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Metallurgical calculations,
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METALLURGICAL CALCULATIONS
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Metallurgical Calculations
BY
JOSEPH W. RICHARDS, A.C., Ph.D.
Pr.ofessor of Metallurgy in Lehigh University,
Secretary {and Past-President) of the American Electrochemical Society,
Vice-President, American Institute of Mining Engineers, Mem-
ber of the U. S. Naval Consulting Board, Author of
"Aluminium, Its Metallurgy, etc."
ONE VOLUME EDITION
PART I
Introduction, Chemical and Thermal Principles
Problems in Combustion, and
Radiation and Conduction of Heat
PART II
Applications to the Metallurgy of Iron and Steel-
PART .
Applications to Other Metals (Non-ferrous Metals)
Second Imibbssion
McGRAW-HILL BOOK COMPANY, Inc.
239 WEST 39TH STREET. NEW YORK
LONDON: HILL PUBLISHING CO., Ltd.
6 & 8 BOUVERIE ST., E. C.
1918
Copyright, 1910, 1915, 1917, 1918, by the
McGraw-Hill Book Company, Inc.
Copyright, 1906, 1907, 1908, by the
McGraw Publishing Company.
fi-'^lp^^K.
THE AlAPLXl PRESS X O R K PA
PREFACE
While issuing Parts I, II and III separately, for the use of
students, it is thought that most persons, particularly practical
metallurgists, will prefer to have the whole work in one volume.
The writer has endeavored to correct all the mistakes which
crept into former editions, and to add such new and useful phys-
ical and chemical data as have appeared to date. Some new
features are the tables of thermochemical constants (which will
assist in predicting undetermined heats of formation), the esti-
mation of many latent heats of fusion and vaporization of the
elements (which data can be used pro tern, as first approxima-
tions), and the evaluation of vapor tension formulas for the ele-
ments in both liquid and solid states (likewise to be used as
first approximations). My thanks are due to Mr. Leonard Buck
and Mr. Y. Taldkawa, students in metallurgy, for considerable
assistance given in calculating and completing the new tabular
data.
The author is grateful to many friends whose kindly criticisms
have pointed out mistakes or suggested improvements. He ap-
preciates the careful translations which have been made into
Italian by Sr. Remo Catani, into Russian by Engineer Koshkin,
into German by Prof. Neuman and Engineer Brodal, and regrets
that the breaking out of the great war has interrupted transla-
tions already begun into French and Spanish. These inter-
national recognitions are simply proofs that the day of Quanti-
tative Metallurgy is dawning all over the world, and that the
application of Metallurgical Calculations is the path to Metallurgical
Efficiency.
Joseph W. Richards.
Lehigh University,
January 15, 191 8.
CONTENTS.
Page
Introduction. — Scope of the Treatise ' iii-xi
Chapter I. — The Chemical Equation 1-12
Atomic Weights 1
Relative Weights 2
Relative Voliunes of Gases 2
Exact Weights and Exact Volumes 3
Weights and Volumes of Gases 4
Corrections for Temperature 5
Corrections for Pressure 7
Corrections for Temperature and Pressure 7
Problem 1. — Combustion of Coal 8
Problem 2. — Combustion of Natural Gas 10
Problem 3. — Oxidation in a Bessemer Converter.. 11
Chapter II. — The Applications of Thermochemistry . 13-40
Thermochemical Nomenclature 13
Thermochemical Data 18-40
Oxides 18
Hydrates 19
Sulfides 21
Selenides, Tellurides 22
Arsenides, Antimonides, Phosphides, Nitrides ... 23
Hydrides, Hydrocarbons 24
Carbides 25
Silicides, Fluorides 26
Chlorides 27
Carbonates 28
Bi-Carbonates 29
Nitrates 30
Silicates 31
Sulphates 32
Bi-'sulphates, Phosphates, Arsenates, Tungstates. 33
Borates, Molybdates, Titanates 34
Manganates, Aluminates, Cyanides 35
Cyanates, Metallo-cyanides 36
vii
VIU CONTENTS.
Amalgams, Alloys 37
Thermochemical Constants of Bases and Acids. 38
Chapter III. — The Use of the Thermochemical Data . 41-60
Simple Combinations — Complex Combinations - 41
Double Decompositions 43
Calorific Power of Fuels 46
Problem 4. — Combustion of Natural Gas 47
Dulong's Formula 48
The Theoretical Temperature of Combustion 50
Specific Heats of Gases Produced 51
Combustion with Heated Fuel or Air 52
Problem 5. — Calorific Intensity of Natural Gas . 54
The Eldred Process of Combustion 55
Temperatures in the "Thermit Process" 58
Chapter IV. — The Thermochemistry of High Tempera-
tures .• 61-117
Reduction of Zinc Oxide by Carbon 63
General Remarks 68
Reduction of Iron Oxide by Hydrogen 69
Specific Heats of the Elements 70
Latent Heats of Fusion of the Elements 71
Latent Heats of Vaporization of the Elements 73
Thermophysics of the Elements 74
Tables of Thermophysics of the Elements 75-103
Efficiency of Furnaces 104
Problem 6. — Efficiency of a Rockwell Furnace. . 106
Problem 7. — Efficiency of an Amalgam Retort. . 107
Problem 8. — Efficiency of a Zinc Distilling Fur-
nace 108
Thermophysics of Alloys 109
Tables of Thermophysics of Alloys 109-113
Problem 9. — Efficiency of a Steel Melting Fur-
nace 114
Problem 10. — Efficiency of a Siemen's Furnace. . 114
Problem 11. — Efficiency of a Foundry Air-fur-
nace 114
Problem 12. — Efficiency of a Foundry Cupola. . 115
Chapter V. — Thermophysics of Chemical Compounds 118-144
Oxides 118
CONTENTS. ix
Problem 13. — Efficiency of Alumina Melting Fur-
nace 127
. Problem 14. — Reduction of Iron Oxides by CO
gas 128
Problem 15. — Reduction of Iron Oxides by Car-
bon 129
Chlorides ' 131
Bromides, Iodides 133
Fluorides, Sulfides 134
Compound Sulfides 135
Arsenides, Antimonides, Selenides, Tellurides 136
Cyanides, Hyposulfites, Sulfates 136
Carbonates, Nitrates 137
Borates, Phosphates, Chromates, Arsenates 138
Chlorates, Aluminates, Titanates, etc 139
Silicates ' 139
Miscellaneous Materials 142
Slags 144
Chapter VI. — Artificial Furnace Gas 145-184
Simple Producers 146
Problem 16. — Investigation of a Simple Pro-
ducer 149
Problem 17. — Drjdng of Moist Producer Gas . . . 154
Mixed Gas Producers 158
Problem 18. — Investigation of a Morgan Pro-
ducer , 163
Mond Gas 169
Problem 19. — Calculations on a Mond Producer . 169
Reactions on Pre-heating Mond Gas 175
Water Gas "177
Problem 20.— On a Dellwick-Fleischer Plant. ... 179
Problem 21.— Blast Required for D-F. Plant ... 182
Chapter VII. — Chimney Draft and Forced Draft . . 185-199
Chimney Draft 185
Available Head 191
■ Problem 22. — Design of Puddling Furnace Chim-
ney 192
Problem 23. — Raising Steam by Waste Heat . . . 194
X CONTENTS.
Chapter VIII. — Conduction and Radiation op Heat. 186-200
Principles of Heat Conduction 200
Table of Heat Conductivities of Metals 203
Principles of Heat Transfer 206
Problem 24.— Cooling of Hot Air in a Tube .... 208
Table of Heat Conductivities of Insulating Ma-
terials 211
Radiation 213
Table of Radiating Capacity (Emissivity) 215
Appendix. 217-231
Problem 25. — Combustion of Hydrocarbons 217
Problem 26.— Calorific Values of Coal, Oil and Gas.. 219
Problem 27. — Calorific Values of Various Gases 219
Problem 28. — Combustion of Coal under a Boiler. . . 220
Problem 29. — Combustion of Anthracite Coal 221
Problem 30. — ^Air Required for Powdered Coal 221
Problem 31. — Efficiency Test on a Boiler 222
Problem 32. — Temperature of Powdered-coal Flame. 223
Problem 33. — Temperature at Tuyeres in Blast Fur-
nace 223
Problem 34. — Efficiency of Powdered Coal Firing . . '. 224
Problem 35. — Calculation of Chimney Draft 224
Problem 36. — Calorimeter Test of Temperature 224
Problem 37. — Calorimeter Test for Specific Heat 225
Problem 38. — Economy of a Feed-water Heater 225
Problem 39. — ^Working of Gas-engine 226
Problem 40. — Efficiency of a Gas Producer 226
Problem 41. — Temperature of Producer-gas Flame. . 227
Problem 42. — Drying KUn for Peat 227
Problem 43. — Power from Waste Coke-oven Gases. . 228
Problem 44. — Efficiency of By-product Coke Ovens. 229
Problem 45 . — Decomposition of Steam in a Producer . 229
Problem 46. — Chimney for a Gas Furnace 230
Problem 47. — ^Loss of Heat through Sides of Chimney 231
CONTENTS TO PART II.
Page
Chapter I. — Balance Sheet of the Blast Furnace. .235-249
Fuel 236
Ore 239
Flux 243
Blast 244
Problem 51. — Balance of a Swedish Furnace. . . 245
Chapter II. — Calculation op a Furnace Charge. .250-267
Flux and Slag 251
Problem 52. — Calculation of Flux needed 259
Comparison of Values of Fuels 261
Comparison of Values of Fluxes 263
Comparison of Values of Ores 265
Problem 53. — Cost of Pig Iron calculated 267
Chapter III. — Utilization of Fuel in the Blast Fur-
nace 268-280
Problem 54. — Generation of Heat in Blast Fur-
nace 268
Problem 55.— Efficiency of Hot Blast Stoves 270
Problem 56. — Power from Blast Furnace Gases. 272
Griiners Ideal Working 274
Minimum Carbon necessary in the Furnace 277
Chapter IV. — Heat Balance Sheet of Blast Fur-
nace 281-304
Heat received and developed 281
Combustion of Carbon. — Illustration 281
Sensible Heat in Hot Blast. — Illustration 284
Formation of Pig Iron 285
Formation of Slag 286
xi
xii CONTENTS TO PART II.
Page
Heat absorbed and disbursed 287
Heat in Waste Gases 287
Heat in Slag 288
Heat in Pig Iron 289
Heat conducted away and radiated 289
Heat in Cooling Water 290
Heat for drying and dehydrating charges 290
Heat for decomposing carbonates 291
Heat for reduction of iron oxides 292
Heat for reduction of other oxides 292
Decomposition of Moisture of Blast 293
Problem 57.— Complete Heat Balance Sheet. ... 294
Chapter V. — Rationale of Hot-Blast and Dry-Blast 305-312
Problem 58. — (for practice) Heat Balance Sheet . 305
Hot Blast. — Temperatures obtained 308
Dried Blast. — ^Increased Temperatures obtained 309
Illustration: Temperatures in a specific case. . . . 310
Table of temperatures under various conditions. 312
Chapter VI. — Production, Heating and Drying of
Blast 313-332
Production of blast. Work required 313
Problem 59. — ^Work of a blowing engine 315
Measurement of pressure of blast. Indicator cards . . 317
Heating of the blast 318
Problem 60. — Efficiency of an iron-pipe stove . . . 320
Problem 61. — Efficiency of fire-brick stoves 322
Drying air blast 325
Illustration: Increased efficiency of blowing en-
gines 326
Illustration: Refrigeration required 326
Problem 62. — Efficiency of drying apparatus. . . 327
Illustrations: Calculation of saturation tempera-
ture 331
Problem 63. — Cooling compressed air by river
water 331
Chapter VII. — The Bessemer Process 333-350
Air required 333
Problem 64. — Maximum and minimum blast. . . 334
CONTENTS TO PART II. xiii
Page
Air received , 336
Problem 65. — ^Volume efficiency of blowing plant 336
Blast pressure 340
Illustration: Back pressure in converter 341
Problem 66. — Distribution of blast pressure. . . . 343
Flux and Slag 346
Illustration: Lime needed in a basic blow 347
Recarburization 348
Problem 67. — Recarburization balance sheet 349
Chapter VIII. — Thermo-Chemistry of the Bessemer
Process 351-367
Elements consumed 351
Heat balance sheet — litems 353
Heat developed by oxidation 356
Heat of formation of slag 357
Illustration: Heat of formation of basic slag. . . . 358
Heat in escaping gases 360
Heat conducted to the air 362
Heat radiated 363
Problem 68. — Complete heat balance sheet 363
Chapter IX. — The Temperature Increment in the
Bessemer Converter 368-380
Combustion of Silicon 369
Combustion of Manganese 371
Combustion of Iron ' 372
Combustion of Titanium 373
Combustion of Aluminum, Nickel, Chromiuni. 374
Combustion of Carbon '. . . . 375
Combustion of Phosphorus 377
R&um6 379
Chapter X. — The Open-Hearth Furnace 381-395
Gas producers 381
Flues to fiuTiace; regenerators 382
Problem 69. — Sizes of regenerators 384
Valves and ports 388
Problem 70. — ^Areas of gas and air ports 389
Laboratory of furnace 391
xiv CONTENTS TO PART tl.
PagB
\ Problem 71. — Efficiency of laboratory 393
Chimney flues and chimney, miscellaneous 395
Chapter XI. — Thermal Efficiency of Open-Hearth
Furnaces 396-428
Heat in warm charges; in gas used 397
Heat in air used 398
Heat of combustion; of oxidation of the bath 399
Heat of formation of slag; heat in melted steel 399
Reduction of iron ore; decomposition of carbonates. . 400
Heat in slag 401
Loss by imperfect combustion; in chimney gases. . . . 402
Loss by conduction and radiation 403
Problem 72. — Heat balance sheet 403
Problem 73. — Thermal efficiencies 411
Problem 74. — New Siemens' furnace 418
Problem 75. — The Monell process 424
Chapter XII. — The Electrometallurgy of Iron and
Steel 429-451
Electrothermal reduction of iron ores 429
Problem 76. — Production of pig-iron electrically . 430
Problem 77. — Prdduction of nickeliferous pig. . . 435
Production of Steel 440
, Problem 78. — The induction electric furnace. . . 442
Problem 79. — The electric " pig and ore " process 443
Problem 80. — Electrolytic refining; Burgess'
process 446
Appendix: Problems for Practice 452-465
Problem 81. — Diameter of tuyeres for a blast fur-
nace 452
Problem 82. — Blast furnace slag calculation 452
Problem 83. — Carbon burned before the tuyeres 453
Problem 84. — Calculation of blast furnace charge. . . 454
Problem 85. — Slag calculation for various slags 454
Problem 86. — Weight of flux needed 455
Problem 87. — Volume of blast, and cooling of same
by expansion 456
Problem 88. — Output of a blast-furnace 456
CONTENTS TO PART II. XV
Page
Problem 89. — Efficiency of the blowing engines. . . . 456
Problem 90. — Power producible from blast-furnace
gases m 457
Problem 91. — Proportion of gases required by hot-
blast stoves 457
Problem 92. — Complete balance sheet of a blast fur-
nace 458
Problem 93. — Working of a foundry cupola 460
Problem 94. — The Monell process in an open-hearth
furnace 460
Problem 95. — Volume of blast required by a Besse-
mer converter 461
Problem 96. — ^Volume of blast received by a Besse-
mer converter 461
Problem 97. — ^Loss of weight of a Bessemer charge. 462
Problem 98. — Balance sheet of a complete Bessemer
blow 462
Problem 99. — ^Lime needed in a basic Bessemer blow 464
Problem 100. — Power of blowing engines required for
a converter 465
Index 466
CONTENTS TO PART III.
Page
Chapter I. — The Metallurgy of Copper 469-577
Roasting and Smelting of Copper Ores 469
The Roasting Operation 475
Problems 101-104.
Pyritic Smelting 483
Fundamental Principles 485
Theoretical Temperature at the Focus 486
Use of Auxiliary Coke . 488
Rate of Smelting 489
Problems 105, 106.
Smelting of Copper ores 499
Reverberatory Smelting 501
Problems 107, 108.
"Bringing Forward" of Copper Matte 514
The Welsh Process 514
Problem 109.
"Blister Roasting" or Roasting-Smelting 523
"Bessemerizing" Copper Matte 525
Theoretical Temperature Rise 527
Problem 110.
The Electrometallurgy of Copper 534
Electrolytic Processes 536
Treatment of Ores by Electrolysis 536
Problem 111.
Electrolytic Treatment of Matte 539
Use of Matte as Anode 539
Problem 112.
Use of Matte as Cathode 545
Problem 113.
Extraction from Solutions 548
(1) Use of Iron Anodes 548
xvii
xviii CONTENTS TO PART III.
Page
Problems 114, 115.
(2) Use of Insoluble Anodes 551
Problems 116, 117, 118
Electroljrtic Refining of Imptire Copper. 558
Energy Absorbed by Electrolyte 559
Problem 119.
Energy Lost in Contacts 562
Problem 120.
Energy Lost in Conductors. . . ' 567
Problem 121.
Electric Smelting 572
Problem 122.
Chapter II. — The Metallurgy of Lead 578-604
The Volatility of Lead 583
Roasting of Lead Ores 586
Problem 123.
Reduction of Roasted Ore 591
Problem 124.
The Electrometallurgy of Lead 598
Problems 125, 126.
Chapter III. — The Metallurgy of Silver and Gold. 605-619
Electrolytic Refining of Silver Bullion 605
Problems 127, 128.
The Volatilization of Silver and Gold 614
Chapter IV. — The Metallurgy op Zinc 620-657
Roasting of Sphalerite , 620
Problems 129, 130.
Reduction of Zinc Oxide 627
Thermochemical Considerations 628
Heating up the Charge 628
Problem 131.
Distillation of the Charge 632
Problems 132, 133, 134.
Electric Smelting of Zinc Ores 637
Zinc Vapor 643
CONTENTS TO PART HI. jdx
Page
Problem 135.
Condensation of Zinc and Mercury 649
Metallic Mist or Fume Vapor 656
Chapter V. — Metallurgy of Aluminium 658-661
Electrolytic furnace reduction of alumina 658
Index 663
METALLURGICAL CALCULATIONS.
INTRODUCTION.
The making of calculations respecting the quantitative work-
ing of any process, furnace or piece of apparatus used in metal-
lurgical operations is of the greatest importance for estimating
the real efficiency of the process, for determining avenues of
waste and possible lines of improvement, and for obtaining the
best possible comprehension of the real principles of operation
involved.
The possibility of making such calculations respecting any
process, furnace or apparatus depends on skill in collecting such
necessary data as can be obtained by observation or measure-
ment, the insight or intuition to see the further use which can
be made of said data when once obtained, and, finally on the
possession of a working knowledge of the fundamental chem-
ical, physical and mechanical principles involved in the calcu-
lations. The highest desideratum, all in all, however, is a
plain analytical, common sense mind, capable of clear, logical
thinking. It is the writer's conviction that no study of details,
or even observation of plants in actual operation, can supply
the insight into metallurgical processes and principles, such
as is gained by these calculations, in addition to the high grade
of mental training involved.
Scope of the Treatise
Discussion of the chemical equation.
Weights and volumes of gases.
Correction of gas volumes for temperature and pressure.
Combustion of commercial fuels.
Heat of chemical combination; of combustion.
Theoretical flame temperatures:
With pure oxygen.
With ordinary air.
xxi
xxu INTRODUCTION.
With diluted air: Farley's system.
With hot air and cold gas.
With hot air and hot gas.
Effect of excess air.
Calculation of furnace efficiencies.
Chimney draft.
Water gas.
Producer gas: Efficiency, effect of drying.
Mixed gas:
Use of steam in producers.
Increased efficiency.
Maximum steam permissible.
Transmission of heat through metals, brick, etc.
Regenerative gas furnace:
Proportioning of gas and air regenerators.
Efficiency of regenerators.
Heat balance sheet.
Theoretical temperatures under different conditions.
Gas engines:
Calculation of temperature in cylinder.
Efficiency; balance sheet.
Cupolas: Amount of blast required.
Efficiency of running.
Blast furnaces:
Balance sheet of materials.
Calculation of blast received.
Efficiency of blowing engines.
Power and dimensions of blowing engines.
Carbon consumed at tuyeres.
Effect of atmospheric changes.
Effect of the moisture in the blast.
Calculation of the temperature.
Effect of hot-blast.
Heat balance sheet of the furnace.
Power producible from the waste gases.
Hot-blast stoves: Theory of iron-pipe and fire-brick stoves.
Efficiency.
Bessemer Converters:
Blast required and time of operation
Balance sheet of materials.
INTRODUCTION. xxiii
Heating efficiency of various ingredients of bath.
Heat balance sheet.
Theoretical rise in temperature.
Conversion of copper matte.
Open-hearth Furnaces:
Pig and ore process; calculation of charge.
Heat evolved or absorbed in bath reactions.
Efficiency of furnaces; of furnaces and producers.
Heat balance sheet.
Pyritic smelting.
Electric furnaces:
Working temperatures.
Heat balance sheet.
Efficiency.
Electrolytic furnaces:
Absorption of heat in chemical decompositions.
Equilibrium of temperature attained.
Ampere and energy efficiency.
Electrolytic refining:
Calculation of plant and output.
Power requirements; temperature of baths.
Ampere and energy efficiency.
Condensation of metallic vapors:
Principles involved.
Application to condensation of zinc and mercury.
CHAPTER I.
THE CHEMICAL EQUATION.
The calculation of the quantitative side of many metallur-
gical processes depends upon the correct understanding of
chemical equations. Every chemical equation is capable of
giving three most important sets of data concerning the pro-
cess which it represents; its shows the relative weights of the
reacting substances, their relative volumes, when in the gas-
eous state, and the surplus or deficit of energy involved in the
reaction, when the heats of formation of the substances con-
cerned are known.
Atomic Weights.
These are the basis of all quantitative chemical calculations.
For metallurgical purposes we may use them in round num-
bers as:
Hydrogen H 1
Lithitun Li 7
Beryllium Be 9
• Boron. B 11
Carbon C 12
Nitrogen N 14
Oxygen O 16
Fluorine F 19
Sodium Na 23
Magnesium Mg 24
Aluminium Al 27
Silicon Si 28
Phosphorus P 31
Sulphur S 32
Chlorine CI 35.5
Potassium K 39
Calcium Ca 40
Arsenic As 75
Selenium Se 79
Bromine Br 80
Strontium. Sr 87
Zirconium Zr 90
Columbium Cb 94
Molybdenum Mo 96
Palladium...' Pd 106
Silver Ag 108
Cadmium Cd 112
Tin Sn 118
Antimony Sb 120
Tellurium Te 126
Iodine I 127
Barium ; Ba 137
Tantalum Ta 183
Tungsten W 184
2 METALLURGICAL CALCULATIONS.
Titanium Ti 48 •Iridium Ir 193
Vanadium V 51 Platinum .Pt 195
Chromium Cr 52 Gold Au 197
Manganese Mn 55 Mercury Hg 200
Iron Fe 56 Thallium Tl 204
Nickel Ni 58.5 Lead Pb 207
Cobalt Co 59 Bismuth Bi 208
Copper Cu 63.6 Thorium Th 232
Zinc Zn 65 Uranium U 238
Relative Weights.
Writing any chemical equation between elements or their
compounds, the relative weights concerned in the reaction are
obtained directly from using these atomic weights, which are.
themselves, of course, only relative. E.g., the slagging of iron
in Bessemerizing copper matte:
2FeS + 30^ + 2Si02 = 2(FeO.Si02) + 2S0'
176+ 96+ 120= 264 +128
These relative weights may be called kilograms or tons,
pounds, ounces or grains; whatever units of weight we may
be working in. In most metallurgical work we use kilograms
or pounds as the convenient weight units.
Relative Volumes of Gases.
Where gases are involved, the relative number of molecules,
of the gaseous substance concerned in the reaction stands for
the relative volume of that gas concerned in the reaction. It is
usual and convenient to designate these relative volumes by
Roman numerals, placed above the formulae. The following
are some examples:
Complete combustion of carbon:
I I
C + 0^ = CO^
Incomplete combustion of carbon:
I II
2C + 02 = 2C0
Combustion of marsh gas:
I II I II
CH, + 202 = C0, + 2H,0
THE CHEMICAL EQUATION. 3
Production of water gas:
I I I
C + H'O = CO + H^
In each case above, the volume of a solid or liquid cannot be
stated, but the relative volumes of all the gases taking part in a
reaction are derived simply from the number of molecules of
each gas concerned. These relative volumes may be called so
many cubic meters or liters, or cubic feet, or whatever measure
is wanted or being used. In most metallurgical calculations it
is convenient to use cubic meters or cubic feet.
Exact Weights and Exact Volumes.
If we specify or fix the weights used, as, for instance, so
many kilograms of each substance as the numbers representing
the relative weights, then we can, by using one constant factor,
convert all the relative volumes into the real or absolute vol-
umes corresponding to the weights used. If, for instance, we
take the equation of the production of water gas:
I I I
12+ 18 = 28 + 2
With the relative weights written beneath and the relative
volwmes above, then if we fix the weights as kilograms, the
relative volumes can be converted into actual volumes in cubic
meters by multiplying by 22.22. A cubic meter of hydrogen
gas (under standard conditions) weighs 0.09 kilogram, and
thence 2 kilograms- will have a volume of 2-4-0.09 = 22.22 cubic
meters. But the relative volumes show that the CO and
H^O gas are the same in volume as the hydrogen, and it, there-
fore, follows that each Roman I stands for 22.22 cubic meters
of gas, if the weights underneath are called kilograms. The
consideration of these relations is very advantageous, because,
by means of this factor (22.22) we pass at once from the
weight of a gas to its volume; each molecule or molecular
weight of a gas, in kilograms (or, briefly, each kilogram-mole-
cule), represents 22.22 cubic' meters of that gas.
The conversion from weight to volume is quite as simple
using English measures; and, by a strange coincidence, the
same factor can be used as in the metric system. The coin-
4 METALLURGICAL CALCULATIONS.
cidence alluded to is the fact (which the writer, as far as he
can discover, was the first to notice) that there happens to be
the same numerical relation between an ounce (av.) and a kilo-
gram, as there is between a cubic foot and a cubic meter; in
short, there are 35.26 ounces (av.) in a kilogram, and 35.31
cubic feet in a cubic meter. The diflference is only one-seventh
of one per cent., which can be ignored, and we can therefore say
that if the relative weights in an equation are called ounces
(av.), each molecule of gas in the equation represents 22.22
cubic feet.
Example. — The production of acetylene from calcium carbide :
I
CaC=-l-H20 = CaO + C2H'
64 + 18 = 56 + 26
Interpreting by weights, and calling the relative weights
ounces, we can call the I molecule of C^H' gas 22.22 cubic feet,
so that, theoretically, 64 ounces of pure carbide, acting on 18
ounces of water, produce 56 ounces of lime and 26 ounces of
C^H'' gas, the volume of which is 22.22 cubic feet.
Weights and Volumes of Gases.
The weight of one cubic meter of dry air, under standard con-
ditions (at 0° Centigrade and at a pressure of 760 millimeters
of mercury), is 1.293 kilograms. The composition of air is:
By Weight. By Volume.
Oxygen 3 21
Nitrogen 10 80
or, in percentages.
Oxygen 23.1 20.8
Nitrogen 76.9 79.2
While these may not represent the absolutely accurate aver-
age composition of dry air, yet the variations are such that the
above simple ratios, 3 to 10 and 21 to 80, are close enough for
all practical purposes in metallurgy.
The weight of one cubic foot of dry air is 1.293 ounces (av.).
The weight of one cubic meter of hydrogen gas, at standard
conditions, is 0.09 kilogram (1 cubic foot, 0.09 ounces). The
formula of hydrogen gas is H^, is molecular weight 2; and
since the densities of all gases are found experimentally to
jlecula
/■eight.
r Density Referred
to Hydrogen.
Weight of 1
Cubic Meter.
2
1
0.09 kilos.
18
9
0.81 "
28
14
1.26 "
32
16
1.44 "
28
14
1.26 "
44
22
1.98 •■
16
8
0.72 "
THE CHEMICAL EQUATION. 5
be proportional to their molecular weights, it follows that the
density of any gas referred to hydrogen is expressed numer-
ically by one-half its molecular weight. But the weight of a
cubic meter of gas is the weight of a cubic meter of hydrogen
multiplied by the density of the gas referred to hydrogen; thus,
is obtained the weight of a cubic meter of any gas whose formula
is known. Examples follow:
Formula.
Hydrogen H^
Water vapor H^O
Nitrogen N^
Oxygen O''
Carbon monoxide . . . CO
Carbon dioxide CO^
Marsh gas CH*
Etc., etc.
In the case of water vapor, a particular explanation is neces-
sary. It cannot exist under standard conditions, but condenses
to liquid at 100° C. if under 760 millimeters pressure. It does
exist at lower temperatures than 100°, but only under partial
pressures of fractions of an atmosphere; thus, at a pressure of
1-50 atmosphere (when it forms 1-50 of a mixture of gases)
it can exist uncondensed at ordinary temperatures (15° C. or
60° F.). The above weight for a cubic meter of water vapor
(0.81 kilos, per cubic meter at standard conditions) is, there-
fore, only a hypothetical value, but it is extremely useful, be-
cause it enables us to calculate, by the principles to be ex-
plained further on, the weight of a cubic meter of water vapor
under any conditions of temperature and pressure at which it
is possible for it to exist.
Corrections for Temperature.
The volumes of all permanent gases increase uniformly for
uniform increase of temperature, so that, starting with a given
volume at 0° C, it is found that their volume increases 1/273
for every degree Centigrade rise in temperature. Thus, at 273°
C, the volume is just double the volume at 0°. Stating this
fact in another way, we may say that the gas acts as if it would
have no volume at — 273° C, and would increase uniformly
6 METALLURGICAL CALCULATIONS.
in volume from this point up to all measurable temperatures,
the increment being, for each degree, 1-273 of the volume
which the gas has at 0° C. A still briefer statement is that the
volimie of a gas is proportional to its temperature above — 273°
C, or to its absolute temperature — the latter being its tem-
perature in C° + 273.
The converse of these principles is, that the density of a gas;
that is, the weight of a unit volume, varies inversely as its
absolute temperature.
In Fahrenheit degrees, we can say that a gas expands 1-490
(1-273 X 5-9) for every degree rise above 32" C. ; or that the
volume is proportional to the absolute temperatures, i.e., to
the F°. -32-1-490 ( = F.°-l-458).
• These principles are in constant use in metallurgical calcula-
tions. Thus, one kilogram of coal will need 8 cubic meters of
air to bum it, at 0° and 760 millimeters pressure. What
volume will that be at 30° C. and the same pressure? Since
30° C. is 30 + 273 = 303° absolute, the two temperatures will
be 273 and 303, and the
Volume at 30° C. = volume at 0° C.X^
It is always to be recommended to make such calculations in
the above form.; that is, to first put down the known volume,
and then to multiply it by a fraction, the numerator and de-
nominator of which are the two absolute temperatures. A
moment's reflection will show which way the fraction must be
written; if the new volume must be greater than the old, the
value of the fraction must be greater than unity, the higher
temperature must be in the numerator; if the fraction were
inverted, we know that the result would be less than the start-
ing volume instead of greater, which would be wrong.
Taking an example in Fahrenheit degrees: What is the vol-
ume under standard conditions of 175 cubic feet of gas meas-
ured at 90° F. and standard pressure (29.93 inches of mer-
cury)? Since 32° F. is 490° absolute and 90° F. is 548° abso-
lute, and the new volume must be less than the starting vol-
ume, we have
Volume at 32° F. = 175X^ = 156.5 cubic feet
THE CHEMICAL EQUATION. 7
Corrections for Pressure.
The principle is that the volumes of a gas are inversely as
the pressure upon it, so that doubling the pressure halves the
volume, etc. Since the practical problems almost always pre-
sent the pressure as two numbers, all that is necessary is to
mtdtiply the original volume by a fraction whose numerator
and denominator are the two pressures concerned, and ar-
ranged with the numerator the larger or the smaller of the two
numbers, according as to whether the final volume should be
greater or less than the starting one. Putting the solution in
this manner avoids the primary school method of making a
proportion, which is so apt to be expressed upside down, and
absolutely avoids error with the minimum exercise of brain
power.
Examples. — What is the volume of 100 cubic meters of any
gas, if the pressure is changed to 700 millimeters?
Answer: lOOX^z^jj; = 108.6 cubic meters.
What is the volume at standard pressure of 150 cubic feet of
gas measured at 28.50 inches of mercury?
28 50
Answer: ISOX^qqo = 142.8 cubic feet.
Corrections for Temperature and Pressure.
These can be both allowed for, by simply correcting first for
one, and then for the other. Actually, the simplest statement
is to put down the original volume, then to multiply it by one
fraction, which corrects for temperature, and again by another
fraction correcting for pressure, thinking out carefully for each
fraction the proper way of expressing it, i.e., whether it should
increase or decrease the volume.
Examples. — What does 100 cubic meters of air at standard
conditions become at 50° C. and 780 millimeters pressure?
crj I 0*7*3 7fi0
Solution: 100 X' ^^g X^ = 115.3 cubic meters.
What is the weight of one cubic meter of hydrogen at 1000°
8 METALLURGICAL CALCULATIONS.
C. and 250 millimeters pressure, its weight at standard condi-
tions being 0.09 kilograms?
Solution: O.OQXjoog^X^ - 0.00637 kilograms.
"What weight of oxygen is in 1500 cubic feet of dry air af
100° F. and at 28.50 inches of mercury? (Refer to weight of
air at standard conditions, and percentage composition.)
Solution : 1 . 293 X — X ||? X ||4§ X 1 500 = 374 ounces.
What is the weight of 50 cubic meters of wa^er vapor at a
temperature of 30° C. and a pressure of 31.6 millimeters?
Solution: 0.81 x|^Xyg^X50 = 1.517 kilograms.
or50X3^XYgQ-X0.81 = 1.517 kilograms.
The first expression calculates the weight of a cubic meter
of water vapor at the assumed conditions, and multiplies by
50; the second calculates the hypothetical volume of the 50
cubic meters if reduced to standard conditions, and multiplies
by the hypothetical weight of a cubic meter at those condi'
tions.
Problems Illustrating Preceding Principles.
Problem 1.
A bituminous coal contains on analysis:
Carbon 73.60 Moisture 0.60
Hydrogen 5.30 Ash 8.05
Nitrogen 1 . 70
Sulphur 0.75. 100.00
Oxygen 10.00
It is powdered and blown into a cement kiln by a blast of air.
Required: 1. The volume of dry air, at 80° F. and 29 inches
barometric pressure theoretically required for the perfect com-
bustion of one pound of the coal.
2. The volume of the products of combustion, using no excess
THE CHEMICAL EQUATION. 9
of air, at 550° F. and 29 inch^ barometer, and their percentage,
composition.
Solution: The reactions of the combustion are:
C + O^ = CO^ S + 0^ - SO^'
12 32 44 32 32 64
2FP + 02 = 2H='0
4 32 36
Requirement (]):
The oxygen required for burning one pound of coal is:
Oxygen for carbon = 0.7360X32/12 = 1.963 pounds.
Oxygen for hydrogen = 0.0530x32/4 = 0.424
Oxygen for sulphur = 0.0075x32/32 = 0.0075
Total required 2.3915
Oxygen in coal 0.1000
Oxygen to be supplied 2.2945
Nitrogen accompanying 7.6483
Air necessary. .- 9.9428
= 159.08 ounces (av.).
Volume of air necessary (standard conditions)
Volume of air necessary at 80° F. and 29 inches barometer =
123.03X?^^^X?^ - 139.4 cubic feet. (1)
Requirement (2): Pounds.
The -weight of CO^ formed is 0.7360+1.963 = 2.699
The weight of H'O formed is 0.0530 + 0.424 = 0.477
The weight of moisture is 0.006
The weight of SO^ formed is '. . .0.0075 + 0.0075= 0.015
The weight of nitrogen altogether is 7.6483 + 0.0170 =
7.6653 pounds. Converting these weights into ounces, and
dividing each by the weight of a cubic foot of each gas in ounces,
we have the volume of these theoretical products at standard
conditions :
10 METALLURGICAL CALCULATIONS.
Volume CO' = 2.699X16^198 ='
43.1844-1.98 = 21.80 cubic feet.
Volume H^O = 0 483X164-0.81 =
7 7284-0.81. = . 9 54 "
Volume SO' = 0 015X16 4-2.88 =
0.2400-^2.88= 0.08 "
Volume W = 7 665X16 4-1.26 =
122.645 4-1.26= 97.34 "
Total volume at standard conditions = 128.76
Volume at 550° F. and 29 inches barometer =
128.76X^^^X^^-273.4 " "
The percentage composition .by volume follows from the
above volumes as:
CO' 17.0 per cent. N' 75. 5 per cent.
TJ2Q 7 4 " —
so"v.;::::::: 0:1 ■■ loo.oo
Problem 2.
Natural gas in the Pittsburg district contains:
Marsh gas CH* 60.70 per cent.
Hydrogen H' 29.03
Ethane C'H" 7.92
Olefiant gas C'H' 0.98
Oxygen O' 0.78
Carbonous oxide CO 0.58
Required:
(1) The volume of air necessary to bum it.
(2) The volume of the products of combustion.
Reactions:
CH'-|-20' = CO' -h2H'0
2H' f O', = 2H'0
2C'H' + 70" = 4C0' -1-611=0
C'H' + 30' = 2CO'-t-2H'0
2C0-I- O' = 2C0'
Solution:
Oxygen required for CH* = 0.6070 X 2 = 1.2140 parts.
Oxygen required for H' = 0.2903X i = 0.1451 "
THE CHEMICAL EQUATION. H
Oxygen required for C^H*. . . . = 0.0792x7/2 = 0.2772 parts.
• Oxygen required for C^H^ = 0.0098 X 3 = 0.0294 "
Oxygen required for CO. = 0.0058 X \ =0.0029 "
1.6686 "
Deduct oxygen already present 0.0078
Leaves oxygen to be supplied 1.6608 "
Corresponding to air V^^ ^ '^'^^^ " ^^^
Volumes of products of combustion:
CO^ H^O N^
From CtP 0.6070 1.2140
From W 0.2903
From C^H^ 0.1584 0.2376
From C^H* 0.0196 0.0196
From CO 0.0058
From air 6. 3242
Total products 0 . 7908 1 . 761 5 6 . 3242 (2)
The above solution is entirely in relative volumes, which
may be all considered cubic feet or cubic meters, and are true
for equal conditions of temperature and pressure.
Problem 3.
A Bessemer converter contains 10 metric tons of pig iron
of the following composition:
Carbon 3 . 00 per cent.
Manganese 0 . 50
Silicon 1.50
Iron... ....95:00
On being blown, one-third the carbon bums to CO^, the rest
to CO; 5 per cent, of iron is oxidized, and no free oxygen escapes
from the converter. Blast is assumed to be dry.
Requirements:
(1) What weight of oxygen is needed during the blow.
(2) How many cubic meters of air, at standard conditions,
will be needed.
(3) What will be the average composition of the gases.
12
METALLURGICAL CALCULATIONS
Reactions:
C + O^ = CO'
12 32 'J4
20 + 0^= 2C0
21 32 56
2Mn + 02 = 2MiiO
110 32 142
Si4 0» = SiO»
28 32 60
2Fe + 0'' = 2FeO
112 32 144
Oxygen needed-
C to C0=. . .
C to CO
Mn to MnO.
Si to SiO' .
. 100 kilos. X 32/ 1 2 = 266 7 kilos.
,200 ■ X 32/24 =266 7
50 " X 32/1 10 = 14.5
,150 " X 32/28 = 171.4
Fe to FeO. ..500
X 32/1 12 --= 142.8
Total 862.1
Nitrogen accompanying this = 2873.7
Air needed. 3735 8
Volume of air.
3735 8
'1.293
Volume of products of combustion:
CO' = 100+266.7 = 366.7 kilos = ^^-p^ = 185.2 cu. m
CO = 200 + 266.7 -■= 466.7 kilos = ^^^ = 370.4 cu. m.
N2
1.26
2873 . 7
1 26
= 2280. T "
Total volume 2836.3 "
Percentage composition by volume:
CO' - . 6.5 per cent.
CO .
N' ...
.13.1
.80.4
(1)
= 2889.3 cubic meters (2)
CHAPTER II.
THE APPLICATIONS OF THERMOCHEMISTRY.
The ordinary interpretation of the chemical equation by
weight gives us the quantitative relations governing the re-
actions of substances upon each other, when the reaction pro-
ceeds to a finish. Unfortunately, much chemical instruction,
as given in our elementary schools, and even in some of the
higher ones, stops with the consideration of the weight re-
lations and does not proceed to those equally important rela-
tions, the energetics of chemical reactions. In most of the
reactions which occur in practical metallurgy, the quantity of
the combustible' used, or, more broadly, the amount of energy
in the form of heat or electrical energy, necessary for pro-
ducing the reactions desired, is the controlling factor regulat-
ing the practicability or impracticability, the commercial success
or failure, of the process.
The relative values of fuels, the manufacture and utilization
of gas, the principles of the regenerative furnace, the Bessemer
process, electric reduction, and a host of metallurgical pro-
cesses, depend essentially on the realization and utilization
of chemical energy, and the only way to become conversant
with the amounts of energy involved or evolved in these opera-
tions is to' understand thoroughly the thermochemistry of the
reactions concerned.
Thermochemical Nomenclature
The heat evolved when compounds are formed from the
elements (in a few cases heat is absorbed) is determined ex-
perimentally by the use of the calorimeter. This branch is
sometimes called " chemical calorimetry; " it is practically a
department of experimental physics. The data obtained give
the heat evolved for the total change from the components
at room temperature to the resulting products, at room tem-
13
14 METALLURGICAL CALCULATIONS.
perature (or very near to it), and may be expressed per unit
of weight of either component or of the substance formed.
Thus, if carbon is burned in a calorimeter to carbonic acid
gas, CO', the heat evolved may be reported or expressed as
8100 gram calories per gram of carbon burnt.
Or, 3037 gram calories per gram of oxygen used.
Or, 2209 gram calories per gram of CO' formed.
Of these three methods of expressing the results, the first
is the more often used, especially by the physicist.
The chemist, however, finds it often more convenient and
logical to express these heats of combination per formula
weight of the substances combining and of product formed.
E.g., In the case of CO', which contains 12 parts of car-
bon and 32 of oxygen in 44 of the gas, the chemist would write,
(C, 0') = 97,200,
meaning thereby that when 12 grams of carbon is burnt by 32
grams of oxygen, forming 44 grams of carbon dioxide, there is
evolved 97,200 gram-calories. These are the laboratory units;
for practical purposes, we call the weights kilograms and the
heat units kilogram-calories (gram-calories are abbreviated to
" cal. "; kilogram-calories to " Cal."). Ostwald, in his thermo-
chemical tables writes (C, 0') = 972 K, where K represents a
unit 100 times as large as a gram-calorie, if weights are taken
as being grams. Berthelot writes (C, O') = 97.2, where the
heat units are kilogram-calories, if the weights concerned are
taken as grams. Both these methods of expression are liable
to cause confusion ; the writer prefers to follow the older thermo-
chemists (Hess, Naumann) and to use the larger number, e.g.,
97,200, which then means gram-calories, if weights are taken
in grams (laboratory units), and kilogram-calories, if weights
are taken in kilograms (practical units).
If it is desired to work in " British Thermal Units " (1 B. T.
U. is the heat needed to raise one pound of water one degree
Fahrenheit), the weights represented by the formula may be
called pounds, and then the expression is as follows:
(C, 0') = [97,200X9/5] = 174,960 B. T. U.
The factor 9/5 is simply the relation of 1° C. to 1° F.; the
APPLICATIONS OF THERMOCHEMISTRY. la
reasoning is, of course, that if the combustion of 12 kilograms
of carbon evolves 97,200 kilogram calories, or would heat
97,200 kilograms of water 1° C, that the combustion of 12
pounds of carbon would heat 97,200 pounds of water 1° C, or
174,960 pounds 1° F. From, the equation as thus expressed
and interpreted, the heat of combination per pound of carbon
burnt, or of oxygen used, or of product formed, may be found
in B. T. U. by dividing 174,960 by 12, 32 or 44, respectively.
Since it is very inconvenient, as well as unscietitific, to have
the two unrelated heat units, with their resulting double sets
of experimental data, I strongly recommend the use of the
metric data and metric Centigrade heat unit. It is, however,
sometimes convenient, when all the data of a problem are
given in English weights, to use as the unit of heat the " pound
— 1° C," or the heat required to raise the temperature of one
pound of water 1° C. This may be called the " pound cal.,"
as distinguished from the B. T. U. The advantage of using it
is that all the experimental data of the metric system units
are at once transferable to the English weights. E.g., (C, O^)
= 97,200 pound cal., if the weights concerned in the formula
(12, 32, 44) are called pounds.
_The thermochemist gives all his experimental data in the
form above explained, and we will now give all the important
thermochemical data known which are useful in metallurgical
calculations, the data being for the reactions beginning and ending
at 15° C. (60° F.):
INTRODUCTION TO THERMOCHEMICAL TABLES.
In the thermochemical tables following, the heats of formation
or combination are given first for the molecular weights repre-
sented by the formulas. What these molecular weights are, can
be found by adding up the atomic weights of the atoms repre-
sented in the formula. The atomic weight table on pages 1 and
2 supplies the necessary data. The formula given represents,
thermochemically, the union of the constituent parts separated
by commas, the whole reaction being enclosed in parentheses,
to distinguish it from the ordinary chemical formula. Thus,
while (Ca, 0) represents the union of 40 parts of calciimi with
16 parts of oxygen to form 56 parts of calcitmi oxide, and (Ca,
Si, O3) represents the combination of 40 parts of calcium with
16 METALLURGICAL CALCULATIONS.
28 parts of silicon and 48 parts of oxygen to form 116 parts of
calcium silicate, (CaO, SiOa) represents the combination of 56
parts of lime with 60 parts of siHca to form 116 parts of calcium
silicate.
In order to facilitate the use of these data in metallurgical
calctdations, the succeeding coltrams give the heat evolution
per. unit weight of the metal or base involved, of the acid element
or acid radical involved, and of the compound formed. This
will save much calculation, for while the heat evolution per
formula weight is of prime use in discussing the heat energy of
chemical reactions, the heats of combination per unit weight of
constituents or of the product are most convenient in ordinary
calculations.
The colimin headed "To Dilute Solution" gives the total heat
evolution inclusive of the heat of solution in a large excess of
water. The differences between correspond.ing coltimns under
this heading and the heading "Anhydrous," are the heat of solu-
tion, per formula weight, per unit of base, of acid, or of compound.
The order in which the elements are arranged in each table is
the order of their heats of combination with unit weight of the
acid element or radical, which expresses the real order of affinity
of the metals in. each case. The order is given for the "An-
hydrous" condition, because this is more frequently concerned
in metallurgy than "To Dilute Solution." Theoretically, the
order of arrangement "To Dilute Solution" is the more uniform
and logical, besides being invariable, but it is less useful in metal-
lurgy, except where dilute solutions are concerned.
In order that an estimate may be made of heats of formation
so far undetermined, there is given at the end of the tables a
list of the " Thermochemical Constants of the Elements" per
chemical equivalent or per unit of valence with which they enter
into combination. These are based on the known law that the
heats of formation of salts to dilute solution are additive functions,
being simply the sum of the thermochemical combining power
of the basic element and that of the acid element or radical.
Assuming arbitrarily that hydrogen has zero for its thermo-
chemical constant, the constants for those elements can be cal-
culated for which any thermochemical data, even a single heat
of formation, exist. Thus, referring to the table, on page 38,
57,200 for Na means that 23 parts of sodium going into combina-
APPLICATIONS OF THERMOCHEMISTRY . 17
tion contributes 57,200 calories towards the heat of formation
of the compound formed, irrespective of what it is combining
with. Similarly, 39,400 for CI means that 35.5 parts of chlorine
going into combination contributes 39,400 calories .towards the
heat of formation of any chloride, irrespective of the element it
is combining with. The sum of these, 96,600, is the heat of
formation of (Na, CI) to dilute solution. By means of the table
of constants given, simple addition gives the heat of formation
of a large number of salts, from their elements (taken from their
usual physical condition at room temperature) per chemical
equivalent of base and acid, or of compound, involved. This is
not per formula weight, except for monovalent elements, but is
per chemical equivalent weight; it must be mtiltiplied by the
valency of the base in the formula, that is, by the number of
chemical equivalents represented by the formula, to get the heat
of formation per formula weight. Thus, J^Al 40,100 plus
CI 39,400 gives 79,500 as the heat of formation of 3^A1 CI3;
from which we have 238,500 as the heat of formation of Al CI3,
per formula weight.
While this table has considerable uses, it does not give the heats
of formation of anhydrous compounds; its values must be cor-
rected by the heats of solution, where known, to give the values
for the anhydrous condition. Yet, these are frequently correc-
tions of a minor order, and leave the table highly useful for
getting first approximations to undetermined thermochemical
data. Names of some elements are introduced whose thermo-
chemical constants are unknown, being placed, at a guess, in
their most probable position in the table. All quantities are
positive, except these indicated negative. The -f- exponent
after a symbol represents one positive valence, the — exponent
one negative valence. A more detailed explanation of the matter
can be found in an article by the writer in the Trans. American
Electrochemical Society, 1903, Vol. iv, page 142.
18
MET ALL URGICAL CALC ULA TIONS.
Heat of Formation of Oxides
Anhydrous
To Dilute Solution
Formula
Pit Unit Weight of
Per Unit Weight of
•
Molecu-
lar
Molecu-
lar
Metal
Oxygen
Oxide
Metal
Oxygen
Oxide
(Th, Oj)
326.000
1.364
10,188
1,235
(Ml!, 0.)'
472,000
1.655
9,850
1,417
(Lai, Os)
444,700
1.602
9.265
1,364
(Nd2, 0.)
435.100
1,506
9.006
1,291
(Mg, 0)
143.400
5,975
8.963
3,586
148.800
6,200
9.300
3.720
(Pn, 0.)
412,400
1,467
8,592
1,297
(Lij, 0
135,800
9,700
8,488
4.627
167.000
11,929
10,437
5,667
(Ba, 0)
133,400
974
8.338
872
161.500
1.179
10,938
1.051
(Ca. 0)
131,500
3,288
8.219
2,348
149,600
3,740
9,350
2.671
(Sr. 0)
131,200
1,508
8.200
1.274
168,400
1,821
9,900
1,538
(Ah, Oi)
392,600
7,270
8.179
3,849
(V2, 0,)
353,200
3,463
7.358
2,355
(Ce, OO
224,600
1,603
7,019
1,308
(Ti. O2)
218,400
4,550
6,825
2,730
(U., Os)
845.200
1.184
6,603
1,004
(V, 0)
104,300
2,045
6.519
1,557
(U. 0.)
303.900
1.277
6,341
1,063
(Naj, 0)
100,400
2.183
6,275
1.620
155,900
3,389
9,744
2,516
(K2. 0)
98,200
1.259
6.138
1.045
166,200
2,118
10,325
1.758
(Si. 0.)
196,000
7.000
6,125
3,267
(Mn, 0)
90,900
1.653
6.681
1.280
(B.. 0.)
272,600
12.391
6,679
3,894
279,900
12.723
6,831
3,999
(Vj. 00
447.000
4.324
5,588
2,423
(Zr, Oi)
177,500
1.972
5.547
1.455
(Zn, 0)
84,800
1.306
5.300
1.058
(Mn,, 00
328,000
1.988
6,125
1.434
(Crz. 0.)
243,900
2.345
5,081
1,605
(Li,, 00
152,650
10.904
4.833
3,318
169,840
11.417
4.995
3,474
(Pj, Ot)
365.300
5.895
4.566
2.572
400,900
6.466
5,013
2,823
(Ba. 0=)
145,500
1,062
4,547
861
(Mo, Oj)
142,800
1.488
4,463
1,116
(Sn, 0)
70,700
599
4,419
527
(Sn. Oi)
141,300
1.197
4.416
942
(CO, 0) (gas)
68,040
2,430
4.253
1,546
73,940
2.641
4,621
1.680
solid
70.400
35,200
4.400
3.911
(Hs, 0) liquid
69.000
34.500
4.313
3.833
gas
68,060
29,030
3.629
3.226
(Pe.. 0.)
270,800
1,612
4.231
1.167
(Cd. 0)
66,300
592
4,144
518
(Pe. 0)
65,700
1,173
4.106
913
(W. Oi)
131,400
714
4.106
620
(W, 0,)
196,300
1,067
4.089
846
(Pes. 0.)
195,600
1,746
4.075
1.223
(Co. 0)
64.100
1,086
4.006
855
(Mn, O2)
125,300
2,278
3.916
1.440
' Mi = Misohmetal— a mixture of rare earth metals sold commercially, containing
cerium, thorium, yttrium, etc.
APPLICATIONS OF THERMOCHEMISTRY .
19
Heat of Formation of Oxides. (Continued)
Anhydrous
To Dilate Solution
Formula
Molecu-
lar
Per Unit Weight of
Molecu-
lar
Per Unit Weight of
Meal
Oxygen
Oxide
Metal
Oxygen Oxide
(Ni, 0)
61,500
1,051
3,844
826
(Naz, O2)
119,800
2,604
3,744
1,536
(Mo, 0.)
175,000
1,823
3,646 ■
1,215
(Sbj, Oa)
166,900
695
3,479
580
(Sb2, O4)
209,800
874
3,278
690
(As!, Oj)
156,400
1,043
3,258
790
148,900
993
3,102
752
(Pb. 0)
50,800
245
3,175
228
(C. 00 (gas)
97,200
8,100
3,038
2,209
103,100
8,592
3,222
2,343
(Cr, O3)
140,000
2,692
2,917
1,400
(Bis, 0,)
139,200
335
2,900
300
(Sb2, Os)
231,200
963
2,890
.723
(As2, Of)
219,400
1,463
2,743
954 :
225,400
1,503
2,818
980
(Cu2. 0)
43,800
344
2,738
306
(TI2. 0)
42,800
105
2,675
101
39,700
97
2,481
94
(Te. 00
78,300
624
2,447
497
(Cu, 0)
37,700
593
2,356
474
(S, O2)
69,260
2,196
2,196
1,098
77,600
2,425
2,425
1,213
(Pb, Os)
. 63,400
306
1,981
265
(S, Oa) (gas)
91,900
2,872
1,915
1,149
141,000
4,406
2,938
1,763
(TI2, 0.)
87,600
215
1,825
192
(C, O) (gas)
29,160
2,430
1,823
1,041
(H2, O2)
47,300
23,650
•1,478
1,391
(Hg2, 0)
22,200
56
1,388
53
(Hg, 0)
21,500
108
1,344
100
(Pd, O)
21,000,
198
1,313
172
(Pt, 0)
17,000
87
1,063
81
(Ag2, 0)
7,000
32
438
30
(Au2, Oa)
-11,500
-29
-240
-26
Heat OF Formation of Hydrates
A. From the Elements
Formula
Anhydrous
To Dilute Solution
and
Per Formula
Per Unit Weight
Per Formula
Per Unit Weight
Weight
of Metal
Weight
of Metal
(Li, 0, H)
112,200
16,030
118,000
16,860
.(Mg, 0,, HO
217,800
9,075
(Sr, 0„ HO
217,300
2,498
227,400
2,614
(Ca, O2, H,)
215,600
5,390
219,500
5,488
(K, 0, H)
104,600
2,682
117,100
3,003
(Na, 0, H)
102,700
4,465
112,450
4,889
(Al, O3, Ha)
301,300
11,160
(N, H,, 0, H)
88,800
90,000
5,294
(Zn, O2, HO
151,100
2,325
solid
70,400
(H, 0, H)
liquid
gas
69,000
58,060
(Tl, 0, H)
57,400
260
54,300
246
(Bi, O3. Ha)
171,700
825
20
METALLURGICAL CALCULATIONS.
Heat of Formation of Hydrates-
A. From the Elements
-Continued
- Formula
Anhydrous
To Dilute Solution
and
Reaction
Per Formula
Weight
Per Unit Weight
of Metal
Per For-nula
Weight
Per Unit Weight
of Metal
(Te, O4, HO
(Te, O3, HO
(Se, O3, H2)
(Se, O4, HO
(Tl, Oa, Ha)
145,600
128,220
78,700
1,156
1,623
386
166,740
124,600
145,020
1,323
1,576
1,836
B. From Metal, Oxygen and Water
Anhydrous
To Dilute Solution
Formula
and
Reaction
Per
Formula
Weight
Per Unit
Weight of
Metal
Per Unit
Weight of
Oxygen
Per
Formula
Weight
Per Unit
Weight of
Metal
Per Unit
Weight of
Oxygen
{U„ 0, H2O)
155,400
11,100
9,710
167,000
11,930.
10,440
(Mg, 0, H2O)
148,800
6,200
9,300
(Sr, 0, H2O
148,300
1,705
9,280
158,400
1,822
9,900
(Ca, 0, H2O)
146,600
3,665
9,162
150,500
3,763
9,406
(K2, 0, H2O)
140,200
1,797
8,762
165,200
2,118
10,325
(Naa, 0, H,0)
136,400
2,965
8,525
155,900
3,389
9,744
(AU, O3, 3H2O)
395,600
7,326
8,242
(Zn, 0, HaO)
82,100
1,263
5,131
(TU, 0, H2O)
45,800
112
2,863 .
39,600
97
2,475
(Bia, Oa, 3H2O)
136,400
328
2,842
(Te, O2, H2O)
76,600
6,079
2,394
(Te, O3, H2O)
97,740
776
2,036
(Se, O2, H2O)
■ 65,500
703
1,735
(Se, Oa, H2O
59,220
750
1,234
76,020
962
1,583
(Tl,, O3, 3H2O)
86,400
212
1,800
C
From
Metallic Oxide and HiO
Anhydrous
To Dilute Solution
Formula
and .
Per
Formula
Weiiht
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Reaction
Metallic
Oxide
H2O
Hydrate
Metallic
Oxide
Com-
bined
H,0
Dis-
solved
Hydrate
(Li! 0, HiO)
19,600
■ 653
1,089
408
31,200
1,040
1,733
650
CMgO, HiO)
5,400
135
300
93
(SrO, H2O)
17,100
166
950
141
27,200
264
1,511
225
(CaO, H2O)
15,100
270
839
204
19,000
339
1,056
257
(KsO, H2O)
42,000
447
2,333
375
67,000
713
3,722
598
(NajO, H2O)
36,000
581
2,000
450
55,500
895
3.083
694
(N H.)zO, H^O
(AUOj, 3H2O)
3,000
30
56
19
(ZnO, HjO)
-2,700
-33
-150
-27
(TI2O, H2O)
3,000
7
167
7
-3,200
-7
-178
-7
(BizOs, 3H2O)
-2,800
-7
-62
-6
(TI2OJ, SH'O)
-1,200
-3
-22
-2
APPLICATIONS OF THERMOCHEMISTRY.
21
Heat of
Formation of Sulfides
Anhydrous
To Dilute Solution of
Formula
Per
Per Unit Weight of
Per
Per Unit Weight of
rorfnulo
' . FoTTnuld
Weight
Metal
Sulfur
Sulfide
Weight
Metal
Sulfur
Sulfide
(Lis, S)
(Kj. S)
115,400
8,243
3,606
2,509
103,500
1,327
3,234
941
113,500
1,455
3,547
10,319
(Ba, S)
102,900
751
3,216
609
109,800
802
3,431
474
(Sr, S)
99,300
1,141
3,103
835
106,700
1,225
3,334
897
(Naa, S)
89,300
1,941
2,791
1,145
104,300
2,267
3,269
1,337
(Nd2, S3)
285,900
993
2,977
744
(Ca, S)
94,300
2,358
2,947
1,310
100,600
2,515
3,144
1,397
(Mg, S)
79,400
3,308
2,481
1,418
(Mn, S)
45,600
829
1,425
524
(Zn, S)
43,000
662
1,344
443
(Ab, Sa)
126,400
2,341
1,316
843
(N, Hs, S)
40,000
2,105
1,250
776
36,700
1,931
1,147
720
(Cd, S)
34,400
307
1,075
239
(K, S.)
59,300
1,521
927
576
59,700
1,531
933
579
(Na, S,)
49,500
2,152
773
556
54,400
2,365
850
611
(B', S3)
75,800
3,445
790
642
(Pe, S)
24,000
428
760
273
(Co, S)
21,900
371
685
241
(TI2. S)
21,600
106
675
92
(CU2, S)
20,300
160
634
127
(Pb, S)
20,200
98
631
85
(Si, S2)
40,000
1,429
625
435
(Ni, S)
19,500
333
609
215
(Sb2, S3)
34,400
143
358
102
(Hg, S)
10,600
63
331
46
(Cu, S)
10,100
159
316
106
(H2, S)
(gas)
4,800
2,400
150
141
9,500
4,750
297
279
(Ag2, S)
3,000
14
94
12
(C, Sj)
(gas)
-25,400
-2,117
-397
-334
(liquid)
-19,000
-1,583
-297
-250
(I,S)
0
0
0
0
22
METALLURGICAL CALCULATIONS
Heat of Formation of Selenides
Formula
Anhydrous
Per
Formula
Weight
Per Unit Weight of
Metal
Selen- Selen-
ide
To Dilute Solution
Per
Formula
Weight
Per Unit Weight of
Metal
Selen' Selen-
ide
(Lii, Se)
(Ks, Se)
(Ba, Se)
(Sr, Se)
(Na2, Se)
(Ca, Se)
(Zn, Se)
(Cd, Se)
(Mn, Se)
(N, Ht. Se)
(Cu, Se)
(Pb, Se)
(Fe, Se)
(Ni, Se)
(Co, Se)
(Th, Se)
(Cu2, Se)
(Hg, Se)
(Ag!, Se)
(Hj, Se) (gas)
(N, Se) :
83,000
79,600
69,900
67,600
60,900
58,000
30,300
23,700
22,400
17,800
17,300
17,000
15,200
14,700
13,900
13,400
8,000
6,300
2,000
-25,100
-42,300
5,929
1,021
510
777
1,324
1,450
466
212
407
937
272
82
262
251
236
33
63
32
93
- 12,550
-3,021
1,051
1,008
885
856
720
727
384
300
284
225
219
215
192
186
178
170
101
80
25
-318
-534
892
507
324
408
487
487
210
124
167
184
114
59
111
107
101
21
39
23
7
-310
-455
93,700
87,900
78,600
12,800
6,693
1,127
1,709
1,186
1,113
995
1,008
560
629
674
- 15,800
-7,900
-200
-195
Heat of Formation of Tellurides
Anhydrous
Formula
Per
Per Unit Weight <
>/
Formula
Weight
Metal
Tellurium
Telluride
(Zn, Te)
31,000
477
246
162
(Cd, T^)
16,600
148
132
70
(Co, Te)
13,000
220
103
70
(Pe. Te)
12,000
214
95
66 .
(Ni, Te)
11,600
198
92
63
(Tl., Te)
10,600
26
84
20
(Cu2, Te)
8,200
64
65
32
(Pb, Te)
6,200
30
49
19
(Hj, Te) (gas)
-34,900
-17,450
-277
-272
APPLICATIONS OF THERMOCHEMISTRY.
23
Heat of Formation of Arsenides
Per
Formula
Weight
Per Unit Weight of
Metal
Arsenic
Arsenide
(H3, As) (gas)
-44,200
-14,733
-689
-567
Heat of Formation of Antimonides
Per
Formula
Weight
Per Unit Weight of
Metal
Antimony
Aniimonide
(H3, Sb) (gas)
-86,800
-28,933
-723
-706
Heat of Formation of Phosphides
Per
Formula
Weight
Per Unit Weight of
Metal
Phosphorus
Phosphide
(Mna, P2)
(Ha, P) (gas)
(Fe, P)
70,900
4,900
+ 0
430
1,633
±0
1,144
158
±0
312
144
±0
Heat of Formation of Nitrides
Anhydrous
To Dilute Solution
Formula
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
Nitrogen
Nitride
Base
Acid
Com-
pound
(Bas, Ni)
149,400
363
5,336
340
(Mga, Nj)
119,700
1,662
4.275
1.197
(Cas, NO
112,200
935
4,007
758
(Lis, N)
49,500
2,357
3,536
1,414
(Al, N)
45.450
1,683
3,246
1,109
(Li, N)
18,750
2,679
1,339
• 893
(Hs, N) (gas)
12,200
4.067
871
718
21,000
7,000
1,500
1,235
(liquid)
16,600
5,533
1,186
976
(P., N.)
81,500
876
1,164
500
24
METALLURGICAL CALCULATIONS.
Heat of Formation of Hydrides
Per
Per Unit Weight of
Per
Per Unit Weight of
Formula
Formula
Weisht
Fromula
Weight
Metal
Hydro-
gen
Hy-
dride
Base
Acid
Com-
pound
(Ca, H2)
46,200
1,155
23,100
1,100
(Li, H)
21,600
3,086
21,600
2,700
(Sr, HO
38,400
" 441
19,200
431
(Ba, H2)
37,500
274
18,750
271
(Ptit, H)
16,950
6
16,950
6
(Ptio, H)
14,200
7
14,200
7
(P12, Hs)
53,400
144
8,900
141
(Pdi6, H)
4,600
3
4,600
3
<P, H3) (gas)
+4,900
158
+ 1,633
144
(Si, HO (gas)
-6,700
-239
-1,675
-209
(Ni, HO
-19,000
-339
-4,750
-317
-26,100
.-466
-6,525
-435
(As, HO
-44,200
-589
-14,730
-567
(Sb, HO
-86,800
-723
-28,930
-706
Heat of Formation of Hydrocarbons
{From Amorphous Carbon, to Gas, Unless Otherwise Specified)
Formula
Name
State
Per
Formula
Weight
Per Unit Weight of
Reaction
Hydro-
Com-
gen
pound
(C, HO
Methane (Marsh gas)
Gas
+21,700
+ 1,808
+5,425
+ 1,356
(C2, HO
Acetylene
Gas
-52,300
-2,179
-26,150
-2,016
(Ci, HO
Ethylene (olefiant gas)
Gas
-8,700
-363
-2,175
-311
(C2, HO
Ethane (ethylene hydride)
Gas
+29,100
+ 1,213
+4,850
+970
(Ca. HO
AUylene
Gas
-44,000
-1,222
-11,000
-1,100
(C3, HO
Propylene
Gas
-700
-22
-133
-19
(C HO
Propane (propylene hydride)
Gas
+39,200
+ 1,089
+4,900
+891
(Ci, HO
Benzene
Gas
+7,200
+ 100
+ 1,200
+92
Liquid
+ 14,400
+200
+ 2,400
+ 185
(Ct, ho
Toluol
Liquid
+21,900
+261
+ 2,738
+238
(Cio, HO
Naphthalene
Liquid
+ 9,100
+76
+ 1,138
+ 71
Solid
+ 13,700
+ 114
+ 1,713
+ 107
(Cio, HiO
Turpentine
Gas
+23,800
+ 197
+ 1,488
+ 175
Liquid
+33,200
+277
+2,075
+244
(Ci4, HiO
Anthracene
Sohd
-2,600
-16
-260
— 15
(C, H4,0)
Methyl alcohol (wood alcohol)
Gas
+56,000
4,667
14,000
1,750
Liquid
+65,050
5,420
16,260
2,033
(C2, Hb, 0)
Ethyl alcohol (grain alcohol)
Gas
+64,200
2,675
10,700
1,396
Liquid
+74,900
3,120
12,480
1,628
(Cj, Ho, 0)
Acetone
Gas
+ 63,150
1,754
10,525
1,089
Liquid
+71,300
1,980
11,880
1,229
APPLICATIONS OF THERMOCHEMISTRY.
25.
Heat
OF Combustion
OF Hydrocarbons
Formula
Name
state
Per Formula Weight
Calor-
ies per
Calor-
ies per
Kg
B.T.U.
per
ft.'
B.T.U.
Calorime-
ter Value
Practical
Value
Water formed not condensed
CHi
Methane
Gas
213,500
191,620
8,623
11,976
970
21,555
CaH2
Acetylene
Gas
315,700
304,760
13,714
11,722
1,543
21,000
CjH4
Ethylene
Gas
341,100
319,220
14,365
11,400
1,616
20,520
C2H6
Ethane
Gas
372,300
339,480
15,277
11,315
1,719
20,365
CaH.
Allylene
Gas
473,600
451,720
20,327
11,295
2,287
20,150
CsHc
Propylene
Gas
499,300
466,480
20,992
11,105
2,362
19,995
CaH.
.Propane
Gas
528,400
484,640
21,812
11,015
2,464
19,825
CsH.
Benzene
Gas
783,000
750,180
33,758
9,618
3,798
17,312
Liquid
775,800
742,980
9,525
17,145
C7H8
Toluol
Liquid
934,500
890,740
9,682
17,430
CioHs
Naphthalene
Liquid
1,238,900
1,195,150
9,337
16,805
Solid
1,234,300
1,190,550
9,300
16,740
CioHi.
Turpentine
Gas
1,500,200
1,412,680
63,570
10,385
7,152
18,695
Liquid
1,490,800
1,403,280
10,318
18,570
CuHio
Anthracene
Solid
1,708,400
1,653,700
9,290
16,545
CHiO
Methyl alcohol
Gas
179,200
157,320
7,079
4,919
786
8,854
Liquid
170,150
148,270
4,633
8,340
C2H(iO
Ethyl alcohol
Gas
337,200
304,380
13,700
6,617
1,541
11,910
Liquid
326,500
293,680
6,384
11.490
CsHeO
Acetone
Gas
435,450
402,630
18,120
6,942
2,038
12,495
Liquid
427,300
394,480
6,801
12,240
Heat of Formation of Carbides
Per Unit Weight 0
f
Per Formula
Weight
Formula
Metal
Carbon
Carbide
(AI4, Cs)
232,000
2,148
6,446
1,815
(Si, C)
26,520
947
2,210
663
(Mns, C) .
9,900
60
825
67
(Pea, C)
8,460
50
705
47
(Ca, C^)
-7,250
181
302
113
(Na, C)
-4,400
191
367
126
(Li, C)
-5,750
821
479
303
(Ag, C)
-43,575
403
3,631
363
(N2, C2)
-73,000 (gas)
-2,607
-3,042
-1,404
.26
METALLURGICAL CALCULATIONS
Heat of Formation of Silicides
Per Formula
Weight
Per Unit Weight of
Metal
Silicon
suicide
Mn,, Siz
47,400
123
1,693
115
Heat of Formation of Fluorides
Anhydrous
To Dilute Solution
Formula
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
Fluorine
Com-
pound
Metal
Fluorine
Com-
pound
(Sr. P,)
(Ba, Fd
(Li, Pi)
(K.?)
(Ca, Pz)
(Mg, Pa)
(Na, P)
(N, Hi, F)
(Al. Fa)
(B, Ps)
(Mn, Fa)
(Zn, Fa)
(Si, Pi)
224,020
224,000
2,575
1,635
5,898
5,895
1,792
1,280
221,500
116,880
113,600
1.617
16.696
2,913
5,829
6,151
5,979
1,266
4,496
110,000
216,450
209,500
109,720
101.250
2,821
5,411
8,729
4,770
5,625
5,789
5,696
5,513
5,775
5,329
1,897
2,775
3,379
2,612
2,736
1.959
109,120
99,750
275,220
219,345
153,310
138,220
4,744
5,542
10,193
19,940
2,787
2,127
5,743
5,250
4,828
3,848
4.034
3,637
2,598
2,696
3 276
234,800
21,345
4,119
3.453
3,226
1 648
1,342
275,920
9,854
3,631
2,653
(Pe, Fi)
125,220
121,720
120,340
118,980
164.940
54,405
2.236
1.087
2.040
2.034
2,947
267
3,295
3,203
3,167
3,131
2,894
2,863
.1,332
811
1,241
1,233
1.4?0
(Cd, Pz)
(Co, FO
(Ni, Ps)
(Fe, Fi)
(Tl, P)
(Pb, P,) • '
101,600
38,500
491
38,500
2,674
2,026
415
1,925
(H, F)
(Sb. Ps)
50,300
136,680
88,160
25,470
50,300
1,139
1,386
236
2,647
2,398
2.320
1.341
2,515
772
868
201
(Cu, Fj)
(Ag. P)
22,070
204
1,162
174
APPLICATIONS OF THERMOCHEMISTRY
Heat of Formation of Chlorides
27
Anhydrous
To Dilute Solution
Formula
Per
Per Unit Weight of
Per
Per Unit Weight 0/
Formula
Weight
Formula
Weight
Metal
Chlo-
rine
Chloride
Metal
Chlo-
rine
Chloride
(Cs, CI)
109,860
826
3,095
652
105,110
790
2.961
624
(Rb, CI)
105,940
1,239
2,983
875
101,210
1,184
2.851
836
(K, CI)
105,700
2,710
2,977
1,419
101,200
2,595
2,879
1.358
(Be, Ch)
155,000
17,222
2,183
1,938
199,500
22,167
2,810
2.494
(fia, CI2)
197,100
1,439
2,776
948
198,300
1,447
2,793
953
(Na, CI)
97,900
4,257
2,758
1,671
96,600
4,200
2,721
1,651
(Li. CI)
93,900
13,414
2,645
2,209
102,300
14,614
2,882
2,407
(Sr, CU)
184,700
2,123
2,601
1,169
195,850
2.251
2,758
1,240
(Ca, CU)
169,900
4,248
2,393
1,531
187,400
4,685
2,639
1,688
(Nd, Cli)
249,500
1,738
2,344
998
284,900
1,985
2,675
1.140
(N, H4, CI)
76,800
4.267
2,192
1,209
72,800
4,044
2,051
1.146
(Mg, CU)
151,200
6,300
2,129
1,592
187,100
7,796
2,635
1,969
(Th, CI.)
■300,200
1,294
2,114
803
(V, CU)
186,900
3.665
1,755
1,187
(Al. CU)
161,800
5,993
1.519 ■
1.212
238,100
8,819
2,236
1,784
(Mn, CU)
112,000
2,037
1.577
889
128,000
2,327
l',803
1,016
(Zn, Ch)
97,400
1,498
1,372
716
113,000
1.7.S8
1.592
831
(Tl, CI)
48,600
238
1,369
203
38,400
188
1.082
1,060
(Cd, CU)
93,700
837
1,320
512
96,400
861
1,358
527
(Pb. CU)
83,900
405
1,182
302
77,900
376
1,097
280
(Fe, CU)
82,200
1,468
1,158
647
100,100
1.788
1,410
788
(Sn. CU)
80,900
686
1,139
428
(Co, CU)
76,700
1,300
1,080
590
95,000
1.610
1,338
731
(Ni, CU)
74,700
1.277
1,052
577
93,900
1.605
1,323
725
(Cu. CI)
35,400
557
997
358
(Sn, CU)
129,8001
1;100
914
499
(Si, CU)
128,800"
4,600
907
758
(Fe, CU)
96,150
1,717
903
592
127,850
2,283
1,200
787
(Hg, CI)
31,320
157
882
133
(Sb, CU)
91,400
762
858
404
(Bi, CU)
90,800
437
853
289
(B, CU)
89,100
8,100
837
758
(Ag, CI)
29,000
269
817
202
(Hg, CU)
53,300
267
751
197
50.300
252
708
185
(Cu, CU)
51,400
80S
724
382
62,500
983
880
464
(As, CU)
71,500'
953
671
394
(H. CI)
22,0002
22,000
620
603
39,400
39,400
1,110
1,080
(Sb. CU) ;
104,500'
871
589
351
(Pd, CU)
40,500
382
570
229
(Te, CU)
77,380
614
545
289
(Pt, CU)
60,200
308
424
179
79,800
409
562
237
(Se, CU)
46,160
584:
325
209
(Se, CU)
22.150'
280'
312
148
(Sj, Ch)
17.600'
10,900>
275
170
248
154
: 130
81
(Aa. CU) .
22,800
116
216
75
27,200
138
255
90
(Au, CI)
5,800
29
163
25
i
' Liquid.
2 Gas.
28
METALLURGICAL CALCULATIONS.
Heat of Formation of Carbonates
A. From the Elements
Anhydrous
To Dilute Solution
Per Formula
Per Unit
Per Formula
Per Unit
Weight
Weight of Metal
Weight
Weight of Metal
(Ba, C, O3)
286,300
2,090
(K2, C, O3)
282,100
3,617
288,600
3,700
(Sr, C, Os)
281,400
3,234
(Ca, C, O3)
273,850
6,846
(Na2, C, O3)
273,700
5,950
279,300
6,072
(Mg, C, 0,)
269,900
11,245
(Mn, C, O3)
210,300
3,824
(N, Hs, C, Os)
208,600
10,980
202,300
10,640
(Zn, C, O3)
197,500
3,038
(Fe, C, O3)
187,800
3,354
(Cd, C, Os)
183,200
1,636
(Pb, C, O3)
170,000
821
(Cu, C, Os)
146,100
2,301
(Aga, C, O3)
123,800
593
B. From Metal, Oxygen and CO^
To Dilute Solution
Formula
and
Reaction
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
Oxygen
Metal
Oxygen
(Ba, 0, CO2)
(Kj, 0, CO2)
189,100
184,900
184,200
176,650
176,500
172,700
113,100
111,400
100,300
90,600
86,000
72,800
48,900
26,600
1,380
2,370
2,117
4,416
3,837
7,195
2,056
5,863
1,543
1,618
768
352
770
123
191,400
182,100
105,100
2,454
3,959
5,532
(Sr, 0, CO2)
(Ca, 0, CO.)
(Naa, 0, CO2)
(Mg, 0, CO2)
(Mn, 0, CO2)
(N, Hs, 0, CO2)
fZn, 0, CO2)
(Pe, 0, CO2)
(Cd, 0, CO2)
(Pb, 0, CO2)
(Cu, 0, CO2)
(Ag2, 0, CO2)
APPLICATIONS OF THERMOCHEMISTRY.
29
C. From Metallic Oxide and CO2
Anhydrous
To Dilute Solution
and
Per Unit Weight of
Per Unit Weight of
Reaction
Per
Formula
Per
Formula
Weight
Metallic
COz
Car-
Weight
Metallic
Car-
Oxide
bonate
Oxide
COi
bonate
(BaO, CO2)
55,700
364
1,266
282
(KsO, CO2)
86,700
922
1,970
628
93,200
992
2,118
675
(SrO, CO!)
53,000
515
1,205
361
(CaO, CO2)
45,150
806
1,026
451
(NazO, CO2)
76,100
1,219
1,718
713
81,700
1,318
1,857
771
(MgO, CO2) .
29,300
733
666
349
(MnO, CO2)
22,200
313
505
193
(NH5, 0, C02)
22,600
646
514
286
16,300
466
370
206
(ZnO, C02)
15,500
191
352
124
(FeO, CO2)
24,900
346
566
215
(CdO, CO2)
19,700
154
448
115
(PbO, CO2)
22,000
99
500
83
(CuO, CO2)
11,200
140
255
91
(Ag20, CO2)
19,600
82
445
71
Heat of Formation of Bicarbonates
A, From the Elements
Formula
Anhydrous
To Dilute Solution
and
Reaction
Per Formula
Weight
Per Unit
Weight of
Metal
Per Formula
Weight
Per Unit
Weight of
Metal
(K, H, C, O3)
(Na, H, C, O3)
(N, Hi, H. C, O3)
233,300
227,000
205,300
5,987
9,870
228,000
222,700
199,000
5,846
9,683
B. From Metal, Oxygen, H2O and COi
Anhydrous
To Dilute Solution
Formula and
Per
For-
mula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
CO2
Metal
CO2
Com-
pound
(K2, 0, H20, 2CO2)
(Naj, 0, H2O, 2CO2)
203,200
190,600
2,605
4,143
2,309
2,166
1,016
1,135
192,600
182,000
2,469
3,957
2,189
2,036
963
1,083
C. From Metallic Oxide, H2O and COi
Anhydrous
To Dilute Solution
Formula
Per
For-
mula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metallic
Oxide
CO2
Com-
pound^
Metallic
Oxide
CO,
Com-
pound
(K2O, H20, 2CO2)
(Na20, HiO, 2CO2)
105,000
90,200
1,117
1,961
1,193
1,025
525
537
94,400
81,600
1,004
1,774
1,073
927
472
486
30
METALLURGICAL CALCULATIONS.
Heat of Formation of Nitrates
A. From the Elements
Anhydrous
To DUute
Solution
Per Formula
Per Unit Weight
Per Formula
Per Unit Weight
Weight
of Metal
Weight
of Metal
(K, N, Ob)
119,000
3,051
■
110,700
2,582
(Ba. Nz, 0.)
227,200
1,658
217,800
1,590
(Li, N, Os)
111,610
15,945
111,910
15,985
(Na, N. O3)
110,700
4,813
106,000
4,609
(Sr. Nz, Oe)
219,800
2,726
214,700
2,468
(Ca, Na, Oe)
202,000
5,050
206,000
5,150
(Mg, N2, 0()
206,300
8,596
(Mn, Nj, Ob)
146,900
2,671
(Zn, N2, Ob)
131,700
2,022
(Tl, N, Os)
58,150
285
48,180
236
(Fe, fl2, Ob)
119,000
2,125
(Co, N2, Ob)
115,720
1,961
(Ni, Ni, Ob)
113,230
1,936
(Pb", N2, Ob)
105,400
509
98,200
474
(Cu, N2, Ob)
82,230
1,295
(Fe, Ns, Os)
157,150
2,S06
(H, N, Oa) (gas)
34,400
34,400
48,800
48,800
(Hg, N, Ob)
28,900
145
(Hg, N2, Oe
54,000
270
(Ag, N, O3)
28,700
275
23,000
213
Heat of Formation of Nitrates
B. From Metallic Oxide and NiOb
'
Anhydrous
To Dilute Solution
Formula
and
Reaction
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metallic
Oxide
2V2OB
Nitrate
Metallic
Oxide
NiOi
Nitrate
(K2O, N2OB)
(BaO, N2OB)
(Li20, NjOb)
(Na20, NsOi)
(SrO, N2O6)
(CaO, N2OB)
(MgO, NiOb)
(MnO, N2OB)
(ZnO, N20t)
(TI2O, N2OB)
(FeO, N2OB)
(CoO, N2OB)
(NiO, N2OB)
(PbO, N2O5)
(CuO, NiOs)
(Fe203, 3N20b)
(H2O, NjOt)
(HgiO, N2OB)
<HgO. N2O.)
(Ag20,NiOB)
141,000
92,600
86,220
119,800
87,400
69,300
72,300
53,400
-1,400
49,200
1,50C
60.
2,87^
l,93i
84!
1,23!
171
24(
-7!
21!
1,306
857
798
1,109
809
642
669
494
-13
456
69E
354
58:
70£
414
42C
ise
16]
-1
14!
>
124,400
83,200
86,820
110,400
82,300
73,300
61,600
54,800
45,700
52,360
52,100
50,420
50,530
46,200;
43,330
115,100
27,400
34,400
31,300
37,800
1,323
544
2,894
1,781
799
1,309
1,540
772
564
124
724
672
678
207
545
719
1,522
827
145
163
1,152
770
804
1,022
762
679
570
507
423
485
482
467
468
428
401
1,066
254
319
290
350
616
319
587
649
390
447
416
306
242 .
98
274
276
277
140
231
428
214
656
97
111
APPLICATIONS OF THERMOCHEMISTRY.
31
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32
METALLURGICAL CALCULATIONS.
Heat of Formation of Sulfates
A. From the Elements
Formula
(Kj, s, oo
(Ba, S, Oi)
(Lia, S, OO
(Sr, S, Oi)
(Naa, S, OO
{Ca,S, OO
(Mg, S, OO
(Ala, Sa, Oii)
(Na. Ha, S, OO
(Mn, S, OO
(Zn, S, 00
(Pe, S, OO
(Co, S, OO
(Ni, S, OO
(Fez, Ss, Oia)
(Tla, S, OO
(Cd, S, OO
(Pb, S, OO
(Ha, S, OO
(Cu, S, OO
(Hga, S, OO
(Aga, S, OO
(Hg, S, OO
Anhydrous
Per Formula
Weight
344,300
339,400
333,500
330,200
328,100
317,400
300,900
283,500
249,400
229,600
221.800
219,900
215,700
192,200
181,700
175,000
167.100
165,100
Per Unit Weight
of Metal
4.414
2.477
23.820
3.795
7,133
7,935
12,540
7.875
4.535
3.532
544
1.963
1.042
96,100
2,861
438
774
826
To Dilute Solution
Per Formula
Weight
337,700
339,600
328,500
321,800
321,100
879,700
281,100
263,200
248,000
234,900
228,900
228,700
650,500
213,500
231.600
210,200
197,500
162,600
Per Unit Weight
of Metal
4.340
24.260
7,141
8,045
13.380
16.290
7.808
4,785
3.815
4.195
3.880
3.919
5.808
523
2,068
105,100
3,110
753
Heat of Formation of Sulfates
B. From Metallic Oxide and SOs
Pormula
and
Reaction
Anhydrous
To Dilute Solution
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metallic
Oxide
SOa
Sulfate
Metallic
Oxide
SOa
Sulfate
(KaO, SOa)
(BaO, SOs)
(LiaO, SOs)
(SrO, SO,)
(NaaO, SOs)
(CaO, SOs)
(MgO, SOa)
(AlaOs, 3SOs)
(MnO, SOs)
(ZnO, SOs)
(PeO, SOs)
(CoO, SOs)
(NiO, SOs)
(PeaOs, 3SOs)
(TlaO, SOa)
(CdO, SOa)
(PbO. SOs)
(HaO, SOs)
(CuO, SOs)
(HgaO, SOs)
(AgaO, SOO
(HgO. SOs)
154,200
114,100
105,800
107,100
135,800
94,000
65.600
66.600
52.900
87.100
61.700
73.000
31.300
52,100
60,900
68,200
51,700
1,640
745
3.527
1.040
2,190
1,679
1,640
938
653
205
482
328
1,739
655
146
294
239
1.928
1.426
1.322
1.339
1.698
1.175
820
832
661
1.089
759.
912
391
651
761
852
646
886
490
962
585
956
692
547
441
328
173
297
241
319
326
123
219
175
147.600
111,900
136,200
98.400
85,800
211,400
80,400
71,300
77,300
72,900
75,300
179,200
78,800
73,400
49,300
67,900
63,700
1,570
3,730
2.197
1.757
2,145
2,072
1,132
880
1,074
972
1,011
1,120
186
573
2,739
853
275
1,845
1,399
1,702
1,230
1.072
881
1.005
891
966
911
941
747
985
918
616
849
796
848
1,017
959
723
715
618
532
443
509
470
487
448
156
353
503
425
204
APPLICATIONS OF THERMOCHEMISTRY.
Heat of Formation of Bi-sulfates
33
Anhydrous
To Dilute Solution
Formula
and
Reaction
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
Bi-sulfate
Metal
Bi-sulfate
(K, H, S, 0.)
(Na, H, S, O4)
(N, Hi, S, GO
276,100
269,100
244,600
7,080
11,700
13,588
2,030
2,243
2,127
272,900
268,300
245,100
6,974
11,665
13,617
2,006
2,236
2,131
Heat of Formation of Phosphates
Anhydrous
A. Fiom the Elements
B. From Metallic Oxide and PiOs
Formula and
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Reaction
Metallic
Oxide
PzOt
Phos-
phate .
(Bas, P2, Os)
(Srs, P2, Os)
(Cas, P2. Os)
(Mgs. P, Os)
(Naa, P. 00
(Mna, P2, Oe)
951,700
945,000
919,200
910,600
452,400
737,500
2,316
7,130
7,660
12',647
6,557
4,470
(3BaO, P2O6)
(3SrO, P2O6)
(3CaO, P2OS)
(3MgO, P2O5)
(3Na20, P2OS)
(3MnO, P206)
186,200
186,100
159,400
115,100
238,300
99,500
406
602
949
959
1,281
467
1,311
1,311
1,123
811
1,678
701
310
413
514
439
727
280
Heat of Formation of Arsenites and Arsenates
A. From the Elements
Anhydrous
To Dilute Solution
Formula and
Per Forjmula
Per Unit Weight
Per Formula
Per Unit Weight
Reaction
Weight
of Metal
Weight
of Metal
(Na, As, O2)
158,050
6,872
(Nas, As, O4)
360,800
5,229
381,500
6,529
(Ka, As, Oi)
396,200
3,386
(Sra, Asa, Os)
761,000
2,915
(Caa, AS2, Os)
732,800
6,107
(Mgj, As2, Os)
712,600
9,897
(Baa, As!, Os)
629,200
1,531
Heat of Formation of Tungstates
Anhydrous
A. From the Elements
B. From Metallic Oxide and WOa
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Formula and
Reaction
Metallic
Oxide
W0>
Tung-
state
(Na2, W, 00
391,400
8,509
(Na20, WOa)
94,700
1,527
408
322
34
METALLURGICAL CALCULATIONS.
B. From Metallic Oxide and AsiOs or As20s
Anhydrous
To Dilute Solution
Formula
and
Reaction
Per
For-
mula
Weight
Per Unit Weight of
Per
For-
mula
Weight
Per Unit Weight of
Metallic
Oxide
AsiOa or
ASiOi
Com-
pound
Metallic
Oxide
AJ20» 01-
AssOb
Com-
pound
(NajO, AsjOs)
(3Na20, AszOs)
(3K2O, AssOe)
(3SrO, AszOb)
(3CaO, AsjOa)
(3MgO, AsjOe)
(3Ba!0, AsiOs)
874
643
502
274
42
483
302
299
151
14
59,300
242,400
278,400
9,565
1,303
987
300
1.054
1,210
227
201,000
148,000
118,900
63,000
9,600
/
1,081
567
708
525
21
583
544
Heat of
Formation of Borates. A. From the Elements
Anhydrous
To Dilute Solution
and
Reaction
Per Formula
Weight
Per Unit Weight
of Metal
Per Formula
Weight
Per Unit Weight
of Metal
(Naa, Bi. O7)
748.100
16,263
758,300
16,485
B. From Metallic Oxide and B2O3
Anhydrous
To Dilute Solution
Formula
and
Per
For-
mula
Weight
Per Unit Weight of
Per
For-
mula
Weight
Per Unit Weight of
Metallic
Oxide
BiO,
Borate
Metallic
Oxide
B203
Borate
(NazO, 2B20a)
102,500
1,653
732
507
112,700
1,818
805
558
Heat of Formation of Molybdates. Anhydrous
A. From the Elements
B. From Metallic Oxide and MoOs
Formula and
Reaction
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Metallic
Oxide
MoOi
Molyb-
date
(Na2, Mo, 0.)
363,800
7,909
(Na20, MoOs)
88,400
1,426
612
429
Heat of Formation of Titanates. Anhydrous
a. From the Elements
B. From Metallic Oxide and Ti02
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Reaction
Metallic
Oxide
TiOi
Titan-
ate
(Naj, Ti, 0.)
388,500
8,446
(NasO, TiO.)
69,700
1,124
871
491
APPLICATIONS OF THERMOCHEMISTRY.
35
Heat of Formation of Manganates and Per-Manganates
Anhydrous
A. From the Elements
B. From Metallic Oxide and MnsO?
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaaion
Per
Formula
Weight
Per Unit Weight o/
Reaction
Metallic
Oxide
MmO,
MniOi
Com-
pound
(Nas, Mn, Oi)
(K, Mn, OO
269,400
205,050
5,857
5,129
(NasO, MnO.)
(KjO, MniO,)
Heat of Formation of Aluminates
Anhydrous
A. From the Elements
B. From Metallic Oxide and AlzOi
Formula and
Reaction
Per
Formula
Weight
Per Unit
Weight
of Metal
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Metallic
Oxide
AhOt
Alumin-
ate
(Na, Al, O2)
(Ca. Ah, 0.)
(Caj, Ah, Os)
(Cai, Ah, 0.)
533,000
524,550
668,900
780,050
11,587
13,114
8,236
6,500
(NajO, AI2O.)
(CaO, Ah03)
(2CaO, AhO.)
(3CaO, AhOj)
40,000
450
3,300
-7,050
645
8
29
-42
394
4
32
-69
244
3
15
-26
Heat of Formation of
Cyanides
Anhydrous
To Dilute Solution
Formula and
Reaction
Per
Formula
Per Unit Weight of
Per
Formula
Per Unit Weight of
Weight
Metal
CN
Cyanide
Weight
Metal
CN
Cyan-
ide
(Ca, C2, NO
(K, C, N)
41,650
30,250
1,041
776
801
1,163
453
33,450
858
1,287
515
465
(Na, C, N)
25,950
1,128
998
530
25,450
1,107
979
519
(Sr, C2, N2)
(Ba, C2, N2)
47,000
540
904
338
48,300
353
929
256
(Ca, Ci, N2)
41,650
1,041
801
453
(Mg, Ca, N2)
34,000
1,417
654
447
(Ni, C2, N2)
-23,400
-400
-450
-212
(Zn, C2, N2)
-24,550
-378
-472
-210
(Fe7, Cis, Nis)
-256,700
-655
-549
-298
(Cd, C2, N2)
-31,850
-284
-612
-194
(Cu, C, N)
-20,375
-320
-784
-227
(Pd, Cj, N2)
-49,250
(gas)
-465
-947
-312
(H, C, N)
-27,150
-27,150
-1,044
-1,006
-21,050
-21,050
-810
-781
(Hg, C2, N2)
-59,150
-296
-1,137
-235
(Ag, C, N)
-34,000
-315
-1,308
-254
36
METALLURGICAL CALCULATIONS.
Heat of Formation of Cyanates
A. From the Elements
Anhydrous
To Dilute Solution
Formula and
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
• Cyanate
Metal
Cyanate
(K, C, N, 0)
(Na, C, N, 0)
(Ag, C, N, 0)
(Hg, C2, N2, O2)
105,850
105,0p0
26,450
-62.900
2,714
4,667
245
-315
1,307
1,616
176
-221
100,650
100,250
2,581
4,359
1,242
1,542
B. From Cyanide and Oxygen
Anhydrous
To Dilute Solution
Formula and
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Cyan-
ide
Oxygen
Cyanate
Cyan-
ide
Oxygen
Cyanate
(KCN, 0)
(NaCN, 0)
(AgCN, 0)
(HgCjNj, O2)
72,400
79,100
60,450
-3,750
1,116
1,614
451
-15
4,525
4,944
3,781
-117
894
1,217
403
-13
67,200
74,300
1,034
1,516
4,200
4,644
830
1,143
Heat of Formation of Metallo-cyanides
A. From the Elements
Anhydrous
To Dilute Solution
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Per
Formula
Weight
Per Unit Weight of
Metal
CN
Cyanide
Metal
CN
Cyanide
(Ki, Fe, C, Ne)
(H4, Fe, Cj, Ns)
(Ka, Fe, Cc, N.)
(Hs, Fe, Cs, Ne)
157,300
-102,000
129,600
1,007
-25,500
1,108
1,007
-652
831
428
-472
394
69
145,300
-101,600
100,800
-127,400
5,350
931
-25,375
862
-42,467
137
931
-650
647
817
103
395
-470
307
593
27
(K, Ag, C2, Nj)
13,700
351
264
B. From the Constituent Cyanides
Anhydrous
To Dilute Solution
Formula and
Reaction
Per
Formula
Weight
Per UMt Weight of
Per
Formula
Weight
Per Unit WHght of
1
Cyan-
ide I
Cyan-
ide 11
Com-
pound
Cyan-
ide I
Cyan-
ide II
Com'
pound
(KCN, AgCN)
-86,100
-1,325
-642
-433
-94',460
-1,452
-704
-474
APPLICATIONS OF THERMOCHEMISTRY .
37
Heat of Formation of Amalgams
To Formula Given
Formula and
Reaction
Per
Formula
Weight
Per Unit Weight of
Mercury
Metal
Amalgam
(Hg, Na)
(Hg2, Na)
(Hg4, Na)
(Hg8, Na)
(Hg., Na)
(Hg„ K)
(Hg4, K)
(Hgs, K)
(Hgi2,K)
(Hg., K)
(Hg., Ag)
(Hg., All)
(Hg3.4i, Zn) \
8.7% Zn 1
(Hgo.94, Zn \
25.9% Zn /
10,300
17,800
20,000
21,900
19,000
20,000
29,700
33,000
34,600
25,600
2,470
2,580
-1,345
-703
52
45
25
18
50
37
18
14
-2
-3.8
448
774
870
952
826
513
762
846
887
656
23
13
-21
-U
46
42
24
18
46
35
18
14
-1.8
-2.8
Heat
OF Formation
OF Alloys
Formula
or
Composition
To Formula or Composition Given
Percentage Compos
Per
Formula
Given
Per Unit Weight of
Metal I
Metal II
Alloy
Metal I
Metal 11
Per
Cent.
Per Cent.
(Cu, Zm)
10,143
160
78
52
Cu
32.8
Zn
67.2
(Cu, Zn)
5,783
91
89
45
Cu
49.4
Zn
50.6
(Cus, Al)
26,910
141
997
124
Cu
87.6 ■
Al
12.4
(Cuz. Al)
21,278
167
788
138
Cu
82.5
Al
17.5
(Cua. Al!)
17,395
91
322
71
Cu
77.9
Al
22.1
(Cu, Al)
1,887
30
70
21
Cu
70.2
Al
29.8
(Cua, Al.)
10,196
80
126
49
Cu
61.1
Al
38.9
(Cu, Ah)
31,900
502
591
271
Cu
54.0
Al
46.0
(Naz, K) (liquid)
-2,930
-64
-75
-34
Na
54.1
K
45.9
(Na, Kb) (liquid)
1,940
84
25
19
Na
22.77
K
77.23
(Na, Ks) (liquid)
1,160
50
10
8
Na
16.43
K
83.57
(Pbij, Zn)
23,188
6
357
5.8
Pb
98.4
Zn
1.6
(Pb, Zn)
952
4.6
15
+3.5
Pb
76.1
Zn
23.9
(Pbi.26, Bi)
- 1,782
-6.9
-8.6
-3.8
Pb
55.6
Bi
44.4
(Snt.i, Zn)
-4,789
-6.6
-74
-6.1
Sn
91.7
Zn
8.3
(Pb, Sm5.7)
-8,034
-39
-4.3
-3.9
Pb
10.0
Sn
90.0
(Pb, Sm.8)
-1,028
-5.0
-3.1
-1.9
Pb
38.2
Sn
61.8
(Pbj.i, Sn)
450
1.0
3.5
0.8
Pb
79.0
Sn
21.0
{Pbio.8, Sn)
2,594
1.2
22
1.1
Pb
95.0
Sn
5.0
(PbM. Sn)
5,914
1,0
50
1.0
Pb
98.0
Sn
2.0
(Mg, Zns)
24,900
1,038
192
162
Mg
15.6
Zn
84.4
(Mgi, All)
164,800
1,717
204
93
Mg
54.2
Al
45.8
(Mg, Cd)
17,700
738
158
130
Mg
17.6
Cd
82.4
(Na, Cdz)
30,800
1,339
137
125
Na
9.3
Cd
90.7
(Na, CdB)
60,600
2,635
108
104
Na
3.9
Cd
96.1
(Ca, ZnO
56,600
1,390
214
185
Ca
13.3
Zn
86.7
(Ca, Znio)
199,100
4,978
306
29
Ca
5.8
Zn
94.2
(Cu!, Cds)
47,700
376
142
103
Cu
27.4
Cd
72.6
38
METALLURGICAL CALCULATIONS.
Thermochemical Constants of Bases to Dilute Solution."
{Per Chemical Equivalent of Base.)
*Por explanation of nature and use of these tables, see page 16.
Caesium Cs+ +81,240
Rubidium Rb+ 76,265
Lithium Li+ 62,900
Potassium K+ 61,900
Beryllium (Glucinum) HBe++ 60,350
Barium J^Ba++ 59,950
Thorium i^Th++++
Strontium J^Sr++ 58,700
Sodium Na+ 57,200
Lanthanum J^La"'"++
Neodymium i^Nd+++ 55,600
Calcium HCa++ 54,400
Praesodymium }^Pr"''++
Magnesium J^Mg++ 54,300
Aluminium • HA\+++ 40,100
Ammonium (from the elements) (N, H4)+ 33,400
Vanadium J^V+++
Cerium i4Ce++++
Titanium J^Ti++++
Uranium y£^+++++
Vanadium J^V^^
Vanadium i^V++++++
Silicon }4Si^+'*-+
Boron HB+++
Vanadium i^y++++-i-
Zirconium 3^Zr''"''"++
Manganese KMn++ 24,900
Zinc i^Zn++ 17,200
Chromium J^Cr"''++
Phosphorus i^p++++-i-
Lithium (per-salts) }4U.'^^
Barium (per-salts) HBa++++
Iron (ferrous) HFe++ 10,900
Tungsten (Wolfram) KW++++
Tungsten (Wolfram) j^\Y++++++
Cadmium i^Cd++ 9,000
Cobalt HCo++ 8,200
Manganese J4Mn++++
Sodium (per-salts) J^Na++
Nickel i^Ni++ ' 7,700
Antimony J^Sb ''"+■'"
Arsenic J^As"''++
Sulfur i^S++++++ ' 2,800
APPLICATIONS OF THERMOCHEMISTRY.
39
Iron (ferric)
Tin (stannous)
Tin (stannic)
Lead
Carbon
Chromium
Bismuth
Antimony
Arsenic
Copper (cuprous)
Hydrogen
Tellurium
Tellurium
Sulfur
'Selenium
Selenium
Tellurium
Copper (cupfic)
Lead (per-salts)
Thallium
Carbon
Hydrogen (per-salts)
Mercury (mercurous)
Palladium
Mercury (mercuric)
Platinum
Silver
Gold (auric)
Gold (aurous)
HSn++
MSn++++
3^Pb++
i^Cr++++++
HBi+++
J^Sb+++++
J^As+
Cu+
H+
Te+
MTe+
MS++++
3^Se++
HSe+++
3^Te++
HCu++
JiPb++++
HT1.+++
J^C++
^H++
Hg+
KHg++
MPt++++
Ag+
HAU+++
Au+
27,30
1,900
400
+ 0
-900
-1,300
-7,900
-14,250
-19,450
-25,300
-30,300
Thermochemical Constants of Acids to Dilute Solution
{Per Chemical Equivalent of Acid Element or Radical)
Fluorine
Chlorine
Bromine
Oxygen
Iodine
Sulfur-
Selenium
Tellurium
Elements
F-
ci-
Br-
I-
J^s-
MSe-
KTe-
52,900
39,400
27,500
20,700
13,200
-5,100
-17,900
Acid Radicals {from the elements)
Hydrate
Sulf-hydrate
(O, H)-
(S, H)-
55,200
3,400
40 METALLURGICAL CALCULATIONS.
Selen-hydrate (Se, H)" 19,100
Hypochlorite (CI, O)- 27,500
Chlorate (CI, O,)- 21,900
Per-chlorate (CI, O4)- 39,400
Hypobromite (Br, O)- 22,800
Bromate (Br, Os)" 6,700
lodate (I, Os)- 60,470
Per-iodate (I, O4)- 48,070
Hypo-sulfite J^(S2, Os)— 71,750
Sulfite 3^(S, Os)— 75,100
Bi-sulfite (H, S, Os)" 149,400
Pyro-sulfite ^^(Sj, Os)— 115,200
Sulfate J^(S, O4)— 107,000
Bi-sulfate (H, S, Ot)- 211,100
Per-sulfate 1^(82, Os)~ 158,100
Di-thionate ^(82, Oe)"" 138,350
Tri-thionate , ^^(Ss, Oe)— 136,500
Tetra-thionate ' J^(S4, Oe)— 130,600
Penta-'thionate ^(85, Oe)"" 133,100
Selenite H(Se, Os)— 60,060
Selenate Ji(Se, O4)— 72,800
Hypo-nitrite (N, O)" -3,800
Nitrite (N, O2)- 27,000
Nitrate (N, O3)- 48,800
Phosphate 1^(P, O4) 99,300
Mono-H-Phosphate J^(H, P, O4)— 152,750
Di-H-Phosphate (H2, P, O4)- 307,700
Arsenite }^(As2, O4)— 102,150
Arsenate 3^ (As, O4) 70,200
Mono-H- Arsenate }^(H, As, O4)"- 108,050
Di-H-Arsenate (H2, As, O4)- 217,200
Cyanide (C, N)" -34,900
Cyanate (C, N, O)- 37,100
Sulfo-cyanate (C, N, 8)- -18,100
Ferro-cyanide Ji(Pe, Ce, Ne) -25,600
Ferri-cyanide K(Pe, Ce, Ne) -52,800
Carbonate i^(C, Os)" 82,450
Bi-carbonate (H, C, Os)" 169,100
Formate (C, H, O2)- 104,600
Acetate (C2, H,, O2)- 120,500
Oxalate >^(Ca, 04)~ 99,800
CHAPTER III.
THE USE OF THE THERMOCHEMICAL DATA.
Simple Combinations.
If the problem is the simple calculation of how much heat
is evolved in the combination of a given weight of one element
with another, the factors needed, the heat evolved per unit
weight of substance combining, are obtained by a simple divi-
ion from the thermochemical data given. Thus, suppose the
question to be the total heat evolved in Problem 3, in. the
oxidation in a Bessemer converter of
100 kilos, of carbon to carbonic oxide.
200 kilos, of carbon to carbonous oxide.
50 kilos, of irianganese to MnO.
150 kilos, of silicon to SiC.
500 kilos, of iron to FeO.
The solution would be
lOOX ^^^^ = 100X8100 = 810,000 Calories.
200X^^^=200X2430= 486,000
QO QOO
50 X ;r = 60X1653= 82,650
55
150X^^5^°= 150X7000 = 1,050,000
Act
500 X -i^^^ = 500X1173 = 586,500
^ Total 3,015,150 "
Complex Combinations.
If the problem is a step more complex, that is, includes the
combination of compoi^nds with each other to form a more
41
42
METALLURGICAL CALCULATIONS.
complex compound, the molecular weights are still our guide,
together with the thermochemical data given. If, for in-
stance, the question above solved is complicated by the further
requirement, to add the heat evolved by the formation of the
slag, that is, of the MnO and FeO with SiO^ to form silicate,
we may calculate the heat of union of FeO and MnO with
SiO' as follows:
(Mn, Si, 0') ,= 276,300 Calories.
But (Mn, O) = 90,900
and (Si, O^) = 180,000
therefore (MnO, SiO') = 5.400
Similarly
5 400
or, per kilo, of MnO = — = 76 Calories.
71
(Fe, Si, 0') = 254,600 Calories
(Fe, 0) = 65,700 "
(Si, 0==) = 180,000 "
(FeO, 810==) = 8,900
• ., , -- ^ 8,900
per kilo, of FeO =
72
= 124 Calories.
Therefore the additional heat of formation of the slag may be
Wt. MnO = 64.5 kilos. X 76 = 4,902 Calories.
Wt. FeO =642.8 " X 124 = 79,707
Sum = 84,609
Similar principles of calculation apply to all the oxygen-
containing salts. Thus, if from the heat of formation of any
sulphate we subtract the heat of formation of SO', and also of
the metallic oxide present, the residue is the heat of combina-
tion of the metallic oxide with SO', — the weights involved
being always those represented by the formulae. Thus, calling
MO any metallic oxide, we may express the principle as follows:
(MO, SiO^) = (M, Si, O') — (M, 0)— (Si, O'')
(MO, SO') = (M, S, 0') — (M, O)— (S, 0')
(MO, CO^) = (M, C, O') — (M, O)— (C, 0')
(3M0, P^O') = (M', P^ 0») — 3(M, 0)— (P^ 0=)
etc. etc.
THERMOCHEMICAL DATA. 43
Example. — What is the heat required to calcine limestone?
(CaO. CO') = (Ca, C, 0')— (Ca, 0) — (C, O')
= 273,850 —131,500—97,200
= 45,150 Calories.
This is the heat required to split up CaCO' (100 parts) into
CaO (56) and CO^* (44) ; the heat required is therefore
45,150 H- 100 = 451.5 Calories per kilo, of CaCO' decomposed.
45,150-^ 44 = 1026. " " " CO^ driven off.
45,150-^ 56 = 806. " " " CaO remaining.
And either of these quantities may be used, according to
convenience in working the problem.
Double Decompositions.
If the thermochemical problem involves the simultaneous
decomposition of one or more substances and formation of one
or more others, then the chemical equation should be written,
and a thermochemical interpretation given of all the energy
involved in the passage from starting compounds to products.
Every chemical equation can be thus interpreted, if the heats
of formation of all the compounds represented in it are known.
We obtain the net energy of the reaction by assuming that all
the substances used are resolved into their elements, and that
all the products are formed from their elements; the first item
is therefore to add together the heats of formation of all the
substances used, starting with their elements, and by changing
the algebraic sign of this sum we have the heat necessary to
decompose the substances used into their elements; the second
item is similarly found by adding together the heats of forma-
tion of all the substances formed; the difference between these
two items is the net energy of the reaction. In making these
summations, regard must, of course, be paid to the number of
molecules of each substance concerned, as shown in the re-
action (because the tabulated data are the heats of formation
of one molecule only) and to the algebraic sign of the heats of
formation.
Examples. — (a) What heat is evolved when dry ferric oxide
is reduced by aluminium, in the Goldschmidt " Thermite "
process ?
44 METALLURGICAL CALCULATIONS.
Fe='0' + 2A1 = AP0' + 2Fe
—195,600
+ 392,600
+ 197,000 Calories.
(b) Write the equation showing the reaction of metallic
goditim on water in excess:
2H20 + 2Na = 2NaOH +HH aq. (excess water)
-2(69,000)
-138,000
+ 2(112,500)
+ 225,000
+ 87,000 Calories.
(c) "What heat is evolved in the reaction of water on cal-
cium carbide to form acetylene gas?
CaC^ +2^0 = Ca02tP+ C^H^
— (—6250)— 138,000
—131,750
+ 215,600+ (—54,750)
+ 160,850
+ 29,100
The last case leaves out the very small heat of solution of
the calcium hydrate which would normally go into solution.
If such a large excess of water was used that all the calcium
hydrate formed could dissolve, the heat of formation of the
hydrate in dilute solution, 219,500 Calories, would be used, and
the final result would be 33,000 Calories. The case is also very
instructive, because it contains two compounds which are
endothermic, that is, their heat of formation is negative, and
therefore, as in the case of CaC^, heat is given out when it
decomposes, while in the case of C^tP heat is absorbed when it
is formed.
(d) What are the heats of combustion of the gaseous hydro-
carbons which form constitutents of common fuels, expressed
per cubic meter of gas burning, and with the water formed
assumed remaining as vapor, uncondensed? To calculate these
we write the equations of combustion, with water as gas, and
find the sum total of heat evolved; since everj'' molecule of gas
burning represents 22.22 m' of gas, we can, by a simple division,
obtain the value sought.
THERMOC HEMIC AL DATA. 46
Methane:
CHH20' = C0H2tP0
—22,250
+ 97,200 + 2(58,060)
Ethane:
+ 191,070 Calories. = 8,598 per m" CH«.
2C='H» + 70' = 4CO=' + 6H='0
—2(26,650)
+ 4(97,200) +6(58,060)
+ 683,860 Calories. =15,387 perm'C'H'.
Propane:
C'H» + 502 = 3CO' + 4H'0
— 33,850
+ 3(97,200) +4(58,060)
+ 489,990 Calories. = 22,050 per ni' C»H«.
By applying these principles to all the hydrocarbons whose
heat of formation has been given in the tables, we obtain the
following useful table, water being considered as uncondensed
and cold:
Molecular Heat of
Combustion . Heat of Combustion of
Hydrocarbon. Calories. 1 m' (Calories) 1 ft' (b.t.u.)
Methane (gas)..; 191,070 8,598 966
Ethane (gas) 341 ,930 15,387 1728
Propane (gas) 489,990 22,050 2477
Ethylene (gas) 321 ,770 14,480 1627
Propylene (gas) 471 ,830 21 ,232 2385
Toluene (liquid) 906,990
(liquid 758,130
Benzene -j ^^^ 765,330 34,440 3869
(liquid 1,428,930
lurpentme | ^^^ 1,438,330 64,725 7271
(solid 1,223,690 ......
JNapntnaiine | lig^i^ _ i 228,290 55,273 6209
Anthracine (solid) 1,690,150
Acetylene (gas) 307,210 13,825 1555
,,,,,, ,j liquid.. 148,270
Methyl alcohol I g^^ ^^gg^O 7050 799
T. , , , , , (liquid... 295,330
Ethyl alcohol I g^^ 305^30 13^44
1544
46 METALLURGICAL CALCULATIONS.
Molecular Heat of
Combustion. Heat of Combustion of
Acetone -1 "'l^'^ ^^^'^^°
Acetone ^ ^^^ 403,630 18,163 2040
(To the above we will add data for three other common
gaseous fuels.)
Carbonous oxide (gas) 68,040 3,062 344
Hydrogen (gas) 58,060 2,613 293 . 5
Hydrogen sulphide 122,520 5^513 619
Calorific Power of Fuels.
By using the principles explained we can calculate the cal-
orific power of any combustible, with the aid of one or two
simple assumptions. Regarding the water formed, we will con-
sider in all ordinary metallurgical calculations that it remains
uncondensed, thus putting it on the same footing as the other
products of combustion. This amounts virtually to assuming
that the latent heat of condensation of the vapor formed has
not been generated in the furnace, an assumption which is
quite justified if, as is almost always the case, the water formed
inevitably escapes as vapor. If, in any special case, the pro-
ducts of combustion are in reality cooled down so far that the
water does condense, then it would be proper to assume the
higher calorific powers for the combustion of the hydrogen-
containing materials, and thus a more accurate heat-balance of
the furnace operation could be constructed. An instance of
the latter might be the case of the partial combustion of a fuel •
in a gas producer, where the fuel gas is afterwards cooled be-
fore using, in order to condense from it ammonia water, etc.
(Mond's producer). In this case, it is possible that some
water formed by the partial combustion would be afterwards
condensed, and to obtain a correct heat-balance-sheet of the
producer it would be necessary to assume that this heat had
also been generated. Such cases occur very rarely in metal-
lurgical practice, and it is therefore recommended to always
use the lower calorific values, assuming water vapor uncon-
densed and its latent heat of condensation not to have been
generated — ^unless the conditions of the problem point plainly
to the opposite course as correct. It is no more correct to
charge against a furnace the latent heat of condensation of the
THERMOCHEMICAL DATA. 47
water vapor formed, than it would be to charge against it the
latent heat of condensation of the carbonic oxide formed, if the
conditions of practice are such that it is impossible to utilize
in any useful manner either one of them. Those who persist
in assuming that the latent heat of vaporization of the water
formed by combustion is generated in and chargeable against
the furnace, and then, of necessity, charge the same quantity
against the heat carried off in the products of combustion, are
in the great majority of cases adding to both sides of the balance
sheet a quantity which cannot possibly be utilized, and are
therefore distorting all the factors of heat generation and
distribution. By thus abnormally increasing the sensible heat
in the waste gases, they abnormally decrease the proportionate
value of other items of heat distribution and utilization. I
have dwelt on this matter at length, because it is of importance
in furnaces using fuel, such as gas, rich in hydrogen; where, in
some cases, the counting of the latent heat of condensation of
the water vapor formed in the escaping gases, would perhaps
double the apparent chimney loss, and give distorted values
to the whole heat balance sheet. What we are after, in every
case, are the real facts and figures as to the working of a furnace
or operation, and we must therefore judge the operation by
a theoretically perfect standard, it is true, but not by a theo-
retically impossible standard; i.e., our standard must be the
theoretically possible one under the practical conditions neces-
sarily prevailing. A radical reform is very much needed in
just this respect, — in the treatment of the latent heat of vapor-
ization of the water formed — ^by writers on metallurgical pro-
cesses and problems of Combustion.
Problem 4.
, A natural gas from Kokomo, Ind., contained by analysis:
CH* 94.16, W 1.42, C^H* 0.30, CO^ 0.27, CO 0.55, O^ 0.32, N:"
2.80; H^'S 0.18 per cent.
Required:
(1) What is its metallurgical calorific power per cubic meter
and per cubic foot?
(2) Comparing it with coal, having a calorific power of 8,000
Calories, how much gas gives the same generation of heat as
a metric ton of coal?
48 METALLURGICAL CALCULATIONS.
(3) If the natural gas costs $0.15 per 1,000 cubic feet, at
what price per metric ton would the coal furnish heating power
at the same cost for fuel ?
Solution. — Requirement (1) Heat of combustion of 1 cubic
meter:
CH^0.9416X 8,598 = 8095.9 Calories.
H2 0.0142X 2,613 = 37.1
C^H^ 0.0030 X 14,480 = 43.4
CO 0.0055 X 3,062 = 16.8
H^S 0.0018 X 5,513 = 9.9
Total = 8203.1
In British thermal units per cubic foot we have
8,203.1X3.967 -i- 35.314 =
8,203.1X0.11233 = 921.5 b. t. u. perft' (1),
Requirement (2):
8000X1000
8,203.1 -975 m
= 34.450 ft».
If the ton of coal be taken as 2240 pounds, instead of the
metric ton of 2204 pounds, the equivalent volume is
2940
34;450xg^ = 35,013 ft'.
Requirement (3):
34,450
1000
or, per ton of 2240 pounds.
X0.15 = $5.16f per metric ton.
2240
$5.16fX 2204 = $5.25 per long ton.
Dulong's Formula.
When a solid or liquid carbonaceous fuel is to be burned,
its calorific power may either be determined directly in a calori-
meter (which is the best way, if carefully done), or may be
calculated from its analysis. In case it is determined calori-
THERMOCHEMICAL DATA. 49
metrically, the weight of water produced per unit of fuel should
be determined either in the same or by a separate experi-
ment, and the latent heat of vaporization of this water sub-
tracted from the calorimetric value of the fuel, in order to get
its metallurgical, or practical, calorific power.
Example. — A coal gave 9215 calories per gram in the calori-
meter, and its products of combustion gave up 0.45 grams of
condensed water. What is its practical calorific power?
Taking the products cold, the heat of condensation per gram
of water vapor may be taken as 606.5 calories (Regnault), and
we therefore have obtained in the calorimeter
0.45X606.5 = 273 Calories
more than if the products were cold and water uncondensed.
The practical calorific power is therefore
9215 — 573 = 8942 Calories per kilogram.
Dulong's simple formula, or method of calculation, is the state-
ment that the calorific power of a fuel can be calculated from
the amount of carbon, hydrogen, and oxygen it contains by
assttming all the carbon free to bum, the oxygen to be all
combined with hydrogen in the proportions of 0^ to 2H^ (32
to 4), and that the rest of the hydrogen is free to bum.
Example. — Taking the bituminous coal of Problem 1, con-
taining carbon 73.60, hydrogen 5.30, nitrogen 1.70, sulphur 0.75,
oxygen 10.00, moisture 0.60, ash 8.05, its calculated calorific
power is per kilogram:
Carbon 0.7360 X 8100 = 5962 Calories.
Hydrogen 0.0530
0.0125
0.0405 X 34,500 = 1397 " (water liquid)
Sulphur 0.0075 X 2164 = ^
Total 7376
This calculation is made, however, on the assumption that
all the water in the products is condensed to liquid, i.e., not
only the water formed by combustion of the free hydrogen,,
but also the already-formed H'O containing the oxygen of the
coal, as well as the moisture present to start with. To obtain
60 METALLURGICAL CALCULATIONS.
the practical calorific power by calculation we must subtract
from the above-generated heat the latent heat of all the vapor
of water in the products, as follows:
Moisture in coal 0 . 0060 kilos.
Moisture formed by combined hydrogen . . 0 . 1125 "
Moisture formed by free hydrogen 0.3645 "
Total. 0.4830 "
Latent heat of vaporization = 606.5x0.4830 = 293 Calories.
Practical calorific power, water all as vapor = 7083 Calories.
The values thus calculated are found to agree satisfactorily
with the laboratory and the practical calorific powers of the
fuels.
The Theoretical Temperature of Combustion.
If we start with a cold fuel and cold air the maximum tem-
perature which can be obtained by the combustion is simply
that temperature to which the heat generated can raise the
products of combustion, assuming that all that heat resides
primarily in the products as sensible heat. This is then a prob-
lem in physics, in which we have a known amount, of heat
available, and the question is to what temperature it will
raise the products of combustion. The problem is at once
soluble when we know the mean specific heats of the products
of combustion and their respective quantities.
Until a few years ago these specific heats at high tempera-
tures were unknown, and very false results were obtained by
assimiing constant specific heats from ordinary temperaltures
up, and values were thus calculated which were known to be
hundreds, and in some cases, thousands, of degrees too high.
The most satisfactory values for the specific heats of the fixed
gases and water vapor and carbonic oxide are those deter-
mined by Mallard and Le Chatelier, and their proper use has
entirely solved the question of theoretical temperatures of com-
bustion, and removed a positively disgraceful discrepancy
between theory and practice. The specific heats of these gases
increase with temperature, so that the actual specific heat of a
cubic meter (measured at standard conditions) is, at the tem-
perature t, —
THERMOCHEMICAL DATA. 51
for N^ 0^ H', CO — S = 0.303 + 0.000054t
" ' CO^— S = 0.37 +0.00044t
" " (vapor) H^O— S = 0.34 +0.00030t
For calculating the quantity of heat needed to raise the gas
from 0° to t°, however, we need the mean specific heats, Sm,
between 0° and t° These are, of course, the above constants
plus half the increase, viz.:
for N^ 0^ H^ CO— Sm = 0.303 + 0.000027t
CO^*— Sm = 0.37 +0.00022t
" (vapor) H^iO— Sm = 0.34 +0.00015t
and the quantity of heat needed to raise 1 cubic meter (at
standard conditions) from 0° to t° is
for N^ O' H^ CO— Q (o-t) = 0.303t + 0.000027t2
CO''— Q (o-t) = 0.37t +0.00022t=
H='0— Q (o-t) = 0.34t -h 0.0001 St"
or, finally, the quantity of heat needed to raise 1 cubic meter
(measured at standard conditions) from t to t' is
for W, 0^ H^ CO— Q (t'-t) =
0.303 (t'-t)-j- 0.000027 (t'^-t^)
00==- Q (t'-t) =
0.37 (t'-t) -1-0.00022 (t"-t=')
H^'O- Q (t'-t) =
0.34 (t'-t) -f 0.00015 (t"-t=')
And the mean specific heat, Sm, between any two tempera-
tures is
for N* 0^ H^ CO — Sm (t'-t) =
0.303 -^ 0.000027 (t'-l-t)
CO'' — Sm (t'-t) =
0.37 -t- 0.00022 (t'-f-t)
H'O — Sm (t'-t) =
0.34 -fO.00015 (t'-f-t)
Examples:
(1) What is the temperature of the hottest part of the oxy-
hydrogen blowpipe flame?
According to the equation 2H'' + 0'' = 2I-P0, the hydrogen
gas forms an equal volume of water vapor. (Equal numbers
of molecules.) The heat of combustion of one cubic meter of
hydrogen is 2613 Calories. The question is, therefore: "To
52 MEIALLURGICA I. CA IjCULATlONS.
what temperature will 2613 Calories raise one cubic meter of
water vapor? " The answer is, using the data above,
Q (o-t) = 0.34t + 0.00016t2 = 2613
Whence t = 3191°
(2) What is the maximuni temperature of the hydrogen
flame burning in dry air?
The heat evolved is, as before, 2613 Calories, and there is
formed also one cubic meter of water vapor, but the products
will contain also the nitrogen which accompanied the 0.5 cubic
0.5
meter of oxygen necessary for combustion, viz.: „ —0.5
= 1.9 cubic meters, and this is heated as well as the water
vapor. Therefore,
Q (o-t) = (0.34t-|-0.00015t2)-l-1.9 (0.303t-|-0.000027t') = 2613
Whence 0.916t + 0.0002013t=' = 2613
And t = 2010°
(3) What is the temperature of the air-hydrogen blowpipe,
if 25 per cent, excess of air is used, above that required ?
All the data are the same as above, except that to the pro-
ducts will be added 0.25 (0.5-M.9) = 0.6 cubic meters of
unused air, which has the same specific heat as nitrogen, and
therefore the equation becomes
Q (o-t) = (0.34t-|-0.00015t2)-f2.5 (0.303t-F0.000027t=') = 2613
Whence t = 1764°
This calculation brings out very clearly the uselessness and
ineffectiveness of using more air than is theoretically neces-
sary; any excess simply passes unused into the products of
combustion, and thus reduces their maximum possible tem-
perature. The obtaining of the maximum possible temperature
depends upon accurately proportioning the supply of oxygen
or air to the quantity of gas burned ; an excess or a deficiency
will result in a lower, temperature.
Combustion with Heated Fuel or Air.
If the fuel itself or the air which bums it is preheated, the
sensible heat in either one or in both is simply added to the
heat generated by the combustion, to give the total amount of
THERMOCHEMICAL DATA . 53
heat which must be present as sensible heat in the products of
combustion. The effect is exactly the same as if the heat
developed by combustion had been increased by the sensible
heat in the fuel or air used.
What is the calorific intensity (theoretical maximum tem-
perature) obtained by burning carbonous oxide gas? (a)
Cold, -with cold air; (b) cold, with hot air at 700° C; (c) hot,
with hot air, both at 700° C.
(a) Take one cubic meter of carbonous oxide. Calorific
power 3,062 Calories; product one cubic meter of carbonic
oxide, and 1.904 cubic meters of nitrogen.
Let t be the temperature attained; then
Heat in the 1 m' carbonic oxide = 0.37t +0.00022t=
1.904 m' nitrogen gas = 0.577t + 0.0000514t'
" products = 3,062 Cal's. = 0.947t + 0.0002714t'
Whence t = 2050°
(b) If the 2.404 cubic meters of air needed are preheated
to 700° C, they will.bring in as serisible heat
Q (O-700) = 2.404 [0.303 (700) +0.000027 (700)'
= 552 Calories.
The total heat in the products will be 3062 + 552 = 3614
Calories, and therefore 0.947t + 0.000271 4t2 = 3614
Whence t = 2189°
(c) If the gas itself is also preheated it brings in
Q (O-700) = 0.303 (700) +0.000027 (700)'
= ' 225 Calories.
The total heat in the products will be 3614 + 225 = 3839
Calories, and therefore 0.947t + 0.0002714t' = 3839
Whence t = 2284°
Heating both gas and air to 700° before they burn thus
raises the theoretical temperature from 2050 to 2284 or 234°.
Problem 5.
Statement— The natural gas of Kokomo, Ind., contains by
analysis: Methane (marsh gas) 94.16, ethylene (olefiant gas)
54
METALLURGICAL CALCULATIONS.
0.30, hydrogen 1.42, carbonous oxide 0.55, carbonic oxide 0.27,
oxygen 0.32, nitrogen 2.80, hydrogen sulphide 0.18, — ^in per-
centages, by volume.
Required:
(1) The maximum flame temperature, if burned cold with
the theoretical amount of cold, dry air necessary.
(2) The calorific intensity, if burned cold, with the requisite
air preheated to 1000° C.
(3) The calorific intensity, if burned cold, with 25 per cent,
more air than theoretically necessary, preheated to 1000°.
Solution :
[The practical calorific power of this gas has been already
calculated in Problem 4 as 8203 Calories per cubic meter.
The gas itself is always burned cold, because, if preheated, it
decomposes and deposits carbon in the regenerators.]
The products of combustion of the gas, say per cubic meter,
must first be found, using the formulae for combustion.
Requirement (1):
1 Cubic Meter
of Gas.
Oxygen
Needed.
m'
CO'
Products.
H^O
SO'
N'
CH«"
0.9416
1.8832
0.9416
1.8832
C^H*
0.0030
0.0090
0.0060
0.0060
H'
0.0142
0.0071
• . . .
0.0142
CO
0.0055
0.00275
0.0055
> • . ■
CO'
0.0027
....
0.0027
0»
0.0032
0.0032
....
* • • ■
N»
0.0280
. . . ■
....
• • • •
.... 0
0280
H'S
0.0018
0.0027
....
0.0018
1.9052
0.0018
....
1.90155
0.9558
0.0018 0
0280
Air re
quired
9.14
(carrying
in N') = 7
Total N' 7
238
266
The heat generated will exist as sensible heat in the CO',
H'O, SO' and N' of the products. The mean specific heat of
SO' per cubic meter is 0.444 at ordinary temperatures; what
it is at high temperature has not been determined; we will
give it the same index of increase as the analogous gas CO',
THERMOCHEMICAL DATA. 55
and take for it Sm (o-t) = 0.444 + 0.00027t, or Q (o-t) =
0.444t + 0.00027t^
At the temperature attained by combustion, t, the products
will contain the following amounts of heat:
W = 7.266 (0.303t + 0.000027t')
H^O = 1.9052 (0.34t +0.00015t=')
CO^ = 0.9558 (0.37t +0.00027t2)*
SO^ = 0.0018 (0.444t + 0.00027t')
Total = 3.2044t + 0.00074057t2 = 8203 Calories.
From which t = 1806° (1)
Requirement (2) :
Heat in 1 m^ of air at 1000° = 0.303 (1000) +0.000027 (1000)^
= 330 Calories.
" 9.14 m» " 1000° = 3016 "
• Therefore: 3.2044t + 0.000740571^ = 8203 + 3016
Whence t = 2288°
Requirement (3):
Excess air used = 9.14x0.25.= 2.285m'.
Heat in 11.425 m' at 1000° = 11.425x330 = 3770 Calories,
The heat capacity of the excess air must be added to the heat
in the products at temperature t, viz.:
Excess air = 2.285 (0.3031 + 0.0000271==)
making the total heat in the products (adding to previous ex-
pression):
3.8968t + 0.000802271' = 8203 + 3770
Whence 1 = 2134°
The Eldred Process of Combustion.
A means of regulating the temperature of the flame has
been proposed by Eldred, and described by Mr. Carlton Ellis
in the December, 1904, issue of El. Chem. Ind. The proposition
is simply to mix with the air used for combustion a certain
* These and some succeeding problems were worked out using this
specific heat. Later the author has adopted a slightly different value for
CO*, viz., (0.37 + 0.00022t), which he now considers more nearly correct.
56 METALLURGICAL CALCULATIONS.
proportion of the products of combustion themselves. The
principle is easily understood when the requisite calculations
of the theoretical flame temperatures are made. When solid
fuel is burnt the temperature is often too high locally, and re-
sults in burning out grate bars or overheating the brick work
of the fire-place, or overheating locally the material which is
mixed with the fuel. If the air is diluted with products of
combustion the initial theoretical temperature is lowered, and
the above evils may be obviated. Using solid fuel, the heat in
the fuel before the air actually bums it must be added to the
heat generated by combustion to get the actual temperature in
the hottest part of the fire.
Examples: (1) What will be the highest temperature in a
charcoal fire fed by air, assuming complete combustion without
excess of air?
Assuming the charcoal to be pure carbon, and to be heated
to the maximum temperature t before it bums (by the com-
bustion of the preceding part), the heat available is:
Heat of combustion of 1 kilo of carbon 8100 Cal's.
Sensible heat in the carbon at t° 0.5t — 120 "
Total heat available to raise the temperature . . . 7980 -1- 0.5t "
Products of combustion CO^ 1.85 m'
. .W 7.04 "
Heat of product at t° CO^ 1.85 (0.37t +0.00022t2)
W 7.04 (0.303t + 0.000027t2)
Sum 2.81t -1-0. 0006012
therefore 2.81t-f-0.00060t2 = 7980-|-0.5t
whence t= 2199°
(2) What will be the temperature in the same case, if the
air used is diluted with an equal volume of the products of
combustion ?
The heat available is the same as before, 7980-l-0.5t; but
since the mixed air for combustion contains only half as much
oxygen per cubic meter, the products will be exactly doubled in
amount for a unit weight of carbon bumt, and w? therefore
have directly
2 (2.8lOt-(-0.00060t=') = 7980 + 0.5t
whence t = 1213''
THERMOCHEMICAL DATA. 57
It is therefore evident that the maximixm temperature of the
hot gases at their moment of formation is nearly halved by the
procedure stated.
(3) Taking the cases cited by Mr. Ellis, where the products
contained originally 15 per cent, oxygen and 6 per cent, carbon
dioxide, and after mixing the air used with half its volume of
the chimney gases, 9 per cent, oxygen and 12 per cent, carbon
dioxide (the gas-air mixture entering containing 15 per cent,
oxygen and 6 per cent, carbon dioxide), what are the maximum
temperatures obtained in the two cases?
Case 1 : The heat available is as before ; the products of com-
bustion are CO^ 1.85 m', O^ 4.62 m^ N^ 24.36 m', and their
sensible heat at temperature t —
Heat in CO^ = 1.85 (0.37t +0.00027t2)
O' + N^ = 28.98 (0.303t-|-0.000027t^)
Sum = 9.46t + 0.00128t2
therefore 9.46t+ 0.001 280t2= 7980-|-0.5t
whence t = 800°
Case 2: The products, per kilogram of carbon burnt, will
be CO' 3.70 m', O' 2.77 m^ N' 24.36 m', and we have
Heat in CO' = 3.70 (0.37t +0.00027t')
O'+N' = 27.13 (0.303t + 0.000027t')
Sum = 9.09t-|-0.00173t'
therefore 9.09t + 0.00173t' = 7980-t-0.5t
whence t = 764°
Conclusions: The calculations regarding the Eldred process
bring out what was not stated in the printed description of
the method, viz.: that if a small excess, or no excess of oxygen
escapes, to the chimney, the temperature of the flame will be
greatly reduced by the dilution of the air used, because more
gas-air mixture will be required per unit of fuel burnt; but if
there is any large amount of unused oxygen escaping in the
first instance, the dilution practiced will scarcely affect the
temperatiire of the flame at all, because it makes very little
difference whether the fuel is heating up unused oxygen or
carbon dioxide dilutant substituted for some of it, as long
58 METALLURGICAL CALCULATIONS.
IS there is more than enough oxygen to bum the fuel. The
fact of the specific heat of carbon dioxide being greater than
that of oxygen, is the only reason for the small calculated dif-
ference of 36°, in cases 1 and 2. It is true that the dilution
practiced will result in less heat being lost in the waste gases,
in the Example (3), but the same economy could be obtained
by simply using less excess of air in the first instance.
Temperatures in the " Thermit " Process.
The Goldschmidt porcess of reducing metallic oxides by
powdered aluminium, igniting the cold mixture, is only a spe-
cial case of our general rule, as far as concerns the calcula-
tion of the theoretical temperatures obtained. In any case,
the total heat available is the surplus evolved in the chemical
reaction, and the temperature sought is that to which this
quantity of heat will raise the products of the reduction. The
products are alumina and the reduced metal. The heat in the
latter, in the melted state, is well known in many cases; it
is most clearly expressed by the sum of the heat in such melted
metal just at its melting point (which is easily determined
calorimetrically, and is well known for many metals), plus
the heat in the melted metal from the melting point to its
final temperature, which is equal to t, minus the melting tem-
perature, multiplied by the specific heat in the melted condition
These data are known for many metals; for some they may
have to be assumed from some general laws correlating these
values. The heat in melted alumina has not been determined
calorimetrically, to my knowledge. Its sensible heat solid is
0.2081t + 0.0000876t=' (determination made in author's labora-
tory), which, evaluated for the probable melting point, 2200'
C, would give the heat in it at that temperature 881.8 Calories;
the latent heat of fusion of molecular weight is very probably
2.1 T, where T is the absolute temperature of the melting point,
making the latent heat of fusion per kilogram 2.1 (2200 +
273) -H 102 = 5193 -^ 102 = 50.9 Calories. The specific heat in
the melted state is probably equal to the specific heat of the
solid at the melting point, viz.: 0.208H- 0.0001752 (2200) =
0.5935. We, therefore, have for the heat in melted alumina at
terftperature t —
|0
THERMOCHEMICAL DATA. 59
Heat in solid alumina to the melting point 881 .8 Calories
Latent heat of fusion 50.9 "
Heat in liquid alumina to its setting point 0 . 5935 (t-2200)
Total .932.7 + 0.5935 (t-2200)
or, for a molecular weight, 102 kilograms: 95,135 + 60.54 (t-2200).
Examples:
(1) If black cupric oxide is reduced by powdered aluminium
what is the temperature attained?
The reaction is
3CuO + 2Al = ATO' + 3Cu
—3(37,700) j + 392,600
+ 279,500
The products must therefore be raised to such a tempera-
ture that they contain 279,500 Calories. The heat in molecular
weight of alumina at temperature t has already been found;
that in copper at t° is, for one kilogram (using the author's
determinations) :
Heat in melted co'pper at setting point = 162 Calories.
Heat in melted copper above 1065° = 0.1318 (t— 1065)
Calories.
Total = 162 + 0.1318 (t— -1065).
Per atomic weight (63.6 kilos.) = 10303 + 8.3825 (1—1065)
From these data there follows the equation:
95,135 + 60.54 (t— 2200) + 3[10,303 +
8.3825 (t— 1065)] = 279,500
whence t = 3670°
A little further calculating will show that approximately one-
third of all the heat generated exists in the copper, and two-
thirds in the melted alumina.
(2) If pure ferric oxide is reduced by the " Thermit " process,
what is the temperature of the resulting iron and melted alumina ?
The reaction and the heat evolved have been already given
in the preceding discussion of these calculations (page 44.)
They show that per molecular weight of alumina formed there,
are two atomic weights (112 kilos.) of iron formed, and there
is disposable altogether 197,000 Calories. The heat in a kilo-
60 METALLURGICAL CALCULATIONS.
gram of pure iron at its melting point (1600°) is 300 Calories,
the latent heat of fusion approximately 69 Calories, and the
specific heat in the melted condition 0.25. The total heat in a
kilogram of melted iron is therefore 369 + 0.25 (t— 1600) Cal-
ories, or per atomic weight = 20,664+14 (t — 1600) Calories.
We, therefore, can write the equation:
95,135 + 60.54 (t— 2200) + 2[20,664+14 (t— 1600)] = 197,000
whence t = 2694°
A similar calculation made for the reduction of MnO by the
theoretical amount of aluminium, shows a reduction tem-
perature less than the melting point of alumina. This would
mean that the melting down of the mass to a fused slag of
pure alumina could not take place. What happens in the re-
duction of manganese is that an excess of manganous oxide is
used, whereby all the aluminium is consumed, and none at all
gets into the reduced manganese, and, furthermore, the excess
of manganous oxide unites with the alumina to form a slag
of manganous aluminate, which is fusible at the temperature
attained. Without the latter arrangement no fused slag could
result.
Similar calculations, made with silicon as the reducing
agent, show similar difficulties regarding the theoretical tem-
peratures attainable, when iron or manganese are reduced.
By using an excess of the oxides of these metals, however,
calculation shows that temperatures sufficient to fuse the
metals and the manganous silicate, or ferrous silicate produced,
ought to be obtained.
CHAPTER rv.
THE THERMOCHEMISTRY OF HIGH TEMPERATURES.
The problem is: Knowing the heat evolved (or absorbed)
in the formation of a compound, or in a double reaction, start-
ing with the reacting materials cold, and ending with the pro-
ducts cold, what is the heat evolved, (or absorbed) in either of
the following cases?
(a) Starting with the reagents cold and ending with the
products hot.
(b) Starting with the reagents hot and ending with the
products hot.
(c) Starting with the reagents hot and ending with the pro-
ducts cold.
Of these three cases (b) is the most general form of the prob-
lem, and occurs frequently in metallurgical practice, particu-
larly in electrometallurgy; (a) is a more limited fomi, and
requires less data for its calculation, while it is very frequently
the desideratum in discussing the thermochemistry or heat re-
quirements of a metallurgical process; (c) is derivable at once
if the data exist for calculating (b), and is of such rare occur-
rence in practice that we can dispense with its lengthy dis-
cussion.
The thermochemical data already given and described in pre-
ceding sections enable us to calculate the heat of any chemical
reaction starting with cold reagents and ending with the pro-
ducts cold. For instance: (Zn, 0) = 84,800 Calories means
that if we take 65 kilograms of solid zinc, at room tempera-
ture, and 16 kilograms of oxygen, as gas at room temperature,
ignite them, and after the reaction cool the 81 kilograms of
zinc oxide formed down to the same starting temperature,
there will be developed a total of 84,800 Calories. Similarly,
using the datum (C, 0) = 29,160 Calories, we can construct
61
62 METALLURGICAL CALCULATIONS.
9.nd interpret the reduction reaction, starting with cola ma-
terials and finally ending with cold materials, as follows:
ZnO + C = Zn + CO
—84,800
+ 29,160
—55,640
This reaction, as interpreted, stands for none of the above
cases (a), (6) or (c); in fact, it represents only a calorimetric
determination in the laboratory, and does not correspond to
either of the three cases actually taking place in practice, that
is, it is not directly applicable to practical conditions, without
being modified by the conditions actually arising in practice.
Case (a): If we, in practice, start with the reagents cold,
and the products pass away from the furnace hot, at some de-
termined temperature t, the total heat energy necessary to
cause this transformation is calculable in two ways. The first
way is to follow the course of the' reaction, and to say that
the total heat absorbed is that necessary to heat the reacting
bodies to the temperature t, plus the heat of the chemical re-
action assumed as starting and finishing at that temperature.
The first of these items can be obtained if we know the sensi-
ble heat in the reacting bodies up to the temperature t; it is
a question of specific heats of the reacting bodies up to t (in-
cluding any physical changes of state occurring in them be-
tween o and t); the second item requires a knowledge of the
heat of formation of all the compounds involved, starting with
their ingredients at t, and ending with the products at t. But
the latter item is the general question of the heat of a reaction
starting with the ingredients hot and ending with the^ pro-
ducts hot; it is the most general case, which we have desig-
nated as Case (6), and which will be discussed later. This
way of solving Case (a), therefore, really includes the solu-
tion of Case (b), and we will defer its consideration for the
present. The second method of solving Case (a), and one
which does not involve the more general solution, is to take
the heat of the reaction, starting with the ingredients cold and
ending with the products cold — the ordinary heat of the re-
action from ordinary thermochemical data — and to add to this
THERMOCHEMISTRY OF HIGH TEMPERATURES. 63
the amount of heat which would be required to raise the pro-
ducts from zero to the temperature t. This, of course, does
not actually represent the sequence in which the reactions take
place in practice, but it accurately represents the heat involved
or evolved in passing from the cold reagents to the hot pro-
ducts, and is, therefore, exactly the practical quantity which
we are endeavoring to find. Moreover, it involves a knowledge
of only the specific heats of the products, and not that of the
ingredients or substances reacting.
Illustration: Starting with a mixture of zinc oxide anTl car-
bon in the proportions Zn O and C, at ordinary temperature,
and shoveling them cold into a retort, how much heat is- ab-
sorbed in converting them into Zn vaipor and CO gas, issuing
from the retort at 1300° C. ?
If we start to calculate this quantity, by first finding the heat
necessary to heat Zn O and C up to 1300°, that is obtained by
midtiplying the weights of each by their mean specific heats
from 0° to 1300°, and then by 1300, as follows:
Calories.
81 kilos. ZnOX*[0.1212 (1300) +0.0000315 (1300)^] = 17,075
12 " C X*[0.5 (1300)— 120] = 6,360
Sum = 23,435
To this must then be added the heat of the reaction ZnO +
C = Zn + CO, starting with the reacting bodies at 1300°, and
ending with the products at the same temperature. This can
only be found by solving the general Case (b) for this par-
ticular reaction, which we will find involves a knowledge of the
heat required to raise Zn, O, C. ZnO and CO from 0° to t.
Anticipating such a solution, we may say that the heat of the
reaction starting and ending at 1300° is — 80,166 Calories,
making the sum total of energy required 103,601 Calories.
The solution is usually much simpler if we take the second
method, and add to the heat of the reaction, starting and
ending at zero ( — 55,640 Calories), the heat required to raise
65 kilos, of zinc and 28 kilos, (22.22 cubic meters) of carbonous
*Heat in 1 kilo of carbon, for temperature above 1,000°, 0.5 t - 120
(deduction from Weber's results) ; zinc oxide 0.1212 t + 0.0000315 1' (de-
termination by the author).
64 METALLURGICAL CALCULATIONS.
oxide, from their ordinary condition at zero to their normal
condition at 1300°. The calculation for the CO gas is simply:
22.22X[0.303 (1300) +0.000027 (1300)'] = 9,766 Calories.
For the zinc, the calculation is more complicated:
Heat in solid zinc to the melting point
(420°) :
. 65 X [0.09058 (420) +0.000044 (420)=*] = 2,977 Calories.
Latent heat of fusion 65X22.61=1,470 «
Heat in melted zinc, 420° to boiling point
(930°) :
65X[0.0958 + 0.000088 (420)]X (930—420) = 4,228
Latent heat of vaporization (Trouton's rule)
28 X (930 + 273) = 27,670 "
Heat in zinc vapor (monatomic) 5X (1300
—930) = 1,850 Calories.
Sum = 38,195 "
The total sensible heat required is, therefore, 9,766 + 38,195
= 47,961 Calories, which, added to the 55,640 absorbed in the
chemical reaction, if it started and ended at zero, makes a total
heat requirement of 103,601 Calories for the practical carrying
out of this reaction, starting with the reagents cold and ending
with the hot products at 1300°.
[To be absolutely accurate, regard should be paid in the
above case to the fact that the above calculations are based
on the substances being all at atmospheric pressure, while in
the mixture of Zn vapor and CO gas each is under only 0.5
atmospheric tension. Since each of these represents a molecular
weight, the outer work which has been included in the calcu-
lations is 2X2T = 2X2 (1300 + 273) = 6292 Calories, whereas
it should really be only half that much, or 3146 Calories. The
corrected heat required is, therefore, 103,601 — 3,146 = 100,455
Calories, or 1,545 Calories per kilogram of zinc. This datum
is exactly the net heat requirement on which calculations of.
the net electrical energy required to produce zinc from its oxide,
or calculations of the net efficiency of an ordinary zinc furnace,
would be based.]
THERMOCHEMISTRY OF HIGH TEMPERATURES. 65
Case (b) : To calculate the heat of a chemical reaction start-
ing and finishing at any temperature t, two methods are avail-
able; The most general solution, and that easiest to understand,
is to calculate for each compound involved the heat of its
formation at the temperature t, that is, the heat evolved if
the elements start at t and the product is cooled to t. Having
these heats of formation at t, they are used in the equation
in just the same manner as the heats of formation at zero
are ordinarily used in obtaining the heat of the reaction start-
ing and ending at zero. The calculations are based on this
general principle: The heat evolved when the cold elements
unite to form the hot product at temperature t equals the heat
of union at zero, minus the heat necessary to raise the product
from zero to t ; if to this difference we add the heat which would
be necessary to heat the uncombined elements from zero to t,
the sum is the desired heat of formation at t°.
