# Full text of "Engineering Thermodynamics"

## See other formats

Google This is a digital copy of a book that was preserved for generations on Hbrary shelves before it was carefully scanned by Google as part of a project to make the world's books discoverable online. It has survived long enough for the copyright to expire and the book to enter the public domain. A public domain book is one that was never subject to copyright or whose legal copyright term has expired. Whether a book is in the public domain may vary country to country. Public domain books are our gateways to the past, representing a wealth of history, culture and knowledge that's often difficult to discover. Marks, notations and other maiginalia present in the original volume will appear in this file - a reminder of this book's long journey from the publisher to a library and finally to you. Usage guidelines Google is proud to partner with libraries to digitize public domain materials and make them widely accessible. Public domain books belong to the public and we are merely their custodians. Nevertheless, this work is expensive, so in order to keep providing this resource, we liave taken steps to prevent abuse by commercial parties, including placing technical restrictions on automated querying. We also ask that you: + Make non-commercial use of the files We designed Google Book Search for use by individuals, and we request that you use these files for personal, non-commercial purposes. + Refrain fivm automated querying Do not send automated queries of any sort to Google's system: If you are conducting research on machine translation, optical character recognition or other areas where access to a large amount of text is helpful, please contact us. We encourage the use of public domain materials for these purposes and may be able to help. + Maintain attributionTht GoogXt "watermark" you see on each file is essential for informing people about this project and helping them find additional materials through Google Book Search. Please do not remove it. + Keep it legal Whatever your use, remember that you are responsible for ensuring that what you are doing is legal. Do not assume that just because we believe a book is in the public domain for users in the United States, that the work is also in the public domain for users in other countries. Whether a book is still in copyright varies from country to country, and we can't offer guidance on whether any specific use of any specific book is allowed. Please do not assume that a book's appearance in Google Book Search means it can be used in any manner anywhere in the world. Copyright infringement liabili^ can be quite severe. About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http : //books . google . com/| of tbe lanlvcrett^ ot TWli»con»in ENGINEERING THERMODYNAMICS ikimiiiM' miMlMiiK B ¥"*'*'" ^ ™p™ McGraw-Hill Book. Company Ne-^v Yorlc iSucce4.Sor« to tke&ookDeporlinent« of tke ' McGmwPublbhing Company Hill Publishing Company l^lUlaher5 of I^ooLs for Electrical World TkeEngineeritig And Mining Journol Engineering Record Americon MacKinist Electric Railway JounniJ Coal Age MetaDuigical and Chemical Engineering R>wer ENGINEERING THERMODYNAMICS BY CHARLES EDWARD LUCKE, PH.D., Prtf€990T of Mechanical Engineering in Columbia Uniwreity, New York CUy McGBAW-HILL BOOK COMPANY 239 WEST 39TH STREET, NEW YORK 6 BOUVERIJB STREET, LO^DO^, E. C. 1912 Copyright 1912, by thb MoGRAW-HILL BOOK COMPANY THK SCICNTIPIC PflKSS ROBKflT DRUMMONO AND COMPANY ■ROORLYN. N. Y. 172771 MAW 3 1 1913 PREFACE ^3^^^ V Calcttlations about heat as a form of energy, and about work, another related form, both of them in connection with changes in the condition of all sorts of substances that may give or take heat, and perform or receive work while changing condition, constitute the subject matter of this book. The treatment of the subject matter of this text is the result of personal experience in professional engineering practice and teaching students of engineering at Columbia University. Even a brief examination of the conditions surrounding changes in sub- stances as they gain or lose heat, do work or have work done on them, and of the corresponding relations between heat and work as forms of energy independ- ent of substances, will convince any one that the subject is one of great com- plexity. Accordingly the simplicity needed for practical use in the industries can be reached only by a consideration of a great mass of sub-topics and data. That the doing of work, and the changes in heat content of substances were related phenomena, and that these relations when formulated, would con- stitute a branch of science, was conceived about a half century ago, and the science was named Thermodynamics. The Engineer Rankine, who helped to create it, defined thermodynamics as " the reduction of the laws according to which such phenomena took place to a physical theory or corrected system of principles." Since Rankine's. time thermodynamics has become a very highly developed science and has proved of great assistance in the formu- lation of modern physical chemistry, and to those branches of engineering that are concerned with heat. Unfortunately, as thermodynamics developed as a separate subject it did not render proportionate service to engineering, which itself developed even more rapidly in the same period under the guidance of men whose duty it was to create industrial apparatus and make it work properly, and who had little or no time to keep in touch with purely scientific advances or to interpret such advances for utilitarian ends. Thermodynamics proper is concerned with no numerical quantities nor with any particular substance nor for that matter with any actual substances whatever, but it is a physical theory of energy in relation to matter as a branch of natural philosophy. Engineering, however, is concerned with real substances, such as coal, steam, and gases and with nimierical quantities, horse-powers of engines, temper- atures of steam, the heats of combustion of oils, so that alone, the principles of thermod3mamic philosophy will not yield a solution of a practical problem, vi PREFACE be it one of design or one of analysis of test performance of actual heat machine or thermal apparatus. It is the province of engineering thermodynamics to guide numerical computation on thermal problems for real substances being treated in real apparatus. Its field, while including some of that of pure thermodynamics, extends far beyond the established provinces of that subject and extends to the interpretation of all pertinent principles and facts for purely useful purposes. Engineering thermodynamics, while using whatever prin- ciples of pure thermodynamics may help to solve its problems, must rely on a great mass of facts or relations that may not yet have risen to the dignity of thermodynamic laws. The workers in shops, factories, power plants or laboratories engaged in designing or operating to the best advantage machines and apparatus using heat with all sorts of substances, have developed great quantities, of rules, methods and data that directly contribute to the ends sought. While for each class or type of apparatus there has grown up a separate set of data and methods in which much is common to several or all groups, not nearly so much assistance is rendered by one to another as should be by a proper use of engineering thermodynamics, which applies methods, principles and conclusion to all related problems. Classes of apparatus about which such groups of methods of analysis or synthesis, or collections of special data have developed and which it is province of engineering thermodjmamics to unify so far as may be, include air compressors, and compressed air engines, reciprocating steam engines, steam turbines, steam boUers, coal-, oil- and gas- fired furnaces, gasifiers of coal and oil, gas producers, gas engines, complete steam or gas power plants, mechanical refrigeration and ice-making plants and chemical factory equipment, or more generally, machinery and apparatus for heating and cooling, evaporating and condensing, melting and freezing, moisten- ing and drying, gasification and combustiom. The nature of the subject and its division are better indicated by the classes of problems to be solved by its aid or the contributions expected of it than by the kinds of apparatus to which they apply. Probably its broadest contribution is the establishment of limits of possible performance of heat apparatus and machines. These limits will show what might be expected of a steam engine, gas engine or refrigerating machine when its mechanism is quite perfect and thus they become standards of reference with which actual per- formance can be compared, and a measure of the improvements yet possible. These same methods and practices are applicable to the analysis of the operat- ing performance of separate units and complete plants to discover the amount of energy being lost, how the total amount is divided between the different elements of the apparatus, which of the losses can be prevented and how, and finally which are absolutely unavoidable. This sort of analysis of the per- formance of thermal apparatus is the first step to be taken by the designer or manufacturer to improve the machine that he is creating for sale, and is essential to the purchaser and user of the machine, who cannot possibly keep it in the best operating condition without continually analyzing its performance and comparing results with thermodynamic possibilities. PREFACE vii The subject naturally divides into three general parts, the first dealing with the conditions surrounding the doing of work without any consideration of heat changes, the second heat gains and losses by substances without reference to work involved and the third, transformation of heat into work or work into heat in conjunction with changes in the condition of substances. The first part applies to the behavior of fluids in the cylinders of compressors and engines. The second part is concerned with the development of heat by combustion, its transmission from place to place, and the effect on the physical condition of solids, liquids, gases with their mixtures, solutions and reactions. The third part is fundamental to the efficient production of power by gases in internal combustion gas engines or compressed-air engines and by steam or other vapors in steam engines and tiu'bines, and likewise as well to the production of mechanical refrigeration by ammonia, carbon dioxide and other vapors. Accordingly, the six chapters of the book treat these three parts in order. The first three chapters deal with work without any particular reference to heat, the second two with heat, without any particular reference to work, while the last is concerned with the relation between heat and work. After establishing in the first chapter the necessary units and basic principles for fixing quantities of work, the second chapter proceeds at once to the determina- tion of the work done in compressor cylinders and the third, the available work in engine cylinders, in terms of all the different variables that may determine the work for given dimensions of cylinder or for given quantities of fluid. There is established in these first three chapters a series of formulas directly applicable to a great variety of circumstances met with in ordinary practice. All are deiived from a few simple principles and left in such form as to be readily available for numerical substitution. This permits of the solution of niunerical problems on engine and compressor horse-power, fluid consumption or capacity with very little labor or time, although it has required the expansion of the subject over a conmderable number of pages of book matter. A similar pro- cedure is followed in the succeeding chapters, formulas and data are developed and placed always with a view to the maximum clearness and utility. The essential unity of the entire subject has been preserved in that all the important related subjects are treated in the same consistent manner and at sufficient length to make them clear. When no general principles were available for a particular solution there has been no hesitation in reverting to specific data. The subject could have been treated in a very much smaller space with less labor in book writing but necessitating far greater labor in numerical work on the part of the user. This same aim, that is, the saving of the user's time and facilitating the arrival at numerical answers, is responsible for the insertion of a very considerable number of large tables, numerous original diagrams and charts, all calculated for the purpose and drawn to scale. These, however, take a great deal of room but are so extremely useful in everyday work as to justify any amount of space thus taken up. For the sake of clearness all the steps in the derivation of any formula used are given, and numerical examples are added to illustrate their meaning viii PREFACE and application. This also requires a considerable amount of space but with- out it the limitations of the formulas would never be clear nor could a student learn the subject without material assistance. Similarly, space has been used in many parts of the book by writing formulas out in words instead of express- ing them in symbols. This saves a great deal of time and labor in hunting up the meaning of symbols by one who desires to use an unfamiliar formula involv- ing complex quantities, the meaning of which is often not clear when it is entirely symbolic. Thus, in the discussion of steam boiler capacity and efficiency, a dozen or more pages are taken up with formulas that could have been con- centrated in a single page were symbols used entirely, but only at the sacrifice of clearness and utiUty. Where in the derivation of a new formula or in the treatment of a new subject, reference to an old formula or statement is needed and imp)ortant, repetition is resorted to, rather than mere reference, so that the new topic may be clear where presented, witlu)ut constantly turning the pages of the book. It will be found, therefore, that while the size of the book is unusually large it will be less difficult to study than if it were short. As a text the book may be used for courses of practically any length, but it is not intended that in any course on the subject every page of the book shall be used as assigned text. In the new graduate course in mechanical engineering at Columbia University, about three-fourths of the subject matter of the book will be so used for a course of about one hundred and twenty periods of one hour each. All of the book matter not specifically assigned as text or reference in a course on engineering thermodynamics in any school may profitably be taken up in courses on other subjects, serving more or less as a basis for them. It is therefore adapted to courses on gas power, compressed air, steam turbines, steam power plants, steam engine design, mechanical refrigeration, heating and ventilation, chemical factory equipment, laboratory practice and research. Whenever a short course devoted to engineering thermodynamics alone is desired, the earlier sections of each chapter combined in some cases with the closing sections, may be assigned as text. In this manner a course of about thirty hours may be profitably pursued. This is a far better procedure than using a short text to fit a short course, as the student gets a better perspective, and may later return to omitted topics without difficulty. The preparation of the manuscript involves such a great amoimt of labor, that it would never have been undertaken without the assurance of assistance by Mr. E. D. Thurston, Jr., a fellow instructor at Columbia, in checking text and tables, calculating diagrams, writing problems and working examples. This help has been invaluable and is gratefully acknowledged. Recognition is also due for material aid rendered by Mr. T. M. Gimn in checking and in some cases deriving formulas, more especially those of the first three chapters. In spite, however, of all care to avoid errors it is too much to expect complete success in a new work of this character, but it is hoped that readers finding errors will point them out that future editions may be corrected. C £. Li. CoLXTMBiA University, New York, September, 1912, CONTENTS Chapter I. Work and Power. General Principles PAOK 1. Work defined 1 2. Power defined 2 3. Work in terms of pressure and volume 3 4. Work of acceleration and resultant velocity 6 5. Graphical representation of work 8 6. Work by pressure volume change 10 7. Work of expansion and compression 13 8. Vahies of exponent 8 defining special cases of expansion or compression 20 9. Work phases and cycles, positive, negative and net work 24 10. Work determination by mean effective pressure 31 11. Relation of pressure-volume diagrams to indicator cards 34 12. To find the clearance 37 13. Measurement of areas of PV diagrams and indicator cards 43 14. Indicated horse-power 44 15. Effective horse-power, brake horse-power, friction horse-power, mechanical efficiency, efficiency of transmission, thermal efficiency 47 16. Specific displacement, quantity of fluid per hour, or per minute per I.H.P 49 17. Velocity due to free expansion by PV method 62 18- Weight of flow through nozzles by PV method 55 19. Horse-power of nozzles and jets, by PV method 67 Chapter II. Work of Compressors. Horse-power and Capacity of Air, Gas and Vapor Compressors, Blowing Engines and Dry Vacuum Pumps 1. General description of structures and processes 73 2. Standard reference diagrams or PV cycles for compressors and methods of analysis of compressor work and capacity 75 3. Single-stage compressor,, no clearance, isothermal compression. Cycle I. Work, capacity, and work per cubic foot in terms of pressures and volumes 81 i. Single-stage compressor with clearance, isothermal compression, Cycle II. Work, capacity, and work per cubic foot in terms of pressures and volumes 85 5. Single-stage compressor, isothermal compression. Capacity, volumetric effioiency, work, mean effective pressure, hor&e-power and horse-power per cubic foot of substance, in terms of dimensions and cylinder clearance 87 6. Single-«tage compressor, no clearance, exponential compression. Cycle Illi Work, capacity and w<»rk per cubic foot, in terms of pressures and volumes 91 7. Single-stage compressor with clearance, exponential compression, Cycle IV. Work, capacity, and work per cubic foot in terms of pressures and voliunes 96 8. Single-stage compressor, exponential compression. Relation between capacity, volumetric efficiency, work, mean effective pressure, horse-power and horse- power per cubic foot of substance and the dimensions of cylinder and clearance. . 98 ix X CONTENTS PAGE 9. Two-stage oompressor, no clearance, perfect interoooling, exponential compression, best-receiver pressure, equality of stages. Cvcle V. Work and capacity in terms of pressures and volumes 103 10. Two-stage compressor, with clearance, perfect intercooling ex^ < ncntial compres&ion, beet-receiver pressure, equality of stages. Cycle VI. Work and capacity in terms of pressures and volumes 109 11. Two-stage compressor, any receiver pressure, exponential compression. Capacity, volumetric efficiency, work, mean effective pressure and horse-power, in terms of dimensions of cylinders and clearances 113 12. Two-stage compressor, vnlh best^eceiver pressure, exponential compression. Capacity, volumetric efficiency, work, mean effective pressure and horse-power in terms of dimensions of cylinders and clearances 120 13. Three-stage compressor, no clearance, perfect intercooling exponential compree- bion, best two receiver pressiu-es, equality of stages. Cycle VII. Work and capacity, in terms of pressures and volumes 125 14. Three-stage compressor with clearance, perfect intercooling exponential compression, best-receiver pressures, equality of stages. Cycle VIII. Work and capacity in terms of pressures and volumes ICl 15. Three-stage compres^r, any receiver pressure, exponential compression. Capacity, volumetric efficiency, work, mean effective pressure, and horse-power in terms of dimensions of cylinders and clearances 135 16. Three-stage compressor vnlh best-receiver pressures, exponential compression. Capac- ity, volumetric efficiency, work, mean effective pressure and horse-power in terms of dimensions of cylinders and clearances 143 17. Comparative economy or efficiency of compressors 148 18. Conditions of maximum work of compressors 151 19. Compressor characteristics 153 20. Work at partial capacity in compressors of variable capacity 160 21. Graphic solution of compressor problems 168 Chapter III. Work op Piston Engines. Horse-power and Consumption op Piston Engines Using Steam, Compressed Air, or any other Gas or Vapor tJNDER Pressure 1. Action of fluid in single cylinders. General description of structure and processes. . 187 2. Simple engines. Standard reference cycles or PV diagrams for the work of expansive fluids in a single cylinder 192 3. Work of expansive fluid in single cyUnder without clearance. Logarithmic expan- sion. Cycle I. Mean effective pressure, horse-power and consumption of simple engines 197 4. Work of expansive fluid in single cylinder without clearance. Exponential expan- sion, Cycle II. Mean effective pressure, horse-power and consumption of simple engines 205 5. Work of expansive fluid in single cyhnder with clearance. Logarithmic expansion and compression; Cycle III. Mean effective pressure, horse-power, and con- sumption of simple engines 208 6. Work of expansive fluid in single cylinder with clearance; exponential expansion and compression. Cycle IV. Mean effective pressure, horse-power and consumption of simple engines 219 7. Action of fluid in multiple-expansion cylinders. General description of structure ai)d processes 225 8. Standard ref^'ence cycles or PF diagrams for the work of expansive fluids in two- cylinder compound engines 236 t CONTENTS xi PAGB 9. Compound engine with infinite receiver. Logarithnic law. No clearance, Cycle V. General relations between pressures, dimensions, and \york 256 10. Compound engine with infinite receiver. Exponential law. No clearance, Cycle VI. General relations between pressures, dimensions and work 268 11. Compound engine with finite receiver. Logarithmic law. No clearance. Cycle. VII. General relations between dimensions and work when H.P. exhaust and L.P. admission are not coincident 274 12. Compound engine with finite receiver. Exponential law, no clearance. Cycle VIII. General relations between pressures, dimensions, and work, when high pressure Exhaust cjid low-pressure admission are independent 287 13. Compound engine without receiver. Logarithmic law. No clearance. Cycle IX. General relations between dimensions aiid work when high-pressure exhaust and low-pressure admission are coincident 292 14. Compound engine without receiver. Exponential law. No clearance, Cycle X. General relations between dimensions and work when high-pressure exhaust and low-pressure admission are coincident 301 15. Compound engine with infinite receiver. Logarithmic law, with clearance and compression, Cycle XI. General relations between pressures, dimensions and work 306 16. Compound engine with infinite receiver. Exponential law, with clearance and compression, Cycle XII. General relations between pressures, dimensions and work 319 17. Compound engine with finite receiver. Logarithmic law, with clearance and com- pression, Cycle XIII. General relations between pressures, dimensions, and work when H.P. exhaust and L.P. admission are independent 325 18. Compound engine with finite receiver. Exponential law, with ch^rance and com- pression, Cycle XIV. General relations between pressiu-es, dimensions, and work when H.P. exhaust and L.P. admission are independent 335 19. Compound engine without receiver. Logarithmic law, with clearance and com- pression. Cycle XV. General relations between pressures, dimensions, and work when H.P. exhaust and L.P. admission are coincident 339 20. Compound engine without receiver. Exponential law, with clearance and compres- sion, Cycle XVI. General relations between pressures, dimension, and work, when H.P. exhaust and I^.P. admission are coincident 346 21. Triple-expansion engine with infinite receiver. Logarithmic law. No clearance. Cycle XYII. General relations between pressures, dimensions and work 349 22. Multiple-expansion engine. General case. Any relation between cylinders and receiver. Determination of pressure-volume diagram and work, by graphic methods 357 23. Mean effective pressure, engine horse-power, and work distribution and their vari- ation with valve movement and initial pressure. Diagram distortion and diagram factors. Mechanical efficiency 363 24. Consumption of steam engines and its variation with valve movement and initial pressure. Best cut-off as affected by condensation and leakage 371 25. Variation of steam consmnption with engine load. The Willans hne. Most eco- nomical load for more than one engine and best load division 381 26. Graphical solution of problems on engine horse-power and cylinder sizes 387 3di CONTENTS Chapter IV. Heat and Matteb. Qualitatzve and Quantitative Relations BETWEEN Heat Content of Substances and Phtsical-Chemical State FAQS 1. Substances and heat effects important in engineering ■. . 398 2. Classification of heating processes. Heat addition and abstraction with or without temperature change, qualitative relations 401 3. Thermometry based on temperature change, heat effects. Thermometer and abso- lute temperature scales 407 4. Caiorimetry based on proportionality of heat effects to heat quantity. Units of heat and mechanical equivalent 415 5. Temperatiu^ change relation to amount of heat for solids, liquids, gases and vapors not changing state. Specific heats 419 6. Volume or density variation with temperature of solids, liquids, gases and vapors not changing state. Coefficients of expansion. Coefficients of pressure change for gases and vapors 435 7. Pressure, volume, and temperature relations for gases. Perfect and real gases 438 8. Gas density and specific volume and its relation to molecular weight and gas constant . 446 9. Pressure and temperature relations for vapor of liquids or solids. Vaporization, sublimation and fusion curves. Boiling- and freezing-points for pure liquids and dilute solutions. Saturated and superheated vapors 451 10. Change of state with amount of heat at constant temperature. Latent heats of fusion and vaporization. Total heats of vapors. Relation of specific volume of liquid and of vapor to the latent heat 467 11. Gas and vapor mixtures. Partial and total gas and vapor pressures. Volume, weight and gas constant relations. Saturated mixtures. Humidity 481 12. Absorption of gases by liquids and by solids. Relative volumes and weights with pressure and temperature. Heats of absorption and of dilution. Properties of aqua ammonia 493 13. Combustion and related reactions. Relative weights and volumes of substances and elements before and after reaction 506 14. Heats of reaction. Calorific power of combustible elements and of simple chemical compounds. B.T.U. per pound and per cubic foot 516 15. Heat transmission processes. Factors of internal conduction, surface resietanoc, radiation and convection 528 16. Heat transmission between separated fluids. Mean temperature differences, coeffi- cients of transmission , 538 17. Variation in coefficient of heat transmission due to kind of substance, character of separating wall and conditions of flow 555 Chapter V. Heating by Combustion. Fuels, Furnaces, Gas Producers and Steam Boilers 1. Origin of heat and transformation to useful form. Complexity of fuels as sources of heat. General classification, solid, liquid, gaseous, natural and artificial 644 2. Natural solid fuels, wood peat, lignite, bituminous and anthracite coals. Chemical and physical properties. Classifications based on ultimate and proximate analysis and on behavior on heating 640 3. Calorific power of coals and the combustible of coals. Calculation of calorific power from ultimate and proximate analyses. Calorific power of the volatile 662 4. Mineral oil and natural gas fuel. Chemical and physical properties. Calorific power direct and as calculated for oils from ultimate analysis or from density, and for gas from sum of constituent gases 670 CONTENTS xiii PAGB 5. Charoo&l, coke, ooke oven and retort ooal gas as products of heating wood and coal. Chemical, phjrsical, and calorific properties per pound. Calorific power of gases per cubic foot in terms of constituent gases. Yield of gas and coke per pound coal 676 6. Distillate oils, kerosene and gasoline, residue oils, and oil gas as products of heating mineral oils. Chemical, physical, and calorific properties. Calorific power of frac- tionated oils in terms of, (a) carbon and hydrogen; (6) density per pound, and estimated value per cubic foot vapor. Calorific power of oil gas per poimd and per cubic foot in terms of constituent gases. Yield of distillates and oil gas 685 7. Gasification of fixed carbon and coke by air-blast reactions, producing air gas, and blast-furnace gas. Comparative yield per pound coke and air. Sensible heat and heat of combustion of gas. Relation of constituents in gas. Efficiency of gasification 695 8. Gasification of fixed carbon, coke, and coal previously heated, by steam-blast reac- tions, producing water gas. Composition and relation of constituents of water gas, yield per pound of steam and coal. Heat of combustion of gas and limitation of yield by negative heat of reaction 710 9. Gasification of coals by steam and air blasts, resulting in producer gas. Composition and relation of constituents of producer gas, yield per pound of fixed carbon, air and steam. Modification of composition by addition of volatile of coal. Heat of combustion of gas, sensible heat, and efficiency of gasification. Horse-power of gas producers 719 10. Combustion effects. Final temperature, volume and pressure for explosive and non-explosive combustion. Estimation of air weights and heat suppression due to CO in products from volumetric analysis 740 11. Temperature of ignition and its variation with conditions. Limits of proportion air gas neutral, or detonating gas and neutral, for explosive combustion of mix- tures. Limits of adiabatic compression for self-ignition of mixtures 758 12. Rate of combustion of solid fuels with draft. Propagation rates, uniform and detonating for explosive gaseous mixtures 765 13. Steam4x>iler evaporative capacity and horse-power. Horse-power imits, equivalent rates of evaporation and of heat absorption. Factors of evaporation. Relation between absorption rates and rates of heat generation. Lifluence of heating and grate surface, calorific power of fuels, rates of combustion and furnace losses 773 14. Steam-boiler efficiency, furnace and heating-surface efficiency. Heat balances and variation in heat distribution. Evaporation and losses per pound of fuel 796 Chafteb VI. Heat and Work. General Relations between Heat and Work. Thermal Efficibnct of Steam, Gab, and Compressed-Air Engines. Flow OF Expansive Fluids. Performance of Mechanical Refrigerating Systems 1. General heat and work relations. Thermal cycles. Work and efficiency deter- mination by heat differences and ratios. Graphic method of temperature entropy heat diagram 874 2. General energy equation between heat change, intrinsic energy change, and work done. Derived relations between physical constants for gases and for changes of state, solid to Uquid, and liquid to vapor 882 3. Quantitative relations for primary thermal phases, algebraic, and graphic to PF, and T4 coordinates. Constancy of PV, and T for gases and vapors, wet, dry and superheated 892 4. Quantitative relations for secondary thermal phases. Adiabatics for gases and vapors. Constant quality, constant total heat, and logarithmic expansion Unes for steam 904 xiv CONTENTS PAQB 6. Thermal cycles representative of heat-engine iprocesses. Cyclic efficiency. A reference standard for engines and fuel-burning power systems. Classification of steam cycles 927 6. The Rankine cycle. Work, mean effective pressure, jet velocity, water rate, heat consumption and efficiency of steam Cycle I. Adiabatio expansion, constant pressure, heat addition and abstraction, no compression 936 7. Carnot cycle and derivatives. Work, mean effective pressure, water rate, heat consumption and efficiency of steam Cycle II. Adiabatic expansion and com- pression, constant pressure heat addition and abstraction 957 8. Gas cycles representative of ideal processes and standards of reference for gas engines 970 9. Brown, Lenoir, Otto and Langen non-compression gas-cycles. Work, mean effective pressure, volume and pressure ranges, efficiency, heat and gas con- sumption 978 10. Stirling and Ericsson gas cycles. Work, efficiency and derived quantities for isothermal compression with and without regenerators 993 11. Otto, Atkinson, Brayton, Diesel, and Carnot gas cycles. Work, efficiency and derived quantities for adiabatic compression gas cycles 1006 12. Comparison of steam and gas cycles with the Rankine as standard for steam, and with the Otto and Diesel, as standard for gas. Relations of Otto and Diesel to Rankine cycle. Conditions for equal efficiency 1031 13. Gas cycle performance as affected by variability of the specific heats of gases, apphed to the Otto cycle 1035 14. Actual performance of Otto and Diesel gas engines, and its relation to the c^'clic. Diagram factors for mean effective pressure and thermal efficiency. Effect of load on efficiency. Heat balance of gas engines alone an^ with gas producers 1042 15. Actual performance of piston steam engines and steam turbines at their best load and its relation to the cyclic. Effect on efficiency of initial pressure, vacuum, superheat, jacketing, and reheating. Heat balances of steam power plants .... 1062 16. Flow of hot water, steam and gases through orifices and nozzles. Velocity, weight per second, kinetic energy, and force of reaction of jets. Nozzle friction and reheating and coefficient of efflux. Relative proportions of series nozzles for turbines for proper division of work of expansion 1083 17. Flow of expansive fluids under small pressure drops through orifices, valves, and Venturi tubes. Relation between loss of pressure and flow. Velocity heads and quantity of flow by Pitot tubes 1097 18. Flow of gases and vapors in pipes, flues, ducts, and chimneys. Relation between quantity of flow and loss of pressure. Friction resistances. Draught and capacity of chimneys 1111 19. Thermal efficiency of compressed-air engines alone and in combination with air compressors. Effect of preheating and reheating. Compressor suction heating, and volumetric efficiency. Wall action 1127 20. Mechanical refrigeration, general description of processes and structures. Thermal cycles and refrigerating fluids. Limiting temperatures and pressures 1142 21. Performance of mechanical refrigerating cycles and systems. Quantity of fluid circulated per minute per ton refrigeration, horse-power, and heat supplied per ton. Refrigeration per unit of work done and its relation to thermal effi- ciency of the system 1157 LIST OF TABLES Ma FAQB 1. Ck>nyer8ion table of units of distance 62 2. Conversion table of units of surface 62 3. Conversion table of units of volume 62 4 Conversion table of units of weights and force 63 5. Conversion table of units of pressure 63 6. Conversion table of units of work 64 7. Conversion table of units of power 64 8. Units of velocity 64 9. Barometric heights, altitudes and pressures 65 10. Values of 8 in the equation PV'» constant for various substances and conditions . . 67 11. Horse-power per pound mean effective pressure 68 12. Ratio of cut-offs in the two cylinders of the compound engine to give equal work for any receiver volume 284 13. Piston positions for any crank angle 395 14. Values for z for use in Heck's formula for missing water 396 15. Some actual steam engine dimensions 396 16. Fixed tempcrattu'es 411 17. Fahrenheit temperatures by hydrogen and mercury thermometers 414 18. Freezing-point of calcium chloride brine 425 19. Specific heat of sodium chloride brine 427 20. Specific heat and gas constants, 431 21. The critical point 453 22. Juhlin's data on the vapor pressure of ice 456 23. Tamman's value on fusion pressure and temperature of water-ice 456 24. Lowering of freezing-points 465 25. Berthelot's data on heat for complete dilution of ammonia solutions 500 26. Air required for combustion of various substances 515 27. Badiation coefficients 535 28. Coefficients of heat transfer 550 29. Temperatures, Centigrade and Fahrenheit 571 30. Heat and power conversion table 573 31. Specific heat of solids 574 32. Specific heats of liquids 576 33. Baum6-specific gravity scale 577 34. Specific heats of gases 578 35. Coefficient of linear expansion of solids 580 36. Coefficient of cubical expansion of solids 581 37. Coefficient of volimietric expansion of gases and vapors at constant pressure 582 XV xvi LIST OF TABLES wo, PAQB 38. Coefficient of pressure rise of gases and vapors at constant volume 683 39. Compressibility of gases by their isothermals 584 40. Values of the gas constant R 684 41. Density of gases 685 42. International atomic weights 686 43. Melting- or freezing-points 586 44. Boiling-points 688 45. Latent heats of vaporization 690 46. Latent heats of fusion 691 47. Properties of saturate steam 692 48. Properties of superheated steam 596 49. Properties of saturated ammonia vapor 603 60. Properties of saturated carbon dioxide vapor 618 51. Relation between pressure, temperature and per cent NHj in solution 628 62. Values of partial pressure of ammonia and water vapors for various temperatures and per cents of ammonia in solution 632 63. Absorption of gases by liquids 634 64. Absorption of air in water 635 66. Heats of combustion of fuel elements and chemical compounds 636 66. Internal thermal conductivity 639 57. Relative thermal conductivity 642 58. General classification of fuels 648 69. Comparison of cellulose and average wood compositions 650 60. Classification of coals by composition 652 61. Classification of coals by g&s and coke qualities 654 62. Composition of peats 655 63. Composition of Austrian lignites 666 64. Composition of English coking coals 658 65. Wilkesbarre anthracite coal sizes and average ash content 659 66. Density and calorific power of natural gas 673 67. Products of wood distillation 676 68. Products of peat distillation 678 69. Products of bituminous coal distillation 680 70. Gas yield of English cannel coals 682 71. Comparison of coke oven and retort coal gas 682 72. Relation between oxygen in coal and hydrocarbon in gas 684 73. Density and calorific power of coke oven gas 684 74. Average distillation products of crude mineral oils 686 76. American mineral oil products 687 76. U.S. gasolene and kerosene bearing crude oils 688 77. Calorific power of gasolenes and kerosenes 691 78. Properties of oil-gas 693 79. Yield of retort oil gas 694 80. Density and calorific power of oil gas 694 81. Boudouard's equihbrium relations for. CO and COt with temperature 697 82. Change of Oj in air to CO and CO2 at 1472*^ F 699 83. Composition of hypothetical air gas, general 704 84. Composition of h>TX)thetical air gas, no CO2 and no CO 705 85. Density and calorific power of blast furnace gas 708 86. Water gas characteristics with bed temperature 710 87. Composition of hypothetical water gas, general 714 88. Composition of hypothetical water gas, no COj and no CO 715 LIST OF TABLES xvii KO. PAOB 89. Density and calorific power of water gas 718 90. Composition of hypothetical producer gas from fixed carbon 725 91. Density and calorific power of producer gas 737 92. Characteristics of explosive mixtures of oil gas and air 747 93. Calculated ignition temperatures for producer gas 761 94. Compressions commonly used in gas engines 762 95. Ignition temperatures : 763 96. Variation of ignition temperature of charcoal with distillation temperature 763 97. Per cent detonating mixture at explosive limits of proportion 764 98. Velocity of detonating or explosive waves » . . . 766 99. Rate of propagation, feet per second, for different proportions in gaseous mixtures . 768 100. Rates of combustion for coal 769 101. Constants of proportion for rate of coal combustion for use in Eq. (848) 771 102. Boiler efficiency sunmiaries 799 103. Three examples of heat balance for boilers 800 104. Composition and calorific power of characteristic coals. 818 105. Combustible and volatile of coals, lignites and peats 826 106. Paraffines from Pennsylvania petroleums 835 107. Calorific power of mineral oils by calorimeter and calcidation by density formula of Sherman and Kropff 836 108. Properties of mineral oils 838 109. Composition of natural gases 841 110. Composition of coke oven and retort coal gas 842 111. Composition of U. S. coke 846 112. Fractionation tests of kerosenes and petroleums 847 113. Fractionation tests of gasolenes 851 1 14. Composition of blast-furnace gas and air gas 853 115. Rate of formation of CO from COi and carbon 855 116. Composition of water gas 857 117. Composition of producer gas 858 118. Gas producer tests 864 119. Composition of oil producer gas 866 120. Composition of powdered coal producer gas 866 121. Calorific powers of best air-gas mixtures 867 122. Composition of boiler-fiue gases 868 123. Limits of proportions of explosive air-gas mixtures 869 124. Rate of combustion of coal with draft 870 125. Rate of combustion of coal 871 126. Values of 8 for adiabatic expansion of steam 912 127. Values of s for adiabatic expansion of steam determined from initial and final volumes only 913 128 1042 129. Diagram factors for Otto cycle gas engines 1046 130. Mechanical efficiencies of gas engines 1050 131. Allowable compression for gas engines 1050 132. Mean effective pressure factors for Otto cycle engines 1053 133. GQldner's values of Otto engine real volumetric efficiency with estimated mean suction resistances 1055 134. Comparative heat balances of gas producer and engine plants 1057 135. Heat balances of gas producer plants 1060 136. Heat balances of gas and oil engines 1060 137. Steam plant heat balances 1063 xvm LIST OF TABLES NO. PAOB 138. Efficiency factors for reciprocating Bteam engines and turbines 1064 130. Steam turbine efficiency and efficiency factors with varying vacuum and with steam approximately at constant initial pressure 1071 140. Efficiency factors for low-pressure steam in piston engines 1074 141. Ck)efficient of discharge for various air pressure and diameters of orifice (Durley). 1101 142. Values of C for air flow (Weisbach) 1101 143. Flow* change resistance factors Fr (Reitschel) 1121 TABLE OF SYMBOLS A »area in square feet. » constant, in formula for most economical load of a steam engine. Chapter III. = constant, in pipe flow formula, Chapter VI. » excess air per pound of coal, Chapter V. B pounds of ammonia dissolved per pound of weak liquor, Chapter IV. a » area in square inches. » coefficient of linear expansion. Chapter IV. » constant in equation for the ratio of cylinder sizes for equal work distribution in com- pound engine, Chapter III. = constant in equation for change in intrinsic energy. Chapter VI. -constant in equation for specific heat. Chapter IV. = cubic feet of air per cubic foot of gas in explosive mixtures, Chapter V. = effective area of piston, square inches. Chapter I. B« constant in equation for the most economical load of the steam engine. Chapter III. » constant in equation for flow in pipes, Chapter VI. Bd. »Baum6. B.H.P. = brake horse-power, Chapters III and VI. = boiler horse-power, Chapter V. B.T.U.= British thermal unit. bs constant in equation for change in intrinsic energy, Chapter VI. « constant in equation for specific heat. Chapter IV. (bk.pr.) =back pressure in pounds per square inch. C= Centigrade. « circumference or perimeter of ducts in equations for flow. Chapter VI. » constant. =heat suppression factor. Chapter V. s ratio of pressure after compression to that before compression in gas engine cycles, Chapter VI. = specific heat, Chapter IV. Cc» per cent of ammonia in weak hquor. Chapter VI. Cy= specific heat at constant pressure. Cfi=per cent of ammonia in rich liquor. Chapter VI. C,«q>ecific heat of water. Chapter VI. CffS specific heat at constant volume. C<= clearance expressed in cubic feet. c» clearance expressed as a fraction of the displacement -constant. cu.ft. s cubic foot. cu.in.s cubic inch. i)= constant in equations for pipe flow, Chapter VI. =» density, Chapter IV. = diameter of pipe in feet, Chapter VI. « displacement in cubic feet. Z)f» specific displacement, Chapter I. xix XX TABLE OF SYMBOLS d«oongtant in equation for change in intrinsic energy, Chapter VI. a diameter of a cylinder in inches, Chapter I. s diameter of pipe in inches, Chapter VI. » differential. (del.pr.)-> delivery pressure in pounds per square inch, Chapter II. ^B constant in equation for pipe flow, Chapter VI. » external latent heat, Chapter IV. s thermal efficiency. Chapter VI. ^ir— thermal efficiency referred to brake horse-power, Chapter III. ^ft= boiler efficiency, Chapter V. £/= furnace efficiency, Chapter V. £/= thermal efficiency referred to indicated horse-power. Chapter III. ^ms mechanical efficiency, Chapter III. ^f= heating surface efficiency. Chapter V. ^r^ volumetric efficiency (apparent), Chapter VI. Ev'^ volumetric efficiency (true), Chapter VI. e »as a subscript to log to designate base e. s constant in equation for change in intrinsic energy. Chapter VI. ei » ratio of true volumetric efficiency to hypothetical. Chapter II. ei"B ratio of true volumetric efficiency to apparent, Chapter III. eis ratio of true indicated horse-power to hypothetical, Chapter U. F-> constant in equation for pipe flow, Chapter VI. "diagram factor for gas engine indicator cards, Chapter VI. "Fahrenheit, "force in pounds. Ff" friction factor, F^X velocity head "loss due to friction, Chapter VI. -Fa "resistance factor, F«X velocity head "loss due to resistances, ChapteT VI. Ffi "Special resistances to flow in equations for chimney draft, Chapter VI. /"Constant in equation for changes in intrinsic energy. Chapter VI. "function, ft. "foot, ft.-lb. "foot-pound. G" constant in equation for pipe flow, Chapter VI. "Weight of gas per hour in equation for chimney flow, Chapter VI. Crm^niaximum weight of gases in equation for chimney flow. Chapter VI. G. S. "grate surface. y" acceleration due to gravity, 32.2 (approx.) feet per second, per second. fl"" as a subscript to denote high pressure cylinder, "heat per pound of dry saturated vapor above 32® F. "heat per cubic foot gas. "heat transmitted. Chapter IV. "height of column of hot gases in feet, Chapter VI. "pressure or head in feet of fluid, Chapter VI. Ha "difference in pressure on two sides of an orifice in feet of air, Chapter VI. i^o** equivalent head of hot gases, Chapter VI. Hm = pressure in feet of mercury, Chapter VI. Hit "pressure in feet of water, Chapter VI. H.P. "high pressure. "horse-power. Chapter I. TABLE OF SYMBOLS xxi H.S. "'heating surface. (H.P.cap.) i^high pressure cylinder capacity, Chapter III. h »heat of superheat. Hm » difference in pressure on two sides of an orifice in inches of mercury, Chapter VI. Afps difference in pressure on two sides of an orifice in inches of water, Chapter VI. /»as a subscript to denote intermediate cylinder, Chapter III. I. H. P. s indicated horse-power. in. » inch, (in. pr.)» initial pressure in pounds per square inch. J s Joule's equivalent » 778 (approx.) foot-pounds per B.T.U. X» coefficient of thermal conductivity. Chapter IV. » constant, s proportionality coefficient in equation for draft. Chapter VI. /r«» engine constant » ^^^ in expression for horse-power. Chapter IlL If a as a subscript to denote low-pressure cylinder, a distance in feet, —latent heat, Chapters IV and VI. —length of stroke in feet. Chapter I. L —per cent of heat in fuel lost in furnace. Chapter V. L.P.— low pressure. (L.P.Cap.)— low-pressure capacity, Chapter II. Z— constant, Chapter III. = length. Chapter IV. lb. —pound. log— logarithm to the base 10. logs —logarithm to the base e. Af — mass. (M.E.P.) —mean effective pressure, pounds per square foot. • m— constant, Chapter III. area —mean hydraulic radius = — : . perimeter —molecular weight, Chapter IV. —ratio of initial pressure to that end of expansion in Otto and Langen gas cycle. Chapter VI. (m.b.p.) —mean back pressure in pounds per square inch, (m.e.p.) —mean effective pressure in pounds per square inch. (m.f.p.) —mean forward pressure in pounds per square inch. .Vs constant, Chapter III. —revolutions per minute. n— cycles per minute. —constant, Chapter III. —cubic foot of neutral per cubic foot of gaseous mixture. Chapter V. ' —number of degrees exposed on thermometer stem, Chapter IV. —ratio of volume after expansion to volume before in Atkinson gas cycle, Chapter IV. dfic volume of dry saturated steam, Chapter VI. xxii TABLE OF SYMBOLS = volume of receiver of compound engine in cubic feet, Chapter III. P= draft in pounds per square foot, Chapter VL =load in kilowatts, Chapter III. = pressure in pounds per square foot. Pf= static pressure in poimds per square foot lost in wall friction, Chapter VT. Pa » static pressure in pounds per square foot lost in changes of cross-section, etc., Chapter VI. Pv —velocity head in pounds per square foot. p= pressure in pounds per square inch. Pe==mean exhaust pressure. Chapter VI. Pf =>mean suction pressure, Chapter VI. Pv = partial pressure of water vapor in air. Chapter VI. Q= quantity of heat or energy in B.T.U. gained or lost by a body passing from one state to another. Qi'— heat added from fire in Stirling and Ericsson cycles, Chapter VI. Qi" =heat added from regenerator in Stirling and Ericsson cycles, Chapter VI. Qt'»heat abstracted by water jacket in Stirling and Ericsson cycles, Chapter VI. Qt"— heat abstracted by regenerator in Stirling and Ericsson cycles, Chapter VI. 9 » quantity of heat per pound of liquid above 32° F. 72= ratio of heating surface to grate surface, Chapter V. * «gas constant. ^72(7= ratio of cylinder sizes in twoHStage air compressor or compound engine. Chapters II and III. /2j7= ratio of expansion in high-pressure cylinder, Chapter III. 122;= ratio of expansion in low-pressure cylinder, Chapter III. Rp^T&iio of initial to back pressure. Chapters III and VI. /2pB ratio of delivery to supply pressure. Chapter II. Rv » ratio of larger volume to smaller volume. r =rate of flame propagation in explosive mixtures, Chapter V. rp« pressure differences (maximum— minimmn) in gas cycles, Chapter VI. fK=volume differences (maximum — minimum) in gas cycles, Chapter VL (rec.pr.) = receiver pressure in pounds per square inch, Chapter IIL (rel.pr.) —release pressure in pounds per square inch. Chapter III. « iS=per cent of ammonia in solution. Chapter IV. «: piston speed. Chapter I. » pounds of steam per pound of air in producer blast, Chapter V. = specific heat. Chapter IV. » specific heat of superheated steam. Chapter VI. (Sup.Vol.) = volume of steam supplied to the cylinder per stroke. Chapter III. 8= general exponent of F in expansion or compression of gases. sp.gr. = specific gravity, sp.ht. = specific heat, sq.ft. = square foot, sq.in. ^square inch, (sup.pr.) = supply pressure, in pounds per square inch. r=* temperature, degrees absolute. Tc— temperature of air. Chapter VI. TiT™ temperature of gases in chinmey, Chapter VL ^""temperature in degrees scale. TABLE OF SYMBOLS xxiii (/»rate of heat transfer in B.T.U. per square foot per hour per degree difference in tem- perature, Chapter IV. IT'S intrinsic energy. Chapter VI. ii= velocity in feet per second. UiB» velocity in feet per minute, Chapter VI. r= volume in cubic feet. Va. =» cubic feet per pound air, Chapter VI. V(7= cubic feet per pound, gas, Chapter VI. Yl —volume of Uquid in cubic feet per pound. Vs = volume of solid in cubic feet per pound. Vy = volume of vapor in cubic feet per pound. V =«volume, Chapter IV. IT = work in foot-pounds. W.R. s= water rate. tr= pounds of water per pound of ammonia in solution, Chapter IV. » weight in pounds, u^js ^pounds of rich Uquor per pound of ammonia, Chapter VI. A' = compression in the steam engine as a fraction of the stroke, Chapter III. - , heat added = 1-1 ; , Chapter VI. temperature at beginning of addition X specific heat at constant volume x» constant in the expression for missing water, Chapter III. = fraction of liquid made from solid or vapor made from liquid, Chapter VI. =per cent of carbon burned to CO2, Chapter V. =per cent of nozzle reheat. Chapter VI. =per cent of steam remaining in high-pressure cylinder of compound engine at any point of the exhaust stroke, Chapter III. —quantity of heat added in generator of absorption system in addition to the amount of heat of absorption of 1 lb. of ammonia. Chapter VI. -ratio of low-pressure admission voliune to high-pressure admission voliune, Chapter III. K = total steam used per hour by an engine, Chapter III. heat added «. „, = 1 + — — , Chapter VI. temperature at beginning of addition X specific heat at constant pressure 1/ =per cent of vane reheat Chapter VI. » ratio of the volume of receiver to that of the high-pressure cylinder of the compound engine, Chapter III. Zs fraction of the stroke of the steam engine completed at cut-off. Chapter III. - . heat added from regenerator ^, ^ ,., = 1H ~ ; , Chapter VI. temperature at beginning of addition X specific heat at constant volume Z' « hypothetical best value of Z, - . heat added from regenerator _, ^ _._. =s 1 -I — ; — ; ZB ^ Chapter VI. temperature at begmnmg of addition X specific heat at constant pressure ;B= ratio of R.P.M. to cycles per minute. as an angle, Chapter I. =ooefficient of cubical expansion. Chapter III. BCODstaot in the equation for latent heat, Chapter VI. xxiv TABLE OF SYMBOLS B constant in equation for variable specific heat at constant volume, Chapter VL oe'= constant in equation fOr variable specific heat at constant pressure, Chapter VI. ^« constant in equation for latent heat, Chapter VI. » fraction of fuel heat available for raising temperature, Chapter V. Y=> constant in equation for latent heat. Chapter VI. » ratio of crossHsection to perimeter. Chapter IV. ... sp. ht. at const, press. «= special value for « for adiabatic expansion or compression » ; sp. ht. at const, vol. y's ratio of specific heat at constant pressure to specific heat at constant volume when each is a variable, Chapter VI. A B increment. 8 » density in pounds per cubic foot, dcf— density in cold gases in equations for chimney draft, Chapter VI. 8/r» density of hot gases in equations for chimney draft. Chapter VI. t^» coefficient of friction, Chapter VI. (Immaterial coefficient in heat transfer expression, Chapter IV. p» internal thermal resistance, Chapter IV. Z» summation. o» surface thermal resistance, Chapter IV. T— time in seconds. *= entropy. Chapter VI. <!>= entropy. Chapter VI. Note. A small letter when used as a subscript to a capital in general refers to a point on a diagram, e.g.. Pa designates pressure at the point A. Two small letters used as sub- scripts together, refer in general to a quantity between two points, e.g., Wab designates work done from point A to point B. ENGINEERING THERMODYNAMICS CHAPTER I WORK AND POWER. GENERAL PRINCIPLES. 1. Work Defined. Work, in the popular sense of performance of any labor, is not a sufficiently precise term for use in computations, but the analytical mechanics has given a technical meaning to the word which is definite and which is adopted in all thermodynamic analysis. The mechanical definition of work is mathematical inasmuch as work is always a product of forces opposing motion and distance swept through, the force entering with the product being limited to that acting in the direction of the motion. The unit of distance in the English system is the foot, and of force the pound, so that the imit of work is the foot-pound. In the metric system the distance unit is the meter and the force unit the kilogramme, making the work unit the kilogrammeter. Thus, the lifting of one pound weight one foot requires the expenditure of one foot- pound of work, and the falling of one pound through one foot will perform one foot-pound of work. It is not only by lifting and falling weights that work is expended or done; for if any piece of mechanism be moved through a distance of one foot, whether in a straight or curved path, and its movement be resisted by a force of one pound, there will be performed one foot-pound of work against the resistance. It is frequently necessary to transform work from one sys- tem of units to the other, in which case the factors given at the end of this Chapter are useful. Work is used in the negative as well as in the positive sense, as the force considered resists or produces the motion, and there may be both positive and negative work done at the same time; similar distinctions may be drawn with reference to the place or location of the point of application of the force. Con- sider, for example, the piston rod of a direct-acting pump in which a certain force acting on the steam end causes motion against some less or equal force acting at the water end. Then the work at the steam end of the pump may be considered to be positive and at the water end negative, so far as the move- ment of the rod is concerned; when, however, this same movement causes a movement of the water, work done at the water end (although negative with reference to the rod motion^ since it opposes that motion) is positive with refer- 2 ENGINEERINa THERMODYNAMICS ence to the water, since it causes this motion. It may also be said that the steam does work on the steam end of the rod and the water end of the rod does work on the water, so that one end receives and the other delivers work, the rod acting as a transmitter or that the work performed at the steam end is the input and that at the water end the output work. (See the end of Chapter I for Tables I, II, III, IV, and VI, Units of Distance, of Surface, of Volimie, of Weight and Force, and of Work.) Example. An elevator weighing 2000 lbs. is raised 80 ft. How much work is done in foot-pounds? Foot-pounds »force xdistance =2000 X80 = 160,000 ft-lbs. Ans. 160,000 ft.-lbs. Prob. 1. A pump lifts 150 gallons of water to a height of 250 ft. How much woikdoes it do? ^ Prob. 2. By means of a jack a piece of machinery weigjiing 10 tons is raised f in. What is the work done? Prob. 3. A rifle bullet weighing 2 oz. travels vertically upward l{ miles. What work was done in foot-pounds? Prob. 4. A cubic foot of water falls 50 ft. in reaching a water-wheeL How much work can it do? Prob. 5. A piston of an elevator is 12 ins. in diameter and has acting on it a pressure of 80 lbs. per square inch. What work is done per foot of travel? Prob. 6. It has been found that a horse can exert 75 lbs. pull when going 7 miles per hour. How much work can be done per minute? ^^ Prob. 7. How much work is done by an en^e which raises a 10-ton casting 50 ft.? Prob. 8. The pressure of the air on front of a train is 50 lbs. per square foot when the speed is 50 miles per hour. If the train presents an area of 50 sq.ft., \(4iat work is done in overcoming wind resistance? ^ * . .wv • v • Prob. 9. The pressure in a 10-inch gun diiring ike time of firing, is 2000 lbs. per square inch. How much work is done in ejecting the projectile if the gun is 33 ft. 4 ins. long? 2. Power Defined. Power is defined as the rate of working or the work done in a given time interval, thus introducing a third unit of mechanics, time, so that power will always be expressed as a quotient, the niunerator being a prod- uct of force and distance, and the denominator time. This is in opposition to the popular use of the word, which is very hazy, but is most often applied to the capability of performing much work] or the exertion of great force, thus, popularly, a powerful man is one who is strong, but in the technical sense a man would be powerful only when he could do much work continuously and rapidly. An engine has large power when it can perform against resistance many foot- pounds per minute. The unit of power in the English system is the horse-power, or the performance of 550 foot-pounds per second or 33,000 foot-poimds per minute, or 1,980,000 foot-pounds per hour. In the metric system the horse- power is termed cheval-vapeur, and is the performance of 75 killogrammeters =542i foot-pounds per second, or 4500 kilograjnmeters= 32,549 foot-pounds WORK AND POWER. 3 per minute, or 270,000 kilograinmeters= 1,952,932 foot-pounds per hour. Table VII at the end of Chapter I gives conversion factors for power units. Example. The piston of a steam engine travels 600 ft. per minute and the mean force of steam acting upon it is 65,000 lbs. What is the horse-power? TT foot-pounds per minute Horse-power '^ 33 qg tune " 33,000 ^65,000X60^ 33,000 Prob. 1. The draw-bar pull of a locomotive is 3000 lbs. when the train is traveling 50 miles per hour. What horse-power is being developed? Prob. 2. Amine cage weighing 2 tons is lifted up a 2000-ft. shaft in 40 seconds. What horse-power will be required if the weight of the cable is neglected? Prob. 3. By direct pull on a cable it is found possible to lift 4 tons 20 ft. per second. With a diffoential pulley 40 tons may be lifted 3 ft. per second. What is the difference in power required? Prob. 4. A horse exerts a pull of 100 lbs. on a load. How fast must the load be moved to develop one horse-power? Prob. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 lbs. What horse-power must be available to maintain this speed? (One knot is a speed of one nautical mile per hour.) /^ . * ; - '' jju „ Ptob. 6. It is estimated that 100,000 cu. ft. of water gp over a fall 60 ft. hi^ every second. What horse-power is going to waste? Prob. 7. The force acting on a piston of a pump is 80,000 lbs. If the piston speed is 150 ft. per minute, what is the horse-power? Prob. 8. To draw a set of plows 2i miles per hour requires a draw-bar pull of 10,000 lbs. What must be the horse-power of a tractor to accomplish this? Prob. 9. The horse-power to draw a car up a grade is the sum of the power necessary to pull it on a level and that necessary to lift it vertically the same number of feet as it rises on the grade. What will be the horse-power required to draw a car 20 miles per hour up a 12 per cent grade if the car weighs 2500 lbs. and the draw-bar pull on the level is 250 lbs.? 3. Work in Terms of Pressure and Volume. Another of the definitions of mechanics fixes pressure as force per unit area so that pressure is always a quotient, the numerator being force and the denominator area, or length to the second jjower. If, therefore, the pressure of a fluid be known, and accord- ing to hydromechanics it acts equally and normally over all surface in contact with it, then the force acting in a given direction against any surface will be the product of the pressure and the projected area of the surface, the projection being on a plane at right angles to the direction considered. In the case of pis- tons and plungers the line of direction is the axis of the cylinder, and the pro- jected area is the area of the piston less the area of any rod passing completely through the fluid that may be so placed. When this plane area moves in a 4 ENGINEERING THERM0t>YNAMIC8 (fireclion perpendicular to itself, the product of its ai^ and the distance will be the volume swept through, and if a piston be involv^ the volume is technically the displacement of the piston. Accordingly, work may be expressed in three ways, as follows: Work = force X distance ; Work = pressure X area X distance ; Work = pressure X volume. The product should always be in foot-pounds, but will be, only when appro- priate units are chosen for the factors. These necessary factors are given as ffdOows: , W<xk in foot-pounds = force in lbs. X distance in ft. = pressure in lbs. per sq.ft. X area in sq.ft. X distance in ft = pressure in lbs. per sq.ui. X area in sq.in. X distance in ft Wv, , ^ =pressure in lbs, per sq.ft. X volume in cu.ft. * -«*^ r= pressure in lbs. per sq.in. X 144 X volume in cu.f t. As pressures are in practice expressed in terms not only as above, but alsa in heights of columns of common fluids and in atmospheres, both in English and metric systems, it is convenient for calculation to set down factors of equivalence as in Table V, at the end of the Chapter. In thermodynamic computations the pressure volume product as an expres- sion for work is most useful, as the substances used are always vapors and gases, which, as will be explained later in more detail, have the valuable property of changing volume indefinitely with or without change of pressure according to the mode of treatment. Every such increase of volmne gives, as a conse- quence, some work, since the pressure never reaches zero, so that to derive work from vapors and gases they are treated in such a way as will allow them to change volmne considerably with as much pressure acting as possible. It should be noted that true pressures are always absolute, that is, measured above a perfect vacuum or counted from zero, while most pressure gages and other devices for measuring pressure, such as indicators, give results measured above or below atmospheric pressure, or as commonly stated, above or below atmosphere. In all problems involving work of gases and vapors, the absolute values of the pressures must be used; hence, if a gage or indicator measure- ment is being considered, the pressure of the atmosphere found by means of the barometer must be added to the pressure above atmosphere in order to obtain the absolute or true pressures. When the pressures are below atmosphere the combination with the barometric reading will depend on the record. If a record be taken by an indicator it will be in pounds per square inch below atmosphere and must be subtracted from the barometric equivalent in the same units to give the absolute pressure in pounds per square inch. When, however, a vacuum gage reads in inches of mercury below atmosphere, as such gages do, the difference between its reading and the barometric gives the absolute WORK AND POWER. 5> pressure 5n inches of mercury directly, which can be converted to the desiredl units by the proper factors. While it is true that the barometer is continually fluctuating at every place, it frequently happens that standards for various altitudes enter into calculations, and to facUitate such work, values are given for the standard barometer at various aU,itudes with equivalent pressures in pounds per square inch in Table IX. Frequently in practice, pressures are given without a definite statement of what units are used. Such a custom frequently leads to ambiguity, but it is often possible to interpret them correctly from a knowledge of the nature of the problem in hand. For instance, steam pressures stated by a man in ordinary prac- tice as being 100 lbs. may mean 100 lbs. per square inch gage (above atmaephere), but may be 100 lbs. per square inch absolute. Steam pressures are then most commonly stated per square inch and should be designated as either gage or abso- lute. Pressures of compressed air are commonly expressed in the same units la steam, either gage or absolute, though sometimes in atmospheres. Steam pressures; below atmosphere may be stated as a vacuum of so many inches of mercury,, meaning that the pressure is less than atmosphere by that amoimt, or may/ be given as a pressure of so many inches of mercury absolute, or as so many pounds per square inch absolute. The pressures of gases stored in tanks under high pressure are frequently recorded in atmospheres, due to the convenience of computation of quantities on this basis. Pressures ot air obtained by blowers or fans are visually given by the manufacturers of such apparatus in ounces per square inch above (or below) atmosphere. Such pressures and also differ- ences of pressure of air due to chimney draft, or forced draft, and the pressure: of illuminating gas in city mains, are commonly stated in inches of water, each inch of water being equivalent to 5.196 lbs. per square foot. The pressure of water in city mains or other pressure pipes may be stated either in pounds per square inch or in feet of water head. Example. A piston on which the mean pressure is 60 lbs. per square inch sweeps through a Yohime of 300 cuit. What is the work done? W^PxVj where 7= cuit. and P^lbs. per sq.ft. .-. }r«60xl44x300 =2,592,000 ft.-lbs. Prob. 1. The mean pressure acting per square inch when a mass of air changes in volume from 10 cu.ft. to 50 cuit. is 40 lbs. per square inch. How much work is done? Prob. 2, An engine is required to develop 30 HP. If the volimie swept through per minute is 150 cuit., what must the mean pressure be? Prob. 3. The mean effective pressure in compressing air from one to five atmospheres is 2S.7 ibs. per square inch. How many horse-power are required to compress 1000 cu.ft. of free air per minute? Prob. 4. At an altitude of 4100 ft. a pressure gage showed the pressure on one side of a piston to be 50 lbs. per square inch while the pressure on the opposite side is 3 lbs. per square inch absolute. What pressure was tending to move the piston? Prob. 6. At an altitude of 1 mUe the mean pressure in a gas engine cylinder during the suction stroke was found to be 12 lbs. per square inch absolute. What work was done by the enpne to draw in a charge if the cylinder was 5 ins. in diameter md the stroke 6 ins.? 6 ENGINEERING THERMODYNAMICS Prob. 6, After explosion the piston of tbe above engme was forced out 80 that the gas volume wa£ five times that at the beginning of the stroke. What must the M.E.P. have been to get 20,000 ft.-lbs. of work? Prob. 7, On entering a heating oven cold air expands to twice its volume. TVTiat work is done per cubic foot of air? Prob. 8. A projectile is forced from a gun by a constant air pressure of 1000 lbs. per square inch. Before it begins to move there is i a cu.ft. of air in the barrel, and at the instant it leaves the barrel the volume is 10 cuit. What work was done on the projectile? Prob. 9. Water is forced from a tank against a head of 75 ft. by filling the tank with compressed air. How much work is done in emptying a tank containing 1000 cu.ft.? 4. Work of Acceleration and Resultant Velocity. When a force acting on a mass is opposed by an equal resistance there may be no motion at all, or there may be motion of constant velocity. Any differences, however, between the two opposing forces will cause a change of velocity so long as the difference lasts, and this difference between the two forces may be itself considered as the only active force. Observations on imresisted falling bodies show that they increase in velocity 32.16 ft. per second for each second they are free to fall, and this quantity is universally denoted by g. If then, a body have any velocity, wi, and be acted on by a force equal to its own weight in the direction of its motion for a time, t seconds, it will have a velocity U2 after that time. U2=ui+g'z (1) It may be that the force acting is not equal to the weight of the body, in which case the acceleration will be different and so also the final velocity, due to the action of the force, but the force producing any acceleration will be to the weight of the body as the actual acceleration is to the gravitational acceleration. So that Actual accelera tio n force actual acce leration Weight of body or gravitational force gravitational acceleration {gY and Actual accelerating force = rr-r* — y \ — zr- rr X actual acceleration. gravitational acceleration (jg) or change of velocity Force = mass X acceleration = mass X time of change F=MX^^^^^^ (2) The work performed in accelerating a body is the product of the resistance met into the distance covered, L, while the resistance, or the above-defined force, acts, or while the velocity is being increased. This distance is the product t)f the time of action and the mean velocity, or the distance in feet, X=^t% (3) WORK AND POWER. 7 The work is the product of Eqs. (2) and (3), or, work of acceleration is ^ T ^~2 where w is the weight in pounds. Exactly the same result will be obtained by the calculus when the acceleration is variable, so that Eq. (4) is of universal application. The work performed in ctcceleraiing a body depends on nothing hut its mass and the initial and final velocities, and is in every case equal to the product of half the mass and the difference between the squares of the initial arid final velocities^ or the product of the weight divided by 64-4 ^^ ^ difference between the squares of the initial and final velocities. It frequently happens that the velocity due to the reception of work is desired, and this is the case with nozzle flow in injectors and turbines, where the steam performs work upon itself and so acquires a velocity. In all such cases the velocity due to the reception of the work energy is M2 where W is work in foot-pounds and w, as before, is weight in pounds. Or if the initial velocity be zero, as it frequently is, U2 =J»=J^l32l. (6) \ w yi w For conversion of velocity units, Table VIII, at the end of the Chapter, is useful. * Example. A force of 100 lbs. acts for 5 seconds on a body weighing 10 lbs. ; if the original velocity of the body was 5 ft. per second, what will be the final velocity, the distance traveled and the wor]r done? 100- 10 ("'-5) ■ Uj = 1615 ft. per second; 5=(j^')x=4050ft. yf^M{u^-u^ „405^(X)0 ft.-lbs. Ptob. 1. A stone weighing \ lb. is dropped from a height of 1 mile. With what veloc- ity and in what length of time will it strike if the air resistance is zero? Prob. 2. A car moving 20 miles per hour and weighing 25 tons is brought to rest in 500 ft. What is the negative acceleration, the time required to stop, and the work done? 8 ENGINEERING THERMODYNAMICS Prob. 3. Steam escapes through an opening with a velocity of half a mile per second. How many foot-pounds of energy were imparted to each pound of it to accomplish this? Prob. 4. A weight of 100 lbs. is projected upward with a constant force of 200 lbs. How much further will it have gone at the end of 10 seconds than if it had been merely falling under the influence of gravity for the same period of time? Prob. 6. A projectile weighing 100 lbs. is dropped from an aeroplane at the height of i mile. How soon will it strike, neglecting air resis'ance? Prob. 6. A water-wheel is kept in motion by a jet of water impinging on flat vanes. The velocity of the vanes is one-half that of the jet. The jet discharges 1000 lbs. of water per minute with a velocity of 200 ft. per second. Assuming no losses, what is amount of the work done? Prob. 7. With the wind blowing 30 miles per hour, how much work could a 12-ft. windmill perform if 25 per cent of the available work were utilized. Note. The weight of a cubic foot of air may be taken as .075 lb. Prob. 8. An engine has a piston speed of 600 ft. per minute and runs at 150 R.P.M. If the reciprocating parts weigh 500 lbs., how much work is done in accelerating the piston during each stroke? Prob. 9. A flywheel with rim 10 ft. in diameter to center of section and weighing 5 tons, revolves at a rate of 150 R.P.M.; 100,000 ft.-lbs. of work are expended on it. How much will the speed change? 6. Grap^cal Representation of Work. As work is always a product of force and distance or pressure and volume, it may be graphically expressed by B • c 5 r • * B 8 1 IT. r> **• 2 * 1 D / { 1 t i e 1 3 i i 5 Distaaces Id Feet Fig. 1. — Constant Force, Work Diagram, Force-Distance CJoordinates. an area on a diagram having as coordinates the factors of the product. It is customary in such representations to use the horizontal distances for volumes and the vertical for pressures, which, if laid off to appropriate scale and in proper units, will give foot-pounds of work directly by the area enclosed. Thus in Fig. 1, if a force of 5 lbs. (AB) act through a distance of 5 ft. (BC) there will be performed 25 foot-pounds of work as indicated by the area of the WORK AND POWER rectangle A BCD, which encloses 25 unit rectangles, each representing one foot- pound of work. If a steam cylinder piston suffers a displacement of 5 cu.ft. under the steam pressure (absolute) of 5 lbs. per square foot then the operation which results in the performance of 25 foot-pounds of work is represented by the diagram Fig. 2, ABCD. I 5 § C A 00 * u & '3 3 c 3 O B 2 B c D I 2 8 4 5 • Volumes in Cubic Eeet Fig. 2. — Constant Pressure Work Diagram, Pressure- Volume Coordinates. Prob. 1. Following the method given for Fig. 1, draw a diagram for the example of Section 3 Prob. 2. By means of a diagram, show that the work done by a pressure of 1000 lbs. per square foot traversing a distance of 10 ft. is 10,000 ft.-lbs. Prob. 3. Draw a diagram for the case of a volume change from 1 to 10 cu.ft. while the pressure acting is 20 lbs. per square inch. Prob. 4, Draw a pressure volume diagram for the case of forcing a piston out of a cylinder by a water pressure of 15,000 lbs. per square foot, the volume of the cylinder at the start is i cu.ft. and at the end 6 cu.ft. Make a diagram to scale and report work per square inch of diagram. Prob.6. A pump draws in water at a constant suction pressure of 14 lbs. and dis- charges it at a constant delivery pressure of 150 lbs. per sq.in. Considering the pump barrel to be empty at beginning of suction and end of delivery and to contain 3 cu.ft. when full, draw the diagram for this case and find the foot-pounds of work done. Prob. 8. In raising a weight a man pulls on a rope with a constant force of 80 lbs. If the weight is lifted 40 ft., find from a diagram the work done. Prob. 1. In working a windlass a force of 100 lbs. is applied at the end of a 6-ft. le\'er, the drum of th windlass being 1 ft. in diameter. Draw a work diagram for work applied and for work done in lifting if there be no loss in the windlass. Prob. 8. The steam and water pistons of a pump are on the same rod and the area of the former is twice that of the latter, the stroke being 3 ft. Show by a diagram that the work done in the two cylinders is the same if losses be neglected. Prob. 9. An engine exerts a draw-bar pull of 8000 lbs. at speed of 25 miles an hour. A change in grade occurs and speed increases to 40 miles per hour and the pull decreases to 5000 lbs. Show by a diagram the change in horse-power. 10 ENGINEEEING THERMODYNAMICS 6. Work by Pressure Volume Change. Suppose that instead of being constant the pressure were irregular and, being measured at intervals of 1 cu.ft. displacement, found to be as follows: Pressure. Displacement Lbs. per Sq.Ft. Volume Cu Ft. 100 125 1 150 2 100 3 75 4 50 5 12 3 4 5 6 Volumes In Cubic Feet Fig. 3. — ^Work Diagram, Pressure-Volume Coordinates. Discontinuous Pressure- Volume Relations. This condition might be plotted as in Fig. 3, A, S, C, D, Ey F, G, H. The work done will be the area under the line joining the observation points. In the absence of exact data on the nature of the pressure variations bet^'^een the two observation points A and S, a variety of assumptions might be made as to the precise evaluation of this area, as follows: (a) The pressure may have remained constant at its original value for the first cubic foot of displacement, as shown dotted A-B/ and then suddenly have risen to B, In this case the work done for this step would be 100 foot-pounds. (6) Immediately after the measurement at A the pressure may have risen to A' and remained constant during displacement A' to B, in which case the work done would be 125 foot-pounds. (c) The pressure may have risen regularly along the solid line AB^ in which case the work area is a trapezoid and has the value ^ XI = 112.5 foot- pounds. WORE AND POWER 11 It thus appears that for the exact evaluation of work done by pressure volume change, contmuous data are necessary on the value of pressure with respect to the volume. If such continuous data, obtained by measurement or otherwise, be plotted, there will result a continuous line technically termed the pressure-volume curve for the process. Such a curve for a pressure volmne change starting at 1 cu.ft. and 45 lbs. per square foot, and ending at 7 cu.ft., and 30 lbs. per square foot, is represented by Fig. 4, Ay B, C, D, E, The work done during this displacement under continuously varying pressure isUkewise the area between the curve and the horizontal axis when pressures are laid oS vertically, and will be in foot-pounds if the scale of pressures is pounds per square foot and volumes, cubic feet. Such an irregular area can be divided into small vertical rectangular strips, each so narrow that the pressure is sensibly constant, however much it may differ in dififerent strips. The area of the rectangle is PA V, each having the width AV and the height P, and the work ^ ^ --- ^ y \ V A / \ \ \ s, c • "^ '^ E D ' 1 I \ 9 K ! 5 # ' VolumeB In Cubic Feet Fig. 4. — Work Diagram, Preesure-Volume Coordinates. Gontinuoua Pressure-Volume Relations. area will be exactly evaluated if the strips are narrow enough to fulfill the conditions of sensibly constant pressure in any one. This condition is true only for infinitely narrow strips having the width dV and height P, so that each has the area PdV and the whole area or work done is =/ Tr= I PdY. (7) This is the general algebraic expression for work done by any sort of continuous pressure volume change. It thus appears that whenever there are available sufficient data to plot a continuous curve representing a pressure volume change, the work can be found by evaluating the area lying under the curve and bounded by the curve coordinates and the axis of volumes. The work done may be found by actual measurement of the area or by algebraic solution of Eq. (7), which can be integrated only when there is a known algebraic relation between the pressure and the corresponding volume of the expansive fluid, gas or vapor. Prob* 1. Draw the diagrams for the following cases: (a) The pressure in a cylinder 12 ins. in diameter was found to vary at different parts of an 18-in. stroke as follows: 12 ENGINEERING THERMODYNAMICS Pressure in Pounds Per Cent of per Sq.Ia. Stroke. 100 100 10 100 .30 100 50 83.3 60 71.5 70 62.5 80 55.5 90 60.0 100 (6) On a gas engine diagram the following pressures were found for parts of stroke. In Out V P V P V P 0.25 cu.ft. 14.7 0.1 45.2 0.13 146.2 0.20 '* 19.5 0.102 79.7 0.15 116.7 0.14 " 29.7 0.104 123.2 0.17 95.7 0.10 *• 45.2 0.106 157.7 0.19 80.7 0.108 181.7 0.21 68.7 0.11 188.2 0.23 58.7 0.12 166.2 Prob. 2. Steam at a pressure of 100 lbs. per square inch absolute is admitted to a cylinder containing .1 cu.ft. of steam at the same pressure, until the cylinder contains 1 cu.ft., when the supply valve closes and the volurtie increases so that the product of pressure and volume is constant imtil a pressure of 30 lbs. is reached. The exhaust valve is opened, the pressure drops to 10 lbs. and steam is forced out until the volume becomes 1 cu.ft., when the exhaust valve closes and the remaining steam decreases in volume so that product of pressure and volume is constant until the original point is reached. Draw the pressure volume diagram for this case. Prob. 3. Diu'ing an air compressor stroke the pressures and volumes were as follows: Volume in Prcf59ure in Lbs. Cu.Ft. Sq.In. 2.0 14.0 1.8 15.5 1.6 17.5 1.4 20.0 1.2 23.3 1.0 28.0 0.8 28.0 0.4 28.0 0.0 28.0 Draw the diagram to a suitable scale to give work area in foot-pounds directly. Prob. 4. Draw the diagrams for last two problems of Section 3. WORK AND POWER 13 7. Work of Expansion and Compression. Any given quantity of gas or vapor confined and not subject to extraordinary thermal changes such as explosion, will suffer regular pressure changes for each unit of volume change, or conversely, suffer a regular volume change for each imit of pressure change, so that pressure change is dependent on volume change and vice versa. When the volume of a mass of gas or vapor, Vi, is allowed to increase to V2 by the movement of a piston in a cylinder, the pressure will regularly increase or decrease from Pi to P2, and experience has shown that no matter what the gas or vapor or the thermal conditions, if steady, the voliunes and pressures will have the relation for the same mass, PiFi'=P2F2^=if, (8) or the product of the pressure and s power of the volume of a given mass will always be the same. The exponent « may have any value, but usually lies between 1 and 1.5 for conditions met in practice. The precise value of « for any given case depends on (a) The substance. (6) The thermal conditions surroimding expansion or compression, « being different if the substance receives heat from, or loses heat to, external sur- roundings, or neither receives nor loses. (c) The condition of vapors as to moisture or superheat when vapors are under treatment. Some commonly used values of s are given in Table X at the end of this chapter for various substances subjected to different thermal conditions dur- ing expansion or compression. Not only does Eq. (8) express the general law of expansion, but it likewise 'expresses the law of compression for decreasing volimiies in the cylinder with jcorresponding rise in pressure. Expansion in a cylinder fitted with a piston is called balanced expansion because the pressure over the piston area is balanced by resistance to piston movement and the mass of gas or vapor is subtitaatially at rest, the work of expaasion being imparted to the J)iston and resisting mechanism attached to it. On the other hand when the gas or vapor under pressure passes through a nozzle orifice to a region of lower pressure the falling pressure is accompanied by increasing volumes as before, but the work of expansion is imparted not to a piston, because there is none, but to the fluid itself, accelerating it until a velocity has been acquired as a resultant of the work energy received. Such expansion is termed free expansion and the law of Eq. (8) applies as well to free as to balanced expansion. This equation, then, is of very great value, as it is a convenient basis for computations of the work done in expansion or compression in cylinders and nozzles of all sorts involv- ing every gas or vapor substance. Some expansion curves for different values of 8 are plotted to scale in Fig. 5, and the corresponding compression curves in fig. 6, in which Curve A has the exponent s= Curve B '' ^V s= .5 ENGINEERINa THERMODYNAMICS Curve C has the exponent «= 1.0 Curve D Curve E Curve F Curve G Curve K Volumea In Cubic Feet Fig. 6. — ComparisoD of ExpanHion Lines having Different Values of *, The volume after expansion is given by (9) SO that the final volume depends on the original volume, on the ratio of the two pres.sures and on the value of the exponent. Similarly, the pressure after expansion ''"-^■(f;)' ("» WORK AND POWER 15 depends on the original pressure, on the ratio of the two volumes and on the exponent. The general equation for the work of expansion or compression can now be integrated by means of the Eq. (8), which fixes the relation between pressures and volumes. From Eq. (8), Fig. 6. — Comparison of Compression Curves having Different Values of 8. which, substituted in Eq.(7), gives ^KdV W -p ya f but as /^ is a constant, W -/I?- The integral of Eq. (11) will have two forms: (1) When 8 is equal to one, in which case PiVi=P2V2=K^; (2) When s is not equal to one. (11) 16 ENGINEERING THEEM0DYNA14ICS Taking first the case when « is equal to one, •VtdV W rv, Whence W=K' log, Y^ =PiViloe» = P2F2l0g, Yl Yl Vi (a) (&) W (d) (e) When 8 - 1 . (12) =i^log.g =PiFilog.^^ = P2V2l0g.g (/) Eqs. (12) are all equal and set down in different forms for convenience in computation; in them F2= largest volume = initial vol. for compression = final vol. for expansion. P2 = smallest pressure = initial pres. for compression = final pres. for expansion. Fi = smallest volume = final vol. for compression = initial vol. for expansion. Pi = largest pressure = final pres. for compression = initial pres. for expansion. These Eqs. (12) all indicate that the work of expansion and compression of this class is dependent only on the ratio of pressures or volumes at the beginning! and end of the process, and the PV product at either beginning or end, this product being of constant value. When the exponent s is not equal to one, the equation takes the form, i-sL Vi^-'-Yx^-" •] As s is greater than one, the denominator and exponents will be negative, so, changing the form to secure positive values, s-1VFi«-» F2'-V This can be put in a still more convenient form. Multiplying and dividing by 1 1 vv-i or Y.»-v --f-r.M^r -^'^uH^<ir\ WORK AND POWER 17 Substituting the value of X=P2F2*=PiFi', Whence -^/mm"'-^]-^^mHW} ^-mivr-^] <«' . . Whens5^1. . . (13) V2 P2 Vi Pi Eqs. (13) gives the work for this class of expansion and compression in terms of pressure ratios and volume ratios and in them largest volume = initial vol. for compression = final vol. for expansion; • smallest pressure = initial pres. for compression = final pres. for expansion ; = smallest volume = final vol. for compression = initial vol. for expansion; -largest pressure = final pres. for compression = initial pres. for expansion. The work of expansion or compression of this class is dependent according to Eqs. (13), upon the ratio of pressures or volumes at beginning and end of the process, the exponent, and on the pressure volume product appropriately taken. It should be remembered that for the result to be in foot-pounds appropriate units should be used and all pressures taken absolute. Examination of Eqs. (12) and (13), for the work done by expansion or compression of both classes, shows that it is dependent on the initial and final values of pressures and volumes and on the exponent 8, which defines the law of variation of pressure with volume between the initial and final states. Example 1. Method of calculating Diagram, Figs. 5 and 6. Consider the curve for which 8 = 1 A as typical of the group. Assumed Data, Then Fi «1.0 cu.ft. Pi =20,000 lbs. per square foot. « = 1.4. Pi7i* =»i2: =20,000 Xl^-* =20,000. For any other value of P, V was found from the relation. 18 ENGINEERING THERMODYNAMICS LetPa:«6000, then or ^- ©'^-[s]-">^"' log 3.33 =.5224 .715 X .5224 = .373 =log Vt. .'. 7* =2.36. A series of points, as shown below, were found, through which the curve was drawn. p 20.000 P ' , 20,000 log— y— . 1 .20,000 s^^ P • V 18000 1.111 0.0453 0.032 1.08 14000 1.430 0.1653 0.111 1.30 10000 2.000 0.3010 0.214 1.64 6000 3.330 0.5224 0.373 2.36 2000 10.000 1.0000 0.714 5.18 1000 20.000 1.3010 0.930 8.51 Curves for other values of 8 were similarly drawn. Starting at a coi^mon volume of 20 cu. ft. the compression curves of Fig. 6 were determined by the same methods. Example 2. A pound of air at 32^ F. and under atmospheric pressure is compressed to a pressure of five times the original. What will be the final volume and the work done if 8»1 and if 8 = 1 A? The voliune of 1 lb. of air at 32^ F. and one atmosphere is 12.4 cu.ft. approx. For« = l, P.' 7, = 12.4cu.ft.; 12.4 =5, whence Fi= 2.48 cu.ft. For « = 1.4, W^P.Vtloge^ = 2116X12.4 log* 5; Pi =2116X12.4X1.61=42,300 foot-pounds. f:-©" 12.4 .-.11 ...-71 _- = (5) = (5) . 5 may be raised to the .71 power by means of logarithms as follows: (5)"^^ is equal to the number whose logarithm is .71 log 5. WORK AND POWER 19 Log 5 =.699, .71 X.699 = .4963, and number of which this is the logarithm is 3.13, hence, 71 = 7,^-3.13 or Fi=3.96; = ^^^^><12.4 ^ ggg ^ 3g 200 f t.-lbs. .4 The value of W can also be found by any other form of equation (13) such as. -Si'-©-]- The value of Yx being found as before, the work expression becomes after numerical substitution ^ 10,580 X3.96r, /3.96V'^1 ^= A L^"VT2l/ J- As the quantity to be raised to the .4 power is less than one, students may find it oasier to use the reciprocal as follows: . /3.96\ -4 ^ 1 ^ 1 !_ ^ goo Vl2.4/ /IMV* (^-^^^^ ^-^ \3.96/ Hence ^^10^X3^^^ -.632) =38,200 f t.