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MetaDuigical and Chemical Engineering R>wer 



ENGINEERING 
THERMODYNAMICS 



BY 



CHARLES EDWARD LUCKE, PH.D., 

Prtf€990T of Mechanical Engineering in Columbia Uniwreity, New York CUy 



McGBAW-HILL BOOK COMPANY 

239 WEST 39TH STREET, NEW YORK 
6 BOUVERIJB STREET, LO^DO^, E. C. 

1912 



Copyright 1912, by thb 
MoGRAW-HILL BOOK COMPANY 



THK SCICNTIPIC PflKSS 
ROBKflT DRUMMONO AND COMPANY 
■ROORLYN. N. Y. 



172771 

MAW 3 1 1913 

PREFACE 



^3^^^ 



V 



Calcttlations about heat as a form of energy, and about work, another 
related form, both of them in connection with changes in the condition of all 
sorts of substances that may give or take heat, and perform or receive work 
while changing condition, constitute the subject matter of this book. The 
treatment of the subject matter of this text is the result of personal experience 
in professional engineering practice and teaching students of engineering at 
Columbia University. 

Even a brief examination of the conditions surrounding changes in sub- 
stances as they gain or lose heat, do work or have work done on them, and of 
the corresponding relations between heat and work as forms of energy independ- 
ent of substances, will convince any one that the subject is one of great com- 
plexity. Accordingly the simplicity needed for practical use in the industries 
can be reached only by a consideration of a great mass of sub-topics and data. 

That the doing of work, and the changes in heat content of substances were 
related phenomena, and that these relations when formulated, would con- 
stitute a branch of science, was conceived about a half century ago, and the 
science was named Thermodynamics. The Engineer Rankine, who helped 
to create it, defined thermodynamics as " the reduction of the laws according 
to which such phenomena took place to a physical theory or corrected system 
of principles." Since Rankine's. time thermodynamics has become a very 
highly developed science and has proved of great assistance in the formu- 
lation of modern physical chemistry, and to those branches of engineering 
that are concerned with heat. Unfortunately, as thermodynamics developed 
as a separate subject it did not render proportionate service to engineering, 
which itself developed even more rapidly in the same period under the guidance 
of men whose duty it was to create industrial apparatus and make it work 
properly, and who had little or no time to keep in touch with purely scientific 
advances or to interpret such advances for utilitarian ends. Thermodynamics 
proper is concerned with no numerical quantities nor with any particular 
substance nor for that matter with any actual substances whatever, but it is a 
physical theory of energy in relation to matter as a branch of natural philosophy. 
Engineering, however, is concerned with real substances, such as coal, steam, 
and gases and with nimierical quantities, horse-powers of engines, temper- 
atures of steam, the heats of combustion of oils, so that alone, the principles 
of thermod3mamic philosophy will not yield a solution of a practical problem, 



vi PREFACE 

be it one of design or one of analysis of test performance of actual heat machine 
or thermal apparatus. It is the province of engineering thermodynamics to 
guide numerical computation on thermal problems for real substances being 
treated in real apparatus. Its field, while including some of that of pure 
thermodynamics, extends far beyond the established provinces of that subject 
and extends to the interpretation of all pertinent principles and facts for purely 
useful purposes. Engineering thermodynamics, while using whatever prin- 
ciples of pure thermodynamics may help to solve its problems, must rely on 
a great mass of facts or relations that may not yet have risen to the dignity 
of thermodynamic laws. The workers in shops, factories, power plants or 
laboratories engaged in designing or operating to the best advantage machines 
and apparatus using heat with all sorts of substances, have developed great 
quantities, of rules, methods and data that directly contribute to the ends sought. 
While for each class or type of apparatus there has grown up a separate set 
of data and methods in which much is common to several or all groups, not 
nearly so much assistance is rendered by one to another as should be by a proper 
use of engineering thermodynamics, which applies methods, principles and 
conclusion to all related problems. Classes of apparatus about which such 
groups of methods of analysis or synthesis, or collections of special data 
have developed and which it is province of engineering thermodjmamics to 
unify so far as may be, include air compressors, and compressed air engines, 
reciprocating steam engines, steam turbines, steam boUers, coal-, oil- and gas- 
fired furnaces, gasifiers of coal and oil, gas producers, gas engines, complete 
steam or gas power plants, mechanical refrigeration and ice-making plants and 
chemical factory equipment, or more generally, machinery and apparatus for 
heating and cooling, evaporating and condensing, melting and freezing, moisten- 
ing and drying, gasification and combustiom. 

The nature of the subject and its division are better indicated by the 
classes of problems to be solved by its aid or the contributions expected of it 
than by the kinds of apparatus to which they apply. Probably its broadest 
contribution is the establishment of limits of possible performance of heat 
apparatus and machines. These limits will show what might be expected of 
a steam engine, gas engine or refrigerating machine when its mechanism is quite 
perfect and thus they become standards of reference with which actual per- 
formance can be compared, and a measure of the improvements yet possible. 
These same methods and practices are applicable to the analysis of the operat- 
ing performance of separate units and complete plants to discover the amount 
of energy being lost, how the total amount is divided between the different 
elements of the apparatus, which of the losses can be prevented and how, and 
finally which are absolutely unavoidable. This sort of analysis of the per- 
formance of thermal apparatus is the first step to be taken by the designer 
or manufacturer to improve the machine that he is creating for sale, and is 
essential to the purchaser and user of the machine, who cannot possibly keep 
it in the best operating condition without continually analyzing its performance 
and comparing results with thermodynamic possibilities. 



PREFACE vii 

The subject naturally divides into three general parts, the first dealing 
with the conditions surrounding the doing of work without any consideration 
of heat changes, the second heat gains and losses by substances without reference 
to work involved and the third, transformation of heat into work or work into 
heat in conjunction with changes in the condition of substances. The first 
part applies to the behavior of fluids in the cylinders of compressors and engines. 
The second part is concerned with the development of heat by combustion, 
its transmission from place to place, and the effect on the physical condition of 
solids, liquids, gases with their mixtures, solutions and reactions. The third 
part is fundamental to the efficient production of power by gases in internal 
combustion gas engines or compressed-air engines and by steam or other vapors 
in steam engines and tiu'bines, and likewise as well to the production of 
mechanical refrigeration by ammonia, carbon dioxide and other vapors. 

Accordingly, the six chapters of the book treat these three parts in order. 
The first three chapters deal with work without any particular reference to 
heat, the second two with heat, without any particular reference to work, 
while the last is concerned with the relation between heat and work. After 
establishing in the first chapter the necessary units and basic principles for 
fixing quantities of work, the second chapter proceeds at once to the determina- 
tion of the work done in compressor cylinders and the third, the available work 
in engine cylinders, in terms of all the different variables that may determine 
the work for given dimensions of cylinder or for given quantities of fluid. There 
is established in these first three chapters a series of formulas directly applicable 
to a great variety of circumstances met with in ordinary practice. All are 
deiived from a few simple principles and left in such form as to be readily 
available for numerical substitution. This permits of the solution of niunerical 
problems on engine and compressor horse-power, fluid consumption or capacity 
with very little labor or time, although it has required the expansion of the 
subject over a conmderable number of pages of book matter. A similar pro- 
cedure is followed in the succeeding chapters, formulas and data are developed 
and placed always with a view to the maximum clearness and utility. The 
essential unity of the entire subject has been preserved in that all the important 
related subjects are treated in the same consistent manner and at sufficient 
length to make them clear. When no general principles were available for 
a particular solution there has been no hesitation in reverting to specific data. 
The subject could have been treated in a very much smaller space with less 
labor in book writing but necessitating far greater labor in numerical work 
on the part of the user. This same aim, that is, the saving of the user's 
time and facilitating the arrival at numerical answers, is responsible for the 
insertion of a very considerable number of large tables, numerous original 
diagrams and charts, all calculated for the purpose and drawn to scale. 
These, however, take a great deal of room but are so extremely useful in 
everyday work as to justify any amount of space thus taken up. For 
the sake of clearness all the steps in the derivation of any formula used 
are given, and numerical examples are added to illustrate their meaning 



viii PREFACE 

and application. This also requires a considerable amount of space but with- 
out it the limitations of the formulas would never be clear nor could a student 
learn the subject without material assistance. Similarly, space has been used 
in many parts of the book by writing formulas out in words instead of express- 
ing them in symbols. This saves a great deal of time and labor in hunting up 
the meaning of symbols by one who desires to use an unfamiliar formula involv- 
ing complex quantities, the meaning of which is often not clear when it is entirely 
symbolic. Thus, in the discussion of steam boiler capacity and efficiency, a 
dozen or more pages are taken up with formulas that could have been con- 
centrated in a single page were symbols used entirely, but only at the sacrifice 
of clearness and utiUty. Where in the derivation of a new formula or in the 
treatment of a new subject, reference to an old formula or statement is needed 
and imp)ortant, repetition is resorted to, rather than mere reference, so that 
the new topic may be clear where presented, witlu)ut constantly turning the 
pages of the book. It will be found, therefore, that while the size of the book 
is unusually large it will be less difficult to study than if it were short. 

As a text the book may be used for courses of practically any length, but 
it is not intended that in any course on the subject every page of the book shall 
be used as assigned text. In the new graduate course in mechanical engineering 
at Columbia University, about three-fourths of the subject matter of the book 
will be so used for a course of about one hundred and twenty periods of one 
hour each. All of the book matter not specifically assigned as text or reference 
in a course on engineering thermodynamics in any school may profitably be 
taken up in courses on other subjects, serving more or less as a basis for them. 
It is therefore adapted to courses on gas power, compressed air, steam turbines, 
steam power plants, steam engine design, mechanical refrigeration, heating 
and ventilation, chemical factory equipment, laboratory practice and research. 
Whenever a short course devoted to engineering thermodynamics alone is 
desired, the earlier sections of each chapter combined in some cases with the 
closing sections, may be assigned as text. In this manner a course of about 
thirty hours may be profitably pursued. This is a far better procedure than 
using a short text to fit a short course, as the student gets a better perspective, 
and may later return to omitted topics without difficulty. 

The preparation of the manuscript involves such a great amoimt of labor, 

that it would never have been undertaken without the assurance of assistance 

by Mr. E. D. Thurston, Jr., a fellow instructor at Columbia, in checking text 

and tables, calculating diagrams, writing problems and working examples. 

This help has been invaluable and is gratefully acknowledged. Recognition 

is also due for material aid rendered by Mr. T. M. Gimn in checking and in 

some cases deriving formulas, more especially those of the first three chapters. 

In spite, however, of all care to avoid errors it is too much to expect complete 

success in a new work of this character, but it is hoped that readers finding 

errors will point them out that future editions may be corrected. 

C £. Li. 
CoLXTMBiA University, New York, September, 1912, 



CONTENTS 



Chapter I. Work and Power. General Principles 

PAOK 

1. Work defined 1 

2. Power defined 2 

3. Work in terms of pressure and volume 3 

4. Work of acceleration and resultant velocity 6 

5. Graphical representation of work 8 

6. Work by pressure volume change 10 

7. Work of expansion and compression 13 

8. Vahies of exponent 8 defining special cases of expansion or compression 20 

9. Work phases and cycles, positive, negative and net work 24 

10. Work determination by mean effective pressure 31 

11. Relation of pressure-volume diagrams to indicator cards 34 

12. To find the clearance 37 

13. Measurement of areas of PV diagrams and indicator cards 43 

14. Indicated horse-power 44 

15. Effective horse-power, brake horse-power, friction horse-power, mechanical efficiency, 

efficiency of transmission, thermal efficiency 47 

16. Specific displacement, quantity of fluid per hour, or per minute per I.H.P 49 

17. Velocity due to free expansion by PV method 62 

18- Weight of flow through nozzles by PV method 55 

19. Horse-power of nozzles and jets, by PV method 67 

Chapter II. Work of Compressors. Horse-power and Capacity of Air, Gas and 
Vapor Compressors, Blowing Engines and Dry Vacuum Pumps 

1. General description of structures and processes 73 

2. Standard reference diagrams or PV cycles for compressors and methods of analysis 

of compressor work and capacity 75 

3. Single-stage compressor,, no clearance, isothermal compression. Cycle I. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 81 

i. Single-stage compressor with clearance, isothermal compression, Cycle II. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 85 

5. Single-stage compressor, isothermal compression. Capacity, volumetric effioiency, 

work, mean effective pressure, hor&e-power and horse-power per cubic foot of 

substance, in terms of dimensions and cylinder clearance 87 

6. Single-«tage compressor, no clearance, exponential compression. Cycle Illi Work, 

capacity and w<»rk per cubic foot, in terms of pressures and volumes 91 

7. Single-stage compressor with clearance, exponential compression, Cycle IV. Work, 

capacity, and work per cubic foot in terms of pressures and voliunes 96 

8. Single-stage compressor, exponential compression. Relation between capacity, 

volumetric efficiency, work, mean effective pressure, horse-power and horse- 
power per cubic foot of substance and the dimensions of cylinder and clearance. . 98 

ix 



X CONTENTS 

PAGE 

9. Two-stage oompressor, no clearance, perfect interoooling, exponential compression, 
best-receiver pressure, equality of stages. Cvcle V. Work and capacity in terms 
of pressures and volumes 103 

10. Two-stage compressor, with clearance, perfect intercooling ex^ < ncntial compres&ion, 

beet-receiver pressure, equality of stages. Cycle VI. Work and capacity in terms 
of pressures and volumes 109 

11. Two-stage compressor, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure and horse-power, in terms of 
dimensions of cylinders and clearances 113 

12. Two-stage compressor, vnlh best^eceiver pressure, exponential compression. Capacity, 
volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 120 

13. Three-stage compressor, no clearance, perfect intercooling exponential compree- 

bion, best two receiver pressiu-es, equality of stages. Cycle VII. Work and 
capacity, in terms of pressures and volumes 125 

14. Three-stage compressor with clearance, perfect intercooling exponential compression, 

best-receiver pressures, equality of stages. Cycle VIII. Work and capacity in 
terms of pressures and volumes ICl 

15. Three-stage compres^r, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure, and horse-power in terms 
of dimensions of cylinders and clearances 135 

16. Three-stage compressor vnlh best-receiver pressures, exponential compression. Capac- 

ity, volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 143 

17. Comparative economy or efficiency of compressors 148 

18. Conditions of maximum work of compressors 151 

19. Compressor characteristics 153 

20. Work at partial capacity in compressors of variable capacity 160 

21. Graphic solution of compressor problems 168 

Chapter III. Work op Piston Engines. Horse-power and Consumption op 
Piston Engines Using Steam, Compressed Air, or any other Gas or Vapor 
tJNDER Pressure 

1. Action of fluid in single cylinders. General description of structure and processes. . 187 

2. Simple engines. Standard reference cycles or PV diagrams for the work of expansive 

fluids in a single cylinder 192 

3. Work of expansive fluid in single cyUnder without clearance. Logarithmic expan- 

sion. Cycle I. Mean effective pressure, horse-power and consumption of simple 
engines 197 

4. Work of expansive fluid in single cylinder without clearance. Exponential expan- 

sion, Cycle II. Mean effective pressure, horse-power and consumption of simple 
engines 205 

5. Work of expansive fluid in single cyhnder with clearance. Logarithmic expansion 

and compression; Cycle III. Mean effective pressure, horse-power, and con- 
sumption of simple engines 208 

6. Work of expansive fluid in single cylinder with clearance; exponential expansion 

and compression. Cycle IV. Mean effective pressure, horse-power and consumption 

of simple engines 219 

7. Action of fluid in multiple-expansion cylinders. General description of structure 

ai)d processes 225 

8. Standard ref^'ence cycles or PF diagrams for the work of expansive fluids in two- 

cylinder compound engines 236 



t 



CONTENTS xi 

PAGB 

9. Compound engine with infinite receiver. Logarithnic law. No clearance, Cycle 

V. General relations between pressures, dimensions, and \york 256 

10. Compound engine with infinite receiver. Exponential law. No clearance, Cycle 

VI. General relations between pressures, dimensions and work 268 

11. Compound engine with finite receiver. Logarithmic law. No clearance. Cycle. 

VII. General relations between dimensions and work when H.P. exhaust and 
L.P. admission are not coincident 274 

12. Compound engine with finite receiver. Exponential law, no clearance. Cycle VIII. 

General relations between pressures, dimensions, and work, when high pressure 
Exhaust cjid low-pressure admission are independent 287 

13. Compound engine without receiver. Logarithmic law. No clearance. Cycle IX. 

General relations between dimensions aiid work when high-pressure exhaust 
and low-pressure admission are coincident 292 

14. Compound engine without receiver. Exponential law. No clearance, Cycle X. 

General relations between dimensions and work when high-pressure exhaust 
and low-pressure admission are coincident 301 

15. Compound engine with infinite receiver. Logarithmic law, with clearance and 

compression, Cycle XI. General relations between pressures, dimensions and 
work 306 

16. Compound engine with infinite receiver. Exponential law, with clearance and 

compression, Cycle XII. General relations between pressures, dimensions and 
work 319 

17. Compound engine with finite receiver. Logarithmic law, with clearance and com- 

pression, Cycle XIII. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are independent 325 

18. Compound engine with finite receiver. Exponential law, with ch^rance and com- 

pression, Cycle XIV. General relations between pressiu-es, dimensions, and 
work when H.P. exhaust and L.P. admission are independent 335 

19. Compound engine without receiver. Logarithmic law, with clearance and com- 

pression. Cycle XV. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are coincident 339 

20. Compound engine without receiver. Exponential law, with clearance and compres- 

sion, Cycle XVI. General relations between pressures, dimension, and work, 
when H.P. exhaust and I^.P. admission are coincident 346 

21. Triple-expansion engine with infinite receiver. Logarithmic law. No clearance. 

Cycle XYII. General relations between pressures, dimensions and work 349 

22. Multiple-expansion engine. General case. Any relation between cylinders and 

receiver. Determination of pressure-volume diagram and work, by graphic 
methods 357 

23. Mean effective pressure, engine horse-power, and work distribution and their vari- 

ation with valve movement and initial pressure. Diagram distortion and diagram 
factors. Mechanical efficiency 363 

24. Consumption of steam engines and its variation with valve movement and initial 

pressure. Best cut-off as affected by condensation and leakage 371 

25. Variation of steam consmnption with engine load. The Willans hne. Most eco- 

nomical load for more than one engine and best load division 381 

26. Graphical solution of problems on engine horse-power and cylinder sizes 387 



3di CONTENTS 



Chapter IV. Heat and Matteb. Qualitatzve and Quantitative Relations 
BETWEEN Heat Content of Substances and Phtsical-Chemical State 

FAQS 

1. Substances and heat effects important in engineering ■. . 398 

2. Classification of heating processes. Heat addition and abstraction with or without 

temperature change, qualitative relations 401 

3. Thermometry based on temperature change, heat effects. Thermometer and abso- 

lute temperature scales 407 

4. Caiorimetry based on proportionality of heat effects to heat quantity. Units of 

heat and mechanical equivalent 415 

5. Temperatiu^ change relation to amount of heat for solids, liquids, gases and vapors 

not changing state. Specific heats 419 

6. Volume or density variation with temperature of solids, liquids, gases and vapors 

not changing state. Coefficients of expansion. Coefficients of pressure change 
for gases and vapors 435 

7. Pressure, volume, and temperature relations for gases. Perfect and real gases 438 

8. Gas density and specific volume and its relation to molecular weight and gas constant . 446 

9. Pressure and temperature relations for vapor of liquids or solids. Vaporization, 

sublimation and fusion curves. Boiling- and freezing-points for pure liquids 
and dilute solutions. Saturated and superheated vapors 451 

10. Change of state with amount of heat at constant temperature. Latent heats of 

fusion and vaporization. Total heats of vapors. Relation of specific volume 
of liquid and of vapor to the latent heat 467 

11. Gas and vapor mixtures. Partial and total gas and vapor pressures. Volume, 

weight and gas constant relations. Saturated mixtures. Humidity 481 

12. Absorption of gases by liquids and by solids. Relative volumes and weights with 

pressure and temperature. Heats of absorption and of dilution. Properties 
of aqua ammonia 493 

13. Combustion and related reactions. Relative weights and volumes of substances 

and elements before and after reaction 506 

14. Heats of reaction. Calorific power of combustible elements and of simple chemical 

compounds. B.T.U. per pound and per cubic foot 516 

15. Heat transmission processes. Factors of internal conduction, surface resietanoc, 

radiation and convection 528 

16. Heat transmission between separated fluids. Mean temperature differences, coeffi- 

cients of transmission , 538 

17. Variation in coefficient of heat transmission due to kind of substance, character of 

separating wall and conditions of flow 555 

Chapter V. Heating by Combustion. Fuels, Furnaces, Gas Producers and 
Steam Boilers 

1. Origin of heat and transformation to useful form. Complexity of fuels as sources 

of heat. General classification, solid, liquid, gaseous, natural and artificial 644 

2. Natural solid fuels, wood peat, lignite, bituminous and anthracite coals. Chemical 

and physical properties. Classifications based on ultimate and proximate analysis 
and on behavior on heating 640 

3. Calorific power of coals and the combustible of coals. Calculation of calorific power 

from ultimate and proximate analyses. Calorific power of the volatile 662 

4. Mineral oil and natural gas fuel. Chemical and physical properties. Calorific 

power direct and as calculated for oils from ultimate analysis or from density, 
and for gas from sum of constituent gases 670 



CONTENTS xiii 

PAGB 

5. Charoo&l, coke, ooke oven and retort ooal gas as products of heating wood and coal. 

Chemical, phjrsical, and calorific properties per pound. Calorific power of gases 
per cubic foot in terms of constituent gases. Yield of gas and coke per pound 
coal 676 

6. Distillate oils, kerosene and gasoline, residue oils, and oil gas as products of heating 

mineral oils. Chemical, physical, and calorific properties. Calorific power of frac- 
tionated oils in terms of, (a) carbon and hydrogen; (6) density per pound, and 
estimated value per cubic foot vapor. Calorific power of oil gas per poimd and per 
cubic foot in terms of constituent gases. Yield of distillates and oil gas 685 

7. Gasification of fixed carbon and coke by air-blast reactions, producing air gas, and 

blast-furnace gas. Comparative yield per pound coke and air. Sensible heat 
and heat of combustion of gas. Relation of constituents in gas. Efficiency 
of gasification 695 

8. Gasification of fixed carbon, coke, and coal previously heated, by steam-blast reac- 

tions, producing water gas. Composition and relation of constituents of water 
gas, yield per pound of steam and coal. Heat of combustion of gas and limitation 
of yield by negative heat of reaction 710 

9. Gasification of coals by steam and air blasts, resulting in producer gas. Composition 

and relation of constituents of producer gas, yield per pound of fixed carbon, air 
and steam. Modification of composition by addition of volatile of coal. Heat 
of combustion of gas, sensible heat, and efficiency of gasification. Horse-power 
of gas producers 719 

10. Combustion effects. Final temperature, volume and pressure for explosive and 

non-explosive combustion. Estimation of air weights and heat suppression 
due to CO in products from volumetric analysis 740 

11. Temperature of ignition and its variation with conditions. Limits of proportion 

air gas neutral, or detonating gas and neutral, for explosive combustion of mix- 
tures. Limits of adiabatic compression for self-ignition of mixtures 758 

12. Rate of combustion of solid fuels with draft. Propagation rates, uniform and 

detonating for explosive gaseous mixtures 765 

13. Steam4x>iler evaporative capacity and horse-power. Horse-power imits, equivalent 

rates of evaporation and of heat absorption. Factors of evaporation. Relation 
between absorption rates and rates of heat generation. Lifluence of heating and 
grate surface, calorific power of fuels, rates of combustion and furnace losses 773 

14. Steam-boiler efficiency, furnace and heating-surface efficiency. Heat balances and 

variation in heat distribution. Evaporation and losses per pound of fuel 796 

Chafteb VI. Heat and Work. General Relations between Heat and Work. 
Thermal Efficibnct of Steam, Gab, and Compressed-Air Engines. Flow 
OF Expansive Fluids. Performance of Mechanical Refrigerating Systems 

1. General heat and work relations. Thermal cycles. Work and efficiency deter- 

mination by heat differences and ratios. Graphic method of temperature 
entropy heat diagram 874 

2. General energy equation between heat change, intrinsic energy change, and work 

done. Derived relations between physical constants for gases and for changes 

of state, solid to Uquid, and liquid to vapor 882 

3. Quantitative relations for primary thermal phases, algebraic, and graphic to PF, 

and T4 coordinates. Constancy of PV, and T for gases and vapors, wet, dry 
and superheated 892 

4. Quantitative relations for secondary thermal phases. Adiabatics for gases and 

vapors. Constant quality, constant total heat, and logarithmic expansion Unes 

for steam 904 



xiv CONTENTS 

PAQB 

6. Thermal cycles representative of heat-engine iprocesses. Cyclic efficiency. A 
reference standard for engines and fuel-burning power systems. Classification 
of steam cycles 927 

6. The Rankine cycle. Work, mean effective pressure, jet velocity, water rate, heat 

consumption and efficiency of steam Cycle I. Adiabatio expansion, constant 
pressure, heat addition and abstraction, no compression 936 

7. Carnot cycle and derivatives. Work, mean effective pressure, water rate, heat 

consumption and efficiency of steam Cycle II. Adiabatic expansion and com- 
pression, constant pressure heat addition and abstraction 957 

8. Gas cycles representative of ideal processes and standards of reference for gas 

engines 970 

9. Brown, Lenoir, Otto and Langen non-compression gas-cycles. Work, mean 

effective pressure, volume and pressure ranges, efficiency, heat and gas con- 
sumption 978 

10. Stirling and Ericsson gas cycles. Work, efficiency and derived quantities for 

isothermal compression with and without regenerators 993 

11. Otto, Atkinson, Brayton, Diesel, and Carnot gas cycles. Work, efficiency and 

derived quantities for adiabatic compression gas cycles 1006 

12. Comparison of steam and gas cycles with the Rankine as standard for steam, and 

with the Otto and Diesel, as standard for gas. Relations of Otto and Diesel 

to Rankine cycle. Conditions for equal efficiency 1031 

13. Gas cycle performance as affected by variability of the specific heats of gases, 

apphed to the Otto cycle 1035 

14. Actual performance of Otto and Diesel gas engines, and its relation to the c^'clic. 

Diagram factors for mean effective pressure and thermal efficiency. Effect 

of load on efficiency. Heat balance of gas engines alone an^ with gas producers 1042 

15. Actual performance of piston steam engines and steam turbines at their best load 

and its relation to the cyclic. Effect on efficiency of initial pressure, vacuum, 
superheat, jacketing, and reheating. Heat balances of steam power plants .... 1062 

16. Flow of hot water, steam and gases through orifices and nozzles. Velocity, weight 

per second, kinetic energy, and force of reaction of jets. Nozzle friction and 
reheating and coefficient of efflux. Relative proportions of series nozzles for 
turbines for proper division of work of expansion 1083 

17. Flow of expansive fluids under small pressure drops through orifices, valves, and 

Venturi tubes. Relation between loss of pressure and flow. Velocity heads 
and quantity of flow by Pitot tubes 1097 

18. Flow of gases and vapors in pipes, flues, ducts, and chimneys. Relation between 

quantity of flow and loss of pressure. Friction resistances. Draught and 
capacity of chimneys 1111 

19. Thermal efficiency of compressed-air engines alone and in combination with air 

compressors. Effect of preheating and reheating. Compressor suction heating, 
and volumetric efficiency. Wall action 1127 

20. Mechanical refrigeration, general description of processes and structures. Thermal 

cycles and refrigerating fluids. Limiting temperatures and pressures 1142 

21. Performance of mechanical refrigerating cycles and systems. Quantity of fluid 

circulated per minute per ton refrigeration, horse-power, and heat supplied 
per ton. Refrigeration per unit of work done and its relation to thermal effi- 
ciency of the system 1157 



LIST OF TABLES 



Ma FAQB 

1. Ck>nyer8ion table of units of distance 62 

2. Conversion table of units of surface 62 

3. Conversion table of units of volume 62 

4 Conversion table of units of weights and force 63 

5. Conversion table of units of pressure 63 

6. Conversion table of units of work 64 

7. Conversion table of units of power 64 

8. Units of velocity 64 

9. Barometric heights, altitudes and pressures 65 

10. Values of 8 in the equation PV'» constant for various substances and conditions . . 67 

11. Horse-power per pound mean effective pressure 68 

12. Ratio of cut-offs in the two cylinders of the compound engine to give equal work 

for any receiver volume 284 

13. Piston positions for any crank angle 395 

14. Values for z for use in Heck's formula for missing water 396 

15. Some actual steam engine dimensions 396 

16. Fixed tempcrattu'es 411 

17. Fahrenheit temperatures by hydrogen and mercury thermometers 414 

18. Freezing-point of calcium chloride brine 425 

19. Specific heat of sodium chloride brine 427 

20. Specific heat and gas constants, 431 

21. The critical point 453 

22. Juhlin's data on the vapor pressure of ice 456 

23. Tamman's value on fusion pressure and temperature of water-ice 456 

24. Lowering of freezing-points 465 

25. Berthelot's data on heat for complete dilution of ammonia solutions 500 

26. Air required for combustion of various substances 515 

27. Badiation coefficients 535 

28. Coefficients of heat transfer 550 

29. Temperatures, Centigrade and Fahrenheit 571 

30. Heat and power conversion table 573 

31. Specific heat of solids 574 

32. Specific heats of liquids 576 

33. Baum6-specific gravity scale 577 

34. Specific heats of gases 578 

35. Coefficient of linear expansion of solids 580 

36. Coefficient of cubical expansion of solids 581 

37. Coefficient of volimietric expansion of gases and vapors at constant pressure 582 

XV 



xvi LIST OF TABLES 

wo, PAQB 

38. Coefficient of pressure rise of gases and vapors at constant volume 683 

39. Compressibility of gases by their isothermals 584 

40. Values of the gas constant R 684 

41. Density of gases 685 

42. International atomic weights 686 

43. Melting- or freezing-points 586 

44. Boiling-points 688 

45. Latent heats of vaporization 690 

46. Latent heats of fusion 691 

47. Properties of saturate steam 692 

48. Properties of superheated steam 596 

49. Properties of saturated ammonia vapor 603 

60. Properties of saturated carbon dioxide vapor 618 

51. Relation between pressure, temperature and per cent NHj in solution 628 

62. Values of partial pressure of ammonia and water vapors for various temperatures 

and per cents of ammonia in solution 632 

63. Absorption of gases by liquids 634 

64. Absorption of air in water 635 

66. Heats of combustion of fuel elements and chemical compounds 636 

66. Internal thermal conductivity 639 

57. Relative thermal conductivity 642 

58. General classification of fuels 648 

69. Comparison of cellulose and average wood compositions 650 

60. Classification of coals by composition 652 

61. Classification of coals by g&s and coke qualities 654 

62. Composition of peats 655 

63. Composition of Austrian lignites 666 

64. Composition of English coking coals 658 

65. Wilkesbarre anthracite coal sizes and average ash content 659 

66. Density and calorific power of natural gas 673 

67. Products of wood distillation 676 

68. Products of peat distillation 678 

69. Products of bituminous coal distillation 680 

70. Gas yield of English cannel coals 682 

71. Comparison of coke oven and retort coal gas 682 

72. Relation between oxygen in coal and hydrocarbon in gas 684 

73. Density and calorific power of coke oven gas 684 

74. Average distillation products of crude mineral oils 686 

76. American mineral oil products 687 

76. U.S. gasolene and kerosene bearing crude oils 688 

77. Calorific power of gasolenes and kerosenes 691 

78. Properties of oil-gas 693 

79. Yield of retort oil gas 694 

80. Density and calorific power of oil gas 694 

81. Boudouard's equihbrium relations for. CO and COt with temperature 697 

82. Change of Oj in air to CO and CO2 at 1472*^ F 699 

83. Composition of hypothetical air gas, general 704 

84. Composition of h>TX)thetical air gas, no CO2 and no CO 705 

85. Density and calorific power of blast furnace gas 708 

86. Water gas characteristics with bed temperature 710 

87. Composition of hypothetical water gas, general 714 

88. Composition of hypothetical water gas, no COj and no CO 715 



LIST OF TABLES xvii 

KO. PAOB 

89. Density and calorific power of water gas 718 

90. Composition of hypothetical producer gas from fixed carbon 725 

91. Density and calorific power of producer gas 737 

92. Characteristics of explosive mixtures of oil gas and air 747 

93. Calculated ignition temperatures for producer gas 761 

94. Compressions commonly used in gas engines 762 

95. Ignition temperatures : 763 

96. Variation of ignition temperature of charcoal with distillation temperature 763 

97. Per cent detonating mixture at explosive limits of proportion 764 

98. Velocity of detonating or explosive waves » . . . 766 

99. Rate of propagation, feet per second, for different proportions in gaseous mixtures . 768 

100. Rates of combustion for coal 769 

101. Constants of proportion for rate of coal combustion for use in Eq. (848) 771 

102. Boiler efficiency sunmiaries 799 

103. Three examples of heat balance for boilers 800 

104. Composition and calorific power of characteristic coals. 818 

105. Combustible and volatile of coals, lignites and peats 826 

106. Paraffines from Pennsylvania petroleums 835 

107. Calorific power of mineral oils by calorimeter and calcidation by density formula 

of Sherman and Kropff 836 

108. Properties of mineral oils 838 

109. Composition of natural gases 841 

110. Composition of coke oven and retort coal gas 842 

111. Composition of U. S. coke 846 

112. Fractionation tests of kerosenes and petroleums 847 

113. Fractionation tests of gasolenes 851 

1 14. Composition of blast-furnace gas and air gas 853 

115. Rate of formation of CO from COi and carbon 855 

116. Composition of water gas 857 

117. Composition of producer gas 858 

118. Gas producer tests 864 

119. Composition of oil producer gas 866 

120. Composition of powdered coal producer gas 866 

121. Calorific powers of best air-gas mixtures 867 

122. Composition of boiler-fiue gases 868 

123. Limits of proportions of explosive air-gas mixtures 869 

124. Rate of combustion of coal with draft 870 

125. Rate of combustion of coal 871 

126. Values of 8 for adiabatic expansion of steam 912 

127. Values of s for adiabatic expansion of steam determined from initial and final volumes 

only 913 

128 1042 

129. Diagram factors for Otto cycle gas engines 1046 

130. Mechanical efficiencies of gas engines 1050 

131. Allowable compression for gas engines 1050 

132. Mean effective pressure factors for Otto cycle engines 1053 

133. GQldner's values of Otto engine real volumetric efficiency with estimated mean 

suction resistances 1055 

134. Comparative heat balances of gas producer and engine plants 1057 

135. Heat balances of gas producer plants 1060 

136. Heat balances of gas and oil engines 1060 

137. Steam plant heat balances 1063 



xvm 



LIST OF TABLES 



NO. PAOB 

138. Efficiency factors for reciprocating Bteam engines and turbines 1064 

130. Steam turbine efficiency and efficiency factors with varying vacuum and with 

steam approximately at constant initial pressure 1071 

140. Efficiency factors for low-pressure steam in piston engines 1074 

141. Ck)efficient of discharge for various air pressure and diameters of orifice (Durley). 1101 

142. Values of C for air flow (Weisbach) 1101 

143. Flow* change resistance factors Fr (Reitschel) 1121 



TABLE OF SYMBOLS 



A »area in square feet. 

» constant, in formula for most economical load of a steam engine. Chapter III. 

= constant, in pipe flow formula, Chapter VI. 

» excess air per pound of coal, Chapter V. 

B pounds of ammonia dissolved per pound of weak liquor, Chapter IV. 
a » area in square inches. 

» coefficient of linear expansion. Chapter IV. 

» constant in equation for the ratio of cylinder sizes for equal work distribution in com- 
pound engine, Chapter III. 

= constant in equation for change in intrinsic energy. Chapter VI. 

-constant in equation for specific heat. Chapter IV. 

= cubic feet of air per cubic foot of gas in explosive mixtures, Chapter V. 

= effective area of piston, square inches. Chapter I. 



B« constant in equation for the most economical load of the steam engine. Chapter III. 
» constant in equation for flow in pipes, Chapter VI. 
Bd. »Baum6. 
B.H.P. = brake horse-power, Chapters III and VI. 

= boiler horse-power, Chapter V. 
B.T.U.= British thermal unit. 

bs constant in equation for change in intrinsic energy, Chapter VI. 
« constant in equation for specific heat. Chapter IV. 
(bk.pr.) =back pressure in pounds per square inch. 



C= Centigrade. 
« circumference or perimeter of ducts in equations for flow. Chapter VI. 
» constant. 

=heat suppression factor. Chapter V. 
s ratio of pressure after compression to that before compression in gas engine cycles, 

Chapter VI. 
= specific heat, Chapter IV. 
Cc» per cent of ammonia in weak hquor. Chapter VI. 
Cy= specific heat at constant pressure. 
Cfi=per cent of ammonia in rich liquor. Chapter VI. 
C,«q>ecific heat of water. Chapter VI. 
CffS specific heat at constant volume. 
C<= clearance expressed in cubic feet. 
c» clearance expressed as a fraction of the displacement 
-constant. 
cu.ft. s cubic foot. 
cu.in.s cubic inch. 



i)= constant in equations for pipe flow, Chapter VI. 

=» density, Chapter IV. 

= diameter of pipe in feet, Chapter VI. 

« displacement in cubic feet. 
Z)f» specific displacement, Chapter I. 

xix 



XX TABLE OF SYMBOLS 

d«oongtant in equation for change in intrinsic energy, Chapter VI. 
a diameter of a cylinder in inches, Chapter I. 
s diameter of pipe in inches, Chapter VI. 
» differential. 
(del.pr.)-> delivery pressure in pounds per square inch, Chapter II. 



^B constant in equation for pipe flow, Chapter VI. 
» external latent heat, Chapter IV. 
s thermal efficiency. Chapter VI. 
^ir— thermal efficiency referred to brake horse-power, Chapter III. 

^ft= boiler efficiency, Chapter V. 

£/= furnace efficiency, Chapter V. 

£/= thermal efficiency referred to indicated horse-power. Chapter III. 
^ms mechanical efficiency, Chapter III. 
^f= heating surface efficiency. Chapter V. 
^r^ volumetric efficiency (apparent), Chapter VI. 
Ev'^ volumetric efficiency (true), Chapter VI. 

e »as a subscript to log to designate base e. 
s constant in equation for change in intrinsic energy. Chapter VI. 

ei » ratio of true volumetric efficiency to hypothetical. Chapter II. 

ei"B ratio of true volumetric efficiency to apparent, Chapter III. 

eis ratio of true indicated horse-power to hypothetical, Chapter U. 



F-> constant in equation for pipe flow, Chapter VI. 
"diagram factor for gas engine indicator cards, Chapter VI. 
"Fahrenheit, 
"force in pounds. 
Ff" friction factor, F^X velocity head "loss due to friction, Chapter VI. 
-Fa "resistance factor, F«X velocity head "loss due to resistances, ChapteT VI. 
Ffi "Special resistances to flow in equations for chimney draft, Chapter VI. 
/"Constant in equation for changes in intrinsic energy. Chapter VI. 
"function, 
ft. "foot, 
ft.-lb. "foot-pound. 



G" constant in equation for pipe flow, Chapter VI. 
"Weight of gas per hour in equation for chimney flow, Chapter VI. 
Crm^niaximum weight of gases in equation for chimney flow. Chapter VI. 
G. S. "grate surface. 

y" acceleration due to gravity, 32.2 (approx.) feet per second, per second. 



fl"" as a subscript to denote high pressure cylinder, 
"heat per pound of dry saturated vapor above 32® F. 
"heat per cubic foot gas. 
"heat transmitted. Chapter IV. 
"height of column of hot gases in feet, Chapter VI. 
"pressure or head in feet of fluid, Chapter VI. 
Ha "difference in pressure on two sides of an orifice in feet of air, Chapter VI. 
i^o** equivalent head of hot gases, Chapter VI. 
Hm = pressure in feet of mercury, Chapter VI. 
Hit "pressure in feet of water, Chapter VI. 
H.P. "high pressure. 

"horse-power. Chapter I. 



TABLE OF SYMBOLS xxi 

H.S. "'heating surface. 
(H.P.cap.) i^high pressure cylinder capacity, Chapter III. 
h »heat of superheat. 
Hm » difference in pressure on two sides of an orifice in inches of mercury, Chapter VI. 
Afps difference in pressure on two sides of an orifice in inches of water, Chapter VI. 



/»as a subscript to denote intermediate cylinder, Chapter III. 
I. H. P. s indicated horse-power. 

in. » inch, 
(in. pr.)» initial pressure in pounds per square inch. 



J s Joule's equivalent » 778 (approx.) foot-pounds per B.T.U. 



X» coefficient of thermal conductivity. Chapter IV. 
» constant, 
s proportionality coefficient in equation for draft. Chapter VI. 

/r«» engine constant » ^^^ in expression for horse-power. Chapter IlL 



If a as a subscript to denote low-pressure cylinder, 
a distance in feet, 
—latent heat, Chapters IV and VI. 
—length of stroke in feet. Chapter I. 
L —per cent of heat in fuel lost in furnace. Chapter V. 
L.P.— low pressure. 
(L.P.Cap.)— low-pressure capacity, Chapter II. 
Z— constant, Chapter III. 
= length. Chapter IV. 
lb. —pound. 

log— logarithm to the base 10. 
logs —logarithm to the base e. 



Af — mass. 

(M.E.P.) —mean effective pressure, pounds per square foot. 

• m— constant, Chapter III. 

area 

—mean hydraulic radius = — : . 

perimeter 

—molecular weight, Chapter IV. 

—ratio of initial pressure to that end of expansion in Otto and Langen gas cycle. 
Chapter VI. 
(m.b.p.) —mean back pressure in pounds per square inch, 
(m.e.p.) —mean effective pressure in pounds per square inch. 
(m.f.p.) —mean forward pressure in pounds per square inch. 



.Vs constant, Chapter III. 

—revolutions per minute. 
n— cycles per minute. 

—constant, Chapter III. 

—cubic foot of neutral per cubic foot of gaseous mixture. Chapter V. ' 

—number of degrees exposed on thermometer stem, Chapter IV. 

—ratio of volume after expansion to volume before in Atkinson gas cycle, Chapter IV. 
dfic volume of dry saturated steam, Chapter VI. 



xxii TABLE OF SYMBOLS 

= volume of receiver of compound engine in cubic feet, Chapter III. 
P= draft in pounds per square foot, Chapter VL 
=load in kilowatts, Chapter III. 
= pressure in pounds per square foot. 
Pf= static pressure in poimds per square foot lost in wall friction, Chapter VT. 
Pa » static pressure in pounds per square foot lost in changes of cross-section, etc., 

Chapter VI. 

Pv —velocity head in pounds per square foot. 

p= pressure in pounds per square inch. 

Pe==mean exhaust pressure. Chapter VI. 

Pf =>mean suction pressure, Chapter VI. 

Pv = partial pressure of water vapor in air. Chapter VI. 



Q= quantity of heat or energy in B.T.U. gained or lost by a body passing from one state to 

another. 
Qi'— heat added from fire in Stirling and Ericsson cycles, Chapter VI. 
Qi" =heat added from regenerator in Stirling and Ericsson cycles, Chapter VI. 
Qt'»heat abstracted by water jacket in Stirling and Ericsson cycles, Chapter VI. 
Qt"— heat abstracted by regenerator in Stirling and Ericsson cycles, Chapter VI. 
9 » quantity of heat per pound of liquid above 32° F. 



72= ratio of heating surface to grate surface, Chapter V. 
* «gas constant. 
^72(7= ratio of cylinder sizes in twoHStage air compressor or compound engine. Chapters II 

and III. 
/2j7= ratio of expansion in high-pressure cylinder, Chapter III. 
122;= ratio of expansion in low-pressure cylinder, Chapter III. 
Rp^T&iio of initial to back pressure. Chapters III and VI. 
/2pB ratio of delivery to supply pressure. Chapter II. 
Rv » ratio of larger volume to smaller volume. 

r =rate of flame propagation in explosive mixtures, Chapter V. 
rp« pressure differences (maximum— minimmn) in gas cycles, Chapter VI. 
fK=volume differences (maximum — minimum) in gas cycles, Chapter VL 
(rec.pr.) = receiver pressure in pounds per square inch, Chapter IIL 
(rel.pr.) —release pressure in pounds per square inch. Chapter III. « 



iS=per cent of ammonia in solution. Chapter IV. 
«: piston speed. Chapter I. 

» pounds of steam per pound of air in producer blast, Chapter V. 
= specific heat. Chapter IV. 
» specific heat of superheated steam. Chapter VI. 
(Sup.Vol.) = volume of steam supplied to the cylinder per stroke. Chapter III. 
8= general exponent of F in expansion or compression of gases. 
sp.gr. = specific gravity, 
sp.ht. = specific heat, 
sq.ft. = square foot, 
sq.in. ^square inch, 
(sup.pr.) = supply pressure, in pounds per square inch. 



r=* temperature, degrees absolute. 
Tc— temperature of air. Chapter VI. 
TiT™ temperature of gases in chinmey, Chapter VL 
^""temperature in degrees scale. 



TABLE OF SYMBOLS xxiii 

(/»rate of heat transfer in B.T.U. per square foot per hour per degree difference in tem- 
perature, Chapter IV. 
IT'S intrinsic energy. Chapter VI. 
ii= velocity in feet per second. 
UiB» velocity in feet per minute, Chapter VI. 



r= volume in cubic feet. 
Va. =» cubic feet per pound air, Chapter VI. 
V(7= cubic feet per pound, gas, Chapter VI. 
Yl —volume of Uquid in cubic feet per pound. 
Vs = volume of solid in cubic feet per pound. 
Vy = volume of vapor in cubic feet per pound. 

V =«volume, Chapter IV. 



IT = work in foot-pounds. 
W.R. s= water rate. 

tr= pounds of water per pound of ammonia in solution, Chapter IV. 
» weight in pounds, 
u^js ^pounds of rich Uquor per pound of ammonia, Chapter VI. 



A' = compression in the steam engine as a fraction of the stroke, Chapter III. 

- , heat added 

= 1-1 ; , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

x» constant in the expression for missing water, Chapter III. 

= fraction of liquid made from solid or vapor made from liquid, Chapter VI. 

=per cent of carbon burned to CO2, Chapter V. 

=per cent of nozzle reheat. Chapter VI. 

=per cent of steam remaining in high-pressure cylinder of compound engine at any point 

of the exhaust stroke, Chapter III. 
—quantity of heat added in generator of absorption system in addition to the amount of 

heat of absorption of 1 lb. of ammonia. Chapter VI. 
-ratio of low-pressure admission voliune to high-pressure admission voliune, Chapter III. 



K = total steam used per hour by an engine, Chapter III. 

heat added «. „, 

= 1 + — — , Chapter VI. 

temperature at beginning of addition X specific heat at constant pressure 

1/ =per cent of vane reheat Chapter VI. 

» ratio of the volume of receiver to that of the high-pressure cylinder of the compound 

engine, Chapter III. 



Zs fraction of the stroke of the steam engine completed at cut-off. Chapter III. 

- . heat added from regenerator ^, ^ ,., 
= 1H ~ ; , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

Z' « hypothetical best value of Z, 

- . heat added from regenerator _, ^ _._. 
=s 1 -I — ; — ; ZB ^ Chapter VI. 

temperature at begmnmg of addition X specific heat at constant pressure 
;B= ratio of R.P.M. to cycles per minute. 



as an angle, Chapter I. 
=ooefficient of cubical expansion. Chapter III. 
BCODstaot in the equation for latent heat, Chapter VI. 



xxiv TABLE OF SYMBOLS 

B constant in equation for variable specific heat at constant volume, Chapter VL 
oe'= constant in equation fOr variable specific heat at constant pressure, Chapter VI. 



^« constant in equation for latent heat, Chapter VI. 
» fraction of fuel heat available for raising temperature, Chapter V. 



Y=> constant in equation for latent heat. Chapter VI. 

» ratio of crossHsection to perimeter. Chapter IV. 

... sp. ht. at const, press. 

«= special value for « for adiabatic expansion or compression » ; 

sp. ht. at const, vol. 

y's ratio of specific heat at constant pressure to specific heat at constant volume when each 

is a variable, Chapter VI. 



A B increment. 

8 » density in pounds per cubic foot, 
dcf— density in cold gases in equations for chimney draft, Chapter VI. 
8/r» density of hot gases in equations for chimney draft. Chapter VI. 



t^» coefficient of friction, Chapter VI. 



(Immaterial coefficient in heat transfer expression, Chapter IV. 



p» internal thermal resistance, Chapter IV. 



Z» summation. 

o» surface thermal resistance, Chapter IV. 



T— time in seconds. 

*= entropy. Chapter VI. 
<!>= entropy. Chapter VI. 



Note. A small letter when used as a subscript to a capital in general refers to a point 
on a diagram, e.g.. Pa designates pressure at the point A. Two small letters used as sub- 
scripts together, refer in general to a quantity between two points, e.g., Wab designates 
work done from point A to point B. 



ENGINEERING THERMODYNAMICS 



CHAPTER I 
WORK AND POWER. GENERAL PRINCIPLES. 

1. Work Defined. Work, in the popular sense of performance of any labor, 
is not a sufficiently precise term for use in computations, but the analytical 
mechanics has given a technical meaning to the word which is definite and which 
is adopted in all thermodynamic analysis. The mechanical definition of work is 
mathematical inasmuch as work is always a product of forces opposing motion 
and distance swept through, the force entering with the product being limited 
to that acting in the direction of the motion. The unit of distance in the 
English system is the foot, and of force the pound, so that the imit of work is 
the foot-pound. In the metric system the distance unit is the meter and the 
force unit the kilogramme, making the work unit the kilogrammeter. Thus, 
the lifting of one pound weight one foot requires the expenditure of one foot- 
pound of work, and the falling of one pound through one foot will perform one 
foot-pound of work. It is not only by lifting and falling weights that work is 
expended or done; for if any piece of mechanism be moved through a distance 
of one foot, whether in a straight or curved path, and its movement be resisted 
by a force of one pound, there will be performed one foot-pound of work against 
the resistance. It is frequently necessary to transform work from one sys- 
tem of units to the other, in which case the factors given at the end of this 
Chapter are useful. 

Work is used in the negative as well as in the positive sense, as the force 
considered resists or produces the motion, and there may be both positive and 
negative work done at the same time; similar distinctions may be drawn with 
reference to the place or location of the point of application of the force. Con- 
sider, for example, the piston rod of a direct-acting pump in which a certain 
force acting on the steam end causes motion against some less or equal force 
acting at the water end. Then the work at the steam end of the pump may be 
considered to be positive and at the water end negative, so far as the move- 
ment of the rod is concerned; when, however, this same movement causes a 
movement of the water, work done at the water end (although negative with 
reference to the rod motion^ since it opposes that motion) is positive with refer- 



2 ENGINEERINa THERMODYNAMICS 

ence to the water, since it causes this motion. It may also be said that the 
steam does work on the steam end of the rod and the water end of the rod does 
work on the water, so that one end receives and the other delivers work, the rod 
acting as a transmitter or that the work performed at the steam end is the input 
and that at the water end the output work. 

(See the end of Chapter I for Tables I, II, III, IV, and VI, Units of 
Distance, of Surface, of Volimie, of Weight and Force, and of Work.) 

Example. An elevator weighing 2000 lbs. is raised 80 ft. How much work is done in 
foot-pounds? 

Foot-pounds »force xdistance 

=2000 X80 = 160,000 ft-lbs. 

Ans. 160,000 ft.-lbs. 

Prob. 1. A pump lifts 150 gallons of water to a height of 250 ft. How much woikdoes 
it do? ^ 

Prob. 2. By means of a jack a piece of machinery weigjiing 10 tons is raised f in. What 
is the work done? 

Prob. 3. A rifle bullet weighing 2 oz. travels vertically upward l{ miles. What work was 
done in foot-pounds? 

Prob. 4. A cubic foot of water falls 50 ft. in reaching a water-wheeL How much work can 
it do? 

Prob. 5. A piston of an elevator is 12 ins. in diameter and has acting on it a pressure of 
80 lbs. per square inch. What work is done per foot of travel? 

Prob. 6. It has been found that a horse can exert 75 lbs. pull when going 7 miles per 
hour. How much work can be done per minute? ^^ 

Prob. 7. How much work is done by an en^e which raises a 10-ton casting 50 ft.? 

Prob. 8. The pressure of the air on front of a train is 50 lbs. per square foot when the 
speed is 50 miles per hour. If the train presents an area of 50 sq.ft., \(4iat work is done in 
overcoming wind resistance? ^ * . .wv • v • 

Prob. 9. The pressure in a 10-inch gun diiring ike time of firing, is 2000 lbs. per square 
inch. How much work is done in ejecting the projectile if the gun is 33 ft. 4 ins. long? 

2. Power Defined. Power is defined as the rate of working or the work 
done in a given time interval, thus introducing a third unit of mechanics, time, 
so that power will always be expressed as a quotient, the niunerator being a prod- 
uct of force and distance, and the denominator time. This is in opposition 
to the popular use of the word, which is very hazy, but is most often applied to 
the capability of performing much work] or the exertion of great force, thus, 
popularly, a powerful man is one who is strong, but in the technical sense a man 
would be powerful only when he could do much work continuously and rapidly. 
An engine has large power when it can perform against resistance many foot- 
pounds per minute. The unit of power in the English system is the horse-power, 
or the performance of 550 foot-pounds per second or 33,000 foot-poimds per 
minute, or 1,980,000 foot-pounds per hour. In the metric system the horse- 
power is termed cheval-vapeur, and is the performance of 75 killogrammeters 
=542i foot-pounds per second, or 4500 kilograjnmeters= 32,549 foot-pounds 



WORK AND POWER. 3 

per minute, or 270,000 kilograinmeters= 1,952,932 foot-pounds per hour. 
Table VII at the end of Chapter I gives conversion factors for power units. 

Example. The piston of a steam engine travels 600 ft. per minute and the mean force 
of steam acting upon it is 65,000 lbs. What is the horse-power? 

TT foot-pounds per minute 
Horse-power '^ 33 qg 

tune 

" 33,000 

^65,000X60^ 
33,000 

Prob. 1. The draw-bar pull of a locomotive is 3000 lbs. when the train is traveling 50 
miles per hour. What horse-power is being developed? 

Prob. 2. Amine cage weighing 2 tons is lifted up a 2000-ft. shaft in 40 seconds. What 
horse-power will be required if the weight of the cable is neglected? 

Prob. 3. By direct pull on a cable it is found possible to lift 4 tons 20 ft. per second. With 
a diffoential pulley 40 tons may be lifted 3 ft. per second. What is the difference in power 
required? 

Prob. 4. A horse exerts a pull of 100 lbs. on a load. How fast must the load be moved to 
develop one horse-power? 

Prob. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 lbs. What 
horse-power must be available to maintain this speed? (One knot is a speed of one nautical 
mile per hour.) /^ . * ; - '' jju „ 

Ptob. 6. It is estimated that 100,000 cu. ft. of water gp over a fall 60 ft. hi^ every 
second. What horse-power is going to waste? 

Prob. 7. The force acting on a piston of a pump is 80,000 lbs. If the piston speed is 150 
ft. per minute, what is the horse-power? 

Prob. 8. To draw a set of plows 2i miles per hour requires a draw-bar pull of 10,000 
lbs. What must be the horse-power of a tractor to accomplish this? 

Prob. 9. The horse-power to draw a car up a grade is the sum of the power necessary to 
pull it on a level and that necessary to lift it vertically the same number of feet as it rises on 
the grade. What will be the horse-power required to draw a car 20 miles per hour up a 12 per 
cent grade if the car weighs 2500 lbs. and the draw-bar pull on the level is 250 lbs.? 

3. Work in Terms of Pressure and Volume. Another of the definitions 
of mechanics fixes pressure as force per unit area so that pressure is always a 
quotient, the numerator being force and the denominator area, or length to 
the second jjower. If, therefore, the pressure of a fluid be known, and accord- 
ing to hydromechanics it acts equally and normally over all surface in contact 
with it, then the force acting in a given direction against any surface will be 
the product of the pressure and the projected area of the surface, the projection 
being on a plane at right angles to the direction considered. In the case of pis- 
tons and plungers the line of direction is the axis of the cylinder, and the pro- 
jected area is the area of the piston less the area of any rod passing completely 
through the fluid that may be so placed. When this plane area moves in a 



4 ENGINEERING THERM0t>YNAMIC8 

(fireclion perpendicular to itself, the product of its ai^ and the distance will be 
the volume swept through, and if a piston be involv^ the volume is technically 
the displacement of the piston. Accordingly, work may be expressed in three 
ways, as follows: 

Work = force X distance ; 

Work = pressure X area X distance ; 

Work = pressure X volume. 

The product should always be in foot-pounds, but will be, only when appro- 
priate units are chosen for the factors. These necessary factors are given as 
ffdOows: 

, W<xk in foot-pounds = force in lbs. X distance in ft. 

= pressure in lbs. per sq.ft. X area in sq.ft. X distance in ft 
= pressure in lbs. per sq.ui. X area in sq.in. X distance in ft 
Wv, , ^ =pressure in lbs, per sq.ft. X volume in cu.ft. 

* -«*^ r= pressure in lbs. per sq.in. X 144 X volume in cu.f t. 

As pressures are in practice expressed in terms not only as above, but alsa 
in heights of columns of common fluids and in atmospheres, both in English and 
metric systems, it is convenient for calculation to set down factors of equivalence 
as in Table V, at the end of the Chapter. 

In thermodynamic computations the pressure volume product as an expres- 
sion for work is most useful, as the substances used are always vapors and gases, 
which, as will be explained later in more detail, have the valuable property of 
changing volume indefinitely with or without change of pressure according 
to the mode of treatment. Every such increase of volmne gives, as a conse- 
quence, some work, since the pressure never reaches zero, so that to derive work 
from vapors and gases they are treated in such a way as will allow them to change 
volmne considerably with as much pressure acting as possible. 

It should be noted that true pressures are always absolute, that is, measured 
above a perfect vacuum or counted from zero, while most pressure gages and 
other devices for measuring pressure, such as indicators, give results measured 
above or below atmospheric pressure, or as commonly stated, above or below 
atmosphere. In all problems involving work of gases and vapors, the absolute 
values of the pressures must be used; hence, if a gage or indicator measure- 
ment is being considered, the pressure of the atmosphere found by means of the 
barometer must be added to the pressure above atmosphere in order to obtain the 
absolute or true pressures. When the pressures are below atmosphere the 
combination with the barometric reading will depend on the record. If a record 
be taken by an indicator it will be in pounds per square inch below atmosphere 
and must be subtracted from the barometric equivalent in the same units to 
give the absolute pressure in pounds per square inch. When, however, a 
vacuum gage reads in inches of mercury below atmosphere, as such gages 
do, the difference between its reading and the barometric gives the absolute 



WORK AND POWER. 5> 

pressure 5n inches of mercury directly, which can be converted to the desiredl 
units by the proper factors. 

While it is true that the barometer is continually fluctuating at every place, 
it frequently happens that standards for various altitudes enter into calculations, 
and to facUitate such work, values are given for the standard barometer at various 
aU,itudes with equivalent pressures in pounds per square inch in Table IX. 

Frequently in practice, pressures are given without a definite statement 
of what units are used. Such a custom frequently leads to ambiguity, but it 
is often possible to interpret them correctly from a knowledge of the nature of the 
problem in hand. For instance, steam pressures stated by a man in ordinary prac- 
tice as being 100 lbs. may mean 100 lbs. per square inch gage (above atmaephere), 
but may be 100 lbs. per square inch absolute. Steam pressures are then most 
commonly stated per square inch and should be designated as either gage or abso- 
lute. Pressures of compressed air are commonly expressed in the same units la 
steam, either gage or absolute, though sometimes in atmospheres. Steam pressures; 
below atmosphere may be stated as a vacuum of so many inches of mercury,, 
meaning that the pressure is less than atmosphere by that amoimt, or may/ 
be given as a pressure of so many inches of mercury absolute, or as so many 
pounds per square inch absolute. The pressures of gases stored in tanks under 
high pressure are frequently recorded in atmospheres, due to the convenience 
of computation of quantities on this basis. Pressures ot air obtained by blowers 
or fans are visually given by the manufacturers of such apparatus in ounces 
per square inch above (or below) atmosphere. Such pressures and also differ- 
ences of pressure of air due to chimney draft, or forced draft, and the pressure: 
of illuminating gas in city mains, are commonly stated in inches of water, each 
inch of water being equivalent to 5.196 lbs. per square foot. The pressure of 
water in city mains or other pressure pipes may be stated either in pounds per 
square inch or in feet of water head. 

Example. A piston on which the mean pressure is 60 lbs. per square inch sweeps through 
a Yohime of 300 cuit. What is the work done? 

W^PxVj where 7= cuit. and P^lbs. per sq.ft. 
.-. }r«60xl44x300 =2,592,000 ft.-lbs. 

Prob. 1. The mean pressure acting per square inch when a mass of air changes in 
volume from 10 cu.ft. to 50 cuit. is 40 lbs. per square inch. How much work is done? 

Prob. 2, An engine is required to develop 30 HP. If the volimie swept through per 
minute is 150 cuit., what must the mean pressure be? 

Prob. 3. The mean effective pressure in compressing air from one to five atmospheres is 
2S.7 ibs. per square inch. How many horse-power are required to compress 1000 cu.ft. of 
free air per minute? 

Prob. 4. At an altitude of 4100 ft. a pressure gage showed the pressure on one side of 
a piston to be 50 lbs. per square inch while the pressure on the opposite side is 3 lbs. per 
square inch absolute. What pressure was tending to move the piston? 

Prob. 6. At an altitude of 1 mUe the mean pressure in a gas engine cylinder during the 
suction stroke was found to be 12 lbs. per square inch absolute. What work was done 
by the enpne to draw in a charge if the cylinder was 5 ins. in diameter md the stroke 6 ins.? 



6 ENGINEERING THERMODYNAMICS 

Prob. 6, After explosion the piston of tbe above engme was forced out 80 that the gas 
volume wa£ five times that at the beginning of the stroke. What must the M.E.P. have 
been to get 20,000 ft.-lbs. of work? 

Prob. 7, On entering a heating oven cold air expands to twice its volume. TVTiat 
work is done per cubic foot of air? 

Prob. 8. A projectile is forced from a gun by a constant air pressure of 1000 lbs. per 
square inch. Before it begins to move there is i a cu.ft. of air in the barrel, and at the instant 
it leaves the barrel the volume is 10 cuit. What work was done on the projectile? 

Prob. 9. Water is forced from a tank against a head of 75 ft. by filling the tank with 
compressed air. How much work is done in emptying a tank containing 1000 cu.ft.? 

4. Work of Acceleration and Resultant Velocity. When a force acting 
on a mass is opposed by an equal resistance there may be no motion at all, or 
there may be motion of constant velocity. Any differences, however, between 
the two opposing forces will cause a change of velocity so long as the difference 
lasts, and this difference between the two forces may be itself considered as the 
only active force. Observations on imresisted falling bodies show that they 
increase in velocity 32.16 ft. per second for each second they are free to fall, 
and this quantity is universally denoted by g. If then, a body have any 
velocity, wi, and be acted on by a force equal to its own weight in the direction 
of its motion for a time, t seconds, it will have a velocity U2 after that time. 

U2=ui+g'z (1) 

It may be that the force acting is not equal to the weight of the body, in which 
case the acceleration will be different and so also the final velocity, due to the 
action of the force, but the force producing any acceleration will be to the 
weight of the body as the actual acceleration is to the gravitational acceleration. 
So that 

Actual accelera tio n force actual acce leration 

Weight of body or gravitational force gravitational acceleration {gY 

and 

Actual accelerating force = rr-r* — y \ — zr- rr X actual acceleration. 

gravitational acceleration (jg) 

or 

change of velocity 



Force = mass X acceleration = mass X 



time of change 



F=MX^^^^^^ (2) 

The work performed in accelerating a body is the product of the resistance 
met into the distance covered, L, while the resistance, or the above-defined force, 
acts, or while the velocity is being increased. This distance is the product t)f 
the time of action and the mean velocity, or the distance in feet, 

X=^t% (3) 



WORK AND POWER. 7 

The work is the product of Eqs. (2) and (3), or, work of acceleration is 

^ T ^~2 

where w is the weight in pounds. Exactly the same result will be obtained by 
the calculus when the acceleration is variable, so that Eq. (4) is of universal 
application. 

The work performed in ctcceleraiing a body depends on nothing hut its mass 
and the initial and final velocities, and is in every case equal to the product of 
half the mass and the difference between the squares of the initial arid final 
velocities^ or the product of the weight divided by 64-4 ^^ ^ difference between 
the squares of the initial and final velocities. 

It frequently happens that the velocity due to the reception of work is desired, 
and this is the case with nozzle flow in injectors and turbines, where the steam 
performs work upon itself and so acquires a velocity. In all such cases the 
velocity due to the reception of the work energy is 



M2 






where W is work in foot-pounds and w, as before, is weight in pounds. Or if 
the initial velocity be zero, as it frequently is, 



U2 



=J»=J^l32l. (6) 

\ w yi w 



For conversion of velocity units, Table VIII, at the end of the Chapter, 
is useful. * 

Example. A force of 100 lbs. acts for 5 seconds on a body weighing 10 lbs. ; if the 
original velocity of the body was 5 ft. per second, what will be the final velocity, the 
distance traveled and the wor]r done? 

100- 10 ("'-5) ■ 

Uj = 1615 ft. per second; 
5=(j^')x=4050ft. 

yf^M{u^-u^ „405^(X)0 ft.-lbs. 

Ptob. 1. A stone weighing \ lb. is dropped from a height of 1 mile. With what veloc- 
ity and in what length of time will it strike if the air resistance is zero? 

Prob. 2. A car moving 20 miles per hour and weighing 25 tons is brought to rest in 
500 ft. What is the negative acceleration, the time required to stop, and the work done? 



8 



ENGINEERING THERMODYNAMICS 



Prob. 3. Steam escapes through an opening with a velocity of half a mile per second. 
How many foot-pounds of energy were imparted to each pound of it to accomplish this? 

Prob. 4. A weight of 100 lbs. is projected upward with a constant force of 200 lbs. 
How much further will it have gone at the end of 10 seconds than if it had been merely 
falling under the influence of gravity for the same period of time? 

Prob. 6. A projectile weighing 100 lbs. is dropped from an aeroplane at the height of 
i mile. How soon will it strike, neglecting air resis'ance? 

Prob. 6. A water-wheel is kept in motion by a jet of water impinging on flat vanes. 
The velocity of the vanes is one-half that of the jet. The jet discharges 1000 lbs. of 
water per minute with a velocity of 200 ft. per second. Assuming no losses, what is 
amount of the work done? 

Prob. 7. With the wind blowing 30 miles per hour, how much work could a 12-ft. 
windmill perform if 25 per cent of the available work were utilized. 

Note. The weight of a cubic foot of air may be taken as .075 lb. 

Prob. 8. An engine has a piston speed of 600 ft. per minute and runs at 150 R.P.M. 
If the reciprocating parts weigh 500 lbs., how much work is done in accelerating the 
piston during each stroke? 

Prob. 9. A flywheel with rim 10 ft. in diameter to center of section and weighing 5 
tons, revolves at a rate of 150 R.P.M.; 100,000 ft.-lbs. of work are expended on it. How 
much will the speed change? 

6. Grap^cal Representation of Work. As work is always a product of 
force and distance or pressure and volume, it may be graphically expressed by 





B 






• 














c 




5 

r 








































• 








* 


B 8 

1 

IT. r> 


























**• 2 
















* 










1 






















D 




/ 


{ 




1 


t 

i 


e 


1 


3 




i 


i 


5 





Distaaces Id Feet 
Fig. 1. — Constant Force, Work Diagram, Force-Distance CJoordinates. 



an area on a diagram having as coordinates the factors of the product. It is 
customary in such representations to use the horizontal distances for volumes 
and the vertical for pressures, which, if laid off to appropriate scale and 
in proper units, will give foot-pounds of work directly by the area enclosed. 
Thus in Fig. 1, if a force of 5 lbs. (AB) act through a distance of 5 ft. (BC) 
there will be performed 25 foot-pounds of work as indicated by the area of the 



WORK AND POWER 



rectangle A BCD, which encloses 25 unit rectangles, each representing one foot- 
pound of work. 

If a steam cylinder piston suffers a displacement of 5 cu.ft. under the steam 
pressure (absolute) of 5 lbs. per square foot then the operation which results 
in the performance of 25 foot-pounds of work is represented by the diagram 
Fig. 2, ABCD. 



I 5 
§ 

C A 
00 * 

u 

& 

'3 3 

c 

3 
O 

B 2 



B 




















c 
























































































































D 





I 



2 8 4 5 

• Volumes in Cubic Eeet 

Fig. 2. — Constant Pressure Work Diagram, Pressure- Volume Coordinates. 

Prob. 1. Following the method given for Fig. 1, draw a diagram for the example of 
Section 3 

Prob. 2. By means of a diagram, show that the work done by a pressure of 1000 lbs. 
per square foot traversing a distance of 10 ft. is 10,000 ft.-lbs. 

Prob. 3. Draw a diagram for the case of a volume change from 1 to 10 cu.ft. while the 
pressure acting is 20 lbs. per square inch. 

Prob. 4, Draw a pressure volume diagram for the case of forcing a piston out of a 
cylinder by a water pressure of 15,000 lbs. per square foot, the volume of the cylinder at 
the start is i cu.ft. and at the end 6 cu.ft. Make a diagram to scale and report work 
per square inch of diagram. 

Prob.6. A pump draws in water at a constant suction pressure of 14 lbs. and dis- 
charges it at a constant delivery pressure of 150 lbs. per sq.in. Considering the pump 
barrel to be empty at beginning of suction and end of delivery and to contain 3 cu.ft. 
when full, draw the diagram for this case and find the foot-pounds of work done. 

Prob. 8. In raising a weight a man pulls on a rope with a constant force of 80 lbs. 
If the weight is lifted 40 ft., find from a diagram the work done. 

Prob. 1. In working a windlass a force of 100 lbs. is applied at the end of a 6-ft. 
le\'er, the drum of th windlass being 1 ft. in diameter. Draw a work diagram for 
work applied and for work done in lifting if there be no loss in the windlass. 

Prob. 8. The steam and water pistons of a pump are on the same rod and the area 
of the former is twice that of the latter, the stroke being 3 ft. Show by a diagram 
that the work done in the two cylinders is the same if losses be neglected. 

Prob. 9. An engine exerts a draw-bar pull of 8000 lbs. at speed of 25 miles an hour. 
A change in grade occurs and speed increases to 40 miles per hour and the pull decreases 
to 5000 lbs. Show by a diagram the change in horse-power. 



10 



ENGINEEEING THERMODYNAMICS 



6. Work by Pressure Volume Change. Suppose that instead of being 
constant the pressure were irregular and, being measured at intervals of 1 cu.ft. 
displacement, found to be as follows: 



Pressure. 


Displacement 


Lbs. per Sq.Ft. 


Volume Cu Ft. 


100 





125 


1 


150 


2 


100 


3 


75 


4 


50 


5 




12 3 4 5 6 

Volumes In Cubic Feet 

Fig. 3. — ^Work Diagram, Pressure-Volume Coordinates. Discontinuous Pressure- Volume 

Relations. 

This condition might be plotted as in Fig. 3, A, S, C, D, Ey F, G, H. The 
work done will be the area under the line joining the observation points. In 
the absence of exact data on the nature of the pressure variations bet^'^een the 
two observation points A and S, a variety of assumptions might be made as to 
the precise evaluation of this area, as follows: 

(a) The pressure may have remained constant at its original value for the 
first cubic foot of displacement, as shown dotted A-B/ and then suddenly have 
risen to B, In this case the work done for this step would be 100 foot-pounds. 

(6) Immediately after the measurement at A the pressure may have risen 
to A' and remained constant during displacement A' to B, in which case the 
work done would be 125 foot-pounds. 

(c) The pressure may have risen regularly along the solid line AB^ in which 

case the work area is a trapezoid and has the value ^ XI = 112.5 foot- 
pounds. 



WORE AND POWER 



11 



It thus appears that for the exact evaluation of work done by pressure 
volume change, contmuous data are necessary on the value of pressure with 
respect to the volume. If such continuous data, obtained by measurement or 
otherwise, be plotted, there will result a continuous line technically termed the 
pressure-volume curve for the process. Such a curve for a pressure volmne 
change starting at 1 cu.ft. and 45 lbs. per square foot, and ending at 7 cu.ft., 
and 30 lbs. per square foot, is represented by Fig. 4, Ay B, C, D, E, 

The work done during this displacement under continuously varying pressure 
isUkewise the area between the curve and the horizontal axis when pressures are 
laid oS vertically, and will be in foot-pounds if the scale of pressures is pounds 
per square foot and volumes, cubic feet. Such an irregular area can be divided 
into small vertical rectangular strips, each so narrow that the pressure is sensibly 
constant, however much it may differ in dififerent strips. The area of the 
rectangle is PA V, each having the width AV and the height P, and the work 









































^ 


^ 


--- 


^ 




















y 












\ 


V 














A 


/ 














\ 


\ 






























\ 


s, 






c 






















• 




"^ 


'^ 






































E 
























D 





' 




1 
I 


\ 




9 


K 


! 




5 






# 


' 



VolumeB In Cubic Feet 

Fig. 4. — Work Diagram, Preesure-Volume Coordinates. Gontinuoua Pressure-Volume 

Relations. 

area will be exactly evaluated if the strips are narrow enough to fulfill the 
conditions of sensibly constant pressure in any one. This condition is true only 
for infinitely narrow strips having the width dV and height P, so that each has 
the area PdV and the whole area or work done is 



=/ 



Tr= I PdY. 



(7) 



This is the general algebraic expression for work done by any sort of continuous 
pressure volume change. It thus appears that whenever there are available 
sufficient data to plot a continuous curve representing a pressure volume change, 
the work can be found by evaluating the area lying under the curve and bounded 
by the curve coordinates and the axis of volumes. The work done may be 
found by actual measurement of the area or by algebraic solution of Eq. (7), 
which can be integrated only when there is a known algebraic relation between 
the pressure and the corresponding volume of the expansive fluid, gas or vapor. 

Prob* 1. Draw the diagrams for the following cases: (a) The pressure in a cylinder 
12 ins. in diameter was found to vary at different parts of an 18-in. stroke as follows: 



12 



ENGINEERING THERMODYNAMICS 



Pressure in Pounds 


Per Cent of 


per Sq.Ia. 


Stroke. 


100 





100 


10 


100 


.30 


100 


50 


83.3 


60 


71.5 


70 


62.5 


80 


55.5 


90 


60.0 


100 



(6) On a gas engine diagram the following pressures were found for parts of stroke. 



In 




Out 




V 


P 


V 


P 


V 


P 


0.25 cu.ft. 


14.7 


0.1 


45.2 


0.13 


146.2 


0.20 '* 


19.5 


0.102 


79.7 


0.15 


116.7 


0.14 " 


29.7 


0.104 


123.2 


0.17 


95.7 


0.10 *• 


45.2 


0.106 


157.7 


0.19 


80.7 






0.108 


181.7 


0.21 


68.7 






0.11 


188.2 


0.23 


58.7 






0.12 


166.2 







Prob. 2. Steam at a pressure of 100 lbs. per square inch absolute is admitted to a 
cylinder containing .1 cu.ft. of steam at the same pressure, until the cylinder contains 
1 cu.ft., when the supply valve closes and the volurtie increases so that the product of 
pressure and volume is constant imtil a pressure of 30 lbs. is reached. The exhaust 
valve is opened, the pressure drops to 10 lbs. and steam is forced out until the volume 
becomes 1 cu.ft., when the exhaust valve closes and the remaining steam decreases in 
volume so that product of pressure and volume is constant until the original point 
is reached. Draw the pressure volume diagram for this case. 

Prob. 3. Diu'ing an air compressor stroke the pressures and volumes were as 
follows: 



Volume in 


Prcf59ure in Lbs. 


Cu.Ft. 


Sq.In. 


2.0 


14.0 


1.8 


15.5 


1.6 


17.5 


1.4 


20.0 


1.2 


23.3 


1.0 


28.0 


0.8 


28.0 


0.4 


28.0 


0.0 


28.0 



Draw the diagram to a suitable scale to give work area in foot-pounds directly. 
Prob. 4. Draw the diagrams for last two problems of Section 3. 



WORK AND POWER 13 

7. Work of Expansion and Compression. Any given quantity of gas or 
vapor confined and not subject to extraordinary thermal changes such as 
explosion, will suffer regular pressure changes for each unit of volume change, 
or conversely, suffer a regular volume change for each imit of pressure change, 
so that pressure change is dependent on volume change and vice versa. When 
the volume of a mass of gas or vapor, Vi, is allowed to increase to V2 by the 
movement of a piston in a cylinder, the pressure will regularly increase or 
decrease from Pi to P2, and experience has shown that no matter what the gas 
or vapor or the thermal conditions, if steady, the voliunes and pressures will 
have the relation for the same mass, 

PiFi'=P2F2^=if, (8) 

or the product of the pressure and s power of the volume of a given mass 
will always be the same. The exponent « may have any value, but usually 
lies between 1 and 1.5 for conditions met in practice. 

The precise value of « for any given case depends on 

(a) The substance. 

(6) The thermal conditions surroimding expansion or compression, « being 
different if the substance receives heat from, or loses heat to, external sur- 
roundings, or neither receives nor loses. 

(c) The condition of vapors as to moisture or superheat when vapors are 
under treatment. 

Some commonly used values of s are given in Table X at the end of this 
chapter for various substances subjected to different thermal conditions dur- 
ing expansion or compression. 

Not only does Eq. (8) express the general law of expansion, but it likewise 
'expresses the law of compression for decreasing volimiies in the cylinder with 
jcorresponding rise in pressure. Expansion in a cylinder fitted with a piston 
is called balanced expansion because the pressure over the piston area is 
balanced by resistance to piston movement and the mass of gas or vapor is 
subtitaatially at rest, the work of expaasion being imparted to the J)iston and 
resisting mechanism attached to it. On the other hand when the gas or vapor 
under pressure passes through a nozzle orifice to a region of lower pressure the 
falling pressure is accompanied by increasing volumes as before, but the work 
of expansion is imparted not to a piston, because there is none, but to the fluid 
itself, accelerating it until a velocity has been acquired as a resultant of the 
work energy received. Such expansion is termed free expansion and the law of 
Eq. (8) applies as well to free as to balanced expansion. This equation, then, 
is of very great value, as it is a convenient basis for computations of the work 
done in expansion or compression in cylinders and nozzles of all sorts involv- 
ing every gas or vapor substance. Some expansion curves for different values 
of 8 are plotted to scale in Fig. 5, and the corresponding compression curves in 
fig. 6, in which 

Curve A has the exponent s= 
Curve B '' ^V s= .5 



ENGINEERINa THERMODYNAMICS 



Curve C has the exponent «= 1.0 

Curve D 

Curve E 

Curve F 

Curve G 

Curve K 



Volumea In Cubic Feet 
Fig. 6. — ComparisoD of ExpanHion Lines having Different Values of *, 



The volume after expansion is given by 



(9) 



SO that the final volume depends on the original volume, on the ratio of the two 
pres.sures and on the value of the exponent. Similarly, the pressure after 
expansion 

''"-^■(f;)' ("» 



WORK AND POWER 



15 



depends on the original pressure, on the ratio of the two volumes and on 
the exponent. 

The general equation for the work of expansion or compression can now be 
integrated by means of the Eq. (8), which fixes the relation between pressures 
and volumes. From Eq. (8), 




Fig. 6. — Comparison of Compression Curves having Different Values of 8. 

which, substituted in Eq.(7), gives 

^KdV 



W 



-p 



ya f 



but as /^ is a constant, 



W 



-/I?- 



The integral of Eq. (11) will have two forms: 

(1) When 8 is equal to one, in which case PiVi=P2V2=K^; 

(2) When s is not equal to one. 



(11) 



16 



ENGINEERING THEEM0DYNA14ICS 



Taking first the case when « is equal to one, 

•VtdV 



W 



rv, 



Whence 



W=K' log, Y^ 



=PiViloe» 



= P2F2l0g, 



Yl 

Yl 
Vi 



(a) 



(&) 



W 



(d) 



(e) 



When 8 - 1 . 



(12) 



=i^log.g 
=PiFilog.^^ 

= P2V2l0g.g (/) 

Eqs. (12) are all equal and set down in different forms for convenience in 
computation; in them 

F2= largest volume = initial vol. for compression = final vol. for expansion. 

P2 = smallest pressure = initial pres. for compression = final pres. for expansion. 

Fi = smallest volume = final vol. for compression = initial vol. for expansion. 

Pi = largest pressure = final pres. for compression = initial pres. for expansion. 

These Eqs. (12) all indicate that the work of expansion and compression of 
this class is dependent only on the ratio of pressures or volumes at the beginning! 
and end of the process, and the PV product at either beginning or end, this 
product being of constant value. 

When the exponent s is not equal to one, the equation takes the form, 






i-sL 



Vi^-'-Yx^-" 



•] 



As s is greater than one, the denominator and exponents will be negative, so, 
changing the form to secure positive values, 



s-1VFi«-» F2'-V 



This can be put in a still more convenient form. Multiplying and dividing by 
1 1 



vv-i 



or 



Y.»-v 



--f-r.M^r -^'^uH^<ir\ 



WORK AND POWER 



17 



Substituting the value of X=P2F2*=PiFi', 



Whence 



-^/mm"'-^]-^^mHW} 



^-mivr-^] <«' 






. . Whens5^1. . . (13) 



V2 
P2 
Vi 
Pi 



Eqs. (13) gives the work for this class of expansion and compression in terms 
of pressure ratios and volume ratios and in them 

largest volume = initial vol. for compression = final vol. for expansion; 
• smallest pressure = initial pres. for compression = final pres. for expansion ; 
= smallest volume = final vol. for compression = initial vol. for expansion; 
-largest pressure = final pres. for compression = initial pres. for expansion. 

The work of expansion or compression of this class is dependent according 
to Eqs. (13), upon the ratio of pressures or volumes at beginning and end of the 
process, the exponent, and on the pressure volume product appropriately taken. 
It should be remembered that for the result to be in foot-pounds appropriate 
units should be used and all pressures taken absolute. Examination of Eqs. 
(12) and (13), for the work done by expansion or compression of both classes, 
shows that it is dependent on the initial and final values of pressures and volumes 
and on the exponent 8, which defines the law of variation of pressure with 
volume between the initial and final states. 

Example 1. Method of calculating Diagram, Figs. 5 and 6. Consider the curve 
for which 8 = 1 A as typical of the group. 



Assumed Data, 



Then 



Fi «1.0 cu.ft. Pi =20,000 lbs. per square foot. 
« = 1.4. 
Pi7i* =»i2: =20,000 Xl^-* =20,000. 



For any other value of P, V was found from the relation. 



18 



ENGINEERING THERMODYNAMICS 



LetPa:«6000, 



then 



or 



^- ©'^-[s]-">^"' 



log 3.33 =.5224 



.715 X .5224 = .373 =log Vt. 
.'. 7* =2.36. 

A series of points, as shown below, were found, through which the curve was drawn. 



p 


20.000 
P ' 


, 20,000 
log— y— . 


1 .20,000 
s^^ P • 


V 


18000 


1.111 


0.0453 


0.032 


1.08 


14000 


1.430 


0.1653 


0.111 


1.30 


10000 


2.000 


0.3010 


0.214 


1.64 


6000 


3.330 


0.5224 


0.373 


2.36 


2000 


10.000 


1.0000 


0.714 


5.18 


1000 


20.000 


1.3010 


0.930 


8.51 



Curves for other values of 8 were similarly drawn. Starting at a coi^mon volume 
of 20 cu. ft. the compression curves of Fig. 6 were determined by the same methods. 

Example 2. A pound of air at 32^ F. and under atmospheric pressure is compressed 
to a pressure of five times the original. What will be the final volume and the work 
done if 8»1 and if 8 = 1 A? The voliune of 1 lb. of air at 32^ F. and one atmosphere 
is 12.4 cu.ft. approx. 



For« = l, 



P.' 



7, = 12.4cu.ft.; 






12.4 



=5, whence Fi= 2.48 cu.ft. 



For « = 1.4, 



W^P.Vtloge^ = 2116X12.4 log* 5; 

Pi 

=2116X12.4X1.61=42,300 foot-pounds. 



f:-©" 



12.4 .-.11 ...-71 
_- = (5) = (5) . 



5 may be raised to the .71 power by means of logarithms as follows: (5)"^^ is equal to 
the number whose logarithm is .71 log 5. 



WORK AND POWER 19 

Log 5 =.699, .71 X.699 = .4963, and number of which this is the logarithm is 3.13, 

hence, 

71 = 7,^-3.13 or Fi=3.96; 

= ^^^^><12.4 ^ ggg ^ 3g 200 f t.-lbs. 
.4 

The value of W can also be found by any other form of equation (13) such as. 



-Si'-©-]- 



The value of Yx being found as before, the work expression becomes after numerical 
substitution 



^ 10,580 X3.96r, /3.96V'^1 
^= A L^"VT2l/ J- 

As the quantity to be raised to the .4 power is less than one, students may find it 
oasier to use the reciprocal as follows: 

. /3.96\ -4 ^ 1 ^ 1 !_ ^ goo 

Vl2.4/ /IMV* (^-^^^^ ^-^ 

\3.96/ 
Hence 



^^10^X3^^^ -.632) =38,200 f t.-lbs. 

Prob. 1. Find Yx and W for Example 2 if s = 1.2 and 1.3. 

Prob. 2. If a pound of air were compressed from a pressure of 1 lb. per square inch 
absolute to 15 lbs. per square inch absolute ifind Yx and W when « = 1 and 1.4. F, = 180 
QXi.iU What would be the H.P. to compress 1 lb. of air per minute? 

Prob. 3. Air expands so that 8 = 1. If Pi = 10,000 lbs. per square foot, Yx = 10 cu.ft. 
and r^ «100 cu.ft. and the expansion takes place in 20 seconds, wha'. is the H.P. devel- 
oped? 

Prob. 4. 100 cu.ft. of air at atmospheric pressure is compressed in a cylinder to a pres- 
sure of 8 atmospheres and then expelled against this constant pressure. Find graphically 
and by calculation the foot-pounds of work done for the case where s = 1 and for the case 
where « = 1.4. 

Prob. 5. At an altitude of 4000 ft., a r is compressed to a pressure of 60 lbs. per 
sq.in. gage. Find he H.P. required to compress 1000 cu.ft. of free air per minute. 

Prob. 6. From the algebraic equation show how much work is done for a volume 
change of 1 to 4, provided pressure is originally 1000 lbs. per square foot when 

(a) FY^^Kx, 
(6) VY^K^, 
(c) PF«=i^3. 



20 ENGINEERING THERMODYNAMICS 

Prob. 7. A vacuum pump compresses air from 1 lb. per square inch absolute to 15 
lbs. per square inch absolute and discharges it. An air compressor compresses air from 
atmosphere to 15 atmospheres and discharges it. Compare the work done for equal 
initial volumes, « = 1.4. 

Prob. 8. For steam expanding according to the saturation law, compare the work 
done by 1 lb. expanding from 150 lbs. per square inch absolute to 15 lbs. per square inch 
absolute with, the work of the same quantity expanding from 15 lbs. to 1 lb. per square 
inch absolute. 

Note. 1 lb. of steam occupies 3 cu.ft. at 150 lbs. per square inch absolute. 

Prob. 9. Two air compressors of the same size compress air adiabatically from atmos- 
phere to 100 lbs. gage and discharge it. One is at sea level, the other at 10,000 ft. ele- 
vation. Compare the work in the two cases. 

8. Values of Exponent s Defining Special Cases of Expansion or Compres- 
sion. There are three general methods of finding s for the definition of particular 
cases of expansion or compression to allow of the solution of numerical problems. 
The first is experimental, the second and third thermodynamic. If by measure- 
ment the pressures and volumes of a series of points on an expansion or com- 
pression curve, obtained by test with appropriate instruments, for example, 
the indicator, be set down in a table and they be compared in pairs, values of 
8 can be found as follows: Calling the points A, J8, C, etc., then, 

PaVa' = PbVb', 

and 

log Pa+S log Fa^log Pb+S log 76, 

or 

. S(l0g Vb - log Va) = log Pa - log Pb, 

hence 

logPg-logP , . 
"^^logn-logF, ^""^ 



_ _ a 

or 






(14) 



According to Eq. (14a), if the difference between the logarithms of the pressures 
at B and A be divided by the differences between the logarithms of the volumes 
at A and B respectively, the quotient will be s. According to Eq. (146), the 
logarithm of the ratio of pressures, Bio Ay divided by the logarithm of the ratio 
of voliunes, AtoB respectively will also give s. It is interesting to note that 
if the logarithms of the pressures be plotted vertically and logarithms of volumes 
horizontally as in Fig. 7, then the line AC equal to the intercept on the horizontal 
axis represents the difference between the logarithms of volumes or, 



CA = logya~^ogy6, 



WORK AND POWER 



21 



and similarly 



Hence 



CB=logPb- logP, 



CB , 
CA 



or the slope of the line indicates the value of s. This is a particularly valuable 
method, as it indicates at a glance the constancy or variability of s, and there 
are many cases of practice where s does vary. Should 8 be constant the line 
will be straight; should it be variable the line will be curved, but can generally 




.6 Jd 1. 

• L(yg. V 

Fig. 7. — Graphic Method of Finding s, from Logarithms of Pressures and Volumes. 

be divided into parts, each of which is substantially straight and each will 
have a different 8. It is sometimes most convenient to take only the beginning 
and end of the curve and to use the value of s corresponding to these points, 
neglecting intermediate values. 

A second method for finding s for a given compression or expansion line 
by means of areas is indicated in a note in Section 17 of this Chapter that 
is omitted here because it depends on formulas not yet derived. It is by this 
sort of study of experimental data that most of the valuable values of s have 
been obtained. There is, however, another method of finding a value for 
s by purely thermodynamic analysis based on certain fimdamental hypo- 
theses, and the value is as useful as the hypotheses are fair or true to the 
facts of a particular case. 

One of the most common hypotheses of this sort is that the gas or vapor 
undergoing expansion or compression shall neither receive any heat from, 
nor give up any to bodies external to itself during the process, and such a process 
is given the name adiabatic. Whether adiabatic processes are possible in actual 
cylinders or nozzles does not affect the analysis with which pure thermody- 
namics is concerned. By certain mathematical transformations, to be carried 
out later, and based on a fundamental thermodynamic proposition, the adia- 



22 ENGINEERING THERMODYNAMICS 

batic hypothesis will lead to a value of s, the use of which gives results valuable 
as a basis of reference, and which when compared with an actual case will per- 
mit of a determination of how far the real case has departed from the adiabatic 
condition^ and how much heat has been received or lost at any part of the 
process. The particular value of s which exists in an adiabatic change is repre- 
sented by the symbol y. 

Another common hjrpothesis on which another value of s can be derived, 
is that gases in expansion or compression shall remain at a constant temperature, 
thus giving rise to the name isothermal. This is generally confined to gasej? 
and superheated vapors, as it is difficult to conceive of a case of isothermal or con- 
stant temperature expansion or compression of wet vapors, as will be seen later. 

In the study of vapors,' which, it must be understood, may be dry or wet, 
that is, containing liquid, a common hypothesis is that during the expansion or 
compression they shall remain just barely dry or that they shall receive or lose 
just enough heat to keep any vapor from condensing, or but no more than 
sufficient to keep any moisture that tends to form always evaporated. Expan- 
sion or compression according to this hypothesis is said to follow the saturation 
law, and the substance to remain saturated. It will appear from this thermal 
analysis later that the value of s for the isothermal hypothesis is the same for all 
gases and equal to one, but for the adiabatic hj^othesis s=y will have a 
different value for different substances, though several may have the same 
value, while for vapors y will be found to be a variable for any one, its value 
depending not only on the substance, but on the temperatures, pressures and 
wetness. • 

When gases or vapors are suffered to expand in cylinders and nozzles or 
caused to compress, it is often difficult and sometimes impossible or perhaps 
undesirable to avoid interference with the adiabatic conditions for vapors 
and gases, with the isothermal for gases or with the saturation law for vapors, 
yet the wqrk to be done and the horse-power developed cannot be predicted 
without a known value of s, which for such cases must be found by experi- 
ence. A frequent cause of interference with these predictions, which should 
be noted, is leakage in cylinders, which, of course, causes the mass under 
treatment to vary. 

According to these methods those values of s have been found which are 
given in Table X, at the end of the Chapter. Mixtures of common gases such 
as constitute natural, producer, blast furnace or illuminating gas, alone or 
with air or products of combustion, such as used in internal combustion engines, 
have values of 5 that can be calculated from the elementary gases or measured 
under actual conditions. 

All vapors, except those considerably overheated, have variable exponents 
for adiabatic expansion and compression. This fact makes the exact soluticii 
of problems of work for wet vapors, expanding or compressing, which form the 
bulk of the practical cases, impossible by such methods as have been described. 
This class of cases can be treated with precision only by strictly thermal 
methods, to be described later. 



WORK AND POWER 



23 



Prob. 1. By plotting the values for the logarithms of the following pressures and vol- 
umes, see if the value for s is constant, and if not find the mean value in each case. 

(a) Gas Engine Comfbession 



V p V p V 
10 45.2 13 32.2 18 


P 
21.0 


11 39.7 14 29.7 20 


19.5 


12 35.7 16 24.7 25 


14.7 


(6) Gas Engine Expansion 




V p V p V 

13 146.2 19 


P 
80.7 


11 188.2 15 116.7 21 


68,7 


12 166.2 17 65.7 23 


58.7 


(c) Steam Expansion 




V p V p 
2.242 203.3 7.338 52.5 




2.994 145.8 12.44 28.8 




4.656 89.9 22.68 14.7 





Prob. 2. By plotting the values for the logarithms of the volumes and pressures on 
the expansion and compression curves of the following cards, find value for a. 



»10H 

0- 
flO- 

60- 
40- 

38- 

r- 

10- 

0- 
600 -j 

900-^ 

20O 
JOO- 

0- 



Atmagphero 




Atmospheto 





Atinoei>here> 



125- 

IQD- 

76- 

60- 

26- 

0- 

120-1 

90 

60- 

«)- 

0- 




(Mmom 




Atmoaphere 



24 



ENGINEERING THERMODYNA^nCS 



160 

140 

120 

100 

80 

eo 

40 
20 


80 




Atinosphei'e 




Atmosphere 



Prob. 3. From the steam tables at the end of Chapter IV. select the pressures and 
volumes for dry-saturated steam and find the value of s between 

(a) 150 lbs. per square inch and 1 lb. per square inch. • 

(6) 15 

Prob. 4. Find for superheated steam at 150 lbs. per square inch and with 100^ 
of superheat expanding to 100 lbs. per square inch without losing any superheat, the 
corresponding value of 8, using tabular data. 

Prob. 6. From the ammonia table data for dry-saturated vapor find the value of s 

between 

(a) 160 lbs. per square inch and 1 lb. per square inch. 

(6) 15 



(( 



It 



it 



9. Work Phases and Cycles, Positive and Negative and Net Work. Accord- 
ing to the preceding it is easy to calculate or predict numerically the work of 
expansion or compression whenever the conditions are sufficiently definite to 
permit of the selection of the appropriate s. It very seldom happens, however, 
that the most important processes are single processes or that the work of 
expansion or compression is of interest by itself. For example, before expansion 
can begin in a steam cylinder steam must be first admitted, and in air com- 
pressors air must be drawn in before it can be compressed. Similarly, aftei* 
expansion in a steam cylinder there must be an expulsion of used vapor before 
another admission and expansion can take place, while in the air compressor 
after compression the compressed air must be expelled before more can enter 
for treatment. The whole series of operations is a matter of more concern 
than any one alone, and must be treated as a whole. The effect can be most 
easily found by the summation of the separate effects, and this method of 
summation will be found of universal application. 

The whole series of processes taking place and involving pressure volume 
changes is called a cycle, any one of them a phase. It is apparent that there 
can be only a limited number of phases so definite as to permit of the mathe- 



WORK AND POWER 25 

matical treatment necessary for prediction of work, but it is equally clear that 
there may be a far greater number of combinations of phases constituting 
cycles. Before proceeding to analyze the action of steam or gas in a cylinder 
it is necessary first to determine on structural, thermal or any other logical 
grounds, what series of separate processes will be involved, in what order, and 
the pressure volume characteristics of each. Then and then only, can the 
cycle as a whole be treated. These phases or separate and characteristic proc- 
esses affecting the work done or involving pressure volume changes are divisible 
into two classes so far as the causes producing them are concerned, the first 
thermal and the second mechanical. It requires no particular knowledge of 
thermodynamics to realize that if air be confined in a cylinder with a free piston 
and is heated, that the volume will increase while pressure remains constant, 
since the piston will move out with the slightest excess of pressure inside over 
what is outside. This is a pressure constant, volume increasing, phase, and 
is thermal since it is a heat effect. If an ample supply of steam be available 
from a boiler held at a constant pressure by the manipulation of dampers and 
fires by the fireman and the steam be admitted to a cylinder with a piston, 
the piston will move out, the pressure remaining constant and volume increas- 
ing. This is also a pressure constant, volume increasing phase, exactly as before, 
but is mechanical because it is due to a transportation of steam from the boiler 
to the cylinder, although in another sense it may be considered as thermal if 
the boiler, pipe and cylinder be considered as one part during the admission. 
A sioiilar constant pressure phase will result when a compressor piston is 
forcibly drawn out, slightly reducing the pressure and permitting the outside 
atmosphere to push air in, to follow the piston, and again after compression of 
air to a slight excess, the opening of valves to storage tanks or pipe lines having 
a constant pressure will allow the air to flow out or be pushed out of the cylinder 
at constant pressure. These two constant pressure phases are strictly mechan- 
ical, as both Represent transmission of the mass. If a cylinder contain water 
and heat be applied without permitting any piston movement, there will be 
a rise of pressure at constant volume, a similar constant volume pressure rise 
phase will result from the heating of a contained mass of gas or vapor under the 
same circumstances, both of these being strictly thermal. 

However much the causes of the various characteristic phases may differ, 
the work effects of similar ones is the same and at present only work effects 
are under consideration. For example, all constant volume phases do no 
work as work cannot be done without change of volume. 

The consideration of the strictly thermal phases is one of the. principal 
problems of thermodynamics, for by this means the relation between the work 
done to the heat necessary to produce the phase changes is established, and a 
basis laid for determining the ratio of work to heat, or eflSciency. For the 
present it is sufficient to note that the work effects of any phase will depend 
jonly on the pressure volume changes which characterize it. 
I Consider a cycle Fig. 8, consisting of {AB)y admission of 2 cu.ft of steam at 
k constant pressure of 200,000 lbs. per square foot, to a cylinder originally 



26 ENGINEERING THERMODYNAMICS 

coittainmg nothing, followed by (BC), expansion with s=l, to a pressure of 
20,000 lbs. per square foot; (C^, constant volume change of pressure, and 
(FG), constant pressure exhaust at 10,000 lbs, per square foot. These opera- 
tions are plotted to scale. Starting at zero volume, because the cylinder 



Volumes In Cublo Pert 



ori^nally contains nothing, and at a pressure of 200,000 lbs. per square foot, 
the line AB, ending at volume 2 cu.ft., represents admission and the cross- 
hatched area under AB represents the 400,000 ft.-lba. of work done during 
admission. At B the admission ceases by closure of a valve and the 2 cu.ft. 
of ateam at the original pressure expands with lowering pressure according 
'CO the law 



So that when 



PaFa=nF6=200,OOOx2 = 400,000 ft.-lbs., 
y = 4 cu.ft., P = —^*^=100,000 lbs. per sq.ft.; 

V = 5 cu.ft.. p=M^= 80,000 lbs. per sq.ft.; 

V = 10cu.ft., P=^^^;^= 40,000 lbs. per sq.ft. 



lbs. per square foot, and the work done during expansion is the cross-hatched 
area JBCD under the expansion curve BC, the value of which can be found by 
measuring the diagram or by using the formula Eq. (12), 



WOEK AND POWER 27 

which on substitution gives 

Trbc=400,000 log. 10=400,000X2.3; 
=920,000 ft.-lbs. 

This completes the stroke and the work for the stroke can be found by addition 
of the numerical values, 

TFafc=400,000 ft.-lbs. , • 
TF»c= 920,000 ft.-lbs.; 
Tra5+TF6c= 1,320,000 ft.-lbs. 

It is often more convenient to find an algebraic expression for the whole, 
which for this case will be, 



TF6c=PftFJoge^'; 



(l+log.^), 



=400,000(H-lo& 10) =400,000X3.3 = 1,320,000 ft.-lbs. 

On the return of the piston it encounters a resistance due to a constant pressure 
of 10,000 lbs. per square inch, opposing its motion; it must, therefore, do work 
on the steam in expelling it. Before the return stroke begins, however, the 
pressure drops by the opening of the exhaust valve from the terminal pressure 
of the expansion curve to the exhaust or hack pressure along the constant volume 
line, CF, of course, doing no work, after which the return stroke begins, the 
pressure volimie line being FG and the work of the stroke being represented by 
the cross-hatched area DFGHy 

TF/a=P/F/= 10,000X20=200,000 ft.-lbs. 

This is negative work, as it is done in opposition to the movement of the piston. 
The cycle is completed by admission of steam at constant zero volume, raising 
the pressure along GA. The net work is the difference between the positive 
and negative work, or algebraically 



Tr=Pftn(l+loge y^ -PfVf, 

= l,320,000-200,000 = l,120,000^ft.-lbs. 



28 ENGINEERING THERMODYNAMICS 

. Consider now a cycle of an an compressor, Fig. 9. AdmissioD or Buction 
18 represented by AB, compression by BC, delivery by CD and constant volume 
drop in pressure after delivery by DA, The work of admission is represented 
by the area ABFE or algebraically by 

W,a = PbVb, 

the work of compression by the area FBCG, or algebraically since 8=1,4 by 



^m^r 



the work of delivery by the area CDEG, or algebraically 

The positive work is that assisting the motion of the piston durii^ suction; 
the area ABFE or algebraically PbVh. The negative work, that in opposi- 
tion to the motion, is the sum of the compression and delivery work, the area 
FBCDE, or algebrajcally, 



-«+---.''-^'[0"'-']- 



The net work is the difference and is negative, as such a cycle is mainly resistant, 
and to execute it the piston must be driven with expenditure of work on the 
gas. The value of the net work is, 

W=Wb,+W,d-W<a, 



WORK AND POWER 29 

an expression which will be simplified in the chapter on compressors. This 
net work is represented by the area ABCD, which is the area enclosed by the 
cycle itself independent of the axes of coordinates. 

It might seem from the two examples given as if net work could be 
obtiuned without the tedious problem of summation, and this is in a sense true 
if the cycle is plotted to scale or an algebraic expression be available, but 
these processes are practically equivalent to summation of phase results. It 
might also seem that the work area would always be that enclosed by the 
cycle, and this is true with a very important limitation, wiiich enters when 
tbe cycle has loops. If, for example, as in Fig. 10, steam admitted A to B, 
expanded along BC to a pressure C, then on opening the exhaust the pressure 
instead of falling to the back pressure or exhaust line as in Fig. 8, would here 



Volumes In Cubic Feet 
Fio, 10. — AnalyBiB of Work Diagram for Engine with Over-expansion Negative Work Loop. 

rise along CD, as the back pressure is higher thaa the terminal expansion pres- 
sure, after which exhaust will take place at constant back pressure along DE. 
The forward stroke work is that under AB and BC or ABCEG, the return 
stroke work is the area DEGH and the net work is 

Area ABC£(?-Area DEGH. 

A.S the area HGECX is common to both terms of the difference, the net work 
may be set down as equal to 

krf^&ABXH-CDX, 

It may be set down then in general for looped cycles that the net work area 
is the difference between that of the two loops. If, however, the method laid 
down for the treatment of any cycle be adhered to there need not be any dis- 
tinction drawn between ordinary and looped cycles, that is, in finding the work 



30 



ENGINEERING THERMODYNAMICS 



of a cycle divide it into characteristic ptiases and group (hem into positive and 
negativcj find the work for each and lake the algebraic sum. 

Special cases of cycles and their characteristics for steam compressors and 
gas engine cylinders, as well as nozzle expansion, will be taken up later in more 
detail and will constitute the subject matter of the next two chapters. 

Example 1. Method of calculating Diagram, Fig. 8. 



Assumed data 



To obtain pomt C. 



Fe=2 " 



PcVc-^PbV, or Vr= 



Pa =200,000 lbs. per square foot. 

Pt^Pa 

Pc =20,000 " " 

P/= 10,000 
I Pe^Pf. 



tt 
It 



tt 



P&V^^ 200,000X2 20 
Pr 20,000 " ' 



A Fc=20 and Pc =20,000. 

Intermediate points B to C are obtained by assuming various pressures and 
finding the corresponding volumes as for Vc. 

Example 2. Method of calculating Diagram, Fig. 9. 



Assumed data 



f Fa =Ocu.ft. 1 
Fj,=20 *' 
Fd=0 

8 = 1.4 



n 



' Pa =2116 lbs. per square foot. 

Pb^Pa 

Pc = 14,812 " " 



tt 



tt 



To obtain point C, 

P*F,i4 = p^FM or V,==V,-r^(^Y^^V,-^(^py 



716 



P 
P 



1 

r=7, log 7 =.845 ; and .715 X.845=log^~y* =.6105, 



or 
Therefore, 



F.=4.02. 



Fc =4.02, and Pc = 14,812. 



Intermediate values BtoC may be found by assuming pressures and finding volumes cor- 
responding as for Vc. 

Prob. 1. Steam at 150 lbs. per square inch absolute pressure is admitted into a cylin- 
der in which the volume is originally zero until the volume is 2 cu.ft., when the valve is 
closed and expansion begins and continues until the volume is 8 cu.ft., then exhaust 
valve opens and the pressure faUs to 10 lbs. absolute and steam is entirely swept out. 
Draw the diagram and find the net work done. 



WORK AND POWER 31 

Prob. 2. A piston moving forward in a cylinder draws in 10 cu.ft. of CO2 at a pressure 
of .9 of an atmosphere at sea level and then compresses it adiabatically until the pressure 
rises to 9 atmospheres and discharges it at constant pressure. Draw the diagram and 
find net work done. 

Prob. 3. A cylinder 18 ins. in diameter and 24 ins. piston stroke receives steam at 
100 lbs. per square inch absolute pressiu^ for i of the stroke. It then expands to the 
end of the stroke and is exhausted at atmospheric pressure. Draw the diagram and 
find the H.P. if the engine makes 100 strokes per minute. 

Prob. 4. Two compressors without clearance each with a cylinder displacement of 
2 cu.ft. draw in air at 14 lbs. per square inch absolute and compress it to 80 lbs. per 
square inch absolute before delivery. Find the difference in H.P. per 1000 cu.ft. of 
free air per minute if one is compressing isothermally and the other adiabatically. 
Draw diagram for each case. 

Prob. 6. A quantity of air 5 cu.ft. in volume and at atmospheric pressure is compressed 
in a cylinder by the movement of a piston until the pressure is 50 lbs. per square inch 
gage. If the air be heated the pressure will rise, as in an explosion. In this case the 
piston remains stationary, while the air is heated until the pressure reaches 200 lbs. per 
square inch gage. It then expands adiabatically to the original volume when the 
pressure is reduced to atmosphere with no change in volume. Draw the diagram, and 
find the work done. 

Prob. 6. The Braytan cycle is one in which gas is compressed adiabatically and then, 
by the addition of heat, the gas is made to expand without change of pressure. Adi- 
abatic expansion then follows to original pressure and the cycle ends by decrease in volume 
to original amount without change of pressure. Draw such a cycle startmg with 5 cu.ft. 
of air at atmospheric pressure, compressing to 4 atmospheres, expanding at constant 
pressure to 5 cu.ft., expanding adiabatically to original pressure and finally ending at 
original point. Find also, work done. 

Prob. 7. In the Ericsson cycle air is expanded at constant temperature, cooled at 
constant pressure, compressed at constant temperature and receives heat at constant 
volume. Draw a diagram for the case where 5 cu.ft. at atmospheric pressure are com- 
pressed to 1 cu.ft., heated until volume is 8 cu.ft., expanded to atmosphere and then 
cooled to original volume. Find the work. 

Prob. 8, In the Stirling cyde constant volume heating and cooling replace that at 
constant pressure in the Ericsson. Draw diagram starting with 5 cu.ft. and atmsopheric 
pressure compressing to 1 cu.ft. and then after allowing the pressure to double, expand 
to original volume and cool to atmosphere. Find the work. 

Prob. 9. The Joule cyde consists of adiabatic compression and expansion and con- 
stant pressure heating and cooling. Assuming data as in last problem draw the 
diagram and find the work. 

Prob. 10. The Camot cycle consists of isothermal expansion, adiabatic expansion, 
isothermal compression and adiabatic compression. Draw the diagram for this cycle 
and find the work. 

10. Work Determination by Mean Eflfective Pressure. While the methods 
already described are useful for finding the work done in foot-pounds for a defined 
cycle with known pressure and volume limiis^ they are not, as a rule, convenient 
for the calculation of the work done in a cylinder of given dimensions. As 
work done can always be represented by an area, this area divided by its length 
will give its mean height. If the area be in foot-pounds with coordinates 



32 ENGINEERING THERMODYNAMICS 

pounds per square foot, and cubic feet, then the division of area in foot-pounds 
by length in cubic feet will give the mean height or the mean pressure in 
pounds per square foot. Again, dividing the work of the cycle into forward- 
stroke work and back-stroke work, or the respective foot-pound areas divided 
by the length of the diagram in cubic feet, will give the mean forward pressure 
and the mean back pressure. The difference between mean forward pressure 
and mean back pressure will give the m,ean effective pressure, or that average 
pressure which if maintained for one stroke would do the same work as the 
cycle no matter how many strokes the cycle itself may have required for its 
execution, which is very convenient considering the fact that most gas engines 
require four strokes to complete one cycle. The mean eflFective pressure may 
also be found directly from the enclosed cycle area, taking proper account of 
loops, as representative of net work by dividing this net work area by the length 
of the diagram in appropriate units. This method is especially convenient 
when the diagram is drawn to odd scales so that areas do not give foot-pounds 
directly, for no matter what the scale the mean height of the diagram, when 
multiplied by the pressure scale factor, represents the mean effective pressure. 
This mean height can always be found in inches for any scale of diagram by 
finding the area of the diagram in square inches and by dividing by the length 
in inches, and this mean height in inches multiplied by the scale of pressures 
in whatever imits may be used will give the mean effective pressure in the same 
units. 

Mean pressures, forward, back or effective, are found and used in two general 
ways; first, algebraically, and second graphically and generally in this case 
from test records. By the first method, formulas, based on some assumed 
laws for the phases, can be found, and the mean effective pressure and its value 
predicted. This permits of the prediction of work that may be done by a given 
quantity of gas or vapor, or the work per cycle in a cylinder, or finally the horse- 
power of a machine, of which the cylinder is a part, operating at a given speed 
and all without any diagram measurement whatever. By the second method, 
a diagram of pressures in the cylinder at each point of the stroke can be obtained 
by the indicator, yielding information on the scale of pressures. The net work 
area measured in square inches, when divided by the length in inches, gives 
the mean height in inches, which, multiplied by the pressure scale per inch 
of height, gives the mean effective pressure in the same imits, which are 
usually pounds per square inch in practice. 

As an example of the algebraic method of prediction, consider the cycle 
represented by Fig. 8. The forward work is represented by 



Forward work =PbVb 



(i+iog.f;), 



the length of the diagram representing the volume swept through in the per- 
formance of this work is Vcy hence 

Mean forward pressure = —vr- ( 1+logc ^^ ) . 



WORE AND POWER 



But PbVh^PcVe by the law of this particular expansion curve, hence 



33 



Mean forward pressure =Pc{ bflogc -^Y 



As the back pressiu^ is constant its mean value is this constant value, hence 

C!onstant (mean) back pressure =P/. 
By subtraction 

Mean effective pressure —Pc\ 1 +loge ^\ — Pf 

= 3.3Pc-Pf; 

= 3.3X20,000-10,000; 

=66,000-10,000=56,000 lbs. per sq.ft. 

The work done in foot-pounds is the mean effective pressure in poimds per 
square foot, multiplied by the displacement in cubic feet. 

F=56,000X20= 1,120,000 ft.-lbs. as before. 




Fio. 11.- 



!^iaA-£jiginc Indicator Card. For Determination of Mean Effective Pressure 

without Volume Scale. 



As an example of the determination of mean effective pressure from a test 
or indicator diagram of unknown scale except for pressures, and without axes 
of coordinates, consider Fig. 11, which represents a gas engine cycle in four 
strokes, the precise significance of the lines being immaterial now. The 
pressure scale is 180 lbs. per square inch, per inch of height. 

By measurement of the areas in square inches it is found that 

Large loop area CDEXC =2.6 sq.in. 

Small loop area ABXA =0.5 sq.in. 

Net cycle area =2.1 sq.in. 

Length of diagram =3.5 in. 

Mean height of net work cycle =0.6 in. 

Mean effective pressure = 120X.06 = 72 lbs. per square inch. 



34 ENGINEERING THERMODYNAMICS 

It is quite immaterial whether this diagram were obtained from a large or 
a small cylinder; no matter what the size, the same diagram might be secured 
and truly represent the pressure volume changes therein. If this particular 
cylinder happened to have a diameter of 10 ins. and a stroke of 12 ins. the 
work per stroke can be foimd. The area of the cylinder will be 78.54 sq.ins., 
hence the average force on the piston is 72 lbs. per square inch X 78.54 sq.ins. = 
5654.88 lbs., and the stroke being 1 ft. the work per stroke is 5654.88 ft.-lbs. 
Both of these methods are used in practical work and that one is adopted in 
any particular case which will yield results by the least labor. 

Prob. 1. An indicator card from an air compressor is foimd to have an area of 3.11 
sq. ins., while the length is 2^ ins. and scale of spring is given as 80 lbs. per square inch 
per inch height. What is m.e.p. and what would be the horse-power if the compressor 
ran with a piston speed of 250 ft. per minute and had a piston 9 ins. in diameter? 

Prob. 2. For the same machine another card was taken with a 60-lb. spring and had 
an area of 4.12 sq.ins. How does this compare with first card, the two having the same 
length? 

Prob. 3. A steam engine having a cylinder 18 ins. in diameter and a stroke of 24 ins., 
takes in ^ cu.ft. of steam at 100 lbs. absolute, allows it to expand and exhausts at atmos- 
pheric pressure. An indicator card taken from the same engine showed a length of 3 
ins., an area of .91 sq.in. when an 80-lb. spring is used. How does the actual m.e.p. 
compare with the computed? 

Prob. 4. Find m.e.p. by the algebraic method of prediction for, 

(a) Brayton cycle; 

(6) Camot cycle; 

(c) Stirling cycle; 

(d) Ericsson cycle; 

(e) Joule cycle. 

(See problems following Section 9). 

11. Relation of Pressure-Volume Diagrams to Indicator Cards. The 
Indicator. When a work cycle or diagram of pressure volume changes is drawn 
to scale with pressures and volumes as coordinates, it is termed a pressure 
volume or PV diagram, and may be obtained by plotting point by point from 
the algebraic expression for the law of each phase or by modifying the indicator 
card. The indicator card is that diagram of pressures and stroke obtained 
by appljdng the indicator to a cylinder in operation. This instrument consists 
essentially of a small cylinder in which a finely finished piston moves freely 
without appreciable friction, with a spring to oppose its motion, a pencil mechan- 
ism to record the extent of the motion, and a drum carrying paper which is moved 
in proportion to the engine piston movement. The indicator cylinder is open 
at the bottom and fitted with a ground union joint for attachment to the main 
cylinder through a special cock, which when open permits all the varying 
pressures in the main cylinder to act on the indicator piston, and when closed 
to the main cylinder opens the indicator cylinder to the atmosphere. The 



WORK AND POWER 35 

upper side of the indicator piston being always open to the atmosphere, its 
movement will be the result of the diflferenee between the pressure in the main 
cylinder and atmospheric pressure. A helical spring, carefully calibrated and, 
therefore, of known scale, is fixed between the indicator piston and open cap or 
head of its cylinder, so that whenever the pressure in the main cylinder exceeds 
atmosphere the indicator piston moves toward the open head of the indicator 
cylinder, compressing the spring. Pressures in the main cylinder if less than 
atmosphere will cause the indicator piston to move the other way, extending 
the spring. This compression and extension of the spring is found in the 
calibration of the spring to correspond to a definite nmnber of pounds per 
square inch above or below atmosphere per inch of spring distortion, so that 
the extent of the piston movement measures the pressure above or below atmos- 
phere. A piston rod projects outward through the cylinder cap and moves a 
series of levers and links carrjdng a pencil point, the object of the linkage being 
to multiply the piston movement, but in direct proportion, giving a large 
movement to the pencil for a small piston movement. A cylinder drum carry- 
ing a sheet of paper is pivoted to the cylinder frame so that the pencil move- 
ment will draw on the paper a straight line parallel to the axis of the drum, if 
drum is stationary, or perpendicular to it if drum rotates and pencil is sta- 
tionary. The height of such lines then above or below a zero or datum line, 
which is the atmospheric line drawn with the cock closed, measures the pressure 
of the fluid under study. The springs have scale nmnbers which give the 
pressure in pounds per square inch per inch of pencil movement. This paper- 
carrying drum is not fixed, but arranged to rotate about its axis, being pulled 
out by a cord attached to the piston or some connecting part through a pro- 
portional reducing motion so as to draw out the cord an amount slightly less 
than the circumference of the drum no matter what the piston movement. 
After having been thus drawn out a coiled spring inside the drum draws it back 
on the return stroke. By this mechanism it is clear that, due to the combined 
movement of the pencil up and down, in proportion to the pressure, and that 
of the drum and paper across the pencil in proportion to the piston movement, 
a diagram will be drawn whose ordinates represent pressures above and below 
atmosphere and abscissse, piston stroke completed at the same time, or dis- 
placement volume swept through. It must be clearly understood that such 
indicator diagrams or cards do not give the true or absolute pressures nor the 
true volumes of steam or gas in the cylinder, but only the pressures above or 
below atmosphere and the changes of volume of the fluid corresponding to the 
piston movement. Of course, if there is no gas or steam in the cylinder at the 
beginning of the stroke, the true volume of the fluid will be always equal to the 
displacement, but no such cylinder can be made. 

While the indicator card is sufficient for the determination of mean eflFective 
pressure and work per stroke, its lack of axes of coordinates of pressure and 
volume prevents any study of the laws of its curves. That such study is 
important must be clear, for without it no data or constants such as the exponent 
« can be obtained for prediction of results in other similar cases, nor can the 



36 



ENGINEEEING THERMODYNAMICS 



presence of leaks be detected, or the gain or loss of heat during the variou-s 
processes studied. In short, the most valuable analysis of the operations is 
impossible. 

To convert the indicator card, which is only a diagram of stroke or displace- 
ment on which are shown pressures above and belpw atmosphere into a pres- 
sure volume diagram, there must first be found (a) the relation of true or abso- 
lute pressures to gage pressures, which involves the pressure equivalent of 
the barometer, and (h) the relation of displacement volumes to true volumes? 
of vapor or gas present, which involves the clearance or inactive volume of the 
cylinder. The conversion of gage to absolute pressures by the barometer 
reading has already been explained, Section 3, while the conversion of displace- 
ment volumes to true fluid volumes is made by adding to the displacement 
volume the constant value in the same units of the clearance, which is usually 
the result of irregularity of form at the cylinder ends dictated by structural 
necessities of valves, and of linear clearance or free distance between the pis- 
ton at the end of its stroke and the heads of the cylinder to avoid any possi- 
bility of touching due to wear or looseness of the bearings. 



150 tl60(h 




Ftg. 12. — Ammonia Compressor Indicator Card with Coordinates of Pressures and Volumes 

Added to Convert it into a Pressure- Volume Diagram. 



Let ABCD, Fig. 12, represent an indicator card from an ammonia compressor 
on which EF is the atmospheric line. The cylinder bore is 14 ins., stroke 
22 ins., and the scale of the indicator spring 100, barometer 28 ins., and measured 
clearance 32 cu.in. According to Table IX, 28 ins. of mercury corresponds to 
13.753 lbs. per square inch, and as 100 lbs. per square inch, according to the spring 
scale, corresponds to 1 in. of height on the diagram, 1 lb. per square inch cor- 
responds to 0.01 in. of height, or 13.75 lbs. per square inch atmospheric pres- 
sure to .137 in. of height. The zero of pressures then on the diagram must 
lie .137 in. below the line EF. Lay off then a line Af/f, this distance below 
EF, This will be the position of the axis of volume coordinates. 

Actual measurement of the space in the cylinder with the piston at the end 
of its stroke gave the clearance volume of 32 cu. ins. As the bore is 14 
ins. the piston area is 153.94 sq. ins. which in connection with the stroke 



WORK AND POWER 37 

of 22 ins. gives a displacement volume of 22X153.94 = 3386.68 cu. ins. 

32 
Compared with this the ctearance volume is ooo^ ^o =.94 per cent of the 

3386.68 

displacement. It should b^ noted here that clearance is generally expressed 
in per cent of displacement volume. Just touching the diagram at the ends 
drop two lines at right angles to the atmospheric line intersecting the axis of 
volumes previously found at G and H. The intercept GH then represents the 
displacement, or 3386.68 cu.in. or 1.96 cu.ft. Lay oflf to the left of G, .0094, or 
in round numbers 1/100 of GH, fixing the point Af , MG representing the clearance 
to scale, and a vertical through M the axis of pressures. The axes of coordinates 
are now placed to scale with the diagram but no scale marked thereon. The 
pressure scale can be laid off by starting at M and marking off inch points 
each representing 100 lbs. per square inch. Pounds per square foot can also 
be marked by a separate scale 144 times as large. As the length of the diagram 
is 2.94 ins. and displacement 1.96 cu.ft., 1 in. of horizontal distance corresponds 
to .667 cu.ft. or 1 cu.ft to 1.50 ins. of distance. Lay off then from M dis- 
tances of 1,60 ins. and mark the first 1 cu.ft. and the second 2 cu.ft., dividing 
the intervals into fractions. A similar scale of volumes in cubic inches might 
also be obtained. 

By this proceas any indicator card may be converted into a pressure volume 
diagram for study and analysis, but there will always be required the two factors 
of true atmospheric pressure to find one axis of coordinates and the clearance 
volume to find the other. 

