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ENGINEERING THERMODYNAMICS
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ENGINEERING
THERMODYNAMICS
BY
CHARLES EDWARD LUCKE, PH.D.,
Prtf€990T of Mechanical Engineering in Columbia Uniwreity, New York CUy
McGBAWHILL BOOK COMPANY
239 WEST 39TH STREET, NEW YORK
6 BOUVERIJB STREET, LO^DO^, E. C.
1912
Copyright 1912, by thb
MoGRAWHILL 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 subtopics 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, horsepowers 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 icemaking 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 compressedair 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 horsepower, 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 threefourths 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 pressurevolume diagrams to indicator cards 34
12. To find the clearance 37
13. Measurement of areas of PV diagrams and indicator cards 43
14. Indicated horsepower 44
15. Effective horsepower, brake horsepower, friction horsepower, 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. Horsepower of nozzles and jets, by PV method 67
Chapter II. Work of Compressors. Horsepower 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. Singlestage compressor,, no clearance, isothermal compression. Cycle I. Work,
capacity, and work per cubic foot in terms of pressures and volumes 81
i. Singlestage compressor with clearance, isothermal compression, Cycle II. Work,
capacity, and work per cubic foot in terms of pressures and volumes 85
5. Singlestage compressor, isothermal compression. Capacity, volumetric effioiency,
work, mean effective pressure, hor&epower and horsepower 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. Singlestage compressor with clearance, exponential compression, Cycle IV. Work,
capacity, and work per cubic foot in terms of pressures and voliunes 96
8. Singlestage compressor, exponential compression. Relation between capacity,
volumetric efficiency, work, mean effective pressure, horsepower and horse
power per cubic foot of substance and the dimensions of cylinder and clearance. . 98
ix
X CONTENTS
PAGE
9. Twostage oompressor, no clearance, perfect interoooling, exponential compression,
bestreceiver pressure, equality of stages. Cvcle V. Work and capacity in terms
of pressures and volumes 103
10. Twostage compressor, with clearance, perfect intercooling ex^ < ncntial compres&ion,
beetreceiver pressure, equality of stages. Cycle VI. Work and capacity in terms
of pressures and volumes 109
11. Twostage compressor, any receiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure and horsepower, in terms of
dimensions of cylinders and clearances 113
12. Twostage compressor, vnlh best^eceiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure and horsepower in terms
of dimensions of cylinders and clearances 120
13. Threestage compressor, no clearance, perfect intercooling exponential compree
bion, best two receiver pressiues, equality of stages. Cycle VII. Work and
capacity, in terms of pressures and volumes 125
14. Threestage compressor with clearance, perfect intercooling exponential compression,
bestreceiver pressures, equality of stages. Cycle VIII. Work and capacity in
terms of pressures and volumes ICl
15. Threestage compres^r, any receiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure, and horsepower in terms
of dimensions of cylinders and clearances 135
16. Threestage compressor vnlh bestreceiver pressures, exponential compression. Capac
ity, volumetric efficiency, work, mean effective pressure and horsepower 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. Horsepower 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, horsepower and consumption of simple
engines 197
4. Work of expansive fluid in single cylinder without clearance. Exponential expan
sion, Cycle II. Mean effective pressure, horsepower 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, horsepower, 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, horsepower and consumption
of simple engines 219
7. Action of fluid in multipleexpansion 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 lowpressure admission are independent 287
13. Compound engine without receiver. Logarithmic law. No clearance. Cycle IX.
General relations between dimensions aiid work when highpressure exhaust
and lowpressure admission are coincident 292
14. Compound engine without receiver. Exponential law. No clearance, Cycle X.
General relations between dimensions and work when highpressure exhaust
and lowpressure 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 pressiues, 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. Tripleexpansion engine with infinite receiver. Logarithmic law. No clearance.
Cycle XYII. General relations between pressures, dimensions and work 349
22. Multipleexpansion engine. General case. Any relation between cylinders and
receiver. Determination of pressurevolume diagram and work, by graphic
methods 357
23. Mean effective pressure, engine horsepower, 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 cutoff 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 horsepower and cylinder sizes 387
3di CONTENTS
Chapter IV. Heat and Matteb. Qualitatzve and Quantitative Relations
BETWEEN Heat Content of Substances and PhtsicalChemical 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 freezingpoints 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 airblast reactions, producing air gas, and
blastfurnace 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 steamblast 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. Horsepower
of gas producers 719
10. Combustion effects. Final temperature, volume and pressure for explosive and
nonexplosive 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 selfignition 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 horsepower. Horsepower 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. Steamboiler efficiency, furnace and heatingsurface 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 CompressedAir 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 heatengine iprocesses. Cyclic efficiency. A
reference standard for engines and fuelburning 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 noncompression gascycles. 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 compressedair 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, horsepower, 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. Horsepower per pound mean effective pressure 68
12. Ratio of cutoffs 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. Freezingpoint 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 waterice 456
24. Lowering of freezingpoints 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. Baum6specific 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 freezingpoints 586
44. Boilingpoints 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 oilgas 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 blastfurnace 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 airgas mixtures 867
122. Composition of boilerfiue gases 868
123. Limits of proportions of explosive airgas 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 lowpressure 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 horsepower, Chapters III and VI.
= boiler horsepower, 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 horsepower, Chapter III.
^ft= boiler efficiency, Chapter V.
£/= furnace efficiency, Chapter V.
£/= thermal efficiency referred to indicated horsepower. 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 horsepower 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. "footpound.
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.
"horsepower. 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 horsepower.
in. » inch,
(in. pr.)» initial pressure in pounds per square inch.
J s Joule's equivalent » 778 (approx.) footpounds 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 horsepower. Chapter IlL
If a as a subscript to denote lowpressure 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.)— lowpressure 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 crosssection, 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 highpressure cylinder, Chapter III.
122;= ratio of expansion in lowpressure 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 footpounds.
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
= 11 ; , 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 highpressure 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 lowpressure admission voliune to highpressure 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 highpressure cylinder of the compound
engine, Chapter III.
Zs fraction of the stroke of the steam engine completed at cutoff. 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 footpound. 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
footpound 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 footpound 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 directacting 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
footpounds?
Footpounds »force xdistance
=2000 X80 = 160,000 ftlbs.
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 footpounds?
Prob. 4. A cubic foot of water falls 50 ft. in reaching a waterwheeL 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 10ton 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 10inch 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 horsepower,
or the performance of 550 footpounds per second or 33,000 footpoimds per
minute, or 1,980,000 footpounds per hour. In the metric system the horse
power is termed chevalvapeur, and is the performance of 75 killogrammeters
=542i footpounds per second, or 4500 kilograjnmeters= 32,549 footpounds
WORK AND POWER. 3
per minute, or 270,000 kilograinmeters= 1,952,932 footpounds 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 horsepower?
TT footpounds per minute
Horsepower '^ 33 qg
tune
" 33,000
^65,000X60^
33,000
Prob. 1. The drawbar pull of a locomotive is 3000 lbs. when the train is traveling 50
miles per hour. What horsepower is being developed?
Prob. 2. Amine cage weighing 2 tons is lifted up a 2000ft. shaft in 40 seconds. What
horsepower 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 horsepower?
Prob. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 lbs. What
horsepower 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 horsepower 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 horsepower?
Prob. 8. To draw a set of plows 2i miles per hour requires a drawbar pull of 10,000
lbs. What must be the horsepower of a tractor to accomplish this?
Prob. 9. The horsepower 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 horsepower required to draw a car 20 miles per hour up a 12 per
cent grade if the car weighs 2500 lbs. and the drawbar 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 footpounds, but will be, only when appro
priate units are chosen for the factors. These necessary factors are given as
ffdOows:
, W<xk in footpounds = 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 horsepower 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 = rrr* — 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 abovedefined 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 644 ^^ ^ 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 footpounds 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 footpounds 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 waterwheel is kept in motion by a jet of water impinging on flat vanes.
The velocity of the vanes is onehalf 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 12ft.
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, ForceDistance 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 footpounds 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 footpounds 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 footpounds 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 footpounds 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 6ft.
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 drawbar 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 horsepower.
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, PressureVolume 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 AB/ and then suddenly have
risen to B, In this case the work done for this step would be 100 footpounds.
(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 footpounds.
(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
pressurevolume 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 footpounds 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, PreesureVolume Coordinates. Gontinuoua PressureVolume
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 18in. 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 footpounds 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,
isL
Vi^'Yx^"
•]
As s is greater than one, the denominator and exponents will be negative, so,
changing the form to secure positive values,
s1VFi«» F2'V
This can be put in a still more convenient form. Multiplying and dividing by
1 1
vvi
or
Y.»v
fr.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 footpounds 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 footpounds.
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 footpounds 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
logPglogP , .
"^^lognlogF, ^""^
_ _ 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 horsepower 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
1201
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 drysaturated 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 drysaturated 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 crosshatched
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(Hlo& 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 crosshatched 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,320,000200,000 = 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,dW<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 Overexpansion 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^&ABXHCDX,
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 footpounds 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 footpounds with coordinates
32 ENGINEERING THERMODYNAMICS
pounds per square foot, and cubic feet, then the division of area in footpounds
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 backstroke work, or the respective footpound 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 footpounds
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
= 3.3PcPf;
= 3.3X20,00010,000;
=66,00010,000=56,000 lbs. per sq.ft.
The work done in footpounds 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 horsepower 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 60lb. 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 80lb. 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 PressureVolume 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 various
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 crosssection 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
crosssection 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 lefthand 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 stripwidths 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 footpounds 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 wellknown instrument
specially designed for direct measurements of area.
14. Indicated Horsepower. Work done by the fluid in a cylinder, because it
is most often determined by indicator card measurements, measures the indicated
horsepower, 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 footpounds 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 horsepower 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 horsepower 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 doubleacting 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 horsepower 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 horsepower 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 horsepower 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. doubleacting steam engine runs at 250 R.P.M. and the
mean effective pressme 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 singleacting 2cylinder engine has a piston 10 ins. in diameter with
a 2in. 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 horsepower 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 Horsepower, Brake Horsepower, Friction Horsepower,
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 horsepower output for the
system, which may be compared to the horsepower 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 horsepower 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 horsepower 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 horsepower 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 Horsepower. The difference between
brake and indicated horsepower of engines is the friHion horsepower of the
engine and the ratio of brake to indicated horsepower is the mechanical efficiency
of the engine. For an engine, then, the effective horsepower or useful horse
power is the brake horsepower. When the power cylinders drive in one machine
a pump or an air compressor, the friction horsepower of the machine is the
difference between the indicated horsepower 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 porsepower to indicated horsepower
of the power cylinders. Whether the indicated horsepower 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 horsepowers 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 horsepower 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 measmes 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 footpound 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 horsepower 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 100ft. 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 waterwheel 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 overall
efficiency?
Prob. i. Perry ^ves a rule for the brake horsepower of steam engines as being
equal to .95 I.H.P. — 10. On this basis find the mechanical efficiency of a 500H.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 horsepower 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 horsepower, termed the specific
displacement, is the basis of computations on the steam consumption of steam
engines, the horsepower per cubic feet of free air per minute for air compressors,
the horsepower per ton refrigeration for refrigerating machines and the con
sumption of fuel per hour per horsepower 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«,
^ ^ 60LXjjjXiV ^ 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 Horsepower 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 horsepower delivered
by a 56x72in. blowing engine with 4 per cent clearance and mean effective pressure of
10 lbs. per square inch?
Prob. 2. An 18x22in. 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 horsepower?
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. — PressureVolume Diagram for Nozzle Expansion Measuring the Acceleration
Velocity and Horsepower of Jets.
But
PeVe' = PeVeVe'^=PbVb' = PbVbVb»'';
81
.\ PeVe = PbVb
m"^^^w
81
81
Whence
r..K..S[.(ft)]P.K.(';=)
al
.^^"'[•(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 overexpansion, 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?
.1X144X100X4.36[1Q"]
=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 pressuredrop
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
^^Mmmf
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. Horsepower of Nozzles and Jets. Although, strictly speaking, nozzles
can have no horsepower, 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 footpounds 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 horsepower developed by the jet, or the nozzle horsepower.
Accordingly,
1
HP^^'«*=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 horsepower 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 horsepower 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 horsepower?
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 horsepower 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) horsepower 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 threequarters 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
waterpower 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 waterwheels 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 onequarter
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 horsepower. (6) Consider steam to be admitted onehalf stroke without
other change. How will the horsepower vary? (c) What will be the horsepower for
onequarter admission if the exhaust pressure is 15 lbs. per square inch absolute? (d)
What will be the horsepower 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 horsepower?
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 horsepower of the engine?
If the above installation were replaced with an airdriven 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 horsepower of this engine?
10. Show by a P7 diagram, assuming any convenient scales, that the quantity of air
discharged byacompressor and the horsepower, both decrease asthe altitude increases, and
that the horsepower 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 onequarter of the stroke and the initial
pressure raised sufficiently to maintain the same horsepower, what would be the new
initial pressure and the new value of the steam used per horsepower 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 motorfire 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 horsepower if the same
amount of air is taken in and how many cubic feet per minute will be discharged?
(c) What will be the horsepower 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 horsepower?
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
wellmade 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 horsepower 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 horsepower 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
81.3 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 crosssection 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 horsepower in the two cases.
23. The power from a hydroelectric 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 horsepower 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 waterwheel 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 threequarters 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) horsepower 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) horsepower per pound m.e.p.?
26. A waterpower site has available at all times 3500 cu.ft. of water per min
ute at a 100ft. fall. Turbines of 70 per cent efficiency are installed which take the place
of two doubleacting 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:
p0.4912a32) 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.
Footpounds.
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
Footpounds per
Second.
Footpounds per
Minute.
Horsepower.
ChevalVapeur.
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
HORSEPOWER 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
91
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«Vi
Indicated horsepower = (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^,=8.024^P,7l(^)~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, HORSEPOWER 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 directconnected 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 gasdriven blowing engines,
showing approximately twice the economy of steamdriven 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 gasdriven blowing engines to equal the steamdriven.
Some lowpressure 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 turbocompressors, is being rapidly adapted to such higher pres
sures as have heretofore required piston compressors.
When highpressure 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 lowpressure 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, noncondensible 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. Naturalgas compressors, some steam and
some gasengine 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 lowpressure gases from a given place, or the delivery
of a quantity of higherpressure gas to another place or both, but all include
compression as an intermediate step between constantpressure admission and
constantpressure 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 wetvapor or vvetgas 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 highpressure 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 reexpansion 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 pressurevolume 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 heatexchange 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 constantpressure 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 multistage with various
amounts of cooling in the intercooler, but in multistage 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 gasvapor 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 pressurevolume
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 pressurevolume cycles defined by their phases,
as shown in Fig. 23, four for singlestage 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.
Singlestage Compression Reference Cycles or PV Diagrams
a
Cycle 1. Singlestage Isothermal Compression without Clearance.
Phase (a) Constant pressure supply.
" (6) Isothermal compression.
" (c) Constant pressure delivery.
" (d) Constant zerovolume pressure drop.
Cycle 2. Singlestage Isothermal Compression with Clearance.
Phase (a) Constant pressure supply.
" (6) Isothermal compression,
(c) Constant pressure delivery.
id) Isothermal reexpansion.
Cycle 3. Singlestage Exponential Compression without Clearance.
Phase (a) Constant pressure supply.
(6) Exponential compression.
(c) Constant pressure delivery.
(d) Constant zerovolume pressure drop.
Cycle 4. Singlestage Exponential Compression with Clearance.
Phase (a) Constant pressure supply.
" (6) Exponential compression.
" (c) Constant pressure delivery.
" (d) Exponential reexpansion.
it
It
WORE OF COMPRESSORS
\
*
1
1' rt

S
"T"'

1 1
8
s
.
%\
if
ti~
:S^
JX
mn^T
i' s
S 3
Ljl bgBdvpmimfli t*.
9
ll
IJ ^ Md vpuOij (q w w^
T
±
.?
±
1 ;
±
f' i
T
±
~iL \
A
%
' *
t ~'''X
1 „iJ»
i
a;
il:i
I I i I I
u'^'^iv<»e>i'n •MHid
; 78 ENGINEERING THERMODYNAMICS
MuiynSTAQE Comprbssion
The phases making up multistage 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
stantpressure decreasing volimie, delivery line whose length is \ of the whole
diagram, exactly as in singlestage 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 constantpressure 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 constantpressure 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
pressurevolume 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.
Twostage Compression Reference Cycles or PV Diagrams
«
Cycle 5. Twostage 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 zerovolume pressure drop.