-lbs. Prob. 1. Find Yx and W for Example 2 if s = 1.2 and 1.3. Prob. 2. If a pound of air were compressed from a pressure of 1 lb. per square inch absolute to 15 lbs. per square inch absolute ifind Yx and W when « = 1 and 1.4. F, = 180 QXi.iU What would be the H.P. to compress 1 lb. of air per minute? Prob. 3. Air expands so that 8 = 1. If Pi = 10,000 lbs. per square foot, Yx = 10 cu.ft. and r^ «100 cu.ft. and the expansion takes place in 20 seconds, wha'. is the H.P. devel- oped? Prob. 4. 100 cu.ft. of air at atmospheric pressure is compressed in a cylinder to a pres- sure of 8 atmospheres and then expelled against this constant pressure. Find graphically and by calculation the foot-pounds of work done for the case where s = 1 and for the case where « = 1.4. Prob. 5. At an altitude of 4000 ft., a r is compressed to a pressure of 60 lbs. per sq.in. gage. Find he H.P. required to compress 1000 cu.ft. of free air per minute. Prob. 6. From the algebraic equation show how much work is done for a volume change of 1 to 4, provided pressure is originally 1000 lbs. per square foot when (a) FY^^Kx, (6) VY^K^, (c) PF«=i^3. 20 ENGINEERING THERMODYNAMICS Prob. 7. A vacuum pump compresses air from 1 lb. per square inch absolute to 15 lbs. per square inch absolute and discharges it. An air compressor compresses air from atmosphere to 15 atmospheres and discharges it. Compare the work done for equal initial volumes, « = 1.4. Prob. 8. For steam expanding according to the saturation law, compare the work done by 1 lb. expanding from 150 lbs. per square inch absolute to 15 lbs. per square inch absolute with, the work of the same quantity expanding from 15 lbs. to 1 lb. per square inch absolute. Note. 1 lb. of steam occupies 3 cu.ft. at 150 lbs. per square inch absolute. Prob. 9. Two air compressors of the same size compress air adiabatically from atmos- phere to 100 lbs. gage and discharge it. One is at sea level, the other at 10,000 ft. ele- vation. Compare the work in the two cases. 8. Values of Exponent s Defining Special Cases of Expansion or Compres- sion. There are three general methods of finding s for the definition of particular cases of expansion or compression to allow of the solution of numerical problems. The first is experimental, the second and third thermodynamic. If by measure- ment the pressures and volumes of a series of points on an expansion or com- pression curve, obtained by test with appropriate instruments, for example, the indicator, be set down in a table and they be compared in pairs, values of 8 can be found as follows: Calling the points A, J8, C, etc., then, PaVa' = PbVb', and log Pa+S log Fa^log Pb+S log 76, or . S(l0g Vb - log Va) = log Pa - log Pb, hence logPg-logP , . "^^logn-logF, ^""^ _ _ a or (14) According to Eq. (14a), if the difference between the logarithms of the pressures at B and A be divided by the differences between the logarithms of the volumes at A and B respectively, the quotient will be s. According to Eq. (146), the logarithm of the ratio of pressures, Bio Ay divided by the logarithm of the ratio of voliunes, AtoB respectively will also give s. It is interesting to note that if the logarithms of the pressures be plotted vertically and logarithms of volumes horizontally as in Fig. 7, then the line AC equal to the intercept on the horizontal axis represents the difference between the logarithms of volumes or, CA = logya~^ogy6, WORK AND POWER 21 and similarly Hence CB=logPb- logP, CB , CA or the slope of the line indicates the value of s. This is a particularly valuable method, as it indicates at a glance the constancy or variability of s, and there are many cases of practice where s does vary. Should 8 be constant the line will be straight; should it be variable the line will be curved, but can generally .6 Jd 1. • L(yg. V Fig. 7. — Graphic Method of Finding s, from Logarithms of Pressures and Volumes. be divided into parts, each of which is substantially straight and each will have a different 8. It is sometimes most convenient to take only the beginning and end of the curve and to use the value of s corresponding to these points, neglecting intermediate values. A second method for finding s for a given compression or expansion line by means of areas is indicated in a note in Section 17 of this Chapter that is omitted here because it depends on formulas not yet derived. It is by this sort of study of experimental data that most of the valuable values of s have been obtained. There is, however, another method of finding a value for s by purely thermodynamic analysis based on certain fimdamental hypo- theses, and the value is as useful as the hypotheses are fair or true to the facts of a particular case. One of the most common hypotheses of this sort is that the gas or vapor undergoing expansion or compression shall neither receive any heat from, nor give up any to bodies external to itself during the process, and such a process is given the name adiabatic. Whether adiabatic processes are possible in actual cylinders or nozzles does not affect the analysis with which pure thermody- namics is concerned. By certain mathematical transformations, to be carried out later, and based on a fundamental thermodynamic proposition, the adia- 22 ENGINEERING THERMODYNAMICS batic hypothesis will lead to a value of s, the use of which gives results valuable as a basis of reference, and which when compared with an actual case will per- mit of a determination of how far the real case has departed from the adiabatic condition^ and how much heat has been received or lost at any part of the process. The particular value of s which exists in an adiabatic change is repre- sented by the symbol y. Another common hjrpothesis on which another value of s can be derived, is that gases in expansion or compression shall remain at a constant temperature, thus giving rise to the name isothermal. This is generally confined to gasej? and superheated vapors, as it is difficult to conceive of a case of isothermal or con- stant temperature expansion or compression of wet vapors, as will be seen later. In the study of vapors,' which, it must be understood, may be dry or wet, that is, containing liquid, a common hypothesis is that during the expansion or compression they shall remain just barely dry or that they shall receive or lose just enough heat to keep any vapor from condensing, or but no more than sufficient to keep any moisture that tends to form always evaporated. Expan- sion or compression according to this hypothesis is said to follow the saturation law, and the substance to remain saturated. It will appear from this thermal analysis later that the value of s for the isothermal hypothesis is the same for all gases and equal to one, but for the adiabatic hj^othesis s=y will have a different value for different substances, though several may have the same value, while for vapors y will be found to be a variable for any one, its value depending not only on the substance, but on the temperatures, pressures and wetness. • When gases or vapors are suffered to expand in cylinders and nozzles or caused to compress, it is often difficult and sometimes impossible or perhaps undesirable to avoid interference with the adiabatic conditions for vapors and gases, with the isothermal for gases or with the saturation law for vapors, yet the wqrk to be done and the horse-power developed cannot be predicted without a known value of s, which for such cases must be found by experi- ence. A frequent cause of interference with these predictions, which should be noted, is leakage in cylinders, which, of course, causes the mass under treatment to vary. According to these methods those values of s have been found which are given in Table X, at the end of the Chapter. Mixtures of common gases such as constitute natural, producer, blast furnace or illuminating gas, alone or with air or products of combustion, such as used in internal combustion engines, have values of 5 that can be calculated from the elementary gases or measured under actual conditions. All vapors, except those considerably overheated, have variable exponents for adiabatic expansion and compression. This fact makes the exact soluticii of problems of work for wet vapors, expanding or compressing, which form the bulk of the practical cases, impossible by such methods as have been described. This class of cases can be treated with precision only by strictly thermal methods, to be described later. WORK AND POWER 23 Prob. 1. By plotting the values for the logarithms of the following pressures and vol- umes, see if the value for s is constant, and if not find the mean value in each case. (a) Gas Engine Comfbession V p V p V 10 45.2 13 32.2 18 P 21.0 11 39.7 14 29.7 20 19.5 12 35.7 16 24.7 25 14.7 (6) Gas Engine Expansion V p V p V 13 146.2 19 P 80.7 11 188.2 15 116.7 21 68,7 12 166.2 17 65.7 23 58.7 (c) Steam Expansion V p V p 2.242 203.3 7.338 52.5 2.994 145.8 12.44 28.8 4.656 89.9 22.68 14.7 Prob. 2. By plotting the values for the logarithms of the volumes and pressures on the expansion and compression curves of the following cards, find value for a. »10H 0- flO- 60- 40- 38- r- 10- 0- 600 -j 900-^ 20O JOO- 0- Atmagphero Atmospheto Atinoei>here> 125- IQD- 76- 60- 26- 0- 120-1 90 60- «)- 0- (Mmom Atmoaphere 24 ENGINEERING THERMODYNA^nCS 160 140 120 100 80 eo 40 20 80 Atinosphei'e Atmosphere Prob. 3. From the steam tables at the end of Chapter IV. select the pressures and volumes for dry-saturated steam and find the value of s between (a) 150 lbs. per square inch and 1 lb. per square inch. • (6) 15 Prob. 4. Find for superheated steam at 150 lbs. per square inch and with 100^ of superheat expanding to 100 lbs. per square inch without losing any superheat, the corresponding value of 8, using tabular data. Prob. 6. From the ammonia table data for dry-saturated vapor find the value of s between (a) 160 lbs. per square inch and 1 lb. per square inch. (6) 15 (( It it 9. Work Phases and Cycles, Positive and Negative and Net Work. Accord- ing to the preceding it is easy to calculate or predict numerically the work of expansion or compression whenever the conditions are sufficiently definite to permit of the selection of the appropriate s. It very seldom happens, however, that the most important processes are single processes or that the work of expansion or compression is of interest by itself. For example, before expansion can begin in a steam cylinder steam must be first admitted, and in air com- pressors air must be drawn in before it can be compressed. Similarly, aftei* expansion in a steam cylinder there must be an expulsion of used vapor before another admission and expansion can take place, while in the air compressor after compression the compressed air must be expelled before more can enter for treatment. The whole series of operations is a matter of more concern than any one alone, and must be treated as a whole. The effect can be most easily found by the summation of the separate effects, and this method of summation will be found of universal application. The whole series of processes taking place and involving pressure volume changes is called a cycle, any one of them a phase. It is apparent that there can be only a limited number of phases so definite as to permit of the mathe- WORK AND POWER 25 matical treatment necessary for prediction of work, but it is equally clear that there may be a far greater number of combinations of phases constituting cycles. Before proceeding to analyze the action of steam or gas in a cylinder it is necessary first to determine on structural, thermal or any other logical grounds, what series of separate processes will be involved, in what order, and the pressure volume characteristics of each. Then and then only, can the cycle as a whole be treated. These phases or separate and characteristic proc- esses affecting the work done or involving pressure volume changes are divisible into two classes so far as the causes producing them are concerned, the first thermal and the second mechanical. It requires no particular knowledge of thermodynamics to realize that if air be confined in a cylinder with a free piston and is heated, that the volume will increase while pressure remains constant, since the piston will move out with the slightest excess of pressure inside over what is outside. This is a pressure constant, volume increasing, phase, and is thermal since it is a heat effect. If an ample supply of steam be available from a boiler held at a constant pressure by the manipulation of dampers and fires by the fireman and the steam be admitted to a cylinder with a piston, the piston will move out, the pressure remaining constant and volume increas- ing. This is also a pressure constant, volume increasing phase, exactly as before, but is mechanical because it is due to a transportation of steam from the boiler to the cylinder, although in another sense it may be considered as thermal if the boiler, pipe and cylinder be considered as one part during the admission. A sioiilar constant pressure phase will result when a compressor piston is forcibly drawn out, slightly reducing the pressure and permitting the outside atmosphere to push air in, to follow the piston, and again after compression of air to a slight excess, the opening of valves to storage tanks or pipe lines having a constant pressure will allow the air to flow out or be pushed out of the cylinder at constant pressure. These two constant pressure phases are strictly mechan- ical, as both Represent transmission of the mass. If a cylinder contain water and heat be applied without permitting any piston movement, there will be a rise of pressure at constant volume, a similar constant volume pressure rise phase will result from the heating of a contained mass of gas or vapor under the same circumstances, both of these being strictly thermal. However much the causes of the various characteristic phases may differ, the work effects of similar ones is the same and at present only work effects are under consideration. For example, all constant volume phases do no work as work cannot be done without change of volume. The consideration of the strictly thermal phases is one of the. principal problems of thermodynamics, for by this means the relation between the work done to the heat necessary to produce the phase changes is established, and a basis laid for determining the ratio of work to heat, or eflSciency. For the present it is sufficient to note that the work effects of any phase will depend jonly on the pressure volume changes which characterize it. I Consider a cycle Fig. 8, consisting of {AB)y admission of 2 cu.ft of steam at k constant pressure of 200,000 lbs. per square foot, to a cylinder originally 26 ENGINEERING THERMODYNAMICS coittainmg nothing, followed by (BC), expansion with s=l, to a pressure of 20,000 lbs. per square foot; (C^, constant volume change of pressure, and (FG), constant pressure exhaust at 10,000 lbs, per square foot. These opera- tions are plotted to scale. Starting at zero volume, because the cylinder Volumes In Cublo Pert ori^nally contains nothing, and at a pressure of 200,000 lbs. per square foot, the line AB, ending at volume 2 cu.ft., represents admission and the cross- hatched area under AB represents the 400,000 ft.-lba. of work done during admission. At B the admission ceases by closure of a valve and the 2 cu.ft. of ateam at the original pressure expands with lowering pressure according 'CO the law So that when PaFa=nF6=200,OOOx2 = 400,000 ft.-lbs., y = 4 cu.ft., P = —^*^=100,000 lbs. per sq.ft.; V = 5 cu.ft.. p=M^= 80,000 lbs. per sq.ft.; V = 10cu.ft., P=^^^;^= 40,000 lbs. per sq.ft. lbs. per square foot, and the work done during expansion is the cross-hatched area JBCD under the expansion curve BC, the value of which can be found by measuring the diagram or by using the formula Eq. (12), WOEK AND POWER 27 which on substitution gives Trbc=400,000 log. 10=400,000X2.3; =920,000 ft.-lbs. This completes the stroke and the work for the stroke can be found by addition of the numerical values, TFafc=400,000 ft.-lbs. , • TF»c= 920,000 ft.-lbs.; Tra5+TF6c= 1,320,000 ft.-lbs. It is often more convenient to find an algebraic expression for the whole, which for this case will be, TF6c=PftFJoge^'; (l+log.^), =400,000(H-lo& 10) =400,000X3.3 = 1,320,000 ft.-lbs. On the return of the piston it encounters a resistance due to a constant pressure of 10,000 lbs. per square inch, opposing its motion; it must, therefore, do work on the steam in expelling it. Before the return stroke begins, however, the pressure drops by the opening of the exhaust valve from the terminal pressure of the expansion curve to the exhaust or hack pressure along the constant volume line, CF, of course, doing no work, after which the return stroke begins, the pressure volimie line being FG and the work of the stroke being represented by the cross-hatched area DFGHy TF/a=P/F/= 10,000X20=200,000 ft.-lbs. This is negative work, as it is done in opposition to the movement of the piston. The cycle is completed by admission of steam at constant zero volume, raising the pressure along GA. The net work is the difference between the positive and negative work, or algebraically Tr=Pftn(l+loge y^ -PfVf, = l,320,000-200,000 = l,120,000^ft.-lbs. 28 ENGINEERING THERMODYNAMICS . Consider now a cycle of an an compressor, Fig. 9. AdmissioD or Buction 18 represented by AB, compression by BC, delivery by CD and constant volume drop in pressure after delivery by DA, The work of admission is represented by the area ABFE or algebraically by W,a = PbVb, the work of compression by the area FBCG, or algebraically since 8=1,4 by ^m^r the work of delivery by the area CDEG, or algebraically The positive work is that assisting the motion of the piston durii^ suction; the area ABFE or algebraically PbVh. The negative work, that in opposi- tion to the motion, is the sum of the compression and delivery work, the area FBCDE, or algebrajcally, -«+---.''-^'[0"'-']- The net work is the difference and is negative, as such a cycle is mainly resistant, and to execute it the piston must be driven with expenditure of work on the gas. The value of the net work is, W=Wb,+W,d-W<a, WORK AND POWER 29 an expression which will be simplified in the chapter on compressors. This net work is represented by the area ABCD, which is the area enclosed by the cycle itself independent of the axes of coordinates. It might seem from the two examples given as if net work could be obtiuned without the tedious problem of summation, and this is in a sense true if the cycle is plotted to scale or an algebraic expression be available, but these processes are practically equivalent to summation of phase results. It might also seem that the work area would always be that enclosed by the cycle, and this is true with a very important limitation, wiiich enters when tbe cycle has loops. If, for example, as in Fig. 10, steam admitted A to B, expanded along BC to a pressure C, then on opening the exhaust the pressure instead of falling to the back pressure or exhaust line as in Fig. 8, would here Volumes In Cubic Feet Fio, 10. — AnalyBiB of Work Diagram for Engine with Over-expansion Negative Work Loop. rise along CD, as the back pressure is higher thaa the terminal expansion pres- sure, after which exhaust will take place at constant back pressure along DE. The forward stroke work is that under AB and BC or ABCEG, the return stroke work is the area DEGH and the net work is Area ABC£(?-Area DEGH. A.S the area HGECX is common to both terms of the difference, the net work may be set down as equal to krf^&ABXH-CDX, It may be set down then in general for looped cycles that the net work area is the difference between that of the two loops. If, however, the method laid down for the treatment of any cycle be adhered to there need not be any dis- tinction drawn between ordinary and looped cycles, that is, in finding the work 30 ENGINEERING THERMODYNAMICS of a cycle divide it into characteristic ptiases and group (hem into positive and negativcj find the work for each and lake the algebraic sum. Special cases of cycles and their characteristics for steam compressors and gas engine cylinders, as well as nozzle expansion, will be taken up later in more detail and will constitute the subject matter of the next two chapters. Example 1. Method of calculating Diagram, Fig. 8. Assumed data To obtain pomt C. Fe=2 " PcVc-^PbV, or Vr= Pa =200,000 lbs. per square foot. Pt^Pa Pc =20,000 " " P/= 10,000 I Pe^Pf. tt It tt P&V^^ 200,000X2 20 Pr 20,000 " ' A Fc=20 and Pc =20,000. Intermediate points B to C are obtained by assuming various pressures and finding the corresponding volumes as for Vc. Example 2. Method of calculating Diagram, Fig. 9. Assumed data f Fa =Ocu.ft. 1 Fj,=20 *' Fd=0 8 = 1.4 n ' Pa =2116 lbs. per square foot. Pb^Pa Pc = 14,812 " " tt tt To obtain point C, P*F,i4 = p^FM or V,==V,-r^(^Y^^V,-^(^py 716 P P 1 r=7, log 7 =.845 ; and .715 X.845=log^~y* =.6105, or Therefore, F.=4.02. Fc =4.02, and Pc = 14,812. Intermediate values BtoC may be found by assuming pressures and finding volumes cor- responding as for Vc. Prob. 1. Steam at 150 lbs. per square inch absolute pressure is admitted into a cylin- der in which the volume is originally zero until the volume is 2 cu.ft., when the valve is closed and expansion begins and continues until the volume is 8 cu.ft., then exhaust valve opens and the pressure faUs to 10 lbs. absolute and steam is entirely swept out. Draw the diagram and find the net work done. WORK AND POWER 31 Prob. 2. A piston moving forward in a cylinder draws in 10 cu.ft. of CO2 at a pressure of .9 of an atmosphere at sea level and then compresses it adiabatically until the pressure rises to 9 atmospheres and discharges it at constant pressure. Draw the diagram and find net work done. Prob. 3. A cylinder 18 ins. in diameter and 24 ins. piston stroke receives steam at 100 lbs. per square inch absolute pressiu^ for i of the stroke. It then expands to the end of the stroke and is exhausted at atmospheric pressure. Draw the diagram and find the H.P. if the engine makes 100 strokes per minute. Prob. 4. Two compressors without clearance each with a cylinder displacement of 2 cu.ft. draw in air at 14 lbs. per square inch absolute and compress it to 80 lbs. per square inch absolute before delivery. Find the difference in H.P. per 1000 cu.ft. of free air per minute if one is compressing isothermally and the other adiabatically. Draw diagram for each case. Prob. 6. A quantity of air 5 cu.ft. in volume and at atmospheric pressure is compressed in a cylinder by the movement of a piston until the pressure is 50 lbs. per square inch gage. If the air be heated the pressure will rise, as in an explosion. In this case the piston remains stationary, while the air is heated until the pressure reaches 200 lbs. per square inch gage. It then expands adiabatically to the original volume when the pressure is reduced to atmosphere with no change in volume. Draw the diagram, and find the work done. Prob. 6. The Braytan cycle is one in which gas is compressed adiabatically and then, by the addition of heat, the gas is made to expand without change of pressure. Adi- abatic expansion then follows to original pressure and the cycle ends by decrease in volume to original amount without change of pressure. Draw such a cycle startmg with 5 cu.ft. of air at atmospheric pressure, compressing to 4 atmospheres, expanding at constant pressure to 5 cu.ft., expanding adiabatically to original pressure and finally ending at original point. Find also, work done. Prob. 7. In the Ericsson cycle air is expanded at constant temperature, cooled at constant pressure, compressed at constant temperature and receives heat at constant volume. Draw a diagram for the case where 5 cu.ft. at atmospheric pressure are com- pressed to 1 cu.ft., heated until volume is 8 cu.ft., expanded to atmosphere and then cooled to original volume. Find the work. Prob. 8, In the Stirling cyde constant volume heating and cooling replace that at constant pressure in the Ericsson. Draw diagram starting with 5 cu.ft. and atmsopheric pressure compressing to 1 cu.ft. and then after allowing the pressure to double, expand to original volume and cool to atmosphere. Find the work. Prob. 9. The Joule cyde consists of adiabatic compression and expansion and con- stant pressure heating and cooling. Assuming data as in last problem draw the diagram and find the work. Prob. 10. The Camot cycle consists of isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. Draw the diagram for this cycle and find the work. 10. Work Determination by Mean Eflfective Pressure. While the methods already described are useful for finding the work done in foot-pounds for a defined cycle with known pressure and volume limiis^ they are not, as a rule, convenient for the calculation of the work done in a cylinder of given dimensions. As work done can always be represented by an area, this area divided by its length will give its mean height. If the area be in foot-pounds with coordinates 32 ENGINEERING THERMODYNAMICS pounds per square foot, and cubic feet, then the division of area in foot-pounds by length in cubic feet will give the mean height or the mean pressure in pounds per square foot. Again, dividing the work of the cycle into forward- stroke work and back-stroke work, or the respective foot-pound areas divided by the length of the diagram in cubic feet, will give the mean forward pressure and the mean back pressure. The difference between mean forward pressure and mean back pressure will give the m,ean effective pressure, or that average pressure which if maintained for one stroke would do the same work as the cycle no matter how many strokes the cycle itself may have required for its execution, which is very convenient considering the fact that most gas engines require four strokes to complete one cycle. The mean eflFective pressure may also be found directly from the enclosed cycle area, taking proper account of loops, as representative of net work by dividing this net work area by the length of the diagram in appropriate units. This method is especially convenient when the diagram is drawn to odd scales so that areas do not give foot-pounds directly, for no matter what the scale the mean height of the diagram, when multiplied by the pressure scale factor, represents the mean effective pressure. This mean height can always be found in inches for any scale of diagram by finding the area of the diagram in square inches and by dividing by the length in inches, and this mean height in inches multiplied by the scale of pressures in whatever imits may be used will give the mean effective pressure in the same units. Mean pressures, forward, back or effective, are found and used in two general ways; first, algebraically, and second graphically and generally in this case from test records. By the first method, formulas, based on some assumed laws for the phases, can be found, and the mean effective pressure and its value predicted. This permits of the prediction of work that may be done by a given quantity of gas or vapor, or the work per cycle in a cylinder, or finally the horse- power of a machine, of which the cylinder is a part, operating at a given speed and all without any diagram measurement whatever. By the second method, a diagram of pressures in the cylinder at each point of the stroke can be obtained by the indicator, yielding information on the scale of pressures. The net work area measured in square inches, when divided by the length in inches, gives the mean height in inches, which, multiplied by the pressure scale per inch of height, gives the mean effective pressure in the same imits, which are usually pounds per square inch in practice. As an example of the algebraic method of prediction, consider the cycle represented by Fig. 8. The forward work is represented by Forward work =PbVb (i+iog.f;), the length of the diagram representing the volume swept through in the per- formance of this work is Vcy hence Mean forward pressure = —vr- ( 1+logc ^^ ) . WORE AND POWER But PbVh^PcVe by the law of this particular expansion curve, hence 33 Mean forward pressure =Pc{ bflogc -^Y As the back pressiu^ is constant its mean value is this constant value, hence C!onstant (mean) back pressure =P/. By subtraction Mean effective pressure —Pc\ 1 +loge ^\ — Pf = 3.3Pc-Pf; = 3.3X20,000-10,000; =66,000-10,000=56,000 lbs. per sq.ft. The work done in foot-pounds is the mean effective pressure in poimds per square foot, multiplied by the displacement in cubic feet. F=56,000X20= 1,120,000 ft.-lbs. as before. Fio. 11.- !^iaA-£jiginc Indicator Card. For Determination of Mean Effective Pressure without Volume Scale. As an example of the determination of mean effective pressure from a test or indicator diagram of unknown scale except for pressures, and without axes of coordinates, consider Fig. 11, which represents a gas engine cycle in four strokes, the precise significance of the lines being immaterial now. The pressure scale is 180 lbs. per square inch, per inch of height. By measurement of the areas in square inches it is found that Large loop area CDEXC =2.6 sq.in. Small loop area ABXA =0.5 sq.in. Net cycle area =2.1 sq.in. Length of diagram =3.5 in. Mean height of net work cycle =0.6 in. Mean effective pressure = 120X.06 = 72 lbs. per square inch. 34 ENGINEERING THERMODYNAMICS It is quite immaterial whether this diagram were obtained from a large or a small cylinder; no matter what the size, the same diagram might be secured and truly represent the pressure volume changes therein. If this particular cylinder happened to have a diameter of 10 ins. and a stroke of 12 ins. the work per stroke can be foimd. The area of the cylinder will be 78.54 sq.ins., hence the average force on the piston is 72 lbs. per square inch X 78.54 sq.ins. = 5654.88 lbs., and the stroke being 1 ft. the work per stroke is 5654.88 ft.-lbs. Both of these methods are used in practical work and that one is adopted in any particular case which will yield results by the least labor. Prob. 1. An indicator card from an air compressor is foimd to have an area of 3.11 sq. ins., while the length is 2^ ins. and scale of spring is given as 80 lbs. per square inch per inch height. What is m.e.p. and what would be the horse-power if the compressor ran with a piston speed of 250 ft. per minute and had a piston 9 ins. in diameter? Prob. 2. For the same machine another card was taken with a 60-lb. spring and had an area of 4.12 sq.ins. How does this compare with first card, the two having the same length? Prob. 3. A steam engine having a cylinder 18 ins. in diameter and a stroke of 24 ins., takes in ^ cu.ft. of steam at 100 lbs. absolute, allows it to expand and exhausts at atmos- pheric pressure. An indicator card taken from the same engine showed a length of 3 ins., an area of .91 sq.in. when an 80-lb. spring is used. How does the actual m.e.p. compare with the computed? Prob. 4. Find m.e.p. by the algebraic method of prediction for, (a) Brayton cycle; (6) Camot cycle; (c) Stirling cycle; (d) Ericsson cycle; (e) Joule cycle. (See problems following Section 9). 11. Relation of Pressure-Volume Diagrams to Indicator Cards. The Indicator. When a work cycle or diagram of pressure volume changes is drawn to scale with pressures and volumes as coordinates, it is termed a pressure volume or PV diagram, and may be obtained by plotting point by point from the algebraic expression for the law of each phase or by modifying the indicator card. The indicator card is that diagram of pressures and stroke obtained by appljdng the indicator to a cylinder in operation. This instrument consists essentially of a small cylinder in which a finely finished piston moves freely without appreciable friction, with a spring to oppose its motion, a pencil mechan- ism to record the extent of the motion, and a drum carrying paper which is moved in proportion to the engine piston movement. The indicator cylinder is open at the bottom and fitted with a ground union joint for attachment to the main cylinder through a special cock, which when open permits all the varying pressures in the main cylinder to act on the indicator piston, and when closed to the main cylinder opens the indicator cylinder to the atmosphere. The WORK AND POWER 35 upper side of the indicator piston being always open to the atmosphere, its movement will be the result of the diflferenee between the pressure in the main cylinder and atmospheric pressure. A helical spring, carefully calibrated and, therefore, of known scale, is fixed between the indicator piston and open cap or head of its cylinder, so that whenever the pressure in the main cylinder exceeds atmosphere the indicator piston moves toward the open head of the indicator cylinder, compressing the spring. Pressures in the main cylinder if less than atmosphere will cause the indicator piston to move the other way, extending the spring. This compression and extension of the spring is found in the calibration of the spring to correspond to a definite nmnber of pounds per square inch above or below atmosphere per inch of spring distortion, so that the extent of the piston movement measures the pressure above or below atmos- phere. A piston rod projects outward through the cylinder cap and moves a series of levers and links carrjdng a pencil point, the object of the linkage being to multiply the piston movement, but in direct proportion, giving a large movement to the pencil for a small piston movement. A cylinder drum carry- ing a sheet of paper is pivoted to the cylinder frame so that the pencil move- ment will draw on the paper a straight line parallel to the axis of the drum, if drum is stationary, or perpendicular to it if drum rotates and pencil is sta- tionary. The height of such lines then above or below a zero or datum line, which is the atmospheric line drawn with the cock closed, measures the pressure of the fluid under study. The springs have scale nmnbers which give the pressure in pounds per square inch per inch of pencil movement. This paper- carrying drum is not fixed, but arranged to rotate about its axis, being pulled out by a cord attached to the piston or some connecting part through a pro- portional reducing motion so as to draw out the cord an amount slightly less than the circumference of the drum no matter what the piston movement. After having been thus drawn out a coiled spring inside the drum draws it back on the return stroke. By this mechanism it is clear that, due to the combined movement of the pencil up and down, in proportion to the pressure, and that of the drum and paper across the pencil in proportion to the piston movement, a diagram will be drawn whose ordinates represent pressures above and below atmosphere and abscissse, piston stroke completed at the same time, or dis- placement volume swept through. It must be clearly understood that such indicator diagrams or cards do not give the true or absolute pressures nor the true volumes of steam or gas in the cylinder, but only the pressures above or below atmosphere and the changes of volume of the fluid corresponding to the piston movement. Of course, if there is no gas or steam in the cylinder at the beginning of the stroke, the true volume of the fluid will be always equal to the displacement, but no such cylinder can be made. While the indicator card is sufficient for the determination of mean eflFective pressure and work per stroke, its lack of axes of coordinates of pressure and volume prevents any study of the laws of its curves. That such study is important must be clear, for without it no data or constants such as the exponent « can be obtained for prediction of results in other similar cases, nor can the 36 ENGINEEEING THERMODYNAMICS presence of leaks be detected, or the gain or loss of heat during the variou-s processes studied. In short, the most valuable analysis of the operations is impossible. To convert the indicator card, which is only a diagram of stroke or displace- ment on which are shown pressures above and belpw atmosphere into a pres- sure volume diagram, there must first be found (a) the relation of true or abso- lute pressures to gage pressures, which involves the pressure equivalent of the barometer, and (h) the relation of displacement volumes to true volumes? of vapor or gas present, which involves the clearance or inactive volume of the cylinder. The conversion of gage to absolute pressures by the barometer reading has already been explained, Section 3, while the conversion of displace- ment volumes to true fluid volumes is made by adding to the displacement volume the constant value in the same units of the clearance, which is usually the result of irregularity of form at the cylinder ends dictated by structural necessities of valves, and of linear clearance or free distance between the pis- ton at the end of its stroke and the heads of the cylinder to avoid any possi- bility of touching due to wear or looseness of the bearings. 150 tl60(h Ftg. 12. — Ammonia Compressor Indicator Card with Coordinates of Pressures and Volumes Added to Convert it into a Pressure- Volume Diagram. Let ABCD, Fig. 12, represent an indicator card from an ammonia compressor on which EF is the atmospheric line. The cylinder bore is 14 ins., stroke 22 ins., and the scale of the indicator spring 100, barometer 28 ins., and measured clearance 32 cu.in. According to Table IX, 28 ins. of mercury corresponds to 13.753 lbs. per square inch, and as 100 lbs. per square inch, according to the spring scale, corresponds to 1 in. of height on the diagram, 1 lb. per square inch cor- responds to 0.01 in. of height, or 13.75 lbs. per square inch atmospheric pres- sure to .137 in. of height. The zero of pressures then on the diagram must lie .137 in. below the line EF. Lay off then a line Af/f, this distance below EF, This will be the position of the axis of volume coordinates. Actual measurement of the space in the cylinder with the piston at the end of its stroke gave the clearance volume of 32 cu. ins. As the bore is 14 ins. the piston area is 153.94 sq. ins. which in connection with the stroke WORK AND POWER 37 of 22 ins. gives a displacement volume of 22X153.94 = 3386.68 cu. ins. 32 Compared with this the ctearance volume is ooo^ ^o =.94 per cent of the 3386.68 displacement. It should b^ noted here that clearance is generally expressed in per cent of displacement volume. Just touching the diagram at the ends drop two lines at right angles to the atmospheric line intersecting the axis of volumes previously found at G and H. The intercept GH then represents the displacement, or 3386.68 cu.in. or 1.96 cu.ft. Lay oflf to the left of G, .0094, or in round numbers 1/100 of GH, fixing the point Af , MG representing the clearance to scale, and a vertical through M the axis of pressures. The axes of coordinates are now placed to scale with the diagram but no scale marked thereon. The pressure scale can be laid off by starting at M and marking off inch points each representing 100 lbs. per square inch. Pounds per square foot can also be marked by a separate scale 144 times as large. As the length of the diagram is 2.94 ins. and displacement 1.96 cu.ft., 1 in. of horizontal distance corresponds to .667 cu.ft. or 1 cu.ft to 1.50 ins. of distance. Lay off then from M dis- tances of 1,60 ins. and mark the first 1 cu.ft. and the second 2 cu.ft., dividing the intervals into fractions. A similar scale of volumes in cubic inches might also be obtained. By this proceas any indicator card may be converted into a pressure volume diagram for study and analysis, but there will always be required the two factors of true atmospheric pressure to find one axis of coordinates and the clearance volume to find the other. Prob. 1. If in cards Nos. 1 and 2, Section 8, the clearances are 5 per cent and 3 per cent respectively of the displacement, convert the cards to P7 diagrams on the same base to scales of 4 ins. to 1 cu.ft. and 1 in. to 1000 lbs. per square foot, for cylinders 9J ins. and 14 J ins. respectively in diameter and stroke 12 ins. Prob. 2. Do the same for cards Nos. 3 and 4, if clearances are 7 per cent and 4 per cent respectively, for cylinders 10 ins. and 17 ins. respectively in diameter and stroke 12 ins. Prob. 8. Do the same for cards Nos. 7 and 8 if clearances are 12 per cent and 8 per cent respectively and cylinders 12 ins. and 19 ins. in diameter and stroke 24 ins. 12. To Find the Clearance. There are two general methods for the find- mg of clearance, the first a direct volumetric measurement of the space itself by filling with measured liquid and the second a determination by algebraic or graphic means from the location of two points on the expansion or com- pression curves of the indicator card based on an assumed law for the curves. The first method of direct measurement is the only one that offers even a promise of accuracy, but even this is difficult to carry out because of the tendency of the measuring liquid to leak past piston or valves, which makes the result too large if the liquid be measured before the filling of the clearance space and too small if the liquid be measured after filling and drawing off. There is also a tendency in the latter case for some of the liquid to remain inside the space, besides the possibility in all cases of the failure to completely fill the space due to air pockets at high places. i .100 A' A' V 80 A'X A \ \ ^i\ e:, i \ N.« 00 B' BU \ V V 40 \ > 2 4 6 Displaoement Fig. 13. 8 Displacement Fig. 14. 10 • ft) A-' a\a \ \ S \ 80 B" B' "^ \, :*.B --- 10 A* A'--. ,a^ ~,^ _« _/ B B _ B n I 4 z Displacement Fig. 15. Diagrams Illustrating Location of Clearance Line from Expansion or Compression Lines of Known Laws. It WORK AND POWER 39 By the second general method any two points, A and B, on an expansion or compression curve. Figs. 13, 14, 15, may be selected and horizontals drawn to the vertical line indicating the beginning of the stroke. The points A' and B' are distant from the milocated axis an amount A'A"^B'B'\ representing the clearance. Let the clearance volume ^Cl,\ " the displacement up to A = Da; the displacement up to B^D^; the whole displacement =i); " « be the exponent in PF*= constant, which defines the law of the curve. Then in general, But and hence or / 1 l\ 1 1 whence the clearance in whatever units the displacement may be measured will be 1 1 1 . 1 ' PJ -Pbf or „ ^••^-(r)'^-] '^-iW"- 1 1 1 m-^\ iw- P^ and CI, in per cent of the whole displacement will be Db_/PayDa Clearance as a fraction of displacement = c = ^^—^ . When fi= 1 this takes the form Db_(Pa\Da Clearance in fraction of displacement =c= ^^—^ . . . (15) m- 40 ENGINEERING THERMODYNAMICS « To use such an expression it is only necessary to measure off the atmospheric pressure below the atmospheric line, draw verticals at ends of the diagram and use the length of the horizontals and verticals to the points in the formula, each horizontal representing one D and each vertical a P. Graphic methods for the location of the axis of pressures, and hence the clearance, depend on the properties of the curves as derived from analytics. For example, when «=1, PaVa^PhVh, which is the equation. of the equilateral hyperbola, a fact that gives a common name to the law, i.e., hyperbolic expansion or hyperbolic compression. Two common characteristics of this curve may be used either separately or together, the proof of which need not be given here, first that the diagonal of the rectangle having two opposite comers on the curve when drawn through the other two comers will pass through the origin of coordinates, and second, that the other diagonal drawn through two points of the curve and extended to intersect the axes of coordinates will have equal intercepts between each point and the nearest axis cut. According to the first principle, lay off. Fig. 16, the vacuum line or axis of volume XY and selecting any two points A and B, construct the rectangle ACBD. Draw the diagonal CDS and erect at E the axis of pressures EZ, then will EZ and EY be the axes of coordinates. According to the second principle, proceed as before to locate the axes of volumes XY and select two points, A and B, Fig. 17. Draw a straight line through these points, which represents the other diagonal of the rectangle ACBD, producing it to inter- sect XY Bit M and lay off AN—BM. Then will the vertical NE be the axis of pressures. It should be noted that these two graphic methods apply only when 8=^1; other methods must be used when s is not equal to 1. A method of finding the axis of zero volume is based upon the slope of the exponential curve, pr=c. Differentiation with respect to V gives or whence Ps y= {-%) -(-S) (-) WORK AND POWER 41 In other words, the true volume at any given point on the known curve may be found by dividing the product of P and 8 by the tangent or the slope of the line at the given point, ^ith the sign changed. This method gives results dependent for their accuracy upon the determination of the tangent to the curve, which is sometimes difficult. 70 00 Z . \ • |40 ao ao \ \ A c v / V / ^ / B Atmospheric Line 10 X 9 / '^ . / r " • — Y 4 i 3 i 5 < ' 1 I) 11 15 Displaoement Fig. 16. TO z Itl V V- c 1 \s ^y y' • St\ l\ V X \.' y 0^ «XI \ >^ ^ X <w • ,^ "^s \ A tmoBi >beric Line 10 X y x\y ^ ^^ — . B L^sT- » wm^im MM •s"^ Y 1 L 1 \ { \ i } ) 1 1 1 3 Displaoement Fig. 17. Graphic Methods of Locating the Clearance Line for Logarithmic or Hyperbolic Expansion and Compression Curves. The following graphical solution is dependent upon the principle just given, and while not mathematically exact, gives results so near correct that the error is not easily measured. The curve ACBj Fig. 18, is first known experi- mentally or otherwise and therefore the value of s, and the axis FV from which pressures are measured is located. Assume that the axis of zero volume, 42 ENGINEERING THERMODYNAMICS KP, is not known but must be found. Selecting any two convenient points, A and B, on the curve, complete the rectangle AHBG with sides parallel and perpendicular to FV. The diagonal HG cuts the curve at C and the horizontal axis at E. From C drop the perpendicular CD. If now the distance DE be multiplied by the exponent s, and laid oflF DK, and the vertical KP erected, this may be taken as the zero volume axis. It cannot be too strongly stated that methods for the finding of clearance or the location of the axes of pressures from the indicator card, much as they have been used in practice, are inaccurate and practically useless unless it is positively known beforehand just what value s has, since the assumed value Fig. 18. — Graphia Method of Locating the Clearance Line for Exponential Expansion and Compression Curves. of 8 enters into the work, and s for the actual diagram, as already explained, is affected by the substance, leakage, by moisture or wetness of vapor and by all heat interchange or exchange between the gas of vapor and its container. Prob. 1. If in card No. 6, Section 8, compression follows the law PF* «=A', where s = 1.4 and the barometer reads 29.9 ins. of mercury, locate the axes algebraically and graphically. Prob. 2. If in card No. 3, Section 8, expansion follows the law PF» =A', where s «1.37 and the barometer reads 29.8 ins. of mercury, locate the axes algebraically and graphically. Prob. 3. If in card No. 5, Section 8, expansion follows the law P7* =A', where s = l and the barometer reads 27.5 ins. of mercury, locate axes algebraically and graphically. WORK AND POWER 43 13. Measurement of Areas of PV Diagrams and Indicator Cards. Areas of pressure volume diagrams or indicator cards must be evaluated for the determination of- work or mean effective pressure, except when calculation by formula and hypothesis is possible. There are two general methods applicable to both the indicator card and PV diagram, that of average heights, and the planimeter measiu'e, besides a third approximate but very useful method, especially applicable to plotted curves on cross-section paper. The third method assumes that the diagram may be divided into strips of equal width as in Fig. 19, which is very easily done if the diagram is plotted on cross-section paper. At the end of each strip, a line is drawn perpendicular to the axis of the strip, such that the area intercepted inside the figure is apparently equal to that outside the figure. If this line is correctly located, the area of the rectangular strip will equal the area of the strip bounded by the irregular lines. Volumee FiQ. 19. — Approximate Method of Evaluating Areas and Mean Effective Pressures of Indicator Cards and P.V. Diagrams. If the entire figmre has irregular ends it may be necessary to subdivide one or both ends into strips in the other direction, as is done at the left-hand side of Fig. 19. The area of the entire figure will be equal to the summation of lengths of all such strips, multiplied by the common width. This total length may be obtained by marking on the edge of a strip of paper the successive lengths in such a way that the total length of the strip of paper when measured will be the total length of the strips. The mean height will be the total length of such strips divided by the num- ber of strip-widths in the length of the diagram. By a little practice the proper location of the ends of the strips can be made with reasonable accuracy, and consequently the results of this method will be very nearly correct if care is exercised. By dividing the diagram into equal parts, usually ten, and finding the length of the middle of each strip, an approximation to the mean height of each strip 44 ENGINEERING THERMODYNAMICS will be obt&in^d; these added together and divided by the number will give the mean height in inches from which the mean efiFective pressure may be found by multiplying by the scale as above, or the area in square inches by multiply- ing by the tength in inches, which can be converted into work by multiplying by the foot-pounds per square inch of area as fixed by the scales. As the pres- sures usually vary most, near the ends of the diagram a closer approximation can be made by subdividing the end strips, as is done in Fig. 20, which repre- sents two steam engine indicator cards taken from opposite ends of the same cylinder and superimposed. The two diagrams are divided into ten equal spaces and then each end space is subdivided. The mean heights of the sub- divisions are measured and averaged to get the mean height of the whole end division, or average pressure in this case for the division. The average heights of divisions for diagram No. 1 are set down in a column on the left, while those 30 40 50 00 70 Displacement In Per Cent of Stroke Fig. 20. — Simpaon's Method for Finding Mean Effective Pressure of Indicator Cards. for No. 2 are ma the right; the sum of each column divided by ten and multiplied by the sprinf* scale gives the whole m.e.p. The heights of No. 1 in inches marked off eMtinuously on a slip of paper measured a total of 11.16 ins. and for No. 2, 10.79 ins., these quantities divided by 10 (number of strips) and multi- plied by the spring scale, 50, gives the m.e.p., as before. This method is often designated as Simpson's rule. The best and most commonly used method of area evaluation, whether for work or m.e.p. determination, is the planimeter, a well-known instrument specially designed for direct measurements of area. 14. Indicated Horse-power. Work done by the fluid in a cylinder, because it is most often determined by indicator card measurements, measures the indicated horse-power, but the term is also applicable to work that would be done by the execution of a certain cycle of pressure volume changes carried out at a specified rate. The mean effective pressure in pounds per square inch, whether of an indicator card or PV cycle, when multiplied by piston area in square inches, WORK AND POWER 45 gives the average force acting on the piston for one stroke, whether the cycle required one, two or x strokes for its execution, and this mean force multiplied by the stroke in feet gives the foot-pounds of work done by the cycle. Therefore, Let in.e.p.=mean eflfective pressure in pounds per square inch for the cycle referred to one stroke; " a = eflfective area of piston in square inches; L = length of stroke in feet; n = number of equal cycles completed per minute; JV = number of revolutions per minute; iS=mean piston speed = 2L AT feet per minute; iV 2= number of revolutions to complete one cycle = — . Then will the indicated horse-power be given by, T TT p _ (m.e.p.)Lan , ' i.n.r.- gg^QQ W __ (m.e .p.)LaAr ... 33000z~" ^ ^ _ (m.e.p.)aS . . ""33000X22 ^^^ (17) When there are many working chambers, whether in opposite ends of the same C3'linder or in separate cylinders, the indicated horse-power of each should be found and the sum taken for that of the machine. This is important not only because the eflfective areas are often unequal, as, for example, in opposite ends of a double-acting cylinder with a piston rod passing through one side only or with two piston rods or one piston rod and one tail rod of unequal diameters, but also because unequal valve settings which are most common will cause diflferent pressure volume changes in the various chambers. It is frequently useful to find the horse-power per pound mean effective pressfiire, which may be symbolized by Ke, and its value given by j^ _ Lan _ LaN ^"33000" 330007 Using this constant, which may be tabulated for various values of n, stroke and bore, the indicated horse-power is given by two factors, one involving cylinder dimensions and cyclic speed or machine characteristics, and the other the resultant PV characteristic, of the fluid, symbolically, I.H.P. = iiLc(m.e.p.). These tables of horse-power per pound m.e.p. are usually based on piston speed rather than rate of completion of cycles and are, therefore, directly applicable 46 ENGINEERING THERMODYNAMICS when z=i or n=2iV, which means that the two cycles are completed in one revolution, in which case, S=2Li\r=Ln, and " 33000' whence LH.P.=ir,(m.e.p.)=-^5||^?^ (18) Table XI at the end of this chapter gives values of (H.P. per lb. m.e.p.) or Ke for tabulated diameters of piston in inches and piston speeds in feet per minute. Tables are frequently given for what is called the engine constant, which is variously defined as either (a) ob?)7)Q> which must be multiplied by m.e.p. Xn to obtain H.P., or (6) QQQQQ ? which must be multiplied by m.e.p. XL Xn to obtain H.P. For an engine which completes two cycles per revolution, this is the same as multiplying by m.e.p. XS. Before using such a table of engine constants it must be known whether it is computed as in (a) or in (6). Example. A 9 in. Xl2 in. double-acting steam engine runs at 250 R.P.M. and the mean effective pressm-e is 30 lbs. What is H.P. per pound m.e.p. and the I.H.P.? j^ Lan 1X63.6X500 33000 33000 I.H.P. = .9636 X 30 =28.908. Prob. 1- A pump has a piston speed of 250 ft. per minute; piston diameter is 24 ins. What is the H.P. per pound mean effective pressure? . Prob. 2. A simple single-acting 2-cylinder engine has a piston 10 ins. in diameter with a 2-in. rod and a stroke of 15 ins. It runs with a mean effective pressure of 45 lbs. per square inch at a speed of 220 R.P.M. What is the H.P.? Prqb. 3. A gas engine has one working stroke in every four. If the speed is 300 R.P.M. what must be the m.e.p. to give 6 H.P. when the cylinder has a diameter of 6 ins. and a stroke of 12 ins.? Prob. 4. An air compressor is found to have a mean effective pressure of 50 lbs. If the cylinder is double acting and is 12 ins. diameter and 16 ins. stroke, what H.P. will be needed to drive it at 80 R.P.M.? Prob. 5. A gasoline engine has an engine constant (a) of .3. What must be the m.e.p. to give 25 H.P.? Prob. 6. A blowing engine has an m.e.p. of 10 lbs. Its horse-power is 500. What is the H.P. per pound m.e.p.? WORK AND POWER 47 Prob. 7. Two engmes of the same size and speed are so run that one gives twice the power of the other. How will the engine constants and m.e.p. vary? Prob. 8. From the diagrams following Section 9 what must have been the H.P. per pound m.e.p. to give 300 H.P. in each case? Prob. 9. How will the H.P. per pound m.e.p. vary in two engines if the speed of one is twice that of the other, if the stroke is twice, if the diameter of piston is twice? 16. Effective Horse-power, Brake Horse-power, Friction Horse-power, Mechanical Efficiency, Efficiency of Transmission, Thermal Efficiency. Work is done and power developed primarily in the power cylinder of engines, and is transmitted through the mechanism with friction loss to some point at which it is utilized. There is frequently a whole train of transmission which may involve transformation of the energy into other fonns, but always with some losses, including the mechanical friction. For example, a steam cylinder may drive the engine mechanism which in turn drives a dynamo, which transforms mechanical into electrical energy and this is transmitted to a distance over wires and used in motors to hoLst a cage in a mine or to drive electric cars. There is mechanical work done at the end of the system and at a certain rate, so that there will be a certain useful or effective horse-power output for the system, which may be compared to the horse-power primarily developed in the power cylinders. A similar comparison may be made between the primary power or input and the power left after deducting losses to any intermediate point in the system. For example, the electrical energy per minute delivered to the motor, or motor input, is, of course, the output of the transmission line. Again, the electrical energy delivered to the line, or electrical transmission inpuij is the same as dj'^namo output, and mechanical energy delivered to the dynamo Ls identical with engine output. The comparison of these measure- ments of power usually takes one of two forms, and frequently both; first, a comparison by diflfcrences, and second, a comparison by ratios. The ratio of any horse-power measurement in the system to the I.H.P. of the power cylin- der is the e^Hciency of the power system up to that point, the difference between the two is the horse-power loss up to that point. It should be noted that, as both the dynamo and motor transform energy from mechanical to electrical or vice versa, the engine mechanism transmits mechanical energy and the wires electrical energy, the system is made up of parts which have the function of (a) transmission without change of form, and (Jb) transformation of form. The ratio of output to input is always an efficiency ^ so that the efficiency of the power system is the product of all the efficiencies of transformation and of transfer or transmission, and the power loss of the system is the sum of trans- fonnation and transmission losses. Some of these eflSeiencies and losses have received names which are generally accepted and the meaning of which is gen- erally understood by all, but it is equally important to note that others have no names, simply because there are not names enough to go around. In dealing with eflSciencies and power losses that have accepted names these names may with reason be used, but in other cases where names are differently under- stood in different places or where there is no name, accurate description must be 48 ENGINEERING THERMODYNAMICS relied on. As a matter of fact controversy should be avoided by definition of the quantity considered, whether descriptive names be used or not. Effective horse-power is a general term applied to the output of a machine, or power system, determined by the form of energy output. Thus, for an engine it is the power that might be absorbed by a friction brake applied to the shaft, and in this case is universally called Brake Horse-power. The difference between brake and indicated horse-power of engines is the friHion horse-power of the engine and the ratio of brake to indicated horse-power is the mechanical efficiency of the engine. For an engine, then, the effective horse-power or useful horse- power is the brake horse-power. When the power cylinders drive in one machine a pump or an air compressor, the friction horse-power of the machine is the difference between the indicated horse-power of the power cylinders and that for the pump or compressor cylinders, and the mechanical eflBciency is the ratio of pump or air cylinder indicated porse-power to indicated horse-power of the power cylinders. Whether the indicated horse-power of the air or pump cylinders can be considered a measure of useful output or not is a matter of difference of opinion. From one point of view the machine may be as considered built for doing work on wat^r or on air, in which case these horse-powers may properly be considered as useful output. On the other hand, the power pump is more often considered as a machine for moving water, in which case the useful work is the product of the weight of water moved into its head in feet, and includes all friction through ports, passages and perhaps even in pipes or conduits, which the indicated horse-power of the pump cylinder does not include, especially when leakage or other causes combine to make the pump cylinder displacement differ from the volume of water actually moved. With compressors the situation is still more complicated, as the air compressor may be considered useful only when its discharged compressed air has performed work in a rock drill, hoist or other form of an engine, in which case all sorts of measm-es of use- ful output of the compressor may be devised, even, for example, as the purely hypothetically possible work derivable from the subsequent admission and complete expansion of tha compressed air in a separate air engine cylinder. Too accurate a definition, then, of oviput and inpvt energy in machines and power systems is not possible for avoidance of misunderstanding, which may affect questions both of power losses and efficiency of transmission and trans- formation whether in a power system or single machine. It is interesting to note here that not only is the indic/ited work of the power cylinder always con- sidered the measure of power input for the system or machine, but, as in the other cases, it is itself an output or result of the action of heat on the vapor or gas and of the cycle of operations carried out. The ratio of the indicated power or cylinder work, to the heat energy both in foot-pound units, that was expended on the fluid is the thermal efficiency of the engine referred to indicated horse- power or the efficiency of heat transformation into work, the analysis of which forms the bulk of the subject matter of Chapter VI. Similarly, the ratio of any power measurement in the system to the equivalent of the heat supplied is the thermal efficiency of so much of the system as is included. WORK AND POWER 49 Example.. It has been found that when the mdicated horse-power of an engine is 250, a generator is giving out 700 amperes at 220 volts. At end of a transmis ion line is a motor using the output of the generator. This motor on test gave out 180 brake horse- power. Assuming no losses in the transmission line, what was the efficiency of the motor, of the generator, of the engine, and of the system? Motor efficieacy-^-^-87^ 746 Note: Volts Xamperes= watts, and, watts -^ 746 =H.P. 220x700 Engine and generator efficiency = — ^j- — a82.4%. 250 180 Efficiency of system = — • =72% or 82.4 x87.2 -72%. iuOU Prob. 1. An engine is belted to a pmnp; the I.H.P. of the engine is 50, of the pump iO, and the pump delivers 1200 gallons water per minute against 100-ft. head. What is the efficiency of each part and of the entire system. Prob. 2. An engine is geared to air compressor. Upon test, the efficiency of the engine alone, glaring alone and compressor alone were each 80 per cent. When the com- pressor H.P. was 100 what was that of the engine? Prob. 3* A water-wheel is run by the discharge from a pump. The B.H.P. of wheel is foimd to be 20 when the pump is delivering 45 gallons of water per minute at a head of 1000 lbs. per square inch. The water I.H.P. of the pump is 30 and the steam I.H.P. is 40. What are the efficiencies of each part of the system and the over-all efficiency? Prob. i. Perry ^ves a rule for the brake horse-power of steam engines as being equal to .95 I.H.P. — 10. On this basis find the mechanical efficiency of a 500-H.P. engine from 200 to 500 H.P. Show results by a curve with B.H.P. and per cent efficiency as coordinates. Prob. 6- Perry gives a rule for the efficiency of an hydraulic line bsH = ,7I —25 where // is the useful power of the pump and / is the indicated. Find / for values of H from 100 to 300 and plot a curve of results. Prob. 6. An engine gives one I.H.P. for every 3 lbs. of coal per hour. One pound of coal contains 9,500,000 ft.-lbs. of energy. What is the thermal efficiency? Prob. 7. A gas engine has a mechanical efficiency of 70 per cent when delivering power to a generator which in turn has an efficiency of 90 per cent. If the engine uses 15 cu.ft. of gas per indicated horse-power hour and the gas contains 700,000 ft.-lbs. per cubic foot, what is the net thermal efficiency of the system? 16. Specific Displacement, Quantity of Fluid per Hour or per Minute per IJ5.P. It has been shown that the work done in cylinders by pressure volume changes of the vapor or gas depends on the mean effective pressure and on the displacement, or that there is a relation between I.H.P. and displacement. The quantity of fluid used also depends on the displacement and may be expressed 50 ENGINEERING THERMODYNAMICS in cubic feet per minute at either the low pressure or high pressure con- dition when the work is done between two definite pressure limits, or in terms of pounds per minute or hour, which involves the application of fluid densities to volumes and which eliminates the double expression for the two conditions of pressure. The displacement per hour per horse-power, termed the specific displacement, is the basis of computations on the steam consumption of steam engines, the horse-power per cubic feet of free air per minute for air compressors, the horse-power per ton refrigeration for refrigerating machines and the con- sumption of fuel per hour per horse-power for gas and oil engines. It is, therefore, a quantity of great importance in view of these applications. Apply- ing the s3anbols already defined to displacement in one direction of one side of a piston Displacement in cu.ft. per stroke =LXyt7J Displacement in cu.ft. per minute ^LXYrrXiV; Displacement in cu.ft. per hour =60LXYjTXiV. T J- X J 1 (m.e.p.)Lan (m.e.p.)LaN Indicated horsepower 33^^^^^ SSOOOT"- Whence expressing displacement per hour per I.H.P. or specific displacement in one direction for one side of a piston by D«, ^ ^ 60LXj-jjXiV ^ gQ^33QQQ^ ^ 1 3750g ^ - .^g. * (p^-e .pQLaJV 144(m.e.p.) (m.e.p.) ~3300(te From Eq. (19) it appears that (he specific displacement is equal to zy,13,750 divided by the mean effective pressure in pounds per square inch. If two points, A and B, be so located on the indicator card. Fig. 21, as to have included between them a fluid transfer phase, either admission to, or expulsion from the cylinder, then calling 8a = pounds per cubic foot or density at point A J and 8^= pounds per cubic foot or density at point B, the weight of fluid present at A is, {Da+Cl)da lbs., and weight of fluid present at B is {Db+Cl)db lbs., whence the weight that has changed places or passed in and out per stroke is, {Db+Ct)8b-{Da+Cr)da lbs. per stroke. WORK AND POWER 51 If both A and B lie on the same horizontal as A and B', ^a = ^6=^, the density of fluid at the pressure of measurement, whence the weight of fluid used per stroke, will be and the voliune per stroke used at density d is Db' — Da cu.ft., which compared to the displacement is Dt'-Da D ' P 1 \ 1 \ \ \ > \ \ M \ \ A 4Hl»Mi^< >a^ ^^B^fe B' -^ \ \i- v^ ^ \ s -1 \ Fig. 21. — Determination of Consumption of Fluid per Hour per Indicated Horse-power from the Indicator Card. This is the fraction of the displacement representing the volume of fluid pass- ing through the machine at the selected pressiu*e. Multiplying the specific displacement by this, there results, Cu.ft. of fluid per hr. at density (d) per I.H.P. = ^ ^ — ^^ — -y (m.e.p.) D and Lb6.offluidperhr.perLH.P=7-^^^f^^^^V (20) (m.e.p.) \ D / ^ ^ More generally, that is, when A and B are not taken at the same pressures Lbs. fluid per hr. per IM.F.-- J^^JJ(Db+Cl)^J,'^-(Da+Cl)^a^ . (21) The particular forms which this may take when applied to special cases will be examined in the succeeding chapters. 62 ENGINEERING THERMODYNAMICS Example. An air compressor whose cylinder is 18x24 ins. (18 ins. in diameter and stroke 2 ft.) runs at 60 R.P.M. and is double acting. The mean effective air pressure is 50 lbs. per square inch. What is the specific displacement? Cu.ft. per hour =^^^ = 60 X2 X^^4 Xl20 =25,600. 144 144 TTTP _m.e .p.Lan _50 x2x254.5xl20 _^oo ^•^•^- "^^OOO 2Sm ^^''^'' Cu.f t. per hour ^ 25,600 ^gy- I.H.P. 92.3 ' or by the formula directly, 13,750 13,750 ^^ m.e.p. 50 Prob. 1. What will be the cubic feet of free air per hour per horse-power delivered by a 56x72-in. blowing engine with 4 per cent clearance and mean effective pressure of 10 lbs. per square inch? Prob. 2. An 18x22-in. ammonia compressor works with a mean effective pressure of 45 lbs. per square inch. What is the weight of NHt per I.H.P. hour if the speed is 50 R.P.M. and compressor is double acting having a volumetric efficiency of 90 per • cent? Use tabular NHz densities. Prob. 3. A steam engine whose cylinder is 9 X 12 ins. runs at a speed of 300 R.P.M. and is double acting. If the m.e.p. is 60 lbs. and the density of steam at end of the stroke is .03, how many pounds of steam are used per hour per indicated horse-power? 17. Velocity Due to Free Expansion by PV Method. All the cases examined ior the work done by P7 cycles have been so far applied only to their execution in cylinders, but the work may be developed in nozzles accelerating the gas or vapor in free expansion, giving, as a consequence, a high velocity to the fluid. It was noted that for cylinders many combinations of phases might be found worthy of consideration as, typical of possible actual conditions of practice, but this is not true of proper nozzle expansion, which has but one cycle, that of Fig. 22. That this is the cycle in question is seen from the following considera- tions. Consider a definite quantity of the gas or vapor approaching the nozzle from a source of supply which is capable of maintaining the pressure. It pushes forward that in front of it and work will be done, ABCD^ equal to the admission of the same substance to a cylinder, so that its approach AB may be considered as a constant- pressure, volume- increasing phase for which the energy comes from the source of supply. This same substance expanding to the lower pres- sure will do the work CBEF; but there will be negative work equivalent to the pushing awary or displacing of an equivalent quantity of fluid at the low pres- sure, or FEGDj making the work cycle ABEG, in which AG is the excess of WORK AND POWER 53 initial over back pressure or the effective working pressure, remaining constant during approach and lessening regularly during expansion to zero excess at E. The work done will be from Eq. (13), w=p.v.+i^[i-{^yyp.v.. A B » ■ \ ■ \ \ L * \ \ \ \ X, G ^v . F Fig. 22. — Pressure-Volume Diagram for Nozzle Expansion Measuring the Acceleration Velocity and Horse-power of Jets. But PeVe' = PeVeVe'-^=PbVb' = PbVbVb»-''; 8-1 .\ PeVe = PbVb m"-^^^w-- 8-1 8-1 Whence r-..K..S[.-(ft)-]-P.K.(';=) a-l -.-^^"-'[•-(ft) ■ ] <'^' 54 ENGINEERING THERMODYNAMICS Aissuming the initial velocity to be zero, and the work of Eq. (22) to be done on 1 lb., the final or resultant velocity will be according to Ekj. (6), T"^ Y - u=V2gW • ••••• V^^/ or This velocity is in feet per second when pressures are in pounds per square foot and volumes in cubic feet of 1 lb. of substance, and is known as Zeuner's equation for the velocity of a gas or vapor expanding in a nozzle. It is generally assiuned that such expansion, involving as it does very rapid motion of the fluid past the nozzle, is of the adiabatic sort, as there seems to be no time for heat exchange between fluid and walls. As already noted, the value of s for adiabatic expansion of vapors is not constant, making the correct solution of problems on vapor flow through orifices practically impossible by this method of pressure volume analysis, but as will be seen later the thermal method of solution is exact and comparatively easy. Note. A comparison of Eqs. (22) and (13) and the figures corresponding will show that the area under the process curve, which is the same as the work done during the compression or expansion,- if multiplied by 8 will equal the area to the left of the process curve, which in turn represents, as in Fig. 21, for engines, the algebraic sum of admission, complete expansion, and exhaust work areas, or as in Fig. 9 for compressors, the algebraic sum of suction, compression and delivery work areas. This statement must not be thought to refer to the work area of such a cycle as that of Fig. 8, where expansion is incomplete, nor a case of over-expansion, Fig. 10. Example. In Fig. 22 assume the initial pressure at 100 lbs. per square inch absolute, back pressure at atmosphere, and expansion as being adiabatic! What will be the work per pound of steam and the velocity of the jet, if F^ is 4.36 cu.ft. and s = 1.3 for superheated steam? .1|X144X100X4.36[1-Q"] =27,206 X. 608 = 16,541 ft.-lbs.; tt = V^=8.02Vl6,541; = 1028 ft. per second. WORK AND POWER 55 Prob. 1. Taking the same pressure range as above, find W and u for adiabatic expan- sion of air, also for isothermal expansion. Prob. 2. How large must the effective opening of the suction valve be, in an air compressor 18x24 ins. to aUow the cylinder to properly fill if the mean pressure-drop through the valve is 1 lb. per square inch and the compressor runs at 80 R.P.M.? Prob. 3. What must be the diameter of the inlet valve in a gas engine to fill a cylinder of i cu.ft. capacity if the lift of the valve is | ins., allowing a pressure drop of 1 lb. per square inch? Engine makes 150 working strokes per minute. Prob. i. It has been found from experiment that the velocity of air issuing from a hole in plate orifice is 72 per cent of what would be expected from calculation as above when the absolute pressure ratio is 2 to 1, and 65 per cent when the absolute pressure ratio is 1} to 1. What will be the actual velocity for air flowing from a tank to atmosphere for these pressure ratios? Prob. 5. COs stored in a tank is allowed to escape through an orifice into the air. What will be the maximum velocity of the jet if the pressure on the tank be 100 lbs. per square inch gage? ' . Note: 1 lb. CO* at pressure of 100 lbs. per square inch gage occupies 1.15 cu.ft. Prob. 6. If ammonia gas and hydrogen were allowed to expand from the same pres- sure, how would their maximum velocities compare? Vol. of 1 lb. of NH| at 50 lbs. per square inch gage is 4.5 cu.ft. Vol. of 1 lb. of H at same pressure is 77.5 cu.ft. 18. Weight of Flow through Nozzles. Applying an area factor to the velocity equation will give an expression for cubic flow per second which becomes weight per second by introducing the factor, density. Let the area of an orifice at the point of maximum velocity, u,he A sq.ft., then will the cubic feet per second efflux be Au, Assume the point of maxi- mum velocity, having area A, to be that part of the nozzle where the pressure has fallen to P«, Fig. 22, and the gas or vapor to have the density 80 pounds per cubic fcx)t. Then will the nozzle flow in pounds per second be But the weight per cubic foot is the reciprocal of the cubic feet per pound, Fe, which it has already been assumed, is the final volume, of one pound of the fluid. Hence, ilA This may be put in terms of initial gas or vapor conditions for, 1 /PA Whence -<w- 1 uA uA/Pe\» (f) 66 ENGINEERING THERMODYNAMICS Substituting in this the value of u from Eq. (24), -»"^.(ft)"'"{;^i''''''['-(K)~]r"»'»'"--- « This weight will be a maximum for a certain value of the pressure ratio, depend- ing on the value of 8 only, and this value can be found by placing the first dif- ferential coeflScient of w with respect to f p^j equal to Kft) =0. To accomplish this, rearrange Eq. (25) as follows: w=A ^^Mm-mf- But as the other factors do not enter to effect the result so long as Pb does not varv. t& is a maximum when the bracket vary, t& is a maximum when the bracket 2 « + i or is a maximum or when (rKr""-H-»- But as ( p^j* cannot be equal to zero in practice, then which gives the condition that w is a maximum when or (Pe\~r 8 + 1 \PbJ ~_ 2 ' WORK AND POWER 57 or mayitniim flow for given iiutial pressure occurs when i^H^r ,<»' Pe For air expanding adiabatically 8 '=1.407. Maximum flow occurs when = .528 and for most conunon values of s it will be between .50 and .60 Pb This result is quite remarkable and is verified by experiment reasonably closely. It shows that, contrary to expeciaUori, the weight of effliix from nozzles mil not continuously and regularly increase with increasing differences in pressure, bid for a given initial pressure the weight discharged per second will have reached its limit when the final pressure has been diminished to a certain fraction of the im'tial, and any further decrease of the discharge pressure vrill not increase the flow tiirough an orifice of a given area. The subject of flow in nozzles will be treated more completely in Chapter VI. Prob. 1« For the following substances under adiabatic expansion determine the pressure ratio for maximum flow and find the rate of flow per square inch of orifice under this condition when flow is into a vacuum of 10 ins. of mercury with standard barometer: (a) Carbon dioxide. (b) Nitrogen. (c) Hydrogen. (d) Ammonia. (e) Dry steam according to saturation law. 19. Horse-power of Nozzles and Jets. Although, strictly speaking, nozzles can have no horse-power, the term is applied to the nozzle containing the orifice through which flow occurs and in which a certain amount of work is done per minute in giving to a jet of gas or vapor initially at rest a certain final velocity, and amount of kinetic energy. The foot-pounds of work per pound of fluid multiplied by the pounds flowing per second will give the foot- pounds of work developed per second within the nozzle, and this divided by 550 will give the horse-power developed by the jet, or the nozzle horse-power. Accordingly, 1 H-P-^^'«*=-556-^-560X7;y ^" •• • ^^^ W A/P 550 F, w^ W . (6) -"<(&) •{i^''.''.['-(g)f]}*.<^ (27) 58 ENGINEERING THERMODYNAMICS where the expression in the bracket is the work done per pound of substance. The pressures are expressed in pounds per square foot, areas in square feet and volumes in cubic feet. Etample. A steam turbine operates on wet steam at 100 lbs. per square inch abso- lute pressure which is expanded adiabatically to atmospheric pressure. What must be the area of the nozzles if the turbine is to develop 50 H.P. ideally? Note: 1 cu.ft. of steam at 100 lbs. = .23 lb. / 2 \«-i By Eq. (26), maximum flow occurs when the pressure ratio is ("77) » or, for 1.11 this case when the pressure is 100^ (oTt) ~^^ ^^' ^^ square inch absolute. As the back pressure is one atmosphere, the flow will not be greater than for the above critical pressure. Substituting it in Eq. (25) will give the flow weight to, and using the actual back pressiu^ in Eq. (22) will give the work W. .11 T. T. /ooxTTT /lll\ /14400\r, / 2116 \ 111] ByEq.(22),W.(— )x(-— )[l-(^^) J = 110000 ft.-lbs. per pound of steam. ByEq.(25),«;=8.02Ax.23x(.58)rnjl^X^^|l-(.58)rri^ P = 198A lbs. of steam per second. By Eq. (27a), Wxw_ 110000 X198A 550 550 Whence Prob. 1. What will be the horse-power per square inch of nozzle for a turbine using hot gases if expansion follows law PVs—kf when s = 1.37, the gases being at a pressure of 200 lbs. per square inch absolute and expanding to atmosphere. Let the volume per pound at the high pressure be 2 cu.ft. Prob. 2. What will be the horse-power per square inch of nozzle for the problems of Section 17? Prob. 3. Suppose steam to expand according to law PFs= A:, where s = 1.111, from atmosphere to a pressure of 2 lbs. per sq. inch absolute. How will the area of the ori- fice compare with that of the example to give the same horse-power? Note: 76=-26.4. Prob. 4. Suppose steam to be superheated in the case of the ex£[mple and of the last problem, how will this affect the area of nozzle? Note: Let 7^ =5 and 32 respectively. Prob. 6. How much work is done per inch of orifice if initial pressure is 100 lbs. absolute on one side and final 10 lbs. absolute on other side of a valve through which air is escaping? WORK AND POWER 59 GENERAL PROBLEMS ON CHAPTER I 1. An air compressor is required to compress 500 cu.ft. of free air per minute to a pressure of 100 lbs. per square inch gage; the compressor is direct connected to a steam engine. The mechanical efficiency of the machine is 80 per cent. What will be the steam horse-power if compression is (a) isothermal; (6) adiabatic? 2. A mine hoist weighing 10 tons is raised 4500 ft. in one minute. In the first 20 seconds it is accelerated from rest to a speed of 6000 ft. per minute; during the next 30 seconds speed is constant at this value, and during last 10 seconds it is brought to rest. What will be (a) work of acceleration for each period; (6) work of lift for each period; (c) total work supplied by engine; (d) horse-power during constant velocity period? 3. The en^e driving the above hoist is driven by compressed air. If air is supplied at a pressure of 150 lbs. per square inch gage and is admitted for three-quarters of the stroke, then expanded adiabatically for the remainder of the stroke and exhausted to the atmosphere find (a) what must the piston displacement be to lift the hoist, the work of acceleration being neglected? (6) To what value could the air pressure be reduced if air were admitted full stroke? 4. It is proposed to substitute an electric motor for the above engine, installing a water-power electric plant at a considerable distance. The type of wheel chosen is one in which a jet of water issuing from a nozzle strikes against a series of revolving buckets. The available head at the nozzle is 1000 ft. Assuming the efficiency of the motor to be 85 per cent, transmission 80 per cent, generators 90 per cent, and water-wheels 60 per cent, what will be the cubic feet of water per minute? 6. A steam turbine consists of a series of moving vanes upon which steam jets issuing from nozzles impinge. It is assumed that for best results the speed of the vanes should be half that of the jets. The steam expands from 100 lbs. per square inch gage to 5 lbs. per square inch absolute, (a) What must be the best speed of vanes for wet steam where 8 = 1.111? (&) If 55 per cent of the work in steam is deUvered by the wheel what must be the area of nozzle per 100 H.P., and weight of steam per hour 100 H.P? Note: 1^6=3.82. 6. It has been found that a trolley car uses a ciurent of 45 amperes at 550 volts when running 12 miles per hour. If the motor is 80 per cent efficient, what is tractive effort? Note : Volts X amperes = watts, and watts -i- .746 = H.P. 7. An engine whose cylinder is 18x24 ins. (diameter = 18 ins., stroke =24 ins.) is double acting and runs at a speed of 125 R.P.M. Steam is admitted for one-quarter stroke at a pressure of 125 lbs. per square inch gage, allowed to expand for the rest of the stroke and then exhausted into a vacuum of 28.25 ins. of Hg. (a) Draw a PV diagram of the cycle and find the m.e.p. from the diagram and by calculation, and then find the horse-power. (6) Consider steam to be admitted one-half stroke without other change. How will the horse-power vary? (c) What will be the horse-power for one-quarter admission if the exhaust pressure is 15 lbs. per square inch absolute? (d) What will be the horse-power if the steam pressure be made 150 lbs. per square inch absolute, other conditions as in (a)? (e) Suppose all conditions as in (a) but the speed lowered to 75 R.P.M. What will be the horse-power? 8. Assuming that 50 per cent of the work in the jet is transformed to useful work, what must be the total area of the nozzles of a steam turbine to develop the same horse- 60 ENGINEERING THERMODYNAMICS power as the engine in problem (7a), the pressure range being the same and 8 being 1.3? 76=3.18. 9. Water is being pumped from the bottom of a shaft 700 ft. deep at the rate of 1000 gallons per minute by an electrically driven pump. Efficiency of pump is 70 per cent, motor 90 per cent, transmission line 95 per cent, generator 85 per cent, and mechanical efficiency of engine 80 per cent. What will be the indicated horse-power of the engine? If the above installation were replaced with an air-driven pump of 05 per cent efficiency, efficiency of transmission being 100 per cent, and that of the compressor and engine 80 per cent, what would be the horse-power of this engine? 10. Show by a P7 diagram, assuming any convenient scales, that the quantity of air discharged byacompressor and the horse-power, both decrease asthe altitude increases, and that the horse-power per cubic foot of air delivered increases under the same condition. 11. A centrifugal pump is driven by a steam engine directly connected to it. The pump is forcing 1000 gallons of water per minute against a head of 250 ft. and runs at a speed of 450 R.P.M. The engine is double acting and its stroke equals the diameter of the cylinder. Steam of 100 lbs. per square inch gage is admitted for half stroke, allowed to expand the rest of the stroke so that s = 1, and is then exhausted to atmosphere. What must be the size of the engine if the pump efficiency is 65 per cent and the engine efficiency 75 per cent? 12. (a) What will be the pounds of steam used by this engine per hour per horse- power? (b) If the steam were admitted but one-quarter of the stroke and the initial pressure raised sufficiently to maintain the same horse-power, what would be the new initial pressure and the new value of the steam used per horse-power per hour? Note: Weight of steam per cubic foot for (a) is .261; for (6) is .365. 13. If it were possible to procure a condenser for the above engine so that the exhaust pressure could be reduced to 2 lbs. per square inch absolute, (o) how much would the power be increased for each of the two initial pressures already given? (6) How would the steam consumption change? 14. A motor-fire engine requires a tractive force of 1300 lbs. to drive it 30 miles per hour, its rated speed. The efficiency of engine and transmiasion is 80 per cent. When the same engine is used to actuate the pumps 70 per cent of its power is expended on the water. What wiU be the rating of the engine in gallons per minute when pumping against a pressure of 200 lbs. per square inch? 16. A compressor when compressing air at sea level from atmosphere to 100 lbs. per square inch absolute, expends work on the air at the rate of 200 H.P., the air being com- pressed adiabatically. (a) How many cubic feet of free air are being taken into the compressor p6r minute and how many cubic feet of high pressure air discharged? Compressor is moved to altitude of 8000 ft. (6) What will be the horse-power if the same amount of air is taken in and how many cubic feet per minute will be discharged? (c) What will be the horse-power if the same number of cubic feet are discharged as in case (a) and what will be the number of cubic feet of low pressure air drawn in? (d) Should superheated ammonia be substituted for air at sea level, what would be the necessary horse-power? 16. A car weighing 40 short tons is at rest and is struck by a train of four cars, each of the same weight as the first, (a) Upon impact the single car is coupled to the train and all move off at a certain velocity. If the original velocity of the train was 3 miles per hour, what will it be after attachment of tlie extra car? (6) If instead of coupling, the extra car after impact moved away from the train at twice the speed the train was then moving, what would be the speed of train? WOHK AND POWER 61 17. To drag a block of stone along the ground requires a pull of 1000 lbs. If it be placed on rollers the pull will be reduced to 300 lbs., while if it be placed on a wagon with well-made wheels, the pull will be but 200 lbs. Show by diagram how the work required to move it 1000 ft. will vary. 18. 100 cu. ft. of air, at atmospheric pressure, is compressed to six times its original pressure, (a) What will be the difference in horse-power to do this in 45 seconds isothermally and adiabaticafly at an elevation of 8000 ft. (6) What will be the final vol- umes? (c) What will be the difference in horse-power at sea level? {d) What will be the final volumes? 19. An engine operating a hoLst is run by compressed air at 80 lbs. per square inch gage. The air is admitted half stroke, then expanded for the rest of the stroke so that 8-1.3 and then exhausted to atmosphere. The engine must be {K>werful enough to lift a ton cage and bring it up to a speed of 2000 ft. per minute in 15 seconds. What will be' the necessary displacement per minute? 20. CJonstnict P7 diagrams for Probs. 1, 11, TS and 15, showing by them that the work of admission, compression or expansion, and discharge or exhaust, is equal to that found algebraically. 21. The elongation of wrought iron under a force F is equal to the force times the length of the piece divided by 25,000,000 times the cross-section of metal in the piece. A 4i in. pipe has 3.75 sq. ins. of metal; a line of this pipe 1 mile long is running full of water with a velocity of 600 ft. per minute. This is stopped in 1 second by closing a valve. Assuming pip>e did not burst, what would be the elongation? 22. Two steam turbines having nozzles of equal throat areas are operating on a steam pressure of 150 lbs. per square inch gage. One is allowing steam to expand to atmosphere the other to 2 lbs. per square inch absolute, both cases having an exponent for expansion of 1.11. Find the relation of the horse-power in the two cases. 23. The power from a hydro-electric plant is transmitted some distance and then used to drive motors of various sizes. At the time of greatest demand for current it has been found that 1000 horse-power is given out by the motors. Taking the average efficiency of the motors as 70 per cent, transmission efficiency as 85 per cent, generator efficiency as 85 per cent, and water-wheel efficiency as 70 per cent, how many cubic feet of water per second will the plant require if the fall is 80 ft.? 24. A small engine used for hoisting work is run by compressed air. Air is admitted for three-quarters of the stroke and then allowed to expand for the rest of the stroke in such a way that 8 »1.4 and finally exhausted to atmosphere. For the first part of the hoisting, full pressure (80 lbs. per square inch gage) is applied, but after the load has been accelerated the pressure is reduced to 30 lbs. per square inch gage. If the engine has two cylinders each 9 Xl2 in., is double acting and runs at 200 R.P.M., what will be the (a) horse-power in each case, (6) the specific displacement? 25. A steam engire has a cylinder 12x18 in., is double acting, and runs at 150 R.P.M. (a) What is the engine constant, and (6) horse-power per pound m.e.p.? 26. A water-power site has available at all times 3500 cu.ft. of water per min- ute at a 100-ft. fall. Turbines of 70 per cent efficiency are installed which take the place of two double-acting steam engines whose mechanical efficiencies were 85 per cent. The speed of the engines was 150 R.P.M. , m.e.p. 100 lbs. per square inch, and stroke was twice the diameter. What was the size of each engine? 27. Assuming the frictional losses in a compressor to have been 15 per cent, how many cu.ft. of gas per minute could a compressor operated by the above engines compress from atmosphere to 80 lbs. per square inch gage if s = 1.35? 