Illustration: The heat of formation of ZnO at zero is 84,800,
starting with cold Zn and O and ending with cold ZnO. If
we started with cold Zn and 0 and ended with hot ZnO, say
at 1300°, the heat evolved altogether would be less than 84,800
by the sensible heat in the 81 kilograms of ZnO at 1300°, which
has already been calculated (see previous illustration) to be
17,075 Calories. The difference is 67,725 Calories, and repre-
sents the transformation from cold Zn and O to hot ZnO.
If the Zn and 0 were heated to 1300° before combining, they
would contain as sensible heat the following quantities:
Heat in 65 kilograms of Zn (0 to 1300°)
. already calculated" 38,195 Calories
Heat in 16 kilograms of O = 11.11 m'X
[0.303 (1300) -F 0.000027 (1300)2] = 4,884
Sum 43,079
And the heat evolved in passing from the hot reagents to the
hot product at 1300° must be 67,725 plus this sensible heat,
or 110,804 Calories. We can express this datum as follows:
(Zn, O)'""" = 110,804, which means that when the zinc and
oxygen taken in their normal state at 1300° combine to form
ZnO at 1300°, 110,804 Calories are evolved.
66 METALLURGICAL CALCULATIONS.
A similar calculation for CO is as follows :
(C, O) = 29,160 Calories
Sensible heat in CO (0 to 1300°) already cal-
culated = 9,766
Heat evolved when cold C and 0 form CO
at 1300° = 19,394
Sensible heat in C (0 to 1300°) = 6360
« 0 ( " ) = 4884 11,244 Calories
Heat evolved hot C and 0 to hot CO = Sum = 30,638
Or (C, O)""" = 30,638
Uniting the two data found, for the heats of formation of
ZnO and CO at 1300°, in the equation of reduction, we have
ZnO + C = Zn + CO
atl300 ° = —110,804 | +30,638
—80,166
The actual reduction is thus seen to absorb 24,526 Calories
more at 1300° than at zero, an increase of over 42 per cent.
If to this heat of reaction at 1300° we add the heat neces-
sary to raise the ZnO and C to 1300° (already calculated under
Case (a) as 23,435 Calories), we will have a total heat require-
ment of 103,601 Calories, which would be required practically
if we started with cold ZnO and C and ended with the hot
Zn and CO. This agrees, as it indeed must, with the sum
of the heat absorbed in the reaction at zero, 55,640, increased
by the sensible heat in Zn and CO at 1300°, already found to
be 47,961, or a total of 103,601.
Another method of calculating the heat of the reaction
ZnO + C = Zn4-C0 at 1300°, without using the heat of forma-
tion of ZnO and CO at 1300°, is based on the following gen-
eral principle: If from the heat of any reaction, starting and
ending cold, there be subtracted the heat necessary to raise
the products from 0 to t, the difference is the heat of the trans-
formation from cold reagents to hot products; if to this be
added the heat which would be contained in the reagents if
they were heated to t, the sum is the heat of transformation
from heated reagents to heated products, all at the tempera-
ture.
THERMOCHEMISTRY OF HIGH TEMPERATURES. &1
Illustration: The heat of the reaction we have been study-
ing, starting and ending cold, is — 55,640 Calories
Sensible heat in Zn at 1300° =38,195
" CO " = 9,766 47,961
Difference— 103,601
Sensible heat in ZnO at 1300° = 17,075
" " C " = 6,360 23,435
Sum— 80,166 "
The above is the simplest way of calculating the heat of any
chemical reaction at any desired temperature, since it involves
the knowledge of the heat capacities of only those substances
which occur individually in the reaction, and not that of the
elements of which the compounds present are composed. The
above calculation, for instance, involves the heat capacities
of ZnO, C, Zn, and CO, but not that of oxygen, which does not
occur free in the reaction.
Case {c) : This hardly needs discussion, because if we have
the data for calculating Case (b) , this case can be easily worked.
If to the heat of the reaction at t° we add the heat given out
by the products in cooling from t to 0, the sum is the total
heat evolved in passing from the hot reagents to the cold pro-
ducts. This solution of the problem presupposes, however,
the solution of Case (b) and requires the maximum amount
of data, but it has the advantage of following and representing
the logical course of the reaction. A simpler solution is to add to
the heat of the cold reaction at zero, the heat necessary to heat
the ingredients to t, and the sum will be the quantity required.
Illustration: Taking the same case as before:
Heat of reaction at 1300° = —80,166
Sensible heat in products at 1300° = 47,961
or. Sensible heat in reagents at 1300°
Heat of reaction at zero
The reasoning involved in the above is simply that, starting
with the reagents hot and ending with the products cold, the
Sum =
—32,205
=
23,435
—55,640
Sum =
—32,205
68 METALLURGICAL CALCULATIONS.
heat evolution must be the same whether we suppose the sys-
tem to pass along the one path or the other.
General Remarks.
It will be evident from this enunciation of principles and the
illustrations adduced, that for many practical purposes the
calculation according to Case (a) will suffice for finding the
net energy involved in many metallurgical operations; it gives
the sum total of energy necessary to be supplied to pass from
the cold reagents to the hot products as they come from the
furnace, but the two items of which this sum is tomposed do
not actually represent the two items of the division of this
total in the furnace, i.e., into heat necessary to heat the re-
agents to the reacting temperature and heat actually absorbed
as the reaction takes place. The latter item can only be calcu-
lated by one of the two methods explained under Case (h),
and then the former item can be obtained by difference, or by
direct use of the heat capacities of the reacting substances.
It is, of course, obvious that this whole subject of calcu-
lating the thermochemistry of high temperatures (as distin-
guished from the ordinary zero thermochemistry) necessitates
the use of all available data regarding specific heats in the
solid, liquid and gaseous states of both elements and com-
pounds, and a:lso in many cases of their latent heats of fusion
and vaporization. When, however, the necessary physical data
are known, or can be assumed with approximate accuracy, the
way is open to make many calculations of the greatest value in
practical metallurgy and chemistry. The exact heats of forma-
tion of cherriical compounds at a certain temperature, which
may be, and often are, very different from these values at zero,
and, having frequently different relations to each other, will
explain in many cases many hitherto little understood and ap-
parently contradictory reactions. The calculations also enable
us to understand many reactions taking place only at high tem-
peratures, to calculate whether they are really exothermic or
endothermic at the temperatures at which they occur, and to
compare these data with the data as to the heat of such re-
actions obtained by studying the chemical equilibrium of such
reactions and deducing the heat of the reaction from the rate
of displacement of the chemical equilibrium. One such ex-
THERMOCHEMISTRY OF HIGH TEMPERATURES. 69
ample may gufBce to suggest the large field here open to the
scientific metallurgist.
Example: G. Preuner (Zeitschrift fur Physikalische Chemie<
March 15, 1904, p. 385) reports a long and careful investigation
of the equilibrium of the reaction
Fe'O* + 4H2 <"=: 3Fe + iWO
in which he finds the equilibrium constant for different tem-
peratures, and calculates from its value at 960°, by the use of
Van't Hofi's formula, that the heat value of the reduction at
that temperature is — 11,900 Calories, and noticing the great
difference between this value and the value derived from the
ordinary heats of formation (—270,800 + 4(58,060) = —38,560),
he concludes that the Van't Hoff formula gives wrong re-
sults when applied to this class of reactions. Now, the facts
are that Preuner's observations were good, Van't Hoff's for-
mula is correct, and applies, but the thermochemical value of
the reaction, — 38,560 Calories is the correct value only for
the reaction beginning and ending at ordinary temperature. Our
thermochemical principles enable us to calculate the heat of
the reaction at 960°, as follows:
Heat of the reaction beginning and ending at
zero = — 38,560 Calories
Heat in products at 960°:
3Fe = 3(56) X [0.21 8 (960) — 39] Pionchon =
28,560
4H='0 = 4(22.22) X[0.34(960) +0.00015
(960)2] _ 41 300
69,860 Calories
Heat in reagents at 960°:
Fe'0< = 232X[0.1447(960) +0.0001878
(960)2]* _ 72_384
4H2 = 4(22.22) X[0.303(960) +0.000027
(960)2] = 28,075
100,459
The heat of the reaction at 960° is, therefore, according to
our method of Case (6),— 38,560— 69,860 + 100,459 = —7,961
* Approximate determination of heat in Fe'O*, made recently in
author's laboratory, Q-= 0.1447t +0.00018781*.
70 METALLURGICAL CALCULATIONS. ~
Calories. It thus appears that Preuner's dilemma was mostly
caused by his thinking that the thermochemical value of the
reaction at zero should be a constant for any temperature: a
proper thermochemical calculation removes the dilemma.
Since the whole treatment of the subject of the thermo-
chemistry of high temperatures requires a knowledge of data
concerning the heat capacity of elements and compounds in the
solid, liquid and gaseous states, as well as of their latent heats
of fusion and vaporization, the next instalment of these cal-
culations will supply these data as far as they are known, and
discuss a number of applications of these principles to various
metallurgical processes.
In order to apply the principles explained in the preceding
discussion, two sets of data are necessary; first, the thermo-
chemical data, such as are ordinarily obtained by laboratory
experiments at laboratory temperatures; second, physical data
concerning the specific heats and latent heats of fusion, and
volatilization of elements and compounds. The first have
been given, at least for all important compounds met with in
metallurgy, in a previous place (p. 18), the latter will now
be discussed and the data presented as far as they have been
determined.
Specific Heats of the Elements.
Dulong and Petit's law announces the fact that the specific
heat of atomic weight of a soHd metal is nearly constant, the
value varying between 6 and 7, and averaging 6.4. This gen-
eralization was made chiefly upon the basis of the specific heats
of the metals, as determined in the range 100° C. to 10° or 15°
C, such as in Regnault's accurate experiments. About the only
notable exceptions to this rule are carbon, boron and silicon,
and it has been naively remarked by more modem physicists
that these exceptions to the rule disappear if we find the spe-
cific heat of these three elements at high temperatures, that,
for instance, the specific heat of carbon above 1000° C. is 0.5.
making its atomic specific heat 0.5X12 = 6, and therefore
the exceptions to the rule are all accounted for. Now, the
rule is an important one, and has done good service, but the
exceptions just noted and their behavior at high temperatures
really prove that the rule must be made more general, or else
THERMOCHEMISTRY OF HIGH TEMPERATURES. 71
abandoned altogether. The fact is, that the specific heats of
almost all the solid elements increase with the temperature
at a rate equal to an increase of about 0.04 per cent, of their
value for each degree centigrade, so that the atomic specific
heat of the majority of the elements, which is about 6.4 at
ordinary temperatures, becomes about 40 per cent, greater at
1000° C. for such elements as are not melted at that tempera-
ture. Therefore, while we may say that at ordinary tem-
peratures the specific heat of atomic weight of a solid element
is 6.4, and its specific heat per unit of weight is 6.4, divided
by the atomic weight, yet it will be more accurate, if actual
determinations have not been made, to assume that the
actual specific heat increases 0.04 per cent, for every degree
rise in temperature, and mean specific heat to zero half that
fast.
The specific heat in the liquid state has not been determined
for many elements. It is in general higher than the specific
heat of the solid at ordinary temperatures; in fact, it appears
to be more nearly equal to the specific heat of the solid element
just before fusion, and may be so assumed if no determina-
tions have been made. It, is found, furthermore, not to change
perceptibly with rise of temperature, so that it may be as-
sumed constant.
The specific heat of the gaseous elements has been deter-
mined only for those- which are gaseous at low temperatures.
For the metals which, as far as known, have monatomic vapors,
i.e., vapors in which the atoms exist alone and uncoupled with
each other, the specific heat of atomic weight, occupying 22.22
cubic meters at standard conditions, should be theoretically 5.0
Calories at constant pressure, or 0.225 per cubic meter. We
may thus estimate the specific heat of metallic vapors which
have not been determined.
Latent Heats op Fusion of the Elements.
The passage from the solid to the liquid state is in all cases
accompanied by an absorption of heat, which in amount varies
from one or two Calories up to 100 Calories per unit of weight..
This quantity has been most frequently determined by finding
calorimetrically how much heat is given out by unit weight
of the melted element just at its setting point, in cooling to
72 - METALLURGICAL CALCULATIONS.
ordinary temperatures, and subtracting from this the heat in
unit weight of the solid substance at the melting point, a"
determined most accurately by exterpolating the value of
mean specific heat of the soHd up to the melting point. In
this manner the latent heat of fusion for a number of elements
has been directly determined.
If a crucible full of melted metal is allowed to cool, the
temperature falls regularly until the melting point is reached,
and then stays constant, or nearly so, for some time, while
the metal is setting. A comparison of the rate of cooling
before and after setting, with the length of time during
which the temperature was constant, gives the relative value
of the latent heat of fusion in terms of the specific heat of
the melted metal and of the solid metal near to the melting
point.
While the latent heat of fusion per unit weight shows no
perceptible regularities, it is found that as soon as the latent
heat of fusion is expressed per atomic weight of the element
(analogous to specific heat of atomic weight) that notable
regularities appear. The elements with high melting points
have high atomic heats of fusion, and vice versa, so that if
the elements are arranged in the order of their melting points
their latent heats of fusion per atomic weight of each are in
the same order, and very nearly proportionately so. If, for
instance, a chart or diagram is made, using the melting points
as abscissas, and latent heats of fusion of atomic weights as
ordinates, the latter will lie very nearly in a straight line.
Numerically, if the melting points be expressed in degrees of
absolute temperature (centigrade temperatures plus 273), the
latent heats of fusion of atomic weights average about 2.1
times the temperature of the melting point. This rule may be
used to predict an undetermined latent heat of fusion.
In addition to the above general rule, another one bearing on
the same question was also discovered and applied by the
writer. (See Journal of the Franklin Institute, May, 1897.)
According to this observation, the continued product of the
latent heat of fusion of atomic weight and the coefficient of
expansion and the cube root of the atomic volume (atomic
weight divided by specific gravity) is a constant. If the co-
efficient of linear expansion between 0° and 100° C. is used.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 73
the constant is 0.095, or if the actual linear expansion of unit
length from 0° to 100° C. is used, the constant is 9.5. This
rule, applied to all elements whose latent heat of fusion is
known, gives satisfactory agreements, and enables us therefore
to predict the latent heat of fusion of nearly a dozen other
elements for which the coefficient of expansion is known.
Latent Heats of Vaporization of the Elements.
This datum has been determined for but a very few elements.
Some of the metalloids, like sulphur, phosphorous and arsenic,
are known to become complex vapors immediately above their
boiling point, corresponding to such formulae as S°, P*, As*;
the metals, as far as they have been tested, pass into monatomic
vapors, such as Na, K, Hg, Zn and Cd, in which each atom
represents a molecule. In the latter cases the following gen-
eralization may be made: The latent heat of vaporization of
atomic weight is proportional to the absolute temperatures of
the boiling point at atmospheric pressure, and is numerically
equal to about twenty-three times that temperature (twenty-
one times, if the outer work of overcoming the atmospheric
pressure be not included). From this rule it is possible to
estimate the amount of heat necessary to vaporize any metal
whose boiling point under atmospheric pressure is known.
Examples: The boiling point of carbon under atmospheric
pressure is 3,700° C, and if its vapor is monatomic, the latent
heat of vaporization is, for an atomic weight of carbon (C =
12), 23X (3700 + 273) = 92,080 Calories, equal to 7,673 Calories
per kilogram, of carbon. If the vapor is diatomic, and its for-
mula C, then the above latent heat is for twenty-four parts of
carbon, and for one part by weight is 3,837 Calories. Other
considerations, from thermochemistry, make the latter value
the more probable one.
The boiling point of cadmium is 772° C, and its vapor is
known to be monatomic, what is its latent heat of vaporiza-
tion? The atomic weight being 112, the latent heat of vapor-
ization of this quantity is 23 X (772 + 273) = 24,035 Calories,
which is 215 Calories per kilogram.
In making such calculations it must be strictly observed that
the boiling point under atmospheric pressure is to be used, and
not any temperature at which vapors may appear at partial
74 METALLURGICAL CALCULATIONS.
tensions which may be only small fractions of atmospheric
pressure.
Thermophysics of the Elements.
Having laid down the laws and the empirical rules which
appear to govern these phenomena, we will now discuss the
data for the common elements, giving both what is known and
what may be assumed as probably true wherever actual deter-
minations have not been made. The elements will be taken
up in the order of their atomic weights, — the only scientifically
logical order.
In all cases, the actually measured mean specific heats, Sm,
will be given. In the case of gases, these will be the mean
specific heats under constant pressure ; if they are desired under
constant volume the amount of outer work must be calculated
in Calories (two Calories per degree for a molecular weight
of a gas, 0.09 Calories per degree for 1 cubic meter, and 0.09-7-
weight of 1 cubic meter for a kilogram of gas), and subtracted
from the specific heat at constant pressure. The specific
heat of 1 cubic foot is, in pound Calories, the specific heat per
cubic meter divided by 35.32 and multiplied by 2.204, or, in
brief, multiplied by 0.0624; in British thermal units it will
be the same as in pound Calories, since t is then Fahrenheit
degrees.
If, from the data given, it is desired to find the mean specific
heat between any two temperatures, t and t', instead of the
mean specific heat from t to 0°, as given directly by the for-
mula, it need only be observed that Sm from t' to t is obtain-
able by finding Sm from 0° to (t'-l-t). If, for instance, Sm
(o to t) = 0.303-|-0.000027t, then Sm (t to t') = 0.303-1-
0.000027 (t'-ft). Furthermore, if the actual specific heat at
any temperature t is desired, it is equal to the mean specific
heat from 0° to 2t-, e.g., in the above cases, S (at any tempera-
ture t) = 0.303-1-0.000027 (2t) = 0.303 + 0.000054 (t).
Temperatures will be always given and represented in centi-
grade degrees, except where the specific heat is given in British
thermal units, in which case I represents, of course, Fahrenheit
degrees, and will also be printed in italics, t, to further dis-
tinguish it from t in centigrade degrees. Absolute tempera-
tures, if used, will be designated as T, and are equal to t4-273.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 75
The vapor tensions of all the elements are expressed in the
form log p = — ^ + B , where p is used in millimeters of mercury, T
is the corresponding temperature in absolute measure (K°), and
A and B are constants for the elements in question, which can be
found if merely two values of p and T are known. The latent
heat of vaporization of molecular volume is, thermodynamically,
4.57 A. B is found nearly constant for all elements, averaging
7.9 to 8.4 in the case of the liquid element and 8.85 for the solid
element. If only one value of p and T are known, then B is
assumed either 7.9 for elements of low boiling point or 8.4 for
elements of high boiling point. A little study of the question
will show that these values correspond to using 23 or 25.1 for
Trouton's constant (KT) expressing the latent heat of vapori-
zation of molecular volume at p = 760 mm. (A/T = Trouton's
constant/4.56). The vapor tension at the melting point is cal-
culated by putting T = M. P., at which point the liquid and solid
have the same vapor tension. This gives one value of p and T
A'
in the analogous formula for the solid state, (log. p= — -;p-|-B'),
where A' can be calculated from A and the latent heat of fusion,
and B' thus becomes known. The vapor tension down to 0° C.
is then calculated from this formula.
Hydrogen.
From the experiments of Mallard and LeChatelier we deduce:
Sm (0-t) 1' kilogram (up to 2000° C.) =3.370+0.0003t
Calories.
1 pound (up to 2000° C.) =3.370-|-0.0003t
I pound Calories.
1 pound (up to 3600° F.) =3.370+0.00017i
B. T. U.
1 cu. meter (up to 2000° C.) = 0.303 -|-0.000027t
Calories.
1. cu. foot (up to 2000° C.) =0.0189-f-0.0000017t
pound Calories.
1 cu. foot (up to 3600° F.) =0.0189-)-0.0000009i
B. T. U.
For higher temperatures, such as electric furnace heats be-
76
METALLURGICAL CALCULATIONS.
tween 2000° and 4000° C..(3600° to 7200° F.), Berthelot and
Vielle have made experiments which give us:
Sm (0 — ^t) 1 kilogram
1 pound
1 pound
1 cubic meter
1 cubic foot
1 cubic foot
Sm (0— t)
S (at M. P. = 179°)
Q (solid, at M. P.)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
Q (liquid at B. P. ,60 = 1450°)
L. H. Vaporization (1450°)
Q (vapor; 1450°)
S (vapor) per cubic meter
per kilo
Vapor tension, liquid :
at M. P.
solid:
at 0° C.
= 2.75 +0.0008t Calories.
= 2.75 +0.0008t pound Cal.
= 2.75 +0.00044* B. T. U.
= 0.2575 +0.000072t Calories.
= 0.0161 +0.0000045t pound Cal.
= 0.0161 +0.0000025* B. T. U.
Lithium.
= 0.7895 +0.0024t+
0.0000063t=' (Bernini).
= 2.255.
= 254 Cal.
= 136 Cal. (calculated by 2.1
Trule).
= 173 Cal. (calculated by
second rule).
= 390 Cal. (calculated).
= 2.255 (assumed).
= 3483 Cal. (calculated).
= 5660 Cal. (Trouton's rule,
K = 23).
= 9140 Cal. (calculated).
= 0.225 (assumed).
= 0.714 (calculated).
1 8690 , „ „„
log p = - ^^ +7.92
(B = 7.92, assumed).
= 4.9 X 10-" mm. (calculated).
logp = -«?0%8.38.
= e.OXlO-^^mm. (calculated).
Beryllium (Glucinum).
Sm (0-t) = 0.38+0.0004t (Nilson & Pettersson).
S (at M. P. = 1430°) = 1.52.
Q (solid, at M. P.) = 1358 Cal.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 77
L. H. Fusion = 400 cal. (calculated by 2.1 T rule).
Q (liquid, at M. P.) = 1758 Cal.
S (liquid) = 1.52 (assumed).
Boron.
Sm (0— t) (t under 233°) =0.22 + 0.00035t (Weber).
S (solid, at 500°) = 0.57 (by extension of above
formula) .
S (solid, above 500°) = 0.57 (assumed to remain
constant).*
88
Sm (0— t) (t over 500°) = 0.57 — ^•
Q (solid, at M. P. = 2500°) = 1337 Cal. (calculated).
L. H. Fusion = 265 Cal. (calculated by 2.1
T rule, and molecule B2).
Q (liquid, at M. P.) = 1602 Cal (calculated).
S (liquid) = 0.57 (assumed).
Q (liquid at B. P.760 = 3700°) = 2285 Cal.
L. H. Vaporization (3700°) = 4155 Cal. (Trouton's rule,
K = 23, and molecule B2).
Q (vapor, 3700°) = 6440 Cal.
S (vapor) per cubic meter = 0.315 (assumed for B2).
per kilo = 0.318 (calculated).
19 950
Vapor tension, liquid : log p = - -~ 1-7.92 (B = 7.92
assumed).
at M. P. =5 mm.
,., , 21,230 , „ ,,
solid: log p = „ 1-8.36.
atO°C. = 3.9 X 10-" mm. (calculated).
Carbon.
{Amorphous and Graphitic.)
Sm (0°— t°) (up to 250°) = 0.1567+0.00036t.
(0°— 1°) (250°— 1000°) = 0.2142 + 0.000166t.
S at 1000° = 0.528 (VioUe); 0.546 (Weber).
Q at 1000° = 380 Cal.
106.5
Sm (0-t) (t over 1000°) = 0.4430-t-0.0000425t -^
(VioUe).
*This figure would give atomic specific heat = 6.27, and we assume this
constancy by analogy with similar behavior of carbon.
78
METALLURGICAL CALCULATIONS.
Q (0— t) (t over 1000°) = 0.4430t+0.0000425t2- 106.5.
Q (solid) (atB. P.veo = 3700°) = 2114 Cal.
L. H. Sublimation (3700°) = 4140 Cal. (Richard's rule,
K = 2.1, and Trouton's rule, K = 23;forC2).
Q (vapor) (3700°)
S (solid, at M. P. = 4400°)
S (vapor) per cubic meter
per kilo
Q (solid) (at M. P. 4400°)
L. H. Fusion (4400°)
Q (liquid) (4400")
S (liquid)
Q (vapor) (4400°)
Vapor tension, solid:
= 6254.
= 0.777 (Violle's equation).
= 0.315 (assumed).
= 0.292 (calculated for C2).
= 2666 Cal. (by Violle's equation).
= 409 Cal. (by 2.1 T rule for C2).
= 396 Cal. (by second rule, for C2).
- = 3066 Cal. (calculated).
= 0.777 (assumed).
= 6806 Cal. (calculated).
logp= -?^+9.0 (B' = 9.0,
assumed).
= 6310 mm. = 8.3 atmospheres.
at M. P.
f -A 1 22,195 , „ ^
liquid: logp= 'j^ 1-8.5.
at 0°C. = 1 X 10-8" mm. (calculated) .
Nitrogen.
Sm (0— t) 1 kilogram (up to 2000° C.)
1 pound (up to 2000° C.)
1 pound (up to 3600° F.)
= 0.2405 +0.0000214t
Cal.
= 0.2405 +0.0000214t
pound Cal.
= 0.2405+0.0000119i
B. T. U.
1 cubic meter (up to 2000° C.) = 0.303 +0.000027t
Cal.
1 cu. foot (up to 2000° C.) = 0.0189+0.0000017t
pound Cal.
1 cu. foot (up to 3600° F.) = 0.0189 +0.0000009*
B. T. U.
For temperatures between 2000° and 4000° C. the following
are the values of the mean specific heats to zero :
Sm (0— t) 1 kilogram = 0.2044+0.000057t Cal.
1 pound . = 0.2044+0.000057t pound Cal.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 79
1 pound = 0.2044+0.000032* B. T. U.
1 cubic meter = 0.2575 +0.000072t Cal.
1 cubic foot = 0.1601+0.0000045t pound Cal.
1 cubic foot = 0.0161+0.0000025* B. T. U.
Oxygen.
Sm (0— t) 1 kilogram (up to 2000° C.) = 0.2104+0.0000187t
Cal.
1 pound (up to 2000° C.) = 0.2104+0.0000187t
pound Cal.
1 pound (up to 3600° F.) = 0.2104+0.0000104*
B. T. U.
1 cu. meter (up to 2000° C.) = 0.303 +0.000027t Cal.
1 cu. foot (up to 2000° C.) = 0.0189+0.0000017t
pound Cal.
1 cu. foot (up to 3600° F.) = 0.0189+0.0000009*
B. T. U.
Sm (0— t) 1 cu. meter (2000° -4000°C) =0.2575 +0.000072t Cal.
leu. foot (2000° -4000° C.) = 0.0161 +0.0060045t
pound Cal.
leu. foot (3600° - 7200° F.) = 0.0161+0.0000025*
B. T. U.
1 kilogram (2000°-4000° C) = 0.1788+0.00005t Cal.
1 pound (2000°-4000° C.) = 0.1788+0.00005t
pound Cal.
1 pound (3600°-7200° F.) = 0.1788+0.00003t
B. T. U.
Sodium.
Sm (0— t) ' = 0.2932+0.00019t (Bernini).
Q (solid, at M. P. = 98°) = 30.6 Cal. (by formula).
L. H. Fusion ' = 27.2 (Rengade).
Q (liquid, at M. P.) = 57.8 Cal.
S (liquid, 98°— 100°) = 0.333 (Bernini).
Q (liquid at B. P.jeo = 877°) =317 Cal. (calculated).
L. H. Vaporization (877°) = 905 Cal. (from vapor tension
constants).
Q (vapor, at 877°) = 1222 Cal.
S (vapor) for cubic meter = 0.225 (assumed),
per kilo = 0.218 (calculated).
80
METALLURGICAL CALCULATIONS.
4565
Vapor tension, liquid : logp = ^p- +6.85 (from Gebhardt's
data)
at M. P. = 3.55X10-5 mm. (calculated),
solid: log. p =- ^+7.30.
at 0° C. = 8.1 X 10"' mm. (calculated).
Magnesium.
Sm (0— 1°)
Q (solid, at M. P. = 650°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
Q (liquid, at B. P. 750 = 1120°)
L. H. Vaporization (1120°)
Q (vapor, at 1120°)
S (vapor) per cubic meter
kilo
Vapor tension, liquid : log p
at M. P.
solid : logp
at 0° C.
= 0.2372+0.000093t+
0.0000000685^ (Stucker).
= 212 Cal. (by formula).
= 70 Cal. (Roos).
= 282 Cal.
= 0.445 (assumed; same as solid
at M. P.).
= 491 Cal. (calculated).
= 1443 Cal. (Trouton's rule,
K = 25.2).
= 1934 Cal.
= 0.225 (assumed).
= 0.208 (calculated).
= ™-+8.4 (constants from
Trouton's rule).
= 1.15 mm. (calculated).
--^ + 8.81.
= 5.25X10-'"' mm. (calculated).
Aluminium.
Sm(0— 1°)
Q (solid, at M. P. = 657°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
= 0.2220+0.00005t (Richards).
= 167.4 Cal. (from formula).
= 90.9 Cal. (Richards).
= 258.3 Cal. (Richards).
= 0.308 (Pionchon).
Q (liquid, at B. P.jeo = 2200°) = 733.5 Cal. (calculated).
L. H. Vaporization (2200°) = 2305 Cal. (Trouton's riile,
K = 25.2).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 81
Q (vapor, at 2200°) = 3039 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
kilo = 0.185 (calculated).
Vapor tension, liquid : log p = t^ 1-8.4 (constants from
Trouton's rule).
at M. P. = 5.25 X 10-' mm. (calculated).
solid : logp = -^!^^^+8.98.
at 0° C. = 1.0X10-*' mm. (calculated).
Silicon.
Sm(0— t°)(up to 234°) = 0.17+0.00007t (Weber).
Q (solid, at M. P. = 1430°) = 386 Cal. (by extension of for-
mula).
L. H. Fusion = 128 Cal. (by 2.1 T rule).
= 108 Cal. (by second rule).
Q (liquid, at 1430°) = 494 Cal.
S (liquid) = 0.37 (assumed, same as solid
at M. P.).,
Q (liquid, at B. P.reo = 2300°) = 816 Cal. calculated).
L. H. Vaporization (2300°) = 1153 Cal. (Trouton's rule, K
= 25.2, vapor = Sia).
Q (vapor, at 2300°) = 1969 Cal.
S (vapor) per cubic meter = 0.315 (assumed, for Si2).
per kilo = 0.125 (calculated).
14 200
Vapor tension, liquid : log p = ^+8.4 (constants from
Trouton's rule).
at M. P. = 1.32 mm. (calculated).
solid: logp= -i^+9.00.
at 0° C. = 8X10-" mm. (calculated).
Phosphorus.
Sm(-20to7°) =0.1788.
Q (solid, at M. P. = 44°) = 8 Cal.
L. H. Fusion (44°) = 5 Cal. (Person).
Q (liquid, at 44°) = 13 Cal.
S (liquid, 44° to 98°) = 0.205.
Q (liquid at B. P.760 = 287°) = 63 Cal.
82
METALLURGICAL CALCULATIONS.
L. H. Vaporization (287°) = 130 Cal.
Q (vapor, at 287°) = 193 Cal.
S (vapor) per cubic meter = 0.405 (assumed, for Ps).
per kilo = 0.097 (calculated).
Vapor Tension, liquid : log p = ^^-+7.0.
at M. P. = 1.63 mm.
1-j 1 2260 , _ „ .
solid : log p = Tfi — 1-7.34.
at 0° C.
T
= 0.115 mm.
Sulfur.
Sm(15°— 97°)
Q(solid, at M. P. = 113°)
L. H. Fusion =
Q (liquid, at M. P.)
Sm(0—t°) (liquid, 114°--445°) =
Q (liquid at B. P.jeo = 444°.5) =
L. H. Vaporization (444°.5)
Q (vapor, at 444°.5) =
S (vapor) per cubic meter =
per cubic meter =
per cubic meter, =
per kilo =
Vapor tension, liquid : log p
at M. P.
solid : log p
at 0° C.
0.18 (Regnault).
20 Cal.
9.4 (Person).
29.4 Cal.
0.338+0.000187t-
0.00000058t2.
126 Cal.
72 Cal.
198 Cal.
0.855 for Sg, close to 445°.
0.674 for Ss, up to 500°.
0.315 for Sa, over 800°.
0.074 for Ss, close to 445°.
0.078 for Se, up to 500°.
0.109 for Ss, over 800°.
0.028 mm.
-iy°+9.i5.
IX 10-8 mm.
Chlorine.
Sm (gas) per cubic meter
per kilo
= 0.40 (13°— 202°) (Regiiault).
= 0.1241 (Regnault).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 83
Potassium.
Sm(0— t°) = 0.1858+0.00008t (Bernini).
Q (solid, at M. P. = 60°) = 11.4 Cal.
L. H. Fusion = 13.6 Cal. (Bernini).
Q (liquid, at M. P.) = 25.0 Cal.
Sm (liquid) (0— 1°) = 0.1422+0.00067t (Rengade).
Q (liquid, at B. P.yeo = 757°) = 181.5 Cal.
L. H. Vaporization (757°) = 607 Cal. (Trouton's rule,
k = 23).
Q (vapor, at 757") = 789 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.128 (calculated).
5190
Vapor Tension, liquid : logp = „ — [-7.92 (constants from
Trouton's rule).
at M. P. = 2.25X10-8 mm.
solid: logp= -^+8.26.
at 0°C. = 7. IX 10-" mm.
Calcium.
Sm(0°— 100°) = 0.1704 (Bunsen).
Q (soHd, at M. P. = 800°) = 136.3 Cal.
L. H. Fusion = 56.3 Cal. (calculated by 2.1
T rule).
Q (liquid, at M. P.) = 192.6.
Titanium.
S (solid, at 0°) = 0.0978 (Nilson and
Pettersson).
Sm (0— 1°) = 0.0978+0.0000035t (assum-
ing atomic specific heat = 10
at M. P.).
Q (solid, at M. P. = 1795°) = 187 Cal. (from above formula).
L. H. Fusion = 90 Cal. (from 2.1 T rule).
Q (liquid, at M. P.) = 277 Cal.
S (liquid) = 0.2084 (assumed, same as
solid at M. P.).
Q (Uquid, at B. P.jeo = 2475°) = 319 Cal.
84 METALLURGICAL CALCULATIONS.
L. H. Vaporization (2475°) = 143 Cal. (from Trouton's rule.
K = 25).
Q (vapor, at 2475°) = 462 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.104 (calculated).
Vapor tension, liquid : log p = ^p 1- 8.4.
at M. P. = 11.75 mm.
solid : log p = - ^^ + 8.85.
atO°C. =8.3X10-" mm.
Vanadium.
Sm (0— 1°) = 0.105+0.00003t.
Q (solid, at M. P. = 1710°) = 267 Cal.
L. H. Fusion = 82 Cal. (by 2.1 T rule).
Q (liquid, at M. P.) = 349 Cal.
S (liquid) = 0.208 (assumed, same as solid
at M. P.).
Chromium.
Sm (0— 1°) (to 600°) = 0.1039+0.00000008t2 (Adler).
Q (solid, at M. P. = 1489°) = 352 Cal.
L. H. Fusion = 71 Cal. (calculated by 2.1 T
(rule).
Q (liquid, at M. P.) = 423 Cal.
S (liquid) =0.24 (estimated).
Q (liquid at B. P. 750 = 2200°) = 594 Cal.
L. H. Vaporization (at 2200°) = 1197 Cal. (Trouton's rule,
K = 25.1).
Q (vapor, at 2200°) = 1791 Cal.
S (vapor), per cubic meter = 0.225 (asstmied).
per kilo = 0.096
Vapor tension, liquid : log p. = m 1-8.4 (constants from
Trouton's rule, K = 25.1).
at M. P. = 4.47 mm.
14 460
solid : log p. = - i^^ + 8.86.
at0°C. = 7.8X10-«mm,
THERMOCHEMISTRY OF HIGH TEMPERATURES. 85
Manganese.
Sm (0— t") = 0.1088+0.000103t.
Q (solid, at M. P. = 1207°) = 281 Cal.
L. H. Fusion = 43 Cal. (calculated by second
rule.)
Q (liquid, at M. P.) = 324 Cal.
S (liquid) = 0.357 (assumed from S solid at
M. P.).
Q (liquid at B. P.760 = 1900°) = 572 Cal.
L. H. Vaporization (1900°) = 995 Cal. (Trouton's rule,
K = 25.1).
Q (vapor, at 1900°) = 1567 Cal.
S (vapor) per cubic meter = 0.225 (asstmied).
per kilo. = 0.091 (calculated).
11 995
Vapor tension, liquid : log p = k h 8.4 (Constants
from Trouton's rule, K = 25.1).
at M. P. = 1.98 mm. Hg.
solid : log p. = - ^^ + 8.85.
at 0° C. = 3.2X10-'8mm. Hg.
Iron.
•Sm(0— 1°) up to 660° = 0.11012+0. 000025t+
0.0000000547t'' (Pionchon;
also Oberhoffer),
Q at 660° = 96.1 Cal.
Q at 750° = 125.6 Cal.
L.H. Change of state (730°) =5.3 Cal. (Pionchon); 5.0
(Leschtschenko) .
S above 750° = 0.1675 (approximately con-
stant.)
Q (0_t°) t = 750° to 1500° = 0.1675t - 51 Cal.
L. H. Change of state (900°) = 6.0 Cal. (Pionchon); 6.1
(Leschtschenko).
Q (solid, at M. P. = 1535°) = 256 Cal.
L. H. Fusion = 66 Cal. (calculated by 2.1 T
rule).
= 69 Cal. (calculated by second
rule).
(4.3% C.) • = 59 Cal. (Schmidt).
86 METALLURGICAL CALCULATIONS.
Q (liquid, at 1535°) = 322 Cal.
S (liquid) = 0.20 (estimated).
Q (liquid at B. P.yeo = 2450°) = 505 Cal.
L. H. Vaporization (2450°) = 1224 Cal. (Trouton's rule,
K = 25.1).
S (vapor) per cubic meter = 0.225 (assumed),
per kilo. = 0.089 (calculated).
Vapor tension, liquid : log p. = — '—Fp" — h 8.4 (Constants
from Trouton's rule, K = 25.1).
at M. P. ■= 1.20 mm. Hg.
solid : log p. = - ^^^ + 8.84.
= at 0° C. = 6.9 X 10-5" mm.
Nickel.
Sm (0— t") up to 230° = 0. 10836 +0.00002233t
(Pionchon).
L. H. Change of State = 4.64 Cal., 230° to 400°,
(Pionchon).
= 3.11 Cal. 363° (Leschtschenko).
ft KK
Sm (0— 1°) t = 440° to 1050° = 0.099 +0.00003375t + ^
(Pionchon).
Q (solid, at M. P. = 1450°) = 221 Cal.
L. H. Fusion = 68 Cal. (calculated by second
rule).
Q (liquid, at 1450°) = 289 Cal.
S (liquid) = 0.197 (estimated).
Q (liquid, at B. P.jeo = 2150°) = 427 Cal.
L. H. Vaporization (2150°) = 1042 Cal. (Trouton's rule,
K= 25.1).
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.086 (calculated).
Vapor tension, liquid : log p. = 'y^ 1- 8.4 (Constants
from Trouton's rule, K = 25.1) .
at M. P. = 4.36 mm. Hg.
1-^ 1 14,245 , „„
solid : log p. = '^ 1- 8.9.
at 0° C. = 5.25 X lO"** mm. Hg.
THERMOCHEMISTRY OF HIGH TEMPERATURES.
Cobalt.
87
Sm (0— 1°) up to 900°
Sm (0— t°)t over 900°
Q (solid, at M. P. = 1490°)
L. H. Fusion
Q (liquid, at 1490°)
S (liquid)
Sm (0— t)
= 0.10584+0.00002287t+
0.000000022t2 (Pibnchon).
= 0.124+0.00004t-^
(Pionchon).
= 259 Cal. (extension of above
formula).
= 68 Cal. (calculated by second
rule).
= 327 Cal.
= 0.243 (assumed, same as solid
at M. P.).
Copper.
= 0.0939+0.00001778t (Frazier
and Richards).
Q (solid, at M. P. = 1083°) = 118.7 Cal.
L. H. Fusion = 43.3 Cal. (Richards).
Q (liquid, at M. P.) = 162.0 Cal. (Frazier & Richards).
S (liquid) ' = 0.156 (Glaser).
Q (liquid, at B. P.reo = 2310°) = 353 Cal. ,
L. H. Vaporization (2310°) = 1087 Cal. (calculated, from va-
por tension curve).
Q (vapor, at 2310°) = 1440 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.079 (calculated).
Vapor tension, liquid : log p. = ij. H 8.75 (from Green-
wood's data),
at M. P. = 3.5 X 10-' mm. Hg.
VA 1 15,765
solid : log p. = T^ 1- 9-17.
at 0° C. = 2.6 X 10-« mm. Hg.
Zinc.
Sm (0— 1°)
Q (solid, at M. P. = 419°)
^= 0.0906+0.000044t.
= 45.2 Cal.
88
METALLURGICAL CALCULATIONS.
L. H. Fusion = 22.6 Cal. (Richards).
Q (liquid, at M. P.) = 67.8 Cal.
S (liquid) = 0.179 (Glaser).
Q (liquid, at B. P.reo = 930°) = 159 Cal.
L. H. Vaporization (930°) = 446 Cal. (calculated from vapor
tension curve).
Q (vapor at 930°) = 605 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
= 0.077 (calculated),
, 6365 ,„,_/,
log p = „ — h 8.17 (from vanous
observations).
per kilo
Vapor tension, liquid :
at M. P,
solid :
at 0° C.
Sm (12° to 23°)
Q (solid, at M. P. = 30°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid, 30°— 119°)
= 9. 3X10- ''mm. Hg.
log p. = - ^5 + 8.63.
= 1.03X10-"mm. Hg.
Gallium.
= 0.079 (Berthelot).
= 2.4 Cal.
= 19.3 Cal. (Berthelot).
= 21.7 Cal.
= 0.084 (Berthelot).
Arsenic.
Sm (21° to 65°) amorphous = 0.0758 (Bettendorf and
Wiillner).
crystalline = 0.083 (Bettendorf and
Wvillner).
Q (solid, at S. P.veo = 616°) = 47 Cal.
L. H. Sublimation (616°)
Q (vapor, at 616°)
S (vapor) per cubic meter
per kilo
Q solid, at M. P. = 827°)
L. H. Fusion (827°)
Q (liquid at 827°)
L. H. Vaporization, 827°)
Q (vapor, at 827°)
= 114 Cal. (calculated from vapor
tension curve, for Ass) .
= 161 Cal.
= 0.405 (assumed for Ass).
= 0.040 (calculated).
= 63 Cal. '
= 10 Cal.
= 73 Cal.
= 104 Cal.
= 177 Cal.
under 10,000 mm.
pressure.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 89
Vapor tension, solid : log p. = ^ + 9.10 (from experi-
mental data),
at M. P. = 10,000 mm. Hg.
5115
liquid : log p. = ^ + 8.65 (from experi-
mental data.)
at 0° C. = 2.8 X 10-" mm. Hg.
Selenium.
Sm (60°— 200°) = 0.084 (Bettendorf and Wullner).
Q (solid, at M. P. = 217°) = 18 Cal.
L. H. Fusion = 4 Cal.
Q (liquid at M. P.) = 22 Cal.
S (liquid) = 0.12 (assumed).
Q (liquid, at B. P.reo = 690°) = 79 Cal.
L. H. Vaporization (690°) = 97 Cal. (calculated from vapor
tension curve, for Seg).
Q (vapor, at 690°) = 176 Cal.
S (vapor) per cubic meter = 0.405 (assumed for Sea),
per kilo = 0.038 (calculated).
5025
Vapor tension, liquid :log p = ^ — 1-8.10 (from experi-
mental data),
at M. P. = 0.71 mm. Hg.
solid : log. p = -^-1-8.56
atO°C. = 2.1X10-"mm.Hg.
Bromine.
Q (solid, at M. P. = -7°) = -16.9 Cal.
L. H. Fusion (-7°) = 16.2 Cal. (Regnault).
Q (liquid, at M. P.) = -0.7 Cal.
Sm (liquid, -6° to -|-58°) = 0.105-|-0.0011t
Q (liquid, at B. P. 760 = 58°) = 10 Cal.
L. H. Vaporization (58°) = 43.7 Cal. (BerthelotandOgier).
Q (vapor, at 58°) = 53.7 Cal.
S (vapor) per cubic meter = 0.40
per kilo = 0.0555 (Regnault).
Vapor tension, liquid : log p = ^-|-7.8 (from experimental
data),
at M. P. = 45 mm. Hg.
at 0° C. = 66 mm. Hg.
90 METALLURGICAL CALCULATIONS.
Rubidium. !
Sm (20° to 35°) = 0.0792+O.OOOOlt (Deusz).
Q (solid, at M. P. = 38°) = 3 Cal.
L. H. Fusion = 6 Cal. (E. Duess).
Q (liquid, at 38°) = 9 Cal.
S (liquid) = 0.080 (assumed).
Q (liquid, at B. P. 760 = 696°) = 62 Cal.
L. H. Vaporization (696°) = 261 Cal. (from Trouton's rule;
K = 23).
Q (vapor, at 696°) = 323 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.059 (calculated).
Vapor tension, liquid : log. p = jp — 1-7.90 (constants from
Trouton's rule; K = 23).
at M. P. = 1.5X10-8 mm. Hg.
solid: logp ^+8.28.
at 0° C. = 6.7 X 10-" mm. Hg.
Strontium.
Sm (0°— 100°) = 0.0735 (assumed by Dulong and
Petit's Law).
Q (solid, at M. P. = 800°) = 59 Cal.
L. H. Fusion = 26 Cal. (calculated by 2. 1 T rule) .
Q (liquid, at 800°) = 85 Cal.
S (liquid) = 0.092 (assumed, from at. sp.
(heat = 8.0).
Zirconium.
Sm (0°— 100°) = 0.0662 (Mixter and Dana) .
Q (solid, at M. P. = 2350°) = 155 Cal.
L. H. Fusion = 61 Cal. (calculated by 2.1 T rule).
Q (liquid, at 2350°) = 216 Cal.
S (liquid) =0.11 (assumed, from at. sp.
heat = 10).
CoLUMBiuM (Niobium).
Sm (0—100°) = 0.068 (assumed, from Dulong
Q (solid, at M. P. = 1950°) = 133 Cal. and Petit's Law).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 91
L. H. Fusion = 50 Cal. (calculated by2.1 T rule).
Q (liquid, at 1950°) = 183 Cal.
S (liquid) = 0.107 (assumed from at. sp.
heat = 10).
Molybdenum.
Sm (0°— t°) = 0.0655+0.00002t (Defacqz and
(Guichard).
Q (solid, at M. P. = 2500°) = 289 Cal.
L. H. Fusion = 61 Cal. (calculated by 2.1 T rule).
Q (liquid, at 2500°) = 350 Cal.
S (liquid) = 0.165 (assumed, from extension
of formula for S (solid).
Q (liquid at B. P.jeo = 3600°) = 522 Cal.
L. H. Vaporization (3600°) = 1015 Cal. (calculated from
Trouton's rule; K = 25).
Q (vapor, at 3600°) ^ 1537 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.052 (calculated).
21 380
Vapor tension, liquid : log p = 7^ — |- 8.4 (constants from
Trouton's rule, K = 25).
at M. P. = 4.9 mm. Hg.
soUd : log p = -??^+8.85.
at 0° C. = 7.6 X 10-" mm. Hg.
Ruthenium.
Sm (0°— 100°) = 0.0611 (Bunsen).
Sm (0°— t°) = 0.0601 +0.00001t (function of t
assvuned).
Q (solid, at M. P. = 1950°) = 155 Cal. (extension of above
formula).
L. H. Fusion = 46 Cal. (calculated by second
rule).
Q (liquid, at M. P.) = 201 Cal.
S (liquid) = 0.099 (assumed same as solid
at M. P.).
Q (liquid at B. P.760 = 2520°) = 257 Cal.
L. H. Vaporization = 689 Cal. (calculated from
Trouton's rule, K = 26).
92 METALLURGICAL CALCULATIONS.
Q (vapor, at 2520°) = 946 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.049 (calculated).
15 420
Vapor tension, liquid : log p = 7^ — 1-8.4 (constants from
Trouton's rule; K = 25).
at M. P. = 28.8 mm. Hg.
... . 16,450 , „ _.
solid: logp = Tf. — h-8.86.
at 0° C. = 4X10-'" mm. Hg.
Rhodium,
Sm (10°— 97°) = 0.0580 (Regnault).
Sm (0°— t°) = 0.0574-l-O.OOOOlt (function of t
assumed).
Q (solid, at M. P. = 1970°) = 152 Cal.
L. H. Fusion = 53 .Cal. (calculated by second
rule).
Q (liquid, at M. P.) = 205 Cal.
S (liquid) = 0.097 (assumed same as solid at
M. P.
Q (liquid at B. P.760 = 2500°) = 256 Cal.
L. H. Vaporization (2500°) = 677 Cal. (calculatied from
Trouton's rule, K = 25).
Q (vapor, at 2500°) = 933 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.049 (calculated).
Vapor tension, liquid : log p = 7p — f-8.4.
atM. P. =38mm. Hg.
,. , , 16,500 , „ - .
solid : log p = ^ — f-8.94.
at 0° * = 3.2 X lO-''' mm. Hg.
Palladium.
Sm (0°— 1°) = 0.0582+0.00001t (VioUe).
Q (solid, at M. P. = 1549°) = 110 Cal. (VioUe).
L. H. Fusion = 36 Cal. (Violle).
Q (liquid, at M. P.) = 146 Cal. (Violle).
S (liquid) = 0.089 (assumed same as solid at
M. P.).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 93
Q (liquid, atB. P. 750 = 2538°) = 238 Cal.
L. H. Vaporization (2535°) = 667 Cal. (calculated from Trou-
ton's rule; K = 25).
Q (vapor, at 2535°) = 905 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.047 (calculated).
15 500
Vapor tension, liquid : log p = ki [-8.4 (constants from
Trouton's rule, K = 25.1).
at M. P. = 7.8 X 10-1 mm. Hg.
VA 1 16,340 , „„,
solid : log p = 7^ h 8.85
at 0° ■ = 1 .4 X lO-'" mm. Hg.
Silver.
Sm (0°— 1°) up to 400° = 0.0555-|-0.00000943t
(Pionchon) .
over 400° = 0.05758+0.0000044t+
0.000000006^! (Pionchon).
Q (solid, at M. P. = 962°) = 64.8 Cal.
L. H. Fusion = 24.35 Cal. (Pionchon).
Q (liquid, at M. P.) = 89.15 Cal. (Pionchon).
S (liquid, 962° to 1020°) = 0.0748 (Pionchon).
Q (liquid at B. P.jeo = 2040°) = 170 Cal.
L. H. Vaporization (2040°) = 490 Cal. (calculated from
vapor tension data).
Q (vapor, at 2040°) = 660 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.046 (calculated).
Vapor tension, liquid : log p. = 7^ 1- 7.9 (from experi-
mental data),
at M. P. = 3.2X10-=' mm. Hg.
, 12,180
solid : log p = 7^ h 0.6b.
at 0° C. = 5.6 X 10-" mm. Hg.
Cadmium.
Sm (0°— 1°) = 0.0546-|-0.000012t (Naccari).
Q (solid, at M. P. = 321°) = 18.8 Cal.
L. H. Fusion = 13.0 Cal. (Person).
94 METALLURGICAL CALCULATIONS.
Q (liquid, at M. P.) = 31.8 (Person).
S (liquid) = 0.0623 (calculated same as
solid at M. P.).
Q (liquid, at B. P.760 = 778°) = 60.3 Cal.
L. H. Vaporization (778°) = 251 Cal. (calculated from ,
vapor tension curve).
Q (vapor, at 778°) = 311 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo. = 0.045 (calculated).
fii ^1
Vapor tension, liquid : log p = t^ + 8.74 (from experi-
mental data).
atM. P. =2. 6X10-2 mm.
solid : log p = - ^ + 9.28.
at 0° C. = 3.8 X 10-1* mm. Hg.
Tin.
Sm (0°— 1°) = 0.0560-|-0.000044t (Bede,
combined with Regnault.).
Q (solid, at M. P. = 232°) = 14.34 Cal.
L. H. Fusion = 13.82 Cal. (Richards).
Q (liquid, at M. P.) = 26.16 Cal. (Richards).
Sm (0°— t°)(t = 232° to 1000°) = 0.06129- 0.00001047t-l-
14 37
0.00000001035^ + -j-
(Pionchion).
S (liquid) att = 232° to 1000° = 0.06129-0.00002094t+
0.00000003105t'' (Pionchon).
Q (liquid, at B. P.jeo = 2250°) = 157 Cal.
L. H. Vaporization (2250°) = 538 Cal. (calculated from
Trouton's rule, K = 25.1).
Q (vapor, at 2250°) = 695 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.042 (calculated).
13 930
Vapor tension, liquid : log p = ^j 1-8.4 (Constants from
Trouton's rule, K = 25.1).
at M. P. = 6.3 X 10-2» mm. Hg.
,., , 14,290 , „,
solid : log p = ^ h 9.1.
at 0° C. = 5.6 X 10-" mm. Hg.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 95
Antimony.
= 0.04864+0.0000084t
Sm (0°— 1°)
Q (solid, at M. P. = 630°)
L. H. Fusion
Q (liquid, at M. P. )
S (liquid, 632° to 830°)
Q (liquid, at B. P.^eo = 1500°) = 127 Cal.
= 34 Cal.
= 40.3 Cal. (Richards).
= 74.3 Cal. (Richards).
= 0.0605 (Richards).
(Naccari).
L. H. Vaporization (1500°)
Q (vapor, at 1500°)
S (vapor) per cubic meter
per kilo
Vapor tension, liquid : log p = —
= 124 Cal. (calculated from
Trouton's rule for Sba, K =
25.1).
= 251 Cal.
= 0.405 (assumed, for vapor =
Sba).
= 0.025 (calculated for Sbs).
9790
+ 8.4 (constants from
at M. P.
solid : . log p =
at 0° C.
Trouton's rule, K = 25.1).
3.55 X 10-' mm. Hg
-^^V9.58.
= 6. 5 X 10-" mm. Hg.
Iodine.
Sm (9°— 98°)
Q (solid, at M; P. = 114°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
Q (liquid, at B. P.jeo = 185°)
L. H. Vaporization (185°)
Q (vapor, at 185°)
S (vapor) per kilo (206° to 377°)
per cubic meter
per cu. meter (> 1200°)
per kilo
L. H. Change of State (h = 21)
= 0.05412 (Regnault).
= 6.2 Cal.
= 11.7 Cal. (Person).
= 17.9 Cal.
= 0.0676 (assumed).
- 22.7 Cal.
= 23.95 Cal. (Fabre and Silber-
mann).
= 46.65 Cal. per kg.
= 0.034 (Strecker).
= 0.389 (calculated).
= 0.225 (assmned for I gas).
= 0.039 (calculated).
= 145 Cal. per kilo (calculated
from vapor tension curve).
= 1658 Cal. per cubic meter I2.
96
METALLURGICAL CALCULATIONS.
Q (vapor, at 1200° = I vapor) = 226 Cal. per kilo.
2390
Vapor tension, liquid : log p = Tf, — h 8 . 10 (from experi'
mental data).
at M. P. = 89 mm. Hg.
solid : log p ^+10.43.
at0° = 2X10-2 mm. Hg.
Tellurium.
= 0.0613 (Richards).
= 0.0495+0.000026t (Richards).
= 27.3 Cal.
= 19.0 Cal. (Richards).
= 46.3 Cal. (Richards).
= 0.0731 (calculated same as
solid at M. P.)
Q (liquid at B. P.^eo = 1390°) = 115 Cal.
Sm (0°— 455°)
Sm (0°— 1°)
Q (solid, at M. P. = 455°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
L, H. Vaporization (1390)
Q (vapor, at 1390°)
S (vapor) per cubic meter
per kilo
Vapor tension, liquid :
at M. P.
solid :
at 0° C.
= 166 Cal. (calculated by Trou-
ton's rule for Tea;
K = 25.1).
= 281 Cal.
= 0.315 (assumed for Tea)
= 0.028 (calc. for Te2).
log p = ~ — 1-8.4 (constants from
Trouton's rule, K = 25.1).
= 6.3X10-5 mm. Hg.
logp=-i»+9.84.
T
= 8.1X10-
' mm. Hg.
Caesium.
Sm (0°— 26°)
Q (solid, at M. P. = 29°)
L. H. Fusion
Q (liquid, at M. P.)
L. H. Vaporization (670°)
= 0.0482 (Eckhardt and Graefe).
= 1.4 Cal.
= 4.8 Cal. (calculated by 2.1 T
rule).
= 6.2 Cal.
= 163 Cal. (calculated by Trou-
ton's rule; K = 23).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 97
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.038 (calculated).
Vapor tension, liquid : log p = =r-+7.92 (constants from
Trouton's rule; K = 23).
at M. P. = 1.2X10-8 mm. Hg.
4Q0r)
solid : log p = -=Y^+8.35.
at 0° C. = 2.5 X 10-1" mm. Hg.
Barium.
Sm (0°— 100°) = 0.05 (Mendeleeff).
Q (solid, at M. P. = 850°) = 42.5 Cal.
L. H. Fusion = 17.0 Cal. (calculated by 2.1 T
rule).
Q (liquid, at M. P.) = 59.5 Cal.
Cerium.
Sm (0°— 100°) = 0.0448 (Hillebrand).
Q (solid, at M. P. = 623°) = 27.9 Cal.
L. H. Fusion = 13.4 Cal. (calculated by 2.1 T
rule).
Q (liquid, at M. P.) = 41.3 Cal.
S (liquid) = 0.07 (assumed).
Tantalum.
Sm. (0° — 100°) = 0.0354 (assumed, from at. sp.
heat = 6.4).
Sm (0°— 1°) = 0.0338+0. OOOOiet (function
of t assumed).
Q (soUd, at M. P. = 2900°) = 233 Cal.
L. H. Fusion = 37 Cal. (calculated by 2.1 T
rule).
Q (liquid, at M. P.) = 270 Cal..
Tungsten.
Sm (0°— t°)(to423°) = 0.033+O.OOOOllt (Defacqz
and Geuchard).
Q (solid at M. P. = 3350°) = 234 Cal.
L. H. Fusion = 40 Cal. (calculated by 2.1 T
rule).
98 METALLURGICAL CALCULATIONS.
Q (liquid, at M. P.) = 274 Cal.
S (liquid) = 0.10 (assumed, same as solid
at M. P.).
Q (liquid, at B. P.jeo = 5164°) = 455 Cal.
L. H. Vaporization (5164°) = 743 Cal. (B. P. according to
Langmiur).
Q (vapor, at 5164°) = 1198 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.027 (calculated).
Vapor tension, liquid : log p = ^ +8.4 (constants from
Trouton'srule, K = 25.1).
at M. P. = 1.3 mm. Hg.
solid : log p = -5i^°+8.86.
at 0° C. = 7.4X lO-ios mm. Hg.
Osmium.
Sm (19°— 98°) = 0.03113 (Regnault).
Sm (0°— 1°) = 0.0305+0.000006t (function of
t assumed).
Q (solid, at M. P. = 2200°) = 96 Cal.
L. H. Fusion = 36 Cal. (calculated by second
rule).
Q (liquid, at M. P.) = 132 Cal.
S (liquid) = 0.057 (assumed from sp. ht.
solid at M. P.).
Q (liquid, B. P. 760 = 2600°) = 155 Cal.
L. H. Vaporization (2600°) = 379 Cal. (calculated from
Trouton's rule, K = 25.1).
Q (vapor, at 2600°) = 534 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.026 (calculated).
1 Pi R7n
Vapor tension, liquid : log p = — -~7p — 1-8.4 (constants from
Trouton's rule; K = 25.1).
at M. P. = 97.7 mm. Hg.
soUd: logp= -1M80+8.81.
at 0° C. = 9.6 X 10-" mm. Hg.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 99
Iridium.
Sm (0°— 1°) = 0.0317 +0.000006t (VioUe).
Q (solid, at M. P. = 2360°) = 108 Cal.
L. H. Fusion = 28 Cal. (calculated by second
rule).
Q (liquid, at M. P.) = 136 Cal.
S (liquid) = 0.060 (assumed, same as solid
at M. P.).
Q (liquid, at B. P. ,60 = 2550°) = 147 Cal.
L. H. Vaporization (2550°) = 368 Cal. (calculated from
Trouton's rule; K = 25.1).
Q (vapor, at 2550°) = 515 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.020 (calculated).
15 580
Vapor tension, liquid : log p = ts 1-8.4 (constants from
Trouton's rule; K = 25.1).
at M. P. = 302 mm. Hg.
solid : log p = -^^^+8.81.
at 0° C. = 5.6 X 10-" mm. Hg.
Platinum,
Sm (0°— 1°) = 0.0317 +0.000006t (VioUe).
Q (solid, at M. P. = 1755°) = 75.2 Cal. (Violle).
L. H. Fusion = 27.2 Cal. (Violle).
Q (liquid at M. P.) • = 102.4 Cal. (Violle).
S (liquid) = 0.053 (assumed, same as solid
at M. P.).
Q (liquid, at B. P.jeo = 2450°) = 139 Cal.
L. H. Vaporization (2450°) = 351 Cal. (calculated from
Trouton's rule; K = 25.1).
Q (vapor, at 2450°) = 490 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.026 (calculated).
15,040 , „ , , , , c
Vapor tension, liquid : log p = ^ +8.4 (constants from
Trouton's rule; K = 25.1).
at M. P. = 9.55 mm. Hg.
solid: logp= -^V9.0.
at 0° C. = 3.0 X 10-" mm. Hg.
100
METALLURGICAL CALCULATIONS.
S (0°— 600°)
Sm (0°— 1°)600° to 1064°
Gold.
= 0.0316 constant (Violle).
= 0.0289 +0.0000045t-' ^^'^^
Q (solid, at M. P. = 1064°)
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
t
(Violle).
= 34.63 Cal. (from VioUe's equa-
tion).
= 16.30 Cal.
= 50.93 Cal. (Roberts-Austen).
= 0.0358 (assumed, same as solid
at M. P.).
Q (liquid, at B. P.tso = 2530°) = 103.4 Cal.
L. H. Vaporization (2530°) = 358 Cat (calculated from
Trouton's rule; K = 25.1).
Q (vapor, at 2530°) = 461 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
= 0.025 (calculated).
log p = 7p — \-9>A (constants from
Trouton's rule, K = 25.1).
= 6.8 X 10-* mm. Hg.
16,175
logp = =--
per kilo
Vapor tension, liquid
at M. P.
solid :
at 0° C.
S (solid) at -59°
Q (solid) at M.P. -39°
L. H. Fusion (-39°)
Q (liquid, at M. P.)
Sm (liquid, -36° to 0°)
Sm (0°— 1°) up to 250°
Q (liquid, at B. P. 7,0 = 357°)
L. H. Vaporization (357°)
(357°)
(0°)
-8.93.
= 4.8 X 10-" mm. Hg.
Mercury.
= 0.0319 (Regnault).
4.1 Cal.
= 2.84 Cal. (Person).
= -1.30 Cal.
= 0.0333 (Pettersson).
= 0.03337 -0.0000027t+
0.00000000055^! (Naccari).
= 117 Cal.
= 67.8 Cal. (Kurbatoff).
= 66.8 Cal. ] (Calc. thermo-
= 77.14 Cal.
L.H.V. at critical temperature
(1139°) = 0.0 Cal.
dynamically,
from vapor ten-
sion curve).
1139° is calcu-
lated critical
temperature.
THERMOCHEMISTRY OF HIGH TEMPERATURES. 101
Q (vapor, at 357°) = 185 Cal.
(vapor, at 1139° = critical temp.) = 205 Cal.
S (vapor) per cubic meter. = 0.225 (assumed).
per '.
kilo
= 0.025 (calailated).
f\ C% i^ ^x ^\
Vapor tension,
, liquid :
log
3330.6 , , „„ , ^
P = ^ + 1.75 log. T-
at M. P.
0.00221T+4.6544 (from vapor
tensions at 357°, 260° and 140°) .
= 5.13X10-«mm. Hg.
solid :
log
p = Y^+8.26 (8.26 assumed).
Thallium.
Sm (17°- 100°)
Sm (0°— t°)
Q (solid, at M. P. = 303°;
L. H. Fusion
Q (liquid, at M. P.)
S (liquid)
)
= 0.08355 (Regnault).
= 0.030+0.00002t.
= 10.9 Cal. (calculated from for-
mula).
= 7.2 Cal. (Robertson).
= 18.1 Cal.
= 0.042 (assumed, from solid at
M. P.).
Q (liquid, atB. P.reo = 1700°) = 77 Cal.
L. H. Vaporization (1700°) = 221 Cal. (calculated by Trouton's
rule; K = 23).
Q (vapor, at 1700°) = 298 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.025 (calculated).
9 940
Vapor tension, liquid : log. p = — tj h 7.92 (constants from
Trouton's rule; K = 23).
at M. P. = 4.7 X 10-" mm. Hg.
10,260
solid: logp = Ty l-o.o.
at 0° C. = 8.3 X 10-'" mm. Hg.
Lead.
Sm (0°— 1°) = 0.02925 + 0.000019t(Bede, com-
bined with Regnault).
Q (solid, at M. P. = 327°) =11.6 Cal. (Le Verrier).
L. H. Fusion = 6.0 Cal. (Richards).
Q (liquid, at M. P.) =17.6 Cal. (Person) .
S (liquid, 335° to 430°) = 0.0402 (Person).
102 METALLURGICAL CALCULATIONS.
Q (liquid, at B.P.760 = 1580°) = 68 Cal.
L. H. Vaporization = 209 Cal. (calculated from vapor
tension curve).
Q (vapor, at 1580°) = 277 Cal.
S (vapor) per cubic meter = 0.225 (assumed) .
per kilo = 0.024 (calculated).
9500
Vapor tension, liquid: log. p = — ?^ — \- 8.0 calculated from
Greenwood's data).
at M. P. = 1.5X10-8 mm. Hg.
9744
solid : log p
= -^+8.41.
at 0° C.
= 4.9XlO-''»mm. Hg.
Bismuth.
Sm (0°— 1°)
= 0.0285 +0.00002t (Bede, com-
bined with Regnault).
Q (solid, at M. P. = 269°)
= 9.0 Cal.
L. H. Fusion
= 12.0 Cal.
Q (liquid, at M. P.)
= 21.0 Cal. (Person).
Sm (liquid, 280° to 360°)
= 0.0363 (Person).
Q (liquid, atB. P. 750 = 1435°)
= 63 Cal.
L. H. Vaporization (1435°) = 208 Cal. (calculated from vapor
tension curve).
Q (vapor, at 1435°) = 271 Cal.
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.024 (calculated).
9460
Vapor tension, liquid : log p = ~ — 1-7.45 (from experimen-
tal data),
at M. P. = iXlO-i" mm. Hg.
... , 10,000 , „ ._
solid : log p = T^ h 8.45.
at 0° C. = 6.6 X lO-''' mm. Hg.
Radium.
Sm 0° = 0.027 (from Dulong and Petit's
Law).
Sm (0°-t°) = 0.027-|-0.000015t (coefficient of
t assumed).
Q (solid, at M. P. = 700°) = 26.3 Cal.
L. H. Fusion = 9.0 Cal. (calculated by 2.1 T
rule).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 103
Q (liquid, at M.P.) = 35.3 Cal.
S (liquid) = 0.048 (assumed, same as solid
.at M. P.).
S (vapor) per cubic meter = 0.225 (assumed),
per kilo = 0.022 (calculated).
Thorium.
Sm(0°-100°) = 0.0276 (Nilson).
Sm (0°-t°) = 0.0270+0.000008t
(coefficient of t assumed).
Q (solid, at M. P. = 1690°) = 68.5 Cal.
L. H. Fusion = 18.0 Cal. (calculated by 2.1 T
rule).
Q (liquid, at M. P.) = 86.5 Cal.
S (liquid) = 0.054 (assumed, same as solid
at M. P.).
S (vapor) per cubic meter = 0.225 (assumed)
per kilo = 0.022 (calculated).
Uranium.
Sm (O^-gS") - 0.028 (Blumcke).
Sm (0°-t°) = 0.0275+0.000007t (coefficient
of t assumed).
Q (solid, at M. P. = 2000°) = 83 Cal.
L. H. Fusion = 20 Cal. (calculated by 2.1 T
rule).
Q (liquid, at M. P.) = 103 Cal.
S (liquid) = 0.055 (assumed, same as solid
at M. P.).
Q (liquid, at B. P.760 = 2900°) = 152 Cal.
L. H. Vaporization (2900°) = 333 Cal. (calculated from
Trouton'srule; K = 25.1).
Q (vapor, at 2900°) = 485 Cal.
S (vapor) per cubic meter = 0.225 (assumed).
per kilo = 0.021 (calculated).
17 400
Vapor tension, liquid : log p = k^ h 8.4 (constants from
Trouton's rule; K = 25.1).
at M. P. = 5.5 X 10-« mm. Hg.
solid: logp= -^^^+8.85
at0° = lX10-"mm. Hg.
104 METALLURGICAL CALCULATIONS.
In several of the preceding instalments we have given the
heats of formation of alloys and compounds and the thermo-
physics of the elements. Before passing to the thermophysics
of alloys and compounds and problems involving their use,
we will consider a few simple cases of the application of data so
far given. Such include operations in which metals are melted
or volatilized, or amalgams retorted. A few words may be in
order, to clear the ground, regarding what is to be regarded as
the efficiency of a furnace.
Efficiency of Furnaces. .
Under this term we must distinguish a generic sense and a
specific sense, the first referring to furnaces in which the object
is to maintain a certain temperature for a certain time with
the minimum constraiption of fuel, the second, in which the
object is to perform a certain thermal operation with the small-
est consumption of fuel. In the first case, one furnace may be
compared with another, and thus comparative efficiencies cal-
culated; in the second case real or absolute efficiencies can be
also calculated. A few examples will illustrate this difference,
which is an essential difference as far as making calculations is
concerned.
Cases of Specific Efficiency: Whenever it is desired to melt
a metal for the purpose of casting it, a certain definite amount
of heat must be imparted to the metal, and the ratio between
this efficiently utilized heat and the heating power of the fuel
consumed, is the efficiency of the furnace. If the furnace is
electric the theoretical heat value of the electric energy used
is the divisor. If, in addition to the heat required to raise
the substances to the desired temperature, there is also heat
absorbed in chemical reactions, this amount can be added in
as usefully applied heat, and the sum of this and the heat in
the final products be regarded as the total efficiently applied
heat. If a blast furnace takes iron ore and furnishes us melted
pig iron, the sum of the heat absorbed in the chemical decom-
position of the iron oxide and the sensible heat in the melted
pig iron is the efficiently applied heat, because it is the neces-
sary theoretical minimvmi required; all other items are more
THERMOCHEMISTRY OF HIGH TEMPERATURES. 105
or less susceptible of reduction, but these are necessary items
and, therefore, measure the net efficiency. If my dwelling re-
quires 200 cubic feet of hot air per minute at 150° F. to keep
it at 65° F., while the outside air is at 0° F., the ratio of the
heat required to warm the 200 cubic feet of air from 0° F. to
150° F., to the calorific power of the fuel used per minute,
measures the specific efficiency of the " heater; " the question
whether this amount of hot air keeps the temperature of the
rooms at 65° F. is a question of the general efficiency of the
construction of the house.
Cases of Generic Efficiency: Such are those in which prac-
tically all the heat generated eventually leaves the furnace by
radiation or conduction, or useless heat in waste gases; this is
the case when a certain temperature has to be continuously
maintained for a given time, and where the time element is the
controlling one, and not any definite amount of thermal work
is to be done. Examples are numerous: An annealing furnace,
where steel castings, let us say, are to be kept at a red heat
for two days, or a brick kiln, where several days slow burning
are required, or a puddling furnace, where the melted iron
must be held one to two hours to oxidize its impurities. In
all these cases we may say that one furnace keeps its con-
tents at the right heat for the right time with so much fuel,
another does the same work with 10 or 25 per cent, less fuel,
and is, therefore, 10 or 25 per cent, more efficient ; but we can-
not, in the nature of the case, speak of the absolute or specific
efficiency of the furnace, because there is no definite term,
expressible in calories, to compare with the thermal power of
the fuel.
In many cases the two efficiencies are mixed in the same
process or operation, and then the calculation of absolute or
specific efficiency can be made for that portion of the operation
wherein a certain definite amount of thermal work is done.
Thus, in an annealing kiln, 50 tons of castings may be brought
up to annealing heat in 24 hours, starting cold, and the heat
absorbed by the castings compared with the calorific power of
the coal burnt during this period, is a measure of the real effi-
ciency of this part of the operation. During the rest of the
operation, while the castings are simply kept at annealing heat.
106 METALLURGICAL CALCULATIONS.
there can be no calculation of the absolute or specific efficiency
of the furnace, because one of the terms necessary for the
comparison has disappeared, in that part of the process we
can only speak of relative efficiency compared to some other
furnace doing a similar operation.
It goes, almost without saying, that we can, of course, apply
the conception of efficiency in its relative or general sense to
the whole operation or to any part of it
Problem 6.
The Rockwell Engineering Co. state in their current adver-
tisements that their regenerative oil-burning furnace melts 100
pounds of copper with the consumption of less than 1.5 gallons
of oil. Assume that 1.5 gallons of oil is used, and that the
copper is heated from 25° C. to melted metal 100° C. above its
melting point.
Required: The "efficiency" of the furnace; i.e., its specific
efficiency as calculated from the net heat utilized.
Solution: One gallon of fuel oil averages in weight 7.5 pounds,
and its calorific power 11,000 Calories per kilogram, or 11,000
pounds Calories per pound. The calorific power of the fuel
used in melting 100 pounds of copper is therefore:
Heat generated 11,000X7.5X1.5 = 82,500X1.5 = 123,750
pound Calories.
The heat imparted to the copper is as follows, taking the
data from Article V. of these calculations: (-p. 68.)
Heat in 1 lb. melted copper at melting point = 162 lb. Cal.
Heat in 1 lb. solid copper at 25° C. = 2
Heat required to just melt the copper = 160 "
Heat to superheat liquid copper 100° C.
0.133X100= 13 "
Total heat expended on each pound of copper = 173 "
Heat usefully applied per 100 pounds = 17,300 "
17 300
Net efficiency of furnace = ' ^ = 0.14 = 14%
It is proper to remark that although this efficiency appears
low, yet it is considerably greater than is attained in simple
melting holes or wind furnaces, and yet the calculations show
THERMOCHEMISTRY OF HIGH TEMPERATURES. 107
what a large margin for improvement and greater efficiency
exists in even some of the best and relatively most efficient
metallurgical furnaces.
Problem 7.
In the distillation of silver amalgam in iron retorts, 1000
kilos, of amalgam, containing 200 kilos, of silver, is retorted
with the consumption of 550 kilos, of wood, the mercury vapor
passes off at an average, temperature of 450° C, and the silver
is raised towards the end of the operation to 800° C, in order
to expel the last of the mercury. Assume the calorific power
of the wood 3000 Calories.
Required: The net efficiency of the furnace.
Solution: The heating power of the wood is 550X3000 =
1,650,000 Calories.
The heat utilized is that absorbed in separating the silver
from the mercury plus the sensible heat in the mercury vapor
at 450°, plus the sensible heat in the silver at 800°. These are
calculated as follows:
Heat to decompose amalgam = 2470 Calories per 108 kilos,
of silver = 200 X (2470 ^ 108) = 200X22.9 = 4,580 Calories.
Heat in silver at 800°, using Pionchon's formula, is 800
[0.05758 + 0.0000044 (800) +0.000000006 (800)^] X 200 = 10,390
Calories.
Heat in 800 kilos, of mercury vapor at 450° is
(a) heat to boiling point (Naccari)
357 [0.03337—0.00000275 (357) +
0.0000000667 (357)'] X 800 = 11,677 Cal.
(6) heat to vaporize 72.5X800 = 68,000 "
(c) heat in vapor at 450° =
0.025 X (450— 357) X 800 = 1,880 "
Total = 71,567 "
Heat usefully applied :
In decomposing amalgam = 4,580 "
In mercury vapor, as sensible heat = 71 ,557 "
In silver, as sensible heat = 10,390 "
Total = 86,527 «
Efficiency of furnace = ^ ^^^ ^^^ = 0.052 = 5.2%.
108 METALLURGICAL CALCULATIONS.
Problem 8.
In a zinc works, impure zinc is refined by redistillation in
fire-clay retorts, a bank of retorts distilled 970 kilos, of zinc
with the expenditure of 912 kilos, of small anthracite coal.
Assume that the zinc vapors pass out of the muffles at the
boiling point (93:0°).
Required: (1) The net efficiency of the furnace.
(2) The electrical power which would be required, in horse-
power-hours, to do the same work, assuming the heating effi-
ciency of the electric furnace is 75 per cent.
Solution: (1) The small anthracite may be assumed to have
a calorific power of 7850 Calories; therefore, the total heat
which should be developed is 7850X912 = 7,159,200 Calories.
The heat in 1 kilo, of zinc in the state of vapor at its boiling
point can be calculated from the thermophysical data supplied
for zinc as:
(a) In solid zinc to melting point (420°) 45.20 Cal.
(6) Latent heat of fusion 22 . 61
(c) Heat in melted zinc to boiling point (930°) .... 65 . 05
(d) Latent heat of vaporization 425.00
Total 557.86
Heat required for 970 kilos. = 541,124
541 124
Efficiency of furnace = ' = 0.075 = 7.5%.
(2) One electric horse-power-hour = 644 . 0 Cal.
Efficiently applied heat = 644 X. 75 = 483.0 "
4g9 q95
Electric horse-power-hours required = — r^^ = 1013 E.H.P hours
483
One metric ton of zinc requires ^ ^„^ = 1044 E.H.P. hours.
0.9/0
Cost of power, at $20.00 per E. H. P. year = ?^^ x 1044 =
8766
0.00228X1044 = $2.38.
This cost of electric
metric tons of small anthracite, equal to a cost of $2.53 for
0 Q12
This cost of electric power would replace the use of "^ -
THERMOCHEMISTRY OF HIGH TEMPERATURES. 109
electric power sufficient to replace a metric ton of coal for this
purpose.
Many other examples could be given of the technical use of
the thermophysical data concerning the elements, but the
problems given illustrate the methods of calculation.
When one is acquainted with some of the ordinary metal-
lurgical operations, such as melting and distilling the metals,
it is surprising, to notice how little is known or thought of the
efficiency or lack of efficiency of the furnaces used. One man
melts 100 pounds of metal by the use of 150 pounds of coal, he
builds a new furnace and does it more cheaply by using 100
pounds of coke, which is certainly relatively more efficient;
but it is seldom that the operator knows that in one case he
is getting probably only 7 per cent, efficiency from his fuel
and in the other case only 10 per cent. It is the knowledge
of these absolute efficiencies which tells the practical man just
what he is accomplishing, and shows him how much room
there still remains for improvement.
Thermophysics of Alloys.
There does not exist, in technical literature, much data of
this nature concerning alloys. There is here a wide and inter-
esting field for metallurgical research, whose cultivation would
yield results both of high practical and high theoretical inter-
est, and yet it is comparatively untouched. What is wanted
is complete data concerning the specific heat of solid and liquid
alloy, and latent heat of fusion. These, combined with the
determination of the heat evolved in the alloying, would fur-
nish a sound basis for a practical theory of alloys, besides
enabling workers with these alloys to control the efficiency of
their furnaces and, in general, to know with scientific exactness
what they are accomplishing.
Alloys of Tin and Lead.
Per Cent, of Tin. Sm. Latent Heat of Fusion.
4.8 (Pb^'Sn) 5.5 (Mazotto)
10.2 (Pb'Sn) 8.0 at 307° (Spring)
12.5 CPb*Sn) 8.3 at 292° (Spring)
110 METALLURGICAL CALCULATIONS.
Per Cent, of Tin. Sm. Latent Heat of Fusion.
16.0 (Pb'Sn) 9.1 at 289° (Spring)
22.2 (Pb^Sn) ........ 9.5 at 270° (Spring)
7.9 (Mazotto)
36.3 (PbSn) 0.04073 (12°— 99°) 11.6 at 241° (Spring)
(Regnault)
9.4 (Mazotto)
50.0 = total heat to 0° in 1 kilo.
melted metal 18.0 from 202° (Ledebur)
53.3 (PbSn^) 0.04507 (10°— 99°) 10.5 at 197° (Mazotto)
(Regnault)
63.1 (PbSn») 15.5 at 179° (Spring)
69.5 (PbSn*) 17.0 at 188° (Spring)
83.0 = total heat to 0° in 1 kilo.
melted metal 21.5 from 205° (Ledebur)
90.1 (PbSn") 12.9 (Mazotto)
Alloys of Tin and Bismuth.
Per Cent, of Tin. Sm. Latent Heat of Fusion.
6.7 (Bi'Sn) 11.4 Cal. (Mazotto)
22.1 (Bi^Sn) 11.2 Cal. (Mazotto)
36.2 (BiSn) 0.0400 (20°— 99°) 11.6 Cal. (Mazotto)
(Regnault)
53.1 (BiSn') solid, 0.0450 11.6 (Cal. Mazotto)
(Regnault)
liquid, 0.0454
(146°— 275°) Person
69.1 (BiSn*) 11.1 Cal. at 140°
(Mazotto)
82.7 (BiSn«) 12.6 Cal. (Mazotto)
90.1 (BiSn") 12.8 Cal. (Mazotto)
Alloys op Tin and Zinc.
Per Cent, of Tm. Sm. Latent Heat of Fusion.
78.4 (ZnSn') 23.5 Cal. (Mazotto)
92.7 (ZnSn') 16.2 Cal. at 197°
(Mazotto)
95.6 (ZnSn«) , . . , 16.3 Cal. (Mazotto)
97.3 (ZnSn='») 15.1 Cal. (Mazotto) :,
THERMOCHEMISTRY OF HIGH TEMPERATURES. -HI
Alloys of Tin and Copper.
Bell metal (20 per cent, tin) Sm (14°— 98°) = 0.0862 (Reg-
nault).
Bronze (15 per cent. tin).
Total heat in melted metal (to 0°) = 130 Cal. (Ledebur).
If strongly superheated = 143.5 Cal. (Ledebur).
Alloys of Tin and Antimony.
Britannia metal (90 per cent, tin ) requires to melt it, starting
cold, 28.0 Calories per kilogram (melting point 236°) ; with 82
per cent, of tin, 25.7 Calories; melting point 205° (Ledebur).
Alloys of Tin, Bismuth and Antimony.
BiSn'Sb (bismuth 34.3, tin 41.9, antimony 23.8 per cent.).
Sm (15°— 100°) = 0.0462 (Regnault).
Alloy of Tin, Bismuth, Antimony and Zinc.
BiSn='SbZn' (bismuth 29.8, tin 34.0, antimony 17.3, zinc 18.9
per cent.).
Sm (15°— 100°) = 0.0566 (Regnault).
Alloys of Lead and Bismuth.
Per Cent, of Lead. * Sm. Latent Heat of Fusion.
11.1 (PbBi») 10.2 (Mazotto)
33.2 (PbBi^) 6.4 (Mazotto)
39.9 (Pb^Bi') solid, 0.03165
(16°— 99°) Person,
liquid, 0.03500
(144o_358°) Person
42.7 (Pb'Bi*) 4.7 at 127° (Mazotto)
49.9 (PbBi) • 4.0 (Mazotto)
66.6 (Pb^Bi) 3.6 (Mazotto) '
88.8 (Pb'Bi) 4.9 (Mazotto)
Alloys of Lead and Antimony.
With 63.0 per cent of lead, Sm (10°— 98°) = 0.0388 (Regnault)
With 82.0 per cent, of lead, heat in 1 kilo, melted metal =
15.6 Calories (Ledebur).
With 90.0 per cent, of lead, total heat in 1 kilo, "melted metal
= 13.8 Calories (Ledebur).
112 METALLURGICAL CALCULATIONS.
Alloys of Lead, Tin and Bismuth.
D'Arcet's Alloy, containing 32.5 lead, 18.5 tin, 48.7 bismuth
Sm solid (5°— 65°) = 0.0372 (Mazotto)
Sm solid (12°— 50°) = 0.049 (Person).
Sm solid (14°— 80°) = 0.060 (Person).
Sm liquid (107°— 136°) = 0.047 (Person).
Sm liquid (120°— 150°) = 0.0399 (Mazotto).
Sm liquid (136°— 300°) = 0.0360 (Person).
Latent heat of fusion = 5.96 Cal. at 96° (Person).
= 5.77 Cal. at 99° (Mazotto).
Rose's Alloy, containing 24.0 lead, 27.3 tin, 48.7 bismuth:
Sm solid (5°— 65°) = 0.375 (Mazotto).
Sm fluid (119°— 338°) = 0.0422 (Person).
Latent heat of fusion = 6.85 Cal. at 99° (Mazotto).
Fusible Alloy, containing 31.8 lead, 36.2 tin, 32.0 bismuth:
Sm solid (18°— 52°) = 0.0423 (Person).
Sm solid (11°— 98°) = 0.0448 (Regnault).
Sm fluid (143°— 330°) = 0.0460 (Person).
Latent heat of fusion = 7.63 Cal. at 145° (Person).
Wood's Alloy, containing 25.8 lead, 14.7 tin, 52.4 bismuth:
7 cadmium:
Sm solid (5°— 50°) = 0.0352 (Mazotto).
Sm fluid (100°— 150°) = 0.0426 (Mazotto).
Latent heat of fusion = 7.78 Cal. at 75° (Mazptto).
Lipowitz's Alloy, containing 25.0 lead, 14.2 tin, 50.7 bismuth,
10.1 cadmium:
Sm solid (5°— 50°) = 0.0345 (Mazotto).
Sm fluid (100°— 150°) = 0.0426 (Mazotto).
Latent heat of fusion = 8.40 Cal. at 75° (Mazotto).
Alloys of Copper and Zinc.
Red Brass S at 0° = 0.0899 (Lorenz)
S at 50° = 0.0924 (Lorenz)
(Copper 85%) S at 75° = 0.0940 (Lorenz)
Yellow Brass S at 0° = 0.0883 (Lorenz)
at 50° = 0.0922 (Lorenz)
(Copper 65%) at 175°= 0.0927 (Lorenz)
Heat in 1 kilo, of melted, somewhat superheated, brass =
130 Calories (Ledebur).
THERMOCHEMISTRY OF HIGH TEMPERATURES. 113
Alloys of Copper, Zinc and Nickel.
German Silver:
(74Cu. 20Zn. 6Ni) Sin(0— t) = 0.0941 +0.0000053t
(Tomlinson)
Alloys of Copper and Aluminium.
Copper 88.7% Sm (20°— 100°) =0.10432 (Luginin)
Alloys of Silver and Platinum.
Silver 66.7% Sm (0— t) = 0.04726 + 0.00001 38t (Tomlinson). •
Alloys of Mercury and Tin.
HgSn (37.1% Sn) Sm (—30°— 15°) = 0.04083 (Schuz)
(—25°— 15°) = 0.04218 (Schuz)
( 22°— 99°) = 0.07294 (Regnault)
HgSn (54.1% Sn) Sm ( 25°— 99°) = 0.06591 (Regnault)
HgSn=(74.7% Sn) Sm (—16°— 15°) = 0.05039 (Scliuz)
Alloys of Mercury and Lead.
Pb Hg (50.9% Pb) Sm (—69°— 20°) - 0.03458 (Schxiz)
Sm ( 2.3°— 99^=) = 0.03827 (Regnault)
Pb'Hg (67.4% Pb) Sm (—72°— 20°) = 0.03348 (Schtiz)
Alloys of Cadmium and Tin.
CdSn= (67.8% Sn) Sm (—77°— 20°) = 0.05537 (Schuz)
whence we have Sm ( 0 — t) = 0.0557 +
0.00000366t (Schuz)
Alloys of Iron and Carbon.
Soft Steel (0.15% carbon) Sm (20°— 98°) = 0.1165 (Regnault).
Hard Steel (1.00% carbon) Sm (20°— 98°)= 0.1175 (Regnault).
Total heat in 1 kilo melted steel at 1350° = 300 Calories
(Ledebur).
Cast Iron (4.0% carbon) Sm (0—1200°) = 0.175.
Sm (0— 1°) = 0.12 + 0.000046t
Total heat in 1 kilo, melted at 1200° = 245 Calories (Ledebur).
Total heat in 1 kilo, coming from blast furnace = 250 to 325
Calories (Akermann).
114 METALLURGICAL. CALCULATIONS.
Problem 9.
A steel-melting crucible contains 110 pounds of steel, which
is melted in a wind furnace with the use of 150 pounds of coke.
Assume the coke to be 90 per cent, fixed carbon, and the steel
to be superheated 100° C. above its melting point.
Required: The net efficiency of the furnace.
Solution: The calorific power of the coke may be assumed
as 90 per cent, that of pure carbon, and therefore:
= 1 SOX (8100X0.90) = 150x7290 = 1,093,500 1b. Cal.
Heat in steel at melting point:
110X300 (Ledebur) = 33,000
Heat to superheat 100°:
110X100X0.15 (assumed) = 1,650
Total 34,650 lb. Cal.
Efficiency of furnace == r-j^^^ = 0.032 = 3.2%
Problem 10.
A Siemen's regenerative furnace holds eighteen steel cruci-
bles, each containing 100 pounds of steel. Assume that the
efficiency of utilization, of the heat for melting the steel is 5
per cent., and that the furnace is fed by natural gas, having a
calorific power of 512-pound Calories per cubic foot.
Required: The number of cubic feet of natural gas required
per furnace heat of eighteen crucibles = 1800 pounds of cast
steel.
Solution: Heat in steel = 1800x315 = 567,000 lb. Cal.
Heating power of gas required = -^"7^=— =11,340,000 lb, Cal.
Cubic feet of gas required = — '-—-^ = 22,150 cubic feet.
Gas required per ton of steel, 2000 lbs. = 24,610 cubic feet.
Cost of gas, at $0.08 per 1000 cubic feet = $1.97.
Problem 11.
In a malleable- casting foundry the pig iron is melted in a
reverbatory air furnace, 3000 kilos, being melted in two hours
THERMOCHEMISTRY OF HIGH TEMPERATURES. 115
by the. combustion of 1200 kilos, of bituminous coal, having a
calorific power of 8500 Calories.
Required; The melting efficiency of the furnace.
Solution: Calorific power of coal used:
1200X8500 = 10,200,000 Calories.
Heat in melted iron at foundry heat:
3000X250 (Ledebur) = 750,000 Calories.
Efficiency of furnace = ^q^q^^q = 0.0735 = 7.35%
Problem 12.
In an iron foundry cupola 14 metric tons of pig iron are
melted in one hour, using 1.5 tons of coke (90 per cent, car-
bon). The gases .passing away contain by volume CO 13 per
cent., CO' 13 per cent., nitrogen 74 per cent., and leave the cupola
at 500° C. The body of the cupola is 1.5 meters in diameter
outside and 4 meters high.
Required:
(1) The net melting efficiency of the cupola.
(2) The proportion of the calorific power of the coke lost,
(a) By the sensible heat of the hot gases escaping.
(6) By the imperfect combustion of the coke.
(c) By radiation from bottom and walls of the cupola.
(3) The amount of heat in Calories radiated, on an average,
from each square meter of outside surface per minute.
Solution :
(1) Calorific power of the coke;
1500X0.90X8100. = 10,935,000 Calories.
Heat in melted iron:
14,000X250 (Ledebur) = 3,500,000 Calories.
Efficiency of-melting = ^^^ = 0.32 = 32%
(2, o) Weight of carbon escaping = 1500X0.90 = 1350 kilos.
116 METALLURGICAL CALCULATIONS.
Volume of CO and CO^ escaping = ^jg = 2500 m*
(Because 1 m' of either gas carries 0.54 kilos. C.)
Volume of escaping gas = q 134,0 13 ^ 9615m'
Volume of nitrogen (by difference) = 7115 m*
Sensible heat of nitrogen and CO
(7115 + 1250) X [0.303 (500) +
0.000027 (500)^] = 1,323,760 Cal.
Sensible heat of CO'
1250 X [0.37 (500) +0.00022 (500)*] = 300,000 "
Total sensible heat in gases = 1,623,760 "
Proportion of calorific power of the fuel thus lost:
= raSl = »»^= ''■^^'-
(2, b) Volume of CO escaping = 1250 m»
Calorific power of this gas = 1250 X
3062 = 3,827,500 Cal.
Proportion of calorific power of the fuel thus lost:
3,827,500
10.935.000 ~ "-^^ - ^^^0
(2, c) Heat in pig iron = 3,500,000 Cal,
Heat in waste gases = 1,623,760 "
Lost by imperfect combustion = 3,827,500 "
Accounted for 8,951,260 "
Calorific power of the coke «= 10,935,000 "
Difference, loss by radiation = 1,983,740 "
Proportion of calorific power of the fuel thus lost:
1,983,740 n-,o-,r: noirn^
= 10.935.000 = °-l^l^ = ^^■^^'^0
(3) Area of bottom of cupola = (1.5)'' X 0.7854 = 1.77 m«
Area of sides f cupola = 1.5X3,14X4 = 18.85 m*
Total radiating area = 20.62 m*
Heat radiated per sq. m. per hour = qff'If" = 96,200 Cal.
Heat radiated per sq. m. per min. -» -^^ - 1.603 "
THERMOCHEMISTRY OF HIGH TEMPERATURES. 117
In general, we may state that the net efficiency in melting,
etc., of metals is very various, from, say, 2 or 3 per cent, in a
crucible steel-melting wind furnace to about 10 or 15 per cent.
in reverberatory furnaces, 20 to 30. per cent in regenerative
open-hearth furnaces, 30 to 50 per cent, in shaft furnaces, where
material to be heated and fuel burned are in direct contact with
each other, 50 to 75 per cent, in steam boilers and hot-blast
stoves, and 60 to 85 per cent, in large electrical furnaces.
CHAPTER V.
THERMOPHYSICS OF CHEMICAL COMPOUNDS.
The metallic oxides are the most important compovinds
occurring in metallurgy, together with the oxides of hydrogen,
carbon and of sulphur, arsenic and antimony, which are formed
during combustion and in roasting operations. "We do not
include as oxides those combinations of two or more oxides
in chemical proportions, such as of silica with metallic oxides,
wliich form oxygen salts or ternary oxygen compotinds. They
will be separately discussed. In the following lists Sm, as
before, means the mean specific heat per kilogram, in large
Calories (or per pound in lb. Cal.) in the range of temperature
given; S means actual specific heat at the temperature given,
and Q is the quantity of heat absorbed in fusion or volatiliza-
tion or in passing through any designated range of tempera-
ture. Temperatures are given only in centigrade degrees, ex-
cepting that t signifies Fahrenheit temperature; volumes of
gases are understood as being at 0° C. and 760 m.m. pressure,
\anless distinctly stated otherwise.
Oxides.
Hydrogen Oxide, WO:
Ice Sm (— 78°— 0°) = 0.463 (Regnault).
(— 30°— 0°) = 0.505 (Person).
Water— Sm (0°— 100°) = H-0.00015t (Pfaundler).
S (at 15°) = 1.0045 (Pfaundler).
S •(atO°) = 1.0000 .(assumed).
In all metallurgical calculations it will be sufficiently exact
to assvime the specific heat of water at ordinary temperatures
as unity. In heating to the boiling point, from zero, 101.5
Calories are absorbed.
118
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 119
Gas Sm (0° to t°)
1 m' up to 2000° = 0.34 + 0.00015t (Mallard and Le
Chatelier).
1 m' 2000°— 4000° = 0.48 + 0.00017t (Vielle). '
1 ft' up to 2000° = 0.021 + 0.000009t— lb. Cal.
1 ft' 2000°— 4000° = 0.030 + 0.000011t— lb. Cal.
1 kilo, up to 2000° = 0.42 + 0.000185t— Cal.
1 kilo. 2000°— 4000° = 0.59 + 0.00021t Cal.
1 lb. up to 2000° = 0.42 + 0.000185t— lb. Cal.
1 lb. 2000°— 4000° = 0.59 + 0.00021t— lb. Cal.
1 ft' up to 3600° F. = 0.021 + 0.000005^ B. T. U.
1 ft' 3600°— 7200° F. = 0.030 + 0.000006i B. T. U.
1 lb. up to 3600° F. = 0.42 + 0.000103* B. T. U.
1 lb. 3600°— 7200° F. = 0.59 + 0.00012* B. T. U.
Above figures are for water vapor as it exists in almost all
metallurgical problems, as true gas far removed from its max-
imum tension at any given temperature. When the water
vapor exists as saturated steam under its maximum tension,
as in a steam boiler, it is recommended to consult tables giving
the total heat in steam at any temperature condensing to liquid
water at 0°, or to use Regnault's formula for the same:
Q = 606.5 + 0.305t
which expresses the amount of .heat required to convert 1 kilo-
gram of water, liquid at 0°, into steam at its maximum pressure
at the temperature t. A little reflection will show that the
above formula amounts to taking the specific heat of a kilo-
gram of saturated steam, under a constantly increasingpressure,
as Q = 0.305t
or under increasing pressure Sm = 0.305
or per cubic meter (0.81 kilos.) Sm = 0.247
These figures apply, to boiler practice only. In cases where water
is evaporated at or below 100° C, and converted afterwards
into steam at a high temperature, I recommend first calculating
the heat required to convert the water into vapor at 0° (606.5
Cal.) .and then treating the vapor afterwards as true gas, bj'
the formulas of Mallard and Le Chatelier. This puts water
vapor at once, from the beginning, on the same footing as all
the other gases, and greatly simplifies all subsequent calcula-
tions, without introducing any unnecessary errors.
120 METALLURGICAL CALCULATIONS, j
Latent heat of fusion = 80 Cal. (Bunsen).
Latent heat of vaporization = 606.5 Cal. at 0° (Regnault).
= 600.0 Cal. at 10°.
= 537.0 Cal. at 100° (to vapor at
760 mm).
Total heat in saturated
steam = 606.5 + 0.305t (Regnault).
Most metallurgical problems which have to do with convert-
ing moisture in fuel, ores, etc., into vapor, are concerned with
the evaporation of the water far below 100° by means of dry
gases, into which the water vapor enters at a partial pressure,
which is only a small fraction of atmospheric pressure. The
problem is, in these cases, to find how much heat is necessary
to convert the water into vapor at the low temperature corre-
sponding to such low pressure. By finding the temperature
corresponding to said partial pressure, as a maximum tension,
applying Regnault 's formula, we get the heat absorbed in pro-
ducing the vapor at that temperature. We can then add to
this the heat required to raise the vapor up to the higher tem-
perature at constant pressure, using Mallard and Le Chatelier's
formulae.
Example: Wet peat is dried in a kiln by hot air, the issuing
air being at 50° C. and the tension of the moisture in it 25
millimeters. How much heat is absorbed in latent heat, and
.bow much as sensible heat per 1 kilogram of water evaporated?
Solution: The tension, 25 millimeters, is the maximum
tension of aqueous vapor at 26° (Tables). To change 1 kilo-
gram of water at zero to saturated vapor at 26° requires, ac-
cording to Regnault's formula, 606.5+0.305 (26) = 614 Calories.
(The real latent heat of vaporization at 26° is, therefore, 614 — 26
= 588 Calories.) The heat required to evaporate 1 kilogram of
water under the conditions of the peat kiln is therefore:
Heat to produce vapor at 25 m.m. (26°) = 614.0 Calories.
Heat to raise vapor from 26° to 50° (constant
pressure) = [0.42 + 0.000185 (50 + 26)]X
(50—26) = jioj
Total = 624.4
In most cases of drying or evaporation the results will be
sufficiently accurate by assuming the water first evaporated at
THERMOPHYSICS OF CHEMICAL COMPOUNDS.
121
0°, with latent heat of evaporation 606.5 Calories, and then
raised as gas to the end temperature. The difference between
this method and the above will be small, where the temperature
of the issuing gas is below 100° and the proportion of vapor in
it small; where the temperature of issuing gas is high and the
amount of vapor in it large, the more exact method should be
used. To facilitate this, the maximum tension of aqueous
vapor for temperatures up to 100° is here given:
Max. Tension.
Max.
Tension,
Temperature C°.
mm. of Hg.
Temperature C°.
mm
. of Hg.
0°
4.6
50°
92.0
5°
6.5
55°
117.5
10°
9.1
60°
148.9
15°
12.7
65°
187.1
20°
17.4
70°
233.3
25°
23.5
75°
288.8
30°
31.5
80°
354.9
35°
41.8
85°
433.2
40°
54.9
90°
525.5
45°
71.4
95°
633.7
50°
92.0
100°
760.0
Beryllium Oxide, Be^O'
Sm (0°— 100°)
Boric Oxide, B^'O':
Sm (16°— 98°)
Carbonous Oxide, CO :
Sm (0° to t°) 1 m'
2000°
1 m' 2000°— 4000°
1 ft' up to 2000°
1 ft' 2000°— 4000°
1 kilo, up to 2000°
1 kilo. 2000°— 4000°
1 lb. up to 2000°
1 lb. 2000°— 4000°
1 ft.' up to 3600° F.
1 ft. 3600—7200° F.
1 lb. up to 3600° F.
1 lb. 3600°— 7200° F.
= 0.2471 (Nilson and Pettersson).
= 0.2374 (Regnault).
up to
= 0.303 + 0.000027t (Mallard and
Le Chatelier).
= 0.2575 + 0.000072t (Berthelot).
= 0.0189 + 0.0000017t lb. Cal.
= 0.0161 +0.0000045t lb. Cal.
= 0.2405 + 0.0000214t Cal.
= 0.2044+ 0.000057t Cal.
= 0.2405 + 0.0000214t lb. Cal.
= 0.2044 + 0.000057t lb. Cal.
= 0.0189 + 0.0000009* B. T. U.
= 0.0161 + 0.0000025* B. T. U.
= 0.2405 + 0.0000119* B. T. U.
= 0.2044 + 0.000032* B. T. U.
122 METALLURGICAL CALCULATIONS.
Carbonic Oxide, CO' :
There has been much doubt about the specific heat of this
gas, and several experimenters have given it particular atten-
tion. Direct combustion of CO to CO' in air has recently
given Mallard and Le Chatelier a directly-observed temperature
of 2050°, while the formula, which has so far been considered
by us as the most rehable (Sm = 0.37-l-0.00027t), leads to
1947°. After renewed consideration of the whole subject the
writer considers the best values those given below, because by
accepting these they will agree with the observed temperature
of combustion, 2050°- "We will hereafter use these values
instead of the one just above. Above 2000° the values of
Berthelot and Vielle are the only ones to be used.
Sm (0° to t°) 1 m» up to
2050° = 0.37 + 0.00022t (calculated by
Richards).
1 m' 2000°— 4000° = 0.815 + 0.0000675t (Vielle).
1 ft' up to 2050° = 0.023 -F0.000014t lb. Cal.
1 ft' 2000°— 4000° = 0.051 + 0.0000042t lb. Cal. ,
1 kilo, up to 2050° = 0.19 + O.OOOllt Cal.
1 kilo. 2000°— 4000° = 0.42-|-0.000034t Cal.
1 lb. up to 2050° = 0.19 + O.OOOllt lb. Cal.
1 lb. 2000°— 4000° = 0.41+0.000034t lb. Cal.
1 ft' up to 3700° F. = 0.023-I-0.000008* B. T. U.
1 ft' 36000 —7200° F. = 0.051 + 0.0000023i B. T. U.
1 lb. up to 3700° F. = 0.19 + 0.00006i B. T. U.
1 lb. 3600°— 7200° F. = 0.4H-0.000019< B. T. U.
In all the above formulae, Q from t° to 0° is equal to Sm
multiplied by t; e.g., Q (0° to t°) 1 m' to 2050° = 0.37t-l-
0.00022t'.
Nitrogen Peroxide, NO':
Sm. (27°— 280°) 1 m' = 1.35 (Berthelot and Ogier).
1 kilo. = 0.65 Cal.
Magnessium Oxide, MgO :
Sm (24°— 100°) = 0.2440 (Regnault).
Sm (0°— t°) = 0.2420 + 0.000016t (assumed).
Mg (OH') Sm (19°— 50°) = 0.312 (Kopp).
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 123
Aluminium Oxide, Al'O' (alumina, corundum):
Sm(0°tot°) = 0.2081 + 0.0000876t (constant by
Regnault, coefficient of t by
Richards, on a corundum crystal
tested up to 1200°).
Melting point = about 2050°.
Latent heat of fusion = 2.1T = 4,878 Cal. for Al^O^.
4,878
r^ = 48 Cal. per kilo.
Heat in solid at 2050° = 795 Cal. by formula.
Total heat in liquid to 0° = 843 Cal.
Specific heat, liquid = 0.567 = specific heat of solid at
M. P.
Silicon Oxide, SiO' (silicon, quartz):
Sm(0°tot°) = 0.1833 + 0.000077t (constant by
Regnault, coefficient of t by
■Richards, on clear quartz up to
1200°).
Latent heat of fusion (at
1750°) = 135 Cal. (Vogt).
Melting point = 1900° (Boudouard).
Latent heat of fusion = 2.1T = 4,563 Cal. for SiO'.
= 3^ = 76.1 Cal. per kilo.
= 135 Cal. (Vogt).
Heat in solid at 1900° = 626 Cal. by formula.
Total heat in liquid to 0° = 761 Cal.
Specific heat, liquid = 0.476 = specific heat of solid at
M. P.
Sulphurous Oxide, SO':
Sm (16°— 202°) 1 m' = 0.4447 (Regnault).
1 kilo. = 0.1,544.
If we assume that the molecular specific heat of SO^' is 8
Calories (from the rule that the molecular specific heat of a
gas at constant pressure is 5+n, where n is the number of
atoms in the molecule), we would have
S (at 0°) Ira? ' = -^^ = 0.36 Cal.
124 METALLURGICAL CALCULATIONS.
Combining this with Regnault's value of Sm, we would get the
formula
Sm (0° to t°) 1 m» = 0.36 + 0.0003t
1 kilo. = 0.125 + O.OOOlt
These values are probably accurate enough for furnace
calculations, and are very useful in pyritic smelting and the
roasting of sulphide ores. While it is always desirable to have
direct determinations of such important quantities, yet when
they have never been made it is allowable to work out and
use the most probable values.
Sulphuric Oxide, SO':
The specific heat of this important gas has not been measured.
As an approximation we may assume
S (atO°) per molecule = 9.0 Cal. (assumed),
per m' = 0.405 Cal.
per kilogram =0.11 Cal.
For getting an approximation to Sm we may assume the
coefficient of t the same as in NH', which contains the same
number of atoms and is of analogous formula, and we then
have
Sm (0° to t°) per molecule = 9.0 +0.0036t
perm' = 0.405 + 0.00017t
per kilogram = 0. 100 + 0.00004t
Calcium Oxide, CaO (lime):
This important datum has not been determined, but since
it is so closely analogous to MgO, an approximation may be
obtained by assuming that the molecular specific heats of the
two compounds are alike. That for MgO is 0.2440X40 = 9.76
Calories. That for CaO would therefore be
Sm (24°— 100°) = ^|5 = 0.1743
and assuming a usual coefficient of t; for compounds of this
type:
Sm (0°— 1°) = 0. 1715 + 0 00007t
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 125
Titanic Oxide, TiO':
S (up to 200°)
Q (0° to 200°)
Q (0° to t°) for t over 200° =
Chromium Oxide, Cr^O':
Sm (10°— 99°)
Manganese Oxides:
MnO Sm (13°— 98°)
Mn^O' Sm (15°— 99°)
(Manganite) Mn='0'H^O
Sm (21°— 52°)
(Pyrolusite) MnO^
Sm (17°— 48°)
Iron Oxides, Fe^O':
Sm (0°— 1°)
Fe'O* Sm (0°— 1°)
0.1790 (Nilson and Pettersson).
35.8 Cal.
35.8 + 0.1790 (t— 200) + .000055
(t— 200)2
0.1796 (Regnault).
0.1570 (Regnault).
0.1620 (Oeberg).
0.1760 (Kopp).
0.1590 (Kopp).
0.1456 + 0.0001 88t (Regnault and
Richards).
0.1447 + 0.000188t (Regnault and
Richards).
= 0.1588 (Regnault)
= 0.1110 (Oeberg).
= 0.1420 (Regnault).
Nickel Oxide, NiO:
Sm (13°— 98°)
Copper Oxides:
Cu^O Sm (19°— 51°)
CuO Sm (12°— 98°)
Zinc Oxide, ZnO:
Sm (0°— 1°) (tot=1000°)= 0.1212 + 0.0000315t (Richards)
Arsenious Oxide, As'O':
Sm (13°— 97°)
Zirconium Oxide, ZrO^
Sm (0°— 100°)
Columbic Oxide, Cb^O"
Sm (0°— 1°)
= 0.1276 (Regnault).
0.1076 (Nilson and Pettersson).
Molybdic Oxide, MoO':
•Sm (21°— 52°)
= 0.1037 + 0.000070t(Kruss and Nil-
son).
- 0.1540 (Kopp).
126
METALLURGICAL CALCULATIONS.
Tin Oxide, SnO^*:
Sm (16°— 98°)
Sm (U— t)
Antimonious Oxide, Sb'O':
Sm (18°— 100°)
Tungstic Oxide, WO':
Sm (8°— 98°)
Mercuric Oxide, HgO:
Sm (5°— 98°)
Lead Oxide, PbO:
Sm (22°— 98°)
Bismuth Oxide, Bi^O':
Sm (20°— 98°)
Thoric Oxide, Th^O':
Sm (0°— 100°)
0.0936 (Regnault).
0.1050+ O.OOOOOet (Richards).
= 0.0927 (Neumann).
= 0.0798 (Regnault).
= 0.0518 (Regnault).
= 0.0512 (Regnault).
= 0.0605 (Regnault).
= 0.0548 (Nilson and Pettersson).
For those metallic oxides whose specific heat has not been
determined, an' approximation to the specific heat can be
found by assuming that the metallic atoms in a molecule of
oxide have each a specific heat 6.4 Calories, and each atom of
oxygen in an oxide molecule has an atomic specific heat of 3.6.
As a rough approximation, we can further assume that the
mean specific heat from 0° to t° increases 0.04 per cent, for
each degree of temperature, so that the above calculated spe-
cific heat being from 0° to 100°, we can figure out S at zero
and the rate of increase of S or Sm with t.
Example: What is the most probable value of the specific
heat of CaO? Since molecule weight is 56, and the molecule
may be assumed to have a molecular specific heat of 6.4 + 3.6
= 10.0 Calories, the mean specific heat per kilogram (0° — 100°)
is 10.0-^56 = 0.1786. S at zero will then be 0.1786-r-1.04 =
0.1717, and we would have the formulae:
Sm = 0.1717 +0.000071t
S = 0.1717 +0.000142t
Q = 0.1717t + 0.000071t*
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 127
Problem 13.
The Jacobs process of producing artificial emery consists in
melting down impure bauxite in an electric furnace, and letting
the mass cool. Assume the bauxite to be calcined before use,
and to contain Per cent.
Alumina. 88
Ferric oxide (Fe^O') . 5
Silica (SiO^).... .- 5
Titanic oxide (TiO^) 2
Required:
(1) The heat necessary to just melt a metric ton of this
material at 2000°.
(2) The net electric power theoretically required.
(3) The total electric power actually required.
Solution :
(1) Because of the presence of the impurities the alumina
melts easier than it otherwise would. We can, therefore,
calculate the heat in the alutnina melted at 2000°, and the
heat in the others melted at their proper melting points, because
they absorb their latent heats of fusion at the lower tempera-
ture. Take 1000 kilos, of material.
Heat in 1 kilo, of alumina:
Q (0°— 2000°) = 0.2081 (2000)4-0.0000876 (2000)^ =767 Cal.
L.H.F. at 2000° = 2.1 (2000 + 273) -h 102 (mol. wt.
Al^O') =_^ "
Total =814 "
Heat in 1 kilo, of ferric oxide
Q (0°— 1600°) = 0.1456 (1600)4-0.000188 (1600)^ (assumed)
= 713 "
L.H.F. at 1600° = 2.1 (1600 + 273)^160 =25 "
Heat in Hquid = (2000— 1600) X 0.75 =aOO_ "
Total =1038 "
Heat in 1 kilo, of silica:
Q (00—1900°) = 0.1833 (1900) + 0.000077 (1900)2 =626 "
L.H.F. (Voigt) =135 •'
Q (1900°— 2000°) liquid = 0.476x100 = 48 "
Total =809 "
128 METALLURGICAL CALCULATIONS.
Heat in 1 kilo, of titanic oxide :
Q (0°— 2000°) = 35.8+0.1790 (1800) +0.000055
(1800)' = 536 Cal.
L.H.F. =2.1T = 2.1 (2000+273) -=-80 = _59 «
Total = 595 "
Total heat to fuse 1000 kilos. :
Heat in alumina 880X814 = 716,320 Cal.
Heat in ferric oxide 50X1038 = 51,900 "
Heat in silica 50X809 = 40,450 "
Heat in titanic oxide 20X595 = 11.900 "
Total = 820,570 "
(2) One kilowatt equals 859 Calories per hour.
820,570
Kilowatt-hours absorbed by melt = — ocg — = 956.3 kw-hr.
(3) Assume an efficiency of 50 per cent., requires 1911 kw-hrs.
If the dynamos work at 95 per cent., efficiency, the working
power of the water turbine, gas engines or steam engines would
1Q11
be ^ = 2000 kw. hr.
Problem 14.
The ferric oxide charged into a blast furnace is reduced at
900° C. by carbonous oxide gas, CO, according to the reaction:
Fe='0'4-9CO-l- (17.15N') = FeH 300' -|- 6C0 + (17.15N=').
Required:
(1) The thermal value of the reaction at 900°.
(2) What would be the temperature of the resulting products
after the reaction had occurred?
Solution :
(1) The 6C0 and 17.15N' occurring on both sides of the equa-
tion need not be considered. The heat value of the equation
at ordinary temperatures would therefore be:
Decomposition of Fe^'O* = —195,600 Cal.
Decomposition of 3C0 = — 87,480 "
Formation of 300' = +291,600 "
Thermal value of reaction = + 8,520 "
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 129
At 900°, the value is changed as follows:
Added:
Heat in Fe^'O' (160 kilos.) at 900° = 160 [0.1456
(900) +0.000188(900)2] = + 45,200 Cal.
Heat in SCO (= 66.66 m') at 900° = 66.66
[0.303(900) + 0.000027(900)='] = + 19.640 "
Total to be added = + 64,840 "
Subtracted:
Heat in Fe=' (112 kilos.) at 900° = 112
X(0.218t— 39) = 17,585 "
Heat in 300^ (66.66 m») at 900° = 66.66
[0.37(900) +0.00022(900)=*] = 34,080 "
Total to be subtracted 51,665 "
Heat of reaction at 900° = 8,520 + 64,840—
51,665 = + 21,695 "
(2) The above heat would serve to raise the temperature of
all the products, viz. : Fe^ 3C0^ 6C0, 17.15N2 (from blast).
Their thermal capacity, 900° to t, is:
Fe^ = 112 [0.218(t— 900)]
300^ = 66.66 [0.37(t— 900)+0.00022(t^— 900=)]
6C0 = 133.33 [0.303(t— 900) +0.000027 (t^— 900=*)]
17.15N2 = 17.15 [ " " ]
Sum = 0.018517t2 + 94.68t- 100.383 = 21,695 Calories.
Whence t = 1065°.
It follows from the preceding discussion that the reduction of
iron oxide by CO, at about 900°, is an exothermic reaction,
heat being evolved, and that the reaction proceeds to a finish
when once started. The relations for FeO would be probably
similar, but we do not have the specific heat of FeO to use in
making the exact calculations.
Problem 15.
If Fe^O' charged into a blast furnace were reduced by solid
carbon, at 900° C, by the reaction
4Fe='0' + 9C = 3CO=' + 6CO+4Fe''.
Required:
(1) The thermal value of the reaction at 900° C
130 METALLURGICAL CALCULATIONS.
(2) The temperature of the resultant products after the reac-
tion.
Solution :
(1) Value of the reaction at zero =
- 4(195,600) Cal.
+ 3(97,200)
+ 6(29,160)
Add: =
— 315,840
Heat in 4Fe='0' at 900°
+ 180,800 "
Heat in 9C at 900° = 9(12) [0.2142(900) +
0.000166(90.0^)]
+ 35,340 "
Subtract:
Heat in 4Fe= at 900° =
— 70,340 "
Heat in 3C0^' at 900°
— 34,080 "
Heat in 6C0 at 900° = 6 (22.22) [0.303 (900)
+ 0.000027(900')]
— 39,275 "
Algebraic sum =
— 243,395 "
Absorbed per molecule of Fe^O' =
— 60,850 "
This reaction is therefore strongly endothermic, and therefore
tends to check itself constantly by the resultant cooling effect.
(2) If the Fe^O' and C were at 900° to start with; the, pro-
ducts would be below 900° after the reaction, since they would
have to furnish the deficit of heat.
The products would give out in cooling from 900° to t°.
4Fe' = 4(112) [0.218(900— t)]
SCO^" = 66.66 [0.37(900— t)+0.00022(9002—t2)]
6C0 = 133.33 [0.303(900— 1) + 0.000027 (900^— t^")]
Sum = 0.0183t=' 162.73t+ 161,250,= 243,395 Cal.
Whence t = — 537°.
This result is, of course, due to the hypothetical assumption
that this endothermic reaction would go on until completed,
without checking itself. The impossible result obtained means
that if this single reaction really goes on, it will soon check
itself on account of the decrease of temperature. •:
There are a few other compounds than those previously
mentioned whose thermophysics has been investigated, but
their number is in reality infinitesimal compared with the num-
ber of those , which have not been touched. There are many
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 131
compounds of great importance in metallurgy whose specific
heats are not known, to say nothing of their latent heats of
fusion, etc. The wide introduction of the electric furnace has
rendered desirable the latent heats of vaporization and specific
heats at high temperatures, but these are altogether lacking;
we can only estimate their values. It is to be hoped that many
metallurgical laboratories may be incited to take up this very
neglected field, and thus bring forward numerical data which
would be of the. greatest theoretical and practical value. One
example of such work, which deserves special commendation,
is the quite recent publication of Prof. J. H. L. Vogt, of Chris-
tiania, on the " SilikatschmelzlQsungen," i.e., on " Melted
Silicate Solutions," in which determinations of the melting
points and latent heats of fusion of many simple and complex
silicates are given for the first time; we venture to predict
that the data and conclusions of Prof. Vogt will be of great
value to the physicist, chemist, metallurgist, and geologist, in
both a practical as well as a theoretical sense.
In collating the remaining available data, it is rather diflficult
to decide as to what has immediate metallurgical interest and
what has not. « Electrometallurgy, in particular, is busying
itself with the treatment and decomposition of so many com-
pounds heretofore considered outside of the metallurgist's
sphere of interest, that almost all of the common chemical
compounds now have either a present or a prospective interest
to the metallurgist.
THERMOPHYSICS OF CHLORIDES
Substance
HCl (gas)
HCl (liquid)
RbCl (fused)
Li CI (fused)
KCl
KCl
NaCl
NaCl
NH4CI
Tern- L. H.
peratuire Specific Heai Fusion
22-214° 0.1867
0.3067(lm«)
-83
16-45 0.112
13-97 0.2821
14-99 0.1730
772 86.0
16-98 0.2140
804 123.5
23-100 0.3908
L. B.
Vapari' Authority
zation
Regnault
97.5 Elliott &.
Mcintosh
. Kopp
. Regnault
. Regnault
. Plato
. Regnault
. Plato
. Neumann
132
METALLURGICAL CALCULATIONS.
Tem-
L.B.
L. H,
Substance
peralure
St'cifie Beat
Fusion
Vapori- Aulhorily
tation
NH4C1
350°
709 Marignao
BaCls (fused)
14r-98
0.0896
Regnault
BaCU (fused)
969
27.8
Plato
SrCl2
13-98
0.1199
Regnault
SrClj
872
25.6
Plato
CaCla (fused)
23-99
0.1642
Regnault
CaCU (fused)
774
54.6
Plato
MgCU (fused)
24^100
0.1946
Regnault
AICI3
-22-15
0.188
Band
TiCU (solid)
13-99
0.1881 •
Regnault
TiCli (gas)
163-271
0.1290
Regnault
SiCU (liquid)
10-15
0.1904
Regnault
SiCl, (gas)
90-234
0.1322
Regnault
SiCU (gas)
90-234
1.0113(lm»)
Regnault
SiCU (liquid)
?
37.3 Ogier
ecu (liquid)
20°
0.207
Timofejew
ecu (Uquid)
0°
52.0 Regnault
MnCU
15-98(?)
0.1425
Regnault
MnCU (liquid)
?
49.4(?)Ogier.
ZnCU (fused)
21-99
0.1362
Regnault.
TlCl (solid)
0-t
r 0.0525
\ +0.000007t
:::::.
Russell,
Goodwin &
TlCl (at M. P.)
427°
0.0585
^H* ^.^V^^J. IT AAA \JU
TlCl (solid)
427°
16.6
XxCLliU Uo
TlCl (liquid)
427-530
0.0590
CrClj
?
0.1430
Kopp
PCI, (solid)
11-98
0.2092
Regnault
PCI3 (liquid)
78.5
51.42 Andrews
PCI, (gas)
111-246
0.135
Regnault
AsCl, (liquid)
14-98
0.1760
Regnault
AsCl, (liquid)
69.74 Regnault
AsCl, (gas)
159-268
0.1122
Regnault
SnCU (fused)
20-99
0.1016
Regnault
SnCU (liquid)
10-15
0.1904
Regnault
SnCU (liquid)
112
46.84 Regnault
SnCU (gas)
149-273
0.0939
Regnault
HgCl (solid)
7-99
0.0521
Regnault
HgCU (solid)
13-98
0.0689
Regnault
CuCi (solid)
17-98
0.1383
Regnault
PbCU (solid) 0-t
PbCU (at M. P.) 498
r 0.0630
\+0.000021t
0.0839
Lindner,
Goodwin &
Kalmus
PbCU (solid)
498
18.5
PbCla (liquid)
> 498
0.1035
Ehrhardt
AgCl (fused)
O-t
f 0.0897
l+0.0000125t
.....'
Regnault
, Goodwin,
THERMOPHYSICS OF CHEMICAL COMPOUNDS.
133
Substance
AgCl (at M. P.)
AgCl (solid)
AgCl (liquid)
Tem-
peratur
455°
455
455-533
Specific Heal
0.1011
L. H. L. H.
Fusion Vapori-
zation
30.5
HBr (gas)
HBr (gas)
HBr (liquid)
KBr (fused)
NaBr
AlBrj
SbBrs
AsBra
SnBri
TlBr
PbBrj (fused)
PbBrj (at M. P.)
PbBrj (at M. P.)
PbBra (at M. P.)
PbBrj (liquid)
AgBr (fused)
AgBr (at M. P.)
AgBr (at M. P.)
AgBr (liquid)
HI (gas)
HI (gas)
HI (liquid)
KI (solid)
Nal (solid)
Cul (solid)
HgJ (red)
Hgal (yellow)
Hgsl (liquid)
Hgl (solid)
Hglj (solid)
Hglj (at M. P.)
0.129
THERMOPHYSICS OF BROMIDES
Authority
Robertson
&
Kalmus
11-100
11-100
-70
16-98
0-96
93°
95
31
30
460
0.0820
0.2989 (perm')
0.1132
0.1170
0-t
488
488
488
488-587
0-t
430
430
430-563
1+0.
0526
OOOOOOt
0.0585
10.47
9.76
8.93
6.26
12.7
9.9
12.34
0.0780
0.0736
+0.0000025t
0.0757
0.0760
12.6
Strecker
Strecker
48.68 Estreigher &
Schnerr
Regnault
Koref
Kablukow
Telloczko
Telloczko
Telloczko
Goodwin &
Kalmus
Goodwin &
Kalmus
from formula
G. & K.
Ehrhardt
G. & K.
Regnault,
Goodwin &
Kalmus.
Calculated
G&K.
G. &K.
THERMOPHYSICS OF lODIDES
21-100
21-100
-37
13-48
16-99
18-99
0-100
0-247
250-327
17-99
18-99
250
0.0550
0.3168 (permO
0.0766
0.0868
0.0687
0.0406
0.0446
0.0554
0.0395
0.0420
35.00
9.79
Strecker
Strecker
r Elliott &
\ Mcintosh
Nernst
Regnault
Regnault
Guinchan'
Guinchant
Guinchant
Regnault
Regnault
Guinchant
134
METALLURGICAL CALCULATIONS.
Tem-
L. H.
L.H.
Substance
perature
Specific Heat
Fusion
Vapori- Authority
zation
Pblj (soUd)
0-375°
0.0430
Ehrhardt
Pbia (at M. P.)
375
11.50
Ehrhardt
Pbis (liquid) >
375
0.0645
Ehrhardt
Agl (solid)
15-264
0.0577
Bellati &
Romanese
Pblz.AgI
10-124
0.0476
Bellati &
Romanese
Thermophysics of Fluorides
HP (liquid)
?
360 Guntz
KF
-76-0
0.1930
Koref
KP (M. P.)
860
108
Plato
NaP
15-53
0.2676
Band
SNaP.AlF,
16-99
0.2522
Oeberg
(Cryolite)
CaP^
15-99
0.2164
Regnault
AIP3
15-53
0.2294
Band
AlP3.7HjO
15-53
0.342
Band
PbPj
(K34
0.0722
Schottky
Thermophysics of Sulfides
nS (gas)
20-206
0.2451
Regnault
HjS (gas)
20-206
0.3750(lm»)
Regnault
HjS (gas).
0-t
(lm3)0.34
H-0.00015t
Richards
CSs (liquid)
14-29
0.2468
Person
CSa (gas)
86-190
0.1596
Regnault
CS» (gas)
86-190
0.5458(lm')
Regnault
CS2 (liquid)
46
83.8
Wirtz
MnS
10-100
0.1392
Sella
ZnS
16-98
0.1230
Regnault
CdS
26
0.0908
Russell
PeS
17-98
0.1367
Regnault
PejSa
20-100
0.1602
Regnault
(Pyrrhotite)
Pes,
19-98
0.1301
Regnault
PeSa
565
(Gives off S)
Richards
Cos
15-98
0.1261
Regnault
NiS
15-98
0.1281
Regnault
M0S2
20-100
0.1233
Regnault
(Molybdenite)
CujS
9-97
0.1212
Regnault
(Chalcocite)
103 Transformation Point
f BeUati &
\ Lussana
190
0.1454
THERMOPHYSICS OF CHEMICAL COMPOUNDS.
135
Substance
CuS
(Covellite)
AsS
(Realgar)
As2Sa
(Orpiment)
(Orpiment) (gas)
SbsSs
(Stibnite)
SnS
SnSj
PbS
PbS (liquid)
BiiS,
HgS
AgjS
AgsS
(fused)
Tem-
perature
25°
20-100
20-100
0-t
23-99
13-98
12-95
16-98
1050
11-99
14-98
7-98
O^t
Specific Heat
0.1243
0.1111
0.1132
(1 m»)0.34
+0.00015t
0.0840
0.0837
0.1193
0.0509
L. H.
Fusion
50
(104 to 0°)
0.0600
0.0512
0.0746
0.0685
+0.00005t
L. H.
Vapori- Authority
zation
Russell
Naumann
Naumann
Richards
Regnault
Regnault
Regnault
Regnault
Richards
Regnault
Regnault
Regnault
Bellati &
Lussana
THERMOPHYSICS OF COMPOUND SULFIDES
SCujS.PesS,
10-100
0.1177
... Sella
(Bornite)
Cu2S.Fe2Sj
14-98
0.1310
. . . Kopp
(Chalcopyrite)
(Chalcopyrite)
Cu Matte
720
0-t
(Gives off S)
0.2110
. . . Richards
. . . Landis
(47% Cu)
Cu Matte
1000°
-0.0000366t
30 .
. . . Landis
(47% Cu)
Cu Matte (liquid)
1000°
(204
to0°)
. . . Landis
4Cu2S.Sb8S8
10-100
0.0987
... Sella
(Tetrahedrite)
2PbS.Cu2S.Sb2S,
10-100
0.0730
... Sella
(Boumonite)
FeS2.FeAs2
10-100
0.1030
... Sella
.(Arsenopyrite)
C0S2.C0AS2
15-99
0.0991
... Sella
(Cobaltite)
3Ag2S.As2Ss
3Ag2S.Sb2S,
10-100
10-100
0.0807
0.0757
... Sella
... Sella
136
METALLURGICAL CALCULATIONS.
Thermophysics of Selenides and Tellurides
Substance
Cu2Se
AgjSe
Tem-
perature
20-200°
37-187
specific Heat
0.1047
0.0684
L. H. L. H.
Fusion Vapori- Authority
sation
:::::}
Bellati &
Lussana
Thermophysics of Arsenides and Antimonides
FeAss
(Lollingite)
CoAsz
(Smaltite)
CusAs
(Domeykite)
AgsSb
(Dyscrasite)
K4Fa(CN)6
K2Zn(CN)i!
Hg(CN)2(cryst)
NajSaOs
BaS208
PbS203
10-100
10-100
10-100
10-100
0.0864
0.0830
0.0949
0.0558
Sella
.. .. . Sella
Sella
Sella
Thermophysics of Cyanides
1-46
14^46
11-46
0.2142
0.241
0.100
Thermophysics of Hyposulfites
20-100
25-100
17-100
15-100
0.197
0.221
0.163
0.092
Thermophysics of Sulfates
H2SO4
H2SO4
H2SO4
K2SO4
KHSO4
NaiSO,
NaaSOi-lOHsO
BaS04
SrSOi
CaS04
MgS04
Al2K2(S04)a.24H20
(NH4)2SOi.l2H20
ZnS04
MnSOi
Cr2K2(S04)4.24H20
10.5
5-22
326
15-98
19-51
17-98
31
10-98
21-99
13-98
25-100
15-52
13-45
22-100
21-100
19-51
0.332
0.1901
0.2440
0.2312
0.1128
0.1428
0.1965
0.2250
0.3490
0.3500
0.1740
0.1820
0.3240
26
61
122
Nernst
Kopp
Kopp
. . Pape
. . Pape
, . Pape
. . Pape
. . Bronsted
. . Cattanes
. 1 Person
. . Regnault
. . Kopp
. . Regnault
. . Cohen
. . Regnault
. . Regnault
. . Regnault
. . Pape
. . Band
. . Kopp
. . Pape
. . Pape
. . Kopp
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 137
Substance
Tem-
perature
Specific Heat
r IT
FusU>'n Va'porl Authority
zatton
PeSO^JHjO
9-16°
0.3460
.. Kopp
CoSOi-THaO
16-80
0.3430
. . Kopp
NiSOi
15-100
0.1840
. . Pape
CUSO4
23-100
0.1840
. . Pape
PbS04
20-99
0.0872
. . Regnault
Hg2S04
0-34
0.0624
.. Schottky
THERMOPHYSICS OF
Nitrates
HNOa
-47
9.64 ..
. . Berthelot
HNOs
86
11£
.1 Berthelot
LiNOj
169-260
0.387
. . Goodwin
& Kalmus
LiNOa
250
88.6 ..
. . Goodwin
& Kalmus
LiNOs (liquid)
260-302
0.390
. . Goodwin
& Kalmus
KNOs
13-98
0.239
• • • • ■ •
. . Regnault
KNO, .
334°
48.9 ..
. . Person
KNO3 (liquid)
350-435
0.332
■ ■ • > > • •
. . Person
NaNOa
14-98
0.278
. . Regnault
NaNOa
306
64.9 ..
. . Person
NaNOa (liquid)
320-430
0.413
« • • • * > •
. . Person
KjNa(N0a)2
15-100
0.235
. . Person
NHiNOa
14-31
0.455
. . Kopp
BaNOa
13-98
0.1523
. . Regnault
SrNOa
17-47
0.181
. . Kopp
CaNOa.4H20
42
33.5 ..
. . . Pickering
AgNOa
16-99
0.1435
. . Regnault
AgNOa
209
17.6 ..
. . Guinchant
AgNOa (liquid)
208-281
0.187
. . Guinchant
PbNOa
17-100
0.1173
. , Neumann
THERMOPHYSICS OF CARBONATES
PbsCOa (fused)
18-47
0.123
. . Kopp
K2CO3
23-99
0.2162
. . Regnault
NaaCO,
16-98
0.2728
. . Regnault
BaCOa
11-99
0.1104
. . Regnault
SrCOa
8-98
0.1475
. . Regnault
CaCOa (calcite)
20-100
0.2086
. . Regnault
CaCOa (argonite)
18-99
0.2085
. . Regnault
CaCOa (marble)
23-98
0.2099
. . Regnault
CaMg(C03)2
20-100
0.2179
. . Regnault
(Dolomite)
Mg7Pe2(C03)9
17-100
0.2270
. . Neumann
(Brown magnesite)
138
METALLURGICAL CALCULATIONS:
Substance
p^Zre 5i,.«Ji. He
ZnCO,
l+O.OOOlt
CuC0,.Cu(0H)2
15-99 0.1763
(Malachite)
FeCOs
9-98 0.1935
PbCOs
16-47
0.0791
Lindner
Oeberg
Regnault
Kopp
Thermophysics of Chromates
KjCrO«
KjCrjOT
KaCrsOj
KjCraOi (liquid)
NajCr04
PeCr04
(Chromite)
PbCrOi
19-98
0-t
397
397-484
23
19-50
19-50
0.1851
0.1803
+0.00008t
0.335
0.1690
0.0900
Regnault
(Regnault, &
Goodwin &
Kalmus
29.8 Goodwin&K.
Goodwin & K.
39.2 Berthdot
Kopp
Kopp
Thermophysics of Borates
K2B204
16-98
0.2048
Regnault
K2B407
18-99
0.2198
Regnault
Na2B204
17-97
0.2571
Regnault
Na2B40,
16-98
0.2382
Regnault
PbB204
15-98
0.0905
Regnault
PbB407
16-98
0.1141
Regnault
Thermophysics of Phosphates
H3P04
18
25.71 ..
. . . Thomsen
K4P207
17-98
0.1910
, .
. . . Regnault
Na4P20,
17-98
0.2283
. . . Regnault
CaP206
15-98
0.1992
. . . Regnault
3CaaP208.CaP2
15-99
0.1903
. . Oeberg
(Apatite)
PbjPjO, (fused)
11-98
0.0821
. . . Regnault
PbsPaOs
11-98
0.0798
. . . Regnault
Ag3P04
19-50
0.0898
. . . Kopp
Thermophysics of
Arsenates
PbjAssOe (fused)
13-97
0.0728
. . . Regnault
KjAsjOe (fused)
17-99
0.1563
. . . Regnault
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 139
Thermophysics of Chlorates and Per-chlorate3
Substance
Tem-
perature
KCIO, (fused) 16-98°
KCIO4 14^45
NaClOa (fused) 184-255
NaClOs (fused) 255
NaClOs (liquid) 255-299
Specific Heat
0.2096
0.190
0.320
0.325
Regnault
Kopp
Goodwin
& Kalmus
49.6 Goodwin
& Kalmus
Goodwin
& Kalmus
MgAUOi
(Spinel)
BeAljOi
(Chrysoberyl)
FeTiOs
PbMo04
(Wulfenite)
CaWOi
(Scheelite)
Fe(Mn)W04
(Wolframite)
KMn04
Mg2Si04
(Olivin)
Thermophysics of Aluminates
15-47 0.1940
0-100 0.2004
Thermophysics of Titanates
15-50 0.177
Thermophysics of Molybdates
15-50 0.083
•Thermophysics of Tungstates
15-50 0.097
15-60 0.098
Thermophysics of Manganates
15-50 0.179
Thermophysics of Silicates
0-100 0.2200
Kopp
Nilson &
Petersson
Kopp
Kopp
, . . Kopp
. . . Kopp
Kopp
Vogt
140
METALLURGICAL CALCULATIONS.
Substance
MgzSiOi
(Olivin)
MgzSiOi (liquid)
MgSiOa
(Enstatite)
MgSiOs
(Enstatite)
MgSiOs (liquid)
CaSiOs
(Wollastonite)
CaSiOs
(Wollastonite)
CaSiOs (liquid)
CaMgSijOe
CaMgSijOe
CaMgSijOc (liquid) 1225
0-t
Tem-
perature
1400°
1400
0-t
1300
1300
0-t
1250
1250
0-t
1225
Specific Heat
L. H.
L. H.
CasMgSiiOij
(Malacolite)
CasMgSiiOu 1200
CaaMgSiiOia (liquid) 1200
HjAUSiaOs 20-98
(Kaolin)
Al2Si(F)06 12-100
(Topaz)
KAlSisOa 20-100
(Orthoclase)
KAlSiaOs 1200
(Orthoclase)
KAlSisOs 20-100
(Microcline)
KAlSiaOs 1170
(Microcline)
CaAlaSijOs 0-t,
(Anorthite)
CaAljSijOs 1220
(Anorthite)
CaAljSiaOs (liquid) 1220
0.1974
+0.000086t
130
0.1690
+0.0001t
0.186
+0.00008t
0.179
+0.00007t
0.2243
0.1997
0.1877
0.197
0.1790
+0.0001t
[520
toO°]
125
[403
toO°]
100
[360
toO°]
100
[344
toO°]
94
[319
toO"]
100
83
100
[358
toO°l
Authority
Vogt
Vogt
Vogt
Vogt
Vogt
Vogt
Vogt
Vogt
Vogt
Vogt
Vog
Vog
Vogt
Vogt
UWch
Joly
Oeber
Vogt
Bogajaw-
lensky
Vogt
Vogt
Vogt
Vogt
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 141
Substance
Tem-
perature
Specific Heat
ZrSiOi
15-100°
0.1456
(Zircon)
BeAljSijOs
12-100
0.2066
(Beryl)
FesAljSiaOia
16-100
0.1758
(Iron Garnet)
AlaSiOs
0-100
0.1684
(Audalusite)
Asbestos
20-98
0.1947
Serpentine
16-98
0.2586
Talc (soapstone)
20-98
0.2092
Potash mica
20-98
0.2080
Sodium mica
20-98
0.2085
Magnesia mica
20-98
0.2061
Oligoclase
20-98
0.2048
Spodumene
20-100
0.2176
Labradorite
20-98
0.1949
Hypersthene
20-98
0.1914
Augite
20-98
0.1931
Hornblende
20-98
0.1952
Granite
20-524
0.2290
Basalt
0-t
0.1937
+0.000086t
Basalt
?
Lava (Etna)
0-500
0.1820
+0.000155t
Lava (Etna)
500-800
0.270
Pumice
10-100
0.240
Ca, K, Glass
14-99
0.198
Ca, K, Glass
O-300
0.1.90
Flint Glass
10-50
0.117
PUnt Glass
18-100
0.082
Crown Glass
10-50
0.161
Mirror Glass
10-50
0.186
Thermometer glass
0.1869
(French, hard)
Jena Glass
18-99
0.2182
Thermometer glass
19-100
0.1988
(Normal, German)
Porcelain
0-t
0.2826
-0.0000264t
Cement clinker
28-40
0.186
Portland cement
28-30
0.271
Humus (soil)
20-98
0.443
Quartz sand
20-98
0.1910
h H t,. H.
Fusion ^°'^°^*- Authority
zation
Regnault
Joly
Oeberg
Lindner
Ukich
Oeberg
Ulrich
Ulrich
Uhrich
Ukich
Ulrich
Schulz
Ulrich
Ulrich
Ulrich
Ulrich
Bartoli
Bartoli
130 Tamman
Bartoli
Bartoli
Dunn
Regnault
Dulong&Petit
H. Meyer
Winkelmann
H. Meyer
H. Meyer
Zonboff
, Winkelmann
, Winkelmann
, Harker
Hartl
Hartl
Ukich
Ukich
142
METALLURGICAL CALCULATIONS.
Substance
Sandstone
Slag
Enamel slag
Bessemer slag
Tem-
perature
0-100°
14-99
15-99
14-99
Specific Heat
0.174
0.1888
0.1865
0.1691
T w" ■^' ^'
Authority
.... Hecht
. . . . Oeberg
. . . . Oeberg
. . . . Oeberg
Miscellaneous Materials
Paraffin (solid)
0-t
0.532
+0.0012t
Paraffin (liquid)
52-63
0.706
Beeswax (solid)
26-42
0.82
Beeswax (liquid)
65-100
0.50
Soft Para Rubber
0-100
0.48
Vulcanite
20-100
0.331
Anthracite
0-12
0.312
Oak Wood
0.57
Fir Wood "
0.65
Charcoal
0.20
Red Brick
0.22
Ether
25
0.54
Ether
35
Alcohol
20-26
0.58
Alcohol (liquid)
78
Benzol (liquid)
10
0.40
Benzol (solid)
5
Benzol (liquid)
80
Glycerin
0.58
Glycerin
13
Machine oil
0.40 '
Petroleum
0.50
Turpentine
0.42
Turpentine
159
Stearic acid
64
Ammonia (gas)
0-t
0.38
+0.00016t
Ammonia (solution) 18
1.00
Ammonia (liquid)
16
Methane (CH,)
0-t
0.38
+0.00022t
Ethelyne (CsHi)
0-t
0.46
+0.0003t
30
42.5
47.6
. . Battelli
. . Battelli
. . Person
. . Person
. . Gee & Terry
. . A. M. Mayer
. . Hecht
Regnault
90 Brix
Bose
206 Schall
Pickering
J. Meyer
94 Tyrer
Berthelot
74 Brix
Bruner
LeChatelier
Thomsen
297 Regnault
LeChatelier
For silicates in general, the specific heat is found to agree
fairly well with what would be calculated from their percent-
THERMOPHYSICS OF CHEMICAL COMPOUNDS. 143
age composition, giving each oxide constituent its proper
specific heat. Those oxides whose specific heat at 0° are known
are as follows:
SiO^ = 0.1833 (Richards)
'APO' = 0.2081 (Richards).
Fe^O' = 0.1456 (Richards).
MgO = 0.2420 (Regnault).
CaO = 0.1779 (calculated).
Those which are not known give good results if assumed to
be as follows:
FeO = 0.1460 (Vogt).
MnO = 0.1511 (Vogt).
K^O = 0.1390 (\'ogt).
Na^O = .0.2250 (Vogt).
Li'O = 0.4430 (Vogt).
Using these data, S at 0° is calculated from the percentage
composition of the silicate. For S at higher temperatures it
can be assumed with considerable approximation to the truth,
that S 'increases 0.078 per cent, for each degree, and so, calling
So the specific heat at zero, we would have
S = S„ (H-0.00078t)
Sm = S„ (H-0.00039t)
Q (0°— 1°) = SmXt
Besides the above generalizations, which enable one to cal-
culate the heat in the solid silicate at the melting point (when
the latter is known), Vogt has shown that if the heat neces-
sary to heat the silicate from— 273° to the melting point is
calculated, the latent heat of fusion may be taken as being 20
to 25 per cent, of this quantity, say an average of 22.5 per cent,
and thus the heat required for fusion may be approximately
calculated.
Akerman has determined for many metallurgical silicate
slags the total heat contained per kilogram of melted, slag at
the melting point. These quantities vary from 347 to 530
Calories, depending principally on the elevation of the melting
point of the slag. A brief tabulation of Akerman's results are
as follows, arranged according to the amount of heat in the
just-melted slag:
144
METALLURGICAL CALCULATIONS.
Calories. %SiO='
%CaO
%APO»
347 \
59
36
5
39
42
19
'63
35
2
58
35
■ 7
58
37
5
350
53
37
10
'
41
42
17
38
47
15
39
43
19
37
40
23
66
32
2
59
38
3
48
42
10
360
40
48
12
34
48
18
31
37
32
46
37
17
58
32
10
f58
27
15
62
37
1 -
380
38
52
10
25
34
41
44
33
23
'60
20
20
65
35
0
400
41
52
7
37
53
10
21
32
47
43
30
27
The above includes most varieties of acid and basic iron
blast-furnace slags. Data for other slags made in the metal-
lurgy of iron and of other metals, are almost altogether lack-
ing. A wide field is here open for metallurgical experiments;
data thus obtained would be immediately useful in practical
calculations.
CHAPTER VI.
ARTIFICIAL FURNACE GAS.
There are many different forms of producers for making artifi-
cial furnace gas. For the purposes of making calculations upon
them they may be conveniently divided into four classes, as
follows :
1. Simple producers, those which use ordinary fuels, such as
wood, peat, lignite, bituminous coal or anthracite, and in which
no water or water vapor is introduced other than the water in
the fuel itself and the normal moisture of the air used.
2. Mixed gas producers, in which water vapor or steam is
introduced with the air for combustion, in such amount as to
be entirely decomposed in passing through the fuel.
3. Mond gas producers, in which, for a special purpose, more
steam is introduced than can be decomposed in the producer,
thus producing very wet gas.
4. Water gas producers, in which air and steam alone are
alternately fed to the producer, the former for heating up, the
latter for producing water gas.
As far as the calculations are concerned, the essential dif-
ference between these classes is the varying amount of water
vapor or steam introduced under the fuel bed while producing
the gas, from all air in Class 1 to all steam in Class 4.
The calculations which it is of immediate interest to make, and
the results of which are of immediate value to the metallurgist,
are those concerned with the volume of gas produced per unit of
fuel, its calorific power compared to that of the fuel from which
it is produced, the items of the heat losses during the operation
of transforming the solid fuel into gaseous fuel, the ftmction
of steam in the producer, the limits up to which the use of steam
is permissible, the increase of efficiency of the gas by subsequently
drying it, the advantages as to final efficiency which are gained
by gasifying the fuel over burning solid fuel directly.
145
146 METALLURGICAL CALCULATIONS.
For information as to the construction and operation of gas
producers, reference may be made to treatises such as Sexton's
or Wyer's " Producer Gas," Groves and Thorp's " Chem-
ical Technology, Vol. V., Fuels." Trade pamphlets and cata-
logues, such as those of R. D. Wood and Co. on " Gas Pro-
ducers," of the De la Vergne Machine Co. on " Koerting Gas
Engines and Gas Producers," etc., contain a great deal of
. exact and useful information, and may be usually had for the
asking. The monograph of Jtiptner and Toldt on " Genera-
toren und Martinofen " (Felix, Leipzig, 1900), is concerned
wholly with calorimetric calculations concerning the produc-
tion of gas and its utilization in regenerative gas furnaces.
. 1. — Simple Producers.
In these a deep bed of fuel is burnt by air or fan blast, in-
troducing no more moisture than happens to be in the atmo-
sphere at the time being. The fuel fed into the producer is first
dried by the hot gases, then is heated and distilled or coked,
and finally is oxidized by the incoming air. The residue is
the ash of the coal, which is ground out at the bottom or drops
through the grate, containing more or less unbumt fixed car-
bon. Great loss of efficiency sometimes occurs from the ashes
being rich in carbon. The escaping gases issue at tempera-
tures of 300° up to 1000° C, carrying much sensible heat out
of the producer.