Prob. 1. If in cards Nos. 1 and 2, Section 8, the clearances are 5 per cent and 3 per 
cent respectively of the displacement, convert the cards to P7 diagrams on the same 
base to scales of 4 ins. to 1 cu.ft. and 1 in. to 1000 lbs. per square foot, for cylinders 
9J ins. and 14 J ins. respectively in diameter and stroke 12 ins. 

Prob. 2. Do the same for cards Nos. 3 and 4, if clearances are 7 per cent and 4 per 
cent respectively, for cylinders 10 ins. and 17 ins. respectively in diameter and stroke 
12 ins. 

Prob. 8. Do the same for cards Nos. 7 and 8 if clearances are 12 per cent and 8 per 
cent respectively and cylinders 12 ins. and 19 ins. in diameter and stroke 24 ins. 

12. To Find the Clearance. There are two general methods for the find- 
mg of clearance, the first a direct volumetric measurement of the space itself 
by filling with measured liquid and the second a determination by algebraic 
or graphic means from the location of two points on the expansion or com- 
pression curves of the indicator card based on an assumed law for the curves. 

The first method of direct measurement is the only one that offers even a 
promise of accuracy, but even this is difficult to carry out because of the 
tendency of the measuring liquid to leak past piston or valves, which makes 
the result too large if the liquid be measured before the filling of the clearance 
space and too small if the liquid be measured after filling and drawing off. 
There is also a tendency in the latter case for some of the liquid to remain 
inside the space, besides the possibility in all cases of the failure to completely 
fill the space due to air pockets at high places. 



i 



.100 






























A' 


A' 












V 














80 


A'X 


A 












\ 






























\ 












^i\ 


e:, 


i 














\ 


N.« 










00 


B' 


BU 
















\ 


V 




V 




40 




\ 


















> 



































2 



4 6 

Displaoement 
Fig. 13. 



8 

Displacement 



Fig. 14. 



10 














• 


















ft) 


A-' 








a\a 


























\ 


\ 




























S 


\ 














80 


B" 








B' 






"^ 


\, 


:*.B 
































--- 




10 


A* 








A'--. 


,a^ 


~,^ 
















_« 








_/ 






B 








B 








_ 


B 







n 








































I 






4 


z 



Displacement 
Fig. 15. 

Diagrams Illustrating Location of Clearance Line from Expansion or Compression Lines 

of Known Laws. 



It 



WORK AND POWER 39 

By the second general method any two points, A and B, on an expansion or 
compression curve. Figs. 13, 14, 15, may be selected and horizontals drawn to 
the vertical line indicating the beginning of the stroke. The points A' and B' 
are distant from the milocated axis an amount A'A"^B'B'\ representing the 
clearance. 

Let the clearance volume ^Cl,\ 
" the displacement up to A = Da; 
the displacement up to B^D^; 
the whole displacement =i); 
" « be the exponent in PF*= constant, which defines the law of the curve. 

Then in general, 
But 
and 
hence 

or 

/ 1 l\ 1 1 

whence the clearance in whatever units the displacement may be measured will be 



1 1 



1 . 1 ' 
PJ -Pbf 
or 



„ ^••^-(r)'^-] '^-iW"- 



1 1 

1 



m-^\ iw- 



P^ 
and CI, in per cent of the whole displacement will be 

Db_/PayDa 
Clearance as a fraction of displacement = c = ^^—^ . 

When fi= 1 this takes the form 

Db_(Pa\Da 

Clearance in fraction of displacement =c= ^^—^ . . . (15) 



m- 



40 ENGINEERING THERMODYNAMICS 

« 

To use such an expression it is only necessary to measure off the atmospheric 
pressure below the atmospheric line, draw verticals at ends of the diagram 
and use the length of the horizontals and verticals to the points in the formula, 
each horizontal representing one D and each vertical a P. 

Graphic methods for the location of the axis of pressures, and hence the 
clearance, depend on the properties of the curves as derived from analytics. 
For example, when «=1, 

PaVa^PhVh, 

which is the equation. of the equilateral hyperbola, a fact that gives a common 
name to the law, i.e., hyperbolic expansion or hyperbolic compression. Two 
common characteristics of this curve may be used either separately or together, 
the proof of which need not be given here, first that the diagonal of the rectangle 
having two opposite comers on the curve when drawn through the other two 
comers will pass through the origin of coordinates, and second, that the other 
diagonal drawn through two points of the curve and extended to intersect the 
axes of coordinates will have equal intercepts between each point and the 
nearest axis cut. 

According to the first principle, lay off. Fig. 16, the vacuum line or axis 
of volume XY and selecting any two points A and B, construct the rectangle 
ACBD. Draw the diagonal CDS and erect at E the axis of pressures EZ, 
then will EZ and EY be the axes of coordinates. According to the second 
principle, proceed as before to locate the axes of volumes XY and select two 
points, A and B, Fig. 17. Draw a straight line through these points, which 
represents the other diagonal of the rectangle ACBD, producing it to inter- 
sect XY Bit M and lay off AN—BM. Then will the vertical NE be the axis 
of pressures. It should be noted that these two graphic methods apply only 
when 8=^1; other methods must be used when s is not equal to 1. 

A method of finding the axis of zero volume is based upon the slope of the 
exponential curve, 

pr=c. 

Differentiation with respect to V gives 

or 

whence 

Ps 



y= 



{-%) 



-(-S) (-) 



WORK AND POWER 



41 



In other words, the true volume at any given point on the known curve 
may be found by dividing the product of P and 8 by the tangent or the slope 
of the line at the given point, ^ith the sign changed. This method gives 
results dependent for their accuracy upon the determination of the tangent to 
the curve, which is sometimes difficult. 



70 
00 


Z 




























. 


































\ 






• 






















|40 

ao 
ao 




\ 






























\ 


A 




c 






















v 


/ 








V 


/ 


























^ 


/ 




B 












Atmospheric 


Line 












10 
X 9 




/ 






'^ 




. 


















/ 


r 
















" 




• 




— 


Y 






4 




i 


3 


i 


5 


< 


' 


1 


I) 


11 


15 



Displaoement 
Fig. 16. 



TO 


z 






























Itl 
































V 


V- 














c 
















1 


\s 








^y 


y' 












• 




St\ 




l\ 


V 


X 


\.' 


y 


0^ 


















«XI 






\ 


>^ 


^ 


X 


<w 


• 






















,^ 


"^s 






\ 








A 


tmoBi 


>beric 


Line 




10 

X 


y 


x\y 


^ 






^^ — . 







B 














L^sT- 








» wm^im MM 








•s"^ 










Y 






1 


L 


1 


\ 


{ 


\ 


i 


} 




) 


1 


1 


1 


3 



Displaoement 
Fig. 17. 

Graphic Methods of Locating the Clearance Line for Logarithmic or Hyperbolic Expansion 

and Compression Curves. 



The following graphical solution is dependent upon the principle just given, 
and while not mathematically exact, gives results so near correct that the 
error is not easily measured. The curve ACBj Fig. 18, is first known experi- 
mentally or otherwise and therefore the value of s, and the axis FV from 
which pressures are measured is located. Assume that the axis of zero volume, 



42 



ENGINEERING THERMODYNAMICS 



KP, is not known but must be found. Selecting any two convenient points, 
A and B, on the curve, complete the rectangle AHBG with sides parallel and 
perpendicular to FV. The diagonal HG cuts the curve at C and the horizontal 
axis at E. From C drop the perpendicular CD. If now the distance DE 
be multiplied by the exponent s, and laid oflF DK, and the vertical KP erected, 
this may be taken as the zero volume axis. 

It cannot be too strongly stated that methods for the finding of clearance 
or the location of the axes of pressures from the indicator card, much as they 
have been used in practice, are inaccurate and practically useless unless it is 
positively known beforehand just what value s has, since the assumed value 




Fig. 18. — Graphia Method of Locating the Clearance Line for Exponential Expansion and 

Compression Curves. 

of 8 enters into the work, and s for the actual diagram, as already explained, 
is affected by the substance, leakage, by moisture or wetness of vapor and by 
all heat interchange or exchange between the gas of vapor and its container. 



Prob. 1. If in card No. 6, Section 8, compression follows the law PF* «=A', 
where s = 1.4 and the barometer reads 29.9 ins. of mercury, locate the axes algebraically 

and graphically. 

Prob. 2. If in card No. 3, Section 8, expansion follows the law PF» =A', 
where s «1.37 and the barometer reads 29.8 ins. of mercury, locate the axes algebraically 
and graphically. 

Prob. 3. If in card No. 5, Section 8, expansion follows the law P7* =A', 
where s = l and the barometer reads 27.5 ins. of mercury, locate axes algebraically and 
graphically. 



WORK AND POWER 



43 



13. Measurement of Areas of PV Diagrams and Indicator Cards. Areas 
of pressure volume diagrams or indicator cards must be evaluated for the 
determination of- work or mean effective pressure, except when calculation by 
formula and hypothesis is possible. There are two general methods applicable 
to both the indicator card and PV diagram, that of average heights, and the 
planimeter measiu'e, besides a third approximate but very useful method, 
especially applicable to plotted curves on cross-section paper. 

The third method assumes that the diagram may be divided into strips of 
equal width as in Fig. 19, which is very easily done if the diagram is plotted on 
cross-section paper. At the end of each strip, a line is drawn perpendicular to 
the axis of the strip, such that the area intercepted inside the figure is apparently 
equal to that outside the figure. If this line is correctly located, the area of the 
rectangular strip will equal the area of the strip bounded by the irregular lines. 




Volumee 
FiQ. 19. — Approximate Method of Evaluating Areas and Mean Effective Pressures of 

Indicator Cards and P.V. Diagrams. 



If the entire figmre has irregular ends it may be necessary to subdivide one or 
both ends into strips in the other direction, as is done at the left-hand side of 
Fig. 19. The area of the entire figure will be equal to the summation of lengths 
of all such strips, multiplied by the common width. This total length may be 
obtained by marking on the edge of a strip of paper the successive lengths in 
such a way that the total length of the strip of paper when measured will be the 
total length of the strips. 

The mean height will be the total length of such strips divided by the num- 
ber of strip-widths in the length of the diagram. By a little practice the proper 
location of the ends of the strips can be made with reasonable accuracy, and 
consequently the results of this method will be very nearly correct if care is 

exercised. 

By dividing the diagram into equal parts, usually ten, and finding the length 
of the middle of each strip, an approximation to the mean height of each strip 



44 



ENGINEERING THERMODYNAMICS 



will be obt&in^d; these added together and divided by the number will give 
the mean height in inches from which the mean efiFective pressure may be found 
by multiplying by the scale as above, or the area in square inches by multiply- 
ing by the tength in inches, which can be converted into work by multiplying 
by the foot-pounds per square inch of area as fixed by the scales. As the pres- 
sures usually vary most, near the ends of the diagram a closer approximation 
can be made by subdividing the end strips, as is done in Fig. 20, which repre- 
sents two steam engine indicator cards taken from opposite ends of the same 
cylinder and superimposed. The two diagrams are divided into ten equal 
spaces and then each end space is subdivided. The mean heights of the sub- 
divisions are measured and averaged to get the mean height of the whole end 
division, or average pressure in this case for the division. The average heights 
of divisions for diagram No. 1 are set down in a column on the left, while those 




30 40 50 00 70 

Displacement In Per Cent of Stroke 

Fig. 20. — Simpaon's Method for Finding Mean Effective Pressure of Indicator Cards. 



for No. 2 are ma the right; the sum of each column divided by ten and multiplied 
by the sprinf* scale gives the whole m.e.p. The heights of No. 1 in inches 
marked off eMtinuously on a slip of paper measured a total of 11.16 ins. and 
for No. 2, 10.79 ins., these quantities divided by 10 (number of strips) and multi- 
plied by the spring scale, 50, gives the m.e.p., as before. This method is 
often designated as Simpson's rule. 

The best and most commonly used method of area evaluation, whether for 
work or m.e.p. determination, is the planimeter, a well-known instrument 
specially designed for direct measurements of area. 

14. Indicated Horse-power. Work done by the fluid in a cylinder, because it 
is most often determined by indicator card measurements, measures the indicated 
horse-power, but the term is also applicable to work that would be done by the 
execution of a certain cycle of pressure volume changes carried out at a specified 
rate. The mean effective pressure in pounds per square inch, whether of an 
indicator card or PV cycle, when multiplied by piston area in square inches, 



WORK AND POWER 45 

gives the average force acting on the piston for one stroke, whether the cycle 
required one, two or x strokes for its execution, and this mean force multiplied 
by the stroke in feet gives the foot-pounds of work done by the cycle. Therefore, 

Let in.e.p.=mean eflfective pressure in pounds per square inch for the cycle 

referred to one stroke; 

" a = eflfective area of piston in square inches; 

L = length of stroke in feet; 

n = number of equal cycles completed per minute; 

JV = number of revolutions per minute; 

iS=mean piston speed = 2L AT feet per minute; 

iV 
2= number of revolutions to complete one cycle = — . Then will the 

indicated horse-power be given by, 

T TT p _ (m.e.p.)Lan , ' 

i.n.r.- gg^QQ W 






__ (m.e .p.)LaAr ... 

33000z~" ^ ^ 

_ (m.e.p.)aS . . 

""33000X22 ^^^ 



(17) 



When there are many working chambers, whether in opposite ends of the 
same C3'linder or in separate cylinders, the indicated horse-power of each should 
be found and the sum taken for that of the machine. This is important not 
only because the eflfective areas are often unequal, as, for example, in opposite 
ends of a double-acting cylinder with a piston rod passing through one side 
only or with two piston rods or one piston rod and one tail rod of unequal 
diameters, but also because unequal valve settings which are most common 
will cause diflferent pressure volume changes in the various chambers. 

It is frequently useful to find the horse-power per pound mean effective pressfiire, 
which may be symbolized by Ke, and its value given by 

j^ _ Lan _ LaN 
^"33000" 330007 

Using this constant, which may be tabulated for various values of n, stroke 
and bore, the indicated horse-power is given by two factors, one involving 
cylinder dimensions and cyclic speed or machine characteristics, and the other 
the resultant PV characteristic, of the fluid, symbolically, 

I.H.P. = iiLc(m.e.p.). 

These tables of horse-power per pound m.e.p. are usually based on piston speed 
rather than rate of completion of cycles and are, therefore, directly applicable 



46 ENGINEERING THERMODYNAMICS 

when z=i or n=2iV, which means that the two cycles are completed in one 
revolution, in which case, 

S=2Li\r=Ln, 
and 

" 33000' 
whence 

LH.P.=ir,(m.e.p.)=-^5||^?^ (18) 

Table XI at the end of this chapter gives values of (H.P. per lb. m.e.p.) or 
Ke for tabulated diameters of piston in inches and piston speeds in feet per 
minute. Tables are frequently given for what is called the engine constant, 
which is variously defined as either 

(a) ob?)7)Q> which must be multiplied by m.e.p. Xn to obtain H.P., or 
(6) QQQQQ ? which must be multiplied by m.e.p. XL Xn to obtain H.P. 

For an engine which completes two cycles per revolution, this is the same as 
multiplying by m.e.p. XS. Before using such a table of engine constants it 
must be known whether it is computed as in (a) or in (6). 

Example. A 9 in. Xl2 in. double-acting steam engine runs at 250 R.P.M. and the 
mean effective pressm-e is 30 lbs. What is H.P. per pound m.e.p. and the I.H.P.? 

j^ Lan 1X63.6X500 



33000 33000 

I.H.P. = .9636 X 30 =28.908. 

Prob. 1- A pump has a piston speed of 250 ft. per minute; piston diameter is 24 ins. 
What is the H.P. per pound mean effective pressure? 

. Prob. 2. A simple single-acting 2-cylinder engine has a piston 10 ins. in diameter with 
a 2-in. rod and a stroke of 15 ins. It runs with a mean effective pressure of 45 lbs. 
per square inch at a speed of 220 R.P.M. What is the H.P.? 

Prqb. 3. A gas engine has one working stroke in every four. If the speed is 300 
R.P.M. what must be the m.e.p. to give 6 H.P. when the cylinder has a diameter of 6 ins. 
and a stroke of 12 ins.? 

Prob. 4. An air compressor is found to have a mean effective pressure of 50 lbs. If 
the cylinder is double acting and is 12 ins. diameter and 16 ins. stroke, what H.P. will 
be needed to drive it at 80 R.P.M.? 

Prob. 5. A gasoline engine has an engine constant (a) of .3. What must be the 
m.e.p. to give 25 H.P.? 

Prob. 6. A blowing engine has an m.e.p. of 10 lbs. Its horse-power is 500. What 
is the H.P. per pound m.e.p.? 



WORK AND POWER 47 

Prob. 7. Two engmes of the same size and speed are so run that one gives twice the 
power of the other. How will the engine constants and m.e.p. vary? 

Prob. 8. From the diagrams following Section 9 what must have been the H.P. per 
pound m.e.p. to give 300 H.P. in each case? 

Prob. 9. How will the H.P. per pound m.e.p. vary in two engines if the speed of 
one is twice that of the other, if the stroke is twice, if the diameter of piston is twice? 

16. Effective Horse-power, Brake Horse-power, Friction Horse-power, 
Mechanical Efficiency, Efficiency of Transmission, Thermal Efficiency. Work 
is done and power developed primarily in the power cylinder of engines, and is 
transmitted through the mechanism with friction loss to some point at which 
it is utilized. There is frequently a whole train of transmission which may 
involve transformation of the energy into other fonns, but always with some 
losses, including the mechanical friction. For example, a steam cylinder may 
drive the engine mechanism which in turn drives a dynamo, which transforms 
mechanical into electrical energy and this is transmitted to a distance over 
wires and used in motors to hoLst a cage in a mine or to drive electric cars. 
There is mechanical work done at the end of the system and at a certain rate, 
so that there will be a certain useful or effective horse-power output for the 
system, which may be compared to the horse-power primarily developed in 
the power cylinders. A similar comparison may be made between the primary 
power or input and the power left after deducting losses to any intermediate 
point in the system. For example, the electrical energy per minute delivered 
to the motor, or motor input, is, of course, the output of the transmission line. 
Again, the electrical energy delivered to the line, or electrical transmission 
inpuij is the same as dj'^namo output, and mechanical energy delivered to the 
dynamo Ls identical with engine output. The comparison of these measure- 
ments of power usually takes one of two forms, and frequently both; first, 
a comparison by diflfcrences, and second, a comparison by ratios. The ratio 
of any horse-power measurement in the system to the I.H.P. of the power cylin- 
der is the e^Hciency of the power system up to that point, the difference between 
the two is the horse-power loss up to that point. It should be noted that, 
as both the dynamo and motor transform energy from mechanical to electrical 
or vice versa, the engine mechanism transmits mechanical energy and the 
wires electrical energy, the system is made up of parts which have the 
function of (a) transmission without change of form, and (Jb) transformation 
of form. The ratio of output to input is always an efficiency ^ so that the efficiency 
of the power system is the product of all the efficiencies of transformation and of 
transfer or transmission, and the power loss of the system is the sum of trans- 
fonnation and transmission losses. Some of these eflSeiencies and losses have 
received names which are generally accepted and the meaning of which is gen- 
erally understood by all, but it is equally important to note that others have no 
names, simply because there are not names enough to go around. In dealing 
with eflSciencies and power losses that have accepted names these names may 
with reason be used, but in other cases where names are differently under- 
stood in different places or where there is no name, accurate description must be 



48 ENGINEERING THERMODYNAMICS 

relied on. As a matter of fact controversy should be avoided by definition 
of the quantity considered, whether descriptive names be used or not. 

Effective horse-power is a general term applied to the output of a machine, 
or power system, determined by the form of energy output. Thus, for an engine 
it is the power that might be absorbed by a friction brake applied to the shaft, 
and in this case is universally called Brake Horse-power. The difference between 
brake and indicated horse-power of engines is the friHion horse-power of the 
engine and the ratio of brake to indicated horse-power is the mechanical efficiency 
of the engine. For an engine, then, the effective horse-power or useful horse- 
power is the brake horse-power. When the power cylinders drive in one machine 
a pump or an air compressor, the friction horse-power of the machine is the 
difference between the indicated horse-power of the power cylinders and that 
for the pump or compressor cylinders, and the mechanical eflBciency is the 
ratio of pump or air cylinder indicated porse-power to indicated horse-power 
of the power cylinders. Whether the indicated horse-power of the air or pump 
cylinders can be considered a measure of useful output or not is a matter of 
difference of opinion. From one point of view the machine may be as considered 
built for doing work on wat^r or on air, in which case these horse-powers may 
properly be considered as useful output. On the other hand, the power pump 
is more often considered as a machine for moving water, in which case the 
useful work is the product of the weight of water moved into its head in feet, 
and includes all friction through ports, passages and perhaps even in pipes or 
conduits, which the indicated horse-power of the pump cylinder does not include, 
especially when leakage or other causes combine to make the pump cylinder 
displacement differ from the volume of water actually moved. With compressors 
the situation is still more complicated, as the air compressor may be considered 
useful only when its discharged compressed air has performed work in a rock 
drill, hoist or other form of an engine, in which case all sorts of measm-es of use- 
ful output of the compressor may be devised, even, for example, as the purely 
hypothetically possible work derivable from the subsequent admission and 
complete expansion of tha compressed air in a separate air engine cylinder. 

Too accurate a definition, then, of oviput and inpvt energy in machines and 
power systems is not possible for avoidance of misunderstanding, which may 
affect questions both of power losses and efficiency of transmission and trans- 
formation whether in a power system or single machine. It is interesting to 
note here that not only is the indic/ited work of the power cylinder always con- 
sidered the measure of power input for the system or machine, but, as in the 
other cases, it is itself an output or result of the action of heat on the vapor or 
gas and of the cycle of operations carried out. The ratio of the indicated power 
or cylinder work, to the heat energy both in foot-pound units, that was expended 
on the fluid is the thermal efficiency of the engine referred to indicated horse- 
power or the efficiency of heat transformation into work, the analysis of which 
forms the bulk of the subject matter of Chapter VI. Similarly, the ratio of 
any power measurement in the system to the equivalent of the heat supplied 
is the thermal efficiency of so much of the system as is included. 



WORK AND POWER 49 

Example.. It has been found that when the mdicated horse-power of an engine is 
250, a generator is giving out 700 amperes at 220 volts. At end of a transmis ion line is a 
motor using the output of the generator. This motor on test gave out 180 brake horse- 
power. Assuming no losses in the transmission line, what was the efficiency of the 
motor, of the generator, of the engine, and of the system? 

Motor efficieacy-^-^-87^ 

746 

Note: Volts Xamperes= watts, and, watts -^ 746 =H.P. 

220x700 
Engine and generator efficiency = — ^j- — a82.4%. 

250 

180 
Efficiency of system = — • =72% or 82.4 x87.2 -72%. 

iuOU 

Prob. 1. An engine is belted to a pmnp; the I.H.P. of the engine is 50, of the pump 
iO, and the pump delivers 1200 gallons water per minute against 100-ft. head. What is 
the efficiency of each part and of the entire system. 

Prob. 2. An engine is geared to air compressor. Upon test, the efficiency of the engine 
alone, glaring alone and compressor alone were each 80 per cent. When the com- 
pressor H.P. was 100 what was that of the engine? 

Prob. 3* A water-wheel is run by the discharge from a pump. The B.H.P. of wheel 
is foimd to be 20 when the pump is delivering 45 gallons of water per minute at a 
head of 1000 lbs. per square inch. The water I.H.P. of the pump is 30 and the 
steam I.H.P. is 40. What are the efficiencies of each part of the system and the over-all 
efficiency? 

Prob. i. Perry ^ves a rule for the brake horse-power of steam engines as being 
equal to .95 I.H.P. — 10. On this basis find the mechanical efficiency of a 500-H.P. 
engine from 200 to 500 H.P. Show results by a curve with B.H.P. and per cent efficiency 
as coordinates. 

Prob. 6- Perry gives a rule for the efficiency of an hydraulic line bsH = ,7I —25 where 
// is the useful power of the pump and / is the indicated. Find / for values of H from 
100 to 300 and plot a curve of results. 

Prob. 6. An engine gives one I.H.P. for every 3 lbs. of coal per hour. One pound 
of coal contains 9,500,000 ft.-lbs. of energy. What is the thermal efficiency? 

Prob. 7. A gas engine has a mechanical efficiency of 70 per cent when delivering 
power to a generator which in turn has an efficiency of 90 per cent. If the engine uses 
15 cu.ft. of gas per indicated horse-power hour and the gas contains 700,000 ft.-lbs. per 
cubic foot, what is the net thermal efficiency of the system? 

16. Specific Displacement, Quantity of Fluid per Hour or per Minute per 
IJ5.P. It has been shown that the work done in cylinders by pressure volume 
changes of the vapor or gas depends on the mean effective pressure and on the 
displacement, or that there is a relation between I.H.P. and displacement. 
The quantity of fluid used also depends on the displacement and may be expressed 



50 ENGINEERING THERMODYNAMICS 

in cubic feet per minute at either the low pressure or high pressure con- 
dition when the work is done between two definite pressure limits, or in terms 
of pounds per minute or hour, which involves the application of fluid densities 
to volumes and which eliminates the double expression for the two conditions 
of pressure. The displacement per hour per horse-power, termed the specific 
displacement, is the basis of computations on the steam consumption of steam 
engines, the horse-power per cubic feet of free air per minute for air compressors, 
the horse-power per ton refrigeration for refrigerating machines and the con- 
sumption of fuel per hour per horse-power for gas and oil engines. It is, 
therefore, a quantity of great importance in view of these applications. Apply- 
ing the s3anbols already defined to displacement in one direction of one side of 
a piston 

Displacement in cu.ft. per stroke =LXyt7J 
Displacement in cu.ft. per minute ^LXYrrXiV; 

Displacement in cu.ft. per hour =60LXYjTXiV. 

T J- X J 1 (m.e.p.)Lan (m.e.p.)LaN 
Indicated horsepower 33^^^^^ SSOOOT"- 

Whence expressing displacement per hour per I.H.P. or specific displacement 
in one direction for one side of a piston by D«, 

^ ^ 60LXj-jjXiV ^ gQ^33QQQ^ ^ 1 3750g ^ - .^g. 

* (p^-e .pQLaJV 144(m.e.p.) (m.e.p.) 

~3300(te 

From Eq. (19) it appears that (he specific displacement is equal to zy,13,750 
divided by the mean effective pressure in pounds per square inch. 

If two points, A and B, be so located on the indicator card. Fig. 21, as to 
have included between them a fluid transfer phase, either admission to, or 
expulsion from the cylinder, then calling 8a = pounds per cubic foot or density 
at point A J and 8^= pounds per cubic foot or density at point B, the weight 
of fluid present at A is, 

{Da+Cl)da lbs., 

and weight of fluid present at B is 

{Db+Cl)db lbs., 

whence the weight that has changed places or passed in and out per stroke is, 

{Db+Ct)8b-{Da+Cr)da lbs. per stroke. 



WORK AND POWER 



51 



If both A and B lie on the same horizontal as A and B', ^a = ^6=^, the 
density of fluid at the pressure of measurement, whence the weight of fluid 
used per stroke, will be 

and the voliune per stroke used at density d is 

Db' — Da cu.ft., 
which compared to the displacement is 

Dt'-Da 

D ' 



















P 


















1 






\ 










1 






\ 


\ 








\ 






> 


\ 








\ 






M 


\ 








\ 


A 


4Hl»Mi^< 







>a^ ^^B^fe 


B' 


-^ 


\ 


\i- 










v^ 


^ 




\ 


s 










-1 




\ 













































Fig. 21. — Determination of Consumption of Fluid per Hour per Indicated Horse-power from 

the Indicator Card. 

This is the fraction of the displacement representing the volume of fluid pass- 
ing through the machine at the selected pressiu*e. Multiplying the specific 
displacement by this, there results, 

Cu.ft. of fluid per hr. at density (d) per I.H.P. = ^ ^ — ^^ — -y 



(m.e.p.) D 



and 



Lb6.offluidperhr.perLH.P=7-^^^f^^^^V (20) 

(m.e.p.) \ D / ^ ^ 

More generally, that is, when A and B are not taken at the same pressures 

Lbs. fluid per hr. per IM.F.-- J^^JJ(Db+Cl)^J,'^-(Da+Cl)^a^ . (21) 

The particular forms which this may take when applied to special cases will be 
examined in the succeeding chapters. 



62 ENGINEERING THERMODYNAMICS 

Example. An air compressor whose cylinder is 18x24 ins. (18 ins. in diameter and 
stroke 2 ft.) runs at 60 R.P.M. and is double acting. The mean effective air pressure 
is 50 lbs. per square inch. What is the specific displacement? 

Cu.ft. per hour =^^^ = 60 X2 X^^4 Xl20 =25,600. 

144 144 

TTTP _m.e .p.Lan _50 x2x254.5xl20 _^oo 
^•^•^- "^^OOO 2Sm ^^''^'' 



Cu.f t. per hour ^ 25,600 ^gy- 
I.H.P. 92.3 ' 



or by the formula directly, 



13,750 13,750 ^^ 
m.e.p. 50 

Prob. 1. What will be the cubic feet of free air per hour per horse-power delivered 
by a 56x72-in. blowing engine with 4 per cent clearance and mean effective pressure of 
10 lbs. per square inch? 

Prob. 2. An 18x22-in. ammonia compressor works with a mean effective pressure 
of 45 lbs. per square inch. What is the weight of NHt per I.H.P. hour if the speed is 
50 R.P.M. and compressor is double acting having a volumetric efficiency of 90 per 
• cent? Use tabular NHz densities. 

Prob. 3. A steam engine whose cylinder is 9 X 12 ins. runs at a speed of 300 R.P.M. 
and is double acting. If the m.e.p. is 60 lbs. and the density of steam at end of the 
stroke is .03, how many pounds of steam are used per hour per indicated horse-power? 

17. Velocity Due to Free Expansion by PV Method. All the cases examined 
ior the work done by P7 cycles have been so far applied only to their execution 
in cylinders, but the work may be developed in nozzles accelerating the gas 
or vapor in free expansion, giving, as a consequence, a high velocity to the fluid. 
It was noted that for cylinders many combinations of phases might be found 
worthy of consideration as, typical of possible actual conditions of practice, 
but this is not true of proper nozzle expansion, which has but one cycle, that of 
Fig. 22. That this is the cycle in question is seen from the following considera- 
tions. Consider a definite quantity of the gas or vapor approaching the nozzle 
from a source of supply which is capable of maintaining the pressure. It pushes 
forward that in front of it and work will be done, ABCD^ equal to the admission 
of the same substance to a cylinder, so that its approach AB may be considered 
as a constant- pressure, volume- increasing phase for which the energy comes 
from the source of supply. This same substance expanding to the lower pres- 
sure will do the work CBEF; but there will be negative work equivalent to the 
pushing awary or displacing of an equivalent quantity of fluid at the low pres- 
sure, or FEGDj making the work cycle ABEG, in which AG is the excess of 



WORK AND POWER 



53 



initial over back pressure or the effective working pressure, remaining constant 
during approach and lessening regularly during expansion to zero excess at E. 
The work done will be from Eq. (13), 



w=p.v.+i^[i-{^yyp.v.. 



A 


B 






















» 














■ 










\ 






■ 
















\ 






















\ 


L 














* 








\ 






















\ 


\ 






















\ 


X, 












G 










^v 










. 
























F 





Fig. 22. — Pressure-Volume Diagram for Nozzle Expansion Measuring the Acceleration 

Velocity and Horse-power of Jets. 



But 



PeVe' = PeVeVe'-^=PbVb' = PbVbVb»-''; 



8-1 



.\ PeVe = PbVb 



m"-^^^w-- 



8-1 



8-1 



Whence 



r-..K..S[.-(ft)-]-P.K.(';=) 



a-l 



-.-^^"-'[•-(ft) ■ ] <'^' 



54 ENGINEERING THERMODYNAMICS 

Aissuming the initial velocity to be zero, and the work of Eq. (22) to be done 
on 1 lb., the final or resultant velocity will be according to Ekj. (6), T"^ Y - 



u=V2gW 



• ••••• V^^/ 



or 



This velocity is in feet per second when pressures are in pounds per square foot 
and volumes in cubic feet of 1 lb. of substance, and is known as Zeuner's 
equation for the velocity of a gas or vapor expanding in a nozzle. It is 
generally assiuned that such expansion, involving as it does very rapid motion 
of the fluid past the nozzle, is of the adiabatic sort, as there seems to be no 
time for heat exchange between fluid and walls. As already noted, the value 
of s for adiabatic expansion of vapors is not constant, making the correct 
solution of problems on vapor flow through orifices practically impossible by 
this method of pressure volume analysis, but as will be seen later the thermal 
method of solution is exact and comparatively easy. 

Note. A comparison of Eqs. (22) and (13) and the figures corresponding 
will show that the area under the process curve, which is the same as the work 
done during the compression or expansion,- if multiplied by 8 will equal the area 
to the left of the process curve, which in turn represents, as in Fig. 21, for 
engines, the algebraic sum of admission, complete expansion, and exhaust work 
areas, or as in Fig. 9 for compressors, the algebraic sum of suction, compression 
and delivery work areas. This statement must not be thought to refer to the 
work area of such a cycle as that of Fig. 8, where expansion is incomplete, nor a 
case of over-expansion, Fig. 10. 

Example. In Fig. 22 assume the initial pressure at 100 lbs. per square inch absolute, 
back pressure at atmosphere, and expansion as being adiabatic! What will be the 
work per pound of steam and the velocity of the jet, if F^ is 4.36 cu.ft. and s = 1.3 for 
superheated steam? 



.1|X144X100X4.36[1-Q"] 

=27,206 X. 608 = 16,541 ft.-lbs.; 
tt = V^=8.02Vl6,541; 
= 1028 ft. per second. 



WORK AND POWER 55 

Prob. 1. Taking the same pressure range as above, find W and u for adiabatic expan- 
sion of air, also for isothermal expansion. 

Prob. 2. How large must the effective opening of the suction valve be, in an air 
compressor 18x24 ins. to aUow the cylinder to properly fill if the mean pressure-drop 
through the valve is 1 lb. per square inch and the compressor runs at 80 R.P.M.? 

Prob. 3. What must be the diameter of the inlet valve in a gas engine to fill a 
cylinder of i cu.ft. capacity if the lift of the valve is | ins., allowing a pressure drop of 
1 lb. per square inch? Engine makes 150 working strokes per minute. 

Prob. i. It has been found from experiment that the velocity of air issuing from a hole 
in plate orifice is 72 per cent of what would be expected from calculation as above when 
the absolute pressure ratio is 2 to 1, and 65 per cent when the absolute pressure ratio 
is 1} to 1. What will be the actual velocity for air flowing from a tank to atmosphere 
for these pressure ratios? 

Prob. 5. COs stored in a tank is allowed to escape through an orifice into the air. 
What will be the maximum velocity of the jet if the pressure on the tank be 100 lbs. 
per square inch gage? ' . 

Note: 1 lb. CO* at pressure of 100 lbs. per square inch gage occupies 1.15 cu.ft. 

Prob. 6. If ammonia gas and hydrogen were allowed to expand from the same pres- 
sure, how would their maximum velocities compare? Vol. of 1 lb. of NH| at 50 lbs. per 
square inch gage is 4.5 cu.ft. Vol. of 1 lb. of H at same pressure is 77.5 cu.ft. 

18. Weight of Flow through Nozzles. Applying an area factor to the 
velocity equation will give an expression for cubic flow per second which 
becomes weight per second by introducing the factor, density. 

Let the area of an orifice at the point of maximum velocity, u,he A sq.ft., 
then will the cubic feet per second efflux be Au, Assume the point of maxi- 
mum velocity, having area A, to be that part of the nozzle where the pressure 
has fallen to P«, Fig. 22, and the gas or vapor to have the density 80 pounds 
per cubic fcx)t. Then will the nozzle flow in pounds per second be 

But the weight per cubic foot is the reciprocal of the cubic feet per pound, 
Fe, which it has already been assumed, is the final volume, of one pound of the 
fluid. Hence, 

ilA 



This may be put in terms of initial gas or vapor conditions for, 

1 
/PA 

Whence 



-<w- 



1 
uA uA/Pe\» 



(f) 



66 ENGINEERING THERMODYNAMICS 

Substituting in this the value of u from Eq. (24), 

-»"^.(ft)"'"{;^i''''''['-(K)~]r"»'»'"--- « 

This weight will be a maximum for a certain value of the pressure ratio, depend- 
ing on the value of 8 only, and this value can be found by placing the first dif- 
ferential coeflScient of w with respect to f p^j equal to 



Kft) 



=0. 



To accomplish this, rearrange Eq. (25) as follows: 



w=A 



^^Mm-mf- 



But as the other factors do not enter to effect the result so long as Pb does not 
varv. t& is a maximum when the bracket 



vary, t& is a maximum when the bracket 



2 « + i 



or 



is a maximum or when 

(rKr""-H-»- 

But as ( p^j* cannot be equal to zero in practice, then 
which gives the condition that w is a maximum when 



or 



(Pe\~r 8 + 1 
\PbJ ~_ 2 ' 



WORK AND POWER 57 

or mayitniim flow for given iiutial pressure occurs when 

i^H^r ,<»' 



Pe 



For air expanding adiabatically 8 '=1.407. Maximum flow occurs when 
= .528 and for most conunon values of s it will be between .50 and .60 



Pb 

This result is quite remarkable and is verified by experiment reasonably closely. 
It shows that, contrary to expeciaUori, the weight of effliix from nozzles mil not 
continuously and regularly increase with increasing differences in pressure, bid 
for a given initial pressure the weight discharged per second will have reached 
its limit when the final pressure has been diminished to a certain fraction of the 
im'tial, and any further decrease of the discharge pressure vrill not increase the 
flow tiirough an orifice of a given area. 

The subject of flow in nozzles will be treated more completely in 
Chapter VI. 

Prob. 1« For the following substances under adiabatic expansion determine the 
pressure ratio for maximum flow and find the rate of flow per square inch of orifice under 
this condition when flow is into a vacuum of 10 ins. of mercury with standard barometer: 

(a) Carbon dioxide. 

(b) Nitrogen. 

(c) Hydrogen. 

(d) Ammonia. 

(e) Dry steam according to saturation law. 

19. Horse-power of Nozzles and Jets. Although, strictly speaking, nozzles 
can have no horse-power, the term is applied to the nozzle containing the 
orifice through which flow occurs and in which a certain amount of work is 
done per minute in giving to a jet of gas or vapor initially at rest a certain 
final velocity, and amount of kinetic energy. The foot-pounds of work per 
pound of fluid multiplied by the pounds flowing per second will give the foot- 
pounds of work developed per second within the nozzle, and this divided by 
550 will give the horse-power developed by the jet, or the nozzle horse-power. 
Accordingly, 

1 
H-P-^^'«*=-556-^-560X7;y ^" •• • ^^^ 



W A/P 



550 F, 



w^ 



W . (6) 



-"<(&) •{i^''.''.['-(g)f]}*.<^ 



(27) 



58 ENGINEERING THERMODYNAMICS 

where the expression in the bracket is the work done per pound of substance. 
The pressures are expressed in pounds per square foot, areas in square feet 
and volumes in cubic feet. 

Etample. A steam turbine operates on wet steam at 100 lbs. per square inch abso- 
lute pressure which is expanded adiabatically to atmospheric pressure. What must be 
the area of the nozzles if the turbine is to develop 50 H.P. ideally? 



Note: 1 cu.ft. of steam at 100 lbs. = .23 lb. 



/ 2 \«-i 
By Eq. (26), maximum flow occurs when the pressure ratio is ("77) » or, for 

1.11 

this case when the pressure is 100^ (oTt) ~^^ ^^' ^^ square inch absolute. As 

the back pressure is one atmosphere, the flow will not be greater than for the above 
critical pressure. Substituting it in Eq. (25) will give the flow weight to, and using the 
actual back pressiu^ in Eq. (22) will give the work W. 

.11 

T. T. /ooxTTT /lll\ /14400\r, / 2116 \ 111] 
ByEq.(22),W.(— )x(-— )[l-(^^) J 

= 110000 ft.-lbs. per pound of steam. 

ByEq.(25),«;=8.02Ax.23x(.58)rnjl^X^^|l-(.58)rri^ P 

= 198A lbs. of steam per second. 
By Eq. (27a), 

Wxw_ 110000 X198A 



550 550 

Whence 

Prob. 1. What will be the horse-power per square inch of nozzle for a turbine using 
hot gases if expansion follows law PVs—kf when s = 1.37, the gases being at a pressure of 
200 lbs. per square inch absolute and expanding to atmosphere. 

Let the volume per pound at the high pressure be 2 cu.ft. 

Prob. 2. What will be the horse-power per square inch of nozzle for the problems 
of Section 17? 

Prob. 3. Suppose steam to expand according to law PFs= A:, where s = 1.111, from 
atmosphere to a pressure of 2 lbs. per sq. inch absolute. How will the area of the ori- 
fice compare with that of the example to give the same horse-power? 

Note: 76=-26.4. 

Prob. 4. Suppose steam to be superheated in the case of the ex£[mple and of the last 
problem, how will this affect the area of nozzle? 

Note: Let 7^ =5 and 32 respectively. 

Prob. 6. How much work is done per inch of orifice if initial pressure is 100 lbs. 
absolute on one side and final 10 lbs. absolute on other side of a valve through which 
air is escaping? 



WORK AND POWER 59 



GENERAL PROBLEMS ON CHAPTER I 

1. An air compressor is required to compress 500 cu.ft. of free air per minute to a 
pressure of 100 lbs. per square inch gage; the compressor is direct connected to a steam 
engine. The mechanical efficiency of the machine is 80 per cent. What will be the 
steam horse-power if compression is (a) isothermal; (6) adiabatic? 

2. A mine hoist weighing 10 tons is raised 4500 ft. in one minute. In the first 20 
seconds it is accelerated from rest to a speed of 6000 ft. per minute; during the next 30 
seconds speed is constant at this value, and during last 10 seconds it is brought to rest. 
What will be (a) work of acceleration for each period; (6) work of lift for each period; 
(c) total work supplied by engine; (d) horse-power during constant velocity period? 

3. The en^e driving the above hoist is driven by compressed air. If air is supplied at 
a pressure of 150 lbs. per square inch gage and is admitted for three-quarters of the 
stroke, then expanded adiabatically for the remainder of the stroke and exhausted to the 
atmosphere find (a) what must the piston displacement be to lift the hoist, the work of 
acceleration being neglected? (6) To what value could the air pressure be reduced if 
air were admitted full stroke? 

4. It is proposed to substitute an electric motor for the above engine, installing a 
water-power electric plant at a considerable distance. The type of wheel chosen is one in 
which a jet of water issuing from a nozzle strikes against a series of revolving buckets. 
The available head at the nozzle is 1000 ft. Assuming the efficiency of the motor to be 
85 per cent, transmission 80 per cent, generators 90 per cent, and water-wheels 60 per 
cent, what will be the cubic feet of water per minute? 

6. A steam turbine consists of a series of moving vanes upon which steam jets issuing 
from nozzles impinge. It is assumed that for best results the speed of the vanes should be 
half that of the jets. The steam expands from 100 lbs. per square inch gage to 5 lbs. 
per square inch absolute, (a) What must be the best speed of vanes for wet steam where 
8 = 1.111? (&) If 55 per cent of the work in steam is deUvered by the wheel what must 
be the area of nozzle per 100 H.P., and weight of steam per hour 100 H.P? 

Note: 1^6=3.82. 

6. It has been found that a trolley car uses a ciurent of 45 amperes at 550 volts when 
running 12 miles per hour. If the motor is 80 per cent efficient, what is tractive effort? 

Note : Volts X amperes = watts, and watts -i- .746 = H.P. 

7. An engine whose cylinder is 18x24 ins. (diameter = 18 ins., stroke =24 ins.) is 
double acting and runs at a speed of 125 R.P.M. Steam is admitted for one-quarter 
stroke at a pressure of 125 lbs. per square inch gage, allowed to expand for the rest of 
the stroke and then exhausted into a vacuum of 28.25 ins. of Hg. (a) Draw a PV 
diagram of the cycle and find the m.e.p. from the diagram and by calculation, and then 
find the horse-power. (6) Consider steam to be admitted one-half stroke without 
other change. How will the horse-power vary? (c) What will be the horse-power for 
one-quarter admission if the exhaust pressure is 15 lbs. per square inch absolute? (d) 
What will be the horse-power if the steam pressure be made 150 lbs. per square inch 
absolute, other conditions as in (a)? (e) Suppose all conditions as in (a) but the 
speed lowered to 75 R.P.M. What will be the horse-power? 

8. Assuming that 50 per cent of the work in the jet is transformed to useful work, 
what must be the total area of the nozzles of a steam turbine to develop the same horse- 



60 ENGINEERING THERMODYNAMICS 

power as the engine in problem (7a), the pressure range being the same and 8 being 1.3? 
76=3.18. 

9. Water is being pumped from the bottom of a shaft 700 ft. deep at the rate of 1000 
gallons per minute by an electrically driven pump. Efficiency of pump is 70 per cent, 
motor 90 per cent, transmission line 95 per cent, generator 85 per cent, and mechanical 
efficiency of engine 80 per cent. What will be the indicated horse-power of the engine? 
If the above installation were replaced with an air-driven pump of 05 per cent efficiency, 
efficiency of transmission being 100 per cent, and that of the compressor and engine 
80 per cent, what would be the horse-power of this engine? 

10. Show by a P7 diagram, assuming any convenient scales, that the quantity of air 
discharged byacompressor and the horse-power, both decrease asthe altitude increases, and 
that the horse-power per cubic foot of air delivered increases under the same condition. 

11. A centrifugal pump is driven by a steam engine directly connected to it. The 
pump is forcing 1000 gallons of water per minute against a head of 250 ft. and runs at a 
speed of 450 R.P.M. The engine is double acting and its stroke equals the diameter of 
the cylinder. Steam of 100 lbs. per square inch gage is admitted for half stroke, allowed 
to expand the rest of the stroke so that s = 1, and is then exhausted to atmosphere. What 
must be the size of the engine if the pump efficiency is 65 per cent and the engine 
efficiency 75 per cent? 

12. (a) What will be the pounds of steam used by this engine per hour per horse- 
power? (b) If the steam were admitted but one-quarter of the stroke and the initial 
pressure raised sufficiently to maintain the same horse-power, what would be the new 
initial pressure and the new value of the steam used per horse-power per hour? 

Note: Weight of steam per cubic foot for (a) is .261; for (6) is .365. 

13. If it were possible to procure a condenser for the above engine so that the exhaust 
pressure could be reduced to 2 lbs. per square inch absolute, (o) how much would the 
power be increased for each of the two initial pressures already given? (6) How would 
the steam consumption change? 

14. A motor-fire engine requires a tractive force of 1300 lbs. to drive it 30 miles per 
hour, its rated speed. The efficiency of engine and transmiasion is 80 per cent. When 
the same engine is used to actuate the pumps 70 per cent of its power is expended on 
the water. What wiU be the rating of the engine in gallons per minute when pumping 
against a pressure of 200 lbs. per square inch? 

16. A compressor when compressing air at sea level from atmosphere to 100 lbs. per 
square inch absolute, expends work on the air at the rate of 200 H.P., the air being com- 
pressed adiabatically. (a) How many cubic feet of free air are being taken into the 
compressor p6r minute and how many cubic feet of high pressure air discharged? 
Compressor is moved to altitude of 8000 ft. (6) What will be the horse-power if the same 
amount of air is taken in and how many cubic feet per minute will be discharged? 

(c) What will be the horse-power if the same number of cubic feet are discharged as in 
case (a) and what will be the number of cubic feet of low pressure air drawn in? 

(d) Should superheated ammonia be substituted for air at sea level, what would be the 
necessary horse-power? 

16. A car weighing 40 short tons is at rest and is struck by a train of four cars, each 
of the same weight as the first, (a) Upon impact the single car is coupled to the train and 
all move off at a certain velocity. If the original velocity of the train was 3 miles 
per hour, what will it be after attachment of tlie extra car? (6) If instead of coupling, 
the extra car after impact moved away from the train at twice the speed the train was 
then moving, what would be the speed of train? 



WOHK AND POWER 61 

17. To drag a block of stone along the ground requires a pull of 1000 lbs. If it be 
placed on rollers the pull will be reduced to 300 lbs., while if it be placed on a wagon with 
well-made wheels, the pull will be but 200 lbs. Show by diagram how the work required 
to move it 1000 ft. will vary. 

18. 100 cu. ft. of air, at atmospheric pressure, is compressed to six times its 
original pressure, (a) What will be the difference in horse-power to do this in 45 seconds 
isothermally and adiabaticafly at an elevation of 8000 ft. (6) What will be the final vol- 
umes? (c) What will be the difference in horse-power at sea level? {d) What will be the 
final volumes? 

19. An engine operating a hoLst is run by compressed air at 80 lbs. per square inch 
gage. The air is admitted half stroke, then expanded for the rest of the stroke so that 
8-1.3 and then exhausted to atmosphere. The engine must be {K>werful enough to lift 
a ton cage and bring it up to a speed of 2000 ft. per minute in 15 seconds. What will be' 
the necessary displacement per minute? 

20. CJonstnict P7 diagrams for Probs. 1, 11, TS and 15, showing by them that the 
work of admission, compression or expansion, and discharge or exhaust, is equal to that 
found algebraically. 

21. The elongation of wrought iron under a force F is equal to the force times the 
length of the piece divided by 25,000,000 times the cross-section of metal in the piece. 
A 4i in. pipe has 3.75 sq. ins. of metal; a line of this pipe 1 mile long is running full of 
water with a velocity of 600 ft. per minute. This is stopped in 1 second by closing a 
valve. Assuming pip>e did not burst, what would be the elongation? 

22. Two steam turbines having nozzles of equal throat areas are operating on a steam 
pressure of 150 lbs. per square inch gage. One is allowing steam to expand to atmosphere 
the other to 2 lbs. per square inch absolute, both cases having an exponent for expansion 
of 1.11. Find the relation of the horse-power in the two cases. 

23. The power from a hydro-electric plant is transmitted some distance and then 
used to drive motors of various sizes. At the time of greatest demand for current it has 
been found that 1000 horse-power is given out by the motors. Taking the average 
efficiency of the motors as 70 per cent, transmission efficiency as 85 per cent, generator 
efficiency as 85 per cent, and water-wheel efficiency as 70 per cent, how many cubic feet 
of water per second will the plant require if the fall is 80 ft.? 

24. A small engine used for hoisting work is run by compressed air. Air is admitted 
for three-quarters of the stroke and then allowed to expand for the rest of the stroke in 
such a way that 8 »1.4 and finally exhausted to atmosphere. For the first part of the 
hoisting, full pressure (80 lbs. per square inch gage) is applied, but after the load has been 
accelerated the pressure is reduced to 30 lbs. per square inch gage. If the engine has 
two cylinders each 9 Xl2 in., is double acting and runs at 200 R.P.M., what will be the 
(a) horse-power in each case, (6) the specific displacement? 

25. A steam engire has a cylinder 12x18 in., is double acting, and runs at 150 
R.P.M. (a) What is the engine constant, and (6) horse-power per pound m.e.p.? 

26. A water-power site has available at all times 3500 cu.ft. of water per min- 
ute at a 100-ft. fall. Turbines of 70 per cent efficiency are installed which take the place 
of two double-acting steam engines whose mechanical efficiencies were 85 per cent. 
The speed of the engines was 150 R.P.M. , m.e.p. 100 lbs. per square inch, and stroke was 
twice the diameter. What was the size of each engine? 

27. Assuming the frictional losses in a compressor to have been 15 per cent, how many 
cu.ft. of gas per minute could a compressor operated by the above engines compress 
from atmosphere to 80 lbs. per square inch gage if s = 1.35? 



62 



ENGINEERINa THERMODYNAMICS 



Table I 



CONVERSION TABLE OF UNITS OF DISTANCE 



Meters. 1 


Kilometers. 


Inches. Feet. 


SUtute Miles. 


Nautical Miles. 


1 

1000 
0.0254 
0.304801 

1609.35 

1853.27 


0.001 

1 

0.0000254 
0.0003048 
1.60935 
1.85327 


39.37 
39370.1 
1 

12 
63360 
72963.2 


3.28083 
3280.83 

0.083333 

1 
5280 
6080.27 


0.000621370 

0.62137 

0.0000157828 

0.000189394 

1. 

1 . 15157 


0.000539587 

0.639587 

0.0000137055 

0.000164466 

0.868382 

1. 



1 In aocordance with U. S. Standards (see Smithsonian Tables). 



Table II 



CONVERSION TABLE OF UNITS OF SURFACE 



Sq. Meters. 


Sq. Inches. 


Sq. Feet. 


Sq. Yards. 


Acres. 


Sq. Miles. 


1 

.000645 
.0929 
.8361 
4046.87 


1550.00 

1 

144 

1296 


10.76387 

.00694 

1 

9 
43560 
27878400 


1 . 19599 

.111 
1 
4840 
3097600 


.000247 

.000206 

1 
640 


001562 


2589999 




1 









Table III 



CONVERSION TABLE OF UNITS OF VOLUME 



Cu. Meters. 


Cu. Inches. 


Cu. Feet. 


Cu. Yards. 


Lities 
(1000 Cu. Cm.) 


GaUons (U.S.) 


1 


61023.4 

1 

1728 

46656 

61.023 

231 


35.3145 
.000578 
1 

27 

.035314 
.13368 


1.3079 


1000 

.016387 
28.317 


264.170 
00433 


• .028317 
.76456 


.03704 
1 
.001308 
.004951 


7.4805 
201.974 


.001 
.003785 


1 
3.7854 


.26417 
1 



TABLES 



63 



Table IV. 



CONVERSION TABLE OF UNITS OF WEIGHT AND FORCE 



1 

Kilocrammea. Metric Tons. 


Pounds. 


U. S* or Short Tons. 


British or Long Tons. 


1. 

1000. 

0.453593 
907.186 
1016.05 


0.001 

1. 

0.000453593 

0.907186 

1.01605 


2.20462 
2204.62 

1. 
2000. 
2240. 


0.00110231 
1 . 10231 
0.0005 
1. 
1.12000 


0.000984205 

0.984205 

0.000446429 

0.892957 

1. 



Table V 



CONVERSION TABLE OF UNITS OF PRESSURE 



One lb. per sq. ft 

One lb. per sq. in 

One ounce per sq. in 

One atmoephere (standard at sea 
level) 

One kilogramme per square meter . . 

One gramme per square millimeter . 

One kilogramme per square centi- 
meter 

FLUID PRESSURES 

One ft. of water at 39.1° F. (max. 
dens.) 

One ft. of water at 62° F 

One in. of wat«r at 62° F 

One in. of mercury at 32° F. (stand- 
ard) * 

One centimeter of mercury at 0° C. . 

One ft. of air at 32° F., one atmos. 
press 

One ft. of air, 62° F 



Pounds i>er 
Square Foot. 



1 
144. 
9. 

2116.1 
20.4817 
204.817 

2048.17 



62.426 

62.355 

5.196 

70.7290 
27.8461 

0.08071 
0.07607 



Pounds per 
Square Inch. 



0.006944 

1. 

0.0625 

14.696 
0.142234 
1.42234 

14.2234 



0.43350 
0.43302 
0.036085 

0.491174 
0.193376 

0.0005604 
0.0005282 



Inches of 

Mercury at 

32° F. 



0.014139 

2.03594 

0.127246 

29.924 
0.289579 
2.89579 

28.9579 



0.88225 
0.88080 
0.07340 

1. 
0.393701 

0.0011412 
0.0010755 



Atmospheres 

(Standard at 

Sea Level). 



0.0004724 

0.06802 

0.004252 

1. 

0.009678 

0.09678 

0.9678 



0.029492 
0.029460 
0.002455 

0.033416 
0.013158 

0.00003813 
0.00003594 



> pR£ASCRE8 M KARUBBD BT THB Mkrcurt Column. For temperatures other than 32^ F.. the density 
of mercury, pounds per cubic inch, and hence the pressure, pounds per square inch, due to a column of mercury 
1 inch hicn, is given with sufficient accuracy by the following formula: 

p-0.4912-a-32) XO.OOOl. 

The mercurial barometer is commonly made with a brass scale which has its standard or correct Length 
at 62** F, and a linear coefficient of expansion of about 0.000001 for each degree Fahrenheit. Hence, to 
correct the standard of mercury at 32° F., the corrected reading will be 



where Hf ia the obeerved height at a temperature of t^ F. 



64 



ENGINEERING THERMODYNAMICS 



Table VI 



CONVERSION TABLE OF UNITS OF WORK ' 



Kilogrammeten. 


Foot-pounds. 


Foot Tons (Short Tons). 


Foot Tons (Long Tons). 


1. 

0.138255 
276.510 
309.691 


7.23300 

1. 
2000. 
2240. 


0.00361650 
0.000500 
1. 
1.12000 


0.00322902 
0.000446429 
0.892857 
1. 



1 See also more complete table of Units of Work and Energy in Chapter IV on Work and Heat. 



Table VII 



CONVERSION TABLE OF UNITS OF POWER 



Foot-pounds per 
Second. 


Foot-pounds per 
Minute. 


Horse-power. 


Cheval-Vapeur. 


Kilogrammeters per 
Minute. 


1. 

0.0166667 
550.000 
542.475 

0.120550 


60. 
1. 
33000. 
32548.5 

7.23327 


0.00181818 

0.000030303 

1. 

0.986319 

0.000219182 


0.00184340 

0.0000307241 

1.01387 

1. 

0.000222222 


8.29531 
0.138252 

4562.42 

4500.00 
1. 



Table VIII 



UNITS OF VELOCITY 



One foot per second 

One foot per minute 

One statute mile per hour 

One nautical mile per hour = 1 knot. 

One kilometer per hour 

One meter per minute , 

One centimeter per second ......... 



Feet per Minute. 


Feet per Second. 


60. 


1. 


1. 


0.016667 


88. 


1.4667 


101.338 


1.6890 


54.6806 


0.911344 


3.28084 


0.054581 


2.00848 


0.032808 



TABLES 



63 



Table IX 
TABLE OF BAROMETRIC HEIGHTS, ALTITUDES, AND PRESSURES 

(Adapted from Smithsonian Tables) 

Barometric heights are given in inches and millimeters of mercury at its standard density 
(32** F.). 

Altitudes are heights above mean sea level in feet, at which this barometric height is 
standard. (See Smithsonian Tables for corrections for latitude and temperature.) 

Pressures given are the equivalent of the barometric height in lbs. per sq. in. and per 
sq. ft. 



Standard Barometer. 


Altitude, Feet above 
Sea Level. 


Pressure, Pounds per 


Inches. 


Centimeters. 


Square Inch. 


Square Foot. 


17.0 
17.2 
17.4 
17.6 
17.8 


43.18 
43.69 
44.20 
44.70 
45.21 


15379 
15061 
14746 
14435 
14128 


8.350 
8.448 
8.546 
8.645 
8.742 


1202.3 
1216.6 
1230.7 
1244.8 
1259.0 


18.0 
18.2 
18.4 
18.6 
18.8 


45.72 
46.23 
46.73 
47.24 
47.75 . 


13824 
13523 
13226 
12931 
12640 


8.840 
8.940 
9.038 
9.136 
9.234 


1273.2 
1287.3 
1301.4 
1315.6 
1329.7 


19.0 
19.2 
19.4 
19.6 
19.8 


48.26 
48.77 
49.28 
49.78 
50.29 


12352 
12068 
11786 
11507 
11230 


9.332 
9.430 
9.529 
9.627 
9.726 


1343.8 
1357.9 
1372.1 
1386.3 
1400.4 


20.0 
20.2 
20.4 
20.6 
20.8 


50.80 
51.31 
51.82 
52.32 
52.83 


10957 
10686 
10418 
10153 
9890 


9.825 - 

9.922 
10.020 
10.118 
10.217 


1414.6 
1428.7 
1442.9 
1457.0 
1471.2 


21.0 
21.2 
21.4 
21.6 
21.8 


53.34 
53.85 
54.36 
54.87 
55.37 


9629 
9372 
9116 
8863 
8612 


10.315 
10.414 
10.511 
10.609 
10.707 


1485.3 
1499.4 
1513.6 
1527.7 
1541.8 


22 
22.2 
22.4 
22.6 
22.8 


55.88 
56.39 
56.90 
57.40 
57.91 


8364 
8118 
7874 
7632 
7392 


10.806 
10.904 
11.002 
11.100 
11.198 


1556.0 

1570.1 
1584.3 
1598.4 
1612.6 


23.0 
23.2 
23.4 
23.6 
23.8 


58.42 
58.92 
59.44 
59.95 
60.45 


7155 
6919 
6686 
6454 
6225 


11.297 
11.395 
11.493 
11.592 
11.690 


1626.7 
1640.8 
1655.0 
1669.3 
1683.3 


24.0 
24.2 
24.4 
24.6 
24.8 


60.96 
61.47 
61.98 
62.48 
62.99 


5997 
5771 
5547 
5325 
5105 


11.788 
11.886 
11.984 
12.083 
12.182 


1697.4 
1711.6 
1725.7 
1739.9 
1754.0 


25.0 
25.2 
25.4 
25.6 


63.50 
64.01 
64.52 
65.02 
65.53 


4886 
4670 
4455 
4241 
4030 


12.280 
12.377 
12.475 
12.573 
12.671 


1768.2 
1782.3 
1796.5 
1810.7 
1824.8 



66 



ENGINEERING THERMODYNAMICS 



Table IX — Continued 





Altitude, Feet above 
Sea Level. 


Pressure, Pounds per 


Inches. 


Centimeters. 


Square Ineh. 


Square Foot. 


26.0 
26.1 
26.2 
26.3 
26.4 


65.04 
66.30 
66.55 
66.80 
67.06 


3820 
3715 
3611 
3508 
3404 


12.770 
12.819 
12.868 
12.918 
12.967 


1838.9 

1846.0 
1853.1 
1860.2 
1867.3 


26.5 
26.6 
26.7 
26.8 
26.9 


67.31 
67.57 
67.82 
68.08 
68.33 


3301 
3199 
3097 
2995 
2894 


13.016 
13.065 
13.113 
13 163 
13.212 


1874.3 
1881.4 
1888.5 
1895.5 
1902.6 


27.0 
27.1 
27.2 
27.3 
27.4 


68.58 
68.84 
69.09 
69.34 
69.60 


2793 
2692 
2592 
2493 
2393 


13.261 
13.310 
13.359 
13.408 
13.457 


1909.7 
1916.7 
1923.8 
1930.9 
1938.0 


27.6 
27.6 
27.7 
27.8 
27.9 


69.85 
70.10 
70.35 
70.61 
70.87 


2294 
2195 
2097 
1999 
1901 


13.507 
13.556 
13.605 
13.654 
13.704 


1945.1 
1952.1 
1959.2 
1966.3 
1973.3 


28.0 
28.1 
28.2 
28.3 
28.4 


71.12 
71.38 
71.63 
71.88 
72.14 


1804 
1707 
1610 
1514 
1418 


13.753 
13.802 
13.850 
13.899 . 
13.948 


1980.4 
1987.5 
1994.5 
2001.6 
2008.7 


28.5 
28.6 

28.7 
28.8 
28.9 


72.39 
72.64 
72.90 
73.15 
73.40 


1322 
1227 
1132 
1038 
943 


13.998 
14.047 
14.096 
14.145 
14.194 


2015.7 
2022.8 
2030.0 
2037.0 
2044.1 


29.0 
29.1 
29.2 
29.3 
29.4 


73.66 
73.92 
74.16 
74.42 
74.68 


849 
756 
663 
570 
477 


14.243 
14 . 293 
14.342 
14.392 
14.441 


2051.2 
2058.2 
2065.3 
2072.4 
2079.4 


29.5 
29.6 
29.7 
29.8 
29.9 


74.94 
75.18 
75.44 
75.69 
75.95 


384 
292 
261 
109 

+18 


14.490 
14.539 
14.588 
14.637 
14.686 


2086.5 
2093.6 
2100.7 
2107.7 
2114.7 


29.92 


76.00 





14.696 


2116.1 


30.0 
30.1 
30.2 
30.3 
30.4 


76.20 
76.46 
76.71 
76.96 
77.22 


- 73 
-163 
-253 
-343 
-433 


14.734 
14.783 
14.833 
14.882 
14.931 


2121.7 
2128.8 
2135.9 
2143.0 
2150.1 


30.5 
30.6 
30.7 
30.8 
30.9 


77.47 
77.72 
77.98 
78.23 
78.48 


-522 
-611 
-700 

-788 
-877 


14.980 
15.030 
15.078 
15.127 
15 . 176 


2157.2 
2164.2 
2171.3 
2178.4 
2185.5 


31.0 


78.74 


-965 


15.226 


2192.6 



TABTiES 67 
Table X 

VALUES OF s IN THE EQUATION PV =CONg TANT FOR VARIOUS SUBSTANCES 

AND CONDITIONS 


Substanee. 




8 


Remarks or Authority. 


Ml eaws ,.,..., 


Isothermal 
Constant pressure 

Isothermal 

Constant volume 

Adiabaiic 

Compressed in cylinder 

Adiabatic, wet 

Adiabatic, superheated 

Adiabatic 

Adiabatic 

Adiabatic 

Adiabatic 
Adiabatic 

Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 

Adiabatic 

Adiabatic 

Adiabatic 

Expanding in cylinder 

Saturation Law 


1 ] 




00 

1.4066 

1.4 

1.1 

1.3 

1.293 

1.300 

1.403 

1.200 
1.323 

1.106 
1.029 
1.410 
1.276 
1.316 
1.410 
1.291 
1.24 
1.26 
1.300 
Variable 

1.111 
1 + 14 X% moist. 
1.035 H-.1X% moist. 
1. 
1.0646 




All gases and vapors . . 
All saturated vapors . . 
All gases and vapors . . 
Air 


Accepted thermody- 
namic law 

Smithsonian Tables 


Air 


Experience 


Ammonia (NHs) 

Ammonia (NHs) 

Brnmine 


Average 

Thermodynamics 

Strecker 


Carbon dioxide (COi) . 
Carbon monoxide (CO) 
Carbon disulphide 
(OS,) 


Rontgen, Wullner 
Cazin, Wullner 

Beyne 


Chlorine (Q) 

Chloroforuu 

(CCl,CH(OH),).... 
Ether (CiHiOCH,) . . . 

Hydrogen (H,) 

Hydrogen sulph . (HsS) 

Methane (OH4) 

Nitrogen (N«) 

Nitrous oxide (NOj) . . 
Pintsch gas 


Strecker 

Be3me, Wullner 

MDller 

Cazin 

MUller 

Mttller 

Cazin 

Wullner 

Pintsch Co. 


Sulphide diox (SOs) .. . 
Steam, superheated . .. 


Cazin, Mttller 
Smithsonian Tables 
(From less than 1 to 




more than 1.2) 
Rankine 


Steam, wet 


Perry 


Steam, wet 


Gray 


Steam, wet 


Average from practice 


Stwun. dry 


Regnault 







68 



ENGINEERING THERMODYNAMICS 



Table XI 
HORSE-POWER PER POUND MEAN EFFECTIVE PRESSURE. 

aJS 





• 






\Jt\lj\JCj 


"' '^'" 33000 








Diameter 

nf 


Speed of Piston in Feet per Minute. 


Cylinder, 
Inches. 




















100 


200 


300 


400 


500 


600 


700 


800 


900 


4 


0.0381 


0.0762 0.1142 


0.1523 


0.1904 


0.2285 


0.2666 


0.3046 


0.3427 


41 


0.0482 


0.0964 


0.1446 


0.1928 


0.2410 


0.2892 


0.3374 


0.3856 


0.4338 


5 


0.0592 


0.1190 0.1785 


0.2380 


0.2975 


0.3570 


0.4165 


0.4760 


0.5355 


51 


0.0720 


0.1440 0.2160 


0.2880 


0.3600 


0.4320 


0.5040 


0.5760 


0.6480 


6 


0.0857 


0.1714 0.2570 


0.3427 


0.4284 


0.5141 


0.5998 


0.6854 


0.7711 


61 


0.1006 


0.2011 0.3017 


0.4022 


0.5028 


0.6033 


0.7039 


0.8044 


0.9050 


7 


0.1166 


0.2332 0.3499 


0.4665 


0.5831 


0.6997 


0.8163 


0.9330 


1.O490 


71 


0.1339 


0.2678 0.4016 


0.5355 


0.6694 


0.8033 


0.9371 


1.0710 


1.2049 


8 


0.1523 


0.3046 0.4570 


0.6093 


0.7616 


0.9139 


1.0662 


1.2186 


1.3709 


81 


0.1720 


0.2439 0.5159 


0.6878 


0.8598 


1.0317 


1 . 2037 


1.3756 


1.5476 


9 


0.1928 


0.3856 


0.5783 


0.7711 


0.9639 


1 . 1567 


1 . 3495 


1 . 5422 


1.7350 


9i 


0.2148 


0.4296 


0.6444 


0.8592 


1.0740 


1.2888 


1.5036 


1.7184 


1 1.9532 


10 


0.2380 


0.4760 


0.7140 


0.9520 


1.1900 


1 . 4280 


1.6660 


1.9040 


2.1420 


11 


0.2880 


0.5760 


0.8639 


1.1519 


1.4399 


1 . 7279 


2.0159 


2.3038 


2.5818 


12 


0.3427 


0.6854 


1.0282 


1.3709 


1.7136 


2.0563 


2.3990 


2 . f 418 


3.0845 


13 


0.4022 


0.8044 


1.2067 


1.6089 


2.0111 


2.4133 


2.8155 


3.2178 


3.6200 


14 


0.4665 


0.9330 


1.3994 


1.8659 


2.3324 


2 . 7989 


3.2654 


3.7318 


4.1983 


15 


0.5355 


1.0710 


1.6065 


2.1420 


2.6775 


3.2130 


3 . 7485 


4.2840 


4.8195 


16 


0.6093 


1.2186 


1.8278 


2.4371 


3.0464 


3.6557 


4.2650 


4.8742 


5.4835 


17 


0.6878 


1 . 2756 


1.9635 


2.6513 


3.3391 


4.0269 


4.6147 


5.4026 


6.1904 


18 


0.7711 


1.5422 


2.3134 3.0845 


3 . 8556 


4.6267 


5.3987 


6.1690 


6.4901 


19 


0.8592 


1.7184 


2.5775 3.4367 


4.2858 


5.1551 


6.0143 


6.8734 


7.7326 


20 


0.9520 


1.9040 


2.8560 


3.8080 


4.7600 


5.7120 


6.6640 


7.6160 


8.5680 


21 


1.0496 


2.0992 


3.1488 


4.1983 


5.2475 


6.2975 


7.3471 


8.3966 


9.4462 


22 


1.1519 


2.3038 


3.4558 


4.6077 


5.7596 


6.9115 


8.0643 


9.2154 


10.367 


23 


1.2590 


2.5180 


3.7771 


5.0361 


6.2951 


7.5541 


8.8131 


10.072 


11.331 


24 


1.3709 


2.7418 


4.1126 


5.4835 


6.8544 


8.2253 


9.5962 ; 


10.967 


12.338 


25 


1.4875 


2.9750 


4.4625 


5.9500 


7.4375 


8 . 9250 


10.413 ! 


11.900 


13.388 


26 


1.6089 


3.2178 


4.8266 


6.4355 


8.0444 


9.6534 


11.262 


12.871 


14.480 


27 


1.7350 


3.4700 


5.2051 


6.9401 


8.6751 


10.410 


12.145 


13.880 


15.615 


28 


1.8659 


3.7318 5.5978 


7.4637 


9.3296 


11.196 


13.061 


14.927 


16.793 • 


29 


2.0016 


4.0032,6.0047 


8.0063 


10.008 


12.009 


14.011 


16.013 


18.014 


30 


2 . 1420 


4.2840 


6.4260 


8.5680 


10.710 


12 . 852 


14.994 


17.136 


19.278 


31 


2.2872 


4.5744 


6.8615 9.1487 


11.436 


13 . 723 


16.010 


18.287 


20.585 


32 


2.4371 


4.8742 


7.3114 9.7485 


12.186 


14.623 


17.060 


19.497 


21.934 


33 


2.5918 


5.1836 7.7755! 


10.367 


12.959 


15.551 


18.143 


20.735 


23.326 


34 


2.7513 


5.5026 


8.2538 


11.005 


13 . 756 


16.508 


19.259 


22.010 


24.762 


35 


2.9155 


5.8310 8.7465 


11.662 


14.578 


17.493 


20.409 


23.224 


26.240 


36 


3.0845 6.1690 9.2534 


12.338 


15.422 •' 


18.507 


21.591 


24.676 


27.760 


37 


3.2582 6.5164 9.7747 


13.033 


16.291 


19.549 


22.808 


26.066 


29.324 


38 


3.4367 6.8734,10.310 13.747 


17.184 


20.620 


24.057 


27.494 


30.930 


39 


3.6200 


7.2400 


10.860 14.480 


18.100 


21.720 


25.340 


28.960 


32.680 























TABLES 



69 



Table XI — Cordinued 



Diameter 

of 


Speed of Piaton in Feet per Minute. 


Cylinder, 
Inchea. 




















100 


200 


300 


400 


600 


600 


700 


800 


900 


40 


3.8080 


7.6160 


11.424 


15.232 


19.040 


22.848 


26.656 


30.464 


34.272 


41 


4.0008 


8.0016 


12.002 


16.003 


20.004 


24.005 


28.005 


32.006 


36.007 


42 


4.1983 


8.3866 


12.585 


16.783 


20.982 


25.180 


29.378 


33.577 


37.775 


43 


4.4006 


8.8012 


13.202 


17.602 


22.003 


26.404 


30.804 


35.205 


39.606 


44 


4.6077 9.2154 


13.823 


18.431 


23.038 


27.646 


32.254 


36.861 


41.469 


45 


4.819519.6390 


14.459 


19.278 


24.098 


28.917 


33.737 


38.556 


43.376 


46 


5.0361 10.072 


15.108 


20.144 


25.180 


30.216 


35.253 


40.289 


45.325 


47 


5.2574 10.515 


15.772 


21.030 


26.287 


31.545 


36.802 


42.059 


47.317 


48 


5.3845 10.967 


16.451 


21.934 


27.418 


32.901 


38.385 


43.868 


49.352 


49 


5.7144 1 11.429 


17.143 


22.858 


28.572 


34.286 


40.001 


45.715 


51.429 


50 


5.9500 11.900 


17.850 


23.800 


29.750 


35.700 


41.650 


47.600 


53.550 


51 


6.1904 12.381 


18.571 


24.762 


30.952 


37.142 


43.333 


49.523 


55.713 


52 


6.4355 12.871 


19.307 


25 . 742 


32 . 178 


38.613 


45.049 


61.484 


57.920 


53 


6.68541 13.371 


20.056 


26.742 


33.427 


40.113 


46.798 


53.483 


60.169 


54 


6.9401 


13.880 


20.820 


27.760 


34.700 


41.640 


48.581 


55.521 


62.461 


55 


7.1995 


14.399 


21.599 


28.798 


35.998 


43.197 


50.397 


57.696 


64.796 


56 


7.4637 


14.927 


22.391 


29.855 


37.318 


44.782 


52.246 


69.709 


67.173 


57 


7.7326 


15.465 


23.198 


30.930 


38.663 


46.396 


64.128 


61.861 


69.594 


58 


8.0063 


16.013 


24.019 


32.025 


40.032 


48.038 


56.044 


64.051 


72.057 


59 


8.2849 


16.570 


24.854 


33 . 139 


41.424 


49.709 


57.993 


66.278 


74.563 


60 

• 


8.5680 


17.136' 25.704 

I 


34.272 


42.840 


51.408 


59.976 


68.544 


77.112 



70 ENGINEERING THERMODYNAMICS 



GENERAL FORMULA RELATING TO PRESSURE- VOLUME 
CALCULATIONS OF WORK AND POWER 

Work = TT = Force X Distance =F xL; 

= Pressure X Area X Distance = P XA xL; 
= Pressure X Volume change =P X( 7,— FJ. 

Force of acceleration ^massXacceleration. 

g dx 
Work of acceleration -^ ^^"^ X difference of (velocity)*, 

2 g^ ^ 64.32^ '^ 

' Velocity due to work of acceleration = square root of the sum of (initial velocity)' 
plus 2g Xwork per pound of substance accelerated. 



' w 

or if initial velocity is zero, 

M,«V64.32-=8.02J^. 
' w ^ w 

Pressure Volume Relation for Expansion or Compression 
PV =PiFi» ^P^Vi' =Ky a constant. (See Table VIII for values of «.) 

log (yj) 
(Note graphical method for finding s when variable, see text.) 



FORMULA RELATING TO PRESSURE- VOLUME CALCULATIONS 71 



Work done during a pressure volume change represented by the equation (PF* —if) 
between points represented by 1 and 2 in figure =area under curve, Wi. 



If «°1, 



PdV 'K I ~. 



W, -P.7, log. ?^ 



=PtVt log. 



F. 



P.V, log. ^' 



=P,F,log. 



P, 

Pa J 



8 



= 1. 



If s is not equal to 1, 



^"'-^my-'-A 1 



«— 1 






9-1 



-S[-©"] 



89^1. 



Work of admission, complete expansion and exhaust for engines » area to left of 
curve (see figure) Wt, Same for admission, compression and expulsion for com- 
pressors. Both cases without clearance. 

(When « = 1, TTj is same as area xmder curve, Wi (see above). 



wh«.^iir..^,p.v,[(f;)'"'-i] 



»-l 



--.".^'-fn 



=«xTFi. 



Clearance, expressed as a fraction of displacement =c, as a voliune =CT. 



72 ENGINEERING THERMODYNAMICS 

liPaVa^PbVt', 

Db _Da (Pa\ 7 

D D \PJ 



CI 



t«V-i 



Indicated horse-power = (mean effective pressure, pounds per square inch Xeffecti ve 
area of piston, square inches X length of stroke, feet X number of working cycles per- 
formed per minute) divided by 33,000. 

■ 

^ • • "" 33.000 • 
Specific displacement B displacement in one direction for one side of a piston, in 

cuit. per hour per H.P. =D« = 1?,750, r, 

(m.e.p./ 

where « is the number of strokes required o conipl ' e one cycle. 

Velocity of a jet due to its own expansion - the square root of the product of 2g X work 

done by admission, complete expansion and exhaust of 1 lb. of the substance. 

ti = V2^,=8.02|4^P,7|l-(^)~j| . 

Weight of flow through nozzle or orifices, pounds per second « to = (velocity, feet per 
second Xarea, square feet, of orifice) 4- (volume per pound of substance at section where 
area is measiu^). 



uA 

"-v 



7-^<.W\7h^H'-(^f]\' 



Maximum discharge w for a given initial pressure occurs when 

a 

p. 



ill ( ^ \*-' 



CHAPTER II 

WORK OF COMPRESSORS, HORSE-POWER AND CAPACITY OF AIR, GAS AND 
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS. 

1. General Description of Structure and Processes. There is quite a 
large class of machines designed to receive a cylinder full of some gas at one 
constant pressure and after the doing of work on the gas through decreasing 
volumes and rising pressures, to discharge the lesser volume of gas against a 
constant higher pressure. These machines are in practice grouped into sub- 
classes, each having some specific distinguishing characteristic. For example, 
blowing engines take in air at atmospheric pressure or as nearly so as the 
valve and port resistance will permit, and after compression deliver the air 
at a pressure of about three atmospheres absolute for use in bla^ furnaces. 
These blowing engines are usually very large, work at low but variable speeds, 
but always deliver against comparatively low pressures; they, therefore, have 
the characteristics of large but variable capacity and low pressures. A great 
variety of valves and driving gears are used, generally mechanically moved 
suction and automatic spring closed discharge valves, but all valves may be 
automatic. The compressor cylinder is often termed the blowing tub and the 
compressed or blast air frequently is spoken of as wind by furnace men. They 
are all direct-connected machines, an engine forming with the compressor one 
machine. The engine formerly was always 9f the steam type, but now a change 
is being affected to permit the direct internal combustion of the blast furnace 
waste gases in the cylinders of gas engines. These gas-driven blowing engines, 
showing approximately twice the economy of steam-driven machines, will in 
time probably entirely displace steam in steel plants, and this change will take 
place in proportion to the successful reduction of cost of repairs, increase of 
reliability and life of the gas-driven blowing engines to equal the steam-driven. 
Some low-pressure blowers are built on the rotary plan without reciprocating 
pistons, some form of rotating piston being substituted, and these, by reason 
of greater leakage possibilities, are adapted only to such low pressures as 5 lbs. 
per square inch above atmosphere or thereabouts. These blowers are coming 
into favor for blasting gas producers, in which air is forced through thick coal 
beds either by driving the air or by drawing on the gas produced beyond the 
bed. They are also used for forcing illuminating gas in cities through pipes 
otherwise too small, especially when the distances are long. In general very 
low pressures and large capacities are the characteristics of the service whether 
the work be that of blowing or exhausting or both. For still lower pressures, 
measurable by water or mercury columns, fans are used of the disk or propeller 

73 



74 ENGINEERING THERMODYNAMICS 

or centrifugal type. These fans are most used for ventilation of buildings and 
mines, but a modification, based on the principles of the steam turbine reversed, 
and termed turbo-compressors, is being rapidly adapted to such higher pres- 
sures as have heretofore required piston compressors. 

When high-pressure air is required for driving rock drills in mines and for 
hoisting engines, for tools, as metal drills, riveters, chipping chisels, for car 
air' brakes, the compressors used to provide the air are termed simply air 
compressors. These compressors usually take in atmospheric air and compress 
it to the desired pressure, the capacity required being usually adjustable; 
they have valves of the automatic type throughout commonly, but in large 
sizes frequently are fitted with mechanically operated suction valves to 
decrease the resistance to entrance of air and so increase economy, a com- 
plication not warranted in small machines. When the pressures of delivery 
are quite high the compression is done in stages in successive cylinders, the 
discharge from the first or low-pressure cylinder being delivered through a 
water cooler or intercooler to the second cyUnder and occasionally to a third 
in turn. This staging with intermediate or intercooling results in better 
economy, %s will be seen later in detail, and permits the attainment of the 
desired quantity of cool compressed air for subsequent use with the expenditure 
of less work, the extra comphcation and cost being warranted only when 
machines are large and final pressures high. 

In the operation of large steam condensers, non-condensible gases will 
collect and spoil the vacuum, which can be maintained only by the continuous 
removal of these gases, consisting of air, carbon dioxide and gases of animal 
and vegetable decomposition originally present in the water. When these gases 
are separately removed the machine used is a special form of compressor termed 
a dry vacuiun pmnp which, therefore, receives a charge at the absolute pres- 
sure corresponding to the vacuum, or as nearly so as the entrance resistance 
permits, and after compression discharges •into the atmosphere at a pressure 
in the cylinder above atmosphere equivalent to discharge resistance. Natural- 
gas wells near exhaustion can sometimes be made to flow freely by the applica- 
tion of a compressor capable of drawing a charge at a pressure below atmosphere, 
but whether the charge be received below atmospheric pressure or above as 
in normal wells, the compressor will permit the delivery of the gas to distant 
cities or points of consumption even 250 miles away through smaller pipes 
than would be otherwise possible. Natural-gas compressors, some steam- and 
some gas-engine driven, are in use for both these purposes, compressing natural 
gas from whatever pressure may exist at the well to whatever is desired at the 
beginning of the pipe line. 

In the preparation of liquid ammonia or carbonic acid gas for the market, 
as such, or in the operation of refrigerating machinery, wet or dry vapor is com- 
pressed into a condenser to permit liquefaction by the combined eflPect of high 
pressure and cooling. One form of refrigerating machine merely compresses 
air, subsequently expanding it after preliminary cooling by water, so that 
after expansion is complete it will become extremely cold. 



WORK OF COMPRESSORS 75 

All these compresEdng machines have, as a primary pmpose, either the 
removal of a quantity of low-pressure gases from a given place, or the delivery 
of a quantity of higher-pressure gas to another place or both, but all include 
compression as an intermediate step between constant-pressure admission and 
constant-pressure discharge as nearly as structure may permit. They will all 
involve the same sort of physical operations and can be analyzed by the same 
principles except the wet-vapor or vvet-gas compressors, in which condensation 
or evaporation may complicate the process and introduce elements that can 
be treated only by thermal analysis later. Safe compressors cannot be built 
with zero cylinder clearance, hence at the end of delivery there will remain in 
the clearance space a volume of high-pressure gases equal to the volume of the 
clearance space. On the return stroke this clearance volume will expand until 
the pressure is low enough to permit suction, so that the new charge cannot 
enter the cj'^linder until some portion of the stroke has been covered to permit 
this re-expansion of clearance gases. 

It is quite impossible to study here all the effects or influences of structure 
as indicated by the compressor indicator cards, but a quite satisfactory treat- 
ment can be given by the establishment of reference diagrams as standards of 
comparison and noting the nature of the differences between the actual cases and 
the standard reference diagram. These standard reference diagrams will really 
be pressure-volume diagrams, the phases of which correspond to certain hypoth- 
eses capable of mathematical expression, such as constant pressure, constant 
volume, expansion, and compression, according to some law, or with some 
definite value of s fixing either the heat-exchange character of the process or 
the substance, as already explained. 

2. Standard Reference Diagrams or PV Cycles for Compressors and Methods 
of Analysis of Compressor Work and Capacity. All the standard reference 
diagrams will include constant-pressure lines corresponding to delivery and 
supply at pressures assumed equal to whatever exists outside the.^Qylinder on 
either delivery or suction side, that is, assiuning no loss of pressure on delivery 
or suction. The compression may be single or multi-stage with various 
amounts of cooling in the intercooler, but in multi-stage compression 
the standard reference diagram will be assumed to involve intermediate 
cooling of the gases to their original temperature, so that the gases 
entering all cylinders will be assumed to have the same temperature and 
to maintain it constant during admission. Another difference entering into 
the classification of standard reference diagrams is the laws of compression 
as defined by the exponent s. Integration of the differential work expres- 
sion will take a logarithmic form for s = l, and an exponential form for all 
other values, thus giving two possible reference compression curves and two 
sets of work equations. 

(a) The isothermal for which 5=1, no matter what the gas, and which is 
the consequence of assuming that all the heat liberated by compression is con- 
tinuously carried away as fast as set free, so that the temperature cannot rise 
at all. 



76 ENGINEERING THERMODYNAMICS 

(6) The exponential for which s has a value greater than one, generally 
different for every gas, vapor or gas-vapor mixture, but constant for any one 
gas, and also for dry vapors that remain dry for the whole process. Wet vapors 
having variable values of 8 cannot be treated by the simple pressure-volume 
analysis that suffices for the gases, but must be analyzed thermally. The 
adiabatic value of s is a consequence of assuming no heat exchange at all 
between the gas and anything else and is a special case of the general exponen- 
tial class. 

Just why these two assumptions of thermal condition should result in the 
specified values of s will be taken up imder the thermal analysis part of this 
work, and is of no interest at this time. 

As a consequence of these phase possibilities there may be established eight 
standard reference diagrams or pressure-volume cycles defined by their phases, 
as shown in Fig. 23, four for single-stage compression and two each for two and 
three stages. These might be extended by adding two more for four stages 
and so on, but as it seldom is desirable, all things being considered, to go beyond 
three, the analysis will stop with the eight cycles or reference diagrams shown. 

Single-stage Compression Reference Cycles or PV Diagrams 

a 

Cycle 1. Single-stage Isothermal Compression without Clearance. 
Phase (a) Constant pressure supply. 
" (6) Isothermal compression. 
" (c) Constant pressure delivery. 
" (d) Constant zero-volume pressure drop. 
Cycle 2. Single-stage Isothermal Compression with Clearance. 
Phase (a) Constant pressure supply. 
" (6) Isothermal compression, 
(c) Constant pressure delivery. 
id) Isothermal re-expansion. 
Cycle 3. Single-stage Exponential Compression without Clearance. 
Phase (a) Constant pressure supply. 
(6) Exponential compression. 

(c) Constant pressure delivery. 

(d) Constant zero-volume pressure drop. 
Cycle 4. Single-stage Exponential Compression with Clearance. 

Phase (a) Constant pressure supply. 

" (6) Exponential compression. 

" (c) Constant pressure delivery. 

" (d) Exponential re-expansion. 






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; 78 ENGINEERING THERMODYNAMICS 



Muiyn-STAQE Comprbssion 

The phases making up multi-stage compression cycles may be considered in 
two wnys, first, as referred to each cylinder and intercooler separately, or 
second, as referred to the pressure volume changes of the gases themselves 
regardless of whether the changes take place in cylinders or intercoolers. 

For example, if 10 cu.-ft. of hot compressed air be delivered from the first 
cylinder of 50 cu.-ft. displacement, the phase referred to this cylinder is a con- 
stant-pressure decreasing volimie, delivery line whose length is \ of the whole 
diagram, exactly as in single-stage compression. If this 10 cu.-ft. of air deliv- 
ered to an intercooler became 8 ft. at the same constant pressure as the first 
cylinder delivery, the phase would be indicated by a constant-pressure volume 
reduction line 2 cu.-ft long to scale, or referred to the original volume of air 
admitted to the first cylinder, a line jV of its length. Finally, admitting this 
8 cu.-ft of cool air to the second cylinder and compressing it to i of its vol- 
ume would result in a final delivery line at constant pressure of a length of i of 
the length of the second cylinder diagram, but as this represents only 8 cu.-ft., 
the final delivery will represent only ^X8 = 1.6 cu.-ft. This 1.6 cu.-ft. will, 
when referred to the original 50 cu.-ft. admitted to the first cylinder, be repre- 

1.6 
sented by a constant-pressure line, —=.032, of the whole diagram length, 

ou 

which in volume is equivalent to i of the length of the second cylinder 
diagram. It should be noted also that three volume change operations take 
place at the intermediate pressure; first, first cylinder delivery; second, 
volume decrease due to intercooling; third, second cylinder admission, the 
net effect of which referred to actual gas volumes, regardless of place where 
the changes happen, is represented by the volume decrease due to inter- 
cooling only. A diagram of volumes and pressures representing the resultant 
of all the gas processes is called in practice the combined PV diagram for the 
two cylinders, or when plotted from actual indicator cards with due regard 
for the different clearances of each cylinder the combined indicator diagrams. 
It is proper in the study of the whole process of compression to consider the 
cycle consisting of phases referred to true gas volumes rather than phases 
referring to separate cylinder processes, which is equivalent to imagining the 
whole cycle carried out in one cylinder. 

Intercooling effects measured by the amount of decrease of volume at 
constant pressure will, of course, depend on the amount of cooling or reduc- 
tion of temperature, but in establishing a standard reference diagram some 
definite amount capable of algebraic . description must be assumed as an 
intercooling hypothesis. 

It has already been shown, Fig. 6, Chapter I, that from any original state 
of pressure and volume the exponential and isothermal could be drawn, diverging 
an amount depending on the difference between the defining exponent, s. 
If, after reaching a given state on the exponential curve, the gas be cooled at 



WORK OF COMPRESSORS 79 

constant pressure to its original temperature, the point indicating its condition 
will lie by definition on the other curve or isothermal and the cooling process 
is represented by a horizontal joining the two curves. Such ihtercooling as 
this will be defined as perfect inUrcooling, for want of a better name, and its 
pressure-volume effects can be treated by the curve intersections. It is now 
possible to set down the phases for the standard reference diagrams of multi- 
stage compression, if in addition to the above it be admitted, as will be proved 
later, that there is a best or most economical receiver pressure. 



Two-stage Compression Reference Cycles or PV Diagrams 

« 

Cycle 5. Two-stage Exponential Compression without Clearance, Perfect 

Intercooling at Best Receiver Pressxire. 
Phase (a) Constant pressure supply. 
' ' (6) Exponential compression to best receiver pressure. 
' ' (c) Constant pressure perfect intercooling of delivered gas. 
* ' (d) Exponential compression from best receiver pressure. 
* ' (e) Constant pressure delivery. 
* * (/) Constant zero-volume pressure drop. 
Cycle 6. Two-stage Exponential Compression with Clearance, Perfect 

Intercooling at Best Receiver Pressure. 
Phase (a) Constant pressure supply. 
* ' (6) Exponential compression to best receiver pressure. 
* ' (c) Constant pressure perfect intercooling of delivered gas. 
' ' (d) Exponential re-expansion of first stage clearance, 
(e) Exponential compression from best receiver pressure. 
(J) Constant pressure delivery. 
ig) Exponential re-expansion of second stage clearance. 



i t 
ti 



Three-stage Compression Reference Cycles or PV Diagrams 

Cycle 7. Three-stage Exponential Compression, without Clearance, Per- 
fect Intercooling at Best Two Receiver Pressxires. 
Phase (a) Constant pressure supply. 

(6) Exponential compression to first receiver pressure, 
(c) Perfect intercooling at best first receiver presssure. 
{d) Exponential compression from best first to best second 
receiver pressure.' 
' ' (c) Perfect intercooling at best second receiver pressure. 
' ' (/) Exponential compression from best second receiver pressure. 
" ig) Constant pressure delivery. 
** (A) Constant zero-volume pressure drop. 



i t 



80 ENGINEERING THERMODYNAMICS 

Cycle 8. Three-stage Adiabatic Compression with Clearance, Perfect 

Intercooling at Best Two Receiver Pressures. 
Phase (a) Constant pressure admission. 
' ' (6) Exponential compression to best first receiver pressure. 
' ' (c) Perfect cooling of delivered gas at best first receiver pressure. 
' ' (d) Exponential re-expansion of first stage clearance. 
'' (e) Exponential compression from best first to best second 

receiver pressure. 
* ' (/) Perfect intercooling of delivered gas at best second receiver 

pressure. 
** (g) Exponential re-expansion of second stage clearance. 
* ' (h) Exponential compression from best second receiver pressure. 
* ' (i) Constant pressure delivery. 
' ' (j) Exponential re-expansion of third stage clearance. 

It should be noted that cycles 6 and 8 may be sub-divided mto any number 
of cases, of which some of the most characteristic are shown: (a) where the 
clearance volume in each cylinder bears the same ratio to the displacement 
of that cylinder, and commonly called equal clearances; (6) where the clearances 
are such that the volume after re-expansion in the higher-pressure cylinder 
is equal to the voliune of clearance in the next lower-pressure cylinder, causing 
the combined diagram to have a continuous re-expansion line, a case which 
may be called proportionate clearance; and (c) the general case in which there 
is no particular relation between clearances in the several cylinders. 

By means of these definitions or their mathematical equivalents in symbols 
it will be possible to calculate work as a function of pressures and volumes 
and by various transformations of a general expresssion for work of a reference 
cycle to calculate the horse-power corresponding to the removal of a given 
volume of gas per minute from the low-pressure supply or to the delivery of 
another volume per minute to the high-pressure receiver or per unit weight 
per minute. It will also be possible to calculate the necessary cylinder size 
or displacement per unit of gas handled, and the horse-power necessary to 
drive the compressing piston at a specified rate and further to calculate the 
work and horse-power of cylinders of given size and speed. In order that these 
calculations of a numerical sort may be quickly made, which is quite necessary 
if they are to be useful, the formulas must be definite and of proper form, the 
form being considered proper when little or no algebraic transformation is 
necessary before niunerical work is possible. While special expressions for 
each case are necessary to facilitate numerical work, it is equally important, 
if not more so, to make clear the broad general principles or methods of attact, 
because it is quite impossible to set down every case or even to conceive at the 
time of writing of all different cases that must in future arise. The treat- 
ment, then, must be a combination of general and special, the general methods 
being applied successively, to make them clear and as a matter of drill, not 
to every possible case, but only to certain characteristic or type forms 



WORK OF COMPRESSORS 81 

of cases, such as are here set down as standard reference diagrams. 
Individual cases may be judged by comparison with these and certain factors 
of relation established which, being ratios, may be and are called efficiencies. 
Thus, if a single-stage compressor should require two horse-power per cubic 
foot of free air compressed per minute, and Cycle I should for the same 
pr^sure limits require only one horse-power for its execution, then the efficiency 
of the real compression would be 50 per cent referred to Cycle I, and similar 
factors or efficiencies for other compressors similarly obtained; a com- 
parison of the factors will yield information for a judgment of the two 
compressors. 

In what follows on the work and gas capacity of compressors two methods 
of attack will be used. 

1. General pressure-volume analysis in terms of gas pressures and volumes 
resulting in the evaluation of work per cubic foot of low- or high- pressure 
gaseous substance. 

2. Transformation of results of (1) to yield volumetric efficiencies, mean 
effective pressures, work, horse-power, and capacity in terms of dimensions 
of cylinders and clearances. 

3. Single-stage Compressory No Clearance, Isothennal Compression 
(Cycle 1), Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. The standard reference diagram is represented by Fig. 24, on 
which the process {A to B) represents admission or supply at constant pressure; 
{B to C) compression at constant temperature; (C to D) delivery at constant 
pressure; and (D to A) zero-volume. 

Let Fft = The number of cubic feet of low pressure gas in the cylinder aftel* 

admission, represented to scale on the diagram by AB and equal 

to the volume at B; 
Fc= volume in cubic feet of the gas in cylinder when discharge begins, 

represented by DC, which is the volume at C; 
P»= absolute pressure in pounds per square foot, at which supply 

enters cylinder = (Sup.Pr.) = pressure at B; 
p&=Pft-^ 144 = absolute supply pressure in pounds per square inch = 

(sup.pr.) ; 
Pc= absolute pressure in pounds per square foot, at which delivery 

occurs = (Del. Pr.) = pressure at C; 

Pc=Pc-j- 144 = absolute delivery pressure in pounds per square inch 

= (del.pr.); 
p 
Rp = 5^= ratio of delivery pressure to supply pressure; 

W = foot-pounds work done for the cycle; 

(H. P. Cap.) = volume of gas delivered in cubic feet per cycle, at 

temperature same as that of supply; 
(L. P. Cap.) = volume of gas drawn into cylinder, cubic feet per cycle. 

For this no clearance case (L. P. Cap.) = Vft. 



a 



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82 ENGINEERING THERMODYNAMICS 

Referring to Fig. 24, the work for the cycle is the sum of compression and 
delivery work, less admission work, or by areas 

Net work ^4 BCD = compression work £BCG+delivery work GCDF 

— admission work BBAF. 
Algebrucally this is equivalent to 



I 



Fig. 24. — One-stage Compreseor Cycle 1, No Clearance, iBothermal. 
But since PrVc=PbVb the expression becomes 

W = F^niog.^, 



which is the work for the execution of the cycle when pressures and volumes 
are in pounds per square foot, and cubic feet. The equivalent expression 
for pounds peT square inch and cubic feet is 

^ = 144 ptFJog,^' (29) 



WOKK OF COMPRESSORS 83 

Since, when there is no clearance the volume taken into the cylinder for 
each cycle (L. P. Cap.) is equal to the volume at B, Vb, the expression Eq. (29) 
may be stated thus, symbolic form. 

Tr=144(sup.pr.)(L. P. Cap.) logeJKp (30) 

The work per cubic foot of low pressure gas, foot-pounds, will be the above 
expression divided by (L. P. Cap.), or 

I W 

^^-^-^^=144(sup.pr.)log.i?, (31) 

The work per cubic foot of high-pressure gas delivered will be 

W 

= 144 (sup.pr.) Rp log* Rp, .... (32) 





(H.P.Cap.) ' 


since 


PiV, = PcVc 


or 


^b 



which expressed symbolically is 

(L.P.Cap.) = (H.P.Cap.)Xffp (33) 

Expressions (31) and (32) for the isothermal compressor are especiallj'^ useful 
as standards of comparison for the economy of the compressors using methods 
other than isothermal. It will be found that the work per cubic foot of either 
low pressure or cooled high-pressure gas is less by the isothermal process ^han 
by any other process discussed later, and that it is the limiting case for the 
economy of multi-stage compressors with a great number of stages. The fact 
that this process of isothermal compression is seldom if ever approached in 
practice does not make it any the less a suitable basis for comparison. 

Example 1. Method of calculating Diagram Fig. 24. 
Assumed Data. 

Pa -Pb^ 2116 lbs. per square foot. Va = F^ =0. 

Pc—Pd- 18,000 lbs. per square foot. Capacity = 5 cu.ft. 

s = l. 
To obtain pwint C, 

.•. Vc^.SQ, Pc =18,000. 



84 ENGINEERING THERMODYNAMICS 

Intermediate points B to C are obtained by assuming various pressures and finding 
corresponding volumes as for Vc. 

Example 2. To compress and deliver 5 cu.ft. of air from atmospheric pressure (2117 
lbs. per square foot) to 8.5 atmospheres (18,000 lbs. per square foot) isothennally 
without clearance, how much work is necessary? 

P6=2116 Pc- 18,000 

Fft-5 

Vb Pc 

Work of admission ^-PtVb -2116 X5 - 10,585 ft.-lbs. 

p 
Work of compression =P6 Ft log* ~ - 10,585 Xlogr 8.5 =22,600 ft.lbs. 

Work of delivery =^PeVe = 10,585 ft.-lbs. 

Total work « 10,585 + 22,600 - 10,585 « 22,600 ft.-lbs. 

Or by the general formula, 

W = (sup.pr.)(L.P.Cap.) log* Rp =21 16 X5 xlog. 8.5 =2116 X5 X2.14 =22,652 ft.-lbs. 

Prob. 1. How many cubic feet of free air may be compressed and delivered i>er 
minute from 14 lbs. absolute to 80 lbs. per square inc^', absolute per horse-power in a 
compressor with zero clearance if compression is isothermal? 

Prob. 2. Oasis being forced through mains at the rate of 10,000 cu.ft. per minute 
under a pressure of 5 lbs. per square inch above atmosphere. The gas is taken into the 
compressor at atmospheric pressure and compression is isothermal. What horse- 
power will be needed at sea level and at an elevation of 5000 feet? 

Prob. 3. Natural gas is drawn from a well, compressed isothennally and forced through 
a main at the rate of 200,000 cu.ft. per hour measured at the pressure on the suction 
side. What steam horse-power will be required to operate the compressor if the 
mechanical efficiency be 80 per cent? Suction pressure is 8 lbs. per square inch 
absolute, delivery pressure 60 lbs. per square inch absolute. 

Prob. 4. A vacuum cleaning pump is required to maintain a pressure of 14 lbs. per 
square inch absolute, move 500 cu.ft. of free air per minute and discharge it against 
an atmospheric pressure of 15 lbs. per square inch absolute. What horse-power will 
be required (isothermal)? 

Prob. 6. A blower 6imishes 45 cu.ft. of air a minute at a pressure of 5 ins. of mercur3^ 
above atmosphere. Assuming compression to be isothermal and supply pressure to be 
atmospheric, what horse-power will be needed? 

Prob. 6. A compressor has a piston displacement of 3 cu.ft. At what speed can it be 
run if air be compressed isothennally from 1 to 10 atmospheres and the horse-power 
supplied is 100? 

Prob. 7. A tank of 1000 cu.ft. capacity contains air at atmospheric pressure. A 
compressor taking air from atmosphere compresses it isothermally and discharges it into 
the tank until the pressure reaches 100 lbs. per square inch gage. What horse-power 
will be required to fill tank at this pressure in ten minutes? 

Prob. 8. A compressor receives air at atmosphere and compresses it isothennally to 
five atmospheres above atmosphere. It takes in 1000 cu.ft. of free air per minute. 
How much would the capacity increase if the discharge pressure dropped to 3 atmos- 
pheres and the horse-power remained the same? 



WORK OF COMPBESSOES 85 

Prob. 9. Suppose that the pressure in the above problem were raised to 8 atmos- 
phrrea. How much would the capacity decrease if the horse-power remained the same 
and liow much more power would be required to keep the capacity the same? 

Prob. 10. By means of suitable apparatus, the water from the side of a waterfall is 
diverted to a vertical shaft, and in falling 126 ft. compresses air from atmospheric pressure 
to a value equal to 90 per cent of the head of the water. To deliver 1000 cu.ft. of com- 
pressed air per hour, how much wat«r is required if the work of falling water is 80 per 
ceut useful in compressing the air? 

4. Single-stage CompreBsor with Clearance, Isothennal Compression, 
(Cycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. 



"'jr " ^ 

Fia. 25. — One-Stage Compressor Cycle 2, Clearance, Isothermal. 

Referring to Fig, 25, the work of the cycle is, by areas. 
Net work &tc& = EBCG+G CDF 
-HAD F-EB AH 
=Area ABCD. 

It is easily seen that this area is also equal to {JBCL) — {JADL), both 
of which are areas of the form evaluated in the preceding section. Accordingly 



Net work a.TeA=JBCL-JADL, 



86 ENGINEERING THERMODYNAMICS 

Algebraically, 

W = P,V, log. ^-PaVa l0&^ 



=P»(F„-r,)log.^' (34) 



b 



which is the general expression for the work of the cycle in foot-pounds when 
pressures are in pounds per square foot, and volumes in cubic feet. Substituting 
the symbolic equivalents and using pressures in pounds per square inch, there 
results, since (Vt— Va) = (L. P. Cap.), 

Work = 144 (sup.pr.)(L. P. Cap.) log* Rp, . . . ' . (35) 

which is identical with Eq. (30), showing that for a given low-pressure capacity 
the work of isothermal compressors is independent of clearance. The value of 
the low-pressure capacity (Vt—Va) maj^ not be known directly, but may 
be found if the volume before compression, Vb, the clearance volume befor^ 
re-expansion, Vd, and the ratio of delivery to supply pressure, ftp, are known, 
thus 

from which 

(L. P. Cap.) = (F,-Ftfflp) (36) 

From Eq. (35) the work per cubic foot of low-pressure gas is, in foot-pounds, 

W 
(L.RCap.) = ^^^^^PP^'^^^g-^'- (3^^ 

and the work per cubic foot of high-pressure gas delivered, ft.-lbs. 

W 

^^p-^— r=144(sup.pr.)fiplogeffp (38) 

By comparison, Eqs. (37) and (38) are found to be identical with (31) and 
(32) respectively, since clearance haSy as found above, no effect on the work done 
for a given volume of ga^ admitted^ however much it mxiy affect the work of the 
cycle between given volume limits or work per unit of dispUicement, 

It is interesting to note that the work areas of Figs. 24 and 25 are equal 

when plotted on equal admission lines AB or delivery lines CD and any 
horizontal intercept xy will be equal in length on both if drawn at the same 

pressure. 

In what precedes, it has been assumed that AB represents admission volume 
and CD represents delivery volume which is true for these established cycles 



WORK OF COMPRESSORS 87 

of reference, but it is well to repeat that for real compressors these are only 
apparent admission and delivery lines, as both neglect heating and cooling 
effects on the gas during its passage into and out of the cylinder. Also that 
in real compressors the pressure of the admission line cannot ever be as high 
as the pressure from which the charge is drawn and the delivery pressure must 
be necessarily higher than that which receives the discharge, in which cases 
the volume of gas admitted, as represented by AB, even if the temperature 
did not change, would not equal the volume taken from the external 
supply, because it would exist in the cylinder at a lower pressure than it 
originally had, and a similar statement would be true for delivered gas. 

Problems. Repeat all the problems of the last sectioD, assuming any numerical 
value for the clearance up to 10 per cent of the displacement. 



6. Single-stage Compressor Isothermal Compression. Capacityi Volu- 
metric Efficiency, Work, Mean Effective Pressure, Horse-power and Horse- 
power per Cubic Foot of Substance, in Terms of Dimensions of Cylinder and 
Clearance. 

Consider first the case where clearance is not zero. Then Fig. 25 is the 
reference diagram. 

Let D= displacement = volume, in cubic feet, displaced by piston in one 
stroke = area of piston in sq.ft. X stroke in ft. = (V6— Vd). 
" (H. P. Cap.) = high pressure capacity = vol. cu.ft. of gas delivered per 

cycle at temperature equal to that of supply = (Fc—yd); 
*' (L. P. Cap.) = low pressure capacity = vol. in cu.ft. of gas entering 
cylinder per cycle = (F&— Va) ; 



ii 



„ , * • a: • L.P. Cap. Vt-Va 
E, = volumetnc efficiency = j;: — — = ^7 — W > 

U Vtt — Yd 



" CZ = volume of clearance, cubic feet= Fd 

" c = clearance volume expressed as a fraction of the displacement; 

whence Cl = cD) 



CI Vi 



W 
" M.E.P. = mean eflFective pressure, lbs. per square foot = y^; 

W 
" m.e.p. = mean eflFective pressure, lbs. per square inch = tttj^] 

" iV = number of revolutions per minute; 

" n= number of cycles per minute; 

N 
" 2= number of revolutions per cycle =—; 

n 

" I.H.P. = indicated horse-power of compressor; 



88 ENGINEERING THERMODYNAMICS 

The low-pressure capacity of the single-stage isothermal compressor with 
clearance is, 

(L.P.Cap.) = (Vft-ya), but Va=VaX^. 

Whence (L. P. Cap.) = { V'ft— Fd— I for which may be substituted the symbols 
for displacement and clearance volumes, thus 

(L. P. C8Lp.)=D+cD-'cDRp, 
• =D(l+c-cRp) (39) 

For convenience the term, Volumetric Efficiency, Ev is introduced. Since 
this is defined as the ratio of the low-pressure capacity to the displacement, 

E,JJ±F:J^-^^l+c-cR, (40) 

Referring to Eq. (35) it is seen that the value of (L. P. Cap.) can be 
substituted from Eq. (39) and the result is: 

Work per q/cfe, foot-pounds^ in terms of supply pressure^ pound spersq uare 
inch, displacement cubic feet, clearance as a fraction of displacement, and ratio 
of delivery to supply pressure is, 

TF = 144 (sup.pr.)D(H-c-cfip)log,/?,„ .... (41) 

or in the terms of the same quantities omitting clearance and introducing 
volumetric efficiency, Ev, 

W = lU{sup,i>T.) DEAogeRp, (42) 

To obtain the mean effective pressure for the cycle, the work done per cycle 
is divided by displacement, D. 

Mean eflFective pressure, pounds per square inch, 

W 

(m.e.p.) = 



1442>' 
whence 

(m.e.p.) = (sup.pr.)(H-c— cBp) log* JKp (43) 

or 

(m.e.p.) = (sup.pr.) J5» log« /2p . (44) 



WORK OF COMPRESSORS 89 

The indicated horse-power of the isothermal compressor is equal to the 

work per minute, in ft.-Ibs. divided by 33,000. If n cycles are performed 
per minute, then 



Wn _ 144n 
33000 ""33000 



I.H.P. = ^^t^Fip: = 5^K?^ (sup.pr.)i)(l +C - cRp)l0ge Rp 



= ^^gg^i>(l+c-ci2,)loge/2, (45) 

-^^^^DE. log. Rp^ (46) 

Introducing the effective area of the piston, in square inches, a, and the piston 
speed S, feet per minute, then since 

144 144 22 288z' 

'•«•''- w1^-'°^«' »'> 

The same expression for the indicated horse-power may be derived by the 
substitution of the value of (m.e.p.) Eq. (44) in the following general expres- 
sion for indicated horse-power. 

( m.e.p.)a ^ 
^•^•^- 33000X2«' 

Example !• Method of calculating Diagram Fig. 25. 
Assumed Data: 

Pa -Pb =2116 lbs. per square foot; 
Pd "Pc - 18,000 lbs. per square foot; . 
c =3 per cent. L.P. Cap. =5 cu.ft. « » 1. 

To obtain point D: 

From formula Eq. (39), L.P. Cap. -2)(l+c-cftp) or 5=Z)(l+.03-.03x8.5), 
or 

D =6.5 cu.ft. and CI. -.03 X6.5 = .195 or approximately .2 cu.ft. 

■ 

.'. Fd = .2 cu.ft., Pd = 18,000 lbs. sq.ft. 

To obtain point A : 

PaVa-PdVi or Fa=^7d-8.5X.2 = 1.7, 
.*. Fa -1.7 cu.ft., Pa =2116 lbs. sq.ft. 



# 



90 ENaiNEERING THERMODYNAMICS 

Intermediate points Z) to A are obtained by assuming various pressures and finding the 
corresponding volumes as for 7o. 
To obtain point B: 

Vt = Fa +5 = 6.7 cu.f t. Pb = 21 16 lbs. sq.ft. 

To obtain point C: 

DT7 DT7 T/ ^^^^ 2116X6.7 _ ,^ 

PcVc ^PbVby or Vc = -p— = ~Y8060~ " ^^ 
.*. Vc = .79 cu.ft., Pc = 18,000 lbs. sq.ft. 

Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 
atmospheres isothermally in a compressor having 4 per cent clearance. What must 
be the displacement work, per 100 cu.ft. of supplied and delivered air, and horse-power 
of machine? Speed is 150 R.P.M., compressor is double acting and stroke » 1 .5 diameters. 
Neglect piston rods. 

Z)=(L.P.Cap.)-5-J^, and ^, = (1 +.04 -.04x8.5) =.7, 
.'. D = 1000 ^ .7 = 1428 cu.ft. per minute. 

Work per cu.ft. of supplied air = (sup.pr.) 144 log« ftp = 144 X 14.7 loge Rp = 4530 ft.-lbs. ; 
Hence the work per 100 cu.ft. =453,000 ft.-lbs. 

Work per cu.ft. of deUvered air - 144(sup.pr.)Hp loge ftp = 144 X 14.7 x8.5 X2.14 =38,550 

ft.-lbs. 

Hence the work per 100 cu.ft. =3,855,000 ft.-lbs. 

■■H.P..^-.37.. 

^ = ^c/xwo = ^-76 cu.ft. per stroke. 

IoUa^ 

D = LX A = 1.5dXTCP = 1.18d3 =4.76. 

4 



-©*=■■■ 



Hence cylinder diameter = ( — -- ) =1.59 feet = 19.1 inches. 



Prob. 1. How many cubic feet of free air per minute may be compressed isother- 
mally to 100 lbs. per square inch absolute in a compressor having 6 per cent clearance 
if the horse-power supplied is 60? 

Prob. 2. A compressor has a cyUnder 18x24 ins., clearance 4 per cent, is double 
acting and runs at 150 R.P.M. If it compresses air from atmosphere to 100 lbs. per 
square inch gage, what will be its high- and low-pressure capacity, its horge-power, and 



WORK OF COMPRESSORS 91 

how will the horse-power and the capacity compare with these quantities in a 
hypothetical conipressor of the same size but having zero clearance? How will 
the horse-power per cubic foot of delivered air compare? 

Prob. 3. A manufacturer gives for a 10jxl2 in. double-acting compressor running 
at 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50-100 lbs. 
per square inch gage. What clearance does this assume for the lowest and highest 
pressure if the compression is isothermal? The horse-power is given as from 23 to 35. 
Check this. 

Prob. 4. Air enters a compressor cylinder at 5 lbs. per square inch absolute and is 
compressed to atmosphere (barometer =30i ins.). Another compressor receives air 
at atmosphere and compresses it to 3 atmospheres. If each has 6 per cent clearance 
what must be the size of each to compress 1000 cu.ft. of free air per minute, how will 
the total work compare in each machine, and how will the work per cubic foot of high 
and low pressure air compare in each? Assimie compression to be isothermal. 

Prob. 5. 1800 cu.ft. of free air per minute is to be compressed isothermally to a pres- 
sure of 100 lbs. per square inch gage. What must be the displacement and horse-power 
of a hypothetical zero clearance compressor, and how will they compare with those of a 
compressor with 6 per cent clearance? 

Prob. 6. Consider a case of a compressor compressing air isothermally from atmos- 
phere to 100 lbs. per square inch gage. Plot curves showing how displacement and 
horse-power will vary with clearance for a 1000 cu.ft. free air per minute capacity 
taking clearances from 1 per cent to 10 per cent. 

Prob. 7. Two compressors of the same displacement, namely 1000 cu.ft. per min- 
ute, compress air isothermally from 50 lbs. per square inch gage to 150 lbs. per square 
inch gage. One has 5 per cent clearance, the other 10 per cent. How will their capaci- 
ties and horse-power compare with each other and with a no clearance compressor? 

Prob. 8. A 9 Xl2 in. compressor is compressing air from atmosphere to 50 lbs. gage. 
How much free air will it draw in per stroke, and how much compressed air will it dis- 
charge per stroke for each per cent clearance? 

Prob. 9. The volumetric efficiency of a compressor is .95 as found from the indicator 
card. It is double acting and has a cylinder 18 X24 ins. What will be its capacity and 
required horse-power for 100 lbs. per square inch gage delivery pressure? 

6. Single-stage Compressor, No Clearance Exponential Compression, 
(Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures 
and Volumes. The cycle of the single-stage exponential compressor without 
clearance is represented by Fig. 26. Referring to work areas on this diagram, 



Net work 45CZ> = compression work EBCG 

-h delivery work GCDF 



—admission work FA BE. 



Algebraically, 



<— 1 



92 



ENGINEERING THERMODYNAMICS 



But 
or 



and 



PcVcVc'~'^P,V,V,'-\ 



•-1 



Pcv,=p,v,(^y '=p,v,(p^ ' 



»-i 



PcVc-P,V,=P,V,[{j,^ ' -l]. 



(HP. Cap 1 




Oold) •Vu.l'.t*! 




Hot) 






J - M 


'^ 1 ! 1 1 ' M 1 


1 


j 1 


-» ■ -4 - —^ 




r> i i/ i /^ 


1 1 1 


lomn t' i K_ _ ' C ^ i 


i_. , _ i 




< 1 1 1 ^ 


1 A 


' ' 1 




_L JL ' i 


I ; 1 ' 




il_ t ijui iir ^41 ■ 


11 III 1 




L JLV— 1 ; 


i ' I 




V^- u ' 


1 ' 1 ■ ' 




IT. ti-i ^41 






V I \ ' ' ' ! 


1 ! ~^ 1 ' 




1 \ ! 


i 1 ( 




1UiVl I l\ ' 


-. _^ ._ _^ _ 




^**^ \ \ 






t-l V \ 






^* Ji . JL, \]^ J 


' , , _. 




5 4J Ip \ -^ 


1 ^ -L- 






; 1 




1 IV ' V ' ' 


-.-. i . ^ ' J L 




8 ■ r\ - - - \ r- U 


. ' ' 1 




a \v -1 ^-4- 1 


I ' 




5 - l-V \>i. 


^ ' 1 




5*ifi¥¥i ; i \ 11 Wj 


1 1 




QQilXMU W \^ 


i 1 ^ 


fe WT (^ IT 


! 1 ' ' 


u 

• 


8. y ;_ W: 


! ' ' P 


■ ii X^ A. 




■9 _ X Jmk w i 


' i "< 


' I vi V 


t ' c 


9 Tx4- ^% ^ 


1 . 


£ 1 V^ aIIu 


i 1 




^ iJ-W. .. >^_t 


i 




a • k ' V t 


-r ^ -• -h 




8 T^ ^ -K-^ 


_i 11' 






1 ' 




H J '^' ^k. 


1 




m _iL ^ __r^ 


1 




1 H - - v,^ J- 


> t 1 






^y 




£ lF_ _ "'^^ 


s 




^ _4 -. 


. Ni^ 




1 
1. 


"^ "*>- "^^s 




I, 


'^■^ "**^ i ! 




iJOUU 1 ti 


'''T>4 "^">.J 




1 






-^i U T ' ■ 


1 ''t*t»4». ^]h*«»^.^ 




A .,.=. ^,„j^i .1-1 r -,-t;^- -j- 


L L ' Ll_ 1 ^^^ -Tg» M^ R 




^^ .1. ! 1 I 1 i 






1 1 1 


\- ~u' I 4- I -L i 




• i i 1 


r i± — ^ 




, t . ; 


1 I ',3 




r ir ^ ' 


J m T 1 cs 




F X-lS-i-j L— 


J-X t.y , 


1 


1 2 

Volumes 


345 

in Cubic Feet 





r*- 



D 



L.P. Cap- 



Fia. 26. — One-Stage Compressor Cycle 3, No Clearance, Exponential. 



Substituting above 



a-l 



»-i 



'^-f:S[©"-]+''-^-[©""-'] 



»-i 



■ -^.>'.[(&)"-]CV)- 



WORK OF COMPRESSORS 93 

Whence 



«-i 



'^%4iW[(g) • -l]. ....... (48) 

Eq. (48) gives the work in foot-pounds for the execution of the cycle when 
])ressures are in pounds per square foot, and volumes in cubic feet. 
The equivalent expression for pressures in pounds per square inch is 

W-l^^,J.y\{^y'-l] (49) 

When there is no clearance, as before, Vb represents the entire vol* 
ume of displacement, which is also here equal to the volume admitted 

(L. P, Cap.), Pb is the supply pressure (sup.pr.) pounds per square inch 

Vc 

absolute and — is the ratio of delivery to supply pressure, Rp. 

Accordingly, the work of an exponential, singlenstage compressor with no 
clearance is 



• -1 



}r=144^(sup.pr.)(L.P.Cap.)(ft/ -l) (50) 

The work per cubic feet of low pressure gas, foot-pounds is 

• -1 



W 



—^ = 144^ (sup.pr.) (ft/ -l) (51) 



(L. P. Cap 

Before obtaining the work per cubic foot of high-pressure gas, it is neces- 
sary to describe two conditions that may exist. Since the exponential com- 
pression is not isothermal, it may be concluded that a change in temperature 
will take place during compression. This change is a rise in temperature, 
and its law of variation will be presented in another chapter. 

1. If the compressed air is to be used immediately, before cooling takes 
place, the high-pressure capacity or capacity of delivery will be equal to the 
volume at C, Ve and may be represented by (H. P. Cap. hot). 

2. It more commonly occurs that the gas passes to a constant-pressure 
holder or reservoir, in which it stands long enough for it to cool approximately 
to the original temperature before compression, and the volume available 
after this cooling takes place is less than the actual volume discharged from 
the cylinder in the heated condition. Let this volume of discharge when 
reduced to the initial temperature be represented by (H. P. Cap. cold) which 
is represented by Vt, Fig. 26. 



94 ENGINEERING THERMODYNAMICS 

Since B and C in Fig. 26 lie on the exponential compression line, P^Vb' =P«V/, 



-<w- 



or 

(L. P. Cap.) = (H. P. Cap. hot) {Rp)T (52) 

Hence, the work in foot-pounds per cubic foot of hot gas delivered from 
compressor is 

On the other hand, B and K lie on an isothermal and PbVh^PtVt, or 
since Ph—Pc, 

whence 

(L. P. Cap.) = (H. P. Cap. cold)flp. * (M) 

The work foot-pounds per cubic foot of gas cooled to its original tempera- 
ture is, therefore, 

^jj_^____. = 144--(8up.pr.)iep(^/e, . -Ij, . . . . (55) 

or 

W s / ml \ 

This last equation is useful in determining the work required for the storing 
or supplying of a given amount of cool compressed air or gas, under conditions 
quite comparable with those of common practice. 

Example 1. Method of calculating Diagram Fig. 26. 
Assumed Data: 

Pa = Pft = 21 1 6 lbs. per square foot ; Pe=Pd^ 18,000 lbs. per square foot. 
CZ=0; Va = Vd=0; L, P. Capacity =5 cu.ft.; « = 1.4 (adiabatic value of s). 