62 ENGINEERINa THERMODYNAMICS Table I CONVERSION TABLE OF UNITS OF DISTANCE Meters. 1 Kilometers. Inches. Feet. SUtute Miles. Nautical Miles. 1 1000 0.0254 0.304801 1609.35 1853.27 0.001 1 0.0000254 0.0003048 1.60935 1.85327 39.37 39370.1 1 12 63360 72963.2 3.28083 3280.83 0.083333 1 5280 6080.27 0.000621370 0.62137 0.0000157828 0.000189394 1. 1 . 15157 0.000539587 0.639587 0.0000137055 0.000164466 0.868382 1. 1 In aocordance with U. S. Standards (see Smithsonian Tables). Table II CONVERSION TABLE OF UNITS OF SURFACE Sq. Meters. Sq. Inches. Sq. Feet. Sq. Yards. Acres. Sq. Miles. 1 .000645 .0929 .8361 4046.87 1550.00 1 144 1296 10.76387 .00694 1 9 43560 27878400 1 . 19599 .111 1 4840 3097600 .000247 .000206 1 640 001562 2589999 1 Table III CONVERSION TABLE OF UNITS OF VOLUME Cu. Meters. Cu. Inches. Cu. Feet. Cu. Yards. Lities (1000 Cu. Cm.) GaUons (U.S.) 1 61023.4 1 1728 46656 61.023 231 35.3145 .000578 1 27 .035314 .13368 1.3079 1000 .016387 28.317 264.170 00433 • .028317 .76456 .03704 1 .001308 .004951 7.4805 201.974 .001 .003785 1 3.7854 .26417 1 TABLES 63 Table IV. CONVERSION TABLE OF UNITS OF WEIGHT AND FORCE 1 Kilocrammea. Metric Tons. Pounds. U. S* or Short Tons. British or Long Tons. 1. 1000. 0.453593 907.186 1016.05 0.001 1. 0.000453593 0.907186 1.01605 2.20462 2204.62 1. 2000. 2240. 0.00110231 1 . 10231 0.0005 1. 1.12000 0.000984205 0.984205 0.000446429 0.892957 1. Table V CONVERSION TABLE OF UNITS OF PRESSURE One lb. per sq. ft One lb. per sq. in One ounce per sq. in One atmoephere (standard at sea level) One kilogramme per square meter . . One gramme per square millimeter . One kilogramme per square centi- meter FLUID PRESSURES One ft. of water at 39.1° F. (max. dens.) One ft. of water at 62° F One in. of wat«r at 62° F One in. of mercury at 32° F. (stand- ard) * One centimeter of mercury at 0° C. . One ft. of air at 32° F., one atmos. press One ft. of air, 62° F Pounds i>er Square Foot. 1 144. 9. 2116.1 20.4817 204.817 2048.17 62.426 62.355 5.196 70.7290 27.8461 0.08071 0.07607 Pounds per Square Inch. 0.006944 1. 0.0625 14.696 0.142234 1.42234 14.2234 0.43350 0.43302 0.036085 0.491174 0.193376 0.0005604 0.0005282 Inches of Mercury at 32° F. 0.014139 2.03594 0.127246 29.924 0.289579 2.89579 28.9579 0.88225 0.88080 0.07340 1. 0.393701 0.0011412 0.0010755 Atmospheres (Standard at Sea Level). 0.0004724 0.06802 0.004252 1. 0.009678 0.09678 0.9678 0.029492 0.029460 0.002455 0.033416 0.013158 0.00003813 0.00003594 > pR£ASCRE8 M KARUBBD BT THB Mkrcurt Column. For temperatures other than 32^ F.. the density of mercury, pounds per cubic inch, and hence the pressure, pounds per square inch, due to a column of mercury 1 inch hicn, is given with sufficient accuracy by the following formula: p-0.4912-a-32) XO.OOOl. The mercurial barometer is commonly made with a brass scale which has its standard or correct Length at 62** F, and a linear coefficient of expansion of about 0.000001 for each degree Fahrenheit. Hence, to correct the standard of mercury at 32° F., the corrected reading will be where Hf ia the obeerved height at a temperature of t^ F. 64 ENGINEERING THERMODYNAMICS Table VI CONVERSION TABLE OF UNITS OF WORK ' Kilogrammeten. Foot-pounds. Foot Tons (Short Tons). Foot Tons (Long Tons). 1. 0.138255 276.510 309.691 7.23300 1. 2000. 2240. 0.00361650 0.000500 1. 1.12000 0.00322902 0.000446429 0.892857 1. 1 See also more complete table of Units of Work and Energy in Chapter IV on Work and Heat. Table VII CONVERSION TABLE OF UNITS OF POWER Foot-pounds per Second. Foot-pounds per Minute. Horse-power. Cheval-Vapeur. Kilogrammeters per Minute. 1. 0.0166667 550.000 542.475 0.120550 60. 1. 33000. 32548.5 7.23327 0.00181818 0.000030303 1. 0.986319 0.000219182 0.00184340 0.0000307241 1.01387 1. 0.000222222 8.29531 0.138252 4562.42 4500.00 1. Table VIII UNITS OF VELOCITY One foot per second One foot per minute One statute mile per hour One nautical mile per hour = 1 knot. One kilometer per hour One meter per minute , One centimeter per second ......... Feet per Minute. Feet per Second. 60. 1. 1. 0.016667 88. 1.4667 101.338 1.6890 54.6806 0.911344 3.28084 0.054581 2.00848 0.032808 TABLES 63 Table IX TABLE OF BAROMETRIC HEIGHTS, ALTITUDES, AND PRESSURES (Adapted from Smithsonian Tables) Barometric heights are given in inches and millimeters of mercury at its standard density (32** F.). Altitudes are heights above mean sea level in feet, at which this barometric height is standard. (See Smithsonian Tables for corrections for latitude and temperature.) Pressures given are the equivalent of the barometric height in lbs. per sq. in. and per sq. ft. Standard Barometer. Altitude, Feet above Sea Level. Pressure, Pounds per Inches. Centimeters. Square Inch. Square Foot. 17.0 17.2 17.4 17.6 17.8 43.18 43.69 44.20 44.70 45.21 15379 15061 14746 14435 14128 8.350 8.448 8.546 8.645 8.742 1202.3 1216.6 1230.7 1244.8 1259.0 18.0 18.2 18.4 18.6 18.8 45.72 46.23 46.73 47.24 47.75 . 13824 13523 13226 12931 12640 8.840 8.940 9.038 9.136 9.234 1273.2 1287.3 1301.4 1315.6 1329.7 19.0 19.2 19.4 19.6 19.8 48.26 48.77 49.28 49.78 50.29 12352 12068 11786 11507 11230 9.332 9.430 9.529 9.627 9.726 1343.8 1357.9 1372.1 1386.3 1400.4 20.0 20.2 20.4 20.6 20.8 50.80 51.31 51.82 52.32 52.83 10957 10686 10418 10153 9890 9.825 - 9.922 10.020 10.118 10.217 1414.6 1428.7 1442.9 1457.0 1471.2 21.0 21.2 21.4 21.6 21.8 53.34 53.85 54.36 54.87 55.37 9629 9372 9116 8863 8612 10.315 10.414 10.511 10.609 10.707 1485.3 1499.4 1513.6 1527.7 1541.8 22 22.2 22.4 22.6 22.8 55.88 56.39 56.90 57.40 57.91 8364 8118 7874 7632 7392 10.806 10.904 11.002 11.100 11.198 1556.0 1570.1 1584.3 1598.4 1612.6 23.0 23.2 23.4 23.6 23.8 58.42 58.92 59.44 59.95 60.45 7155 6919 6686 6454 6225 11.297 11.395 11.493 11.592 11.690 1626.7 1640.8 1655.0 1669.3 1683.3 24.0 24.2 24.4 24.6 24.8 60.96 61.47 61.98 62.48 62.99 5997 5771 5547 5325 5105 11.788 11.886 11.984 12.083 12.182 1697.4 1711.6 1725.7 1739.9 1754.0 25.0 25.2 25.4 25.6 63.50 64.01 64.52 65.02 65.53 4886 4670 4455 4241 4030 12.280 12.377 12.475 12.573 12.671 1768.2 1782.3 1796.5 1810.7 1824.8 66 ENGINEERING THERMODYNAMICS Table IX — Continued Altitude, Feet above Sea Level. Pressure, Pounds per Inches. Centimeters. Square Ineh. Square Foot. 26.0 26.1 26.2 26.3 26.4 65.04 66.30 66.55 66.80 67.06 3820 3715 3611 3508 3404 12.770 12.819 12.868 12.918 12.967 1838.9 1846.0 1853.1 1860.2 1867.3 26.5 26.6 26.7 26.8 26.9 67.31 67.57 67.82 68.08 68.33 3301 3199 3097 2995 2894 13.016 13.065 13.113 13 163 13.212 1874.3 1881.4 1888.5 1895.5 1902.6 27.0 27.1 27.2 27.3 27.4 68.58 68.84 69.09 69.34 69.60 2793 2692 2592 2493 2393 13.261 13.310 13.359 13.408 13.457 1909.7 1916.7 1923.8 1930.9 1938.0 27.6 27.6 27.7 27.8 27.9 69.85 70.10 70.35 70.61 70.87 2294 2195 2097 1999 1901 13.507 13.556 13.605 13.654 13.704 1945.1 1952.1 1959.2 1966.3 1973.3 28.0 28.1 28.2 28.3 28.4 71.12 71.38 71.63 71.88 72.14 1804 1707 1610 1514 1418 13.753 13.802 13.850 13.899 . 13.948 1980.4 1987.5 1994.5 2001.6 2008.7 28.5 28.6 28.7 28.8 28.9 72.39 72.64 72.90 73.15 73.40 1322 1227 1132 1038 943 13.998 14.047 14.096 14.145 14.194 2015.7 2022.8 2030.0 2037.0 2044.1 29.0 29.1 29.2 29.3 29.4 73.66 73.92 74.16 74.42 74.68 849 756 663 570 477 14.243 14 . 293 14.342 14.392 14.441 2051.2 2058.2 2065.3 2072.4 2079.4 29.5 29.6 29.7 29.8 29.9 74.94 75.18 75.44 75.69 75.95 384 292 261 109 +18 14.490 14.539 14.588 14.637 14.686 2086.5 2093.6 2100.7 2107.7 2114.7 29.92 76.00 14.696 2116.1 30.0 30.1 30.2 30.3 30.4 76.20 76.46 76.71 76.96 77.22 - 73 -163 -253 -343 -433 14.734 14.783 14.833 14.882 14.931 2121.7 2128.8 2135.9 2143.0 2150.1 30.5 30.6 30.7 30.8 30.9 77.47 77.72 77.98 78.23 78.48 -522 -611 -700 -788 -877 14.980 15.030 15.078 15.127 15 . 176 2157.2 2164.2 2171.3 2178.4 2185.5 31.0 78.74 -965 15.226 2192.6 TABTiES 67 Table X VALUES OF s IN THE EQUATION PV =CONg TANT FOR VARIOUS SUBSTANCES AND CONDITIONS Substanee. 8 Remarks or Authority. Ml eaws ,.,..., Isothermal Constant pressure Isothermal Constant volume Adiabaiic Compressed in cylinder Adiabatic, wet Adiabatic, superheated Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Adiabatic Expanding in cylinder Saturation Law 1 ] 00 1.4066 1.4 1.1 1.3 1.293 1.300 1.403 1.200 1.323 1.106 1.029 1.410 1.276 1.316 1.410 1.291 1.24 1.26 1.300 Variable 1.111 1 + 14 X% moist. 1.035 H-.1X% moist. 1. 1.0646 All gases and vapors . . All saturated vapors . . All gases and vapors . . Air Accepted thermody- namic law Smithsonian Tables Air Experience Ammonia (NHs) Ammonia (NHs) Brnmine Average Thermodynamics Strecker Carbon dioxide (COi) . Carbon monoxide (CO) Carbon disulphide (OS,) Rontgen, Wullner Cazin, Wullner Beyne Chlorine (Q) Chloroforuu (CCl,CH(OH),).... Ether (CiHiOCH,) . . . Hydrogen (H,) Hydrogen sulph . (HsS) Methane (OH4) Nitrogen (N«) Nitrous oxide (NOj) . . Pintsch gas Strecker Be3me, Wullner MDller Cazin MUller Mttller Cazin Wullner Pintsch Co. Sulphide diox (SOs) .. . Steam, superheated . .. Cazin, Mttller Smithsonian Tables (From less than 1 to more than 1.2) Rankine Steam, wet Perry Steam, wet Gray Steam, wet Average from practice Stwun. dry Regnault 68 ENGINEERING THERMODYNAMICS Table XI HORSE-POWER PER POUND MEAN EFFECTIVE PRESSURE. aJS • \Jt\lj\JCj "' '^'" 33000 Diameter nf Speed of Piston in Feet per Minute. Cylinder, Inches. 100 200 300 400 500 600 700 800 900 4 0.0381 0.0762 0.1142 0.1523 0.1904 0.2285 0.2666 0.3046 0.3427 41 0.0482 0.0964 0.1446 0.1928 0.2410 0.2892 0.3374 0.3856 0.4338 5 0.0592 0.1190 0.1785 0.2380 0.2975 0.3570 0.4165 0.4760 0.5355 51 0.0720 0.1440 0.2160 0.2880 0.3600 0.4320 0.5040 0.5760 0.6480 6 0.0857 0.1714 0.2570 0.3427 0.4284 0.5141 0.5998 0.6854 0.7711 61 0.1006 0.2011 0.3017 0.4022 0.5028 0.6033 0.7039 0.8044 0.9050 7 0.1166 0.2332 0.3499 0.4665 0.5831 0.6997 0.8163 0.9330 1.O490 71 0.1339 0.2678 0.4016 0.5355 0.6694 0.8033 0.9371 1.0710 1.2049 8 0.1523 0.3046 0.4570 0.6093 0.7616 0.9139 1.0662 1.2186 1.3709 81 0.1720 0.2439 0.5159 0.6878 0.8598 1.0317 1 . 2037 1.3756 1.5476 9 0.1928 0.3856 0.5783 0.7711 0.9639 1 . 1567 1 . 3495 1 . 5422 1.7350 9i 0.2148 0.4296 0.6444 0.8592 1.0740 1.2888 1.5036 1.7184 1 1.9532 10 0.2380 0.4760 0.7140 0.9520 1.1900 1 . 4280 1.6660 1.9040 2.1420 11 0.2880 0.5760 0.8639 1.1519 1.4399 1 . 7279 2.0159 2.3038 2.5818 12 0.3427 0.6854 1.0282 1.3709 1.7136 2.0563 2.3990 2 . f 418 3.0845 13 0.4022 0.8044 1.2067 1.6089 2.0111 2.4133 2.8155 3.2178 3.6200 14 0.4665 0.9330 1.3994 1.8659 2.3324 2 . 7989 3.2654 3.7318 4.1983 15 0.5355 1.0710 1.6065 2.1420 2.6775 3.2130 3 . 7485 4.2840 4.8195 16 0.6093 1.2186 1.8278 2.4371 3.0464 3.6557 4.2650 4.8742 5.4835 17 0.6878 1 . 2756 1.9635 2.6513 3.3391 4.0269 4.6147 5.4026 6.1904 18 0.7711 1.5422 2.3134 3.0845 3 . 8556 4.6267 5.3987 6.1690 6.4901 19 0.8592 1.7184 2.5775 3.4367 4.2858 5.1551 6.0143 6.8734 7.7326 20 0.9520 1.9040 2.8560 3.8080 4.7600 5.7120 6.6640 7.6160 8.5680 21 1.0496 2.0992 3.1488 4.1983 5.2475 6.2975 7.3471 8.3966 9.4462 22 1.1519 2.3038 3.4558 4.6077 5.7596 6.9115 8.0643 9.2154 10.367 23 1.2590 2.5180 3.7771 5.0361 6.2951 7.5541 8.8131 10.072 11.331 24 1.3709 2.7418 4.1126 5.4835 6.8544 8.2253 9.5962 ; 10.967 12.338 25 1.4875 2.9750 4.4625 5.9500 7.4375 8 . 9250 10.413 ! 11.900 13.388 26 1.6089 3.2178 4.8266 6.4355 8.0444 9.6534 11.262 12.871 14.480 27 1.7350 3.4700 5.2051 6.9401 8.6751 10.410 12.145 13.880 15.615 28 1.8659 3.7318 5.5978 7.4637 9.3296 11.196 13.061 14.927 16.793 • 29 2.0016 4.0032,6.0047 8.0063 10.008 12.009 14.011 16.013 18.014 30 2 . 1420 4.2840 6.4260 8.5680 10.710 12 . 852 14.994 17.136 19.278 31 2.2872 4.5744 6.8615 9.1487 11.436 13 . 723 16.010 18.287 20.585 32 2.4371 4.8742 7.3114 9.7485 12.186 14.623 17.060 19.497 21.934 33 2.5918 5.1836 7.7755! 10.367 12.959 15.551 18.143 20.735 23.326 34 2.7513 5.5026 8.2538 11.005 13 . 756 16.508 19.259 22.010 24.762 35 2.9155 5.8310 8.7465 11.662 14.578 17.493 20.409 23.224 26.240 36 3.0845 6.1690 9.2534 12.338 15.422 •' 18.507 21.591 24.676 27.760 37 3.2582 6.5164 9.7747 13.033 16.291 19.549 22.808 26.066 29.324 38 3.4367 6.8734,10.310 13.747 17.184 20.620 24.057 27.494 30.930 39 3.6200 7.2400 10.860 14.480 18.100 21.720 25.340 28.960 32.680 TABLES 69 Table XI — Cordinued Diameter of Speed of Piaton in Feet per Minute. Cylinder, Inchea. 100 200 300 400 600 600 700 800 900 40 3.8080 7.6160 11.424 15.232 19.040 22.848 26.656 30.464 34.272 41 4.0008 8.0016 12.002 16.003 20.004 24.005 28.005 32.006 36.007 42 4.1983 8.3866 12.585 16.783 20.982 25.180 29.378 33.577 37.775 43 4.4006 8.8012 13.202 17.602 22.003 26.404 30.804 35.205 39.606 44 4.6077 9.2154 13.823 18.431 23.038 27.646 32.254 36.861 41.469 45 4.819519.6390 14.459 19.278 24.098 28.917 33.737 38.556 43.376 46 5.0361 10.072 15.108 20.144 25.180 30.216 35.253 40.289 45.325 47 5.2574 10.515 15.772 21.030 26.287 31.545 36.802 42.059 47.317 48 5.3845 10.967 16.451 21.934 27.418 32.901 38.385 43.868 49.352 49 5.7144 1 11.429 17.143 22.858 28.572 34.286 40.001 45.715 51.429 50 5.9500 11.900 17.850 23.800 29.750 35.700 41.650 47.600 53.550 51 6.1904 12.381 18.571 24.762 30.952 37.142 43.333 49.523 55.713 52 6.4355 12.871 19.307 25 . 742 32 . 178 38.613 45.049 61.484 57.920 53 6.68541 13.371 20.056 26.742 33.427 40.113 46.798 53.483 60.169 54 6.9401 13.880 20.820 27.760 34.700 41.640 48.581 55.521 62.461 55 7.1995 14.399 21.599 28.798 35.998 43.197 50.397 57.696 64.796 56 7.4637 14.927 22.391 29.855 37.318 44.782 52.246 69.709 67.173 57 7.7326 15.465 23.198 30.930 38.663 46.396 64.128 61.861 69.594 58 8.0063 16.013 24.019 32.025 40.032 48.038 56.044 64.051 72.057 59 8.2849 16.570 24.854 33 . 139 41.424 49.709 57.993 66.278 74.563 60 • 8.5680 17.136' 25.704 I 34.272 42.840 51.408 59.976 68.544 77.112 70 ENGINEERING THERMODYNAMICS GENERAL FORMULA RELATING TO PRESSURE- VOLUME CALCULATIONS OF WORK AND POWER Work = TT = Force X Distance =F xL; = Pressure X Area X Distance = P XA xL; = Pressure X Volume change =P X( 7,— FJ. Force of acceleration ^massXacceleration. g dx Work of acceleration -^ ^^"^ X difference of (velocity)*, 2 g^ ^ 64.32^ '^ ' Velocity due to work of acceleration = square root of the sum of (initial velocity)' plus 2g Xwork per pound of substance accelerated. ' w or if initial velocity is zero, M,«V64.32-=8.02J^. ' w ^ w Pressure Volume Relation for Expansion or Compression PV =PiFi» ^P^Vi' =Ky a constant. (See Table VIII for values of «.) log (yj) (Note graphical method for finding s when variable, see text.) FORMULA RELATING TO PRESSURE- VOLUME CALCULATIONS 71 Work done during a pressure volume change represented by the equation (PF* —if) between points represented by 1 and 2 in figure =area under curve, Wi. If «°1, PdV 'K I ~. W, -P.7, log. ?^ =PtVt log. F. P.V, log. ^' =P,F,log. P, Pa J 8 = 1. If s is not equal to 1, ^"'-^my-'-A 1 «— 1 9-1 -S[-©"] 89^1. Work of admission, complete expansion and exhaust for engines » area to left of curve (see figure) Wt, Same for admission, compression and expulsion for com- pressors. Both cases without clearance. (When « = 1, TTj is same as area xmder curve, Wi (see above). wh«.^iir..^,p.v,[(f;)'"'-i] »-l --.".^'-fn =«xTFi. Clearance, expressed as a fraction of displacement =c, as a voliune =CT. 72 ENGINEERING THERMODYNAMICS liPaVa^PbVt', Db _Da (Pa\ 7 D D \PJ CI t«V-i Indicated horse-power = (mean effective pressure, pounds per square inch Xeffecti ve area of piston, square inches X length of stroke, feet X number of working cycles per- formed per minute) divided by 33,000. ■ ^ • • "" 33.000 • Specific displacement B displacement in one direction for one side of a piston, in cuit. per hour per H.P. =D« = 1?,750, r, (m.e.p./ where « is the number of strokes required o conipl ' e one cycle. Velocity of a jet due to its own expansion - the square root of the product of 2g X work done by admission, complete expansion and exhaust of 1 lb. of the substance. ti = V2^,=8.02|4^P,7|l-(^)~j| . Weight of flow through nozzle or orifices, pounds per second « to = (velocity, feet per second Xarea, square feet, of orifice) 4- (volume per pound of substance at section where area is measiu^). uA "-v 7-^<.W\7h^H'-(^f]\' Maximum discharge w for a given initial pressure occurs when a p. ill ( ^ \*-' CHAPTER II WORK OF COMPRESSORS, HORSE-POWER AND CAPACITY OF AIR, GAS AND VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS. 1. General Description of Structure and Processes. There is quite a large class of machines designed to receive a cylinder full of some gas at one constant pressure and after the doing of work on the gas through decreasing volumes and rising pressures, to discharge the lesser volume of gas against a constant higher pressure. These machines are in practice grouped into sub- classes, each having some specific distinguishing characteristic. For example, blowing engines take in air at atmospheric pressure or as nearly so as the valve and port resistance will permit, and after compression deliver the air at a pressure of about three atmospheres absolute for use in bla^ furnaces. These blowing engines are usually very large, work at low but variable speeds, but always deliver against comparatively low pressures; they, therefore, have the characteristics of large but variable capacity and low pressures. A great variety of valves and driving gears are used, generally mechanically moved suction and automatic spring closed discharge valves, but all valves may be automatic. The compressor cylinder is often termed the blowing tub and the compressed or blast air frequently is spoken of as wind by furnace men. They are all direct-connected machines, an engine forming with the compressor one machine. The engine formerly was always 9f the steam type, but now a change is being affected to permit the direct internal combustion of the blast furnace waste gases in the cylinders of gas engines. These gas-driven blowing engines, showing approximately twice the economy of steam-driven machines, will in time probably entirely displace steam in steel plants, and this change will take place in proportion to the successful reduction of cost of repairs, increase of reliability and life of the gas-driven blowing engines to equal the steam-driven. Some low-pressure blowers are built on the rotary plan without reciprocating pistons, some form of rotating piston being substituted, and these, by reason of greater leakage possibilities, are adapted only to such low pressures as 5 lbs. per square inch above atmosphere or thereabouts. These blowers are coming into favor for blasting gas producers, in which air is forced through thick coal beds either by driving the air or by drawing on the gas produced beyond the bed. They are also used for forcing illuminating gas in cities through pipes otherwise too small, especially when the distances are long. In general very low pressures and large capacities are the characteristics of the service whether the work be that of blowing or exhausting or both. For still lower pressures, measurable by water or mercury columns, fans are used of the disk or propeller 73 74 ENGINEERING THERMODYNAMICS or centrifugal type. These fans are most used for ventilation of buildings and mines, but a modification, based on the principles of the steam turbine reversed, and termed turbo-compressors, is being rapidly adapted to such higher pres- sures as have heretofore required piston compressors. When high-pressure air is required for driving rock drills in mines and for hoisting engines, for tools, as metal drills, riveters, chipping chisels, for car air' brakes, the compressors used to provide the air are termed simply air compressors. These compressors usually take in atmospheric air and compress it to the desired pressure, the capacity required being usually adjustable; they have valves of the automatic type throughout commonly, but in large sizes frequently are fitted with mechanically operated suction valves to decrease the resistance to entrance of air and so increase economy, a com- plication not warranted in small machines. When the pressures of delivery are quite high the compression is done in stages in successive cylinders, the discharge from the first or low-pressure cylinder being delivered through a water cooler or intercooler to the second cyUnder and occasionally to a third in turn. This staging with intermediate or intercooling results in better economy, %s will be seen later in detail, and permits the attainment of the desired quantity of cool compressed air for subsequent use with the expenditure of less work, the extra comphcation and cost being warranted only when machines are large and final pressures high. In the operation of large steam condensers, non-condensible gases will collect and spoil the vacuum, which can be maintained only by the continuous removal of these gases, consisting of air, carbon dioxide and gases of animal and vegetable decomposition originally present in the water. When these gases are separately removed the machine used is a special form of compressor termed a dry vacuiun pmnp which, therefore, receives a charge at the absolute pres- sure corresponding to the vacuum, or as nearly so as the entrance resistance permits, and after compression discharges •into the atmosphere at a pressure in the cylinder above atmosphere equivalent to discharge resistance. Natural- gas wells near exhaustion can sometimes be made to flow freely by the applica- tion of a compressor capable of drawing a charge at a pressure below atmosphere, but whether the charge be received below atmospheric pressure or above as in normal wells, the compressor will permit the delivery of the gas to distant cities or points of consumption even 250 miles away through smaller pipes than would be otherwise possible. Natural-gas compressors, some steam- and some gas-engine driven, are in use for both these purposes, compressing natural gas from whatever pressure may exist at the well to whatever is desired at the beginning of the pipe line. In the preparation of liquid ammonia or carbonic acid gas for the market, as such, or in the operation of refrigerating machinery, wet or dry vapor is com- pressed into a condenser to permit liquefaction by the combined eflPect of high pressure and cooling. One form of refrigerating machine merely compresses air, subsequently expanding it after preliminary cooling by water, so that after expansion is complete it will become extremely cold. WORK OF COMPRESSORS 75 All these compresEdng machines have, as a primary pmpose, either the removal of a quantity of low-pressure gases from a given place, or the delivery of a quantity of higher-pressure gas to another place or both, but all include compression as an intermediate step between constant-pressure admission and constant-pressure discharge as nearly as structure may permit. They will all involve the same sort of physical operations and can be analyzed by the same principles except the wet-vapor or vvet-gas compressors, in which condensation or evaporation may complicate the process and introduce elements that can be treated only by thermal analysis later. Safe compressors cannot be built with zero cylinder clearance, hence at the end of delivery there will remain in the clearance space a volume of high-pressure gases equal to the volume of the clearance space. On the return stroke this clearance volume will expand until the pressure is low enough to permit suction, so that the new charge cannot enter the cj'^linder until some portion of the stroke has been covered to permit this re-expansion of clearance gases. It is quite impossible to study here all the effects or influences of structure as indicated by the compressor indicator cards, but a quite satisfactory treat- ment can be given by the establishment of reference diagrams as standards of comparison and noting the nature of the differences between the actual cases and the standard reference diagram. These standard reference diagrams will really be pressure-volume diagrams, the phases of which correspond to certain hypoth- eses capable of mathematical expression, such as constant pressure, constant volume, expansion, and compression, according to some law, or with some definite value of s fixing either the heat-exchange character of the process or the substance, as already explained. 2. Standard Reference Diagrams or PV Cycles for Compressors and Methods of Analysis of Compressor Work and Capacity. All the standard reference diagrams will include constant-pressure lines corresponding to delivery and supply at pressures assumed equal to whatever exists outside the.^Qylinder on either delivery or suction side, that is, assiuning no loss of pressure on delivery or suction. The compression may be single or multi-stage with various amounts of cooling in the intercooler, but in multi-stage compression the standard reference diagram will be assumed to involve intermediate cooling of the gases to their original temperature, so that the gases entering all cylinders will be assumed to have the same temperature and to maintain it constant during admission. Another difference entering into the classification of standard reference diagrams is the laws of compression as defined by the exponent s. Integration of the differential work expres- sion will take a logarithmic form for s = l, and an exponential form for all other values, thus giving two possible reference compression curves and two sets of work equations. (a) The isothermal for which 5=1, no matter what the gas, and which is the consequence of assuming that all the heat liberated by compression is con- tinuously carried away as fast as set free, so that the temperature cannot rise at all. 76 ENGINEERING THERMODYNAMICS (6) The exponential for which s has a value greater than one, generally different for every gas, vapor or gas-vapor mixture, but constant for any one gas, and also for dry vapors that remain dry for the whole process. Wet vapors having variable values of 8 cannot be treated by the simple pressure-volume analysis that suffices for the gases, but must be analyzed thermally. The adiabatic value of s is a consequence of assuming no heat exchange at all between the gas and anything else and is a special case of the general exponen- tial class. Just why these two assumptions of thermal condition should result in the specified values of s will be taken up imder the thermal analysis part of this work, and is of no interest at this time. As a consequence of these phase possibilities there may be established eight standard reference diagrams or pressure-volume cycles defined by their phases, as shown in Fig. 23, four for single-stage compression and two each for two and three stages. These might be extended by adding two more for four stages and so on, but as it seldom is desirable, all things being considered, to go beyond three, the analysis will stop with the eight cycles or reference diagrams shown. Single-stage Compression Reference Cycles or PV Diagrams a Cycle 1. Single-stage Isothermal Compression without Clearance. Phase (a) Constant pressure supply. " (6) Isothermal compression. " (c) Constant pressure delivery. " (d) Constant zero-volume pressure drop. Cycle 2. Single-stage Isothermal Compression with Clearance. Phase (a) Constant pressure supply. " (6) Isothermal compression, (c) Constant pressure delivery. id) Isothermal re-expansion. Cycle 3. Single-stage Exponential Compression without Clearance. Phase (a) Constant pressure supply. (6) Exponential compression. (c) Constant pressure delivery. (d) Constant zero-volume pressure drop. Cycle 4. Single-stage Exponential Compression with Clearance. Phase (a) Constant pressure supply. " (6) Exponential compression. " (c) Constant pressure delivery. " (d) Exponential re-expansion. 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 ± ~i-L \ A- -% ' * t ~'''X 1 „iJ» i a; il:i I I i I I -u'^'^iv<»e>i'n •MHid ; 78 ENGINEERING THERMODYNAMICS Muiyn-STAQE Comprbssion The phases making up multi-stage compression cycles may be considered in two wnys, first, as referred to each cylinder and intercooler separately, or second, as referred to the pressure volume changes of the gases themselves regardless of whether the changes take place in cylinders or intercoolers. For example, if 10 cu.-ft. of hot compressed air be delivered from the first cylinder of 50 cu.-ft. displacement, the phase referred to this cylinder is a con- stant-pressure decreasing volimie, delivery line whose length is \ of the whole diagram, exactly as in single-stage compression. If this 10 cu.-ft. of air deliv- ered to an intercooler became 8 ft. at the same constant pressure as the first cylinder delivery, the phase would be indicated by a constant-pressure volume reduction line 2 cu.-ft long to scale, or referred to the original volume of air admitted to the first cylinder, a line jV of its length. Finally, admitting this 8 cu.-ft of cool air to the second cylinder and compressing it to i of its vol- ume would result in a final delivery line at constant pressure of a length of i of the length of the second cylinder diagram, but as this represents only 8 cu.-ft., the final delivery will represent only ^X8 = 1.6 cu.-ft. This 1.6 cu.-ft. will, when referred to the original 50 cu.-ft. admitted to the first cylinder, be repre- 1.6 sented by a constant-pressure line, —=.032, of the whole diagram length, ou which in volume is equivalent to i of the length of the second cylinder diagram. It should be noted also that three volume change operations take place at the intermediate pressure; first, first cylinder delivery; second, volume decrease due to intercooling; third, second cylinder admission, the net effect of which referred to actual gas volumes, regardless of place where the changes happen, is represented by the volume decrease due to inter- cooling only. A diagram of volumes and pressures representing the resultant of all the gas processes is called in practice the combined PV diagram for the two cylinders, or when plotted from actual indicator cards with due regard for the different clearances of each cylinder the combined indicator diagrams. It is proper in the study of the whole process of compression to consider the cycle consisting of phases referred to true gas volumes rather than phases referring to separate cylinder processes, which is equivalent to imagining the whole cycle carried out in one cylinder. Intercooling effects measured by the amount of decrease of volume at constant pressure will, of course, depend on the amount of cooling or reduc- tion of temperature, but in establishing a standard reference diagram some definite amount capable of algebraic . description must be assumed as an intercooling hypothesis. It has already been shown, Fig. 6, Chapter I, that from any original state of pressure and volume the exponential and isothermal could be drawn, diverging an amount depending on the difference between the defining exponent, s. If, after reaching a given state on the exponential curve, the gas be cooled at WORK OF COMPRESSORS 79 constant pressure to its original temperature, the point indicating its condition will lie by definition on the other curve or isothermal and the cooling process is represented by a horizontal joining the two curves. Such ihtercooling as this will be defined as perfect inUrcooling, for want of a better name, and its pressure-volume effects can be treated by the curve intersections. It is now possible to set down the phases for the standard reference diagrams of multi- stage compression, if in addition to the above it be admitted, as will be proved later, that there is a best or most economical receiver pressure. Two-stage Compression Reference Cycles or PV Diagrams « Cycle 5. Two-stage Exponential Compression without Clearance, Perfect Intercooling at Best Receiver Pressxire. Phase (a) Constant pressure supply. ' ' (6) Exponential compression to best receiver pressure. ' ' (c) Constant pressure perfect intercooling of delivered gas. * ' (d) Exponential compression from best receiver pressure. * ' (e) Constant pressure delivery. * * (/) Constant zero-volume pressure drop. Cycle 6. Two-stage Exponential Compression with Clearance, Perfect Intercooling at Best Receiver Pressure. Phase (a) Constant pressure supply. * ' (6) Exponential compression to best receiver pressure. * ' (c) Constant pressure perfect intercooling of delivered gas. ' ' (d) Exponential re-expansion of first stage clearance, (e) Exponential compression from best receiver pressure. (J) Constant pressure delivery. ig) Exponential re-expansion of second stage clearance. i t ti Three-stage Compression Reference Cycles or PV Diagrams Cycle 7. Three-stage Exponential Compression, without Clearance, Per- fect Intercooling at Best Two Receiver Pressxires. Phase (a) Constant pressure supply. (6) Exponential compression to first receiver pressure, (c) Perfect intercooling at best first receiver presssure. {d) Exponential compression from best first to best second receiver pressure.' ' ' (c) Perfect intercooling at best second receiver pressure. ' ' (/) Exponential compression from best second receiver pressure. " ig) Constant pressure delivery. ** (A) Constant zero-volume pressure drop. i t 80 ENGINEERING THERMODYNAMICS Cycle 8. Three-stage Adiabatic Compression with Clearance, Perfect Intercooling at Best Two Receiver Pressures. Phase (a) Constant pressure admission. ' ' (6) Exponential compression to best first receiver pressure. ' ' (c) Perfect cooling of delivered gas at best first receiver pressure. ' ' (d) Exponential re-expansion of first stage clearance. '' (e) Exponential compression from best first to best second receiver pressure. * ' (/) Perfect intercooling of delivered gas at best second receiver pressure. ** (g) Exponential re-expansion of second stage clearance. * ' (h) Exponential compression from best second receiver pressure. * ' (i) Constant pressure delivery. ' ' (j) Exponential re-expansion of third stage clearance. It should be noted that cycles 6 and 8 may be sub-divided mto any number of cases, of which some of the most characteristic are shown: (a) where the clearance volume in each cylinder bears the same ratio to the displacement of that cylinder, and commonly called equal clearances; (6) where the clearances are such that the volume after re-expansion in the higher-pressure cylinder is equal to the voliune of clearance in the next lower-pressure cylinder, causing the combined diagram to have a continuous re-expansion line, a case which may be called proportionate clearance; and (c) the general case in which there is no particular relation between clearances in the several cylinders. By means of these definitions or their mathematical equivalents in symbols it will be possible to calculate work as a function of pressures and volumes and by various transformations of a general expresssion for work of a reference cycle to calculate the horse-power corresponding to the removal of a given volume of gas per minute from the low-pressure supply or to the delivery of another volume per minute to the high-pressure receiver or per unit weight per minute. It will also be possible to calculate the necessary cylinder size or displacement per unit of gas handled, and the horse-power necessary to drive the compressing piston at a specified rate and further to calculate the work and horse-power of cylinders of given size and speed. In order that these calculations of a numerical sort may be quickly made, which is quite necessary if they are to be useful, the formulas must be definite and of proper form, the form being considered proper when little or no algebraic transformation is necessary before niunerical work is possible. While special expressions for each case are necessary to facilitate numerical work, it is equally important, if not more so, to make clear the broad general principles or methods of attact, because it is quite impossible to set down every case or even to conceive at the time of writing of all different cases that must in future arise. The treat- ment, then, must be a combination of general and special, the general methods being applied successively, to make them clear and as a matter of drill, not to every possible case, but only to certain characteristic or type forms WORK OF COMPRESSORS 81 of cases, such as are here set down as standard reference diagrams. Individual cases may be judged by comparison with these and certain factors of relation established which, being ratios, may be and are called efficiencies. Thus, if a single-stage compressor should require two horse-power per cubic foot of free air compressed per minute, and Cycle I should for the same pr^sure limits require only one horse-power for its execution, then the efficiency of the real compression would be 50 per cent referred to Cycle I, and similar factors or efficiencies for other compressors similarly obtained; a com- parison of the factors will yield information for a judgment of the two compressors. In what follows on the work and gas capacity of compressors two methods of attack will be used. 1. General pressure-volume analysis in terms of gas pressures and volumes resulting in the evaluation of work per cubic foot of low- or high- pressure gaseous substance. 2. Transformation of results of (1) to yield volumetric efficiencies, mean effective pressures, work, horse-power, and capacity in terms of dimensions of cylinders and clearances. 3. Single-stage Compressory No Clearance, Isothennal Compression (Cycle 1), Work, Capacity, and Work per Cubic Foot in Terms of Pressures and Volumes. The standard reference diagram is represented by Fig. 24, on which the process {A to B) represents admission or supply at constant pressure; {B to C) compression at constant temperature; (C to D) delivery at constant pressure; and (D to A) zero-volume. Let Fft = The number of cubic feet of low pressure gas in the cylinder aftel* admission, represented to scale on the diagram by AB and equal to the volume at B; Fc= volume in cubic feet of the gas in cylinder when discharge begins, represented by DC, which is the volume at C; P»= absolute pressure in pounds per square foot, at which supply enters cylinder = (Sup.Pr.) = pressure at B; p&=Pft-^ 144 = absolute supply pressure in pounds per square inch = (sup.pr.) ; Pc= absolute pressure in pounds per square foot, at which delivery occurs = (Del. Pr.) = pressure at C; Pc=Pc-j- 144 = absolute delivery pressure in pounds per square inch = (del.pr.); p Rp = 5^= ratio of delivery pressure to supply pressure; W = foot-pounds work done for the cycle; (H. P. Cap.) = volume of gas delivered in cubic feet per cycle, at temperature same as that of supply; (L. P. Cap.) = volume of gas drawn into cylinder, cubic feet per cycle. For this no clearance case (L. P. Cap.) = Vft. a 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. — One-stage Compreseor Cycle 1, No Clearance, iBothermal. But since PrVc=PbVb the expression becomes W = F^niog.^, which is the work for the execution of the cycle when pressures and volumes are in pounds per square foot, and cubic feet. The equivalent expression for pounds peT square inch and cubic feet is ^ = 144 ptFJog,^' (29) WOKK OF COMPRESSORS 83 Since, when there is no clearance the volume taken into the cylinder for each cycle (L. P. Cap.) is equal to the volume at B, Vb, the expression Eq. (29) may be stated thus, symbolic form. Tr=144(sup.pr.)(L. P. Cap.) logeJKp (30) The work per cubic foot of low pressure gas, foot-pounds, will be the above expression divided by (L. P. Cap.), or I W ^^-^-^^=144(sup.pr.)log.i?, (31) The work per cubic foot of high-pressure gas delivered will be W = 144 (sup.pr.) Rp log* Rp, .... (32) (H.P.Cap.) ' since PiV, = PcVc or ^b which expressed symbolically is (L.P.Cap.) = (H.P.Cap.)Xffp (33) Expressions (31) and (32) for the isothermal compressor are especiallj'^ useful as standards of comparison for the economy of the compressors using methods other than isothermal. It will be found that the work per cubic foot of either low pressure or cooled high-pressure gas is less by the isothermal process ^han by any other process discussed later, and that it is the limiting case for the economy of multi-stage compressors with a great number of stages. The fact that this process of isothermal compression is seldom if ever approached in practice does not make it any the less a suitable basis for comparison. Example 1. Method of calculating Diagram Fig. 24. Assumed Data. Pa -Pb^ 2116 lbs. per square foot. Va = F^ =0. Pc—Pd- 18,000 lbs. per square foot. Capacity = 5 cu.ft. s = l. To obtain pwint C, .•. Vc^.SQ, Pc =18,000. 84 ENGINEERING THERMODYNAMICS Intermediate points B to C are obtained by assuming various pressures and finding corresponding volumes as for Vc. Example 2. To compress and deliver 5 cu.ft. of air from atmospheric pressure (2117 lbs. per square foot) to 8.5 atmospheres (18,000 lbs. per square foot) isothennally without clearance, how much work is necessary? P6=2116 Pc- 18,000 Fft-5 Vb Pc Work of admission ^-PtVb -2116 X5 - 10,585 ft.-lbs. p Work of compression =P6 Ft log* ~ - 10,585 Xlogr 8.5 =22,600 ft.lbs. Work of delivery =^PeVe = 10,585 ft.-lbs. Total work « 10,585 + 22,600 - 10,585 « 22,600 ft.-lbs. Or by the general formula, W = (sup.pr.)(L.P.Cap.) log* Rp =21 16 X5 xlog. 8.5 =2116 X5 X2.14 =22,652 ft.-lbs. Prob. 1. How many cubic feet of free air may be compressed and delivered i>er minute from 14 lbs. absolute to 80 lbs. per square inc^', absolute per horse-power in a compressor with zero clearance if compression is isothermal? Prob. 2. Oasis being forced through mains at the rate of 10,000 cu.ft. per minute under a pressure of 5 lbs. per square inch above atmosphere. The gas is taken into the compressor at atmospheric pressure and compression is isothermal. What horse- power will be needed at sea level and at an elevation of 5000 feet? Prob. 3. Natural gas is drawn from a well, compressed isothennally and forced through a main at the rate of 200,000 cu.ft. per hour measured at the pressure on the suction side. What steam horse-power will be required to operate the compressor if the mechanical efficiency be 80 per cent? Suction pressure is 8 lbs. per square inch absolute, delivery pressure 60 lbs. per square inch absolute. Prob. 4. A vacuum cleaning pump is required to maintain a pressure of 14 lbs. per square inch absolute, move 500 cu.ft. of free air per minute and discharge it against an atmospheric pressure of 15 lbs. per square inch absolute. What horse-power will be required (isothermal)? Prob. 6. A blower 6imishes 45 cu.ft. of air a minute at a pressure of 5 ins. of mercur3^ above atmosphere. Assuming compression to be isothermal and supply pressure to be atmospheric, what horse-power will be needed? Prob. 6. A compressor has a piston displacement of 3 cu.ft. At what speed can it be run if air be compressed isothennally from 1 to 10 atmospheres and the horse-power supplied is 100? Prob. 7. A tank of 1000 cu.ft. capacity contains air at atmospheric pressure. A compressor taking air from atmosphere compresses it isothermally and discharges it into the tank until the pressure reaches 100 lbs. per square inch gage. What horse-power will be required to fill tank at this pressure in ten minutes? Prob. 8. A compressor receives air at atmosphere and compresses it isothennally to five atmospheres above atmosphere. It takes in 1000 cu.ft. of free air per minute. How much would the capacity increase if the discharge pressure dropped to 3 atmos- pheres and the horse-power remained the same? WORK OF COMPBESSOES 85 Prob. 9. Suppose that the pressure in the above problem were raised to 8 atmos- phrrea. How much would the capacity decrease if the horse-power remained the same and liow much more power would be required to keep the capacity the same? Prob. 10. By means of suitable apparatus, the water from the side of a waterfall is diverted to a vertical shaft, and in falling 126 ft. compresses air from atmospheric pressure to a value equal to 90 per cent of the head of the water. To deliver 1000 cu.ft. of com- pressed air per hour, how much wat«r is required if the work of falling water is 80 per ceut useful in compressing the air? 4. Single-stage CompreBsor with Clearance, Isothennal Compression, (Cycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures and Volumes. "'jr " ^ Fia. 25. — One-Stage Compressor Cycle 2, Clearance, Isothermal. Referring to Fig, 25, the work of the cycle is, by areas. Net work &tc& = EBCG+G CDF -HAD F-EB AH =Area ABCD. It is easily seen that this area is also equal to {JBCL) — {JADL), both of which are areas of the form evaluated in the preceding section. Accordingly Net work a.TeA=JBCL-JADL, 86 ENGINEERING THERMODYNAMICS Algebraically, W = P,V, log. ^-PaVa l0&^ =P»(F„-r,)log.^' (34) b which is the general expression for the work of the cycle in foot-pounds when pressures are in pounds per square foot, and volumes in cubic feet. Substituting the symbolic equivalents and using pressures in pounds per square inch, there results, since (Vt— Va) = (L. P. Cap.), Work = 144 (sup.pr.)(L. P. Cap.) log* Rp, . . . ' . (35) which is identical with Eq. (30), showing that for a given low-pressure capacity the work of isothermal compressors is independent of clearance. The value of the low-pressure capacity (Vt—Va) maj^ not be known directly, but may be found if the volume before compression, Vb, the clearance volume befor^ re-expansion, Vd, and the ratio of delivery to supply pressure, ftp, are known, thus from which (L. P. Cap.) = (F,-Ftfflp) (36) From Eq. (35) the work per cubic foot of low-pressure gas is, in foot-pounds, W (L.RCap.) = ^^^^^PP^'^^^g-^'- (3^^ and the work per cubic foot of high-pressure gas delivered, ft.-lbs. W ^^p-^— r=144(sup.pr.)fiplogeffp (38) By comparison, Eqs. (37) and (38) are found to be identical with (31) and (32) respectively, since clearance haSy as found above, no effect on the work done for a given volume of ga^ admitted^ however much it mxiy affect the work of the cycle between given volume limits or work per unit of dispUicement, It is interesting to note that the work areas of Figs. 24 and 25 are equal when plotted on equal admission lines AB or delivery lines CD and any horizontal intercept xy will be equal in length on both if drawn at the same pressure. In what precedes, it has been assumed that AB represents admission volume and CD represents delivery volume which is true for these established cycles WORK OF COMPRESSORS 87 of reference, but it is well to repeat that for real compressors these are only apparent admission and delivery lines, as both neglect heating and cooling effects on the gas during its passage into and out of the cylinder. Also that in real compressors the pressure of the admission line cannot ever be as high as the pressure from which the charge is drawn and the delivery pressure must be necessarily higher than that which receives the discharge, in which cases the volume of gas admitted, as represented by AB, even if the temperature did not change, would not equal the volume taken from the external supply, because it would exist in the cylinder at a lower pressure than it originally had, and a similar statement would be true for delivered gas. Problems. Repeat all the problems of the last sectioD, assuming any numerical value for the clearance up to 10 per cent of the displacement. 6. Single-stage Compressor Isothermal Compression. Capacityi Volu- metric Efficiency, Work, Mean Effective Pressure, Horse-power and Horse- power per Cubic Foot of Substance, in Terms of Dimensions of Cylinder and Clearance. Consider first the case where clearance is not zero. Then Fig. 25 is the reference diagram. Let D= displacement = volume, in cubic feet, displaced by piston in one stroke = area of piston in sq.ft. X stroke in ft. = (V6— Vd). " (H. P. Cap.) = high pressure capacity = vol. cu.ft. of gas delivered per cycle at temperature equal to that of supply = (Fc—yd); *' (L. P. Cap.) = low pressure capacity = vol. in cu.ft. of gas entering cylinder per cycle = (F&— Va) ; ii „ , * • a: • L.P. Cap. Vt-Va E, = volumetnc efficiency = j;: — — = ^7 — W > U Vtt — Yd " CZ = volume of clearance, cubic feet= Fd " c = clearance volume expressed as a fraction of the displacement; whence Cl = cD) CI Vi W " M.E.P. = mean eflFective pressure, lbs. per square foot = y^; W " m.e.p. = mean eflFective pressure, lbs. per square inch = tttj^] " iV = number of revolutions per minute; " n= number of cycles per minute; N " 2= number of revolutions per cycle =—; n " I.H.P. = indicated horse-power of compressor; 88 ENGINEERING THERMODYNAMICS The low-pressure capacity of the single-stage isothermal compressor with clearance is, (L.P.Cap.) = (Vft-ya), but Va=VaX^. Whence (L. P. Cap.) = { V'ft— Fd— I for which may be substituted the symbols for displacement and clearance volumes, thus (L. P. C8Lp.)=D+cD-'cDRp, • =D(l+c-cRp) (39) For convenience the term, Volumetric Efficiency, Ev is introduced. Since this is defined as the ratio of the low-pressure capacity to the displacement, E,JJ±F:J^-^^l+c-cR, (40) Referring to Eq. (35) it is seen that the value of (L. P. Cap.) can be substituted from Eq. (39) and the result is: Work per q/cfe, foot-pounds^ in terms of supply pressure^ pound spersq uare inch, displacement cubic feet, clearance as a fraction of displacement, and ratio of delivery to supply pressure is, TF = 144 (sup.pr.)D(H-c-cfip)log,/?,„ .... (41) or in the terms of the same quantities omitting clearance and introducing volumetric efficiency, Ev, W = lU{sup,i>T.) DEAogeRp, (42) To obtain the mean effective pressure for the cycle, the work done per cycle is divided by displacement, D. Mean eflFective pressure, pounds per square inch, W (m.e.p.) = 1442>' whence (m.e.p.) = (sup.pr.)(H-c— cBp) log* JKp (43) or (m.e.p.) = (sup.pr.) J5» log« /2p . (44) WORK OF COMPRESSORS 89 The indicated horse-power of the isothermal compressor is equal to the work per minute, in ft.-Ibs. divided by 33,000. If n cycles are performed per minute, then Wn _ 144n 33000 ""33000 I.H.P. = ^^t^Fip: = 5^K?^ (sup.pr.)i)(l +C - cRp)l0ge Rp = ^^gg^i>(l+c-ci2,)loge/2, (45) -^^^^DE. log. Rp^ (46) Introducing the effective area of the piston, in square inches, a, and the piston speed S, feet per minute, then since 144 144 22 288z' '•«•''- w1^-'°^«' »'> The same expression for the indicated horse-power may be derived by the substitution of the value of (m.e.p.) Eq. (44) in the following general expres- sion for indicated horse-power. ( m.e.p.)a ^ ^•^•^- 33000X2«' Example !• Method of calculating Diagram Fig. 25. Assumed Data: Pa -Pb =2116 lbs. per square foot; Pd "Pc - 18,000 lbs. per square foot; . c =3 per cent. L.P. Cap. =5 cu.ft. « » 1. To obtain point D: From formula Eq. (39), L.P. Cap. -2)(l+c-cftp) or 5=Z)(l+.03-.03x8.5), or D =6.5 cu.ft. and CI. -.03 X6.5 = .195 or approximately .2 cu.ft. ■ .'. Fd = .2 cu.ft., Pd = 18,000 lbs. sq.ft. To obtain point A : PaVa-PdVi or Fa=^7d-8.5X.2 = 1.7, .*. Fa -1.7 cu.ft., Pa =2116 lbs. sq.ft. # 90 ENaiNEERING THERMODYNAMICS Intermediate points Z) to A are obtained by assuming various pressures and finding the corresponding volumes as for 7o. To obtain point B: Vt = Fa +5 = 6.7 cu.f t. Pb = 21 16 lbs. sq.ft. To obtain point C: DT7 DT7 T/ ^^^^ 2116X6.7 _ ,^ PcVc ^PbVby or Vc = -p— = ~Y8060~ " ^^ .*. Vc = .79 cu.ft., Pc = 18,000 lbs. sq.ft. Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 atmospheres isothermally in a compressor having 4 per cent clearance. What must be the displacement work, per 100 cu.ft. of supplied and delivered air, and horse-power of machine? Speed is 150 R.P.M., compressor is double acting and stroke » 1 .5 diameters. Neglect piston rods. Z)=(L.P.Cap.)-5-J^, and ^, = (1 +.04 -.04x8.5) =.7, .'. D = 1000 ^ .7 = 1428 cu.ft. per minute. Work per cu.ft. of supplied air = (sup.pr.) 144 log« ftp = 144 X 14.7 loge Rp = 4530 ft.-lbs. ; Hence the work per 100 cu.ft. =453,000 ft.-lbs. Work per cu.ft. of deUvered air - 144(sup.pr.)Hp loge ftp = 144 X 14.7 x8.5 X2.14 =38,550 ft.-lbs. Hence the work per 100 cu.ft. =3,855,000 ft.-lbs. ■■H.P..^-.37.. ^ = ^c/xwo = ^-76 cu.ft. per stroke. IoUa^ D = LX A = 1.5dXTCP = 1.18d3 =4.76. 4 -©*=■■■ Hence cylinder diameter = ( — -- ) =1.59 feet = 19.1 inches. Prob. 1. How many cubic feet of free air per minute may be compressed isother- mally to 100 lbs. per square inch absolute in a compressor having 6 per cent clearance if the horse-power supplied is 60? Prob. 2. A compressor has a cyUnder 18x24 ins., clearance 4 per cent, is double acting and runs at 150 R.P.M. If it compresses air from atmosphere to 100 lbs. per square inch gage, what will be its high- and low-pressure capacity, its horge-power, and WORK OF COMPRESSORS 91 how will the horse-power and the capacity compare with these quantities in a hypothetical conipressor of the same size but having zero clearance? How will the horse-power per cubic foot of delivered air compare? Prob. 3. A manufacturer gives for a 10jxl2 in. double-acting compressor running at 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50-100 lbs. per square inch gage. What clearance does this assume for the lowest and highest pressure if the compression is isothermal? The horse-power is given as from 23 to 35. Check this. Prob. 4. Air enters a compressor cylinder at 5 lbs. per square inch absolute and is compressed to atmosphere (barometer =30i ins.). Another compressor receives air at atmosphere and compresses it to 3 atmospheres. If each has 6 per cent clearance what must be the size of each to compress 1000 cu.ft. of free air per minute, how will the total work compare in each machine, and how will the work per cubic foot of high and low pressure air compare in each? Assimie compression to be isothermal. Prob. 5. 1800 cu.ft. of free air per minute is to be compressed isothermally to a pres- sure of 100 lbs. per square inch gage. What must be the displacement and horse-power of a hypothetical zero clearance compressor, and how will they compare with those of a compressor with 6 per cent clearance? Prob. 6. Consider a case of a compressor compressing air isothermally from atmos- phere to 100 lbs. per square inch gage. Plot curves showing how displacement and horse-power will vary with clearance for a 1000 cu.ft. free air per minute capacity taking clearances from 1 per cent to 10 per cent. Prob. 7. Two compressors of the same displacement, namely 1000 cu.ft. per min- ute, compress air isothermally from 50 lbs. per square inch gage to 150 lbs. per square inch gage. One has 5 per cent clearance, the other 10 per cent. How will their capaci- ties and horse-power compare with each other and with a no clearance compressor? Prob. 8. A 9 Xl2 in. compressor is compressing air from atmosphere to 50 lbs. gage. How much free air will it draw in per stroke, and how much compressed air will it dis- charge per stroke for each per cent clearance? Prob. 9. The volumetric efficiency of a compressor is .95 as found from the indicator card. It is double acting and has a cylinder 18 X24 ins. What will be its capacity and required horse-power for 100 lbs. per square inch gage delivery pressure? 6. Single-stage Compressor, No Clearance Exponential Compression, (Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures and Volumes. The cycle of the single-stage exponential compressor without clearance is represented by Fig. 26. Referring to work areas on this diagram, Net work 45CZ> = compression work EBCG -h delivery work GCDF —admission work FA BE. Algebraically, <— 1 92 ENGINEERING THERMODYNAMICS But or and PcVcVc'~'^P,V,V,'-\ •-1 Pcv,=p,v,(^y '=p,v,(p^ ' »-i PcVc-P,V,=P,V,[{j,^ ' -l]. (HP. Cap 1 Oold) •Vu.l'.t*! Hot) J - M '^ 1 ! 1 1 ' M 1 1 j 1 -» ■ -4 - —^ r> i i/ i /^ 1 1 1 lomn t' i K_ _ ' C ^ i i_. , _ i < 1 1 1 ^ 1 A ' ' 1 _L JL ' i I ; 1 ' il_ t ijui iir ^41 ■ 11 III 1 L JLV— 1 ; i ' I V^- u ' 1 ' 1 ■ ' IT. ti-i ^41 V I \ ' ' ' ! 1 ! ~^ 1 ' 1 \ ! i 1 ( 1UiVl I l\ ' -. _^ ._ _^ _ ^**^ \ \ t-l V \ ^* Ji . JL, \]^ J ' , , _. 5 4J Ip \ -^ 1 ^ -L- ; 1 1 IV ' V ' ' -.-. i . ^ ' J L 8 ■ r\ - - - \ r- U . ' ' 1 a \v -1 ^-4- 1 I ' 5 - l-V \>i. ^ ' 1 5*ifi¥¥i ; i \ 11 Wj 1 1 QQilXMU W \^ i 1 ^ fe WT (^ IT ! 1 ' ' u • 8. y ;_ W: ! ' ' P ■ ii X^ A. ■9 _ X Jmk w i ' i "< ' I vi V t ' c 9 Tx4- ^% ^ 1 . £ 1 V^ aIIu i 1 ^ iJ-W. .. >^_t i a • k ' V t -r ^ -• -h 8 T^ ^ -K-^ _i 11' 1 ' H J '^' ^k. 1 m _iL ^ __r^ 1 1 H - - v,^ J- > t 1 ^y £ lF_ _ "'^^ s ^ _4 -. . Ni^ 1 1. "^ "*>- "^^s I, '^■^ "**^ i ! iJOUU 1 ti '''T>4 "^">.J 1 -^i U T ' ■ 1 ''t*t»4». ^]h*«»^.^ A .,.=. ^,„j^i .1-1 r -,-t;^- -j- L L ' Ll_ 1 ^^^ -Tg» M^ R ^^ .1. ! 1 I 1 i 1 1 1 \- ~u' I 4- I -L i • i i 1 r i± — ^ , t . ; 1 I ',3 r ir ^ ' J m T 1 cs F X-lS-i-j L— J-X t.y , 1 1 2 Volumes 345 in Cubic Feet r*- D L.P. Cap- Fia. 26. — One-Stage Compressor Cycle 3, No Clearance, Exponential. Substituting above a-l »-i '^-f:S[©"-]+''-^-[©""-'] »-i ■ -^.>'.[(&)"-]CV)- WORK OF COMPRESSORS 93 Whence «-i '^%4iW[(g) • -l]. ....... (48) Eq. (48) gives the work in foot-pounds for the execution of the cycle when ])ressures are in pounds per square foot, and volumes in cubic feet. The equivalent expression for pressures in pounds per square inch is W-l^^,J.y\{^y'-l] (49) When there is no clearance, as before, Vb represents the entire vol* ume of displacement, which is also here equal to the volume admitted (L. P, Cap.), Pb is the supply pressure (sup.pr.) pounds per square inch Vc absolute and — is the ratio of delivery to supply pressure, Rp. Accordingly, the work of an exponential, singlenstage compressor with no clearance is • -1 }r=144^(sup.pr.)(L.P.Cap.)(ft/ -l) (50) The work per cubic feet of low pressure gas, foot-pounds is • -1 W —^ = 144^ (sup.pr.) (ft/ -l) (51) (L. P. Cap Before obtaining the work per cubic foot of high-pressure gas, it is neces- sary to describe two conditions that may exist. Since the exponential com- pression is not isothermal, it may be concluded that a change in temperature will take place during compression. This change is a rise in temperature, and its law of variation will be presented in another chapter. 1. If the compressed air is to be used immediately, before cooling takes place, the high-pressure capacity or capacity of delivery will be equal to the volume at C, Ve and may be represented by (H. P. Cap. hot). 2. It more commonly occurs that the gas passes to a constant-pressure holder or reservoir, in which it stands long enough for it to cool approximately to the original temperature before compression, and the volume available after this cooling takes place is less than the actual volume discharged from the cylinder in the heated condition. Let this volume of discharge when reduced to the initial temperature be represented by (H. P. Cap. cold) which is represented by Vt, Fig. 26. 94 ENGINEERING THERMODYNAMICS Since B and C in Fig. 26 lie on the exponential compression line, P^Vb' =P«V/, -<w- or (L. P. Cap.) = (H. P. Cap. hot) {Rp)T (52) Hence, the work in foot-pounds per cubic foot of hot gas delivered from compressor is On the other hand, B and K lie on an isothermal and PbVh^PtVt, or since Ph—Pc, whence (L. P. Cap.) = (H. P. Cap. cold)flp. * (M) The work foot-pounds per cubic foot of gas cooled to its original tempera- ture is, therefore, ^jj_^____. = 144--(8up.pr.)iep(^/e, . -Ij, . . . . (55) or W s / ml \ This last equation is useful in determining the work required for the storing or supplying of a given amount of cool compressed air or gas, under conditions quite comparable with those of common practice. Example 1. Method of calculating Diagram Fig. 26. Assumed Data: Pa = Pft = 21 1 6 lbs. per square foot ; Pe=Pd^ 18,000 lbs. per square foot. CZ=0; Va = Vd=0; L, P. Capacity =5 cu.ft.; « = 1.4 (adiabatic value of s). To obtain point C: X PcVc^-^^PtVt^'^ or Fc = F6-^(^)^■* Pc/P* = 8.5; loge8.5=.929, and .71 log. 8.5 = .665; (Pc/Pt)!* =4.6, WORK OF COMPRESSORS 95 hence Fc =5 4-4.6 = 1.09 cu. ft. Pc = 18,000 lbs. per sq.ft. Intermediate points B to C are obtained by assuming various pressiues and finding the corresponding volumes as for 7c. Example 2. To compress 5 cu.ft.of air from atmospheric pressure (2116 lbs. per square foot) to 8.5 atmospheres (18,000 lbs. per square foot) adiabatically and with no clearance requires how many foot-pounds of work? Pft =2116 lbs. sq.ft., Pe = 18,000 lbs. sq.ft., 7^=5 cu.ft. Vb 7^= 7P\ .71 =5^4.57 = 1.092 cu.ft. Work of admission is PftFft =2116 X5 = 10,585 ft.-lbs. Work of compression, using y to represent the adiabatic value of s is, F^l (fI) "^ -l] --^[(«-5)" -11 =^^ ><•««' -22,350 ft.-lbe. Work of delivery is PeVe = 18,000 X 1 .092 = 19,650 ft.-lbs. Total work -19,650+22,350 -10,585 =31,425 ft.-lbs., or by the formula Eq. (50) directly W = 144^(sup. pr.) (L. P. Cap.) {Rp~ - 1.) = 144+3.46x2116x5x[(8.5)-2»-l]; = 144+3.46 X21 16 X5 X. 86 =31,450 ft.-lbs. Prob. 1. A single-stage zero clearance compressor compresses air adiabatically from 1 to 6 atmospheres. How many cubic feet of free air per minute can be handled if the compressor is supplied with 25 H.P. net? Prob. 2. The same compressor is used for superheated ammonia under the same pressure conditions. For the same horse-power will the capacity be greater or less and how much? Prob. 3. A dry-vacuum pump receives air at 28 ins. of mercury vacuum and delivers it against atmospheric pressure. What will be the work per cubic foot of low-pressure air and per cubic foot of high-pressure air hot? Barometer reads 29.9 ins. 96 ENGINEERING THERMODYNAMICS Prob. 4. The manufacturer gives for a lOl X 12 in. double acting compreesor run- ning at 160 R.P.M., a capacity of 177 cu.ft. of free air per minute and a horse-power of 25 to 35 when delivering against pressures from 50 to 100 lbs. Check these figures. Prob. 6. A set of drills, hoists, etc., are operated on compressed air. For their opera- tion 3000 cu.ft. of air at 70 lbs. gage pressure are required per minute. What must be the piston displacement and horse-power of a compressor plant to supply this air if compression is adiabatic and there is assumed to be no clearance? Prob. 6. Air is compressed from atmosphere to 60 lbs. per square inch gage by a compressor having a 12x18 in. cylinder and running at 100 R.P.M. Find its capacity and horse-power at sea level and loss in capacity and horse-power if operated at an alti- tude of 10,000 ft. for zero clearance. Prob. 7. 10,000 cu.ft. of free air per minute at a pressure of 15 lbs. above atmos- phere are compressed and delivered by a blowing engine. Find the horse-power required to do this and find how much free air could be delivered by same horse-power if the pressure were tripled. Prob. 8. In a gas engine the mixture of air and gas is compressed in the cylinder before ignition. If the original pressure is 14 lbs. per square inch absolute^ final pressiu'e 85 lbs. absolute and compression is adiabatic, what will be the work of compression only, per pound of mixture? Note: Weight per cubic foot may be taken as .07 and y as 1-38. Prob. 9. A vacuum pump is maintaining a 25-in. vacuum and discharging the air removed against atmospheric pressure. Compare the work per cubic foot of low pres- sure air with that of a compressor compressing from atmosphere to 110 lbs. above atmos- phere. 7. Single-stage Compressor with Clearance^ Exponential Compression, (Cycle 4). Work, Capacity, and Work per Cubic Foot in Terms of Pressures and Volumes. When clearance exists in the cylinder, it is evident that a volume equal to the clearance, Va, will not be expelled during the delivery of compressed gas, and this volume will expand with fall in pressure as the piston returns, causing pressure-volume changes represented by the line DA on the diagram, Fig. 27. Until the pressure has fallen to that of supply, the admission valve will not open, so that while the total volume in the cylinder at end of admission is Vbj the volume Va was already present by reason of the clearance, and the volume taken in is (Vb^Va) which is the low-pressure capacity (L. P. Cap.). The work area of the diagram is ABCD, which may be expressed as Work QTeA^^JBCL-JADL, which areas are of the form evaluated in Section 6. Hence, the above expression in algebraic terms is WOEK OF C0MPEESS0B9 97 This is the general expression for the work of the cycle, in foot-pounda, when the pressures are expressed in pounds per square foot, and volumes in cubic feet. Using symbolic equivalents W = 144-4T{8uppr}(L-P-Cap.)[(ffp)^'-l,l .... (58) Fig. 27. — One-Stage Compressor Cycle 4, Clearance, E^xponential. Eq. (58) is identical with Eq. (50), showing that for adiahatic as for 'sothermal compressors, Ike wort ■ done for a given low-pressure capcuHly is inde- pendent of clearance. Due to this fact, the expressions derived for the expo- nptitial compressor without clearance will hold for that with clearance: Work, in foot-pounds per cubic foot of low-pressure gas is, W (L:prcap.)-"ta(^"PP^-^^'^'' -i> ■ Work, in foot-pounds per cubic foot of hot gas delivered is, (H. P. Cap. hot) ■144^(sup.pr.)fi; (r,' -l). 98 ENGINEERING THERMODYNAMICS Work, in foot-pounds per cubic foot of cooled gas to its original temperature Is, • -1 W (H. P. Cap. cold) = 144^(del.pr.)(ftp' -l) (61) The relation of high-pressure qapacity either hot or cold to the low-pressure capacity is also as given for the case of no clearance, as will be shown. In Fig. 27, the high-pressure capacity, hot, is DC=Ve'-Va* The low- — — 11 1 i pressure capacity is AB = Vb— Fa, but VcPe* — F^P^• and FdPd* = FoPa», or 1 1 Fft = VcRp" and Va = VJip' . ' 1 Hence (L. P. Cap.) = (H. P. Cap. hot)ftp« (62) If the delivered gas be cooled to its original temperature, then the volume after delivery and cooling will be (H. P. Cap. cold) = ^^- P- ^^P-) , or (L. P. Cap.) = (H. P. Cap. cold)B„ .... (63) From the work relations given above, it is seen that in general, the work per unit of gas, or the horse-power per unit of gas per minute is independent of clearance. 8. Single-stage Compressor Exponential Compressor. Relation between Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, Horse- power and H.P. per Cubic Foot of Substance and the Dimensions of Cylinder and Clearance. As indicated on Fig. 27,. for the single-stage exponential com- pressor with clearance, the cylinder displacement D, is {Vb—Vd)> The low- pressure capacity per cycle is (L. P. Cap.) = (F&— Fa). The actual volume of gas or vapor delivered by the compressor is (H. P. Cap. hot) = (Fc— F^). This is, in the case of a gas at a higher temperature than during supply, but if cooled to the temperature which existed at B will become a less volume. This delivered volume after cooling is symbolized by (H. P. Cap. cold) and is /T ^ i-. X (sup.pr.) (L. P. Cap.) , ^ . , equal to (L. P. Cap.) X -77-; : or where Rp is the ratio of delivery (del.pr.) Rp ^ pressure to supply pressure. Volumetric efficiency, Evj already defined as the ratio of low-pressure capacity to displacement is „ Vb-V g (L. P. Cap.) WORK OF COMPRESSORS 99 Clearance, e, expressed as a fraction of the displacement is the ratio of clearance volume, CI, to displacement, D, and is, CI Va c= Mean effective pressure, pounds per square foot (M.E.P.), is the mean height of the diagram or the work area W, divided by displacement, D. If expressed in pounds per square inch the mean effective pressure will be indicated by (m.e.p.) = jj4^. Let (I.H.P.) be indicated horse-power of the compressor; N the number of revolutions per minute; n the number of cycles per minute and z the number of revolutions per cycle, whence nXz^N. a Then, the low-pressure capacity is (L. P. Cap.) = (F»-7«). But Va^VaX w- since the re-expansion DA is exponential and similar to compression as to value of «, whence J 1 (L. P. Cap.) = (V*-Va) = y»-7*Bp' ; 1 ^D+Cl-VdRp^; =D+cD-cDRp^ ; or (L. P. Cap.) = D (l + c - cBp"« ) (64) From this, by definition, the volumetric eflBciency is Referring to Eq.(67), in which may be substituted the value Eq. (64) for (Fi— Fd), the work of the single-stage exponential compressor in terms of dis- iOO ENGINEERING THERMODYNAMICS placement, clearance (as a fraction of displacement), and pressures of supply and delivery in pounds per square foot is, "^-^M'+'-wwr -'] <*' or using pressures, pounds per square inch, and inserting the symbols, this may be stated in either of the following forms: TF=144~(sup.prOD(l+c-c72pr)rft/-^-ll. . . . (67) = 144-^ (sup.pr.)D£'«[ftpV-l] (68) The mean effective pressure in pounds per square foot is this work divided by the displacement, in cubic feet, and may be converted to pounds per square inch by dividing by 144, whence Mean effective pressure, pounds per square inch, (m.e.p.) = - — (sup.pr.)(l +c—c/?p« )hRp • — 1 > • . (69) 5 5-1 (sup.pr.)J^Jflp^-ll (70) The indicated horse-power of the single-stage exponential compressor from (67) is, T TT P - ^'^ - ^_ (suP-P rQnPE , r ^^__ J /7,x ^•^•^•""33000""s~l 229.2 " [^'^ ' ij . . . K^ i) Where n is the number of cycles per minute, or in terms of piston speed S and effective area of piston, square inches, and z the number of revolutions per cycle, I-H.P.-^— 1 ^^^^^ [iep • -IJ (72) Since it was found in Section 7, that the work per unit volume of gas is the same with clearance as without clearance, the horse-power per cubic foot per minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ). Horse-power per cubic foot of gas supplied per minute I.H.P. n(L. P. Cap __ s isup.pjg r izi_ ] .)~s~i 229.2 L ^J • • • • ^^^^ WORK OF COMPRESSORS 101 The horse-power per cubic foot of hot gas delivered per minute is IH.P. s_ Isup.prO^ 1 r 1^1 1 . n(H.P.Cap.hot)~8-l 229.2 ^ [^' ' ^J • • • U*; Horse-power per cubic foot of gas delivered and cooled is I.H.P. « (sup.pr.) r o Izi [fip • -l|, . . . (75) n(H. P. Cap .cold) s-l 229.2 ^'^ ^' ' _ » (d el.pr.) [ff^ -] «„v -pi~"229:2~L^ ~ J ^ ^ In the above formulae (del.pr.) and (sup.pr.) indicate delivery pressure and supply pressure, in pounds per square inch. Example 1. Method of calculating Diagram, Fig. 27. Assumed data: Pa =^6 «*2116 lbs. per square foot. Pc=Pd- 18,000 lbs. per square foot. CI. =3.5 per cent. L. P. Capacity =5 cu.ft. « = 1.4. To obtain point D: L. P. Cap.=i>(l+c-c/^p7) or 5 =Z)(l +.035 -.035(8.5)) ; Hence D -5 -5- (1 +.035 -.035 X4.6) =5.72 cu.ft. and CI = .035 X5.72 = .2 cu.ft /. Vd = .2 cu.ft. ; Pa = 18,000 lbs. sq.ft. .716 To obtain point A : -m^'y- = 4.6X.2=.92; .'. 7a = .92 cu.ft.; Pa =2116 lbs. sq.ft. Intermediate pK)ints i) to A are obtained by assuming various pressures and finding the corresponding volumes as for Va- To obtain point B: Vo = Va+L, p. Cap. = .92 +5 =5.92, /. Vt =5.92 cu.ft. ; P* = 2116 lbs. sq.ft. 102 ENGINEERING THERMODYNAMICS To obtain point C: Fc-n AT =5.92 +4.6 = 1.29 cu.ft. .". Fc = 1 .29 cu.ft. ; Pc = 18,000 lbs. sq.ft. Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 atmospheres absolute so that «» 1.4, in a compressor having 4 per cent clearance. . What must be the displacement of the compressor, work per 100 cu.ft. of supplied and delivered air, hot and cold, and horse-power of machine? Speed is 150 R.P.M., compressor is double acting and stroke = 1.5 diameters. D = L. P. Cap. •^^», and Ev = (l -f c -cfi,7 J . .'. ^,= (l+.04-.04x(8.5)-^^) =.86; .-. D = 1000 + .86 = 1 162 cu.ft. per min. 8 — Work per cubic foot of supplied air = 144 (sup.pr)[/2p « ■" ^l* = 144 X3.46 X 14.7 X.86 =6300 ft.-lbs. .*. Work per 1000 cu.ft. = 6,300,000 ft.-lbs. Work per cubic foot of delivered air cold is Rp times work per cubic foot of supplied air, hence work per 100 cu.ft. of delivered cooled air is 5,350,000 ft.-lbs. i. Work per cubic foot of delivered air hot \a Rp t times work per cubic foot of suppUed air, hence work per 100 cu.ft. of hot delivered air is 2,800,000 ft.-lbs. «^ --r -n. (m.e.p.)ajS , , .. s—l. ._ r_ — =■ ^1 I.H.P.- r^ ; 2-i; (m.e.p.) — - (8up.pr.)S,[B, *-^-l,\ 66,0002 ' " ' *^' « or (m.e.p.) -3.46x14.7 x.86 X.86 -37.7 lbs. per square inch. a-f ; ^-150x2xif ; ^-^-1162; . .*. ii«=5690 or <i = 17.85. o =260 sq-inches. S =670 ft. per min. .'. I.H.P. = 191, WORK OF COMPRESSORS 103 Prob. 1. A denBe-air ice machine requires that 4000 cu.ft. of air at 50 lbs. per square inch absolute be compressed each minute to 150 lbs. per square inch absolute. The compression being such that « = 1.4, clearance being 6 per cent, find the work required. What would be the work if clearance were double? Half? Prob. 2. The compressor for an ammonia machine compresses from one atmos- phere to 8 atmospheres absolute. With adiabatic compression and 4 per cent clear- ance, what will be work per cubic foot of vapor at the low pressure and at the high? Assume vapor to be superheated. Prob. 3. On a locomotive an air-brake pump compresses air adiabatically from atmosphere to 80 lbs. per square inch gage. It is required to compress 50 cu.ft. of free air per minute and clearance is 5 per cent. What horse-power must be supplied to it? Prob. 4. In a manufacturing process a tank must be maintained with a vacuum of 29 ins. when barometer reads 30 ins. To do this 100 cu.ft. of carbon dioxide must be removed from it per minute and returned under atmospheric pressure. Compression is adiabatic and clearance 7 per cent. How much power must be supphed to compressor and what should be its displacement? Prob. 5. Two compressors each 12x18 in., double acting, with 8 per cent clear- ance, and running at 150 R.P.M. compress in the one case air, in the other carbon disul- phide. The compression being adiabatic in each case, what (a), is the difference in power required, (6), in low-pressure capacities? Take pressures as 2 and 15 atmos- pheres of 26 inches mercury. Prob- 6. A compressor is supplied with 40 horse-power. If it draws in air from atmosphere to what pressure can if 500 cu.ft. per minute be compressed, when 5 = 1.38 and clearance 10 per cent? Prob. 7. For forcing gas through a main, a pressure of 50 lbs. per square inch gage is required. What is the work done per cubic foot of high-pressure gas, if a compressor having 6 per cent clearance is used, and s for the gas is 1.36? What should be its dis- placement? Prob. 8. A gas compressor 20x22 ins. has a volumetric efficiency cf 90 per cent, supply pressure =4 lbs. per square inch and delivery 110 lbs. per square inch gage. WhsA are its L. P., H. P., hot and cold capacities, and its I.H.P. if single acting at 70 R,P.M. when* = 1.35? 9. Two-Stage Compressor, no Clearance, Perfect Intercooling, Exponential Compression, Best Receiver Pressure, Equality of Stages (Cycle V). Work and Capacity in Terms of Pressures and Volumes. The common assumption in con< sideringthe multi-stage compressor is that in passing from one cylinder to the next, the gas is cooled to the temperature it had before entering the com- pressor, which has already (Section 2), been defined as " perfect intercooling." This condition may be stated in other words by saying that the product of pressure and volume must be the same for gas entering each cylinder. If then the voliune and pressure of gas entering the first stage be determined, fixing the volume entering the second stage will determine the pressure of the gas entering the second stage, or fixing the pressure of the gas entering the second stage will determine the volume that must be taken in. Using subscripts referring to Fig. 28, for the no clearance case, P!,V,=PaVa (77) 104 .ENGINEERING THERMODYNAMICS The net work of the compressor, area ABCDEF=BTesk ABCH first stages- area HDEF second stage. Using the general expression, Eq. (48) for these work areas with appropriate changes in subscripts TT^^P^yl^^^) ' -ll. . (first stage) •-1 8 37j Pdl^d I ( p^ 1 ■" M • • (second stage) But from the above, and since Pc—Pd, •-1 «-i -^Mi^r <9r -^\ (78) which is the general expression for work of a two-stage compressor without clearance, perfect intercooling, and may be restated with the usual s3anboLs as follows; TF = 144-^-:~ (sup.pr.)(L. P. Cap.)r(/2pi)"-~+(/i;,2) V-2I , . (79) in which (Rpi) and Rp2) are the ratios of delivery to supply pressures for the first stage and for the second stage respectively. From Eq. (79), work per cubic foot of gas supplied is, ^j--^»^ = 144--^(8up.pr.)[fl,x-+ff,2--2j . . (80) Work per cubic foot of gas discharge and cooled to its original temperature is W (H. P. Cap. cold) 8 [ 'si} iiJ 1 = 144-^^(sup.pr.)i?p iZpi • +Rp2 • ""2 , = 144^-j(del.pr.)rflpiV+/?p2'' ^-2J (81) The low-pressure capacity stated in terms of high-pressure capacity hot, as actually discharged is 1 (L.P. Cap.) = (H.P. Cap. hot)ftp2'"ftpi, (82) whence Work per cubic foot hot gas discharged W (H. P. Cap. hot) « 1 r J.-1 #-1 1 = 144-^--(sup.pr.)ftp2^/epi ftpi r+Rj,2 • -21 . (83) WORK OF COMPRESSORS 105 Examination of Fig. 28 will show without analysis that there must be some best-receiver pressure at which least work will be required. For if tJie receiver pressure approached Pj, then the compression would approach single stage and ;3 > a o P-t i O o « O O t-4 B o O ^ooj ourenbs jad spunod ^1 SQ^nssaij cc (N the compression line approach BCG. The same would be true as the receiver pressure approached Pg = Pe, whereas at any intermediate point C, intercooling causes the process to follow BCDE with a saving of work over single-stage operation represented by the area DCGE. This area being zero when C is at 106 ENGINEERING THERMODYNAMICS « either B or (7, it must have a maximum value somewhere between, and the pressure at which this least-compressor work will be attained is the best-receiver pressure. By definition the best-receiver pressure is that for which TF is a minimum, or that corresponding to dPc Performing this differentiation upon £q. (78), equating the result to zero, and solving for Pc, (Best rec.pr.) = (P»P.)M(sup.pr.) (del. pr)]* (84) Substituting this value in the general expression for work Eq. (78), notii^ that pr~pr [pj''''^pr[pj jt-i F=2::^A7.[(^;) '• -l], (85) s-r Eq. (85) is the general expression for two-stage work with perfect inter- cooling at best-receiver pressure in terms of pressures and volumes. Sub- stituting the symbols for the pressures and volumes and noting that as in Cycle 1, 75=(L. P. Cap.) and 7.= (H. P. Cap. hot) and using.(/2,) for (^\ W = 288—% (sup.pr.) (L. P. Cap.) (ftpV - 1) . . . ' . . (86) This equation gives the same value as Eq. (85), but in terms of different units. It should be noted here that the substitution of best-receiver pressure in the expressions for the two stages preceding Eq. (78), mil show tfiat the work done in the two cylinders is equal. The work per cubic foot of low-pressure gas, from Eq. (86) is, W (L.P.Cap.) 7) =288—^ (sup.pr.) [/jpSr-l] (87) To transform Eq. (85) into a form involving delivery volmnes, use the rela- tion from the diagram, 1 1 y-<W-<%{W- WORK OF COMPRESSORS 107 Whence -'-<mw- which for the best-receiver pressure becomes «+i Fft=FJ2p2*. bubstituting in Eq. (85), W^2j^P,V.(Rp)^[RM-i\ (88) Introducing the S3miboIs, TF=288~(sup.pr.)(H.P. Cap. hot)flp^rftpV--ll, . . (89) • and W (H. P. Cap. hot) g a + ir .-1 -I =288-^(sup.pr.)ffp~2r[iJ,-27-lJ ... (90) The voliune of gas discharged at the higher pressure when reduced to its original temperature will become such that (L.P/Cap.) ^P, (H. P. Cap. cold) Pr^' or (sup.pr.) (L. P. Cap.) = (del.pr.) (H. P. Cap. cold), . . . (91) which may be substituted in Eq. (86), Tr= 288-^ (del.pr.) (H.P. Cap. cold) 1 /ZpV - 1, 1 . . . . from wnich the work per cubic foot of gas delivered and cooled is, (92) IF (H.P. Cap. cold) = 288j^ (del.pr.) j/2pV^ - 1 1 (93) 108 ENGINEERING THERMODYNAMICS Example 1. Method of calculating diagram, Fig. 28. Assumed data: Fa =0 cu.ft. Pa =2116 lbs. per square foot. F/=0 cu.ft. Pc^Pa^VPaPe =6172 lbs. sq.ft. F6=5 cu.ft. •P/=P,=P^= 18,000 lbs. sq.ft. « = 1.4. To obtain point C: or To obtain point Z>: To obtain point E\ 1 Fc = 7* + {-) ^'^ =2.36 cu.ft.; .*. Vc = 2.36 cu.f t. Pc = 6172 lbs. sq.ft -_ _, Pb - 2116 ^ -_, ,^ Fd = Fft X o- =5 X - - = 1.71 cu.ft. re Ol7J /. Fd = 1.71 cu.ft. Pd =6172 sq.ft. v,^v..(y;)h but by definition (£)"-(?t)"— . hence, Fe = 1.71 H-2.14 =.8 cu.ft. Pe = 18000 lbs. sq.ft. Example 2. To compress 5 cu.ft. of air from one atmosphere (2116 lbs. per square foot) to 8.5 atmospheres (18,050 lbs. per square foot) in two stages with best-receiver pressure and perfect intercooling requires how much work? FT =288 --(sup.pr.)(L. P. Cap.)(/2p ^' -1), 5 — 1 (sup.pr.) « 14.7. (L. P. Cap.) =5. Rp =8.5. .'. TF=288x3.463xl4.7x5x(8.5~2^-l) =26,800 ft.-lbs. WORK OF COMPRESSORS 109 Prob. L Air at 14 lbs. per square inch absolute is compressed to 150 lbs. per square inch absolute by a two-stage compressor. What will be the work per cubic foot of air delivered? What will be the work per cubic foot if the air be allowed to cool to the original temperature, and how will this compare with the work per cubic foot of sup- plied air? Best receiver-pressure and perfect intercooling are assumed for the above compressor, s = 1.4. Prob. 2. A compressor receives air at atmosphere and compresses it to half its vol- ume, whereupon the air is discharged to the cooler and its temperature reduced to the original point. It then enters a second cylinder and is compressed to 80 lbs. absolute. What will be the work per cubic foot of supplied air in each cylinder and how will the work of compressing a cubic foot to the delivery pressure compare with the work (lone if compression were single stage, compression being adiabatic. Prob. 3. Air is to be compressed from 15 lbs. per square inch absolute to 10 times this pressure. What would be the best-receiver pressure for a two-stage compressor? How many more cubic feet may be compressed per minute in two stage than one stage by the same horse-power? Prob. 4. A manufacturer sells a compressor to run at best-receiver pressure when (sup.pr.) is 14 lbs. per square inch absolute and (del.pr.) 100 lbs. per square inch absolute. What will be the work per cubic foot of supply-pressure air done in each cylin- der? Another compressor is so designed that the receiver pressure for same supply pressure and delivery pressure is 30 lbs. per square inch absolute, while a third is so designed that receiver pressure is 50 lbs. per square inch absolute. How will the work done in each cylinder of these machines compare with that of first machine? Prob. 5. For an ice machine a compressor works between 50 and 150 lbs. per square inch absolute. It is single stage. Would the saving by making compression two stage at best-receiver pressure amount to a small or large per cent of the work in case of single stage, how much? Prob. 6. A compressor has been designed to compress 1000 cu.ft. of carbon dioxide per minute from 15 to 150 lbs. per square inch absolute. What horse-power will be required at best-receiver pressure? Should delivery pressure change to 200 lbs., what power would be required? To 100 lbs. what power? Prob. 7. A -gas-compressing company operates a compressor which has to draw COs gas from a spring and compress it to 150 lbs. per square inch gage. In the morning pressure on the spring is 10 lbs. gage, while by evening it has dropped to 5 lbs. absolute. If the compressor was designed for the first condition, how will the high-pressure capacity cold and horse-power per cubic foot of high-pressure gas at night compare with corresponding values in morning? Assume a barometric reading. Prob. 8. On a mining operation a compressor is supplying a number of drills and hoists with air at 150 lbs. per square inch absolute, the supply pressure being 14 lbs. Wliat will be the difference in horse-power per cubic foot of delivered air at compressor and per cubic foot received at drills if air is a long time in reaching drills? Prob. 9. With a best-receiver pressure of 40 lbs. per square inch absolute and a supply pressure of 14 lbs. per square inch absolute, what horse-power will be required to compress and deliver 1000 cu.ft. of high-pressure air per minute at the delivery pressure for which compressor is designed and what is that delivery pressure? 10. TVo-stage Compressor^ with Clearance, Perfect Intercooling Expo- nential Compression, Best-receiver Pressure, Equality of Stages, (Cycle 6). Work and Capacity in Terms of Pressures and Volumes. The two-stage expo- no ENGINEERING THEEM0DTNAJUC8 nential compressor with clearance aad perfect intercooling is represented by the PV diagrams Figs. 29, 30, 31, which are clearly made up of two aii^e-stage compression processes, each with clearance. ij looj SMnoe Ma oq-i □! gMnww J Applying Eq. (57) to the two stages and supplying proper subscript, referrii^ to Fig. 29, WORK OF COMPRESSORS lU '^^1^'^^^''^^'[{f) ' "^]* • (second stage) If the condition of perfect intercooling be imposed, it is plain that since the weight of gas entering the second stage must equal that entering the first stage, and the temperature in each case is the same, and noting also that Pc^Pd, w.^^p^v.-y.my\{^y' -2], . ■ ■ (M) Eq. (94) is the general expression for the work of two-stage exp>onential compressor with perfect intercooling, Pc being the receiver pressure. p As in Section 9, let (Rpi) be the pressure ratio -^ for the first stage eLnd{Rp2) Pb p the pressure ratio -^ for the second stage and using instead of P» its equivalent Pc 144 (sup.pr.) lbs. per square inch. Tr=144^(sup.pr.)(L.P.Cap.) [(fl,i)VV(iJp2V-2,l . . (95) which is identical with (79), showing that far tvxhstage compressors vrith perfect intercooling {as for single stagey Section 7), the work for a given low-pressure capacity is independent of clearance. The work per cubic foot of gas supplied is given by Eq. (80); per cubic foot of cold gas delivered by Eq. (81) and per cubic foot hot gas delivered by Eq. (83). The reasoning regarding best-receiver pressure followed out in Section 9, will dW hold again in this case, and by putting -f^^O in Eq. (94), and solving for Pe a±c it will again be found that best-receiver pressure will be (best-rec. pr.) = (P6P,)* (96) Substitution of this value for Pe in Eq. (94), gives the following expression forwork of the two-stage exponential compressor with best-receiver pressure, Tr-2^P»(7»-F.)[(g)''-l], (97) 112 ENGINEERING THEBMODYNAMICS which may be exprf^ssed in terms of supply pressure, pounds per square inch low-pressure capacity, cubic feet, and ratio of compression, 1 S. o I 1oo£ oraniiB lad wfi ui eainBeajj which is the same as Eq. (86). Substitution of the value of best-receiver pressure in the expression for the wore: of compressors 113 work of the two stages separately will show the equality of work done in the respective stages for this case with clearance. Work per cubic foot gas supplied to compressor is . (L7^) = 288^j(sup.pr.)[7i,'ii-'-l] (99) Work per cubic foot of high-pressure gas hot is W (H. P. Cap :h^) = 288^^(sup.pr.)ft, 2. l/2p2. -ij. . . (100) The work per cubic foot of air delivered and cooled to its original tem- perature iSy Due to the fact that clearance has no effect upon the work per cubic foot of substance, as previously noted, Eqs. (99), (100) and (101) are identical with (87), (90) and (93). 11. Two-Stage Compressor, any Receiver Aressure, Exponential Compres- sion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and Horse-power, in Terms of Dimensions of Cylinders and Clearances. Referring to Fig. 29, let Di be the displacement of the first stage cylinder in cubic feet = (Vft— Ft), i>2 the displacement of the cylinder of the second stage in cubic feet = (Fd— V/), ci the clearance of the first stage, stated as a fraction of the displacement of that cylinder, so that the clearance of the first stage cubic feet = ciDi, and that of the second stage = C22)2. ^ The low-pressure capacity of the first stage (L. P. Cap.) in cubic feet is {W—Va)f and, as for the single-stage compressor, is expressed in terms of dis- placement, clearance and ratio of compression of the first stage as follows, see Eq. (64) : (L.P.Cap.i)=Z>i(l + ci-ciftpi"^)=DiJ5.i (102) For the second stage, the low-pressure capacity (L. P. Cap.2) is and is equal to (L.P.CSLp.2)=D2\l + C2-C2Rp2')=D2Ef^ .... (103) Volumetric efliciency of the first stage is given by E,i = l+ci-ciRpi' (104) 114 ENGINEERING THERMODYNAMICS Volumetric efficiency for second stage E^=l + C2-C2Rp2' ■ (105) 1^ It I! loo^ ajBncrgjaa gqi ii|Bajn«8»j ~ - It may be required lo find the receiver pressure (iocidental to the finding of work or horse-power) for a compressor with given cylinder sizes and deliver}' pressure. The condition as.sumed of perfect intercooling stipulates that (L. P, Cap.i)fsup.pr.) = (L. P. Cap.j) (rec.pr.), WORK OF COMPRESSORS 115 whence f V , .(L.P. Cap.i) (rec.pr.)==(sup.pr.) (j^pc^p^) If the volumetric efficiencies are known or can be sufficiently well approx- imated this can be solved directly. If, however^ £,1 and £,2 are not known, but the clearances are known, since these are both dependent upon the receiver presstire sought, the substitution of the values of these two quantities will give (recpr.) = (sup.pr.)— ^^ \»"P-P''-/ ^ J , . . . (107) i)2|l+C2 ['+-«fif)1 an expression which contains the receiver pressure on both sides of the equation. This can be rearranged with respect to (rec. pr.)i but results in a very complex expression which is difficult to solve and not of sufficient value ordinarily to warrant the expenditure of much labor in the solution. Therefore, the relations are left in the form (107). It may be solved by a series of approximations, the first of which is (rec.pr.) = (sup.pr.)^ approx (108) With this value for the receiver pressure, substitution may be made in the sec- ond member of the Eg. (107), giving a result which will be very nearly correct. If desirable, a third approximation could be made. To find the work of a two-stage exponential compressor in terms of displace- ment of cylinders, supply pressure, receiver pressure and delivery pressure, pounds per square inch, and volimietric efficiency of the first stage, jB,i, from (79) or (94) , « r .-1 .-1 -| jf=144^-^(sup.pr.)Di£?,i fipi"r-|-/2p2~-2 . . . .. (109) in which (rec^ and i?,2=f*^-"^^^ ^ (sup.pr.) (rec.pr.) To solve this the receiver pressure must be found as previously explained and the volimietric efficiency must be computed by Eq. (104) or otherwise be known. It is impracticable to state work for this general case in terms of displace-* ment and clearances directly, due to the difficulty of solving for (rec. pr.) in Eq. 116 ENGINEERING THERMODYNAMICS (107). It may, however, be stated purely in terms of supply and delivery pres- sures, in pounds per square inch, displacement, in cubic feet, and volumetric efficiencies, as follows: From Eq. (106), Rvi = I and D2E92 „ _ del.pr . D2Et2 _j^ D2gg ^^ sup.pr. Z)iA\i""^^Z)i£?,i' Hence «-i «-i ,r-144j-tj(a„p.pr.)/)A[(°J^) " +(«,?£) " -2J . (110) The mean effective pressure of the two-stage compressor referred to the low- pressure cylinder is found by dividing the work of the entire cycle Eq. (110), by the displacement of the first-stage cylinder, and by 144, to give pounds per square inch. m.e.p. referred to first-stage cylinder, pounds per square inch is, «-l 8-1 W .-:-,(sup.pr.)E.,[(||) ■ +(«,gg)-----2]. . a„) 144Di 8-1 It is well to note that this may also be found by multiplying (work done per cubic foot of gas supplied) by (volumetric efficiency of the first stage, Eti), and dividing the product by 144. In terms of the same quantities, an expression for indicated horse-power may be given as follows: where n is the number of cycles completed per minute by the compressor. For n may be substituted the number of revolutions per minute, divided by the revolutions per cycle, N The horse-power per cubic foot of gas supplied per minute is «-i »-i 'I.H.P s (sup.pr.) n(L.P.Cap.) s-1 229.2 [VA^ (gitr-"H«'^)"-4 • »-) WORK OF COMPRESSORS 117 Horse-power per cubic foot of gas delivered and cooled per minute. a-l » 1 I.H.P. n(H. P. Cap. cold) =j^-f'is^'[(l£)^+('^tt-:)"'"-^ »») Horse-power per cubic foot of hot gas delivered per minute *-i I.H.P. s sup.pr./I)iEvi\ » o L /T("H7PrCap. hot) "s-i 229.2 VD2£r2/ ''' *-i mr-i^m^-^- ■ <"« For the case where clearance is zero or negligible, these expressions may be simplified by putting Et? and Eei equal to unity. «-l a- I ■•H"'-.4^,"^^'U[(^;) •+«:)■ -4. . (116, I.H.P. per cubic foot, gas supplied per minute LH.P. _ s (sup.pr.)r/Z)i\ • , / n2\-T^' ] n(L.P.Cap:)~s-l 229:2 [W ^\"'dJ "^J* ' ^"^^ I.H.P. per cubic foot gas delivered and cooled per minute •-1 »- 1 I.H.P. -rli'-^m'-'^Ky-'l '"«) n(H. P. Cap. cold) I.H.P. per cubic foot hot gas delivered per minute a- I » — 1 «— 1 I.H.P. n(H. P. Cap. hot) -ASf(»:) ■ «-'[(&)"+«;)""-^] <"« Example 1. Method of calculating diagram, Figs. 29, 30, 31. Assumed data. Pa = Pft=2116 lbs. per square foot; p^=P^=P;^=Pt =6172 lbs. per square foot. P^=P,=P/ = 18,000 lbs. per square foot. C/(H. P.) = 7.5 per cent; C/(L. P.) 7.5 per cent; « = 1.4; L. P. Capacity =5 cu.ft. 118 ENGINEERING THERMODYNAMICS To obtain point K. From formula Eq. (64), 1 L. P. Cap. = A(l +Ci —CiRpu) 5=Z)i(l+.075-. 