Calculations as to the amount of gas produced per unit of
fuel consumed are to be based entirely on the carbon. The gas
must be carefully analyzed, so that it can be calculated from
this analysis how much carbon, in weight, is contained in a
given volume of gas. (An alternative method is to take a
carefully measured volume of the gas, mix it with excess of
ox^'gen, and explode in a gas burette, determining the amount
of carbon dioxide formed, and from that calculate the weight
of carbon in the volume of gas taken.) Knowing this, the
rest is simple; the carbon in unit weight of fuel minus the
carbon lost in the ashes by poor combustion gives the weight
of carbon gasified; this divided by the weight of carbon in
unit volume of gas produced, gives the volume of the latter
per unit weight of fuel.
Illustration: A fuel used in a gas producer contains 12 per
ARTIFICIAL FURNACE GAS. 147
cent, of ash and 72 per cent, of carbon. The ashes made con-
tain 20 per cent, of unbumt carbon; the gas produced contains
by analysis and calculation 0.162 ounce of carbon per cubic
foot of gas, measured at 60° F. and 29.8 inches barometric
pressure. What volume of gas, measured at above conditions,
is being produced per ton of 2240 pounds of coal used?
Solution: The distinction between ash and askes must be
noted; the former is the analytical expression for the amount
of inorganic material left after complete combustion during the
chemical analysis ; the latter term means the waste matter pro-
duced industrially, and consists, if weighed dry, of the true
ash, plus any tinbumt carbon. The calculations are therefore
as follows:
Lbs.
Ash in 2240 pounds of coal = 2240 XO.12 = 268. 8
Ashes corresponding = 268.8 -h 0.80 = 336.0
(the ashes are 80 per cent, ash)
Carbon in the ashes = 67 . 2
Carbon in the coal = 2240X0.72 = 1612.8
Carbon going into the gas (gasified) '^ 1545.6
Carbon in 1 cubic foot of gas = 0.162 -=-16 = 0.010125.
Volume of gas produced per 2240 poimds of
coal = 1545.6^0.010125 = 152,652 cu. ft
It will be noted that these calculations absolutely require
the percentage of total carbon in the fuel, as determined by
chemical analysis. This is not a difficult analysis, as it consists
in burning the carbon in a heated tube in a stream of oxygen
or air free from carbon dioxide; the products of combustion
are dried and then passed through caustic potash solution to
absorb CO^ gas, the weight of which is obtained by the in-
creased weight of the potash bulb, and the total carbon thus
obtained. The fixed carbon and volatile matter of the coal, as
determined by the ordinary proximate analysis, cannot be used
in this calculation, since while all the fixed carbon is carbon,
the volatile matter is of variable composition, containing such
varying proportions of carbon that no fixed percentage of the
latter in it can be assumed without considerable possible error.
The calorific power of the gas, per cubic meter or cubic
foot, can be calculated from its analysis, using the calorific
powers of the combustible constituents as already given in our
148 METALLURGICAL CALCULATIONS.
tables. This, multiplied by the volume of gas produced per
unit of coal, gives the calorific power of the gas as compared
with that of the coal from which it is made. The difference
is the heat loss in the operation of producing the gas, including
loss by unbumt carbon in the ashes. In fact, we may state that
the heat balance is based on the following equations:
Heating power of the coal, per unit
— Heating power of the gas per unit of coal
= Calorific losses in conversion.
The latter item is composed of:
Loss by unbumt carbon in the ashes.
Sensible heat of the hot gases issuing.
Heat conducted to the ground.
Heat radiated to the air.
These items may be modified as follows: If the air used is
hotter than the normal outside temperature, its sensible heat
above this dattim should be added to the heating power of the
coal, because it~increases the total available heat. If the ashes
are removed hot, and not allowed to be completely cooled by
the incoming air, their sensible heat should be included in the
calorific losses during conversion. If the air used is moist, its
moisture wiH be decomposed to hydrogen and oxygen, but the
heat absorbed in doing this is exactly represented by the
calorific power of this increased amount of hydrogen in the
gases, and the heat absorbed is not lost but really represents so
much saved as available calorific power of the gases. This item
must, therefore, M0< becounted as one of the heat losses during
the operation, as those losses have been defined by us. If the
fuel is wet, considerable heat is required to evaporate the mois-
ture in it, but this heat is not to be reckoned as one of the
losses in conversion, if we have taken as the heating power of
the coal the practical metallurgical value; that is, its value
assuming all the water in its products of combustion to re-
main as vapor and none to condense. If this value has been
so taken, the heat required to vaporize the moisture in the
coal will have already been allowed for. Similarly, it may take
a little heat energy to break up a bituminous coal so as to
expel its volatile matter, but this should not be reckoned in as
ARTIFICIAL FURNACE GAS. 149
a heat loss in the producer, because a little reflection will show
that, whatever this amount may be, it has been properly al'
lowed for in the determination or calculation of the total calo-
rific power of the fuel. Juptner and Toldt call this the " gas-
ifying heat," and use it in all their calculations, but it is doubt-
ful whether it really amounts to an appreciable quantity, for
one thing, and even if it does it should not be reckoned as a
heat loss in the producer.
Problem 16.
Juptner and Toldt ran a gas producer with lignite of the
following composition: (Generator en, p. 49).
Carbon 69 . S3 per cent.
Hydrogen 4 . 33
Nitrogen 0. 50
Oxygen 12.38
Moisture 7 . 25
Ash 5.71
Of this coal, 3214 kilograms was used in 8 hours, 50 minutes,
producing gas which contained, analyzed dry, by volume;
Carbon dioxide, C0=' 5 . 21 per cent.
Carbon monoxide, CO 23.99
Oxygen, 0^ 0.63
Methane, CH^ 0.25
Hydrogen, H^ 10.64
Nitrogen, N^' 59.28
The ashes produced weighed 22.23 kilograms per 100 kilo-
grams of coal used, and contained 68.76 per cent, of unbumed
carbon. The calorific power of the coal, determined in the
calorimetric bomb (in compressed oxygen, moisture resulting
condensed) was 6949 Calories per kilogr. Temperature of
hot gases, 282° C. ; temperature of air used, 9° C. ; humidity,
62 per cent.; barometer, 712 millimeters of mercury.
Required:
1. The volume of gas, measured at 0° and 760 mm. pressure
(and assumed dry), produced per metric ton (1000 kilos. =
2204 pounds) of fuel used.
150 METALLURGICAL CALCULATIONS.
2. The calorific power of the coal, per kilogram, with mois-
ture formed by its combustion assumed uncondensed.
3. The proportion of the calorific power of the coal develop-
able by burning the gas- produced from it.
4. The loss of heat in conversion.
5. The loss of heat by unbumt carbon in the ashes.
6. The loss of heat as sensible heat in the gases.
7. The loss of heat by radiation and conduction, expressed:
(a) Per unit of coal burnt.
(b) Per minute.
8. The volume of air required by the producer, at the con-
ditions of the atmosphere, per kilogram of coal burnt.
Solution :
(1) The gas contains, per cubic meter at standard condi-
tions :
Carbon in CO' 0.0521 X 0.54 kilos.
Carbon in CO 0.2399X0.54 "
Carbon in CH< 0.0025X0.54 "
Total 0.2945X0.54 = 0.1590 kilos.
The carbon gasified from 1000 kilograms of coal is:
Carbon in coal = 698.3 kilos.
Carbon in ashes 222.3X0.6876 =152.8 "
Carbon gasified = 545.5 "
Therefore,
545 5
Gas (dry) produced = = 3430 cubic meters. (1)
0 . 1590
(2) The calorific power of the coal as given must be dimin-
ished by the heat required to vaporize all the moisture formed
by its combustion, leaving such moisture as theoretical mois-
ture at 0° C. There will be formed per kilogram oif coal:
From moisture of coal = 0.0725 kilos.
From hydrogen 0.0433X9 =0.3897 "
Total = 0.4622 "
To evaporate this to theoretical moisture at zero (thus
ARTIFICIAL FURNACE GAS. 151
putting the water vapor on the same footing as the other pro-
ducts of combustion, CO'' and N^) requires:
0.4622X606.5 (Regnault), = 280 Calories,
leaving as the metallurgical or practical calorific power
6949—280 = 6669 Calories, (2)
(3) The calorific power of each cubic meter of gas (meas-
ured dry at standard conditions) is
CO = 0.2399 m'X 3,062 = 734 . 6 Calories
CH* = 0.0025 m'X 8,598 = 21.5
H=' = 0.1064 m^X 2,613 = 278.0
Total = 1034.1
Calorific power of gas from 1 kilogram of coal:
1034.1X3.43 = 3547.0 Calories.
3547 0
which equals s^oq-q = 53.2 per cent. (3)
(4) The loss of calorific power in conversion is 100 — 53.2
= 46.8 per cent, of the calorific power of the coal, or per kilo-
gram of coal:
6669—3547 = 3122 Calories. (4)
(5)
Carbon in ashes = 0.1528X8100 = 1237.7 Cal.
= 18.6 per cent. (5)
(6) The gases produced carry off, per cubic meter measured
dry, the following amounts of heat:
Volume X mean specific heat (0° — 282°) = heat capacity
per 1°-
CO^ 0.0521X0.432 = 0.0225
CH< 0.0025X0.428 = 0.0011
CO
o^
0.9454X0.311 = 0.2940
Sum = 0.3176
Heat carried out = 0.3176X282 = 89.56 Calories.
Per kilogram of coal = 89.56X3.43 = 307 Calories.
307
Proportion of calorific power = ^gq = 4.6 per cent.
152 METALLURGICAL CALCULATIONS.
The above result is, however, subject to a small correction,
because some of the moisture in the coal goes undecomposed
into the gases, and is not represented in the analysis of the
dried gas. The amount of this moisture can be obtained with
sufficient accuracy by finding how much moisture would be
obtained by burning the dried gas from 1 kilogram of coal,
and comparing this with the moisture which would be ob-
tained from 1 kilogram of coal itself; the difference must rep-
resent the moisture accompanying the gas as water vapor,
and which has not been included in the above computation.
Burning 1 cubic meter of gas, the H^O vapor is:
From CH* 0.0025 X 2 = 0.0050 cubic meters.
FromH^ 0.1064X1 = 0.1064 "
0.1114 "
Perkilo. of coal = 0.1114X3.43 =0.3821 "
But the weight of water vapor from burning 1 kilogram
of coal has already been found to be (2) 0.4622 kilograms, the
volume of which is
0.4622-^0.81 = 0.5706 cubic meters.
Leavmg, therefore, 0.5706—0.3821 = 0.1885 cubic meters of
water vapor as such accompanying the 3.43 cubic meters of
(dried) gas from 1 kilogram of coal. This would take out
0.1885X0.382X282 = 20.3 Calories.
Thus increasing the sensible heat in the gases to
307+20.3 = 327.3 Calories = 4.9 per cent. (6)
[In reality, a still further correction should be made; viz.:
to add in the moisture in the air used, because it would also
reappear as moisture on final combustion of the gases. Its
amount is found from the amount of air used, which, if 62 per
cent, saturated at 9° would carry moisture having 0.62 X
8.6 mm. (if saturated) = 5.3 millimeters tension, which repre-
sents, barometer being 712 mm., 0.7 per cent, of the volume
of the air used, or practically 0.9 per cent, of the volume of
nitrogen in the air. Since the nitrogen in the gas represents
almost entirely the nitrogen in the air used, the moisture 'to be
accounted for from the air amounts to 0.5928X0.009 = 0..0053
ARTIFICIAL FURNACE GAS. 153
cubic meters per cubic meter of gas, or = 0.0182 cubic meters
per kilogram of coal burnt. This correction is altogether too
small to affect the results in this case, but should be taken
into account whenever the air used is warm and moist.]
(7) The calorific loss in conversion was 3122 Calories. Of
this we have accounted for:
Lost by unbumt carbon in ashes 1237.7 Calories.
Sensible heat of gases (including moisture) 327.3
Total 1665.0
Loss by radiation and conduction = 1557.0 " (a)
Per 8 hours 5 minutes there is burnt 3214 kilograms of
fuel, making the loss of heat by radiation and conduction per
minute =
1557X3214 n..o n ^ ■ n,^
^^Tj = 9,442 Calories. (b)
(8) At the conditions given, each cubic meter of moist air
used contained 0.7 per cent, of its volume of moisture, making
its percentage composition by volume:
Water vapor 0.70 per cent.
j Oxygen 20.65
^^'" I Nitrogen 78.65
The volume of gas produced per kilogram of coal is 3.43 cubic
meters, of which 59.28 per cent, is nitrogen, equal to 2.0333
cubic meters, and weighing 2.0333x1.26 = 2.562 kilograms.
Of this 0.0050 kilograms came from the coal itself, leaving
2.512 kilograms to come from the air, or 2.512 4-1.26 = 1.9921
cubic meters. This would correspond to 1.9921+0.7855 =
2.5329 cubic meters of moist air if measured at standard con-
ditions. At 9° and 712 mm. pressure the real volume of moist
air used at prevailing conditions, per kilogram of coal burnt, is
2 . 5329 X ^^i^ X 7^ = 2-80 cubic meters. (8)
It must not be thought that the conditions of working in
the above producer represent good practice; they are very
poor practice as far as concerns the utilization of the fuel.
Many producers make gas having 75 to 90 per cent, of the
154 METALLURGICAL CALCULATIONS.
calorific power of the coal from which it is made, so that the
losses by unbumt carbon and radiation and conduction in this
case must be regarded as highly abnormal and very poor prac-
tice. The writer chose this example for calculating, because of
the carefulness with which Jtiptner and Toldt had collected the
necessary data, and because it illustrated so well the principles
to be employed in similar calculations.
Problem 17.
A gas producer run in Sweden uses saw-dust of the following
composition:
Water 27.0 per cent.
Ash 0.5
Carbon 37.0
Hydrogen 4.4
Oxygen 30.6
Nitrogen 0.5
Assume that it is run by dry air and that 0.5 per cent, of
ashes are made. The gas formed, dried before analysis, con-
tains, by volume:
Carbon dioxide, CO' 6.0 per cent.
Carbon monoxide, CO 29.8
Ethylene, C^H* 0.3
Methane, CH^ 6.9
Hydrogen, H^' 6.5
Nitrogen, N=' 50.5
The gas actually produced is partly dried before use by
having its temperature reduced by cold water, in a surface
condenser, to 29° C, in order to increase its calorific intensity
of combustion.
Required:
(1) The proportion of the moisture in the moist gas which
is condensed out.
(2) The calorific intensity of the moist gas, if burned pre-
heated to 800° C. by the theoretical quantity of air preheated
also to 800° C.
ARTIFICIAL FURNACE GAS. 155
(3) The calorific intensity of the dried gas, burnt under
exactly similar conditions.
Solution :
(1) It is first necessary to find the weight or volume of
water vapor accompanying the gas before condensation, next
that accompanying it after passing the condenser. The first
can be calculated best on the basis of the hydrogen present
in the tuel and in the (dried) gas made from it; the difference
is the hydrogen of the moisture removed before analysis, i.e.,
the hydrogen of the nloisture in the wet gas.
The first step is to find the volume of gas (dry) produced
per unit of fuel, as follows:
Carbon in 1 kilo, of fuel = 0.370 kilos.
Carbon in 1 m' of gas = CO=' + CO+CH* +
2C='H* (0.060 + 0.298 + 0.069 + 0.006) X 0.54 =0.2338 "
0 370
Dry gas per kilo, of fuel = ' ^ = 1.5825 rn^
The next step is to calculate the water which would be formed
by the combustion of 1 kilogram of fuel:
Water present in fuel 0. 270 kilos.
Water produced by hydrogen. 0.396 "
Total = 0.666 "
Volume at standard conditions = q' ^„ = 0.8222 m*
From this we subtract the moisture which would be pro-
duced by the combustion of the 1.5825 cubic meters of dry gas,
obtained as follows:
Water from ethylene = 0.003 X 2 X 1.5825 m=
Water from methane = 0.069 X 2 X 1 .5825
Water from hydrogen = 0.065 X 1 X 1.5825
= 0.209 X 1.5825 = 0.3307 m»
Difference = water vapor to 1.5825 m' of dried gas
= 0.8222—0.3307 = 0.4915 m'
= 0.4915 -^ 1.5825 = 0.3106 m' per 1 m«
of dried gas, as analyzed
156 METALLURGICAL CALCULATIONS.
This is to be compared with the amount of moisture ac-
companying the same quantity of (dried) gas as it escapes
from the condenser. This is obtained directly from the fact
that the gas escaping will be saturated with moisture at 29°
C, that the latter will, therefore, have a tension of 30 milli-
meters (tables), and that, assuming the barometer normal
(760 mm.), the partial tensions of moisture and gas proper
are as 30 to 760 — 30, or as 30 to 730. Since their respective
volumes (if both were measured separately at normal pres-
sures) are in the same proportion, it follows that each cubic
meter of (dry) gas is accompanied by 30 -j- 730 = 0.0411 cubic
meters of uncondensed moisture.
The respective quantities of moisture accompanying 1 cubic
meter of dry, uncondensable gas, are, therefore, 0.3106 before
cooling and 0.0411 after cooling, showing that 13.2 per cent, of
all the moisture escapes condensation, and that, therefore,
86.8 per cent, of moisture is condensed. (1)
(2) The wet gas has a calorific power, calculating on the
basis of 1 cubic meter of dried gas analyzed:
CO
0.298 X 3,062 =
912.5 Calories.
C^'H^
0.003X14,480 =
43.4
CH<
0.069 X 8;598 =
593.3
H'
0.065 X 2,613 =
Total =
169.8
1719.0
There is added to this available heat, when burned, the
sensible heat in the gas itself, at 800° C, and also that of the
necessary air, also at 800°.
Heat in gas:
CO,H',N^ = 0.868 m«x 0.3246 = 0.2817 Cals. per 1°
C0=' = 0.060 m'X 0.5460 = 0.0328
CH< = 0.069 m»X 0.4485 = 0.0309
C?W = 0.003 m»X0.50 = 0.0015
H'O = 0.3106 m'X 0.460 =0.1429
Calorific Capacity = 0.4898 "
Total sensible heat = 0.4898X800 = 391.8 Calories.
ARTIFICIAL FURNACE GAS. 157
The air required theoretically is:
For CO 0.298 m« = 0. 1490 m' oxygen
For C='H* 0.003 m' = 0.0090 m»
For CH* 0.069 m' = 0.1380 m' "
For H* 0.065 m' = 0.0325 m'
Sum = 0.3285 m' "
= 1.58 m' air
Heat in this at 800° = 1.58X0.3246X800 = 410.3 Calories,
Sum total of heat going into the products:
Developed by combustion 1719.0 Cals.
Sensible heat in gas 391.8 "
Sensible heat in air 410.3 "
Total 2520.0 "
The products of the combustion are CO^, H'O and N', as
follows: CO^ = 0.060 (in gas) +0.298 (from CO) +0.006 (from
C^HO+ 0.069 (from CH*) = 0.433 m'.
H^'O = 0.311 (with gas) +0.006 (from C'HO +0.138 (from
CHO+ 0.065 (from H^) = 0.520 m'.
N'' = 0.505 (in gas) +[1.58—0.33 = 1.25] (from air) = 1.755
m'.
Since the 2520.0 Calories remains as sensible heat in the
above products, at some temperature t, we have
Heat'capacity of the CO'' = 0.433 (0.37 +0.00022t)
Heat capacity of the H^O = 0.520 (0.34 +0.00015t)
Heat capacity of the N* = 1.755 (0.303 +0.000027t)
Heat capacity of the products = 0.8688 +0.00022065t
and the calorific intensity t must be
2520.0
* ~ 0.8688 + 0.00022065t
whence •
t = 1942°. (2)
(3) When the dried gas is burned under similar conditions,
the only difference is that 0.2695 m' of water vapor are absent
from the gas arid from the products, having been condensed.
158 METALLURGICAL CALCULATIONS.
This reduces the available heat by the sensible heat in this
much water vapor at 800°, viz. :
0.2695X0.460X800 = 99.2 Calories,
and decreases the calorific capacity of the products by
0.2695 (0.34+0.00015t).
Our equation, therefore, becomes
_ 2360.2
0.7772+0.00018023t
whence
t = 2056°
The increased efficiency of the dried gas for obtaining high
temperatures is too evident to need further comment, the
difEerence being in round numbers 150° C, equal to 270° F.,
in favor of the dried gas.
2. Mixed Gas Producers.
This Class of producers are those most commonly used. In
them a moderate amount of steam or vapor of water passes
with the air into the fire, and is decomposed, producing carbon
monoxide and hydrogen gases by the reaction:
I I I
H'O + C = CO + H='
18 12 28 2
which may be read as follows: One voltmie of steam forms
one volume of carbon monoxide and one volimie of hydrogen;
or eighteen parts by weight of water vapor act upon twelve
parts of solid carbon, producing twenty-eight parts of carbon
monoxide and two parts of hydrogen. If we speak of kilo-
grams as the above weights, then we can call each " volume "
spoken of 22.22 cubic meters: or if we call the weights ounces
avoirdupois, each " volume " represents 22.22 cubic feet.
The water vapor is admitted either automatically, as in the
old Siemen's type of producer, where water was run into the
ash pit to be evaporated by the heat radiated from the grate or
by hot ashes falling into it, or as in the modem water-seal
ARTIFICIAL FURNACE GAS. 159
bottom producer, where the ashes rest upon water in a large
pan, and so are continually kept soaked by capillary action;
or, finally, steam is positively blown under the grate, either
as a simple steam jet, or, more economically, by using it in a
steam blower, so as to have it produce by injector action an air
blast sufficient to run the producer. In the latter case the
proportions of air and steam may be regulated with precision,
and the blast action produced makes the production and de-
livery of the gas practically independent of chimney draft. The
use of steam also rots or disintegrates the ashes, preventing
or breaking up masses of clinker, and so facilitating the re-
moval of the ashes.
Steam or water vapor cools down the fire in the producer
so that it runs cooler; at the same time gas is produced which
is rich in hydrogen, and, therefore, of higher calorific power.
This saves unnecessary waste of heat in the producer, and in-
creases the efficiency of the gas in the furnace in which it is
burnt. The scieiitific reasons for these facts are to be found
in a consideration of the thermochemistry of the reaction by
which steam is decomposed.
ffO-l-C = CO-f-ff
—69,000 -1-29,160 = —39,840 Calories.
This would be the deficit in decomposing 18 kilograms of
water if it starts in the Hquid state. If, however, it is used as
steam at 100° C, each kilogram contains 637 Calories of sensi-
ble heat, making 18X637 = 11,466 Calories altogether, leav-
ing the deficit 28,374 Calories, or the deficit
= 1,576 Calories per kilogram of steam decomposed.
= 2,364 Calories per kilogram of carbon thus burnt.
In making this calculation it might be objected that the
steam used is often at three or four atmospheres pressure, and
its temperature, therefore, over 100° C; but it must not be
overlooked that this steam expands suddenly to atmospheric
tension, and that in so doing it cools itself to an amount roughly
proportional to its excess pressure so that the expanded steam
at atmospheric pressure is usually close to 100° C. When
mixed with air, the temperature of the mixture is usually
below 100°, some 40° to 50° C, and in this condition some of
160 METALLURGICAL CALCULATIONS.
the steam is possibly condensed to fog, but it must not be for-
gotten that the heat of condensation thus given out has been
absorbed in raising the temperature of the admixed air, and
therefore goes as sensible heat into the bed of burning fuel.
Fbr the purpose of calculation, we will, therefore, be very
nearly right in assuming that we are dealing in each case with
steam at 100° C, requiring the above calculated deficits to be
made up, in order for the decomposition to proceed..
It will be evident that any heat thus used in the producer
must be sensible heat of the hot carbon, which, if not so used,
would be lost as sensible heat in the gases produced or radi-
ated and conducted away from the producer. The basic pro-
cess of the running of the producer is the burning of carbon
to carbon monoxide, liberating 2,430 Calories per kilogram of
carbon, which is 2,4304-8,100 = 30 per cent, exactly of the
calorific power of the carbon. This heat, if no water vapor
gets into the producer, is lost as sensible heat of the hot gases
and by radiation and conduction, and is thus largely a dead
loss. In Problem 16, for instance, these items figured out 28.25
per cent, of the calorific power of the coal used. Now, the
facts are, that while small producers need that much heat to
keep them up to working temperature, large producers need
very much less, and run far too hot if no steam is admitted
to check the rise of temperature. In the largest producers the
sensible heat in the gases, plus the losses by radiation and
conduction, does not exceed 10 per cent, of the calorific power
of the fuel. Out of the 30 per cent, of the calorific power of the
carbon "inevitably generated, only some 10 per cent, is, therefore,
needed to supply the losses in a large producer, leaving 20 per
cent, applicable to decomposing steam. We therefore have:
Per 1 Kilo, of Carbon Gasified.
Heat generated 2,430 Calories.
Necessary loss in a large producer 810- "
Useful for decomposing steam 1,620 "
Required to decompose 1 kilo, steam at 212° F.
(69,000— 11 ,466) -^ 18 = 3,196
Maximum steam decomposable 1,620-5-3,196 = 0.507 kilo.
Calculation, therefore, shows that about one-half of a unit
weight of steam is the maximum which can be used per unit
ARTIFICIAL FURNACE GAS. 161
weight ot carbon burnt to carbon monoxide, consistent with
keeping the producer at proper working temperature. This
would be equal to
0.08S kilos, of steam per kilo, of air used.
0.088 pounds of steam per potmd of air used.
0.114 kilos, of steam per cubic meter of air used.
0.007 pounds of steam per cubic foot of air used.
The above discussion is on the assumption that the bed of
fuel in the producer is thick enough, and its temperature al-
ways high enough, to bum all the carbon to carbon monoxide.
Such is only an ideal condition, for the irregular charging,
descent and working of the fuel always allow of some carbon
dioxide being produced, and the best regular gas usually con-
tains 1 to 5 per cent, of dioxide, representing some 3 to 20 per
cent, of the total carbon oxidized in the producer.
Assuming that on an average 10 per cent, of the carbon
oxidized inevitably forms carbon dioxide, we can calculate
under these more usual conditions how much steam can be
used, because for each kilogram of carbon oxidized, 0.1 kilo,
then gives us 8,100 Calories per kilogram instead of 2,430 in
the producer, a surplus of
0.1 X (8,100—2,430) = 567 Calories.
over the conditions when only monoxide is formed. Instead
of the 1,620 Calories available for decomposing steam we will
now have 2,187 Calories, which will decompose
9 1R7
|i^= 0.684 kilo, of steam.
which reckoned on the air used would be
0.108 kilos, of steam per kilo, of air used.
0.108 pounds of steam per pound of air used.
0.140 kilos, of steam per cubic meter of air used.
0.009 pounds of steam per cubic foot of air used.
The above proportions are those which cannot practically
be exceeded, if producing gas as low as is usually possible in
carbon dioxide.
[Working the producer comparatively cold, with excess of
162 METALLURGICAL CALCULATIONS.
steam, much larger proportions of carbon dioxide are formed
and correspondingly larger proportions of steam are decom-
posed, but this manner of working is abnormal, is not unusual,
and will be discussed tmder the next heading, treating of Mond
gas.]
Making gas containing any given proportion of CO^ to CO,
by volume, the ratio thus obtained is identical with the pro-
portionate weight of carbon oxidized to CO^ and CO in the
producer, and by the application of the principles just described
it can be calculated how much steam can be used per unit of
carbon oxidized, making proper allowance for losses by radia-
tion, etc.
Illustration: A Siemen's producer with chimney draft pro-
duced gas containing, by volvune, 4. 3 per cent. CO^ and 25.6
per cent. CO. How much steam could be used per pound of
air used, assuming 50 per cent, of the heat generated by the
oxidation of the carbon to be needed to run the producer?
Solution:
4 3
Per cent, of carbon burnt to CO'' = . o/qk = '^^■^ V^^ ^^^^-
Heat generated by C to CO^ = 0.144X8,100 = 1,166 lb. Cal.
Heat generated by C to CO = 0.856X2,430 = 2,080
Heat generated per kilo, of C = 3,246 "
Heat lost by radiation, etc. (50 per cent.) = 1,623 "
Heat available for decomposing steam = 1,623 "
1 623
Steam decomposable = ' = 0.508 pounds.
The above is expressed per kilogram of carbon oxidized, but
the same proportion is true per pound. The air required per
pound of carbon oxidized is fotmd from the oxygen required
to form CO and 00=*:
Oxygen for C to C0= = 0.144X8-3 = 0.384 lbs.
Oxygen for C to CO =0.856X4-3 = 1.141 "
Total = l";525 "
6 608 ■^'^= 6.608 "
Volume of air = q^^j^ = 81.8 cu. feet.
ARTIFICIAL FURNACE GAS. 163
Steam per cubic foot of air = ' = 0.006 lbs.
ol .o
Steam per pound of air = " ^ = 0.077 "
Air required per pound of steam = 13.0 "
It should easily be seen that whatever heat is absorbeddn the
producer in decomposing steam, is entirely recovered when the
hydrogen thus produced is burnt and steam is reproduced. If
then, in any case, it is possible to absorb in the producer, in
the decomposition of steam, an amount of heat equal to say,
20 per cent, of the total calorific power of the fuel, then that
20 per cent, is regained and capable of being utilized when the
hydrogen so produced is burnt in the furnace. In other words,
20 per cent, less of the calorific power of the fuel will be lost
in the process of conversion into gas in the producer, and 20
per cent, more will be obtained in the burning of the gas when
it is used. The great advantages of using steam judiciously
are thus clearly evident.
Problem 18.
R. W. Hunt & Co. report the following tests made of the
running of a Morgan continuous gas producer. Coal used,
" New Kentucky " Illinois coal, run of mine. Composition:
Fixed carbon 50.87 per cent.
Volatile matter 37.32
Moisture 5.08
Ash •■■• 6.73 "-
100.00
The ultimate composition was:
Total carbon 69.72
Hydrogen 5.60
Nitrogen 2.00
Total sulphur 0.94
Oxygen 1100
Moisture 6.08
Inorganic residue (less sulphur) 6.06
The ash, on combustion, contains 1.12 per cent, of its weight
164 METALLURGICAL CALCULATIONS.
of sulphur (as FeS); the ashes obtained from the producer
contain 4.66 per cent, of unbumt carbon.
The gas produced, dried, contained by volume:
Carbon monoxide (CO) 24.5 per cent.
Marsh gas (CH^) 3.6
Ethylene (C='H^) 3.2
Carbon dioxide (CO') 3.7
Hydrogen (H^) 17.8
Oxygen (0^) 0.4
Nitrogen (N^) (by difference) 46.8
[The moisture and sulphur compounds in the gas not having
been determined, we can calculate the former, and, for the
purposes of calculation, are justified in assuming the sulphur
present in the gas as H'S, and in subtracting it from the hydro-
gen. We will also assume that the moisture in the coal goes
unchanged into the gases as moisture, and that all the steam
used is decomposed. The 0.94 per cent, of sulphur in the
coal will furnish 0.86 per cent, to the gases, because 6.73 X
0.0112 = 0.08 per cent, will go into the ash as ferrous sulphide.
The 0.86 pound of sulphur would produce 0.91 pounds of H^S,
equal in volume to 9.56 cubic feet per 100 pounds of coal used,
or (since we will see later that 53.83 cubic feet of gas are pro-
duced per potmd of coal) there will be 0.2 per cent, of H'S in
the gases, leaving 17.6 per cent, of hydrogen.]
On the basis of above data and assumptions:
Required: (1) The volume of gas produced per poimd of
fuel used in the producer.
(2) The weight of steam used per 100 cubic feet of air blown
in, assuming the air dry.
(3) The proportion of the total heat generated in the pro-
ducer which is utilized in decomposing steam.
(4) The percentage of increased economy thus obtained
reckoned on the calorific power of the fuel.
(5) The efficiency lost by unbumt carbon in the ashes.
(6) The efficiency of the gas, burnt cold, compared with the
coal from which it is made.
Solution: (1) Per pound of coal burnt, there remains in
the ashes 0.0673 X (0.0466^0.9534) = 0.0033 pounds of un-
bumt carbon, leaving 0.06972— 0.0033 = 0.6939 pounds gasified.
ARTIFICIAL FURNACE GAS. 166
One cubic foot of gas, at 32° F., contains the following weight
of carbon:
C in CO = 0.245 X 0.54 ounces
CinCH* = 0.036X0.54 "
C in C^H* = 0.032X1.08 "
CinCO^ = 0.037X0.54 "
Total = 0.382X0.54 "
= 0.20628 ounces av
= 0.01289 pounds av.
Gas produced, measured dry, per pound of coal, at 32° F.:
0.6939
0.01289
= 53.83 cubic feet
(2) If the moisture of the coal is assumed to pass unchanged
into the gas, as moisture, then all the hydrogen in the dry
gas, in any form or combination, must have come either from
hydrogen in the coal or in the air blast. The hydrogen in the
coal is given as 5.60 per cent. The hydrogen in the gas is
calculated as follows, per cubic foot:
H= in H^* = 0.176X0.09 ounces.
H^in H^S = 0.002X0.09 "
HMn CR* = 0.036X0.18 "
HMnC'H< = 0.032X0.18 "
Total = 0.314X0.09 "
= 0.02826 ounces av.
= 0.00177 pounds av.
Hydrogen in gas from 1 pound of coal:
0.00177X53.83 = 0.0953 pounds.
Hydrogen from decomposition of steam:
0.0953—0.0560 = 0.0393 pounds.
Weight of steam decomposed per pound of coal used:
0.0393X9 = 0.3537 pounds.
To express this weight relatively to the air blown in, we
must calculate the air used per pound of coal, as follows:
166 METALLURGICAL CALCULATIONS.
Nitrogen in gas per cubic foot
0.468 X (14 X. 09) = 0.5897 oz. av.
= 0.036856 lbs. av.
Per pound of coal = 0.036856X53.83 = 1.9840
Subtract W in 1 pound of coal = 0.0200
Leaves N^ from air = 1.9640
Weight of air = 1.9640 X II =2.5532
Volume of air = 2.5532 X 16 4- 1 .293 = 31.61 cu. ft.
Steam used per 100 cubic feet of air blown in;
0.3537.
31.61
XlOO = 1.119 pounds. (2)
(3) The heat utilized in decomposing steam has been found
to be 3,196-poutid Calories per potmd of steam at 212° F. We
therefore, have the heat so used per pound of fuel used:
0.3537X3,196 = 1,130-pound Calories.
This quantity must now be compared with the total heat gen-
erated in the producer, and the latter quantity can be deter-
minated in two ways: (1) We may subtract from the total
calorific power of the coal the calorific power of the gas pro-
duced and of the unbumt carbon in the ashes; the difference
must be the net heat generated in the producer, i.e., the total
heat generated minus that absorbed in decomposing steam.
The total heat generated is the net heat thus calculated plus
the heat absorbed in decomposing steam. (2) We may calcu-
late the heat of formation of the CO and CO^ in the gas, and
assume that as the total heat generated in the producer. This
method is not so accurate as (1).
The total calorific power of the coal is given as practically
7,747-pound Calories per poimd, water formed being con-
densed, which would be decreased by the latent heat of vapor-
ization, if the latter is assumed uncondensed. The deduction
is 606.5 X [(0.056 X 9) -f 0.0508] = 337-pound Calories, leaving
7,410-pound Calories as the practical metallurgical calorific
power of the fuel.
ARTIFICIAL FURNACE GAS. 167
Calorific power of the gas (dried) per cubic foot:
CO 0.245 X 3,062 = 750.2 ounce Cal.
CH^ 0.036 X 8,598 = 309.5
cm* 0.032X14,480 = 463.4
ff 0.176X 2,613 = 459.9
H^S 0.002 X 5,513 = 11.0
Total = 1994.0
= 124.6 pound Cal.
Calorific power of gas per pound of coal:
124.6X53.83= 6,707 pound Cal.
Calorific power of carbon in ashes:
0.0033X8,100= 27
Sum = 6,734
Calorific power of 1 pound coal = 7,410
Net heat lost in conversion = 676
Used in decomposing steam = 1,130
Gross heat generated in producer = 1,806
Proportion of this utilized in decomposing steam:
1,130
1,806
0.626 = 62.6 per cent. (3)
(4) We can state this result in another way, by saying that
1,806-^7,410 = 24.40 per cent, of the calorific power of the
fuel is generated in the producer, of which 1,130-^7,410 =
15.25 per cent, is utilized to decompose steam, and 9.15 per
cent, is lost by radiation, conduction and sensible heat in the
gases. The calorific power of the gases represents 6,707 -?-
7,410 = 90.50 per cent, of the calorific power of the coal of
which 15.25 per cent., however, is clear gain from the employ-
ment of steam. Reckoning on the total calorific power of the
coal, the increased economy from the use of steam is 15.2
per cent. (4)
(5) The loss of calorific power by the tmbumt carbon in
the ashes is 27-pound Calories, or, on the whole heat available,
27
= - ■ ,.= 0.36 per cent.
168 METALLURGICAL CALCULATIONS.
This loss is exceptionally low, and may be very profitably
compared with the analogous loss of 18.6 per cent, occurring
in the case discussed in Problem 16.
(6) This has already been calculated as
1^=90.50 per cent.
on the assumption that the gases are burnt cold. If they are
burnt hot, say issuing from the producer at 1200° F. (649° C),
and are burnt when at 1,000° F. (548° C.) their sensible heat
at 1,000° F. will be added to their efficient heating power,
and can be calculated with exactness, using the principle ex-
plained and used in requirement (6) of Problem 14. Under
such conditions the total efficiency of the producer, reckoned
on the calorific power of the coal used, approximates 95 per
cent.
To be fair to everybody concerned, however, we must de-
duct from this the coal required to be burnt to raise the steam
used. There is a little over one-third pound of steam used
per pound of coal used in the producer. This requires in
ordinary boiler practice -5-X-^ = kt pound of coal, or about 4
per cent, of the weight of fuel burnt in the producer. The
results of the previous calculation must, therefore, be dimin-
ished in this proportion, if the steam has to be raised by burning
coal under boilers. Under these conditions (6) becomes:
90.50-1-1.04 = 87 per cent, efficiency. (6)
If the steam can be obtained from waste gases of a blast
furnace, or from the hot producer gases themselves (in case
they are going to be burnt cold), then no such deductions need
be made. It is very evident, however, that if 9.15 per cent,
greater efficiency is gained by burning 4 per cent, more coal
to raise the steam, that the net gain would be only 5.15 per
cent. Even under these conditions it pays to use steam, be-
cause of the greater calorific intensity of the richer gas, the
usefulness of the steam for supplying air blast, and the rotting
of the clinkers therewith obtained.
ARTIFICIAL FURNACE GAS. 169
3. MoND Gas.
In the Mond producer, an excess of steam is introduced with
the heated air used for combustion; the producer is thus run
much colder than the ordinary producer, and far more carbon
dioxide is present in the gas. In fact, the gas, compared with
ordinary producer gas, is very high in carbon dioxide (10 to
20 per cent.), very high in hydrogen (20 to 30 per cent.), very
low in carbon monoxide (10 to 15 per cent.), low" in nitrogen
(40 to 50 per cent.), and carries an extraordinary amount of
undecomposed moisture. The fuel used is low-grade bitumin-
ous slack, and the object of using so much steam is to keep
the temperature in the producer so low that a maximum amount
of the nitrogen in the coal is evolved as ammonia. The gas
is cooled to ordinary temperature by contact with water spray,
so that all but a small amount of moisture is. condensed, the
ammonia is removed by dilute sulphuric acid, and the cold
nearly dry gas is then used for gas engines or in furnaces.
The calorific power of the gas is not low, because the high
proportion of hydrogen compensates for the low carbon mon-
oxide, while the great heat evolved by the large formation of
carbon dioxide has been mostly absorbed in decomposing
steam, and is, therefore, potentially present in the gas in the
form of hydrogen.
Problem 19.
Bituminous slack coal used in Mond producers for gener-
ating gas for a gas-engine power plant contained:
Moisture 8.60 per cent.
Carbon 62.69
Hydrogen 4.57
Oxygen 10.89
Nitrogen 1.40
Ash 10.42
Calorific power, determined in a bomb calorimeter, water
condensed, 6,786 Calories per unit of dried fuel. Ashes pro-
duced 268 potmds per ton (2,240 pounds) of moist slack used;
contains 12 per cent, of carbon.
Air used for running is heated to 300° C. by the waste heat
of producer gases, and carries in 2J tons of water as steam (at
170 METALLURGICAL CALCULATIONS.
same temperature) for every ton of fuel burnt. The steam
is generated by a tubular boiler nm by the escape gases from
the gas engines, but is heated from 100° C. to 300° C. by the
waste heat of the producer gases. The latter escape from the
producer at 350° C.
Composition of waste gases passing out of condensers at
15° C:
Carbon monoxide (CO) 11.0 per cent
Hydrogen (H^) 27.5
Marsh gas (CH*) 2.0
Carbonic oxide (CO^) 16.5
Nitrogen (N^") 41.3
Water vapor (H^'O) 1.7
100.0
Assume all the nitrogen of the fuel to form ammonia gas NH'.
Required:
(1) The calorific power of the Mond gas.
'. (2) The volume of gas produced per ton of fuel used.
(3) The efficiency of the producer.
(4) The weight of steam which is decomposed in the pro-
ducer.
(5) The proportion of the calorific power of the fuel saved
to the gas by the decomposition of steam.
(6) The proportion of the heat generated in the producer
which is saved to the gas by the decomposition of
steam.
(7) The proportion of the calorific power of the coal lost
from the producer by radiation and conduction.
Solution: (1) One cubic foot of the gas at 0° C. would gen-
erate the following amounts of heat (water uncondehsed).
CO = 0.110 cubic feet X 3062 = 336.8 ounce Calories.
H= = 0.275 cubic feet X 2613 = 718.6 "
CH* = 0.020 cubic feet X 8598 = 172.0 "
Totar= 1227.4
= 76.7 pound
= 138.1 B. T. U.
Per cubic meter =1227.4 kilo. "
ARTIFICIAL FURNACE GAS. 171
If measured at 15° C. (60° F.) the above values will be re-
duced by the factor 273 h- (273 + 1,5), and become
Per cubic foot = 72.7 pound Calories. ;
Per cubic meter = 1163 kilo. " (J)
(2) Carbon in 1 cubic foot of gaa at 0° C:
In CO O.IIOXO.M ounces r" '
In CH* 0.020X0.54 "
In CO" 0.165X0.54 "
0.295X0.54 " =0.1593 ounces
= 0.009956 pounds
Carbon going into gas per. pound of fuel burnt:
Carbon in fuel 0.6269 pou^ds
268
.. Carbon in ashes g^;gx 0.12 = 0.0144 ' "
Carbon in gas = 0.6125
Volume of gas (at 0° C.) per pound of fuel used:
0.6125
0.009956
61.52 cubic feet
Per ton of 2240 pounds = 137,805 cubic feet
At 15° C. (60° F.) '
= 137,805 X^^ -145,375 " "(2)
(3) The calorific power of the dried fuel is given as 6,786
Calories per pound, water condensed. Since one pound of
wet fuel contains 100 — 8.60 = 91.40 per cent, dried fuel, the
calorific power of 1 pound of wet fuel, moisture condensed, is
6,786X0.9140 = 6,202 pound Calories.
But, 1 pound of wet fuel would produce, on combustion,
0.0860 + 9 (0.0457) =^ 0.497.3 pounds moisture,
which, remaining vaporized at 15°, would retain
0.4973X596 = 296 Calories,
leaving the net metallurgical c^lb'rific power as
6,202—296 = 5,906 Calories.
172 METALLURGICAL CALCULATIONS.
The 61.52 cubic feet of gas produced per pound of moist fuel
will have a calorific power, burnt cold, of
61.52X76.7 = 4,719 Calories.
making the efficiency of the producer, on fuel consumed in it,
4 71Q
Zili^= 0.799 = 79.9 per cent. (3)
The above figure is true only on the assumption that the
steam used is obtained from waste heat, and therefore does
not require the combustion of extra fuel
(4) The gas produced contains, per cubic foot, the following
amount of hydrogen, free and as CH*:
As H^ 0.275 cubic feet X 0.09 =0.02475 ounces.
As CH< 0.040 cubic feet X 0.09 = 0.00360
Sum = 0.02835
= 0.00] 772 pounds
Per pound moist fuel = 0.001772x61.52 = 0.1090*
Present as ammonia gas 0.0140 X (3 h- 14) = 0.0030
Total (not including H as water) =0.1120
Hydrogen in 1 pound of coal = 0.0457 "
Hydrogen from decomposition of steam = 0.0663 "
Water decomposed in the producer =0.5967 "(4)
Since the steam introduced weighs 2.5 pounds for every
pound of fuel burnt, we see that only
^,fy = 0.2387 = 23.87 per cent.
^ . o
of the steam introduced is decomposed.
[In the writer's opinion this unused 76.85 per cent, can only
pass in and pass out carrying out sensible heat, and it seems
a very wasteful method of keeping down the temperature in
the producer. From the standpoint of regarding the 2 poimds
of steam as a mere absorber of sensible heat, it could probably
be replaced by some of the gases of combustion from the gas
engine or open-hearth furnace. The products of combustion
of the above gas would contain approximately
Nitrogen ■. .68.7 per cent.
Carbon dioxide 14.7 "
Water vapor 16.6 "
ARTIFICIAL FURNACE GAS. 173
And if the J pound of water vapor decomposed in the pro-
ducer were thus supplied, it would bring in, per pound of coal
burnt.
Nitrogen 41.2 cubic feet.
Carbon dioxide 8.8 "
Water vapor 9.9 "
And if the CO' thus introduced were reduced to CO, as it prob-
ably would be, the gases would receive from this source
Nitrogen .41.2 cubic feet.
Carbon monoxide 27.5 "
Hydrogen 9.9
thus producing gas quite up to standard as regards combus-
tibles, while the heat absorbed in the reduction of CO^ to CO
would cool the fire down quite as effectually as the extra steam
now used.]
(5) The steam decomposed, 0.5967 pounds per pound of fuel
burnt, may be assumed to be at 100° C. on entering the fire,
and to therefore absorb 3,196-pound Calories per pound of
steam decomposed. The heat absorbed in the producer, and
thus transferred into potential calorific power is, therefore
3196X0.5967 = 1907 pound Calories,
which expressed in per cent, of the calorific power of the fuel is
1907
i^^ = 0.323 = 32.3 per cent (5)
591)0
(6) The heat generated in the producer equals the calorific
power of the coal minus the heat lost by carbon in the ashes,
minus the calorific power of the gases, plus the heat absorbed
in decomposing steam. The loss by carbon in the ashes is
7=^X0.12X8100 = 117 Calories.
2240
The heat generated in the producer is, therefore,
5906—117 — 4719+1850 = 2920 Calories,
equal to 2920-^5906 = 49.4 per cent, of the calorific power of
174 METALLURGICAL CALCULATIONS.
the coal. Of this, 1907 Calories is absorbed in decomposing
1907
steam, which is -^-^ = 66.3 per cent, of the total heat gener-
ated. (6)
(7) There are 2922 Calories generated in the producer, of
which 1907 are absorbed in decomposing steam, leaving 1015
Calories to supply radiation and conduction and as sensible
heat in the hot gases, to which must be added the sensible
heat in the hot air and steam used at 300° C.
The air used per pound of coal is fotmd from the nitrogen
in the gases:
Nitrogen in 1 cubic foot of producer gas = 0.413 cubic foot
Nitrogen in 61.52 cubic foot of producer gas =25.40 "
Air used = 25.4-4-0.792 =32.08
Steam used = (2.5X16) -^ 0.81 =49.40
Heat in steam and air at 300° C. (from 15° C.) :
32.08X0.3116X285 = 2850 ounce Calories
49.40X0.3872X285 = 5450 "
Sum = 8300 "
= 519 poimd Calories
Total heat radiated, conducted and in hot gases:
1015 + 519 = 1534 Calories.
The heat in the hot gases is as follows, per cubic foot of gas
produced :
CO 0.110X0.313X335]
W 0.275X0.313X335 [ = 83.7
W 0.413X0.313X335 J
CO^ 0.165X0.450X335 = 24.9
CH< 0.020X0.460X335 = 3.1
Sum = 111.7 ounce Calories.
= 7.0 pound Calories.
Per pound of coal burnt :
7.0X61.52 = 430 pound Calories
ARTIFICIAL FURNACE GAS 175
To this must be added the heat in undecomposed water vapor
in the gases, as follows:
Steam used per pound of coal = 2.50 pounds
Steam decomposed per pound of coal = 0.60
Steam remaining in gases • =1.90
Moisture in coal = 0.086
Total in gases = 1.986
Volume = (1.986X16)^0.81 =39.2 cubic feet
Heat contained in this at 350° C. :
39.2X0.395X335 = 5187 ounce Calories
= 324 poimd Calories
Total heat in producer gases:
430 + 324 = 754 Calories
Heat lost by radiation and conduction:
1589—754 = 827 Calories
Proportion of calorific power of coal thus lost:
827
^± = 0.140 = 14.0 per cent. (7)
Whea Mond gas is heated in the regenerator of an open-
hearth furnace it changes considerably in composition, as is
shown by the following analyses made by Mr. J. H. Darby, on
gas dried before analysis:
Before After
Regenerator. Regenerator.
Carbonic acid gas, CO^" 17.8 10.5
Carbon monoxide, CO 10.5 21 .6
Ethylene, C^H< 0.7 0.4
Methane, CH* 2.6 2.0
Hydrogen, H'' 24.8 17.7
Nitrogen, W 43.6 47.8
100.0 100.0
The above changes are very interesting, and their discussion
profitable. Before heating, the gas bums with a non-luminous
176 METALLURGICAL CALCULATIONS.
flame; after heating, it bums with a brilliant white flame.
Let us inquire:
(1) What chemical change occurs during the heating?
(2) What are the relative volumes of the gas before and
after heating (excluding water)?
(3) What change in the calorific power is produced by the
heating?
(1) An inspection of the analyses shows undoubtedly that
at a high temperature the CO^ cannot hold all its oxygen in the
presence of such a large amount of hydrogen, and that the
following reaction must occur:
I I 1 I
CO^+H=' = CO + H='0
The figures given do not check exactly, but assimiing that
the above reaction takes place, we can find to what extent, by
expressing the composition of the gas after heating for the
same amount of nitrogen as was in the gas before heating,
since this gas is unchanged:
Before After Loss
Heating. Heating. or Gain.
CO^ 17.8 9.6 —8.2
CO 10.5 19.7 +9.2
H^ 24.8 16.1 —8.7
N=' 43.6 43.6 0.0
Since, according to the reaction written, the volume of CO'
reduced to CO will be equal to the volume of H' thus con<.
sumed, and will produce an equal volume of CO; the above
table proves, within the probable limits of error, that the
reaction written actually takes place.
The separation of luminous carbon is probably due to the
splitting up of C-H^.
The relative voltimes of the heated and unheated gases will
be inversely as the percentage of nitrogen in each (since this
gas is unchanged), viz.: as 47.8 to 43.6, or as 100 to 91.2. The
contraction, 8.8 parts, would again correspond almost exactly
to the amo\mt of water formed in the assumed reaction, which
would be equal to the hydrogen so used, thus giving another
check on the validity of the reaction assumed to take place.
ARTIFICIAL FURNACE GAS.
177
The calorific power of 1 cubic foot of original gas is:
CO
IP
0.105 X 3,062 =
0.007X14,480 =
0.026 X 8,598 =
0.248 X 2,613 =
Sum =
321.5 ounce Calories
101.4 "
223.5 "
648.0 "
Per 100 cubic feet
1294.4 "
80.9 pound
8090 "
The calorific power of 1 cubic foot of heated gas is:
CO 0.216X3,062= 661.4 ounce Calories
0.004X14,480 = 57.9
0.020 X 8,598 = 172.0
0.177X 2,613 = 462.5
Sum
C^'H*
CH<
1353.8 "
= 84.6 pound
For 91.2 cubic feet the calorific power would be 84.6X91.2
= 7715 pound Calories.
The net conclusion is that the heated gas (aside from its
sensible heat) gives less heat by combustion in the furnace
per unit of coal gasified in the producers than the correspond-
ing quantity of unheated gas; but the difference is only some 4
per cent. On the other hand, the calorific power of the heated
gas per cubic foot is soine 5 per cent, greater than that of the
unheated gas, and, therefore, its calorific intensity will have been
increased by the heating — quite aside from the question of its
temperature being higher, and therefore its sensible heat greater.
4. Water Gas.
This gas is made intermittently, by firs't burning part of the
fuel until the fire is very hot, and then introducing steam,
cool or superheated, into the fire. The reaction producing the
gas is:
III
C + H^O = CO + H'
And if the reaction is complete, and only carbon is present as
fuel, the gas produced is theoretically composed of equal parts
by voltome of CO and H^, the . volume of each of these being
equal to that of the steam used (if measured at the same tem-
perature and pressure).
178 MET A LLURGICA L CALCULA TIONS.
Deviations from this ideal composition are caused in practice
by the occurrence of undecomppsed steam in the gas, also of
CO^ and W, which come from the residual gas formed during
the heating up of the fire, some of which will get into the first
portion of the water-gas produced, and also hydrocarbons, tar
and ammonia from the distillation of the fuel. A typical
analysis of water-gas, given by Mr. W. E. Case, is:
Hydrogen (H') , . . .48.0
Carbon monoxide (CO) 38 . 0
Methane (CH^ 2.0
Carbon dioxide (C0=) 6.0
Nitrogen (N") 5.5
Oxygen (0^) 0.5
The* production of water-gas is more expensive than that of
other artificial producer gases, because of the large amount
of steam necessarily required, the heat lost during the heating
up, and the essentially intermittent character of the operation.
Its essential advantages are its very high proportion of com-
bustibles, averaging 90 per cent, and the consequent high
calorific intensity which it is capable of producing. (Urlcar-
buretted, it is well known that water-gas is a valuable domestic
fuel, and illuminant when used in mantle burners; carburetted,
it forms our principal illuminating gas, and as such is manu-
factured on an immense scale. We will treat here only of its
metallurgical uses.)
Discussing the manufacture of the gas, the first step is the
heating up of the fuel. This is accomplished by blowing air
through it. At this point we can distinguish two systems.
The older oiie is that of blowing in air under mdd<irate pres-
sure, which passes through the fire at moderate velocity, and-
produces a fair grade of ordinary producer gas. This gas is'
either wasted, or burned in furnaces requiring such quality of
gas, or burned in regenerators where heat is stored up to be
utilized in the next stage in superheating steam. The newer
system is to blow in air at high pressure, such that a large
percentage of carbon dioxide remains in the gases, thus nearly
completely burning what carbon is oxidized, and. storing upa;
cQiresponding quantity of heat in the remaining carbon. The
incombustible gases produced are passed through a recuperator
ARTIFICIAL FURNACE GAS. 179
or regenerator, where their sensible heat is partly communi-
cated to the steam used, so as to superheat it when forming
water-gas. This system consumes the minimum of carbon
in " heating up " the fuel, saves time in this unproductive period,
and allows the producer to be run longer " on steam." In
practice the fuel is probably raised to an average temperature
of 1500° C. during the heating up.
We can calculate the efficiency of this heating up, that is,
the proportion of the calorific power of the carbon burnt which
is stored up in the remaining fuel, if we know how much fuel
is in the producer, how much air is blown through, the com-
position of the gases produced, and the average rise in tem-
perature of the fuel. Since this operation is, however, only
supplementary to the real formation of water-gas by the de-
composition of steam, we will first make calculations upon
the cooling down period, during which water-gas is made.
During the use of steam the reaction absorbs heat, and the
producer rapidly cools. Steam is passed through until the
temperature of the fuel is 800° C, and must then be stopped,
because between 800° and 600° the reaction is mostly
C + 2ffO = C0H2H''
resulting in a rapid increase of CO' in the gas. The heat to
decompose st-eam is furnished partly by the oxidation of carbon
to CO, and partly by the sensible heat in the carbon itself.
Since the carbon, from temperatures of about 1000° C. up has
a specific heat of 0.5, we can easily calculate how much steam
can be decomposed before the temperature falls to 800°.
Problem 20.
A Dellwick-Fleischer water-gas producer contains 3 tons of
coke (90 per cent, carbon), heated up to 1500° C. Steam,
heated to 300° C, is passed through for 8 minutes, until the
temperature of the gas escaping is 700° C, or 800° at. the fuel
bed. The composition of the gas is (Prof. V. B. Lewes):
Hydrogen 50 . 0
Carbon monoxide 40 . 0
Carbon dioxide 5.0
Oxygen • 10
Nitrogen. 4.0
180 METALLURGICAL CALCULATIONS.
Assume 9000 Calories per minute lost by radiation and con-
duction.
Required: (1) The amount of steam used in the 8 minutes
(2) The volvime of gas produced in the 8 minutes.
(1) The amoimt of steam which could have been used is
limited by the available heat to decompose it. The latter is
furnished by
(a) Oxidation of carbon to monoxide.
(6) Oxidation of carbon to dioxide.
(c) Sensible heat of the carbon and ash of fuel.
(d) Sensible heat of the steam used.
While the items of heat absorption and loss are:
(e) Decomposition of the steam.
(/) Sensible heat in the gases,
(g) Radiation and conduction.
The simplest way to arrive at a solution is to make a few
necessary assumptions, and to then let X represent the weight
of steam used. The assumptions are that the average tem-
perature of the fuel bed falls to 1000° C, that the average
temperature of the escaping gases is (1500 + 700) h- 2 = 1100° C,
that the average specific heat of carbon in the range 1000 — 1500
is 0.5, and of the ash of the fuel 0.25. Then, .casting up g,
heat balance sheet in terms of X, we can finally arrive at an
expression for the heat which was available during the 8 minutes
for decomposing steam, and thus at the weight of steam decom-
posed, and (making this equal to X) at a solution.
(a) The analysis of the gas shows that eight times as much
carbon was burnt to monoxide as to dioxide, making the equa-
tion of combustion:
9C+10H='O = 8CO + COHi0H='
By weight, 8X12 = 96 parts of carbon was burnt to CO per
10 X 18= 180 parts of steam used, or 0.533 parts per one part of
steam. The heat generated in forming monoxide is, therefore :
0.533XXX2430 = 1296 X Calories.
(6) Only one-eighth as much carbon bums to dioxide, giving,
therefore, as the heat evolved:
(0.533h-8)XXx8100 = 540 X Calories.
ARTIFICIAL FURNACE GAS. 181
(c) 3000 kilos, of fuel, representing 2700 kilos, of carbon and
300 kilos, of ash, cool from 1500° to 1000°:
2700X0.50X500 = 675,000 Calories
300X0.25X500 = 37,500
Sum = 712,500
(d) X-t-0.81 will be the volume of the steam used at normal
conditions, which brings in considered only as vapor, at 300°:
(X -h 0.81) X 0.385X300 = 142.6 X Calories.
The sum total of (a) + (b) + (c) + (d) gives the total available
heat, viz.:
1978.6 X + 712,500 Calories.
(e) The steam ' used requires for its decomposition, con-
sidered theoretically as cold steam, producing cold products:
(X -^ 9) X 29,042 = 3227 X Calories
(/) The gas being 50 per cent, hydrogen, and the latter being
equal to the volume of steam used, the volume of gas must be
2X (X-:-0.81), which multiplied by the mean specific heat of
gas of this composition between 0° and 1100° per cubic meter,
and by the temperature, will give the heat thus carried out of
the producer:
2 (X^0.81)X0.347X1100 = 942.5 X Calories
(g) 9000X8 = 72,000 Calories.
The sum total of (e) + (f ) + (g) gives the total heat distribu-
tion, viz.:
4.169.5 X + 72,000 Calories.
Since the heat available equals the heat distributed:
1.978.6 X + 712,500 = 4,169.5 X + 72,000
or X = 292 kilograms (1)
(2) .The volume of this steam, at assumed standard condi-
tions, would be
292-^0.81 = 360 cubic meters,
and of gas, since it is 50 per cent, hydrogen,
360X2 = 720 cubic meters.
m METALLURGICAL CALCULATIONS.
The above figures represent the maximum attainable pro-
duction, on the assumption that sufl&cient steam-generating
power is available to furnish the steam, and that the fuel in
the producer is of small size and the bed so uniform that the
production of gas is regular all over it. In practice, figures con-
siderably below this are attained, but it is always well to know
the possible maximum which is attainable.
Problem 21.
In a Dellwick-Fleischer water-gas producer the heaiting up is
accomplished in 2 mintites by blast from a Root blower, fur-
nishing air through a 9.5 in h pipe at a total water-gauge
pressure of 19 inches of water, temperature of air 15° C. The
gases escaping from the producer analyze: >
Carbon dioxide 17.9 per cent.
Carbon monoxide 1.8 "
Nitrogen 78.6
Oxygen 1.7
Temperature of waste gases 900° C. ; heat lost by radiation
and conduction 9000 Calories per minute; assume producer to
contain 3000 kilos, of fuel, consisting of 90 per cent, carbon and
10 per cent. ash.
Required: (1) The average rise in temperature of the fuel bed.
(2) The proportion of the heat generated which is thus
stored up as useful heat for producing water-gas.
(3) Assuming that the production pf water-gas lasts 8 min-
utes, during which 2500 Calories are absorbed from the fuel
bed per kilogram of steam used, what should be the steam
supply in kilograms per minute ?
(4) The ratio between the volume of air supplied during the
blowing up period and the weight of steam used in the gas-
i making period.
Solution: (1) We miist first find the amount of air fur-
nished by the blower, To do this, we calculate the pressure
head in terms of air at 15° C, instead of water, and then apply
the well-known formula V = V 2g. h. Water is 772 times as
heavy as air at 0° C, and, therefore, Jti inches of water pres-
sure would represent 19X772-^12 = 1222 fee^ of air pressure
(that is, a column of .fluid as light as air, 1222 feet high). But
ARTIFICIAL FURNACE GAS. 183
air at 15° is lighter' still than air at 0°, in the ratio 273 to 288»
so that the air pressure measured in terms of air at 15° will be
1222X2^ = 1290 feet
The velocity of the air supplied will therefore be, in feet per
second:
V=\/ 64X1290
= 287 feet per second,
and the volume delivered, per minute, at 15° C:
287 X 60 X (0.7854 X 9.5 X 0.5 -^ 144) cubic feet
= 17,220X0.492 = 8472 cubic feet,
which in temisof air at 0° C, would be
847 X'i^l = 8050 cubic feet
>- . - Zoo
• i ; = 228 cubic meters ; -, _\
The volume of waste gases produced in the 2 minutes can
be ; found from the relative percentages -of nitrogen in the air
(79.2V and in the gases (78.6), as follows:
.' ■ '■■ ' 79' 2 ' ' J ^ '
228X2X;^-A = 459.0 cubic meters
- -f ,., .
Containing, therefore, from its analysis:, . .
-' ' Carbon dioxides . .-. .-; .-. . . . . : . 82.25 cubic meters
Carbon monoxide 8.27
;. •■•. 'Oxygen... "J. .;.;.:..., .-.......' ?.81 r"":-^ . " - ( )
Nitrogen 361.15 " " "
-. ' : 459.48 i •• ««
Tl\e carbpn burnt to CO^ an(l CO. will be: "
C to CO' 82.25X0.54 =' 44.42 kilos.
• ^CfoXO- 8.^7X0.54-^" 4.47 " "'
Sum = 48.89 "
And the heat thus generated:
' C to CO' = 44.42X8100 = 359,800 Calories
C to CO = 4.47 X 3430 = 10.860 "
,. ,. Sum = 370,660
184 METALLURGICAL CALCULATIONS.
To find the amotint of this heat left in the producer at the
end of the 2 minutes blowing up, we must subtract the 2x
9000 = 18,000 Calories lost by radiation and conduction, and
then, in addition, the heat carried out by the hot gases, at an
average temperature of 900° C, which latter will be:
CO, 0^N' 377.25X0.327X885 = 109,150 Calories
CO'' 82.25X0.571X885 = 41.565
Sum = 150,715
Heat left in the fuel bed:
370,660—168,715 = 201,945 Ca:lories
heat capacity of the fuel bed per 1° C. :
3000X0.9 kilos, carbon XO.5 = 1370 Calories
3000X0.1 kilos, ash XO.25 = 75
Sum = 1445
Average rise in temperature of the fuel bed.
201,945
1445
140° C. (1)
(2) The useful heat thus stored up in the fuel bed amounts
to the following proportion of the total heat generated during
the blowing up:
201,945
370,660
= 0.545 = 54.5 per cent. (2)
(3) Steam which can be decomposed in the gas-producing
period:
201,945 -._,.,
25Q0 = ^^-^ kilograms
= 10.1 kilograms per minute (3)
(4) Air supplied in 2 minutes = 456 cubic meters. Steam
used in 8 minutes = 80.8 kilograms.
80 8
Ratio = -— - = 0.177 kilos, steam per 1 m* air (4)
= 0.177 ovince steam per 1 ft' air
= 1.1 poxtnds steam per 100 ft' air
CHAPTER VII.
CHIMNEY DRAFT AND FORCED DRAFT.
In all problems concerning combustion, we must furnish the
air needed for combustion either by suction .or by pressure.
The original and almost universal method is by chimney draft;
the more positive and reliable method is forced draft. Often
the two are combined with very satisfactory results.
The waste heat from any metallurgical process or furnace
is generally considerable. Most furnaces must be kept above
a red heat, and the gases pass directly out of the furnace into
the chimney. In such cases the chimney is indicated as the
proper source of draft, because it utilizes, although very in-
efficiently, the lascensive force of the hot gases, and thus works
by otherwise wasted energy. In other cases it is practicable
to pass the gases through boilers before they go to the chim-
ney, and thus to raise large amounts of steam. The gases are
then cooled down so far that they enter the chimney too cold
to furnish all the draft needed; in such cases a small fraction
of the steam generated will run a steam engine or steam tur-
bine, and run a fan capable of furnishing all the draft needed.
In this manner considerable steam is available for other pur-
poses, and great economy is effected.
Chimney Draft.
The principles involved are not obscure or complicated.
The total pull, or suction, which a chimney can produce, as-
suming it to be filled with hot air, is simply due to the ascen-
sive force of the hot air inside, and the measure of this is the
difference of weight of the chimney full of hot gases and what
it would be if filled with cold air of the temperature outside.
Illustration: A chimney is 6 feet square inside and 100 feet
high, uniform, with the gases inside at an average tempera-
ture of 500° F,, and specific gravity (air = 1) of 1.06. The
185
186 METALLURGICAL CALCULATIONS.
air outside is at 80° F. What is the ascensive force of the hot
gas inside, in total pounds, in ounces per square inch and.
inches of water gauge ?
The volume of the space in the chimney — chimney volume —
is 100X6X6 = 3600 cubic feet. This volume, filled with air
at 32° F., would weigh
3600X1.293 = 4654.8 oz. av. = 290.9 pounds.
And, filled with gas at 500° F.,
491
290.9X1.06X^00_3,^^Q, = 157.9 pounds.
If filled with outside air, at 80° F., the weight would be
4Q1
290-9X80=32T491 = 265-0 pounds.
We, therefore, see that the hot gases in the chimney, weigh-
ing 157.9 pounds', displace 265.0 pounds of cold air, and the
tendency of the former to rise upwards in this ocean of air
must be
265.0—157.9 = 107.1 pounds.
To put it in another way, if a piston fitted into the chimney
at the bottom, and could move without friction, the piston
would have to be loaded with 107.1 pounds to keep it from
moving up the chimney. The total upward pull of the chimney
is therefore 107.1 pounds.
Since this would be exerted on a piston 6X6 = 36 square
feet in area, the pull or suction per square foot, in pounds, is
107.1 H- 36 = 2.98 pounds, and in ounces per square inch.
,(2,98-^,144) X 16 = 0.331 ounce per square inch.
If. the pull or suction is measured on a gauge, as by water
pressure, the pressure of a 1-foot column of water at ordi-
nary ;teinperatures is 1000 ounces per square foot, or 1 inch of
water is
(1000 -f- 144) -^ 12 = 0.597 ounces per square inch.
The total pull of the chimney is therefore eciuivalent to-
0.331
0.579
0.57 inch of water gauge.
CHIMNEY DRAFT AND FORCED DRAFT. 187
By exactly similar methods of calculation- the theoretical total
suction of a chimney of any given height and temperature of
gases inside and of air outside may be obtained. The suction
expressed in ounces per square inch, or in water gauge is, of
course, independent of the cross-sectional area of the chimney;
it depends only on its height and on the temperatures inside
and outside.
The above calculated total suction (allowing nothing for
friction, etc.) is called the total head of the chimney, and is
usually expressed in terms of cold air (at 0° C.) instead of in
water. Cold air is a fluid, and water is 772 times as heavy as
it; therefore, a gauge pressure or hydrostatic head of 0.57
inch of water is the same as
0.57X772 = 440 inches of air.
= 36.5 feet of air.
What this head really represents is clearly seen from the
above calculation. Its value is to be obtained directly from
the height of the chimney, temperature inside and out and
specific gravity of the chimney gases (air = 1) by the fol-
lowing relations, in which
h^ ^ total head in feet of air at 32° F.
ho = total head in meters of air at 0° C.
t = temperature in the chimney, F°.
t = temperature in the chimney, C°-
t' = temperature of outside air, F°.
t' = temperature of outside air, C°.
D = Specific gravity of chimney gases, air = 1.
H = Height of chimney in feet.
H = Height of chim.ney in meters.
coef = coefficient of gaseous expansion, F° = ^
coef = Coefficient of gaseous expansion, C° = ^=^ = a
_ r(l— D+W [Cf-32)-D (r— 32)]-|
^o ~ L[l+'^«^/ (<'— 32)] [1+coef (/— 32)]J
r(l— D-l-coef (t— DtQ-l
188 METALLURGICAL CALCULATIONS.
The author is not fond of using formulas whenever their
use can be avoided. The above formulas express in the sim-
plest mathematical form the principles which have been so
far explained and used in the calculations, but it is strongly-
urged that the formulas be kept " for exhibition purposes only,"
and that when any specific case is to be worked it be attacked
from the standpoint of the principles involved, as explained in
the case worked. In other words, if one understands properly
and thoroughly the "basic principles, he has no need of the
formula; if one does not understand the principles, the formula
had better be kept forever in " innocuous desuetude."
The total head, obtained as above, is the theoretical head.
It is like the pressure on the piston of a locomotive — ^the total
available force for all purposes. Just as the pressure on the
locomotive piston is used up in friction in the engine and in
moving the engine itself, and the residue is the available pull
on the draw-bar which moves the train, so the total head of
the chimney is partly used up in friction in the chimney itself,
partly in giving velocity to the gases as they pass out of the
chimney, and the residue is the available head which draws or
pulls the gases through fire-grates, furnaces and flues up to
the base of the chimney. If the chimney could be momen-
tarily completely closed at the bottom, except for the gauge
opening, and the gas inside be brought to rest, the gauge would
show the total head ; as soon as dampers are opened connecting
the flues, gas moves up the chimney, and the gauge pressure
is lessened by the head required to move the gases and that
absorbed in the friction in the chimney.
Head Represented in Velocity of Issuing Gases. — This item
always exists when the chimney is working, and depends only
on the velocity of the gases as they escape and their tempera-
ture. The hydraulic head necessary to give any fluid a velocity
V is simply the same as the height which a falling body must
fall in order to acquire that same velocity; i.e.',
in which expression g is the constant acceleration of gravity,
9.8 meters or 32.2 feet, and the velocity is in meters or feet
per second. ' If we kno-v« therefore, the velocity of the gases
CHIMNEY DRAFT AND FORCED DRAFT. 189
issuing from the chimney, or can calculate or assume it, we
can get h. In practice the velocity does not vary within very
wide limits. In small house chimneys it may not exceed 3 feet
per second, in boiler chimneys 6 to 12 feet per second, in fur-
nace chimneys 12 to 20 feet per second. The temperatures of
these issuing gases is, moreover
C° F"
In small chimneys 100 to 200 200 — 350
In boiler chimneys 100 to 300 200° — 550
In furnace chimneys 300 to 1000 550°— 1800°
If the chimney in question has, therefore, a known velocity
of exit of its gases, h can be calculated; but it must not be
forgotten that h will be in terms of the kind of gases which
is escaping; i.e., of hot gas, and to subtract it from or com-
pare it with ho we must reduce it to its equivalent head in
terms of cold gas. This is merely a matter of taking into
account the specific gravities or relative densities of hot gas
and cold air, which are inversely proportional to their absolute
temperatures; that is, if D represents the relative density of
air and chimney gases at the same temperature:
^o '^^l + at 2g l + at
Illustration: Assuming the actual velocity of the gases is-
suing from a furnace chimney to be 15 feet per second, and
their temperature 500° F., density 1.06 (air = 1), what will
be the head represented by the velocity of these gases in terms
of cold air at 32° F. ?
The head represented, in terms of hot gases at 500° F., is
, V 15^ O C ^ 4.
In terms of air at 500° F. is
3.5X1.06 = 3.71 feet.
And in terms of air at 32° F.,
3-7lX5ooZ§+491 = ^-^^^^*-
190 METALLURGICAL CALCULATIONS.
Out of the total head which this chimney produces (say 36.5
feet), 1.9 feet is represented by the velocity of the issuing
gases, or 5.2 per cent, of the whole, leaving 34.6 feet to repre-
sent loss by friction in the chimney and the available head.
We will proceed to discuss the loss of head due to friction in
the chimney.
Head Lost in Friction in the Chimney. -^T^his varies with
the smoothness or roughness of the walls, and has been deter-
mined experimentally for air moving with different velocities.
The manner of expressing the friction loss is, to put it as a
function of the head necessary to give the gases their actual
velocity, assuming there were no friction. Thus, supposing as
in the preceding paragraph, the actual velocity of the hot
gases is 15 feet per second, and the head (in terms of cold
air) necessary to give that velocity, not considering friction,
is 1.9 feet, then the head lost in friction in getting up this
velocity will be
h friction = i.9x^K.
d
That is, it will be proportional to H, the height of the chimney,
inversely as d, the diameter or side, if square, and to a co-
efficient K, determined by experiment. The latter varies, ac-
cording to Grashof's experiments, between 0.05 for a smooth
interior to 0.12 for a rough one, and averages 0.08.
Illustrations: Assuming the height of the chimney, 100 feet,
its section to be 6 feet square, the coefficient of friction K =
0.08, and the head represented by the net velocity of the hot
gases in the chimney to be 1.9 feet of cold air, what is the
head lost in friction in the cliimney?
The ratio of height to side is 100-^6 = 16.67, which mul-
tiplied by K gives 1,33 as the value of the function containing
these three terms. This means that 1.33 times as much head
has been lost in friction as is represented by the net actual
velocity of the gases as they pass up the chimney. Therefore,
h f™*"" = 1.9X1.33 = 2.5 feet cold air.
Another way of looking at this, which is sometimes useful
in considering the height of a chimney, is to say
h ^"^twn = 0.025 H.
CHIMNEY DRAFT AND FORCED DRAFT. 191
Or, that in this case, the head lost in friction amounts numer-
ically to one-fortieth the height of the chimney.
If we subtract the head lost in friction plus that represented
in the net velocity of the gases, from the total gross head, the
residue is that available for doing work external to the chimney.
In the specific case of the preceding illustrations we have
ho = total head = 36.5 feet = 100 per cent.
h velocity = velocity hea'd = 1.9 " = ~5
h f™*»" = friction in chimney = 2.5 " = 7
h available _ available head = 32,1 " = 88 - "
Available Head of a Chimney. — ^This is the part of the total
head which remains after subtracting the head lost in friction
in the chimney and that represented by the velocity of the
issuing gases. In the specific cases considered in the above
illustrations, the net available head amounted to 88 per cent,
of the whole theoretical head. If we assiime limiting con-
ditions as found in practice, we can find the limiting values of
this proportion. Calling the cases I and II, those with mini-
mum and maximvun absorption of head in the chimney itself,
we have
Case I. Case II.
Temperature of issuing gases .100° C. 1000° C.
Velocity of issuing gases per second 1 meter 7 meters
Ratio H to d ' 10 50
Coefficient K 0.05 0.12
Specific gravity of gases (air = 1) 1-00 1.06
Head as velocity of gases (meters of air) . .0.04 m. 0.56 m.
Head as velocity of gases (feet of air) 0.13 1.87
Head absorbed in friction (meters of air) . .0.02 3.36
H^ad absorbed in friction (feet of air) 0.07 11.2
Head used up in chimney (meters). . .... .0.06 to 3.92
Head used up in chimney (feet) . . . . .• ...... 0.20 -to 13.0
Water gauge pressure thus lost, m.m 0.1 to 5.0
Water gauge pressure thus lost, inches' 0.003 to 0.2
The available head, will, therefore, be the theoretical total
head minus a loss in the chimney itself, which may amount
to a maximum of 3.9 meters or 13 feet, representing an ab-
sorption of- water gauge pressure up to 5 millimeters, or 0.2
192 METALLURGICAL CALCULATJONS.
inch at a maximum. Under ordinary conditions half these
quantities would be a rather high chimney loss.
In most conditions which confront the metallurgist, the
question is to determine how high a chimney should be built
in order to supply a certain available draft determined by
practice to be necessary. For instance, to burn a certain
amount of coal per hour on any grate requires a certain amount
of draft. This amount is increased if the draft is increased,
and vice versa. In boilers, 18 pounds of coal burned per square
foot of grate surface per hour is highly economical practice,
and requires a draft of 0.4 inch to 0.8 inch of water gauge,
according to the kind of coal burned. In furnaces where the
amount of coal burned is greater per hour there will be usually
a con-espondingly greater temperature in the chimney. To
calculate the height of chimney required it is necessary to assume
only the temperature in the chimney, the available draft re-
quired and an average chimney loss.
Problem 22.
It is desired to design a chimney for a puddling furnace, the
grate of which is 4 feet by 6 feet, and which shall burn 30 pounds
of bituminous coal per hour per square foot of grate surface.
Temperature of gases entering the chimney 1200° C, at the
top probably 1000° C. Specific gravity of gases 1.03 (air = 1).
Draft required 0.6 inch of water gauge. Outside temperature
30° C.
Solution: We can assume that since the gases will be at
an average temperature of 1100° C. in the chimney, their- ve-
locity will be high, and that at least 0.1 inch of water gauge
pressure will be absorbed by the chimney itself. This makes
a total requirement of 0.7 inches of water for total head, or
h„ = 0.7 X 772 -H 12 = 45 feet of cold air.
Or an unbalanced pressure or ascensive force of
45X1.293-^16 = 3.64 pounds per square foot.
Considering the air outside the chimney, its weight at 30° C.
is equal, per cubic foot,
1.293 X 1^ -Hie = 0.073 pounds.
CHIMNEY DRAFT AND FORCED DRAFT. 193
The gases inside the chimney weigh, per cubic foot,
1.293X1.03Xj3qq^^-h16 = 0.0166 pounds.
The height of the chimney being called H and its cross-
section S, the volume is HxS, and the weight of hot gas inside
it is
(HXS)X 0.0166 pounds.
And of an equal volume of cold air outside
(HXS)X 0.073 pounds,
giving a total ascensive force of
(HXS)X 0.0564 pounds.
But there is needed a total ascensive force of
SX 3.64 pounds,
in order to give the pull of 3.64 pounds per square foot, and,
therefore, of necessity,
HXSX 0.0564 = SX3.64,
from which
H = K^nSr = 64.5 feet.
0 . 0564
Concerning the cross-section of this chimney, it would not
be safe to make it less in diameter than one-fiftieth of its height,
because of lack of stability; in fact, one-twenty-fifth would
be better practice. This consideration would make its in-
ternal diameter 2 ft. 7 inches, area 5.2 square feet. Another
way of arriving at a diameter is to calculate the volume of the
hot gases which must pass up the chimney and assume for them
some maximum velocity in the chimney, such as, let us say,
6 meters (20 feet) per second, and so get the minimim:i area
necessary for filling this condition as follows:
Coal burnt per hour 4 X 6 X 30 = 720 poimds.
Air theoretically necessary, assuming aver-
age bituminous coal (see Prob. 1) = 123
X72Q = 88,560 cubic feet.
194 METALLURGICAL CALCULATIONS-
Products of combustion at standard condi-
tions = 129X720 = 92,880 cubic teet
Volume chimney gases at 1100° C. =
92,880X^i5^§i^ = 467.100 "
2/3
Volume per second = 130 "
Area of chimney, if maximum velocity is 20
feet per second = • 6 . 5 sq. feet
Diameter, if round = 2 ft. 10 in.
This chimney would do its work better, and there wotdd be
much less loss in friction, if the internal diameter were made 25
per cent, greater than the above calculated minimum, say, there-
fore, 3 feet 6 inches, making the area nearly 50 per cent, greater
and cutting down the velocity in the chimney to 13.5 feet per
second.
Problem 23.
In the case of the puddling furnace of Problem 22, assume
that the hot gases, instead of going directly into the chimney,
are passed through the flues of a boiler placed above the fur-
nace, and thence pass into the chimney at a point 15 feet
higher than before. Assume chimney 3 feet 6 inches internal
diameter, 64.5 feet high above the furnace flue, and that the
gases now passing into it 15 feet higher up are at 350° C, and
cool to 250° C. at the top of the chimney. The boiler flues in-
troduce additional frictional resistance equal to 0.1 inch of
water. The boiler raises steam at a net efficiency of 45 per
cent., the steam engine utilizes the steam at a mechanical
efficiency of 20 per cent., and a centrifugal fan supplies the
forced draft needed at a mechanical efficiency of 25 per cent.
Required: (1) The total head of the chimney, when the
furnace discharged directly into it. and the average tempera-
ture of the gases in it was 1100° C, and specific gravity 1.03
(air = 1).
(2) The head absorbed as velocity of the outgoing gases,
their temperature being 1000° C.
, (3) The head lost in friction in the chimney, in this case.
(4) The head which was available to run the puddling furnace.
(5) The total head of the chimney with the gases entering
15 feet above former flue, and average temperature 300° C.
CHIMNEY DRAFT AND FORCED DRAFT. 195
(6) The head absorbed in this case as velocity of outgoing
gases, their temperature being 250° C.
(7) The head lost in friction in the chimney in this case.
(8) The available head to draw gases into the chimney.
(9) The deficit of head which must be made up by forced
blast under the grate of puddling furnace.
(10) The horse-power absorbed by the fan which furnishes
this blast.
(11) The horse-power furnished by the engine using the
steam from the boiler.
(12) The excess of power which is thus saved and available
for other purposes.
Solution:
(1) Volume of gases in chimney:
64.5X3.5X3.5X0.7854 = 620.5 cu. ft.
Weight at 32° F. (0° C):
620.5X(1.293-hl6)'xl.03 = 51.65 lbs.
Weight if temperature is 1100° C:
^l-«^XllO0T273 = ^«-2^^^^-
Weight of equal volume of air outside at 30° C:
273
620.5X(1 -293-1-16) X3Q_^273 ^" ^5.18 lbs
Difference of weight = ascensive force,
45.18—10.27 = 34.91 lbs.
Ascensive force per square foot,
34.91 -f- 9.62 = 3.63 lbs.
Total head in terms of cold air at 0° C,
3.63 -^ (11293 -h 16) = 44.9 ft. - (1)
In terms of water gauge pressure,
44.9 X 12 -H 772 = 0.685 ins. (1)
196 METALLURGICAL CALCULATIONS.
(2) Volume of gases per hour at 0° C,
(Prob. 22) = 92,880 cu. ft.
Volume at 1000° C,
1 000 4- 27"^
= 92.880 X^yg = 433,100 cu. ft.
Velocity per second,
433,100-4- (3600) -^ 9.62 = 12.50 ft.
Head necessary to give this velocity, in terms of hot gases,
at 1000° = (12.50)^-^64.3 (2g) = 2.43 ft.
In terms of gases at 0° C,
" 'l^v — n "i" ft
"•^^^1000+273 "•^""•
In terms of air at 0° C,
0.52X1.03 = 0.55 ft.
(2)
In terms of water gauge pressure,
0.55X12^772 = 0.008 in.
(2)
(3) Assuming K, the coefficient of friction, 0.08, then
h friction = T. ^73_ H
^0 2g 273 + t d^-^-
This is only an abbreviated form of the operations done
under (2), adding the terms which account for the height,
diameter and friction. Now, the velocity per second:
n 00 -1- 97^
V = 92,880 X "JIh- 3600-^9.62 = 13.5 ft.
^7o
Head necessary to give this velocity in terms of air at QP,
Proportion of this velocity head lost in friction =
f K = -||x0.08 = 1.47
CHIMNEY DRAFT AND FORCED DRAFT. 197
Head lost in friction in chimney,
0.58X1.47 = 0.85 ft. (3)
In terms of water gauge pressure,
0.85 X 12 -h 772 = 0.013 in. (3)
(4) Cold Air. Water Gauge.
Total head 44.90 feet 0.685 inch
Absorbed in velocity of gases. ...... 0.55 " 0.008 "
Absorbed in friction in chimney ... . 0.85 " 0.013 "
Available for the furnace 43.50 " "0664 " (4)
(5) Volume of chimney gases,
(64.5— 15) X 3.5X3.5X0.7854 = 475.7 cu. ft.
Weight at 300° C, specific gravity 1.03 (air = 1).
475.7X(1.293^16)X1 OSX^^g^g^p = 18.86 lbs.
Weight of equal volume of outside air at 30° C,
07-}
475.7X(1.293H-16)X^y3_^3Q = 34.68 lbs.
Ascensive force of air per square foot,
(34.68— 18.86) H- 9.62 = 1.64 lbs.
Total head in terms of cold air,
1.64-^(1.293-^16) = 20.3 ft. (5)
In terms of water gauge pressure,
20.3 X 12 H- 772 = 0.32 in. (5)
(6) Velocity of issuing gases, per second, at 250° C,
92,880 X ^^ 2^ ^ ^ 3,600 -^ 9.62 = 5.14 ft
Head as velocity in terms of cold air at 0° C,
273
In terms of water gauge pressure = 0.003 in.
(7) Average velocity of gases in chimney at 300° C,
92,880^ 3,600 X~||^ ^9.62.= 5.63ft.
198 METALLURGICAL CALCULATIONS.
Head lost in friction in terms of cold air,
(5.63)'^.84.3X3QQ^^y3Xl.03X-^^X0.08 = 0.27 ft. (7)
In terms of water gauge pressure = 0.004 in. (7)
(8) Cold Air. Water Gauge.
Total head 20.30 feet 0.320 inch
Absorbed in velocity of gases 0.22 " 0.003 "
Absorbed in friction in chimney 0.27 " 0.004 "
Available to draw gases in 19.81 " 0.313 " (8)
(9)
Available head needed for puddling
(4) 43.50 " 0.664 "
Available head needed for boiler. . . . 6.43 " 0.100 "
Total head needed for both 49.93 " 0.764 "
Available head from chimney (8). . .19.81 " 0.313 "
Deficit, to be supplied by blast 30.12 " 0.451 "
(10) The 0.451 inches of water gauge equals
(0.451^ 12) X 62.5 = 2.35 lbs. per sq. ft.
The volume of air to be supplied is, at 30° C,
27^ -I- ^0
88,560 (Prob. 22)^60X 273 = ^'^^^ ^"- **• P®'" ^°^-
Net work done by the fan,
1,638X2.35 = 3,850 ft. lbs. per min.
Gross power needed by the fan,
3,850^0.25 (efficiency) = 15,400 ft. lb. per min.
Horse-power needed to drive the fan,
15,400 -=-33,000 = 0.47 H. P. (10)
(11) The boiler receives the gases at 1200° C. and discharges
them at. 350°, and 92.880 cubic feet of gases (measured at
standard conditions) pass through per hour. The composition
of these gases is not given, but from the specific gravity we
might conclude that they contain on an average 10 per cent, of
carbon dioxide, since if they contained the maximum amount
CHIMNEY DRAFT AND FORCED DRAFT. 199
of that gas (about 20 per cent.) their specific gravity would be
1.06 (air=l). Assuming them, therefore, to contain
CO' 10 per cent.
H'O 10
CO, N', O' 80
their heat capacity per degree per cubic foot would be, between
350° and 1200°,
CO* 0.1X[0.37 +0.00022 ,(350 + 1200)] = 0.0711 oz. cal.
WO 0.1X[0.34 +0.00015 (350 + 1200)] = 0.0573 "
CO, N^ O' 0.8X[0.303+0.000027 (350 + 1200)] = 0.2759 "
Sum = 0.4043 "
Heat given up per cubic foot,
0.4043X (1200—350) = 343.7 oz. cal.
Heat given up by gases per hour to boiler,
92,880X343.7 = 31,922,850 oz. cal.
= 1,995,000 lb. cal.
Heat in the steam produced per hour,
1,995,000X0.45 (efficiency) = 897,750 lb. cal.
Heat equivalent of mechanical energy of steam engine per
hour,
897,750X0.20 (efficiency) = 179,9:0 lb. cal.
Heat equivalent of 1 hp hour = 635 kg. cal.
= 1,400 lb. cal.
Horse-power generated by the engine,
179,950-^1,400 = 128 H. P. (H)
(12) Net available power after supplying fan,
128—0.5 = 127.5 H. P, (12)
CHAPTER VIII.
CONDUCTION AND RADIATION OF HEAT.
These two factors are of the greatest practical importance
to the metallurgist, yet they are also the subjects of all others
upon which the practical metallurgist is usually the most
poorly informed. It often happens that a furnace is built of
twice the capacity of its predecessor or of its neighbors, and
the manager unexpectedly and most agreeably discovers that
it keeps up a more uniform heat and requires considerably less
than twice the amount of fuel to keep it running. Two reasons
were operative; first, the walls were made thicker to support
the heavier structure, but that made them also poorer conduc-
tors of heat; second, the capacity increased as the cube of a
linear dimension, while the radiating surface increased as its
square, so that radiation was, therefore, less than twice the
primary amount from the furnace of double capacity. Many
practical metallurgists have learned the practical results, but
are entirely ignorant of the principles upon which the results
are obtained. While it is well to be successful, it is better to
be intelligently so.
Principles of Heat Conduction.
It is well known that the ability of metals to conduct heat
is very nearly the same as their ability to conduct electricity.
The order of metals in these two series is almost identical.
Further, the specific conductance for heat is closely analogous
to specific conductance for electricity. Just as we say that the
electrical resistance of a conductor is proportional to its length
and inversely as its cross-section, so we can make the same
statement concerning heat resistance. Just as we can add
electrical resistances when the bodies are in series, and must
add electrical conductances when they are in parallel, so we
200
CONDUCTION AND RADIATION OF HEAT. 201
can add heat resistances when the bodies are in line, and must
add heat conductances when they are hooked up together in
parallel.
The unit of electrical resistance is an ohm, and a substance
has unit resistivity when a cube of it, 1 centimeter on a side,
has that resistance. In such a case a drop of potential of 1
volt from one side to the opposite one sends through the cube
1 coulomb of electricity per second. Since conductivity is
merely the reciprocal quality, actually and mathematically, to
resistivity, the unit of conductivity is defined in exactly the
same way, and the conductivity of anj^- substance may be ex-
pressed as so many reciprocal ohms.
If we read in the preceding paragraph, thermal resistance
for electrical resistance, degrees temperature for volts, and
gram calories for coulombs, we have defined the corresponding
unit of thermal resistivity. A substance has unit capacity
for conducting heat when a cube 1 centimeter on a side trans-
mits 1 gram calorie of heat per second, with a drop of tem-
perature from one surface to the other of 1° C. Many investi-
gators have adopted slightly different units, such as 1-kilo-
gram Calorie per cubic meter per hour per degree difference;
but such are only simple multiples or fractions of the centi-
meter-gram-second unit above defined.
Illustration: If the thermal conductivity of copper is 0.92
imits, what would be the amount of heat passing per hour
through a sheet 1 millimeter thick and 1 meter square, with
a constant difference of 1° C. between the two sides?
Solution: The 0.92 units means that a column of copper 1
centimeter long and 1 square centimeter in cross-section, hav-
ing a difference of temperature at its two surfaces of 1° C.
would allow 0.92 gram-calories of heat to flow through it per
second. For a sheet one4enth as thick, 10,000 times the area
and during 1 hour, the heat passing per 1° C. difference will be
0.92X^X^^^X3600 = 331,200,000 cal.
= 331,200 Cal.
Such calculations as the above are, of course, very useful for
comparing different thicknesses of plates of different material,
as to their relative heat-carrying capacity. With some slight
202 METALLURGICAL CALCULATIONS.
modifications they are applicable to the actual conveyance of
heat through such walls, from a fluid on one side to a fluid on
the other. This introduces an idea analogous to transfer or
contact resistance in electrolytic conduction. Thus, if we call
R the thermal specific resistance (resistivity) in C. G. S. units,
of the material of a partition having a thickness of d centi-
meters, and an area of S square centimeters, then the thermal
resistance of the body of the partition will be
R X d
and its thermal conductance
-^ = k
RX d
Besides this resistance in the body of the material there is
transfer resistance at its two sides, or at its inner and outer
surfaces, which depends on the materials which communicate
heat and receive heat, and their velocity. These resistances
may be expressed as so many units per square centimeter, the
unit being of exactly the same nature as R, except that it
connotes no thickness but merely a surface effect. Calling R'
the specific resistance of transfer from the inner fluid to the
inner surface of the material, and R' the outside specific trans-
fer resistance to the fluid outside, we can consider all three
resistances as being in series, and the total thermal resistance
to transfer from the inside fluid to the outside fluid to be
R X c/ R' R'
5 "'"5 5
or the thermal conductance of the system
S
(RXa!)+R' + R^
= k
The following table gives the thermal resistivity of various
materials, in C. G. S. gram-Calorie units, also the thermal
conductivity in units which are reciprocals of the resistance
units:
CONDUCTION AND RADIATION OF HEAT. 203
, Conductivity Reststivtty
Material. (in reciprocal (inC.GS.
resistance units .) gram-cal. umts.)
Silver (at 0°) IJO 0.91
Copper (0°— 30°) " 0.92 1.09
Copper, commercial 0 . 82 1 . 22
Copper, phosphorized 0 . 72 1 . 39
Magnesium 0.38 2.63
Aluminium (0°) 0 . 34 2 94
Aluminium (100°) 0.36 2.75
Zinc (15°) 0.30 3.33
Brass, yellow (0°) 0 . 20 5 . 00
Brass, yellow (100°) 0 . 25 4 . 00
Brass, red (0°) 0.25 4.00
Brass, red (100°) 0.28 3.57
Cadmium (0°) 0.20 5.00
Tin (15°) 0.15 6.67
Iron, wrought (0°) , 0 . 21 4. 76
Iron, wrought (100°). 0. 16 6.25
Iron, wrought (200°) . 0.14 7.14
Iron, steel, soft , 0.11 9.01
Iron, steel, hard 0.06 16.67
German silver (0°) 0.07 14.28
German silver (100°) 0.09 11.11
Lead (0°) 0 084 11.90
I>ead (100°) 0.076 13.16
Antimony (0°) 0.044 22.73
Mercury (0°) 0.015 66.67
Mercury (50°) 0.019 52.63
Mercury (100°). 0.024 41.67
Bismuth (0°) 0.018 55.55
Wood's alloy (7°) 0.032 31.25
Alloy, ISn. 99 Bi 0.008 125.00
The last alloy, bismuth, with only 1 per cent, of tin, is re-
markable for its extremely low heat conductivity, less than one
hvindredth as good as copper. It has also the lowest electrical
conductivity of any alloy, about one two hundredth that of
copper. These peculiar properties ought to make it of use for
some particular purposes.
The above data enables one to calculate the rate at which heat
204 METALLURGICAL CALCULATIONS.
will pass through a metallic partition or wall when the tempera-
ture of its two surfaces is known. This is a very useful calcu-
lation,- for in many cases the temperature inside a partition or
pipe is known, or can be easily determined, and when the tem-
perature of the outside surface is determined, the calculation
can be made. The temperature of the outside of a partition or
pipe can be found by several methods: one is to lay a very flat
bulb thermometer (made especially for this purpose) against
it; another is to. put the junction of a thermocouple against it,
covering the couple with a little putty or clay; another is to
take small pieces of metals, or alloys of known melting points,
and see which melts against the hot metal. The only uncer-
tainty then in the calculation is the question as to what differ-
ence in the conductivity may be caused by the higher tempera-
ture. The conductivities in most cases decrease as tempera-
ture rises, but in others increase. There is here a large field
for metallurgical experiment, in determining the heat con-
ductivities of metals, alloys and fire-resisting materials at high
temperature.
Illustration: An iron pipe, 6 centimeters in diameter out-
side and 5 centimeters inside,' is filled with water at 10° C, and
surrounded by hot gases at 198° C. From the rise in tempera-
ture of the water it is known that for each square centimeter of
heating surface 0.084-gram calories of heat pass per second.
Assuming the thermal conductivity of the iron (k) = 0.14,
what is the difference of temperature of the two surfaces, inside
and outside, of the pipe?
Solution: The thickness of the walls is (6 — 5) -h 2 = 0.5
centimeter. If the walls were 1 centimeter thick, 1° difference
would transmit 0.14 calorie; but being only half that thick, 1°
difference would transmit 0.28-gram calorie. The actual dif-
ference of temperature of the two surfaces must then be
5-5?f = 03°C
0.28 ""^ ^•
Of course, such calculations can be turned arotuid, and if the
temperature of the inside and outside surfaces is known, the
heat being transmitted can be calculated ; or if the temperature
of these two surfaces is known and the heat being transmitted
is measured, the therm.al conductivity of the partition can be
CONDUCTION AND RADIATION OF HEAT. 205
reckoned; or, again, if the temperature of the two surfaces is
known, and also the thermal conductivity of the partition, and
the temperature of either fluid on either side, the thermal re-
sistance of the transfer from either of the fluids to the sur-
face of the partition can be calculated.
Illustration; In the previous illustration the temperature of
the hot gases was 198°, while that of the pipe in contact with
them was practically 10°. What was the thermal resistivity of
the transfer from gas to metal?
Solution: The thermal resistivity is the reciprocal of the
thermal conductivity, and in the case of this transfer resistance
is the reciprocal of the number of gram calories which will be
transferred to 1 square centimeter of pipe surface from the
hot gases per second per each degree centigrade of difference
of temperature causing the flow. This is a surface or skin
resistance, and, therefore, no linear dimension representing
thickness enters into the calculation. There is 0.084 calorie
per second being transferred to each square centimeter of pipe,
with a difference of temperature acting as propelling force of
198—10 = 188°. The thermal conductivity of this transfer, k,
is, therefore,
0.084 -- 188 - 0.00045
And the thermal resistivity, R, the reciprocal, viz.: 2222. Since
the thermal resistivity of the iron wall is 1-^0.28 = 3.57 units,
it follows that the contact surface offers 2222-^3.57 = 622
times as much resistance to the flow of heat as the metal itself.
In this specific instance, we can conclude that the transfer of
heat from the gases to the pipe is the principal item which
conditions the flow of heat, the passage of the heat through the
wall of the pipe itself taking place 622 times as readily.
The valuable practical conclusion is that the thermal re-
sistance of the walls of the pipe is insignificant as compared
with the whole thermal resistance, and the thickening or thin-
ning of the walls of the pipe, or the substitution of copper
for iron, because of its greater thermal conductivity, is prac-
tically unnecessary for thermal considerations, since such can
.be expected to make practically no change in the thermal re-
sistance of the whole system. •
If the pipe under discussion, however, acquires during use a
206 METALLURGICAL CALCULATIONS.
layer of scale deposited from the water, then the thermal re-
sistance of the pipe or of the system is materially affected. A
study of the thermal resistivity of various boiler scales would
be a very useful and practical subject, but has not been done, as
far as the writer is aware. Assuming a deposit, 0.5 centimeter
thick, of material having the thermal conductivity of .plaster of
paris, for which k = 0.0013, the specific resistance of this ma-
terial is 0.14-7-0.0013 = 108 times that of iron, and therefore
0.5 centimeter of this represents 54 centimeters thickness of
iron. The thermal resistances in this case, neglecting that of
transfer from the water to the pipe or scale, are as follows:
Resistance of transfer, gases to pipe = 2222.0 = 85 per cent.
0 5
Resistance of 0.5 cm. iron = jr-T-r = 3.6 =0 "
0.14
0 5
Resistance of 0.5 cm. scale = ' = 384.4 = 15 "
U.OUliS
Total = 2610.0
It thus appears that a deposit of scale from the water to a
thickness of 5 millimeters increases the total thermal resistance
of the system some 18 per cent, of its original amount, and
would cut down the efficiency of this part of the heating surface
of a boiler by this amount. These considerations are not only
of vital importance to the steam boiler engineer, but they
are the essential principles which condition the efficiency of
steam-heating apparatus, feed-water heaters, hot-air stoves and
ovens, air-cooling of parts of furnaces and the efficiency of
water jackets.
Principles of Heat Transfer.
We have already had to speak of the transfer of heat from
fluids to soUds, or vice versa, and in one specific case we de-
duced the value 2222 for the transfer resistivity from hot gases
to the surface of iron pipe, meaning thereby that for each
degree of temperature difference between the gases and out-
side of the pipe 0.00045 gram calorie passed per second through
each square centimeter of contact surface. A consideration
of the transfer of heat through such contact surfaces, from
gases or liquids to solids and vice versa, has shown that the
CONDUCTION AND RADIATION OF HEAT. 207
transfer resistivity varies with the solid and with the fluid con-
cerned, but much more with the latter than with the former,
and is very largely dependent upon the circulation of the
fluid, that is, upon the rate at which it is renewed, and there-
fore upon its velocity. The conductivity or resistivity of such
a transfer must, therefore, contain a term which includes the
velocity of the fluid. Various tests by physicists have shown
the specific conductance (or conductivity of transfer) to vary
approximately as the square root of the velocity of the fliiid.
From metal to air or similar gases, the mean velocity of
flow being expressed in centimeters per second, and the other
units being square centimeters and gram calories, the transfer
resistivity is approximately
_ 36,000
2 + Vv
and the transfer conductivity of the contact
k = 0.000028 (2 + Vv)
from hot water to metal the relations are similar, but the
conductivity is much better Experiments show values as
follows:
k = 0.000028 (300 + ]80>/7)
36,000
K. =
300-|-180v/v
Illustration: In the preceding case of the iron pipe, calculate
the difference of temperature of the water in the pipe and the
inner surface of the pipe, assuming the water to be passing
through at a velocity of 4 centimeters per second.
Using the above given formula, the heat transfer per 1°
difference would be
0.000028 (300-|-180\/4)"= 0.0185 calories,
and the difference to transfer 0.084 calories per second will be
0.084
0.0185
= 4°.C
The inner surface of the iron pipe will be, therefore, con-
tinuously 4°.6 higher than the water, and, therefore, at 14°.6;
208 METALLURGICAL CALCULATIONS.
the outer surface will be continuously 0°.3 higher, or prac-
tically at 15°.
Illustration: A steam radiator, surface at about 100° C,
caused a current of hot air to rise having a velocity of about
10 centimeters per second, which was insufficient to keep the
room warm. An electric fan was set to blow air against the
radiator, which it did with a velocity of about 300 cm. per
second, and keeping the room comfortably warm. What were
the relative quantities of heat taken from the radiator in the
two cases?
The relative thermal conductivities of transfer were
2 + \/l0 : 2 + V300
or 5 : 16
Showing over three times as much heat taken away per unit
of time in the second instance.
This illustration proves the great efficiency which the metal-
lurgist may attain in air cooling of exposed surfaces, by blow-
ing the air against them instead of merely allowing it to be
drawn away by its ascensive force.
Problem 24.
Dry air of the volume of 33,000 cubic meters per hour passes
through an iron pipe exposed to the air, 30 meters long, 1.5
meters inside diameter, thickness of walls 1 centimeter, and
lined inside with 5 centimeters of fire-brick. Assume the hot
air entering at 1,000°, the outside air to be at 3°, the coefficient
of internal transfer 0.000028 (2-Fv'vy, the conductivity of
the fire-bricks 0.0014, of the iron 0.14, of external transfer
0.000028 (2 + \/^_ the velocity of the wind against the outside
10 kilometers per hour.
Required: The temperature of the hot air leaving the tube.
Solution: The mean temperature in the tube is the factor
which conditions the mean velocity in the tube, and the rate of
flow of heat towards the outside. If, therefore, we let t repre-
sent that mean temperature, the solution can be stated in the
simplest terms. We then have:
Voltune of air per hour, at 0°, = 33,000 cubic m,
Voltime of air per second, at 0°, = 9.167 "
CONDUCTION AND RADIATION OF HEAT. 209
t + 273
Volume of air per second, at t, = 9.167 ^
Cross-section of inside of tube
(1.5—0.1)^X0.7854 =1.54sq. m.
Velocity of air at t, in pipe
Qifi7i±273 . 1 .. ,„- /t+273\
= 5.95 (1 + at) m. p. s.
= 595 (l + at) cm. p. s.
Coefficient of internal transfer =
0.000028 (2 + V595(i + «t))
Coefficient of conductance of 5 cm. of fire-brick lining =
0.0014-^-5 = 0.00028
Coefficient of conductance of 1 cm. of iron =
0.14-^1.= 0.14
Coefficient of external conductance (v = 278 cm. per sec.) =
0.000028 (2-|-\/278y =^ 0.000524
Total fall of temperature of air = 2 (1000 — t)
End temperature of the air = 1000—2 (1000— t)
= 2 t— 1000
Mean specific heat of air per cubic meter between 1000 and end
temperature =
0.303 -F 0.000027 (2t)
Heat given out by the air per hour =
33,000X2(1000— t)X (0.303 -f 0.000027 (2t) ) =
19,998,000—16,434 1— 3.564 1'
This heat is the quantity transferred through the pipe in
kilogram Calories. The outside surface of the pipe may be
taken as the conducting surface, because the larger part of the
resistance to the flow of heat takes place there. If we desired
to be more exact, the mean area of iron, fire-bricks and the
inner surface could be each used separately. The outside
diameter being 150 centimeters, and the length 30 meters, the
outer surface is 1.52X3.1416X30 = 142.3 square meters =
1,423,000 square centimeters, the total driving force is the
inside temperature minus that outside, or t — 3, and the total
210 METALLURGICAL CALCULATIONS.
thermal resistance is the sum of the four thermal resistances in
series, that is, the svim of thermal resistance gas to fire-bricks =
1
0.000028 (2+V595 (l + « t)
thermal resistance of fire-brick lining =
1
0.00028
thermal resistance of iron shell =
1
0.14
thermal resistance of iron to air =
0.000524
The reciprocal of the sum of these four resistances is the
thermal conductance of the system per square centimeter,
which, multiplied by the conducting surface, 1,423,000 square
cm., and by the total difference of temperature, t — 3, gives
the heat transferred per second in gram calories.
We, therefore, have the final equality expressed as the heat
given up by the air, per second, equals the heat transmitted to
the outside air per second; i.e.,
1^ (19,998,000— 16.434 1— 3.564 1'') =
1
1 I i i
0.000028 (2 + v'595 (l + «t) 0.00028 0.14 ' 0.000524
X 1,423,000 X (t— 3)
Whence t = 965°.5
And the temperature of the air at the end of the tube is
2t— 1000 = 931° (1)
It would be interesting to compare the temperatures of the
inside and outer surfaces of the tube. The air inside' is at a
mean temperature of 965°, the heat transmitted per second is
0.1574-gram calories per square centimeter of outer surface,
CONDUCTION AND RADIATION OF HEAT. 211
whose thermal conductivity is 0.000515, and therefore, the out-
side surface must be
0-1574 _
0.000515 ~
hotter than the surrounding air; that is, must be at 309' C.
The extra loss of heat by radiation from this outside surface
would be small at that low temperature, but has not been
allowed for in the working of the problem.
For such metallurgical problems we need the data as to the
thermal conductivity or resistivity of ordinary furnace ma-
terials. These have been determined in but few instances, and
in most cases not at the high temperatures at which they are
practically used. A whole series of physico-metallurgical ex-
periments is needed just upon this point. The following are
probably nearly all that have been determined and the values
published. The unit k is the C. G. S.-gram calories unit, the
same as used for the metals.
k.
Ice (datum useful in refrigerating plants, where pipes
become coated with ice, as in Gayley's method of
drying blast) 0.00500
Snow 0.00050
Glass (10°— 15°) 0.00150
Water.' 0.00120
Quartz sand (18°— 98°) 0.00060
Carborundum sand (18°— 98°) 0.00050
SiHcate enamel (20°— 98°) 0.00040
(Explains the sma41 conductance of enameled iron
ware.)
Fire-brick, dust (20°— 98°) 0.00028
Retort graphite dust (20°— 100°) 0 . 00040
(Datum useful where articles are packed in this
poorly conducting material.)
Lime (20°— 98°) • 0.00029
(Datum would be highly useful for oxyhydrogen
platinum furnaces, if it were only known at high tem-
peratures.)
Magnesia brick, dust (20°— 100°) 0 . 00050
Magnesia calcined, Grecian, granular (20°— 100°) 0.00045
Masonry 0.0036 to 0.0058
Water, uncirculated 0.0012 to 0.0016
212 METALLURGICAL CALCULATIONS.
Magnesia calcined, Styrian, granular (20°— 100°) 0.00034
Magnesia calcined, light, porous (20°— 100°) 0.00016
Infusorial earth (Kieselguhr) (17°— 98°) 0.00013
Infusorial earth (0°— 650°) 0.00038
Clinker, in small grains (0°— 700°) 0.00110
Coarse ordinary brick dust (0°— 100°) 0.00039
Chalk (0°— 100°) 0.00028
Wood ashes (0°— 100°) 0.00017
Powdered charcoal (0°— 100°) ■■ . . .0.00022
Powdered coke (0°— 100°) 0.00044
Gas retort carbon, solid (0°— 100°) ■. 0.01477
Cement (0°— 700°) 0.00017
Alumina bricks (0°— 700°) • .0.00204
Magnesia bricks (0°— 1300°) 0.00620
Fire-bricks (0°— 1300°) 0.00310
Fire-bricks (0°— 500°) 0.00140
Marble, white (0°) 0.0017
Pumice 0.0006
Plaster of paris 0.0013
Felt 0.000087
Paper 0.00040
Cotton 0.000040
Wool 0.000035
Slate 0.00081
Lava. 0.t)0008
Pumice 0.00060
Cork 0.00072
Pine wood 0.00047
Oak wood .' 0.00060
Rubber 0.00047
A study of the above table will in many cases show the
metallurgist how conduction of heat can be checked, and to
what degree. The substance chosen must be able to stand the
temperature without being destroyed; but it is in many cases
possible to use one kind of material inside, where the heat is
greatest, and a material of much poorer conductivity outside,
where the heat will not destroy it. Such compound linings or
coverings may be very advantageous. Infusorial earth is one
of the very best insulators for moderate temperature; above
CONDUCTION AND RADIATION OF HEAT.
213
a bright red it loses efficiency greatly, and is then hardly better
than powdered fire-brick.
Mr. Irving Langmuir has recently conducted important ex-
periments on convection and radiation of heat, recorded in a
paper before the American Electrochemical Society (Trans-
actions (1913) 23, 299-332). His experimental results for
convection alone, from horizontal and vertical surfaces, to still
air at 27° C, are as follows, first in calories per second and next
in watts:
Gram-calories per Sec. per Sq. Cm. — to Still Air at 27° C.
100°
200°
300°
400°
500°
Downward from — Surface
from Surface
Upward from — Surface
0.002
0.005
0.011
0,014
0.029
0.032
0.024
0.050
0.055
0.035
0.074
0.081
0.047
0.099
0.108
Watts per Sq. Cm. — to Still Air at 27° C.
100°
200°
300°
400°
500°
Downward from — Surface
from Surface
Upward from — Surface
0.010
0.020
0.046
0.060
0.120
0.132
0.100
0.210
0.233
0.146
0.308
0.341
0.196
0.416
0.452
These figures are for convection alone, and do not include
radiation. They are different from the formula given on page
179, but are probably more reliable. If data are required for
over 500°, the above values may be plotted and the curves
extrapolated.
If the air is in motion, Mr. Langmuir found that the heat loss
by convection could be expressed as .. / — ^t — times the loss to
\ 35
still air, where V is the velocity of air current in centimeters per
second. Therefore, the velocity being known, evaluate the
above expression and multiply the tabulated values by this
factor.
Radiation.
A body placed in a vacuum, with no ponderable substance in
contact with it, radiates heat to its surroundings. All experi-
ments so far made confirm the accuracy of Stefan's law, that
every hot body radiates energy in proportion to the fourth
power of its absolute temperature, and that the transfer of
heat by radiation is therefore proportional to the difference
214 METALLURGICAL CALCULATIONS.
between the fourth powers of the absolute temperatures of the
hot body and its surroundings, respectively.
Mr. Langmuir uses Stefan's law of radiation expressed in
the following form
Qr = E 5.9 X 10-"(r2^— Ti*) watts per sq. cm.
= E 1.41XlO-"(r2^ — Ti*) cal. per sec. per sq. cm.
These formulae express the fact that the radiation from a body
at absolute temperature Tj to one at Ti, is proportional to the
difference between the fourth powers of the temperatures in
question, and that for each unit of such difference the radiation
loss from the hotter body is 1. 41X10-" calories per square
centimeter per second, if the body has unity radiating power
(black body), or £x 1.41x10-*^ calories if its coefficient of
radiating power (emissivity) is E. If the heat flow is expressed
as energy flow, 1 . 41 calories per second =5.9 watts.
For convenience sake, to avoid large numbers, the absolute
temperatures may be expressed on a scale of 1000°, in which
case 10-'^ disappears, and the formula becomes:
<3r = £X5.9
=£;xi.4i
\1000/ \1000/
Viooo/ \1000/
watts
cal. per sec.
In the above form, this formula is very convenient for cal-
culating radiation losses from any body at absolute temperature
T^ to its surroundings at Ti, provided the emissivity of the hot
body, E, is known. Dull lampblack has an emissivity nearly
1 . 00, but all other materials have emissivities less than unity,
the polished metals having values as low as 0.03. Unfortu-
nately, total energy emissivity is not quite independent of
temperature, and most of the determinations have been made
for low temperatures only.
In the following table, some emissivities have been calculated
from the older experiments on radiation, and some are Mr.
Langmuir's own values.
CONDUCTION AND RADIATION OF HEAT.
Radiating Capacity (Emissivity) = E
215
Theoretical black body
Lampblack
Smoothblack paint ....
Black paper
Cast iron, rusted
Cast iron, new
Cast iron, polished. . , .
Cast iron, liquid
Soft steel, liquid
Iron oxide scale
Iron rust
Russian sheet iron
Ordinary sheet iron. . . .
Leaded sheet iron
Tinned sheet iron
Hematite
Chromite
Cuprous oxide
Graphite, polished
Graphite, mat
Copper, oxidized
Copper, calorized
Copper, polished
Copper, liquid
Aluminium polished. . .
Aluminium paint
Brass, polished
Silver, polished
Gold, polished
Gold, enamel
Platinum, polished ....
Silvered paper
Monel metal, bright. , .
Monel metal, oxidized.
Chromium
Zinc, bright
Nickel, bright
Nickel oxide
Tin, bright
Magnesium
Silicon
Tungsten
Alumina
Zirconia
Magnesia
Lime
Glass
Porcelain
White fire clay
Building stone
Plaster"
Wood
(At al
(SO'
(50=
(50'
(50'
(50'
(50'
(1200'
(1600'
(500'
(440'
(50'
(50'
(50'
(50'
(500'
(50'
(50'
(50-
(50"
(50°;
(50'
(50'
(1075'
(240'
(50'
(50'
(50"
(50"
(50'
(50'
(1160'
(50'
(50'
(50'
(50"
(50"
(50"
(50"
(800"
(50'
(50"
(50'
(3250'
(500'
(50'
(50'
(50'
(50'
(50"
(300"
(50'
(50'
(SO'
(300)° 0.67
(1750°) 0.28
(1000°) 0.88
(468i°) 0.85
(1000°) 0.99
(720°) 0.60
temperatures)
0.95 - 0.98
0.68
0.64
0.56
0.52
0.17
0.28
0.28
0.85
0.91
0.56
0.47
0.11
0.04
0.98
0.99
0.60
0.65
0.90
0.77
0.39
0.03
0.16
0.156
0.67
0.04
0.02
0.03
0.33
0.035
0.14
0.07
0.50
0.50
0.11
0.04
0.05
0.3S
0.68
0.04
0.07
0.72
0.39'
0.10
0.06
0.06
0.10
0.49
0.25
0.73
0.60
0.60
0.60
1.00
(300°) 0.82
(300°) 0.70
(300°) 0.26
(1000°) 0.09
(1175°) 0.15
(360°) 0.19
(127°) 0.45
(300°) 0.03
(500°) 0.07
(126°) 0.37
(300°) 0.061
(1430°) 0.16
(300°) 0.30
(300°) 0.41
(500°) 0.19
(500°) 0.06
(300°) 0.40
(1000°) 0.75
(300°) 0.14
(1000°) 0.72
(Langmuir).
(500°) 0.09
(500°) 0.09
(500°) 0.40
(500°) 0.50
(400°) 0.79
<500)°0.53
(Thwing)
(1200°) 0.89
(946°) 0.60
(400^ 0.92
(500°) 0.73
(500°) 0.23
(1275°) 0.13
(500°) 0.04
(500°) 0.08
(1695°) 0.175
(500°) 0.38
(1000°) 0.37
(1000°) 0.17
(500°) 0.48
(1075)° 0.80
(500°) 0.23
(1170°) 0.61 (Burgess)
(500°) 0.95
(1000°) 0.70
(1000°) 0.20
(1290°) 0.15 Thwing
(1000°) 0.124
(Burgess)
(710°) 0.62
(1300°) 0.88
APPENDIX TO PART I.
Problem 25.
(1) Write the equations showing the relative weights and
relative volumes (of gases) concerned in the combustion of
Methane (marsh gas) C U.*
Acetylene C'' H^
Ethylene (olefiant gas) C R*
Methylene C^ H*
AUylene C H*
Propylene C H«
Propylene hydride C H*
Benzine C H»
Turpentine (liquid) C"H"
Napthaline (liquid) .- CH^
(2) The molecular heats of combustion of the above gases to
CO* and liquid H^O are given by Berthelot as:
for C n* 213,500 Calories
« C* H* 315,700
" C* R* 341,100 "
" cm' 372,300
" G' H* 473,600 "
" C H« 499,300
" C H». 528,400 "
" C»H« 784,100
" C'»H'» (liquid) 1,490,800 "
" C"H' (liquid) 1,241,800
(a) Calculate the molecular lieats of combustion to CO* and
H^O vapor.
(b) Calculate the molecular heats of formation of the hydro-
carbons themselves, using (C, O'') 97,200 and (H^O) 69,000 liquid
or 58,060 gas.
(c) Calculate the heat of combustion, to CO'' and H^O vapor, of
one cubic meter of each gas, and of one cubic foot in B. T, units.
217
218
METALLURGICAL CALCULATIONS.
Answers (1):
I
16
(2)
C H*
C'H*
C'H«
C',H»
C»H«
Ciojjie
I
28
II
2C^H«
60
■ I
cm*
40
II
2C''H«
84
C»H»
44
I
136
I
C10H8
128
II
+ 20*
64
II V
2C='H2 + 50'
52 160
III
+ 30='
96
VII
+ 10^
224
IV
+ 40=*
128
IX
+ 90'
288
+
+
+
V
50'
160
XIV
140'
448
XII
120'
384
(liqtiid)
(gas)
C"H« (liquid)
Molecular Heat
of Formation
(C Amorphous.)
+ 21,700
—52,300
— 8,700
+ 29,100
—44,000
— 700
+ 39,200
+ 6,100
+ 33,200
+ 23,800
+ 6,200
I
CO'
44
IV
4C0'
176
+
II
2H'0
36
II
2H'0
36
II II
= 2C0' + 2H'0
88 36
IV VI
= 4C0' + 6H'0
176 108.
III II
= 3C0' + 2H'0
132 36
VI VI
= 6C0' + 6H'0
264 108
III IV
= 3C0' + 4H'0
132 72
X
lOCO' +
440
VIII
8H'0
144
X IV
= lOCO' + 4H'0
440 72
Molecular Heat
of Combustion
{to IPO Vapor .)
Heat of Combustion of
191,620
304,760
319,220
339,480
451,720
466,480
486,640
751,280
+ 1,403,280
+ 1,412,680
+ 1,198,040
1 m' (Cal.)
+ 8,623
+ 13,714
+ 14,365
+ 15,277
+ 20,327
+ 20,992
+ 21,899
+ 33,808
I ft.'
B.T.U.
'+ 970
+ 1,543
+ 1,616
+ 1,719
+ 2,287
+ 2,362
+ 2,464
+ 3,803
+ 63,570 +7,152
APPENDIX. 219
Problem 26.
The following typical analyses are from Poole's book on
" The Calorific Power of Fuels ":
Bituminous Coal. Fuel Oil. Natural Gas.
"Carnegie, Pa." "Lima,0." "Findlay.O."
Carbon 77.20 80.2 H' 1.64
Hydrogen 5.10 17.1 CR* 93.35
Oxygen 7.22 1.3 C^H^ 0.35
Nitrogen 1.68 1.4 CO' 0.25
Sulphur 1.42 CO 0.41
Moisture 1.45 0=* 0.39
Ash 5.93 N= 3.41
H'S 0.20
Required: (1) The practical, metallurgical calorific powers of
(a) The coal, per kilogram or pound 7,447
(b) The oil, per kilogram or pound 11,393
(c) The gas, per cubic meter 8,143 kg. Cal.
(d) The gas, per 1000 cubic feet 509,000 lb. Cal.
(2) Compare the calorific values of 1 ton (2240 lbs.) of coal,
1 ton (2000 lbs.) of oil, and 1000 cubic feet of natural gas.
32.8:44.8:1.
Problem 27.
The composition of some commercial gases used as fuels is
given by Wyer (Treatise on Producer-Gas and Gas Producers,
p. 50), as in following table:
H' CU* cm* N' CO 0' C0«
Natural gas (Pitts-
burg). 3.0 92.0 3.0 2.0
Oil gas 32.0 48-0 16.5 3.0 0.5
Coal gas, from re-
torts 46.0 40.0 5.0 2.0 6.0 0.5 0.5
Coke-oven gas 50.0 36.0 4.0 2.0 6.0 0.5 1.5
Carburetted water-
gas 40.0 25.0 8.5 4.0 19.0 0.5 3.0
Watergas 48.0 2.0 5.5 38.0 0.5 6.0
Producer gas (hard
coal, using steam) 20.0 49.5 25.0 0.5 5.0
Producer gas (soft
coal) 10.0 3.0 0.5 58.0 23.0 0.5 5.0
2-20 METALLURGICAL CALCULATIONS.
Requirements: For each of the above gases calculate
(1) The volume of air theoretically necessary to bum it.
^2) The volume of the products of combustion.
(3) The caloriific power
(a) per cubic meter, in kilogram Calories.
(b) per cubic foot, in pound Calories.
(c) per cubic foot, in British Thermal Units.
(4) The theoretical temperature, if burned cold with cold
air (calorific intensity).
P«c-«7*f Volume Volume
/veiHtM. of air to of prod- Calorific Power. Calori-
burn I uctsper Kg. Cat. Lb. Cal. B.T.U. fie In.
volume. 1 volume per fn". per ffl. per ffi. tensity.
Natural gas (Pittsburg) 9.35 10.33 8423 526 948 1852°
Oil gas 7.74 8.58 7350 460 828 1915°
Coal gas, from retorts.. 5. 79 6.50 5550 347 625 1896°
Coke-oven gas 5.36 6.08 5159 322 579 1892°
Carburetted water-gas.. 5. 06 5.77 5008 313 563 1914°
Water gas 2.24 2.81 2590 162 292 1928°
Producer gas (with
steam) 1.08 1.86 1290 81 146 1676°
Producergas (ordinary) 1.13 1.98 1300 82 148 1655°
Problem 28.
An anthracite coal containing on analysis
Carbon 89 per cent.
Hydrogen 3
Oxygen. 1 "
Ash 7
is burned under a boiler, and the ashes produced weigh, dry,
10 per cent, of the weight of the coal used. The chimney gases,
dried and then analyzed, contain CO^ 15.3 per cent., 0' 3.5
per cent., N^ 81.2 per cent., and 25.9 grams of water is obtained
for each cubic meter of dry gas collected.
Required: (1) The volume of chimney gas (measured dry)
produced per kilo, of coal burned 10.41 m'.
(2) The volume of air used, measured dry at normal condi-
tions, per kilo, of coal burned 10.67 m*.
(3) The weight of dry air used 13.80 kg.
(4) The volume of dry air theoretically necessary for the
complete combustion of one kilo, of coal 8.72 tn'.
APPENDIX. • 221
(5) The excess of aif used, in percentage of that theoretically
aacessary for the perfect combustion of the coal 22.36%.
(6) The voluirie of chimney gas, at standard conditions, per
kilo, of coal burned 10.74 m^.
Problem 29.
An anthracite coal contains
Carbon 89 per cent.
Hydrogen 3
Oxygen 1 "
Ash 7
It is burned on a grate, using 15 per cent, more air than is
theoretically needed for its perfect combustion. The ashes
weigh 10 kilos, per 100 kilos, of coal burned.
Required: (1) The volume of air (assumed dry and at standard
conditions) used per kilo, of coal burned 9.71 m'.
(2) The number of cubic feet of air per pound of coal 155.4.
(3) The percentage composition (by voltime, of course) of
the chimney gas, assuming it contains no soot or unbumed
gas. N^ 77.9, 00== 16.1, 0=* 2.6, H='0 3.4.
(4) The percentage composition of the same, if first dried
and then analyzed N^* 80.6, CO^ 16.7, 0=* 2.7.
(5) The number of grams of moisture carried per cubic meter
of dried gas measured 28.4.
(6) The number of grains per cubic foot 12.4.
(7) The volume of the chinmey gkses, at standard conditions,
(water assumed uncondensed) per kilo, of coal burned 9.85 m'.
(8) The number of cubic feet of products per pound of coal
157.6.
(9) The volume of the products in cubic meters at 350° C.
and 700 m.m. pressure 24.4.
(10) The volume of the products in cubic feet at 600'' F. and
29 inches pressure 349.6.
Problem 30. '
A bituminous coal contains
Carbon 75 per cent.
Hydrogen 5
Oxygen ; . . 10
Ash ' 10 "
222 METALLURGICAL CALCULATIONS.
It is powdered and injected into a cement kiln with 25 pei
cent, more air than is theoretically necessary for its complete
combustion. The kiln calcines 200 barrels of cement daily,
using 125 pounds of coal per barrel of cement. Assume outside
air at 0° C. Hot gases enter stack at 819° C. Omit from cal-
culation the water vapor and carbonic acid gas expelled from
the charge.
Required: (1) The volume of air per minute which the
blower or fan must furnish. 2,672 ft'.
(2) The volume of the hot products of combustion per min-
ute, as they pass into the stack. 11,044 ft'.
Problem 31.
A bituminous coal (as of Problem 30) contains
Carbon 75 per cent.
Hydrogen 5 "
Oxygen 10
Ash 10
It is burned under a boiler with 25 per cent, more air than
is theoretically necessary for its complete combustion. As-
sume combustion perfect, and no unbumt carbon to remain
in the ashes. The wet steam produced contains 3 per cent,
of water of priming, and 8.82 pounds of water is evaporated per
poiind of coal burnt. Steam pressure 6 atmosphejres effective
pressure. Outside air 0° C, feed water 10° C, stack gases
310° C.
Required: (1) The calorific power of the coal, by calculation,
water formed considered uncondensed, in pound calories per
pound of coal, and British Thermal Units perpound. 7096; 12773.
(2) The percentage of the calorific power of the coal repre-
sented by the heat in the dry steam produced 78.0.
(3) ditto in the water of priming 0.6.
(4) ditto in the sensible heat of the chimney gases 14.5.
(5) ditto lost by radiation and conduction 6.9.
(6) What increase in the net efficiency (requirement 2)
would be obtained by using a feed water heater, which ab-
stracted 60 per cent, of the sensible heat in the chimney gases,
all other conditions remaining constant? 8.6%.
APPENDIX. 223
Problem 32.
A bituminous coal contains (as in Problems 30 and 31):
Carbon 75 per cent.
Hydrogen 5 "
Oxygen . . , 10 "
Ash 10
It is blown into a revolving cylindrical furnace with a high
pressure blast which supplies only 14.9 per cent, of the air neces-
sary for its complete combustion, producing near the burner a
highly luminous flame surroimded by a cylindrical sheath of
auxiliary air drawn in by the injector action of the fuel-air
jet. Assuming that in the body of the jet the hydrogen only
of the coal is consumed, the luminosity being due to vuicon-
sumed particles of carbon, and ash; that the mean specific heat
120
of carbon is 0.5 , of the ash 0.25-
Required: (1) The theoretical temperature of the interior
of the jet of burning fuel 1140°C.
(2) The temperature if the supply of air in the fuel jet is
increased to that theoretically necessary for perfect combustion
of the whole fuel 1945° C.
Problem 33.
Calculate the maximum temperature obtainable in the region
of the tuyers of a blast-furnace.
(1) Using cold, dry air 1683° C.
(2) Using dry air, heated to 700° C. 2272° C.
(3) Using moist air, heated to 700° C, the amount of moisture
present being such as air at 37° C. can carry when saturated
with moisture. 1947° C.
(4) Using cold, moist air of above composition. 1367° C.
Problem 34.
Powdered coal having the following composition is burnt in a
cement kiln:
Carbon 73.60 per cent.
Hydrogen 5.30
Nitrogen... 1.70
Sulphur. 0.75
224 METALLURGICAL CALCULATIONS.
Oxygen 10.00 per cent.
Ash 8.05 "^
Moisture 0.60
The finely-ground coal is burnt by cold air, at 20° C, and
760 m.m. pressure. The gases resulting, together with carbon'
dioxide gas from the charge, pass into the stack at 820° C.
The analysis of the flue gases, by volume, dried before analysis,
is:
Carbon dioxide 25.9 per cent.
Oxygen 3.1
Carbon monoxide 0.2
Sulphur dioxide not determined.
Nitrogen difference.
Of the carbon dioxide in the gases assume 40 per cent, of it
to come from the carbonates in the charge being treated, and
not from the coal. Assunie no water in the furnace charge.
The air used is saturated with moisture, at 20° C, tension of
the moisture 22 m.m. of mercury.
Required: (1) The theoretical calorific power of the coal 7090.
(2) The theoretical temperature of the hottest part of the
flame 1593° C.
(3) The proportion of the -calorific power of the fuel carried
out by the hot gases 60.4%.
(4) The percentage excess of air admitted above that theoret-
ically reqtiired 16.4%.
Problem 35.
Calculate the draft of a chimney, in inches of water gauge
and feet of cold air, if measured at its base, its height
being 120 feet, inside diameter, (round) 6 feet, temperature of
gases inside at bottom 300° C, at top 200° C, specific gravity
(air = 1) 1.03, and taking in 200 cubic feet (measured at 0°
and 760 m.m.) of products of conibustion per second. Section
inside uniform top to bottom, sides fairly smooth, assume
K = 0.04. Outside temperature 0° C. 0.91 in.; 59.25ft.
Problem 36.
A copper cylinder weighing 22.092 grams was placed in a
small iron box on the end of a rod, and held several minutes
in the hot blast main of a blast-furnace. On removal, and drop-
APPENDIX. 225
ping instantly into a calorimeter, containing 301.3 grams of
water, the temperature rose 4°. 183 C. in three minutes. From
previous experiments with this calorimeter, it was known that
in three minutes it would abstract from the water 30 gram
calories for each 1° observed rise of temperature above starting.
The mean specific heat of copper between 0° and t° being
0.09393 + 0.00001778t, what was th§ temperature of the hot-
blast?
[N.B. — Calculate what wotdd theoretically have been the
temperature of the water of the calorimeter if no heat had been
lost to the calorimeter, and work out using this as the final
temperature of the water.] 618° C.
Problem 37.
A piece of tin-stone (Cassiterite, SnO^) from Bolivia was
tested to obtain its specific heat. It was heated to two tem-
peratures determined by a Le Chatelier thermo-electric pyro-
meter, and dropped into a calorimeter. Corrections for calori-
meter losses were made as explained in the Journal of the Frank-
lin Institute, August, 1901. Data of the two tests were as
follows:
Experiment 1. Experiment 2.
Weight of tin-stone used 12.765 gms. 12.765 gms.
Temperature of same .476° C. 1018° C.
Weight of water in calorimeter. . . 299.4 gms. 300.7 gms.
Temperature of same, starting 16°.527 18°.705
Temperature of same, 3 minutes. . 18°.444 22°.986
Water value of calorimeter per 3',
for each 1° rise 30 cal. 30 cal.
Required: A formula of the form S^ = a + /?t, for the particu-
lar piece of Cassiterite used. 0.1050 -1-0.000006 It.
Problem 38.
The chimney gases from a boiler enter the stack at a tem-
perature of 400° C. Their composition is:
Carbon dioxide 15.0 per cent.
Oxygen 5.9
Nitrogen 79.1
What percentage of the total calorific power of the coke burnt
will be saved by using a feed-water heater which reduces the
226 METALLURGICAL CALCULATIONS.
temperature of these gases to 200° C. and sends 75 per cent,
of the heat thus abstracted into the boiler with the feed-water?
7.92%.
Problem 39.
A blast-furnace gas contains, by voltime:
Carbon monoxide 23 per cent.
Carbon dioxide 12
Hydrogen 2
Methane 2
Water vapor 3
Nitrogen 58
Its temperature is 20° C, and it is taken to a gas engine,
mixed with the theoretical amount of dry air needed for perfect
combustion, and compressed to 4 atmospheres tension (total
pressure) before being ignited. Neglect the heating of the
gas-air mixture by compression, i.e., assume the temperature
of the compressed mixture 20° C. before ignited. Assume that
at the instant of ignition the voltmie occupied by the gases
remains constant, so that the specific heat at constant volume
applies. (Specific heat of 1 cubic meter at constant volvmie
equals specific heat at constant pressure — 0.09.)
Required: (1) The calorific power of the gas per cubic meter,
at constant pressure, and at constant voltime. 928.5; 925.5.
(2) The theoretical temperature at the moment after ignition
has taken place. 1592° C.
(3) The maximum pressure exerted. 22§ atmospheres.
(4) The voltime of gas needed per horse-power hour, at a
..nechanico-thermal efficiency of 30 per cent. 2.81 m', at 20° C.
Problem 40.
Bituminous coal containing carbon 78 per cent., hydrogen
5, oxygen 8, ash 8, water 1, is used in a gas producer. Assume
the calorific power of the coal (water formed uncondensed) as
7480 Calories; ashes formed 12 per cent. Gas formed leaves
producer at 600° C; composition:
Carbon monoxide 35 per cent.
Carbon dioxide 5 "
Methane 5 "
Hydrogen 5 "
Nitrogen 50 "
APPENDIX. 227
Required: (1) The-^'olume of gas obtained per kilo, of coal
burnt. 3.045 m'
(2) The calorific power of the gas per cubic meter.
1633 Calories.
(3) The proportion of the calorific power of the fuel obtain-
able on burning the gas. 66.5 per cent.
(4) The proportion of the calorific power of the fuel which
has been sacrificed in making the gas, assuming it burnt cold.
33.5 per cent.
Problem 41.
Producer gas of the following composition:
Carbon monoxide 28 per cent.
Carbon dioxide 4 "
Hydrogen 4 "
Methane 2 "
Water vapor 1 "
Nitrogen 61 "
is burned with 10 per cent, more air than theoretically re-
quired, both air and gas being preheated to 1000° C.
Required: The theoretical maximum temperature of the flame.
2100° C.
Problem 42.
Kiln-dried peat from Livonia contained by analysis;
Carbon .' 49.70 per cent.
Hydrogen 5.33
Oxygen 30.76
Nitrogen 1.01
Ash 13.23
The wet peat, as taken from the ground, carried 75 per cent,
of water, when air-dried 20 per cent., and when kiln-dried
none, as per analysis.
The air-dried peat is dried in a kiln by means of a current of
hot air, which enters the kiln, and comes in contact with the
freshly-charged peat, at a temperature of 150° C, while it
leaves the kiln, near the discharge end, at 50° C. Outside tem-
perature 0° C, outside air dry. The kiln loses by radiation
10 per cent, of the sensible heat of the hot air coming into it,
the rest represents the sensible heat of the warm moist air and
228 METALLURGICAL CALCULATIONS.
dried peat, issuing at 50° C, and the heat necessary to evaporate
the moisture. The air required is heated in a stove where
kiln-dried peat is burned, 75 per cent, of the heat generated
being transferred to the air. Mean specific heat of kiln-dried
peat 0.25.
Required: (1) The practical calorific powers of wet peat, air
dried peat and kiln dried peat. 607, 3278; 4249.
(2) The volume of air, at standard conditions, needed for
drying one metric ton of air-dried peat. 5,122 m^
(3) The percentage degree of saturation, with moisture, of
the issuing air. 35.4.
(4) The amount of kiln-dried peat required to be burned in
the stove per ton of air-dried peat piit through the kiln, and the
percentage of the total fuel necessary for this purpose.
70.5 kg. 9.25%.
Problem 43.
A set of four coke ovens produce 10,000 cubic feet of gas
per hour, measured at 60° F., and having the composition H'
64.3 per cent., CO 20.7, CH* 5.4, C^H^ 0.5, .C«H» 0.5, €0== 2.0,
0^ 1.0, N^ 5.6 per cent. Temperature leaving the ovens 2900° F.
For half of each hour the whole gas is burned by the theoretical
amount of cold air, as it passes into recuperators, whence the
products at 1900° F. pass under steam boilers where their
temperature is reduced to 500° before passing to the stack.
For the second half of each hour the gases pass through the
recuperators unmixed with air, are there heated to 1900° F.,
and at that, temperature pass under boilers where they meet
with the theoretical quantity of air needed for combustion,
are burned, and the products pass to the stack at 500° F.
Required: (1) The horse-power of the boilers during the first
30 minutes, calling 1 horse-power the ability to evaporate 34^
pounds of water per hour at 212° F., and assimiing the boilers
to produce steam representing 50 per cent, of all the heat re-
ceived by them and generated within them (net efficiency 50
per cent.). 23.4 H. P.
(2) The same, for the second 30 minutes. 57.0 H. P.
Problem 44.
A plant of by-product coke ovens uses bituminous coal con-
taining ,76 per cent, of carbon, and having a calorific power
APPENDIX.
229
of 9000, produces coke containing an average of 86 per cent,
of carbon, and having a calorific power of 7000, while the by-
product tar produced contains 20 per cent, of carbon. The
coke weighs 70 per cent, and the tar 5 per cent, of the weight
of the coal used. The average analysis of the gases for the
month of January, 1904 (samples dried before analysis) was:
Carbon dioxide 3.00 per cent.
Oxygen 0.50
Carbon monoxide 5.10
Marsh gas, CH< 35.00
[cm' 2.13
Ill\miinants { C=H» 1.06
IC«H« 1.06
Hydrogen 40.00
Nitrogen 12.15
Required: (1) The volume of by-product gases, at standard
barometric pressure and at 60° F., produced per ton (2000
povmds) of coal used. 16,294 cubic feet.
(2) The proportion of the calorific power of the coal repre-
sented by the calorific powers of the coke and the gases.
54.4%; 27.4%.
(3) Using half the gases produced in gas-engines, at an effi-
ciency of conversion into power of 25 per cent., how many
horse-power-hours could be thus generated per pound of coal
coked? 0.22 H. P. hours.
Problem 45.
A gas producer uses coal which analyzed: total carbon 75.68
per cent., oxygen 12.70 per cent., hydrogen 4.50 per cent.,
ash 7.12 per cent. The ashes produced contain 21.07 per cent,
of tmbumt carbon. The gas carried 30,4 grams of moisture
to each cubic meter of dried gas collected, and the latter shows
on analysis:
Carbon dioxide 5.7 per cent.
Carbon monoxide 22.0
Methane (CH^) 2.6
Ethylene (C^H^ 0.6
Hydrogen 10.5
Oxygen 0.4
Nitrogen 58.2
230 METALLURGICAL CALCULATIONS.
A steam blower furnishes blast, forcing in the outside air which
carries 15 grams of moisture per cubic meter, measured dry
at 20° C.
Required: (1) The volume of gas (dry) produced per kilo,
of coal used. 4.337 m\
(2) The weight of steam used by the blower per kilo, of
coal used. 0.2691 kg.
(3) The weight and volume of air blown in, per kilo, of steam
used by the blower. 15.5 kg. ; 12.95 m'.
(4) The percentage of the steam blown in which is not de-
composed in the producer, assvmiing the moisture in the gas to
represent steam used and not decomposed (assumption not
strictly accurate). 49.3%.
(5) The mechanical efficiency of the blower, asstmiing it
uses steam at 4 atmospheres effective pressure, and produces
10 centimeters of water gauge pressure in the ash pit of the
producer. 4.67%.
Problem 46.
Assume the gas of Problem 42 to be produced by the com-
bustion of 1000 kilos, of coal per hour in the producers, and
to be burned in a furnace with such excess of air (carrying
at 20° C. 15 grams of moisture per cubic meter measured dry)
that the chimney gas, analyzed dry, contains:
Carbon dioxide 12.7 per cent.
Nitrogen. 80.6
Oxygen 6.7
These products of combustion enter the chimney at 500° C.
and leave it at 350° C, their velocity entering at the base is
4 meters per second. Outside air 20°. The efficient draft is
2.5 centimeters of water gauge, measured at the base; assume
this 90 per cent, of the total head of the chimney.
Required: (1) The diameter of the chimney, asstmiing it
round, and its height, assuming its section uniform.
1.52 meters; 41.3 meters.
(2) The work done by the chimney, expressed in horse-
power. 2.6
(3) The energy efficiency of the chimney, i.e., the ratio of the
mechanical work it performs to the mechanical equivalent of the
heat which it receives. 0.079 per cent.
APPENDIX. 231
Problem 47.
Assume the chimney of Problem 46 to be built of fire-brick
of an average thickness of 60 centimeters; that the gases passing
through per hour are 1586 cubic meters of nitrogen and air,
315 cubic meters of carbon dioxide and 234 cubic meters of
water vapor; the temperature at the base is 500° C, at the top
350°; the velocity of the hot gases at the base, 4 meters per
second; diameter 1.52 meters, height 41.3 meters. Assume
further the coefficient of transfer of heat from gas to brick
and brick to air to be „„ ^^ (in C. G. S. units), and the
oo,UuU
velocity of the wind outside to be 20 kilometers per hour.
Required: The coefficient of conductivity of the fire-brick
in C. G. S. units. 0.31
PART II.
IRON AND STEEL.
233
CHAPTER I.
BALANCE SHEET OF THE BLAST FURNACE.
As the most important factor in the production of the most
important metal, the blast furnace is the most important fur-
nace or piece of metallurgical apparatus in the world. It is,
therefore, proper that we should commence a series of articles
on the application of metallurgical principles and calculations
to the metallurgy of iron, by a discussion of the blast furnace;
and since this discussion, to be complete, must include a wide
range of topics, we will commence with the simplest, viz.: the
balance sheet of materials, entering and leaving the furnace.
Later we can discuss the balance sheet of heat entering in,
developed within and leaving the furnace, the reactions taking
place in the furnace, the action of hot and of dried blast, the
calculation of the proper constituents of the charge, the tem-
peratures attained before the tuyeres, unused combustible
energy of the gases, efficiency of the hot-blast stoves, and other
interesting and practically valuable factors in the running of
the furnace.
The blast furnace may be regarded from several points of
view; we will mention two. First, it may be regarded as a
huge gas producer, run by hot, forced blast, in which the in-
combustible portions of the contents are melted down (with a
little tmbumt carbon) to liquid metal and slag, and are nm out
beneath, while the gaseous products pass upwards through 50
to 100 feet of burden, and escape above. The escaping gases
are primarily of the composition of producer gas, with some
of its carbonous oxide changed to CO^ by the oxygen abstracted
from the burden, with some CO^ added from the decomposition
of the carbonates of the charge, and with the usual increment
of moisture from the charge and volatile matter (if any) from
the distillation of the. fuel. From this point of view, the blast
furnace is a huge gas producer, giving a rather inferior quality
of combustible gas in very large quantities, and incidentally
235
23f} METALLURGICAL CALCULATIONS.
reducing to metal and slag the burden of iron ore and' flux
(limestone) which is put in with the fuel. The treatment of
the furnace as a metallurgical problem may then proceed as
the discussion of a gas producer, with the composition of the
gas produced somewhat modified by the amount of oxygen
given up tp the gas by the reducible portions of the charge of
the furnace.
The other viewpoint is to regard the furnace as primarily
an apparatus for deoxidizing or reducing iron ore, for which
purpose the ore is charged with sufficient carbonaceous fuel
to do two things, viz.: to abstract all the oxygen from the
reducible metallic oxides, and to furnish enough heat, or high
enough temperature, to melt down to superheated liquids the
pig iron and slag (combinations of irreducible metallic oxides)
formed. In this view, the fuel must supply the reducing
energy and the melting-down or smelting requirements; the
first by acting upon the metallic oxides at a red to a white heat
and abstracting their oxygen; the second, by being burned at
the foot of the furnace by hot air blast, and there generating
the heat and higher temperatures necessary for the smelting
down of the already reduced materials.
Materials Charged and Discharged.
The materials put into a blast furnace may all be classed
under four heads:
Fuel "I
Iron ore \ Charged at the throat.
Fluxes J
Blast Blown in at the tuyeres.
The materials discharged from the furnace may be classed
under four heads also:
ci^ I Tapped from the crucible.
T-j . > Passing out at the top.
We will discuss the resolution of each of the four materials
charged into the four avenues- of escape.
Fuel.
The fuel used is sometimes charcoal, but in the great ma-
jority of cases coke, with perhaps some raw bituminous coal
BALANCE SHEET OF BLAST FURNACE. 237
or anthracite coal, or in a few cases all raw bituminous coal.
The composition of these fuels consists of moisture, volatile
matter, fixed carbon, sulphur and ash consisting of silica, lime,
iron, alumina, alkalies, etc.
The moisture is driven off near the top, and goes into the
gases as moisture. The volatile matter is expelled near to the
top; almost all of it goes unchanged into the gases, but part
of the hydrocarbons thus expelled may be decomposed and
deposit fixed carbon on the iron oxides, etc., surrounding them.
This carbon, however, will take up oxygen from the charge
lower down in the furnace, and thus eventually pass into the
gases as CO or CO^. We can, therefore, assume without error
that all the volatile constituents of the fuel pass into the gases,
but cannot be certain in exactly what state of combination,
except as regards the moisture. It will be quite exact if we
know the ultimate composition of the volatile matters of the
coal, as so much carbon, hydrogen, oxygen, nitrogen, sulphur,
etc., to charge them thus entirely to the gases.
The fixed carbon all finds its way ultimately into the dust
or the gases, either as CO, CO', CH* or HCN, or alkali cyanides,
excepting the amount represented by the carbon in the pig
iron. Subtracting the carbon in the pig iron from the total
fixed carbon in the fuel, the difference can safely be put down
as entering the gases or being in the dust carried away by the
gases.