Cycle 6. Twostage 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 reexpansion of first stage clearance,
(e) Exponential compression from best receiver pressure.
(J) Constant pressure delivery.
ig) Exponential reexpansion of second stage clearance.
i t
ti
Threestage Compression Reference Cycles or PV Diagrams
Cycle 7. Threestage 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 zerovolume pressure drop.
i t
80 ENGINEERING THERMODYNAMICS
Cycle 8. Threestage 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 reexpansion 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 reexpansion of second stage clearance.
* ' (h) Exponential compression from best second receiver pressure.
* ' (i) Constant pressure delivery.
' ' (j) Exponential reexpansion of third stage clearance.
It should be noted that cycles 6 and 8 may be subdivided 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 reexpansion in the higherpressure cylinder
is equal to the voliune of clearance in the next lowerpressure cylinder, causing
the combined diagram to have a continuous reexpansion 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 horsepower corresponding to the removal of a given
volume of gas per minute from the lowpressure supply or to the delivery of
another volume per minute to the highpressure 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 horsepower necessary to
drive the compressing piston at a specified rate and further to calculate the
work and horsepower 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 singlestage compressor should require two horsepower per cubic
foot of free air compressed per minute, and Cycle I should for the same
pr^sure limits require only one horsepower 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 pressurevolume 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, horsepower, and capacity in terms of dimensions
of cylinders and clearances.
3. Singlestage 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) zerovolume.
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=Pcj 144 = absolute delivery pressure in pounds per square inch
= (del.pr.);
p
Rp = 5^= ratio of delivery pressure to supply pressure;
W = footpounds 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
ti
ti
it
n
it
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. — Onestage 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, footpounds, 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 highpressure 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 highpressure 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 multistage 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
Fft5
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 horsepower 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 horsepower 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 horsepower 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 horsepower 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 horsepower
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 horsepower
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 horsepower 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 horsepower 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. Singlestage CompreBsor with Clearance, Isothennal Compression,
(Cycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures
and Volumes.
"'jr " ^
Fia. 25. — OneStage Compressor Cycle 2, Clearance, Isothermal.
Referring to Fig, 25, the work of the cycle is, by areas.
Net work &tc& = EBCG+G CDF
HAD FEB 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=JBCLJADL,
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 footpounds 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 lowpressure capacity
the work of isothermal compressors is independent of clearance. The value of
the lowpressure capacity (Vt—Va) maj^ not be known directly, but may
be found if the volume before compression, Vb, the clearance volume befor^
reexpansion, 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 lowpressure gas is, in footpounds,
W
(L.RCap.) = ^^^^^PP^'^^^g^' (3^^
and the work per cubic foot of highpressure 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. Singlestage Compressor Isothermal Compression. Capacityi Volu
metric Efficiency, Work, Mean Effective Pressure, Horsepower 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. VtVa
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 horsepower of compressor;
88 ENGINEERING THERMODYNAMICS
The lowpressure capacity of the singlestage isothermal compressor with
clearance is,
(L.P.Cap.) = (Vftya), 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+ccRp) (39)
For convenience the term, Volumetric Efficiency, Ev is introduced. Since
this is defined as the ratio of the lowpressure capacity to the displacement,
E,JJ±F:J^^^l+ccR, (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, footpounds^ 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(Hccfip)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.)(Hc— cBp) log* JKp (43)
or
(m.e.p.) = (sup.pr.) J5» log« /2p . (44)
WORK OF COMPRESSORS 89
The indicated horsepower 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+cci2,)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 horsepower may be derived by the
substitution of the value of (m.e.p.) Eq. (44) in the following general expres
sion for indicated horsepower.
( 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+ccftp) 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 :
PaVaPdVi or Fa=^7d8.5X.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 horsepower
of machine? Speed is 150 R.P.M., compressor is double acting and stroke » 1 .5 diameters.
Neglect piston rods.
Z)=(L.P.Cap.)5J^, 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 horsepower 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 lowpressure capacity, its horgepower, and
WORK OF COMPRESSORS 91
how will the horsepower and the capacity compare with these quantities in a
hypothetical conipressor of the same size but having zero clearance? How will
the horsepower per cubic foot of delivered air compare?
Prob. 3. A manufacturer gives for a 10jxl2 in. doubleacting compressor running
at 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50100 lbs.
per square inch gage. What clearance does this assume for the lowest and highest
pressure if the compression is isothermal? The horsepower 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 horsepower
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
horsepower 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 horsepower 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 horsepower for 100 lbs. per square inch gage delivery pressure?
6. Singlestage Compressor, No Clearance Exponential Compression,
(Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures
and Volumes. The cycle of the singlestage 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
PcVcP,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
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_L JL ' i
I ; 1 '
il_ t ijui iir ^41 ■
11 III 1
L JLV— 1 ;
i ' I
V^ u '
1 ' 1 ■ '
IT. tii ^41
V I \ ' ' ' !
1 ! ~^ 1 '
1 \ !
i 1 (
1UiVl I l\ '
. _^ ._ _^ _
^**^ \ \
tl V \
^* Ji . JL, \]^ J
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5 4J Ip \ ^
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a \v 1 ^4 1
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5  lV \>i.
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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
^ iJW. .. >^_t
i
a • k ' V t
r ^ • h
8 T^ ^ K^
_i 11'
1 '
H J '^' ^k.
1
m _iL ^ __r^
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> t 1
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. Ni^
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'^■^ "**^ i !
iJOUU 1 ti
'''T>4 "^">.J
1
^i U T ' ■
1 ''t*t»4». ^]h*«»^.^
A .,.=. ^,„j^i .11 r ,t;^ j
L L ' Ll_ 1 ^^^ Tg» M^ R
^^ .1. ! 1 I 1 i
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r ir ^ '
J m T 1 cs
F XlSij L—
JX t.y ,
1
1 2
Volumes
345
in Cubic Feet
r*
D
L.P. Cap
Fia. 26. — OneStage Compressor Cycle 3, No Clearance, Exponential.
Substituting above
al
»i
'^f:S[©"]+''^[©""']
»i
■ ^.>'.[(&)"]CV)
WORK OF COMPRESSORS 93
Whence
«i
'^%4iW[(g) • l]. ....... (48)
Eq. (48) gives the work in footpounds 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
Wl^^,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, footpounds is
• 1
W
—^ = 144^ (sup.pr.) (ft/ l) (51)
(L. P. Cap
Before obtaining the work per cubic foot of highpressure 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 highpressure 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 constantpressure
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 footpounds 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 footpounds 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 44.6 = 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 footpounds 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 singlestage 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 horsepower will the capacity be greater or less
and how much?
Prob. 3. A dryvacuum pump receives air at 28 ins. of mercury vacuum and delivers
it against atmospheric pressure. What will be the work per cubic foot of lowpressure
air and per cubic foot of highpressure 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 horsepower 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 horsepower 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 horsepower at sea level and loss in capacity and horsepower 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 horsepower required
to do this and find how much free air could be delivered by same horsepower 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 138.
Prob. 9. A vacuum pump is maintaining a 25in. 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. Singlestage 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 pressurevolume 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 lowpressure
capacity (L. P. Cap.).
The work area of the diagram is ABCD, which may be expressed as
Work QTeA^^JBCLJADL,
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 footpounda, when
the pressures are expressed in pounds per square foot, and volumes in cubic
feet. Using symbolic equivalents
W = 1444T{8uppr}(LPCap.)[(ffp)^'l,l .... (58)
Fig. 27. — OneStage 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 lowpressure 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 footpounds per cubic foot of lowpressure gas is,
W
(L:prcap.)"ta(^"PP^^^'^'' i> ■
Work, in footpounds per cubic foot of hot gas delivered is,
(H. P. Cap. hot)
■144^(sup.pr.)fi; (r,' l).
98 ENGINEERING THERMODYNAMICS
Work, in footpounds 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 highpressure qapacity either hot or cold to the lowpressure
capacity is also as given for the case of no clearance, as will be shown.
In Fig. 27, the highpressure 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 horsepower per unit of gas per minute is independent
of clearance.
8. Singlestage 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 singlestage 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 lowpressure
capacity to displacement is
„ VbV 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 horsepower 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 lowpressure capacity is
(L. P. Cap.) = (F»7«).
But
Va^VaX
w
since the reexpansion DA is exponential and similar to compression as to
value of «, whence J
1
(L. P. Cap.) = (V*Va) = y»7*Bp' ;
1
^D+ClVdRp^;
=D+cDcDRp^ ;
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 singlestage 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+cc72pr)rft/^ll. . . . (67)
= 144^ (sup.pr.)D£'«[ftpVl] (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
51
(sup.pr.)J^Jflp^ll (70)
The indicated horsepower of the singlestage exponential compressor from
(67) is,
T TT P  ^'^  ^_ (suPP 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,
IH.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 horsepower per cubic foot per
minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ).
Horsepower 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 horsepower 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)~8l 229.2 ^ [^' ' ^J • • • U*;
Horsepower 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) sl 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+cc/^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:
Fcn
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 horsepower 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^ ; 2i; (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.
af ; ^150x2xif ; ^^1162; .
.*. ii«=5690 or <i = 17.85.
o =260 sqinches. S =670 ft. per min. .'. I.H.P. = 191,
WORK OF COMPRESSORS 103
Prob. 1. A denBeair 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 airbrake 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 horsepower 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 lowpressure capacities? Take pressures as 2 and 15 atmos
pheres of 26 inches mercury.
Prob 6. A compressor is supplied with 40 horsepower. 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 highpressure 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. TwoStage 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 multistage 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 twostage 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) V2I , .
(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,22j . . (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 lowpressure capacity stated in terms of highpressure 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
bestreceiver 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
Pt
i
O
o
«
O
O
t4
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 singlestage
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 leastcompressor work will be attained is the bestreceiver
pressure.
By definition the bestreceiver 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
jti
F=2::^A7.[(^;) '• l], (85)
sr
Eq. (85) is the general expression for twostage work with perfect inter
cooling at bestreceiver 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 bestreceiver 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 lowpressure gas, from Eq. (86) is,
W
(L.P.Cap.)
7) =288—^ (sup.pr.) [/jpSrl] (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 bestreceiver pressure becomes
«+i
Fft=FJ2p2*.
bubstituting in Eq. (85),
W^2j^P,V.(Rp)^[RMi\ (88)
Introducing the S3miboIs,
TF=288~(sup.pr.)(H.P. Cap. hot)flp^rftpVll, . . (89)
• and
W
(H. P. Cap. hot)
g a + ir .1 I
=288^(sup.pr.)ffp~2r[iJ,27lJ ... (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 H2.14 =.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 bestreceiver
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 twostage 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 receiverpressure 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 bestreceiver pressure for a twostage compressor?
How many more cubic feet may be compressed per minute in two stage than one stage
by the same horsepower?
Prob. 4. A manufacturer sells a compressor to run at bestreceiver 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 supplypressure 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 bestreceiver 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 horsepower will be
required at bestreceiver pressure? Should delivery pressure change to 200 lbs., what
power would be required? To 100 lbs. what power?
Prob. 7. A gascompressing 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 highpressure
capacity cold and horsepower per cubic foot of highpressure 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 horsepower 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 bestreceiver pressure of 40 lbs. per square inch absolute and a
supply pressure of 14 lbs. per square inch absolute, what horsepower will be required
to compress and deliver 1000 cu.ft. of highpressure air per minute at the delivery
pressure for which compressor is designed and what is that delivery pressure?
10. TVostage Compressor^ with Clearance, Perfect Intercooling Expo
nential Compression, Bestreceiver Pressure, Equality of Stages, (Cycle 6).
Work and Capacity in Terms of Pressures and Volumes. The twostage 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^estage
compression processes, each with clearance.
ij
looj SMnoe Ma oqi □! 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 twostage 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(iJp2V2,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 lowpressure
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 bestreceiver 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 bestreceiver pressure will be
(bestrec. pr.) = (P6P,)* (96)
Substitution of this value for Pe in Eq. (94), gives the following expression
forwork of the twostage exponential compressor with bestreceiver pressure,
Tr2^P»(7»F.)[(g)''l], (97)
112 ENGINEERING THEBMODYNAMICS
which may be exprf^ssed in terms of supply pressure, pounds per square inch
lowpressure 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 bestreceiver 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 highpressure 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. TwoStage Compressor, any Receiver Aressure, Exponential Compres
sion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and
Horsepower, 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 lowpressure capacity of the first stage (L. P. Cap.) in cubic feet is
{W—Va)f and, as for the singlestage 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 + ciciftpi"^)=DiJ5.i (102)
For the second stage, the lowpressure capacity (L. P. Cap.2) is
and is equal to
(L.P.CSLp.2)=D2\l + C2C2Rp2')=D2Ef^ .... (103)
Volumetric efliciency of the first stage is given by
E,i = l+ciciRpi' (104)
114 ENGINEERING THERMODYNAMICS
Volumetric efficiency for second stage
E^=l + C2C2Rp2' ■
(105)
1^
It
I!
loo^ ajBncrgjaa gqi iiBajn«8»j ~ 
It may be required lo find the receiver pressure (iocidental to the finding of
work or horsepower) 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.)— ^^ \»"PP''/ ^ J , . . . (107)
i)2l+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 twostage 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
,r144jtj(a„p.pr.)/)A[(°J^) " +(«,?£) " 2J . (110)
The mean effective pressure of the twostage compressor referred to the low
pressure cylinder is found by dividing the work of the entire cycle Eq. (110), by
the displacement of the firststage cylinder, and by 144, to give pounds per
square inch.
m.e.p. referred to firststage cylinder, pounds per square inch is,
«l 81
W
.:,(sup.pr.)E.,[() ■ +(«,gg)2]. . a„)
144Di 81
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 horsepower
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 horsepower per cubic foot of gas supplied per minute is
«i »i
'I.H.P s (sup.pr.)
n(L.P.Cap.) s1 229.2 [VA^
(gitr"H«'^)"4 • »)
WORK OF COMPRESSORS 117
Horsepower per cubic foot of gas delivered and cooled per minute.
al » 1
I.H.P.
n(H. P. Cap. cold)
=j^f'is^'[(l£)^+('^tt:)"'"^ »»)
Horsepower per cubic foot of hot gas delivered per minute
*i
I.H.P. s sup.pr./I)iEvi\ » o L
/T("H7PrCap. hot) "si 229.2 VD2£r2/ '''
*i
mri^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:)~sl 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 = CihD, = 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 horsepower per
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow
ing compressor: Twostage, doubleacting cylinders, 22i and 34^X24 in., running
at 100 R.P.M. Highpressure clearance 6 per cent, lowpressure 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^XTTr 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 horsepower per 1000 cu.ft. of free air per minute is, = 150.
From Eq. (115) horsepower per cubic foot (del.pr.) air, hot = that of (sup.pr.)
air X/^p * \~^~^) or 5.85 times that of (sup.pr.) air.
.'. Horsepower per 1000 cu.ft. of hot (del.pr. air) =150x5.85 =877.
Prob. 1. A twostage doubleacting compressor has volumetric efficiencies as shown
by cards of 98 per cent and 90 per cent for the high and lowpressure 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
horsepower 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 horsepower will be needed to drive a twostage 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 lowpressure 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
horsepower per cubic foot of (sup.pr.) air?
Prob. 4. How many cubic feet of free air can be compressed in twostage compres
sor 18ix30ix24 ins. with 5 per cent clearance in highpressure 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 lowpressure cylinder is known to be 95 per
cent, and of the highpressure cylinder 85 per cent. The cylinder sizes are 15 J X25i X 18
ins. and speed is 120 R.P.M. What horsepower 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 horsepower for the range of discharge pressure, for « = 1.3.
Prob. 7. The volumetric efficiency of the lowpressure cyUnder of a twostage 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
horsepower if the machine runs at 120 R.P.M. and the lowpressure 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 doubleacting 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 electricdriven 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 lowpressure
volumetric efficiency is 90 per cent, s being 1.4?
12. Twostage Compressor with Best Receiver Pressure Exponential
Compression. Capacity, Volumetric Efficiency, Work, Mean EfFective Pres
sures and Horsepower in Terms of Dimensions of Cylinders and Clearances.