To obtain point C: 



X 

PcVc^-^^PtVt^'^ or Fc = F6-^(^)^■* 



Pc/P* = 8.5; loge8.5=.929, and .71 log. 8.5 = .665; (Pc/Pt)!* =4.6, 



WORK OF COMPRESSORS 95 

hence Fc =5 4-4.6 = 1.09 cu. ft. Pc = 18,000 lbs. per sq.ft. 

Intermediate points B to C are obtained by assuming various pressiues and finding 
the corresponding volumes as for 7c. 

Example 2. To compress 5 cu.ft.of air from atmospheric pressure (2116 lbs. per 
square foot) to 8.5 atmospheres (18,000 lbs. per square foot) adiabatically and with no 
clearance requires how many foot-pounds of work? 

Pft =2116 lbs. sq.ft., Pe = 18,000 lbs. sq.ft., 

7^=5 cu.ft. 

Vb 
7^= 7P\ .71 =5^4.57 = 1.092 cu.ft. 



Work of admission is 

PftFft =2116 X5 = 10,585 ft.-lbs. 

Work of compression, using y to represent the adiabatic value of s is, 

F^l (fI) "^ -l] --^[(«-5)" -11 =^^ ><•««' -22,350 ft.-lbe. 

Work of delivery is 

PeVe = 18,000 X 1 .092 = 19,650 ft.-lbs. 

Total work -19,650+22,350 -10,585 =31,425 ft.-lbs., 

or by the formula Eq. (50) directly 

W = 144^(sup. pr.) (L. P. Cap.) {Rp~ - 1.) 

= 144+3.46x2116x5x[(8.5)-2»-l]; 

= 144+3.46 X21 16 X5 X. 86 =31,450 ft.-lbs. 

Prob. 1. A single-stage zero clearance compressor compresses air adiabatically from 
1 to 6 atmospheres. How many cubic feet of free air per minute can be handled if the 
compressor is supplied with 25 H.P. net? 

Prob. 2. The same compressor is used for superheated ammonia under the same 
pressure conditions. For the same horse-power will the capacity be greater or less 
and how much? 

Prob. 3. A dry-vacuum pump receives air at 28 ins. of mercury vacuum and delivers 
it against atmospheric pressure. What will be the work per cubic foot of low-pressure 
air and per cubic foot of high-pressure air hot? Barometer reads 29.9 ins. 



96 ENGINEERING THERMODYNAMICS 

Prob. 4. The manufacturer gives for a lOl X 12 in. double acting compreesor run- 
ning at 160 R.P.M., a capacity of 177 cu.ft. of free air per minute and a horse-power of 
25 to 35 when delivering against pressures from 50 to 100 lbs. Check these figures. 

Prob. 6. A set of drills, hoists, etc., are operated on compressed air. For their opera- 
tion 3000 cu.ft. of air at 70 lbs. gage pressure are required per minute. What must be 
the piston displacement and horse-power of a compressor plant to supply this air if 
compression is adiabatic and there is assumed to be no clearance? 

Prob. 6. Air is compressed from atmosphere to 60 lbs. per square inch gage by a 
compressor having a 12x18 in. cylinder and running at 100 R.P.M. Find its capacity 
and horse-power at sea level and loss in capacity and horse-power if operated at an alti- 
tude of 10,000 ft. for zero clearance. 

Prob. 7. 10,000 cu.ft. of free air per minute at a pressure of 15 lbs. above atmos- 
phere are compressed and delivered by a blowing engine. Find the horse-power required 
to do this and find how much free air could be delivered by same horse-power if the 
pressure were tripled. 

Prob. 8. In a gas engine the mixture of air and gas is compressed in the cylinder 
before ignition. If the original pressure is 14 lbs. per square inch absolute^ final 
pressiu'e 85 lbs. absolute and compression is adiabatic, what will be the work of 
compression only, per pound of mixture? 

Note: Weight per cubic foot may be taken as .07 and y as 1-38. 

Prob. 9. A vacuum pump is maintaining a 25-in. vacuum and discharging the air 
removed against atmospheric pressure. Compare the work per cubic foot of low pres- 
sure air with that of a compressor compressing from atmosphere to 110 lbs. above atmos- 
phere. 

7. Single-stage Compressor with Clearance^ Exponential Compression, 
(Cycle 4). Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. When clearance exists in the cylinder, it is evident that a 
volume equal to the clearance, Va, will not be expelled during the delivery 
of compressed gas, and this volume will expand with fall in pressure as the 
piston returns, causing pressure-volume changes represented by the line DA 
on the diagram, Fig. 27. Until the pressure has fallen to that of supply, the 
admission valve will not open, so that while the total volume in the cylinder 
at end of admission is Vbj the volume Va was already present by reason 
of the clearance, and the volume taken in is (Vb^Va) which is the low-pressure 
capacity (L. P. Cap.). 

The work area of the diagram is ABCD, which may be expressed as 



Work QTeA^^JBCL-JADL, 

which areas are of the form evaluated in Section 6. Hence, the above expression 
in algebraic terms is 



WOEK OF C0MPEESS0B9 97 

This is the general expression for the work of the cycle, in foot-pounda, when 
the pressures are expressed in pounds per square foot, and volumes in cubic 
feet. Using symbolic equivalents 

W = 144-4T{8uppr}(L-P-Cap.)[(ffp)^'-l,l .... (58) 



Fig. 27. — One-Stage Compressor Cycle 4, Clearance, E^xponential. 

Eq. (58) is identical with Eq. (50), showing that for adiahatic as for 
'sothermal compressors, Ike wort ■ done for a given low-pressure capcuHly is inde- 
pendent of clearance. Due to this fact, the expressions derived for the expo- 
nptitial compressor without clearance will hold for that with clearance: 

Work, in foot-pounds per cubic foot of low-pressure gas is, 

W 



(L:prcap.)-"ta(^"PP^-^^'^'' -i> ■ 

Work, in foot-pounds per cubic foot of hot gas delivered is, 



(H. P. Cap. hot) 



■144^(sup.pr.)fi; (r,' -l). 



98 ENGINEERING THERMODYNAMICS 

Work, in foot-pounds per cubic foot of cooled gas to its original temperature Is, 

• -1 



W 



(H. P. Cap. cold) 



= 144^(del.pr.)(ftp' -l) (61) 



The relation of high-pressure qapacity either hot or cold to the low-pressure 
capacity is also as given for the case of no clearance, as will be shown. 

In Fig. 27, the high-pressure capacity, hot, is DC=Ve'-Va* The low- 

— — 11 1 i 

pressure capacity is AB = Vb— Fa, but VcPe* — F^P^• and FdPd* = FoPa», or 

1 1 

Fft = VcRp" and Va = VJip' . 

' 1 
Hence (L. P. Cap.) = (H. P. Cap. hot)ftp« (62) 

If the delivered gas be cooled to its original temperature, then the volume 
after delivery and cooling will be 

(H. P. Cap. cold) = ^^- P- ^^P-) , 

or 

(L. P. Cap.) = (H. P. Cap. cold)B„ .... (63) 

From the work relations given above, it is seen that in general, the work 
per unit of gas, or the horse-power per unit of gas per minute is independent 
of clearance. 

8. Single-stage Compressor Exponential Compressor. Relation between 
Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, Horse- 
power and H.P. per Cubic Foot of Substance and the Dimensions of Cylinder 
and Clearance. As indicated on Fig. 27,. for the single-stage exponential com- 
pressor with clearance, the cylinder displacement D, is {Vb—Vd)> The low- 
pressure capacity per cycle is (L. P. Cap.) = (F&— Fa). The actual volume of 
gas or vapor delivered by the compressor is (H. P. Cap. hot) = (Fc— F^). 
This is, in the case of a gas at a higher temperature than during supply, 
but if cooled to the temperature which existed at B will become a less volume. 
This delivered volume after cooling is symbolized by (H. P. Cap. cold) and is 

/T ^ i-. X (sup.pr.) (L. P. Cap.) , ^ . , 
equal to (L. P. Cap.) X -77-; : or where Rp is the ratio of delivery 

(del.pr.) Rp ^ 

pressure to supply pressure. 

Volumetric efficiency, Evj already defined as the ratio of low-pressure 
capacity to displacement is 

„ Vb-V g (L. P. Cap.) 



WORK OF COMPRESSORS 99 

Clearance, e, expressed as a fraction of the displacement is the ratio of 
clearance volume, CI, to displacement, D, and is, 

CI Va 
c= 



Mean effective pressure, pounds per square foot (M.E.P.), is the mean height 
of the diagram or the work area W, divided by displacement, D. If expressed 
in pounds per square inch the mean effective pressure will be indicated by 

(m.e.p.) = jj4^. 

Let (I.H.P.) be indicated horse-power of the compressor; 
N the number of revolutions per minute; 
n the number of cycles per minute and 
z the number of revolutions per cycle, whence nXz^N. 



a 



Then, the low-pressure capacity is 

(L. P. Cap.) = (F»-7«). 
But 



Va^VaX 



w- 



since the re-expansion DA is exponential and similar to compression as to 
value of «, whence J 

1 
(L. P. Cap.) = (V*-Va) = y»-7*Bp' ; 

1 
^D+Cl-VdRp^; 

=D+cD-cDRp^ ; 



or 



(L. P. Cap.) = D (l + c - cBp"« ) 



(64) 



From this, by definition, the volumetric eflBciency is 



Referring to Eq.(67), in which may be substituted the value Eq. (64) for 
(Fi— Fd), the work of the single-stage exponential compressor in terms of dis- 



iOO ENGINEERING THERMODYNAMICS 

placement, clearance (as a fraction of displacement), and pressures of supply 
and delivery in pounds per square foot is, 

"^-^M'+'-wwr -'] <*' 

or using pressures, pounds per square inch, and inserting the symbols, this 
may be stated in either of the following forms: 

TF=144~(sup.prOD(l+c-c72pr)rft/-^-ll. . . . (67) 
= 144-^ (sup.pr.)D£'«[ftpV-l] (68) 

The mean effective pressure in pounds per square foot is this work divided 
by the displacement, in cubic feet, and may be converted to pounds per square 
inch by dividing by 144, whence 

Mean effective pressure, pounds per square inch, 

(m.e.p.) = - — (sup.pr.)(l +c—c/?p« )hRp • — 1 > • . (69) 



5 
5-1 



(sup.pr.)J^Jflp^-ll (70) 



The indicated horse-power of the single-stage exponential compressor from 
(67) is, 

T TT P - ^'^ - ^_ (suP-P rQnPE , r ^^__ J /7,x 

^•^•^•""33000""s~l 229.2 " [^'^ ' ij . . . K^ i) 

Where n is the number of cycles per minute, or in terms of piston speed S and 
effective area of piston, square inches, and z the number of revolutions per 
cycle, 

I-H.P.-^— 1 ^^^^^ [iep • -IJ (72) 

Since it was found in Section 7, that the work per unit volume of gas is the 
same with clearance as without clearance, the horse-power per cubic foot per 
minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ). 

Horse-power per cubic foot of gas supplied per minute 

I.H.P. 



n(L. P. Cap 



__ s isup.pjg r izi_ ] 

.)~s~i 229.2 L ^J • • • • ^^^^ 



WORK OF COMPRESSORS 101 

The horse-power per cubic foot of hot gas delivered per minute is 

IH.P. s_ Isup.prO^ 1 r 1^1 1 . 

n(H.P.Cap.hot)~8-l 229.2 ^ [^' ' ^J • • • U*; 

Horse-power per cubic foot of gas delivered and cooled is 

I.H.P. « (sup.pr.) r o Izi 



[fip • -l|, . . . (75) 



n(H. P. Cap .cold) s-l 229.2 ^'^ ^' ' 
_ » (d el.pr.) [ff^ -] «„v 

-pi~"229:2~L^ ~ J ^ ^ 

In the above formulae (del.pr.) and (sup.pr.) indicate delivery pressure and 
supply pressure, in pounds per square inch. 

Example 1. Method of calculating Diagram, Fig. 27. 

Assumed data: 

Pa =^6 «*2116 lbs. per square foot. 

Pc=Pd- 18,000 lbs. per square foot. 

CI. =3.5 per cent. L. P. Capacity =5 cu.ft. « = 1.4. 
To obtain point D: 

L. P. Cap.=i>(l+c-c/^p7) or 5 =Z)(l +.035 -.035(8.5)) ; 

Hence 

D -5 -5- (1 +.035 -.035 X4.6) =5.72 cu.ft. and CI = .035 X5.72 = .2 cu.ft 

/. Vd = .2 cu.ft. ; Pa = 18,000 lbs. sq.ft. 



.716 



To obtain point A : 



-m^'y- 



= 4.6X.2=.92; 
.'. 7a = .92 cu.ft.; Pa =2116 lbs. sq.ft. 

Intermediate pK)ints i) to A are obtained by assuming various pressures and finding the 
corresponding volumes as for Va- 

To obtain point B: 

Vo = Va+L, p. Cap. = .92 +5 =5.92, 

/. Vt =5.92 cu.ft. ; P* = 2116 lbs. sq.ft. 



102 ENGINEERING THERMODYNAMICS 

To obtain point C: 



Fc-n 



AT 



=5.92 +4.6 = 1.29 cu.ft. 
.". Fc = 1 .29 cu.ft. ; Pc = 18,000 lbs. sq.ft. 

Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 
atmospheres absolute so that «» 1.4, in a compressor having 4 per cent clearance. . What 
must be the displacement of the compressor, work per 100 cu.ft. of supplied and delivered 
air, hot and cold, and horse-power of machine? Speed is 150 R.P.M., compressor is 
double acting and stroke = 1.5 diameters. 

D = L. P. Cap. •^^», and Ev = (l -f c -cfi,7 J . 

.'. ^,= (l+.04-.04x(8.5)-^^) =.86; 
.-. D = 1000 + .86 = 1 162 cu.ft. per min. 

8 — 

Work per cubic foot of supplied air = 144 (sup.pr)[/2p « ■" ^l* 

= 144 X3.46 X 14.7 X.86 =6300 ft.-lbs. 
.*. Work per 1000 cu.ft. = 6,300,000 ft.-lbs. 



Work per cubic foot of delivered air cold is Rp times work per cubic foot of supplied air, 
hence work per 100 cu.ft. of delivered cooled air is 5,350,000 ft.-lbs. 

i. 
Work per cubic foot of delivered air hot \a Rp t times work per cubic foot of suppUed air, 

hence work per 100 cu.ft. of hot delivered air is 2,800,000 ft.-lbs. 

«^ --r -n. (m.e.p.)ajS , , .. s—l. ._ r_ — =■ ^1 

I.H.P.- r^ ; 2-i; (m.e.p.) — - (8up.pr.)S,[B, *-^-l,\ 



66,0002 ' " ' *^' « 



or 



(m.e.p.) -3.46x14.7 x.86 X.86 -37.7 lbs. per square inch. 

a-f ; ^-150x2xif ; ^-^-1162; . 

.*. ii«=5690 or <i = 17.85. 
o =260 sq-inches. S =670 ft. per min. .'. I.H.P. = 191, 



WORK OF COMPRESSORS 103 

Prob. 1. A denBe-air ice machine requires that 4000 cu.ft. of air at 50 lbs. per square 
inch absolute be compressed each minute to 150 lbs. per square inch absolute. The 
compression being such that « = 1.4, clearance being 6 per cent, find the work required. 
What would be the work if clearance were double? Half? 

Prob. 2. The compressor for an ammonia machine compresses from one atmos- 
phere to 8 atmospheres absolute. With adiabatic compression and 4 per cent clear- 
ance, what will be work per cubic foot of vapor at the low pressure and at the high? 
Assume vapor to be superheated. 

Prob. 3. On a locomotive an air-brake pump compresses air adiabatically from 
atmosphere to 80 lbs. per square inch gage. It is required to compress 50 cu.ft. of free 
air per minute and clearance is 5 per cent. What horse-power must be supplied to it? 

Prob. 4. In a manufacturing process a tank must be maintained with a vacuum of 
29 ins. when barometer reads 30 ins. To do this 100 cu.ft. of carbon dioxide must be 
removed from it per minute and returned under atmospheric pressure. Compression is 
adiabatic and clearance 7 per cent. How much power must be supphed to compressor 
and what should be its displacement? 

Prob. 5. Two compressors each 12x18 in., double acting, with 8 per cent clear- 
ance, and running at 150 R.P.M. compress in the one case air, in the other carbon disul- 
phide. The compression being adiabatic in each case, what (a), is the difference in 
power required, (6), in low-pressure capacities? Take pressures as 2 and 15 atmos- 
pheres of 26 inches mercury. 

Prob- 6. A compressor is supplied with 40 horse-power. If it draws in air from 
atmosphere to what pressure can if 500 cu.ft. per minute be compressed, when 5 = 1.38 
and clearance 10 per cent? 

Prob. 7. For forcing gas through a main, a pressure of 50 lbs. per square inch gage 
is required. What is the work done per cubic foot of high-pressure gas, if a compressor 
having 6 per cent clearance is used, and s for the gas is 1.36? What should be its dis- 
placement? 

Prob. 8. A gas compressor 20x22 ins. has a volumetric efficiency cf 90 per cent, 
supply pressure =4 lbs. per square inch and delivery 110 lbs. per square inch gage. 
WhsA are its L. P., H. P., hot and cold capacities, and its I.H.P. if single acting at 70 
R,P.M. when* = 1.35? 

9. Two-Stage Compressor, no Clearance, Perfect Intercooling, Exponential 
Compression, Best Receiver Pressure, Equality of Stages (Cycle V). Work and 
Capacity in Terms of Pressures and Volumes. The common assumption in con< 
sideringthe multi-stage compressor is that in passing from one cylinder to the 
next, the gas is cooled to the temperature it had before entering the com- 
pressor, which has already (Section 2), been defined as " perfect intercooling." 

This condition may be stated in other words by saying that the product 
of pressure and volume must be the same for gas entering each cylinder. If 
then the voliune and pressure of gas entering the first stage be determined, 
fixing the volume entering the second stage will determine the pressure of the 
gas entering the second stage, or fixing the pressure of the gas entering the 
second stage will determine the volume that must be taken in. 

Using subscripts referring to Fig. 28, for the no clearance case, 

P!,V,=PaVa (77) 



104 .ENGINEERING THERMODYNAMICS 



The net work of the compressor, area ABCDEF=BTesk ABCH first stages- 
area HDEF second stage. Using the general expression, Eq. (48) for these 
work areas with appropriate changes in subscripts 



TT^^P^yl^^^) ' -ll. . (first stage) 



•-1 



8 



37j Pdl^d I ( p^ 1 ■" M • • (second stage) 



But from the above, and since Pc—Pd, 

•-1 «-i 



-^Mi^r <9r -^\ 



(78) 



which is the general expression for work of a two-stage compressor without 
clearance, perfect intercooling, and may be restated with the usual s3anboLs as 
follows; 



TF = 144-^-:~ (sup.pr.)(L. P. Cap.)r(/2pi)"-~+(/i;,2) V-2I , . 



(79) 



in which (Rpi) and Rp2) are the ratios of delivery to supply pressures for the 
first stage and for the second stage respectively. From Eq. (79), work per 
cubic foot of gas supplied is, 

^j--^»^ = 144--^(8up.pr.)[fl,x-+ff,2--2j . . (80) 

Work per cubic foot of gas discharge and cooled to its original temperature is 

W 



(H. P. Cap. cold) 



8 [ 'si} iiJ 1 

= 144-^^(sup.pr.)i?p iZpi • +Rp2 • ""2 , 



= 144^-j(del.pr.)rflpiV+/?p2'' ^-2J 



(81) 



The low-pressure capacity stated in terms of high-pressure capacity hot, 

as actually discharged is 

1 

(L.P. Cap.) = (H.P. Cap. hot)ftp2'"ftpi, (82) 

whence 

Work per cubic foot hot gas discharged 

W 



(H. P. Cap. hot) 



« 1 r J.-1 #-1 1 

= 144-^--(sup.pr.)ftp2^/epi ftpi r+Rj,2 • -21 . (83) 



WORK OF COMPRESSORS 



105 



Examination of Fig. 28 will show without analysis that there must be some 
best-receiver pressure at which least work will be required. For if tJie receiver 
pressure approached Pj, then the compression would approach single stage and 





;3 



> 



a 




o 
P-t 






i 

O 
o 

« 
O 

O 



t-4 

B 
o 

O 



^ooj ourenbs jad spunod ^1 SQ^nssaij 



cc 

(N 



the compression line approach BCG. The same would be true as the receiver 
pressure approached Pg = Pe, whereas at any intermediate point C, intercooling 
causes the process to follow BCDE with a saving of work over single-stage 
operation represented by the area DCGE. This area being zero when C is at 



106 ENGINEERING THERMODYNAMICS 

« 
either B or (7, it must have a maximum value somewhere between, and the 
pressure at which this least-compressor work will be attained is the best-receiver 
pressure. 

By definition the best-receiver pressure is that for which TF is a minimum, 
or that corresponding to 

dPc 

Performing this differentiation upon £q. (78), equating the result to zero, 
and solving for Pc, 

(Best rec.pr.) = (P»P.)M(sup.pr.) (del. pr)]* (84) 

Substituting this value in the general expression for work Eq. (78), notii^ that 

pr~pr [pj''''^pr[pj 

jt-i 
F=2::^A7.[(^;) '• -l], (85) 



s-r 



Eq. (85) is the general expression for two-stage work with perfect inter- 
cooling at best-receiver pressure in terms of pressures and volumes. Sub- 
stituting the symbols for the pressures and volumes and noting that as in 
Cycle 1, 

75=(L. P. Cap.) and 7.= (H. P. Cap. hot) and using.(/2,) for (^\ 

W = 288—% (sup.pr.) (L. P. Cap.) (ftpV - 1) . . . ' . . (86) 

This equation gives the same value as Eq. (85), but in terms of different units. 

It should be noted here that the substitution of best-receiver pressure in the 
expressions for the two stages preceding Eq. (78), mil show tfiat the work done 
in the two cylinders is equal. 

The work per cubic foot of low-pressure gas, from Eq. (86) is, 



W 



(L.P.Cap.) 



7) =288—^ (sup.pr.) [/jpSr-l] (87) 



To transform Eq. (85) into a form involving delivery volmnes, use the rela- 
tion from the diagram, 

1 1 



y-<W-<%{W- 



WORK OF COMPRESSORS 107 

Whence 



-'-<mw- 



which for the best-receiver pressure becomes 



«+i 



Fft=FJ2p2*. 



bubstituting in Eq. (85), 



W^2j^P,V.(Rp)^[RM-i\ (88) 



Introducing the S3miboIs, 



TF=288~(sup.pr.)(H.P. Cap. hot)flp^rftpV--ll, . . (89) 



• and 

W 



(H. P. Cap. hot) 



g a + ir .-1 -I 

=288-^(sup.pr.)ffp~2r[iJ,-27-lJ ... (90) 



The voliune of gas discharged at the higher pressure when reduced to its 
original temperature will become such that 

(L.P/Cap.) ^P, 
(H. P. Cap. cold) Pr^' 

or 

(sup.pr.) (L. P. Cap.) = (del.pr.) (H. P. Cap. cold), . . . (91) 

which may be substituted in Eq. (86), 



Tr= 288-^ (del.pr.) (H.P. Cap. cold) 1 /ZpV - 1, 1 . . . . 
from wnich the work per cubic foot of gas delivered and cooled is, 



(92) 



IF 



(H.P. Cap. cold) 



= 288j^ (del.pr.) j/2pV^ - 1 1 



(93) 



108 ENGINEERING THERMODYNAMICS 

Example 1. Method of calculating diagram, Fig. 28. 
Assumed data: 

Fa =0 cu.ft. Pa =2116 lbs. per square foot. 

F/=0 cu.ft. Pc^Pa^VPaPe =6172 lbs. sq.ft. 
F6=5 cu.ft. •P/=P,=P^= 18,000 lbs. sq.ft. 

« = 1.4. 



To obtain point C: 



or 



To obtain point Z>: 



To obtain point E\ 



1 
Fc = 7* + {-) ^'^ =2.36 cu.ft.; 



.*. Vc = 2.36 cu.f t. Pc = 6172 lbs. sq.ft 



-_ _, Pb - 2116 ^ -_, ,^ 
Fd = Fft X o- =5 X - - = 1.71 cu.ft. 

re Ol7J 

/. Fd = 1.71 cu.ft. Pd =6172 sq.ft. 



v,^v..(y;)h 



but by definition 



(£)"-(?t)"— . 



hence, 

Fe = 1.71 H-2.14 =.8 cu.ft. Pe = 18000 lbs. sq.ft. 

Example 2. To compress 5 cu.ft. of air from one atmosphere (2116 lbs. per square 
foot) to 8.5 atmospheres (18,050 lbs. per square foot) in two stages with best-receiver 
pressure and perfect intercooling requires how much work? 

FT =288 --(sup.pr.)(L. P. Cap.)(/2p ^' -1), 

5 — 1 

(sup.pr.) « 14.7. (L. P. Cap.) =5. Rp =8.5. 
.'. TF=288x3.463xl4.7x5x(8.5~2^-l) =26,800 ft.-lbs. 



WORK OF COMPRESSORS 109 

Prob. L Air at 14 lbs. per square inch absolute is compressed to 150 lbs. per square 
inch absolute by a two-stage compressor. What will be the work per cubic foot of air 
delivered? What will be the work per cubic foot if the air be allowed to cool to the 
original temperature, and how will this compare with the work per cubic foot of sup- 
plied air? Best receiver-pressure and perfect intercooling are assumed for the above 
compressor, s = 1.4. 

Prob. 2. A compressor receives air at atmosphere and compresses it to half its vol- 
ume, whereupon the air is discharged to the cooler and its temperature reduced to the 
original point. It then enters a second cylinder and is compressed to 80 lbs. absolute. 
What will be the work per cubic foot of supplied air in each cylinder and how will 
the work of compressing a cubic foot to the delivery pressure compare with the work 
(lone if compression were single stage, compression being adiabatic. 

Prob. 3. Air is to be compressed from 15 lbs. per square inch absolute to 10 times 
this pressure. What would be the best-receiver pressure for a two-stage compressor? 
How many more cubic feet may be compressed per minute in two stage than one stage 
by the same horse-power? 

Prob. 4. A manufacturer sells a compressor to run at best-receiver pressure 
when (sup.pr.) is 14 lbs. per square inch absolute and (del.pr.) 100 lbs. per square inch 
absolute. What will be the work per cubic foot of supply-pressure air done in each cylin- 
der? Another compressor is so designed that the receiver pressure for same supply 
pressure and delivery pressure is 30 lbs. per square inch absolute, while a third is so 
designed that receiver pressure is 50 lbs. per square inch absolute. How will the 
work done in each cylinder of these machines compare with that of first machine? 

Prob. 5. For an ice machine a compressor works between 50 and 150 lbs. per square 
inch absolute. It is single stage. Would the saving by making compression two stage 
at best-receiver pressure amount to a small or large per cent of the work in case of single 
stage, how much? 

Prob. 6. A compressor has been designed to compress 1000 cu.ft. of carbon dioxide 
per minute from 15 to 150 lbs. per square inch absolute. What horse-power will be 
required at best-receiver pressure? Should delivery pressure change to 200 lbs., what 
power would be required? To 100 lbs. what power? 

Prob. 7. A -gas-compressing company operates a compressor which has to draw 
COs gas from a spring and compress it to 150 lbs. per square inch gage. In the morning 
pressure on the spring is 10 lbs. gage, while by evening it has dropped to 5 lbs. absolute. 
If the compressor was designed for the first condition, how will the high-pressure 
capacity cold and horse-power per cubic foot of high-pressure gas at night compare 
with corresponding values in morning? Assume a barometric reading. 

Prob. 8. On a mining operation a compressor is supplying a number of drills and 
hoists with air at 150 lbs. per square inch absolute, the supply pressure being 14 lbs. 
Wliat will be the difference in horse-power per cubic foot of delivered air at compressor 
and per cubic foot received at drills if air is a long time in reaching drills? 

Prob. 9. With a best-receiver pressure of 40 lbs. per square inch absolute and a 
supply pressure of 14 lbs. per square inch absolute, what horse-power will be required 
to compress and deliver 1000 cu.ft. of high-pressure air per minute at the delivery 
pressure for which compressor is designed and what is that delivery pressure? 

10. TVo-stage Compressor^ with Clearance, Perfect Intercooling Expo- 
nential Compression, Best-receiver Pressure, Equality of Stages, (Cycle 6). 
Work and Capacity in Terms of Pressures and Volumes. The two-stage expo- 



no ENGINEERING THEEM0DTNAJUC8 

nential compressor with clearance aad perfect intercooling is represented by 
the PV diagrams Figs. 29, 30, 31, which are clearly made up of two aii^e-stage 
compression processes, each with clearance. 




ij 



looj SMnoe Ma oq-i □! gMnww J 



Applying Eq. (57) to the two stages and supplying proper subscript, 
referrii^ to Fig. 29, 



WORK OF COMPRESSORS lU 

'^^1^'^^^''^^'[{f) ' "^]* • (second stage) 

If the condition of perfect intercooling be imposed, it is plain that since 
the weight of gas entering the second stage must equal that entering the first 
stage, and the temperature in each case is the same, 

and noting also that 

Pc^Pd, 



w.^^p^v.-y.my\{^y' -2], . 



■ ■ (M) 



Eq. (94) is the general expression for the work of two-stage exp>onential 
compressor with perfect intercooling, Pc being the receiver pressure. 

p 

As in Section 9, let (Rpi) be the pressure ratio -^ for the first stage eLnd{Rp2) 

Pb 
p 

the pressure ratio -^ for the second stage and using instead of P» its equivalent 

Pc 

144 (sup.pr.) lbs. per square inch. 



Tr=144^(sup.pr.)(L.P.Cap.) [(fl,i)VV(iJp2V-2,l . . 



(95) 



which is identical with (79), showing that far tvxhstage compressors vrith perfect 
intercooling {as for single stagey Section 7), the work for a given low-pressure 
capacity is independent of clearance. 

The work per cubic foot of gas supplied is given by Eq. (80); per cubic 
foot of cold gas delivered by Eq. (81) and per cubic foot hot gas delivered by 
Eq. (83). 

The reasoning regarding best-receiver pressure followed out in Section 9, will 

dW 
hold again in this case, and by putting -f^^O in Eq. (94), and solving for Pe 

a±c 

it will again be found that best-receiver pressure will be 

(best-rec. pr.) = (P6P,)* (96) 

Substitution of this value for Pe in Eq. (94), gives the following expression 
forwork of the two-stage exponential compressor with best-receiver pressure, 

Tr-2^P»(7»-F.)[(g)''-l], (97) 



112 ENGINEERING THEBMODYNAMICS 

which may be exprf^ssed in terms of supply pressure, pounds per square inch 
low-pressure capacity, cubic feet, and ratio of compression, 




1 S. 



o I 



1oo£ oraniiB lad wfi ui eainBeajj 



which is the same as Eq. (86). 

Substitution of the value of best-receiver pressure in the expression for the 



wore: of compressors 113 

work of the two stages separately will show the equality of work done in the 
respective stages for this case with clearance. 

Work per cubic foot gas supplied to compressor is 

. (L7^) = 288^j(sup.pr.)[7i,'ii-'-l] (99) 

Work per cubic foot of high-pressure gas hot is 

W 



(H. P. Cap 



:h^) = 288^^(sup.pr.)ft, 2. l/2p2. -ij. . . 



(100) 



The work per cubic foot of air delivered and cooled to its original tem- 
perature iSy 

Due to the fact that clearance has no effect upon the work per cubic foot 
of substance, as previously noted, Eqs. (99), (100) and (101) are identical with 
(87), (90) and (93). 

11. Two-Stage Compressor, any Receiver Aressure, Exponential Compres- 
sion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and 
Horse-power, in Terms of Dimensions of Cylinders and Clearances. Referring to 
Fig. 29, let Di be the displacement of the first stage cylinder in cubic feet = 
(Vft— Ft), i>2 the displacement of the cylinder of the second stage in cubic 
feet = (Fd— V/), ci the clearance of the first stage, stated as a fraction of the 
displacement of that cylinder, so that the clearance of the first stage cubic 
feet = ciDi, and that of the second stage = C22)2. ^ 

The low-pressure capacity of the first stage (L. P. Cap.) in cubic feet is 
{W—Va)f and, as for the single-stage compressor, is expressed in terms of dis- 
placement, clearance and ratio of compression of the first stage as follows, see 
Eq. (64) : 

(L.P.Cap.i)=Z>i(l + ci-ciftpi"^)=DiJ5.i (102) 

For the second stage, the low-pressure capacity (L. P. Cap.2) is 

and is equal to 

(L.P.CSLp.2)=D2\l + C2-C2Rp2')=D2Ef^ .... (103) 

Volumetric efliciency of the first stage is given by 

E,i = l+ci-ciRpi' (104) 



114 ENGINEERING THERMODYNAMICS 

Volumetric efficiency for second stage 



E^=l + C2-C2Rp2' ■ 




(105) 



1^ 



It 

I! 



loo^ ajBncrgjaa gqi ii|Bajn«8»j ~ - 

It may be required lo find the receiver pressure (iocidental to the finding of 
work or horse-power) for a compressor with given cylinder sizes and deliver}' 
pressure. The condition as.sumed of perfect intercooling stipulates that 

(L. P, Cap.i)fsup.pr.) = (L. P. Cap.j) (rec.pr.), 



WORK OF COMPRESSORS 115 

whence 

f V , .(L.P. Cap.i) 

(rec.pr.)==(sup.pr.) (j^pc^p^) 

If the volumetric efficiencies are known or can be sufficiently well approx- 
imated this can be solved directly. If, however^ £,1 and £,2 are not known, 
but the clearances are known, since these are both dependent upon the receiver 
presstire sought, the substitution of the values of these two quantities will give 

(recpr.) = (sup.pr.)— ^^ \»"P-P''-/ ^ J , . . . (107) 

i)2|l+C2 



['+-«fif)1 



an expression which contains the receiver pressure on both sides of the equation. 
This can be rearranged with respect to (rec. pr.)i but results in a very complex 
expression which is difficult to solve and not of sufficient value ordinarily to 
warrant the expenditure of much labor in the solution. Therefore, the relations 
are left in the form (107). It may be solved by a series of approximations, 
the first of which is 

(rec.pr.) = (sup.pr.)^ approx (108) 

With this value for the receiver pressure, substitution may be made in the sec- 
ond member of the Eg. (107), giving a result which will be very nearly correct. 
If desirable, a third approximation could be made. 

To find the work of a two-stage exponential compressor in terms of displace- 
ment of cylinders, supply pressure, receiver pressure and delivery pressure, 
pounds per square inch, and volimietric efficiency of the first stage, jB,i, from (79) 
or (94) , 



« r .-1 .-1 -| 

jf=144^-^(sup.pr.)Di£?,i fipi"r-|-/2p2~-2 . . . .. 



(109) 



in which 



(rec^ and i?,2=f*^-"^^^ 
^ (sup.pr.) (rec.pr.) 



To solve this the receiver pressure must be found as previously explained and 
the volimietric efficiency must be computed by Eq. (104) or otherwise be known. 
It is impracticable to state work for this general case in terms of displace-* 
ment and clearances directly, due to the difficulty of solving for (rec. pr.) in Eq. 



116 ENGINEERING THERMODYNAMICS 

(107). It may, however, be stated purely in terms of supply and delivery pres- 
sures, in pounds per square inch, displacement, in cubic feet, and volumetric 
efficiencies, as follows: 
From Eq. (106), 



Rvi = 



I 



and 



D2E92 



„ _ del.pr . D2Et2 _j^ D2gg 
^^ sup.pr. Z)iA\i""^^Z)i£?,i' 



Hence 



«-i «-i 



,r-144j-tj(a„p.pr.)/)A[(°J^) " +(«,?£) " -2J . (110) 

The mean effective pressure of the two-stage compressor referred to the low- 
pressure cylinder is found by dividing the work of the entire cycle Eq. (110), by 
the displacement of the first-stage cylinder, and by 144, to give pounds per 
square inch. 

m.e.p. referred to first-stage cylinder, pounds per square inch is, 

«-l 8-1 



W 



.-:-,(sup.pr.)E.,[(||) ■ +(«,gg)-----2]. . a„) 



144Di 8-1 

It is well to note that this may also be found by multiplying (work done 
per cubic foot of gas supplied) by (volumetric efficiency of the first stage, Eti), 
and dividing the product by 144. 

In terms of the same quantities, an expression for indicated horse-power 
may be given as follows: 

where n is the number of cycles completed per minute by the compressor. 
For n may be substituted the number of revolutions per minute, divided by the 
revolutions per cycle, 

N 

The horse-power per cubic foot of gas supplied per minute is 

«-i »-i 



'I.H.P s (sup.pr.) 



n(L.P.Cap.) s-1 229.2 [VA^ 



(gitr-"H«'^)"-4 • »-) 



WORK OF COMPRESSORS 117 

Horse-power per cubic foot of gas delivered and cooled per minute. 

a-l » 1 

I.H.P. 



n(H. P. Cap. cold) 



=j^-f'is^'[(l£)^+('^tt-:)"'"-^ »») 



Horse-power per cubic foot of hot gas delivered per minute 



*-i 



I.H.P. s sup.pr./I)iEvi\ » o L 

/T("H7PrCap. hot) "s-i 229.2 VD2£r2/ ''' 



*-i 



mr-i^m^-^- ■ <"« 



For the case where clearance is zero or negligible, these expressions may be 
simplified by putting Et? and Eei equal to unity. 



«-l a- I 



■•H"'-.4^,"^^'U[(^;) •+«:)■ -4. . (116, 

I.H.P. per cubic foot, gas supplied per minute 

LH.P. _ s (sup.pr.)r/Z)i\ • , / n2\-T^' ] 
n(L.P.Cap:)~s-l 229:2 [W ^\"'dJ "^J* ' ^"^^ 

I.H.P. per cubic foot gas delivered and cooled per minute 

•-1 »- 1 



I.H.P. 



-rli'-^m'-'^Ky-'l '"«) 



n(H. P. Cap. cold) 
I.H.P. per cubic foot hot gas delivered per minute 

a- I » — 1 «— 1 



I.H.P. 



n(H. P. Cap. hot) 



-ASf(»:) ■ «-'[(&)"+«;)""-^] <"« 



Example 1. Method of calculating diagram, Figs. 29, 30, 31. 
Assumed data. 

Pa = Pft=2116 lbs. per square foot; 
p^=P^=P;^=Pt =6172 lbs. per square foot. 
P^=P,=P/ = 18,000 lbs. per square foot. 
C/(H. P.) = 7.5 per cent; C/(L. P.) 7.5 per cent; « = 1.4; L. P. Capacity =5 cu.ft. 



118 ENGINEERING THERMODYNAMICS 

To obtain point K. 

From formula Eq. (64), 

1 
L. P. Cap. = A(l +Ci —CiRpu) 

5=Z)i(l+.075-. 075X2.14), hence Z)i = 5.45 cu.ft. 

Ch - 7* - 5.45 X .075 * .41 cu.ft. 

.'. 7k - .4 cu.ft. ; Pk =6172 lbs. sq.ft. 



To obtain point A: 



-<w- 



1 

I 

4X2. 14 = .856 cu.ft. 



.*. 7a = .85 cu.ft.; Pa -2116 lbs. sq.ft. 
To obtain point B: 

7» = 7a+5 = .85+5 =5.85 cu.ft.; Pft=2116. lbs. sqit. 



To obtain point C: 



7c = 75 -5- ^ V* =5.85 -^2.14 =2.73, 

.'. 7c =2.73 cu.ft.; Pc=6172 lbs. sq.ft. 
To obtain point D: 

Volume at Z> is the displacement plus clearance of H. P. cylinder. This cannot be 
found imtil the capacity is known. The capacity is the amount gas which must be 
taken in each stroke and which is also the amount actually delivered by L. P. cylinder 
cooled to original temperature. The amount of cool gas taken in by the second cylinder 
is 

(L. P. Cap.2) = fe'''^/-) (L. P. Cap.i) = |J^X5= 1.7 cu.ft. 

\rec.pr. / dI72 

But 

_ (L. P. Cap.2) _ 1.7 , _^ . 

(, , o -I 1+.075-. 075X2.14 

\1+C2 — C2/Cp2V 

Ci, = 7/=C2D, =.075 Xl.88 = .14 cu.ft. 

7(f = Ci-hD, = 1.88 +.14 =2.02 cu.ft. 

Pd =6172 lbs. sq.ft. 

Other points are easily determined by relations too obvious to warrant setting dowiL 

Example 2. What will be the capacity, volumetric efficiency and horse-power per 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow- 
ing compressor: Two-stage, double-acting cylinders, 22i and 34^X24 in., running 
at 100 R.P.M. High-pressure clearance 6 per cent, low-pressure 4 per cent. Supply 



WORK OF COMPRESSORS 119 

pressure 14 lbs. per sqiiare inch absolute. Delivery pressure 115 lbs. per square inch 
absolute. 

The capacity wOl be the cylinder displacement times the volumetric efficiency. 

Di ^displacement of a 34ix24 in. cylinder, or 12.8 cu.ft. and D2 =displacement of 
a 22 J +24" cylinder or 5.4 cu.ft. To obtain the volumetric efficiency, make use of 
approximation of formula £q. (108), 

D 12 8 

(rec.pr.) « (sup.pr.)^ « 14 X-^-V =33.2 lbs. sq.in, 

Ut 0.4 

and then by Eq. (107) checking, 

,2.8[n-.04-.04(^\TU-| 

(rec.pr.) -l^X-TTr tTk — TliT ~^^'^ ^^^- sq.in. 

E,i = 1 +ci -CiiRp,) • from Eq. (104), 

1 
- 1 +.04 - .04 X (2.5) • =96.8 per cent 

Therefore the capacity will be, 

200 X 12.8 X. 968 =2480 cu.ft. per minute; 

1 
En^l+ct -CtiRpi)* from Eq. (105), 

= 1 +.06 -.06 X (3.28)-^^* =92 per'cent. 
From Eq. (113), I H.P. per cu.ft. (sup.pr.) air per minute is, 

LrJ: <-l 

8 sup.pr. r (DiEn\ B yljy^A •' ol 

"«-l 229.2 l\D^EJ ^\^^D,E~J, "^> 

1.4 14 r/lM_X.968\.286 / 5.4X.92 V^se i 
T^229":2L\ 5.4X.92 / ^ \^'^h2,^^ml ~^J ^'^^' 

Whence horse-power per 1000 cu.ft. of free air per minute -is, = 150. 

From Eq. (115) horse-power per cubic foot (del.pr.) air, hot = that of (sup.pr.) 

air X/^p * \~^~^) or 5.85 times that of (sup.pr.) air. 

.'. Horse-power per 1000 cu.ft. of hot (del.pr. air) =150x5.85 =877. 

Prob. 1. A two-stage double-acting compressor has volumetric efficiencies as shown 
by cards of 98 per cent and 90 per cent for the high- and low-pressure cylinders respect- 
ively. It is nmning at 80 R.P.M. and compressing from atmosphere to 80 lbs. per square 
mch gage. If the cylinders are 15ix25ixl8 ins., and speed is 120 R.P.M., what 



120 ENGINEERING THERMODYNAMICS 

horse-power is being used and how many cubic feet of free and compressed air (hot 
and cold) are being delivered per minute, when « equals 1.41? 

Prob. 2. What horse-power will be needed to drive a two-stage compressor 10 i ins. 
and 16j Xl2 ins., double acting, with 5 per cent clearance in each cylinder at 160 R.P.M. 
when the supply pressure is atmosphere, delivery pressure 100 lbs. per square inch 
gage, when s equals 1.35? 

Prob. 3. A thousand cubic feet of free air per minute must be compressed by a two- 
stage compressor to 80 lbs. per square inch gage from a supply pressure of 10 lbs. per 
square inch absolute. The volumetric efficiencies for the high- and low-pressure cylin- 
4ers are 85 per cent and 95 per cent respectively, and the receiver pressure is 25 lbs. 
per square inch absolute. What will be the displacement of each cylinder and the 
horse-power per cubic foot of (sup.pr.) air? 

Prob. 4. How many cubic feet of free air can be compressed in two-stage compres- 
sor 18ix30ix24 ins. with 5 per cent clearance in high-pressure cylinder and 3 per cent 
in low if (sup.pr.) is atmosphere and (del.pr.) 80 lbs. per square inch gage? How would 
the answer be affected if clearance were taken as zero? Take s = 1.41. 

Prob. 5. The volumetric efficiency of the low-pressure cylinder is known to be 95 per 
cent, and of the high-pressure cylinder 85 per cent. The cylinder sizes are 15 J X25i X 18 
ins. and speed is 120 R.P.M. What horse-power must be supplied to the machine if 
the mechanical efficiency is 80 per cent and the pressure ratio 10 with a (sup.pr.) of one 
atmosphere? 

Prob. 6. A compressor runs at 120 R.P.M. and is double acting. It is compressing 
air from 14 lbs. per square inch absolute to pressures ranging from 70 lbs. per square 
inch gage to 100 lbs. per square inch gage. The cyUnders are 20i X32i x24 ins., and 
clearances 8 per cent and 4 per cent. Find the approximate receiver pressure, capacity 
and horse-power for the range of discharge pressure, for « = 1.3. 

Prob. 7. The volumetric efficiency of the low-pressure cyUnder of a two-stage com- 
pressor is known to be 95 per cent, the receiver pressure as shown by gage is 40 lbs., 
delivery pressure 100 lbs., and supply pressiu^ one atmosphere. What will be the 
horse-power if the machine runs at 120 R.P.M. and the low-pressure cylinder is 18 X 12 in.? 
« = 1.4. 

Prob. 8. An air compressor appears to require more power to run it than should 
be necessary. It ib a double-acting 18x30x24 in. machine running at 100 R.P.M. 
The volumetric efficiencies are 85 per cent and 90 per cent respectively and supply and 
delivery pressures 14 lbs. and 110 lbs. per square inch, both absolute. What would be 
the minimum work per cubic foot of (sup.pr.) air, per cubic foot of (del.pr.) air, hot and 
cold, for adiabatic compression? 

Prob. 9. The efficiency of the driving gear on an electric-driven compressor is 
75 per cent. Power is being supplied at the rate of 150 H.P. How much air should 
be compressed per minute from 4 lbs. per square inch absolute to 100 lbs. per square 
inch gage, if the receiver pressure is 35 lbs. per square inch absolute and the low-pressure 
volumetric efficiency is 90 per cent, s being 1.4? 

12. Two-stage Compressor with Best Receiver Pressure Exponential 
Compression. Capacity, Volumetric Efficiency, Work, Mean EfFective Pres- 
sures and Horse-power in Terms of Dimensions of Cylinders and Clearances. 

For the two-stage exponential compressor with or without clearance, and per- 
fect intercooling, the best-receiver pressure was found to be (Eq. 84), 

(best-rec.pr.) = [(sup.pr.) (del.pr.)]* (120) 



WORK OF COMPUESSORS 121 

This expression Eq. (120) for best-receiver pressure makes it possible to 
evaluate Rpi and Rp2 as follows: 

g,xfor(be8t-rec.pr.) = ^"^*-^"^P'' = ^^^"PP''-^^^^^P^-^^ ' = [^g^1*=Ji!A (121) 

sup.pr. sup.pr. Lsup.pr.J 

and 

D e /u 4. \ (del.pr.) (del.pr.) 

Rp2 for (best-ree. pr.) = 7,-— -- — -. = ,7 ; .^ , ttt- 

(best-rec.pr.) [(sup.pr.) (del.pr.)]* 

-[(tij;)]'-«-'' • <'^> 

The use of these values for Rp\ and Rp2 in the expressions previously given 
for volumetric eflSciency for the general case, Eqs. (104) and (105) results in 

Volumetric eflBciency, first stage 

Je.i = (l+ci-cii2p2i), (123) 

and volumetric efficiency, second stage 

£^ = (1+C2-C2ftp2.) (124) 

The work was found to be represented by Eq. (98), which may be stated 
in tenns of displacement and volumetric efficiency of the first stage, as follows: 

W = 2SS~(mp.pr.)DiEjRp^--l\, .... (125) 

where Rp = 7 — -^-^ and where (sup.pr.) is in pounds per square inch. 

If the clearance is known for the first stage this becomes by the use of 
Eq. (104), 

s 1 r *-^ I 

WS8 ^^ (sup.pr.)Z)i(l+ci-ci«p2,) /2^ 2. - 1 , . . . (126) 

which is a direct statement of the work of a two-stage adiabatic compressor 
with perfect intercooling .in terms of supply pressure and delivery pressure, 
pounds per square inch, displacement, cubic feet and clearance as a fraction 



122 ENGINEERING THERMODYNAMICS 

of displacement, provided the cylinder sizes and clearances are known to be such 
as to give best-receiver pressures. 

The mean effective pressure reduced to firstnstage displacement, in pounds 
per square inch, may be derived from either Eq. (125) or (126) by dividing the 
work by the displacement of the first-stage cylinder, and again dividing by 144. 

w 2s r i^ 1 

m.e.p. = j;p^ =^3Y (8up.pr.)£^i ii!p 2t -1 

2s I -LXr 5-- 1 ^' ' ' ' ^'^^ 

= -j^(8Up.pr.)ll+Ci— Ciflp2« 1 i2p2t —1 

Since the work done is equally divided between the two cylinders when best- 
receiver pressure is maintained, the mean efifective pressure, in pounds per 
square foot, for each cylinder will be, one-half the total work divided by the 
displacement of the cylinder in question, 

w s r 1^ 1 

m.e.p., first stage =2Q8nr'^7Zi;(^^P-P''-)^»'h^^ (128) 

Note that this is one-half as great as the m.e.p. of the compressor reduced 
to first stage, (127), 

m.e.p., second stage = ogsZT " IZT ^^P-P^-^ n~^»i U^p «« — 1 , . . . (129) 



But 



(sup.pr.)^^ = (rec.pr.) = (sup.pr.)(del.pr.) U, 

whence, 

m.e.p., second stage = (sup.pr.)(del.pr.) ♦JS?,2 Lb, ^T— 1. . , . (130) 

It is next necessary to investigate what conditions must be fulfilled to obtain 
the best-reieeiver pressure, the value of which is stated, Eq. (120). The condition 
of perfect intercooling provides that the temperature of the gas entering the 
second stage is the same as that entering the first stage, and hence that the 
product (volume entering second stage) X (pressure when entering second stage) 
must be equal to the product (volume entering first stage) X (pressure of supply- 
to first stage), or 

(L. P. Cap. 2)(rec. pr.) = (L. P. Cap. i) (sup. pr.). , . . (131 



WORK OF COMPRESSORS 123 



Combining with Eq. (120) 



(L 



. P. Cap.i) _ [(8up.pr.)(del.pr.)P _ [" (deLpr.) ]*_ p ^ 
. P. Cap.2) "" (sup.pr.) L (sup.pr.) J " ^ ' 



or 



(1) (2) r (3) 



r (3) J -| 

DA 1+ci— ciRp2t 



■,,_ (I..P.C»p. i)_DiE., — L"" -"^J ,,„, 

From this three-part equation proper values may be found to fulfill require- 
ments of best-receiver pressure for: 

1. The ratio of capacities for a given ratio of pressures, or conversely, the 
ratio of pressures when capacities are known; 

2. The ratio of cylinder displacements for known volumetric efficiencies; 

3. The ratio of cylinder displacements when the clearances and ratio of com- 
pression are known, or conversely, with known displacements and clearances 
the ratio of pressures which will cause best-receiver pressure to exist. This 
last case in general is subject to solution most easily by a series of approxi- 
mations. 

There is, however, a special case which is more or less likely to occur in prac- 
tice, and which lends itself to solution, that of equal clearance percentages. If 
ci=C2 the parenthesis in the numerator of part (3) of Eq. (132) becomes equal 
to the parenthesis in the denominator, and evidently the volumetric efficiency 
of the two cylinders are equal, hence for equal clearance percentages in the two 
tages, 

Wr^' (133) 

A case which leads to the same expression, Eq. (133), is that of zero clearance, 
a condition that is often assumed in machines where the clearance is quite 
small. 

The work per cycle, Eq. (126), when multiplied by the number of cycles 
performed per minute, n, and divided by 33,000, gives 

LH.P. = ^-^ ^^u^l'^ ''^' (^ +^^ - ^^^^^^) (^^^ ~ ^)> • (1^) 

from which are obtained the following: 
I.H.P. per cubic foot supplied per minute 

LH.P. _ . •(sup.pr.)(g^ir^^j)^ _ (J35J 



n(L.P. Cap.) s-1 114.6 
I.H.P. per cubic foot delivered and cooled per minute 



I.H.P. _ « (del.pr.)(^^--jJ_i)^_ ^ _ (j3gj 



n(H. P. Cap. cold) s-l 114.6 



124 [ENGINEERING THERMODYNAMICS 

and I.H.P. per cubic foot delivered hot per minute 

m PP^' K n-° S ^'rr;V-^-(iJ,'^'-l). . (137) 

n(H. p. Cap. hot) s — 1 1 14.6 

These e^qpressions, Eqs. (165), (166) and (167) are all independent of clear- 
ance. 

Example. What will be the capacity, volumetric efficiency and horse-power per 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for tht 
following compressor for a = 1.4? Two-stage, double-acting, cylinders 22i x341 X24 ins., 
running at 100 R.P.M. Low-pressure clearance 5 per cent, high-pressure clearance 
such as to give best-receiver pressure. Supply pressure 15 lbs. per square inch abso- 
lute, delivery pressure 105 lbs. per square inch absolute. 

Capacity will be cylinder displacement times low pressure volumetric efficiency, or, 
200DiX^rt. 

Di« 17.5 cu.ft. 

]_ 
Eti from Eq. (123) = (1 +Ci -CiRp2») 

= 1 + .05 - .05 X 7^^^ = 95 per cent. 

Therefore low pressure capacity =200 X 1 2.8 X. 95 =2430 cu.ft. per minute. 
Horse-power per cubic foot of (sup.pr.) air per minute is from Eq. (135) 

8 sup.pr. i_-J 
7-1 114.6"^^' '' "^^^ 
14. ^^ * 

.4 ^ 105^ ^ 

Therefore, horse-power per 1000 cu.ft. of sup.pr. air = 160. 

Horse-power per cubic foot of (del.pr.) air, hot, is from Eq. (137) 

Rp 2a times power per cu.ft. of (sup.pr.) air, 

hence, 

160 X5.3 =850 = horse-power per 1000 cu.ft. of (del.pr.) air, hot, per minute. 

Problem Note. In the following problems, cylinders are assumed to be proportioned 
with reference to pressures so as to give best-receiver pressure. Where data conflict, 
the conflict must be found and eliminated. 

Prob. 1. Air is compressed adiabatically from 14 lbs. per square inch absolute to 
80 lbs. per square inch gage, in a 20 J X32i X24 in. compressor, running at 100 R.P.M. . 
the low-pressure cyUnder has 3 per cent clearance. What will be horse-power re- 
quired, to run compres or and what will be the capacity in cubic feet of low pressure 
and in cubic feet of (del.pr.) air? 

Prob. 2. What must be the cylinder displacement of a two-stage compressor with 5 
per cent clearance in each cylinder to compress 500 cu.ft. of free air per minute from 
14 lbs. per square inch absolute to 85 lbs. per square inch gage, so that s equals 1.4? 
What will be the horse-power per cubic foot of (del.pr.) air hot and cold? 

Prob. 3. A two-stage compressor is compressing gas with a value of s = 1.2o 
from 10 lbs. per square inch gage to 100 lbs. per square inch gage. The cylinders are 
18jx30tx24 ins., and speed is 100 R.P.M. If the clearance is 5 per cent in the low- 



WORK OF COMPRESSORS 125 

pressure cylinder and 10 per cent in the high, what will be the cubic foot of (sup.pr.) gas' 
handled i>er minute and what will be the horse-power at best receiver pressure? 

Prob. 4. A manufacturer states that his 201^x32^X24 in. double-acting compres- 
sor when running at 100 R.P.M. at sea level will have a capacity of 2390 cu.ft. of free air 
per minute, pressure range being from atmosphere to SO lbs. per square inch gage. At 
best-receiver pressure what clearance must the compressor have, compression being 
adiabatic? 

Prob. 5. The cylinder sizes of a two-stage compressor are given as 10} X 16^x12 
ins., and clearance in each is 5 per cent. What will be the best-receiver pressures when 
operating between atmosphere and following discharge pressures, 60, 70, 80, 90, 100 
and 1 10 lbs. per square inch gage, for s equal 1.4? 

Prob. 6, 1500 cu.ft. of air at 150 lbs. per square inch gage pressure are needed per 
minute for drills, hoists, etc. The air is supplied from 3 compressors of the same size 
and speed, 120 RJ*.M. Each has 4 per cent clearance in each cylinder. What will 
be sizes of cylinders and the horse-power of the plant for best-receiver pressure, when 
s = 1.41? 

Prob. 7. The cards taken from a compressor show volumetric efficiencies of 95 per 
cent and 80 per cent in low- and high-pressure cylinder respectively. What will be 
(del.pr.) for best-receiver pressure if compressor is 151x25^X18 ins., and (sup.pr.) 15 
ll)s. \yeT square inch absolute to 10 lbs. absolute, and what will be the work in each case, 
9 being 1.35? 

Prob. 8. A manufacturer gives a range of working pressure of his lOj Xl6}xl2 in. 
compressor from 80-100 lbs. per square inch page. If clearances are, low 4 per cent, 
high 8 per cent, and (sup.pr.) is atmosphere, find by trial which end of the range comes 
nearest to giving best-receiver pressure? If clearances were equal which would give 
best-receiver pressure? 

Prob. 9 A 16} x25} Xl6 in. compressor is rated at 1205 cu.ft. free air per minute at 
135 R.P.M. at sea level. What would be the clearance if compressor were compressing 
air from atmosphere to 100 lbs. gage at sea level? With same clearance what would be 
the size of a low-pressure cylinder to give the same capacity at altitude of 10,000 ft. with 
the same clearance and the same (del.pr.) , best-receiver pressure always being maintained? 

13. Three-Stage Compressor, no Clearance, Perfect Intercooling Expo- 
nential Compression (Cycle 7), Best Two Receiver Pressures, Equality of 
Stages. Work and Capacity, in Terms of Pressures and Volumes. The three- 
stage exponential compressor cycle with no clearance, perfect intercooling Cycle 
7, is shown in Fig. 32. The net work area, ABCDEFGHJKA, is made up of 
three areas which may be computed individually by the formulae for single stage 
Eq. (48), provided the requisite pressures and volumes are known, as follows: 

«-i 

T^=-l-|P^F^[(^) ' -ij (first Stage) 



8-1 



-_;^PdVa I W ) * — 1 (second stage) 



8 

t-i 



s 



ZTi^rVfU^) ' -A (third stage) 



. (138) 



ENGINEERING THERMODTNAMICS 



But the conditioD of perfect intercooling provides that for no clearance, 
P^Vt=pgV^^PfVf. 



and it may be noted that Pi=P„ and P/^P,. Accordingly, 



Pressures in this expression are in pounds per square foot. 



(140) 



WORK OF COMPKESSORS 127 

Changing the equation to read in terms of supply pressure pounds per square 
inch, low-pressure capacity cubic feet, and ratios of pressures, first stage 
(Rpi)y second stage {Rp2) and third stage {Rpz)^ it becomes 

Work done by three-stage compressor, perfect intercooMng 

W^ = 144^(8up.pr.)(L. P. Cap.) [{Rj-^ + {Rp2)^+{Rpz)^-^]f (141 

From this the following expressions are derived: 
Work per cubic foot supplied 

(L.pSap.) °^^A^^''^'^'''^[^^'''^^'"^ (fii>2p"^ + (gp3/-^-3]]. . . (142) 
Work per cubic foot gas delivered and cooled 

(H.P.cTp.cold) = ^^(^^^-P^4^^-'^'"^"+^^''^'^"+^^''^'"^"4^^^^ 

Work per cubic foot gas, as delivered hot 

•-1 
W 



(H. P. Cap. hot) 



= 144^^(8up.pr.)(/2pi)(/2p2)(fl,8)*[(/2,i) ' 

+(Rp2f^ + {Rp3fr-3\ (144) 



Best Two Receiver Pressures, Referring to Fig. 32, Pe is the pressure in the 
first receiver (1 rec.pr.) and P« is the pressure in the second receiver, (2 rec.pr.). 
It is evident that if either receiver pressure be fixed and the other is varied, 
the work necessary to compress a given initial volume of gas will be varied, 
and will have a minimum value for some particular value of the varying receiver 
pressure. By a variation of both receiver pressures a minumim may be found 
for the work when both receiver pressures have some specific relation to supply 
and delivery pressures. For instance, assume that Pc is fixed. Then a change 
in /*• can change only the work of the second and third stages, and the three- 
stage compressor may be regarded as consisting of 

One single-stage compressor, compressing form P* to Pc. 
One two-stage compressor, compressing from Pc to P^. 

In this two-stage compressor, best-receiver pressure is to exist, accord 
ing to Eq. (84), 

P*= (best 2 rec.pr.) = (PJ>,)*. . ^ (145) 



128 ENGINEERING THERMODYNAMICS 

Similar reasoning, assuming P« fixed and making Pc variable, would show 
that 

Pc=(best 1 rec.pr.) = (PeP6)* (146) 

Eliminate Pc from Eq. (145) and the expression becomes, 

P.= (best 2 rec.pr.) = {PJPg^) = [(sup.pr.)(del.pr.)2] * (147) 

Similarly, from Eq. (146) 

Pc= (best 1 rec.pr.) = {P^^P,) = r(sup.pr.)2(del.pr.)l* (148) 

From these expressions may be obtained. 



Pc^P.^P.^/P.y 

P, Pc ~Pe \Pj 

or 

Rp 1 = Rp2 ~ Rpi = -Bp* 



(149) 



Substitution in Eq. (140) gives, 

Work, three-stage, best-receiver pressure no clearance 

IF=3^^P.n[@'^'-l] (150) 

Arranging this equation to read in terms of supply pressure, pounds per square 
inch, low-pressure capacity, cubic feet, and ratio of pressures 

Work, three-stage best-receiver pressure 

Tr=432-^(sup.pr.)(L.P. Cap.)(i2pV-l), . . . (151) 

The work of the compressor is equally divided between the three stages 
when best-receiver pressures are maintained, which may be proven by substitu- 
tion of Eq. (149) in the three parts of Eq. (138), and 

Work of any one stage of three-stage compressor with best-receiver pressure. 

Wi = W2 = W^ = U4-^(sup,pT.){L. P. Cap.)(/2/"ir-l). . (152) 



WORK OF COMPRESSORS 129 

From Eq.(151), may be derived the expressions for work per unit of capacity. 
^ork per cubic foot low-pressure gas is, 



(L 



-^^=432^-^(sup.pr.)[fi,^-l] (153) 



$ince 



(L. P. Cap.) = (H. P. Cap. cold)i2p. 



RTork per cubic foot cooled gas delivered is, 



^ Tr=432-*-(sup.pr.)/2p(i2,V-l). . . . (154) 



(H. P. Cap. cold) 5-1 



\gain, from Fig. 32, 



^'(?f -"'-<;)- ^'-'©(^f' 



which is to say that, when best-receiver pressures are maintained, 

(L. P. Cap.) = (H. P. Cap. hot)Rj,JRprB, 



or 



2g + l 

= (H. P. Cap. hot)fl,"3r-, .... (155) 



hence 

Work per cubic foot hot gas delivered 



=432-^(sup.pr.)i2,"3r-(i2p-8r-l). . . (156) 



(H.P. Cap. hot) — 5-1 

Example 1. Method of calculating Diagram, Fig. 32. 
Assumed dcUa, 

Pa =Pft =2116 lbs. per sq.ft. 

Pc =Pd =best first-receiver pressure =PjPg^ =4330. 

Pe =P/=best second-receiver pressure =Pi,^Pg^ =8830. 

Pg^Pn = 18,000 lbs. per sq.ft. 

7a = 7A =0 cu.ft. 7& =5 cu.ft. s -1.4. 



130 ENGINEERINQ THERMODYNAMICS 

To obtain point C: 

F,-F»-i-(^)^*-5+1.67=3cuit. 



'-"-(§)■' 



A Fc - 3 cu.ft. Pe =4330 lbs. sq.ft. 

Intermediate points Bix>C may be found by assuming various'pressures and finHmg 
the ccMTesponding volumes as for 7c. 
To obtain point D: 

•TT <rr * * » 2116 ^ .. ». 

7d-7ftX5- -6Xt^^-2.44 cu.ft. 
Pd 4330 



Va -2.44 cu.ft., Pa -4330 lbs. sq.ft. 



To obtain point E: 



1 



by assumption of best-receiver pressure. 

Hence 7* -2.44 -s- 1.67 =1.46 cu.ft., an P.=8830 lbs. sq.ft. 

Intermediate points Dio E may be found by assuming various pressures and finding 
corresponding volimies as for 7«, andsucceeding points are found by similar methods 
to these already used. 

7^ -.72, Pi, = 18,000, 

Example 2. What will be the horse-power required to compress 100 cuit. of free 
air per minute from 15 lbs. per square inch absolute to 90 lbs. per square inch gage in a 
no-clearance, three-stage compressor if compression be adiabatic? What will be the 
work per cubic foot of (del.pr.) air hot or cold? 

From Eq. (153) work per cubic foot of (sup.pr.) air is, 

8 t^ 

432 — -(sup.pr.) (iJp 3. -1), 

8 — 1 

-432X^X16X(709S2..1) ^4500 ft.-lbs., 
.4 

or 

H.P. for 100 cu.ft. per minute - — -13.6. 

OtiJ,UUU 

From Eq. (154) work per cubic foot of (del.pr.) air cold is Rp times that per cubic 
foot of (sup.pr.) air, or in this case is 31,500 ft.-lbs. 

i From Eq. (156) work per cubic foot of (del.pr.) air hot laRp ^ times that per cubic 

foot of (sup.pr.) air, or in this case 5.8x45,000 «46,200 ft.-lbs. 



WORK OF COMPRESSORS 131 

Prob. 1. What work will be required to supply 2000 cu.ft. of air at 200 lbs. per inch 
gage pressure if compressing is done adiabatically by three-stage compressors, taking 
air at atmosphere, neglecting the clearances? 

Prob. 2. A motor is available for running a compressor for compressing gajs, for 
which 8 equals 1.3. If 60 per cent of the input of the motor can be expended on the 
air, to what delivery pressure can a cubic foot of air at atmospheric pressure be com- 
pressed in a zero clearance three-stage machine? ' How many cubic feet per minute 
could be compressed to a pressure of 100 lbs. gage per H.P. input to motor? 

Prob. 3. Two compressors are of the same size and speed. One is compressing 
air so that exponent is 1.4, the other a gas so that exponent is 1.1. Each is three stage. 
Which will require the greater power to drive, and the greater power per cubic foot 
of (sup.pr.) gas, and per cubic foot of (del.pr.) gas, hot, and how much more, neglect- 
ing clearance? 

Prob. 4. How will the work per cubic foot of (sup.pr.) air and per cubic foot of (del. 
pr.) air difiFer for a three-stage compressor compressing from atmosphere to 150 lbs. per 
square inch gage from a single- and a two-stage, neglecting clearance? 

Prob. 6. A table in " Power " gives the steam used per hour in compressing air to 
various pressures single stage. A value for air compressed to 100 lbs. is 9.9 lbs. steam 
per hour per 100 cu.ft. of free air. Using the same ratio of work to steam, find the value 
for the steam if compression had been three-stage, zero clearances to be assumed. 

Prob. 6. A 5 in. drill requires 200 cu.ft. of free air per minute at 100 lbs. per square 
inch gage pressure. What work will be required to compress air for 20 such drills if 
three-stage compressors are used, compared to single-stage for no clearance? 

Prob. 7. What would be the steam horse-power of a compressor delivering 150 
cu.ft. of air per minute at 500 lbs. per square'^inch pressure if compression is three-stage, 
adiabatic, clearance zero, and mechanical efficiency of compressor 80 per cent? 

14. Three-stage Compressor with Clearance, Perfect Intercooling Expo- 
nential Compression (Cycle 8), Best-receiver Pressures, Equality of Stages. 
Work and Capacity in Terms of Pressures and Volumes. The pressure- 
volume diagrams of the three-stage compression is shown in Figs. 33, 34 and 
35, on which the clearance volume and displacements, low-pressure capacity 
and high-pressure or delivery capacity for hot gas are indicated. 

If perfect intercooling exists, as is here assumed, 



(V,- Va)Pi> = (V^- Vi)Pa = {Vf- Vj)Pf 
and also 

(L. P. Cap.)P6 = (H.P. Cap. cold)P^. 

Apply Eq. (57) to the three stages and the entire work done is, 
JF=^P,(F,-7a)[(^;)'"^-l] (first stage) 



(157) 



8 



+» 



~P<,{V^-V,) [ (^^J-r - 1 j (second stage) 
~lPfiyf-Vj) [(^')'"^'-l] (thW stage) 



. . . (168) 



ENGINEERING THERMODYNAMICS ' 



By use of the above conditions of perfect intercooling Eq. (157) this 
expression becomes, 



(159) 
I 



/ A A 

qI £L <^. 

I — -• 



a- S 



15 



V»>& aranDs jaa Epanoj u] 



WORK OF COMPEESSOES 133 

Id terms of supply pressure, pounds per square inch, low-pressure capacity, 
cubic foot and ratios of pressures as above, the worl^ of a three-stage com- 
pressor with perfect intercooling and with clearance is 

H -144-%(sup.pr.)(L.P. Cap.) f(ftpi)^'-|-{ft;.2) V+(/i'^.i)^-3l, (l(iO) 



^ / / 

I - _l 



el 






Si 



which is identical with Eq. (141), ahoufing that dearance kas no effect upon the 
irnrk for a given capadiy. 



ENGINEERING THERMODYNAMICS 



It readily follows that the work per UDJt of gas is independent of clearance, 
and hence Eqs. (142), (143) and (144), will give a correct value for the work 



/ A / 

i — _i 



-ir I-; 



w 

so 






per cubic foot of gas supplied, per cubic feet delivered and cooled, and per 
cubic foot as delivered hot, respectivply. 

Since in two-stage compressors the reasoning leading to the determination 
of best-receiver pressure applies equally well with and without clearance, and 
since the value of best-receiver pressures for three-stage are found by eon- 



WORK OF COMPRESSORS 135 

sidering the threenstage a combination of one- and two stage-compressors, the 
same expressions for best-receiver pressures will hold with clearance as without; 
see Eqs. (147) and (148). 

Pg=(best 2 rec.pr.) = [(sup.pr.)(del.pr.)^l*- 
Pc— (best 1 rec.pr.) = [(sup.pr.)^(del.pr.)l*. 

The use of these expressions for best-receiver pressures leads to the same 
result as for no clearance £q. (150), except for the volumes, 

Work, three-stag^ best-receiver's pressure with clearance 

which is stated below in terms of supply pressure, pounds per square inch low- 
pressure capacity, cubic foot, and ratio of compression Rp, 

Work, three-stage besfr^receiver pressure. 

Tr=432-^(sup.pr.)(L.P.Cap.)(iBp^^-l) . . . (162) 

which is identical with Eq. (151). 

From this may be obtained expressions for the work per cubic foot of low- 
pressure gas supplied to compressor per cubic foot of gas delivered and cooled, 
and per cubic foot of gas as delivered hot from the compressor, when the re- 
ceiver-pressures are best, and these will be respectively identical with E3qs. 
(153), (154), and (156), in the foregoing section. 

16. Three-stage Compressor^ any Receiver-pressure Exponential Com- 
pression. Capacity^ Volumetric Efficiency^ Work, Mean Effective Pressurei 
and Horse-power in Terms of Dimensions of Cylinders and Clearances. 

Dis displacement of the first-stage cylinder, in cubic feet=(F6— Fm); 
2)2= displacement of the second-stage cylinder, in cubic feet = (7d— Fa)> 
D3» displacement of the third-stage cylinder, in cubic feet = (F/— F*). 

ci, C2f cs are the clearances of the first, second and third stages respectively, 
stated as fractions of the displacement, so that, 

Clearance volume, 1st stage, in cubic feet = Fm=ciDi; 
Clearance volume, 2d stage, in cubic feet»^F* = C2l>2; 
Clearance volume, 3d stage, in cubic feet = FA = C3l>3. 



136 ENGINEERING THERMODYNAMICS 

The low-pressure capacity of the first stage, and hence for the compressor 
is (Vb^Va)f and in terms of clearance, ci, and displacement Di of the fir<t 
stage is, according to Eq. (64), 

j_ 
(L.P. Cap.i)=Di(l+ci-ci/epi')=^i^.i (163) 

For the second stage, the low-pressure capacity is (Fd— Vi) and is equal to 

j_ 
(L. P. Cap.2) = 2>2(1 +C2 - C2Rp2 ' ) = D2E,2, .... (164) 

and for the third stage (F/— Vj) or, 

(L. P. Cap.3) = 2>3(1 +C3 - cafips * ) = D^E^z (165 j 

The volumetric efficiency of 1st stage is 

j_ 
jB,i = (l+ci-ciflpiO (166) 

Volumetric efficiency of second stage is 

JB.2 = (1 + C2-C2flp20 (167) 



Volumetric efficiency of third stage is 



1 



S.3=(l+C3-C3flp3'), (168) 

The work of the three-stage compressor with the assistance of Eq. (163) may 
be stated in terms of supply pressure, pounds per square inch, displacement 
of first-stage cylinder, in cubic feet and volumetric efficiency of first stage, and 
also ratios of compression existing in the first, second, and third stages, 



TF=144^(sup.pr.)2>i£.i[(/epi) '~r+ {R,2) V+ (/J^g) V-3 1 . 



(169) 



To make use of this formula for the work of the compressor the two 
receiver pressures must be known, and it is, therefore, important to derive a 
relation between receiver pressures, displacements and clearances or volumetric 
efficiencies. 

The assumption of perfect intercooling which has already been made use 
of in obtaining Eq. (169), regardless of the receiver-pressure, requires that — 
see Eq. (157) : 

(L.P. Cap.i) (sup.pr.) = (L.P. Cap.2) (1 rec.pr.) 

= (L.P. Cap.3)(2 rec.pr.). . (170) 



WORK OF COMPRESSORS 137 

Using values of capacities in Eqs. (163), (164), and (165) and solving for 
first- and second-receiver pressures. 

/■i \ / V (L. P. Cap.i) / .DiEvi /^^^\ 

(lrec.pr.) = (sup.pr.)-^L.P:Cap7)=(^"P-P'->5^£72' " ' ^^^^^ 



and 



(2rec.prO = (sup.prOJ^;-^;g^5;;} = (sup.pr.)^. . . (172) 



Then 



P _(1 rec.pr.) _DiE^ ,, . 

''"'"Isu^prO ~D2£,2' ^"^^ 

P _ (2 rec.pr.) _ Z)2S,2 ,,_.. 

By definition, (del.pr.) =Rp (sup.pr.), 

__ (del.pr . _ (sup.pr.) _ DzE,:^ . . 

^^"(2rec.pr.)'"^''(2rec.pr.)~^''Di£,i ^^^^^ 

The work of the three-stage compressor may then be stated in terms of 
supply pressure, pounds per square inch, displacements, cubic feet, volumetric 
efficiencies, and overall ratio of compression, Rp, as follows: 

^.|44.-^,(s„p.p..)O.E..[(^*)^'+(|g)'-^" 



»-l 



+(«'gt:) ■ -']■ ■ <"« 



In Terms of PressureSf Displacements, and Clearances, an expression can be 
written by substitution of values of Evi, E^ and Evz from Eqs. (166), (167) and 
(168), but it becomes a long expression, further complicated by the fact that 
Rvi, Rp2 and Rpz remain in it. This may be solved by the approximation 
based first upon the assumption that all volumetric efficiencies are equal to each 
other or to unity when 






(If volumetric efficiencies are each equal to each other 
or to unity) (177) 



138 ENaiNEERING THERMODYNAMICS 

This process amounts to the same thing as evaluating Evu E^* <uid Evb from 
Eqs. (166), (167) and (168), making use of the approximation Eq. (177) and 
substituting the values found in Eq. (176). 

Since the above can be done with any expression which is in terms of volu- 
metric efficiencies, the following formulae will be derived from Eq. (176), as it 
stands. 

The mean effective pressure of the three-stage compressor reduced to the first- 
stage cylinder is found by dividing the work of the entire cycle, Eq. (176) by 
displacement of the first stage, and by 144 to reduce to pounds per square inch. 

(m.e.p.) reduced to first stage cylinder, 

j^^=— j(sup.pr.)fi.,[(^^^-j . +(5^3) • 

Note here that this may also be obtained by multiplying (work per cubic foot 
supplied) by (volumetric efficiency of first stage) and dividing the product by 144. 
The indicated horse-power of a compressor performing n cycles per minute 
will be equal to the work per cycle multiplied by n and divided by 33,000, or, 
for the three-stage compressor with general receiver pressures, 

'•H.p-.4T^^''"'.-..[(5s:i)'^'+(Sl!)^" 



t+(«'tfe)"'-4 • ("») 



For n may be substituted the number of revolutions per minute, N, divided 
by the revolutions required to complete one cycle 



N 
n=— . 
z 



The horse-power per cubic foot of gas supplied'per minute is 

I.H.P. 8 (sup.pr:) r/ Dig,i \ii:i / D2E,2 \'-^ 

n(L.P.Cap.) «-l 229.2 [[OiEj \'^\DzEJ ' 



+ 



«l.)^'-4 • (>»«» 



Horse-power per cubic foot gas delivered and cooled per minute is 

I. H.P. ^ 8 (d el.pr.) V/ DiE^A 'T' / D2gr2 \ V 

n(H.P.Cap.cold) s-1 229.2 [\D2Ej '^[DsE.sJ 

8-1 
+ 



«rfe)^""-4 • ('") 



WORK OF COMPRESSORS 139 

Horse power per cubic foot hot gas delivered per minute is 

I.H.P. 8 (sup.pr.) /Pig,i\V Lr/Pig.i\V 

n(H. P. Cap. hot) s-l 229.2 [DsE^j/ '' [\D2E,2) 



+ 



/DiE,2 



»-l • •-! 



^r^^'^r-'i ■ a»^) 



The last equation is obtained by means of the relation 

(L. P. Cap.) = (H. P. Cap. hot) X (^'^^)^X (^-I^^^) 
^ ^ / V *^ ' \2 rec.pr./ \ sup.pr. / 



= (H.P.Cp.ho.)X«^)^(^-jf^) 



»-l 



= (H.P.Cap.hot)Xi?pr(^^) ' . ..... (183) 

// clearance is zero or negligible, these expressions may be rewritten, putting 
Ev, Ev2 and Evz each equal to unity. 



«— 1 J «— 1 «— 1 



•"••"-j^^f^-^^®) ■■+(t) ■ +«) ■ -4 <'«> 



H.P. per cubic foot of gas supplied per minute is 

I.H.P. _ (sup.pr.) r /DA '^ /D2\ ^\ / „ ^\^*_il 
n(L.P. Cap.) 229.2 L\W W \'Di) ^\ 

H.P. per cubic foot delivered and cooled per minute is 

•-1 »-i »-i 



. . (185) 



I.H.P. 



=^Sf^[(S)"+(t)"+(«.|)"-'] ■ ^m 



n(H. P. Cap. cold) 
H.P. i>er cubic foot hot gas delivered per minute is 

I.H.P. ^ 8 (sup.prQ /DA'-f-' if/^X^' /^\^' 

n(H. P. Cap. hot) .«-l 229.2 \Ds) ^Lw W 

8-1 



+ (i2.^) ' -3]. . . (187) 



140 ENGINEERING THERMODYNAMICS 

Example 1. Method of calculating Diagram, Fig. 35. 
Assumed data: 

■Pa =Pft =2116 lbs. per square foot. 

Pc =Pd-Pi-Pm =4330 lbs. per square foot. 

» 

P«=P/=P^=P4=8830 lbs. per square foot. 

Pg^Ph- 18,000 lbs. per square foot. 
ci=7.5 per cent for all cylinders; « = 1.4. 
L.P. capacity 5 cu.ft. 

To obtain point M: 

1 

From formula Eq. (163) L. P. Capi. - A(l +Ci -ci/2,i •) 

or 

5 =r>i(l +.075 -.075 X 1.67) or Z)i =5.3 cu.ft. and clearance volume 

Vm = 5.3 X .075 = .387 cu.ft. 

Therefore, 

Vm = .39 cu.ft. ; Pm =4330 lbs. sq.ft. ; 

To obtain Point A : 



Va^VmX (-^j 1.4. = .39 X 1.67 = .67 cu.ft. 



Additional points M to A may be found by assuming pressures and finding corr& 
sponding volumes as for Va- 

To obtain point B: 

Vb = 7a + (L. p. Capi.) «.67+5 =5.67 cu.ft. i 



Therefore, 



Fft = 5.67 cu.f t. ; P* = 21 16 lbs. sq. ft. ; 



To obtain point C: 



Therefore, 



i_ 
Vc^Vb^ (~-j ^'^ =5.67 -^ 1.67 =3.45 cu.ft. 



Ve = 3.45 cu.f t. ; Pe = 4330 lbs. sq.ft. 



Intermediate points J5 to C may be found by assuming various pressures and findi 
corresponding volumes as for Vb^ 



WORK OF COMPRESSORS 141 



*o obtain point D: 



Volume at Z> is the displacement plus clearance of the intermediate cylinder. This 
umot be found until the capacity is known. Appl3dng the same sort of relations as 
rere used in calculating the diagram for the two-stage case with clearance, 

D2(l +C2 -C2RP2) • =2.44 or D2 =2.57, 

nd clearance volume. 

. Vt = .075 X2.57 = .192 cu.ft., 
ence, 

74=2.57+.19=2.76cu.ft. 
Tierefore, 

Vd =2.76 cu.ft; Pa =4330 lbs. sq.ft. ; 

The rest of the points are determined by methods that require no further explana- 
lon and as pressures were fixed only volumes are to be found. These have the following 
alues, which should be checked: 

% = 1.65; Vr = 1.32; 7^ = .79; Fa = .09; Fy = .15; Fz = .32; 7, = .65; 7^ = 1.23; F, = .14. 

Bxample 2. A three-stage compressor is compressing air from atmosphere to 140 
)s. per sqxiare inch absolute. The low-pressure cyUnder is 32 X 24 ins. and is known to 
ave a clearance of 5 per cent. From gages on the machine it is noted that the first- 
Bceiver pressure is 15 lbs. per square inch gage and the second-receiver pressure is 
5 lbs. per square inch gage. What horse-power is being developed if the speed is 
00 R.P.M. and s = 1.4? From the formula Eq. (169), 

W =144-^(sup.pr.)D,^„ Rprr -\-rJ~b~ -\-RprT~ --Z . 

5 — 1 L J 

From gage readings 

/Cpi=--— Z. /Cj»j=— — Z.,5tJ, ^J»l— yQ-— -6. 

1 

Et, = (1 +Ci -CiRpC* ) from Eq. (166), 



r, 



lence. 



En^{l +.05 -.05X1.65) =67.5 per cent. 



TF = 144x^Xl5Xll.2x.675(1.22+1.28+1.22-3); 
.4 

=59,200 ft.lbs. per stroke or 200 X59,200 ft. =lbs. per minute; 

=358 I.H.P. 



142 ENGINEERING THERMODYNAMICS 

Examples. Another compressor has cylinders 12x20x32x24 in. and it is known 
that the volumetric efficiencies of the high, intermediate and low-pressure cylinders are 
respectively 70 per cent, 85 per cent and 98 per cent. The (del.pr.) is 150 lbs. per square 
inch absolute. What is the horse-power in this case if the speed is 100 R.P.M.? 

From the formula Eq. (176), 

-i«x^.><„..x..[(;-iti)-^.(if^)^ 



+ 



V"ll.2xW ^J 



= (1.309+1.495+1 -3) =66,400 ft.-lbs. per stroke, 

200X66,400 .„ 
Whence I.H.P. = — qoTw^ — =402. 

• 

Prob. 1. What will be the horse-power required to drive a 12 X22 X34 X30 in. three- 
stage compressor with volumetric efficiencies of 75, 85, and 95 per cent in the high, 
intermediate and low-pressure cylinders, at 100 R.P.M. when compressing natural gas 
from 25 lbs. per square inch gage to 300 lbs. per square inch gage, adiabatically? 

Prob. 2. A three-stage compressor for supplying air for a compressed-air locomo- 
tive receives air at atmosphere and delivers it at 800 lbs. per square inch gage. Should 
the receiver pressures be 50 lbs. and 220 lbs. respectively in the first and second and the 
volumetric efficiency of the first stage 90 per cent, what would be its displacement 
and horse-power when compressing 125 cu.ft. of free air per minute, adiabatically? 
What are the cylinder displacements? 

Prob. 3. Find the work done on a gas, the value for s of which is 1.3, in compressing 
it from atmosphere to 7 atmospheres absolute, adiabatically in three stages, the com- 
pressor having a low-pressure cylinder displacement of 60 cu.ft. per minute and a volu- 
metric efficiency of 95 per cent, first receiver pressure being 2 atmospheres absolute, 
and second-receiver pressure 4 atmospheres absolute. If air were being compressed 
instead of the above gas, how would the work vary? 

Prob. 4. The cylinders of a compressor are 8x12x18x24 ins. and clearance such 
as to give volumetric efficiencies of 80, 90 and 98 per cent in the different cylinders in the 
order given. Compressor is double acting, running at 120 R.P.M. and compressing air 
adiabatically from 14 lbs. per square inch absolute to 150 lbs. per square inch gage. 
What is the capacity in cubic feet per minute, work done per cubic feet of (sup.pr.) air, 
(del.pr.) air hot and cold and the horse-power of the compressor? What would be 
the effect on these quantities if the clearances were neglected? 

Prob. 6. If the cyHnders of a compressor are 10x14x20x18 ins., and clearances 
are 8, 5 and 3 per cent, what work is being done in adiabatically compressing air from 
10 lbs. per square inch absolute to 100 bs. per square inch gage? 

Note: Solve by approximate method. 

Prob. 6. For special reasons it is planned to keep the first-receiver pressure of a 
throe-stage compressor at 30 lbs. per square inch absolute, the second-receiver pressure at 
60 lbs. per square inch absolute, and the line pressure at 120 lbs. per square inch absolute 



WORK OF COMPRESSORS 143 

The (sup.pr.) is 14 lbs. per square inch absolute. If the clearances are 4 per cent in 
the low and 8 per cent in the intermediate and high-pressure cylinders, what must be 
the cylinder sizes to handle 500 cu.ft. of free air per minute at 120 R.P.M. and what 
power must be supplied to the compressor on a basis of 80 per cent mechanical effi- 
ciency, for a value of « equal to 1.39? 

Prob. 7. Should the above pressures (Prob. 6) be gage pressures instead of absolute, 
how would the quantities to be found be affected? 

Prob. 8. The receiver pressures on a COt gas compressor are 50 lbs. per square inch 
absolute, and 200 lbs per square inch absolute, the (del.pr.) being 1000 lbs. per square 
inch absolute. The machne has a low-pressure cylinder 8x10 ins. with 3 per cent 
clearance. What horse-power will be required to run it at 100 R.P.M and what would 
be the resultant horse-power and capacity if each pressure were halved? (Sup.pr.) = 14.7 
lbs. per square inch. 

16. Three-stage Compressor with Best-receiver Pressures Exponential 
Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure 
and Horse-power in Terms of Dimensions of Cylinders and Clearances. It 
was found that for the three-stage adiabatic compressor with perfect inter- 
cooling, the work was a minimum if the first and second receivers had pressures 
defined as follows, see Eqs. (147) and (148) : 

(best 1 rec.pr.) = [(sup.pr.)2(del.pr.)]* (188) 

(best 2 rec.pr.) = [(sup.pr.) (del.pr.)^]* (189) 

(best Irec p ro ^/deLpry^^ _ ^ 

(sup.pr.) \sup.pr./ 

(best 2 recp rj ^ / dehprA* ^^^^ 

^ (best 1 rec.pr.) \8up.pr./ 

,prO_ /dehpry^ , (192) 

rec.pr.) \ sup.pr./ 



*''"(best2 



The use of these values in connection with expressions previously given 
for volumetric efficiency, Eqs. (166), (167) and (168), gives. 



Volumetric efficiency of first stage =JS?,i = (l+Ci-Ci/2,3.) .... (193) 



Volumetric efficiency of second stage = -Bt2 = (1+^2— C2fip^*) .... (149) 
Volumetric efficiency of third stage =Bt3 = (l+C3— ^^s/Zp^O .... (195) 



144 ENGINEERING THERMODYNAMICS 

The work of the three-stage compressor with best-receiver pressures, Ekj- 
(162), when expressed in terms of displacement and volumetric efficiency becomes 

TF=432--(sup.pr.)I>iS.i(flp^-l) (196) 

where 

(del.pr.) 



Rp= 



(sup.pr.) 



If clearance is known, the value of Evi may be ascertained by Eq. (193) 
and inserted in Eq. (196). Since this may be so readily done the substitution 
will not here be made. 

The mean effective pressure of the compressor referred to the first stage is 
obtained by dividing the work Eq. (196) by 144 Di: 

(m.e.p.) referred to first-stage cylinder 

14^,=3^-l(sup.pr.)^.i(«,3r-l) (197) 

The mean effective pressures of the respective stages, due to the equality 
of work done in the three stages will be as follows: 



For first stage 



8 *— 1 

(m.e.p.) = T (sup.pr.) J?i,i(Bp3« — 1) (198) 

o J. 



For second stage 



(m.e.p.)=-^(sup.pr.)~J5:.i(i2pV-l) (199) 

5—1 1/2 



For third stage 



(m.e.p.) =-^(sup.pr.)^£„(i?p^-l) (200) 



But 



(sup.pr,) ^-^'- = (1 rec.pr.) = [(sup.pr.)2(del.pr.)]*, 



and also 



(sup.pr.)^-^l^ = (2 rec.pr.) = [(sup.pr.) (del.pr.)2] J. 



WORK OF COMPRESSORS 145 



Hence 

For second stage 



8 -1 



(m.e.p.) = -^[(sup.pr.)2(del.pr.)]*iE?.2(fli,'3r - 1). . (201) 

8 "~" X 



For third stage 



s •-» 



(m.e.p.) = --i[(sup.pr.)(deI.pr.)2]»£,3(«p^-l). ■ ■ (202) 



Conditions to Give Best-Receiver Pressures. All the foregoing discussion 
of best-receiver pressures for the three-stage compressor can apply only to cases 
in which all the conditions are fulfilled necessary to the existence of best-receiver 
pressures. These conditions are expressed by equations (173), (174), (175), 
(190), (191), and (192), which may be combined as follows: 

(1) (-2) (3) (4) 

^^ (L.P.Cap.2) (L.P.Cap.3) D2E,2 ~DsE,3 

(5) ^ ^®^ 1 r- • (203) 

^ Dl{l+Ci—ClRj^») ^ 1)2(1 +C2-C2fiy^ 
D2(l+C2-C2fip-3«) Ds^l+Ci-CsRp^ 

Parts (1) and (2) of this equation state the requirements in terms of 
capacities; (3) and (4) in terms of displacements and volumetric efficiencies; 
(5) and (6) in terms of displacements and clearances. In order, then, that best- 
receiver pressure may be obtained, there must be a certain relation between 
the given ratio of compression and dimensions of cylinders and clearances. 
Since, after the compressor is once built these dimensions are fixed, a given 
multi-stage comprassor can be made to give best-receiver pressures only when 
compressing through a given range, i.e., when Rp has* a definite value. If Rp 
has any other value the receiver pressures are not best, and the methods of the 
previous Section (15) must be applied. 

When clearance 'percentages are equal in all three cylinders, ci=C2=C3, and 
the volumetric efficiencies are all equal then, when best-receiver pressures exist, 
Eq. (203) becomes, 

fip* = ^ = ^ = f or equal clearance per cent. . . (204) 

Evidently this same expression holds if clearances are all zero or negligible. 
What constitutes negligible clearance is a question requiring careful thought 
and is dependent upon the ratio of compression and the percentage of error 
allowable. 



146 ENGINEERma THERMODYNAMICS 

Indicated horse-power of the compressor is found by multiplying the work 
per cycle, Eq. (196) by the number of cycles per minute, n, and dividing the 
product by 33,000. 

IM.V.^j^^^^^nDiEn(Rv^-l) .... (205) 

From this are obtained the following: 
H.P. per cubic foot supplied per minute 



H.P. per cubic foot delivered and cooled per minute 

I.H.P. _ 9 (deLpr.) , !^^ 

n(H. P. Cap. cold) s-1 76.4 ^"*' ''' 

H.P. per cubic foot delivered hot per minute 

I.H.P. 
n(H. P. Cap. hot) 

(See Eq. (156)). 



(207) 



=^^^f^^fi,^(/2,V'-l). . . (208) 



It is useful to notie that these expressions are all independent of clearance, 
which is to be expected, since the multi-^tage compressor may be regarded as a 
series of single-stage compressors, and in single stage such an independence 
was found for work and horse-power per unit of capacity. 

Example. If the following three-stage compressor be run at best-receiver pressures 
what will be the horse-power and the best-receiver pressures? Compressor has low- 
pressure cylinder 32x24 ins. with 5 per cent clearance, is compressing air from atmos- 
phere to 140 lbs. per square mch absolute, so that s equals 1.4 and it runs at 100 R.P.M. 

From the formula Eq.'(196) 



432« / •-! \ 

W = — r(8up.pr.)Di£^,i [RjTzT^I) 



From the formula Eq. (188) 



(best 1 rec.pr.)=[(sup.pr.)*(del.pr.)]i 

= (15)«Xl40]i=31.6. 



WORK OF COMPRESSORS 147 

Prom Eq. (189) 

(best 2 rec.pr.) =[(sup.pr.)(del.pr.)*]* 

= [15 X (140)*]* =66.5. 
From Eq, (193) 



^n = ( l+c,-Ci«/M 



=» 1 +.05 -.05 X (^) ^' =96.5; 
hence, 



TF-432x^Xl5xll.2x96.5x(9.35'®^-l) =59,000 ft-lbs., 



or. 



TTTT. 59,000X200 ,^o 
'•^•*^- 33;000 ^^ 

Prob. 1. There is available for running a compressor 176 H.P. How many cu.ft. 
of free air per minute can be compressed from atmosphere to 150 lbs. per square inch 
gage by a three-stage adiabatic compressor with best-receiver pressures? 

Prob. 2. The low-pressure cylinder of a three-stage compressor has a capacity of 
4i cu.ft. per stroke. If the stroke of all three cylinders is 18 ins., what must be the 
diameters of the intermediate and high to insure best-receiver pressures, if clearance 
be n^lected, and (sup.pr.) be 1 lb. per square inch absolute and (del.pr.) 15 lbs. per 
square inch absolute, 8 being 1.4. 

Prob. 3. The above compressor is used as a dry-vacuum pump for use with a sur- 
face condenser. If 800 cu.ft. of (sup.pr.) gas must be handled per minute what horse- 
power will be needed to run it? What will be the horse-power per cubic foot of atmos- 
pheric air? 

Prob. 4. Will a 15 X22 x34 x24 in. compressor with clearances of 3, 5 and 8 per cent 
in low, intermediate and high-pressure cylinders respectively be working at best-receiver 
pressures when (sup.pr.) is 15 lbs. per square inch absolute and (del.pr.) 150 lbs. per 
square inch absolute? If not, find by trial, the approximate (del.pr.), for which this 
machine is best, with s equal to 1.4? 

Prob. 6. For the best (del.pr.) as found above find the horse-power to run the 
machine at 100 R.P.M. and also the horse-power per cu.ft. of (del. pr.) air cold? 

Prob. 6. Should this compressor be used for compressing ammon'a would tic 
best (del. pr.) change, and if so what would be its value? Also what power would Le 
needed for this case? 

Prob. 7. Compare the work necessary to compress adiabatically in three stages from 
20 lbs. per square inch absolute to 200 lbs. per square inch absolute, the following gases: 

Air; Oxygen; Gas-engine mixtures, for which s = 1.36. 

Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio 
for best-receiver pressure and a pressure ratio of 10? 



148 ENGINEERING THERMODYNAMICS 

Prob. 9. A compressor, the low-pressure cylinder of which la 30 X20 ins. with 5 per 
cent clearance is compressing air adiabatically from atmosphere to 150 lbs. per square 
inch gage, at best-receiver pressure. Due to a sudden demand for air the (del. pr.) 
drops to 100 lbs. per square inch gage. Assuming that the (1 rec. pr.) dropped to 5 lbs. 
per square inch gage and (2 rec.pr.) dropped to 40 lbs. per square inch gage, how much 
would the speed rise if the power supplied to machine was not changed? 

17. Comparative Economy or Efficiency of Compressors. As the prime 

duty of compressors of all sorts is to move gas or vapor from a region of low 
to a region of high pressure, and as this process always requires the expenditure 
of work, the compressor process which is most economical is the one that 
accomplishes the desired transference with the least work. In this sense, then, 
economy of compression means something different than efficiency, as ordi- 
narily considered. Ordinarily, efficiency is the ratio of the energy at one point 
in a train of transmission or transformation, to the energy at another point, 
whereas with compressors, economy of compression is understood to mean the 
ratio of the work required to compress and deliver a unit of gas, moving it 
from a low- to a high-pressure place, to the work that would have been required 
by some other process or hypothesis, referred to as a standard. This economy 
of compression mast not be confused with efficiencyof compressors as machines, 
as it is merely a comparison of the work in the compressor cylinder for an actual 
case or hypothesis to that for some other hyjwthesis taken as a standard. The 
standard of comparison may be any one of several possible, and unfortunately 
there is no accepted practice wdth regard to this standard. It will, therefore, 
be necessary to specify the standard of reference whenever economy of compres- 
sion is under consideration. The following standards have been used with 
some propriety and each is as useful, as it supplies the sort of information really 
desired. 

First Standard, The work per cubic foot of supply gas necessary to com- 
press isoihermally (Cycle 1), from the supply pressure to the delivery pressure 
of the existing compressor and to deliver at the high pressure is less than that of 
any commercial process of compression, and may be taken as a standard for 
comparison. Since, however, actual compressors never depart greatly from 
the adiabatic law, their economy compared with the isothermal standard will 
always be low, making their performance seem poor, whereas they may be as 
nearly perfect as is possible, so that it may appear that some other standard 
would be a l>etter indication of their excellence. 

Second Standard. The work per cubic foot of gas supplied when compressed 
adiabatically in a single stage (Cycle 3), if taken as a standard, will indicate a 
high economy, near unity for single-stage compressors, and an economy above 
unity for most multi-stage compressors. For the purpose of comparison it 
will be equally as good as the first standard,and the excess of the economy over 
unity will be a measure of the saving over single-stage adiabatic compression. 
Since, however, single-stage adiabatic compression is not the most economical 
obtainable in practice for many cases, this standard may give an incorrect 
idea of the perfection of the compressor. 



WORK OF COMPRESSORS 149 

Third Standard, Due to the facts noted above, it may be a better indica- 
tion of the degree of perfection of the compressor to compare the work per 
cubic foot of gas supplied with that computed for the standard adiabatic cycle 
most nearly approaching that of the compressor. This standard is, however, 
open to the objection that a multi-stage compressor is not referred to the same 
cycle as a single-stage compressor, and a multi-stage compressor with other 
than best-receiver pressure is not referred to the same cycle as another operating 
with best-receiver pressure. This is, therefore, not a desirable standard for 
comparing compressors of different types with one another, although it doeS 
show to what extent the compressor approaches the hypothetical best condi- 
tion for its own type and size. 

Other standards might be chosen for special reasons, each having a value 
in proportion as it supplies the information that is sought. 

It is seen from the discussion of the second standard that its only advantage 
over the first is in that it affords a measure of the saving or loss as compared 
with the single-stage adiabatic compressor cycle. 

If the first standard, that of the isothermal compressor cycle, be adopted 
for the purpose of comparison, it at once gives a measure of comparison with 
the isothermal, which is more and more nearly approached as the number 
of stages is increased, though never quite reached, or as the gas is more effect- 
ively cooled during compression. It may be regarded as the limiting case of 
multi-stage compression with perfect intercooling, or the limiting case of con- 
tinuous cooling. 

In order to ascertain how nearly the actual compressor approaches the 
adiabatic cycle most nearly representing its working conditions, the economy of 
of the various reference cycles heretofore discussed may be tabulated or charted, 
and the economy of the cycle as compared with that of the actual performance 
of the compressor will give the required information. The process of com- 
putation by which this information is obtained will depend upon the nature 
of information sought. The economy of actual compressor compared with the 
isothermal may be stated in any of the following ways: 



Computed work per cubic foot supplied, isothermal , . 



Indicated work per cu.ft. actual gas supplied to compressor 

I.H.P. per cubic foot per minute suj>plied, isothermal 
I.H.P. per cubic foot per minute actual supplied 

Single stage 

(m.e.p.) isothermal, pounds per square inc h, no clearance 
(m.e.p.) actual -5- true volumetric efficiency 

Multi-stage 

(m.e.p.) isothermal, no clearance 



(b) 



(209) 



(c) 



(m.e.p.) reduced to first stage -5- first stage vol. eff. 



(d) 



150 ENGINEERING THERMODYNAMICS 

In this connection it is useful to note that for the case of the no-clearance 
cycles, the work per cubic foot of supply is equal to the mean effective pressure 
(M.E.P.) in pounds per square foot, an4 when divided by 144 gives (m.e.p.) 
in pounds per square inch. Also, that in cases with clearance, or even actual 
compressors with negligible clearance, but in which, due to leakage and other 
causes, the true volumetric efficiency is not equal to unity. 

Work per cubic foot gas supplied X^»=144(m.e.p.). . . (210) 

The information that is ordinarilj'^ available to determine the economy 
of the compressor will be in the form of indicator cards from which the (m.e.p.) 
for the individual cylinders may be obtained with ordinary accuracy. The 
volumetric efficiency may be approximated from the indicator cards also, but 
with certain errors due to leakage and heating, that will be discussed 
later. If by this or other more accurate means the true volumetric 
efficiency is foimd, the information required for the use of Eq. (209) (c) 
or (d) is available. Evaluation of the nimierator may be had by Eq. (31), 
which is repeated below, or by reference to the curve sheets found at the end 
of this chapter. (Fig. 50.) 

Mean effective pressure, in pounds per square inch for the isothermal com- 
pressor without clearance is given by 

(m.e.p.) isothermal = (sup.pr.) logs Jip (211) 

The curve sheet mentioned above also gives the economy of adiabatic 
cycles of single stage, also two and three stages with best-receiver pressures. 
The value of 8 will depend upon the substance compressed and its condition. 
The curve sheet is arranged to give the choice of the proper value of s applying 
to the specific problem. 

If it is required to find the economy of an actual compressor referred to 
the third standard, i.e., that hypothetical adiabatic cycle which most nearly 
approaches the actual, then 



Economy by third standard is 

Econ. actual referred to isothermal 



Econ. hypothetical referred to isothermal' 



(212) 



It is important to notice that for a vapor an isothermal process is not one 
following the law PXV = constant. What has, in this section, been called an 
isothermal is correctly so. called only so long as the substance is a gas. Since, 
however, the pressure-volume analysis is not adequate for the treatment of 
vapors, and as they will be discussed under the subject of Heat and Work, 
Chapter VI, it is best to regard this section as referring only to the treatment 
of gases, or superheated vapors which act very nearly as gases. However, 
it must be understood that whenever the curve follows the law PXF= constant, 
the isothermal equations for work apply, even if the substance be a vajwr 
and the process is not isothermal. 



WORK OF COMPRESSORS 151 

18. Conditions of Mazimiim Work of Compressors. Certain types of com- 
pressors are intended to operate with a delivery pressure approximately con- 
stanty but may have a varying supply pressure. Such a case is foimd in pumps 
or compressors intended to create or maintain a vacuum and in pmnping 
natural gas from wells to pipe lines. The former deliver to atmosphere, thus 
having a substantially constant delivery pressure. The supply pressiu'e, 
however, is variable, depending upon the vacuiun maintained. In order that 
such a compressor may have supplied to it a sufficient amount of power to 
keep it running under all conditions, it is desirable to learn in what way this 
power required will vary, and if it reaches a maximum what is its value, and 
under what conditions. 

Examine first the expression for work of a single-stage adiabatic compressor 
with clearance. The work per cycle will vary directly as the mean effective 
pressure. Eq. (69.) 

(:«..p.)..4i(»P.P-.)[l+c-c(^)-][(^f-l]. (21S, 



This will have a maximum value when 

d(m.e.p.) 
d(sup.pr.) 



=0, 



or when 



f^5^y'-'-:f^ri+c-c?^(^?i:Pi^yi==o. . . . (214) 

Vsup.pr./ 1+c L « \sup.pr./ J ^ ^ 



Solving this for the value of supply pressure will give that supply pressure 
at which the work will be a maximum, in terms of a given delivery pressure, 
clearance and the exponent a. 

The assumption most conunonly used is that clearance is zero. If this is 
trae^or the assmnption permissible, the above equation becomes simplified^ 



• 



m^-'" (2'« 

The value of 8 for air, for instance, is 1.406, and hence the ratio of compression 
for maximum work for the hypothetical air compressor is 

(1.406)^*^=3.26 ; (216) 

It may be noted that when s = 1 in the above expression, the value of the 
ratio of compression become indeterminate. To find the supply pressure for 
maximum work in this case, take the expression for mean effective pressure 
for the isothermal compressor (s = 1), Eq. (43), 

(m.e.p.).(,up.pr.)[l+.-.(,^)]:„^^). ... (217) 



152 ENGINEERING THERMODYNAMICS 

DiflFerentiate with respect to (sup.pr.) and place the differential coefficient equal 
to zero. This process results in the expression 



log. (^'^■^'•- 
^ \sup.pr. 



del.pr.\ c /del.prA_j^^j (218) 

/ 1+c \8up.pr./ 



When c=0, this becomes, 

,og.fd?l:Pr.\ = l or (^^'P":- ) = 2.72 (219) 

^ \sup.pr./ \sup.pr./ 

The expressions Eqs. (215) and (219) given are easily solved, but Eqs. (214) 
and (218) are not, and to facilitate computations requiring their solution 
the results of the computation are given graphically on the chart. Fig. 48, at 
the end of this chapter. 

The mean efifective pressure for a compressor operating imder maximum 
work conditions may be found by substituting the proper ratio of compression, 
found as above, in Eq. (213) or (217). In Fig. 48 are found also the results 
of this computation in the form of curves. Note in these curves that the mean 
efifective pressure is expressed as a decimal fraction of the delivery pressure. 

The discussion so far applies to only single-stage compressors. The problem 
of maxiinum work for muUi'Stage compressors is somewhat different, and its 
solution is not so frequently required. Moreover, if the assumption of perfect 
intercooling is made, the results are not of great value, as a still greater amount 
of power might be required, due to the failure for a period of time of the supply 
of cooling water. Consider this case first. 

If intercooling be disconUnued in a multi-stage compressor, the volume 
entering the second stage will equal that delivered from the first, and similarly 
for the third and second stages. The entire work done in all stages will be the 
same as if it had all been done in a single stage. It might be questioned as 
to whether this would hold, when the ratio of compression is much less than 
designed. The first stage will compress until the volume has become as small 
as the low-pressure capacity of the second stage. If the delivery pressure is 
reached before this volume is reached, there is no work left to be done in the 
second or any subsequent stages, and, due to the pressure of the gas, the valves, 
if automatic, will be lifted in the second and higher stages, and the gas ynW 
be blown through, with only friction work. It appears then that under the 
condition of no intercooling the multi-stage compressor acts the sanve as a 
single stage, and the conditions of maximum work will be the same. 

If intercooling is maintained perfect there will still be a range of pressures 
on which all the work of compression is done in the first stage, merely blowing 
the discharge through receivers, valves and cylinders in the upper stages. If 
this range is such that this continues beyond a ratio of pressures, which gives a 
maximum (m.e.p.) for the single stage, then the maximum will have been reached 
while the compressor is operating single stage, and the single-stage formulae 
and curves may be applied to this case also. 



WORK OF COMPRESSORS 



153 



That this condition frequently exists with multi-stage compressors of 
ordinary design is shown by the fact that the ratio of compression in each stage 
is seldom less than 3, and more frequently 3.5, 4 or even more. The ratio 
of compression giving maximum work for single stage, has values from 2.5 
to 3.26, dependent on clearance and the value of 8 for the gas compressed, and 
is, therefore, less for the majority of cases. 

If a curve be drawn, Fig. 36, with ratio of compression as abscissas and 
(m.e.p. -hdel.pr.) as ordinates, so long as the action is single stage, a smooth curve 
will result, but when the ratio of compression is reached above which the second 
cylinder begins to act, the curve changes direction suddenly, falling as the ratio 




Values of Rp. 

Fig. 36. — Curve of Relation between Mean Effective Pressure per Pound of Delivery Pressure 
in Terms of Pressure Ratio for Air, showing Maximum Value. 



of compression increases. Hence, if the ratio of cylinders is such that the single- 
stage maximum is not reached before the second stage begins to operate^ the highest 
point of the curve, or maximum work for a given delivery pressure will occur 
when the ratio of supply and delivery pressures is such as to make first-receiver 
jrressure equal to delivery pressure, 

19. Actual Compressor Characteristics. Air or gas compressors are very 
commonly made double acting, so that for a single cylinder, two cycles will 
be performed during one revolution, one in each end of the cylinder. If a rod 
extends through one of these spaces and not through the other, the displace- 
ment of that end of the cylinder will be less than the other by a volume equal 
to the area of rod multiplied by the stroke. To avoid mechanical shock 
at the end of either stroke, it is necessary to leave some space between the 



154 



ENGINEERING THERMODYNAMICS 



piston and cylinder head. Passages must also be provided, communicating 
with inlet and discharge valves. The total volume remaining in this spact^ 
and in the passages when the piston is at the nearer end of its stroke constitute 
the clearance. The amount of this clearance volume varies from .5 or .6 of 
one per cent in some very large compressors to as much as 4 or 5 per cent of 
the volume of displacement in good small cylinders. 

In order to study the performance of an actual compressor and to compare 
it with the hypothetical cycle, it is necessary to obtain an indicator card, and 
knowing the clearance and barometric pressure to convert the indicator card 
into a pressure volume diagram, by methods explained in Chapter I. Fig. 37, 
is such a diagram for a single-stage compressor. In the pipe leading to the 



Press. 



H.P.Cap Hot (Apparent) 

!G 



iLD (DeL Pr.) 








-L^B€ap-H^l)ethetieal 



-LrPrOap-^-pparent 



Displaoemcnt- 



(Sup. Pr.) 



•Vol. 



Fig. 37. — Compressor Indicator Card Illustrating Departure from Reference Cycle. 



intake valve the pressure is determined and a horizontal line AB is drawn 
on the diagram at a height to represent the supply pressure. Similarly, 
discharge pressure is determined and drawn on the diagram, KE, Consider 
the four phases of this diagram in succession. 

1. Intake Line. At a point somewhat below A the intake vaJve opens, 
say at the point F. This remains open till a point H is reached at or near the 
end of the stroke. The line connecting these two points indicates variations? 
of pressures and volumes throughout the supply stroke. In general this line 
will lie below the supply-pressure line AB due to first, the pressure necessary 
to lift the inlet valve from its seat against its spring and inertia, if automatic, 
and support it, and second, friction in the passages leading to the cylinder from 
the point where the supply pressure was measured. While the former is nearly 
constant the latter varies, depending upon the velocity of gases in the passages. 
The piston attains its highest velocity near the middle of the stroke^ thus 



WORK OF COMPRESSORS 155 

causing the intake line to drop below the supply pressure more at this part 
of stroke. These considerations do not, however, account completely for the 
form of the intake line. Frequently the first portion of the line lies lower than 
the last portion, even at points where the piston velocities are equal. This 
is more prominent on a compressor having a long supply pipe, and is due to 
the forces required to accelerate the aor in the supply pipe while piston velocity 
is increasing, and to retard it while piston velocity is decreasing. In com- 
pressors where the inlet valve is mechanically operated and the supply pipe 
long, it is possible to obtain a pressure at the end of the intake, line H, even 
in excess of the supply pressure. The effect of this upon volumetric efficiency 
will be noted later. 

The apparent fluctuations in pressure during the first part of the intake 
line may be attributed, first, to inertia vibrations of the indicator arm, in which 
case the fluctations may not indicate real variations of pressure; second, the 
indicator card may show true variations of pressure due to inertia of the gases 
in the supply pipe, since a moment before the valve opened at F the gases were 
stationary in the supply pipe. When F is reached the piston is already in motion 
and a very considerable velocity is demanded in the sugply pipe to supply 
the demaQd. This sudden acceleration can be caused only by a difference in 
pressure, which is seen to exist below and to the right of F on the diagram. 
The suddenness of this acceleration may start a surging action which will cause 
rise and fall of pressure to a decreasing extent immediately after. A third cause 
is possible, that is, a vibration of the inlet valve due to its sudden opening 
when it is of the common form, mechanical valves change the conditions. It 
is closed by weight or a spring and opened by the pressure difference. Between 
these forces the valve disk may vibrate, so affecting the pressure. 

2. Compression Line, From the time the inlet valve closes at the point 
H until the discharge valve opens at the point (?, the gases within the cylinder 
are being compressed. The compression is very nearly adiabatic in ordinary 
practice, but due to the exchange of heat between the cylinder walls, at first 
from walls and later from gas to walls, which are cooled by water jacket to 
prevent the metal from overheating, there is a slight departure from the adia- 
batic law almost too small to measure. 

A second factor which influences the form of this curve to a greater extent 
is leakage. This may occur around the piston, permitting gas to escape from 
one end of the cylinder to the other. During the compression process there is 
first an excess of pressure in the other end of the cylinder due to reexpansion, 
tending to increase pressures on the first part of compression. Later, the 
pressure rises and the pressure on the other side of the piston falls to supply 
pressure. During this period leakage past piston tends to decrease successive 
pressures or lowers the compression line. Leakage also occurs through cither 
discharge or inlet valves. The former will raise the compression line, while 
excessive leakage of the inlet valve will lower it. 

It is then evident that unless the nature of the leakage is known, it is 
impossible to predict the way in which it will change this line. It is, however. 



156 ENGINEERING THERMODYNAMICS 

more frequently the case that the piston and inlet leakage are large as compared 
with the discharge valves, in which case the actual compression line has a 
tendency to fall lower than the adiabatic as the volumes are decreased. Com- 
pression lines lowered by leakage are often mistaken for proofs of effective 
cooling, and cases have been known where isothermal compression of air was 
claimed on what proved to be evidence only of bad leakage. 

3. Delivery Line. After the delivery pressure of the compressor has been 
exceeded sufficiently, the discharge valve is opened and the gas is delivered to 
the discharge pipe or receiver till the end of the stroke is reached and at the 
point J the valve closes. The same group of factors influence the form of this 
line as act upon the intake line; spring resistance of discharge valve; friction 
in discharge passages varying with piston velocitj'-; inertia of gases in delivery 
pipe; sudden acceleration of gases in delivery pipe when discharge valve opens, 
and inertia of indicator arm, but in addition a strong tendency for the valve 
to chatter or jump open and shut alternately. I^eakage also occurs through 
tntake valve and past piston during this process, with the result that less gas 
passes through the discharge valve than is shown on the indicator card. 

4. Reexpansion Line. From the time the discharge valve closes, at J, 
till the intake valve opens, at F, the gas which remained in the clearance space 
after delivery expands, due to the advancing of the piston, till the pressure 
has fallen to such an amount that the intake valve will open. The same factors 
influence this line as the compression line. Heat is exchanged with the jacketed 
cylinder walls, at first cooling and later heating the gas as the pressure falls. 
This, for any given volume, changes the pressure. I^eakage occurs inward 
through the discharge vaJve and outward through intake valve and past piston. 
If these last two are in excess, the pressures will fall more rapidly than if the 
expansion were that of a constant quantity of gas. 

Work due to gas friction and inertia, it should be noted, is fully represented 
on the indicator card, and may be regarded as being equal to that extra area 
below the supply-pressure line and above the delivery-pressure line. In the 
combined card of a two-stage compressor there would be an overlapping of 
the diagrams due to this frictional loss. 

Low-pressure Capacity, Referring to the adiabatic compression and expan- 
sion lines, CD and KL^ Fig. 37, it is seen that the low-pressure capcun^y of the 
hypothetical cycle is the volume, LC. 

The apparent low-pressure capacity of the actual compressor, measured at 
the supply pressure is AB, This is not, however, the true volume of gas at 
supply pressure and temperature that is taken in, compressed, and finally 
delivered per cycle. First, the valves, passages and walls are not at the same 
temperature as the entering gas, due to the heat left from the compression of 
the previous charge. This causes the temperature of the gas within the cylin- 
der to be something higher than the supply gas outside. This causes it to 
be less dense, and hence an equivalent weight of gas at supply temperature 
and pressure would occupy a volume somewhat less than AB, Second, the gas 
which occupies the volume AB has not all entered the cylinder through the intake 



WORK OF COMPRESSORS 157 

valve. After reexpansion is completed the intake valve opens and gas enters 
the end of the cylinder under consideration. At the same time compression 
is taking place in the other end, and later deliver3\ During these processes 
whatever gas leaks past the piston tends to fill the end of the cylinder in which 
intake Ls going on. Leakage past the discharge valve also tends to fill the cylin- 
der with leakage gas. Both of these tendencies decrease the quantity of gas 
entering through this intake valve, and its true amount when reduced to external 
supply pressure and temperature is, therefore, less than the volume AB. 

The triAe lovypressure capacity of the compressor is the true volume of gas 
under external supply conditions that enters the cylinder for each cycle. This 
cannot be determined from the indicator card except by making certain assump- 
tions which involve some error at best. It can, however, be ascertained by 
means of additional apparatus, such as meters or calibrated nozzles or receivers, 
by means of which the true amount of gas compressed per unit of time is made 
known. This reduced to the volume per cycle under supply pressure and tem- 
perature will give the true low-pressure capacity. 

Volumetric efficiency is defined as being the ratio of low-pressure capacity 
to displacement. On the diagram. Fig. 37, the displacement is represented 
to the voliune scale by the horizontal distance between verticals through the 
extreme ends of the diagram, K and H. Since there are three ways in which 
the low-pressure capacity may be approximated or determined, there is a 
corresponding number of expressions for volumetric efficiency. 

1. The volumetric efficiency of the hypothetical cycle is 

JT /^i^^^f Ko+inon - (hypothetical L. P. Cap.) . . 

^.(hypothetical) — -(dis^ace^nt)" ~"' " ' ' ' ^^^^ 

and this is evaluated and used in computations in the foregoing sections of 
this chapter. 

2. The apparent volumetric efficiency is 

T? / x\ (apparent L. P. Cap.) ,^oi \ 

^.(apparent) = ^^^ ,. -, - t/ -, (221) 

^^ (displacement) 

and would be very nearly equal to the true volumetric efficiency were it not 
for leakage valve resistance and heating during suction, but due to this may 
be very different from it. 

3. The trv/e volumetric efficiency is 

g.(true)= (y«L.P.Cap O ^222) 

(displacement) 

In problems of design or prediction it is necessary either to find dimensions, 
speeds and power necessary to give certain actual results, or with given dimen- 
sions and speeds to ascertain the probable power and capacity or other 



158 ENGINEERING THERMODYNAMICS 

characteristics of actual performance. Since it is impossible to obtain actual 
performance identical with the hypothetical, and since the former cannot be 
computed, the most satisfactory method of estimate is to perform the computa- 
tions on the hypothetical cycle, as is explained in previous sections of this chap- 
ter, and then to apply to these results factors which have been foimd by 
comparing actual with hypothetical performance on existing machines as nearly 
like that under discussion as can be obtained. This necessitates access to data 
on tests performed on compressors in which not only indicator cards are taken 
and speed recorded, but also some reliable measurement of gas compressed. 
The following factors or ratios will be found of much use, and should be 
evaluated whenever such data is to be had: 

_ S»(true) (true L. P. Cap.) .^^m 

^ E9 (hypothetical) "" (hypothetical L. P. Cap.)* 

= -^i>(true) _ (true L . P. Ca p.) .^ . 

Et(apparent) (apparent L. P. Cap.) 

— true I.H.P. _ true m.e.p. ,^^. 

^ hypothetical I.H.P. hypothetical m.e.p. 



Then 



true work per cu.ft. gas, supplied 
hypothetical work per cu.ft. gas, supplied 

I.H.P. 



true 



_ true I.H.P. per cu.ft. g as supplied (L. P. Cap.) 

hypothetical I.H.P. per cu.ft. supplied , +Vi f 1 I-H.P. 



(L. P. Cap.) 
true m.e.p. . true L. P. Cap. 63 



(226) 



hypothetical m.e.p. ' hypothetical L. P. Cap. ci 

This ratio can be used to convert from hypothetical work per cubic foot 
gas supplied to probable true work per cubic foot. 

Multi-stage Compressors are subject in each stage to all of the characteristics 
described for single stage to a greater or less extent. Valve resistance, friction 
and inertia affect the intake and discharge lines; heat transfer and leakage 
influence the form of compression and reexpansion lines, and the true capacity 
of the cylinder is made different from the apparent due to leakage, pressure 
and temperatures changes. 

In addition to these points it is useful to note one special way in which the 
multi-stage compressor differs from the single stage. The discharge of the first 
stage is not delivered to a reservoir in which the pressure is constant, but a 
receiver of limited capacity. The average rate at which gas is delivered to 
the receiver must equal the average rate at which it passes to the next cylinder. 
The momentary rate of supply and removal is not constantly the same, however. 



WORK OF COMPRESSORS 



159 



thjB causes a rise or fall of pressure. It is evident that this pressure fluctuation 
is greatest for a small receiver. Very small receivers are not, however, used 
3n gas compressors due to the necessity of cooling the gas as it passes from one 
stage to the next. To accomplish this a large amount of cooling surface must 
be exposed, requiring a large chamber in which it can be done. Thus, it is 
seen that the hypothetical cycles assumed for multi-stage cpmpressors do not 
truly represent the actual cycle, but the difference can never be very great, 
due to the large size of receiver which must always be used. 