The sulphur in the fuel has a more varied history. When it
is partly present in the form of iron pyrites, some may go into
the gases as sulphur vapor, and eventually be burned to SO'
when the gases are burned; another part may be oxidized in
the furnace itself to SO', and as such appear in the gases; the
rest, along with organic sulphur, passes either into the slag
or the pig iron. Sulphur passing into the slag seems to do so
as calcium sulphide, CaS, formed by some such reaction as
CaO + C + FeS = CaS + CO + Fe
or, if the sulphur was present in the fuel as gypsum,
CaS0* + 4C = CaS + 4C0
The amount of sulphur going into the iron depends really upon
238 METALLURGICAL CALCULATIONS.
the opporttinity for it to go into the slag. If the temperature
of the furnace at the tuyeres is very high, and especially if the
slag is low in silica, sulphur will keep out of the iron and go
into the slag, to the extent of ten or twenty times as much
being in the slag as in the iron ; but if the temperature is low
and the slag rich in silica the reverse may be the case. A
high temperature and a high percentage of lime in the slag
are the blast furnace manager's means of keeping down the
sulphur in the iron, although high magnesia or high alumina
are also efficacious. In casting up the balance sheet it can be
assumed that when using coke or charcoal all the sulphur of
the fuel goes either into the slag or the iron, and knowing from
the analysis of the pig iron made how much goes into it, the
rest can be calculated as going into the slag as CaS. If raw
coal is used, it is uncertain how much sulphur goes into the
gases, and an exact analysis of either the slag or gases, for sul-
phur, in addition to that of the pig iron, would be necessary
to fix its distribution.
The ash of the ^fuel counts in with the other incombustible
ingredients of the charge. Some of the silica in it may be
reduced to silicon, and some of the CaO to Ca, to form CaS;
while most of the iron will pass into the pig iron. It is in
most cases uncertain whether the silicon in the pig iron comes
at all from the fuel ash, so it is usual to assume it as coming
from the silica of the ore only ; as to the iron, it is best to assume
it all reduced to the metallic state, as is probably always the
case.
Besides all these avenues of escape for the constituents of
the fuel, it is sometimes necessary to take into account the
possibility of some of it, in fine particles, being carried out of
the furnace bodily with the outgoing gases. If the amount
of this in the dust is determined, it must be subtracted in toto
from the fuel charged, and then the remainder distributed as
just discussed.
Illustration. — A blast furnace is charged, per 1,000 kilos, of
pig iron produced, with 925 kilos, of coke, containing by analysis:
Fixed carbon, 86 per cent.; volatile carbon, 2; hydrogen, 1;
oxygen, 0.5; nitrogen, 0.5; sulphur, 1.0; iron, 2; silica, 5; lime,
1; moisture, 1. The pig iron contains 3.5 per cent, of carbon
and 0.1 per cent, of sulphur. The dust carries 15 kilos, of dry
BALANCE SHEET OF BLAST FURNACE.
239
coke per metric ton of pig iron. Required the distribution of
the coke in the furnace per ton of pig iron made:
Charges. Pig Iron. Slag. Gases.
Dust
Fixed C...
Volatile C.
H
. . 15.0
..782.6
.. 18.2
. . 9.1
0
. . 4.5
N
. . 4.5
S
.. 9.1
Fe
. . 18.2
SiO='
CaO.......
WO
. . 45.5
. . 9.1
.. 9.3
C 35.0
C
C
H
O
N
S
Fe
1.0 S 8.1
18.2 ....
. . . SiO^ 45.5
. .. CaO 9.1
747.6
18.2
9.1
4.5
4.5
Dust.
Coke 15.0
WO 9.3
Ore.
Whatever the varieties of ore used they can be averaged to-
gether, so as to get the average composition of the ore charged.
Then, knowing its weight per unit of pig iron made, the dis-
tribution into pig iron, slag, gases and dust can be made.
There may, first of all, be blown out as ore dust up to 25 per
cent, of the ore charged. This must be first deducted as dry
ore and then the rest distributed to pig iron, slag and gases.
The moisture of the ore, also any carbonic acid, may be con-
sidered as going over bodily into the gases. The sulphur may
partly go into the gases if present as iron pyrites (this amovrnt
would have to be checked by an analysis of the gases for sul-
phur or hydrogen sulphide), but mostly into the slag as CaS.
Some of it may be put down as going into the pig iron, if the
sulphur in the fuel does not account for all that appears in the
analysis of the iron. The iron oxides present must be as-
sumed reduced to metallic iron sufficient to furnish the iron in
the pig iron from its analysis; the excess, if any, if put down
as going into the slag as FeO; all the oxygen given off (the
difference between the weight of iron oxide in the ore and
the sum of iron going in the pig iron and ferrous oxide passing
in the slag) goes to the gases. If there is not enough iron in
the ore to account for all in the pig iron, then none is assumed
to go into the slag.
240 METALLURGICAL CALCULATIONS.
The manganese oxides in the ore furnish the manganese in
the pig iron, the excess going into the slag as manganous oxide
MnO, the oxygen (by difference) goes to the gases. The pro-
portion of manganese reduced to metal increases with the tem-
perature at which the furnace is run and as the slag is less
siliceous. The amount reduced is known, however, only by
the analysis of the pig iron.
Zinc in the ore is partly found as ZnO in the slag, and partly
as flakes of white zinc oxide in the gases, which latter partly
deposit in the 'dust catcher and are partly carried by the cur-
rent of gases into the stoves and under the boilers. The rela-
tive amounts going into slag and gases can be best controlled
by analysis of the slag.
Copper, silver, gold, nickel, cobalt, phosphorous, antimony and
arsenic are almost completely reduced into the pig iron ; careful
analysis of the latter will show exactly to what extent, but
without this careful analysis they may be assumed to pass
completely into the iron. Lead is mostly carried out as fume, a
small amount passes into the pig iron, and, if present in quan
tity, a large amount may collect as metallic lead beneath the
pig iron and, if it can, soak into the foundation of the furnace.
Alumina usually passes completely, as such, into the slag.
When present in large amount, producing a slag rich in alu-
mina, and with very hot blast, the pig iron may contain as
much as 1 per cent, of aluminium, the oxygen thereof passing
into the gases. Magnesia may be assumed as passing com-
pletely into the slag; none is reduced. Lime goes into the slag,
except a not-unimportant quantity which is reduced by carbon
in the presence of sulphur compounds, and forms CaS, its
oxygen passing into the gases ; a very small amount may go as
calcium into the pig iron. Alkaline metals partly go into the
slag, while some may pass into the gases as alkaline cyanides.
Titanium oxide, tungsten oxide, chromium oxide and the
oxides of molybdenum, uranium, vanadium are sometimes re-
duced in small amounts, the more the hotter the furnace is run
and the more basic the slag, while the bulk of them passes into
the slag as the lowest oxide which each is capable of forming.
Silica mostly goes into the slag as SiO^ but a portion is al-
ways reduced to silicon in the pig iron. The amotmt reduced
is greater the hotter the furnace is run, the more slowly it is
BALANCE SHEET OF BLAST FURNACE. 241
run:, and the more siliceous the slag. In some cases as much
as one-quarter of all the silica going into a furnace is reduced
to silicon. It is probably reduced only by carbon dissolved in
iron at the lower part of the furnace. The oxygen of the
silica reduced goes into the gases.
Illustration. — 1956^8 kilograms of ore is charged into a fur-
nace per metric ton of pig iron made. The ore analyses: Fe^O',
71.43 per cent.; Si6^ 14.24; CaO, 2.05; MgO, 1.51; MnO^ 4.15;
S0^ 1.40; H^O, 5.00; Cu^O, 0.22 per cent. The pig iron con-
tains 93.03 per cent, iron, 3.27 carbon, 1.20 manganese, 0.08
sulphur, 0.40 copper, 2.02 silicon. Assume the dry ore dust to
weigh 4 per cent, of the weight of ore charged, and that there
is no sulphur found in the gases, but all the sulphur in the pig
iron comes from the fuel. Cast up the distribution of the ore:
Charge. Pig Iron. Slag. Gases. Dust.
Ore...., 1956.8 kg.
Dust 78.1 " Ore 78.1
Fe^O^ 1339.2 " Fe 930.3 FeO 9.1 O 499.8
SiO^ 267.0 " Si 20.2 SiO=' 223.6 0 22.2
MnO' 77.8 " Mn 12.0 MnO 48.0 0 17.8
Cu^O 4.1 " Cu 3.6 .... 0 0.5
^° 3«-*" •■•■ {a°S;io 5.2
MgO 28.3 " .... MgO 28.3 ....
S0» 26.3 " .... S 10.5 0 15.8
WO 97.6 " H^O 97.6
In calculating the above we note that the dust is dry, and
weighs 4 per cent, of the ore, making 78.1 kilos, of dry dust,
representing 82 kilos, of. moist ore, leaving 1874.8 kilos, of
moist ore to be distributed, plus the 3.9 kilos, of water from the
dust, which also goes into the gases. This 1874.8 kilos, contains
the weights given, calculating from its analysis. The 1339.2
kilos, of Fe^O' contains 937.3 kilos, of iron; but there are only
930.3 kilos, in the ton of pig iron, therefore the other 7.0 kilos.
72
must go into the slag as 7.Qy.-rx= 9.1 kilos, of ferrous oxide,
00
while 1339.2- (930.3 H- 9.1) = 499.8, the weight of oxygen
242 METALLURGICAL CALCULATIONS.
going into the gases. The 267.0 kilos, of silica contains 124.6
kilos, of silicon, but there are only 20.2 kilos, in the pig iron,
therefore, 104.4 kilos, must remain unreduced, passing into the
slag as 104.4X^5 = 223.6 kilos, of silica, while 267- (20.2 +
223.6) = 22.2 kilos, of oxygen goes into the gases. Another,
and equally logical procedure, is to start -with the 20.2 kilos,
of silicon in the pig iron, which must have required 20.2 X
1^ = 42.4 kilos, of silica to furnish it, yielding 42.4- 20.2 =
28
22.2 kilos, of oxygen to the gases, and leaving 267.0— 42.4 =
223.6 kilos, of silica unreduced to go into the slag.
The 12.0 kilos, of manganese in the pig iron would be reduced
87
from 12.0 X-^ = 19.0 kilos, of MnO^ furnishing, therefore, 7.0
kilos, of oxygen to the gases, and leaving 77.8 — 19.0 = 58.8
kilos, of MnO^ to go into the slag as MnO. The molecular
weights of MnO and MnO^ being respectively 71 and 87, there is
71
58.8X5= = 48.0 kilos, of MnO going into the slag, while 10.8
87
kilos, more of oxygen will be supplied to the gases, making alto-
gether 10.8 + 7 = 17.8 kilos, of oxygen given up by the MnO'-
The 4.1 kilos, of Cu^'O contain 3.6 kilos, of copper, all of which
enters the pig iron and contributing 0.5 kilos, of oxygen to the
slag.
The 38.4 kilos, of CaO must supply enough Ca to form CaS
32
with the S of the SO'. The latter quantity is 26.3 X on =
40
10.5 kilos., to supply which there is needed 10.5 X^ = 13.1
56
kilos, of calcium. The latter will be supplied by 13.1X7^ =
18.3 kilos, of CaO, furnishing 5.2 kilos, of oxygen to the gases,
and leaving 38.4— 18.3 = 20.1 kilos, of CaO unreduced to go
into the slag. The MgO in the ore goes directly into the slag.
The oxygen of the SO' goes into the gases. The 5 per cent, of
BALANCE SHEET OF BLAST FURNACE. 243
water of the whole quantity of ore charged goes into the gases,
as vapor.
Flux.
The flux is used for the purpose of making a fusible slag
with the slag-forming ingredients, contributed by the ore and
fuel. If we consider the distribution of ore and fuel given in
the preceding illustrations we see that the chief material to be
fluxed is silica, with smaller quantities of FeO, MnO, CaO,
MgO and CaS. The cheapest and most available material to
flux silica is limestone, the slag formed being a silicate of lime,
magnesia and alumina, with CaS and smaller quantities of
other basic oxides. We will not discuss at present the con-
siderations governing the amotint of flux used, since this is a
calculation requiring separate treatment, as the proper working
of the furnace depends fundamentally upon it. We may re-
mark here that enough flux must be used to make an easily
fusible fluid slag, rich enough in lime, magnesia or alumina to
carry away satisfactorily the bulk of the sulphur, and so pro-
duce good pig iron.
The flux usually contains CaO, MgO, APO', SiO^ FeO, CO^
and H^O. Its H^O and 00=" are driven off in the upper third
of the furnace, and may be put down as going as such into the
gases. The FeO may be reduced if the slag is very clean, but
under ordinary conditions may be put down as all going into
the slag, unless in quite large amount, because the iron in ore
and fuel usually supplies the total weight of iron in the pig iron.
The silica and alumina may be carried over bodily into the
slag. The magnesia can be put at once into the slag, but the
lime cannot in many cases be treated that way, because quite
frequently some is needed to supply calcium for the sulphur
of the fuel. In the fuel previously illustrated, for instance,
there is not enough CaO present to furnish Ca for the S, whence
it follows that some CaO from the flux will be needed to make
up the deficit. We may, in such a case, either consider all the
CaO of the fuel to form CaS with part of the sulphur, and then ■
take enough CaO from the flux to unite with the remainder,
or, it is equally permissible to take all the CaO necessary to
furnish Ca to all the sulphur of the fuel, and to let the CaO
of the fuel figure as passing entirely into the slag. The latter
reqviires a little less calculation.
244
METALLURGICAL CALCULATIONS.
Illustration. — A blast furnace receives 503 kilos, of lime-
stone flux per metric ton of pig iron made, which analyses CaO,
29.68 per cent.; MgO, 20.95; SiO^ 3.07; KVO\ 2.66; FeO, 0.48;
C0^ 42.66; WO, 0.50 per cent. Assume 8.1 kilos, of sulphur
in the fuel, for which the flux must provide calcium. Required
the distribution of the flux, assuming it to make no dust:
Charge
.
Pig Iron. S
lag.
Gases.
Flux. .
503.0 kg.
CaO...
149.3 "
(Ca
■■ tCaO
10.1
135.1
0 4.1
MgO..
105.4 "
.... MgO
105.4
....
sio^..
15.4 "
.... SiO^
15.4
....
Al^O'. .
13.4 "
2.4 "
.... APO'
.... FeO
13.4
2.4
FeO...
• . • >
C0\ . .
214.6 "
. . .
00=! 214.6
WO...
2.5 "
H^O 2.5
The only calculation needed above is that 8.1 kilos, of sulphur
40
require 8.1x^77 = 10.1 kilos, of calcium, which would be
5o
14.2 kilos, of lime, leaving 4.1 kilos.
32
furnished by -10.1X^7^
■' 40
of oxygen to go into the gases and 135.1 of lime to go into the
slag.
Blast.
The remaining item needed to complete the balance sheet is
the amount of blast. This may be roughly estimated by ob-
taining the piston displacement of the blowing engines, and
assuming a coefficient of delivery into the furnace. This is
very rough, because the efficiency is not known, and may vary
anywhere between 0.5 and 0.95. Another rough approximation
may be obtained by observing the pressure of the blast, its
temperature, the back pressure in the furnace, and knowing
the area of the tuyeres, and assuming a coefficient of contrac-
tion of the hot air jet as it emerges from the tuyeres. Here,
again, are several tmcertain factors, and the coefficient may
vary between 0.9 and 0.98. Calculations on this basis are very
rough.
BALANCE SHEET OF BLAST FURNACE. 24f
The only satisfactory way to determine the blast is to care-
fully analyze the gases, determining carefully all the carbon,
oxygen and nitrogen which they contain. Since the carbon
comes only from the charges, the amount of gases produced per
tmit of pig iron made becomes known, and thence the oxygen
and nitrogen contained in them. These, minus the oxygen and
nitrogen coming from the solid charges, leave the oxygen and
nitrogen which must have come from the blast. The oxygen
3
in the blast, minus — the nitrogen, gives the oxygen entering
as water vapor; but this last calculation is not so satisfactory
as to observe the atmospheric conditions, and calculate the air
and moisture on the basis of the contained nitrogen.
The blast contains oxygen, nitrogen and moisture. All its
constituents pass into the gases, being put down as so much
oxygen, nitrogen and hydrogen. Just how much of that hydro-
gen gets into the gases as free hydrogen and how much as
water vapor is not known. Argon and other rare gases in the
blast are counted and treated as nitrogen. The carbonic acid
of the air is present relatively in such a small amount that it
can be neglected, as far as all ordinary calculations are con-
cerned.
Problem 51.
A blast furnace at Herrang, Sweden, is run on ore briquettes
made by pressing and calcining fine concentrates. The analyses
of briquettes, charcoal and limestone flux are as follows (see
Journal Iron and Steel Institute, I., 1904):.
Briquettes.
Fe^O' 85.93
FeO 3.96
SiO^ 5.50
MnO 0.63
APO' 0.76
CaO 2.23
MgO 0.97
P^O' 0.006
S 0.010
Cu 0.007
Limestone.
Charcoal.
0.18
0.32
C
80.31
3.14
0.19
N
0.08
0.32
0
3.54
53.74
0.89
0.17
0.10
0.006
0.0068
0.001
0.017
COM2. 42
H^O
14.04
K^O
0.50
246 METALLURGICAL CALCULATIONS.
The pig iron contains phosphorus, 0.012 per cent.; sulphur,
0.007; manganese, 0.025; silicon, 0.60; carbon, 2.70; iron,
96.656. There is used in charging the furnace:
Briquettes 1,190 pounds
Limestone 90
Charcoal 530
And the fuel consumption is 682 pounds of charcoal per 1,000
pounds of pig iron made.
The gases at the throat (dried) analyze: N^ 57.3 per cent.;
CO, 23.1; C0^ 14.8; H^ 4.3; CHS 0.5 (Rinman). Assume
blast dry. Dust in gases neglected.
Required: (1) A balance sheet of materials entering and leav-
ing the furnace, per 1,000 pounds of pig iron made. (2) The
percentages of iron, manganese, silicon, sulphur and phosphorus
going into the furnace, which go into the pig iron.
Solution: — (1) See table opposite page.
(2) The total iron in the charge is 969.2 kilos., while that in
the pig iron is 966.6; the efficiency of the reduction of iron is
therefore 99.7 per cent.
55
The total manganese in the charge is 9.6Xwt = 7.4 kilos.,
of which only 0.25 gets into the pig iron, or 3.4 per cent.
The total silica charged is 89.1 kilos., representing 41.6 kilos,
of silicon, of which 6.0 kilos, enters the pig iron, or 14.4 per
cent.
The sulphur charged is 0.270 kilos., of which the pig iron
contains 0.07, or 25.9 per cent.
The phosphorus charged is 0.063 kilos., while the analysis of
the pig iron shows in it 0.12 kilos. It is thus evident that all.
the phosphorus goes into the pig iron; for while the analysis
shows more phosphorus in the pig iron than was put into the
furnace, yet the divergence is evidently due to segregation or
concentration of phosphorus in the sample taken, and the prac-
tical conclusion is that all the phosphorus in the charge finds its
way into the pig iron.
Notes on the Balance Sheet.
The Fe^O' of the ore is assumed all reduced, because the
920.4 kilos, of iron in it is less than the 966.6 kilos, of iron
known to be in the 1,000 kilos, of pig iron from its analysis.
BALANCE SHEET OF BLAST FURNACE. 247
Solution.- (1)
Balance Sheet (Per 1000 op Pig Iron).
Charges Pig Iron Slag Gases
Ore 1530.2
Fe'O* 1314.9 Fe 920.4 O 394.5
FeO 60.6 Fe 46.2 FeO 1.2 O 13.2
SiO^ 84.2 Si 6.0 SiO^ 69.6 O 8.6
MnO 9.6 Mn 0.25 MnO 9.3 O 0.1
AlW 11.6 A1='0' 11.6
CaO 34.1 CaO 34.1 6 0.03
MgO 14.8 MgO 14.8
P^O' 0.092 P 0.04 O 0.05
S 0.153 S 0.07 CaS 0.19
Cu 0.11 Cu 0.11 O 0.01
Limestone
115.8
Fe^O*
0.2
SiO='
3.6
APO»
0.4
CaO
62.2
MgO
0.2
0.007
S
0.001
CO^
49.1
Charcoal
682.0
C
547.7
N
0.5
0
24.1
Fe^O*
2.2
SiO^*
1.3
CaO
6.1
MgO
p205
0.7
0.046
S
0.116
K^'O
3.4
H^O
95.8
Blast
2416.8
0^
557.7
N2
1859.1
FeO 0.2 O 0.02
SiO^ 3.6 ....
AlW 0.4
CaO 62.2
MgO 0.2
0.003 O 0.00
CaS 0.0
CO^ 49.1
27.0
0.02
, ,
C
520.7
N
0.5
....
0
24.1
FeO
2.0
0
0.2
SiO^
1.3
CaO
5.9
o
0.06
MgO
0.7
o
0.03
CaS
0.25
K^O
3.4
H^O 95.8
O 557.7
N= 1859.1
Totals 4744.0 1000.0 220.8 3543.7
248 METALLURGICAL CALCULATIONS.
The FeO, however, cannot be assumed all reduced, because It
would furnish 47.1 kilos, of iron, and there is only 966.6—
920.4 = 46.2 kilos, of iron yet to be supplied. "We, therefore,
put down 46.2 kilos, of iron as going to the pig iron, thus fur-
nishing all the iron in the pig iron, and leaving 0.9 kilos, of
iron to go over into the slag as 1.2 kilos, of FeO. Having thus
allowed for all the iron in the pig iron, the Fe^O' in the lime-
stone and fuel must be assumed as passing entirely into the
slag as FeO.
The 6 kilos, of silicon in the pig iron is put down as coming
entirely from the SiO^ of the ore, of which 14.6 kilos, is thus
used up, leaving 15.6 kilos, to- go into the slag. The SiO^ of
flux and fuel must then be regarded as passing entirely into
the slag.
The 0.25 of manganese in the pig iron comes from the MnO
of the .ore, requiring 0.35 of MnO, and leaving 9.3 of MnO to
go into the slag.
The APO' and MgO of ore, flux and fuel go bodily into the
slag.
The sulphur in the ore, 0.153 kilos., is more than enough to
supply the 0.07 kilos in the pig iron. We, therefore, put down
0.07 kilos, as going into the pig iron, supplying all the latter
contains, and calculate the remaining 0.083 kilos, to CaS going
into the slag. The CaO necessary to furnish this calcium is
56 for every 32 of sulphur (CaO = 56, S = 32), or 0.14 kilos.,
which, therefore, must be deducted from the 34.1 kilos, of CaO
present in the ore. The oxygen of this 0.14 kilos, of CaO finds
its way into the gases.
The 0.092 kilos, of P^O' present in the ore contain only 0.04
kilos, of phosphorus, and since the pig iron contains, from its
analysis, 0.12 kilos., we may assume all of this going into the
pig iron. The same remarks are true of the P^0° in flux and
fuel; altogether, they come somewhat short of supplying all
the phosphorus in the pig iron, and are, therefore, considered as
completely reduced. The copper goes entirely into the pig iron,
although not given in the analysis.
The Fe^O' of the limestone must be transferred entirely as
FeO to the slag, since all the iron needed for the pig iron has
been already provided. The same is true of the Fe^O' of the
fuel; and an analogous statement applies to the SiO* and sul-
BALANCE SHEET OF BLAST FURNACE 249
phur of both flux and fuel. The sulphur of the fuel does not
produce an amount of CaS which counts in significant figures,
and the CaO required is likewise insignificant, as is also the
oxygen thus furnished the gases. In such cases, instead of
ignoring the item altogether, or putting down wholly insignifi-'
cant quantities, the amounts are expressed as 0 00, denoting no
significant amount.
The fixed carbon of the fuel, only, furnishes the carbon in
the pig iron, the rest going into the gases. The blast is calcu-
lated as follows:
12
Carbon in CO^ of flux = 49.1 X^r = 13 . 39 kilos.
44
Carbon in gases from fuel = 520 . 70 "
Carbon in gases altogether ■= 534 . 09 "
Carbon in 1 cu. meter of gas (0.231 + 0.148 +
0.005) X 0.54 =0.20736 "
Volume of gas per 1,000 of pig iron 534.09 -J-
0.20736 = 2575.6 m'
Nitrogen in this gas 2575.6X0.573 = 1475.9 "
Weight of nitrogen 1475.9X1.26. = 1859.6 kg.
Nitrogen from fuel = 0.5 "
Nitrogen from blast = 1859 . 1
Oxygen from blast 1859.1X0.3 = 557.7 "
CHAPTER II.
CALCULATION OF THE CHARGE OF THE BLAST
FURNACE.
In the last chapter we discussed the balance sheet of ma-
terials entering and leaving the furnace, showing the distribu-
tion of the ingredients of the charge and the blast into the
various products and by-products of the furnace. We did not
there go into the question as to how the proportions of the
charge are determined by the metallurgist in charge of the
furnace. There are, however, very few factors of the charge
which can be controlled at will. The blast furnace reduces
practically all of the iron present in the ore into the pig iron,
and, therefore, if the ore contains 50 per cent, of metallic iron
and the pig iron 90 per cent., it will take 0.90 -h 0.50 = 180
parts of ore to furnish the iron in 100 parts of pig iron. The
amount of ore to be used per unit of pig iron made is there-
fore fixed by the richness or poverty of the ore, and is not
capable of variation. The amount of fuel used per unit of
pig iron made is not fixed a priori, as is the amount of ore,
but is governed by the calorific requirements of the furnace
while in operation. If the pig iron and slag run colder than
they should, it is evident that more heat must be put in or
developed within the furnace, which the manager promptly
proceeds to accomplish by increasing the temperature of the
air blast (if he can), or by relatively increasing the amount of
fuel in the charge (which he always can). The ratio of the
weight of the ore and flux in the charge to the weight of fuel
used is called the burden of the furnace, and in practice it is
usual to charge at one time a fixed weight of fuel, and to vary
the burden according to the heat requirements of the furnace.
Changing the burden is therefore only another expression for
changing the relative amount of fuel used, and this is varied
simply from observation of the temperatures of iron and slag
and the conclusions therefrom as to whether the burden is too
250
CALCULATION OF THE CHARGE. 251
heavy or unnecessarily light. The amount of blast used is
another factor relatively fixed per unit of pig iron produced.
Blow more blast, and more pig iron is produced; blow no blast
(bank the furnace), and no pig iron is made; the ratio is not
quite exact, but it is quite nearly true that, other conditions
being equal, the output of pig iron is nearly proportional to the
amount of blast blown. Variations in the temperatures of the
blast produce important changes, which will be separately dis-
cussed.
The amount of flux used is really the one factor in which the
manager has the greatest freedom of action. The amount of
this indispensable substance used is determined by many fac-
tors, and can be varied between quite wide limits without
fundamentally deranging the furnace. It is here a question of
using sufficient flux in the charge to make with the unreduced
constituents of ore and ash of the fuel a slag which shall be
well-fused at the temperature of the furnace, shall carry off
considerable sulphur, if much is present, and shall not corrode
the lining of the furnace. These considerations are so im-
portant, and often so little understood, that we will discuss
them more at length.
Calculation of the Flux and Slag.
From the balance sheet of problem 51 it will be seen that the
metallurgist running the Swedish blast furnace at Herrang,
used, per 1,000 of pig iron made, 1530.2 of ore, 115.8 of lime-
stone flux, 682 of charcoal, and blew in 2416.8 parts by weight
of blast. It may with safety be believed that the amount of
charcoal used was the minimum which he found by experience
necessary to keep his furnace at proper temperature; and most
American blast furnace managers will wonder how he could
get along with so little. The amount of ore used was the
necessary proportion to furnish the iron. The amount of blast
wa? probably all that could be gotten out of the blowing ap-
paratus with which the furnace was provided. Finally, the
amottnt of flux was that amount necessary to make a proper
slag. Let us investigate the question as to how its amount was
determined and the characteristics of the proper slag.
The ingredients of the slag produced in the case in question
are, from the balance sheet:
252 METALLURGICAL CALCULATIONS.
From Ore. From Flux. From Fuel. Total
SiO^ 69.6 3.6 1.3 74.5
APO' 11.6 0.4 .... 12.0
CaO 34.0 62.2 5.9 102.1
MgO 14.8 0.2 0.7 15.7
FeO 1.2 0.2 2.0 3.4
MnO 9.3 .... .... 9.3
K^O .... 3.4 3.4
CaS 0.2 .... 0.25 0.45
140.7 66.6 13.55- 220.85
And the percentage composition of the slag:
SiO^ = 33.73 per cent. FeO = 1.54 per cent.
APO' = 5.43 " MnO = 4.21
CaO =26.23 " K^O = 1.54 "
MgO =7.11 " CaS =0.20 "
The crucial question now presents itself, " What guided the
metallurgist in choosing the quantity of lime stone used, and
in making a slag of the above composition? " The answer to
this will develop the whole practice of fluxing.
Primarily, the fundamental guide in this matter is previous
experience, as revealed in recorded analyses of slags which
have worked properly. Such analyses have been made for a
hundred years, and freely published; they show what com-
positions of slag have been found practicable and suitable in
blast furnace practice. As reported by the chemists, the analyses
show the varying percentages of SiO'', APO', CaO, MgO, FeO,
MnO, etc., found in actual slags made in successful practice,
and this information would be the metallurgist's guide in calcu-
lating the amount of flux to use to produce a good slag. On
studying these analyses, we find that silica and lime are the pre-
dominating constituents of all blast furnace slags, the reason
being that silica is the principal material to be fluxed, and that
lime (from limestone) is the cheapest material which will flux
it, and form a fusible slag.
Analyses of numerous blast furnace slags show the following
variations of composition:
SiO' 25 to 65 per cent. FeO 0 to 6 per cent.
APO' 3 " 30 " MnO 0 " 14 "
TiO' 0 " 10 " K^O \ n <■ Q ..
CaO 12 " 50 " Na^O / " "^
MgO 0 " 18 " CaS 0 " 9 "
CALCULATION OF THE CHARGE. 263
The above limits are not reached simultaneously by one and
the same slag. The ordinary variations may be summarized
as follows (according to Ledebur) :
SiO= APO' CaO + MgO
Producing ^ray iron, using char-
coal 45—65 10— 5 45—25
Producinggrayiron, using coke. 30 — 35 15 — 10 50 — 55
Producing white iron, using char-
coal 45—50 10— 5 45—
Producing white iron, using coke 30 — 40 10—5 60 — 55
Producing spiegeleisen, using
coke 30 10 55—45
In the latter case there is present 5 — 15 per cent, of MnO.
Among these varied compositions, however, there are various
degrees of fusibility, and of suitability to the blast furnace's
needs. The most easily fusible slags are those (considering
only the main ingredients) which contain 35 per cent, of lime,
and in which, if alumina is present, each 1 per cent, of alumina
is balanced by the presence of 0.5 per cent, additional lime.
This rule gives good slags up to about 65 per cent, of lime and
alumina cotmted together. Another observation is, that with
33 .to 40 per cent, of lime in the slag, the amount of silica and
alumina together being some 60 to 70 per cent., it is possible to
change the relations of silica to alumina within large limits
{i.e., from 40 to 50 silica and 20 alumina to 25 or 35 silica and
35 alumina) with very little effect upon the fusibility of the
slag. On the other hand, with a low proportion of silica in the
slag, say 30 to. 40 per cent., it is equally possible to change the
relations of lime to alumina within large limits {i.e., from 50
lime and 15 alumina to 35 lime and 35 alumina) with very little
eflect upon the fusibility of the slag.
The blast furnace manager usually decides upon the kind
of slag he will make, on one of three or four assumptions:
(1) If there is little alumina present, and practically no
magnesia, he usually assumes some ratio between the weights
of silica and lime which he desires his slag to possess, and cal-
culates the weight of flux necessary to make slag conforming
to that condition. In charcoal furnaces, where there is prac-
254 METALLURGICAL CALCULATIONS.
tically no sulphur in the charge, the silica may be 1.5 to 2.0
times the lime present; in coke furnaces, where it is necessary
to eliminate much sulphur, this ratio is usually 0.5 to 1.0.
Illustration. — If a ton of iron ore carries 300 pounds of silica,
how much limestone, which is practically pure calcium car-
bonate, will be required to slag it, neglecting flux necessary
to slag the ash of the fuel?
Solution. — Pure limestone is CaCO', or CaO . CO^ and car-
ries 56 per cent, of lime, which will go into the slag, and 44
per cent, of carbonic oxide (CO^), which goes into the gases.
Using X pounds of limestone, the weight of lime going into the
slag is 0.56 x. If we assume the ratio of silica to lime = 1 for
a coke furnace, and 1.75 for a charcoal furnace, we get the two
equations and corresponding values of x:
= 1 whence x = 536 pounds.
0.56 a;
300
0.56 a;
= 1.75 whence x = 307
(2) If considerable magnesia is present it Is usual to either
count it simply as so much lime, or else to calculate the weight
of lime to which it is chemically equivalent, and add this to the
lime, calling the sum the " summated lime." The ratio of silica
to lime is then used as the ratio of silica to lime plus magnesia,
or of silica to summated lime. When there is considerable
magnesia present the chemical summation should always be
made. Small amounts of MnO, FeO and K^O or Na^O are
also chemically summated as lime. The chemical summation is
based on the fact that one molecule of a base containing one
atom of oxygen is considered the equivalent in fluxing power
of any other similar molecule; e.g., CaO, MgO, FeO, MnO,
K'O, Na^O are considered as equivalent; but as these mole-
cules weigh differently, we have equivalent fluxing powers in
the following weights:
CaO 56
MgO 40
CALCULATION OF THE CHARGE. 255
FeO 72
MnO 71
K^O 94
Na^O 62
from which we conclude that since 40 parts by weight of mag-
nesia is equivalent to 56 parts of lime, that, therefore, the
56
lime equivalent of any weight of magnesia is — of the weight
of the magnesia. Similarly, we get the lime equivalent of these
bases as follows:
56
CaO equivalent of any weight of MgO = •;-X weight MgO
FeO = §X " FeO
?! " MnO = ^X " MnO
K^O = ^X " K^'O
" " Na^'O = i^X " Na'O
Illustration. — An iron ore contributes to the slag 350 pounds
of silica, 12 of FeO and 60 of MnO. It is desired to flux it by
using limestone containing 38.1 per cent, lime, 13.6 magnesia,
3.4 silica, and 44.9 carbonic oxide (CO^). How much flux
must be used to produce a ratio of silica to summated lime
= 0.8?
Solution. — Letting x be the pounds of limestone used, the
ingredients of the slag will be
SiO'' = 350 + 0.034 a;
CaO = 0.381 X
MgO = 0.136 ic
FeO = 12
MnO = 60
256 METALLURGICAL CALCULATIONS.
The lime equivalents of the MgO, FeO and MnO are:
.56
CaO equivalent of MgO " ■*"" "'
MgO = 0.136 :kX^ = 0.1904^
FeO = 12
X|= 9.33
MnO = 60
Xf^ = 47.32
CaO = 0.381 XX 1 = 0.381 x
Summated CaO = 0.5714 a; + 56.65
The ratio of silica to summated lime is therefore:
350 + 0.034 «: .„
56.65 + 0.5714 a;
Whence x = 720 pounds.
It is easily seen that this method of solution, calling x the
weight of flux used, and then getting expressions for the weight
of each ingredient in the slag and the weight of the whole slag,
is a very general solution which is applicable to any kind of
assumed composition to which it is desired that the slag shall
conform.
(3) If alumina is present in the slag-forming constituents of
the ore it also may be reckoned with in several ways. It may
be reckoned as so much by weight, and added in as such to the
silica or the lime, or it may be calculated to its silica or its
lime equivalent, and added into the summated silica or the
summated lime. Here we touch on a question which has
agitated blast furnace managers and theorists for a generation:
Should the alumina be reckoned with the bases or with the
acids; summated as silica or as lime? It would be presump-
tuous to set forth a dictum on a subject which has been so
long and so ably discussed by some of the best iron metal-
lurgists, but we will assume, as somewhere near the truth, that
as far as the elimination of sulphur in the slag is concerned.
CALCULATION OF THE CHARGE. 257
alumina acts in slags low in silica as though it were lime, not
in the proportions of its lime equivalent
= IrKo X weight of alumina)
but rather in about the proportions of its simple weight. As fSlr
as fusibility is concerned, in high silica slags alumina increases
the fusibility up to a certain point, above which it decreases it.
It acts in these, therefore, like lime, and may be classed with
the bases. In low silica slags, below 45 per cent., alumina acts
like silica when considerable is present, and like lime when less
is present; for instance, in a low lime, high alumina slag, alu-
mina and silica may be substituted one for the other within
wide limits, without materially affecting the fusibility of the
slag; in a high lime, high alumina slag, alumina and lime may
be substituted for each other within wide limits without sen-
sibly changing the fusibility. To state the matter as succinctly
as possibe, in slags low in silica (30 to 35 per cent.), alumina
reinforces the bases in the elimination of sulphur; in regard to
fusibility, it acts like silica in a slag low in both silica and lime,
and like lime in all other blast furnace slags.
Illustration. — ^An iron ore carries 10 per cent, of its weight of
silica and 6 per cent, of alumina. The lime stone on hand con-
tains 37.3 per cent, lime, 13.3 magnesia, 3.3 silica, 44 carbonic
oxide (CO^), and 2.1 per cent, alumina. How much flux is
required per 1,000 parts of ore to make (a) a slag with 49 per
cent, of silica plus alumina; (b) a slag with 33 per cent, silica;
(c) a slag with summated silica = the summated lime?
Solution.
(a) Weight of SiO' in slag = 100 + 0.033 x
Weight of APO^ in slag = 60 + 0.021 x
Weight of CaO in slag = 0.373 x
Weight of MgO in slag = 0.133 a;
Total weight of slag = 160 + 0.560 »
therefore,
160 + 0.054 s; = 0.49 (160 + 0.560 a;)
whence
x = 370
258 METALLURGICAL CALCULATIONS.
(b) 100 + 0.033 a; = 0.33 (160 + 0.560 a:)
whence
X =_313
(c) Silica = 100 +0.033 a;
180
_^^^ Silica equivalent of alumina = -rr-r (60 + 0.021 a;)
^i* 204 '
Summated silica =153
+0.0515 a;
Lime =
0.373 X
Lime equivalent of magnesia = -nr
40
(0.133 s;)
Summated lime =
0.5592 X
therefore. 153 + 0.0515 « = 0.5592 a:
whence x = 300
Returning to the slag resulting from the proportions of flux
chosen in Problem 51, and containing:
SiO^" 33.74 per cent. FeO 1.54 per cent.
APO' 5.43 " MnO 4.21
CaO 46.23 " K^O 1.54 "
MgO 7.11 " CaS 0.20
We see that its percentage of silica is low, therefore it is adapted
to produce pig iron low in sulphur; its percentage of alumina
is low, and therefore its presence increases the fusibility of
the slag, which would otherwise be rather deficient, because
of the high lime and somewhat considerable amount of other
bases. In such a slag, alumina would be summated with the
bases.
33 74
The ratio of silica to bases is „„" ^ = 0.516
66.06
33 74
The ratio of silica to lime plus magnesia is H5~qq ~ 0.632
OO ■ do
The summated lime = 46.23
+ CaO equivalent of kVQP = ^ (5.43) = 8.94
CALCULATION OF THE CHARGE. 259
56
+ CaO equivalent of MgO = -— - (7.11) = 9.95
+ CaO equivalent of FeO = ||'(1.54) = 1.20
+ CaO equivalent of MnO = ~ (4.21) = 3.32
+ CaO equivalent of K^'O = -^ (1.54) = 0.92
94
Ratio silica to summated lime
70.56
vAv 33.74
'^^^'"^^ 70.56
= 0.478
Problem 52.
In a blast furnace charge, consisting of 1530.2 pounds of ore
and 682 pounds of charcoal, per 1,000 pounds of pig iron made,
it is known from the balance sheet of above materials that they
will contribute to the slag the following slag-forming ingre-
dients. (See balance sheet, Problem 51) :
SiO=' 70.9 pounds. FeO 3.2 pounds.
APO' 11.6 " MnO 9.3
CaO 39.9 " K^O 3.4
MgO 15.5 " CaS 0.4
The limestone at hand contains:
CaO 53.74 per cent. APO» 0.32 per cent.
MgO .0.17 " Fe^O' 0.18 "
SiO^ 3.14 " CO^' 42.42 "
Required. — The weight of limestone to be used to make:
(1) A slag containing 33.74 per cent, of silica. (2) A slag in
which the ratio of silica to bases is 0.516. (3) A slag in which
the ratio silica to summated lime 's 0.478.
Solution. — This problem embodies the conditions which con-
260 METALLURGICAL CALCULATIONS.
front the metallurgist when desiring to calculate the flux needed-
by any given furnace, and we have assumed certain working
ratios to be aimed at in the slag, in order to elucidate the method
of solution.
SiO^.
APO'.
CaO..
MgO.
FeO.
MnO.
CaS..
Constituents of Slag.
From Ore
and Fuel.
From Flux.
Total.
70.9
0.0314 ic
70.9 + 0.0314.
11.6
0.0032 a;
11.6 + 0.0032.
39.9
0.5374 a;
39.9 + 0.5374.
15.5
0.0017^
15.5 + 0.0017.
3.2
;^J (0.0018.)
3.2 + 0.0016.
9.3
9.3
3.4
3.4
0.4
0.4
154.2
0.5753.
154.2 + 0.5753..
(1) To make a slag with 33.74 per cent, of silica we must
have
70.9 + 0.0314. = 0.3374 (154.2 + 0.5753.)
sirhence . = 116 pounds.
(2) To make a slag witli ratio of silica to bases 0.516 we
must have
70.9 + 0.0314. = 0.516 (82.9 + 0.5439.)
whence . = 116 pounds.
(3) To make a slag with ratio of silica to summated lime
0.478, we must first summate the lime as follows:
Lime = 39.9 + 0.5374.
1 fiS
Lime equiv. of APO' = ■— (11.6 + 0.0032 .) = 19.1+0.0053 .
MgO = -^(15.5 + 0.0017.) =21.7 + 0.0024.
FeO = -yj( 3.2 + 0.0016.) = 2.5 + 0.0012.
CALCULATION OF THE CHARGE. 2B1
MnO ,= ^ ( 9.3) = 7.3
K^O = |i ( 3.4) = 2.0
Summated lime =92.5 + 0.5463*
therefore
70.9 + 0.0314 a; = 0.478 (92.5 + 0.5463 a;)
whence a; = 116 pounds.
Comparison of Fuels, Fluxes and Ores.
By properly utilizing the preceding, principles, it is possible
to compare different varieties of fuels, fluxes or ores with each
other, and thus to determine their relative values to the fur-
nace, as far as can be inferred from their chemical composi-
tion. (Some of the following methods are from a paper by
Mr. F. W. Gordon, Trans. Am. Institute Mining Eng., 1892,
p. 61.)
Comparison of Fuels.
If different qualities of fuel are available it is possible to
calculate which is the most advantageous to use in the furnace.
The fixed carbon only is efficient for the furnace, and not all
of that, because the ash of the fuel needs to be fluxed to slag,
and a certain amount of the fixed carbon will need to be burned
simply to melt this slag. Then the cost of the limestone to be
used to flux this ash must be counted in, and finally part of the
labor costs of running the furnace must be charged against the
slag. We can thus calculate the total charges against the fuel
to supply one part of available carbon, which is the best basis
upon which to compare different fuels.
Illustration. — Two varieties of coke are available for a blast
furnace, analyzing respectively:
No. 1. No. 2.
Fixed carbon '. . . 84 per cent. 90 per cent.
Volatile matter 2 " 1
Moisture 5 " 3 "
Ash 9 " 6 "
And costing respectively $4.50 and $5.50 per ton. The ash of
the fuels analyzes respectively:
262 METALLURGICAL CALCULATIONS.
No. 1. No. 2.
Silica 55 per cent. 25 per cent.
Alumina 25 " 5 "
Lime 15 " 50 "
Magnesia 5 " 10 "
Ferric oxide — " 10
They are to be fluxed with the limestone of preceding prob-
lem, assumed to cost $1.00 per ton, and to make a slag carrying
40 per cent, of silica and alumina together. Assume an average
of 0.228 parts of fixed carbon necessary to melt down 1 part
of slag, and that the manufacturing costs borne by the slag
amount to $1.00 per ton. What are the relative values of the
two fuels in this furnace?
Solution. — The amormts of flux needed to 100 parts of each
fuel burned will be found as follows, letting x be the amount of
flux used:
Slag No. 1. Slag No. 2.
Silica 4. 95+0. 0314 « 1.50 + 0.0314 «
Alumina 2.25 + 0.0032 a; 0.30 + 0.0032*
Lime 1.35 + 0. 5374a; 3. 00 + 0. 5374a;
Magnesia 0.45 + 0. 0017 a; 0.60 + 0.0017*
Ferrous oxide 0.0016* 0.54 + 0.0016*
Total weights 9.00 + 0.5753 * 5.94 + 0.5753*
Therefore, in case No. 1:
7.20 + 0.0346* = 0.40 (9.00 + 0.5753*)
whence * = 18.4
And in case No. 2:
1.80 + 0.0346* = 0.40 (5.94+0.5753*)
whence * = —2.9
The negative value in case No. 2 simply means that the ash
of fuel No. 2 is more basic than the slag, and, therefore, requires
no limestone, but itself acts as a basic flux. Per ton of fuel
burned, there would be required respectively 0.184 and — 0.029
tons of limestone, and the weights of slag would be 0.196 tons
and 0.0427 tons. (Substituting values of * in the total weights
of slags.)
The weights of fixed carbon necessary to smelt these weights
of slag would be:
CALCULATION OF THE CHARGE. 263
No. 1. 0.196 X0.228 = 0.0447 tons.
No. 2. 0.0427X0.228 = 0.0097 "
Leaving as available fixed carbon for the furnace in each case:
No. 1. 0.84- 0.0447 = 0.7953 tons.
No. 2. 0.90- 0.0097 = 0.8903 "
From these figures the cost of 1 ton of available carbon fur-
nished by each fuel, adding in manufacturing cost chargeable
against the slag, is
No. 1.
Cost of coke, $4.50-5-0.7953 = $5,658
Cost of limestone, $1.00X0.184 + 0.7953 \= 0.231
Costs against slag, $1.00x0.196-^0.7953 U 0.246
$6,135
No. 2.
Cost of coke, $5.50-^0.8985 = $6,233
Cost of limestone, $1.00 X (- 0.092) -s- 0.8903 =-0.101
Cost against slag, $1.00X0.0065X0.8903 = 0.007
$6,139
The two fuels, at the prices given, are therefore of almost
exactly the same value to the furnace.
The solution along the lines shown is general, for any de-
sired composition of slag, or for use with any given limestone,
and is a valuable means of comparing the values of different
fuels. The cost of a ton of pure, available carbon, when fur-
nished by any given fuel, is an item which is useful when com-
paring the relative values of different fluxes or ores with each
other.
Comparison of Fluxes.
If different qualities of flux are available it is very desirable
to be able to calculate which is the most economical to use in
the furnace. Any acid ingredients in the flux diminish very
sharply its efficient fluxing power, because they must first be
satisfied from the bases present in the same proportions as acid
to bases in the final slag. The slag thus formed from the im-
purities requires further to be melted, and other costs are
properly chargeable against it. The best comparison is finally
;2^'*
METALLURGICAL CALCULATIONS.
obtained by calculating for each flux available the cost from it
of pure net lime, or net summated lime, analogous to the cal-
culation for pure net carbon in the case of fuels. Any ordinary
condition may be imposed upon the slag which the furnace is
to produce.
Illustration. — There are available for a furnace two qualities
of limestone, containing respectively:
CaO 53.74 per cent. 47.80 per cent.
MgO 0.17 " 4.61
SiO^" 3.14 " 5.12
APO^ 0.32 " 3.36
Fe^O^ O.is--'" 1.10
CO^ ,r..-..A2A2 " 37.55
'"-*■/'
The first costs $1.00 per ton, the second $0.80. Assume them
smelted with fuel furnishing pure available fixed carbon at
$6,135 per ton ; that 0.228 tons of pure fixed carbon is needed to
smelt 1 ton of slag; that manufacturing costs against slag are
$1.00 per ton, and that the slag, to be made in the furnace must
have summated silica equal to summated lime. Compare the
relative values of the two fluxes.
Solution. — We will direct our calculations towards finding
the net cost of 1 ton of pure available summated lime from
each of the two limestones. The summated lime and silica in
each flux are:
No. 1. No. 2.
Summated lime 0.5411 0.5502
Summated silica 0.0342 0.0808
Excess of summated lime 0 . 5069
0.4694
The weights of slag formed from the impurities present in
each limestone will be:
No. 1. No. 2.
CaO (difTerence between amount present
and excess of summated lime found) . .0.0305 0.0086
MgO... 0.0017 0.0461
FeO .0.0016 0.0099
SiO' 0.0314 0.0512
APO» 0.0032 0.0336
Totals 0.0684
0.1494
CALCULATION OF THE CHARGE 265
The cost of 1 ton of pure available lime from each of theS'?
fluxes will therefore be, adding in tosts chargeable against the
slags formed by the impurities present:
No. 1.
Cost of limestone, $1.00^0.5069 = $1,973
Cost of carbon for melting slag, $6,135X0.228X0.0684
-0.5069 ■ =0.188
Costs of running, chargeable against slag, $1.00X0.0684
-0.5069 =0.135
$2,296
No. 2.
Cost of limestone, $0.80 H- 0.4694 = 1.904
Cost of carbon for melting slag, $6,135X0.228X0.1494
-0.4694 = 0.465
Cost of running, chargeable against slag, $1.00X0.1494
-0.4694 = 0.318
$2,487
The conclusion is that the poorer limestone, at $0.20 per ton
less cost, is in reality costing $0,191 per ton more for pure
available lime, or is in reality 8.3 per cent, dearer than the first,
instead of being 20 per cent, cheaper.
The method of calculation here described is quite general for
any compositions of limestone or other flux, and for any as-
sumed conditions which the slag must conform to.
Comparison of Ores.
As the more complicated, we come to the comparison of
various ores which may be at the iron master's disposal. Here
a similar method of procedure is advisable. It can be calcu-
lated first, for a unit weight of ore, how much pure lime would
be required to flux its impurities, how much pure carbon would
be required to melt the slag thus formed, and to the costs of
each of these would be added the handling of the slag. Each
of these can be also expressed per unit of pure oxide of iron
in the ore; and if to their sum we add the cost of ore necessary
to furnish unit weight of pure oxide of iron we obtain the total
costs per unit weight of pure iron oxide (Fe^O'). This is the
basis on which different ores may then be compared. It must
not be forgotten that one ore may, because of higher sulphur
266 METALLURGICAL CALCULATIONS.
content, require the production of a more basic slag, so that the
amounts of flux required and slag formed will be influenced
by the condition necessary to impose on the slag in each case.
Illustration. — The iron ore briquettes (of Problem 51) con-
tained Fe=0', 85.93 per cent.; FeO, 3.96; SiO^ 5.50; MnO, 0.63;
APO^ 0.76; CaO, 2.23; MgO, 0.97 per cent. If these cost $4.40
per ton, and are smelted in a furnace making slag with ratio
of silica to bases 0.516, and assuming 0.3 per cent, of the iron,
82.7 per cent, of the silica, and 96.6 per cent, of the manga-
nese to go into the slag, what is the cost per ton of pure Fe^O'
from this source, charging pure lime for fluxing at $2,296 per
ton, pure carbon for smelting slag at $6,135 per ton, and re-
quiring 0.228 tons of carbon for one of slag, adding also manu-
facturing costs at $1.00 per ton of slag?
Solution. — The slag-forming ingredients from 1 ton of ore
briquettes are:
72
FeO 0.003 x4i X
56
r(o.8593X ~j +(o.0396x||)] = 0.0024 tons
MnO 0.966X0.0063 =0.0061 "
CaO = 0.0223 "
MgO = 0.0097 "
APO' = 0.0076 "
SiO'' = 0.0455 "
and the bases to satisfy the silica present must
be 0.0455 -J- 0.516 =0.0882 "
But, sum of bases already present = 0.0482 "
therefore, pure lime to be added = 0.0417 "
and total weight of slag = 0.1337 "
Efficient Fe^'O' in 1 ton of ore = 0.8593
plus Fe^O' equivalent of reduced FeO
= j^ (0.0396- 0.0024)= 0.0413
Total available Fe^O» = 0.9006 "
METALLURGICAL CALCULATIONS. 267
Cost of 1 ton pure available Fe'O' from these briquettes:
Cost of ore, $4.40 -i- 0.9006 = $4,885
Cost of pure lime, $2,296 X 0.0417 -^ 0.9025 = 0 . 106
Cost of carbon for melting slag, $6,135X0.228X0.1337
-5-0.9025 = 0.207
Costs chargeable against slag, $1.00x0.1337^-0.9025 = 0.148
Total = $5,346
A similar calculation is possible with any ore of any given
composition, and making any assumed quality of slag. The
costs of 1 ton of pure ferric oxide thus calculated will give the
relative costs of the iron obtained from these different sources,
and therefore indicate the relative values of the different ores
to the blast furnace manager.
Problem BS.
Assume a blast furnace manager to use the ore of preceding
illustration, furnishing pure Fe^O' at a net cost of $5,336 per
ton, and to use with it fuel furnishing pure carbon at $6,135 per
ton, there being required for reduction and melting the iron
produced and furnishing it with carbon, 0.66 tons of pure
available carbon per ton of pig iron produced, and the pig
iron containing 96.656 per cent, of iron. The running costs of
the furnace are $3.00 per ton of pig iron produced (costs against
slag not included).
Required: — The cost of the pig iron per ton.
Solution:
Fe'O* required 160 -^ 112 X 0.96656 = 1 . 3808 tons.
Cost of the ore, $5.336 X 1.3808 = $7,368
Costof fuel, $6,135X0.66 = 4.049
Cost of manufacturing (share against pig iron) = 2 . 000
Total cost = $13,417
CHAPTER III.
UTILIZATION OF FUEL IN THE BLAST FURNACE.
The blast furnace, in its simplest terms, may be regarded as
a huge gas producer, producing in the region of the tuyeres
pure producer gas from fixed carbon and heated air; the gas
thus produced is partly oxidized in its ascent through the
furnace by the oxygen abstracted from the charge (which
latter iteni is almost a constant quantity per unit of pig iron
made), and has added to it carbon dioxide from the carbonates
of the charge. 'But, after all, the unoxidized and combustible
ingredients of the gas escaping represent a large part, in fact,
often the largest part, of the total calorific power of the fuel.
Problem 54.
A blast furnace uses 2,240 pounds of coke, containing 90 per
cent, fixed carbon and 350 pounds of limestone, containing 10
per cent, of carbon (as carbonic acid, CO^) to produce a ton of
pig iron containing 4 per cent, of carbon. The gases contain
24 per cent, of carbonous oxide, CO, 12 per cent, of carbonic
oxide, CO^ 2 per cent, of hydrogen, 2 per cent, of methane, and
60 per cent, of nitrogen.
Required. — (1) The volume of gas, as analyzed, produced per
ton of pig iron made.
(2) The calorific power of the gas.
(3) The proportion of the calorific power of the coke which
has been generated in the furnace.
Solution. — (1) The carbon going into the gases will be that
in the coke, less that in the pig iron, plus that in the carbonates
of the charge.
Carbon in coke = 2,240X0.90 = 2,016 pounds
Carbon in carbonates = 350X0.10= 35 "
Carbon charged = 2,051 "
Carbon in pig iron =2,240X0.04= 89.6 "
Carbon going into the gases = 1,961.4 "
268
UTILIZATION OF FUEL IN BLAST FU^ACE. 269
Carbon in 1 cubic foot of gas:
In CO 0.24X0.54
In CO' 0.12X0.54
In CH* 0.02X0.54
Total 0.38X0.54 = 0.2052 ounces av.
= 0.012825 pounds
1 961 4
Gas produced per ton of pig iron = ■ ' ^ = 152,935 cu. ft.(l)
(2) Calorific power of 1 cubic foot of gas:
CO 0.24X3,062 = 734.9 oz. cal.
ff 0.02X2,613 = 52.3 "
CH* 0.02X8,598 = 172.0 "
Sum f= 959.2 "
= 59.95 pound cal.
per 152,935 cubic feet = 9,168,450 pound cal. (2)
(3) The calorific power of the coke considering it to contain
simply 90 per cent, of fixed carbon, would be
■ 8,100X0.90 = 7,290 pound cal. per pound.
The presence of CH* in the gases points, however, to there
being probably some available hydrogen in it, which would in-
crease its calorific power somewhat. A closer approximation
to the calorific power of the coke could, therefore, be obtained
by assuming at least as mach available hydrogen in it as would
correspond to the hydrogen in the CH* in the gas.
CH* in 152,935 cu. ft. of gas = 3,059 cu. ft.
Weight of this = 3,059 X (0.09 X 8) = 2,202 oz. av.
137.7 lbs.
Hydrogen = 137.7 X (4 -H 16) = 34.4 lbs.
Available hydrogen in coke = 34.4
-^ 2,240 = 1.54 per cent.
Calorific power 0.0154X29,030 == 447 lb. cal. per lb.
Total calorific power of the coke = 7,737 " "
Calorific power of coke used per ton
7,737X2,240 = 17,330,880 lb. cal.
Calorific power of the gases = 9,168,450 "
Calorific power generated = 8,162,430 "
= 47.1 per cent. \
270 METALLURGICAL CALCULATIONS.
This figure, however, practically over-charges the furnace
little bit, because the pig iron with its 4 per cent, of carbon
really takes out of the furnace some unbumt fuel, whose heat
of combustion may be utilized outside the furnace as in the
Bessemer converter. The furnace does not generate heat from
this, representing:
2,240X0.04X8,100 = 725,760 lb. cal.
leaving as actually generated in the furnace
8,162,430- 725,760 = 7,436,670 lb. cal.
= 42.9 per cent. (3)
Such an average blast furnace cannot, therefore, be accused
of generating within it over some 43 per cent, of the calorific
power of the fuel put into it, while the heat rejected as poten-
tial energy of combustion of the waste gases amounts to more
than half the calorific power of the fuel.' Fifty years or more
ago, when these waste gases were allowed to bum truly to waste
the blast furnace was indeed a devourer of fuel, but matters
have been improved by the utilization of the waste gases to heat
the blast, and thus one of the largest " leaks " of heat from
the furnace has been patched up to some extent, although yet
far from satisfactorily.
Problem 65.
Assume that in problem 54, one-third of the gases produced
are burnt in hot-blast stoves, preheating the air blown in at the
tuyeres, and that the blast is thus preheated to 450° C.
Required. — (1) The amount of blast blown in per ton of pig
iron made.
(2) The heat in the blast.
(3) The efficiency of the hot-blast stoves.
(4) The increased efficiency of the blast furnace plant as a
whole in generating the calorific power of the fuel, when thus
provided with this hot-blast apparatus.
Solution.
(1) Volume of (dry) gases per ton = 152,935 cu. ft.
Nitrogen present in these (60%) = 91,761 "
Q1 7fi1
Air containing this = - ' ■ = 115,860 "
0.792 ^^j
UTILIZATION OF FUEL IN BLAST FURNACE. 271
This is the volume of the blast per ton of pig iron produced,
assuming no nitrogen to come from the coke used, and the
blast to be dry. If the blast were moist, and its hydrometric
condition known, the volume of moist blast could be calculated.
(2) Assuming the blast dry, it is heated to 450° C, requiring
115,860 X [0.303 + 0.000027 (450)]X450 = 16,430,975 oz. cal.
= 1,026,936 lb. cal.
(3) The hot-blast stoves receive one-third of all the gas pro-
duced, having, therefore, a calorific power of
9,168,450 -H 3 = 3,056,150 lb. cal.
Efficiency of the stoves = ^'^^^^^^ = 0.336 = 33.6 per cent.
3,056,150 /^\
(4) The blast furnace was primarily rejecting unused 57.1
per cent, of the calorific power of the fuel, 4.2 per cent., how-
ever, as a necessary loss, to supply the carbon in the pig iron,
but 52.9 per cent, as combustible power of unbumt waste gases.
If one-third of these gases are completely burnt in hot-blast
stoves, then the combined plant — furnace plus stoves — is
utilizing 17.6 per cent, more of the calorific power of the fuel
than before, or 42.9+17.6 = 60.5 per cent., and, therefore, re-
jecting undeveloped 39.5 per cent, of the calorific power. (4)
[The net efiEect of the use of the hot^blast stove, upon the
heat generation in the furnace, is practically to put back into
the furnace, as sensible heat, 1-3x33.6 = 11.2 per cent, of the
calorific power of the waste gases, equal, therefore, to 0.112X
52.9 = 5.9 per cent, of the total calorific power of the fuel.
This renders available, for the working of the furnace, 42.9 +
5.9 = 48.8 per cent, of the calorific power of the fuel, an in-
5 9
crease of available heat for reducing and smelting of ■ ' =
13.7 per cent, of the former available quantity.]
The practical conclusion is that a blast furnace generates
in itself not much over 40 per cent, of the calorific power of
the fuel used, and rejects nearly 60 per cent. ; by using part of
the waste gases to heat the blast, however, some of this re-
jected heat, to an amount representing net 5 to 10 per cent, of
the calorific power of the fuel used, is returned to and injected
272 METALLURGICAL CALCULATIONS.
bodily into the furnace, thus rendering available for the pur-
poses of running the furnace some 50 per cent, of the calorific
power of the fuel as a maximum. The efficiency with which
the furnace applies this 50 per cent, usefully to the objects of
reducing and smelting, is another question for investigation.
Problem 56.
Assume that at the furnace of Problems 54 and 55 the two-
thirds of the waste gases are burnt tmder boilers, raising steam
which runs the blowing engines, hoists and pumps, and pro-
viding 10 effective horse-power for each ton of pig iron made
per day in the furnace.
Required. — (1) The efficiency of development of the calorific
power of the fuel in the plant (furnace, stoves, boilers, engines)
regarded as a whole.
(2) The thermo-mechanical efficiency of the boiler and engine
plant.
(3) The power which could be generated if gas engines, at
25 per cent, thermo-mechanical efficiency, were used in their
stead.
Solution. — (1) Since the stoves completely bum one-third
of the waste gases, and the boilers the other two-thirds, all the
combustible power of the waste gases is developed in the com-
bined plant, and the only part of the calorific power of the
fuel which is unused is the 4.9 per cent, represented by the
carbon necessarily entering into the composition of the pig
iron. The plant as a whole, therefore, develops or generates
95.1 per cent, of the calorific power of the fuel used.
(2) For each ton of pig iron, the heat developed under the
boilers will be two-thirds of the calorific power of the gases, or
9,168,450X2-3 = 6,112,300 lb. cal.
There is generated thereby 10 effective horse-power days,
equal to
10X33,000X60X24 = 475,200,000 ft. lbs
But 1 lb. cal. = 425X3.2808 = 1394.3 "
Therefore, thermal equivalent of
1 , 475,200,000 ^, „ „
work done = ^^^^ 'g -•« 340,800 lb. cal.
UTILIZATION OF FUEL IN BLAST FURNACE. 273
Thenno-mechanical efficiency of boiler and engine plant:
(3) Gas engines, at 25 per cent, thermo-mechanical efficiency,
would give power representing per ton of pig iron produced :
6,112,300X 0.25 = 1,528,075 lb. cal.
Equal to 1,528,075X1394.3 = 2,130,645,900 ft. lbs.
2,130,645,900 ,,„^ .
33,000X60X24 = 44.9 horse-power days. (3)
A quicker solution is:
10 X 5^ = 44.9 " " (3)
Leaving net surplus power per ton of pig iron produced per
day = 34.9 horse-power.
The preceding problems have elucidated the question of the
small proportion of the calorific power of the fuel which is
generated in a blast furnace, showing it to be, in usual practice
only 40 to, at most, 50 per cent, of the calorific power. The
discussion has not explained " why," but a further consideration
will throw light on this question also.
The proportion of the calorific power of a fuel which is gen-
erated in a blast furnace is solely a question of how much of
it is burned to carbonic oxide, CO^, and how much to carbonous
oxide, CO? If all the carbon were burned to CO^, practically
all the calorific power of the fuel would be generated; if all
2 430
were burned to CO, only „' ■= 0.30 = 30 per cent, of the
0,100
heating power of the carbon Would be generated. If a blast
furnace was filled with nothing but coke, and air blown in as
usual at the tuyeres, carbon would be burned in the furnace
only to CO, and but 30 per cent, of its calorific power be gen-
erated and available for the needs of the furnace. The entire
gain over this percentage is due to the oxidation of COtoCO^
by the oxygen abstracted from the solid charges, that is, by
the act of reduction. In Problem 54 we calculated that under
ordinary conditions, between 40 and 50 per cent, of the calorific
274 METALLURGICAL CALCULATIONS.
power of the fuel is generated in the furnace; the excess of
this above 30 per cent, is due to the oxidation of CO to CO^
during the reduction of the metallic oxides in the charge. From
this standpoint it is advisable to strive to perform the greatest
possible proportion of the reduction in the furnace by CO gas,
because in this case the total generation of heat in the furnace
per imit of fuel charged will tend towards a maximum. Since
no carbon can be burned to CO^ at the tuyeres, it follows that,
from the standpoint of the generation of the maximum quantity
of heat in the furnace, from a given weight of fuel, Griiner was
right in formulating his dictum of the ideal working of a blast
furnace, viz.:
Gruner's " Ideal Working."
All the carbon burnt in the furnace should be first oxidized
at the tuyeres to CO, and all reduction of oxides above the
tuyeres should be caused by CO, which thus becomes CO^.
This dictum is not in Gruner's own words, but expresses their
sense, and from the point of view of the present discussion, it
is the correct principle upon which to obtain the maximum gen-
eration of heat in the furnace from a given weight of fuel. It
practically directs us to generate at the tuyeres 30 per cent.
of the calorific power of the carbon oxidized in the furnace,
and the rest that can be obtained from the carbon is to be
generated during the reduction of the charge.
If we apply this principle to the furnace and data of Problem
54, we should first observe that the carbon oxidized in the
furnace is:
Carbon in coke charged 2016. pounds
Carbon in pig iron produced 89.6 "
Carbon oxidized in furnace 1926. 6 "
Requiring, if all oxidized to CO at the tuyeres
4
1926. 6 X-o- lbs. oxygen = 2569. pounds.
o
But, Problem 52 shows us that there was actually blown
into this furnace 115,860 cubic feet of blast, containing, there-
fore,
115,860X0.208X (1.44-^16) = 2,169 lbs. oxygen,
UTILIZATION OF FUEL IN BLAST FURNACE. 275
capable of oxidizing to CO at the tuyeres
2,169X0.75 = 1,627 lbs. carbon.
Proportion of carbon gasified burnt at the tuyeres,
^^ = 0.844 = 84.4 per cent.
It is, therefore, true of the furnace under discussion, that if
Griiner's ideal working be called standard, this furnace attains
to 84.4 per cent, of that ideal; and it is also true that this fur-
nace generates from the carbon burnt at the tuyeres 84.4 per
cent, of the amount of heat which could have been generated if
Gruner's ideal working iad been attained.
It is always possible to find out for any given blast furnace,
by similar calculations, how much carbon is burned at the
tuyeres, and how much is burned above the tuyeres, and thus
to determine -how closely the furnace running approximates to
Gruner's ideal working. This proportion or percentage will
not necessarily express how efficiently the furnace is rimning,
as regards fuel used per unit of iron made, but it will tell what
proportion of the calorific power of the fuel used is being gen-
erated at the tuyeres, and in possibly nine cases out of ten
this proportion indicates the general efficiency of the furnace
as regards fuel consumption.
It will be next profitable to inquire when and under what
conditions Gruner's ideal working does not correspond to
maximum fuel economy, and why it usually does. The answer
is not difficult to understand: if all the carbon gasified in the
furnace is burned to CO at the tuyeres, 30 percent of the total
calorific power of the carbon burned is there developed, which
is more than half of all the heat generated from carbon in the
furnace. To this must also be added the sensible heat in hot
blast, which may amotmt (as in Problem 52) to some 5.9 per
cent, of the calorific power of the carbon, making, therefore, a
total of 35 per cent, of the calorific energy of the fuel gen-
erated at the tuyeres out of a total of about 50 per cent, de-
veloped in the furnace. If, however, the blast be heated to a
very high temperature, or particularly if it be dried, or if the
ore and fuel are extra pure, so that a smaller quantity of heat
is needed to melt down slag at the tuyeres, then there may not
276 METALLURGICAL CALCULATIONS.
be needed at the tuyeres the generation by combustion of so
much heat as Griiner's ideal working would require and cause
to be produced, and to bum at the tuyeres all the carbon oxi-
dized in the furnace would be wasteful of fuel. In this,case
although less heat would be generated per unit of fuel, by
burning some of it above the tuyeres, yet economy in fuel con-
sumption as a whole would be attained, because of the better
distribution of the heat which was generated from a smaller
total quantity of fuel.
Illustration. — In Problem 55 we assumed that the furnace
ran with blast heated to 450° C, and that this hot blast, burn-
ing at the tuyeres 84.4 per cent, of all the carbon gasified in the
furnace, smelted down pig iron and slag satisfactorily and kept
the tuyere region at proper temperature. If the temperature
of this blast were raised to 900° C, how much greater pro-
portion of heat would be available in the tuyere region?
We have already calculated that the 115,860 cubic feet of blast
used per ton of iron made, brought in at 450° C, 1,026,936
poimd calories of hea,t, equal to 5.9 per cent, of the calorific
power- of the fuel put into the furnace. If the temperature
were 900° C, the heat brought in would be
115;860X [0.303 + 0.000027 (900)]X900 = 37,920,978 oz. cal.
= 2,370,061 lb. cal.
which equals
2,370,061 .^_ ,„_
17,330,880 = °-^^^ = ^^-^ P^'' '=^''*-
of the calorific power of the fuel, a gain of 7.8 per cent,, added
to the heat available in the tuyere region. This causes a very
great increase in the smelting-down power of the furnace,
enabling the same work per ton of ore smelted to be done with
much less consumption of fuel in the tuyere region. An idea
of this increased smelting-power may be obtained from the
following comparison of heat available for smelting purposes
in the tuyere region in the two cases just discussed:
Case 1.
Heat developed by oxidation of carbon per
ton of iron made, 1,627 lbs. X 2,430 = 3,953,610 lb. cal.
Sensible heat in blast at 450° C. = 1,026,936 "
Total heat available = 4,980,546
UTILIZATION OF FUEL IN ^LAST FURNACE. 277
Case 2.
Heat developed by same quantity of air
burning same carbon = 3,953,610 "
Sensible heat in blast at 900° C. = 2,370,061
Total heat available = 6,323,671
It is, therefore, seen that the heat generated and available
at the tuyeres is increased 1,343,125 calories, amounting to 27
per cent, of the amount disposable in Case 1. It follows, there-
fore, that the smelting down power has been increased 27
per cent., and that, if the 4,980,546 calories were sufficient for
satisfactorily smelting down the iron and slag in the first in-
stance, that the extra heat of Case 2 can all be utilized for
smelting down 27 per cent, extra burden. We can, there-
fore, charge 27 per cent, more burden per unit weight of coke
in Case 2, because we have the requisite smelting down power
at the region of the tuyeres, which amounts to saying that
we can charge 22 per cent, less coke for a given weight of pig
iron made.
In actual practice, as the amount of burden is increased and
the temperature of the blast increased, the change causes more
and more of the carbon to be oxidized above the tuyeres, and
a smaller proportion to be oxidized at the tuyeres, thus ob-
taining less service in the furnace from oxidation of carbon as
a whole, but compensating for this by the extra heat in the
hot blast. Or, looking at it in another way, we may say that
the same heat could be made available in the region of the
tuyeres, when using hot blast, by the combustion there of a
smaller quantity of carbon; therefore, we can bum more of it
above the tuyeres and yet work more economically on the
whole,' than we were working in the first instance, with the
colder blast.
Minimum Carbon Necessary in the Furnace.
Many writers have assumed that in the reduction of iron,
oxide, such as Fe^O^, the reaction of its reduction by carbonous
oxide, CO, is expressed as follows:
Fe^0'-|-3C0 = 2Fe + 3CO='
If this wei:e true, there would need to be burnt to CO at the
278 METALLURGICAL CALCULATIONS.
tuyeres only 3C, or 36 parts of carbon, to ensure the reduction
of Fe^O^ representing 112 parts of iron. The reaction does
not, however, progress as shown, because CO^ acts oxidizingly
on Fe to such a degree that when ICO^ is present in the gas
for ICO left unused, the reduction practically stops, even
though the gases are moving slowly through the warm ore.
The real reaction of reduction by CO gas is therefore more
nearly represented by
Fe='0» + 6C0 = 2Fe + 3CO + 3CO^
which shows that 112 parts of iron would require at least 72
parts of carbon to be oxidized at the tuyeres to CO, in order to
produce the gas necessary for its reduction. The presence of
some CO' in the furnace coming directly from carbonates in
the charge would neutralize still more of the reducing power
of the CO gas, and cause still more of it to be theoretically re-
quired for reduction. The minimum amount of carbon neces-
sary to be charged in the furnace will be that necessary to
furnish fixed carbon enough for this reducing gas and for the
carbon in the pig iron. This would be, per 100 parts of pig
iron, containing say 93 iron and 3 carbon, and using coke con-
taining 90 per cent, of fixed carbon:
Carbon burnt at tuyeres = 72x93-^112 =59. 8
Carbon in pig iron = 3.0
Total fixed carbon necessary =62.8
Total coke to supply this = 62.8 -hO.9 =69.8
It results from these calculations that if " Griiner's ideal
working " of a blast furnace were carried out to the practical
extent of reducing all the charge by carbonous oxide, CO,
and oxidizing no carbon at all directly above the tuyeres, that
about 63 parts of fixed carbon would be required per 100 of
pig iron made, requiring from 70 to 80 parts of fuel, according
to its richness in fixed carbon (90 to 80 per cent.). In prac-
tice, as is well known, more than this is commonly used, be-
cause of the larger proportion of unused CO in the gases than
above assumed: and less than this has been regularly used,
showing that economy of fuel can be attained without adher-
ing to " Griiner's ideal working," in fact, by transgressing it as
far as one dares.
UTILIZATIOTSI OF FUEL IN BLAST FURNACE. 279
The principle involved can be best grasped by a calculation
of the amount of carbon which would be required by the fur-
nace, supposing all the heat necessary for melting down the
charge were supplied by electrical means, thus dispensing, for
the purposes of this supposition, with the necessity of blast
and the consequent necessity of oxidizing any carbon by any
other agent than tlie oxygen given up by the ore. In this case
the gases resulting would be, let us assume, of the same com-
position as before, that is, containing equal volumes of CO
and C0^ and since this oxygen is abstracted altogether from
the ore, the reaction is
Fe='0H2C = 2Fe-fCO + C02
This would represent the utilization of carbon in an electrically-
heated furnace, and would require per 100 of pig iron made,
assuming it 3 per cent, carbon and 93 iron :
Carbon for reduction 24 X 93 ^ 112 =19.9
Carbon in pig iron = 3.0
Total fixed carbon necessary = 22 . 9
Or only a little over one-third as much as the minimum re-
quired when the smelting down is done by blast.
Aside from electrical furnace practice, however, this discus-
sion proves that whatever fixed carbon bums or oxidizes above
the region of the tuyeres, in a blast furnace; absorbs oxygen
from the charges with three times the efficiency of carbon first
burnt at the tuyeres. Every pound of oxygen abstracted from
the charges by solid carbon requires the use or intervention of
only one-third as much carbon as that which is abstracted by
CO gas; or, each pound of carbon abstracting oxygen directly
from the charge takes from it three times as much oxygen as
a pound of carbon first burnt to CO at the tuyeres possibly can.
The ordinary furnace produces at the tuyeres, in order to get
heat enough to melt down the charges, more CO gas than is
needed to abstract all the oxygen from the charges; under
these conditions it is imeconomical to oxidize any carbon at
all above the tuyeres. The exceptional furnace, because of
pure ores, small amount of slag, pure fuel, high temperature
of blast, or dry blast, gets heat enough at the tuyeres to melt
280 METALLURGICAL CALCULATIONS.
down the charges without producing enough CO gas to reduce
all the charges; under these conditions more or less reduction
is effected by solid carbon, and with the greatest economy in
quantity of carbon required in the furnace. These are the
conditions tuider which, having passed the turning point, the
greater economy of fuel is attained the farther away one can
get from " Gruner's ideal working." '
CHAPTER IV.
THE HEAT BALANCE SHEET OF THE BLAST FURNACE.
Twenty-eight years ago, Sir Lothian Bell first constructed
a satisfactory heat-balance sheet for a blast furnace. His ob-
servations were largely, and his experience altogether, confined
to the reduction of the argillaceous siderite ores of the Cleve-
land district, England, and although he made numerous at-
tempts to draw general conclusions from the data at hand
applicable to iron smelting in general, yet many of his deduc-
tions remain true only for the particular ores and manner of
working characteristic of the Cleveland district.
No treatment of this subject, however, can be based other-
wise than upon Bell's researches, following the lines laid down,
in his " Principles of the Manufacture of Iron and Steel."
Heat Received and Developed.
The items on this side of the balance sheet are:
(1) Combustion of carbon to carbonous oxide (CO),
(2) Combustion of carbon to carbonic oxide (CO').
(3) Sensible heat of the hot blast.
(4) Heat of formation of the pig iron from its constituents.
(5) Heat of formation of slag from its oxide constituents.
(1) and (2) Combustion of carbon in the furnace. There is
but one satisfactory way to determine with exactness the
amounts under this heading. From the balance sheet, the total
amount of carbon passing iiito the gases is obtained; from the
analysis of the gases, the weight of carbon per unit volume
of gases is calculated; the first divided by the second gives the
volume of gases per unit weight of pig iron produced. The
amount of CO and CO* in these gases is then obtained by use
of the gas analysis, and if from the total CO and CO* in the
gases there be subtracted the CO and CO* contributed as such
by the solid charges, the difference is' the CO and CO* which
have been formed in the furnace. The heat evolved in the
formation of these quantities can then be calculated.
281
282 METALLURGICAL CALCULATIONS.
Illustration. — In Problem 51 it was calculated that per 1,000
kilos, of pig iron produced, 634.09 kg. of carbon went into the
gases; also that the analysis of the gases showed 0.20736 kg.
of carbon in each cubic meter of gas. The quotient indicated,
therefore, 2575.6 cubic meters of gas produced per ton of pig
iron. From the analysis of the gases there was in this volume,
2575.6X0.231 = 595.0 m' of CO
2575.6X0.148 = 381.2 m'' of CO^
whose weights were
595.0X1.26 = 749.7 kg. CO
381.2X1.98 = 754.8 kg. CO^
The balance sheet shows, however, 49. 1 kg. of CO^ contained in the
limestone flux used, which can be assumed as entering the
gases bodily. Subtracting this we have 705.7 kg. of CO'' formed
in the furnace, and 749.7 kg. of CO, containing respectively
705.7X^1 = 192.5 kg. of C in CO^*
12
749. 7X^ = 321.3 kg. of C in CO
The heat generated in the furnace by the oxidation of carbon
is, therefore,
192.5X8100 = 1,559,250 Calories
321.3X2430= 780,760 "
2,340,010
If this carbon could have been entirely burnt to CO^, there
would have been generated
513.8X8100 = 4,161,780 Calories
Showing that only 56 per cent, of the calorific power of the
carbon was developed in the furnace; the other 46 per cent,
exists as potential calorific power in the waste gases, and part
THE HEAT BALANCE SHEET. 283
of it is really put back into the furnace as sensible heat in
the hot blast.
There is a little doubt as to how to consider the CH* in the
gases; that is, whether the heat of its formation should be
reckoned in as developed in the furnace. This would be (C,H*)
= 22,250, or 1,854 Calories per kg. of carbon contained therein.
Its presence in the gas probably results largely from the dis-
tillation of the fuel at a high temperature, and the heat re-
quired to disunite the CH* from the solid fuel is probably as
great as is represented by its heat of formation from carbon and
hydrogen. The item is, therefore, a doubtful one, and as far
as we know, we may be coming about as near to the truth by
omitting it altogether as by counting it in. If we wished to
add it in the illustration just given the calculation would be:
Volume of CH« = 2575.6 X 0.005 = 12.88 m'
Weight of C = 12.88X0.54 = 6.9 kg.
Heat of formation = 6.9 X 1,854 = 12,793 cat.
It should be emphasized that in this calculated heat of oxi-
dation of carbon in the furnace, no account has been taken
whatever of where in the furnace this heat is generated. Above
all, the mistake should not be made of supposing that the
780,760 Calories produced by formation of CO represents the
heat generation at the region of the tuyeres ; nothing could be
further from the truth. A great deal of carbon is burnt to CO
at the tuyeres, and some above the tuyeres, but a goodly pro-
portion of this CO oxidizes by abstracting oxygen from the
charge and becomes CO''. It would not be incorrect, however,
to divide the heat of oxidation of carbon in the furnace into
two parts, viz: to assume all the carbon as first forming CO,
and part of this CO afterwards forming CO^, corresponding to
the amount of the latter formed in the furnace. If this were
done, we would have
513.8 kg. C to CO = 513.8X2430 = 1,248,535 Cal.
449.2 kg. CO to CO'' = 449.2X 2430 = 1,091,475 "
2,340,010
This analysis gives us the information that of the total heat
generated by the oxidation of carbon in the furnace, some-
284 METALLURGICAL CALCULATIONS.
where about one-half is generated by its burning to CO, and
the other half by the further oxidation of CO to CO^; and we
also know that the larger part of the former takes place at the
tuyeres, an d all of the latter takes place during the reduction
of the charges in the upper part of the furnace.
If we know, however, or have calculated the amount of the
blast received by the furnace, or, more properly speaking, the
amount of oxygen in the blast, then the heat generated by
oxidation of carbon at the tuyeres becomes known. In the
previous illustration, taken from Problem 51, we can also take
from the same problem the weight of oxygen in the blast,
557.7 kilos. This would bum 557.7X0.75 = 418.3 kg. of
carbon to CO at the tuyers, generating there
418.3X2430 = 1,016,410 Calories,
or 44 per cent, of all the heat generated by oxidation of carbon
in the furnace, leaving 1,323,610 Calories as generated above the
tuyeres by the agency of the oxygen of the charges. These
figures tell us just where and how the principal items of heat
were generated in this particular furnace, and the similar cal-
culation may be made for any blast furnace for which we have
the necessary data.
(3) Sensible heat in the hot blast. To calculate this item,
we need to know the weight or volume of the different con-
stituents of the blast and their temperature. The question at
once arises, as to what base line of temperature shall be chosen.
It is most convenient to choose 0° C, since that is not over
15° from the average temperature in the largest iron produc-
ing countries. However, any other prevailing temperature
may be taken as the base line, involving merely a little more
calculation, since our specific heats are reckoned from 0° C.
The temperature ought, moreover, to be taken as near to the
tuyeres as possible, to properly take into accoimt the effect
of cooling of the blast in the bustle and feeder pipes, from
radiation and expansion. The blast consists of air p oper
and moisture, the former with a mean specific heat between
0° and t° C of 0.303 -|-0.000027t, in kilogram Calories per cubic
meter, or in ounce calories per cubic foot, the latter with a
similar mean specific heat of 0.34-f 0.00015t. Since the mois-
ture at times amounts to as much as 5 per cent, of the blast,
it should be calculated separately.
THE HEAT BALANCE SHEET. 285
Illustration: With the outside air at 30° C, and saturated
with moisture (raining), calculate the heat carried into a blast
furnace by blast carrying in 1859.1 kilos, of nitrogen, the tem-
perature of the heated blast being 600° C. Barometer 720
millimeters of mercury. Temperature base line 0° C.
Solution: One cubic meter of the moist blast, as taken into
the blowing cylinders, carries all the moisture it can hold, the
tension of which is therefore 31.5 millimeters. The tension of
the air proper present is therefore 720—31.5 = 688.5 milli-
meters, and each cubic meter of moist air carries
31 5
= 0.0438 cubic meter of moisture, and
720
688.5
720
= 0.9562 cubic meter of air proper.
Whatever the temperature of the blast, the moisture and air
proper will be in this same proportion whenever its tempera-
ture is over 30° C. If the temperature were 0° C. the moisture
would be mostly condensed, but for the purposes of calculating
the heat brought in we may assume the moist air to be at 0°
C, with its moisture uncondensed. That volume of blast
which would be 1 cubic meter at 0° and 760 mm. pressure,
would, therefore, bring in, at 600° C, the following quantity
of heat :
ffO 0.0438X[0.34 +0.00015 (600)] X600 = 11.3 Calories
Air 0.9562 X [0.303 + 0.000027 (600)] X600 = 179.7 "
Total 191.0 "
Since the nitrogen present in this is
0.9562X1.293X^ = 0.9511 kg.
the heat brought in per 1859.1 kg. of nitrogen is
191.0 X^St = 373,344 Calories.
An amount equal to over one-third of all the heat generated
by combustion of carbon at the tuyeres.
(4) Heat of formation of pig iron from its constituents. The
286 METALLURGICAL CALCULATIONS.
pig iron contains several per cent., some 5 to 10 altogether,
of carbon, silicon, manganese, phosphorus, sulphur and other
elements. The energy of their combination with the iron is a
somewhat indefinite quantity, and in no case can be consid-
able. Berthelot states the energy of combination of carbon
with iron as (Fe',C) = 8,460, which would be 705 Calories per
kilogram of carbon, and another investigator (Ponthiere)
states the heat of combination of phosphorus with iron to be
zero. In the present state of imcertainty it is hardly allowable
to add in any other than the heat of combination of the car-
bon in the iron, and leave out that of the other elements.
(5) Heat of formation of the slag from its constituent oxides.
Here we touch upon a quantity of more than insignificant
proportions, yet which is not yet quantitatively known with
satisfactory accuracy. The main constituents of the slag
are SiO^ APO*, CaO, MgO and CaS, which are provided by
clay, limestone and iron sulphide. If we allow, on the other
side of the balance sheet, for the heat necessary to de-hydrate
clay, drive carbonic acid off carbonates, and break up iron sul-
phide and enough CaO to furnish Ca for CaS, we are then en-
titled, on the other hand, to place in the heat evolution column
the heat of combination of aluminum silicate with lime and
magnesia, the heat of formation of CaS and its heat of so-
lution in the silicate slag. The heat of formation of CaS is
94,300 Calories, or 2,947 Calories per kilogram of sulphur;
its heat of combination with a silicate slag is unknown. The
heat of combination of lime with aluminum silicate has been
determined only for the proportions 3CaO to APSi^O^, that is,
for 168 parts of CaO uniting with 222 parts of aluminum sili-
cate. This has been determined in Le Chatelier's laboratory as
(SCaO, APSi^O') = 33,500 Calories which is 200 Calories per
vmit of CaO combining, or 150 Calories per unit weight of APO'
-fSiO^. The calculation would be made on the basis of the
amount of lime (plus lime equivalent of magnesia present),
if it were present in a smaller ratio than 168 to 222 of silica
and alumina, and on the basis of the silica and alumina, if
their ratio to the summated lime were lesfe than 222 to 168.
It is probable that in the near future these quan^ties will be
known more accurately.
One item of heat received by the furnace has not been men-
THE HEAT BALANCE SHEET. 287
tioned, because of its usual absence, viz.: heat in hot charges.
Very rarely roasted ore comes hot to the furnace, in which
case its sensible heat must be counted in, else the thermal
sheet of the furnace will be that much out of balance.
Heat Absorption and Disbursement.
The items on this side of the balance sheet are:
(1) Sensible heat in waste gases, including water vapor only
as vapor.
(2) Sensible heat in outflowing slag.
(3) Sensible heat in outflowing pig iron.
(4) Heat conducted to the ground.
(5) Heat conducted and radiated to the air.
(6) Heat abstracted by cooling water, tuyeres, etc.
(7) Heat for de-hydrating the charges.
(8) Heat for vaporizing water from charges.
(9) Heat absorbed by decomposition of carbonates.
(10) Heat absorbed in reduction of iron oxides.
(11) Heat absorbed in reduction of other metallic oxides.
(12) Heat absorbed by decomposition of moisture of the
blast.
(1) Sensible heat in waste gases^ The amount of these gases
is known only from the carbon contained in unit volume, by
analysis, and the known weight of carbon entering and leav-
ing the furnace. If there is much fine coke carried over by
the blast, allowance must be made for the carbon in it, be-
cause this would not be represented in the gas analysis. The
analysis of completely dried gas is that usually obtained, be-
cause if the gas is measured without drying, an uncertain
amount of moisture is condensed, and, therefore, it is usual
to dry before measuring and analyzing. The amount of mois-
ture in the gases is either assumed as that driven off from the
charges, as shown by the balance sheet, or else is determined
direcstly by drawing the gases through a calcium chloride tube
or other dessicating apparatus. Several tests should be made
to get a fair average, because much more will be in the gases
immediately after charging than immediately before. Dust
must be excluded from the drying tube by filtering the gases
through dry asbestos. The average temperature of the gases
should be known over a considerable period; a thermo-couple
288 METALLURGICAL CALCULATIONS.
in the down-comer gives this most accurately and more uni-
formly than if inserted above the stock line in the furnace.
The weight of moisture per unit volume of dry gas is then
converted into volume at standard conditions by dividing by
0.81 (Icubic meter = 0.81 kg; 1 cubic foot = 0.81 ounce
avoirdup.). The sensible heat of the gases is then calculated,
using 0° C. as the base line, and the proper mean specific heats
of the gases per unit of volume. The water vapor will here
be considered simply as a gas, and its sensible heat above water
vapor at 0° only calculated. This leaves the latent heat of
vaporization of this water to be considered as a separate item
(606.5 Calories), that is, as heat absorbed by reactions in the
furnace, thus putting it on exactly the same footing as the
CO^ in the gases which has been expelled from carbonates in
the furnace. By so proceeding much uncertainty as to the
heat in the water vapor is avoided.
If the amount of flue dust is considerable its quantity should
be ascertained, and the heat in it also calculated, and added in
to the heat in the moist gases. Its specific heat may be ap-
proximated as so much carbon, iron oxide and silica, the pro-
portions of each of these present being known.
(2) Sensible heat in outflowing slag. The weight of slag
produced is seldom taken directly, but can be reckoned up
with all needful accuracy from the balance sheet of materials
entering and leaving. Its temperature and specific heat, solid
and liquid, melting point and latent heat of fusion, are un-
fortunately almost always unknown factors. The one datum
which is needful, however, is the total heat in a unit weight of
liquid slag as it flows from the furnace, and this is not a diffi-
cult quantity to obtain. A rough calorimeter with a reliable
thermometer and containing a carefully weighed quantity of
water, may be easily constructed. Some liquid slag is nm
directly into it, and by observing the rise of temperature and
afterwards filtering out, drying and weighing the granulated
slag, a satisfactory determination can be arrived at. This is
corrected to 0° C. by using an approximate specific heat, say
0.20, for the range of final calorimeter temperature .to zero.
In this connection it is important to note that the calorimetric
determinations of Akerman on blast-furnace slags, give the
heat in the just-melted slag, whereas slags flowing out of a
THE HEAT BALANCE SHEET. 289
furnace are considerably, some 200° to 500° C, above their
melting point, and therefore contain some 50 to 150 Calories
more heat than that given by Akerman for a slag of similar com-
position. Since Akerman 's values run from 350 to 400 Calories,
the actual heat in the outflowing slag may be between 400
and 550 Calories. Akerman himself states that an average
of twenty-seven Swedish furnaces gave 530 Calories as the
actual heat in unit weight of outflowing slag, and Bell uses
550 in most of his calculations on Cleveland (England) furnaces.
(3) Heat in outflowing pig iron. The heat in just-melted
pig iron is evidently too small a quantity to use in this con-
nection. The heat in the outflowing pig iron at 200° to 500°
above its melting point will be 50 to 100 Calories greater. The
former quantity is about 245 Calories; the latter will be 300
to 350. Bell takes 330 for Cleveland furnaces; Akerman states
250 to 325 for Swedish furnaces. We may conclude then,
to use 300 Calories for a coke furnace running cool, and up
to 350 Calories for a very hot furnace.
(4) Heat conducted to the ground. ' This is a very uncertain
quantity. It varies with the kind of ground, and is more
nearly a constant per day than per unit of pig iron produced.
It is, therefore, expressed per unit of iron produced, larger
for small furnaces run slowly than for large furnaces run fast.
It is less when running rich ores and greater with poor ores,
other things being equal. As nearly as can be assumed we
would put this item as lying between 60 and 200 Calories per
unit of pig iron made. Bell uses 169 on one Cleveland furnace,
but it is certainly less than 100 in some charcoal furnaces using
pure ores and fuel, and consequently with a small heat re-
quirement.
(5) Heat conducted and radiated to the air. This item is
likewise more nearly a constant quantity per day for a given
furnace, and is therefore less per unit of iron produced the
faster the furnace is r\m. It may vary between 60 and 250
Calories per unit of pig iron, the former in furnaces of low heat
requirement per unit of iron produced, the latter in those
of high heat requirement. If the amount were calculated it
would figure out as a time fimction, and would require the
temperature of the. outside shell, that of the air, the velocity
of the wind, and the total outside surface, in order to calculate
290 METALLURGICAL CALCULATIONS.
by the principles of heat radiation and conduction, the amount
radiated per day. No one has done this yet for any one fur-
nace, and, in brief, items (4) and (5) of this schedule are usually
grouped together and determined simply by difference, their
sum aggregating from 100 to 500 Calories per unit of pig iron,
averaging 100 to 150 for charcoal furnaces of low heat
requirement, 200 to 450 for Cleveland furnaces (Bell), and
300 to 500 for large, modem furnaces with thin walls and great
height.
(6) Heat abstracted by cooling water. In the old-fashioned
heavy masonry, cold blast furnace, this item was zero. With
the advent of hot blast, the water needed for cooling the tuyeres
entered as a heat abstracting factor. It is greater the harder
a furnace is blown, but does not increase proportionately with
the output. The heat lost by tuyere-cooling water may be
50 to 100 Calories per tinit of pig iron made. That for cooling
of bosh plates and the outside of the crucible in modem fur-
naces, may vary all the way up to 200 Calories. These two
items are very large in a modem furnace, but are necessary
expenditures of heat energy in order to preserve the lines of
the furnace during fast rtuining. For any particular furnace
they may be determined with all needful accuracy by mea-
suring the amoimt of water pumped or used for these purposes
and its temperature before and after using.
(7) and (8) Drying and de-hydrating charges. Water goes
into the furnace as moisture and as combined water of the
charges. To convert the moisture, such as is evaporated by a
current of moderately warm air, into vapor requires 606.5
Calories per unit of water. This allows merely for its vapor-
ization in the furnace, and not fojr any sensible heat which it
may carry out of the furnace at the temperature of the waste
gases. This latter item is properly considered in with the
sensible heat of the waste gases. The common practice of
saying that it takes 637 Calories to evapprate the moisture of
the charges is wrong, because this amoimt would convert water
at 0° to vapor at 100°, and, therefore, would include part of
what is properly the sensible heat of the waste gases. On the
other hand, it is equally wrong to say that this heat of vapor-
ization should be counted in as sensible heat in the hot gases;
it would be just as logical or, rather, equally illogical, to count
the latent heat of vaporization of CO^ as sensible heat in the gases.
THE HEAT BALANCE SHEET. 291
To drive off water of hydration from hydrated minerals in
the charge requires an additional amount of chemically-ab-
sorbed heat. As far as is known, this is small for the water
driven off hydrated iron oxides, so small as to be a doubtful
quantity and safely left out ; but if it comes from clay the large
amount of 611 Calories is absorbed in merely separating it
from its chemical combination (2H='0, APSi^O') = 22,000 Calories,
which would require 611 -f 607 = 1,218 Calories to put into
the state of vapor each unit weight of water entering the fur-
nace chemically combined in clay. (This does not concern
the ordinary moisture in moist qlay, expelled at 100° C, but
only the combined water in the dry clay). Where much clay
occurs in the ores this quantity becomes important, and its
amount explains some of the difficulties met in working clayey
charges, particularly since a large part of this chemically com-
bined water is expelled only at a red heat, and, therefore,
cools greatly the hotter zones of the furnace.
(9) Decomposition of carbonates. Raw limestone, or dolo-
mite, is the usual flux of the blast furnace, and its carbonic
acid is evolved at temperatures between 600° and 800°. Whether
some of this is subsequently decomposed by contact with
carbon and reduced to CO, is immaterial to the balance sheet,
because more than enough CO^ escapes from the furnace to
represent the CO^ of the flux, and we charge the furnace only
with the formation of the CO and CO^ actually found in the
gases, less the CO* from fluxes. Bell charges up the heat
absorbed also in the assumed reaction CO*-^C = 2C0, but
this is an error, because it is doubtful how much of the CO*
is thus decomposed, and the question, in its last analysis,
is one of heat distribution in the furnace, and does not concern
the totals of heat absorbed or evolved. We can, therefore,
omit the item of decomposition of this CO* (as likewise, and
for analogous reasons, the heat evolved in carbon deposition in
the upper part of the furnace— 2C0 = C + CO^), and need
consider only the heat required to expel the CO* from car-
bonates. This is :
(CaO, CO*) = 45,150 Calories = 1,026 Calories per kg. CO*
(MgO, CO*) = 29,300 " = 666 "
(MnO, CO*) = 22,200 " = 500 "
(FeO, CO*) = 24,900 " = 566 "
(ZnO, CO*) = 15,500 " = 352 "
292 METALLURGICAL CALCULATIONS.
By using the above figures, in connection with the known
composition of ore and fluxes, the heat required to decompose
carbonates can be correctly calculated.
(10) Reduction of iron oxides. The heats of formation of
the various oxides of iron are:
(Fe,0) = 65,700 Calories = 1,173 Calories per kg. iron
(Fe',0*) = 270,800 " = 1,671
(Fe^O') = 195,600 " = 1,746
And, therefore, just these quanities of heat are required per
unit weight of iron reduced from these compounds. If the
ore is a carbonate the heat absorbed in driving of CO" from
FeCO' can be first allowed for, and then the heat required
for reduction of the FeO calculated on the weight of the re-
duced iron. If some FeO goes into the slag it will be as FeO,
and if the ore was Fe^O', or Fe'O^, the reduction of unit weight
of iron from the state of Fe^O', or Fe'O*, to the state of FeO,
absorbs respectively 573 or 498 Calories, as may be readily
deduced from the heats of formation of the three oxides con-
cerned. If FeS is present its heat of formation is
(Fe, S) = 24,000 Calories = 429 Calories per kg. Fe.
If the iron is charged partly as silicate, such as mill or tap
ciftder, an additional amount of heat will be required for re-
duction, equal to that needed to separate the iron oxides from
their combination with silica. The heat of formation of the
bi-silicate slag only has been determined.
(FeO. SiO^) = 8,900 Calories = 148 Calories per kg. SiO'.
And since the cinders concerned contain relatively more iron
than this, we can best make allowance for the heat required to
set free the silica. It is necessary, therefore, to take the amount
of silica in the iron cinder charged, and allow, as necessary
for its decomposition into FeO and SiO", 148 Calories for each
unit weight of SiO" contained.
(11) Reduction of non-ferrous oxides. Silicon is usually
present in pig iron, its reduction from silica requiring:
(Si, 0==) = 180,000 Calories = 6,413 Calories per kg. Si.
There is a little doubt about this (Berthelot's) figure; more
THE HEAT BALANCE SHEET 293
recent determinations, not yet published, point rather to 196,000
and 7,000 Calories respectively.
Manganese is often present in the ores as Mn^'O', Mn'O*, or
MnO^ and going partly into the slag as MnO. The heat ab-
sorbed ia reduction to manganese is :
(Mn, O) = 90,900 Calories = 1,653 Calories per kg.Mn
(Mn», O*) = 328,000 " = 1,988 "
(Mn, 0=") = 125,300 " = 2,278 "
For the MnO produced and going into the slag, the reduc-
tion from Mn'O*, or MnO^, per imit weight of contained man-
ganese, absorbs 335 or 625 Calories respectively.
Sulphur generally comes from the reduction of FeS, re-
quiring 667 Calories per kg. of sulphur ; but care must be taken
not to allow for this heat twice, since if reckoned once tmder
the head of iron reduction it must not be reckoned on the
balance sheet a second time under sulphur. One reduction of
FeS liberates both constituents.
Phosphorus may be reduced in large quantity in making basic
iron. It probably comes mostly from calcium phosphate, in
which case we must reckon on not only the heat of oxidation
of phosphoric oxide but also its heat of combination with lime:
(P^ 0=) = 365,300 Cal. = 5,892 Cal. per kg. of P.
(3Ca0, P^O') = 159,400 " = 2,410 " " contamed.
Making a total heat requirement of 8,302 Calories to separate
unit weight of phosphorus from phosphate of lime, and leave
free lime. In a furnace making a pig iron with several per
cent, of phosphorus, this item becomes quite large.
Calcium occurs in the slag as CaS, its reduction from lime
requiring
(Ca, O) = 130,500 Calories = 3,263 Calories per kg. Ca.
Other elements than the above rarely occur in pig iron in
notable quantity. If rare ones occur, their heat of reduction
can be sought in thermochemical tables. Those of tungsten,
titanium, molybdendum and chromium are, however, not at
present known.
(12) Decomposition of moisture in the blast. This is to be
294 METALLURGICAL CALCULATIONS.
counted as vapor of water, the heat required to decompose,
which is:
(H^, O) vapor = 58,060 Cal. = 3,226 Cal. per kg. H»0.
= 29,030 " " W.
It is not correct to allow here for the sensible heat in this
water vapor coming in with the hot blast, because that heat
should go on the other side of the balance sheet as heat de-
livered to the furnace. Neither is it correct to subtract the
heat of combination of the oxygen of this moisture T^th car-
bon to form CO at the tuyeres ; because, although that com-
bination actually does take place, yet the heat thereby evolved
properly belongs also on the other side of the balance sheet,
and is there properly taken care of as part of the heat of oxida-
tion of carbon in the furnace.
Problem 57.
(Data partly from paper by the author. Transactions Ameri-
can Institute of Mining Engineers, 1905).
A blast furnace running on Lake Superior ore has the fol-
lowing charges, per 100 of pig iron produced :
Hematite ore: 177.6; composition WO — 10.0 per cent.
SiO^ —10.0
APO* — 3.5
Fe^O^ —76.5
Limestone : 44.4 ; composition SiO^ — 5.0
MgO — 4.8
CaO —47.6
CO' —42.6
Coke: 95.8 ;composition SiO' — 5.3
CaO — 5.3
WO — 1.0
C —88.0
Pig iron produced: 100; composition Si — 1.0
C — 4.0
Fe —95.0
Gases produced: composition dry: CO^ — 13.0
CO —22.3
N» —64.7
THE HEAT BALANCE SHEET. 295
Blast used: Contained 5.66 grains of moisture per cubic foot
of dry air, at 24° C. = 75° F.
Charges in pounds: Coke, 10,200; ore, 20,000; stone 5,000.
Product per day: Pig iron, 358 tons = 801,920 pounds.
Coke used per day : 768,626 pounds.
Temperature of blast 720° F. = 382° C.
Temperature of waste gases = 538° F. = 281° C.
Displacement of blowing engines = 40,000 ft.' per min.
Heat in one unit of pig iron = 325 Calories.
Heat in one unit of slag = 525 Calories.
Cooling water, per day, heated 50° C. = 300,000 gallons.
Required: (1) The volume of gases per 100 kg. of pig iron
(2) A balance sheet of materials entering and leaving the
furnace, per 100 imit of pig iron.
(3) The volume and weight of blast per 100 kg. of pig iron.
(4) The efficiency of the blowing plant.
(5) The heat balance sheet of the furnace.
(6) The proportion of the fixed carbon of the fuel burnt at
the tuyeres.
(7) The proportion of the whole heat generated at the
tuyeres.
(8) The proportion' of the iron reduced in the furnace which
is reduced by solid carbon from FeO.
(9) The theoretical maximum of temperature at the tuyeres.
(10) The theoretical maximum if all the moisture were
removed from the blast.
Solution: (1) To find the volume of the gases :
Carbon in the coke used 95.8X0.88 = 84.3 kg.
3
Carbon in the limestone 44.4 X0.426X jj = ^.2 "
Total carbon entering furnace = 89.5
Carbon in 100 of pig iron = 4.0
Carbon entering gases =85.5
Carbonous and carbonic oxides in gases = 35.3 per cent
Carbon in 1 m' of dried gas = 0.54X0.353 , = 0.19062 kg.
85 5
Dry gas per 100 kg. of pig iron = ^^ ^qqq^ = 448.5 m'. (1)
296 METALLURGICAL CALCULATIONS.
Dry gas per 100 oz. of pig iron , = 448.5 ft.'
Dry gas per 100 pounds of pig iron = 448.5
X16 ■ =7176.0"
Dry gas per 2240 povinds of pig iron = 7176.0
X22.4 = 160,742"
„ . ^ 160.742X358 (ton^) qq q«o «
Dry gas per minute = qox24: "" 39,962
[2) Balance Sheet
OF
Materials, per 100
OF Pig-iron
CO
Charges
Pig
Iron
Slag
Gases
Fe^O'
135.6
Fe 95.0
0
40.7
?
H^iO
17.8
,
....
H^O
7.8
b
SiO'
17.8
Si
1.0
SiO=' 15.7
O
1.1
SiO='
CaO
2.2
21.1
•
SiO^ 2.2
CaO 21.1
i
....
1
MgO
2.1
MgO 2.1
,
Ix
00="
19.0
—
CO'
19.0
00
C
84.3
c
4.0
C
80.3
s
SiO'
5.3
SiO^" 5.3
£
CaO
5.3
CaO 5.3
H^O
0.9
H'O
0.9
C4
0^
96.4
O
96.4
^
N2
321.3
N2
321.3
1
H^O ■
4.5
.
• • • •
{o
0.5
m
4.0
Total
740.0
100.0
68.0
582.0
The charges are calculated simply from the weights of ore,
flux and fuel used, and their percentage composition. The
blast is calculated as given in solution of requirement (3).
The 1.0 of silicon in the pig iron requires 1.0 X (60^28)
= 2.1 parts of SiO' to furnish it, leaving 15.7 of unreduced
SiO' to go into the slag, and 1.1 of oxygen to the gases.
(3) The balance sheet shows that the solid charges furnish
to the gases 41.8 of 0, 190 of 00=" and 18.7 of ffO. The WO
goes as such into the gases, and therefore its oxygen is not
present in the sample of dry gas analyzed. The oxygen in
CO' is 19.0X(32-h44) = 13.8 kg., which added to the 41.8
gives 55.6 of oxygen getting into the gases as CO or CO', from
the solid charges.
THE HEAT BALANCE SHEET. 2d7
The oxygen in the CO and C0= of the gases is to be calcu-
lated from the oxygen in unit volume of gas and the total
volume of gas produced. The oxygen in 1 cubic meter of gas
will be:
OinCO = 0.223X (o.OQxf) X^ = 0.16056kg.
OinCO= = 0.130 X (o.OQXy) X j| =0.18720 "
0.34776 "
A quicker solution is to note that CO^ represents 0^ its own
volilme of oxygen, and CO represents O, or half its volume of
oxygen, and since 1 cubic meter of oxygen weighs 0.09 X (32 -h 2)
= 1.44 kilograms, the weight of oxygen required is
(0 22^ \
-^2-- + 0.13J X 1.44 = 0.34776 kg.
The total oxygen in 448.5 m' of gases is therefore
448.5X0.34776
subtracting that furnished by the solid charges =
there remains oxygen furnished by blast
If the blast were perfectly dry air, its nitrogen would
be 100.4 X (10-7-3)
and the weight of dry blast
and its volume at 0° C. 435.1-^1.293
The blast, however, is not dry, and the 100.4 kilos, of oxygen
includes that brought in by the moisture. The moisture weighs
5.66 grains per cubic foot of measured dried air; it is, there-
fore, simplest to calculate the oxygen entering the furnace per
unit volume of air proper entering. Since 5.66 grains = 5.66
4-437.5 = 0.01294 ounces av., the calculation can be made in
ounces per cubic foot or kilograms per cubic meter in identical
figures as follows:
c
Oxygen in 0.01294 parts of water = 0.01294 X-g- = 0.0115
Oxygen in 1 volume of air proper, at 24° C.
= 1-293 Xj3X273T^ =0-2743
156
kg.
55.6
((
100.4
tt
334.7
(<
435.1
If
336.5
m?
Sum = 0.2858
298 METALLURGICAL CALCULATIONS.
The blast received, per 100.4 kg. of oxygen thus received, is
therefore, measured dry at 24° C:
>. „"-o = 351.8 cubic meters,
U . zooo
and in cubic feet, per 100.4 pounds of oxygen:
100.4X16
0.2858
5628.8 cubic feet.
The volume of the moist air containing this will be the volume
of assumed dried air plus the volume of the water vapor. The
latter is, per unit' volume of dried air:
0.01294 4- (o.OQXy X ^^|^-^) = 0.01738
and the volume of moist air received, at 24°, is, therefore,
351 . 8 X 1 . 01738 = 357 . 9 cubic meters.
Or 5628 . 8 X 1 . 01738 = 5726 . 6 cubic feet.
The weights of water, oxygen and nitrogen received per 100
of pig iron are (as already given on the balance sheet) :
WO 0.01294X351.8= 4.52 kg.
0=! 0.2743 X351.8 = 96.4 "
N^ 0.9143X351.8 = 321.3 "
And the volumes of these, considered theoretically at 0° C. and
standard pressure, per 100 kg. of pig iron made:
H^O vapor 4 . 52 -4- 0 . 81 = 5.6 cubic meters.
Air 417.7 -f-1.293 = 322.8 "
Sum = 328.4 "
There are several other modifications of this solution which
will suggest themselves to anyone who studies out the question,
but while half a dozen ways may be equally correct, that one
is logically to be preferred which is most easily understood
and is the shortest. One solution, however, based on volume
relations, is well to know. Water vapor, H^O, represents half
THE HEAT BALANCE SHEET. 299
its volume of contained oxygen, while air has 0.208 oxygen.
The cubic foot of dried air at 24° C. was accompanied by
0 ■ 01294 H-0. 81 Xo'T^ = 0.0174 cubic foot
of moisture, which was removed in making the test. The
oxygen per cubic foot of dry blast was, therefore, at 24° C. :
; 0 as H^O = 0.0174-^2 = 0.0087 cubic foot.
O^ as air =1.0000X0.208 = 0.2080 "
0.2167 "
Weight of this oxygen :
1.44X^^3^^^X0.2167 = 0.2868
And volume of dry blast, at 24°,' per 100 of pig iron:
100.4
0.2868
350.1 cubic meters.
The difference of about 0.4 per cent between this and the pre-
viously-Calculated result is due to the figures not being carried
out to a sufficient number of decimal places.
(4) The efficiency of the blowing plant is found by calcu-
lating the volume of air and moisture received by the furnace
per minute, calculateii to 24° C. = 75° F., and comparing this
with the piston displacement of 40,000 cubic feet per minute.
Volume of moist air received, at 24° C, per
100 lbs. of pig iron made ' = 5726.6 ft.'
Volume per day = 5726.6X8019.2 = 45,922,750 ft.'
Volume per minute = 45,922,750-^-1440 = 31,891"
31 890
Efaciency of blowing plant ' - = 79.7 per cent. (4)
(5) The heat balance sheet is based for most of its data upon
the balance sheet of materials, the calculations already made
and the additional data given in the statement.
The balance sheet shows 80.3 kilos, of carbon oxidized in
the furnace. Of this, the amoimt oxidized to CO and reinain-
300 METALLURGICAL CALCULATIONS.
ing as such in the gases is obtained from the amount of CO in
the gases, as follows:
C in CO = 448.5X0.223X0.54 = 54.0 kg.
C oxidized to CO^ is tlierefore 80.3 - 54 = 26.3 kg.
The heat in hot blast is calculated from its volume assumed
to be at 0° C, and already calculated, viz.,; 322.8 cubic meters
of air proper and 5.6 cubic meters of water vapor, with mean
specific heats per cubic meter between 0° and 382° C. of 0.313
and 0.40 respectively.
The heat of formation of the pig iron may be taken from
the amount of carbon in the iron, as 705 Calories per kilogram
of carbon, and that of silicon omitted from consideration.
The heat of formation of the slag, since it contains 29.5 of
silica and alumina to 28.5 of lime and magnesia, may be taken
as 150 Calories per unit of silica and alumina.
The total heat received and generated, and to be accovinted
for, in the furnace, is therefore, per 100 kg. of pig iron:
Carbon to CO 54.0X2430 = 131,220 Calories.
Carbon to CO^" 26.3X8100 = 213,030
In ffO vapor 5.6X0.40 X382\ _ on ooe
In air proper 322.8 X 0 . 313 X 382 / ~ 6\),6Jio
Solution of C in iron 4X705 '= 2,820
Formation of slag 29.5X150 = 4,425
Total = 390,880
The items of heat distribution are 325 Calories in each kilo-
gram of pig iron, 525 Calories in each kilogram of slag, heat
in the waste gases at 281°, heat in cooling water, lost by radia-
tion and conduction (by difference), evaporation of the mois-
ture in charges, expulsion of CO^ from carbonates, decom-
position of moisture of blast, reduction of silicon and iron.
Reduction of Fe
95kg.X 1,746
= 165,870 Calories
Reduction of Si
1 " X 7,000 = 7,000
Expulsion of CO''
from
CaCO»
16.7 " XI. 026
= 18,666 "
Expulsion of CO^
from
MgCO*
2.3 " X 666
Evaporation of H^O
18.7 " X606.6
- 11,342 "
THE HEAT BALANCE SHEET. 301
X281° = 43,836
Heat in waste gases:
N and CO 400 m^xO.3106
CO^ 58.3 m^X 0.446
ffO 23.1 m'X 0.382
Decomposition of moisture of blast:'
H2O4.5kg.X(29,040H-9) = 14,511
Heat in slag 58 kg. X 525 = 30,450
Heat in pig iron 100X325 = 32,500
Heat in cooling water (300,000 gallons per
diem) 300,000X8.3 (lbs.) X 50° -=-8019.2 = 15,525
Loss by radiation and conduction (differ-
ence) = 51,180
Total = 390,880 " (5)
(6) The proportion of the fixed carbon burned at the tuyeres
is obtained by calculating the weight of carbon which the
oxygen entering at the tuyeres could oxidize to CO, as follows:
Oxygen entering at the tuyeres = 100.4 kg.
12
Carbon burned to CO = 100.4 XtI = 75.3"
lb
Fixed carbon charged = 84.3 "
Proportion iDumed at tuyeres = 89.3 per cent.
A more just proportion is that between the carbon burned
at the tuyeres and the total fixed carbon oxidized, because the
fixed carbon which carbonizes the iron cannot be oxidized tinder
any circumstances. This proportion in this furnace is:
93.8 per cent. (6)
84.3-4.0
indicating a very fair approximation to Griiner's ideal work-
ing. If we make the further allowance, that the silicon in the
pig iron is necessarily reduced by solid carbon, and that there-
fore the solid carbon needed to reduce the 1 kilogram of silicon
[lX(24-h28)] is in no case available for combustion at the
tuyeres, we have the approach to Gruner's ideal working meas-
ured by the ratio
75.3
80.3- 0.9
= 94.8 per cent.
302 METALLURGICAL CALCULATIONS.
in spite of which, however, the furnace is not doing very good
work.
(7) The proportion of the heat requirement generated or
available at or about the tuyeres is deternained as follows:
Combustion of C to CO = 75.3x2430 = 182,979 Calories.
Heat in hot blast = 39,385
Formaton of pig iron = 2,820 "
Formation of slag = 4,425 "
Total = 229,609
Against which must be charged heat required
to decompose moisture of blast = 14,511 "
Leaving net heat available = 215,098 "
which is 215,098-7-390,880 = 55 per cent,
of the total heat generated and available in the furnace.
Another way. in which it is sometimes put, is that the oxida-
tion of carbon to CO or CO^ furnishes a certain amount of
heat to the furnace (346,250 Calories in this case), of which
a certain amount is generated by combustion at the tuyeres
(182,979 Calories), making the ratio thus considered, in this
case, 53 per cent., — ^which is almost the same figures as above,
but not so significant, since it is illogical to consider the heat
brought in by the hot blast as not being available heat for
doing work in the tuyere region.
(8) The proportion of iron reduced by solid carbon is found
by finding how much carbon is used up in that reduction.
Fixed carbon charged = 84.3 kg.
Fixed carbon carbonizing the pig iron = 4.0
Fixed carbon oxidized in the furnace = 80.3
Fixed carbon oxidized by the blast = 75.3
Fixed carbon oxidized above the tuyeres = 5.0
Carbon needed to reduce 1 kg. silicon = 0.9
Carbon used for reducing FeO = 4.1
56
Amount of Fe, thus reduced = 4.1 Xy^ = 19.1
Proportion of the total Fe, thus reduced
19 1
= -g^ =20 per cent. (8)
THE HEAT BALANCE SHEET. 303
(9) The maximum temperature of the gases in the region
of the' tuyeres is that temperature to which the heat there
available will raise the products of combustion. This question
is best resolved by simply considering the combustion of 1
kilogram of carbon, evolving 2430 calories, while the heat in
the hot carbon itself just before it is burnt, and that in the
hot blast required, will also exist as sensible heat in the pro-
ducts,— ^the whole diminished by the heat necessary to decom-
pose the water vapor blown in.
Since, per 75.3 kg. of carbon burned at the tuyeres there
enter 5.6 m* of water vapor and 322.8 m' of air proper, meas-
ured at 0°, the volume of blast per kilogram of carbon oxidized
at the tuyeres is
WO 5.6-^-75.3 = 0.0738 m' = 0.0598 kg.
Air 322.84-75.3 = 4.2869 m'
The products of the combustion are, per kg. of carbon burned:
CO 22.22-4-12 = 1.8519 cubic meters.
N^ 321.3^1.26-4-75.3 =3.3865 "
H' equal to H^O decomposed = 0.0738 "
Total = 5.3122 "
The heat available to raise their temperature is:
Heat of combustion of 1 kg. carbon = 2430 Calories.
Heat in hot blast = 39,385-4-75.3 = 523
Heat in hot carbon at t° (or very nearly) = 0.5t— 120 "
Less heat absorbed in decomposing steam
14,511^75.3 = 193
Net heat available in gaseous products =2640-f 0.5tCal
Calorific capacity of gaseous products
= 5.3122 (0.303t-f 0.000027^")
Therefore 5.3122 (0.303t-|-0.000027t^) = 2640-f 0.5t
Whence t = 1910° (9)
This represents the absolute maximum of temperature which
the gaseous products formed at the tuyeres can possess.
(10) If all the moisture were removed from the blast, the
heat available would be:
304 METALLURGICAL CALCULATIONS.
By combustion of 1 kg. carbon = 2430 Calories.
Heat in 4.4685 m' of dry air at 382° C.
= 4.4685X0.313X382 = 574
Heat in 1 kg. of carbon at t° = 0.5t- 120"
Net heat available in gaseous products = 0.5t + 2884 Cal.
Calorific capacity of gaseous products
= 5.3976 (0.303t + 0.000027t=)
Therefore 5.3976 (0.303t + 0.000027t=) = 2884 + 0.5t
Whence t = 2018° (10)
It is to be noted that this is 108° C. = 194° F. higher than
with moist blast ; and while the slag and iron in passing through
this zone of high temperature will not reach this maximum
temperature, yet they would be heated approximately 100° C.
higher when using dry blast, if all other conditions were kept
constant.
N. B. — In working this problem, the metric measurements
and English measures have been purposely used interchange-
ably, in order that readers may understand better that if weights
are taken in pounds and the heat xmit is the pound Calorie
(1° C.) , the same numbers represent a solution in either system.
When volumes are concerned, cubic feet and ounces, or ounce
calories, have practically the same numerical expression as
cubic meters and kilograms or kilogram calories.
CHAPTER V.
THE RATIONALE OF HOT-BLAST AND DRY-BLAST.
Problem 58 (for practice).
The blast furnace of Problem 57 had its blast dried before
using, to the extent of leaving on an average 1.75 grains of
moisture per cubic foot of air proper, at — 5° C, in the air
pumped. The composition of the ore, limestone and coke used
was tmchanged, also that of the pig iron made. The weights
charged per 100 of pig iron were: Ore 177.6, flux 44.4, fuel 77.0,
and the blast used calculates out oxygen 76.5, nitrogen 255.0,
moisture 1.0. Analysis of gases: CO 19.9, CO^* 16, N^ 64.1 per
cent. Product per day 447 tons. Temperature of gases 191°
C, of blast 465° C. Piston displacement (air at - 5° C.) 34,000
cubic f«et per minute. Assume heat in pig iron and slag same
as before, in cooling water 20 per cent, greater per day.
Required. — (1) The volume of gases per 100 kg. of pig Iron
made.
Answer. — Measured dry, 355.9 m'.
(2) A balance sheet of materials entering and leaving the
furnace, per 100 units of pig iron. (See table on page 281.)
(3) The volume of blast per 100 kg. of pig iron.
Answer.— 252.9 m^ at- 5° C.
(4) The efficiency of the blowing plant.
Answer. — 82.3 per cent.
(5) The heat balance sheet of the furnace, per 100 of pig iron.
Developed.
Carbon to CO 92,950 Calories.
Carbon to CO^ 206,955
Heat in hot blast 37,850
Solution of carbon in iron 2,820 "
Formation of slag 4,260
Total 344,835
305
306 METALLURGICAL CALCULATIONS.
Distributed.
Reduction of iron 165,870 Calorics.
Reduction of silicon 7,000
Expulsion of carbonic oxide (CO^) 18,666
Evaporation of moisture of charges 11,342
Heat in waste gases 23,799
Decomposition of moisture of blast 3,225
Heat in slag 29,820
Heat in pig iron • • 32,500
Heat in cooling water 14,922
Lost by radiation and conduction (difi.) . 37,791
344,835
(6) Compare the heat items which are substantially different
for the furnace run by moist and dried blast.
Moist Blast. Dried Blast.
Combustion of C to CO 131,220 92,950
Heat in waste gases 43,836 23,799
Decomposing moisture in blast 14,511 3,225
Loss by radiation and conduction 51,180 37,791
■ It may be noticed, that using moist blast too much carbon
was burnt to CO at the tuyeres; the chief item of economy
with dried blast is the ability to get along with less. The
smaller total amoimt of gases, particularly nitrogen, accounts
for the lower temperature of the waste gases, with dried blast.
The direct saving by reason of decomposition of the moisture
is the smallest item of economy. The reduced losses by radia-
tion £Cnd conduction are mainly because of the faster rate of
running, these losses being nearly constant per day. The ratio
of these losses is found to be 0.74, whereas, the inverse ratio
of the pig iron productions per day was 0.79.
(7) Calculate the carbon burnt at the tuyeres, the propor-
tion of the carbon available thus consumed. Compare these
items with those of the furnace on moist blast.
Moist Blast. Dried Blast.
Carbon burnt at tuyeres.... 75.3 58.05
Total fixed carbon charged.. ... .-. 84.3 67.8
Proportion burnt at tuyeres 89 . 3 85 . 6
Fixed carbon really available 79 . 4 62 . 9
Proportion burnt at tuyeres 94 . 8 92 , 3
RATIONALE OF HOT-BLAST AND DRY-BLAST
307
(In some charcoal furnaces of low heat requirement, i.e.,
with pure ores and fuel, as little as 37 parts of carbon is burned
at the tuyeres per 100 of pig iron produced, which represents,
moreover, only 70 to 75 per cent, of the available fixed carbon
in these furnaces.)
Charges.
Pig Iron.
Slag.
Gases.
(D
FeW 135.7
H^O 17.8
SiO' 17.8
Al'O' 6.3
Fe 95.0
Si 1.0
O 40.7
I-H
H^O 17.8
1
Si02 15.7
AlW 6.3
0 1.1
SiO^ 2.2
CaO 21 . 1
MgO 2.1
CO^ 19.0
SiO^ 2.2
CaO 21.1
MgO 2.1
5
S
5
CO^ 19 0
C 67.8
SiO' 4.2
CaO 4.2
H^O 0.8
C 4.0
■
C 63.8
f:
SiO^ 4.2
CaO 4.2
•s
d
H^O 0 . 8
lO
O* 76.5
N^ 255.0
H^O 1.0
O 76.5
IN
N 255.0
■s
/ H 0.1
m
\ O 0.9
Totals 631.5
100.0
55.8
475.7
(8) The proportion of the whole heat requirement available
in the region of the tuyeres.
Answer. — 53 per cent.
(9) The proportion of the iron in the furnace which is re-
duced by solid carbon from FeO.
Answer. — 23.8 per cent.
(10) The theoretical maximum of temperature at the tuyeres,
Answer,— 1966° C.
308
METALLURGICAL CALCULATIONS.
Hot Blast.
For several centuries blast furnaces were run by charcoal as
fuel and with cold blast. How great the variations in tem-
perature of the cold blast may be, can be inferred from the
experience of a furnace manager in the Urals, Russia, who
noted temperatures of the air nearly 40° C. in the summer
and — 60° C. in the winter. Assuming an average temperature
of 0° C. for unheated blast, burning charcoal to CO, the theo-
retical maximum of temperature before the tuyeres can be
calculated as follows:
Heat generated by combustion
Heat in hot carbon being burnt
Volume of CO and N^ formed
2430 Calories
0.5t- 120
5.3795 cubic meters
2310-l-0.5t
Temperature ^ ^^^^ (0.303 -f-0.000027t)
Whence t = 1678°
This does not mean that the pig iron and slag will be car-
ried to this temperature, any more than if a locomotive could
run alone 90 miles an hour it could therefore pull a train of
cars that fast. The hot gases, CO and N^, are at the start at
this temperature, and as they ascend and come in contact with
the descending iron and slag, these are raised to temperatures
approximating towards, but always lower than, the above.
In fact, the temperature to which the iron and slag is raised
depends on the relative quantities of iron and slag to fuel burnt,
and the speed with which the furnace is worked.
If the blast is heated, its sensible heat is simply added to the
numerator of the above expression. We can easily find out
how much sensible heat the 4.4685 cubic meters of air brings
in, at any temperature desired [Q = 4.4685 (0.303t-f0.000027t=')],
and then solve the quadratic anew. We thus find
Temp, of Blast. Heat in Blast. Theoretical Temp.
0°C.
— Calories
1678° C.
100° '
137
1762° "
200° '
276
1845° "
300° *
417
1929° "
400° '
561
2012° "
500° '
707
2096° "
600° '
866
2180° "
RATIONALE OF HOT-BLAST AND DRY-BLAST
309
Temp, of Blast. Heat in Blast. Theoretical Temp.
700° C. 1007 Calories. 2265° C.
800° " 1160 " 2350° "
900°" 1316 " 2435° "
1000° " 1475 " 2520° "
The heating of the blast thus raises the maximum tempera-
ture available some 85° C. for each 100° C. to which the blast
is heated. It not only increases the temperature available, but
also the number of heat units, thus increasing both the quan-
tity and the intensity of the heating in the tuyere region. Of
these two items of increase, that of the intensity factor is the
most important, since it regulates the rapidity of transfer of
heat to the charge and the efficiency and speed of the smelting
action of the furnace.
Dried Blast.
Each kilogram of water vapor decomposed absorbs 29040 -s-
9 = 3227 Calories, which would not be needed if the 0.67 kilo.
, of carbon thus employed was oxidized by air instead of by
moisture. Per kilogram of carbon oxidized by water vapor,
there is absorbed 58,080-^12 = 4,840 Calories, while this kilo-
gram of carbon can only furnish 2430 Calories in becoming
CO, leaving a net absorption of 2410 Calories per kilogram of
carbon thus burned, against which, however, can be credited
the heat in the kilo, of carbon burned and the sensible heat in
the water vapor itself; the former is 0.5t— 120, and the latter
can be calculated from the volume of thp water vapor, 1.8519
cubic meters. We thus have, per kilogram of carbon thus
oxidized :
Temperature
of Water Vapor.
Heat in Moisture.
Heat in Products.
100°
66 Calories
0.5t - 2460
Calories
200°
137
0.5t- 2393
(
300°
214
0.5t-2316
(
400°
296
0.5t-2234
(
500°
384
0.5t-2146
t
600°
478
0.5t-2042
t
700°
577
0.5t- 1953
i
800°
682
0.5t- 1848
i
900°
792
0.5t-1738
c
1000"
907
0.5t- 1623
t
310 METALLURGICAL CALCULATIONS.
Since the heat in: the carbon burnt (0.5t— 120) can never
equal numerically 1623, it is seen that under no circumstances
can the water vapor do anything but diminish the 'quantity of
heat available at the tuyeres, while the products of its decom-
position CO and HI increase the volume of the products and
so diminish still further the intensity of temperature attainable.
The best way to determine the amount of moisture in the air
is to draw it through a tube containing concentrated sulphuric
acid, measure the quantity of dry air drawn through, and
Weigh the amount of moisture caught by the tube. This gives
the weight of moisture accompanying unit volume of dry air
(not weight of moisture in unit volume of moist air). De-
terminations by the wet and dry bulb thermometers, the whirled
psychrometer, humidity gauges, etc., are none of them so re-,
liable as the above described method, which is direct, simple
and accurate. The results may be obtained in grains per
cubic foot of dry air or milligrams per liter. The best way to
express thfem, for further use, is in ounces avoirdupois per
cubic foot, or kilograms per cubic meter. The first is obtained
by dividing the number of grains by 437.5, the second by divid-
ing the milligrams per liter by 1000. The numbers thus ob-
tained are practically identical in the two systems.
The theoretical temperatures attainable with moist blast of
varying degrees of moisture and heated', to various tempera-
tures are obtained by applying the previously explained prin-
ciples. We have already calculated the temperature obtained
by burning carbon with dry air of various temperatures. We
have also calculated a table of the deficit of heat available pro-,
duced by the entrance of 1.5 kilos, of water vapor (which would
oxidize 1 kilo, of carbon and produce 1.8519 cubic meters of
CO and H^). We are, therefore, prepared to calculate a table
of the maximum temperature attainable using blast of any
degree of humidity heated to any practical temperature. .Before
giving the table we will run through the details of one calcula-
tion, to make clear the method employed.
Illustration. — ^What is the theoretical maximum tempera-
ture using air which carries normally 10 grams of moisture per
cubic meter of dry air, reduced to standard conditions (i.e. j 10
grams per 1.293 kilograms of dry air)", and heated to 500° C?
It takes 3.5275 cubic meters of dry air at standard: condi-
RATIONALE OF HOl -Bi^.iST AXD DRY-BLAST 311
tions to bum 1 kilogram of carbon. If dry, and at 506°-, there
is 2430 Calories generated by combustion, 707 Calories in the
dry air used, 0.5t— 120 Calories in the hot carbon, and the
total heat thus at hand raises the 5.3944 cubic meters of pro-
ducts to the temperature of 2096° C, as is determined by solving
the equation
^ 2430 + 707+ (O.St- 120) _
5.3944 (0.303 + 0.000027t) ~ ^^
As modifiedby the moisture, the 4.4685 cubic meters of dry
air would be accompanied by 44.685 grams of moisture, or
0.044685 kg., which would oxidize two-thirds of its weight of
carbon, or 0.02979 kg. of carbon, which would contribute to the
heat available 0.02979 (0.5t— 2146) Calories, making a con-
tribution of 0.015t— 64 Calories to the numerator of above
equation. The products of combustion will be increased by H^
and CO equal in volume to twice the volume of the moisture,
or 2X0.044685-^0.81 = 0.1102 cubic meters, the mean specific
heat of which goes into the denominator. We then have
3017 + 0.5t + 0.0-15t-64 .^
~ (5.3944 + 0.1102) (0.303 + 0.000027t)
Another m.ethod of solution, not using the previously calcu-
lated tables but based entirely on first' principles, is to base the
whole calculation on the use of 1 cubic meter of dry air, with
its accompanying moisture, as follows:
Oxygen present in 1 m' dry air, = 1,293x3/13 = 0.2984kg.
Oxygen present in the moisture = 0.010X8/9- =0.0089 "
• : ' , Total = 0.3073 "
Carbon consumed = 0.3073X0.75 .' =0.2305"
Volume of moisture = 0.010-^0.81 = 0.0123 m'
Volume of oxygen in dry air = 0.2078
Volume of products from dry air = 1.0000 + 0.2078= 1.2078 "
Volume.of .products from moisture = ,2X0.0123 =0.0246
Total volume of products = 1^2324
Heat of combustion of carbon = 0.2305X2430 = 560 Cal.
Heatin carbon att° = 0.2305X (0.5t- 120) = 0.1152t- 28 "
312 METALLURGICAL CALCULATIONS.
Heat in dry air at 500° = IX [0.303 + 0.000027
(500)] 500 = 158 Cal.
Heat in moisture at 500° = 0.0123 [0.34 + 0.00015
(500)] = 3 "
Heat absorbed in decomposing moisture = 0.0123
X2614 = -32 "
Whence results the equation
_ 0.1152t + 661 ^ o
0 . 2324 (0 . 303 + 0 . 000027t)
By applying either of the two methods of calculation ex-
plained, the temperatures in the following table are obtained:
Theoretical Temperatures Before the Tuyeres
{ Grams per cubic meter of dry air reduced to 0° C.
Moisture. | q^.^^^ p^j. (^y^ic foot of dry air reduced to 0° C.
Grams 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00
Grains. 2.19 4.38 6.57 8.75 10.94 13.13 15.32 17.50
Blast.
40° 1678° 1647° 1615° 1673° 1536° 1507° 1471° 1443°
100° 1723 1692 1666 1627 1587 1548 1526 1496
200° 1807 1776 1751 1712 1673 1636 1612 1584
300° 1892 1861 1837 1800 1760 1725 1700 1673
400° 1976 1945 1921 1885 1846 1813 1786 1760
600° 2061 2030 2007 1970 1933 1902 1874 1848
600° 2146 2115 2093 2056 2020 1991 1962 1936
700° 2232 2201 2178 2144 2108 2080 2060 2025
800° 2318 2287 2264 2227 2195 2169 2138 2114
900° 2404 2373 2361 2313 2282 2257 2226 2203
1000° 2490 2459 2437 2399 2369 2345 2314 2292
The calculations show that the temperature before the tuyeres
may vary as much as 235° C. = 423° F., from change in the
moisture of the air, from dryness to warm air saturated with
moisture.
CHAPTER VI.
PRODUCTION, HEATING AND DRYING OF AIR BLAST.
This is a subject intimately related to the running of iron
blast furnaces, and incidentally related more or less to all
classes of metallurgical operations. The principles involved
are those of physics — mechanical and thermal — and when once
thoroughly understood can be used for the most various classes
of metallurgical problems.
Production of Blast.
There are two different principles upon which air is com-
pressed, represented by the fan and the piston machines. The
first is constant in its operation, the second intermittent, the
first draws in and expels a continuous current of air, the second
draws in a given quantity of air, compresses it and expels it.
Measuring the work done by the difference in static conditions
of the compressed and of the tmcompressed air, the work actu-
ally done is a fixed and calculable quantity, independent of
what type of machine performs it. During the compression,
heat is generated, and the mechanical work done includes the
mechanical equivalent of the heat thus generated. This is
allowed for in the well-known formula for adiabatic com-
pression :
in which
J- = the ratio of the specific heat of air at constant pres-
sure to that at constant volume = 1.408.
Vo = the volume of the uncompressed air.
po = the tension of the uncompressed air.
Pi = the tension of the compressed air.
If we use the. value of 7- = 1.408, the coefiicient — ^ be-
comes 3.45, and ^^^-~ =0.29. If we then use the other di-
r
313
314 METALLURGICAL CALCULATIONS.
mensions in, feet and pounds, the resulting work done will be
in foot-pounds; if we use meters and kilograms, in kilogram-
meters. The element of time does not enter into this equation,
the work actually done is the same for compressing a given
amount of air, and is independent of the time. If I lift a kilo-
gram 1 meter high, the same amount of work is done whether
I lift it in 1 minute or in 1 second, the rate of doing work would
be, however, sixty times as great in the last instance as in the
first, but the actual amotmt of work accomplished is the same
in each case. If, therefore, we use in the formula the volume
of air entering the compressor per minute, the result will give
the work done per minute; if we use the volume per second
we get the work done per second. If we wish to transpose
the work done into horse-power, we bear in mind that a horse-
power is understood in English units as 33,000 foot-pounds of
work done per minute, and in the metric system as 75 kilogram-
meters of work done per second.
In applying the formula we may further notice that — ex-
Po
presses the ratio of compression; that is, by how many times
the tension of the original air is increased. If air at one at-
mosphere tension (such as the air we usually have to breathe)
is compressed to two atmospheres tension, the ratio of com-
pression is 2; if air entered a machine at two atmospheres ten-
sion and was therein compressed to four atmospheres, the
ratio of compression would be likewise 2, and the work done
(for a given quantity of air) the same as before. The effective
pressure, as shown on a gauge, is the difference between these
two tensions: it is not pj. It is highly important that this
point be held clearly in mind. The tension of the uncompressed
air is its barometric pressure, as m-easured against a vacuum;
the tension of the compressed air is likewise its pressure as
measured against a vacuum; the effective pressure of the com-
pressed air is the difference between these two tensions; the
tension of the compressed air is, therefore, (by transposition)
equal to the barometric pressure of the tmcompressed air plus
the effective pressure of the compressed air, i.e., plus the gauge
pressure. Never use the effective pressure of the compressed
air as pi, but always add to it the barometric pressure of the
uncompressed air pg for the correct value of p^.
. PRODUCING HEATING AND DRYING BLAST. 315
Regarding the volume of uncompressed air, V^ is its volume
measured at the pressure P„, i.e., the actual volume of uncom-
pressed air at its actual tension. If more convenient, however,
we may use the volume of this air measured at standard baro-
metric pressure, if we multiply it by p^, the standard pres-
sure. Thus, if we know the volume of tmcompressed air meas-
ured at one atmosphere standard tension, we may use for the
po immediately following it the standard pressure 10.334 (kilos,
per square meter), or 2,120 (pounds per square foot). By the
law of the reciprocal nature of volume and pressure, the pro-
duct thus obtained is the same as would be found by multi-
plying the volume of the same air at any other pressure by
that pressure (Vq Po = Vj Pi = V p^, etc.). Whatever pres-
sure we use in getting the product Vj Po, however, we must
use only the actual tension of the uncompressed air as the
denominator in getting the ratio of compression (PiH-p„).
Problem 59.
A blast furnace requires 2,615 cubic meters of air, measured
at — 5° C, for every metric ton of pig iron made. Assuming
outside barometric pressure 735 millimeters of mercury column,
the efficient pressure of the compressed blast to be 1.0 kilo-
gram per square centimeter (as measured in a blast regulating
reservoir), a mechanical efficiency of the blowing engine of
90 per cent., of delivery of blast 82.3 per cent., and output 447
tons per day:
Required:
' (1) The ratio of compression of the blast.
(2) The piston displacement per ton of pig iron made.
(3) The work of the blowing cylinders,' per ton of iron made.
(4) The gross or indicated horse-power of the steam cylinders.
Solution:
(1) The effective pressure being 1.0 kilogram per square
centimeter, represents ^" „„.X760 = 735 millimeters of mer-
cury. But the uncompressed air is at 735 mm. barometric
tension, therefore, the total tension of the compressed air, pj, is
1470 mm. of m-ercury, and the
Ratio of compression = -=^ = 2.0 (1)
316 METALLURGICAL CALCULATIONS.
(2) Piston displacement per metric ton of pig iron:
2615
0.823
3165 cubic meters. (2)
= 111,560 cubic feet.
(3) The volume displaced must be the. basis of calculating
the power required, since subsequent losses of air (100— 82.3
= 17.7 per cent.) do not affect the work done in the blowing
cylinder. This volume, is 3,165 cubic meters, representing that
volume of uncompressed air at — 5° C. and 735 mm. pressure.
The initial temperature of the air does not affect the work to
be done, and we can either use its volume at 735 mm. tension
or correct this to 760 mm. tension, just as we chose. If we
consider the volume at 735 mm. tension, then the product of
volume and pressure is
Vp = 3,165 X (l0,334 X ^~\ = 31,755,000
If we first change the volume to standard pressure, we have
VoPo = (3,165 X ^) X 10,344 = 31,755,000
The work done in compression, in the blowing cylinder is,
therefore,
W = 3.45 X 3,073 X 10,334 X[2<'-=»- 1]
= 109,554,750X[1. 2225-1]
= 24,326,800 kilogrammeters. (3)
(4) The steam in the steam cylinders does more mechanical
work than would be represented by the compression of air in
the blowing cylinders. In this case, we assume 90 per cent,
mechanical efficiency, or 10 per cent, mechanical loss.. The
gross work of the steam is, therefore,
24,326,800 -h 0.90 = 27,029,800 kilogrammeters,
or, in horse-power,
27,029,900X447
1440X4500
1,866 horse-power.
This amounts to 4.17 h.p. per metric ton of pig iron made
per day.
PRODUCING, HEAriNG AND DRYING BLAST. 317
Measurement of Pressure.
In making calculations of the work done in furnishing blast,
as in the preceding problem, it is important to note how the
pressure of the compressed air is measured. The real pressure
is measured correctly by a pressure gauge only where the air
is comparatively at rest, as on a pressure-equalizing reservoir.
If the pipe communicating with a gauge is connected with a
blast main, in which the velocity of the air is considerable, the
pressure recorded will vary greatly with the direction of the
end of this tube relative to the direction of the air current.
The total pressure of the air in motion is the pressure which it
exerts against the sides of the main, plus the pressure which
has been used in giving it velocity. If the pressure gauge
tube is cut off at right angles to the flow of air in the main,
the pressure recorded is even less than the actual static pres-
sure of the air against the sides of the main (because of suction
effect), and does not include at all the pressure represented
by the velocity. The only way to measure correctly in a
main the total pressure which has been impressed upon the
blast is to bend the end of the pressure tube until it is parallel
with the axis of the main and faces the current of air. The
gauge then records the static pressure, plus the velocity head,
and gives the proper pressure to use in calculating the work done
by the blowing engine. However, it is still preferable to put
the gauge on a blast reservoir, if there is one, where the air
is nearly at rest and velocity head approximates zero — for
the reason that the velocity of air passing through a tube is
greatest in the center and least against the walls, and it is diffi-
cult to place the pressure tube so as to measure the mean ve-
locity. An approximation ■ to the mean value is found by
placing the end of the pressure tube not in the center but at
one-third of the radius of the main from the center.
Indicator Cards.
If the blowing engine cylinder can be tested by pressure
indicator cards, then a different method of obtaining the work
of producing the blast is. furnished. The integration of the
diagram gives the mean pressure on the piston during the
stroke. Expressing this in kilograms per square meter or
pounds per square foot, and multiplying by the area of the
318 METALLURGICAL CALCULATIONS.
piston and the length of the cylinder, we get the work done
per stroke; again multiplying by the number of strokes per
minute we have the work done per minute, from which is readily
obtained the horse-power. These operations are usually com-
bined in the formula:
Work = PXLXAXN
If we observe that LxAxN is the piston displacement per
minute, the formula becomes:
Work = PX Piston displacement per minute,
in which, it must not be forgotten, P represents mean pres-
sure on the piston during the stroke, and is not to be con-
founded with gauge pressure of the compressed air. Such a
formula is, therefore, totally inapplicable to finding the work
done by a fan or rotary blower, in which only the final pressure
of the compressed air is known. The mistake of so applying
it is often made. When only the final pressure of the compressed
air is known, the formula for adiabatic compression is the only
correct one to use.
It may not be useless to call attention to the fact that when
using the formula for adiabatic compression, the raising of the
ratio of compression to the 0.29 power has to be done by logar-
ithms:
'*[(?-n = "^(i;).x''-^
If a table of logarithms is not at hand, a satisfactory ap-
proximation may be made by taking the cube root of the ratio,
since
Heating of the Blast.
Air blast is commonly heated either continuously, by direct
transmission of heat through metal or earthenware pipes, or
discontinuously by the heating up of fire-brick surfaces, which
are subsequently cooled by the blast. It is not within the prov-
ince of these calculations to enter into a description of the
various types of such hot-blast stoves, but to indicate the
principles upon which the metallurgist can base calculations as
to the efficiency of such stoves, and thus be prepared to find
PRODUCING, HEATING AND DRYING BLAST. 319
out which do the best work, and wherein lie the principal ad-
vantages or disadvantages of each type.
The efficiency of a hot-blast stove is measured by the ratio
of the heat imparted to the blast to that contained ia the air
and fuel used and generated by combustion. It is a furnace
whose useful effect is the heat imparted to the air, and all other
items of heat distribution are more or less necessary losses.
The problem is simplest when the stove is a recuperator (con-
tinuous type), using solid fuel. Then the item of heat genera-
tion is perfectly definite, since the amount of fuel used in a
given time can be definitely determined. When the hot-blast
stove receives gaseous fuel, however, from a blast furnace, the
amount of gas received by the stoves is usually a very tmcertain
quantity, since only part of all the gas produced by the furnace
is used by the stoves, and the question of what fraction is thus
used is difficult to determine. In such cases, the usual method
of comparing the sizes of the pipes leading gas to the stoves
and those carrying gas to boilers, etc., affords but a very tm-
certain determination, since the draft and consequent velocity
of the gas may be very different in the two sets of pipes. In
such cases, knowing the total volume of gas produced by the
furnace, not only the relative sizes of the hot-blast stove pipes
should be noted, but also the relative velocities of the gas
currents in the two sets of pipes. Another method is to de-
termine the quantity of chimney gas passing away from the
stoves into the chimney flues, by measuring the size of the
chimney flue, temperature of the gases and average velocity;
given in addition the chemical analyses of the chimney gas and
of the furnace gas, the quantity of the latter being used can be
calculated, using the carbon in the two gases as the basis of
calculation.
Problem 60.
Statement: Air for drying peat in a kiln is heated from
0° C. to 150° C. by an iron pipe stove, the latter consuming
^ dried peat, whose ultimate analysis is:
Carbon 49.70 per cent.
Hydrogen 5.33
Oxgyen 30,76
Nitrogen 1.01
Ash 13.23 "
320 METALLURGICAL CALCULATIONS.
The calorific power of this peat is 4249, and by the com-
bustion of 92.5 kilograms, 5122 cubic meters of air is heated
to the required temperature. The products of combustion pass
away from the stove at 200° C, and contain (analyzed dry)
14.8 per cent, of CO^, and no CO or CH^ or other gas containing
carbon.
Required:
(1) The heat generated in the stove.
(2) The heat in the hot air, and the net efficiency of the
stove.
(3) The volume and composition of the products of com-
bustion in the stove.
(4) The heat carried out in the chimney of the stove.
(5) The heat lost by radiation and combustion.
(6) The excess of air used in burning the peat.