For the twostage exponential compressor with or without clearance, and per
fect intercooling, the bestreceiver pressure was found to be (Eq. 84),
(bestrec.pr.) = [(sup.pr.) (del.pr.)]* (120)
WORK OF COMPUESSORS 121
This expression Eq. (120) for bestreceiver pressure makes it possible to
evaluate Rpi and Rp2 as follows:
g,xfor(be8trec.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 (bestree. pr.) = 7,—  — . = ,7 ; .^ , ttt
(bestrec.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+cicii2p2i), (123)
and volumetric efficiency, second stage
£^ = (1+C2C2ftp2.) (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+cici«p2,) /2^ 2.  1 , . . . (126)
which is a direct statement of the work of a twostage 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 bestreceiver 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 firststage 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, onehalf the total work divided by the
displacement of the cylinder in question,
w s r 1^ 1
m.e.p., first stage =2Q8nr'^7Zi;(^^PP'')^»'h^^ (128)
Note that this is onehalf as great as the m.e.p. of the compressor reduced
to first stage, (127),
m.e.p., second stage = ogsZT " IZT ^^PP^^ 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 bestreieeiver 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 threepart equation proper values may be found to fulfill require
ments of bestreceiver 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 bestreceiver 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.) s1 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) sl 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 horsepower 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? Twostage, doubleacting, cylinders 22i x341 X24 ins.,
running at 100 R.P.M. Lowpressure clearance 5 per cent, highpressure clearance
such as to give bestreceiver 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.
Horsepower per cubic foot of (sup.pr.) air per minute is from Eq. (135)
8 sup.pr. i_J
71 114.6"^^' '' "^^^
14. ^^ *
.4 ^ 105^ ^
Therefore, horsepower per 1000 cu.ft. of sup.pr. air = 160.
Horsepower 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 = horsepower 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 bestreceiver 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 lowpressure cyUnder has 3 per cent clearance. What will be horsepower 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 twostage 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 horsepower per cubic foot of (del.pr.) air hot and cold?
Prob. 3. A twostage 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 horsepower at best receiver pressure?
Prob. 4. A manufacturer states that his 201^x32^X24 in. doubleacting 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
bestreceiver pressure what clearance must the compressor have, compression being
adiabatic?
Prob. 5. The cylinder sizes of a twostage compressor are given as 10} X 16^x12
ins., and clearance in each is 5 per cent. What will be the bestreceiver 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 horsepower of the plant for bestreceiver 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 highpressure cylinder respectively. What will be
(del.pr.) for bestreceiver 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 80100 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 bestreceiver pressure? If clearances were equal which would give
bestreceiver 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 lowpressure cylinder to give the same capacity at altitude of 10,000 ft. with
the same clearance and the same (del.pr.) , bestreceiver pressure always being maintained?
13. ThreeStage 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^=lP^F^[(^) ' ij (first Stage)
81
_;^PdVa I W ) * — 1 (second stage)
8
ti
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, lowpressure capacity cubic feet, and ratios of pressures, first stage
(Rpi)y second stage {Rp2) and third stage {Rpz)^ it becomes
Work done by threestage 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^ + {Rp3fr3\ (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 singlestage compressor, compressing form P* to Pc.
One twostage compressor, compressing from Pc to P^.
In this twostage compressor, bestreceiver 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, threestage, bestreceiver pressure no clearance
IF=3^^P.n[@'^'l] (150)
Arranging this equation to read in terms of supply pressure, pounds per square
inch, lowpressure capacity, cubic feet, and ratio of pressures
Work, threestage bestreceiver pressure
Tr=432^(sup.pr.)(L.P. Cap.)(i2pVl), . . . (151)
The work of the compressor is equally divided between the three stages
when bestreceiver 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 threestage compressor with bestreceiver pressure.
Wi = W2 = W^ = U4^(sup,pT.){L. P. Cap.)(/2/"irl). . (152)
WORK OF COMPRESSORS 129
From Eq.(151), may be derived the expressions for work per unit of capacity.
^ork per cubic foot lowpressure 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,Vl). . . . (154)
(H. P. Cap. cold) 51
\gain, from Fig. 32,
^'(?f "'<;) ^''©(^f'
which is to say that, when bestreceiver 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(i2p8rl). . . (156)
(H.P. Cap. hot) — 51
Example 1. Method of calculating Diagram, Fig. 32.
Assumed dcUa,
Pa =Pft =2116 lbs. per sq.ft.
Pc =Pd =best firstreceiver pressure =PjPg^ =4330.
Pe =P/=best secondreceiver 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 ^ .. ».
7d7ftX5 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 bestreceiver 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 horsepower 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
noclearance, threestage 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 threestage 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 threestage 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 threestage compressor compressing from atmosphere to 150 lbs. per
square inch gage from a single and a twostage, 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 threestage, 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
threestage compressors are used, compared to singlestage for no clearance?
Prob. 7. What would be the steam horsepower of a compressor delivering 150
cu.ft. of air per minute at 500 lbs. per square'^inch pressure if compression is threestage,
adiabatic, clearance zero, and mechanical efficiency of compressor 80 per cent?
14. Threestage Compressor with Clearance, Perfect Intercooling Expo
nential Compression (Cycle 8), Bestreceiver Pressures, Equality of Stages.
Work and Capacity in Terms of Pressures and Volumes. The pressure
volume diagrams of the threestage compression is shown in Figs. 33, 34 and
35, on which the clearance volume and displacements, lowpressure capacity
and highpressure 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,) [ (^^Jr  1 j (second stage)
~lPfiyfVj) [(^')'"^'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, lowpressure capacity,
cubic foot and ratios of pressures as above, the worl^ of a threestage 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 twostage compressors the reasoning leading to the determination
of bestreceiver pressure applies equally well with and without clearance, and
since the value of bestreceiver pressures for threestage are found by eon
WORK OF COMPRESSORS 135
sidering the threenstage a combination of one and two stagecompressors, the
same expressions for bestreceiver 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 bestreceiver pressures leads to the same
result as for no clearance £q. (150), except for the volumes,
Work, threestag^ bestreceiver'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, threestage 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
ceiverpressures are best, and these will be respectively identical with E3qs.
(153), (154), and (156), in the foregoing section.
16. Threestage Compressor^ any Receiverpressure Exponential Com
pression. Capacity^ Volumetric Efficiency^ Work, Mean Effective Pressurei
and Horsepower in Terms of Dimensions of Cylinders and Clearances.
Dis displacement of the firststage cylinder, in cubic feet=(F6— Fm);
2)2= displacement of the secondstage cylinder, in cubic feet = (7d— Fa)>
D3» displacement of the thirdstage 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 lowpressure 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+cici/epi')=^i^.i (163)
For the second stage, the lowpressure 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+ciciflpiO (166)
Volumetric efficiency of second stage is
JB.2 = (1 + C2C2flp20 (167)
Volumetric efficiency of third stage is
1
S.3=(l+C3C3flp3'), (168)
The work of the threestage compressor with the assistance of Eq. (163) may
be stated in terms of supply pressure, pounds per square inch, displacement
of firststage 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) V3 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 receiverpressure, 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 secondreceiver pressures.
/■i \ / V (L. P. Cap.i) / .DiEvi /^^^\
(lrec.pr.) = (sup.pr.)^L.P:Cap7)=(^"PP'>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 threestage 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 threestage 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 horsepower 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 threestage 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 horsepower 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 • (>»«»
Horsepower 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) s1 229.2 [\D2Ej '^[DsE.sJ
81
+
«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) sl 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
81
+ (i2.^) ' 3]. . . (187)
140 ENGINEERING THERMODYNAMICS
Example 1. Method of calculating Diagram, Fig. 35.
Assumed data:
■Pa =Pft =2116 lbs. per square foot.
Pc =PdPiPm =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 twostage 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 threestage compressor is compressing air from atmosphere to 140
)s. per sqxiare inch absolute. The lowpressure 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 secondreceiver pressure is
5 lbs. per square inch gage. What horsepower 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+1.223);
.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 lowpressure 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 horsepower 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 horsepower 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 lowpressure 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 threestage compressor for supplying air for a compressedair 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 horsepower 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 lowpressure 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 secondreceiver 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 horsepower 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 firstreceiver pressure of a
throestage compressor at 30 lbs. per square inch absolute, the secondreceiver 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 highpressure 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 lowpressure cylinder 8x10 ins. with 3 per cent
clearance. What horsepower will be required to run it at 100 R.P.M and what would
be the resultant horsepower and capacity if each pressure were halved? (Sup.pr.) = 14.7
lbs. per square inch.
16. Threestage Compressor with Bestreceiver Pressures Exponential
Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure
and Horsepower in Terms of Dimensions of Cylinders and Clearances. It
was found that for the threestage 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+CiCi/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 threestage compressor with bestreceiver 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 firststage cylinder
14^,=3^l(sup.pr.)^.i(«,3rl) (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(i2pVl) (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 BestReceiver Pressures. All the foregoing discussion
of bestreceiver pressures for the threestage compressor can apply only to cases
in which all the conditions are fulfilled necessary to the existence of bestreceiver
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 +C2C2fiy^
D2(l+C2C2fip3«) Ds^l+CiCsRp^
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
multistage comprassor can be made to give bestreceiver 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 bestreceiver 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 horsepower 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) s1 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 singlestage compressors, and in single stage such an independence
was found for work and horsepower per unit of capacity.
Example. If the following threestage compressor be run at bestreceiver pressures
what will be the horsepower and the bestreceiver 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,
TF432x^Xl5xll.2x96.5x(9.35'®^l) =59,000 ftlbs.,
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 threestage adiabatic compressor with bestreceiver pressures?
Prob. 2. The lowpressure cylinder of a threestage 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 bestreceiver 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 dryvacuum 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 horsepower 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 highpressure cylinders respectively be working at bestreceiver
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 horsepower to run the
machine at 100 R.P.M. and also the horsepower 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; Gasengine mixtures, for which s = 1.36.
Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio
for bestreceiver pressure and a pressure ratio of 10?
148 ENGINEERING THERMODYNAMICS
Prob. 9. A compressor, the lowpressure 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 bestreceiver 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 highpressure 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 singlestage compressors, and an economy above
unity for most multistage 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 singlestage adiabatic compression.
Since, however, singlestage 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 multistage compressor is not referred to the same
cycle as a singlestage compressor, and a multistage compressor with other
than bestreceiver pressure is not referred to the same cycle as another operating
with bestreceiver 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 singlestage 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
multistage 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
Multistage
(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 noclearance
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 bestreceiver 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 pressurevolume 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 singlestage adiabatic compressor
with clearance. The work per cycle will vary directly as the mean effective
pressure. Eq. (69.)
(:«..p.)..4i(»P.P.)[l+cc(^)][(^fl]. (21S,
This will have a maximum value when
d(m.e.p.)
d(sup.pr.)
=0,
or when
f^5^y'':f^ri+cc?^(^?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 singlestage 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 multistage 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 lowpressure 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 multistage 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 singlestage formulae
and curves may be applied to this case also.
WORK OF COMPRESSORS
153
That this condition frequently exists with multistage 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 firstreceiver
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 singlestage compressor. In the pipe leading to the
Press.
H.P.Cap Hot (Apparent)
!G
iLD (DeL Pr.)
L^B€apH^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 supplypressure 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 supplypressure line and above the deliverypressure line. In the
combined card of a twostage compressor there would be an overlapping of
the diagrams due to this frictional loss.
Lowpressure Capacity, Referring to the adiabatic compression and expan
sion lines, CD and KL^ Fig. 37, it is seen that the lowpressure capcun^y of the
hypothetical cycle is the volume, LC.
The apparent lowpressure 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 lowpressure capacity.
Volumetric efficiency is defined as being the ratio of lowpressure 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 lowpressure 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 IH.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.
Multistage 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
multistage 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 multistage 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 multistage compressor
commonly differs from assumptions made in the foregoing discussions is in
Fio. 38. — ^Effect of Loss of Intercooling in Twostage Compressors on Receiverpressure 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, twostage (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 fourstage 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, twostage compressors are used, especially for larger
sizes, up to ratios of 34 to 51 (500 to 750 lbs. gage). Some threestage 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 fourstage
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,
gasengine, oilengine or waterpower 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
siue during inlet and delivery is not so great, and hence the mean effective pres
sure is not quite so great. If the compressor is multistage, 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
mvlMistagej 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 fullload lowpres
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 multistage compressor may be illustrated as
in Fig. 40, by considering the twostage 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 bestreceiver 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
\
\
\
\
%
•
\
\
\
\
\
\
\
\
\
\
\
\
\
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 Onestjige 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 cutofiF, 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 Multistage Compressors on Receiverpressure, 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 cutofiF 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 cutoflF 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 cutoflFs may be so
adjusted as to give and maintain bestreceiver pressiu'e.
,
p
•
1 \
)
\
\

\
\
■ ■ \
\
\
\
V
^
\
\
\
\
\
N
\
\
>
\.
\
D
,
.^^
(Su
?. Pr.
\
V
••^
G
A
^
B
A
B
p
D
^
rf
^
1
t
\
\
\
\
\
\
\
V
\
"n
\
\
SE
^1
^^
">.
B
A^
V
F
(A)
D
G
c
\
\
1
\
^
\
\
\
I.
\
\
N,
\
\
F
^^.
^^^
V
^^*>>^
'ite \t
y
(B)
C D
Fig. 41. — Control of Compressor Capacity by A. Open Inlet Valve; B. Closed Inlet Valve;
C. Suction Cutofif; Z). Delayed Suction Closure.
Since the lowpressure 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+ccJBp« ).
(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
; — \"
\ \
\
\^
\
\
.
^^
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 lowpressure 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 reexpansion 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
\
\
V
\
>
\
\
\
r
\
\
A
\
\
\
\
V
^
\
\
\
\
\
j
\
\
<
>
\
— <
\
^N
^
\
\
"n
s
^N,
^
.^
0'
1
'A
r
1
•
A'
1
1
B
^
W—' — I
i.P. Cap Part Load
uirl
*■
h
—CD
"Ui: "
.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 twostage 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 bestreceiver
pressure can be found,
[ici(jep2li)]Di
C2 = i i (229)
(Rj^*l)Rp2sD2
For every value of firststage clearance there is a corresponding clearance of
second ^stage that will give bestreceiver pressure, found by this equation. Sim
ilar reasoning can be applied to three or fourstage 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 horsepower 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 horsepower 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 bestreceiver 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 horsepower 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 horsepower 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 horsepower 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
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•tqv 'ni 'bS
8S
CO CO CO A w '^
6
ENGINEERING THEEIMODYNAMICS
111
'sqv 'QI ''>I>8 'isil JV <''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 horsepower 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 multistage compressors with bestreceiver pressure and perfect inter
»oling, the (m.e.p.) of each cylinder may be found by considering each cylinder
is a singlestage 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 multistage compressors with perfect intercooling and bestreceiver
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.) GuasUl
Fiti. 47. — Mean EffecUve Preseuro of Compreswra, One, Two, and Threeetagea.
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 ThreeBtages.
176 ENGINEERING THERMODYNAMICS
each to be a singlestage 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 lowpressure cylinder is found by taking work per cubic feet of (sup.pr.)
gas and multiplying by volumetric efficiency of lowpressure 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
bestreceiver 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 lowpressure cylinder and is 36.5. From the — ^^ —
33,000
formula, horsepower 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 righthand 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.
81 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 tanaix«K
XiOjunininrrsjiioj 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 ciuTes
laid off to their proper values referred to these axes. Through points as plotted
the clearance lines were drawn. The righthand 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 lefthand 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 lefthand 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 lefthand 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. doubleacting compressor running
200 R.P.M., having 5 per cent clearance and delivering against 45 lbs. per square
inch gage. Also the horsepower. Compression such that 8 = 1.3.
WORK OF COMPRESSOKS
Projecting from the value 1.3 for s on the lefthand 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 righthand 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 ThmcGtage 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 twostage or threestage compression with
bestrreceiver 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 multistage to onestage work. This value
gives per cent power needed for one stage that will be required to compress
the same gas multistage. Saving by multistage 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 twostage 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 '^
81
{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 noclearance 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+cci2p«).
The righthand 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 lefthand 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 lefthand diagram are
drawn lines of piston speed, and on lefthand 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 lowpressure 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
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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 noclearance on&«tage
compressor;
(h) Adiabatically in a twostage, noclearance compressor;
(c) Adiabatically in twostage compressor each cylinder having 5 per cent clearance;
Prob. 2. Air is being compressed in three plants. One is singlestage, the second is
twostage, and the third is threestage 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 horsepower required and cylinder
sizes in each case?
Prob. S* A twostage compressor with 5 per cent clearance n the h gh and 3 per
cent in the lowpressure cylinder is compressing air from 14 lbs. per square inch al>
solute to 125 lbs. per square inch gage. What is the bestreceiver 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 singlestage 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 doubleacting 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 twostage compressor is to be used, the clearance in low pres
sure of which is 3 per cent. What must be the displacement of the lowpressure cylin
der and what will be the horsepower of the compressor?