Another way in which the performance of this multi-stage compressor 
commonly differs from assumptions made in the foregoing discussions is in 




Fio. 38. — ^Effect of Loss of Intercooling in Two-stage Compressors on Receiver-pressure and 

Work Distribution in the Two Cylinders. 

r^ard to intercooling. It seldom occurs that the gases enter all stages at 
the same temperature. In the several stages the temperature of the gases 
will depend on the amount of compression, on the cooling surface and on the 
amount and temperature of cooling water. The effect of variations in tem- 
perature upon the work and receiver pressures will be taken up later. It may 
be noted now, however, that if all cooling water is shut off, the gas passes from 
one cylinder to the next without cooling j there is no decrease in volmne in the 
receiver. For simplicity take the case of zero clearance, two-stage (Fig. 38). 
let ABCDEF be the cycle for perfect intercooling. AB and KD are the low- 
pressure capacities of the first and second stages respectively. If now, inter- 
cooling ceases, the gas will no longer change volume in the receiver. The 
receiver gas, in order to be made sufficiently dense to occupy the same 



160 ENGINEERING THERMODYNAMICS 

volume {KD) as it did before, must be subjected to a greater pressure in the 
first stage. The new receivei^ line will be K'D'. The wo rk of the first 
stage will therefore be ABD'K'', of the second stage K'D'GF, and the total 
work in the new condition is greater than when intercoooling was perfect by an 
amount represented by the area DCGE. 

In the case where clearance is considered, the effect is the same, except that 
the increasing receiver pressure, increasing the ratio of compression of the first 
stage, causes the volumetric efficiency of the first stage to become less, and 
hence lessens the capacity of the compressor. The efifect on work per unit of 
capacity is the same as without clearance. 

The question as to how many stages should be used for a ^ven compressor 
is dependent upon the ratio of compression largely, and so is due, first, to con- 
siderations of economy, which can be imderstood from the foregoing sections; 
second, for mechanical reasons, to avoid high pressures in large cylinders; third, 
for thermal reasons, to avoid such high temperatures that the lubrication of 
the cylinders would be made difficult, or other dangers, such as explosions, 
involved. 

Practice varies very widely as to the limiting pressures for single, two, 
three or four-stage compressors. Air compressors of a single stage are com- 
monly used for ratios of compression as high as 6 or 7 (75 to 90 lbs. gage). For 
ratios greater than these, two-stage compressors are used, especially for larger 
sizes, up to ratios of 34 to 51 (500 to 750 lbs. gage). Some three-stage com- 
pressors are used for ratios as low as 11 or 14 (150 or 200 lbs. gage), although 
installations of this nature are rare, and are warranted only when power is costly 
and the installation permanent and continuously used to warrant the high 
investment cost. As a minimum ratio for three stages, 11 (160 lbs. gage) j 
is used for large units, while a few small units compress as high as 135 or even | 
170 atmospheres (2000 or 2500 lbs. gage). A notable use for the four-stage 
compressor is for charging the air flask of automobile torpedoes used by the 
various navies, which use pressures from 1600 to 3000 lbs. per square inch 
(110 to 200 atmospheres). These require special design of valves, cylinders 
and packings to withstand the extremely high pressures, small clearances, and 
special precautions against leakage, due to the great loss of voliunetric efficiency 
and economy that would otherwise result. 

20. Work at Partial Capacity in Compressors of Variable Capacity. I 
is seldom that a gas compressor is run continuously at its full capacity. I 
the duty of the compressor is to charge storage tanks, it may be made to run 
at its full capacity until the process is completed and then may be stopped 
entirely, by hand. Even where the compressed gas is being used continuously 
it is common practice to have a storage reservoir into which the compressor may 
deliver. This enables the compressor to deliver a little faster or slower thaa 
the demand for a short period without a great fluctuation pressure in tha 
reservoir. For many purposes hand regulation is not sufficient or is too 
expensive, hence the demand for automatic systems of capacity regulation 
These systems may be classified in a general way in accordance with the method 



WORK OF COMPRESSORS 161 

of driving. Some methods of power application permit of speed variations while 
others require constant speed. The former provides in itself a means of regulat- 
ing capacity within certain limits, while, if the compressor must run at constant 
speed, some additional means of gas capacity control must be provided. 

Ck)mpressors driven by an independent steam engine, or steam cylinders 
constituting part of the same machine may be made to nm at any speed required 
within a very wide range and still kept low enough for safety. If driven 
by gear, belt, rope, chain or direct drive from a source of power whose speed 
is constant, the speed of the compressor cannot be varied. Electric motor, 
gas-engine, oil-engine or water-power drives are subject to only limited speed 
alteration and may, therefore, be placed in the constant speed class. 

Regulation of Capacity by Means of Speed Change. If the speed of the com- 
pressor is decreased below normal: 

1. Displacement of piston is decreased in proportion to the speed. 

2. Mean effective pressure, as to hypothetical considerations, is the same, 
but due to the decrease of velocities in gas passages, the frictional fall of pres- 
siu-e during inlet and delivery is not so great, and hence the mean effective pres- 
sure is not quite so great. If the compressor is multi-stage, since a smaller 
quantity of gas is passing through the intercooler, it is probable that the inter- 
cooling is more nearly perfect, thus decreasing the mean effective pressures in 
the succeeding stages. 

3. The volumetric eflSciency is changed, due first to the fact that leakage 
is about the same in total amount per minute as at full speed, but the total 
quantity of gas being less, leakage is a larger percentage of the total; second, 
the inertia of gases in the supply pipe, as well as their friction, has been decreased. 
The former tends to decrease vulmetric efficiency, while the latter may tend 
to increase or decrease it. It may be expected that the true voliunetric eflSciency 
will be somewhat greater at fractional speed than at full speed. 

For any compressor there is a speed of maximum economy above and below 
which the economy is less, though it may be that this most economical speed is 
greater than any speed of actual operation. 

It is not desirable at this point to discuss the effect of speed variation upon 
the economy of the engine or other motor supplying the power. The reasoning 
above applies to the term economy as applied to the compression effect obtained 
per unit of power applied in the compression cylinder. It might be noted here, 
however, that the decrease of speed has little effect upon the mechanical efficiency 
of the compressor as a machine, since frictional resistance between solid parts 
remains nearly constant, and, therefore, power expended in friction will vary 
as the speed, as does approximately also the power to drive the compressor. 
The ratio of frictional power to total may then be expected to remain nearly 
constant. 

Regulation of Capacity at Constant Speed may be accomplished in a nimiber 
of ways: 

1. Intermittent running; 

2. Throttling the supply to compressor; 



162 ENGINEEEING THERMODYNAMICS 

3. Periodically holding open or shut the intake valve; 

4. Closing intake valve before end of intake stroke, or holding intake valve 

open until compression stroke has been partially completed; 

5. Large clearance; 

6. Variable clearance, 

The first necessitates some means for stopping and starting the compressor, 
which is simple with electric drive, and may be accomplished in other cases by 
means of a detaching clutch or other mechanical device. The pressure in the 
reservoir is made to control this stopping and starting device by means of a 
regulator. This arrangement is made to keep the pressure in the reservoir 
between certain iBxed limits, but does not maintain a constant pressure. The 
economy of compression in this case is evidently the same as at full speed 
continuous running, provided there is no loss in the driving system due to 
starting and stopping, which may not be the case. This method of 
regulation is used mainly for small compressors in which inertia is not 
great, such as supply the air brakes on trolley cars. The sudden change 
of load on the driving machinery would be too great if large compressors 
were arranged in this way. 

If the compressor whose capacity is regulated by intermittent running is 
mvlMi-stagej the constant supply of water to the intercoolers while the compres- 
sor is stopped will lower the temperature of the cooling surface, causing naore 
nearly perfect intercooling when the compressor is started. Leakage, on the 
other hand, will permit the loss of pressure to a greater or less extent in the 
receivers while the compressor is stationary, which must be replaced after 
starting before effective delivery is obtained. 

Throttling the gas supply to the compressor has certain effects that may 
be studied by referring to Fig. 39, which represents the hypothetical cycles most 
nearly approaching this case. In order to reduce from the full-load low-pres- 
sure capacity, ilB, to a smaller capacity, AE^ the supply pressure is decreased 
by throttling to the pressure of B\ such that B' and E lie on the same adiabatir. 
The work area A'B'EA is entirely used up in overcoming the throttle resistance 
and is useless friction, so that economy is Seriously reduced by this method 
of regulation. Such compressors may use almost as much power at partial 
as at full capacity. 

It is easily seen that this method of regulation would be undesirable, its 
only advantage being simplicity. 

The effect of throttling upon a multi-stage compressor may be illustrated as 
in Fig. 40, by considering the two-stage compressor cycle without clearance, 
ABCDEF. The ratio of compression of the first cylinder is determined with 

perfect intercooling by the ratio of displacements Pc — Pb(jr\ - When the 

supply pressure is throttled down to Pc', the new receiver pressure will be 

Pc =Pb \fr) > a pressure much lower than Pc. Hence the receiver pressure 

is decreased, less work done in the first stage, and far more than half the work 



WORK OF COMPRESSORS 



163 



of compression done in the second stage. If best-receiver pressiire existed at 
normal capacity, it does not exist in the throttled condition. 

The intake valve may be held wide open or completely closed during one 
or more revolutions, thereby avoiding the delivery of any gas during that period. 
If the intake valve is held wide open, the indicator card would be as shown in 
Fig. 41A in full lines, ABCD, the dotted lines showing the cycle performed when 



D 


c 


c 














* 




\ 

\ 

\ 

\ 

\ 

\ 


\ 
















1 


\ 

\ 
\ 

\ 






% 




• 










\ 

\ 
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\ 

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\ 




\ 

\ 

\ 

\ 

\ 

\ 
















1 \ 


\ 


\ 

\ 

\ 

\ 

\ 






\ 












\ 




\ 
\ 

\ 

N 


E 








B 






K 

1 N 

1 \ 

\ 

1 > 


\ 






■----, 












1 


N 

N 


A' 










B' 
















\^ 


ft -L.P.Cap Throttled *> 


T V f^awx a4 TNilI ▼ -^^^A ... ->- 






1^ 

1 


14. t^ .Cap at Villi 


XJUliU 






v^ 






J 


J 






^ 





Fig. 39. — Effect of Throttling the Suction of One-stjige Compressors, on Capacity and 

Econonij'. 



normal operation is permitted. With the inlet valve open in this way there is 
a loss of power due to friction of the gas in passage during both strokes, measured 
by the area within the loop. 

Closing the inlet valve and holding it shut will give an indicator card of the 
form EFG, Fig. 41 S, which will be a single line retraced in both directions 
except for probable leakage effects. If leakage is small, there will be but little 



164 



ENGINEERING THERMODYNAMICS 



area enclosed between the lines. At a high speed this might be expected to 
incur less lost power than the former plan. 

Certain types of compressors are made with an intake valve controlled 
by a drop cut-ofiF, much like the steam valve of the Corliss engine. The effect 
of this is to cut oflF the supply of gas before the end of the stroke, after which 
time the gas must expand hjrpothetically according to the adiabatic law. The 
return stroke causes it to compress along the same line continued up to the 
delivery pressure, as indicated by the line FEGy Fig. 41C. There is little work 




FiQ. 40. — ^Effect of Throttling Multi-stage Compressors on Receiver-pressure, Work Distri- 
bution, Capacity and Economy. 



lost in the process, none, if the line is superimposed as in the figure, and hence 
the process is the same as if only the cycle AEGD were performed. 

The same quantity of gas might have been entrapped in the cylinder by 
holding the intake valve open until the end of the stroke and on the return till 
the point E, Fig. 41D, was reached, then closing it. The same compression 
line EG will be produced. The line AB will not coincide with BE, due to 
friction of the gas in passages, and hence will enclose between them a small 
area representing lost work, which may be no larger than that lost in the process 
EFE. 



WORK OF COMPRESSORS 



165 



If such an automatic cut-ofiF were applied only to the first stage of a mvMi- 
singe compressor, the eflFect would be to lower receiver pressures as in the throt- 
tling process. To avoid this, the best practice is to have a similar cut-oflF to act 
on the supply to all of the stages. If this is properly adjusted, the receiver 
pressures can be maintained the same as at full load. An additional advantage 
of this system is that even if the compressor is to be used for a delivery pres- 
sure for which it was not originally designed, the relative cut-oflFs may be so 
adjusted as to give and maintain best-receiver pressiu'e. 



, 
















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(Su 


?. Pr. 


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'ite \t 




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(B) 













C D 

Fig. 41. — Control of Compressor Capacity by A. Open Inlet Valve; B. Closed Inlet Valve; 

C. Suction Cut-ofif; Z). Delayed Suction Closure. 

Since the low-pressure capacity per cycle of a compressor involves clearance 
and ratio of compression as two of its variables, it is possible to change capacity 
by changing either the clearance or the ratio of compression. 



(L. P. Cap.)=Z)£f,=D(l+c-cJBp« ). 



(227) 



Assuming that clearance is a fixed amount and not zero, it is evident that an 
increase in the ratio of compression decreases the capacity, and when it has 



166 



ENGINEERING THERMODYNAMICS 



reached a certain quantity will make the capacity zero. If the clearance is 

large, making the coefficient of fip* large in the equation the efifect of a change 
in that factor is increased. Fig. 42 indicates the hjrpothetical performance 
of a compressor with large clearance. When the pressure of delivery is low 
(say Pe) the capacity is large, AB. The cycle is then ABCD. An increase 
of the delivery pressure to P/ changes the cycle to A' BCD' and the low pres- 
sure capacity is A'B, If the compressor is delivering to a receiver from which 
no gas is being drawn, the delivery pressure will continue to rise and the capacity 





m 














• 
















0' 


V 
















\ \ 














D 


\ \ 

\ 

\ 


\c 














; — \" 
\ \ 

\ 




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^^ 












A 




A' 




B 

• 


^ L 1^ ■% 


^ -rx «^ 




^"C XJ ^ 






±j 






y 





Fia. 42. — ^Variation of Compressor Capacity with Rise of Delivery Pressure, Fixed Clearance, 

Pressure for Zero Delivery. 

to decrease till the capacity approaches zero as the delivery pressure approaches 
the pressure Pe as a limit. 

(limiting del.pr.)=(sup.pr.)( ?j (228) 

When the limiting condition has been reached and the capacity has become 
zero, the compression and reexpansion lines coincide and enclose zero area 
between them; hence, the njcan effective pressure and the indicated horse- 
power are zero, for the hypothetical case. Leakage will prevent a perfect 
coincidence of the lines and cause some power to be required in addition to that 
of friction. 

Such a simple method of regulation as this is used for some small com- 
pressors driven constantly from some source of power used primarily for 
other purposes. When it is not necessary to have a constant delivery pressure, 



WORK OP COMPRESSORS 



167 



but only to keep it between certain limits, this may be made use of, especially 
if the limits of pressure are quite wide. 

The expression for low-pressure capacity Eq. (227), suggests the possi- 
bility of decreasing capacity by the increase of clearance. The effect of this is 
shown in Fig. 43. The original compression cycle (full capacity) is shown by 
ABCD, with a clearance volume of cD, so that the axis of zero volume is OP. 
Increasing the clearance to c'jD causes a smaller volume CD to be delivered 
and due to the more sloping re-expansion DA\ a smaller volume of gas is 
taken in, A'B, 

It has been shown in previous sections that clearance has no effect upon the 
economy of a compressor so far as hypothetical considerations are regarded. 
In practice it is found that a slight loss of economy is suffered at light load. 





p 


D 




c 




C 


















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\ 
V 




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\ 


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r 


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— < 












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^N 


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"n 


s 






^N, 


^ 


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0' 




1 


'A 

r 

1 
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1 
1 












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W—' — I 


i.P. Cap Part Load 

uirl 


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h 


—C-D 


"U-i: " 


.r. VAip 


-© — 


IlKl 


» 





Fig. 43. — ^Variation of Compressor Capadty by Changing Clearance. 

as might be expected, due to greater leakage per unit of capacity. The addi- 
tional clearance is provided in the form of two or more chambers connected to the 
clearance space of the compressor by a passage in which is a valve automatically 
controlled by the receiver pressure. 

In the muUi'Stage compressor, decreasing the capacity of the first stage by 
an increase of its clearance would evidently permit a decrease of receiver pres- 
sures unless the capacity of each of the various stages is decreased in the same 
proportion. Eq. (132) gives the condition which must be fulfilled to give best 
receiver pressure for a two-stage compressor. 



RJ = 



M 



1 + Cl — Cl/?|,2« 



[ 



I>2 1+C2 — C2/2p2« 



Is 



168 ENGINEERING THERMODYNAMICS 

Since Di, D29 and Rp remain fixed, for any chosen value of clearance of the 
first stage, a, the clearance of the second stage, C2, to give best-receiver 
pressure can be found, 

[i-ci(jep2l-i)]Di 

C2 = i i (229) 

(Rj^*-l)Rp2sD2 

For every value of first-stage clearance there is a corresponding clearance of 
second ^stage that will give best-receiver pressure, found by this equation. Sim- 
ilar reasoning can be applied to three- or four-stage compressors. 

21. Graphic Solution of Compressor Problems. In order to obviate the 
necessity of working out the formulas given in this chapter each time a prob- 
lem is to be solved, several of them have been worked out for one or more 
cases and results arranged to give a series of answers graphically. By the 
use of the charts made up of these curves many problems may be solved 
directly and in many others certain steps may be shortened. A description of 
each chart, its derivation and use is given in subsequent paragraphs. 

Chart, Fig. 44. This chart gives the work required to compress and deliver 
a cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute if the ratio of pressure (del.pr.)-f- (sup.pr.), the value 
of s and the (sup.pr.) are known, and compression occurs in one stage. The 
work or H.P. for any number of cubic feet is directly proportional to number 
of feet. The curves are dependent upon the formulas, Eq. (31), for the case 
when 8=1, and Eq. (51) for the case when s is not equal to 1. These formulas 
are: 

Eq. (31), W per cu.ft. = 144 (sup.pr) log« Rp] 
" (51), W per cu.ft. = 144--^— (sup.pr.) (fip"^-lV 

These equations are difficult to solve if an attempt is made to get a relation 
between the work and ratio of pressures. This relation may, however, be 
worked out for a number of values of pressure ratios and results plotted to 
form a curve by which the relation may be had for any other ratio within 
limits. This has been done in this figure in the following manner: 

On a horizontal base various values of Rp are laid off, starting with the value 
2 at the origin. The values for work were then found for a number of values 
of Rp with a constant value of (sup.pr.) and s. A vertical work scale was 
then laid off from origin of Rp and a curve drawn through the points found 
by the intersection of horizontal lines through values of work, with vertical 
lines through corresponding values of Rp, The process was then repeated for 
other values of s and curves similar to the first, drawn for the other values 
of 8. From the construction so far completed it is possible to find the work per 
cubic foot for any pressure ratio and any value of s for one (sup.pr.) by pro- 



WORK OF COMPRESSORS 



atnpwqY -01 bg wd "gqi (■y -iiofl) 



170 ENGINEERING THERMODYNAMICS 

jecting up from the proper value of Rp to the curve of value of s and then hor- 
izontally to the scale of work. It will be noted from these formulas, however, 
that the work may be laid off on the horizontal base and a group of lines drawn 
so that the slope of the line equals ratio of work for any supply pressure to that 
for the (sup.pr.) originally used. For convenience, in order that the group of 
8 curves and the latter group may be as distinct as possible, the origin of the 
latter group is taken at the opposite end of the base line. If from the point 
for work originally found, a projection is made horizontally to the proper 
(sup.pr.) curve, the value for work with this (sup.pr.) will be found directh- 
below. It will be noted that from point of intersection of the vertical from the 
Rp value with the s curve, it is only necessary to project horizontally far enough 
to intersect the desired (sup.pr.) curve, and since no information of value will 
be found by continuing to the work scale for the original (sup.pr.) this is omitted 
from the diagram. 

In brief, then, the use of this chart consists in projecting upward from the 
proper value of Rp to the proper s curve, then passing horizontally to the value of 
(sup.pr.) and finally downward to the work scale. As an example of use of the 
curve Ex. 2 of Section (8) may be solved directly. This is to find the work 
to compress 1000 cu.ft. of free air from 1 to 8J atmospheres adiabatically. 
On the curve project upward from iiJp = 8.5 to curve of s = 1.406, then over 
to 14.7 (sup.pr.) curve and down to read work = 6,300,000 as found, for exam- 
ple, by use of formulas in Section (8). 

Chartj Fig. 45. This gives the work required to compress and deliver a 
cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute if the ratio of pressures, the value of s and (sup.pr.) 
are known and if compression occurs in two stages with best-receiver pressure 
and perfect intercooling. The work or H.P. for any other number of cubic feet 
may be found by multiplying work per foot by the -number of feet. The 
method of arriving at this chart was exactly the same as that for one stage. 

As an example of the use of the chart-. Example 2 of Section (9) may be 
solved directly. This problem calls for the work to compress 5 cu.ft. of free 
air from 1 to 8| atmospheres adiabatically in two stages. Project upward from 
ftp = 8.5 to curve 5 = 1.406, then over to 14.7 curve and down to read 5320 ft.-lbs. 
per cubic foot, which is same as found from the formula in Section (9). 

Chartj Fig. 46. This chart gives the work necessary to compre&s and deliver 
a cubic foot of (sup.pr.) air or horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute, if the ratio of pressures, the value of s, and the 
(sup.pr.) are known and if the compression occurs in three stages with best- 
receiver pressures and perfect intercooling. The work or horse-power for any 
other number of cubic feet may be found by multiplying the work for one 
foot by the number of feet. 

As an example of use of this chart, Example 2 of Section (13) may be 
solved directly byit. This calls for the horse-power to compress 100 cu.ft. free 
air per minute adiabatically in three stages from 15 lbs. per square inch abs. 
to 90 lbs. per square inch gage. From Rp = 7f project to curve of s = 1.4 then 



WORK OF COMPRESSORS 



171 



«\ -i 




I 



O 



& 



o o 
V 8 






o 

W 

S 
I 

c 

o 



H5 

O 



k ^ 



•tqv 'ni 'bS 



8S 



CO CO CO A w '^ 



6 



ENGINEERING THEEIMODYNAMICS 



111 



'sqv 'QI ''>I>8 'isil J|V <''i<I'<Ii>9) 



-IS 



(,2.5 ■ -c - 

l» Its 

SI i| 

— i i 's i 






i ■ 



WORK OF COMPRESSOES 173 

)ver to (sup.pr.) = 15 and down, and the horse-power will be found to be 13.6 
\s before by use of formulas. 

Chart, Fig, 47. This chart is for finding the (m.e.p.) of compressors. In the 
lase of multi-stage compressors with best-receiver pressure and perfect inter- 
»oling, the (m.e.p.) of each cylinder may be found by considering each cylinder 
is a single-stage compressor, or the (m.e.p.) of the compressor referred to the 
[j.P. cylinder may be found. 

The chart depends on the fact that the work per cubic foot of (sup.pr.) gas 

s equal to the (m.e.p.) for the no clearance case and that the (m.e.p.) with 

:learance is equal to the (m.e.p.) for no clearance, times the volumetric 

fficiency. Diagrams 1, 2 and 4 are reproductions of Figs. 45, 46 and 47 to a 

imaller scale and hence need no explanation as to derivation. Their use may 

)e briefly given. From the proper ratio of pressures project upward to the 

>roper curve, then horizontally to the (sup.pr.) and downward to read work per 

*ubic feet of (sup.pr.) gas. 

The volumetric eflSciency diagram was drawn in the following manner: 

1 

From Eq. (65) vol. eflf. = (l+c— cfip* ), showing that it depends upon three 
variables, Rp, c and s. A horizontal scale of values of Rp was laid off. Values 

)f Rp* were found and a vertical scale of this quantity laid off from the same 

jrigin as the Rp values. Through the intersection of the verticals from various 

iralues of Rp with the horizontals drawn through the corresponding values of 
1 

[Rp)' for a known value of s, a curve of this value of s was drawn. In a similar 

ifrav curves of other values of s were drawn. From the construction so far 

1 

completed it is possible to find the value of (Rp) * by projecting upward from any 

value oi Rp to the curve of s and then horizontally to the scale of (Rp)' . Values 

1 

rf volumetric efficiencies found for various clearances and the values of {Rp)' 

ire laid off on a horizontal base, with the origin at the opposite end of scale 

from that of Rp values, in order that clearance curves and s curves might be 

as distant as possible. These clearance curves were drawn through the inter- 
im 

?ection of horizontals through the (Rp)' values and of verticals through the 

vulmetric efficiency values corresponding to them for the particular clearance 

in question. 

To find volumetric efficiency then it is merely necessary to project from value 

Df Rp to the correct s curve, then across to the proper clearance and finally 

1 

down to volumetric efficiency. As the value of (Rp)' is not desired, the hori- 
zontal projection is carried only to the intersection with the clearance curve 
and not to the edge of the diagram. To find the (m.e.p.) for single stage, the 
work per cubic foot is found from the diagram and then the volumetric efficiency, 
both as described above. The product is (7n,e,p,) 

For multi-stage compressors with perfect intercooling and best-receiver 
presssure, as stated above, the (m.e.p.) of each cylinder may be found, considering 



ENGINEERING THERMODYNAMICS 



II ItuUootPressurp i I I 1 J 

« et 77 7D SS SB 49 4^! 3S S8 21 It t 

Work ptr Cu. Ft. Of (Sup. I'r.) Guas-Ul 

Fiti. 47. — Mean EffecUve Preseuro of Compreswra, One-, Two-, and Three-etagea. 



WORE OF COMPRESSOES 
Initial Pi«isUTe Lba. per Sq. In, Ati«. 



Hulio of Pressures 
nsi7TTD<BSeiaU35e8El ul ( 

Work per Cu. Pi. at (Sup. IT.) Oii3-^l*4 

Fio. 47. — Mean Effective Pressure of ComprcBsora, One-, Two-, and Three-Btages. 



176 ENGINEERING THERMODYNAMICS 

each to be a single-stage compressor and remembering that (1 rec.pr.) becomes 
(sup.pr.) for second stage, and (del.pr.) for first stage and that (2 rec.pr.) 
becomes (sup.pr.) for third stage, (del.pr.) for second stage. The (m.e.p.) reduced 
to low-pressure cylinder is found by taking work per cubic feet of (sup.pr.) 
gas and multiplying by volumetric efficiency of low-pressure cylinder. 

To illustrate the use of this curve the example of Section (16) may be 
solved. Projecting upward from the pressure ratio of 9.35 to the line of a = 1.4 
and then over to (sup.pr.) = 15 in diagram 4, since compression is three stage 
and from 15 lbs. per square inch to 140 lbs. per square inch, work per cubic 
foot or (m.e.p.), is found for no clearance to be 37.8 abs. per square inch. Since 
best-receiver pressure assumed is 31.6, which gives a ratio of 2.1 for the low- 
pressure cylinder. From diagram 3, by projecting upward from Rp = 2.1 and 
over to the 5 per cent clearance line volumetric efficiency is 96.5. The product 

gives (m.e.p.) reduced to low-pressure cylinder and is 36.5. From the — ^^ — 

33,000 

formula, horse-power is found to be 358 as before. 

Charty Fig. 48. As mentioned in Section 18, there is one (sup.pr.), which 

for a given (del.pr.) will give the maximum work of compression. The chart, 

Fig. 48, originated by Mr. T. M. Gunn, gives a graphical means of finding thi\ 

value of (sup.pr.) when the (del.pr.), clearance and value of s are known. It also 

gives on the right-hand of the chart a means for finding the (m.e.p.) for this 

condition. The figure was drawn by means of Eqs. (214) and (218). For the 

value of s = 1 the ratio of (del.pr.) to (sup.pr.) was found for cases of clearances 

from per cent to 15 by means of Eq. (218) by trying values of this ratio which 

would fulfill the condition of the equation, log^^p^T^) = ^""1^1^ (sup.pr.) * 

For values of s not 1, Eq. (214) was used, and a set of values of Rp found 
for the values of s= 1.4 and 1.2 by trial, the correct value of Rp being that which 
satisfied the equation. 



8-1 1 

/del.pr. \ « ^ fi I ^~ 1 /del.pr. 






\sup.pr./ 1+c 



L s Vsup.pr./ J 



As an example the work foi* the case where s = 1.2 and c= 10 per cent is given. 
Try Rp=2.6, then, R,T^ =j~ri+.l-. 1X^^X2.6*331 

= 1.091(1.1 - .01667X2.218), 



= 1.161 



WORK OF COMPRESSORS 



omwMjXiaAnsaoKiI'aiii] ]0 oiisg tanai|x«K 



X-iOjunininrrsjiioj fy 'Bajnwai j jo odbh 



178 ENGINEEEING THEEMODYNAMICS 

Rp — (1.161)^ =2.45, which shows the value of 2.6 to be incorrect. For a second 
trial take 2.45, and then, 

/?p*, = 1.091(l.l-.01667X2.45-^), 

= 1.1627 

= (1.1617)6 = 2.458, 
which is sufficiently close. Therefore the value of R^^ for s = 1.2 and clearance = 
10 per cent, is 2.45; that is, maximum work will occur for a given (del.pr.) 

when (sup.pr.) is g^ times the (del.pr.) 

When the values for iJp had been obtained a horizontal axis of values of % 
and a vertical one of Rp values, were laid off and the points for clearance ciuT-es 
laid off to their proper values referred to these axes. Through points as plotted 
the clearance lines were drawn. The right-hand diagram was plotted in a 
similar manner from Eqs. (213) and (217), for s not equal to 1 and equal to 
1 respectively. 

The latter formula was rearranged in the form 

/m.e.p.\ logei2p„ 
Vdel.pr./ Rp ^'' 

the last term being foimd from curve of Fig. 45. The value of Rp for each 
value of the clearance was taken from the left-hand diagram, and substituted 

in the above expression to obtain ( j *, ' - jfor the case of s= 1. EJq. (213) was 

put in form 

/m.e.p.V s /p'— .^jp 

and values of Rp for each value of the clearance found in the left-hand diagram 

were substituted, together with Ej, values from Fig. 45 and the value of ( , \ ' j 

found for each case of clearance when s = 1.4. When the points for 8 = 1 and 
8 = 1,4 had been found, a horizontal axis of values of 8 and a vertical one of 
values of Rp were laid off, and points for the clearance curves plotted as for 
the left-hand diagram and the curves drawn in. 

To find the (sup.pr.) to give maximum work for any (del.pr.) it is only 
necessary to project from the proper value of 8 to the proper clearance curve, 
and then horizontally to read the value of Rp. The (del.pr.) divided by this 
gives the (sup.pr.) desired. To obtain the (m.e.p.) project upward from the 

value of 8 to the clearance curve, then horizontally to read the ratio {ttV"^") 

The {del.pr,) times this quantity gives the m.e.p. 

As an example of the use of this chart let it be required to find the (sup.pr.) 
for the case of maximum work for 9X12 in. double-acting compressor running 
200 R.P.M., having 5 per cent clearance and delivering against 45 lbs. per square 
inch gage. Also the horse-power. Compression such that 8 = 1.3. 



WORK OF COMPRESSOKS 



Projecting from the value 1.3 for s on the left-hand diagram to the line of 

60 
5 per cent clearance find flp to be 2.8, hence (sup.pr.) =2g=21.4 lbs. per square 

inch absolute = 6.4 lbs. per square inch gage. Again, projecting from value 1.3 



for s on right-hand dia^am to line of 5 per cent clearance find that^ 
bcnce fm.e.p). = 23 and 



THP -23X1X64X400 ,„„ 
LH.P.---33^j^^— =17.8. 



( m.e.p .) ^ 
Cdei.pr.)" 



Fig. 49. — R«lativc Work of Two- and Thmc-Gtage Compressors Compared to Single Stapc. 

Chart, Fig. 49. This chart is designed to show the saving in work done in 
compressing and delivering gases by two-stage or three-stage compression with 
bestr-receiver pressure and perfect intcrcooling over that required for com- 
pressing and delivering the same ga.s between the -same pressures in one stage. 



180 ENGINEERING THERMODYNAMICS 

The chart was made by laying ofT on a horizontal base a scale of pressure 
ratios. From the same origin a scale of work for two or three stage divided by 
the work of one stage was drawn vertically. For a number of values of if, 
the work to compress a cubic foot of gas was found for one, two and three stage 
for each value of s. The values found by dividing the work of two or three 
stage by the work of single stage were plotted above the proper Rp values and 
opposite the proper ratio values and curves drawn through all points for one 
value of s. To find the saving by compressing in two or three stages project 
from the proper Rp value to the chosen s curve for the,desired number of stages, 
then horizontally to read the ratio of multi-stage to one-stage work. This value 
gives per cent power needed for one stage that will be required to compress 
the same gas multi-stage. Saving by multi-stage as a percentage of single 
stage is one minus the value read. 

To illustrate the use of this chart, find the per cent of work needed to 
compress a cubic foot of air adiabatically from 1 to 8i atmospheres in two 
stages compared to doing it in one stage. From examples under chart Nos. 
44 and 46 it was found that work was 6300 ft.-lbs. and 5320 ft.-lbs. respec- 
tively, for one- and two-stage compression, or that two stage was 84.5 per cent 
of one stage. From Rpj 8J project up on Fig. 49 to 8 = 1.406 for two stage and 
over to read 84.6 per cent, which is nearly the same. 

Chart, Fig, 50. This chart, designed by Mr. T. M. Gunn, shows the 
economy compared to isothermal compression. 

The chart was drawn on the basis of the following equation: 

•n /• xi_ i\ m.e.p. isothermal (no clearance) 

Economy (isothermal) = '^ - -— -^— - — l 

^ m.e.p. actual -r^, actual 

(sup .pr.) log e fip 



8-\ 

logeiZ 



(sup.pr.) (/2j, B — 1) 



s '-^ 



8-1 



{Rp—-\) 



Values of this expression were worked out for each exponent, for assumed values 
of Rp, A scale of values of Rp was laid off horizontally and from the same 
origin a vertical scale of values of the ratio of isothermal to adiabatic. The 
results found were then plotted, each point above its proper Rp and opposite its 
ratio value. Curves were then drawn through all the points found for the 
same value of s. In a similar way a set of curves for two stage and a set for 
three stage were drawn. 

This chart Is also useful in obtaining the (m.e.p.) of th^ cycle if the (sup.pr.) 
and the volumetric efficiency of the cylinder be known. A second horizontal 
scale laid off above the Rp scale shows the (m.e.p.) per pound of (sup.pr. for) 
the isothermal no-clearance cycle. This is found to be equal to log« Rp, since 



WORK OF COMPEESSOES 



r 



'pjBpireiB |viiuaii)osi iqju pa]Bdiua3 ^luouosa. 



182 ENGINEERING THERMODYNAMICS 

the (m.e.p.) for no clearance is equal to the work per cubic foot of (sup.pr.) ga=, 
which, in turn, for the isothermal case is (sup.pr.) log* Rp or log« Rp when 

(sup.pr.) = 1. 

Knowing the ratio of pressures, economy compared to isothermal can be 
found as explained above. Also knowing Rp the (m.e.p.) per pound initial is 
found from the upper scale. 

Since the latter quantity is assumed to be known, by multiplying it by- 
factor just found there is obtained (m.e.p.) isothermal. Since volumetric efficiency 
is assumed known, all the factors are known for the first equation given above 
which, rearranged, reads 

, N X 1 m.e.p. isothermal (no clearance) 

(m.e.p.) actual = -r- = — tt jr — ^ , 

(economy isothermal) -^£r, 

Chart, Fig, 51. This chart is drawn to give the cylinder displacement for a 
desired capacity, with various values of /2p, s and clearance. From the formula 
Eq. (64) 

1^ 
(L. P. Cap.)=Z)(l+c-ci2p«). 

The right-hand portion of the diagram is for the purpose of finding values 

of (/2p) « for various values of Rp and s, and is constructed as was the similar 

curve in Fig. 45. The values of the lower scale on the left-hand diagram give 

i. 

values of D = (L. P. Cap.)-4-(l+c— c/?p*)i where capacity is taken at 100 cu.ft., 

this scale was laid out and the clearance curves points found by solving the 

above equation for various values of (Rp) * for each value of c. To obtain the 
displacement necessary for a certain capacity with a given value of Rp^ c and 
5, project upward from Rp to the proper s curve across to the c curve and down to 
read displacement per hundred cubic feet. Also on the left-hand diagram are 
drawn lines of piston speed, and on left-hand edge a scale of cylinder areas 
and diameters to give displacements found on horizontal scale. To obtain 
cylinder areas or approximate diameters in inches project from displacement to 
piston speed line and across to read cylinder area or diameter. Figures given 
are for 100 cu.ft. per minute. For any other volume the displacement and 
area of cylinder will be as desired volume to 100 and diameters will be as 

v^desired volume to 100. 

As an example, let it be required to find the low-pressure cylinder size for a 
compressor to handle 1500 cu.ft. of free air per minute. Receiver pressure to 
be 45 lbs. per square inch gage and (sup.pr.) to be atmosphere. Piston speed 
limited to 500 ft. per minute. Compression to be so that s=L4 and clear- 
ance =4 per cent. Projecting upward from /2p=4 tos==1.4, across to c = 4%, 
and down to piston speed = 500, find the diameter of a cylinder for 100 cu.ft. 

per minute is 6.3. For 1500 cu.ft. diameter will be as Vl5X6.3 = 3,9X6.3=24 
ins. 



WORK OF COMPRESSORS 



183 




o 

08 

& 

O 
c 
> 

O 

O 



o 

8 



c 
O 

I 

u 

a 



.9 



if 



o 






o 



r-t "^S 



o 






00 






5 



184 ENGINEERING THERMODYNAMICS 



GENERAL PROBLEMS ON CHAPTER II. 

Prob. 1. One hundred cubic feet of HjS are compressed from 15 lbs. per square 
nch absolute to 160 lbs. per square inch absolute. 

(a) Find work done if compression occurs isothermaUy in a no-clearance on&«tage 
compressor; 

(h) Adiabatically in a two-stage, no-clearance compressor; 

(c) Adiabatically in two-stage compressor each cylinder having 5 per cent clearance; 

Prob. 2. Air is being compressed in three plants. One is single-stage, the second is 
two-stage, and the third is three-stage Considering the compressors to have no clear- 
ance and 1000 cu.ft. of free air compressed adiabatically per minute from atmosphere 
to 160 lbs. per square inch gage, what will be the horse-power required and cylinder 
sizes in each case? 

Prob. S* A two-stage compressor with 5 per cent clearance n the h gh and 3 per 
cent in the low-pressure cylinder is compressing air from 14 lbs. per square inch al> 
solute to 125 lbs. per square inch gage. What is the best-receiver pressing and what 
must be the size of the cylinders to handle 500 cu.ft. of free air per minute? 

Prob. 4. A manufacturer gives a capacity of 970 cu.ft. for a 22i x24 in. single- 
stage compressor running at 142 R.P.M. when working pressures are 50 to 100 lbs. per 
square inch gage. What would be the clearance for each of these pressures assuming 

8 = 1.4? 

Prob. 5. The card taken from a single-stage compressor cylinder showed an appar- 
ent volmetric efficiency of a 95 per cent and a mean value for s of 1.3. What is the 
clearance and what would be the (m.e.p.) for the ratio of pressures of 6? 

Prob. 6. A compressor with double-acting cylinder 12x14 ins., having 6 per cent 
clearance, is forcing air into a tank. Taking the volumetric efficiency as the mean of 
that at the start and the end, how long w 11 it take to build up 100 lbs. per square inch 
gage pressing in a tank of 1000 cu.ft. capacity if the speed is 100 M.R.P. and compres 
sion is isothermal, and how long wi 1 it take to do it if the compression is adiabatic and 
the air in the tank does not cool during filling? What is the ma>ximum attainable pressure? 

Prob. 7. It is desired to supply 1000 cu.ft. of air per minute at a pressure of 80 lbs. 
per square inch gage. A two-stage compressor is to be used, the clearance in low pres- 
sure of which is 3 per cent. What must be the displacement of the low-pressure cylin- 
der and what will be the horse-power of the compressor? 

Prob. 8. The low-pressure cylinder of a compressor is 18x24 ins. and has a clear- 
ance of 4 per cent. The receiver pressure is 60 lbs. per square inch absolute. The high- 
pressure cylinder has a 5 per cent clearance. What must be its diameter, stroke being 
same as low, so that compressor will operate at its designed receiver pressure? 

Prob. 9. The discharge pressure of a two-stage compressor is 120 lbs. per square 
inch absolute and the supply pressure is 15 lbs. per square inch absolute The compressor 
is 10x16x12 ins. The clearance in the low-pressure cylinder is 3 per cent. What 
must be the clearance in the high-pressure cylinder for the machine to operate at best- 
receiver pressure? Compression is adiabatic. 

Prob. 10. If the clearance in the high-pressure cylinder of Prob. 9 were reduced 
to 3 per cent, would the receiver pressure increase or decrease, how much and why? 

Prob. 11. If the discharge pressure in Prob. 9 fell to 100 lbs. per square inch absolute, 
what would be the new best-receiver pressure and why? Would the original clearance 
allow the new best-receiver pressure to be maintained? 



PROBLEMS ON CHAPTER II 185 

Prob. 12. The discharge pressure for which a 20jx32jx24 in. compressor is 
designed, is 100 lbs. per square inch gage, supply pressure being 14 lbs. per square inch 
absolute. The d scharge pressure is raised to 125 lbs. per square inch gage. The 
clearance on the high-pressure cyUnder can be adjusted. To what value must it be 
changed to enable the compressor to carry the best-receiver pressure for the new 
dischai^ pressure? Low-pressure clearance is 5 per cent at all times and com- 
pression being adiabatic. 

Prob. 13. A manufacturer builds his 15jx25ixl8 in. compressors with low-pres- 
sure cylinders of larger diameter for high altitude work. What would be the diameter 
of a special cylinder for this compressor to work at an altitude of 10,000 ft. and what 
would be the horse-power per cubic foot of low-pressure air in each case? 

Prob. 14. A three-stage compressor has 4 per cent clearance in all the cylinders. The 
low-pressure cylinder is 34x36 ins., delivery pressure 200 lbs. per square inch gage, 
supply pressure 14 lbs. per square inch absolute. What must be the size of the other 
cylinders for the machine to operate at best-receiver pressure. 

Prob. 16. The cylinders of a two-stage compressor are given as lOJ and 16J ins., 
the stroke being 12 ins. The machine has a capacity of 440 cu.ft. per minute at 160 
R.P.M., the supply pressure is 14 lbs. per square inch absolute and the dehvery 
pressure 100 lbs. per square inch gage, \^^lat is the clearance of each cylinder? 

Prob. 16. To perform the required useful refrigerating effect in an ice plant, it is 
necessary to compress 250 cu.ft. of ammonia vapor per minute from 30 lbs. per square 
inch gage to 150 lbs. per square inch gage. What must be the size of the compressor 
to handle this at 75 R.P.M. if it be double acting and the clearance be 5 per cent? 

Prob. 17. The maximum deUvery pressure of a type of compressor is controlled by 
making the clearance large so that the volumetric efficiency will decrease as the pressure 
rises and become zero at the desired pressure. What must be the clearance for a single- 
stage compressor where the supply pressure is 14 lbs. per square inch absolute and the 
maximum deUvery pressure 140 lbs. per square inch absolute? What will be the volu- 
metric efficiency of the same machine at a delivery pressure of J the maximum? At J? 

Prob. 18. A three-stage compressor has a clearance of 5 per cent in each cylinder. 
What must be the cylinder ratios for the best-receiver pressures when the machine is 
compressing to 170 lbs. per square inch gage from atmosphere? 

Prob. 19. Show why it was very essential to keep the clearance low in cylinders of 
three-stage compressor used for compressing air for air-driven cars, where the delivery 
pr^sure carried was 2500 lbs. per square inch, by assuming numerical data and 
calculating numerical proof. 

Prob. 20. With water falling 150 ft. and used to compress air directly, how many 
cubic feet of air could be compressed per cubic foot of water? 

Prob. 21. Air is compressed from atmosphere to 150 lbs. per square inch absolute, 
iaothermally in one stage. How much more work would be required per cubic foot if 
compression were adiabatic? How much of this excess would be saved by compressing 
two stage? Three stage? 

Prob. 22. 150 I.H.P. is delivered to the air cylinders of a 14i X22i Xl8 in. compres- 
sor, running at 120 R.P.M. The supply pressure is 15 lbs. per 'square inch absolute. 
The volumetric efficiency as found from the indicator card is 95 per cent. What was 
the discharge pressure? 

Prob. 23. The clearance in the high-pressure cylinder of a compressor is 5 per cent, 
which allows the compressor to run with the best-receiver pressure for a discharge of 
100 lbs. per square inch absolute when the compressor is at sea-level. What would the 



186 ENGINEERING THERMODYNAMICS 

clearance be if the discharge pressure were kept the same and the altitude were 10,000 
ft. to keep the best-receiver pressure? 

Prob. 24. How many cubic feet of supply-pressure air may be compressed per minute 
from 1 to 8 atmospheres absolute by 100 horse-power if the compression in all cases k 
adiabatic? 

(6) Three stage, no clearance; 

(c) Two stage with 5 per cent clearance; 

(d) Single stage with 5 per cent clearance; 
(a) Two stage, no clearance 

Prob. 25. The capacity of a 14jx22jxl4 in. compressor when running at 140 
R.P.M. is said to be 940 cu.ft. for working pressures of 80 to 100 lbs. per square inch 
gage and atmospheric supply at sea level. Check these figures. 

Prob. 26. What horse-power would be required by an 18jx30jx24 in. compressor 
operating at 100 P.P.M. and on a working pressure of 100 lbs. per square inch gage if 
the clearance in low-pressure cylinder is 4 per cent? What would be the capacity? 

Prob. 27. By means of water jackets on a compressor cylinder the va ue for s of com- 
pression curve in single-stage machine is lowered to 1.3. Compare the work to com- 
press 1000 cu.ft. of air from 1 to 8 atmospheres absolute with this condition with that 
required for isothermal and adiabatic compression. 

Prob. 28. What must be the size of cylinders in a three-stage compressor for com- 
pressing gas from 50 lbs. per square inch absolute to 600 lbs. per square inch absolute 
when 8 equals 1.3 and each cylinder has 3 per cent clearance. Compressor to run at 
100 R.P.M. to be double acting and handle 1000 cu.ft. of gas per hour? 

Prob. 29. How many cu.ft. of free air could be compressed per minute with an avail- 
able horse-power of 1000 H.P. from atmosphere to 150 lbs. per square inch gage; (a) 
if compression is isothermal; (6) if compression be single-stage adiabatic; (c) if com- 
pression be three-stage adiabatic? 

Prob. 30. A single-stage compressor is compressing air adiabatically at an altitude 
of 6000 ft. to a pressure of 80 lbs. per square inch gage. The cylinder has 2 per cent 
clearance. What must be the s'.ze of the cylinder to compress 2000 cu.ft. of free air per 
minute if the piston speed is limited to 600 ft. per minute? What, if the clearance be 
zero? 

Prob. 31. What would be the size of the two-stage compressor for same data as in 
Prob. 30? 



CHAPTER III 

WORK OF PISTON ENGINES. HORSE POWER AND CONSUMPTION OF 
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER 
GAS OR VAPOR UNDER PRESSURE. 

1. Action of Fluid in Single Cylinders. General Description of Structure 
and Processes. The most commonly used class of engines is that in which 
the operation is dependent on the pushing action of high-pressure fluids on 
pistons in cylinders, and this includes all piston steam engines of the recipH 
rocating or straight-hne piston path group as well as the less common 
rotary group, having pistons moving in curved and generally circular paths. 
In these same engines there may be used compressed air as well as steam, 
and equally as well the vapors of other substances or any other gases, without 
change of structure,' except perhaps as to proportions, providing only that the 
substance to be used be drawn from a source of supply under high pressure, 
be admitted to the cylinder, there used and from it discharged or exhausted 
to a place of lower pressure. This place of lower pressure may be the open 
air or a closed chamber; the used fluid may be thrown away and wasted or 
used again for various purposes without in any way affecting the essential 
process of obtaining work at the expense of high-pressure gases or vapors. 
It is evident that, regarding a piston as a movable wall of a cylinder, when- 
ever a fluid acts on one side with greater pressure than on the other, the 
piston will move toward the lower pressure end of the cylinder, and in so 
moving can exert a definite force or overcome a definite resistance, measured 
by the difference in pressure on the two sides and the areas exposed to the 
pressure. It is not so evident, but just as true, that the piston may be made 
to move from one end of the cylinder to the other when the average pressure 
on one side is greater than the average pressure on the other, and also do work 
even if the excess of pressure should reverse in direction during the stroke 
and provided only some energy storage device is added. In the common 
steam or compressed-air engine this is a flywheel with the usual connecting 
rod and crank mechanism, uniting the reciprocating piston movement with 
the continuous rotary movement of the flywheel mass. In certain forms 
of pumps the energy is stored in extra cylinders at times of excess and given 
out at times of deficiency in the path of the piston, so that its motion from 
end to end of cylinder may not be interrupted even if the pressure on the 
driving side should fall below that on the resisting side, assuming, of course, 
the average pressure for the whole stroke to be greater on the driving side 
than on the resisting side. 

187 



188 ENGINEERING THERMODYNAMICS 

It appears, therefore, that piston movement in dhgines of the common 
form and structure, and the doing of work by that movement is not a 
question of maintaining a continuously greater pressure on one side than 
on the other. On the contrary, the process is to be studied by examination 
of the average pressure on the driving side and that on the resisting side, 
or by comparing the whole work done on one side with the whole work 
done on the other side by the fluid. The work done by the fluid on one 
side of a piston may be positive or negative, positive when the pressures art- 
assisting motion, negative when they are resisting it. It is most con- 
venient to study the action of fluids in cylinders by considering the 
whole action on one side from the beginning of movement at one end to 
the end of movement at the same point, after the completion of one complete 
forward and one complete return stroke. All the work done by the pressure 
of the fluid on the forward stroke on the side of the piston that is apparently 
moving away from the fluid is positive work, all the work done by the pres- 
sure of the fluid on the same side of the piston dming the return stroke is 
negative, and for this stroke the side of the piston under consideration is 
apparently moving toward the fluid or pushing it. 

For the complete cycle of piston movement covering the two strokes 
the work done on one side is the algebraic sum of the forward stroke 
work, considered positive, and the back stroke work, considered negative. 
During the same time some pressures are acting on the other side of the 
piston, and for them also there will be a net work done equal to the cor- 
responding algebraic sum. The work available for use during the complete 
two strokes, or one revolution, will be the sum of the net work done by the 
fluid on the two sides of the piston during that time, or the algebraic smn of 
two positive and two negative quantities of work. Methods of analysis of the 
work of compressed fluids in cylinders are consequently based on the action in 
0716 end of a cylinder, treated as if the other end did not exist. 

Just how the high-pressure fluid from a source of supply such as a boiler 
or an air compressor is introduced into one end of a cylinder, how it is treated 
after it gets there, and how expelled, will determine the natiu*e of the varia- 
tion in pressure in that end that acts on that side of the piston, and theso 
are subjects to be studied. To determine the work done in the cylinder 
end by the fluid, it is necessary to determine laws of pressure change with 
stroke, and these are fixed first by valve action controlling the distribution 
of the fluid with respect to the piston and second by the physical properties 
of the fluid in question. 

It is necessary that the cylinder be fitted with a valve for getting fluid 
into a cylinder, isolating the charge from the source of supply and getting it 
out again, and it may be that one valve will do, or that two or even more 
are desirable but this is a structural matter, knowledge of which is assumed 
here and not concerned with the effects under investigation. The first step 
in the process is, of course, admission of fluid from the source of supply to the 
cylinder at one end, which may continue for the whole, or be limited to a part 



• WORK OF PISTON ENGINES 189 

of the stroke. When admission ceases or supply is cut off before the end 
of the stroke there will be in the cylinder an isolated mass of fluid which 
will, of course, expand as the piston proceeds to the end. Thus the forward 
stroke, considering one side of the piston only, always consists of full pressure 
admission followed by expansion, the amount of which may vary from zero 
to a very large amount; in fact the final volume of the fluid after expansion 
may be hundreds of times as great as at its beginning, when supply was cut 
off. 

At the end of this forward stroke an exhaust valve is opened, which 
permits communication of the cylinder with the atmosphere in non-con- 
densing steam and most compressed-air engines, or with another cylinder, 
or with a storage chamber, or with a condenser in the case of a steam engine 
in which the pressure approximates a perfect vacuum. If at the moment of 
exhaust opening the cylinder pressing is greater or less than the back pressure, 
there will be a more or less quick equalization either up or down before the 
piston begins to return, after which the return or exhaust stroke will proceed 
with some back-pressiu*e resistance acting on the piston, which is generally 
though not always constant. This may last for the whole back stroke or 
for only a part, as determined by the closure of the exhaust valve. When 
the exhaust valve closes before the end of the return stroke the unexpelled 
steam will be trapped and compressed to a pressure depending partly on the 
point of the stroke when closure begins and the pressure at the time, and 
partly on the clearance volume of the cylinder into which the trapped steam 
is compressed. Of course, at any time near the end of the stroke the admission 
valve may be opened again., and this may occur, 1st, before compression is 
complete, which will result in a sudden pressure rise in the cylinder before 
the end of the stroke to equalize it with the source of supply; 2d, just at the 
end of the stroke, which may result in a rise or a fall to equalize, or no 
change at all, depending on whether compression has raised the cylinder 
pressure not quite to supply pressure, or to something greater than it, or to 
a value just equal to it; 3d, after the end of the stroke, which will result in 
a reexpansion of the steam previously under compression, and then a sudden 
rise. It may be said in general that in cylinders there are carried out with 
more or less variation the following processes: 

Forward stroke, constant-pressure admission followed by expansion. 

Back strokCj constant-pressure exhaust followed by compression, while at 
both ends of the stroke there may or may not be a vertical line on a pressure- 
volume diagram representing a constant-volume change of pressure. 

These processes will result in a cycle of pressure- volume changes which will 
be a closed curve made of more or less accurately defined phases, and the work 
of the cycle will be the area enclosed by the cyclic curve. Of course, there are 
causes of disturbance which make the phases take on peculiar characteristics. 
For example, the valve openings and closures may not take place as desired 
or as presupposed with respect to piston positions; leakage may occur, steam 
may condense during the operations in the cylinder, and water of condensation 



190 



ENGINEERING THERMODYNAMICS 



may evaporate; the resistance through valves will always make the cylinder 
pressure during admission less than in the supply chamber and greater during 
exhaust than the atmosphere or than in exhaust receiver or condenser and mav 
through the valve movements make what might have been a constant-pit^ 
sure straight line become a curve. There will, by reason of these influences, 
encountered in real engines, be an almost infinite variety of indicator cards 
or pressure-volume cycles for such engines. 



































A 


■ 






B 
















B 

j0^ 




A' 












\ 




AB STEAM ADMITTED 

B-C •• EXPANDED 

C-D •• EXHAUSTED 
D-A •• COMPRESSED 




/ 


/^ 
















\ 






/ 



















\ 










y 


ACTUAL CARD FROM 
CORU6S ENGINE 












> 


N 


e 


c 










.,/ 




A 














^ 












D 








\ 




















, 




C 








\ 


\ 




























\ 


^ 
































D""" 


















C 







Fig. 52. — Diagram to Indicate Position of Admission, Cut-off, Release, Compression (»n 

Engine Indicator Card. 



The various points of the stroke at which important events occur, 
important in their pressure-volume significance, have names, as do also the 
lines between the points, and these names are more or less commonly accepted 
and generally understood as follows: letters referring to the diagram Fig. 52. 

Point Names: Events of Cycle. 

A. Admission is that point of the stroke where the supply valve Is 

opened. 

B, Cut-off is that point of the stroke where the supply valve is closed. 

C, Release is that point of the stroke where the exhaust valve is opened. 

D. Compression is that point of the stroke where the exhaust valve 

is clovsed. 
Names of Lines, or Periods: 

i-B, Admission or steayn line joins the points of admission and cut-off. 

B-C, Expansion line joins the points of cut-off and release. 

C-D, Exhaust line joins the points of release and compression if there 

is any, or admission if there is not. 
D-A. Compression line joins the points of compression and admission. 



WORK OP PISTON ENGINES 



191 



By reason of the interferences discussed, these points on actual indicator 
cards may be difficult to locate, one line merging into the next in so gradual 
a manner as to make it impossible to tell where the characteristic point lies, 
as will be apparent from Fig. 53, in which is reproduced a number of actual 
indicator cards. In such cases equivalent points must be located for study 











Fig. 53. — ^Actual Steam Engine Indicator Cards Showing Distortion of Lines and Uncertain 

Location of Characteristic Points. 



These same terms, which it appears sometimes refer to points and some- 
times to lines, are also .used in other senses, for example, cutroff is com- 
monly used to mean the fraction of stroke completed up to the point of 
cut-off, and compression that fraction of stroke remaining incomplete at the 
point of compression, while compression is also sometimes used to express 
the pressure attained at the end of the compression line. In general, there 



192 



ENGINEERING THERMODYNAMICS 



is nothing in the use of these words to indicate just which of the various 
meanings is intended except the text, and experience will soon eliminate most 
of the possible chances of confusion. 

Prob. 1. Draw a card in which admission and exhaust are late. Draw a card in 
which there is no compression, in which compression is very early, so that compression 
pressure is equal to admission pressure. Draw a card with per cent cut-off, and cui- 
off = 100 per cent. Draw cards with same cut-off but with varying initial pressures. 

Prob. 2. The following diagrams, Fig. 54, are reproductions of indicator cards 
actually taken from engines. Explain what features are peculiar to each and if 
possible give an explanation of the cause. 











Fig. 64. — Indicator Diagram from Steam Engines with Improperly Set Valve Gear. 

For example, in No.' 1 a line of pressure equalization between the end of the com- 
pression line and the beginning of the admission line inclines to the right instead of 
being perpendicular, as in a perfect diagram. This is due to the fact that admis- 
sion does not occur until after the piston has begun to move outward, so that pres- 
sure rise does not occur at constant volume, but during a period of increasing volume. 

2. Standard Reference Cycles or PV Diagrams for the Work of Expansive 
Fluids in a Single Cylinder. Simple Engines. To permit of the derivation 
of a formula for the work of steam, compressed air, or any other fluid in a 



WORK OF PISTON ENGINES 193 

ylinder, the various pressure volume, changes must be defined algebraically, 
rhe first step is, therefore, the determination of the cycle or pressure-volume 
liagram representative of the whole series of processes and consisting of a 
lumber of well-known phases or single processes. These phases, ignoring all 
orts of interferences due to leakage or improper valve action, will consist of 
K^nstant-pressure and constant-volume lines representing fluid movement 
nto or from the cylinder, combined with expansion and compression lines 
representing changes of condition of the fluid isolated in the cylinder. These 
*xpansion and compression lines represent strictly thermal phases, laws for 
p^hich will be assumed here, but derived rigidly later in the part treating of 
the tbermal analysis; however, all cases can be represented by the general 
expression 

PV' = C, 

in w^hich the character of the case is fixed by fixing the value of s. For all 
gases and for vapors that do not contain liquid or do not form or evaporate 
any during expansion or compression, i.e., continually superheated, the exponent 
8 may have one of two characteristic valiies. The first is isoOiermal expansion 
and compression, and for this process s is the same for all substances and 
equal to xmity. The second is for exponential expansion or compression and 
for this process s will have values peculiar to the gas or superheated vapor 
under discussion, but it is possible that more than one substance may have the 
same value, as may be noted by reference to Section 8, Chapter I, from 
which the value s= 1.406 for air and 5 = 1.3 for superheated steam or ammonia 
adiabatically expanding are selected for illustration. 

When steam or any other vapor not so highly superheated as to remain 

free from moisture during treatment is expanded or compressed in cylinders 

different values of s must be used to truly represent the process and, of coiu^e, 

there can be no isothermal value, since there can he no change of pressure of 

wet vapors mithoiU a change of temperature. For steam expanding adiabatically 

the value of s is not a constant, as will be proved later by thermal analysis, so 

that the exact solution of problems of adiabatic expansion of steam under 

ordinary conditions becomes impossible by pressure-volume analysis and can 

be handled only by thermal analysis. However, it is sometimes convenient 

or desirable to find a solution that is approximately correct, and for this a 

sort of average value for s may be taken. Rankine's average value is 5=1. 111 = ^ 

for adiabatic expansion of ordinarily wet steam, and while other values have 

l:)een su^ested from time to time this is as close as any and more handy than 

most. The value s = 1.035+.14X(the original dryness fraction), is ^ven by 

Perry to take account of the variation in original moisture. 

Steam during expansion adiabatically, tends to make itself wet, the 
condensation being due to the lesser heat content by reason of the work done; 
but if during expansion heat be added to steam originally just dry, to keep it so 
continuously, as the expansion proceeds, it may be said to follow the saturation 
law of steam, for which s = 1.0646. This is a strictly experimental value found 



194 



ENGINEERING THERMODYNAMICS 



A 



d{ 



m 



^'€ 



Opmp. 

















s 




Exp. 






B 


V 


V. 


m 




































"V 



P 














• 


























-* 


-( 


:i 


E 















































\ 




■Exp. 


4' 


F 


^ 


V. 



































U-»i 


«1 

-Exp. 






< v 








Vr-Ooi 


^>^ 

















M 



Ck>mi >. 



i 



r' 




















—s= 


• 






^ 


V 


Q 


• 






>> 








G<>mp. 
















V 




Fig. 55. — Standard Reference Cycles or Pressure-volume Diagrams for Eiq>anfiive Fluid 

in Simple Engines. 



WORK OF PISTON ENGINES 



195 



by studying the volume occupied by a poimd of just dry steam at various 
pressures, quite independent of engines. 

Direct observation of steam engine indicator cards has revealed the fact 
that while, in general, the pressure falls faster at the beginning of expansion 
and slower at the end than would be the case if 5 = 1, yet the total work is 
about the same as if « had this value all along the curve. This law of expansion 
and compression, which may be conveniently designated as the logarithmic law, 
iS almost universally accepted as representing about what happens in actual 
>t3am engine cylinders. Later, the thermal analysis will show a variation of wet- 
ness corresponding to «=1, which is based on no thermal hjrpothesis what- 
ever, but is the result of years of experience with exact cards. Curiously 




Fig. 56. — Simple Engine Reference, Cycles or PF Diagrams. 



enough, this value of 8 is the same as results from the thermal analysis of con- 
stant temperature or isothermal expansion for gases, but it fails entirely to 
represent the case of isothermal expansion for steam. That «= 1 for isothermal 
gas expansion and actual steam cylinder expansion is a mere coincidence, a 
fact not understood by the authors of many books often considered standard, 
as m them it is spoken of as the isothermal curve for steam, which it most 
positively is not. This discussion of the expansion or compression laws indicates 
that analysis falls into two classes, first, that for which s=l, which yields a 
logarithmic expression for work, and second, that for which 5 is greater or 
less than one, which yields an exponential expression for work, and the former 
will be designated as the logarithmic and the latter as the exponential laws, for 
convenience. 



196 



ENGINEERING THERMODYNAMICS 



The phases to be considered then may be summed up as far as this aoalysis 
is concerned as: 

1. Admission or exhaust , pressure constant, P= const. 

2. Admission or exhaust, volume constant, F= const. 

3. Expansion, Py= const., when « = 1. 

4. Expansion, PF* = const., when 8 is greater or less than I. 

5. Compression, PF = const., when 8=1. 

6. Compression, PF*= const., when s is greater or less than I. 

Considering all the possible variations of phases, there may result any or 
the cycles represented by Fig. 55. These cycles have the characteristics indicated 
by the following table, noting the variation in the law of the expansion or 
compression that may also be possible. 



1 

Cycle. Clearance. 


Eipanaion. 


Compression. 


A 
B 
C 
D 


Zero 
Zero 
Zero 
Zero 


Zero 

Little 

Complete 

Over^xpansion 


Zero 
Zero 
Zero 
Zero 


E 
F 
G 
H 


Little 
Little 
Tiittle 
Little 


Zero 

Little 

Complete 

Over^xpansion 


Zero 
Zero 
Zero 
Zero 


I 
J 
K 
L 


Little 
Little 
Little 
Little 


Zero 

Tiittle 

Complete 

Over-expansion 


little 
Little 
Little 
Little 


M 

N 

P 


Little 
Little 
Little 
Little 

Little 
Little 
Little 
Little 


Zero 

Little 

Complete 

Over-expansion 


Complete 
Complete 
Complete 
Complete 


Q 
R 
S 
T 


Zero 

Little 

Complete 

Over-expansion 


Too much 
Too much 
Too much 
Too much 



It is not necessary, however, to derive algebraic expressions for all these 
cases, since a few general expressions may be found involving all the variables 
in which some of them may be given a zero value and the resulting expression 
will apply to those cycles in which that variable does not appear. The result- 
ing cycles, Fig. 56, that is is convenient to treat are as follows: 



SIMPLE ENGINE REFERENCE CYCLE OR P7 DIAGRAMS 

Cycle 1. Simple Engine, Logarithmic Expansion without Clearance. 
Phase (a) Constant pressure admission. 

(6) Expansion PF =const. (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent) . 

(d) Constant pressure exhaust. 

(e) Constant (zero) volume admission equalization of pressure with supply. 



(( 



(I 



<t 



u 



WORK OF PISTON ENGINES 



197 



U 



« 



«( 



<( 



Cycle II. Simple Engine, Exponential Expansion without Clearance. 
Phase (a) Constant pressure admission. 

(6) Ebcpansion PP = const, (may be absent). 

(c) Ck>nstant- volume equalization of pressure with exhasut (may be absent). 

(cO Constant-pressure exhaust. 

(e) Constant (zero) volume admission equalization of pressure with supply. 

Cycle III. Simple Engine, ^garithmic Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

(6) Expansion P7b const, (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent). 

(d) Constant pressure exhaust. 

(e) Compression PF -const, (may be absent). 
(/) Constant volume admission, equalization of pressure with supply (may 

be absent). 

Cycle IV. Simple Engine, Exponential Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

(6) Expansion PV=const. (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent). 

(d) Constant pressure exhaust. 

(e) Compi^ession PF'= const, (may be absent). 

(f) Constant admission, equalization of pressure with supply (may be 
absent). 



<i 



«f 



€t 



€1 



It 



«< 



If 



If 



IC 



t€ 



p 


^ 




7n 








B 




















A 














(ln.pr) 
















\ 
































V 


\ 










• 






















S 


\ 


































X 


^^ 


































^^ 


>^ 




Q 


(rel.pr.) 


• 






































































E 




























D_ 


(bk.pr.) 
















































n 




















6 












F 














• 


iL 





Fig. 57. — ^Work of Expansive Fluid in Single Cylinder with No Clearance. Logarithmic 

Expansion for Cycle I. Exponential for Cycle II. 



3. Work of Eicpansive Fluid in Single Cylinder without Clearance. Loga- 
rithmic Expansion, Cycle I. Mean Eflfective Pressure, Horse-power and 
Consunqition of Simple Engine. Referring to the diagram, Fig. 57, the net 
work, whether expansion be incomplete, perfect, or excessive, is the sum of 



198 



ENGINEERING THERMODYNAMICS 



admission and expaneion work less the back-pressure work, or by areas, 
net work area, ABCDE^Budmiasion work area ABFG 



+expansion work area FBCH 



or algebraically, 



—back-pressure work area GEDH 



lF=P»7»-|-P»7»log,§-P*7«, • 



(230) 



Work of cycle in foot-pounds is 



=P.7il+log. ^) -P^Vs 
=P»7»(l-|-log.^^)-P*7<, 
=P«Fyi-|-log, ^j -P^Vt 



(a) 



(6) 



(c) 



(d) 



. (231) 



As Vi or Ve represent the whole displacement, the mean effec^ve presmare 
will be obtained by dividing Eqs. (231), by Vt or F,, givmg. 



M 



.E.P.=Pyi-|-log.^)-P<, (a) 
=P.(l+log.^^)-P, 



(6) 



(c) 



.e.p. = pe ( 1 + log, ^ j - Pa 

=pJl+Iog,^j-Pd (d) 

'=pt(l+log,y)(Y)-pa (c) 



(232) 



Similarly, dividing the Eq. (232) by the volume of fluid admitted, F», will 
give the work per cubic foot, which is a good measure of economy, greatest 
economy being defined by maximum work per cubic foot, which, it may be noted, is 
the inverse of the compressor standard. 



Work per cu.ft. supplied = P»( 1+loge ^ j —Pd^ (a) 

=p»(i+iog.^)-p.f; (6)_ 



(233) 



WORK OF PISTON ENGINES ' 199 

According to Eq. (19), Chapter I, the piston displacement in cubic feet 

13 750 
per hour per I.H.P. is for 2=l,y— ^ — , and this multiplied by the fraction of 

rhole displacement occupied in charging the cylinder or representing admission, 
ffhich is 

7, 7, 

Va °' 7.' 

ill give the cubic feet oj high pressure fluid supplied per hour per I.H.P., 
ence 



will give 
hence 

CuA^pplied per hr.per I.H.P. « ^3,750 V^ .. 

(m.e.p.) Vc 

__ 13,750 Pfl .j^v 

(m.e.p.) Pb 



(234) 



Introducing a density factor, this can be transformed to weight of fluid. If 
then d\ is the density of the fluid as supplied in pqunds per cubic foot, 



Lbs. fluid supplied per hr. per I.H.P. = -. — - — r X^X Bi (a) 

\m.e.p.j V c 

^ 13^ g, g 
(m.e.p.) Pft 



. •. (235) 



All these expressions, Eqs. (230) to (235), for the work of the cycle, the 
mean effective pressure, work per cubic feet of fluid supplied, cubic feet and 
pounds of fluid supplied per hour per I.H.P., are in terms of diagram point 
conditions and must be transformed so as to read in terms of more generally 
defined quantities for convenience in solving problems. The first step is to 
introduce quantities representing supply and back pressures and the amount 
of expansion, accordingly: 

Let (in.pr.) represent the initial or supply pressure p» expressed in pounds 

per square inch; 
" (rel.pr.) represent the release pressure pc, in pounds per square inch; 
" (bk.pr.) represent the back pressure pd, in pounds per square inch; 
" Ry represent the ratio of expansion defined as the ratio of largest to 

smallest volume on the expansion line \w\^^\w) which is, of 

course, equal to the ratio of supply to release pressure ( )i when 

the logarithmic law is assumed; 
D represent the displacement in cubic feet which is Vd or Vc when no 
clearance is assumed; 



200 



ENGINEEMNa THERMODYNAMICS 



Let Z represent the fraction of stroke or displacement completed up to 
cut-off so that ZD represents the volume Vh admitted to the cylinder. 

In this case when clearance is zero, Z=-^ . 

Mxy 



Work of the cycle in foot-pounds 



W 



= 144 (in.pr. 



l+logefir 



R 



vj 



-.(bk.pr.)l/) (a) ' 

J }• . . 

= 144[(reLpr.) (1 +loge Af) - (bk.pr.)]/) (6) 

« 

m.e.p. = (rel.pr.) (1 +loge Rv) — (bk.pr.) (a) 
= (m.pr.)(^^'')-(bk.pr.) (6) 

= (m.pr.)z(l+log,^)-(bk.pr.) (c) 

Work per cu.ft. supplied = 144[ (in.pr.) (1 +log, Rv) — (bk.pr.)fiv] (a) ' 

= 144[(in.pr.)(l+Iog.|)-^] (6) _ 



Cu.ft. supplied per hr. per I.H.P. = — ^ — 5- (a) 

m.e.p. Mxy 



13,750 
(mie.p.) 



Z (6) 



Lbs. supplied per hr. per I.H.P. 



13,750 Si . . 



(m.e.p.) Ry 
13,750 



(m.e.p.) 



ZSi(6) 



(236) 



(237) 



(238) 



(239) 



(240) 



The indicated horse-power may be found by multiplying the work of the 
cycle, Eq. (236), by the number of cycles performed per minute n and divid- 
ing the product by 33,000. 

l+Iog,/2v 

Ry 

or 

Z)n(m.e.p.) 



^^'^^^"^'^^^k 



J -(bk.pr.) , . . . 



(241) 



I.H.P.= 



229.2 



(242) 



In any of these expressions where Ry is the ratio of greatest to smallest volume 
diuing expansion, either flp, ratio of greater to smaller pressures, or — , the 



WOKK OF PISTON ENGINES 



201 



reciprocal of the cut-off, may be substituted, since the expressions apply only 
to the logarithmic law, and clearance is assumed equal to zero. When 
clearance is not zero; it is shown later that the cut-off as a fraction of stroke 
is not the reciprocal of Rp or Rv 

These expressions are perfectly general, but convenience in calculation 
will be served by deriving expressions for certain special cases. The first of 
these is the case of no expansion at all, the second that of complete expansion 
without over-expansion. This latter gives the most economical operation from 
the hypothetical standpoint, because no work of expansion has been left 
unaccomplished and at the same time no negative work has been introduced 
by over-eiq>ansion. 



A 


























B 


III. pr.^ 






















































■ 












































• 
























•1 




































































































































D^ 


)k. pr.) 


E 






















































H_ 





\/ 



Fig. 58. — ^First Special Case of Cycles I and II. Expansion = zero. Full Stroke Admission. 

First Special Case. If there is no expansion, together with the above assump- 
tion of no clearance, the diagram takes the form (Fig. 58), and 

Tr=144Z>[(in.pr.)-(bk.pr.)] (243) 



m.e.p. = (in.pr.) — (bk.pr.) . . 



(244) 



Work per cu.ft. supplied = 144[(in.pr.) — (bk.pr.)] (245) 



Cu.ft. supplied per hr. per I.H.P.=7; 



13,750 



(in.pr.) — (bk.pr.) 



(246) 



Lbs. supplied per hr. per I.H.P. = 77 



13,7505i 



(in.pr.) — (bk.pr.) 



(247) 



202 



ENGINEERING THERMODYNAMICS 



Second Special Case. When the expansion is complete without over-expan- 
sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) =^ (bk.pr.), 

hence Ry=Rp=^ 7 hlr \ ~ 7' ^^* value of cul-off^ Z, is known as best cutoff, as 
ii is thai which uses all the available energy of the fluid by expansion. 



»r« I44D(in.pr.) ^5&^= 144Z) (bk.pr.) log. Rp. 

JtCp 



, \ n \ 10g« Rp 

(m.e.p.) « (m.pr.) — ^ — 



Work per cuit. supplied = 144(in.pr.) log. Rp. 



. (248) 



(249) 
(250) 



A 




B(ln 


pr) 


























\ 














• 














\ 






■ 






















\ 


s, 




























\ 




























N 


\ 




























V 


V 






























^ 


^-^ 














E 






















, 




D(i 
































• 

H 






















1 

If 














• 














V 



>k.pr) 



Fig. 59. — Second Special Case of Cycles I and II. Complete Expansion Without Over 

Expansion Case of ^eet Cut-off. 



13,750 



Cu.ft. supplied per hr. per I.H.P. = 7r v , r» • 
^^ ^ ^ (in.pr.) loge iZp 

Lbs. supplied per hr. per I.H.P. = ;; — - . . ^ ^ . 
*^*^ ^ ^ (m.pr.) log, iZp 



(251) 



(252) 



Bzample 1. Method of calculating diagrams. Fig. 57 and Fig. 59. 
Assumed data for Fig. 57. 

Pa *Py =90 lbs. per sq.in. abs. Va = F« =0 cu.ft. 
P^ =P« = 14 lbs. per sq.in. abs. yc'^Vd- 13.5 cuit. 

Fft=6cu.ft. 



WOEK OF PISTON ENGINES 203 



To obtain point C: 



Vb 6 

Pc "^Pb X— =90 Xttz -40 lbs. per sq-in. abs. 

Vc lo.o 



Assumed data for Fig. 59. 



Pa ''Pit =90 lbs. per sq-in. Va = Ve =0 cuit. 
P^ =.p, B 14 lbs. per sq.in. Vd = 13.5 cu.ft. 



To obtain point B: 



Pd 14 

Vi, = Vd X^ = 13.5 X;^ =2.1 cu.ft. 



Example 2* A simple double-acting engine admits steam at 100 per square inch 
absolute for \ stroke, allows it to expand to the end of the stroke and then exhausts it 
against a back pressure of 5 lbs. per square inch absolute. If the engine has no 
clearance, a 7 x9-in. cylinder and runs at 300 R.P.M., what is the horse-power and steam 
consumption when steam is expanding according to the logarithmic law? Note: 
1 cuit. steam at 100 lbs. per square inch absolute weighs .2258 lb. 

From Eq. (237ft), 

m.e.p. = (m.pr.) ( ^ — I - (bk.pr.) 

-100 ^^ +log. ^) -5 ,100(l±iJgg> -5 -54.7: 



(m.e.p.)Lon 54.7 X. 75x38.5x600 
33,000 33,000 



or directly from Eq. (242) 

Z)n(m.e.p.), 



I.H.P. - 



229.2 
.2X600X54.7 



229.2 

Tu X TTTT» 13,750 8i 
Lbs. steam per I.H.P. = — x 



=28, 



m.e.p. Rv 

13750 .2258 _ ,„ 
X--7-= 14.13. 



54.7 

Therefore, steam per hour used by engine = 14.15x28 =396 lbs. 

Prob. 1. A steam engine has a cylinder 12x18 ins. with no clearance. It runs at 
200 R.P.M. and is double-acting. If the steam pressure be fixed at 100 lbs. per 
square inch absolute, and the back pressure at 10 lbs. per square inch abs., show how 
the horse-power and steam consumption will vary as cut-off increases. Take cut-off 
from i to f by eighths. Plot 



204 ENGINEERING THERMODYNAMICS 

Prob. 2. Two engines of the same size and design as above are running on a steam 
pressure of 100 lbs. per square inch absolute, but one exhausts through a long pipe to 
the atmosphere, the total back pressure being 20 lbs. per square inch absolute, 
while the other exhausts into a condenser in which the pressure is but 3 Ibe. per 
square inch absolute. If the cut-ofif is in each case f , how will the I.H.P. and steam 
used in the two cases vary? 

Prob. 3. By finding the water rate and the horse-power in the two following cases, 
show the saving in steam and loss in power due to using steam expansively. A pump 
having a cylinder 9 X 12 ins. admits steam full stroke, while an engine of same size admits 
it but i of the stroke; both run at the same speed and have the same back pressure. 

Prob. 4. Steam from a 12x24 in. cylinder is exhausted at atmospheric pressure 
(15 lbs. per square inch absolute) into a tank, from which a second engine takes steam. 
Neither engine has clearance. The first engine receives steam at 100 lbs. per square 
inch absolute and the cut-off is such as to give complete expansion. The second engine 
exhausts into a 24 in« vacuum and its cut-off is such that complete expansion occurs in 
its cylinder. Also the cylinder voliune up to cut-off equals that of the first cylinder 
at exhaust. If the stroke is the same in both engines and the speed of each is 200 
R.P.M., what is the diameter of the larger cylinder, the total horse-power developed, 
the total steam used, and the work per cubic foot of steam admitted to the first 
cylinder, the water rate of each engine and the total horse-power derived from each 
pound of steam? 

Prob. 6. The steam pressure for a given eng^e is changed from 80 lbs. per square 
inch gage to 120 lbs. per square inch gage. If the engine is 12x16 ins., running 250 
R.P.M. with a fixed cut-off of 25 per cent and no clearance, the back pressure being 
15 lbs. per square inch absolute, what will be the horse-power and the water rate in 
each case? 

Note: 1 cuit. of steam at 80 and 120 lbs. weighs .216 and .3 lb. respectively. 

Prob. 6. By trial, find how much the cut-off should have been shortened to 
keep the H.P. constant when the pressure was increased and what effect this would 
have had on the water rate. 

Prob. 7. A certain tjrpe of automobile engine uses steam at 600 lbs. per square 
inch absolute pressure. The exhaust is to atmosphere. For a cut-off of i and no 
clearance, what would be the water rate? 

Note: for 600 lbs. ^i =1.32. 

Prob. 8. Engines are governed by throttling the initial pressure or shortening the 
cut-off. The following cases show the effect of light load on economy. Both engines, 
12x18 ins., running at 200 R.P.M., with 125 lbs. per square inch absolute. Initial 
pressure and back pressure of 10 lbs. per square inch absolute. The load is sufficient 
to require full steam pressure at 5 cut-off for each engine. Load drops to a point 
where the throttle engine requires but 50 lbs. per square inch absolute initial pressure 
with the cut-off still fixed at i. What is the original load and water rate, and new 
load and water-rate for each engine? 

Note: d for 125 lbs. absolute = .279 and for 50 lbs. =.117 lb. 

Prob. 9. The guarantee for a simple engine 18x24 ins., running at 200 R.P.M., 
states that the I.W.R. when cut-off is i will not exceed 15 lbs. if the initial pressure be 
100 lbs. per square inch gage, and back pressure 5 lbs. per square inch absolute. 
If engine has no clearance, see if this would be possible. 

4. Work of Expansive Fluid in Single Cylinder without Clearance. Erpo- 
nential Expansion Cycle 11. Mean Effective Pressure, Horse-power and 



WORK OF PISTON ENGINES 



205 



Constunptioii of Simple Engines. Referring to the diagram^ Fig. 57, the 
work is given by the same areas as for Cycle I, but its algebraic expression 
is different because s is greater than 1 and an exponential expansion results on 
integration instead of a logarithmic one. 

In general, from Eq. (13a), Section 7, Chapter I, 

Tr=P,F6+^^[(^y"'-l]-P.7, (253) 



Putting this in terms of initial conditions by the relations 



Also 



there results 






VJt 



b^V* 



p„v> 



bVb 



ir=p»7»+ .-_:'j^^"-, iRv*-'- 1) -PaV^v 

1 



=P»7» H 



[ 



-PaViRv. 



(«-l)i2v^x(i2v--l) 

= 1442) \z (in.pr .) ^^j— - (bk.pr.)l 

which is the general equation for work of this cycle. 
Dividing by Vt, the volume of fluid supplied, 

Work per cu .ft. supplied = P^ ( -—7 — / .^vp ,_i j — PaRv (a) 



(a) 



(b) 



(c) 



. (254) 



= 144(in.pr.)(-^-^i-^— 



(&) 



. . (255) 



Similarly, the mean effective pressure results from dividing the work by the 
displacement, Vd=VJiv 



or 



M.E.P. = -^ ( r— 7 TTjy ,_i ) ~Pd 

Rv\s—l (a— l)ftv' V 
, . (in.pr.)Y s 1 \ /ui \ 

= Z(m.pr.) { ^~_^ ) - (bk.pr.) 



(a) 



(b) 



(c) 



. (256) 



206 



ENGINEERING TELERMODYNAMICS 



First special case of no expansion^ when i^F^ly results the same diagram 
as in the previous section, Fig. 58, and exactly the same set of formulas. 

Second special case when the expansion is complete without over-expansion, 
is again represented by Fig. 59 and for it 



Whence 






or 



Work for complete expansion is 



W=PtVt 



*^tO"57^V 



(a) 



,,^(in^ _^(,__L^>) (^ 



• . . (257) 



which is the general equation for the work of Fj or I p- 1 cubic feet of fluid 

when the economy is best or for best cut-ofif. 

The work per cubic foot of fluid supplied for this case of complete expansion 
gives the maximum value for Eq. (255) and is obtained by dividing Ek]. (257) 



. Max. work per cu.ft. supplied =Pft-3Y(l""p;7^i ) 

= 144(m.pr.)^(l-^.) 



(a) 



(b) 



, ..(258) 



which is the general equation for maximum work per cubic foot of fluid supplied. 
The expression for mean effective pressure becomes for this case of best cut-off, 



M.E.P.= 



or, 



Rv s 

( 



-iV Rv'-^j 



, X _ (in.pr.) s /, \_\ 

(m.e.p.)^ r;- jri(^l-:R;rrij 



(a) 



(6) 



. . . (259) 



It is convenient to note that in using Eqs. (257), (258) and (259) it may be 
desirable to evaluate them without first finding Rv- Since 



^"^rrw""^'*' 



WORK OF PISTON ENGINES 207 

this substitution may be made, and 

Ry ^ Rp » 

• 

Bzample. Compare the horse-power and the steam conswnption of a 9x12 in. 
simple double-acting engine with no clearance and running at 250 R.P.M. when initial 
pressure is 100 lbs. per square inch absolute and cut-off is }, if 

(a) steam remains diy and saturated throughout expansion, 

(6) remains superheated throughout expansion, and 

(c) if ori^nally dry and suffers adiabatic expansion. 

Back pressure is 10 lbs. per square inch absolute. 



<-'-'-^*fe-(:rik^)-<^->- 



100/1.0646 

4 \.0646 .0646x4 



For case (o) «- 1.0646 and (m.e.p.) =-7- 1 -7^7^- r^. a ^.064e ) -10-48.6; 






" (6) » - 1.3 and (m.e.p.) -^ (^ ~T^) ~ ^° "^"^' 



" (c)«-l.lll and (m^.p.)-^(^- ^^^^^.i„ ) -10-47^. 



iJl.r . — (m.e,p,)ljan « ^^ — .yoo m.e.p. 

A I JI.P. for case (a) -46.9, 

" (b) -42.0, 
" " (c)-45.8. 

From Eq. (240), lbs. steam per hour per I.H.P.-» * X-^ 

m.e.p. Rv 

1? / X X 1. AOf^ 13,750 8i 

.'. For case (a) steam per hr. -46.9 X '^ X-r ; 
- 4 4o.o 4 

" (b) steam per hr. -42 X-^^ X-r; 

4o.o 4 

a/\_x 1. Af a 13,750 8i 

(c) steam per hr. -45.8 X~-=- X-7. 

47.5 4 

Prob. 1. On starting a locomotive steam is admitted full stroke, while in running 
the valve gear is arranged for f cut-off. If the engine were 18 x30 ins., initial pressure 
150 lbs. per square inch absolute, back pressure 15 lbs. per square inch absolute, what 
would be the difference in horse-power with the gear in normal running position and 
in the starting position for a speed of 20 miles per hour with 6-ft. driving wheels? Con- 
sider the steam to be originally dry and expanding adiabatically. What would be the 
difference in steam per horse-power hour for the two cases and the difference in total 
steam? Clearance neglected. 

Prob. 2. Consider a boiler horse-power to be 30 lbs. of steam per hour; what must be 
the horse-power of a boiler to supply the following engine? Steam is supplied in a super- 



208 ENGINEERING THERMODYNAMICS 

heated state and remains so throughout expansion. Initial density of steam « .21 lbs. per 
cubic foot. Engine is 12x20 ins., double-acting, 200 R.P.M., no clearance, initial 
pressure 125 lbs. per square inch absolute, back pressure a vacuum of 26 ins. of 
mercury. Cut-ofif at maximum load J, no load, A. What per cent of rating o^ boiler 
will be required by the engine at no load? 

Prob. 3. While an engine driving a generator is running, a short circuit occurs 
putting full load on en^ne, requiring a J cut-off. A moment later the circuit-breaker 
opens and only the friction load remains, requiring a cut-off of only -fj. The engine 
being two-cylinder, double-acting, simple, 12 X 18 ins., running at 300 R.P.M., and having 
no clearance, what will be the rate at which it uses steam just before and just after 
circuit-breaker opens if the steam supplied is at 125 lbs. per square inch absolute and is 
just dry, becoming wet on expanding, and back pressure is 3 lbs. per square inch 
absolute? 

Prob. 4. A pmnping engine has two double-acting steam cylinders each 9x12 ins. 
and a fixed cut-off of ). It runs at 60 R.P.M. on 80 lbs. per square inch absolute steam 
pressure and atmospheric exhaust. Cylinder is jacketed so that steam stays dry 
throughout its expansion. How much steam will it use per hour? Neglect clearance. 

Prob. 6. If an engine 10x14 ins. and running 250 R.P.M. has such a cut-off that 
complete expansion occurs for 90 lbs. per square inch absolute initial pressure, and at 
atmospheric (15 lbs. absolute) exhaust, what will be the horse-power and steam used per 
hour, steam being superheated at all times, and what would be the value for the horse- 
power and steam used if full stroke admission occurred? 

Prob. 6. The steam consumption of an engine working under constant load is 
better than that of a similar one working under variable load. For a 16 X24 ins. engine 
running at 250 R.P.M. on wet steam of 125 lbs. per square inch absolute and atmospheric 
exhaust, find the horse-power and steam used per horse-power per hour for best con- 
dition and by taking two hghter and three heavier loads, show by a curve how steam 
used per horse-power per hour will vary. 

Prob. 7. For driving a shop a two cylinder single-acting engine, 6x6 ins., running 
at 430 R.P.M., is used. The cut-off is fixed at J and intitial pressure varied to control 
speed. Plot a curve between horse-power and weight of steam per hour per horse- 
power for 20, 40, 60, 80, 100, 120 lbs. per square inch absolute initial pressure and 
atmospheric exhaust. Steam constantly dry. Clearance zero. 

Note: ^i for above pressures equals .05, .095, .139, .183, .226, and .268 lbs. per 
cubic foot, respectively. 

Prob. 8. Taking the loads found in Prob. 7, find what cut-off would be required 
to cause the engine to run at rated speed for each load if the initial steam pressure 
were 100 lbs. per square inch absolute, and the back pressure atmosphere, and a plot 
curve between horse-power and steam used per horse-power hour for this case. 

Prob. 9. For working a mine hoist a two-cylinder, double-acting engine is used 
in which compressed air is admitted f stroke at 125 lbs. per square inch absolute and then 
allowed to expand adiabatically and exhaust to atmosphere. If the cylinders are 18 x24 
ins. and speed is 150 R.P.M., find the horse-power and cubic feet of high pressure air 
needed per minute. 

6. Work of Expansive Fluid in Single Cylinder with Clearance. Loga- 
rithmic Expansion and Compression; Cycle m. Mean Eflfective Pressure, 
Horse-power, and Consumption of Simple Engines. As in previous cycles, 
the net work of the cycle is equal to the algebraic sum of the positive work 



WORK OF PISTON ENGINES 



209 



done on the forward stroke and the negative work on the return stroke. By 
areas, Fig. 60, this is 



Work BxeA^JABN+NBCW-WDEO-OEFJ. 
Expressed in terms of diagram points this becomes 



1F= 



P*(n-Fa)+P.F, log, :^ 
-Pa{V,-V,)-PeVel0ge] 



f ) 



(260) 



P CD) 


f 


apply 


yolun 
7r\ 


e-, — 




B 




[ la.pr. 












n 


J 


1 ' 






f 


r A 






cu^ 




^1 


i 












\ 


















\ 












S 


\ 
















\ 

\ 














S 














M - 


IF 


















\ 


^ 








Wl 


n 




























C(. 




i\ 
1 \ 
































1 > 

1 


V 




























G- 


1 
-.1. 


\ 


























5L 


1 


\ 




■ 
























t. 


1 












r* 












--» 


w 


L/ m 




3 


1 








N 


D 












K 


M 


«y-iJ 


























V 



il.pr) 



Fig. 60. — ^Work of Expansive Fluid in Single Cylinder with Clearance. Logarithmic Expan- 
sion for Cycle III. Exponential for Cycle IV. 



Expressing this in terms of displacement, in cubic feet D; clearance as a frac- 
tion of displacement, c; cut-oif as a fraction of displacement, Z\ compres- 
sion as a fraction of displacement, X; initial pressure, in pounds per square 
inch (in.pr.), and exhaust or back pressure, in pounds per square inch (bk.pr): 



P5 = 144(in.pr.); 
Pi =144 (bk.pr.); 
(V»-F,)=ZZ); 

7» D{,Z+c) Z+c' 



(F<,-7.) = (1-X)Z); 
7»=Z)(Z+c); 
F.=I>(X+c). 
V, D(X+c) X+c 



Vt 



Dc 



210 ENGINEERING THERMODYNAMICS 

Whence 
Work in ft.-lbs. per cycle is 



W= 



144i> [ (in.pr.) [z+ (Z+c) loge |t^] 
-(bk.pr.) [^(l«X)+(X+c)log.^] j 



- (261) 



From Eq. (261), the mean effective pressure, pounds per square inch, follows 
by dividing by 144D: 



(m.e.p.)= - 



(in.pr.) Z+(Z+c) loge^TT- • • (i»ean forward press.) 
-(bk.pr.)r(l-X)+(X+c)loga^t?. 1 (meanbk.pr.) 



(262) 



This is a general expression of very broad use in computing probable mean 
effective pressure for the steam engine with clearance and compression, or for 
other cases where it is practicable to assim[ie the logarithmic law to hold. Fig. 
117, at the end of this chapter, will be found of assistance in evaluating this 
expression. 

Indicated horse-power, according to expressions ah^ady given, may be 
found by either of the following equations: 

y TT p _ (m.e,p,)Lan _ 144(m.e.p.)Dn _ (m.e.p.)Dn 
33,000 33,000 ~ 229.2""' 

where L is stroke in feet, a is efifective area of piston, square inches, n is the 
number of cycles performed per minute and D the displacement, cubic feet. 

It might seem that the work per cubic foot of fluid supplied could be 
found by dividing Eq. (261) by the admission volume, 

but this would be true only when no steam is needed to build up the pressure 
from F to A. This is the case only when the clearance is zero or when com- 
pression begins soon enough to carry the point F up to point A, i.e., when by 
compression the pressure of the clearance fluid is raised to the initial pressure. 
It is evident that the fluid supplied may perform the two duties: first, 
building up the clearance pressure at constant or nearly constant volume, 
and second, filling the cylinder up to cut-oflf at constant pressure. To measure 
the steam supplied in terms of diagram quantities requires the fixing of the 
volume of live steam necessary to build up the pressure from F to A and adding 
it to the apparent admission volume (Vi,-'Va)- This can be done by producing 



WORK OF PISTON ENGINES 211 

the compression line EF to the initial pressure Q, then LQ is the volume that 
the clearance steam would have at the initial pressure and QA the volume of 
live steam necessary to build up the pressure. The whole volume of steam 
admitted then is represented by QB instead of AB or by (Vb—Vq) instead^ of 
by {Vb'-V^)y and calling this the supply volume, 



But 



(Sup.Vol.) = (Fft-7,.) 



Pq (m.pr.) (m.pr.) 



Hence 



(Sup.Vol.)=Z)[(Z+c)-(X+c)|g^], .... (263) 



which is the cubic feet of fluid admitted at the initial pressure for the dis- 
placement of D cubic feet by the piston. Dividing by D there results 

(Sup Vol.) ^ _ (;i.+,)jbk-prj (264) 

D V / V ^(m.pr.) ^ ^ 

which is the ratio of admission volume to displacement or cubic feet of live 
steam admitted per cubic foot of displacement. 

Dividing the work done by the cubic feet of steam supplied gives the 
economy of the simple engine in terms of volumes, or 

W 
Work per cu.ft. of fluid supplied = 



(Sup.Vol.) 



(in.pr.)[z+(Z+c)lo& M -(bk.pr.)[(l-X)+(X+c)lo& ^''l 
=144 L Z+^i I £_J. (265) 

(Z+c)-(X+c)^^) 
^ ^ ^ ^(m.pr.) 

It is more common to express economy of the engine in terms of the weight 

of steam used per hour per horse-power or the " water rate," which in more 

general terms may be called the consumption per hour per LH.P. 

Let di be the density or weight per cubic foot of fluid supplied, then the 

weight per cycle is (Sup.Vol.) 5i, and this weight is capable of performing 

W 
W foot-poirnds of work or (Sup.Vol.) ^i lbs. per minute will permit of «^«^^^^ 

horse-power. But (Sup.Vol.) di lbs. per mintue corresponds to 60 (Sup.Vol.) 
i\ lbs. per hour, whence the number of pounds per hour per horse-power is 

60(Sup.Vol.)d^ 



1^733,000 



212 . ENGINEERING THERMODYNAMICS 

which is the pounds consumption per hour per I.H J*., whence 

n +• • IK u TXTi> 60X33,000(Sup.VoL) ^i ,^.. 

Consumption m lbs. per hr. per I.H.P. = — 2^ ^ ^— , . . . (266; 

which is the general expression for consumption in terms of the cubic feet of 
fluid admitted per cycle, ^i initial density, and the work per cycle. 

As work is the product of mean efiFective pressure in pounds per square foot, 
(M.E.P.,) and the displacement in cu.ft. or TF=(M.E.P.)D, or in termB of 
mean effective pressure pounds per square inch W = 144 (m.e.p.)D, the 
consumption may also be written 

n *• • lu u TXJT> 60X33,000 (Sup.Vol.)ai 

Consumption m lbs.,per hr. per I.H.P.= '. — *-r^r — - — 

x^rff ^m.e.p.^xy 

13,750 (Sup.Vol.)ai 
(m.e.p.) D 

-^^Uz+c)-(X+c)^^]3^ (267) 

(m.e.p.) L '(m.pr.) J "^ 

which gives the water rate in terms of the mean effective pressure, cut-off, 
clearance, compression, initial and back pressures and initial steam density. 
It is sometimes more convenient to introduce the density of fluid at the back 
pressure <^2i which can be done by the relation (referring to the diagram). 



whence 



n^a ^'•^'or^bk.pr.) V, d2' 



(bk.pr.) 
(m.pr.) 



This on substitution gives 
Consumption in lbs., per hr. per I.H.P. 

-^[^^+'^'^-^''+'^''] (268) 

Since the step taken above of introducing <?£ has removed all pressure or 
volume ratios from the expression, Eq. (268) is general, and not dependent 
upon the logarithmic law. It gives the consumption in terms of mean effective 
pressure, cut-off, clearance, compression and the density of steam at initial 
and back pressure, which is of very common use. 

It cannot be too strongly kept in mind thai all the preceding is true only when 
no steam forms from moisture water during expansion or compression or no steam 
condenses, which assumption is known to he untrue. These formulae are, there- 
fore, to be considered as merely convenient approximations, although they 



WORK OF PISTON ENGINES 



213 



Eire almost universally used in daily practice. (See the end of this chapter for 
diagrams by which the solution of this expression is facilitated.; 

Special Cases. First, no expansion and no compression would result in 

:• 61. For it 



Tr= 144Z>[(in.pr.) - (bk.pr.)], 



(m.e.p.) = (in.pr.) — (bk.pr.) 



(269) 
(270^ 



pf 

LL 


311 












oliime 














1 














'*»«» — ' 














B 


i\n^T%T.\ 




7 




































































































































i 


































i 


































"1 


\ 

1 
































N . 






























D 


Cbk.pr.) 




_ 

r 


























1 

—tr* 




K _ 


p- 














"br- 












— ► 








t^ 


























V 


/ \ 


1/ 



Fig. 61. — ^First Special Case of Cycles III and IV. Expansion and Compression both Zero, 

but Clearance Finite. 

The voliune of fluid supplied per cycle is QB, or from Eq. (263) it is 

(Sup.Vo1.)=d[i+c-c^^^^^^] (271) 

Consumptioninlbs. perhr. per I.H.P.= 7; v ' .iT- A 1+c— c,. -*^-/- ^ 

(m.pr.) — (bk.pr.) L (m.pr.) J (; 



(272) 



or in terms of initial and final densities. 



13 750 
Consumption in lbs. per hr. per I.H. P. = ,. \^/^ki; — \[(l+c)^i— C(?2] (273) 

The second special case is thai of complete expansion and compression^ as 
indicated in Fig. 62. Complete expansion provides that the pressure at the 



214 



ENGINEERING THERMODYNAMICS 



end of expansion be equal to the back pressure, and complete compression that 
the final compression pressure be equal to the initial pressure. 

Here 











V, 


V, 


1+c 


X+c 


(in.pr.) 












F» r. Z+c c (bk.pr.)' 




and hence 




yr-^rSr.-' ^"^^ (^' ^.)=z>[i+c- 


/in.pr.y 
\bk.pr./J 


and 






(y,-7^)=ZD. 






P 

L 


Aj;^-ZD- 


5b 


















^^^^^"' 






Cln.pr.) 








i 












. 














eD 


i 




A 


\ 


























i 


i 




\ 


\ 






■ 




















V 


1 

1 

1 






V 


X, 






















\ 


1 

1 

1 

1 








^S 




^^ 














G 


_! — 





A 


















. 





-P-^ 


Q>k.Dr^ 


























S-^ 




1 







——— 




t)- — 








~^ 




•-W 






K 


Is 




i" 

























W 





Fig. 62. — Second Special Case of Cycles III and IV. Perfect Expansion and Perfect 

Compression with Clearance. 

Hence by substitution 



I I (bly)r.)J^_ (in.pr.) 

"(bk.pr.) ' 



from which 



ZD 



^-<'+«) w$- 



(274) 



Again, 



_ V,- Va [ Vhk.pr./ J _ I /jn.prA _ 



'] 



. . (275) 



WORK OF PISTON ENGINES 215 

Eq. (274) gives the cut-oflF as a fraction of the displacement necessary to give 
complete expansion, while Eq. (275) gives the compression as a fraction of dis- 
placement to give complete compression, both in terms of clearance, initial 
pressure and back pressure, provided the logarithmic law applies to expansion 
and compression. 

Substitution of the values given above in Eq. (261) gives, after simplifi- 
cation, 

F=144Z)[(l+c)(bk.pr.)-c(in.pr.)]lo&|^^j. . . (276) 

(m.e.p.) = [(l+c)(bk.pr.)-c(in.pr.)]log.|™^^ (277) 

In this case the volume supplied is exactly equal to that represented by the 
admission line AB, and is equal to 

(Sup.Vol.)=ZD (278) 

Hence, the consumption, in pounds fluid per hpur per I.H.P. in terms of 
initial density, is 

Consumption in lbs. per hr. per I.H.P.=-- — r Zdi, 



but 



(bk^) ^ 
(m.pr.) 



m 



.e.p. ,. . r,. , .(bk.pr.) "I, (in.pr. ) .. v, (in.pr.) 



hence 



13,750^ 



Consumption in lbs. fluid per hr. per I.H.P.= — ,. r-. . . (279) 

This last equation is interesting in that it shows the consumption (or water 
rate, if it is a steam engixie) is independent of clearance, and dependent only 
upon initial density, and on the initial and final pressures. 

An expression may also be easily derived for the consumption in terms of 
iiiitial and final density, but due to its limited use, will not be introduced here. 

Example 1. Method of calculating Diagrams Fig. 60 and Fig. 62. 
Assumed data for Fig. 60: 

Pq ^Pa «p6 =90 lbs. per square inch abs. Fa = V/ = .6 cu.ft. 
Pg «Pe =Pif = 14 lbs. per square inch abs. Vd'^Ve- 135 cu.ft. 
P/ =60 lbs. per square inch abs. F> =6 cu.ft. 



216 ENGINEERING THERMODYNAMICS 

To obtain point C: 



To obtain point E: 



To obtain point Q: 



Vh 6 

P« =Pft Xtt =90 X— -T =40 lbs. per 8q.in. abs. 

Ve 16.0 



F, = y;.x^=.5X7^=1.78 cu.ft. 



P, KA 

F, = 7/X^«.6x^«.278cu.ft. 



Intermediate points from B to C and J^ to Q are found by assuming volumes 
and computing the corresponding pressures by relation PxVx==PbVb or PxVx=-PeVc^ 

Clearance is ^ — -i^s- =^ =3.8 per cent, 
Cut-ofif is -=7^ — ~ =-^ =42.3 per cent, 

Yd — r o lo 

Compression is -jz — =7- =-7^ =9.9 per cent. 

Assumed data for Fig. 62. 

p^ =P5 «90 lbs. per square inch absolute. Va = .5 cu.ft. 
Pg==Pc- 14 lbs. per square inch absolute. Va = 13.5 cu.ft. 



To obtain point B: 



To obtain point E: 



F^ = 7c§ =13.5X^=2.11 cu.ft. 

jt ft » yu 



p 00 

F. = Fa X^ = .5 Xt. =3.2 cu.ft. 

x^« 14 



Intermediate points from B to C and from A to £? are to be found by assuming 
. various volumes and finding the corresponding pressures from relation PzVx =PaVa or 

PxVx^PbVi,, 

Example. 2. What will be the horse-power of, and steam used per hour by the 
following engine: 

(a) cut-off 50 per cent, compression 30 per cent, 

(6) complete expansion and compression, 

(c) no expansion or compression. 

Cylinder, 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial pressure 
85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, and 
logarithmic expansion and compression. 



WORK OF PISTON ENGINES 217 

Note: d for 85 lbs. gage = .23, ^i for 15 lbs. absolute = .038 cu.ft. 
(a) From Eq. (262) 

(m.e.p.) =(in.pr.)[z -[-(Z+c) log, |~^1 -(bk.pr.) f'(l-X)+(Z+c)log,^j, 

= 10o[.5+(.6+.07)log,|±|^]-15[(l-.3)-|-(.3+.07) log, — J^] 

=86-20=66 lbs. sq.in. 
(m.e. p.) Lan _66 Xl.5^X llil X400 _ 

From Eq. (267) steam per hour per I.H.P. in pounds is 

,750 [,„ , ,,^ ,/bk.pr.\l 

^ r (Z+c)-(X-|-c) -*^ 8i, 

.e.p.)L \m.pr./J 

= -^^[(.5+.07) -(.3+.07) X^l X.23=25 lbs. 

Hence total steam per hour =25x135 =3380 lbs. 
(6) From Eq. (277) 

(m.e.p.) =[(1 +c)(bk. pr.) -c(in. pr.)] log, f g^~ )» 
=[(1 +.07) X15-.07 XlOO] log, 6.67 =17.2 lbs. sq.in. 



13 
(m 



^„^ 17.2X1.5X113.1X400 ^^ , 
I.H.P. = ^-^ =35.4. 



From Eq. (279) 



a. TTn> I, 13,7505i 13,750X.23 _. 

Steam per I.H.P. per hour = /. \ = ^,^^-/^-^ =16.6, 

»»7 



,. ^, /in.pr.\ 100 XL! 



Total steam per hour = 16.6x35.4 =588 lbs. 

(c) From Eq. (270) (m.e.p.) =(in.pr.)- (bk.pr.) =100-15=85 lbs. sq.in. 

. I HP 85X1.5 X13 .1X400 

•• ^•^•^•~ 33,000 -^^^•^• 

From Eq. (273) 

Steam per I.H.P. per hour = — ^ r[(l +c)di -c^a] ^^[1 .07x .23 - .07x .038] =35.4. 

^m.e.p.^ 00 

Total steam per hour = 174.5x35.4 =6100 lbs. 

Pr6b. 1. What will be the horse-power and water rate of a 9x12 in. simple engine 
having 5 per cent clearance, running at 250 R.P.M. on 100 lbs. per square inch abso- 



218 ENGINEERING THERMODYNAMICS 

lute initial pressure and 5 lbs. per square inch absolute back pressure when the cut-off is 
I, \f and If expansion follows the logarithmic law, and there is no compression? 

Note : B for 100 lbs. absolute = .23, 5 for 5 lbs. absolute = .014. 

Prob. 2. Will a pump with a cylinder 10 Xl5 ins. and 10 per cent clearance give the 
same horse-power and have the same water rate as a piunp with cylinders of the same 
size but with 20 per cent clearance, both taking steam full stroke? Solve for a case 
of 125 lbs. per square inch absolute initial pressure, atmospheric exhaust and a speed of 
50 double strokes. No compression. 

Note: 5 for 125 lbs. absolute = .283, 8 for 15 lbs. absolute = .039. 

Prob. 3. Solve the above problem for an engine of the same size, using steam expan- 
sively when the cut-off is J and R.P.M. 200, steam and exhaust pressure as in Prob. 2 
and compression zero. 

Prob. 4. Two engines, each 9x15 ins., are running on an initial pressure of 90 lbs. 
per square Jnch absolute, and a back pressure of one atmosphere. One has no clearance, 
the other 8 per cent. Each is cutting off so that complete expansion occurs. The speed 
of each is 200 and neither has any compression. What will be the horse-power and 
water rate? 

Note: 8 for 90 lbs. -.24, 8 for 15 lbs. =.039. 

Prob. 6. By finding the horse-power and water rate of a 12x18 in. double-acting 
engine with 8 per cent clearance, running at 150 R.P.M. on an initial steam pressure 
of 90 lbs. per square inch absolute and atmosphere exhaust for a fixed cut-off of } and 
variable compression from to the point where the pressure at the end of compression 
is equal to 125 per cent of the initial pressure, plot the curves between compression and 
horse-power, and compression and water rate to show the effect of compression on 
the other two. 

Note: 8 for 90 lbs. = .21, 8 for 15 lbs. = .039. 

Prob. 6. A steam engine is running at such a load that the cut-off has to be f at 
a speed of 150 R.P.M. The engine is 14 x20 ins. and has no clearance. Initial pressure 
100 lbs. per square inch absolute and back pressure 5 lbs. per square inch absolute. 
What would be the cut-off of an engine of the same dimensions but with 10 per cent 
clearance under similar conditions? 

Prob. 7. The steam pressure is 100 lbs. per square inch gage and the back 
pressure is 26 ins. of mercury vacuum when the barometer is 30 ins. For a 16x22 
in.engine with 6 per cent clearance running at 125 R.P.M., cut-off at § and 30 per cent 
compression, what will be the horse-power and- the water rate? Should the steam 
pressure be doubled what would be the horse-power and the water rate? If it should 
be halved? 

Note: 8 for 100 lbs. gage = .2017, 8 for 26 ins. Hg.=.0058. 

Prob. 8. While an 18 x24 in. simple engine with 4 per cent clearance at speed 
of 150 R.P.M. is running with a i cut-off and a compression of ^ on a steam pressure of 
125 lbs. per square inch gage, and a vacuum of 28 ins. of mercury, the condenser fails 
and the back pressure rises to 17 lbs. per square inch absolute. What will be the change in 
the horse-power and water rate if all other factors stay constant? What would the new 
cut-off have to be to keep the engine running at the same horse-power and what 
would be the water rate with this cut-off? 

Note: 8 for 125 lbs. gage = .315, 8 for 28 in. Hg. = .0029, 8 for 17 lbs. absolute 
= .0435. 

Prob. 9. Under normal load an engine has a cut-off of |, while imder light load 
the cut-off is but A. What per cent of the steam used at normal load will be used 



WORK OF PISTON ENGINES 



219 



at light load for the following data? Cylinder 10x18 ins.; 7 per cent clearance; 200 
R.P.M.; initial pressure 120 lbs. per square inch gage; back pressure 2 lbs. per square 
inch absolute; compression at normal load 5 per cent; at light load 25 per cent. 
Note : B for 120 lbs. gage = .304, i for 2 lbs. absolute = .0058. 

6. Work of Expansive Fluid in Single Cylinder with Clearance ; Exponential 
Expansion and Compression, Cycle IV. Mean Effective Pressure, Horse- 
power and Consumption of Simple Engines. As pointed out in several places, 
the logarithmic expansion of steam only approximates the truth in real engines 
and is the result of no particular logical or physically definable hypothesis as to 
the condition of the fluid, moreover its equations are of little or no value 
for compressed air or other gases used in engine cylinders. All expansions 
that can be defined by conditions of physical state or condition of heat, 
including the adiabatic, are expressible approximately or exactly by a definite 
value of 8, not unity, in the expression PF* = const. All these cases can 
then be treated in a group and expressions for work and mean effective 
pressure foimd for a general value of «, for which particular values belonging 
to, or following from any physical hypothesis can be substituted. The area 
under such expansion curves is given by Eq. (13) Chapter I, which applied to 
the work diagram, Fig. 60, in the same manner as was done for logarithmic 
expansion, gives the net work: 



W=P,{V,-Va)+j~\l-(p) 1 (area JABCWJ) 



«-i 



-Pd(F, - Ve) -^' [ (^;) - 1] (area WDEFJW) 



(280) 



Introducing the symbols, 

Pft = 144(in.pr.), 



Ptf = 144(bk.pr.), 
(F»-Fa)=ZA 

(Ftf-F.) = Z)(l-X), 



Ve=D(X+c). 



/VA_Z+c 
\VJ~l+c 

/V,\_X±c 
\VJ~ c • 



W=14AD 



-(M:.p..,[(I-X,+|t'[(^-±-')-'-.]] 



(281) 



Eq. (281) pves the work in foot-pounds for D cubic feet of displacement in a 
cylinder having any clearance c, cut-off Z, and compression X, between two 



220 ENGINEERING THERMODYNAMICS 

pressures, and when the law of expansion is PV* = const, and s anything except 
unity, but constant. 

The mean efifective pressure, pounds per square inch, is obtained by divid- 
ing the expression for work by 144D, giving 

(m.e.p.) = (in.pr.) J Z-\ — — r- 1 — ( ^TT ) [ (mean for'd pr.) 



— (bk.pr.) 



(l-X)+^'[(^y '-l]j (mean bk.pr.) 



, (282) 



which is the general expression for mean effective pressure for this cycle. 

It was pointed out in Section (5) that the cubic feet of fluid admitted at the 
initial pressure was not represented by AB, Fig. 60, but by QB, and the same 
is true for this case, so that the 

^Sup.Vol.) = 76-F«. 

But when the expansion and compression laws have the form PV*=^c 



.K.(g)--C(X+«)(g^-)^ 



Whence 



1 

(Sup.Vol.)=Z)[(Z+c)-(X+c)(g^-y] (283) 

Eq. (283) gives for these cases what was given by Eq. (263) for the logarithmic 
law, the cubic feet of fluid supplied at the initial pressure for the displacement 
of D cubic feet in terms of cut-off, clearance, compression and the pressures. 
From this, by division there is found the volume of high pressure fluid per 
cu.ft. of displacement 

The consumption is given by the general expression already derived, 
Eq. (34), from which is obtained. 

Consumption lbs. per hr. per I.H.P. 

^13J50r ^ ^)/bk^.\h ^285) 

(m.e.p.) L ^\m.pr. / J 



WORK OF PISTON ENGINES 221 

Eq. (285) ^ves the water rate or gas consumption in terms of mean effective 
pressure, initial and back pressure, cut-off, clearance, compression and initial 
fluid density. Introducing the density at the back pressure by the relation, 

X g r (J = ±e y e ) 

1 



F,*52"Vbk:pF:/ •' 

_1_ 

_ / in.pr. \ . 

-^^bk:^.; ' 



there results 

Consumption lbs. per hr. per I.H.P. = iM^ \ (z+c) h - (X+c) 82I , . (286) 

^m.e.p.; [_ J 

which is identical with Eq. (268) and is, as previously observed, a general expres- 
sion, no matter what the laws of expansion and compression, in terms of 
mean effective pressure, cut-off, clearance, compression and the initial and 
final steam density. 

The first special case of full admission, no compression mig)it at first thought 
appear to be the same as in the preceding section, where the logarithmic law 
was assumed to hold, and so it is as regards work and mean effective pressure, 
Eq. (269) and (270), but referring to Fig. 61 it will be seen that since the expo- 
nential law is now assumed instead of the logarithmic, the point Q will be dif- 
ferently located (nearer to A than it was previously if s is greater than 1), and 
hence the supply volume QB is changed, and its new value is 

1 
(Sup.Vol.)=Z)[l+c-c(^)'] (287) 

Hence, consumption stated in terms of initial density of the fluid §1, is 
Consumption lbs. per hr. per I.H.P. 

13,750 r, . /bk.pr.\i^ 1 ^ ,^^^^ 

=7: r- Sn ^ll+c-cli — ^) 81. ... (288) 

(m.pr.) — (bk.pr.) [ \m.pr. /J ^ ^ 

Stated in terms of initial and final densities, the expression is as before, Eq. (273). 
The second special case, complete expansion and compression is again repre- 
sented by Fig. 62. From the law of expansion it is evident that 



222 ENGINEERING THERMODYNAMICS 

or stated in symbolic form, 

1 

Z>(Z+c)=Z)(l+c)(g^)', 



whence 



\m.pr. / 
Again referring to Fig. 62, 

Y^ ^*"^ *^- Vbk.pr./ 

Vc-Va D 

Whence 

^-'[(fe^)'-'] <^> 

Eq. (289) gives the cut-off as a fraction of displacement necessary to g^ve 
complete expansion, and (290), the compression fraction to give complete com- 
pression, both in terms of clearance, initial and back pressures, and the exponent 
s, in the equation of the expansion or compression line, PF*= const. 

The work of the cycle becomes for this special case, by substitution in Eq. 
(281), 

W-.UD (u,.p,.,.4i[(i+<^)'-][i-(^)"']. . m) 

and the mean effective pressure, lbs. per sq.in., is 

The volume of fluid supplied is, 

(Sup.Vol.)=ZD (293) 

hence 

13 750 
Consumption, lbs. per hr. per I.H.P.= — - — Z8i, 

Xfl .CD. 

but 



iM^f -«] 



s— IL^ \m.pr. / -J L \m.pr. / J 



1 



( 



•■'P'•)7-^['-t^)■■']. 



WORK OF PISTON ENGINES 223 

whence 

Consumption lbs. fluid per hr. per I.H.P. is, 

!^750XJj__^ (2^j 

the expression for smallest consumption (or water rate if steam) of fluid for 
the most economical hypothetical cycle, which may it be noticed, is again in- 
dependent of clearance. 

The expressions for work and mean effective pressure are not, however, 
independent of clearance, and hence, according to the hypothetical cycles here 
considered, it is proved that large clearance decreases the work capacity of a 
a cylinder of given size, but does not afifect the economy, provided complete 
expansion and compression are attained, a conclusion similar to that in regard 
to clearance effect on compressor capacity and economy. Whether the actual 
performance of gas or steam engines agrees with this conclusion based only 
on hypothetical reasoning, will be discussed later. 

Example 1. What will be the honse-power of and steam used per hour by the 
following engine: 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial 
pressure 85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, 
and expansion such that 8 ===1.3. 

(a) cut-oflf=50 per cent; compression =30 per cent; 

(6) complete expansion and compression; 

(c) no expansion or compression; 

Note: ^ for 85 lbs. gage =.18; for 15 lbs. absolute. = .03. 
(a) From Eq. (282) 

,„.e.p,.<i..p,.,{.4!-;[.-(f2)-'] 

-»-){<' -«-f^l(^?)""-']} 

-10o{;i+f[,-(j|)']}-.6{.7+f[(f)*-,]}-59.81b.„.i„. 

_„_ 59.8X1.5X113.1X400 ^^^ 
LH.P. ^^-^ -122. 

From Eq. (286) 
Steam per hour per LH.P. 

. ^.^^^Lz+c)^,^{X+c)^,] =-f^[(.57)X.18-(.37)X.03] =20.9 lbs. 
(m.e.p.)L J 59.8 

/. Steam per hour = 122 X20.9 =2560 lbs. 



224 ENGINEERING THERMODYNAMICS 

(6) From Eq. (292) 



«-i 



(„.e.p.,.a..p.,,-i;[a..,(^^)^-«][,-(^)^. 

_„_ 26.2X1.5X113.1X200 ^, 

I.H.P. = — — =64 

3000 

From Eq. (294) steam used I.H.P. per hour is, 

13,750§i 13,750 X.18 



«— 1 .3 



|— « 16.5 lbs., 



hence total steam per hour = 16.5 X64 = 1060 lbs. 

(c) From Eq. (270) which holds for any value of s, m.e.p. =100 — 15 «85 lbs. sq.in. 

A Txrr* 85x1.5x113.1x400 ,_ ^ 
and I.H.P. = ^^-^ 174.5, 

From Eq. (288) steam per I.H.P. hour 

1 1 

_13,750Bir, /bk.pr.\«1 13,750x.l8r, ^^ ^^ /ISXOI ^^ ^ ,^ 

^-r rh+^-^h— ^ = c"^ h+-07-.07xL-7^ =24.51h.<^.. 

(m.e.p.) L \m.pr. /J 85 L \100/ J 

and total steam per hour = 174.5X24.5 =2475 lbs. 

Prob. 1. Two simple engines, each 12 Xl8 ins., with 6 per cent clearance are operated 
at J cut-off and with 20 per cent compression. One is supplied with air at 80 Ihs. 
per square inch gage, and exhausts it to atmosphere; the other with initially dry steam 
which becomes wet on expansion and which also is exhausted to atmosphere. For a speed 
of 200 R.P.M. what is the horse-power of each engine and the cubic feet of stuff supplied 
per horse-power hour ? 

Prob. 2. A crank-and-flywheel two-cylinder, double-acting, pumping engine is 
supplied with dry steam and the expansion is such that it remains dry until exhaust. 
The cylinder size is 24x36 ins., cut-off to give perfect expansion, clearance 5 per cent, 
compression to give perfect compression, initial pressure 50 lbs. per square inch al>- 
solute. back pressure 5 lbs. per square inch absolute. What is the horse-power and 
water rate? What would be the horse-power and water rate of a full-stroke pump of the 
same size ahd clearance but having no compression, running on the same pressure range 
and quahty of steam. 

Note: S for 50 lbs. absolute =.12, § for 15 lbs. absolute = .0387. 

Prob. 3. Should the cylinder of the following engine be so provided that the 
steam was always kept dry, would there be any change in the horse-power developed as 



WORK OF PISTON ENGINES 225 

compared with steam expanded adiabatically, and how much? Cylinder 20 x24 ins., 
initial pressure 125 lbs. per sq. in. gage, back pressure 26 ins. vacuum, standard barom- 
eter, clearance 6 per cent, cut-off f, compression 10 percent, and speed 125 R.P.M. 
Prob. 4. What will be the total steam used per hour by a 20x28-in. double-acting 
engine running at 150 R.P.M. if the initial pressure be 125 lbs. per square inch absolute, 
back pressure one atmosphere, clearance 8 per cent, compression zero, for.cut-off 
i, i, f , and i, if steam expands adiabatically and is originally dry and saturated? 
Note: 8 for 125 lbs. absolute = .283, B for 15 lbs. absolute = .0387. 
Prob. 6. An engine which is supplied with superheated steam is said to have an 
indicated water rate of 15 lbs. at } cut-off and one of 25 lbs. at i cutr-off. See if 
this is reasonable for the following conditions: engine is 15 X22 ins., 7 per cent clearance, 
no compression, initial pressure 100 lbs. per square inch gage., back pressure 28-in. 
vacuum, barometer 30 ins. and speed 180 R.P.M. 

Note: B for 100 lbs. gage -.262, 5 for 28 in. Hg = .0029. 

Prob. 6. The tank capacity of a locomotive is 4500 gals. The cylinders are 
18x36 ins., initial pressure 200 lbs. per square inch gage, exhaust atmospheric, cut-off i, 
clearance 7 per cent, speed 200 R.P.M., no compression. The stream is dry at 
start and expansion adiabatic, how long will the water last if 40% condenses during 
admission? 

Note: 8 for 200 lbs. gage =.471, 8 for 15 lbs. absolute = .0387. 
Prob. 7. To drive a hoist, an air engine is used, the air being supplied for i 
stroke at 80 lbs. per square inch gage expanded adiabatically and exhausted to atmos- 
sphere. If the clearance is 8 per cent and there is no compression how many cubic 
feet of air per hour per horse-power will be needed? What, with complete compression? 
Prob. 8. A manufacturer rates his 44 x42-in. double-acting engine with a speed 
of 100 R.P.M. at 1000 H.P. when running non-condensing, initial pressure 70 lbs. 
per square inch gage and cut-off i. No clearance is mentioned and nothing said about 
manner of expansion. Assuming s = 1.0646, find on what clearance basis this rating is 
made. 