(1) Heat generated in the stove:
4249X92.5 = 393,033 Calories (1)
(2) Heiat imparted to the air:
[0.303 + 0.000027 (150)] X 150 X
5122 = 235,907
Net efficiency = 60.0 per cent. (2)
(3) Volume of products of combustion:
Weight of carbon in fuel burnt— 92.5
X 0.4970 = 45.97 kilos.
Volume of CO^ thus produced— 45.97
-f-0.54 = 85.13 cub. meters
Volume of (dry) gas produced— 85.13
-hO.148 = 575.20 "
Volume of N^ and 0^ in products —
575.2-85.13 = 490.07 "
Volume of water vapor in products —
92.5 X 0.0533 X 9 H- 0.81 = 54.78 "
Percentage composition of products of combustion:
Moist. Dried.
CO^" 13.5 14.8
N^andO'' 77.8 85.2
H^'0 8.7
PRODUCING, HEATING AND DRYING BLAST. 321
To separate the nitrogen and oxygen would necessitate con-
siderable calculation, and is not required just here, because
the two gases have the same heat capacity per cubic meter,
and therefore can thermally be reckoned together:,
(4) Heat in the chimney gases at 200°:
CO^ 85.13 m'x0.4140 = 35.24
N^ and OM90.07m'X 0.3084 = 151.14
WO 54.78 m'X 0.3700 = 20.27
206.65X200 = 41,330 Cal.
Proportion thus lost = 10.5 P. C. (4)
(5) Loss by radiation and conduction
100- (60.0 + 10.5) =29.5P. C. (5)
(6) The separation of N" and 0' in the products is not easy.
It is best based on the consideration that part of these con-
sists of the nitrogen of the air which was necessary for com-
bustion (plus the nitrogen of the fuel) and of the excess air.
The first can be calculated, and thus the latter obtained by
difference, and then the percentage of air in excess determined.
Oxygen necessary for combustion:
C to CO^ = 45.97X^1 = 122.59 kilos.
H to H^O = 4.93X8 = 39.44 "
Oxygen present in peat 92.5X0.3076
Oxygen to be supplied
Nitrogen accompanying
Air necessary
Nitrogen from coal, 92.5X0.0101
Total nitrogen from coal and necessary air
Volume = 446.2-^1.26
Total N'' and 0=* in products
162.03
((
=
28.45
(C
__
133.58
(<
=
445.27
f(
=
578.85
((
=
447.7
m"
=
0.93 kilos.
=
446.2
It
=
354.1
m'
=
490.1
((
Therefore, excess air = 136.0
P.I
(6)
Percentage of excess air . .-"- = 0.303 = 30.3 P. C.
a22 METALLURGICAL CALCULATIONS.
Problem 61.
A blast furnace has three hot-blast stoves, two of which are
always on gas and one on blast. Per long ton of pig iron pro-
duced there issues from the furnace (reduced to standard tem-
perature and pressure) :
Nitrogen 81,763 cubic feet
Carbon monoxide 25,383
Carbon di-oxide 20,409 "
Water vapor 8,230 "
For the same quantity of pig iron made, 92,330 cubic feet
(standard conditions) of air is heated in the stoves from —5°
C. to 465° C. The hot gases reach the stoves at 175° C, are
there burned by 10 per cent, excess of air at 0°, and the chimney
gases resulting (assume perfect combustion) pass out of the
stoves at 120° C.
Required:
(1) The net efficiency of the stoves, assuming they receive
25, 30, 35, 40, 45 or 50 per cent, of the gas produced by the
furnace.
(2) Assuming that each stove radiates and loses to the
ground one-third as much heat as the blast furnace itself (the
heat balance sheet of the furnace showed 846,518 pound Cal-
ories thus lost per long ton of pig iron produced), what pro-
portion of the whole gas goes to the hot-blast stoves, and what
is the net efficiency of the stoves?
Solution:
(1) We will first calculate the efficiency of the stoves, as-
suming that they received all the gas produced, from which
datum can then be obtained the efficiency, supposing any as-
sumed fraction of the gas goes to the stoves.
Heat in the blast (-5° to 465°):
92,330 X [0.303 -1-0.000027 (465 -5)]X[465 -( -5)]
= 92,330 X 0.31542X470 = 13,687,683 ounce cal.
= 855,480 pound Cal.
Sensible heat in the hot gases (0° to 175°) :
N^ and CO 107,146X0.3077 = 32,969
C0=' 20,409X0.4085 = 8,337
WO 8,230X0.3763 = 3,097
Heat capacity per 1° = 44,403 ounce cal.
= 2,775 pound CaL
Heat capacity per 175.. 2,775X175° = .485,625 "
PRODUCING, HEATING AND DRYING BLAST. 323
Heat generated by combustion:
CO to 00^25,383X3062
= 77,722,746 ounce cal.
4,857,672 pound Cal.
= 5,343,297 "
If the gas were all used in the stoves the efficiency of the
latter would be only
Total heat available
855,480
5,343,297 '
0.160 = 16.0 per cent.
With smaller proportions of the whole gas used the efficiency
of the stoves calculates out as follows:
Using 50 per cent, of the gas Efficiency 32.0 per cent.
45
40
35
30
25
16
36.0
40.0
45.7
53.3
64.0
100.0
(1)
All that the above analysis tells us is, that certainly over
16 per cent, of the gas produced by the furnace must be used
by the stoves ; but if we can deduce any probable value for the
percentage of the gas actually used, such as by measuring the
several gas mains and the velocities of the gas in each, we can
then reckon out the value of the efficiency of the stoves, with
about the same degree of probability. Blast furnaces use
from 33 to 60 per cent, of the gas, they produce in the stoves.
If we assume 50 per cent., in this case, the efficiency of the
stoves would be 32 per cent.
(2) There is another way of solving the problem, which is
to either measure, calculate or assume the heat lost by radia-
tion and conduction from the stoves; adding to this the heat
going out in the products of combustion, and the net heat in
hot blast, the sum is the total heat which has been brought
into and generated within the stoves. But, all the gas would
bring in and generate in the stoves 5,343,297 Calories; we can,
therefore, find easily what proportion of all the gas is being
used in the stoves. In requirement (2) we are told to as-
sume that the three stoves lost by radiation and conduction
324
METALLURGICAL CALCULATIONS.
pig iron made, an amount
furnace itself. The heat
thus calculated:
846,518 pound Calories per long ton of
equal to that similarly lost by the
in the chimney gases, at 120°. can be
Oxygen required l-^ CO)
Air required
Excess of air used
Nitrogen of required air
Nitrogen already in gas
Nitrogen in these two items
Chimney products:
CO'
H^'O
w ■
Excess air
Heat in chimney gases :
CO^ 45,792'X 0.3964
H^O 8,230X0.3580
N^ and air 136,190X0.3064
Heat capacity per 1°
Heat capacity per 120°
If all the furnace gas were burnt in the stoves, 471,122 pound
Calories would go up the stove chimneys. But, since only a
part of the gas is so used, only a fraction of this amount of
heat is lost to the chimneys. If we call X the proportion of
the gases used, then 471,122 X will represent the chimney loss,
and the total heat requirements of the stoves will be:
Heat in air blast = 855,480 lb. Cal.
Heat in chimney gases = 471,122 X "
Radiation and conduction = 846,518 "
=
12,692 cubic feet
=
61,017
(( It
=
6,102
(( n
=
48,325
tt tt
=
81,763
It ((
=
130,088
tt It
=
45,792
tt tt
=
8,230
tt tt
=
130,088
tt tt
=
6.102
tt tt
_
18,152 (
Dtince cal.
=
2,946
tt tt
=
41,729
tt tt
=
62,827
t( ((
=
3,927 pound Cal,
=
471.122
(1 tt
Total = 1,701,998 + 471,122 X lb. Cal.
The proportion of the total gases needed to supply this, X, is
1,701,998 + 471,122 X
= X
whence
5,343,297
X = 0.349 = 34.9 per cent.
(2)
PRODUCING, HEATING AND DRYING BLAST. 325
and the net efficiency of the stoves is
855,480 855,480
5,343,297X0.349 1,864,812
= 45.9 per cent. (2)
Arithmetically, the finding of X can be simplified, perhaps,
by considering that the chimney loss represents in any case
471,122-^5,343,297 = 0.088 = 8.8 per cent, of the total heat
received by the stoves, leaving 91.2 per cent, to be applied to
heating the blast and for radiation and conduction loss. The
last two items, however, must amount to 1,701,998 Calories,
and, therefore, the total heat requirement of the stoves is
1,701,998
0.912
requiring
1,866,200 pound Calories,
i|||5= 0.349. 34.9 per cent. (2)
of all the gas produced by the furnace.
In the above solution the only uncertain factor in the calcu-
lation is the radiation and conduction loss from the stoves, and
this uncertainty does not largely affect the reliability of the
result obtained, allowing that we have assumed an approxi-
mately correct value for this loss. All uncertainty could be .
dispelled, however, were the radiation and conduction losses
measured directly. This could be accomplished by finding
experimentally the temperature of the outside shells of the
stoves, and calculating the external losses of heat by the laws
of radiation and condiiction; but in order to do this satisfac-
torily, it would be necessary to divide the shells of the stoves
into zones, and determine the temperature and calculate the
losses from each zone separately — a rather long operation, but
one worth doing.
Drying Air Blast.
The advantage of dried blast has been already discussed at
length in these calculations. The advantage is due primarily
to the higher temperature obtained when the moisture has been
removed.
The means adopted commercially for dr)ring the air have
326 METALLURGICAL CALCULATIONS.
been those of refrigerating the uncompressed air, before its
entrance into the blowing cylinders. This has the great ad-
vantage of furnishing the cylinders with cold air, and, there-
fore, of greatly increasing their blowing capacity, since the
weight or quantity of air blown is proportional to the absolute
temperature of the air entering the cylinders.
Illustration. Outside air being at 30° C, what increase in
the amount of air furnished by blowing engines will result if
the temperature of the air is artificially reduced to —5° C?
How much slower can the engines be driven, in the second
case, in order to furnish the same weight of air as before?
The two temperatures are respectively 273 + 30 and 273—5
absolute, that is, 303 and 268. The engines, if run at uniform
speed, would furnish 303^-268 = 1.13 times as much air in
the second case, or 13 per cent. more. If the engines were
slowed down to 268 -v- 303 = 0.884 of their former speed they
would furnish the same amount of air; that is, they could be
run 11.6 per cent, slower. In fact, they could be rtm more
than 11.6 per cent, slower in the second case, and yet supply
the same quantity of air, because at the 'slower running the
delivery efficiency is somewhat higher.
The disadvantage of cooling the uncompressed blast is that
it must be cooled much more, to eliminate from it a given per-
centage of its moisture, than if it were first compressed, and
to obtain nearly dry air the moisture must be practically frozen
out.
Illustration: Air at 30° C. and 85 per cent, humidity is to
be cooled until 95 per cent, of its moisture is eliminated, with-
out compression; to what temperature must it be cooled?
From the tables of aqueous tension, we find that the maxi-
mum tension which aqueous vapor can exert at 30° C. is 31.5
31.5
millimeters, which means practically that ' of a cubic
meter of moisture accompanies „„ of a cubic meter of air
proper. If the humidity is 85 per cent., then this same quantity
of air is accompanied by
31 5
-=^X0.85 = 0.0352 cubic meters
7o0
PRODUCING, HEATING AND DRYING BLAST. 327
of moisture, or, per cubic meter of air measured dry —
0.0352-T-^^ = 0.0368 cubic meters.
7dU
If 95 per cent, of this is removed by cooling, then the mois-
ture left, per cubic meter of air measured dry, is
0.0368X0.05 = 0.00184 cubic meters.
And the relative tensions of air and moisture, to attain this
dryness, must have been reduced to
1.0000:0.00184.
Since the sum of these tensions is always 760 mm. in the un-
compressed air, the actual tensions of air and moisture will be
758.6:1.4,
that is, the temperature must be reduced until the moisture
present can exert only, 1.4 mm. pressure. This, on examining
the tables of tension of aqueous vapor is found to be less than
0° C, in fact -15° C.
Problem 62.
Air at 30° C, carrying 85 per cent, of its possible amount' of
moisture, is cooled to 0° C, and the moisture condensed to
liquid water at that temperature. Barometer 760 mm.
Required:
(1) The percentage of the moisture condensed.
(2) The amount of heat to be abstracted from each cubic
meter of original moist air (refrigerating effect).
(3) The percentage of moisture which would be condensed
if the temperature were reduced to —6° C.
(4) The total heat to be abstracted, per cubic meter of orig-
inal air, in the latter case.
Solution :
(1) From the preceding illustration we can take the figures
that in the moist air taken, each cubic meter of air proper is
accompanied by 0.0368 cubic meter of moisture. In the cooled
328 METALLURGICAL CALCULATIONS.
air at 0°, the relative volumes of air proper and residual moisture
will be as their relative tensions, viz.:
755.4:4.6
or
1:0.0061.
The moisture accompanying the given quantity of air is, there-
fore, reduced from 0.0368 to 0.0061, showing a condensation
of 0.0307, equal to
^^ = 0.834 = 83.4 per cent. (1)
(2) The air at 30° contains, as before figured out, air proper
and moisture in the proportions
1:0.0368.
or, in 1 cubic meter of moist air, in the proportions
0.9645:0.0355.
We have, therefore, to calculate the heat to be extracted from
0.9645 cubic meter of air proper, falling 30° to 0°, and from
0.0355 cubic meter of water vapor, falling 30° to 0° and 83.4
per cent, of the latter condensing to liquid at 0°. The figures
are, remembering that the volumes heretofore handled are
at 30°:
Air proper:
273
0.9645X303X0.3038x30 = 7.920 Calories
Moisture, if all uncondensed:
273
0.0355X3^X0.3445x30 = 0.332 "
Heat of condensation:
273
0.0355X0.834X3^X0.81X606.5 = 13.105 "
Total = 21.357 " (2)
(3) If the temperature were reduced to —5° C, the tension
of the residual moisture would be 3.4 mm. of mercury, and
PRODUCING, HEATING AND DRYING BLAST. 329
that of the air proper 756.6 mm., their relative volumes would
be in the same ratio, viz.:
756.6:3.4
or
1:0.0045.
showing that out of 0.0368 cubic meter of moisture originally-
accompanying 1 cubic meter of air proper, 0,0323 had been
condensed, or
n 032*?
^-:^ = 0.878 = 87.8 per cent. (3)
(4) The condensation is seen to be 87.8—83.4, or 4.4 per
cent, more, if the air is cooled to —5° C. A considerably larger
amount of refrigeration, however, is required in this case,
because the moisture would all be frozen. The 1 cubic meter
of moist air at 30°, containing, as before calculated, 0.9645
cubic meter of air proper and 0.0355 cubic meter of moisture,
would have 87.8 per cent, of the latter condensed to ice, or
0.03117 cubic meter, and, therefore, 0.00433 cubic meter re-
maining uncondensed.
Air proper, 30° to -5° C:
0. 9645 Xip^X 0.3037X35 = 9.237 Calories
Moisture, if all uncondensed, 30° to —5° C. :
97^
0.0355X^X0.3438X35 . = 0.385 "
Condensation of 0.03177 m' to liq. at - 5° C. :
0. 03177 X 1^X0.81X605 = 14.025 "
Freezing of same, at — 5° G. :
0.03177x|§X0.8lX80 = 1-855 "
= 25.502 " (4)
The conclusion is, that an increased condensation of 4.4 per
cent, has been obtained by an increase in the refrigerating
requirement of 25.502-21.357 = 4.145 Calories, or nearly
330 METALLURGICAL CALCULATIONS.
20 per cent. Another way of stating the comparison, is that
when not freezing the moisture condensed, about 4 per cent,
of the moisture was condensed from each cubic meter of air
for one Calorie of refrigeration, whereas, the removal of ad-
ditional moisture by cooling below zero requires nearly one
Calorie refrigeration for each additional per cent, of moisture
elin:iinated.
A practical 'conclusion is, that the expense of refrigeration
might easily be justified down to 0° C, and yet be unjustified
by the results when cooling below 0°.
Mr. James Gayley has, I believe, patented the scheme of
refrigerating the air in two stages; first, nearly to zero, re-
moving the moisture thus condensed as liquid, and then cooling
the nearly dry air further, eliminating more moisture as ice.
In this manner, the amount of refrigeration required is reduced
by the latent heat of solidification of the larger part of the
moisture, and cooling below zero becomes profitable. The
saving in the above example would be 80 Calories per kilogram
on all the moisture condensed at 0°, viz.:
0.0216X80 = 1.728 Calories,
cutting down the refrigeration required from 25.502 to 23.774
Calories, that is, enabling 4.4 per cent, additional drying to be
produced for 2.4 Calories additional refrigeration. This scheme
of Mr. Gayley is fovinded on sound scientific as well as prac-
tical considerations.
The idea of cooling the compressed blast by moderately
cool water, recently proposed,- is also a very practical idea
and founded on sound scientific principles. At a given tem-
perature moisture cannot exert more than a maximum vapor
tension. It follows that if we start with air saturated with
moisture, at a given temperature, and compress it to double its
initial tension, keeping its temperature constant, about half of
its moisture must condense out. If, at the same time, it is
artificially cooled, then more than half of its moisture will
condense as liquid.
If we start with air not saturated with moisture, the com-
pression, temperature being constant, will increase the tension
of the moisture present until the air becomes saturated, after
which increased pressure will cause condensation.
PRODUCING, HEATING AND DRYING BLAST. 331
Illustration: Air at 30° C, 85 per cent, saturated with mois-
ture, is compressed. At what effective pressure does it become
saturated with moisture, temperature remaining constant?
The moisture present is exerting 85 per cent, of its maximum
vapor tension at this temperature. Therefore, the tension
must be increased in the ratio of 85 to 100, to make the air
saturated, viz.: in the ratio 1 to 1.177. The effective pressure
necessary to be applied is, therefore, 1.177 — 1.000 = 0.177
atmospheres (2.6 pounds per square inch).
Illustration: If air at 30° C, 85 per cent, saturated with
moisture, is compressed to one atmosphere effective pressure
and its temperature kept constant, what proportion of its
moisture will condense?
Before compression, the tension of the moisture being 31.5 X
0.85 = 26.8 mm., the relative volumes of air proper and mois-
ture are as
733.2 : 26.8
or, as
1 : 0.0367
After compression, the tension am the mixture being two
atmospheres (1520 mm.), and the tension of the uncondensed
moisture being it maximum, or 31.5 mm., the relative volumes
of air and uncondensed moisture will be as
1488.5 : 31.5,
or, as
1 : 0.0212.
The proportion of the original moisture remaining uncondensed
is therefore,
^1^ = 0.578 = 57.8 per cent.
and the amotint condensed out = 42.2 per cent.
Problem 63.
Air at 30° C, and 85 per cent, saturated with moisture is
compressed to one atmosphere effective pressure (760 mm. of
mercury), and simultaneously cooled by river water to 10° C.
Barometer 730 mm.
Required:
(1) The percentage of the original moisture, condensed.
332 METALLURGICAL CALCULATIONS.
(2) The weight of moisture remaining in the air, expressed
in grams per cubic meter of dry air at standard conditions
(i.e., per 1.293 kilograms of air proper).
Solution: (1)
Tension of uncompressed moist air = 730 . 0 mm.
Tension of nioisture present 31.5X0.85 = 26.8 "
Tension of air proper, tui compressed = 703.2 "
Relative volumes of air proper and moisture = 1 : 0.0381
Tension of compressed moist air 730 + 760 =1490.0 "
Tension of uncondensed moisture (maximum
tension at 10°) = 9.1 "
Tension of air proper, when compressed = 1480.9 "
Relative volumes of air proper and uncondensed
moisture = 1 : 0.0061
Proportion of moisture condensed:
0.0381-0.0061 ,.oA oAn + /in
__ = 0.84 = 8.40 per cent. (1)
(2) The relative volumes of air proper and uncondensed
moisture are, as found above,
1 : 0.0061,
And the relative specific gravities of air proper and moisture
are as the standard weights of 1 cubic meter of each, viz.: as
1.293 : 0.81.
It follows, therefore, that 1.293 kilos, of air proper (1 cubic
meter at standard conditions) must be accompanied by,
0.0061X0.81 = 0.0049 kg. of moisture,
or = 4.9 grams. (2)
The original moist air contained, similarly considered,
0.0381X0.81 = 0.0309 kg.
= 30.9 grams.
CHAPTER VII.
THE BESSEMER PROCESS. '
The outlines of this famous process are known to every
educated person; the mechanical and most of the chemical
details are familiar to most technologists; the exact way to run
the converter is the source of income to himdreds of superin-"
tendents of works, and yet the quantitative side of the chemical
and physical operations involved is mastered by very few.
To state the case briefly, melted pig iron is put into the con-
verter, numerous air jets are blown through, the impurities of
the iron — carbon, silicon, manganese and, in a special case,
phosphorus — oxidize relatively faster than the iron, and the
final product is usually nearly pure iron. This is recarburized
to steel by spiegel-eisen. During the blow very little free
oxygen escapes from the converter, and the gases produced are
principally nitrogen, carbon monoxide and some carbon di-
oxide, while some hydrogen may come from the decomposition
of the moisture of the air. The silicon, manganese, phosphorus
and iron form silica, manganous oxide (MnO) principally, fer-
rous oxide (FeO) principally, phosphorus pentoxide, P^O',
which go into the slag, while a little Fe^'OS Mn^O* and SiO^
escape as fume.
The applications of calculations to this process are numerous
and important. They include such subjects as the amount of
air theoretically required per ton of iron, the dimen.sions and
power of the blowing engines, the weight of slag produced, the
balance sheet of materials, the balance sheet of heat evolved
and distributed, the radiation losses, the discussion of the
efficiency of the various impurities as heating agents in the
process.
Air Required,
Basing our calculations on the analysis of the pig Iron used,
and assuming it to be blown to pure iron, we must next as-
sume the probable loss of iron itself in the blow. This varies
333
334 METALLURGICAL CALCULATIONS.
considerably, from 1 to 10 per cent, on the pig iron used in
ordinary practice, but as much as 20 to 25 per cent, in some
carelessly run " Baby " Bessemers in steel-casting foimdries.
The silicon all oxidizes to SiO^; iron mostly to FeO, and a
small part, say not over one-tenth, to Fe^O'; manganese mostly
to MnO, a small part, up to one-fourth, may form Mn^O^;
phosphorus forms only P^O'; carbon bums mostly to CO, but
from one-fifth, at times nearly one-half, bums to CO^. When
all the calculated oxygen has been foimd, the blast to contain
it can be calculated, if it is assumed that no free oxygen es-
capes from the converter; at times, however, up to one-third
of all the oxygen blown in may thus escape, but this is very
exceptional, ordinarily less than one-fifth thus escapes, and
often none at all.
Problem 64.
Pig iron containing 3.10 per cent, carbon, 0.98 silicon, 0.40
manganese, 0.101 phosphorus and 0.06 sulphur is blown in an
acid-lined converter, to metal practically free from carbon,
silicon and manganese, but no sulphur or phosphorus is elimi-
nated. To get the minimum and the maximum amounts of
air which could be needed, make the following assumptions:
Case 1. Case 2.
Per cent, of iron lost by oxidation 1 15
Proportion of iron oxidized to Fe^'O* none one-tenth
Proportion of Mn oxidized to Mn^O' none one-fifth
Proportion of C oxidized to CO^ one-fifth one-half
Proportion of 0^ escaping in the gases none one-third
Requirement: (1) Find the weight of dry air needed per
metric ton of pig iron blown, in each case, and its volu;ne at
0° C. Express the results also in povinds and cubic feet per
ton of 2,000 pounds.
Solution :
Case 1.
Oxygen needed per 1,000 kg. of pig iron:
•JO
C to CO^ 6.2 kg.xg = 16.53 kg.
CtoCO 24.8 " xj|= 33.07 "
THE BESSEMER PROCESS. 335
SitoSiO^ 9.8 " x|= 11.20kg.
Mn to MnO 4.O " X— = 1.16 "
55
FetoFeO lO.o " x^ = 2.86 "
00
64.82 "'
N^ accompanying = 216.07 "
Air needed = 280.89 "
Volume at 0° C = 217.2 m'
Volume needed per 1000 oz. Av = 217.2 ft'
Volume needed per 2000 lbs. Av = 6,950 ft*
Case 2.
Oxygen needed per 1000 kg. of pig iron:
C to CO' 15.5 kg.x|| = 41.33 kg.
CtoCO 15.5 " X^= 20.67 "
SitoSiO' 9.8 " X^= 11.20 "
MntoMnO 3.2 " X^ = 0.93 "
48
Mn to Mn'O* 0.8 " X y^= 0.34 "
FetoFeO 135.0 "X^ = 38.57 "
56
Fe to Fe^O» 15.0 " X~^= 6.43' "
0' unused (one-half sum of above) = 59.73 "
179.20 "
N^ accompanying = 597.33 "
Air used = 776.53 "
Volume at 0° C = 600 m'
Volume used per 1000 oz. Av = 600 ft*
Volume used per 2000 lbs. Av = 19,200 ft'
336 METALLURGICAL CALCULATIONS.
For temperatures of the air above 0° C, a corresponding
increase in the volume used would be found. Since this is net
air received by. the converter an allowance for loss of 10 to
25 per cent, (in exceptional cases 50 per cent.) would be needed
to get the piston displacement of the blowing engines. The
above figures are the maximum and minimum for this quality
of pig iron only, blown in an acid-lined converter; other quali-
ties of pig iron might require a little more or less, and if blown
in a basic-lined converter considerably more, to oxidize the
phosphorus. The detailed calculations can be made in each
specific case.
Air Received.
The converse of the preceding proposition is to take an
actual case, in a Bessemer converter, and to calculate how
much air is being received. This will serve as a check on the
blowing engines, since the volume received, divided by the
piston displacement, gives the volume efficiency of the blowing
plant. To make the calculation we need to know the weight
and analysis of the pig iron and the analysis of the blown metal,
in order to find the weights of impurities oxidized, also the
average composition of the escaping gases, to find the propor-
tion of carbon burning to CO^ and of unused oxygen; also the
composition of the slag, to get therefrom the weight of iron
oxidized and the weight of slag, if practicable, but this can
sometimes be calculated; also the weight and composition of
the fume, if it is considerable. The temperature of the air
entering the blowing cylinders its hygrometric condition and
the barometric pressure should also be noted.
Problem 65.
At the South Chicago works of the Illinois Steel Co. (see
paper by Howe, in Trans. American Institute of Mining Engi-
neers, XIX. [1890-91], p. 1127), the charge weighed 22,500
poimds, and contained 2.98 per cent, carbon, 0.94 silicon, 0.43
manganese, 0.10 phosphorus, and 0.06 sulphur. After blowing
9 minutes 10 seconds the bath contained 0.04 per cent, carbon,
0.02 silicon, 0.01 manganese, 0.11 phosphorus and 0.06 sulphur.
The slag formed contained 63.56 per cent, silica, 3.01 alumina,
21.39 FeO, 2.63 YeO\ 8.88 MnO, 0.90 CaO and 0.36 MgO, of
THE BESSEMER PROCESS. 337
which the APO', Cal and MgO and part of the SiO' come from
the lining. The gases, analyzed dry, show an average com-
position during the blow of
C0^ 5.20 per cent.
CO 19.91
IP 1.39
N^ 73.50
and were free from fume.
The piston displacement during the. blow was 190,406 cubic
feet, air in engine room 36° C, humidity 50 per cent., barometer
756 millimeters.
Requirements: (1) The weight of oxygen acting on the bath
during the blow, and the theoretical volume of air at standard
conditions to which this would correspond, per 2,000 pounds
of metal blown.
(2) The volume of moist air at the conditions of the engine
room, received by the converter during the blow, and the vol-
ume efificiency of the blowing machine and connections.
(3) The weight of slag produced and the loss in weight of
the lining by corrosion during the blow.
Solution: (1) The percentages of impurities left in the bath
are so small that we can take them as equivalent to the same
percentages reckoned on the original weight of the bath If
they had been larger it would be necessary to assume an ap-
proximate loss of iron during the blow, find the final weight
of the bath and reckon the percentages on this revised weight.
Making this assumption, we know at once the weights of
carbon, silicon and manganese oxidized, but we do not know
the weight of iron lost. That follows, however, from a con-
sideration of the slag, for the manganese and iron in the slag
are derived only from the metallic bath, and the analysis of the
slag practically gives us the relation between the weights of
manganese and iron in it; since we know the weight of the
former oxidized, the weight of iron lost can be calculated.
Thus the slag contains :
MnO 8.88 per cent = 6.88 per cent. Mn
FeO 21.39 " = 16.64 " Fe as FeO
Fe='0= 2.63 " = 1.84 " Fe as Fe='0»
338 METALLURGICAL CALCULATIONS.
The loss of manganese being 0.42 per cent. == 94.5 pounds, the
loss of iron is :
94.5 X-^^ = 228.6 pounds Fe as FeO
O.oo
1 84
94.5 X^^ = 25.3 pounds Fe as Fe='0»
D.Oo
The remaining item still undetermined is the weight of carbon
oxidizing to CO^ and to CO. The gas analysis shows 5.20 per
cent. CO^ to 19.91 per cent. CO, and _since equal volumes of
each of these gases contain equal weights of carbon, it follows
5.20
that " of the total carbon is present in the gas as CO^,
and the rest as CO. Since the total carbon oxidized is 22,500 X
0.0294 = 661.5 pounds, we have
520
661.5 X;^ = 137.0 pounds C burning to CO''
aoXX
= 524.5 " " CO
Weight of oxygen absorbed by the bath :
C to CO^ 137.0 pounds X 8/3 = 365.3 pounds
CtoCO 524.5
SitoSiO=' 207.0
Mn to MnO 94.5
FetoFeO 228.6
FetoFe^'O' 25.3
X 4/3 = 699.3
X 32/28 = 236.6
X 16/55 = 27.5
X 16/56 = 65.3
X 48/112 = 10.8
1404.8
N^ corresponding. = 4682.7
Dry air corresponding = 6087.5 "
Volume at 0° C = 75,483 ft'
Weight of 0== per 2000 pounds = 124.9 Ibs.(l)
Volume of air per 2000 poimds = 6,710 ft'(l)
(This result is for comparison with data of Problem 64).
THE BESSEMER PROCESS. 339
(2) The nitrogen in the gases can be obtained from its vol-
ume relation to the carbon, and from this we can calculate the
real volume of blast used.
Weight of carbon in 1 cubic foot of gases :
(0.0520 + 0.1991) X0.54 = 0.1356 oz. Av.
Volume of gases produced at standard conditions:
Volume of nitrogen at standard conditions:
78,057X0.7350 = 57,372 ft'
Weight = 57,372X1.26 = 72,289 oz. Av.
= 4,518 lbs.
The question now is, how much nitrogen is contained in each
cubic foot of air in the engine room. Knowing that, we are
prepared to calculate the volume of this actually received by
the converter:
Barometric pressure = 756 m.m.
Tension of moisture (44X0.5) = 22 "
Tension of air present = 734 "
Tension of nitrogen present (734X0.792) =. 580 "
Weight of nitrogen in 1 cubic foot:
^•2«XW36XW = 0.8495 OZ.AV
Volume of air actually received :
4,518X16
0.8495
Volume efficiency of machinery:
85,095
= 85,095 ft.3 (2)
190,406
= 0.447 = 44.7p. c. (2)
(3) The slag contains 8.88 per cent, of MnO, equal to 6.88
per cent, of Mn, as already calculated. But 94.5 pounds of
340 METALLURGICAL CALCULATIONS.
manganese is oxidized, therefore, the weight of slag produced
is:
94.5-^0.0688 = 1374 pounds. (3)
Weight of SiO^'in slag (1374 X 0.6356) = 873 pounds
SiO^ from the Si of bath (207.0 + 236;6) =444
SiO^ corroded from the lining = 429 "
CaO, APO' and MgO (1374X0.0427) = 59 "
Loss in weight of lining = 488 " (3)
Blast Pressure.
It is necessary to use sufficient blast pressure to overcome
the static pressure of the metallic bath, plus that of the slag
formed, also the back pressure in the converter, to give the
necessary velocity to the air in the tuyeres and to overcome
friction in the same. When the tuyeres are near to the sur-
face of the bath, pressures of 1 or 2 pottnds will rtm the small
converter, but the ordinary converter with bottom tuyeres re-
quires from 15 to 30 pounds pressure per square inch (1.054 to
2.108 kg. per square cm.). We will consider the latter, the
more frequent and the more complex case to discuss.
The metal lies 12 to 24 inches (30 to 60 cm.) deep in the
converter. Since its specific gravity melted is about 6.88 (Rob-
erts and Wrightspn), the ferro-static pressure which it exerts
is practically 0.25 pounds per square inch for each inch depth
of metal, or 0.00688 kilos, per square centimeter for each centi-
meter depth.
The slag lying on the metal has a specific gravity melted
of approximately half that of the metal. Its amount may vary
from 5 to 10 per cent, of the weight of the metal treated, in an
acid-lined converter, up to from 15 to 35 per cent, in a basic-
lined vessel. Taking into account its lower specific gravity,
its depth in the converter may be, therefore 10 to 20 per cent,
the depth of metal in an acid-lined vessel, and 30 to 70 per
cent, in a basic-lined converter; but the static pressure exerted
would be only in direct pioportion to the relative weights;'
i. e., 5 to 10 or 15 to 35 per cent, of that exerted by the metal.
The static pressure of the slag may, therefore, be reckoned as
0.125 pounds per square inch for each inch in depth, or 0.00344
THE BESSEMER PROCESS. 341
kilos, per square centimeter for each centimeter depth, and the
depth of slag as lying between the following extremes :
Acid Lined. Basic Lined.
Depth of metal (^^^es. 12 to 24 12 to 24
^ \ Centimeters. . 30 to 60 30 to 60
^»or^+ll rif clotr / Inchcs 1.2 to 4.8 3.6 to 16.8
ueptn oi siag ^ Centimeters. . 3.0 to 12..0 9.0 to 42.0
The probable depth of slag can be calculated in any par-
ticular case, when the composition of the metal to be blown is
known, its approximate depth in the vessel, and the approxi-
mate composition of the slag to be formed.
The back pressure of gases in the converter itself, that is,
their static pressure, will vary, with the shape of the converter
and the size of the free opening for their escape into the air.
A measurement at the Pennsylvania Steel Go's, works gave
0.275 pounds per square inch, but it is not stated just how
the measurement was made. If we know approxima;tely the
volume of gas which must escape from the converter and from
its temperature aiid the time and the size of the outlet calculate
its velocity, the static pressure giving- it this velocity can be
calculated, as
in which, if V is in feet per second, 2g = 64.3 and the resultant
pressure is in feet of the hot gas; if V is in meters per second,
2g = 19.6, and h is in meters of the hot gas. Knowing the
approximate specific gravity of the hot gas (weight of 1 cubic
foot in poimds or of 1 cubic meter in kilograms) the static
pressure is obtainable in pounds per square foot or kilograms
per square meter.
Illustration: The gases escaping from a converter are 78,057
cubic feet (standard conditions), and weigh 0.0801 pounds per
cubic foot (standard conditions). They escape from the con-
verter at an average temperature of 1,500° C, and the opening
is 24 inches in diameter. What is the gaseous back pressure in
the converter? Time of blow 9 minutes 10 seconds.
342 METALLURGICAL CALCULATIONS.
Volume of gas at 1,500°:
78,057 X ^^""+1^^ = 506,940 ft^"
Volume per second:
506,940^550 = 921.7 "
Area of outlet :
2X2X0.7854 = 3.1416 ft^
Velocity (assuming 0.9 coefficient) :
921.7^-0.9-^3.1416 = 326 ft. per second
Head of hot gas giving velocity:
326 X 326
64.3
Pressure of this column per square foot :
1,653 ft.
1.653X^^X0.0801 = 20.4 pounds
Per square inch = 0.14 "
This solution omits one consideration; the velocity of the
gases in the body of the converter is neglected. This is some-
what counterbalanced by the great friction of the gases against
the sides of the converter, so that the one item tends to neu-
tralize the other. If the interior were 8 feet in diameter, the
velocity of the gases therein would average only some 20 feet
per second, showing the above corrections to be practically
negligible, since the pressure thus represented would be only
0.4 per cent, of the total obtained above.
The pressure necessary to force the blast through the tuyeres
is calculable on principles similar to the above; the differences
are that the blast, at temperatures varjdng from 100° C. in the
blast box to possibly 200° at its entrance into the metal, is
divided up into fifty or 150 streams of approximately 1 centi-
meter (0.4 inch) in diameter, the length of tuyeres being some
50 centimeters (20 inches). The formula similar to that used
THE BESSEMER PROCESS. 343
for chinmey draft, or rather, frictional resistance in a chimney,
applies to this case.
, V^" KL
in which h is the head in terms of the air passing, V is its velocity,
2g the gravitation constant, L the length of the tuyere, D its
diameter, and K the coefficient of friction, which latter is for
relatively smooth flues 0.05 (Grashof) , and may be so assumed
here.
Problem 66.
In the converter mentioned in Problem 65, where 22,500
pounds of metal was blown in 9 minutes 10 seconds, using
as therein calculated 85,095 cubic feet of air, at 36° C, and
producing 1,374 pounds of slag, assume the inside diameter of
the converter as 7 feet, and that the bottom contains fourteen
tuyere blocks, each containing eleven openings of 0.5 inch
diameter each; blocks 24 inches long. Assume back pressure
in converter 0.14 pounds per square inch, total blast pressure
in equalizing reservoir 27 potmds per square inch. Temper-
ature of air in the tuyeres 150° C.
Required: (1) The pressure needed to overcome the head
of metal and slag.
(2) The pressure absorbed in friction in the tuyeres.
(3) The pressure represented by the velocity of the blast
in the tuyeres.
(4) The loss of pressure from the reservoir to the blast-box.
(5) The distribution of the total pressure.
(6) The length of the blow if the blast pressure were reduced
to 20 pounds.
(7) The length of the blow if the pressure were maintained
at 27 pounds, but twenty-one tuye/e blocks (each with eleven
^-inch holes) were used.
Solution: (1) At the start there is 22,500 pounds of melted
metal, the volume of which will be
22.500 ^ jgjOO ^ 52.3 cubic feet
6.88X62.5 430
344 METALLURGICAL CALCULATIONS.
The depth of metal, the inside diameter being 7 feet, is
= — ;; — " -„, . = 7^7^— z = 1.356 feet
7X7X0.7854 38.5
= 16.4 inches
Static pressure = 16.4 X 0.25 = 4.1 lbs. 'per sq. in.
The slag, formed during the first, half of the blow, weighs
1,374 pounds, and has a volume of
= 6.4 cubic feet
3.44X62.5 215
The depth of slag, at its maximum, will be
6.4-r38.5 = 0.167 feet
= 2.0 inches
Static pressure 2.0X0.125 = 0.25 lbs. per sq. in.
The static pressure during the blow will, therefore, be 4.1
pounds to start with, increasing during the first half of the
blow to 4.35 pounds, and stajHing practically constant at that,
and, therefore, will average
4.1 + 4.35,4.35 ... , ,,,
— 2^2 "^~2~ ^ pounds (1)
(2) Each of the 14X11 = 154 tuyeres receives 85,095 -f-
154 -h 550 = 1.005 cubic feet of air per second, measured at
36° C. At 150° C. this volume is
1.005X^2?3 + 3T " -^-^^^ ''''^'*' ^®^*
And the velocity in the tuyere:
, „„. 0.5X0.5X0.7854 .„„ , , ,
1.375-; —- = 1009 feet per second
144
The head absorbed in friction in the 24-itich tuyeres will be
, 1009X1009^0.05X2 „^ „„„ , ^
^ = 64.3 X-o:04l7 = ^^'^^^ ^^^*
Changing this pressure of air at 150° C. to pounds per square
inch we have:
THE BESSEMER PROCESS.
345
Weight of 1 cubic foot of air at 0° =
Weight of 1 cubic foot of air at 150° =
Weight of air column = 37,893X0.0522 =
Pressure in pounds per square inch =
(3) The pressure absorbed as velocity-
expressed in getting the friction in the tuyeres
head is simply:
V^ _ 1009X1009
2g"
0.0808 pounds
0.0522 "
1978 - "
13.7 " (2)
has already been
The velocity
h =
64.3
= 15,835 feet
which becomes in pressure
15,835X0.0522 = 826.6 pounds per square foot
= 5.73 pounds per square inch (3)
(4) The remaining part of the 27 poxmds pressure used is
lost between the blast reservoir and the entrance to the tuyeres.
It is
27.00- (13.70 + 5.73 + 4.29 + 0.14). = 3.14 pounds. (4)
(5) Distribution of blast pressure:
Fall between reservoir and tuyeres
Absorbed in friction in the tuyeres
Absorbed in velocity in the tuyeres
Static head of liquid bath
Velocity head of issuing gases
3. 14 pounds =
11.6%
13.70 "
50.7%
5.73 " =
21.2%
4.29
15.9%
0.14 "
0.6%
27.00
= 100.0%
(6) All the items of absorption of pressure are proportional
to the square of the velocity of the gases, excepting the static
pressure of the bath. It remains constant at 4.29 potonds. If
the total pressure were reduced to 20 pounds, there would be
only 20— 4.29 = 15.71 pounds pressure to give velocity and
overcome friction, instead of 27— 4.29 = 22.71 pounds. The'
relative quantities of air blown through in a given time in the
two instances would be practically proportional to the square
roots of the two effective pressures, i.e.:
V22?n : ^15.71 = 1 : 0.832
and the times of the blows inversely as the latter:
550 sec. -^0.832 = 673 seconds
= 11 min. 13 sec.
(6)
346 METALLURGICAL CALCULATIONS.
(7) If the tuyere area were increased 50 per cent., then the
velocity of the air in the tuyeres would be decreased one-third,
assuming the amount of air passing to be unchanged. This
would decrease the pressure absorbed in friction, and in giving
velocity in the tuyeres to (0.67)^ = 0.444 of its former amount.
The 19.43 potmds previously absorbed in these two items
would then become 19.43X0.444 = 8.61 pounds, and the total
pressure needed to rtm the converter just as fast as before
would be 27— (19.43— 8.61) = 16.18 pounds per square inch.
If, however, the pressure were maintained at 27 pounds, giving
still 27— 4.29 = 22.71 pounds to overcome frictional resist-
ances and to give velocity, then the velocity and consequent
amotmt of air blown through by this 22.71 pounds pressure
would increase in proportion to the square root of these two
available pressures; i.e., be as
Vie.lS- 4.29 : V27-4.29 = 1 : 1.38
The duration of the blow would be just that much shorter ; i.e. :
550 sec. -5- 1.38 = 398 seconds
= 6 min. 38 sec. (7)
Flux and Slag.
No flux is used in the acid-lined converter, and the silica,
iron oxides and manganese oxide formed in the converter
unite to a silicate slag which corrodes the lining and thus takes
up more silica. The slag being analyzed, its weight is ob-
tained by considering the percentage of manganese which it
contains, because the weight of manganese oxidized is known
definitely from the analysis of the bath; it is usually all oxi-
dized. Calculation of the weight of slag cannot be based upon
the silica, because an vinknown amotmt comes from the lining;
nor upon the iron, because the weight of iron left in the con-
verter is not definitely known. Having the weight of the slag,
analysis tells us the total weight of silica in it, as also the amount
of iron. The silica in the slag minus that formed from silicon
in the pig iron, gives silica corroded from the lining.
In the basic Bessemer converter, phosphorus is nearly en-
tirely eliminated from the metal, so that, assuming none to be
volatilized, the amotmt going into the slag is known, and using
the slag analysis the weight of slag can be calculated. In
THE BESSEMER PROCESS. 347
this process the lining is mainly dolomite, containing CaO and
MgO, in proportion easily determined by analysis. The weight
of slag beiag known, the amotint of corrosion of the lining
can be determined from the percentage of magnesia therein,
which may be assumed as practically coming entirely from
the lining ; it cannot be told from the CaO in the slag, because
nearly pure CaO is added during the blow, and some of it,
a variable amount, gets blown out of the converter. For
the same reason it is not possible to base a good calculation
of the weight of slag on the lime alone which is added, be-
cause of the indefinite proportion of it which is blown out.
The weight of slag may also be gotten from the silica or man-
ganese oxide in it, assuming these to come almost entirely from
the oxidation of silicon or manganese.
Lime must be added as flux, in the- basic converter, to pro-
tect the lining and to make the slag so basic that the percentage
of silica in it is below 15 per cent., phosphoric acid below 20
per cent., and lime over 50 per cent. These considerations must
be balanced in each particular case.
Illustration: Pig iron blown in a basic-lined converter con-
tained 1.22 per cent, silicon, 2.18 phosphorus, 1.03 manganese
I and 3.21 carbon. It is blown until all of these and 2.00 per
cent, of iron are oxidized, and burnt lime is added to form
slag during the blow. Composition of the burnt lime: MgO,
1.00 per cent.; SiO^ 2.00 per cent.; CaO, 97 per cent. How
much lime should be added per 10 metric tons of pig iron charged?
The slag-forming ingredients from the oxidation of the bath,
and the addition of X kilos, of lime, are
SiO^ 10,000X0. 0122 X 60/28 = 261.4 kg.
P^O' 10,000X0.0218X142/62 = 499.3"
MnO 10,000X0. 0103 X 71/55 = 133.0 "
FeO 10,000X0. 0200 X 72/56 = 257.1 "
rCaO XXO.9700 = 0.97 X
MgO XXO.OIOO = 0.01 X
[siO^ XX 0.0200 = 0.02 X
Weight for slag = X + 1150.8 kg.
Corrosjon of the lining will undoubtedly increase this weight,
SO some allowance should be made, say to increase it 5 per
348
METALLURGICAL CALCULATIONS.
cent., probably an outside figure. Of this 5 per cent., half can
be considered lime and half magnesia. The total weight of
slag will then be 1.05 X + 1208.3, and of the ingredients prin-
cipally in question:
SiO^ = 261. kg. + 0.02 X
MgO = 28.7 " +0.035 X
CaO = 28.7 " +0.995 X
To make our slag 50 per cent. CaO wall require the addition
of enough to make'
28.7 + 0.995 X'= 0.50 (1.05 X+ 1208. 3)
X = 1224 kg.
To make a slag with at most 15 per cent, of SiO^ reqtdres
261.4 + 0.02 X =
0.15 (1.05X + 1208.3)
X = 583 kg.
To make a slag with at most 20 per cent, of P^O' reqtdres
499.3 = 0.20 (1.05 X + 1208.3)
X = 1227 kg.
The larger of these three amounts would be used, with 10
per cent, added to cover lime dust blown out, making 1350 kg.
added, and the calculated composition of the slag:
1250 kg. = 50.1 percent.
723 " = 2.9
286 " = 11.4
499 " = 20.0
257 " = 10.3
133 " = 5.3
CaO
MgO
SiO^
pZQS
FeO
MnO
Total 2497 "
Recarburization.
When the bath has been blown to nearly pure iron, melted
spiegeleisen is run in, to add the necessary carbon and man-
ganese. Knowing the approximate composition and weight of
the bath, and the composition of the melted spiegel, a simple
arithmetical calculation would give the amount of the latter
to be added, assuming no loss of carbon or manganese in the
THE BESSEMER PROCESS. 349
operation. But experience shows that there is some loss, and
that the carbon and manganese in the finished metal are always
lower than the calculated amount. An interesting field is open
here for calculating the loss of manganese and carbon and the
amovmt of oxygen which must have been in the metal to cause
these losses. A tabulation of many such calculations gives
the metallurgist the necessary data for assuming the average
amounts of carbon and manganese lost during recarburization,
under different conditions of working, such as letting the metal
stand before pouring or pouring at once, turning on the blast
5 or 10 seconds to mix up the bath, etc.
Problem 67.
At the end of the blowing the converter of Problem 65 con-
tained 21,283 pounds of metal of the composition 0.04 per
cent, carbon, 0.02 silicon, 0.01 manganese, 0.11 phosphorus, 0.06
sulphur, an unknown amount of oxygen (probably < 0.3 per
cent.) and the rest iron. There is added to it 2,500 pounds
of spiegeleisen, containing 4.64 per cent, carbon, 0.035 silicon,
14.90 manganese, and 0.139 phosphorus. The finished metal
contained 0.45 per cent, of carbon, 0.038 silicon, 1.15 manganese,
0.109 phosphorus and 0.059 sulphur. Assume no iron oxidized.
Required: (1) A balance sheet of materials before and after
recarburizing.
(2) The proportions of carbon and manganese going into the
finished metal.
Solution: (1)
Blown Gases
Metal Spiegel. Steel, or Slag.
C 8.5 116.0 106.5 18.0
Si 4.3 0.9 9.0 -3.8
Mn 2.1 372.5 271.2 103.4
P 23.4 3.5 25.8 1.1
S 12.8 14.0 - 1.2
Fe 21,232 2007 23,239
The differences in the sulphur, phosphorus and silicon are
within the limits of error of the data, but there is no doubt as
to the loss of carbon and manganese.
(2) The proportions of the two elements in question going
into the finished steel are:
350 METALLURGICAL CALCULATIONS.
Carbon 106.5-^124.5 = 0.85 = 85 per cent.
Manganese 271.2-^374.6 = 0.72 = 72
The calculated percentages in the finished steel should have
been, and actually were:
Carbon 0.53- 0.45, loss = 0.08 per cent.
Manganese 1.58- 1.15, loss = 0.43
Concerning oxygen removed, if we assume the loss of carbon
and manganese to be due to their combining with oxygen dis-
solved in the bath, to form CO and MnO, the percentage of
oxygen thus absorbed is:
By carbon 0.08x16/12 = 0.11 per cent.
By manganese 0.43x16/55 = 0.12
0.23
[The next chapter will consider the thermo-chemistry of the
Bessemer process.]
CHAPTER VIII.
THERMO-CHEMISTRY OF THE BESSEMER PROCESS.
The feature of the Bessemer operation which strikes the
observer as most wonderful, is that cold air is blown in great
quantity through melted pig iron, and yet the iron is hotter at
the end than at the beginning. If the observer will reflect a
moment, however, he can see that if nothing but fuel, on fire,
was in the converter, it would certainly be made much hotter
by the air blast ; in similar manner, the oxidation or combustion
of part of the ingredients of the pig iron furnishes all the heat
required for the process. Ten tons of pig iron contains, for
example, some 350 kilograms, of carbon which is all burnt out
in the bessemerizing, furnishing heat equal to the combustion
of some 400 kilograms of coke — a not insignificant quantity,
since it is burned and its heating power utilized in a very few
minutes.
Elements Consumed.
The usual ingredients of pig iron are:
Iron 90.0 to 95.0 per cent
Carbon 2.5
Manganese 0.5
Silicon 0.5
Phosphorus 0.01
Sulphur 0.01
Some of the unusual constituents are nickel, chromium, titan-
ium, aluminium, vanadium, timgsten, copper and zinc; all of
these are rare, and there is seldom present as much as 0.5 per
cent, of any one except in unusual cases.
In the Bessemer operation, carried out with the usual silica
lining, iron, carbon, manganese and silicon are freely oxidized,
but phosphorus and sulphur remain practically unchanged. In
the basic Bessemer, lined with burnt dolomite and tar, phos-
351
to 95.0
4.5
4.0
4.0
4.0
0.5
352 METALLURGICAL CALCULATIONS.
phorus is also freely oxidized at the end of the operation, but
sulphur is only slightly diminished — ^the more, the more man-
ganese is in the slag. After all the oxidizable impurities are
eliminated, iron itself oxidizes in much larger quantity, oc-
casioning great loss if the blast is permitted to continue.
Iron oxidizes during the blow mostly to FeO, which enters
the slag as ferrous silicate, and partly to Pe^O', The brown
fume which escapes in large amount if the blow is contiaued
too long contains iron as Fe^O'.
Fe to FeO 1173 Calories per kg. of Fe
FetoFe=0» 1746
FetoFe^O* 1612
The amount of iron oxidized can be gotten from the weight
and perc^tage composition of the slag; also from the com-
parison of the weights of materials used and weight of ingots
produced, knowing the weight of other impurities oxidized.
Carbon oxidizes mostly to CO gas, and partly, especially in
the early part of the blow, to CO^ gas. The proportion of each
of these formed can only be known by analyzing the gases
produced at various stages of the blow. The proportiona.te
volumes of CO and CO^ express the proportionate amoimts
of carbon forming each respective gas. A shallow bath allows
more CO' to pass, oh essentially the same principle that a deep
layer of fuel on a grate favors the production of CO. The
heat evolved by oxidation of carbon is :
C to CO 2430 Calories per kg. of C
CtoCO' 8100
Manganese oxidizes quickly and mostly to MnO. If the
metal is a little overblown, Mn'O^ in small amount is found in
the slag, while Mn^O* is also present in the fumes. The heat
evolved in these oxidations is :
Mn to MnO 1653 Calories per kg. of Mn
MntoMn'O' .2480(?)"
The last figure is estimated; it has not yet been determined
experimentally.
Silicon oxidizes rapidly and early in the blow to SiO^, form-
ing silicate slag with the metallic oxides formed. Its heat
THE BESSEMER PROCESS. 353
of oxidation has been usually taken as 7830 Calories per kilo-
gram, but recent investigations have thrown doubt on this
figure, Berthelot having found as low as 6414 Calories, and Mr.
H. N. Potter, 7595 for crystalline sihcon, equal to 7770 for
amorphous silicon. Under these circumstances the best course
is probably to use the middle value ad interim, and consider
Si to SiO=' 7000 Calories per kg. of Si
We hope that the exact figure will soon be determined.
Phosphorus oxidizes to P^O^ and only towards the end of
the blow. It is practically completely eliminated, going into
the slag as calcium phosphate :
P to P^Qs 5892 Calories per kg. of P
Sulphur is not eliminated at all in the acid-lined converter.
In the basic converter it is reduced in amoimt in the last few
minutes, while phosphorus is disappearing, and partly escapes,
mostly as SO^ in the gases. The presence of a very basic slag
is necessary, but the' sulphur, while possibly going into the slag,
does not remain there, but passes into the gases. The heat of
oxidation of the unusual elements sometimes present are, as
far as known,
Ni to NiO 1051 Calories per kg. of Ni
TitoTiO^ 4542 " " Ti ,
AltoAPO' 7272 " " Al
ZntoZnO ..1305 " " Zn
VtoV^O= 4324 " " V
WtoWO' 1047 " " W .
CrtoCr'O'... 2344 " " Cr
Some of the above values are a little uncertain, and none of
them include the heat of combination of the oxide formed with
the slag.
Heat Balance Sheet.
Taking o° C. (32° F.) as a base line, we may express the
total heat contents of the pig iron, steel, gases, slag, blast, etd.,
354 METALLURGICAL CALCULATIONS.
from this temperature. This method is simpler than to take the
bath at any one high temperature and to reckon from there.
The items of heat available during the blow are:
Heat in the body of converter at starting.
Heat in the melted pig iron used.
Heat in the spiegleisen or ferro-manganese added.
Heat in hot lime added (sometimes in basic process).
Heat in the blast, if warm on entering.
Heat developed by oxidation of the bath.
Heat developed by formation of the slag.
The items of heat distribution are:
Heat in the "body of the converter at finishing.
Heat in the steel poured.
Heat in the slag at finishing.
Heat in the gases escaping.
Heat in the fume.
Heat in the slag or metal blown out.
Heat absorbed in decomposing moisture of the blast.
Heat to separate the constituents of the bath.
Heat conducted away by supports, blast pipe, etc.
Heat conducted to the air in contact with converter.
Heat radiated during the blow.
These two columns should balance each other if all the items
of each are correctly determined.
Heat in Converter Body at Starting.
If the converter were quite cold when the pig iron was run
in and the blow started, this item would be zero. But such a
case would result disastrously, since the heat absorbed by the
converter during the blow would be more than any ordinary
heat could afford to lose. It is, therefore, customary, when
starting for the first time, to build a fire in the converter and
turn on a little air blast, so as to bring the inside up to bright
redness, say 900° to 1000° C. The outside shell wotdd, under
these conditions, be at about 200°, and the mean temperature of
the converter lining, say 400°- If the converter were in regular
operation, one charge being introduced as soon as the other
was finished, then the heat in the body of the converter at start-
ing could be practically regarded as equal to the heat in the
THE BESSEMER PROCESS. 355
same at finishing, assuming the heats are running regularly.
An estimate of the heat in the converter body at starting is
therefore only necessary when the converter is first started up,
or when it is allowed to stand some time between blows.
Illustration : . Assume a converter weighing, without charge,
25 tons, of which 5 tons is iron work and 20 tons silica lining.
The mean specific heat of iron (for low temperatures) being
0.11012 + 0.000025t + 0.0000000547t^ and for silica 0.1833 +
0.000077t, calculate the heat contained in the body of the con-
verter.—
(1) When the temperature of the outside shell is 200° and
that of the lining averages 400° (converter warmed up for
starting).
(2) When the temperature of the outside shell is 300° and
that of the lining averages 725° (converter empty at end of a
blow).
Solution: (1)
Heat in 5000 kg. of iron work:
0.11012 + 0.000025 (200) +0.0000000547 (200)' = 0.11731
0.11731X200X5000 = 117,310 Calories.
Heat in 20,000 kg. of silica lining:
0.1833 + 0.000077 (400) = 0.2141
0.2141 X 400 X 20,000 = 1,712,800 Calories.
Total heat in converter body = 1,830,110 „
It may be remarked that this would require the consumption
of at least 250 kilograms of coke, with a calorific power of
8,000 Calories, to warm the converter to this extent.
(2) Heat in 5,000 kg. of iron work:
0.12354X300X5,000 = 185,310 Calories.
Heat in 20,000 kg. of siUca lining:
0.2391 X 725X20,000 = 3,466,950
Total = 3,652,260
This condition is assumed as representing the .converter when
just emptied and immediately refilled. In this case, in regular
working, the heat in the converter body is practically the same
at the beginning and at the end of a blow, and the heat losses
through it are only those due to conduction to the air and
groimd and radiation.
35e METALLURGICAL CALCULATIONS.
Heat in Melted Pig Iron Used.
This quantity depends on the temperature at which the pig
iron is run into the converter. If the iron is high in silicon,
which would tend to produce a hot blow, it may be run in
somewhat cool ; but if low in silicon it should be rtm in much
hotter. The minimum quantity of heat contained in a kilogram
of melted pig iron may be put at 250 Calories; the maximum,
for very hot pig iron, 350 Calories; about 300 Calories would
be a usual average figure. This may be easily determined ex-
perimentally in any given case by granulating a sample in
water in a rough calorimeter.
Heat in Metallic Additions.
Melted spiegeleisen is usually not very hot when run into the
converter. It may contain 250 to 300 Calories per kilogram;
say an average of 275. Ferro-manganese is sometimes added
red-hot, at 800° to 900° C. At this temperature it would con-
tain 120 to 135 Calories per kilogram, assuming a mean specific
heat of 0.15.
Heat in Preheated Lime.
Taking the mean specific heat of CaO as 0.1715 -|-0.00007t,
it would contain 154 Calories per kilogram at 700°, and 211
Calories at 900° The heat content can be calculated for any
known temperature at which the lime is used.
Heat in Warm Blast.
No Bessemer converters are r\m by hot blast, but the air
pressure used is so great (20 to 35 pounds per square inch)
that the blast is warmed by compression in the cylinders. If
air at 1 atmosphere tension (ordinary air) be compressed to 2
or to 3 atmospheres tension, giving effective blast pressures of
1 to 2 atmospheres, the air is heated 60° or 103°, respectively,
above its initial temperature. While some of this heat may be
lost in the cylinder and conduits, yet the air, unless artificially
cooled, passes to the tuyeres at 25° to 50° C. above the outside
temperature, and thus imparts some heat to the converter.
Heat Developed by Oxidation.
We have already discussed the thermochemical data required
for this calculation. To use the data we must find the weights
THE BESSEMER PROCESS. 357
of each ingredient oxidized (not forgetting the iron itself) and
the nature of the oxide it forms. This is deduced from the
analysis of slag, gases and steel produced, as compared with
those of the pig iron and additions used.
Heat of Formation of Slag.
The slag is a mechanical mixture or mutual solution of sili-
cates of iron and manganese, with a little alumina (up to 5
per cent.)' and lime and magnesia from nothing up to 5 per
cent. In the basic process a large amotmt of calcium phos-
phate is present, representing over half of the entire slag, while
magnesia is present in considerable amount, and alumina is
almost absent.
Having, in the preceding column, calculated the heat of
oxidation of all the metals oxidized in the converter, it re-
mains to calculate the heat of combination of these to form
slag. The silicate of alumina contributes nothing, since it
occurred combined in the lining or lime added. The same can
be said of the lime and magnesia in the acid process slags.
The FeO and MnO are then to be considered, and the only
thermochemical data we have are :
(MnO, SiO=) = 5,400 Calories.
(FeO, SiO^) = 8,900
These data are for 71 or 72 parts of MnO or FeO respectively,
uniting with 60 parts of SiO^ Since in acid slags there is
always proportionately less of the bases, we should utilize
the above heats of formation by expressing them per unit
weight of MnO or FeO going into combination. These figures
are: . ; .
Per kg. of MnO = 5,400-^71 = 76 Calories.
Per kg. of FeO = 8,900-h72 = 124
From these figures the heat of formation of the slag can be
calculated : Any Fe^O'- present in the slag would be calculated
as its equivalent weight of FeO (by multiplying by 144-^160).
- Illustration: A Bessemer slag contained by analysis:
SiO^ 47.25 per cent.
APO« 3.45
FeO 15.43
MnO 31.89
CaO, MgO 1.84
358 METALLURGICAL CALCULATIONS.
What is its heat of formation per kilogram?
Solution: The Al^O', CaO and MgO are to be neglected,
for reasons already given. The heat of combination, per kilo-
gram of slag, is therefore,
FeO uniting with SiO= = 0.1543X124 = 19.1 Cal.
MnO imiting with SiO^ = 0.3189X 76 = 23.2 "
Total = 42.3 "
In basic Bessemer slags the conditions are much more com-
plicated. The content of P^O' is not under 14 per cent., runs
as high as 25 per cent., and averages 19 per cent.; the CaO
averages 45 per cent., limits 35 to 55; the silica is usually below
12 per cent., and averages 6 to 8 per cent.; magnesia is present
from 1 to 7 per cent., average about 4 per cent. In such slags
we should first of all assume the P^O' to be combined with
CaO as 3CaO. P^O', containing 168 of CaO to 142 of P^O',
(1,183 to 1) and having a heat of formation from CaO and
P^O' of 159,400 Calories per molecule, or 1123 Calories per
unit weight of P^O*- Next the iron and manganese present
may be calculated to FeO and MnO respectively, and treated
as to their combination with SiO^, the same as in an acid slag.
Alumina may be considered in such basic slags as an acid, and
equivalent to 120/102 of its weight of silica. These allowances
will leave considerable lime and either an excess or deficiency
of silica; if an excess, we can assume it combined with lime
and magnesia, with a heat evolution equal to 476 Calories per
kilogram of silica; if a deficiency, we let the calculations stand
without further modification.
Illustr.ation: Slag made at a Rhenish works contained:
SiO' 7.73 per cent.
P^O' 21.90
A1=0» 3.72
Fe^'O' 1.00
FeO ; 4.73
MnO 2.05
CaO 50.76
MgO 4.00
CaS 1.71
What is its heat of formation per unit of slag?
THE BESSEMER PROCESS. 359
Solution: The 21.90 parts of P'0= would be combined with
21.90X168/142 = 25.91 parts of CaO. This leaves 50.76—
25.91 = 24.85 of CaO as either free, dissolved CaO or partly
combined, in the slag. The 2.05 MnO would correspond to
2.05x60/71 = 1.73 SiO^ the 4.73 FeO to 4.73x60/72 =
3.94 SiO^; the 1.00 Fe^'O^ to 1.00x60/80 = 0.75 SiO^; a total
SiO^ requirement for these three bases of 6.42 per cent. The
SiO^ present is 7.73 per cent., adding to which the SiO^ equiva-
lent to the Al^O' present (3.72x120/102 = 4.38), we have
12.11 per cent, of summated silica. The ratio of summated
FeO to summated SiO^ is considerably below the ratio 72 to
60, we can, therefore, consider the- summated FeO as all com-
bined with silica. The summated FeO is :
FeO = 4.73 per cent.
FeO equivalent of MnO = 2.08 "
FeO equivalent of Fe^'O' = 0.90 "
Total = 7.71 "
And the SiO^ combining with this as FeO.SiO^ is
7.71x60/72 = 6.42 per cent.
The excess of summated silica free to combine with lime is
12.11 — 6.42 = 5.69 per cent. As there is 24.85 of CaO and
4.00 of MgO for it to combine with, the heat of this combina-
tion must be calculated on the SiO^ going into this combination.
We then have the formation heat of the slag as:
P'O^ to 3CaO . P^O' 21.90X 1123 = 24,594 Cal.
MnO to MnO . SiO'' 2.05 X 76= 156"
FeO to FeO . SiO^' 5.63X 124= 698 "
SiO^* to SCaO . SiO^ 5.69X 476 = 2,708 "
28,156 "
This equals 281.6 Calories per unit weight of slag, forming
a very important item in the heat balance sheet, particularly
when the slag is large in amount.
Heat in Converter Body at Finishing.
This item reaches its maximum at the end of the blow, and
would be equal to that calculated for the beginning of the blow,
360 METALLURGICAL CALCULATIONS.
■with the exception that some of the lining, silica or dolomite
has been corroded and passed into the slag, carrying with it its
sensible heat.
Heat in Finished Steel.
This should be determined experimentally in each particular
case. If not so determined an average value may be assumed,
based on the following considerations: The finishing tempera-
ture averages 1650° C. ; at this heat average Bessemer steel
will contain a total of 350 Calories of heat per kilogram. If the
temperature is determined by a pyrometer, a correction of 1/5
Calorie can be made for every degree hotter or colder than
1650°.
Heat in Slag.
Some experimental data are badly needed, concerning the
heat in slags of different composition at different temperatures.
At present it is necessary to make guesses, wherever the heat
in the slag produced is not directly determined. At a finishing
temperature of 1650° it is likely that the slag contains 550
Calories per kilogram; with a variation of 1/4 Calorie for each
degree hotter or colder than 1650°.
Heat in Escaping Gases.
The amount of these gases can only be determined satis-
factorily from their analysis and the known weights of carbon
oxidized. Direct estimation from the piston displacement is of
much less exactness, because the slip and leakage at these
high pressures may reach 25 to 50 per cent, or even more. The
temperature of the gases is only slightly less than that of the
bath, that is, some 1350° at starting and 1650° at finishing.
Outside in the air the Bessemer flame may be much hotter than
, this, but that is due to further combustion of CO to CO^ out-
side the converter, and should be disregarded. Where extra air
is blown upon the surface of the bath, as in " baby " converters,
it is quite possible that the gases in the converter and the es-
caping gases may be considerably hotter than the bath itself
These variations must be taken into consideration. The most
satisfactory condition is to insert a pyrometer tube into the
opening of the converter and measure the temperature directly.
The heat carried out by the gases can then be calculated ac-
THE BESSEMER PROCESS. 361
curately, using the proper mean specific heats to these high tem-
peratures already used in these calculations.
Heat in 'Escaping Fume.
This is mostly oxideS of iron and manganese, with some-
times silica. It is relatively small in amount. Its quantity
being known, consider it at the same temperature as the gases,
with a mean specific heat of 0.40 if free from silica, and 0.35
if siliceous.
Heat in Slag or Metal Blown Out.
These can be coimted as equal to the heat in an equal quantity
of slag or metal at the finishing temperature.
Heat Absorbed in Decomposing Moisture.
Knowing the amount of moisture blown in with the blast, a
proper allowance is 29,040^-9 = 3,227 Calories for every kilo-
gram of moisture thus blown in. It is probable that this
moisture is all decomposed, its hydrogen appearing in the
gases. In the absence of data as to the hygrometric condition
of the blast, the amount of moisture entering may be inferred
and calculated from the amount of hydrogen in the gases.
Heat to Separate Constituents of Bath.
We here meet the question, how much heat of combination
exists between the bath and the various ingredients which are
removed — carbon, silicon, manganese, phosphorus, sulphur.
Le Chatelier believes manganese to exist in the bath as Mn^C,
requiring 80 Calories per kilogram of manganese to decompose
it. Silicon and carbon have, as far as has at present been
determined, no sensible heat of combination with iron. Sulphur
requires 750 Calories to separate each kilogram from iron.
Phosphorus requires, according to Ponthiere, 1397 Calories to
separate each kilogram from iron, but the reliability of this
datum is doubtful. Until more reliable tests are made it is
perhaps better to omit this item than to use it, although it must
be of great importance if as large as Ponthiere states it to be.
Heat Conducted Away by Supports.
This is a very difficult quantity to determine, being con-
ditioned by the size of the supports, their cooling surface and
362 METALLURGICAL CALCULATIONS.
the kind of connection they have with other objects. The heat
which would pass to the blast pipe is practically returned to the
converter by the incoming blast. The heat passing into the
supports is perhaps best found by taking their temperature at
different places, and calculating the heat loss from their sur-
face by radiation and conduction to the air. This amounts
practically to considering them as part of the outer cooling
surface of the converter; the calculation of these surface losses
is given in the two following paragraphs :
Heat Conducted to the Air.
This is a function of the extent of outside surface, its tem-
perature, the temperature of the air, and the velocity of the
air current. Measurement will give the extent of surface in
contact with the air and the average velocity of the air cur-
rent; the surface being rough iron, the coefficient of transfer
conductivity may be taken as k = 0.000028 (2 + \/ v). where
V is the air velocity in cm. per second, and k is the heat con-
ducted per second in gram-calories, from each square cm.
of surface per 1° difference of temperature. The temperature
of the outer surface should be carefully measured, so that a
reliable average is obtained, and the air velocity likewise aver-
aged, since it has considerable influence on the heat lost to
the air.
Illustration: A converter has an outside surface of 50 square
meters, at an average temperature during the blow of 200° C.
the average air current being 1 meter per second, and out-
side air 30° C. What is the heat loss by conduction to the
air in kilogram Calories per minute?
Solution :
Coefficient of transfer conductivity :
0.000028 (2+ VTOO) = 0.000336 .
Heat loss per 1° difference, per second, in gram-calories:
50X10,000X0.000336 = 168 calories.
Heat loss per 170° difference, per minute, in kg.-calories:
168 X 170 X 60 -T- 1000 = 1714 calories.
THE BESSEMER PROCESS. 363
Heat Radiated During the Blow.
This is a function of the temperature of the outside shell, the
mean temperature of the surroundings of the converter and
the nature of the metallic surface. As the surface is oxidized
iron, it would lose about 0.0141 gram-calories from each square
centimeter per second, if at a temperature of 100° and the
surroundings at 0°; or practically 1 gram-calorie per second
from each square meter, for every 100,000,000 of numerical dif-
ference between the fourth powers of the absolute tempera-
ture of the radiating surface and its surroundings. (See Metal-
lurgical Calculations, Part 1., p. 185).
Illustration: Assuming the surroundings of the converter at
30°, in the preceding illustration, what amount of heat is radiated
per minute in large Calories?
Solution: The absolute temperatures in question are 273 -f
30 = 303, and 273-1-200 = 473. The difference of their fourth
powers is :
473*— 303* = 41,626,500,000,
which, divided by 100,000,000, gives 416.265 gram-calories
lost per second per each square meter of radiating surface.
The radiation loss for the whole surface per minute is, therefore:
416.265 X 50 X 60 -^ 1000 = 1249 kg.-Calories.
While the assumption made as to the temperature of the out-
side shell is doubtless only approximate, yet if the tempera-
ture of the same is carefully determined the radiation loss can
be accurately calculated. If the outside of the converter were
polished, this radiation loss might be reduced nearly nine
tenths.
Problem 68.
From the data and results of calculation of Problem 65 (see
pages 310, 317 and 323,) we see that 22,500 pounds of pig
iron and 2,500 potinds of spiegeleisen produced 24,665 pounds
of steel', there being eliminated during the blow and recar-
burization :
Carbon 679.5 pounds (140.7 to CO^)
Silicon 203.2
Manganese 197.9 "
Iron 253.9 " (25.3 to Fe^O')
864 METALLURGICAL CALCULATIONS.
The gases contain :
00=* 5.20 per cent.
CO 19.91
W 1.39
N^* 73.50
The slag contains :
SiO\...' 63.56
APO' 3.01
FeO 21.39
Fe^O' 2.63
MnO 8.88
CeO 0.90
MgO 0.36
Make average assumptions for requisite data not given.
Required: A balance sheet of' heat evolved and distributed.
Solution: The items of this balance sheet have already been
discussed in detail. We will apply them to this specific case:
Heat in Body of Converter at Starting: Assuming that this
is a blow in regular running, the heat may be taken at any
reasonably approximate quantity, because the same quantity
with only a slight deduction will be allowed as contained in
the same on finishing. We will, therefore, take a figure already
calculated, 8,034,970-pound Calories, as the heat in the con-
verter body at starting.
Heat in Melted Pig Iron: We will take it at 300 Calories
per pound, or a total of 300X22,500 = 6,750,000 Calories.
Heat in Spiegeleisen: 2,500 X 300 = 750,000 Calories.
Heat in Blast: This may safely be considered as warmed
by compression and entering the .converter at 60° C. The
amotmt of blast received altogether is calculated thus:
Carbon oxidized = 679.5 pounds.
Volume of CO and CO^ formed:
679.5X16-^0.54 = 20,133 cu. ft.
Volume of gases ~^ = 80,179
Volume of nitrogen 80,179 X 0.735 = 58,932
Volume of air proper in blast = 74,408
Volume of hydrogen in gases .
80,179X0.0139= 1,115
Volume of moisture in gases = 1,115
THE BESSEMER PROCESS. 365
Assuming the blast at 60° C, the heat in it is:
Air 74,408X0.3046= 22,665 oz. Cal. per. 1°
ffO 1,115X0.3790 = 423 "
23,088 •'
23,088X60 = 1,385,280 oz. Cal.
86,580 lb. Cal.
Heat of Oxidation:
C to CO'' 140.7X8100 = 1,139,670 Calories
C to CO 538.8X2430 = 1,309,280
Si to SiO=i 203.2X7000 = 1,422,400
MntoMnO 197.9X1653= 327,130
FetoFeO 228.6X1173= 268,150
FetoFe^'O' 25.3X1746= 44,170
4,510,800
Heat of Formation of Slag: The 197.9 pounds of manganese
oxidized forms 255.5 of MnO. Hence the weight of the slag
is 255.54-0.0888 = 2877 pounds. The slag, therefore, con-
tains also 2877X0.6356 = 1829 pounds of SiO^ of which 203.2
X 60/28 = 435 pounds came from the silicon oxidized, and
1794 pounds from the lining. The lining will also have lost
the APO', CaO and MgO in the slag, equal to 2877 X 0.0427
= 123 pounds. The total iron going into the slag, 253.9
pounds, is equivalent to 326.4 pounds of FeO.
The heat of formation of the slag will therefore be:
FeO 326.4X124 = 40,474 Calories.
MnO 255.5 X 76 = 19,418
Total 59,892 "
Heat in Converter at Finishing: This will be the same as at
starting, less the heat in 1794 + 123 = 1917 -pounds of liaing,
which was corroded and entered the slag. Assuming this to
have been on the inner surface, at an average temperature of
1500°, the heat in it, using the mean specific heat of silica, will
have been
1917X0.2988X1500 = 851,200 Calories.
366 METALLURGICAL CALCULATIONS.
And the heat in the converter body at the finish:
8,034,970—851,200 = 7,183,770 pound Calories
Heat in Finished Steel: Taking its temperature as 1650°,
with 350 Calories per unit, we have
24,665X350 = 8,632.750 Calories.
Heat in the Slag:
2877X550 = 1,582,350
Heat in Escaping Gases: These have already been calcu-
lated as consisting of
Nitrogen 58,932 cubic feet.
Hydrogen 1,115 " "
Carbon monoxide 15,964 "
Carbon dioxide 4,169 "
The first three have the same heat capacity per cubic foot, so
assuming their temperature 1550° :
N^ H^ CO 76,011X534.5 = 40,627,900 oz. Cal.
CO^ 4,169X947.1 = 3,948,250
44,576,150
= 2,786,000 lb. Cal.
Absorbed in Decomposing Moisture: The 1,115 cubic feet
of hydrogen in the gases represent so much steam or water
vapor decomposed. Since 1 cubic foot = 0.09 ounces, the
heat absorbed is
1,115X0.09X29,040 = 2,914,160 oz. Cal.
= 182,130 lb. Cal.
Heat Conducted to the Air: Assuniing the conditions worked
out in the illustration tmder this heading,, this item v/ould
be approximately, for 9 min. 10 sees., in lb. Cal.
1,714X2.204X9.167 = 34,630 lb. Cal.
Heat Lost by Radiation: Making similar assumption, we
have
1,249X2.204X9.167 = 25,2401b. Cal.
THE BESSEMER PROCESS. 367
•
Recapitulation.
Lb. Cal.
Heat in converter body at starting 8,034,970
melted pig-iron 6,750,500
spiegeleisen 750,000
blast 86,580
Heat of oxidation 4,510,800
" formation of slag 59,890
Total on hand and developed 20,192,740
Heat in converter body at finish 7,183,770
" finished steel 8,632,750
slag 1,582,350
" escaping gases 2,786,000
Heat absorbed in decomposing moisture 182,130
Heat conducted to the air 34,630
Heat lost by radiation 25,240
Total accounted for ; 20,426,870
Another way of expressing this balance is to itemize the
avenues of heat evolution and utilization, as follows:
Lb. Cal.
Received from converter body 851,200
Received by oxidation 4,510,800
Received by formation of slag 59,890
Total ; . . . 5,421,890
Lb. Cal.
Used, excess of heat in gases over blast 2,699,420
Used, excess of heat in steel and slag over pig iron and
Spiegel 2,714,600
Decomposition of moisture * . . . . 182,130
Radiation and conduction 59,870
Total. 5,656,020
CHAPTER IX.
THE TEMPERATURE INCREMENT IN THE BESSEMER
CONVERTER.
In the preceding chapter, we have studied the generation of
heat in the Bessemer converter, arid its distribution. We saw
in that analysis that in a typical operation, nearly one-half of
the heat generated during the blow is carried out by the hot
gases, about half is represented in the increased temperature of
the contents of the converter, while only about 5 per cent, is
lost by radiation, etc. In the present paper we wish to analyze
still further this question of increased temperature, which is so
vitally necessary for the proper working of the process, and to
calculate the relative efficiency of the various substances oxidized
in causing this rise of temperature.
While at no time in the Bessemer operation is only one sub-
stance being oxidized, yet we can get the best basis for our
computations by assuming a charge in operation and only one
substance oxidized at a time. Whatever substance is in ques-
tion, let us assume one kilogram burnt in a given short period
of time, generating the heat of its combustion. With the
bath at a given temperature, the air, at say 100° C, comes
in contact with it, bearing the necessary oxygen. If nothing
was oxidized, the oxygen and nitrogen would simply be heated
to the temperature of the bath, and pass on and out, while the
bath would be meanwhile losing heat also by radiation. It
is evident then, that unless at least as much heat as the sum
of these two items is generated, the bath will cool, and that
only the excess of heat above this requirement . is available for
increasing the temperature ot the bath and resulting gases.
The proper procedure for us will therefore be to calculate,
in each case, the chilling effect of the air entering, subtract
this from the heat generated, and the residue is net heat avail-
able for raising the temperature of the contents of the con-
368
THE BESSEMER CONVERTER. 369
verter and the gases, and supplying radiation losses. The
latter are proportional to the time, and therefore to the amotmt
of air used, assuming blast constant.
Silicon.
This is burnt out in the first part of the process, during which
the temperature begins low and ends high. "We will there-
fore assume two temperatures, and calculate the thermal
increment at each. We will take 1250° and 1600°. The net
heat is absorbed by the bath, slag and nitrogen.
Oxygen necessary to bum one kilogram of silicon:
1X32:28= 1.143 kg.
Nitrogen accompanying this oxygen = 3.810 "
Weight of air needed = 4.953 "
Volume of air = 4.953-^1.293 = 3.831 m*
Specific.heat, 100° to 1250°, per m' = 0.3395
Specific heat, 100° to 1600°, per m' = 0.3489
Chilling effect of air at 100°, bath at 1250°
= 3.831X0.3395X1150 = 1496 Cal.
Shilling effect of air at 100°, bath at 1600°
= 3.831X0.3489X1500 = 2043 Cal.
Heat generated per kilogram of silicon = 7000 Cal.
This is, however, for cold oxygen and cold silicon burning
to- cold solid silica. Under the conditions prevailing we have
melted silicon at the temperature of the bath, oxidized by hot
oxygen to hot silica, giving a slightly different heat of corri-
bination, calculated as follows:
Heat in melted silicon at 1250° = 480 Cal.
" " oxygen required, at 1250° = 334 "
" " silica, at 1250° = 750 "
Heat of oxidation at 1250°
= 7,000 + 480 + 334—750 = 7,064 "
The difference from 7,000 is so small as to be within the
possible error of the 7,000 itself, and we can therefore make
the calculations with all the accuracy they allow, taking the
ordinary heats of oxidation from the tables.
One factor of heat generation has, however, not been men-
tioned, viz.: the heat of combination of silica with oxides of
370 METALLURGICAL CALCULATIONS.
iron and manganese to form slag. This is 148 Calories per kg.
of silica when forming iron silicate, and 90 when forming man-
ganese silicate, which quantities would become 317 and 193
Calories respectively when calculated per kg. of silicon oxi-
dizing. The question arises, however, whether it is fair to
credit all this to the silica, because this generation of heat by
slag formation is really a mutual affair, chargeable to the credit
of both silica and the other oxides; we should, therefore, not
charge it all up to the credit of silica formation, and we do
not know what part to charge to the credit of silica if we do
not charge it all. In this dilemma it may be well to remem-
ber that silicon is probably oxidized before the iron and man-
ganese, and that the heat of formation of the slag is therefore
more properly considered as being generated afterwards, and
therefore may be practically credited entirely to the oxidation
of iron and manganese.
Resume for oxidation of 1 kilo, of silicon:
Cal.
Heat generated = 7,000
Chilling effect of the blast, 100° to 1250° = 1,496
Chilling effect of the blast, 100° to 1600° = 2,043
Available heat, bath at 1250° = 5,504
Available heat, bath at 1600° = 4,957
If we assume a radiation loss proportional to the length of
the blow, i. e., proportional to the air blown in, we can find
out from average blows that this amounts to about 50 Calories
per cubic meter of blast used. The radiation loss during com-
bustion of 1 kilogram of silicon would therefore be
Cal.
3.831X50 = 192
Leaving net available heat at 1250° = 5,312
At 1600° = 4,765
The above quantity of heat is expended in raising the tem-
perature of 99 kg. of bath, 2.143 kg. of silica, and 3.810 kg. of
nitrogen gas, from their initial temperature. At the tempera-
tures of 1250° and 1600°, respectively, the heat capacity of
these products of the operation will be, per 1° C. rise:
THE BESSEMER CONVERTER. 371
Specific Heat Heat Capacity
Products. at 1250° at 1600° at 1250° at 1600°
Bath. 99 kg 0 25 0.25 24.8 24.8
Si0^2.14kg. 0.37 0.43 0.8 0.9
N^3.02m» 0.37 0.39 1.1 1.2
Totals. 26.7 26.9
Theoretical rise of temperature :
5,312 -^ 26.7 = 199° (bath at 1250°)
4,7654-26.9 = 177° (bath at 1600°)
Average rise, per average
1% of silicon = 188° C.
Manganese.
As high as 4 per cent, of manganese may be oxidized during
the blow, and therefore this heat of combustion is sometimes
important. We will calculate the net heat available for rais-
ing temperature and the rise of temperature per 1 per cent, of
manganese oxdized, i. e., for 1 kilo, of manganese per 100
kilos, of bath.
Oxygen necessary 1X16/55
Nitrogen accompanying this
Weight of air used
Volume of air used
Chilling effect of air at 100° on bath at
1250° =.0.975X0.3395X1150
Heat generated per kg. of manganese
Heat of formation of MnO.SiO^'
1.291kg. MnOx76
Total heat developed
Heat available =*= 1751 — 381
Radiation loss = 0.975X60
Net available heat
Heat capacity of 99 kg. of bath per 1°
Heat capacity of 2.4 kg. of slag
Heat capacity of 0.8 m' nitrogen
Heat capacity of products per 1°
=
0.291 kg.
=
0.970 "
=
1.261 "
=
0.975 m^
^_
381 Cal
=
1653 "
^
98 "
—
1751 "
=
1370 "
=
49 "
=
1321 "
=
24.8 "
=
0.7 "
=
0.3 "
=
25.8 "
372 METALLURGICAL CALCULATIONS.
Theoretical rise of temperature :
1321-^25.8 = 51° C.
This is. in round numbers, one-fourth the efficiency of silicon.
Iron.
While it is not desired to oxidize iron, and while it is rela-
tively less oxidizable than silicon or manganese, yet some
is always oxidized because of the great excess of iron present
in the converter. The amount of iron thus lost is variable, and
some of it, towards the end of the blow, may be oxidized to
Fe^O' instead of FeO, the larger part, however, oxidizes to
FeO. We will make the calculations for both oxides, per kilo-
gram of iron.
Formation of FeO.
Oxygen necessary 1 X 16/56 = 0.286 kg.
Nitrogen accompanying this = 0.953 "
Weight of air used = 1.239 "
Volume of air used = 0.958 m'
Chilling effect of air at 100° on bath at
1250° = 0.958X0.3395X1150 - 374 Cal.
Chilling effect of air at 100° on bath at
1600° = 0.958 X. 3489X1500 = 501 "
Heat generated per kg. of iron * = 1,173 "
Heat of formation of FeO.SiO^ = 1.286 kg.
FeO X 124 = 159 "
Total heat developed = 1,332 "
Net heat available at 1250° = 1332—374 = 958 "
Net heat available at 1600° = 1332—501 = . 821 "
Radiation losses = 0.958X50 = 48 "
Net available heat at 1250° = 910 "
Net available heat at 1600° = 773 "
Heat capacity of 99 kg. of bath per 1° = 24.8 "
Heat capacity of 2.4 kg. of slag per 1° - = 0.7 "
Heat capacity of 0.8 m^ of nitrogen = 0.3 "
.Heat capacity of products per 1° = 25.8 "
Theoretical rise of temperature:
910 -H 25.8 = 36° C.
773^25.8 = 30° C.
This is only about one-sixth as efficient as silicon.
THE BESSEMER CONVERTER. 373
Formation of Fe^O^.
Weight of air used = 1.859 kg.
Volume of air used = 1.438 m'
Chilling effect of air at 100° on bath at
1600° = 1.438X0.3489X1500 = 753 Cal.
Total heat developed by oxidation = 1746
Heat of formation of slag = 159
Total heat developed = 1905
Heat available = 1905—753 = 1152
Radiation losses = 1.438X50 = 72
Net heat available = 1080
Heat capacity of products = 26
Theoretical rise of temperature :
1080-^26 = 42° C.
Titanium.
While titanium is an unusual constituent of pig iron, yet it is
conceivable that titaniferous pig iron might be made and blown
to steel. If so, the following calculation, based on a quite re-
cently determined value for the heat of oxidation of titanium,
will show that the titanium is a valuable heat producing sub-
stance, being, in fact, weight for weight three fourths as efficient
as silicon.
Oxygen needed 1 X 32/48 = 0.667 kg.
Nitrogen accompanying this = 2.222 "
Air used = 2.888
Volume of air needed = 2.250 m'
Chilling effect of air at 100° on bath at
1250° = 2.250X0.3395X1150 = 878 Cal.
Heat generated per kg. of Ti = 4542 "
Heat of formation of slag — imknown
Net heat available = 4542—878 = 3664 "
Deducting 144 for radiation losses = 3520 "
Heat capacity of products per 1° C. = 26.5 "
Theoretical rise of temperature:
3520-3-26.5 => 133° C.
=:
0.889
kg.
=
2.963
It
=
3.852
(C
=
2.964 :
m'
=
1157 Cal,
=
7272
((
=
6115
((
=
5967
ft
=
26.6
<i
374 METALLURGICAL CALCULATIONS..
Aluminium.
This metal is also rarely found in pig iron, yet when present
it would be a powerful heat producer, as the following calcu-
lations show:
Oxygen needed 1X48/54
Nitrogen
Air
Volume of air
Chilling efEect of air at 100° on bath at
1250° = 2.964X0.3395X1150
Heat generated per kg. of Al
Heat of formation of slag — uncertain
Heat available = 7272—1157
Deducting for radiation losses
Calorific capacity of products per 1° C.
Theoretical rise of temperature:
5967 -H 26.6 = 224° C.
If a blow was running cold, and ferro-silicon was not on
hand to add in order to increase its temperature, ferro-alum-
inium, or aluminium itself, would be a good substitute in the
emergency.
Nickel.
It is hardly probable that nickeliferous pig iron would be
blown to steel, because of the waste of valuable nickel in the
slag ; yet if 1 per cent, of nickel were thus oxidized, calculations
similar to the preceding would show a net rise in temperature
of the contents of the bath of about 33° C.
Chromium.
Quite recently some chromiferous pig iron has been blown
in the Bessemer conveijter, and those in charge were hampered
by the lack of data as to how chromium would behave during
the blow and its heat value to the converter, Technical litera-
ture will probably soon contain an account of the practice
which has been developed at the Sparrow's Point works of the
Maryland Steel Company. No thermo-chemist has, as yet,
determined the heat of formation of chromium slag. From
TEE BESSEMER CONVERTER. 375
what we know of the chemical reactions of chromium it is
likely that this heat ■ is considerable. Neglecting the heat of
formation of slag, we have the following approximation to its
heating efficiency, assimiing it to be oxidized when the bath
is near to its maximum temperature:
Oxygen needed 1x48/104 = 0.462 kg.
Nitrogen entering = 1.540 "
Air used . = 2.002 "
Volume of air = 1.548 m*
Chilling effed; of air at 100° on bath at
1600° = 1.548X0.3489X1500 = 810 Cal.
Heat of oxidation = 2,344 "
Net heat = 2344—810 = 1,534 "
Deducting for radiation losses = 1,457 "
Calorific capacity of products per 1° = 26.1 "
Theoretical rise of temperature :
1457 -- 26.1 = 56° C.
Carbon.
This element commences to be oxidized in large amount only
towards the middle of the blow, when the temperature of the
bath is high, because of the previous oxidation of silicon. It
will, be about right, therefore, to estimate the bath at an
average temperature of 1600° during the elimination of carbon.
The product is mostly CO, but partly CO^. We will, therefore,
calculate for each of these possible products separately. The
net heat available, after allowing for average radiation losses,
is used to increase the temperature of the products, i. e., of the
bath, the nitrogen and the CO or CO^.
Oxidation to CO^.
Oxygen required = 1x32/12 = 2.667 kg.
Nitrogen accompanying = 8.889 "
Air used = 11.556 "
Volume of air = 8.937 m'
Chilling effect of air at 100'^ on bath at
1250° = 8,937X0.3395X1150 = 3,489 Cal.
Heat of oxidation = 8,100 "
376 METALLURGICAL CALCULATIONS.
Heat available = 8,100—3,489 = 4,611 Cal.
Radiation losses = 8.937X50 = 447 "
Net heat available = 4,164 "
Heat capacity of 99 kg. bath = 99X0.25= 24.80 "
Heat capacity of 7.05 m^ N^ = 7.05 X 0.37 = 2.61 "
Heat capacity of 1.90m3CO2= 1.90X0.88= 1.70 "
Heat capacity of products, per 1° C. = 29.11 "
Theoretical rise of temperature :
4,164^29.11 =143°C.
Since carbon bums to CO^ principally at the beginning of
the blow, while the bath is cold, we see that carbon thus con-
sumed is about three-quarters as efficient as an equal weight of
silicon in raising the temperature of the bath.
Oxidation to CO.
Oxygen needed 1X16/12
Nitrogen accompanying
Air used
Volume of air
Chilling effect of air at 100° on bath at
1600° = 4.469X0.3489X1500
Heat of oxidation
Heat available = 2,430—2,339
Radiation losses = 4.469X50
Net heat available = 91—233 =—142 "
Heat capacity of 99 kg. of bath = 99X0.26 = 24.8 "
Heat capacity of 3.5 m^ of N^ \ ^ .^^v on _ „ , ■<
Heat capacity of 1.9 m^ of CO / ^-^X^-^^ " ^-^
Heat capacity of products = 26.9 "
Theoretical rise of temperature :
— 142 -f- 26.9 = — 5°C.
The result is, therefore, that when carbon bums, as it mostly
does, to CO, and the temperature of the bath is high, there is
practically no further rise of tempefature, for the heat of oxi-
dation is barely sufficient 'to counteract the chilling effect
of the air and to supply radiation and conduction losses.
In the abpve calculations, no allowance was made for heat
=
1.333 kg.
=
4.444 "
=
5.777 "
=
4.469 m'
=
2,339 Cal,
2,430 "
91 "
233 "
THE BESSEMER CONVERTER. 377
required to separate carbon from its combination with iron,
or for the variation in the heat of combination of carbon with
oxygen from the combination heats at ordinary temperatures.
The former is not known, or perhaps is very nearly zero. The
heat of oxidation of liquid carbon at 1250° to CO^ or at 1600°
to CO is calculated as follows :
Oxidation 1 kg. C to CO' at 0° = 8,100 Cal
Heat to raise 1 kg. C to 1250° = 505 Cal.
Heat to liquefy 1 kg. at 1250° =. 129 "
Heat to raise 2.67 kg. 0= to 1250° = 779 "
Heat to raise reacting substance to 1250° = 1,413 "
Heat in 3.67 kg. CO' at 1250° = 1,493 "
Heat of reaction at 1250° (8,100+1,413-1,493) = 8,020 "
For production of CO, at 1600°, the correction is larger,
as is seen from the following:
Oxidation 1 kg. C to CO at 0° = 2,430 Cal.
Heat to raise 1 kg. to 1600° = 680 Cal.
Heat to liquefy 1 kg. C at 1600° = 156 "
Heat to raise 1.33 kg. O' to 1600° = 554 "
Heat to raise reacting substances to 1600° = 1,390
Heat in 2.33 kg. of CO at 1600° = 1,104 "
Heat of reaction at 1600° (2,430 + 1,390—1,104) = 2,716 "
The use of this corrected value makes the oxidation of C
to CO give a small net heat development, with consequent
•slight rise of temperature, instead of the slight cooling effect
before calculated. The conditions are so nearly even, however,
that a slight increase of temperature or slowing up of the blow
would wipe out the heat excess.
Phosphorus.
This is the last important element to be considered, and is
always eliminated after the carbon, at the maximum bath tem-
perature, which we will assume, for calculation, atl600°. The
heat generated, at ordinary temperatures, is 5892 Calories per
kilogram of solid phosphorus. Per kilogram of liquid phos-
phorus it would be only 5 Calories more, or 5897 Calories. For
the reaction at .1600° we would have a difEerent value, probably
378 METALLURGICAL CALCULATIONS.
some 500 Calories more, but the necessary data concerning the
specific heats of P and P^O' are not known, and we must omit
this calculation. The heat of combination of iron and phos-
phorus is also a doubtful quantity. Ponthiere places it as high
as 1,397 Calories per kilogram of phosphorus, but this ap-
pears altogether improbable since another experimenter could
obtain no heat of combination at all. ^ As concluded in another
place, I advise for the present omitting this questionable
quantity. '
The phosphorus pent-oxide forms 3CaO . P^O' with the lime
added, but since there is always more lime present than cor-
responds to these proportions (BCaO : P^O^ ::168 : 142), the
calculation of heat of formation of the slag must be based on
the amount of P^O' formed (1123 Calories per kilogram of
P^O^). This amounts to a considerable item. On the other
hand, the lime needed for slag is put in, usually preheated, for
the sole purpose of combining with the P^O^- It seems, there-
fore, only right to charge the phosphorus with the heat re-
quired to raise this lime to the temperature of the bath. The
lime added averages three times the weight of P^O' formed,
and is preheated usually to about 600°. Assuming these con-
ditions, the following calculations can be made per kilogram
of phosphorus oxidized :
Oxygen required
Nitrogen accompanying
Air used
Volume of air
Heat of formation of slag,
2.29 kg. P='0'X1123
Heat of oxidation of phosphorus
Total heat developed
Chilling effect of air at 100° on bath at
1600° = 4.32X0.3489X1500 = 2261
Chilling effect of lime (600° to 1600°)
(2.29X3) X 0.328X1000 = 2253
Chilling effect of blast and lime = 4514
Heat available = 8469—4514 = 3955
Radiation losses = 4.32X50 = 216
Net heat available = 3739
1.29 kg.
4.30 "
5.59 "
4.32 m'
2572 Cal,
5897 "
8469 "
THE BESSEMER CONVERTER. 379
Heat capacity 99 kg. of bath = 99 X 0.25 = 24.8 kg.
Heat capacity 3.4 m= of N^ = 3.4X0.39 = 1.3 "
Heatcapacity6.9kg. of slag = 6.9X0.3 = 2.1 "
Heat capacity of products, per 1° = 28.2 "
Theoretical rise of temperature :
3739-^28.2 = 133° C.
If the lime were added cold, its cooling effect would be 883
Calories greater, the net heat available would be 883 Calories
less, and the calculated rise of temperature 31° less, or 102° C.
Using the preheated lime we can regard phosphorus as being
practically two-thirds as efficient, weight for weight, as silicon;
with cold lime, about one-half as efficient.
Resume.
Heat effect of oxidizing 1 kilogram of element.
M -IS, « 5 O^
.o K "§ 'o" S "I 2 a
I ^ 2 "fel I i ~ I
^ I 1 '^l ^^ 1|
^ «>. H^ *«* r^ 'i.i c* to S
^ o o -s^^a-Sto
te! (I, K oa5:§te;E~,K
Silicon 7,000 ... 7,000 1,688 5,312 188°
Manganese ' 1,653 98 1,751 430 1,321 51°
Iron (to FeO) 1,173 159 1,332 422 910 33°
Iron (to FcjOj) 1,746 159 1,905 825 1,080 42°
Titanium 4,642 . . . 4,542 1,022 3,520 133°
Aluminium 7,272 ... 7,272 1,305 5,967 224°
Nickel 1,051 159 1,210 378 832 33°
Chromium 2,344 ... 2,344 887 1,457 56°
Carbon (to COj) 8.100 ... 8,100 3.936 4,164 143°
Carbon (to CO) : . . : .2,430 . . 2,430 2,572 -142 —5°
Phosphorus 5,8972,572 8,469(2,477)3,739 133°
? 2,253*)
* Chilling effect of lime added, preheated to 600°
It must be observed that the above table is for comparison
only, it cannot be used for an actual case, such as when 1 per
cent; of silicon, 3 of iron, 4 of carbon and 2 of phosphorus are
380 METALLURGICAL CALCULATIONS.
oxidized. In such a case, the rise in temperature would be
only very roughly:
From silicon 1 X 188 = 188°
From iron 3X 33 = 99°
From carbon - say = 0°
From phosphorus 2X 133 = 266°
Total = 553° C.
It IS to be recommended that in each specific case the calcu-
lation be made for the specific conditions obtaining, such as
temperature of the metal at starting, temperature of the blast,
time of the blow (as far as this affects radiation and conduc-
tion losses), proportion of carbon burned to CO^, free oxygen
in the gases, moisture in the blast, temperature and quantity
of lime added, corrosion of lining. When all these items and
conditions are taken into account there will be room for only
small discrepancy between the calculated and the observed rise
of temperature. The chief items needing experimental re-
search at present are: The specific heat of the melted bath,
the specific heat of the slag, the heat of combination of various
elements comprising the bath, the heat of formation of the
slag, and the heat of oxidation of some of the rarer elements.
Such establishments' as the Carnegie Institution could not do
the cause of metallurgy better service than to subsidize metallur-
gical laboratories for the determination of such data.
CHAPTER X.
THE OPEN-HEARTH FURNACE.
By the above title we mean to designate not only the re-
generative gas furnaces for maldng steel but also those for
reheating puiposes; in other words, regenerative or recupera-
tive reverberatory furnaces. Prominent among these is the
Siemens-Martin furnace, with complete gas and air preheating
regenerative chambers. In all these furnaces the charge is
heated, or kept hot, partly by direct contact with the gaseous
products of combustion and partly by radiation from the flame
and the sides and roof of the furnace. It was the Siemens
brothers who first insisted on the relatively great importance
of the radiation principle, in distinction to the direct impinge-
ment of the flame on the material, pointing out that a luminous
flame radiates from all parts of its volume, while a hot, solid
body radiates only from its surface, and direct impingement
interferes with the development of perfect combustion and
cotnmunicates heat relatively slowly at best.
Gas Producers.
Gas producers are the usual adjunct for open-hearth fur-
naces, excepting where natural gas or blast furnace gas is
available. They may be placed far from the furnace, when
they deliver cool gas to the regenerators, or close to the fur-
nace, delivering comparatively hot gas to the regenerators, or
even be made part of the furnace itself, delivering their hot
gas immediately to the ports of the furnace. The latter is un-
doubtedly the most economical arrangement where practicable.
The following generalizations concerning the relations of the
gas producer to the open-hearth furnace may be made: Pro-
ducers furnish 4,300 to 4,600 cubic meters of gas per metric
ton of coal used (150,000 to 160,000 cubic feet per short ton) ;
the gas produced runs 3 to 8 per cent. CO*, 5 to 20 per cent. H*.
20 to 30 per cent. CO, and 50 to 60 per cent. W; its calorific
381
382 METALLURGICAL CALCULATIONS.
power is 750 to 1000 Calories per cubic meter (47 to 63-pound
Calories, or 85 to 115 B. T. U. per cubic foot) ; its calorific
power represents 60 to 90 per cent, of the calorific power of the
fuel used; in steel-making processes, the keeping of the furnace
up to proper heat requires the gasifying of 25 to 35 kilograms
(50 to 80 pounds) of coal per hour, in the producers, for each
ton of metal capacity of the furnace; good producers gasify
60 to 65 kilograms of coal per hour per each square meter of
gas-producing area (10 to 15 pounds per hour per each square
foot) ; a furnace therefore requires some 0.4 to 0.6 square meter
(4 to 6.5 square feet) of gas-producing air in the producers
for each ton of metal capacity of the furnace.
Flues to Furnace.
In conducting the gases to the furnace, the flues or conduits
should be of ample size. If too small the gas must pass through
them with high velocity, requiring considerable draft to give
them this velocity, which the chimney, or blower may or may
not be capable of furnishing. Producers are almost always
worked by a steam blower, furnishing mixed air and steam
and a plenum of pressure in the upper part of the producer,
which suffices to send the gas through the conduits under a
slight pressure, and thus avoids any sucking in of air through
crevices in the conduits. With too small conduits the result-
ing friction and high velocity required may give the blower
more work than it can do, and thus entail demands for draft
upon the furnace stack. A reasonable rule is to give the flues
such cross-sectional area that the hot gas which must pass
through them shall have a velocity between 2 and 3 meters per
second.
Regenerators.
The dimensions of the regenerators are of the first import-
ance to the working of the furnace. They should have suf-
ficient length in the direction the gas currents are passing,
so that the gases may be properly cooled or heated ; they should
have sufficient cross-sectional area of free space, so that the
velocity of the gases through them is not too great; they must
bave sufficient thermal capacity, so that they can absorb the
requisite quantity of heat.
Length. — From 4 to 6 meters (13 to 20 feet) is a sidtable
THE OPEN-HEARTH FURNACE. 383
length in the direction of the gas currents. This permits the
hot products to become properly cooled before going to the
chimney, and the gas or air to be properly heated before en-
tering the furnace. The shorter length may be used when the
regenerator is of large cross-sectional area, with slow velocity
of gas currents through the free spaces; the longer when the
regenerator is rather restricted in cross-section and the gas
currents have somewhat high velocity.
Cross-Section. — The free-space sectional area should be such
that the gases where hottest should not have a calculated
velocity of over 3 meters (10 feet) per second, and if calculated
for 2 meters (6.5 feet) will give much better results as regards
transmission of heat to the checker work. In this manner,
knowing how much gas must go through the regenerator and
what its maximum teniperature will probably be, the cross-
section area of free passage space can be calculated. The rela-
tion of this to the cross-section of the entire stove must next
be considered, and this is entirely a question of how the checker
work is built up. If the bricks are stacked close together the
free space may be reduced to as much as one-half the total;
as ordinarily stacked it may be 60 to 80 per cent, of the total;
if perforated bricks are used, as in blast furnace firebrick-
stoves, the area- of free space averages one-half the total ; with
ordinary bricks the average is 70 per cent. This is a question
which is very variously worked out in different furnaces, and
to which not as much scientific thought has been given as should
be. The thickness of the bricks influences greatly the rela-
tive amotmt of free space and filled space, and the rate at
which the generator heats up or cools off. In a regenerator
of given length and cross-section closer packing of the bricks
gives more heat absorbing surface, increases the velocity of the
gases and dinmiishes the cross-sectional area of each passage
and of the sum of all the passages; some of these factors in-
crease the efficiency of the regenerator, others tend to decrease
it, and there are, therefore, several independent variables to be
considered in finding the beSt arrangement for highest effi-
ciency. A numerical solution is indeed a possibility, but is
too involved for an elementary presentation of the subject.
Relative Sizes. — The relative sizes of gas and air regenera-
tors is a question of importance which admits of easy solu-
,38^ METALLURGICAL CALCULATIONS.
tion by calculation. So far we have treated the pair of re-
generators together, and discussed the sum of their cross-sec-
tions as deduced from the volume of products passing through
at an assumed maximum temperature and allowable velocity.
The regenerators at one end of a furnace are, however, usually
divided into a pair or set, one for heating gas and the other
for heating air. This is not usual where natural gas is used
because of the deposition of soot in the regenerator by the
latter when it is heated, but nine out of ten open-hearth fur-
naces preheat their gas as well as the air. The heating capacity
of the regenerators should be divided in proportion to the
calorific capacities of the gas and air simultaneously heated.
The problem is. therefore to find the heat capacity per degree
of the gas and air used, or, more exactly, the total heat capacity
of each of these between the temperature at which they enter
the regenerators and that at which it is desired that they should
enter the furnace.
Problem 69.
An open-hearth furnace uses producer gas containing, by
volume, at it reaches the regenerators:
CO 26.97 per cent.
CO^ 4.37
CW 0.33
ff 13.00
NH' 0.21
ffS 0.10
W 54.01
Air 1.03
Each cubic meter, measured at 20° C. and 720 m. m. baro-
metric pressure, is accompanied by 73.22 grams of moisture, as
determined by draAving thiough a. drying tube and weighing the
moisture. The air used is at 20° C, 720 m.m. barometer, and
three-quarters saturated with moisture. A maximum of 10 per
cent, more air is used than is theoretically necessary to com-
pletely bum the gas (assuming NH' to bum to H^O and N0^
and allowing for the air already present in the gas). The
gas and air may be both assilmed as coming to the regenerators
at 20° C, and to be heated, in the regenerators to 1200° C.
THE OPEN-HEARTH FURNACE. 385
Required. — (1) The relative volumes of gas and air passing
through the gas and air regenerators.
(2) The total amotmts of heat necessary to be furnished to
each, per cubic meter of gas used.
(3) The relative sizes of the two regenerators.
Solution. — (1) Taking 1 cubic meter of the dry gas, as rep-
resented by the analysis, the combustion of its combustible in-
gredients is represented by the equations:
2C0 +02 = 2C02
CH* +202 = C02 + 2ffO
2H2 +02 =2H20
4NH?+702 =6H20 + 4N02
2H2S +302 =2H20 + 2S02
And since molecules represent volumes, each unit volume of
CO, CHS H2, NH* and H2S is seen to require respectively 0.5,
2, 0.5, 1.75, or 1.5 volumes of oxygen. The 1 cubic meter
of dry gas therefore requires oxygen as follows:
CO 0.2697X0.5 = 0.1349 m'
CH* 0.0437X2.0 = 0.0874 "
H2 0.1300X0.5 = 0.0650 "
NH'0.0021X1.75 = 0.0037 "
H^S 0.0010X1.5 = 0.0015 "
Total =. 0.2925 "
Air needed = 0.2925-^0.208 = 1.4062 m«
Add 10 per cent, excess = 1.5468 "
Air present in gas = 0.0103 "
Air to be supplied = 1.5365 "
Each cubic meter of dry gas, at any given conditions of tem-
perature and pressure, would require 1.5365 cubic meters of dry
air at the same conditions of temperature and pressure. This,
however, is not exactly the relation required, for the reason
that the gas is accompanied by considerable moisture, which
in reality adds to its volume, while the air is also moist, add-
ing to its volume. Two corrections must therefore be ap-
plied; first, to calculate the volume of the moisture accom-
ipanjdng 1 cubic meter of (assumed) dried gas; the second, to
calculate the volume of moisture accompanying 1.5365 cubic
386
METALLURGICAL CALCULATIONS.
meters of (assumed) dry air. Assuming our gas and moist
air both at 20° and 720m.m. pressure, the volume of moisture
accompanying 1 cubic meter of (assumed) dry gas is the vol-
ume of 73.22 grams of moisture at these conditions, which is
73.22 -^ 1000-^0.81 X
273 -I- 20 .760
273
X ^ = 0.1024 m=
The air being 0.75 saturated with moisture at 20°, the tension
of this moisture will be 17.4X0.75 = 13 m.m. The air proper
is therefore imder 720 — 13 = 707 m.m. tension instead of
720 m.m. and the volume of the moist air containing 1.5365
cubic meters of (assumed) dry gas is therefore:
720
1.5365 Xsq= = 1.5648 cubic meters.
The relative volumes of actual (moist) gas and actual (moist)
air used are therefore :
1.1024:1.5684 = 1.0000:1.419 (1)
(2) To calculate the heat necessary to raise gas and air from
20° to 1200° per cubic meter of gas used, the best preliminary
is to calculate the composition of the gas, including its mois-
ture. Since 1 cubic meter of (assumed) dry gas is accom-
panied by 0.1024 cubic meter of water vapor, the sum being
1.1024 cubic meters, we can calculate the real percentage com-
position to be:
CO 24.47 per cent.
CO^ 3.96
CH* 0.30
H^" 11.79
NH'.
W...
Air..
H'O.
. 0.19
. 0.09
.48.99
. 0.93
. 9.29
Of the above quantities the (assumed) dry gas in 1 cubic meter
of (actual) moist gas is 0.9071 cm. The air used for its com-'
bustion will therefore be:
THE OPEN-HEARTH FURNACE.
387
0.9071X1.5365 = 1.3938 m' dry air.
0.9071X1.5648 = 1.4080 m^ moist air.
1.4080—1.3938 = 0.0142 m' of moisture.
The heat required by the 1 cubic meter of (actual) moist gas
is fotmd as follows :
Volume XMea,n Specific Heat
20°— 1200°
CO 0.2447
H^ 0.1179
■ X 0.3359 = 0.2895 Calories
N* 0.4899
Air 0.0093
CO^ 0.0396
X 0.6384 = 0.0253
H^O 0.0929
X 0.5230 = 0.0486
CH^ 0.0030
X 0.6484 = 0.0019
NH' 0.0019
X 0.5752 = 0.0011
H^S 0.0009
X 0.5230 = 0.0005
Mean cal. capacity per 1° = 0.3669 "
Total calorific capacity 20°— 1200°
0.3669X1180 = 432.9 Calories.
The calorific capacity of the moist air simultaneously heated
through the same range will be:
Air 1.3938X0.3359 = 0.4682 Calories.
H^O 0.0142X0.5230 = 0.0074
Sum
= 0.4756
Total calorific capacity 20°— 1200°
0.4756X1180 = 561.2 Calories.
The air regenerator should therefore .have' 561.2^-432.9 =
1.30 times the heating power or cross-section of the gas re-
generator; i. e., 30 per cent. more. Or the combined capacity
of the pair of regenerators should be divided so as to give 57
per cent, to the air regenerator and 43 per cent, to the gas
regenerator. In ordinary practice it is usual to allow about 60
388 METALLURGICAL CALCULATIONS.
and 40 per cent, respectively; it is better to calculate ahead for
the specific case in hand, if the composition of the gas to be
used is known.
Valves and Ports.
No very exact rule can be given as to the size of the gas
and air valves, or those leading the products to the chimney.
If made too large they are cumbersome to operate and apt
to warp; if made too small they give undue obstruction to the
flow of gas. A general rule is to calculate the free opening,
such as to give the gases passing through a velocity between
3 and 5 meters (10 and 16 feet) per second, allowing, of course,
for the average temperature of the gas or air products of com-
bustion passing through them. It may be remarked that
while water-seal valves are very convenient, the water is evap-
orated where in contact with gas or air, and diminishes the
heating efficiency of the furnace, the use of a non-volatile
oil or a fine sand would appear perferable to water.
The ports are a very important part of the furnace, and may
be designed in many different styles for various ways in which
a furnace is to be worked. Their cross-section, however, can
be calculated when we know the volume of gas or air leaving
the regenerators and their temperature, or the volume of the
products of combustion entering the regenerators and their
temperature. They should be so designed that the velocity of
the gases through them is not over 20 meters per second, while
10 meters per second is a better velocity to use. A long fur-
nace can admit of higher velocities at the ports than a short
one; but in any case the higher the velocity the farther com-
plete combustion will occur from the ports, and if the velocity
is too high for the length of the furnace combustion may even
be continued in the opposite regenerators and less than the
maximum oecur in the furnace. This is a condition to be
scrupulously avoided if possible.
Problem 70.
Producer gas of the following composition:
CO 24.47 per cent. NH» 0.19 per cent.
C0=' '. . 3.96 " N2 48.99
CH* 0.30 " Air 0.93
W 11.79 " H^'0 9.29
ffS 0.09
THE OPEN-HEARTH FURNACE. 389
is burned with 1.408 times its volume of moist air (see Prob-
lem 69). The furnace treats 50 metric tons bf steel in 12 hours,
using 17.5 tons of coal in the producers, from which 15 tons
of carbon pass into the gas. The gas and air pass out of the
regenerators at 1200°, and the products of combustion (as-
sumed complete) pass into the opposite regenerators at 1400°.
Assume a maximum velocity of the hot gas and air as 10 meters
per second, as they pass through the ports.
Required. — (1) The volume of gas and air at 20° C. and
720 m.m. barometer used by the furnace per second.
(2) The areas of the gas and air ports.
(3) The velocity of the products entering the opposite ports.
Solution. — (1) The carbon in 1 cubic meter of the gas at
standard conditions is
CO 0.2447
CO^ 0.0396
CH* 0.0030
0.2873X0.54 = 0.1551 kg.
Gas used in 12 hours (standard conditions) :
15,000-^0.1551 = 96,710 m'
Per second = 2.24 m'
Gas used at 20° C. and 720 m.m. :
2.24 X ?^|^ X ?^ = 2-53 m» per second. (1)
Air used (standard conditions) :
2.24X1,408 = 3.15 m'
Air used at 20° C. and 720 m.m. :
2.53 X 1,408 = 3.56 m' per second. (1)
(2) The volume of gas used per minute, as it issues from the
ports at 1200°, is
and Of air 3.15X " X " = 17.9 "
and assuming a maximum velocity for each of 10 meters per
second, the areas of the ports must be:
390 METALLURGICAL CALCULATIONS.
Gas ports 1 .28 m'
Air ports." 1 ; 79 "
Sum 3.07 " (2)
(3) There is usually contraction when gases burn, the pro-
ducts having less volume than the gas and air used. In-
specting the equations of combustion of CO, CH^ H^ ffS and
NH' given in Problem 69, we can construct the following table
of relative volumes concerned and the ensuing contraction:
CO CH^ W H'S NH»
Volume used 1.0 1.0 1.0 1.0 1.0
Oxygen used 0.5 2.0 .0.5 1.5 1.75
Gases combining 1.5 3.0 1.5 2.5 2. 75
Volume of products 1.0 3.0 1.0 2.0 2.5
Contraction 0.5 0.0 0.5 0.5 0.25
Using 1 cubic meter of producer gas the contraction resulting
from its combustion with an excess of air is
CO 0.2447X0.5 = 0.12235 m'
CH^ = 0.00000 "
iP 0.1179X0.5 = 0.05895 "
H^S 0.0009X0.5 = 0.00045 "
NH' 0.0019X0.25= 0.00050 "
Total contraction = 0 . 1822 "
Since the volume of gas, plus air used, is 2.408 m', the volume
of the products, at standard conditions, is
2.408-0.182 = 2.226 m'
per cubic meter of gas used under standard conditions. The
volume of products per minute, at standard conditions, is,
therefore,
2.226X2.24 = 4.986 m'.
And at 1400° and 720 m.m. pressure:
4.9Sexi«5_^xl-3.T„..
THE OPEN-HEARTH FURNACE. 391
Since the sum of the area of gas and air ports is 2.91 m^ the
velocity of the products in these ports will be
32.7 -h 3.07 = 10.7 m. per second. (3)
In both the calculations of the size of the ports and the
velocity of the products we have assumed the tension of the
gas, air or products in the ports to be the prevailing atmos-
pheric tension. This may or may not be exactly true, because
the air or gas may be under a slightly less tension, being drawn
into the furnace by the stack draft. If the pressure inside
the furnace, with doors closed, is greater or less than atmos-
pheric pressure, the tension of the gases in the entrance ports
will be correspondingly greater or less than the atmospheric
pressure, while the tension of the products will probably always
be less than atmospheric pressure, because of the stack draft.
Under ordinary conditions these corrections are too small to
need to be taken into consideration.
Laboratory of Furnace.
The laboratory consists of the open space enclosed between
the hearth, sides, ends and roof. Its dimensions vary with the
intended capacity of the furnace and the ideas of the designer.
If a hearth is to contain, say, 50 tons of melted steel, which
weighs some 7 tons per cubic meter (425 pounds per cubic
foot) , there will be contained in the furnace at one time 7 cubic
meters, or 260 cubic feet of steel. The deeper this lies the
more slowly it- will be heated or oxidized by the flame, and there-
fore there is a limiting depth of, say, 50 centimeters, or 20
inches, which it is not advisable to exceed, while a more shallow
bath will result in faster working. Assuming a depth of 40
centimeters (16 inches), the volume divided by the depth will
give the area of the bath:
7-^0.4 = 17.5 square meters
26d-f- 1.25 = 208 square feet.
We can then either choose a convenient width, consistent
with a practicable roof span, and derive the length, or choose
a length and derive the width, or chooge a certain ratio of
length to width, and derive both. If the width is 3 meters the
392 METALLURGICAL CALCULATIONS.
length must be 5.8; if the length is assumed 5 meters, the width
is 3.5; if the ratio of length to width is 2 to 1, the length figures
out 5.92 meters and the width -2.96. These dimensions are
those of the bath of metal, and each should be increased by
at least 1 meter to get the area of the hearth inside the walls,
thus allowing 0.5 meter clear space all around the metal.
If a furnace is short it should be wide and the roof high,
in order to give cross-sectional area and thus diminish the
velocity of the gases over the hearth. The gases attain their
maximum temperature in the laboratory, theoretically, some
1700° to 1900°, and their velocity depends solely on the vertical
cross-sectional area of the laboratory or body of the furnace.
In Problem 70, for instance, about 5 cubic meters of products
of combustion (measured at standard conditions) pass through
the furnace per second. At 1800° this volume would be 38
cubic meters, and if the laboratory were 4.5 meters wide by
1.5 meters high above the level of bath, there would be 7.25
square meters of cross-section, and the velocity of the gases
would be 38-5-7.25 = 5.2 meters per second. This would
allow barely 1 second for the hot gases to pass over the bath,
which would result in a low rate of heating and probable in-
complete combustion, for the gas can only burn as it gets mixed
with air, and it is hardly likely that 100 per cent, of, it would
get mixed with air and consumed in 1 second. Such could
only be attained by sub-division of the gas and air and very
intimate mixture at the ports. Much of the economy undoubt-
edly attained by raising the roof of open-hearth furnaces is due
to the slowing-up of the gas currents in the laboratory, though
it is usually ascribed to avoidance of contact of flame and
bath, increased heating by radiation, etc. In the writer's opinion
the raising of the roof from 1 to 2 meters, let us say, thus doub-
ling the vertical cross-sectional area, cutting in half the velocity
of the gases through the furnace, and doubling the period in
which they are able to combine, and to radiate or impart heat
to the furnace walls and charge — is the principal reason for
the increased economy observed.
An equally important improvement is lengthening the dis-
tance between ports. There is a limit to the width of the
furnace, set by the practicable arch for the roof; there is also
a limit to the height of roof, set by the increasing distance of
THE OPEN-HEARTH FURNACE. 393
the gases from the hearth ; when both these factors have reached
their maximum, further efficiency of utilization of the heat
of combustion can only be secured, as far as the body of the
furnace is concerned, by lengthening the hearth. There is
no mechanical limit, and in every case the distance between
the ports and the velocity of the gases should be such that
complete combustion takes place in the furnace laboratory
before the products pass into the regenerators.
Problem 71.
H- H, Campbell gives in the Transactions of the American
Institute of Mining Engineers, 1890, analyses made at the
Pennsylvania Steel Co.'s works, as follows:
Gas Burned, Products,
Entering Furnace. Leaving Furnace.
CO' 5.5 per cent. 3.1 per cent.
0== 2.3
CO 8.2
CH* ■ 7.3
W 39.8
W 36.9
0.7
7.1
0.0
11.6
77.5
Required. — (1) The proportion of the calorific power of the
fuel developed while passing through the body of the furnace.
(2) The proportion of the air necessary for complete com-
bustion which was used.
Solution.— {!) One cubic meter of the gas contains the fol-
lowing weight of carbon:
(0.055 + 0.0824-0.073) X 0.54 = 0.1134 kg.
One cubic meter of products contains:
(0.071 -h 0.031) X 0.54 = 0.0551 kg.
Therefore, volume of products per 1 cubic meter of gas:
0.1134-^0.0551 = 2.06 m'.
Calorific power of 1 cubic meter of gas:
CO 0.082X3062= 251 Calories
CH* 0.073X8623 = 629
IP 0.398X2613 = 1040
1920
394 METALLURGICAL CALCULATIONS.
Calorific power of 2.06 cubic meters of products:
CO 0.071X3062= 217 Calories.
B? 0.116X2613= 303
520
520X2.06 = 1071 • "
Heat developed in the furnace:
1920 - 1071 =.849 Calories.
Proportion of the possible heat development:
849-^1920 = 0.442 = 44.2 per cent. (1)
(2) The 1 cubic meter of gas needed, to burn its combustible
constituents, the following amount of oxygen:
CO
0.082X0.5
= 0.041 m'
CH«
0.073X2.0
= 0.146 "
W
0.98 X0.5
= 0.199 "
Sum = 0.386
Oxygen present in gas = 0.023
Oxygen needed from air = 0.363 "
Air = 0.363-:- 0.208 = 0.745 "
The 2.06 m' of products of incomplete combustion require for
their combustion:
CO 0.071X2.06X0.5 = 0.0731 m' oxygen,
ff 0.116X2.06X0.5 = 0.1195"
Sum = 0.1926 "
Oxygen present in the products = 0.0070 "
Oxygen needed from the air = 0.1856 "
Air needed to complete combus-
tion = 0.892 "
Total air needed for complete
combustion = 1.745 "
Air supplied in the furnace = 0.853 "
Percentage supplied:
0.853 H- 1.745 = 0.489 = 48.9 per cent. (2)
THE OPEN-HEARTH FURNACE. 395
It is almost needless to remark that with less than half the
air necessary for complete combustion supplied, a high calorific
intensity of flame and a high utilization of the calorific power
of the fuel are impossible. More air should have been used
and the furnace made longer, so as to secure perfect combus-
tion in the furnace, and not have over half the possible de-
velopment left to take place in the regenerators or stack.
Chimney Flues and Chimney.
The gases pass into the chimney flues at from 150° to 450°.
If their volume at an assumed average temperature of, say,
300° is calculated, they can be given an assumed velocity of
2 to 3 meters (5 to 10 feet) per second, aiid thus a suitable cross-
sectional area of the chimney flues obtained. The stack will
work best with a velocity of 5 meters per second, and thus its
cross-sectional area may be calculated. A height of 25 to 30
meters (75 to 100 feet) is sufficient for most furnaces.
Miscellaneous.
Some other data useful in figuring up the dimensions and
running conditions of modem open-hearth steel furnaces are
the following, taken mostly from an article by H. D. Hess, in
the Proceedings Engineering Club of Philadelphia, January,
1904:
Average coal consumption, in pounds, per hour per ton of
metal capacity of the furnace, 55 to 80.
Cubical feet of space in one pair of regenerators, per ton
of metal capacity of the furnace, 30 to 75.
Cubic feet of space in one pair of regenerators per pound
of coal consumed per hour, 0.5 to 1.0.
A correllation and combination of data of this sort, with
details as to the actual working of the furnaces, would point
the way towards a general solution, which would furnish the
best condition for every possible case, with strict consideration
for all the variables involved.
CHAPTER XI.
THERMAL EFFICIENCY OF OPEN-HEARTH FURNACES.
The ordinary open-hearth steel furnace receives cold pig
iron, cold scrap, warm ferro-manganese, cold limestone and
cold ore, it receives cold air and moderately warm producer
gas, and it furnishes melted steel and slag at the tapping heat.
The larger part of the usefully applied heat is that contained
in the melted steel, for it must be melted in order to be cast,
and when olice taken away from the furnace the latter is done
with it. ^
The total heat available for the purposes of the furnace and
which should be charged against it consists of the following
items :
(1) Heat in warm or hot charges.
(2) Heat in warm or hot gas as it reaches the furnace.
(3) Heat in warm or hot air as it reaches the furnace.
(4) Heat which could be generated by complete combustion
of the gas.
(5) Heat of oxidation of those constituents of the charge
which are oxidized in the furnace.
(6) Heat of formation of the slag.
The items of distribution of this total will be as follows:
(1) Heat in the melted steel at tapping.
(2) Heat absorbed in reducing iron from iron ore.
(3) Heat absorbed in decomposing limestone added for flux.
(4) Heat absorbed in evaporating any moisture in the
charges.
These first three items constitute the usefully applied heat,
and their sum measures the net thermal efficiency of the furnace.
(5) Heat absorbed in reducing ferric oxide to ferrous oxide.
(6) Heat in the slag.
(7) Heat lost by imperfect combustion.
396
EFFICIENCY OF OPEN-HEARTH FURNACES. 397
(8) Heat in the chinmey gases as they leave the furnace.
(9) Heat absorbed by cooling water.
(10) Heat lost by conduction to the ground.
(11) Heat lost by conduction to the air.
(12) Heat lost by radiation.
(1) Heat in Warm Charges.
If the pig iron is charged melted instead of cold an immense
amount of thermal work is spared the furnace, and it should
be charged with all the heat (reckoning from 0° C. as a base
line) which is in the melted pig iron as it runs into the fur-
nace. This will average 275 Calories per unit of pig iron,
but should be actually determined calorimetrically in each
specific instance wherever possible. 'The net thermal efficiency
of the furnace will figure out higher with cold charges than
with melted pig iron, because, with a possible flame tempera-
ture of 1,900° C. in the furnace, heat is absorbed much more
rapidly by cold charges than by hot ones, and a larger per-
centage of the available heat will be thus usefully applied.
Scrap is almost always charged cold, but if any of it is hot
its weight and temperature should be known and the amount
of heat thus brought in charged against the furnace. Or a
small piece may be dropped into a calorimeter and its heat
content per unit of weight measured directly, and thus the heat
in all the hot scrap used may be estimated.
Ferro-manganese is often added cold, but -usually is pre-
heated to cherry redness (about 900°) in another small fur-
nace, in order that it may dissolve more quickly in the bath.
Knowing its weight, temperature and specific heat, the heat
which it brings into the furnace can be calculated; a better plan
is to drop a piece into a calorimeter and measure the actual
heat in a sample of it.
Limestotie and ore are almost invariably put into the fur-
nace cold. If used warm the heat in them can be> determined
by the methods just described.
/ (2) Heat in the Gas Used.
By this is meant, not the heat in the gas after it is heated
by the regenerators, but its sensible heat as it reaches the
furnace. This applies only to furnaces where the producers
398 METALLURGICAL CALCULATIONS.
or gas supply are independent of the furnace. Where the pro-
ducers are an integral part of the furnace it is impracticable
to consider them separately from the furnace, and the efficiency
of the whole plant, including the producers, must be consid-
ered together. But where the gas supply from whatever source
comes to the furnace from outside, and reaches the furnace
warm, its sensible heat is to be charged against the furnace as
part of the heat which the furnace must account for. If the
gas comes from producers its amount is satisfactorily found
from the known weight of carbon gasified per hour, or per
furnace charge, and the weight of carbon contained in unit
volume of gas, as calculated from its analysis. If gas comes
from a common main which supplies several furnaces, or is
simply natural gas, its amovint can only be roughly estimated
by measuring the area of the gas supply pipe or flue and meas-
uring the velocity of flow by a pressure gauge or Pitot tube
or anemometer. None of these methods just mentioned are
satisfactorily accurate, and there is great need of simple methods
for determining accurately the flow of gases in flues or pipes.
If the velocity of warm gas is determined suitable correction
for its temperature must be made to reduce it to volume at
standard conditions.
(3) Heat in the Air Used.
If the air coming to the furnace is warm its sensible heat
must be charged against the furnace. If the air is warmed,
however, before it goes into the regenerators by waste heat
from the furnace itself, then its sensible heat should not be
charged against the furnace, because that would amotint to
charging the furnace twice with this quantity of heat. Such
preheating it in reality only a part of- the regenerative princi-
ple, even though it may not be done in regenerators, but, for
instance, by circulating air around a slag-pot or through the
hollow walls of the furnace. If the air used is moist its mois-
ture should not be omitted in the calculation.
The amoimt of air used is best determined by a comparison
of the analyses of gas and chimney products, and a calculation
based on the carbon contents of each and the known volume
of gas used.
EFFICIENCY OF OPEN-HEARTH FURNACES. 399
(4) Heat of Combustion.
Under this head comes the principal item of heat available
for the furnace. In reckoning it we should calculate the total
heat which could be generated by the perfect combustion of
the gas used, to C0^ N^ and H^O vapor. If there is in reality
imperfect combustion, as is shown by analysis of the chimney
gases, that is a defect of operation of the furnace which should
be written down against it. Problem 71 showed an actual case
in which there was very incomplete combustion in the body of
the furnace, but where combustion was afterwards completed
in the regenerators. In such a case the same principle applies;
the furnace must be charged with the total calorific power of
the fuel used, and incomplete combustion can be charged
against the furnace as a whole only on the basis of uncon-
sumed ingredients in the chimney gases — ^the products finally
rejected by the furnace. If there is poor combustion in the body
of the furnace and combustion is only completed in the re-
generators, the furnace will not give as high net thermal effi-
ciency as if combustion were complete above the hearth.
(5) Oxidation of the Bath.
The oxidation of carbon, iron, silicon, manganese and some-
times phosphorus and sulphur, add a not inconsiderable amount
to the heat resources of the furnace. Carbon should be burnt
to CO^, iron is usually oxidized to FeO, manganese to MnO,
silicon to SiO^ phosphorus to P^O', and sulphur 'to SO^^ All
of these oxidations generate heat, and, moreover, heat which
should be very efficiently utilized, being produced in direct
contact with the metallic bath ; it should all be charged against
the furnace as part of its available heat.
(6) , Formation of Slag.
The metallic oxides produced unite with each other, and
with the lime and silica of ore used and lining of the hearth to
produce the slag, the heat of formation of which can be calcu-
lated and counted in as available heat.
(1) Heat in Melted Steel.
This is 9. large item in the work done by the furnace ; in fact,
usually the largest single item. It should be determined
400 METALLURGICAL CALCULATIONS.
calorimetrically when possible; if this is not done its tem-
perature should be known and its composition, in order to
compare it with the calorimetric experiments of others, and
thus derive a probable value for its heat contents. Not many-
experimental values in this line have so far been published,
and a very much needed investigation is one upon the total
heat in melted steels of different compositions at different
temperatures. Values from 275 to 350 Calories per kilogram
have been observed.
(2) Heat of Reduction of Iron from Ore.
This is an item which appears whenever ore is used to facili-
tate oxidation of the bath. The weight and composition
of the charges and the products will easily show how much
iron has been reduced. The ore used is almost always hema-
tite, less frequently magnetite; hydrated iron oxides are not
used for obvious reasons. The heat absorbed is 1,746 Calories
per kilogram of iron reduced from Fe^O' and 1,612 Calories
per kilogram from Fe'O*.
(3) Decomposition of Limestone Flux.
If limestone is charged raw, as is usually done in order to
avoid the dusting caused by using burnt lime, then the furnace
is called upon to bum this limestone in place of the lime kiln.
The heat absorbed may be taken as either
451 Calories per kilogram of CaCO' decomposed.
1,026 Calories per kilogram of CO^ driven off.
806 Calories per kilogram of CaO produced.
(4) Evaporation of Moisture in the Charges.
If the ore, flux, scrap or ore are put into the furnace wet
their moisture must be evaporated. The correct figure for this
evaporation is the latent heat at ordinary temperatures, viz.:
606.5 Calories per kilogram. This allows for the heat re-
quired to convert into cold vapor of water, and puts the H^O
thereafter upon the same basis as all the other gas going out
of the furnace. The chimney gases carry out sensible heat,
and the H^'O in them can be calculated as carrying out a cer-
tain amount of heat as gas, reckoning from zero,* and thus
the correct chimney loss obtained. It is incorrect either to
EFFICIENCY OF OPEN-HEARTH FURNACES. 401
charge the latent heat of vaporization as chimney loss or to
charge the sensible heat of the water vapor in the chinmey
gases to heat absorbed in evaporating water in the furnace.
It is also incorrect to do as is frequently done, viz.: to calcu-
late, the heat required to evaporate the moisture to water
vapor at 100° — 637 Calories — and say that this is the heat to
evaporate the moisture. With almost no moisture in the gases,
the moisture of the charges would commence to evaporate at
once, while they were yet cold, and the moisture is no more
evaporated at 100° or to vapor at 100° than,, it is to 200° or
500°. The only safe course is to confine the evaporation heat
to that necessary to convert the moisture into cold vapor, and
let its sensible heat as it escapes as vapor at any other tem-
perature be reckoned in with the sensible heat of the chimney
gases.
(5)' Reduction- of Ore Into the Slag.
While considerable of the iron in the ore used is reduced to
the metallic state, yet often the larger part is reduced merely
to the state of FeO, and as such goes into the slag. The amount
so reduced can be determined by subtracting the iron reduced
from ore from the total iron in the ore used; the differences
gives the iron from the ore going into the slag as FeO. The
weight of FeO corresponding is then easily calculated. The
heat absorbed in this partial reduction is :
446 Calories per kilogram of FeO reduced from Fe^O'.
341 Calories per kilogram of FeO reduced from Fe'O*.
(6) Heat in Slag.
This is usually a small item in open-hearth practice, but may
amount to a very considerable one in the method of running
with large ore charges, as in the Monell process. The varia-
tions of composition of the slag, and especially in the tempera-
ture at which it is run off, are so large that the heat in the
slag should always be determined calorimetiically for each
specific case. If assumptions have to be made, 450 to 550
Calories per kilogram of slag would be assumed. The weight
of slag is seldom taken, although it could in most cases be
done if desired. If the weight is not known it may be calcu-
lated from the known weight of either iron, manganese or
402 METALLURGICAL CALCULATIONS.
phosphorous going into it, as seen from the balance sheet and
the percentages of these elements in the slag as shown by ana-
lysis.
(7) Loss BY Imperfect Combustion.
This is based upon the unconsumed ingredients of the chim-
ney gases, as shown in an analysis. From this the calorific
power of the unbumt gases in 1 cubic meter can be calculated.
If then we know the volume of chimney gases per unit of charge,
we get the heat loss by imperfect combustion per unit of charge.
The volume of chimney gas is found by means of the carbon
in it, which must all come from the carbon in the gas used,
plus the carbon oxidized out of the charges, plus the carbon
of CO^, driven off raw limestone used as flux. Or, putting
it in another way, the total carbon going into the furnace
in any form, less the carbon in finished steel, must give the
carbon in the chimney gases. This divided by the weight of
carbon in unit volume of chimney gas gives the volume of
the latter, per whatever unit of charge is used as the basis
of calculations. This volume times the calorific power of unit
volume of chimney gas, gives the total heat lost by imperfect
combustion.
(8) Sensible Heat op Chimney Gases.
The temperature of these gases should be taken as they
enter the chimney flue. Their amount -is determined as ex-
plained under the previous heading. The water vapor con-
tained should not be overlooked, being reckoned simply as
vapor or gas in exactly the same category as the other gases.
The analysis of the chimney gases being usually given on dried
gas, a separate determination of the moisture carried per unit
volume of such dried gas is necessary. If this is not done an
approximation can be made by considering all the hydrogen
in the gas burned to form water vapor, and add in the moisture
of the air used and the moisture in the charge.
(9) Heat Lost in Cooling Water.
This is a very variable amount, and must be determined for
each furnace by measuring the amount of water used per unit
of time and its temperature before reaching and after leaving
EFFICIENCY OF OPEN -HEARTH FURNACES. 403
the furnace. Doors are frequently water-cooled, also ports,
where the heat is fiercest, and sometimes a ring around the
hearth at the slag line.
(10) Loss BY Conduction to the Ground.
This is a quantity extremely difficult to measure or to .esti-
mate. If a closed 'vessel filled with water were put into the
foundations the rate at which its temperature rose might give
some idea of the rate at which heat passed in that direction per
unit of surface contact. At present, lacking all reliable data,
we must put this item in the " by difference " class.
(11) Loss BY Conduction to the Air.
This is an amount which can be observed and calculated with
some approach to satisfaction. The sina qua non for this par-
pose is a F^ry radiation pyrometer, by which the temperature
of the outside of the furnace at different parts can be accurately
determined. Then the velocity of the air blowing against the
furnace, if it is in a current of air, is observed with a wind
gauge, and its temperature before reaching the furnace. With
these data and by the methods of calculation before explained
in these calculations (Part I, Chap. VIII) the heat lost to the
air may be calculated.
(12) Radiation Loss.
Having determined the temperature of the outer surface of
the furnace and measured its extent, as above explained, the
radiation loss can also be calculated, knowing the mean tem-
perature of the surroundings, by the principles of radiation,
having due regard to the nature of the radiating surface. Tables
of specific radiation capacity of different substances (fire-brick,
stone, iron) will be found at the reference just given above.
Problem 72.
Jiiptner and Toldt (Generatoren und Martinofen, p. 73) ob-
served the following data with regard to an open-hearth steel
furnace charge:
Weight of cold charges, at 26° C 3,745 kg.
Weight of hot charges, at 700° C 1,700 "
Total weight of charge 5,445 "
404 METALLURGICAL CALCULATIONS.
Average composition of charge C = 1.07 per cent.
Si = 0.50 "
Mn = 1.33
Coal gasified in producers 1,980 kg.
Carbon gasified from coal 47.13 per cent.
Average composition of dried gas used CO^ 3.81 "
■ O' 0.98
CO 23.82
CW 0.42
W 8.75 "
N^ 62.22
Moisture accompanying each m* of gas = 82 grams.
Temperature of gas reaching furnace 165° C.
Temperature of air used 26° C
Moisture accompanying each m' of air 12 grams.
Barometer 717 m.m.
Steel produced 5,191 kg.
Composition of steel C = 0.12 per cent.
Si = 0.04 "
Mn = 0.19 "
Temperature of steel when tapped 1410° C.
Heat in 1 kg. steel, by calorimeter (to 0° C.) . . 277 Cal.
Composition of slag SiO^ 45.65 .per cent.
FeO 33.60
MnO 18.21
CaO 2.54 "
Weight of slag 425 kg.
Temperature of slag when issuing. .' 1,410° C.
Heat in 1 kg. slag, by calorimeter (to 0°) . 560 Cal.
Composition of the stack gases (dry) CO^ ll.l^j per cent.
0^ 6.78 "
N^ 82.10 "
Temperature of the gases in chimney flue 500° C.
Requirements :
(1) A balance sheet of materials entering and leaving the
furnace.
(2) A balance sheet of the heat available and its distribution.
(3) What excess of air above what is necessary for complete
combustion is used, and what per cent, of all the available
heat of the furnace is thereby lost?
EFFICIENCY OF OPEN-HEARTH FURNACES.
405
(4) What is the thermal efficiency of the furnace?
Shlution: (1)
Charges.
Metal,
C,
Si.
Mn,
Fe,
Limestone,
CaO,
C,
O,
Hearth,
SiO'
Gas,
C,
O,
H,
N,
Air,
0,
N,
H,
Totals,
Balance Sheet op Materials.
5,445 kg.
68 "
27 "
72 "
5,288 "
20 "
11 "
2.5 "
8.5 "
132 "
7,884 "
933 "
2,003 "
118 "
4,830 "
16,026 "
3,812 "
12,195 "
19 "
Steel.
6
2
10
5,173
Slag.
25
62.
115
11
132
29,507
"5,191
80
425
Gases.
52
2.5
6.5
933
2,003
118
4,830
3,732
12,195
17
23,891
Notes on the Balance Sheet.
The distribution of carbon, silicon, manganese and iron is
governed by the known amounts of these elements present in
the steel, the rest of the carbon going into the gases (as CO^),
and the manganese, silicon and iron passing into the slag (as
MnO, SiO^ and FeO, respectively.)
The amount of limestone used was not stated, but was de-
duced from the fact that the slag was said to weight 425 kilos,
and to contain 2.54 per cent, of CaO, which makes the CaO
11 kilos, and assuming pure limestone, this would bring in 9
Idlos. of CO^, which appears on the balance sheet as 2.5 kilos,
of carbon and 6.6 kilos, of oxygen, contributed to the gases.
406 METALLURGICAL CALCULATIONS.
The weight of silica contributed by the hearth is deduced
from the fact that the slag must contain the silicon, manganese
and iron oxidized from the charge, as SiO^ MnO and FeO.the
CaO of the flux and the SiO^ contributed by the hearth, and its
total weight is 425 kilos. The ingredients of the slag must,
therefore, be
Kg.
SiO=' 25x60/28 = 53.5
MnO 62X71/55 = 80.0
FeO 115X72/56 = 147.9
CaO = 11.0
Sum 293
From hearth (difference) 132
Total slag 425
The gas used we find by starting with the fact that 1,980
kilos, of coal is used, from which 47.13 per cent, of carbon
enters the gases. This makes carbon in the gases 933 kilos.
Each cubic meter of dry gas, as analyzed, contains 0.2805
cubic meter of CO*, CO and CH* added together, and there-
fore, 0.2805X0.54 = 0.1515 kilos, of carbon. The volume of
dry producer gas used is therefore, at standard conditions,
933-^0.1515 = 6,160 cubic meters, containing by its analysis:
CO'' 6,160X0.0381= 235 m' = 465 kg.
O' " X 0.0098 = 60m'= 86
CO " X 0.2382 = 1,467 m' = 1,840
CH< " X 0.0042 = 26m'= 19
H* " X 0.0875 = 539 m' = 49
N' " X 0.6222 = 3,833 m' = 4,830
6,160 m' = 7,297
The moisture is 82 grams per each cubic meter of gas, meas-
ured at 26° and 717 m.m. pressure; but the 6,160 cubic meters
of gas at standard conditions would be 7,175 cubic meters at
those conditions of temperature and pressure, and therefore
be accompanied by
7,175X82 = 588,350 grams
= 588 kg. of H'O vapor.
EFFICIENCY OF OPEN-HEARTH FURNACES. 407
We can now enter the gas on the balance sheet either as so
much CO^, CO, H^O, etc., or else resolve it into its essential
constituents C, H, O and N, which course we have followed
on the balance sheet. The carbon in the gas, is 933 kg. by-
assumption, the oxygen is 8/11 the C0^ all the 0^ 4/7 the CO
and 8/9 the H^O, a total amounting to 2,003 kilos; the hydro-
gen is 4/16 the CHS all the W and 1/9 the H='0, a total of
118 kilos.
The air supplied is best found from the volume of the chim-
ney gases. The total carbon entering these is 52-1-2.5 + 933
= 987.5 kilos., as seen from the balance sheet. Each cubic
meter of dry chimney gas contains 0.1112 m' of CO^, carrying
0.1112X0.54 = 0.0600 kilos, of carbon. The volume of dry
chimney gas at standard conditions is therefore 987.5-^0.0600
= 16,458 m». This contains 16,458X0.8210 = 13,512 m' of
N^, and since 3,833 cubic meters came in with the gas 9,679
m' must have come in with the air, corresponding to 12,220 m'
of dry air at standard conditions. This would consist of 12,195
kilos, of W and 3,660 kilos, of O^. To find the moisture present
the volume of this air at 26° and 717 m.m. pressure would be
14,230 m^ and would be therefore accompanied by
14,230X12 = 170,760 grams
= 171 kg. of ffO.
This consists of 19 kilos, of hydrogen and 152 kilos, of oxygen,
the latter increasing the total oxygen in the air used to 3 660 +
152 = 3,812 kilos.
(2) Heat Balance Sheet.
Heat Available.
Per-
Cal. centages.
Heat in the warm charges 189,210= 2.45
Sensible heat of air used 99,480 = 1.29
Sensible heat of gas used 360,550= 4.68
Heat of combustion of gas 6,202,300 = 80.44
Heat of oxidation of the bath 833,600 = 10 . 81
Heat of formation of slag 24,200 = 0. 31
Total 7,709.340 = 100.00
408 METALLURGICAL CALCULATIONS. ■
Heat Distribution.
In melted steel at tapping 1,437,900 = 18. 65
Decomposition of limes;tone 9,200 = 0. 12
Sensible heat of slag 238,000= 3.09
Sensible heat of chimney gases 3,118,460 = 40.45
All other losses, not classified 2,905,790 = 37. 69
Total 7,709,340 = 100.00
Notes on the Heat Balance Sheet.
The warmed charges weighed 1,700 kilos, at 700°, and the
cold charges 3,745 kilos at 26°. Taking 0° C. as the base line
the sensible heat in these is
Cal.
1,700X0.15X700 = 178,500
3,745 X 0.11 X 26 = 10,710
Sum = 189,210
The air used contains, at standard conditions, 12,220 m' of
air and 171-^0.81 = 211 m^ of water vapor. These carry at
26°, heat as follows:
Cal.
12,220X0.3037X26 = 98,490
211X0.3439X26= 990
Sum = 99,480
The gas used, coming in at 165° C, carries in heat as follows:
Cal.
0^ CO, W, W 5,899 m'X 0.3075 = 1,814
CQ^ 235 m'X 0.4063 = 95
CH< 26m'x0.4163= 11
H^O 726 m'X 0.3648 = 265
Average calorific capacity per 1° = 2,185
Heat content 2,185 X 165 = 360,525
The heat of combustion is that of the combustible ingredi-
ents of the gas used to CO^ and H'O vapor:
EFFICIENCY OF OPEN-HEARTH FURNACES. 409
Cal.