Prob. 8. The lowpressure 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 twostage 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 lowpressure cylinder is 3 per cent. What
must be the clearance in the highpressure cylinder for the machine to operate at best
receiver pressure? Compression is adiabatic.
Prob. 10. If the clearance in the highpressure 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 bestreceiver pressure and why? Would the original clearance
allow the new bestreceiver 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 highpressure cyUnder can be adjusted. To what value must it be
changed to enable the compressor to carry the bestreceiver pressure for the new
dischai^ pressure? Lowpressure clearance is 5 per cent at all times and com
pression being adiabatic.
Prob. 13. A manufacturer builds his 15jx25ixl8 in. compressors with lowpres
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 horsepower per cubic foot of lowpressure air in each case?
Prob. 14. A threestage compressor has 4 per cent clearance in all the cylinders. The
lowpressure 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 bestreceiver pressure.
Prob. 16. The cylinders of a twostage 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 threestage compressor has a clearance of 5 per cent in each cylinder.
What must be the cylinder ratios for the bestreceiver 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
threestage compressor used for compressing air for airdriven 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 highpressure cylinder of a compressor is 5 per cent,
which allows the compressor to run with the bestreceiver pressure for a discharge of
100 lbs. per square inch absolute when the compressor is at sealevel. 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 bestreceiver pressure?
Prob. 24. How many cubic feet of supplypressure air may be compressed per minute
from 1 to 8 atmospheres absolute by 100 horsepower 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 horsepower 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 lowpressure 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 singlestage 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 threestage 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 horsepower of 1000 H.P. from atmosphere to 150 lbs. per square inch gage; (a)
if compression is isothermal; (6) if compression be singlestage adiabatic; (c) if com
pression be threestage adiabatic?
Prob. 30. A singlestage 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 twostage 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 highpressure fluids on
pistons in cylinders, and this includes all piston steam engines of the recipH
rocating or straighthne 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 highpressure 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 compressedair 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 highpressure 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 noncon
densing steam and most compressedair 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 backpressiu*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, constantpressure admission followed by expansion.
Back strokCj constantpressure 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 constantvolume 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 constantpit^
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 pressurevolume cycles for such engines.
A
■
B
B
j0^
A'
\
AB STEAM ADMITTED
BC •• EXPANDED
CD •• EXHAUSTED
DA •• COMPRESSED
/
/^
\
/
\
y
ACTUAL CARD FROM
CORU6S ENGINE
>
N
e
c
.,/
A
^
D
\
,
C
\
\
\
^
D"""
C
Fig. 52. — Diagram to Indicate Position of Admission, Cutoff, Release, Compression (»n
Engine Indicator Card.
The various points of the stroke at which important events occur,
important in their pressurevolume 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, Cutoff 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:
iB, Admission or steayn line joins the points of admission and cutoff.
BC, Expansion line joins the points of cutoff and release.
CD, Exhaust line joins the points of release and compression if there
is any, or admission if there is not.
DA. 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
cutoff, 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 cutoff, and cui
off = 100 per cent. Draw cards with same cutoff 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 pressurevolume
liagram representative of the whole series of processes and consisting of a
lumber of wellknown phases or single processes. These phases, ignoring all
orts of interferences due to leakage or improper valve action, will consist of
K^nstantpressure and constantvolume 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 pressurevolume 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
VrOoi
^>^
M
Ck>mi >.
i
r'
—s=
•
^
V
Q
•
>>
G<>mp.
V
Fig. 55. — Standard Reference Cycles or Pressurevolume 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
Overexpansion
little
Little
Little
Little
M
N
P
Little
Little
Little
Little
Little
Little
Little
Little
Zero
Little
Complete
Overexpansion
Complete
Complete
Complete
Complete
Q
R
S
T
Zero
Little
Complete
Overexpansion
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 Constantpressure 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, Horsepower 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 backpressure work, or by areas,
net work area, ABCDE^Budmiasion work area ABFG
+expansion work area FBCH
or algebraically,
—backpressure work area GEDH
lF=P»7»P»7»log,§P*7«, •
(230)
Work of cycle in footpounds is
=P.7il+log. ^) P^Vs
=P»7»(llog.^^)P*7<,
=P«Fyilog, ^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.=Pyilog.^)P<, (a)
=P.(l+log.^^)P,
(6)
(c)
.e.p. = pe ( 1 + log, ^ j  Pa
=pJl+Iog,^jPd (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
cutoff 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 footpounds
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 horsepower 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 cutoff, 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 cutoff 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 overexpansion. 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 overeiq>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 overexpan
sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) =^ (bk.pr.),
hence Ry=Rp=^ 7 hlr \ ~ 7' ^^* value of culoff^ 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 Cutoff.
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 sqin. abs.
Vc lo.o
Assumed data for Fig. 59.
Pa ''Pit =90 lbs. per sqin. 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 doubleacting 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 x9in. cylinder and runs at 300 R.P.M., what is the horsepower 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 _ ,„
X7= 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 doubleacting. 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 horsepower and steam consumption will vary as cutoff increases. Take cutoff
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 cutofif 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 horsepower 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 cutoff is such as to give complete expansion. The second engine
exhausts into a 24 in« vacuum and its cutoff is such that complete expansion occurs in
its cylinder. Also the cylinder voliune up to cutoff 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 horsepower 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 horsepower 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 cutoff of 25 per cent and no clearance, the back pressure being
15 lbs. per square inch absolute, what will be the horsepower 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 cutoff 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 cutoff 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
cutoff. 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 cutoff for each engine. Load drops to a point
where the throttle engine requires but 50 lbs. per square inch absolute initial pressure
with the cutoff still fixed at i. What is the original load and water rate, and new
load and waterrate 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 cutoff 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, Horsepower 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(i2vl)
= 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 overexpansion,
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 cutofif.
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 =Pft3Y(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 cutoff,
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 horsepower and the steam conswnption of a 9x12 in.
simple doubleacting engine with no clearance and running at 250 R.P.M. when initial
pressure is 100 lbs. per square inch absolute and cutoff 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 ) 1048.6;
" (6) »  1.3 and (m.e.p.) ^ (^ ~T^) ~ ^° "^"^'
" (c)«l.lll and (m^.p.)^(^ ^^^^^.i„ ) 1047^.
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 '^ Xr ;
 4 4o.o 4
" (b) steam per hr. 42 X^^ Xr;
4o.o 4
a/\_x 1. Af a 13,750 8i
(c) steam per hr. 45.8 X~= X7.
47.5 4
Prob. 1. On starting a locomotive steam is admitted full stroke, while in running
the valve gear is arranged for f cutoff. 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 horsepower with the gear in normal running position and
in the starting position for a speed of 20 miles per hour with 6ft. driving wheels? Con
sider the steam to be originally dry and expanding adiabatically. What would be the
difference in steam per horsepower hour for the two cases and the difference in total
steam? Clearance neglected.
Prob. 2. Consider a boiler horsepower to be 30 lbs. of steam per hour; what must be
the horsepower 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., doubleacting, 200 R.P.M., no clearance, initial
pressure 125 lbs. per square inch absolute, back pressure a vacuum of 26 ins. of
mercury. Cutofif 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 cutoff. A moment later the circuitbreaker
opens and only the friction load remains, requiring a cutoff of only fj. The engine
being twocylinder, doubleacting, 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
circuitbreaker 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 doubleacting steam cylinders each 9x12 ins.
and a fixed cutoff 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 cutoff that
complete expansion occurs for 90 lbs. per square inch absolute initial pressure, and at
atmospheric (15 lbs. absolute) exhaust, what will be the horsepower 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 horsepower and steam used per horsepower per hour for best con
dition and by taking two hghter and three heavier loads, show by a curve how steam
used per horsepower per hour will vary.
Prob. 7. For driving a shop a two cylinder singleacting engine, 6x6 ins., running
at 430 R.P.M., is used. The cutoff is fixed at J and intitial pressure varied to control
speed. Plot a curve between horsepower 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 cutoff 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 horsepower and steam used per horsepower hour for this case.
Prob. 9. For working a mine hoist a twocylinder, doubleacting 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 horsepower 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,
Horsepower, 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+NBCWWDEOOEFJ.
Expressed in terms of diagram points this becomes
1F=
P*(nFa)+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
«yiJ
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; cutoif 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.) = (1X)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(lX)+(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 horsepower, 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 cutoflf 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.) = (Fft7,.)
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.+,)jbkprj (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.)[(lX)+(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 horsepower 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 footpoirnds of work or (Sup.Vol.) ^i lbs. per minute will permit of «^«^^^^
horsepower. But (Sup.Vol.) di lbs. per mintue corresponds to 60 (Sup.Vol.)
i\ lbs. per hour, whence the number of pounds per hour per horsepower 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, cutoff,
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, cutoff, 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+cc^^^^^^] (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 cutoflF 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,
Cutofif 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 horsepower of, and steam used per hour by the
following engine:
(a) cutoff 50 per cent, compression 30 per cent,
(6) complete expansion and compression,
(c) no expansion or compression.
Cylinder, 12xl8in. doubleacting, 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'(lX)+(Z+c)log,^j,
= 10o[.5+(.6+.07)log,±^]15[(l.3)(.3+.07) log, — J^]
=8620=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)(Xc) *^ 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.) =10015=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 horsepower 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 cutoff 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 horsepower 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 cutoff 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 horsepower and
water rate?
Note: 8 for 90 lbs. .24, 8 for 15 lbs. =.039.
Prob. 6. By finding the horsepower and water rate of a 12x18 in. doubleacting
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 cutoff 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
horsepower, 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 cutoff 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 cutoff 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., cutoff at § and 30 per cent
compression, what will be the horsepower and the water rate? Should the steam
pressure be doubled what would be the horsepower 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 cutoff 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 horsepower and water rate if all other factors stay constant? What would the new
cutoff have to be to keep the engine running at the same horsepower and what
would be the water rate with this cutoff?
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 cutoff of , while imder light load
the cutoff 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
(FtfF.) = Z)(lX),
Ve=D(X+c).
/VA_Z+c
\VJ~l+c
/V,\_X±c
\VJ~ c •
W=14AD
(M:.p..,[(IX,+t'[(^±')'.]]
(281)
Eq. (281) pves the work in footpounds for D cubic feet of displacement in a
cylinder having any clearance c, cutoff 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.)
(lX)+^'[(^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.) = 76F«.
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 cutoff, 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, cutoff, 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, cutoff, 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+cc(^)'] (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+ccli — ^) 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./
VcVa D
Whence
^'[(fe^)''] <^>
Eq. (289) gives the cutoff 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 honsepower of and steam used per hour by the
following engine: 12xl8in. doubleacting, 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) cutoflf=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+07.07xL7^ =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 cutoff 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 horsepower of each engine and the cubic feet of stuff supplied
per horsepower hour ?
Prob. 2. A crankandflywheel twocylinder, doubleacting, pumping engine is
supplied with dry steam and the expansion is such that it remains dry until exhaust.
The cylinder size is 24x36 ins., cutoff 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 horsepower and
water rate? What would be the horsepower and water rate of a fullstroke 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 horsepower 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, cutoff f, compression 10 percent, and speed 125 R.P.M.
Prob. 4. What will be the total steam used per hour by a 20x28in. doubleacting
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.cutoff
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 } cutoff and one of 25 lbs. at i cutroff. 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 28in.
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, cutoff 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 horsepower will be needed? What, with complete compression?
Prob. 8. A manufacturer rates his 44 x42in. doubleacting engine with a speed
of 100 R.P.M. at 1000 H.P. when running noncondensing, initial pressure 70 lbs.
per square inch gage and cutoff 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 directacting noncondensing
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 crankandflywheel pump with 6 per cent clearance to give the same
horsepower on the same steam and exhaust pressures with a cutoff 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 Multipleezpansion 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 multipleexpansion 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 pressurevolimie 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 threecylinder compound. Similarly, tripleexpansion 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 pressurevoliune 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 lowpres
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 highpressure cylinder discharging into it will be a constantpressure line,
as will also the admission line of the lowpressure cylinder. When, however,
the receiver is of finite size highpressure 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 lowpressure 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 highpressure exhaust and low
pressure admission with, of course, a constant average value. The greatest
possible change of pressure during highpressure exhaust and lowpressure 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 highand lowpressure 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 noreceiver 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 multipleexpansion 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 highpressure exhaust and out of it, by lowpressure 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 highpressure 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 highpressure 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.
Multipleexpansion 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 cutoflf
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 lowpressure 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 pressiues 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 mjiiltipleexpansion 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 lowpressure steam will be cooler than highpressure
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 multipleexpansion
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 compressedair leakage in a siinple engine is a direct loss, whereas in a
compjound highpressure cylinder leakage has at least a chance to do some
work in the lowpressure 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 multipleexpan
sion engines are capable of greater economy than simple. The thermal analysis
must also consider the influence of the reheating receiver, the steamjacketed
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 multipleexpansion engines some indi
cator cards are given for a few typical cases in Figs. 63 to 66, together
with the combined diagrams of pressurevolume 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 noreceiver
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 lowpressure 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 cutoff 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 lowpressure
cylinder.
In set A the highpressure 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 cutoff 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 highpressure cylinder occurred and the admission
valve of the lowpressure cylinder opened, the steam dropping in pressure until
the pressure in both high and lowpressure 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 Noreceivei
Compomid Steam Engine.
WORK OF PISTON ENGINES 231
in the passages between the high and the lowpressure cylinders. At D the
highpressure exhaust valve closed and compression of the steam trapped in
the highpressure cylinder occurred to point A, thus closing the cycle. From
point D' in the lowpressure cylinder, which corresponds to D in the high
pressure, no more steam was admitted to the lowpressure 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 highpres
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 cutofiF is made earlier. In the lowpressure 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 highpressure
cylinder being equal to admission pressure and above it in the lowpressure
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 crosssection
paper and then combined. Cards taken from the different cylinders of a
multipleexpansion 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 highpressure
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 zeropressure 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
highpressure diagram. A number of points A'B^C, etc., were then chosen
on the lowpressure card, and the corresponding points a'6'c', etc., plotted by
making the distances of a\b\ etc., from the zerovolume 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 lowpressure card was formed. The highpressure card was then redrawn
Fig. 64. — Diagram to Show Method of Combining the High and Lowpressure Cylinder
Indicator Cards of the Noreceiver 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 zerovolume 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 crosssection paper for
ease in combining, and with the zerovolume axis added. On the highpressure
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 lowpressure 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 lowpressure card, admission occurred at A' and continued to B' at
constant volume, the piston being on dead center as from A to B in highpressure
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 highpres
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 tripleexpansion 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 crosssection paper. On
the highpressure 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 cutoff. The pressure rise is slight, however, due to the size of the receiver,
which is large compared to the cylinder. At twothirds 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 Tripleexpansion 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 cutoff occurred and steam expanded to D', where release
took place, and the steam was exhausted. As in the case of the highpressure
cylinder the back pressure was rising for twothirds 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 twothirds of the stroke
the lowpressure 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 lowpressure 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 lowpressure 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 cutoff
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 singlevalve engine, 12x20x12 ins., with 33 per cent clearance in
highpressure cylinder and 9 per cent in low.
No. 3. From a fourvalve Corliss engine 22x44x60 ins., with 2 per cent clearance
in the highpressure cylinder and 6 per cent in low.
No. 4. From a single valve engine 18 X30xl6 ins., with 30 per cent clearance in the
highpressure 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 doublevalve engine 14x28x24 ins., with 3.5 per cent clearance in
the highpressure cylinder and 6.5 per cent in the low.
Prob. 2. In Fig. 68 are shown four sets of indicator cards from tripleexpansion
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 steamship, 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 steamship 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 tripleexpansion 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 Twocylinder 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
101
No.l
No. 2
lOi
No. 8
ICM
rlOO
60
80
plSO
flO
"0
p40
20
0
i10
100
00
20
100
60
20
40
20
0
10
fuo
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 Tripleexpansion 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
1201
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 horsepower equiv
alence between the simple and compound, if, in the latter case the steam
WORK OF PISTON ENGINES
239
admitted up to cutoff may be assimied to be acting only in the lowpressure
cylinder, that is, ignoring the highpressure 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 Tripleexpansion Engines.
cutoff 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 highpressure cylinder and that below in the low. This is merely equiv
alent to saying that a volume of steam AB Is admitted to the highpressure
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 lowpressure 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
\
n
*x
'^
^
HP.