Prob. 9. The water supply of a town is supplied by a direct-acting non-condensing 
pump with two cyUnders, each 24x42 ins., with 10 per cent clearance, and no com- 
pression, initial pressure being 100 lbs. per square inch gage. What must be the size of the 
steam cylinder of a crank-and-flywheel pump with 6 per cent clearance to give the same 
horse-power on the same steam and exhaust pressures with a cut-off of J? Speed in 
each case to be 50 R.P.M. 

Note: B for 100 lbs. gage = .262, 5 for 15 lbs. abs. = .0387. 

7. Action of Fluid in Multiple-ezpansion Cylinders. General Description 
of Structure and Processes. When steam, compressed air, or any other high 
pressure working fluid is caused to pass through more than one cylinder in 
series, so that the exhaust from the one is the supply for the next, the engine 
is, in general, a multiple-expansion engine, or more specifically, a compound 
when the operations are in two expansion stages, triple for three, and quadruple 
for four stages. It must be imderstood that while a compound engine is one 
in which the whole pressure-volimie change from initial to back pressure takes 
place in two stages, it does not necessarily follow that the machine is a two-, 
cylinder one, for the second stage of expansion may take place in two cylinders, 
in each of which, half of the steam is put through identical operations; this 



226 ENGINEERING THERMODYNAMICS 

would make a three-cylinder compound. Similarly, triple-expansion engines, 
while they cannot have less than three may have four or five or six cylindenj 
Multiple expansions engine, most of which are compound, are of two classe? 
with respect to the treatment and pressure-voliune changes of the steam, first 
without receiver^ and second, mth receiver. A receiver is primarily a chamber 
large in proportion to cylinder volumes, placed between the high- and low-pres- 
sure cylinders of compounds or between any pair of cylinders in triple or 
quadruple engines, and its purpose is to provide a reservoir of fluid so that 
the exhaust from the higher into it, or the admission to the lower from it, will 
be accomplished without a material change of pressure, and this will be accom- 
plished as its volume is large in proportion to the charge of steam received 
by it or delivered from it. With a receiver of infinite size the exhaust line of 
a high-pressure cylinder discharging into it will be a constant-pressure line, 
as will also the admission line of the low-pressure cylinder. When, however, 
the receiver is of finite size high-pressure exhaust is equivalent to increasing 
the quantity of fluid in the receiver of fixed volume and must be accompanied 
by a rise of pressure except when a low-pressure cylinder may happen to be 
taking out fluid at the same rate and at the same time, which in practice never 
happens. As the receiver becomes smaller in proportion to the cylinders, the 
pressure in it will rise and fall more for each high-pressure exhaust and low- 
pressure admission with, of course, a constant average value. The greatest 
possible change of pressure during high-pressure exhaust and low-pressure admis- 
sion would occur when the receiver is of zero size, that is when there is none at all, 
in which case, of course, the high-and low-pressure pistons must have synchronous 
movement, both starting and stopping at the same time, but moving either in 
the same or opposite directions. When the pistons of the no-receiver compound 
engines move in the same direction at the same time, one end of the high- 
pressure cylinder must exhaust into the opposite end of the low; but with 
oppositely moving pistons, the exhaust from high will enter the same end of the 
low. It is plain that a real receiver of zero volume is impossible, as the connect- 
ing ports must have some volume and likewise that an infinite receiver is equally 
impracticable, so that any multiple-expansion real engine will have receivers 
of finite volume with corresponding pressure changes during the period when a 
receiver is in communication with a cylinder. The amount of these pressure 
changes will depend partly on the size of the receiver with respect to the cylin- 
ders, but also as well, on the relation between the periods of flow into receiver, 
by high-pressure exhaust and out of it, by low-pressure admission, which 
latter factor will be fixed largely by crank angles, and partly by the settings 
of the two valves, relations which are often extremely complicated. 

For the purpose of analysis it is desirable to treat the two limiting cases 
of no receiver and infinite receiver, because they yield formulas simple enough to 
be useful, while an exact simple solution of the general case is impossible. These 
simple expressions for hypothetical cases which are very valuable for estimates 
and approximations are generally close to truth for an actual engine especially 
if intelligently selected and used. 



WORK OF PISTON ENGINES 227 

Receivers of steam engines may be simple tanks or temporary storage 
chambers or be fitted with coils or tubes to which live or high-pressure steam 
is supplied and which may heat up the lower pressure, partly expanded steam 
passing from cylinder to cylinder through the receiver. Such receivers are 
refiecUing receivers, and as noted, may heat up the engine steam or may evapo- 
rate any moisture it might contain. As a matter of fact there can be no 
heating of the steakn before all moisture is first evaporated, from which 
it appears that the action of such reheating receivers may be, and is quite 
complicated thermally, and a study of these conditions must be postponed 
till a thermal method of analysis is established. This will introduce no 
serious diflBculty, as such reheating receivers assist the thermal economy 
of the whole system but little and have little effect on engine power, likewise 
are now little used. Reheating of air or other gases, as well as preheating 
them before admission to the high-pressure cylinder is a necessary practice, 
when the supply pressure is high,' to prevent freezing of moisture by the gases, 
which get very cold in expansion if it be carried far. This is likewise, however, 
a thermal problem, not to be taken up till later. 

Multiple-expansion engines are built for greater economy than is possible 
in simple engines and the reasons are divisible into two classes, first mechanical, 
and second thermal. It has already been shown that by expansion, work is 
obtained in greater amounts as the expansion is greater, provided, of course, 
expansion below the back pressure is avoided, and as high initial and low 
back pressures permit essentially of most expansion, engines must be built 
capable of utilizing all that the steam or compressed gas may yield. If 
steam followed the logarithmic law of expansion, pressure falling inversely 
with volume increase, then steam of 150 lbs. per square inch absolute 
expanding to 1 lb. per square inch absolute would require enough ultimate 
cylinder space to allow whatever volume of steam was admitted up to cut-oflf 
to increase 150 times. This would involve a valve gear and cylinder structure 
capable of admitting Thr = -0067 of the cylinder volume. It is practically 
impossible to construct a valve that will accurately open and close in this 
necessarily short equivalent portion of the stroke. This, however, is not the 
worst handicap even mechanically, because actual cylinders cannot be made 
without some clearance, usually more than 2 per cent of the displacement and 
in order that any steam might be admitted at all, the clearance in the example 
would have to be less than .67 per cent of the total volume, which is quite 
impossible. These two mechanical or structural limitations, that of admission 
valve gear and that of clearance limits, supply the first argimient for multiple- 
expansion engines, the structure of which is capable of utilizing any amount 
of expansion that high boiler pressure and good condenser vacuum make 
available. For, if neglecting clearance, the low-pressure cylinder had ten 
times the volume of the high, then the full stroke admission of steam to the high 
followed by expansion in the low would give ten expansions, while admission 
to the high for ^ of its stroke would give 16 expansions in it, after which this 
final volume would increase in the low ten times, that is, to 160 times the original 



228 ENGINEERING THERMODYNAMICS 

volume, and cylinder admission of ^ or 6.7 per cent is possible with ordinary 
valve gears, as is also an initial volimie of 6.7 per cent of a total cylinder 
volume, even with clearance which in reasonably la^-ge engines may be not 
over 2 per cent of the whole cylinder volume. 

It is evident that the higher the initial and the lower the back pressures 
the greater the expansion ratio will be for complete expansion, and as in steam 
practice boiler pressiu-es of 225 lbs. per square inch gage or approximately 240 
lbs. per square inch absolute with vacuum back pressures as low as one or even 
half a pound per square inch are in use, it should be possible whether desirable 
or not, to expand to a final volume from 250 to 600 times the initial in round num- 
bers. This is, of course, quite impossible in simple engine cylinders, and as it 
easy with multiple expansion there is supplied another mechanical argument for 
staging. Sufficient expansion for practical purposes in locomotives and land 
engines under the usually variable load of industrial service is available for even 
these high pressures by compounding, but when the loads are about constant, 
as in waterworks pumping engines, and marine engines for ship propul.sion, 
triple expansion is used for pressures in excess of about 180 lbs. gage. 

Use of very high initial and very low back pressures will result in simple 
engines, in a possibility of great unbalanced forces on a piston, its rods, pins 
and crank, when acting on opposite sides, and a considerable fluctuation in tan- 
gential turning force at the crank pin. Compounding will always reduce the 
unbalanced force on a piston, and when carried out in cylinders each of which 
has a separate crank, permits of a very considerable improvement of turning 
effort. So that, not only does multiple expansion make it possible to utilize 
to the fullest extent the whole range of high initial and low back pressures, 
but it may result in a better force distribution in the engine, avoiding shocks, 
making unnecessary, excessively strong pistons, and rods and equalizing turn- 
ing effort so that the maximum and minimum tangential force do not depart 
too much from the mean. 

The second or thermal reason for bothering with mjiiltiple-expansion com- 
plications in the interest of steam economy is concerned with the prevention 
of steam loss by condensation and leakage. It does not need any elaborate 
analysis to show that low-pressure steam will be cooler than high-pressure 
steam and that expanding steam in a cylinder has a tendency to cool the 
cylinder and piston walls, certainly the inner skin at least, so that after 
expansion and exhaust they will be cooler than after admission; but as 
admission follows exhaust hot live steam will come into contact with cool 
walls and some will necessarily condense, the amount being smaller the less 
the original expansion; hence in any one cylinder of a multiple-expansion 
engine the condensation may be less than a simple engine with the same range 
of steam pressures and temperatures. Whether all the steam condensation 
during admission added together will equal that of the simple engine or not 
is another question. There is no doubt, however, that as the multiple expan- 
sion engine can expand usefully to greater degree than a simple engine, and 
so cause a lower temperature by expansion, that it has a greater chance to 



WORK OF PISTON ENGINES 229 

reevaporate some of the water of initial condensation and so get some work 
out of the extra steam so evaporated, which in the simple engine might have 
remained as water, incapable of working until exhaust opened and lowered 
the pressure, when, of course, it could do no good. It is also clear that steam 
or compressed-air leakage in a siinple engine is a direct loss, whereas in a 
compjound high-pressure cylinder leakage has at least a chance to do some 
work in the low-pressure cylinder. The exact analysis of the thermal reasons 
for greater economy is compUcated and is largely concerned with a study of 
steam condensation and reevaporation, but the fact is, that multiple-expan- 
sion engines are capable of greater economy than simple. The thermal analysis 
must also consider the influence of the reheating receiver, the steam-jacketed 
working cylinder, and the use of superheated steam, their effects on the pos- 
sible work per poimd of steam and the corresponding quantity of heat expended 
to secure it, and for air and compressed gas the parallel treatment of pre- 
heating and reheating. 

To illustrate the action of steam in multiple-expansion engines some indi- 
cator cards are given for a few typical cases in Figs. 63 to 66, together 
with the combined diagrams of pressure-volume changes of the fluid in all 
cylinders to the same scale of pressures and volumes, which, of course, makes 
the diagram look quite different, as indicator cards are usually taken to the 
same base length, fixed by the reducing motion, and to different pressure 
scales, to get as large a height of diagram as the paper will permit. Fig. 63 
shows fdur sets of cards taken from an engine of the compound no-receiver 
type, namely, a Vauclain compound locomotive. In this machine there are 
two cylinders, one high pressure and one low, on each side, the steam from 
the high pressure exhausting directly into the low-pressure cylinder so that 
the only receiver space is made up of the clearance and connecting passages 
between the cylinders. Starting with set A, the cards show a decreasing 
high pressure cut-off of 76 per cent in the case of set A to 54 per cent in the 
case of set D. The letters A, B, C and D refer in each case to admission, cut- 
off, release and compression, the use of primed letters denoting the low-pressure 
cylinder. 

In set A the high-pressure admission line AFB may be considered as made 
up of two parts, the part AF representing pressure rise at constant volimie, 
wh^ch is the admission of steam to the clearance space at dead center to raise the 
pressure from that at the end of compression to that of boiler pressure. From 
F to B admission occiured at constant pressure, steam filling the cylinder 
volume as the piston moved outward. At B cut-off or closure of the steam 
valve occurred and the steam in the cylinder expanded. At C, release or open- 
of the exhaust valve of the high-pressure cylinder occurred and the admission 
valve of the low-pressure cylinder opened, the steam dropping in pressure until 
the pressure in both high- and low-pressure clearance became equal, and then 
expanding in both cylinders, as the exhaust from the high and admission to 
the low occurred, the exhaust line CD of the high pressure and the admission 
line F'B' of the low pressure being identical except for the sUght pressure drop 



230 



ENGINEERING THEEM0DYNAM1C8 







(A) 



(B) 







(C) (D) 

FiQ. 63. — Set of Indicator Cards from Vauclain Locomotive Illustrating the No-receivei 

Compomid Steam Engine. 



WORK OF PISTON ENGINES 231 

in the passages between the high- and the low-pressure cylinders. At D the 
high-pressure exhaust valve closed and compression of the steam trapped in 
the high-pressure cylinder occurred to point A, thus closing the cycle. From 
point D' in the low-pressure cylinder, which corresponds to D in the high 
pressure, no more steam was admitted to the low-pressure cylinder. What 
steam there was in the low expanded to the point C when the exhaust valve 
opened and the pressure dropped to the back pressure and the steam was 
exhausted at nearly constant back pressure to Z)', when the exhaust valve closed 
and the steam trapped in the cylinder was compressed to A', at which point 
steam was again admitted and the cycle repeated. 

In set B the cycle of operation is exactly the same as in set A. In set C 
the cycle is the same as in A, but there are one or two points to be especially 
noted, as they are not present in set A. The admission line of the high-pres- 
sure cylinder is not a constant pressure, but rather a falling pressure one, due to 
throttling of the steam, or ''wire drawingy^ as it is called, through the throttle 
valve or steam ports, due to the higher speed at which this card was taken. 
It will also be noticed that the compression pressure is higher in this case, due 
to earlier closing of the exhaust valve, which becomes necessary with the type of 
valve gear used, as the cut-ofiF is made earlier. In the low-pressure card it will 
be seen that the compression pressure is greater than the admission pressure 
and hence there is a pressure drop instead of rise on admission. In set D the 
peculiarities of C are still more apparent, the compression in high-pressure 
cylinder being equal to admission pressure and above it in the low-pressure 
cylinder. The wire drawing is also more marked, as the speed was still 
higher when this set of cards was taken. 

In Fig. 64, one set of the cards of Fig. 63 is redrawn on cross-section 
paper and then combined. Cards taken from the different cylinders of a 
multiple-expansion engine will in nearly all cases have the same length, the great- 
est that can be conveniently handled by the indicator, and will be to two different 
pressures scales, in as much as that indicator spring will be chosen for each 
cylinder which will give the greatest height of card consistent with safety to 
the instrument. To properly compare the cards they must be reduced to the 
same pressure scale, and also to the same volume scale. As the lengths 
represent volumes, the ratio of the two volume scales will be as that of the 
cylinder volumes or diameters squared. Hence, the length of the high-pressure 
card must be decreased in this ratio or the low increased. As a rule it is found 
more convenient to employ the former method. When the cards have been 
reduced to a proper scale of pressures and volumes the clearance must be 
added to each in order that the true volume of the fluid may be shown. 
The cards may now be placed with these atmospheric lines and zero volume 
lines coinciding and will then appear in their true relation. In this case the 
cylinder ratio was 1.65, the indicator springs 100 lbs. and 70 lbs. respectively 
and clearance 5 per cent in each cylinder. 

The steps in combining the cards were as follows: The zero volimie lines were 
first drawn perpendicular to the atmospheric line and at a distance from the end of 



232 



ENGINEERING THERMODYNAMICS 



the card equal to the length of the card times the clearance. PV axes were laid 
oflf and a line drawn parallel to the zero-pressure line at a distance above it equal to 
14.7 lbs. to scale of combined diagram. This scale was taken to be that of the 
high-pressure diagram. A number of points A'B^C, etc., were then chosen 
on the low-pressure card, and the corresponding points a'6'c', etc., plotted by 
making the distances of a\b\ etc., from the zero-volume line equal to thote 
of A^B'y etc., and the distances of the new points above the atmosphere .7 
the distances of the original. By joining the points as plotted, the new diagram 
for the low-pressure card was formed. The high-pressure card was then redrawn 




Fig. 64. — Diagram to Show Method of Combining the High- and Low-pressure Cylinder 

Indicator Cards of the No-receiver Compound Engine. 



by taking a number of points A, B, C, etc., and plotting new points a, &, c, 

etc., so that the distances of a, 6, c, etc., from the zero-volume line were t-— the 

distances of A, B, C, etc., while the distances of new points above the atmos- 
pheric line were the same as for the original points. 

In Fig. 65 are shown two cards from a compound steam engine with receiver. 
Diagram A shows the cards as taken, but transferred to cross-section paper for 
ease in combining, and with the zero-volume axis added. On the high-pressure 
card admission occurred practically at constant volume, piston being at rest 
at dead center, at A, bringing the pressures in the cylinder up to the initial 
pressure at B. Admission continued from J? to C at nearly constant pressure, 
the piston moving slowly with correspondingly small demand for steam and 
consequently little wire drawing. From C to D the piston is moving more 
rapidly and there is in consequence more wire drawing, admission being no 
longer at constant pressure. At D the steam valve closes and expansion occurs, 
to E, where release occurs, the pressure falling to that in the receiver. From 
F to G exhaust occurs with increase of pressure due to the steam being forced 
into the receiver, (receiver + decreasing H.P. cyl.vol.) while from G to H the 



WORK OF PISTON ENGINES 



233 



pressure falls, due to the low-pressure cylinder taking steam from the receiver 
and consequently voliune of receiver, (receiver + increasing L.P. cyl.voL+ 
decreasing H.P. cyl.vol.) increasing. At H exhaust closes, a very slight com- 
pression occurring from H U> A. 

On the low-pressure card, admission occurred at A' and continued to B' at 
constant volume, the piston being on dead center as from A to B in high-pressure 
cylinder. From B' to C admission occurred with falling pressure due to increase 
in receiver volume, (receiver+increasing L.P. cyl.vol.), and from C to D' 
admission still took place, but with less rapidly falling pressure, as high-pres- 
sure cylinder is now exhausting and receiver volume, (receiver+increasing 
L.P. cyl.vol. +decreasing H.P. cyl.vol.) was receiving some steam as well as 




Fig. 65. — Indicator Cards from a Compound Engine with a Receiver, as Taken and as 

Combined. 



delivering. At ZX admission ceased and expansion took place to E' where 
release occurred, the pressure falling to the back pressure and continuing from 
F' to G', where the exhaust valve closed and compression took place to A' , 
thus completing the cycle. At H' leakage past the exhaust valve was so great 
as to cause the curve to fall off considerably from H' to A', instead of con- 
tinuing to be a true compression curve, ending at /, as it should have done. 
The combined diagrams are shown in B. 

In Fig. 66 are shown a set of three cards from a triple-expansion pumping 
engine with large receivers and cranks at 120°. In diagram A the cards are 
shown with the same length and with different pressure scales as taken, but 
with the zero volume line added and transferred to cross-section paper. On 
the high-pressure card admission occurred at A^ causing a constant volume 
pressure rise to -B, the piston being at rest with the crank at dead center. From 
B to C admission occurred at nearly constant pressure to C, where steam was 



234 



ENGINEERING THERMODYNAMICS 



cut off and expansion took place to D. At this point release occurred^ the 
pressure dropping at constant volume to E with the piston at rest. From E 
exhaust took place with slightly increasing pressure, since the intermediate 
cylinder was taking no steam, the intermediate piston being beyond the point 
of cut-off. The pressure rise is slight, however, due to the size of the receiver, 
which is large compared to the cylinder. At two-thirds of the exhaust stroke, 
point F, the back pressure became constant and then decreased, for at this 
point the speed of the intermediate piston increased and the receiver pressure 
fell. At G exhaust closed and a slight pressure rise occurred to A, due to the 
restricted passage of the closing exhaust valve. On the intermediate card 





















































c 














p 


h 






• 
















B 






V 














T 




























\ 


\, 




































A 
















D 




1 
























G 




F 












Afvn 




























X 
















AUU 


























F* 


■^ 


— c 


V 






































A' 






\ 


X^ 












.) 


\ « 






















^ 






— 


^v 






Atn 


i. 
/ 

n' 







id 
\e 




















■ 


a" 














<u 

Atm 






\cf 






















"^ 




-j 










t 


a' 




V 




















A 

V 
















D" 




r^ 




\ 


s^d' 


C" 
















v.. 
















e;' 




\£ 


' 













'd 






V 












I 


^^^ 






^„^^,^ 








1- 




- 



® 

Fig. 66. — Indicator Cards from a Triple-expansion Engine with Receiver as Taken and 

Combined. 



admission occurred at A\ the pressure rising to B'. From B' the admission 
was at nearly constant pressure to X while the piston speed was low and then 
at a falling pressure to C. Pressure was falling, since the steam was supplied 
from a finite receiver into which no steam was flowing during intermediate 
admission. At C cut-off occurred and steam expanded to D', where release 
took place, and the steam was exhausted. As in the case of the high-pressure 
cylinder the back pressure was rising for two-thirds of the stroke, since the 
steam was being compressed into the receiver or rather into a volume made 
up of receiver and intermediate cylinder volume, which is, of course, a decreas- 
ing one, since the cylinder volume is decreasing. At two-thirds of the stroke 
the low-pressure cylinder begins to take steam and the receiver volume is 



WORK OF PISTON ENGINES 235 

now increased, inasmuch as it was made up of the receiver portion of the inter- 
mediate cylinder and a portion of the low-pressure cylinder, and the low- 
pressure cylinder volume increased faster than intermediate decreased for the 
same amount of piston travel. At G' exhaust closed and a slight compression 
occurred to A', thus completing the cycle. 

On the low-pressure card admission occurred at A" and the pressure rose at 
constant volume to B"y and then admission continued first at constant pressure 
and then falling, as in the intermediate cylinder, to the point of cut-off 
at C". From here expansion took place to Z)". At this point the exhaust 
valve opened, the pressure fell nearly to back pressure at E", and the steam 
was exhausted at practically constant back pressure to G"f where the exhaust 
valve closed and there was compression to A", thus completing the cycle. 
The combined diagram is shown iii B. 

Prob. 1. In Fig. 67 are shown six sets of indicator cards from compound en^es. 
The cylinder sizes and clearances are given below. Explain the cylinder events and the 
shaf>e of lines for each card and form a combined diagram for each set. 

No. 1. From a four- valve Corliss engine, 26x48 ins., with 3 per cent clearance in 
each cylinder. 

No. 2. From a single-valve engine, 12x20x12 ins., with 33 per cent clearance in 
high-pressure cylinder and 9 per cent in low. 

No. 3. From a four-valve Corliss engine 22x44x60 ins., with 2 per cent clearance 
in the high-pressure cylinder and 6 per cent in low. 

No. 4. From a single- valve engine 18 X30xl6 ins., with 30 per cent clearance in the 
high-pressure cylinder and 8 per cent in the low. 

No. 5. From a single- valve engine 11^X18^X13 ins., with 7 per cent clearance in 
the high and 10 per cent in the low. 

No. 6. From a double-valve engine 14x28x24 ins., with 3.5 per cent clearance in 
the high-pressure cylinder and 6.5 per cent in the low. 

Prob. 2. In Fig. 68 are shown four sets of indicator cards from triple-expansion 
marine engines. The cylinder sizes and clearances are given below. Explain the cylinder 
events and the shape of the lines of each card and form a combined diagram of each set. 

No. 1. From the engine of a steam-ship, cylinders 21.9x34x57 ins.x39 ins. with 
6 per cent clearance in each and fitted with simple slide valves. 

No. 2. From an engme 20x30x50x48 ins. 

No. 3. From the engine of a steam-ship with cylinders 22x35x58x42 ins., 
assume clearance 7 per cent in each cylinder. 

No. 4. From the engine of the steamer "Aberdeen," with cylinders 30x45 X 70x54 
ins., and with 4 per cent clearance in the high, 7 per cent in the intermediate, and 8 per 
cent in the low. 

Prob. 3. In Fig. 69 are shown some combined cards from compound engines. 
Explain the cylinder events and the shape of the lines and reproduce the indicator cards. 

Prob. 4. In Fig. 70 are shown some combined cards from triple-expansion engines. 
Draw the individual cards and explain the cylinder events and shape of lines. 

8. Standard Reference Cycles or PV Diagrams for the Work of Expansive 
Fluids in Two-cylinder Compound Engines. The possible combinations of 
admission with all degrees of expansion for forward strokes and of exhaust with 



236 



ENGINEERING THERMODYNAMICS 



all degrees of compression for back strokes, with and without clearance, id 
each of the two cylinders of the compound engine, that may have any volume 
relation one to the other and any size of receiver between, and finally, any sort 
of periodicity of receiver receipt and discharge of fluid, all make possible a 




5-| 

0- 

5- 

10-1 



No.l 



No. 2 



lO-i 



No. 8 



ICM 



rlOO 



-60 



-80 





plSO 
-flO 

"0 




p40 
-20 

-0 

i-10 




-100 

-00 

20 





-100 

-60 

-20 





40 

-20 

-0 
10 





f-uo 



-80 



-40 



LO 



No. 6 




Fig. 67. — Six Sets of Compound Engine Indicator Cards. 



very large number of cycles. In order that analysis of these conditions of work- 
ing may be kept within reasonable space, it is necessary to proceed as was 
done with compressors and simple engines, concentrating attention on such 
type forms as yield readily to analytical treatment and for which the formulas 



WORK OF PISTON ENGINES 



237 



are simple even if only approximate with respect to. actual engines, but, of 
course, keeping in mind the possible value of the formulas, as those that teach 




150- 

100- 
50- 





I5-| 

30- 
26- 



140- 

70- 

0- 





10 lao 





No. 2 




No. 4 




25- 





FiQ. *68. — Four Sets of Triple-expansion Engine Indicator Cards. 

no principles or fail to assist in solving problems must be discarded as useless. 
The work that fluids under pressure can do by losing that pressure is no 



238 



ENGINEERING THERMODYNAMICS 



different in compound than in simple engines, if the fluid has a chance to do 
what it can. Provided the structure is such as will not interfere with the com- 
pleteness of the expansion, and no fluid is wasted in filling dead spaces without 



120-1 



80- 



40- 




100- 




300- 



100- 




No. 8 




Fia. 69. — Ck)mbiiied Diagrams of Compound Engines. 

working, then the work per cubic foot or per pound of fluid is the same for simple, 
compound and triple engines. Furthermore, there is a horse-power equiv- 
alence between the simple and compound, if, in the latter case the steam 



WORK OF PISTON ENGINES 



239 



admitted up to cut-off may be assimied to be acting only in the low-pressure 
cylinder, that is, ignoring the high-pressure cylinder except as it serves as a 



250- 



aeo- 



150- 



100- 



60- 




1J50- 



lOO 



50- 



No. 1 



No. 2 



100 



60 





No. 8 




No. 5 




Fig. 70. — Combined Diagrams of Triple-expansion Engines. 

cut-off means or meter. This should be clear from a comparison of Figs. 71, 
A and B. In Fig. 71 A, representing the case of the simple engine without 



240 



ENGINEERING THERMODYNAMICS 



clearance and with complete expansion, the volume admitted, ABy expands 
to the back pressure on reaching the full cylinder volume DCt and exhauFtt 
at constant back pressure, the work represented by the area ABCD- It should 
be clear that no difference will result in the work done if a line be drawn across 
the work area as in Fig. 7 IB, all work done above the line HG to be developed 
in the high-pressure cylinder and that below in the low. This is merely equiv- 
alent to saying that a volume of steam AB Is admitted to the high-pressure 
cylinder expands completely to the pressure at G on reaching the full higb- 
pressure cylinder volume, after which it exhausts at constant pressure (into 
a receiver of infinite capacity), this same amount being subsequently admitted 
without change of pressure to the low-pressure cylinder, when it again expands 
completely. Thus, it appears that the working of steam or compressed air 



p 


rr 


Vol. 


Idmi 


ttedi 


oCyl 


1 

inder 






P\ 


fZ 


Vol.. 


^dmi 


;tedt 


oH.P.Cyl 


nder 




A 




U 
















A 




U 


























• 


































, 


















\ 
















• 


1 








A 


















B 










\ 


















\ 
















\ 














M 


( r 


u 


-V 


olum 


eAdi 


litted to I 


.P.Q 


rl. 






\ 




^ 


n 










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n 


*-x 


'^ 


^ 


HP. 


Displ 


icezn 


mt 












V 


















[V 












n 










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> 




C 


D 
















C 












— > 






1 T» T^i- »— ' 












1 I 


^ispia 


i:-^''iiii: 


lit > u 


tiilluc 






L.p.r 


jispia 


cem€| 


nt — 





Fig. 71. — Diagram to Show Equality of Work for Expansion in One-cylinder Simple and in 
Two-cylinder Compound Engines for the Same Rate of Expansion. 

in two successive cylinders instead of one will in no way change the maximum 
amount of work a cubic foot supplied can do, the compounding merely making 

it easier to get this maximum. In simple engine cases, Fig. TlA, the cut-off in per 

AB 
cent of stroke is 100 Xy^, which is a very small value, leaving but little time to 

open and close the admission valve, whereas in the compound case the per cent 

AB , ^ 
cut-off in the high-pressure cylinder is 100X7T7;;,and in the low-pressure cylinder, 

HC 
100 X -^t:;, which are quite large enough ratios to be easily managed with ordinary 

valve gears. 

Compounding docs, however, introduce possibilities of loss not present in 
the single-stage expansion, if the dimensions or adjustments are not right, 
which may be classed somewhat improperly as receiver losses, and these are 



WORK OF PISTON ENGINES 



241 



of two kinds, one due to incomplete 
expaxLsion in the high and the other 
to over-expansion there. Thus, in 
Fig. 72, if ABCEFGDA represent a 
combined compound diagram for 
the case of complete expansion in 
the high-pressure cylinder continued 
in the low without interruption but 
incomplete there, DC represents the 
volume in the low-pressure cylinder 
at cat-off^ and at the same time the 
total high-pressure cylinder volume. 

If now, the low-pressure cut-off 
be made to occur later. Fig. 73, 
then the volume that the steam 

would occupy when expansion began c.,^ ^o t^- *. au n * t ' 

. . 1. , . ^*^' '2. — Diagram to Show Correct Low-pressure 

m the low-pressure cylmder is rep- 
resented by D'C. This adjust- 























A 




B 




















\ 




















\ 




















\ 


V 




















\ 














D 








V 


sX 










__, 




- . T\ 






.A»^^w, 






F 

F 








•L>1 






^ ^ 




G 
































1 

1 

1 




L^ T^ 




^ 


Y 


1 






jj 


^2 









Cut-off for No Receiver Loss. 



ment could not, of course, change the high-pressure total volume DC, so 

that at release in the high-pressure 

cylinder the pressure would drop 

abruptly to such a value in the 

receiver as corresponds to filling 

the low pressure up to its cut-off, 

and work be lost equal to area 

CCW^. 

. A shortening of the low-pressure 

cut-off will have an equally bad 

effect by introducing negative work 

as indicated in Fig. 74, in which 

the L.P. cut-off volimie is reduced 

from DC to D'C Expansion in the 

high pressure proceeds as before till 

the end of the stroke, at which time 

it has a volume DC greater than. 

the low pressure can receive D'C, 

hence the receiver pressure must 

rise to such a value as will reduce 

the volume the required amount, 

introducingthe negative work CC'C". 

Changes of low-pressure cut-off may, 

therefore, introduce negative work, 

P. -o T^• A au !:!«•* rr ^u -^ chaugc thc rcccivcr pressure and. 

Fig. 73. — Diagram to Show Effect of Lengthening ^ v j. .i . 

L.P. Cut-off Introducing a Receiver Loss Due to 01 COUrse, modlfiy the distribution 
Incomplete High-pressure Cylinder Expansion, of work between high and low, but 

























A 




B 






















\ 






















\ 






















\ 
















■ 






\ 






















\ 


V 






















\ 
















.5... 


■«»^^>^« 


XV 


.\ 


c 














1" 


















Di- 
















%-. 


„' 


















v'/i^'///^ 


^///hi 


K 










D' 








c 




















^ 




E 




G 


















F 












TV 






















-Dj 



































242 



ENGINEERING THERMODYNAMICS 



these saxae effects might also have resulted from changes of high-pressure 
cut-off or of cylinder ratio. 

For such conditions as have been assumed it seems that compounding 
does not increase the work capacity of fluids, but may make it easier to realize 
this capacity, introducing at the same time certain raiher rigid relations between 
ciU-offs and cylinder volumes as necessary conditions to its attainment. It 
can also be shown that the same proposition is true when there are clearance and 
compression, that is, in real cylinders and when the receiver is real or not 
infinite in size, or when the exhaust of high and admission of low, are not con- 

stant^pressure lines. The former 
needs no direct proof, as inspec- 
tion of previous diagrams makes 
it clear, but the latter requires 
some discussion. 

A real receiver of finite size 
is at times in communication 
with the high-pressure cylinder 
during its exhaust, and at other 
times with the low-pressure 
cylinder during admission, and 
these two events may take place 
at entirely independent times, be 
simultaneous as to time, or over- 
lap in all sorts of ways. Suppose 
the periods to be independent 
and there be no cylinder dear- 
ance, then at the beginning of 
high-pressure exhaust two sepa- 
rate volimies of fluid come 
together, the contents of both 
the high-pressure cylinder and the 
receiver, and this double volume 
is compressed by the H.P. piston 
into the receiver, in which case the 
high-pressure exhaust would take 
place with rising pressure. Follow- 
ing this will come low-pressure admission, during which the volume of fluid in the 
receiver expands into the low-pressure cylinder up to its cut-off, and if the same 
volume is thus taken out of the receiver as entered it previously, low-pressure 
admission will take place with falling pressure, the line representing it exactly 
coinciding with that for the high-pressure exhaust. Independence of H.P. 
cylinder exhaust and L.P. cylinder admission, as to time, may result in a cycle 
such as is represented in Fig. 75 for the case of no cylinder clearance. On this 
diagram the receiver line is DC, an expansion or compression line referred to a 
second axis of volumes XJ, placed away from the axis of purely cylinder volumes 



A 




B 






















i 






















\ 






















\ 


















D' 




\. 


:' 


C" 


















m 


















• 




















D 








c 


















Tk 




^ 


















■Dj- 














\ 


V 






















\ 


\ 




















¥ 




■^ 


^ 


E 


— 




G 


















F 












■^ 








^ 





























Fig. 74. — Diagram to Show Effect of Shortening L.P. 
Cut-off, Introducing a Receiver Loss Due to 
• Over-expansion in the High-pressure Cylinder. 



WORK OF PISTON ENGINES 



243 



by the distance LD, equal to the receiver volume to scale. All diagram points 
are referred to the axis A I except those on the line DC. 

This same case of time independence of H.P. exhaust and L.P. admission 
yields quite a different diagram when the cylinder clearance is considered. 
Such a case is represented by the diagram, Fig. 76, which also serves to illustrate 
the eflFect of incomplete expansion and compression as to equalization of 
receiver with cylinder pressures. At high-pressue release the volume of fluid 
in the H.P. cylinder is ML and its pressure iaJJi. This is about to come 
into communication with the receiver volume ON from which the low-pres- 
sure cylinder finished taking fluid and which is, therefore, at the same pressure 



d 


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eterr 






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Fig. 75. — Diagram to Illustrate Variable Receiver Pressure for the Case of Independent 
High-pressure Exhaust and Low-pressure Admission with Zero Clearance. 



as the L.P. cylinder cut-off KS- The question, therefore, is — ^what will be 
the pressure at P in both H.P. cylinder and receiver when LM cu.ft. of fluid 
at LR pressure cpmbines with ON cu.ft. at pressure KS, and together occupy 
a volume ON+LM- By hypothesis the pressure after mixture is 

(first volume Xits pressure) + (second voliuneXits pressure) 

simi of volumes 



From this or the graphic construction following, the point P is located. If the 
high-pressiire expansion had continued to bring LQ to the receiver pressure 
KS, it would reach it at X, At this hypothetic point there would be a volume 
NX in the H.P. cylinder to add to the volume ON in the receiver at the same 
pressure, resultmg in OX cu.ft. This fluid would have a higher pressure at the 
lesser volume of receiver and H.P. cylinder and the value is found by a compres- 



244 



ENGINEERING THERMODYNAMICS 



sion line through X, XPAT referred to the receiver axis. This same line is also 
the exhaust of the H.P. cylinder from P to ^4 . A similar situation exists at admis- 
sion to the L.P. cylinder as to pressure equalization and location of admission 
line. At the end of the L.P. compression there is in the L.P. cylinder FE cu.ft. 
at pressure EH, to come into communication with the receiver volume CB cu.ft. 
at pressure BGy that at which H.P. exhaust ended. Producing the L.P. com- 
pression line to /, the volume BI is found, which, added to receiver, results 
in no pressure change. An expansion line, referred to the receiver axis through /, 
fixes the equalized pressure J and locates the L.P. admission line JK, which, it 
must be observed, does not coincide with the H.P. exhaust. 

So far only complete iridependence of the time of H.P. exhaust and L.P. 
admission have been considered, and it is now desirable to consider the diagram- 

















































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Fia. 76. — Diagram to Illustrate Variable Receiver Pre.ssiire for the Case of Independent Higb- 

pressure Exhaust and Low-pressure Admission with Clearance. 

matic representation of the results of complete coincidence. Such cases occur 
in practice with the ordinarj*^ tandem compound stationary steam engine and 
twin-cylinder single-crosshead Vauclain compound steam locomotive. In the 
latter structure both pistons move together, a single valve controlling both 
cylinders, exhaust from high taking place directly into low, and for exactly 
equal coincident time periods. The diameter of the low-pressure cylinder being 
greater than the high, the steam at the moment of release suffers a drop in 
pressure in filling the low-pressure clearance, unless, as rarely happens, the pres- 
sure in the low is raised by compression to be just equal to that at H.P. release. 
After pressure equalization takes place, high-pressure exhaust and low-pres- 
sure admission events are really together a continuation of expansion, the 
volume occupied by the steam at any tinui being equal to the difference between 



WORK OF PISTON ENGINES 245 

tlie two cylinder displacements and clearances up to that point of the stroke. 
Before this period of communication, that is, between high-pressure cut-off and 
release, the volume of the expanding fluid is that of the high-pressure displace- 
ment up to that point of the stroke, together with the high-pressure clearance. 
After the period of commimication the volume of the expanding fluid is that 
of the low pressiu-e cylinder up to that point of the stroke, together with the low 
pressure clearance, plus the high-pressure displacement not yet swept out, 
and the high-pressure clearance. 

These fluid processes cannot be clearly indicated by a single diagram, 
because a diagram drawn to indicate volumes of fluid will not show the volumes 
in the cylinders without distortion. If there be no clearance, Fig. 77 will 
assist in showing the way in which two forms of diagram for this purpose 
are derived. Referring to Fig. 77 A, the volume AB admitted to the high 
pressure cylinder expands in it until it occupies the whole H.P. cylinder volume 
DC At this point expansion proceeds in low and high together, with decreasing 
volumes in high and increasing in low until the low-pressure cylinder volume is 
£^ttained at E. The line BCE then indicates the pressures and volumes of the 
fluid expanding, but does not clearly show the volume in either cylinder between 
C and Ey with the corresponding pressure. It is certain, however, that when 
the volume in the H.P. cylinder becomes zero the pressure must have fallen 
to a value the same as that in the low when the fluid completely fills it, 

orP/=Pe. 

As the high-pressure piston returns, on the exhaust stroke, the low- 
pressure piston advances an equal distance, on its admission stroke, sweep- 
ing through a greater volume than the high pressure, in the ratio of low-pres- 
sure to high-pressure displacements. If at any point of the stroke the volume 
remaining in the high-pressure cylinder be x, and the high- and low-pressiu-e 
displacements be respectively Di, and D2, then (Dx—x) is the volume swept 

out by high-pressiu'e piston, x the volume remaining in it, and -^^(Di— x) 

Di 

the volume swept in by the low-pressure piston. Then the total volume 
still in the two cylinders is, for a point between C and E, 



7=x+^(Di-x)=Z)2-a:(~?-l). 



Since the equation of the curve CE is, PV = PcVc = PcDif the value of V may 
1 c substituted, giving P \D2 — xi-~ — l] \= PcDi, = constant. Dividing by 



^— 1 jthis becomes P 



D2 

X 



i®-') 



= PN.-_-^ — x =a new constant, so 



246 



ENGINEERING THERMODYNAMICS 

from the axis GP anv 



that if a new axis LM be laid off on B, GV=i n n i 



point on FC will be distant from the new axis LM an amount f ^ ^^ — \x\ 

as the product of this distance and the pressure P, is constant, the curve PC 
is an equilateral hyperbola referred to the axis LM, Therefore Fig. 77 B is 




COMBINED DIAGRAM TO ONE SCALE 

Fig. 77. — Diagram to Illustrate the Compound Engine Cycle with No-receiver, and Exact 
Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance. 

the pressure-volume representation of the entire cycle of the high-pressure 
cylinder. 

In Fig. 77C is shown the corresponding pressure-volume diagram for the low- 
nressure cylinder, for which it may similarly be shown that the curve DE may be 



WORK OF PISTON ENGINES 247 

plotted to an axis NO at a distance to the left of the axis GP equal to the same 
quantity, 

("'-S^k w 

These diagrams, 77-4, B and C may be superposed, as in Fig. 77 E, giving 
one form of combined diagram used for this purpose, and the one most nearly 
comparable with those ah-eady discussed. In this diagram, the area ABC FA 
represents the work of the high-pressure cylinder, and DEIHD, the work of 
the lo w-pressure cylinder. Together, they equal the work of the enclosing 
figure ABEIHA, and hence the work of the low-pressure cylinder must also be 
represented by the area FCEIHK 

It is not difficult to show that if a vertical, CJ, be dropped from the point 
C to the exhaust line HJ, the figure CFHJ, in Fig. 77D is similar to DEIH, 
in Fig. 77E, reversed, but drawn to a different horizontal scale. Here the 
length of the low-pressure diagram is made equal to the length of the high- 
pressure diagram. The two scales of volumes are shown above and below the 
figure. While this appears to be a very convenient diagram, it will be found 
to be less so when clearance and compression are considered. 

It may be noted that since it has been shown that the curves CF and DE 
are of the same mathematical form (hyperbolic) as the expansion line CE, 
they may be plotted in the same way after having in any way found the axis. 
The location of this axis may be computed as given above, or may be found 
graphically by the method given in connections with the subject of clearance. 
Chapter I, and shown in Fig. 77B, Knowing two points that lie on the curve, 
C and Fj the rectangle CDFK is completed. Its diagonal, DK, extended, cuts 
the horizontal axis GV in the point Af , which is the base of the desired axis ML. 

If now the axis NO and the figure DEIHj part of which is referred to this 
axis, be reversed about the axis GP, Fig. 77C, NO will coincide with AfL, 
Fig. 77D, and Fig. 78 results. Note that the axis here may be found graphically, 
in a very simple way. Draw the vertical CK from C to the axis GV and 
the horizontal DJ to the vertical IE extended. DC is then the high-pressure 
displacement and DJ the low-pressure displacement. Draw the two diagonals 
DK and JG, extending them to their intersection X. By similar triangles it 
may be shown that a horizontal line, UW, will have an intercept between these 
two lines, JG and DK, equal to the volume of fluid present in the two cylinders 
combined. The intersection X is the point at which this volume would 
become zero if the mechanical process could be continued unchanged to that 
point, and is, therefore, on the desired axis ML extended. T being the inter- 
section of WU with the axis GP, when the volume UT is present in the high- 
pressure cylinder, TW gives the volimie in the low-pressure cylinder. 

When clearance and compression are considered, this diagram is changed in 
many respects, and is shown in Fig. 79. The axes OP, OV and OV are laid out, 
with OZ equal to the clearance and ZK, the displacement, of the high-pressure 



248 



ENGINEERING THERMODYNAMICS 



cylinder, and OQ and QF, clearance and displacement of the low-pressure cylinder. 

It is necessary to know high-pressure cut-oflf, ;z^l high-pressure compression, 
Zli. 

: and low-pressure compression, =^, in addition to the initial and back 
ZK YQ 

pressures, in order to lay out the diagram. The drop in pressure CD at releasee 

is due to the coming together of (volume Vg at pressure P<.), with (volume T'^ 

at pressure P,). If the volimie Vj (measured from axis OP) were decreased 

sufficiently to raise the pressure in the low-pressure clearance to the pressure 




Fig. 78. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound 
Engine at any Point of th'e Stroke for the Case of No-clearance and Coincidence of 
H.P. Exhaust with L.P. Admission. 

at C, the voliune woxild become F„ as indicated at the point S, and the volumes 
now combined in the hypothetical condition would occupy the volume SC 
Increasing this volume to D'D, that actually occupied after communication, 
the pressure would fall along the curve SD\ which is constructed on KV and KT 
as axes. The pressure of D is then laid out equal to the pressure at D\ To 
find the axis, MLy for the cui^es DE and £>'£', from any convenient point A' 
on ZAy draw the line NK extended to X. Extend HG to A, at a height equal 
to that of N, and draw RQX, and through the intersection draw the desired axis 
XML, The fraction of stroke completed at E' in the low pressure at cut-ofif must 
be equal to that completed at E in the high pressure at compression, and may be 
laid out graphically by projecting up from E to the point U on the line NK and 
horizontally from t7 to TF on the line RQ. Projecting down from W to the 
curve at £' locates the point of effective cut-ofif in the low-pressure cylinder. 



WORK OF PISTON ENGINES 



249 



After the supply from the high-pressure cylinder has been cut off at E', the 
expansion is that of the volume in the low-pressure cylinder and its clearance, and 
hence the curve E'G is constructed on OP as an axis. 

While in this last case a zero receiver volume has been assumed, there is 
nothing to prevent a receiver volume being interposed between H.P. and L.P. 
so that common expansion takes place with a volume greater than assumed 
by so much as this volume, the effect of which is to decrease the slope of DE 
and D'E\ Such receivers usually consist of the spaces included in a valve 
body and connecting passages and may be treated generally as increased L.P. 
clearances. 

The most conamon of all relations between H.P. exhaust and L.P. admis- 
sion is, of course, that of partial coincidence of periods, as it is thus with all cross- 




-u 



p. Cut Oft due 



L.P. Diaptacctpent 



Displaceinenr*'^^ 



\ 



-:s 



^ 



Fig. 79. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compoimd 
Engine at any Point of the Stroke for the Case of Clearance and Coincidence of H.P. 
Exhaust with L.P. Admission. 



compound and triple-expansion engines having separate cranks for the 
individual cylinders. For these there is no simple fixed relation between the 
periods, for, while crank angles are generally fixed in some comparatively simple 
relation, such as 90°, 180° and 270° for conipounds and 120° for triples, they 
are sometimes set at all sorts of odd angles for better balance or for better 
turning effort. Even if the angles were known the receiver line would have 
to be calculated point by point. When the H.P. cylinder begins to discharge into 
a receiver for, say, a cross compound with cranks at 90°, steam is compressed 
into the receiver, and so far the action is the same as already considered for 
independence of periods, but at near mid-stroke the low-pressure admission 
opens while high-pressure exhaust continues. This will cause the receiver 
pressure to stop rising and probably to fall until the low pressure cuts off, which 



250 



ENGINEERING THERMODYNAMICS 



may occur before the H.P. exhaust into the receiver ceases. If it does, the receiver 
pressure will again rise. Exact determination of such complex receiver lines is 
not often wanted, and when needed is best obtained graphically point by point. 
Its value lies principally in fixing exactly the toork distribution between cylin- 
ders, which is not of great importance except for engines that are to work at 
constant load nearly all the time, such as is the case with city water worfc? 
pumping, and marine engines. While equations could be derived for these 
cases, they are not worth the trouble of derivation, because they are too cumber- 
some, and graphic methods are to be substituted or an approximation to be made. 
Four kinds of approximation are available, as follows, all of which ignore 
partial coincidence of periods: 

1. Receiver pressure constant at some mean value and clearance ignored. 

2. Receiver pressure constant at some constant value and clearance con- 
sidered with compression zero or complete. 

3. Receiver pressure fluctuates between fixed limits as determined by an 
assumed size, clearance ignored. 

4. Receiver pressure fluctuates between fixed limits as determined by an 
assumed size, clearance considered, with compression zero or complete. 

These are not all of equal difficulty in solution, and the one to be used is 
that nearest the truth as to representation of conditions, which is usually the 
most difficult, provided time permits or the information is worth the trouble. 
Quickest work is accomplished with assumption (1) and as this is most often 
used in practical work it indicates that its results are . near enough for most 
purposes. 

This discussion leads, therefore, to the analytical study of the following cycles: 

Infinite Receiver, Zero Cylinder Clearance. 

CYCLES V, AND VI (Fig. 80). 

' (a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cyl. (may be zero) by law PV -c lor {V)\ 

Pr^c for (VI). "^ 
(c) Equalization of H.P. cyl. pressure with receiver pressure 

at constant volume (may be zero). 
{d) Exhaust into infinite receiver at constant pressure from 

H.P. cylinder, 
(e) Equalization of H.P. cylinder pressure with supply pres.sure 

at constant zero volume. 
' (/) Admission from receiver at constant receiver pressure to 

L.P. cylinder. 
{g) Expansion in L.P cylinder (may be zero) by law PV = c for 

(V); Pr=c for (VI). 
(A) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero), 
(i) Exhaust at constant back pressure for L.P. cylinder. 
(j) Equalization of L.P. cylinder pressure with receiver pressure 

at constant zero volume. 



H.P. Cylinder 
Events 



L.P. Cylinder 
Events 



WORK OF PISTON ENGINES 



251 





























































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\ Cycles VII & VIII 






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Cycle* V & VI 
Cycle Y PV«C Cycle VI PV«-C 














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ACycle VII PV-C Cycle VIII PV«.C 






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Cycles IX & X 
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Fig. 80. — Compound Engine Standard Reference Cycles or PK Diagrams. 



252 



ENGINEERING THERMODYNAMICS 



Relations 
between h.p. 

AND L.P. CyLINDEB 

Events 



(1) H.P. exhaust and L.P. admission independent as to time, 

coincident as to representation (except as to length). 

(2) H.P. expansion line produced coincides as to representation 

with L.P. expansion line. 

(3) Tlie length of the constant pressure receiver line up to the 

H.P. expansion line produced is equal to the length of 
the L.P. admission line, 



H.P. Cylinder 
Events 



Finite Receiver, Zero Cylinder Clearance. 
CYCLES VII, AND VIII, (Fig. 80). 



(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV=c 
for (VII); Pr*=c for (VIII). 

(c) Equalization of H.P. cylinder pressure with receiver pressure 

at constant volume (may be zero) with a change of receiver 
pressure toward that at H.P. cylinder release (may be 
zero). 

(d) Exhaust into finite receiver from H.P. cylinder at rising 

pressure equivalent to compression of fluid in H.P. cylinder 
and receiver into receiver by law P7=c for (VII) and 
P7*=c for (VIII). 

(e) Equalization of H.P. cylinder pressure with supply pressure 

at constant zero volume. 



L.P. Cylinder 
Events 



Relation 

between H.P. AND 

L.P. Cylinder 
Events 



' (f) Admission from receiver to L.P. cylinder at falling pressure 

equivalent to expansion of fluid in receiver into receiver 

and L.P. cylinder together by law P7=c for (VII), PV*-^c 

for (VIII). 
ig) Expansion in L.P. cylinder (may be zero) by law PV=c 

for (VII); PV*-c for (VIII). 
(A) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero), 
(i) Exhaust at constant back pressure for L.P. cylinder. 
(t) Equalization of L.P. cylinder pressure with receiver pressure 

at constant zero volume to value resulting from HJP. 

exhaust. 

' (1) H.P. exhaust and L.P. admission independent as to time, 
coincident as to representation, except as to length. 

(2) H.P. expansion line produced coincides as to representation 
with L.P. expansion line. 

(3) The length of the receiver pressure line up to the H.P. 
expansion line produced is equal to the length of the 
L.P. admission line. 



WORK OF PISTON ENGINES 



253 



No Receiver, Zero Cylinder Clearance. 
CYCLES IX, AND X, (Fig. 80). 



H^. Cylinder 
Events 



Both H.P. and L.P. 
Simultaneously 

H.P. Cylinder 
Events 



L.P. Cylinder 
Events 



(a) Admission at constant supply pressure to H.P. Cylinder. 

(b) Expansion in H.P. cylinder (may be zero) by law PV'^c 
for (IX); PF'=c for (X). 

' (c) Transference of fluid from H.P.to L.P. cylinder with simul- 
taneous continuation of expansion until all fluid is in 
L.P. cylinder and expanded to its full volume by law 
PV =c for (IX) ; PV' =c for (X). 

r (d) Equalization of H.P. cylinder pressure to the pressure of 
I supply. 

' (e) Equahzation of L.P. cylinder pressure with back pressure 

at constant volume (may be zero). 
(/) Exhaust at constant back pressure for L.P. cylinder. 
(g) Equalization of L.P. cylinder pressure to the pressure in 

H.P. cylinder at the end of its expansion. 



H.P. Cylinder 
Events 



L.P. Cylinder 

Events 



Infinite Receiver, with Cylinder Clearance. 

CYCLES XI, AND XII, (Fig. 80). 

(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV^c 
for (XI); Pr=c for (XII). 

(c) Equahzation of H.P. cylinder pressure with receiver pressure 
at constant volume (may be zero) pressure. 

(d) Exhaust into infinite receiver at constant pressure from H.P. 
cylinder. 

(c) Compression in H.P. cyhnder to clearance volume (may be 
zero) by law PV =c for (XI) ; P7* -c for (XII). 

(/) Equalization of H.P. cyhnder pressure with supply pressure 
at constant clearance volume, may be zero. 

' (g) Admission from receiver at constant-receiver pressure to 
L.P. cyhnder. 

(h) Expansion in L.P. cyhnder (miay be zero) by law PV ==c 
for (XI); P7'=c for (XII). 

(t) Equahzation of L.P. cyhnder pressure with back pressure 
at constant volume (may be zero). 

(j) Exhaust at constant back pressure for L.P. cyhnder. 

(k) Compression in L.P. cyhnder to clearance volume by law 
PV^c for (XI) ; PV =c for (XII) (may be zero). 

(0 Equahzation of L.P. cyhnder pressure with receiver pres- 
sure at constant clearance volume without change of 
receiver pressure (may be zero). 



254 



ENGINEERING THERMODYNAMICS 



Relations 

between h.p. and 

L.P. Cylindeb 

Events 



' (1) H.P. exhaust and L.P. admission independent as to time, 
coincident as to representation except as to length. 

(2) L.P. expansion line does not coincide as to representation 

with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The length of the constant-receiver pressure line intercepted 

between H.P. compression line and H.P. expansion line 
produced is equal to the same intercept between L.P. 
expansion line and L.P. compression line produced. This 
is equivalent to the condition that the volume taken in bj 
low is equal to expelled by the high reduced to the same 
pressure. 



Finite Receiver, with Cylinder Clearance. 



CYCLES XIII, AND XIV, (Fig. 80). 



H.P. Cylinder 

Events 



L.P. Cylinder 
Events 



' (a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV ^c 
for (XIII) ; PV* =c for (XIV). 

(c) Equalization of H.P. cylinder pressure with receiver pres- 
sure at constant volume (may be zero) toward that at 
H.P. cyhnder release (may be zero). 

[d) Exhaust into finite receiver from H.P. cyUnder at rising 
pressure equivalent to compression of fluid in H.P. cylinder 
and receiver into receiver by law P7 =c for (XIII) \ PV* -c 
for (XIV). 

(c) Compression in H.P. cylinder to clearance volume (may be 
zero) by law PV =c for (XIII) ; PV* -c for (XIV). 

(/) Equalization of H.P. cylinder pressure with supply pressure 
at constant clearance volume. 

{g) Admission from receiver to L.P. cylinder at falling pressure 
equivalent to expansion of fluid in receiver into receiver 
and L.P. cylinder together by PV =c for (XIII); PV* ^c 
for (XIV). 

(A) Expansion in L.P. cylinder (may be zero) by law PV^c 
for(V); PV =cfor(VI). 

(i) EquaUzation of L.P. cylinder pressure with back pressure 
at constant volume (may be zero). 

( j) Exhaust at constant back pressure for L.P. cylinder. 

(A;) Compression in L.P. cylinder to clearance volume by law 
PV^c for (XI) ; PV* =c for (XII) (may be zero). 

(0 Equalization of L.P. cylinder pressure with receiver pressure 
at constant clearance volume with change of receiver 
pressure in direction of L.P. compression pressure (may 
be zero). 



WORK OF PISTON ENGINES 



255 



Relations 

between h.p. and 

L.P. Cylinder 

Events 



(1) H.P. exhaust and L.P. admission independent as to time, 
representation and length. 

(2) L.P. expansion line does not coincide as to representation 
with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The high-pressure exhaust and low-pressure admission lines 
do not coincide as to representation by reason of clearance 
influences. 

(4) There is a relation between the lengths of the L.P. admission 
and H.P. exhaust lines, but not a simple one. 



No Receiver, with Cylinder Clearance. 
CYCLES XV, AND XVI, (Fig. 80). 



H.P. 

Cylinder 

Events 



Also 

L.P. 

Event 



LP. 

Cylinder 
Events 



(o) Admission at constant-supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) according to law 
PV^c for (XV); PV' =c for (XVI). 

(c) Equalization of pressures in H.P. cylinder after expansion 

with that in L.P. after compression at constant volume 
(may be zero). 

(d) Transference of fluid from H.P. to L.P. cylinder imtil all 

fluid is in L.P. cylinder and expanded to its full volume 
by same law as (b) . 

(e) Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV^c for (XV) ; P7* -c for (XVI). 

(f) Equalization of pressure in H.P. cylinder with supply at 

constant-clearance volume (may be zero). 

(g) Expansion in L.P. cylinder may be zero by law PV^c for 

(XV); P7'=c for fXVI). 
(h) Equalization of pressure in L.P. cylinder with back pressure, 

at constant volume (may be zero). 
{i) Exhaust at constant pressure for L.P. cylinder. 
(j) Compression in L.P. cylinder to clearance, may be zero by 

law PV =c for (XV) ; PV =c for (XVI). 
{k) Equahzation of L.P. cylinder pressure with H.P. cyhnder 

pressure. 



Cycle XVII, Fig. 81, for the triple expansion is defined in the same way 
as the corresponding case of compounds Cycle V, with appropriate alterations 
in wording to account for a third or intermediate cylinder between high- and low- 
pressure cylinders and an additional receiver. Thus, high- pressure cylinder- 
exhausts into first, and intermediate cylinder into second receiver: inter- 
mediate cylinder receives its supply from first,"^and low-pressure cylinder from 
second receiver. This being the case, it is imnecessary to write out the cylin- 
der events, noting their relation to the corresponding compound case. 



256 



ENGINEERING THERMODYNAMICS 



9. Compound Engine with Infinite Receiver, Logarithmic Law. No Clear- 
ance, Cycle V. General Relations between Pressures, Dimensions, and Work 

It must be understood that the diagrams representing this cycle, Fig. 82, 
indicating ^A) incomplete expanmon and {B) over-expansion in both cylinders. 

P 

























A 


B 




















M 








































e 




\ 



















\ 


C 
Dp 


[\ 


G " 














\ 

c 












E 


d 














f 




A 






















• 

J 
















1 




■ 




h 


H 


J 


• 












A: 


— _ 






>j 




K 













in 






M 








n 








L 

























Fig. 81. — Triple-expansion Engine Standard Reference Diagram or FV Cycle for Infinite 

Receiver. 



may just as well stand for over, complete or incomplete expansion in all possible 
combinations in the two cylinders. Applying the principles already derived 
for calculating the work areas, 



High-pressure cylinder work 



Wh^F^\{\+\o^, J'^^-PtfFtf, 



(296) 



Low-pressure cylinder work 



W,.^l\\\[ 



1+iog. J/)-7>,F„ 



(297) 



WORK OF PISTON ENGINES 257 

'otal work 

TF=P»n(l+log, ^^) +Pe7e(l+log. ^^) -P,7^-P,F,, ^ . (298) 

>re8sure being in pounds per square foot, and voliunes in cubic feet. 

In theses expressions the receiver pressure P,= Puis imknown, but determinate 
IS it is a function of initial pressure and certain volumes; giving it the value, 

is merely satisfying the condition that the point E at which expansion begins 
in the low-pressure cylinder must lie in the expansion line of the high. Sub- 
stituting this value there results 

W^P^V^{l+\o^y^ +P^V^(l+\o^ ^^ -P,vi^-P,V, 

=P5n[2+log,^;+loge^;-^]-P,F,. . (299) 

This is a perfectly general' expression for the work of the fluid expanding to 
any degree in two cylinders in succession when the clearance is zero and receiver 
volume infinite, in terms of initial and back pressures, pounds per square foot, 
the volumes occupied by the fluid in both cylinders at cut-off, and at full 
stroke in cubic feet. Dividing this by the volume of the low-pressure cylinder 
Vg gives the mean effective pressure referred to the low-pressure cylinder, 
from which the horse-power may be determined without considering the high- 
pressure cylinder at all. Hence, in the same imits as are used for P» and P,, 

(M.E.P. referred to L.P.) ^P.^'h+log, ^'+log, ^-^1 -Pi,- (300) 

Proceeding as was done for simple engines, the work per cubic feet of fluid 
supplied is found by dividing Eq. (299) by the volume admitted to the high- 
pressiire cylinder F*, whence. 



Work per cu.ft. supplied = Ph 2+loge tt + lo& i/"" p- "" ^oijf' • • • 
Also applying the consumption law with respect to horse-power, 



(301) 



13 750 Vb 
Cu.ft. supplied per hour per I.H.P. = ^^^^ ^;^ ^ ^^^^ y. (302) 

Lbs. supplied per hr. per IH.P. = (^^p^^;Jft^ j^^^ ^fi (303) 



258 



ENGINEERING THERMODYNAMICS 



These last five equations, (299), (300), (301), (302), (303), while character- 
istic, are not convenient for general use in their present form, but are ren- 




(atm.pr.) 
G(bk.pr,) 






1_1 



I 



N 



:1...4bk.pr.) 

f— -(Pel.pr.)L 
(atm.pr.) 



H V 



H.P. 





INOICATOII CAMM OF 

EQUAL BAM AND 

HEIOHT 





INMCATOfl CARDS OP 
EQUAL BASE ANO 
HEIQHT 



Fia. 82. — Work of Ejqjansive Fluid in Ck)mpound Engine with Infinite Receiver, Zero 
Clearance. Cycle V, Logarithmic Expansion and Cycle VI Exponential Expansion. 

dered so by substituting general symbols for initial and back pressures 
displacement, cut-off, and amount of expansion for each cylinder. 



WORK OF PISTON ENGINES 



259 



et (in.pr.) = initial or supply pressure, pounds per square inch = ^52? 
(rel.pr.)j5r = release pressure, in H.P. cylinder pounds per square inch = 



(rel.pr.)j!i = release pressure in L.P. cylinder, pounds per square inch 

(rec.pr.) =receiver pressure, pounds per square inch=r^=yv~; 

P 

*(bk.pr.) = back pressure, pounds per square inch=Y24J 

y 
/?// = ratio of expansion in high-pressure cylinder = y-*^ 5 



144' 
144' 



Fe' 



jBl= ratio of expansion in low-pressure cylinder = 

V, 
J2v= ratio of expansion for whole expansion ==:fd^; 

Vb 

Z)h= displacement of high-pressure cylinder =Fd = Fc; 

Z)l = displacement of low-pressure cylinder = F/= Vg\ 

-Re = cylinder ratio = 7^ =t/> 

l^H Vd 

Zh = fraction of displacement completed up to cut-off in high-pressure 

cylinder, so that ZhDh = F^ = -^; 
Zl= fraction of displacement completed up to cut-oflf in low-pressure 

Rl 



cylinder, sq that ZlDl = F« = ^ 



Substitution of these general symbols in Eqs. (299), (300), (301), (302), and 
(303) gives another set of five equations in useful form for direct substitu- 
tion of ordinary data as follows: 



Work of cycle 

= 144Dir(in.pr.) r^(2+loge fi//+loge Rl) ~£^ 

= 144D. { (in.pr.)|(2+log. A+iog, J_ ._i_) .(bk.pr.) 
(in.pr.)^^(2+lo& R/r+loge Rl-^^ " (bk.pr.) 



-144(bk.pr.)i)L (a) 



(b) 



= IUDl 



(m.e.p.) lbs. per sq.in. referred to L.P. cyl. 



(c) 



. (304) 






= (in.pr.)g^(2+lofe R„+\oe, Rl-^ - (bk.pr.) 
= (in.pr.)^( 2+lofe z~„'^^^^'Z',~ ZTRc) ~ ^^^-^^-^ 



(a) 



(b) 



. (305) 



260 



ENGINEERING THERMODYNAMICS 



Work per cu.ft. supplied 

= 144r(in.pr.) (2+loge Rh +log, Bl-^) - (bk.prOJKcfiJ 

= 144[(in.pr.)(2+loge ^+lo& ^-^) -(bk.pr.)|^] 

Cu.ft. supplied per hr. per I.H.P. 

13,750 1 



(a) 



(6) 



(306) 



(m.e.p. ref. to L.P.) RhRc 

13,750 Z„ 

(m.e.p. ref. to L.P.) Re 



(a) 



(b) 



(307) 



J 



From this, of course, the weight in pounds supplied per I.H.P. results directly 
from multiplication by the density of the fluid. 

To these characteristic equations for evaluating work, mean pressure, 
economy and consumption in terms of the initial and final pressures and cylin- 
der dimensions there may be added a series defining certain other general rela- 
tions of value in fixing the cycle for given dimensions and initial and final 
pressures, and in predicting dimensions for specified total work to be done and 
its division between high- and low-pressure cylinders. 

Returning to the use of diagram points and translating into the general 
symbols as each expression is derived, there results. 



Receiver pressure =Pd—Pe = Pbjr- 



:. (rec.pr.) = On.pr.)^^^^ = (m.pr.)^-^ 



= (in.pr.) 



Rl 



RcRh 



(a) 



(6) 



(308) 



(rel.pr.)w = (in.pr.)- 



High-pressure cylinder release pressure =Pc=^»-Tr> 

1^ 

Rh 

= (in.pr.)Zj5r 

y 

Low-pressure cylinder release pressure =P/=P,.:j:3r. 

yf 

1 



(a) 
(6) 



(309) 



(rel.pr.)i, = (in.pr.) 



RcRh 



Zh 

Re 



= (in.pr.) 

_ (in.pr.) 
Rv 

_ (rel.pr.)ir 
"ftc" 



(o) 



(b) 



(c) 



(d) 



' • 



(310) 



WORK OF PISTON ENGINES 261 

Division of work between cylinders may be made anything for a given load by 
suitably; proportioning cylinders, and equations giving the necessary relations 
to be fulfilled can be set down. It is quite common for designers to fix on 
equal division of work for the most commonly recurring or average load or that 
corresponding to some high pressure cut-off or low-pressure terminal pressure^ 
generally the latter. Therefore, a general expression for dimensional rela- 
tions to be fulfilled for equal division of work is useful. On the other hand, 
for an engine the dimensions of which are determined, it is often necessary to 
find the work division for the imposed conditions, so that the following equa- 
tions are of value. 

Frwn Eqs. (296) and (297), noting that Pd=P«=P5^, 

'9 



Low-pressure cylinder work ^^^^^ _^^^^ 



High-pressure cylinder work _ 



('+'<* ft) - 

(■+■-«• rrr) 



P»7i 



1+log. Rb~ 
\m.pr. / 



. . . . (311) 



This is a general expression for work division between the cylinders in 
terms of (a) ratio of expansion in each cylinder, initial and back-pressure 
ratio and cylinder ratio, or, in terms of (b) cut-off in each, associated with cylinder 
and pressure ratios. 

This expression Eq. (311) is less frequently used in its general form as above, 
than in special forms in which the work of the two cylinders in made equal, 
or the expression made equal to imity. The conditions thus found for equal 
division of work between cylinders may be expressed either (a) in terms of initial 
and back pressures, release pressm*e of low-pressure cylinder and ratio of L.P. 
admission volume to H.P. displacement, and cylinder ratio, or (b) cut-off in 
high- and low-pressure cylinders, initial and back pressures and cylinder ratio. 
Still more special conditions giving equality of work may be found (c) when the 



262 ENGINEERING THERMODYNAMICS 

cylinder ratio is made such that equality of work is obtained at all loads^ by 
equalizing high and low cut-offs. 

(a) To find the first set of conditions, equate Eqs. (296) and (297) from the 
first part of this section, and by simplification there results, 



or 






•• F.F/ ^ 

Introducing the usual symbols and putting in addition 

Low-pressure admission volume _Ve_ _j p _ (rel.pr.)g . 
High-pressure displacement volume ~Fe ~ ^ <^~ (rec.pr.) ' 

Therefore, 

r(bk.pr.) 1^-1 

rL(rerpr.),-,-J (inj,r) J» 

L (rel.pr.)i, J 

This is of value when a given release pressure is to be reached in the low 
pressure cylinder and with a particular value of low-pressure cut-off volume 
as fixed by x in terms of high-pressure cylmder displacement. 

(b) Again for equal division of work, make Eq. (311) equal to unity, 
whence, 

1.1 1 1 1.1 1 (bk.pr.) Re 

or 

|^'j^'^+ifclog.|-^-^=0, 
Lh (m.pr.) Lh ItL 

which may be reduced to the following, solving for Rc^ 

^ "2 lbk.pr .) • • ' • ^^^^^ 

Zif(in.pr.) 

Equal division of work for given initial and back pressure is to be obtained 
by satisfying these complex relations Eqt (313) between the two cut-offs, or their 
equivalent ratios of expansion in connection with a given cylinder ratio, or the 
relation between pressures and volumes in Eq. (312) equally complex. 



WORK OF PISTON ENGINES 



263 



(c) An assumption of eqiuil cut-off in both cylinders gives results which are 
of interest and practical value, although it is a special case. Eq. (313) then 
becon^es, when Zh^Zl or Rh—Rl, 



Ri 



_ / in.pr. Y^ 
\bk.pr./ 



(314) 



1 

X 



As would be expected, this may also be derived from Ea. (312), since 

7 / \ and x=-,rT^—T under these conditions. 
(rel.pr.)L (bk.pr.) 

The receiver pressure imder these conditions is constant, and is, from Eq. (308) 

, . (in.pr.) (in.pr.) f.. .... .1. .^.^. 

(rec.prO=^^--^=y^;£y^ . . .(315) 

\bk.pr./ 



The high-pressure release pressure is not affected by any change in the low- 
pressure cut-ofif, and hence Eq. (309) gives the value of high-pressure release, 
pressure for the case. Low-pressure release pressure Eq. (310) may be expressed 
for the case of equal cut-ofF, 



(tel.pr.);-= fepirff = ^*[(ii^-P') (bk.pr.)]* 
\bk.pr./ 



(a) 



=^[(in-pr.)(bk.pr.)J (6) . 



(316) 



The foregoing equations up to and including Eq. (311), are perfectly general, 
and take special forms for special conditions, the most important of which is 
that of complete expansion in both cylinders^ the equations of condition for which 
are, referring to Fig. 82. 

Pc — Pd't 

which, when fulfilled, yield the diagram. Fig. 83. These equations of condi- 
tion are equivalent to fixing a relation between the cut-off in both the high- 
and low-pressure cylinders, and the volume of high-pressure cylinder with respect 
to the low-pressure volume, so that 



p 
ytf—Vf^y or symbolically, 

«-i(iS:) <" 



(317) 



264 



ENGINEERING THERMODYNAMICS 



Similarly the low-pressure cylinder cut-ofif volume must equal the high- 
pressure displacement volume or Dh^ZiJ)l, 



Z), 






"- Di. Re 



(a) 
(b)\ 



(318) 



indicating that low-pressiire cut-off is the reciprocal of the cylinder ratio. Making 
the necessary substitution there result the following equations for this cyde 
which, it must be noted, is that for most economical use of fluid in compound 

























A 




B 








-<-• (ii 


i.Dr.) 










[ 










•H**/ 








^z, 


iDh-* 


\ 






















\ 






















\ 


V 
















N 






\ 


c 





<r 


5c.pr.) 










D„ = 


T Tk- 




V 












ZlI>l 




\ 








1 
1 

i 

1 






X 


^ 










M 




1 

1 
1 

1 










"^ 






E(rel 






1 






r» 






















^L 




















1 












H 





Fig. 83. — Special Case of Cycles V and VI, Complete Expansion in both Cylinders of 
Compound Engine with Infinite Receiver and Zero Clearance. 

cylinders without clearance and with infinite receiver, and in which the same 
work is done as in Cycle I, for simple engines at best cut-oflF. 
From Eq. (308) 

R, _(in.pr.) Re ^^^(bk.pr.). . . (319) 



(recpr.) = (in.pr.)^^ = -^ 1 (in.pr.) 



Re (bk.pr.) 



From Eq. (309), 



(rel.pr Oh = ^-""^^-^ = (^^-P^') -^^-J^^ = RcQ^k.pv,) = (in.pr .)i. . (320) 



WORK OF PISTON ENGINES 



265 



From Eq. (310), 



(rel.pr.)x,--^^-^ 1 (in.pr.) "(bk-pr.) 



(321) 



Re (bk.pr.) 



Frran Eq. (311), 
High pressure cylinder work _ 



H-log.«/r-|^ 

lie 



Low pressure cylinder work - , , ^ (bk.pr.) •, •, 

1 +log, Rl - ;. T. / flcflg 

(in.pr.; 



log, fix, 



(a) 



loge 



1 



z 



J7 



lo& 



(fc) 



(322) 



Z 



For the case of most economical operation, that of complete and perfect 
expansion in both cylinders, there may be set down the four characteristic 
Ek|s. (304), (305), (306), (307) with suitable modifications to meet the case. 
These become 



Work of cycle = 144(in.pr.)Di,- 



/ in.pr. \ 
^ Vbk.pr./ _ 
/ in.pr. \ 
\bk.pr./ 



= 144(in.pr.)Dz, 



loggfii 
Rv 



(323) 



(324) 



1 / ^P-pr- \ 

/ \ f t i. T ■D\ ^ r \ \bk.pr./ ,. xlogrfir 

(m.e.p.) (ref.toL.PO-i44^^=(m.prO-yj^p^ = (m.pr.)-^. . 

\bk.pr./ 
Work per cu.ft. supplied = 144(in.pr.) log, [ '^ ' ) = 144(in.pr.) loge Rv (325) 

13,750 / bk.pr. X 

r.P.)\in.pr./ ^^^ 



Cu.ft. per hr. per I.H.P.== 



(m.e.p. ref . to L, 

13,750 J_ 

(m.e.p.ref. to L.P.) Rv 



(b) 



. a 



(226) 



For equal division of work with complete expansion in both cylinders, the 
ratios of Eqs. (317) and (318) becomes 






(327) 



and this is evidently a case to which Eqs. (314) and (315) apply without change. 



266 ENGINEERING THERMODYNAMICS 

Szample. 1. Method of calculating diagram, Fig. 82. 
Afisumed data for Case A: 

Pa=Pb- 100 lbs. per sq-in. abs. Fa « F» = Fm = cu.ft. 

Pn-Pd^Pe- 50 lbs. per sqin. abs. Vc-Vd^ .6 cu.ft. 

Pm^Pg- 10 lbs. per sq.in. abs. Vf='Vg^ 2 cu.ft. 

Pc = 60 lbs. per sq.in. abs. Vg = .8 cu.ft. 



To obtain point B: 



To obtain point F: 



P 60 

n - 7, x^ = .6 X-TTi^ = .36 cu.ft. 



V 8 

Pf--Pg X^ =50 Xiz =20 lbs. per sq.in. 
Vf z 



To construct the indicator cards: 

Lay off ND of the PY diagrams to equal the length of the card, and t^A peq)en- 
dicular to it at "N to equal the height of the card. Cut off equals AB-^ND, From .4 
on card lay off this ratio times the length of the card. From D on the card lay ofif 
a perpendicular equal to CD of the PV diagram reduced by the same proportion 
as AN of the card is to AN of the diagram. Join the points B and C by a curve 
through points located from intermediate points on the PV diagram. The low- 
pressure card is constructed in same manner. 

Example. 2. A 12- and 18 x24-in. steam engine without clearance run8» on 150 
lbs. per square inch absolute initial pressure, 10 lbs. per square inch absolute back pressure, 
and has a speed of 125 R.P.M. What will be (o) the horse-power for | cut-off in H.P. 
cylinder, (b) poimds of steam per I. H.P. hour, (c) terminal pressures, (d) L.P. cut-oflf 
for continuous expansion^ (e) work done in each cylinder. 

Nons: 8 for 150 lbs. =.332. 

(a) From Eq. (305) 

(m.e.p.) referred to L.P. cylinder is 

(in.pr.) ( 2 -floge fl/f +log« fli - ^ ) - (bk.pr.) . 

iChiCc\ iCc/ 

In this case 

Rh-^2, /ec = (J|y=2.25, «x.=2.25, 

since vol. of L.P. cyl. at cut-off must be equal to the entire volume of the, high for 
continuous expansion, hence 

(m.e.p.)=150Xr— r-T3X(2+.69+.81-l)-10=73.3 lb. sq. inch, 

and 

(m.e.p.) Lan 

^•"•^•- 33,000 ^^^• 
(6) From Eq. (307) 

n ,* , -„-, 13,750 Zh 13,750 .5 ,, _ 

Cu.ft. per hour per LH.P.^^^;^;^ ______ x^, 41. 7, 



WORK OF PISTON ENGINES 267 

(c) Emm Eq. (308) 

(rel.pr.)«^ = (m.pr.)ZH, 

-. •«■■■■ ■■>■• 

= 150X§-751bs.Bqin, 
and from Eq. (310) we have • 

(rel.pr.)i, = — ^ , 

75 
= r-rr « 33.3 lbs. sq.in. 
^.25 



(e) From Eq. (311) 



^''"2.25""^V 



l+lo&^ ^ 



H.P. work ^Zh RcZl 



L.P. work , , , 1 (bk.pr.) Re ' 
l+loge 



Zl (in.pr.) Zh 



1+.69 



2.25 X. 44 .69 ,^^ 

.450, 



1+.81-Jix2-^ ^-^^ 



150 .5 

or 

H.P,work =.456XL.P.work, 

also 

H.P. woik +L.P. work =282 I.H.P. 

Hence 

H.P. work =88 I.H.P. 
and 

L.P. work = 194 I.H.P. 

■ 

ProtK 1. What must be the cylinder diametens of a cross compound engine to run 
on 100 lbs. per wjuare inch absolute steam pressure, 18 ins. of mercury vacuum and to 
develop 150 H.P. at a speed of 200 R.P.M. with J cut-off in each cyhnder, if cylinder 
ratio is 3 and stroke is 18 ins.? Engine is double-acting and assumed to have no 
clearance. 

Prob. 2. What will be the release pressure in each cylinder and the receiver 
pressure of the engine of Prob. 1? If cut-off were reduced to J in H.P. cyhnder, 
how would these pressures be affected and to what extent? How would the horse- 
p>ower change? 

Prob. 3. A 15- and 22 x30-in. infinite receiver engine has no clearance, a speed of 
loO R.P.M., initial pressure 125 lbs. per square inch gage. What will be the horse- 
power and steam consumption for a H.P. cut-off of }, i, }, }, and that value which 
will give complete expansion in high-pressure cylinder? Low-pressure cut-off to be 
fixed at |. ' 

Note: 6 for 150 lbs. gage = .363. 



268 ENGINEERING THERMODYNAMICS 

Ptob. 4. What will be the release and receiver pressures, and the woik done in each 
cylinder for Prob. 3? 

Prob. 6. An 18 and 24x30-in. infinite receiver engine is to be operated so as 
to give complete expansion in both cylinders. What will be the cut-off to acoomplish 
this and what horse-power will result if the initial pressure is 100 lbs. and back pressure 
10 lbs. per square inch absolute? 

Prob. 6. Draw the PV diagram for following cases. Cylinder ratio 1 to 2.5, 
(in.pr.), 100 lbs. per square inch absolute, (bk.pr.), 20 lbs. per square inch absolute, 
H.P. cut-off (a)-i, (6) -J, (c)=l. L.P. cut-off (a) -I, (6) -A, (c)=f. 

Prob. 7. For the following conditions find the horse-power, steam used per hour, 
receiver pressure and release pressures. Engine, 10- and 15x24-in. 150 R.P.M., 
125 lbs. per square inch gage initial pressure, 2 lbs. per square inch absolute, back 
pressure, i cut-off in high-pressure cylinder, A cut-in low-pressure cylinder with 
infinite receiver. 

Note : B for 125 lbs. = .31 1 . 

Prob. 8. An infinite receiver engine is to develop 150 H.P. at 200 R.P.M. when 
initial pressure is 150 lbs. per square inch absolute. Cylinder ratio is 1 to 3 and 
back pressure is one atmosphere. What must be its size if the stroke is equal to 
the low-pressiue cylinder diameter for i cut-off in the high-pressure cylinder and 
J cut-off in the low-pressure cylinder? 

Prob. 9. Find by trial the cut-offs at which work division will be equal for an 
infinite receiver engine with a cylinder ratio of 2.5, an initial pressure of 100 lbs. per 
square inch absolute and a back pressure of 5 lbs. per square inch absolute? 

10. Compound Aigine with Infinite Receiveri B^Kmential Law. No 
Clearance, Cycle VI. General Relations between Pressuresi Dunendonsy 
and Work. Again referring to Fig. 82, which may be used to represent this 
cycle also, the work of each cylinder may be expressed as follows, by the 
assistance of Eq. (254) derived in Section 4. 



Fi = 144Z)j,rZ^(rec.pr.)(?^^|^')-(bk.pr.)L • - . 



(328) 



(329) 



Vh 
where Zh is the cut-off in the high pressure, = v^ and Zl, low-pressure cut-off 

y 
= :=^. In combining these into a single equation for the total work, the term 

for receiver pressure (rec.pr.) should be eliminated. Referring to Fig. 82, 



hence 



(rec.pr.) =P*=P.=P»(^;)'^=(in.pr.)(^y, . . . 



(330) 



r..««J,..p,.>r(|)(i^)-X(|-)- 



+^'(^)"C-i^)]-*''-'"'l' • <'"» 



WORK OF PISTON ENGINES 269 

a rather complex expression which permits of little simplification, but offers 
no particular difficulty in solution. 

Mean effective pressure referred to the low-pressure cylinder is 

(m..p. rrf. to l.P.,.a..pr.)[(|)(i:^)-i(^j- 

Work per cubic feet fluid supplied may be foimd by dividing Eq. (331) by 
the supply volume, which in terms of low-pressure displacement is 

(Sup.Vol.)=Z>i|^ (333) 

The consumption of fluid, cubic feet per hoiu* per indicated horse-power is 

Consimiption cu.ft. per hr. per I.H.P.=7 \ . t t> \ t^i • • • (334) 

'^ x- x- (m.e.p. ref . to L.P.) Re 

which is the same expression as for the logarithmic law. Multiplying this by 
li, the initial density of the fluid, pounds per cubic feet, gives cansumptUm, 
pounds i)er fluid hour per I.H.P. 

The receiver pressure has already been determined in Eq. (330). 

Release pressure of the high-pressure cylinder is 

(rel.pr.)^=(in.pr.)Z^*, (335) 

and for the low-pressure cylinder, 

(rel.pr.)L = (in.pr.) \^\ (a) 
_ (in.pr.) 



fi • 



(ft) 



(336) 



where Ry is the ratio of maximum volume in the low pressure, to, volume at 

R 
cut-off in the high, and equals -^. 

The distrubition of work between the high- and low-pressiu*e cylinders may 
be found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by 
means of Eq. (330) 

l s-Zn^-^ \ (Zh\ 
L \RcZl/ \ s-l J \m.pr. /J 



270 



ENGINEERING THERMODYNAMICS 



Equality of work in the two cylinders will be obtained if this expression 
is equal to unity, giving a complex relation between high- and low-pressurt 
cut-offS; cylinder ratio and ratio of initial and back pressures, to be satisfied. 
It is found at once that the simple conditions for equality in the case of logarith- 
mic law will not give equality of work for the exponential law. There is, how- 
ever, a case under this law which yields itself to analysis, that of complete expan- 
sion in both cylinders, without over-expansion. The conditions for equality 
of work for this case will be treated after deriving work and mean effective 
pressure for it. 

Complete expansion, without over-expansion, in lx>th cylinders may be 
represented by Fig. 83. 



„ AB „ NC 

NC MEf 



and since 



' NC=Dh and ME=Di., 



^ D„ NC Zl. 



ME 1 72/^ 
The true ratio of expansion =jBr== =--—=—-, but this is also equal to 

AB ^H^L liH 

1 



/ in.pr. \ « 
\bk.pr./ 



due to the law of the curve, PbVb—PtVe. 



By means of Eq. (257) in Section (4) the work of the two cylinders may be 
evaluated, 



but since 



Wh = 144(m.pr.)Di,Z,r^ (l - Z«« '^\ 
«I44(m.pr.)Z).|^4^(l-Z..-i) 

Tr. = 144(bk.pr.)D.^4j(^- 1), 



ia) 



(P) 



Zl^ 



8 






Wl = 144(bk.pr.)i)L ~{R<f'^-l) 



(a) 



= lU(m.pT.)DLj^ (^) \Rc-'-1) (6) 



(338) 



.... (339) 



WORK OF PISTON ENGINES 



271 



The total work is evidently the same as that of a cylinder equal in size to 

the low-pressure cylinder with a cut-ofif equal to ^, working between the 

tic 

given (in.pr.) and (bk.pr.) and may be stated by reference to Eq. (257), 
Section 4, or by taking the sum of Wh and Wl given above, 

TF=144(in.prOI)L|^^ J (l-Zir'-0 + (^y" W 



which reduces to 



TF = 144(in.pr.)Dx, 



Zh s 



Rcs-l 



!'-(in 



(340) 



For this case of complete expansion in both cylinders, the ratio of high- to 
low-pressure work is given by division and cancellation, 



Wl 



1-Zb-' 



w. (1).-. 



(/ic-^-1) 






(341) 



Ek}uality of work, obtained by placing this expression equal to unity, pro- 
vides the condition that 



^--' = (l)'"'<HSr)'-'. 



or 



«-i 



Rc^ 



\bk.pr./ 



+1 



1 



(342) 



for equal work and complete expansion, and 



«-i 



1^ 
p _ / bk.pr A • 

'^"■\in.pr. /; 



' / bk.pr A « 

XVC— 7^ 



+ 1 



1 
• -1 



(343) 



Since Zl—i^- for complete expansion,Tand (in Fig. 83) PcVc—PfVf\ the 
receiver pressure, P^ is 

(rec.pr.) = (bk.pr.) (^'y=(bk.pr.)/?c', (344) 

in which Re will have the value given above if work is equally distributed. 

Example 1. What will be (a) the horse-power, (6) consumption, (c) work ratio, 
{d) receiver and release pressures for the following conditions? Engine 12 and 
18x24 ins., running at 125 R.P.M. on initial air pressure of 150 lbs. per square 
inch absolute, and back pressure of 10 lbs. per square inch absolute, with \ cut-ofiF 
in high-pressure cylinder and continuous expansion in low-pressure cylinder. Exponent 
of expansion curve;«1.4 for compressed air, infinite receiver. 



272 



ENGINEERING THERMODYNAMICS 



(a) From Eq. (332) 
(m.e.p.) -(in.pr.)[(g) ('-^-^) -i(^)' 



H-iT)'irf^)] -a>--) 



which, on substituting values from above, gives for (m.e.p.) 63 lbs. per sq. inch. 
Hence, the indicated horse-power "242. 
(6) From Eq. (334) 

Compressed air per hour per I.H.P. = — - — -^g- cu.ft., 

m.e.p. tic 



which, on substitution, gives 



13,750 .5 ,„^ ^^ 



(c) From Eq. (337) 









which gives 



Wl 



•<->(=?.) 



1.4 



2.25 



1 



.5 



1.4/1 



-G-ii)' 



.294. 



10 



2.251 ^ «, 1 
2.25 X 



.4 



150 



2.21 



and 



Hence 



and 



TrH+Trz.-242I.H.P. 



Trir=56 I.H.P. 



Wl = 184 I.H.P. 



(d) From Eq. (330) 

(rec.pr.) =(in.pr.) (^f^j 



= 150i 



.5 



1.4 



2.25 X 



_1_ 
2.25. 



=57 lbs. per sq.in. 



From Eq. (335) 



(rel.pr.)^ = (in.pr.)Zi5r*, 



= 150 X (.5)^ * -57 lbs. per sq.in. 



WORK OF PISTON ENGINES 273 

Fiom £q. (336) 

(rel.pr.)i. = (in.pr.) -^R'vi 

= 150 -i-21.85 -6.85 lbs. per sq.m. 

These values may be compared with those of Ex. 1, Section 9, which were for the 
same data with logarithmic expansion. 

Prob. 1. What will be the horse-power and steam used per hour by the follow- 
ing engine imder the conditions given? Cyhnders 18 and 30x48 ins., speed 100 
R.P.M.y initial pressure 150 lbs. per square inch absolute, back-pressure 10 lbs. per 
square inch absolute, steam continually dry. Cut-off at first } in high-pressure and 
^ in low, and then i in each infinite receiver. 

Prob. 2. The very large receiver of a compound pumping engine is fitted with safety 
valve which is to be set to blow at 25 per cent above ordinary pressure. The cylinder 
ratio is 1 to 3.5, and cut-offs are i in high and ^ in low. If initial pressure is 125 
lbs. per square inch gage, for what must valve be set? What vacuum must be carried 
in the condenser to have complete expansion in low-pressure cylinder? Superheated 
steam. 

Prob. 3. A compoimd engine is to be designed to work on superheated steam of 
125 lbs. per square inch absolute, initial pressure and on an 18-inch vacuum. The 
load which it is to carry is 150 horse-power and piston speed is to be 500 ft. per 
minute at 200 R.P.M. Load is to be equally divided between cylinders and there is 
to be complete expansion in both cylinders. What must be cylinder sizes, and 
what cut-offs will be used for an infinite receiver? 

Prob. 4. How will the economy of the two following engines compare? Each is 
14 and 20x24 ins., runs at 200 R.P.M., on compressed air of 100 lbs. per square 
inch gage pressure, and 15 lbs. per square inch absolute exhaust pressure. Low-pres- 
sure cut-off of each is } and high pressure of one is i, the other, i. Infinite 
receivers. 

Prob. 6. A compound engine 12 and 18 X24 ins. is running at 200 R.P.M. on 
superheated steam of 100 lbs. per square inch absolute pressxue and exhausting to a 
condenser in which pressure is 10 lbs. per square inch absolute. The cut-off is i in 
high-pressure cylinder and i in low-pressure cylinder. Compare the power and steam 
consumption under this condition with corresponding values for wet steam imder 
same conditions of pressure and cut-off and infinite receivers. 

Prob. 6. The initial pressure of an engine is 150 lbs. per square inch absolute, the 
back pressure one atmosphere, the cylinder ratio 3. As operated, both cut-offs are at i. 
What will be the receiver pressure, high-pressure release pressure, and low-pressure 
release pressure? What will be the new values of each if (a) high-pressure cut-off is 
made f, (6) i, without change of an3rthing else, (c) low pressure cut-off is made i, 
(d) f, without change of anything else? Infinite receiver, s = 1.3. 

Prob. 7. In the above problem for i cut-off in each cylinder how will the release and 
receiver pressures change if (a) initial pressure is raised 25 per cent, (6) lowered 25 
per cent, (c) back pressure raised 25 per cent, (d) lowered 25 per cent? 

Prob. 8. How many pounds of initially dry steam per hour will be required to 
supply an 18-in. and 24x30-in. engine running at i cut-off in each cylinder if speed 



274 



ENGINEERING THERMODYNAMICS 



be 200 R.P.M., initial pressure 100 lbs. per square inch gage and back preasore 5 
lbs. per square inch absolute? Expansion to be adiabatic and receiver infinite. 
Note: d for 100 lbs. =.26. 



-fiup. Vol. 




OVER EXPANSION 

Fig. 84. — Work of Expansive Fluid in Compound Engines with Finite Receiver, Zero Clear- 
ance. Cycle VII, Logarithmic Expansion; Cycle VIII, Exponential Expansion. 

11. Compound Engine with Finite Receiver. Logarithmic Law. No 
Clearance, Cycle Vn. General Relations between Dimensdons and Work 
when H.P. Exhaust and L.P. Admission are not Coincident. The diagrams, 
Fig. 84, while showing only two degrees of expansion, that of over and imder 
in both cylinders, suffice for the derivation of equations applicable to all 
degrees in either cylinder. Voliunes measured from the axis AL are those 



WORK OF PISTON ENGINES 275 

occiipied by the fluid in either cylinder alone, while fluid voliunes entirely in the 
receiver, or partly in receiver, and in either cylinder at the same time are meas- 
ured from the axis A'V , No confusion will result if all volumes represented by 
points be designated by the (F) with a subscript, and to these a constant rep- 
resenting the receiver volume be added when part of the fluid is in the receiver. 
Then, 

High-pressure cylinder work is 

Fi,=P^F.(l+lo&pj-P„01oge(^^) (345) 

Low-pressure cylinder work is 

Pr^ = PnO log, (^^^)+PeFel0ge^-P, 7,. . . . (346) 

Total work 

Tr = P,7»(l+l0ge^)+Pn0 l0g-(^^) 

+PJ%log.^;-P„01og.(^^)-P,7,. . (347) 

These expressions include some terms not known as initial data and may 
be reduced by the following relations, 

and 



Fn0^p,{y,-\-O) =p,v,(^^}j 



Hence 



»-P.v.[>+,o.i:.(MO),c.(K^) 

Dividing by the low-pressure cylinder displacement, Vg, the result will be 
the mean eflfective pressure referred to the low-pressure cylinder, 

(M.E.P. ref. to L.P.) 

-4:['+'-f:+m'-(^°)+-f: 



276 ENGINEERING THERMODYNAMICS 

A similar division but with the volume supplied, V^ as the divisor, gives 
Work per cu.ft. supplied 

-m'-m]-4:-----<»«' 

Also as in previous cases 
Cu.ft. supplied per hr. per I.H.P. 

- 13'^«^ .X& (351) 



(m.e.p. ref. to L.P.) Vg 



Of course, the weight per hour per I.H.P. follows from Eq. (361) by introduc- 
ing the density as a multiplier. 

While the last four equations can be used for the solution of problems, it is 
much better to transform them by introducing dimensional relations as in the 
previous cases developed. 

Let (rec.pr.)i= maximum receiver pressure P», which is also the initial admis- 
sion pressure for the low-pressure cylinder; 
(rec.pr.) 2= minimum receiver pressiu^ Pe, which is the terminal admission 

pressure for the low-pressure cylinder and that at which 
expansion begins there; 



<( 



_ receiver volume _2. — ^ a Q ^ ^bV _ y 

^""high-pressure cyl. displ. Vc Dh ' Dl", Dl " Re 



Other symbols necessary are unchanged from the meaning imposed in Section (9). 
Substitution in Eqs. (348), (349), (350), and (351) gives the following set 
in a form for direct substitution of ordinary data: 

Work of cycle 

= 144(in.pr.)Z«Z),, 1 1+log. ^+log. ^ 

+ (^+zfe)h 0+-f )-^^^(^+^)] } -l^(bk.pr.)i>. (a) 
= 144(in.pr.)|^ j 1+loge /2/.+log. Rl+ (i+^) [log. (l+^) 



-loge (l+-)l ) -144(bk.pr.)i5, 



(6) 



. (35) 



WORK OF PISTON ENGINES 



277 



(m.e.p. ref. to L.P.) 

= (in-pr .)^ ( 1 +log. ^+loge ^4 

•tic [ iiH ^L 






-log.(l+^)]j-(bk.pr.) 



(6) 



(353) 



Work per cu.ft. supplied 

- 144(m.pr.) j 1+log, ^+log, ^ 



- 144(m.pr.) 1 1 +log. ie^+log. i2^+ (l +^) [log. (l +^) 

-log. (l +^)] j - 144(bk.pr.)i?cii:i/ (6) 



(354) 



Cu.ft. supplied per hr. per I.H.P. 

13,750 



X 



(m.e.p. ref. to L.P.) Re 



13,750 



X 



(m.e.p. ref. to L.P.) RhRc 



(a) 



% 



(355) 



It is desirable at this point to introduce a series of expressions fixing the 
relations between the dimensions, the cycle that may follow, and the fluctua- 
tions in the receiver pressure, and for the selection of cylinder and receiver 
dimensions for a required output of work and division of it between cylinders. 

In doing this it will be convenient to start with diagram points and finally 
substitute general symbols in each case. There will first be established the 
nummum and minimum receiver pressures and the fiuduations. 

Maximum receiver pressure 






= (in.pr.) ^^ 



■1 



A 



^ir2/ RffRc 



) 



) 



(b) 



(356) 



278 



ENGINEERING THERMODYNAMICS 



Minimum receiver pressure 



P =Pk— • 



/. (rec.pr.)2 = (m.pr.) Yj)^ "" ^^'^^^Y'jMc ^ 



= (m.pr.)5-p- 



ib) 



• ■ 



(357) 



Fluctuation in receiver] pressure =(P»— Pe)=ft-7r. 

.*- (rec.pr.) i — (rec.pr.)2 = (m.pr.) — ;^— = (m.pr.) — (a) 







y 



= (in.pr.)^-- (5) 



Rky 



(358) 



It is interesting to note that the rninimum receiver pressure is exactly the same 
as the value of the constant-receiver pressure for infinite receiver y so that limit- 
ing the size of receiver does not affect the point E, but only raises point N higher, 
tending to throw more work on the L.P. cylinder for the same valve setting. 

The two release pressures Pc and P/ can be evaluated as in the case of the 
infinite receiver, as both these points lie on the common expansion line, which 
is not at all affected by the receiver-pressure changes, and the values are the 
same as for the infinite receiver, and are here reproduced from Eqs. (309) and 
(310) with new numbers to make the set of equations complete: 



(rel. pr.)^ = (in.pr.) 



(a) 



Rh 

= (in.pr.)Z« (6) ^ 



(369) 



(rel.pr.)L = (in.pr.) 



RcRh 



= (m.pr.)^ 
(in.pr.) 



Rv 

^ (rel.pr.) ff 
Re 



(a) 



(h) 



(c) 



(d) 



(360) 



where Rv is the ratio of maximimi volume in the low- to the volume at cut-off 
in the high-pressure cylinder. 

Division of work between the cylinders cannot, as pointed out, be the same 
as for the infinite receiver, the tendency being to throw more work on the low 
as the receiver becomes smaller, assuming the cut-off to remain the same. As, 
therefore, equal division was obtainable in the case of infinite receiver with 



WORK OF PISTON ENGINES 279 

equal cut-offs when the cylinder ratio was equal to the square root of initial 
over back pressure, it is evident that a finite receiver will require unequal cut- 
offs. As increase of low-pressure admission period or cut-off fraction lowers 
the receiver pressure and reduces the low-pressiu*e work, it follows that with 
the finite receiver the low-pressure cut-off must be greater than the high for 
equal work division, and it is interesting to examine by analysis the ratio 
between them to determine if it should be constant or variable. 

For equal toork division Eqs. (345) and (346) should be equal, hence by 
diagram points 

P,V,{l+log,^^-PnO\ogs (^-)=P„0 loge (^^^)+P.F.l0g.y^-P,7,. 

-p. y'{^yf) lofc (^) +f .n log, y-P.V.- 
hence for equal division of work, the following relations must be satisfied: 

_/bk^.\^ ^^ 

\m.pr. / ^ ' 

It will be shown later that when expansion is complete in both cylinders 
and work equal that the high-pressure cylinder cut-off or the equivalent ratio 
of expansion bears a constant relation to that of the low, according to 

^=o^ (362) 

in which a is a constant depending only on the size of the receiver. It will 
also be shown that the cylinder ratio is a constant function of the initial and 
back pressures and the receiver volume for equal division of work, according to 

•^-Km^y (»') 

in which (a) is the same constant as in Eq. (362). It is impo tant to know if 
these same values will also give equal division for this general case. Substi- 
tuting them in Eq. (361) 



21og.a=/H 







aRi 



/in.pr. 
\bk.pr. 



1. 



2S0 



ENGINEERING THERMODYNAMICS 



Here there is only one variable, R^y the evaluation of which can be made by 
inspection, for if 

Rl 



_1 / in.pr. 
a\bk.pr/ 
the equation will become 

21ogea = (l+y) loge (l+^)'-2 = 2(l+y) loge(l+i)-2, ' 



or 



^^^[(i+i/)io«.(i+M-ij, 



(3W) 



which 18 a constant. 



LOO 








1 


II 


/ 


/ 




/ 


/ 




■■■ 


/ 








71 










1 U JL 


J 




/ 


J 


/ 




/ 








7 








4 






'?/ ^ 


f/ 


1 


\ 


<0 


^J 


7 




^ 


/ 




b> 


f 










7 




iff 




7 


7 


r 


> 


7^ 












.80 












' 


f 


J 


J 


f 




/ 






/ 


















: 1 




1 




/ 


/ 




/ 






7 






















1 


f 


/ 


/ 




/ 






7 












MJ aik 






/ 


/// 


\i 




/> 


/ 


/ 


f 




/ 
















|.eo 






/ 


// 




T 


/ 


/ 


i 


/ 




7 


















1 




////' > 


/ 


V 


r 


/ 




7 




















I 




/// 


/ 


/ 


/ 


7 




/ 
























^ / 


/ 


/ 


/ 




7 


























///// 


/ 


r 


/ 


/ 








RECEIVER VOLUME EQUALS 
y X H.P. DISPLACEMENT 










m^ 


/. 


/ 


/ 
































m 


/ 


/ 


/ 
































jw/ 


^ 


A 


/ 
































jao 


IwlL 


V 


/ 








« 




























\m/£ 


7 






































\m 








































E 









































.». 



.40 .eo 

Hiffh Brenore Gut Oft 



«80 



.100 



Fig. 85.- 



-Diagram to Show Relation of High- and Low-Pressure Cut-offs for Equal Work in 
the Two Cylinders of a Finite-receiver Compound Engine with Zero Clearance 
and Logarithmic Law. 



As only one constant value of low-pressiu-e ratio of expansion or cut-off 
satisfies the equation for equal division of work when there is a fixed ratio 
between the values for high and low, that necessary for equal division with 
complete expansion in both, it is evident that equal division of work between 
the two cylinders cannot be maintained at all values of cut-ofif by fixing the 
ratio between them. As the relation between these cut-offs is a matter of some 
interest and as it cannot be derived by a solution of the general equation it is 
given by the curve, Fig. 85, to scale, the points of which were calculated. 



WORK OF PISTON ENGINES 



281 



A special case of this cycle of sui&cient importance to warrant derivation 
of equations because of the simplicity of their form and consequent value 
in estimating when exact solutions of a particular problem are impossible, 
18 the caee of complete and perfect expansion in both cylinders. For it the 
following equations of condition hold, referring to Fig. 84, 




Fig. 86. — Special Case of Cycles VII and VIII Ck>mplete Expansion, in both Cylinders of the 

Finite Receiver Compound Engine. Zero Clearance. 

which when fulfilled yield the diagram. Fig. 86. These equations of conditions 
are equivalent to fixing the cut-off in both high- and low-pressure cylinders, 
and the volume of the high- with respect to the low-pressure volume. Accordingly, 



Vt=Ve 



Pe 



^-"^^M w 



/2j5r = 



Re Vbk.pr./ Re 



{b) 



. . . . (365) 



Also for the low-pressure cylinder the cut-off volume must equal the whole 
ligh-pressure volume, or Db'=ZiJ)l- Therefore, 



Zl=^ (a) 

IhC 

Rt,^Rc {b) 



(366) 



282 ENGINEEMNQ THERMODYNAMICS 

Substituting these equations of condition in the characteristic set Eqs. (352), 
(353), (354), and (355), there results the following for most economical operation: 

Work of cycle 

r=.44(in.pr.)/fc(g^>-{l+U*[(2^.)i]+.*R, 

+a+y) log. ^1+i-) -log,(l+M I -144(bk.pr.)Z>i; 



= 144(bk.pr.)i)Jog.(^) (o) 

1 / 'P-pr- \ 
= 144(in.pr.)Z)x ^^^ = 144(m.pr.)D^ ^^" (6) 

\bk.pr./ 



(367) 



W 

(m.e.p. ref. to L.P.) = 144: 



144i)i 



/ui \i /in.pr. \ ,. . ^Vbk.pr./ ,. .log^Rv 
= (bk.pr.) log. (^j = (m.pr.) ^^^^ - (m.pr.) -|^ . 

\bk.pr./ 



(368j 



W 
Work per cu.ft. supplied =«-yr- 



= 144 

.pr 



, (bk.pr.)D L I /m.pr. \ ..... ., /m.pr. \ 

t \i| \ loge I VI I = 144(m.pr.) log, (rr-^— ) 

p / bk.pr . \ ^ '^\bk.pr./ ^ *^ ' ^Xbk.pr./ 

\in.pr. / ^ 

= 144(in.pr.) log, /Jr. . (369) 
Cu.ft. supplied per hr. per I.H.P. 

13,750 ^/bk.pr.\ 13,750 ^^ 1 

(m.e.p. ref. to L. P.) \in.pr. / (m.e.p. ref. to L. P.) Rv* 

For this special case of best economy the receiver and release pressures, 
of course, have special values obtained by substituting the equations of 
condition Eqs. (365) and (366), in Eqs. (356), (357), (358), (359), (360). 

(rec.pr.) , = (in.pr.) (^+^1-) = (m.pr.) (^+g) 

(rec.pr.)2 = (in.pr.) ^r^ = (in.pr .)^- = (bk.pr .)i2c. . . . (372) 



WORK OF PISTON ENGINES 283 

Therefore 

(rec.pr.)i — (rec.pr.)2 = (in.pr.)p — ; (373) 

(rel.pr.)H=(in-pr.)D"~0>'^-Pr-)J'2c=(rec.pr.)2; .... (374) 
(rel.pr.)x,=(in.pr.)p-n- = (bk.pr.) (375) 

•These last two expressions might have been set down at once, but are 
KTorked out as checks on the previous equations. 

For equal division of work in this special case the general Eq. (361) becomes 



)r 



log.g=2[(l+.)log.(l+i)-l]. 



Therefore 



r 

|H = ^<l+»)lo..(l+i-)-l] = o2 (376) 



This term, a, has already been used in previous discussions of equality of 
^orky while the derivation of its value has not been made up to this point. 

This indicates that ratio of cut-offs or individual ratios of expansion is a 
Hmction of the receiver size for equal division of work. 

From Eq. (376) the cylinder ratio can be found in terms of a, and the ratio 
)f expansion. Referring to Fig. 86, 



Rh V, Y? 



Rl Vf V,Vf 



=a2, 



F.-aVlvT/, 



^"^^'"""^/^aWl^WS^^aV^^^ • • • ^^^^^ 



v. aW, 



whence the q/linder ratio is equal to a constant depending on the receiver sizcy 
muUiplied by the valvs for the infinite receiver, i.e., the square root of the initial 
divided by back pressure. 



284 ENQINEEBINQ THERMODYNAMICS 

The high-pressure cylinder ratio of expansion is 

and the corresponding value for the low-pressure cylinder is 



. (378) 



iJ.=^'= 






For convenience in calculation Table, XII of values of a and o' is added for 
various size of receivers. 

Table XII. 



Receiver Vol. 
H.P.Cyl. Dwp."'' 


^^j;(i+if)iog,(i+i.)-i] 


o* 




.5 


1.016 


2.64 


.76 


1.624 


3.67 


1.0 


1.474 


2.17 


1.6 


1.322 


1.76 


2.0 


1.243 


1.66 


2.5 


1.198 


1.437 


3.0 


1.164 


1.369 


4.0 


1.1223 


1.262 


6.0 


1.0973 


1.204 


7.0 


1.0690 


1.143 


10.0 


1.0478 


1.008 


14.0 


1.0366 


1.068 


20.0 


1.0228 


1.046 


Infinite 


1.0 


1.0 



At the end of this chapter there is presented a chart which gives the relation 
between cylinder and receiver volumes, cylinder ratio, and high- and low-pressure 
cut-offs graphically. 

The corresponding values of maximum and minimum receiver pressure 
for equal division of work for this case of best economy are 

(»c.pr.),.(bk.p,.,l^(l+l)-^^&SSS^l+l) . (», 



/~«r.,^ -^K^«r^l /in.pr. V(in.pr.)(bk.pr.) 
(rec.pr.)2 = (bk.pr.)-^y^ 

(jec.pr.h-{rec.pr.h = ^^^^^^^^. . . 

ay 



(381) 



(382) 



WORK OF PISTON ENGINES 286 

Example 1. Method of calculating Diagram, Fig. 84. 
Assumed data for case A: 

Pa'^Pt" 120 lbs. per 8q.in. aba. 7. - V* - F« -0 cu.ft. 
Pm ^Pf « 10 lbs. per 8q.in. abs. Vb * .4 cu.ft. 

7i-l cu.ft. 

0-1.2cu.ft. 

7r- .8cu.ft. 



To find point C: 



To obtain point E: 



Pe "Pb-TT- « — 5-—- -60 lbs. per sq. inch. 



^ ^Vb 120 X. 4 ..„ . , 

Pb "PvtT - — :; — -48 lbs. per sq. inch* 
' • 1 



To obtain point D: 



Pe(7*+0)-Pd(7c+0) or Pd-^^^-53 lbs. per sq. inch. 



To obtain point N: 



48 v2 2 
P»(0+V»)-P.(F.+0) or P» -^^^jy^«88 lbs. per sq. inch. 



To obtain point F: 



P/-— Jr-^ - — - — -24 lbs. per sq. inch. 
Vf Z 



Example SL Find (a) the horse-power, (6) steam used per hour^ (c) the release 
and receiver pressures for a 12- and 18 x24-in. engine with receiver twice as large as 
the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-ofiFs \ in 
high-pressure and such a value in the low pressure as to give complete expansion, and 
[ learances zero. 

(a) Frtnn Eq. (353) 



(m.e.p.) «(in.pr.) 



9 

1 



RhRc 



l+log, Bff +log, «£ 



+ 



e^)h('-«^)-'-K)]}— ■ 



ffrhich on substituting the above values gives 



(m.e.p.) 



2x2.25 



l+.69+.81+(l+2-'^^'' 



2.25 



:«?(.8-.8) 



-10 -73.3 lbs. 



hence I.H.P. -282. 

(6) From Eq. (355) we have 

n e. ^ u u 13,750 1 13,760 1 

Cu.ft. steam per hour per ho^e-power - ^^^^^ Xj^ --^3- X^^^^^ -41.7, 



286 ENGINEERING THERMODYNAMICS 

(c) From Eqs. (356) and (357) for maximuin and minimnm receiver preflsures 
respectively: 



( 



^•P'-K^+^) *^^ ^^P'-^fe' 



maximum receiver 



pre88ure-150^— — +~-;25) "^^'^ ^^^' ^^ ^' ^^^ 



2.25 
minimmn receiver pressure = 150 X^--v^= 75 lbs. per sq. inch. 

From Eqs. (359) and (360) for release pressures 

[m.pr. )liM and , 

high pressure cylinder release pressure » 150 X. 5 « 75 lbs., per sq. inch. 

150 
low pressure cylinder release pressure —-777 «33.9 lbs. per sq. inch. 

.444 



These results may be compared with those of Example 1 of Sections 9 and 10, which 
are derived for same engine, with data to fit the special cycle described in the particular 
section. 

Note: In all the following problems clearance is to be neglected. 

Prob. 1. A 12- and 18 x24-in. engine has a receiver equal to 5 times the volume of 
the high-pressure cylinder. It is running on an initial pressure of 150 lbs. per square 
inch gage and exhausts to the atmosphere. It has a speed of 150 R.P.M. and the cut-o£^ 
are iV and \ in high- and low-pressure cyUnders respectively. What is the horse-power 
and the steam used in cubic feet per hour? 

Prob. 2. What will be the release pressures, and yariation of receiver pressure ia 
an engine in which the cylinder ratio is 3, cut-offs | and I, in high and low, initial pres- 
sure is 100 lbs. per square inch absolute, and receiver 2 times low-pressure cylinder 
volume? 

Prob. 3. Show whether or not the following engine will develope equal cylinder 
work for the conditions given. Cylinder diameters, 15 and 22 in., initial pressure 135 
lbs. per square inch gage, back pressure 10 lbs. per square inch absolute, cut-offs I 
and {, receiver volume 4 times high-pressure cylinder, strokes equal. 

Prob. 4. For the same conditions as above, what low-pressure cut-off would give 
equal work? 

Prob. 6. What will be the most economical load for a 16- and 24x30-in. engine 
running at 125 R.P.M. on 150 lbs. per square inch absolute initial pressure and at- 
mospheric backpressure? What will be the economy at this load? 

Prob. 6. What will be the release and receiver pressures for the above engine if 
^he receiver has a volume of 15 cu.ft.? 



WORK OF PISTON ENGINES 287 

Prob. 7. F!nd the cut-offs and cylinder ratio for equal work division and complete 
expansion when initial pressure is 150 lbs. per square inch absolute and back 
pressure is 10 lbs. per square inch absolute, receiver four H.P. volumes. 

Prob. 8. WiU a 14- and 20 x20-in. engine, with a receiver volume equal to 5 times 
the H.P. cylinder and running on { cut-off on the high-pressure cylinder and I cut-off 
on the low, with steam pressure of 100 lbs. per square inch gage and back pressure 
of 5 lbs. per square inch absolute, have complete expansion and equal work distri- 
bution? If not, what changes must be made in the cut-off or initial pressure? 

Prob. 9. What must be the size of an engine to give 200 1.H.P. at 150 R.P.M. on an 
initial steam pressure of 150 lbs. per square inch absolute, and 10 lbs. per square inch 
absolute back pressure, if the piston speed is limited to 450 ft. per minute and complete 
expansion and equal work distribution is required? Receiver is to be 6 times the volume 
of high-pressure cylinder and H.P. stroke equal to diameter. 

12. Compoimd Engine with Finite Receiver. Exponential Law^ No 



Clearance. Cycle Vm. General Relations between Pressures, Dimensions, 
and Work, when EBgh Pressure Exhaust and Low-pressure Admission are 
Independent. The diagram Fig. 84 may be used to represent this cycle, as well 
as cycle VII, by conceiving a slight change in the slope of the expansion and 
receiver lines. Using the same symbols as those of the preceding section, 
and the expression for work as found in Section 7, Chapter I, 

.iuD.Uia.^.)z.+^Ji^(i-z.--y^^y[i.(^)--']l 



but 



(rec.pr.)i=P«=P»(prl [—q-J . 
or 

(rec.pr.), = (m.pr.)(^ ['^-) 

and the last term in the equation for Wh within the bracket may therefore be 
written 



or 



288 ENGINEERlNa THERMODYNAMICS 

and hence by simplifying the first two terms also, 

•^.-«<'-p')'^'{--.--(^-)'l«k+')'['-fe)'"']}-« 

Work of the low-pressure cylinder may be expressed in tenns of pressure 
and volumes at N, E, and G, but it is convenient to use instead of the pressure 
at N or at E, its equivalent in terms of the point B, The pressure at iV^ is 

(rec.pr.). = (in.pr.)(^)'(^+l)' 

and when multiplied by the receiver volume vDh, it becomes 
(rec.pr.).yD,-(in.pr.)D.Z.(X)(^)'(^^+l)- 

= (in.pr.)D.Z.(^y-^(^^+l)' 
At E the product of pressure and volume is 

(rec.pr.)2 xZlDl = (in.pr.)^HZ)ir (^f J 

Using these quantities, the following equation gives the work of the low- 
pressure cylinder: 

+ (^y~*[l-z/"*]|-144(bk.pr.)Z)i„ . (384) 
and the total work is, by adding (Ws) and (Wl), 

— <-»''!^-^('---+ff)-(«k+')"[c-fir" 

-C-+ferJ'"]+(fe)'"'('-^''"')}-'"«''''"'"'' • • "*" 

This Eq. (385) is the general expression for work of the zero clearance com- 
pound engine with exponential expansion, no clearance, and finite receiver. 
From this the following expressions are derived: 

(m.e.p. ref. to L.P.) 

-^i?r^|{.---'+(t)-'(A-)'[C-^0'"' 



WORK OF PISTON ENGINES 289 

Work per cu.ft. supplied is 

-(H^&rJ"1+(o;)'""-^'-'))-'«»''-'"-'fe • ««'> 

Cu.ft. supplied per hr. per I.H.P. 

13,750 Zu. .ggg. 

(m.e.p. ref. to L. P.) Re 

(rec.pr.)2 = (in.pr.)(^Y; (389) 

(rec.pr.)i = (in.pr.)(^)'(l+^)'; .... (390) 

(rel.pr.),r=(in.pr.)ZH*; (391) 

(rel.pr.)i,=(rec.pr.)2Zj,* = (in.pr.)f^y (392) 

If work is equally divided between the cylinders, Wh, Eq- (383), and Wl, 
Eq. (384), will become equal, henoe 

-— (^)-(^.-)'[-fen-(f)-(ife«)' 

This equation shows conditions to be fulfilled in order that an equal division 
of work may be obtained. It does not yield directly to a general solution. 
When expansion is complete in both cylinders, 

Z, 4 and (^ = (|^)*. 
Rc \in.pr. / \Rc/ 

Introducing these values in the general expression Eq. (385) for work of 
this cycle, it may be reduced to the following: 

F = 144(in.pr.)Z^D^^[l-(|^y"'] (394) 

From which are obtained 

(m.e.p. ref.toL.P.) = (in.pr.)|^^[l-(^y""n. . . (395) 

Work per cu.ft. supplied = 144(in.pr. -3^ l — (^j (396) 



s 



290 ENGINEERING THERMODYNAMICS 

Cu.ft. supplied per hr. per I.H.P. = 7 * . y t^\ (p^) 

^^ ^ ^ (m.e.p. ref . to L. P.) \/2cv 



13>750 (bk:PL-)\ (397) 

(m.e.p. ref. to L. P.) \m.pr. / 



If work is equally divided and complete expansion is maintained in both 
cylinders Eq. (381) becomes 



8 



-.-- .-^['-(,-^.n 






which may be simplified to the form. 



2Zr-^ 



'H 



>+^hfer]}-^ — <- 



where Rv is the ratio of maximum low-pressure volume, to the high-pressure 
volume at cut-ofif, 

hence 

ZRc 

Jttv 

and the value of Rv may be found from original data. 



«'- (S£/ « 



Eq. (398) may easily be solved for Zh, from which the required cylinder 
ratio may be found by, 

Rc=ZhRv (400) 

This is the cylinder ratio which gives equal work in the two cylinders and 
complete expansion in both, when used with the value found for the high- 
pressure cut-oflf Zffy the assumed initial and back pressures, and the assumed 
ratio, y, of receiver volume to high-pressure displacement. 



WORK OF PISTON ENGINES 



291 



Example. Find (a) the horse-power, (b) steam used per hour, (c) the release 
and receiver pressures of a 12^ and 18x24-in. engine, with a receiver twice as large as 
the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-offs i in the 
high and such a value in the low as to give complete expansion. Exponent for ex- 
pansion curve = 1.4. 



(o) From Eq. (386) 



(m.e.p.) = 



( in.pr.) Zh 

8-1 Re 



8 — 






— bk.pr 



which, on substituting above values, gives 



150 ^5_ J 4 

.4 ^2.25 K-"*"-^" "^V4.5/ 



(±y 



4.5 



1.4 



2.25 X 



1 



+1 



2.25 




4.5 



.4" 



4.5 +2.25 X 



2.25. 



.5 



2.25 



X2.25 



{b-(^']]- 



10 



or 



hence 



(m.e.p.) =57.5 lbs. per sq.in., 



I.H.P. =221. 



(6) From Eq. (388) 



Cubic feet of steam per hour per horse-power = 



^3/50Zflr 
m.e.p. Re 



13,750 .5 ^^^ ^, 
-575- ^2:2-5 =^-2 ^"•^^•' 



hence total poimds per hour will be 



53.2 X221X. 332 =3910. 



From Eqs. (389) to (392): 



(rec.pr.)x = (in.pr.) (£|j (l +^^) , 
(rec.pr.)a = (in.pr.) (^-^ , 



(rel.pr.)tf = (in.pr.)Zi/*, 
(rel.pr.)i, = (rec.pr.)iZ£,*, 



292 ENGINEERING THERMODYNAMICS 

These, on substitution of the proper numerical values, become: 

(rec.pr.)i = 150 X l^j I Xl +tt) =75 lbs. per sq. inch, 

(rec.pr.)i = 150X(.5)^* =57 lbs., 

(reLpr.)£r = 150 X (.5)^* =57 lbs, 

(rel.pr.)L = 57 X (^^j ' =32.1 lbs. " 

Note: In all the following problems clearance is assumed to be zero. 

Prob. 1. A 12x18x24 in. engine is running on superheated steam of 150 lbs. per 
square inch absolute pressure, and exhausts to the atmosphere. If the speed is 100 
R.P.M., high-pressure cut-off J, low pressure cut-off i, and receiver volume 10 
cu.ft., what horse-power will be developed and what steam used per hour? 

Prob. 2. What would be the effect on the power and the economy of (a) changJTig 
to wet steam in the above? (6) to compressed air? 

Prob. 3. What would be the receiver and the release pressures for each caae? 

Prob, 4. Will there be equal work distribution between the two cylinders? 

Prob. 5. It is desired to obtain complete expansion in a 14x22x36-in. engine 

running on fluid which gives a value for a of 1.2. Initial pressure is 100 lbs. per 

square inch gage, and back pressure 5 lbs. per square inch absolute. What must be 

the cut-offs and what power will be developed at 500 ft. piston speed? Receiver =3 

XH.P. volume. 

Prob. 6. How large must the receiver be for the above engine in order that the 
pressure in it shall not fluctuate more than 5 lbs. per sq. inch? 

Prob. 7. An engine is to run on steam which will give a value of « = 1.1, and t^ 
develope 500 horse-power at 100 R.P.M. Piston speed is not to exceed 500 ft. per 
minute. Steam pressure, 150 lbs. per square inch absolute, back pressure, 5 lbs. per 
square inch absolute. Complete expansion and equal work distribution, for this load are 
to be accomphshed. What will be the cyUnder sizes and the high-pressure cut-off if the 
receiver is to be 3 times the high-pressure cyhnder volume? 

Prob. 8. What will be the steam used per hour by the engine of Prob. 7, and 
what wnll be the variation in the receiver pressure? 

Prob. 9. If the high-pressure cut-off were halved, how would the power and 
economy be affected? 