075X2.14), hence Z)i = 5.45 cu.ft. Ch - 7* - 5.45 X .075 * .41 cu.ft. .'. 7k - .4 cu.ft. ; Pk =6172 lbs. sq.ft. To obtain point A: -<w- 1 I 4X2. 14 = .856 cu.ft. .*. 7a = .85 cu.ft.; Pa -2116 lbs. sq.ft. To obtain point B: 7» = 7a+5 = .85+5 =5.85 cu.ft.; Pft=2116. lbs. sqit. To obtain point C: 7c = 75 -5- ^ V* =5.85 -^2.14 =2.73, .'. 7c =2.73 cu.ft.; Pc=6172 lbs. sq.ft. To obtain point D: Volume at Z> is the displacement plus clearance of H. P. cylinder. This cannot be found imtil the capacity is known. The capacity is the amount gas which must be taken in each stroke and which is also the amount actually delivered by L. P. cylinder cooled to original temperature. The amount of cool gas taken in by the second cylinder is (L. P. Cap.2) = fe'''^/-) (L. P. Cap.i) = |J^X5= 1.7 cu.ft. \rec.pr. / dI72 But _ (L. P. Cap.2) _ 1.7 , _^ . (, , o -I 1+.075-. 075X2.14 \1+C2 — C2/Cp2V Ci, = 7/=C2D, =.075 Xl.88 = .14 cu.ft. 7(f = Ci-hD, = 1.88 +.14 =2.02 cu.ft. Pd =6172 lbs. sq.ft. Other points are easily determined by relations too obvious to warrant setting dowiL Example 2. What will be the capacity, volumetric efficiency and horse-power per 1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow- ing compressor: Two-stage, double-acting cylinders, 22i and 34^X24 in., running at 100 R.P.M. High-pressure clearance 6 per cent, low-pressure 4 per cent. Supply WORK OF COMPRESSORS 119 pressure 14 lbs. per sqiiare inch absolute. Delivery pressure 115 lbs. per square inch absolute. The capacity wOl be the cylinder displacement times the volumetric efficiency. Di ^displacement of a 34ix24 in. cylinder, or 12.8 cu.ft. and D2 =displacement of a 22 J +24" cylinder or 5.4 cu.ft. To obtain the volumetric efficiency, make use of approximation of formula £q. (108), D 12 8 (rec.pr.) « (sup.pr.)^ « 14 X-^-V =33.2 lbs. sq.in, Ut 0.4 and then by Eq. (107) checking, ,2.8[n-.04-.04(^\TU-| (rec.pr.) -l^X-TTr tTk — TliT ~^^'^ ^^^- sq.in. E,i = 1 +ci -CiiRp,) • from Eq. (104), 1 - 1 +.04 - .04 X (2.5) • =96.8 per cent Therefore the capacity will be, 200 X 12.8 X. 968 =2480 cu.ft. per minute; 1 En^l+ct -CtiRpi)* from Eq. (105), = 1 +.06 -.06 X (3.28)-^^* =92 per'cent. From Eq. (113), I H.P. per cu.ft. (sup.pr.) air per minute is, LrJ: <-l 8 sup.pr. r (DiEn\ B yljy^A •' ol "«-l 229.2 l\D^EJ ^\^^D,E~J, "^> 1.4 14 r/lM_X.968\.286 / 5.4X.92 V^se i T^229":2L\ 5.4X.92 / ^ \^'^h2,^^ml ~^J ^'^^' Whence horse-power per 1000 cu.ft. of free air per minute -is, = 150. From Eq. (115) horse-power per cubic foot (del.pr.) air, hot = that of (sup.pr.) air X/^p * \~^~^) or 5.85 times that of (sup.pr.) air. .'. Horse-power per 1000 cu.ft. of hot (del.pr. air) =150x5.85 =877. Prob. 1. A two-stage double-acting compressor has volumetric efficiencies as shown by cards of 98 per cent and 90 per cent for the high- and low-pressure cylinders respect- ively. It is nmning at 80 R.P.M. and compressing from atmosphere to 80 lbs. per square mch gage. If the cylinders are 15ix25ixl8 ins., and speed is 120 R.P.M., what 120 ENGINEERING THERMODYNAMICS horse-power is being used and how many cubic feet of free and compressed air (hot and cold) are being delivered per minute, when « equals 1.41? Prob. 2. What horse-power will be needed to drive a two-stage compressor 10 i ins. and 16j Xl2 ins., double acting, with 5 per cent clearance in each cylinder at 160 R.P.M. when the supply pressure is atmosphere, delivery pressure 100 lbs. per square inch gage, when s equals 1.35? Prob. 3. A thousand cubic feet of free air per minute must be compressed by a two- stage compressor to 80 lbs. per square inch gage from a supply pressure of 10 lbs. per square inch absolute. The volumetric efficiencies for the high- and low-pressure cylin- 4ers are 85 per cent and 95 per cent respectively, and the receiver pressure is 25 lbs. per square inch absolute. What will be the displacement of each cylinder and the horse-power per cubic foot of (sup.pr.) air? Prob. 4. How many cubic feet of free air can be compressed in two-stage compres- sor 18ix30ix24 ins. with 5 per cent clearance in high-pressure cylinder and 3 per cent in low if (sup.pr.) is atmosphere and (del.pr.) 80 lbs. per square inch gage? How would the answer be affected if clearance were taken as zero? Take s = 1.41. Prob. 5. The volumetric efficiency of the low-pressure cylinder is known to be 95 per cent, and of the high-pressure cylinder 85 per cent. The cylinder sizes are 15 J X25i X 18 ins. and speed is 120 R.P.M. What horse-power must be supplied to the machine if the mechanical efficiency is 80 per cent and the pressure ratio 10 with a (sup.pr.) of one atmosphere? Prob. 6. A compressor runs at 120 R.P.M. and is double acting. It is compressing air from 14 lbs. per square inch absolute to pressures ranging from 70 lbs. per square inch gage to 100 lbs. per square inch gage. The cyUnders are 20i X32i x24 ins., and clearances 8 per cent and 4 per cent. Find the approximate receiver pressure, capacity and horse-power for the range of discharge pressure, for « = 1.3. Prob. 7. The volumetric efficiency of the low-pressure cyUnder of a two-stage com- pressor is known to be 95 per cent, the receiver pressure as shown by gage is 40 lbs., delivery pressure 100 lbs., and supply pressiu^ one atmosphere. What will be the horse-power if the machine runs at 120 R.P.M. and the low-pressure cylinder is 18 X 12 in.? « = 1.4. Prob. 8. An air compressor appears to require more power to run it than should be necessary. It ib a double-acting 18x30x24 in. machine running at 100 R.P.M. The volumetric efficiencies are 85 per cent and 90 per cent respectively and supply and delivery pressures 14 lbs. and 110 lbs. per square inch, both absolute. What would be the minimum work per cubic foot of (sup.pr.) air, per cubic foot of (del.pr.) air, hot and cold, for adiabatic compression? Prob. 9. The efficiency of the driving gear on an electric-driven compressor is 75 per cent. Power is being supplied at the rate of 150 H.P. How much air should be compressed per minute from 4 lbs. per square inch absolute to 100 lbs. per square inch gage, if the receiver pressure is 35 lbs. per square inch absolute and the low-pressure volumetric efficiency is 90 per cent, s being 1.4? 12. Two-stage Compressor with Best Receiver Pressure Exponential Compression. Capacity, Volumetric Efficiency, Work, Mean EfFective Pres- sures and Horse-power in Terms of Dimensions of Cylinders and Clearances. For the two-stage exponential compressor with or without clearance, and per- fect intercooling, the best-receiver pressure was found to be (Eq. 84), (best-rec.pr.) = [(sup.pr.) (del.pr.)]* (120) WORK OF COMPUESSORS 121 This expression Eq. (120) for best-receiver pressure makes it possible to evaluate Rpi and Rp2 as follows: g,xfor(be8t-rec.pr.) = ^"^*-^"^P'' = ^^^"PP''-^^^^^P^-^^ ' = [^g^1*=Ji!A (121) sup.pr. sup.pr. Lsup.pr.J and D e /u 4. \ (del.pr.) (del.pr.) Rp2 for (best-ree. pr.) = 7,-— -- — -. = ,7 ; .^ , ttt- (best-rec.pr.) [(sup.pr.) (del.pr.)]* -[(tij;)]'-«-'' • <'^> The use of these values for Rp\ and Rp2 in the expressions previously given for volumetric eflSciency for the general case, Eqs. (104) and (105) results in Volumetric eflBciency, first stage Je.i = (l+ci-cii2p2i), (123) and volumetric efficiency, second stage £^ = (1+C2-C2ftp2.) (124) The work was found to be represented by Eq. (98), which may be stated in tenns of displacement and volumetric efficiency of the first stage, as follows: W = 2SS~(mp.pr.)DiEjRp^--l\, .... (125) where Rp = 7 — -^-^ and where (sup.pr.) is in pounds per square inch. If the clearance is known for the first stage this becomes by the use of Eq. (104), s 1 r *-^ I WS8 ^^ (sup.pr.)Z)i(l+ci-ci«p2,) /2^ 2. - 1 , . . . (126) which is a direct statement of the work of a two-stage adiabatic compressor with perfect intercooling .in terms of supply pressure and delivery pressure, pounds per square inch, displacement, cubic feet and clearance as a fraction 122 ENGINEERING THERMODYNAMICS of displacement, provided the cylinder sizes and clearances are known to be such as to give best-receiver pressures. The mean effective pressure reduced to firstnstage displacement, in pounds per square inch, may be derived from either Eq. (125) or (126) by dividing the work by the displacement of the first-stage cylinder, and again dividing by 144. w 2s r i^ 1 m.e.p. = j;p^ =^3Y (8up.pr.)£^i ii!p 2t -1 2s I -LXr 5-- 1 ^' ' ' ' ^'^^ = -j^(8Up.pr.)ll+Ci— Ciflp2« 1 i2p2t —1 Since the work done is equally divided between the two cylinders when best- receiver pressure is maintained, the mean efifective pressure, in pounds per square foot, for each cylinder will be, one-half the total work divided by the displacement of the cylinder in question, w s r 1^ 1 m.e.p., first stage =2Q8nr'^7Zi;(^^P-P''-)^»'h^^ (128) Note that this is one-half as great as the m.e.p. of the compressor reduced to first stage, (127), m.e.p., second stage = ogsZT " IZT ^^P-P^-^ n~^»i U^p «« — 1 , . . . (129) But (sup.pr.)^^ = (rec.pr.) = (sup.pr.)(del.pr.) U, whence, m.e.p., second stage = (sup.pr.)(del.pr.) ♦JS?,2 Lb, ^T— 1. . , . (130) It is next necessary to investigate what conditions must be fulfilled to obtain the best-reieeiver pressure, the value of which is stated, Eq. (120). The condition of perfect intercooling provides that the temperature of the gas entering the second stage is the same as that entering the first stage, and hence that the product (volume entering second stage) X (pressure when entering second stage) must be equal to the product (volume entering first stage) X (pressure of supply- to first stage), or (L. P. Cap. 2)(rec. pr.) = (L. P. Cap. i) (sup. pr.). , . . (131 WORK OF COMPRESSORS 123 Combining with Eq. (120) (L . P. Cap.i) _ [(8up.pr.)(del.pr.)P _ [" (deLpr.) ]*_ p ^ . P. Cap.2) "" (sup.pr.) L (sup.pr.) J " ^ ' or (1) (2) r (3) r (3) J -| DA 1+ci— ciRp2t ■,,_ (I..P.C»p. i)_DiE., — L"" -"^J ,,„, From this three-part equation proper values may be found to fulfill require- ments of best-receiver pressure for: 1. The ratio of capacities for a given ratio of pressures, or conversely, the ratio of pressures when capacities are known; 2. The ratio of cylinder displacements for known volumetric efficiencies; 3. The ratio of cylinder displacements when the clearances and ratio of com- pression are known, or conversely, with known displacements and clearances the ratio of pressures which will cause best-receiver pressure to exist. This last case in general is subject to solution most easily by a series of approxi- mations. There is, however, a special case which is more or less likely to occur in prac- tice, and which lends itself to solution, that of equal clearance percentages. If ci=C2 the parenthesis in the numerator of part (3) of Eq. (132) becomes equal to the parenthesis in the denominator, and evidently the volumetric efficiency of the two cylinders are equal, hence for equal clearance percentages in the two tages, Wr^' (133) A case which leads to the same expression, Eq. (133), is that of zero clearance, a condition that is often assumed in machines where the clearance is quite small. The work per cycle, Eq. (126), when multiplied by the number of cycles performed per minute, n, and divided by 33,000, gives LH.P. = ^-^ ^^u^l'^ ''^' (^ +^^ - ^^^^^^) (^^^ ~ ^)> • (1^) from which are obtained the following: I.H.P. per cubic foot supplied per minute LH.P. _ . •(sup.pr.)(g^ir^^j)^ _ (J35J n(L.P. Cap.) s-1 114.6 I.H.P. per cubic foot delivered and cooled per minute I.H.P. _ « (del.pr.)(^^--jJ_i)^_ ^ _ (j3gj n(H. P. Cap. cold) s-l 114.6 124 [ENGINEERING THERMODYNAMICS and I.H.P. per cubic foot delivered hot per minute m PP^' K n-° S ^'rr;V-^-(iJ,'^'-l). . (137) n(H. p. Cap. hot) s — 1 1 14.6 These e^qpressions, Eqs. (165), (166) and (167) are all independent of clear- ance. Example. What will be the capacity, volumetric efficiency and horse-power per 1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for tht following compressor for a = 1.4? Two-stage, double-acting, cylinders 22i x341 X24 ins., running at 100 R.P.M. Low-pressure clearance 5 per cent, high-pressure clearance such as to give best-receiver pressure. Supply pressure 15 lbs. per square inch abso- lute, delivery pressure 105 lbs. per square inch absolute. Capacity will be cylinder displacement times low pressure volumetric efficiency, or, 200DiX^rt. Di« 17.5 cu.ft. ]_ Eti from Eq. (123) = (1 +Ci -CiRp2») = 1 + .05 - .05 X 7^^^ = 95 per cent. Therefore low pressure capacity =200 X 1 2.8 X. 95 =2430 cu.ft. per minute. Horse-power per cubic foot of (sup.pr.) air per minute is from Eq. (135) 8 sup.pr. i_-J 7-1 114.6"^^' '' "^^^ 14. ^^ * .4 ^ 105^ ^ Therefore, horse-power per 1000 cu.ft. of sup.pr. air = 160. Horse-power per cubic foot of (del.pr.) air, hot, is from Eq. (137) Rp 2a times power per cu.ft. of (sup.pr.) air, hence, 160 X5.3 =850 = horse-power per 1000 cu.ft. of (del.pr.) air, hot, per minute. Problem Note. In the following problems, cylinders are assumed to be proportioned with reference to pressures so as to give best-receiver pressure. Where data conflict, the conflict must be found and eliminated. Prob. 1. Air is compressed adiabatically from 14 lbs. per square inch absolute to 80 lbs. per square inch gage, in a 20 J X32i X24 in. compressor, running at 100 R.P.M. . the low-pressure cyUnder has 3 per cent clearance. What will be horse-power re- quired, to run compres or and what will be the capacity in cubic feet of low pressure and in cubic feet of (del.pr.) air? Prob. 2. What must be the cylinder displacement of a two-stage compressor with 5 per cent clearance in each cylinder to compress 500 cu.ft. of free air per minute from 14 lbs. per square inch absolute to 85 lbs. per square inch gage, so that s equals 1.4? What will be the horse-power per cubic foot of (del.pr.) air hot and cold? Prob. 3. A two-stage compressor is compressing gas with a value of s = 1.2o from 10 lbs. per square inch gage to 100 lbs. per square inch gage. The cylinders are 18jx30tx24 ins., and speed is 100 R.P.M. If the clearance is 5 per cent in the low- WORK OF COMPRESSORS 125 pressure cylinder and 10 per cent in the high, what will be the cubic foot of (sup.pr.) gas' handled i>er minute and what will be the horse-power at best receiver pressure? Prob. 4. A manufacturer states that his 201^x32^X24 in. double-acting compres- sor when running at 100 R.P.M. at sea level will have a capacity of 2390 cu.ft. of free air per minute, pressure range being from atmosphere to SO lbs. per square inch gage. At best-receiver pressure what clearance must the compressor have, compression being adiabatic? Prob. 5. The cylinder sizes of a two-stage compressor are given as 10} X 16^x12 ins., and clearance in each is 5 per cent. What will be the best-receiver pressures when operating between atmosphere and following discharge pressures, 60, 70, 80, 90, 100 and 1 10 lbs. per square inch gage, for s equal 1.4? Prob. 6, 1500 cu.ft. of air at 150 lbs. per square inch gage pressure are needed per minute for drills, hoists, etc. The air is supplied from 3 compressors of the same size and speed, 120 RJ*.M. Each has 4 per cent clearance in each cylinder. What will be sizes of cylinders and the horse-power of the plant for best-receiver pressure, when s = 1.41? Prob. 7. The cards taken from a compressor show volumetric efficiencies of 95 per cent and 80 per cent in low- and high-pressure cylinder respectively. What will be (del.pr.) for best-receiver pressure if compressor is 151x25^X18 ins., and (sup.pr.) 15 ll)s. \yeT square inch absolute to 10 lbs. absolute, and what will be the work in each case, 9 being 1.35? Prob. 8. A manufacturer gives a range of working pressure of his lOj Xl6}xl2 in. compressor from 80-100 lbs. per square inch page. If clearances are, low 4 per cent, high 8 per cent, and (sup.pr.) is atmosphere, find by trial which end of the range comes nearest to giving best-receiver pressure? If clearances were equal which would give best-receiver pressure? Prob. 9 A 16} x25} Xl6 in. compressor is rated at 1205 cu.ft. free air per minute at 135 R.P.M. at sea level. What would be the clearance if compressor were compressing air from atmosphere to 100 lbs. gage at sea level? With same clearance what would be the size of a low-pressure cylinder to give the same capacity at altitude of 10,000 ft. with the same clearance and the same (del.pr.) , best-receiver pressure always being maintained? 13. Three-Stage Compressor, no Clearance, Perfect Intercooling Expo- nential Compression (Cycle 7), Best Two Receiver Pressures, Equality of Stages. Work and Capacity, in Terms of Pressures and Volumes. The three- stage exponential compressor cycle with no clearance, perfect intercooling Cycle 7, is shown in Fig. 32. The net work area, ABCDEFGHJKA, is made up of three areas which may be computed individually by the formulae for single stage Eq. (48), provided the requisite pressures and volumes are known, as follows: «-i T^=-l-|P^F^[(^) ' -ij (first Stage) 8-1 -_;^PdVa I W ) * — 1 (second stage) 8 t-i s ZTi^rVfU^) ' -A (third stage) . (138) ENGINEERING THERMODTNAMICS But the conditioD of perfect intercooling provides that for no clearance, P^Vt=pgV^^PfVf. and it may be noted that Pi=P„ and P/^P,. Accordingly, Pressures in this expression are in pounds per square foot. (140) WORK OF COMPKESSORS 127 Changing the equation to read in terms of supply pressure pounds per square inch, low-pressure capacity cubic feet, and ratios of pressures, first stage (Rpi)y second stage {Rp2) and third stage {Rpz)^ it becomes Work done by three-stage compressor, perfect intercooMng W^ = 144^(8up.pr.)(L. P. Cap.) [{Rj-^ + {Rp2)^+{Rpz)^-^]f (141 From this the following expressions are derived: Work per cubic foot supplied (L.pSap.) °^^A^^''^'^'''^[^^'''^^'"^ (fii>2p"^ + (gp3/-^-3]]. . . (142) Work per cubic foot gas delivered and cooled (H.P.cTp.cold) = ^^(^^^-P^4^^-'^'"^"+^^''^'^"+^^''^'"^"4^^^^ Work per cubic foot gas, as delivered hot •-1 W (H. P. Cap. hot) = 144^^(8up.pr.)(/2pi)(/2p2)(fl,8)*[(/2,i) ' +(Rp2f^ + {Rp3fr-3\ (144) Best Two Receiver Pressures, Referring to Fig. 32, Pe is the pressure in the first receiver (1 rec.pr.) and P« is the pressure in the second receiver, (2 rec.pr.). It is evident that if either receiver pressure be fixed and the other is varied, the work necessary to compress a given initial volume of gas will be varied, and will have a minimum value for some particular value of the varying receiver pressure. By a variation of both receiver pressures a minumim may be found for the work when both receiver pressures have some specific relation to supply and delivery pressures. For instance, assume that Pc is fixed. Then a change in /*• can change only the work of the second and third stages, and the three- stage compressor may be regarded as consisting of One single-stage compressor, compressing form P* to Pc. One two-stage compressor, compressing from Pc to P^. In this two-stage compressor, best-receiver pressure is to exist, accord ing to Eq. (84), P*= (best 2 rec.pr.) = (PJ>,)*. . ^ (145) 128 ENGINEERING THERMODYNAMICS Similar reasoning, assuming P« fixed and making Pc variable, would show that Pc=(best 1 rec.pr.) = (PeP6)* (146) Eliminate Pc from Eq. (145) and the expression becomes, P.= (best 2 rec.pr.) = {PJPg^) = [(sup.pr.)(del.pr.)2] * (147) Similarly, from Eq. (146) Pc= (best 1 rec.pr.) = {P^^P,) = r(sup.pr.)2(del.pr.)l* (148) From these expressions may be obtained. Pc^P.^P.^/P.y P, Pc ~Pe \Pj or Rp 1 = Rp2 ~ Rpi = -Bp* (149) Substitution in Eq. (140) gives, Work, three-stage, best-receiver pressure no clearance IF=3^^P.n[@'^'-l] (150) Arranging this equation to read in terms of supply pressure, pounds per square inch, low-pressure capacity, cubic feet, and ratio of pressures Work, three-stage best-receiver pressure Tr=432-^(sup.pr.)(L.P. Cap.)(i2pV-l), . . . (151) The work of the compressor is equally divided between the three stages when best-receiver pressures are maintained, which may be proven by substitu- tion of Eq. (149) in the three parts of Eq. (138), and Work of any one stage of three-stage compressor with best-receiver pressure. Wi = W2 = W^ = U4-^(sup,pT.){L. P. Cap.)(/2/"ir-l). . (152) WORK OF COMPRESSORS 129 From Eq.(151), may be derived the expressions for work per unit of capacity. ^ork per cubic foot low-pressure gas is, (L -^^=432^-^(sup.pr.)[fi,^-l] (153) $ince (L. P. Cap.) = (H. P. Cap. cold)i2p. RTork per cubic foot cooled gas delivered is, ^ Tr=432-*-(sup.pr.)/2p(i2,V-l). . . . (154) (H. P. Cap. cold) 5-1 \gain, from Fig. 32, ^'(?f -"'-<;)- ^'-'©(^f' which is to say that, when best-receiver pressures are maintained, (L. P. Cap.) = (H. P. Cap. hot)Rj,JRprB, or 2g + l = (H. P. Cap. hot)fl,"3r-, .... (155) hence Work per cubic foot hot gas delivered =432-^(sup.pr.)i2,"3r-(i2p-8r-l). . . (156) (H.P. Cap. hot) — 5-1 Example 1. Method of calculating Diagram, Fig. 32. Assumed dcUa, Pa =Pft =2116 lbs. per sq.ft. Pc =Pd =best first-receiver pressure =PjPg^ =4330. Pe =P/=best second-receiver pressure =Pi,^Pg^ =8830. Pg^Pn = 18,000 lbs. per sq.ft. 7a = 7A =0 cu.ft. 7& =5 cu.ft. s -1.4. 130 ENGINEERINQ THERMODYNAMICS To obtain point C: F,-F»-i-(^)^*-5+1.67=3cuit. '-"-(§)■' A Fc - 3 cu.ft. Pe =4330 lbs. sq.ft. Intermediate points Bix>C may be found by assuming various'pressures and finHmg the ccMTesponding volumes as for 7c. To obtain point D: •TT <rr * * » 2116 ^ .. ». 7d-7ftX5- -6Xt^^-2.44 cu.ft. Pd 4330 Va -2.44 cu.ft., Pa -4330 lbs. sq.ft. To obtain point E: 1 by assumption of best-receiver pressure. Hence 7* -2.44 -s- 1.67 =1.46 cu.ft., an P.=8830 lbs. sq.ft. Intermediate points Dio E may be found by assuming various pressures and finding corresponding volimies as for 7«, andsucceeding points are found by similar methods to these already used. 7^ -.72, Pi, = 18,000, Example 2. What will be the horse-power required to compress 100 cuit. of free air per minute from 15 lbs. per square inch absolute to 90 lbs. per square inch gage in a no-clearance, three-stage compressor if compression be adiabatic? What will be the work per cubic foot of (del.pr.) air hot or cold? From Eq. (153) work per cubic foot of (sup.pr.) air is, 8 t^ 432 — -(sup.pr.) (iJp 3. -1), 8 — 1 -432X^X16X(709S2..1) ^4500 ft.-lbs., .4 or H.P. for 100 cu.ft. per minute - — -13.6. OtiJ,UUU From Eq. (154) work per cubic foot of (del.pr.) air cold is Rp times that per cubic foot of (sup.pr.) air, or in this case is 31,500 ft.-lbs. i From Eq. (156) work per cubic foot of (del.pr.) air hot laRp ^ times that per cubic foot of (sup.pr.) air, or in this case 5.8x45,000 «46,200 ft.-lbs. WORK OF COMPRESSORS 131 Prob. 1. What work will be required to supply 2000 cu.ft. of air at 200 lbs. per inch gage pressure if compressing is done adiabatically by three-stage compressors, taking air at atmosphere, neglecting the clearances? Prob. 2. A motor is available for running a compressor for compressing gajs, for which 8 equals 1.3. If 60 per cent of the input of the motor can be expended on the air, to what delivery pressure can a cubic foot of air at atmospheric pressure be com- pressed in a zero clearance three-stage machine? ' How many cubic feet per minute could be compressed to a pressure of 100 lbs. gage per H.P. input to motor? Prob. 3. Two compressors are of the same size and speed. One is compressing air so that exponent is 1.4, the other a gas so that exponent is 1.1. Each is three stage. Which will require the greater power to drive, and the greater power per cubic foot of (sup.pr.) gas, and per cubic foot of (del.pr.) gas, hot, and how much more, neglect- ing clearance? Prob. 4. How will the work per cubic foot of (sup.pr.) air and per cubic foot of (del. pr.) air difiFer for a three-stage compressor compressing from atmosphere to 150 lbs. per square inch gage from a single- and a two-stage, neglecting clearance? Prob. 6. A table in " Power " gives the steam used per hour in compressing air to various pressures single stage. A value for air compressed to 100 lbs. is 9.9 lbs. steam per hour per 100 cu.ft. of free air. Using the same ratio of work to steam, find the value for the steam if compression had been three-stage, zero clearances to be assumed. Prob. 6. A 5 in. drill requires 200 cu.ft. of free air per minute at 100 lbs. per square inch gage pressure. What work will be required to compress air for 20 such drills if three-stage compressors are used, compared to single-stage for no clearance? Prob. 7. What would be the steam horse-power of a compressor delivering 150 cu.ft. of air per minute at 500 lbs. per square'^inch pressure if compression is three-stage, adiabatic, clearance zero, and mechanical efficiency of compressor 80 per cent? 14. Three-stage Compressor with Clearance, Perfect Intercooling Expo- nential Compression (Cycle 8), Best-receiver Pressures, Equality of Stages. Work and Capacity in Terms of Pressures and Volumes. The pressure- volume diagrams of the three-stage compression is shown in Figs. 33, 34 and 35, on which the clearance volume and displacements, low-pressure capacity and high-pressure or delivery capacity for hot gas are indicated. If perfect intercooling exists, as is here assumed, (V,- Va)Pi> = (V^- Vi)Pa = {Vf- Vj)Pf and also (L. P. Cap.)P6 = (H.P. Cap. cold)P^. Apply Eq. (57) to the three stages and the entire work done is, JF=^P,(F,-7a)[(^;)'"^-l] (first stage) (157) 8 +» ~P<,{V^-V,) [ (^^J-r - 1 j (second stage) ~lPfiyf-Vj) [(^')'"^'-l] (thW stage) . . . (168) ENGINEERING THERMODYNAMICS ' By use of the above conditions of perfect intercooling Eq. (157) this expression becomes, (159) I / A A qI £L <^. I — -• a- S 15 V»>& aranDs jaa Epanoj u] WORK OF COMPEESSOES 133 Id terms of supply pressure, pounds per square inch, low-pressure capacity, cubic foot and ratios of pressures as above, the worl^ of a three-stage com- pressor with perfect intercooling and with clearance is H -144-%(sup.pr.)(L.P. Cap.) f(ftpi)^'-|-{ft;.2) V+(/i'^.i)^-3l, (l(iO) ^ / / I - _l el Si which is identical with Eq. (141), ahoufing that dearance kas no effect upon the irnrk for a given capadiy. ENGINEERING THERMODYNAMICS It readily follows that the work per UDJt of gas is independent of clearance, and hence Eqs. (142), (143) and (144), will give a correct value for the work / A / i — _i -ir I-; w so per cubic foot of gas supplied, per cubic feet delivered and cooled, and per cubic foot as delivered hot, respectivply. Since in two-stage compressors the reasoning leading to the determination of best-receiver pressure applies equally well with and without clearance, and since the value of best-receiver pressures for three-stage are found by eon- WORK OF COMPRESSORS 135 sidering the threenstage a combination of one- and two stage-compressors, the same expressions for best-receiver pressures will hold with clearance as without; see Eqs. (147) and (148). Pg=(best 2 rec.pr.) = [(sup.pr.)(del.pr.)^l*- Pc— (best 1 rec.pr.) = [(sup.pr.)^(del.pr.)l*. The use of these expressions for best-receiver pressures leads to the same result as for no clearance £q. (150), except for the volumes, Work, three-stag^ best-receiver's pressure with clearance which is stated below in terms of supply pressure, pounds per square inch low- pressure capacity, cubic foot, and ratio of compression Rp, Work, three-stage besfr^receiver pressure. Tr=432-^(sup.pr.)(L.P.Cap.)(iBp^^-l) . . . (162) which is identical with Eq. (151). From this may be obtained expressions for the work per cubic foot of low- pressure gas supplied to compressor per cubic foot of gas delivered and cooled, and per cubic foot of gas as delivered hot from the compressor, when the re- ceiver-pressures are best, and these will be respectively identical with E3qs. (153), (154), and (156), in the foregoing section. 16. Three-stage Compressor^ any Receiver-pressure Exponential Com- pression. Capacity^ Volumetric Efficiency^ Work, Mean Effective Pressurei and Horse-power in Terms of Dimensions of Cylinders and Clearances. Dis displacement of the first-stage cylinder, in cubic feet=(F6— Fm); 2)2= displacement of the second-stage cylinder, in cubic feet = (7d— Fa)> D3» displacement of the third-stage cylinder, in cubic feet = (F/— F*). ci, C2f cs are the clearances of the first, second and third stages respectively, stated as fractions of the displacement, so that, Clearance volume, 1st stage, in cubic feet = Fm=ciDi; Clearance volume, 2d stage, in cubic feet»^F* = C2l>2; Clearance volume, 3d stage, in cubic feet = FA = C3l>3. 136 ENGINEERING THERMODYNAMICS The low-pressure capacity of the first stage, and hence for the compressor is (Vb^Va)f and in terms of clearance, ci, and displacement Di of the fir<t stage is, according to Eq. (64), j_ (L.P. Cap.i)=Di(l+ci-ci/epi')=^i^.i (163) For the second stage, the low-pressure capacity is (Fd— Vi) and is equal to j_ (L. P. Cap.2) = 2>2(1 +C2 - C2Rp2 ' ) = D2E,2, .... (164) and for the third stage (F/— Vj) or, (L. P. Cap.3) = 2>3(1 +C3 - cafips * ) = D^E^z (165 j The volumetric efficiency of 1st stage is j_ jB,i = (l+ci-ciflpiO (166) Volumetric efficiency of second stage is JB.2 = (1 + C2-C2flp20 (167) Volumetric efficiency of third stage is 1 S.3=(l+C3-C3flp3'), (168) The work of the three-stage compressor with the assistance of Eq. (163) may be stated in terms of supply pressure, pounds per square inch, displacement of first-stage cylinder, in cubic feet and volumetric efficiency of first stage, and also ratios of compression existing in the first, second, and third stages, TF=144^(sup.pr.)2>i£.i[(/epi) '~r+ {R,2) V+ (/J^g) V-3 1 . (169) To make use of this formula for the work of the compressor the two receiver pressures must be known, and it is, therefore, important to derive a relation between receiver pressures, displacements and clearances or volumetric efficiencies. The assumption of perfect intercooling which has already been made use of in obtaining Eq. (169), regardless of the receiver-pressure, requires that — see Eq. (157) : (L.P. Cap.i) (sup.pr.) = (L.P. Cap.2) (1 rec.pr.) = (L.P. Cap.3)(2 rec.pr.). . (170) WORK OF COMPRESSORS 137 Using values of capacities in Eqs. (163), (164), and (165) and solving for first- and second-receiver pressures. /■i \ / V (L. P. Cap.i) / .DiEvi /^^^\ (lrec.pr.) = (sup.pr.)-^L.P:Cap7)=(^"P-P'->5^£72' " ' ^^^^^ and (2rec.prO = (sup.prOJ^;-^;g^5;;} = (sup.pr.)^. . . (172) Then P _(1 rec.pr.) _DiE^ ,, . ''"'"Isu^prO ~D2£,2' ^"^^ P _ (2 rec.pr.) _ Z)2S,2 ,,_.. By definition, (del.pr.) =Rp (sup.pr.), __ (del.pr . _ (sup.pr.) _ DzE,:^ . . ^^"(2rec.pr.)'"^''(2rec.pr.)~^''Di£,i ^^^^^ The work of the three-stage compressor may then be stated in terms of supply pressure, pounds per square inch, displacements, cubic feet, volumetric efficiencies, and overall ratio of compression, Rp, as follows: ^.|44.-^,(s„p.p..)O.E..[(^*)^'+(|g)'-^" »-l +(«'gt:) ■ -']■ ■ <"« In Terms of PressureSf Displacements, and Clearances, an expression can be written by substitution of values of Evi, E^ and Evz from Eqs. (166), (167) and (168), but it becomes a long expression, further complicated by the fact that Rvi, Rp2 and Rpz remain in it. This may be solved by the approximation based first upon the assumption that all volumetric efficiencies are equal to each other or to unity when (If volumetric efficiencies are each equal to each other or to unity) (177) 138 ENaiNEERING THERMODYNAMICS This process amounts to the same thing as evaluating Evu E^* <uid Evb from Eqs. (166), (167) and (168), making use of the approximation Eq. (177) and substituting the values found in Eq. (176). Since the above can be done with any expression which is in terms of volu- metric efficiencies, the following formulae will be derived from Eq. (176), as it stands. The mean effective pressure of the three-stage compressor reduced to the first- stage cylinder is found by dividing the work of the entire cycle, Eq. (176) by displacement of the first stage, and by 144 to reduce to pounds per square inch. (m.e.p.) reduced to first stage cylinder, j^^=— j(sup.pr.)fi.,[(^^^-j . +(5^3) • Note here that this may also be obtained by multiplying (work per cubic foot supplied) by (volumetric efficiency of first stage) and dividing the product by 144. The indicated horse-power of a compressor performing n cycles per minute will be equal to the work per cycle multiplied by n and divided by 33,000, or, for the three-stage compressor with general receiver pressures, '•H.p-.4T^^''"'.-..[(5s:i)'^'+(Sl!)^" t+(«'tfe)"'-4 • ("») For n may be substituted the number of revolutions per minute, N, divided by the revolutions required to complete one cycle N n=— . z The horse-power per cubic foot of gas supplied'per minute is I.H.P. 8 (sup.pr:) r/ Dig,i \ii:i / D2E,2 \'-^ n(L.P.Cap.) «-l 229.2 [[OiEj \'^\DzEJ ' + «l.)^'-4 • (>»«» Horse-power per cubic foot gas delivered and cooled per minute is I. H.P. ^ 8 (d el.pr.) V/ DiE^A 'T' / D2gr2 \ V n(H.P.Cap.cold) s-1 229.2 [\D2Ej '^[DsE.sJ 8-1 + «rfe)^""-4 • ('") WORK OF COMPRESSORS 139 Horse power per cubic foot hot gas delivered per minute is I.H.P. 8 (sup.pr.) /Pig,i\V Lr/Pig.i\V n(H. P. Cap. hot) s-l 229.2 [DsE^j/ '' [\D2E,2) + /DiE,2 »-l • •-! ^r^^'^r-'i ■ a»^) The last equation is obtained by means of the relation (L. P. Cap.) = (H. P. Cap. hot) X (^'^^)^X (^-I^^^) ^ ^ / V *^ ' \2 rec.pr./ \ sup.pr. / = (H.P.Cp.ho.)X«^)^(^-jf^) »-l = (H.P.Cap.hot)Xi?pr(^^) ' . ..... (183) // clearance is zero or negligible, these expressions may be rewritten, putting Ev, Ev2 and Evz each equal to unity. «— 1 J «— 1 «— 1 •"••"-j^^f^-^^®) ■■+(t) ■ +«) ■ -4 <'«> H.P. per cubic foot of gas supplied per minute is I.H.P. _ (sup.pr.) r /DA '^ /D2\ ^\ / „ ^\^*_il n(L.P. Cap.) 229.2 L\W W \'Di) ^\ H.P. per cubic foot delivered and cooled per minute is •-1 »-i »-i . . (185) I.H.P. =^Sf^[(S)"+(t)"+(«.|)"-'] ■ ^m n(H. P. Cap. cold) H.P. i>er cubic foot hot gas delivered per minute is I.H.P. ^ 8 (sup.prQ /DA'-f-' if/^X^' /^\^' n(H. P. Cap. hot) .«-l 229.2 \Ds) ^Lw W 8-1 + (i2.^) ' -3]. . . (187) 140 ENGINEERING THERMODYNAMICS Example 1. Method of calculating Diagram, Fig. 35. Assumed data: ■Pa =Pft =2116 lbs. per square foot. Pc =Pd-Pi-Pm =4330 lbs. per square foot. » P«=P/=P^=P4=8830 lbs. per square foot. Pg^Ph- 18,000 lbs. per square foot. ci=7.5 per cent for all cylinders; « = 1.4. L.P. capacity 5 cu.ft. To obtain point M: 1 From formula Eq. (163) L. P. Capi. - A(l +Ci -ci/2,i •) or 5 =r>i(l +.075 -.075 X 1.67) or Z)i =5.3 cu.ft. and clearance volume Vm = 5.3 X .075 = .387 cu.ft. Therefore, Vm = .39 cu.ft. ; Pm =4330 lbs. sq.ft. ; To obtain Point A : Va^VmX (-^j 1.4. = .39 X 1.67 = .67 cu.ft. Additional points M to A may be found by assuming pressures and finding corr& sponding volumes as for Va- To obtain point B: Vb = 7a + (L. p. Capi.) «.67+5 =5.67 cu.ft. i Therefore, Fft = 5.67 cu.f t. ; P* = 21 16 lbs. sq. ft. ; To obtain point C: Therefore, i_ Vc^Vb^ (~-j ^'^ =5.67 -^ 1.67 =3.45 cu.ft. Ve = 3.45 cu.f t. ; Pe = 4330 lbs. sq.ft. Intermediate points J5 to C may be found by assuming various pressures and findi corresponding volumes as for Vb^ WORK OF COMPRESSORS 141 *o obtain point D: Volume at Z> is the displacement plus clearance of the intermediate cylinder. This umot be found until the capacity is known. Appl3dng the same sort of relations as rere used in calculating the diagram for the two-stage case with clearance, D2(l +C2 -C2RP2) • =2.44 or D2 =2.57, nd clearance volume. . Vt = .075 X2.57 = .192 cu.ft., ence, 74=2.57+.19=2.76cu.ft. Tierefore, Vd =2.76 cu.ft; Pa =4330 lbs. sq.ft. ; The rest of the points are determined by methods that require no further explana- lon and as pressures were fixed only volumes are to be found. These have the following alues, which should be checked: % = 1.65; Vr = 1.32; 7^ = .79; Fa = .09; Fy = .15; Fz = .32; 7, = .65; 7^ = 1.23; F, = .14. Bxample 2. A three-stage compressor is compressing air from atmosphere to 140 )s. per sqxiare inch absolute. The low-pressure cyUnder is 32 X 24 ins. and is known to ave a clearance of 5 per cent. From gages on the machine it is noted that the first- Bceiver pressure is 15 lbs. per square inch gage and the second-receiver pressure is 5 lbs. per square inch gage. What horse-power is being developed if the speed is 00 R.P.M. and s = 1.4? From the formula Eq. (169), W =144-^(sup.pr.)D,^„ Rprr -\-rJ~b~ -\-RprT~ --Z . 5 — 1 L J From gage readings /Cpi=--— Z. /Cj»j=— — Z.,5tJ, ^J»l— yQ-— -6. 1 Et, = (1 +Ci -CiRpC* ) from Eq. (166), r, lence. En^{l +.05 -.05X1.65) =67.5 per cent. TF = 144x^Xl5Xll.2x.675(1.22+1.28+1.22-3); .4 =59,200 ft.lbs. per stroke or 200 X59,200 ft. =lbs. per minute; =358 I.H.P. 142 ENGINEERING THERMODYNAMICS Examples. Another compressor has cylinders 12x20x32x24 in. and it is known that the volumetric efficiencies of the high, intermediate and low-pressure cylinders are respectively 70 per cent, 85 per cent and 98 per cent. The (del.pr.) is 150 lbs. per square inch absolute. What is the horse-power in this case if the speed is 100 R.P.M.? From the formula Eq. (176), -i«x^.><„..x..[(;-iti)-^.(if^)^ + V"ll.2xW ^J = (1.309+1.495+1 -3) =66,400 ft.-lbs. per stroke, 200X66,400 .„ Whence I.H.P. = — qoTw^ — =402. • Prob. 1. What will be the horse-power required to drive a 12 X22 X34 X30 in. three- stage compressor with volumetric efficiencies of 75, 85, and 95 per cent in the high, intermediate and low-pressure cylinders, at 100 R.P.M. when compressing natural gas from 25 lbs. per square inch gage to 300 lbs. per square inch gage, adiabatically? Prob. 2. A three-stage compressor for supplying air for a compressed-air locomo- tive receives air at atmosphere and delivers it at 800 lbs. per square inch gage. Should the receiver pressures be 50 lbs. and 220 lbs. respectively in the first and second and the volumetric efficiency of the first stage 90 per cent, what would be its displacement and horse-power when compressing 125 cu.ft. of free air per minute, adiabatically? What are the cylinder displacements? Prob. 3. Find the work done on a gas, the value for s of which is 1.3, in compressing it from atmosphere to 7 atmospheres absolute, adiabatically in three stages, the com- pressor having a low-pressure cylinder displacement of 60 cu.ft. per minute and a volu- metric efficiency of 95 per cent, first receiver pressure being 2 atmospheres absolute, and second-receiver pressure 4 atmospheres absolute. If air were being compressed instead of the above gas, how would the work vary? Prob. 4. The cylinders of a compressor are 8x12x18x24 ins. and clearance such as to give volumetric efficiencies of 80, 90 and 98 per cent in the different cylinders in the order given. Compressor is double acting, running at 120 R.P.M. and compressing air adiabatically from 14 lbs. per square inch absolute to 150 lbs. per square inch gage. What is the capacity in cubic feet per minute, work done per cubic feet of (sup.pr.) air, (del.pr.) air hot and cold and the horse-power of the compressor? What would be the effect on these quantities if the clearances were neglected? Prob. 6. If the cyHnders of a compressor are 10x14x20x18 ins., and clearances are 8, 5 and 3 per cent, what work is being done in adiabatically compressing air from 10 lbs. per square inch absolute to 100 bs. per square inch gage? Note: Solve by approximate method. Prob. 6. For special reasons it is planned to keep the first-receiver pressure of a throe-stage compressor at 30 lbs. per square inch absolute, the second-receiver pressure at 60 lbs. per square inch absolute, and the line pressure at 120 lbs. per square inch absolute WORK OF COMPRESSORS 143 The (sup.pr.) is 14 lbs. per square inch absolute. If the clearances are 4 per cent in the low and 8 per cent in the intermediate and high-pressure cylinders, what must be the cylinder sizes to handle 500 cu.ft. of free air per minute at 120 R.P.M. and what power must be supplied to the compressor on a basis of 80 per cent mechanical effi- ciency, for a value of « equal to 1.39? Prob. 7. Should the above pressures (Prob. 6) be gage pressures instead of absolute, how would the quantities to be found be affected? Prob. 8. The receiver pressures on a COt gas compressor are 50 lbs. per square inch absolute, and 200 lbs per square inch absolute, the (del.pr.) being 1000 lbs. per square inch absolute. The machne has a low-pressure cylinder 8x10 ins. with 3 per cent clearance. What horse-power will be required to run it at 100 R.P.M and what would be the resultant horse-power and capacity if each pressure were halved? (Sup.pr.) = 14.7 lbs. per square inch. 16. Three-stage Compressor with Best-receiver Pressures Exponential Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and Horse-power in Terms of Dimensions of Cylinders and Clearances. It was found that for the three-stage adiabatic compressor with perfect inter- cooling, the work was a minimum if the first and second receivers had pressures defined as follows, see Eqs. (147) and (148) : (best 1 rec.pr.) = [(sup.pr.)2(del.pr.)]* (188) (best 2 rec.pr.) = [(sup.pr.) (del.pr.)^]* (189) (best Irec p ro ^/deLpry^^ _ ^ (sup.pr.) \sup.pr./ (best 2 recp rj ^ / dehprA* ^^^^ ^ (best 1 rec.pr.) \8up.pr./ ,prO_ /dehpry^ , (192) rec.pr.) \ sup.pr./ *''"(best2 The use of these values in connection with expressions previously given for volumetric efficiency, Eqs. (166), (167) and (168), gives. Volumetric efficiency of first stage =JS?,i = (l+Ci-Ci/2,3.) .... (193) Volumetric efficiency of second stage = -Bt2 = (1+^2— C2fip^*) .... (149) Volumetric efficiency of third stage =Bt3 = (l+C3— ^^s/Zp^O .... (195) 144 ENGINEERING THERMODYNAMICS The work of the three-stage compressor with best-receiver pressures, Ekj- (162), when expressed in terms of displacement and volumetric efficiency becomes TF=432--(sup.pr.)I>iS.i(flp^-l) (196) where (del.pr.) Rp= (sup.pr.) If clearance is known, the value of Evi may be ascertained by Eq. (193) and inserted in Eq. (196). Since this may be so readily done the substitution will not here be made. The mean effective pressure of the compressor referred to the first stage is obtained by dividing the work Eq. (196) by 144 Di: (m.e.p.) referred to first-stage cylinder 14^,=3^-l(sup.pr.)^.i(«,3r-l) (197) The mean effective pressures of the respective stages, due to the equality of work done in the three stages will be as follows: For first stage 8 *— 1 (m.e.p.) = T (sup.pr.) J?i,i(Bp3« — 1) (198) o J. For second stage (m.e.p.)=-^(sup.pr.)~J5:.i(i2pV-l) (199) 5—1 1/2 For third stage (m.e.p.) =-^(sup.pr.)^£„(i?p^-l) (200) But (sup.pr,) ^-^'- = (1 rec.pr.) = [(sup.pr.)2(del.pr.)]*, and also (sup.pr.)^-^l^ = (2 rec.pr.) = [(sup.pr.) (del.pr.)2] J. WORK OF COMPRESSORS 145 Hence For second stage 8 -1 (m.e.p.) = -^[(sup.pr.)2(del.pr.)]*iE?.2(fli,'3r - 1). . (201) 8 "~" X For third stage s •-» (m.e.p.) = --i[(sup.pr.)(deI.pr.)2]»£,3(«p^-l). ■ ■ (202) Conditions to Give Best-Receiver Pressures. All the foregoing discussion of best-receiver pressures for the three-stage compressor can apply only to cases in which all the conditions are fulfilled necessary to the existence of best-receiver pressures. These conditions are expressed by equations (173), (174), (175), (190), (191), and (192), which may be combined as follows: (1) (-2) (3) (4) ^^ (L.P.Cap.2) (L.P.Cap.3) D2E,2 ~DsE,3 (5) ^ ^®^ 1 r- • (203) ^ Dl{l+Ci—ClRj^») ^ 1)2(1 +C2-C2fiy^ D2(l+C2-C2fip-3«) Ds^l+Ci-CsRp^ Parts (1) and (2) of this equation state the requirements in terms of capacities; (3) and (4) in terms of displacements and volumetric efficiencies; (5) and (6) in terms of displacements and clearances. In order, then, that best- receiver pressure may be obtained, there must be a certain relation between the given ratio of compression and dimensions of cylinders and clearances. Since, after the compressor is once built these dimensions are fixed, a given multi-stage comprassor can be made to give best-receiver pressures only when compressing through a given range, i.e., when Rp has* a definite value. If Rp has any other value the receiver pressures are not best, and the methods of the previous Section (15) must be applied. When clearance 'percentages are equal in all three cylinders, ci=C2=C3, and the volumetric efficiencies are all equal then, when best-receiver pressures exist, Eq. (203) becomes, fip* = ^ = ^ = f or equal clearance per cent. . . (204) Evidently this same expression holds if clearances are all zero or negligible. What constitutes negligible clearance is a question requiring careful thought and is dependent upon the ratio of compression and the percentage of error allowable. 146 ENGINEERma THERMODYNAMICS Indicated horse-power of the compressor is found by multiplying the work per cycle, Eq. (196) by the number of cycles per minute, n, and dividing the product by 33,000. IM.V.^j^^^^^nDiEn(Rv^-l) .... (205) From this are obtained the following: H.P. per cubic foot supplied per minute H.P. per cubic foot delivered and cooled per minute I.H.P. _ 9 (deLpr.) , !^^ n(H. P. Cap. cold) s-1 76.4 ^"*' ''' H.P. per cubic foot delivered hot per minute I.H.P. n(H. P. Cap. hot) (See Eq. (156)). (207) =^^^f^^fi,^(/2,V'-l). . . (208) It is useful to notie that these expressions are all independent of clearance, which is to be expected, since the multi-^tage compressor may be regarded as a series of single-stage compressors, and in single stage such an independence was found for work and horse-power per unit of capacity. Example. If the following three-stage compressor be run at best-receiver pressures what will be the horse-power and the best-receiver pressures? Compressor has low- pressure cylinder 32x24 ins. with 5 per cent clearance, is compressing air from atmos- phere to 140 lbs. per square mch absolute, so that s equals 1.4 and it runs at 100 R.P.M. From the formula Eq.'(196) 432« / •-! \ W = — r(8up.pr.)Di£^,i [RjTzT^I) From the formula Eq. (188) (best 1 rec.pr.)=[(sup.pr.)*(del.pr.)]i = (15)«Xl40]i=31.6. WORK OF COMPRESSORS 147 Prom Eq. (189) (best 2 rec.pr.) =[(sup.pr.)(del.pr.)*]* = [15 X (140)*]* =66.5. From Eq, (193) ^n = ( l+c,-Ci«/M =» 1 +.05 -.05 X (^) ^' =96.5; hence, TF-432x^Xl5xll.2x96.5x(9.35'®^-l) =59,000 ft-lbs., or. TTTT. 59,000X200 ,^o '•^•*^- 33;000 ^^ Prob. 1. There is available for running a compressor 176 H.P. How many cu.ft. of free air per minute can be compressed from atmosphere to 150 lbs. per square inch gage by a three-stage adiabatic compressor with best-receiver pressures? Prob. 2. The low-pressure cylinder of a three-stage compressor has a capacity of 4i cu.ft. per stroke. If the stroke of all three cylinders is 18 ins., what must be the diameters of the intermediate and high to insure best-receiver pressures, if clearance be n^lected, and (sup.pr.) be 1 lb. per square inch absolute and (del.pr.) 15 lbs. per square inch absolute, 8 being 1.4. Prob. 3. The above compressor is used as a dry-vacuum pump for use with a sur- face condenser. If 800 cu.ft. of (sup.pr.) gas must be handled per minute what horse- power will be needed to run it? What will be the horse-power per cubic foot of atmos- pheric air? Prob. 4. Will a 15 X22 x34 x24 in. compressor with clearances of 3, 5 and 8 per cent in low, intermediate and high-pressure cylinders respectively be working at best-receiver pressures when (sup.pr.) is 15 lbs. per square inch absolute and (del.pr.) 150 lbs. per square inch absolute? If not, find by trial, the approximate (del.pr.), for which this machine is best, with s equal to 1.4? Prob. 6. For the best (del.pr.) as found above find the horse-power to run the machine at 100 R.P.M. and also the horse-power per cu.ft. of (del. pr.) air cold? Prob. 6. Should this compressor be used for compressing ammon'a would tic best (del. pr.) change, and if so what would be its value? Also what power would Le needed for this case? Prob. 7. Compare the work necessary to compress adiabatically in three stages from 20 lbs. per square inch absolute to 200 lbs. per square inch absolute, the following gases: Air; Oxygen; Gas-engine mixtures, for which s = 1.36. Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio for best-receiver pressure and a pressure ratio of 10? 148 ENGINEERING THERMODYNAMICS Prob. 9. A compressor, the low-pressure cylinder of which la 30 X20 ins. with 5 per cent clearance is compressing air adiabatically from atmosphere to 150 lbs. per square inch gage, at best-receiver pressure. Due to a sudden demand for air the (del. pr.) drops to 100 lbs. per square inch gage. Assuming that the (1 rec. pr.) dropped to 5 lbs. per square inch gage and (2 rec.pr.) dropped to 40 lbs. per square inch gage, how much would the speed rise if the power supplied to machine was not changed? 17. Comparative Economy or Efficiency of Compressors. As the prime duty of compressors of all sorts is to move gas or vapor from a region of low to a region of high pressure, and as this process always requires the expenditure of work, the compressor process which is most economical is the one that accomplishes the desired transference with the least work. In this sense, then, economy of compression means something different than efficiency, as ordi- narily considered. Ordinarily, efficiency is the ratio of the energy at one point in a train of transmission or transformation, to the energy at another point, whereas with compressors, economy of compression is understood to mean the ratio of the work required to compress and deliver a unit of gas, moving it from a low- to a high-pressure place, to the work that would have been required by some other process or hypothesis, referred to as a standard. This economy of compression mast not be confused with efficiencyof compressors as machines, as it is merely a comparison of the work in the compressor cylinder for an actual case or hypothesis to that for some other hyjwthesis taken as a standard. The standard of comparison may be any one of several possible, and unfortunately there is no accepted practice wdth regard to this standard. It will, therefore, be necessary to specify the standard of reference whenever economy of compres- sion is under consideration. The following standards have been used with some propriety and each is as useful, as it supplies the sort of information really desired. First Standard, The work per cubic foot of supply gas necessary to com- press isoihermally (Cycle 1), from the supply pressure to the delivery pressure of the existing compressor and to deliver at the high pressure is less than that of any commercial process of compression, and may be taken as a standard for comparison. Since, however, actual compressors never depart greatly from the adiabatic law, their economy compared with the isothermal standard will always be low, making their performance seem poor, whereas they may be as nearly perfect as is possible, so that it may appear that some other standard would be a l>etter indication of their excellence. Second Standard. The work per cubic foot of gas supplied when compressed adiabatically in a single stage (Cycle 3), if taken as a standard, will indicate a high economy, near unity for single-stage compressors, and an economy above unity for most multi-stage compressors. For the purpose of comparison it will be equally as good as the first standard,and the excess of the economy over unity will be a measure of the saving over single-stage adiabatic compression. Since, however, single-stage adiabatic compression is not the most economical obtainable in practice for many cases, this standard may give an incorrect idea of the perfection of the compressor. WORK OF COMPRESSORS 149 Third Standard, Due to the facts noted above, it may be a better indica- tion of the degree of perfection of the compressor to compare the work per cubic foot of gas supplied with that computed for the standard adiabatic cycle most nearly approaching that of the compressor. This standard is, however, open to the objection that a multi-stage compressor is not referred to the same cycle as a single-stage compressor, and a multi-stage compressor with other than best-receiver pressure is not referred to the same cycle as another operating with best-receiver pressure. This is, therefore, not a desirable standard for comparing compressors of different types with one another, although it doeS show to what extent the compressor approaches the hypothetical best condi- tion for its own type and size. Other standards might be chosen for special reasons, each having a value in proportion as it supplies the information that is sought. It is seen from the discussion of the second standard that its only advantage over the first is in that it affords a measure of the saving or loss as compared with the single-stage adiabatic compressor cycle. If the first standard, that of the isothermal compressor cycle, be adopted for the purpose of comparison, it at once gives a measure of comparison with the isothermal, which is more and more nearly approached as the number of stages is increased, though never quite reached, or as the gas is more effect- ively cooled during compression. It may be regarded as the limiting case of multi-stage compression with perfect intercooling, or the limiting case of con- tinuous cooling. In order to ascertain how nearly the actual compressor approaches the adiabatic cycle most nearly representing its working conditions, the economy of of the various reference cycles heretofore discussed may be tabulated or charted, and the economy of the cycle as compared with that of the actual performance of the compressor will give the required information. The process of com- putation by which this information is obtained will depend upon the nature of information sought. The economy of actual compressor compared with the isothermal may be stated in any of the following ways: Computed work per cubic foot supplied, isothermal , . Indicated work per cu.ft. actual gas supplied to compressor I.H.P. per cubic foot per minute suj>plied, isothermal I.H.P. per cubic foot per minute actual supplied Single stage (m.e.p.) isothermal, pounds per square inc h, no clearance (m.e.p.) actual -5- true volumetric efficiency Multi-stage (m.e.p.) isothermal, no clearance (b) (209) (c) (m.e.p.) reduced to first stage -5- first stage vol. eff. (d) 150 ENGINEERING THERMODYNAMICS In this connection it is useful to note that for the case of the no-clearance cycles, the work per cubic foot of supply is equal to the mean effective pressure (M.E.P.) in pounds per square foot, an4 when divided by 144 gives (m.e.p.) in pounds per square inch. Also, that in cases with clearance, or even actual compressors with negligible clearance, but in which, due to leakage and other causes, the true volumetric efficiency is not equal to unity. Work per cubic foot gas supplied X^»=144(m.e.p.). . . (210) The information that is ordinarilj'^ available to determine the economy of the compressor will be in the form of indicator cards from which the (m.e.p.) for the individual cylinders may be obtained with ordinary accuracy. The volumetric efficiency may be approximated from the indicator cards also, but with certain errors due to leakage and heating, that will be discussed later. If by this or other more accurate means the true volumetric efficiency is foimd, the information required for the use of Eq. (209) (c) or (d) is available. Evaluation of the nimierator may be had by Eq. (31), which is repeated below, or by reference to the curve sheets found at the end of this chapter. (Fig. 50.) Mean effective pressure, in pounds per square inch for the isothermal com- pressor without clearance is given by (m.e.p.) isothermal = (sup.pr.) logs Jip (211) The curve sheet mentioned above also gives the economy of adiabatic cycles of single stage, also two and three stages with best-receiver pressures. The value of 8 will depend upon the substance compressed and its condition. The curve sheet is arranged to give the choice of the proper value of s applying to the specific problem. If it is required to find the economy of an actual compressor referred to the third standard, i.e., that hypothetical adiabatic cycle which most nearly approaches the actual, then Economy by third standard is Econ. actual referred to isothermal Econ. hypothetical referred to isothermal' (212) It is important to notice that for a vapor an isothermal process is not one following the law PXV = constant. What has, in this section, been called an isothermal is correctly so. called only so long as the substance is a gas. Since, however, the pressure-volume analysis is not adequate for the treatment of vapors, and as they will be discussed under the subject of Heat and Work, Chapter VI, it is best to regard this section as referring only to the treatment of gases, or superheated vapors which act very nearly as gases. However, it must be understood that whenever the curve follows the law PXF= constant, the isothermal equations for work apply, even if the substance be a vajwr and the process is not isothermal. WORK OF COMPRESSORS 151 18. Conditions of Mazimiim Work of Compressors. Certain types of com- pressors are intended to operate with a delivery pressure approximately con- stanty but may have a varying supply pressure. Such a case is foimd in pumps or compressors intended to create or maintain a vacuum and in pmnping natural gas from wells to pipe lines. The former deliver to atmosphere, thus having a substantially constant delivery pressure. The supply pressiu'e, however, is variable, depending upon the vacuiun maintained. In order that such a compressor may have supplied to it a sufficient amount of power to keep it running under all conditions, it is desirable to learn in what way this power required will vary, and if it reaches a maximum what is its value, and under what conditions. Examine first the expression for work of a single-stage adiabatic compressor with clearance. The work per cycle will vary directly as the mean effective pressure. Eq. (69.) (:«..p.)..4i(»P.P-.)[l+c-c(^)-][(^f-l]. (21S, This will have a maximum value when d(m.e.p.) d(sup.pr.) =0, or when f^5^y'-'-:f^ri+c-c?^(^?i:Pi^yi==o. . . . (214) Vsup.pr./ 1+c L « \sup.pr./ J ^ ^ Solving this for the value of supply pressure will give that supply pressure at which the work will be a maximum, in terms of a given delivery pressure, clearance and the exponent a. The assumption most conunonly used is that clearance is zero. If this is trae^or the assmnption permissible, the above equation becomes simplified^ • m^-'" (2'« The value of 8 for air, for instance, is 1.406, and hence the ratio of compression for maximum work for the hypothetical air compressor is (1.406)^*^=3.26 ; (216) It may be noted that when s = 1 in the above expression, the value of the ratio of compression become indeterminate. To find the supply pressure for maximum work in this case, take the expression for mean effective pressure for the isothermal compressor (s = 1), Eq. (43), (m.e.p.).(,up.pr.)[l+.-.(,^)]:„^^). ... (217) 152 ENGINEERING THERMODYNAMICS DiflFerentiate with respect to (sup.pr.) and place the differential coefficient equal to zero. This process results in the expression log. (^'^■^'•- ^ \sup.pr. del.pr.\ c /del.prA_j^^j (218) / 1+c \8up.pr./ When c=0, this becomes, ,og.fd?l:Pr.\ = l or (^^'P":- ) = 2.72 (219) ^ \sup.pr./ \sup.pr./ The expressions Eqs. (215) and (219) given are easily solved, but Eqs. (214) and (218) are not, and to facilitate computations requiring their solution the results of the computation are given graphically on the chart. Fig. 48, at the end of this chapter. The mean efifective pressure for a compressor operating imder maximum work conditions may be found by substituting the proper ratio of compression, found as above, in Eq. (213) or (217). In Fig. 48 are found also the results of this computation in the form of curves. Note in these curves that the mean efifective pressure is expressed as a decimal fraction of the delivery pressure. The discussion so far applies to only single-stage compressors. The problem of maxiinum work for muUi'Stage compressors is somewhat different, and its solution is not so frequently required. Moreover, if the assumption of perfect intercooling is made, the results are not of great value, as a still greater amount of power might be required, due to the failure for a period of time of the supply of cooling water. Consider this case first. If intercooling be disconUnued in a multi-stage compressor, the volume entering the second stage will equal that delivered from the first, and similarly for the third and second stages. The entire work done in all stages will be the same as if it had all been done in a single stage. It might be questioned as to whether this would hold, when the ratio of compression is much less than designed. The first stage will compress until the volume has become as small as the low-pressure capacity of the second stage. If the delivery pressure is reached before this volume is reached, there is no work left to be done in the second or any subsequent stages, and, due to the pressure of the gas, the valves, if automatic, will be lifted in the second and higher stages, and the gas ynW be blown through, with only friction work. It appears then that under the condition of no intercooling the multi-stage compressor acts the sanve as a single stage, and the conditions of maximum work will be the same. If intercooling is maintained perfect there will still be a range of pressures on which all the work of compression is done in the first stage, merely blowing the discharge through receivers, valves and cylinders in the upper stages. If this range is such that this continues beyond a ratio of pressures, which gives a maximum (m.e.p.) for the single stage, then the maximum will have been reached while the compressor is operating single stage, and the single-stage formulae and curves may be applied to this case also. WORK OF COMPRESSORS 153 That this condition frequently exists with multi-stage compressors of ordinary design is shown by the fact that the ratio of compression in each stage is seldom less than 3, and more frequently 3.5, 4 or even more. The ratio of compression giving maximum work for single stage, has values from 2.5 to 3.26, dependent on clearance and the value of 8 for the gas compressed, and is, therefore, less for the majority of cases. If a curve be drawn, Fig. 36, with ratio of compression as abscissas and (m.e.p. -hdel.pr.) as ordinates, so long as the action is single stage, a smooth curve will result, but when the ratio of compression is reached above which the second cylinder begins to act, the curve changes direction suddenly, falling as the ratio Values of Rp. Fig. 36. — Curve of Relation between Mean Effective Pressure per Pound of Delivery Pressure in Terms of Pressure Ratio for Air, showing Maximum Value. of compression increases. Hence, if the ratio of cylinders is such that the single- stage maximum is not reached before the second stage begins to operate^ the highest point of the curve, or maximum work for a given delivery pressure will occur when the ratio of supply and delivery pressures is such as to make first-receiver jrressure equal to delivery pressure, 19. Actual Compressor Characteristics. Air or gas compressors are very commonly made double acting, so that for a single cylinder, two cycles will be performed during one revolution, one in each end of the cylinder. If a rod extends through one of these spaces and not through the other, the displace- ment of that end of the cylinder will be less than the other by a volume equal to the area of rod multiplied by the stroke. To avoid mechanical shock at the end of either stroke, it is necessary to leave some space between the 154 ENGINEERING THERMODYNAMICS piston and cylinder head. Passages must also be provided, communicating with inlet and discharge valves. The total volume remaining in this spact^ and in the passages when the piston is at the nearer end of its stroke constitute the clearance. The amount of this clearance volume varies from .5 or .6 of one per cent in some very large compressors to as much as 4 or 5 per cent of the volume of displacement in good small cylinders. In order to study the performance of an actual compressor and to compare it with the hypothetical cycle, it is necessary to obtain an indicator card, and knowing the clearance and barometric pressure to convert the indicator card into a pressure volume diagram, by methods explained in Chapter I. Fig. 37, is such a diagram for a single-stage compressor. In the pipe leading to the Press. H.P.Cap Hot (Apparent) !G iLD (DeL Pr.) -L^B€ap-H^l)ethetieal -LrPrOap-^-pparent Displaoemcnt- (Sup. Pr.) •Vol. Fig. 37. — Compressor Indicator Card Illustrating Departure from Reference Cycle. intake valve the pressure is determined and a horizontal line AB is drawn on the diagram at a height to represent the supply pressure. Similarly, discharge pressure is determined and drawn on the diagram, KE, Consider the four phases of this diagram in succession. 1. Intake Line. At a point somewhat below A the intake vaJve opens, say at the point F. This remains open till a point H is reached at or near the end of the stroke. The line connecting these two points indicates variations? of pressures and volumes throughout the supply stroke. In general this line will lie below the supply-pressure line AB due to first, the pressure necessary to lift the inlet valve from its seat against its spring and inertia, if automatic, and support it, and second, friction in the passages leading to the cylinder from the point where the supply pressure was measured. While the former is nearly constant the latter varies, depending upon the velocity of gases in the passages. The piston attains its highest velocity near the middle of the stroke^ thus WORK OF COMPRESSORS 155 causing the intake line to drop below the supply pressure more at this part of stroke. These considerations do not, however, account completely for the form of the intake line. Frequently the first portion of the line lies lower than the last portion, even at points where the piston velocities are equal. This is more prominent on a compressor having a long supply pipe, and is due to the forces required to accelerate the aor in the supply pipe while piston velocity is increasing, and to retard it while piston velocity is decreasing. In com- pressors where the inlet valve is mechanically operated and the supply pipe long, it is possible to obtain a pressure at the end of the intake, line H, even in excess of the supply pressure. The effect of this upon volumetric efficiency will be noted later. The apparent fluctuations in pressure during the first part of the intake line may be attributed, first, to inertia vibrations of the indicator arm, in which case the fluctations may not indicate real variations of pressure; second, the indicator card may show true variations of pressure due to inertia of the gases in the supply pipe, since a moment before the valve opened at F the gases were stationary in the supply pipe. When F is reached the piston is already in motion and a very considerable velocity is demanded in the sugply pipe to supply the demaQd. This sudden acceleration can be caused only by a difference in pressure, which is seen to exist below and to the right of F on the diagram. The suddenness of this acceleration may start a surging action which will cause rise and fall of pressure to a decreasing extent immediately after. A third cause is possible, that is, a vibration of the inlet valve due to its sudden opening when it is of the common form, mechanical valves change the conditions. It is closed by weight or a spring and opened by the pressure difference. Between these forces the valve disk may vibrate, so affecting the pressure. 2. Compression Line, From the time the inlet valve closes at the point H until the discharge valve opens at the point (?, the gases within the cylinder are being compressed. The compression is very nearly adiabatic in ordinary practice, but due to the exchange of heat between the cylinder walls, at first from walls and later from gas to walls, which are cooled by water jacket to prevent the metal from overheating, there is a slight departure from the adia- batic law almost too small to measure. A second factor which influences the form of this curve to a greater extent is leakage. This may occur around the piston, permitting gas to escape from one end of the cylinder to the other. During the compression process there is first an excess of pressure in the other end of the cylinder due to reexpansion, tending to increase pressures on the first part of compression. Later, the pressure rises and the pressure on the other side of the piston falls to supply pressure. During this period leakage past piston tends to decrease successive pressures or lowers the compression line. Leakage also occurs through cither discharge or inlet valves. The former will raise the compression line, while excessive leakage of the inlet valve will lower it. It is then evident that unless the nature of the leakage is known, it is impossible to predict the way in which it will change this line. It is, however. 156 ENGINEERING THERMODYNAMICS more frequently the case that the piston and inlet leakage are large as compared with the discharge valves, in which case the actual compression line has a tendency to fall lower than the adiabatic as the volumes are decreased. Com- pression lines lowered by leakage are often mistaken for proofs of effective cooling, and cases have been known where isothermal compression of air was claimed on what proved to be evidence only of bad leakage. 3. Delivery Line. After the delivery pressure of the compressor has been exceeded sufficiently, the discharge valve is opened and the gas is delivered to the discharge pipe or receiver till the end of the stroke is reached and at the point J the valve closes. The same group of factors influence the form of this line as act upon the intake line; spring resistance of discharge valve; friction in discharge passages varying with piston velocitj'-; inertia of gases in delivery pipe; sudden acceleration of gases in delivery pipe when discharge valve opens, and inertia of indicator arm, but in addition a strong tendency for the valve to chatter or jump open and shut alternately. I^eakage also occurs through tntake valve and past piston during this process, with the result that less gas passes through the discharge valve than is shown on the indicator card. 4. Reexpansion Line. From the time the discharge valve closes, at J, till the intake valve opens, at F, the gas which remained in the clearance space after delivery expands, due to the advancing of the piston, till the pressure has fallen to such an amount that the intake valve will open. The same factors influence this line as the compression line. Heat is exchanged with the jacketed cylinder walls, at first cooling and later heating the gas as the pressure falls. This, for any given volume, changes the pressure. I^eakage occurs inward through the discharge vaJve and outward through intake valve and past piston. If these last two are in excess, the pressures will fall more rapidly than if the expansion were that of a constant quantity of gas. Work due to gas friction and inertia, it should be noted, is fully represented on the indicator card, and may be regarded as being equal to that extra area below the supply-pressure line and above the delivery-pressure line. In the combined card of a two-stage compressor there would be an overlapping of the diagrams due to this frictional loss. Low-pressure Capacity, Referring to the adiabatic compression and expan- sion lines, CD and KL^ Fig. 37, it is seen that the low-pressure capcun^y of the hypothetical cycle is the volume, LC. The apparent low-pressure capacity of the actual compressor, measured at the supply pressure is AB, This is not, however, the true volume of gas at supply pressure and temperature that is taken in, compressed, and finally delivered per cycle. First, the valves, passages and walls are not at the same temperature as the entering gas, due to the heat left from the compression of the previous charge. This causes the temperature of the gas within the cylin- der to be something higher than the supply gas outside. This causes it to be less dense, and hence an equivalent weight of gas at supply temperature and pressure would occupy a volume somewhat less than AB, Second, the gas which occupies the volume AB has not all entered the cylinder through the intake WORK OF COMPRESSORS 157 valve. After reexpansion is completed the intake valve opens and gas enters the end of the cylinder under consideration. At the same time compression is taking place in the other end, and later deliver3\ During these processes whatever gas leaks past the piston tends to fill the end of the cylinder in which intake Ls going on. Leakage past the discharge valve also tends to fill the cylin- der with leakage gas. Both of these tendencies decrease the quantity of gas entering through this intake valve, and its true amount when reduced to external supply pressure and temperature is, therefore, less than the volume AB. The triAe lovypressure capacity of the compressor is the true volume of gas under external supply conditions that enters the cylinder for each cycle. This cannot be determined from the indicator card except by making certain assump- tions which involve some error at best. It can, however, be ascertained by means of additional apparatus, such as meters or calibrated nozzles or receivers, by means of which the true amount of gas compressed per unit of time is made known. This reduced to the volume per cycle under supply pressure and tem- perature will give the true low-pressure capacity. Volumetric efficiency is defined as being the ratio of low-pressure capacity to displacement. On the diagram. Fig. 37, the displacement is represented to the voliune scale by the horizontal distance between verticals through the extreme ends of the diagram, K and H. Since there are three ways in which the low-pressure capacity may be approximated or determined, there is a corresponding number of expressions for volumetric efficiency. 1. The volumetric efficiency of the hypothetical cycle is JT /^i^^^f Ko+inon - (hypothetical L. P. Cap.) . . ^.(hypothetical) — -(dis^ace^nt)" ~"' " ' ' ' ^^^^ and this is evaluated and used in computations in the foregoing sections of this chapter. 2. The apparent volumetric efficiency is T? / x\ (apparent L. P. Cap.) ,^oi \ ^.(apparent) = ^^^ ,. -, - t/ -, (221) ^^ (displacement) and would be very nearly equal to the true volumetric efficiency were it not for leakage valve resistance and heating during suction, but due to this may be very different from it. 3. The trv/e volumetric efficiency is g.(true)= (y«L.P.Cap O ^222) (displacement) In problems of design or prediction it is necessary either to find dimensions, speeds and power necessary to give certain actual results, or with given dimen- sions and speeds to ascertain the probable power and capacity or other 158 ENGINEERING THERMODYNAMICS characteristics of actual performance. Since it is impossible to obtain actual performance identical with the hypothetical, and since the former cannot be computed, the most satisfactory method of estimate is to perform the computa- tions on the hypothetical cycle, as is explained in previous sections of this chap- ter, and then to apply to these results factors which have been foimd by comparing actual with hypothetical performance on existing machines as nearly like that under discussion as can be obtained. This necessitates access to data on tests performed on compressors in which not only indicator cards are taken and speed recorded, but also some reliable measurement of gas compressed. The following factors or ratios will be found of much use, and should be evaluated whenever such data is to be had: _ S»(true) (true L. P. Cap.) .^^m ^ E9 (hypothetical) "" (hypothetical L. P. Cap.)* = -^i>(true) _ (true L . P. Ca p.) .^ . Et(apparent) (apparent L. P. Cap.) — true I.H.P. _ true m.e.p. ,^^. ^ hypothetical I.H.P. hypothetical m.e.p. Then true work per cu.ft. gas, supplied hypothetical work per cu.ft. gas, supplied I.H.P. true _ true I.H.P. per cu.ft. g as supplied (L. P. Cap.) hypothetical I.H.P. per cu.ft. supplied , +Vi f 1 I-H.P. (L. P. Cap.) true m.e.p. . true L. P. Cap. 63 (226) hypothetical m.e.p. ' hypothetical L. P. Cap. ci This ratio can be used to convert from hypothetical work per cubic foot gas supplied to probable true work per cubic foot. Multi-stage Compressors are subject in each stage to all of the characteristics described for single stage to a greater or less extent. Valve resistance, friction and inertia affect the intake and discharge lines; heat transfer and leakage influence the form of compression and reexpansion lines, and the true capacity of the cylinder is made different from the apparent due to leakage, pressure and temperatures changes. In addition to these points it is useful to note one special way in which the multi-stage compressor differs from the single stage. The discharge of the first stage is not delivered to a reservoir in which the pressure is constant, but a receiver of limited capacity. The average rate at which gas is delivered to the receiver must equal the average rate at which it passes to the next cylinder. The momentary rate of supply and removal is not constantly the same, however. WORK OF COMPRESSORS 159 thjB causes a rise or fall of pressure. It is evident that this pressure fluctuation is greatest for a small receiver. Very small receivers are not, however, used 3n gas compressors due to the necessity of cooling the gas as it passes from one stage to the next. To accomplish this a large amount of cooling surface must be exposed, requiring a large chamber in which it can be done. Thus, it is seen that the hypothetical cycles assumed for multi-stage cpmpressors do not truly represent the actual cycle, but the difference can never be very great, due to the large size of receiver which must always be used. Another way in which the performance of this multi-stage compressor commonly differs from assumptions made in the foregoing discussions is in Fio. 38. — ^Effect of Loss of Intercooling in Two-stage Compressors on Receiver-pressure and Work Distribution in the Two Cylinders. r^ard to intercooling. It seldom occurs that the gases enter all stages at the same temperature. In the several stages the temperature of the gases will depend on the amount of compression, on the cooling surface and on the amount and temperature of cooling water. The effect of variations in tem- perature upon the work and receiver pressures will be taken up later. It may be noted now, however, that if all cooling water is shut off, the gas passes from one cylinder to the next without cooling j there is no decrease in volmne in the receiver. For simplicity take the case of zero clearance, two-stage (Fig. 38). let ABCDEF be the cycle for perfect intercooling. AB and KD are the low- pressure capacities of the first and second stages respectively. If now, inter- cooling ceases, the gas will no longer change volume in the receiver. The receiver gas, in order to be made sufficiently dense to occupy the same 160 ENGINEERING THERMODYNAMICS volume {KD) as it did before, must be subjected to a greater pressure in the first stage. The new receivei^ line will be K'D'. The wo rk of the first stage will therefore be ABD'K'', of the second stage K'D'GF, and the total work in the new condition is greater than when intercoooling was perfect by an amount represented by the area DCGE. In the case where clearance is considered, the effect is the same, except that the increasing receiver pressure, increasing the ratio of compression of the first stage, causes the volumetric efficiency of the first stage to become less, and hence lessens the capacity of the compressor. The efifect on work per unit of capacity is the same as without clearance. The question as to how many stages should be used for a ^ven compressor is dependent upon the ratio of compression largely, and so is due, first, to con- siderations of economy, which can be imderstood from the foregoing sections; second, for mechanical reasons, to avoid high pressures in large cylinders; third, for thermal reasons, to avoid such high temperatures that the lubrication of the cylinders would be made difficult, or other dangers, such as explosions, involved. Practice varies very widely as to the limiting pressures for single, two, three or four-stage compressors. Air compressors of a single stage are com- monly used for ratios of compression as high as 6 or 7 (75 to 90 lbs. gage). For ratios greater than these, two-stage compressors are used, especially for larger sizes, up to ratios of 34 to 51 (500 to 750 lbs. gage). Some three-stage com- pressors are used for ratios as low as 11 or 14 (150 or 200 lbs. gage), although installations of this nature are rare, and are warranted only when power is costly and the installation permanent and continuously used to warrant the high investment cost. As a minimum ratio for three stages, 11 (160 lbs. gage) j is used for large units, while a few small units compress as high as 135 or even | 170 atmospheres (2000 or 2500 lbs. gage). A notable use for the four-stage compressor is for charging the air flask of automobile torpedoes used by the various navies, which use pressures from 1600 to 3000 lbs. per square inch (110 to 200 atmospheres). These require special design of valves, cylinders and packings to withstand the extremely high pressures, small clearances, and special precautions against leakage, due to the great loss of voliunetric efficiency and economy that would otherwise result. 20. Work at Partial Capacity in Compressors of Variable Capacity. I is seldom that a gas compressor is run continuously at its full capacity. I the duty of the compressor is to charge storage tanks, it may be made to run at its full capacity until the process is completed and then may be stopped entirely, by hand. Even where the compressed gas is being used continuously it is common practice to have a storage reservoir into which the compressor may deliver. This enables the compressor to deliver a little faster or slower thaa the demand for a short period without a great fluctuation pressure in tha reservoir. For many purposes hand regulation is not sufficient or is too expensive, hence the demand for automatic systems of capacity regulation These systems may be classified in a general way in accordance with the method WORK OF COMPRESSORS 161 of driving. Some methods of power application permit of speed variations while others require constant speed. The former provides in itself a means of regulat- ing capacity within certain limits, while, if the compressor must run at constant speed, some additional means of gas capacity control must be provided. Ck)mpressors driven by an independent steam engine, or steam cylinders constituting part of the same machine may be made to nm at any speed required within a very wide range and still kept low enough for safety. If driven by gear, belt, rope, chain or direct drive from a source of power whose speed is constant, the speed of the compressor cannot be varied. Electric motor, gas-engine, oil-engine or water-power drives are subject to only limited speed alteration and may, therefore, be placed in the constant speed class. Regulation of Capacity by Means of Speed Change. If the speed of the com- pressor is decreased below normal: 1. Displacement of piston is decreased in proportion to the speed. 2. Mean effective pressure, as to hypothetical considerations, is the same, but due to the decrease of velocities in gas passages, the frictional fall of pres- siu-e during inlet and delivery is not so great, and hence the mean effective pres- sure is not quite so great. If the compressor is multi-stage, since a smaller quantity of gas is passing through the intercooler, it is probable that the inter- cooling is more nearly perfect, thus decreasing the mean effective pressures in the succeeding stages. 3. The volumetric eflSciency is changed, due first to the fact that leakage is about the same in total amount per minute as at full speed, but the total quantity of gas being less, leakage is a larger percentage of the total; second, the inertia of gases in the supply pipe, as well as their friction, has been decreased. The former tends to decrease vulmetric efficiency, while the latter may tend to increase or decrease it. It may be expected that the true voliunetric eflSciency will be somewhat greater at fractional speed than at full speed. For any compressor there is a speed of maximum economy above and below which the economy is less, though it may be that this most economical speed is greater than any speed of actual operation. It is not desirable at this point to discuss the effect of speed variation upon the economy of the engine or other motor supplying the power. The reasoning above applies to the term economy as applied to the compression effect obtained per unit of power applied in the compression cylinder. It might be noted here, however, that the decrease of speed has little effect upon the mechanical efficiency of the compressor as a machine, since frictional resistance between solid parts remains nearly constant, and, therefore, power expended in friction will vary as the speed, as does approximately also the power to drive the compressor. The ratio of frictional power to total may then be expected to remain nearly constant. Regulation of Capacity at Constant Speed may be accomplished in a nimiber of ways: 1. Intermittent running; 2. Throttling the supply to compressor; 162 ENGINEEEING THERMODYNAMICS 3. Periodically holding open or shut the intake valve; 4. Closing intake valve before end of intake stroke, or holding intake valve open until compression stroke has been partially completed; 5. Large clearance; 6. Variable clearance, The first necessitates some means for stopping and starting the compressor, which is simple with electric drive, and may be accomplished in other cases by means of a detaching clutch or other mechanical device. The pressure in the reservoir is made to control this stopping and starting device by means of a regulator. This arrangement is made to keep the pressure in the reservoir between certain iBxed limits, but does not maintain a constant pressure. The economy of compression in this case is evidently the same as at full speed continuous running, provided there is no loss in the driving system due to starting and stopping, which may not be the case. This method of regulation is used mainly for small compressors in which inertia is not great, such as supply the air brakes on trolley cars. The sudden change of load on the driving machinery would be too great if large compressors were arranged in this way. If the compressor whose capacity is regulated by intermittent running is mvlMi-stagej the constant supply of water to the intercoolers while the compres- sor is stopped will lower the temperature of the cooling surface, causing naore nearly perfect intercooling when the compressor is started. Leakage, on the other hand, will permit the loss of pressure to a greater or less extent in the receivers while the compressor is stationary, which must be replaced after starting before effective delivery is obtained. Throttling the gas supply to the compressor has certain effects that may be studied by referring to Fig. 39, which represents the hypothetical cycles most nearly approaching this case. In order to reduce from the full-load low-pres- sure capacity, ilB, to a smaller capacity, AE^ the supply pressure is decreased by throttling to the pressure of B\ such that B' and E lie on the same adiabatir. The work area A'B'EA is entirely used up in overcoming the throttle resistance and is useless friction, so that economy is Seriously reduced by this method of regulation. Such compressors may use almost as much power at partial as at full capacity. It is easily seen that this method of regulation would be undesirable, its only advantage being simplicity. The effect of throttling upon a multi-stage compressor may be illustrated as in Fig. 40, by considering the two-stage compressor cycle without clearance, ABCDEF. The ratio of compression of the first cylinder is determined with perfect intercooling by the ratio of displacements Pc — Pb(jr\ - When the supply pressure is throttled down to Pc', the new receiver pressure will be Pc =Pb \fr) > a pressure much lower than Pc. Hence the receiver pressure is decreased, less work done in the first stage, and far more than half the work WORK OF COMPRESSORS 163 of compression done in the second stage. If best-receiver pressiire existed at normal capacity, it does not exist in the throttled condition. The intake valve may be held wide open or completely closed during one or more revolutions, thereby avoiding the delivery of any gas during that period. If the intake valve is held wide open, the indicator card would be as shown in Fig. 41A in full lines, ABCD, the dotted lines showing the cycle performed when D c c * \ \ \ \ \ \ \ 1 \ \ \ \ % • \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ N E B K 1 N 1 \ \ 1 > \ ■----, 1 N N A' B' \^ ft -L.P.Cap Throttled *> T V f^awx a4 TNilI ▼ -^^^A ... ->- 1^ 1 14. t^ .Cap at Villi XJUliU v^ J J ^ Fig. 39. — Effect of Throttling the Suction of One-stjige Compressors, on Capacity and Econonij'. normal operation is permitted. With the inlet valve open in this way there is a loss of power due to friction of the gas in passage during both strokes, measured by the area within the loop. Closing the inlet valve and holding it shut will give an indicator card of the form EFG, Fig. 41 S, which will be a single line retraced in both directions except for probable leakage effects. If leakage is small, there will be but little 164 ENGINEERING THERMODYNAMICS area enclosed between the lines. At a high speed this might be expected to incur less lost power than the former plan. Certain types of compressors are made with an intake valve controlled by a drop cut-ofiF, much like the steam valve of the Corliss engine. The effect of this is to cut oflF the supply of gas before the end of the stroke, after which time the gas must expand hjrpothetically according to the adiabatic law. The return stroke causes it to compress along the same line continued up to the delivery pressure, as indicated by the line FEGy Fig. 41C. There is little work FiQ. 40. — ^Effect of Throttling Multi-stage Compressors on Receiver-pressure, Work Distri- bution, Capacity and Economy. lost in the process, none, if the line is superimposed as in the figure, and hence the process is the same as if only the cycle AEGD were performed. The same quantity of gas might have been entrapped in the cylinder by holding the intake valve open until the end of the stroke and on the return till the point E, Fig. 41D, was reached, then closing it. The same compression line EG will be produced. The line AB will not coincide with BE, due to friction of the gas in passages, and hence will enclose between them a small area representing lost work, which may be no larger than that lost in the process EFE. WORK OF COMPRESSORS 165 If such an automatic cut-ofiF were applied only to the first stage of a mvMi- singe compressor, the eflFect would be to lower receiver pressures as in the throt- tling process. To avoid this, the best practice is to have a similar cut-oflF to act on the supply to all of the stages. If this is properly adjusted, the receiver pressures can be maintained the same as at full load. An additional advantage of this system is that even if the compressor is to be used for a delivery pres- sure for which it was not originally designed, the relative cut-oflFs may be so adjusted as to give and maintain best-receiver pressiu'e. , 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 Cut-ofif; Z). Delayed Suction Closure. Since the low-pressure capacity per cycle of a compressor involves clearance and ratio of compression as two of its variables, it is possible to change capacity by changing either the clearance or the ratio of compression. (L. P. Cap.)