Displ
icezn
mt
V
[V
n
^■" — '
>
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 Onecylinder Simple and in
Twocylinder 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 cutoff 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 , ^
cutoff in the highpressure cylinder is 100X7T7;;,and in the lowpressure 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 singlestage 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 overexpansion there. Thus, in
Fig. 72, if ABCEFGDA represent a
combined compound diagram for
the case of complete expansion in
the highpressure cylinder continued
in the low without interruption but
incomplete there, DC represents the
volume in the lowpressure cylinder
at catoff^ and at the same time the
total highpressure cylinder volume.
If now, the lowpressure cutoff
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 Lowpressure
m the lowpressure 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
Cutoff for No Receiver Loss.
ment could not, of course, change the highpressure total volume DC, so
that at release in the highpressure
cylinder the pressure would drop
abruptly to such a value in the
receiver as corresponds to filling
the low pressure up to its cutoff,
and work be lost equal to area
CCW^.
. A shortening of the lowpressure
cutoff will have an equally bad
effect by introducing negative work
as indicated in Fig. 74, in which
the L.P. cutoff 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 lowpressure cutoff 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. Cutoff Introducing a Receiver Loss Due to 01 COUrse, modlfiy the distribution
Incomplete Highpressure Cylinder Expansion, of work between high and low, but
A
B
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^///hi
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242
ENGINEERING THERMODYNAMICS
these saxae effects might also have resulted from changes of highpressure
cutoff 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
ciUoffs 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 highpressure cylinder
during its exhaust, and at other
times with the lowpressure
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
highpressure exhaust two sepa
rate volimies of fluid come
together, the contents of both
the highpressure cylinder and the
receiver, and this double volume
is compressed by the H.P. piston
into the receiver, in which case the
highpressure exhaust would take
place with rising pressure. Follow
ing this will come lowpressure admission, during which the volume of fluid in the
receiver expands into the lowpressure cylinder up to its cutoff, and if the same
volume is thus taken out of the receiver as entered it previously, lowpressure
admission will take place with falling pressure, the line representing it exactly
coinciding with that for the highpressure 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
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:'
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m
•
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c
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^
■Dj
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E
—
G
F
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Fig. 74. — Diagram to Show Effect of Shortening L.P.
Cutoff, Introducing a Receiver Loss Due to
• Overexpansion in the Highpressure 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 highpressue 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 lowpres
sure cylinder finished taking fluid and which is, therefore, at the same pressure
d
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— —
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ne
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Fig. 75. — Diagram to Illustrate Variable Receiver Pressure for the Case of Independent
Highpressure Exhaust and Lowpressure Admission with Zero Clearance.
as the L.P. cylinder cutoff 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
highpressiire 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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— J^^
\
L
t t^^^m ^iM
~4
L
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1
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me_.
F E
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1
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H
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s
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Fia. 76. — Diagram to Illustrate Variable Receiver Pre.ssiire for the Case of Independent Higb
pressure Exhaust and Lowpressure 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
twincylinder singlecrosshead 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 lowpressure cylinder being
greater than the high, the steam at the moment of release suffers a drop in
pressure in filling the lowpressure 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, highpressure exhaust and lowpres
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 highpressure cutoff and
release, the volume of the expanding fluid is that of the highpressure displace
ment up to that point of the stroke, together with the highpressure clearance.
After the period of commimication the volume of the expanding fluid is that
of the low pressiue cylinder up to that point of the stroke, together with the low
pressure clearance, plus the highpressure displacement not yet swept out,
and the highpressure 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 lowpressure 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 highpressure 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 lowpres
sure to highpressure displacements. If at any point of the stroke the volume
remaining in the highpressure cylinder be x, and the high and lowpressiue
displacements be respectively Di, and D2, then (Dx—x) is the volume swept
out by highpressiu'e piston, x the volume remaining in it, and ^^(Di— x)
Di
the volume swept in by the lowpressure piston. Then the total volume
still in the two cylinders is, for a point between C and E,
7=x+^(Dix)=Z)2a:(~?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 Noreceiver, and Exact
Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance.
the pressurevolume representation of the entire cycle of the highpressure
cylinder.
In Fig. 77C is shown the corresponding pressurevolume 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, 774, 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 aheady discussed. In this diagram, the area ABC FA
represents the work of the highpressure cylinder, and DEIHD, the work of
the lo wpressure cylinder. Together, they equal the work of the enclosing
figure ABEIHA, and hence the work of the lowpressure 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 lowpressure 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 highpressure
displacement and DJ the lowpressure 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 lowpressure 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 highpressure
248
ENGINEERING THERMODYNAMICS
cylinder, and OQ and QF, clearance and displacement of the lowpressure cylinder.
It is necessary to know highpressure cutoflf, ;z^l highpressure compression,
Zli.
: and lowpressure 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 lowpressure clearance to the pressure
Fig. 78. — Diagram to Show Volume of Steam in the Cylinders of the Noreceiver Compound
Engine at any Point of th'e Stroke for the Case of Noclearance 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 cutofif 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 cutofif in the lowpressure cylinder.
WORK OF PISTON ENGINES
249
After the supply from the highpressure cylinder has been cut off at E', the
expansion is that of the volume in the lowpressure 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 Noreceiver 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 tripleexpansion 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 midstroke the lowpressure admission
opens while highpressure 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
p
P
a
I
"l
I
\ Cycles VII & VIII
\
Cycle* V & VI
Cycle Y PV«C Cycle VI PV«C
e
ACycle VII PVC Cycle VIII PV«.C
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Cycles IX & X
Cycle IX PV«C Cycle X PV««C
f
\ Cycles XIII 8c XIV
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\Cycle XIII PV«C Cycle XIV PV».C
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CyclM XI <i XII
Cycte XI PV=C Cycle XII PV»=C
j
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Cycles XV & XVI
Cycle XV PV=C Cycle XVI PV»aC
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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 constantreceiver 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 constantreceiver 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 highpressure exhaust and lowpressure 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 constantsupply 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
constantclearance 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 lowpressure 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) overexpansion 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. — Tripleexpansion 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,
Highpressure cylinder work
Wh^F^\{\+\o^, J'^^PtfFtf,
(296)
Lowpressure 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 lowpressure 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 cutoff, and at full
stroke in cubic feet. Dividing this by the volume of the lowpressure cylinder
Vg gives the mean effective pressure referred to the lowpressure cylinder,
from which the horsepower 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 horsepower,
(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, cutoff, 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 highpressure cylinder = y*^ 5
144'
144'
Fe'
jBl= ratio of expansion in lowpressure cylinder =
V,
J2v= ratio of expansion for whole expansion ==:fd^;
Vb
Z)h= displacement of highpressure cylinder =Fd = Fc;
Z)l = displacement of lowpressure cylinder = F/= Vg\
Re = cylinder ratio = 7^ =t/>
l^H Vd
Zh = fraction of displacement completed up to cutoff in highpressure
cylinder, so that ZhDh = F^ = ^;
Zl= fraction of displacement completed up to cutoflf in lowpressure
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 lowpressure 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.)
Highpressure cylinder release pressure =Pc=^»Tr>
1^
Rh
= (in.pr.)Zj5r
y
Lowpressure 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 cutoff or lowpressure 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
Lowpressure cylinder work ^^^^^ _^^^^
Highpressure 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 backpressure
ratio and cylinder ratio, or, in terms of (b) cutoff 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 lowpressure cylinder and ratio of L.P.
admission volume to H.P. displacement, and cylinder ratio, or (b) cutoff in
high and lowpressure 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 cutoffs.
(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
Lowpressure admission volume _Ve_ _j p _ (rel.pr.)g .
Highpressure 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 lowpressure cutoff volume
as fixed by x in terms of highpressure 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 cutoffs, 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 cutoff 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 highpressure release pressure is not affected by any change in the low
pressure cutofif, and hence Eq. (309) gives the value of highpressure release,
pressure for the case. Lowpressure release pressure Eq. (310) may be expressed
for the case of equal cutofF,
(tel.pr.);= fepirff = ^*[(ii^P') (bk.pr.)]*
\bk.pr./
(a)
=^[(inpr.)(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 cutoff in both the high
and lowpressure cylinders, and the volume of highpressure cylinder with respect
to the lowpressure volume, so that
p
ytf—Vf^y or symbolically,
«i(iS:) <"
(317)
264
ENGINEERING THERMODYNAMICS
Similarly the lowpressure cylinder cutofif volume must equal the high
pressure displacement volume or Dh^ZiJ)l,
Z),
" Di. Re
(a)
(b)\
(318)
indicating that lowpressiire cutoff 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 cutoflF.
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.) "(bkpr.)
(321)
Re (bk.pr.)
Frran Eq. (311),
High pressure cylinder work _
Hlog.«/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
Eks. (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 / ^Ppr \
/ \ f t i. T ■D\ ^ r \ \bk.pr./ ,. xlogrfir
(m.e.p.) (ref.toL.POi44^^=(m.prOyj^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 sqin. abs. Fa « F» = Fm = cu.ft.
PnPd^Pe 50 lbs. per sqin. abs. VcVd^ .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 XTTi^ = .36 cu.ft.
V 8
PfPg 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 x24in. 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 horsepower for  cutoff in H.P.
cylinder, (b) poimds of steam per I. H.P. hour, (c) terminal pressures, (d) L.P. cutoflf
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 = (Jy=2.25, «x.=2.25,
since vol. of L.P. cyl. at cutoff must be equal to the entire volume of the, high for
continuous expansion, hence
(m.e.p.)=150Xr— rT3X(2+.69+.81l)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
= rrr « 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+.81Jix2^ ^^^
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 cutoff in each cyhnder, if cylinder
ratio is 3 and stroke is 18 ins.? Engine is doubleacting 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 cutoff 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 x30in. 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. cutoff of }, i, }, }, and that value which
will give complete expansion in highpressure cylinder? Lowpressure cutoff 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 24x30in. infinite receiver engine is to be operated so as
to give complete expansion in both cylinders. What will be the cutoff to acoomplish
this and what horsepower 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. cutoff (a)i, (6) J, (c)=l. L.P. cutoff (a) I, (6) A, (c)=f.
Prob. 7. For the following conditions find the horsepower, steam used per hour,
receiver pressure and release pressures. Engine, 10 and 15x24in. 150 R.P.M.,
125 lbs. per square inch gage initial pressure, 2 lbs. per square inch absolute, back
pressure, i cutoff in highpressure cylinder, A cutin lowpressure 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 lowpressiue cylinder diameter for i cutoff in the highpressure cylinder and
J cutoff in the lowpressure cylinder?
Prob. 9. Find by trial the cutoffs 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 cutoff in the high pressure, = v^ and Zl, lowpressure cutoff
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()
+^'(^)"Ci^)]*'''"'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 lowpressure 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 lowpressure displacement is
(Sup.Vol.)=Z>i^ (333)
The consumption of fluid, cubic feet per hoiu* per indicated horsepower 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 highpressure cylinder is
(rel.pr.)^=(in.pr.)Z^*, (335)
and for the lowpressure 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
cutoff in the high, and equals ^.
The distrubition of work between the high and lowpressiu*e cylinders may
be found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by
means of Eq. (330)
l sZn^^ \ (Zh\
L \RcZl/ \ sl 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 lowpressurt
cutoffS; 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 overexpansion. The conditions for equality
of work for this case will be treated after deriving work and mean effective
pressure for it.
Complete expansion, without overexpansion, 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^(lZ..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 lowpressure cylinder with a cutofif 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 (lZir'0 + (^y" W
which reduces to
TF = 144(in.pr.)Dx,
Zh s
Rcsl
!'(in
(340)
For this case of complete expansion in both cylinders, the ratio of high to
lowpressure work is given by division and cancellation,
Wl
1Zb'
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 horsepower, (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 \ cutofiF
in highpressure cylinder and continuous expansion in lowpressure 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(^)'
HiT)'irf^)] a>)
which, on substituting values from above, gives for (m.e.p.) 63 lbs. per sq. inch.
Hence, the indicated horsepower "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
Gii)'
.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 i21.85 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 horsepower 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, backpressure 10 lbs. per
square inch absolute, steam continually dry. Cutoff at first } in highpressure 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 cutoffs 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 lowpressure 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 18inch vacuum. The
load which it is to carry is 150 horsepower 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 cutoffs 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. Lowpres
sure cutoff 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 cutoff is i in
highpressure cylinder and i in lowpressure cylinder. Compare the power and steam
consumption under this condition with corresponding values for wet steam imder
same conditions of pressure and cutoff 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 cutoffs are at i.
What will be the receiver pressure, highpressure release pressure, and lowpressure
release pressure? What will be the new values of each if (a) highpressure cutoff is
made f, (6) i, without change of an3rthing else, (c) low pressure cutoff is made i,
(d) f, without change of anything else? Infinite receiver, s = 1.3.
Prob. 7. In the above problem for i cutoff 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 18in. and 24x30in. engine running at i cutoff 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,
Highpressure cylinder work is
Fi,=P^F.(l+lo&pjP„01oge(^^) (345)
Lowpressure 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 lowpressure cylinder displacement, Vg, the result will be
the mean eflfective pressure referred to the lowpressure 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 lowpressure cylinder;
(rec.pr.) 2= minimum receiver pressiu^ Pe, which is the terminal admission
pressure for the lowpressure cylinder and that at which
expansion begins there;
<(
_ receiver volume _2. — ^ a Q ^ ^bV _ y
^""highpressure 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.)
= (inpr .)^ ( 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.)5p
ib)
• ■
(357)
Fluctuation in receiver] pressure =(P»— Pe)=ft7r.
.* (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 constantreceiver 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 receiverpressure 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 cutoff
in the highpressure 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 cutoff to remain the same. As,
therefore, equal division was obtainable in the case of infinite receiver with
WORK OF PISTON ENGINES 279
equal cutoffs 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 lowpressure admission period or cutoff fraction lowers
the receiver pressure and reduces the lowpressiu*e work, it follows that with
the finite receiver the lowpressure cutoff 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, yP.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 highpressure cylinder cutoff 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+Mij,
(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 LowPressure Cutoffs for Equal Work in
the Two Cylinders of a Finitereceiver Compound Engine with Zero Clearance
and Logarithmic Law.
As only one constant value of lowpressiue ratio of expansion or cutoff
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 cutofif by fixing the
ratio between them. As the relation between these cutoffs 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 cutoff in both high and lowpressure cylinders,
and the volume of the high with respect to the lowpressure volume. Accordingly,
Vt=Ve
Pe
^"^^M w
/2j5r =
Re Vbk.pr./ Re
{b)
. . . . (365)
Also for the lowpressure cylinder the cutoff volume must equal the whole
lighpressure 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 / 'Ppr \
= 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=(inpr.)D"~0>'^Pr)J'2c=(rec.pr.)2; .... (374)
(rel.pr.)x,=(in.pr.)pn = (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 cutoffs 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 highpressure cylinder ratio of expansion is
and the corresponding value for the lowpressure 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 lowpressure
cutoffs 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.
7il cu.ft.
01.2cu.ft.
7r .8cu.ft.
To find point C:
To obtain point E:
Pe "PbTT « — 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 horsepower, (6) steam used per hour^ (c) the release
and receiver pressures for a 12 and 18 x24in. engine with receiver twice as large as
the lowpressure 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 cutofiFs \ in
highpressure 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^epower  ^^^^^ 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
pre88ure150^— — +~;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 x24in. engine has a receiver equal to 5 times the volume of
the highpressure 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 cuto£^
are iV and \ in high and lowpressure cyUnders respectively. What is the horsepower
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, cutoffs  and I, in high and low, initial pres
sure is 100 lbs. per square inch absolute, and receiver 2 times lowpressure 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, cutoffs I
and {, receiver volume 4 times highpressure cylinder, strokes equal.
Prob. 4. For the same conditions as above, what lowpressure cutoff would give
equal work?
Prob. 6. What will be the most economical load for a 16 and 24x30in. 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 cutoffs 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 x20in. engine, with a receiver volume equal to 5 times
the H.P. cylinder and running on { cutoff on the highpressure cylinder and I cutoff
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 cutoff 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 highpressure 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 Lowpressure 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^(iz.y^^y[i.(^)']l
but
(rec.pr.)i=P«=P»(prl [—qJ .
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 lowpressure 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~*[lz/"*]144(bk.pr.)Z)i„ . (384)
and the total work is, by adding (Ws) and (Wl),
— <»''!^^('+ff)(«k+')"[cfir"
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 lowpressure volume, to the highpressure
volume at cutofif,
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 cutoflf Zffy the assumed initial and back pressures, and the assumed
ratio, y, of receiver volume to highpressure displacement.