CO 1,467 m»X 3,062 = 4,492,000
CH* 26 m'X 8,598 = 223,500
W 539 m»X 2,613 - 1 486,800
Sum = 6,202,300
The heat of oxidation of the bath is from the various sub-
stances oxidized:
Cal.
C to CO^ 62X8,100 = 421,200
Si to SiO^* 25X7,000 - 175,000
Mn to MnO 62x1,653 = 102,500
Fe to FeO 115X 1,173 = 134,900
Sum = 833,600
The heat of formation of the slag is the heat of combination
of 80 kilos, of MnO,- 148 kilos, of FeO and 11 kilos, of CaO,
with 186 kilos, of SiO^. This will be, since the bases are largely
in excess of the silica:
186X130 = 24,200 Cal.
The figure 130 is an average for the heat of combination of
1 kilo, of SiO' with about 2 parts of FeO to 1 part MnO.
The heat in the steel at tapping is simply its weight multi-
plied by its heat contents per kilo. (5,191X277).
The heat in the slag is similarly obtained (425 X 560) .
The decomposition of limestone represents 9 kilos, of CO'
liberated, and the heat absorbed is 9X1,026.
The sensible heat in the chimney gaseS is obtained by first
noting that their volume (measured dry) , as already obtained,
is 16,458 cubic meters. The CO^, 11.12 per cent., becomes,
therefore, 1,843 m^ the 0^ 1,116 m^ N^ 13,512 m', while the
H^O accompanying this will be 9 tipies the weight of hydrogen
passing into the gases, which is 9X (1184-17) = 1,215 kilos.
= 1,500 m'. A simpler way to get the volume of the water
vapor is to observe that it is always equal to the volume of
hydrogen going into it, and, therefore, in this case would be
(118 + 17) H- 0.09 = 1,500 m». The heat carried out by these
gases would therefore be;
410 METALLURGICAL CALCULATIONS.
N' + O' = 14,628 m'X 0.3165 = 4629.8 Cal. per 1°
CO' = 1,843 m'X 0.4800 = 984.4 "
H=0 = 1,500 m'X0.4150 = 622.5 "
•^o
Calorific capacity = 6236.9 "
Total capacity = 3,118,450 " per 500°
The heat balance sheet as a whole discloses the fact that in
this furnace the fuel only supplies some 80.5 per cent, of the
total heat available, and that about 10.8 per cent, is furnished
by the oxidation of the bath itself. On the other hand, the
melted steel accounts for 18.6 per cent., while chemical reactions
absorb almost none, giving a net thermal efiiciency of slightly
under 19 per cent. The other important items are 40.5 per cent,
of the total heat lost up the chimney, and 38 per cent, lost by
radiation and conduction. Such data as these point the way .
to avenues of possible saving or avoidance of waste of heat.
(3) The excess of air is obtained directly from the chimney
gases. The 1,116 m' of oxygen, unused, represents 1,116-;-
20.8 = 5,365 m' of air in excess, which leaves 16,458 — 5,365
= 11,093 m* of air which came in and was used. The per-
centage of air used in excess of that which was necessary was:
5,365h- 11,093 = 0.4845 = 48.5 per cent. (3)
No properly run open-hearth regenerative gas furnace should
ever have such a large excess of air, for it cuts down the tem-
perature of the flame and leads to high chimney losses.
The air used brought in 171 kilos of water vapor, and there-
fore the excess air brought in
48 5
171X,.o r = 56 kilos, of water,
140.6
the volume of which, at standard conditions, would be
56 -t- 0.81 = 61 cubic meters.
The excess air, with its accompanying water, going into the
chimney at 500°, carried out heat as follows:
Cal.
5,365X0.3165X500 = 849,000
61X0.4150X500= 12,650
861,650
EFFICIENCY OF OPEN-HEARTH FURNACES. 411
Representing 861,6504-7,709,340 = 0.112 = 11.2 per cent. (3)
(4) The thermal efficiency has been already added up as
18.65 + 0.12 = 18.77 per cent. (4)
Problem 73.
In the open-hearth furnace of the preceding problem,
assume that the calculations therein made showed that, per
heat of steel produced, 5,191 kilograms, there entered and
left the furnace the following volumes of gases, measured at
standard conditions:
Producer Gas.
Air.
Chimney Gases.
C0= 235 m'
1,830 m'
0=" 60 m^
3,833 m»
1,116 m'
CO 1,467 m'
CH^ 26 m'
H= 539 m'
N^! 3,834 m'
9,679 m'
13,512 m'
H^O 726 m'
211 m^
1,500 ID?
The temperature of air used was 26°, of producer gas 165°,
of chimney gases 400°. The excess of air used was 48.5 per
cent. The gas and air entered the laboratory of the furnace
preheated to 1,100°, and the products of combustion entered
the regenerators at 1,450. Items of heat balance sheet:
Available.
Calories
In warm charges 189,210
Sensible heat of air used at 26° 90,480
Sensible heat of gas used at 165° 360,550
Heat of combustion of the gas 6,202,300
Heat of oxidation of the bath 833,600
Heat of formation of slag 24,200
7,700,340
Distribution.
In melted steel at tapping at 1,410° 1,437,900
In slag at tapping at 1,410° 238,000
Decomposition of limestone 9,200
Sensible heat in chiflmey gases at 400° 3,065,350
All other losses, not classified 2,949,890
7,700.340
412 METALLURGICAL CALCULATIONS.
Required:
(1) The thermal efficiency of the regenerators.
(2) The thermal efficiency of the laboratory of the furnace.
(3) The temperature of the flame.
(4) The change in (3) if only the theoretical amount of air for
combustion were used.
Solution:
(1) The products of combustion, entering the regenerators
at 1,450°, carry into them the following amounts of heat:
Calories.
00== 1,830X0.689 = 1,261
0= + N^ 14,628X0.342 = 5,003
H'O 1,500X0.557 = 836
Calorific capacity per 1° = 7,100
Calorific capacity per 1,450° = 10,295,000 Calories.
The same gases entering the chimney flue at 400° carry
with them the following amounts :
Calories.
CO^ 1,830X0.458 = 838.1
O^ + N^ 14,628X0.314 = 4,593.2
ffO 1,500X0.400 = 600.0
Calorific capacity per 1° = 6,031.3
Calorific capacity per 400° = 2,412,500 Calories.
The gas used for combustion, entering the regenerators at
165° and leaving it at 1,100°, carried into the regenerators
360,550 Calories, as' already given in the balance sheet, and
carried out at 1,100° the following:
Calories.
CO^ 235X0.612 = 143.8
CH* . 26X0.620= 16.1
0^ N^ W, CO 5,899X0.333 = 1,964.4
WO 726X0.505 = 366.6
Calorific capacity per 1° = 2,490.9
Calorific capacity per 1,100° = 2,780,000 Calories.
Heat abstracted from regenerators:
2,780,000—360,550 = 2,419,450 Calories.
EFFICIENCY OF OPEH-HEARI H FURNACES: 413
The air used, entering the regenerators at 26°, carries in as
sensible heat 90,480 Calories, as already given in the' balance
sheet, and issuing from them at 1,100° carries out the fol-
lowing:
Calories.
0^+W 13,512X0.333 = 4,499.5
H^O 211X0.405= 85.5
Calorific capacity per 1° = 4,585.0
Calorific capacity per 1,100° = 5,043,500 Calories.
Heat abstracted from regenerators :
5,043,500—90,500 = 4,953,000 Calories.
The thermal efficiency of the regenerators may now be cal-
culated from three standpoints. There is no doubt that the
gas and air take from the regenerators, and return to the body
of the furnace 2,419,450 + 4,953,000 = 7,372,450 Calories. This
is, therefore, the usefully returned heat, and the ratio of this to
the heat received by the regenerators measures their efficiency
qua regenerators. The three figures obtained for this efficiency
depend on what is to be considered as the heat chargeable
against the regenerators. Are they to be charged with all the
heat in the hot products at 1,450° (10,295,000 Calories), or only
with the heat left in the regenerators by these products leaving
•at 400° (10,295,000—2,412,500 = 7,882,500 Calories), or per-
haps only with the heat carried in less a certain assumed amount
representing the minimum temperature to which it is desirable
to cool the products before they enter the chimney?
If we charge the regenerators with all the heat brought in
by the products their thermal efficiency figures out:
^g™ = 0.r2=72pe.c.„.. (1)
If we charge them with the heat left in the regenerators, by
the products, their efficiency is :
7,372,450 r.nA aA * /1^
7^882:500 = 0-^*=^*P^'"^"^*- ^^^
414 METALLURGICAL CALCULATIONS.
leaving 7 per cent, of the heat chargeable against them lost by
radiation from their walls and conduction to the ground.
If we think that the first calculation gives too low an effi-
ciency, because the gases, must leave the regenerators hot, in
order to be used for chimney draft, and therefore some or all
of the heat in the chimney gases should not be charged against
the regenerators, on the other hand, the second calculation may
represent too high ah efficiency, because the gases may be dis-
carded to the chimney at a higher temperature than is neces-
sary to provide the requisite chimney draft, and this excess of
chimney temperature and consequent heat loss is a defect of
the regenerators which they should be charged with. If we
assume that a chimney gives very nearly its maximum drawing
capacity with the gases entering it at 300°, it would be per-
fectly proper to charge the regenerators with all the heat which
could be given out by the products in cooling from 1,450°
to 300°, heat which they should have entirely absorbed. In
the case in hand, the products at 300° would contain (by calcu-
lations similar to those already made) 1,777,400 Calories, leaving
10,295,000—1,777,400 = 8,517,600 Calories chargeable against
the regenerators, as the heat which they should have absorbed
or intercepted. Measured by this standard, their efficiency is:
gf;|5 = 0.87 =87 per cent. (1)
The losses from the regenerators in this view would be 7 per*
cent, (about) by radiation and 7 per cent, by unnecessary chim-
ney loss.
Comparing these three methods of considering efficiency,
the third appears to the writer as the fairest, and the one which
gives the metallurgist the most reliable criterion of the real
work which his regenerators are doing for him, and the best
basis of comparison when considering the work of different
regenerators or of the same regenerators under different con-
ditions.
(2) The laboratory of the furnace, the space enclosed be-
tween the hearth, roof, side walls and ports, receives from the
entering preheated gas and air their sensible heat, and the heat
generated by combustion. In the case in point these items
total, as already calculated :
EFFICIENCY OF OPEN-HEARTH FURNACES. 415
Calories.
Sensible heat in preheated gas 2,722,300
Sensible heat in preheated air 5,043,500
Heat generated by the combustion 6,202,300
Total 13,968,100
Hot products at 1,450° take out 10,295,000
Heat left in the laboratory 3,673,100
These figures show that the laboratory of the furnace ap-
propriates to its own purposes 3,673,100 Calories out of the
13,968,100 Calories poured into it. This part of the furnace,
therefore, qua laboratory, has an efficiency in this respect of
?gg50_, 0.26 = 26 per cent. (2,
This is the datum which would be useful to the metallurgist
in comparing the efficiencies of differently shaped laboratories,
such as those with differently shaped hearths, differently shaped
roof, differently arranged ports, etc. This conception of effi-
ciency is that taken by Damour, and repeated by Queneau
in his book on " Industrial Furnaces." We must be careful
here not to compare the heat left in the laboratory with the
heat of combustion alone'. This would give
3,673,100 „ en =n
6:202:306 = '■'' = "" p^'" ^^"*-
But this is not the heat absorption of the laboratory alone, but
is a function of the furnace as a whole, and depends largely on
the efficiency of the regeneration accomplished by the regen-
erators. If we wish to obtain an idea of the perfection wjth
which the laboratory of the furnace appropriates the heat pass-
ing into and ^through it, we must simply compare what it ap-
propriates with what was sent into it. The percentage of this
appropriatiori measures the efficiency of the laboratory for
abstracting heat for the purpose of heating itself— a datum
highly useful to know if the furnace is used simply for keeping
a given working space up to a given temperature for an in-
416 METALLURGICAL CALCULATIONS.
definite time. If the heating of the furnace charge to the furnace
heat is only a minor part of the useful work of the furnace,
and keeping it at that temperature is the chief function of the
furnace, then the relation of the heat thus appropriated and
utilized to the total heat available to the laboratory, is a measure
of the perfection of construction of the laboratory. This
is the view of Damour and Queneau, and seems in some re-
spects plausible.
However, if we are examining this question thus in detail,
it appears to the writer that the view just explained and the
conclusions derived thereform may really be the very opposite
of the truth of the matter, and lead to entirely erroneous con-
clusions. If two exactly siniilar open-hearth furnaces are
constructed, with the sole difference that the body of one is
thicker walled than the body of the other, it is perfectly true
that maintaining the same temperature in each will require
more gas in the thin-walled furnace, but that it will abstract
and radiate a considerably larger proportion of the heat passing
through it than the thick- walled furnace. Yet, according to
Damour and Queneau's principle, the thin-walled furnace
would be considered as being the more efficient laboratory of
the two. The truth is, we must either com'pute the heat left
in the laboratories per unit of time per cubic meter of stock
space in order to compare different furnaces, or else to get
absolute efficiency compute the ratio of the heat passing into
the laboratory with that absorbed usefully by the charge, viz.:
in this case:
1,447,100 ^ jj^Q^ _ ^Q^ p^^ ^^^^
13,968,100 (2)
(3) The temperature of the flame is that to which the pro-
ducts of combustion can be raised by the maximum heat which
they contain. When perfectly consumed they contain the heat
pre-existing as sensible heat in the hot gas and air, plus the
heat of combustion. We have already calculated this total as
13,968,100 Calories. The. mean heat capacity of the products
of, combustion, per 1° between 0° and t°, is:
-EFFICIENCY OF OPEN-HEARTH FURNACES. 417
00== 1,830 m« (0.37 +0.00022t) = 677 + 0.4026t
O^' + N^ 14,628 m' (0.303 + 0.000027t) = 4,428 + 0.3950t
H^O 1,500 m' (0.34 +0.00015t) = 510 + 0.2250t
Therefore,
Whence,
Sum = 5,615+ 1.0226t
13,968,100
5,615+ 1.0226t
t = 1,749° (3)
(4) If the excess air were not .used, the 1,116 cubic meters of
oxygen in the products, together with its corresponding quan-
tity of nitrogen, 4,249 cubic meters, would be absent from the
products of combustion, as would also some of the 211 cubic
meters of water vapor. Since the oxygen in the air was 3,833
m' and the imused excess 1,116, the proportion of the 211 m'
of water vapor belonging to the excess air was
and the final products are CO^ 1,830 m^ N^ 9,263 m* and
H^O 1,439 m^
The heat available will be the same as before from com-
bustion, the same from preheated gas, but less from preheated
air, because of the absence of the excess air. Since the air
formerly used brought in 5,043,500, the proportion of this
carried by the excess air is
5,043,500 X ^^= 1,468,500,
leaving heat brought in by air in the second case the difference,
viz.: 3,575,000, and the total heat in the products 13,968,100—
1,468,500 = 12,499,600 Calories.
The mean heat capacity of the diminished quantity of pro-
ducts, per 1°, is, by the same methods as previously used,
3,973 + 0.8685t, and, therefore,
12,499,600 g . ,go ,4.
* = 3,973 + 0.8685t- ^'^*^ ^^^
418 METALLURGICAL CALCULATIONS.
a gain of nearly 400° in maximum temperature, by cutting off
the 40 per cent, excess of air which was being used.
Problem 74.
The new style Siemens furnace for small plants has the gas
producers built in as part of the furnace, between the two re-
generators. The furnace has only one regenerative chamber
at each end for preheating the air used for burning the gas,
while the latter comes to the ports, hot directly from the pro-
ducers. This plant is compact, economical of fuel, allows high
temperatures to be reached, and occupies little floor space. As
compared with the old style separate furnace and producer
plant, it occupies about half the floor space, costs about 60
per cent, as much to build and uses about 60 per cent, as much
coal. They are largely used abroad for melting steel for cast-
ings in small foundries and for melting pig iron for castings to
be subsequently annealed to malleable castings.
Such a furnace melted a charge of 3,000 kilograms of cast
iron in 4 hours, using 750 kilograms of coal. The cast iron
was charged cold, at 0°, and run out melted at 1,450°, con-
taining 300 Calories of sensible heat per kilogram. The coal
and gases were of the following compositions:
Coal— C 75, H 5, 0 12, WO 2; ash 6.
Producer Gas— CO 20, CO^ 5, CH^ 2, W 16, N^ 57. •
Chimney Gas— CO^ 19, 0^ 1.8, N^" 79.2.
The air coming to the furnace is at 0° and dry; a steam blower
is used to run the producer, using 1 kilo, of steam per 6 kilos,
of air blown in, and the raising of this steam requires 0.1 kilo
of coal used under boilers. The ashes produced weigh 75 kilos,
per heat run. Temperature of the hot producer gas entering
the furnace laboratory 600°, of preheated air 1,000°, of pro-
ducts leaving laboratory 1,400°, of products entering chimney
flue 350°.
Required:
(1) A heat balance sheet of the furnace as a whole.
(2) The net thermal efficiency of the furnace.
(3) The net thermal efficiency of the laboratory of the fur-
nace.
(4) The net thermal efficiency of the regenerators.
EFFICIENCY OF OPEN-HEARTH FURNACES. 419
(5) The theoretical flame temperature in the furnace.
Solution :
(1) Heat balance sheet per charge of 3,000 kilos.
Heat Available.
Calories.
Calorific power of coal used in producers 5,248,500
' under boilers 398,900
tt ti tt
5,647,400
Heat Distribution.
In melted cast iron 900,000
In chimney products 768,950
Loss by tmburnt carbon in ashes 243,000
Used for raising steam 398,900
Loss by radiation and conduction 3,336,550
5,647,400
Calculation of the Heat Balance.
The heat of combustion of a kilogram of coal is not given,
therefore must be calculated from its composition:
Calories.
C 0.75X8,100 = 6,075
(»-^)
H (0.05--^^) X 34,500 = 1,208
Calorific power to liquid water = 7,283
Vaporization heat of water formed
(0.02 + 0.45) X 606.5 = 285
Calorific power to vapor of water = 6,998
Of 750 kilos, in the producers = 5,248,500
The coal used under the boilers can only be found by first
finding how much steam was used, which in its turn can be
gotten from the air blown in, and the nitrogen of this can be
found from the total nitrogen in the producer gas. The volume
of producer gas can be gotten from its carbon content per cubic
meter and the known weight of carbon gasified. Or, turning
this chain of reasoning the other way, if we subtract the carbon
in the ashes from the carbon in the coal, the difference is the
carbon entering the gases; this divided by the carbon in 1
420 METALLURGICAL CALCULATIONS.
cubic meter of chimney gas (calculated from its analysis) gives
the volume of chimney gas, and by the carbon in 1 cubic meter
of producer gas gives the volume of this gas used, from which
the nitrogen in it can be calculated, which divided by 0.792
gives the volume of air used, from which its weight is obtained,
thence the amount of steam used, and finally from this the
amount of coal necessary to burn under boilers to raise this
steam.
[In working most metallurgical problems the difficulty of
finding the connection or succession of steps connecting the
requirements with the data given is often easiest overcome by
starting with the requirement in mind and noting from what
other figure its value may be calculated, and thus passing
backwards from one figure to another we finally arrive at one
which can be found directly from the data given. The logical
sequence of operations is thus disclosed and the problem is in
reality solved; the following of the thread in the reverse direc-
tion, from data to requirement, involves calculations only, not
hard thinking, and is usually a matter of the simplest arith-
metic]
The operations for calculating the steam used are as follows:
Carbon in coal in producers = 750x0.75
Carbon in ashes from producers = 75 — 45
Carbon in gases from producers
Carbon in producer gas per 1 m' = 0.27x0.54
Producer gas per heat = 532.5 -r- 0.1458
Nitrogen in producer gas = 3,652x0.57
Air used in the producer = 2,082 -^ 0.792
Carbon in chimney gas per 1 m' = 0.19X0.54
Chimney gas per heat = 532.5^-0. 1026
Nitrogen in chimney gas = 5,190X0.792
Nitrogen from air used = 4,110 — 2,082
Air used to burn gas = 2,028-^0.792
Weight of this air = 2,561 X 1.293
Weight of air used in producer
Weight of steam used in producer = 3,399-^6
Weight of boiler coal used = 567X0.1
Calorific power of this coal = 6,998x57
=
562.5 kg.
=
30.0 "
=
532.5 "
=
0.1458 "
=
3,652 m»
=
2,082 "
=
2,629 "
=
3,399 kg.
=
0.1026 "
=
5,190 m'
=
4,110 "
=
2,028 "
=
2,561 "
=
3,311 "
=
3,399 "
=
567 "
=
57 "
=
398,900 Cal,
EFFICIENCY OF OPEN-HEARTH FURNACES. 421-
The heat in the melted cast iron is simply its weight times
300 = 300X3,000 = 900,000 Calories.
(2) The net thermal efficiency follows directly from this, as
900,000 „ ,^ <»
= 0.16 = 16 per cent. (2)
5,647,400.
(3) The gaseous products consist dry, as analyzed, of
00== 5,190X0.19 = 986 m'
O^ 5,190X0.018 = 93 "
W 5,190X0.792 = 4,110 "
And in addition all the -^ater formed from the coal, plus the
steam used, a total of: ,
Water from coal = 0.47X750 = 352.5 kg.
Steam = 566.5 "
Total water vapor in gases = 919.0 "
Volume at standard conditions = 919-^0.81 = 1,135 m'
At 350° these products carry out of the furnace heat as fol-
lows:
Calories.
0^ + W 4,203X0.312 = 1,311
CO* 986X0.447 = 441
H^'O 1,135X0.392 = 445
Average caloric capacity per 1° =2,197
Capacity per 350° = 768,950 Calories.
These same products would carry heat out of the laboratory
of the furnace at 1,400° as follows:
Calories.
02 + N'' 4,203X0.341 = 1,433
CO* 986X0.678 = 669
H*0 1,135X0.550 = 624
Average caloric capacity per 1° = 2,726
Capacity per 1,400° = 3,816,400 Calories.
The heat developed in the laboratory of the furnace by the
combustion of the 3,652 m' of producer gas is
422 METALLURGICAL CALCULATIONS.
Calories.
CO 0.20X3062 = 612
CH* 0.02X8,598 = 172
W 0.16X2,613 = 418
Per 1 m» = 1,202
Per 3,652 m' = 4,389,700 Calories.
The 3,652 m' of gas comes into the laboratory of the furnace
at 600°, and therefore carries in as sensible heat the following:
CO' 0 . 05 X 0 . 502 = 0 . 0251 Cal. per 1°
CH^ 0.02X0.512 = 0.0102 " "
CO, N^ H' 0.93X0.319 =* 0.2967 " "
0.3320X600 =199.2 Cal.
Per 3,652 m' = 199.2X3,652 = 727,480 Calories.
This does not count, however, the water vapor accompanying
the producer gas. This is best determined from the fact that
the total hydrogen in the coal and steam used in the producer
must appear as hydrogen in the producer gas in some form or
other, and whatever is not present in the dry producer gas
must be in the water vapor accompanying it. The amount of
this and the heat it carries into the laboratory of the furnace
are thus determined:
Hydrogen in 750 kg. coal = 750 X 0.0522 = 39. 15 kg.
Hydrogen in steam used = 567 -^ 9 = 63.90 "
Hydrogen going into producer gases = 102.15 "
Hydrogen in dry producer gases,
3,652X0.20X0.09= 65.74 "
Hydrogen in producer gas as vapor of
water = 36.41 "
Water vapor in producer gas = 327.7 "
Volume of this vapor = 405 "
Heat in this at 600°, 405 X 0.43 X 600 = 104,500 Cal.
Total heat in moist producer gas,
727,480 + 104,500 = 831,980 "
We must also calculate the heat brought into the laboratory
of the furnace by the preheated air, at 1,000°, used for com-
bustion. This is
2,661 m«X 0.330X1,000 = 845,130 Calories.
EFFICIENCY OF OPEN -HEARTH FURNACES. 423
It follows from the above calculations that the laboratory of
the furnace receives per heat of steel:
Calories.
From hot producer gas at 600° = 831,980
From preheated air at 1,000° = 845,130
From combustion = 4,389,700
Total = 6,066,810
Of this total there is rejected in the hot products at 1,400°,
.3,816,400 Calories, leaving in the laboratory of the furnace
2,250,410 Calories.
According to the view of Damour and Queneau (Industrial
Furnaces, page 56) the ratio of the heat thus left in the body
of the furnace to the calorific power of the fuel used, measures
the thermal efficiency of the furnace. According to that view
this new style Siemen's furnace has an efficiency of
|g|L«= 0.400 = 40.0 per cent.
This figure, however, is an illusive one. It enables us to
compare two furnaces and see which regenerates the waste
heat best, or which furnace laboratory is designed best so as
to catch and retain most of the heat furnished to it; but the
real question is the comparison of different laboratories as to
their net melting efficiency. This laboratory abstracts from
the heat supply furnished to it 2,250,410 Calories, and fur-
nishes to the steel 900,000 Calories. Its proportion of usefully
applied heat to heat appropriated is, therefore:
QQO'QQ" = 0.400 = 40.0 per cent. (3)
2,250,410
The laboratory of the furnace therefore loses by radiation
60 per cent, of the heat which it abstracts from the gases, or
2,250,410-900,000 = 1,350,410 Calories. This is 24 per cent,
of the calorific power of the coal used.
(4) The regenerators receive 3,816,400 Calories from the
laboratory of the furnace, and return to it the heat in the pre-
heated air at 1,000°, viz.: 845,130 Calories. If we call the
ratio of these two the efficiency of the regenerators we have
j^^'^^^ = 0.222 = 22.2 per cent.
3,816,400
424 METALLURGICAL CALCULATIONS. .
If, however, we charge them only with the heat actually left
in them by the hot products, entering 1,400° and leaving at
350°, we have an efficiency of
845,130 „„^„ „„„
3,816,400-768,950 = ^^^^ = 27-7 per cent. ^^^
As long as the chimney gases are near in temperature to 300°,
we can use the latter form of calculation as representing the
real efficiency of the regenerator. The regenerators, there- .
fore, lose by radiation and conduction to the air 2,202,320
Calories, which is 72.3 per cent, of the heat left in them by the
hot gases and 39 per cent, of the calorific power of the coal
used.
(5) The flame temperature is that to which the 6,066,810
Calories available in the laboratory of the furnace will raise
the products of combustion. The average mean specific heat
of the latter per 1° is :
C02 986m'X(0.37 +0.00022t) = 365 + 0.2169t
02 + N2 4,203 m'X(0.303 + 0.000027t)= l,274 + 0.1135t
H^'O 1,135 m'X (0.34 +0.00015t) = 386 +0.703t
Therefore t =
Sum = 2,025 + 0.5008t
6,066,810
2,025 +0.5008t
Whence t = 2,003 (5)
Problem 75.
In an basic lined open-hearth furnace using the Monell pro-
cess, 50 short tons of melted pig iron, at 1,300°, is run upon
30,000 pounds of Lake Superior iron ore (90 per cent. Fe^O',
10 per cent. SiO^) lying on the hearth and previously heated to
1,300°. There is 2,000 pounds of burnt lime lying on the ore
to help to form slag. The ore reacted quickly, almost vio-
lently, on the melted pig iron, so that at the close of the re-
action, in 20 minutes, the slag could be nm off. The pig iron
contained,
EFFICIENCY OF OPEN-HEARTH FURNACES. 425
On Running in. After the Reaction
Carbon 3.50 3.OO
Silicon 2.00 0.00
Phosphorus 0.75 0.00
Manganese 0.50 0.00
Iron 93.25 97.00
The oxidation of C, Si, P and Mn may be assumed as being
produced first by reducing all Fe'O^ present to FeO, and com-
pleted by the reduction of some FeO to Fe. The carbon is
assumed oxidized in the reaction to CO, although this CO may
be subsequently partly burned to ■ CO^ by excess air in the
furnace.
Required:
(1) The amount of iron reduced into the bath.
(2) The weight and percentage composition of the slag
formed.
(3) The items of heat evolved and absorbed in the reaction.
Solution :
(1) We cannot make a simple, direct solution, because the
weight of the metal after the reaction, analysis of which is
given, is not known. The inexperienced calculator might be
tempted to say that there was 3.50 — 3.00 = 0.50 per cent, of
carbon oxidized, 2.00 per cent, of silicon, 0.75 per cent, of phos-
phorus and 0.50 per cent, of manganese, calculate the weights
of these, reckon up the oxygen they would absorb in being
oxidized, and thus get at the amount of Fe^O^ reduced to FeO
and FeO reduced to Fe. This would be correct as far as silicon,
manganese and phosphorus are concerned, because they are
entirely oxidized, but incorrect for the weight of carbon, be-
cause the 3.00 per cent, not oxidized is per cent, of the final
bath, which is of different weight from the original one, and,
therefore, we are in error in subtracting 3.00 from 3.50. Being
confronted with this dilemma we can see, however, that if
we only knew the weight of the bath after the reaction we
could calculate correctly how much carbon is in it, thence get
the correct weight of carbon oxidized, then the cor"ect amount
of oxygen absorbed in the reaction, and from this the weight
of iron reduced into the bath. We could also get the latter
quantity more simply by finding the iron present in the bath
426 METALLURGICAL eALC'ULATIONS.
after the reaction (97 per cent.), and subtracting from it the
iron in the original bath. We thus have two ways of getting
the same requirement if we only know the weight of the bath.
In such a case the mathematical key to the situation is, of
course, to let X represent the weight of the bath, and work out
the amount of iron reduced into the bath by the two methods,
getting the result expressed in each case in terms of X. Since
the two expressions represent the same quantity, we put them
equal to each other and solve for X. Having obtained X, we
substitute it in either expression and get the result asked for.
Let X be the weight of the bath after the reaction, in pounds;
the bath before the reaction weighs 100,000 pounds.
Oxidized out:
Carbon (100,000 X 0.035)— X 0.03 = 3,500—0.03 X pounds.
Silicon 100,000X0.02 = 2,000
Phosphorus 100,000X0.0075 = 750
Manganese 100,000X0.005 = 500
Oxygen required:
For carbon (3,500—0.03 X) 16/12 = 4,677—0.04 X pounds.
For silicon 2,000X32/28 =2,286
For phosphorus 750X80/62 = 968
For manganese 500X16/55 = 145 "
Sum = 8,076—0.04 X pounds.
Oxygen supplied:
If all Fe^O' of ore (27,000 pounds) were reduced to
FeO = 27,000X16/160 = 2,700 pounds.
If all Fe^O' of ore were reduced to Fe 27,000X48/160 =
8,100 pounds.
We thus see that we will certainly require more oxygen than
the Fe'^O' can give up in becoming all FeO, but not as much
as would be given up if it all became Fe. If all the Fe^O' were
considered first reduced by the reaction to FeO, giving up
2,700 pounds of oxygen, the reaction will still require the fur-
nishing of
(8,076—0.04 X)— 2,700 = 5,376—0,04 X pounds.
of oxygen, which would have to be furnished by FeO becoining
EFFICIENCY OF OPEN-HEARTH FURNACES. 427
Fe. In that reduction, however, 72 parts of FeO gives up 16
of oxygen in becoming 56 Fe. The reduced Fe will be, there-
fore, 56/16 of the weight of oxygen thus furnished, and therefore,
Reduced Fe = 56/16 (5,376—0.04 X) pounds.
= 18,816— 0.14 X pounds.
The same quantity is obtained more directly as follows:
Fe in original bath 100,000X0.9325 = 93,250 pounds.
Fe in bath after reaction = 0.97 X pounds.
Therefore, Fe reduced = 0.97 X— 93,250 lbs.
These two expressions represent the same thing, and, therefore,
18,816— 0.14 X = 0.97 X— 93,250
Whence X = 100,960
And the reduced iron = 0.97 X— 93,250 = 4,681 pounds. (1)
(2) The ore used contains altogether
Fe = 27,000 X 112/160 = 18,900 lbs.
Fe reduced to metallic state = 4,681 "
Fe remaining in slag as FeO =14,219 "
"Weight of FeO = 14,219x72/56 = 18,282 " = 61.1 per cent.
WeightofP^O' = 750 + 968 = 1,718 " = 5.7
Weight of MnO= 500+145 = 645 " = 2.1
Weight of CaO = 2,000 " = 6.7
WeightofSiO^ =3,000 + 4,286 = 7,286 " =24.4
Total weight of slag = 29,931 " (2)
(3) The physical data available do not permit of calculating
the actual heat of the reaction at 1,300°, since many of the
specific heats needed are lacking. We will therefore foot up
the items of heat evolution and absorption uncorrected for
temperature, which is the only thing to be done under the cir-
cumstances.
Heat Evolution. Calories
SitoSiO'' 2,000X7,000 =14,000,000
PtoP^'O' 750X5,892 = 4,419,000
MntoMnO 500X1,653 = 826,500
CtoCO 471X2,430 = 1,144,500
SiOHo FeO. SiO» 7,286 X 144 = 1,049,200
CaO to SCaO. P^O' 2,000 X 949 = 1,898,000
Total = 23.337,200
428 METALLURGICAL CALCULATIONS.
Heat Absorption.
Calories.
Fe^O'toFeO 18,900 X 573 =10,829,700
FeOtoFe 4,681X1,173 = 5,490,800
Fe'CtoFe' + C 471 X 705 (?) = 332,000 (?)
FeSitoFe + Si 2,000 X 931 (??) = 1,682,000 (??)
Fe'P to FeHP 750X1,400 (??) = 1,050,000 (??)
Sum = 19,564,500
These calculations, therefore, show a minimum surplus of
heat in the reaction of 23,337,200 - 19,564,500 ^ 3,772,700 Cal,
an amount which would raise the temperature of the slag and
resulting metal approximately 100° C. above the 1,300° at which
the reacting materials came together. The quantity above
marked (?) is a little doubtful in amount, but those marked
(??) are very doubtful, perhaps may not exist at all. If these
are omitted from the heat absorption the surplus heat is in-
creased some 50 per cent, of its value, and the rise in tem-
perature might be in the neighborhood of 150°. Further, some
of the CO formed by the oxidation of carbon may be burned to
CO^ close to the surface of the bath, by free oxygen in the furnace,
still further increasing the rise in temperature.
The conclusion from these calculations and discussions is
that the pig iron and ore reaction in the Monell process is a
heat evolving reaction, which, independently of the hpating
effect of the fuel used by the furnace, could increase the tem-
perature of the contents of the furnace at least 100°, and pos-
sibly 150°. It would be highly interesting to have a typical
heat such as this followed closely with a good reliable pyro-
meter, so as to check the indications of the thermochemical
study of the process.
CHAPTER XII.
THE ELECTROMETALLURGY OF IRON AND STEEL.
Electrical methods may enter into the extraction of a metal
from its ores either as electrolytic or as electrothermal pro-
cesses. Electrolytic processes are those in which the electric
current is used for its electrolytic action, i.e., for its electrical
decomposing and depositing properties; electrothermal pro-
cesses are such as use the current merely as a source of heat,
to furnish the sensible heat and high temperature necessary
for melting materials or for bringing about chemical reactions.
So far electrolytic processes have entered the metallurgy of
iron only as used by Burgess for electrolytically refining nearly
pure iron in an aqueous electrolyte and depositing chemically
pure iron; the electrolysis of fused iron salts has not been prac-
tically utilized. Up to the present, electrothermal processes
are in commercial use for melting together wrought iron and
cast iron to make steel, also for keeping cast iron melted while
its impurities are being extracted by oxidation ; the electro-
thermal reduction of iron ores to cast iron has been proved
technically possible, and may in some places prove commer-
cially practicable.
Electrothermal Reduction of Iron Ores.
If the electric current is used to furnish the heat energy
necessary to reduce iron ore, it cannot displace the reducing
agent — carbon. In ordinary blast furnace practice the carbon
is first burned to provide the heat necessary to smelt down the
pig iron and slag, and the product of this incomplete combus-
tion— CO — abstracts oxygen from the ore. The two equations
are practically:
At the tuyeres —
30C + 150^+57.15 N'' = 30CO + 57.15 N^
Reduction :
4Fe='O'-l-30CO-F57.15 N* = 12CO=' + 18CO + 57.15 NH8Fe
429
430 METALLURGICAL CALCULATIONS.
These equations show us that to produce 8Fe = 448 parts,
30C = 360 parts of carbon is the minimum necessary, which is
first burned to CO at the tuyures, and then the producer gas
thus formed (N^ and CO) reduces the iron oxides above. If
the heat for fusion is furnished electrically, the first combus-
tion is unnecessary, all blowing in of air is dispensed with and
the reaction taking place is:
HFe^O' + aOC = 12CO=' + 18CO + 28Fe
And we have 28Fe = 1,568 parts reduced by 30C = 360 parts,
a consumption of less than one-third as much carbon as is
required in the blast furnace. The operation consists, therefore,
in mixing iron ore and carbon so that for every part of iron
present about 0.25 parts of carbon is present, using the proper
quantity of limestone or other material to flux the gangue of
the ore to a fusible slag, and then furnishing electrically the
heat necessary to cause the chemical reaction, melt down the
resulting iron and slag, and supply radiation losses. The
gases resulting from this electrical reduction are combustible,
just as the gases from the blast furnace, and since there is no
blast to be heated they can very well be utilized to preheat the
charges coming into the furnace, and thus save some of the
electrical energy needed.
Problem 76.
A magnetite ore contains:
Per Cent. Per Cent.
Fe^O' 60.74 MgO 5.50
FeO 17. 18 P^O' 0.04
SiO^ 6.60 S 0.57
AW 1.48 CO^ 2.05
CaO 2.84 H^'O 3.00
It is to be mixed with pure carbon (charcoal fines) and a
suitable flux; the fixed carbon being 0.25 per cent, of the iron
present; the flux silica sand, so as to make a slag with 33 per
cent, of SiO^ Neglect the ash and assume 10 per cent, of
moisture in the charcoal. Assume also :
(a) The pig iron to contain 4 per cent. C, 3.5 per cent. Si,
92.4 per cent. Fe.
ELECTROMETALLURGY OF IRON AND STEEL. 431
(b) The slag and pig iron to contain at tapping 600 and
400 Calories of heat respectively.
(c) The heat losses by radiation, etc., to be 30 per cent, of
the total heat requirement of the furnace.
(d) The hot gases to escape at 300° C.
(e) The iron to be completely reduced into the pig iron.
(/) The sulphur to go entirely into the slag as CaS.
Requirements :
(1) The weights of ore, flux and charcoal dust needed per
1,000 kg. of pig iron produced'.
(2) A balance sheet of materials entering and leaving the
furnace.
(3) A heat balance sheet of the furnace.
(4) The number of kilowatt days of electrical energy re-
quired per metric ton of pig iron produced.
Solution :
(1) The ore must supply 924 kg. of iron. But 100 parts of
ore contains iron as follows:
Kg.
In Fe='0» 60.74x112/160 ^ 42.52
In FeO 17.18X 56/72 = 13.36
Sum = 55.88 (1)
Ore required per 1,000 kg. of pig iron:
924 H- 0.5588 = 1,654 kg.
The slag-forming ingredients from this amovint of ore are
as follows:
Kg.
A1^0n,654X 0.0148 = 24.6
Mg01,654X 0.0550 = 910
CaO 1,654X (0.0284— O.0057X 56/32)
= 1,654X0.0184 = 30.4
810^(1,654X0.0660)— (35x60/28) = 109.2—75 = 34.2
CaS 1,654 X (0.0057x72/32) = ^1.2
Sum = 201.4
If X parts of SiO' sand are added to these, the total weight
432 METAUJJRGICAL CAlJCVLATlONS.
of slag will be 201.4+ a;, of SiO^ in it 34. 2+ a;. And since the
SiO^ is to be 33 per cent, of the weight of slag, then
34.2 + a: = 0.33 (201.4 + ic)
Whence x= 48 kg. (1)
The charcoal dust used must contain fixed carbon equal to
0.25 of the iron present, *'. e.,
924X0.25 = 231 kg.
And since it is 90 per cent, fixed carbon the dust required is :
231-^0.90 = 257 kg.
Charges. Pig Iron. Slag. Gases.
Fe^O'
...1004.6
284.2
....109.2
24.6
46.9
91.0
0.6
9.4
Fe....
Fe....
Si....
...703.2
...221.0
35.0
0....
0
0
..301.4
FeO
. . .63.2
SiO^
APO'
SiOj....
AI.O3. .
CaO....
MgO...
Ca
( S
...34.2
...24.6
...30.4
...91.0
...11.8
9.4
...40.0
CaO
0....
4 7
MgO
P.....
0.3
P'O'
s
0....
....0.3
C02
33.9
co^..
HjO..
33 9
H^O ... .
49.6
...48 Kg.
48.0
. .257 Kg.
....231.0
26.0
.49.6
Flux
SiO^
SiOjj....
...48.0
Charcoal . .
C
C
40.0
C....
HjO..
191 0
H^O
26 0
1959.0 999.5 249.4 118.1
(3) Heat available is the heat of oxidation of carbon plus
that furnished by the electric current. The charge gives up
to the carbon, as shown on the balance sheet, 301.4 + 63.2 +
40.0 + 4.7 + 0.3 = 409.6 kg. of oxygen. The 191 kg. of carbon
burned would take 191x16/12 = 254.7 kg. of oxygen to bum
it to CO, leaving 154.9 kg. of oxygen to bum CO to CO^ This
would bum 154.9x28/16 = 271.1 kg. of CO to CO".
The heat of formation of the slag may be taken as approxi-
ELECTROMETALLURGY OF IRON AND STEEL. 433
mately' 150 Calories per kg. of contained SiO^ + APO'- The
heat of combination of carbon with the iron is a doubtful quan-
tity, which may be taken at 705 Calories per kilogram of car-
bon. The formation of CaS gives 2,947 Calories per kilogram
of sulphur.
Letting the heat furnished by the electric current = x,
and neglecting the heat of oxidation of the small amount of
electrode carbon consumed, we have :
Heat Available.
Calories.
Supplied by electric current :
X
Oxidation of C to CO 191 X 2,430 =
434,130
Oxidation of CO to 00=* 271.1 X 2,430 =
658,770
Formation of silicate slag 106.8 X 150 =
16,020
Formation of CaS 9.4 X 2,947 =
27,700
Formation of Fe^C 40. X 705 =
28,200
Sum =
1,194,820 + a;
Heat Distribution.
Calories,
Reduction of Fe from Fe^'O'
703.2X1,746 =1,229,790
Reduction of Fe from FeO
221.0X1,173 = 259,230
Reduction of Si from SiO^
35.0X7,000 = 245,000
Reduction of P from P^O'
0.3X5,892 = 1,770
Reduction of Ca from CaO
11.8X3,288 = 38,800
Expulsion of CO^ from ore
33.9X1,026 = 34,780
Evaporation of H^O from charges
75.6X606.5 = 45,850
Sensible heat in gases, at 300°
CO 174.6 kg. = 138 m»
X0.311 = 42.9
434 METALLURGICAL CALCULATIONS.
C0= 459.9 kg. = 232 m'
X 0.436 = 101.1
H^O 75.6 kg. = 93 m'
X 0.385 = 35.8
179.8X300 = 53,940
Sensible heat in slag 249.4X600 = 149,640
Sensible heat in pig iron 1 ,000 X 400 = 400 ,000
Loss by radiation, etc., 0.30 (1,194,820 + ac) = 358,450 + 0. 3a;
Sum total = 2,817,250 + 0.3i(;
Equating the heat available and accounted for we have:
1,194,820 + a; = 2,817,250 + 0.3 x.
Whence x = 2,317,760 Calories.
And the sum total of heat requirement is
3,512,580 Calories,
of which the electric current supplies
§gg2 = 0.66 = 66 per cent. (3)
(4) A kilowatt-day of electricity is equal to
0.239X60X60X24 = 20,650 Calories.
There is, therefore, required per ton of pig iron produced:
2,317,760
20,650
112 kilowatt days. (4)
This figure might be materially reduced by using the waste
gases to warm up the charges entering the furnace. It is also
possible that in a properly designed shaft the gases passing out
might contain nearly equal volumes of CO and CO^, instead of
3C0 to 2C0, as assumed in this problem, from best ordinary
blast furnace practice. Any greater utilization of the heat-
producing power of the carbon would decrease the electrical
energy required; the above calculated value is given as a safe
figure for this ore on which to base working calculations.
ELECTROMETALLURGY OF IRON AND STEEL.
Problem 77.
435
At Sault Sainte Marie, Canada, roasted pyrrhotite ore was
smelted •with the addition of limestone and charcoal fines. The
mixture used contained 400 pounds of ore, 110 pounds" charcoal
dust and 85 potinds limestone. The analyses of each of these
materials was:
Roasted Ore.
Fe^O' 65.43
CuO 0.51
NiO 2.84
SiO^ 10.96
ATO' 3.31
CaO 3.92
MgO 3.53
80= 3.90
p20^ 0.03
H^O 6.57
Limestone.
CaO 52.00
MgO 2.10
FeW 0.60
Al^O^ 0.21
Si02 1.71
P20' 0.01
SO' 0.13
CO^ 43.15
Charcoal Dust.
Fixed C 55.90
Vol. matter 28.08
Moisture 13 . 48
Ash 2.54
Using 165.65 kilowatts effective electric energy for 56 hours
20 minutes, there was produced 7,336 pounds of nickeliferous
pig iron and 5,062 potinds of slag, having the following average
compositions:
Slag.
SiO' 16.44
Al^O' 13.86
CaO 42.87
MgO 8.80
CaS 13.34
FeO 0.84
Undetermined 3.85
c
Pig Iron.
3.05
Si
5.24
s
0.01
p
0.05
Cu
0.81
Ni
3.94
Fe
86.90
Requirements:
(1) A balance sheet of materials entering and leaving the
furnace.
(2) A heat balance sheet of the furnace, making necessary
assumptions where data is not furnished.
(3) The thermal efiSciency of the furnace.
Solution:
(1) The amount of ore used may be calculated either from
the iron, the nickel or the copper. The nickel and copper are
the easiest to use, because there is supposed to be none of
436 METALLURGICAL CALCULATIONS.
them in the slag, but they are the least reliable, because present
in such small amount. The pig-metal contains 7,336X0.81 —
59.4 pounds of copper, and since the roasted ore contains 0.51
X „„ g = 0.41 per cent, copper, the weight of ore used, on
this basis, would be 59.4-^0.0041 = 14,493 pounds. As for
nickel, the pig-metal contains 7,336X0.0394 = 289 pounds of
59
nickel, and since the roasted ore contained 2.84 X=f = 2.23
75
per cent. , the weight of this used should have been 289 -h 0.0223 =
12,960 pounds. These figures differ so much that we will
make the calculation on the basis of the iron. There is iron
present as follows :
Pounds.
In the pig iron 7,336X0.8690 = 6,375
In the slag 5.062 X 0.0084 X 56/72 = 33
Total in products = 6,408
In 100 pounds of ore 65.43X0.7 = 45.80
In 21 potinds limestone 21. X 0.0060 X0!7 = .08
In 27.5 poimds of charcoal 27.50 X 0.0025
X0.7 = .04
In charge, per 100 pounds of ore used = 45.92
Therefore, ore necessary to supply the iron in products:
Pounds.
6,408 -^ 0.459 = 13,961
with which will be used :
Charcoal = 13,961x110/400 = 3,839
Limestone = 13,961x85/400 = 2,967
We are now ready to construct the balance sheet as soon
as we assume probable values for the composition of the vola-
tile matter and ash of the charcoal. The ash might be taken
as containing on an average: K^O 15 per cent., CaO 40, MgO
20, MnO 15, and Fe='0' 10 per cent. The volatile matter is due
to insuflBicient charring, and the gases given off on heating
may be assumed as, by volume, CO' 25 per cent., CO 15, H'
ELECTROMETALLURGY OF IRON AND STEEL.
437
50, CH* 10. This would make the volatile matter to contain, in
per cents, by weight, carbon 33.7, oxygen 58.5, hydrogen 7. 8
per cent. (The verification of this last statement is a nice
little exercise in chemical arithmetic.) The full statement of
the elementary composition of the charcoal, for use in making
the balance sheet, is therefore:
Per Cent,
Fixed carbon 55.90
Volatile carbon 9.46
Volatile hydrogen 2.19 28.08%
Volatile oxygen 16.43
Moisture 13.48
K^O..
CaO..
MgO.,
MnO.,
0.381
1.02
0.51
0.38
0.25
Charges.
Roasted Ore
Balance Sheet, per 7,336 Lhs. Pig-metal.
Pig-metal. Slag.
.13,961
\ 2.54%
Gases.
Fe^O'.
NiO..,
9,135 Fe 6,375 FeO 25 O 2,375
CuO.
397 Ni.
71 Cu.
.289 NiO 28 O.
.59
O.
SiO' 1,530 Si 384 SiO^ 1,091 O.
.80
.12
.55
CaO..
MgO..
SO'..,
H^O..
461 ..
547 ..
493 ..
545 S.
4 P.
778 ..
AlW 461
CaO 166 O . . . .
MgO 493
CaS 490 O...,
O...,
H^O.
.109
.327
.. 0
.778
Limestone. . . 2,967
O.
CaO 1,543 CaO 1,542 O.
MgO
FeW
AlW
SiO^
P^O^. .
SO^..
62
18
6
51
0
4
00^ 1,280
MgO.
FeO..
AlW.
SiO^..
P='0^.
CaS...
.62
.16
. 6
.51
. 0
. 4
CO
.1280
Charcoal dust 3,839
438 METALLURGICAL CALCULATIONS.
FixedC 2,146 C 224 C 1,922
Vol. C 363 C 363
Vol. H 84 H 84
Vol.0 631 0 631
H^O 518 HjO 518
K^O 15 K'0 15
CaO 39 CaO 39
MgO 20 MgO 20
MnO 15 MnO 15
Fe^O^ 10 FeO 9 0 1
Electrode 66
C 86 C 66
20,828 7,336 4,533 8,962
There is a lack of close correspondence between the weights
and compositions of slag, as observed and as calculated, due
evidently to inaccurate sampling and analyses of the roasted
ore and slag.
(2) From the balance sheet we can deduce the heat evolved
and absorbed in the chemical reactions in the furnace. The
more involved items are calculated as follows:
Oxidation of carbon to CO : All the carbon put into the fur-
nace as fixed carbon goes out as either CO or CO^, except that
going into the pig iron. The carbon burnt to CO in the fur-
nace may be, therefore, taken as 1,922 + 66 = 1,988 pounds,
evolving 1,988X2,430 = 4,830,800 pound-Calories.
Oxidation of CO to CO'^: There is given up in the furnace,
by the reductions accomplished, 3,021 pounds of oxygen, of
which 1,988x16/12 = 2,651 pounds would burn fixed carbon
to CO, as above shown, leaving 370 pounds to burn CO to CO'.
This would oxidize 370x28/16 = 648 pounds of CO to C0^
which would evolve 648X2,430 = 1,574,300 po'und-Calories.
Heat energy of electric current: One kilowatt-second is 0.239
kilogram Calories, or 0.527 pound-Calories. The current being
on 56 hours, 20 minutes, or 202,800 seconds, the heat equiva-
lent of the current used is:
0.527X202,800X165.65 = 17,704,000 pound-Calories.
Heat in escaping gases: We are here confronted with the
fact that no observation of the temperature of these was given.
ELECTROMETALLURGY OF IRON AND STEEL.
439
There is no essential reason why they should escape very hot
from the furnace, if properly run and conducted, so we will
assume a maximum temperature of 500° C. The gases would
consist of 0,958 + 2,651 - 648 = 4,061 pounds of CO and 648 +
370 = 1,018 pounds of CO^, from the oxidation of fixed carbon
in the furnace; plus 1,280 of CO^ from the limestone, and 666
pounds C0^ 254 pounds CO, 97 pounds CH* and 60 pounds
of H^ from the volatile matter of the charcoal. To these must
be added 1,290 pounds of water vapor. The heat is therefore
CO 4,315 lbs. = 54,800 ft' X 0.304 = 16,650 oz. Cal.
2,964 lbs. = 23,9butt''X0.480= 11,500
97 lbs. = 2,150 ft=X 0.490= 1,050
60 lbs. = 10,700 ft=X 0.304= 3,250
1,290 lbs. = 25,500 ft' X 0.415 = 10,550
CO^
CH«
H^'O
Sum =117,100 ft'
per 1°
per 500
43,000
= 21,500,000
= 1,343,750 Ib.-Cal.
The other items of the heat balance sheet are almost self-
explanatory, and the complete balance is as follows:
Heat Available.
Energy of the electric current
Oxidation of C to CO
Oxidation of CO to CO^
Combination of C with Fe' 224X705
Combination of Ca with S 220X2,947
Formation of silico-aluminate slag
(SiO' + APO') 1609X150
Heat Distribution.
Pound-Cal.
= 17,704,000
= 4,830,800
= 1,574,600
157,900
= .648,300
241,400
Total = 25.157.000
Reductions:
Fe'O' to Fe
NiO to Ni
CuO to Cu
SiO' to Si
SO' to S
P'O' to P
CaO to Ca
Fe^O' to FeO
6,375X1,746
289X1,051
59 X 593
384X7,000
218X2,872
4X5,892
274X3,288
50 X 446
11,130,750
303,750
35,000
2,688,000
626,100
23,550
900,900
22.300
440 METALLURGICAL CALCULATIONS.
Expulsion of CO^ from flux:
1,280X1,026 =
1,313,300
Evaporation of H'O
1,290X606.5 =
783,400
Sensible heat in gases
=
1,343,750
Sensible heat in pig iron
7,336 X 400 =
2,934,400
Heat in slag
4,533 X 600 =
2,719,800
Loss by radiation, conduction
., etc. =
Total =
332,000
25,157,000
(2)
(3) The essential work done by the furnace is the reduc-
tions, evaporation of moisture and decomposition of carbon-
ates. The heat in slag, iron, gases and radiation loss are all
susceptible of diminution or of being more or less returnable
to the furnace. The usefully applied heat is, therefore, 17,-
827,050 pound-Calories. To produce this there was consumed
the 25,157,000 pound-Calories actually generated, and there
was wasted 13,390,000 pound-Calories, the calorific power of
the gases escaping from the furnace, which should have been
generated or might be utilized, making a total of 38,527,000
pound-Calories disposable.
The working thermal efficiency over all was therefore:
17,827,050 r,.a Aa 4. ri\
38.527.000 = "^-^^ = *6 P^^ '=^"*- (^)
Production of Steel.
There are three methods of producing steel electrically which
are practicable. First, the electric furnace may replace the
crucible simply as a melting apparatus, in producing a cast
steel from cemented bars; second, the electric furnace may re-
place the crucible or open-hearth furnace as an apparatus in
which to melt together wrought iron and pure cast iron, such
as washed pig metal, although in this operation, the electric
furnace is more like the crucible method, in that there is not
necessarily any oxidation of the metal or of its impurities in
the operation; third, the electric furnace may be used to melt
or keep melted cast iron, while its impurities are oxidized out
by additions of iron ore, in this operation resembling the Ucha-
tius method of making crucible steel or the " pig and ore "
process of making steel in the open-hearth furnace.
ELECTROMETALLURGY OF IRON AND STEEL. 441
The particular advantages possessed by the electric furnace
processes are, compared with the crucible process, the larger
quantities in which the steel can be made in one operation, the
absence of carbon in the furnace lining, thus controlUng better
the carbon and silicon in the steel, and the higher temperature
enabling a more basic slag to be kept fluid and the sulphur to
be better eliminated; the advantages compared with the open-
hearth furnace are the absence of gases of combustion in con-
tact with the metal, and the higher- temperature available,
which permits of very basic, refractory slags being made and
kept thinly fluid, and thus gives better control of sulphur and
phosphorus. In addition to these, in both cases, may be
mentioned the commercial advantages, for the saving in cru-
cibles alone makes the electric furnace superior in this respect
to the crucible process, and the electric furnace can compete
successfully as regards cost with the regenerative open-hearth
furnace wherever water power costs less than $10.00 per horse-
power-year and coal costs over $5.00 per ton.
A particular point to be noted is, that when heating by com-
bustion is used the efficiency of the absorption of heat by the
charges decreases very rapidly as the temperature gets higher.
For instance, if a cold ingot of iron is placed in a furnace the
temperature of which is 1,500°, the iron absorbs heat with very
great rapidity from the start up to, say, 1,000°; but with de-
creasing rapidity thereafter. The rate of transfer of heat
from the gases to the iron is proportional to the difference of tem-
perature, and is some fifteen times as fast when the iron is
at 0° as when it is at 1,400° If the efficiency of the heating
by furnace gases is, say, 25 per cent, in bringing metal up to
1,500°, it is likely that the distribution of this efficiency is pro-
portioned about as follows:
Heating from 0° to 500° 45 per cent, efficiency.
Heating from 500° to 1,000° 27 per cent, efficiency.
Heating from 1,000° to 1,500° 3 per cent, efficiency.
On the other hand, the conversion of electrical energy into
heat in the substance of the material to be heated is not a
contact or transfer phenomenon, but a thermodynamically
frictionless transfer of 100 per cent, efficiency, and equally so at
the highest as at the lowest temperatures. Only radiation and
442 METALLURGICAL CALCULATIONS. ■
conduction losses need be considered, the problem is not how
much heat can you get into a body but how much can you
keep in; it is already all in, 100 per cent, of it, to start with.
In a large, properly designed electric furnace the radiation and
conduction losses of heat, even working to the highest tem-
peratures, can be kept at 15 to 25 per cent, of the total heat
generated, giving an efficiency of 75 to 85 per cent.
It may very well be, that there are places where the relative
prices of coal and electric power are such that coal is the cheaper
for heating to 500°, at 45 per cent, efficiency, or to 1,000° at
36 per cent, efficiency, or even to 1,500°, at 25 per cent, effi-
ciency, but that the combination of heating by fuel to 1,000°
at 36 per cent, efficiency and then by electricity from 1,000° to
1,500°, at 70 per cent, efficiency, would be the cheaper plan,
or even by fuel to 500°, at 45 per cent, efficiency, and then
by electricity from 500° to 1,500°, at, say, 75 per cent, efficiency,
would be commercially advantageous.
Illustration: Steel bars are to be melted in an electrical
furnace. It takes 300 Calories effective in the steel per kilo-
gram to heat it to a tapping heat; the electric furnace supplies
this at a net thermal efficiency of 75 per cent. To heat the bars
to 750°, cherry red, without melting them, requires, 88 Calories,
or 29 per cent, of the total. If the bars were heated in a coal
furnace to 750°, and then transferred to the electric furnace,
some 25 per cent, of the electrical power might be saved. If
this heating were done by coal having a calorific power of
8,500, at a thermal efficiency of 25 per cent., there would be
needed 40 grams of coal. The question is, therefore, the rela-
tive cost of 88-^0.75 =. 117 Calories delivered electrically, and
40 grams of coal. The former requires 117-5-0.239 = 490
kilowatt-seconds = 8.2 kilowatt-minutes = 0.14 kilowatt-hours.
At $10.00 per kilowatt-year (8,760 hours) this would cost 0.14
X 0.114 = 0.016 cents. At $5.00 per metric ton the coal would
cost 0.040X0.5 = 0.020 cent. Under such assumed conditions
of cost of power and of coal the electrical heating, even up
to 750°, would be the cheaper.
Problem 78.
In an induction electric furnace of 170 kilowatts capacity,
4.7 tons of steel is made per day by melting together colij-
ELECTROMETALLURGY OF IRON AND STEEL. 443
washed pig iron and scrap iron, the melted steel carrying 350
Calories per kilogram.
Required:
(1) The electric energy in kilowatt hours required per ton
of steel produced.
(2) The thermal efficiency of the furnace.
(3) If one-third the material used were put into the furnace
melted, carrying 275 Calories per kilogram, what would be the
production per day and the power required per ton of steel?
Solution:
(1) Energy for 4.7 tons = 170 kw-days.
Energy for 1.0 ton = 36.2 "
Energy for 1.0 ton = 0.10 kw-year
Energy for 1.0 ton = 869 kw-hours.
(1)
(2) Ikw-hour = 0.239X60X60 = 860 Calories
869 kw-hours = 747,340
Heat in 1 ton of steel = 350 X 1 , 000 = 350, 000 "
TU ^ cc ■ 350,000
Thermal efficiency = ^^^ ^^^ = 0.47 = 47 per cent. (2)
(3) Heat in melted material used per kilogram of steel pro-
duced = 275x1/3 = 92 Calories.
Heat to be supplied by the current 350 — 92 = 258 Calories.
Production per day under these conditions:
. » 350
4.7X253 = ®"* *°''^- ^^)
Relative times for the heats = 1 to 0.74.
Energy required per ton = 170-^6.4 = 26.6 kw-days.
= 0.07 kw-year.
= 638 kw-hours. (3)
Problem 79.
An electric steel furnace running at full heat, and contain-
ing about 2,500 kg. of steel, loses by radiation, etc., 250,000
Calories per hour; 2,500 kg. of melted pig iron is run into the
hot furnace, carrying 250 Calories per kilogram, and it is treated
with 500 kg. of iron ore, previously heated to 500° C, and 50
444 METALLURGICAL CALCULATIONS.
kg. of limestone added cold. The steel produced carries 400
Calories per kilogram and the slag 600 Calories. The opera-
tion lasts 1 hour. Assume the following composition of ma-
terials used and made:
Pig Iron.
Iron Ore.
Limestone.
Steel.
Fe.
96.666
Fe203....85.93
CaO 53.74 Fe.
99.60
r
2.700
0.600
FeO 3.96
SiOj 5.50
MgO 0.17 C.
0.11
Si..
SiOj 3.14 Si..
0.11
Mn,
0.025
MnO 0.63
Pe^Oj.... 0.18 Mn,
0.15
S..
0.007
AljO^ 0.76
AI2O3 0.32 S..
0.02
P..
0.012
CaO 2.23
MgO 0.97
PjO., 0.006 P..
S. 0.001
0.01
The bath was treated by the final addition of 10 kg. of cold
ferro-manganese, carrying 80 manganese, 16 iron and 4 car-
bon. The steel obtained weighed 2,630 kg.
Required :
(1) A balance sheet of materials entering and leaving the
furnace.
(2) The weight and percentage composition of the slag.
(3) A balance sheet of the heat received and distributed.
(4) The net power required to run the furnaces and the cost
of power per ton of steel made, at $25.00 per kilowatt-year.
Balance Sheet.
Charges. Steel. Slag. Gases,
Pig Iron.... (2600 Kg.)
Fe 2416.4 2416.4
C 67.6 2.5 C 65.0
Si 16.0 2.9
Mn 0.6
S 0.2 0.5
P 0.3 0.3
Si02
... 24.9
MnO....
... 0.8
FeO
...125.6
FeO
... 19.8
Si02
... 27.5
MnO....
... 3.2
AlW...
.... 3.8
CaO
... 11.2
MgO....
... 4.8
Ore (500 Kg.)
FeW 429.7 203.1 FeO 125.6 0 87.2
FeO 19.8 ....
SiO^ 27.6 .:..
MnO 3.2
Al^O' 3.8 ....
CaO. 11.2
MgO 4.8 ....
ELECTROMETALLURGY OF IRON AND STEEL
445
CaO..
26.9
0.1
1.6
0.1
0.2
21.2
manganese (10 K$
1.6
■J
CaO....
MgO....
SiO*....
FeO
AlW...
... 26.9
... 0.1
... 1.6
... 0.1
... 0.2
MgO..
sio^.
Fe^O'.
AlW.
CO*..
COjj....
21.2
Ferro-
Fe...
',■)
1
0
3
6
4
9
C ..
0.4
Mn .
8.0
MnO....
... 5.3
3060.0
2631
6
255.8
173.4
(2) The slag contains :
Per Cent.
SiO=' 54.0 pounds = 21.1
APO' 4.0
CaO 38.1
MgO 4.9
FeO 145.5
MnO 9.3
255.8
= 1.5
= 14.9
= 1.9
= 56.9
= 3.6
= 99.9 (2)
Heat Available.
Electric current :
Oxidation of C to CO
Oxidation of CO to CO'
Oxidation of Si to SiO'
Oxidation of Mn to MnO
Formation of slag
Heat in melted pig iron
Heat in iron ore
Calories.
X
65.0X2,430 = 157,950
0.9X2,430 = 2,190
12.1X7:000 = 84,700
4.7X1,653 = 7,770
22.6 X 150 = 3,390
2,500 X 250 = 625,000
500 X 77 = 38,500
Sum = a;+ 919,500
Heat Distribution.
Heat in melted steel
Heat in melted slag
Reduction of Fe'0» to FeO
2,630 X
256 X
386.7 X
400 =
600 =
446 =
Calories
1,052,000
153,600
172,470
446 METALLURGICAL CALCULATIONS.
Reduction of FeO to Fe 203.1 X 1,173 = 238,240
Separation of carbon from iron 65 X 705= 45,800
Heat in gas at 1,500°:
CO = 151.7 kg. = 120.4 in^X0.32x500 = 19,260
CO^ = 1.4 kg. = 0.7 m'X 0.60X500 = 200
Loss by radiation, etc. = 250,000
Total = 1,931,570
Heat to be supplied by current:
X = 1,931,570—919,500 = 1,012,070 Calories. (3)
(4) One kilowatt furnishes per hour 860 Calories, therefore
the power required to run the furnace is:
1,012,070-^860 = 1,177 kilowatts. (4)
At $25.00 per kilowatt-year a kilowatt-hour would cost
$25.00^8,760 = 0.2854 cents.
And the power to run the furnace 1 hour would cost
0.002854X1,177 = $3.36.
And the cost per ton of steel :
$3.36 -J- 2.630 = $1.32.
Problem 80.
It is desired to design a plant for the electro-deposition of
pure iron by the Burgess process (see Transactions American
Electrochemical Society, Vol. V., p. 201). The desiderata and
data are as follows:
Output, 25 metric tons per day.
Current density, 110 amps, per square meter.
Anodes, 0.75X0.5 meters immersed X3m.m. thick.
Cathodes, 0.75X0.5 meters immersed Xl m.m. thick at
starting.
Cathodes to be run until deposit is 1.5 cm. thick on each side.
Anodes run until 0.9 consumed.
Tanks, 1.00 meter deep, 0.6 m. wide, 2 m. long inside, filled to
withiii 0.10 meter of top with electrolyte.
ELECTROMETALLURGY OF IRON AND STEEL. 447
Working distance between anode and cathode 6 centimeters
at starting.
Electrolyte contains 10 per cent. FeSO^. 7H^0, and 5 per cent.
(NH*)^SOS specific gravity 1.1, electrical resistivity 20 ohms
per centimeter cube.
Voltage drop in connections and conducting rods 0.3 volt per
tank.
Main conductors carry 2 amps, per m.m. square of section.
Net cost of electrical power, $25.00 per kilowatt-year.
Requirements :
(1) The number of anodes and cathodes per tank and the
number of tanks in the plant and their arrangement.
(2) The weight of anodes and cathode sheets, increasing the
weight of immersed part 10 per cent. Specific gravity of the
rolled iron 7.9
(3) The weight of ferrous sulphate and ammonium sulphate
required to start the plant.
(4) The drop of potential across the electrodes at starting
and at the close of a deposition ; the drop of potential per tank ;
the total voltage needed at starting the plant and when it is
in regular operation.
(5) The cross sectional area of the main conductors.
(6) The time required to consume an anode plate, *'. e., in
dissolving iron away equal to 0.9 of its weight.
(7) The time required to deposit a full cathode plate ; specific
gravity of the deposit 7.6
(8) The electric power required to run the plant and its cost
per ton of iron deposited.
Solution :
(1) 110 amps, per square meter deposits per day:
0.00001036X110X ^X60X60X24 = 2,757 grams Fe.
Therefore, depositing surface required:
25,000^2,757 = 9,140 square meters.
Since one cathode plate has a depositing area on both sides
of 0.75X0.5X2 = 0.75 square meter, the number of cathode
plates required in the whole plant is:
9,140-^0.75 = 12,187
■448 METALLURGICAL CALCULATIONS.
In one tank, if there are x cathodes and x+1 anodes, the
thickness of these plates at starting is, in millimeters, x + 3
(x+1) = 4x + 3 m.m. The number of spaces between anodes
and cathodes being 2x, and each of these being 6 cm. = 60
mm. at starting, the spaces are 120ic m.m. The length of
the tank being, inside, 2,000 m.m. :
124s: + 3 = 2,000
whence x = 16.1.
Each tank will therefore contain 16 cathodes and 17 anodes,
at a distance apart, at starting of
2,000-16-3 (17) (.f.,^^
■ = 60. 4 m.m.
2 (16) ^ 6.04 cm. (1)
Since we need 12,187 cathode plates we need
12,187-^16 = 761.7 tanks
Which means that we would use 762 tanks. (1)
The arrangement of the tanks in series and groups of series
can be best discussed when we know the voltage drop per
tank, grouping them so as to absorb either 110 or 220 volts
per series in one group.
(2) An anode sheet weighs:
75X50X0.3X7.9X1.1 = 9,776 grams,
of which there is immersed 8,888 grams.
The 17 anodes per tank weigh altogether:
9.776X17 = 166.2 kg.,
and the 16 cathodes, which are one-third as thick:
3.259X16 = 521.1kg.
In the whole plant, at starting, the weights will be:
Anode sheets 166.2X762 = 126,644 kg.
Cathode sheets 52.1x762= 39,700 " (2)
(3) Volume of liquid in tank:
(1— 0.1)X0.6X2 , = 1.08 cubic meters.
ELECTROMETALLURGY OF IRON AND STEEL. 449
Weight of solution per tank:
0.18X1,000X1.1 = 1,188 kg.
Weight of dissolved salts:
Copperas = 118.8 "
Ammonium sulphate = 59.4 "
Weight in the whole plant:
Copperas = go. 5 tons.
Ammonium sulphate = 45.3 " (3)
(4) At starting the surface of the electrodes are 6.04 cm.
apart, and 110 amps, passes through each square meter of
electrode surface; therefore, 110X0.375 = 41.25 amps, pass
from each free side of each anode plate to the corresponding
side of a cathode plate. Neglecting the small cross sectional
area of electrolyte outside the plates, the resistance of each
space would be
20X6.04
75X50 = °-^^^ °^"'''
and the drop of voltage across two electrodes:
0.0322X41.25 = 1.33 volts.
If we take into accotmt the 5 cm. free space at the sides
of each electrode, and allow an equal amount as effective be-
neath, the cross-section of the electrolyte may be taken as
(75 + 5) X (50 4- 10) = 4,800 cm^,
and the resistance between two plates:
20x6.04 4-4.800= 0.0252 ohms,
and the drop 0.0252X41.25 = 1.04 volts.
This value is the more probable one of the two.
At the close of a deposition, neglecting the decreased thick-
ness of the thin anode plates, the working distance is decreased
by one and a half centimeters, and the voltage drop between
plates will then be :
"^^^t^ X41.25 = 0.78 volt.
4,800
450 METALLURGICAL CALCULATIONS.
In both cases the voltage drop -in contacts and conductors
being 0.3 volt, the working voltage per tank will be:
Volts.
At starting 1.34
At end 1.08
Average. 1.21
The voltage needed at the generators can only be calculated
when we assume a plan of grouping the 762 tanks. If we as-
sume 110 volts to be desired at the generators, we could run
102 tanks in one series, which would give 7J . series. If we
used 220 volts at the generators 2 series of 190 cells and 2 of
191 would absorb at starting 255 and 256 volts respectively, but
when in regular rimning, with tanks in all stages of deposition,
205 and 206 volts. This would be a reasonable and practicable
arrangement. In reality, at least one if not two additional
tanks would be slipped into each series, for at least that num-
ber would be out of circuit continuously, being cleaned and
made ready for re-starting.
(5) The amperes per tank would be:
0.75X0.5X2X16X110 = 1,320
And the area of the main conductors in each series:
1,3204-2 = 660 sq. m.m. (5)
(6) The part of the anode sheet immersed weighs 8,888
grams, of which 0.9 is 8,000 grams, and if the anode is an in-
termediate one it is corroded on both sides, and receives, there-
fore, 85 amps, of current. This current dissolves, per second:
0.00001036X28X85 = 0 024657 grams.
And therefore the anode sheet will last
8,000X0.024657 = 324,000 seconds.
= 90 hours. (6)
(7) The weight of deposit on both sides of a cathode plate is
75X50X2X1.5X7.6 = 85,500 grams.
And the time required to deposit this, since it is deposited by
85 amps., is
85,500^0.024657 = 3,467,600 seconds.
= 40 days 3 hours. (8)
ELECTROMETALLURGY OF IRON AND STEEL. 451
(8) Each series requires 1,320 amps, at 205 volts, or
1,320X2054-1,000 = 270.6 kilowatts.
The three series therefore require 812 kilowatts, which will
cost
$25.00X812-^365.25 = $55.68 per day.
An average cost of power per ton of iron refined of
$55.68-^25 = $2.23. (8)
The other items of cost in a well conducted refirery will
about equal this sum, making the total cost of refining about
$4.50 per ton of pure iron. With cheap soft steel used as the
raw material, there is a striking possibility of such a process
being commercially practicable for furnishing one of the raw
materials for producing the finest qualities of steel, the other
raw material being washed pig metal of standard quality. We
commend this possibility to the attention of the makers of fine
steel.
APPENDIX TO PART II.
PROBLEMS FOR PRACTICE.
Problem 81.
A blast furnace will require 7000 cubic feet of cold air sup-
plied per minute, and assuming that it will be provided with
5 tuyeres, and that the pressure in the main will be kept at 8
pounds per square inch, that the back pressure in the furnace
will average 2 pounds per square inch, and that the coefificient
of contraction of the jet, for a conical nozzle, is 0.92 —
Required :
(1) The diameter of each nozzle.
Ans. : 3 inches.
Problem 82.
A blast-furnace charge is composed of the following ingre-
dients, in percentage composition:
Ore Coal Ash of Coal Flux
Fe=0' 71.49 fixed C 90 SiO^ 48.00 CaO 52.83
SiO^ 15.73 ash 10 APO^ 41.20 00== 41.51
APO' 4.21 Fe^O' 8.00 SiO^ 5.66
CaO 5.00 CaO 2.80
MgO 3.57
Per ton of pig iron there is used 1.5 tons of coal. Pig iron
contains 94 per cent. Fe, and 2.8 per cent. Si, 99 per cent, of the
iron in the charge goes into the pig iron. Quantivalent ratio
of the slag, calling both Si and Al acids, 1.882.
Required, per ton of pig iron produced:
(1) The weight of ore charged. Ans. : 1.88 tons
(2) " " " flux " Ans.:0A8tons
(3) " " " slagproduced ^m5.: 0.91 tons
(4) The percentage composition of the slag.
452
APPENDIX. 453
Ans.: SiO=' 36.95 per cent.
APO' 15.56
CaO 38.79 "
MgO 7.40 «
FeO 1.30 «
100.00
Problem 83.
The composition of the waste gases of a blast furnace is by
weight (not by volume)
Nitrogen 55.40
Carbonous oxide 28.00
Carbonic oxide .16.50
Hydrogen .• 0 . 10
From analyses of the charges and products it is known that
for every 100 kg. of pig-iron made 79.52 kg. of oxygen enters
the gas from the solid charges and 15.5 kg. of carbon enters the
gases from other sources than the fixed carbon of the fuel.
The fuel contains 90 per cent, fixed carbon; the pig-iron con-
tains 3 per cent, carbon. In 24 hours there is produced 41,400
kg. of pig-iron; the blowing engine works 23 hours. Assume
blast dry.
Required:
(1) Per 100 kg. of pig iron, the weight of fuel charged.
Ans. : 114.4 kg.
(2) " " " " the carbon burned at the tuyeres.
Ans. : 87.4 kg.
(3) " " " " the carbon otherwise consumed.
Ans. : 12.6 kg.
(4) " " " " the weight of the gases.
Ans : 700 kg.
(5) " " " " the weight of the gases.
Ans. : 504 kg.
(6) The volume of blast received per minute.
Ans. : 117 m^
(7) The dimensions of the blast cylinder.
Ans.: 1=1.5 m, i=1.88 m., 20 r.p.m.
454 METALLURGICAL CALCULATIONS.
Problem 84.
Four varieties of ore are used in a blast-furnace, containing
by analysis:
A B C D
Fe 63.25 60.10 64.35 62.35
SiO' 5.86 4.20 5.30 6.58
•APO^ 1.48 1.98 1.96 1.87
CaO 1.04 0.16 0.90
MgO 0.75 0.09 0.51
P 0.019 0.107 0.019 0.015
The flux, fuel and pig iron contained:
Flux Fuel Pig Iron
SiO^ 6.46 5.64 Fe 95.24
APO' 1.53 3.74 Si i.40
CaO 47.00 0.56
MgO 3.60 0.60
P 0.010 0.020
FeS 1.32
The ores are mixed in the proportions J A, J B, | C, -J D.
A " round " of fuel consists of 12 barrows of coke containing
620 pounds per barrow. 1885 pounds of coke is used in pro-
ducing a " ton " of 2300 pounds of pig-iron. Assume -jV
of the sulphur to go into the slag, and -jV of the phosphorus
to go into the pig-iron.
Required:
(1) A corrected fluxing table, showing for each substance
(100 parts) the weights of ingredients furnished to the slag,
the quantivalence of each element thus contributed, and the net
available acid or basic quantivalence assuming the slag to be a
bi-silicate, counting Si and Al as the acid elements in the slag.
(2) The weight of each ingredient of the burden per round of
coke, Ans.: Ore 11,640 lb., flux 1090, coke 6240.
(3) The weight of slag produced per 2300 lbs. of pig-iron.
(4) The percentage composition of the slag.
(5) The percentage of S and P in the pig-iron.
Problem 85.
The flux used in a blast-furnace and the slag produced by its
use have the following compositions:
APPENDIX. 465
Flux
Slag
CaCO»
80.36
CaO 29.0
MgCO'
15.75
MgO 4.8
SiO'
3.00
810=^ 53.4
H=0
0.89
APO' 9.0
FeO 2.6
Na^'O 1.2
Per 100 kg. of pig iron produced there is used 28.7 kg. of flux,
and there is made 50.15 kg. of slag.
Required:
(1) The quantivalent ratio of the slag.
Ans.: 2.96
(2) How much flux would be needed to produce a uni-silicate
slag. Ans.: 105 kg.
(3) What would be the weight of this slag.
Ans. : 93 kg.
(4) What would be its percentage composition; check the
quantivalent ratio from this composition.
Problem 86.
A blast furnace works on the following materials:
Ore
Coke
Flux
Pig Iron
Gases
Fe^O^ -85
Fixed C
87
CaO 52.50
Fe 94.8
CO 28.41
SiO' 9
SiO^
6
810== 6.25
8i 1.4
CO^ 10.85
APO» 1
APO»
5
CO^ 41.25
C 3.8
N^ 60.78
CaO 2.
8
H^O
2
WO 2.
2
100 kg. of coke and 160 kg. of ore are used per 100 kg. of pig-
iron produced.
Required: (1) The weight of flux charged, per 100 of coke,
to make a slag with the quantivalent ratio:
Quant. Si + Quant. Al ^ , yg
Quant. Ca " ' Ans.: 39.1kg.
(2) The weight of carbon burned at the tuyeres per 100 of
pig-iron. ^w.: 70.1 kg.
(3) The weight of carbon consumed above the tuyeres per 100
of pig-iron. Ans.: 13.1kg.
456 METALLURGICAL CALCULATIONS.
Problem 87.
The blast-furnace of preceding problem has 7 tuyeres, each
5 inches in diameter. Temperature of blast in blast main 819°,
indicated pressure on blast-main gauge 22.08 lbs. per sq. inch,
the back pressure in the furnace in the region of the tuyeres is
3.69 lbs. per sq. inch. The air jet contracts in issuing until the
coefficient of contraction of area of the jet — /i — is 0.9.
Required: (1) The actual temperature of the blast entering
the furnace, assuming it to be cooled only by expansion. As-
sume the mean specific heat of air under the above conditions
as 0.40 (oz. cal. per cubic foot or Cal. per cubic meter).
Ans.: 743°
(2) The volume of blast per minute received by the furnace,
at standard conditions of measurement. Ans.: 48,400 ft'.
Problem 88.
Taking the calculated data of Problems 86 and 87, and com-
bining them, how many tons of pig-iron, at 2240 lbs. per ton,
are produced per day, assuming the blast to be on the furnace
23 hours 30 minutes? Ans.: 700 tons.
Problem 89.
A blast furnace makes 330,000 kg. of pig iron in 24 hours.
The blowing engine runs 1400 minutes per day, 3 cylinders,
double acting, 2 m. diameter by 2 m. stroke, piston rod 0.1 m.
diameter and passing through one end of the cylinder only; 40
strokes per minute. Temperature in engine room 27.3°; assume
air dry.
Charges per 100 kg. of pig iron: Ore 155.0, flux 45.2, coke
84.0. Ore contains 62 per cent, of Fe, as Fe^O*, 5 per cent.
H^O and the rest SiO^ The flux is 3 per cent. SiO^" and the
rest CaCO*. The coke contains 89 per cent, fixed carbon, 1
per cent. H^O and the rest half SiO'' and half APO'. The pig-
iron carries 3.5 carbon, 1 silicon, 94.6 per cent. iron. The waste
gases contain 2.3 volumes of CO to 1 volume of CO^. Assume
all the oxygen in the CO and CO^ to come from the blast, re-
duction of Fe^'O' and SiO^ and the CO^ of the flux.
Required :
(1) The weight of carbon burned at the tuyeres, per 100 kg. of
pig-iron produced. Ans. : 57.8 kg.
APPENDIX. 457
(2) The weight of carbon burned above the tuyeres, in direct
reduction, per 100 of pig-iron. Ans. : 13.5 kg.
(3) The volume of blast, at standard conditions, received by
the furnace per minute. Ans. : 608.5 m'
(4) The same, per ton of pig-iron. Ans. : 2582 m'
(5) The efficiency of the blowing plant, i.e., the ratio of air
received to the piston displacement. Ans.: 88.9 per cent.
Problem 90.
A blast-furnace produces 300 metric tons of pig-iron per day,
and is charged, per 100 kg. of pig-iron, with
156 kg. iron ore
50 " limestone
90 " coke
The percentage compositions of these materials and of the pig-
iron and gases produced are:
Ore
Limestone
Coke
Pig Iron
Gases
Fe^O='
85
CaO 51 . 66
Fixed C 88
Fe 94.00
CO 26.50
SiO^
8
MgO 2.50
SiO^ 8
Si 2.10
CO^ 13.25
APO'
4
SiO^ 2.50
FeS 2
C 3.75
N^.60.25
H=0
3
CO^ 43.34
H^O 2
S 0.10
Required:
(1) The volume of gas, measured dry, at 0° and 760 m.m.
pressure, produced*per 100 kg. of pig-iron made.
Ans. : 379 m'
(2) The volume of blast, assumed dry and at 0° and 760 m.m.,
received per 100 kg. of pig-iron made. Ans. : 288 m*
(3) The weight of carbon consumed by the blast, at the tuy-
eres, per 100 kg. of pig-iron made. Ans. : 66 kg.
(4) The weight and percentage composition of the slag.
Ans. : 49 kg.
Composition: SiO=' 32.6, APO^ 12.4, CaO 49.3, MgO 2.4, FeO 1.2,
CaS 2.4.
(5) The horse-power producible from one-half of the gases,
assuming them used in gas-engines at 25 per cent, net thermo-
mechanical efficiency. Ans. : 7570 h.p.
Problem 91.
A blast-furnace makes 300 metric tons of pig-iron daily,
producing per 100 kg. of pig iron made 464.5 m' of gas of the
458 METALLURGICAL CALCULATIONS.
following composition: CO 24, CO^ 12, N' 64, and using 352.5 m"
of blast, at 700° C. The blowing engines consume 1,000 indi-
cated horse-power, the lift and pumps 150 h.p. The. blast (as-
sumed dry at 0°) is heated to 700° in fire-brick stoves, at an
efficiency of 50% on the heat of combustion of the gas used in
them.
Required:
(1) The proportion of the gas made by the furnace required
to be used by the stoves. Ans. : 46.6 per cent.
(2) The rest of the gas being burned under boilers, and just
raising the power required by the furnace, what is the net
thermo-mechanical efficiency of the boilers and engines together?
Ans. : 3.2 per cent.
(3) If these gases were used in gas engines at an efficiency
of 25 per cent., what surplus power would the blast-furnace
produce above its own requirements? Ans.: 7830 horse-power.
Problem 92.
An Alabama blast furnace uses and produces materials of the
following compositions :
Ore
Dolomite
Coke
Pig Iron
Fe^O' 72.85
CaO
30.24
C 86.00
Fe
93.54
SiO^ 9.00
MgO
20.48
SiO^ 4.Q0
C
3.50
APO' 3.75
Fe^'O'
0.50
APO^ 2.00
Si
2.25
CaO 0 75
APO'
0.50
Fe^O' 2.00
s
0.06
P'0= 0.90
SiO^
2.00
S 1.00
p
0.65
SO^ 0.25
CO^
46.28
WO 5.00
CO^ 0.60
ffO 11.90
Assume: Burden = 2.00 {i.e., ore plus dolomite = twice
coke). All the iron in the charge reduced into the pig iron.
Sulphur not in pig iron goes into slag as CaS.
Phosphorus not in pig iron goes into slag as P^O'.
Slag to contain 34.00 per cent. SiO^
Eighty-nine per cent, of the fixed carbon of the coke will be
consumed at the tuyeres, by dry blast.
Required: A balance sheet of materials entering and leaving
the furnace.
APPENDIX,
Solution
x + y
[N.B. — Let X = ore used, y = dolomite, then
coke, and the balance sheet thus constructed leads easily
to a solution conforming to above conditions.]
Balance Sheet (per 1000 of pig iron).
Charges
Pig Iron
Slag
Gases
Ore
1792
Fe='03
1305.5
Fe
913.9
0
391.6
SiO='
161.3
Si
22.5
Si02
113.1
0
25.7
Ol'O'
67.2
APO'
67.2
CaO
13.4
i
fCa
^CaO
1.5
11.3
0
0.6
p205
16.1
P
6.5
pZQS
1.3
0
8.4
SO'
4.5
S
0.6
S
1.2
C02
10.8
CO^
10.8
H^O
213.2
H^O
213.2
Flux
411
CaO
124.3
1
fCa
[CaO
13.8
105.0
0
5.5
MgO
84.2
MgO
84.2
Fe^O'
2.1
Fe
1.5
O
0.6
AP03
2.1
Al^O'
2.1
SiO^
8.2
SiO^
8.2
CO^
190.2
co^ -
190.2
Coke
1101
C
946.9
G
35.0
c
911.9
SiO^
44.0
SiO^
44.0
Al^O'
22.0
APO'
22.0
Fe^O'
22.0
Fe
15.4
O
6.6
S
11.0
S
11.0
H^O
55.1
H^O
55,1
Blast
4004 '
02
924.
O
924.0
N^
3080.
N
3080.0
.7308.
1000.0
485.9
5824.2
Charges
Slag
Coke
1000
SiO^
34.
0%
CaS
5.7
Ore
1628
APO' 18.
8%
9%
pZQS
0.3
Flux
373
CaO
^a.
100.2
Pig Iron 908 MgO 17.6%
460
METALLURGICAL CALCULATIONS.
Problem 93.
A foundry cupola melts 6,000 kg. of pig iron per hour, using
480 kg. of coke containing 85 per cent, fixed carbon. Average
composition of issuing gases, by volume, N^ 75, CO^ 16, CO 4.
Some Fe, Mn and Si are oxidized, forming the silicate Fe° Mn
SiO*. No carbon is oxidized out of or absorbed by the pig-
iron. Assume the blast dry and neglect the ash of the coke
and corrosion from the lining.
Requirements :
(1) The volume of blast received by the cupola per minute.
Ans. : 69.7 m'
(2) The percentage loss in weight of the pig-iron, assuming
it to lose only Fe, Mn and Si. Ans. : 4.74 per cent.
(3) The proportion of the whole heat generated which is due
to the oxidation of carbon and to the oxidation of the pig-iron.
Ans. : 84.5 and 15.5 per cent.
Problem 94.
Dichmann, in Stahl und Eisen, 1 December, 1905, gives the
following data respecting the Monell open-hearth operation, as
practised by him.
Charge: 1000 kg. limestone, CaO 54.1, SiO^" 1.65, MgO 0.68.
3276 " hematite ore, Fe^O^ 95.0, SiO^ 3.7.
These were put on a basic hearth and heated nearly to melt-
ing. There was then run upon them 20,300 kg. of melted pig-
iron. Starting at 2.30 p.m., the subsequent analyses showed:
P.M.
2.30
3.00
5.40
5.45
6.15
Metal
C Si
4.61 0.84
4.66 0.19
1.47 0.05
P
0.15
0.05
0.03
Mn
2.20
0.45
0.63
(819 kg. more ore added)
0.43 0.05 0.03 0.49
Slag
FeO Fe='0' .Mn P^O^ SiO='
47.88 6.10 15.22 2.36 17.68
10.36 3.23 12.67 2.35 23.00
11.44 2.37 12.04 2.03 22.90
Required:
(1) The weight of iron reduced into the bath during the slag-
forming period — 2.30 to 3.00 p.m.
Ans. : 1219 kg. = 56% of the iron in the ore.
APPENDIX. 461
(2) The weight of iron reduced into the bath during the boil —
3.00 to 5.40 P.M.
Ans. : 705 kg. = 32.4% of the iron in the ore.
(3) The weight of iron reduced from the additional ore added
—5.40 to 6.15 PM
Ans. : 509 kg. = 93% of the iron in the ore added.
(4) The proportion of oxidation produced by the ore and by
the furnace gases in each of these periods.
Ans. : 1st period 100 and 0.
2d " 30 and 70.
3d « 73 and 27.
Problem 95.
A Bessemer converter is charged with 10,000 kg. of pig-iron,
from which is oxidized
Si 2.8 per cent. *
C 3.3 " 1/5 to CO^
Fe 1.12 "
Length of blow 15 minutes, no free oxygen escapes from the
converter. Temperature in engine room 0°, barometer 760
m.m., mean pressure on piston of blowing cylinder during the
stroke 1 kg. per sq. centimeter. No moistiu:e in the air. As-
sume a coefficient of delivery of 0.60.
Required:
(1) The volume of blast received per minute.
Ans. : 196 m»
(2) The dimensions and speed of the blowing cylinder.
Ans. : d 2.4 m., / 2 m., 18 r.p.m.
(3) The horse-power exerted by the blowing cylinder.
Ans. : 727 h.p.
(4) If the speed of the blowing cylinder is kept constant, but
the temperature in the engine room becomes 30° and saturated
with moisture, how much longer will the blow last?
Ans. : 46 seconds.
Problem 96.
A Bessemer converter contains 10,000 kg. of pig-iron, from
which there is oxidized by the blast: Si 1, Fe 2, C as CO^ 0.7,
C as CO 2.8 per cent., while no free oxygen or H^O vapor escapes
462 METALLURGICAL CALCULATIONS.
from the converter. The blowing engines have a piston dis-
placement of 326.5 m' per minute. The air in the engine rooni
is at 27.3°, barometer 741 m.m., air saturated with moisture.
The blow lasts 9 minutes 10 seconds, up to the end of the boil.
Required:
(1) The coefficient of useful delivery of the blowing cylinders
ajid conduits. Ans. : 66 per cent.
(2) The average composition of the gases produced.
Ans.: CO 5.18, 00== 20.73, iP 3.38, N^ 70.80 per cent.
Problem 97.
The gases from a Bessemer converter had the following av-
erage composition during the slag-forming period and the boil,
respectively :
First 3 minutes. Last 2 minutes.
0= '. 0.6 0.1
CO^ 8.4, 4.3
CO 5.2 26.1
W 0.6 0.6
N^" ; 85.2 68.9
Duration of blow 5 minutes. Weight of charge 2500 kg.
Weight of carbon oxidized 3.9 per cent, of the weight of the
charge. Assume the blast constant per minute throughout,
and that the Gomposition of the slag formed from the Fe, Mn
and Si, oxidized is Fe Mn Si^O". Neglect the moisture of the
blast.
Required:
(1) The volume of the blast per minute. Ans.: 168.5 m'
(2) The proportion of the total carbon oxidized passing off
as CO^- Ans. : 30.9 per cent.
(3) The percentage loss in weight of the charge from the
oxidation of its ingredients. Ans.: 9.30 per cent.
Problem 98.
A Bessemer converter is charged with 8000 kg. of pig-iron.
The slag-forming period lasts 6 '20", the boil 4' 10". At the
end of the boil 800 kg. of spiegeleisen is added, and the con-
verter momentarily turned up. The steel ingots obtained weigh
7900 kg. Hydrogen in gases comes from moisture in the blast.
APPENDIX. 463
Temperature of air in engine room 30° C, barometer 760 m.m.
Assume blast constant per minute. Blast cylinder 1.5 m. in-
ternal diameter, 2 m. stroke, double acting, 90 strokes per min-
ute, piston rod through both ends, diameter 0.15 m. Ef-
fective pressure of the blast, 2 . atmospheres.
Analyses :
Pig Iron End of End of Spiegel Steel
\st period 2d period
C 4.00 2.63 0.04 5.00 0.49
Si 2.00 0.26 0.03 3.00 0.10
Mn 1.40 0.42 0.01 15.00 0.78
P 0.05 0.05 0.06 0.12 0.06
S 0.05 0.05 0.06 0.05
Fe 92.50 96.59 99.80 76.88 98.52
Average composition of gases:
In \st period In 2d period Final Slag
CO 7.19 27.45 SiO^ 48.62
CO^ 7.19 2.60 MnO 21.72
0=' 2.60 FeO' 29.66
W 2.81 2.36
N^ 80.21 67.59
Required :
(1) A balance sheet, showing constituents of the bath at each
stage of the blow.
(2) The voltmie of blast per minute, at the engine rooni tern?
perature; the weight of blast per ton of pig-iron treated.
(3) .The relative volumes of gas issuing per minute in each
period.
(4) The proportion of the total carbon burned in the blow
which is burned to CO and CO^.
(5) The proportions of the carbon, silicon and manganese of
the spiegeleisen lost during re-carbonization. Express these
also as percentages of the weight of steel produced
(6) The total weight of the slag ; the weight of silica corroded
from the lining.
(7) The volume efificiency of delivery of the blowing engine.
(8) The net effective horse-power of the blowing cylinder.
484
METALLURGICAL CALCULATIONS.
Solution:
(1)
Balance Sheet
Pig Iron
Loss
End
Loss
End
Spiegel Loss
Steel
1st period
2d period
C
320
120
200
197
3
40
4
39
Si
160
140
20
18
2
24
18
8
Mn
112
80
32
31
1
120
59
62
P
4
4
0
4
1
0
5
S
4
4
0
4
0
0
4
Fe
7400
60
7340
124
7216
615
49
7782
Total
(2)
(3)
(4)
8000
400
7600 370 7230 800 130 7900
Volume received by converter per minute at 30°:
287 m'
Weight per ton of pig-iron treated : 436 kg.
Relative volumes of gases per minute : 1 to 1.20
Proportion of carbon burned to CO : 75.7 per cent.
(5) Loss in per cent, of Spiegel: in per cent, of steel
C
Si
Mn
0.5
2.3
7.4
0.05
0.23
0.75
(6) Weight of slag: 1005 kg. Loss of lining: 112 kg.
(7) Efficiency of delivery = 82.5 per cent.
(8) Net effective horse-power = 900; gross, 1000.
Problem 99.
Ten metric tons (10,000 kg.) of pig-iron is charged into a
basic lined (Thomas) Bessemer converter, and blown 12 min-
utes. The lining is burnt dolomite, composition practically
CaO, MgO. Analyses showed:
Metal at end of blow Slag
Pig Iron
c
3.05
Mn
0.41
Si
0.83
S
0.33
P
1.37
Fe
94.01
C
0.1687
SiO^
5.98
Mn
0.0973
MnO
1.39
Si
S
CaS
pjQS
2.49
10.08
0.0540
P
0.0650
FeO
7.14
Fe
99.6150
Fe^O"
1.20
CaO
67.05
MgO
4.67
APPENDIX. 465
Assume all the Fe, Si and Mn oxidized to be removed in the
first period, all the carbon in the second, \ to CO^ and | to
CO, and all the P and S in the third period, and no free oxygen
to escape from the converter.
Requirements:
(1) The weight of slag produced. AM5.:2975kg.
(2) « « " steel " A«5. :9250 kg.
.(3) " " " lining corroded away during the blow.
Ans. : 333 kg.
(4) " "^ " lime added during the blow.
Ans. : 1858 kg.
(5) The duration of each period, assumed as sharply defined.
Ans.: 2' 27", 6' 50", 2' 43".
Problem 100.
In a basic-lined Bessemer converter, there is oxidized during
the blow:
Silicon 0.84 per cent.
Carbon 3.00 « 1/5 to CO''
Phosphorus ...1.86 "
Manganese 0.55 "
Iron 2.80
Bath weighs at starting 8000 kg., blow lasts 18 minutes, no
free oxygen escapes from the converter. Effective pressure of
the blast 1.5 atmospheres, outside air at 0° and 760 m.m.
Requirements:
(1) The volume of the blast per minute. Ans. : 134 m'
(2) The net horse-power of the blowing engine.
Ans. : 313 h.p.
(3) If the slag has the formula: 3 Ca*P^O». MnFe'Ca'^ Si'O^*,
how much lime must be added during the blow and what is the
weight and percentage composition of the slag?
Ans. : 1075 kg. CaO
1905 kg. slag.
, p^o* 17.89 per cent.
SiO^ 7.56
CaO 56.46 «
MnO 2.98 "
FeO 15.11 «
PART III.
THE METALS OTHER THAN IRON.
(NON-FERROUS METALS.)
467
CHAPTER T.
THE METALLURGY OF COPPER.
Roasting and Smelting Copper Ores.
The facts with regard to the metallurgy of copper may be
found in condensed form, very clearly stated, in Schnabel's
" Handbook of Metallurgy," Vol. I; they may be found dis-
cussed at greater length in Dr. Peters' " Modern Copper Smelt-
ing " and " Principles of Copper Smelting," in Eissler's " Hydro-
metallurgy of Copper," and in Ulke's " Modem Electrolytic
Copper Refining." *In the following presentation it will be pos-
sible to merely enumerate the different processes concerned, the
principles embodied in each and the methods of calculating
quantitatively the nature of the reactions involved.
The chief ore of copper is chalcopyrite, CuFeS^ which con-
tains when pure approximately 30 per cent iron and 35 per cent
each of copper and sulphur. It is most frequently foimd mixed
with silica, SiO^, as gangue, although various other gangue
materials are sometimes present. This mineral is fusible, and
if it were merely melted out of the enclosing rock, in an atmos-
phere which did not act upon it, it would simply lose one-fourth
of its sulphur and melt to a fluid double sulphide of iron and
copper, according to the following reaction:
2CuFeS='= Cu='S.2FeS-FS.
The melted sulphide of copper and iron resulting is called
" matte," and, in fact, any mixture of Cu^'S and FeS in any pro-
portions is called matte, and in practice may contain small
quantities of PbS,, ZnS, BaS, NiS, As, Sb, Te and even of Fe'OS
which is slightly soluble in these fused sulphides. In the major-
ity of cases the matte is practically near enough to a mixture
of Cu^S and FeS to regard it as containing only those two sub-
stances.
*More recently, in Prof. H. O. Hofman's splendid treatise on "The
Metallurgy of Copper."
469
470 METALLURGICAL CALCULATIONS.
Since the first operation in the extraction of copper from its
usual sulphide ores is invariably the concentration to matte,
we will first study the composition of this substance. Assum-
ing matte to consist of Cu^S and FeS in varying proportions,
we may note that if it is all Cu^S it contains 2X63.6 = 127.2
parts of copper to 32 parts of sulphur, or practically 4 copper
to 1 sulphur, or is 80 per cent copper. For every 1.0 per cent
of copper sulphide present there is, therefore, 0.8 per cent of
copper in the matte; and, conversely, for every 1 per cent of
copper in the matte there is 1.25 per cent of copper sulphide in
it. The rest of the matte being iron sulphide, which contains
56 parts iron to 32 sulphur, it can be seen that the per cent
of iron can be calculated for any per cent of copper; in fact,
the per cent of copper fixes both that of iron and sulphur.
Illustration: A matte contains 40 per cent of copper. How
much iron and sulphur does it contain? How much copper
sulphide and iron sulphide?
40 per cent copper=40X5/4 = 50 per cent Cu^S.
100—50 = 50 " FeS.
50 per cent FeS = 50X56 -h88 = 31.8 " Fe.
100— (40 + 31.8)= 28.2 " S.
We may generalize this solution and say that if X represents
the percentage of copper in a matte, that the composition of the
matte is as follows:
Per cent of copper = X
Per cent of Cu^S = 1.25 X
Per cent of FeS = 100 — 1.25 X
Per cent of Fe= (100 - 1.25 X) 56/88 = 63.6 - 0.795 X
Per cent of S = 100 - X - Fe =36.4 — 0.205 X
Similarly, if Y represents the percentage of iron in a matte
its composition is:
Per cent of Fe = Y
Per cent of FeS = 1.57 Y
Per cent of Cu^'S = 100 — 1.57 Y
Per cent of Cu = 80 — 1.26 Y
Per cent of S = 20 + 0.26 Y
THE METALLURGY OF COPPER. 471
Finally, if Z represents the percentage of sulphur in a matte
its composition is:
Per cent of S =2
Per cent of Fe = 3.39 z _ 77 g
Per cent of Cu = I77 g _ 4 gg z
Per cent of FeS = 6.1 1 Z - 122.3
Per cent of Cu^S = 222.3 - 6.11 Z
The following tables may then be drawn up, and will be
found useful for reference:
Percentages
',
Cu.
Fe.
S.
Cu^S.
FeS.
80
0.0
20.0
100.0
0.0
75
4.0
21.0
93.8
6.2
70
8.0
22.0
87.5
12.5
65
12.0
23.0
81.3
18.7
■60
15.9
24.1
75.0
25.0
55
19.9
25.1
68.8
31.2
50
23.9
26.1
62.5
37.5
45
27.9
27.1
56.3
43.7
40
31.8
28.2
50.0
50.0
35
35.8
29.2
43.8
56.2
80
39.8
30.2
37.5
62.5
25
43.8
31.2
31.3
68.7
20
47.7
32.3
25.0
75.0
15
51.7
33.3
18.8
81.2
10
55.7
34.3
12.5
87.5
5
59.7
35.3
6.3
93.7
0
63.6
36.4
0.0
100.0
For regular use in a smelting works it is easy to calculate
a table similar to the above for every one per cent of copper,
and keep it under a glass cover for constant reference. Or, if
preferred, a large piece of cross-section paper can be taken,
and a diagram prepared with the percentages of copper as
abscissas, from 0 to 80, and with ordinates running up to 100,
A line drawn from an ordinate 36.4 at abscissa 0 to ordinate
' 20 -on abscissa 80, will represent the sulphur content ; a line
from ordinate 63.6 on abscissa 0 to ordinate 0 on abscissa 80,
472 METALLURGICAL CALCULATIONS.
will represent the iron content; a line from 0 to ordinate 100
on abscissa 80 will represent the Cu^S content; and a line from
ordinate 100 on abscissa 0 to ordinate 0 on abscissa 80 will rep-
resent FeS content.
It has been already shown that if a chalcopyrite ore were
smelted down, as in a shaft furnace, to a matte, that about a
35 per cent matte is as rich as could be made by simple fusion.
In fact, a much poorer matte would usually result, because
chalcopyrite is often accompanied by pyrite, FeS^, which on
simple heating becomes FeS, and thus dilutes the matte still
further.
Illustration: A chalcopyrite copper ore contains 30 per cent
by weight of chalcopyrite and 25 per cent of iron pyrite. What
grade of matte would result from a simple fusion of this ore,
without roasting, in a reducing atmosphere?
Thirty per cent CuFeS^ contains:
30 X 160 -=- 368 = 13 per cent Cu^^S. .
30X176^368 = 14 " FeS.
25 per cent of FeS^* will produce
25 X 88:^120 = 18 " FeS.
Total matte = 45 " of the ore.
Composition of matte:
Cu^S = 29 per cent = 23 per cent copper.
FeS = 71
It is thus seen that the presence of iron pyrites in the ore
tends to lower the grade of the matte produced.
The essential principle of the metallurgy of copper is now
before us. It is this fact: that if some of the sulphur in the
ore be first removed by roasting, and this operation be followed
by smelting, the copper will first take enough of the sulphur
to form Cu^S, and then what sulphur is left over will form FeS.
The amount of FeS which . will accompany the Cu'S into the
matte is entirely a question of how much sulphur is left to com-
bine with iron after all the copper has been satisfied, and this
amount of sulphur can be exactly controlled by the preliminary
roasting.
Illustration: Taking the ore mentioned in the previous illus-"
tration, containing 30 per cent of chalcopyrite (= 10.4 per cent
THE METALLURGY OF COPPER. 473
of copper) and 25 per cent- of iron pyrites, what per cent of
sulphur does it contain, and what would be the quaUty of 'the
matte produced by fusion in a reducing atmosphere if the sulphur
were previously roasted down to J or i or J of its original
amount?
Sulphur in 30 per cent chalcopyrite :
30 X 128 -f- 368 = 10.5 percent.
Sulphur in 25 per cent pyrites:
25 X 64^120 = 13.3 "
Total = 23.8 "
If the sulphur were reduced to J, i or ^ its original amount
there would remain, out of the 23.8 per cent, either 11.9, 5.95
or 2.97 parts of sulphur per 100 of original ore; i.e., per 10.4
parts of the copper. The 10.4 of copper requires 2.6 of
sulphur to form Cu^S, leaving the following amounts of sulphur
in excess to form FeS:
Sulphur left in ore 11.90 5.95 2.97
Sulphur needed for Cu^S 2.60 2.60 2 60
Sulphur left to form FeS 9.30 3.35 0.37
FeS which will be formed 25.61 9.21 1.02
Cu^S formed 13.00 13.00 13.00
Matte formed 38,61 22.21 14.02
Per cent of Cu in matte 27.0 46.8 74.2
It is thus evident that the degree of the previous roasting
determines absolutely the quality or grade of the matte formed
on the subsequent smelting.
It can be readily seen that since partial roasting virtually
oxidizes a portion or fraction of the ore, that it is the practical
equivalent of roasting if we can get oxide ore to mix with sulphide
ore. The raw oxide ore acts exactly as so much roasted ore,
and thus enriches the matte on subsequent smelting.
Illustration: What proportion of a cuprite (copper oxide)
ore, free from sulphur and containing 28 per cent of copper, can
be mixed with the chalcopyrite ore of the previous illustrations,
to produce a matte on subsequent reducing smelting containing
60 per cent of copper?
474 METALLURGICAL CALCULATIONS.
One hundred parts of the chalcopyrite ore contain 10.4 parts
Cu; 20.8 parts Fe, 23.8 parts S.
The sulphur, however, is not all available for making matte,
because one-fourth of the sulphur in chalcopyrite and one-half
that in pyrite is given off by simple heating. There would be
lost, therefore, and not available for matte, sulphur as follows:
S in chalcopyrite 10.5 Xi = 2.6 not available.
Sin pyrite 13.3 Xi = 6.7 "
9.3 "
Available sulphur =23.8 — 9.3 = 14.5 parts per 100 of ore.
In a 50 per cent matte there is (see previous table) 26.1 per
cent of sulphur; 14.5 parts of sulphur available for matte is there-
fore capable of producing 14.5-^0 261 = 55.5 parts of 50 per
cent matte, which would contain 27.75 parts of copper. But
there is only 10.4 of copper in the chalcopyrite ore, leaving there-
fore 16.35 of copper to be supplied by the cuprite ore, and there-
fore capable of taking up 16.35-^0.28 = 58 parts of the oxide
ore per 100 of the sulphide ore. A mixture of 100 of the sulph-
ide ore and 58 parts of the oxide ore would therefore smelt
down in a reducing atmosphere to a 50 per cent matte.
We have carefully specified a reducing atmosphere as the con-
dition for smelting down to produce the calculated results,
because under these conditions the oxygen of the roasted
ore or of the raw oxide ore added cannot combine with the sul-
phur of the ore and go off as S0^ but is taken up by the carbon
or CO in the smelting furnace. In smelting in a shaft furnace
with carbon as fuel these conditions exist, and the sulphur in
the charge can be counted on as practically all forming matte.
If the smelting is done in a reverberatory furnace, with a neutral
atmosphere, much sulphur will be removed from the charge by
the oxygen of the roasted ore or of the raw ore added, and a
richer matte results. The reactions of this elimination of sul-
phur are mostly
Cu2S-)-2CuO = 4CU + S02
Cu2S-|-2Cu20 = 6CuH-S0^
On an average, a matte 10 per cent richer in copper will be
obtained from a given roasted ore by smelting it down in a
reverberatory furnace, where these reactions between sulphide
THE METALLURGY OF COPPER.
475
and oxide. of copper, can occur, than in shaft furnace smelting
with coke, in which the oxygen of the charge is taken up by
the carbon.
There is a third variety of smelting which is in reality a
combined roasting and smelting-down operation. We refer to
what is called pyritic smelting, where a sulphide ore is roasted
in a blast of air so rapidly that the heat generated melts down
the charge and produces a concentrated matte. This operation
is done on pure sulphides, however, without preliminary roast-
ing, and will be subsequently discussed by itself.
The Roasting Operation.
When an ore is roasted it loses some sulphur and takes up
oxygen. The roasted ore will therefore have a different weight
from the unroasted ore. Thus:
100 parts of Cu^S become
100 parts of Cu^S become
100 parts of Cu^S become
100 parts of FeS^ become
100 parts of FeS^ become
100 parts of FeS^ become
90.0 parts of Cu^O.
100.0
"
CuO.
100.0
a
CuSO*
50.0
It
CuO.
64.5
u
Fe'O^
66.7
It
Fe^O»
126.7
11
FeSO<
Problem 101.
A pyritous copper ore from Ely, Vt. (see Peters' " Modem
Copper Smelting," p. 133), contained before and after roasting,
in percentages:
Before. After.
Sulphur...., 32.6 7.4
Copper.... 8.2 9.1
The condition of the copper in the roasted sample was de-
termined as
Per Cent
Copper as CuSO* 1-3
Copper as CuO 2.1
Copper as Cu^S 5.7
Required:
(1) The weight of roasted ore per 100 of raw ore.
(2) The proportion of the sulphur removed by roasting.
476 METALLURGICAL CALCULATIONS.
(3) The grade of matte which would be formed by smelting
down in a reducing atmosphere in a shaft furnace.
(4) The grade of matte which would be formed by smelting
down in a neutral atmosphere in a reverberatory furnace.
Solution :
(1) Since no copper is lost, there will be produced per 100
of raw ore:
8 2
100 Xq^ = 90.1 of roasted ore. (1)
Parts.
(2) Sulphur in 100 of raw ore =32.6
Sulphur in 90.1 of roasted ore = 90.1X0.074 = 6.7
Sulphur eliminated = 25 . 9
(2)
Proportion removed:
M = «-^^^ = ■'^•* p^^
cent.
(3) Per 100 of roasted ore we have:
Kg.
Sulphur to form Cu^S = 9.1X0.25
= 2.3
Sulphur to form FeS = 7.4 - 2.3
= 5.1
FeS formed = 5.1X88/32
= 14.0
Cu^'S
= 11.4
Matte formed = 25.4
9 1
Per cent of copper = ^i^ = 0.358 = 35.8 per cent. (3)
(4) In
CuSOHCu='S = 3CU + 2S02
every 1 part of copper as CuSO^ reduces two parts of copper as
Cu^S, and causes the eUmination as gas of 1 part of sulphur.
In
2CUO + CU2S = 4CU + S02
1 part of copper as CuO reduces 1 part of copper as Cu'S, and
THE METALLURGY OF COPPER. 477
causes the elimination of one-fourth its weight of sulphur.
Therefore, the sulphur eliminated by the reactions given is:
Parts.
By reaction of sulphate on sulphide 1.3X1 = 1.3
By reaction of oxide on sulphide 2.1 Xi =0.5
Total = 1.8
Sulphur remaining to form matte = 7.4 — 1.8 = 5.'6
Sulphur needed to form Cu^S = 2.3
Sulphur left to form FeS = 5.6 — 2.3 =3.3
FeS formed = 3.3X88/32 = 9.1
Cu^S formed =11.4
Matte formed = 20 . 5
9 1
Per cent of copper = ' = 0.444 = 44.4 per cent.
(4)
In the roasting of ores a large amount of heat is set free,
since both the sulphur and the metal are usually oxidized.
Problem 102.
In the ore of Problem 101, taking the data given for its
weight before and after roasting, how much heat was generated
in its roasting per kilogram of ore treated.
Solution: Starting with 100 parts of original ore it contained
32.6 parts of sulphur and 8-2 of copper. The latter corresponds
to 2.05 of sulphur as Cu^S, which would mean 8.2 of sulphur
altogether as chalcopyrite, leaving 24.4 of sulphur present in
the ore as iron pyrites.
In 100 parts of roasted ore we have 5.7 parts of copper as
Cu^S, therefore unchanged, 2.1 parts changed to CuO and 1.3
parts changed to CuSO*. These contain sulphur as follows:
Parts.
Sulphur for 5.7 Cu as Cu^S = 1.4
Sulphur for 1.3 Cu as CuSO* = 0.7
Sulphur for copper compounds = 2.1
Sulphur for FeS = 7.4 - 2. 1 =5.3
FeS = 5.3X88/32 =14.6
478 METALLURGICAL CALCULATIONS.
Per 90.1 of roasted ore there is present:
Cu as CuSO<
=
1.17
Cu as Cu^'S
=
5.14
Cu as CuO
=
1.89
Fe asFeS = 9.3 X. 901
=
8.38
Sulphur
=
6.67
We therefore have:
Heat to Decompose Raw Ore.
Calories.
8.20CuasCu2S = 8.20X160
=
1,312
18.85 Fe as sulphide = 18.85X429
=
8,087
Sum =
9,399
Heat of Formation of Products.
S.HCuasCu^S = 5.14 X 160 =
822
1.17 Cu as CuSO^ = 1.17X2,857 =
3,343
1.89 Cu as CuO = 1.89 X 593.=
1,121
25.93 S to SO^ =25.93X2,164 =
56,112
9.55 Fe to Fe'O^ = 9.55X1,612 =
15,395
9.30FetoFeS = 9.30 X 429 =
3,990
Sum =
80,783
Net heat evolved = 80,783 - 9,399 = 71,384 Calories.
Per 1 kg. of ore = 713
In the roasting of copper ore the heat generated by the oxida-
tion of the ore is an important item in the heat required by the
furnace. In the above case, for instance, each kilogram of ore
developed about one-tenth as much heat as a kilogram of
coke, and with proper arrangements for conserving heat, such
as thick walls and compact shape of furnace, combined with
regeneration of heat from the cooling ore, there is no reason
why such an ore should not be made to roast itself. Such a
scheme, when practicable, is highly economical in districts
where fuel is dear. A long-bedded calciner, with its single
hearth and large radiating surface, cannot meet these condi-
tions, but furnaces having superposed hearths and arrange-
ments for discharging cold ore, cooled by the incoming air,
such as the Spence and MacDougal furnaces, can work on
many ores without using other fuel.
THE METALLURGY OF COPPER. 479
Problem 103.
An improved Spence' furnace used at Butte, Mont., calcines
90,000 pounds of concentrates in 24 hours, of the following
composition :
Per CenU
Copper 9.8
Iron 33.8
Silica 13.3
Sulphur 41.2
Four-fifths of the sulphur is removed, and the roasted ore
is practically discharged cold; 1,500 pounds of slack coal, cal-
orific power 6,500; is used per day. The furnace gases contain:
Per Cent.
CO^ 0.6
S0= 7.2
ffO 0.6
N^ 81.3
O^ 10.3
100.0
and escape into the chimney at 200° C. Ore charged and dis-
charged cold. Furnace has an outside radiating surface of
5,000 square feet.
Required:
(1) The heat balance sheet of the furnace.
(2) The proportion of the heat generated by the roasting of
the ore and by the combustion of fuel.
(3) The heat radiated and conducted to the air per square
foot of outside surface per minute.
Solution :
(1) Per 100 of ore used the unroasted and roasted ore will
contain, respectively:
Copper 9.8 9.8
Iron 33.8 33.8
Silica .13.3 13.3
Sulphur .41.2 8.2
Assuming that the copper remains in the roasted ore, two-
thirds as Cu^S, one-quarter as CuO and one-tenth as CuSO*,
while theiron takes the rest of the sulphur to form FeS, the ex-
480 METALLURGICAL CALCULATIONS.
cess of iron forming half Pe^'O^ and half Fe^O*. We have the
composition of the roasted ore as :
Cu^S 8.2
CuO 3.1
CuSO< 2.0
FeS 17.0
Fe^O^ 16.4
Fe'O* 15.9
SiO^., 13.3
75.9
The copper and iron existed in the unroasted ore virtually
as Cu^S, FeS and FeS^ and therefore the 8.2 of Cu^S and the
17.0 of FeS in the roasted ore can be considered as unchanged.
The actual change in the roasting has been the formation of
3.1 CuO and 2.0 CuSO^ from Cu^S, and the formation of 16.4
Fe^O' and 15.9 Fe'O* from FeS. If we, therefore, subtract
from the heat of formation of these amounts of CuO, CuSO*,
Fe^O' and Fe'OS and the SO^ formed, the heat required to
resolve the Cu^S, FeS and FeS^ into their constituents, we
shall get the net heat evolution in the roasting process. The
heats of formation involved are:
Molecular Heat.
(Cu2, S) = 20,300 Calories = 127 Cal. per kg. of product.
(Cu, 0) = 37,700 " = 471 "
(Cu, S, 0*) = 181,700 " = 1,136 "
(Fe, S) = 24,000 " = 273 "
(Fe^ 0=) = 195,600 " = 1,223 "
(Fe^ 0<) = 270,800 " =1,167 "
(S, O^) = 69,260 " = 1,082 "
We then have the heat evolved and absorbed as follows:
Evolved. Calories.
Formation of CuO 3.1 X 471 = 1,460
Formation of CuSO^ 2.0X1,136 = 2,272
Formation of Fe^O' 16.4X1,223 = 20,057
Formation of Fe'O* 15.9X1,167 = 18,555
Formation of SO^ 66.0X1,082 = 71,412
Total = 113,756
THE METALLURGY OF COPPER. 481
Absorbed.
Decomposition of Cu^S 4.1 X 127 = 521
Decomposition of FeS 36.1 X 273 = 9,855
Total = 10,376
Net heat evolution per 100 of concentrates:
113,756 - 10,376 = 103,380 Calories.
The gases contain 33 pounds of sulphur for every 100 of ore
roasted, and the analysis shows there is 0.072 cubic foot of
SO^ in each cubic foot of chimney gas. This weighs 0.072 X
2.88 = 0.2074 ounces, and contains just half its weight =
0.1037 ounces = 0.00674 pounds of sulphur. There is therefore
produced, per 100 pounds of ore, 33 h-0. 00674 = 4,896 cubic
feet of chimney gas.
This contains, from its analysis:
00=" 29 cub. ft.
S02 353 " "
WO 29 " "
0== 504 " "
W 3,981 " "
and at 200° C. carries heat up the chimney as follows:
S« (0 - 200°)
00=* 29X0.414 = 12 oz. cal. per 1°
SO^ 353X0.420 = 148
WO 29X0.370 =11 " "
2! I 4,485X0.308 = 1,381
N^ )
Sum = 1,552
= 97 lb. Cal. per 1°.
= 19,400 lb. Cal. per 200°.
The fuel evolves 6,500X1,500 = 9,750,000 pound Calories
per day = 10,830 pound Calories per 100 of ore roasted.
(1) Balance Sheet per 100 of Ore Roasted.
Lb. Calories.
Evolved by the fuel 10,830
Evolved by the roasting operation 103,380
Sum 114,210
482 METALLURGICAL CALCULATIONS.
Loss in chimney gases 19,400
Loss by radiation and conduction 94,810
Sum 114,210
(2) The proportion of the total heat generated by the ore
itself is:
i^= 0.90 = 90 per cent.
by the fuel 10
The heat lost by radiation and conduction per day is:
Lb. Calories.
94,810X900 = 85,329,000
Per minute = 59,000
Loss per square foot of surface per
. ^ 59,000 _ - ,„.
^^^^*^ =5:000= "-^ (^)
Problem 104.
The Evans-Klepetko cylindrical roaster used at Butte, Mont.,
is 19 feet high, 18 feet in diameter, and roasts 80,000 pounds
of concentrates daily from 35 per cent sulphur down to 7 per
cent (Dr. Peters). The evolution of heat in the roasting opera-
tion is sufficient to supply all the heat needed. The stirrer
arms are cooled by water circulation, 100 pounds of water
being used per minute, and raised in temperature 50° C. As-
sume the evolution of heat to be 90 per cent as great as in the
roasting of ore in Problem 103, and the chimney loss to be
correspondingly smaller.
Required:
(1) The heat balance sheet per 100 of ore roasted.
(2) The I9SS by radiation and conduction per square foot of
outside surface of the furnace.
Solution:
(1) Heat Evolved.
Lb. Calories.
90 per cent of 103,380 = 93.040
THE METALLURGY OF COPPER. 483
Heat Distribution.
Heat in chimney gases = 17,460
Heat in cooling water = 9 qoO
Heat lost by radiation and conduction = 66,580
Sum = 93,040 (1)
(2) The outside surface consists of the top and bottom and
sides. Their area is as follows:
Sq. Feet.
Sides 18X3.14X19 = 1,074
Bottom 18X18X0.78 = 253
Top = 253
Total = 1,580
Loss by radiation and conduction Lb. Calories
Per 100 of ore = 66,580
Per day ^ = 53,264,000
Per minute . = 37,000
Per square foot surface per minute
37,000
1,580
23.4
(2)
It is interesting to note, as comparing this compact cylindri-
cal furnace with the rectangular furnace of Problem 103, that
although no fuel is used and cooling water is used, and radia-
tion losses per square foot are greater because of thinner walls,
yet the much smaller radiating surface more than counter-
balanced these considerations, and permitted the furnace to
run more economically.
Thick walls and minimum radiating surface are the requisites
for economy of fuel in general, and the sine qua non for roasting
copper sulphide ores by their own self-generated heat of oxida-
tion.
Pyritic Smelting.
This method of smelting is sometimes called pyrite smelting,
because usually practiced on ores rich in pyrite; it is, however,
just as applicable to ores rich in pyrrhotite (magnetic pyrites)
or chalcopyrite (copper pyrites) , and therefore the more general
term pyritic smelting is really more proper and descriptive of
the range of process.
484 METALLURGICAL CALCULATIONS.
For a full description of pyritic smelting we would refer the
reader to the volume " Pyrite Smelting,"' a symposium of in-
formation contributed by nearly forty metallurgists, and edited
by T. A. Rickard, or to Sticht's monograph, " Ueber das Wesen
les Pyrit Schmelzverfahrens,"^ cTr to Dr. Peters' " Principles
of Copper Smelting,"' which contains a 125-page chapter upon it.
A brief statement of the principles of the process is as follows:
Given an ore containing considerable silica in the free state
and a good percentage of pyrite, pyrrhotite or chalcopyrite,
it is possible to smelt it Aown. in a shaft furnace to a ferrous
silicate slag and a matte by means of cold blast and without
using any carbonaceous fuel. Whether fuel is cheap or dear the
possibility of dispensing with it when smelting down certain
sulphide ores is highly important ; yet it is only within a very few
years that the principles involved have been well enough under-
stood to make the process a recognized success. As far as the
modus operandi is concerned, the ore is cliarged into a shaft fur-
nace, water-jacketed at the boshes and smelting zone, with enough
flux to furnish 10 to 20 per cent of CaO or similar alkaline earth
base to the slag, and a high pressure of blast is employed, such
as would correspond to a high rate of driving in ordinary smelt-
ing (up to 3^ 'pounds pressure per square inch is used). The
matte and slag either collect in the well of the furnace or run
continuously into an external settler, where they separate as
in ordinary smelting. The heat necessary to run the furnace
is all supplied by the oxidation of sulphur and iron in the fur-
nace, the former escaping as SO^ in the gases and the latter as
silicate of FeO in the slag. The process may be regarded as
a very quick partial roasting of the ore, accompanied by sim-
ultaneous formation of melted slag and matte, because of the
high temperature generated by the oxidation itself.
There is hardly any process in the whole of metallurgy which
invites so strongly to quantitative calculations, such as we
are endeavoring to encourage in this treatise; and there are
fewer processes which present such a lack of data upon which
to base the calculations. Not only are the physical and chemi-
cal (thermochemical) data scarce, but the ordinary industrial
1 Engineering and Mining Journal, New York, 1905.
2 Wilhelm Knapp, Halle, 1906. Reprinted from " Metallurgie."
3 Hill Publishing Company, New York, 1907.
THE METALLURGY OF COPPER. 485
data, careful details of the running of the furnaces, weights
and compositions of charges and products, analyses and tem-
perature of gases, are very largely lacking — most that we
have of value are those furnished as recently as 1906 by Robert
Sticht, director of the Mount Lyell furnaces in Tasmania, and
these apply only to his particular furnaces and operations.
Fundamental Principles.
Taking the simplest case, and supposing FeS^ and SiO^ to be
charged into a pyritic smelting furnace, we know without a
doubt that the FeS^ becomes approximately FeS at a red heat,
and the sulphur thus evolved is driven off at a part of the
furnace where there is no free oxygen, so remains in the furnace
gases as sulphur vapor. At a temperature of 1400° Sticht
has shown experimentally that the FeS becomes something
like Fe^S', and it is known that the sulphur in the matte is
often less than can correspond to FeS. But the actual tem-
perature of combustion attained at the moment of oxidation in
pyritic smelting is higher even than 1,400°, and it is almost
certain that as FeS oxidizes, it is heated so intensely that it
really goes through two stages, first decomposing to Fe^S and
then oxidizing to FeO:
2FeS = Fe^S-FS
Fe^S + SiO^ + 20^ = SO^ + 2FeO . SiO^
or, putting it all together:
2FeS-hSiOH20^ = S + SO^-h2FeO. SiO^
If we cast up the thermochemistry of this reaction we find
a very considerable evolution of heat, easily comparable with
the heat of oxidation of carbon, and accounting for the run-
ning of the furnace. The thermochemical analysis of the above
reaction is :
Absorbed. Calories.
Decomposition of 2FeS 48,000
Evolved.
Formation of 2FeO = 131,400
80^ = 69,260
2FeO. SiO^* = 8,900
209,560
Excess of heat evolved = 161,560
486 METALLURGICAL CALCULATIONS.
The FeS and SiO^ come into the zone of oxidation already
heated to a high temperature by their contact with the rising
hot gases. They come to this zone, or focus, heated to at least
1,000°, and perhaps hotter, before they begin themselves to
oxidize. The air comes in cool, and we will not credit it with any
sensible heat at the moment oxidation begins.
Theoretical Temperature at the Focus.
With these data, and a few reasonable assumptions, we can
calculate how hot the focus will become at its hottest point.
This calculation is similar to that for the calorific intensity of
combustion of a fuel, or the theoretical temperature before the
tuyeres of a blast furnace. Making the calculation for the
quantities represented by the equation, and assuming
Heat in FeS at 1000° melted 200 Calories
" " SiO=' at 1000° soUd 260
" " Slag (-|- weight of FeO): 0.27 (t- 1100) +300
" " Matte (i weight of slag) : 0.14 (t-1000) +200
we can calculate the theoretical temperature t, to which the
slag, matte and gases will be raised by the heat at hand, which
latter is the sensible heat in FeS and SiO^ at, say, 1,000°, plus
the heat evolved in the reaction, plus sensible heat in matte at
1000°:
Calories.
Heat in 2FeS at 1000°: 176X200 = 35,200
" " SiO^ at 1000°: 60X260 = 15,600
" of the reaction =161,560
in matte: 80 X 200 = 16,000
Total heat available =228,360
Calorific capacity of the products:
80== = 22.22 (0.36t + 0,0003t=')
N=' = 170 (0.303t + 0.000027t2)
Slag = 240 [300+ (t-llOO) 0.27]
Matte = 80 [200 + (t-1 000) 0.14]
Sum = 0.0113t2+135.5t + 5,520 =228,360
Whence t = 1465°C.
THE METALLURGY OF COPPER. 487
The result of our calculation is to show that a temperature
sufficient to run the furnace is theoretically obtainable if the
weight of slag made, carrying a given weight of FeO, is not too
great. Supposing the silica and other slag-forming ingredients
are so heavy in comparison to the amount of FeO formed by
oxidation that the slag contains only 50 per cent of such FeO,
then the weight of slag above would be 288 instead of 240, and
the temperature attained only 1,365° instead of 1,465°. This
reduction, however, is getting perilously near the temperature
necessary to keep the furnace in operation. The formation
temperatures of copper blast-furnace slags are 1,100° to 1,200°,
and an over-heating of at least 50° is necessary, for practical
operation of the furnace. Suppose we put 1,200° as the mini-
mum theoretical temperature which will run the furnace, then
the maximum weight of slag can be calculated in the above
solution for t, and thence the mininium percentage of FeO.
Let X be the maximum weight of slag, for t a minimum
of 1.200°. Then we have:
Calories.
Heat in SO^ at 1200°
= 19,248
" " N^ at 1200°
= 68,422
" " Slag at 1200°
330X
" " Matte at 1200°
= 18,240
Whence
105,940 -f- 330X = 228,360
and
X= 371kg,
Since this maximum weight of slag must contain the 2FeO =
144 kg. of FeO, the minimum percentage of FeO produced by
oxidation and going into slag in pure pyritic smelting, must be
144-7-371 = 0.39 = 39 per cent.
The margin for working pyritic smelting is ordinarily so
small that variations of the temperature of the blast and its
humidity must exercise a large influence in the running of the
furnace. This coincides with experience as far as the heating
of the blast is concerned. At La Lustre Smelter, Santa Maria
del Oro, Mr. Koch says that " a warm blast of 200° C. is a
sine qua non with, us; it spelled success ; cold blast meant failure."
This is, however, an extreme position ; many other metallurgists
working other ores have run entirely with cold blast, but there
488 METALLURGICAL CALCULATIONS.
is no doubt that smelting is easier, faster and more regular
when using warm blast. No one has yet tried drying the blast,
but there can be no doubt that under some circumstances this
would contribute greatly to the regularity of running and would
increase the theoretical temperature obtainable at the focus
or smelting zone.
Use of Auxiliary Coke.
When the amount of slag-forming material in the charges
is high, pyritic smelting using cold, ordinary blast may become
impossible for lack of sufficient calories to melt the slag and
matte. In this case, which often occurs, some coke may be
used, the combustion of which to CO^ at the focus increases
materially the heat available and the temperature. If the
temperature desired is, let us say, 1,400°, carbon at 1,000° can
bum to CO^, yielding CO^ and N^ as products, and give a large
surplus of heat to the charge at this temperature.
One kilogram of coke, containing 0.9 kg. of carbon, will
form 3.3 kg. of CO^ (1.67 cubic meters) and 6.35 cubic meters
of N^. The heat generated is 7,290, and adding in the heat in
hot carbon at 1,000° (342) we have 7,632 Calories generated.
But in the products at 1,400" we have:
Calories.
COU.67 [0.37 + 0.00022 (1400)] X 1400 =1582
N^ 6.35 [0.303 + 0.000027 (1400)] X 1400 =3038
4620
Surplus, to help the fusion
7632 - 4620 = 3012 Calories.
Each kilogram of coke used will therefore melt down, at 1400°,
3012 -H 330 = 9 kg.
of slag, and thus relieves the situation materially. At Mount
Lyell, 0.5, 1.0 and 1.5 per cent of coke (reckoned on the charge)
is found necessary, according to the nature of the ore worked.
From this we have increasing quantities of coke used up to
several per cent, according to the exigencies of the case, and
we pass by insensible gradations through partial pyritic smelt-
ing to ordinary smelting. As the carbon is increased the sul-
phides get less and less of the oxygen blown in, and therefore
the degree of oxidation of sulphides is lower and the concentra-
THE METALLURGY OF COPPER. 489
tion poorer. The best concentration, using unroasted sulphide
ore, is obtained by pure pyritic smelting, if there are enough
sulphides present to generate the requisite heat and temperature.
Rate of Smelting.
In all that precedes it has been assumed that the furnace was
driven fast enough. Other things being equal, the harder a
furnace is blown the more material can be put through it, and
the smaller the heat losses by radiation and conduction when
expressed per unit of charge treated or of product obtained.
It is, therefore, possible to increase the temperature in the
focus simply by harder blowing, and thus to decrease the amount ■
of auxiliary coke needed. A similar effect is produced by height-
ening the umace shaft, since this incrieases the regeneration of
heat by the charges, they coming to the hot zone more intensely
preheated by the ascending gases. The present tendency in
pyritic smelting is undoubtedly to increase the rate of driving,
heighten the furnace, and thus decrease the auxiliary coke
needed and dispense with hot blast, which is somewhat of a
complication.
The possibility of smelting in any manner depends on being
able to generate the theoretical temperature necessary for run-
ning the furnace, and then keeping the rate of smelting per
minute per unit area of the smelting zone as high as prac-
ticable. This achieves two practical results, viz. : makes the
actual temperature of melted-down slag and matte approximate
closer to the theoretical temperature of the focus, and gets the
largest tonnage through the furnace. As we gradually increase
the smelting rate we increase the temperature of the focus,
because of decreased radiation losses; but, on the other hand,
we tend to decrease the relative amount of oxidation, because
of the decreased time that the charge is subjected to oxidation;
there must be a certain rate of driving which will attain the
maximum of oxidation, i.e., of concentration, and past which
increased tonnage is put through at the cost of decreased con-
centration.
Problem 105.
W. H. Freeland {Engineering and Mining Journal, May 2,
1903), at Isabella, Tenn., smelted Ducktown pyrrhotite ore in
a water-jacketed Herreshoff furnace, having a cross-sectional
490 METALLURGICAL CALCULATIONS.
area at the tuyeres of 21.7 square feet. The analyses of the
materials used and the products are as follows:
Charges.
Ore. Quartz. Slag. Coke.
Cu 2.744 .... 0.73
Fe 36.519 1.45 39.20 2.30
S 24.848 0.32 1.75 1.58
SiO^ 18.548 96.79- 30.90 8.41
CaO 7.294 0.23 8.51
MgO 2.672 .... 2.71
Zn : 2.556 .... 2.88
APO^ 0.911 0.32 1.90 3.56
Mn 0.770 .... 0.85
O ■ .... 0.38 11.37 1.00
C -. 83.86
CO^ 3.138
Loss 0.39
Products.
Matte. Flue Dust. Slag.
Cu 20.00 2.20 0.37
Fe 47.15 30.80 38.84
S 24.00 16.51 1.74
SiO^ 0.44 23.92 32.60
CaO 0.10 4.45 8.24
MgO 1.38 3.44
Zn 2.05 2.98 1.54
ATO3 0.82 1.94 1.50
Mn 0.53 0.15 0.80
0 4.91 15.26 10.88
The charges and products per 24 hours and per 1,000 pounds
of ore used were
Charges :
Ore 68.0 tons 1000 lbs.
Quartz 5.4 " 80 "
Slag 9.8 " 145 "
Coke 2.3 " 34 "
Products :
Matte 8.34 tons 122.65 lbs.
Flue dust 1 . 75 " 25 . 71 "
Slag 63.81 " 938.24 "
THE METALLURGY OF COPPER.
491
Blast applied, 4,500 cubic feet displacement per minute, at
17-ounce pressure. Assume, temperature of gases 450° C, and
that they contain no CO, SO^ or free O^ (no analyses are given).
Assume matte and slag issuing from the furnace at 1,300° (no
temperature given).
Required :
(1) A balance sheet of everything entering and leaving the
furnace.
(2) The volume efHciency of the blowing plant.
(3) The heat generated per minute, per square foot of cross-
section, in the focus of the furnace.
(4) The theoretical temperature at the focus.
(5) The proportion of the heat generated in the focus by the
combustion of carbon and by the oxidation of sulphides.
(6) If hot blast were used what should be its temperature
to be able to dispense with the coke charged? Assume that
the pressure was increased so as to keep the delivery to the
furnace constant per minute.
Per 1000 of Ore Smelted.
Charges.
Ore (1000).
Matte. Flue Dust. Slag. Gases.
Cu
.... 27.44
24.53-
0.57
2.34
Fe
....365.19
57.83
7.92
299.44
S
. ..248.48
29.44
4.24
16.33
SiO^
....185.48
0.54
6.15
178.79
CaO
. ... 72.94
0.12
1.14
71.68
MgO
.... 26.72
0.35
26.37
Zn
. ... 25.56
2.51
0.77
22.28
APO^
.... 9.11
1.00
0.64
0.50
0.14
7.61
Mn
. . . 7.70
6.92
CO'
. ... 31.38
....
....
....
Quartz (80).
Fe
.... 1.16
1.16
S
. ... 0.26
SiO^"
... 77.43
77.43
CaO
... 0.18
0.18
AW
. ... 0.26
0.26
H'O
... 0.71
98
31
47
38
.26
0.71
492
METALLURGICAL CALCULATIONS
Charges.
Slags (145).
Cu
... 1.06
Matte.
FlueL
')ust. Slag.
1.06
Gases.
Fe
... 56.84
. 56.84
■ . . ■
S
... 2.54
2.54
SiO^
... 44.81
. 44.81
• . . .
CaO
... 12.34
12.34
■ . • •
MgO
Zn
... 3.93
... 4.18
3.93
4.18
APO^
... 2.76
2:76
1.23
Mn
... 1.23
■ ■ . >
0
... 15.31
6.04
3.93 5.34
....
Coke (34)..
Fe
... 0.78
0.78
S
... 0.54
....
. . .
. ....
0.54
SiO^
... 2.86
2.86
C
... 28.51
• . .
. ....
28.51
APO^
... 1.21
1.21
0.10
0
... 0.10
Blast (1,191).
0^
...274.91
. 97.92
176.99
N^
...916.37
916.37
2450
122.65
25.71 946.16
1355.77
Notes on the Balance Sheet.
The oxygen of the blast goes as SO^ and CO^ into the gases
and as oxides into the slag. The amount required for the
gases is
Oin SO^' =SX
32
32
= 100.96
OinCO= =Cxfi
= 76.03
0 for Fe = 330.68 X^
00
176.99
94.48
THE METALLURGY OF COPPER. 493
O for Zn = 26.46 X i| = 6.51
bo
OforMn = 8.15X^ = 2.37
O in slag = 103 . 36
Contributed by charges = 5 . 44
Contributed by blast = 97 . 92
To bum sulphides and carbon at the focus = 176 . 99
Total from blast = 274.91
(2) The furnace receives 1,191 pounds of blast per 1,000 of
ore smelted. This represents at 0° C. :
1191X16 = 14 738 cubic feet.
1.293
Per 2000 lbs. ore = 29,475 "
Per 68 tons ore = 2,004,300 " " per day
= 83,510 " " « hour.
1,400 " " " minute.
At50°C. = 1,477 " " "
Blower displacement = 4,500 " " "
1 477
Efficiency of blower =-t-^ = 0-328 = 32.8 per cent. (2)
[It is high time that practical furnace men should stop giv-
ing the mechanical displacement of their blower as the amount
of air received by their furnace. They still keep doing that,
although the furnace receives only from 85 per cent down
to 25 per cent of the given volume. What per cent of the
piston displacement the furnace is actually receiving is a highly
important datum, but is more often unknown than known to
the practical men. It can usually only be found by calculation,
as is illustrated in the preceding case. We urge upon practical
furnace managers, for their own information and use, to cease
being satisfied with piston displacement, and to get deeper into
the real inwardness of their furnace processes by calculating
494 METALLURGICAL CALCULATIONS.
the air actually received by their furnaces. The two things
are enough different to make it always " worth while."]
(3) To calculate the heat generated at the focus we will
assume that all the fixed carbon of the coke is there burned to
CO^, and that the rest of the oxygen blown in produces the
reaction characteristic of pure pyritic smelting. We have
treated per minute
68X2000 c^A A^u t
— T-TT^r — = 94.4 lbs. of ore.
1440
and the carbon burned per 1000 of ore is 28.51, generating
28.51X8100 = 230' 930 lb. Calories.
this absorbs
28.51X8/3 = 76.03 lbs. of oxygen.
leaving 274.91 — 76.03 = 198.88 lbs. of oxygen to oxidize sul-
phides.
Since, now, 20^ generates by the pyritic smelting reaction
161,560 calories, we have generated per pound of oxygen thus
used:
161,560^64 = 2,5241b. Calories.
and per 1000 of ore smelted we will have
2,524 X198.88 = 502,000 lb. Calories.
total heat generated
502,000 + 230,930 = 732,9301b. Calories.
Since this is per 1000 of ore smelted, per minute we have
732,930-^1000x94.4 = 77,640 lb. Calories.
and per square foot of smelting zone area
77,640-^-21.7 = 3,575 lb. Calories. (3)
This last figure is extremely useful in determining the rela-
tive activity of different furnaces, and in most cases will be a
reliable index of the rate at which the furnace is capable of
being driven.
THE METALLURGY OF COPPER.
495
(4) Taking as the basis of calculation 1,000 pounds of ore
smelted, there is generated at the focus 732,930 pound Calories,
there is used 1,190 pounds of blast, and there arrives at the
focus all of the charges except the flue dust, CO^ of ore, H^O of
quartz, and approximately one-quarter of the sulphur. We
therefore, have arriving at the focus about
28.51 lbs. of fixed carbon.
526.50 lbs. of sulphides.
621.20 lbs. of inert slag-forming material.
These, assuming them to reach the focus at 1,000°, would
bring back into it the following amounts of heat:
Carbon 28.51 X380 = 10,834 lb. Calories.
Sulphides, melted 526.50 X200 = 105,300 "
Slag-forming material 621.20X174 = 108,089 "
224,223 "
Heat generated at the focus = 732,930 "
Total heat available at the focus = 957,153 " "
Letting t be the theoretical temperature at the focus, then
we have
Heat in Slag 946 [300 -j- (t - 1 100) 0.27]
" "Matte 123[200-|-(t- 1000) 0.14]
" " S vapor 76 [179-1- (t - 445) 0.11]
" " SO^ 35[0.36t-|-0.0003t2]
" " CO^ 53[0.37t-F0.00022t2]
" " N^ 727[0.303t-|-0.000027t=]
[The 76 pounds of sulphur vaporized at the focus is the
total sulphur charged, less the 25 pounds assumed as vaporized
above the focus, and less the 100.96 pounds oxidized at the
focus; the expression is the total heat in 1 pound of sulphur
vapor at t, in pound-Calories. The 70 of SO^ is the weight of
sulphur dioxide formed, in pounds, divided by 2.88; the 53 of
CO^ is the weight of CO^ formed, in pounds, divided by 1.98;
the 727 of N^ is the M^eight of N^ divided by 1.26. To be strictly
logical the weights of SO^, CO^ and N^ should be first multiplied
by 16 to get ounces, and the resulting expression in ounce-
496 METALLURGICAL CALCULATIONS.
calories divided by 16 to get pound-Calories. We have simply
dispensed with these two numerical operations.]
The sum of the calorific capacity of the products at t° is:
20,098 + 530.5t-l-0.0425t=' = 957,153
whence t = 1569° (4)
It will be noted that the temperature is much higher than
could have been attained by attempting to smelt this mixture
without coke. If the coke were omitted from the charge we
would be without the 28.51 pounds of fixed carbon and some
5.5 pounds of slag-forming material; the products would be
without the CO^ and the N^ corresponding to it, and the heat
available would be decreased 230,930 pound-Calories from the
absence of carbon. If these corrections are made, and it is
still assumed that the materials come down into the smelting
zone at 1,000°, the theoretically calculated temperature is
t = 1447°
While this temperature is theoretically sufficient, yet varia-
tions in ore quality and temperature and humidity of the air
would make running under these conditions much more pre-
carious than with the coke.
(5) This has already been calculated. We found 230,930
pound-Calories to be due to the formation of CO^ and 502,000
to the oxidation of sulphides, making of the total
31 per cent from oxidation of carbon
69 per cent from oxidation of sulphides. (5)
(6) If coke were dispensed with, the theoretical temperature
previously attained, 1,569°, might be reached if the air used
were heated. With cold blast we have, under (4), calculated a
theoretical temperature of 1,447°. Under these conditions the
heat available was altogether 725,266 pound-Calories. But to
heat the products, whose heat capacity was
20,083 -I- 451.7tH-0.0247t2
to t = 1,569° would require (evaluating)^ 789,600 pound-Cal-
ories, which is 64,334 pound-Calories more than are available.
To supply this difference by heating the blast we reckon first
THE METALLURGY OF COPPER. 497
the amount of blast per 1,000 of ore smelted, which will be 862
pounds and its heat capacity:
•j^ (0.303t + 0.000027t2)
making this equal to the heat to be suppUed, 64,334 lb. Cal. ViQ
have
202t + 0.01805t2 = 64,334
whence
t = 313°. (6)
Problem 106.
At Mount Lyell, Tasmania, R. Sticht analyzed the gases at
different depths of the shaft in a pyritic smelting furnace, and
found them of nearly uniform composition for 6 feet down,
leaving out the sulphur vapor, which condensed in taking the
samples. At 6 to 7 feet below the top, close to the focus of the
furnace, the mean of 5 analyses gave by volume
H^
0.00
SO^
0.00
S0='
7.90
00==
3.56
CO
0.00
02
0.88
N^ (difference)
87.66
The CO^ comes from coke, which it was stated was used to the
extent of 1.5 per cent of the charge. The slag contained 53
per cent FeO, 30 per cent SiO^
Required:
(1) The percentage of the heat generated in the furnace
coming from the oxidation of sulphides and the combustion of
carbon.
(2) The heat developed in the furnace per unit of slag formed.
(3) The volume of blast per 1,000 kg. of slag formed.
Solution:
(1) We will assume the reaction of oxidation of sulphides
to be
498 METALLURGICAL CALCULATIONS.
Fe2S + 202 = SCP + 2FeO
and, in fact, the above analyses are our chief justification for
so doing. The air blown in is approximately 20.8 per cent of
oxygen by volume, and according to the above equation the
oxygen going to form FeO is equal to that going tp form SO^.
We have then for the oxygen used
0^ to form 7.90 SO^ 7.90
O^ to form FeO 7.90
0^ to form 3.56 CO^ 3.56
02 to form 0.88 02 0.88
Sum 20.24
on Q
O^ corresponding to W 87.66 X^^ = 23.00
There is thus a small deficit of oxygen, but if the sulphide
oxidizing were assumed to be FeS the reaction would be
2FeS + 302 = 2S02 + 2FeO
and the oxygen accounted for from the gases would be only
0=' to form 7.90 SO^ 7.90
O^ to form FeO 3 . 95
02 to form 3.56 CO^ 3.56
0= to form 0.88 0^ 0.88
16.29
which barely accounts for two-thirds of the oxygen which
must have accompanied the nitrogen.