13. Compound Engine without Receiver, Logarithmic Law. No Clear- 
ancei Cycle IX. General Relations between Dimensions and Work when 
High-Pressure Exhaust and Low-Pressure Admission are Coincident Such 
a peculiar case as this admits of but little modification of the cycle compared 
with the receiver cases, b; cause the low-pressure expansion is necessarily a direct 
continuation of the high pressure without any possible break. There can be no 
over-expansion in the high nor can expansion there be incomplete, as there is, 
properly speaking, no back pressure with which to compare the high-pressure 
cylinder terminal pressure. There may, however, be over and incomplete 
expansion in the low-pressure cylinder. It might appear that the high-pres- 
sure cylinder negative work was equal to the low-pressure admission work, as 
each is represented by the area below DC, Fig. 874, but this is not the case, since 



WORK OF PISTON ENGINES 



293 



le diagram is drawn to two different scales of volumes, showing the pressure- 
roke relation between high and low. This is apparent from the diagram, Fig. 
rC showing fluid volumes in each cylinder to a single scale on which A BCD 
the wor k done in the high-pressure cylinder, ABD'EF the whole work, whence 
CD'EF is the part done in the low-pressure cylinder. There is, of course, 
) low-pressure cut-off or even admission as ordinarily considered. The cyde, 
» far as the work to be done is concerned, is the same as for a simple engine, 





B.P.C]rL Voh. 

1 i 






































A 




B. 












A 






,8 


















A 




^ 








\ 












































\ 


















\ 


















H.P. V^ 








\ 


















\ 
























\ 


















\ 























H.P.W 

1 


ork 


V 


















V 
















D 










\ 


c 
















\ 


ic 














C 










) 


v* 














b 




\ 














F 










J 


V 






















\ 




















v 


























Vs 


**-> 








L P iH' 


D 


^ 




























• 






^ 


^ 


D 


L..r . ^ 


E 

L 


E 




^P. 


Woi 


k 


r 








c 






















E- 


IDICATOR CARDS OF EQUAI 


^^ 








r 








r 






















BASE AND HEIGHT 






L.1 


*.Qr 


LV, 


4.. 






iXSE 






D 


lAGfl 


AM( 


►F F 


.UID 


WOI 

Dte- 


MR 
RIBl 


IGAF 

mo 


OLE 


)6 


























DM 


ORAM ON 

UNEQUA 


EQUAL i 
L VOLUM 


rfROKEI 
E6CALE 








1 


>FC 


YLIK 


OER 


4 
























qI 


^ 


A 






R 








































A 






n 














\ 








































M 




r 








9. 










\ 








































Q 






\ 


















\ 














































\ 








45 










\ 














































\ 








•< 


C 








\ 






































-< 








V 


^^ 




1 




c 


H.I 


*.Cy 


.W. 


r^ 


P 




































cl 






^-^ 


\ 


c> 








\ 






J 




































/I 


> 


-^^ 


••^ 




A 


/I 














> 


^1 




s 






























^ 




--^ 


Q 




y 


V 

/ 


'R" 












^ 


/^ 


"T^ 




"v 


*--, 




















^ 


-^ 


.^^ 


^' 








y 


y 


r — 1 










D 


-^ 


^ 


,— 


1 

._!_ 


•«M« 




:=a 




D' 




D' 


— 






,<^ 


,^' 






b 
nw^ 




a. 


^-^ 

I 
( 


f 












F 






1 


;p.jcyi 


DderW 


»rk 






r 




— C- 






^ 


^-^ 




























. 






r 










L 




l^jT 


^ 


















'/ 








^ 









INDtVlCUAL CYUNKR WORK SMOWN TO SAME 
SCALE OF PRESSURE AND VOLUME 



N 



K L 

HIGH AND LOW PRESSURE CYLINDER DIAGRAMS PLOTTED TO SAME /M9 



"iQ. 87. — ^Work of Expansion in the No-receiver Compound Engine, Zero Clearance, Cycle 
IX, Logarithmic Expansion, Cycle X, Exponential Coincident Piston Movement. 

nd the only reason for introducing formulas for overall work, work per cubic 
3et supplied, {m.e,p. referred to low), and fluid consumption, is to put them into 
orm for immediate substituiion of dimensional relations. Because of the absence 
f cut-off in the low, the distribution of work between high and low will 
lepend solely on the cylinder ratio and high-pressure cut-off, for, the earlier the 
ligh-pressure cut-off, and the larger the high-pressure cylinder, the greater 
he fraction of the total work that will be done there, as there is only a fixed 
ffnount available, and the less there will be left to be done in the low 



294 ENGINEERING THERMODYNAMICS 

The diagrams of the two cylinders are plotted to combined axes in 
Fig. 87D. The points Q and R at equal heights. KN is the L. P. displace- 
ment, and KG that of the H.P. It has been shoVn in Section 8, that 
the expansion lines CD and CD' may be plotted to the axes LN and LXM, 
the point X being the intersection of NQ and KR extended, and that the distance 



and 



also 



^^-w^.-'>--D^r'''sh- ■ ■ ■ « 



^=^+'''-5^- <*»' 



Hence the work area under CD is 

W„ =GLXPc log. p = 144(rel.pr.)ir77^4r 'o& ^ ■ 

but 

(rel.pr.V = (m.pr.)Z^, 



hence 



Tr^ = 144(in.pr.)ZirDH|l+loge^-^^-^logeiec|. . (404) 



Again the work area under C'T^ is 



W^t. =KLXPc log.£i=144(reI.pr.)/r7^^ log. ^, 
hence 

TrL==144(in.pr.)Z^DH(^^1log. ftc-144(bk.pr.)D^, . (4C5:) 
and the total work, 



Tr = 144(in.pr.)ZHDH] 1+loge ^^ R^i^og, Re 



Re 



+^3l^^&^ 



-144(bk.pr.)Di 



= 144(in.pr.)Z^Dtf 



l+log.i^+(^^-.^Jloge/2cl -144(bk.pr.)D^. 



WORK OF PISTON ENGINES 296 

But 



\Rc-l Rc-l) ' 



and 

1 R 

lo& -^+l0g. iJc = loge, -^ = loge Ry, 

SO that 

Tr-144(m.pr.)^ifZ)ir(n-Iog.|^)-144(bk.pr.)DL, . . (406) 

which shows by its similarity to the work o| the simple engine that, as before 
stated, the total work is the same for this cycle as if the entire expansion were 
made to take place in a single cylinder. 

This same result could have been attained in another way sufficiently 
interesting to warrant se tting it down. Since the low-pressure work is repre- 
sented truly to scale by C'D'EFj Fig. 87C, the mean eflfective pressure of the 
low-pressure cylinder is given by the area divided by F«. By contracting all 
volumes proportionately, CD' takes the position CD' and C'F the position 
CF\ hence 

area CiyEF 



F,-F/. 



-P. 



represents the mean effective pressure in the low-pressure cylinder just as 
truly. Therefore, 

L.P. cylinder work = ( y ^y P. j 7, 

As the high-pressure work is (total— low), 
H.P.cylmderwork=P»F»(l+log. y\-P»V»-PcVly^^\og,^^ ^P^V, 

Introducing symbols 
L.P. cylinder work=144(in.pr.)ZirZ)ir(p^)log,i2c-144(bk.pr.)Z)ji. (407) 

H.P. cylinder work=144(in.pr.)^fl2>H[l+log,|^-g%| log. fie] 

= 144(in.pr.)ZKD«ri+log.^-^5rilo8'.^]- • (*08) 
which check with Eqs. (404) and (405). 



296 



ENGINEERING THERMODYNAMICS 



Dividing the total work by the low-pressure cylinder volume and the hi^- 
pressure admission volume in turn, 



(m.e.p. ref . to L.P.) = (in.pr.)-^{ 1+loge y-j" (bk.pr.) 



(a) 



= (in.pr.) pVf 1 +I0& iRnRc)] - (bk.pr.) (6) 



. . (409) 



Work per cu.ft. supplied = 144( 



in.pr.)ZH( 



1+lofe 



Rc 
Zi 



| — (bk.pr. 



)Rc (a) 



1 



= 144(m.pr.):^[l+log.(ftiiBc7)]- (bk.pr.)ifc (6) 



. (410; 



Cuit. supplied per hr. per I.H.P. = 7 ' . ^ p\ X-tt («) 

^^ ^ ^ (m.e.p. ref. to L.P.) Rc 

13,750 1 



(m.e.p. ref. to L.P.) RhRc 



(b) 



'. . (411) 



For equal division of work there can obviously be only one setting of the 
high-pressure cut-off for a given cylinder ratio and any change of load to be met 
by a change of initial pressure or of high-pressure cut-off will necessarily unbalance 
the work. Equating the high-pressure and low-pressure work expressions, 
Eqs. (404) and (405), 



l+loge^j — P — = 



loge/2c = 



Rc 



Rc-l 



logeRi 



/ bk.pr. \ Rc 
\in.pr. / Zh 



or 



,,, 1 , (bk.pr.) Rc Rc+l . j, ^ 
Another relation exists between Zu and Rc, namely, that 

Zji fRc 

where Rv is the ratio of volumetric expansion. Then 



but 






R^ 



2R„ f 



hence 



Rc^c-"^ 



Rc^c-^ 



2R, 



log.«A-^ 
Rv 



= l+/bk:P£:)fl^ (412) 

\m.pr. / ^ ' 



WORK OF PISTON ENGINES 



297 



With this formula it is possible to find the necessary ratio of cylinder 
displacements for given initial and back pressures and for given ratio of 
expansion Rv 

For convenience in solving this, a curve is given in Fig. 88 to find value 



2R 



of Re when Rc^c"^ has been found. 



1 






































9 


























^ 


^ 










«% mm 






































i 


















^ 
































^ 




























































> 






^ 


/ 






























8 




/ 


/ 
































1 




T 






































26 




6 







76 




U 


K) 


nn- 


126 




160 







2Kc 

Values of (Rc)"K«^ 



2A. 



Fio. 88. — Curve to Show Relation between Values of Re and {JRk)^c~^ for Use in Solving 
Eq. (412), Giving Cylinder Ratio in Terms of Ratio of Expansion for the No-receiver 
Compound Engine without Clearance. 

The complete expansion case of this cycle results from the condition 
Pd=P« or (rel.pr.)L=(bk.pr.) or ^^=(vX^)> 

which when applied to Fig. 87, transforms the diagrams to the form Fig. 89. 
It also follows that 



(bk.pr.)Dx = (in.pr.)ZjyDjy 



and 



Re 



— /hupj[i\ 
Vbk.pr./ 



These conditions will, of course, reduce the total work Eq. (406) to the 
common value for all cycles with logarithmic expansion and likewise those 
for mean effective pressure, work per cubic foot supplied, and consumption. 
For the equal division of work imder this condition, Eq. (412), becomes 



2R. 



R 



R -1 



C C 



= 7.39/2, (413) 



since (. — ^J/2r=l and R may represent ratio of expansion or ratio of 
\m.pr. / 



298 



ENGINEERING THERMODYNAMICS 



initial to back preasures, these being equal. Fig. 90 gives a curve showing the 
relation between cylinder ratio and ratio of expansion established by the above 
condition. 



p 




H J. Cyl. Volg. 

2 


A 






B 






A 






I 












\ 












\ 










® 


\ 














\ 












\ 


c 










) 










/ 


/ 








y 


/ 






D 


^^^ 






F 


L-J^ 


_L- 


J_ 



P 
























A 






B 
























\ 
























\ 
























\ 






















(D 


\ 


k 
























V 
























\ 


c 
























\ 
























> 


N 


























"^ 


>v^ 








F 






















n 














^^^ 


1 


~T^ 





6 4 8 2 1 
L.P. Cyl. Vols. 



p 
























A 






J 
















































\ 
























\ 
























\ 






















© 




\ 
















C 


\ 






\ 


c 














V 




) 


v 


















X 


> 


/ 


> 


\ 
















y 


/ 






,^ 


"^v 


^-. 


1 — 






D 


rf<^ 


^ 










— — 


^"^, 


fc^ 


E 























Fig. 89. — Special Case of Cycles IX and X. Complete Expansion in both Cylinders of the 

No-Receiver Compound Engine, Zero Cleamace. 



Example 1. Method of calculating Diagram, Fig. 87. 

A, As described in the text this diagram is drawn to two- volume scales, so that 
there may be two volumes for one point. 



WORK OF PISTON ENGINES 

Assumed data: 

Pa =P» - 120 Iba. per sq.in. abs. Fa - F« - F< - F, - F/ =0 ou.ft. 
P.^Pf- 10 lbs. per Bq.in. abs. F.-l cu.ft. 

Fc=2cuit. 
Fi-F.-ScuJt. 



To locate point C: 



To locate point D: 









To locat« intermediate pcunts from C to D. The volume at any intermediate 
point is (the volmne of low-pressure cylinder up to that point) + (volume ot lii^- 



S 



I 



Fio. 90. — Curve to Show Relation between Values of Rct the Cylinder Ratio, and R the Ratio 
of Initial to Back Preaeure tor Complete Expanaion in the No-receiver Corapound 
Bogine without Clearance (Bq. (413).) 

preaauie cylinder from that point to end of stroke), e.g., at f stroke the volume in 
low, is .75 X5, and the volume in the high is .25 X2, or total 4.25, and the pressure 
at that point is found by the PV relation as above. 

B. ABBumed data: 



Pa 
P, 


=P.- 


120 lbs. per sq.in 
10 lbs. per sq.iii. 


.abs. 
ahs. 


Va 


•"F/ = 
= F.- 


=0 cu.ft. 
■5 cu.ft. 
'\ cu.ft. 


To locate pmnt D 
















-^ 


120 X 


1_, 


24 lbs. 


per sq.in. 




5 



Intermediate points from B to D found by assuming volumes and computing 
pressures from the PV relation as above. 

C. Figure ABCD constructed as in A. Figure C'D'EF is figure CDEF of A to the 
same pressure scale but to a volume scale 2.5 times as large. 

D. Figures constructed as in C. 



300 ENGINEERING THERMODYNAMICS 

To draw indicator cards. The volume and pressure scales are chosen and from 
diagram A^ a distance AB is laid off to the volume scale, AD is then laid off equal to 
AD of diagram A to the pressure scale. Point C is located to these scales and joined 
to B and D by drawing curves through the intermediate points plotted from the FY 
diagram to the scales of the card. For the low-pressure card EF is laid off to the 
volume scale, and FC" and ED' to pressure scale. C and U are then joined in same 
manner as C and D for high-pressure card. 

Example 2. Find (a) the horse-power, and (6) steam used per hour for a 12 Xl8 x24 
in. engine with no clearance when initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-off in 
the high-pressure cylinder is i, there being no receiver. 

(a) From Eq. (409) we have 

(m.e.p.)=(in.pr.)^5-^-[l +loge(/^///2c)] -(bk.pr.), 

= 150X^ ], — X(l+.8)-10=60 1bs. sq.in. 
JX^.^0 

hence 

I.H.P.=192. 

(6) From Eq. (411) we have 

1 o rr c r\ i 

Cubic feet of steam per I.H.P. per hour =7 — ^ rX 



(m.e.p.) RhRc 

_ 13,750 J 

~ 50 ^2X2.25" ' 

hence the weight of steam used per hour will be 

61 .2 X .332 X 1 92 = 3890 pounds. 

Example 3. What will be the cylinder ratio and the high-pressure cut-off to give 
equal work distribution for a ratio of expansion of 6, an initial pressure of 150 lbs. per 
square inch absolute and back pressure of 10 lbs. per square inch absolute? 

Ratio of baek to initial pressures is .067 and 



hence from Eq. (412) 



or 



Rv =6, 

2Rc 

lofo%^ 1.40, 



R^R^-i =24.36, 



and from Fig. 88 



Re =2.8. 



p o Q 

From the relation Z/f = h" = high-pressure cut off =-^ =.446. 

tiv o 



WORK OF PISTON ENGINES 301 

Prob. 1. A compound locomotive has no receiver and runs on an initial pressure of 
175 lbs. per square inch gage and atmospheric exhaust. The cylinders are 18 and 
30x42 in. The steam pressure may be varied, as may also the cut-off to a limited 
degree. For a speed of 200 R.P.M. and a constant cut-off of f, find how the power 
will vary with initial pressure and for constant initial pressure equal to boiler pres- 
sure show how the power at the same speed will vary from i cut-off to full stroke. 

Prob. 2. Show how the steam used per horse-power hour will vary in above 
problem. 

Note: 8 for 175 lbs. =.416. 

Prob- 3. With the cut-off at f , what should the initial pressure be to give equal 
work distribution? 

Prob. 4. With full boiler pressure and f cut-off what would be tenninal pressure 
in the low-pressure cylinder? 

Prob. 6. What must be size of cylinders for a tandem compound engine with 
negligable receiver volume to run at 125 R.P.M. with complete expansion and equal 
work distribution on an initial pressure of 125 lbs. per square inch gage and a back 
pressure of 5 lbs. per square inch absolute, when carrying a load of 500 horse-power, 
the piston speed to be less than 500 ft. per minute? 

Prob. 6. What will be the steam used by the above engine in pounds per hour? 

Note: 5 for 125 lbs. -.311. 

Prob. 7. A builder gives following data for a tandem compound steam engine. 
Check the horse-power and see if the work is equally divided at the rated load. 
Cylinders 10 ins. and 17 J X 15 ins., initial pressure 125 lbs., speed 250 R.P.M., horse- 
power 155. Neglect the receiver volume. 

Prob. 8. Another manufacturer gives for his engine the following, check this: 
Cylinders 20 and 32x18 ins., initial pressure 100 lbs., atmospheric exhaust, speed 
200 R.P.M., horse-power 400. Neglect the receiver volume. 

14. Compound Engine without Receiver, Exponential Law, Cycle X. 
General Relations between Dimensions and Work wlien High-pressure 
Exhaust and Low-pressure Admission are Coincident. Referring to Fig. 
S7D it is desirable first to evaluate the work areas CDKG and C'D'NK, As 
before, 



>/f 5— -J and KL = Dh ^^ _^ , 



hence 



Wc^ = 



PcXGL 



r['-in-""^[^v^j-an. 



8 

but 

(rel.pr.)if = (in.pr.)Z»', 

so that 



W^H = 144(in.pr.) 



DjtZh 

s-1 



-^»--^«-rM~) 



, . . . (414) 



^ _PcXKL\ /XL\-> 1 _ (rel.pr.)A, D^iJcf, / 1 V"'! 



302 
whence 



ENQDIEERING THERMODYNAMICS 



Wl = 144(in.pr .) ^^Rc 



1- L '-' 

Re 
Rc-l 



-144(bk.pr.)Di;. 



(415) 



It is to be expected that the sum of high- and low-pressure work will be 
of a form similar to that which would be obtained if all work were performed 
in a single cylinder of a displacement equal to that of the low pressure, adding, 



W = Wa-\- Wl = 144(in.pr.) 






s-Z^-^-Z^-i^^^^ 



+Z, 



--•44ff 



-144(bk.pr.)Di, 



= 144(in.pr.) 



DaZa 

8-1 



«-ZH»-i+Zir*-M l-(^y~n I -144(bk.pr.)Dx> 



Z 1 

whence, substituting ^^=-5- 



Tr=144(in.pr.)^[8-(^)'"']-144(bk.pr.)Dii; (416) 



. . (417) 



(m.e.p.ref.toL.P.)=^|[«-(l)'"]-(bk.pr.) . • . 

Work per cu.ft. supplied = 144^^^^ \a- (^"^\ - 144(bk.pr.)|^ . (418) 



Cu.ft. supplied per hr. per I.H.P.= 



13,750 



(m.e.p. ref. to L.P.) Re 

13,750 



(m.e.p. ref. to L.P.) Ry 



rs-. . (419) 



Conditions for equal division of work between high- and low-pressure 
cylinders may be obtfuned by equating Eqs. (414) and (415), 



s-Z«— i-Z«'-' 



-(i)-' 

Rc-l 



= Za—'Rr 



Rc-l 



_{bk.pr.\Rc, 



WORK OF PISTON ENGINES 303 

Rearranging 

The last term in the first member of this equation may be expressed as 



\m.pr. / 



and the relation 



Re 

w 



V 



exists between Zh and Re, hence, making these substitutions, 



Re 

Re 



'^^[Rcf-^-l\+R<f-^^Rv'-^]^s+(^ . (420) 



which is not a simple relation, but can be solved by trial. 

The assumption of complete expansion in the low-pressure cylinder (it is 
always complete in high, for this cycle), leads to this following relations: 



(!§.)=«'•. 



hence 



144(bk.pr.)I>i, = 144(in.pr.)5-V>H, 

llv 



and from Eq. (414), 

^.=.«(i.,.,?^"[,-(±).-.-i'^-i.i 



but 



Re 



1 

„ . RvS—1 8—\ . 1 /bk.prAi 

Tr=144(in.pr.)M«;:iy[l-(g^-) ']. ... (421) 
(m.e.p. ref. to L.P.)=in.pr.^ ITtU"!- — ) I* ' ' i^*^^) 



304 



ENGINEERING THERMODYNAMICS 



The expression for egualUyofwork Eq. (420) becomes, for this case of complete 
expansion, 



Rc+l 
Rc-\ 



(i2c?-i - 1) +fi^-» = 8ftF*-» +(«- 1), 



• • 



. (423) 



by which it is not diflSicult to find the ratio of expansion Rvy which gives equality 
of work for given values of «, and /Ec, the cylinder ratio. Values for Rv for 
various values of Re and a are given by the curves of Fig. 91. 



10 



& 
8 

















■ 










1.6 


1.4 


Valuta of 
If 112 


S 


LI 


























y. 


i^ 


^ 
^ 


1> 








'^ 
























yy 


g 


;^ 


X^ 


^ 




























y. 


^ 


^ 


'y' 




^ 




























i^ 


t 


^ 
































V 


4 


^ 


>" 
































-A 


4 


■y^ 
































/' 


^ 


>* — 


































^ 


y 




































^ 


7^- 













































































10 
ValttM of Rv 



15 



Fig. 91. — Curves to Show Relation between Re the Cylinder Ratio, and Ry the Ratio of 
Expansion, for Various Values of (s), Applied to the No-receiver Compound Engine 
without Clearance, when the Expansion is not Logarithmic. 



Example 1. Find (a) the horse-power, and (6) the steam used per hour for a 
12- and 18 x24-in. engine with no receiver when the initial pressure is 150 lbs. per 
square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 
R.P.M., cut-off in high-pressure cylinder is J, there being no receiver and steam 
having expansion, such that $ = 1.3. 

From Eq. (417) 

, , (in.pr.)Z^r /I \— 1-1 ... . 

(^.e.p.) .-_ _ _-|^,> ( _) j ^(bk.pr.), 

which, on substituting the above values, becomes 



T>^2l5[^-^-(i^) ]-10-C3.31bs.per8q.in. 



hence the indicated horse-power -=243. 



WORK OF PISTON ENGINES 305 

(6) From Eq. (419) the steam used per hour in cu.ft. per horse-power is 

13J50Zj^ 
m.e.p. Re* 

which, for the data given above, becomes 

13,750 .5 ,^^ ^^ 

or pounds per hour total, is, 48.2 X243X. 332 =3880. 

Example 2. What will be the high-pressure cut-off and cylinder ratio to give 
equal work distribution and complete expansion for an initial pressure of 150 lbs. per 
square inch absolute, and back pressure of 10 lbs. per square inch absolute? 



(in.pr.X 
bk!^./; 



From relation fiF'^Irr-— ), iBr«"6.9 and from this, by the curve of Fig. 91, 



/2c =5.4. 

For complete expansion 



Rv 6.9 



Prob. 1. A tandem compound engine without receiver has cylinders 18- and 
30x42-ins. and runs at 200 R.P.M. What will be the horse-power developed at 
this speed if the initial pressure is 175 lbs. per square inch gage, back pressure 
atmosphere, high-pressure cut-off }, and 8 has a value of (a) 1.1, (6) 1.3? Compare 
the results with Prob. 1 of Sec. 13. 

Prob. 2. What will be the weight of steam used per horse-power per hour for 
the two cases of the above problem? Compare these results with those of Prob. 2, 
Sec. 13. 

Note: B=.416. 

Prob. 3. What must be the cut-off in a 10- and 15 x20-in. compressed air engine 
running on 100 lbs. per square inch gage initial pressure and atmospheric back pres- 
sure, to give complete expansion, and what will be the horse-power per 100 ft. per 
minute piston speed, a being 1.4? 

Prob. 4. It is desired to run the following engine at its most economical load. 
What will this load be and how much steam will be needed per hour? 

Cylinders 20- and 32x18 ins., speed 150 R.P.M., steam pressure 100 per square 
inch gage, atmospheric exhaust, dry saturated steam. 

Prob. 6. Should the load increase 50 per cent in Prob. 4, how would the cut-off 
change and what would be th effect on the amount of steam used? 

Prob. 6. Vf hat would be gain in power and the economy of the engine of Prob. 4 
were superheated steam used, for which s = 1.3? 

Prob. 7. In a 14- and 20x24-in. engine will the work be equally divided 
between the cylinders for the following conditions? If not, what per cent will be done 
in each? Steam pressure' 100 lbs. per square inch absolute, back pressure 10 lbs. per 
square inch absolute, s = 1.2, cut-off = J. 

Prob. 8. What would be the work and steam used by the above engine if there 
were complete expansion and equal distribution? 



306 



ENGINEERING THERMODYNAMICS 



16. Compound Engine with Infinite Receiver. Logarithmic Law. With 
Clearance and Compression, Cycle XI. General Relations between Pressures, 
Dimensions and Work. In terms of point pressures and volumes, Fig. 92, the 

P 




INDICATOR CARDS OP EQUAL BASE 
ARO HEIQHT FOR CASE OF INCOM- 
PLETE EXPANSION AND COMPRESMCN. 



INCjOMP.LETE EXfANSION ANDf COMPRESSION. Y 





H.P. 




^^ 



--(bk.pr.) 



• ( rcl.pr.) 




INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT FOR CASE OF OVER 
EXPANSION AND COMPRESSION. 



OVER EXPANSION AND COMPRESSION. 



Fig. 92. — ^Work of Expansive Fluid in Compound Engines with Infinite Receiver, with Clearance, 
Cycle X Logarithmic, and Cycle IX Exponential Expsuision, and Compression. 

work of the two cylinders may be written down at once as if each were inde- 
pendent of the other, the connection between them being fixed first by making 
the back pressure of the high equal to the initial pressure of the low, or to 



WORK OF PISTON ENGINES 307 

the receiver pressure, and second by making the volume admitted to the low 
equal to that discharged from the high reduced to the same pressure. This 
last condition may be introduced in either of two ways, 



(a) EM--NH, 

(6) [{PV) on H.P. expansion line— (P 7) on H.P. comp. line]> 
= [(P7) on L.P. expansion line— (PV) on L.P. comp. line]. 

Without introducing the last relation 
IFH=P5F»(l+loge^^)-P/7/(l+log.^')-(Pa-P/)7a-P,(F,-7.); (424) 

TfL=P*V,(l+loge^)-P,7,(l+log.^J-(P,-Pi)F,-PXr,-F») . (425) 

W^PtV, loge ^+PnVH lOge P - PeVe loge ^^-P,V, logc ^ 

Vb Vh Vf Vi 

+PbVt-PrVf-PaVa+PfVj-PaVa+P.V. . 
+PAF,-P,7,-P,7,+P,7,-P,7,+P*Ft. 
The second condition is 

or 

P»F»+P.y,+P.y.+P»y»=2(P»F»+P*7») (426) 

Substituting 



W=P{V„ log. ^+PhV^ log. ~PcV. log. ^-P.y* log. ^ ^ 

yb Vh Vf Vi 

+ 2{P,V,+PtVu)-PaVa-PoV,-PaV^-P,Vs 



. (417; 



This expression, Eq. (427) contains, however, the receiver pressure which is 
related to the release pressure by 

/ \r>nDDD^</i \ L.P. max.vol. 

(rec.pr.)=P,=P.=P-=P»=P.^=(reI.pr.).L:p— ^r^g-^. 

Introducing this 



Tr=P»niog.^+P,y, log. ^-pL-V. log. ^'-PtF* log. ^ ^ 

•'» Vh Vh Vf Vi 

+2(P»F»+Pt7t)-Pay.-pX* V, -PXrVi-P,V, 

V h V h 



■ . (428) 



308 



ENGINEERING THERMODYNAMICS 



Introducing the usual symbols in Eqs. (427) and (428) and in additicri 
the following: 

Z = cut-off as fraction of stroke, so that ZhDh is the displacement volume 

up to cut-off. 
c= clearance volume divided by displacement, so that Ch Db is the clearance 

Volume and (Zh+Ch)Dh is the volume in the high-pressure cylinder 

at cut-off. 
X=that fraction of the stroke during which compression is taking place so 

that {Xh-\-Ch)Dh is the volume in the high-pressure cylinder when 

compression begins. 

Applying the general symbols to Eq. (427), 



l+cg\ 



W^IU 



1 1-^ 
[m.pr.)(Z^H-Ciir)Z)j5rloge \^ 



4-(rec.pr.)(ZL+CL)Z)Lloge (^^J^) 
-(rec.pr.)(X^+c^)I>fflog« ^— ^ — ^j 



Ch 

Xl+ci^ 



-.(bk.pr.)(X^ +Ci,)Z)i, lege (^^) 

H-2(in.pr.) {Zh +ch)Dh - (in.pr.)c^Dir - (rec.pr.) (1 '\-Ch)Dh 
4-2(bk.pr.) {Xl +cl)Dl - (rec.pr.)ci/>i, - (bk.pr.) (1 -hCz.)Z)i, J 



(4^9) 



This expression gives the work in terms of initial, receiver and back pressures, 
the valve periods, cut-off and compression, the clearances and cylinder dis- 
placements. 

Substitution of the symbols in Eq. (428) will give another equivalent expres- 
sion in terms of the same quantities except that low-pressure cylinder release 
pressure will take the place of receiver pressure. This is 

{m.pT,)(ZH+CH)D„ log.f^''-) +(rel.pr.)L(l +Cl)Z)l log. (Pr^) 

-<""-''(ii.t)<^'+-"'»'°«-(^") 

-(bk.pr.)(Xi, +c^)Z)i, lege (^^'') 



Pr = 144 



-f 2(in.pr.)(Z^ ■^-Cu)Dh - (in.pr.)c^Z)jy - (^^-PrOLU^^- j (1 +Ch)Dh 
+2(bk.prO(Xjr,H-CL)Z)L-(rel.prOi,f -^ Ui^l- (bk.pr.) (1 H-C£.)Dr. 



(430^ 



WORK OF PISTON ENGINES 



309 



It is sometimes more convenient to involve the cylinder ratio and low- 
pressure displacement than the two displacements as involved in Eq. (430) 
and the ratios of expansion instead of (iut-offs. This may be done by 












Rl-'^^ 






(431) 



and it should be noted here that the ratio of expansion in each cylinder is no 
longer the reciprocal of its cutroff, as was the case when clearance was zero, nor 
is the whole ratio of expansion equal to the product of the two separate ones 
because the low-pressure cylinder expansion line is not a continuation of that 
in the high. Making these substitutions for cylinder and expansion ratios, 
Eq. (430) becomes, 






(432) 



(in.pr.)(l +ch)-^-^ lege /2ir +(reLpr.)z.(l +Cz.) loge Rl 

KhKc 

- (rel.pr.)i(ZH +<'*)f^log.(^^^) - (bk.pr.) (X^ +cl) log.(^-^^ 

+2(in.pr.)(l -^-ch):^-^ -(in.pr.)^— (rel.pr.)L5^(l +ch) 
^ +2(bk.pr.)(Xz,4-CL) -(rel.pr.)z72i,ri, -(bk.pr.) (1 -\-Cl) 



It is interesting to npte that this reduces to Eq. 304 of Section 9, by making 
clearance and compression zero. 

From any of the expressions for work, but more particularly (430) and 
(432), the usual expressions for (m.e.p.) referred to low-pressure cylinder, work 
per cubic foot supplied, and consumption per hour per I.H.P. can be found, 
but as these are long they are not set down, but merely indicated as follows: 



(m.e.p. ref . to L.P.) = 



W 



144Z>z.* 



Work per cu.ft. supplied = 



W 



DnUi 



h-\-Ch) — {Xh-\-Ch) 



/rec.pr.Xl 
\in.pr./J 



. (433) 



. (434) 



Cu.ft. sup. per hr. per I.H.P. 



13,750 



(m.e.p. ref. to L 



_[(Z.+,,)_(X.+,,)(-H:)]^. 



13,760 



(m.e.p. ref. to L 



^[(Z.+.)-(X.+.)(|^)](^i). (435, 



310 



ENGINEERING THERMODYNAMICS 



As the receiver pressure is related to the initial and back pressures and to 
the relation between the amount taken out of the receiver to that put in, which 
is a function of the compression as well as the cut-off and cylinder ratio, it is 
expressed only by a complicated function which may be derived from the 
equivalence of volumes in the high and low, reduced to equal pressure. 



Therefore, 



Ph=^P 



b 



+p. 



V, 



v^+v» ' 'v^+v: 



Introducing symbols 



Hence 
(rec.pr.)-(m.pr.)(^^^^^)^^^(^^^^^)+(bk.pr.)(^:^-j^-:p^^^^p^. (436) 

This Eq. (436) gives the receiver pressure in terms of initial and back pressures, 
the two clearances and compressions, the cylinder ratio and the cut-off in each 
cylinder. 

Proceeding in a similar way, the release pressures can be found in terms of 
initial data, 

V, 



P'-Poy^. 



or 



(rel.pr.)a - (in.pr.) ( - "jp^ j (a) 



(in.pr.)5-- 
Kb 



And 



Pi = P»y" = JP» 






.\14 



V, 



+P>\ 






(437) 



14 



7. 



or 



(rel.prOi = (in.pr.) 



, = (in.pr.) 



\{\+Cl )Rc } 

, , {Xh + Cii) 
r /1+C«\ 1 

\1 + ClIRhRc 



+(bk.pr.) 



1 + 



1 + 



{X„+ci,)R i. 
a+Ct.)Rc 



+ (bk.pr.) 



1+ 



(Xi.+Cr.) -, 

(Xh+Ch) 
(Zl+Cl)Rc 

(A'^ + c^) 
{Xh+Ch)Rl 



(1+Ci,)Rq -■ 



va) 



(6) 



(438) 



WORK OF PISTON ENGINES 



311 



These three pressures all reduce to those of Eqs. (308), (309), (310), Section 
9, when clearance and compression are zero. 

Equal work in both cylinders is, of course, possible, but it may be secured 
by an almost infinite variety of combinations of clearance, compression and 
cut-oflF in the two cylinders for various ratios of expansion; it is, therefore, not 
worth while setting down the equation of condition to be satisfied, but reference 
may be had to Eqs. (424) and (425), which must be made equal to each other, the 
result of which must be combined with the equation of cylinder relations. 

























A 




B 


















1 




y 






















\ 






















\ 


















1 


y 




\ 
















G 


V 




"N 


c 














\ 






\ 




• 


V 






\ 
















\ 








^ 


^-, 














SK 














■.» . 1 















































INDICATOR CARDS OF EQUAL 
BAIE AND HEMHT 



Fio. 93. — Special Case of Cycles XI and XII Complete Expansion and Compression in both 
Cylinders, of Compoimd Engine with Clearance and Infinite Receiver. 

There are certain special cases of this cycle for which equations expressing 
important relations are simpler, and they are for that reason worth investigat- 
ing. Those that will be examined are 

(a) Complete expansion and compression in both cylinders. Fig. 93. 

(b' Complete expansion m both cylinders with no compression, any clearance, 
Fig. 94. 

(c) Any amount of expansion and compression but equal in both cylinders, 
equal clearance percentages and a cylinder ratio equal to the square root of 
the ratio of initial to back pressures, Fig. 95. 

Case (a) When both expansion and compression are complete in both cylin- 
ders, Fig. 93, 

TF^ = 144(in.prOZ,,D,,log. (^^"^^^ .... (439) 
Tr., = 144(rec.pr.)Zx^^loge ([§^^7^^^) .... (440) 



(in.pr. ) 
(bk.pr.)' 



312 ENGINEERING THERMODYNAMICS 

but 

and 

log.(Ji^PI:))+log. (5^) =log. (J-M^4 ^S^) =log. 
^\(rec.pr.)/ ^ \(bk.pr.)/ \(rec.pr.) (bk.pr.)/ 

hence 

Tr=144(in.pr.)Z^D^ log. |gg^ (441) 

(m.e.p.ref.toLJO = (m.pr.)|^loge(g^) (442) 

Work per cu.ft. supplied =144(in.pr.) logf (^^-^j (443) 

Consumption, cu.ft. per hr. per I.H.P. = 7 \ ^ ^ T>^ tt-- • • • (444) 

^ ' *- x- (m.e.p. ret. to L.P.) Re 

EqualUy of work in high- and low-pressure cylinders is obtained by making 

/ (in.pr. ) \ ^ /(rec.pr.)\ ^ /(in .pr. ) \ * 

\(rec.pr.)/ \(bk.pr.)/ \(bk.pr.)/ ' 
or 

(rec.pr.) = [(in.pr.) (bk.pr.)]* (445) 

It is desirable to know what clearances and displacements will permit of 
equal work and complete expansion and compression. 

(rec.pr.) = (m.pr.)|^ = (m.pr.) ( ^^^^ 1 

= (bk.pr.)^ =(bk.pr.)(^j), 

hence 

/( in.pr. )\ ^ 1 +Ch ^ /(in .pr. ) ^\ 

\(rec.pr.)/ Zh+ch \(bk.pr.) /' 

/ (rec.pr.) \ ^ 1+Cz. ^ / (in.pr. ) \ * 

\( bk.pr.)/ Zz,+cx, \(bk.pr.)/ ' 

or calling 

/(in.pr. )\ _ 

\(bk.pr'0/ ^^''^ 

^ l+c^-cgfip* , „ 1+Cl- ClRp^ /..^v 

Zk = ^-^1 , and, Zx, = ^^ . . . . (446) 



Equating discharge of high and intake of low-pressure cylinders, 

ZffDeRp^^ZJ)^ or ^=Rc=^RpK 



$ 



WORK OF PISTON ENGINES 



313 



Inserting in this the values just found for Zh and Z^y 

1+Cu — ChRp^ 



«c-j4.;^^^^^^r^p*, 



(447) 



which is the required relation between cylinder sizes, clearances and ratio of 
pressures, which, together with cut-offs given in Eq. (446),. will give equal work 
and complete expansion and compression. The compression in the high- 
pressure cylinder is such that 

Xh=ch{Rp^-1) 

(448) 



and for L.P. cylinder. 



XL=ci,(Rpi-l). 




A 


B 






A 


\ 


G 


H 




K 


\ 


^^ • 



INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



Fig. 94. — Special Case of Cycles XI and XII. Complete Expansion and Zero Compression 
in both Cylhders of Compounds Engine with Clearance and Infinite Receiver. 

Case (6) With complete expansion and no compression, both cylinders, any 
clearance, Fig. 94, 

WH = 14^H^{m.pr.)iZH+CH)\og, ^^^^^j -c,.[(in.pr.)-(rec.pr.)]] (449) 

Tri,=144Z)i,r(rec.pr.)(Zi,+cz,)log, ([^t.'^j) " ^^^ (^50) 

with the added requiremen t tha t the high-pressure discharge volume, EC = low 
pressure admission volume FH, or 



D^=2)i,| 



Zl + Cl 



V(rec.pr.)/J' 



(451) 



314 ENGINEERING THERM0DYNAMI08 

and 

(rec.pr.) = (bk.pr.)^'=(bk.pr.)^^, (452) 

hence 

r (bk.pr. )] ^£t+ct 

|_(rec.pr.)J l+Cr ' 
which substituted in Eq. (451) and rearranging gives 

Re ^^ ,. ^ , (463) 



Zl+c. 



/Zl+ClV 



Eq. (453) indicates that for this special case of complete expansion and no com- 
pression the cylinder ratio required to give this case, is determined entirely by 
the L.P. cut-off and clearance. If the cylinder ratio and clearance are fixed, the 
required cut-oflf in the L.P. cylinder can be found by solving Eq. (453) for Zl, 

Z^=i^-Ci„ (454) 



and from Eq. (452), 

r 1+Cr, 

(rec.pr.) = (bk.pr.) 



— Cl-tCl 



Ri 



= (bk.pr .)ii;c. . • (455) 



Cut-oflf in the high-pressure cylinder is determined by clearance, initial 
pressure and receiver pressure, which in turn depends on low-pressure cut-oflf 
and clearance Eq.(452), or may be reduced to cylinder ratio and low-pressure 
clearance by Eq.(454), as follows 

Vc^ 1+cg ^ / ( in.pr . )\ ^ /(in .pr. ) \ Zl+Cl 

Vt Zh+ch V(rec.pr.)/ \(bk.pr.)/ 1+Cl ' 
hence 

^^^ Rp{Z^+c^) ~'^- 



Elimmate Zl by Eq. (454), 



(456) 



Since the high- and low-pressure cut-offs are functions of cylinder and clearance 
dimensions, and of Rpy the rato of initial and back pressures, the work of high- 
and low-pressure cylinders may be expressed entirely in terms of these quantities. 



TF^ = 144Z)H(in.pr.)j(l+c^)|^loge(|^)-c,,(l-|^)|. . . (457) 



TFi, = 144DH/ec(bk.pr.)[(l+cz,)loge/2c-c,.(/2c-l)]. . . . (458) 



WORK OF PISTON ENGINES 315 

Hence^ total work by addition is 



TF=144Z)i7(in.pr.)i^ 



/•+'•> "*(l)-''(l-i) 



+(l+Cjr,)logfiBc-Ci(i2e-l) . . . (459) 

» 

Expressions might be easily written for mean effective pressure referred 
to the L.P. cylinder, work per cubic foot fluid supplied, and consumption, but 
will be omitted for brevit y. I t is important to note, however, the volume of 
fluid used per cycle is not AB, but is LB, Fig. 94, and is, 

{Sup.YoI)^Dh[{Zh+Ch)-cJ^^^^=Dh[(^^^ . (460) 

W 
(m.e.p.ref.toL.P.) =j^^^^-g (461) 

W 
(Work per cu.ft. supplied) =^^ — vHT (462) 

is 750 r 

Consumption cu.ft perhr. per LH.P.=^^^^j-^^-p-^-|^(Z«+c«) 

Itpjltc 

Equality of work, secured by equating Eqs. (457) ajnd (458) gives 

(l+Cir)loge(|^)-CH(|^-l) = (l+c,,)logeiec-c,,(/?c-l). . (464) 

This equation may be satisfied in an infinite number of ways. One case 
worth noting is that of equal clearances, when it is evident that if 

Cs==CLf and -^^Rc, or Rc—^Rp 

JtCc 

the Eq. (464) is satisfied. This last condition is the same as that which satisfied 
Case (a) with complete compression. 
CcLse (c), Fig. 95, assumes that 

and 



316 



ENGINEERING THERMODYNAMICS 



and corresponds to the first special case considered in Section 9, which lead 
in the no-clearance case to equality of high- and low-pressure work. 




Fig. 95. — Special Case of Cycles XI and XII, Equal Per Cent Clearance in Each Cylinder of 
Compound Engine with Infinite Receiver and Cylinder Ratio Equal to the Square Root 
of Initial Divided by Back Pressure. 

The assumptions already made are sufficient to determine the receiver pres- 
sure. By Eq. (436) 



(rec.pr.) = (in.pr.) 



Z+c 






f(bk.pr.) 






+X+C 



+X+C 



[(in.pr.) (bk.pr.)]* 



= [(m.pr.)(bk.pr.)]» (465) 

The work of the high-pressure cylinder may now be evaluated. 

War 144D£r(in.pr.) | Z |^1 +log. (^^ j - < 

- 144Dtf[(in.pr.) (bk.pr.)]* { (X+c) log. (^^^\+l-X ■ . . (466) 



WORK OF PISTON ENGINES 317 

The low-pressure cylinder work may be similarly stated, 

Trz,«1441)4(in.prO(bk.prO]*|z[l+log.(^^)^^ (467) 

-144i)^(bk.pr.) |(X+c) log. ^^)+l-Z ., 

but 

Dz.[(in.pr.) (bk.pr.)]* = D^fic[(in.pr.) (bk.pr.)]* 

=^ir(^^)*[(in.pr.)(bk.pr.)l* 

=D^(in.pr.),l 

and similarly, 

Z),,(bk.pr.)=2)£r[(in.pr.)(bk.pr.)]* • 

With these substitutions the value of low-pressure ie7orfc,Trx,,Eq. (467), becomes 
equal to high pressure work, Eq. (466), hence the total work 

W=2Wh = 2Wi^ (468) 

Example 1. Method of calculating Diagrams, Fig. 92. 
Assumed data: 

Pq ^Pa '^Pb = 120 lbs. per square inch abs. Va = F/= .12 cu.ft. 
Pn =P# —Pe -Pd^Ph- 50 lbs. per square inch abs. Vh = .4 cu.ft. 

Pk ^i'i = 10 lbs. per square inch abs. Vc^Vd^ .8 cu.ft. 

Vg^ Fi = .16cu.ft. 

Vi-'Vi^ 2cu.ft. 

F«- .2cu.ft. 

(Va-7«)=(F„-F,). Fit- .4cu.ft. 

The above may be expressed in initial pressure, etc., and in terms of cut-off, etc., 
but as the relation of the lettered points to these terms is shown on the diagram values 
for cut-off, etc., they will not be given here, as they may readily be found from values of 
the lettered points. 



To locate point C: 



To locate point F: 



- -Fft 120X.4 _„ 
Pe=Pb-77-^ — — — -w lbs. per sq.m. 

V c -O 



P, re »'^< 50 X .2 _ _ _ -, 
/=— ir-= — -^r- =83.3 lbs. per sq.m. 

Vf AJt 



To locate point Q: 



Pe7e^^ 

P« 120 



Va — ^' = 7:^=.083cu.ft. 



318 ENGINEERING THERMODYNAMICS 

To locate point L: 



_ PtVk IPX. 4 „,, 
Pi = — rr- = — TT— =26 lbs. per sq.m. 
Vi .16 



To locate point N: 



F„=^=^-.08cuit. 



To locate point H: 

(FA-F„) = (Fm-F.), or F*-F«+Fn-Fe = .96+.08-.2 = .84cu.ft., 
since 



To locate point 7: 



48 
P«F« = P,V,, = F« =g - .96 cu.ft. 



■n PhVh 50X.84 _^ ,, 
Pi = —.— « — - — =21 lbs. per sq.m. 
Vi 2 



Example 2. Find (a) the horse-power, (6) steam used per hour, and (c) receiver 
and release pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per 
cent clearance in high-pressure cylinder, and 4 per cent clearance in low-pressure cylinder, 
when initial pressure is 150 lbs. per square inch absolute* back pressure 10 lbs. i>er 
square inch absolute, speed 125 R.P.M., cut-ofif in high-pressure cylinder is }, low- 
pressure cut-off is such as to give complete H.P. expansion, and compression is 15 per 
cent in high and complete in low. 

(a) For complete high-pressure expansion the receiver pressure must be equal to the 
high-pressure release, and to maintain the receiver pressure constant the low-pressure 
cylinder must take as Liuch steam per stroke as the high-pressure discharges. With 
initial pressure and cut-off as given, the release pre3&ure for the high-pressure cylinder 
may be found from the relation (inpr.)(cj5r+ZH)=(rel.pr.)/r(cj5f+Z)i5r) or 150x(.56) 
-(rel.pr.)»(1.04), or (rel.pr.)iy=79.3 lbs. Since there is 15 per cent compression in 
high-pressure cylinder there is exhausted each stroke 85 per cent of its volume. Also 
since compression in low-pressure cylinder is complete, the low-pressure clearance is 
full of steam at the receiver pressure at the beginning of the stroke. Hence the 
low-pressure displacement up to cut-off must equal MDa or L.P. cut-off =.85Dj, 
divided by cylinder ratio, or .85 4-2.26 =.378. As compression is complete, the 
per cent compression may be found from the relation CLX(rec.pr,)=(cL-f^L)(bk.pr), 
or .04x79.3 = (.04-fXi.)10, or Zi, = .28. 

From Eq. (432), (m.e.p.) referred to low-pressure cylinder is obtained by dividing 
by 144 Dl, and on substituting the above values it becomes, 

150(.1+.06)(^|-) log. 2+30(1 +.04) log,2.64-30(.15+.06) (|~)log, {^^^) 

-10(.28+.04)log. (-^^) +2xl50(l+.06)^y .150x^ 

" 2 64 
-30X;r~|(l+.06)+2xl0(.28+.04)-30x2.64x.04-10(l+.04)=60.5 lbs. per. sq.in.. 

hence 

I.H.P.=235. 



WORK OF PISTON ENGINES 



319 



(6) From Eq. (435) by substituting the above values 

Cuit. steam per hour per horse-power = i|^ \({M+ .04) - (.28 + M)^) ^1 =46.5, 

oU.o L \ 79.0/ 160 J 

or pounds per hour will be 3550. 

(c) Release pressure for high-pressure cylinder has been shown to be 79.3 lbs. 
and may be cheeked by Eq. (437), as follows: 

(rel.pr.)zr = 15o(^^^) =79.3 lbs. 

Receiver pressure has already been shown to be equal to this quantity and may 
be checked by Eq. (436) 



(rec.pr.)j/ = 



150 X (.5 +.06) 



10 X (.28 +.04)2.25 



(.378 +.04)2.25 +(.28 +.04) ' (.378 +.04)2.25 + (.28 +.04) 



79.3 lbs. 



Low-pressure release pressure is found from Eq. (438) to be 



(reLpr.)i.=150 



1+.06 



X 



1 +.04 2x2.25 



1 + 



(.15 +.06)2.64 



U (1 +.04)2.25 J 



+10 



.28 +.04 
1+.04 



(.15 +.06)2.64 



H 



L (1 +.04)2.25 J 



=30 lbs. 



Prob. 1. What will be the horse-power and steam used by the following engine 
for the data as given? 

Engine 20 and 28x36 ins., running at 100 R.P.M., clearance 5 per cent in high 
pressure. 3 per cent in low. From cards H.P. cut-off = .3, L.P. =.4, H.P. compression, 
.1, L.P., .2. Gages show (in.pr.) to be 150 lbs., (r c.pr.) 60 lbs., (bk.pr.) 26 ins. 
Hg. (barometer ==30 ins.). 

Prob. 2, What must be ihe cut-offs and the cylinder ratio of an engine to give 
equal work and complete expansion and ompression for 200 lbs. per square inch 
absolute initial pressure and atmospheiic exhaust, if clea ance is 5 per cent in the 
high and 3 per cent in the low-pressure cylinder? What will the horse-power for an 
engine with a low-pressure cylinder 24 X36 ins., running at 100 R.P.M. for this case? 

Prob. 3. Should there be no compression, how would the results of Prob. 2 be 
altered? 

Prob. 4. What will be the total steam used by engines of Probs. 2 and 3? 

Prob. 6. For an 11- and 19x24-in. engine with 5 per cent clearance in each 
cylinder, i cut-off in each cyhnder, and 20 per cent compression in each cylinder, what 
will be the horse-power and the steam consumption when the speed is 125 R.P.M., the 
initial pressure 150 lbs. per square inch gage, and back pressure at atmosphere? 



16. Compound Engine with Infinite Receiver. Exponential Law, with 
Clearance and Compression, Cycle Xn. General Relation between Pressures, 
Dimensions and Work. Referring to Fig. 92, of the preceding section, which 



320 ENGINEERING THERMODYNAMICS 

will represent this cycle by a slight change of slope of the expansion and com- 
pression lines, the high-pressure work may be expressed in terms of dimensions, 
ratios and pressures. Since this must contain receiver pressure as a factor, 
and since that is not an item of original data, it is convenient first to state 
receiver pressure in terms of fundamental data: 



F«-F.= F,-Fn. 



But 



1^ 2 

7„=77iE£!LV and Fn = F*/bk^V 
Vrec.pr./ \rec.pr./ 



Whence, 



1 



whence 



\rec.pr./ \rec.pr./ 



/ \ r \\ \m.pr. / 

(rec.pr. ) = (m.pr.) * 



Vn+V. 



or in terms of dimensions and pressures, 



(rec.pr.) = (in.pr.)l r^(^z,+c,)+X.-\-'c^ ' j ' ' ^^^^ 

The high-pressure work may be stated as follows: 
Tr«=144D«{ (in.pr.)^^i^^[«- (^^^^)'"'] -(rec.pr.) (1-X«) 

-(rec.pr.)^J:^[(^^)'"'l-]-(ia.pr.)cH 1- . (470) 

Zu+CH+Rc{Xt.+cM^^y 
^ _ ,. . , Vm.pr. / 

-144D«(m.pr.)\^ r^^Z,+c,)-^X.+c„- 

'^;±;?[(^^)'"l-]+l-X«j. (471) 



WORK OF PISTON ENGINES 



321 



W,=lUDJ(Tec.pv.)^^f±^[s- (-Ag-y-'j _(bk.pr.)(l-X^) 

-(bk.pr.)^^^[(^^)*"-l]-(rec.pr.Kl. . (472) 



= 144D«fic(in.pr.)' 



Zll+CB+Rc(Xi,+Ci. 



>/bk.pr.yY 
\in.pr. / I 



Rc(Zt.+CL)+Xa+eB 



(473) 



The expression for total work need not be written here, as it is simply the sum 
of Eqs. (470) and (472) or of (471) and (473), the former containing receiver pres- 
sure and the latter containing only dimensions, initial and back pressures and 
both, the expK>nent of the expansion, 8, 

The volume of high-pressure fluid supplied per cycle is QB, Fig. 92, which 
may be expressed either in terms of high pressure or of low pressure points, 
thus; 



(Sup.Vol.) = Db \Zb+Cb - {Xb+Cb) (^^) ' ] 



(a) 



-^^[^'M^y -'^^Miw ^'^ 



(474) 



The following quantities will be indicated, and may be evaluated by sub- 
stitution from the preceding: 



W W 

(m.e.p. ref. to ^;P)-iuDl'T4iD^, 



W 



Work per cu-ft. fluid supplied = t^ — y-p; 

Consumption cu.ft. per hr. per I.H.P. 

13,750 



(475) 



(476) 



(Sup.Vol.) 



(m.e.p. ref. to L.P.) Dl 



(477) 



Eqtud division of work between high- and low-pressure cylinders requires 
that Eqs. (470) and (472), or (471) and (i73) be made equal, The latter will 



322 ENGESTEERINQ THERMODYNAMICS 

give an expression showing the required relation between dimensions and initial 
and final pressures, cut-off and compression in high- and low-pressure cylinders. 
In this expression there are so many variables that an infinite number of arnir 
hinaiions may be made to give equality of work. 

It is desirable to examine the results of assmning special conditions such as 
those of the previous section, the most important of which is that of complete 
expansion and compression in both cylinders, which is represented by Fig. 93. 



9-1 

TF„=144(in.pr.)Z.D^[l-(^) ' ]. ... (4: 



'8) 



but 



Tr. = 144(rec.pr.)Z^.^[l-(^^;)-]. . . . (479) 

\rec.pr./ 



hence 



,r=.44(in.pr)^,Z^|[l-(^)'-"] 



L in? > 

/re c.pr. X / in.pr. y j" _ / bk.pr. \ * "I 1 

\ in.pr. / \rec.pr. / [ \rec.pr./ J J 



= 144C ^-^ - * 



+ 
•-1 



in.pr .)D^Zhj—y J ^ / rec.pr. \ « 

[ \ in.pr. / 



. / rec.pr. \ « / rec.p r.X « / bk.pr. \ « I 
\ in.pr. / \ in.pr. / \rec.pr./ J 



= 144(in.pr.)Z>.Z^[l-(gi^) ' ] (480) 

The receiver pressure may be found as follows. In Fig. 93, EC=GH: 

Equating 

l+c.-c.(H:P?:i)^=ficri+c.-c.(^h(^F_V- . (481) 
Vrec.pr./ [ \bk.pr./ J \rec.pr./ ^^^ ^ 



WORK OF PISTON ENGINES 



323 



When this is solved for receiver pressure it results in an equation of the 
lecond degree, which is somewhat cumbersome, and will not be stated here. 
5q. (481) is, however, used later to find Ro- 
lf work is to be equally distributed between high- and low-pressure cylinders, 
rom Eqs. (478) and (479), 

«-l 8-1 •-! *~1 

«» { ^^^'P ^'\ • _ / rec.prA « __ / rec.prA « /bk.pr. \ « 
\ in.pr. / \ in.pr. / \ in.pr. / \rec.pr./ ' 



•r 



«-i 



«-i 



/rec£r.\ . =i+/bk^r.\ ' , 
\m.pr. / \m.pr. / 







































































































.6 


1 
















































I 


















































1 
































































































^i 




i\ 


















































v\ 
















































A 


k 












































§ 




A 


^ 


^ 










































p 

1 






\ 


§ 


^ 




^ 










































\ 


^ 


^ 










- 




































\ 


<: 




^ 


^ 










"■"' 




— 








^ 


L6 
•LI 


9 






IS 


t* 












"v 


-"^ 








"^ 








— 




~ 




— 




s 
























— 














~ 




~^ 


































~ 









— 






a.o 






































































































c 










z 


If 

> 








B 


O 








71 










u 


X) 









s. 96. — Curves to Show Receiver Pressure to Give Equal Work Distribution when Expansion 
and Compression are Complete in both Cylinders of the Compound Engjne with Infinite 
Receiver, with Clearance when Expansion and Compression are not Logarithmic. 



nee, for equal division of work, 



(rec.pr.) = (in.pr.) 



\ in.pr. / 



s-l 



. . . . (482) 



lich, if satisfied, will give equality of work in the two cylinders, for this case 
perfect conipression and expansion. 

In Fig. 96, is given a set of curves for use in determining the value of the 
tio of (rec.pr.) to (in.pr.) as expressed by Eq. (482). 



324 



ENGINEERING THERMODYNAMICS 



When (rec.pr.) has been found by Eq. (482) it is possible by means of (481) 
and the clearances to find Re* The events of the stroke must have the follow- 
ing values to maintain complete and perfect compression and expansion. 



Zh^(1+ch) 



Zl={1+Cl) 



1 

/rec.pr A r 

Vrec.pr./ 



(4831 



(4&t) 






X 



Example. Find (a) the horse-power, (6) compressed air used per hour, and (e) receiver 
and relea.e pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per cent 
clearance in the high-pressure cylinder, and 4 per cent in the low-pressure cylinder, when 
initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square 
inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder }, low-pressure 
cut-off such as to give complete expansion in high-pressiu^ cylinder, compression in 
high-pressure cylinder 15 per cent, and complete in low. Expansion such that «»1.4. 

(6) As in example of Section 15, receiver pressure equal high-pressure release 
pressure, and low-pressure volume at cut-off must equal volume of steam exhausted 
from high pressure. Release pressure may be found from relation (in.pr.Xc£r+^2r)* 
^(relpr.) -(ch+DuY: or 160(.06+.6)^**(rel.pr.)-(.06+l)^-*, or(rel.pr.)-601bs. As 
in the previous example, the low-pressure cut-off is .38, and the low-pressure compressioQ 
may be found from the relation Cz,*(rec.pr.) =*(ci.+Xii)'(bk.pr.),or(.04)^'^X60 = 
(M+Xl)^"^ (10), or Zl = .09. 

From the siun of equations (471) and (473) divided by 144Diyl2c, and with the 
proper values substituted the following expression for (m.e.p.) results: 



150 
2.25 



-4 -['•'- Vl^-loe-j J '•^) -2:25 






.5+.06+2.25(.15+.06) 



/JO \ -71 1 1 ♦ 
\150/ 



2.25(.38+.04)+.15+.06 

.5-h.06+2.25(.15-h.04)f ^j 
. 2.25(.38+.04)+.15+.06 



-.10- 
hence the horse-power is 214. 



'.09 + .04r/.09H-.04\* 1., ) ^^ ,^ 
4~~ \ — 04 ~ ) ""M ~ I " ^^ ^ 



WORK OF PISTON ENGINES 325 



From Eq. (477) with proper values substitute, 

Cu.ft. per I.H.P. hr--^^|pX r(.38+.04)(~y^' -(.09+.04)^^^l'^' =50, 

r total steam per hour will be 

60X214 = 10700 cu.ft. 

(c) Release for the high-pressiuie cylinder has already been given as 60 lbs. and the 
■ceiver pressure the same. The latter quantity may be checked by equation (469) 
Qd will be found to be the same. The low-pressure release pressure may be found 
•cm the relation (rec.pr.)(ZL-fci,)^*«»(rel.pr.)L(l+CL)**, which on proper substitution 

ives 



/ .38+.04 \i.4 

\ 1.04 ; 



(reLpr.)i,=«60( — ' 1 =27 lbs. per sq. inch 

Prob. 1. What will be the horse-power and steam used per hour by an 18- and 
4x30-in. engine with 5 per cent clearance in each cylinder and with infinite receiver 
iinning on 100 lbs. per square inch gage initial pressure, and 5 lbs. per square inch 
bsolute back pressure, when the speed is 100 R.P.M. and the cut-ofif in high-pressure 
vlinder is ^ and in low -^7 

Note: » = 1.3 and S=.2. 

Prob. 2. What must be the receiver pressure for equal work distribution when the 
litial pressure has the following values for a fixed back pressure of 10 lbs. per 
quare inch absolute? 200, 175, 150, 125, 100, and 75 lbs. per square inch gage? 

Prob. 3. For the case of 150 lbs. per square inch gage initial pressure and 14 lbs. 
ler square inch absolute back pressure, what will be the required high-pressure cyhnder 
izeforan air engine with a low-pressure cylinder 18x24 ins., to give equality of work, 
learance in both cylinders being 5 per cent? 

Prob. 4. What will be the horse-power and air consumption of the above engine 
rhen running at a speed of 150 R.P.M., and under the conditions of perfect expansion 
nd compression? 



17. Compound Engine with Finite Receiver. Logarithmic Law, with 
3earance and Compression, Cycle Xm. General Relations between Pressures, 
)imensions, and Work when H.P. Exhaust and L.P. Admission are Inde- 
pendent As this cycle, Fig. 97, is made up of expansion and compression lines 
eferred to the different origins together with constant pressure, and constant 
rolume lines, the work for high- and low-pressure cylinders and for the cycle can 
)e set down at once. These should be combined, however, with the relation 
loted for the case of infinite clearance which might be termed the condition 
or a steady state 

[{PV) on H.P. expansion line] — [(PF) on H.P compression line] 

=[(Py) on L.P. expansion line— PV on L.P. compression line,] 

W, Pi,Vi,-PeVe=^PnVn-PtVt (487) 



ENGINEERING THERMODYNAMICS 




OVER EXPANSION AND COMPRESSION 



Fig. 97. — Work of Expansive Fluid in Compound Engine with Finite Receiver and with 
Clearance. Cycle XIII, Logarithmic Cycle XIV, Exponential Expansion, and Com- 
pression. 



Also 



WORK OF PISTON ENGINES 327 

Beside this there is a relation between H.P. exhaust and L.P. admission 
pressures, corresponding to the equality that existed for the infinite receiver, 
that may be set down as follows: 

PmV^^P.V,, and Pn.{V„,+0)^Pa{Va+0); 
.-. Pn^V^=Pi>V,=Pa{V, +0)-Pn,0, and P^^Ph; 

PnVn=P,V,, and Pn{Vn+0)=P,{V,+0); 

/. PnVn^PtVu^P,{V,+0)-PnO, and Pn^Pe, 

^^= vl+o' ^^^^ 

These two expressions for the pressure at D and at G are not available in 
their present form, since they involve two unknown pressures— those at H and 
E, but two other equations of relation can be set down from which four equa- 
tions, the four unknown pressures P«, Pa, Pg and Pn, can be found. These 
other equations are 

Pd(F<i+0)=Pe(F,+0), or P,=P,(^±^), . , 
and 

P,(n+0)=P.(7.+0), or P,=p,^^±g). . . (491) 

Equating (488) to (490), 

V^ + ^'^^\Vi + 0)' ^^ Pl>V!>+PkO-=^Pe{Ve+0), 

and 

p _ P,V,+PnO 

Ve+0 ' 

Equating (489) to (491) 

and 

P*(F»+0)-PtF» 



(490) 



P.= 







Therefore „ ^Pi>yi,±Pj^ ^ P*(F>+0)-P«Ft 

V,+0 

P,V>0+P,(P =P»(7»+0) iV»+0) -PtV,iVe+0) ; 

PH[{v^+o)(y,+o)-o^] =PtVtO+PtVt{y,+o), 

p ^ PtV,0 +P,Vt{V.+0)P„V,^+ PtVtiVe+ 0) 



328 



ENGINEERING THERMODYNAMICS 



Therefore 



P»= 






(a) 



Substitution will give 



P.= 



P»V,(V>,+0)+PtVtO 
(V.+0)(Vh+0)-(P 



'"L v,{v,+o)+v/) \lv,+o\ 

> = \ p*y»(yk+ o)+PkVtOi [ F.+Q i 

' L(^.+0)(t^»+0)-02JLF,+0j 



(&) 



(c) 



(d) 



(492) 



It will be found that the use of these pressures is equivalent to the applica- 
tion of the equation of condition given in Eq. (487), for substitution of them 
reduces to an identity, therefore the work of the two cylinders can be set down 
by inspection in terms of point pressures and volumes and the above preasures 
substituted. The result will be the work in terms of the pressures and cylinder 
dimensions. 



(P.-P,)7. 



WH--P>vJl+loe,Y^ -PfVr(l+log, ^^)-P.F«logc 

-p,y»(i+iog. ^) -p,v,-p,v, log. Y-P.v. og, ^i^-PaF.+p.y. 

=P.n(l+lo„ ^) -P.F. log. (^) (^) -P.F.. 



Va+0 

v.+o 

+0 



Therefore 



(4<- 



Trir=P»F.^l+log, Y^ -P.Va 

_ \ P»V^(V^+0)+PtVtO l „ , (V.\ ( Va+0 \ 

TFi,=P,Fil+log.^±^)+P»r»log.^;-P,F, (l+log. ^*)-(p,-P,)r.. 

-PjiVj-V,) 
=P,V, log. (^±§) +PhVh log. y-PtV, log. ^-PsiVj- F.). 

Therefore 



^ _ r p>F>o+p.F.(F.+o) /z»±o\] J (n±q\ ) 



+ 



r PtF>0-|-P>Ft(F.+0 )] y .Vi 

[ v^(y,+o)+v,o J'^*'°*'F» 



-P»F* log. ^-P/F,- Ft). 



• • 



(494) 



WORK OF PISTON ENGINES 



329 



Adding Wb and Wl 

P»F»(l+l0g. ^) -P.y, log. ^^ -PaVa-Pj{V,-V^) 

_ r p»7t(y>+o)+p«n oi „ , (VA m+o \ 

[ (7.+o)(y»+o)-02 J "^^ '°^ \vJ\v.+o} 



Tr= 



+ 



+ 



fftF, 

L F 

[ 



+o)(y»+o) 



*(F.+0)/F»+0\"| y»+o 



■»(F.+0)+F.O \V,+0 
P ,V>0+P,Vt(V, 



v,+ o 



p^-J F» log. ^ 



(495) 



While this Eq. (495) for the cyclic work is in terms of initial data, it is not 
of very much value by reason of its complex form. To show more clearly 
that only primary terms are included in it, the substitution of the usual symbols 
will be made. 



F-144X 



(in.pr.)(Za+C£f)Z)H 1 +log. „ i 



1+ch 1 
Zh 



- (bk.pr.)(Zi, +cl)Dl log. 



Cl 



— (in.pr.)c£ri)a -(bk.pr.)(l —Xi.)Di. 

_ ( in.pr. ) (ZH+CH)DnliZi, +ci,)Dj, +0] + (bk.pr.) (X z,+CL)Z?i,0 
' [{Xa+CH)Da+0](iZi.+c,^)DL+0]-0* 

(X«+c«)D.log. (— -j |^^^^_--^^-^J 

(in.pr.) {Zu +ch)DhO + (bk.pr.) {Xl +cl)Dl [(Jh +ch)Dh +0] 



X 



+ 



(2i +Ci,)Z)i[(Xif +ch)£)j/ +0] + (Xh +Cfl)i)flO 



X 



[ 



(Zl+cl)Dl+0 



] 



ClDl loge 



[ 



{Zl+cl)Dl+0 



] 



I 



(in.pr.) (Zg +Ciy)D/rQ -f (bk.pr.) [(Xz, -f C£.)Z)l][X^ +c//)Z)^ +0] 
(Zl +cl)DlKXh +cji)DH +0] +{Xh +ch)DhO 

14-cz. 



X 



(^.+a)Z).log.(j^) 



• . (496) 



Such equations as this are almost, if not quite, useless in the solution of 
yrMms requiring numerical answers in engine design, or in estimation of engine 
V^ormance, and this fact justifies the conclusion thai in cases of finite receivers 
graphic methods are to be used rather than the analytic for aU design work. When 
estimates of power of a given engine are needed, this graphic work is itself seldom 
justifiable, as results of sufficient accuracy for all practical engine operation 
problems can be obtained by using the formulas derived for infinite receiver when 
reasonably, large and zero receivers when small and the pistons move together. 



330 



ENGINEERING THERMODYNAMICS 



It *mighi also be possible to derive an expression for work with an equivalevi 
constant-receiver pressure^ that woyld give the same total work and approxi- 
mately th^ same work division as for this ca^, but this case so seldom arises that 
it is omitted here. 

Inspection of the work equations makes it clear that any attempt to find 
equations of condition for equal division of work for the general case must be 
hopeless. It is, however, worth while to do this for one special case, that of 
complete expansion and compression in both stages, yielding the diagram Fig. 
98. This is of value in drawing general conclusions on the influence of receiver 
size by comparing with the similar case for the infinite receiver. 

By referring to Fig. 98, it will be seen by inspection that cylinder sizes, 
clearances and events of the stroke must have particular relative values in order 




Fio. 98. — Special Cajse of Cycles XIII and XIV, Complete Expansion and Compression in 
both Cylinders of Compound Engine with Clearance and Finite Receiver. 



to give the condition assumed, i.e., complete expansion and compression. It 
is, therefore, desirable to state the expressions for work in terms which may 
be regarded as fundamental. For this purpose are chosen, initial pressure 
(in.pr.), back pressure (bk.pr.); high-pressxu-e displacement, D^; cylinder 
ratio, Re] high-pressure clearance, Ch] and ratio of receiver volume to high- 
pressure displacement, y. Call 



( 






It will be convenient first to find values of maximum receiver pressure 
(rec.pr.)i, and minimum (rec.pr.)2; high-pressure cut-ofiF Zj?, and compression 
Xh\ low-pressure clearance cl, cut-off Zl, and compression, Xlj in terms of 
these quantities. Nearly all of these are dependent upon the value of cl and 
it will, therefore, be evaluated first. 



WORK OF PISTON ENGINES 331 

From the points C and /, Fig. 98, 



From A and E^ 



trec.pr.)2=(bk.pr.)^^^^^ (497) 



(rec.pr.)i = (in.pr.)^^, (498) 



and from E and C, 

(rec.prQi 1+c^+y ,^^. 

(rec.pr.)2 RcCl+V • • • v ; 

Dividing Eq. (498) by Eq. (497) and equating to Eq. (499), 

RpChJI+Ch) ^ I+Ch+V 
R(^cl(1+Cl) RcCl+v ■ 

Multiplying out and arranging with respect to cl, the relation to be 
fulfilled in order that complete expansion and compression may be possible is, 

CL^[Rd'{l+ca+y)]+CL{R<f^{l+CH+y)-RcRpCH{l+CH)] 

-[2/fipC/,(l+C/,)]=0. . (500) 

This is equivalent to 

C]?l+CLm,—n—Qy (501) 

and the value of cl is 

^^ Jf^+Un)^-m ^^2) 

It is nauch simpler in numerical calculation to evaluate Z, m, and n and 
insert their values in Eq. (502) than to make substitutions in Eq. (500), which 
would make a very cumbersome formula. (rec.pr.)2 and (rec.pr.)i may 
now be evaluated from Eqs. (497) and (498) by use of the now known value of Cl.. 

High-pressure cut-ofiF Zh may be found from the relation of points B and 
/, Fig. 98, 

Rp(Zh+Cu)^Rc{1+Cl) 
or 

Zh^^{1'\-Cl)-Ch (503) 

tip 



Low-pressure cut-off, Zl from, 

Rc{Zl+Cl)^I^Ch, 



or 

Zl^^^-Cl (504) 

tic 

High-pressiu^ compression, Xh^ 

Xh^clRc-Ch (505) 



332 ENGINEERING THERMODYNAMICS 

Low-pressure compression, Xl, by the use of points A and K^ 

Rc(Xl + Cl) = RpCh9 * 



or 

Re 



R 
Xl= j<~Ch—Cl (506) 



If Cl is regarded as being part of the original data, though it is related to 
Re, Rpt Ch and y as indicated in Eq. (488), the expressions for high- and low-pres- 
sure work and may be stated as follows: 



W^H=144(in.pr.)2)^ 



(|a+„,)[.+,^|fl±g] 



lliTS''+"+"'"*^S7--h'*'f]}- ■ <»"'* 



Tri;=144(m.pr.)I>if 



Re (I+Cl) [f. I ^ ,„si^„1±£e+V 



+ (1 +Ch) log. ficj^l - [1^(1 +&) - cJ - CB log. fj I • (508) 
Adding these two equations gives the total work of the cycle as follows: 

TF=144(in.pr.)Z)Hfic(l+c^) {^^[l+lo& |j [j~^)] +|; log. Rc\ 

l+cu+y , 1+cg+y . l+CH+y , 1+cy+ y 
fip(l+Ci,) '^^ ficCi^+y "*'ftp(l+cii) '""^ i?cci:+y J 

-144(m.pr.)Z)^{cH[l+ log, ^^j+c^loge ^+^/^+Ci:)-c^ 



+Cl 
+Cb 



This, however, may be greatly simplified, 



and 



'<*iTffS + "*<S-'*''" 



logc^^^-^+loge bTT =lo& ^P' 
Ch iCcCl 



Hence 



TF^144(in.pr.)Z>^r|^(l+CL)-cJlog, Rp. 

= 144(in.pr.)Z>HZ/flogfti> (609) 



WORK OF PISTON ENGINES 333 

From this may be obtained mean effective pressure referred to the low- 
pressure cylinder, work per cubic foot supplied, and consumption per hour 
per indicated horse-power, all leading to the same results as were foimd for the 
case of complete expansion and compression with infinite receiver (Section 15,) 
and will not be repeated here. 

To find the conditions of equal dwisum of work between cylinders, equate 
Eqs. (507) and (508). 

which may be simplified to the form, 

1+ch ^ RcCL+y ^ Rc{1+cl) Rc(1+cl)1 ^ RcCl J 

(510) 

This equation reduces to Eq. (376) of Section 11, when ch and cl are put 
equal to zero. In its present form, however, Eq. (510) it is not capable of 
solution, and it again becomes apparent that for such cases the graphical solution 
of the problem is most satisfactory. 

Example 1. Method of calculating Diagram, Fig. 97. 

Assumed data: 

Pa =P» = 120 lbs. per square inch abe. F; = Fi = 2 cu.ft. 

P^^Pj^^ 30 lbs. per square inch abs. Vc^Vd^Vt^ .8 cu.ft. 

P>= 10 lbs. per square inch abs. Vg^Vi= .24 cu.ft. 

Fe= .2 cu.ft. 

Va^Vf^ .12cu.ft. 

V 0^1.2 cu.ft. 

Vb= .4 cu.ft. 



To locate point C: 



To locate point M\ 



_ P^n 120 X. 4 

Pc =-7>— - — 5 — =60 lbs. per 8q.m. 



_ P^n 120 X. 4 ,. ^ 
Vm^-rs—^ — ^ — =1.6cu.ft. 



To locate point D\ 



Pd(Ftf+0)«Pm(Fm+0>, or Ptf«30^-^^=421b8.persq.in. 



334 ENGINEERING THERMODYNAMICS 

To locate point E: 

Pe^{Ve+0)^P„iVm+0), or P.=r^^ X30-60 Ibs. per sq.iii. 



To locate point F: 



To locate point L: 



P "e K« 60 X .2 , /WN 11 
Vf .u 



Pi =— Tr" ^ — ^";~33.3 lbs. per sq.in. 
Vi .Z4 



^ P»7t IPX .8 ,- ,^ 



To locate point N: 

since Pn-Pe 

To locate point G: 

P,{V,+0)=Pn{Vn+0) or. P, = 6o|^||^J =55.6 lbs. per 8q.in. 

To locate point H: 

Ph(.Vh+0)=P,(V,+0) v.^^-Xl+MllM 

or 

.24X55.5+55.5X1.2-30X1.2 



^^ ^_- — — - . 1.46 cuit. 



To locate point /: 



J, P*F» 1.46x30 „. ^ ,, 

Pi ^—Tr- *= 5 — =21.7 lbs. per sq.m. 

Vi ^ 



Example 2. Find the horse-power of a 12- and lSx24-]n. engme, running at 125 
R.P.M., with a receiver volume twice as large as the low-pressure cylinder, 6 per cent 
clearance in the high-pressure cylinder, 4 per cent in the low, when the initial pressure 
is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high- 
pressure cut-ofF J, low-pressure f , high-pressure compression 10 per cent, low 3 p)er cent. 

From Eq. (496) divided by 144Di„ and with the values as given above, the (m.e.p.) 

is equal to following expression: 

150x.56X2^(l+log.^|)-10X.34 log. |j-150X.06X2~^ 

150 X .56[.79 X2.25 +4.5] 4-10(.34)2.25 X4.5 .^ _i_ , ^6 1.06x2.25+4.5 
[.16 +4.511.79X2.25 +4.5] -(4.5)« ^'^^^2.25 ^^ .06^.16 X2.25+4.5 

.79x2.25+4.5 
04x2.25+4.5 



150 X .56 X4.5 + 10 X .34 X2.25[.16 +2.25] [.79 X2.25+4.5] ^ o 25 1 ' 
79 X2.25[.16+2.25] +(.16X2.25) [.04x2.25 +4.5 J ' ^ ^8« j 



+ — ^ 



. 150x.56x4.5 + 10(.34x2.25)[.16+2.25]r^^, 1.04"! ^^^,, 
-^ .79x2.25[.16+4.5] + (.16x 2:25) L"^' ^^«^ li^\ ^^^'^ ^^''^ ^' ^•^^- 

h(^r^cQ the horse-power will be 191. 



WORK OF PISTON ENGINES 335 

Prob. 1. Find the work done in the high-pressure cyKnder and in the low-pressure 
Under of the following engine under the conditions given. 

Engine 14 and 30 X28 ins., 100 R.P.M., 5 per cent clearance in each cylinder, 
gh-pressure cut-off iV, low-pressure cut-off A, high-pressure compression A, low- 
lessure compression iV, initial pressure 100 lbs. per square inch gage, back pressure 

lbs. per square inch absolute, and receiver volume 3 times the high-pressure 
splacement. Logarithmic expansion. 

Prob. 2. The following data are available: initial pressure 200 lbs. per square inch 
)solute, back pressure 10 lbs. per square inch absolute, engine 10x15x22 ins., with 
per cent clearance in the high- and low-pressure cylinders, speed 100 R.P.M. What will 
3 the cut-offs, and compression percentages to give complete expansion and compression, 
ogarithmic expansion? 

Prob. 3. What will be the work done by the above engine working under these 
)nditions? 

Prob. 4, What must be the low-pressure clearance, cut-offs, and compression 
ereentages, to give complete expansion and compression for a similar engine work- 
ig under the same conditions as those of Prob. 2, but equipped with a receiver twice 
3 large as the high-pressure cylinder? 

18. Compound Engine with Finite Receiver, Exponential Law, with 
leaiance and Compression, Cycle XIV. General Relations between 
tessures, Dimensions, and Work when HJP. Exhaust and L.P. Admission 
re Independent. It cannot be expected that the treatment of this cycle by 
jrmulas will give satisfactory results, since even with the logarithmic expansion 
iw, Cycle XIII gave formulas of unmanageable form. For the computation of 
wk done during the cycle, however, and for the purpose of checking pressures and 
ork determined by graphical means, it is desirable to have set down the relations 
f dimensional proportions, initial and final pressures, and valve adjustments, to 
le receiver pressures, release pressures and work of the individual cylinders. 

The conditions of a steady state, explained previously, require that (Fig. 97) 

1 1 

n-v,=v.{^y -v.{^y , (511) 

hich is the same as to say, that the quantity of fluid passing per cycle in the 

gh-pressure cylinder must equal that passing in the low. Expressed in terms 

dimensions, 

1 

L (in.pr.) J 

, . ^ (in.pr.) 
, rearrangmg, and usmg Rp = /|^^j; x • 

1 

Up, L (in.pr.) J 1 

+fl.(c.+Z.)r(''^*:^?P'--H', .(512) 

L (in.pr.) J 



336 ENGINEERING THERMODYNAMICS 

an equation which contains two unknown pressures (rec.pr.)i and (cut ofif pr.)i. 
To evaluate either, another equation must l>e found: 

where Pn=Pe, so that 

1 

V.'RcD„(c.+X.)[^J' (613) 

Hence 

Vn+0 \ 

or 

1 



=p«(^ 



(cut-off pr.)i= (rec.pr.)i 



^L(rec.pr.)iJ 
y+Rc{cL+ZL) 

_ r y(rec.pr.) M+ R c(cL+Xi,)(bk.pr. )* ] ' . 

L y+Rc{cL+ZL) J, ■ ^^ ^ 



1-.. 



which constitutes a second equation between (cut-off pr.)L and (rec.pr.)i, which, 
used with fk). (512) makes it possible to solve for the imknown. By substitu- 
tion in Eq. (512) and rearranging, 



1 



r..n r.. ^ fhV r.r ^ \ «/»* fa + ^n) [y+/Zc(CL + Zi:)]+ficy(CL+Xx) ]' ,,. ,. 

(rec.pr.)i = (bk.pr.)[ ^^^+x.)[y^Rc(c,+Z,)]+Rcy{c.+Z,) [ ' ^^^^^ 

This expression is of great assistance even in the graphical construction 
of the diagram, as otherwise, with all events known a long process of trial and 
error must be gone through with. It should also be noted that when s = l this 
expression does not become indeterminate and can, therefore, be used to solve 
for maximum receiver pressure for Cycle XIII, as well as Cycle XIV. 

Cut-off pressure of the low-pressure cylinder, which is same as the pressure at 
H or at My Fig. 97, is now found most easily by inserting the value found bj 
Eq. (515) for (rec.pr.)i in Eq. (5i4). 

Enough information has been gathered now to set down the expressions for 
work. 

W„ = 144D« I (in.pr.) '-^f [s - ('fj^)* "] -c.(m.pr.) 

(ch+Xh) y . [ /ch+XsV"^ 



8-1 



JV+Ch+Xh) 

s-1 






WORK OF PISTON ENGINES 337 

ir..l44D,|(cut^prO/-±-'^i-±?^[(t!^«^?iJ)'-'-l] 

-(bk.pr.)-'^^^^±^^[(^^)'"'-l]-(bk.pr.)(l-X^^ficj. . (517) 

Addition of these two Eqs. (516) and (517) gives an expression for the total 
work Wf and equating them gives conditions which must be fulfilled to give 
equality of work in the high- and low-pressure cylinders. Since these equations 
so obtained cannot be simplified or put into more useful form, there is no object in 
inserting them here, but if needed for any purpose they may be easily written. 
In finding the conditions of equal work, the volumes of (rec.pr.)i and (cut-off pr.)L 
must be inserted from Eqs. (514) and (515) in (516) and (517), in order to have 
terms in the two equations consist of fimdamental data. This, however, increases 
greatly the complication of the formula. 

After finding the total work of the cycle, the mean effective pressure referred 
to the low pressure is obtained by dividing by 144XDi». 

To assist in finding the work per cubic foot supplied and consumption, 
and the cubic feet or pounds per hour per I.H.P. it is important to know the 
volume of fluid supplied per cycle, 

r - 

(Snp^lol) ^QB=DH\{cH+ZH)-{cH+XB)y^^^ . (518) 

Example. Find the horse-power of and compressed air steam used by a 12- and 
18x24-in' engine running at 125 R.P.M., with a receiver volume twice as large as the 
low-pressure cylinder, 6 per cent clearance in the high-pressure cylinder, 4 per cent in 
the low, when the initial pressure is 150 lbs. per square inch absolute, back pressure 
10 lbs. per square inch absolute, high-pressure cut-off J, low-pressure cut-off }, 
high-pressure compression 10 per cent, low-pressure compression 30 per cent, and 
expansion and compression follow the law PV^'^=c. 

From £q. (515) (rec.pr.)i is found to-be as follows when values for this problem 
are substituted: 

X ,^ri5'7(.56)[4.5-|-2.25(.79)]-h4.6x2.25(.34)1« ^, ^ „ . ^ . 

(rec.pr.).^10[. ^e [4.5-h2.25X.79]4-4.5x2.25x.79 J '^^'^ ^^«- ^^^^^^ ^^^^^^' 

and by using this in Eq. (514) 

.71 .71 « 

, ^ re ^ / 4.5X81.7 +2 .25X.34X10 \ „„ • u w 

(cut-off pr.)L = ( 4 5-f-2 2 5x 79 / '^ ^^^- ^'^^' absolute. 



338 ENGINEERING THERMODYNAMICS 

It is now possible by use of Eqs. (516) and (517) by addition and division by 
14ADl, to obtain (m.e.p.). Substituting the values found above and carrying out 
the process just mentioned. 

'»«-'-2^{'™xfx['*-(ia)1-"''<'«'-f ><»»[(:«)■*-■] 

- -^X82|^l-|^^j^-p^j J+53X ^ LU5+2:25T.04; "^J 

,^^^^ 4.5-h2.25x.79 r, / 4.5+2.25x.79 \ *] ,^^^ 2.25x.34 r/.34\ -* 1 
+53X ^i^l—-——-J J_iox— ^— |^(^-j -Ij 

-lOx2.25x.7l =51.5 lbs. 8q.k 

hence the horse-power will be 200. 

By means of Eq. (518) the supply volume may be found. This gives upon 
substituting of the proper values: 

. (Sup.Vol.) =Z)^[.56-16x(^^y'n =.462>^. 

i-u-rx 1. TTTT> 13,750 ^^Sup.Vol. 

Cubic feet per hour per LH.P. = r- rX — ~ , 

(ta.e.p.) Dl 

^13^50 ^^ 
51.5 ^2.25"^*^' 

hence the total volume of air per hour will be 

64.5X200 = 10900 cu.ft. 

Prob. 1. What will be the receiver pressure and L.P. cut-off pressure for a 
cross-compound compressed air engine with 5 per cent clearance in each cylinder, run- 
ning on 100 lbs. per square mch gage mitial pressure and atmospheric exhaust, when 
the high-pressure cut-off is i, low-pressure f , high-pressure compression 15 per cent, 

low 25 per cent, and 5 = 1.4. Receiver volume is twice the high-pressure cylinder 
volume. 

Prob. 2. Find the superheated steam per hour necessary to supply a 14- and 21 x28- 
m. engine with 5 per cent clearance m each cylinder and a receiver twice' the aze 
of the high-pressure cylinder when the initials pressure is 125 lbs. per square inch gage, 
back pressure 7 lbs. per square inch absolute, speed 100 R.P.M., high-pressui^ 
cut-off i, low-pressure A, high-pressure compression 15 per cent, low pressure 40 per 
cent and s = 1.3. 

Note: B = .3. 

Prob. 3. If the high-pressure cut-off is changed to } without change of any other 
factor in the engine of Prob. 2, how will the horse-power, total steam per hour, 
and steam per horse-power per hour be affected? If it is changed to i ? 

Prob. 4. A boiler capable of supplying 5000 lbs. of steam per hour at rated load fur- 
nishes steam for a 12- and 18 x24-in. engine with 5 per cent clearance in each cyhnder and 
running at 125 R.P.M. The receiver is three times as large as the high-pressure cylinder, 



WORK OF PISTON ENGINES 



339 



the initial pressure 150 lbs. per square inch gage, back pressure 5 lbs per square inch 
absolute, the low-pressure cut-off fixed at i and low-pressure compression fixed at 
30 per cent. At what per cent of its capacity will boiler be working for these follow- 
ing cases, when »1.2 for all and 20% of the steam condenses during admission ? 
(a) high-pressure cut-off i, high-pressure compression 80 ) er cent, 
(6) high-pressure cut-off J, high-pressure compression 20 per cent, 
(c) high-pressure cut-off i, high-pressure compression 10 per cent. 
Note: 8 =.33; 




Fig, 99. — ^Work of Expansion in Compound Engine without Receiver and with Clearance. 
Qrcle XV, Logarithmic Expansion; Cycle XVI, Exponential, High-pressure Exhaust 
and Low-pressure Admission Coincident. 

19. Compound Engine without Receiver. Logariflunic Law, with Clear- 
ance and Compression, Cycle XV. General Relations between Pressures, 
Dimensions, and Work when H.P. Exhaust and L.P. Admission are Coinci- 
dent The graphical construction for this cycle has been described to some 
extent in connection with the first description of the cycle, given in Section 8, of 
this chapter, and is represented here by Fig. 99 in more detail. 

To show that the expansion from D to £ is the same as if volumes were 
measured from the axis ML, consider a point Y on DE, If the hypothesis 
18 correct 



P4XKM = FyX{KM+rK) 



(519) 



340 ENGINEERING THERMODYNAMICS 

The true volume when the piston is at the end D of the stroke, i^ 
2)^(1 +Cjj+i2cCi,), and at F, the true volume is 

where y is the fraction of the return stroke that has been completed in both 
cylinders when the point Y has been reached. Then 

Dividing though by (fie— 1), 

^This equation may be observed to be similar in form to Eq. (519). More- 
over, the last term within the bracket, D^y, is equal to the corresponding term 
Y'K, in Eq. (519), hence, 

\\-\-Ch-\-RcCl 



■\-RcCl'\ 



KM=DBy-^^£^'-^\ (52i: 



SimUarly, the distance QM, or equivalent volume at 1/ is 



QM 



=i)Jl±gt«^^l (522) 

The following quantities will be evaluated preparatory to writing the expres- 
sions for work: 

(rel.pr.)ir=(in.pr.)^p^; (523) 

(bk.pr.) 1+c.^ ^ 



(rel.pr.)jy (m.pr.; 



Zh+Ch 



SC 
(in.pr.)L = Pd' = (rel.pr.)/f^g; 



_-^. „x £H+Cff L (m.pr.) £b+cu I 

-^m.pr.j[^ j^^^ J Di,[1+Cb+RcCl] 

ZM+CH+Rc{Xr,+ci.)^^ 
= (in.pr.)— ^^^^^^-^ (^25) 

(cutoff pr.). = (m.pr.). ^-^ = (^-pr.). \y^^^ ^^^^^ (^^^.-ij^iHx^) J 



= (in.pr.) 



(m.pr.) 



.l+CH+ficCL+(fic-l)(l-X«). 



(52(.: 



WORK OF PISTON ENGINES 



341 



The ratio of expansion from E' to G is equal to 

(cut-off pr.)L _ 1 +cl 
(rel.pr.)L ~ (1 -Xh) +Cl 



• • • • • 



. • (527) 



Hence 



F^ = 144Dir(in.pr.) 



(Z.+c.)[l+lo«.^] 



(in.pr. ) 



Rc-l 



Zi^+Ci,+2ee(Xi,+Ci,)^''- P'-^ 



log. 



[' 



^-ch^RcCl+{Rc-1){X -Xn) 



1-\-ch-\'RcCl 



] 



(in.pr. ) 



\l-\-CH'\'RcCL-{-{fic-l){l-XH) 



t . IT \ 1 ^h-\-Xh 
(Ch+Xh) log* Ch 



ch 



(528) 



TrL=144Dx,(iii.pr.) 



Z.+c.+i2c(X.+c.)j^PI:] 



Rc-1 



loga 



14- C^+/2cCl+ (fie- 1) (1 -Z 



1+c^+ficC^ 



.)] 



+ 



(m.pr. ) 

11+Cb+RcClh+ (Rc- 1) (1 - Xb) J 

-144Dz,(bk.pr.) 



(l-fC.-X.)l0ge(-j^g^J 



l-ZL+(Xz.+Cz.)log,^^^^. . . (529) 



Cl 



The totoZ iDork found by adding TFh and Wl as given above, leads to the 
following: 



)F = 144i>^(in.pr.) 



(2^H+CH)l0g6 






+Z.+[zH+c.+fieCX.-fc.)^]loge [i±^ 



-|-CH+ficCi.+(ffc-l)(l -X 



+C/f-|-ficCL 



. 



+ 



ZH+CH'\-RciXL-^CL)$^'^^'^ 



1-hCH^RcCLHRc 



-i)(i-£) J [^^fi+^^-^^) X 



'*(RS3f;)-'-«'>'*Hf')]l 



-144Dx,(bk.pr.) j 1 -Xi,+(Xl+cl) log. ^ 






. . (530) 



342 



ENGINEERING THERMODYNAMICS 



This is the general expression for the work of the compound engine without 
receiver, with clearance and compression^ when high-pressure exhaust and low- 
pressure admission are simvUaneous and expansion and compression logartthmic, 
in terms of fundamental data regarding dimensions and valve periods. 

From this the usual expressions for mean eflfective pressure, work per 
cubic foot supplied, and consumption per hour i>er I.H.P., may be easily 
written, provided the supply volume is known. This is given by 



(Sup. Vol.) =^A'B=dJ(Zh+Ch) - fa+X^) fr"^.'^^P!''H 

L Cm.pr.) J 



=Dh Zh+Ch-(ch+Xh) 



Zh+Ch+Rc(Xl+cl) 



(bk.pr.) 
(in.pr. ) 



1+Ch+RcCl+(Rc-IKI-Xh)] 



(631) 



To find the conditions which must be fulfilled to give equal work in the two 
cylinders, equate Eqs. (528) and (529). 



^.+(Z.+c.)log.(^) 



ft< 



fi< 



:^[ 



Zn-^CH+RciXL-^CL) 



(bk 
( 



Zh -{-Ch +RciXL -f cl)7t-^4 

{ in.pr.; 



?^i log, r^ 



-hCM-\-RcCLHRc-l)(l-X 



I-^ch-^-RcCl 



'] 



ll+CH-^RcCLHRc-mi-XH)] 



Rc{l-\-CL-XH)l0gfi 



\1+Cl-XhI 



+{chXh) logs 



ch-\-Xh 

Ch 



+«c 



(bk.p r.) 
( in.pr.) 



1 -Xl + {Xl+cl) log«(^^^) I =0. (532) 



These expressions are perfectly general for this cycle, and expressed in 
terms of fundamental data, but are so complicated that their use is ver}^ 
limited, as in the case of some of the general expressions previously derived 
for other cycles. 

As in other cycles, it is desirable to investigate a special case, that of 
complete expansion and compression in both cylinders, Fig. 100. First it is 
necessary to determine what are regarded as fundamental data in this case, and 
then to evaluate secondary quantities in terms of these quantities. The following 
items are assumed to be known: (in.pr.), (bk.pr.), which is equal to (rel.pr.)2^, 
Rcy Ch, and Cl, and Dy, which are dimensions, and it is known that the 
pressure at the end of compression in L.P. is equal to (rel.pr.)^. 

Referring to the diagram, displacements, clearances, and the axis for the 
common expansion, ML, can all be laid out, and the location of the points 
A and G determined. 

The points E and E' are at the end of the common expansion within the 
two cylinders, and at beginning of high-pressure compression and of separate 
low-pressure expansion, hence p«=p«'. 



WORK OF PISTON ENGINES 



343 



From the points A and E: 



p.=(in.pr.)y-j— 



From the points G and B', 



p«'=p«=(bk.pr.) 



1+Ci 



I+Cl-Zj^' 




M V 



Fig. 100. — Special Case of Cycles XV and XVI. Complete Expansion and Compression in 
both Cylinders of Compound Engine without Receiver and with Clearance High- 
Pressure Exhaust and Low-Pressure Admission Coincident 



and equating, 



1+Ci 






whence 



where 



Xh = Ch 



1+Cl + CuRp ' 



(533) 



Ri 



_ / in.pr. \ 
\bk.pr./ 



Substituting the value of X^ in either of the expressions for pet which is 
the low-pressure cut-oflf pressure, 

+ Cl + RpCh1 ^ /KOA\ 



(cut-oflf pr.)L=Ptf = (bk.pr.) — 

L A 



344 ENGINEERINa THERMODYNAMICS 

It may be noted here that the cylinder ratio does not enter into this, but 
only clearances and pressures. In the no-clearance case, it may be remembered 
that the point E or -E' was not present, as it coincided with G, 

Next, to find the high-pressure release pressure, pa, by means of points £ 
and D, and their relation to the axis ML, Fig. 100. 

l+CH+RcCL-i 
+Cl + CuRc 

1+Cl+Ch 



(rel.pr.)^ ^Pd^p, =^ = (bk.pr.) I -^ 



1''Xh+ 



Re- 1 



I+cu+RcCl 
Rc-l' 



= (bk.pr.) i^,^^jg^^ (53oi 

Knowing the release pressure of the high-pressure cylinder, it is possible to 
find the high-pressure cut-off and compression necessary to give the required 
performance. 

7 _/i , ^ .(relprO^ ^ _(1+Ch)[ Rc(1+Cl)+RpCh ^ ^ ,„.^ 

Y ^ (rel.pr.)H ^ ^ [Rc(1+cl)+RpCh J ,„.. 

^^='"^7bk:^y~"^=^H"T4^^+ft^^"^J- • • • (^'^ 

The work of the two cylinders is as follows: 

w i^n fir. «, ^ f (1+c^) r ^c(l+Ci.)+fii 'Cw1 f, , . „ Rpjl+cg+RcC j.)'] ^ 
TFH=144DH(m.pr.) |-^ [ l+CK+fieC.-JL^+'^«'i2o(l+c.)+22.^.J -'" 

_ n+CH+RcCL ] \ RcO.+Ci,)-\-RpCa ] J_ , \ Rcil+CL)+RpCB ] r 1+Cx+ Cjr 1 
L fic-1 JL 1+Cir+iecCt J/ip^L l+CB+RcCt. \ll+Ci.+RpCa\ 

_ i4An rt„ «, ^ [ (l+<'g) fic(l+Ci,)+gpCg r, , Rp O.+Ch-\-RcCl '\ ^ 
L i?p(i?c-l) J ^'L l+to+ficCt \\\+Cl-^RpCh\ 



Tr.=144Dx(bk.pr.) 1 1^ ^^3j-^ J log. |^-j^-^^-__J |^- 



«<7(l+&)+fii<ill 






WORK OF PISTON ENGINES 345 

These expressions, when added and simplified, give the following for total 
work per cycle, 

W^lW)^{hk,pT.)ll^Cr.[-f^^^^ . (540) 

in which of course DhRc may be used instead of Dl and —^— instead of 

ftp 

(bk.pr.) and then 

TF = 144Z)^Zj^(in.pr.)loge/ep, . ... (541) 

Zjt, having the value of Eq. (536). 

IJquality of work the in high- and low-pressure cylinders results, if W 
Eq. (538) equals W^, Eq. (539), or if 

2W„=W, or 2Fl = F, 

all of which lead to equivalent expressions. Simplification of these expres- 
sions, however, does not lead to any direct solution, and hence the equations 
will not be given here. 

Example. Find (a) the horse-power and (6) steam used per hour for a 12- and 18 x24-in. 
tandem compound engine with no receiver, 6 per cent clearance in the high-pressure 
cylinder, and 4 per cent in the low, when the initial pressure is 150 lbs. per square inch 
absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., high- 
pressure cut-off J, high-pressure compression 15 per cent and low-pressure compression 
is complete. 

(a) Since the low-pressure compression is complete, the pressure at end of compression 
must be equal to the release pressure of the high. This latter quantity may be found 
rom the relation (in.pr.)(ZH-f-c^)=(rel.pr.)H(l +Cii), or 

(rel.pr.)^ = 150 T^- =79.3 lbs. per sq.in. absolute. 

l.Oo 

Low-pressure compression may be found from the relation (rel.pr.)^(cL) = (bk.pr.) 
(cl-{-Xl)j or Xi, = .28. (m.e.p.) may be found from Eq. (540) divided by 144DL,which 
on substitution gives 



150 
2.25 



/ 



1 08 

(.5 -f .06) log, i^ 



+ 



K^\ K^na^ooKfOQ^i^A^ ^^li r i-i-0^+2.25X .04-H.25X.85l 
.5+ |^.5-K.06-h2.25(.28+.04) -Jloge [ i^.06-H2:25x':04" J 



.5 -f .06 +2.25(.28 + .04) ^ 
1-f. 06+2.25 X.04-|-1.75-f-.85 

.06 +.15 



2.25(L04-.15)loge-^-^^ 



(.06+.15)loge 



-10 



.06 
and the horse-power will be 271, 



1.28 +(.28 +.04) loge — tir \ =69.7. lbs. sq. in. 

.04 



346 ENGINEERINa THERMODYNAMICS 

(b) Since the consumption in cubic feet per hour per horse-power is equal to 

13,750 Sup.Vol. 



(m.e.p.) Dl 
and supply volume is given by Eq. (531), this becomes 



13,750 1 
X: 



69.7 2.25 



.56+2.25(.32)^ 
.56-.2ir ^^ 



1.06+.09-I-1.25X.85, 



=44, 



hence the consumption per hour will be 

44 X271 X.332 =4000 pounds. 

Prob. 1. A Vauclain compound locomotive has cylinders 18 and 30x42 ins., with 5 
pec cent clearance in each and runs on a boiler pressure of 175 lbs. per square inch gage 
and atmospheric exhaust. The steam pressure may be varied as may also the cut-ofif to a 
limited degree. For a speed of 200 R.P.M., a cut-off f and 10 per cent compression 
in each cylinder, find how the horse-power will vary with the initial pressures of 175, 
150, 125, and 100 lbs. gage. 

Prob. 2. When the cut-off is reduced to i in the above engine compression 
increases in tlie high-pressure cylinder to 20 per cent. For the case of 175 lbs. gage 
initial pressure find the change in horse-power. 

Prob. 3. Find the steam used by the engine per hour for the first case of 
Prob. 1 and for Prob. 2. 

Prob, 4. It is desired to run a 12- and 18x24-in. no-receiver engine with 5 per cent 
clearance in each cylinder, under the best possible hypothetical economy conditions for an 
initial pressure of 200 lbs. per square inch absolute and atmospheric exhaust. To 
give what cut-off and compression must the valves be set and what horse-power 
wUl result for 100 R.P.M.? 

20. Compound Engine Without Receiver. Exponential Law, with Clear- 
ance and Compression, Cycle XVI. General Relations between Pressures, 
Dimensions, and Work, when H.P. Exhaust and L.P. Admission are Coin- 
cident. Again referring to Fig. 99, it may be observed that reasoning 
similar to that in Section 19 but using the exponential law, would show that 
the same formulas and graphical constructions will serve to locate the 
axes of the diagram, hence, as before, 

KM=Dh ^-j , (542) 

and 

QM=D^±f±^ (543) 



Release pressure in the high-pressure cylinder is 



(rel.pr.)H = (in.pr.)f-^-^j (544) 



WORK OF PISTON ENGINES 347 

Immediately after release the pressure is equalized in the high-pressure 
cylinder and the low-pressure clearance. The pressure after equalization, 
«rmed (iii.pr.)i;i is found by the relation of the volume at S and that at D\ 
Pig. 99, measured from the axis KT. 



(in.pr.)L = (rel.pr.)^ 



'+-+«'(-+^')(s^)' 



which, by means of Eq. (544) becomes 



« 



1 ^ « 
(m.pr.)^=(bk.pr.)[— ^:p-^-;;:p^^;^^ J (545) 

The expansion of the fluid goes on as it passes from the high^pressure 
cylinder to the greater volume in the low-pressure, as indicated .by. liD'JS' 
and DEy and when the communicating valve closes, the pressure has become 



(cut-oflf pr.)L = (in.pr.)L 



Rc-l 



l+C^+ficCL_^(j_^^) 



Rc-l 



which, by means of Eq. (545) reduces to 

1 



(ou.^ pr.).-(bk.pr.)[ , «;'^^^;> ±'^^g4 ]-. . . (546) 

After cut-oflf in the low pressure, expansion goes on in that cylinder alone 
to the end of the stroke, when release occurs at a pressure 

(rel.pr.)L= (cut-oflf pr.)z,(?^±^^y 

or by substitution from equation (546), 

(rel.pr.).-(bk.pr.)[ i+c«+i2^,+(i— jQ(^,;- i) ( l+c, )J " (^47) 

In terms of these quantities the work of the high- and low-pressure 
cylinders can be written out as follows: 

TTh = 144D« { (in.pr.)'-^ [a- (^--±£i?)' "] -c«(in.pr.) 
(bk 



:.pr.) r g«p(cg+Z«)+gc(cf,+.Yx.) ]'r / \+ch+RcCl V-H 

-1 L fic-1 JL \l+c„+RcCL+0-X„){Rc-l)) J 



1 



-^<-«4rS^^"^Sf«!^)]TC-^)-'-'] j (-) 



348 ENGINEERING THERMODYNAMICS 

and 



FFL=144i>z, 



1 • 

(bk. 



8 



-1 L Rc-i J 



^ L \l+c,,+i2c7C/^-f-(l-X)(l2c-l)y J 



1 



, (bk.pr.) ,, , ^ ^ J R'p(ch+Zh)+Rc{ cl+Xl) ]'L / 1+C£,-X„ \'-M 
+ ,_1 ^^+<^^-^")[i+c„+RcCj.+ (l^X^){Rc-T)\ L^"V 1+c^ / J 

_^k£i^.J:Z.)|^^ (549) 

These are general expressions for work of high- and low-pressure cylinders 
for this cycle, and from them may be obtained the total work of the cycle, 
mean elBfective pressure referred to the low-pressure cylinder, and by equating 
them may be obtained the relation which must exist between dimensions, 
events, and pressures to give equal division of work. It would, however, 
be of no advantage to state these in full here, as they can be obtained from 
the above when needed. 

The supply volume, cubic feet per cycle, is represented by A'B, Fig. 99, 
and its value is found by referring to points B and E as follows: 

iSnp.Yol)^D„[(c.+Z.)-{c„+X,)(^^^^f]' 

=z)«^c«+Zh -^—\i+cs+r^+{i-x^){r:^T)}\- ^^^ 

Rs p 

Work per cubic foot supplied is foimd from Eqs. (548), (549), and (550). 

Work per cu.ft. suppUed = -r^^-^y^p:. (551) 

Consiunption, cubic feet per hour per I.H.P., is found from mean elBfective 
pressure referred to L.P. cyl. and supply volume as follows: 

Consiunption, cu.ft. per hr. per I.H.P. 

13,7.50 (Sup.Vol.) ^552) 

(m.e.p. ref. to L.P.) Dl 

This will give pounds consumption by introducing the factor of density. 
Further than this, it will be found more practicable to use graphical 
methods instead of computations with this cycle. 



WORK OF PISTON ENGINES 349 

Bzample. Find (a) the horse-power and (6) consumption of a 12- and 18x24-in. 
no-receiver engine having 6 per cent clearance in the high pressure cyUnder and 4 per 
cent in the low when the initial pressure is 150 lbs. per square inch absolute, back 
pressure 10 lbs. per square inch abzolute, speed 125 R.P.M., high-pressure cut-off i, 
high-pressure compression 15 per cent, and low-pressure compresaion is complete. 

(a) The per cent of low-pressure compression may be found as in the Example of 
Section 19, using the value of « in this case of 1.4. Then 

(in.pr.)(ZH+c/f)** « (rel.pr.)^ (1 -\-ch)^'\ 
or 

(rel.pr.)j5r=61.5 lbs. sq. inch absolute, 
and 

(rel.pr.)^ Xcl^* = (bk.pr.) (clH-Xz,)^*, 
or 

Zl = .11 

From the sum of Eqs. (548) and (549) divided by 144Dx, and with proper 

values substituted, (m.e.p.) =48.5 lbs.; hence the horse-power is 189. 

13 750 
(6) From Eqs. (550) the value for (Sup. Vol.), which when multiplied by — ■' — , 

m.e.p. 

and divided by Dl, gives cubic feet air per hour per I.H.P. 

I?i750 y J_r nfi_. ^ _ ^21 15-7(.56)+2 .25(.15) 1 « .63 cu.ft. per 

48,5 ^2.25L "^ (15)-^ H-.06+2.25x.04+(l -.15)(2.25-1)J hourper I.H.P. 

Prob, 1. If the locomotive of Prob. 1, Section 19, should be equipped with superheater 
so that the steam expanded in such a way that 8 = 1.3, what would be the effect upon 
the horsu-power for conditions of that problem and on the cylinder event pressures? 

Prob. 2. A 30- and 42 x54-in. no receiver steam pumping engine runs at 30 R.P.M. 
and has 3 per cent clearance in the high-pressure cylinder and 2 per cent in low. There 
is no compression in either cylinder. Initial pres ure is 120 lbs. per square inch gage, 
and back pressure 28 ins. of mercury (barometer reading 30 ins.). The steam is such 
that the expansion exponent is 1.25. What will be the horse-power of, and the steam 
used by the engine when the cut-off in the high is i? 

Prob. 3. By how much would the power change if the cut-off were shortened to | 
and then to i, and what would be the effect of these changes on the economy? 

21. Triple-Expansion Engine with Lafinite Receiver. Logarithmic Law. 
No Clearance, Cycle XVU. General Relations between Pressures, Dimen- 
sions and Work. Fig. 101 represents the cycle of the triple-expansion engine 
with infinite receiver, no clearance, showing one case of incomplete expansion 
in all cylinders, and another where overexpansion takes place in all cylinders. 



350 



ENGINEERING THERMODYNAMICS 



The reasoning which follows applies equally well to either case, and to any 
combination of under or overexpansion in the respective cylinders. 

It is desired to express the work of the respective cylinders and the 
total work in terms of dimensions, initial and back pressures, and the cut-offs 
of the respective cylinders. To do this, it is convenient first to express the 




bk.K-) 



INCOMPLETE EXPANSION 



OVER EXPANSION 



CYCLE XVll 



Fig. 101. — Work of Expansive Fluid in Triple-Expansion Engine with Infinite Receiver and 

Zero Clearance. Cycle XVII, Logarithmic Expansion. 

first receiver pressure (1st rec.pr.) and second receiver pressure (2d rec.pr.) in 
terms of these quantities. The subscript I refers to the intermediate cylinder. 



or 



and 



/I i. \ /■ \ZhDh 

(1st rec.pr.) = (m.pr.)-^-p:— , 






(553) 



or 



(2d rec.pr.) = (in.pr.) 



Z„Dh 
ZlDl' 



• • • 



. . (654) 



WORK OF PISTON ENGINES 



Work of high-pressure cylinder is 



= 144(in.pr.)Z)i, | Z« (l +log. ^) - ^^ 



Work of intermediate cylinder is 



= 144(in.pr.)i)/f | Z^^l +log. ^) - 
- lU(m.pr.)DH I Za^l+log. ^) - 



DhZJ)i, 



Z„D,\ 
ZlDl 



Work of low-pressure cyUnder is 



The total work by addition is 



W = 144(in.pr.)Z)HZ;, I (l +I0& ^) + (l +log. ~j + U -hlog. ^\ 



= 144(in.pr.)Z)aZjif 



3+loge 



D 



H 



D: ] 



ZhZiZj^ Z,Dt ZJ)i 



351 



(555) 



(556) 



= 144(in.pr.)i)i,Z«U+lo&^\-144(bk.pr.)Z)i,. . . 



(557) 



-144(bk.pr.)Z)i. . (558) 



Mean effective pressure referred to the low-pressure cylinder is found by 
dividing W by 144D/,, and is therefore 



(m.e.p. ref. to L.P.) 



Z)i 



= (in-pr. )Zh^ 



3-hlog. 



1 



D 



H 



^1 

ZuZiZt. ZiDi ZlDl 



-(bk.pr.). . (559) 



352 ENGINEERING THERMODYNAMICS 

Work done per cubic foot supplied is equal to W divided by the supply 
volume AB or ZhDu, 

Work per cu.ft. supplied 



= 144(in.pr.) 



3+I0&: 



ZhZjZl ZiDj ZlDl 



► -144(bk.pr.)^^. (560) 



The volume of fluid supplied per hour per indicated horse-power is 

Consumption, cu.ft. per hr. per I.H.P. 

13,750 ZhBh 



(m.e.p. ref. to L.P.) Dl 



(561) 



The wdqhi of fluid used per hour hour per indicated horse-power is of 
course found by multiplying this volume Eq. (561) by the density of the fluid 
used. 

The conditions which will provide for eqiwl division of work between the 
three cylinders may be expressed in the following ways: 

which is equivalent to, first: 

WH=Wly 



or 



hence 



, 1 1/Dh\ , 1 l/Dj\ 

''^T,-zXDj)=^''^zrYAD-j' 

^k-m)-m) <«») 



Similarly from Wh = Wl, 

^ Zh Zi \Di I (m.pr.) \DhI Zi ^ 

These two equations, (562) and (563), show the necessary relations between 

in order that work shall be equally divided. Since there are six independent 
quantities entering (as above) and only two equations, there must be jorar 
of these quantities fixed by conditions of the problem, in order that the other 



WORK OF PISTON ENGINES 353 

two may be found. For instance, if the cylinder ratios, the pressure ratio, 
and one cut-off are known, the other two cut-offs may be found, though the 
solution is difficult. 

Again, if cutroffs are equal, and the ratio of initial to back pressure is known, 
it is possible to find the cylinder ratios. This forms a special case which is of 
sufficient importance to require investigation. 

If ZH=Zj^ZLy Eq. (562) becomes 



and Eq. (563) reduces to 



/>/ 


Bl 


Da 


Dj' ' 


D,Dl 
D^ 


(in.pr.) 
" (bk.pr.)' 



(564) 



(565) 



but from Eq. (564), 



Di. Di^ ^Di. ^ (DtY 
Dj Dm Dm \Dm) ' 



and therefore 



DJhjDA 
D^ \Dm) 



3 



and 



%'wA^y <^) 



which, along with the condition assumed that 

Zh^Zj^Zl (567) 

constitute one set of conditions that mil make work equal in the three cylinders 
This is not an uncommon method of design, since by merely maintaining 
equal cut-offs, the work division may be kept equal. 

The work done in any one cylinder imder these conditions Eqs. (566) 

and (567) is then 

Tf^^}F, = Tr^ = 144(in.pr.)D^{z(l-hloge|)-(^^)*} . (568) 
and the total work 

ir=432(m.pr.)Z)„{z(l+logc|)-(^-) I (569) 

in which Z represents the cut-off in each cylinder, all bemg equal. 



354 



ENGINEERING THERMODYNAMICS 



A special case of the triple-expansion engine with infinite receiver and no 
clearance which demands attention is that of complete expansion in all 
cylinders, represented by Fig. 102. Here 

^.=^=#^; ....... (570) 

Dl (2drec.pr.) 



^ ^Dg^ (2d rec.pr.) , 
' Di (1st rec.pr.)' 




(571) 



Fig. 102.— Special Case of Cycle XVIII Complete Expansion in Triple-expanaon Engine 

with Infinite Receiver, Zero Clearance, Logarithmic Expansion. 



and 



7 _ (1st rec.pr.) _ / bk.pr. X D^ ^^^^2) 

^ (in.pr.) \in.pr. /Z)/f' 



hence the receiver pressures are as follows: 

(1st rec.pr.) = (bk.pr.)yp, 



(573) 



WORK OF PISTON ENGINES 355 

and 

(2drec.pr.)=(bk.pr.)^ (574) 

The work of the respective cylinders, expressed in terms of initial and 
back pressures and displacements is then, 



-l«(bk.pr.)C.l<*(|g^|«) (675) 

Similarly, 

Wj = lUihk.pr.)DL\og,(^^, (576) 



and 



iri=144(bk.pr.)Z)iloge^ (577) 



Total work, by addition, is 

W. 144(bk.pr.)i,4l* ({H|j^ ^^)+lo^ l+log. f ] 

- 144(bk.pr.)Di log, |j^ (578) 

If for this special case of complete expansion equality of work is to be 
obtained, then from Eqs. (575), (576), and (577), 

(inj)r 0^ Dh^D^^Dl ,--q. 

"(bk.pr.)i)L Dh ,D/ ^""'^^ 

wfiich is readily seen to be the same result as was obtained when all cut-offs 
were equalized, Eqs. (564) and (565). This case of complete expansion and 
equal work in all cylinders is a special case of that previously discussed where 

cut-offs are made equal. Hence for this case cut-offs are equal, 

* 

„ „ „ Bn Di Dl (bk.pr.) . . 

Dl Dl Dh (m*prO 

Bzample. A triple-expansion engine 12- and 18- and 27x24-ins., with infinite 
receiver and no clearance, runs at 125 R.P.M. on an initial pressure of 150 lbs. 
per square inch absolute, and a back pressure 10 lbs. per square inch absolute. 



356 ENGINEERING THERMODYNAMICS 

If the cut-offs in the different cylinders, beginning with the high, are i, i, and i, 
what will be (a) the horse-power, (6) steam consumed per hour, (c) release and 
receiver pressures? 
(a) From Eq. (669) 



= 150X.5X- 



1 f. - ... 8 8 



5.06 
hence 



3 +log, 14.2 -T-;rrr -^ « «, \ -10 »39 lbs. pei sq.iii, 
^ 3X2.26 3X2.25' 



}- 



39X2X573X250 
^•^•*^' 33;000 ^^ 

(6) From Eq. (561). 

Cubic feet per hoise-power per hour= , — - — ^v -~-^' 

(m.e.p.j Dl 

13,750 1 

X .o X - -^yj ^34«9| 



39 5.06 

hence total pounds per hour will be, 

34.9 X.338X.332 -3920. * 



(c) From Eq. (553) 



1st (rec.pr.)«(in.pr.) 



T>hZh 
ZiDi' 



5 

150 X o>Tg o ;7? = 89 lbs. per Bq.in absolute. 
•o7oX^u&5 



From Eq. (554) 



(rec.pr.) =(m.pr.)^-^, 



.5 
= 160 X »^, ' , ^^ =3.75. per sq.in absolute. 
.375x5.06 

High-pressure release pressure may be found from relation (in.pr.)Zfii)ir 
= (peI.pr.)£fZ)/^ .'. (reLpr.)/f =75 lbs. Similarly 1st (rec.pr .)ZiDi = (reLpr.)/!)/, or 
(rel.pr.)/=33.4. Similarly 2d (rec.pr.)Zi/)i, = (rel.pr.)ii>L, or (rel.pr.)z, = 14.8. 

Prob. 1. What would be the horse-power and steam usfed per hour by a 10- and 

16- and 25x20-in. infinite receiver, no-clearance engine, running at 185 R.P.M. 

on an initial pressure of 180 lbs. per square inch gage and atmospheric exhaust. 

Cut-offs .4, .35^ and .3. 

Prob. 2. The following data are reported for a test of a triple engine: 

Size 20 x33 X52 x42 ins., speed 93 R.P.M., initial pessure 200 lbs. per square inch 

gage, back pressure one atmosphere, H.P. cut-off ,5, horse-power 1600, steam per haore. 



WORK OF PISTON ENGINES 357 

power per hour 17 lbs. Check these results, using cut-oflfs in other cylinders to give 
approximately even work distribution. 

Prob. 3. What change in cyhnder sizes would have to be made in the above engine 
to have equal work with a cut-off of J in each cylinder, keeping the high pressiu-e the 
same size as before? 

Prob. 4. What would be the horse-power of a triple-expansion engine whose low- 
pressure cylinder was 36x3 ins., when running on 150 lbs. per square inch absolute 
initial pressure and 10 lbs. per square inch absolute back pressure, with a cut-off in 
each cylinder of .4 and equal work distribution? Make necessary assumptions. 

Prob. 6. A triple engine 18x24x36x30 ins., running at 100 R. P.M. on an initial 
pressure of 200 lbs. per square inch absolute and back pressure of 20 lbs. per 
square inch absolute, is to be run at such cut-offs as will give complete expansion in 
all cylinders. What will these be, what receiver pressures will result, what horse- 
power can be produced under these conditions, and how much steam will be needed 
per hour? 

Note: 8 for 200 lbs. -.437. 

22. Multiple-Expansion Engine. General Case. Any Relation between 
Cylinder and Receiver. Determination of Pressure Volume-Diagram and 
Work, by Graphic Methods. It is possible to arrange multiple-expansion 
engines in an almost infinite variety of ways with respect to the pressure-volume 
changes of the fluid that take place in their cylinders and receivers. There 
may be two or three cylinder compounds of equal or unequal strokes, pistons 
moving together by connection to one piston rod, or separate piston rods 
with a common cross-head or even with completely independent main parts 
and cranks at 0**, 180**, displaced with either one leading, or the pistons may 
not move together, being connected to separate cranks at any angle apart, 
and any order of lead. Moreover, there may be receivers of large or small 
size, and there may be as a consequence almost any relation between H.P. 
discharge to receiver and low-pressiu-e receipt from it, any amount of fluid 
passing to correspond to engine load demands and consequently any relation 
of cut-offs, compressions, and receiver-pressure fluctuations. Triple and 
quadruple engines oflfer even greater varieties of combination of related factors, 
so that problems of practical value cannot be solved by anal3rtical methods 
with anything like the same ease as is possible by graphic means, and in 
some cases not at all. 

These problems that demand solution are of two classes. 

1. To find the work distribution and total work for cylinders of given 
dimensions, clearances, receiver voliunes and mechanical connection or move- 
ment relation, with given initial and back pressures, and given valve gear at 
any setting of that valve gear or at a variety of settings. 

2. To find the cylinder relations to give any proportion of the total work 
in any cylinder at any given valve setting or any fraction of initial pressure 
or any value of release pressure or total number of expansions. 

The essential differences between these two classes of problems is that in the 
first the cylinder dimensions are given, while in the second they are to be found. 



358 ENGINEERING THERMODYNAMICS 

In general,* however, the same methods will do for both with merely a 
change in the order, and in what follows the dimensions of cylinders, 
valve periods, receiver volimie, initial and back pressures will be assumed 
and the diagrams found. By working to scale these diagrams will give the 
work by evaluation of their area, by means either of cross-section paper directly, 
on which strips can be measured and added, or by the planimeter. Thus 
will high- and low-pressure work be evaluated through the foot-pound equiva- 
lent per square inch of diagram, and the total work or the equivalent mean 
eflfective pressure found by the methods of mean ordinates referred to the 
pressure scale of ordinates. 