WORK OF PISTON ENGINES
291
Example. Find (a) the horsepower, (b) steam used per hour, (c) the release
and receiver pressures of a 12^ and 18x24in. engine, with a receiver twice as large as
the lowpressure 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 cutoffs 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
81 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 horsepower =
^3/50Zflr
m.e.p. Re
13,750 .5 ^^^ ^,
575 ^2:25 =^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., highpressure cutoff J, low pressure cutoff i, and receiver volume 10
cu.ft., what horsepower 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 14x22x36in. 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 cutoffs 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 horsepower 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 highpressure cutoff if the
receiver is to be 3 times the highpressure 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 highpressure cutoff 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
HighPressure Exhaust and LowPressure 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 lowpressure expansion is necessarily a direct
continuation of the high pressure without any possible break. There can be no
overexpansion in the high nor can expansion there be incomplete, as there is,
properly speaking, no back pressure with which to compare the highpressure
cylinder terminal pressure. There may, however, be over and incomplete
expansion in the lowpressure cylinder. It might appear that the highpres
sure cylinder negative work was equal to the lowpressure 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 highpressure cylinder, ABD'EF the whole work, whence
CD'EF is the part done in the lowpressure cylinder. There is, of course,
) lowpressure cutoff 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 Noreceiver 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 cutoff in the low, the distribution of work between high and low will
lepend solely on the cylinder ratio and highpressure cutoff, for, the earlier the
lighpressure cutoff, and the larger the highpressure 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.)ZirDHl+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. ftc144(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
\Rcl Rcl) '
and
1 R
lo& ^+l0g. iJc = loge, ^ = loge Ry,
SO that
Tr144(m.pr.)^ifZ)ir(nIog.^)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 lowpressure work is repre
sented truly to scale by C'D'EFj Fig. 87C, the mean eflfective pressure of the
lowpressure 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 lowpressure cylinder just as
truly. Therefore,
L.P. cylinder work = ( y ^y P. j 7,
As the highpressure 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,i2c144(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 lowpressure cylinder volume and the hi^
pressure admission volume in turn,
(m.e.p. ref . to L.P.) = (in.pr.)^{ 1+loge yj" (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\ Xtt («)
^^ ^ ^ (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
highpressure cutoff for a given cylinder ratio and any change of load to be met
by a change of initial pressure or of highpressure cutoff will necessarily unbalance
the work. Equating the highpressure and lowpressure work expressions,
Eqs. (404) and (405),
l+loge^j — P — =
loge/2c =
Rc
Rcl
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 Noreceiver
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
LJ^
_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
NoReceiver 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.
FiF.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 lowpressure 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 Noreceiver 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 lowpressure 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 highpressure card.
Example 2. Find (a) the horsepower, 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 cutoff in
the highpressure 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 highpressure cutoff 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" = highpressure 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 cutoff to a limited
degree. For a speed of 200 R.P.M. and a constant cutoff 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 cutoff to full stroke.
Prob. 2. Show how the steam used per horsepower hour will vary in above
problem.
Note: 8 for 175 lbs. =.416.
Prob 3. With the cutoff at f , what should the initial pressure be to give equal
work distribution?
Prob. 4. With full boiler pressure and f cutoff what would be tenninal pressure
in the lowpressure 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 horsepower,
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 horsepower 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., horsepower 400. Neglect the receiver volume.
14. Compound Engine without Receiver, Exponential Law, Cycle X.
General Relations between Dimensions and Work wlien Highpressure
Exhaust and Lowpressure 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^jan.
8
but
(rel.pr.)if = (in.pr.)Z»',
so that
W^H = 144(in.pr.)
DjtZh
s1
^»^«rM~)
, . . . (414)
^ _PcXKL\ /XL\> 1 _ (rel.pr.)A, D^iJcf, / 1 V"'!
302
whence
ENQDIEERING THERMODYNAMICS
Wl = 144(in.pr .) ^^Rc
1 L ''
Re
Rcl
144(bk.pr.)Di;.
(415)
It is to be expected that the sum of high and lowpressure 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.)
sZ^^Z^i^^^^
+Z,
•44ff
144(bk.pr.)Di,
= 144(in.pr.)
DaZa
81
«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 lowpressure
cylinders may be obtfuned by equating Eqs. (414) and (415),
sZ«— iZ«''
(i)'
Rcl
= Za—'Rr
Rcl
_{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 lowpressure cylinder (it is
always complete in high, for this cycle), leads to this following relations:
(!§.)=«'•.
hence
144(bk.pr.)I>i, = 144(in.pr.)5V>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 Noreceiver Compound Engine
without Clearance, when the Expansion is not Logarithmic.
Example 1. Find (a) the horsepower, and (6) the steam used per hour for a
12 and 18 x24in. 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., cutoff in highpressure cylinder is J, there being no receiver and steam
having expansion, such that $ = 1.3.
From Eq. (417)
, , (in.pr.)Z^r /I \— 11 ... .
(^.e.p.) ._ _ _^,> ( _) j ^(bk.pr.),
which, on substituting the above values, becomes
T>^2l5[^^(i^) ]10C3.31bs.per8q.in.
hence the indicated horsepower =243.
WORK OF PISTON ENGINES 305
(6) From Eq. (419) the steam used per hour in cu.ft. per horsepower 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 highpressure cutoff 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
30x42ins. and runs at 200 R.P.M. What will be the horsepower developed at
this speed if the initial pressure is 175 lbs. per square inch gage, back pressure
atmosphere, highpressure cutoff }, 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 horsepower 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 cutoff in a 10 and 15 x20in. 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 horsepower 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 cutoff
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 20x24in. 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, cutoff = 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) EMNH,
(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.^')(PaP/)7aP,(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
+PbVtPrVfPaVa+PfVjPaVa+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)PaVaPoV,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. ^pLV. log. ^'PtF* log. ^ ^
•'» Vh Vh Vf Vi
+2(P»F»+Pt7t)Pay.pX* V, PXrViP,V,
V h V h
■ . (428)
308
ENGINEERING THERMODYNAMICS
Introducing the usual symbols in Eqs. (427) and (428) and in additicri
the following:
Z = cutoff as fraction of stroke, so that ZhDh is the displacement volume
up to cutoff.
c= clearance volume divided by displacement, so that Ch Db is the clearance
Volume and (Zh+Ch)Dh is the volume in the highpressure cylinder
at cutoff.
X=that fraction of the stroke during which compression is taking place so
that {Xh\Ch)Dh is the volume in the highpressure cylinder when
compression begins.
Applying the general symbols to Eq. (427),
l+cg\
W^IU
1 1^
[m.pr.)(Z^HCiir)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 (^^)
H2(in.pr.) {Zh +ch)Dh  (in.pr.)c^Dir  (rec.pr.) (1 '\Ch)Dh
42(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, cutoff 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 lowpressure 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,HCL)Z)L(rel.prOi,f ^ Ui^l (bk.pr.) (1 HC£.)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 (iutoffs. 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 lowpressure 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,4CL) (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 lowpressure 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 cutoff 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 cutoff 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
cutoflF 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. (JM^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 lowpressure 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 lowpressure 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^
«cj4.;^^^^^^r^p*,
(447)
which is the required relation between cylinder sizes, clearances and ratio of
pressures, which, together with cutoffs 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,(Rpil).
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 highpressure 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. cutoff and clearance. If the cylinder ratio and clearance are fixed, the
required cutoflf 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.)
— CltCl
Ri
= (bk.pr .)ii;c. . • (455)
Cutoflf in the highpressure cylinder is determined by clearance, initial
pressure and receiver pressure, which in turn depends on lowpressure cutoflf
and clearance Eq.(452), or may be reduced to cylinder ratio and lowpressure
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 lowpressure cutoffs are functions of cylinder and clearance
dimensions, and of Rpy the rato of initial and back pressures, the work of high
and lowpressure 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/2cc,.(/2cl)]. . . . (458)
WORK OF PISTON ENGINES 315
Hence^ total work by addition is
TF=144Z)i7(in.pr.)i^
/•+'•> "*(l)''(li)
+(l+Cjr,)logfiBcCi(i2el) . . . (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,,)logeiecc,,(/?cl). . (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 noclearance case to equality of high and lowpressure 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 highpressure cylinder may now be evaluated.
War 144D£r(in.pr.)  Z ^1 +log. (^^ j  <
 144Dtf[(in.pr.) (bk.pr.)]* { (X+c) log. (^^^\+lX ■ . . (466)
WORK OF PISTON ENGINES 317
The lowpressure cylinder work may be similarly stated,
Trz,«1441)4(in.prO(bk.prO]*z[l+log.(^^)^^ (467)
144i)^(bk.pr.) (X+c) log. ^^)+lZ .,
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 lowpressure 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.
(Va7«)=(F„F,). Fit .4cu.ft.
The above may be expressed in initial pressure, etc., and in terms of cutoff, etc.,
but as the relation of the lettered points to these terms is shown on the diagram values
for cutoff, 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=Pb77^ — — — 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:
(FAF„) = (FmF.), or F*F«+FnFe = .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 horsepower, (6) steam used per hour, and (c) receiver
and release pressures for a 12 and 18x24in. engine with infinite receiver, 6 per
cent clearance in highpressure cylinder, and 4 per cent clearance in lowpressure 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., cutofif in highpressure cylinder is }, low
pressure cutoff is such as to give complete H.P. expansion, and compression is 15 per
cent in high and complete in low.
(a) For complete highpressure expansion the receiver pressure must be equal to the
highpressure release, and to maintain the receiver pressure constant the lowpressure
cylinder must take as Liuch steam per stroke as the highpressure discharges. With
initial pressure and cutoff as given, the release pre3&ure for the highpressure 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
highpressure cylinder there is exhausted each stroke 85 per cent of its volume. Also
since compression in lowpressure cylinder is complete, the lowpressure clearance is
full of steam at the receiver pressure at the beginning of the stroke. Hence the
lowpressure displacement up to cutoff must equal MDa or L.P. cutoff =.85Dj,
divided by cylinder ratio, or .85 42.26 =.378. As compression is complete, the
per cent compression may be found from the relation CLX(rec.pr,)=(cLf^L)(bk.pr),
or .04x79.3 = (.04fXi.)10, or Zi, = .28.
From Eq. (432), (m.e.p.) referred to lowpressure 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,2.6430(.15+.06) (~)log, {^^^)
10(.28+.04)log. (^^) +2xl50(l+.06)^y .150x^
" 2 64
30X;r~(l+.06)+2xl0(.28+.04)30x2.64x.0410(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 horsepower = 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 highpressure 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.
Lowpressure 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 horsepower 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. cutoff = .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 cutoffs 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 lowpressure cylinder? What will the horsepower for an
engine with a lowpressure 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 19x24in. engine with 5 per cent clearance in each
cylinder, i cutoff in each cyhnder, and 20 per cent compression in each cylinder, what
will be the horsepower 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 highpressure 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 highpressure work may be stated as follows:
Tr«=144D«{ (in.pr.)^^i^^[« (^^^^)'"'] (rec.pr.) (1X«)
(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]+lX«j. (471)
WORK OF PISTON ENGINES
321
W,=lUDJ(Tec.pv.)^^f±^[s (Agy'j _(bk.pr.)(lX^)
(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 highpressure 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 cuft. fluid supplied = t^ — yp;
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 lowpressure 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, cutoff and compression in high and lowpressure 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.
91
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 lowpressure cylinders,
rom Eqs. (478) and (479),
«l 81 •! *~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. /
sl
. . . . (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 horsepower, (6) compressed air used per hour, and (e) receiver
and relea.e pressures for a 12 and 18x24in. engine with infinite receiver, 6 per cent
clearance in the highpressure cylinder, and 4 per cent in the lowpressure cylinder, when
initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square
inch absolute, speed 125 R.P.M., cutoff in highpressure cylinder }, lowpressure
cutoff such as to give complete expansion in highpressiu^ cylinder, compression in
highpressure cylinder 15 per cent, and complete in low. Expansion such that «»1.4.
(6) As in example of Section 15, receiver pressure equal highpressure release
pressure, and lowpressure volume at cutoff 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 lowpressure cutoff is .38, and the lowpressure 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^loej J '•^) 2:25
.5+.06+2.25(.15+.06)
/JO \ 71 1 1 ♦
\150/
2.25(.38+.04)+.15+.06
.5h.06+2.25(.15h.04)f ^j
. 2.25(.38+.04)+.15+.06
.10
hence the horsepower 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 highpressiuie 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 lowpressure release pressure may be found
•cm the relation (rec.pr.)(ZLfci,)^*«»(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 horsepower and steam used per hour by an 18 and
4x30in. 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 cutofif in highpressure
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 highpressure cyhnder
izeforan air engine with a lowpressure cylinder 18x24 ins., to give equality of work,
learance in both cylinders being 5 per cent?
Prob. 4. What will be the horsepower 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 lowpressure 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=^PnVnPtVt (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.
WHP>vJl+loe,Y^ PfVr(l+log, ^^)P.F«logc
p,y»(i+iog. ^) p,v,p,v, log. YP.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..
PjiVjV,)
=P,V, log. (^±§) +PhVh log. yPtV, log. ^PsiVj F.).
Therefore
^ _ r p>F>o+p.F.(F.+o) /z»±o\] J (n±q\ )
+
r PtF>0P>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. ^^ PaVaPj{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.
F144X
(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
14cz.
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
constantreceiver 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.); highpressxue displacement, D^; cylinder
ratio, Re] highpressure 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; highpressure cutofiF Zj?, and compression
Xh\ lowpressure clearance cl, cutoff 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..
Highpressure cutofiF 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
Lowpressure cutoff, Zl from,
Rc{Zl+Cl)^I^Ch,
or
Zl^^^Cl (504)
tic
Highpressiu^ compression, Xh^
Xh^clRcCh (505)
332 ENGINEERING THERMODYNAMICS
Lowpressure 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 lowpres
sure work and may be stated as follows:
W^H=144(in.pr.)2)^
(a+„,)[.+,^fl±g]
lliTS''+"+"'"*^S7h'*'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 horsepower, 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^^ X3060 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 PnPe
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+55.5X1.230X1.2
^^ ^_ — —  . 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 horsepower of a 12 and lSx24]n. engme, running at 125
R.P.M., with a receiver volume twice as large as the lowpressure cylinder, 6 per cent
clearance in the highpressure 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 cutofF J, lowpressure f , highpressure 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. j150X.06X2~^
150 X .56[.79 X2.25 +4.5] 410(.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 horsepower will be 191.
WORK OF PISTON ENGINES 335
Prob. 1. Find the work done in the highpressure cyKnder and in the lowpressure
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,
ghpressure cutoff iV, lowpressure cutoff A, highpressure 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 highpressure
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 lowpressure cylinders, speed 100 R.P.M. What will
3 the cutoffs, 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 lowpressure clearance, cutoffs, 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 highpressure 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
nv,=v.{^y v.{^y , (511)
hich is the same as to say, that the quantity of fluid passing per cycle in the
ghpressure 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«(^
(cutoff 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 (cutoff 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.
Cutoff pressure of the lowpressure 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"^
81
JV+Ch+Xh)
s1
WORK OF PISTON ENGINES 337
ir..l44D,(cut^prO/±'^i±?^[(t!^«^?iJ)''l]
(bk.pr.)'^^^^±^^[(^^)'"'l](bk.pr.)(lX^^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 lowpressure 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 (cutoff 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 horsepower of and compressed air steam used by a 12 and
18x24in' engine running at 125 R.P.M., with a receiver volume twice as large as the
lowpressure cylinder, 6 per cent clearance in the highpressure 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, highpressure cutoff J, lowpressure cutoff },
highpressure compression 10 per cent, lowpressure compression 30 per cent, and
expansion and compression follow the law PV^'^=c.
From £q. (515) (rec.pr.)i is found tobe as follows when values for this problem
are substituted:
X ,^ri5'7(.56)[4.52.25(.79)]h4.6x2.25(.34)1« ^, ^ „ . ^ .
(rec.pr.).^10[. ^e [4.5h2.25X.79]44.5x2.25x.79 J '^^'^ ^^« ^^^^^^ ^^^^^^'
and by using this in Eq. (514)
.71 .71 «
, ^ re ^ / 4.5X81.7 +2 .25X.34X10 \ „„ • u w
(cutoff pr.)L = ( 4 5f2 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
,^^^^ 4.5h2.25x.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 horsepower 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)^[.5616x(^^y'n =.462>^.
iurx 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. cutoff pressure for a
crosscompound 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 highpressure cutoff is i, lowpressure f , highpressure compression 15 per cent,
low 25 per cent, and 5 = 1.4. Receiver volume is twice the highpressure 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 highpressure 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., highpressui^
cutoff i, lowpressure A, highpressure compression 15 per cent, low pressure 40 per
cent and s = 1.3.