The conclusion must therefore be that the sulphide which is
being oxidized is Fe^S, and not FeS, for otherwise the gas
analyses would show impossible conditions.
The oxidation of the sulphides gives for each O^ thus used
161,560^2 = 80,780 Calories, and the oxidation of C to CO^
97,200. From the above analyses we infer that for 3.56 volumes
of oxygen used for CO^ 15.80 volumes were used in oxidizing
sulphides. The relative amounts of heat thus generated are
therefore :
THE METALLURGY OF COPPER.
499
By carbon 97,200 X 3.56 = 346,000 = 21.3 per cent.
By sulphides 80,780X15.80 = 1,276,300 = 78.7 « " (1)
(2) The production by oxidation of 2FeO (144 kilograms)
is accompanied by the production of SO^ (64 kilograms) and
the evolution of 161,560 Calories. The slag being 53 per cent
FeO, the slag formed by the reaction is 144 + 0.53 = 272 kg.
The 64 kg. of SO^* would be, in volume, 22.22 cubic meters, and
would be accompanied in the gases by
22.22X~= lOm'ofCO^.
which contains 10X0.54 = 5.4 kg. of C.
whose heat of oxidation to CO^ is
5.4X8100= 43,740 Cal.
but, heat from sulphides = 161,560 "
total heat available 205,300 "
per kg. of slag 205,300 ^ 272 = 755 Cal. (2)
Per unit weight of charge this heat would be slightly lower.
(3) The oxygen used per 272 kg. of slag is
for SO^ 22 . 22 m^
" FeO 22.22 "
" CO^ 10.00 "
Total 54.44 "
per kg. of slag 0 . 20 "
" 1000 kg. of slag 200. «
Volume of air per 1000 kg. of slag made
200 + 0.208 - 960 cubic meters (3)
SMELTING OF COPPER ORES.
The smelting down to matte is done either in reverberatory
furnaces or in shaft furnaces. The charges are usually com-
posed of roasted ore, roasted matte or speiss, mixed with un-
roasted sulphides, usually concentrates, and with siKceous rock
SCO METALLURGICAL CALCULATIONS.
or limestone as flux. The important reaction during the smelt-
ing down is the formation of Cu^S by the copper present, the
formation of other sulphides, mostly FeS, by the larger part
of the sulphur left over, and the slagging of the elements which
do not enter the matte. If much lead is present metallic lead
will separate out, carrying the bulk of the precious metals, but
this phenomenon will be considered under the metallurgy of
lead. The important subjects for calculation are: the propor-
tions of roasted and unroasted materials to be used, and the
proportions of flux to use to make a satisfactory slag.
Some data of importance for calculations on copper smelting ,
have been determined in the author's laboratory by Prof. Walter
S. Landis.
A copper matte containing 47.3 copper, 26.2 iron and 23.6
sulphur, was found to have the following thermo-physical
characteristics :
Melting point 1,000° C.
Mean specific heat, 0 - 1 = Sm = 0.21104 — 0.0000366t
Actual specific heat, at t = S = 0.21104 — 0.0000732t
Heat content, solid, at 1,000° = 174 Cal.
Heat content, Hquid, at 1,000° = 204 Cal.
Latent heat of fusion, at 1,000° = 30 "
Specific heat, at 1,000° = 0.138
Heat of formation from Cu^S and FeS not satisfactorily de-
termined.
A copper blast furnace slag containing 35.5 SiO^ 39.7 FeO,
1.0 MnO, 11.4 CaO, 2.7 MgO, 9.2 ATO', 0.42 Cu, 0.42 S, was
found to have the following characteristics:
Melting point 1,114° C.
Mean specific heat, 0-t = Sm = 0.20185 -t-0.0000302t
Actual specific heat, at t = S = 0.20185 +0.0000604t
Heat content, solid, at 1,114° "= 262 Cal.
Heat content, liquid, at 1,114° = 302 "
Latent heat of fusion, at 1,114° = 40 "
Specific heat, at 1,114° = 0.269- "
Heat of formation from its constituent oxides = 133 Calories
per kg. of slag.
THE METALLURGY OF COPPER. 501
In the case of the slag, its melting point and latent heat of
fusion are not nearly as sharply defined as those of the matte,
since the matte melts sharply, but the slag goes through a pasty
or viscous stage. The values given are the best approxima-
tions which could be gotten from the experiments. As regards
heat of formation, the constituent oxides were carefully weighed,
mixed with a known weight of carbon, and ignited in a Berthelot
bomb calorimeter. Several experiments gave satisfactory con-
cordant results, so that the heat of combination of the con-
stituents of this slag may be taken as not far from 133 Calories
per kilo of slag. By following this scheme it is possible to
measure experimentally the heat of formation of any slag
whose analysis is known.
I. Reverberator y Smelting.
In the reverberatory furnace the atmosphere is in no case
strongly reducing; it may be varied from weakly reducing to
strongly oxidizing, and in the smelting operation usually aver-
ages about neutral. The consequence of this is that copper
oxides or sulphate in the charge are not deoxidized by carbon,
as in the shaft furnace, nor is Fe^O' reduced to FeO by carbon,
but the oxygen thus contained in the charge largely goes off
with sulphur as S0^ thus decreasing the amount of sulphur
left to form matte, and therefore increasing the percentage of
copper in the matte formed. The reactions are mainly:
2Cu04-2FeS-f 2Fe20*-t-6Si02 = SO^ -f Cu^S -F 6FeSiO^
The slag FeSiO^ is, however, heavy and viscous; it would
contain 45 per cent of silica, but runs better and gives a cleaner
separation from matte if some hme is added to it. The re-
placement of 10 per cent of FeO in this siUcate by 10 per cent
of CaO lowers its melting point from 1,110° C. to 1,010°, and
therefore makes a slag that is much easier to keep fluid and of
greater fluidity at any given furnace temperature.
Problem 107.
Peters {Modern Copper Smelting, p. 446) gives the com-
position of the average mixture smelted in the reverberatory
furnaces at Argo, Col., as
502 METALLURGICAL CALCULATIONS.
Per Cent.
SiO^^ 33.9
Iron 10.8
BaSO^ 15.5
APO= 5.6
CaCO^ 8.5
MgCO' 5.8
ZnO 6.1
Copper 2.0
Sulphur 5.1
Oxygen 6.4
99.7
The furnace smelts 50 tons of this mixture (charged hot, at
350° C.) in 24 hours, using 13.5 tons of bituminous coal and
producing a matte with 40 per cent of copper. Outside dimen-
sions of furnace 20 x 40 x 6 feet. Area of stack, fire-box and
hearth 16, 32.5 and 481 square feet, respectively. Temperature,
of stack gases 1,000° C. Composition of the coal: moisture
1.40 per cent, fixed carbon 54.90, volatile • matter 32.90, ash
10.80 per cent; assunie 10 per cent more air used than necessary
for theoretical combustion. Temperature of slag and matte
1,200°.
Required :
(1) The weight of matte produced, assuming the slag to
carry 0.2 per cent of copper (as intermingled matte), and the
matte 40 per cent.
(2) The loss of copper in the slag, expressed in per cent of
the total copper present.
(3) The percentage of the calorific power of the fuel exist-
ing in the stack gases, the slag and the matte.
(4) The heat lost by radiation and conduction in pound-
Calories per minute per square foot of furnace surface.
(5) The velocity of the hot gases at the base of the stack.
(6) The horse-power theoretically obtainable by passing the
hot gases through a boiler which reduces their temperature to
200° C, and which, together with the steam engine, gives a
total thermomechanical efficiency of 7.5 per cent upon the heat
furnished to (entering) the boiler.
THE METALLURGY OF COPPER. 503
Solution :
(1) The entire matte formed, that free plus thatintermingled
with the slag, will be per 100 of ore mixture
2.0 -=-0.40 = 5.0 pounds.
Pounds.
FeS in this matte = 5.0- 2.5 = 2.5
Fe in this matte = 2.5X56/88 = i.e
Fe going to slag as PeO = 10.8-1.6 = 9.2
FeO in slag = 9.2x72/56 =11.8
Constituents of slag:
SiO' = 33.9
FeO = 11.8
BaO = 15.5 X 153/233 =10 2
.APO' = 5.6
CaO = 8.5x56/100 = 4.8
MgO = 5.8X40/84 = 2.8
ZnO = 6.1
Weight of slag = 75.2
Copper in slag as intermingled matte = 0. 15
Intermingled matte lost in slag = 0.37
Free matte obtained = 5.0—0.37 = 4.63 (1)
(2) 0 15
' = 0.075 = 7.5 per cent. (2)
(3) The calorific power of the fuel may be calculated from
its proximate analysis by the method of Goutal {Electro-
chemical and Metallurgical Industry, April, 1907, p. 145), as
follows :
Pure fuel 1.0000 - 0. 1220 = 0.8780
0.3290
Per cent of this volatile f." „„„ „ = 37.5 per cent
Calorific power of volatile matter = 8,650 Cal.
Calorific power (to liquid H^O) :
Carbon 0.5490X8,100 = 4,447 "
Vol. matter 0.3290X8,650 = 2,846 "
Sum = 7,293
504 METALLURGICAL CALCULATIONS.
Water formed = 0.45
Latent heat of vaporization of water
= 0.45X606.5 = 273 Cal.
Metallurgical calorific power = 7,020 "
Assuming the volatile matter to be 15 per cent hydrogen, 40
per cent oxygen c:aC 45 per cent carbon, the coal contains:
Hydrogen 0.329x0.15 = 0.049
Volatile carbon 0.329X0.45 = 0.148
Fixed carbon = 0.549
Total carbon = 0.697
and the air necessary for combustion and products thei'efrom
are per unit weight of fuel used :
OxygenforC= 0.697x8/3
Oxygen for H = 0.049 X 8
Sum =
Oxygen in coal = 0.329X0.40
Oxygen needed from air
.. , , 2.119X13/3X16
Air needed = ^-±^-
Nitrogen therein = 113.7X0.792
10 per cent surplus air
17 1 inr.2 2.556X16 „„„
Volume of CO^ = • — .- = 20.7
1.9o
T7 1 ixsin 0.455X16
Volume of ffO = — p^-^- — = 9.0
U.ol
Assuming the tons mentioned are of 2,000 pounds each, the
heat generated per 27,000 pounds of fuel used per day is
27,000X7,020= 189,540,000 pound-Calories,
and, therefore, per 100 pounds of ore smelted:
189,540,000 -h 1,000 = 189,540 pound-Calories.
The fuel used per 100 pounds of ore being 54 pounds, the
products of combustion are 54 times the above calculated vol-
umes, to which must be added
1.859
0.392
potmds.
2.251
0.132
2.119
(1
113.7
:u. ft.
90.0
11.4
(1
THE METALLURGY OF COPPER. 505
7.4 pounds SO^ = (7.4X16) -f- 2.88 = 41 cubic feet.
6.0 pounds SO' = (6.0X16) -f-3.60 = 27 " "
6.7 pounds SO^* = (6.7X 16) h- 1.98 = 54 " "
found later to be driven off the charge. The stack gases and
the heat carried out by them are as follows :
C0=' 613X0.59X1,000 =
H^O 243X0.49X1,000 =
^.^ j- 2,738X0.33X1,000 =
SO^ 41X0.66X1,000 =
SO' 27X0.58X1,000 =
1,427,000 "
= 89,200 pound-Cal.
Proportion of the calorific power of the fuel in the hot gases:
89,200,
361.670 (
3unce-cal.
119,070
(t tc
903,540
" a
27,060
a u
15,660
it «
189,540
= 0.47 = 47 per cent. (3)
To .find the heat in the slag tapped, taking Landis's deter-
mination of 302 Calories in slag melted at its melting point,
1,114°, and assuming a specific heat in the melted state of 0.27,
we have the heat in it per unit weight at 1,200°:
302+ (0.27X86) = 325 Cal.
Heat in total slag:
325X75.2 = 24,440 Cal.
Proportion of calorific power of fuel in hot slag:
24,440
189,540
= 0.129 = 12.9 per cent. (3)
To find the heat in the matte, we may take Landis's deter-
mination of 204 Calories in matte just melted at 1,000°, and
assuming a specific heat of 0.14 we have heat in unit weight at
1,200°:
204 -KO. 14X200) = 232 Cal.,
and, therefore, heat in the matte formed:
4.65X232 = 1,080 Cal.
506 METALLURGICAL CALCULATIONS.
Proportion of calorific power of fuel in nielted matte;
1,080
189.540
= 0.006 = 0.6 per cent. (3)
(4) The heat lost by radiation and conduction equals all the
heat brought in and generated by combustion and other chem-
ical reactions in the furnace, minus that absorbed by chemical
reactions in the furnace, and minus that issuing as sensible
heat in the stack gases, slag and matte. The ore mixture
being charged hot, at 350° C, and having an assumed specific
heat of 0.15, its sensible heat is
Calories.
100X0.15X350 = 5,250
Add heat of combustion of coal = 189,540
Heat available = 194,790
The heats of chemical reaction of sulphates and oxides on
sulphides are quite complex, and we will show the calculation
in detail at the end of the problem. Suffice here to give the end
results of the calculation:
Calories.
Chemical reactions, net absorption 16,480
Heat of formation of slag; evolved + 9,430
Net heat absorbed in reactions and combinations 7,050
The heat balance sheet will therefore show:
Available.
Calories.
Heat in hot charges 5,250
Heat of combustion of coal 189,540
194,790
Distribution.
In chimney gases 89,200
In liquid slag, tapped out 24,440
In liquid matte, tapped out 1,080
Absorbed in reactions and combinations 7,050
Loss by radiation and conduction 77,020
194,790
THE METALLURGY OF COPPER. 607
The loss by radiation and conduction per day will be:
Calories.
77,020 X 270 = 20,795,000
per minute 20,795,0004-1,440 = 14,400
The whole outside area of the furnace, including the base, is
2 (20X40) +2 (6X40) +2 (6X20) = 2,320 sq. ft.
Loss in pound-Calories per square foot of surface per minute :
14,400^-2,320 = 6.2 pound-Cal. (4)
(5) The volume of stack gases, at 0° C, has been found to
be 3,662 cubic feet per 100 pounds of fuel used, or 3,662 X
270 = 988,740 cubic feet per day = 11.4 cubic feet per second.
At 1.000° this volume would be
11.4X (1,000-1-273) ^273 = 53.2 cubic feet.
And since the area of stack cross-section is 16 square feet the
velocity of the hot gases entering the stack is
53.2^ 16 = 3.3 feet per second. (5)
This is a low velocity and shows good design, which will
result in better economy of fuel than if high velocity were used.
(6) The gases were found to contain 89,200 pound-Cal. per
100 of ore smelted, or per day:
89,200X1,000 = 89,200,000 pound-Cal.
per hour = 3,717,000 "
Since 1 hp. equals 1,400 pound-Cal. per hour, we have:
Horse-poWer at 100 per cent efficiency:
3,717,000^1,400 = 2,650
Horse-power at 7.5 per cent efficiency:
2,650X0.075= 199 hp. (6)
In connection with requirement (3) of above problem it will
be interesting and instructive to discuss the chemical and es-
pecially the thermochemical' phenomena accompanying the
smelting down of the ore mixture." The discussion will be
clearest if we take the actual figures of the problem in question.
Aside from the SiO^ APO^ etc., the ore mixture contains:
508 METALLURGICAL CALCULATIONS.
Per Cent.
Fe 10.8
Cu 2.0
S 5.1
O. 6.4
and these four ingredients are present either as FeS, FeO,
Fe^O', FeSO^ Cu=S, Cu^O, CuO or CuSO*. With the above
quantities of the four elements in question, however, the man-
ner in which they are combined is fixed within rather narrow
limits. Each of the four binds each of the others, and it can
be found, by some patience and trying out, that the four ele-
ments, constituting 24.3 per cent of the ore mixture, must be
combined about as follows in order to be present in the quan-
tities given:
Per Cent
CuO 1.0
CuSO' 3.0
-FeS 7.3
FeSO< 8.9
FeO 0.4
Fe^'O' 3.7
The best proof of this statement is to resolve the weights of
these compounds into their components, and to thus prove
that they agree with the premises.
On melting this down to 40 per cent matte we produce
Cu^S 2.5 per cent in matte,
FeS 2.5 per cent in matte,
SO^ 7.4 per cent in gases,
FeO 11.8 per cent to slag.
The heat represented by the formation of the products must
be subtracted from the heat of formation of the materials
reacting to get the net heat of the reaction. We will multiply
the amount of each material by the heat of formation of unit
weight from its elements as follows:
Cal.
CuO 1.0X( 37,700^ 79.6) = l.OX 473 = 473
CuSO< 3.0X (181,700^ 159.6) = 3.0X1,139 = 3,417
FeS 7.3X( 24,000 -^ 88 ) = 7.3X 273 = 1,993
THE METALLURGY OF COPPER. 609
FeSO^ 8.9 X (214,500^152 )= 8.9X1,411 =12,558
FeO 0.4X( 65,700 -H 72 ) = 0.4X 913 = 365
Fe^O' 3.7 X (195.600^ 160 ) =3.7X1,223 = 4,525
Sum= 23,331
Heat of formation of products:
Cu'S 2.5 X( 20,300 -H 159) = 2.5 X 128 = 320
FeS 2.5X( 24,0004- 88) = 2.5 X 273 = 683
SO^ 7.4X( 69,260^- 64) = 7.4X1,082 = 8,007
FeO 11.8X( 65,700h- 72) = 11.8X 913 =10,773
Sum = 19,782
Difference, heat absorbed = 3,549
We have, therefore, a deficit, or heat to be supplied. This
deficit is increased by the heat required to decompose BaSO<
into BaO and SO', CaCO' into CaO and CO^ and MgCO' into
MgO and CO^; while it is decreased by the heat of combination
of Cu^S.and FeS to form matte, and the heat of combination
of SiO^ with BaO, APO', CaO, MgO, ZnO and FeO to form
slag.
The driving off of SO' and CO^ absorbs
Calories.
SO' from BaO, SO' 6.0X1,189 = 7,134
COMrom CaO, C0=' 3.7x1,026 = 3,796
CO^ from MgO, CO^ 3.0X 666 = 1,998
Sum = 12,928
The net result of all these reactions, leaving out the forma-
tion heat of the slag from its oxide constituents, is
Lb. Cal.
Absorbed 23,331 + 12,928 = 36,259
Evolved 19,782
Deficit 16,477
And even if we credit the heat of formation of the slag:
Lb. Cal.
70.9X133 = 9,430
there remains a net deficit of 7,047
510 METALLURGICAL CALCULATIONS.
The reaction of the ore mixture to form matte and slag is
therefore an endothermic. operation, in spite of the fact that
much sulphur goes off as SO^. The prime reason for this is
that the bulk of the sulphur in the roasted ore was present as
sulphate and not as sulphide.
Problem 108.
A copper blast furnace has at its disposal materials of the
following compositions in percentages:
Cu. Fe. S. SiO\
Selected raw ore 15 20 35 25
Roasted concentrates 25 35 10 12
Refinery slag 50 10 — 25
Limestone CaO — 50 1
It is desired to make a matte with 50 per cent copper and
a slag containing 35 per cent SiO^, 40 per cent FeO and 15
per cent CaO, or with these ingredients in those proportions.
Required :
(1) The proportions of charge.
(2) A balance sheet showing distribution of these materials.
Solution: The foregoing conditions are those which fre-
quently confront the copper smelter. He has at hand raw ore
and roasted concentrates, whose relative quantities he can
usually vary at will by simply concentrating and roasting more
or less material. He has return slags from the refinery fur-
naces which, however, are not unlimited in quantity, but bear a
general relation to the weight of matte made and sent on to
the further operations. Finally, the limestone can be varied at
will. It will readily be seen that if the charge is fixed at a
certain quantity of raw ore to start with, that there are then
three variables, the quantities of the other three materials, and
to fix these there are practically only two conditions to be
fulfilled, the two ratios between three components of the slag.
In order to make a solution possible, it is necessary to make
an assumption which will practically reduce the number of
variables by one, and on looking over the ground it is seen
that the most rational assumption which can be made is to
assume the refinery slag to bear a given relation to the weight
of matte produced. Such an assumption eliminates the weight
THE METALLURGY OF COPPER. 511
of refinery slag as a variable, and leaves us with only two vari-
ables and two conditions to fill, which makes a solution pos-
sible. Assuming the charges to be based on 100 of roasted
ore, we can call the weight of roasted concentrates used X, the
weight of refinery slag Y, and the weight of limestone Z, and
then figure out the weight of matte produced in terms of X,
Y and Z. Assuming then that the refinery- slag is, say, 0.25
of the weight of matte, we have Y = -^ (expression for weight
of matte), and thus one equation between X, Y and Z. The
ingredients of the slag being figured out in terms of X, Y and
Z, the assumed relations between FeO, CaO and SiO^ in the
slag give us two more equations between X, Y and Z, and thus
all three quantities can be determined.
The provisional balance sheet, based on 100 of ore and X,
Y and Z of other ingredients of the charge, will be as follows:
Balance Sheet.
Slag.
0.14Y FeO 16.5- C.1.8Y
0.03X-I-0.26Y
SiO^ 25.0
FeO 0.30X
SiO='0.12X
Ore. (100).
Matte.
Cu 15
Cul5
Fe 20
Fe 1.2 +
S 35
S 7.8 +
SiO^ 25
R'st'd
Cone. (X).
Cu 0.25X
Cu 0.25X
Fe 0.35X
Fe 0.12X
S O.IOX
S O.IOX
SiO^ 0.12X
Rej.
Slag. (Y).
Cu 0.50Y
Cu 0.50Y
Fe O.IOY
Fe O.IOY
SiO^ 0.25Y
Lime-
stone. (Z).
CaO 0.50Z
SiO^ O.OIZ
SiO^ 0.25Y
CaO 0.50Z
SiO^ O.OIZ
Notes on above balance sheet:
Copper in the matte 15 -^0.25X-f0.50Y
Weight of matte (50 per cent Cu) 30 -F0.50X-I-1.00Y
612 METALLURGICAL CALCULATIONS.
Weight of S in matte 7.8 + 0.13X + 0.26Y
Weight of Fe in matte 7.2 + 0.12X + 0.24Y
These weights of S and Fe are therefore provided for under
the column " matte," taking them from the various materials
charged.
The refinery slag being assumed 0.25 of the matte, we have
at once
Y = 0.25 (30 + 0.50X + 1.00Y)
= 7.5 + 0. 13X + 0.25Y
whence Y = 10 + 0.17X
That is, the refinery slag equals in weight 0.1 the ore plus 0.17
the roasted concentrates. This practically amounts to leaving
the roasted concentrates and Umestone as the only variables.
The problem can now be solved by summing up the ingredi-
ents of the slag as follows:
FeO = 16.5-0.18Y + 0.30X
SiO^ = 25.0 + 0.25YH-0.12X + 0.01Z
CaO = 0.5Z
or substituting Y =10 +0.17X
FeO = 14.7 + 0.27X
SiO' = 27.5 + 0.16X + 0.01Z
CaO = 0.5Z
And since the requirements of the slag are that
and
we have:
SiO^ = ^ FeO
40
CaO = ^ FeO
40
27.5 + 0.16X + 0.01Z = 1^ (14.7 + 0.27X)
0.50Z = i^(14.7 + 0.27X)
whence
X = 195 Z = 48
and, therefore, Y = 43
THE METALLURGY OF COPPER.
513
The final balance sheet then becomes
Balance Sheet.
Ore.
(100).
Matte.
Slag.
Cu
15
Cul5
Fe
20
Fe 20
S
35
S 35
SiO^
25
SiO^ 25
Jis't'd Cone.
(195).
Cu
49
Cu49
Fe
68
Fel7
Fe0 66
S
20
S 10
SiO='
23
SiO^ 23
Ref. Slag.
(43).
Cu
22
Cu22
Fe
4
Fe 4
SiO^
Limestone.
11
(48).
SiO^ 11
CaO
24
CaO 24
SiO^
1
SiO^ 1
Matte. - Per Cent.
Cu 86 = 50
Fe 41 = 24
S 45 = 26
172
172 150
Slag. Ratio.
SiO^ 60 = 36
FeO 66 = 40
CaO 24 = 14
150 90
Furnace managers are usually afraid of X, Y and Z, and in
regular running there is usually little need for an algebraic
solution, yet many occasions arise when a judicious use of
algebra solves a problem in an exact manner which hardly
any amount of guessing or approximating can attain efficiently.
In bringing forward this solution we may lay ourselves open to
being called pedantic or academic, but the fact is that the
conditions of this problem were proposed to the writer as a
difficult nut to crack by a practical copper smelter, and the
algebraic solution furnished was characterized by him as the
most satisfactory solution of this class of problems which he
had yet seen.
514 METALLURGICAL CALCULATIONS.
" Bringing Forward " of Copper Matte.
The above term is the old Welsh expression for the further
treatment of mattes which gradually ehminates iron and sul-
phur and finally results in " crude " or " blister " copper, usually
90 to 99 per cent pure. The matte obtained by the first smelt-
ing operation varies considerably in richness, between 20 and
50, or even up to 60 per cent, of copper. The reason it is not
always made high in copper is that the richer the matte the
more copper goes into the slag ; one of the expert metallurgist's
best accomplishments in copper smelting is to make a rich
matte and poor slag. The making of slag of the best compo-
sition, and the use of extra large settlers or fore-hearths, helps
most to this end. Having then this first " matte " the problem
of the metallurgist is to get from it the metallic copper which
it contains.
The Welsh Process.
The principle employed is to partially roast the matte and
then to smelt it down to a richer matte and a ferruginous slag.
The grade of matte formed in this smelting depends entirely
upon the amount of roasting to which the matte has been sub-
jected, the more sulphur eliminated by roasting the less can be
present to form matte and the richer the matte. ,
Illustration: A matte containing 21.36 per cent of copper
and 22.95 per cent of sulphur is roasted until its sulphur con-
tents is two-thirds eliminated. What grade of matte can be
expected on the subsequent smelting.
The sulphur left is 22.95 h- 3 = 7.65. The 21.36 of copper
requires 5.34 of sulphur to form Cu^S, leaving 7.65—5.34 =
2.31 of sulphur to form FeS. This forms 2.31X88/32 = 6.35
FeS. The resulting matte cannot contain more FeS than this;
its composition will therefore be approximately:
Copper 21.36 = 64 per cent
Iron 4.04 = 12
Sulphur 7.65 = 23
33.05
In practical work a richer matte than this will be formed
in reverberatory furnace smelting and a slightly poorer matte
by shaft furnace smelting, because in the former there is some
THE METALLURGY OF COPPER. 515
reaction between the oxides and sulphides of the roasted ore,
resulting in an expulsion of SO^ during smelting, and in the
latter, using carbonaceous fuel, the temperature is so high and
reducing action so strong that metalhc iron is formed and
enters the matte, practically acting as if the matte contained
some Fe^S, and thus diluting the matte.
This roasting is a slow operation unless the matte is crushed
and roasted fine. Lump matte is very little pervious to gases;
it roasts slowly. By breaking a lump in half we, on an aver-
age, increase its surface 50 per cent, and therefore the exposed
surface available for oxidation increases, for a given weight
of material, very quickly as its size is decreased. For auto-
roasting by its own self-generated heat of oxidation the finer
the better, and roasting thus in lump form is impracticable
because of the slowness of the operation.
Pjritic smelting of matte is probably altogether out of ques-
tion as far as pure pyritic smelting without carbonaceous fuel
is concerned. Using some coke, however, an oxidizing smelt-
ing analogous to partial pyritic smelting is possible, as was
proved by Mr. Freeland, at Isabella, Tenn. Such a concen-
trating pyritic smelting is more feasible with a low-grade matte
than with a high grade, because there is more iron and sul-
phur to oxidize. It will be profitable to calculate closely the
details of this matte smelting to rich matte, and to compare it
with the details of the smelting of raw ore in the same fur-
nace— the subject of Problem 105, page 463.
Problem 109.
W. H. Freeland at Isabella, Tenn. (see Engineering and Min-
ing Journal, May 2, 1903), smelted a low-grade matte without
preliminary roasting in a water-jacketed Herreshoff furnace,
having a free area at the tuyeres of 21.7 square feet. The
analyses of materials used and the products are as follows:
Charges.
Raw Laboratory
Matte. Ore. Samplings. Slags. Quartz. Coke.
Cu 20.00 2.79 2.45 0.73
Fe 47.15 43.26 31.07 39.20 1.45 2:'30
S 24-.0(y 29.18 14.84 1.75 0.32 1.58
sio^ 0.44' 10.01 22.66 30.90 96.79 8.41
516
METALLURGICAL CALCULATIONS.
Matte.
Ore. 5
>amplings.
SI
2gs. Qua:
rtz.
Coke.
CaO
. 0.10
6.32
5.71
8
.51 0.23
MgO
....
1.39
2.03
2
.71
. .
Zn
2.05
2.56
2.05
2
.88
0.00
APO»
. 0.82
1.00
1.15
1
.90 0.
32
3.56
Mn
. 0.53
0.69
0.75
0.85
0.00
0
. 4.91
....
3.39
11
.37 0.
38
1.00
C
....
....
13.90
.
. . .
83.86
C0^ etc. . .
2.80
....
.
. . .
. .
....
Loss
0.
39
Products.
MaUe.
Flue Dust.
Slag.
Cu
. 49.63
2.49
0.60
Fe
. 25.24
24.79
43.99
s
. 23.00
. 0.26
8.91
31.43
3.31
1.19
SiO^
33.72
CaO
2.03
MgO
1.18
0.57
Zn
1.53
3.81
2.12
Aro'
0.39
3.93
0.30
2.16
Mn
0.50
0
....
3.97
12.86
C
....
15.88
> > • *
The charges and products per 24 hours and per 1,000 of
matte used were:
Charges.
Matte
Raw Ore
Laboratory samplings
Slag
Quartz
Coke
Products :
Matte
Flue dust
Slag
47.5 tons
1,000 lbs.
8.1 "
170 «
1.6 "
34 "
7.6 "
160 "
15.7 «
330 "
4.5 "
95 «
19.1 tons
401.6 lbs,
0.6 "
12.0 "
56.1 «
1182.2 "
Blast applied, 4,500 cubic feet displacement, at 17 ounces
pressure. Assume temperature of gases 450° C, and that they
THE METALLURGY OF COPPER.
517
contain no CO, SO' or free O^ (no analyses are given). As-
sume matte and slag issuing from the furnace at 1,300° C. (no
temperature is given).
Required:
(1) A balance sheet of everything entering and leaving the
furnace.
(2) The volume efficiency of the blowing plant.
(3) The heat generated per minute per square foot of cross-
section in the focus of the furnace.
(4) The theoretical temperature at the focus.
(5) The proportion of the heat generated in the focus by the
combustion of carbon and by the oxidation of sulphides.
(6) The concentration effected in this smelting.
(1) Balance Sheet (per 1000 of matte smelted).
Charges.
Matte. Flue Dust.
Slag.
Gas
Matte (1,000)
Cvi
200.00
199
31 0.30
0.39
Fe
471.50
101
36 2.97
367.17
. .
S
240.00
92
18
147.
SiO^
4.40
1
04
3.36
. .
CaO
1.00
,
1.00
, .
Zn
20.50
6.
14
14.36
APO'
.8.20
,
8.20
. .
Mn
5.30
1
57
3.73
. .
0
49.10
• «
49.10
••
Ore (170)
Cu
4.74
• . . •
4.74
Fe
73.54
. . . > • >
73.54
S
49.60
1.07
14.15
34.
SiO^
17.02
3.77
13.25
CaO
10.74
0.40
10.34
MgO
2.36
0.14
2.22
Zn
4.35
0.46
3.89
*
APO^
1.70
0.47
1.23
Mn
1.17
0.04
1.13
0
2.40
....
2.40
CO'
2.38
• f . ■
2.
.82
38
518
METALLURGICAL CALCULATIONS.
Charges.
Samplings (34]
Cu
0.83
Fe
10.56
S
5.05
SiO^
7.70
CaO
1.94
MgO
0.69
Zn.
0.70
APO'
0.39
Mn
0.26
0
1.15
c
4.73
Slags (160)
Cu
1.17
Fe
62.72
S
2.80
SiO'
49.44
CaO
13.62
MgO
4.34
Zn
4.61
APO»
3.04
Mn
1.36
0
16.90
Quartz (330)
Fe
4.78
S
1.06
SiO^
319.41
CaO
0.76
APO^
1.06
WO
2.93
Coke (95)
Fe
2.19
S
1.50
SiO^
7.99
APO'
3.38
C
79.67
H^O
0.27
Matte. Flue Dust. Slag.
0.83
10.56
7.70
1.94
0.69
0.70
0.39
0.26
0.67
48
91,
1.17
62.72
49.44
13.62
4.34
4.61
3.04
1.36
16.90
4.78
319.41
0.76
1.06
2.19
Gases.
05
2.82
80
06
93
1.50
7.99
....
3.38
■ > • .
79.67
• • • •
0.27
THE METALLURGY OF COPPER. 519
Charges.
Matte.
Flue Dust.
Slag.
Gases.
Blast (2,065)
0== 476.63
81.24
395.39
N^ 1588.77
1588.77
(3854) 401.60 12.01 1176.05 2264.84
Notes on the Balance Sheet.
The sulphur and carbon of the charge passing into the gases
are, from the balance sheet, 192.61 and 82.49, respectively. We
can assume all the carbon in the gases as CO^ and all the sul-
phur as SO^ except, say, half of that from the ore. This leaves
192.61 — 17.19 = 175.42 of. sulphur to be oxidized at the focus:
32
Oxygen for sulphur = 175 . 42 X ^ = 175 . 42
38
Oxygen for carbon = 82.49Xyq= 219.97
Total to bum sulphur and carbon at focus 395 . 39
1 A
Oxygen f or Fe in slag = 499 . 3 1 X ^g = 1 42 . 66
1 p.
Oxygen for Mn in slag = 6.48X ^^ = 1-88
Oxygen for Zn in slag = 23.56X gg = 5.77
Total in sla§ 150.31
Total oxygen in gas and slag = 545 . 70
Oxygen furnished by soUd charges 69 . 07
Oxygen furnished by the blast 476 . 63
Nitrogen furnished by the blast 1588 . 77
(2) The furnace receives 2,065 pounds of blast per 1,000 of
matte concentrated. This represents at 0°:
^"^^^^^ = 25,550 cubic feet.
1 . 293
520 METALLURGICAL CALCULATIONS.
Per 2000 lbs. matte = 51,100 "
Per 47.5 tons matte = 2,427,250 " " per day.
= 101,135 " " " hour.
1,686 " " " minute.
At 50° C. = 1,958 " " "
Blower displacement = 4,500 " " " "
Efficiency of blower = j^ = 0.435 = 43.5 per cent. (2)
(3) To calculate the heat generated at the focus we will
assume that all the fixed carbon of the coke is there burned
to CO^ and. that the rest of the oxygen blown in produces the
reaction characteristic of pure pyritic smelting. There is
treated per minute:
47.5X2000 „„,,• . ^^
— -777 — = 66 lbs. of matte,
1440
and the carbon burned per 1000 of matte is 82.49 lbs., generating
82.49 X 8100 = 668.169 Ib.-Calories.
and absorbing 219.97 lbs. of oxygen. This leaves 476.63-
219.97 = 256.67 lbs. of oxygen to oxidize sulphides.
Since by the " pyritic smelting " reaction, 20^ generates 161,560
calories, there is generated per pound of oxygen thus used
161,5604-64 = 2,524 Ib.-Calories.
and per 1000 lbs. of matte smelted we have
2,524X256.67 = 677,820 Ib.-Calories.
but heat generated by carbon = 668,170 " "
Total ^ 1,345,990 " "
Since this is per 1000 lbs. of matte smelted, we have, per minute
1,345,990 -f- 1,000X66 = 88,835 Ib.-Calories,
and per square foot of smelting zone area, per minute
88,835^21.7 = 4,094 Ib.-Calories. (3)
Comparing this with the 732,930 pound-Cal. generated at the
focus in smelting 1,000 pounds of ore and the 3,575 pound-Cal.
THE METALLURGY OF COPPER. 521
there generated per square foot per minute, we see that the
matte smelting requires more heat per unit of charge, princi-
pally because the . smelting is done more slowly. Radiation
losses are over twice as great per unit of charge treated when
concentrating matte than when ore concentrating. This points
to the utility of hard driving when smelting matte, as the
direction likely to yield greatest economy.
(4) Taking as a basis of calculation 1,000 pounds of matte
treated, there is generated at the focus 1,345,900 pound-Cal.,
there is used 2,065 pounds of blast, and there arrives at the
focus all of the charges except the flue dust, CO^ and H^O
of charges, and approximately one-half of the sulphur con-
tained in the raw ore. We therefore have arriving at the focus
approximately :
82.5 lbs. of fixed carbon,
1133.0 lbs. of sulphides.
545.0 lbs. of inert slag-forming material.
These come to the focus preheated by the ascending gases,
and assuming them to be preheated to 1,000°, we can find the
correction to be used for their sensible heat:
Carbon 82 . 5 X 380 = 3 1 ,350 lb. Calories.
Sulphides, melted 1133.0X200= 226,000 "
Slag-forming material 545.0X174= 94,850 "
352,200 "
Heat generated at the Focus 1,345,990 " "
Total in the hot products, at Focus 1,698,190 "
Letting t be the theoretical temperature at the focus, then we
have:
Heat in Slag 1176 [300+ (t - 1100)0.27]
" "Matte 402 [200+ (t- 1000)0.14]
" « S vapor 17 [179+ (t- 445)0.11]
" " SO^ -2§i[0.36t + 0.0003t2]
" " ^°' p^[0.37t + 0.00022t=']
" " ^' — ^[0.303t + 0.000027t2]
522 METALLURGICAL CALCULATIONS.
The calorific capacity of these products at t° is therefore
29,858 + 858.7t + 0.1046t='
and making this equal to the heat available at the focus, we have
0.1046t=' + 858.7t + 29,858 = 1,698,190
whence t = 1622°. (4)
This is 53° higher than we calculated for the same furnace
smelting ore, with only one-third as much coke. If the coke
were omitted from this charge the theoretical temperature
would be:
t = 1388°.
Such a theoretical temperature at the focus would not suffice
to form slag, supply radiation and conduction losses, and see
the matte and slag safely out of the furnace, except with very
hard driving and very fast running iii a large furnace.
(5) The heat generated by oxidation at the focus has already
been calculated. The proportion to credit to carbon is
668,170 _ Q^gg ^ ^g g p^^ ^^^^_
1,345,990
and to oxidation of sulphides 60.4 " (5)
In the ore smelting these figures were found to be 31 and 69
per cents, respectively.
(6) The ratio of concentration is usually found by com-
paring the per cent of copper in the material treated with that
in the material produced. This would give in this case:
20.00 ~ ^-^ ^^^
A more reasonable factor, however, is the ratio of weight of
fresh copper-bearing materials treated to weight of copper-
bearing materials produced which must be further treated.
This would be in this case:
THE METALLURGY OF COPPER. 523
414
whereas in the ore smelting it was
1080
148
7.3
"Blister-Roasting" or "Roasting-Smelting."
When a matte has been concentrated to somewhere between
70 and 80 per cent of copper it is called " white metal," and is
in shape for reduction to metallic copper:
Cu^S. FeS.
With 70 per cent copper, matte contains. . . 87.5 12.5
With 80 per cent copper, matte contains. . . 100.0 0.0
The problem being to get metallic copper, the operation is
entirely one of oxidation; first, slow melting down of the lumps
of matte in a highly oxidizing atmosphere on the hearth of a
reverberatory furnace; second, continued oxidation tmtil all
the sulphur is removed and the bath remains as metallic copper,
saturated with Cu^O and Cu^S.
During the melting down the matte is oxidized superficially
to such an extent that when fusion is finally complete the cop-
per oxides have reacted upon the iron sulphides sufficiently to
eliminate all the iron from the matte. The reactions and phe-
nomena of this period are exactly those of the matte concen-
tration processes.
After the melting down sulphur continues to be oxidized
with formation of copper oxide, and then this latter reacts with
more of the sulphide to set free metallic copper.
(a) 2Cu2S + 30^ = 2Cu20 + 2S0^-
(b) Cu='S-}-2Cu20 = eCu-hSO^-
(c) Cu^'S-t-O^ = Cu-hSO^
Reactions (a) and (b) are probably consecutive, but assuming
them simultaneous we have equation (c).
524 METALLURGICAL CALCULATIONS.
Thermochemically, equation (a) analyzes as follows:
2(Cu^ S) = 2(20,300) = 40,600 Calories absorbed.
2(Cu^0) = 2(43,800) = 87,600 " evolved.
2(S, 0^) = 2(69,260) = 138,520 "
Algebraic sum = 185,520 " "
Per kilo, of Cu=S = 580 "
Similarly, equation (b) :
(Cu^ S) = 20,300 Calories absorbed.
2(Cu^ O) = 2(43,800) = 87,600
(3,0^) = 69,260 " evolved.
Algebraic sum = 38,640 Calories absorbed.
Per kilo, of Cu^S = 243
Equation (c) gives:
(Cu^ S) = 20,300 Calories absorbed.
(S, O^) = 69,260 " evolved.
Algebraic sum = 48,960 " "
Per kilo, of Cu^S = 308
Per kilo, of copper liberated = 385 " "
The net result of these figures is to show that the first re-
action evolves a large amount of heat, that the second absorbs
considerable, but that the two together constitute a highly exo-
thermic reaction. When we reflect that a kilogram of melted
Cu^S only carries some 250 Calories, and a kilogram of melted
copper not over 200 at any furnace temperature, the excess
of heat above calculated shows up very strikingly.
In a reverberatory furnace the rate of oxidation is so slow
that but minor advantage is taken of the heat of oxidation of
the Cu^S.
Illustration: In a reverberatory furnace, 8 tons of " white
metal " is smelted to blister copper in 48 hours, using 5 tons
of coal. About what proportion does the. heat of oxidation of
the charge bear to the heat of combustion of the coal?
Assuming the tons to be 1,000 kilos., the white metal to be
THE METALLUR(?y OF COPPER. 525
nearly pure Cu^'S, and the calorific power of the coal 6,000, we
have:
Heat of oxidation of the bath 8000 X 308 = 2,464,000 Calories.
" of combustion of coal 5000X6000 = 30,000,000 "
" requirement for 48 hours 32,464,000 "
" requirement per hour 676,333 "
Proportion of heat requirement furnished by oxidation of
bath:
2,464,000 ^„^„ „^ ■
= 0.759 = 7.6 per cent.
34,920,000
The above figures teach us, however, that when the charge is
once melted, if the oxidation of the bath could be per-
formed quickly enough it alone would keep the furnace up to
heat. For example, the furnace requires an average of 727,500
Cal. per hour to keep it up to heat. The heat of oxidation
would therefore supply this for
. 2,464,000 .,, ,
727,500 =3i hours,
which means that if the bath, when melted, could be oxidized
in that time all exterior firing would be unnecessary after the
charge had been once melted. Some companies force air onto
the surface of the bath, and one company blows compressed
air through wrought iron pipes into the bath, producing the
required oxidation in one-fifth the time usually required, and
saving greatly in coal.
Bessemerizing Copper Matte.
John HoUaway, in 1878, patented the process of oxidizing
matte to metallic copper in an apparatus similar to the Bes-
semer steel converter; in 1880, Manh&s, in France, was suc-
cessful in accomplishing this result, and in 1884 ran the first
commercial plant at the Parrot works in Butte. While the
principles are similar the details are very different from the
blowing of pig iron to steel.
In oxidizing pig iron the carbon silicon, manganese, etc.,
which are to be oxidized out rarely exceed 10 per cent, while
526 METALLURGICAL CALCULATIONS.
the iron itself is oxidizable, and some 3 to 20 per cent may be
lost. In oxidizing matte some 40 to 70 per cent of the whole
charge is to be oxidized, a very voluminous slag results, and
the operation lasts five to fifteen times as long. Towards the
end of a " steel " blow, the pure iron formed is itself oxidizable,
and is not chilled but really heated by th^ passing of the blast
through it; in a " matte " blow the pure copper separating at the
end is not oxidizable under the prevailing conditions, and
therefore is only chilled by the blast if the latter strikes it.
On the latter account, tuyeres in the bottom are impracticable
when Bessemerizing matte, since they become filled with chilled
copper; it is imperative to use lateral tuyeres which, by the
swinging of the converter, are kept always below the surface of
the matte but above the pool of copper as it separates out and
sinks to the bottom.
The two best treatises for details of bessemerizing copper
matte are *Jannettaz's "Les Convertisseurs pour le Cuivre" and
Dr. F. Mayr's " Das Bessemem von Kupf ersteinen ; " very satis-
factory information can be found in Dr. Peters' " Modem
Copper Smelting " and " Principles of Copper Smelting," in
Hixon's " Notes on Lead and Copper Smelting and Copper
Converting," and in Schnabel's " Handbook of Metallurgy,"
last edition, Vol. I.
Assume that a converter is just emptied, is at a bright red
heat (1,100°), and that a charge of melted matte at, say, 1,100°
(setting point 1,000°), is poured into it. The lining is some
60 cm. thick, the outside shell of the converter is at about
200° C, and the surface is losing heat at a nearly steady rate
of, say, 50 Cal. per square meter of outside surface per minute
(10 pound-Cal. per square foot). A converter of 25 square
meters outside surface would thus lose 1,250 Cal. per minute
if standing still. If it had 3,000 kg. of liquid matte in it, with
a specific heat of 0.14, the latter would give out 3,000X0.14 =
420 Calories for every degree which it cooled, and would there-
fore cool off about 1,250-^420 = 3° per minute. If poured
in 100° above its setting point it might be some 30 minutes in
cooling to its setting point and 70 minutes in setting, assuming
no radiation losses through the throat, i.e., that the throat
were tightly covered.
* Mem. de la Soc. Ing. Civile, 1902, (I), 268-319.
THE METALLURGY OF COPPER. 527
Under such conditions, if blast is turned on, the reaction is
2FeS + 30^ + 2Si02 (lining) = 2(FeO, SiO^") +280='
and the heat evolved is
Decomposition of 2FeS = 2(24,000) = - 48,000 Calories.
Formation of 2FeO =2(65,700) =+131,400 "
Union of 2FeO with 2SiO=' = 2(8,900) = + 17,800
Formation of 280=' = 2(69,260) = + 138,520 "
Total =+239,720 "
Heat evolution per 1 kg. Fe8 = 1,362 Calories.
Heat evolution per 1 kg. 0= = 2,497
Theoretical Temperature Rise.
Supposing we oxidize an amount of FeS equal to 1 per cent
of the weight of matte. Let us calculate the theoretical rise
in temperature of the contents of the converter.
The oxygen required is 30^ to 2FeS, or per kilo, of FeS.
Oxygen required = 96-^176 = 0.545 kg.
N^ accompanying = 1.818 "
Air required = 2.363
Volume = 2.363-^1.293 = 1.827 m'.
In being raised from say 50° C. to 1100°, the assumed tempera-
ture of the matte, the air will absorb
1.827[0.303 + 0.000027(1 150)] X 1050 = 641 Calories.
This leaves available, for raising the temperature of the pro-
ducts of the reaction:
1,362-641 = 721 Calories.
The immediate products are:
99. kg. of unoxidized matte.
1.50 " " liquid slag.
0.73 " " SO^gas.
1.82 " " N^gas.
528
METALLURGICAL CALCULATIONS.
The heat capacities of these, at 1100°, are as follows, per 1° C.
Matte
Slag
80="
N2
99X0.14 = 13.86 Calories.
1.50X0.27= 0.41
0.73-^2.88X1.02= 0.25 "
1.818H-1.26X0.36 = 0.52 "
Sum = 15.04 "
Theoretical temperature rise
721 -=-15.04 = 47.9° C.
The above rise represents the rate at which the temperature
tends to rise at the beginning of the blow. Supposing this
amount of FeS (1 per cent) is oxidized in 1 minute, the tem-
perature at the end of 1 minute would rise 47.9° less the cooling
off in 1 minute, which latter might be from 2° to 10°, according
to the size of the converter and its charge, thiclcness of lining,
.etc. For practical purposes the cooling-off rate may be as-
sumed as nearly constant, but the heating-up rate is quite
variable. As the FeS disappears the amount of slag increases,
while that of the matte decreases; but since 1.50 kilos, of slag
with a calorific capacity of 0.41 Cal. per degree takes the place
of 1 kilo, of matte with a calorific capacity of 0.14 Cal. per 1°,
the heat capacity of the products is increasing 0.27 Cal. for
each 1 per cent, of FeS oxidized, while the available heat for
raising temperature is slightly decreasing, because of the higher
temperature of the bath and therefore the greater chilling effect
of the air. Assuming that the bath cools 5° during the time
that 1 per cent of FeS is being oxidized out, we can calculate
the following table:
Cahrific
Temp.
Heat
Net
Capacity
FeS'
at
Absorbed
Heat
of
Temp.
Oxidized.
Start.
by Air.
Available.
Products.
Rise.
Oto 1%
1100°
641
715
15.04
42.9°
1 " 2%
1142.5°
670
686
15.31
40°
2 " 3%
1182°
694
662
15.58
37°
3 « 4%
1219°
722
634
15.85
35°
4 " 5%
1254°
742
614
16.12
33°
5 « 6%
1287°
768
588
16.39
31°
Calorific
Temp.
Heat
Net
Capacity
FeS
at
Absorbed
Heat
of
Oxidized.
Start.
by Air.
Available.
Products.
6 " 7%
1318°
789
567
16.66
7 " 8%
1347°
809
547
16.93
8 " 9%
1374°
823
533
17.20
9 " 10%
1405°
847
509
17.47
At 10%
1429°
THE METALLURGY OF COPPER. 629
Temp.
Rise.
29°
27°
26°
24°
There is no object in extending above table, because it is
constructed on the particular assumption that radiation losses
would amount to 5° during the burning out of 1 per cent of
FeS. This quantity would evidently vary with the size of the
converter, the amount of matte being treated and the speed of
blowing, because as the contents become hotter their rate of
cooling would be increased. The object of the above table, so
far as it went, was to show that 47.5° was the theoretical rise
for the first 1 per cent, but that the theoretical rises for suc-
ceeding per cents would be less and less; in such manner, in-
stead of rising (47.9 - 5) X 10 = 429° for an oxidation of 10
per cent of FeS the actual rise figures out only 329°.
One of the most necessary data which is badly needed for
these converters, and in fact for Bessemer converters in gen-
eral, is the heat loss by radiation and conduction, in other
words, how much heat would be lost by the charge if simply
standing still, how much would the temperature of a given
charge fall per minute if standing still. This could be easily
obtained by running in a charge of pretty hot matte, and fol-
lowing its temperatue curve as it cools, without any necessity
of letting it freeze, but merely starting up the blast when the
rate of cooling has been satisfactorily determined. If these
determinations were coupled with details as to the temperature
of the outside air, its velocity of impingement against the con-
verter, the temperature of the outside shell, the area of radiating
surface and the thickness of the lining, we would soon get data
with which to render entirely definite and exact the whole
thermal investigation of a " Bessemerizing " operation. This
should be supplemented by analyses of the gases during the
blow and the temperature curve of the contents as the blow
530 METALLURGICAL CALCULATIONS.
progresses. Scientific metallurgists must, at present, simply
assume many of these data, because of lack of them. We are
looking to the metallurgical directors of copper plants for some
of this badly-needed technical data.
Problem 110.
W. Randolph Van Liew (Trans. Am. Inst. Mining Eng.,
1904, p. 418) gives the following analyses of a charge of matte
blown to blister copper in a Bessemer converter:
Cu. Fe. S. Zn.
Matte. 49.72 23.31 21.28 1.19
10 minutes 50.20 23.15 20.95 1.20
20 " 56.88 17.85 19.74 0.84
30 « 64.60 10.50 18.83 0.70
40 " (last slag skimmed) 76. 37 2.40 16.30 0.45
70 " (blister copper) 99.120 0.038 0.159 0.09
As. Sb. Ag. Au.
Matte 0.11 0.14 0.152 0.00055
lOminutes 0.09 0.12 0.147 0.00048
20 " 0.08 0.10 0.176 0.00069
30 " 0.08 0.13 0.191 0.00083
40 " (last slag skimmed) 0.08 0.13 0.240 0.00110
70 " (blister copper) 0.0012 0.006 0.312 0.00111
The percentage composition does not exhibit clearly the
relative time and amount of the elimination of impurities, be-
cause of the varying weight of the bath.
Required:
(1) Assuming 1,000 kilograms of matte to be treated, find
the weight of the matte at each period of the blow.
(2) The loss of each constituent of the bath during each
period.
(3) The heat evolution during each period.
(4) The loss of heat per minute due to radiation and con-
duction, assuming that the heat starts with matte at 1,100° and
ends with blister copper at 1,200°, and that 1 per cent of copper
(reckoned on the matte) is oxidized during the last period.
THE METALLURGY OF COPPER. 531
(1) The first question is to find some constituent of the
matte whose weight does not vary during the blow, to serve
as a basis for the calculations. If the analyses be taken as re-
liable in all details (we can make no other assumption) we
see that there is very little loss of anything in the first 10 min-
utes, evidently because of the low temperature of the matte.
Afterwards, iron falls off rapidly, also sulphur and ziiic, while
silver and gold increase in percentage, because of the falling off
in weight of the bath as a whole. The slag up to the last skim-
ming, contains usually but very little copper; in the last period
we are told to assume a loss of 10 kilos of copper by oxidation.
The most rational basis for calculating the weight of the bath
is to assume the copper contents constant for the first 40 minutes.
Copper in 1000 kg. of matte at starting = 497.2 kg.
Weight of matte at starting 1000.0 "
"■ 10 minutes = 497.2 -h 0.5020 = 990.4 "
« " « 20 " = 497.2-^0.5688 = 874.1 "
« 30 " = 497.24-0.6460 = 769.7 "
« " « 40 " = 497.2 -=-0.7637 = 651.0 kg.
« " metal 70 " = 487.2^-0.9912 = 491.5 "
From the analyses given, aiid calling the shortage of per-
centage in the anal^^ses oxygen, we have the following table
of weights and eliminations: (2)
Per 1000 kg. of Original Matte.
Cu. Fe. S. 0.
Start 497.2 233.1 212.8- 41.0
Eliminated 0.0 3.8 5.3 0.1
End 10' 497.2 229.3 207.5 40.9
Eliminated 0.0 73.3 35.0 2.9
'End20' 497.2 156.0 172.5 38.0
Eliminated: 0.0 75.2 27.6 —0.3
End 30' 497.2 80.8 144.9 38.3
Eliminated 0.0 65.2 38.8 12.0
End 40' •■• 497.2 15.6 106.1 26.3
Eliminated 10.0 15.4 105.3 24.9
End 70' 487.2 0.2 0.8 1.4
532 METALLURGICAL CALCULATIONS.
Zn.
^5.
Sb.
Ag.
Au.
11.9
1.1
1.4
1.52
0.0055 =
1000.0
0.0
0.2
0.2
0.06
0.0007 =
9.6
11.9
0.9
1.2
1.46
0.0048 =
990.4
4.6
0.2
0.3
—0.08
—0.0012 =
116.3
7.3
0.7
0.9
1.54
0.0060 =
874.1
1.9
0.1
—0.1
0.07
—0.0004 =
104.4
5.4
0.6
1.0
1.47
0.0064 =
769.7
2.6
0.1
0.2
—0.09
—0.0008 =
118.7
2.9
0.5
0.8
1.56
0.0072 =
651.0
2.5
0.5
0.8
0.03
0.0017 =
159.6
0.4
0.0
0.0
1.53
0.0055 =
491.5
Our table is not absolutely accurate, as can be seen from an
apparent gain in both silver and gold in the periods 10' — 20'
and 30' — 40'. This is caused by a loss of copper during those
periods, but the amounts of gold and silver are too small to
serve as a base for revising the table. If we were to assume
the gold as constant it would make a smaller weight of bath
at the ends of those periods, but the figures are not reliable
enough to make this correction worth while.
The very small total loss in the first 10 minutes and the
loss of time thus occasioned, could in all probability be obviated
by putting the matte hotter into the converter at the start.
A graphic representation of the course t)f a blow is usually
made by plotting the percentages of each element in the bath
at given periods. This plan is largely misleading; the diagram
should be made by plotting the calculated weights of each
element present at the given period, such as are found in the
above table.
(3) The heat evolution is to be found by taking the weights
of each element oxidized out, calculating the heat of its oxida-
tion and formation of slag, and subtracting the heat necessary
to break up the equivalent amount of its sulphide. The heat
necessary to separate the sulphides from each other is unknown.
Period 7.— Start to 10'.
Heat of oxidation: Cal. Cal.
FetoFeO.SiO^ 3.8X1332 = 5,062
S " SO^* 5.3X2164 =11,469
THE METALLURGY OF COPPER. 533
Cal. Cal.
As " As^O' 0.2X1043 = 209
Sb « Sb^O' 0.2X695 = 139
—■ 16,879
Denomposition of sulphides:
FefromFeS 3.8X429 = 1630
As " As'S' 0.2X2000 (?)= 400
Sb " Sb^S^ 0.2X1433 = 287
2,317
Net heat evolution 14,562
The other periods are similarly calculated, and yield the fol-
lowing results:
Start-IQ' 10'-20' 20'-30' 30'-40' 40'-70' | last.
Heat of oxida-
tion 16,879 179,797 162,476 174,316 257,046 85,682
Decomp. of sul-
phides,.... 2,317 35,321 33,719 30,113 10,410 3,470
Net heat evolu-
tion 14,562 144,476 128,757 144,203 246,636 82,212
The last colunin is added for comparison, being the heat
evolution per average 10' in the last period. (3)
(4) The loss of heat by radiation and conduction can be
found, as a whole, by assuming the converter body to contain
the same heat at finishing as at starting, which is a likely as-
sumption, since the lining loses somewhat in weight but ends
up at a higher temperature. Then we know how much heat
was in the original matte at 1,100°, how much was generated,
how much is in the slag and copper, and can calculate ap-
proximately how much is carried out in the gases. These
enable us to find, by difference, the loss by radiation and con-
duction which can then be averaged up per minute.
Calories.
Heat in 1000 of matte at 1100° = 214,000
Net heat generated in the blow = 678,635
Total available = 892,635
534 METALLURGICAL CALCULATIONS.
Heat in 491.5 Copper, at 1200° = 85,995
" « 424.0 S0^ " 1000° = 97,020
" " 947. N^ " 1000° = 248,160
" " 550. Slag, " 1250° = 187,000
Heat accounted for ■618,175
Loss by radiation and conduction = 274,460
Loss per minute, per 1000 kg. matte = 3,920 (4)
It is thought that the principles explained and the methods
of calculation illustrated will suffice to show to copper metal-
lurgists how important information is obtainable by applying
the principles of thermochemistry to copper smelting, and to
indicate the lines along which experiment and results of meas-
urement and observation are greatly to be desired.
THE ELECTROMETALLURGY OF COPPER.
The electric current is used in metallurgy either for its elec-
trolytic effect or for its electrothermal action. In the former
the property of the current utiUzed is its ability when- pass-
ing in one direction through an electrolyte, of causing at the
cathode a reducing action, such as the separation of metal
from the electrolyte or the reducing of a ferric salt to a ferrous
satlt, and at the anode a perducing effect, the direct opposite
chemically of reducing, such as the taking of a metal into the
electrolyte or the perducing of a ferrous or cuprous salt to a
ferric or cupric condition. In such a process heat is inevitably
generated to some extent by the passage of the current through
the ohmic resistance offered by the electrolyte. The extent to
which heat is thus generated, coincident with the electrolytic
action of the current, is of no significance whatever upon the
nature of the process, which remains essentially electrolytic
as long as the current is used for and performs its electrolytic
decomposing function. However, when the heat thus coin-
cidently and inevitably generated in the operation of a process
essentially electrolytic, is sufficient to keep melted an electrolyte
which is not liquid at ordinary temperatures, the apparatus as
a whole may not improperly be regarded as a furnace, and is
very properly classed as an electrolytic furnace.
THE METALLURGY OF COPPER.
535
Electrothermal processes are those characterized by the
absence of electrolysis, as shown by the arrangement and
working of the apparatus, use of alternating current, etc., and
in which the current is used solely for its heating effects. Such
processes are carried on in electric furnaces, and are essentially
processes in which chemical reactions or physical changes are
brought about in the charge solely by the action of the tem-
perature maintained by the assistance of the electric current.
Metallurgically, the furnaces used are resistance furnaces, arc
furnaces and combinations of the two. In resistance furnaces
the heat operating the furnace is produced through the agency
of the resistance of the substance or charge itself, or of a
solid, liquid or granular resistor, intermingled with, in contact
with, or placed in the neighborhood of the material to be heated.
In arc furnaces the current jumps a gap between two poles,
and generates locally the very high temperature of the arc,
which is utilized by feeding the material to be treated into
it or by bringing in close proximity to it. In the combined
arc-resistance furnace, the material being treated forms one or
both poles of the arc, and is therefore heated by the arc itself
as well as by the passage of the current through its own substance.
In electrolytic processes the output of material, or com-
mercial efSciency of the process, is essentially dependent upon
the amperage of the current, since electrolytic effects are pro-
portional to the amperes passing through the electrolyte. In
electrothermal processes the commercial efficiency of output
will vary with the total energy dropped by the current in the
furnace, i. e., will be proportional to the watts Of current used
up, not to its amperage or voltage, but to the product of these.
Direct currents only are employed in electrolytic processes;
direct or alternating may be used in electrothermal processes,
but alternating are preferred, because of the complete absence
of one-sided electrolytic effects when they are used.
Assuming familiarity with the ordinary non-electric metal-
lurgy of copper, we may divide the electrometallurgy of copper
into the following classes:
I. Electrolytic processes —
1. Direct treatment of ores.
2. Treatment of matte.
536 METALLURGICAL CALCULATIONS.
3. Extraction from solutions.
4. Refining of impure copper.
II. Electrothermal processes —
1. Direct smelting of ores.
2. Melting and casting of copper.
I.
Electrolytic Processes.
Copper has an atomic weight of 63.6. It occurs chemically
as cuprous compounds, formulae CuA' , or cupric compounds,
formulae CuA" , where A' is a univalent or monad acid radicle
and A" a bivalent or dyad acid radicle. As a monad atom^
copper has a chemical equivalent of 63.6, as a dyad clement
31.8. The amounts of copper dissolved into or deposited from
a cupric or cuprous salt are proportional to the chemical equi-
valent of copper in these two states and to the amperes flowing.
Assuming that 1 ampere liberates electrolytically 0.00001036
grams of hydrogen per second, we will have as the amount
of copper concerned ifi the passage of 1 ampere through one
tank:
Cuprous Compounds. Cupric Compounds.
1 Ampere, per second . . 0.0006589 grams 0.0003295 grams
per minute . . 0.03953 " 0.01977
per hour 2.372 " 1.186
per day 56.93 " 28.46
per year 20.78 kilograms. 10.39 kilograms
A useful datum to remember, if one is used to working in
EngUsh measures, is that one ampere deposits practically one
ounce avoirdupois (28.35 grams) of copper per day in each cell
using cupric compounds, and 2 ounces for cuprous compounds,
or, respectively, one-sixteenth and one-eighth of a pound per
day. These figures, are of course, the theoretical figures for an
anipere efficiency of 100 per cent.
I. 1. — Treatment of Ores by Electrolysis.
There are no native ores of copper susceptible of being melted
and electrolyzed — in the manner for instance, that sodium nitrate
(Chili saltpeter) can be melted and electrolyzed for the pro-
duction of sodium. The most abundant ore of copper, its
THE METALLURGY OF COPPER. 537
sulphide, occurs mostly mixed with several times its weight
of foreign material, and even if picked out pure and melted it
redissolves copper at the cathode so actively that no electrolysis
is possible. Faraday showed that melted cupric oxide is de-
composed by the current ; and if the oxides of copper were found
anywhere in sufficient purity and quantity they could be
melted and decomposed electrolytically, but hardly in any
case as cheaply as they can be reduced by carbonaceous
material.
Cupric chloride is found in nature as atacamite, with the
formula CuCP. 3Cu^0. 2H^0, in a comparatively pure condi-
tion, but in small quantity, in Chili. Such material will dis-
solve in considerably quantity in melted salt (NaCl), and can
then be electrolyzed. The dissolved copper salt is first re-
duced by the current, with evolution of chlorine, to CuCl, and
then this decomposed. The quantity of this material is at
present insignificant, but there is a possibility of an electrolytic
process along this line.
Cuprous chloride does not occur in nature, but can be readily
made from other copper-bearing material. Chlorine gas,
for instance, converts the common copper sulphide, Cu^S,
into CuCl (Ashcroft's process). When melted, cuprous chloride
conducts the current very well, copper separating out as fine
leaves. The melt cannot be heated to the melting point of
copper and the copper obtained liquid, because it vaporizes
too easily. It is a good conductor, its resistivity at 50° above
its melting point being only 6 ohms (per centimeter cube).
Problem 111.
In electrolyzing a bath of melted cuprous chloride, using an
unattackable anode, the electrodes are 4 centimeters apart, and
a current density of 0.5 amperes per square centimeter is used.
Required:
(1) The voltage required for running the bath.
(2) The proportion of the energy of the current converted
into heat.
(3) The volume of chlorine, at 0°, liberated per minute by
a current of 1,000 amperes.
(4) The output of copper in kilograms per kilowatt-hour of
electric energy employed.
538 METALLURGICAL CALCULATIONS.
Solution :
(1) The voltage required includes that necessary to over-
come the ohmic resistance of the electrolyte plus that absorbed
in chemical decomposition. With a resistivity of 6 ohms, cur-
rent density 0.5 amperes per square centimeter, and distance be-
tween electrodes 4 cm., the first item is
V^ = 6X0.5X4= 12
The second item comes from the heat of formation, by divid-
ing that heat expressed per chemical equivalent by 23,040.
Since (Cu, CI) = 35,400 Calories, we have
y = 35,400-^23,040 = 1.5
The total voltage required would be
V^ +V'^ = 13.5 volts. (1)
This is independent of drop of voltage at contacts of electrodes
with the conductors. That may amount to 0.5 or 1.0 volt,
unless the contacts are very closely looked after.
(2) The current drops as sensible heat:
12 -h 13.5 = 0.89 = 89 per cent of its energy. (2)
(3) Assuming 100 per cent efficiency of liberation of chlorine,
we have:
Hydrogen gas liberated by 1 ampere, in 1 sec.= 0.00001036 gr.
Chlorine gas liberated by 1 ampere, in 1 sec.
0.00001036X35.5 (Chem. eq. wt.) = 0.0003677 "
Chlorine gas per 1000 amperes in 1 minute = 22.06 "
Volume of chlorine, at 0° C.
22.06-^(0.09X35.5) = 6.9 litres. (3)
(4) One kilowatt, at 13.5 volts, gives 1000-^13.5 = 74 am-
peres. This current, in one hour, will furnish, at 100 per cent
efficiency
2.372X74 = 175 grams of copper. (4)
Lower current densities would absorb less voltage and give
a greater power-factor output.
-THE METALLURGY OF COPPER. 639
I. 2 — Electrolytic Treatment op Matte.
Matte is a mixture of Cu^S with FeS, and frequently im-
pure with Pb, Zn, Ag, Au, As, Sb, Ba, etc. It melts sharply
at about 1,000° C, to a thin liquid; it sets quickly to a hard,
stony mass. When melted it can dissolve copper rapidly, so
that electrolysis results in the matte being re-formed as quickly
as it tends to be decomposed, and it apparently conducts the
current without decomposition. If it could be dissolved in
certain other fused sulphides in small amount, such as infused
sodium sulphide, it could conceivably be electrolyzed continu-
ously, but no process has as yet been developed along these lines.
If it could be dissolved in aqueous solutions of alkaline or
other - sulphides, it might be electrolyzed in such solvents,
but no successful solvent of this nature has been found. It. is
not impossible that some aqueous solutions may be found to
answer this purpose.
Solid matte is conducting, and may be used as an anode or
a cathode. . When so used it is acted upon by the electric cur-
rent, reducingly as cathode and perducingly as anode; that is,
used as cathode it tends to be reduced in situ to me^Uic cop-
per and iron, if there is any base present capable of uniting
with and carryiiig away the sulphur; used as anode its copper
tends to pass into combination with the acid radicle of the
electrolyte, leaving the sulphur behind.
Use- of Matte as Anode.
This has appeared to many persons a hopeful application of
electrometallurgy to copper. The matte is cast around some
strips of copper, or copper netting is better, these giving the piece
strength, preventing its disintegrating too quickly, and serving
as conductors. Used in acidulated copper sulphate solution,
the tendency is to dissolve out copper as copper sulphate and
leave the sulphur as a residue. The latter is a non-conductor,
forming an insulating layer, which increases greatly the voltage
required to run the bath. Attempts to scrape off this layer are
fruitless, because of the irregular, cavernous corrosion of the
anodes. Marchesi, an Italian, installed a plant to work this
process near Genoa in 1882; a plant was also erected in Stol-
berg, Germany, Both were subsequently abandoned as im-
oracticable.
540 METALLURGICAL CALCULATIONS.
The theoretical reaction is for purest matte
Cu^S = 2Cu + S
so that for four chemical equivalents of copper deposited one
molecule of .Cu^S is broken up. The voltage corresponding to
the chemical work done is, therefore, •
20,300 ^4 - „„ ,,
23,040 = 0-22 ^°lt-
The voltage required by the cell is this plus that absorbed in
overcoming the ohmic resistances of the circuit.
If the matte were only FeS, the reaction would consist in
the solution of iron, sliming of sulphur, and deposition of an
equivalent amount of copper:
FeS + CuSO%q. = Cu + FeSO^aq. + S
and the energies involved are
(Fe, S) = 24,000 Calories absorbed
(Cu, S, OS aq.) = 197,500 "
(Fe, S, OS aq.) = 234,900 " evolved
Sum = 13,400 "
and the voltage contributed to the circuit is
13,400 ^ 2
23,040
= 0.29 volt.
If the matte is, as it usually is, part Cu^S and part FeS, then
when it is uniformly corroded there occurs a combination of
the above two reactions. If the matte corresponded, for in-
stance, to the formula Cu^S.FeS, and these were simultaneously
acted upon, the current would be divided in the proportions
required by the preceding reactions; that is, twice as much to
handle the Cu^S as for the FeS. The voltages concerned will
enter into the calculation in the proportions 2 to 1, and the
calculated voltage of decomposition will be
0.22 X f = 0. 147 volt absorbed.
0.29Xi = 0.097 " contributed.
Sum = 0.05 " absorbed.
THE METALLURGY OF COPPER. 541
The calculation can be made for any proportions of Cu^S
and FeS, remembering that every 159.2 parts of Cu^S require
twice as much current as 88 parts of FeS.
Problem 112.
Marchesi erected and operated for some time an electrolytic
plant using copper matte as anodes. The matte used contained
30 per cent copper, 30 sulphur and . 40 iron, and the anodes
measured 800 x 800 x 30 mm., and weighed 125 kilograms.
The cathodes were 700 x 700 x 0.3 mm. The tanks were
lead lined, with interior dimensions 2,000 x 900 mm. x 1,000
mm. deep. Electrolyte an acid solution of copper and iron
sulphates. Resistivity assumed at 6 ohms. The plant con-
tained 120 tanks, arranged in ten groups of twelve each, each
group being run by a separate dynamo. Current density, 30
amperes per sq. meter of cathode surface; each tank contained
twenty anodes and twenty-one cathodes. Conductors, 30 mm.
diameter; total length to one group 10 meters.
Required:
(1) The electrical current required by each group of twelve
tanks and the motive power to run its dynamo.
(2) The rate at which the anodes lost weight per day.
(3) The rate at which FeSO* accumulates in the bath, in
per cent of the weight of the bath.
(4) The output of copper per day.
(5) The length of time necessary to plate 1 cm. thickness
of copper on one side of each cathode plate.
Solution :
(1) The first point to be solved is the voltage required per
tank, and that consists of voltage absorbed by ohmic resistance
and that of chemical decomposition. The first must be found
from the ohmic resistance of the baths, the second from the
chemical reactions involved.
With twenty anodes, each 30 mm. thick, and twenty-one
cathodes (the end ones against the ends of the tank) 0.3 mm.
thick, the thickness of electrodes in a tank is
(20X30) -[-(21X0.3) =606.3 mm,
= 60.6 cm.
542 METALLUR6I'6AL CALCULATIONS.
and the free space between, in the -length of the tank, is
200.0 - 60.6 = 139.4 cm. '. ,
Since this is divided into 40 spaces, the free space is
7 139.4^-40 = 3.5 cm.
The total anode surface is, assuming them entirely immersed,
80.0X80.0X2X20 = 256,000 sq. cm.'
and of the cathodes,
70.0X70.0X2X20 = 196,000 sq. cm.
giving current passing through a tank
19.6X30 = 588 amperes.
Since the tank is wider and deeper than the plates, the ef-
fective cross-sectional area of electrolyte is 25.6 sq. m. at the
anode surface, 19.6 sq. m. at the cathode surface, and greater
than either (by spreading of current lines) in between. It
will not be far, wrong, under these conditions, to take the ef-
fective area of the electrolyte as about that of the larger elec-
trode, viz., at 25.6 sq. m. The resistance of the tank thus
becomes
^ = iSlsb = «-«000^2 ohm.
and the voltage absorbed in overcoming ohmic resistance
0.000082X588 = 0.048 volt.
During active corrosion, if copper -and iron are dissolved in
the proportions 30 to 40, this would be, in. chemical equivalent
proportions as 304-31.8 to 40-^28
or as 0.943 to 1.429
since the current divides, in doing mixed electrolysis, in pro-
portion to the number of chemical equivalents dissolved or
deposited; It thus results that 0.943 -h (0.943 -1-1.429) =0.4 =
40 per cent of the current is dissolving copper and 60 per cent
dissolving iron. The corresponding voltage of the chemical
work is. therefore.
THE METALLURGY OF COPPER. 543
0.22X0.40= 0.088
0.29X0.60 = -0.174
Sum = - 0.086 volt.
The sum of these two voltages is
V^ = 0.048 volt
V = - 0.086 "
V = - 0.038 "
The conclusion is, that as long as the surface of the matte
is clean, and no resistance offered by the non-conducting sul-
phur slime, the bath will really require no outside current
to run it, but will practically run itself. The resistance of con-
nections would absorb the small excess of voltage produced.
This is as far as calculation can go. Practical experience with
the bath records that the voltage required quickly rose to 1
volt, and after a few days running reached 5 volts. It is prac-
tically certain that this was caused entirely by the non-conducting
film formed on the anodes.
(2) With 588 amps, passing, and 40 per cent dissolving cop-
per, or 235 amps., the copper dissolved per day in one tank is
28.46X235 = 6688 grams
= 6.688 kilograms
and the weight of matte dissolved per day
6.688-^0.30 = 22.3 kilograms.
Since the 20 anodes in a tank weigh 20X125 = 2500 kilograms,
their loss of weight per day is
; 22.3-^2500 = 0.009 = 0.9 per cent (2)
a,nd to lose their whole weight would theoretically require
J,;,,. lOO-hO.9 =111 days.
P/actically, the plates went to, pieces in about half that time,
when about one-half of their weight had been dissolved.
(3) The iron dissolved is 40 per cent of the matte decom-
posed, ox ,8. 9' kilograms per day. This forms
544 METALLURGICAL CALCULATIONS.
8.9X^ = 24.2 kg. FeSO^
oo
The tank is not quite full of solution, but say to within 3 cm. of
the top. This gives a space of
200X90X97 = 1,746,000 cc.
= 1.746 cu. m.
From this would be deducted the volume of the electrodes :
80 X 80 X 3 X 20 = 384,000 cc.
70X70X0.3X21 = 34,300 "
Sum = 418,300 "
= 0.418 cu.m.
Volume of solution = 1.328 "
Assuming its specific gravity as 1.2, the weight of solution in a
tank is
1.328X1.2 = 1594 kilograms
and its content of FeSO* is increased per day
24.2 -=-1594 = 0.015 = 1.6 per cent. (3)
The total salts present in the electrolyte, however, do not
increase quite that fast, for since 16.73 kg. of copper are de-
posited per day and only 6.67 kg. are dissolved, the electrolyte
loses copper at the rate of 10.06 kg. per day, equal to 25.24 kg.
of CuSO^ per day, which is 1.6 per cent of the weight of the
electrolyte. We may, therefore, say that the electrolyte would
lose 1.6 per cent of its weight of CuSO* per day and gain 1.5
per cent of FeSO* — amounting to a virtual displacement of
CuSO* by FeSO* until the copper was all removed. This would
result in not many days running before the electrolyte would
have to be replaced.
(4) We have previously calculated for one tank, that 16.73
kg, of copper is deposited per day. This amounts for the whole
plant to
16.73X120 = 2008 kilograms. (4)
Of this amount, however, only 6.67X120 = 800 kg. came
from the matte, and the rest from the solution. Ample means
THE METALLURGY OF COPPER. 545
must therefore be provided to supply fresh CuSO* solution,
which was obtained by Marchesi from the roasting and leaching
of ore.
(5) Precipitated copper has a density of 8.9. One amp.
precipitates per day 28.46 grams. One amp. per square centi-
meter would therefore deposit in a day
28.46 - „ , .
o Q = 3.2 cubic cm.
= 3.2 cm. thickness.
But the current density used for depositing by Marchesi was
only 30 amps, per square meter, = 0.003 amps., per square
centimeter, which would deposit a layer per day of
3.2X0.003 = 0.0096 cm.
= 0.096 mm.
To deposit a layer 1 cm. thick would therefore require
1.0-^0.0096 = 104 + days. (5)
This is a much slower rate of deposition than is used at pres-
ent in copper refining, where 100 to 500 amps, per square meter
(9 to 45 amps, per square foot) are used. Low-current density
deposits purer copper and absorbs less power, but gives a
smaller output for a given installation, with consequent higher
interest charges, labor costs, amortisation and general ex-
penses.
Use of Matte as Cathode.
Pedro G. Salom describes the use of this principle, called by
him cathodic reduction, and has applied it on a large scale to
the reduction of PbS concentrates, used as cathode in dilute
sulphuric acid, to spongy lead. If the process is applied to
granulated copper matte, assumed as nearly pure Cu'S to
simplify this discussion, the anode product is 0^ gas, and the
cathode product H^ gas mixed with varying quantities of H^S.
A. T. Weigh tman {Transactions American Electrochemical So-
ciety, II (1902), p. 76) gives details of such an experiment.
546 METALLURGICAL CALCULATIONS.
Problem 113.
Fifteen grams of Cu^S were used as cathode in a 5 per cent
solution of ffSO*, using an unattackable antimonial-lead
anode. A current of 3.6 amps, was sent through for 3 hours
(voltage not given). Area of electrodes 50 sq. cm. each, dis-
tance apart 4 cm. Gases contained H' and H^S in the follow-
ing proportions:
At 5' 42.4 57.6
At 180' 90.8 9.2
Average 0-180' 79.5 20.5
Required:
(1), The voltage required to run the cell, if H^S alone were
liberated, 100 per cent pure.
(2) The voltage required to run the cell at the beginning, at
the end, and the average throughout the run.
(3) The proportion of the Cu^S reduced to Cu during the run.
Solution :
(1) The voltage to overcome ohmic resistance requires first
the resistivity of the electrolyte, which for 5 per cent solution
is 4.8 ohms. The ohmic resistance of the cell is, therefore,
4.8X4H-50 = 0.38 ohm
and the voltage drop to send 3.6 amperes through this
0.38X3.6=1.37 volts.
If ffS were liberated pure, in which case the solution would
be saturated with H^S, the chemical reactions are
Cu^S + H^O = 2Cu + ffS (gas)+0
and the energy involved
(Cu^ S) = 20,300 absorbed
(ff, O) .= 69,000
(ff , S) = 4,800 evolved
Sum = 84,500 absorbed.
The voltage absorbed in producing this chemical reaction is:
84,500^2 , ^- ,^
23.040 = ^-^^ ^°^*'
THE METALLURGY OF COPPER. S47
The total voltage necessary to run the cell is
1.37 + 1.83 = 3.20 volts. (1)
(2) If no H^S were formed, and the current liberated only
H^ gas, there would be absorbed in decomposition 1.50 volts,
and by the cell as a whole 2.87 volts. Since a molecule of H^S
contains the same amount of hydrogen as a molecule of H^,
equal volumes of H^S or H^ would be liberated by an equal
electric current. It follows, therefore, that at 5 minutes 57.6
per cent of the current was performing reduction, hberating
H^S, and 42.4 per cent liberating hydrogen.
The principle here involved in calculating the voltage dropped
corresponding to the chemical work done is that of the com-
position and resolution of voltages, explained by the writer
in Transactions American Electrochemical Society V. (1904),
p. 89. The numerical result desired may be reached in several
ways, for instance,
1.83X0.576 = 1.054
1.50X0.424 = 0.636
V^ = 1.69 volts,
but, for conduction, V'^ = 1.37 "
therefore, total voltage V = 3 . 06
(2)
At the end of the run we have similarly
• 1.83X0.092 = 0.168 volts '
1.50X0.908 = 1.362 "
V^ = 1.53
V^ = 1.37
V = 2.90 "
(2)
For the average running
1.83X20.5 = 0.375 volts
1.50X79.5 = 1.193 "
V^ = 1.57
V' = 1.37 "
V = 2.94 " (2)
548 METALLVRGICAL CALCULATIONS.
(3) Since 20.5 per cent of the current reduced copper during
the run, the weight of copper reduced must have been
1.186X3.6X0.205 = 0.88 grams.
But, the 15 grams Cu^S contained, if pure,
127 2
^^><159:2=^2^'^'°'-
Proportion reduced in the three hours
0.88
12
0.073 = 7.3 percent. (3)
This method of reduction of matte would need radical im-
provement before there could be any possibility of its com-
mercial application.
I. 3 — Electrolytic E-xtraction from Solutions.
This branch of the subject covers some promising processes
which have not, however, been as yet practically successful.
Their consideration, however, from a quantitative point of
view, is not devoid of interest or lacking in instruction.-
The waters of many copper mines carry copper sulphate,
produced from the weathering and leaching of copper sulphide
ores. The solutions are generally too dilute to allow of
concentration and crystallizing out to blue vitriol. The stand-
ard method of treatment is to pass the solutions over pig iron
or scrap iron, thus precipitating the copper as a sort of metallic
mud called " cement copper," which contains much iron, be-
sides the impurities of the iron used, so that it is sometimes
only 90 per cent down to 60 per cent copper. This precipitate
needs a strong refining to bring it up to merchant copper, with
considerable loss of copper in the operation.
Two electrolytic methods are applicable to the treatment of
such solutions: 1. The use of soluble iron anodes. 2. The use
of insoluble anodes.
(1). Use of Soluble Iron Anodes.
If iron in plates or sheets, or bundles of scrap iron in a crate
or holder are immersed in copper sulphate solution and simul-
taneously connected electrically with a copper plate to serve
THE METALLURGY OF COPPER. 649
as cathode, no copper precipitates on the iron, but all is pre-
cipated on the copper. The iron acts as a soluble anode, going
into solution as ferrous sulphate, while the copper is deposited
out passive, dense, and practically chemically pure, on the
cathode sheet. Since more energy is developed by the solution
of the iron than is absorbed in the deposition of the copper,
there is electromotive force generated by the chemical action,
which will run the tank hke a short-circuited battery cell, if
the iron and copper are simply connected by a low resistance
wire. This electromotive force will send a current of a certain
amount through the cell, depending on its internal ohmic resis-
tance plus the resistance of the external conductor. If it is
desired to force matters, and to precipitate the copper faster
than this auto-precipitation, an impressed electromotive force
from a dynamo may be put on and the cell made to work faster.
Problem 114.
A copper sulphate solution whose resistivity is 50 ohms is-
run for precipitation through tanks, each of which contains
fifteen anodes of cast iron 40 x 80 cm. in size, and sixteen
sheets of copper of similar size, the distance between aver-
aging 5 cm. The anodes and cathodes are short-circuited by
resting on a triangular copper distributing bar of negligible
resistance. Assume resistance of contacts to be such that 0.1
volt will be lost at them.
Required:
(1) The electromotive force generated by the chemical action.
(2) The total current operative in each tank, and the current
density.
(3) The weight of copper deposited in each tank per day.
(4) The weight of iron dissolved in each tank per day.
Solution:
(1) From the thermochemical tables (Metallurgical Cal-
culations, Part I. p. 24) we have:
(Fe, S, 0\ aq.) = 234,900 Calories
(Cu, S, O^ aq.) = 197,500 «
Excess of anode energy = 37,400 "
Since this is generated for each atom of copper deposited
and of iron dissolved, the excess energy per chemical equiva-
550 METALLURGICAL CALCULATIONS.
lent concerned is 37,400-^2 = 18,700 Calories, and the total
electromotive force developed is
18,700-^23,040 = 0.81 volt. (1)
(2) The loss of voltage at the contacts being 0.1 volt, there,
is 0.71 volt operative to overcome the ohmic resistance of the
solution. From the data given, the fifteen anodes operating
on both sides, sandwiched between sixteen cathodes, give
15X2X40X80 = 96,000 sq. cm.
active electrode surface, and taking this as the cross-section
of the electrolyte between, we have its resistance as
50 -f- 96,000X5 = 0.0026 ohm,
the total current passing in the tank
0.71-^0.0026 = 273 amperes, (2)
and the current density
273-^9.6 = 28.4 amperes per square meter (2)
= 2.6 amperes per square foot.
(3) Copper deposited in the tank per day:
28.46X273 = 7770 grams
= 7.77 kilograms
= 17.13 pounds. (3)
(4) The iron going into solution will be to the copper dis- .
solved as 56 to 63.6. If the iron is cast iron, this may be only
90 to 93 per cent of the loss of weight of the anodes, because
they contain only that percentage of iron. If wrought iron or
steel sheets are used, the iron dissolved would represent 99
to 99.8 per cent of the loss of weight of the anodes. The iron
dissolved per day is
7 . 77 X 56 -;- 63 . 6 = 6.84 kilograms (4)
= 16.08 pounds.
Problem 115.
One hundred tanks of the kind described in Problem 114 are
arranged in a series, the anodes in each tank, however, con-
nected as one large anode, and the cathodes in each as one
THE METALLURGY OF COPPER. 551
large cathode. A dynamo is connected to the series capable
of maintaining 110 volts across its terminals. The bus-bars
provided are altogether 280 meters long, and are 1x4 cm. in
cross-section, of pure copper. All other details of the tanks
are as before: resistance of contacts being 0.1 volt -h 273 amps. =
0.0004 ohm.
Required:
(1) The ohmic resistance of the bus-bars.
(2) The current operative in the circuit, and the current
density.
(3) The weight of copper deposited in each tank per day.
Solution :
(1) The conductivity of copper, in reciprocal ohms, is 600,000;
its resistivity 1-h 600,000 = 0.00000167 ohms. This is its
resistance per centimeter cube. The resistance of the bus-bars
is therefore.
0 . 00000167 X 28000 ^ 4 = 0.0117 ohm. (1)
(2) The total voltage operative in the circuit is Ihe self -gen-
erated voltage of 100 tanks, plus the 110 volts from the dynamo,
or
(0.81X100) + 110 = 191 volts.
The total resistance of the circuit is that of the 100 tanks,
plus 100 sets of connections, plus that of the bus-bars
(0. 0026 X 100) -h (0.0004X100) +0.01 17 = 0.3117 ohm.
The total current flowing will therefore be
191 -^ 0 . 3117 = 618 amperes (2)
and the current density
618 -^ 9.6 = 64.4 amperes per square meter (2)
= 5.9 amperes per square foot.
(3) 28.46X618-4-1000 = 17.59 kilograms per day (3)
= 38.79 pounds per day.
(2). Use of Insoluble Anodes.
If insoluble anodes are used, the copper may be extracted
en masse from such solutions, and its acid left behind. In
such operations, the choice of an anode is not easy. Graph-
652 METALLURGICAL CALCULATIONS.
itized carbon plates are unattacked, and have practically no
back electromotive force; high-silicon iron plates are also said
to be very resistant, and pur 3 silicon itself resists still better,
but introduces considerable ohmic resistance. Sheet lead an-
odes become coated with a brown coating of PbO^, which is
permanent and evolves oxygen freely. Many other substances
may possibly be used as anodes; not many practical results of
such tests have been published. When operating thus, the
electrolyte offers practically the same ohmic resistance as be-
fore, averaging probably a little less, and the chemical reac-
tion absorbs energy. The resulting acidified solution may be
utilized for dissolving easily-attacked copper compounds from
fresh quantities of raw or roasted ore.
Problem 116.
An electrolytic depositing plant treats dilute copper sulphate
solution containing 318 grams of copper per cubic meter as
CuSO^, and its copper is deposited by passing through tanks
having insoluble anodes, electrodes 3 centimeters apart, and
current density of 20 amps, per square meter. With good cir-
culation, copper is precipitated with no evolution of hydrogen
until the precipitation is practically complete. By having many
tanks in series the current passing is kept very nearly constant
at 300 amps. Each tank contains 1.5 cu. m! of electrolyte.
Required:
(1) The voltage absorbed in the chemical reaction when
precipitating copper, and when the copper is all precipitated.
(2) The voltage absorbed in overcoming the ohmic resistance
of the electrolyte at starting and at the end of the precipitation.
(3) The total voltage across the electrodes of a tank, at
the beginning of the precipitation, just before the last of the
copper is precipitated and after all copper is precipitated.
(4) The time required to precipitate the copper from a batch
of solution.
(5) The output of copper per average kilowatt-hour of
electric energy used, adding 0.2 volts to the potential across
the electrodes for loss in contacts and bus-bars per tank.
Solution :
(1) When precipitating copper, we destroy CuSO*aq. and
form the corresponding H^SO^aq. Their heats of formation
being
THE METALLURGY OF COPPER. 553
(Cu, S, OS aq.) = 197,500 Calories
(H^ S, OS aq.) = 210,200 «
and since the reaction is
CuSO%q. + ffO = ffSO%q. + 0 + Cu
we have to supply
(Cu, S, 0\ aq.) = 197,500 Calories
(H^ O) = 69,000
Total = 266,500
and get back (H^, S, 0\ aq.) = 210,200
therefore supplying net = 66,300
and voltage absorbed In decomposition
56,300-^2
23,040
= 1.22 volts. (1)
When the copper is all deposited, then only H^ and O appear
at the electrodes, so that the voltage absorbed in decomposition
becomes
69,000-^-2 . „ ,^
23,040 =1-50 volts. (1)
(2) At starting, the solution contains
318X-^^^= 798 grams
of CuSO^ per cubic meter, and since the cubjc meter weighs at
16° practically the same as water, i.e., 999.1 kg., our solution is
798-^1000 . _„ , -„ „„,
■ — = 0 . 08 per cent CuSO^
and also equals 0.798-; ^^- = 0.01 normal.
At the end of the electrolysis the 0.01 normal CuSO* solution
becomes a 0.01 normal H^SO^ solution. The resistivities of
both these solutions can be found at once from electrochemical
554 METALLURGICAL CALCULATIONS.
tables {e.g., Kohlrausch, in Landolt-Bomstein-Meyerhoffer's
Tabellen).
With a current density of 20 amps, per square meter (0.002
per square centimeter) and a working distance of 3 cm., the
corresponding voltages absorbed in overcoming these ohmic
resistances will be
1395X3X0.002 = 8.37 volts
325X3X0.002 = 1.95 " (2)
(3) The total voltage across the electrodes, assuming the
solution to contain no free acid at starting, will be
at start, precipitating copper: 8.37+1.22 = 9.59 volts
at close, precipitating last copper: 1.95+1.22 = 3.17 "
at close, evolving hydrogen: 1.95+1.50 = 3.45
(4) Copper present in a tank: 318x1.5 = 477 grams.
Time required, 300 amperes
477 ^- (300X0.0003295) = 4826 seconds
= 1 hr. 20.5 min. (4)
(4) The average voltage consumed in overcoming ohmic
resistance will be considerably less than the mean of 8.37 and
1.95 = 5.16 volts, because when the electrolysis is only half
completed, and the solution is 0.005 normal in each ingredient,
its resistance will be 487 ohms, and not (1,395 + 325) -h 2 =
860 ohms, and the corresponding voltage 2.92. This is because
sulphuric acid is a so much better conductor than CuSO*, that
as soon as acid forms the solution becomes much better con-
ducting. If we calculate the resistances and corresponding
voltages absorbed in several steps, we would get more ac-
curate results. Without going into the detailed calculation we
will take 3.52 as the average voltage, and we then have the
average voltage required per tank:
Decomposition 1 . 22 volts
Resistance 3 . 52 "
Contacts 0 . 20 "
Sum 4.94
Kilowatts used per tank
4.94X300
1000
= 1.482 kw.
THE METALLURGY OF COPPER. 555
Output per tank per hour
300X1.186 = 355.8 grains.
Output per kilowatt-hour
355 . 8 H- 1 . 482 = 240 grams. (4)
This is, of course, a small output compared with that of a
refining process, but it must be remembered that this is an
extraction process, and its cost is not properly comparable with
simple refining but with that of the processes which it replaces,
such as precipitation by iron, and with cheap power and ex-
pensive iron there may be localities where this method would be
economical.
Siemens and Halske invented a combined wet-extraction and
electrolytic process, consisting in leaching the roasted copper
ore or matte with a solution of ferric sulphate acidulated with
sulphuric acid, producing thus a solution of cupric and ferrous
sulphates, and then electrolyzing this solution in a cell having
an insoluble carbon anode and a diaphragm between the elec-
trodes. The solution is run first through the series of cathode
compartments, ■yhere its copper is extracted, and back through
the anode compartments, where its ferrous sulphate is per-
duced to ferric sulphate, ready to be used again in leaching ore.
The chemical reactions in leaching ore are
Cu^S + 2Fe2(SO*)'aq. = 2CuS0%q. + 4FeS0*aq. -i- S.
3CuO-hFe2(SO^)=aq. = 3CuS0%q. -F Fe^O^
CuO-f-ffSO%q. = CuSG^aq.-t-H^G.
In the electrical precipitation we have:
in the cathode compartments : CuSO* = Cu -^ SO^
in the anode compartments: 2FeS0HS0^ = Fe^CSO^)
altogether: 2FeS0^-fCuS0^ = Cu + Fe^(SO0
Confining our calculations to the electrolysis, we have CuSO<
destroyed and 2FeS0^ simultaneously perduced to Fe2(S0^)*,
which is then available for re-use in the leaching tanks. The
chemical work involved is:
(Cu, S, O*, aq.) = 197,500 Calories absorbed
2(Fe, S, OS aq.) = 469,800 "
(Fe^ S^ 0»^ aq.) = 650,500 " evolved
Sum = 16,800 " absorbed
3
3
556 METALLURGICAL CALCULATIONS.
*
Voltage thus absorbed = ' " — = 0.36 volt.
In addition to this, the voltage necessary to overcome the
ohmic resistance of the cell must be counted in.
Problem 117.
An electrolytic tank for working the Siemens-Halske method
was of wood, lead lined, 220 cm. long, 100 cm. wide and 100
cm. deep. Each cell contained fifteen anodes and sixteen
cathodes, the latter 80 x 80 cm. x 1 mm. The anodes were
of carbon rods arranged as a grid 45 cm. x 100 cm. x 1 cm.
thick. A porous partition 1 cm. thick was used, the equivalent
effective free area of which was 0.05. Resistivity of electrolyte
in cathode compartment 5 ohms, in anode compartment 8
ohms, partition in middle. Current density 100 amperes per
square meter.
Required:
(1) The distance between the diaphragm and each anode
and cathode surface, and the resistance of the cell, taking into
consideration the contraction of current path" due to the dia-
phragm.
(2) The voltage necessary to run the tank.
Solution:
(1) 15 anodes XI = 15.0 cm.
16 cathodes X 0,1 = 1.6 "
• Sum =16.6 "
Space, anode to cathode
220-16.6 _g
30r- =6.8 cm.
Space, anode or cathode, to diaphragm
^^.= 2.9 cm. (1)
Assume the solution in the diaphragm to have mean re-
sistivity of 6.5 ohms, the equivalent free area being 100 X
95X0.05 = 475 sq. cm. (area of diaphragm a trifle less than
cross-section of tank). Mean area of electrolyte
THE METALLURGY OF COPPER. 557
cathode compartment (6400 + 9500) -^2 = 7950 sq. cm.
anode compartment (4500 + 9500) -^ 2 = 7000 " "
Resistance of
anode compartment 8X2.9-^7000 = 0.0033 ohm.
cathode compartment 5 X 2.9 h- 7950 =0.0018 "
diaphragm 6.5X1-^475 = 0.0137 "
0.0188 « (1)
Voltage absorbed, for 64 amperes (to 0.64 sq. m. depositing
surface)
0.0188X64 = 1.20 volts
V-* = 0.36 "
1.56 " (2)
Carl Hoepfner devised a process along similar lines to the
above, but based on the use of the two chlorides of copper. A
solution of CuCP in brine is used to act upon the ores or roasted
matte, dissolving the copper with the formation of CuCl.
2CuCP + Cu2S = 4CuCl + S.
The cuprous chloride stays in solution in the brine, and is elec-
trolyzed in tanks containing diaphragms, half the solution
going through the cathode compartments, and half through the
anode compartments. In the former, copper is precipitated;
in the latter, the equivalent amount of chlorine converts the
CuCl present into CuCP, which can be used over, mixed with
the depleted cathode solution. The total reaction is
CuCl + CuCl = Cu + CuCP
and the energy involved
2(Cu, CI) = 70,800 absorbed
(Cu, CP) = 62,500 evolved
Sum = 8,300 absorbed
= 0.18 volt.
Coehn devised a cell for carrying on this process without a
diaphragm. The cathode is only half as long as the carbon
anode, and the CuCP formed at the latter is so heavy that it
sinks to the bottom and is drawn off therefrom, while fresh
CuCl solution is quietly poured in above. Since the cathode
558 METALLURGICAL CALCULATIONS.
does not touch the cupric solution, it is not redissolved thereby,
and the two solutions are kept separate.
A great advantage of this process is that Cu is monovalent
in CuCl, and therefore twice as much copper is deposited as
from CuSO* by a given number of amperes.
Problem 118.
Coehn (Jahrbuch der Electrochemie, 1895, p. 155) describes
his improved partitionless Hoepfner apparatus as containing a
copper cathode 100 cm. wide by 50 cm. deep, at a distance of
12 cm. from a carbon anode; solution, 10 per cent NaCl, plus
varying amounts of CuCl and CuCP. Current density 20 amps,
per square meter.
Required :
(1) The voltage required to run the cell.
(2) The weight of copper deposited therein per day.
(3) The weight of copper per kilowatt hour electric power
used.
Solution :
(1) The resistivity of 10 per cent NaCl solution is 8.5 ohms.
R = ^'^r^n^ = 0-0204 ohm
V , for ohmic resistance = 0.0204 X 10= 0.20 volt
V, for chemical reactions 0.18 "
V = Sum = 0.38 " (1)
(2) 28.46X2X10= 569 grams per day (2)
(3) 569 1000 =6210 grams
24 ^10X0.38 = 6.21kg. (.3)
I. 4. Electrolytic Refining of Impure Copper.
This subject is a large one. It lends itself very well to cal-
culation. It is also comparatively simple in principle. Given
a nearly pure copper plate or slab as an anode, in a solution of
a copper salt, and a suitable conducting cathode, the metal is
simply dissolved and deposited, or is " plated over." The pro-
cess really consists in extracting copper from the solution at
the cathode, and sending it and other soluble impurities into
the solution at the anode. The mechanical transportation of
the copper, in space, through a distance equal to the space be-
tween anode and cathode, is sometimes emphasized as the chief
THE METALLURGY OF COPPER. 559
mechanical work done. It is a real fact that the metal is taken
out of the solution at a different spot from where it goes in,
but the electric current does not do that transportation — dif-
fusion and circulation are the agents which transport the copper
entering the electrolyte at the surface of the anode plate to
the surface of the cathode plate, and electric energy does nothing
more than create the differences of concentration, which are
thus physically neutralized.
The chief expenditure of electrical energy in the refining
bath is that converted into sensible heat in overcoming the
ohmic resistance of the electrolyte. The other items of elec-
trical work are the overcoming of the ohmic resistance of hangers,
rods, bus-bars, clamps and connectors, and the not unimportant
resistance of contacts; also the difference of chemical work
done in depositing all copper and dissolving part copper and
part other impurities; also the chemical work of separating
the ingredients of the impure copper, which may be alloyed
and have some heat of combination which must be furnished
by the current in order to separate them.
Energy Absorbed by the Electrolyte.
The electrolyte is a conductor and absorbs energy according
to Ohm's law whenever current passes through it. The solu-
tion used in copper refining is almost invariably copper sulphate
acidulated with sulphuric acid. The resistivity of copper sul-
phate, iron sulphate and sulphuric acid solutions, at 20° C, is
as follows, in ohms per centimeter cube and ohms per inch
cube:
Percent CuSOK FeSO*. H^SO*.
in
Q per
Q per
Q per
Q per
Q per
Q per
Solution.
cm?
inch^.
cm^.
inch^.
cw?.
inch^.
2.5
92
37
. .
....
....
5.
53
21
.
4.8
1.9
7.5
..
, .
65
26
....
10.
31
12
. .
2.5
1.0
15.
24
10
34
14
1.8
0.7
17.5
22
9
....
20.
• m
, .
. •
1.5
0.6
25.
• «
• !•
. .
1.4
0.56
30.
• *
M «
25
10
1.37
0.55
560 METALLURGICAL CALCULATIONS.
The plain copper sulphate solution is thus seen to be a rather
poor conductor, iron sulphate is not so good ; sulphuric acid is
highly conducting. It is therefore evident that the ohmic re-
sistance of the bath is, for practical purposes, dependent upon
its content in free acid; hence the great economic importance
of keeping the bath well acidulated. As the bath grows foul
with iron sulphate and loses sulphuric acid by formation of
insoluble sulphates, chemical solution of slimes, etc., the bath
becomes poorer conducting and the energy absorbed in it much
greater. A solution in good condition, with 15 to 20 per cent
of CuSO^ and 5 to 10 per cent free H2S0^ may have a resist-
ivity of only 2 to 5 ohms at starting, which may increase to
20 or 25 ohms as the solution becomes impure and the free acid
disappears. These facts are of immense importance in the eco-
nomics of refining.
A comprehensive, systematic study of the resistivities of
solutions of various strengths of copper sulphate with various
percentages of iron sulphate present and various content of
sulphuric acid is greatly needed, and would not be a difficult
research to carry out.
Problem 119.
A refining bath is run with a current density of 250 amperes
per square meter, with electrodes 4 cm. apart, and starting with
electrolyte 15 per cent CuSO^ and 10 per cent H^SO^. It is
run until the sulphuric acid has decreased to 5 per cent.
Required :
(1) The probable voltage drop across the electrodes at start-
ing.
(2) The same when the second condition is reached.
(3) The rate at which the solution would start to rise in
temperature at the beginning, ignoring the electrodes.
(4) The rate, assuming anodes 5 cm. thick and copper cath-
odes 0.5 cm. thick.
Solution :
(1) Lacking the experimentally determined datum of the
resistivity of the primary solution, we know that it will be
somewhere about that of the 10 per cent sulphuric acid solution,
viz. : 2.5 ohms. It should be slightly less, because of the copper
salt present, and if we assume the conductivity of the solutions
THE METALLURGY OF COPPER.
561
as additive, we would have a conductivity of the mixed solu-
tion, in reciprocal ohms as
C = 2:5+^ = 0.40 + 0.04 = 0.44
Rip = Q~^ = 2.3 ohms.
Using this resistivity, the voltage drop across the electrodes,
at starting, will be
V"^ = 2.3X4X0.0250 = 0.23 volts. (1)
(0.0250 amperes goes across each square centimeter and the
plates are 4 centimeters apart.)
(2) With H^SO* only 5 per cent, and copper contents, say,
20 per cent, because of solution of copper by free acid, we would
have
C =- 4^ + 2^ = 0.21+0.05 = 0.26
Rjp = Q-gg = 3.6 ohms.
Using this resistivity, the voltage drop is
V^ = 3.6X4X0.0250 = 0.36 volts. (2)
In actual practice an increase of 0.01 volt above this might be
expected,, because of the formation of a film of slime on the
surface of the anode.
(3) The solution at starting has a specific gravity of 1.20
and a specific heat of 0.875. (Combination of data from
Landolt-Bomstein-Meyerhofler's Tabellen.) The water value
as heat absorber of a column of liquid 1 cm.^ and 4 cm. long,
between the electrodes, is therefore
1.20X4X0.875 = 4.2 calories per 1° C.
Heat equivalent of the electric energy expended
0.23X0.0250X0.2389 = 0.00138 calories per 1".
562 METALLURGICAL CALCULATIONS.
Rate of rise of temperature of solution
0 . 00138 -^ 4 . 2 = 0.00033° per second
= 0.0198° " minute
= 1.2'= " hour (3)
= 28.8° " day
(4) With half the thickness of anode and cathode to be sup-
plied with heat, the copper having a specific gravity of 8.9 and
specific heat of 0.093, the water value of the copper concerned
per square centimeter area of electrolyte section is
/^2 4. 2^^8.9X0.093 = 2.276 calories per 1°
Rate of rise of temperature of solution plus electrodes:
0. 00138 -f- (2.276 + 4.2) = 0.00021° per second
= 0.013° " minute
= 0.78° " hour
= 18.7° " day (4)
Energy Lost in Contacts.
This is an extremely variable and yet important factor in the
economics of copper refining. No one item in copper refining
will at the present time better repay close attention and study
of methods of improvement. A proper contact should cause a
very small voltage drop to begin with, and should be capable
of being kept efficient — that is the chief requirement. Con-
tacts are often made on the hit or miss style, and arranged in
such position as to receive spattering or drippings from the
electrolyte; such may cause immense losses of electrical energy
as they rapidly pass from bad to worse.
B. Magnus {Electrochemical and Metallurgical Industry, De-
cember, 1903), and L. Addicks {idem., January, 1904) have
given us valuable information on this point. The weight of
the plates, when resting on the contacts, improves the contact;
in such cases the anode contacts are best when the tank is first
set up and the cathode contacts worst. In such cases a screw
clamp, to put several hundred pounds of mechanically applied
pressure on the joint, should be adopted, so as to make the
contacts " best " all the time.
THE METALLURGY OF COPPER. 563
Figures given by Magnus as a carefully determined average
of results in a western refinery are:
Current 4,000 amperes
Current density (per sq. ft.). .11 "
Total voltage per tank 0.230 volts
Energy .used per tank 0.920 kw. Per
Drop bus-bar to anode rod. 0.0270 volts
" anode rod to anode
hanger 0.0060
" anode hook to anode. 0.0008
" anode to cathode 0.1782 " =0.1782= 77.5
" cathode to cathode
rod 0.0045 " V
" cathode rod to bus- I = 0.0180 = 7.8
bar 0.0135 "
Cent.
= 0.0338 = 14.7
0.2300 100.0
It thus appears that in this tank the loss in contacts was
some 20 per cent of the total voltage employed (assuming the
anode rods to be of such design that their resistance was neg-
ligible). Magnus fitted up a similar tank with mercury cup
contacts, and found a drop from bus-bar to anode rod of 0.0050
volt instead of 0.0270, an'd cathode rod to bus-bar 0.0050 instead
of 0.0135 volt. If it were possible in the case of above tank to
make these improvements alone, the saving on voltage per tank
would be 0.0305 volt, or 13.3 per cent of the voltage used,
which means a saving of 13.3 per cent of the power required to
run the tank.
Problem 120.
In an electrolytic refinery system of 200 tanks the resistance
of the contacts was, on an average, 0.0368 volt per tank, and
0.2567 volt across two electrodes; 3,800 amperes was passed
through the series, and the voltage at the terminals of the
dynamo was 67 volts. Power costs $157 per kilowatt-year,
Efficiency of deposition of copper 82 per cent.
Required:
(1) The cost of power per ton (2,000 pounds) of copper
refined.
(2) The cost per ton of copper produced, of the power lost
564 METALLURGICAL CALCULATIONS.
by (a), resistance of the conductor bars (b), by the contacts and
(c) , consumed in the electrolyte itself.
(3) If interest on capital is 6 per cent, what expenditure
would be justified on means for (a), reducing the conductor
losses (b), reducing the contact losses (c), reducing the elec-
trolyte losses.
(4) If the ampere efficiency of deposition could be increased
5 per cent by reducing the current density 20 per cent, would it
pay?
Solution :
(1) Copper deposited in 1 tank per day
3800X0.82 = 3118 oz. = 195 lbs.
In 200 tanks
200X195 = 39,000 "
Power used
3800x67-^1000= 255 kw.
Cost of power per day
255 X $157 -5- 365 = $109.70
Cost of power per 2,000 lbs. copper
$109. 70 -f- 19. 5 = $5.63 (1)
(2) The voltage lost in the conductors is
67 - 200(0.0368 + 0.2567) = 67 - 58.7 = 8.3 volts
and the power thus lost
3800 X 8. 3 -f- 1000 = 31.54 kw.
Cost per year. 31.54X $157 =$4,952.00
Cost per day $4952 -h 365 = $13.57
Cost per ton of copper produced = $0.70 (2a)
Power lost in the contacts
200 X 0 . 0368 X 3800 -f- 1000 =. 27.97 kw.
Cost, per year = $4,391.00
Cost, per day = $12.03
Cost, per ton = $0.62 (26)
Power consumed in the electrolyte
200 X 0 . 2567 X 3800 ^ 1000 = . 195 kw.
Cost, per year = $30,615.00
Cost, per day = $83.88
Cost, per ton = $4.30 (36)
THE METALLURGY OF COPPER. 565
(3) The cost of the power lost per year, divided by 0.06,
will give the capital investment representing that yearly outlay.
This would be
Cost of Power Capitalized.
Conductor resistances $4,952 $82,530
Contact resistances 4,391 73,180
Electrolyte resistances 30,615 510,250
In the specific cases in point each 1 per cent of the power
consumption which could be saved by improvements of a per-
manent character, not subject to depreciation, would • justify
capital expenditures in the plant up to the following amounts:
Conductor resistances $825
Contact resistances 732
Electrolyte resistances 5,100
If the improvements were liable to depreciation in use, such
as larger tanks, contact clamps, etc., the depreciation must be
added to the interest to find the justifiable capital expenditure.
Assuming depreciation 10 and 20 per cent, respectively, this
plus interest amounts to 16 and 26 per cent, and the justified
expenditures must be less than the following amounts for 1 per
cent saving of power in each direction:
10% _ 20% _
Depreciation. Depreciation.
Conductor resistances $309 . 50 $190 . 50
Contact resistances 274 . 50 169 . 00
Electrolyte resistances 1,913 . 50 736 . 00
There is here a wide field for experiment and improvement.
In the case in point, making all allowances, one would be justi-
fied in expending in permanent plant for each tank at least
$1.00 for every 1 per cent reduction which could be made in
the power loss in the conductors, $0.85 for every 1 per cent
gained in reducing contact resistances and $3.68 for each 1
per cent reduction in the resistance of the electrolyte. To be
more specific, assume the main conductors to be proportioned
to carry 2,000 amperes per square inch of cross-section. Their
resistance in the case in point is 8.3-^3,800 = 0.0022 ohm,
cross-section 1.9 square' inches, resistance per running inch
0.00000067-^1.9 = 0.00000035 ohm, and total length 0.0022^
0.00000035 = 6,286 inches = 524 feet.
566 METALLURGICAL CALCULATIONS.
The weight, at 62.5X8.9^1,728 = 0.322 lbs. per cubic inch is
6,286X1.9X0.322 = 3,846 lbs.
which at $0.20 per lb. costs
3,846X0.20 = $769.20
Since there is practically no depreciation (except change in
market value) in these copper conductors, if we increased the
cross-section of the conductors 100, 200, 300, and 400 per cent,
respectively, we would decrease their resistance 50, 67,, 75 and
80 per cent, respectively, and arrive at the following figures:
Per Ton of Copper Produced.
Per Cent Interest
Per Cent Decreased on
Increased Resistance Increased Increased Decreased
Size of of Cost of Cost of Cost of
Conductors. Conductors. Plant. Plant. Power. Gain.
100 50 $769 $0.0065 $0.35 $0.35
200 67 1,538 0.013 0.47 0.46
300 75 2,307 0.019 0.52 0.50
400 80 3,077 0.026 0.56 0.53
900 90 6,923 0.058 0.63 0.57
1,900 95 14.613 0.123 0.67 0.54
We thus see that for this plant, with power costing $157 a
kilowatt-year and copper bar $0.20 per pound, a great reduc-
tion in the power costs up to $0.57 per ton of copper produced
can be made by making the main conductors ten times as large;
that is, by making them carry only 200 amperes per square
inch of section instead of 2,000.
At a locality like Great Falls, Montana, where the water
power does not cost over one-tenth the above price, let us say
$15.70 per kilowatt-year, the decreased cost of power, in the
above table would be one-tenth the above figures — there would
be only $0.07 per ton of copper to save if all were saved, and
it can easily be seen that the gains would be
for 100 per cent increased size $0,029
"200 " " "....-. 0.034
"300 " " " 0.033
"400 " " " 0.030
THE METALLURGY OF COPPER. 567
We reach the maximum saving in this case by increasing the
size 200 per cent, giving a current density of 660 amperes per
square inch of conductor section.
It can easily be seen that the cost of copper bars, rate of in-
terest, and cost of power are all three factors in determining
the economic size to be given the conductors. In general
they are made much smaller than they should be because of
lack of capital to invest in them. As soon as a refinery is run-
ning and paying dividends part of its profits should be ap-
plied to permanent investment in heavier copper conductors in
order to reduce the running costs.
(4) If the ampere efficiency were increased 5 per cent and
the current density 20 per cent, all other data being constant
except those affected by these changes, the output of copper
would be 87 per cent of the theoretical output of 3,040 amperes
instead of 82 per cent of the theoretical output of 3,800 am-
peres. The output of copper per day would therefore be, in
place of 39,000 pounds:
3,040X0.87X2004-16 = 33,060 lbs. per day.
The power cost will be reduced 20 per- cent by the lower
amperage used, that is, be $5.63X0.80 = $4.50 per ton — a sav-
ing of $1.13. Whether it will pay to do this depends of what
the other fixed charges of the plant are. Decreasing the out-
put 15 per cent increases the fixed charges per ton of copper
produced 12 per cent. It is then a question whether 12 per
cent of the fixed charges in the first case is greater or less than
$1.13. If the fixed charges amounted to, say, a usual figure
of $5.00 per ton, decreasing the output 15 per cent would in-
crease this charge per ton of copper produced 12 per cent,
or $0.60; but if there is a power saving of $1.13 there is a mar-
gin of $0.53 per ton in favor of making the change. If the
power cost were only $0.56 per ton of copper it would just
barely pay to make the change. Everything depends upon the
relative cost of power and of the other fixed charges.
Energy Lost in Conductors.
We have already touched on this subject, but it deserves
still greater attention. The resistivity of copper is 0.00000167
ohm per centimeter cube, or 0.00000067 ohm per inch cube.
S68 METALLURGICAL CALCULATIONS.
Given a conductor of certain length its resistance in ohms is
0.00000167 multipUed by its length in centimeters and divided
by its cross sectional area in square centimeters. The similar
calculation can be made, for the inch units. The drop in po-
tential in the bus-bars is their, resistance multiplied by the
amperes passing, and the power consumption is the voltage
drop into the amperes passing or the resistance in ohms into the
square of the amperes passing. The watts thus expended
may be expressed in kilowatts, or transposed into horse-power
by dividing by 746. The heat generated in the conductors is,
per second, in gram calories, the number of watts expended in
them multiplied by 0.2389. The rate of rise of temperature of
the conductors, leaving out radiation and conduction to the
air, will be, per second, the gram calories generated in the con-
ductors divided by the heat capacity of the conductors per de-
gree, which in the case of copper is volume in cubic centi-
meters into 0.837. This rise in temperature will continue until
by radiation and conduction to the air the conductors lose as
much heat per second as the current generates in them. This
will vary with the shape of the cross section of the conductors,
because the loss of heat is proportional to the surface exposed,
and the area of exposed surface for a given cross section is
greater the thinner the conductor and is a minimum for a
round conductor. The amount of heat thus lost can be cal-
culated by the principles explained in Metallurgical Calcula-
tions, Part I., pp. 172-186. If it is desired to get along with
the minimum heating of the conductors, they should be as flat
a,nd thin as possible; if heating can be ignored they may be of
the cheap round shape. It should not be forgotten also that
the resistivity of copper increases as it gets hotter in direct
proportion to its absolute temperature (C° -1-273), so that
the heating effect really aggravates itself from this increased
resistance.
Problem 121.
A copper conductor is desired to carry 4,000 amperes, 300
meters. Round bars sell at 18 cents per pound. Interest rate
10 per cent. Cost of electric power as delivered $50 per kilo-
watt year. Average temperature of surrounding air 20° C. ;
air still. Conductivity of the copper 600,000 reciprocal ohms
per centimeter cube. Specific gravity of copper 8.9, specific
heat 0.094.
THE METALLURGY OF COPPER. 569
Aluminium conductors to replace same have conductivity of
375,000 reciprocal ohms, specific gravity 2.60, specific heat
0.230; cost of round bars 36 cents per pound.
Required:
(1) The cross-sectional area of the conductors of copper
which will give the minimum cost of power and interest on the
investment.
(2) The same for aluminium.
(3) The running temperature of the conductors of copper
above the room temperature.
(4) The same for aluminium.
Solution :
(1) Let S = cross section of conductor, in sq. cm.
Resistance of copper conductor
0.00000167X(300X100)-^S = ^1^21 ohm.
Power expended in the conductor:
0.0501 ,,. „„„,, 801,600 ,,
— g — X (4,000)=' = g watts.
Cost of power thus expended, per year
^5^.1,000X50 = i5fO dollars.
Weight of copper in conductor:
(300 X 100) X S X 8 . 9 . 1 ,000 = 267 X S kilograms.
Cost of conductor, if round
267. SX 0.18X2.205 = 106 XS dollars.
Interest on cost of conductor, at 10% = 10.6XS dollars.
Sum of cost of power and interest on investment:
i2^+(io.6xS)
Solving for S a minimum, we find S = 61.5 sq. cm. (1)
Total costs, substituting S above
$652 (for power) +$652 (interest) = $1,304 per year.
570 METALLURGICAL CALCULATIONS.
(2) Resistance of aluminium conductor
0.00000267X(300X100)-^S = ^1^2^ ohm.
Power expended in the conductor
0.0801 ^,. „„,. 1,281,600 ,,
— g — X (4,000)=' = ~ — g watts.
Cost of power thus expended, per year
1,281,600 i„o^^Kn 64,080,,,
- — ^ ■ -^ 1,000X50 = — q — dollars.
Weight of aluminium in conductor
(300X 100) X 8X2.6^1,000 = 78xS. kilograms.
Cost of conductor, if round
78. S.X 0.36X2.205 = 62xS. dollars.
Interest on cost of conductor, at 10% = 6.2xS. dollars.
Sum of cost of power and interest on investment:
^+(6.2XS.)
Solving for S = a minimum : S = 101.2 sq. cm. (2)
Total costs, substituting S above
$633 (for power) +$627 (interest) = $1,260 per year.
(3) Power expended in copper conductor
801,600 .on-^K +4.
—prr-i^ — = 13,035 watts.
61.5
Heat generated in conductor, per second
13,035X0.2389 = 3,114 gm. cal.
Area of conductor surface
(300X 100) X 3.14 (v/61.5h- 0.7854) =
= 30,000X27.8 (cm. circumference) = 834,000 sq. cm.
THE METALLURGY OF COPPER. 571
Loss of heat per second by contact with still air, per 1° difference
0.000056X834,000 = 46.7 calories.
Rise of temperature, excluding radiation loss
3,114-^46.7 = 67°.
The radiation loss must, however, be considered. From
copper it is 0.00068 gram calories per second per sq. centi-
meter, if the surroundings are at 0° and the hot body at 100°,
this is approximately 0.0000068 gram calories per each degree
in this range. We have the combined conduction and radia-
tion losses per 1° per second
0.000056 + 0.000007 = 0 . 000063 calories
and the rise in temperature of the conductor
3,114^(0.000063X834,000) = 59°.
The conductors would therefore run at a temperature of
20 + 59 = 79° C. (3)
= 174° F.
A correction of the second order would be to take into con-
sideration the fact that at 79° the copper would have less con-
ductivity than at 20°, its resistivity at 79° being
97."^ -I- 7Q
0. 00000167 X 1^1^ = 0.0000020 ohm.
This would make the required section for minimum expense
68 sq. cm. instead of 61.5, and would, by increasing the sur-
face, make a different rise in temperature. The power ex-
pended would be the equivalent of 3,373 gram calories per
second, and the revised temperature of the conductor would be
(3,373^52.54) +20 = 64 + 20 = 84° C.
(4) Power expended in the aluminium conductor
1,281,600
101.2
= 12,665 watts.
Heat generated in conductor, per second
12,665X0.2389 = 3,026 calories.
572 METALLURGICAL CALCULATIONS.
Area of conductor in square centimeters = 1,074,000 sq. cm.
Conduction and radiation losses, per 1°
(0.000056 + 0.000010) X 1,074,000 =70.88 grm. cal.
Rise in temperature of conductor
3, 026 -^ 70. 88 = 43°.
Working temperature of conductor
43 + 20 = 63° C.
The aluminium conductor thus is seen to work cooler than
the copper, and to admit of a lower minimum of working costs.
Electric Smelting of Copper Ores.
The use of electrothermal processes for smelting copper ores
has been proposed in recent years, and in a few cases experi-
mentally tested. The whole question depends on the recog-
nition and appreciation of a few fundamental facts. By far
the greater bulk of all copper ores are sulphide ores, and in many
cases the sulphur and iron present are sufficient heat producers
to allow of the smelting of the ore simply by the heat of their
oxidation, if the operation is skillfully conducted. Such is the
basic principle of pyritic smelting, aiid whenever it can be
applied it is very economical to do so, and electric processes
have no chance of an application.
Other ores of copper contain so small an amount of the
metal that the result of the smelting down of the whole to a
fluid mass, by any method of fusion, would not repay the cost
of the operation. Some such ores can be treated by aqueous
and other methods not involving fusion, with a margin of
profit to pay for the operation.
The only field for electrothermic processes appears to be in
the smelting down of ores carrying too little sulphur to be
" pyxitically " smelted, and which require, as usually treated,
the use of carbonaceous fuel to assist the fusion. Wherever such
fuel is expensive, and electric power may be obtained at a low
price, an electrothermic process might be theoretically possible
and profitable.
Such conditions may easily occur in the vicinity of copper
mines. Situated frequently among the mountains, remote from
THE METALLURGY OF COPPER. 573
railways and cheap fuel, they frequently are near large water
powers which would furnish cheap electric power. Some may
be even so situated that concentration of a very lean sulphide
ore to matte by use of fuel, or by mechanical concentration,
would be unprofitable, and yet a concentration or simple melt-
ing to matte by electrically-generated heat be profitable.
There are, so far as the writer knows, no electrothermal
processes yet in commercial operation on copper ores, yet they
have been tested experimentally with promising results. Vat-
tier, for instance, conducted experiments at the Livet works of
the Compagnie Electrothermique Keller et Leleux, on April 23,
1903, in the presence of several distinguished metallurgists,
such as Mr. Stead, of Middlesborough ; Mr. Allen, of Sheffield,
and Mr. Saladin, of the Creusot works. The ores were Chilian
ores sent by the Chilian Government to test the possibility of
electric smelting, taken from the " Vulcan " mine, owned by G.
Denoso, and low-grade ore from Santiago. At the mines, coke
costs $20 per ton, but the slopes of the Andes afford a fine
opportunity for developing large water powers cheaply; it is
estimated that the total cost of electric power delivered should
not exceed $6 per kilowatt-year. A detailed account of these
tests can be found in the appendix to the Report of the Can-
adian Commission on Electric Smelting, 1904.
Problem 122.
Vattier mixed his rich and poor Chilean ores to a charge
consisting of
Per Cent.
Copper 5.10
Sulphur 4.13
Iron 28.50
Manganese 7. 64
Silica 23.70
Alumina 4.00
Carbonic acid 4.31
Lime 7.30
Magnesia 0 . 33
Phosphorus 0 . 05
This was charged directly into a shaft furnace 1.8 meters
long, 0.9 meter wide and 0.9 meter high. The melted material
574 METALLURGICAL CALCULATIONS.
ran out into a forehearth 1.2 meters long, 0.6 meter wide and
0.6 meter high. Two carbon electrodes, 30 centimeters square
by 170 centimeters long, were used in the smelting chamber,
and two electrodes, 25 ceiitimeters square by 100 centimeters
long, were put in the forehearth to reheat it for tapping.
The matte obtained analyzed:
Per Cent.
Silica 0.80
Alumina 0 . 50
Iron 24.30
Manganese 1 . 40
Sulphur 22.96
Copper 47.90
The slag contained:
Silica 27.20
Alumina 5.20
Lime 9.90
Magnesia 0 . 39
Iron 32.50
Manganese 8 . 23
Sulphur 0.57
Phosphorus 0 . 06
Copper 0.10
The current used was 4,750 amperes at 119 volts; power
factor = 0.9; 8,000 kilograms of ore mixture were smelted in
8 hours; consumption of electrodes 50 kilograms.
Required :
(1) A balance sheet of materials entering and leaving the
furnace per hour.
(2) A balance sheet of the heat development and distribution
in the furnace.
(3) The saving in cost of treatment per ton of ore, assuming
electrodes to cost 4 cents per kilogram, coke $20 per metric
ton, and that when coke is used one-third of its calorific power
leaves the furnace in the escaping hot gases. Assume all other
costs similar, except that air blast costs $0.10 per ton of ore
smelted additional. Electric power $6 per kilowatt-year.
(4) If the resulting slag were smelted electrically for ferro-
THE METALLURGY OF COPPER. 575
silicon, at an expenditure of 0.75 electric horse power-year per
ton of ferro-silicon obtained, and other smelting costs were
$10 per ton, how much ferro-silicon would be obtained per ton
of copper ore treated, and what would be its cost ?
Solution :
(1) Balance sheet per 1,000 kg. of ore.
Charges Matte Slag Gases
Ore: 1,000 kg.
Silica 237 kg. 1 236
Alumina 40 " 0.5 39.5
Iron 285 " 25 260
Manganese 76 " 1.5 74.5
Sulphur 41 " 24 5 12
Copper 51 " 50 1
Carbonic acid 43 " .... .... 43
Lime 73 " .... 73
Magnesia 3 " .... 3 ....
Oxygen 137 " 2 74 61
Electrode :
Carbon 6 " .... .... 6
104 766 122
Composition of Slag.
Calculated. Analyzed.
Silica 29 . 5per cent. 27 . 2 per cent.
Alumina 5.0 " 5.2
Lime 9.2 « 9.9 "
Iron 32.5 « 32.5 «
9Q " S 9 "
^^^^^^^^ -^
Sulphur 0.6 " 0.6
Heat Development.
Electric energy per hour:
4,750 X 119 X 0 . 9 = 509 kilowatt-hours.
'509X860 = 437,750 Cal. per hour.
Oxidation of carbon :
6X4,300 = 25,800 « " "
463,550 " " "
676 METALLURGICAL CALCULATIONS.
Heat Distribution.
Heat in matte:
104X270 = 28,080 Calories.
Heat in slag:
796X400 = 318,400
Heat in gases = 30,000 "
Loss by radiation
and conduction = 87,070
463,550
The radiation and conduction loss is 19 per cent of the total
heat generated in the furnace — a satisfactory showing. The
useful effect of the furnace is nearly 75 per cent, reckoning as
usefully applied heat that contained as sensible heat in the
melted matte and slag. Nearly 90 per cent of this usefully
applied heat is contained in the slag.
(3) If coke werq used instead of electric heat, enough coke
would have to be used to furnish 433,550 Calories plus the heat
in the gases. Since the latter is assumed to be one-third of its
calorific power, the coke must have a calorific power of 650,325
Calories, which at 7,000 Calories per kilogram would require
93 kg. of coke per 1,000 of ore smelted. The costs of coke and
blowing engine would therefore be
Coke, 93X0.02 = $1.96
Engine, 0.10
Sum $2.06
The cost of electrodes and electric energy, which would
replace coke an4 blowing engine, would be:
Electrodes, 6 kg. X $0.04 = $0.24
Power 509kw.X ^„?„= 0.35
Sum, $0.59
Saving, $1.47
(4) The resulting slag contains, according to the balance
sheet, 260 kg. of iron, 74.5 kg. of manganese, and 236 kg. of
silica equal to 126 kg. of siHcon. It should produce, if com-
THE METALLURGY OF COPPER. 677
pletely reduced, 460 kg. of alloy (omitting the carbon it might
contain) of the approximate composition:
Fe
260 kg.
= 56 per cent.
Mn
74 kg.
= 13
Si
126 kg.
460 kg.
= 28
Cost of power
$6.00X1 =
$4 . 50 per ton.
Other costs
10.00 "
Total costs per ton $14 . 50
This cost is very low, because of the very low figure assumed
for power cost. However, the whole calculation points to
the possibility of great saving in some localities in smelting
down copper ores by electrically applied heat, and the possi-
bility of cheaply producing, at the same spot, valuable ferro-
silicon as a by-product.
CHAPTER II.
THE METALLURGY OF LEAD.
The extracting of lead from its ores and the refining of the
crude metal to commercial lead, constitutes the metallurgy of
lead. The chief ore is galena, lead sulphide, PbS; but the
carbonate, cerussite, PbCO^, and the sulphate, Anglesite, PbSO*,
occur at some places in important quantities; the silicate, phos-
phate, molybdate, tungstate, chloride and native lead are rare
minerals.
The reduction of the oxidized lead ores, such as often occur
at the outcrops of sulphide veins, is very simple; carbon, the
cheapest and most universal reducing agent, reduces them
satisfactorily at a red heat, with some loss of lead by volatiliza-
tion. The sulphide of lead, however, is not reducible by carbon;
it requires other treatment. If we tabulate the most common
sulphides according to their thermochemical heats of formation,
expressed per unit weight of sulphur held in combination (32
kilograms), we find the common metals arranged as follows:
Calories.
Potassium (K^ S) 103,500
Calcium (Ca, S) 94,300
Sodium (Na^ S) 89,300
Manganese (Mn, S) 45,600
Zinc (Zn, S) 43,000
Cadmium (Cd, S) 34,400
Iron (Fe, S) 24,000
Cobalt (Co, S) .- 21,900
Copper (Cu^ S) 20,300
Lead (Pb, S) 20,200
Nickel (Ni, S) 19,500
Mercury (Hg, S) 10,600
Hydrogen (H^ S) 4,800
Silver (Ag^ S) 3,000
578
THE METALLURGY OF LEAD. 579
A glance at this table shows us that some metals unite with
sulphur more energetically than lead does, and others less
energetically, and points to the theoretical possibility of de-
composing lead sulphide by the agency of copper, cobalt, iron,
cadmium, zinc, etc. Of these agents, iron is, of course, the
cheapest and most available, and can be used to reduce lead
sulphide with formation of iron sulphide.
PbS + Fe = FeS + Pb
- 20,200 + 24,000
Showing an excess heat development per 207 parts of lead
set free, of 24,000 - 20,200 = 3,800 Calories. If lead sulphide
and iron are brought together at a red heat, at which heat the
sulphide is molten, they react energetically with evolution of
heat — enough theoretically to raise the temperature of the
207 kg. of lead and 88 kg. of iron sulphide some 280° C. The
reaction is fairly complete if sufficient reducing agent is present
and time is allowed, so that it can be used in assaying to deter-
mine lead by fire assay, but in commercial practice more or less
undecomposed lead sulphide always remains in the iron sul-
phide, forming a sort of double sulphide or iron-lead matte.
It is interesting to note, en passant, that lead vigorously
reduces silver sulphide, the liberated silver alloying with the
excess of lead. This reaction is accompanied by a large evolu-
tion of heat, as the table shows, is the basis of the ordinary
assaying methods for silver ores, and is used commercially to
extract silver from its ores wherever lead is plentiful.
The affinity of lead for oxygen is a no less interesting sub-
ject, since it concerns not only the reduction of oxide ores
but also the roasting of sulphide ores to oxide, and very largely
controls the refining of impurities from lead by methods in-
volving oxidation. The heat of combination of some of the
more common elements with oxygen, expressed per unit weight
of 16 kilograms of oxygen held in combination, is as follows:
Calories.
Magnesium (Mg, O) 143,400
Calcium (Ca, O) 131,500
Aluminium ^ (AP, O^) 130,870
Sodium (Na^ O) 100,900
Silicon J (Si, O^) 98,000
580 METALLURGICAL CALCULATIONS.
Calories.
Manganese (Mn, O) 90,900
Zinc(Zn, O) 84,800
TinKSn, 0^) 70,650
Iron (Fe, 0) 65,700
IronJ(Fe^O') 65,200
Nickel (Ni, 0) 61,500
Hydrogen (H^ 0) 58,060
Antimony ^(Sb^ O^) 55,630
Arsenic \{As\ 0') 52,130
Lead (Pb, 0) 50,800
Carbon \{Q,, 0^) 48,600
Bismuth iCBi^, 0=) 46,400
Copper (Cu^ 0) 43,800
Sulphur KS, O^) 34,630
Sulphur ^(S, 0=) 30,630
Carbon (C, O) 29,160
Mercury (Hg, O) 21,500
Silver (Ag^ O) 7,000
Reduction of Lead Oxide. — An inspection of above table
shows that lead oxide is a weak oxide, weaker than the oxides
of most of the common metals. It is reduced to metallic lead
by many reagents with evolution of heat. It is also reduced
by some weaker reagents with absorption of heat, provided
that the oxide formed is gaseous and the necessary heat energy
is supplied from outside. For instance,
PbO + C = Pb + CO
- 50,800 +29,160
involves an absorption* of 21,640 Calories, or several times as
much heat as is necessary to raise the reacting substances,
PbO and C, to the reacting temperature, a red heat. The
reaction is endothermic, absorbing heat, and therefore only
progresses in measure as the necessary calories are supplied
from outside. The fact that all the substances concerned are
liquid or solid except the product CO, gives a predisposing
cause which facilitates the reaction, that is, the reaction can
only go one way, since the CO gas escapes from the sphere of
action as soon as formed.
THE METALLURGY OF LEAD. 681
Oxidation of Lead Sulphide. — -When PbS is roasted, that is,
heated to redness with free access of air, both the sulphur and
the lead tend to oxidize, generating a large heat of oxidation,
against which there is absorbed only the comparatively feeble
heat of decomposition of PbS. The equation may be discussed
as
2PbS + 302 = 2PbO + 2SO='
- (20,200) + 2(50,800) +2(69,260)
The net heat evolved in the equation is 240,120 — 40,400 =
199,720 Calories, which is 33,290 Calories per unit weight (16
kg.) of oxygen used, or 418 Calories for each kilogram of lead
sulphide oxidized. This great heat of oxidation, evolved in
roasting, is quite sufficient, in fact more than sufficient, to
provide the heat required for self-roasting without the use of
any other fuel, the chief difficulty is really to keep the charge
from getting too hot and melting everything down to a liquid
before the roasting is anywhere near complete. In the ordinary
hand-worked reverberatory roaster, the oxidation is so slow
that the fire on the grate really controls the temperature on the
hearth, and the temperature of the roasting ore can be regu-
lated accordingly. Where the roasting is done quickly, by an
air blast, as in " Pot Roasting," the temperature is kept down
somewhat by previously roasting off part of the sulphur, or by
liberally wetting the charge, or by using limestone in with the
ore, to absorb by its decomposition into CaO and CO^ a large
part of the excess heat — while the melting down of the charge
is prevented by the mechanical interference presented by inter-
mixed, inert and infusible material, such as siUca, lime, etc.,
which simply prevents the really melted globules of sulphide from
running together and thus melting down to a liquid mass.
An interesting variation is the roasting of lead sulphide to
sulphate; some of this alwaj^s forms, due to the high formation
heat and difficult decomposabiUty of the sulphate.
PbS + 202=PbSO«
-(20,200) +(215,700)
The net heat evolved is 215,700 - 20,200 = 195,500 Calories,
or 48,875 Calories per unit weight (16 kg.) of oxygen used, or
818 Calories for every kilogram of lead sulphide so oxidized.
682 METALLURGICAL CALCULATIONS.
This is a higH heat of oxidation, and explains the great ten-
dency to form sulphate observed during roasting. If the con-
ditions could be found whereby only this reaction occurred, the
sulphate roasting could be easily made automatic without out-
side fuel. There is needed, at the present, time, a careful lab-
oratory investigation of the conditions,, mechanical, physical,
chemical and thermal, for the roast exclusively to sulphate —
just as a bit of badly needed metallurgical information.
Double Reactions.— A. large part in the metallurgy of lead is
played by double reactions, such as the following:
PbS + PbSO* = 2Pb +2S02
PbS + 2PbO = 3Pb +80==
PbS + 3PbS0^ = 4PbO + 4S02
On roasting lead sulphide for a short time, either lead or
lead sulphate or mixtures of these are formed, according to the
temperature and excess of air provided. With large excess of
air and low temperature, and especially in presence of infusible
materials which act as catalyzers {i.e., which promote the
union of SO^ with O, and consequent formation of SO' and
PbSO^), sulphate may be formed almost exclusively. If the
temperature is then raised, the tendency of PbSO* to react
upon the undecomposed PbS rapidly gets stronger, until at an
orange heat this takes place rapidly and fairly completely,
forming the one single gaseous product, SO^.
PbS + PbSO< = 2Pb + 2S02
- (20,200) - (215,700) -1-2(69,260)
Net deficit. 97,380 Calories.
PbS-|-2Pb0 = 3Pb-|-S02
- (20,200) - 2(50,800) + (69,260)
Net deficit, 52,540 Calories.
In both these cases the well-known phenomena of an endo-
thermic reaction are manifest — the high temperature and
strong firing necessary, and the formation of a single gaseous
product from non-gaseous materials, assisting the reaction.
The reaction:
PbS + 3PbS0< = 4Pb0 + 4S0='
- 20,200 - 3(215,700) -F 4(50,800) 4-4(69,260)
Net deficit, 187,060 Calories,
THE METALLURGY OF LEAD. 683
is supposed to take place in " pot roasting " where excess of air
is blown through the finely divided material, but it is too endo-
thermic a reaction to take place in a pot-roasting operation to
more than a very subsidiary extent.
Oxidation Refining. — Impure lead is refined or " softened " by
oxidation at a red heat. "We must expect a great deal of lead
to be oxidized in this operation, simply because of its pre-
ponderating mass, but the impurities present will oxidize rela-
tively faster or slower than lead in proportion as their affinity
for oxygen is relatively greater or less. The skimmings or
slags obtained during softening are always principally com-
posed of lead oxide, PbO, but they come off containing, in
order, zinc, tin, antimony, arsenic, bismuth and small quantities
of silver. .During cupellation down to silver, which is con-
tinued oxidation until all the lead is oxidized, bismuth oxide is
concentrated in the last parts of lead oxide formed, which may
also carry silver in small amount; before this happens, how-
ever, the lithage formed is almost chemically pure.
The converse of these oxidation reactions also holds, viz.:
differential reduction. Taking a softening skimming rich in
antimony, for instance, it is possible by mixing it with a small
amount of reducing agent, such as carbon, to reduce out of it
all the silver and considerable of the lead which it contains
without reducing much antimony. This leaves the remaining
unreduced material desilverized and poorer in lead, or richer
in antimony, and on subsequent reduction of this by excess of
reducing agent a rich antimony-lead alloy is obtained. The
easy reducibility of lead oxide is complementary to the slow
oxidation of lead ; both facts are clear from the heat of formation
of the various metallic oxides, and both are extensively utilized
in the metallurgy of lead.
The Volatility of Lead.
The melting point of lead is 326°, its mean specific heat in
the solid state 0.02925 -|-0.000019t, heat in melted lead at its
melting point 11.6 Calories, latent heat of fusion 4.0 Calories,
heat in just melted lead 15.6 Calories, specific heat in the Hquid
state 0.042 — and approximately constant — boiling point at
normal atmospheric pressure about 1,800° C, latent heat of
vaporization, calculated by Trouton's rule (23 T) 47,680
584 METALLURGICAL CALCULATIONS.
Calories per molecular weight = 230 Calories per kilogram,
assuming the vapor monatomic, specific gravity of vapor 103.5,
referred to hydrogen gas at the same temperature and pres-
sure, or, theoretically, 9.315 kg. per cubic meter at 0° C. and
760 mm. pressure, as a standard datum.
The question of the volatility of lead at other temperatures
than 1,800° C. is highly important in the smelting of lead ores,
yet is practically an unknown quantity. The following is an at-
tempt to calculate these data, so important in practical metal-
lurgy.
The vapor tension curve of mercury is known for very low
and up to comparatively high pressures. A rule has been
observed between mercury and water vapor, in that the absolute
temperatures at which these two substances have the same vapor
tensions are found to stand in the ratio 1.7 to 1 through a
large range of temperatures. Since lead vapor is in all prob-
ability monatomic, like mercury vapor, we will deduce the
vapor tension curve of lead from that of mercury, using the
constant ratio derived from the two temperatures at which
their respective vapors have atmospheric tension, viz.:
TPb ^ 1,8004-273 _ 2,073 _
THg 357 -F 273 630
The following table gives the most reliable data for the
vapor tension curve of mercury, and the corresponding data
calculated for lead, assuming the constant ratio 3.3 between the
absolute temperatures at which they have the same vapor
tension :
Tension of Vapor
mm. of Hg.
0.0002
0.004
0.045
0.28
1.47
5.73
18.25
50.
106.
lercury
Lead
C°
C°
0
625
33
735
67
844
100
954
133
1,064
167
1,173
200
1,283
233
1,393
267
1,502
THE METALLURGY OF LEAD. 585
Tension
of Vapor
Mercury
Lead
mm.
ofHg.
C°
C°
242.
300
1,612
484.
Atmospheres.
333
1,722
760.
= 1.0
357
1,800
849.
= 1.1
367
1,841
1588.
= 2.1
400
1,951
4.3
450
2,116
8.0
500
2,280
13.8
550
2,445
22.3
600
2,609
34.
650
2,774
50.
700
2,938
72.
750
3,103
102.
800
3,267
137.5
850
3,436
162.
880
3.525
From the above table a vapor pressure curve of mercury and
lead can be constructed. An examination of the data shows
that lead is certainly volatile to a minute extent at a low red
heat, and that a current of inert gas passing across the surface
of melted lead at that temperature certainly carries away
vapor of lead; at the melting point of silver the tension is only
about one-quarter of a millimeter, or one three thousandth of
an atmosphere, yet this means that each cubic meter of inert gas
carries off one three-thousandth of its volurhe, or one-thirtieth
of Iper cent of its volume, of lead vapor. At 1,300°, the tem-
perature of a commercial zinc retort, or of a lead smelting
furnace, the tension is about one-fortieth of an atmosphere,
which would mean that any other gas or vapor could carry
2.5 per cent of its volume of lead vapor with it. It must be
remembered, moreover, that such gas saturated with lead
vapor if suddenly cooled does not deposit its excess of lead
vapor as Hquid lead, but that a suspension similar to hoar-
frost almost inevitably results, the so-called " lead fume,"
which is simply molecularly divided Uquid or sohd lead carried
in suspension by a current of gas. The lead vapor is thus al-
most entirely carried out of the furnace without condensation
and deposition.
Looking at the higher pressures, we can understand why an
586 METALLURGICAL CALCULATIONS.
explosive reaction results when PbO is reduced by finely di-
vided aluminium, the " aluminothermic " reaction. The im-
mense heat liberated in the reaction,
3PbO + 2Al = 3Pb + AP0',
220,000 Calories, raises the products to an electric furnace tem-
perature, to some 3,000° C, at which temperature the lead
vapor has a maximum tension, according to our table, of 60
atmospheres. No wonder, then, that when Tissier tried this
test for the first time, in 1857, using a piece of sheet aluminium
weighing less than 3 grams (0.1 ounce), " le creuset a ^t^ hns6
en mille pieces et les portes du foumeau projet^es au loin " —
the crucible was broken into a thousand pieces and the doors
of the furnace blown to a distance.
Roasting of Lead Ores.
The principal operations in the metallurgy of lead are the
roasting of the ore, its reduction to metal and the refining of
the crude metal.
The chief ore of lead being galena, PbS, the operation of
roasting it in air converts it partly into PbO and partly into
PbSO*. Since PbS is easily fusible, and is volatile per se at
a yellow heat, it is necessary to roast carefully and slowly,
avoiding high temperatures, which first make the ore sticky
or pasty and afterwards fuse it, thus practically stopping the
roasting reaction. The only manner in which rapid roasting
can be done is by having the ore mixed with so much infusible
inert matter, like lime, that the globules of melted sulphide
cannot run together, but are kept isolated and continue to
oxidize on their surfaces, the mass being meanwhile kept " open "
or porous by the inert, infusible material, to allow the rapid
passage of air through the mass. This is the principle of " pot
roasting " — the most radical improvement of recent years in
the metallurgy of lead.
Problem 123.
A Savelsberg " pot roaster " treats 5 tons of ore mixture,
consisting of 100 parts lead ore, 10 parts quartzose silver ore,
10 parts spathic iron ore, 19 parts limestone. The lead ore is
galena ore, containing 78 per cent of lead and 15 per cent of
THE METALLURGY OF LEAD. 587
sulphur; the silver ore may be called silica sand; the spathic
iron ore FeCO^ the limestone CaCO^ The mixture is mois-
tened with 5 per cent of its weight of water. Air blast is kept
constant at 7 cubic meters per minute, blower displacement at
15° C, and the operation lasts 18 hours, leaving 2 per cent of
sulphur in the product, which may be assumed as undecom-
posed PbS. Gases produced contain approximately 10 per
cent of SO^ and 5 per cent free oxygen; 50 kg. of charcoal is
used to start the operation, and is assumed to be pure carbon.
Required:
(1) The weight of product and its percentage composition.
(2) The complete analysis of the gases escaping.
(3) The efficiency of delivery of the blower.
(4) The proportion of the heat generated by the oxidation
of the ore which is absorbed in the decomposition of the lime-
stone.
(5) The proportion of the heat generated in the pot used
in evaporating the moisture of the charge.
(6) The proportion of the heat generated in the pot carried
away by the gases at an average temperature of 300° C.
(7) Make a heat balance sheet of the whole operation.
Solution :
(1) The components of the 5-ton (=5,000 kilos) charge are:
Galena 3597 kg. = Pb 2806 kg. S : 540 kg.