=Z)£f,=D(l+c-cJBp« ). (227) Assuming that clearance is a fixed amount and not zero, it is evident that an increase in the ratio of compression decreases the capacity, and when it has 166 ENGINEERING THERMODYNAMICS reached a certain quantity will make the capacity zero. If the clearance is large, making the coefficient of fip* large in the equation the efifect of a change in that factor is increased. Fig. 42 indicates the hjrpothetical performance of a compressor with large clearance. When the pressure of delivery is low (say Pe) the capacity is large, AB. The cycle is then ABCD. An increase of the delivery pressure to P/ changes the cycle to A' BCD' and the low pres- sure capacity is A'B, If the compressor is delivering to a receiver from which no gas is being drawn, the delivery pressure will continue to rise and the capacity m • 0' V \ \ D \ \ \ \ \c ; — \" \ \ \ \^ \ \ -.- ^^ A A' B • ^ L 1^ ■% ^ -rx «^ ^"C XJ ^ ±j y Fia. 42. — ^Variation of Compressor Capacity with Rise of Delivery Pressure, Fixed Clearance, Pressure for Zero Delivery. to decrease till the capacity approaches zero as the delivery pressure approaches the pressure Pe as a limit. (limiting del.pr.)=(sup.pr.)( ?j (228) When the limiting condition has been reached and the capacity has become zero, the compression and reexpansion lines coincide and enclose zero area between them; hence, the njcan effective pressure and the indicated horse- power are zero, for the hypothetical case. Leakage will prevent a perfect coincidence of the lines and cause some power to be required in addition to that of friction. Such a simple method of regulation as this is used for some small com- pressors driven constantly from some source of power used primarily for other purposes. When it is not necessary to have a constant delivery pressure, WORK OP COMPRESSORS 167 but only to keep it between certain limits, this may be made use of, especially if the limits of pressure are quite wide. The expression for low-pressure capacity Eq. (227), suggests the possi- bility of decreasing capacity by the increase of clearance. The effect of this is shown in Fig. 43. The original compression cycle (full capacity) is shown by ABCD, with a clearance volume of cD, so that the axis of zero volume is OP. Increasing the clearance to c'jD causes a smaller volume CD to be delivered and due to the more sloping re-expansion DA\ a smaller volume of gas is taken in, A'B, It has been shown in previous sections that clearance has no effect upon the economy of a compressor so far as hypothetical considerations are regarded. In practice it is found that a slight loss of economy is suffered at light load. p D c C \ \ 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 —C-D "U-i: " .r. VAip -© — IlKl » Fig. 43. — ^Variation of Compressor Capadty by Changing Clearance. as might be expected, due to greater leakage per unit of capacity. The addi- tional clearance is provided in the form of two or more chambers connected to the clearance space of the compressor by a passage in which is a valve automatically controlled by the receiver pressure. In the muUi'Stage compressor, decreasing the capacity of the first stage by an increase of its clearance would evidently permit a decrease of receiver pres- sures unless the capacity of each of the various stages is decreased in the same proportion. Eq. (132) gives the condition which must be fulfilled to give best receiver pressure for a two-stage compressor. RJ = M 1 + Cl — Cl/?|,2« [ I>2 1+C2 — C2/2p2« Is 168 ENGINEERING THERMODYNAMICS Since Di, D29 and Rp remain fixed, for any chosen value of clearance of the first stage, a, the clearance of the second stage, C2, to give best-receiver pressure can be found, [i-ci(jep2l-i)]Di C2 = i i (229) (Rj^*-l)Rp2sD2 For every value of first-stage clearance there is a corresponding clearance of second ^stage that will give best-receiver pressure, found by this equation. Sim- ilar reasoning can be applied to three- or four-stage compressors. 21. Graphic Solution of Compressor Problems. In order to obviate the necessity of working out the formulas given in this chapter each time a prob- lem is to be solved, several of them have been worked out for one or more cases and results arranged to give a series of answers graphically. By the use of the charts made up of these curves many problems may be solved directly and in many others certain steps may be shortened. A description of each chart, its derivation and use is given in subsequent paragraphs. Chart, Fig. 44. This chart gives the work required to compress and deliver a cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. of (sup.pr.) air per minute if the ratio of pressure (del.pr.)-f- (sup.pr.), the value of s and the (sup.pr.) are known, and compression occurs in one stage. The work or H.P. for any number of cubic feet is directly proportional to number of feet. The curves are dependent upon the formulas, Eq. (31), for the case when 8=1, and Eq. (51) for the case when s is not equal to 1. These formulas are: Eq. (31), W per cu.ft. = 144 (sup.pr) log« Rp] " (51), W per cu.ft. = 144--^— (sup.pr.) (fip"^-lV These equations are difficult to solve if an attempt is made to get a relation between the work and ratio of pressures. This relation may, however, be worked out for a number of values of pressure ratios and results plotted to form a curve by which the relation may be had for any other ratio within limits. This has been done in this figure in the following manner: On a horizontal base various values of Rp are laid off, starting with the value 2 at the origin. The values for work were then found for a number of values of Rp with a constant value of (sup.pr.) and s. A vertical work scale was then laid off from origin of Rp and a curve drawn through the points found by the intersection of horizontal lines through values of work, with vertical lines through corresponding values of Rp, The process was then repeated for other values of s and curves similar to the first, drawn for the other values of 8. From the construction so far completed it is possible to find the work per cubic foot for any pressure ratio and any value of s for one (sup.pr.) by pro- WORK OF COMPRESSORS atnpwqY -01 bg wd "gqi (■y -iiofl) 170 ENGINEERING THERMODYNAMICS jecting up from the proper value of Rp to the curve of value of s and then hor- izontally to the scale of work. It will be noted from these formulas, however, that the work may be laid off on the horizontal base and a group of lines drawn so that the slope of the line equals ratio of work for any supply pressure to that for the (sup.pr.) originally used. For convenience, in order that the group of 8 curves and the latter group may be as distinct as possible, the origin of the latter group is taken at the opposite end of the base line. If from the point for work originally found, a projection is made horizontally to the proper (sup.pr.) curve, the value for work with this (sup.pr.) will be found directh- below. It will be noted that from point of intersection of the vertical from the Rp value with the s curve, it is only necessary to project horizontally far enough to intersect the desired (sup.pr.) curve, and since no information of value will be found by continuing to the work scale for the original (sup.pr.) this is omitted from the diagram. In brief, then, the use of this chart consists in projecting upward from the proper value of Rp to the proper s curve, then passing horizontally to the value of (sup.pr.) and finally downward to the work scale. As an example of use of the curve Ex. 2 of Section (8) may be solved directly. This is to find the work to compress 1000 cu.ft. of free air from 1 to 8J atmospheres adiabatically. On the curve project upward from iiJp = 8.5 to curve of s = 1.406, then over to 14.7 (sup.pr.) curve and down to read work = 6,300,000 as found, for exam- ple, by use of formulas in Section (8). Chartj Fig. 45. This gives the work required to compress and deliver a cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. of (sup.pr.) air per minute if the ratio of pressures, the value of s and (sup.pr.) are known and if compression occurs in two stages with best-receiver pressure and perfect intercooling. The work or H.P. for any other number of cubic feet may be found by multiplying work per foot by the -number of feet. The method of arriving at this chart was exactly the same as that for one stage. As an example of the use of the chart-. Example 2 of Section (9) may be solved directly. This problem calls for the work to compress 5 cu.ft. of free air from 1 to 8| atmospheres adiabatically in two stages. Project upward from ftp = 8.5 to curve 5 = 1.406, then over to 14.7 curve and down to read 5320 ft.-lbs. per cubic foot, which is same as found from the formula in Section (9). Chartj Fig. 46. This chart gives the work necessary to compre&s and deliver a cubic foot of (sup.pr.) air or horse-power to compress and deliver 1000 cu.ft. of (sup.pr.) air per minute, if the ratio of pressures, the value of s, and the (sup.pr.) are known and if the compression occurs in three stages with best- receiver pressures and perfect intercooling. The work or horse-power for any other number of cubic feet may be found by multiplying the work for one foot by the number of feet. As an example of use of this chart, Example 2 of Section (13) may be solved directly byit. This calls for the horse-power to compress 100 cu.ft. free air per minute adiabatically in three stages from 15 lbs. per square inch abs. to 90 lbs. per square inch gage. From Rp = 7f project to curve of s = 1.4 then WORK OF COMPRESSORS 171 «\ -i I O & o o V 8 o W S I c o H5 O k ^ •tqv 'ni 'bS 8S CO CO CO A w '^ 6 ENGINEERING THEEIMODYNAMICS 111 'sqv 'QI ''>I>8 'isil J|V <''i<I'<Ii>9) -IS (,2.5 ■ -c - l» Its SI i| — i i 's i i ■ WORK OF COMPRESSOES 173 )ver to (sup.pr.) = 15 and down, and the horse-power will be found to be 13.6 \s before by use of formulas. Chart, Fig, 47. This chart is for finding the (m.e.p.) of compressors. In the lase of multi-stage compressors with best-receiver pressure and perfect inter- »oling, the (m.e.p.) of each cylinder may be found by considering each cylinder is a single-stage compressor, or the (m.e.p.) of the compressor referred to the [j.P. cylinder may be found. The chart depends on the fact that the work per cubic foot of (sup.pr.) gas s equal to the (m.e.p.) for the no clearance case and that the (m.e.p.) with :learance is equal to the (m.e.p.) for no clearance, times the volumetric fficiency. Diagrams 1, 2 and 4 are reproductions of Figs. 45, 46 and 47 to a imaller scale and hence need no explanation as to derivation. Their use may )e briefly given. From the proper ratio of pressures project upward to the >roper curve, then horizontally to the (sup.pr.) and downward to read work per *ubic feet of (sup.pr.) gas. The volumetric eflSciency diagram was drawn in the following manner: 1 From Eq. (65) vol. eflf. = (l+c— cfip* ), showing that it depends upon three variables, Rp, c and s. A horizontal scale of values of Rp was laid off. Values )f Rp* were found and a vertical scale of this quantity laid off from the same jrigin as the Rp values. Through the intersection of the verticals from various iralues of Rp with the horizontals drawn through the corresponding values of 1 [Rp)' for a known value of s, a curve of this value of s was drawn. In a similar ifrav curves of other values of s were drawn. From the construction so far 1 completed it is possible to find the value of (Rp) * by projecting upward from any value oi Rp to the curve of s and then horizontally to the scale of (Rp)' . Values 1 rf volumetric efficiencies found for various clearances and the values of {Rp)' ire laid off on a horizontal base, with the origin at the opposite end of scale from that of Rp values, in order that clearance curves and s curves might be as distant as possible. These clearance curves were drawn through the inter- im ?ection of horizontals through the (Rp)' values and of verticals through the vulmetric efficiency values corresponding to them for the particular clearance in question. To find volumetric efficiency then it is merely necessary to project from value Df Rp to the correct s curve, then across to the proper clearance and finally 1 down to volumetric efficiency. As the value of (Rp)' is not desired, the hori- zontal projection is carried only to the intersection with the clearance curve and not to the edge of the diagram. To find the (m.e.p.) for single stage, the work per cubic foot is found from the diagram and then the volumetric efficiency, both as described above. The product is (7n,e,p,) For multi-stage compressors with perfect intercooling and best-receiver presssure, as stated above, the (m.e.p.) of each cylinder may be found, considering ENGINEERING THERMODYNAMICS II ItuUootPressurp i I I 1 J « et 77 7D SS SB 49 4^! 3S S8 21 It t Work ptr Cu. Ft. Of (Sup. I'r.) Guas-Ul Fiti. 47. — Mean EffecUve Preseuro of Compreswra, One-, Two-, and Three-etagea. WORE OF COMPRESSOES Initial Pi«isUTe Lba. per Sq. In, Ati«. Hulio of Pressures nsi7TTD<BSeiaU35e8El ul ( Work per Cu. Pi. at (Sup. IT.) Oii3-^l*4 Fio. 47. — Mean Effective Pressure of ComprcBsora, One-, Two-, and Three-Btages. 176 ENGINEERING THERMODYNAMICS each to be a single-stage compressor and remembering that (1 rec.pr.) becomes (sup.pr.) for second stage, and (del.pr.) for first stage and that (2 rec.pr.) becomes (sup.pr.) for third stage, (del.pr.) for second stage. The (m.e.p.) reduced to low-pressure cylinder is found by taking work per cubic feet of (sup.pr.) gas and multiplying by volumetric efficiency of low-pressure cylinder. To illustrate the use of this curve the example of Section (16) may be solved. Projecting upward from the pressure ratio of 9.35 to the line of a = 1.4 and then over to (sup.pr.) = 15 in diagram 4, since compression is three stage and from 15 lbs. per square inch to 140 lbs. per square inch, work per cubic foot or (m.e.p.), is found for no clearance to be 37.8 abs. per square inch. Since best-receiver pressure assumed is 31.6, which gives a ratio of 2.1 for the low- pressure cylinder. From diagram 3, by projecting upward from Rp = 2.1 and over to the 5 per cent clearance line volumetric efficiency is 96.5. The product gives (m.e.p.) reduced to low-pressure cylinder and is 36.5. From the — ^^ — 33,000 formula, horse-power is found to be 358 as before. Charty Fig. 48. As mentioned in Section 18, there is one (sup.pr.), which for a given (del.pr.) will give the maximum work of compression. The chart, Fig. 48, originated by Mr. T. M. Gunn, gives a graphical means of finding thi\ value of (sup.pr.) when the (del.pr.), clearance and value of s are known. It also gives on the right-hand of the chart a means for finding the (m.e.p.) for this condition. The figure was drawn by means of Eqs. (214) and (218). For the value of s = 1 the ratio of (del.pr.) to (sup.pr.) was found for cases of clearances from per cent to 15 by means of Eq. (218) by trying values of this ratio which would fulfill the condition of the equation, log^^p^T^) = ^""1^1^ (sup.pr.) * For values of s not 1, Eq. (214) was used, and a set of values of Rp found for the values of s= 1.4 and 1.2 by trial, the correct value of Rp being that which satisfied the equation. 8-1 1 /del.pr. \ « ^ fi I ^~ 1 /del.pr. \sup.pr./ 1+c L s Vsup.pr./ J As an example the work foi* the case where s = 1.2 and c= 10 per cent is given. Try Rp=2.6, then, R,T^ =j~ri+.l-. 1X^^X2.6*331 = 1.091(1.1 - .01667X2.218), = 1.161 WORK OF COMPRESSORS omwMjXiaAnsaoKiI'aiii] ]0 oiisg tanai|x«K X-iOjunininrrsjiioj fy 'Bajnwai j jo odbh 178 ENGINEEEING THEEMODYNAMICS Rp — (1.161)^ =2.45, which shows the value of 2.6 to be incorrect. For a second trial take 2.45, and then, /?p*, = 1.091(l.l-.01667X2.45-^), = 1.1627 = (1.1617)6 = 2.458, which is sufficiently close. Therefore the value of R^^ for s = 1.2 and clearance = 10 per cent, is 2.45; that is, maximum work will occur for a given (del.pr.) when (sup.pr.) is g^ times the (del.pr.) When the values for iJp had been obtained a horizontal axis of values of % and a vertical one of Rp values, were laid off and the points for clearance ciuT-es laid off to their proper values referred to these axes. Through points as plotted the clearance lines were drawn. The right-hand diagram was plotted in a similar manner from Eqs. (213) and (217), for s not equal to 1 and equal to 1 respectively. The latter formula was rearranged in the form /m.e.p.\ logei2p„ Vdel.pr./ Rp ^'' the last term being foimd from curve of Fig. 45. The value of Rp for each value of the clearance was taken from the left-hand diagram, and substituted in the above expression to obtain ( j *, ' - jfor the case of s= 1. EJq. (213) was put in form /m.e.p.V s /p'— .^jp and values of Rp for each value of the clearance found in the left-hand diagram were substituted, together with Ej, values from Fig. 45 and the value of ( , \ ' j found for each case of clearance when s = 1.4. When the points for 8 = 1 and 8 = 1,4 had been found, a horizontal axis of values of 8 and a vertical one of values of Rp were laid off, and points for the clearance curves plotted as for the left-hand diagram and the curves drawn in. To find the (sup.pr.) to give maximum work for any (del.pr.) it is only necessary to project from the proper value of 8 to the proper clearance curve, and then horizontally to read the value of Rp. The (del.pr.) divided by this gives the (sup.pr.) desired. To obtain the (m.e.p.) project upward from the value of 8 to the clearance curve, then horizontally to read the ratio {ttV"^") The {del.pr,) times this quantity gives the m.e.p. As an example of the use of this chart let it be required to find the (sup.pr.) for the case of maximum work for 9X12 in. double-acting compressor running 200 R.P.M., having 5 per cent clearance and delivering against 45 lbs. per square inch gage. Also the horse-power. Compression such that 8 = 1.3. WORK OF COMPRESSOKS Projecting from the value 1.3 for s on the left-hand diagram to the line of 60 5 per cent clearance find flp to be 2.8, hence (sup.pr.) =2g=21.4 lbs. per square inch absolute = 6.4 lbs. per square inch gage. Again, projecting from value 1.3 for s on right-hand dia^am to line of 5 per cent clearance find that^ bcnce fm.e.p). = 23 and THP -23X1X64X400 ,„„ LH.P.---33^j^^— =17.8. ( m.e.p .) ^ Cdei.pr.)" Fig. 49. — R«lativc Work of Two- and Thmc-Gtage Compressors Compared to Single Stapc. Chart, Fig. 49. This chart is designed to show the saving in work done in compressing and delivering gases by two-stage or three-stage compression with bestr-receiver pressure and perfect intcrcooling over that required for com- pressing and delivering the same ga.s between the -same pressures in one stage. 180 ENGINEERING THERMODYNAMICS The chart was made by laying ofT on a horizontal base a scale of pressure ratios. From the same origin a scale of work for two or three stage divided by the work of one stage was drawn vertically. For a number of values of if, the work to compress a cubic foot of gas was found for one, two and three stage for each value of s. The values found by dividing the work of two or three stage by the work of single stage were plotted above the proper Rp values and opposite the proper ratio values and curves drawn through all points for one value of s. To find the saving by compressing in two or three stages project from the proper Rp value to the chosen s curve for the,desired number of stages, then horizontally to read the ratio of multi-stage to one-stage work. This value gives per cent power needed for one stage that will be required to compress the same gas multi-stage. Saving by multi-stage as a percentage of single stage is one minus the value read. To illustrate the use of this chart, find the per cent of work needed to compress a cubic foot of air adiabatically from 1 to 8i atmospheres in two stages compared to doing it in one stage. From examples under chart Nos. 44 and 46 it was found that work was 6300 ft.-lbs. and 5320 ft.-lbs. respec- tively, for one- and two-stage compression, or that two stage was 84.5 per cent of one stage. From Rpj 8J project up on Fig. 49 to 8 = 1.406 for two stage and over to read 84.6 per cent, which is nearly the same. Chart, Fig, 50. This chart, designed by Mr. T. M. Gunn, shows the economy compared to isothermal compression. The chart was drawn on the basis of the following equation: •n /• xi_ i\ m.e.p. isothermal (no clearance) Economy (isothermal) = '^ - -— -^— - — l ^ m.e.p. actual -r^, actual (sup .pr.) log e fip 8-\ logeiZ (sup.pr.) (/2j, B — 1) s '-^ 8-1 {Rp—-\) Values of this expression were worked out for each exponent, for assumed values of Rp, A scale of values of Rp was laid off horizontally and from the same origin a vertical scale of values of the ratio of isothermal to adiabatic. The results found were then plotted, each point above its proper Rp and opposite its ratio value. Curves were then drawn through all the points found for the same value of s. In a similar way a set of curves for two stage and a set for three stage were drawn. This chart Is also useful in obtaining the (m.e.p.) of th^ cycle if the (sup.pr.) and the volumetric efficiency of the cylinder be known. A second horizontal scale laid off above the Rp scale shows the (m.e.p.) per pound of (sup.pr. for) the isothermal no-clearance cycle. This is found to be equal to log« Rp, since WORK OF COMPEESSOES r 'pjBpireiB |viiuaii)osi iqju pa]Bdiua3 ^luouosa. 182 ENGINEERING THERMODYNAMICS the (m.e.p.) for no clearance is equal to the work per cubic foot of (sup.pr.) ga=, which, in turn, for the isothermal case is (sup.pr.) log* Rp or log« Rp when (sup.pr.) = 1. Knowing the ratio of pressures, economy compared to isothermal can be found as explained above. Also knowing Rp the (m.e.p.) per pound initial is found from the upper scale. Since the latter quantity is assumed to be known, by multiplying it by- factor just found there is obtained (m.e.p.) isothermal. Since volumetric efficiency is assumed known, all the factors are known for the first equation given above which, rearranged, reads , N X 1 m.e.p. isothermal (no clearance) (m.e.p.) actual = -r- = — tt jr — ^ , (economy isothermal) -^£r, Chart, Fig, 51. This chart is drawn to give the cylinder displacement for a desired capacity, with various values of /2p, s and clearance. From the formula Eq. (64) 1^ (L. P. Cap.)=Z)(l+c-ci2p«). The right-hand portion of the diagram is for the purpose of finding values of (/2p) « for various values of Rp and s, and is constructed as was the similar curve in Fig. 45. The values of the lower scale on the left-hand diagram give i. values of D = (L. P. Cap.)-4-(l+c— c/?p*)i where capacity is taken at 100 cu.ft., this scale was laid out and the clearance curves points found by solving the above equation for various values of (Rp) * for each value of c. To obtain the displacement necessary for a certain capacity with a given value of Rp^ c and 5, project upward from Rp to the proper s curve across to the c curve and down to read displacement per hundred cubic feet. Also on the left-hand diagram are drawn lines of piston speed, and on left-hand edge a scale of cylinder areas and diameters to give displacements found on horizontal scale. To obtain cylinder areas or approximate diameters in inches project from displacement to piston speed line and across to read cylinder area or diameter. Figures given are for 100 cu.ft. per minute. For any other volume the displacement and area of cylinder will be as desired volume to 100 and diameters will be as v^desired volume to 100. As an example, let it be required to find the low-pressure cylinder size for a compressor to handle 1500 cu.ft. of free air per minute. Receiver pressure to be 45 lbs. per square inch gage and (sup.pr.) to be atmosphere. Piston speed limited to 500 ft. per minute. Compression to be so that s=L4 and clear- ance =4 per cent. Projecting upward from /2p=4 tos==1.4, across to c = 4%, and down to piston speed = 500, find the diameter of a cylinder for 100 cu.ft. per minute is 6.3. For 1500 cu.ft. diameter will be as Vl5X6.3 = 3,9X6.3=24 ins. WORK OF COMPRESSORS 183 o 08 & O c > O O o 8 c O I u a .9 if o o r-t "^S o 00 5 184 ENGINEERING THERMODYNAMICS GENERAL PROBLEMS ON CHAPTER II. Prob. 1. One hundred cubic feet of HjS are compressed from 15 lbs. per square nch absolute to 160 lbs. per square inch absolute. (a) Find work done if compression occurs isothermaUy in a no-clearance on&«tage compressor; (h) Adiabatically in a two-stage, no-clearance compressor; (c) Adiabatically in two-stage compressor each cylinder having 5 per cent clearance; Prob. 2. Air is being compressed in three plants. One is single-stage, the second is two-stage, and the third is three-stage Considering the compressors to have no clear- ance and 1000 cu.ft. of free air compressed adiabatically per minute from atmosphere to 160 lbs. per square inch gage, what will be the horse-power required and cylinder sizes in each case? Prob. S* A two-stage compressor with 5 per cent clearance n the h gh and 3 per cent in the low-pressure cylinder is compressing air from 14 lbs. per square inch al> solute to 125 lbs. per square inch gage. What is the best-receiver pressing and what must be the size of the cylinders to handle 500 cu.ft. of free air per minute? Prob. 4. A manufacturer gives a capacity of 970 cu.ft. for a 22i x24 in. single- stage compressor running at 142 R.P.M. when working pressures are 50 to 100 lbs. per square inch gage. What would be the clearance for each of these pressures assuming 8 = 1.4? Prob. 5. The card taken from a single-stage compressor cylinder showed an appar- ent volmetric efficiency of a 95 per cent and a mean value for s of 1.3. What is the clearance and what would be the (m.e.p.) for the ratio of pressures of 6? Prob. 6. A compressor with double-acting cylinder 12x14 ins., having 6 per cent clearance, is forcing air into a tank. Taking the volumetric efficiency as the mean of that at the start and the end, how long w 11 it take to build up 100 lbs. per square inch gage pressing in a tank of 1000 cu.ft. capacity if the speed is 100 M.R.P. and compres sion is isothermal, and how long wi 1 it take to do it if the compression is adiabatic and the air in the tank does not cool during filling? What is the ma>ximum attainable pressure? Prob. 7. It is desired to supply 1000 cu.ft. of air per minute at a pressure of 80 lbs. per square inch gage. A two-stage compressor is to be used, the clearance in low pres- sure of which is 3 per cent. What must be the displacement of the low-pressure cylin- der and what will be the horse-power of the compressor? Prob. 8. The low-pressure cylinder of a compressor is 18x24 ins. and has a clear- ance of 4 per cent. The receiver pressure is 60 lbs. per square inch absolute. The high- pressure cylinder has a 5 per cent clearance. What must be its diameter, stroke being same as low, so that compressor will operate at its designed receiver pressure? Prob. 9. The discharge pressure of a two-stage compressor is 120 lbs. per square inch absolute and the supply pressure is 15 lbs. per square inch absolute The compressor is 10x16x12 ins. The clearance in the low-pressure cylinder is 3 per cent. What must be the clearance in the high-pressure cylinder for the machine to operate at best- receiver pressure? Compression is adiabatic. Prob. 10. If the clearance in the high-pressure cylinder of Prob. 9 were reduced to 3 per cent, would the receiver pressure increase or decrease, how much and why? Prob. 11. If the discharge pressure in Prob. 9 fell to 100 lbs. per square inch absolute, what would be the new best-receiver pressure and why? Would the original clearance allow the new best-receiver pressure to be maintained? PROBLEMS ON CHAPTER II 185 Prob. 12. The discharge pressure for which a 20jx32jx24 in. compressor is designed, is 100 lbs. per square inch gage, supply pressure being 14 lbs. per square inch absolute. The d scharge pressure is raised to 125 lbs. per square inch gage. The clearance on the high-pressure cyUnder can be adjusted. To what value must it be changed to enable the compressor to carry the best-receiver pressure for the new dischai^ pressure? Low-pressure clearance is 5 per cent at all times and com- pression being adiabatic. Prob. 13. A manufacturer builds his 15jx25ixl8 in. compressors with low-pres- sure cylinders of larger diameter for high altitude work. What would be the diameter of a special cylinder for this compressor to work at an altitude of 10,000 ft. and what would be the horse-power per cubic foot of low-pressure air in each case? Prob. 14. A three-stage compressor has 4 per cent clearance in all the cylinders. The low-pressure cylinder is 34x36 ins., delivery pressure 200 lbs. per square inch gage, supply pressure 14 lbs. per square inch absolute. What must be the size of the other cylinders for the machine to operate at best-receiver pressure. Prob. 16. The cylinders of a two-stage compressor are given as lOJ and 16J ins., the stroke being 12 ins. The machine has a capacity of 440 cu.ft. per minute at 160 R.P.M., the supply pressure is 14 lbs. per square inch absolute and the dehvery pressure 100 lbs. per square inch gage, \^^lat is the clearance of each cylinder? Prob. 16. To perform the required useful refrigerating effect in an ice plant, it is necessary to compress 250 cu.ft. of ammonia vapor per minute from 30 lbs. per square inch gage to 150 lbs. per square inch gage. What must be the size of the compressor to handle this at 75 R.P.M. if it be double acting and the clearance be 5 per cent? Prob. 17. The maximum deUvery pressure of a type of compressor is controlled by making the clearance large so that the volumetric efficiency will decrease as the pressure rises and become zero at the desired pressure. What must be the clearance for a single- stage compressor where the supply pressure is 14 lbs. per square inch absolute and the maximum deUvery pressure 140 lbs. per square inch absolute? What will be the volu- metric efficiency of the same machine at a delivery pressure of J the maximum? At J? Prob. 18. A three-stage compressor has a clearance of 5 per cent in each cylinder. What must be the cylinder ratios for the best-receiver pressures when the machine is compressing to 170 lbs. per square inch gage from atmosphere? Prob. 19. Show why it was very essential to keep the clearance low in cylinders of three-stage compressor used for compressing air for air-driven cars, where the delivery pr^sure carried was 2500 lbs. per square inch, by assuming numerical data and calculating numerical proof. Prob. 20. With water falling 150 ft. and used to compress air directly, how many cubic feet of air could be compressed per cubic foot of water? Prob. 21. Air is compressed from atmosphere to 150 lbs. per square inch absolute, iaothermally in one stage. How much more work would be required per cubic foot if compression were adiabatic? How much of this excess would be saved by compressing two stage? Three stage? Prob. 22. 150 I.H.P. is delivered to the air cylinders of a 14i X22i Xl8 in. compres- sor, running at 120 R.P.M. The supply pressure is 15 lbs. per 'square inch absolute. The volumetric efficiency as found from the indicator card is 95 per cent. What was the discharge pressure? Prob. 23. The clearance in the high-pressure cylinder of a compressor is 5 per cent, which allows the compressor to run with the best-receiver pressure for a discharge of 100 lbs. per square inch absolute when the compressor is at sea-level. What would the 186 ENGINEERING THERMODYNAMICS clearance be if the discharge pressure were kept the same and the altitude were 10,000 ft. to keep the best-receiver pressure? Prob. 24. How many cubic feet of supply-pressure air may be compressed per minute from 1 to 8 atmospheres absolute by 100 horse-power if the compression in all cases k adiabatic? (6) Three stage, no clearance; (c) Two stage with 5 per cent clearance; (d) Single stage with 5 per cent clearance; (a) Two stage, no clearance Prob. 25. The capacity of a 14jx22jxl4 in. compressor when running at 140 R.P.M. is said to be 940 cu.ft. for working pressures of 80 to 100 lbs. per square inch gage and atmospheric supply at sea level. Check these figures. Prob. 26. What horse-power would be required by an 18jx30jx24 in. compressor operating at 100 P.P.M. and on a working pressure of 100 lbs. per square inch gage if the clearance in low-pressure cylinder is 4 per cent? What would be the capacity? Prob. 27. By means of water jackets on a compressor cylinder the va ue for s of com- pression curve in single-stage machine is lowered to 1.3. Compare the work to com- press 1000 cu.ft. of air from 1 to 8 atmospheres absolute with this condition with that required for isothermal and adiabatic compression. Prob. 28. What must be the size of cylinders in a three-stage compressor for com- pressing gas from 50 lbs. per square inch absolute to 600 lbs. per square inch absolute when 8 equals 1.3 and each cylinder has 3 per cent clearance. Compressor to run at 100 R.P.M. to be double acting and handle 1000 cu.ft. of gas per hour? Prob. 29. How many cu.ft. of free air could be compressed per minute with an avail- able horse-power of 1000 H.P. from atmosphere to 150 lbs. per square inch gage; (a) if compression is isothermal; (6) if compression be single-stage adiabatic; (c) if com- pression be three-stage adiabatic? Prob. 30. A single-stage compressor is compressing air adiabatically at an altitude of 6000 ft. to a pressure of 80 lbs. per square inch gage. The cylinder has 2 per cent clearance. What must be the s'.ze of the cylinder to compress 2000 cu.ft. of free air per minute if the piston speed is limited to 600 ft. per minute? What, if the clearance be zero? Prob. 31. What would be the size of the two-stage compressor for same data as in Prob. 30? CHAPTER III WORK OF PISTON ENGINES. HORSE POWER AND CONSUMPTION OF PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER GAS OR VAPOR UNDER PRESSURE. 1. Action of Fluid in Single Cylinders. General Description of Structure and Processes. The most commonly used class of engines is that in which the operation is dependent on the pushing action of high-pressure fluids on pistons in cylinders, and this includes all piston steam engines of the recipH rocating or straight-hne piston path group as well as the less common rotary group, having pistons moving in curved and generally circular paths. In these same engines there may be used compressed air as well as steam, and equally as well the vapors of other substances or any other gases, without change of structure,' except perhaps as to proportions, providing only that the substance to be used be drawn from a source of supply under high pressure, be admitted to the cylinder, there used and from it discharged or exhausted to a place of lower pressure. This place of lower pressure may be the open air or a closed chamber; the used fluid may be thrown away and wasted or used again for various purposes without in any way affecting the essential process of obtaining work at the expense of high-pressure gases or vapors. It is evident that, regarding a piston as a movable wall of a cylinder, when- ever a fluid acts on one side with greater pressure than on the other, the piston will move toward the lower pressure end of the cylinder, and in so moving can exert a definite force or overcome a definite resistance, measured by the difference in pressure on the two sides and the areas exposed to the pressure. It is not so evident, but just as true, that the piston may be made to move from one end of the cylinder to the other when the average pressure on one side is greater than the average pressure on the other, and also do work even if the excess of pressure should reverse in direction during the stroke and provided only some energy storage device is added. In the common steam or compressed-air engine this is a flywheel with the usual connecting rod and crank mechanism, uniting the reciprocating piston movement with the continuous rotary movement of the flywheel mass. In certain forms of pumps the energy is stored in extra cylinders at times of excess and given out at times of deficiency in the path of the piston, so that its motion from end to end of cylinder may not be interrupted even if the pressure on the driving side should fall below that on the resisting side, assuming, of course, the average pressure for the whole stroke to be greater on the driving side than on the resisting side. 187 188 ENGINEERING THERMODYNAMICS It appears, therefore, that piston movement in dhgines of the common form and structure, and the doing of work by that movement is not a question of maintaining a continuously greater pressure on one side than on the other. On the contrary, the process is to be studied by examination of the average pressure on the driving side and that on the resisting side, or by comparing the whole work done on one side with the whole work done on the other side by the fluid. The work done by the fluid on one side of a piston may be positive or negative, positive when the pressures art- assisting motion, negative when they are resisting it. It is most con- venient to study the action of fluids in cylinders by considering the whole action on one side from the beginning of movement at one end to the end of movement at the same point, after the completion of one complete forward and one complete return stroke. All the work done by the pressure of the fluid on the forward stroke on the side of the piston that is apparently moving away from the fluid is positive work, all the work done by the pres- sure of the fluid on the same side of the piston dming the return stroke is negative, and for this stroke the side of the piston under consideration is apparently moving toward the fluid or pushing it. For the complete cycle of piston movement covering the two strokes the work done on one side is the algebraic sum of the forward stroke work, considered positive, and the back stroke work, considered negative. During the same time some pressures are acting on the other side of the piston, and for them also there will be a net work done equal to the cor- responding algebraic sum. The work available for use during the complete two strokes, or one revolution, will be the sum of the net work done by the fluid on the two sides of the piston during that time, or the algebraic smn of two positive and two negative quantities of work. Methods of analysis of the work of compressed fluids in cylinders are consequently based on the action in 0716 end of a cylinder, treated as if the other end did not exist. Just how the high-pressure fluid from a source of supply such as a boiler or an air compressor is introduced into one end of a cylinder, how it is treated after it gets there, and how expelled, will determine the natiu*e of the varia- tion in pressure in that end that acts on that side of the piston, and theso are subjects to be studied. To determine the work done in the cylinder end by the fluid, it is necessary to determine laws of pressure change with stroke, and these are fixed first by valve action controlling the distribution of the fluid with respect to the piston and second by the physical properties of the fluid in question. It is necessary that the cylinder be fitted with a valve for getting fluid into a cylinder, isolating the charge from the source of supply and getting it out again, and it may be that one valve will do, or that two or even more are desirable but this is a structural matter, knowledge of which is assumed here and not concerned with the effects under investigation. The first step in the process is, of course, admission of fluid from the source of supply to the cylinder at one end, which may continue for the whole, or be limited to a part • WORK OF PISTON ENGINES 189 of the stroke. When admission ceases or supply is cut off before the end of the stroke there will be in the cylinder an isolated mass of fluid which will, of course, expand as the piston proceeds to the end. Thus the forward stroke, considering one side of the piston only, always consists of full pressure admission followed by expansion, the amount of which may vary from zero to a very large amount; in fact the final volume of the fluid after expansion may be hundreds of times as great as at its beginning, when supply was cut off. At the end of this forward stroke an exhaust valve is opened, which permits communication of the cylinder with the atmosphere in non-con- densing steam and most compressed-air engines, or with another cylinder, or with a storage chamber, or with a condenser in the case of a steam engine in which the pressure approximates a perfect vacuum. If at the moment of exhaust opening the cylinder pressing is greater or less than the back pressure, there will be a more or less quick equalization either up or down before the piston begins to return, after which the return or exhaust stroke will proceed with some back-pressiu*e resistance acting on the piston, which is generally though not always constant. This may last for the whole back stroke or for only a part, as determined by the closure of the exhaust valve. When the exhaust valve closes before the end of the return stroke the unexpelled steam will be trapped and compressed to a pressure depending partly on the point of the stroke when closure begins and the pressure at the time, and partly on the clearance volume of the cylinder into which the trapped steam is compressed. Of course, at any time near the end of the stroke the admission valve may be opened again., and this may occur, 1st, before compression is complete, which will result in a sudden pressure rise in the cylinder before the end of the stroke to equalize it with the source of supply; 2d, just at the end of the stroke, which may result in a rise or a fall to equalize, or no change at all, depending on whether compression has raised the cylinder pressure not quite to supply pressure, or to something greater than it, or to a value just equal to it; 3d, after the end of the stroke, which will result in a reexpansion of the steam previously under compression, and then a sudden rise. It may be said in general that in cylinders there are carried out with more or less variation the following processes: Forward stroke, constant-pressure admission followed by expansion. Back strokCj constant-pressure exhaust followed by compression, while at both ends of the stroke there may or may not be a vertical line on a pressure- volume diagram representing a constant-volume change of pressure. These processes will result in a cycle of pressure- volume changes which will be a closed curve made of more or less accurately defined phases, and the work of the cycle will be the area enclosed by the cyclic curve. Of course, there are causes of disturbance which make the phases take on peculiar characteristics. For example, the valve openings and closures may not take place as desired or as presupposed with respect to piston positions; leakage may occur, steam may condense during the operations in the cylinder, and water of condensation 190 ENGINEERING THERMODYNAMICS may evaporate; the resistance through valves will always make the cylinder pressure during admission less than in the supply chamber and greater during exhaust than the atmosphere or than in exhaust receiver or condenser and mav through the valve movements make what might have been a constant-pit^ sure straight line become a curve. There will, by reason of these influences, encountered in real engines, be an almost infinite variety of indicator cards or pressure-volume cycles for such engines. A ■ B B j0^ A' \ AB STEAM ADMITTED B-C •• EXPANDED C-D •• EXHAUSTED D-A •• COMPRESSED / /^ \ / \ y ACTUAL CARD FROM CORU6S ENGINE > N e c .,/ A ^ D \ , C \ \ \ ^ D""" C Fig. 52. — Diagram to Indicate Position of Admission, Cut-off, Release, Compression (»n Engine Indicator Card. The various points of the stroke at which important events occur, important in their pressure-volume significance, have names, as do also the lines between the points, and these names are more or less commonly accepted and generally understood as follows: letters referring to the diagram Fig. 52. Point Names: Events of Cycle. A. Admission is that point of the stroke where the supply valve Is opened. B, Cut-off is that point of the stroke where the supply valve is closed. C, Release is that point of the stroke where the exhaust valve is opened. D. Compression is that point of the stroke where the exhaust valve is clovsed. Names of Lines, or Periods: i-B, Admission or steayn line joins the points of admission and cut-off. B-C, Expansion line joins the points of cut-off and release. C-D, Exhaust line joins the points of release and compression if there is any, or admission if there is not. D-A. Compression line joins the points of compression and admission. WORK OP PISTON ENGINES 191 By reason of the interferences discussed, these points on actual indicator cards may be difficult to locate, one line merging into the next in so gradual a manner as to make it impossible to tell where the characteristic point lies, as will be apparent from Fig. 53, in which is reproduced a number of actual indicator cards. In such cases equivalent points must be located for study Fig. 53. — ^Actual Steam Engine Indicator Cards Showing Distortion of Lines and Uncertain Location of Characteristic Points. These same terms, which it appears sometimes refer to points and some- times to lines, are also .used in other senses, for example, cutroff is com- monly used to mean the fraction of stroke completed up to the point of cut-off, and compression that fraction of stroke remaining incomplete at the point of compression, while compression is also sometimes used to express the pressure attained at the end of the compression line. In general, there 192 ENGINEERING THERMODYNAMICS is nothing in the use of these words to indicate just which of the various meanings is intended except the text, and experience will soon eliminate most of the possible chances of confusion. Prob. 1. Draw a card in which admission and exhaust are late. Draw a card in which there is no compression, in which compression is very early, so that compression pressure is equal to admission pressure. Draw a card with per cent cut-off, and cui- off = 100 per cent. Draw cards with same cut-off but with varying initial pressures. Prob. 2. The following diagrams, Fig. 54, are reproductions of indicator cards actually taken from engines. Explain what features are peculiar to each and if possible give an explanation of the cause. Fig. 64. — Indicator Diagram from Steam Engines with Improperly Set Valve Gear. For example, in No.' 1 a line of pressure equalization between the end of the com- pression line and the beginning of the admission line inclines to the right instead of being perpendicular, as in a perfect diagram. This is due to the fact that admis- sion does not occur until after the piston has begun to move outward, so that pres- sure rise does not occur at constant volume, but during a period of increasing volume. 2. Standard Reference Cycles or PV Diagrams for the Work of Expansive Fluids in a Single Cylinder. Simple Engines. To permit of the derivation of a formula for the work of steam, compressed air, or any other fluid in a WORK OF PISTON ENGINES 193 ylinder, the various pressure volume, changes must be defined algebraically, rhe first step is, therefore, the determination of the cycle or pressure-volume liagram representative of the whole series of processes and consisting of a lumber of well-known phases or single processes. These phases, ignoring all orts of interferences due to leakage or improper valve action, will consist of K^nstant-pressure and constant-volume lines representing fluid movement nto or from the cylinder, combined with expansion and compression lines representing changes of condition of the fluid isolated in the cylinder. These *xpansion and compression lines represent strictly thermal phases, laws for p^hich will be assumed here, but derived rigidly later in the part treating of the tbermal analysis; however, all cases can be represented by the general expression PV' = C, in w^hich the character of the case is fixed by fixing the value of s. For all gases and for vapors that do not contain liquid or do not form or evaporate any during expansion or compression, i.e., continually superheated, the exponent 8 may have one of two characteristic valiies. The first is isoOiermal expansion and compression, and for this process s is the same for all substances and equal to xmity. The second is for exponential expansion or compression and for this process s will have values peculiar to the gas or superheated vapor under discussion, but it is possible that more than one substance may have the same value, as may be noted by reference to Section 8, Chapter I, from which the value s= 1.406 for air and 5 = 1.3 for superheated steam or ammonia adiabatically expanding are selected for illustration. When steam or any other vapor not so highly superheated as to remain free from moisture during treatment is expanded or compressed in cylinders different values of s must be used to truly represent the process and, of coiu^e, there can be no isothermal value, since there can he no change of pressure of wet vapors mithoiU a change of temperature. For steam expanding adiabatically the value of s is not a constant, as will be proved later by thermal analysis, so that the exact solution of problems of adiabatic expansion of steam under ordinary conditions becomes impossible by pressure-volume analysis and can be handled only by thermal analysis. However, it is sometimes convenient or desirable to find a solution that is approximately correct, and for this a sort of average value for s may be taken. Rankine's average value is 5=1. 111 = ^ for adiabatic expansion of ordinarily wet steam, and while other values have l:)een su^ested from time to time this is as close as any and more handy than most. The value s = 1.035+.14X(the original dryness fraction), is ^ven by Perry to take account of the variation in original moisture. Steam during expansion adiabatically, tends to make itself wet, the condensation being due to the lesser heat content by reason of the work done; but if during expansion heat be added to steam originally just dry, to keep it so continuously, as the expansion proceeds, it may be said to follow the saturation law of steam, for which s = 1.0646. This is a strictly experimental value found 194 ENGINEERING THERMODYNAMICS A d{ m ^'€ Opmp. s Exp. B V V. m "V P • -* -( :i E \ ■Exp. 4' F ^ V. U-»i «1 -Exp. < v Vr-Ooi ^>^ M Ck>mi >. i r' —s= • ^ V Q • >> G<>mp. V Fig. 55. — Standard Reference Cycles or Pressure-volume Diagrams for Eiq>anfiive Fluid in Simple Engines. WORK OF PISTON ENGINES 195 by studying the volume occupied by a poimd of just dry steam at various pressures, quite independent of engines. Direct observation of steam engine indicator cards has revealed the fact that while, in general, the pressure falls faster at the beginning of expansion and slower at the end than would be the case if 5 = 1, yet the total work is about the same as if « had this value all along the curve. This law of expansion and compression, which may be conveniently designated as the logarithmic law, iS almost universally accepted as representing about what happens in actual >t3am engine cylinders. Later, the thermal analysis will show a variation of wet- ness corresponding to «=1, which is based on no thermal hjrpothesis what- ever, but is the result of years of experience with exact cards. Curiously Fig. 56. — Simple Engine Reference, Cycles or PF Diagrams. enough, this value of 8 is the same as results from the thermal analysis of con- stant temperature or isothermal expansion for gases, but it fails entirely to represent the case of isothermal expansion for steam. That «= 1 for isothermal gas expansion and actual steam cylinder expansion is a mere coincidence, a fact not understood by the authors of many books often considered standard, as m them it is spoken of as the isothermal curve for steam, which it most positively is not. This discussion of the expansion or compression laws indicates that analysis falls into two classes, first, that for which s=l, which yields a logarithmic expression for work, and second, that for which 5 is greater or less than one, which yields an exponential expression for work, and the former will be designated as the logarithmic and the latter as the exponential laws, for convenience. 196 ENGINEERING THERMODYNAMICS The phases to be considered then may be summed up as far as this aoalysis is concerned as: 1. Admission or exhaust , pressure constant, P= const. 2. Admission or exhaust, volume constant, F= const. 3. Expansion, Py= const., when « = 1. 4. Expansion, PF* = const., when 8 is greater or less than I. 5. Compression, PF = const., when 8=1. 6. Compression, PF*= const., when s is greater or less than I. Considering all the possible variations of phases, there may result any or the cycles represented by Fig. 55. These cycles have the characteristics indicated by the following table, noting the variation in the law of the expansion or compression that may also be possible. 1 Cycle. Clearance. Eipanaion. Compression. A B C D Zero Zero Zero Zero Zero Little Complete Over^xpansion Zero Zero Zero Zero E F G H Little Little Tiittle Little Zero Little Complete Over^xpansion Zero Zero Zero Zero I J K L Little Little Little Little Zero Tiittle Complete Over-expansion little Little Little Little M N P Little Little Little Little Little Little Little Little Zero Little Complete Over-expansion Complete Complete Complete Complete Q R S T Zero Little Complete Over-expansion Too much Too much Too much Too much It is not necessary, however, to derive algebraic expressions for all these cases, since a few general expressions may be found involving all the variables in which some of them may be given a zero value and the resulting expression will apply to those cycles in which that variable does not appear. The result- ing cycles, Fig. 56, that is is convenient to treat are as follows: SIMPLE ENGINE REFERENCE CYCLE OR P7 DIAGRAMS Cycle 1. Simple Engine, Logarithmic Expansion without Clearance. Phase (a) Constant pressure admission. (6) Expansion PF =const. (may be absent). (c) Constant volume equalization of pressure with exhaust (may be absent) . (d) Constant pressure exhaust. (e) Constant (zero) volume admission equalization of pressure with supply. (( (I <t u WORK OF PISTON ENGINES 197 U « «( <( Cycle II. Simple Engine, Exponential Expansion without Clearance. Phase (a) Constant pressure admission. (6) Ebcpansion PP = const, (may be absent). (c) Ck>nstant- volume equalization of pressure with exhasut (may be absent). (cO Constant-pressure exhaust. (e) Constant (zero) volume admission equalization of pressure with supply. Cycle III. Simple Engine, ^garithmic Expansion and Compression with Clearance. Phase (a) Constant pressure admission. (6) Expansion P7b const, (may be absent). (c) Constant volume equalization of pressure with exhaust (may be absent). (d) Constant pressure exhaust. (e) Compression PF -const, (may be absent). (/) Constant volume admission, equalization of pressure with supply (may be absent). Cycle IV. Simple Engine, Exponential Expansion and Compression with Clearance. Phase (a) Constant pressure admission. (6) Expansion PV=const. (may be absent). (c) Constant volume equalization of pressure with exhaust (may be absent). (d) Constant pressure exhaust. (e) Compi^ession PF'= const, (may be absent). (f) Constant admission, equalization of pressure with supply (may be absent). <i «f €t €1 It «< If If IC t€ p ^ 7n B A (ln.pr) \ V \ • S \ X ^^ ^^ >^ Q (rel.pr.) • E D_ (bk.pr.) n 6 F • iL Fig. 57. — ^Work of Expansive Fluid in Single Cylinder with No Clearance. Logarithmic Expansion for Cycle I. Exponential for Cycle II. 3. Work of Eicpansive Fluid in Single Cylinder without Clearance. Loga- rithmic Expansion, Cycle I. Mean Eflfective Pressure, Horse-power and Consunqition of Simple Engine. Referring to the diagram, Fig. 57, the net work, whether expansion be incomplete, perfect, or excessive, is the sum of 198 ENGINEERING THERMODYNAMICS admission and expaneion work less the back-pressure work, or by areas, net work area, ABCDE^Budmiasion work area ABFG +expansion work area FBCH or algebraically, —back-pressure work area GEDH lF=P»7»-|-P»7»log,§-P*7«, • (230) Work of cycle in foot-pounds is =P.7il+log. ^) -P^Vs =P»7»(l-|-log.^^)-P*7<, =P«Fyi-|-log, ^j -P^Vt (a) (6) (c) (d) . (231) As Vi or Ve represent the whole displacement, the mean effec^ve presmare will be obtained by dividing Eqs. (231), by Vt or F,, givmg. M .E.P.=Pyi-|-log.^)-P<, (a) =P.(l+log.^^)-P, (6) (c) .e.p. = pe ( 1 + log, ^ j - Pa =pJl+Iog,^j-Pd (d) '=pt(l+log,y)(Y)-pa (c) (232) Similarly, dividing the Eq. (232) by the volume of fluid admitted, F», will give the work per cubic foot, which is a good measure of economy, greatest economy being defined by maximum work per cubic foot, which, it may be noted, is the inverse of the compressor standard. Work per cu.ft. supplied = P»( 1+loge ^ j —Pd^ (a) =p»(i+iog.^)-p.f; (6)_ (233) WORK OF PISTON ENGINES ' 199 According to Eq. (19), Chapter I, the piston displacement in cubic feet 13 750 per hour per I.H.P. is for 2=l,y— ^ — , and this multiplied by the fraction of rhole displacement occupied in charging the cylinder or representing admission, ffhich is 7, 7, Va °' 7.' ill give the cubic feet oj high pressure fluid supplied per hour per I.H.P., ence will give hence CuA^pplied per hr.per I.H.P. « ^3,750 V^ .. (m.e.p.) Vc __ 13,750 Pfl .j^v (m.e.p.) Pb (234) Introducing a density factor, this can be transformed to weight of fluid. If then d\ is the density of the fluid as supplied in pqunds per cubic foot, Lbs. fluid supplied per hr. per I.H.P. = -. — - — r X^X Bi (a) \m.e.p.j V c ^ 13^ g, g (m.e.p.) Pft . •. (235) All these expressions, Eqs. (230) to (235), for the work of the cycle, the mean effective pressure, work per cubic feet of fluid supplied, cubic feet and pounds of fluid supplied per hour per I.H.P., are in terms of diagram point conditions and must be transformed so as to read in terms of more generally defined quantities for convenience in solving problems. The first step is to introduce quantities representing supply and back pressures and the amount of expansion, accordingly: Let (in.pr.) represent the initial or supply pressure p» expressed in pounds per square inch; " (rel.pr.) represent the release pressure pc, in pounds per square inch; " (bk.pr.) represent the back pressure pd, in pounds per square inch; " Ry represent the ratio of expansion defined as the ratio of largest to smallest volume on the expansion line \w\^^\w) which is, of course, equal to the ratio of supply to release pressure ( )i when the logarithmic law is assumed; D represent the displacement in cubic feet which is Vd or Vc when no clearance is assumed; 200 ENGINEEMNa THERMODYNAMICS Let Z represent the fraction of stroke or displacement completed up to cut-off so that ZD represents the volume Vh admitted to the cylinder. In this case when clearance is zero, Z=-^ . Mxy Work of the cycle in foot-pounds W = 144 (in.pr. l+logefir R vj -.(bk.pr.)l/) (a) ' J }• . . = 144[(reLpr.) (1 +loge Af) - (bk.pr.)]/) (6) « m.e.p. = (rel.pr.) (1 +loge Rv) — (bk.pr.) (a) = (m.pr.)(^^'')-(bk.pr.) (6) = (m.pr.)z(l+log,^)-(bk.pr.) (c) Work per cu.ft. supplied = 144[ (in.pr.) (1 +log, Rv) — (bk.pr.)fiv] (a) ' = 144[(in.pr.)(l+Iog.|)-^] (6) _ Cu.ft. supplied per hr. per I.H.P. = — ^ — 5- (a) m.e.p. Mxy 13,750 (mie.p.) Z (6) Lbs. supplied per hr. per I.H.P. 13,750 Si . . (m.e.p.) Ry 13,750 (m.e.p.) ZSi(6) (236) (237) (238) (239) (240) The indicated horse-power may be found by multiplying the work of the cycle, Eq. (236), by the number of cycles performed per minute n and divid- ing the product by 33,000. l+Iog,/2v Ry or Z)n(m.e.p.) ^^'^^^"^'^^^k J -(bk.pr.) , . . . (241) I.H.P.= 229.2 (242) In any of these expressions where Ry is the ratio of greatest to smallest volume diuing expansion, either flp, ratio of greater to smaller pressures, or — , the WOKK OF PISTON ENGINES 201 reciprocal of the cut-off, may be substituted, since the expressions apply only to the logarithmic law, and clearance is assumed equal to zero. When clearance is not zero; it is shown later that the cut-off as a fraction of stroke is not the reciprocal of Rp or Rv These expressions are perfectly general, but convenience in calculation will be served by deriving expressions for certain special cases. The first of these is the case of no expansion at all, the second that of complete expansion without over-expansion. This latter gives the most economical operation from the hypothetical standpoint, because no work of expansion has been left unaccomplished and at the same time no negative work has been introduced by over-eiq>ansion. A B III. pr.^ ■ • •1 D^ )k. pr.) E H_ \/ Fig. 58. — ^First Special Case of Cycles I and II. Expansion = zero. Full Stroke Admission. First Special Case. If there is no expansion, together with the above assump- tion of no clearance, the diagram takes the form (Fig. 58), and Tr=144Z>[(in.pr.)-(bk.pr.)] (243) m.e.p. = (in.pr.) — (bk.pr.) . . (244) Work per cu.ft. supplied = 144[(in.pr.) — (bk.pr.)] (245) Cu.ft. supplied per hr. per I.H.P.=7; 13,750 (in.pr.) — (bk.pr.) (246) Lbs. supplied per hr. per I.H.P. = 77 13,7505i (in.pr.) — (bk.pr.) (247) 202 ENGINEERING THERMODYNAMICS Second Special Case. When the expansion is complete without over-expan- sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) =^ (bk.pr.), hence Ry=Rp=^ 7 hlr \ ~ 7' ^^* value of cul-off^ Z, is known as best cutoff, as ii is thai which uses all the available energy of the fluid by expansion. »r« I44D(in.pr.) ^5&^= 144Z) (bk.pr.) log. Rp. JtCp , \ n \ 10g« Rp (m.e.p.) « (m.pr.) — ^ — Work per cuit. supplied = 144(in.pr.) log. Rp. . (248) (249) (250) A B(ln pr) \ • \ ■ \ s, \ N \ V V ^ ^-^ E , D(i • H 1 If • V >k.pr) Fig. 59. — Second Special Case of Cycles I and II. Complete Expansion Without Over Expansion Case of ^eet Cut-off. 13,750 Cu.ft. supplied per hr. per I.H.P. = 7r v , r» • ^^ ^ ^ (in.pr.) loge iZp Lbs. supplied per hr. per I.H.P. = ;; — - . . ^ ^ . *^*^ ^ ^ (m.pr.) log, iZp (251) (252) Bzample 1. Method of calculating diagrams. Fig. 57 and Fig. 59. Assumed data for Fig. 57. Pa *Py =90 lbs. per sq.in. abs. Va = F« =0 cu.ft. P^ =P« = 14 lbs. per sq.in. abs. yc'^Vd- 13.5 cuit. Fft=6cu.ft. WOEK OF PISTON ENGINES 203 To obtain point C: Vb 6 Pc "^Pb X— =90 Xttz -40 lbs. per sq-in. abs. Vc lo.o Assumed data for Fig. 59. Pa ''Pit =90 lbs. per sq-in. Va = Ve =0 cuit. P^ =.p, B 14 lbs. per sq.in. Vd = 13.5 cu.ft. To obtain point B: Pd 14 Vi, = Vd X^ = 13.5 X;^ =2.1 cu.ft. Example 2* A simple double-acting engine admits steam at 100 per square inch absolute for \ stroke, allows it to expand to the end of the stroke and then exhausts it against a back pressure of 5 lbs. per square inch absolute. If the engine has no clearance, a 7 x9-in. cylinder and runs at 300 R.P.M., what is the horse-power and steam consumption when steam is expanding according to the logarithmic law? Note: 1 cuit. steam at 100 lbs. per square inch absolute weighs .2258 lb. From Eq. (237ft), m.e.p. = (m.pr.) ( ^ — I - (bk.pr.) -100 ^^ +log. ^) -5 ,100(l±iJgg> -5 -54.7: (m.e.p.)Lon 54.7 X. 75x38.5x600 33,000 33,000 or directly from Eq. (242) Z)n(m.e.p.), I.H.P. - 229.2 .2X600X54.7 229.2 Tu X TTTT» 13,750 8i Lbs. steam per I.H.P. = — x =28, m.e.p. Rv 13750 .2258 _ ,„ X--7-= 14.13. 54.7 Therefore, steam per hour used by engine = 14.15x28 =396 lbs. Prob. 1. A steam engine has a cylinder 12x18 ins. with no clearance. It runs at 200 R.P.M. and is double-acting. If the steam pressure be fixed at 100 lbs. per square inch absolute, and the back pressure at 10 lbs. per square inch abs., show how the horse-power and steam consumption will vary as cut-off increases. Take cut-off from i to f by eighths. Plot 204 ENGINEERING THERMODYNAMICS Prob. 2. Two engines of the same size and design as above are running on a steam pressure of 100 lbs. per square inch absolute, but one exhausts through a long pipe to the atmosphere, the total back pressure being 20 lbs. per square inch absolute, while the other exhausts into a condenser in which the pressure is but 3 Ibe. per square inch absolute. If the cut-ofif is in each case f , how will the I.H.P. and steam used in the two cases vary? Prob. 3. By finding the water rate and the horse-power in the two following cases, show the saving in steam and loss in power due to using steam expansively. A pump having a cylinder 9 X 12 ins. admits steam full stroke, while an engine of same size admits it but i of the stroke; both run at the same speed and have the same back pressure. Prob. 4. Steam from a 12x24 in. cylinder is exhausted at atmospheric pressure (15 lbs. per square inch absolute) into a tank, from which a second engine takes steam. Neither engine has clearance. The first engine receives steam at 100 lbs. per square inch absolute and the cut-off is such as to give complete expansion. The second engine exhausts into a 24 in« vacuum and its cut-off is such that complete expansion occurs in its cylinder. Also the cylinder voliune up to cut-off equals that of the first cylinder at exhaust. If the stroke is the same in both engines and the speed of each is 200 R.P.M., what is the diameter of the larger cylinder, the total horse-power developed, the total steam used, and the work per cubic foot of steam admitted to the first cylinder, the water rate of each engine and the total horse-power derived from each pound of steam? Prob. 6. The steam pressure for a given eng^e is changed from 80 lbs. per square inch gage to 120 lbs. per square inch gage. If the engine is 12x16 ins., running 250 R.P.M. with a fixed cut-off of 25 per cent and no clearance, the back pressure being 15 lbs. per square inch absolute, what will be the horse-power and the water rate in each case? Note: 1 cuit. of steam at 80 and 120 lbs. weighs .216 and .3 lb. respectively. Prob. 6. By trial, find how much the cut-off should have been shortened to keep the H.P. constant when the pressure was increased and what effect this would have had on the water rate. Prob. 7. A certain tjrpe of automobile engine uses steam at 600 lbs. per square inch absolute pressure. The exhaust is to atmosphere. For a cut-off of i and no clearance, what would be the water rate? Note: for 600 lbs. ^i =1.32. Prob. 8. Engines are governed by throttling the initial pressure or shortening the cut-off. The following cases show the effect of light load on economy. Both engines, 12x18 ins., running at 200 R.P.M., with 125 lbs. per square inch absolute. Initial pressure and back pressure of 10 lbs. per square inch absolute. The load is sufficient to require full steam pressure at 5 cut-off for each engine. Load drops to a point where the throttle engine requires but 50 lbs. per square inch absolute initial pressure with the cut-off still fixed at i. What is the original load and water rate, and new load and water-rate for each engine? Note: d for 125 lbs. absolute = .279 and for 50 lbs. =.117 lb. Prob. 9. The guarantee for a simple engine 18x24 ins., running at 200 R.P.M., states that the I.W.R. when cut-off is i will not exceed 15 lbs. if the initial pressure be 100 lbs. per square inch gage, and back pressure 5 lbs. per square inch absolute. If engine has no clearance, see if this would be possible. 4. Work of Expansive Fluid in Single Cylinder without Clearance. Erpo- nential Expansion Cycle 11. Mean Effective Pressure, Horse-power and WORK OF PISTON ENGINES 205 Constunptioii of Simple Engines. Referring to the diagram^ Fig. 57, the work is given by the same areas as for Cycle I, but its algebraic expression is different because s is greater than 1 and an exponential expansion results on integration instead of a logarithmic one. In general, from Eq. (13a), Section 7, Chapter I, Tr=P,F6+^^[(^y"'-l]-P.7, (253) Putting this in terms of initial conditions by the relations Also there results VJt b^V* p„v> bVb ir=p»7»+ .-_:'j^^"-, iRv*-'- 1) -PaV^v 1 =P»7» H [ -PaViRv. («-l)i2v^x(i2v--l) = 1442) \z (in.pr .) ^^j— - (bk.pr.)l which is the general equation for work of this cycle. Dividing by Vt, the volume of fluid supplied, Work per cu .ft. supplied = P^ ( -—7 — / .^vp ,_i j — PaRv (a) (a) (b) (c) . (254) = 144(in.pr.)(-^-^i-^— (&) . . (255) Similarly, the mean effective pressure results from dividing the work by the displacement, Vd=VJiv or M.E.P. = -^ ( r— 7 TTjy ,_i ) ~Pd Rv\s—l (a— l)ftv' V , . (in.pr.)Y s 1 \ /ui \ = Z(m.pr.) { ^~_^ ) - (bk.pr.) (a) (b) (c) . (256) 206 ENGINEERING TELERMODYNAMICS First special case of no expansion^ when i^F^ly results the same diagram as in the previous section, Fig. 58, and exactly the same set of formulas. Second special case when the expansion is complete without over-expansion, is again represented by Fig. 59 and for it Whence or Work for complete expansion is W=PtVt *^tO"57^V (a) ,,^(in^ _^(,__L^>) (^ • . . (257) which is the general equation for the work of Fj or I p- 1 cubic feet of fluid when the economy is best or for best cut-ofif. The work per cubic foot of fluid supplied for this case of complete expansion gives the maximum value for Eq. (255) and is obtained by dividing Ek]. (257) . Max. work per cu.ft. supplied =Pft-3Y(l""p;7^i ) = 144(m.pr.)^(l-^.) (a) (b) , ..(258) which is the general equation for maximum work per cubic foot of fluid supplied. The expression for mean effective pressure becomes for this case of best cut-off, M.E.P.= or, Rv s ( -iV Rv'-^j , X _ (in.pr.) s /, \_\ (m.e.p.)^ r;- jri(^l-:R;rrij (a) (6) . . . (259) It is convenient to note that in using Eqs. (257), (258) and (259) it may be desirable to evaluate them without first finding Rv- Since ^"^rrw""^'*' WORK OF PISTON ENGINES 207 this substitution may be made, and Ry ^ Rp » • Bzample. Compare the horse-power and the steam conswnption of a 9x12 in. simple double-acting engine with no clearance and running at 250 R.P.M. when initial pressure is 100 lbs. per square inch absolute and cut-off is }, if (a) steam remains diy and saturated throughout expansion, (6) remains superheated throughout expansion, and (c) if ori^nally dry and suffers adiabatic expansion. Back pressure is 10 lbs. per square inch absolute. <-'-'-^*fe-(:rik^)-<^->- 100/1.0646 4 \.0646 .0646x4 For case (o) «- 1.0646 and (m.e.p.) =-7- 1 -7^7^- r^. a ^.064e ) -10-48.6; " (6) » - 1.3 and (m.e.p.) -^ (^ ~T^) ~ ^° "^"^' " (c)«-l.lll and (m^.p.)-^(^- ^^^^^.i„ ) -10-47^. iJl.r . — (m.e,p,)ljan « ^^ — .yoo m.e.p. A I JI.P. for case (a) -46.9, " (b) -42.0, " " (c)-45.8. From Eq. (240), lbs. steam per hour per I.H.P.-» * X-^ m.e.p. Rv 1? / X X 1. AOf^ 13,750 8i .'. For case (a) steam per hr. -46.9 X '^ X-r ; - 4 4o.o 4 " (b) steam per hr. -42 X-^^ X-r; 4o.o 4 a/\_x 1. Af a 13,750 8i (c) steam per hr. -45.8 X~-=- X-7. 47.5 4 Prob. 1. On starting a locomotive steam is admitted full stroke, while in running the valve gear is arranged for f cut-off. If the engine were 18 x30 ins., initial pressure 150 lbs. per square inch absolute, back pressure 15 lbs. per square inch absolute, what would be the difference in horse-power with the gear in normal running position and in the starting position for a speed of 20 miles per hour with 6-ft. driving wheels? Con- sider the steam to be originally dry and expanding adiabatically. What would be the difference in steam per horse-power hour for the two cases and the difference in total steam? Clearance neglected. Prob. 2. Consider a boiler horse-power to be 30 lbs. of steam per hour; what must be the horse-power of a boiler to supply the following engine? Steam is supplied in a super- 208 ENGINEERING THERMODYNAMICS heated state and remains so throughout expansion. Initial density of steam « .21 lbs. per cubic foot. Engine is 12x20 ins., double-acting, 200 R.P.M., no clearance, initial pressure 125 lbs. per square inch absolute, back pressure a vacuum of 26 ins. of mercury. Cut-ofif at maximum load J, no load, A. What per cent of rating o^ boiler will be required by the engine at no load? Prob. 3. While an engine driving a generator is running, a short circuit occurs putting full load on en^ne, requiring a J cut-off. A moment later the circuit-breaker opens and only the friction load remains, requiring a cut-off of only -fj. The engine being two-cylinder, double-acting, simple, 12 X 18 ins., running at 300 R.P.M., and having no clearance, what will be the rate at which it uses steam just before and just after circuit-breaker opens if the steam supplied is at 125 lbs. per square inch absolute and is just dry, becoming wet on expanding, and back pressure is 3 lbs. per square inch absolute? Prob. 4. A pmnping engine has two double-acting steam cylinders each 9x12 ins. and a fixed cut-off of ). It runs at 60 R.P.M. on 80 lbs. per square inch absolute steam pressure and atmospheric exhaust. Cylinder is jacketed so that steam stays dry throughout its expansion. How much steam will it use per hour? Neglect clearance. Prob. 6. If an engine 10x14 ins. and running 250 R.P.M. has such a cut-off that complete expansion occurs for 90 lbs. per square inch absolute initial pressure, and at atmospheric (15 lbs. absolute) exhaust, what will be the horse-power and steam used per hour, steam being superheated at all times, and what would be the value for the horse- power and steam used if full stroke admission occurred? Prob. 6. The steam consumption of an engine working under constant load is better than that of a similar one working under variable load. For a 16 X24 ins. engine running at 250 R.P.M. on wet steam of 125 lbs. per square inch absolute and atmospheric exhaust, find the horse-power and steam used per horse-power per hour for best con- dition and by taking two hghter and three heavier loads, show by a curve how steam used per horse-power per hour will vary. Prob. 7. For driving a shop a two cylinder single-acting engine, 6x6 ins., running at 430 R.P.M., is used. The cut-off is fixed at J and intitial pressure varied to control speed. Plot a curve between horse-power and weight of steam per hour per horse- power for 20, 40, 60, 80, 100, 120 lbs. per square inch absolute initial pressure and atmospheric exhaust. Steam constantly dry. Clearance zero. Note: ^i for above pressures equals .05, .095, .139, .183, .226, and .268 lbs. per cubic foot, respectively. Prob. 8. Taking the loads found in Prob. 7, find what cut-off would be required to cause the engine to run at rated speed for each load if the initial steam pressure were 100 lbs. per square inch absolute, and the back pressure atmosphere, and a plot curve between horse-power and steam used per horse-power hour for this case. Prob. 9. For working a mine hoist a two-cylinder, double-acting engine is used in which compressed air is admitted f stroke at 125 lbs. per square inch absolute and then allowed to expand adiabatically and exhaust to atmosphere. If the cylinders are 18 x24 ins. and speed is 150 R.P.M., find the horse-power and cubic feet of high pressure air needed per minute. 6. Work of Expansive Fluid in Single Cylinder with Clearance. Loga- rithmic Expansion and Compression; Cycle m. Mean Eflfective Pressure, Horse-power, and Consumption of Simple Engines. As in previous cycles, the net work of the cycle is equal to the algebraic sum of the positive work WORK OF PISTON ENGINES 209 done on the forward stroke and the negative work on the return stroke. By areas, Fig. 60, this is Work BxeA^JABN+NBCW-WDEO-OEFJ. Expressed in terms of diagram points this becomes 1F= P*(n-Fa)+P.F, log, :^ -Pa{V,-V,)-PeVel0ge] f ) (260) P CD) f apply yolun 7r\ e-, — B [ la.pr. n J 1 ' f r A cu^ ^1 i \ \ S \ \ \ S M - IF \ ^ Wl n C(. i\ 1 \ 1 > 1 V G- 1 -.1. \ 5L 1 \ ■ t. 1 r* --» w L/ m 3 1 N D K M «y-iJ V il.pr) Fig. 60. — ^Work of Expansive Fluid in Single Cylinder with Clearance. Logarithmic Expan- sion for Cycle III. Exponential for Cycle IV. Expressing this in terms of displacement, in cubic feet D; clearance as a frac- tion of displacement, c; cut-oif as a fraction of displacement, Z\ compres- sion as a fraction of displacement, X; initial pressure, in pounds per square inch (in.pr.), and exhaust or back pressure, in pounds per square inch (bk.pr): P5 = 144(in.pr.); Pi =144 (bk.pr.); (V»-F,)=ZZ); 7» D{,Z+c) Z+c' (F<,-7.) = (1-X)Z); 7»=Z)(Z+c); F.=I>(X+c). V, D(X+c) X+c Vt Dc 210 ENGINEERING THERMODYNAMICS Whence Work in ft.-lbs. per cycle is W= 144i> [ (in.pr.) [z+ (Z+c) loge |t^] -(bk.pr.) [^(l«X)+(X+c)log.^] j - (261) From Eq. (261), the mean effective pressure, pounds per square inch, follows by dividing by 144D: (m.e.p.)= - (in.pr.) Z+(Z+c) loge^TT- • • (i»ean forward press.) -(bk.pr.)r(l-X)+(X+c)loga^t?. 1 (meanbk.pr.) (262) This is a general expression of very broad use in computing probable mean effective pressure for the steam engine with clearance and compression, or for other cases where it is practicable to assim[ie the logarithmic law to hold. Fig. 117, at the end of this chapter, will be found of assistance in evaluating this expression. Indicated horse-power, according to expressions ah^ady given, may be found by either of the following equations: y TT p _ (m.e,p,)Lan _ 144(m.e.p.)Dn _ (m.e.p.)Dn 33,000 33,000 ~ 229.2""' where L is stroke in feet, a is efifective area of piston, square inches, n is the number of cycles performed per minute and D the displacement, cubic feet. It might seem that the work per cubic foot of fluid supplied could be found by dividing Eq. (261) by the admission volume, but this would be true only when no steam is needed to build up the pressure from F to A. This is the case only when the clearance is zero or when com- pression begins soon enough to carry the point F up to point A, i.e., when by compression the pressure of the clearance fluid is raised to the initial pressure. It is evident that the fluid supplied may perform the two duties: first, building up the clearance pressure at constant or nearly constant volume, and second, filling the cylinder up to cut-oflf at constant pressure. To measure the steam supplied in terms of diagram quantities requires the fixing of the volume of live steam necessary to build up the pressure from F to A and adding it to the apparent admission volume (Vi,-'Va)- This can be done by producing WORK OF PISTON ENGINES 211 the compression line EF to the initial pressure Q, then LQ is the volume that the clearance steam would have at the initial pressure and QA the volume of live steam necessary to build up the pressure. The whole volume of steam admitted then is represented by QB instead of AB or by (Vb—Vq) instead^ of by {Vb'-V^)y and calling this the supply volume, But (Sup.Vol.) = (Fft-7,.) Pq (m.pr.) (m.pr.) Hence (Sup.Vol.)=Z)[(Z+c)-(X+c)|g^], .... (263) which is the cubic feet of fluid admitted at the initial pressure for the dis- placement of D cubic feet by the piston. Dividing by D there results (Sup Vol.) ^ _ (;i.+,)jbk-prj (264) D V / V ^(m.pr.) ^ ^ which is the ratio of admission volume to displacement or cubic feet of live steam admitted per cubic foot of displacement. Dividing the work done by the cubic feet of steam supplied gives the economy of the simple engine in terms of volumes, or W Work per cu.ft. of fluid supplied = (Sup.Vol.) (in.pr.)[z+(Z+c)lo& M -(bk.pr.)[(l-X)+(X+c)lo& ^''l =144 L Z+^i I £_J. (265) (Z+c)-(X+c)^^) ^ ^ ^ ^(m.pr.) It is more common to express economy of the engine in terms of the weight of steam used per hour per horse-power or the " water rate," which in more general terms may be called the consumption per hour per LH.P. Let di be the density or weight per cubic foot of fluid supplied, then the weight per cycle is (Sup.Vol.) 5i, and this weight is capable of performing W W foot-poirnds of work or (Sup.Vol.) ^i lbs. per minute will permit of «^«^^^^ horse-power. But (Sup.Vol.) di lbs. per mintue corresponds to 60 (Sup.Vol.) i\ lbs. per hour, whence the number of pounds per hour per horse-power is 60(Sup.Vol.)d^ 1^733,000 212 . ENGINEERING THERMODYNAMICS which is the pounds consumption per hour per I.H J*., whence n +• • IK u TXTi> 60X33,000(Sup.VoL) ^i ,^.. Consumption m lbs. per hr. per I.H.P. = — 2^ ^ ^— , . . . (266; which is the general expression for consumption in terms of the cubic feet of fluid admitted per cycle, ^i initial density, and the work per cycle. As work is the product of mean efiFective pressure in pounds per square foot, (M.E.P.,) and the displacement in cu.ft. or TF=(M.E.P.)D, or in termB of mean effective pressure pounds per square inch W = 144 (m.e.p.)D, the consumption may also be written n *• • lu u TXJT> 60X33,000 (Sup.Vol.)ai Consumption m lbs.,per hr. per I.H.P.= '. — *-r^r — - — x^rff ^m.e.p.^xy 13,750 (Sup.Vol.)ai (m.e.p.) D -^^Uz+c)-(X+c)^^]3^ (267) (m.e.p.) L '(m.pr.) J "^ which gives the water rate in terms of the mean effective pressure, cut-off, clearance, compression, initial and back pressures and initial steam density. It is sometimes more convenient to introduce the density of fluid at the back pressure <^2i which can be done by the relation (referring to the diagram). whence n^a ^'•^'or^bk.pr.) V, d2' (bk.pr.) (m.pr.) This on substitution gives Consumption in lbs., per hr. per I.H.P. -^[^^+'^'^-^''+'^''] (268) Since the step taken above of introducing <?£ has removed all pressure or volume ratios from the expression, Eq. (268) is general, and not dependent upon the logarithmic law. It gives the consumption in terms of mean effective pressure, cut-off, clearance, compression and the density of steam at initial and back pressure, which is of very common use. It cannot be too strongly kept in mind thai all the preceding is true only when no steam forms from moisture water during expansion or compression or no steam condenses, which assumption is known to he untrue. These formulae are, there- fore, to be considered as merely convenient approximations, although they WORK OF PISTON ENGINES 213 Eire almost universally used in daily practice. (See the end of this chapter for diagrams by which the solution of this expression is facilitated.; Special Cases. First, no expansion and no compression would result in :• 61. For it Tr= 144Z>[(in.pr.) - (bk.pr.)], (m.e.p.) = (in.pr.) — (bk.pr.) (269) (270^ pf LL 311 oliime 1 '*»«» — ' B i\n^T%T.\ 7 i i "1 \ 1 N . D Cbk.pr.) _ r 1 —tr* K _ p- "br- — ► t^ V / \ 1/ Fig. 61. — ^First Special Case of Cycles III and IV. Expansion and Compression both Zero, but Clearance Finite. The voliune of fluid supplied per cycle is QB, or from Eq. (263) it is (Sup.Vo1.)=d[i+c-c^^^^^^] (271) Consumptioninlbs. perhr. per I.H.P.= 7; v ' .iT- A 1+c— c,. -*^-/- ^ (m.pr.) — (bk.pr.) L (m.pr.) J (; (272) or in terms of initial and final densities. 13 750 Consumption in lbs. per hr. per I.H. P. = ,. \^/^ki; — \[(l+c)^i— C(?2] (273) The second special case is thai of complete expansion and compression^ as indicated in Fig. 62. Complete expansion provides that the pressure at the 214 ENGINEERING THERMODYNAMICS end of expansion be equal to the back pressure, and complete compression that the final compression pressure be equal to the initial pressure. Here V, V, 1+c X+c (in.pr.) F» r. Z+c c (bk.pr.)' and hence yr-^rSr.-' ^"^^ (^' ^.)=z>[i+c- /in.pr.y \bk.pr./J and (y,-7^)=ZD. P L Aj;^-ZD- 5b ^^^^^"' Cln.pr.) i . eD i A \ i i \ \ ■ V 1 1 1 V X, \ 1 1 1 1 ^S ^^ G _! — A . -P-^ Q>k.Dr^ S-^ 1 ——— t)- — ~^ •-W K Is i" W Fig. 62. — Second Special Case of Cycles III and IV. Perfect Expansion and Perfect Compression with Clearance. Hence by substitution I I (bly)r.)J^_ (in.pr.) "(bk.pr.) ' from which ZD ^-<'+«) w$- (274) Again, _ V,- Va [ Vhk.pr./ J _ I /jn.prA _ '] . . (275) WORK OF PISTON ENGINES 215 Eq. (274) gives the cut-oflF as a fraction of the displacement necessary to give complete expansion, while Eq. (275) gives the compression as a fraction of dis- placement to give complete compression, both in terms of clearance, initial pressure and back pressure, provided the logarithmic law applies to expansion and compression. Substitution of the values given above in Eq. (261) gives, after simplifi- cation, F=144Z)[(l+c)(bk.pr.)-c(in.pr.)]lo&|^^j. . . (276) (m.e.p.) = [(l+c)(bk.pr.)-c(in.pr.)]log.|™^^ (277) In this case the volume supplied is exactly equal to that represented by the admission line AB, and is equal to (Sup.Vol.)=ZD (278) Hence, the consumption, in pounds fluid per hpur per I.H.P. in terms of initial density, is Consumption in lbs. per hr. per I.H.P.=-- — r Zdi, but (bk^) ^ (m.pr.) m .e.p. ,. . r,. , .(bk.pr.) "I, (in.pr. ) .. v, (in.pr.) hence 13,750^ Consumption in lbs. fluid per hr. per I.H.P.= — ,. r-. . . (279) This last equation is interesting in that it shows the consumption (or water rate, if it is a steam engixie) is independent of clearance, and dependent only upon initial density, and on the initial and final pressures. An expression may also be easily derived for the consumption in terms of iiiitial and final density, but due to its limited use, will not be introduced here. Example 1. Method of calculating Diagrams Fig. 60 and Fig. 62. Assumed data for Fig. 60: Pq ^Pa «p6 =90 lbs. per square inch abs. Fa = V/ = .6 cu.ft. Pg «Pe =Pif = 14 lbs. per square inch abs. Vd'^Ve- 135 cu.ft. P/ =60 lbs. per square inch abs. F> =6 cu.ft. 216 ENGINEERING THERMODYNAMICS To obtain point C: To obtain point E: To obtain point Q: Vh 6 P« =Pft Xtt =90 X— -T =40 lbs. per 8q.in. abs. Ve 16.0 F, = y;.x^=.5X7^=1.78 cu.ft. P, KA F, = 7/X^«.6x^«.278cu.ft. Intermediate points from B to C and J^ to Q are found by assuming volumes and computing the corresponding pressures by relation PxVx==PbVb or PxVx=-PeVc^ Clearance is ^ — -i^s- =^ =3.8 per cent, Cut-ofif is -=7^ — ~ =-^ =42.3 per cent, Yd — r o lo Compression is -jz — =7- =-7^ =9.9 per cent. Assumed data for Fig. 62. p^ =P5 «90 lbs. per square inch absolute. Va = .5 cu.ft. Pg==Pc- 14 lbs. per square inch absolute. Va = 13.5 cu.ft. To obtain point B: To obtain point E: F^ = 7c§ =13.5X^=2.11 cu.ft. jt ft » yu p 00 F. = Fa X^ = .5 Xt. =3.2 cu.ft. x^« 14 Intermediate points from B to C and from A to £? are to be found by assuming . various volumes and finding the corresponding pressures from relation PzVx =PaVa or PxVx^PbVi,, Example. 2. What will be the horse-power of, and steam used per hour by the following engine: (a) cut-off 50 per cent, compression 30 per cent, (6) complete expansion and compression, (c) no expansion or compression. Cylinder, 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial pressure 85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, and logarithmic expansion and compression. WORK OF PISTON ENGINES 217 Note: d for 85 lbs. gage = .23, ^i for 15 lbs. absolute = .038 cu.ft. (a) From Eq. (262) (m.e.p.) =(in.pr.)[z -[-(Z+c) log, |~^1 -(bk.pr.) f'(l-X)+(Z+c)log,^j, = 10o[.5+(.6+.07)log,|±|^]-15[(l-.3)-|-(.3+.07) log, — J^] =86-20=66 lbs. sq.in. (m.e. p.) Lan _66 Xl.5^X llil X400 _ From Eq. (267) steam per hour per I.H.P. in pounds is ,750 [,„ , ,,^ ,/bk.pr.\l ^ r (Z+c)-(X-|-c) -*^ 8i, .e.p.)L \m.pr./J = -^^[(.5+.07) -(.3+.07) X^l X.23=25 lbs. Hence total steam per hour =25x135 =3380 lbs. (6) From Eq. (277) (m.e.p.) =[(1 +c)(bk. pr.) -c(in. pr.)] log, f g^~ )» =[(1 +.07) X15-.07 XlOO] log, 6.67 =17.2 lbs. sq.in. 13 (m ^„^ 17.2X1.5X113.1X400 ^^ , I.H.P. = ^-^ =35.4. From Eq. (279) a. TTn> I, 13,7505i 13,750X.23 _. Steam per I.H.P. per hour = /. \ = ^,^^-/^-^ =16.6, »»7 ,. ^, /in.pr.\ 100 XL! Total steam per hour = 16.6x35.4 =588 lbs. (c) From Eq. (270) (m.e.p.) =(in.pr.)- (bk.pr.) =100-15=85 lbs. sq.in. . I HP 85X1.5 X13 .1X400 •• ^•^•^•~ 33,000 -^^^•^• From Eq. (273) Steam per I.H.P. per hour = — ^ r[(l +c)di -c^a] ^^[1 .07x .23 - .07x .038] =35.4. ^m.e.p.^ 00 Total steam per hour = 174.5x35.4 =6100 lbs. Pr6b. 1. What will be the horse-power and water rate of a 9x12 in. simple engine having 5 per cent clearance, running at 250 R.P.M. on 100 lbs. per square inch abso- 218 ENGINEERING THERMODYNAMICS lute initial pressure and 5 lbs. per square inch absolute back pressure when the cut-off is I, \f and If expansion follows the logarithmic law, and there is no compression? Note : B for 100 lbs. absolute = .23, 5 for 5 lbs. absolute = .014. Prob. 2. Will a pump with a cylinder 10 Xl5 ins. and 10 per cent clearance give the same horse-power and have the same water rate as a piunp with cylinders of the same size but with 20 per cent clearance, both taking steam full stroke? Solve for a case of 125 lbs. per square inch absolute initial pressure, atmospheric exhaust and a speed of 50 double strokes. No compression. Note: 5 for 125 lbs. absolute = .283, 8 for 15 lbs. absolute = .039. Prob. 3. Solve the above problem for an engine of the same size, using steam expan- sively when the cut-off is J and R.P.M. 200, steam and exhaust pressure as in Prob. 2 and compression zero. Prob. 4. Two engines, each 9x15 ins., are running on an initial pressure of 90 lbs. per square Jnch absolute, and a back pressure of one atmosphere. One has no clearance, the other 8 per cent. Each is cutting off so that complete expansion occurs. The speed of each is 200 and neither has any compression. What will be the horse-power and water rate? Note: 8 for 90 lbs. -.24, 8 for 15 lbs. =.039. Prob. 6. By finding the horse-power and water rate of a 12x18 in. double-acting engine with 8 per cent clearance, running at 150 R.P.M. on an initial steam pressure of 90 lbs. per square inch absolute and atmosphere exhaust for a fixed cut-off of } and variable compression from to the point where the pressure at the end of compression is equal to 125 per cent of the initial pressure, plot the curves between compression and horse-power, and compression and water rate to show the effect of compression on the other two. Note: 8 for 90 lbs. = .21, 8 for 15 lbs. = .039. Prob. 6. A steam engine is running at such a load that the cut-off has to be f at a speed of 150 R.P.M. The engine is 14 x20 ins. and has no clearance. Initial pressure 100 lbs. per square inch absolute and back pressure 5 lbs. per square inch absolute. What would be the cut-off of an engine of the same dimensions but with 10 per cent clearance under similar conditions? Prob. 7. The steam pressure is 100 lbs. per square inch gage and the back pressure is 26 ins. of mercury vacuum when the barometer is 30 ins. For a 16x22 in.engine with 6 per cent clearance running at 125 R.P.M., cut-off at § and 30 per cent compression, what will be the horse-power and- the water rate? Should the steam pressure be doubled what would be the horse-power and the water rate? If it should be halved? Note: 8 for 100 lbs. gage = .2017, 8 for 26 ins. Hg.=.0058. Prob. 8. While an 18 x24 in. simple engine with 4 per cent clearance at speed of 150 R.P.M. is running with a i cut-off and a compression of ^ on a steam pressure of 125 lbs. per square inch gage, and a vacuum of 28 ins. of mercury, the condenser fails and the back pressure rises to 17 lbs. per square inch absolute. What will be the change in the horse-power and water rate if all other factors stay constant? What would the new cut-off have to be to keep the engine running at the same horse-power and what would be the water rate with this cut-off? Note: 8 for 125 lbs. gage = .315, 8 for 28 in. Hg. = .0029, 8 for 17 lbs. absolute = .0435. Prob. 9. Under normal load an engine has a cut-off of |, while imder light load the cut-off is but A. What per cent of the steam used at normal load will be used WORK OF PISTON ENGINES 219 at light load for the following data? Cylinder 10x18 ins.; 7 per cent clearance; 200 R.P.M.; initial pressure 120 lbs. per square inch gage; back pressure 2 lbs. per square inch absolute; compression at normal load 5 per cent; at light load 25 per cent. Note : B for 120 lbs. gage = .304, i for 2 lbs. absolute = .0058. 