In the finding of the pressure-volume diagram point by point there is but one 
common principle to be applied, and that is that for a given mass the product 
of pressure and volume is to be taken as constant (for nearly all steam prob- 
lems, which is the almost sole appUcation of this work) and when two masses 
come together at originally different pressures and mix, the prodiict of the remlt- 
ing pressure and the new volume, is equal to the sum of the PV products of the 
two parts before mixture. At the beginning of operations in the high-pressure 
cylinder, a known volmne of steam is admitted at a given pressure and its 
pressure and volume are easily traced up to the/ time when it communicates 
with 'the receiver in which the pressure is unknown, and there difficulty is 
encoimtered, but this can be met by working from the other end of the 
series of processes. The low-pressure cylinder, having a known compression 
volume at the back pressure, there will be in it at the time of opening to 
receiver a known voliune, its clearance, at a known compression pressure. 
The resulting receiver pressure will then be that for the mixture. These two 
receiver pressures are not equal ordinarily, but are related by various compres- 
sions and expansions, involving high- and low-pressure cylinder partial 
displacements, grouped with receiver volumes in various ways. 

Take for an illustrative example the case of a two-cylinder, single-acting, 
cross-compoimd engine with slide valves, cylinder diameters 12f and 20 ins. 
with 24 ins. stroke for both. High-pressure clearance is 10 per cent, low- 
pressure clearance 8 per cent. Receiver volume 4000 cu.ins. High-pressure 
crank following by 90**. Find the mean effective pressure for the high- and low- 
pressure cylinders, for a cut-off of 50 per cent in the high, and 60 per cent 
in the low, a compression of 10 per cent in the high and 20 per cent in the low, 
itiitial pressure 105 lbs. per square inch gage, back pressure 5 lbs. per 
square inch absolute, expansion according to logarithmic law. 

On a horizontal line SZ, Fig. 103, lay off the distances 

r[7= low-pressure cylinder displacement volume in cubic inches to scale. 
[77= low-pressure cylinder clearance volume in cubic inches to scale, 
ypl/'ss receiver voliune in cubic inches to scale. 

TFX= high-pressure cylinder clearance volume in cubic inches to scale. 
XF= high-pressure cylinder displacement volume in cubic inches to scale. 



WORK OF PISTON ENGINES 



359 



Through these points draw verticals produced above and below, f'T"^ 
V'U", Y'Y", WW", X'X'', Tr\ Then will WW and WZ be PV 
coordinates for the high pressure diagram in the quadrant WWZ, and V'V 




Fig. 103. — Graphical Solution of Compound Engine with Finite Receiver and with Clearance 
Illustrating General Method of Procedure for any Multiple Expansion Engine. 

and VS the PV coordinates for the low pressure diagram in the reversed 
quadrant V'VS. 

Lay o£F AB to represent the high-pressure admission at a height XA rep- 



360 ENGINEERING THERMODYNAMICS 

resenting absolute initial pressure; lay off LM at a height TL representing 
low-pressure exhaust at a constant absolute back pressure to the same scale. 

Locate point B at the cut-off point AB^,50XY on the initial pressure 
line, and drop a vertical BB^ and draw similar verticals JJ^, GG^, MAP, at 
suitable fractional displacements to represent L.P. cut-off, H.P. and L.P. 
compression volumes resprectively. 

This operation will fix two other points besides the points A and L, B the 
H.P. cut-off at the initial pressure and M the low-pressure compression at the 
back pressure. Through the former draw an expansion line BC and through 
the latter a compression line MN, locating two more points, C and N, at the 
end of the outstroke of the high and instroke of the low. 

At point C the H.P. cylinder steam releases to the receiver of imknown 
pressure, and at N, the L.P. cylinder steam is opened to both the receiver and 
high-pressure cylinder at unknown pressure and volume. 

To properly locate these pressures and volumes from the previously known 
pressures and volumes in a simple manner, the construction below the line 
SZ is used. 

Lay off on WW" the high-pressure crank angles 0-360°, and to the right 
of each lay off from the clearance line XX'* the displacement of the piston at the 
various crank angles for the proper rod to crank ratio, locating ^the curve 
A^B^C^E^F^Om^. This is facilitated by Table XIII at the end of the 
chapter, but may be laid out graphically by drawing the crank circle and 
sweeping arcs with the connecting rod as radius. 

Opposite H.P. crank angle 270° locate L.P. crank angle 0°=360° and 
draw to left of the low-pressure clearance line UU" the crank angle dis- 
placement cmrve for that piston. 

It will be noted that steam volumes are given in the lower diagram by the 
distances from either of these curves toward the other as far as circum- 
stances call for open valves. Thus H.P. cylinder volumes are distances from 
the H.P. displacement curve to WW", but when H.P. cylinder is in communi- 
cation with receiver, the volume of fluid is the distance from H.P. displacement 
curve to VV", and when H.P. cylmder, receiver and L.P. cylinder are all 
three in communication the volimie is given by the distance from H.P. 
displacement curve to L.P. displacement curve. This pair of displacement 
curves located one with respect to the other as called for by the crank angle 
relations, which may be made to correspond to any other angular relation, 
by sliding the low up or down with respect to high-pressure curve, serve 
as an easy means of finding and indicating the volumes of fluid occupying 
any of the spaces that it may fill at any point of either stroke. 

On each curve locate the points corresponding to valve periods by the 
intersection of the curve with verticals to the upper diagram, such as BB^. 
These points being located, the whole operation can be easily traced. 

At H.P. cut-off (fi) the volume of steam is B^B^. During H.P. expansion 
(B to C) steam in the high increases in volume from B^B^ to C^C^. 

During H.P. release (C to D) the volume of steam in the high C'C'^ is 



WORK OF PISTON ENGINES 361 

idded to the receiver volume C^C, making the total volume C^C^. During 
H.P. exhaust (D to E) the steam volume C^C^ in H.P. and rec. is com- 
pressed to volume PE^. 

At L.P. admission (AT) in low and (E) in high, the volume PP is added, 
making the total volume PE^ in high, low, and receiver. 

During (/ to Q) in low and (F to G) in high the volume PE^ in high, low 
and receiver, changes volume until it becomes Q^G^ in high, low, and receiver. 

At H.P. compression, G in the high, the steam divides to Q^G^ in low 
and receiver, while G^G^ remains in high and is compressed to H^H^, at the 
beginning of admission in the high. The former volume Q^G^, in low and 
receiver, expands to PJ^, at the moment cut-off occurs in the low, which 
divides the volume into, PJ^ in receiver, which remains at constant volume 
bill high-pressure release, and the second part, PP in the low, which expands 
In that cylinder to K^K^. 

After low-pressure release the volume in low decreases from K^K^ to 
M^M^f when the exhaust valve closes and low-pressure compression begins. 

During compression in low, the volume decreases from M^M^ to PP which 
is the volimie first spoken of above, which combines with PE^j causing the drop 
in the high-pressure diagram from E to F. 

The effects upon pressures, of the various mixings at constant volume 
between high, low, and receiver steam and the intermediate common expan- 
sions and compressions may be set down as follows: 

At C, steam in high, at pressure Pe, mixes with steam in receiver at 
pressure Py, resulting in high and receiver volume at pressure Pd. 

From D to E there is compression in high and receiver resulting in 
pressure P«. 

At E steam in high and receiver at pressure P« mixes with steam in low, 
at P», locating points / in low and F in high at same pressure. 

From (P to G in high) and (/ to Q in low) there is a common compression- 
mansion in high, low, and receiver, the pressures varying inversely as the total 
rolume measured between the two displacement curves. At G in there begins 
compression in high alone to H. 

In the low and receiver from Q to J there is an expansion and consequent 
fall in pressure from Pq to Pj. 

After low-pressure cut-off at J the expansion takes place in low-pressure 
cylinder alone, to pressure P*, when release allows pressure to fall (or rise) 
to exhaust pressure Pi. 

When cut-off in low occurs at J the volume PP is separated off in the 
receiver, where it remains at constant pressure P^ imtil high-pressure release 
it point C. 

At the point M compression in low begins, increasing the pressure in low 
ilone from Pm to Pn- 

There are, it appears, plenty of relations between the various inter- 
nediate and common points, but not enough to fix them unless one be first 
established. One way of securing a starting point is to assume a compression 



362 ENGINEERING THERMODYNAMICS 

pressure Pg for the begiiming of H.P. compression and draw a compre^ion 
line HG through it, produced to some pressure line a/, cutting low-presure 
compression line at d. Then the H.P, intercept {e-f) must be equal to the 
low-pressure intercept (d-c); this fixes (c) through which a Py = const, line 
intersects the L.P. cut-off voliune at J. 

Now knowing by this approximation the pressure at J, the pressure may 
by foimd at D, E, f , and at G, The pressure now found at G may differ 
considerably from that assumed for the point. If so, a new assiuuption for 
the pressure at G may be made, based upon the last figure obtained, and 
working aroimd the circuit of pressures, /, D, B, F, and back to G should 
give a result fairly consistent with the assumption. If necessary, a third 
approximation may be made. 

It might be noted that this is much the process thai goes on in the receiver 
when the engine is being started^ the receiver pressure rising upon each relea^i 
from the high, closer and closer to the limiting pressure that is completely reachai 
only after running some time. 

These approximations may be avoided by the following computation, 
representing point pressures by P with subscript and volumes by reference 
to the lower diagram. Pj is the unknown pressure in receiver before high- 
pressure release and after low-pressure cut-oflf. 

Pressure after mixing at D is then 

Pj(C^(^) + Pc(C^C^) _-p 
The pressure at F, after mixing is 

(PE^) 
This pressure multiplied by 7Q2pk gives P^, and this in turn multiplir 

by j2j^ will give Pj. . 
Writing this in full. 



Solving for Pj, 






{Q'G^) {.PJ*) - {C*C^) (Q2g;3) 



WORK OF PISTON ENGINES 363 

which is in terms of quantities all of which are measurable from the diagram. 
While this formula applies to this particular case only, the manner of obtain- 
ing it is indicative of the process to be followed for other cases. 

When there are three successive cylinders the same constructions can be 
jsed, the intermediate diagrams taking the position of the low for the com- 
[X)und case, while the low for the triple may be placed \mder the high and oflf-set 
irom the intermediate by the volume of the second receiver. In this case 
It is well to repeat the intermediate diagram. Exactly similar constructions 
ipply to quadruple expansion with any crank angle relations. 



Prob. 1. By means of graphical construction find the horse-power of a 12- and 18 X24- 
n. single-acting cross-compound engine with 6 per cent clearance in each cylinder, 
f the receiver volume is 5 cu.ft., initial pressure 150 lbs. per square inch gage, back 
3ressure 10 lbs. per square inch absolute, speed 125 R.P.M., high-pressure compression 
}0 per cent, low pressure 20 per cent, high pressure cut-off 50 per cent, low pressure 
10 per cent, high-pressure crank ahead 70°, logarithmic expansion, and ratio of red 
*o crank 4. - 

Prob. 2. Consider the above engine to be a tandem rather than a cross-compound 
ind draw the new diagrams for solution. 

Prob. 3. A double-acting, 15- and 22 x24-in. compound lengine has the high- 
jressure crank ahead by 60°, and has 5 per cent clearance in the low-pressure cylinder, 
10 per cent in the high, and a receiver 4 times as large as the high-pressure cylinder. 
tVhat will be the horse-power when the speed is 125 R.P.M., initial pressure 150 lbs. per 
iquare inch absolute, back pressure 5 lbs. per square inch absolute, high-pressure cut- 
)ff ij low-pressure i, high-pressure compression 20 per cent, low-pressure 30 per cent, 
ind ratio of rod to crank 5. Determine graphically the horse-power in each cylinder. 

Prob. 4. Consider the engine of Prob. 3 to be a tandem compound and repeat 
he solution. 



23. Mean Effective Pressure, Engine Power, and Work Distribution and 
heir Variation, with Valve Movement and Initial Pressure* Diagram Dis- 
x>rtion and Diagram Factors. Mechanical Efficiency. The indicated power 
ieveloped by a steam engine is dependent upon three principal factors — piston 
Usplacement, speed, and mean effective pressure. The first, piston displacement, 
s dimensional in character, and, fixed for a given engine. Speed is limited by 
Jteam and inertia stresses, with which the present treatment is not concerned, 
)r by losses due to fluid friction in steam passages, a subject that will be 
urther considered under steam flow. Mean effective pressure is a third factor 
irhich is to be investigated, most conveniently by the methods laid down in 
;he foregoing sections. 

In these formulas for mean effective pressure, it will be observed that 
:he terms entering are (a) initial pressure, (6) back pressure, (c) cut-off or 
'atio of expansion, (d) clearance, and (e) compression, for the single-cylinder 
mgine. It is desirable to learn in what way the mean effective pressure 
iraries upon changing any one of these factors. 



364 ENGINEEEING THERMODYNAMICS 

Referring to Section 5, Eq. (262) for logarithmic expansion 

(ra.e.p.) = (iu-pr.)^+(^+c)log. ™q^ (mean forward pressure) 
— (bk.pr.) 1 ~X+{X+c) log, (mean back pressure) 



. (582) 



it is seen that the mean effective pressure la the difference between a mean 
forward pressure and a mean back pressure. The former depends un 
initial pressure, cut-oflf, and clearance, and the latter on back pressure. 



Fio. 104. — Curve to Show Variation of Mean Forward and Mean Back Pressure for Logarithmic 
Expanmon and Coinpreasion in a Single Cylinder Engine with CleMance. 

compression, and clearance. To study the effect of varying these terms it 
is most convenient to draw curves such as are shown in Fig. 104, and examine 
mean forward pressure and mean backward pressure separately. 

Mean forward pressure is seen by inspection to vary in direct proportion 
to initial pressure. Cut-off, when short, gives a low mean forward pressure, 
but it is to be noted that zero cut-off will not give zero mean effective 
pressure so long as there is clearance, due to expansion of steam in the 
clearance space. Increasing the length of cut-off, or period of admi^on. 
increases mean forward pressure, but not in direct proportion, the (m.f.p. 
approaching initial pressure as a limit as complete admission is approached. 



WORK OF PISTON ENGINES 



365 



];iearance has the tendency as it mcreases, to increase the mean forward 
)ressure, though not to a great extent, as indicated by the curve Fig. 104. 

Mean back pressure is usually small as compared to initial pressure, though 
i great loss of power may be caused by an increase of back pressure or com- 
)ression. Back pressure enters as a direct factor, hence the straight line through 
he origin in the figure. So long as compression is zero, back pressure and 
nean back pressure are equal. When compression is not zero, there must 
)e some clearance, and the ratio of (mean bk.pr.) to (bk.pr.) depends on both 
clearance and compression, being greater for greater compressions and for 
smaller clearances. 















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COMPRESSION .80 


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Fig. 105. — Curves to Show Variation of Mean EflFective Pressure for Logarithmic Expansion 
. and Compression in a Single Cylinder Engine with Clearance. 



The mean efifective pressures obtained by subtracting mean back from 
mean forward pressures in Fig. 104 are shown in curve form in Fig. 105. 

The inuUiple expansion engine can not be so simply regarded. In a general 
way each cylinder may be said to be a simple engine, and subject to variations 
of mean effective pressure due to change in its own initial pressure and back 
pressure, clearance, cut-off and compression, which is true. At the same time 
these factors are interrelated in a way that does not exist in the simple 
engine. Consider, for instance, the high-pressure cylinder of a compound 
engine with infinite receiver, with clearance. An increase of high-pressure 
compression tends first to raise the mean back pressure according to the 
reasoning on simple engine, but at the same time the change has decreased 
the volume of steam passing to receiver. No change having been made 



366 ENGINEERING THERMODYNAMICS 

in the low-pressure cylinder, the volume admitted to it will remain the same 
as before, and the receiver pressure will fall, decreasing mean back pressure 
by a greater amount than compression increased it, and mean forward being 
the same as before, the increase of high-pressure compression has increased 
the mean effective pressure of the high-pressure cylinder. The only eflFect 
upon the low-pressure cylinder is that resulting from lowering its initial 
pressure, i.e., the receiver pressure. This results in a decrease of low^ressure 
mean effective pressure. Computation will show that the assmned increase 
of high-pressure compression decreases low-pressure work more than it 
increases high-pressure work, or in other words, decreased mean effective 
pressure referred to the low. 

It is impracticable to describe all results of changing each of the 
variables for the multiple-expansion engine. Initial pressure and cut-off 
in the respective stages have, however, a marked influence upon receiver 
pressures and work distribution which should be noted. Power regulation 
is nearly always accomplished by varying initial pressure, i.e., throttling, 
or by changing cut-off in one or more cylinders. 

The effect of decreasing initial pressure is to decrease the pressures on 
the entire expansion line and in all no clearance cycles, to decrease absolute 
receiver pressures in direct proportion with the initial pressure. Since back 
pressure remains constant, the result is, for these no-clearance cycles, that 
the mean effective pressures of all but the low-pressure cylinder are decreased 
in direct proportion to the initial pressure, while that of the low-pressure is 
decreased in a greater proportion. The same is true only approximately 
with cycles having clearance and compression. 

The conditions giving equal work division have been treated in connection 
with the individual cycles, it may here be noted in a more general way that 
if higb-pressure cut-off is shortened, the supply capacity of that cylinder is 
decreased, while that of the next cylinder remains unchanged. The result 
is that the decreased supply volume of steam will be allowed to expand to 
a lower pressure before it can fill the demand of the next cylinder than it 
did previously, i.e., the receiver pressure is lowered. Similarly, shortening 
cut-off in the second cylinder will tend to increase receiver pressure. To 
maintain constant work division, there must be a certain relation between 
cut-offs of the successive cylinders, which relation can only be determined 
after all conditions are known, but then can be definitely computed and 
plotted for reference in operation. 

So far, in discussing the steam engine, cycles only have been treated. 
These cycles are of such a nature that they can be only approached in 
practice, but since all conclusions have been arrived at through reasoning 
based on assumed laws or hjrpotheses, the term hypothetical may be applied 
to all these cycles. It is desirable to compare the actual pressure-volume 
diagram, taken from the indicator card of a steam engine, and the hyix)thetical 
diagram most nearly corresponding with the conditions. 

In Fig. 106 is shown in full lines a pressure- volume diagram which has 



WORK OF PISTON ENGINES 



367 



>en produced from an actual indicator card taken from a simple non-con- 
msing, four-valve engine having 5 per cent clearance. 

Finding the highest pressure on the admission line A'B' and the lowest 
ressure on the exhaust line £)'£', these pressures are regarded as (initial 
ressure) and (back pressure) and a hypothetical diagram constructed cor- 
ssponding to Cycle III, with cut-off and compression at the same fraction of 
broke as in the actual engine. 

The first difference between the hypothetical and actual PV diagrams 
5 that the point of release C is not at the end of stroke, as was assumed for the 
lypothetical release, C, a difference which is intentional, since it requires time 
or pressure to fall after release to the exhaust pressure. This same fact may 
lause the comer of the diagram to be roimded instead of sharp as at D. 
Similarly, the point of admission f is before the end of the return stroke has 




Fig. 106. — Diagram to Illustrate Diagram Factors. 

been reached, and for a similar reason the comer A' may be roimded, though 
if release and admission are made sufficiently early the comers D' and A' will be 
sharp, approaching the hypothetical, H and A. ' 

These differences, however, have little effect upon the area of the actual 
diagram, which is seen to be much smaller than the hypothetical. This 
deficiency of area is the net result of a large number of influences, only a few 
of which can be fully eicplained in connection with the pressure volume 
discussion. 

Beginning with the pomt of admission, F', the line F'A'B' represents 
the period of admission. The rounding at A' has been explained; the inclina- 
tion of the line from A^ toward B' is due in part to the frictional loss of 
pressure as the fluid passes at high velocity through ports and passages from 
steam chest to cylinder. As the stroke progresses, the linear velocity of the 
piston increases toward' mid-stroke, requiring higher velocities in steam 
passages. The greater consequent friction causes pressure to fall in the 
cylinder. The resistance of pipes and valves leading to the engine have 



368 ENGINEERING THERMODYNAMICS 

an effect on the slope of this line. As cut-off is approached, this pressure 
fall becomes more rapid, due to the partial closure of the admission valve. 

From B', the point of cut-off, to C\ the point of release, is the period 
of expansion, during which the pressures are much lower than during the 
hypothetical expansion line BC, due principally to the lower pressure at the 
point of cut-off B' than at B. Hence, the frictional fall in pressure during 
admission has a marked effect upon the work done^during expansion. The 
curve BV rarely follows the law PF= const, exactly, though it commonly 
gives approximately the same work area. During the first part of expansion, 
the actual pressure commonly falls below that indicated by this curve, but 
rises to or above it before the expansion is complete. This is largely due 
to condensation of steam on the cylinder walls at high pressures, and its 
reevaporation at lower pressure, to be studied in connection with a thermal 
analysis of the cycle. The curve of expansion may also depart from this 
very considerably, due to leakage, either inwardly, through the admissioii 
valve, or by piston from a region of higher pressure, or outwardly, through 
exhaust valve, or by piston into a region of lower pressure, or by drain, 
indicator, or relief valves, or by stuffing-boxes. 

From the opening of the exhaust valve at the point of release, C, till 
its closure at compression S', is the period of exhaust. Pressures during this 
period, as during admission, are affected by frictional losses in the passages 
for steam, in this case the pressure in the cylinder being greater than that 
in*exhaust pipe due to friction, by an increasing amount, as the velocity ol 
the piston increases toward midnstroke. Thus the line DE' rises above tiie 
line DE until the partial closure of the exhaust near the point of compres- 
sion causes it to rise more rapidly. i - 

At the point of compression E' the exhaust valve is completely closed 
and the period of compression continues up to admission at F', Leakage, 
condensation, and reevaporation affect this line in much the same way a< 
they do the expansion, and often to a more marked degree, due to the fact 
that the volume in cylinder is smaller during compression than during 
expansion, and a given weight condensed, reevaporated, or added or removed 
by leakage will cause a greater change in pressure in the small weight 
present than if the change in weight had occurred to a large body of steam. 

In the compound engine all these effects are present in each cylinder in 
greater or less degree. In addition, there are losses of pressure or of volume 
in the receivers themselves between cylinders, due to friction or conden- 
sation, and where especially provided for, reevaporation by means of 
reheating receivers. The effect of these changes in receivers is to cause a 
loss of work between cylinders, and to makQ the discharge volume of one 
cylinder greater or less than the supply volmne of the next, while theee 
were assumed to be equal in the hypothetical cases. 

The effect of all of these differences between the actual and hypothetical 
diagrams is to make the actual indicated work of the cylinder something less 
than that represented by the hypothetical diagram. Since these effects are 



WORK OF PISTON ENGINES 369 

not subject to numerical calculations from data ordinarily obtainable, they 
are commonly represented by a single coefficient or diagram factor which is 
a ratio, derived from experiment, between the actual work and that indicated 
by h3rpothesis. 

It is at once evident that there may be more than one hypothetical 
diagram to which a certain engine performance may be referred as a standard 
of comparison. When the heat analysis of the steam engine is taken up, 
a standard for comparison will be foimd there which is of great use. For 
determination of probable mean eflfective pressure, however, no method of 
calculation has been devised which gives better results than the computa- 
tion of the hypothetical mean eflfective pressure from one of the standard 
h3rpothetical diagrams, and multiplying this by a diagram factor obtained 
by experiment from a similar engine, under as nearly the same conditions 
as can be obtained. 

Such diagram factors are frequently tabulated in reference books on the 
steam engine, giving values for the factor for various types and sizes, under 
various conditions of running. Unfortimately, however, the exact standard 
to which these are referred is not stated. In this text it will be assumed, 
unless otherwise stated, that the diagram factor for an actual engine is the 
ratio of the mean effective pressure of the actual engine to that computed 
for Cycle I, without clearance or compression, logarithmic law, with cut-off 
at the same fraction of stroke as usual, initial pressure equal to maximum 
during admission in actual, and back pressure equal to minimiun during exhaust 
of the actual engine. 

This is selected as the most convenient standard of comparison for mean 
effective pressures, as it is frequently impossible to ascertain the clearance 
in cases where data are supplied. When it is possible to do so, however, closer 
approximation may be made to the probable performance by comparing 
the actual with that hjrpothetical diagram most nearly approaching the cycle, 
using same clearance, cut-off, and compression as are found in the actual. 

Commercial cut-off is a term frequently used to refer to the ratio of the 
volume AK to the displacement, Fig. 106, in which the point K is found on 
the initial pressure line AB, by extending upward from the true point of cut-oflf 
B' a curve PF = const. 

While the diagram factor represents the ratio of indicated horse-power 
to hypothetical, the output of power at the shaft or pulley of engine is less 
than that indicated in the cylinders, by that amount necessary to overcome 
mechanical friction among engine parts. If this power output at shaft or 
pulley of engine is termed brake horse-power (B.H.P) then the ratio of this to 
indicated horse-power is called the mechanical efficiency, Emy of the engine 

The difference between indicated and shaft horse-power is the power 
consumed by friction (F.H.P.). Friction under running conditions consists 



370 



ENGINEERING THERMODYNAMICS 



of two parts, one proportional to load, and the other constant and inde- 
pendent of load, or 

(F.H.P.) =JV[(const.) x(m.e.p.)+(const.)2], 

where N is speed, revolutions per minute. But NX (const.) (m.e.p.) = {IMJP.)Ki 



and 



(F.H.P.) = (I.H.P.)iri+iV(const.)2, 



(584) 



where K'l and (const.)2 are constants to be determined for the engine, whose 
values will change as the conditions of the engine bearing-surfaces or lubri- 
cation alters. This value for (F.H.P.) may be used to evaluate Em, 



^ _ (I.H.P .)-( RH.PO _ , ^ iV(const.)2 
^~" (I.H.P.) "^ ^^ (LH.P.) ' '• 



(585) 



but indicated horse-power divided by speed is proportional to mean effective 
pressure, so that 

K2 



En, = l-Ki- 



(m.e.p.) 



(586) 




0;i 4 6 8 10 12U16l8 90SSiMnS038348888AO 

Mean Effective Pressure 
Fio. 107. — ^Diagram to Show Relation of Mechanical Efficiency and Mean Effective Pressure. 

From this expression, speed has been eliminated, which agrees with general 
observation, that mechanical efficiency does not vary materially with speed. 
Values of the constants jRli and K2 may be ascertained if (m.e.p.) and Em 
are known for two reliable tests covering a sufficient range, by inserting their 
values forming two simultaneous equations. 

The numerical values of Ki foimd in common practice are between .02 
and .05, and for K2 between 1.3 and 2, in some cases passing out of this range. 
In Fig. 107 is shown the form of mechanical efficiency curve when plotted 
on (m.e.p.) as abscissas, using Ki = .04, iC2 = 1.6. It may be noted that at 



WORK OF PISTON ENGINES 371 

higher (m.e.p.) the curve does not approach unity, but the value (l—Ki) 
as a limit. The mechanical efficiency becomes zero for this case, at a mean 
efifective pressure of about 1.67 lbs. per square inch, which is that just 
sufficient to keep the engine running under no load. For a given speed and size 
of cylinders, the abscissas may be converted into a scale of indicated 
horse-power. 

Prob. 1. Assuming a back pressure of 10 lbs. per square inch absolute, a clearance 
of S per cent, a cut-off of 40 per cent, and compression of 20 per cent, show how 
(m.e.p.) varies with initial pressure over a range of 200 lbs., starting at 25 lbs. 

Prob. 2. • For an initial pressure of 150 lbs. per square inch absolute, show how 
(m.e.p.) varies with back pressure over a range of 30 lbs., starting at J lb. per square 
inch absolute, keeping other quantities as in Prob. 1. 

Prob. 3. For values of initial pressure, back pressure, etc., as given in Probs. 1 
and 2, show how (m.e.p.) varies with clearance from 1 per cent to 15 per cent. 

Prob. 4, For values of initial pressure, etc., as given in Probs. 1 and 2, show how 
(m.e.p.) will vary with cut-off from to 1. 

Prob. 6. For values of initial pressure, etc., as given in Probs. 1 and 2, show how 
(m.e.p.) will vary with compression for values from to 50 per cent. 

Prob. 6. A certain engine developing 675 I.H.P. at a speed of 151 R.P.M., 
deUvered at the shaft 606 H.P. measured by an absorption dynamometer. A second 
lost at 100 R.P.M. gave 150 I.H.P., and 114 shaft H.P. If this engine is to deliver 
500 H.P. at the shaft at a speed of 150 R.P.M., what will be the I.H.P.and the mechan- 
ical efficiency? 

Prob. 7. A compound Corliss engine, 25 and 52 ins. diameters, 60 ins. stroke, 
double-acting, was dosigne^^fpr 650 I. HP. at 63 R.P.M. It was found that at this 
sjK^cd and I.H.P. the mechanical efficiency was 91 per cent. When running with no 
load, the cylinders indicated 38.1 I.H.P. at 65 R.P.M. Find the probable mechanical 
efficiency when developing 300 I.H.P. at a speed of 64 R.P.M. 

24. Consumption of the Steam Engine and its Variation wifh Valve Move- 
ment and Initial Pressure. Best Cut-off as Affected by Condensation and 
Leakage. The weight of steam used by a steam engine per hour divided by 
the indicated horse-power is said to be the water rate or steam consumption 
of that engine. It is almost needless to say that this is not a constant for 
a given engine, since it will change with any change of initial pressure, back 
pressure, or valve setting, leakage, or temperature conditions. Since there 
are at least two other uses of terms water rate or consumption, this may be 
termed the actual water rate, or actual consumptiorij the latter being a more 
general term which may refer to the weight of fluid used per hour per 
indicated horse-power, whatever the fluid may be, steam, air, carbon dioxide, 
or any other expansive fluid. The present discussion has special reference 
to steam*. 

From the hypothetical diagram, by computations such as are described 
for the various foregoing cycles, may be obtained a quantity representing 
the weight of fluid required to develop one horse-power for one hour, by the 



372 ENGINEERING THERMODYNAMICS 

performance of the hypothetical cycle. This may be termed the hypothetical 
consumption or for steam cycles the hypothetical water rate. 

By the use of the actual indicator card, may be obtained, by methods still 
to be described, the weight of fluid accoimted for by volumes and pressures 
known to exist in the cylinder, this being called the indicated consumption of 
the engine or indicated water rate if the fluid be steam. 

The heat analysis of the steam-engine cycle will lead to another standard 
of comparison which is of the greatest importance as a basis of determining 
how nearly the actual performance approaches the best that could be 
obtained if the engine were to use all available energy possessed by the steam. 
At present the object is to compare the actual and indicated performance with 
that hypothetically possible with cylinders of the known size. Accordingly 
attention will be confined first to hypothetical consumption, and the quantities 
upon which it is dependent. 

For Cycle III, which is the most general for the single-expansion engine, 
logarithmic law, the expression for consumption in pounds fluid per hour per 
indicated horse-power, found in Section 5, Elq. (267), is as follows: 

Hypothetical consumption, lbs. per hr. per I . H.P. 

= /3.750 r /bk.pr.\j 

(m.e.p.)L^ vn.pr. /J ' 

in which the value of mean effective pressure itself depends upon (in.pr.), 
(bk.pr.), c, Z, and X, The density of the fluid at initial pressure, ^i, is to be 
ascertained from tables of the properties of steam or of whatever fluid is used. 

In Fig. 108 are the results of computations on the hypothetical steam con- 
sumption, using mean effective pressures as plotted in Fig. 105. For each curve, 
conditions are assumed to be as stated on the face of the diagram, varying only 
one of the factors at a time. 

Other conditions remaining unchanged, it may be noted that consumption 
decreases for an increase of initial pressure, though not rapidly in the higher 
pressure range. 

Cut-off has a marked effect upon consumption, the minimum occurring 
when cut-off is such as to give complete expansion. This occurs when 

1+c (in.pr.) 



Z'+c (bk.pr.)' 
or 

which may be termed hypothetically best cut-off. In the case assmned in the 
diagram, 

Z' = (1 + . 1)^1 -.1 = . 065. 



WORK OF PISTON ENGINES 



373 



If clearance be varied, maintaining constant compression and cut-off, large 
clearance will give high consumption due to an excessive quantity of fluid 
required to fill the clearance space. Extremely small clearance leads to a high 
pressure at the end of compression, causing a loss of mean-eflfective pressure, 
and consequent high consumption. Between, the consumption has a minimum 
point, which is dependent for its location on both cut-off and compression. 

Decreasing back pressure has a beneficial effect upon mean effective pressure 
and consmnption. This would be still more marked in the figure if a case had 
been selected with a very short cut-off. 

Compression, throughout the ordinary range of practice, has but slight effect 
upon consumption, indicated by the flat middle portion of the curve in Fig. 











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10 15 20 25 80 .1 .2 .3 .4 .5 .6 .7 .8 .0 1.0 

Fig. 108. — Curves to Show the Variation of Hypothetical Steam Consumption of Simple 

Engines, Logarithmic Expansion and Compression. 

108. Very small or zero compression permits too much high-pressure steam to 
be admitted to the clearance space without doing work, and excessively large 
clearance causes pressures during compression to rise very high, thereby de- 
creasing mean effective pressure; hence this curve of consumption rises at 
both ends. 

Hypothetically, the best attainable consumption for given initial and back 
pressures is obtained when both expansion and compression arc complete. 

The indicated consumption, or, as it is frequently called for the steam engine, 
" steam accounted for by the indicator card " " or '* indicated water rate," is 
determined from the indicator card as follows. Let Fig. 109 represent an indicator 
diagram. The points of cut-off and compression are located from the form of 



374 



ENGINEERING THERMODYNAMICS 



the line, at the highest point on the expansion line and the lowest point on the 
compression line respectively. The fraction of the card lengths completed at 
cut-oflf, 

AB 
and the fraction of card length from point of compression to end of stroke, 



X= 



AC 



AD' 



are determined, the pressure at cut-off and compre&sion measured by the 
proper vertical scale, and the corresponding densities, 81, and I2 respectively, 
are ascertained from steam tables for dry saturated steam. Clearance, CI is 
known or ascertained by the form of the compression curve (Chap. I, Section 12). 

P 



Cutoff 




{^Atm* 



D V 



Fia. 109. — Diagram to Illustrate Method of Determining Indicated Water Rate of fcitoam 

Engine. 

At the point of cut-off, the weight of dry saturated steam present in thi> 
cylinder is D(Z+c)8i, and at compression the weight prcvSent is D(X+r):.>. 
on the assumption that the steam in the cylinder is of density 5i and h at 
these two instants. Accepting this assumption, the weight of steam used per 
cycle is 



Wt. steam per cycle=w = [{Z+c)h-(X+c)^2\D. 



(o^S 



The work per cycle 



TF = 144D(m.e.p.), 



and for n cycles per minute the indicated horse-power is 

^•^•^•" 33;000 



WORK OF PISTON ENGINES 375 

The indicated consumption is then, in pounds per hour per I.H.P. 

wn _ 60X33,0OOXD[(Z+c)$i-(X+c)a2]n 
^I.H.P. ' 144nD(m.e.p.) ' 

or, 

Ind. consrunption, lbs. per hr. per I.H.P. 

13,750 



(m.e.p.) 



[(Z+c)5i-(Z+c)B2], (589) 



which is the expression used to find indicated consumption for either simple- 
or multiple-expansion engines. In applying this to the mvltiple-expansion 
engine the terms Z, X and c are found for any one cylinder, and the mean eflfective 
pressure is referred to that cylinder. There may be, therefore, as many computa- 
tions as there are expansion stages. For a compound engine, for instance, 
indicated consmnption according to high-pressure card is found by inserting 
in formula Z, X and c for the high-pressure card, 8i and 82 for corresponding 
pressures, and for (m.e.p.) use 

(m.e.p. ref. to H.P.) = (m.e.p.)iy+(m.e.p.)Lpp. . . . (590) 

m 

If the computation is done by means of events on the low-pressure card, 
the (m.e.p.) must be referred to the low. 

(m.e.p. ref. to L.P.) = (m.e.p.) /r5^+(m.e.p.)L (591) 

In general for a multiple-expansion engine 

(m.e.p. ref. to cyl. A) = 2 (m.e.p.) YT (592) 

It is often difficult and sometimes impossible to determine the point of 
cut-oflf and of compression on the indicator card. The expansion and compres- 
sion lines are of very nearly hyperbolic in form and are usually recognizable. 
The highest point on the hyperbolic portion of the expansion line is regarded as 
cut-ofiF, and the lowest point on the hjrperbolic portion of the compression line, 
as the point of compression. It must be understood that by reason of the 
condensation and re-evaporation of steam in cylinders the weight of steam 
proper is not constant throughout the stroke, so that calculations like the 
above will give different values for every different pair of points chosen. The 
most correct results are obtained when steam is just dry and these points are 
at release and compression most nearly. 



376 ENGINEERING THERMODYNAMICS 

When under test of actual engines the steam used is condensed and weighed 
and the indicated horse-power determined, then the actual steam consumption 
or water rate can be found by dividing the weight of water used per hour in 
the form of steam by the indicated horse-power. This actual water rate is 
always greater than the water rate computed from the equation for indicate 
consumption. The reasons for the difference have been traced to (c) leakage 
in the engine, whereby steam weighed has not performed its share of work, to 
(6) initial condensation, whereby steam supplied became water before it could 
do any work, (c) variations in the water content of the steam by evaporation 
or condensation during the cycle, whereby the expansion and compression laws 
vary in unpredictable ways, affecting the work. 

Estimation of probable water rate or steam consimiption of engines cannot, 
therefore, be made with precision except for engines similar to those which have 
been tested, in all the essential factors, including, of course, their condition, 
and for which the deficiencies between actual and indicated consumptions 
have been determined. This difference is termed the missing vxtter, and end- 
less values for it have been found by experiments, but no value is of any use 
except when it is found as a function of the essential variable conditions that 
cause it. No one has as yet found these variables which fix the form for an 
empiric formula for missing water nor the constants which would make such a 
formula useful, though some earnest attempts have been made. This is no 
criticism of the students of the problem, but proof of its elusive nature, and the 
reason is probably to be found in the utter impossibility of expressing by a 
formula the leakage of an engine ui imknown condition, or the effect of its 
condition and local situation on involuntary steam condensation and evapora- 
tion. It is well, however, to review some of these attempts to evaluate 
missing water so that steam consumption of engines may be estimated. 
After studying the many tests, especially those of Willans, Perry announced 
the following for non-condensing engines, in which the expansion is but little 

Missing wat^i^^^^_^ 

Indicated steam dVN* 

where d is the diameter of the cylinder in inches and N the number of revolutions 
per minute. This indicates that the missing steam or missing water has been 
found to increase with the amount of expansion and decrease with diameter of 
cylinder and the speed. Thermal and leakage conditions are met by the use 
of difference values of m, for there are given 

rn = 5 for well-jacketed, well-drained cylinders of good construction with 
four poppet valves, that is, with minimum leakage and condensation. 

m = 30 or more for badly drained unjacketed engines with slid, valves, thai 
is, with high leakage and condensation possibilities. 

m = 15 in average cases. 



WORK OF PISTON ENGINES 377 

For condensing engines Perry introduces another variable — ^the initial pres- 
ire pounds per square inch absolute, p giving 



_120(l+|) 



Missing watei^ _ ^ «, /ka/in 



Indicated steam dVnpi 

t might seem as if such rules as these were useless, but they are not, especially 
rhen a given engine or line of engines is being studied or two different engines 
ompared; in such cases actual conditions are being analyzed rather than 
iredictions made, and the analysis will always permit later prediction of con- 
iderable exactness, if the constants are fixed in a formula of the right empiric 
Drm, Similar study by Heck has resulted in a different formula involving 
iflFerent variables and constants, but all on the assumption that the dis- 
repancies are due to initial condensation. He proposes an expression equi- 
valent to 



Missing steam _ .27 /<S(x2— xi) {kqk\ 

Indicated steam "" ^JV\ Pi^ ' 

Q which iV = R.P.M. of the engine; 
d= diameter in inches; 
L = stroke in feet; 

<S=the ratio of cylinder displacement surface in square feet to dis- 
placement in cubic feet. 



^1^2)' 



XL 



K^) 



The term (X2— «i) is a constant supposed to take into account the amount 
)f initial condensation dependable on the difference between cylinder wall 
md live-steam temperature and is to be taken from a table found by trial as 
:he difference between the x for the high pressure and x for the low pressure, 
yoih absolute, see Table XIV at the end of the Chapter. 

In discussing the hypothetical diagrams, it was found that best economy was 
)btained with a cut-off which gives complete expansion. For other than 
lypothetical diagrams this is not true, which may be explained most easily by 
reference to the curves of indicated, and actual consumption, and missing 
steam, Fig. 110. 

The curve ABC is the hypothetical consumption or water rate for a certain 
steam engine. Its point of best economy occurs at such a cut-off, B, that expan- 



378 



ENGINEERING THERMODYNAMICS 



sion is complete. The curve GHI is computed by Heck's formula for missiiif 
water. The curve falls off for greater cut-oflfs. Adding ordinates of these tvo 
curves, the curve DEF for probable consumption is found. The minimuic 
point in this curve, jE, corresponds to a longer cut-off than that of -4BC. 
Since cut-off B gave complete expansion, cut-off E must give incomplete expan- 
sion. In other words, due to missing steam, the condition which really gives 
least steam consiunption per hour per indicated horse-power corresponds to a 
release pressure, which is greater than the back pressure. 

It should be noted that the minimum point mentioned above will not 1« 
best cut-off, for the output of the engine is not indicated, but brake horse-power. 



ao 






30 



10 





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Per Cent Cut off 

Fig. 110. — Diagram to Show Displacement of Best Cut-off Due to Effect of Missing Waier 
from Point B for the Hypothetical Cycle to Some Greater Value E, 

In Fig. ill on cut-off as abscissa are plotted {EFG) consumption pounds 
per hour per I.H.P., and for the case assumed, (OD) the curve of mechanical 
efficiency, based on cut-off, 



(lbs. steam per hr.) . B.H.P. 
I.H.P. "^I.H.P. 

or, in other words, 

Consumption, lbs. per hr. per I.H.P. 



(lbs. steam per hr.) 

B.H.P: 



E. 



-'—^ = Consumption, lbs. per hr. per B .H.P. (59tj ! 



m 



Due to the increasing value of Em for greater cut-offs, the minimum point 5 
corresponds to a cut-off still longer than for the curve EFG, which itself wa> 
found in Fig. 110 to give a longer cut-off than that of the hypothetical curve. 

Hence the best cut-off for economy of steam, where the net power at the 
shaft is regarded as the output, will be such as to give incomplete expansion, 
or a release pressure above back pressure, this effect being caused by l)olli 
missing steam and by frictional losses. 

Prediction of actual consumption of steam engines as a general proposition 
is almost hopeless if any degree of accuracy worth while is desired, though the 



WORK OF PISTON ENGINES 



379 



jflfect on steam consumption of changing the value of any one variable can be 
)retty well determined by the previous discussion qualitatively, that is, in kind, 
;hough not quantitatively in amount. Probably the best attempt is that 
)f Hrabak in German, which takes the form of a large number of tables 
developed from actual tests though not for engines of every class. These tables 
are quite extensile, being in fact published as a separate book and any abstrac- 
tion is of no value. 

There is, however, a sort of case of steam consumption prediction that can be 
carried out with surprising precision and that is for the series of sizes or line 
of engines manufactured by one establishment all of one class, each with 
about the same class of workmanship and degree of fit, and hence having 
leakage and cylinder condensation characteristics that vary consistently through- 
out the whole range. For such as these tables and curves of missing water 







A 








Uecl 


> Eff.- 


1 










40 




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Lmpei 


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■ 


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n.p 


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per Hoor^ 


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fe 


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1 


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.8 



Per Cent Cut off 
Fig. 111. — Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cut-off. 



can be made up and by the best builders are, for making guarantees of steam 
consumption for any service conditions that their engines are able to meet. 

The practice of one firm making what is probably the best line of stationary 
engine in this country is of suflScient interest to warrant description. The 
primary data are curves of indicated water rate plotted to mean effective 
pressure for clearances of three or four per cent, and that mean effective pressure 
is chosen in any one specific case that will give the horse-power desired at the 
fixed speed for some one set of cylinder sizes available. To this indicated 
water rate a quantity is added constituting the missing water which is made 
up of several parts as follows: The first is an addition representing condensa- 
tion which is plotted in curve form as a function of (a) boiler pressure, (6) 
superheat in the steam, (c) piston speed, (d) the class of engine simple, com- 
pound or triple, with jacketed or unjacketed cylinders, and for cylinder ratios 
from 4 to 1, to 6 to 1 in the case of compounds. It is therefore a most complex 
quantity, the nature of the variations in which can only be indicated here. 



380 ENGINEERING THERMODYNAMICS 

For example, increase of piston speed decreases the condensation loss a- 
does multiple expansion, and also jacketing, while increase of superheat in the 
steam also decreases it, but superheat has less effect in triple than in com- 
pounds and less in compound than in simple engines. 

The next factor of correction is that covering leakage losses, also additive 
to indicated water rate and which with it and the condensation loss make 
up the probable steam consumption. The leakage decreases regularly with 
increase of piston speed, is le^s for large than for small engines, the changf. 
being rather fast from 50 to 200 horse-power and much slower later, being 
scarcely anything at all over 2000 horse-power. 

Example 1. What cut-off will give the lowest indicated water rate for a 9x12- 
in. engine, with 5 per cent clearance and no compression when running non-condensing 
on an initial pressure of 100 lbs. per square inch gage, and what will be the value 
of this water rate? What steam will be used per hour per brake horse-power 
hypothetically? From Eq. (587) 

Z' = (l+c)y^-c, 
in.pr. 

15 
= (1 +.05)-7L - .05 =8.7 per cent, 
115 



and 



(m.e.p.) =115| .087x(.087+.05) log*^ | -15=27.2 lbs. sq.in. 



Hence 



13 750 r 15 1 

Steam per hour per I.H.P. =-^— .137-.05X— X.262 = 17.2 lbs. 

From the curve of Fig. 107, assuming it to apply to the engine, for this value of 
(m.e.p.) mechanical efficiency is 90 per cent, hence from Eq. (596) the weight of 
steam per shaft horse-power per hour will be 19.1 poimds. 

Prob. 1. Draw diagram similar to Fig. 108 for following case: 

Initial pressure, 135 lbs. per square inch gage, back pressure 10 lbs. per square 
inch absolute, clearance 5 per cent, cut-off 30 per cent, compression 25 per cent. 

Prob. 2. From indicator diagram shown in Fig. 106 find the indicated water rate 
of the engine from which it was taken. 

Prob. 3. The indicated water rate of a 9Xl2-in. jacketed engine when running 
non-condensing at a speed of 250 R.P.M. with an initial pressure of 100 lbs. per 
square inch gage and J cut-ofiF is 50 lbs. Using Perry's formula what will be the 
probable actual steam used by engine per horse-power hour * 

Prob. 4. A 24 x48-in. engine in good condition is found to have an indicated water 
rate of 25 lbs. when cut-off is i, initial pressure 100 lbs. per square inch gage, back 
pressure 10 lbs. per square inch absolute, and speed of 125 R.P.M. What will be the 
miesing water, and the rate as found by Perry's formula and by Heck's? 



WORK OF PISTON ENGINES 381 

Prob. 5. What will be the probable amount of steam used per hour by a 36 >c48- 
in. engine with 5 per cent clearance running at 100 R.P.M. on an initial pressure of 
150 lbs. per square inch gage a back pressure of 5 lbs. per square inch absolute, i cut- 
off and 10 per cent compression? 

Prob. 6. How will the amount of steam of Prob. 5 compare with that used by 
a 15X22x36-in engine with 5 per cent clearance in each cylinder, running at 100 
R.P.M. on same pressure range with J cut-off in high-pressure cylinder, J cut-off in 
low, and 10 per cent compression in each cylinder? 

25. Variatioii of Steam Consumptioii with Engine Load. The Willans 
Line. Most Economical Load for More than One Engine and- Best Load 
Division. However valuable it may be to the user of steam engines to have 
an engine that is extremely economical at its best load which, it should be 
noted, may have any relation to its rated horse-power, it is more important 
usually that the form of the economy load curve should be ap flat as possible 
and always is this case when the engine must operate under' a wide range of 
load. This being the case it is important to examine the real performance 
curves of some typical engines all of which have certain characteristic 
similarities as well as differences. 

From the discussion of hypothetical and indicated water rates it appears 
that the curve of steam consumption (vertical) to engine load (horizontal) 
is always concave upward and always has a minimum point, not at the maxi- 
mum load. Actual consumption curves are similar in general form, but as 
has been pointed out, the load at which the water rate is least corresponds 
to some greater mean effective pressure than that for the hypothetical, so 
the whole curve is displaced upwards and to the right by reason of cylinder 
condensation and leakage losses. This displacement may be so great as to 
prevent the curve rising again beyond the minimum point, in which case the 
least steam consumption corresponds to the greatest load. Just what form 
the actual water rate-load curve will take depends largely on the form of valve 
geai and type of governing method in use, by throttling initial pressure with 
3, fixed cut-off or, by varying cut-off without changing initial pressure, with or 
without corresponding changes in the other valve periods. 

Whenever the control of power is by throttling of the supply steam the 
curve is found to be almost exactly an hypjerbola, so that (water rate X horse- 
power) plotted to horse-power is a straight line which being characteristic 
is much used in practical work and is known as the WiUans line. All other 
engines, that is, those that govern on the cut-off, have Willans lines that 
are nearly straight, such curvature as exists being expressed by a second 
degree equation instead of one of the first degree. 

Equations for Willans lines can always be found for the working range 
of load, that is, from about half to full load, though not for the entire range, 
except in unusual cases, and these equations are of very great value in pre- 
dicting the best division of load between units, which is a fundamental step 
in deciding, how many and what sizes of engine to use in carrying a given 
load in industrial power plants. 



382 



ENGINEERING THERMODYNAMICS 























































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WORK OF PISTON ENGINES 



3S3 



Before taking up the derivation of equations some actual test curves will 
examined and a number of these are grouped in Fig. 112 for engines of 
.nous sizes, simple and compound, up to 10,000 H.P., on which vertical 
stances represent pounds of steam per' hour, per I.H.P. and horizontal 
Ei.P. To show the essential similarity of the curves for engines of different 
;e more cleariy, these are re-plotted in Fig. 113 to a new load scale based 
I best load of eachj which is taken as unity. This is evidently a function of 
mn effective pressure, just what sort of function does not matter here. In 




75 ICD 125 

Pecoentttgpe ot most economical load 



200 



1. 113. — ^Typical Water Rate — Lead Curves for Steam Engines Plotted to Fractional Loads. 



Bry case the Willans line is also plotted in Fig. 112, each line being num- 
red to correspond to its water rate curve. 

As there is a corresponding similarity of form for the water rate and 
illans line of steam turbines, though the reasons for it will be developed 
er, it must be understood that the mathematical analysis that follows applies 
both turbine and piston steam engines, and finally it makes no difference 
lat units are used for load, whether I.H.P., or B.H.P. or K.W. of a direct- 
inected electric generator. 

In Fig. 114 is shown the water-rate curve to a K.W. base for the 10,000 K.W. 
irtis steam turbine at the Chicago Edison, Fiske Street Station for which the 
lowing equation fits exactly: 

V 17 09 

^=''-;;'^+10.54+.156P, 



384 



ENGINEERING THERMODYNAMICS 



where y= pounds of steam per hour -5- 1000; 
load (in this case in K.W.)-^1000; 
pounds of steam per K.W. hr. 



P 
Y 
P 



A similar equation fits fairly well the curve of Fig. 115, representing th 
7000 H.P. piston engines of the Interborough Railway, Fifty-ninth Stm 
station, as well as the combined piston engine and low-pressure steam turbia 




7000 9000 

1000 P IF Kilowatts 

Fia. 114.— Performance of a 10000-K.W. Steam Turbine. 



taking its exhaust steam, in the same station, but with different numeric 
constants, as below: 

Piston engine, -p = ' + .6-|- 1 .85P, 

Y 89.4 
Combined piston engine and turbine, p = ~p — 2.90+.713P. 

A third case of smaller size is shown in Fig. 116, representing the jut 
formance of a 1000-K.W. Corliss piston engine driving a generator for whi 
the equation is 

Y 9.8 



WORK OF PISTON ENGINES 



385 




10000 11060 12000 laooo 14000 15000 laooo 

Kilowatts - 1000 P 

Fia. 115. — Performance of a 7000-H.P. Piston Engine alone and with a Low-pressure 

Steam Turbine, 



26 24 



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800 



600 900 

Kilowatts»1000P 



1200 



1500 



Fig. 116.— Performance of a 1000-K.W. Steam Turbine. 



386 ENGINEERING THERMODYNAMICS 

These illustrations could be multiplied indefinitely, but those given will 
suffice to establish the fact that the two following equations are fundamental 
over ihe working range of any steam engine of whatever type: 

Water rate line, ^=^-+B+CP (597 

Water per hour, Willans line, Y=A+BP+CI^, (598 

in which Y is the weight of steam per hour and P the engine load whether 
expressed in indicated or brake horse-power, or in kilowatts. 

At the most economical load the water rate is a minimum, so that 



i(?)=<'-^(p+«+'^'') 



dP 
whence the most economical load is 

i^' = >J|. (599) 

Where the Willans line is straight, C=0, and the most economical load 
is the greatest load. 

Two engines carrying the same load must divide it and some one pro- 
portion may be best. To find out, consider first any number of similar engiius. 
that is, engines that have the same constants A, B, and C, denoting each ca^e 
by subscripts. Then 

Let P= total load; 

'' Pu P2, Pd) etc. = individual engine loads; 
*' Y = total water per hour; 
'* Yi, Y2y F3 = water per hour for each engine. 

Then 

r=Fi+F2+F3+. . .+r„ 

= nA+B{Pi+P2 + P3 + . . . Pn)+C(Pl^ + P2' + P3'+. . .+Pn') 

==nA+BP+C{Pi^+P2^+P3^+. . .+Pn2). 
Only the last term is variable and this is a minimum when 

Pl=P2 = P3 = P»/ 

Therefore for similar engines^ ihe best division of load is an equal division. 



WORK OF PISTON ENGINES 387 

When the engines are dissimilar it is convenient to first consider the case 
of straight Willans lines for which C=0. Then for two such engines 

Y=Ai+A2+BiPi+B2P2 

= (Ai+A2) + Bi(P-P2)+B2P2 

= {Ai+A2)+BiP+{B2-Bi)P2. 

At any ^ven load P the first two terms together will be constant, and the 
ivater per hour will be least when the last term is least. As neither factor 
jan be zero, this will occur when P2 is least. 

Therefore for two dissimilar engines the best division of load is thai which 
Tuts the greatest possible share on the one with the smaller value of B, in its equation, 
yrmded each has a straight WiUans line. 

Two dissimilar engines of whatever characteristics yield the equation, 

Y==Ai+A2+BiPi+B2P2+CiPi^ + C2P2^ 

= (Ai+A2+B,P+CiP2) 

+ (B2-2PCi-Bi)P2 

+ {Cl + C2)P2^. 

DiflFerentiation with respect to P2, and solving for P2, the load for the second 
(ngine that makes the whole steam consumption least, gives, 

= constant+constant XP. 

Therefore, fJie load division must be linear and Eq. (600) gives the numerical 
alue, when any two engines share a given load. 

This sort of analysis can be carried much further by those interested, but 
pace forbids any extension here. It is proper to point out, however, that 
y means of it the proper switch-in points for each imit in a large power station 
an be accurately found, to give most economical operation on an increasing 
tation load. 

26. Graphical Solution of Problems on Horse-power and Cylinder Sizes . 

Tie diagram for mean effective pressures in terms of initial and back 
ressure, clearance, compression and cut-off, Fig. 117, facilitates the solution 
f Eq. (262) in Section 5. The mean effective pressure is the difference 
etween mean forward and mean back pressure. The former is dependent 
XK)n clearance, cut-off and initial pressure. In the example shown on the 
gure by letters and dotted lines, clearance is assumed 5 per cent, shown at 
L. Project horizontally to the point P, on the contour line for the assumed 
ut-off, 12 per cent. Project downward to the logarithmic scale for " mean 



388 



ENGINEERING THERMODYNAMICS 







0) 



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'5) 

c 

pa 

C 



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WORK OF PISTON ENGINES 389 

forward pressure in terms of initial pressure " to the point G. On the scale 
for " initial pressure " find the point iT, representing the assumed initial pres- 
sure, 115 lbs. absolute. Through G and H a straight line is passed to the point 
K on the scale for *'mean forward pressure," where the value is read, 
m.f.p.=49.5 lbs. absolute. 

Mean back pressure is similarly dependent upon clearance, compression 
and back pressure, and the same process is followed out by the points A, B, 
Cy D and £, reading the mean back pressure, 3.2 lbs. absolute at the point E. 
Then by subtraction, 

(m.e.p.) = (m.f .p.) - (m.b.p.) =49.5-3.2=46.3 lbs. 

Fig. 118 is arranged to show what conditions must be fulfilled in order to 
obtain equal work with complete expansion in both cylinders in a compound 
engine, finite receiver, logarithmic law, no clearance, Cycle VII, when low- 
pressure admission and high-pressure exhaust are not simultaneous. This is 
discussed in Section 11, and the diagram represents graphically the conditions 
expressed in Eqs. (376), (377), (378), (379). 

To illustrate its use assume that in an engine operating on such a cycle, 

the volume of receiver is 1.5 times the high-pressure displacement, 1,5 = y, then 

1 

~= .667. Locate the point A on the scale at bottom of Fig. 118, corresponding 

y 

to this value. Project upward to the curve marked ''ratio of cut-offs " and at 
the side, C, read ratio of cut-offs 

l^=.672. 

Next extending the line 45 to its intersection D, with the curve (7/f, the point 
D is found. From D project horizontally to the contour line representing the 
given ratio of initial to back pressure. In this case, initial pressure is assumed 
ten times back pressure. Thus the point E is located. Directly above E at 
the top of the sheet is read the cylinder ratio, at /^, 



D 



L 



fic=n-=2.4. 

If cylinder ratio and initial and final pressures are the fimdamental data of 
the problem, the ratio of cut-offs and ratio of high-pressure displacements 
to receiver volume may be found by reversing the order. 



390 



ENaiNEERING THERMODYNAMICS 




to to 



m 



JO ORBH 



WORK OF PISTON ENGINES 391 



GENERAL PROBLEMS ON CHAPTER IH. 

Prob. 1, How much steam will be required to run a 14 xl8-in. ^double-acting 
engine with no clearance at a speed of 200 R.P.M. when initial pressure is 100 lbs. 
per square in. gage, back pressure 28 ins. of mercury (barometer reading 30 ins ), and 
cut-off is i? What will be horse-power imder these conditions? 

Note: 8 for 100 lbs. =.26, for 28 ins. =.0029. 

Prob. 2. Draw the indicator cards and combined diagram for a compound steam 
engine without receiver, and with 3 per cent clearance in low pressure and 5 per cent in 
high, when initial pressure is 100 lbs. per square inch absolute, back pressure 10 lbs. 
per square inch absolute, high-pressure cut-off J, high-pressure compression iV, and 
low-pressure compression i, 

Prob. 3. A simple double-acting engine, 18x24 ins., is running at 100 R.P.M. 
on compressed air, the gage pressure of which is 80 lbs. The exhaust is to atmosphere. 
If the clearance is 6 per cent and cut-off f , and compression 10 per cent, what horse- 
power is being developed, the expansion being adiabatic, and how long can engine be 
run at rated load on 1000 cu.ft. of the compressed air? 

Prob. 4. Will the work be equally distributed in a 12xl8x24-in. engine with 
infinite receiver and no clearance when cut-off is J in high pressure cyhnder, and f in 
low, expansion being logarithmic, initial pressure 150 lbs. per square inch absolute 
and back pressure atmosphere? What will be work in each cylinder? 

Prob. 6. The receiver of a 15X20x22 in. engine is 4 times aa large as high- 
pressure cylinder. What will be the horse-power, steam used per hour, and variation 
in receiver pressure for this engine, if clearance be considered, zero and initial pressure 
is 125 lbs. per square inch gage, back pressure 5 lbs. per square inch absolute, cut-offs 
J and I in high- and low-pressure cylinders respectively, and piston speed is 550 ft. per 
minute? 

Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. 

Prob. 6. At no load an engine having 7 per cent clearance, cuts off at 4 per cent 
of its stroke, while at full load it cuts off at 65 per cent of its stroke. At no load, 
compression is 40 per cent and at fuU load 5 per cent. What percentage of full-load 
horse-power is required to overcome friction, and what percentage of steam used at 
full load, is used on friction load, if initial pressure is constant at 100 lbs. per square 
inch gage, back pressure constant at 5 lbs. per square inch absolute, and expansion 
is logarithmic? 

Note: 8 for 100 lbs. =.262, for 5 lbs. =.014. • 

Prob. 7. The initial pressure on which engine is to run is 115 lbs. per square inch 
gage, and steam is superheated and known to give a value of s = 1.3. For an engine 
in which clearance may be neglected, work is to be equal, and expansion complete 
in both cylinders, when back pressure is 10 lbs. per square inch absolute. What must 
be the cut-offs and cylinder ratio to accomplish this when receiver is 3i times high 
pressure cylinder volume? 

Prob. 8. A 12-in. and 18x24 ins. double-acting engine with zero clearance and 
infinite receiver operates on an initial pressure of 150 lbs. per square inch gage, and 



392 ENGINEERING THERMODYNAMICS 

a back pressure of 5 lbs. per square inch absolute. What will be the release and receiver 
pressures, horse-power, and steam consumption when speed is 150 R.P M., expansion 
logarithmic, and cut-off i in each cylinder? 

Note: 8 for 150 lbs. =.367, for 5 lbs. =.014. 

Prob. 9. If a third cylinder 24 ins. in diameter were added to engine of Prob. 8 
and cut-off in this made i, how would horse-power, steam consumption, receiver and 
release pressures change? 

Prob. 10. What would have to be size of a single cylinder to give iSame horse-power 
at same revolutions and piston speed as that of engine of Prob. 8 under same conditions 
of pressure and cut-off? 

Prob. 11. With the higli-pressure cut-off at f, and low and intermediate cut-offs 
at A, what will be the horse-power, water rate and receiver pressures of a 30 X 48 X 77 X 72- 
in. engiae running at 102 R.P.M. on an initial pressure of 175 lbs. per square inch gage 
and a back pressure of 26 fins, of mercury (barometer reading 30 ins.), if the 
receiver be considered infinite and expansion logarithmic, clearance zero? What change 
in intermediate and low-pressure cut-offs would be required to give equal work distribu- 
tion? 

Note: 8 for 175 lbs. =.419, for 26 ins. =.0058. 

Prob. 12. If it had been intended to have all the cut-offs of the engine of Prob. 
11, equal to i, what should have been the size of the intermediate and low-pre^ure 
cylinders to give equal work for same pressure range and same high-pressure cylinder? 
• Prob. 13. To attain complete expansion in all cylinders of the engine of Prob. 11, 
with the initial and back pressures as there given, what cut-offs would be required 
and what receiver pressures would result? 

Prob. 14. A compoimd locomotive has no receiver, the high- pressure clearance 
is 8 per cent, and low-pressure clearance 5 per cent. The cylinders are 22 and 
33x48 ins., high-pressure cut-off |, high- and low-pressure compression each 10 per 
cent, initial pressure 175 lbs. per square inch gage, back pressure one atmosphere, and 
expansion and compression logarithmic. What will be the horse-power at a speed 
of 40 miles per hour, the engine having 7-ft. driving wheels? At this speed, how 
long will a tank capacity of 45,000 gallons last? 

Note: 5 for 175 lbs. =.419, for 15 lbs. =.038. 

Prob. 16. A superheater has been installed on engine of Prob. 14 and expansion 
and compression, now follow the law PV ^c, when s = 1.2. What effect will this have 
on the horse-power and steam consumption? 

Prob. 16. What will be the maximiun receiver pressure work done in each cylinder 
and total work for a cross-compoimd engine 36 and 66x48 ins., running at 100 R.P.M. 
on compressed air of 100 lbs. per square inch gage pressure, exhausting to atmosphere 
if the high pressure cut-off is J, clearance 6 per cent, compression 20 per cent, low- 
pressure cut-off is f , clearance 4 per cent, compression 15 per cent, and receiver volume 
is 105 cu.ft.? 

Prob. 17. A manufacturer gives the horse-power of a 42x64x60-in. engine as 
2020, when run at 70 R.P.M. on an initial pressure of 110 lbs. per square inch gage, 
atmospheric back pressure, and .4 cut-off in high-pressure cylinder. How does this 
value compare with that found on assumption of 5 per cent clearance in high, 4 per 
cent in low, and complete expansion and compression is each cj^linder? 

Prob. 18. A mine hoisting engine is operated on compressed air with a pressure of 
150 lbs. per square inch absolute and exhausts to atmosphere. The cylinder sizes are 
26X48x36 ins., and clearance is 5 per cent in each. At the start the high-pressure 



WORK OF PISTON ENGINES 393 

cut-off 18 1 and low pressure i, while normally both cut-offs are J. The exhaust 
from high-pressure cylinder is into a large receiver which may be considered infinite. 
The compression is zero at all times. Considering the exponent of expansion to be 
1.4, what will be the horse-power imder the two conditions of cut-off given, for a speed 
of 100 R.P.M.? 

Prob. 19. What must be ratio of cylinders in the case of a compound engine with 
infinite receiver, to give equal work distribution complete expansion and com- 
pression if the least clearance which may be attained is 5 per cent in the high- 
pressure cylinder, and 3 per cent in the low-pressure. The engine is to run non- 
condensing on an initial pressure of 125 lbs. per square inch gage, with expansion 
exponent equal to 1.3? What must be the cut-offs and compressions to satisfy these 
conditions? 

Prob. 20. Assuming 7 per cent clearance in high-pressure cylinder and 5 per cent 
in low, infinite receiver, and no compression, how will the manufacturer's rating of 
2100 H.P. check, for a 36X6x48-in. engine running at 85 R.P.M. on an initial 
pressure of 110 lbs. per square inch gage, and a back pressure of 26 ins. vacuum, with 
.3 cut-off in high pressure cylinder? 

Prob. 21. For a 25X40x36-in. engine, with 5 per cent clearance, i cut-off 
and 20 per cent compression in each cylinder, what will be horse-power for an initial 
pressure of 100 lbs. per square inch gage, and a back pressure of 17.5 lbs. per square inch 
absolute, with logarithmic expansion and compression? 

Prob. 22. What must be the cylinder ratio and cut-off to give complete expansion 
in a no-clearance, 14 and 22 x24-in. engine with no receiver and logarithmic expansion, 
when initial pressure is 100 lbs. per square inch gage, and back pressure 10 lbs. per 
square inch absolute? What will be the horse-power and steam used for these conditions 
at a speed of 150 R.P.M.? 

Note: 8 for 100 lbs. =.262, for 10 lbs. =.026. 

Prob. 23. A 24X20x24 in. engine with no receiver or clearance, runs on com- 
pressed air of 120 lbs. per square inch gage pressure, and exhausts to atmosphere. 
When running at a speed of 125 R.P.M., with high-pressure cut-off f, what horse- 
power will be developed and how many cubic feet of compressed air per minute will 
be required to run the engine, the expansion being adiabatic? WiU the work be equally 
divided between the two cylinders? 

Prob. 24. It is desired to run the above engine as economically as possible. What 
change in cut-off wiU be required, and will this cause a decrease or increase in horse- 
power and how much? How will the quantity of air needed be affected? 

Prob. 26. A mill operates a cross-compound engine with a receiver 3 times as large 
as high-pressure cylinder, on an initial pressure of 125 lbs. per square inch gage, and a 
back pressure of 10 lbs. per square inch absolute. The engine may be considered as 
without clearance, and the expansion as logarithmic. As normally run the cut-off in 
high-pressure cylinder is | and in low, \, It has been found that steam is worth 25 
cents a thousand pounds. What must be charged per horse-power day (10 hours) 
to pay for steam if the missing water follows Heck's formula? 

Note 8 for 125 = .315, for 10 = .026. 

Prob. 26. By installing a superheater the value of 8 in Prob. 25 could be changed 
to 1.3. The cost of st am would then be 30 cents a thousand pounds. From the effect 
on value of s alone would the installation of the superheater pay? 

Prob. 27. When a 26X48x36-in. cross-compound engine with a receiver volume 
of 35 cu.ft. and zero clearance, is being operated on ste^frn of 125 lbs. per square inch 



394 ENGINEERING THERMODYNAMICS 

gage initial pressure, and atmospheric exhaust, is the work distribution equal, when 
high-pressure cut-off is i and low-pressure cut-off J? For these cut-offs what i- 
fluctuation in receiver pressure and what steam will be used per horse-power hour? 

Note; 5 for 125 =.315, for 15 lbs. =.038. 

Prob. 28. To operate engine of Prob. 27 under most economical conditions, what 
values must be given to the cut-offs, and what values will result for receiver pressims, 
horse-power, and eteam used per hour? 

Prob. 29. What will be the horse-power and steam used by a 20x30x36-m. 
engine with infinite receiver and no clearance, if expansion be such, that 8 = li5, 
high-pressure cut-off i, low-pressure cut-off i, initial pressure 100 lbs. per square IlcIi 
gage, back pressure 3 lbs. per square inch absolute, and speed 100 R.P.M. 

Note: 8 for 100 lbs. =.262, for 3 lbs. =.0085. 

Prob. 30. The following engine with infinite receiver and no-clearance, runs od 
steam which expands according to the logarithmic law. Cylinders 9, and 13X1S 
ins., initial pressure 125 lbs. per square inch gage, back pressure 5 lbs. per square incli 
absolute, high-pressure cut-off f, low-pressure |, speed 150 R.P.M. What will be 
horse-power and steam consumption hypothetical and probable? 

Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. 

Prob. 31. By graphical means find the (m.e.p.), of a 15X22x30-in. cto^ 
compound engine, with 5 per cent clearance in each cylinder, if the receiver volume is 
8 cu.ft., initial pressure 125 lbs. per square inch absolute, back pressure 10 lbs. per 
square inch absolute, high-pressure cut-off J, low-pressure A, high-pressure compres- 
sion 40 per cent, low-pressure 20 per cent, high-pressure crank following 90°, logarithmic 
expansion. 

Prob. 32. Show by a series of curves, assuming necessary data, the effect on 
(m.e.p.) of cut-off, back pressure, clearance, and compression. 

Prob. 33. Show by curves, how the indicated, and actual water rate, of an 18 X24- 
in. engine with 5 per cent clearance, and running at 125 R.P.M. on an initial pressure 
of 125 lbs. per square inch gage, and a back pressure of 10 lbs. per square inch absolute, 
may be expected to vary with cut-off from A to }. 



TABLES 



395 



Table XIII 
PISTON POSITIONS FOR ANY CRANK ANGLE 

^ROM BilGINNINO OF STROKE AWAY FROM CrANK ShAFT TO FiND PiSTON POSITION FROM 

Dead-Center Multiply Stroke by Tabular Quantity 



Crank 
Angle. 


-'-4 

r 


-^-4.6 
r 


r 


-=5.5 

r 


r 


-?=7 
r 


^=8 

r 


r 


5 


.0014 


.0015 


.0015 


.0016 


.0016 


.0016 


.0017 


.0019 


10 


.0057 


.0059 


.0061 


.0062 


.0063 


.0065 


.0067 


.0076 


15 


.0128 


.0133 


.0137 


.0140 


.0142 


.0146 


.0149 


.0170 


23 


.0228 


.0237 


.0243 


.0248 


.0253 


.0260 


.0266 


.0302 


25 


.0357 


.0368 


.0379 


.0388 


.0394 


.0405 


.0413 


.0468 


30 ! 


.0513 


.0531 


.0545 


.0556 


.0566 


.0581 


.0592 


.0670 


35 


.0698 


.0721 


.0740 


.0754 


.0767 


.0787 


.0801 


.0904 


40 


.0910 


.0939 


.0962 


.0981 


.0997 


.1022 


.1041 


.1170 


45 


.1152 


.1187 


.1215 


.1237 


.1256 


.1286 


.1308 


.1468 


50 


.1416 


.1458 


.1491 


.1518 


.1541 


.1576 


.1607 


.1786 


55 


.1713 


.1769 


.1828 


.1827 


.1863 


.1892 


.1922 


.2132 


60 


.2026 


.2079 


.2122 


.2167 


.2186 


.2231 


.2296 


.2500 


65 


.2374 


.2431 


.2477 


.2514 


.2546 


.2594 


.2630 


.2886 


70 


.2730 


.2794 


.2844 


.2885 


.2929 


.2973 


.3013 


.3290 


75 


.3123 


.3187 


.3239 


.3282 


.3317 


.3372 


.3414 • 


.3706 


80 


.3516 


.3586 


.3642 


.3687 


.3725 


.3784 


.3828 


.4132 


85 


.3944 


.4013 


.4068 


.4113 


.4151 


.4210 


.4264 


.4664 


90 


.4365 


.4437 


.4495 


.4547 


.4680 


.4641 


.4686 


.5000 


95 


.4816 


.4885 


.4940 


.4985 


.5022 


.6081 


.5126 


.6436 


100 


.5253 


.5323 


.5378 


.5424 


.5461 


.5520 


.6564 


.5868 


105 


.5711 


.5775 


.5828 


.6870 


.5905 


.5961 


.6002 


.6294 


110 


.6150 


.6214 


.6265 


.6306 


.6340 


.6393 


.6630 


.6710 


115 


.6600 


.6657 


.6703 


.6740 


.6771 


.6820 


.6866 


.7113 


120 


.7026 


.7080 


.7122 


.7167 


.7186 


.7231 


.7265 


.7500 


125 


.7449 


.7495 


.7533 


.7563 


.7588 


.7628 


.7668 


.7868 


130 


.7844 


.7885 


.7920 


.7947 


.7969 


.8004 


.8030 


.8214 


135 


.8223 


.8258 


.8286 


.8308 


.8327 


.8367 


.8379 


.8535 


140 


.8570 


.8600 


.8623 


.8642 


.8668 


.8682 


.8703 


.8830 


145 


.8889 


.8913 


.8931 


.8946 


.8968 


.8978 


.8993 


.9096 


150 


.9173 


.9191 


.9204 


.9216 


.9226 


.9241 


.9262 


.9330 


155 


.9420 


.9432 


.9452 


.9451 


.9457 


.9468 


.9476 


.9631 


160 


.9625 


.9633 


.9640 


.9645 


.9650 


.9656 


.9661 


.9698 


165 


1 .9787 


.9792 


.9796 


.9799 


.9802 


.9806 


.9809 


.9829 


170 


.9905 


.9908 


.9909 


.9911 


.9912 


.9913 


.9916 


.9924 


175 


.9976 


.9977 


.9977 


.9977 


.9978 


.9978 


.9979 


.9981 


180 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 



396 



ENGINEERING THERMODYNAMICS 



Table XIV 
VALUES OF z FOR USE IN HECK'S FORMULA FOR MISSING WATER 



Absolute 




Absolute 




Absolute 




Steam Preaeure. 


X 


Steam Pressure. 


X 


Steam Pressure. 


% 





170 


70 


297.5 


165 


393 


1 


175 


75 


304 


170 


397 


2 


179 


80 


310 


180 


405 


3 


183 


85 


316 


185 


409 


4 


186 


90 


321.5 


190 


413 


6 


191 


95 


327 


irs 


4165 


8 


196 


100 


332.5 


200 


420 


10 


200 


105 


338 


210 


427 


15 


210 


110 


343 


220 


431 


20 


220 


115 


348 


230 


441 


25 


229 


120 


353 


240 


447.5 


30 


238 


125 


358 


250 


454 


35 


246 


130 


362.5 


260 


460.5 


40 


254 


135 


367 


270 


467 


45 


262 


140 


371.5 


280 


473 


50 


269.5 


145 


376 


290 


479 


55 


277 


150 


380.5 


300 


485 


60 


284 


155 


385 






65 


291 

1 


160 


389 




• 



Table XV 
SOME ACTUAL ENGINE DIMENSIONS 

Simple 



7X9 


7iX15 


16 X18 


151X24 


24 X36 


8X9 


8iXl5 


16JX18 


16 X24 


26 X36 


9X9 


12 X15 


17 X18 


18 X24 


26iX38 


5iX10 


13 X15 


17iX18 


20 X24 


28 X36 


6iX10 


14 X15 


18 X18 


22 X24 


14 X42 


8 XIO 


14iX15 


19 X18 


24 X24 


15 X42 


9 XIO 


15 X15 


20 X18 


16iX27 


16 X42 


10 XIO 


16 X15 


29 X19 


17iX27 


18 X42 


11 XIO 


17JX15 


12 X20 


10 X30 


20 X42 


OiXlOJ 


11 X16 


14 X20 


12 X30 


22 X^2 


lOiXlOJ 


12 X16 


18 X20 


16 X30 


24 X42 


7fX12 


13 X16 


19 X20 


18 X30 


26 X42 


8 X12 


14iXl6 


28 X20 


18iX30 


28 y.^1 


8iX12 


15 X16 


21 X20 


20 X30 


18 X48 


9 X12 


151X16 


22 X20 


24 X30 


20 X48 


10 X12 


16 X16 


12 X21 


22 X33 


22 X48 


11 X12 


17 X16 


13 X21 


24 X33 


24 X48 


lliXl2 


18 X16 


18iX21 


10 X36 


26 X48 


12 X12 


18iX17 


20 X21 


12 X36 


28 X48 


12iXl2 


23 X17 


20 X22 


14 X36 


24 X54 


13 X12 


26 X17 


18 X24 


16 X36 


26 X54 


14 X12 


10 X18 


10 X24 


18 X36 


28 X54 


10 X14 


11 X18 


12 X24 


20 X36 


28 X60 


11 X14 


15 X18 


14i X24 


22 X36 





TABLES 



397 



Table XV. — Cordinued 



Compound 



NoTSs: 1 to run condenBins or 

2 to run condenmng or 

3 to run condensing or 

4 to run condensing or 

5 to run condensing or 

6 to run condensing or 

7 to run condensing or 



non-condensing 
non-condensing 
non-condensing 
non-condensing 
non-condensing 
non-condensing 
non-condensing 



on initial 
on initial 
on initial 
on initial 
on initial 
on initial 
on initial 



pressure 
pressure 
pressure 
pressure 
pressure 
pressure 
pressure 



of 100-160. 
of 100. 
of 126. 
of 90-100. 
of 110-130. 
of 140-160. 
of 126 



4i- 8 X 6 

6 -10 X 6 

7 -13 X 8 

6 -12 XIO 

7 -12 X 10 

8 -12 XIO 
8i-15jX10 

7 -14 XIO 

8 -14 XIO 
9i-15 Xll 
7H3iXl2 

9 -15JX12 
19 -14 X 12 
10 -16 XX2 
10-18 X12 
U -16 X12 

9 -18 X14 

10 -18 X14 

10 -17iX14 

11 -19 X14 

11 -18 X14 

12 -18 X14 
12 -20 X14 



1 
1 
1 

1 
1 
2 
3 
3 
1 
1 
1 
1 
1 
1 
3 
3 
1 
1 
1 



13 -18 
13 -20 
7M3i 
9 -15J 
11 -19 
13 -19 
7i-13i 
9 -15J 

10 -17i 

11 -19 

11 -22 

12 -21 

13 -22 

13 -22J 

14 -22 
14J-25 

15 -22 
15i-26J 

16 -25 
13 -23 
15 -26 
16i-29 

9 -15J 



X14 


1 


X14 


1.2 


X15 


3 ' 


X15 


3 1 


X15 


4 ! 


X15 


5 


X16 


3 


X16 


3 


X16 


3 


X16 


3 ! 


X16 


1 1 


X16 


3 


X16 


1 


X16 


3 • 


X16 


2 ' 


X16 


3 


X16 


1 


X16 


3 ; 


X16 


2 1 


X17 


4,6 


X17 


4,6 


X17 


4 ; 


X18 


3 



11 -19 X18 

12 -21 X18 

13 -22JX18 

14 -24 X18 
15i-26JX18 
16 -24 X18 
16 -26 X18 
161-28iX18 

8 -12 X20 

9 -14 X20 

16 -28 X20 

17 -30 X20 

18 -28 X20 

19 -30 X20 

19 -30 X22 
9 -15JX21 

12 -21 X21 

13 -22JX21 
14i-26iX21 
15i-28JX21 
18i-32i X21 

20 -36 X21 
13 -23 X22 



3 
3 
1 
3 
1 

3 
7 
7 
1 
1 
1 
1 
1 
3 
3 
3 
3 
3 
3 
3 
5 



14i-26 X22 
18 -32 X22 

10 -17JX24 

11 -19 X24 

12 -18 X24 

13 -20 X24 

14 -22 X24 
16J-28JX24 
17i-30JX24 
22 -38 X24 
24 -A2 X24 

12 -21 X27 

13 -22 J X 27 
16J-28i X27 
17J-30iX27 
14i-25 X30 
15l-26iX30 
18i-32iX30 
20 -36 X30 
28i-50 X30 
30 -54 X30 
16i-28JX33 

17iX30iX33 



4 
5 
3 
3 
7 
7 
7 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 



22 -38 X33 
24 -42 X33 
181-32 J X 36 
20 -36 X36 
26J-46 X36 
281-50 X36 
14i-25 X42 
15i-26iX42 
18J-32JX42 
20 -36 X42 
16 J-28i X48 
17i-30iX48 
22 -38 X48 
24 -A2 X48 
18J-321X54 
20 -36 X64 
26i-46 X54 
28i-50 X54 
22 -38 X60 
24 -42 X60 
30 -54 X60 
32i-57 X60 



3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 
3 



34 -%Q X60j 3 



Triple 

Note: All condensing and to run of initial pressure as given. 



Size. 


V 


Size. 


P 


Size. 


V 


10 -151-26X15 


200 


27 -43. (41x39 


180 


30-50-82X48 


180 


11 -18 -30X20 


250 




25-41-68X48 


190 


12 -20 -34X24 


180 


18 -281-48X40 


180 


27-45-75X54 


190 


12 -19 -32X24 


190 


22 -37 -63X42 


180 


28-45-72X54 


185 




175 


22 -38 -64X42 


185 


1 28-46-75X54 


180 


12^22 -36X24 


180 


321-53 -{5[x48 


265 


' 29-47-83X54 


160 


14 -23 -28X26 


190 


1 32-52-92X50 


200 


18 -29 -47X30 
161-24 -41X30 
18 -30 -50X30 
16 -25M3X30 


200 
180 
200 
190 


35-67-/^X48 
36 -57 -{76X48 


265 
295 


' 34-56-100X60 
35-58 |^qX60 
34-57-104X63 


200 
190 
200 


161-24 -41 X30 


180 


28 -45 -72X48 


180 







CHAPTER IV 

HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS 
BETWEEN THE HEAT CONTENT OF SUBSTANCES AND THEIR PHYSICAL- 
CHEMICAL STATE, 

1. Substances and Heat Effects Important in Engineering. It has been 
shown in preceding chapters concerned with work in general and with the deter- 
mination of quantity of work that may be done in the cyUnders by, or on 
expansive fluids that 

(a) Fluids originally at low may be put in a high-pressure condition by 
the expenditure of work; 

(b) Fluids under high pressure may do work in losing that pressure. 

That work may be done, fluids under pressure are necessary and thai the 
greatest amount of work mxiy be done per unit of fluid the fluid itself must be 
expansive, that is, it mv^t be a gOrS or a vapor. Gases or vapors under pressure 
are, therefore, prerequisites to the economical use of fluids for the doing of work, 
and that this work may be done at the expense of heat or derived from heat, it 
is only necessary that the heat be used to create the necessary primary con- 
dition of high pressure in vapors and gases. There are two general ways of 
accomplishing this purpose — first, to apply the heat to a boiler supplied with 
liquid and discharging its vapor at any pressure as high as desired or as high 
as may be convenient to manage; second, to apply the heat to a gas confined 
in a chamber, raising its pressure if the chamber be kept at a fixed volume, 
which is an intermittent process, or increasing the fluid volume if the size of the 
chamber be allowed to increase, the fluid pressure being kept constant or not, 
and this latter process may be intermittently or continuously carried out. 

These two processes are fundamental to th^ steam and gas engines that are 
the characteristic prime movers or power generators of engineering practice, 
utilizihg heat energy, and with the exception of water-wheels the sole commer- 
cially useful sources of power of the industrial world. Thus, the heating of 
gases and the evaporation of liquids are two most important thermal processes 
to be examined together with their inverse, cooling and condensation, and 
necessarily associated in practical apparatus with the heating and cooling of 
solid containers or associated Uquids. From the power standpoint, the effects 
of heat on solids, liquids, gases and vapors, both without change of state and with 
change of state are fundamental, and the substances to be studied as heat carriers 
do not include the whole, known chemical world, but only those that are cheap 
enough to be used in engineering practice or otherwise essential thereto. These 
substances of supreme importance are, of course, air and water, with all their 

398 



HEAT AND MATTER 399 

physical and chemical variations, next the fuels, coal, wood, oil, alcohol and 
ombustible gases, together with the chemical elements entering into them 
ind the chemical compounds which mixed together may constitute them. 

Probably next in importance from the standpoint of engineering practice 
^re the substances and thermal processes entering into mechanical refriger- 
ation and ice making. There are but three substances of commercial importance 
lere — ammonia, pure and in dilute aqueous solution, carbonic acid and air. 
The process of heating or cooling solids, liquids, gases and vapors, together 
vith solidification of water into ice, evaporation and condensation, fundamental 

power problems, are also of equal importance here, but there is added an 
idditional process of absorption of ammonia vapor in water and its discharge 
rom the aqueous solution. 

Many are the practical applications of heat transfer or transmission, some 
){ which call into play other substances than those named. In the heating 
)f buildings there is first combustion with transfer of heat to water in boilers, 
iow of the hot water or steam produced to radiators and then a transfer of heat 
.0 the air of the room; in feed-water heaters, heat of exhaust steam warms 
^ater on its way to the boilers; in economizers, heat of hot flue gases is trans- 
'erred to boiler feed water; in steam superheaters, heat of hot flue gases is trans- 
ferred to steam previously made, to raise its temperature, steam pipes, boiler 
surfaces and engine cylinders transfer heat of steam to the air which is opposed 
by covering and lagging, in steam engine condensers heat of exhaust steam is 
transferred to circulation water; in cooling cold storage rooms and making ice, 

1 solution of calcium or sodium chloride in water is circulated through pipes 
and tanks and is itself kept cool in brine coolers in which the brine transfers 
the heat absorbed in the rooms and tanks, to the primary substance ammonia 
3r carbonic acid and evaporates it. 

While evaporation and condensation as processes are fundamental to the 
machinery and apparatus of both power and refrigeration, they also are of 
importance in certain other industrial fields. In the concentration of 
s:olutions to promote crystallization such, for example, as sugar, evapora- 
tion of the solution and condensation of the distillate are primary processes 
as also is the case in making gasolene and kerosene from crude oil, in the making 
of alcohol from a mash, and many other cases found principally in chemical 
manufacture. These are examples of evaporation and condensation in which 
little or no gases are present with the vapor but there are other cases in which 
a gas is present in large proportion, the thermal characteristics of which are 
different as will be seen later. Among these processes are: the humidification 
or moistening of air with water in houses and factories to prevent excessive- 
skin evaporation of persons breathing the air, excessive shrinkage of wood-work 
and to facilitate the manufacturing processes like tobacco working and thread 
spinning. Conversely, air may be too moist for the purpose, in which case it 
is dried by cooling it and precipitating its moisture as rain or freezing it out as 
ice, and this is practiced in the Gayley process of operating blast furnaces, where 
excess of moisture will on dissociating absorb heat of coke combustion and reduce 



400 ENGINEERING THERMODYNAMICS 

the iron output per ton of coke, and in the factories where, for example, collodion 
is worked, as in tb'^ manufacture of photographic films, with which moisture 
seriously interferes*. Of course, himiidification of air by water is accomplished 
only by evaporation of water, and evaporation of water is only to be accomplished 
by the absorption of heat, so that humidification of air by blowing it over water 
or spraying water into it must of necessity cool the water, and this is the prin- 
ciple of the cooling tower or cooling pond for keeping down the temperature 
of condenser circulating water, and likewise the principle of the evaporative 
condenser, in which water cooler and steam condenser are combined in one. 
The same process then, may serve to cool water if that is what is wanted, or to 
moisten air, when dry air is harmful, and may also serve to remove moisture 
from solids like sand, crystals, fabrics, vegetable or animal matter to be reduced 
to a dryer or a pulverized state. 

There are some important examples of humidification in which the substances 
are not air and water, and one of these is the humidification of air by gasolene 
or alcohol vapor to secure* explosive mixtures for operating gas engines. Here 
the air vaporizes enough of the fuel, humidifjring or carburetting itself to serve 
the purpose, sometimes without heat being specifically added and sometimes 
with assistance from the hot exhaust. A somewhat similar action takes place 
in the manufacture of carburetted water gas when the water gas having no 
illuminating value is led to a hot brick checkerwork chamber supplied with a 
hydrocarbon oil, the vapor of which humidifies the gas, the heat of vaporization 
being supplied by the hot walls and regularly renewed as the process is inter- 
mittent. Of course, in this case some of the vapors may really decompose 
into fixed gases, a peculiar property of the hydrocarbon fuels, both liquid and 
gaseous, and frequently leaving residues of tar, or soot, or both. 

Finally, among the important processes there is to be noted that of gasifica- 
tion of solid and liquid fuels in gas producers and vaporizers, a process also 
carried on in blast furnaces in which it is only an accidental accompaniment 
and not the primary process. Some of the actions taking place in gas producers 
are also common to the manufacture of coal gas, and coke, in retorts, beehive 
and by-product ovens. 

From what has been said it should be apparent that engineers are concerned 
not with any speculations concerning the nature of heat but only with the kind 
and quantity of effect that heat addition to, or abstraction from, substances 
may be able to produce and not for all substances either. While this interest 
is more or less closely related to philosophic inquiry, having for its object the 
development of all embracing generalizations or laws of nature, and to the 
relation of heat to the chemical and physical constitution of matter, subject 
matter of physical chemistry, the differences are marked, and a clearly defined 
field of application of laws to the solution of nurtierical problems dealing with 
identical processes constitutes the field of engineering thermodynamics. 

It is not possible or desirable to take up and separately treat every single 
engineering problem that may rise, but on the contrary to employ the scientific 
methods of grouping thermal processes or substance effects into types. 



HEAT AND MATTER 401 

Prob. 1. Water is forced by a pump through a feed-water heater and economizer 
to a boiler where it is changed to steam, which in turn passes through a superheater 
to a cylinder from which it is exhausted to a condenser. Which pieces of apparatus 
have to do with heat effects and which with work? Point out similarities and 
differences of process. 

Prob. 2. Air is passed over gasolene in a carburetter; the mixture is compressed, 
burned and allowed to expand in a gas engine cylinder. Which of the above steps have 
to do with heat effects and which with work effects? 

Prob. 3. In certain types of ice machines liquid ammonia is allowed to evaporate, 
the vapor which is formed being compressed and condensed agmn to liquid. Which of 
these steps is a work phase and which a heat phase? Compare with Problem 1. 

Prob. 4. When a gun is fired what is the heat phase and what is the work 
phase? Are they separate or coincident? 

Prob. 5. Air is compressed in one cylinder, then it is cooled and compressed to 
higher pressure and forced into a tank. The air in the tank cools down by giving 
up heat to the atmosphere. From the tank it passes through a pipe hne to a heater 
and then to an engine from which it is exhausted to the atmosphere. Which steps 
in the cycle may be regarded as heat and which as work phases? Compare with 
Problem 2. 

2. Classification of Heating Processes. Heat Addition and Abstraction 
withy or without Temperature Change. Qualitative Relations. That heat 
will pass from a hot to a less hot body if it gets a chance is axiomatic, so that a 
body acquiring heat may be within range of a hotter one, the connection between 
them being, either inmiediate, that is they touch each other, or another body 
may connect them acting as a heat carrier, or they may be remote with no more 
provable connection than the hypothetic ether as is the case with the sun and 
earth. A body may gaiii heat in other ways than by transfer from a hotter 
body, for example, the passage of electrical current through a conductor will 
heat it, the rubbing of two solids together will heat both or perhaps melt one, 
the churning of a liquid will heat it, the mixing of water and sulphuric acid will 
produce a hotter liquid than either of the components before mixture, the absorp- 
tion by water of ammonia gas will heat the liquid. All these and many other 
similar examples that might be cited have been proved by careful investigation 
partly experimental, and partly by calculation based on various hypotheses to 
be examples of transformation of energy, mechanical, electrical, chemical, into 
the heat form. While, therefore, bodies may acquire heat in a great many 
different concrete ways they all fall under two useful divisions: 

(a) By transfer from a hotter body; 

(6) By transformation into heat of some other energy manifestation. 

One body may be said to be hotter than another when it feels so to the 
sense of touch, provided neither is too hot or too cold for injury to the tissues, 
or more generally, when by contact one takes heat from the other. Thus, 
ideas of heat can scarcely be divorced from conceptions of temperature and the 
definition of one will involve the other. As a matter of fact temperature as 
indicated by any instrument is merely an arbitrary number located by some- 
body on a scale, which is attached to a substance on which heat has some visible 



402 ENGINEERING THERMODYNAMICS 

effect. Temperature is then a purely arbitrary, though generally accepted, 
number indicating some heat content condition on a scale, two points of which 
have been fixed at some other conditions of heat content, and the scale space 
between, divided as convenient. Examination of heat effects qualitatively 
will show how thermometers might be made or heat measured in terms of any 
handy effect, and will also indicate what is hkely to happen to any substaHce 
when it receives or loses heat. Some of the more common heat effects of various 
degrees of importance in engineering work are given below: 

Expansion of Free Solids. Addition of heat to free solids will cause them to 
expand, increasing lengths and volumes. Railroad rails and bridges are longer 
in siunmer than winter and the sunny side of a building becomes a little higher 
than the shady side. Steam pipes are longer and boilers bigger hot, than cold, 
and the inner shell of brick chinmeys must be free from the outer to permit 
it to grow when hot without cracking the outer or main supporting stack body. 
Shafts running hot through lack of lubrication or overloading in comparatively 
cool bearing boxes may be gripped tight enough to twist off the shaft or merely 
score the bearing. 

Stressing of Restrained Solids. A solid being heated may be restrained 
in its tendency to expand, in which case there will be set up stresses in the mate- 
rial which may cause rupture. Just as with mechanically applied loads, bodies 
deform in proportion to stress up to elastic limit, as stated by Hooke's law, so 
if when being heated the tendency to expand be restrained the amount of 
deformation that has been prevented determines the stress. A steam pipe 
rigidly fixed at two points when cold will act as a long column in compression 
and buckle when hot, the budding probably causing a leak or rupture. If fixed 
hot, it will tend to shorten on cooUng and being restrained will break something. 
Cylinders of gas engines and air compressors are generally jacketed with water 
and becoming hot inside, remaining cold outside, the inner skin of the metal 
tends to expand while the outer skin does not. One part is, therefore, in tension 
and the other in compression, often causing cracks when care in designing is 
not taken and sometimes in spite of care in large gas engines. 

Expa^^^n of Free Liquids. Heating of liquids will cause them to expand 
just as do solids, increasing their volume. Thus, alcohol or mercury in glass 
tubes will expand and as these liquids expand more than the glass, a tube which 
was originally full will overflow when hot, or a tube of very small bore attached 
to a bulb of cold liquid will on heating receive some liquid; the movement of 
liquid in the tube if proportional to the heat received will serve as a thermometer. 
If the solid containing the Uquid, expanded to the same degree as the liquid 
there would be no movement. Two parts of the same liquid mass may be 
unequally hot and the hotter having expanded will weigh less per cubic foot, 
that is, be of less density. Because of freedom of movement in liquids the lighter 
hot parts will rise and the cooler heavy parts fall, thus setting up a circulation, 
the principle of which is used in hot water heating systems, the hot water from 
the furnace rising to the top of the house through one pipe and cooling on its 
downward path through radiators and return pipe. In general then, liquids 



HEAT AND MATTER 403 

crease in density on heating and increase in density on cooling, but a most 
iportant exception is water, which has a point of maximum density just 
ove the freezing-point, and if cooled below this becomes not heavier but lighter. 
)nsequently, water to be cooled most rapidly should be cooled at first from 
e top and after reaching this point of maximum density, from the bottom, if it 
to h6 frozen. 

Rise of Pressure in Confined Liquids. When liquids are restrained from 
panding under heating they suffer a rise of pressure which may burst the 
ntaining vessel. For this reason, hot water heating systems have at their 
gbest point, open tanks, called expansion tanks, which contain more water 
len the system is hot than when cold, all pipes, radiators and furnaces being 
nstantly full of water. Should this tank be shut off when the water is cold 
mething would burst, or joints leak, before it became very hot. 

Expansion of Free Gases. Just as solids and liquids when free expand under 
mating, so also do gases and on this principle chinmeys and house ventilation 
stems are designed. The hot gases in a chimney weigh less per cubic foot 
lan cooler atmospheric air; they, therefore, float as does a ship on water, 
le superior density of the water or cold gas causing it to flow under and 
t the ship or hot gas, respectively. Similarly, hot-air house furnaces and 
^ntilating systems having vertical flues supplied with hot air can send it upward 
f simply allowing cold air to flow in below and in turn being heated flow up 
id be replaced. 

Rise of Pressure in Confined Gases. Gases when restrained from expanding 
ider heat reception will increase in pressure just as do liquids, only over greater 
inges, and as does the internal stress increase in solids when heated under 
straint. It is just this principle which lies at the root of the operation of 
ins and gas engines. Confined gases are rapidly heated by explosive combus- 
on and the pressure is thus raised sufficiently to drive projectiles or pistons 
I their cylinders. 

Melting of Solids. It has been stated that solids on being heated expand 
at it should be noted that this action cannot proceed indefinitely. Continued 
^ting at proper temperatures will cause any solid to melt or fuse, and the pre- 
iously rising temperature will become constant during this change of state, 
hus, melting or fusion is a process involving a change of state from soUd 
) liquid and takes place at constant temperature. The tanks or cans of ice- 
laking plants containing ice and water in all prc^ortions retain the same 
?mperature until all the water becomes ice, provided there is a stirring or cir- 
ilation so that one part communicates freely with the rest and provided also 
le water is pure and contains no salt in solution. Impure substances, such as 
quid solutions, may suffer a change of temperature at fusion or solidification, 
or pure substances, melting and freezing, or fusion and solidification, are 
onstant temperature heat effects, involving changes from solid to liquid, or 
quid to solid states. 

Boiling of Liquids. Ebuilition. Continued heating of solids causes fusion, 
nd similarly continued heating of liquids causes boiling, or change of state from 



404 ENGINEERING THERMODYNAMICS 

liquid to vapor, another constant temperature process — ^just what temj>erati 
will depend on the pressure at the time. So constant and convenient is tl 
temperature pressure relation, that the altitude of high mountains can be foi 
from the temperature at which water boils. The abstraction of heat from 
vapor will not cool it, but on the contrary cause condensation. Steam boi 
and ammonia refrigerating coils and coolers are examples of evaporating apps 
tus, and house heating radiators and steam and ammonia condensers of c«3 
densing apparatus. 

Evaporation of Liquids; Humidification of Gases, When dry winds b!oi 
over water they take up moisture in the vapor form by evaporation at 
temperature. This sort of evaporation then must be distinguished from ebi 
lition and is really a heat effect, for without heat being added, liquid c{ 
change into vapor; some of the necessary heat may be supplied by the watt 
and some by the air. This process is general between gases and liquids and 
the active principle of cooling towers, carburetters, driers of solids like woa 
kilns. The chilling of gases that carry vapors causes these to condense in part 
As a matter of fact it is not necessary for a gas to come into contact to produa 
this sort of evaporation from a liquid, for if the Uquid be placed in a vacuui 
some will evaporate, and the pressure finally attained which depends on the t^m 
perature, is the vapor pressure or vapor tension of the substance, and the amoun 
that will so evaporate is measured by this pressure and by the rate of remova 
of that which formed previously. 

Evaporation of Solids. Sublimation. Evaporation, it has been shown, mai 
take place from a liquid at any temperature, but it may also take place directli 
from the solid, as ice will evaporate directly to vapor either in the presena 
of a gas or alone. Ice placed in a vacuum will evaporate until the vapor tensioi 
is reached, and it is interesting to note that the pressure of vapors above thei 
solids is not necessarily the same as above their liquids at the same temperature 
though they merge at the freezing-point. This is the case with ice-wata" 
water vapor. 

Change of Viscosity. Heating of liquids may have another effect measured 
by their tendency to flow, or their viscosity. Thus, a thick oil will flow eaae 
when heated, and so also will any liquid. If, therefore, the time for a giva 
quantity to flow through a standard orifice under a given head or pressure U 
measured, this time, which is the measure of viscosity, will be less for any Mquk 
hot, than cold, for the same liquid. Viscosity then decreases with heat additioi 
and temperature rise. 

Dissociation of Gases. When gases ndi simple are heated and the heatinf 
continued to very high temperatures, they will split up into their elements oi 
perhaps into other compound gases. This may be called decomposition or, bett€f, 
dissociation, and is another heat effect. Thus, the hydrocarbon C2H4 will 
split up with solid carbon soot C and the other hydrocarbon CH4 and steam 
H2O into hydrogen and oxygen. This is not a constant temperature process, 
but the per cent dissociated increases as the temperature rises. 

Dissociation of Liquids. Similar to the dissociation of gases receiving heat al 



HEAT AND MATTER 405 

^ temperature is the decomposition of some liquids in the liquid state, notably 
^ fuel and lubricating oils, or hydrocarbons which are compounds of H and C 
various proportions, each having different properties. Sometimes these 
wges of H and C groupings from the old to the new compounds under the 
luenee of heating will be at constant and at other times at varying tempera- 
■es; sometimes the resulting substances remain liquid and sometimes soot 
C separates out, and this is one of the causes for the dark color of some 
inder oils. 

Absorption of Gases in Liquids. Liquids will absorb some gases quite freely; 
IS, water will absorb very large quantities of ammonia, forming aqua anmionia. 
idition of heat will drive off this gas so that another heat effect is the expul- 
n of gases in solution. Use is made of this industrially in the absorption 
{tem of ammonia refrigeration. 

SolvbilUy of Solids in Liquids, The heating of liquids will also affect their 
ubility for solid salts; thus, a saturated solution of brine will deposit crystals 
heat abstraction and take them back into solution on heat addition, 
rtain scale-forming compounds are thrown down on heating the water in- 
ided for boilers, a fact that is made use of in feed-water heating purifiers; 
' these salts increase of temperature reduces solubility. In general then heat 
dition affects the solubility of liquids for solid salts. 

Chemical Reaction, Combustion. If oxygen and hydrogen, or oxygen and 
rbon, be heated in contact, they will in time attain an ignition temperature at 
lich a chemical reaction will take place with heat liberation called combus- 
•n, and which is an exothermic or heat-freeing reaction. Another and 
ferent sort of reaction will take place if CO2 and carbon be heated together, 
• these will together form a combustible gas, CO, under a continuation of heat 
leption. This is an endothermic or heat-absorbing reaction. Neither of 
ese will take place until by heat addition the reaction temperature, called 
lition temperature for combustion, has been reached. 
Electrical and Magnetic Effects. Two metals joined together at two separate 
ints, one of which is kept cool and the other heated, will be found to carry 
electric current or constitute a thermo-electric couple. Any conductor 
nrjing an electric current will on changing temperature suffer a change of 
flstance so that with constant voltage more or less current will flow; this is 
second electrical heat effect and like the former is useful only in instru- 
jnts indicating temperature condition. A fixed magnet will suffer a change 
magnetism on heating so that heat may cause magnetic as well as electric 
ects. 

These heat effects on substances as well as some others of not so great engi- 
ering importance may be classified or grouped for further study in a variety 
ways, each serving some more or less useful purpose. 

Reversible and Non-reversible Processes, There may be reversible and 

in-reversible thermal processes, when the process may or may not be con- 

iered constantly in a state of equilibrium. For example, as heat is applied 

boiling water there is a continuous generation of vapor in proportion to the 



r" 



406 ENGINEERING THERMODYNAMICS 

heat received; if at any instant the heat application be stopped the evap^ 
ration will cease and if the flow of heat be reversed by abstraction, condens* 
tion will take place, indicating a state of thermal equilibrium in which lh 
effect of the process follows constantly the direction of heat flow and is co> 
stantly proportional to the amoimt of heat numerically, and in sign, of directitjs, 
As an example of non-reversible processes none is better than combustion, 
which the chemical substances receive heat with proportional temperature 
until chemical reaction sets in, at which time the reception of heat has no fur- 
ther relation to the temperatures, because of the liberation of heat by coai- 
bustion which proceeds of itself and which cannot be reversed by h 
abstraction. Even though a vigorous heat abstraction at a rate greater th 
it is freed by combustion may stop combustion or put the fire out, no amo 
of heat abstraction or cooling will cause the combined substances to chaoa 
back into the original ones as they existed before combustion. The effect d 
heat in such cases as this is, therefore, non-reversible. 

Constant and Variable Volume or Density. When gitses, liquids or soKii 
are heated they expand except when prevented forcibly from so doing, and ai 
a consequence they suffer a reduction of density with the increase of volume 
this is, of course, also true of changing liquids to their vapors. It should 
noted that all such changes of volume against any resistance whatever, oc 
with corresponding performance of some work, so that some thermal proce* 
may directly result in the doing of work. Heating accompanied by no vol 
change and during which restraints are applied to keep the volume invariable 
cannot do any work or suffer any change of density, but always results in chan^ 
of pressure in liquids, gases and vapors and in a corresponding change of intemi 
stress in solids. 

Constant and Variable Temperature Processes, Another useful division, aa 
that most valuable in the calculation of relations between heat effect and hei 
quantity, recognizes that some of the heating processes and, of course, coolki^ 
occur at constant temperature and others with changing temperature- Fi 
example, the changes of state from liquid to solid, and solid to liquid, or freeziu 
and fusion, are constant temperature processes in which, no matter how mud 
heat is supplied or abstracted, the temperature of the substance changing st.at4? i 
not affected, and the same is true of ebullition and condensation, or the changin 
of state from liquid to vapor, and vapor to liquid. These latter constant 
temperature processes must not be confused with evaporation, which m*] 
proceed from either the solid or liquid state at any temperature whether constan 
or not. 

Prob. 1. From the time a fire is lighted under a cold boiler to the time steaa 
first comes off, what heat effects take place? 

Prob. 2. What heat effects take place when a piece of ice, the temperature o 
which is 20*^ F., is thrown onto a piece of red-hot iron? 

Prob. 8. What heat effects must occur before a drop of water may be evaporate 
from the ocean, and fed back into it as snow? 

Prob. 4. What heat changes take place when soot 13 formed from coal or oil? 



HEAT AND MATTER 407 

Prob. 6. In a gas producer, coal is burned to COi, which is then reduced to CO. 
Steam is also fed to the producer, and H and formed from it. Give all the heat 
effects which occur. 

Prob. 6. By means of what heat effects have you measured temperature changes, 
or have known them to be measured? 

Prob. 7. When the temperature changes from 40° F. to 20° F., give a list of all 
heat effects you know that commonly occur for several common substances. Do the 
same for a change in the reverse direction. 

Prob. 8. If a closed cyUnder be filled with water it will burst if the temperature 
be lowered or raised sufficiently. What thermal steps occur in each case? 

Prob. 9. If salt water be lowered sufficiently in temperature, a cake of fresh ice 
and a rich salt solution will be formed. State the steps or heat effects which occur 
during the process. 

3. Thennometiy Based on Temperature Change Heat Effects. Ther- 
mometer and Absolute Temperature Scales. Those thermal proceeses in which 
heat addition or abstraction is followed as a result by a corresponding and more 
or less proportional temperature change, are quite numerous and important 
both in engineering practice and as furnishing a means for thermometer-mak- 
ing, and temperature definition and measurement. . According to Sir William 
Thomson " every kind of thermoscope must be foimded on some property 
of matter continuously varying with the temperature " and he gives the fol- 
lowing: 

(a) Density of fluid under constant pressure. 

(6) Pressure of a fluid under a constant volume envelope. 

(c) Volume of the liquid contained in a solid holder (ordinary mercury or 

spirit thermometer). 

(d) Vapor pressure of a solid or liquid. 

(e) Shape or size of an elastic solid under constant stress. 
(/) Stress of an elastic solid restrained to constant size. 
ig) Density of an elastic solid under constant stress. 

{h) Viscosity of a fluid, 
(t) Electric current in a thermo-couple. 
(j) Electric resistance of a conductor. 
{k) Magnetic moment of a fixed magnet. 

Any, or all of these — ^pressure, volume, shape, size, density, rate of flow, 
magnetic or electrical effects, may be measured, and their measure constitutes 
a measure of temperature indirectly, so that instruments incorporating these 
temperature effects to be measured, are also thermometers. 

Any temperature-indicating device may be called a thermometer, though 
those in use for high temperatures are generally called pyrometers, which 
indicates the somewhat important fact that no thermometer is equally useful 
for all ranges of temperature. Practically all thermometers in use for tempera- 
tures short of a red heat, depend on certain essential relations between the density 
or volume, the pressure and temperature of a fluid, though metals are used in 
some little-used forms in which change of size is measured, or change of shape 



408 ENGINEERING THERMODYMAMICS 

of a double metallic bar, often brass and iron, consisting of a piece of each 
fastened to the other to form a continuous strip. The two metals are expanded 
by the temperature different amounts causing the strip to bend under heating. 
There are also in use electric forms for all temperatures, and these are the 
only reliable ones for high temperatures, both of the couple and resistance 
types except one dependent on the color of a high temperature body, black 
when cold. That most useful and common class involving the interde- 
pendence of pressure and temperature, or volume and temperature, of a fluid 
is generally found in the form of a glass bulb or its equivalent, to which 
is attached a long, narrow glass tube or stem which may be open or closed at 
the end; open when the changes of fluid voliune at constant pressure are to 
be observed and closed when changes of contained fluid pressure at constant 
restrained volume are to be measured as the effect of temperature changes. 
For the fluid there is used most commonly a liquid alone such as mercury, or 
a gas alone such as air; .though a gas may be introduced above mercury and 
there may be used a liquid with its vapor above. When the fluid is a liquid, 
such as mercury, in the common thermometer, the stem is closed at the end so 
that the mercury is enclosed in a constant-volume container or as nearly so a? 
the expansion or deformation of the glass will permit, which is not filled with 
mercury, but in which a space in the stem is left at a vacuum or filled with a 
gas under pressiu'e, such as nitrogen, to resist evaporation of the mercur}-^ at 
high temperatures. Gas-fill6d mercury thermometers, as the last form is called, 
are so designed that for the whole range of mercury expansion the pressure 
of the gas opposing it does not rise enough to offer material resistance to the 
expansion of mercury or to unduly stress the glass container. It should be 
noted that mercury thermometers do not measure the expansion of mercun- 
alone, but the difference between the voliune of mercury and the glass envelope, 
but this is of no consequence so long as this difference is in proportion to the 
expansion of the mercury itself, which it is substantially, with proi)er glass 
composition, when the range is not too great. Such thermometers indicate 
temperature changes by the rise and fall of merciu'y in the stem, and any numeri- J 
cal value that may be convenient can be given to any position of the mercur}^ 
or any change of position. Common acceptance of certain locations of the scale 
number, however, must be recognized as rendering other possible ones unneces- 
sary and so undesirable. Two such scales are recognized, one in use with metric 
units, the centigrade, and the other with measurements in English units, the 
Fahrenheit, both of which must be known and familiar, because of the frequent 
necessity of transformation of numerical values and heat data from one sj'stem 
to the other. To permit of the making of a scale, at least two points must be 
fixed with a definite nmnber of divisions between them, each called one degree. 
The two fixed points are first, the position of the mercury when the thermometer 
is in the vapor of boiling pure water at sea level, or under the standard atmos- 
pheric pressure of 29.92" = 760 mm. of mercury absolute pressure, and 
second, the position of the mercury when the thermometer is surrounded by 
melting ice at the same pressure. These are equivalent to the boiling- or con- 



HEAT AND MATTER 



409 



densation, and melting- or freezing-points, of pure water at one atmosphere 
pressure. The two accepted thermometer scales have the following character- 
istics with respect to these fixed points and division between them: 

THERMOMETER SCALES 





Pure Water 
Freeiiag-poiDt. 
at one aim. pr. 


Pure Water 
Boiling-point, 
at one atm. pr. 


Number of Equal Divisions 
Between Freesing and BoUing. 


Centigrade scale 



32 


100 
212 


100 


Fahrenheit scale 


180 







From this it appears that a degree of temperature change is on the centigrade 
scale, ^l^ of the linear distance between the position of the mercury surface 
at the freezing- and boiling-^ints of water, and on the Fahrenheit scale, yf^ of 
the same distance. From this the relation between a degree temperature change 
for the two scales can be given. 



One degree temperature 
change centigrade 



180^9 
100 5 



of one degree temperature 
change Fahrenheit; 



or 



One degree temperature 
change Fahrenheit 



of one degree temperatv/re 
change centigrade. 



It is also possible to set down the relation between scale readings, for when the 
temperature is 0^ C, it is 32° F., and when it is 100° C. it is (180+32) =212° F., 
so that 

9 
Temperature Fahrenheit =32+— (Temperature centigrade), 



or 



Temperature centigrade = q (Temperature Fahrenheit— 32). 



For convenience of numerical work tables are commonly used to transform 
temperatures from one scale to the other and such a transformation is shown 
in a curve. Fig. 119, and in Table XXIX at end of the Chapter. 

By reason of the lack of absolute proportionality between temperature 
and effect, other fixed points are necessary, especially at high temperatures, 
and the following of Table XVI have been adopted by the U. S. Bureau of Stand- 
ards and are considered correct to within 5° C, at 1200° C. 



ENGINEEHINa THERMODYNAMICS 



Dcsrcea CoutlBnule 

FiQ. 119. — Graphical Relation between Centigrade and Fahrenheit Thermometer Scalea. 



HEAT AND MATTER 



411 



TABLE XVI 
FIXED TEMPERATURES 

U. S. BUREAU OP STANDARDS 



Temperftture, 

•c. 


Temponture, 


Determined by the Point at which 


232 


449 


Liquid tin solidifies 


327 


621 


Liquid lead solidifies 


419.4 


787 


Liquid zinc solidifies 


444.7 


832.6 


Liquid sulphur boils 


630.5 


1167 


Liquid antimony solidifies ^ 


658 


1216 


Liquid aluminum, 97.7% pm'e, solicUfies 


1064 


1947 


Solid gold melts 


1084 


1983 


Liquid copper solidifies 


1435 


2615 


Solid nickel melts 


1546 


2815 


Solid palladium melts 


1753 


3187 


Solid platinum melts 



Thermometers in which a liquid and its vapor exist together, depend on a 
property to be noted in detail later, the relation of vapor pressure to tempera- 
ture and its independence of the volume of vapor. So long as any vapor exists 
above the liquid the temperature will depend only on the pressure of that vapor 
so that such thermometers will indicate temperature by the pressure measure- 
ment, after experimental determination of this pressure-temperature relation 
of vapors. Conversely, temperature measurements of vapors by mercury ther- 
mometers will lead to pressure values, and at the present time some steam 
plants are introducing mercury thermometers on the boilers and pipe lines, in 
place of the proverbially inaccurate pressure gages. 

Gas thermometer, is the name generally applied to the class in which the 
fluid is a gas, whether air, hydrogen, nitrogen or any other, and whether the 
pressure is measured for a fixed contsdned volume, or the volume measured 
when acted on by a constant pressure. These gas thermometers are so bulky 
as to be practically useless in ordinary engineering work and are only employed 
as standards for comparison and for tests of extraordinary delicacy in investi- 
gation work. They give much larger indications than mercury thermometers 
because the changes of gas volume under constant pressure are far greater 
than for mercury or any other liquid. Regnault was the first to thoroughly 
investigate air thermometers and reported that the second form, that of constant 
gas volume with measurement of pressure, was most useful. 

Using the centigrade scale, fixing freezing point at 0^ C, and making the 
corresponding pressure po, atmospheric at this point, and reading at 100** C. 
another pressure pioo, he found experimentally a relation between these two 
pressiures and the temperature corresponding to any other pressure p, as 
given by the empiric formula. 



<=100-P=PO-. 
Pioo—po 



(601) 



412 ENGINEERING THERMODYNAMICS 

He also determined the pressure at the boiling-point to be related to the pressure 
at the freezing-point, by 

Pioo= 1.3665 po, 
which on substitution gives 

This constant, 272.85, the reciprocal of which is .003665, is, of course, the pressure 
increase factor per degree C. rise of temperature for a gas held at constant 
volume, received extended investigation and it vxxs found that it had about 
the same value applied to the other type of thermometer in which gas volumes 
are measured ai constant pressure. This was true even when the pressure 
used was anything from 44 to 149 cm. of mercury, though it is reported 
that for 44 cm. pressure the value 272.98, and for 149 cm. pressure, the value 
272.7, seemed closer. For hydrogen it was found that the constant was sub- 
stantially the same as for air, while for carbonic acid it was 270.64, and while 
the hydrogen thermometers agreed with the air over the whole scale, showing 
proportional efifects, this was hardly true of carbonic acid. Such uncertainty 
in the behavior of these thermometers and in the fixing of the constants was 
traced to the glass in some cases, but there still remained differences charge- 
able only to the gases themselves. Comparison of the air with mercury ther- 
mometers showed that there was not a proportional change with the temperature 
and that temperatures on the two, consistently departed. 

Examination of Eq. (602), giving the relation between two temperatures and 
the corresponding gas pressures, will show a most important relation. If in Eq. 
(602), the pressure be supposed to drop to zero and it is assumed that 
the relations between pressure and temperature hold, then when p=0, 
i=— 272.85. This temperaiure has received the name of the absolute zero, 
and may be defined as the temperature at which pressure disappears or becomes 
zero at constant volume, and correspondingly, at which the volume also dis- 
appears, since it was foimd that similar relations existed between volume and 
temperature at constant pressure. Calling temperature on a new scale begin- 
ning 272.85** below the centigrade zero by the name absolute temperatures, 
then 

Absolute temperature 1 ^.o^o or_i f Scale temperature 
centigrade J ' [ centigrade 

As this constant or absolute temperature of the centigrade scale zero, is an 
experimental value, it is quite natural to find other values presented by differ- 
ent investigators, some of them using totally different methods. One of these 
methods is based on the temperature change of a gas losing pressure without 
doing work, generally described as the porous plug experiment, and the results 



HEAT AND MATTER 413 

as the Joule-Thomson effect, and another is based on the coefficient of expansion 
of gases being heated. Some of these results agreed exactly with Regnault's 
value for hydrogen between 0° C. and 100° C. for which he gave —273° C. = 
—491.4° F. Still other investigations continued down to the last few years 
yielded results that tend to change the value slightly to between — 491.6°F., 
and —491.7° F., and as yet there is no absolute agreement as to the exact value.. 
In engineering problems, however, it is seldom desirable or possible to work 
to such degrees of accuracy as to make the uncertainty of the absolute zero a 
matter of material importance, and for practical purposes the following values 
may be used with sufficient confidence for all but exceptional cases which are 
to be recognized only by experience. 



I Centigrade =273] 



Absolute Temperature (T) j Fahrenheit =460 



+Scale Temperature (0 



When great accuracy is important it is not possible at present to get a better 
Fahrenheit value than 459.65, the mean of the two known limits of 459.6 and 
459.7, though Marks and Davis in their Steam Tables have adopted 459.64, which 
is very close to the value of 459.63 adopted by Buckingham in his excellent 
Bulletin of U. S. Bureau of Standards and corresponding to 273.13 on the 
centigrade scale. 

These experiments with the gas thermometers, leading to a determination 
of temperature as a function of the pressure change of the gas held at constant 
volume, or its voliune change when held at constant pressure, really supply a 
definition of temperature which before meant no more than an arbitrary number, 
and furnished a most valuable addition to the generalization of relations between 
heat content of a body and its temperature or physical state. 

A lack of proportionality between thermometer indication and temperature, 
has already been pointed out, and it is by reason of this that two identical ther- 
mometers, or as nearly so as can be made, with absolute agreement between 
water boiling- and freezing-points, will not agree at all points between, nor will 
the best constructed and calibrated mercury thermometers agree with a similarly 
good gas thermometer. 

The temperature scale now almost universally adopted as standard is that 
of the constant volume hydrogen gas thermometer, on which the degree F. 
is one one-himdred-and-eightieth part of the change in pressure of a fixed 
voliune of hydrogen between melting pure ice, and steam above boiling pure 
water, the initial pressure of the gas at 32° being 100 cm. = 39.37 ins. Hg. A 
mercury in glass thermometer indication is, of course, a measure of the proper- 
ties of the mercury and glass used, and its F. degree of temperature is defined 
in parallel with the above as one one-hundred-and-eightieth part of the volume 
of the stem between its indications at the same two fixed points. A comparison 
of the hydrogen thermometer and two different glasses incorporated in mercury 
thermometers is given below, Table XVII, from the Bulletin of the U. S. Bureau 
of Standards, v. 2, No. 3, by H. C. Dickinson, quoting Mahlke, but it must be 



414 



ENGINEERING THERMODYNAMICS 



remembered that other glasses will give different results ana even different 
thermometers of the same glass when not similarly treated. 

Table XVII 

FAHRENHEIT TEMPERATURES BY HYDROGEN AND MERCURY 

THERMOMETERS 



Temporsture by 

Hydrogen 
Thermometer. 



32 
212 
302 
392 
428 
464 
500 
536 
672 



Difference in 

Reading bv 

Mercury m Jena 

50'' Glass. 






+ 1.3 



Difference in 

Reading by 

Mercury in 18" 

Jena Glass. 






- .18 
+ .072 
+ .39 
-h .83 
+1.79 
+2.4 
+3.53 



Temperature by 

Hydrogen 
Thermometer. 



617 
662 
707 
752 
797 
842 
887 
932 



Difference in 

Reading bv 

Mercury in Jena 

59" Glass. 



+10.6 
+ 16.6 
+18.7 
+24.6 
+28.2 
+38.3 
+41.4 
+50.0 



Difference in 

Reading by 

Mercury in 16" 

Jena Glass. 



Useful comparisons of air, hydrogen, nitrogen, and other gases, with alcohol 
and mercury, in various kinds of glass, are given in the Landolt-Bomstein- 
Meyerhoffer, and in the Smithsonian Physical Tables, but are seldom needed 
for engineering work. 

One sort of correction that is often necessary in mercury thermometer 
work is that for stem immersion. Thermometers are calibrated as a rule with 
the whole stem immersed in the melting ice or the steam, but are ordinarily 
used with part of the stem exposed and not touching the substance whose tem- 
perature is indicated. For this condition, the following correction is recom- 
mended by the same Bureau of Standards Bulletin: 



When n 
t 
h 



Stem correction = .000088 n{t-tiyF- 

number of degrees exposed ; 

temperature indicated Fahrenheit degrees; 

=mean temperature of emergent stem itself, which must necessarily 

be estimated and most simply by another thermometer next to 

it, and entirely free from the bath. 