Silver ore 360 " = SiO^:
Iron ore 360 " = FeO:
Limestone 683 " = CaO :
360 "
224 " CO^: 136
382 " CO^: 301
Approximate composition of product:
Pbo 2^°^><lo? = ^°^^ ^^■
SiO^ = 360 "
FeO = 224 "
CaO = 382 "
3989
But, since product contains 2 per cent of sulphur as unde-
composed PbS, the above weight is only 99 per cent of the
588 METALLURGICAL CALCULATIONS.
weight of the roasted ore, because the lead combined with this
sulphur has been calculated above to PbO, and the O in PbO
is only half the weight of the S in PbS. The above weight is
short, therefore, by an amount equal to one-half the weight of
the sulphur present, and since the latter is 2 per cent of the
roasted ore the above weight is 1 per cent short. The cor-
rected weight of the roasted ore is, therefore,
3,989-4-0.99 = 4,029 kg.,
and its sulphur content 81 kg., corresponding to 521 kg. of
lead, or 602 kg. of PbS. This leaves present, as PbO, 2,806-
521 = 2,285 kg. of lead, equal to 2,285X223/207 = 2,461 kg.
of PbO.
Revised composition of product:
PbO 2461 kg. = 61 per cent
PbS 602 " = 15
SiO^ 360 " = 9
FeO 224 " = 5 "
CaO 382 " = 9 "
4029 " (1)
(2) The charge gives to the gases
ffO 5000X0.05 = 250 kg.
CO^ (50X3.67) -f 136 -f 301 = 620 "
S 540 - 81 = 459 "
and the blast gives to the charge
0 2461 - 2285 = 176 kg.
The volume of SO^ produced from 459 kg. of sulphur, which
makes 918 kg. of SO^ will be, at standard conditions,
918-f- ^0. 09 X y) = 319 cubic meters.
The free oxygen in the gases is half the volume of the SO',
and, therefore, will be exactly one-quarter of its weight.
The volumes of H'O and CO' present in the gases, at standard
conditions, are
THE METALLURGY OF LEAD. 589
H^O 250-H ^O.OQXy) = 309 cubic meters.
CO^ 620 H- ^.09Xy) = 313 "
The nitrogen present is that in the air used, or 10/3 the
weight of oxygen in the air. The latter is the weight of oxygen
in the PbO of the roasted charge, plus the oxygen necessary to
bum the 50 kg. of charcoal, plus the oxygen oxidizing sulphur
to SO^ and the free oxygen in the gases.
O in PbO = 176 kg.
OtobumC ^0Xl2 = ^^^ "
OtobumS ^^9><32 = ^^^ "
Free O' in the gases = 230
Total 998 "
N^ corresponding = 3327
Composition of escaping gases:
Per Cent.
Nitrogen 3327 kg. = 2609 cu. meters = 70.3
Oxygen 230 " = 159 " " = '4.3
S02 918 " = 319 " " = 8.6
CO^ 620 « = 313 « " =8.5
H'O vapor 250 " = 309 « " = 8.4
3709 100.
(2)
(3) Oxygen delivered by blast 998 kg.
Nitrogen in the blast 3327
Air delivered 4325
Volume at 0° = 4325^ 1.293 = 3345 cu. meters.
Volume at 15° = 3345 X g| = 3529 " "
590 METALLURGICAL CALCULATIONS.
Displacement of blower at 15°
7X60X18 = 7560 " «
3529
Efficiency of blower =^^ = 0.467 = 46.7 per cent. (3)
(4) The heat generated in the oxidation of the ore consists
of the heat of formation of the PbO formed, plus that of the
SO^, and less that of the PbS decomposed. This is not the
complete heat balance of the whole operation, but simply
that concerned with the oxidation of the galena.
Heat of oxidation of Pb to PbO
2285 kg. Pb.X 245 =+ 559,825 Cal.
Heat of oxidation of S to SO^
459 kg. SX2164 =+ 993,275 "
Heat of formation of PbS decomposed
2285 kg. PbX 98 =- 223,930 "
Roasting heat = +1,329,170 "
The decomposition- of the limestone, producing 301 kg. of
CO^, absorbs
301X1,026 = 308,825 Cal.
The proportion of the roasting heat thus absorbed is
Of)Q pOK
^^^^2^ = 0.232 = 23.2 per cent. (4)
(5) The 5-ton charge is moistened with 5 per cent of its
weight of water, or 250 kgs., in order to keep down the tem-
perature. This absorbs, simply in becoming vapor,
250X606.5 = 151,625 Cal.,
Which, on the heat generated in the roasting operation, is
151,625
1,329,170
= 0.114 = 11.4 per cent. (5)
(6) If the gases pass away at an average temperature of 300°,
and the air comes in at an average temperature of 15°, the
THE METALLURGY OF LEAD. 591
gases cool off the pot to the extent of the difference between
these two heat capacities:
Heat in the air entering:
3345m^X[0.303 + 0.000027(15)]X15 = 15,225 Cal.
Heat in gases escaping:
W 2609 m^) ,
Q2 jggj^aj X[0.303 + 0.000027(300)] = 861.1 Cal. per 1°.
S,0^ 319 m= X[0.36 + 0.0003(300)] = 143.6
C0=' 313 m^ X[0.37 + 0.00022(300)] = 136.5
ffO 309 m^ X[0.34 + 0.00015(300)] = 119.0
1260.2
Total = 1260.2X300 = 378,060 Cal.
Proportion of the total roasting heat thus carried out:
T^^O = «-280 = 28.0 per cent. (6)
(7) Heat Available. Calories.
Sensible heat of air blast, at 15° 15,225
Combustion of charcoal 405,000
Net heat generated by the roasting 1,329,170
Total 1,749,395
Heat Distribution.
Sensible heat in the hot gases 378,060 = 22 per cent.
Decomposition of carbonates :
Calcium carbonate 308,825 = 18
Iron carbonate. . .'. 76,980 = 4
Evaporation of moisture 151,625 =9 "
Sensible heat of product, at 400°. . . . 400,000 = 23 "
Loss by radiation and conduction. . . 433,905 = 25 "
1,749,395
Reduction of Roasted Ore.
If the roasted ore is reduced in reverberatory furnaces, the
chances are that the undecomposed sulphide it contains will
react with the oxide or sulphate, forming metal and SO^, and
592 METALLURGICAL CALCULATIONS.
thus eliminating almost all of the sulphur left in the roasted
ore; this results in the formation of no matte, or at most of
very little rich matte. If the roasted ore is reduced in shaft
furnaces, the carbonaceous fuel and strong reducing atmos-
phere of the same, tend to reduce oxides to metal and sul-
phates to sulphides before they reach the temperature at which
they can react on sulphide, thus cutting out very largely the
double reaction, ajid throwing into matte the larger part of
the sulphur left in the roasted ore. This is highly advan-
tageous if the ore has copper in it, even if only in small amount,
as the matte will concentrate the copper in it, is easily sep-
arated from the metal and the slag, and forms a valuable by-
product. Since most lead ores contain some copper, enough
to make it pay to save it, the roasting of these is never pushed
to completion, and the reduction in low shaft furnaces, run
at moderate temperature, is pursued with the object of pro-
ducing a copper-iron matte. Any arsenic or antimony left in
the roasted ore is likely also to form spiess, a compound of
arsenic and antimony with iron, copper, nickel, cobalt, lead
and silver, which is heavier than matte, but lighter than lead,
and separates out in the cooling pots between these. It is a
highly undesirable product, because of its complexity and the
difficulty of satisfactorily and cheaply separating its constitu-
ents; it is always to be advised to remove arsenic and anti-
mony as completely as possible in the roasting operation, even
at the expense of removing too much sulphur, and then to supply
the sulphur needed in the smelting operation by mixing with
the dead-roasted ore some raw sulphide ore free from arsenic
and antimony, if such can be obtained.
In calculating the charge for such a smelting operation, the
roasted ore must be charged with such amounts of iron ore and
limestone as will, together with the gangue of the ore and the
ash of the fuel used, make an easily fusible slag. To produce
such a slag is of fundamental importance, since it must melt
and become thinly liquid at the moderate temperature neces-
sarily prevailing in a lead furnace. Experience has shown
that a typical lead slag will contain about 30 per cent SiO^ 40
per cent FeO and 20 per cent CaQ, with 10 per cent allowed
for other ingredients, such as APO', etc., or in more general
terms, that SiO^ FeO and CaO are best present in the proper-
THE METALLURGY OF LEAD. 693
tions 30 : 40 : 20. The other limitations are that a certain
amount of hand-picked slag is always returned to the furnace
for re-smelting and purification from matte, and that the fuel,
coke, cannot melt more than a certain amount, say seven
times its weight of inert material. With these conditions in
mind, we will work out a typical smelting problem, the data
for which are taken from Hofman's Metallurgy of Lead.
Problem 124.
A mixture of lead carbonate ore and galena (which repre-
sents pret£y closely a partially roasted sulphide ore) is smelted
with the addition of iron ore and limestone. The charges for
the furnace consist of 1,000 pounds of burden (material to be
smelted) and 150 pounds of coke (containing 10 per cent ash),
and the 1,000 pounds consist of 100 pounds return slags and
900 pounds of lead ore, iron ore and limestone. The percentage
composition of these materials is :
Lead Ore. Iron Ore. Limestone. Ash of Cake.
SiO' 32.6 4.3 2.7 40.3
FeO 14.8 72.4 4.5 26.5
MnO 4.3 1.7
CaO 2.2 3.1 37.3 6.9
MgO 5.3 .... 11.9 2.4
APO' 2.5 .... .... 20.4
BaO 1.5
ZnO 2.4 The iron present in these mate-
S 4.4 rials is really present mostly as
As 0.5 Fe^O^ and only partly as FeO, but
Pb 20.7 the analysis is expressed usually as
Cu 2.9 FeO in order to facilitate the slag
Ag 0.17 calculations. The analyses, as given,
are not expected to add up to 100.
In making the calculations, assume that the slag to be pro-
duced contains SiO^:. FeO: CaO in the ratio 30 : 40 : 20; that
the FeO here expressed includes the MnO, calculated to its
FeO equivalent, and the CaO includes the MgO, BaO, and
ZnO calculated to their CaO equivalent. Assume also that
all the ZnO of the charge goes into the slag, all the sulphur
into matte, all the arsenic into speiss, of the formula Fe^As,
all the silver and lead into the lead bullion.
594 METALLURGICAL CALCULATIONS.
Requirements :
(1) The proportions of lead ore, iron ore and limestone to
be used in the 900 pounds of these to a charge.
(2) The weight and composition of the slag formed.
(3) The weight and composition of matte formed.
(4) The weight and composition of speiss formed.
(5) The weight and composition of lead bullion formed.
(6) A balance sheet of the essential materials entering and
leaving the furnace.
Solution :
(1) There are really only two unknown weights to be de-
termined, for the sum of the three weights required is to be
900. Similarly, the ratio 30 : 40 : 20 really gives two condi-
tions to be fulfilled, since there are practically two ratios to
be worked to. The simplest method of solution is, undoubt-
edly, the algebraic one, letting x and y represent two of the
ores, 900— (x-|-y) the other, and then working out the weights
of SiO^, FeO and CaO in the slag, expressed in terms of x and y.
The ratio 30 : 40 : 20 then gives us two independent equations,
containing the two unknowns, and the problem is solved.
Let X = weight of lead ore.
y = weight of iron ore.
900 — X — y = weight of liniestone.
15 = weight of ash of coke.
100 = weight of return slags.
Calculation of FeO going into Slag.
Sulphur in 100 parts ore
Copper in 100 parts ore
Sulphur united with Cu to form Cu^S
Sulphur left over to form FeS
56
Iron needed to form FeS = 3.7 X^ = 6.5
72
FeO corresponding to this Fe = 6.5X-?^ =8.4
Arsenic in 100 parts of ore 0 . 5
280
Iron required to form Fe^As = 0.5 X -=,-=- = 1.9
4.4 lbs,
2.9
tl
0.7
U
3.7
it
'I
THE METALLURGY OF LEAD. 595
72
FeO corresponding to this Fe = 1.9 X-^TT = 2.4 lbs
56
Total FeO disappearing in matte and speiss =10.8 "
FeO left over to go into slag = 14.8— 10.8 = 4.0 "
Therefore x parts of lead ore contributes to the slag:
SiO^" 0.326 X
FeO 0.040 X
MnO 0.043 x = 0.044 x FeO.
CaO 0.022 X
MgO 0.053 X = 0.074 x CaO.
APO= 0.025 X
BaO 0.015 X = 0.006 x CaO.
ZnO 0.024 X = 0.017 x CaO.
The FeO equivalent of MnO is 72/71 the MnO; the CaO
equivalent of MgO is 56/40 the MgO, of BaO is 56/141 the
BaO, of ZnO is 56/81 the ZnO. These ratios are the chem-
ically equivalent values of these oxides, as taken from their
molecular weights. Adding all these together, our lead ore
contributes to the slag the equivalent of
SiO^ 0.326 X, FeO 0.084 x, CaO 0.119 x.
The y parts of iron ore similarly contributes to the slag:
SiO^ 0.043 y, FeO 0.741 y, CaO 0.031 y.
The 900 — x — y parts of limestone contributes likewise;
SiO^* 0.027(900- x-y)
FeO 0.045(900- x-y)
CaO 0.373(900- x-y)
MgO 0.119(900 -x-y) = 0.167(900 -x-y) of CaO.
Therefore, total contribution of the limestone to the slag:
SiO^ 0.027(900 -x-y), FeO 0.045(900 -x-y),
CaO 0.540(900- x-y).
Contribution of ash of fuel to the slag, in similar manner:
SiO^ 6.1 FeO 4.0 CaO 1.4
596
METALLURGICAL CALCULATIONS.
Adding all these together, we have in the slag:
SiO^ 30.4 + 0.299 x + 0.016 y
FeO 44.5 + 0.039 x + 0.696 y
CaO 487.6 -0.421 x -0.509 y
According to assumption, these ingredients as thus summated
should bear the relations 30 : 40 : 20. We therefore have
30.4 + 0.299 x + 0.016 J = ^ (487.6 - 0.421 x- 0.509 y)
40
44.5 + 0.039 x + 0.696 y = |^ (487.6- 0.421 x- 0.509 y)
Whence x = lead ore = 623 lbs.
y = iron ore = 274 "
900 - X - y = limestone = 103 " (1)
(2) With the weights of materials used, as calculated, we can
calculate the weights of the ingredients of the slag as:
Lead Ore.
Iron Ore.
Limestone.
Coke Ash.
Total.
SiO''
.... 170.5
11.8
2.8
6.1
191.2
FeO
.... 20.9
198
4
4.6
4.0
227.9
MnO....
.... 22.5
4
7
27.2
CaO
.... 11.5
8
5
38.4
1.0
59.4
MgO
27.7
. .
.
12.3
0.4
40.4
APO'
13.1
•
3.1
16.2
BaO
7.8
7.8
ZnO
.... 12.6
•
12.6
582.7
= 4.8 per cent F_eO.
Percentage Composition and Check on Ratio.
SiO' 32.8 per cent
FeO 39.0
MnO 4.7
CaO 10.2
MgO 6.9
APO' 2.8
BaO 1.3
ZnO 2.2
Summated SiO^ : FeO : CaO
= 32.8 : 43.8 : 21.9 = 30 :
= 9.7
= 0.5
= 1.5
CaO.
CaO.
CaO.
40 :20
THE METALLURGY OF LEAD. 597
(3) Sulphur in 523 lbs. of lead ore = 23 0 lbs
Copper in 523 lbs. of lead ore = 152 " '
Sulphur to form Cu^S with this Cu = 3^3 "
Sulphur to form FeS = 23.0 - 3.8 = 19.2 «
Fe to form FeS = 19.2 X ^ = 33.6 «
FeS in matte =19.2 + 33.6 =52.8 "
Cu^S in matte = 15.2 + 3.8 =19.0 "
Total weight of matte = 71.3
(3)
Composition of matte
Cu^S = 26.5 per cent = Cu =21 per cent.
FeS = 73.5 « Fe = 47
S =22 " (3)
(4) Arsenic in 523 lbs. of lead ore = 2.6 "
FetoformFe'As =2.6X 7^ = 9 7 «
75
Weight of speiss formed = 12.3 "
Composition: Fe 79 per cent
As 21 " (4)
(5) Lead in 523 lbs. of lead ore = 108.3 lbs.
Silver in 523 lbs. of lead ore = 0.9 "
109.2 "
Composition : Pb 99.2 per cent.
Ag 0.8 " (5)
(6) Balance sheet of materials entering and leaving per 1,000
lbs. of burden:
Charges. Bullion. Matte. Speiss. Slag. Gases,
Lead Ore 623
SiO^ 170.5 170.5
FeO 77.4 ... Fe 33.6 Fe 9.7 FeO20.9 0 13.2
MnO 22.5 22.5
CaO 11.5 11.5
598
METALLURGICAL CALCULATIONS.
Charges.
MgO
27.7 ..
APO^
13.1 ..
BaO
7.8 ..
ZnO
12.6 ..
S
23.0 ..
As
2.6 ..
Pb
108.3 108
Cu
15.2 ..
Ag
0.9 0.
Iron Ore
274
SiO^ .
11.8 ..
FeO
198.4 ..
MnO
4.7 ..
CaO
8.5 ..
Limestone
103
SiO^
2.8 ..
FeO
4.6 ..
CaO
38.4 ..
MgO
12.3 ..
Ash of Fuel
15
SiO^
6.1 ..
FeO
4.0 ..
CaO
1.0 ..
MgO
0.4 ..
APO'
3.1 ..
Return Slag
100 ..
Bullion. Matte. Speiss. Slag. Gases.
27.7
13.1
7.8
12.6
23
15
11.8
198.4
4.7
8.5
2.8
4.6
38.4
12.3
6.1
4.0
1.0
0.4
3.1
100.0
109.2 71.8
12.3 682.7
13.2
The Electrometallurgy of Lead.
As far as the present, the use of electrical methods in the
metallurgy of lead has been confined to the Salom process of
cathodic reduction of galena and the Betts process of refininj
crude lead bullion. Electro.thermic methods of smelting, and
even of roasting, are among future possibiUties, as are also the
leaching by suitable solvents and electrical precipitation of
lead from the solutions, but they have not been commercially
practiced.
The Salom process puts- the powdered galena on a " hard-
THE METALLURGY OF LEAD. 699
lead " plate, in a solution of dilute sulphuric acid, and using
a " hard-lead " anode, passes a strong current through, reducing
PbS in situ to Pb with formation of H^S (with some hydro-
gen) at the cathode and oxygen at the anode. It was run
commercially on a fair scale at Niagara Falls. For further
details reference may be made to Transactions American Elec-
trochemical Society, Vol. I, p. 87; II, p. 65; IV, p. 101.
The Betts process consists in using impure lead as anode
in a solution of lead fluo-silicate, PbF^. SiF*, strongly acid
with HF. The refining plant and its operation are quite sim-
ilar to a copper refining plant. The cathodes are sheet steel,
greased, and the dense sheet lead deposited is stripped off from
time to time.
Problem 125.
In a Salom apparatus powdered galena, density of powder
3.5, is placed in a layer 0.5 mm. thick on a revolving lead table
having an effective treatment area of 2.25 square meters.
Current density 330 amps, per square meter of cathode surface.
Time of treatment of the layer 90 minutes. Electrolyte dilute
H^SO*. Resistance of cell 0.001 ohm. Heats of formation:
(Pb, S) 20,300; (H^ S) 4,800 (as gas); (ff, O) 69,000. Assume
reduction to Pb complete.
Required :
(1) The efficiency of application of the amperes passing to
the reduction of the galena, and the percentage of ampere
efficiency lost by the evolution of hydrogen.
(2) The average composition of the gases coming from the
cell and their volume per hour, at normal pressure and 20° C,
assuming the electrolyte saturated with H^S, W and 0^ at
starting.
(3) The working voltage absorbed, if the efficiency of re-
duction of galena were 100 per cent.
(4) The working voltage if the cell were kept running after
all the galena was reduced.
(5) The average working voltage of the cell in actual opera-
tion and the distribution of this.
(6) The proportion of the power required by the cell which
could be generated by burning the gases produced in a gas engine
(if such were possible) at a thermo-mechanical efficiency of 100
per cent.
600 METALLURGICAL CALCULATIONS.
(1) Current used 2.25X330 = 742 amperes.
Lead theoretically reducible in 90 minutes
0.00001035 X 103.5X60X90X742.5 = 4295 grams.
Galena in the apparatus:
2.25X10,000X0.05X3.5 = 3938 "
Lead under treatment:
907
3938 X~ = 3410 "
Ampere efficiency of the treatment:
3410
= 0.794 = 79.4 per cent.
4295
Per cent of amperes evolving hydrogen = 20.6 " (1)
(2) If all the current evolved H^S, the gases evolved would bes
2 parts 1 part.
If all the current evolved H^, the gases would be:
2ff O^
• 2 parts 1 part.
If 79.4 per cent of the current evolves H^S and 20.6 per cent
hydrogen, the average gases evolved will be:
H^S 1.588 parts = 52.93 per cent.
W 0.412 " = 13.73 "
02 1.000 « = 33.33 " (2)
The volume evolved per hour is found as follows:
Oxygen produced by 1 ampere hour
0.00001035X8X60X60 = 0.29808 grams.
Produced by 742.5 amperes, in one hour
0.29808X742.5 = 221.3
Volume at 760 mm. and 0° C.
221.3-1-1.44 = 153.7 Utres.
Volume of gases produced per hour
153.7X3 = 461. "
= 0.461 cubic meter
Volume at 20°
0.461X^^1^= 0.495 « (2)
THE METALLURGY OF LEAD. 601
(3) Voltage for ohmic resistance
0.001X742 = 0.742 volts.
Chemical work done if only H^'S is formed, per formula
PbS + ffO = ffS + O + Pb
- 20,300 - 69,000 +4,800 = - 84.500 Cai.
Per chemical equivalent concerned = — 42,250 "
Voltage absorbed in chemical work x
42,250 + 23,040 = 1.83 volts
Total voltage drop in the cell 1.83 + 0.74 =2.59 " (3)
(4) Chemical work done if only H^ is liberated = 69,000 Cal.
Voltage absorbed in chemical work
^ + 23,040= 1.49 volts.
Total voltage drop in the cell = 2.23 (4)
(5) Voltage absorbed in chemical work if 79.4 per cent of the
current evolves H^S and 20.6 per cent H^:
(1.83 X 0.794) + (1.49 X 0.206) = 1.76 volts.
Total voltage drop, in average running = 2.50 " (5)
Logically, the 1.76 volts absorbed in chemical decomposition in
average running is properly calculated thus, from the energy
evolved :
(42,250X0.794) + (34,500X0.206) . „^
23:040 ^ ^-^^
(6) Heat of combustion of 1 cubic meter of the average gas :
WS 0.5293 m^X 5,513 = 2918 Cal.
W 0.1373 m'X2,614 = 359 "
Sum = 3277 "
Calorific power of gas produced per hour:
3277X0.461 = 1511 Calories.
Watt hours producible from this at 100 per cent thermo-mechan-
ical efficiency :
1511-^0.860 = 1757 watts.
602 METALLURGICAL CALCULATIONS.
Current used by the cell, average running:
742.5X2.50 = 1856 watts.
Proportion theoretically regainable:
1757
1856
0.95 = 95 per cent. (6)
The reason why more power is theoretically regainable than
is used in doing chemical work in the cell, is that the gas engine
bums the sulphur of the PbS ultimately to S0^ and this is
the source of great energy, while only a relatively small amount
of energy was necessary to convert the PbS into H^S ready for
combustion.
Problem 126.
Lead bullion refined by the Betts process is 96.73 per cent
lead, and the refined lead is practically pure. The anodes are
1.5 inches thick, and weigh, with lugs for suspension, 275 pounds.
They are left in the tanks an average of nine days, running
with 135 amps, per plate, whose active surface is 1,700 square
inches. Space between anode and cathode 1\ inches, specific
resistance of solution 10 ohms. Each tank has twenty-two
anodes and twenty-three cathodes, and produces an average
of 545 pounds of lead per day. Power costs $50 per electric
horsepower-year, as delivered to the tanks. Drop per tank,
due to resistance of contacts, 0.15 volt; specific gravity of pure
lead, 11.35.
Required :
(1) The weight of lead theoretically deposited by the cur-
rent per day.
(2) The ampere efficiency of the deposition.
(3) The voltage drop per tank.
(4) The amount of anode scrap to be remelted in percent
of the total anode weight.
(5) The average thickness of the cathode deposit per day.
(6) The power costs per ton of impure lead refined.
Solution:
(1) Lead theoretically deposited by 1 ampere per day
0 . 00001035 X 103 . 5 X 60 X 60 X 24 = 92.55 grams.
THE METALLURGY OF LEAD. 603
Per anode per day, dissolved
92 . 55 X 135 -^ 1000 = 12.5 kg.
= 27.5 lbs.
Per tank, per day, deposited
27.5X22= 605 " (1)
(2) Efficiency of deposition
gg| = 0.90 = 90 per cent. (2)
(3) Amperes used per plate = 135
Current density per square inch
= 135 -f- 1700= 0.0794 amperes
Current density per square centimeter
= 0.0794-^6.25 = 0.0127
Distance of plates apart = 1.17 inches
= 2.93 cm.
Resistance of 1 cm. cube of electrolyte = 10 ohms.
Voltage drop in electrolyte
10 X 2 . 93 X 0 . 0127 = 0.37 volt.
Voltage drop in connections = 0.15
Total voltage drop per tank = 0.52 " (3)
(4) Dissolved from anode in 9 days
545-h22X9 = 223 lbs.
Anode corroded in 9 days, assuming slime to fall off it
223 -h 0.9673 = 231 lbs.
Anode scrap = 275-231 = 44
44
Percentage of anode scrap 75=^ = 0.16 = 16 per cent.
(5) Deposit of lead one side of a cathode, per day
545^44 = 12.4 lbs.
= 5.636 kg.
Area of 1 side of a cathode = 850 sq. inches.
= 5312 sq. cm.
Deposit, per day, per sq. cm.
5636-^5312 = 1.06 grams.
604 METALLURGICAL CALCULATIONS.
Volume of this deposit 1 . 06 h- 1 1 . 35 = 0.093 cu. cm.
Therefore thickness deposited per day = 0.93 mm.
(6) Bullion refined, per tank, per day
545-^0.9673 = 563
lbs.
Power required per tank
Cost of power per year
per day
per ton of lead treated
2970X0.52 = 1544 watts.
= 2.06 e.h.p.
= 50X2.06 = 103. dollars.
= 103 H- 365
2000
0.28X-
563
0.28
0.99
(5)
(6)
CHAPTER III.
THE METALLURGY OF SILVER AND GOLD.
The processes most used for the extraction of silver and
gold from their ores are the smelting of the ores with copper
ores to matte and eventually blister copper, followed by electro-
lytic refining of the latter, or smelting with lead ores, followed
by desilverization of the resulting lead bullion, cupellation of
the rich lead, or electrolytic refining of the same. In all of these
cases the silver or gold is a comparatively minute constitutent
of the ore, intermediate products and final metal, so that the
metallurgy of silver or gold, after these methods, up to the
production of impure silver or gold, or silver or gold bullion, is
practically the metallurgy of copper or lead — both of which
subjects we have considered at length.
When we obtain products in which the silver or gold is the
chief constituent, we approach a condition in which calculations
upon the precious metal contained may be based, but here
again there are not the same conditions of economy so promi-
nent in the metallurgy of lead or copper. It does not matter
greatly, for instance, whether we use 50 cents' worth of one
fuel or a dollar's worth of another in melting 1,000 ounces
of silver worth over $500; the question of which fuel is the
cleanest or most convenient is of greater importance than the
cost of the fuel.
The object of this introduction is to show that, until it comes
to handling nearly pure silver or gold, close calculations as to
amounts of reagents, running of furnaces, etc., are not prac-
ticable for silver or gold, per se.
Electrolytic Refining of Silver Bullion.
The parting of gold and silver by acids is rapidly giving
place to the cleaner and less expensive electrolytic separation.
The bullion may be of quite variable composition. The follow-
ing analyses of two samples give an idea of this variation.
605
606 METALLURGICAL CALCULATIONS.
Cupelled Silver. Impure Silver.
Ag 98.69 74.08
Pb 1.09 3.71
Cu 0.12 20.23
Fe 0.09 1.01
Au 0.0023 0.05
The electrolytic refining proceeds easiest with the nearly
pure silver, because the impurities going into solution accumu-
late much more slowly, and the solution therefore requires
purifying less frequently. On the other hand, the solution of
a large amount of copper, iron or lead adds electromotive force
to the circuit, since silver only is deposited, and thus decreases
the electrical work to be done.
Problem 127.
Two varieties of silver bullion are refined electro lytically
carrying
No. 1. No. 2.
Ag 98.69 74.08
Pb 1.09 3.71
Cu 0.12 20.23
Fe 0.09 1.01
Au 0.01 0.05
The slimes from each carry in percentages :
No. 1. No. 2.
Ag 60 55
Pb 10 5
Cu 5 15
Fe 3 5
Au 22 20
In each case all the gold goes into the slimes. A current
density of 200 amperes per square meter of electrode surface is
used, and the plates are 2 centimeters thick. Specific gravity
of No. 1, 10.15; of No. 2, 9.79. There is anode scrap equal to
12 per cent of the weight of the plates. The working space
between the electrodes is 4 cm. total current 220 amperes per
tank, specific resistance of electrolyte 20 ohms. Deposited
silver, 19.85 kg. per tank per day.
THE METALLURGY OF SILVER AND GOLD. 607
Required :
(1) The ampere efficiency of the deposition of silver.
(2) The weight of anode corroded in a tank per day.
(3) The deficit in weight of silver in the electrolyte in each
tank per day.
(4) The electromotive force contributed to the circuit by the
solution of impurities in the case of each bullion.
(5) The total drop of potential across each tank, adding 10
per cent for loss in contacts.
(6) The horse power-hours required per kilogram of silver
deposited in each case.
Solution :
(1) Silver deposited theoretically by one ampere, per day
0.00001035X108X60X60X24 = .96.58 grms.
220 amperes deposit 96.58X220 =21,248 "
= 21.248 kg.
Ampere efficiency ' = 0.934 — 93.4 per cent. (1)
This means that 6.6 per cent of all the silver which the cur-
rent is capable of depositing is prevented from depositing by
the nitric acid present, forming AgNO'. The electrolyte is
silver nitrate with copper nitrate, and contains always about
1 per cent of free nitric acid, which thus acts chemically upon
the deposited silver (or acts to prevent its deposition to this
extent, whichever way we wish to look at it).
(2) The weight dissolved, per 100 grams of anode corroded,
is, assuming all the gold to appear in the slimes:
No. 1.
Anode. Slimes. Dissolved.
Ag 98.69 0.03 98.66
Pb 1.09 .... 1.09
Cu 0.12 ... 0.12
Fe 0.09 ... 0.09
Au 0.01 0.01
No. 2.
Anode. Slimes. Dissolved.
Ag 74.08 0.14 77.94
Pb 3.71 0.01 3.70
608 METALLURGICAL CALCULATIONS.
Anode. Slimes. Dissolved.
Cu 20.23 0.04 20.19
Fe 1.01 0.01 1.00
Au 0.05 0.05
The amount of current necessary to dissolve these weights
will be the sum of the current which would be necessary to
dissolve each constituent separately. These will be found in
ampere-hours by dividing each by its electrochemical chemical
equivalent X 3,600.
No. 1.
Ampere-Hours.
Ag 98. 66H-0. 001118X60X60 = 24.51
Pb 1.09-hO. 00107 X60X60^ = 0.28
Cu 0.12-hO. 00033 X60X60 = 0.12
Fe 0.09-hO. 00029 X60X60 = 0.09
26.00
No. 2.
Ag 77.94-^4.025 = 19.36
Pb 3.70^3.856 = 0.96
Cu 20.19-Hl.185 = 17.04
Fe 1.00-^1.044 = 0.96
38.32
Since there is disposable 220 X 24 = 5,280 ampere-hours in
a tank per day, there will be corroded the following weights
of anodes per day in tanks containing Nos. 1 and 2, respectively;
No. 1 100 X^^ = 21,120 Grams.
No. 2 lOOxfli = ^3,779 "
(3) There are dissolved in a tank, per day, the following weights
of silver:
No. 1 anodes: 21,120X0.9866 = 20,837 Grams
No. 2 anodes: 13,779X0.7794 = 10,739 "
Deposit in each tank = 19,850 "
Surplus in No. 1 tank = 987 "
Deficit in No. 2 tank = 9,111 " (3)
THE METALLURGY OF SILVER AND GOLD. 609
(4) The heats of formation of the acid and salts concerned
are:
(Ag, N, 0^ Ag) = 23,000 cal. = 213 cal. per gram Ag
(Pb, N^ 0\ Ag) = 98,200 " = 474 « « Pb
(Cu, N^ 0«, Ag) = 81,300 " = 1,278 " " Cu
(Fe, N^ 0», Ag) = 43,900 " = 784 " « Fe.
(H, N, 0^ Ag) = 48,800 " = 48,800 " « H
The heat balance per 100 grams of anodes No. 1 is:
Solution of Ag, 98.66 X 213 = 21,015 cal.
Pb, 1.09 X 474 = siy "
Cu, 0.12X 1,278 = 153 "
Fe, 0.09 X 784 = 71 "
Heat evolved = 21,756 cal.
DepositionofAg, 93.99 X 213=20,020 "
Liberation of W 0.06X48,800 = 2,928 "
Heat absorbed = 22,948 "
Heat deficit = 1,192 "
Since- this is per 25 ampere-hours [see (2)] and 1 ampere
hour at 1 volt = 860 calories, the voltage which this deficit of
energy will absorb in the case of anodes No. 1 is:
1,192-:- 25 ^860 = 0.06 volt. (4)
The similar calculation for anodes No. 2 gives per 100 grams
corroded :
Solution of Ag, 77.94 X 213 = 16,601 cal.
Pb, 3.70 X 474 = 1,754 "
Cu, 20.19X1,278 = 25,803 "
Fe, 1.00 X 784 = 784 "
Heat evolved = 44,942 cal.
Deposition of Ag, 98.11 X 213 = 20,897 "
Liberation of ff 0.064X8,800= 2,928 "
Heat absorbed 23,826
Surplus heat = 21,117
Voltage generated = 0.98 volt. (4)
1.82 votts.
0.18
((
2.00
Ct
0.78
u
0.18
a
610 METALLURGICAL CALCULATIONS.
(5) Resistance of each element of electrolyte of 1 square
cm. area is 20X4 = 80 ohms. The current passing through
this is 220-^10,000 = 0.022 amperes. The drop of voltage,
due to the resistance of the electrolyte, is, therefore,
0.022X80 = 1.76 volts.
Adding decomposition voltage we have
In case of No. 1 anodes, 1.76 + 0.06
Add for resistance of contacts
"Working voltage 2.00 " (5)
In case of No. 2 anodes, 1.76 - 0.98
Add for resistance of contacts
"Working voltage 0.96 " (5)
(6) Horsepower required per tank:
. , -, , 220X2.00 .._,
Anodes No. 1 : j^^^r = 0.59 hp.
75(J
Anodes No. 2: ^^"^""^^ =0.28 «
760
Power per kilogram of silver deposited:
Anodes No. 1, 0.59^-19.85 = 0.020 hp-days
= 0.48 hp-hours. (6)
Anodes No. 2, 0.28-^-19.85 = 0.014 hp-days.
0.34 hp-hours. 6)
Problem 128.
The "Wohlwill process for refining gold bullion ^operates with
AuCP solution, strongly acid with HCl. Design a plant for
refining bullion having the analysis:
Au 60.3 Fe 2.2
Ag 7.0 Ni 2.0
Cu 6.5 Pb 7.0
Zn 15.0
Assume the following data to start with:
Current density, 1,000 amperes per square meter.
Use ten tanks in series.
Tanks available, 500 mm. X 500 mm. X300 mm. deep, inside.
THE METALLURGY OF SILVER AND GOLD. 611
Electrodes about 4 cm. apart.
Starting sheets 1 mm. thick.
Anodes 20 mm. thick.
Specific gravity of anodes 17.5.
Gold present in solution 50 grams per liter.
Specific resistance of electrolyte 5 ohms.
Specific gravity of the electrolyte 1.15.
Loss of electromotive force at contacts one-half the loss due
to the resistance of the solution.
Anode products, AuCP, AgCl, CuCl, ZnCP, FeCP, NiCP,
PbCP.
Required:
(1) The size and dimensions of electrodes and number in a
tank.
(2) Current used and total voltage of the system.
(3) Gold deposited per day.
(4) Anodes corroded per day, assuming corrosion uniform.
(5) Deficit of gold in the tanks per day.
(6) If anodes are left in until 9/10 dissolved, how long will
they last?
(7) If cathodes are removed every 24 hours, what is the
average time of treatment?
(8) At 6 per cent per annum, what is the interest charge
per kilogram of gold under treatment in the plant, gold being
worth 72.9 cents per gram.
Solution:
(1) Allowing 1 cm. clearance at the sides, the plates must
not be over 48 cm. in width across the tank. The electrolyte
must not be nearer than 2 cm. to the top of the tank, and the
plates ought to be 8 centimeters above the bottom, to allow
space for slimes to accumulate. The immersed depth of plates
is, therefore, 40 cm., and the superficial area 960 sq. cm. on a
side.
With spaces 4 cm., anodes 2 cm. thick and cathodes 0.1 cm.,
using one more cathode than anode, we have the spaces equal
in number to twice the anodes, and, therefore,
(2 X anodes) + (anodes + 1) 0.1 + (2 anodes + 4) = 50, whence the
number of anodes figure out 4.9. (1)
612 METALLURGICAL CALCULATIONS.
Choosing the nearest whole number, 5, there will be 6 cath-
odes, and the working spaces will be 10, and their width
50-5(2)^-6(0.1)^3 3^^^^ (J)
(2) Ignoring the edges of the plates, the working surface
per tank is 960X2X5 f= 9,600 sq. cm., and the current 9,600 -f-
10,000X1,000 = 960 amperes. (2)
The voltage to overcome ohmic resistance is
Add 50 per cent loss at contacts = 0 . 98 "
Sum = 2.95 "
The voltage corresponding to the chemical action occurring
can be calculated from the energy of formation of the AuCP
decomposed minus the energy of formation of the salts formed.
More simply, although not so logically, it may be derived from
the voltages of decomposition of these compounds, allowing for
the proportions of each formed and decomposed. These are:
(Au, CP)
= 27,200 H
r3H
i- 23,040
= 0.393 volt.
(Ag, CI)
= 29,000-^
-1-;
1-23,040
= 1.257 •'
(Cu, CI)
= 35,400 H
r-l-
h 23,040
= 1.536 "
(Zn, CV)
= 113,000-;
■r2-
h 23,040
= 2.448 «
(Fe, CP)
= 100,100-;
r2-
;- 23,040
= 2.170 "
(Ni, CP)
= 93,900 H
■r-2-
H 23,040
= 2.037 «
(Pb, CP).
= 77,900^
■r2-
h 23,040
= 1.690 «
These are the voltages which would be generated at the
anode in case each one of these salts was the only salt being
formed. But, since we know the exact composition of our
anode dissolved, we can calculate the corresponding voltage
generated at the anode:
Au, 0.603X0.393 = 0.337 volt.
Ag, 0.070X1.257 = 0.088 "
Cu, 0.065X1.536 = 0.101 "
Zn, 0.150X2.448 = 0.367 «
Fe, 0.022X2.170 = 0.048 "
THE METALLURGY OF SILVER AND GOLD. 613
Ni, 0.020X2.037 = 0.041 volt.
Pb, 0.070X1.690 = 0.118 "
Sum 1.000
Absorbed at the cathode 0.393
Total voltage generated 0.607
Absorbed in electrolyte and
connections 2.95
Working voltage per tank 2.34 "
Total for the system 23.4 " " (2)
(3) Au deposited by 960 amperes in ten tanks per day:
IQ7
0.00001035X^ X60X60X24X960X10 = 563,729 gr.
= 563.729 kg.
(4) One gram of anode requires for its corrosion:
Au, 0.603-^197 -^ 3-^0.00001035
Ag, 0.603-^108 -=-0.00001035
63.6 -H 0.00001035
Cu, 0.065
Zn, 0.150
Fe, 0.022
Ni, 0.020
Pb, 0.070
65 4-2-^0.00001035
56 -h 2-^0.00001035
59 H- 2-^0.00001035
207 -^2-^0.00001035
= 0.01703 -=-0.00001035 = 1,645 amp. seconds.
Current available per day :
960 X 60 X 60 X 24 X 10 = 829,440,000 amp. seconds.
Anodes corroded per day:
829,440,000 H- 1,645 = 504,220 grams.
= 504.220 kilograms (4)
(5) Gold deposited = 563.729
Gold dissolved, = 504.220X0.603 = 304.045
Deficit of gold in the plant per day = 259.684 " (5)
(6) Weight of anodes:
960X4X17.5X5X10 = 8,360,000 grams.
= 3,360 kilograms.
One-tenth scrap = 336
Weight corroded = 3,024
614 METALLURGICAL CALCULATIONS.
Days to corrode them:
^ 094.
504 22'" 5.93 = practically 6 days. (6)
(7) Average time of treatment, removing one-sixth each day:
(1+2 + 3 + 4 + 5 + 6)^-6 = 3.5 days. (7)
(8) Value of gold in plant is, in regular running, value of
cathodes for all the time plus value of anodes for 3.5 -i- 6 = 58
per cent of the time.
Value of cathodes :
960X0.1X19.2 (sp. gr. gold) X6X 10X0.729 = $80,578
Value of gold in anodes:
3,360,000X0.603X0.729 = 1,477,012
Average value of anode gold under treatment:
1,477,012 X (3.5 -H 6) = 861,590
Sum of. cathode gold continually in stock and average value
of anode gold under treatment:
$80,578 + $861,590 = $942,168
Interest per annum at 6 per cent = 56,530
Interest charge per day = 155
Interest charge per kg. of anode refined:
$155 ^ 504 . 22 = $0.31 per kg.
Interest charge per kg. of gold under treatment :
$155-4-304.045= $0.51 per kg. (8)
The Volatilization op Silver and Gold.
It is known that both of these metals can be vaporized; it
can easily be done in the electric arc. If they, are placed in
a vacuum, in quartz vessels, they show signs of metal vapor-
izing at 680° and 1,070°, respectively, although what vapor
tension this corresponds to we do not exactly know, but we
can surmise it to be a small fraction of a millimeter. If heated
still higher in a vacuum, they show the phenomenon of ebul-
lition, or boil, at 1,360° and 1,800°, respectively, although
here again we do not know what pressure this corresponds to,
THE METALLURGY OF SILVER AND GOLD. 615
except it is the pressure of the melted metals above the spot
where the vapor commences to form beneath the surface,
which might be 5 to 10 millimeters of melted metal, say 10
millimeters of mercury, if the bath of metal were shallow.
From these data, obtained by SchuUer, Krafft and Bergfeld, it
is estimated that the boiUng points of the two metals under
atmospheric pressure are 2,040° and 2,530°, respectively, on
the assumption that the temperature interval between first
signs of vaporization in a vacuum and boiling in a vacuum is
equal to the interval between the latter temperature and the
ordinary boiling point. (This is said to be true of mercury.)
0. P. Watts has estimated the boiling point of silver and
gold as 1,850° and 2,800°, respectively, meaning probably at
atmospheric pressure; but his estimate is based on such as-
sumptions that they cannot lay claim to as much accuracy
as the data and estimates above cited.
If we go back to the very probable rule, that at equal frac-
tions of the normal boiling points, expressed in absolute tem-
peratures, metals have the same vapor tensions, we can com-
pare silver and gold with mercury, whose vapor .tension is
known through a wide range, and get quite probable values
for the vapor tensions of these metals through a large range
of temperature. The following table is from nearly zero ten-
sion to about the temperature at which ebullition is noticeable
in a vacuum:
Tension
of Vapor. Mercury. Lead. Silver. Gold.
mm. of Hg. C C C G
0.0002 0 625 729 942
0 0005 10 658 766 987
0 0013 20 691 802 1,031
0 0029 30 724 839 1,075
0 0063 40 757 . 876 1,120
0 013 50 790 913 1,165
0 026 60 822 949 1,209
0 050 70 855 986 1,254
0 093 80 888 1,023 1,298
0 165 90 921 1,059 1,343
0 285 100 954 1,096 1,387
616 METALLURGICAL CALCULATIONS.
Tension
of Vapor.
Mercury.
Lead
Silver.
Gold.
mm. of Hg.
.C°
. c°
C°
C°
0.478
110
987
1,133
1,432
0.779
120
1,020
1,169
1,476
1.24
130
1,053
1,206
1,520
1.93
140
1,086
1,243
1,565
2.93
150
1,119
1,280
1,611
4.38
160
1,151
1,316
1,654
6.41
170
1,184
1,353
1,699
9.23
180
1,217
1,390
1,743
Inspecting the above table, we see that apparently about
0.0002 mm. of mercury tension is sufficient to make a metal
show signs of vaporization. The corresponding temperatures
at which silver and gold show 0.0002 mm. tension are, from
the above table, 729° and 942°, respectively, whereas it is said
to have been observed for these metals at 680° and 1,070°.
The divergence is not wide, considering the lack of exact data
involved.
Mercury is said to show ebullition in a vacuum at- 180°, lead
at 1,250°, silver at 1,360° and gold at 1,800°. From the above
table we see these metals having the same vapor tension at
180°, 1,217°, 1,390°, and 1,743°, respectively. The agreement
is encouraging.
The table in continuation is for temperatures from those
causing ebullition in a vacuum to those required to boil the
metals at atmospheric pressure:
Tension
,
of Vapor.
Mercury.
Lead.
Silver.
Gold.
mm,, of Hg.
C°
C°
C°
C°
9.23
180
1,217
1,390
1,743
14.84
190
1,250
1,427
1,788
19.90
200
1,283
1,463
1,832
26.35
210
1,316
1,500
1,877
34.70
220
1,349
1,537
1,921
45.35
230
1,882
1,574
1,965
58.82
240
1,415
1,610
2,010
75.75
250
1,448
1,647
2,055
96.73
260
1,480
1,684
2,099
THE METALLURGY OF SILVER AND GOLD, 617
Tension
of Vapor.
Mercurv.
Lead.
Silver.
Gold.
mm. of Hg.
C°'
C°
C°
C°
123.
270
1,513
1,720
2,144
155.
280
1,546
1,757
2,188
195.
290
1,579
1,794
2,233
242.
'300
1,612
1,830
2,277
300.
310
1,645
1,867
2,322
369.
320
1,678
1,904
2,366
451.
330
1,711
1,941
2,410
548.
340
1.744
1,977
2,455
663.
350
1,777
2,014
2,500
760.
357
1,800
2,040
2,630
For tensions of metallic vapors over 1 atmosphere, such as
may easily occur in high temperature work, particularly in
electric furnaces, the following data may be useful:
Tension
of Vapor.
Mercury.
Lead.
Silver.
Gold.
Atmospheres.
C°
C°
C°
C°
1.0
357
1,800
2,040
2,530
2.1
400
1,951
2,197
2,722
4.25
450
2,116
2,380
2,945
8.
500
2,280
2,564
3,167
13.8
550
2,445
2,747
3,390
22.3
600
2,609
2,931
3,612
34.0
650
2,774
3,114
3,835
50.
700
2,938
3,298
4,057
72.
750
3,103
3,481
4,280
102.
800
3,267
3,665
4,502
137.5
850
3,436
3,848
4,725
162.
880
3,525
3,958
4,858
The last table
may not be
of much
immediate
practical
value, but it is
interesting scientific information.
The pre-
ceding tables are, however, technically and commercially
important. They point particularly to the wastefulness of
melting silver or gold in open furnaces Where furnace gases
pass over the metal. Such gases, at temperatures above 700°
for silver and 950° for gold, i.e., even passing over unmelted
metal, can cause volatilization and loss of weight, because they
618 METALLURGICAL CALCULATIONS.
evaporate the metal on exactly the same principle that a current
of dry air evaporates ice or water. They also explain why
silver volatilizes with lead in the cupellation operation. At
1,000° C. lead has a vapor tension of about 0.62 mm. of mer-
cury, and silver about 0.07 mm., and therefore, a considerable
loss of silver is sure to occur with the lead vapors passing off.
Gold at the same temperature has only 0.0007 mm. tension, and,
therefore, proportionately less of it is lost, say only one-fiftieth
as much as the silver loss, reckoning from the proportionate
tensions and the relative densities of the two vapors.
In calculating such metallic vapor losses it is important to
remember that the metals are monatomic when in the state
of vapor, and that the molecular weights of their vapors are
simply their atomic weights. The hypothetical weight of a
cubic meter of such metallic vapor at standard conditions, 0°
C. and 760 mm. pressure, is therefore,
0.09kg.X^'°"^^^7^^g^*,
and from this theoretical datum the weight of any volume at
any temperature and any tension can be calculated.
Illustration: Calculate the weight of silver vapor contained
in 1 cubic meter of furnace gases, if saturated with silver vapor
and at 1,100° C.
Solution: Since the tension of silver at this temperature
is 0.28 mm., the question resolves itself into this: What is
the weight of 1 cubic meter of silver vapor at 1,100° and 0.28
mm. pressure. .One cubic meter at standard conditions would
be 0.09 X (108 4- 2) = 4.86 kilograms; therefore, at these given
conditions, it contains:
^•««XT:iS27-3Xy6f = 0-0035 kg.
= 3.5 grams.
It is true that the silver vapor, being heavy, mixes slowly
with the furnace gases, but, nevertheless, the gases rushing
over and coming in contact with the silver may become nearly
saturated with this vapor, and carry considerable away. In
THE METALLURGY OF SILVER AND GOLD. 619
the Bessemerizing of copper matte to blister copper, by blowing
air directly through it, as much as 30, per cent of all the silver
present may be thus vaporized from the bath' as soon as the
silver has taken the metallic form.
Every smelter and refiner of the precious metals should be
familiar with these facts, and draw conclusions from them
useful to the conduct of his business.
The specific heat of these metallic vapors is 0.225 per cubic
meter (calculated to standard conditions), and the latent heat
of vaporization is approximately twenty-three tirhes the normal
boiling point expressed in absolute temperature for one atomic
weight of the metal — as explained in Part I of these calculations.
For silver and gold we would have the latent heats :
Ag, 23 (2,040 + 273) = 53,200 Cal. per 108 kg.
= 493 " 1 kg.
Au, 23(2,536 + 273) = 64,470 " 197 kg.
= 327 " 1 kg.
The above latent heats of vaporization are for boiling at
normal atmospheric pressure ; for vaporization at other pressures
and corresponding temperatures, correction must be made for
the difference in specific heats of the metallic vapor and liquid
metal.
CHAPTER IV.
THE METALLURGY OF ZINC.
(Including Cadmium and Mercury.)
At the present time, the metallurgy of zinc is briefly com-
prehended in the following statements: The chief ore is zinc
sulphide, ZnS, infusible at ordinary furnace heats, non-volatile,
easily roasted to ZnO; the roasting is done principally in me-
chanically stirred furnaces, the ore being in small pieces, because
it is non-porous, compact, and roasts slowly; the roasted ore,
principally ZnO, is mixed with an excess of carbon as a re-
ducing agent, and heated in closed fire-clay retorts having
condensers attached; zinc vapors begin to come off at 1033° C,
and come off rapidly at the working temperature of the charge,
say 1200° to 1300°; the zinc vapor and carbon monoxide pass
into the condensers, and as they cool deposit the zinc, some
in the form of fine dust (Uke hoar frost), most of it as liquid
drops; the cadmium in the ore and some lead, if present, distil
over with the zinc, constituting its chief impurities. Arsenic
is sometimes present in the condensed product. Iron does
not distil over, but some is absorbed from ladles and moulds
in which the liquid zinc may be handled and cast.
The chief operations with which calculations may be con-
cerned are the roasting, reduction by carbon, condensation of
the vapors, possibility of blast-furnace extraction, of electric
furnace reduction, electrolytic extraction, electrolytic refining.
Roasting of Sphalerite.
Before roasting, the crude ore is crushed and concentrated.
Pure ZnS contains 67 per cent zinc and 33 per cent sulphur;
its specific gravity is 3.9 and it has fine cleavage, so that it
crushes easily, producing much fines or slimes. In the Joplin
district, in Missouri, the largest zinc field in America, the ore
as mined averages some 4.3 per cent of zinc, equal to 6.4 per
cent of pure blende, ZnS, and is concentrated by jigging to
620
THE METALLURGY OF ZINC. 621
heads carrying about 60 per cent of zinc (90 per cent of ZnS),
containing some 70 per cent of the zinc in the ore, and tails
carrying about 1.35 per cent of zinc (2 per cent of ZnS), repre-
senting 30 per cent of the zinc in the ore. The concentrates
average 5 per cent of the weight of the ore ; concentration ratio
20 to 1. Cost of such crushing and concentration 20 to 40
cents per ton of ore milled. (Ingalls.) The average com-
position of these concentrates is:
Zinc 60 per cent = 90 per cent ZnS.
Iron 2 " = SpercentFeS.
Silica 7 "
Sulphur 31 "
100
Zinc sulphide begins to be oxidized by air at a dull red heat,
say, 600° C, and if the supply of air is kept up the oxidation
is rapid, generating a large amount of heat and consequent
high temperature.
2ZnS + 30=' = 2ZnO-|-2S02
The question as to how high a temperature would theoret-
ically result if sphalerite were thus burned is an interesting one.
Ingalls quotes HoUaway as giving 1992° C. We will investigate
this point as being of interest and value in connection with
practical roasting.
Problem 129.
Pure ZnS is oxidized by air. The heats of formation con-
cerned are:
(Zn, S) = 43,000
(Zn, O) = 84,000
(S, 02) = 69,260
The specific heats of the materials concerned are:
Sm to 0° .
ZnS = 0.120 +0.00003t
Zn O = 0.1212 + 0.0000315t
S02(lm=) = 0.36 +0.0003t
02 QvW { " ) = 0.303 +0.000027t
Assuming that the blende is first heated to 600°, to start the
roasting, and then the air supply put on, the exposed roasting
622 METALLURGICAL CALCULATIONS.
surface being sufficient for rapid oxidation, and the tempera-
ture too high to form SO':
Required:
(1) The theoretical temperature at the roasting surface at
starting, if all the oxygen passing is utilized.
(2) The, same, when the operation has continued to its max-
imum temperature.
(3) The same, if three-fourths the oxygen passing is utilized.
(4) The same, if half the oxygen passing is utilized.
(5) The same, if the resulting gases contain only 5 per cent
of S(y gas.
Solution:
(1) The reaction gives, thermally:
Decomposition of ZnS —43,000 Cal.
Formation of ZnO -)- 84,800 "
" SO^ -t- 69,260 "
Net heat evolution 111 , 060 Cal.
This would be concerned in the oxidation of 97 kg. of ZnS
to 81 kg. of ZnO and 64 kg. of SO^ (= 22.22 m=), and requiring
3X16 = 48 kg. of 02 = 208 kg. of air (containing 160 kg. =
127 cubic meters of N^).
Heat in 97 kg. of ZnS at 600° = 8,032 Cal.
Total heat in the products = 119,092 "
Mean heat capacity of the products per 1°, from 0° to t°:
81 kg. ZnO = 9.8172 -I- 0.002552t
22.22 m'SO== = 8.0000 -f-0.006667t
127 m' N^ = 38.4810 -|- 0.003429t.
Sum = 56.2982 -I- 0.012648t
Therefore, theoretical surface temperature, at starting,
_ 119^092 _
56.2982 -I- 0.012648t ^ '
(2) When the surface of the oxidizing material has attained
its maximum temperature, it is practically at t°, and the pro-
ducts of combustion will contain 111,060 Calories plus the
heat in 97 kg. of ZnS at t°. We then have
THE METALLURGY OF ZINC. 623
^ lll,060 + 11.24t + 0.00291t^ _
56.2982 + 0.012648t ^'°" " "-"^^
It must be emphasized that this is the theoretical maximum
under the most favorable conditions as to oxidation and exact
air supply, conditions never realized in practice.
(3) The excess of oxygen would be 48 X J = 16 kg. = 11.11 m^
This would be the most favorable possible proportion for mak-
ing sulphuric acid from the gases, since it would just serve to
oxidize the SO^ to SO' in the acid chambers, the complete re-
actions being:
Roasting -ZnS + 202 = ZnO -I- SO^ -|- 0
Acid chambers -SO^ + O = SO^
The 11.11m' of oxygen corresponds to 53.4 m' of excess
air, whose mean heat capacity is 16.1903 + 0.001443t. Adding
this to the mean heat capacity of the products we have:
. _ 111,060 + 11.24t + 0.00291t^ _
72.4885 + 0.014091t ' ^ '
(4) This is more nearly the practical conditions, and apply-
ing the principles above explained, we have:
_ 111,060 + 11.24t + 0.00291t^_
104.8691 +0.016977t ^'
(5) If the gases contain 5 per cent of SO^ gas, their total
volume per formula weight of SO^ produced, in kilograms, is:
22.22 -=-0.05 = 444.4 m'
and the volume of excess air they contain is:
444.4 - 22.2 - 127 = 295.2 m'
the heat capacity of this excess air, per average 1° is:
89.4456 + 0.007970t
and
_ 111,060 + 11.24t + 0.00291t^ _ ,g-
* ~ 145.7438 + 0.024947t *• '
It follows from this analysis and these results, that effective
auto-roasting of zinc sulphide is practicable provided that the
ore is finely divided, so as to expose large surface, oxidize
quickly, and utiUze the oxygen of the air efficiently. Without
these conditions, auto-roasting is impracticable.
624 METALLURGICAL CALCULATIONS.
It ought to roast satisfactorily by pot roasting, if the proper
conditions as to size of ore, speed of blast, thickness of pot
walls, etc., can be found.
Problem 130.
Average blende concentrates, containing 90 per cent ZnS,
3 per cent FeS, 7 per cent SiO^, are roasted by the use of 30
per cent of coal, in a furnace with a hearth 135 feet effective
length, the unroasted sphalerite remaining being 2.6 per cent
of the roasted material. The^ iron is roasted to Fe^O'. The
coal carries 75 per cent of carbon, 5 per cent of hydrogen,
8 per cent of oxygen, 1 per cent of sulphur and 11 per cent
of ash. The chimney gases average 2 per cent of SO^ and
escape from the furnace at 300° C. Charge is drawn from
the furnace at .1000° C. Furnace roasts 40,000 pounds per
day. Width of furnace, 12 feet; height, 8 feet.
Required :
(1) The composition of the roasted ore.
(2) The heat generated by the roasting of 2,000 pounds of
the ore.
(3) The heat generated by the combustion of the fuel.
(4) The heat in the hot charge as drawn.
(5) The heat in the chimney gases.
(6) The heat lost by radiation and conduction.
(7) The heat loss calculated to pound-calories per square
foot of outer surface per minute.
Solution :
(1) The 2.6 per cent of ZnS in the roasted ore is per cent
of it, and not of the raw ore. The simplest method of getting
the composition of the roasted ore is to represent by x the
quantity of ZnS remain unoxidized per 100 of raw ore. We
then have:
Weight of ZnS oxidized, per 100 raw ore = 90 — x
" " ZnO formed = (90 - x) X 3^ = 75.2 - 0.835 x
" " Fe^O' « = 3 X ^ =2.7
176
THE METALLURGY OF ZINC.
Weight of Si02 remaining = 7.0
roasted ore = 84.9 - 0.835 x
also = x-H 0.026
Therefore
84.9 - 0.835 X = XH- 0.026
Whence
X = 2.2
Percentage of the ZnS oxidized
90-2.2
90
626
0.976 = 97.6 per cent.
Composition and weight of the roasted ore :
ZnS = 2.2 = 2.6 per cent.
ZnO = 73.4 = 86.1
Fe^O^ = 2.7 =3.1
SiO^ = 7.0 = 8.2 "
Weight 85.3
Zinc contents = 60.0 = 70.3 per cent. (1)
(2) Zinc sulphide oxidized to ZnO
2000 X (0 . 90 - 0 . 022) = 1 , 756 lbs.
Heat of oxidation of 97 lbs. of ZnS to
ZnO and SO^ (from Prob. 129) = 111,060 Ib.-Cal.
Heat of oxidation of 1 lb. = 1,145 " "
" « " " 1,756 lbs. = 2,010,620 " « (^,
(3) Heat of combustion of 1 lb. of fuel to CO^ and, H^O vapor:
C to 00^0.75X8100 = 6075 Ib.-Cal.
Available Hydrogen
0.05 - 0.01 = 0.04
H to H^O condensed
0.04X34,500 = 1380 " "
Calorific power to H^O condensed = 7455
Water formed
0.05X9 = 0.45
Latent heat of condensation
0.45X606.5 = 273
Calorific power to H^O vapor = 7182
626 METALLURGICAL CALCULATIONS
Coal used per 2000 lbs. of ore = 600 lbs.
Calorific power = 600X7182 = 4,309,200 Ib.-cal.
(3)
Total heat generated in the ftimace
2,010,620 + 4,309,200 = 6,319,820 " "
Proportion of total generated by the roasting
2,010,620^6,319,820 = 0.325 = 32.5 per cent.
(4) Charge, as drawn, per 2000 Ibsv of raw ore:
ZnS 44 lb.
ZnO 1468 "
Fe^'O' 54 "
SiO^ 140 "
1706 '
Heat in this, at 1,000° C.
ZnS 44X0.150 = 6.6
ZnO 1468X0.153 = 224.6
Fe^O' 54X0.344 = 18.0
SiO^" 140X0.260 = 36.4
285.6X1000 = 285,600 Ib.-Cal.
(4)
(5) Sulphur going into the gases:
S from ZnS =1756X32/97 = 579.4 lb.
S " FeS = 60X32/88 = 21.8 "
S " coal = 600 X 0.01 = 6.0 "
Weight of S = 607.2 «
" " 0 for SO^ = 607.2 "
" SO^ 1214.4 "
Volume of SO^—
(1214.4X16) -H 2.88 = 6,747 cubic feet.
Volume of chimney gas —
6,747^-0.02 = 337,350 " "
Volume of CO^ in it—
(600 X 0.75 X 16) -^ 1.98 = 3,636 " "
Volume of H^O in it—
(600X0.45x16)^-0.81 = 5,333 " "
Volume of N^ and excess air = 321,634 « «
THE METALLURGY OF ZINC. 627
Heat in these gases at 300° C.
CO^ 3,636X0.436= 1,585 oz.-cal. per 1°
H^O 5,333X0.385 = 2,057 " "
S02 6,747X0.450 = 3,036 " "
N^andO^ 321,634X0.311 =100,028 « "
106,706 " '
= 6,669 Ib.-Cal.
= 2,000,700 "
300° (5)
(6) Summary of heat distribution :
Heat available, per 2,000 lb. ore = 6,319,820 Ib.-Cal.
Heat in hot ore drawn, 235,600 "
" chimney gases, 2,000,700 "
Loss by radiation and conduction, 4,083,620 "
6,319,820 " " (6)
(7) Approximate periphery of furnace :
12 + 12 + 8 + 8 = 40 feet.
Approximate outer surface, including base:
40X135 = 5,400 sq., feet.
Ore roasted per hour :
40,000 -f- 24 = 1,6671b.
Heat radiated and conducted away per hour:
^'2^000^^^ ^'^^'' = 3,360,000 Ib.-Cal.
Per square foot outside area, per hour:
3,360,000-7-5,400 = 622 Ib.-Cal.
per minute = 10.5 " " (7)
Reduction of Zinc Oxide.
Since metallic zinc boils under atmospheric pressure at 930° C,
and carbon does not begin to reduce zinc oxide until 1033° is
reached, the zinc reduced is necessarily obtained in the state
of vapor. To make the reaction proceed fast, a temperature
of the charge inside the retorts of 1100° to 1300° C. at the end
628 METALLURGICAL CALCULATIONS.
is necessary. A large &,mount of heat is absorbed in the reac-
tion, which must be supplied as a current or flow of heat through
the walls of the retort in order to keep the reaction and re-
duction going. Since the walls of the retort are fire-clay, 'aver-
aging 4 centimeters (1.6 inch) thick, there must be a consider-
able difference of temperature (temperature head) between
the inside and outside of the retort in order to keep up the
flow of heat inward.
Thermochemical Considerations.
The heat of formation of zinc oxide, at ordinary tempera-
tures, is 84,800 Calories for a molecular weight, 81 kilos, of
oxide, as determined by thermochemical experiment. This
means that cold zinc uniting with cold oxygen, and the hot
product cooled down to the same starting temperature, results
in the evolution of heat stated. Cold carbon uniting with cold
oxygen to cold carbonous oxide, CO, gives similarly 29,160
Calories per molecular weight, 28. kilograms, of gas formed.
These facts are expressed thermochemically as
(Zn, O) = 84,800
(C, O) = 29,160
The equation of reduction becomes
ZnO + C = Zn + CO
- 84,800 +29,160 = - 55,640
This means that if we start with cold zinc oxide and cold
carbon, the heat absorbed is 55,640 Calories ending up with cold
products, zinc and CO at ordinary temperatures.
What is actually done in practice, however, is to first heat
the ZnO and C to a high temperature, at least to 1033°, and
then to supply the heat of the reaction at that temperature, pro-
ducing zinc vapor and CO gas, both at 1033° The process of
reduction, therefore, resolves itself plainly into two steps; (1)
heating the charge up to the reacting. temperature, (2) supply-
ing the latent heat of the chemical change at that temperature.
Heating up the Charge.
The charge usually contains more carbon than is theoretically
necessary for reduction of the zinc oxide, because some is con-
sumed by air in the retort, some in reducing iron oxides and
THE METALLURGY OF ZINC. 629
some is left behind unused; it is cheaper to lose carbon than
unreduced zinc oxide. However, making the calculation on
the theoretical amount of carbon only (anyone can modify it
for any excess of carbon used in any specific case) we need to
know the heat necessary to raise these reacting substances to
the temperature of reaction.
The heat in 1 kilogram of zinc oxide at various temperatures was
determined in the writer's laboratory as 0.1212t + 0.0000315t2.
The heat in 1 kilogram of carbon at temperatures above 1000° is
represented by 0.5t - 120. We have the heat necessary to raise
our 81 kilograms of ZnO and 12 kilograms of carbon to 1033° as
ZnO: 0.1212 (1033) +0.0000315 (1033)^X81
= 159X81 = 12,879 Cal.
C : [0.5 (1033) - 120] X 12 = 396 X 12 = 4,752 "
Sum = 17,631 "
This quantity represents the heat per 81 of oxide and 12 of
carbon, or 93 of charge containing 65 of zinc. Per 1000 kilo-
grams of oxide, containing 800 kilos of zinc, it would be
17,631 X^^ = 217,670 Calories.
It should be noted that this quantity is actually proportional
to the weight of the charge, ore plus reducing agent, and inde-
pendent of the amount of zinc in it. A ton of poor ore will
practically require as much heat to bring it up to the reaction
temperature as a ton of rich ore, so that these costs are propor-
tional to the weight of charge treated and not to the weight of
zinc it contains.
If the charge is heated electrically, the amount of electric
power needed for heating up can be calculated, assuming an
average loss of 10 to 30 per cent of the total heat by radiation
and conduction from the furnace.
Illustration :
A retort contains 300 kilograms of charge mixture and is
heated electrically to 1033°, the reaction temperature at an effi-
ciency of 75 per cent, by an electric current of 250 horse-power.
How long will the heating-up period last.
630 METALLURGICAL CALCULATIONS.
Solution :
Heat needed in the charge, 217,670X0.3 = 65,300 Calories.
Heat to be supplied = 65,300-^0.75 = 87,070 Calories; Power
applied supphes 635X250 = 158,750 Calories per hour. Time
required 87,070^158,750 = 0.55 hour = 33 minutes.
If the charge is heated by furnace gases outside the retort we
must take into consideration that the fire-clay is a poor con-
ductor, that the rate of transmission of heat through the walls
falls off as the charge becomes hot, and that the outside sur-
face of the retort must be kept well above 1033° in order to
get the charge to that temperature in a reasonable time. The
conductance of fire-clay for high temperature is 0.0031; that
is, 0.0031 gram-calories pass through each square centimeter
per second, if one centimeter thick, per 1° C. difference of tem-
perature. When the charge is cold, the inner surface of the
retort may be reduced in temperature to a low red heat, say,
500°, and towards the end of the heating-up period its tempera-
ture becomes at least 1033°, while the outer surface is kept
continually at 1200°, let us say, by the furnace gases. During
the heating-up period there is a difference of temperature
producing heat flow of 700°, for a short time, down to, say,
200°, which we may average up as 300° heat difference. The
heat conductivity of the material of the retort and the thick-
ness of its walls have everything to do with the rate at which
the heat can get through and the charge be heated up to the
reaction temperature.
Problem 131.
An oval zinc retort is 130 centimeters long, 30 centimeters
diameter outside, one way, and 15 centimeters the other, and
its walls are 3 centimeters thick. It is charged with 30 kilo-
grams of ore and 12 kilograms of reduction material. The
temperatures of the charge in the retort and of the gases out-
side the retort were as follows:
In Retort
Outside
DifEerence
At starting.
0°C.
1067°C.
1067°C.
In 0.5 hour
350 ,
1067
717
« 1.0 "
600
1067
467
"1.5 hours
781
1067
286
" 2.0 "
814
1100
286
THE METALLURGY OF ZINC. 631
In Retort
Outside
Difference
2.5 "
869
1100
231
3.0 «
924
1110
187
3.5 "
957
1155
198
4.0 "
935
1166
231
4.5 "
935
1138
203
5.0 "
946
1144
198
5.5 "
946
1155
209
6.0 "
979
1166
187
6.5 "
1001
1177
176
7.0 "
1034
1177
143
Average, 319
Take 159 Calories per kg. as the heat required to bring the
ore to 1034° and 396 for the reduction material.
Required:
The average heat conductivity in C. G. S. units of the mate-
rial of the retorts, in the range given, assuming the inner sur-
face of the retort to be at the same temperature as the charge.
Solution :
Heat passing through the retort walls in 7 hours :
159X30 = 4770 Calories
396X12 = 4752 "
9522
Per second = 0.378 "
= 378 gram-cal.
Surface of retort:'
Periphery V 30X15X3.14 = 66 cm.
Area of sides, 66 X 130 = 8580 sq. cm.
Heat passing through each square cm. per second
378^8580 = 0.044 cal.
Heat passing per 1° difference
0.044-^319 = 0.00014 cal.
Since thickness is actually 3 centimeters, conductance in
C. G. S. units is:
0.00014X3 = 0.00042.
632 METALLURGICAL CALCULATIONS.
Correction:
This coefficient of conductance is far too low. The reason
is that the inner temperature, the temperature of the charge, is
always lower than the temperature of the inside surface of the
retort, and the difference of temperature between the outer and
inner walls of the retort must have .been far less than the aver-
age, 319°. If we take the coefficient of conductance as deter-
mined by experiment for firebrick, viz., 0.0031, then the average
difference of temperature between the inner and the outer
walls of the retort would be:
0.00042
"^^^^ 0.0030" *^ ^•
This is more likely, than that the conductance of the retort
material should be so extraordinarily low. In fact, we are led
to the conclusion that the poor heat conductivity of the charge
itself is the chief obstacle to its rapid heating, and that pre-
heating of the charge would be very advisable if it could be
done by some of the waste heat of the gases leaving the furnace.
Distillation of the Charge.
The driving off of the zinc is altogether a different operation
from the heating up of the charge to the stated reduction tem-
perature. It is an endothermic chemical operation, comparable
to the boiling of water at a constant temperature, the latent
heat of the chemical reaction is exactly comparable to the
latent htat of vaporization. The temperature must be kept
up to the temperature of reduction in order for the reaction
to take place at all, and then heat-calories must be supplied at
that temperature to keep the reaction going, and the reduction
proceeds pari passu with the quantity of heat supplied at that
constant temperature. The question now is: what is the latent
heat of this reaction at the reaction temperature.
We must for this purpose know the heat of formation of the
substances involved, ZnO and CO, at 1033°. The method of
calculating these is too long to insert here, but may be learned
from the writer's book on Metallurgical Calculations, Part I,
p. 51. We have the heats of formation from the elements as
they exist at 1033°:
THE METALLURGY OF ZINC. 633
(Zn, 0)i»== = 112,580
(C, 0)««' = 30,091 ■
and the reaction at 1033°
ZnO + C = CO + Zn ,
- 112,580 + 30,091 = -82,489
This is seen to be nearly 50 per cent greater than the heat
of the reaction calculated to ordinary temperatures, from the
ordinary heats of formation as taken from thermochemical
tables. The metallurgist should understand this difference, for
it is one of the utmost importance in thermochemical calcula-
tions, if we want our calculations to represent actual conditions
and to check up with practice.
This amount of heat is proportional to the zinc oxide re-
duced or to the zinc distilled from the charge, but not to the
weight of ore charge itself. This requirement will, therefore,
be greater, per retort full of material, the richer the charge is
in zinc. It is a constant requirement for a given output of
zinc, and not per given weight of ore treated.
The heat required for the reduction, therefore, as distin-
guished from the heating-up period, is
??i^ = 1,018 Calories per kg. of ZnO reduced
81
8^'^^^ = 1 269 " " " " Zn distilled off
65
= 1,269,000- " " ton " " " "
Furnishing this reduction heat, at the high temperature re-
quired, is the larger part of the heat needed in the whole pro-
cess. We see that it is some 4.7 times the amount of heat
needed to raise the materials to the reduction temperature.
Problem 132.
The retort charge of Problem 131 was kept at the reduction
temperature for 14 hours, the temperature outside the retort
being gradually raised to 1300° and the average difference in
temperature between the gases and the charge being 147°, and
there being distilled from the charge in that time 18 kilograms
of zinc.
634 METALLURGICAL CALCULATIONS.
Required:
The average heat conductance in C. G. S. units of the mate-
rial of the retort from the above data and assumptions.
Solution: ■
The heat furnished in the 14 hours was as follows:
1,269X18 = 22,842 Calories.
Heat furnished per hour:
22,842^14 = 1,632
Heat furnished per second:
= 0.453 "
= 453 calories
Heat passing each sq. cm. of retort surface per second:
453 + 8,580= 0.053 calories.
Per 1° difference:
0.053 -f- 147 = 0.00036 "
Conductance, in C. G. S. units:
0.00036X3 = 0.00108
Remarks: This conductance calculates out 2.5 times as great
as from the data on the heating-up period, the reason of the
higher value being that the charge is at nearly uniform tem-
perature during this reduction period, and therefore the dif-
ference between the temperature taken in the middle of the
charge and the true temperature of the inner walls of the retort
is less than before, and the error from this source less. If we
make the same assumption as in the correction to the previous
problem, i.e., take the conductance of the retort material as
0.0030, the difference in temperature of the outer and inner
walls of the retort during this period would calculate out
147X0.00108 _
0.0030
while the center of the charge was then 147 — 53 = 94° cooler,
on an average, than the inner wall of the retort from which it
was deriving its heat.
These figures appear reasonable, and the writer would con-
THE METALLURGY OF ZWC. 635
elude, from the data so far available, that the conductance
0.003 probably represents a good approximation to the correct
value for zinc retort material, but that, in order to use it, we
should have more experimental data as to the average difference
between the temperature of the inner walls of the retort and
the temperature of the charge at various points in the retort.
Problem 133.
A zinc ore containing 50 per cent of zinc is mixed with 40
per cent of its weight of small anthracite coal, and retorted
in a Belgian furnace. The recovery of zinc was 82 per cent
of the zinc content of the ore. The consumption of anthracite
to heat the furnace was 2.25 tons per 10 of ore. The anthracite
contained 90 per cent of carbon and 10 per cent of ash, and
had a calorific power of 7500.
Required :
(1) The total consumption of fuel per 1000 of zinc obtained.
(2) The efficiency of transfer of heat from the furnace gases
to the charge.
Solution :
(1) Zinc charged, per 1000 of zinc obtained
1000^0.82 = 1220
Ore charged, per 1000 of zinc obtained
1220 -H 0.50 = 2440
Coal charged with ore
2440X0.40 = 976
Coal burned on grate
2440X2.25 = 5490
Total coal used, per 1000 of zinc obtained
976 + 5490 = 6466 (1)
(2) Calorific power of coal burned
5490X7500 = 41,170,000
Heat required to raise ore to reduction point
2440 X 159 = 387,960
63d METALLURGICAL CALCULATIONS.
Heat required to raise fuel to reduction point
Ash 98 X 159 = 15,580
Carbon 878 X 396 = 347,690
Total to raise charge to reduction point
751,230
Heat absorbed in distilling away 1000 of zinc
1,269,000
Total heat utilized
= 2,020,230
Thermal efficiency of utilization of the fuel
2,020,230 _„,,„_,„ ,
41,170,000
— W.WICr I.C JJCl ^^^llb. \£i^
Problem 134.
Natural gas from lola
, Kan., has the following composition
CH*
93 per cent.
H^'
2
CO
1
C^H^
1
w
3
It is used in a zinc retort furnace, at an efficiency of transfer
of heat to the charge of 4.9 per cent, as calculated in Prob.
133, working a charge which absorbed 2,000,000 Calories per
1000 kilograms of zinc distilled off.
Required: The volume of natural gas required to be used
to displace the anthracite fuel used per 1000 kilograms of zinc
produced. The cubic feet of gas per 1000 lbs. of zinc produced.
Solution:
Calorific power of the gas
CH* 0.93 X 8623 = 8009
H^* 0.02X29030 = 581
C^H< 0.01X14365 = 144
CO 0.01 X 3062 = 31
8765
THE METALLURGY OF ZINC. 637
This result may be called Calories per cubic meter of gas, or
ounce-calories (1° C.) per cubic foot according to whether it
is desired to work in metric units or the English system.
Cubic meters of gas required, per 1000 ' kilograms of zinc
produced :
2,000,000
— oTmo — -5-8765 = 4,700 cubic meters.
Per 1000 lbs. of zinc produced:
^' 0^049°^ X 16 -^8765 = 76,200 cubic feet.
Electric Smelting of Zinc Ores.
In an interesting communication to the American Electro-
chemical Society (Vol. XII, p. 117), Gustave Gin calculates the
electric power necessary for the smelting of several varieties
of zinc ore, assuming that the zinc vapor and other gases pass
out of the furnace at the usual high temperature — 1200° C.
Mr. Gin makes his calculations of heat required on the basis
of molecular weight of zinc compound reduced; that is, for 81
parts of ZnO and 65 parts of Zn. Calculating to 1200°, he finds
the heat in the products Zn and CO to be
In 65 parts Zn 26,720 Cal.
"28 " CO 8,280 "
Sum 35,000 Cal.
whereas we calculate for the same quantities
In 65 parts Zn 37,695 Cal.'
"28 " CO 8,945 "
46,640 Cal.
The difference between these numbers is principally in the
latent heat of vaporization of zinc, which Gin assumes as 15,370
Calories, but which by a method of evaluation used by the
writer figures out 27,670 Calories, and almost exactly the same
value has been obtained by a diflPerent method by W. McA.
Johnson.
Assume then, that we start with cold materials and end with
638 METALLURGICAL CALCULATIONS.
the products of the reaction leaving the retort at 1200°, the
sum total of usefully applied heat is the heat of the reaction
calculated at ordinary temperature plus the sensible heat of
the products at 1200°, or
Heat absorbed by reaction, ordinary temperature. . . .84,800 Cal.
Heat in necessary products, at 1200° 46,640
Total 131,440 "
To this must be added the sensible heat in the residue left
in the retort, to get the total heat which has been applied to
the charge. If the charge were pure ZnO, with the theoretical
amount of carbon, the residue would be nil, but in practice
there is always a residue of gangue with unused carbon.
Mr. Gia, having calculated the heat requirement on the above
basis, then makes his calculations of power required per ton
of ore smelted in the following ingenious way: The weight
of each component of 1000 kg. of ore is divided by its mole-
cular weight, and thus the number of molecular weights of
material in one ton of ore determined, which, so to speak,
gives a kind of chemical formula for the ore. An example will
make this clear.
Composition of a calcined calamine zinc ore:
ZnO 40.50 per cent.
ZnSiO' 38.07
Fe^'O^ 9.60
APSiO' 8.10
CaO 2.80 "
If we divide the weight of each compound present by its
molecular weight we get the relative number of molecules
present in the ore, and the formula for the ore:
KG. IN
1000 KG.
ZnO 405 4- 81 = 5.0 molecular weights.
ZnSiO' 380.7 ^ 141 = 2.7
Fe^O' 96. -f- 140 = 0.6 "
APSiO= 81 -^ 162 = 0.5 «
CaO 28 -^ 56 = 0.1 "
THE METALLURGY OF ZINC. 639
The ore may therefore be represented by the formula
5ZnO + 2.7ZnSiO' + O.eFe^O^ + O.SAPSiO' + 0. ICaO.
Assuming that the Fe^O^ becomes FeSiO', and that there is
to be added enough CaO to form CaSiO^ with the rest of the
SiO^ of the zinc silicate, we will need 1.5 CaO to do it, and
since there is 0.1 CaO present, 1.4 CaO must be added, which
represents 1.4X56 = 79 kg. of CaO. To form CO with the
oxygen combined with the zinc and iron, reducing the latter
to FeO, will require:
for 5 ZnO
5.0 C.
« 2.7ZnSiO*
2.7 "
" 0.6 Fe^O^
0.6 "
Sum 8.3 C. = 100 kg.
If we use twice the theoretical amount of carbon needed for
reduction we have the following balance sheets; weight in kilo-
grams being enclosed:
ORE
ADDED
CHARGE
5 ZnO
(405)
....
5 ZnO (405)
2.7 ZnSiO^
(381)
2.7 ZnSiO' (381)
0.6 Fe^O'
(96)
....
0.6 Fe^O' (96)
0.5 APSiO'
(81)
....
0.5 APSiO' (81)
0.1 CaO
(28)
1.4 CaO (79)
1.5 CaO (79)
16.6 C (200)
16.6 C (200)
The distribution of this charge will be as follows:
CHARGE
GASES.
RESIDUE.
5 ZnO
(405)
5 Zn (325)
....
2.7 ZnSiO^
(381)
2.7 Zn (175)
0.6 Fe^O'
(96)
1.2 FeSiO' (158)
0.5 APSiO= (81)
0.5 APSiO= (81)
1.5 CaO
(79)
1.5CaSiO= (174)
16.6 0
(200)
8.3 CO (232)
8.3 C (100)
The essential reactions involved in the reduction are expressed
by the reaction formula:
.5ZnO + 2.7ZnSiO^ + 0.6Fe2O=+1.5CaO + 8.3C
= 7.7Zii + 1.2FeSi03 + 1.5CaSiOH8.3CO
640 METALLURGICAL CALCULATIONS
The heat requirements per ton of ore treated may therefore
be figured out as:
Calories.
Decomposition of 7.7ZnO = 84,800X7.7 = 652,960
2.7ZnSiO'
into ZnO and SiO=' = 10,000X2.7 = 27.000
Decomposition of 0.6Fe^O'
into FeO and O = 64,200X0.6 = 38,520
Heat absorbed in decompositions = 718,480
Formation of 1.2 FeSiO^
from FeO and SiO^ =8,900X1.2 = 10,680
Formation of 1.5 CaSiO'
from CaO and SiO^ =17,850X1.5= 26,770
Formation of 8.3 CO = 29,160X8.3 =222,030
Heat evolved in formation heats 259,480
Net heat of chemical reactions absorbed 459,000
Sensible heat in 7.7 Zn =46,640X7.7 =359,130
8.3 CO = 8,945X8.3 = 74,240
" " 8.3 C = 5,760X8.3 = 47,810
1.2 FeSi'O'
■ 0.5 APSiO'
1.5 CaSiO'
= 413 kilograms = 460X413 = 200,000
slag
Sensible heat in products and residue = 681,180
Add heat absorbed in chemical reactions = 459,000
Total heat requirement of the charge = 1,140,180
Important principle : The heat absorbed in an electric furnace
by chemical Reactions is electric energy utilized at an efificiency
of 100 per cent. It is only part of the sensible heat of the
materials being treated which is lost by radiation and con-
duction. Heat losses by radiation and conduction in electric
furnace processes should be expressed upon the total energy of
the current diminished by the heat absorbed in chemical reac-
tions, and not upon the total energy of the current. This
principle has been expressed mOst clearly by Mr, F. T, Snyder,
THE METALLURGY OF ZINC. 641
of Chicago, the pioneer electric furnace zinc metallurgist of
America.
Expressed thus, electric furnaces give higher efficiencies the
greater the absorption of heat in chemical reactions ' taking
place within them. In mere physical processes, such as melting,
electric furnaces on a large scale give 75 per cent net efficiency,
with 25 per cent loss by radiation and conduction. In the
above case, figured out for 1000 kg. of zinc ore, 40 per cent
of the total heat requirement is absorbed as chemical heat at
100 per cent efficiency, and 60 per cent is needed as sensible
heat. This 60 per cent then represents the net sensible heating
effect, and the loss of heat by radiation and conduction must
be calculated as one-third of this quantity, and not one-third
of the total net heat requirement. We therefore have :
Calories.
Net heat requirements for chemical reactions 459,000
" sensible heat 681,180
Loss by radiation and conduction 227,060
Gross heat requirement of the furnace 1,367,240
Kilowatt hours of current required
860
Mr. Gin's figures have been modified by the writer, as ex-
plained above, and anyone consulting his paper on this subject
may make similar modifications to the other cases cited in that
paper.
Mr. F. T. Snyder, of the Canada Zinc Company, at Van-
couver, B. C, has operated the first practical electric zinc
smelting furnace in America. Treating mixed lead and zinc
ores Mr. Snyder uses the furnace and process protected in his
United States patents of July 2, 1907. (See Electrochemi-
cal and Metallurgical Industry, V. 323, 489.) The lead is
obtained liquid, and the zinc also condensed to the Uquid state
before leaving the furnace proper, the heat of condensation
being partly absorbed by water cooling the condensers and
partly by the descending charges, which carry it back into the
focus of the furnace. Under these circumstances, Mr. Snyder
reports that he has attained unexpectedly low results as to
642 METALLURGICAL CALCULATIONS.
power requirement. It will be recalled, that in discussing the
question of the electric smelting of zinc ores in general, and
Mr. Gin's process in particular, we assumed the zinc vapor and
CO gas to escape from the furnace at a minimum of 1033° C.
If, as in Mr. Snyder's furnace, they escape at about 500°, the
zinc liquid, the heat in these hot products is reduced very con-
siderably, particularly that in the zinc. Mr. Snyder states in
discussing Mr. Gin's paper (loc. cit.) that he has smelted pure
zinc oxide at an expenditure of 1050 kilowatt hours per 1000
kilograms of oxide. Let us see how this coincides with the
theoretical figures:
Calories
Heat value of 1050 kw-hours =1050X860 = 903,000
Heat of the chemical reactions, assumed to take place
at ordinary temperatures 1000X687 = 687,000
Sensible heat in products, at 500°
Zinc: 800 X 80 = 64,000
CO: 342X152 = 52,000
= 116,000
Leaving, by difference for radiation, conduction and
cooling water = 100,000
Mr. Snyder does not give details as to the exact working of
his furnace, but his claim to reduce a ton of zinc ore with 1000
kw-hours is seen to be a possibility, if he can remove the zinc
as liquid zinc from the furnace, not lose too much heat in cool-
ing water, and get most of the heat of condensation of the zinc
vapor usefully returned by the descending charges into the
working focus of the furnace.
Snyder's furnace, working as claimed, would show a useful
efficiency of
803,000 nco CQ
g^PQQ- = 0.88 = 88 per cent.
on the current used, with 12 per cent losses. The real heat
losses, however, should be expressed upon the energy used less
that absorbed in chemical reactions, or as 100,000 loss on
216,000 used for sensible heat,
100,000 r^Aa Aa
2^g^QQ^ = 0.46 = 46 per cent.
THE METALLURGY OF ZINC. 643
This shows that the furnace loses by radiation, conduction
and cooHng water 46 per cent of the energy developed as sen-
sible heat in the furnace, but the latter item is only 24 per
cent of the total energy applied to the furnace.
In short, about three-quarters of all the energy consumed by
the furnace is utilized in the heat of the chemical reactions
produced; of the other one quarter, half of it is represented
by the sensible heat of the products leaving the furnace and
half is lost by radiation, conduction and cooling water.
Zinc Vapor.
The condensation of zinc from the state of vapor follows the
same laws as regulate the condensing of all gases. If it is by
itself in a cooler space, it condenses by contact with the walls
as finely divided zinc dust (blue powder) if the walls are cold,
as liquid zinc if the walls are hot, and it keeps on condensing
until the tension of the remaining vapor is equal to the maxi-
mum tension of zinc vapor at the temperature of the condensing
vessel. This condition having been reached, no more will con-
dense until the condenser is cooled. At any temperature, there-
fore, an amount of zinc remains uncondensed corresponding to
the maximum vapor tension of zinc at that temperature.
If other indifferent gases are present, the zinc vapor sustains
only part of the atmospheric pressure, or pressure on the whole
mixture, so that condensation cannot begin until a lower tem-
perature is reached; that is, a temperature at which the maxi-
mum vapor tension of the zinc equals its partial vapor tension
in the gas mixture. At this point condensation begins and con-
tinues as the temperature falls, the gas mixture always being
saturated with zinc vapor. The phenomenon is precisely simi-
lar to the production of rain by the cooling of air containing
moisture.
As to what the vapor tension of zinc is, the data are scanty.
The boiling point under atmospheric pressure is 930° C, where
its vapor tension is 760 mm. of mercury column. According to
Barus, the maximum tension increases 6.67 mm. for each 1° C.
increase of temperature. This is a difHcult determination, but
corresponds satisfactorily with the calculated increase from an-
alogy to water and mercury.
Water, boiling at 100° C. (373° absolute), varies 1° in boiling
644 METALLURGICAL CALCULATIONS.
point for 27.20 mm. variation in vapor tension ; mercury, boiling
at 357° C. (630° absolute) varies 1° in boiling point for 12.66
mm. variation in vapor tension. The ratio of the two varia-
tions is seen to be somewhere in the inverse ratio of the two
boiling points.
^-^7 27.20
373 ''^•' 12.66 T^-^
If we compare mercury with zinc, with the two boiling points
357° and 930° (630° and 1203° absolute), and the observed va-
riations in vapor tension at the normal boiling point of 12.66
mm. and 6.67 mm. per 1° rise in boiling temperature, the two
ratios are
630 ^-"^ 6.67 ^^"
This coincidence justifies us completely in assuming that the
vapor tension curve of zinc may be calculated directly from
that of mercury, by assigning for any given vapor tension an
absolute temperature to the zinc vapor of 1.91 times that of
mercury vapor. For cadmium we may use the similar ratio
of the normal boiling points in absolute degrees:
273-1-780 _ 1053 ^
2734-357 630
and thus calculate the cadmium curve from the mercury curve.
From incipient vaporization to ebullition in a vacuum.
Tension
of vapor
Mercury
Cadmium
Zinc
mm. of Hg.
C°.
C°.
C°.
0.0002
0
183
248
0.0005
10
200
267
0.0013
20
216
286
0.0029
30
233
305
0.0063
40
250
324
0.013
50
267
344
0.026
60
283
363
0.050
70
300
382
THE METALLURGY OF ZINC. 645
Tension
of vapor
Mercury
Cadmium
Zinc
mm. of Hg.
C°-
C°.
C°.
0.093
80
317
401
0.165
90
333
420
0.285
100
350
439
0.478
110
367
458
0.779
120
383
477
1.24
130
400
496
1.93
140
417
516
2.93
150
433
535
4.38
160
450
554
6.41
170
467
573
9.23
180
483
592
Before proceeding further, we would remark that cadmium
melts at 320° and zinc at 419°; that both can, therefore, show
vaporizing phenomena nearly 150° below their melting points.
At 9 mm. tension, 180° C, mercury begins to show the phenom-
ena of ebullition, and we would, therefore, expect cadmium
to " simmer " at 483° and zinc at 592°. Neither of these obser-
vations has as yet been experimentally investigated, so far as
the author knows.
From ebullition in vacuum to normal boiling point.
Tension
of vapor
Mercury
Cadmium
Zinc
mm. of Hg.
C°.
C°.
C°.
9.23
180
483
592
14.84
190
500
611
19.90
200
517
630
26.35
210
533
649
34.70
220
550
668
45.35
230
567
687
58.82
240
584
706
75.75
250
600
726
96.73
260
617
745
123.
270
634
764
155.
280
650
783
195.
290
667
802
242.
300
684
821
300.
310
700
840
646 METALLURGICAL CALCULATIONS.
Tension
of vapor
Mercury
Cadmium
Zinc
mm. of Hg.
C°.
C°.
C°.
369.
320
717
859
451.
330
734
878
648.
340
750
897
663.
350
767
915
760.
357
780
930
The above table contains the more important data for the
actual condensation of zinc, cadmium and mercury vapors in
practice. Thus, if a mixture of zinc or cadmium vapor with
an equal volume of indifferent gas goes into a condenser, the
partial pressure of the metallic vapor being in this case only
half an atmosphere, or 380 mm., no metal will commence to
condense until the temperature of the gases is reduced to 862°
for zinc and 720° for cadmium. If mercury vapor from a
roaster is mixed with 19 times its volume of other gas, so that
it only forms 5 per cent of the mixture, its partial tension will
be only 5 per cent of 760 mm. = 38 mm., and no mercury will
commence to condense until the ternperature of the gas mixture
is reduced to 224°. These temperatures are exactly analogous
to the phenomenon of the dew point of moist air, the tempera-
ture at which rain forms.
From normal boihng point to high pressures.
Tension
of vapor
Mercury
Cadmiurn
Zinc
atmospheres.
C°.
C°. '
C°.
1.0
357
780
930
2.1
400
851
1012
4.25
450
934
1107
8.
500
1018
1203
13.8
550
1101
1298
22.3
600
1185
1394
34.0
650
1268
1489
50.
700
1352
1585
72.
750
1435
1680
102.
800
1519
1776
137.5
850
1602
1871
162.
880
1652
1928
THE METALLURGY OF ZINC. 647
The latter table may have several special applications. If
zinc, for instance, were placed in a closed electric furnace filled
with indifferent gas and capable of supporting 100 atmospheres
pressure, zinc could be kept in the liquid state up to some
1700° C, and its alloying with other metals greatly facilitated.
In the making of brass, for instance, much zinc is lost by vola-
tilization at the melting point of the copper used, yet a pressure
on the furnace of four atmospheres would entirely prevent
loss of zinc at the melting point of copper. Induction electric
furnaces could easily be run inside a pressure vessel at this
temperature, and the pressure be relieved gradually as the
temperature of the melted alloy was reduced. To keep zinc
from boiling, or to keep' it in the liquid state, at 1300°, not a
high temperature, would require a pressure of nearly 14 at-
mospheres to prevent it from boiling ad libitum.
Problem 135.
Roasted zinc sulphide ore, consisting practically of pure
ZnO, is reduced by carbon, and the reduction gases passed into
a condenser. Assume the reaction to be
ZnO-l-C = Zn + CO.
Required :
(1) The temperature at which zinc commences to condense
from the above gas mixture.
(2) The proportion of the zinc condensed out for each 1°
reduction of temperature below this " dew point."
(3) The proportion , of the zinc escaping uncondensed, if the
gas escapes from the condenser at 600° C.
(4) How would these data be affected, if the reduction was
taking place at Denver, 1500 meters above sea level, barometer
560 mm?
Solution :
(1) Since zinc vapor has been shown to have a specific grav-
ity of 32.5 referred to hydrogen gas at the same temperature
and pressure, and the molecular weight of hydrogen gas, for-
mula B?, is 2, the molecular weight of zinc vapor must be
2X32.5 = 65. This coincides with the atomic weight of zinc,
and therefore zinc vapor is monatomic and the symbol of its
molecule is Zn. The gas mixture in the above equation con-
648 METALLURGICAL CALCULATIONS.
tains, therefore, one molecule each of its constituents, and,
therefore, each gas supports half the atmospheric pressure.
This is ordinarily expressed by saying that the gas is half one
constituent and half the other, meaning that if the two gases
were separated and each measured under the prevailing normal
pressure, the volume of each would be half the volume of the
gas mixture. The statement that each supports half the nor-
mal pressure is the more scientific and logical, for in a gas
mixture each gas certainly possesses the volume of the mixture,
but under only a fraction of the pressure on the mixture, said
fraction being identical with the fraction of the volume of the
whole which it would constitute, if measured separately under
the pressure supported by the mixture'.
Consulting the table, we find zinc vapor to have a pressure of
380 mm. at 862°. This would, therefore, be the dew point, at
which the zinc would commence to condense. (1)
(2) At this condensing temperature, a difference of 19° in
the temperature corresponds to 82 mm. difference in the maxi-
mum vapor tension, and, therefore, in this gas mixture at this
temperature, a reduction of 1° in its temperature will reduce
the tension of the zinc vapor 82-h 19 = 4.3, or practically 4 mm.
4
This will result in the condensation of -5-- th. of the zinc, be-
cause its tension was 380 and was reduced to 376, and, there-
fore, ^^ th. of it remained uncondensed. The proportion con-
densed for the reduction of 1° in temperature is, therefore, a
trifle over 1 per cent. (2)
(3) Leaving the condenser at 600°, the tension of the zinc
vapor escaping would be, from the table, 11.6 mm. The ten-
sion of the CO gas would, therefore, be 760— 11.6 = 748.4 mm.
For every cubic meter of mixed gases, at 862°, containing a
cubic meter of zinc vapor at that temperature and 380 mm.
pressure, there will escape a fraction of a cubic meter of mixed
gas at 600°, containing zinc vapor at 11.6 mm. pressure. The
fractional volume can be calculated from the tension on the CO.
1 m^ CO at 862° and 380 mm. pressure, reduced to 600° and
748.4 mm, pressure, becomes
THE METALLURGY OF ZINC 649
,^600 + 273 380
^^862 + 273^748:4 = 0.381 m^
This is, then, the actual volume of gas 'mixture escaping, for
each actual 1 m' of gas mixture in the condenser at the " dew-
point," 862°.
The uncondensed zinc vapor is, therefore, 0.381 m' at 600°
and 11.6 mm. tension. This would be, at 862° and 380 mm.
tension
n .,01.. 862 + 273^,11.6 __,. ,
^•^^^>< 600 + 273^-380- = °-°^^^-
There, therefore, escaped uncondensed
-I nr^n = 0.015 = 1.5 per cent of the zinc. (3)
A quicker solution, not so easy to understand, however, is
11 CK
„■„" ■ = 0.015 = 1.5 per cent uncondensed.
748 . 4 ^
(4) If the operation takes place at such elevation above sea-
level that the barometer stands normally at 560 mm., then the
partial pressure of the zinc vapor in above case would be 280
mm. instead of 380 mm. and the " dew point " or temperature
at which condensation commences would be 835° instead of 862°,
as under previous conditions.
At this temperature, a difference of 1° in the temperature
corresponds to a difference of 3 mm. in the tension of the zinc
vapor, producing therefore a condensation of the zinc present,
per r fall of temperature, of 3^-280 = 0.011 = 1.1 per cent,
instead of 1.2 per cent.
If the temperature of the escaping gases is 600°, with the
saturated zinc vapor at 11.8 mm. tension and the carbonous
oxide, CO, at 560—11.8 = 548.2 mm., the proportion of the
original quantity of zinc escaping uncondensed is 11.8^-548.2 =
0.022 = 2.2 per cent, instead of 1.5 per cent.
Condensation of Zinc and Mercury Vapors.
The temperature at which a metallic vapor, hke zinc or mer-
cury, commences to condense, depends upon the vapor tension
curve of the metal and the amount of other uncondensable gas
650 METALLURGICAL CALCULATIONS.
with which it is mixed. These intermixed gases reduce the
pressure upon the metallic vapor, and lower the temperature at
which the vapor becomes saturated vapor. The phenomenon is
exactly analogous to the " dew point " of air and the precipita-
tion of rain.
Problem 136.
A roasted zinc ore contains 15 per cent. Fe^ 0' and 70 per cent.
Zn O. It is reduced by excess of carbon, in a retort.
Required:
(1) The average composition, by volume, of the gas mixture
entering the condenser, assuming no CO^ in it.
(2) The " dew point " of the mixture at which it begins to
deposit zinc.
(3) The percentage of zinc escaping condensation, if the gases
leave the condenser at 600° C.
Solution:
(1) Per kilogram of zinc there is used, assuming complete
reduction :
1-^(0.70X65/81) = 1.78 kg.
Oxygen in 1.78 kg. of ore:
As Zn O = 1X16/65 = 0.246 kg.
As Fe^O^ = 1.78X0.15X48/160 = 0.080 "
Sum = 0.326 " '
Co formed = 0.326X28/16 = 0.57 "
Volume of CO = 0.57^1.26 = 0.45 m»
Volume of Zn = 1.00-h2.93 = 0.34 m^
Sum = 0.79 m»
Percentage composition of gases :
CO = 57 per cent.
Zn = 43 " (1)
(2) If the barometric pressure is assumed normal, then the
zinc vapor supports
760X0.43 = 327 mm
and the temperature at which the mixture becomes saturated
with zinc vapor is found from the table on page 620 to be 847°.
This is 83° below the normal boiling point of zinc.
THE METALLURGY OF ZINC. 651
(3) From 847° down, the vapor in the retort will be always
saturated, its tension being found from the tables. The propor-
tion of original zinc condensed, .or remaining uncondensed,
cannot be inferred §imply from the vapor tension at any tem-
perature. For instance, the vapor tension of the zinc at 600° C.
is 12 mm of mercury column. The CO gas leaving the con-
denser at this temperature will, therefore, be at a tension of
760—12 = 748 mm, and will be accompanied by 12/748 of its
volume of zinc vapor; as it came into the condenser it was ac-
companied by 327/433 of its volume of zinc vapor. The ratio
of these two quantities or proportions represents the real frac-
tion of the zinc escaping uncondensed; that is, the proportion
escaping condensation, in terms of the total zinc concerned, is
. 12/748-7-327/423 = 0.016 -^ 0.773 == 0.008 = 0.8 per cent.
The same result can be calculated in several ways. One is
based on the datum- that at any given temperature zinc vapor
is 65-i-28 = 2.3 times as heavy as CO gas.
Problem 137.
Numerous attempts have been made to reduce Zn O continu-
ously in a blast furnace, and condense the zinc from the gases.
In such a case, the gases will consist of zinc vapor, carbon
monoxide and nitrogen, and the deficit of reduction heat must be
obtained by the burning of carbon to CO at the region of the
tuyeres. A low charge column must be used in order that the
gases pass out hot enough to carry the zinc. Assume the gases
to escape at 900° C, and the furnace to lose by radiation and
conduction, etc., an amount of heat equal to the sensible heat in
the blast used.
Required:
(1) The amount of fixed carbon to be charged into the fur-
nace, per 2000 lb. of zinc.
(2) The composition of the gases leaving the furnace.
(3) The temperature at which these gases will begin to de-
posit zinc or become saturated with zinc vapor.
(4) The proportion of the zinc they carry which will escape
from the condensers, as vapor, at 600° C.
(5) The pressure necessary to apply to the furnace to keep
652 METALLURGICAL CALCULATIONS.
the zinc in the melted state at the minimum temperature of re-
duction, 1033° C.
Solution:
(1) If the reaction is assumed as practically
ZnO + C = Zn + CO
there is needed for reduction, per 2000 lb. of zinc
2000 X (12/65) = 369 1b.
There is in this reaction a deficit of heat, which is per 65 parts
of zinc concerned simply
(Zn, 0)-(C, O) = 84,800-29,160 = 55,640.
This deficit can only be made up by burning more C to CO
right at the tuyeres. If two more atoms of carbon were burned,
we would have a little more than enough heat. We will, there-
fore, assume the reaction at the tuyere region as the using of
one atom of carbon for reduction and of double as much for
supplying the heat deficit, and since one volume of oxygen is
accompanied by 3.81 volumes of nitrogen, the whole reaction
will be
Zn 0 + 3 C + 02 + 3.81 N^ = Zn + 3 CO + 3.81 N^
The amount of fixed carbon necessary is approximately three
times that necessary for reduction only and amounts, per
2000 lb. of zinc, to
36X^= 11081b. (1)
DO
(2) The fumacb gases, therefore, contain, by volume,
Zinc vapor = 1 volume = 12.8 per cent.
Carbon monoxide =3 " = 38.4
Nitrogen =3.81 " = 48.8 " (2)
(3) The gases will be saturated with zinc vapor when the
partial pressure of the zinc, which is
760X0.128 = 97 mm
is equal to the maximum tension of zinc at the temperature
attained. This is at 745° C, as seen from inspecting the vapor
tension tables, or 185° below the normal boiling point of zinc. (3)
THE METALLURGY OF ZINC. 653
(4) The relative volume of zinc vapor to uncondensable gas
in the original gases is
In the gases issuing from the condenser at 600° it would be
12.0
748
= 0.016
The proportion of the zinc escaping condensation under these
conditions would, therefore, be
—j^ = 0.109 = 10.9 per cent. (4)
By cooling the gases to the melting point of zinc, it would
be possible to leave uncondensed only
^^^-^0.147 = 0.0014 = 0.14 per cent.
(5) If it was a question only of keeping pure zinc from vaporiz-
ing at 1033°, in the absence of other gases, we could at once
consult the vapor tension table, and find 2.2 atmospheres
as the actual tension of zinc vapor at 1033°, corresponding to
1,2 atmospheres effective pressure upon the furnace. But in
the gas mixture coming from the furnace the zinc is practically
one-eighth, by volume, of the gases, which means that it sus-
tains one-eighth of the pressure upon the whole. It would,
therefore, take a tension of
2.2X8 = 17.6 atmospheres
to keep all this zinc in the liquid state, corresponding to an
effective pressure of 16.6 atmospheres. (5)
This is an entirely impracticable working condition and
shows the practical impossibility of the schemes upon which
much money have been spent for reducing zinc ore to liquid
zinc in a blast furnace run under high pressure.
Problem 138.
In the metallurgy of mercury, the average ore contains 2
per cent, of Hg S and is roasted by the use of 10 per cent, of it5
654 METALLURGICAL CALCULATIONS.
weight of wood having an average composition of water 20
per cent., carbon 32, hydrogen 5.3, oxygen 42.7. Assume that
just enough air enters to completely bum the sulphur and car-
bon, that the ore and air used are dry.
Required:
(1) The percentage composition by volume of the roaster
gases.
(2) The temperature at which the mercury will commence
to condense.
(3) The proportion of the mercury present escaping as
vapor if the gases pass out of the condensers at 100° C, at
50° C, at 15° C.
Solution:
(1) Per 1000 kg. of ore, containing 20 kg. of Hg S, there
must be burned
20 X 32/232 = 2 . 8 kg of sulphur.
100X0.32 = 32 " " carbon.
requiring of oxygen :
2.8X1
=
2,
.8:
kg.
32 X 32/12 =
Sum =
85.
,3
u
88,
,1
u
Nitrogen accompanying = S
Air = i
!93,
,7
u
i81,
,8
a
Composition of the gases:
Hg =
17.2 kg = 17.2
H-9.00
= 1.91
m»
=
0.5
S0= =
5.6
" = 5.6
■^2.88
= 1.91
l(
=
0.5
W0 =
68.0
" = 68.0
-^0.81
= 83.95
It
=
22.1
w =
293.7
" = 293.7
-4-1.26
= 233.10
it
=
61.3
C0'' =
117.3
" = 117.3
-M.98
= 59.24
It
=
15.6
per cent.
380.11 " = 100.00 (1)
(2) In reality, each of these gases possesses the volume of the
gas mixture, and is at the above percentage of the ■ normal
barometric pressure upon the mixture. The tensions of the
different constituents of the mixture are, therefore,
THE METALLURGY OF ZINC. 655
Hg 760 X 0.005 = 3.8 mm
S02 760 X 0.005 = 3.8 "
H^O 760 X 0.221 = 168.0 "
N^ 760 X 0.613 = 465.9 "
CO'' 760X0.156 = 118.5 "
760.
The mercury will, therefore, commence to condense when the
temperature of the gas mixture is that of mercury vapor at a
maximtun tension of 3.80 mm. From the vapor tension table
of mercury we find this to be 156° C, or 201° below the nor-
mal boiling point of mercury. Above this temperature the gas
mixture is not saturated with Hg vapor and no condensation
can occur. (2)
(3) In the gas mixture, the mercury vapor is equivalent to
3.8/760 of the volume of the whole, or 3.8/756.2 of the volume
of the other gases. This latter proportion is 0.00503. At
100° C, the uncondensed mercury vapor can have a tension of
0.285 mm, leaving 759.715 to the other gases, and giving the
ratio of mercury vapor to other gases 0.000375. Since the other
gases have remained constant in amount, the proportion of
mercury remaining condensed is
0.000375 -^ 0.00503 = 0.075 = 7.5 per cent.
If the temperature of the mixture is reduced to 50° C, the
mercury vapor remaining can exert a tension of only 0.013 mm
and the water vapor only 92 mm. The other gases present will
therefore sustain 760- (92 + 0.013) = 668 mm. In the origi-
nal mixture the Hg and WO vapors are, respectively, equiAa-
lent to
Hg 3.8 ^ 587.2 = 0.0065
H^O 168.0 ^ 587.2 = 0.2860
of the volume of N'', CO^ and SO^ together.
In the gas escaping at 50° the Hg and H='0 will be equiva-
lent to
Hg 0.013-^668 = 0.00002
H20 92. -=-668 = 0.1377
of the volume of the same gases.
The proportions of the Hg and H^O escaping uncondensed
at 50° will, therefore, be
656 METALLURGICAL CALCULATIONS.
Hg 0.00002 -5- 0.0065 = 0.0031 = 0.31 per cent.
H^O 0.1377 -^ 0.2860 = 0.4815 = 48.15 per cent.
By exactly similar reasoning, using the maximum tensions of
Hg and H^O at 15° (0.0009 mm and 13 mm, respe,ctively), the
proportions of each to the N^ + CO^ + SO'' are
Hg 0.0009 H- 747 = 0.0000012
Hi'O 13. -^ 747 = 0.0174
and the proportions escaping condensation would be
Hg 0.0000012 -^ 0.0065 = 0.00018 = 0.018 per cent.
H^O 0.0174 -=- 0.2860 = 0.06064 = 6.064 per cent. (4)
The differences between these percentages and 100 will be
the proportions of the^e materials condensed to the liquid state.
Metallic Mist or Fume.
When zinc or mercury are in the state of vapor their atoms
are each separate from other atoms. Chemically speaking,
their vapor molecules are mon-atomic. On condensing to the
liquid state we do not know whether the atoms come together
into compound molecules or whether they simply come closer
together. In either case, the molecules of the liquid metal, as
the temperature is reduced low enough to form them, exist at
first as isolated molecules, constituting the liquid metal in its
finest possible state of sub-division, so small that it is for the
time being practically still a gas. If, however, we give these
liquid molecules the possibility of uniting to liquid masses, we
get the latter. This is a matter principally of reducing the sur-
face tension which holds the liquid molecules each to itself as a
sphere. Contact with a rubbing surface, with dust particles,
filtration through cloth, or even electrification (perhaps a mag-
netic field) reduce this surface tension and cause agglomera-
tion into liquid masses which precipitate.
The amount of liquid or solid particles thus held in suspen-
sion and escaping with the current of uncondensable gas is en-
tirely independent of the quantity escaping as true vapor, which
has been calculated above, except in so faf as they are both
functions of the amount of said uncondensable gas. The veloc-
ity of the escaping current is a large factor in the amount of
mist, because the carrying power of gas for mist particles varies
THE METALLURGY OF ZINC. 657
probably with the cube of its velocity, so that halving the
velocity of the issuing gas will diminish greatly this source
of loss. Large settling chambers in which the mist can de-
posit, because of stagnation of the gas current, and large rub-
bing surfaces, are effective means for catching or depositing
the mist. Filtration through bags is still more effective.
In the case of zinc, the sohdifying point is 420°, while con-
densation from the state of vapor begins at, say, 860°. In this
interval only can the condensed particles, as a mist, collect
together to liquid zinc. If the temperature passes quickly
through this range, the particles have Httle chance to agglomer-
ate into liquid masses, and the proportion which chills into
soUdified mist ^particles is increased. These solidified mist
particles, analogous to " hoar frost " in nature, form the well
known " blue powder " of the zinc works. It settles in large
settling chambers, and is in extremely fine particles, mostly
under 0.01 mm in size. It is quite easy to collect all the zinc
in this form if the vapor is chilled suddenly and the resulting
gas settled properly or filtered.
The loss of mercury or zinc as mist or hoar frost is there-
fore, to be considered entirely apart from the loss as true vapor;
it may be smaller than the latter and it may be larger ; its amount
varies particularly with the nature of the settling apparatus
and the velocity of the gases as they escape. The theory as
to its amount under any given set of conditions would be very
difficult to elaborate, and the data for doing so with exactness
are practically unknown. The best that can be done at present
is to study the loss as mist or hoar frost in actual practice,
experimentally, and tabulate the actual results as a guide for
future metallurgical use.
CHAPTER V.
METALLURGY OF ALUMINIUM.
The two essential principles here involved are " differential
reduction " as used in the electric furnace purification of alumina,
and " electrolytic furnace operation," as illustrated in the
decomposition of the alumina by electrolysis in the manner
usually practiced. Only the latter problem will be covered here.
Electrolytic Furnace Reduction of Alumina.
The Hall process is beautifully simple and technically admir-
able. AP O^ is found to dissolve in melted alkaline-aluminium
double fluorides; it is as pretty a case of solution, so far as ap-
pearances go, as dissolving a spoonful of powdered sugar in a
glass of distilled water. The melting point of the fused fluorides
is reduced by the solution of alumina, just as that of water is
reduced by dissolving salt. In passing the electric current the
constituents of the dissolved alumina appear at the electrodes,
oxygen at the anode and aluminium at the cathode. The best
practical arrangement is to use a carbon-lined pot, with molten
aluminium in the bottom as the cathode, upon it a few inches
depth of the bath, and dipping into this the carbon anodes. The
writer has given most of the technical details of this operation
in his treatise on "Aluminium," and Prof. Haber has published
extensive laboratory studies of the process in the Zeitschrift
fur Elektrochemie.
With a solvent salt consisting of melted sodium fluoride and
aluminium fluoride, such as called for in one of the Hall patents,
with alumina dissolved therein, and using carbon anodes, the
electrolytic elements of the process are simplicity itself. The
bath contains sodium, aluminium, fluorine and oxygen, and the
anode is carbon. Under these conditions those elements or com-
pounds will form at the electrodes which cannot further react
upon the bath material; in other words, those materials most
stable in contact with the bath material or electrodes. A mo-
658
THE METALLURGY OF ALUMINIUM. 659
merit's reflection explains what happens, and what must happen.
At the cathode, sodium cannot be liberated because metallic
sodium reacts chemically on this bath, separating out alumin-
ium; therefore, the electrolytic reducing tendency at the sur-
face of the cathode can only expend itself in separating out
aluminium. At the anode, fluorine cannot be liberated because
fluorine acts strongly upon alumina even when cold, converting
it into fluoride and expelling its oxygen; therefore, the electro-
lytic perducing tendency at the surface of the anode will tend
to set free oxygen. But oxygen cannot be set free at a carbon
surface at a cherry-red heat because of its inevitable tendency
to unite with the carbon to form CO. The electrolytic agency
at the surface of the carbon anode must, therefore, cause the
formation of CO. The whole electrolytic operation results in
the removal of APO^ from the bath and the formation of alumin-
ium and carbon monoxide.
The thermochemical relations, as far as known, agree abso-
lutely with the above explained experimental results. The
materials in presence of each other are sodium fluoride, alumin-
ium fluoride, aluminium oxide and carbon. The heats of
formation of these, per molecule and per chemical equivalent
concerned, are as follows :
(Na, F) = 109,720 = 109,720 per chemical equivalent.
(Al, F^) =275,220= 91,740 "
(AP, O^) = 392,600 = 65,430 "
(C, 0) = 29,160 = 14,580 "
The heat of formation of carbon tetra-fluoride is unknown,
but is probably small, since it is so difficult to form.
From the last column, which largely governs the work done
by the current, we see that the current does far the least work
when it decomposes alumina; in fact, it does still less, by 14,580
calories, because of the assistance rendered by carbon uniting
with the oxygen. Now, although the electrical current does not
consistently adhere to the doctrine of " least work," yet it does
in this case because forced to do so by the chemical relations
of sodium, aluminium, fluorine, oxygen and carbon at the tem-
perature of the bath, as explained in the preceding analysis.
It is probable that in electrolysis the chemical relations of the
possible products control what the current does rather. than the
660 METALLURGICAL CALCULATIONS.
thermochemical relations alone, but in most cases the two condi-
tions or controlling circumstances coincide in their influence
and lead to identical results. This is the case in the process
in question ; it is absolutely normal and is explainable by either
method of reasoning.
If the operation is run with a small vessel and correspond-
ingly small current, the heat necessarily evolved by the passage
of the current is small compared to the conduction and radia-
tion losses and must be supplemented by outside heating to
keep the contents at proper temperature. If the size of the
operation is increased in all its items and' dimensions, the neces-
sarily generated internal " resistance " heat will suffice to keep
the bath at the requisite temperature, when the enlargernent is
done on a certain scale. If enlarged past this point, too much
internal heat is unavoidably generated and means must be used
to artificially cool the pot.
Problem 139.
An electrolytic vessel is. composed of a block of carbon 25 cm
cube, with a cavity 10 cm square by 10 cm deep inside. The
cavity has a round carbon, 6 cm diameter, dipping into it. The
vessel weighs 30 kilograms; the fused bath in the cavity 2 kg;
the carbon immersed in it 0.1 kg. The specific heat of the car-
bon is 0.5, of the bath 0.3, at the running temperature. An ex-
periment showed that the bath material cooled off, at the work-
ing temperature, at the rate of 10° C. per minute, the walls of
the vessel at an average rate of 2° C. per minute, when left to
cool by themselves.
Required:
(1) The number of watts which must be converted into heat
in the vessel in order to maintain it at the working temperature.
(2) Assuming 75 per cent, of the theoretical ampere efficiency
to be obtained, what amperes passed through the pot will keep
it at working temperature if the working voltage is kept at 10
volts ?
Solution:
(1) The 30 kg of vessel material losing heat at the rate of
2° per minute, with specific heat of 0.5, gives a heat loss per
minute of
30X0.5X2 = 30 Calories.
THE METALLURGY OF ALUMINIUM. 661
Similarly, the immersed carbon in the cavity and the bath
material itself lose
0.1X0.5X10 = 0.5 Calories.
2.0X0.3X10 = 6.0
The total heat loss is, therefore, 36.5 Calories per minute.
To maintain the temperature constant the current must fur-
nish this heat, and since 1 watt is 0.239 gram calories per sec-
ond, the watt energy thus converted into heat must be
36.5X1000 -.._ ^^ ,,,
0.239X60 = 2645 watts (1)
(2) If all the amperes passing through separated out metal
the voltage absorbed in decomposition in the bath would be
from the thermochemical heats of formation of chemical equiva-
lent quantities of AP O^ and CO :
65,430-14,580 „ „ u
= 2.2 volts.
23,040
If 75 per cent, of the amperes are efficient, the voltage thus ab-
sorbed is
2.2X0.75 = 1.65 volts.
The voltage disappearing in overcoming ohmic resistance will
then be
10-1.65 = 8.35 volts
and the current which when passed will keep the pot at working
temperature will be
— -- = 306 amperes. (2)
The above-used principles are applicable to any kind of elec-
trolytic-furnace operation.
INDEX.
Addicks, L., on energy loss at electrical contacts 562
Air, composition of 4
Air, received by converter 336
Air, required by converter 333
Alloys, heats of formation 37
Alloys, thermophysics of 109
Alumina, action in blast furnace slag 256
Alumina, behavior in the blast furnace 240
Alumina, reduction of 668
Alumina, thermophysics of 123
Aluminates, heats of formation of 35
Aluminium alloys, thermophysics of 113
Aluminium compounds, heats of formation of 18, 37
Aluminium compounds, thermophysics of 139
Aluminium conductors, econdmical size of 570
Aluminium conductors, heating of, in use 672
Aluminium, electrical conductivity of 569
Aluminium, heat of oxidation of 353, 374
Aluminium, metallurgy of ^ 658
Aluminium, reduction of lead oxide by 586
Aluminium, thermophysics of 80
Amalgam furnace, efficiency of 107
Amalgams, heats of formation of 37
Ammonia gas, mean specific heat of 142
Anode, insoluble, use of 551
Anode, soluble, use of iron as 548
Anode, use of matte as 539
Anthracite coal, combustion of 220, 221
Antimonides, heats of formation of 23
Antimonides, therniophysics of 136
Antimony alloys, thermophysics of 101
Antimony, behavior in the blast furnace 240
Antimony, compounds, heats of formation of 23
Antimony, thermophysics of 05
Aqueous vapor, maximum tension of Ill
Argo, Col., reverberatory smelting at 501
Arsenic, behavior in the blast furnace 240
Arsenic compounds, heats of formation of .... 33
Arsenic compounds, thermophysics of 136, 138
Arsenic, thermophysics of °°
Arsenides, heats of formation of j^^
Ashcroft's process for copper ores °^
Atacamite, formula of °**j
Atomic weights '■
Balance sheet of blast furnace 235
Barium compounds, heats of formation of q_i44.
Barium compounds, thermophysics of H** 1^
Barium, thermophysics of ., ^i
Bauxite, efficiency of furnace, in meltmg ^|'
Beeswax, thermophysics of „„-
Belgian furnace, distillation of zinc ore m a "^o
Bergfeld, on the boiling of silver and gold o^o
663
664 INDEX
Beryllium compounds, heats of formation of 18-40
Beryllium compounds, thermophysics of 113-144
Beryllium, thermophysics of 76
Bessemerizing copper matte 525
Bessemer process, problem concerning 11
Bessemer process, the 333
Betts' refining process for lead 598, 599, 602
bi-carbonates, heats of formation of 29
Bismuth alloys, thermophysics of 110
Bismuth compounds, heats of formation of 18-40
Bismuth compounds, thermophysics of 113-144
Bismuth, thermophysics of 102
Bi-sulfates, heats of formation of 33
Bituminous coal, combustion of 219, 221, 222, 223, 226
Blast, amount used in blast furnace 244
Blast, dry, Gayley process of producing 330
Blast, dry, the rationale of 309
Blast engines, efficiency of 493, 590
Blast furnace, balance sheet of 235
Blast furnace, carbon needed in the 277
Blast furnace gas, in gas engines 226
BlEist furnace gases, surplus power from 272
Blast furnace, heat balance sheet of 281
Blast furnace, reduction of zinc ores 651
Blast furnace, temperature in 223
Blast furnace, utilization of fuel in the 268
Blast, heating of, apparatus for. .'. 318
Blast, hot, the function of 308
Blast pressure required in Bessemer process 340
Blast, warm, use in pyritic smelting 487
Blast, work done in producing 313
Blister roasting of copper matte 523
Blowing engines, efficiency of 299
Boiling of metals in a vacuum 614
Boiling of zinc 641
Borates, heats of formation of 34
Boron compounds, heats of formation of 18-40
Boron compounds, thermophysics of 138
Boron, thermophysics of 77
Brass, thermophysics of 112
Bringing forward of copper matte 514
Bromides, thermophysics of 133
Bromine, thermophysics of 89
Bronze, thermophysics of Ill
Bullion, gold, electrolytic refining of 606
Bullion, silver, electrolytic refining of 605
Burgess process of refining iron- 446
Butte, Mont., copper concentrates from 479
Butte, Mont., roasting furnace at 482
Cadmium alloys, thermophysics of 113
Cadmium compounds, heats of formation of 18-40
Cadmium compounds, thermophysics of 131-144
Cadmium, thermophysics of 93
Cadmium, vapor tension of 644
Caesium compounds, heats of formation of 18-40
Csesium, thermophysics of 96
Calcium compounds, heats of formation of 18-40
Calcium compounds, thermophysics of 131-144
Calcium oxide, heat absorbed in reducing 293
INDEX 665
Calcium oxide, thermophysics of 124
Calcium, thermophysics of 83
Calorific power of fuels, calculation of [[ g03
Calorimeter, test for specific heat by 225
Calorimeter, test for temperature by 224
Campbell, H. H., furnace data by . . . 393
Canada, electrical processes in 435
Canada Zinc Co., operations of the 641
Carbides, heats of formation of 25
Carbon, amount needed in the blast furnace 277
Carbon, heat of oxidation of 352, 375
Carbon, thermophysics of 77
Carbonates, heat absorbed in decomposing 291
Carbonates, heats of formation of 28
Carbonates, thermophysics of 137
Carbonic oxide, thermophysics of 122
Carbonous oxide, thermophysics of 121
Cathodes, use of copper matte as 545
Cathodes, use of lead sulfide as 598
Cement copper, composition of 548
Cerium compounds, heats of formation of 18, 40
Cerium, thermophysics of 97
Chalcopyrite, composition of 474
Charge, blast furnace, calculation of 250
Chimney draft 185
Chimney draft, problems concerning 192, 194, 224, 230, 231
Chimney for open-hearth furnace 395
Chlorates, thermophysics of 139
Chlorides, heats of formation of 27
Chlorides, thermophysics of 131
Chlorine, thermophysics of 82
Chromates, thermophysics of 138
Chromium, heat of oxidation of 353, 374
Chromium, thermophysics of 84
Coal, combustion of powdered 8
Cobalt, behavior in the blast furnace 240
Cobalt compounds, heats of formation of 18-40
Cobalt compounds, thermophysics of 131-144
Cobalt, thermophysics of 87
Coehn, improved Hoepfner apparatus 558
Coke-oven gases, combustion of abb
Coke, use of, in pyritic smelting 488
Columbium, thermophysics of «^
Compression of blast, work done in ^1^
Concentration of zinc ores, cost of <-aq eel
Condensation of metallic vapors d49. 00^
Conduction of heat, principles of orvQ 01 1
Conductivity for heat, tables of ''"^. i^^
Conductors, electrical, economic size of obi
Conductors, electrical, heating of, in use sos
Constants, thermochemical ^r9
Contacts, electric, energy loss at ^oj
Copper alloys, thermophysics of ^^
Copper, behavior in the blast furnace ^?^"
Copper, calculations on metallurgy of *"»■ °i»
Copper chloride, electrolysis of ■ iS-4n
Copper compounds, heats of formation of 131-144
Copper compounds, thermophysics of ,„_
Copper, electrical conductivity of ego
Copper, electrolytic refinery of
666 INDEX
Copper matte, composition of 469, 471
Copper, principles of the metallurgy of 472
Copper ores, electric smelting of 572
Copper solutions, physical properties of 611
Copper solutions, resistivity of 659
Copper, thermophysics of -. 87
Crushing of zinc ores, cost of 621
Cupola, thermal efficiency of 115
Cyanates, heats of formation of 36
Cyanides, heats of formation of 35, 36
Cyanides, thermophysics of 136
Damour, on furnace efficiency 416, 423
Decomposition of moisture of blast 293
Dehydrating charges, heat absorbed in 290
Dellwick-Fleischer gas producer 179
Distillation of zinc from ore 632
Dried blast, rationale of action of 309
Dried blast, temperatures attained with 312
Drying air blast commercially 325
Drying charges, heat absorbed in 290
Drying of wet peat 227
Ducktown pyrrhotite, smelting of 489
Dulong and Petit's laws of specific heats 70
Dulong's laws of heat of combustion of coals 48
Efficiency of electric furnaces 640
Efficiency of electric heating 441
Efficiency of furnaces, discussion of 104
Efficiency of open-hearth furnaces 396
Eissler's hydrometallurgy of copper 469
Eldred process of combustion 55
Electric furnaces, efficiency of 640
Electric melting of steel 440
Electric reduction of iron ore 429
Electric refining of iron 446
Electric smelting of copper ores 572
Electric smelting of zinc ores 637
Electrical contacts, energy loss at 562
Electrolytic furnace, definition of 634
Electrolytic furnace for reducing aluminium 658
Electrolytic refining of copper 558
Electrolytic refining of gold 610
Electrolytic refining of iron 446
Electrolytic refining of lead 698
Electrolytic refining of silver 605
Electrolytic refining, principles of 536
Electrometallurgy of copper 534
Electrometallurgy of iron and steel 429
Electrometallurgy of lead ■. . . 698
Electrothermal processes, definition of 535
Ely, Vermont, copper ore from 475
Ethylene, mean specific heat of 142
Evans-Klepetko furnace, calculations on 482
Feed-water heater, utility of 225
Ferro-cyanides, heats of formation of 36
Fire-clay, heat conductivity of 630
Flues, size of 382
Fluorides, heats of formation of 26
INDEX 667
Fluorides, thermophysics of I34
Flux, amount needed in the blast furnace. .................'. 251
Flux, behavior in the blast furnace 243
Flux, comparison of values of 263
Flux, required in Bessemer converter 346
Freeland, W. H., on smelting pyrrhotite . . . . 489, 515
Fuels, behavior of, in the blast furnace ' 236
Fuels, calorific power of , 4g
Fuels, comparison of values of 261
Fuels, utiUzation of, in the blast furnace 268
Fume, metallic g5g
Furnaces, efficiency of 104
Furnaces, electric, efficiency of 640
Furnaces, electrolytic, definition of 534
Furnaces, electrolytic, for reducing alumina 658
Fusion of latent heats of, discussion 71
Fusion of latent heats of, tables 118-144
Fusion of matte 500
Fusion of sla^ 500
Galena, roasting of 581
Gallium, thermophysics of 88
Gas, artificial 145
Gas, engin6 operation 226
Gas, mixed, producers for 158
Gas, natural, combustion of 10, 47, 53, 219
Gas, natural, from lola, Kansas 636
Gas producers 381
Gases, composition of, from smelting furnace 497
Gases, pressure corrections for 7
Gases, relative volumes of 2
Gases, temperature corrections for 5
Gases, thermophysics of 141
Gases, waste, heat in 287
Gayley process of drying blast 330
Genoa, Marchesi's process at 539
Gin, G., in electrometallurgy of zinc 637
Glass, thermophysics of 141
Gold, behavior in the blast furnace 240
Gold bullion, refining of 610
Gold compounds, heats of formation of 18-40
Gold compounds, thermophysics of 131-144
Gold, metallurgy of 605
Gold, thermophysics of 100
Gold, vapor tension of 616
Gordon, F. W., slag calculations by 261
Goutal's method for calorific power of coal 503
Great Falls, Montana, copper refining at 566
Griiner's ideal working of a blast furnace 274
Haber, on reducing alumina 658
Hall, C. M., on reducing alumina 658
Halske and Siemens' copper process 555
Heat balance sheet of Bessemer converter a5d
Heat balance sheet of blast furnace ^81
Heat balance sheet of electrical furnace 4dd
Heat consumed in reductions •'^^
Heat of formation of chemical Compounds cnn
Heat of formation of slag 5UU
Heat of formation of some nitrates °^"
668 INDEX
Heat of formation of some oxides 579
Heat of formation of some sulfides 678
Heat properties of matte 600
Heat properties of slag '. 600
Heat units 14
Heats of formation 18
HerreshofI furnace, calculations on 489, 515
Hess, H. D., on open-hearth furnaces 395
Hixon's notes on matte smelting 526
Hoepfner's copper chloride process 657, 558
Hofman's Metallurgy of Lead 593
HoUaway, J., on bessemerizing matte 625
HoUaway, J., on oxidation of sphalerite 621
Hot blast, sensible heat in 285
Howe, H. M., notes on a bessemer charge 336
Hydrates, heats of formation of 19
Hydrides, heats of formation of 24
Hydrocarbons, combustion of 45, 217
Hydrocarbons, heats of formation of 24, 217
Hydrogen compounds, heats of formation of 18-40
Hydrogen compounds, thermophysics of 131-144
Hydrogen, thermophysics of 75
Hyposulfites, thermophysics of 136
Ice, specific heat of 119
Ideal working of a blast furnace, Griiner's 274
Ingalls, on cost of treating zinc ores 621
Ingalls, on oxidation of sphalerite 621
Insoluble anodes, use of 651
Iodides, thermophysics of 133
Iodine, thermophysics of 95
lola, Kansas, natural gas from 636
Iridium, thermophysics of 99
Iron alloys, thermophysics of 113
Iron compounds, heats of formation of , 18-40
Iron compounds, thermophysics of 125, 128-130
Iron, electric refining of 446
Iron, heat of oxidation of 352, 366
Iron ores, electrical reduction of 429
Iron oxide, discussion of reduction of 69
Iron oxide, problems concerning reduction of 128, 129
Iron oxides, heat absorbed in reducing 292
Iron persulfate process for copper ores 555
Iron sulfate solutions, resistivity of 559
Iron sulfide,, heat absorbed in reducing 293
Iron, thermophysics of 85
Iron, use as soluble anode 648
Isabella, Tenn., pyritic smelting at 489
Isabella, Tenn., smelting of matte at 616
Jannettaz, "Les Cenvertisseurs pour le Cuivre" 526
Joplin, zinc ore from 620
Jtiptner, data on furnace charges 403
Koch, on warm blast in smelting 487
Krafft, on the boiling of silver and gold 614
Laboratory of open-hearth furnaces 391
Laboratory of furnaces, efficiency of 414
La Lustre smelter, warm blast at 487
INDEX 669
Landis, W. S., laboratory determinations by 500
Latent heats of fusion, discussion of 71
Latent heats of fusion, tables of '.' 75^103 118
Latent heats of vaporization, discussion of ' 73
Latent heats of vaporization, tables of .76-163 118-144
Lead alloys, thermophysics of \ 109
Lead compounds, heats of formation of 18-40
Lead compounds, thermophysics of ' . 131-144
Lead, electrometallurgy of 598
Lead, metallurgy of [ [_ 573
Lead, physical constants of 583
Lead, refining of [ . . 533
Lead, thermophysics of . . . . 101
Lead, use as insoluble anode 551
Lead, vapor tension of 584
Lime, amount needed in Bessemer converter 347
Lime, heat in preheated 356
Lithium compounds, heats of formation of 18-40
Lithium compounds, thermophysics of 131-144
Lithium, thermophysics 76
Magnesium compounds, heats of formation of 18-40
Magnesium compounds, thermophysics of 131-144
Magnesium, thermophysics of 80
Magnus, B., on energy losses at electrical contacts 562
Manganese, behavior in the blast furnace 240
Manganese compounds, heats of formation of 18-40
Manganese compounds, thermophysics of 131-144
Manganese, heat of oxidation of ^. 371
Manganese, oxides, heat absorbed in reducing 293
Manganese, thermophysics of 85
Manhes, P., on bessemerizing matte 525
Marchesi, on electrolytic treatment of matte 539, 541
Marsh gas, mean specific heat of 142
Martin process 381
Matte, bessemerizing of 525
Matte, blister roasting of 523
Matte, bringing forward of 514
Matte, copper, composition of 4.69
Matte, electric furnace production of 574
Matte, electroljrtic treatment of 539
Matte, grade obtained in smelting 474
Matte, thermophysical properties of 500
Maximum tension of aqueous vapor 121
Mayer, P., Das Bessemem von Kupf ersteinen " 526
Mercury alloys, thermophysics of 113
Mercury compounds, heats of formation of 18-40
Mercury compounds, thermophysics 131-144
Mercury, metallurgy of 653
Mercury, thermophysics of 100
Mercury, vapor tension of 584
Metallic mist 656
Methane, mean specific heat of 142
Mine water, extraction of copper from 548
Mist, metallic °56
Moisture, heat absorbed in decomposing ^93
Molybdenum compounds, heats of formation of o, 7; V
Molybdenum compounds, thermophysics of 131-144
Molybdenum, thermophysics of 91
Mond gas 1"°
670 INDEX
Mond gas, superheating of 175
Monell process, problem concerning 424
Morgan producer, problem concerning 163
Mount Lyell, gases of furnace at 497
Mount Lyell, pyritic smelting at 488
Natural gas, combustion of 10, 47, 53, 219
Natural gas, from lola, Kansas 636
Niagara Falls, Salem process as used at 599
Nickel alloys, thermophysics of 113
Nickel, behavior in the blast furnace 240
Nickel compounds, heats of formation of 18-40
Nickel compounds, thermophysics of 131-144
Nickel, heat of oxidation of 353, 374
Nickel, thermophysics of 86
Nickeliferous pyrrhotite, reduction of 435
Nitrates, heats of formation 30, 609
Nitrates, thermophysics of 137
Nitrides, heats of formation of 23
Nitrogen, thermophysics of 78
Oil, fuel, combustion of 219
Oil furnace, efficiency of 106
Open-hearth furnaces 381
Ore, behavior of, in the blast furnace 239
Ore, comparison of values of 265
Ore, electrical reduction of 429
Ores, copper, direct treatment by electrolysis 536
Osmium, thermophysics of 98
Oxides, heat absorbed in reducing 292
Oxides, heats of formation of 18, 579
Oxides, thermophysics of 118
Oxygen, thermophysics of 79
Palladium compounds, heats of formation of 18-40
Palladium, thermophysics of 92
Paraffin, thermophysics of 142
Parrot Works, Butte, Bessemer process at 525
Peat, kiln drying of 227
Peters, "Modern Copper Smelting" 469, 526
Peters, "Principles of Copper Smelting" 469, 484, 566
Phosphates, heats of formation of 33
Phosphates, thermophysics of 138
Phosphides, heats of formation of 23
Phosphorus, behavior in the blast furnace 240
Phosphorus, heat of oxidation of 353, 377
Phosphorus, oxide, heat absorbed in reducing 293
Phosphorus, thermophysics of 81
Pig-iron, heat of formation of 285
Pig-iron, heat in melted 289
Platinum alloys, thermophysics of 113
Platinum compounds, heats of formation of 18-40
Platinum compounds, thermophysics of 131-144
Platinum, thermophysics of 99
Ports of an open-hearth furnace 388
Potassium compounds, heats of formation of 18-40
Potassium compounds, thermophysics of 131-144
Potassium, thermophysics of 83
Pot roasting, applicability to sphalerite 621
Pot roasting, principle of 586
INDEX
671
Potter, H. N., on heat of formation of silica
Powdered coal, combustion of
Power obtainable from furnace gases
Power required to compress blast
Power, surplus, from blast furnace gases
Pressure of blast, methods of measuring
Pressure of blast, required in the Bessemer
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Producer gas, combustion of 219, 227, 228, 229
Producers, gas 246, 381
Pyritic smelting ^°^
672 INDEX
Pyritic smelting, reaction at the focus 498
Pyritic smelting, slag produced by 487
Pjrritic smelting, temperature attained in 485
Pyrrhotite, electrical reduction of 435
Pyrrhotite, smelting of 489
QuENEAU, on furnace eflBciency 416, 423
Radiation, heat 213
Radiation, from a Bessemer converter 533
Radiation, from a blast furnace 289
Radiation, from an electrolytic furnace 660
Radiation, from a roasting furnace 479, 482, 627
Radiation, from a smelting furnace ' 506
Radiation, heat, tables of 213
Radiation, heatjng by 392
Radium, thermophysics of 102
Reactions, double, in copper metallurgy 474
Reactions, double, in lead metallurgy 682
Recarburization, in the Bessemer converter 348
Reduction of iron ores electrically 429
Reduction of roasted lead ore 591
Reduction of zinc oxide 627
Refining, electrolytic, of copper 558
Refining, electrolytic, of gold 610
Refining, electrolytic, of iron 446
Refining, electrolytic, of lead 698
Refining, electrolytic, of silver 605
Refining, electrolytic, .principles of 536
Refining, of impure lead 583
Regenerators, dimensions of 382
Regenerators, efficiency of 412
Resistivity of some solutions 559
Reverberatory smelting 501
Rhodium, thermophysics of 92
Richard, T. A., on pyrite smelting 484
Roasting, Bessemer, of lead ores 686
Roasting, heat generated in 477, 479
Roasting of lead sulfide 581, 586
Roasting of sphalerite 620
Roasting, removal of sulfur in. . . . .- 473
Roasting, smelting of copper matte 523
Rubber, thermophysics of 142
Rubidium, thermophysics of; 90
Ruthenium, thermophysics of 91
Salom, p. G., process for copper matte 545
Salom, P. G., process for reducing galena 598
Saulte Ste. Marie, electrical reduction at 435
Savelsberg pot-roasting operation 586
Schnabel "Handbook of Metallurgy" 469, 526
Schuller, on the boiling of silver and gold 615
Selenides, heats of formation of 22
Selenium, thermophysics of 89
Siemens and Halske's copper process 555
Siemens-Martin process 381
Siemens new-style furnace 418
Silica, behavior of, in the blast furnace 240
Silicates, heats of formation of 31
INDEX 673
Silicates, thermophysics of I39
Silicides, heats of formation of ........'. 26
Silicon compounds, heats of formation of '. 18-40
Silicon compounds, thermophysics of 139
Silicon, heat of oxidation of 353, 369
Silicon, thermophysics of ; ' 81
Silver alloys, thermophysics of 113
Silver, analyses of impure 606
Silver, behavior in the blast furnace 240
Silver, compounds, heats of formation of 18-40
Silver compounds, thermophysics of 131-144
Silver, metallurgy of 605
Silver, thermophysics of 93
Silver, vapor tensions of 616
Slag, amount produced in the blast furnace 251
Slag, blast furnace, composition of 252
Slag, blast furnace, heat in 289
Slag formed in the Bessemer converter 346
Slag, fusion of, in the copper furnace 500
Slag, heat of formation of ■ 286, 357
Slag, summation of ingredients of 254
Slag, thermophysics of 144
Smelting electric of copper ores 572
Smelting of lead ores 591
Smelting of zinc ores , . 637
Smelting, ordinary, of copper ores 499
Smelting, pyritic, rate of ,. 489
Snyder, F. T., on efficiency of electric furnaces 640
Snyder, F. T., on electric zinc smelting 641
Sodium compounds, heats of formation of 18-40
Sodium compounds, thermophysics of 131-144
Sodium, thermophysics of 79
Solutions, extraction of copper from 548
Specific heats of chemical compounds 131-144
Specific heats of products of combustion 51
Specific heats of the elements 76-103
Speiss, formation of ; 592
Spence furnace, cailculations on a 479
Sphalerite, roasting of 620
Steel, electrical production of 440
Steel furnace, efficiency of 114
Steel, thermophysics of 113
Sticht, analysis of furnace gases by 497
Sticht, on pyritic smelting 484
Stolberg, Marchesi's process at 539
Strontium compounds, heats of formation of o 7:5
Strontium compounds, thermophysics of on
Strontium, thermophysics of 90
Sulfates, heats of formation of o^
Sulfates, thermophysics of 1^6
Sulfides, heats of formation of 1 q5
Sulfides, thermophysics of lo4
Sulfur, available in ordinary smelting 4/4
Sulfur, available in pyritic smeltmg 4St.
Sulfur, behavior in blast furnace • • • ■ jAt
Sulfur, condition of, in roasted ore. 475, 479, 507
Sulfur, heat absorbed in separation of ^a^
Sulfur, removal by partial roasting 4/d
Sulfur, thermophysics of • °^
Sulfuric acid solutions, resistivity of oo»
43
674 INDEX
Tantalum, thermophysics of 97
Tellurides, heats of formation of 22
Tellurium compounds, heats of formation of 18-40
Tellurium, thermophysics of 96
Temperature, theoretical, of combustion 50
Temperature, thermochemistry of high 61
Temperatures in the Bessemer converter 368
Temperatures in the blast furnace . . ., 308
Temperatures, using dry blast 312
Tension, majdmum, of aqueous vapor 121
Thallium compounds, heats of formation of 18-40
Thallium compounds, thermophysics of 131-144
Thallium, thermophysics of 101
Thermochemical constants of bases and acids 38
Thermochemistry, applications of 13
Thermochemistry, calculations in 41
Thermochemistry of the Bessemer process 351
Thermophysics of alloys 109-1 13
Thermophysics of chemical compounds 118-144
Thermophysics of the elements 75-103
Thermit process, calculations 43
Thermit process, temperatures in 58
Thorium, thermophysics of 103
Tin alloys, thermophysics of 109
Tin compounds, heats of formation of 18-40
Tin compounds, thermophysics of 131-144
Tin, thermophysics of 94
Tissier, reduction of lead oxide by aluminium 585
Titanates, heats of formation of 34
Titanium, heat of oxidation of 353, 373
Titanium, thermophysics of 83
Toldt, data on furnace charge 403
Tungstates, heats of formation of 33
Tungsten compounds, thermophysics of 131-144
Tungsten, heat of oxidation of 353
Tungsten, thermophysics of 97
Ulke, modern electrolytic copper refining 469
Units, heat, definition of 14
Uranium, thermophysics of 103
Vacuum, volatility of gold and silver in a 614
Valves for open-hearth furnaces 388
Vanadium, heat of oxidation of 353
Vanadium, thermophysics of 84
Vancouver, B. C, electric zinc smelting at 641
Van Liew, on bessemerizing matte 530
Vapor, aqueous, maximum tension of 121
Vaporization, latent heats of, discussion 73
Vaporization, latent heats of, tables 75-103, 118-144
Vapor, metallic, calculation of weight of 618
Vapor, tension of cadmium 644
Vapor, tension of gold 615
Vapor, tension of lead 583
Vapor, tension of mercury • 583
Vapor, tension of silver 615
Vapor, tension of zinc 643
Vattier, on electric smelting of copper ores 573
Volatility of lead 583
Volatility of silver and gold 614
INDEX 675
Voltages of decomposition 647
Vulcanite, thermophysics of 142
Water, cooling, heat in 290
Water gas 177
Water gas, combustion of 219
Water, heat of formation of 16
Water, thermophysics of 119
Water vapor, maximum tension of 121
Watts, O. P., on boilingpoints of silver and gold 615
Weightman, A. T., on treatment of copper matte 545
Welch process of copper concentration 514
Wohlwill process of refining gold bullion 610
Work done by blowing engines 313
Zinc alloys, thermophysics of 110
Zinc, behavior in the blast furnace 240
Zinc compounds, heats of formation of , 18-40
Zinc compounds, thermophysics of 131-144
Zinc, distillation of 632
Zinc, distilling furnace, efficiency of 108
Zinc, electric reduction of 637
Zinc, heat of oxidation of 353
Zinc, metallurgy of 620
Zinc oxide, blast-furnace reduction of 651
Zinc oxide, electric reduction of 626
Zinc oxide, reduction of 63, 627
Zinc oxide, specific heat of 621
Zinc, reduction in a blast furnace 651
Zinc sulfide, roasting of 620
Zinc sulfide, specific heat of 621
Zinc sulfide, temperature of roasting 621
Zinc, thermophysics of 87
Zinc, vapor of °*^
Zirconium, thermophysics of 91*
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