6. Work of Expansive Fluid in Single Cylinder with Clearance ; Exponential Expansion and Compression, Cycle IV. Mean Effective Pressure, Horse- power and Consumption of Simple Engines. As pointed out in several places, the logarithmic expansion of steam only approximates the truth in real engines and is the result of no particular logical or physically definable hypothesis as to the condition of the fluid, moreover its equations are of little or no value for compressed air or other gases used in engine cylinders. All expansions that can be defined by conditions of physical state or condition of heat, including the adiabatic, are expressible approximately or exactly by a definite value of 8, not unity, in the expression PF* = const. All these cases can then be treated in a group and expressions for work and mean effective pressure foimd for a general value of «, for which particular values belonging to, or following from any physical hypothesis can be substituted. The area under such expansion curves is given by Eq. (13) Chapter I, which applied to the work diagram, Fig. 60, in the same manner as was done for logarithmic expansion, gives the net work: W=P,{V,-Va)+j~\l-(p) 1 (area JABCWJ) «-i -Pd(F, - Ve) -^' [ (^;) - 1] (area WDEFJW) (280) Introducing the symbols, Pft = 144(in.pr.), Ptf = 144(bk.pr.), (F»-Fa)=ZA (Ftf-F.) = Z)(l-X), Ve=D(X+c). /VA_Z+c \VJ~l+c /V,\_X±c \VJ~ c • W=14AD -(M:.p..,[(I-X,+|t'[(^-±-')-'-.]] (281) Eq. (281) pves the work in foot-pounds for D cubic feet of displacement in a cylinder having any clearance c, cut-off Z, and compression X, between two 220 ENGINEERING THERMODYNAMICS pressures, and when the law of expansion is PV* = const, and s anything except unity, but constant. The mean efifective pressure, pounds per square inch, is obtained by divid- ing the expression for work by 144D, giving (m.e.p.) = (in.pr.) J Z-\ — — r- 1 — ( ^TT ) [ (mean for'd pr.) — (bk.pr.) (l-X)+^'[(^y '-l]j (mean bk.pr.) , (282) which is the general expression for mean effective pressure for this cycle. It was pointed out in Section (5) that the cubic feet of fluid admitted at the initial pressure was not represented by AB, Fig. 60, but by QB, and the same is true for this case, so that the ^Sup.Vol.) = 76-F«. But when the expansion and compression laws have the form PV*=^c .K.(g)--C(X+«)(g^-)^ Whence 1 (Sup.Vol.)=Z)[(Z+c)-(X+c)(g^-y] (283) Eq. (283) gives for these cases what was given by Eq. (263) for the logarithmic law, the cubic feet of fluid supplied at the initial pressure for the displacement of D cubic feet in terms of cut-off, clearance, compression and the pressures. From this, by division there is found the volume of high pressure fluid per cu.ft. of displacement The consumption is given by the general expression already derived, Eq. (34), from which is obtained. Consumption lbs. per hr. per I.H.P. ^13J50r ^ ^)/bk^.\h ^285) (m.e.p.) L ^\m.pr. / J WORK OF PISTON ENGINES 221 Eq. (285) ^ves the water rate or gas consumption in terms of mean effective pressure, initial and back pressure, cut-off, clearance, compression and initial fluid density. Introducing the density at the back pressure by the relation, X g r (J = ±e y e ) 1 F,*52"Vbk:pF:/ •' _1_ _ / in.pr. \ . -^^bk:^.; ' there results Consumption lbs. per hr. per I.H.P. = iM^ \ (z+c) h - (X+c) 82I , . (286) ^m.e.p.; [_ J which is identical with Eq. (268) and is, as previously observed, a general expres- sion, no matter what the laws of expansion and compression, in terms of mean effective pressure, cut-off, clearance, compression and the initial and final steam density. The first special case of full admission, no compression mig)it at first thought appear to be the same as in the preceding section, where the logarithmic law was assumed to hold, and so it is as regards work and mean effective pressure, Eq. (269) and (270), but referring to Fig. 61 it will be seen that since the expo- nential law is now assumed instead of the logarithmic, the point Q will be dif- ferently located (nearer to A than it was previously if s is greater than 1), and hence the supply volume QB is changed, and its new value is 1 (Sup.Vol.)=Z)[l+c-c(^)'] (287) Hence, consumption stated in terms of initial density of the fluid §1, is Consumption lbs. per hr. per I.H.P. 13,750 r, . /bk.pr.\i^ 1 ^ ,^^^^ =7: r- Sn ^ll+c-cli — ^) 81. ... (288) (m.pr.) — (bk.pr.) [ \m.pr. /J ^ ^ Stated in terms of initial and final densities, the expression is as before, Eq. (273). The second special case, complete expansion and compression is again repre- sented by Fig. 62. From the law of expansion it is evident that 222 ENGINEERING THERMODYNAMICS or stated in symbolic form, 1 Z>(Z+c)=Z)(l+c)(g^)', whence \m.pr. / Again referring to Fig. 62, Y^ ^*"^ *^- Vbk.pr./ Vc-Va D Whence ^-'[(fe^)'-'] <^> Eq. (289) gives the cut-off as a fraction of displacement necessary to g^ve complete expansion, and (290), the compression fraction to give complete com- pression, both in terms of clearance, initial and back pressures, and the exponent s, in the equation of the expansion or compression line, PF*= const. The work of the cycle becomes for this special case, by substitution in Eq. (281), W-.UD (u,.p,.,.4i[(i+<^)'-][i-(^)"']. . m) and the mean effective pressure, lbs. per sq.in., is The volume of fluid supplied is, (Sup.Vol.)=ZD (293) hence 13 750 Consumption, lbs. per hr. per I.H.P.= — - — Z8i, Xfl .CD. but iM^f -«] s— IL^ \m.pr. / -J L \m.pr. / J 1 ( •■'P'•)7-^['-t^)■■']. WORK OF PISTON ENGINES 223 whence Consumption lbs. fluid per hr. per I.H.P. is, !^750XJj__^ (2^j the expression for smallest consumption (or water rate if steam) of fluid for the most economical hypothetical cycle, which may it be noticed, is again in- dependent of clearance. The expressions for work and mean effective pressure are not, however, independent of clearance, and hence, according to the hypothetical cycles here considered, it is proved that large clearance decreases the work capacity of a a cylinder of given size, but does not afifect the economy, provided complete expansion and compression are attained, a conclusion similar to that in regard to clearance effect on compressor capacity and economy. Whether the actual performance of gas or steam engines agrees with this conclusion based only on hypothetical reasoning, will be discussed later. Example 1. What will be the honse-power of and steam used per hour by the following engine: 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial pressure 85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, and expansion such that 8 ===1.3. (a) cut-oflf=50 per cent; compression =30 per cent; (6) complete expansion and compression; (c) no expansion or compression; Note: ^ for 85 lbs. gage =.18; for 15 lbs. absolute. = .03. (a) From Eq. (282) ,„.e.p,.<i..p,.,{.4!-;[.-(f2)-'] -»-){<' -«-f^l(^?)""-']} -10o{;i+f[,-(j|)']}-.6{.7+f[(f)*-,]}-59.81b.„.i„. _„_ 59.8X1.5X113.1X400 ^^^ LH.P. ^^-^ -122. From Eq. (286) Steam per hour per LH.P. . ^.^^^Lz+c)^,^{X+c)^,] =-f^[(.57)X.18-(.37)X.03] =20.9 lbs. (m.e.p.)L J 59.8 /. Steam per hour = 122 X20.9 =2560 lbs. 224 ENGINEERING THERMODYNAMICS (6) From Eq. (292) «-i („.e.p.,.a..p.,,-i;[a..,(^^)^-«][,-(^)^. _„_ 26.2X1.5X113.1X200 ^, I.H.P. = — — =64 3000 From Eq. (294) steam used I.H.P. per hour is, 13,750§i 13,750 X.18 «— 1 .3 |— « 16.5 lbs., hence total steam per hour = 16.5 X64 = 1060 lbs. (c) From Eq. (270) which holds for any value of s, m.e.p. =100 — 15 «85 lbs. sq.in. A Txrr* 85x1.5x113.1x400 ,_ ^ and I.H.P. = ^^-^ 174.5, From Eq. (288) steam per I.H.P. hour 1 1 _13,750Bir, /bk.pr.\«1 13,750x.l8r, ^^ ^^ /ISXOI ^^ ^ ,^ ^-r rh+^-^h— ^ = c"^ h+-07-.07xL-7^ =24.51h.<^.. (m.e.p.) L \m.pr. /J 85 L \100/ J and total steam per hour = 174.5X24.5 =2475 lbs. Prob. 1. Two simple engines, each 12 Xl8 ins., with 6 per cent clearance are operated at J cut-off and with 20 per cent compression. One is supplied with air at 80 Ihs. per square inch gage, and exhausts it to atmosphere; the other with initially dry steam which becomes wet on expansion and which also is exhausted to atmosphere. For a speed of 200 R.P.M. what is the horse-power of each engine and the cubic feet of stuff supplied per horse-power hour ? Prob. 2. A crank-and-flywheel two-cylinder, double-acting, pumping engine is supplied with dry steam and the expansion is such that it remains dry until exhaust. The cylinder size is 24x36 ins., cut-off to give perfect expansion, clearance 5 per cent, compression to give perfect compression, initial pressure 50 lbs. per square inch al>- solute. back pressure 5 lbs. per square inch absolute. What is the horse-power and water rate? What would be the horse-power and water rate of a full-stroke pump of the same size ahd clearance but having no compression, running on the same pressure range and quahty of steam. Note: S for 50 lbs. absolute =.12, § for 15 lbs. absolute = .0387. Prob. 3. Should the cylinder of the following engine be so provided that the steam was always kept dry, would there be any change in the horse-power developed as WORK OF PISTON ENGINES 225 compared with steam expanded adiabatically, and how much? Cylinder 20 x24 ins., initial pressure 125 lbs. per sq. in. gage, back pressure 26 ins. vacuum, standard barom- eter, clearance 6 per cent, cut-off f, compression 10 percent, and speed 125 R.P.M. Prob. 4. What will be the total steam used per hour by a 20x28-in. double-acting engine running at 150 R.P.M. if the initial pressure be 125 lbs. per square inch absolute, back pressure one atmosphere, clearance 8 per cent, compression zero, for.cut-off i, i, f , and i, if steam expands adiabatically and is originally dry and saturated? Note: 8 for 125 lbs. absolute = .283, B for 15 lbs. absolute = .0387. Prob. 6. An engine which is supplied with superheated steam is said to have an indicated water rate of 15 lbs. at } cut-off and one of 25 lbs. at i cutr-off. See if this is reasonable for the following conditions: engine is 15 X22 ins., 7 per cent clearance, no compression, initial pressure 100 lbs. per square inch gage., back pressure 28-in. vacuum, barometer 30 ins. and speed 180 R.P.M. Note: B for 100 lbs. gage -.262, 5 for 28 in. Hg = .0029. Prob. 6. The tank capacity of a locomotive is 4500 gals. The cylinders are 18x36 ins., initial pressure 200 lbs. per square inch gage, exhaust atmospheric, cut-off i, clearance 7 per cent, speed 200 R.P.M., no compression. The stream is dry at start and expansion adiabatic, how long will the water last if 40% condenses during admission? Note: 8 for 200 lbs. gage =.471, 8 for 15 lbs. absolute = .0387. Prob. 7. To drive a hoist, an air engine is used, the air being supplied for i stroke at 80 lbs. per square inch gage expanded adiabatically and exhausted to atmos- sphere. If the clearance is 8 per cent and there is no compression how many cubic feet of air per hour per horse-power will be needed? What, with complete compression? Prob. 8. A manufacturer rates his 44 x42-in. double-acting engine with a speed of 100 R.P.M. at 1000 H.P. when running non-condensing, initial pressure 70 lbs. per square inch gage and cut-off i. No clearance is mentioned and nothing said about manner of expansion. Assuming s = 1.0646, find on what clearance basis this rating is made. Prob. 9. The water supply of a town is supplied by a direct-acting non-condensing pump with two cyUnders, each 24x42 ins., with 10 per cent clearance, and no com- pression, initial pressure being 100 lbs. per square inch gage. What must be the size of the steam cylinder of a crank-and-flywheel pump with 6 per cent clearance to give the same horse-power on the same steam and exhaust pressures with a cut-off of J? Speed in each case to be 50 R.P.M. Note: B for 100 lbs. gage = .262, 5 for 15 lbs. abs. = .0387. 7. Action of Fluid in Multiple-ezpansion Cylinders. General Description of Structure and Processes. When steam, compressed air, or any other high pressure working fluid is caused to pass through more than one cylinder in series, so that the exhaust from the one is the supply for the next, the engine is, in general, a multiple-expansion engine, or more specifically, a compound when the operations are in two expansion stages, triple for three, and quadruple for four stages. It must be imderstood that while a compound engine is one in which the whole pressure-volimie change from initial to back pressure takes place in two stages, it does not necessarily follow that the machine is a two-, cylinder one, for the second stage of expansion may take place in two cylinders, in each of which, half of the steam is put through identical operations; this 226 ENGINEERING THERMODYNAMICS would make a three-cylinder compound. Similarly, triple-expansion engines, while they cannot have less than three may have four or five or six cylindenj Multiple expansions engine, most of which are compound, are of two classe? with respect to the treatment and pressure-voliune changes of the steam, first without receiver^ and second, mth receiver. A receiver is primarily a chamber large in proportion to cylinder volumes, placed between the high- and low-pres- sure cylinders of compounds or between any pair of cylinders in triple or quadruple engines, and its purpose is to provide a reservoir of fluid so that the exhaust from the higher into it, or the admission to the lower from it, will be accomplished without a material change of pressure, and this will be accom- plished as its volume is large in proportion to the charge of steam received by it or delivered from it. With a receiver of infinite size the exhaust line of a high-pressure cylinder discharging into it will be a constant-pressure line, as will also the admission line of the low-pressure cylinder. When, however, the receiver is of finite size high-pressure exhaust is equivalent to increasing the quantity of fluid in the receiver of fixed volume and must be accompanied by a rise of pressure except when a low-pressure cylinder may happen to be taking out fluid at the same rate and at the same time, which in practice never happens. As the receiver becomes smaller in proportion to the cylinders, the pressure in it will rise and fall more for each high-pressure exhaust and low- pressure admission with, of course, a constant average value. The greatest possible change of pressure during high-pressure exhaust and low-pressure admis- sion would occur when the receiver is of zero size, that is when there is none at all, in which case, of course, the high-and low-pressure pistons must have synchronous movement, both starting and stopping at the same time, but moving either in the same or opposite directions. When the pistons of the no-receiver compound engines move in the same direction at the same time, one end of the high- pressure cylinder must exhaust into the opposite end of the low; but with oppositely moving pistons, the exhaust from high will enter the same end of the low. It is plain that a real receiver of zero volume is impossible, as the connect- ing ports must have some volume and likewise that an infinite receiver is equally impracticable, so that any multiple-expansion real engine will have receivers of finite volume with corresponding pressure changes during the period when a receiver is in communication with a cylinder. The amount of these pressure changes will depend partly on the size of the receiver with respect to the cylin- ders, but also as well, on the relation between the periods of flow into receiver, by high-pressure exhaust and out of it, by low-pressure admission, which latter factor will be fixed largely by crank angles, and partly by the settings of the two valves, relations which are often extremely complicated. For the purpose of analysis it is desirable to treat the two limiting cases of no receiver and infinite receiver, because they yield formulas simple enough to be useful, while an exact simple solution of the general case is impossible. These simple expressions for hypothetical cases which are very valuable for estimates and approximations are generally close to truth for an actual engine especially if intelligently selected and used. WORK OF PISTON ENGINES 227 Receivers of steam engines may be simple tanks or temporary storage chambers or be fitted with coils or tubes to which live or high-pressure steam is supplied and which may heat up the lower pressure, partly expanded steam passing from cylinder to cylinder through the receiver. Such receivers are refiecUing receivers, and as noted, may heat up the engine steam or may evapo- rate any moisture it might contain. As a matter of fact there can be no heating of the steakn before all moisture is first evaporated, from which it appears that the action of such reheating receivers may be, and is quite complicated thermally, and a study of these conditions must be postponed till a thermal method of analysis is established. This will introduce no serious diflBculty, as such reheating receivers assist the thermal economy of the whole system but little and have little effect on engine power, likewise are now little used. Reheating of air or other gases, as well as preheating them before admission to the high-pressure cylinder is a necessary practice, when the supply pressure is high,' to prevent freezing of moisture by the gases, which get very cold in expansion if it be carried far. This is likewise, however, a thermal problem, not to be taken up till later. Multiple-expansion engines are built for greater economy than is possible in simple engines and the reasons are divisible into two classes, first mechanical, and second thermal. It has already been shown that by expansion, work is obtained in greater amounts as the expansion is greater, provided, of course, expansion below the back pressure is avoided, and as high initial and low back pressures permit essentially of most expansion, engines must be built capable of utilizing all that the steam or compressed gas may yield. If steam followed the logarithmic law of expansion, pressure falling inversely with volume increase, then steam of 150 lbs. per square inch absolute expanding to 1 lb. per square inch absolute would require enough ultimate cylinder space to allow whatever volume of steam was admitted up to cut-oflf to increase 150 times. This would involve a valve gear and cylinder structure capable of admitting Thr = -0067 of the cylinder volume. It is practically impossible to construct a valve that will accurately open and close in this necessarily short equivalent portion of the stroke. This, however, is not the worst handicap even mechanically, because actual cylinders cannot be made without some clearance, usually more than 2 per cent of the displacement and in order that any steam might be admitted at all, the clearance in the example would have to be less than .67 per cent of the total volume, which is quite impossible. These two mechanical or structural limitations, that of admission valve gear and that of clearance limits, supply the first argimient for multiple- expansion engines, the structure of which is capable of utilizing any amount of expansion that high boiler pressure and good condenser vacuum make available. For, if neglecting clearance, the low-pressure cylinder had ten times the volume of the high, then the full stroke admission of steam to the high followed by expansion in the low would give ten expansions, while admission to the high for ^ of its stroke would give 16 expansions in it, after which this final volume would increase in the low ten times, that is, to 160 times the original 228 ENGINEERING THERMODYNAMICS volume, and cylinder admission of ^ or 6.7 per cent is possible with ordinary valve gears, as is also an initial volimie of 6.7 per cent of a total cylinder volume, even with clearance which in reasonably la^-ge engines may be not over 2 per cent of the whole cylinder volume. It is evident that the higher the initial and the lower the back pressures the greater the expansion ratio will be for complete expansion, and as in steam practice boiler pressiu-es of 225 lbs. per square inch gage or approximately 240 lbs. per square inch absolute with vacuum back pressures as low as one or even half a pound per square inch are in use, it should be possible whether desirable or not, to expand to a final volume from 250 to 600 times the initial in round num- bers. This is, of course, quite impossible in simple engine cylinders, and as it easy with multiple expansion there is supplied another mechanical argument for staging. Sufficient expansion for practical purposes in locomotives and land engines under the usually variable load of industrial service is available for even these high pressures by compounding, but when the loads are about constant, as in waterworks pumping engines, and marine engines for ship propul.sion, triple expansion is used for pressures in excess of about 180 lbs. gage. Use of very high initial and very low back pressures will result in simple engines, in a possibility of great unbalanced forces on a piston, its rods, pins and crank, when acting on opposite sides, and a considerable fluctuation in tan- gential turning force at the crank pin. Compounding will always reduce the unbalanced force on a piston, and when carried out in cylinders each of which has a separate crank, permits of a very considerable improvement of turning effort. So that, not only does multiple expansion make it possible to utilize to the fullest extent the whole range of high initial and low back pressures, but it may result in a better force distribution in the engine, avoiding shocks, making unnecessary, excessively strong pistons, and rods and equalizing turn- ing effort so that the maximum and minimum tangential force do not depart too much from the mean. The second or thermal reason for bothering with mjiiltiple-expansion com- plications in the interest of steam economy is concerned with the prevention of steam loss by condensation and leakage. It does not need any elaborate analysis to show that low-pressure steam will be cooler than high-pressure steam and that expanding steam in a cylinder has a tendency to cool the cylinder and piston walls, certainly the inner skin at least, so that after expansion and exhaust they will be cooler than after admission; but as admission follows exhaust hot live steam will come into contact with cool walls and some will necessarily condense, the amount being smaller the less the original expansion; hence in any one cylinder of a multiple-expansion engine the condensation may be less than a simple engine with the same range of steam pressures and temperatures. Whether all the steam condensation during admission added together will equal that of the simple engine or not is another question. There is no doubt, however, that as the multiple expan- sion engine can expand usefully to greater degree than a simple engine, and so cause a lower temperature by expansion, that it has a greater chance to WORK OF PISTON ENGINES 229 reevaporate some of the water of initial condensation and so get some work out of the extra steam so evaporated, which in the simple engine might have remained as water, incapable of working until exhaust opened and lowered the pressure, when, of course, it could do no good. It is also clear that steam or compressed-air leakage in a siinple engine is a direct loss, whereas in a compjound high-pressure cylinder leakage has at least a chance to do some work in the low-pressure cylinder. The exact analysis of the thermal reasons for greater economy is compUcated and is largely concerned with a study of steam condensation and reevaporation, but the fact is, that multiple-expan- sion engines are capable of greater economy than simple. The thermal analysis must also consider the influence of the reheating receiver, the steam-jacketed working cylinder, and the use of superheated steam, their effects on the pos- sible work per poimd of steam and the corresponding quantity of heat expended to secure it, and for air and compressed gas the parallel treatment of pre- heating and reheating. To illustrate the action of steam in multiple-expansion engines some indi- cator cards are given for a few typical cases in Figs. 63 to 66, together with the combined diagrams of pressure-volume changes of the fluid in all cylinders to the same scale of pressures and volumes, which, of course, makes the diagram look quite different, as indicator cards are usually taken to the same base length, fixed by the reducing motion, and to different pressure scales, to get as large a height of diagram as the paper will permit. Fig. 63 shows fdur sets of cards taken from an engine of the compound no-receiver type, namely, a Vauclain compound locomotive. In this machine there are two cylinders, one high pressure and one low, on each side, the steam from the high pressure exhausting directly into the low-pressure cylinder so that the only receiver space is made up of the clearance and connecting passages between the cylinders. Starting with set A, the cards show a decreasing high pressure cut-off of 76 per cent in the case of set A to 54 per cent in the case of set D. The letters A, B, C and D refer in each case to admission, cut- off, release and compression, the use of primed letters denoting the low-pressure cylinder. In set A the high-pressure admission line AFB may be considered as made up of two parts, the part AF representing pressure rise at constant volimie, wh^ch is the admission of steam to the clearance space at dead center to raise the pressure from that at the end of compression to that of boiler pressure. From F to B admission occiured at constant pressure, steam filling the cylinder volume as the piston moved outward. At B cut-off or closure of the steam valve occurred and the steam in the cylinder expanded. At C, release or open- of the exhaust valve of the high-pressure cylinder occurred and the admission valve of the low-pressure cylinder opened, the steam dropping in pressure until the pressure in both high- and low-pressure clearance became equal, and then expanding in both cylinders, as the exhaust from the high and admission to the low occurred, the exhaust line CD of the high pressure and the admission line F'B' of the low pressure being identical except for the sUght pressure drop 230 ENGINEERING THEEM0DYNAM1C8 (A) (B) (C) (D) FiQ. 63. — Set of Indicator Cards from Vauclain Locomotive Illustrating the No-receivei Compomid Steam Engine. WORK OF PISTON ENGINES 231 in the passages between the high- and the low-pressure cylinders. At D the high-pressure exhaust valve closed and compression of the steam trapped in the high-pressure cylinder occurred to point A, thus closing the cycle. From point D' in the low-pressure cylinder, which corresponds to D in the high pressure, no more steam was admitted to the low-pressure cylinder. What steam there was in the low expanded to the point C when the exhaust valve opened and the pressure dropped to the back pressure and the steam was exhausted at nearly constant back pressure to Z)', when the exhaust valve closed and the steam trapped in the cylinder was compressed to A', at which point steam was again admitted and the cycle repeated. In set B the cycle of operation is exactly the same as in set A. In set C the cycle is the same as in A, but there are one or two points to be especially noted, as they are not present in set A. The admission line of the high-pres- sure cylinder is not a constant pressure, but rather a falling pressure one, due to throttling of the steam, or ''wire drawingy^ as it is called, through the throttle valve or steam ports, due to the higher speed at which this card was taken. It will also be noticed that the compression pressure is higher in this case, due to earlier closing of the exhaust valve, which becomes necessary with the type of valve gear used, as the cut-ofiF is made earlier. In the low-pressure card it will be seen that the compression pressure is greater than the admission pressure and hence there is a pressure drop instead of rise on admission. In set D the peculiarities of C are still more apparent, the compression in high-pressure cylinder being equal to admission pressure and above it in the low-pressure cylinder. The wire drawing is also more marked, as the speed was still higher when this set of cards was taken. In Fig. 64, one set of the cards of Fig. 63 is redrawn on cross-section paper and then combined. Cards taken from the different cylinders of a multiple-expansion engine will in nearly all cases have the same length, the great- est that can be conveniently handled by the indicator, and will be to two different pressures scales, in as much as that indicator spring will be chosen for each cylinder which will give the greatest height of card consistent with safety to the instrument. To properly compare the cards they must be reduced to the same pressure scale, and also to the same volume scale. As the lengths represent volumes, the ratio of the two volume scales will be as that of the cylinder volumes or diameters squared. Hence, the length of the high-pressure card must be decreased in this ratio or the low increased. As a rule it is found more convenient to employ the former method. When the cards have been reduced to a proper scale of pressures and volumes the clearance must be added to each in order that the true volume of the fluid may be shown. The cards may now be placed with these atmospheric lines and zero volume lines coinciding and will then appear in their true relation. In this case the cylinder ratio was 1.65, the indicator springs 100 lbs. and 70 lbs. respectively and clearance 5 per cent in each cylinder. The steps in combining the cards were as follows: The zero volimie lines were first drawn perpendicular to the atmospheric line and at a distance from the end of 232 ENGINEERING THERMODYNAMICS the card equal to the length of the card times the clearance. PV axes were laid oflf and a line drawn parallel to the zero-pressure line at a distance above it equal to 14.7 lbs. to scale of combined diagram. This scale was taken to be that of the high-pressure diagram. A number of points A'B^C, etc., were then chosen on the low-pressure card, and the corresponding points a'6'c', etc., plotted by making the distances of a\b\ etc., from the zero-volume line equal to thote of A^B'y etc., and the distances of the new points above the atmosphere .7 the distances of the original. By joining the points as plotted, the new diagram for the low-pressure card was formed. The high-pressure card was then redrawn Fig. 64. — Diagram to Show Method of Combining the High- and Low-pressure Cylinder Indicator Cards of the No-receiver Compound Engine. by taking a number of points A, B, C, etc., and plotting new points a, &, c, etc., so that the distances of a, 6, c, etc., from the zero-volume line were t-— the distances of A, B, C, etc., while the distances of new points above the atmos- pheric line were the same as for the original points. In Fig. 65 are shown two cards from a compound steam engine with receiver. Diagram A shows the cards as taken, but transferred to cross-section paper for ease in combining, and with the zero-volume axis added. On the high-pressure card admission occurred practically at constant volume, piston being at rest at dead center, at A, bringing the pressures in the cylinder up to the initial pressure at B. Admission continued from J? to C at nearly constant pressure, the piston moving slowly with correspondingly small demand for steam and consequently little wire drawing. From C to D the piston is moving more rapidly and there is in consequence more wire drawing, admission being no longer at constant pressure. At D the steam valve closes and expansion occurs, to E, where release occurs, the pressure falling to that in the receiver. From F to G exhaust occurs with increase of pressure due to the steam being forced into the receiver, (receiver + decreasing H.P. cyl.vol.) while from G to H the WORK OF PISTON ENGINES 233 pressure falls, due to the low-pressure cylinder taking steam from the receiver and consequently voliune of receiver, (receiver + increasing L.P. cyl.voL+ decreasing H.P. cyl.vol.) increasing. At H exhaust closes, a very slight com- pression occurring from H U> A. On the low-pressure card, admission occurred at A' and continued to B' at constant volume, the piston being on dead center as from A to B in high-pressure cylinder. From B' to C admission occurred with falling pressure due to increase in receiver volume, (receiver+increasing L.P. cyl.vol.), and from C to D' admission still took place, but with less rapidly falling pressure, as high-pres- sure cylinder is now exhausting and receiver volume, (receiver+increasing L.P. cyl.vol. +decreasing H.P. cyl.vol.) was receiving some steam as well as Fig. 65. — Indicator Cards from a Compound Engine with a Receiver, as Taken and as Combined. delivering. At ZX admission ceased and expansion took place to E' where release occurred, the pressure falling to the back pressure and continuing from F' to G', where the exhaust valve closed and compression took place to A' , thus completing the cycle. At H' leakage past the exhaust valve was so great as to cause the curve to fall off considerably from H' to A', instead of con- tinuing to be a true compression curve, ending at /, as it should have done. The combined diagrams are shown in B. In Fig. 66 are shown a set of three cards from a triple-expansion pumping engine with large receivers and cranks at 120°. In diagram A the cards are shown with the same length and with different pressure scales as taken, but with the zero volume line added and transferred to cross-section paper. On the high-pressure card admission occurred at A^ causing a constant volume pressure rise to -B, the piston being at rest with the crank at dead center. From B to C admission occurred at nearly constant pressure to C, where steam was 234 ENGINEERING THERMODYNAMICS cut off and expansion took place to D. At this point release occurred^ the pressure dropping at constant volume to E with the piston at rest. From E exhaust took place with slightly increasing pressure, since the intermediate cylinder was taking no steam, the intermediate piston being beyond the point of cut-off. The pressure rise is slight, however, due to the size of the receiver, which is large compared to the cylinder. At two-thirds of the exhaust stroke, point F, the back pressure became constant and then decreased, for at this point the speed of the intermediate piston increased and the receiver pressure fell. At G exhaust closed and a slight pressure rise occurred to A, due to the restricted passage of the closing exhaust valve. On the intermediate card c p h • B V T \ \, A D 1 G F Afvn X AUU F* ■^ — c V A' \ X^ .) \ « ^ — ^v Atn i. / n' id \e ■ a" <u Atm \cf "^ -j t a' V A V D" r^ \ s^d' C" v.. e;' \£ ' 'd V I ^^^ ^„^^,^ 1- - ® Fig. 66. — Indicator Cards from a Triple-expansion Engine with Receiver as Taken and Combined. admission occurred at A\ the pressure rising to B'. From B' the admission was at nearly constant pressure to X while the piston speed was low and then at a falling pressure to C. Pressure was falling, since the steam was supplied from a finite receiver into which no steam was flowing during intermediate admission. At C cut-off occurred and steam expanded to D', where release took place, and the steam was exhausted. As in the case of the high-pressure cylinder the back pressure was rising for two-thirds of the stroke, since the steam was being compressed into the receiver or rather into a volume made up of receiver and intermediate cylinder volume, which is, of course, a decreas- ing one, since the cylinder volume is decreasing. At two-thirds of the stroke the low-pressure cylinder begins to take steam and the receiver volume is WORK OF PISTON ENGINES 235 now increased, inasmuch as it was made up of the receiver portion of the inter- mediate cylinder and a portion of the low-pressure cylinder, and the low- pressure cylinder volume increased faster than intermediate decreased for the same amount of piston travel. At G' exhaust closed and a slight compression occurred to A', thus completing the cycle. On the low-pressure card admission occurred at A" and the pressure rose at constant volume to B"y and then admission continued first at constant pressure and then falling, as in the intermediate cylinder, to the point of cut-off at C". From here expansion took place to Z)". At this point the exhaust valve opened, the pressure fell nearly to back pressure at E", and the steam was exhausted at practically constant back pressure to G"f where the exhaust valve closed and there was compression to A", thus completing the cycle. The combined diagram is shown iii B. Prob. 1. In Fig. 67 are shown six sets of indicator cards from compound en^es. The cylinder sizes and clearances are given below. Explain the cylinder events and the shaf>e of lines for each card and form a combined diagram for each set. No. 1. From a four- valve Corliss engine, 26x48 ins., with 3 per cent clearance in each cylinder. No. 2. From a single-valve engine, 12x20x12 ins., with 33 per cent clearance in high-pressure cylinder and 9 per cent in low. No. 3. From a four-valve Corliss engine 22x44x60 ins., with 2 per cent clearance in the high-pressure cylinder and 6 per cent in low. No. 4. From a single- valve engine 18 X30xl6 ins., with 30 per cent clearance in the high-pressure cylinder and 8 per cent in the low. No. 5. From a single- valve engine 11^X18^X13 ins., with 7 per cent clearance in the high and 10 per cent in the low. No. 6. From a double-valve engine 14x28x24 ins., with 3.5 per cent clearance in the high-pressure cylinder and 6.5 per cent in the low. Prob. 2. In Fig. 68 are shown four sets of indicator cards from triple-expansion marine engines. The cylinder sizes and clearances are given below. Explain the cylinder events and the shape of the lines of each card and form a combined diagram of each set. No. 1. From the engine of a steam-ship, cylinders 21.9x34x57 ins.x39 ins. with 6 per cent clearance in each and fitted with simple slide valves. No. 2. From an engme 20x30x50x48 ins. No. 3. From the engine of a steam-ship with cylinders 22x35x58x42 ins., assume clearance 7 per cent in each cylinder. No. 4. From the engine of the steamer "Aberdeen," with cylinders 30x45 X 70x54 ins., and with 4 per cent clearance in the high, 7 per cent in the intermediate, and 8 per cent in the low. Prob. 3. In Fig. 69 are shown some combined cards from compound engines. Explain the cylinder events and the shape of the lines and reproduce the indicator cards. Prob. 4. In Fig. 70 are shown some combined cards from triple-expansion engines. Draw the individual cards and explain the cylinder events and shape of lines. 8. Standard Reference Cycles or PV Diagrams for the Work of Expansive Fluids in Two-cylinder Compound Engines. The possible combinations of admission with all degrees of expansion for forward strokes and of exhaust with 236 ENGINEERING THERMODYNAMICS all degrees of compression for back strokes, with and without clearance, id each of the two cylinders of the compound engine, that may have any volume relation one to the other and any size of receiver between, and finally, any sort of periodicity of receiver receipt and discharge of fluid, all make possible a 5-| 0- 5- 10-1 No.l No. 2 lO-i No. 8 ICM rlOO -60 -80 plSO -flO "0 p40 -20 -0 i-10 -100 -00 20 -100 -60 -20 40 -20 -0 10 f-uo -80 -40 LO No. 6 Fig. 67. — Six Sets of Compound Engine Indicator Cards. very large number of cycles. In order that analysis of these conditions of work- ing may be kept within reasonable space, it is necessary to proceed as was done with compressors and simple engines, concentrating attention on such type forms as yield readily to analytical treatment and for which the formulas WORK OF PISTON ENGINES 237 are simple even if only approximate with respect to. actual engines, but, of course, keeping in mind the possible value of the formulas, as those that teach 150- 100- 50- I5-| 30- 26- 140- 70- 0- 10 lao No. 2 No. 4 25- FiQ. *68. — Four Sets of Triple-expansion Engine Indicator Cards. no principles or fail to assist in solving problems must be discarded as useless. The work that fluids under pressure can do by losing that pressure is no 238 ENGINEERING THERMODYNAMICS different in compound than in simple engines, if the fluid has a chance to do what it can. Provided the structure is such as will not interfere with the com- pleteness of the expansion, and no fluid is wasted in filling dead spaces without 120-1 80- 40- 100- 300- 100- No. 8 Fia. 69. — Ck)mbiiied Diagrams of Compound Engines. working, then the work per cubic foot or per pound of fluid is the same for simple, compound and triple engines. Furthermore, there is a horse-power equiv- alence between the simple and compound, if, in the latter case the steam WORK OF PISTON ENGINES 239 admitted up to cut-off may be assimied to be acting only in the low-pressure cylinder, that is, ignoring the high-pressure cylinder except as it serves as a 250- aeo- 150- 100- 60- 1J50- lOO 50- No. 1 No. 2 100 60 No. 8 No. 5 Fig. 70. — Combined Diagrams of Triple-expansion Engines. cut-off means or meter. This should be clear from a comparison of Figs. 71, A and B. In Fig. 71 A, representing the case of the simple engine without 240 ENGINEERING THERMODYNAMICS clearance and with complete expansion, the volume admitted, ABy expands to the back pressure on reaching the full cylinder volume DCt and exhauFtt at constant back pressure, the work represented by the area ABCD- It should be clear that no difference will result in the work done if a line be drawn across the work area as in Fig. 7 IB, all work done above the line HG to be developed in the high-pressure cylinder and that below in the low. This is merely equiv- alent to saying that a volume of steam AB Is admitted to the high-pressure cylinder expands completely to the pressure at G on reaching the full higb- pressure cylinder volume, after which it exhausts at constant pressure (into a receiver of infinite capacity), this same amount being subsequently admitted without change of pressure to the low-pressure cylinder, when it again expands completely. Thus, it appears that the working of steam or compressed air p rr Vol. Idmi ttedi oCyl 1 inder P\ fZ Vol.. ^dmi ;tedt oH.P.Cyl nder A U A U • , \ • 1 A B \ \ \ M ( r u -V olum eAdi litted to I .P.Q rl. \ ^ n \ 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 One-cylinder Simple and in Two-cylinder Compound Engines for the Same Rate of Expansion. in two successive cylinders instead of one will in no way change the maximum amount of work a cubic foot supplied can do, the compounding merely making it easier to get this maximum. In simple engine cases, Fig. TlA, the cut-off in per AB cent of stroke is 100 Xy^, which is a very small value, leaving but little time to open and close the admission valve, whereas in the compound case the per cent AB , ^ cut-off in the high-pressure cylinder is 100X7T7;;,and in the low-pressure cylinder, HC 100 X -^t:;, which are quite large enough ratios to be easily managed with ordinary valve gears. Compounding docs, however, introduce possibilities of loss not present in the single-stage expansion, if the dimensions or adjustments are not right, which may be classed somewhat improperly as receiver losses, and these are WORK OF PISTON ENGINES 241 of two kinds, one due to incomplete expaxLsion in the high and the other to over-expansion there. Thus, in Fig. 72, if ABCEFGDA represent a combined compound diagram for the case of complete expansion in the high-pressure cylinder continued in the low without interruption but incomplete there, DC represents the volume in the low-pressure cylinder at cat-off^ and at the same time the total high-pressure cylinder volume. If now, the low-pressure cut-off be made to occur later. Fig. 73, then the volume that the steam would occupy when expansion began c.,^ ^o t^- *. au n * t ' . . 1. , . ^*^' '2. — Diagram to Show Correct Low-pressure m the low-pressure cylmder is rep- resented by D'C. This adjust- A B \ \ \ V \ D V sX __, - . T\ .A»^^w, F F •L>1 ^ ^ G 1 1 1 L^ T^ ^ Y 1 jj ^2 Cut-off for No Receiver Loss. ment could not, of course, change the high-pressure total volume DC, so that at release in the high-pressure cylinder the pressure would drop abruptly to such a value in the receiver as corresponds to filling the low pressure up to its cut-off, and work be lost equal to area CCW^. . A shortening of the low-pressure cut-off will have an equally bad effect by introducing negative work as indicated in Fig. 74, in which the L.P. cut-off volimie is reduced from DC to D'C Expansion in the high pressure proceeds as before till the end of the stroke, at which time it has a volume DC greater than. the low pressure can receive D'C, hence the receiver pressure must rise to such a value as will reduce the volume the required amount, introducingthe negative work CC'C". Changes of low-pressure cut-off may, therefore, introduce negative work, P. -o T^• A au !:!«•* rr ^u -^ chaugc thc rcccivcr pressure and. Fig. 73. — Diagram to Show Effect of Lengthening ^ v j. .i . L.P. Cut-off Introducing a Receiver Loss Due to 01 COUrse, modlfiy the distribution Incomplete High-pressure Cylinder Expansion, of work between high and low, but A B \ \ \ ■ \ \ V \ .5... ■«»^^>^« XV .\ c 1" Di- %-. „' v'/i^'///^ ^///hi K D' c ^ E G F TV -Dj 242 ENGINEERING THERMODYNAMICS these saxae effects might also have resulted from changes of high-pressure cut-off or of cylinder ratio. For such conditions as have been assumed it seems that compounding does not increase the work capacity of fluids, but may make it easier to realize this capacity, introducing at the same time certain raiher rigid relations between ciU-offs and cylinder volumes as necessary conditions to its attainment. It can also be shown that the same proposition is true when there are clearance and compression, that is, in real cylinders and when the receiver is real or not infinite in size, or when the exhaust of high and admission of low, are not con- stant^pressure lines. The former needs no direct proof, as inspec- tion of previous diagrams makes it clear, but the latter requires some discussion. A real receiver of finite size is at times in communication with the high-pressure cylinder during its exhaust, and at other times with the low-pressure cylinder during admission, and these two events may take place at entirely independent times, be simultaneous as to time, or over- lap in all sorts of ways. Suppose the periods to be independent and there be no cylinder dear- ance, then at the beginning of high-pressure exhaust two sepa- rate volimies of fluid come together, the contents of both the high-pressure cylinder and the receiver, and this double volume is compressed by the H.P. piston into the receiver, in which case the high-pressure exhaust would take place with rising pressure. Follow- ing this will come low-pressure admission, during which the volume of fluid in the receiver expands into the low-pressure cylinder up to its cut-off, and if the same volume is thus taken out of the receiver as entered it previously, low-pressure admission will take place with falling pressure, the line representing it exactly coinciding with that for the high-pressure exhaust. Independence of H.P. cylinder exhaust and L.P. cylinder admission, as to time, may result in a cycle such as is represented in Fig. 75 for the case of no cylinder clearance. On this diagram the receiver line is DC, an expansion or compression line referred to a second axis of volumes XJ, placed away from the axis of purely cylinder volumes A B i \ \ D' \. :' C" m • D c Tk ^ ■Dj- \ V \ \ ¥ ■^ ^ E — G F ■^ ^ Fig. 74. — Diagram to Show Effect of Shortening L.P. Cut-off, Introducing a Receiver Loss Due to • Over-expansion in the High-pressure Cylinder. WORK OF PISTON ENGINES 243 by the distance LD, equal to the receiver volume to scale. All diagram points are referred to the axis A I except those on the line DC. This same case of time independence of H.P. exhaust and L.P. admission yields quite a different diagram when the cylinder clearance is considered. Such a case is represented by the diagram, Fig. 76, which also serves to illustrate the eflFect of incomplete expansion and compression as to equalization of receiver with cylinder pressures. At high-pressue release the volume of fluid in the H.P. cylinder is ML and its pressure iaJJi. This is about to come into communication with the receiver volume ON from which the low-pres- sure cylinder finished taking fluid and which is, therefore, at the same pressure d K'fl > 8 A B <*4 € n \ is 7^ \ « 1 1 eterr \ L ^- ■a- -— — — — PT D > V V , N X \ -Bm eixec .Volu: ne ^ \ s^ \ M ^ !^ ......^ E F G ^ I H V Fig. 75. — Diagram to Illustrate Variable Receiver Pressure for the Case of Independent High-pressure Exhaust and Low-pressure Admission with Zero Clearance. as the L.P. cylinder cut-off KS- The question, therefore, is — ^what will be the pressure at P in both H.P. cylinder and receiver when LM cu.ft. of fluid at LR pressure cpmbines with ON cu.ft. at pressure KS, and together occupy a volume ON+LM- By hypothesis the pressure after mixture is (first volume Xits pressure) + (second voliuneXits pressure) simi of volumes From this or the graphic construction following, the point P is located. If the high-pressiire expansion had continued to bring LQ to the receiver pressure KS, it would reach it at X, At this hypothetic point there would be a volume NX in the H.P. cylinder to add to the volume ON in the receiver at the same pressure, resultmg in OX cu.ft. This fluid would have a higher pressure at the lesser volume of receiver and H.P. cylinder and the value is found by a compres- 244 ENGINEERING THERMODYNAMICS sion line through X, XPAT referred to the receiver axis. This same line is also the exhaust of the H.P. cylinder from P to ^4 . A similar situation exists at admis- sion to the L.P. cylinder as to pressure equalization and location of admission line. At the end of the L.P. compression there is in the L.P. cylinder FE cu.ft. at pressure EH, to come into communication with the receiver volume CB cu.ft. at pressure BGy that at which H.P. exhaust ended. Producing the L.P. com- pression line to /, the volume BI is found, which, added to receiver, results in no pressure change. An expansion line, referred to the receiver axis through /, fixes the equalized pressure J and locates the L.P. admission line JK, which, it must be observed, does not coincide with the H.P. exhaust. So far only complete iridependence of the time of H.P. exhaust and L.P. admission have been considered, and it is now desirable to consider the diagram- \ \ \^ \ c — J^^ \ L t t^^^m ^iM ~4 L \ 1 -<%, ^ p\ "1 1 ^ X \ \ 1 s ^^ .-Re :elTei Jir.olu me_. F E \^ 1 ^^ ..^ V 1 1 1 1 D G H F < s V Fia. 76. — Diagram to Illustrate Variable Receiver Pre.ssiire for the Case of Independent Higb- pressure Exhaust and Low-pressure Admission with Clearance. matic representation of the results of complete coincidence. Such cases occur in practice with the ordinarj*^ tandem compound stationary steam engine and twin-cylinder single-crosshead Vauclain compound steam locomotive. In the latter structure both pistons move together, a single valve controlling both cylinders, exhaust from high taking place directly into low, and for exactly equal coincident time periods. The diameter of the low-pressure cylinder being greater than the high, the steam at the moment of release suffers a drop in pressure in filling the low-pressure clearance, unless, as rarely happens, the pres- sure in the low is raised by compression to be just equal to that at H.P. release. After pressure equalization takes place, high-pressure exhaust and low-pres- sure admission events are really together a continuation of expansion, the volume occupied by the steam at any tinui being equal to the difference between WORK OF PISTON ENGINES 245 tlie two cylinder displacements and clearances up to that point of the stroke. Before this period of communication, that is, between high-pressure cut-off and release, the volume of the expanding fluid is that of the high-pressure displace- ment up to that point of the stroke, together with the high-pressure clearance. After the period of commimication the volume of the expanding fluid is that of the low pressiu-e cylinder up to that point of the stroke, together with the low pressure clearance, plus the high-pressure displacement not yet swept out, and the high-pressure clearance. These fluid processes cannot be clearly indicated by a single diagram, because a diagram drawn to indicate volumes of fluid will not show the volumes in the cylinders without distortion. If there be no clearance, Fig. 77 will assist in showing the way in which two forms of diagram for this purpose are derived. Referring to Fig. 77 A, the volume AB admitted to the high pressure cylinder expands in it until it occupies the whole H.P. cylinder volume DC At this point expansion proceeds in low and high together, with decreasing volumes in high and increasing in low until the low-pressure cylinder volume is £^ttained at E. The line BCE then indicates the pressures and volumes of the fluid expanding, but does not clearly show the volume in either cylinder between C and Ey with the corresponding pressure. It is certain, however, that when the volume in the H.P. cylinder becomes zero the pressure must have fallen to a value the same as that in the low when the fluid completely fills it, orP/=Pe. As the high-pressure piston returns, on the exhaust stroke, the low- pressure piston advances an equal distance, on its admission stroke, sweep- ing through a greater volume than the high pressure, in the ratio of low-pres- sure to high-pressure displacements. If at any point of the stroke the volume remaining in the high-pressure cylinder be x, and the high- and low-pressiu-e displacements be respectively Di, and D2, then (Dx—x) is the volume swept out by high-pressiu'e piston, x the volume remaining in it, and -^^(Di— x) Di the volume swept in by the low-pressure piston. Then the total volume still in the two cylinders is, for a point between C and E, 7=x+^(Di-x)=Z)2-a:(~?-l). Since the equation of the curve CE is, PV = PcVc = PcDif the value of V may 1 c substituted, giving P \D2 — xi-~ — l] \= PcDi, = constant. Dividing by ^— 1 jthis becomes P D2 X i®-') = PN.-_-^ — x =a new constant, so 246 ENGINEERING THERMODYNAMICS from the axis GP anv that if a new axis LM be laid off on B, GV=i n n i point on FC will be distant from the new axis LM an amount f ^ ^^ — \x\ as the product of this distance and the pressure P, is constant, the curve PC is an equilateral hyperbola referred to the axis LM, Therefore Fig. 77 B is COMBINED DIAGRAM TO ONE SCALE Fig. 77. — Diagram to Illustrate the Compound Engine Cycle with No-receiver, and Exact Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance. the pressure-volume representation of the entire cycle of the high-pressure cylinder. In Fig. 77C is shown the corresponding pressure-volume diagram for the low- nressure cylinder, for which it may similarly be shown that the curve DE may be WORK OF PISTON ENGINES 247 plotted to an axis NO at a distance to the left of the axis GP equal to the same quantity, ("'-S^k w These diagrams, 77-4, B and C may be superposed, as in Fig. 77 E, giving one form of combined diagram used for this purpose, and the one most nearly comparable with those ah-eady discussed. In this diagram, the area ABC FA represents the work of the high-pressure cylinder, and DEIHD, the work of the lo w-pressure cylinder. Together, they equal the work of the enclosing figure ABEIHA, and hence the work of the low-pressure cylinder must also be represented by the area FCEIHK It is not difficult to show that if a vertical, CJ, be dropped from the point C to the exhaust line HJ, the figure CFHJ, in Fig. 77D is similar to DEIH, in Fig. 77E, reversed, but drawn to a different horizontal scale. Here the length of the low-pressure diagram is made equal to the length of the high- pressure diagram. The two scales of volumes are shown above and below the figure. While this appears to be a very convenient diagram, it will be found to be less so when clearance and compression are considered. It may be noted that since it has been shown that the curves CF and DE are of the same mathematical form (hyperbolic) as the expansion line CE, they may be plotted in the same way after having in any way found the axis. The location of this axis may be computed as given above, or may be found graphically by the method given in connections with the subject of clearance. Chapter I, and shown in Fig. 77B, Knowing two points that lie on the curve, C and Fj the rectangle CDFK is completed. Its diagonal, DK, extended, cuts the horizontal axis GV in the point Af , which is the base of the desired axis ML. If now the axis NO and the figure DEIHj part of which is referred to this axis, be reversed about the axis GP, Fig. 77C, NO will coincide with AfL, Fig. 77D, and Fig. 78 results. Note that the axis here may be found graphically, in a very simple way. Draw the vertical CK from C to the axis GV and the horizontal DJ to the vertical IE extended. DC is then the high-pressure displacement and DJ the low-pressure displacement. Draw the two diagonals DK and JG, extending them to their intersection X. By similar triangles it may be shown that a horizontal line, UW, will have an intercept between these two lines, JG and DK, equal to the volume of fluid present in the two cylinders combined. The intersection X is the point at which this volume would become zero if the mechanical process could be continued unchanged to that point, and is, therefore, on the desired axis ML extended. T being the inter- section of WU with the axis GP, when the volume UT is present in the high- pressure cylinder, TW gives the volimie in the low-pressure cylinder. When clearance and compression are considered, this diagram is changed in many respects, and is shown in Fig. 79. The axes OP, OV and OV are laid out, with OZ equal to the clearance and ZK, the displacement, of the high-pressure 248 ENGINEERING THERMODYNAMICS cylinder, and OQ and QF, clearance and displacement of the low-pressure cylinder. It is necessary to know high-pressure cut-oflf, ;z^l high-pressure compression, Zli. : and low-pressure compression, =^, in addition to the initial and back ZK YQ pressures, in order to lay out the diagram. The drop in pressure CD at releasee is due to the coming together of (volume Vg at pressure P<.), with (volume T'^ at pressure P,). If the volimie Vj (measured from axis OP) were decreased sufficiently to raise the pressure in the low-pressure clearance to the pressure Fig. 78. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound Engine at any Point of th'e Stroke for the Case of No-clearance and Coincidence of H.P. Exhaust with L.P. Admission. at C, the voliune woxild become F„ as indicated at the point S, and the volumes now combined in the hypothetical condition would occupy the volume SC Increasing this volume to D'D, that actually occupied after communication, the pressure would fall along the curve SD\ which is constructed on KV and KT as axes. The pressure of D is then laid out equal to the pressure at D\ To find the axis, MLy for the cui^es DE and £>'£', from any convenient point A' on ZAy draw the line NK extended to X. Extend HG to A, at a height equal to that of N, and draw RQX, and through the intersection draw the desired axis XML, The fraction of stroke completed at E' in the low pressure at cut-ofif must be equal to that completed at E in the high pressure at compression, and may be laid out graphically by projecting up from E to the point U on the line NK and horizontally from t7 to TF on the line RQ. Projecting down from W to the curve at £' locates the point of effective cut-ofif in the low-pressure cylinder. WORK OF PISTON ENGINES 249 After the supply from the high-pressure cylinder has been cut off at E', the expansion is that of the volume in the low-pressure cylinder and its clearance, and hence the curve E'G is constructed on OP as an axis. While in this last case a zero receiver volume has been assumed, there is nothing to prevent a receiver volume being interposed between H.P. and L.P. so that common expansion takes place with a volume greater than assumed by so much as this volume, the effect of which is to decrease the slope of DE and D'E\ Such receivers usually consist of the spaces included in a valve body and connecting passages and may be treated generally as increased L.P. clearances. The most conamon of all relations between H.P. exhaust and L.P. admis- sion is, of course, that of partial coincidence of periods, as it is thus with all cross- -u p. Cut Oft due L.P. Diaptacctpent Displaceinenr*'^^ \ -:s ^ Fig. 79. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compoimd Engine at any Point of the Stroke for the Case of Clearance and Coincidence of H.P. Exhaust with L.P. Admission. compound and triple-expansion engines having separate cranks for the individual cylinders. For these there is no simple fixed relation between the periods, for, while crank angles are generally fixed in some comparatively simple relation, such as 90°, 180° and 270° for conipounds and 120° for triples, they are sometimes set at all sorts of odd angles for better balance or for better turning effort. Even if the angles were known the receiver line would have to be calculated point by point. When the H.P. cylinder begins to discharge into a receiver for, say, a cross compound with cranks at 90°, steam is compressed into the receiver, and so far the action is the same as already considered for independence of periods, but at near mid-stroke the low-pressure admission opens while high-pressure exhaust continues. This will cause the receiver pressure to stop rising and probably to fall until the low pressure cuts off, which 250 ENGINEERING THERMODYNAMICS may occur before the H.P. exhaust into the receiver ceases. If it does, the receiver pressure will again rise. Exact determination of such complex receiver lines is not often wanted, and when needed is best obtained graphically point by point. Its value lies principally in fixing exactly the toork distribution between cylin- ders, which is not of great importance except for engines that are to work at constant load nearly all the time, such as is the case with city water worfc? pumping, and marine engines. While equations could be derived for these cases, they are not worth the trouble of derivation, because they are too cumber- some, and graphic methods are to be substituted or an approximation to be made. Four kinds of approximation are available, as follows, all of which ignore partial coincidence of periods: 1. Receiver pressure constant at some mean value and clearance ignored. 2. Receiver pressure constant at some constant value and clearance con- sidered with compression zero or complete. 3. Receiver pressure fluctuates between fixed limits as determined by an assumed size, clearance ignored. 4. Receiver pressure fluctuates between fixed limits as determined by an assumed size, clearance considered, with compression zero or complete. These are not all of equal difficulty in solution, and the one to be used is that nearest the truth as to representation of conditions, which is usually the most difficult, provided time permits or the information is worth the trouble. Quickest work is accomplished with assumption (1) and as this is most often used in practical work it indicates that its results are . near enough for most purposes. This discussion leads, therefore, to the analytical study of the following cycles: Infinite Receiver, Zero Cylinder Clearance. CYCLES V, AND VI (Fig. 80). ' (a) Admission at constant supply pressure to H.P. cylinder. (6) Expansion in H.P. cyl. (may be zero) by law PV -c lor {V)\ Pr^c for (VI). "^ (c) Equalization of H.P. cyl. pressure with receiver pressure at constant volume (may be zero). {d) Exhaust into infinite receiver at constant pressure from H.P. cylinder, (e) Equalization of H.P. cylinder pressure with supply pres.sure at constant zero volume. ' (/) Admission from receiver at constant receiver pressure to L.P. cylinder. {g) Expansion in L.P cylinder (may be zero) by law PV = c for (V); Pr=c for (VI). (A) Equalization of L.P. cylinder pressure with back pressure at constant volume (may be zero), (i) Exhaust at constant back pressure for L.P. cylinder. (j) Equalization of L.P. cylinder pressure with receiver pressure at constant zero volume. H.P. Cylinder Events L.P. Cylinder Events WORK OF PISTON ENGINES 251 p P a I "l I \ Cycles VII & VIII \ Cycle* V & VI Cycle Y PV«C Cycle VI PV«-C e ACycle VII PV-C Cycle VIII PV«.C \ ^ e 2 s: \ 0^ V \ k \, d \ c \ -^ Rec i dTerVol Bine -♦ S c \ \ d \ \^ / V m 3 N ^ m 3 V* "*^ e t h • h V V p a i P M \ Cycles IX & X Cycle IX PV«C Cycle X PV««C f \ Cycles XIII 8c XIV \ \Cycle XIII PV«C Cycle XIV PV».C 5^ i : tr d V V \ V \ \ \ \ </ \ t Reo rivei ■Vol ime -► ^ < c ^> / p ^ y ^ •v ^•., \ --^ ^ ^^ 5.^ Vf "S L. e N ^ i j V V p P c 'i M \ CyclM XI <i XII Cycte XI PV=C Cycle XII PV»=C j 1 \ Cycles XV & XVI Cycle XV PV=C Cycle XVI PV»aC \ \ / A A V V \ \ \ , — . ^ l* c I c 1 d V \ K s,^ 'N [/ <r -^ I ^ r^ V ■ ii ^^ ^^ ■ i \ ^ • A V V Fig. 80. — Compound Engine Standard Reference Cycles or PK Diagrams. 252 ENGINEERING THERMODYNAMICS Relations between h.p. AND L.P. CyLINDEB Events (1) H.P. exhaust and L.P. admission independent as to time, coincident as to representation (except as to length). (2) H.P. expansion line produced coincides as to representation with L.P. expansion line. (3) Tlie length of the constant pressure receiver line up to the H.P. expansion line produced is equal to the length of the L.P. admission line, H.P. Cylinder Events Finite Receiver, Zero Cylinder Clearance. CYCLES VII, AND VIII, (Fig. 80). (a) Admission at constant supply pressure to H.P. cylinder. (6) Expansion in H.P. cylinder (may be zero) by law PV=c for (VII); Pr*=c for (VIII). (c) Equalization of H.P. cylinder pressure with receiver pressure at constant volume (may be zero) with a change of receiver pressure toward that at H.P. cylinder release (may be zero). (d) Exhaust into finite receiver from H.P. cylinder at rising pressure equivalent to compression of fluid in H.P. cylinder and receiver into receiver by law P7=c for (VII) and P7*=c for (VIII). (e) Equalization of H.P. cylinder pressure with supply pressure at constant zero volume. L.P. Cylinder Events Relation between H.P. AND L.P. Cylinder Events ' (f) Admission from receiver to L.P. cylinder at falling pressure equivalent to expansion of fluid in receiver into receiver and L.P. cylinder together by law P7=c for (VII), PV*-^c for (VIII). ig) Expansion in L.P. cylinder (may be zero) by law PV=c for (VII); PV*-c for (VIII). (A) Equalization of L.P. cylinder pressure with back pressure at constant volume (may be zero), (i) Exhaust at constant back pressure for L.P. cylinder. (t) Equalization of L.P. cylinder pressure with receiver pressure at constant zero volume to value resulting from HJP. exhaust. ' (1) H.P. exhaust and L.P. admission independent as to time, coincident as to representation, except as to length. (2) H.P. expansion line produced coincides as to representation with L.P. expansion line. (3) The length of the receiver pressure line up to the H.P. expansion line produced is equal to the length of the L.P. admission line. WORK OF PISTON ENGINES 253 No Receiver, Zero Cylinder Clearance. CYCLES IX, AND X, (Fig. 80). H^. Cylinder Events Both H.P. and L.P. Simultaneously H.P. Cylinder Events L.P. Cylinder Events (a) Admission at constant supply pressure to H.P. Cylinder. (b) Expansion in H.P. cylinder (may be zero) by law PV'^c for (IX); PF'=c for (X). ' (c) Transference of fluid from H.P.to L.P. cylinder with simul- taneous continuation of expansion until all fluid is in L.P. cylinder and expanded to its full volume by law PV =c for (IX) ; PV' =c for (X). r (d) Equalization of H.P. cylinder pressure to the pressure of I supply. ' (e) Equahzation of L.P. cylinder pressure with back pressure at constant volume (may be zero). (/) Exhaust at constant back pressure for L.P. cylinder. (g) Equalization of L.P. cylinder pressure to the pressure in H.P. cylinder at the end of its expansion. H.P. Cylinder Events L.P. Cylinder Events Infinite Receiver, with Cylinder Clearance. CYCLES XI, AND XII, (Fig. 80). (a) Admission at constant supply pressure to H.P. cylinder. (6) Expansion in H.P. cylinder (may be zero) by law PV^c for (XI); Pr=c for (XII). (c) Equahzation of H.P. cylinder pressure with receiver pressure at constant volume (may be zero) pressure. (d) Exhaust into infinite receiver at constant pressure from H.P. cylinder. (c) Compression in H.P. cyhnder to clearance volume (may be zero) by law PV =c for (XI) ; P7* -c for (XII). (/) Equalization of H.P. cyhnder pressure with supply pressure at constant clearance volume, may be zero. ' (g) Admission from receiver at constant-receiver pressure to L.P. cyhnder. (h) Expansion in L.P. cyhnder (miay be zero) by law PV ==c for (XI); P7'=c for (XII). (t) Equahzation of L.P. cyhnder pressure with back pressure at constant volume (may be zero). (j) Exhaust at constant back pressure for L.P. cyhnder. (k) Compression in L.P. cyhnder to clearance volume by law PV^c for (XI) ; PV =c for (XII) (may be zero). (0 Equahzation of L.P. cyhnder pressure with receiver pres- sure at constant clearance volume without change of receiver pressure (may be zero). 254 ENGINEERING THERMODYNAMICS Relations between h.p. and L.P. Cylindeb Events ' (1) H.P. exhaust and L.P. admission independent as to time, coincident as to representation except as to length. (2) L.P. expansion line does not coincide as to representation with H.P. expansion line produced by reason of clearance influence except in one special and unusual case. (3) The length of the constant-receiver pressure line intercepted between H.P. compression line and H.P. expansion line produced is equal to the same intercept between L.P. expansion line and L.P. compression line produced. This is equivalent to the condition that the volume taken in bj low is equal to expelled by the high reduced to the same pressure. Finite Receiver, with Cylinder Clearance. CYCLES XIII, AND XIV, (Fig. 80). H.P. Cylinder Events L.P. Cylinder Events ' (a) Admission at constant supply pressure to H.P. cylinder. (6) Expansion in H.P. cylinder (may be zero) by law PV ^c for (XIII) ; PV* =c for (XIV). (c) Equalization of H.P. cylinder pressure with receiver pres- sure at constant volume (may be zero) toward that at H.P. cyhnder release (may be zero). [d) Exhaust into finite receiver from H.P. cyUnder at rising pressure equivalent to compression of fluid in H.P. cylinder and receiver into receiver by law P7 =c for (XIII) \ PV* -c for (XIV). (c) Compression in H.P. cylinder to clearance volume (may be zero) by law PV =c for (XIII) ; PV* -c for (XIV). (/) Equalization of H.P. cylinder pressure with supply pressure at constant clearance volume. {g) Admission from receiver to L.P. cylinder at falling pressure equivalent to expansion of fluid in receiver into receiver and L.P. cylinder together by PV =c for (XIII); PV* ^c for (XIV). (A) Expansion in L.P. cylinder (may be zero) by law PV^c for(V); PV =cfor(VI). (i) EquaUzation of L.P. cylinder pressure with back pressure at constant volume (may be zero). ( j) Exhaust at constant back pressure for L.P. cylinder. (A;) Compression in L.P. cylinder to clearance volume by law PV^c for (XI) ; PV* =c for (XII) (may be zero). (0 Equalization of L.P. cylinder pressure with receiver pressure at constant clearance volume with change of receiver pressure in direction of L.P. compression pressure (may be zero). WORK OF PISTON ENGINES 255 Relations between h.p. and L.P. Cylinder Events (1) H.P. exhaust and L.P. admission independent as to time, representation and length. (2) L.P. expansion line does not coincide as to representation with H.P. expansion line produced by reason of clearance influence except in one special and unusual case. (3) The high-pressure exhaust and low-pressure admission lines do not coincide as to representation by reason of clearance influences. (4) There is a relation between the lengths of the L.P. admission and H.P. exhaust lines, but not a simple one. No Receiver, with Cylinder Clearance. CYCLES XV, AND XVI, (Fig. 80). H.P. Cylinder Events Also L.P. Event LP. Cylinder Events (o) Admission at constant-supply pressure to H.P. cylinder. (6) Expansion in H.P. cylinder (may be zero) according to law PV^c for (XV); PV' =c for (XVI). (c) Equalization of pressures in H.P. cylinder after expansion with that in L.P. after compression at constant volume (may be zero). (d) Transference of fluid from H.P. to L.P. cylinder imtil all fluid is in L.P. cylinder and expanded to its full volume by same law as (b) . (e) Compression in H.P. cylinder to clearance volume (may be zero) by law PV^c for (XV) ; P7* -c for (XVI). (f) Equalization of pressure in H.P. cylinder with supply at constant-clearance volume (may be zero). (g) Expansion in L.P. cylinder may be zero by law PV^c for (XV); P7'=c for fXVI). (h) Equalization of pressure in L.P. cylinder with back pressure, at constant volume (may be zero). {i) Exhaust at constant pressure for L.P. cylinder. (j) Compression in L.P. cylinder to clearance, may be zero by law PV =c for (XV) ; PV =c for (XVI). {k) Equahzation of L.P. cylinder pressure with H.P. cyhnder pressure. Cycle XVII, Fig. 81, for the triple expansion is defined in the same way as the corresponding case of compounds Cycle V, with appropriate alterations in wording to account for a third or intermediate cylinder between high- and low- pressure cylinders and an additional receiver. Thus, high- pressure cylinder- exhausts into first, and intermediate cylinder into second receiver: inter- mediate cylinder receives its supply from first,"^and low-pressure cylinder from second receiver. This being the case, it is imnecessary to write out the cylin- der events, noting their relation to the corresponding compound case. 256 ENGINEERING THERMODYNAMICS 9. Compound Engine with Infinite Receiver, Logarithmic Law. No Clear- ance, Cycle V. General Relations between Pressures, Dimensions, and Work It must be understood that the diagrams representing this cycle, Fig. 82, indicating ^A) incomplete expanmon and {B) over-expansion in both cylinders. P A B M e \ \ C Dp [\ G " \ c E d f A • J 1 ■ h H J • A: — _ >j K in M n L Fig. 81. — Triple-expansion Engine Standard Reference Diagram or FV Cycle for Infinite Receiver. may just as well stand for over, complete or incomplete expansion in all possible combinations in the two cylinders. Applying the principles already derived for calculating the work areas, High-pressure cylinder work Wh^F^\{\+\o^, J'^^-PtfFtf, (296) Low-pressure cylinder work W,.^l\\\[ 1+iog. J/)-7>,F„ (297) WORK OF PISTON ENGINES 257 'otal work TF=P»n(l+log, ^^) +Pe7e(l+log. ^^) -P,7^-P,F,, ^ . (298) >re8sure being in pounds per square foot, and voliunes in cubic feet. In theses expressions the receiver pressure P,= Puis imknown, but determinate IS it is a function of initial pressure and certain volumes; giving it the value, is merely satisfying the condition that the point E at which expansion begins in the low-pressure cylinder must lie in the expansion line of the high. Sub- stituting this value there results W^P^V^{l+\o^y^ +P^V^(l+\o^ ^^ -P,vi^-P,V, =P5n[2+log,^;+loge^;-^]-P,F,. . (299) This is a perfectly general' expression for the work of the fluid expanding to any degree in two cylinders in succession when the clearance is zero and receiver volume infinite, in terms of initial and back pressures, pounds per square foot, the volumes occupied by the fluid in both cylinders at cut-off, and at full stroke in cubic feet. Dividing this by the volume of the low-pressure cylinder Vg gives the mean effective pressure referred to the low-pressure cylinder, from which the horse-power may be determined without considering the high- pressure cylinder at all. Hence, in the same imits as are used for P» and P,, (M.E.P. referred to L.P.) ^P.^'h+log, ^'+log, ^-^1 -Pi,- (300) Proceeding as was done for simple engines, the work per cubic feet of fluid supplied is found by dividing Eq. (299) by the volume admitted to the high- pressiire cylinder F*, whence. Work per cu.ft. supplied = Ph 2+loge tt + lo& i/"" p- "" ^oijf' • • • Also applying the consumption law with respect to horse-power, (301) 13 750 Vb Cu.ft. supplied per hour per I.H.P. = ^^^^ ^;^ ^ ^^^^ y. (302) Lbs. supplied per hr. per IH.P. = (^^p^^;Jft^ j^^^ ^fi (303) 258 ENGINEERING THERMODYNAMICS These last five equations, (299), (300), (301), (302), (303), while character- istic, are not convenient for general use in their present form, but are ren- (atm.pr.) G(bk.pr,) 1_1 I N :1...4bk.pr.) f— -(Pel.pr.)L (atm.pr.) H V H.P. INOICATOII CAMM OF EQUAL BAM AND HEIOHT INMCATOfl CARDS OP EQUAL BASE ANO HEIQHT Fia. 82. — Work of Ejqjansive Fluid in Ck)mpound Engine with Infinite Receiver, Zero Clearance. Cycle V, Logarithmic Expansion and Cycle VI Exponential Expansion. dered so by substituting general symbols for initial and back pressures displacement, cut-off, and amount of expansion for each cylinder. WORK OF PISTON ENGINES 259 et (in.pr.) = initial or supply pressure, pounds per square inch = ^52? (rel.pr.)j5r = release pressure, in H.P. cylinder pounds per square inch = (rel.pr.)j!i = release pressure in L.P. cylinder, pounds per square inch (rec.pr.) =receiver pressure, pounds per square inch=r^=yv~; P *(bk.pr.) = back pressure, pounds per square inch=Y24J y /?// = ratio of expansion in high-pressure cylinder = y-*^ 5 144' 144' Fe' jBl= ratio of expansion in low-pressure cylinder = V, J2v= ratio of expansion for whole expansion ==:fd^; Vb Z)h= displacement of high-pressure cylinder =Fd = Fc; Z)l = displacement of low-pressure cylinder = F/= Vg\ -Re = cylinder ratio = 7^ =t/> l^H Vd Zh = fraction of displacement completed up to cut-off in high-pressure cylinder, so that ZhDh = F^ = -^; Zl= fraction of displacement completed up to cut-oflf in low-pressure Rl cylinder, sq that ZlDl = F« = ^ Substitution of these general symbols in Eqs. (299), (300), (301), (302), and (303) gives another set of five equations in useful form for direct substitu- tion of ordinary data as follows: Work of cycle = 144Dir(in.pr.) r^(2+loge fi//+loge Rl) ~£^ = 144D. { (in.pr.)|(2+log. A+iog, J_ ._i_) .(bk.pr.) (in.pr.)^^(2+lo& R/r+loge Rl-^^ " (bk.pr.) -144(bk.pr.)i)L (a) (b) = IUDl (m.e.p.) lbs. per sq.in. referred to L.P. cyl. (c) . (304) = (in.pr.)g^(2+lofe R„+\oe, Rl-^ - (bk.pr.) = (in.pr.)^( 2+lofe z~„'^^^^'Z',~ ZTRc) ~ ^^^-^^-^ (a) (b) . (305) 260 ENGINEERING THERMODYNAMICS Work per cu.ft. supplied = 144r(in.pr.) (2+loge Rh +log, Bl-^) - (bk.prOJKcfiJ = 144[(in.pr.)(2+loge ^+lo& ^-^) -(bk.pr.)|^] Cu.ft. supplied per hr. per I.H.P. 13,750 1 (a) (6) (306) (m.e.p. ref. to L.P.) RhRc 13,750 Z„ (m.e.p. ref. to L.P.) Re (a) (b) (307) J From this, of course, the weight in pounds supplied per I.H.P. results directly from multiplication by the density of the fluid. To these characteristic equations for evaluating work, mean pressure, economy and consumption in terms of the initial and final pressures and cylin- der dimensions there may be added a series defining certain other general rela- tions of value in fixing the cycle for given dimensions and initial and final pressures, and in predicting dimensions for specified total work to be done and its division between high- and low-pressure cylinders. Returning to the use of diagram points and translating into the general symbols as each expression is derived, there results. Receiver pressure =Pd—Pe = Pbjr- :. (rec.pr.) = On.pr.)^^^^ = (m.pr.)^-^ = (in.pr.) Rl RcRh (a) (6) (308) (rel.pr.)w = (in.pr.)- High-pressure cylinder release pressure =Pc=^»-Tr> 1^ Rh = (in.pr.)Zj5r y Low-pressure cylinder release pressure =P/=P,.:j:3r. yf 1 (a) (6) (309) (rel.pr.)i, = (in.pr.) RcRh Zh Re = (in.pr.) _ (in.pr.) Rv _ (rel.pr.)ir "ftc" (o) (b) (c) (d) ' • (310) WORK OF PISTON ENGINES 261 Division of work between cylinders may be made anything for a given load by suitably; proportioning cylinders, and equations giving the necessary relations to be fulfilled can be set down. It is quite common for designers to fix on equal division of work for the most commonly recurring or average load or that corresponding to some high pressure cut-off or low-pressure terminal pressure^ generally the latter. Therefore, a general expression for dimensional rela- tions to be fulfilled for equal division of work is useful. On the other hand, for an engine the dimensions of which are determined, it is often necessary to find the work division for the imposed conditions, so that the following equa- tions are of value. Frwn Eqs. (296) and (297), noting that Pd=P«=P5^, '9 Low-pressure cylinder work ^^^^^ _^^^^ High-pressure cylinder work _ ('+'<* ft) - (■+■-«• rrr) P»7i 1+log. Rb~ \m.pr. / . . . . (311) This is a general expression for work division between the cylinders in terms of (a) ratio of expansion in each cylinder, initial and back-pressure ratio and cylinder ratio, or, in terms of (b) cut-off in each, associated with cylinder and pressure ratios. This expression Eq. (311) is less frequently used in its general form as above, than in special forms in which the work of the two cylinders in made equal, or the expression made equal to imity. The conditions thus found for equal division of work between cylinders may be expressed either (a) in terms of initial and back pressures, release pressm*e of low-pressure cylinder and ratio of L.P. admission volume to H.P. displacement, and cylinder ratio, or (b) cut-off in high- and low-pressure cylinders, initial and back pressures and cylinder ratio. Still more special conditions giving equality of work may be found (c) when the 262 ENGINEERING THERMODYNAMICS cylinder ratio is made such that equality of work is obtained at all loads^ by equalizing high and low cut-offs. (a) To find the first set of conditions, equate Eqs. (296) and (297) from the first part of this section, and by simplification there results, or •• F.F/ ^ Introducing the usual symbols and putting in addition Low-pressure admission volume _Ve_ _j p _ (rel.pr.)g . High-pressure displacement volume ~Fe ~ ^ <^~ (rec.pr.) ' Therefore, r(bk.pr.) 1^-1 rL(rerpr.),-,-J (inj,r) J» L (rel.pr.)i, J This is of value when a given release pressure is to be reached in the low pressure cylinder and with a particular value of low-pressure cut-off volume as fixed by x in terms of high-pressure cylmder displacement. (b) Again for equal division of work, make Eq. (311) equal to unity, whence, 1.1 1 1 1.1 1 (bk.pr.) Re or |^'j^'^+ifclog.|-^-^=0, Lh (m.pr.) Lh ItL which may be reduced to the following, solving for Rc^ ^ "2 lbk.pr .) • • ' • ^^^^^ Zif(in.pr.) Equal division of work for given initial and back pressure is to be obtained by satisfying these complex relations Eqt (313) between the two cut-offs, or their equivalent ratios of expansion in connection with a given cylinder ratio, or the relation between pressures and volumes in Eq. (312) equally complex. WORK OF PISTON ENGINES 263 (c) An assumption of eqiuil cut-off in both cylinders gives results which are of interest and practical value, although it is a special case. Eq. (313) then becon^es, when Zh^Zl or Rh—Rl, Ri _ / in.pr. Y^ \bk.pr./ (314) 1 X As would be expected, this may also be derived from Ea. (312), since 7 / \ and x=-,rT^—T under these conditions. (rel.pr.)L (bk.pr.) The receiver pressure imder these conditions is constant, and is, from Eq. (308) , . (in.pr.) (in.pr.) f.. .... .1. .^.^. (rec.prO=^^--^=y^;£y^ . . .(315) \bk.pr./ The high-pressure release pressure is not affected by any change in the low- pressure cut-ofif, and hence Eq. (309) gives the value of high-pressure release, pressure for the case. Low-pressure release pressure Eq. (310) may be expressed for the case of equal cut-ofF, (tel.pr.);-= fepirff = ^*[(ii^-P') (bk.pr.)]* \bk.pr./ (a) =^[(in-pr.)(bk.pr.)J (6) . (316) The foregoing equations up to and including Eq. (311), are perfectly general, and take special forms for special conditions, the most important of which is that of complete expansion in both cylinders^ the equations of condition for which are, referring to Fig. 82. Pc — Pd't which, when fulfilled, yield the diagram. Fig. 83. These equations of condi- tion are equivalent to fixing a relation between the cut-off in both the high- and low-pressure cylinders, and the volume of high-pressure cylinder with respect to the low-pressure volume, so that p ytf—Vf^y or symbolically, «-i(iS:) <" (317) 264 ENGINEERING THERMODYNAMICS Similarly the low-pressure cylinder cut-ofif volume must equal the high- pressure displacement volume or Dh^ZiJ)l, Z), "- Di. Re (a) (b)\ (318) indicating that low-pressiire cut-off is the reciprocal of the cylinder ratio. Making the necessary substitution there result the following equations for this cyde which, it must be noted, is that for most economical use of fluid in compound A B -<-• (ii i.Dr.) [ •H**/ ^z, iDh-* \ \ \ V N \ c <r 5c.pr.) D„ = T Tk- V ZlI>l \ 1 1 i 1 X ^ M 1 1 1 1 "^ E(rel 1 r» ^L 1 H Fig. 83. — Special Case of Cycles V and VI, Complete Expansion in both Cylinders of Compound Engine with Infinite Receiver and Zero Clearance. cylinders without clearance and with infinite receiver, and in which the same work is done as in Cycle I, for simple engines at best cut-oflF. From Eq. (308) R, _(in.pr.) Re ^^^(bk.pr.). . . (319) (recpr.) = (in.pr.)^^ = -^ 1 (in.pr.) Re (bk.pr.) From Eq. (309), (rel.pr Oh = ^-""^^-^ = (^^-P^') -^^-J^^ = RcQ^k.pv,) = (in.pr .)i. . (320) WORK OF PISTON ENGINES 265 From Eq. (310), (rel.pr.)x,--^^-^ 1 (in.pr.) "(bk-pr.) (321) Re (bk.pr.) Frran Eq. (311), High pressure cylinder work _ H-log.«/r-|^ lie Low pressure cylinder work - , , ^ (bk.pr.) •, •, 1 +log, Rl - ;. T. / flcflg (in.pr.; log, fix, (a) loge 1 z J7 lo& (fc) (322) Z For the case of most economical operation, that of complete and perfect expansion in both cylinders, there may be set down the four characteristic Ek|s. (304), (305), (306), (307) with suitable modifications to meet the case. These become Work of cycle = 144(in.pr.)Di,- / in.pr. \ ^ Vbk.pr./ _ / in.pr. \ \bk.pr./ = 144(in.pr.)Dz, loggfii Rv (323) (324) 1 / ^P-pr- \ / \ f t i. T ■D\ ^ r \ \bk.pr./ ,. xlogrfir (m.e.p.) (ref.toL.PO-i44^^=(m.prO-yj^p^ = (m.pr.)-^. . \bk.pr./ Work per cu.ft. supplied = 144(in.pr.) log, [ '^ ' ) = 144(in.pr.) loge Rv (325) 13,750 / bk.pr. X r.P.)\in.pr./ ^^^ Cu.ft. per hr. per I.H.P.== (m.e.p. ref . to L, 13,750 J_ (m.e.p.ref. to L.P.) Rv (b) . a (226) For equal division of work with complete expansion in both cylinders, the ratios of Eqs. (317) and (318) becomes (327) and this is evidently a case to which Eqs. (314) and (315) apply without change. 266 ENGINEERING THERMODYNAMICS Szample. 1. Method of calculating diagram, Fig. 82. Afisumed data for Case A: Pa=Pb- 100 lbs. per sq-in. abs. Fa « F» = Fm = cu.ft. Pn-Pd^Pe- 50 lbs. per sqin. abs. Vc-Vd^ .6 cu.ft. Pm^Pg- 10 lbs. per sq.in. abs. Vf='Vg^ 2 cu.ft. Pc = 60 lbs. per sq.in. abs. Vg = .8 cu.ft. To obtain point B: To obtain point F: P 60 n - 7, x^ = .6 X-TTi^ = .36 cu.ft. V 8 Pf--Pg X^ =50 Xiz =20 lbs. per sq.in. Vf z To construct the indicator cards: Lay off ND of the PY diagrams to equal the length of the card, and t^A peq)en- dicular to it at "N to equal the height of the card. Cut off equals AB-^ND, From .4 on card lay off this ratio times the length of the card. From D on the card lay ofif a perpendicular equal to CD of the PV diagram reduced by the same proportion as AN of the card is to AN of the diagram. Join the points B and C by a curve through points located from intermediate points on the PV diagram. The low- pressure card is constructed in same manner. Example. 2. A 12- and 18 x24-in. steam engine without clearance run8» on 150 lbs. per square inch absolute initial pressure, 10 lbs. per square inch absolute back pressure, and has a speed of 125 R.P.M. What will be (o) the horse-power for | cut-off in H.P. cylinder, (b) poimds of steam per I. H.P. hour, (c) terminal pressures, (d) L.P. cut-oflf for continuous expansion^ (e) work done in each cylinder. Nons: 8 for 150 lbs. =.332. (a) From Eq. (305) (m.e.p.) referred to L.P. cylinder is (in.pr.) ( 2 -floge fl/f +log« fli - ^ ) - (bk.pr.) . iChiCc\ iCc/ In this case Rh-^2, /ec = (J|y=2.25, «x.=2.25, since vol. of L.P. cyl. at cut-off must be equal to the entire volume of the, high for continuous expansion, hence (m.e.p.)=150Xr— r-T3X(2+.69+.81-l)-10=73.3 lb. sq. inch, and (m.e.p.) Lan ^•"•^•- 33,000 ^^^• (6) From Eq. (307) n ,* , -„-, 13,750 Zh 13,750 .5 ,, _ Cu.ft. per hour per LH.P.^^^;^;^ ______ x^, 41. 7, WORK OF PISTON ENGINES 267 (c) Emm Eq. (308) (rel.pr.)«^ = (m.pr.)ZH, -. •«■■■■ ■■>■• = 150X§-751bs.Bqin, and from Eq. (310) we have • (rel.pr.)i, = — ^ , 75 = r-rr « 33.3 lbs. sq.in. ^.25 (e) From Eq. (311) ^''"2.25""^V l+lo&^ ^ H.P. work ^Zh RcZl L.P. work , , , 1 (bk.pr.) Re ' l+loge Zl (in.pr.) Zh 1+.69 2.25 X. 44 .69 ,^^ .450, 1+.81-Jix2-^ ^-^^ 150 .5 or H.P,work =.456XL.P.work, also H.P. woik +L.P. work =282 I.H.P. Hence H.P. work =88 I.H.P. and L.P. work = 194 I.H.P. ■ ProtK 1. What must be the cylinder diametens of a cross compound engine to run on 100 lbs. per wjuare inch absolute steam pressure, 18 ins. of mercury vacuum and to develop 150 H.P. at a speed of 200 R.P.M. with J cut-off in each cyhnder, if cylinder ratio is 3 and stroke is 18 ins.? Engine is double-acting and assumed to have no clearance. Prob. 2. What will be the release pressure in each cylinder and the receiver pressure of the engine of Prob. 1? If cut-off were reduced to J in H.P. cyhnder, how would these pressures be affected and to what extent? How would the horse- p>ower change? Prob. 3. A 15- and 22 x30-in. infinite receiver engine has no clearance, a speed of loO R.P.M., initial pressure 125 lbs. per square inch gage. What will be the horse- power and steam consumption for a H.P. cut-off of }, i, }, }, and that value which will give complete expansion in high-pressure cylinder? Low-pressure cut-off to be fixed at |. ' Note: 6 for 150 lbs. gage = .363. 268 ENGINEERING THERMODYNAMICS Ptob. 4. What will be the release and receiver pressures, and the woik done in each cylinder for Prob. 3? Prob. 6. An 18 and 24x30-in. infinite receiver engine is to be operated so as to give complete expansion in both cylinders. What will be the cut-off to acoomplish this and what horse-power will result if the initial pressure is 100 lbs. and back pressure 10 lbs. per square inch absolute? Prob. 6. Draw the PV diagram for following cases. Cylinder ratio 1 to 2.5, (in.pr.), 100 lbs. per square inch absolute, (bk.pr.), 20 lbs. per square inch absolute, H.P. cut-off (a)-i, (6) -J, (c)=l. L.P. cut-off (a) -I, (6) -A, (c)=f. Prob. 7. For the following conditions find the horse-power, steam used per hour, receiver pressure and release pressures. Engine, 10- and 15x24-in. 150 R.P.M., 125 lbs. per square inch gage initial pressure, 2 lbs. per square inch absolute, back pressure, i cut-off in high-pressure cylinder, A cut-in low-pressure cylinder with infinite receiver. Note : B for 125 lbs. = .31 1 . Prob. 8. An infinite receiver engine is to develop 150 H.P. at 200 R.P.M. when initial pressure is 150 lbs. per square inch absolute. Cylinder ratio is 1 to 3 and back pressure is one atmosphere. What must be its size if the stroke is equal to the low-pressiue cylinder diameter for i cut-off in the high-pressure cylinder and J cut-off in the low-pressure cylinder? Prob. 9. Find by trial the cut-offs at which work division will be equal for an infinite receiver engine with a cylinder ratio of 2.5, an initial pressure of 100 lbs. per square inch absolute and a back pressure of 5 lbs. per square inch absolute? 10. Compound Aigine with Infinite Receiveri B^Kmential Law. No Clearance, Cycle VI. General Relations between Pressuresi Dunendonsy and Work. Again referring to Fig. 82, which may be used to represent this cycle also, the work of each cylinder may be expressed as follows, by the assistance of Eq. (254) derived in Section 4. Fi = 144Z)j,rZ^(rec.pr.)