Prob. 1. What will be the centigrade scale and absolute temperatures, for the 
following Fahrenheit readings? -25^, 25^, llO*', 140°, 220*^, 263° scale, and 300°, 
460°, 540°, 710°, 2000° absolute. 

Prob. 2. What will be the Fahrenheit scale and absolute temperatures, for the follow- 
ing centigrade readings? -20°, 10°, 45°, 80°, 400°, 610° scale, and 200°, 410°, 650°, 
810°, 2500° absolute. 

Prob. 8. By the addition of a certain amount of heat the temperature of a 
quantity of water was raised 160° F. How many degrees C. was it raised? 



HEAT AND MATTER 415 

Prob. 4. To bring water from 0° C. to its boiling-point under a certain pressure 
equired a temperature rise of 150** C. What was the rise in Fahrenheit degrees? 

Prob. 6. For each degree rise Fahrenheit, an iron bar will increase .00000648 of 
is length. How much longer will a bar be at 150** C. than at 0** C? At 910** C. 
bsolute than at 250** C. absolute? 

Prob. 6. The increase in pressure for SO2 for a rise of 100° C. is given as .3845 at 
onstant volimie. What would have been absolute zero found by Regnault had he 
sed SOs rather than air? 

Prob. 7. A thermometer with a scale from 40° F. to 700° F. is placed in a thermome- 
sr well so that the 200° mark is just visible. The temperature as given by the 
[lerraometer is 450°. If the surrounding temperature is 100° F., what is true tempera- 
are in the well? 

4. Calorimetiy Based on Proportionality of Heat E£fects to Heat Quantity, 
rnits of Heat and Mechanical Equivalent. Though it is generally recognized 
rom philosophic investigations extending over many years, that heat is one 
lanifestation of energy capable of being transformed into other forms such 
s mechanical work, electricity or molecular arrangement, and derivable from 
hem through transformations, measurements of quantities of heat can be made 
rithout such knowledge, and were made even when heat was regarded as a 
ubstance. It was early recognized that equivalence of heat effects proved 
fleets proportional to quantity; thus, the melting of one pound of ice can cool 
» poimd of hot water through a definite range of temperature, and can cool 
wo pounds through half as many degrees, and so on. The condensation of 
, pound of steam can warm a definite weight of water a definite number of 
legrees, or perform a certain niimber of pound-degrees heating effect in water, 
lo that taking the pound-degree of water as a basis the ratio of the heat liberated 
>y steam condensation to that absorbed by ice melting can be found. Other 
ubstances such as iron or oil may suffer a certain number of poimd degree 
hanges and affect water by another number of pound-degrees. The unit 
if heat quantity might be taken as that which is liberated by the condensation 
►f a pound of steam, that absorbed by the freezing of a pound of water, that to 
aise a poimd of iron any number of degrees or any other quantity of heat 
iffoct. The heat unit generally accepted is, in metric measure, the calorie, 
»r the amount to raise one kilogramme of pure water one degree centigrade, 
)T in English units, the British thermal unit, that necessary to raise one pound 
if water one degree Fahrenheit. Thus, the calorie is the kilogramme degree 
lentigrade, and the British thermal unit the poimd degree Fahrenheit, and the 
atter is used in engineering, usually abbreviated to B.T.U. There is also 
iccasionally used a sort of cross unit called the centigrade heat unit, which is 
he pound degree centigrade. 

The relation between these is given quantitatively by the conversion table 
it the end of this Chapter, Table XXX. 

All the heat measurements are, therefore, made in terms of equivalent 
^ater heating effects in pound degrees, but it must be understood that a water 
)ound degree is not quite constant. Careful observation will show that the 



416 ENGINEERING THERMODYNAMICS 

melting of a pound of ice will not cool the same weight of water from 200** F, 
to 180° F., as it will from 60° F. to 40° F., which indicates that the heat capacity 
of water or the B.T.U. per pound-degree is not constant. It is, thereforr. 
necessary to further limit the definition of the heat unit, by fixing on som* 
water temperature and temperature change, as the standard, in addition to the 
selection of water as the substance, and the pound and degree as units of capacity. 
Here there has not been as good an agreement as is desirable, some using 
4° C. = 39.4° F. as the standard temperature and the range one-half degree 
both sides; this is the point of maximum water density. Others have used 15' 
C. = 59° F. as the temperature and the range one-half degree both sides; still 
others, one degree rise from freezing point 0° C. or 32° F. There are gtx 
reasons, however, for the most common present-day practice which will prol)- 
ably become universal, for taking as the range and temperatures, freezing- 
point to boiling-point, and dividing by the niunber of degrees. The heat unit 
so defined is properly named the mean calorie or mean British thermal unit; 
therefore, 

Mean calorie =t7v7j (amount of heat to raise 1 Kg. water from 0° C. to 100° C). 
Mean B.T.U. =— (amount of heat to raise 1 lb. water from 32° F. to 212° F.i. 



In terms of the heat unit thus defined, the amount of heat per degree tem 
perature change is variable over the scale, but only in work of the most accurate 
character is this difference observed in engineering calculations, but in accurate 
work this difference must not be neglected and care must be exercised in using 
other physical constants in heat units reported by different observers, to be sure 
of the unit they used in reporting them. It is only by experience that judgment 
can be cultivated in the selection of values of constants in heat units reported 
for various standards, or in ignoring differences in standards entirely. The 
great bulk of engineering work involves uncertainties greater than these differ- 
ences and they may, therefore, be ignored generally. 

By various experimental methods, all scientifically carried out and exetnding 
over sixty years, a measured amount of work has been done and entirely eon- 
verted into heat, originally by friction of solids and of liquids, for the deter- 
mination of the foot-pounds of work equivalent to one B. T. U., when the 
conversion is complete, that is, when all the work energy has been converted mto 
heat. This thermo-physical constant is the mechanical equivalent of kat 
Later, indirect methods have been employed for its determination by cakula- 
tion from other constants to which it is related. All of these experiments 
have led to large number of values, so that it is not surprising to find doubt as 
to the correct value and different values are used even by recognized authori- 
ties. The experiments used include: 



HEAT AND MATTER 417 

(a) Compression and expansion of air; Joule. 

(6) Steam engine experiments, comparing heat in supplied and exhausted 
steam; Him. 

(c) Expansion and contraction of metals; Edlund and Haga. 

(d) Specific volume of vapor; Perot. 

(e) Boring of metals; Rumford and Him. 
(/) Friction of water; Joule and Rowland. 
(g) Friction of mercury; Joule. 

(A) Friction of metals; Him, Puluj, Sahulka. 

(i) Crushing of metals; Him. 

0) Heating of magneto electric currents; Joule. 

(k) Heating of disk between magnetic poles; Violle. 

(I) Flow of liquids (water and mercury) under pressure; Him, Bartholi. 

(m) Heat developed by wire of known absolute resistance; Quintus Icilius, 

Weber, Lenz, Joule, Webster, Dieterici. 
(n) Diminishing the heat contained in a battery when the current produces 

work; Joule, Favre. 
(o) Heat developed in, and voltage of Daniell cells; Weber, Boscha, Favre 

and Silberman, Joule, 
(p) Combination of electrical heating and mechanical action by stirring 

water; Griffiths. 
{q) Physical constants of gases. 

The results of all of these were studied by Rowland in 1880, who himself 
experimented also, and he concluded that the mechanical equivalent of heat 
was nearly 

778.6 ft.-lbs. = 1 B.T.U., at latitude of Baltimore, 
or 

774.5 ft.-lbs. = 1 B.T.U., at latitude of Manchester. 

with the following corrections to be added for other latitudes. 

Latitude 0** 10*» 20'' 30'' 40'' 50** eO** 70** 80* 90* 

Ft.-lbs 1.62 1.50 1.15 .62 .15 -.75 -1.41 -1.93 -2.30 -2.43 

Since that time other determinations have been made by Re3molds and 
Morby, using mechanical, and Griffiths, Schuste and Gannon, Callendar and 
Barnes, using electrical transformation into heat. Giving these latter deter- 
minations equal weight with those of Joule and Rowland, the average is 

1 small calorie at 20® C. (nitrogen thermometer) =4.181X10^ ergs. ^ 



418 ENGINEERING THERMODYNAMICS 

On the discussion of these results by Smith, Marks and Davis accept and use 
the mean of the results of Reynolds, and Morby and Barnes, which is 

1 mean calorie = 4. 1834 X 10^ ergs, 

= 3.9683 B.T.U. 
1 mean B.T.U. = 777.52 ft.-lbs., 

when the gravitational constant is 980.665 cm. sec^, which corresponds to 32.174 
lbs., and is the value for latitude between 45*^ and 46**. 

For many years it has been most common to use in engineering calculations, 
the round number 778, and for most problems this round number is still the 
best available figure, but where special acciu'acy is needed it is likely that no 
closer value can be relied upon than anything between 777.5 and 777.6 for the 
above latitude. 

^ Example. To heat a gallon of water from 60^ F. to 200^ F. requires the heat 
equivalent of how many foot-pounds? 

1 gallon =8.33 lbs.,' 

200^ F. -60** F. =140^ F. rise, 

8.33 X 140 = 1665 pound-degrees, 

= 1665 B.T.U. 

=778X1665, ft.-lbs. 

-90,800 ft.-lb8. 

Prob. 1. A feed-water heater is heating 5000 gallons of water per hour from 
40* F. to 200** F. What would be the equivalent energj'^ in horse-power units? 

Prob. 2. A pound of each of the following fuels has the heating values as given. 
Change them to foot-pounds. 

Average bituminous coal, 14,000 B.T.U. per lb., Average kerosene, 18,000 B.T.U. per lb. 
" small anthracite, 12,000 B.T.U. per lb., " alcohol, 10,000 B.T.U. per lb. 

Average gasolene, 20,000 B.T.U. per lb. 

Prob. 8. A cubic foot of each of the following gases yields on combusture, the 
number of heat units shown. Change them to foot-pounds. 

Natural gas (average), 880 B.T.U. per cu.ft., Carburetted water gas, 700 B.T.U. per cu.ft. 
Coal gas, 730 B.T.U. per cu.ft., Mond gas, 150 B.T.U. per cu.ft. 

Blast fimiace gas, 100 B.T.U. per cu.ft. 

Prob. 4. A pool contains 20,000 cu.ft. of water and must be warmed from 40** F. 
to 70** F. How much work might be done with the equivalent energy? 






HEAT AND MATTER 419 

Prob. 6. How many calories and how many centigrade heat units would be 
required in Prob. 4? 

Prob. 6. In the course of a test a man weighing 200 lbs. goes up a ladder 25 ft. 
high, every 15 minutes. If the test lasted 12 hours how many B.T.U. did he expend? 

Prob. 7. A reservoir contains 300 billion gallons of water which are heated each 
year from 39® F. to 70° F. What is the number of foot-poimds of work equivalent? 

Prob. 8. A pound of water moving at the rate of 450 ft. per second is brought 
to rest, so that all of its energy is turned into heat. What will be the temperature rise? 

Prob. 9. For driving an automobile 30 horse-power is being used. How long 
will a gallon of average gasolene, sp.gr. = .7, last, if 10% of its energy is converted 
into work? 

Prob. 10. Power is being absorbed by a brake on the flywheel of an engine. 
If the engine is developing 50 horse-power how many B.T.U. per minute must be 
carried off to prevent burning of the brake? 

6. Temperature Change Relation to Amount of Heat, for Solids, Liquids, 
Gases, and Vapors, not Changing State. Specific Heats. Provided gases do not 
decompose, vapors condense, liquids freeze or evaporate, and solids melt, under 
addition or abstraction of heat, there will always be the same sort of relation 
between the quantity of heat gained or lost and the temperature change 
for all, differing only in degree. As the reception of heat in each case 
causes a temperature rise proportional to it and to the weight of the sub- 
stances, this constant of proportionality once determined will give numerical 
relations between any temperature change and the corresponding amount of 
heat. Making the weight of the substance unity, which is equivalent to the 
consideration of one pound of substance, the constant of proportionality may 
be defined as the quantity of heat per degree rise, and as thus defined is the 
specific heat of the substance. Accordingly, the quantity of heat for these cases 
is equal to the product of specific heat, temperature rise and weight of substance 
heated. 

The heat, as ahready explained, may be added in two characteristic ways: 
(o) at constant volume or density, or (6) at constant pressure. It might be 
expected that by reason of the increase of volume and performance of work 
under constant pressure heating, more heat must be added to raise the tempera- 
ture of one pound, one degree, than in the other case where no such work is done, 
and both experimental and thermodynamic investigations confirm this view. 
There are, therefore, two specific heats for all substances, capable of definition: 

(a) The specific heat at constant volume, and 

(6) The specific heat at constant pressure. 
These two specific heats are quite different both for gases and for vapors, which 
suffer considerable expansion imder constant pressure heating, but for solids 
and liquids, which expand very little, the difference is very small and is to be 
neglected here. As a matter of fact, there are no cases of common engineering 
practice involving the specific heat of liquids and solids under constant volume, 
and values for the specific heats of liquids and solids are always without further 
definition to be understood as the constant pressure values. 



420 ENGINEERING THERMODYNAMICS 

Let C, be the specific heat of solids and liquids suffering no change of state. 

C,, be the specific heat of gases and vapors at constant pressure and 
suffering no change of state. 

C„ be the specific heat of gases and vapors at constant volume and suf- 
fering no change of state. 

<2 and^i, be the maximum and min i mum temperatures for the process. 

w, be the weight in pounds. 






Then will the heat added, be given by .the following equation, if the tempera- 
ture rise is exactly proportional to the quantity of heat, or in other words, 
ij the specific heat is constant. 

Q=Cw(t2^ti)y for solids and liquids (603) 

Q=sC.io(b— <i), for gases and vapors (not near condensation) when 

voliune is constant (6(H) 

Q=Cpti?(fe— <i), for gases and vapors (not near condensation) when 

pressure is constant (605) 

When, however, the specific heat is variable, as is the case for many sub- 
stances, probably for all, the above equation cannot be used except when 
the specific heat average value, or mean specific heat is used. If the variation 
is irregular this can be found only graphically, but for some substances the 
variation is regular and integration will give the mean value. It has been 
the custom to relate the specific heat to the temperature above the freez- 
ing-point of water, expressing it as the siun of the value at 32^ F., and 
some fraction of the temperatm^ above this point to the first and second 
powers, as in Eq. (606). 

Specific heat at temperature (0 =a+b(t-Z2)+c{t-32)^ . (606) 

In this equation a is the specific heat at 32^, while b and c are constants, 
different for different substances, c being generally zero for liquids. 

When this is true, the heat added is related to the temperature above 
32^ by a differential expression which can be integrated between limits 



Wt-32 

Q= I [a+6(i-32)+C(i-32)2](ft 

Jtt -32 



= a[(fe-32)-(<i-32)]+|[(fe-32)2-(ii-32)2]+|[(fe-32)3-(<i-32)3]. (607) 

Usually the heats are calculated above 32® so that the heats between any 
two temperatures will be the difference between the heats from 32® to those 
two temperatures. In this case /i=32®, and, t2 = t, whence 

B.T.U. per lb., from 32® to <,= ra+|(i-32)+|(^-32)2l(/-23). . . (608) 



HEAT AND MATTER 421 

''or this range, of temperature 32** to t, the quantity of heat may be ex- 
tressed as the product of a mean specific heat and the temperature range 

Heat from 32^ to t = (mean sp. heat from 32° to r) X (<-32). . (609) 
::omparing Eq. (608) with Eq. (609) it follows that 



Mean specific heat 
from 32° P. tor P. 



} =a+|(<-32)+|(<-32)2 (610) 



The coefficient of (1—32) in the mean specific heat expression, is half that in 
he expression for specific heat at (, and the coefficient of (i— 32)^, is one-third, 
rhis makes it easy to change from specific heat at a given temperature 
ibove 32°, to the mean specific heat from 32° to the temperature in question. 

The specific heats of some substances are directly measured, but for some 
others, notably the gases, this is too difficult or rather more difficult than cal- 
culation of values from other physical constants to which they are related. 
• It often happens that in engineering work the solution of a practical 
problem requires a specific heat for which no value is available, in which case 
the general law of specific heats, known as the law of Dulong and Petit, for 
definite compounds may be used as given in Eq. (611). 

(Specific heat of solids) X (atomic weight) = 6.4. . . . (611) 

This is equivalent to saying that all atoms have the same capacity for 
heat, and while it is known to be not slrictly true, it is a useful relation in 
the absence of direct determinations. Some values, experimentally determined 
for the specific heats of solids, are given in Table XXXI at the end of this Chap- 
ter, together with values calculated from the atomic weights to show the degree 
of agreement. The- atomic weights used are those of the International Com- 
mittee on Atomic Weights (Jour. Am. Chem. Soc, 1910). When the specific 
heat of a solid varies with temperature and several determinations are avail- 
able, only the maximum and minimum are given with the corresponding tem- 
peratures, as these usually suffice for engineering work. 

To illustrate this variability of specific heat of solids, the values deter- 
mined for two samples of iron are given in Figs. 120 and 121, the former 
showing the variation of the mean specific heat as determined by OberhofiFer 
and Harker from 500** F. up, and the latter the amount of heat per pound of 
iron at any temperature above the heat content at 500® F., which is gen- 
erally called its total heat above the base temperature, here 500** F. 

It is extremely probable that the specific heats of liquids all vary irregularly 
with temperature so that the constant values given in Table XXXII at the end 
of the Chapter must be used with caution. This is certainly the case for water, 
and is the cause of the difficulty in fixing the unit of heat, which is best solved 
by the method of means. In Fig. 122 are shown in curve form the values for the 



422 



INGINEERINQ THERMODYNAMICS 



specific heats of water at temperatures from 20** F. to 600® JF., as accepted 
by Marks and Davis after a critical study of the experimental results of 



•18 



.'16 



X 



|.14 

OQ 



S 



.12 



.10 







































■ 






















(a) 
(6) 


Oberi 
Hark 


loffer 
jr 






/ 




V 


*^^m^^^^ 


— — 


— 


>^^ aia^M 


IS) 














f 

1/ 


y 


X 


^.. 


(6) 


















y} 


'/ 






















y 


-A 


'y 






















-;? 


■:>^ 


























y 



















































1000 aooo 

^Temperature in Degrees Fahr. 



Fig. 120. — Mean Specific Heat of Iron above 500** F., Illustrating Irregular Variations not 

Yielding to Algebraic Expression. 

Barnes and Dieterici and adjustment of the differences. The integral curv^e 
is plotted in Fig. 123 which, therefore, gives the heat of water from 32®F. to any 



>'30O 

> 

< 



paoo 



o 

a 
§ 

glOO 

I 



(6) 



From 



Oberh 



t 



ffer Data 



Harker 



^^ 



y 



fiOO 



1000 



./^ 



y 



l^-'- 



^^ 



:'(6' 



^Td) 



2300 



laoo 2000 

Temperature la Degrees Fabr. 

Fio 121. — ^Total Heat of Iron above 500" F., Illustrating its Approximation to a Straight 
Line Relation in Spite of Wide Variation in Specific Heat Given in Fig. 120. 



temperature up to the highest used in steam practice and which is designated 
in steam tables, summarizing all the properties of water and steam, as the 



HEAT AND MATTER 



423 





































































































> 


















































> 


/ 
















































> 


A 
















































/^ 


^ 
















































X 














































X" 






























































































^ 


^ 


•^ 
























> 


















^ 


^ 






































. 




















































It 


» 






» 


N) 






» 


X) 






400 






600 













1.15 



I 
B-LIO 

«3 



I 



rni^OS 



LOO 



Temperature Degrees F&hr. 
Fig. 122. — Specific Heat of Water at Various Temperatures. 




iOO 400 

Temperature In Degrees Fabr. 

Fig. 123.— Total Heat of Water from 32*^ F., to any Temperature, tfie Heal of the Liquid at 

that Temperature above 32** F. 



424 



ENGINEERING THERMODYNAMICS 



heat of the liquid. For the purpose of comparison, the mean specific heat 
of water is given in Fig. 124 from 32** F. to any temperature which is obtained 
from the heat of the liquid above 32® F. by dividing it by the temperature 
above 32** F. 

In the table of specific heats of liquids there is a column giving the value 
calculated from the atomic weights to show at a glance the degree with 
which liquids satisfy the Dulong and Petit law. 

Variability of specific heat is especially noticeable in liquids that are solu- 
tions with different amounts of dissolved substance, in which case the specific 
heat varies with the density and temperature. Problems of refrigeration 
involve four cases of this kind: (a), calcium, and (b), sodium chloride, 




XO 400 

Temperature in Degrees Fahrenheit 

Fig. 124. — Mean Specific Heat of Water from 32** to any Temperature. 

brines, the densities of which vary considerably but which are used with 
but little temperature range, .seldom over 20® F. and often not over 5® F., 
(c), anhydrous ammonia and (d)y carbonic acid. 

As the density of brines is often reported on the Baum6 scale and liquid 
fuels always so, a comparison of this with specific gravities is given in Table 
XXXIII in connection with the specific heat tables at the end of this Chapter 
to facilitate calculation. 

One of the best-known solutions so far as accuracy of direct experimental 
data is concerned, is calcium brine, results for which, from 35® C. to 20® C. 
given below, are from U. S. Bureau of Standards BuUetui by Dickinson, 
Mueller and George, for densities from 1.175 to 1.250. • For chemically pure 



HEAT AND MATTER 



425 



calcium chloride in water, it was found that the following relation be- 
tween density D, and specific heat (7, at 0® C, 

Z)=2.8821-3.6272C+1.7794C2, (612) 

and these results plotted in Fig. 125 show the specific heat variation with 
temperature to follow the straight line law very nearly. This being the case 
the mean specific heat for a given temperature range is closely enough the 
arithmetical mean of the specific heat at the two limiting temperatures. To 
the figure are also added dotted, the specific heats for some commercial brines, 
not pure calciiun chloride, but carrying magnesium and sodium chloride of 
density 1.2. 

It might be conveniently noted here that the relation between freezing- 
point and density for pure calciiun chloride by the same bulletin is given 
in Table XVIII below: 

Table XVIII 

FREEZING-POINT OF CALCIUM CHLORIDE . 
U. S. Bureau of Standards 



Density of Solution. 


Per cent CaClt by Wt. 


Freesing-point, 


Freeaing^point, 


1.12 


14.88 


- 9 


16.8 


1.14 


16.97 


-13 


8.6 


1.16 


19.07 


-16 


3.2 


1.18 


21.13 


-20 


- 4.0 


1.20 


23.03 


-24 


-11.2 


1.22 


24.89 


-29 


-20.2 


1.24 


26.77 - 


-34 


-29.2 


1.26 


28.55 


-40 


-40.0 



Other values for the specific heats of brines as commonly used are given 
m Table XIX, the accuracy of which is seriously in doubt and whigh 
may be checked by more authoritative values at different points where deter- 
minations have been made. 

Anhydrous ammonia liquid, has a variable specific heat with temperature, 
but the exi)erimental values are too few to make its value and law quite certain. 
Several formulas have been proposed, however, that tend to give an impression 
of accuracy not warranted by the facts though quite convenient in preparing 
tables. 

Specific heat of NH« liquid at t^ F. 

1.0135+.00468 (<-32) (o) ) 
1.118 +.001156 (t-32) (6) 



Authority 

Zeuner 
Dieterici 
Wood 
Ledoux 



1.1352+.00438 (^-32) (c) 
1.0057+.00203 (t-32) (d) 



(613) 



426 



ENGINEERING THEBMODYNAMIO0 




Temperature In Degrees Fahr. 

Fig. 125. — Specific Heat of Calcium Chloride Brine of Various Densities D at Temperatures 

-10*' F. to +70^ F. 



HUAT AND MATTER 



427 



Table XIX 
SPECIFIC HEAT OF SODIUM CHLORIDE BRINE 



Density, B6 


8p.gr. 


1 
Per cent NaCl 
byWi. 


Sp. Heat. 


Temp. F. 


Authority. 


1 


1.007 


1 

1.6 
4.9 
5.0 
10.0 
10.3 
10.3 
11.5 
12.3 
15.0 
18.8 
18.8 
20.0 
24.3 
24.5 
25 


.992 

.978 

.995 

.960 

.892 

.892 

.912 

.887 

.871 

.892 

.841 

.854 

.829 

.7916 

.791 

.783 


-0 
64.4 

66-115 

-0 

-0 
59-120 
59-194 
61-126 

64.4 

-0 
63-125 
68-192 

64-68 
64 


Common 
Thomsen 






WinVelmftnn 


5 
10 


1.037 
1.073 


Common 
Common 
Teudt 






Teudt 






Marignac 
Winlf«'n)ftnn 






15 


1.115 


Common 
Teudt 






Teudt 


19 


1.150 


Common 

WinkAlrnann 






Thomsen 


'23 " 


1.191 


Common 



From these expressions the mean specific heat follows by halving the coefficient 
of (<— 32) F., and these were determined and plotted to scale^ together with some 
direct experimental values of Drewes, in Fig. 126. Giving greatest weight 
to Drewes and Dieterici, a mean curve shown by the solid line is located as 
the best probability of the value for liquid anhydrous and it has the Eq. (614). 



trSSfL'm^^^^^^ ! =1.07+.00056(.-32).. 



(614) 



From this value the heat of liquid ammonia above 32° F. has been determined 
and is presented graphically in Fig. 127 from which, and the equation, the 
tabular values at the end of the Chapter were determined. 

Anmionia dissolved in water, giving an aqueous solution as used in the absorp- 
tion refrigerating sjrstem, has a nearly constant specific heat so closely approxi- 
mating unity as shown by Thomsen, who gives 



3 per cent NH3 in water solution, sp.ht., 

1.8 per cent NH3 in water solution, sp.ht., 

.9 per cent NH3 in water solution, sp.ht.. 



.997, at 66° F. 
.999, at 66° F. 
.999, at 66° F., 



that it is customary in these calculations to ignore any departure from unity, 
the value for water. 

Liquid carbonic acid, another important substance in engineering, especially 
in mechanical refrigeration, is less known as to its specific heat than is ammonia, 
and that is much too uncertain. There is probably nothing better available 
at present for the necessary range than the results of Amagat and Mollier, 
reported by Zeuner for the heat of the liquid, which are reproduced in Fig. 
128, and used in the table at the end of this Chapter. 



428 



ENGINEERING THERMODYNAMICS 



It is, however, with gases that the most complex situation exists with respect 
to specific heats. As iias already been pointed out, gases may be heated at 



Li 



tz; 

I 
I 



L2 



1.0 



99 



.8 




B 



-60 



AA-Zeuner 

BB-Ledouz 

CO -Wood 

DD-Dieterioi 

E£ Drewes 

FF - Mean Used in Book 



Temperature in Degrees Fah^ 



150 



Fig. 126. — Mean Specific Heat of Liquid Anhydrous Ammonia from —50*" F. to 150"* .F 




GO 100 

Temperature in Degrees Fahr . 

Fig. 127. — Heat of Liquid Anhydrous Ammonia above — 50** F. 

constant volume, doing no external work while being heated, or at constant 
pressure, in which latter case work is done by expansion of the gas against the 
resisting constant pressure. Therefore, there must be two difiFerent specific 



HEAT AND MATTER 



429 



heats for each gas, one Cp at constant pressure and the other C« at constant 
volume, the difference between them representing the heat equivalent of the 





















































■ 






























1 




50 


















































/ 


























1 




























/ 


















» 










/ 


• 




fa' 

o 
< 




















/ 


























/ 








a* 


















/ 










o 
















/ 


/ 










w 














/ 


/ 

























/ 


/ 






















/ 


/ 
























/ 


/ 
























/ 


/ 
























y 


/ 






















-a 


/ 


/ 
























26 









s 


s 








* 


6 







% Temperature in Degreea Fahr. 

Fig. 128.— Heat of Liquid Carbonic Acid above 32** F. 

work of expansion done during the rise of temperature. Most experimental 
determinations of the specific heats of gases have been made at constant pressure 



430 ENGINEERING THERMODYNAMICB 

and the constant volume value found from established relations between it 
and other physical constants. These relations most commonly used are two, 
Eq. (615) connecting the difiFerence with a constant R and the other Eq. (616) 

777.52(Cp-C.) = fi, (615) 

§f-T (616) 

connecting their ratio to a constant y. These constants have each 
a special significance that may be noted here and proved later, thus R is the 
ratio of the PV product of a pound of gas to the absolute temperature, and t the 
particular value taken by the general exponent s in PV'^Cj when the expansion 
represented takes place with no heat addition or abstraction, i.e., adiabatic, 
it is also a function of the velocity of sound in gases. Table XXXIV at the end 
of this Chapter gives some authentic values, with those adopted here designated 
by heavy type. 

YanabiUty of specific heats of gases and vapors is most marked and of some 
engineering importance, because so many problems of practice involve highly 
heated gases and vapors, the most common being superheated steam and the 
active gases of combustion in furnaces, gas producers and explosive gas engines. 
In f act^ with regard to the latter it may be regarded as quite impossible with 
even a fair degree of accuracy to predict the temperature that will result in the 
gaseous products from the liberation of a given amount of heat of combustion. 
The first fairly creditable results on the variability of the specific heats of gases 
of combustion at high temperatures were announced by Mallard and LeChatelier, 
Vieille and Berthelot, all of whom agree that the specific heat rises, but who 
do not agree as to the amount. A general law was proposed by LeChatelier, 
giving the specific heat as a function of temperature by an equation of the 
following form: 

Specificheatatr F., (F=C), = C,=a+b(«-32), (617) 

in which a = specific heat at constant volume at 32** F. This yields, 



Mean-specific heat from 
32.^F.,tor F., (F=C), 



^=C'. = a+|(i-32) (619) 



The specific heat at constant pressure is obtained by adding a constant 
to the value for constant volume according to 



HEAT AND MATTEB 



431 



whence 

Specific heat at t" F., (P=C), = C,=a4 



R 



777.52 



f6(<-32), (621) 



B.T.U. per lb. from 
32°F.,to«''F., (P=(7), 



=Q 



32 to t 



'[' 



f 



^^+|(<-32)](<-32) . . (622) 

[Mean specific heat from \ ^r' —n\ ^ \^ff Q9^ rao'Vi 

i32^F., to <^F.,(P-0, J ''"■^777.52+2^^"^^^ ^^^"^^ 

The values of these constants have been determmed by LeChatelier, Clerk, 
Callender, and Holbom and Austm, from which the following values are 
selected. 

Table XX 
SPECIFIC HEAT CONSTANTS, GASES, 



Gm. 


o 


. R 
*'"*"777.52 


h 


b 
2 


Authority. 


CO, 


.1477 


.1944 


.000097 


.0000484 


LeChatelier 


CO, 




.2010 


.0000824 


.0000412 


Holborn and Austin 


N, 


.170 


.2404 


.0000484 


.0000242 


LeChatelier 


N, 




.2350 


.000021 


.0000106 


Holborn and Austin to 2606* F. 


N, 




.2350 


.0000208 


.0000104 


Callender 1644* F. to 2440* F. 


0, 


.1488 


.2126 


.0000424 


.0000212 


LeChatelier 


H,0 


.3211 




.000122 


.000061 


LeChatelier 


Air 




.2431 


.000135 


.0000675 


CaUcnder (1544* F. to 2440* F.) 



For purposes of comparison the following curves are plotted, showing all 
these results of specific heat at constant volume, at temperature t^ F., the total 
heat above 32^ F. per poxmd of gas, and the mean specific heat from 32^ F. to 
r F. in Fig. 129. 

Probably there is now more known of the specific heat of superheated steam 
than of any conunon gaseous substance, and it is likely that other substances 
will be found in time to have somewhat similar characteristics. Pure computa- 
tion from the laws of perfect gases indicates that the specific heat of gases or 
superheated vapors must be either a constant, or a function of temperature 
only, and this is what prompted the form of the LeChatelier formula. Bold 
experimentation on steam, disregarding the law, or rather appreciating that 
superheated steam is far from a perfect gas, principally by Knobloch and 
Jacob and by Thomas, showed its specific heat to be a function of both pres- 
sure and temperature. Results were obtained that permitted the direct solu- 
tion of problems of heat of superheat, or the heat per pound of vapor at any 
temperature above that at which it was produced, or could exist in contact with 
the liquid from which it came. Critical study of various results by Marks and 
Davis led them to adopt the values of Knobloch and Jacob with slight modifi- 



432 



ENGINEERING THERMODYNAMICS 




■«n"X"a ^] 88 ©AOQB SBO JO punoi jad ^toh 




3 



a 
o 

"S 



o 
CO 

a 

2 






O 



o 

CO 

s 

o 



«5 
o 

to 

c 

§ 






O 



eS 

eg 

o 

■ 



HEAT AND- MATTER 



433 




•gfl'X'arnonBJTHiig aAoqB niBois JO pnno^i Jaj ^bsh 




a «^ 



ts s :i| 













§ 3 

i - 



» g 






OS 

eg 



1 
T 



s. 






S '^. s 



434 ENGINEERING THERMODYNAMICS 

cationS; for which evidence was in existence, raising the specific heats at low 
pressures and temperatures, and their conclusions are adopted in this work. 
In Fig. 130 is shown (A) the Marks and Davis modification of the Cp curve 
of Knobloch and Jacobs, the integral of which (C) gives the heat of superheat 
from any temperature of steam generation to actual steam temperature, while 
(B) shows the values for the mean specific heat above the temperature of satura- 
tion for the particular pressure in question. 

When substances of the same class are mixed so that toi, W2, wz^ etc., lbs. 
of the diflFerent substances having specific heats Ci, C2, C3, etc., or Cpi, C,2, 
C^f etc., or Cfi, C92, C,3, etc., then the specific heat of the mixture is given by 

wi+W2+wz+eic. ' 

^ ^ C9iWi+C^ W2+C^W3 +etc. ,^25A 

wi+v>2+wz+eiQ, ' 

^ • C piwi + Cp2W2 + Cpzws +et c. ,^^^. 

tp= ; : j — 7 [oZb) 

Example. If 5 lbs. of olive oil at a temperature of 100° F., 10 lbs. of petroleum 
at a temperature of 150° F., and 50 lbs. of water at 50° F. are mixed together, what 
will be the resultant temperature and how much heat will be required to heat the miA- 
ture 100° above this temperature? 

Sp. ht. of olive oil « .4, 
Sp. ht. petroleum » .511, 
Sp. ht. water =1.000. 

Let x-the final temp. The heat given up by the substances falling in tempera- 
ture is equal to that gained by those rising, hence 

50(x-50) XI -5(100-0:) X.4+10(150-x) X.511, 

50x -2500 -200 -2a: +766 -5.11a:, 

57.11a: =3466, or, a: =60.7° F., 

Sp.ht. of nuxture- ; , from Eq. (611), 

5X.4-H0x.511-f50Xl 57.11 ._. 

^ z z~z z~^ ~" A_ ^ .o7oO« 

5-1-10-1-50 65 

whence the heat required will be 65X.8786 = 57 B.T.U. 

Prob. 1. To change a pound of water at 32° F. to steam at 212° F. requires 
1150.4 B.T.U's. If the same amount of heat be given to a cubic foot of each of 
the following subsfahces at 32° F., what will be final temperature in each case? (a) cop- 
per; (6) iron; (c) mercury; (d) clay; (e) stone. 



HEAT AND MATTER 435 

Prob, 2. How many pounds of the following substances could be wanned 10® F. 
)y the heat required to raise 100 lbs. of water from 40° F. to 200** F.? 
(a) Ethyl alcohol from 100** F.; 
(6) Sea water from 60° F., (density = 1.045); 

(c) Glycerine from 60° F; 

(d) Tin from 480° F. 

Prob. 3. If 150 lbs. of water at 200° F. are added to a tank containing 200 lbs 
)f petroleum at 70° F., what will be the resultant temperature, neglecting any heat 
ibsorbed or given up by the tank itself? 

Prob. 4. To melt 1 lb. of ice requires 144 B.T.U. How much would this lower 
:he temperature of 1 lb. of the following substances (1) at constant pressure; (2) at 
»>nstant volume; (a) air; (6) oxygen; (c) ammonia; (d) hydrogen; (e) nitrogen? 

Prob. 6. What would be the specific heats of the following mixture? Hydrogen 
) lbs., oxygen 1 lb., nitrogen 7 lbs., carbon dioxide 20 lbs., carbon monoxide 10 lbs.? 

Prob. 6. Air is approximately 77 per cent N2, and 23 per cent O2 by weight. By 
means of the specific heats of the components, find its specific heats at constant pres- 
sure, and at constant volume. 

Prob. 7. By means of the specific heats, find the values of R and y most correct 
at atmospheric temperature (60° F.) for, hydrogen, air, carbon dioxide, carbon monoxide 
and nitrogen. 

Prob. 8, How much water could be heated from 40° F. to 60° F. by the heat 
needed to superheat 10 lbs. of steam at 200 lbs. per square inch absolute to 700° F.? 

Prob. 9. A building containing 250,000 cu. ft. of space is heated by a hot-water 
system. Considering the air to change eight times per hour, how many pounds of 
water per hour must be circulated if the drop in temperature of the water is from 
200° to 100° and the temperature of the outside air is 30° F. while that of the room 
is 60° F. neglecting wall conducted heat? 

Prob. 10. How much heat would be required to warm a pound of liquid COi from 
zero to 80° F.? Compare with water and ammonia. 

6. Volume or Density Variation with Temperature of Solids, Liquids, Gases 
and Vapors, Not Changing State. Coefficients of Expansion. Coefficients 
of Pressure Change for Gases and Vapors. Solids increase in length or in any 
linear dimension, a certain fraction of their original length for each degree 
temperature rise and the expansion is usually assumed to be in proportion 
to temperature rise. The relation between original and final length can be 
set down in an equation involving the coefficient of expansion. 

Let a = coefficient of linear expansion = fractional increase in length per 
degree. 

'^ h and <i= original length or any other linear dimension and the cor- 
responding temperature; 

'' I2 and <2= length which h becomes after heating and the corresponding 
temperature. 

Then 

Increase in length = Z2—ii = aZi(te—<i), (627) 

New length Z2 = ii+aZi(fa — ii), 

= li[l+a{t2-ti)] (628) 



436 ENGINEERING THERMODYNAMICS 

Solids, of course, expand eubicaJly and the new volume will be to the old 
as the cubes of the linear dimension. 

Let a = coefl5eient of volumetric expansion; 
t;i= original volume; 
t;2= final volume after heating. 
Then when the temperature rises one degree, 



^=(|y = (l+a)3 = l+3a+3a2+a3 = l+a . . . . 



(629) 



If a is small, and it is generally less than j^j then the square and cube eao 
be neglected in comparison with the first power, whence 

l+a = l+3a and a = 3a. 

so that the coefficient of volumetric expansion may be taken as sensibly 
equal to three times the coefficient of linear expansion, and similarly, the 
coefficient of surface expansion as twice the coefficient of linear expansion. 

Liquids, by reason of the fact that they must always be held in solid 
containers, may be said to have no linear expansion, and therefore, although 
the expansion may be one direction only, the amount is due to the total 
change of volume rather than the change of length along the direction of 
freedom to expand. The same is true of gases, so that for gases and liquids 
only coefficients of volumetric expansion are of value and these are given in the 
Tables XXXV, XXXVI, XXXVII, and XXXVIII at the end of this Chapter. 
With liquids and gases it is usual to take the volume at 0® C. or 32° F. and 
29.92 ins. Hg pressure as a standard, and the coefficient gives the increase sls 
a fraction of this, per degree departure from the freezing-point. This is the 
universal practice with gases. 

It appears that the coefficients of expansion for solids are quite different 
from one another, ranging from over 15X10"* for wax, to.085XlO~* for 
Jena normal glass, a range of over two hundred and sixty times. Detennina- 
tions of the value at various temperatures for any one substance indicate a 
variation with temperature, which proves that proportionality of increase of 
dimensions to temperature rise, does not hold true, a fact which has led to 
formulas of the form 

I2^lill+x{t2-ti)+y(t2-ti)^, 

the value of which is dependent on the determination of the constant and veri- 
fication of correctness of form, which has not by any means been conclusively 
done. For most engineering work the constant values nearest the temperature 
range will suffice except for certain liquids, vapors, and gases. A more marked 
tendency to follow such a law of variation with temperature is found with 
liquids and coefficients for some are given in the standard physical tables. 



HEAT AND MATTER 437 

The two important liquids, mercury and water^ have been separately 
studied in greater detail and the latter exhibits a most important exception 
to the rule. For mercury, according to Broch 

t;2 = vi(l+.00O455e+54Xl0-i2i2+602Xl0-i*^), . . . (630) 

which exhibits a refinement of value only in instnunent work such as barometers 
and thermometers. Water, as already mentioned, has its maximum density 
at 39.1^ F. and expands with both fall and rise of temperature. Its expansion 
is given by a similar formula by Scheel, as follows: 

t;2=t;i(l-.03655Xl0"^<+2.625Xl0^®^2_ii61^). . . (631) 

Most commonly the expansion of water is not considered in this^way, but by 
comparing densities at varying temperatures, and all sets of physical tables 
contain values which in this work are significant only as affecting the change 
of volume in turning water to steam and siLch values as are needed are 
incorporated in the steam tables later. 

The study of the expansion of gases and vapors at constant pressure, and 
rise of pressure at constant volume, per degree has perhaps been fairly com- 
plete and is of greatest significance, because from it most of the important laws 
of thermodynamics have been derived. This work may be said to have 
started with the Regnault air and gas thermometer work, already described. 
Some of the authentic values collected in the Landolt, Bomstein, Myerhoflfer, 
and Smithsonian Physical Tables, are given at the end of this Chapter, 
where oe, is the coeflicient of pressure change at constant volume, and a, the 
coefficient of expansion, or volume change at constant pressure. 

The remarkable thing about the coefficients for these gases and vapors is the 
approach to constancy for most of the gases, not only of the coefficients of expansion 
for P = c nor the similar constancy of the coefficients of pressure rise for F=c, but 
more remarkable than either of these is the similarity of the two constant coeffir- 
dents. These facts permit of the generalizing of effect when P^c, 
and when F=c, and of the announcement of a law by means of which 
ail such problems can be solved instead of applying separate coefficients for 
every substance and every different temperature necessary for solids and 
liquids where, for example, the maximum coefficient was over 260 times as 
great as the least. The average coefficient for all gases, applying both to 
pressures and volumes, is the same as enters into the gas thermometer work 
and its best value is found to be 



a = VqYko ^ -^2034, per degree F. 



a = 27^13 = 003661 , per degree C. 



• , .... (632) 



438 ENGINEERING THERMODYNAMICS 

and approximately 

a = 2^2 = .00203, per degree F. 

a = 070 == .00366, per degree C. 



. (633) 



These are the same as the reciprocals of the absoliUe temperature of i}te 
ice-meUing point, and are but expressions of conditions for reduction of the 
volume and pressure at the ice-melting temperature to zero by constant 
pressure and constant volume abstraction of heat respectively, and by 
stating the amount of reduction per degree give by implication the nimibcr 
of degrees for complete reduction, j 

Example. The rails on a stretch of railroad are laid so that they just touch when 
the temperature is 120** F. How much total space will there be between the rails 
per mile of track at 0** F.? 

For wrought iron a will be nearly the same for Bessemer steel = .00000648. 

Hence the linear reduction in 5280 ft. for a change of 120° F. will be 

5280 X 120 X. 00000648*4.1 ft. 

Prob. 1. A steam pipe is 700 ft. long when cold (60** F.), and is anchored at one 
end. How much will the other end move, if steam at a temperature of 560** F. is 
turned into the pipe? 

Prob. 2. A copper sphere is one foot in diameter at 50** F. What must be the 
diameter of a ring through which it will pass at a temperature of 1000° F.? 

Prob. 3. A hollow glass sphere is completely filled with mercury at 0° F. What 
per cent of the mercury will be forced out if the temperature rises to 300° F.? 

Prob. 4. A room 100 ft.XoO ft.XlO ft. is at a temperature of 40° F. The tem- 
perature rises to 70° F. How many cubic feet of air have been forced from the room? 

Prob. 6. The air in a pneumatic tire is at a pressure of 90 lbs. per square inch 
gage and at a temperature of 50° F. Due to friction of the tire on the ground in 
running, the temperat re rises to 110° F. What will be the pressure? 

Prob. 6. A brick lighthouse is approximate y 200 ft. high. Should it be exactly 
this at 0° F., what would it be at 100° F.? 

Prob. 7. Show that if a glass tube is rigidly held at each end by brackets attached 
to an iron tank it will break if the tank is warmed. 

Prob. 8. From Eq. (618) find the density of water at 60° F., 100° F., 212° 
F., and compare with the values in the steam tables. 

Prob. 9. A drum containing COs gas at a pressure of 250 lbs. per square inch 
gage is raised 100° F. above its original temperature. What will be the new pressure? 

7. Pressure, Volume and Temperature Relations for Gases. Perfect and 
Real Gases. Formulating the relations between the pressure change at constant 
volume and the volume change at constant pressure, 

Let P and V be the simultaneous pressure and volume of gas; 
*' < be its scale temperature at the same time, F.; 
" r be its absolute temperature at the same time, F. ^ 



HEAT AND MATTER 439 

Then at constant volume the pressure reached at condition (a) after heating 
from 32** F. is given by 

Pa — -P32* = 409^32 X (4i — 32) . 



Pa = -P32* 



L ^ 492 I 



Similaxly for another temperature U, the pressure will be 



Pt=Pz2' 



L^ 492 J- 



Whence 



or 



Similarly 



1 4.??r_32 

Pa^ 49 2 ^492--32+<a^<a+460 
P* ti,-S2 492-32+fe fo+460' 

■*■ 492 



§5=^, for 7 constant, (634) 

^b -lb 



V T 

7~=^, for P constant (635) 

Vb lb 



Both EJqs. (634) and (635) are true, for no gas all the time, but very nearly 
true for all, under any range of change, and a hypothetical gas is created for which 
it is exactly true all the time^ known as a perfect gaSy about which calculations 
can be made as would be impossible for real gases and yet the results of which 
are so close to what would be the result with real gases, as to be good enough 
for engineering practice. Therefore, with a mental reservation as a guard 
against too great confidence in the work, all real gases will be assumed perfect 
and to follow Eqs. (634) and (635) except when experience shows the results 
are too far wrong to be useful. 

These laws, known by the names both of Charles and Gay-Lussac, are closely 
associated with another also doubly named as Boyle's or Mariotte's and like- 
wise an idealization of experimental observations known to be nearly true for 
all gases. This is to the effect that so long as temperatures are kept constant 
the pressures of gases vary inversely as their volume, or that. 

Pa Vb 

-^^T7, and, PaVa'^PbVb^^Qon&i^iiiy for T constant . (636) 

i^b y a 

Study of the PY product, for various gases has revealed a good deal on the 
general properties of matter, especially as to the transition from one state to 
another. This is most clearly shown by curves which may be plotted in two 



440 



ENQINEEBINa THERMODYNAMICS 













































































1 
































































































































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a 








































































































i W/l 




















































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9 












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=S 


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r 








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Ml3 


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218 












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Job 



HEAT AND MATTER 441 

ways. To coordinates of pressure and volume a family of equilateral hyper- 
bolas one for each temperature, would represent the true PF=C or isoiAermaZ 
relation and any variation in the constancy of the product would be shown by 
its departure from the hyperbola. Still more clearly, however, will the depart- 
ure appear when the product PV is plotted against pressures, constancy of 
product would require all lines to be straight and inconstancy appear by 
departures from straight lines. To illustrate, the data from Young for car- 
bon dioxide are plotted both ways in Fig. 131, from 32° F. to 496° F., the values 
of PV at 32° and 1 atm. are taken as unity on one scale. It appears th^t up to 
the temperature of 88° F, known as the critical temperature, each isothermal 
plotted to P and PV coordinates consists of three distinct parts: 
(a) a curved line sloping to the right and upwards; 
(6) a straight line nearly or exactly horizontal; 
(c) a nearly straight line sloping upward rapidly and to the left. 
In this region then the isothermals are discontinuous, and this is caused by 
the liquification or condensation of the gas, during which increase of pressure, 
produces no change of volume, provided the temperature is low enough. It 
also appears that each PV line has a minimum point and these minima joined 
result in a parabola. At the end of this Chapter are given in Table XXXIX the 
values of PV at three different temperatures and various pressures for oxygen, 
hydrogen, carbon dioxide and ammonia, in terms of the values at 32° and 1 
atm. for further comparison and use. Further study along these hues is not 
profitable here and the topic while extremely interesting must be dropped with 
the observation, that except near the point of condensation or liquefaction, 
gases or vapors, which are the same thing except as to nearness to the critical 
state, follow the Boyle law closely enough for engineering purposes. 

None of these approximate laws, Eqs. (634), (635) and (636) can be con- 
sidered as general, because each assumes one of the variables to be constant, but a 
general law inclusive of both of these follows from further investigation of a 
fixed mass of gas suffering all sorts of pressure volume and temperature changes, 
such as occur in the cylinders of compressors and gas engines. A table of 
simultaneous experimental values of pressure, volume, and temperature, for any 
gas will reveal the still more general relation inclusive of the preceding three as 
follows: 

Paya _ Pbyb _ Py _f^ /AQTN 

m "~ m ~" /T7 ~^^> woi; 

in which Cg is approximately constant for any one gas and assumed constant 
for perfect gases in all calculations. For twice the weight of gas at the same 
pressures and temperatures Cg would be twice as large, so that taking a constant 
R for one pound, and generally known as the ** gas constant," and introducing 
a weight factor u;, the general characteristic equation for the perfect gas is, 

PV^wRT (638) 



442 



ENGINEERING THERMODYNAMICS 



This general law may be derived from the three primary laws by imagining 
in Fig. 132, two points, A and B, in any position and representing any two states 
of the gas. Such points can always be joined by three lines, one constant 

































A 




X 














\ 


















.^ 


Y 














B 

















pv Y 

Diaamim to derive Law ^=Cq 

Fig. 132. — Curve of ContinuouB Relation between P, F, and T for Gases. 

pressure A to X, one constant temperature X to F, and the other constant 
volume F to B. For these the following relations hold, passing from A to B 



But 



and 



whence 



or 



Passing to B, 



y w" y xm • 



la 



F,= 






r,=r, 



Vf 



Va- 


_ y P» To 

" ^'Pa Ty' 


PaVa 


_P,V, 

- Ty ■ 


P,= 


.pTy 



HEAT AND MATTER 443 

But 

T 

" Ta Ty Ty Th ' 

PV 

-jp- = constant. = wR 



or in general 



when the weight of gas is w lbs. 

For numerical work, the values of R must be fixed experimentally by direct 
measurement ojf simultaneous pressure, volume, and temperature, of a known 
weight of gas or computed from other constants through established relations. 
One such relation already mentioned but not proved is 

fl = 777.52(Cp-C.) (639) 

It is extremely imlikely that the values of R f oimd in both ways by a multi- 
tude of observers under all sorts of conditions should agree, and they do not, 
but it is necessary for computation work that a reasonable consistency be attained 
and that judgment in use be cultivated in applying inconsistent data. In the 
latter connection the general rule is to use that value which was determined by 
measurement of quantities most closely related to the one being dealt with. 
Thus, if jB is to be used to find the state of. a gas as to pressure, volume, and tem- 
perature, that value of R determined from the first method should be selected, but 
the second when specific heats or Joule's equivalent are involved. Of coiu^e, a 
consistency could be incorporated for a perfect gas, but engineers deal with real 
gases and must be on guard against false results obtained by too many hypoth- 
eses or generalizations contrary to the facts. Accordingly, two values of R are 
given in Table XL, at the end of this chapter, one obtained from measure- 
ments of specific heats at constant pressure and determinations of the ratio 
of specific heats unfortunately not always at the same temperature and gen- 
erally by different people, and the other by direct measure of gas volume at 
standard 32° F. temperature and 1 atm. pressure. These measurements are 
separately reported in Sections (5) and (8), respectively. 

If a gas in condition A, Fig. 133, expand in any way to condition B, then 
it has been shown that 

in which « has any value and which becomes numerically fixed only when the 
process and substance are more definitely defined. Comparing the temperatures 
at any two points A and jB, it follows that 



1 ^E^ and T.=^*Z? 
« wR' ^"^ ^' wR' 



4AA 

X. X X 

whence 



ENGINEERING THERMODYNAMICS 



n p.n 



But 
and 



1 a -to ^ a 






whence 



Ta~\Vj ' 



(640) 



















A 










• 




I 














v 














\ 


V 






• 








« 


-^ 




B 






























V 



Fio. 133. — Expansion or Compression of Gas between A and B, Causing a Change of Tem- 

peratiure. 

and 






(641) 



Eqs. (640) and (641) give the relation between temperatures and volumes 
But 

Vg _ TgPo TbPg 

which, substituted in above, gives 



or 



©■-©■"■• 



HEAT AND MATTER 445 

and 

rArf <««> 



or 



k-iW'"- ■ • • <'^> 

Eqs. (629) and (630), give the relation between pressures and temperatures. 
It is convenient to set down the volume and pressure relations again to 
complete the set of three pairs of most important ga& equations. 



Pa \Vj 



(645) 



These are perfectly general for any expansion or compression of any gas, but 
are of value in calculations only when s is fixed either by the gas itsel/ or by 
the thermal process as will be seen later. 

Example. A pound of air has a volume of 7.064 cu.ft. at a pressure of two atmos-. 
pheres and a temperature of 100** F. Find the value of R for air from the data; 
also the final volume and temperature if expansion bccurs so that «*«1.4 until the 
pressure becomes i an atmosphere. 

PV^wRT, or 2116X2X7.064 = 1 Xi2x560, or fi«63.38, 

»-l .4 



^(g) •-<«"-'•«• 



.-. T, = ^1 -^1.49 = ^7^ =352 abs. = -108** F. 

1.49 



HW-^^- 



h^[^y =2.7, or 7, =2.7 7i =19.1 cu.ft. 



Prob. 1. A perfect gas is heated in such a way that the pressure is held constant. 
If the original volume was 10 cu.ft. and the temperature rose from 100^ F. to 400** 
F., what was the new volume? 

Prob. 2. The above gas was under a pressure of 100 lbs. per square inch gage at 
the beginning of the heating. If the volume had been held constant what would have 
been the pressure rise? 

Prob. 3. A quantity of air, 5 lbs. in weight, was found to have a volume of 50 cu.ft. 
and a temperature of 60** F. What was the pressure? 

Prob. 4. A cylinder holding 12 cu.ft. has a pressure of 250 lbs. per square 
inch gage, and the temperature is 50*^ F. What would be the weight of its contents 
were it filled with (a) CO,; (6) NH,; (c) Oxygen; (d) Hydrogen? 



Cp. 


C 


3.409 


2.412 


.217 


.1535 


.2175 


.1551 


.2438 


.1727 



446 ENGINEERING THERMODYNAMICS 

Prob. 6. At a pressure of 14.696 lbs. per square inch and a temperature of 
melting ice, one pound of air has a volume of 12.387 cu.ft. From the data find the 
value of R for air. The specific heats of air are given by one authority as Cp = .2375 
and C» = .1685. Find R from the data and see how the two values obtained compare. 

Prob. 6. At the same pressure and temperature as in Prob. 5, a pound of the 
following substances has a volume as shown. From the data and the valu^ of 
specific heats, find R by the two methods. 

Substance. Cu.ft. per lb. 

Hydrogen 178.93 

Carbon dioxide ..... 8.15 

Oxygen 11.21 

Nitrogen 12.77 

Prob. 7. 5 cu.ft. of gas at a pressure of 3 atmospheres absolute and a tempera- 
ture of 50** F. expand to atmospheric pressure. What will be the final volume and 
temperature, if s = 1.35? 

Prob. 8. 1000 cu.ft. of gas at atmospheric pressure and 60** F. are compressed 
into a tank of 100 cu.ft. capacity. What will be the pressure in the tank and the 
temperature of the gas at the end of the process, if the gas is COj and the com- 
pression adiabatic? 

Prob. 9. What will be the final volume, pressure and temperature, if a pound of 
air at atmospheric pressure (14.7 lbs. per square inch) and a temperature of 60® F. 
be compressed adiabatically until its absolute temperature is six times its original 
value? 

8. Gas Density and Specific Volume and its Relation to Molecular Weight 
and Gas Constant. The density of a gas is best stated for engineering 
purposes as the weight of a cubic foot, but as this becomes less on rise of 
temperature or decrease of pressure it is necessary to fix a standard condition 
for reporting this important physical constant. It is best to take one atmosphere 
760 mm. or 29.92 ins. of mercury as the pressure, and 0® C. = 32°F. as the 
standard temperature, though it is in some places customary in dealing with 
commercial gases, such for example as those used for illumination, to take the 
temperature at 60^ F. and illuminating gas at this condition is often known 
among gas men as standard gas. In this work, however, the freezing-point 
and standard atmosphere will be tmderstood where not specifically mentioned, 
as the conditions for reporting gas density and its reciprocal, the specific volume 
of gases or the cubic feet per pound. The chart, Fig. 134, shows the relation 
of volume and density at any pressure and temperature to the volume and 
density under standard conditions. 

These constants have been pretty accurately determined by many investi- 
gators, whose figures, to be sure, do not agree absolutely, as is always the 
case in experimental work, but the disagreement is found only in the last 
significant figures. Some selected values of reliable origin are reported at 
the end of this Chapter in Table XLI for the important gases and these 
may be used in computation work. 



HEAT AND MATTER 



447 



It often happens in dealing with gases and especially superheated vapors 
that a value is needed for which no determination is available, so that general 



Pressure in Founds Per Sq. In. Abs. 
U 13 la 11 





4) 



1 


1 1 


1 M» 


1 1 


1 1 


1 1 


u 


1.1 


1 



Temperature « uegreeB Fahr. 
I I I 1 1 1-90 I I I I 1^ I 



I I I I t I 



I I I I 



.7 



I I I I I II 



I 

'.A 



I 1^8 

I I I I 



^ ^„, ^^.. Density at 82 A 29.W^^ _ _ , „ ^, Volume at 32"^ 29.98'' 

Upper Soale =Batlo Density at any T » P I^^er Sc^le=Ratio v,.imi»^ atanv T^p 

Inner Scale - Ratio Volu5^aJ_321R ^^^^ ^^^ ^ ^^^.^ Den«lt_y at^2^ 

Volume at any T Density at any T 

Fig. 134. — Equivalent Gas Densities At Different Pressures and Temperatures. 

laws of density or specific volumes of substances are necessary to permit the 
needed constant to be estimated. These relations may be applied to vapors 



448 ENGINEERING THERMODYNAMICS 

as well as to gases even though the standard conditions are those for the 
liquid state, on the assumption that all gases and vapors will expand under 
temperature, or contract imder pressure rise, to the same degree, retaining 
the same relative relations between all substances as exist at the standard 
atmosphere and freezing-point. A vapor thus reported below its point of 
condensation and assumed to have reached that condition from one of higher 
temperature at which it exists as vapor is often called steam gas, or alcohol 
gas, for example in the case of water and alcohol. 

Such general relations between the densities of gases as are so desirable 
and useful in practical work have been found by studying the manner in which 
gases chemically combine with respect to the volume relations before and after 
the reaction. Following several experimenters, who reported observed rela- 
tions, Gay-Lussac stated a general law, as follows: 

When two or more gaseous substances combine to form a compound, the vol- 
umes of the combining gases bear a simple raiio to each other and also to 
thai of the compound when it is also a gas. 

He also attempted to derive some relation between this law and Dalton's atomic 
combining law, which states that, in combining chemically, a simple numerical 
relation exists between the number of atoms of different elements which unite 
to form a compound. This was not successful, but Avagadro later foim^d the 
expected relation by assuming that it is a particle, or a number of atoms, or 
a molecule, that is important in combining, and the law stated is as follows: 

Equal volumes of different gases Tneasured at the same pressure arid tempera- 
ture contain the same number of molecules. 

It is possible by analysis of these two laws to get a relation between the volumes 
of gases and the weights of their molecules because the molecular relation of 
Avagadro, combines with the combining law of Gay-Lussac to define the rela- 
tion between the number of combining molecules. At the same time the weight 
relations in chemical reactions, based on atomic weights, may be put into a 
similar molecular form, since the weight of any one substance entering is the 
product of the number of its molecules present and the weight of the molecule. 
Applying the relation between the number of molecules derived previously, 
there is fixed a significance for the weight of the molecule which for simple gases 
like hydrogen and oxygen is twice the atomic weight and for compound gases, 
like methane and carbon dioxide, is equal to the atomic weight. Appl\dng 
this to the Avagadro law, the weights of equal volumes of different gases must 
be proportional to their molecular weights, as equal volumes of all contain the 
same number of molecules. 

Putting this in symbolic form and comparing any gas with hydrogen, as to 
its density, because it is the lightest gas of all and has well determined charac- 



HEAT AND MATTER 449 

teristics, requires the following symbols, denoting hydrogen values by the 
subscript h. 

Let m= molecular weight of a gas, 

8 = density in lbs. per cu.ft. =— , 
then 

81 Vfll 



8.-m,' (^6> 

and 

• ^=^ (647) 

But as the molecular weight of hydrogen is for engineering purposes equal to 2 
closely enough and hydrogen weighs .00562 lb. per cu.ft. = Bjsr, at 32® F and 
29.92 ins. Hg, 

Lbs. percu.ft. = 8i = .00281mi (648) 

To permit of evaluation of Eq. (648) it is necessary that there be available 
a table of molecular weights of gases and the atomic weights of elements from 
which they are derived, and the values given at the end of this Chapter in Table 
XLII are derived from the international table. As atomic weights are 
purely relative they may be worked out on the basis of any one as imity, and 
originally chemists used hydrogen as imity, but for good reasons that are of no 
importance here, the custom has changed to ^ the value for oxygen as unity. 
These atomic weights are not whole numbers but nearly so, therefore, for con- 
venience and suflScient accuracy the nearest whole number will be used in 
this work and hydrogen be taken as imity except where experience shows it 
to be undesirable. 

The reciprocal expression to Eq. (648) can be set down, giving the specific 
volume of a gas or its cubic feet per pound at 32® F. and 29.92 ins. Hg., as 
follows: 

Cu.ft.perlb.=^ = ;^l28r"i=^ (649) 

This is a most important and useful conclusion as applied to gases and vapors 
for which no better values are available, and in words it may be stated as follows: 

The cubic feet per pound of any gas or vapor at 32^ and 29,92 ins, Hg, is 
equal to 855,87 divided by its molecular weight, 

or 

The molecular weight of any gas or vapor in pounds j will occupy a volume of 
355,87 cu,ft. at 32"* and 29,92 ins, Hg, 

The approach to truth of these general laws is measured by the values 
given for specific volume and density at the end of this Chapter (a) experiment- 
ally derived and, (b) as derived from the hydrogen value by the law. 



450 ENGINEERING THEHMODYNAMICS 

Another and very useful relation of a similar nature is derivable from what 
has been established, connecting the gas constant R with molecular weights. 

The general law PV^wRT when put in the density form by making 8=- 
becomes 

J = ftr (650 



Whence, comparing gases with each other and w^th hydrogen at the same 
pressure and temperature 

ll = h=^^ (651, 

P2 §1 R2 

T2 

— = ^, and oi=-p— (652) 

which indicate that the densities of gases are irwersely proportional to (hi 
ga^ cansiarUs, or the density of any gas is equal to the density of hydrogen times 
the gas constant for hydrogen divided by its own. 

Inserting the values of density at 32^ and 29.92 ins. Hg and of the gas con- 
stant for hydrogen, it follows that for any gas 

Lbs. per cu.ft. = Bi=-^5 — , (653) 

the reciprocal of which gives the specific volume at 32** F. and 29.92 ins. Hg, or 

Cu.ft. perlb. = Fi=^^ (654) 

Example. 1. Explanation and use of Chart, Fig. 134. This diagram is for the 
purpose of finding the cubic feet per pound, or pounds per cubic foot, of a gas at 32*^ 
F. and a pressiue of 29.92 ins. of Hg, if its volume or weight per cubic foot be known 
at any pressiire and temperature. The curves depend upon the fact that the pounds 
per cubic foot (8) vary directly as the pressure and inversely as the temperature. 
That is 

„ ^ T 29.92 
032 , 29.92 = °5rp;^ "T"' 

The line of least slope is so drawn that for any temperature on the horizontal scale 
its value when divided by 492 may be read on the vertical scale. The group of lines 
with the greater slope is so drawn that for any value on the vertical scale this quantity 

29 92 
times — '- — may be used on the horizontal scale. That is, the vertical scale gives the 



HEAT AND MATTER 451 

ratio of densities as affected by temperature for constant pressure, while horizontal 
scale gives the ratio as affected by both temperature and pressure. A reciprocal 
scale is given in each case for volume calculations. 

To find the pounds per cubic foot of gas at 32** F. and 29.92 ins. of mercury when 
its value is known for 90^ and 13 lbs. per sq.in. On the temperature scale, pass 
vertically until the temperature line is reached, then horizontally imtil the curve 
for 13 lbs. absolute is reached. The value on the scale below is found to be 1.265, 
so that the density imder the standard conditions is 1.265 of the value under known 
conditions. Had it been required to find the cubic feet per pound the process would be 
precisely the same, the value being taken from the lower scale, which for the example 
reads .79, or, the cubic feet per pound imder standard conditions is 79 per cent of 
the value under conditions assumed. 

Example 2. By means of the molecular weight find the density of nitrogen at 
32^ F. and 29.92 ins. Hg, and the cubic feet per pound for these conditions. 
From Eq. (646) 

li mi ^ 28 X. 00562 

— =— , or 8i"= . 

Oh f^H J 

Hence 8 for nitrogen = .07868 pounds per cu.ft. and, 

1 1 



8 .07868 



12.709. cu.ft. perlb. 



Prob. 1. Taking the density of air from the table, find the value of R for air, by 
means of Eq. (653) and compare its value with that found in Section 7. 

Prob. 2. Compare the density of carbon monoxide when referred to 32° F. and 
60° F. as the standard temperature, as foimd both ways. 

Prob. 3. By means of their molecular weights find the density of oxygen, nitrogen 
and carbon dioxide at 32° F. and 29.92 ins. Hg. 

Prob. 4. What are the cubic feet per pound of acetylene, methane and ammonia 
at 32° F. and 29.92 ins. Hg? 

Prob. 6. An authority gives the following values for R. Compare the densities 
found by this means with the densities for the same substance found by the use of 
the molecular weights. 

Oxygen 48.1 

Hydrogen 764.0 

Carbon monoxide 55.0 

Prob. 6. What will be the volume and density under standard conditions, of a 
gas which contains 12 cu.ft. per pound at a temperature of 70° F. and a pressure of 
16 lbs. per square inch absolute? 

Prob. 7. What will be the difference in volume and density of a gas when con- 
sidered at 60° and 29.92 ins. of Hg, and at 32° F. and 29.92 ins. of Hg? 

9. Pressure and Temperature Relations for Vapor of Liquids or Solids. 
Vaporization^ Sublimation and Fusion Curves. Boiling- and Freezing-points 
for Pure Liquids and Dilute Solutions. Saturated and Superheated Vapors. 

Substances may exist in one of three states, solid, liquid or gas, the latter being 
generally called vapor when, at ordinary temperatures the conmion state is that 



452 ENGINEEBING THERMODYNAMICS 

of liquid or solid, or when the substance examined is near the point of lique- 
faction or condensation, and just which state shall prevail at any time dep>end5 
on thermal conditions. Within the same space the substance may exist in two of 
these three states or even all three at the same time under certain special condi- 
tioas. These conditions may be such as to gradually or rapidly make that part 
in one state, turn in to another state, or may be such as to maintain the relative 
amounts of the substance in each state constant; conditions of the latter sort are 
known as conditions of equilibrium. These are experimental conclusions, but 
as in other cases they have been concentrated into general laws of which they 
are but special cases. The study of the conditions of equilibrium, whether of 
physical state or chemical constitution, is the principal function of ph^^ical 
chemistry, in the pursuit of which the Gibbs phase rule is a controlling prin- 
ciple. According to this rule each possible state is called a phase, and the 
number of variables that determine which phase shall prevail or how many 
phases may exist at the same time in equilibrium for one chemical substance 
like water, is given by the following relation, which is but one of the conclusions 
of this general principle of equilibrium. 

Number of undefined variables =3— number of phases. 

Now it is experimentally known that if water be introduced into a vacuum 
chamber some of it will evaporate to vapor and that, therefore, water and its 
vapor may coexist or the number of phases is two, but this does not state how 
or when equilibrium will be attained. The rule above, however, indicates that 
for this case there can be but one undefined or independent variable and, of 
course, since the pressure rises more when the temperature is high than when 
low, the two variables are pressure and temperature, of which accordingly only 
one is free or independent, so that fixing one fixes the other. In other words 
when a vapor and its liquid are together the former will condense or the latter 
evaporate until either pressure or temperature is fixed, and fixing the one the 
other cannot change, so that the conditions of equilibrium are indicated by 
a curve to coordinates P and T, on one side of which is the vapor state and 
on the other that of liquid. Such a curve is the vapor pressure-temperature curve 
of the substance, sometimes called its vapor tension curve, and much experi- 
mental information exists on this physical property of substances, all obtained 
by direct measurement of simultaneous pressiu'es and temperatures of a vapor 
above its liquid, carefully controlled so that the pressiu'e or the temperature is 
at any time uniform throughout. 

The conditions of equilibrium between vapor and liquid, defined by the vapor 
tension curve extend for each substance over a considerable range of pressure 
and temperature, but not indefinitely, nor is the range the same for each. At 
the high-pressure and temperature end a peculiar interruption takes place due 
to the expansive eflfect of the temperature on the Uquid and the compressive 
eflfect of the pressure on the vapor, the former making Uquid less dense and the 
latter making vapor more dense, the two densities become equal at some 
pressure and temperature. The point at which this occurs is the " critical point " 
at which the equilibrium between liquid and vapor that previously existed, 



HEAT AND MATTER 



453 



ends and there is no longer any difference between vapor and liquid. This 
point is a most important one in any discussion of the properties of matter, 
and while difficult to exactly locate, has received much experimental attention, 
and some of the best values are given below in Table XXI for the pressure, 
density, and temperature defining it, for the substances important in engineering 

Table XXI 
THE CRITICAL POINT 



Substance. 



Hydrogen 

Ozysen 

Nitrogen 

Amntonia. 

Ammonia 

Carbon dioxide. . . 
Carbon dioxide. . . 

Water 

Water 

Water 

Water 

Water 

Water 



Symbol. 



Ht 
Ot 

Nt 

NHs 

NH; 

CO« 
COa 



HiO 
HsO 
HsO 

HsO 

HsO 

HsO 



Critical Temp. 



0«C. 



-243.5 
-118.1 

-146. » 

+130.0 
+131.0 

+ 31.35 
+ 30.021 



+358.1 
+364.3 
+365.0 

+374. 

+374.6 

+374.5 



O'^F. 



-390.1 
-180.4 

-232.8 

266. 
267.8 

88.43 
87.67 



676.4 
687.7 
680. 

705.2 

706.3 

706.1 



Critical Pres- 
sures. 



Atm. 



20 
601 

35.1 

115. 
113. 

72.9 
77.1 



194.61 
200.5 



Lbs. 
per 
Sq.in. 



294 
735 

515 

1690 
1660 

1070 
1130 



2859 
2944 



3200 



3200 



Critical 

Density 

Water 

at 
4«»C-1. 



■ • • • 

65* 
44S 



464 
45> 



.429 



Authority. 



Olssewski 
1 Wroblewski 
*Dewar 
1 Olssewski 
> Wroblewski 
Dewar 
Vincent and 
Chappuis 
Amagat 
1 Andrews 
* Cailletet And 
Mathias 
Nadejdini 
Batteli 
Cailletet and 
Colardeau 
Traube and 
Teichner 
Holborn and 
Baumann 
Marks 



Criti- 
cal vol. 
Cu.ft. 
per Lb. 



26.8 
13. 



Authority. 



Nadejdini 
Batteli 



To illustrate this discussion there is presented the vapor tension curves of 
water, ammonia and carbon dioxide to a large scale in chart form derived 
from the tabular data both at the end of this Chapter, while a small scale dia- 
gram for water is given in Fig. 136. These data are partly direct experimental 
determinations and partly corrections obtained by passing a smooth 
curve representing an empiric equation of relation between pressure and tem- 
perature, through the major part of the more reliable experimental points. 
These pressure-temperature points are very accurately located for water, the 
first good determinations having been made by Renault in 1862 and the last 
by Holborn and Bamnann of the German Bureau of Standards in the last yoar. 
The data presented are those of Regnault corrected by various investigations 
by means of curve plotting, and empiric equations by Wiebe, ThieGsen and 
Schule, and those of various later observers, including Battelli, Holborn, Hen- 
ning, Baumann, Ramsay and Young, Cailletet and Colardeau, somes eparately, 
but all together as xinified by Marks and Davis in their most excellent steam 



454 



ENGINEERING THERMODYNAMICS 



tables, and later by Marks alone for the highest temperatures 400** F. to the 
critical point, which he accepts as being located at 706.1® F. and 3200 lbs. square 



2800 



S400 



i 

a 



2000 



QQ 

f leoo 

I 

I 



1200 



800 



400 



























































































































































































































































/ 




















































/ 




















































/ 






, 














































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/ 






































































































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/ 




















































/ 




















































' 






































































































/ 




















































/ 




















































/ 


















































/ 
















































\ 


/ 




























• 




















• 




/ 
















• 




. 
































/ 


















































i 


i 






. 




















































































































































1 


















































J 












































w 


ate 


I* 
• 




/ 




V 


ape 


r 




























. 














i 


t 


















































/ 




















































/ 


















































/ 


















































i 


/ 


















































/ 


















































/ 


f 
















































y 


/ 
















































[^ 


^^^ 
























































































u 


K) 








» 


X) 








U 


iO 








« 


iT 













Temperature in Degrees Ffthr. 
Fig. 136. — ^Vapor of Water, Pressure-temperature Curve over Liquid (Water). 

inch. In calculations the values of Marks and Davis, and Marks, will be 
accepted and used. 



HEAT AND MATTER 



455 



Carbon dioxide and ammonia are by no means as well known as steam, 
md the original data plotted, while representing the best values obtainable, must 
ye accepted with some uncertainty. A smooth curve Figs. (139) and (140) 
las been drawn for each through the points at locations that seem most fair, 
or both these substances and the values obtained from it are to be used in 
calculations; these curves have been located by the same method as used by 
Marks in his recent paper and described herein later. The equalized values 
ire given in the separate table at the end of the Chapter with others for latent 



1 
























































1 








/ 




















/ 




• 

cr 

CO 


















/ 




• 
OD 

;3.05 

a 

•pi* 

00 

2 








• 






Ice 


1 


/ 


















/ 







CO 














/ 


A. 


ipor 


















/ 












• 








y 

















.^ 


-^ 


'^ 












-e 







-36 




-1( 


) 




+15 




440 



Temperature in Degrees Fahr. 
Fig. 136. — ^Vapor of Water, Pressure-temperature Curve over Solid (Ice). 

heats and volumes, but while consistent each with the other are probably but 
little more correct than values reported by others which are inconsistent. 

The curves and the equivalent tabular data are most useful in practical 
work, as they indicate the temperature at which the vapor exists for a given 
pressure, either as formed during evaporation or as disappearing during con- 
densation, or the other way round, they indicate the pressure which must be 
maintained to evaporate or condense at a given temperature. 

Just as the vapor-Uquid curves indicate the conditions of equilibrium between 



456 



ENGINEERING THERMODYNAMICS 



vapor and its liquid, dividing the two states and fixing the transition pressure 
or temperature from one to the other, so also does a similar situation exist with 
respect to the vapor-solid relations. In this case the curve is that of " siA- 
limadon " and indicates the pressure that will be developed above the solid 
by direct vaporization at a given temperature in a closed chamber. In Fig. 
• 136 is plotted a curve of sublimation of vapor-ice, based on Juhlin's data, 
Table XXII, which indicates that the line divides the states of ice from that 
of vapor so that at a .constant pressure, decrease of temperature will caus* 
vapor to pass directly to ice and at constant temperature a lowering of pres- 
sure will cause ice to pass directly to vapor. 

Table XXII 

JUHLIN'S DATA ON VAPOR PRESSURE OF ICE 



Temperature. 

• 


Pressure. 


C. 


F. 


Min. Hg. 


Lbs. sq.in. 

• 


-60 


-58. 


.050 


.001 


-40 


-40. 


.121 


.0023 


-30 


-22. 


.312 


.006 


-20 


- 4. 


.806 


.0156 


-15 


5. 


1.279 


.0247 


-10 


14. 


1.999 


.0386 


- 8 


17.6 


2.379 


.0459 


- 6 


21.2 


2.821 


.0544 


- 4 


24.8 


3.334 


.0643 


- 2 


28.4 


3.925 


.0758 


- 


32. 


4.602 


.0888 



Likewise the liquid, water, may pass to the solid, ice, by lowering temperature 
at a fixed pressure as indicated in Fig. 137, plotted from data by Tamman, 
Table XXIII, and which becomes then the curve of " fusion.^' 

Table XXIII 

TAMMAN^S DATA ON FUSION PRESSURE AND 
TEMPERATURE OF WATER-ICE 



Temperature. 


Pressure. 


C. 


F. 


Kg. sq.cm. 


Lbs. sq.in. 





32. 


1 


1423 


- 2.5 


27.5 


336 


4779. 


- 5. 


23. 


615 


8747.4 


- 7.5 


18.5 


890 


13658.8 


-10.0 


14. 


1155 


16428. 


-12.5 


9.5 


1410 


20055. 


-16. 


5. 


1625 


23113. 


-17.6 


.5 


1835 


26100. 


-20. 


- 4. 


2042 


27044. 


-22.1 


- 7.8 


2200 


31291. 



HEAT AND MATTER 



457 



80000 



aeooo 



22000 



S1800Q 



m 
o 

Suooo 



10000 



eooo 



2000 




Temperature in Degrees Fahrenheit 
Fig, 137. — Water — Ice, Pressure-temperature Curve. 



458 



ENGINEERING THERMODYNAMICS 



These three curves plotted to the same scale meet at a point located at a pressure 
of 4.6 mm. Hg. = .18 ins. Hg., and temperature +.0076*' C. =32.01^ F.,ordi. 
narily taken at 32^ F., which point is named the triple point, as indicated in 
Fig. 138. The fact that the vapor pressure for water extends below freezing- 
point and parallels more or less that of ice indicates the condition of supercooled 



.2 



«0 

< 



u 
o 

CQ 

'^ 1 

9 
O 



I 

u 

Pi4 




Ice 




Triple Point 



50 



Fig. 138.— Water Vapoi 



Temperature Degrees Fahr. 

-Water — loe, Combined Curves of Prebsure-temperature Rela- 
tion. The THpU Paint. 



water, one of unstable equilibrium instantly dispelled by the introduction of 
a little ice at the proper stable state for this temperature. 

Ordinary en^neering work is not concerned with the entire range indicated 
in Fig. 138 for any substance, but with the higher temperature ranges for some 
and the low for others, with transition from solid to liquid state for metals 
and similar solids and the transition from liquid to vapor for a great many, of 
which water comes first in importance, then the refrigerating fluids, ammonia 



HEAT AND MATTER 459 

and carbon dioxide^ and last certain fuels like alcohol and the petroleum oils 
with their distillates and derivatives. 

MeUing-paintSy or the fusion temperature of such solids as are important, 
are usually given for only one pressure, the standard atmosphere, as in ordinary 
practice these substances are melted only at atmosphere pressure, and some 
such values are given at the end of the Chapter in Table XLIII. 

This is not the case, however, for boHing-pointa, which must be defined 
a little more closely before discussion. The vapor pressure curves indicate 
that as the temperature of a liquid rises, the pressure rises also if the substance 
is enclosed, but if the pressiu'e were relieved by opening the chamber to a region 
of lower pressure and kept constant, then the temperature would no longer 
rise and boiling or ebullition would take place. The boiling-point then is the 
highest temperature to which the liquid and its vapor could rise under the 
existing pressure. When not otherwise defined the term boilmg-point must 
be taken to mean the temperature of ebullition for atmospheric pressure of 
29.92 ins. Hg, and values for several substances are given at the end of this 
Chapter in Table XLIV. 

Vapor having the temperatiu'e required by the pressure of the pressure- 
temperature curve is known as aaturcUed vapor, and this may be defined as 
vapor having the lowest temperature at which it could exist as vapor, under 
the given pressure. Vapors may, however, be superheated, that is, have 
higher temperatures than satiu'ated vapors at the same pressure, but cannot 
so exist for long in the presence of liquid. Superheating of vapors, therefore, 
implies isolation from the liquid, and the amount of superheat is the number of 
degrees excess of temperature possessed by the vapor over the saturation 
temperature for the pressure. In steam power plant work, especially with 
turbines, it is now customary to use steam with from 75® F. to 150® F. of 
superheat, and it might be noted that all so-called gases like oxygen and 
nitrogen are but superheated vapors with a great amount of superheat. 

It has abeady been mentioned that the saturated vapor pressure-temperature 
curve of direct experiment is seldom accurate as foimd, but must be corrected 
by empiric equations or smooth average curves, and many investigators have 
sought algebraic expressions for them. These equations are quite useful also 
in another way, since they permit of more exact evaluation of the rate of change 
of pressure with temperature, which in the form of a differential coefficient 
is found to be a factor in other physical constants. One of these formulas 
for steam as adopted by Marks and Davis in the calculation of their tables 
is given in Eq. (655), the form of which, was suggested by Thiessen: 

(i-l-459.6) Iog^ = 5.409(i-212®)-3.71X10-I0[(689-0*-477^1, • (655) 

in which i= temperature F.; and p= pressure lbs. sq.in. 

This represents the truth to within a small fraction of one per cent up to 400® 
F., but having been found inaccurate above that point Professor Marks has 



460 ENGINEERING THERMODYNAMICS 

very recently developed a new one, based on Holbom and Baumann's 
high temperature measurements, which fits the entire range, its agreement 
with the new data being one-tenth of 1 per cent, and with the old below 100® 
F., about one-fifth of 1 per cent, maximum mean error. It appears to be the 
best ever found and in developing it the methods of the physical chemists have 
been followed, according to which a pressure is expressed as a fraction of the 
critical pressure and a temperature a fraction of the critical temperature. 
This gives a relation between reduced pressures and temperatures and makes use 
of the principle of corresponding states according to which bodies having the same 
reduced pressure and temperature, or existing at the same fraction of their 
critical are said to be in equivalent states. The new Marks formula is given 
in Eqs. (656) and (657), the former containing symbols for the critical 

1 X m i_ r and the latter giving to them their numerical values, 

temperature I e abs. J ^ « 

in pressure pounds per square inch, and temperature absolute F. 

log 2l = 3.006854 (y-l) Tn-. 0505476^-1- .629547 (-^-.7875^1, . (656) 

log p = 10.515354 - 4873.71 r- 1 - .00405096^-1- .000001392964T2. . (657) 

As the method used in arriving at this formula is so rational and scientific, 
it has been adopted for a new determination, from old data to be sure, of the 
relations between p and T for ammonia and carbon dioxide, so important as 
substances in refrigeration, especially the former. According to this method 
if pc and Tc are the critical pressures and temperatures, both absolute, and 
p and T those corresponding to any other point, then according to Van der 
Waals, 



Iog^=/(^'-lj (658) 

Accordingly, the logarithm of the critical divided by any other pressure, is 
to be plotted against the quantity [(critical temperatur edivided by the tem- 
perature corresponding to the pressure) — 1], and the form of curve p>ennits of 
the determination of the fimction, after which the values of the critical 
point are inserted. This has been done for NH3 and CO2 with the result for 
NH3 

log £^ = . 045+2.75^1?- 1) +.325^1?- 1^, .... (659) 



which on inserting the critical constants, 
Pc=114 atm. = 1675.8 lbs. per square inch 

rc== 727.4^ F. absolute 



which are the Vincent and Chap- 
puis values, 



HEAT AND MATTER 461 

becomes, 

logp=5.60422-1527.54r"i-17196ir-2 .... (660) 

For CO2 it was found that 

log2?=.038+2.65^p-l) + 1.8(~^-iy, .... (661) 

which on inserting the critical constants, 

Pe=77 atm. = 1131.9 lbs. per square inch 

' which are Andrews' values, 
rc=547.27^F. abs. 

becomes, 
log p = 7.46581-4405.765r"i+1617501.366r"2-257086165.8706r"3. (662) 

Curves showing the relation of reduced and actual temperatures and pressures 
are given in Fig. 139 for ammonia and in Fig. 140 for carbon doixide. 

For the past half century far more time and effort have been devoted to 
making other formulas of relation of p to T for saturated vapor not only for 
steam, but also for other vapors, than would have sufficed for accurate exper- 
imental determination, and as these help not at all they are omitted here. Equa- 
tions of physical relations can be no better than the data on which they are based, 
and for the substances ammonia and carbonic acid the charts or formulas must 
be used with a good deal of suspicion. 

In all engineering calculations requiring one of these constants even for steam 
no one is justified in using a formula; the nearest tabular or chart value must 
be employed and it will be as accm*ate as the work requires. Time is at 
least as important as accuracy, if not more so, for if too much time is required 
to make a calculation in commercial work, it will not be made because of the 
cost, indirect and approximate methods being substituted. 

It is sometimes useful in checking the boiling-point of some substance 
little known, to employ a relation between boiling-points of different substances 
at the same pressure appUed to a substance well-known. 

Let Ta and T^ be absolute temperatures of boiling for substances A and B under 

same pressure; 
Ta and Th be absolute temperature of boiling for substances A and B under 

some other pressure. 
Then, 

^! = f;+c(n'-n) (663) 



Such equations as this are useful in finding the saturation curve of other sub- 
stances from that for water, which is now so well established, when enough 
points are known for the other substances to establish, the constant c. Also 

Ta 
the ratio =^ plotted against the temperature difference Tb—Ti, should give 

TjT 



462 



ENGINEKRlNa THERMODYNAMICS 

















































































.. 








I 


/ 




















• 






















> 


/ 










































> 


/^ 








































/ 


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From Values of Wood 

" »» »* DietericI • 

** " Begiiault + 

»• *• " Ledoux ' 










.6 

1 
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.3 












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• 40 •O '^ •«> 

^Critical Temperature Divided by any other Temperature)—! 















































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• 

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From Values of W.ood 
'* »• - Bletedci • 
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Temperatures in Degrees Fahr. 
Fig. 139. — ^Ammonia Pressure-temperature Relations, for Saturated Vapor. 



HEAT AND MATTER 



463 




.2 .3 .4 .5 

(Critical Temperatare Divided by any otlier Temperature)-! 




—100 



50 50 

lemperatureB in Degrees Fahr. 



FiQ. 140. — Carbon Dioxide Pres6ure-temperature Relatipng for Saturaled Vapor. 



464 



ENGINEERING THERMODYNAMICS 



a straight line, and if the line is not straight the experimental values may k 
wrong or the law untrue. This procedure has been followed in Fig. 141, in 
checking the curves for CO2 and NH3 against those for water, but it is impos- 
sible to say whether the discrepancies for CO2 are due to a failure of the law or 
bad experimental values, probably both, as the law holds poorly for water itself. 



250 



200 



10 



CO 



^150 

0) 





I 



100 



50 



o 



O 



.5 



.6 

Values of 






Fig. 141.— Curves for CO, and NHi to Check the Linear Relation Eq. (663). 

All of the preceding refers, of course, to pure substances, but in practid 
work there are frequently encountered problems on solutions where large 
differences may exist compared to the pure liquids. Thus, for salts in water, it 
is well known that addition of a salt lowers the freezing-point, that more salt 
lowers it more, and it was first thought that the depression was in proportion 
to the amount dissolved. This being found to be untrue, recourse was had 



HEAT AND MATTER 



465 



again to molecular relations by Raoult, who announced the general law that 
the molecular depression of the freeztng-point is a constant 



Atm 



Molecular lowering of freezing-point J?'= = const., . . (664) 



w 



in which 



Af = depression of freezing-point in degrees F.; 
ti7= weight dissolved in 100 parts of solvent; 
m= molecular weight of substance dissolved. 

From Eq. (664) the freezing-point for brines may be found as follows: 

w 

Freezing-point of aqueous solutions = 32°— (const.) X — . . . (665) 

m 

As examples of the degree of constancy of the " constant " the following values 
Table XXIV, taken from Smithsonian Tables are given: 

Table XXIV 

LOWERING OF FREEZING POINTS 



Salt. 


g. Mol. 
1000 g. H«0* 


Molecular 
Lowering. 


AuthoritioB. 


NaCl 


.004 
.01 


3.7 
3.67 


■ 








.022 


3.55 


Jones 




.049 


3.51 


Loomis 




.108 
.232 
.429 


3.48 
3.42 
3.37 


Abegg 
Roozeboon 




.7 


3.43 




NH4CI 


.01 


3 6 






.02 


3.56 






.035 
.1 


3.5 
3.43 


Loomis 




.2 


3.4 






.4 


3.39 


• 


CaCli 


.01 
.05 


5.1 

4.85 










.1 


4.79 






.508 


5.33 




• 


.946 

2.432 

3.469 

3.829 

.048 

.153 


5.3 
8.2 
11.5 
14.4 
5.2 
4.91 


Arrhenius 

Joneff^letman 

Jones-Chambers 

Loomis 

Roozeboon 




.331 


5.15 






.612 


5.47 


• 




.788 


6 34 





466 ENGINEERING THERMODYNAMICS 

Just as the pressure of dissolved substances in liquids lowers the freezing- 
point, so also does it lower the vapor pressure at a given temperature or raii^e 
the boiling-point at a given pressure. Investigation shows that a similar 
formula expresses the general relation: 

Mm 
Molecular rise of boiling-point = S = = constant = 5.2, . (666) 

when water is the solvent. 

From Eq. (666) the rise of the boiling-point is found to be 

Rise of boiling-point =5.2 — (667) 

When liquids are mixed, such as is the case with all fuel oils and ^ith 
denatured alcohol, the situation is different than with salts in solution, and 
these cases fall into two separate classes: (a) liquids infinitely miscible like 
alcohol and water or like the various distillates of petroleum with each other, 
and (&) those not miscible, like gasolene and water. 

The vapor pressure for miscible liquid mixtures is a function of the pressure 
of each separately and of the molecular per cent of one in the other when there 
are two. This rule, which can be symbolized, is no use in engineering work, 
because in those cases where such mixtures must be dealt with there will be 
generally more than two liquids, the vapor pressure characteristic and molec- 
ular per cent of each, or at least some of which will be imknown. 

When, however, the two liquids in contact or in fact any number are 
non-miscible they behave in a Very simple manner with respect to each other, 
in fact are quite independent in action. Each liquid will evaporate imtil it.< 
own vapor pressure is established for the temperature, as if the other were not 
there, and the vapor pressure for the mixture will be the sum of all the separate 
ones. On the other hand the boiling-point will be the temperature at which 
all the vapor pressures together make up the pressure of say the atmosphere, 
and this is necessarily lower than the highest and may be lower than the 
lowest value for a single constituent. This action plays a part in vaporizers 
and carburettors using alcohol and petroleum products. To permit of some 
approximations, however, a few vapor tension curves for hydrocarbons and alco- 
hols are given later in the Section on vapor-gas mixtures, and data on the vapor 
pressure and temperature relations of ammonia-water solution are given in the 
section on the solution of gases in liquids. 

Example 1. Through how many degrees has ammonia vapor at a pressure of 
50 lbs. per square inch absolute been superheated, when it is at the temperature at 
which steam is formed under a pressure of 100 lbs. per square inch absolute? 

From the curve of pressure and temperature of steam the temperature is 328* F. 
for the pressure of 100 lbs. From the similar curve for anmionia vaporization occurs 
under a pressure of 50 lbs. at a temperature of 22^ F. Hence, superheat «=32S- 
22«306^F, 



HEAT AND MATTER 467 

Prob. 1. Three tanks contain the foUowing liquids together: water, ammonia, 
and carbon dioxide respectively, and at a temperature of 30^ F. What pressure 
exists in each tank? If the temperature rises to 70^ F. how much will the pressure 
rise in each? 

Prob. 2. The pressure exerted by water vapor in the atmosphere when saturated, 
is that due to the temperature and is independent of the pressure of the air. The 
total pressure read by a barometer is the sum of the air pressure and the water vapor 
pressure. What is the pressure due to each under a saturated condition for tem- 
peratures of 50** F., 100** F., 150** F., and 200* F., the barometer m each case being 
29.92 inches of Hg? 

Prob. 3. In order to secure a sufficiently high rate of heat transfer the steam in 
a radiator must be at a much higher temperature than the room to be warmed. If it is to 
btf 150** above room temperature what must be its pressure for room temperatures 
of 50** F., 60** F., 70** F., 80** F., and 125** F.? 

Prob. 4. In one tj^e of ice machine ammonia gas is condensed at a high pressure 
and evaporated at a low pressure. What is the least pressure at which gas may be 
condensed with cooling water of 70° F., and what is the highest pressure which may 
be carried in the evaporating coils to maintain a temperature in them of 0** F.? 

Prob. 5. Should carbon dioxide be substituted in the above machine what pressures 
would there be in the condensing coils, and in the evaporating coils? 

Prob. 6. How many degrees of superheat have the vapors of water, anunonia and 
carbon dioxide at a pressure of 15 atmospheres and a temperature of 500** F.? 

Prob. 7. Change the following pressures in pounds per square inch absolute xo 
reduced pressures for water, ammonia, and carbon dioxide, 15 lbs., 50 lbs., 100 lbs., 
500 lbs. 

Prob. 8. At the temperature of melting ice what will be the vapor pressure of ammonia 
and carbon dioxide? At the temperature of melting tin what will be the pressure of 
water vapor? At this same temperature how many degrees of superheat would 
ammonia vapor under 100 lbs. pressure have, and how many degrees superheat would 
carbon dioxide vapor have under 1000 lbs. pressure? 

Prob. 9. If 10 lbs. of common salt, NaCl, be dissolved in 100 lbs. of water, what 
will be the boiling point of the solution at atmospheric pressure, what the freezing-point? 

m 

10. Change of State with Amount of Heat at Constant Temperature. Latent 
Heats of Fusion and Vaporization. Total Heats of Vapors. Relation of Spe- 
cific Volume of Liquid and of Vapor to the Latent Heat As previously explained, 
a liquid boils or is converted into a vapor at constant temperature when the 
pressure on the surface is constant. Then during the change of state the amount of 
heat added is indirect proportion to the amount of vapor formed. The amount of 
vapor to convert a pound of liquid into vapor at any one steady tempera- 
ture, is the latent heat of vaporization some values for which are given at the 
end of this chapter in Table XLV, and it must be understood that this 
latent heat is also the amount given up by the condensation of a pound of vapor. 
Latent heat is not the same for different pressures or temperatures of vapori- 
sation but is intimately associated with the volume change in the transition 
From the liquid to the vapor state. That this should be so, is clear on purely 
rational grounds because there is necessarily external mechanical work done 



468 ENGINEERING THERMODYNAMICS 

in converting the liquid to the vapor, since this is accompanied by a change 
of volume against the resisting pressure at which the conversion takes place. 
Thus, if 

Vv= specific volume of the vapor in cubic feet per pound; 
Vl = specific volume of the liquid in cubic feet per pound ; 
P= pressure of vaporization lbs. per sq.ft. absolute. 

Then 

Mechanical external work done dur- ] „,-, rr \ ^i. ii_ iooc. 

... r 1 lu f =F(7v— yj it.lbs (^8 

mg vaponzation of 1 lb. J 

Of course, at high temperatures the volume of a poimd of liquid is greater 
than at low because of its expansion with temperature rise, and imder the cor- 
responding higher pressures the volimie of a pound of vapor is less, because 
of the compressional effect of the pressure, than at low pressures, so that as 
pressures and temperatures rise the difference Vv—Vl becomes less and dis- 
appears at the critical point where it is zero. The latent heat being thus asso- 
ciated with a factor that becomes less in the higher ranges of temperature 
and pressure may be expected, likewise to become less unless some other factor 
tends to increase. All the energy of vaporization making up the latent hea: 
may be said to be used up in (a) doing external work as above, or (6) overcom- 
ing attraction of the* molecules for each other. As at the critical point there 
is no molecular change and no external work, the latent heat becomes zero 
at this point. 

This relation between latent heat and volume change was formulated by 
Clausius and Clapeyron, but Eq. (669) is generally known as ClapeyronV 
equation: 

Let L = latent heat ; 

" J = mechanical equivalent of heat = 778, or better 777.52, in such 

cases as this; 

" T = absolute temperature of vaporization ; 

dP 
" -7^= rate of increase of vapor pressure per degree change of cone 

sponding temperature. 
Then 

^=77?52S(^-^^) (^■' 

This formula is used to calculate latent heat from the specific volumes of vapor 
and liquid and from the curvature of the saturation curve when they are known, 
but as these volumes are especially diflBcult to measure, direct experimental 
determination of the latent heat should be depended upon to get numerical 
values wherever possible. The formula will then be useful for the inverse 
process of calculating specific volumes from latent heats or as a means of 



HEAT AND MATTER 469 

checking experimental values of both, one against the other. It is, however, 

dP 
just as useful to calculate latent heats from the specific volumes, and -— of 

dT 

the vapor curve, when the latent heats are less positively determined than 

the volumes or densities. 

Another simpler relation of a similar general character exists and is useful 

in estimating latent heats approximately for some little known substances 

like, for example, the liquid fuels, and in the use of which accurate physical 

data are badly needed. Despretz announced that 



Vv-Vi 



is nearly constant for all substances, and this was simplified by Ramsay and 
Trouton on the assumption, first, that the voliune of the liquid is very small 
at ordinary temperatures and may be neglected, in comparison with the volume 
of the vapor, and second, that the volume of the vapor is inversely proportional 
to the molecular weight m and directly proportinal to absolute temperatures 
so that (Trouton's law) 



m-jf = constant = C 



(670) 



or 

m 

the constant c is given the following values by Young: 

CO2 c = 21.3 

NH3 c = 23.6 

Hydrocarbons c = 20.21 

Water and alcohols c = 260 

For such substances as water and steam, the properties of which must be 
accurately known, general laws like the above are of no value compared with, 
direct experimental determination except as checks on its results, and even 
these checks are less accurate than others that are known. 

These experimental data are quite numerous for water, but as generally 
made include the heat of liquid water from some lower temperature to the 
boiling-point. The amount of heat necessary to warm a pound of liquid from 
temperature 32° F. to some boiling-point, and to there convert it entirely into 
vapor is designated as the total heat of the dry saturated vapor above the origi- 
nal temperature. This is, of course, also equal to the heat given up by the con- 
densation of a pound of dry saturated vapor at its temperature of existence and 
by the subsequent cooling of the water to some base temperature taken univer- 
sally now as 32° F. in engineering calculations. 



470 ENGINEERING THERMODYNAMICS 

From observations by Regnault and formulated by him in 1863 the present 
knowledge of the total heat of water may be said to date. He gave the 
expression, Eq. (671), in which the first term is the latent heat at 32^ and 
one atmospheric pressure: 

Total heat per pound dry saturated steam^H = 1091.7+.305(<-32). (671) 

This was long used as the basis of steam calculations, but is now to be discarded 
in the light of more recent experimental data, the best of it based on indirect 
measurements by Grindley, Griessmann, Peake, who observed the behavior of 
steam issuing from an orifice, together with the results of Knobloch and Jacob 
and Thomas on specific heats of superheated steam, and in addition on direct 
measurements by Dieterici, Smith, GriflSths, Henning, Joly. All this work 
has been recently reviewed and analyzed by Davis, who accepts 1150.3 
B.T.U. as the most probable value of the total heat under the standard atmos- 
phere and the following formula as representing total heats from 212^ up to 

400° F. 

i3r=1150.3+.3746(<~212)-.000550(«-212)2 . . . (672) 

The Davis curve containing all the important experimental points ^nd the 
accepted line, extended dotted from 212° to 32°, is presented in Fig. 142. 

From the total heats given by this formula the latent heat is obtainable 
by subtraction, according to the relation. 

Latent heat (L)= total heat of vapor above 32° F. (If)— heat of 

liquid from 32° F. to boiling point (A), (673) 

in which the heat of the liquid is computed from a mean curve between Dieterici's 
and Regnault's values, having the equation A = .9983-. 0000288 («- 32) + 
.0002133(^—32)2. This is the basis of the values for latent and total heats in the 
Marks and Davis steam tables referred to, and accepted as the best obtain- 
able to-day. From these tables a pair of charts for latent heat and total heat of 
dry saturated steam are given at the end of this Chapter. 

The specific volume and density of dry saturated steam, given in the charts 
and table are calculated, as this seems to promise more exact results than direct 
experiment, the method of calculation involving three steps:* 

dp 
(a) From the pressure-temperature equation the ratio of — is foimd by 

differentiation as follows: 

log p = 10.516354 -4873.71 r~i-. 004050967+ .000001392964r2, 
whence 

^= f^^%^ -.00405096+. 0000027859287) p. 



dT \ T^ 






HEAT AND MATTER 



471 



(b) From the latent heats the diflFerence between specific volume of vapor 
and liquid, {Vv—Vl) is calculated by substituting (a) in Clapeyron's equation. 

(c) From the Landolt, Bomstein, Myerhoffer tables for density of water 
the volume Vl is taken, whence by addition the volume of the vapor Vv is found, 

For ammonia and carbonic acid there are no data available on total heats 
by either direct measure or by the orifice expansion properties, and very few 




a8° 50° eS*' 86^ 104° 122° 140** 158° 176'' IW" 212" 230° ti»° 266° 284'* JJ08° 820" 838" 356° 874° 888' 

Temperature In. Degrees K 

Fig. 142.— Total Heat of Dry Saturated Steam above 32° F. (Davis). 



determinations of the latent heat itself, so that the process that has proved so 
satisfactory with steam cannot be directly followed with these substances. 
Accordingly, a process of adjustment has been used, working from both ends, 
beginning with the pressure temperature relations on the one hand and specific 
volumes of liquid and vapor on the other, the latent heat is determined by 
Clapeyron's equation and where this does not agree with authentic values an 
adjustment of both latent heat and specific volume is made. 



472 ENGINEERING THERMODYNAMICS 

This process is materially assisted by the so-called Cailletet and Mathia> 
law of mean diameter of the curves of density of liquid and vapor, which are 
given in Figs. 143, 144 and 145, for water, ammonia and carbon dioxide, ol 
which the points are marked to indicate the source of information. 

On each of these curves the line BD is the line of mean density, its abscissa 
being given by the following general equation, 



8 



= i{^+yij=(i+bt+cfi (6: 



Of course, this mean density line passes through the critical volume B. For 
these three cases this Ekj. (674) is found to have the form. 



For water s = 28.7 -.015(<- 300) -.000015(<- 300)2. (a) 

For ammonia s = 20 - .022(< - 30) . (6) 

For carbonic acid.. s = 33.1-.0219(/+20)-.00016(<-20)2. (c) 



(675 - 



A more exact equation for water has been determined by Marks and Da%Ts 
in their steam tables and is 

s = 28.424 -.01650(^-320) -.0000132(^-320)2. . . . (676) 

From the smooth curve, which has the above equation, the volumes and densi- 
ties of liquid and vapor that are accepted have been derived, and are presented 
in chart form on a large scale and in tabular form at the end of the Chapter, 
the values for water being those of Marks and Davis. 

dp 
From these volume differences and the tz, relation the latent heats have 

dT 

been calculated and the newly calculated points are compared with experimental 

values in Fig. 146. 

The total heats are obtained by adding to the latent heat the heat of 
liquid above 32° from —50° F. up to the critical point for CO2 and to 
150° F. for NH3, which include the working range for refrigeration. 
These liquid heats have already been determined in Section 5 in discussing 
specific heats. 

Charts and tables at the end of this Chapter give the final values of total heat, 
heat of liquid, latent heat, specific volume and density of dry saturateti 
vapor based on large-scale plottings, without equations beyond that for the 
pressure-temperature relations for saturated vapor, and the results are be- 
Ueved to be as reliable as it is possible to 'get them without more experimental 
data. 

The properties of dry-saturated steam are given in Table XL VII, and 
charts. A, B, C, D, E, F; the properties of superheated steam, in Table XLVIII; 
dry-saturated ammonia vapor in Table XLIX, and Charts G, H, I, J, K, L; 
and dry-saturated carbon dioxide vapor in Table L, and Charts M, N, 
O, P, Q, R. 



HEAT AND MATTER 



473 




.1 "" .06 .04 

Upper Scale » Lbs. per Cu.Ft, 



.025 '" .08 
Lower Scale =Cu.Ft,per Lb. 




Volume of Vapor in Cu.Ft, 
Fig. 143. — Specific Volume and Density of Liquid and Dry Saturated Vapor of Water. 



474 



ENGINEERING THERMODYNAMICS 



























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Volume of Yapacin Oubio Feet pes VovaaSL, 



Fio. 144. — Specific Volume and Density of Liquid and Dry Saturated Vapor of AmmonU. 



HEAT AND MATTER 



475 




J& .04 
Upper Scales LlM.pei: Cu.Ft, 



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Volume orVapgr io Cu.Ft.per Poon^ , 

Fig. 145. — Specific Volume and Density of Liquid and Dry Saturated Vapor of Carbon 

Dioxide. 



476 



ENGINEERING THERMODYNAMICS 





































































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HEAT AND MATTER 



477 



The volumes of dry-saturated steam determined from the tables when 
compared with their pressiu-es show that there exists an approximate 
relation of the form for steam, 



p{Vv-VLy'^^ = constant.^497y 



(677) 



when pressures are in pounds per square inch and volumes are in cubic feet. 
This curve plotted to PF coordinates is called the saturation curve for the vapor. 
It is useful in approximate calculations of the work that would be done by steam 
expanding so that it remains dry and saturated or the work required to compress 
vapor such as ammonia under the same conditions. But as the specific volume 
of liquid is generally negligible it may be written as one of the general class 

P7'= constant, (678) 



for which «= 1.0646 and constant =497. 

This curve suppUes a means for computing the work for wet vapors (not too wet) 
as well as dry, provided only that they at no time become superheated or change 
their quality, by using for V some fraction of the true specific volume repre- 
senting the dryness. The very fact that a great volume of vapor may .be 
formed from an insignificant volunae of liquid makes the saturation curve a 
useful standard of comparison with actual expansion and compression lines for 
wet vapors. 




Fig. 147. — Comparison of Steam Expansion Line of an Indicator Card with the Saturar 
tion line for both Dry Saturated Vapor and for Vapor Constantly Wet at the Initial Value. 

Plotting the saturation curve beside an actual cylinder expansion or 
compression curve will show the quality of vapor at all times and also give a 
measure of evaporation and condensation taking place during the process. In 
Fig. 147 is shown a set of diagrams taken from a simple Corliss engine, 18X24 
ins. with 4 per cent clearance, a 2J-inch piston rod and tail rod, running at 150 
R.P.M., to which have been added lines of zero pressure and volume by the 
method explained in Chapter I. The discharge from the condenser per hour 
for a constant load of the value to give the above cards was 2600 lbs. Allow- 



478 ENGINEERlAa THERMODYNAMICS 

ing for the rods, the displacement volumes of each end of the cylinder will Ix 
5990 cu.ins., and since the clearance volume is 4 per cent, the steam volume 
will be 239.6 cu.ins. From the left-hand card it will be seen that the cut-off 
was at point C, 16.5 per cent of stroke, hence the volume at C is (.165 X 
5990) +239.6 = 1228 cu.ins. It will also be seen from the card, that the pres- 
sure at C was 73.5 lbs. per square inch absolute. From the curves or the 
tables at the end of the Chapter 1 cu.ft. of dry steam at this pressure weighs 

1228 
.1688 lb. and hence the weight of steam in this end of cylinder was rzrzX.1668 

1728 

or .1185 lb. at cutK)ff. From the card it will also be seen that at the end of 

the exhaust stroke, denoted by the point D, the pressure was 30 lbs., at which 

the weight of 1 cu.ft. of dry steam is .0728 lb., hence the weight of steam in the 

239 6 
cylinder was -=^X. 0728 = .01010 lb., and the amount admitted was . 1185 - 

.0101 = .1084 lb. 

In as much as the two ends of cylinder are identical and as the cards from 
both ends are practically the same, it may be assumed that the same weight 
of steam was in each end, or that .1084X2 = .2168 lb. are accounted for by the 
card per revolution, or .2168X150X60 = 1950 lbs. per hour. There is then 
the difference to otherwise account for, of 2600—1950=650 lbs. per hour, 
which can only have been lost by condensation. 2600 lbs. per hour is 2600 
-5- (150X60X2) = .1442 lb. per stroke, which with the .0101 lb. left from pre- 
vious stroke would make .1543 lb. in the cylinder at cut-off, and if it 
were all steam its volume would be 1581 cu.ins., denoted by point E 
on diagram. The ratio of AC to AE gives the amount of actual 
steam present in the cylinder at cut-off, to the amount of steam and 
water. The saturation curves CF and EG are drawn through C and 
E from tabuUr values and represent in the case of CF the volumes 
which would have been present in the cylinder at any point of stroke had the 
steam and water originally present expanded in such a way as to keep the 
ratio or dryness constant, and in case of EGy volmnes at any point of the stroke 
if all the steam and water originally present had been in form of steam 
and had remained so throughout the stroke. Just as the ratio of AC to AE 
shows per cent of steam present at cut-off, so does the ratio of distances of any 
points Y and Z, from the volmne axis denote the per cent of steam present at 
that particular point of the stroke. By taking a series of points along the 
expansion curve it is possible to tell whether evaporation or condensation is 
occurring during expansion. In this case the ratio, 

= = .795, and = =.86. 
AE XZ 

Hence, it is evident that evaporation is occurring since the percentage of steam 
is greater in the second case. 



HEAT AND MATTER 479 

For some classes of problems it is desirable that the external mechanical 
work be separated from the latent heat, and for this reason latent heat is 
given in three ways: 

(a) Ebctemal latent heat, 

(6) Internal latent heat, 

(c) Latent heat total. 
The external latent heat in foot-pounds is the product of pressure and volume 
change, or expressing pressures in pounds per square inch, 

144 
Extern^ latent heat = ^P(Fk-Fi,) (679) 

This is sometimes reduced by neglecting Yi, as insignificantly small as it really 
is for most problems which are limited to temperatures below 400° for saturated 
vapor, in which case, 

144 
External latent heat -^^Ff (680) 

In all cases 

Internal latent heat =L— (Ext. Lat. Ht.) (a) 



or 



144 
=L-^(7v-Fi)(6) 

144 
~L-^Yv (c) 



(681) 



Fusion and freezing are quite similar to vaporization and condensation 
in that they are constant temperature processes with proportionality between the 
amount of substance changing state and the amount of heat exchangee}. They 
are different in as much as little or no volimie change occurs. As there is so little 
external work done it may be expected that there is little change in their latent 
beats with temperature and pressure, but as a matter of fact it makes very 
little difference in most engineering work just how this may be, because prac- 
tically all freezing and melting takes place under atmospheric pressure. There 
does not appear to be any relation established between heats of fusion like 
those for vaporization that permit of estimates of value from other constants, 
$o direct experimental data must be available and some such are given for a 
■ew substances at the end of this Chapter in Table XL VI. As a matter of fact 
mch laws would be of little use, and this is probably reason enough for their 
ion-discovery. 

Example 1. Pigs of iron having a total weight of 5 tons and a temperature of 
1000® F. are cooled by inmiersing them in open water at a temperature of 60** F. If 
»ne-half of the water is evaporated by boiling, how much must there have been originally? 

The iron must have been cooled to the final temperature of the water, which 
Qust have been 212^ F. Also the heat given up by the iron will be the 



480 ENGINEERING THERMODYNAMICS 

product of its weight, specific heat and temperature difference, or, considering the 
mean specific heat to be .15, 

10,000 X (2000 -212) X.l 5^2,682,000 B.T.U. 

The heat absorbed by the water in being heated, considering its specific heat as unity 
will be its weight times its temperature change and, since one-half evaporates, the 
heat absorbed in evaporating it will be half its weight times the latent heat, or 

W[{212 -60) +i X970] =637»r B.T.U. 

These expressions for heat must be equal, hence 

Do7 

Example 2. A tank of pure water holding 1000 gallons is to be frozen by means 
of evaporating ammonia. The water is originally at a temperature of 60^ F. and the 
ice is finally at a temperature of 20° F. The ammonia evaporates at a pressure 
of one atmosphere and the vapor leaves the coils in a saturated condition. How 
many pounds of ammonia liquid will be needed, how many cubic feet of dry saturated 
vapor will be formed, and how much work will be done in forming the vapor? 

The heat to be removed is the sum of that to cool the water, the latent heat 
of fusion of ice, and that to cool the ice, or for this case 

[(60-32) +144+.5(32-20)]x8333, 

8333 being the weight of 1000 gallons of water. Hence the B.T.U. abstracted 
amount to 1,466,608. 

Each pound of ammonia in evaporating at atmospheric pressure absorbs 594 B.T.U.'s 
as latent heat and, therefore, 2470 lbs. are needed. At this pressure each pound of 
vapor occupies 17.5 cu.ft., hence there will be 43,200 cu.ft. of vapor. At this same 
pressure the volume of a pound of liquid is .024 cu.ft., so that the work done per xx>und 
in evaporating the ammonia is 37,000 ft.-lbs. and the total work is 915x10^ ft.-lbs. 

Prob. 1. How much ice would be melted at 32** F. with the heat necessary to 
boU away 5 lbs. of water at atmospheric pressure, the water being initially at the 
temperature corresponding to the boiling-point at this pressure? 

Prob. 2. What is the work done during the vaporization of 1 lb. of liquid anhydrous 
ammonia at the pressure of the atmosphere? 

Prob. 3. From the tables of properties of anhydrous ammonia check the value of 
the constant in Trouton's law given as 23.6 by Young. 

Prob. 4. As steam travels through a pipe some of it is condensed on account of 
the radiation of heat from the pipe. If 5 per cent of the steam condenses how much 
heat per hour will be given off by the pipe when 30,000 lbs. of steam per hour at a 
pressure of 150 lbs. per square inch absolute is passing through it? 

Prob. 5. Brine having the specific heat of .8 is cooled by the evaporation of ammonia 
in coils. If the brine is lowered 5** F. by ammonia evaporating at a pressure of 20 



HEAT AND MATTER 481 

lbs. per square inch gage, the vapor escaping at brine temperature, how many poimds 
of brine could be cooled per poimd of ammonia? 

Prob. 6. Steam from an engine is condensed and the water cooled down to a 
temperature of 80** F. in a condenser in which the vacuum is 28 ins. of Hg. How many 
pounds of cooling water will be required per pound of steam if the st«am be initially 
10 per cent wet? 

Prob. 7. A poimd of water at a temperature of 60° F. is made into steam at 100 
lbs. per square inch gage pressing. How much heat will be required for this, and what 
will be the volimies at (a) original condition; (b) just before any steam is made; (c) after 
all the water has been changed to steam? 

Prob. 8. A sand mold weighs 1000 lbs. and 100 lbs. of melted cast iron ore poiu^ 
into it. Neglecting any radiation losses and assuming the iron to be practically at 
its freezing temperature how much of the iron will solidify before the mold becomes 
of the same temperature as the iron? 

Prob. 9. How many pounds of ice could be melted by heat given up by freezing 
50 lbs. of lead? 



11. Gas and Vapor Mixtures. Partial and Total Gas and Vapor Pressures. 
Volume, Weighty and Gas Constant Relations. Saturated Mixtures. Humidity. 

One of the characteristic properties of gases distinguishing them from liquids, 
and which also extends to vapors with certain limitations is that of infinite 
expansion^ according to which no matter how the containing envelope or volume 
of the expansive fluid may vary, the space will be filled with it at some pres- 
sure and the weight remain unchanged except when a vapor is brought to 
condensation conditions, or the pressure lowered on the surface of a liquid 
which will, of course, make more vapor. A given weight of gas or vapor (within 
limits) will fill any volume at some pressure peculiar to itself, and two gases, 
two vajwrs, or a vapor and gas, existing together in a given volume, will fill it 
at some new pressure which is the sum of the pressures each would exert sepa- 
rately at the same temperature (if non-misdble). This fact, sometimes des- 
ignated as Dalton's Law, permits of the derivation of equations for the rela- 
tion of any one pressure, partial or total, to any other total or partial, in 
terms of the weights of gas or vapor present, and the gas constants R, It 
also leads to equations for the various constituent and total weights in 
terms of partial and total pressures and gas constants. Such equations sup- 
ply a basis for the solution of problems in humidification and drying of air, in 
carburetion of air for gasolene and alcohol engines, or of water gas for illumina- 
tion, and are likewise useful as check relations in certain cases of gas mixtures 
such as the atmospheric mixture of nitrogen and oxygen, producer gas or gase- 
ous combustibles in general. 

Let wi, W2 and Wx be the respective weights of the constituents of a mixture; 

Wm= 2io be the weight of the mixture; 
r Pif P2, Px be the respective partial pressures of the constituents; 

Pm— 2P be the pressure of the mixture; 

iBi, R2, Rs be the respective gas constants; 

£» be the gas constant for the mixture. 



it 
tt 



482 ENGINEERING THERMODYNAMICS 

Then if wi lbs. of one, and W2 lbs. of another gas or vai)or at temperature 
Tm occupy the volume F» cubic feet together, 

V^Pl^WiRiTn,, (o)] 

and \ (682: 

V,nP2 = W2R2T^, (6) J 

whence 

V^{Pl+P2)=^(WiRi+W2R2)T,n, (683) 

or, in general, 

2;P= X(wR)Tn . . . ' (684) 

Or putting 

sp=p;», (68o: 

and 

X{wR)=R^w^, or /2« = ^^^, ... (686) 

then 

P^V^^W^Rn^Tsny (687) 

so that the mixture will behave thermaUy quite the same as any one gas with 
sixh exceptions as may be due to a different gas constant £». 
Dividing Eq. (682a) by Eq. (683) or (684) gives 

Pi^ wiR i wiRi 

Pm WlRl-lrW2R2 S(«>fi) ^'*^' 

which gives the ratio of any partial pressure to that for the mixture in terms 
of the individual weights and gas constants. Hence 

Pi - wifii . , 

pi;-^^^ (^^^ 

which gives the ratio of any partial pressure to thai for the mixture in terms of 
its oum iveight and gas constant and those for the mixture. 

It is possible to express the ratio of weights as a function of gas constants 
alone which will permit of a third expression for the partial pressures in terms 
of gas constants without involving any weights. For two gases 

Wl-V)m — W2. 

Whence 

^ = 1-^, (690) 

Wm Wni ^ 



HEAT AND MATTER 483 



But from Eq. (686) 



or 



W2 



so that 






1I\» R2\ V)m I 



or 






\ R2/ R2 



and 



tOl R2^Rm 



y>m R2'-'Rl* 



(691) 



11^/ticA 18 the roHo of partial to total weights in terms of gas constants. On sub- 
stitution in Eq. (689), 



Pi 
P, 



'_l^Rl/ R2-'Rm \ ^gg2) 

'p, Rm\R2'-'Rl/' 



which gives the ratio of partial pressures of two gases or vapors to thai for the 
mixture in terms of the individual gas constants and that for the mixture, and 
a similar expression can be found for more than two gases. The ratio of any 
one partial, to the total weight can also be found from Eq. (689) in terms of its 
gas constant and partial pressure, and the mixture gas constant and pressure, 
from Eq. (691) in terms of the gas constants for the constituents and for 
the mixture. This ratio of partial to total mixture weight gives the fractional 
composition by weight. 

It is sometimes necessary to know the volume relations in a mixtiu^ of two 
gases existing at the same pressure or two vax)ors or a vax)or and gas, such, for 
example, as air and water vapor. In this case two different volumes existing 
together at a common temperature and pressure together make up a mixture 
volume equal to their sum. Using similar symbols 

I, (693) 

Pn.V2-^W2R2Tm] 

where Vi and V2 are the volumes occupied by the two constituents respectively 
when at a mixture pressure Pm and temperature Tmt whence for the mixture 

Pm{yi + V2)^{WiRi-^W2R2)Tm (694) 

or 

P..2(T0 = 2(u;/e)r« (695) 



484 ENGINEERING THERMODYNAMICS 

These Eqs. (694) and (695) are identical in form with (683) and (684) except that 
Y replaces P, and F, P, so that all equations just derived also apply to volumes 
as the volume proportion will be identical with pressure prox)ortions. For 
convenience of reference these may be set down. 
From Eq. (688), 

Fi W\K\ taciCC^, 

y;=2^' ^^^^ 

which gives the roiio of any partial volume, to thai for the mixture in terms of 
the individual weights and gas constants. 



From Eq. (689) 

Vi WiRi 

Vm^WmRm' 



(697) 



which gives the ratio of any partial volume to thai of the mixture in terms of its 
own weight and gas constant and those far the mixture. 



From Eq. (692) 

V^ R^{R2-Ri)' 



(698) 



which gives the ratio of any partial volume to that of the mixture in terms of the 
individual gas constants and that for the mixture. 

The volumetric composition of air is given by Eq. (697) or its equal 
numerically Eq. (692), and since the partial pressure of oxygen and nitrogen 
in air are 78.69 per cent and 21.31 per cent, these are its volumetric per cents. 

When one of the constituents is a vapor, all the preceding applies, provided 
the condition of the vapor is such that at the temperatures assumed it is not 
near the condition of condensation, but then the relations become more definite 
since the partial pressure of the vapor is fixed by the temperature. In practical 
work with gas and vapor mixtures the failure^of the perfect gas laws near the 
condensation condition is ignored and they are assumed to be true for the very- 
good reason that there is no other way as good, to get numerical results. 

All liquids, and many, if not all solids will, if placed in a vacuum chamber, 
evaporate until the pressure has reached a certain value depending on the tem- 
perature, at which time the liquid and its vapor are in equilibrium, and evapo- 
ration may be said either to cease or proceed at a rate exactly equal to the rate at 
which vapor condenses, or more precisely, at equilibrium the weight of vapor 
in the vapor form remains constant. The weight of vapor that will rise over a liquid 
in a given space depends on the temperature and pressure which are related 



HEAT AND IdATTER 



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Tempeorature^Deg. Fahr. 

ha. 148. — Vapor Pressure of Hydrocarbons and Light Petroleum Distillates of the Gasolene 

Class. 



486 



ENGINEERING THERMODYNAMICS 



in the so-called vapor tension or vapor pressure tables and curves, such a$ 
shown in Figs. 148, 149 and 160, for some liquid fuels or as given in the pre- 
vious section for water. At any fixed temperature the vapor will con- 
tinue to rise until it exerts its own vapor pressure for the temperature. 

the process being often described as evaporation .without ebullition. If the 
liquid or solid be introduced into a chamber containing dry gas the evapora- 
tion will proceed precisely the same as in the vacuum imtil the pressure has 








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