Note: B = .3.
Prob. 3. If the highpressure cutoff is changed to } without change of any other
factor in the engine of Prob. 2, how will the horsepower, total steam per hour,
and steam per horsepower 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 x24in. 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 highpressure cylinder,
WORK OF PISTON ENGINES
339
the initial pressure 150 lbs. per square inch gage, back pressure 5 lbs per square inch
absolute, the lowpressure cutoff fixed at i and lowpressure 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) highpressure cutoff i, highpressure compression 80 ) er cent,
(6) highpressure cutoff J, highpressure compression 20 per cent,
(c) highpressure cutoff i, highpressure 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, Highpressure Exhaust
and Lowpressure 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+(ficl)(lX«).
(52(.:
WORK OF PISTON ENGINES
341
The ratio of expansion from E' to G is equal to
(cutoff pr.)L _ 1 +cl
(rel.pr.)L ~ (1 Xh) +Cl
• • • • •
. • (527)
Hence
F^ = 144Dir(in.pr.)
(Z.+c.)[l+lo«.^]
(in.pr. )
Rcl
Zi^+Ci,+2ee(Xi,+Ci,)^'' P'^
log.
['
^ch^RcCl+{Rc1){X Xn)
1\ch\'RcCl
]
(in.pr. )
\l\CH'\'RcCL{{ficl){lXH)
t . IT \ 1 ^h\Xh
(Ch+Xh) log* Ch
ch
(528)
TrL=144Dx,(iii.pr.)
Z.+c.+i2c(X.+c.)j^PI:]
Rc1
loga
14 C^+/2cCl+ (fie 1) (1 Z
1+c^+ficC^
.)]
+
(m.pr. )
11+Cb+RcClh+ (Rc 1) (1  Xb) J
144Dz,(bk.pr.)
(lfC.X.)l0ge(j^g^J
lZL+(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.+(ffcl)(l X
+C/fficCL
.
+
ZH+CH'\RciXL^CL)$^'^^'^
1hCH^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 highpressure 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+(RcIKIXh)]
(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\RcCLHRcl)(lX
I^ch^RcCl
']
ll+CH^RcCLHRcmiXH)]
Rc{l\CLXH)l0gfi
\1+ClXhI
+{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 highpressure compression and of separate
lowpressure expansion, hence p«=p«'.
WORK OF PISTON ENGINES
343
From the points A and E:
p.=(in.pr.)yj—
From the points G and B',
p«'=p«=(bk.pr.)
1+Ci
I+ClZj^'
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 LowPressure 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 lowpressure cutoflf pressure,
+ Cl + RpCh1 ^ /KOA\
(cutoflf 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 noclearance case, it may be remembered
that the point E or E' was not present, as it coincided with G,
Next, to find the highpressure release pressure, pa, by means of points £
and D, and their relation to the axis ML, Fig. 100.
l+CH+RcCLi
+Cl + CuRc
1+Cl+Ch
(rel.pr.)^ ^Pd^p, =^ = (bk.pr.) I ^
1''Xh+
Re 1
I+cu+RcCl
Rcl'
= (bk.pr.) i^,^^jg^^ (53oi
Knowing the release pressure of the highpressure cylinder, it is possible to
find the highpressure cutoff 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 fic1 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?cl) 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 lowpressure 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 horsepower and (6) steam used per hour for a 12 and 18 x24in.
tandem compound engine with no receiver, 6 per cent clearance in the highpressure
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 cutoff J, highpressure compression 15 per cent and lowpressure compression
is complete.
(a) Since the lowpressure 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.)(ZHfc^)=(rel.pr.)H(l +Cii), or
(rel.pr.)^ = 150 T^ =79.3 lbs. per sq.in. absolute.
l.Oo
Lowpressure 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 ii0^+2.25X .04H.25X.85l
.5+ ^.5K.06h2.25(.28+.04) Jloge [ i^.06H2:25x':04" J
.5 f .06 +2.25(.28 + .04) ^
1f. 06+2.25 X.041.75f.85
.06 +.15
2.25(L04.15)loge^^^
(.06+.15)loge
10
.06
and the horsepower 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 horsepower 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+.09I1.25X.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 cutofif to a
limited degree. For a speed of 200 R.P.M., a cutoff f and 10 per cent compression
in each cylinder, find how the horsepower will vary with the initial pressures of 175,
150, 125, and 100 lbs. gage.
Prob. 2. When the cutoff is reduced to i in the above engine compression
increases in tlie highpressure cylinder to 20 per cent. For the case of 175 lbs. gage
initial pressure find the change in horsepower.
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 18x24in. noreceiver 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 cutoff and compression must the valves be set and what horsepower
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 highpressure cylinder is
(rel.pr.)H = (in.pr.)f^^j (544)
WORK OF PISTON ENGINES 347
Immediately after release the pressure is equalized in the highpressure
cylinder and the lowpressure 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 lowpressure, as indicated .by. liD'JS'
and DEy and when the communicating valve closes, the pressure has become
(cutoflf pr.)L = (in.pr.)L
Rcl
l+C^+ficCL_^(j_^^)
Rcl
which, by means of Eq. (545) reduces to
1
(ou.^ pr.).(bk.pr.)[ , «;'^^^;> ±'^^g4 ]. . . (546)
After cutoflf 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= (cutoflf 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 lowpressure
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 VH
1 L fic1 JL \l+c„+RcCL+0X„){Rcl)) J
1
^<«4rS^^"^Sf«!^)]TC^)''] j ()
348 ENGINEERING THERMODYNAMICS
and
FFL=144i>z,
1 •
(bk.
8
1 L Rci J
^ L \l+c,,+i2c7C/^f(lX)(l2cl)y J
1
, (bk.pr.) ,, , ^ ^ J R'p(ch+Zh)+Rc{ cl+Xl) ]'L / 1+C£,X„ \'M
+ ,_1 ^^+<^^^")[i+c„+RcCj.+ (l^X^){RcT)\ L^"V 1+c^ / J
_^k£i^.J:Z.)^^ (549)
These are general expressions for work of high and lowpressure cylinders
for this cycle, and from them may be obtained the total work of the cycle,
mean elBfective pressure referred to the lowpressure 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^+{ix^){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 horsepower and (6) consumption of a 12 and 18x24in.
noreceiver 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., highpressure cutoff i,
highpressure compression 15 per cent, and lowpressure compresaion is complete.
(a) The per cent of lowpressure 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.) (clHXz,)^*,
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 horsepower 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_. ^ _ ^21 157(.56)+2 .25(.15) 1 « .63 cu.ft. per
48,5 ^2.25L "^ (15)^ H.06+2.25x.04+(l .15)(2.251)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 horsupower for conditions of that problem and on the cylinder event pressures?
Prob. 2. A 30 and 42 x54in. no receiver steam pumping engine runs at 30 R.P.M.
and has 3 per cent clearance in the highpressure 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 horsepower of, and the steam
used by the engine when the cutoff in the high is i?
Prob. 3. By how much would the power change if the cutoff were shortened to 
and then to i, and what would be the effect of these changes on the economy?
21. TripleExpansion 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 tripleexpansion 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 cutoffs
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 TripleExpansion 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 highpressure 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 lowpressure 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 lowpressure cylinder is found by
dividing W by 144D/,, and is therefore
(m.e.p. ref. to L.P.)
Z)i
= (inpr. )Zh^
3hlog.
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 horsepower 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 horsepower 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)=^''^zrYADj'
^km)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 cutoff are known, the other two cutoffs 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 cutoffs, 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(lhloge)(^^)*} . (568)
and the total work
ir=432(m.pr.)Z)„{z(l+logc)(^) I (569)
in which Z represents the cutoff in each cylinder, all bemg equal.
354
ENGINEERING THERMODYNAMICS
A special case of the tripleexpansion 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 Tripleexpanaon 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* ({Hj^ ^^)+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 cutoffs
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
cutoffs are made equal. Hence for this case cutoffs are equal,
*
„ „ „ Bn Di Dl (bk.pr.) . .
Dl Dl Dh (m*prO
Bzample. A tripleexpansion engine 12 and 18 and 27x24ins., 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 cutoffs in the different cylinders, beginning with the high, are i, i, and i,
what will be (a) the horsepower, (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 hoisepower 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
Highpressure 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 horsepower and steam usfed per hour by a 10 and
16 and 25x20in. infinite receiver, noclearance engine, running at 185 R.P.M.
on an initial pressure of 180 lbs. per square inch gage and atmospheric exhaust.
Cutoffs .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. cutoff ,5, horsepower 1600, steam per haore.
WORK OF PISTON ENGINES 357
power per hour 17 lbs. Check these results, using cutoflfs 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 cutoff of J in each cylinder, keeping the high pressiue the
same size as before?
Prob. 4. What would be the horsepower of a tripleexpansion 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 cutoff 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 cutoffs 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. MultipleExpansion Engine. General Case. Any Relation between
Cylinder and Receiver. Determination of Pressure VolumeDiagram and
Work, by Graphic Methods. It is possible to arrange multipleexpansion
engines in an almost infinite variety of ways with respect to the pressurevolume
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 crosshead 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 lowpressiue receipt from it, any amount of fluid
passing to correspond to engine load demands and consequently any relation
of cutoffs, compressions, and receiverpressure 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 crosssection paper directly,
on which strips can be measured and added, or by the planimeter. Thus
will high and lowpressure work be evaluated through the footpound 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 pressurevolume 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 highpressure
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 lowpressure 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 lowpressure cylinder partial
displacements, grouped with receiver volumes in various ways.
Take for an illustrative example the case of a twocylinder, singleacting,
crosscompoimd engine with slide valves, cylinder diameters 12f and 20 ins.
with 24 ins. stroke for both. Highpressure clearance is 10 per cent, low
pressure clearance 8 per cent. Receiver volume 4000 cu.ins. Highpressure
crank following by 90**. Find the mean effective pressure for the high and low
pressure cylinders, for a cutoff 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= lowpressure cylinder displacement volume in cubic inches to scale.
[77= lowpressure cylinder clearance volume in cubic inches to scale,
ypl/'ss receiver voliune in cubic inches to scale.
TFX= highpressure cylinder clearance volume in cubic inches to scale.
XF= highpressure 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 highpressure admission at a height XA rep
360 ENGINEERING THERMODYNAMICS
resenting absolute initial pressure; lay off LM at a height TL representing
lowpressure exhaust at a constant absolute back pressure to the same scale.
Locate point B at the cutoff 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. cutoff, 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. cutoff at the initial pressure and M the lowpressure 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
highpressure 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 highpressure crank angles 0360°, 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 lowpressure 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 highpressure 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. cutoff (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 cutoff occurs in the low, which
divides the volume into, PJ^ in receiver, which remains at constant volume
bill highpressure release, and the second part, PP in the low, which expands
In that cylinder to K^K^.
After lowpressure release the volume in low decreases from K^K^ to
M^M^f when the exhaust valve closes and lowpressure 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 highpressure 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 lowpressure cutoff at J the expansion takes place in lowpressure
cylinder alone, to pressure P*, when release allows pressure to fall (or rise)
to exhaust pressure Pi.
When cutoff in low occurs at J the volume PP is separated off in the
receiver, where it remains at constant pressure P^ imtil highpressure 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 lowpresure
compression line at d. Then the H.P, intercept {ef) must be equal to the
lowpressure intercept (dc); this fixes (c) through which a Py = const, line
intersects the L.P. cutoff 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 lowpressure cutoflf.
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 oflfset
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 horsepower of a 12 and 18 X24
n. singleacting crosscompound 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., highpressure compression
}0 per cent, low pressure 20 per cent, high pressure cutoff 50 per cent, low pressure
10 per cent, highpressure 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 crosscompound
ind draw the new diagrams for solution.
Prob. 3. A doubleacting, 15 and 22 x24in. compound lengine has the high
jressure crank ahead by 60°, and has 5 per cent clearance in the lowpressure cylinder,
10 per cent in the high, and a receiver 4 times as large as the highpressure cylinder.
tVhat will be the horsepower when the speed is 125 R.P.M., initial pressure 150 lbs. per
iquare inch absolute, back pressure 5 lbs. per square inch absolute, highpressure cut
)ff ij lowpressure i, highpressure compression 20 per cent, lowpressure 30 per cent,
ind ratio of rod to crank 5. Determine graphically the horsepower 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) cutoff or
'atio of expansion, (d) clearance, and (e) compression, for the singlecylinder
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.) = (iupr.)^+(^+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, cutoflf, 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. Cutoff, when short, gives a low mean forward pressure,
but it is to be noted that zero cutoff 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 cutoff, 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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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, cutoff 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 highpressure cylinder of a compound
engine with infinite receiver, with clearance. An increase of highpressure
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 lowpressure 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 highpressure compression has increased
the mean effective pressure of the highpressure cylinder. The only eflFect
upon the lowpressure 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 highpressure compression decreases lowpressure work more than it
increases highpressure 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 multipleexpansion engine. Initial pressure and cutoff
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 cutoff 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 noclearance cycles, that
the mean effective pressures of all but the lowpressure cylinder are decreased
in direct proportion to the initial pressure, while that of the lowpressure 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 higbpressure cutoff 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
cutoff in the second cylinder will tend to increase receiver pressure. To
maintain constant work division, there must be a certain relation between
cutoffs 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 pressurevolume
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 noncon
msing, fourvalve 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 cutoff 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' midstroke, 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 cutoff is approached, this pressure
fall becomes more rapid, due to the partial closure of the admission valve.
From B', the point of cutoff, 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 cutoff 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 stuffingboxes.
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 cutoff
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, cutoff, and compression as are found in the actual.
Commercial cutoff 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 cutoflf
B' a curve PF = const.
While the diagram factor represents the ratio of indicated horsepower
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 horsepower (B.H.P) then the ratio of this to
indicated horsepower is called the mechanical efficiency, Emy of the engine
The difference between indicated and shaft horsepower 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 bearingsurfaces 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 horsepower divided by speed is proportional to mean effective
pressure, so that
K2
En, = lKi
(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
horsepower.
Prob. 1. Assuming a back pressure of 10 lbs. per square inch absolute, a clearance
of S per cent, a cutoff 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 cutoff 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,
doubleacting, 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 Cutoff as Affected by Condensation and
Leakage. The weight of steam used by a steam engine per hour divided by
the indicated horsepower 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 horsepower, 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 horsepower 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 steamengine 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 singleexpansion engine,
logarithmic law, the expression for consumption in pounds fluid per hour per
indicated horsepower, 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.
Cutoff has a marked effect upon consumption, the minimum occurring
when cutoff 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 cutoff. 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 cutoff, 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 meaneflfective pressure,
and consequent high consumption. Between, the consumption has a minimum
point, which is dependent for its location on both cutoff 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 cutoff.
Compression, throughout the ordinary range of practice, has but slight effect
upon consumption, indicated by the flat middle portion of the curve in Fig.
I
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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 highpressure 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 cutoff 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
cutoflf,
AB
and the fraction of card length from point of compression to end of stroke,
X=
AC
AD'
are determined, the pressure at cutoff 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 cutoff, 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 horsepower 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 multipleexpansion engines. In applying this to the mvltipleexpansion
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 highpressure card is found by inserting
in formula Z, X and c for the highpressure 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 lowpressure 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 multipleexpansion 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
cutoflf 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
cutofiF, 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 reevaporation 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 horsepower 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 horsepower. 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 noncondensing 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 welljacketed, welldrained 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 livesteam 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 cutoff 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 cutoff, 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 cutoflfs. 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 cutoff than that of 4BC.
Since cutoff B gave complete expansion, cutoff 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 horsepower 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 cutoff, for the output of the engine is not indicated, but brake horsepower.
ao
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Per Cent Cut off
Fig. 110. — Diagram to Show Displacement of Best Cutoff Due to Effect of Missing Waier
from Point B for the Hypothetical Cycle to Some Greater Value E,
In Fig. ill on cutoff 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 cutoff,
(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 cutoffs, the minimum point 5
corresponds to a cutoff still longer than for the curve EFG, which itself wa>
found in Fig. 110 to give a longer cutoff than that of the hypothetical curve.
Hence the best cutoff 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
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> Eff.
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Per Cent Cut off
Fig. 111. — Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cutoff.
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 horsepower 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 horsepower and much slower later, being
scarcely anything at all over 2000 horsepower.