(?^^|^')-(bk.pr.)L • - . (328) (329) Vh where Zh is the cut-off in the high pressure, = v^ and Zl, low-pressure cut-off y = :=^. In combining these into a single equation for the total work, the term for receiver pressure (rec.pr.) should be eliminated. Referring to Fig. 82, hence (rec.pr.) =P*=P.=P»(^;)'^=(in.pr.)(^y, . . . (330) r..««J,..p,.>r(|)(i^)-X(|-)- +^'(^)"C-i^)]-*''-'"'l' • <'"» WORK OF PISTON ENGINES 269 a rather complex expression which permits of little simplification, but offers no particular difficulty in solution. Mean effective pressure referred to the low-pressure cylinder is (m..p. rrf. to l.P.,.a..pr.)[(|)(i:^)-i(^j- Work per cubic feet fluid supplied may be foimd by dividing Eq. (331) by the supply volume, which in terms of low-pressure displacement is (Sup.Vol.)=Z>i|^ (333) The consumption of fluid, cubic feet per hoiu* per indicated horse-power is Consimiption cu.ft. per hr. per I.H.P.=7 \ . t t> \ t^i • • • (334) '^ x- x- (m.e.p. ref . to L.P.) Re which is the same expression as for the logarithmic law. Multiplying this by li, the initial density of the fluid, pounds per cubic feet, gives cansumptUm, pounds i)er fluid hour per I.H.P. The receiver pressure has already been determined in Eq. (330). Release pressure of the high-pressure cylinder is (rel.pr.)^=(in.pr.)Z^*, (335) and for the low-pressure cylinder, (rel.pr.)L = (in.pr.) \^\ (a) _ (in.pr.) fi • (ft) (336) where Ry is the ratio of maximum volume in the low pressure, to, volume at R cut-off in the high, and equals -^. The distrubition of work between the high- and low-pressiu*e cylinders may be found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by means of Eq. (330) l s-Zn^-^ \ (Zh\ L \RcZl/ \ s-l J \m.pr. /J 270 ENGINEERING THERMODYNAMICS Equality of work in the two cylinders will be obtained if this expression is equal to unity, giving a complex relation between high- and low-pressurt cut-offS; cylinder ratio and ratio of initial and back pressures, to be satisfied. It is found at once that the simple conditions for equality in the case of logarith- mic law will not give equality of work for the exponential law. There is, how- ever, a case under this law which yields itself to analysis, that of complete expan- sion in both cylinders, without over-expansion. The conditions for equality of work for this case will be treated after deriving work and mean effective pressure for it. Complete expansion, without over-expansion, in lx>th cylinders may be represented by Fig. 83. „ AB „ NC NC MEf and since ' NC=Dh and ME=Di., ^ D„ NC Zl. ME 1 72/^ The true ratio of expansion =jBr== =--—=—-, but this is also equal to AB ^H^L liH 1 / in.pr. \ « \bk.pr./ due to the law of the curve, PbVb—PtVe. By means of Eq. (257) in Section (4) the work of the two cylinders may be evaluated, but since Wh = 144(m.pr.)Di,Z,r^ (l - Z«« '^\ «I44(m.pr.)Z).|^4^(l-Z..-i) Tr. = 144(bk.pr.)D.^4j(^- 1), ia) (P) Zl^ 8 Wl = 144(bk.pr.)i)L ~{R<f'^-l) (a) = lU(m.pT.)DLj^ (^) \Rc-'-1) (6) (338) .... (339) WORK OF PISTON ENGINES 271 The total work is evidently the same as that of a cylinder equal in size to the low-pressure cylinder with a cut-ofif equal to ^, working between the tic given (in.pr.) and (bk.pr.) and may be stated by reference to Eq. (257), Section 4, or by taking the sum of Wh and Wl given above, TF=144(in.prOI)L|^^ J (l-Zir'-0 + (^y" W which reduces to TF = 144(in.pr.)Dx, Zh s Rcs-l !'-(in (340) For this case of complete expansion in both cylinders, the ratio of high- to low-pressure work is given by division and cancellation, Wl 1-Zb-' w. (1).-. (/ic-^-1) (341) Ek}uality of work, obtained by placing this expression equal to unity, pro- vides the condition that ^--' = (l)'"'<HSr)'-'. or «-i Rc^ \bk.pr./ +1 1 (342) for equal work and complete expansion, and «-i 1^ p _ / bk.pr A • '^"■\in.pr. /; ' / bk.pr A « XVC— 7^ + 1 1 • -1 (343) Since Zl—i^- for complete expansion,Tand (in Fig. 83) PcVc—PfVf\ the receiver pressure, P^ is (rec.pr.) = (bk.pr.) (^'y=(bk.pr.)/?c', (344) in which Re will have the value given above if work is equally distributed. Example 1. What will be (a) the horse-power, (6) consumption, (c) work ratio, {d) receiver and release pressures for the following conditions? Engine 12 and 18x24 ins., running at 125 R.P.M. on initial air pressure of 150 lbs. per square inch absolute, and back pressure of 10 lbs. per square inch absolute, with \ cut-ofiF in high-pressure cylinder and continuous expansion in low-pressure cylinder. Exponent of expansion curve;«1.4 for compressed air, infinite receiver. 272 ENGINEERING THERMODYNAMICS (a) From Eq. (332) (m.e.p.) -(in.pr.)[(g) ('-^-^) -i(^)' H-iT)'irf^)] -a>--) which, on substituting values from above, gives for (m.e.p.) 63 lbs. per sq. inch. Hence, the indicated horse-power "242. (6) From Eq. (334) Compressed air per hour per I.H.P. = — - — -^g- cu.ft., m.e.p. tic which, on substitution, gives 13,750 .5 ,„^ ^^ (c) From Eq. (337) which gives Wl •<->(=?.) 1.4 2.25 1 .5 1.4/1 -G-ii)' .294. 10 2.251 ^ «, 1 2.25 X .4 150 2.21 and Hence and TrH+Trz.-242I.H.P. Trir=56 I.H.P. Wl = 184 I.H.P. (d) From Eq. (330) (rec.pr.) =(in.pr.) (^f^j = 150i .5 1.4 2.25 X _1_ 2.25. =57 lbs. per sq.in. From Eq. (335) (rel.pr.)^ = (in.pr.)Zi5r*, = 150 X (.5)^ * -57 lbs. per sq.in. WORK OF PISTON ENGINES 273 Fiom £q. (336) (rel.pr.)i. = (in.pr.) -^R'vi = 150 -i-21.85 -6.85 lbs. per sq.m. These values may be compared with those of Ex. 1, Section 9, which were for the same data with logarithmic expansion. Prob. 1. What will be the horse-power and steam used per hour by the follow- ing engine imder the conditions given? Cyhnders 18 and 30x48 ins., speed 100 R.P.M.y initial pressure 150 lbs. per square inch absolute, back-pressure 10 lbs. per square inch absolute, steam continually dry. Cut-off at first } in high-pressure and ^ in low, and then i in each infinite receiver. Prob. 2. The very large receiver of a compound pumping engine is fitted with safety valve which is to be set to blow at 25 per cent above ordinary pressure. The cylinder ratio is 1 to 3.5, and cut-offs are i in high and ^ in low. If initial pressure is 125 lbs. per square inch gage, for what must valve be set? What vacuum must be carried in the condenser to have complete expansion in low-pressure cylinder? Superheated steam. Prob. 3. A compoimd engine is to be designed to work on superheated steam of 125 lbs. per square inch absolute, initial pressure and on an 18-inch vacuum. The load which it is to carry is 150 horse-power and piston speed is to be 500 ft. per minute at 200 R.P.M. Load is to be equally divided between cylinders and there is to be complete expansion in both cylinders. What must be cylinder sizes, and what cut-offs will be used for an infinite receiver? Prob. 4. How will the economy of the two following engines compare? Each is 14 and 20x24 ins., runs at 200 R.P.M., on compressed air of 100 lbs. per square inch gage pressure, and 15 lbs. per square inch absolute exhaust pressure. Low-pres- sure cut-off of each is } and high pressure of one is i, the other, i. Infinite receivers. Prob. 6. A compound engine 12 and 18 X24 ins. is running at 200 R.P.M. on superheated steam of 100 lbs. per square inch absolute pressxue and exhausting to a condenser in which pressure is 10 lbs. per square inch absolute. The cut-off is i in high-pressure cylinder and i in low-pressure cylinder. Compare the power and steam consumption under this condition with corresponding values for wet steam imder same conditions of pressure and cut-off and infinite receivers. Prob. 6. The initial pressure of an engine is 150 lbs. per square inch absolute, the back pressure one atmosphere, the cylinder ratio 3. As operated, both cut-offs are at i. What will be the receiver pressure, high-pressure release pressure, and low-pressure release pressure? What will be the new values of each if (a) high-pressure cut-off is made f, (6) i, without change of an3rthing else, (c) low pressure cut-off is made i, (d) f, without change of anything else? Infinite receiver, s = 1.3. Prob. 7. In the above problem for i cut-off in each cylinder how will the release and receiver pressures change if (a) initial pressure is raised 25 per cent, (6) lowered 25 per cent, (c) back pressure raised 25 per cent, (d) lowered 25 per cent? Prob. 8. How many pounds of initially dry steam per hour will be required to supply an 18-in. and 24x30-in. engine running at i cut-off in each cylinder if speed 274 ENGINEERING THERMODYNAMICS be 200 R.P.M., initial pressure 100 lbs. per square inch gage and back preasore 5 lbs. per square inch absolute? Expansion to be adiabatic and receiver infinite. Note: d for 100 lbs. =.26. -fiup. Vol. OVER EXPANSION Fig. 84. — Work of Expansive Fluid in Compound Engines with Finite Receiver, Zero Clear- ance. Cycle VII, Logarithmic Expansion; Cycle VIII, Exponential Expansion. 11. Compound Engine with Finite Receiver. Logarithmic Law. No Clearance, Cycle Vn. General Relations between Dimensdons and Work when H.P. Exhaust and L.P. Admission are not Coincident. The diagrams, Fig. 84, while showing only two degrees of expansion, that of over and imder in both cylinders, suffice for the derivation of equations applicable to all degrees in either cylinder. Voliunes measured from the axis AL are those WORK OF PISTON ENGINES 275 occiipied by the fluid in either cylinder alone, while fluid voliunes entirely in the receiver, or partly in receiver, and in either cylinder at the same time are meas- ured from the axis A'V , No confusion will result if all volumes represented by points be designated by the (F) with a subscript, and to these a constant rep- resenting the receiver volume be added when part of the fluid is in the receiver. Then, High-pressure cylinder work is Fi,=P^F.(l+lo&pj-P„01oge(^^) (345) Low-pressure cylinder work is Pr^ = PnO log, (^^^)+PeFel0ge^-P, 7,. . . . (346) Total work Tr = P,7»(l+l0ge^)+Pn0 l0g-(^^) +PJ%log.^;-P„01og.(^^)-P,7,. . (347) These expressions include some terms not known as initial data and may be reduced by the following relations, and Fn0^p,{y,-\-O) =p,v,(^^}j Hence »-P.v.[>+,o.i:.(MO),c.(K^) Dividing by the low-pressure cylinder displacement, Vg, the result will be the mean eflfective pressure referred to the low-pressure cylinder, (M.E.P. ref. to L.P.) -4:['+'-f:+m'-(^°)+-f: 276 ENGINEERING THERMODYNAMICS A similar division but with the volume supplied, V^ as the divisor, gives Work per cu.ft. supplied -m'-m]-4:-----<»«' Also as in previous cases Cu.ft. supplied per hr. per I.H.P. - 13'^«^ .X& (351) (m.e.p. ref. to L.P.) Vg Of course, the weight per hour per I.H.P. follows from Eq. (361) by introduc- ing the density as a multiplier. While the last four equations can be used for the solution of problems, it is much better to transform them by introducing dimensional relations as in the previous cases developed. Let (rec.pr.)i= maximum receiver pressure P», which is also the initial admis- sion pressure for the low-pressure cylinder; (rec.pr.) 2= minimum receiver pressiu^ Pe, which is the terminal admission pressure for the low-pressure cylinder and that at which expansion begins there; <( _ receiver volume _2. — ^ a Q ^ ^bV _ y ^""high-pressure cyl. displ. Vc Dh ' Dl", Dl " Re Other symbols necessary are unchanged from the meaning imposed in Section (9). Substitution in Eqs. (348), (349), (350), and (351) gives the following set in a form for direct substitution of ordinary data: Work of cycle = 144(in.pr.)Z«Z),, 1 1+log. ^+log. ^ + (^+zfe)h 0+-f )-^^^(^+^)] } -l^(bk.pr.)i>. (a) = 144(in.pr.)|^ j 1+loge /2/.+log. Rl+ (i+^) [log. (l+^) -loge (l+-)l ) -144(bk.pr.)i5, (6) . (35) WORK OF PISTON ENGINES 277 (m.e.p. ref. to L.P.) = (in-pr .)^ ( 1 +log. ^+loge ^4 •tic [ iiH ^L -log.(l+^)]j-(bk.pr.) (6) (353) Work per cu.ft. supplied - 144(m.pr.) j 1+log, ^+log, ^ - 144(m.pr.) 1 1 +log. ie^+log. i2^+ (l +^) [log. (l +^) -log. (l +^)] j - 144(bk.pr.)i?cii:i/ (6) (354) Cu.ft. supplied per hr. per I.H.P. 13,750 X (m.e.p. ref. to L.P.) Re 13,750 X (m.e.p. ref. to L.P.) RhRc (a) % (355) It is desirable at this point to introduce a series of expressions fixing the relations between the dimensions, the cycle that may follow, and the fluctua- tions in the receiver pressure, and for the selection of cylinder and receiver dimensions for a required output of work and division of it between cylinders. In doing this it will be convenient to start with diagram points and finally substitute general symbols in each case. There will first be established the nummum and minimum receiver pressures and the fiuduations. Maximum receiver pressure = (in.pr.) ^^ ■1 A ^ir2/ RffRc ) ) (b) (356) 278 ENGINEERING THERMODYNAMICS Minimum receiver pressure P =Pk— • /. (rec.pr.)2 = (m.pr.) Yj)^ "" ^^'^^^Y'jMc ^ = (m.pr.)5-p- ib) • ■ (357) Fluctuation in receiver] pressure =(P»— Pe)=ft-7r. .*- (rec.pr.) i — (rec.pr.)2 = (m.pr.) — ;^— = (m.pr.) — (a) y = (in.pr.)^-- (5) Rky (358) It is interesting to note that the rninimum receiver pressure is exactly the same as the value of the constant-receiver pressure for infinite receiver y so that limit- ing the size of receiver does not affect the point E, but only raises point N higher, tending to throw more work on the L.P. cylinder for the same valve setting. The two release pressures Pc and P/ can be evaluated as in the case of the infinite receiver, as both these points lie on the common expansion line, which is not at all affected by the receiver-pressure changes, and the values are the same as for the infinite receiver, and are here reproduced from Eqs. (309) and (310) with new numbers to make the set of equations complete: (rel. pr.)^ = (in.pr.) (a) Rh = (in.pr.)Z« (6) ^ (369) (rel.pr.)L = (in.pr.) RcRh = (m.pr.)^ (in.pr.) Rv ^ (rel.pr.) ff Re (a) (h) (c) (d) (360) where Rv is the ratio of maximimi volume in the low- to the volume at cut-off in the high-pressure cylinder. Division of work between the cylinders cannot, as pointed out, be the same as for the infinite receiver, the tendency being to throw more work on the low as the receiver becomes smaller, assuming the cut-off to remain the same. As, therefore, equal division was obtainable in the case of infinite receiver with WORK OF PISTON ENGINES 279 equal cut-offs when the cylinder ratio was equal to the square root of initial over back pressure, it is evident that a finite receiver will require unequal cut- offs. As increase of low-pressure admission period or cut-off fraction lowers the receiver pressure and reduces the low-pressiu*e work, it follows that with the finite receiver the low-pressure cut-off must be greater than the high for equal work division, and it is interesting to examine by analysis the ratio between them to determine if it should be constant or variable. For equal toork division Eqs. (345) and (346) should be equal, hence by diagram points P,V,{l+log,^^-PnO\ogs (^-)=P„0 loge (^^^)+P.F.l0g.y^-P,7,. -p. y'{^yf) lofc (^) +f .n log, y-P.V.- hence for equal division of work, the following relations must be satisfied: _/bk^.\^ ^^ \m.pr. / ^ ' It will be shown later that when expansion is complete in both cylinders and work equal that the high-pressure cylinder cut-off or the equivalent ratio of expansion bears a constant relation to that of the low, according to ^=o^ (362) in which a is a constant depending only on the size of the receiver. It will also be shown that the cylinder ratio is a constant function of the initial and back pressures and the receiver volume for equal division of work, according to •^-Km^y (»') in which (a) is the same constant as in Eq. (362). It is impo tant to know if these same values will also give equal division for this general case. Substi- tuting them in Eq. (361) 21og.a=/H aRi /in.pr. \bk.pr. 1. 2S0 ENGINEERING THERMODYNAMICS Here there is only one variable, R^y the evaluation of which can be made by inspection, for if Rl _1 / in.pr. a\bk.pr/ the equation will become 21ogea = (l+y) loge (l+^)'-2 = 2(l+y) loge(l+i)-2, ' or ^^^[(i+i/)io«.(i+M-ij, (3W) which 18 a constant. LOO 1 II / / / / ■■■ / 71 1 U JL J / J / / 7 4 '?/ ^ f/ 1 \ <0 ^J 7 ^ / b> f 7 iff 7 7 r > 7^ .80 ' f J J f / / : 1 1 / / / 7 1 f / / / 7 MJ aik / /// \i /> / / f / |.eo / // T / / i / 7 1 ////' > / V r / 7 I /// / / / 7 / ^ / / / / 7 ///// / r / / RECEIVER VOLUME EQUALS y X H.P. DISPLACEMENT m^ /. / / m / / / jw/ ^ A / jao IwlL V / « \m/£ 7 \m E .». .40 .eo Hiffh Brenore Gut Oft «80 .100 Fig. 85.- -Diagram to Show Relation of High- and Low-Pressure Cut-offs for Equal Work in the Two Cylinders of a Finite-receiver Compound Engine with Zero Clearance and Logarithmic Law. As only one constant value of low-pressiu-e ratio of expansion or cut-off satisfies the equation for equal division of work when there is a fixed ratio between the values for high and low, that necessary for equal division with complete expansion in both, it is evident that equal division of work between the two cylinders cannot be maintained at all values of cut-ofif by fixing the ratio between them. As the relation between these cut-offs is a matter of some interest and as it cannot be derived by a solution of the general equation it is given by the curve, Fig. 85, to scale, the points of which were calculated. WORK OF PISTON ENGINES 281 A special case of this cycle of sui&cient importance to warrant derivation of equations because of the simplicity of their form and consequent value in estimating when exact solutions of a particular problem are impossible, 18 the caee of complete and perfect expansion in both cylinders. For it the following equations of condition hold, referring to Fig. 84, Fig. 86. — Special Case of Cycles VII and VIII Ck>mplete Expansion, in both Cylinders of the Finite Receiver Compound Engine. Zero Clearance. which when fulfilled yield the diagram. Fig. 86. These equations of conditions are equivalent to fixing the cut-off in both high- and low-pressure cylinders, and the volume of the high- with respect to the low-pressure volume. Accordingly, Vt=Ve Pe ^-"^^M w /2j5r = Re Vbk.pr./ Re {b) . . . . (365) Also for the low-pressure cylinder the cut-off volume must equal the whole ligh-pressure volume, or Db'=ZiJ)l- Therefore, Zl=^ (a) IhC Rt,^Rc {b) (366) 282 ENGINEEMNQ THERMODYNAMICS Substituting these equations of condition in the characteristic set Eqs. (352), (353), (354), and (355), there results the following for most economical operation: Work of cycle r=.44(in.pr.)/fc(g^>-{l+U*[(2^.)i]+.*R, +a+y) log. ^1+i-) -log,(l+M I -144(bk.pr.)Z>i; = 144(bk.pr.)i)Jog.(^) (o) 1 / 'P-pr- \ = 144(in.pr.)Z)x ^^^ = 144(m.pr.)D^ ^^" (6) \bk.pr./ (367) W (m.e.p. ref. to L.P.) = 144: 144i)i /ui \i /in.pr. \ ,. . ^Vbk.pr./ ,. .log^Rv = (bk.pr.) log. (^j = (m.pr.) ^^^^ - (m.pr.) -|^ . \bk.pr./ (368j W Work per cu.ft. supplied =«-yr- = 144 .pr , (bk.pr.)D L I /m.pr. \ ..... ., /m.pr. \ t \i| \ loge I VI I = 144(m.pr.) log, (rr-^— ) p / bk.pr . \ ^ '^\bk.pr./ ^ *^ ' ^Xbk.pr./ \in.pr. / ^ = 144(in.pr.) log, /Jr. . (369) Cu.ft. supplied per hr. per I.H.P. 13,750 ^/bk.pr.\ 13,750 ^^ 1 (m.e.p. ref. to L. P.) \in.pr. / (m.e.p. ref. to L. P.) Rv* For this special case of best economy the receiver and release pressures, of course, have special values obtained by substituting the equations of condition Eqs. (365) and (366), in Eqs. (356), (357), (358), (359), (360). (rec.pr.) , = (in.pr.) (^+^1-) = (m.pr.) (^+g) (rec.pr.)2 = (in.pr.) ^r^ = (in.pr .)^- = (bk.pr .)i2c. . . . (372) WORK OF PISTON ENGINES 283 Therefore (rec.pr.)i — (rec.pr.)2 = (in.pr.)p — ; (373) (rel.pr.)H=(in-pr.)D"~0>'^-Pr-)J'2c=(rec.pr.)2; .... (374) (rel.pr.)x,=(in.pr.)p-n- = (bk.pr.) (375) •These last two expressions might have been set down at once, but are KTorked out as checks on the previous equations. For equal division of work in this special case the general Eq. (361) becomes )r log.g=2[(l+.)log.(l+i)-l]. Therefore r |H = ^<l+»)lo..(l+i-)-l] = o2 (376) This term, a, has already been used in previous discussions of equality of ^orky while the derivation of its value has not been made up to this point. This indicates that ratio of cut-offs or individual ratios of expansion is a Hmction of the receiver size for equal division of work. From Eq. (376) the cylinder ratio can be found in terms of a, and the ratio )f expansion. Referring to Fig. 86, Rh V, Y? Rl Vf V,Vf =a2, F.-aVlvT/, ^"^^'"""^/^aWl^WS^^aV^^^ • • • ^^^^^ v. aW, whence the q/linder ratio is equal to a constant depending on the receiver sizcy muUiplied by the valvs for the infinite receiver, i.e., the square root of the initial divided by back pressure. 284 ENQINEEBINQ THERMODYNAMICS The high-pressure cylinder ratio of expansion is and the corresponding value for the low-pressure cylinder is . (378) iJ.=^'= For convenience in calculation Table, XII of values of a and o' is added for various size of receivers. Table XII. Receiver Vol. H.P.Cyl. Dwp."'' ^^j;(i+if)iog,(i+i.)-i] o* .5 1.016 2.64 .76 1.624 3.67 1.0 1.474 2.17 1.6 1.322 1.76 2.0 1.243 1.66 2.5 1.198 1.437 3.0 1.164 1.369 4.0 1.1223 1.262 6.0 1.0973 1.204 7.0 1.0690 1.143 10.0 1.0478 1.008 14.0 1.0366 1.068 20.0 1.0228 1.046 Infinite 1.0 1.0 At the end of this chapter there is presented a chart which gives the relation between cylinder and receiver volumes, cylinder ratio, and high- and low-pressure cut-offs graphically. The corresponding values of maximum and minimum receiver pressure for equal division of work for this case of best economy are (»c.pr.),.(bk.p,.,l^(l+l)-^^&SSS^l+l) . (», /~«r.,^ -^K^«r^l /in.pr. V(in.pr.)(bk.pr.) (rec.pr.)2 = (bk.pr.)-^y^ (jec.pr.h-{rec.pr.h = ^^^^^^^^. . . ay (381) (382) WORK OF PISTON ENGINES 286 Example 1. Method of calculating Diagram, Fig. 84. Assumed data for case A: Pa'^Pt" 120 lbs. per 8q.in. aba. 7. - V* - F« -0 cu.ft. Pm ^Pf « 10 lbs. per 8q.in. abs. Vb * .4 cu.ft. 7i-l cu.ft. 0-1.2cu.ft. 7r- .8cu.ft. To find point C: To obtain point E: Pe "Pb-TT- « — 5-—- -60 lbs. per sq. inch. ^ ^Vb 120 X. 4 ..„ . , Pb "PvtT - — :; — -48 lbs. per sq. inch* ' • 1 To obtain point D: Pe(7*+0)-Pd(7c+0) or Pd-^^^-53 lbs. per sq. inch. To obtain point N: 48 v2 2 P»(0+V»)-P.(F.+0) or P» -^^^jy^«88 lbs. per sq. inch. To obtain point F: P/-— Jr-^ - — - — -24 lbs. per sq. inch. Vf Z Example SL Find (a) the horse-power, (6) steam used per hour^ (c) the release and receiver pressures for a 12- and 18 x24-in. engine with receiver twice as large as the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-ofiFs \ in high-pressure and such a value in the low pressure as to give complete expansion, and [ learances zero. (a) Frtnn Eq. (353) (m.e.p.) «(in.pr.) 9 1 RhRc l+log, Bff +log, «£ + e^)h('-«^)-'-K)]}— ■ ffrhich on substituting the above values gives (m.e.p.) 2x2.25 l+.69+.81+(l+2-'^^'' 2.25 :«?(.8-.8) -10 -73.3 lbs. hence I.H.P. -282. (6) From Eq. (355) we have n e. ^ u u 13,750 1 13,760 1 Cu.ft. steam per hour per ho^e-power - ^^^^^ Xj^ --^3- X^^^^^ -41.7, 286 ENGINEERING THERMODYNAMICS (c) From Eqs. (356) and (357) for maximuin and minimnm receiver preflsures respectively: ( ^•P'-K^+^) *^^ ^^P'-^fe' maximum receiver pre88ure-150^— — +~-;25) "^^'^ ^^^' ^^ ^' ^^^ 2.25 minimmn receiver pressure = 150 X^--v^= 75 lbs. per sq. inch. From Eqs. (359) and (360) for release pressures [m.pr. )liM and , high pressure cylinder release pressure » 150 X. 5 « 75 lbs., per sq. inch. 150 low pressure cylinder release pressure —-777 «33.9 lbs. per sq. inch. .444 These results may be compared with those of Example 1 of Sections 9 and 10, which are derived for same engine, with data to fit the special cycle described in the particular section. Note: In all the following problems clearance is to be neglected. Prob. 1. A 12- and 18 x24-in. engine has a receiver equal to 5 times the volume of the high-pressure cylinder. It is running on an initial pressure of 150 lbs. per square inch gage and exhausts to the atmosphere. It has a speed of 150 R.P.M. and the cut-o£^ are iV and \ in high- and low-pressure cyUnders respectively. What is the horse-power and the steam used in cubic feet per hour? Prob. 2. What will be the release pressures, and yariation of receiver pressure ia an engine in which the cylinder ratio is 3, cut-offs | and I, in high and low, initial pres- sure is 100 lbs. per square inch absolute, and receiver 2 times low-pressure cylinder volume? Prob. 3. Show whether or not the following engine will develope equal cylinder work for the conditions given. Cylinder diameters, 15 and 22 in., initial pressure 135 lbs. per square inch gage, back pressure 10 lbs. per square inch absolute, cut-offs I and {, receiver volume 4 times high-pressure cylinder, strokes equal. Prob. 4. For the same conditions as above, what low-pressure cut-off would give equal work? Prob. 6. What will be the most economical load for a 16- and 24x30-in. engine running at 125 R.P.M. on 150 lbs. per square inch absolute initial pressure and at- mospheric backpressure? What will be the economy at this load? Prob. 6. What will be the release and receiver pressures for the above engine if ^he receiver has a volume of 15 cu.ft.? WORK OF PISTON ENGINES 287 Prob. 7. F!nd the cut-offs and cylinder ratio for equal work division and complete expansion when initial pressure is 150 lbs. per square inch absolute and back pressure is 10 lbs. per square inch absolute, receiver four H.P. volumes. Prob. 8. WiU a 14- and 20 x20-in. engine, with a receiver volume equal to 5 times the H.P. cylinder and running on { cut-off on the high-pressure cylinder and I cut-off on the low, with steam pressure of 100 lbs. per square inch gage and back pressure of 5 lbs. per square inch absolute, have complete expansion and equal work distri- bution? If not, what changes must be made in the cut-off or initial pressure? Prob. 9. What must be the size of an engine to give 200 1.H.P. at 150 R.P.M. on an initial steam pressure of 150 lbs. per square inch absolute, and 10 lbs. per square inch absolute back pressure, if the piston speed is limited to 450 ft. per minute and complete expansion and equal work distribution is required? Receiver is to be 6 times the volume of high-pressure cylinder and H.P. stroke equal to diameter. 12. Compoimd Engine with Finite Receiver. Exponential Law^ No Clearance. Cycle Vm. General Relations between Pressures, Dimensions, and Work, when EBgh Pressure Exhaust and Low-pressure Admission are Independent. The diagram Fig. 84 may be used to represent this cycle, as well as cycle VII, by conceiving a slight change in the slope of the expansion and receiver lines. Using the same symbols as those of the preceding section, and the expression for work as found in Section 7, Chapter I, .iuD.Uia.^.)z.+^Ji^(i-z.--y^^y[i.(^)--']l but (rec.pr.)i=P«=P»(prl [—q-J . or (rec.pr.), = (m.pr.)(^ ['^-) and the last term in the equation for Wh within the bracket may therefore be written or 288 ENGINEERlNa THERMODYNAMICS and hence by simplifying the first two terms also, •^.-«<'-p')'^'{--.--(^-)'l«k+')'['-fe)'"']}-« Work of the low-pressure cylinder may be expressed in tenns of pressure and volumes at N, E, and G, but it is convenient to use instead of the pressure at N or at E, its equivalent in terms of the point B, The pressure at iV^ is (rec.pr.). = (in.pr.)(^)'(^+l)' and when multiplied by the receiver volume vDh, it becomes (rec.pr.).yD,-(in.pr.)D.Z.(X)(^)'(^^+l)- = (in.pr.)D.Z.(^y-^(^^+l)' At E the product of pressure and volume is (rec.pr.)2 xZlDl = (in.pr.)^HZ)ir (^f J Using these quantities, the following equation gives the work of the low- pressure cylinder: + (^y~*[l-z/"*]|-144(bk.pr.)Z)i„ . (384) and the total work is, by adding (Ws) and (Wl), — <-»''!^-^('---+ff)-(«k+')"[c-fir" -C-+ferJ'"]+(fe)'"'('-^''"')}-'"«''''"'"'' • • "*" This Eq. (385) is the general expression for work of the zero clearance com- pound engine with exponential expansion, no clearance, and finite receiver. From this the following expressions are derived: (m.e.p. ref. to L.P.) -^i?r^|{.---'+(t)-'(A-)'[C-^0'"' WORK OF PISTON ENGINES 289 Work per cu.ft. supplied is -(H^&rJ"1+(o;)'""-^'-'))-'«»''-'"-'fe • ««'> Cu.ft. supplied per hr. per I.H.P. 13,750 Zu. .ggg. (m.e.p. ref. to L. P.) Re (rec.pr.)2 = (in.pr.)(^Y; (389) (rec.pr.)i = (in.pr.)(^)'(l+^)'; .... (390) (rel.pr.),r=(in.pr.)ZH*; (391) (rel.pr.)i,=(rec.pr.)2Zj,* = (in.pr.)f^y (392) If work is equally divided between the cylinders, Wh, Eq- (383), and Wl, Eq. (384), will become equal, henoe -— (^)-(^.-)'[-fen-(f)-(ife«)' This equation shows conditions to be fulfilled in order that an equal division of work may be obtained. It does not yield directly to a general solution. When expansion is complete in both cylinders, Z, 4 and (^ = (|^)*. Rc \in.pr. / \Rc/ Introducing these values in the general expression Eq. (385) for work of this cycle, it may be reduced to the following: F = 144(in.pr.)Z^D^^[l-(|^y"'] (394) From which are obtained (m.e.p. ref.toL.P.) = (in.pr.)|^^[l-(^y""n. . . (395) Work per cu.ft. supplied = 144(in.pr. -3^ l — (^j (396) s 290 ENGINEERING THERMODYNAMICS Cu.ft. supplied per hr. per I.H.P. = 7 * . y t^\ (p^) ^^ ^ ^ (m.e.p. ref . to L. P.) \/2cv 13>750 (bk:PL-)\ (397) (m.e.p. ref. to L. P.) \m.pr. / If work is equally divided and complete expansion is maintained in both cylinders Eq. (381) becomes 8 -.-- .-^['-(,-^.n which may be simplified to the form. 2Zr-^ 'H >+^hfer]}-^ — <- where Rv is the ratio of maximum low-pressure volume, to the high-pressure volume at cut-ofif, hence ZRc Jttv and the value of Rv may be found from original data. «'- (S£/ « Eq. (398) may easily be solved for Zh, from which the required cylinder ratio may be found by, Rc=ZhRv (400) This is the cylinder ratio which gives equal work in the two cylinders and complete expansion in both, when used with the value found for the high- pressure cut-oflf Zffy the assumed initial and back pressures, and the assumed ratio, y, of receiver volume to high-pressure displacement. WORK OF PISTON ENGINES 291 Example. Find (a) the horse-power, (b) steam used per hour, (c) the release and receiver pressures of a 12^ and 18x24-in. engine, with a receiver twice as large as the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-offs i in the high and such a value in the low as to give complete expansion. Exponent for ex- pansion curve = 1.4. (o) From Eq. (386) (m.e.p.) = ( in.pr.) Zh 8-1 Re 8 — — bk.pr which, on substituting above values, gives 150 ^5_ J 4 .4 ^2.25 K-"*"-^" "^V4.5/ (±y 4.5 1.4 2.25 X 1 +1 2.25 4.5 .4" 4.5 +2.25 X 2.25. .5 2.25 X2.25 {b-(^']]- 10 or hence (m.e.p.) =57.5 lbs. per sq.in., I.H.P. =221. (6) From Eq. (388) Cubic feet of steam per hour per horse-power = ^3/50Zflr m.e.p. Re 13,750 .5 ^^^ ^, -575- ^2:2-5 =^-2 ^"•^^•' hence total poimds per hour will be 53.2 X221X. 332 =3910. From Eqs. (389) to (392): (rec.pr.)x = (in.pr.) (£|j (l +^^) , (rec.pr.)a = (in.pr.) (^-^ , (rel.pr.)tf = (in.pr.)Zi/*, (rel.pr.)i, = (rec.pr.)iZ£,*, 292 ENGINEERING THERMODYNAMICS These, on substitution of the proper numerical values, become: (rec.pr.)i = 150 X l^j I Xl +tt) =75 lbs. per sq. inch, (rec.pr.)i = 150X(.5)^* =57 lbs., (reLpr.)£r = 150 X (.5)^* =57 lbs, (rel.pr.)L = 57 X (^^j ' =32.1 lbs. " Note: In all the following problems clearance is assumed to be zero. Prob. 1. A 12x18x24 in. engine is running on superheated steam of 150 lbs. per square inch absolute pressure, and exhausts to the atmosphere. If the speed is 100 R.P.M., high-pressure cut-off J, low pressure cut-off i, and receiver volume 10 cu.ft., what horse-power will be developed and what steam used per hour? Prob. 2. What would be the effect on the power and the economy of (a) changJTig to wet steam in the above? (6) to compressed air? Prob. 3. What would be the receiver and the release pressures for each caae? Prob, 4. Will there be equal work distribution between the two cylinders? Prob. 5. It is desired to obtain complete expansion in a 14x22x36-in. engine running on fluid which gives a value for a of 1.2. Initial pressure is 100 lbs. per square inch gage, and back pressure 5 lbs. per square inch absolute. What must be the cut-offs and what power will be developed at 500 ft. piston speed? Receiver =3 XH.P. volume. Prob. 6. How large must the receiver be for the above engine in order that the pressure in it shall not fluctuate more than 5 lbs. per sq. inch? Prob. 7. An engine is to run on steam which will give a value of « = 1.1, and t^ develope 500 horse-power at 100 R.P.M. Piston speed is not to exceed 500 ft. per minute. Steam pressure, 150 lbs. per square inch absolute, back pressure, 5 lbs. per square inch absolute. Complete expansion and equal work distribution, for this load are to be accomphshed. What will be the cyUnder sizes and the high-pressure cut-off if the receiver is to be 3 times the high-pressure cyhnder volume? Prob. 8. What will be the steam used per hour by the engine of Prob. 7, and what wnll be the variation in the receiver pressure? Prob. 9. If the high-pressure cut-off were halved, how would the power and economy be affected? 13. Compound Engine without Receiver, Logarithmic Law. No Clear- ancei Cycle IX. General Relations between Dimensions and Work when High-Pressure Exhaust and Low-Pressure Admission are Coincident Such a peculiar case as this admits of but little modification of the cycle compared with the receiver cases, b; cause the low-pressure expansion is necessarily a direct continuation of the high pressure without any possible break. There can be no over-expansion in the high nor can expansion there be incomplete, as there is, properly speaking, no back pressure with which to compare the high-pressure cylinder terminal pressure. There may, however, be over and incomplete expansion in the low-pressure cylinder. It might appear that the high-pres- sure cylinder negative work was equal to the low-pressure admission work, as each is represented by the area below DC, Fig. 874, but this is not the case, since WORK OF PISTON ENGINES 293 le diagram is drawn to two different scales of volumes, showing the pressure- roke relation between high and low. This is apparent from the diagram, Fig. rC showing fluid volumes in each cylinder to a single scale on which A BCD the wor k done in the high-pressure cylinder, ABD'EF the whole work, whence CD'EF is the part done in the low-pressure cylinder. There is, of course, ) low-pressure cut-off or even admission as ordinarily considered. The cyde, » far as the work to be done is concerned, is the same as for a simple engine, B.P.C]rL Voh. 1 i A B. A ,8 A ^ \ \ \ H.P. V^ \ \ \ \ H.P.W 1 ork V V D \ c \ ic C ) v* b \ F J V \ v Vs **-> L P iH' D ^ • ^ ^ D L..r . ^ E L E ^P. Woi k r c E- IDICATOR CARDS OF EQUAI ^^ r r BASE AND HEIGHT L.1 *.Qr LV, 4.. iXSE D lAGfl AM( ►F F .UID WOI Dte- MR RIBl IGAF mo OLE )6 DM ORAM ON UNEQUA EQUAL i L VOLUM rfROKEI E6CALE 1 >FC YLIK OER 4 qI ^ A R A n \ M r 9. \ Q \ \ \ 45 \ \ •< C \ -< V ^^ 1 c H.I *.Cy .W. r^ P cl ^-^ \ c> \ J /I > -^^ ••^ A /I > ^1 s ^ --^ Q y V / 'R" ^ /^ "T^ "v *--, ^ -^ .^^ ^' y y r — 1 D -^ ^ ,— 1 ._!_ •«M« :=a D' D' — ,<^ ,^' b nw^ a. ^-^ I ( f F 1 ;p.jcyi DderW »rk r — C- ^ ^-^ . r L l^jT ^ '/ ^ INDtVlCUAL CYUNKR WORK SMOWN TO SAME SCALE OF PRESSURE AND VOLUME N K L HIGH AND LOW PRESSURE CYLINDER DIAGRAMS PLOTTED TO SAME /M9 "iQ. 87. — ^Work of Expansion in the No-receiver Compound Engine, Zero Clearance, Cycle IX, Logarithmic Expansion, Cycle X, Exponential Coincident Piston Movement. nd the only reason for introducing formulas for overall work, work per cubic 3et supplied, {m.e,p. referred to low), and fluid consumption, is to put them into orm for immediate substituiion of dimensional relations. Because of the absence f cut-off in the low, the distribution of work between high and low will lepend solely on the cylinder ratio and high-pressure cut-off, for, the earlier the ligh-pressure cut-off, and the larger the high-pressure cylinder, the greater he fraction of the total work that will be done there, as there is only a fixed ffnount available, and the less there will be left to be done in the low 294 ENGINEERING THERMODYNAMICS The diagrams of the two cylinders are plotted to combined axes in Fig. 87D. The points Q and R at equal heights. KN is the L. P. displace- ment, and KG that of the H.P. It has been shoVn in Section 8, that the expansion lines CD and CD' may be plotted to the axes LN and LXM, the point X being the intersection of NQ and KR extended, and that the distance and also ^^-w^.-'>--D^r'''sh- ■ ■ ■ « ^=^+'''-5^- <*»' Hence the work area under CD is W„ =GLXPc log. p = 144(rel.pr.)ir77^4r 'o& ^ ■ but (rel.pr.V = (m.pr.)Z^, hence Tr^ = 144(in.pr.)ZirDH|l+loge^-^^-^logeiec|. . (404) Again the work area under C'T^ is W^t. =KLXPc log.£i=144(reI.pr.)/r7^^ log. ^, hence TrL==144(in.pr.)Z^DH(^^1log. ftc-144(bk.pr.)D^, . (4C5:) and the total work, Tr = 144(in.pr.)ZHDH] 1+loge ^^ R^i^og, Re Re +^3l^^&^ -144(bk.pr.)Di = 144(in.pr.)Z^Dtf l+log.i^+(^^-.^Jloge/2cl -144(bk.pr.)D^. WORK OF PISTON ENGINES 296 But \Rc-l Rc-l) ' and 1 R lo& -^+l0g. iJc = loge, -^ = loge Ry, SO that Tr-144(m.pr.)^ifZ)ir(n-Iog.|^)-144(bk.pr.)DL, . . (406) which shows by its similarity to the work o| the simple engine that, as before stated, the total work is the same for this cycle as if the entire expansion were made to take place in a single cylinder. This same result could have been attained in another way sufficiently interesting to warrant se tting it down. Since the low-pressure work is repre- sented truly to scale by C'D'EFj Fig. 87C, the mean eflfective pressure of the low-pressure cylinder is given by the area divided by F«. By contracting all volumes proportionately, CD' takes the position CD' and C'F the position CF\ hence area CiyEF F,-F/. -P. represents the mean effective pressure in the low-pressure cylinder just as truly. Therefore, L.P. cylinder work = ( y ^y P. j 7, As the high-pressure work is (total— low), H.P.cylmderwork=P»F»(l+log. y\-P»V»-PcVly^^\og,^^ ^P^V, Introducing symbols L.P. cylinder work=144(in.pr.)ZirZ)ir(p^)log,i2c-144(bk.pr.)Z)ji. (407) H.P. cylinder work=144(in.pr.)^fl2>H[l+log,|^-g%| log. fie] = 144(in.pr.)ZKD«ri+log.^-^5rilo8'.^]- • (*08) which check with Eqs. (404) and (405). 296 ENGINEERING THERMODYNAMICS Dividing the total work by the low-pressure cylinder volume and the hi^- pressure admission volume in turn, (m.e.p. ref . to L.P.) = (in.pr.)-^{ 1+loge y-j" (bk.pr.) (a) = (in.pr.) pVf 1 +I0& iRnRc)] - (bk.pr.) (6) . . (409) Work per cu.ft. supplied = 144( in.pr.)ZH( 1+lofe Rc Zi | — (bk.pr. )Rc (a) 1 = 144(m.pr.):^[l+log.(ftiiBc7)]- (bk.pr.)ifc (6) . (410; Cuit. supplied per hr. per I.H.P. = 7 ' . ^ p\ X-tt («) ^^ ^ ^ (m.e.p. ref. to L.P.) Rc 13,750 1 (m.e.p. ref. to L.P.) RhRc (b) '. . (411) For equal division of work there can obviously be only one setting of the high-pressure cut-off for a given cylinder ratio and any change of load to be met by a change of initial pressure or of high-pressure cut-off will necessarily unbalance the work. Equating the high-pressure and low-pressure work expressions, Eqs. (404) and (405), l+loge^j — P — = loge/2c = Rc Rc-l logeRi / bk.pr. \ Rc \in.pr. / Zh or ,,, 1 , (bk.pr.) Rc Rc+l . j, ^ Another relation exists between Zu and Rc, namely, that Zji fRc where Rv is the ratio of volumetric expansion. Then but R^ 2R„ f hence Rc^c-"^ Rc^c-^ 2R, log.«A-^ Rv = l+/bk:P£:)fl^ (412) \m.pr. / ^ ' WORK OF PISTON ENGINES 297 With this formula it is possible to find the necessary ratio of cylinder displacements for given initial and back pressures and for given ratio of expansion Rv For convenience in solving this, a curve is given in Fig. 88 to find value 2R of Re when Rc^c"^ has been found. 1 9 ^ ^ «% mm i ^ ^ > ^ / 8 / / 1 T 26 6 76 U K) nn- 126 160 2Kc Values of (Rc)"K«^ 2A. Fio. 88. — Curve to Show Relation between Values of Re and {JRk)^c~^ for Use in Solving Eq. (412), Giving Cylinder Ratio in Terms of Ratio of Expansion for the No-receiver Compound Engine without Clearance. The complete expansion case of this cycle results from the condition Pd=P« or (rel.pr.)L=(bk.pr.) or ^^=(vX^)> which when applied to Fig. 87, transforms the diagrams to the form Fig. 89. It also follows that (bk.pr.)Dx = (in.pr.)ZjyDjy and Re — /hupj[i\ Vbk.pr./ These conditions will, of course, reduce the total work Eq. (406) to the common value for all cycles with logarithmic expansion and likewise those for mean effective pressure, work per cubic foot supplied, and consumption. For the equal division of work imder this condition, Eq. (412), becomes 2R. R R -1 C C = 7.39/2, (413) since (. — ^J/2r=l and R may represent ratio of expansion or ratio of \m.pr. / 298 ENGINEERING THERMODYNAMICS initial to back preasures, these being equal. Fig. 90 gives a curve showing the relation between cylinder ratio and ratio of expansion established by the above condition. p H J. Cyl. Volg. 2 A B A I \ \ ® \ \ \ c ) / / y / D ^^^ F L-J^ _L- J_ P A B \ \ \ (D \ k V \ c \ > N "^ >v^ F n ^^^ 1 ~T^ 6 4 8 2 1 L.P. Cyl. Vols. p A J \ \ \ © \ C \ \ c V ) v X > / > \ y / ,^ "^v ^-. 1 — D rf<^ ^ — — ^"^, fc^ E Fig. 89. — Special Case of Cycles IX and X. Complete Expansion in both Cylinders of the No-Receiver Compound Engine, Zero Cleamace. Example 1. Method of calculating Diagram, Fig. 87. A, As described in the text this diagram is drawn to two- volume scales, so that there may be two volumes for one point. WORK OF PISTON ENGINES Assumed data: Pa =P» - 120 Iba. per sq.in. abs. Fa - F« - F< - F, - F/ =0 ou.ft. P.^Pf- 10 lbs. per Bq.in. abs. F.-l cu.ft. Fc=2cuit. Fi-F.-ScuJt. To locate point C: To locate point D: To locat« intermediate pcunts from C to D. The volume at any intermediate point is (the volmne of low-pressure cylinder up to that point) + (volume ot lii^- S I Fio. 90. — Curve to Show Relation between Values of Rct the Cylinder Ratio, and R the Ratio of Initial to Back Preaeure tor Complete Expanaion in the No-receiver Corapound Bogine without Clearance (Bq. (413).) preaauie cylinder from that point to end of stroke), e.g., at f stroke the volume in low, is .75 X5, and the volume in the high is .25 X2, or total 4.25, and the pressure at that point is found by the PV relation as above. B. ABBumed data: Pa P, =P.- 120 lbs. per sq.in 10 lbs. per sq.iii. .abs. ahs. Va •"F/ = = F.- =0 cu.ft. ■5 cu.ft. '\ cu.ft. To locate pmnt D -^ 120 X 1_, 24 lbs. per sq.in. 5 Intermediate points from B to D found by assuming volumes and computing pressures from the PV relation as above. C. Figure ABCD constructed as in A. Figure C'D'EF is figure CDEF of A to the same pressure scale but to a volume scale 2.5 times as large. D. Figures constructed as in C. 300 ENGINEERING THERMODYNAMICS To draw indicator cards. The volume and pressure scales are chosen and from diagram A^ a distance AB is laid off to the volume scale, AD is then laid off equal to AD of diagram A to the pressure scale. Point C is located to these scales and joined to B and D by drawing curves through the intermediate points plotted from the FY diagram to the scales of the card. For the low-pressure card EF is laid off to the volume scale, and FC" and ED' to pressure scale. C and U are then joined in same manner as C and D for high-pressure card. Example 2. Find (a) the horse-power, and (6) steam used per hour for a 12 Xl8 x24 in. engine with no clearance when initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-off in the high-pressure cylinder is i, there being no receiver. (a) From Eq. (409) we have (m.e.p.)=(in.pr.)^5-^-[l +loge(/^///2c)] -(bk.pr.), = 150X^ ], — X(l+.8)-10=60 1bs. sq.in. JX^.^0 hence I.H.P.=192. (6) From Eq. (411) we have 1 o rr c r\ i Cubic feet of steam per I.H.P. per hour =7 — ^ rX (m.e.p.) RhRc _ 13,750 J ~ 50 ^2X2.25" ' hence the weight of steam used per hour will be 61 .2 X .332 X 1 92 = 3890 pounds. Example 3. What will be the cylinder ratio and the high-pressure cut-off to give equal work distribution for a ratio of expansion of 6, an initial pressure of 150 lbs. per square inch absolute and back pressure of 10 lbs. per square inch absolute? Ratio of baek to initial pressures is .067 and hence from Eq. (412) or Rv =6, 2Rc lofo%^ 1.40, R^R^-i =24.36, and from Fig. 88 Re =2.8. p o Q From the relation Z/f = h" = high-pressure cut off =-^ =.446. tiv o WORK OF PISTON ENGINES 301 Prob. 1. A compound locomotive has no receiver and runs on an initial pressure of 175 lbs. per square inch gage and atmospheric exhaust. The cylinders are 18 and 30x42 in. The steam pressure may be varied, as may also the cut-off to a limited degree. For a speed of 200 R.P.M. and a constant cut-off of f, find how the power will vary with initial pressure and for constant initial pressure equal to boiler pres- sure show how the power at the same speed will vary from i cut-off to full stroke. Prob. 2. Show how the steam used per horse-power hour will vary in above problem. Note: 8 for 175 lbs. =.416. Prob- 3. With the cut-off at f , what should the initial pressure be to give equal work distribution? Prob. 4. With full boiler pressure and f cut-off what would be tenninal pressure in the low-pressure cylinder? Prob. 6. What must be size of cylinders for a tandem compound engine with negligable receiver volume to run at 125 R.P.M. with complete expansion and equal work distribution on an initial pressure of 125 lbs. per square inch gage and a back pressure of 5 lbs. per square inch absolute, when carrying a load of 500 horse-power, the piston speed to be less than 500 ft. per minute? Prob. 6. What will be the steam used by the above engine in pounds per hour? Note: 5 for 125 lbs. -.311. Prob. 7. A builder gives following data for a tandem compound steam engine. Check the horse-power and see if the work is equally divided at the rated load. Cylinders 10 ins. and 17 J X 15 ins., initial pressure 125 lbs., speed 250 R.P.M., horse- power 155. Neglect the receiver volume. Prob. 8. Another manufacturer gives for his engine the following, check this: Cylinders 20 and 32x18 ins., initial pressure 100 lbs., atmospheric exhaust, speed 200 R.P.M., horse-power 400. Neglect the receiver volume. 14. Compound Engine without Receiver, Exponential Law, Cycle X. General Relations between Dimensions and Work wlien High-pressure Exhaust and Low-pressure Admission are Coincident. Referring to Fig. S7D it is desirable first to evaluate the work areas CDKG and C'D'NK, As before, >/f 5— -J and KL = Dh ^^ _^ , hence Wc^ = PcXGL r['-in-""^[^v^j-an. 8 but (rel.pr.)if = (in.pr.)Z»', so that W^H = 144(in.pr.) DjtZh s-1 -^»--^«-rM~) , . . . (414) ^ _PcXKL\ /XL\-> 1 _ (rel.pr.)A, D^iJcf, / 1 V"'! 302 whence ENQDIEERING THERMODYNAMICS Wl = 144(in.pr .) ^^Rc 1- L '-' Re Rc-l -144(bk.pr.)Di;. (415) It is to be expected that the sum of high- and low-pressure work will be of a form similar to that which would be obtained if all work were performed in a single cylinder of a displacement equal to that of the low pressure, adding, W = Wa-\- Wl = 144(in.pr.) s-Z^-^-Z^-i^^^^ +Z, --•44ff -144(bk.pr.)Di, = 144(in.pr.) DaZa 8-1 «-ZH»-i+Zir*-M l-(^y~n I -144(bk.pr.)Dx> Z 1 whence, substituting ^^=-5- Tr=144(in.pr.)^[8-(^)'"']-144(bk.pr.)Dii; (416) . . (417) (m.e.p.ref.toL.P.)=^|[«-(l)'"]-(bk.pr.) . • . Work per cu.ft. supplied = 144^^^^ \a- (^"^\ - 144(bk.pr.)|^ . (418) Cu.ft. supplied per hr. per I.H.P.= 13,750 (m.e.p. ref. to L.P.) Re 13,750 (m.e.p. ref. to L.P.) Ry rs-. . (419) Conditions for equal division of work between high- and low-pressure cylinders may be obtfuned by equating Eqs. (414) and (415), s-Z«— i-Z«'-' -(i)-' Rc-l = Za—'Rr Rc-l _{bk.pr.\Rc, WORK OF PISTON ENGINES 303 Rearranging The last term in the first member of this equation may be expressed as \m.pr. / and the relation Re w V exists between Zh and Re, hence, making these substitutions, Re Re '^^[Rcf-^-l\+R<f-^^Rv'-^]^s+(^ . (420) which is not a simple relation, but can be solved by trial. The assumption of complete expansion in the low-pressure cylinder (it is always complete in high, for this cycle), leads to this following relations: (!§.)=«'•. hence 144(bk.pr.)I>i, = 144(in.pr.)5-V>H, llv and from Eq. (414), ^.=.«(i.,.,?^"[,-(±).-.-i'^-i.i but Re 1 „ . RvS—1 8—\ . 1 /bk.prAi Tr=144(in.pr.)M«;:iy[l-(g^-) ']. ... (421) (m.e.p. ref. to L.P.)=in.pr.^ ITtU"!- — ) I* ' ' i^*^^) 304 ENGINEERING THERMODYNAMICS The expression for egualUyofwork Eq. (420) becomes, for this case of complete expansion, Rc+l Rc-\ (i2c?-i - 1) +fi^-» = 8ftF*-» +(«- 1), • • . (423) by which it is not diflSicult to find the ratio of expansion Rvy which gives equality of work for given values of «, and /Ec, the cylinder ratio. Values for Rv for various values of Re and a are given by the curves of Fig. 91. 10 & 8 ■ 1.6 1.4 Valuta of If 112 S LI y. i^ ^ ^ 1> '^ yy g ;^ X^ ^ y. ^ ^ 'y' ^ i^ t ^ V 4 ^ >" -A 4 ■y^ /' ^ >* — ^ y ^ 7^- 10 ValttM of Rv 15 Fig. 91. — Curves to Show Relation between Re the Cylinder Ratio, and Ry the Ratio of Expansion, for Various Values of (s), Applied to the No-receiver Compound Engine without Clearance, when the Expansion is not Logarithmic. Example 1. Find (a) the horse-power, and (6) the steam used per hour for a 12- and 18 x24-in. engine with no receiver when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder is J, there being no receiver and steam having expansion, such that $ = 1.3. From Eq. (417) , , (in.pr.)Z^r /I \— 1-1 ... . (^.e.p.) .-_ _ _-|^,> ( _) j ^(bk.pr.), which, on substituting the above values, becomes T>^2l5[^-^-(i^) ]-10-C3.31bs.per8q.in. hence the indicated horse-power -=243. WORK OF PISTON ENGINES 305 (6) From Eq. (419) the steam used per hour in cu.ft. per horse-power is 13J50Zj^ m.e.p. Re* which, for the data given above, becomes 13,750 .5 ,^^ ^^ or pounds per hour total, is, 48.2 X243X. 332 =3880. Example 2. What will be the high-pressure cut-off and cylinder ratio to give equal work distribution and complete expansion for an initial pressure of 150 lbs. per square inch absolute, and back pressure of 10 lbs. per square inch absolute? (in.pr.X bk!^./; From relation fiF'^Irr-— ), iBr«"6.9 and from this, by the curve of Fig. 91, /2c =5.4. For complete expansion Rv 6.9 Prob. 1. A tandem compound engine without receiver has cylinders 18- and 30x42-ins. and runs at 200 R.P.M. What will be the horse-power developed at this speed if the initial pressure is 175 lbs. per square inch gage, back pressure atmosphere, high-pressure cut-off }, and 8 has a value of (a) 1.1, (6) 1.3? Compare the results with Prob. 1 of Sec. 13. Prob. 2. What will be the weight of steam used per horse-power per hour for the two cases of the above problem? Compare these results with those of Prob. 2, Sec. 13. Note: B=.416. Prob. 3. What must be the cut-off in a 10- and 15 x20-in. compressed air engine running on 100 lbs. per square inch gage initial pressure and atmospheric back pres- sure, to give complete expansion, and what will be the horse-power per 100 ft. per minute piston speed, a being 1.4? Prob. 4. It is desired to run the following engine at its most economical load. What will this load be and how much steam will be needed per hour? Cylinders 20- and 32x18 ins., speed 150 R.P.M., steam pressure 100 per square inch gage, atmospheric exhaust, dry saturated steam. Prob. 6. Should the load increase 50 per cent in Prob. 4, how would the cut-off change and what would be th effect on the amount of steam used? Prob. 6. Vf hat would be gain in power and the economy of the engine of Prob. 4 were superheated steam used, for which s = 1.3? Prob. 7. In a 14- and 20x24-in. engine will the work be equally divided between the cylinders for the following conditions? If not, what per cent will be done in each? Steam pressure' 100 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, s = 1.2, cut-off = J. Prob. 8. What would be the work and steam used by the above engine if there were complete expansion and equal distribution? 306 ENGINEERING THERMODYNAMICS 16. Compound Engine with Infinite Receiver. Logarithmic Law. With Clearance and Compression, Cycle XI. General Relations between Pressures, Dimensions and Work. In terms of point pressures and volumes, Fig. 92, the P INDICATOR CARDS OP EQUAL BASE ARO HEIQHT FOR CASE OF INCOM- PLETE EXPANSION AND COMPRESMCN. INCjOMP.LETE EXfANSION ANDf COMPRESSION. Y H.P. ^^ --(bk.pr.) • ( rcl.pr.) INDICATOR CARDS OF EQUAL BASE AND HEIGHT FOR CASE OF OVER EXPANSION AND COMPRESSION. OVER EXPANSION AND COMPRESSION. Fig. 92. — ^Work of Expansive Fluid in Compound Engines with Infinite Receiver, with Clearance, Cycle X Logarithmic, and Cycle IX Exponential Expsuision, and Compression. work of the two cylinders may be written down at once as if each were inde- pendent of the other, the connection between them being fixed first by making the back pressure of the high equal to the initial pressure of the low, or to WORK OF PISTON ENGINES 307 the receiver pressure, and second by making the volume admitted to the low equal to that discharged from the high reduced to the same pressure. This last condition may be introduced in either of two ways, (a) EM--NH, (6) [{PV) on H.P. expansion line— (P 7) on H.P. comp. line]> = [(P7) on L.P. expansion line— (PV) on L.P. comp. line]. Without introducing the last relation IFH=P5F»(l+loge^^)-P/7/(l+log.^')-(Pa-P/)7a-P,(F,-7.); (424) TfL=P*V,(l+loge^)-P,7,(l+log.^J-(P,-Pi)F,-PXr,-F») . (425) W^PtV, loge ^+PnVH lOge P - PeVe loge ^^-P,V, logc ^ Vb Vh Vf Vi +PbVt-PrVf-PaVa+PfVj-PaVa+P.V. . +PAF,-P,7,-P,7,+P,7,-P,7,+P*Ft. The second condition is or P»F»+P.y,+P.y.+P»y»=2(P»F»+P*7») (426) Substituting W=P{V„ log. ^+PhV^ log. ~PcV. log. ^-P.y* log. ^ ^ yb Vh Vf Vi + 2{P,V,+PtVu)-PaVa-PoV,-PaV^-P,Vs . (417; This expression, Eq. (427) contains, however, the receiver pressure which is related to the release pressure by / \r>nDDD^</i \ L.P. max.vol. (rec.pr.)=P,=P.=P-=P»=P.^=(reI.pr.).L:p— ^r^g-^. Introducing this Tr=P»niog.^+P,y, log. ^-pL-V. log. ^'-PtF* log. ^ ^ •'» Vh Vh Vf Vi +2(P»F»+Pt7t)-Pay.-pX* V, -PXrVi-P,V, V h V h ■ . (428) 308 ENGINEERING THERMODYNAMICS Introducing the usual symbols in Eqs. (427) and (428) and in additicri the following: Z = cut-off as fraction of stroke, so that ZhDh is the displacement volume up to cut-off. c= clearance volume divided by displacement, so that Ch Db is the clearance Volume and (Zh+Ch)Dh is the volume in the high-pressure cylinder at cut-off. X=that fraction of the stroke during which compression is taking place so that {Xh-\-Ch)Dh is the volume in the high-pressure cylinder when compression begins. Applying the general symbols to Eq. (427), l+cg\ W^IU 1 1-^ [m.pr.)(Z^H-Ciir)Z)j5rloge \^ 4-(rec.pr.)(ZL+CL)Z)Lloge (^^J^) -(rec.pr.)(X^+c^)I>fflog« ^— ^ — ^j Ch Xl+ci^ -.(bk.pr.)(X^ +Ci,)Z)i, lege (^^) H-2(in.pr.) {Zh +ch)Dh - (in.pr.)c^Dir - (rec.pr.) (1 '\-Ch)Dh 4-2(bk.pr.) {Xl +cl)Dl - (rec.pr.)ci/>i, - (bk.pr.) (1 -hCz.)Z)i, J (4^9) This expression gives the work in terms of initial, receiver and back pressures, the valve periods, cut-off and compression, the clearances and cylinder dis- placements. Substitution of the symbols in Eq. (428) will give another equivalent expres- sion in terms of the same quantities except that low-pressure cylinder release pressure will take the place of receiver pressure. This is {m.pT,)(ZH+CH)D„ log.f^''-) +(rel.pr.)L(l +Cl)Z)l log. (Pr^) -<""-''(ii.t)<^'+-"'»'°«-(^") -(bk.pr.)(Xi, +c^)Z)i, lege (^^'') Pr = 144 -f 2(in.pr.)(Z^ ■^-Cu)Dh - (in.pr.)c^Z)jy - (^^-PrOLU^^- j (1 +Ch)Dh +2(bk.prO(Xjr,H-CL)Z)L-(rel.prOi,f -^ Ui^l- (bk.pr.) (1 H-C£.)Dr. (430^ WORK OF PISTON ENGINES 309 It is sometimes more convenient to involve the cylinder ratio and low- pressure displacement than the two displacements as involved in Eq. (430) and the ratios of expansion instead of (iut-offs. This may be done by Rl-'^^ (431) and it should be noted here that the ratio of expansion in each cylinder is no longer the reciprocal of its cutroff, as was the case when clearance was zero, nor is the whole ratio of expansion equal to the product of the two separate ones because the low-pressure cylinder expansion line is not a continuation of that in the high. Making these substitutions for cylinder and expansion ratios, Eq. (430) becomes, (432) (in.pr.)(l +ch)-^-^ lege /2ir +(reLpr.)z.(l +Cz.) loge Rl KhKc - (rel.pr.)i(ZH +<'*)f^log.(^^^) - (bk.pr.) (X^ +cl) log.(^-^^ +2(in.pr.)(l -^-ch):^-^ -(in.pr.)^— (rel.pr.)L5^(l +ch) ^ +2(bk.pr.)(Xz,4-CL) -(rel.pr.)z72i,ri, -(bk.pr.) (1 -\-Cl) It is interesting to npte that this reduces to Eq. 304 of Section 9, by making clearance and compression zero. From any of the expressions for work, but more particularly (430) and (432), the usual expressions for (m.e.p.) referred to low-pressure cylinder, work per cubic foot supplied, and consumption per hour per I.H.P. can be found, but as these are long they are not set down, but merely indicated as follows: (m.e.p. ref . to L.P.) = W 144Z>z.* Work per cu.ft. supplied = W DnUi h-\-Ch) — {Xh-\-Ch) /rec.pr.Xl \in.pr./J . (433) . (434) Cu.ft. sup. per hr. per I.H.P. 13,750 (m.e.p. ref. to L _[(Z.+,,)_(X.+,,)(-H:)]^. 13,760 (m.e.p. ref. to L ^[(Z.+.)-(X.+.)(|^)](^i). (435, 310 ENGINEERING THERMODYNAMICS As the receiver pressure is related to the initial and back pressures and to the relation between the amount taken out of the receiver to that put in, which is a function of the compression as well as the cut-off and cylinder ratio, it is expressed only by a complicated function which may be derived from the equivalence of volumes in the high and low, reduced to equal pressure. Therefore, Ph=^P b +p. V, v^+v» ' 'v^+v: Introducing symbols Hence (rec.pr.)-(m.pr.)(^^^^^)^^^(^^^^^)+(bk.pr.)(^:^-j^-:p^^^^p^. (436) This Eq. (436) gives the receiver pressure in terms of initial and back pressures, the two clearances and compressions, the cylinder ratio and the cut-off in each cylinder. Proceeding in a similar way, the release pressures can be found in terms of initial data, V, P'-Poy^. or (rel.pr.)a - (in.pr.) ( - "jp^ j (a) (in.pr.)5-- Kb And Pi = P»y" = JP» .\14 V, +P>\ (437) 14 7. or (rel.prOi = (in.pr.) , = (in.pr.) \{\+Cl )Rc } , , {Xh + Cii) r /1+C«\ 1 \1 + ClIRhRc +(bk.pr.) 1 + 1 + {X„+ci,)R i. a+Ct.)Rc + (bk.pr.) 1+ (Xi.+Cr.) -, (Xh+Ch) (Zl+Cl)Rc (A'^ + c^) {Xh+Ch)Rl (1+Ci,)Rq -■ va) (6) (438) WORK OF PISTON ENGINES 311 These three pressures all reduce to those of Eqs. (308), (309), (310), Section 9, when clearance and compression are zero. Equal work in both cylinders is, of course, possible, but it may be secured by an almost infinite variety of combinations of clearance, compression and cut-oflF in the two cylinders for various ratios of expansion; it is, therefore, not worth while setting down the equation of condition to be satisfied, but reference may be had to Eqs. (424) and (425), which must be made equal to each other, the result of which must be combined with the equation of cylinder relations. A B 1 y \ \ 1 y \ G V "N c \ \ • V \ \ ^ ^-, SK ■.» . 1 INDICATOR CARDS OF EQUAL BAIE AND HEMHT Fio. 93. — Special Case of Cycles XI and XII Complete Expansion and Compression in both Cylinders, of Compoimd Engine with Clearance and Infinite Receiver. There are certain special cases of this cycle for which equations expressing important relations are simpler, and they are for that reason worth investigat- ing. Those that will be examined are (a) Complete expansion and compression in both cylinders. Fig. 93. (b' Complete expansion m both cylinders with no compression, any clearance, Fig. 94. (c) Any amount of expansion and compression but equal in both cylinders, equal clearance percentages and a cylinder ratio equal to the square root of the ratio of initial to back pressures, Fig. 95. Case (a) When both expansion and compression are complete in both cylin- ders, Fig. 93, TF^ = 144(in.prOZ,,D,,log. (^^"^^^ .... (439) Tr., = 144(rec.pr.)Zx^^loge ([§^^7^^^) .... (440) (in.pr. ) (bk.pr.)' 312 ENGINEERING THERMODYNAMICS but and log.(Ji^PI:))+log. (5^) =log. (J-M^4 ^S^) =log. ^\(rec.pr.)/ ^ \(bk.pr.)/ \(rec.pr.) (bk.pr.)/ hence Tr=144(in.pr.)Z^D^ log. |gg^ (441) (m.e.p.ref.toLJO = (m.pr.)|^loge(g^) (442) Work per cu.ft. supplied =144(in.pr.) logf (^^-^j (443) Consumption, cu.ft. per hr. per I.H.P. = 7 \ ^ ^ T>^ tt-- • • • (444) ^ ' *- x- (m.e.p. ret. to L.P.) Re EqualUy of work in high- and low-pressure cylinders is obtained by making / (in.pr. ) \ ^ /(rec.pr.)\ ^ /(in .pr. ) \ * \(rec.pr.)/ \(bk.pr.)/ \(bk.pr.)/ ' or (rec.pr.) = [(in.pr.) (bk.pr.)]* (445) It is desirable to know what clearances and displacements will permit of equal work and complete expansion and compression. (rec.pr.) = (m.pr.)|^ = (m.pr.) ( ^^^^ 1 = (bk.pr.)^ =(bk.pr.)(^j), hence /( in.pr. )\ ^ 1 +Ch ^ /(in .pr. ) ^\ \(rec.pr.)/ Zh+ch \(bk.pr.) /' / (rec.pr.) \ ^ 1+Cz. ^ / (in.pr. ) \ * \( bk.pr.)/ Zz,+cx, \(bk.pr.)/ ' or calling /(in.pr. )\ _ \(bk.pr'0/ ^^''^ ^ l+c^-cgfip* , „ 1+Cl- ClRp^ /..^v Zk = ^-^1 , and, Zx, = ^^ . . . . (446) Equating discharge of high and intake of low-pressure cylinders, ZffDeRp^^ZJ)^ or ^=Rc=^RpK $ WORK OF PISTON ENGINES 313 Inserting in this the values just found for Zh and Z^y 1+Cu — ChRp^ «c-j4.;^^^^^^r^p*, (447) which is the required relation between cylinder sizes, clearances and ratio of pressures, which, together with cut-offs given in Eq. (446),. will give equal work and complete expansion and compression. The compression in the high- pressure cylinder is such that Xh=ch{Rp^-1) (448) and for L.P. cylinder. XL=ci,(Rpi-l). A B A \ G H K \ ^^ • INDICATOR CARDS OF EQUAL BASE AND HEIGHT Fig. 94. — Special Case of Cycles XI and XII. Complete Expansion and Zero Compression in both Cylhders of Compounds Engine with Clearance and Infinite Receiver. Case (6) With complete expansion and no compression, both cylinders, any clearance, Fig. 94, WH = 14^H^{m.pr.)iZH+CH)\og, ^^^^^j -c,.[(in.pr.)-(rec.pr.)]] (449) Tri,=144Z)i,r(rec.pr.)(Zi,+cz,)log, ([^t.'^j) " ^^^ (^50) with the added requiremen t tha t the high-pressure discharge volume, EC = low pressure admission volume FH, or D^=2)i,| Zl + Cl V(rec.pr.)/J' (451) 314 ENGINEERING THERM0DYNAMI08 and (rec.pr.) = (bk.pr.)^'=(bk.pr.)^^, (452) hence r (bk.pr. )] ^£t+ct |_(rec.pr.)J l+Cr ' which substituted in Eq. (451) and rearranging gives Re ^^ ,. ^ , (463) Zl+c. /Zl+ClV Eq. (453) indicates that for this special case of complete expansion and no com- pression the cylinder ratio required to give this case, is determined entirely by the L.P. cut-off and clearance. If the cylinder ratio and clearance are fixed, the required cut-oflf in the L.P. cylinder can be found by solving Eq. (453) for Zl, Z^=i^-Ci„ (454) and from Eq. (452), r 1+Cr, (rec.pr.) = (bk.pr.) — Cl-tCl Ri = (bk.pr .)ii;c. . • (455) Cut-oflf in the high-pressure cylinder is determined by clearance, initial pressure and receiver pressure, which in turn depends on low-pressure cut-oflf and clearance Eq.(452), or may be reduced to cylinder ratio and low-pressure clearance by Eq.(454), as follows Vc^ 1+cg ^ / ( in.pr . )\ ^ /(in .pr. ) \ Zl+Cl Vt Zh+ch V(rec.pr.)/ \(bk.pr.)/ 1+Cl ' hence ^^^ Rp{Z^+c^) ~'^- Elimmate Zl by Eq. (454), (456) Since the high- and low-pressure cut-offs are functions of cylinder and clearance dimensions, and of Rpy the rato of initial and back pressures, the work of high- and low-pressure cylinders may be expressed entirely in terms of these quantities. TF^ = 144Z)H(in.pr.)j(l+c^)|^loge(|^)-c,,(l-|^)|. . . (457) TFi, = 144DH/ec(bk.pr.)[(l+cz,)loge/2c-c,.(/2c-l)]. . . . (458) WORK OF PISTON ENGINES 315 Hence^ total work by addition is TF=144Z)i7(in.pr.)i^ /•+'•> "*(l)-''(l-i) +(l+Cjr,)logfiBc-Ci(i2e-l) . . . (459) » Expressions might be easily written for mean effective pressure referred to the L.P. cylinder, work per cubic foot fluid supplied, and consumption, but will be omitted for brevit y. I t is important to note, however, the volume of fluid used per cycle is not AB, but is LB, Fig. 94, and is, {Sup.YoI)^Dh[{Zh+Ch)-cJ^^^^=Dh[(^^^ . (460) W (m.e.p.ref.toL.P.) =j^^^^-g (461) W (Work per cu.ft. supplied) =^^ — vHT (462) is 750 r Consumption cu.ft perhr. per LH.P.=^^^^j-^^-p-^-|^(Z«+c«) Itpjltc Equality of work, secured by equating Eqs. (457) ajnd (458) gives (l+Cir)loge(|^)-CH(|^-l) = (l+c,,)logeiec-c,,(/?c-l). . (464) This equation may be satisfied in an infinite number of ways. One case worth noting is that of equal clearances, when it is evident that if Cs==CLf and -^^Rc, or Rc—^Rp JtCc the Eq. (464) is satisfied. This last condition is the same as that which satisfied Case (a) with complete compression. CcLse (c), Fig. 95, assumes that and 316 ENGINEERING THERMODYNAMICS and corresponds to the first special case considered in Section 9, which lead in the no-clearance case to equality of high- and low-pressure work. Fig. 95. — Special Case of Cycles XI and XII, Equal Per Cent Clearance in Each Cylinder of Compound Engine with Infinite Receiver and Cylinder Ratio Equal to the Square Root of Initial Divided by Back Pressure. The assumptions already made are sufficient to determine the receiver pres- sure. By Eq. (436) (rec.pr.) = (in.pr.) Z+c f(bk.pr.) +X+C +X+C [(in.pr.) (bk.pr.)]* = [(m.pr.)(bk.pr.)]» (465) The work of the high-pressure cylinder may now be evaluated. War 144D£r(in.pr.) | Z |^1 +log. (^^ j - < - 144Dtf[(in.pr.) (bk.pr.)]* { (X+c) log. (^^^\+l-X ■ . . (466) WORK OF PISTON ENGINES 317 The low-pressure cylinder work may be similarly stated, Trz,«1441)4(in.prO(bk.prO]*|z[l+log.(^^)^^ (467) -144i)^(bk.pr.) |(X+c) log. ^^)+l-Z ., but Dz.[(in.pr.) (bk.pr.)]* = D^fic[(in.pr.) (bk.pr.)]* =^ir(^^)*[(in.pr.)(bk.pr.)l* =D^(in.pr.),l and similarly, Z),,(bk.pr.)=2)£r[(in.pr.)(bk.pr.)]* • With these substitutions the value of low-pressure ie7orfc,Trx,,Eq. (467), becomes equal to high pressure work, Eq. (466), hence the total work W=2Wh = 2Wi^ (468) Example 1. Method of calculating Diagrams, Fig. 92. Assumed data: Pq ^Pa '^Pb = 120 lbs. per square inch abs. Va = F/= .12 cu.ft. Pn =P# —Pe -Pd^Ph- 50 lbs. per square inch abs. Vh = .4 cu.ft. Pk ^i'i = 10 lbs. per square inch abs. Vc^Vd^ .8 cu.ft. Vg^ Fi = .16cu.ft. Vi-'Vi^ 2cu.ft. F«- .2cu.ft. (Va-7«)=(F„-F,). Fit- .4cu.ft. The above may be expressed in initial pressure, etc., and in terms of cut-off, etc., but as the relation of the lettered points to these terms is shown on the diagram values for cut-off, etc., they will not be given here, as they may readily be found from values of the lettered points. To locate point C: To locate point F: - -Fft 120X.4 _„ Pe=Pb-77-^ — — — -w lbs. per sq.m. V c -O P, re »'^< 50 X .2 _ _ _ -, /=— ir-= — -^r- =83.3 lbs. per sq.m. Vf AJt To locate point Q: Pe7e^^ P« 120 Va — ^' = 7:^=.083cu.ft. 318 ENGINEERING THERMODYNAMICS To locate point L: _ PtVk IPX. 4 „,, Pi = — rr- = — TT— =26 lbs. per sq.m. Vi .16 To locate point N: F„=^=^-.08cuit. To locate point H: (FA-F„) = (Fm-F.), or F*-F«+Fn-Fe = .96+.08-.2 = .84cu.ft., since To locate point 7: 48 P«F« = P,V,, = F« =g - .96 cu.ft. ■n PhVh 50X.84 _^ ,, Pi = —.— « — - — =21 lbs. per sq.m. Vi 2 Example 2. Find (a) the horse-power, (6) steam used per hour, and (c) receiver and release pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per cent clearance in high-pressure cylinder, and 4 per cent clearance in low-pressure cylinder, when initial pressure is 150 lbs. per square inch absolute* back pressure 10 lbs. i>er square inch absolute, speed 125 R.P.M., cut-ofif in high-pressure cylinder is }, low- pressure cut-off is such as to give complete H.P. expansion, and compression is 15 per cent in high and complete in low. (a) For complete high-pressure expansion the receiver pressure must be equal to the high-pressure release, and to maintain the receiver pressure constant the low-pressure cylinder must take as Liuch steam per stroke as the high-pressure discharges. With initial pressure and cut-off as given, the release pre3&ure for the high-pressure cylinder may be found from the relation (inpr.)(cj5r+ZH)=(rel.pr.)/r(cj5f+Z)i5r) or 150x(.56) -(rel.pr.)»(1.04), or (rel.pr.)iy=79.3 lbs. Since there is 15 per cent compression in high-pressure cylinder there is exhausted each stroke 85 per cent of its volume. Also since compression in low-pressure cylinder is complete, the low-pressure clearance is full of steam at the receiver pressure at the beginning of the stroke. Hence the low-pressure displacement up to cut-off must equal MDa or L.P. cut-off =.85Dj, divided by cylinder ratio, or .85 4-2.26 =.378. As compression is complete, the per cent compression may be found from the relation CLX(rec.pr,)=(cL-f^L)(bk.pr), or .04x79.3 = (.04-fXi.)10, or Zi, = .28. From Eq. (432), (m.e.p.) referred to low-pressure cylinder is obtained by dividing by 144 Dl, and on substituting the above values it becomes, 150(.1+.06)(^|-) log. 2+30(1 +.04) log,2.64-30(.15+.06) (|~)log, {^^^) -10(.28+.04)log. (-^^) +2xl50(l+.06)^y .150x^ " 2 64 -30X;r~|(l+.06)+2xl0(.28+.04)-30x2.64x.04-10(l+.04)=60.5 lbs. per. sq.in.. hence I.H.P.=235. WORK OF PISTON ENGINES 319 (6) From Eq. (435) by substituting the above values Cuit. steam per hour per horse-power = i|^ \({M+ .04) - (.28 + M)^) ^1 =46.5, oU.o L \ 79.0/ 160 J or pounds per hour will be 3550. (c) Release pressure for high-pressure cylinder has been shown to be 79.3 lbs. and may be cheeked by Eq. (437), as follows: (rel.pr.)zr = 15o(^^^) =79.3 lbs. Receiver pressure has already been shown to be equal to this quantity and may be checked by Eq. (436) (rec.pr.)j/ = 150 X (.5 +.06) 10 X (.28 +.04)2.25 (.378 +.04)2.25 +(.28 +.04) ' (.378 +.04)2.25 + (.28 +.04) 79.3 lbs. Low-pressure release pressure is found from Eq. (438) to be (reLpr.)i.=150 1+.06 X 1 +.04 2x2.25 1 + (.15 +.06)2.64 U (1 +.04)2.25 J +10 .28 +.04 1+.04 (.15 +.06)2.64 H L (1 +.04)2.25 J =30 lbs. Prob. 1. What will be the horse-power and steam used by the following engine for the data as given? Engine 20 and 28x36 ins., running at 100 R.P.M., clearance 5 per cent in high pressure. 3 per cent in low. From cards H.P. cut-off = .3, L.P. =.4, H.P. compression, .1, L.P., .2. Gages show (in.pr.) to be 150 lbs., (r c.pr.) 60 lbs., (bk.pr.) 26 ins. Hg. (barometer ==30 ins.). Prob. 2, What must be ihe cut-offs and the cylinder ratio of an engine to give equal work and complete expansion and ompression for 200 lbs. per square inch absolute initial pressure and atmospheiic exhaust, if clea ance is 5 per cent in the high and 3 per cent in the low-pressure cylinder? What will the horse-power for an engine with a low-pressure cylinder 24 X36 ins., running at 100 R.P.M. for this case? Prob. 3. Should there be no compression, how would the results of Prob. 2 be altered? Prob. 4. What will be the total steam used by engines of Probs. 2 and 3? Prob. 6. For an 11- and 19x24-in. engine with 5 per cent clearance in each cylinder, i cut-off in each cyhnder, and 20 per cent compression in each cylinder, what will be the horse-power and the steam consumption when the speed is 125 R.P.M., the initial pressure 150 lbs. per square inch gage, and back pressure at atmosphere? 16. Compound Engine with Infinite Receiver. Exponential Law, with Clearance and Compression, Cycle Xn. General Relation between Pressures, Dimensions and Work. Referring to Fig. 92, of the preceding section, which 320 ENGINEERING THERMODYNAMICS will represent this cycle by a slight change of slope of the expansion and com- pression lines, the high-pressure work may be expressed in terms of dimensions, ratios and pressures. Since this must contain receiver pressure as a factor, and since that is not an item of original data, it is convenient first to state receiver pressure in terms of fundamental data: F«-F.= F,-Fn. But 1^ 2 7„=77iE£!LV and Fn = F*/bk^V Vrec.pr./ \rec.pr./ Whence, 1 whence \rec.pr./ \rec.pr./ / \ r \\ \m.pr. / (rec.pr. ) = (m.pr.) * Vn+V. or in terms of dimensions and pressures, (rec.pr.) = (in.pr.)l r^(^z,+c,)+X.-\-'c^ ' j ' ' ^^^^ The high-pressure work may be stated as follows: Tr«=144D«{ (in.pr.)^^i^^[«- (^^^^)'"'] -(rec.pr.) (1-X«) -(rec.pr.)^J:^[(^^)'"'l-]-(ia.pr.)cH 1- . (470) Zu+CH+Rc{Xt.+cM^^y ^ _ ,. . , Vm.pr. / -144D«(m.pr.)\^ r^^Z,+c,)-^X.+c„- '^;±;?[(^^)'"l-]+l-X«j. (471) WORK OF PISTON ENGINES 321 W,=lUDJ(Tec.pv.)^^f±^[s- (-Ag-y-'j _(bk.pr.)(l-X^) -(bk.pr.)^^^[(^^)*"-l]-(rec.pr.Kl. . (472) = 144D«fic(in.pr.)' Zll+CB+Rc(Xi,+Ci. >/bk.pr.yY \in.pr. / I Rc(Zt.+CL)+Xa+eB (473) The expression for total work need not be written here, as it is simply the sum of Eqs. (470) and (472) or of (471) and (473), the former containing receiver pres- sure and the latter containing only dimensions, initial and back pressures and both, the expK>nent of the expansion, 8, The volume of high-pressure fluid supplied per cycle is QB, Fig. 92, which may be expressed either in terms of high pressure or of low pressure points, thus; (Sup.Vol.) = Db \Zb+Cb - {Xb+Cb) (^^) ' ] (a) -^^[^'M^y -'^^Miw ^'^ (474) The following quantities will be indicated, and may be evaluated by sub- stitution from the preceding: W W (m.e.p. ref. to ^;P)-iuDl'T4iD^, W Work per cu-ft. fluid supplied = t^ — y-p; Consumption cu.ft. per hr. per I.H.P. 13,750 (475) (476) (Sup.Vol.) (m.e.p. ref. to L.P.) Dl (477) Eqtud division of work between high- and low-pressure cylinders requires that Eqs. (470) and (472), or (471) and (i73) be made equal, The latter will 322 ENGESTEERINQ THERMODYNAMICS give an expression showing the required relation between dimensions and initial and final pressures, cut-off and compression in high- and low-pressure cylinders. In this expression there are so many variables that an infinite number of arnir hinaiions may be made to give equality of work. It is desirable to examine the results of assmning special conditions such as those of the previous section, the most important of which is that of complete expansion and compression in both cylinders, which is represented by Fig. 93. 9-1 TF„=144(in.pr.)Z.D^[l-(^) ' ]. ... (4: '8) but Tr. = 144(rec.pr.)Z^.^[l-(^^;)-]. . . . (479) \rec.pr./ hence ,r=.44(in.pr)^,Z^|[l-(^)'-"] L in? > /re c.pr. X / in.pr. y j" _ / bk.pr. \ * "I 1 \ in.pr. / \rec.pr. / [ \rec.pr./ J J = 144C ^-^ - * + •-1 in.pr .)D^Zhj—y J ^ / rec.pr. \ « [ \ in.pr. / . / rec.pr. \ « / rec.p r.X « / bk.pr. \ « I \ in.pr. / \ in.pr. / \rec.pr./ J = 144(in.pr.)Z>.Z^[l-(gi^) ' ] (480) The receiver pressure may be found as follows. In Fig. 93, EC=GH: Equating l+c.-c.(H:P?:i)^=ficri+c.-c.(^h(^F_V- . (481) Vrec.pr./ [ \bk.pr./ J \rec.pr./ ^^^ ^ WORK OF PISTON ENGINES 323 When this is solved for receiver pressure it results in an equation of the lecond degree, which is somewhat cumbersome, and will not be stated here. 5q. (481) is, however, used later to find Ro- lf work is to be equally distributed between high- and low-pressure cylinders, rom Eqs. (478) and (479), «-l 8-1 •-! *~1 «» { ^^^'P ^'\ • _ / rec.prA « __ / rec.prA « /bk.pr. \ « \ in.pr. / \ in.pr. / \ in.pr. / \rec.pr./ ' •r «-i «-i /rec£r.\ . =i+/bk^r.\ ' , \m.pr. / \m.pr. / .6 1 I 1 ^i i\ v\ A k § A ^ ^ p 1 \ § ^ ^ \ ^ ^ - \ <: ^ ^ "■"' — ^ L6 •LI 9 IS t* "v -"^ "^ — ~ — s — ~ ~^ ~ — a.o c z If > B O 71 u X) s. 96. — Curves to Show Receiver Pressure to Give Equal Work Distribution when Expansion and Compression are Complete in both Cylinders of the Compound Engjne with Infinite Receiver, with Clearance when Expansion and Compression are not Logarithmic. nee, for equal division of work, (rec.pr.) = (in.pr.) \ in.pr. / s-l . . . . (482) lich, if satisfied, will give equality of work in the two cylinders, for this case perfect conipression and expansion. In Fig. 96, is given a set of curves for use in determining the value of the tio of (rec.pr.) to (in.pr.) as expressed by Eq. (482). 324 ENGINEERING THERMODYNAMICS When (rec.pr.) has been found by Eq. (482) it is possible by means of (481) and the clearances to find Re* The events of the stroke must have the follow- ing values to maintain complete and perfect compression and expansion. Zh^(1+ch) Zl={1+Cl) 1 /rec.pr A r Vrec.pr./ (4831 (4&t) X Example. Find (a) the horse-power, (6) compressed air used per hour, and (e) receiver and relea.e pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per cent clearance in the high-pressure cylinder, and 4 per cent in the low-pressure cylinder, when initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder }, low-pressure cut-off such as to give complete expansion in high-pressiu^ cylinder, compression in high-pressure cylinder 15 per cent, and complete in low. Expansion such that «»1.4. (6) As in example of Section 15, receiver pressure equal high-pressure release pressure, and low-pressure volume at cut-off must equal volume of steam exhausted from high pressure. Release pressure may be found from relation (in.pr.Xc£r+^2r)* ^(relpr.) -(ch+DuY: or 160(.06+.6)^**(rel.pr.)-(.06+l)^-*, or(rel.pr.)-601bs. As in the previous example, the low-pressure cut-off is .38, and the low-pressure compressioQ may be found from the relation Cz,*(rec.pr.) =*(ci.+Xii)'(bk.pr.),or(.04)^'^X60 = (M+Xl)^"^ (10), or Zl = .09. From the siun of equations (471) and (473) divided by 144Diyl2c, and with the proper values substituted the following expression for (m.e.p.) results: 150 2.25 -4 -['•'- Vl^-loe-j J '•^) -2:25 .5+.06+2.25(.15+.06) /JO \ -71 1 1 ♦ \150/ 2.25(.38+.04)+.15+.06 .5-h.06+2.25(.15-h.04)f ^j . 2.25(.38+.04)+.15+.06 -.10- hence the horse-power is 214. '.09 + .04r/.09H-.04\* 1., ) ^^ ,^ 4~~ \ — 04 ~ ) ""M ~ I " ^^ ^ WORK OF PISTON ENGINES 325 From Eq. (477) with proper values substitute, Cu.ft. per I.H.P. hr--^^|pX r(.38+.04)(~y^' -(.09+.04)^^^l'^' =50, r total steam per hour will be 60X214 = 10700 cu.ft. (c) Release for the high-pressiuie cylinder has already been given as 60 lbs. and the ■ceiver pressure the same. The latter quantity may be checked by equation (469) Qd will be found to be the same. The low-pressure release pressure may be found •cm the relation (rec.pr.)(ZL-fci,)^*«»(rel.pr.)L(l+CL)**, which on proper substitution ives / .38+.04 \i.4 \ 1.04 ; (reLpr.)i,=«60( — ' 1 =27 lbs. per sq. inch Prob. 1. What will be the horse-power and steam used per hour by an 18- and 4x30-in. engine with 5 per cent clearance in each cylinder and with infinite receiver iinning on 100 lbs. per square inch gage initial pressure, and 5 lbs. per square inch bsolute back pressure, when the speed is 100 R.P.M. and the cut-ofif in high-pressure vlinder is ^ and in low -^7 Note: » = 1.3 and S=.2. Prob. 2. What must be the receiver pressure for equal work distribution when the litial pressure has the following values for a fixed back pressure of 10 lbs. per quare inch absolute? 200, 175, 150, 125, 100, and 75 lbs. per square inch gage? Prob. 3. For the case of 150 lbs. per square inch gage initial pressure and 14 lbs. ler square inch absolute back pressure, what will be the required high-pressure cyhnder izeforan air engine with a low-pressure cylinder 18x24 ins., to give equality of work, learance in both cylinders being 5 per cent? Prob. 4. What will be the horse-power and air consumption of the above engine rhen running at a speed of 150 R.P.M., and under the conditions of perfect expansion nd compression? 17. Compound Engine with Finite Receiver. Logarithmic Law, with 3earance and Compression, Cycle Xm. General Relations between Pressures, )imensions, and Work when H.P. Exhaust and L.P. Admission are Inde- pendent As this cycle, Fig. 97, is made up of expansion and compression lines eferred to the different origins together with constant pressure, and constant rolume lines, the work for high- and low-pressure cylinders and for the cycle can )e set down at once. These should be combined, however, with the relation loted for the case of infinite clearance which might be termed the condition or a steady state [{PV) on H.P. expansion line] — [(PF) on H.P compression line] =[(Py) on L.P. expansion line— PV on L.P. compression line,] W, Pi,Vi,-PeVe=^PnVn-PtVt (487) ENGINEERING THERMODYNAMICS OVER EXPANSION AND COMPRESSION Fig. 97. — Work of Expansive Fluid in Compound Engine with Finite Receiver and with Clearance. Cycle XIII, Logarithmic Cycle XIV, Exponential Expansion, and Com- pression. Also WORK OF PISTON ENGINES 327 Beside this there is a relation between H.P. exhaust and L.P. admission pressures, corresponding to the equality that existed for the infinite receiver, that may be set down as follows: PmV^^P.V,, and Pn.{V„,+0)^Pa{Va+0); .-. Pn^V^=Pi>V,=Pa{V, +0)-Pn,0, and P^^Ph; PnVn=P,V,, and Pn{Vn+0)=P,{V,+0); /. PnVn^PtVu^P,{V,+0)-PnO, and Pn^Pe, ^^= vl+o' ^^^^ These two expressions for the pressure at D and at G are not available in their present form, since they involve two unknown pressures— those at H and E, but two other equations of relation can be set down from which four equa- tions, the four unknown pressures P«, Pa, Pg and Pn, can be found. These other equations are Pd(F<i+0)=Pe(F,+0), or P,=P,(^±^), . , and P,(n+0)=P.(7.+0), or P,=p,^^±g). . . (491) Equating (488) to (490), V^ + ^'^^\Vi + 0)' ^^ Pl>V!>+PkO-=^Pe{Ve+0), and p _ P,V,+PnO Ve+0 ' Equating (489) to (491) and P*(F»+0)-PtF» (490) P.= Therefore „ ^Pi>yi,±Pj^ ^ P*(F>+0)-P«Ft V,+0 P,V>0+P,(P =P»(7»+0) iV»+0) -PtV,iVe+0) ; PH[{v^+o)(y,+o)-o^] =PtVtO+PtVt{y,+o), p ^ PtV,0 +P,Vt{V.+0)P„V,^+ PtVtiVe+ 0) 328 ENGINEERING THERMODYNAMICS Therefore P»= (a) Substitution will give P.= P»V,(V>,+0)+PtVtO (V.+0)(Vh+0)-(P '"L v,{v,+o)+v/) \lv,+o\ > = \ p*y»(yk+ o)+PkVtOi [ F.+Q i ' L(^.+0)(t^»+0)-02JLF,+0j (&) (c) (d) (492) It will be found that the use of these pressures is equivalent to the applica- tion of the equation of condition given in Eq. (487), for substitution of them reduces to an identity, therefore the work of the two cylinders can be set down by inspection in terms of point pressures and volumes and the above preasures substituted. The result will be the work in terms of the pressures and cylinder dimensions. (P.-P,)7. WH--P>vJl+loe,Y^ -PfVr(l+log, ^^)-P.F«logc -p,y»(i+iog. ^) -p,v,-p,v, log. Y-P.v. og, ^i^-PaF.+p.y. =P.n(l+lo„ ^) -P.F. log. (^) (^) -P.F.. Va+0 v.+o +0 Therefore (4<- Trir=P»F.^l+log, Y^ -P.Va _ \ P»V^(V^+0)+PtVtO l „ , (V.\ ( Va+0 \ TFi,=P,Fil+log.^±^)+P»r»log.^;-P,F, (l+log. ^*)-(p,-P,)r.. -PjiVj-V,) =P,V, log. (^±§) +PhVh log. y-PtV, log. ^-PsiVj- F.). Therefore ^ _ r p>F>o+p.F.(F.+o) /z»±o\] J (n±q\ ) + r PtF>0-|-P>Ft(F.+0 )] y .Vi [ v^(y,+o)+v,o J'^*'°*'F» -P»F* log. ^-P/F,- Ft). • • (494) WORK OF PISTON ENGINES 329 Adding Wb and Wl P»F»(l+l0g. ^) -P.y, log. ^^ -PaVa-Pj{V,-V^) _ r p»7t(y>+o)+p«n oi „ , (VA m+o \ [ (7.+o)(y»+o)-02 J "^^ '°^ \vJ\v.+o} Tr= + + fftF, L F [ +o)(y»+o) *(F.+0)/F»+0\"| y»+o ■»(F.+0)+F.O \V,+0 P ,V>0+P,Vt(V, v,+ o p^-J F» log. ^ (495) While this Eq. (495) for the cyclic work is in terms of initial data, it is not of very much value by reason of its complex form. To show more clearly that only primary terms are included in it, the substitution of the usual symbols will be made. F-144X (in.pr.)(Za+C£f)Z)H 1 +log. „ i 1+ch 1 Zh - (bk.pr.)(Zi, +cl)Dl log. Cl — (in.pr.)c£ri)a -(bk.pr.)(l —Xi.)Di. _ ( in.pr. ) (ZH+CH)DnliZi, +ci,)Dj, +0] + (bk.pr.) (X z,+CL)Z?i,0 ' [{Xa+CH)Da+0](iZi.+c,^)DL+0]-0* (X«+c«)D.log. (— -j |^^^^_--^^-^J (in.pr.) {Zu +ch)DhO + (bk.pr.) {Xl +cl)Dl [(Jh +ch)Dh +0] X + (2i +Ci,)Z)i[(Xif +ch)£)j/ +0] + (Xh +Cfl)i)flO X [ (Zl+cl)Dl+0 ] ClDl loge [ {Zl+cl)Dl+0 ] I (in.pr.) (Zg +Ciy)D/rQ -f (bk.pr.) [(Xz, -f C£.)Z)l][X^ +c//)Z)^ +0] (Zl +cl)DlKXh +cji)DH +0] +{Xh +ch)DhO 14-cz. X (^.+a)Z).log.(j^) • . (496) Such equations as this are almost, if not quite, useless in the solution of yrMms requiring numerical answers in engine design, or in estimation of engine V^ormance, and this fact justifies the conclusion thai in cases of finite receivers graphic methods are to be used rather than the analytic for aU design work. When estimates of power of a given engine are needed, this graphic work is itself seldom justifiable, as results of sufficient accuracy for all practical engine operation problems can be obtained by using the formulas derived for infinite receiver when reasonably, large and zero receivers when small and the pistons move together. 330 ENGINEERING THERMODYNAMICS It *mighi also be possible to derive an expression for work with an equivalevi constant-receiver pressure^ that woyld give the same total work and approxi- mately th^ same work division as for this ca^, but this case so seldom arises that it is omitted here. Inspection of the work equations makes it clear that any attempt to find equations of condition for equal division of work for the general case must be hopeless. It is, however, worth while to do this for one special case, that of complete expansion and compression in both stages, yielding the diagram Fig. 98. This is of value in drawing general conclusions on the influence of receiver size by comparing with the similar case for the infinite receiver. By referring to Fig. 98, it will be seen by inspection that cylinder sizes, clearances and events of the stroke must have particular relative values in order Fio. 98. — Special Cajse of Cycles XIII and XIV, Complete Expansion and Compression in both Cylinders of Compound Engine with Clearance and Finite Receiver. to give the condition assumed, i.e., complete expansion and compression. It is, therefore, desirable to state the expressions for work in terms which may be regarded as fundamental. For this purpose are chosen, initial pressure (in.pr.), back pressure (bk.pr.); high-pressxu-e displacement, D^; cylinder ratio, Re] high-pressure clearance, Ch] and ratio of receiver volume to high- pressure displacement, y. Call ( It will be convenient first to find values of maximum receiver pressure (rec.pr.)i, and minimum (rec.pr.)2; high-pressure cut-ofiF Zj?, and compression Xh\ low-pressure clearance cl, cut-off Zl, and compression, Xlj in terms of these quantities. Nearly all of these are dependent upon the value of cl and it will, therefore, be evaluated first. WORK OF PISTON ENGINES 331 From the points C and /, Fig. 98, From A and E^ trec.pr.)2=(bk.pr.)^^^^^ (497) (rec.pr.)i = (in.pr.)^^, (498) and from E and C, (rec.prQi 1+c^+y ,^^. (rec.pr.)2 RcCl+V • • • v ; Dividing Eq. (498) by Eq. (497) and equating to Eq. (499), RpChJI+Ch) ^ I+Ch+V R(^cl(1+Cl) RcCl+v ■ Multiplying out and arranging with respect to cl, the relation to be fulfilled in order that complete expansion and compression may be possible is, CL^[Rd'{l+ca+y)]+CL{R<f^{l+CH+y)-RcRpCH{l+CH)] -[2/fipC/,(l+C/,)]=0. . (500) This is equivalent to C]?l+CLm,—n—Qy (501) and the value of cl is ^^ Jf^+Un)^-m ^^2) It is nauch simpler in numerical calculation to evaluate Z, m, and n and insert their values in Eq. (502) than to make substitutions in Eq. (500), which would make a very cumbersome formula. (rec.pr.)2 and (rec.pr.)i may now be evaluated from Eqs. (497) and (498) by use of the now known value of Cl.. High-pressure cut-ofiF Zh may be found from the relation of points B and /, Fig. 98, Rp(Zh+Cu)^Rc{1+Cl) or Zh^^{1'\-Cl)-Ch (503) tip Low-pressure cut-off, Zl from, Rc{Zl+Cl)^I^Ch, or Zl^^^-Cl (504) tic High-pressiu^ compression, Xh^ Xh^clRc-Ch (505) 332 ENGINEERING THERMODYNAMICS Low-pressure compression, Xl, by the use of points A and K^ Rc(Xl + Cl) = RpCh9 * or Re R Xl= j<~Ch—Cl (506) If Cl is regarded as being part of the original data, though it is related to Re, Rpt Ch and y as indicated in Eq. (488), the expressions for high- and low-pres- sure work and may be stated as follows: W^H=144(in.pr.)2)^ (|a+„,)[.+,^|fl±g] lliTS''+"+"'"*^S7--h'*'f]}- ■ <»"'* Tri;=144(m.pr.)I>if Re (I+Cl) [f. I ^ ,„si^„1±£e+V + (1 +Ch) log. ficj^l - [1^(1 +&) - cJ - CB log. fj I • (508) Adding these two equations gives the total work of the cycle as follows: TF=144(in.pr.)Z)Hfic(l+c^) {^^[l+lo& |j [j~^)] +|; log. Rc\ l+cu+y , 1+cg+y . l+CH+y , 1+cy+ y fip(l+Ci,) '^^ ficCi^+y "*'ftp(l+cii) '""^ i?cci:+y J -144(m.pr.)Z)^{cH[l+ log, ^^j+c^loge ^+^/^+Ci:)-c^ +Cl +Cb This, however, may be greatly simplified, and '<*iTffS + "*<S-'*''" logc^^^-^+loge bTT =lo& ^P' Ch iCcCl Hence TF^144(in.pr.)Z>^r|^(l+CL)-cJlog, Rp. = 144(in.pr.)Z>HZ/flogfti> (609) WORK OF PISTON ENGINES 333 From this may be obtained mean effective pressure referred to the low- pressure cylinder, work per cubic foot supplied, and consumption per hour per indicated horse-power, all leading to the same results as were foimd for the case of complete expansion and compression with infinite receiver (Section 15,) and will not be repeated here. To find the conditions of equal dwisum of work between cylinders, equate Eqs. (507) and (508). which may be simplified to the form, 1+ch ^ RcCL+y ^ Rc{1+cl) Rc(1+cl)1 ^ RcCl J (510) This equation reduces to Eq. (376) of Section 11, when ch and cl are put equal to zero. In its present form, however, Eq. (510) it is not capable of solution, and it again becomes apparent that for such cases the graphical solution of the problem is most satisfactory. Example 1. Method of calculating Diagram, Fig. 97. Assumed data: Pa =P» = 120 lbs. per square inch abe. F; = Fi = 2 cu.ft. P^^Pj^^ 30 lbs. per square inch abs. Vc^Vd^Vt^ .8 cu.ft. P>= 10 lbs. per square inch abs. Vg^Vi= .24 cu.ft. Fe= .2 cu.ft. Va^Vf^ .12cu.ft. V 0^1.2 cu.ft. Vb= .4 cu.ft. To locate point C: To locate point M\ _ P^n 120 X. 4 Pc =-7>— - — 5 — =60 lbs. per 8q.m. _ P^n 120 X. 4 ,. ^ Vm^-rs—^ — ^ — =1.6cu.ft. To locate point D\ Pd(Ftf+0)«Pm(Fm+0>, or Ptf«30^-^^=421b8.persq.in. 334 ENGINEERING THERMODYNAMICS To locate point E: Pe^{Ve+0)^P„iVm+0), or P.=r^^ X30-60 Ibs. per sq.iii. To locate point F: To locate point L: P "e K« 60 X .2 , /WN 11 Vf .u Pi =— Tr" ^ — ^";~33.3 lbs. per sq.in. Vi .Z4 ^ P»7t IPX .8 ,- ,^ To locate point N: since Pn-Pe To locate point G: P,{V,+0)=Pn{Vn+0) or. P, = 6o|^||^J =55.6 lbs. per 8q.in. To locate point H: Ph(.Vh+0)=P,(V,+0) v.^^-Xl+MllM or .24X55.5+55.5X1.2-30X1.2 ^^ ^_- — — - . 1.46 cuit. To locate point /: J, P*F» 1.46x30 „. ^ ,, Pi ^—Tr- *= 5 — =21.7 lbs. per sq.m. Vi ^ Example 2. Find the horse-power of a 12- and lSx24-]n. engme, running at 125 R.P.M., with a receiver volume twice as large as the low-pressure cylinder, 6 per cent clearance in the high-pressure cylinder, 4 per cent in the low, when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high- pressure cut-ofF J, low-pressure f , high-pressure compression 10 per cent, low 3 p)er cent. From Eq. (496) divided by 144Di„ and with the values as given above, the (m.e.p.) is equal to following expression: 150x.56X2^(l+log.^|)-10X.34 log. |j-150X.06X2~^ 150 X .56[.79 X2.25 +4.5] 4-10(.34)2.25 X4.5 .^ _i_ , ^6 1.06x2.25+4.5 [.16 +4.511.79X2.25 +4.5] -(4.5)« ^'^^^2.25 ^^ .06^.16 X2.25+4.5 .79x2.25+4.5 04x2.25+4.5 150 X .56 X4.5 + 10 X .34 X2.25[.16 +2.25] [.79 X2.25+4.5] ^ o 25 1 ' 79 X2.25[.16+2.25] +(.16X2.25) [.04x2.25 +4.5 J ' ^ ^8« j + — ^ . 150x.56x4.5 + 10(.34x2.25)[.16+2.25]r^^, 1.04"! ^^^,, -^ .79x2.25[.16+4.5] + (.16x 2:25) L"^' ^^«^ li^\ ^^^'^ ^^''^ ^' ^•^^- h(^r^cQ the horse-power will be 191. WORK OF PISTON ENGINES 335 Prob. 1. Find the work done in the high-pressure cyKnder and in the low-pressure Under of the following engine under the conditions given. Engine 14 and 30 X28 ins., 100 R.P.M., 5 per cent clearance in each cylinder, gh-pressure cut-off iV, low-pressure cut-off A, high-pressure compression A, low- lessure compression iV, initial pressure 100 lbs. per square inch gage, back pressure lbs. per square inch absolute, and receiver volume 3 times the high-pressure splacement. Logarithmic expansion. Prob. 2. The following data are available: initial pressure 200 lbs. per square inch )solute, back pressure 10 lbs. per square inch absolute, engine 10x15x22 ins., with per cent clearance in the high- and low-pressure cylinders, speed 100 R.P.M. What will 3 the cut-offs, and compression percentages to give complete expansion and compression, ogarithmic expansion? Prob. 3. What will be the work done by the above engine working under these )nditions? Prob. 4, What must be the low-pressure clearance, cut-offs, and compression ereentages, to give complete expansion and compression for a similar engine work- ig under the same conditions as those of Prob. 2, but equipped with a receiver twice 3 large as the high-pressure cylinder? 18. Compound Engine with Finite Receiver, Exponential Law, with leaiance and Compression, Cycle XIV. General Relations between tessures, Dimensions, and Work when HJP. Exhaust and L.P. Admission re Independent. It cannot be expected that the treatment of this cycle by jrmulas will give satisfactory results, since even with the logarithmic expansion iw, Cycle XIII gave formulas of unmanageable form. For the computation of wk done during the cycle, however, and for the purpose of checking pressures and ork determined by graphical means, it is desirable to have set down the relations f dimensional proportions, initial and final pressures, and valve adjustments, to le receiver pressures, release pressures and work of the individual cylinders. The conditions of a steady state, explained previously, require that (Fig. 97) 1 1 n-v,=v.