Example 1. What cutoff will give the lowest indicated water rate for a 9x12
in. engine, with 5 per cent clearance and no compression when running noncondensing
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 horsepower
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 horsepower 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, cutoff 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 9Xl2in. jacketed engine when running
noncondensing at a speed of 250 R.P.M. with an initial pressure of 100 lbs. per
square inch gage and J cutofiF is 50 lbs. Using Perry's formula what will be the
probable actual steam used by engine per horsepower hour *
Prob. 4. A 24 x48in. engine in good condition is found to have an indicated water
rate of 25 lbs. when cutoff 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 15X22x36in engine with 5 per cent clearance in each cylinder, running at 100
R.P.M. on same pressure range with J cutoff in highpressure cylinder, J cutoff 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 horsepower, 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 rateload 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 cutoff or, by varying cutoff 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 horsepower 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 cutoff, 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 replotted 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 waterrate 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, Fiftyninth Stm
station, as well as the combined piston engine and lowpressure steam turbia
7000 9000
1000 P IF Kilowatts
Fia. 114.— Performance of a 10000K.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 1000K.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 7000H.P. Piston Engine alone and with a Lowpressure
Steam Turbine,
26 24
I
s
S20
a
O
ji
u
s.
3
o
H
u
^20
f
^18
OS
10 16
\
—
—
—
 ■
— ■
— _ .
1
\
\
/
\
•
>
f
\
/
\
1
1/
r
\
^A
/
\
<?^
r
M
/
7
\
»
^
f
\
\
1/
r
Kl
f
1
/
/
\
/
\
\
/
*
\
V
/
^
^
>
^
y^
^
/
y
X
 
:r~=
^^~'
— ■
^
^
800
600 900
Kilowatts»1000P
1200
1500
Fig. 116.— Performance of a 1000K.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 horsepower, 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(PP2)+B2P2
= {Ai+A2)+BiP+{B2Bi)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)
+ (B22PCiBi)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 switchin 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 Horsepower and Cylinder Sizes .
Tie diagram for mean effective pressures in terms of initial and back
ressure, clearance, compression and cutoff, 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, cutoff 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
utoff, 12 per cent. Project downward to the logarithmic scale for " mean
388
ENGINEERING THERMODYNAMICS
0)
o
c
'5)
c
pa
C
I d
.si
GQ I
O C
£3
I I
§
B
O
hi
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.) =49.53.2=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 highpressure 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 highpressure 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 cutoffs " and at
the side, C, read ratio of cutoffs
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 cutoffs and ratio of highpressure 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 xl8in. ^doubleacting
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
cutoff is i? What will be horsepower 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, highpressure cutoff J, highpressure compression iV, and
lowpressure compression i,
Prob. 3. A simple doubleacting 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 cutoff 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 12xl8x24in. engine with
infinite receiver and no clearance when cutoff 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 horsepower, 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, cutoffs
J and I in high and lowpressure 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 fullload
horsepower 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 cutoffs and cylinder ratio to accomplish this when receiver is 3i times high
pressure cylinder volume?
Prob. 8. A 12in. and 18x24 ins. doubleacting 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, horsepower, and steam consumption when speed is 150 R.P M., expansion
logarithmic, and cutoff 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 cutoff in this made i, how would horsepower, steam consumption, receiver and
release pressures change?
Prob. 10. What would have to be size of a single cylinder to give iSame horsepower
at same revolutions and piston speed as that of engine of Prob. 8 under same conditions
of pressure and cutoff?
Prob. 11. With the higlipressure cutoff at f, and low and intermediate cutoffs
at A, what will be the horsepower, 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 lowpressure cutoffs 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 cutoffs of the engine of Prob.
11, equal to i, what should have been the size of the intermediate and lowpre^ure
cylinders to give equal work for same pressure range and same highpressure 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 cutoffs 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 lowpressure clearance 5 per cent. The cylinders are 22 and
33x48 ins., highpressure cutoff , high and lowpressure 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 horsepower at a speed
of 40 miles per hour, the engine having 7ft. 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 horsepower and steam consumption?
Prob. 16. What will be the maximiun receiver pressure work done in each cylinder
and total work for a crosscompoimd 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 cutoff is J, clearance 6 per cent, compression 20 per cent, low
pressure cutoff is f , clearance 4 per cent, compression 15 per cent, and receiver volume
is 105 cu.ft.?
Prob. 17. A manufacturer gives the horsepower of a 42x64x60in. 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 cutoff in highpressure 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 highpressure
WORK OF PISTON ENGINES 393
cutoff 18 1 and low pressure i, while normally both cutoffs are J. The exhaust
from highpressure 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 horsepower imder the two conditions of cutoff 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 lowpressure. 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 cutoffs and compressions to satisfy these
conditions?
Prob. 20. Assuming 7 per cent clearance in highpressure 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 36X6x48in. 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 cutoff in high pressure cylinder?
Prob. 21. For a 25X40x36in. engine, with 5 per cent clearance, i cutoff
and 20 per cent compression in each cylinder, what will be horsepower 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 cutoff to give complete expansion
in a noclearance, 14 and 22 x24in. 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 horsepower 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 highpressure cutoff 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 cutoff 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 crosscompound engine with a receiver 3 times as large
as highpressure 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 cutoff in
highpressure cylinder is  and in low, \, It has been found that steam is worth 25
cents a thousand pounds. What must be charged per horsepower 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 26X48x36in. crosscompound 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
highpressure cutoff is i and lowpressure cutoff J? For these cutoffs what i
fluctuation in receiver pressure and what steam will be used per horsepower 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 cutoffs, and what values will result for receiver pressims,
horsepower, and eteam used per hour?
Prob. 29. What will be the horsepower and steam used by a 20x30x36m.
engine with infinite receiver and no clearance, if expansion be such, that 8 = li5,
highpressure cutoff i, lowpressure cutoff 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 noclearance, 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, highpressure cutoff f, lowpressure , speed 150 R.P.M. What will be
horsepower 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 15X22x30in. 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, highpressure cutoff J, lowpressure A, highpressure compres
sion 40 per cent, lowpressure 20 per cent, highpressure crank following 90°, logarithmic
expansion.
Prob. 32. Show by a series of curves, assuming necessary data, the effect on
(m.e.p.) of cutoff, 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 cutoff 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
DeadCenter 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
noncondensing
noncondensing
noncondensing
noncondensing
noncondensing
noncondensing
noncondensing
on initial
on initial
on initial
on initial
on initial
on initial
on initial
pressure
pressure
pressure
pressure
pressure
pressure
pressure
of 100160.
of 100.
of 126.
of 90100.
of 110130.
of 140160.
of 126
4i 8 X 6
6 10 X 6
7 13 X 8
6 12 XIO
7 12 X 10
8 12 XIO
8i15jX10
7 14 XIO
8 14 XIO
9i15 Xll
7H3iXl2
9 15JX12
19 14 X 12
10 16 XX2
1018 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
7i13i
9 15J
10 17i
11 19
11 22
12 21
13 22
13 22J
14 22
14J25
15 22
15i26J
16 25
13 23
15 26
16i29
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
15i26JX18
16 24 X18
16 26 X18
16128iX18
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
14i26iX21
15i28JX21
18i32i 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
14i26 X22
18 32 X22
10 17JX24
11 19 X24
12 18 X24
13 20 X24
14 22 X24
16J28JX24
17i30JX24
22 38 X24
24 A2 X24
12 21 X27
13 22 J X 27
16J28i X27
17J30iX27
14i25 X30
15l26iX30
18i32iX30
20 36 X30
28i50 X30
30 54 X30
16i28JX33
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
18132 J X 36
20 36 X36
26J46 X36
28150 X36
14i25 X42
15i26iX42
18J32JX42
20 36 X42
16 J28i X48
17i30iX48
22 38 X48
24 A2 X48
18J321X54
20 36 X64
26i46 X54
28i50 X54
22 38 X60
24 42 X60
30 54 X60
32i57 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 15126X15
200
27 43. (41x39
180
305082X48
180
11 18 30X20
250
254168X48
190
12 20 34X24
180
18 28148X40
180
274575X54
190
12 19 32X24
190
22 37 63X42
180
284572X54
185
175
22 38 64X42
185
1 284675X54
180
12^22 36X24
180
32153 {5[x48
265
' 294783X54
160
14 23 28X26
190
1 325292X50
200
18 29 47X30
16124 41X30
18 30 50X30
16 25M3X30
200
180
200
190
3567/^X48
36 57 {76X48
265
295
' 3456100X60
3558 ^qX60
3457104X63
200
190
200
16124 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 highpressure 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 waterwheels 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 feedwater 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 woodwork
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 byproduct 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 feedwater 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 freezingpoint, 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, hotair 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 freezingpoint. This is the case with icewata"
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 scaleforming compounds are thrown down on heating the water in
ided for boilers, a fact that is made use of in feedwater 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 heatfreeing 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 heatabsorbing 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 thermoelectric 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 Nonreversible Processes, There may be reversible and
inreversible 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 nonreversible 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, nonreversible.
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 redhot 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 thermometermak
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 thermocouple.
(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 temperatureindicating 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 littleused 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 constantvolume 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. Gasfill6d 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 freezingpoints, 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
FreeiiagpoiDt.
at one aim. pr.
Pure Water
Boilingpoint,
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 pressuretemperature 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.
<=100P=PO.
Pioo—po
(601)
412 ENGINEERING THERMODYNAMICS
He also determined the pressure at the boilingpoint to be related to the pressure
at the freezingpoint, 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 JouleThomson 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 freezingpoints, 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 onehimdredandeightieth 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 onehundredandeightieth 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 LandoltBomstein
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{ttiyF
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 boilingpoint 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 pounddegrees heating effect in water,
lo that taking the pounddegree 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 pounddegrees. 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 pounddegree 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 onehalf degree
both sides; this is the point of maximum water density. Others have used 15'
C. = 59° F. as the temperature and the range onehalf 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 presentday practice which will prol)
ably become universal, for taking as the range and temperatures, freezing
point to boilingpoint, 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 footpounds 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 thermophysical 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 footpounds?
1 gallon =8.33 lbs.,'
200^ F. 60** F. =140^ F. rise,
8.33 X 140 = 1665 pounddegrees,
= 1665 B.T.U.
=778X1665, ft.lbs.
90,800 ft.lb8.
Prob. 1. A feedwater heater is heating 5000 gallons of water per hour from
40* F. to 200** F. What would be the equivalent energj'^ in horsepower units?
Prob. 2. A pound of each of the following fuels has the heating values as given.
Change them to footpounds.
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 footpounds.
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 footpoimds 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 horsepower 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 horsepower 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
ingpoint 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(tZ2)+c{t32)^ . (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
Wt32
Q= I [a+6(i32)+C(i32)2](ft
Jtt 32
= a[(fe32)(<i32)]+[(fe32)2(ii32)2]+[(fe32)3(<i32)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+(i32)+(^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 onethird,
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
BLIO
«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 bestknown 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)=2.88213.6272C+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
FREEZINGPOINT OF CALCIUM CHLORIDE .
U. S. Bureau of Standards
Density of Solution.
Per cent CaClt by Wt.
Freesingpoint,
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 (t32) (6)
Authority
Zeuner
Dieterici
Wood
Ledoux
1.1352+.00438 (^32) (c)
1.0057+.00203 (t32) (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
66115
0
0
59120
59194
61126
64.4
0
63125
68192
6468
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
AAZeuner
BBLedouz
CO Wood
DDDieterioi
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(CpC.) = fi, (615)
§fT (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,
Meanspecific heat from
32.^F.,tor F., (F=C),
^=C'. = a+(i32) (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.,(P0, 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 xthe final temp. The heat given up by the substances falling in tempera
ture is equal to that gained by those rising, hence
50(x50) XI 5(1000:) X.4+10(150x) 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.4H0x.511f50Xl 57.11 ._.
^ z z~z z~^ ~" A_ ^ .o7oO«
5110150 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 hotwater
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{t2ti)] (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 freezingpoint. 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{t2ti)+y(t2ti)^,
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+54Xl0i2i2+602Xl0i*^), . . . (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
icemeUing point, and are but expressions of conditions for reduction of the
volume and pressure at the icemelting 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 ^49232+<a^<a+460
P* ti,S2 49232+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 GayLussac, 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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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(CpC.) (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
kiW'" ■ • • <'^>
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 freezingpoint
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 1 190 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 freezingpoint. 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, GayLussac 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 GayLussac 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 Freezingpoints
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 pressuretemperature 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 highpressure 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«»C1.
■ • • •
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 pressuretemperature 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
/
/
/
,
/
/
/
/
/
/
'
/
/
/
/
\
/
•
•
/
•
.
/
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, Pressuretemperature 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, Pressuretemperature 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 vaporUquid 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 vaporsolid 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 vaporice, 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 WATERICE
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, Pressuretemperature 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 Prebsuretemperature 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.
MeUingpaintSy 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 boHingpointa, 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 boilingpoint then is the
highest temperature to which the liquid and its vapor could rise under the
existing pressure. When not otherwise defined the term boilmgpoint 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 socalled gases like oxygen and
nitrogen are but superheated vapors with a great amount of superheat.
It has abeady been mentioned that the saturated vapor pressuretemperature
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:
(il459.6) Iog^ = 5.409(i212®)3.71X10I0[(6890*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 onetenth of 1 per cent, and with the old below 100®
F., about onefifth 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 (yl) 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=5.604221527.54r"i17196ir2 .... (660)
For CO2 it was found that
log2?=.038+2.65^pl) + 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 = 7.465814405.765r"i+1617501.366r"2257086165.8706r"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 boilingpoint of some substance
little known, to employ a relation between boilingpoints of different substances
at the same pressure appUed to a substance wellknown.
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
/
•
>
/
>
/^
/
/^
l.4
A
/
r
^^
y
71.2
r
2
^
r
J
y
^
k
y
?r^
■3
c> 8
jH
J
From Values of Wood
" »» »* DietericI •
** " Begiiault +
»• *• " Ledoux '
.6
1
■54
.3
I
f
^
K^
^
^
,<^
/^
y
•^
,
<
• €
J
• T
.5
,€
»
• 40 •O '^ •«>
^Critical Temperature Divided by any other Temperature)—!
1400
;
/
^200
•
"^.1000
>
/
/
/
/
/
a
.
/
'
/
/
in Pou]
/
«
/
—
Pressure
/
/
From Values of W.ood
'* »•  Bletedci •
♦• » »» RefirDault +
u u M Ledoux /
^y
Y
•
K
5<>^
K^^
^^
..^
—4
1
6
1]
IC
K)
ai
LO
2<
X)
Temperatures in Degrees Fahr.
Fig. 139. — ^Ammonia Pressuretemperature 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 Pres6uretemperature 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 freezingpoint, 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 freeztngpoint is a constant
Atm
Molecular lowering of freezingpoint J?'= = const., . . (664)
w
in which
Af = depression of freezingpoint in degrees F.;
ti7= weight dissolved in 100 parts of solvent;
m= molecular weight of substance dissolved.
From Eq. (664) the freezingpoint for brines may be found as follows:
w
Freezingpoint 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
JonesChambers
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 boilingpoint at a given pressure. Investigation shows that a similar
formula expresses the general relation:
Mm
Molecular rise of boilingpoint = S = = constant = 5.2, . (666)
when water is the solvent.
From Eq. (666) the rise of the boilingpoint is found to be
Rise of boilingpoint =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
nonmiscible 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 boilingpoint 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 vaporgas mixtures, and data on the vapor
pressure and temperature relations of ammoniawater 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 freezingpoint?
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
VvVi
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)
mjf = 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
boilingpoint. The amount of heat necessary to warm a pound of liquid from
temperature 32° F. to some boilingpoint, 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 today. 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 pressuretemperature 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 socalled 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 largescale plottings, without equations beyond that for the
pressuretemperature 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 drysaturated steam are given in Table XL VII, and
charts. A, B, C, D, E, F; the properties of superheated steam, in Table XLVIII;
drysaturated ammonia vapor in Table XLIX, and Charts G, H, I, J, K, L;
and drysaturated 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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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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Dioxide.
476
ENGINEERING THERMODYNAMICS
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HEAT AND MATTER
477
The volumes of drysaturated steam determined from the tables when
compared with their pressiues show that there exists an approximate
relation of the form for steam,
p{VvVLy'^^ = 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 2Jinch 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 lefthand card it will be seen that the cutoff
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 cutoff, 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 cutoff, 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 cutoff, 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 footpounds is the product of pressure and volume
change, or expressing pressures in pounds per square inch,
144
Extern^ latent heat = ^P(FkFi,) (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^(7vFi)(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
iondiscovery.
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
»nehalf 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 onehalf 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
[(6032) +144+.5(3220)]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 boilingpoint 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 nonmisdble). 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 WlRllrW2R2 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
WlV)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^{R2Ri)'
(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 socalled 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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