{^y -v.{^y , (511) hich is the same as to say, that the quantity of fluid passing per cycle in the gh-pressure cylinder must equal that passing in the low. Expressed in terms dimensions, 1 L (in.pr.) J , . ^ (in.pr.) , rearrangmg, and usmg Rp = /|^^j; x • 1 Up, L (in.pr.) J 1 +fl.(c.+Z.)r(''^*:^?P'--H', .(512) L (in.pr.) J 336 ENGINEERING THERMODYNAMICS an equation which contains two unknown pressures (rec.pr.)i and (cut ofif pr.)i. To evaluate either, another equation must l>e found: where Pn=Pe, so that 1 V.'RcD„(c.+X.)[^J' (613) Hence Vn+0 \ or 1 =p«(^ (cut-off pr.)i= (rec.pr.)i ^L(rec.pr.)iJ y+Rc{cL+ZL) _ r y(rec.pr.) M+ R c(cL+Xi,)(bk.pr. )* ] ' . L y+Rc{cL+ZL) J, ■ ^^ ^ 1-.. which constitutes a second equation between (cut-off pr.)L and (rec.pr.)i, which, used with fk). (512) makes it possible to solve for the imknown. By substitu- tion in Eq. (512) and rearranging, 1 r..n r.. ^ fhV r.r ^ \ «/»* fa + ^n) [y+/Zc(CL + Zi:)]+ficy(CL+Xx) ]' ,,. ,. (rec.pr.)i = (bk.pr.)[ ^^^+x.)[y^Rc(c,+Z,)]+Rcy{c.+Z,) [ ' ^^^^^ This expression is of great assistance even in the graphical construction of the diagram, as otherwise, with all events known a long process of trial and error must be gone through with. It should also be noted that when s = l this expression does not become indeterminate and can, therefore, be used to solve for maximum receiver pressure for Cycle XIII, as well as Cycle XIV. Cut-off pressure of the low-pressure cylinder, which is same as the pressure at H or at My Fig. 97, is now found most easily by inserting the value found bj Eq. (515) for (rec.pr.)i in Eq. (5i4). Enough information has been gathered now to set down the expressions for work. W„ = 144D« I (in.pr.) '-^f [s - ('fj^)* "] -c.(m.pr.) (ch+Xh) y . [ /ch+XsV"^ 8-1 JV+Ch+Xh) s-1 WORK OF PISTON ENGINES 337 ir..l44D,|(cut^prO/-±-'^i-±?^[(t!^«^?iJ)'-'-l] -(bk.pr.)-'^^^^±^^[(^^)'"'-l]-(bk.pr.)(l-X^^ficj. . (517) Addition of these two Eqs. (516) and (517) gives an expression for the total work Wf and equating them gives conditions which must be fulfilled to give equality of work in the high- and low-pressure cylinders. Since these equations so obtained cannot be simplified or put into more useful form, there is no object in inserting them here, but if needed for any purpose they may be easily written. In finding the conditions of equal work, the volumes of (rec.pr.)i and (cut-off pr.)L must be inserted from Eqs. (514) and (515) in (516) and (517), in order to have terms in the two equations consist of fimdamental data. This, however, increases greatly the complication of the formula. After finding the total work of the cycle, the mean effective pressure referred to the low pressure is obtained by dividing by 144XDi». To assist in finding the work per cubic foot supplied and consumption, and the cubic feet or pounds per hour per I.H.P. it is important to know the volume of fluid supplied per cycle, r - (Snp^lol) ^QB=DH\{cH+ZH)-{cH+XB)y^^^ . (518) Example. Find the horse-power of and compressed air steam used by a 12- and 18x24-in' engine running at 125 R.P.M., with a receiver volume twice as large as the low-pressure cylinder, 6 per cent clearance in the high-pressure cylinder, 4 per cent in the low, when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high-pressure cut-off J, low-pressure cut-off }, high-pressure compression 10 per cent, low-pressure compression 30 per cent, and expansion and compression follow the law PV^'^=c. From £q. (515) (rec.pr.)i is found to-be as follows when values for this problem are substituted: X ,^ri5'7(.56)[4.5-|-2.25(.79)]-h4.6x2.25(.34)1« ^, ^ „ . ^ . (rec.pr.).^10[. ^e [4.5-h2.25X.79]4-4.5x2.25x.79 J '^^'^ ^^«- ^^^^^^ ^^^^^^' and by using this in Eq. (514) .71 .71 « , ^ re ^ / 4.5X81.7 +2 .25X.34X10 \ „„ • u w (cut-off pr.)L = ( 4 5-f-2 2 5x 79 / '^ ^^^- ^'^^' absolute. 338 ENGINEERING THERMODYNAMICS It is now possible by use of Eqs. (516) and (517) by addition and division by 14ADl, to obtain (m.e.p.). Substituting the values found above and carrying out the process just mentioned. '»«-'-2^{'™xfx['*-(ia)1-"''<'«'-f ><»»[(:«)■*-■] - -^X82|^l-|^^j^-p^j J+53X ^ LU5+2:25T.04; "^J ,^^^^ 4.5-h2.25x.79 r, / 4.5+2.25x.79 \ *] ,^^^ 2.25x.34 r/.34\ -* 1 +53X ^i^l—-——-J J_iox— ^— |^(^-j -Ij -lOx2.25x.7l =51.5 lbs. 8q.k hence the horse-power will be 200. By means of Eq. (518) the supply volume may be found. This gives upon substituting of the proper values: . (Sup.Vol.) =Z)^[.56-16x(^^y'n =.462>^. i-u-rx 1. TTTT> 13,750 ^^Sup.Vol. Cubic feet per hour per LH.P. = r- rX — ~ , (ta.e.p.) Dl ^13^50 ^^ 51.5 ^2.25"^*^' hence the total volume of air per hour will be 64.5X200 = 10900 cu.ft. Prob. 1. What will be the receiver pressure and L.P. cut-off pressure for a cross-compound compressed air engine with 5 per cent clearance in each cylinder, run- ning on 100 lbs. per square mch gage mitial pressure and atmospheric exhaust, when the high-pressure cut-off is i, low-pressure f , high-pressure compression 15 per cent, low 25 per cent, and 5 = 1.4. Receiver volume is twice the high-pressure cylinder volume. Prob. 2. Find the superheated steam per hour necessary to supply a 14- and 21 x28- m. engine with 5 per cent clearance m each cylinder and a receiver twice' the aze of the high-pressure cylinder when the initials pressure is 125 lbs. per square inch gage, back pressure 7 lbs. per square inch absolute, speed 100 R.P.M., high-pressui^ cut-off i, low-pressure A, high-pressure compression 15 per cent, low pressure 40 per cent and s = 1.3. Note: B = .3. Prob. 3. If the high-pressure cut-off is changed to } without change of any other factor in the engine of Prob. 2, how will the horse-power, total steam per hour, and steam per horse-power per hour be affected? If it is changed to i ? Prob. 4. A boiler capable of supplying 5000 lbs. of steam per hour at rated load fur- nishes steam for a 12- and 18 x24-in. engine with 5 per cent clearance in each cyhnder and running at 125 R.P.M. The receiver is three times as large as the high-pressure cylinder, WORK OF PISTON ENGINES 339 the initial pressure 150 lbs. per square inch gage, back pressure 5 lbs per square inch absolute, the low-pressure cut-off fixed at i and low-pressure compression fixed at 30 per cent. At what per cent of its capacity will boiler be working for these follow- ing cases, when »1.2 for all and 20% of the steam condenses during admission ? (a) high-pressure cut-off i, high-pressure compression 80 ) er cent, (6) high-pressure cut-off J, high-pressure compression 20 per cent, (c) high-pressure cut-off i, high-pressure compression 10 per cent. Note: 8 =.33; Fig, 99. — ^Work of Expansion in Compound Engine without Receiver and with Clearance. Qrcle XV, Logarithmic Expansion; Cycle XVI, Exponential, High-pressure Exhaust and Low-pressure Admission Coincident. 19. Compound Engine without Receiver. Logariflunic Law, with Clear- ance and Compression, Cycle XV. General Relations between Pressures, Dimensions, and Work when H.P. Exhaust and L.P. Admission are Coinci- dent The graphical construction for this cycle has been described to some extent in connection with the first description of the cycle, given in Section 8, of this chapter, and is represented here by Fig. 99 in more detail. To show that the expansion from D to £ is the same as if volumes were measured from the axis ML, consider a point Y on DE, If the hypothesis 18 correct P4XKM = FyX{KM+rK) (519) 340 ENGINEERING THERMODYNAMICS The true volume when the piston is at the end D of the stroke, i^ 2)^(1 +Cjj+i2cCi,), and at F, the true volume is where y is the fraction of the return stroke that has been completed in both cylinders when the point Y has been reached. Then Dividing though by (fie— 1), ^This equation may be observed to be similar in form to Eq. (519). More- over, the last term within the bracket, D^y, is equal to the corresponding term Y'K, in Eq. (519), hence, \\-\-Ch-\-RcCl ■\-RcCl'\ KM=DBy-^^£^'-^\ (52i: SimUarly, the distance QM, or equivalent volume at 1/ is QM =i)Jl±gt«^^l (522) The following quantities will be evaluated preparatory to writing the expres- sions for work: (rel.pr.)ir=(in.pr.)^p^; (523) (bk.pr.) 1+c.^ ^ (rel.pr.)jy (m.pr.; Zh+Ch SC (in.pr.)L = Pd' = (rel.pr.)/f^g; _-^. „x £H+Cff L (m.pr.) £b+cu I -^m.pr.j[^ j^^^ J Di,[1+Cb+RcCl] ZM+CH+Rc{Xr,+ci.)^^ = (in.pr.)— ^^^^^^-^ (^25) (cutoff pr.). = (m.pr.). ^-^ = (^-pr.). \y^^^ ^^^^^ (^^^.-ij^iHx^) J = (in.pr.) (m.pr.) .l+CH+ficCL+(fic-l)(l-X«). (52(.: WORK OF PISTON ENGINES 341 The ratio of expansion from E' to G is equal to (cut-off pr.)L _ 1 +cl (rel.pr.)L ~ (1 -Xh) +Cl • • • • • . • (527) Hence F^ = 144Dir(in.pr.) (Z.+c.)[l+lo«.^] (in.pr. ) Rc-l Zi^+Ci,+2ee(Xi,+Ci,)^''- P'-^ log. [' ^-ch^RcCl+{Rc-1){X -Xn) 1-\-ch-\'RcCl ] (in.pr. ) \l-\-CH'\'RcCL-{-{fic-l){l-XH) t . IT \ 1 ^h-\-Xh (Ch+Xh) log* Ch ch (528) TrL=144Dx,(iii.pr.) Z.+c.+i2c(X.+c.)j^PI:] Rc-1 loga 14- C^+/2cCl+ (fie- 1) (1 -Z 1+c^+ficC^ .)] + (m.pr. ) 11+Cb+RcClh+ (Rc- 1) (1 - Xb) J -144Dz,(bk.pr.) (l-fC.-X.)l0ge(-j^g^J l-ZL+(Xz.+Cz.)log,^^^^. . . (529) Cl The totoZ iDork found by adding TFh and Wl as given above, leads to the following: )F = 144i>^(in.pr.) (2^H+CH)l0g6 +Z.+[zH+c.+fieCX.-fc.)^]loge [i±^ -|-CH+ficCi.+(ffc-l)(l -X +C/f-|-ficCL . + ZH+CH'\-RciXL-^CL)$^'^^'^ 1-hCH^RcCLHRc -i)(i-£) J [^^fi+^^-^^) X '*(RS3f;)-'-«'>'*Hf')]l -144Dx,(bk.pr.) j 1 -Xi,+(Xl+cl) log. ^ . . (530) 342 ENGINEERING THERMODYNAMICS This is the general expression for the work of the compound engine without receiver, with clearance and compression^ when high-pressure exhaust and low- pressure admission are simvUaneous and expansion and compression logartthmic, in terms of fundamental data regarding dimensions and valve periods. From this the usual expressions for mean eflfective pressure, work per cubic foot supplied, and consumption per hour i>er I.H.P., may be easily written, provided the supply volume is known. This is given by (Sup. Vol.) =^A'B=dJ(Zh+Ch) - fa+X^) fr"^.'^^P!''H L Cm.pr.) J =Dh Zh+Ch-(ch+Xh) Zh+Ch+Rc(Xl+cl) (bk.pr.) (in.pr. ) 1+Ch+RcCl+(Rc-IKI-Xh)] (631) To find the conditions which must be fulfilled to give equal work in the two cylinders, equate Eqs. (528) and (529). ^.+(Z.+c.)log.(^) ft< fi< :^[ Zn-^CH+RciXL-^CL) (bk ( Zh -{-Ch +RciXL -f cl)7t-^4 { in.pr.; ?^i log, r^ -hCM-\-RcCLHRc-l)(l-X I-^ch-^-RcCl '] ll+CH-^RcCLHRc-mi-XH)] Rc{l-\-CL-XH)l0gfi \1+Cl-XhI +{chXh) logs ch-\-Xh Ch +«c (bk.p r.) ( in.pr.) 1 -Xl + {Xl+cl) log«(^^^) I =0. (532) These expressions are perfectly general for this cycle, and expressed in terms of fundamental data, but are so complicated that their use is ver}^ limited, as in the case of some of the general expressions previously derived for other cycles. As in other cycles, it is desirable to investigate a special case, that of complete expansion and compression in both cylinders, Fig. 100. First it is necessary to determine what are regarded as fundamental data in this case, and then to evaluate secondary quantities in terms of these quantities. The following items are assumed to be known: (in.pr.), (bk.pr.), which is equal to (rel.pr.)2^, Rcy Ch, and Cl, and Dy, which are dimensions, and it is known that the pressure at the end of compression in L.P. is equal to (rel.pr.)^. Referring to the diagram, displacements, clearances, and the axis for the common expansion, ML, can all be laid out, and the location of the points A and G determined. The points E and E' are at the end of the common expansion within the two cylinders, and at beginning of high-pressure compression and of separate low-pressure expansion, hence p«=p«'. WORK OF PISTON ENGINES 343 From the points A and E: p.=(in.pr.)y-j— From the points G and B', p«'=p«=(bk.pr.) 1+Ci I+Cl-Zj^' M V Fig. 100. — Special Case of Cycles XV and XVI. Complete Expansion and Compression in both Cylinders of Compound Engine without Receiver and with Clearance High- Pressure Exhaust and Low-Pressure Admission Coincident and equating, 1+Ci whence where Xh = Ch 1+Cl + CuRp ' (533) Ri _ / in.pr. \ \bk.pr./ Substituting the value of X^ in either of the expressions for pet which is the low-pressure cut-oflf pressure, + Cl + RpCh1 ^ /KOA\ (cut-oflf pr.)L=Ptf = (bk.pr.) — L A 344 ENGINEERINa THERMODYNAMICS It may be noted here that the cylinder ratio does not enter into this, but only clearances and pressures. In the no-clearance case, it may be remembered that the point E or -E' was not present, as it coincided with G, Next, to find the high-pressure release pressure, pa, by means of points £ and D, and their relation to the axis ML, Fig. 100. l+CH+RcCL-i +Cl + CuRc 1+Cl+Ch (rel.pr.)^ ^Pd^p, =^ = (bk.pr.) I -^ 1''Xh+ Re- 1 I+cu+RcCl Rc-l' = (bk.pr.) i^,^^jg^^ (53oi Knowing the release pressure of the high-pressure cylinder, it is possible to find the high-pressure cut-off and compression necessary to give the required performance. 7 _/i , ^ .(relprO^ ^ _(1+Ch)[ Rc(1+Cl)+RpCh ^ ^ ,„.^ Y ^ (rel.pr.)H ^ ^ [Rc(1+cl)+RpCh J ,„.. ^^='"^7bk:^y~"^=^H"T4^^+ft^^"^J- • • • (^'^ The work of the two cylinders is as follows: w i^n fir. «, ^ f (1+c^) r ^c(l+Ci.)+fii 'Cw1 f, , . „ Rpjl+cg+RcC j.)'] ^ TFH=144DH(m.pr.) |-^ [ l+CK+fieC.-JL^+'^«'i2o(l+c.)+22.^.J -'" _ n+CH+RcCL ] \ RcO.+Ci,)-\-RpCa ] J_ , \ Rcil+CL)+RpCB ] r 1+Cx+ Cjr 1 L fic-1 JL 1+Cir+iecCt J/ip^L l+CB+RcCt. \ll+Ci.+RpCa\ _ i4An rt„ «, ^ [ (l+<'g) fic(l+Ci,)+gpCg r, , Rp O.+Ch-\-RcCl '\ ^ L i?p(i?c-l) J ^'L l+to+ficCt \\\+Cl-^RpCh\ Tr.=144Dx(bk.pr.) 1 1^ ^^3j-^ J log. |^-j^-^^-__J |^- «<7(l+&)+fii<ill WORK OF PISTON ENGINES 345 These expressions, when added and simplified, give the following for total work per cycle, W^lW)^{hk,pT.)ll^Cr.[-f^^^^ . (540) in which of course DhRc may be used instead of Dl and —^— instead of ftp (bk.pr.) and then TF = 144Z)^Zj^(in.pr.)loge/ep, . ... (541) Zjt, having the value of Eq. (536). IJquality of work the in high- and low-pressure cylinders results, if W Eq. (538) equals W^, Eq. (539), or if 2W„=W, or 2Fl = F, all of which lead to equivalent expressions. Simplification of these expres- sions, however, does not lead to any direct solution, and hence the equations will not be given here. Example. Find (a) the horse-power and (6) steam used per hour for a 12- and 18 x24-in. tandem compound engine with no receiver, 6 per cent clearance in the high-pressure cylinder, and 4 per cent in the low, when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., high- pressure cut-off J, high-pressure compression 15 per cent and low-pressure compression is complete. (a) Since the low-pressure compression is complete, the pressure at end of compression must be equal to the release pressure of the high. This latter quantity may be found rom the relation (in.pr.)(ZH-f-c^)=(rel.pr.)H(l +Cii), or (rel.pr.)^ = 150 T^- =79.3 lbs. per sq.in. absolute. l.Oo Low-pressure compression may be found from the relation (rel.pr.)^(cL) = (bk.pr.) (cl-{-Xl)j or Xi, = .28. (m.e.p.) may be found from Eq. (540) divided by 144DL,which on substitution gives 150 2.25 / 1 08 (.5 -f .06) log, i^ + K^\ K^na^ooKfOQ^i^A^ ^^li r i-i-0^+2.25X .04-H.25X.85l .5+ |^.5-K.06-h2.25(.28+.04) -Jloge [ i^.06-H2:25x':04" J .5 -f .06 +2.25(.28 + .04) ^ 1-f. 06+2.25 X.04-|-1.75-f-.85 .06 +.15 2.25(L04-.15)loge-^-^^ (.06+.15)loge -10 .06 and the horse-power will be 271, 1.28 +(.28 +.04) loge — tir \ =69.7. lbs. sq. in. .04 346 ENGINEERINa THERMODYNAMICS (b) Since the consumption in cubic feet per hour per horse-power is equal to 13,750 Sup.Vol. (m.e.p.) Dl and supply volume is given by Eq. (531), this becomes 13,750 1 X: 69.7 2.25 .56+2.25(.32)^ .56-.2ir ^^ 1.06+.09-I-1.25X.85, =44, hence the consumption per hour will be 44 X271 X.332 =4000 pounds. Prob. 1. A Vauclain compound locomotive has cylinders 18 and 30x42 ins., with 5 pec cent clearance in each and runs on a boiler pressure of 175 lbs. per square inch gage and atmospheric exhaust. The steam pressure may be varied as may also the cut-ofif to a limited degree. For a speed of 200 R.P.M., a cut-off f and 10 per cent compression in each cylinder, find how the horse-power will vary with the initial pressures of 175, 150, 125, and 100 lbs. gage. Prob. 2. When the cut-off is reduced to i in the above engine compression increases in tlie high-pressure cylinder to 20 per cent. For the case of 175 lbs. gage initial pressure find the change in horse-power. Prob. 3. Find the steam used by the engine per hour for the first case of Prob. 1 and for Prob. 2. Prob, 4. It is desired to run a 12- and 18x24-in. no-receiver engine with 5 per cent clearance in each cylinder, under the best possible hypothetical economy conditions for an initial pressure of 200 lbs. per square inch absolute and atmospheric exhaust. To give what cut-off and compression must the valves be set and what horse-power wUl result for 100 R.P.M.? 20. Compound Engine Without Receiver. Exponential Law, with Clear- ance and Compression, Cycle XVI. General Relations between Pressures, Dimensions, and Work, when H.P. Exhaust and L.P. Admission are Coin- cident. Again referring to Fig. 99, it may be observed that reasoning similar to that in Section 19 but using the exponential law, would show that the same formulas and graphical constructions will serve to locate the axes of the diagram, hence, as before, KM=Dh ^-j , (542) and QM=D^±f±^ (543) Release pressure in the high-pressure cylinder is (rel.pr.)H = (in.pr.)f-^-^j (544) WORK OF PISTON ENGINES 347 Immediately after release the pressure is equalized in the high-pressure cylinder and the low-pressure clearance. The pressure after equalization, «rmed (iii.pr.)i;i is found by the relation of the volume at S and that at D\ Pig. 99, measured from the axis KT. (in.pr.)L = (rel.pr.)^ '+-+«'(-+^')(s^)' which, by means of Eq. (544) becomes « 1 ^ « (m.pr.)^=(bk.pr.)[— ^:p-^-;;:p^^;^^ J (545) The expansion of the fluid goes on as it passes from the high^pressure cylinder to the greater volume in the low-pressure, as indicated .by. liD'JS' and DEy and when the communicating valve closes, the pressure has become (cut-oflf pr.)L = (in.pr.)L Rc-l l+C^+ficCL_^(j_^^) Rc-l which, by means of Eq. (545) reduces to 1 (ou.^ pr.).-(bk.pr.)[ , «;'^^^;> ±'^^g4 ]-. . . (546) After cut-oflf in the low pressure, expansion goes on in that cylinder alone to the end of the stroke, when release occurs at a pressure (rel.pr.)L= (cut-oflf pr.)z,(?^±^^y or by substitution from equation (546), (rel.pr.).-(bk.pr.)[ i+c«+i2^,+(i— jQ(^,;- i) ( l+c, )J " (^47) In terms of these quantities the work of the high- and low-pressure cylinders can be written out as follows: TTh = 144D« { (in.pr.)'-^ [a- (^--±£i?)' "] -c«(in.pr.) (bk :.pr.) r g«p(cg+Z«)+gc(cf,+.Yx.) ]'r / \+ch+RcCl V-H -1 L fic-1 JL \l+c„+RcCL+0-X„){Rc-l)) J 1 -^<-«4rS^^"^Sf«!^)]TC-^)-'-'] j (-) 348 ENGINEERING THERMODYNAMICS and FFL=144i>z, 1 • (bk. 8 -1 L Rc-i J ^ L \l+c,,+i2c7C/^-f-(l-X)(l2c-l)y J 1 , (bk.pr.) ,, , ^ ^ J R'p(ch+Zh)+Rc{ cl+Xl) ]'L / 1+C£,-X„ \'-M + ,_1 ^^+<^^-^")[i+c„+RcCj.+ (l^X^){Rc-T)\ L^"V 1+c^ / J _^k£i^.J:Z.)|^^ (549) These are general expressions for work of high- and low-pressure cylinders for this cycle, and from them may be obtained the total work of the cycle, mean elBfective pressure referred to the low-pressure cylinder, and by equating them may be obtained the relation which must exist between dimensions, events, and pressures to give equal division of work. It would, however, be of no advantage to state these in full here, as they can be obtained from the above when needed. The supply volume, cubic feet per cycle, is represented by A'B, Fig. 99, and its value is found by referring to points B and E as follows: iSnp.Yol)^D„[(c.+Z.)-{c„+X,)(^^^^f]' =z)«^c«+Zh -^—\i+cs+r^+{i-x^){r:^T)}\- ^^^ Rs p Work per cubic foot supplied is foimd from Eqs. (548), (549), and (550). Work per cu.ft. suppUed = -r^^-^y^p:. (551) Consiunption, cubic feet per hour per I.H.P., is found from mean elBfective pressure referred to L.P. cyl. and supply volume as follows: Consiunption, cu.ft. per hr. per I.H.P. 13,7.50 (Sup.Vol.) ^552) (m.e.p. ref. to L.P.) Dl This will give pounds consumption by introducing the factor of density. Further than this, it will be found more practicable to use graphical methods instead of computations with this cycle. WORK OF PISTON ENGINES 349 Bzample. Find (a) the horse-power and (6) consumption of a 12- and 18x24-in. no-receiver engine having 6 per cent clearance in the high pressure cyUnder and 4 per cent in the low when the initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch abzolute, speed 125 R.P.M., high-pressure cut-off i, high-pressure compression 15 per cent, and low-pressure compresaion is complete. (a) The per cent of low-pressure compression may be found as in the Example of Section 19, using the value of « in this case of 1.4. Then (in.pr.)(ZH+c/f)** « (rel.pr.)^ (1 -\-ch)^'\ or (rel.pr.)j5r=61.5 lbs. sq. inch absolute, and (rel.pr.)^ Xcl^* = (bk.pr.) (clH-Xz,)^*, or Zl = .11 From the sum of Eqs. (548) and (549) divided by 144Dx, and with proper values substituted, (m.e.p.) =48.5 lbs.; hence the horse-power is 189. 13 750 (6) From Eqs. (550) the value for (Sup. Vol.), which when multiplied by — ■' — , m.e.p. and divided by Dl, gives cubic feet air per hour per I.H.P. I?i750 y J_r nfi_. ^ _ ^21 15-7(.56)+2 .25(.15) 1 « .63 cu.ft. per 48,5 ^2.25L "^ (15)-^ H-.06+2.25x.04+(l -.15)(2.25-1)J hourper I.H.P. Prob, 1. If the locomotive of Prob. 1, Section 19, should be equipped with superheater so that the steam expanded in such a way that 8 = 1.3, what would be the effect upon the horsu-power for conditions of that problem and on the cylinder event pressures? Prob. 2. A 30- and 42 x54-in. no receiver steam pumping engine runs at 30 R.P.M. and has 3 per cent clearance in the high-pressure cylinder and 2 per cent in low. There is no compression in either cylinder. Initial pres ure is 120 lbs. per square inch gage, and back pressure 28 ins. of mercury (barometer reading 30 ins.). The steam is such that the expansion exponent is 1.25. What will be the horse-power of, and the steam used by the engine when the cut-off in the high is i? Prob. 3. By how much would the power change if the cut-off were shortened to | and then to i, and what would be the effect of these changes on the economy? 21. Triple-Expansion Engine with Lafinite Receiver. Logarithmic Law. No Clearance, Cycle XVU. General Relations between Pressures, Dimen- sions and Work. Fig. 101 represents the cycle of the triple-expansion engine with infinite receiver, no clearance, showing one case of incomplete expansion in all cylinders, and another where overexpansion takes place in all cylinders. 350 ENGINEERING THERMODYNAMICS The reasoning which follows applies equally well to either case, and to any combination of under or overexpansion in the respective cylinders. It is desired to express the work of the respective cylinders and the total work in terms of dimensions, initial and back pressures, and the cut-offs of the respective cylinders. To do this, it is convenient first to express the bk.K-) INCOMPLETE EXPANSION OVER EXPANSION CYCLE XVll Fig. 101. — Work of Expansive Fluid in Triple-Expansion Engine with Infinite Receiver and Zero Clearance. Cycle XVII, Logarithmic Expansion. first receiver pressure (1st rec.pr.) and second receiver pressure (2d rec.pr.) in terms of these quantities. The subscript I refers to the intermediate cylinder. or and /I i. \ /■ \ZhDh (1st rec.pr.) = (m.pr.)-^-p:— , (553) or (2d rec.pr.) = (in.pr.) Z„Dh ZlDl' • • • . . (654) WORK OF PISTON ENGINES Work of high-pressure cylinder is = 144(in.pr.)Z)i, | Z« (l +log. ^) - ^^ Work of intermediate cylinder is = 144(in.pr.)i)/f | Z^^l +log. ^) - - lU(m.pr.)DH I Za^l+log. ^) - DhZJ)i, Z„D,\ ZlDl Work of low-pressure cyUnder is The total work by addition is W = 144(in.pr.)Z)HZ;, I (l +I0& ^) + (l +log. ~j + U -hlog. ^\ = 144(in.pr.)Z)aZjif 3+loge D H D: ] ZhZiZj^ Z,Dt ZJ)i 351 (555) (556) = 144(in.pr.)i)i,Z«U+lo&^\-144(bk.pr.)Z)i,. . . (557) -144(bk.pr.)Z)i. . (558) Mean effective pressure referred to the low-pressure cylinder is found by dividing W by 144D/,, and is therefore (m.e.p. ref. to L.P.) Z)i = (in-pr. )Zh^ 3-hlog. 1 D H ^1 ZuZiZt. ZiDi ZlDl -(bk.pr.). . (559) 352 ENGINEERING THERMODYNAMICS Work done per cubic foot supplied is equal to W divided by the supply volume AB or ZhDu, Work per cu.ft. supplied = 144(in.pr.) 3+I0&: ZhZjZl ZiDj ZlDl ► -144(bk.pr.)^^. (560) The volume of fluid supplied per hour per indicated horse-power is Consumption, cu.ft. per hr. per I.H.P. 13,750 ZhBh (m.e.p. ref. to L.P.) Dl (561) The wdqhi of fluid used per hour hour per indicated horse-power is of course found by multiplying this volume Eq. (561) by the density of the fluid used. The conditions which will provide for eqiwl division of work between the three cylinders may be expressed in the following ways: which is equivalent to, first: WH=Wly or hence , 1 1/Dh\ , 1 l/Dj\ ''^T,-zXDj)=^''^zrYAD-j' ^k-m)-m) <«») Similarly from Wh = Wl, ^ Zh Zi \Di I (m.pr.) \DhI Zi ^ These two equations, (562) and (563), show the necessary relations between in order that work shall be equally divided. Since there are six independent quantities entering (as above) and only two equations, there must be jorar of these quantities fixed by conditions of the problem, in order that the other WORK OF PISTON ENGINES 353 two may be found. For instance, if the cylinder ratios, the pressure ratio, and one cut-off are known, the other two cut-offs may be found, though the solution is difficult. Again, if cutroffs are equal, and the ratio of initial to back pressure is known, it is possible to find the cylinder ratios. This forms a special case which is of sufficient importance to require investigation. If ZH=Zj^ZLy Eq. (562) becomes and Eq. (563) reduces to />/ Bl Da Dj' ' D,Dl D^ (in.pr.) " (bk.pr.)' (564) (565) but from Eq. (564), Di. Di^ ^Di. ^ (DtY Dj Dm Dm \Dm) ' and therefore DJhjDA D^ \Dm) 3 and %'wA^y <^) which, along with the condition assumed that Zh^Zj^Zl (567) constitute one set of conditions that mil make work equal in the three cylinders This is not an uncommon method of design, since by merely maintaining equal cut-offs, the work division may be kept equal. The work done in any one cylinder imder these conditions Eqs. (566) and (567) is then Tf^^}F, = Tr^ = 144(in.pr.)D^{z(l-hloge|)-(^^)*} . (568) and the total work ir=432(m.pr.)Z)„{z(l+logc|)-(^-) I (569) in which Z represents the cut-off in each cylinder, all bemg equal. 354 ENGINEERING THERMODYNAMICS A special case of the triple-expansion engine with infinite receiver and no clearance which demands attention is that of complete expansion in all cylinders, represented by Fig. 102. Here ^.=^=#^; ....... (570) Dl (2drec.pr.) ^ ^Dg^ (2d rec.pr.) , ' Di (1st rec.pr.)' (571) Fig. 102.— Special Case of Cycle XVIII Complete Expansion in Triple-expanaon Engine with Infinite Receiver, Zero Clearance, Logarithmic Expansion. and 7 _ (1st rec.pr.) _ / bk.pr. X D^ ^^^^2) ^ (in.pr.) \in.pr. /Z)/f' hence the receiver pressures are as follows: (1st rec.pr.) = (bk.pr.)yp, (573) WORK OF PISTON ENGINES 355 and (2drec.pr.)=(bk.pr.)^ (574) The work of the respective cylinders, expressed in terms of initial and back pressures and displacements is then, -l«(bk.pr.)C.l<*(|g^|«) (675) Similarly, Wj = lUihk.pr.)DL\og,(^^, (576) and iri=144(bk.pr.)Z)iloge^ (577) Total work, by addition, is W. 144(bk.pr.)i,4l* ({H|j^ ^^)+lo^ l+log. f ] - 144(bk.pr.)Di log, |j^ (578) If for this special case of complete expansion equality of work is to be obtained, then from Eqs. (575), (576), and (577), (inj)r 0^ Dh^D^^Dl ,--q. "(bk.pr.)i)L Dh ,D/ ^""'^^ wfiich is readily seen to be the same result as was obtained when all cut-offs were equalized, Eqs. (564) and (565). This case of complete expansion and equal work in all cylinders is a special case of that previously discussed where cut-offs are made equal. Hence for this case cut-offs are equal, * „ „ „ Bn Di Dl (bk.pr.) . . Dl Dl Dh (m*prO Bzample. A triple-expansion engine 12- and 18- and 27x24-ins., with infinite receiver and no clearance, runs at 125 R.P.M. on an initial pressure of 150 lbs. per square inch absolute, and a back pressure 10 lbs. per square inch absolute. 356 ENGINEERING THERMODYNAMICS If the cut-offs in the different cylinders, beginning with the high, are i, i, and i, what will be (a) the horse-power, (6) steam consumed per hour, (c) release and receiver pressures? (a) From Eq. (669) = 150X.5X- 1 f. - ... 8 8 5.06 hence 3 +log, 14.2 -T-;rrr -^ « «, \ -10 »39 lbs. pei sq.iii, ^ 3X2.26 3X2.25' }- 39X2X573X250 ^•^•*^' 33;000 ^^ (6) From Eq. (561). Cubic feet per hoise-power per hour= , — - — ^v -~-^' (m.e.p.j Dl 13,750 1 X .o X - -^yj ^34«9| 39 5.06 hence total pounds per hour will be, 34.9 X.338X.332 -3920. * (c) From Eq. (553) 1st (rec.pr.)«(in.pr.) T>hZh ZiDi' 5 150 X o>Tg o ;7? = 89 lbs. per Bq.in absolute. •o7oX^u&5 From Eq. (554) (rec.pr.) =(m.pr.)^-^, .5 = 160 X »^, ' , ^^ =3.75. per sq.in absolute. .375x5.06 High-pressure release pressure may be found from relation (in.pr.)Zfii)ir = (peI.pr.)£fZ)/^ .'. (reLpr.)/f =75 lbs. Similarly 1st (rec.pr .)ZiDi = (reLpr.)/!)/, or (rel.pr.)/=33.4. Similarly 2d (rec.pr.)Zi/)i, = (rel.pr.)ii>L, or (rel.pr.)z, = 14.8. Prob. 1. What would be the horse-power and steam usfed per hour by a 10- and 16- and 25x20-in. infinite receiver, no-clearance engine, running at 185 R.P.M. on an initial pressure of 180 lbs. per square inch gage and atmospheric exhaust. Cut-offs .4, .35^ and .3. Prob. 2. The following data are reported for a test of a triple engine: Size 20 x33 X52 x42 ins., speed 93 R.P.M., initial pessure 200 lbs. per square inch gage, back pressure one atmosphere, H.P. cut-off ,5, horse-power 1600, steam per haore. WORK OF PISTON ENGINES 357 power per hour 17 lbs. Check these results, using cut-oflfs in other cylinders to give approximately even work distribution. Prob. 3. What change in cyhnder sizes would have to be made in the above engine to have equal work with a cut-off of J in each cylinder, keeping the high pressiu-e the same size as before? Prob. 4. What would be the horse-power of a triple-expansion engine whose low- pressure cylinder was 36x3 ins., when running on 150 lbs. per square inch absolute initial pressure and 10 lbs. per square inch absolute back pressure, with a cut-off in each cylinder of .4 and equal work distribution? Make necessary assumptions. Prob. 6. A triple engine 18x24x36x30 ins., running at 100 R. P.M. on an initial pressure of 200 lbs. per square inch absolute and back pressure of 20 lbs. per square inch absolute, is to be run at such cut-offs as will give complete expansion in all cylinders. What will these be, what receiver pressures will result, what horse- power can be produced under these conditions, and how much steam will be needed per hour? Note: 8 for 200 lbs. -.437. 22. Multiple-Expansion Engine. General Case. Any Relation between Cylinder and Receiver. Determination of Pressure Volume-Diagram and Work, by Graphic Methods. It is possible to arrange multiple-expansion engines in an almost infinite variety of ways with respect to the pressure-volume changes of the fluid that take place in their cylinders and receivers. There may be two or three cylinder compounds of equal or unequal strokes, pistons moving together by connection to one piston rod, or separate piston rods with a common cross-head or even with completely independent main parts and cranks at 0**, 180**, displaced with either one leading, or the pistons may not move together, being connected to separate cranks at any angle apart, and any order of lead. Moreover, there may be receivers of large or small size, and there may be as a consequence almost any relation between H.P. discharge to receiver and low-pressiu-e receipt from it, any amount of fluid passing to correspond to engine load demands and consequently any relation of cut-offs, compressions, and receiver-pressure fluctuations. Triple and quadruple engines oflfer even greater varieties of combination of related factors, so that problems of practical value cannot be solved by anal3rtical methods with anything like the same ease as is possible by graphic means, and in some cases not at all. These problems that demand solution are of two classes. 1. To find the work distribution and total work for cylinders of given dimensions, clearances, receiver voliunes and mechanical connection or move- ment relation, with given initial and back pressures, and given valve gear at any setting of that valve gear or at a variety of settings. 2. To find the cylinder relations to give any proportion of the total work in any cylinder at any given valve setting or any fraction of initial pressure or any value of release pressure or total number of expansions. The essential differences between these two classes of problems is that in the first the cylinder dimensions are given, while in the second they are to be found. 358 ENGINEERING THERMODYNAMICS In general,* however, the same methods will do for both with merely a change in the order, and in what follows the dimensions of cylinders, valve periods, receiver volimie, initial and back pressures will be assumed and the diagrams found. By working to scale these diagrams will give the work by evaluation of their area, by means either of cross-section paper directly, on which strips can be measured and added, or by the planimeter. Thus will high- and low-pressure work be evaluated through the foot-pound equiva- lent per square inch of diagram, and the total work or the equivalent mean eflfective pressure found by the methods of mean ordinates referred to the pressure scale of ordinates. In the finding of the pressure-volume diagram point by point there is but one common principle to be applied, and that is that for a given mass the product of pressure and volume is to be taken as constant (for nearly all steam prob- lems, which is the almost sole appUcation of this work) and when two masses come together at originally different pressures and mix, the prodiict of the remlt- ing pressure and the new volume, is equal to the sum of the PV products of the two parts before mixture. At the beginning of operations in the high-pressure cylinder, a known volmne of steam is admitted at a given pressure and its pressure and volume are easily traced up to the/ time when it communicates with 'the receiver in which the pressure is unknown, and there difficulty is encoimtered, but this can be met by working from the other end of the series of processes. The low-pressure cylinder, having a known compression volume at the back pressure, there will be in it at the time of opening to receiver a known voliune, its clearance, at a known compression pressure. The resulting receiver pressure will then be that for the mixture. These two receiver pressures are not equal ordinarily, but are related by various compres- sions and expansions, involving high- and low-pressure cylinder partial displacements, grouped with receiver volumes in various ways. Take for an illustrative example the case of a two-cylinder, single-acting, cross-compoimd engine with slide valves, cylinder diameters 12f and 20 ins. with 24 ins. stroke for both. High-pressure clearance is 10 per cent, low- pressure clearance 8 per cent. Receiver volume 4000 cu.ins. High-pressure crank following by 90**. Find the mean effective pressure for the high- and low- pressure cylinders, for a cut-off of 50 per cent in the high, and 60 per cent in the low, a compression of 10 per cent in the high and 20 per cent in the low, itiitial pressure 105 lbs. per square inch gage, back pressure 5 lbs. per square inch absolute, expansion according to logarithmic law. On a horizontal line SZ, Fig. 103, lay off the distances r[7= low-pressure cylinder displacement volume in cubic inches to scale. [77= low-pressure cylinder clearance volume in cubic inches to scale, ypl/'ss receiver voliune in cubic inches to scale. TFX= high-pressure cylinder clearance volume in cubic inches to scale. XF= high-pressure cylinder displacement volume in cubic inches to scale. WORK OF PISTON ENGINES 359 Through these points draw verticals produced above and below, f'T"^ V'U", Y'Y", WW", X'X'', Tr\ Then will WW and WZ be PV coordinates for the high pressure diagram in the quadrant WWZ, and V'V Fig. 103. — Graphical Solution of Compound Engine with Finite Receiver and with Clearance Illustrating General Method of Procedure for any Multiple Expansion Engine. and VS the PV coordinates for the low pressure diagram in the reversed quadrant V'VS. Lay o£F AB to represent the high-pressure admission at a height XA rep- 360 ENGINEERING THERMODYNAMICS resenting absolute initial pressure; lay off LM at a height TL representing low-pressure exhaust at a constant absolute back pressure to the same scale. Locate point B at the cut-off point AB^,50XY on the initial pressure line, and drop a vertical BB^ and draw similar verticals JJ^, GG^, MAP, at suitable fractional displacements to represent L.P. cut-off, H.P. and L.P. compression volumes resprectively. This operation will fix two other points besides the points A and L, B the H.P. cut-off at the initial pressure and M the low-pressure compression at the back pressure. Through the former draw an expansion line BC and through the latter a compression line MN, locating two more points, C and N, at the end of the outstroke of the high and instroke of the low. At point C the H.P. cylinder steam releases to the receiver of imknown pressure, and at N, the L.P. cylinder steam is opened to both the receiver and high-pressure cylinder at unknown pressure and volume. To properly locate these pressures and volumes from the previously known pressures and volumes in a simple manner, the construction below the line SZ is used. Lay off on WW" the high-pressure crank angles 0-360°, and to the right of each lay off from the clearance line XX'* the displacement of the piston at the various crank angles for the proper rod to crank ratio, locating ^the curve A^B^C^E^F^Om^. This is facilitated by Table XIII at the end of the chapter, but may be laid out graphically by drawing the crank circle and sweeping arcs with the connecting rod as radius. Opposite H.P. crank angle 270° locate L.P. crank angle 0°=360° and draw to left of the low-pressure clearance line UU" the crank angle dis- placement cmrve for that piston. It will be noted that steam volumes are given in the lower diagram by the distances from either of these curves toward the other as far as circum- stances call for open valves. Thus H.P. cylinder volumes are distances from the H.P. displacement curve to WW", but when H.P. cylinder is in communi- cation with receiver, the volume of fluid is the distance from H.P. displacement curve to VV", and when H.P. cylmder, receiver and L.P. cylinder are all three in communication the volimie is given by the distance from H.P. displacement curve to L.P. displacement curve. This pair of displacement curves located one with respect to the other as called for by the crank angle relations, which may be made to correspond to any other angular relation, by sliding the low up or down with respect to high-pressure curve, serve as an easy means of finding and indicating the volumes of fluid occupying any of the spaces that it may fill at any point of either stroke. On each curve locate the points corresponding to valve periods by the intersection of the curve with verticals to the upper diagram, such as BB^. These points being located, the whole operation can be easily traced. At H.P. cut-off (fi) the volume of steam is B^B^. During H.P. expansion (B to C) steam in the high increases in volume from B^B^ to C^C^. During H.P. release (C to D) the volume of steam in the high C'C'^ is WORK OF PISTON ENGINES 361 idded to the receiver volume C^C, making the total volume C^C^. During H.P. exhaust (D to E) the steam volume C^C^ in H.P. and rec. is com- pressed to volume PE^. At L.P. admission (AT) in low and (E) in high, the volume PP is added, making the total volume PE^ in high, low, and receiver. During (/ to Q) in low and (F to G) in high the volume PE^ in high, low and receiver, changes volume until it becomes Q^G^ in high, low, and receiver. At H.P. compression, G in the high, the steam divides to Q^G^ in low and receiver, while G^G^ remains in high and is compressed to H^H^, at the beginning of admission in the high. The former volume Q^G^, in low and receiver, expands to PJ^, at the moment cut-off occurs in the low, which divides the volume into, PJ^ in receiver, which remains at constant volume bill high-pressure release, and the second part, PP in the low, which expands In that cylinder to K^K^. After low-pressure release the volume in low decreases from K^K^ to M^M^f when the exhaust valve closes and low-pressure compression begins. During compression in low, the volume decreases from M^M^ to PP which is the volimie first spoken of above, which combines with PE^j causing the drop in the high-pressure diagram from E to F. The effects upon pressures, of the various mixings at constant volume between high, low, and receiver steam and the intermediate common expan- sions and compressions may be set down as follows: At C, steam in high, at pressure Pe, mixes with steam in receiver at pressure Py, resulting in high and receiver volume at pressure Pd. From D to E there is compression in high and receiver resulting in pressure P«. At E steam in high and receiver at pressure P« mixes with steam in low, at P», locating points / in low and F in high at same pressure. From (P to G in high) and (/ to Q in low) there is a common compression- mansion in high, low, and receiver, the pressures varying inversely as the total rolume measured between the two displacement curves. At G in there begins compression in high alone to H. In the low and receiver from Q to J there is an expansion and consequent fall in pressure from Pq to Pj. After low-pressure cut-off at J the expansion takes place in low-pressure cylinder alone, to pressure P*, when release allows pressure to fall (or rise) to exhaust pressure Pi. When cut-off in low occurs at J the volume PP is separated off in the receiver, where it remains at constant pressure P^ imtil high-pressure release it point C. At the point M compression in low begins, increasing the pressure in low ilone from Pm to Pn- There are, it appears, plenty of relations between the various inter- nediate and common points, but not enough to fix them unless one be first established. One way of securing a starting point is to assume a compression 362 ENGINEERING THERMODYNAMICS pressure Pg for the begiiming of H.P. compression and draw a compre^ion line HG through it, produced to some pressure line a/, cutting low-presure compression line at d. Then the H.P, intercept {e-f) must be equal to the low-pressure intercept (d-c); this fixes (c) through which a Py = const, line intersects the L.P. cut-off voliune at J. Now knowing by this approximation the pressure at J, the pressure may by foimd at D, E, f , and at G, The pressure now found at G may differ considerably from that assumed for the point. If so, a new assiuuption for the pressure at G may be made, based upon the last figure obtained, and working aroimd the circuit of pressures, /, D, B, F, and back to G should give a result fairly consistent with the assumption. If necessary, a third approximation may be made. It might be noted that this is much the process thai goes on in the receiver when the engine is being started^ the receiver pressure rising upon each relea^i from the high, closer and closer to the limiting pressure that is completely reachai only after running some time. These approximations may be avoided by the following computation, representing point pressures by P with subscript and volumes by reference to the lower diagram. Pj is the unknown pressure in receiver before high- pressure release and after low-pressure cut-oflf. Pressure after mixing at D is then Pj(C^(^) + Pc(C^C^) _-p The pressure at F, after mixing is (PE^) This pressure multiplied by 7Q2pk gives P^, and this in turn multiplir by j2j^ will give Pj. . Writing this in full. Solving for Pj, {Q'G^) {.PJ*) - {C*C^) (Q2g;3) WORK OF PISTON ENGINES 363 which is in terms of quantities all of which are measurable from the diagram. While this formula applies to this particular case only, the manner of obtain- ing it is indicative of the process to be followed for other cases. When there are three successive cylinders the same constructions can be jsed, the intermediate diagrams taking the position of the low for the com- [X)und case, while the low for the triple may be placed \mder the high and oflf-set irom the intermediate by the volume of the second receiver. In this case It is well to repeat the intermediate diagram. Exactly similar constructions ipply to quadruple expansion with any crank angle relations. Prob. 1. By means of graphical construction find the horse-power of a 12- and 18 X24- n. single-acting cross-compound engine with 6 per cent clearance in each cylinder, f the receiver volume is 5 cu.ft., initial pressure 150 lbs. per square inch gage, back 3ressure 10 lbs. per square inch absolute, speed 125 R.P.M., high-pressure compression }0 per cent, low pressure 20 per cent, high pressure cut-off 50 per cent, low pressure 10 per cent, high-pressure crank ahead 70°, logarithmic expansion, and ratio of red *o crank 4. - Prob. 2. Consider the above engine to be a tandem rather than a cross-compound ind draw the new diagrams for solution. Prob. 3. A double-acting, 15- and 22 x24-in. compound lengine has the high- jressure crank ahead by 60°, and has 5 per cent clearance in the low-pressure cylinder, 10 per cent in the high, and a receiver 4 times as large as the high-pressure cylinder. tVhat will be the horse-power when the speed is 125 R.P.M., initial pressure 150 lbs. per iquare inch absolute, back pressure 5 lbs. per square inch absolute, high-pressure cut- )ff ij low-pressure i, high-pressure compression 20 per cent, low-pressure 30 per cent, ind ratio of rod to crank 5. Determine graphically the horse-power in each cylinder. Prob. 4. Consider the engine of Prob. 3 to be a tandem compound and repeat he solution. 23. Mean Effective Pressure, Engine Power, and Work Distribution and heir Variation, with Valve Movement and Initial Pressure* Diagram Dis- x>rtion and Diagram Factors. Mechanical Efficiency. The indicated power ieveloped by a steam engine is dependent upon three principal factors — piston Usplacement, speed, and mean effective pressure. The first, piston displacement, s dimensional in character, and, fixed for a given engine. Speed is limited by Jteam and inertia stresses, with which the present treatment is not concerned, )r by losses due to fluid friction in steam passages, a subject that will be urther considered under steam flow. Mean effective pressure is a third factor irhich is to be investigated, most conveniently by the methods laid down in ;he foregoing sections. In these formulas for mean effective pressure, it will be observed that :he terms entering are (a) initial pressure, (6) back pressure, (c) cut-off or 'atio of expansion, (d) clearance, and (e) compression, for the single-cylinder mgine. It is desirable to learn in what way the mean effective pressure iraries upon changing any one of these factors. 364 ENGINEEEING THERMODYNAMICS Referring to Section 5, Eq. (262) for logarithmic expansion (ra.e.p.) = (iu-pr.)^+(^+c)log. ™q^ (mean forward pressure) — (bk.pr.) 1 ~X+{X+c) log, (mean back pressure) . (582) it is seen that the mean effective pressure la the difference between a mean forward pressure and a mean back pressure. The former depends un initial pressure, cut-oflf, and clearance, and the latter on back pressure. Fio. 104. — Curve to Show Variation of Mean Forward and Mean Back Pressure for Logarithmic Expanmon and Coinpreasion in a Single Cylinder Engine with CleMance. compression, and clearance. To study the effect of varying these terms it is most convenient to draw curves such as are shown in Fig. 104, and examine mean forward pressure and mean backward pressure separately. Mean forward pressure is seen by inspection to vary in direct proportion to initial pressure. Cut-off, when short, gives a low mean forward pressure, but it is to be noted that zero cut-off will not give zero mean effective pressure so long as there is clearance, due to expansion of steam in the clearance space. Increasing the length of cut-off, or period of admi^on. increases mean forward pressure, but not in direct proportion, the (m.f.p. approaching initial pressure as a limit as complete admission is approached. WORK OF PISTON ENGINES 365 ];iearance has the tendency as it mcreases, to increase the mean forward )ressure, though not to a great extent, as indicated by the curve Fig. 104. Mean back pressure is usually small as compared to initial pressure, though i great loss of power may be caused by an increase of back pressure or com- )ression. Back pressure enters as a direct factor, hence the straight line through he origin in the figure. So long as compression is zero, back pressure and nean back pressure are equal. When compression is not zero, there must )e some clearance, and the ratio of (mean bk.pr.) to (bk.pr.) depends on both clearance and compression, being greater for greater compressions and for smaller clearances. 1 — — 1(V\ M .E P. M .E P. M.E •P. lUU - 70 GO 50 40 90 ao 10 ^ ^ i^ ^ ^ y ^ / /^ • .^ — ^ /* y r - i— — "*" y ^ 1 1 J r / y X 1 / t ^ y ■■ t ■ 1 / 1 i ~1 1 — 1 'y ia Wl Presi i. r K ^ ^ A 3u to ff ±j ix jj l_l*|jl IX IX iis i«2L2£ lS. )1 02 40 6 8 K K) ■J* So ) iP •> .3 A 1 . 6 7 .8 .9 "T ( 1 .0 L5 .10 TI T *T^ TT^ 5 100 90 80 70 GO SO 40 30 20 10 3il .£ .P. M .E P. INITIAL PRE '68URE 100 LBS^ABO — ^ -- CLEARANCE .10 CUT Orr .25 COMPRESSION .80 ^ > - 1 "1 *^ ^ - -* r* „^ - 1 1 *^ 1 — ' •v. ,^ 1 "^ . r ■ 1 1 ^ li ^ LL Li ^ _Back.?r $8t » j_ 1 , ^m r^mlbi^ k D < * i 1 I 5 2 2 h I to ( D . 1 .2 .3 .4 • < 5 . 6 .7 .8 .9 1 .0 Fig. 105. — Curves to Show Variation of Mean EflFective Pressure for Logarithmic Expansion . and Compression in a Single Cylinder Engine with Clearance. The mean efifective pressures obtained by subtracting mean back from mean forward pressures in Fig. 104 are shown in curve form in Fig. 105. The inuUiple expansion engine can not be so simply regarded. In a general way each cylinder may be said to be a simple engine, and subject to variations of mean effective pressure due to change in its own initial pressure and back pressure, clearance, cut-off and compression, which is true. At the same time these factors are interrelated in a way that does not exist in the simple engine. Consider, for instance, the high-pressure cylinder of a compound engine with infinite receiver, with clearance. An increase of high-pressure compression tends first to raise the mean back pressure according to the reasoning on simple engine, but at the same time the change has decreased the volume of steam passing to receiver. No change having been made 366 ENGINEERING THERMODYNAMICS in the low-pressure cylinder, the volume admitted to it will remain the same as before, and the receiver pressure will fall, decreasing mean back pressure by a greater amount than compression increased it, and mean forward being the same as before, the increase of high-pressure compression has increased the mean effective pressure of the high-pressure cylinder. The only eflFect upon the low-pressure cylinder is that resulting from lowering its initial pressure, i.e., the receiver pressure. This results in a decrease of low^ressure mean effective pressure. Computation will show that the assmned increase of high-pressure compression decreases low-pressure work more than it increases high-pressure work, or in other words, decreased mean effective pressure referred to the low. It is impracticable to describe all results of changing each of the variables for the multiple-expansion engine. Initial pressure and cut-off in the respective stages have, however, a marked influence upon receiver pressures and work distribution which should be noted. Power regulation is nearly always accomplished by varying initial pressure, i.e., throttling, or by changing cut-off in one or more cylinders. The effect of decreasing initial pressure is to decrease the pressures on the entire expansion line and in all no clearance cycles, to decrease absolute receiver pressures in direct proportion with the initial pressure. Since back pressure remains constant, the result is, for these no-clearance cycles, that the mean effective pressures of all but the low-pressure cylinder are decreased in direct proportion to the initial pressure, while that of the low-pressure is decreased in a greater proportion. The same is true only approximately with cycles having clearance and compression. The conditions giving equal work division have been treated in connection with the individual cycles, it may here be noted in a more general way that if higb-pressure cut-off is shortened, the supply capacity of that cylinder is decreased, while that of the next cylinder remains unchanged. The result is that the decreased supply volume of steam will be allowed to expand to a lower pressure before it can fill the demand of the next cylinder than it did previously, i.e., the receiver pressure is lowered. Similarly, shortening cut-off in the second cylinder will tend to increase receiver pressure. To maintain constant work division, there must be a certain relation between cut-offs of the successive cylinders, which relation can only be determined after all conditions are known, but then can be definitely computed and plotted for reference in operation. So far, in discussing the steam engine, cycles only have been treated. These cycles are of such a nature that they can be only approached in practice, but since all conclusions have been arrived at through reasoning based on assumed laws or hjrpotheses, the term hypothetical may be applied to all these cycles. It is desirable to compare the actual pressure-volume diagram, taken from the indicator card of a steam engine, and the hyix)thetical diagram most nearly corresponding with the conditions. In Fig. 106 is shown in full lines a pressure- volume diagram which has WORK OF PISTON ENGINES 367 >en produced from an actual indicator card taken from a simple non-con- msing, four-valve engine having 5 per cent clearance. Finding the highest pressure on the admission line A'B' and the lowest ressure on the exhaust line £)'£', these pressures are regarded as (initial ressure) and (back pressure) and a hypothetical diagram constructed cor- ssponding to Cycle III, with cut-off and compression at the same fraction of broke as in the actual engine. The first difference between the hypothetical and actual PV diagrams 5 that the point of release C is not at the end of stroke, as was assumed for the lypothetical release, C, a difference which is intentional, since it requires time or pressure to fall after release to the exhaust pressure. This same fact may lause the comer of the diagram to be roimded instead of sharp as at D. Similarly, the point of admission f is before the end of the return stroke has Fig. 106. — Diagram to Illustrate Diagram Factors. been reached, and for a similar reason the comer A' may be roimded, though if release and admission are made sufficiently early the comers D' and A' will be sharp, approaching the hypothetical, H and A. ' These differences, however, have little effect upon the area of the actual diagram, which is seen to be much smaller than the hypothetical. This deficiency of area is the net result of a large number of influences, only a few of which can be fully eicplained in connection with the pressure volume discussion. Beginning with the pomt of admission, F', the line F'A'B' represents the period of admission. The rounding at A' has been explained; the inclina- tion of the line from A^ toward B' is due in part to the frictional loss of pressure as the fluid passes at high velocity through ports and passages from steam chest to cylinder. As the stroke progresses, the linear velocity of the piston increases toward' mid-stroke, requiring higher velocities in steam passages. The greater consequent friction causes pressure to fall in the cylinder. The resistance of pipes and valves leading to the engine have 368 ENGINEERING THERMODYNAMICS an effect on the slope of this line. As cut-off is approached, this pressure fall becomes more rapid, due to the partial closure of the admission valve. From B', the point of cut-off, to C\ the point of release, is the period of expansion, during which the pressures are much lower than during the hypothetical expansion line BC, due principally to the lower pressure at the point of cut-off B' than at B. Hence, the frictional fall in pressure during admission has a marked effect upon the work done^during expansion. The curve BV rarely follows the law PF= const, exactly, though it commonly gives approximately the same work area. During the first part of expansion, the actual pressure commonly falls below that indicated by this curve, but rises to or above it before the expansion is complete. This is largely due to condensation of steam on the cylinder walls at high pressures, and its reevaporation at lower pressure, to be studied in connection with a thermal analysis of the cycle. The curve of expansion may also depart from this very considerably, due to leakage, either inwardly, through the admissioii valve, or by piston from a region of higher pressure, or outwardly, through exhaust valve, or by piston into a region of lower pressure, or by drain, indicator, or relief valves, or by stuffing-boxes. From the opening of the exhaust valve at the point of release, C, till its closure at compression S', is the period of exhaust. Pressures during this period, as during admission, are affected by frictional losses in the passages for steam, in this case the pressure in the cylinder being greater than that in*exhaust pipe due to friction, by an increasing amount, as the velocity ol the piston increases toward midnstroke. Thus the line DE' rises above tiie line DE until the partial closure of the exhaust near the point of compres- sion causes it to rise more rapidly. i - At the point of compression E' the exhaust valve is completely closed and the period of compression continues up to admission at F', Leakage, condensation, and reevaporation affect this line in much the same way a< they do the expansion, and often to a more marked degree, due to the fact that the volume in cylinder is smaller during compression than during expansion, and a given weight condensed, reevaporated, or added or removed by leakage will cause a greater change in pressure in the small weight present than if the change in weight had occurred to a large body of steam. In the compound engine all these effects are present in each cylinder in greater or less degree. In addition, there are losses of pressure or of volume in the receivers themselves between cylinders, due to friction or conden- sation, and where especially provided for, reevaporation by means of reheating receivers. The effect of these changes in receivers is to cause a loss of work between cylinders, and to makQ the discharge volume of one cylinder greater or less than the supply volmne of the next, while theee were assumed to be equal in the hypothetical cases. The effect of all of these differences between the actual and hypothetical diagrams is to make the actual indicated work of the cylinder something less than that represented by the hypothetical diagram. Since these effects are WORK OF PISTON ENGINES 369 not subject to numerical calculations from data ordinarily obtainable, they are commonly represented by a single coefficient or diagram factor which is a ratio, derived from experiment, between the actual work and that indicated by h3rpothesis. It is at once evident that there may be more than one hypothetical diagram to which a certain engine performance may be referred as a standard of comparison. When the heat analysis of the steam engine is taken up, a standard for comparison will be foimd there which is of great use. For determination of probable mean eflfective pressure, however, no method of calculation has been devised which gives better results than the computa- tion of the hypothetical mean eflfective pressure from one of the standard h3rpothetical diagrams, and multiplying this by a diagram factor obtained by experiment from a similar engine, under as nearly the same conditions as can be obtained. Such diagram factors are frequently tabulated in reference books on the steam engine, giving values for the factor for various types and sizes, under various conditions of running. Unfortimately, however, the exact standard to which these are referred is not stated. In this text it will be assumed, unless otherwise stated, that the diagram factor for an actual engine is the ratio of the mean effective pressure of the actual engine to that computed for Cycle I, without clearance or compression, logarithmic law, with cut-off at the same fraction of stroke as usual, initial pressure equal to maximum during admission in actual, and back pressure equal to minimiun during exhaust of the actual engine. This is selected as the most convenient standard of comparison for mean effective pressures, as it is frequently impossible to ascertain the clearance in cases where data are supplied. When it is possible to do so, however, closer approximation may be made to the probable performance by comparing the actual with that hjrpothetical diagram most nearly approaching the cycle, using same clearance, cut-off, and compression as are found in the actual. Commercial cut-off is a term frequently used to refer to the ratio of the volume AK to the displacement, Fig. 106, in which the point K is found on the initial pressure line AB, by extending upward from the true point of cut-oflf B' a curve PF = const. While the diagram factor represents the ratio of indicated horse-power to hypothetical, the output of power at the shaft or pulley of engine is less than that indicated in the cylinders, by that amount necessary to overcome mechanical friction among engine parts. If this power output at shaft or pulley of engine is termed brake horse-power (B.H.P) then the ratio of this to indicated horse-power is called the mechanical efficiency, Emy of the engine The difference between indicated and shaft horse-power is the power consumed by friction (F.H.P.). Friction under running conditions consists 370 ENGINEERING THERMODYNAMICS of two parts, one proportional to load, and the other constant and inde- pendent of load, or (F.H.P.) =JV[(const.) x(m.e.p.)+(const.)2], where N is speed, revolutions per minute. But NX (const.) (m.e.p.) = {IMJP.)Ki and (F.H.P.) = (I.H.P.)iri+iV(const.)2, (584) where K'l and (const.)2 are constants to be determined for the engine, whose values will change as the conditions of the engine bearing-surfaces or lubri- cation alters. This value for (F.H.P.) may be used to evaluate Em, ^ _ (I.H.P .)-( RH.PO _ , ^ iV(const.)2 ^~" (I.H.P.) "^ ^^ (LH.P.) ' '• (585) but indicated horse-power divided by speed is proportional to mean effective pressure, so that K2 En, = l-Ki- (m.e.p.) (586) 0;i 4 6 8 10 12U16l8 90SSiMnS038348888AO Mean Effective Pressure Fio. 107. — ^Diagram to Show Relation of Mechanical Efficiency and Mean Effective Pressure. From this expression, speed has been eliminated, which agrees with general observation, that mechanical efficiency does not vary materially with speed. Values of the constants jRli and K2 may be ascertained if (m.e.p.) and Em are known for two reliable tests covering a sufficient range, by inserting their values forming two simultaneous equations. The numerical values of Ki foimd in common practice are between .02 and .05, and for K2 between 1.3 and 2, in some cases passing out of this range. In Fig. 107 is shown the form of mechanical efficiency curve when plotted on (m.e.p.) as abscissas, using Ki = .04, iC2 = 1.6. It may be noted that at WORK OF PISTON ENGINES 371 higher (m.e.p.) the curve does not approach unity, but the value (l—Ki) as a limit. The mechanical efficiency becomes zero for this case, at a mean efifective pressure of about 1.67 lbs. per square inch, which is that just sufficient to keep the engine running under no load. For a given speed and size of cylinders, the abscissas may be converted into a scale of indicated horse-power. Prob. 1. Assuming a back pressure of 10 lbs. per square inch absolute, a clearance of S per cent, a cut-off of 40 per cent, and compression of 20 per cent, show how (m.e.p.) varies with initial pressure over a range of 200 lbs., starting at 25 lbs. Prob. 2. • For an initial pressure of 150 lbs. per square inch absolute, show how (m.e.p.) varies with back pressure over a range of 30 lbs., starting at J lb. per square inch absolute, keeping other quantities as in Prob. 1. Prob. 3. For values of initial pressure, back pressure, etc., as given in Probs. 1 and 2, show how (m.e.p.) varies with clearance from 1 per cent to 15 per cent. Prob. 4, For values of initial pressure, etc., as given in Probs. 1 and 2, show how (m.e.p.) will vary with cut-off from to 1. Prob. 6. For values of initial pressure, etc., as given in Probs. 1 and 2, show how (m.e.p.) will vary with compression for values from to 50 per cent. Prob. 6. A certain engine developing 675 I.H.P. at a speed of 151 R.P.M., deUvered at the shaft 606 H.P. measured by an absorption dynamometer. A second lost at 100 R.P.M. gave 150 I.H.P., and 114 shaft H.P. If this engine is to deliver 500 H.P. at the shaft at a speed of 150 R.P.M., what will be the I.H.P.and the mechan- ical efficiency? Prob. 7. A compound Corliss engine, 25 and 52 ins. diameters, 60 ins. stroke, double-acting, was dosigne^^fpr 650 I. HP. at 63 R.P.M. It was found that at this sjK^cd and I.H.P. the mechanical efficiency was 91 per cent. When running with no load, the cylinders indicated 38.1 I.H.P. at 65 R.P.M. Find the probable mechanical efficiency when developing 300 I.H.P. at a speed of 64 R.P.M. 24. Consumption of the Steam Engine and its Variation wifh Valve Move- ment and Initial Pressure. Best Cut-off as Affected by Condensation and Leakage. The weight of steam used by a steam engine per hour divided by the indicated horse-power is said to be the water rate or steam consumption of that engine. It is almost needless to say that this is not a constant for a given engine, since it will change with any change of initial pressure, back pressure, or valve setting, leakage, or temperature conditions. Since there are at least two other uses of terms water rate or consumption, this may be termed the actual water rate, or actual consumptiorij the latter being a more general term which may refer to the weight of fluid used per hour per indicated horse-power, whatever the fluid may be, steam, air, carbon dioxide, or any other expansive fluid. The present discussion has special reference to steam*. From the hypothetical diagram, by computations such as are described for the various foregoing cycles, may be obtained a quantity representing the weight of fluid required to develop one horse-power for one hour, by the 372 ENGINEERING THERMODYNAMICS performance of the hypothetical cycle. This may be termed the hypothetical consumption or for steam cycles the hypothetical water rate. By the use of the actual indicator card, may be obtained, by methods still to be described, the weight of fluid accoimted for by volumes and pressures known to exist in the cylinder, this being called the indicated consumption of the engine or indicated water rate if the fluid be steam. The heat analysis of the steam-engine cycle will lead to another standard of comparison which is of the greatest importance as a basis of determining how nearly the actual performance approaches the best that could be obtained if the engine were to use all available energy possessed by the steam. At present the object is to compare the actual and indicated performance with that hypothetically possible with cylinders of the known size. Accordingly attention will be confined first to hypothetical consumption, and the quantities upon which it is dependent. For Cycle III, which is the most general for the single-expansion engine, logarithmic law, the expression for consumption in pounds fluid per hour per indicated horse-power, found in Section 5, Elq. (267), is as follows: Hypothetical consumption, lbs. per hr. per I . H.P. = /3.750 r /bk.pr.\j (m.e.p.)L^ vn.pr. /J ' in which the value of mean effective pressure itself depends upon (in.pr.), (bk.pr.), c, Z, and X, The density of the fluid at initial pressure, ^i, is to be ascertained from tables of the properties of steam or of whatever fluid is used. In Fig. 108 are the results of computations on the hypothetical steam con- sumption, using mean effective pressures as plotted in Fig. 105. For each curve, conditions are assumed to be as stated on the face of the diagram, varying only one of the factors at a time. Other conditions remaining unchanged, it may be noted that consumption decreases for an increase of initial pressure, though not rapidly in the higher pressure range. Cut-off has a marked effect upon consumption, the minimum occurring when cut-off is such as to give complete expansion. This occurs when 1+c (in.pr.) Z'+c (bk.pr.)' or which may be termed hypothetically best cut-off. In the case assmned in the diagram, Z' = (1 + . 1)^1 -.1 = . 065. WORK OF PISTON ENGINES 373 If clearance be varied, maintaining constant compression and cut-off, large clearance will give high consumption due to an excessive quantity of fluid required to fill the clearance space. Extremely small clearance leads to a high pressure at the end of compression, causing a loss of mean-eflfective pressure, and consequent high consumption. Between, the consumption has a minimum point, which is dependent for its location on both cut-off and compression. Decreasing back pressure has a beneficial effect upon mean effective pressure and consmnption. This would be still more marked in the figure if a case had been selected with a very short cut-off. Compression, throughout the ordinary range of practice, has but slight effect upon consumption, indicated by the flat middle portion of the curve in Fig. I r — r * - I 1 T— — ^ \ — - - — A hVp s tef mj Cons. in H.SJ c. •n H.SlC. U - ; lU - - » - - 85 ■ 1 •* ■ / r .,\ " • \ » - / r VJ ■ __ ao - \ AA ■ / r SO . ■ - \ 30 - ■ — 1 / r~ - 25 • s. 23 - / 25 ~ ■ > Sw \ 1 / ^ - J^ IP-- ao - •^ i 30 I V* ^ » ;^ - -= ^ r* _ _ - 1 . \ ^ ^ — 15 " 1 1 15 ^ '* i -- - 1 - 1 10 - Initlai Pi'ess. F j^ Cutoff ' 1 1 ' 1 ! 1 ,.!,, 1 Cloai-aii^el i I t)20 4 u 1) 8 100 120 ^ 1 2 .3 A .5 , 6 7 A ~^ 1 io ' - • 05 .10 .15 -^ .25 J — 1 40 H .s.c an H.SJC. ■ 1 : "1 -■ S5 - tK " - ta - INITIAL PRESSURE 100 LBS.A88 BACK PRESSURE 15 •< '< CLEARANCE .10 CUT OFF M ^— SO - 20 ~ - 25 : 15 - COMPREI »su W .39 - ^ - y ao ■ ^ ^ 10 •- - - ■^ \ 1_ . — , — — =^ - ^ -- T- L' 1 15 ~~i -5 ' \ 1 \— ' ' ' \ i 1 — - ■ 10 ^ li x: LL JA 1 LL 3ac \SIT^I 1 " \ Li . dpmp)^!. >n _ 10 15 20 25 80 .1 .2 .3 .4 .5 .6 .7 .8 .0 1.0 Fig. 108. — Curves to Show the Variation of Hypothetical Steam Consumption of Simple Engines, Logarithmic Expansion and Compression. 108. Very small or zero compression permits too much high-pressure steam to be admitted to the clearance space without doing work, and excessively large clearance causes pressures during compression to rise very high, thereby de- creasing mean effective pressure; hence this curve of consumption rises at both ends. Hypothetically, the best attainable consumption for given initial and back pressures is obtained when both expansion and compression arc complete. The indicated consumption, or, as it is frequently called for the steam engine, " steam accounted for by the indicator card " " or '* indicated water rate," is determined from the indicator card as follows. Let Fig. 109 represent an indicator diagram. The points of cut-off and compression are located from the form of 374 ENGINEERING THERMODYNAMICS the line, at the highest point on the expansion line and the lowest point on the compression line respectively. The fraction of the card lengths completed at cut-oflf, AB and the fraction of card length from point of compression to end of stroke, X= AC AD' are determined, the pressure at cut-off and compre&sion measured by the proper vertical scale, and the corresponding densities, 81, and I2 respectively, are ascertained from steam tables for dry saturated steam. Clearance, CI is known or ascertained by the form of the compression curve (Chap. I, Section 12). P Cutoff {^Atm* D V Fia. 109. — Diagram to Illustrate Method of Determining Indicated Water Rate of fcitoam Engine. At the point of cut-off, the weight of dry saturated steam present in thi> cylinder is D(Z+c)8i, and at compression the weight prcvSent is D(X+r):.>. on the assumption that the steam in the cylinder is of density 5i and h at these two instants. Accepting this assumption, the weight of steam used per cycle is Wt. steam per cycle=w = [{Z+c)h-(X+c)^2\D. (o^S The work per cycle TF = 144D(m.e.p.), and for n cycles per minute the indicated horse-power is ^•^•^•" 33;000 WORK OF PISTON ENGINES 375 The indicated consumption is then, in pounds per hour per I.H.P. wn _ 60X33,0OOXD[(Z+c)$i-(X+c)a2]n ^I.H.P. ' 144nD(m.e.p.) ' or, Ind. consrunption, lbs. per hr. per I.H.P. 13,750 (m.e.p.) [(Z+c)5i-(Z+c)B2], (589) which is the expression used to find indicated consumption for either simple- or multiple-expansion engines. In applying this to the mvltiple-expansion engine the terms Z, X and c are found for any one cylinder, and the mean eflfective pressure is referred to that cylinder. There may be, therefore, as many computa- tions as there are expansion stages. For a compound engine, for instance, indicated consmnption according to high-pressure card is found by inserting in formula Z, X and c for the high-pressure card, 8i and 82 for corresponding pressures, and for (m.e.p.) use (m.e.p. ref. to H.P.) = (m.e.p.)iy+(m.e.p.)Lpp. . . . (590) m If the computation is done by means of events on the low-pressure card, the (m.e.p.) must be referred to the low. (m.e.p. ref. to L.P.) = (m.e.p.) /r5^+(m.e.p.)L (591) In general for a multiple-expansion engine (m.e.p. ref. to cyl. A) = 2 (m.e.p.) YT (592) It is often difficult and sometimes impossible to determine the point of cut-oflf and of compression on the indicator card. The expansion and compres- sion lines are of very nearly hyperbolic in form and are usually recognizable. The highest point on the hyperbolic portion of the expansion line is regarded as cut-ofiF, and the lowest point on the hjrperbolic portion of the compression line, as the point of compression. It must be understood that by reason of the condensation and re-evaporation of steam in cylinders the weight of steam proper is not constant throughout the stroke, so that calculations like the above will give different values for every different pair of points chosen. The most correct results are obtained when steam is just dry and these points are at release and compression most nearly. 376 ENGINEERING THERMODYNAMICS When under test of actual engines the steam used is condensed and weighed and the indicated horse-power determined, then the actual steam consumption or water rate can be found by dividing the weight of water used per hour in the form of steam by the indicated horse-power. This actual water rate is always greater than the water rate computed from the equation for indicate consumption. The reasons for the difference have been traced to (c) leakage in the engine, whereby steam weighed has not performed its share of work, to (6) initial condensation, whereby steam supplied became water before it could do any work, (c) variations in the water content of the steam by evaporation or condensation during the cycle, whereby the expansion and compression laws vary in unpredictable ways, affecting the work. Estimation of probable water rate or steam consimiption of engines cannot, therefore, be made with precision except for engines similar to those which have been tested, in all the essential factors, including, of course, their condition, and for which the deficiencies between actual and indicated consumptions have been determined. This difference is termed the missing vxtter, and end- less values for it have been found by experiments, but no value is of any use except when it is found as a function of the essential variable conditions that cause it. No one has as yet found these variables which fix the form for an empiric formula for missing water nor the constants which would make such a formula useful, though some earnest attempts have been made. This is no criticism of the students of the problem, but proof of its elusive nature, and the reason is probably to be found in the utter impossibility of expressing by a formula the leakage of an engine ui imknown condition, or the effect of its condition and local situation on involuntary steam condensation and evapora- tion. It is well, however, to review some of these attempts to evaluate missing water so that steam consumption of engines may be estimated. After studying the many tests, especially those of Willans, Perry announced the following for non-condensing engines, in which the expansion is but little Missing wat^i^^^^_^ Indicated steam dVN* where d is the diameter of the cylinder in inches and N the number of revolutions per minute. This indicates that the missing steam or missing water has been found to increase with the amount of expansion and decrease with diameter of cylinder and the speed. Thermal and leakage conditions are met by the use of difference values of m, for there are given rn = 5 for well-jacketed, well-drained cylinders of good construction with four poppet valves, that is, with minimum leakage and condensation. m = 30 or more for badly drained unjacketed engines with slid, valves, thai is, with high leakage and condensation possibilities. m = 15 in average cases. WORK OF PISTON ENGINES 377 For condensing engines Perry introduces another variable — ^the initial pres- ire pounds per square inch absolute, p giving _120(l+|) Missing watei^ _ ^ «, /ka/in Indicated steam dVnpi t might seem as if such rules as these were useless, but they are not, especially rhen a given engine or line of engines is being studied or two different engines ompared; in such cases actual conditions are being analyzed rather than iredictions made, and the analysis will always permit later prediction of con- iderable exactness, if the constants are fixed in a formula of the right empiric Drm, Similar study by Heck has resulted in a different formula involving iflFerent variables and constants, but all on the assumption that the dis- repancies are due to initial condensation. He proposes an expression equi- valent to Missing steam _ .27 /<S(x2— xi) {kqk\ Indicated steam "" ^JV\ Pi^ ' Q which iV = R.P.M. of the engine; d= diameter in inches; L = stroke in feet; <S=the ratio of cylinder displacement surface in square feet to dis- placement in cubic feet. ^1^2)' XL K^) The term (X2— «i) is a constant supposed to take into account the amount )f initial condensation dependable on the difference between cylinder wall md live-steam temperature and is to be taken from a table found by trial as :he difference between the x for the high pressure and x for the low pressure, yoih absolute, see Table XIV at the end of the Chapter. In discussing the hypothetical diagrams, it was found that best economy was )btained with a cut-off which gives complete expansion. For other than lypothetical diagrams this is not true, which may be explained most easily by reference to the curves of indicated, and actual consumption, and missing steam, Fig. 110. The curve ABC is the hypothetical consumption or water rate for a certain steam engine. Its point of best economy occurs at such a cut-off, B, that expan- 378 ENGINEERING THERMODYNAMICS sion is complete. The curve GHI is computed by Heck's formula for missiiif water. The curve falls off for greater cut-oflfs. Adding ordinates of these tvo curves, the curve DEF for probable consumption is found. The minimuic point in this curve, jE, corresponds to a longer cut-off than that of -4BC. Since cut-off B gave complete expansion, cut-off E must give incomplete expan- sion. In other words, due to missing steam, the condition which really gives least steam consiunption per hour per indicated horse-power corresponds to a release pressure, which is greater than the back pressure. It should be noted that the minimum point mentioned above will not 1« best cut-off, for the output of the engine is not indicated, but brake horse-power. ao 30 10 D V A \ r _C. \ i"****^ t ^ \ -f-. ^^ ^B4 1 ,—- f— -^— T'"" ! . — ■- •»». _H__ '•^, *^__ "*■*-•-. '"^^. ^-^^wmjt 1 • I \ •I I .3 Per Cent Cut off Fig. 110. — Diagram to Show Displacement of Best Cut-off Due to Effect of Missing Waier from Point B for the Hypothetical Cycle to Some Greater Value E, In Fig. ill on cut-off as abscissa are plotted {EFG) consumption pounds per hour per I.H.P., and for the case assumed, (OD) the curve of mechanical efficiency, based on cut-off, (lbs. steam per hr.) . B.H.P. I.H.P. "^I.H.P. or, in other words, Consumption, lbs. per hr. per I.H.P. (lbs. steam per hr.) B.H.P: E. -'—^ = Consumption, lbs. per hr. per B .H.P. (59tj ! m Due to the increasing value of Em for greater cut-offs, the minimum point 5 corresponds to a cut-off still longer than for the curve EFG, which itself wa> found in Fig. 110 to give a longer cut-off than that of the hypothetical curve. Hence the best cut-off for economy of steam, where the net power at the shaft is regarded as the output, will be such as to give incomplete expansion, or a release pressure above back pressure, this effect being caused by l)olli missing steam and by frictional losses. Prediction of actual consumption of steam engines as a general proposition is almost hopeless if any degree of accuracy worth while is desired, though the WORK OF PISTON ENGINES 379 jflfect on steam consumption of changing the value of any one variable can be )retty well determined by the previous discussion qualitatively, that is, in kind, ;hough not quantitatively in amount. Probably the best attempt is that )f Hrabak in German, which takes the form of a large number of tables developed from actual tests though not for engines of every class. These tables are quite extensile, being in fact published as a separate book and any abstrac- tion is of no value. There is, however, a sort of case of steam consumption prediction that can be carried out with surprising precision and that is for the series of sizes or line of engines manufactured by one establishment all of one class, each with about the same class of workmanship and degree of fit, and hence having leakage and cylinder condensation characteristics that vary consistently through- out the whole range. For such as these tables and curves of missing water A Uecl > Eff.- 1 40 \ ■^^"^ D — E- \ .y" r ^ N Ll>s.< f Ste) Lmpei Shaft ■ / A ^ ^ . B n.p ■ ■■■ i ' per Hoor^ 1 I.P. fe / J —, »er I. 1- F -.Lh k of Steam / p< r Hoi r lA / 10 c ( ■ L 1 I .8 Per Cent Cut off Fig. 111. — Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cut-off. can be made up and by the best builders are, for making guarantees of steam consumption for any service conditions that their engines are able to meet. The practice of one firm making what is probably the best line of stationary engine in this country is of suflScient interest to warrant description. The primary data are curves of indicated water rate plotted to mean effective pressure for clearances of three or four per cent, and that mean effective pressure is chosen in any one specific case that will give the horse-power desired at the fixed speed for some one set of cylinder sizes available. To this indicated water rate a quantity is added constituting the missing water which is made up of several parts as follows: The first is an addition representing condensa- tion which is plotted in curve form as a function of (a) boiler pressure, (6) superheat in the steam, (c) piston speed, (d) the class of engine simple, com- pound or triple, with jacketed or unjacketed cylinders, and for cylinder ratios from 4 to 1, to 6 to 1 in the case of compounds. It is therefore a most complex quantity, the nature of the variations in which can only be indicated here. 380 ENGINEERING THERMODYNAMICS For example, increase of piston speed decreases the condensation loss a- does multiple expansion, and also jacketing, while increase of superheat in the steam also decreases it, but superheat has less effect in triple than in com- pounds and less in compound than in simple engines. The next factor of correction is that covering leakage losses, also additive to indicated water rate and which with it and the condensation loss make up the probable steam consumption. The leakage decreases regularly with increase of piston speed, is le^s for large than for small engines, the changf. being rather fast from 50 to 200 horse-power and much slower later, being scarcely anything at all over 2000 horse-power. Example 1. What cut-off will give the lowest indicated water rate for a 9x12- in. engine, with 5 per cent clearance and no compression when running non-condensing on an initial pressure of 100 lbs. per square inch gage, and what will be the value of this water rate? What steam will be used per hour per brake horse-power hypothetically? From Eq. (587) Z' = (l+c)y^-c, in.pr. 15 = (1 +.05)-7L - .05 =8.7 per cent, 115 and (m.e.p.) =115| .087x(.087+.05) log*^ | -15=27.2 lbs. sq.in. Hence 13 750 r 15 1 Steam per hour per I.H.P. =-^— .137-.05X— X.262 = 17.2 lbs. From the curve of Fig. 107, assuming it to apply to the engine, for this value of (m.e.p.) mechanical efficiency is 90 per cent, hence from Eq. (596) the weight of steam per shaft horse-power per hour will be 19.1 poimds. Prob. 1. Draw diagram similar to Fig. 108 for following case: Initial pressure, 135 lbs. per square inch gage, back pressure 10 lbs. per square inch absolute, clearance 5 per cent, cut-off 30 per cent, compression 25 per cent. Prob. 2. From indicator diagram shown in Fig. 106 find the indicated water rate of the engine from which it was taken. Prob. 3. The indicated water rate of a 9Xl2-in. jacketed engine when running non-condensing at a speed of 250 R.P.M. with an initial pressure of 100 lbs. per square inch gage and J cut-ofiF is 50 lbs. Using Perry's formula what will be the probable actual steam used by engine per horse-power hour * Prob. 4. A 24 x48-in. engine in good condition is found to have an indicated water rate of 25 lbs. when cut-off is i, initial pressure 100 lbs. per square inch gage, back pressure 10 lbs. per square inch absolute, and speed of 125 R.P.M. What will be the miesing water, and the rate as found by Perry's formula and by Heck's? WORK OF PISTON ENGINES 381 Prob. 5. What will be the probable amount of steam used per hour by a 36 >c48- in. engine with 5 per cent clearance running at 100 R.P.M. on an initial pressure of 150 lbs. per square inch gage a back pressure of 5 lbs. per square inch absolute, i cut- off and 10 per cent compression? Prob. 6. How will the amount of steam of Prob. 5 compare with that used by a 15X22x36-in engine with 5 per cent clearance in each cylinder, running at 100 R.P.M. on same pressure range with J cut-off in high-pressure cylinder, J cut-off in low, and 10 per cent compression in each cylinder? 25. Variatioii of Steam Consumptioii with Engine Load. The Willans Line. Most Economical Load for More than One Engine and- Best Load Division. However valuable it may be to the user of steam engines to have an engine that is extremely economical at its best load which, it should be noted, may have any relation to its rated horse-power, it is more important usually that the form of the economy load curve should be ap flat as possible and always is this case when the engine must operate under' a wide range of load. This being the case it is important to examine the real performance curves of some typical engines all of which have certain characteristic similarities as well as differences. From the discussion of hypothetical and indicated water rates it appears that the curve of steam consumption (vertical) to engine load (horizontal) is always concave upward and always has a minimum point, not at the maxi- mum load. Actual consumption curves are similar in general form, but as has been pointed out, the load at which the water rate is least corresponds to some greater mean effective pressure than that for the hypothetical, so the whole curve is displaced upwards and to the right by reason of cylinder condensation and leakage losses. This displacement may be so great as to prevent the curve rising again beyond the minimum point, in which case the least steam consumption corresponds to the greatest load. Just what form the actual water rate-load curve will take depends largely on the form of valve geai and type of governing method in use, by throttling initial pressure with 3, fixed cut-off or, by varying cut-off without changing initial pressure, with or without corresponding changes in the other valve periods. Whenever the control of power is by throttling of the supply steam the curve is found to be almost exactly an hypjerbola, so that (water rate X horse- power) plotted to horse-power is a straight line which being characteristic is much used in practical work and is known as the WiUans line. All other engines, that is, those that govern on the cut-off, have Willans lines that are nearly straight, such curvature as exists being expressed by a second degree equation instead of one of the first degree. Equations for Willans lines can always be found for the working range of load, that is, from about half to full load, though not for the entire range, except in unusual cases, and these equations are of very great value in pre- dicting the best division of load between units, which is a fundamental step in deciding, how many and what sizes of engine to use in carrying a given load in industrial power plants. 382 ENGINEERING THERMODYNAMICS N V \ \l \ \ ^ "K s \ S *^ 1 % c •• lU V c 5 a ■ "Co" ^ * \ « *^ , — 4 ' \ ^ \ \ \ - - > \ ► N! ^ V ^ V \ \ \ N, \ \ ^ i . ■ — - , I ^ 1 % X ^ s « '% • : - — ^ ' *«w "\ — *>> \\ m \ 1 1 M i« S 3 ^ r te & / /■* — — 1 »-i i^ A VVm 1 — / A «— c ?? t^ », S 8 9 S WORK OF PISTON ENGINES 3S3 Before taking up the derivation of equations some actual test curves will examined and a number of these are grouped in Fig. 112 for engines of .nous sizes, simple and compound, up to 10,000 H.P., on which vertical stances represent pounds of steam per' hour, per I.H.P. and horizontal Ei.P. To show the essential similarity of the curves for engines of different ;e more cleariy, these are re-plotted in Fig. 113 to a new load scale based I best load of eachj which is taken as unity. This is evidently a function of mn effective pressure, just what sort of function does not matter here. In 75 ICD 125 Pecoentttgpe ot most economical load 200 1. 113. — ^Typical Water Rate — Lead Curves for Steam Engines Plotted to Fractional Loads. Bry case the Willans line is also plotted in Fig. 112, each line being num- red to correspond to its water rate curve. As there is a corresponding similarity of form for the water rate and illans line of steam turbines, though the reasons for it will be developed er, it must be understood that the mathematical analysis that follows applies both turbine and piston steam engines, and finally it makes no difference lat units are used for load, whether I.H.P., or B.H.P. or K.W. of a direct- inected electric generator. In Fig. 114 is shown the water-rate curve to a K.W. base for the 10,000 K.W. irtis steam turbine at the Chicago Edison, Fiske Street Station for which the lowing equation fits exactly: V 17 09 ^=''-;;'^+10.54+.156P, 384 ENGINEERING THERMODYNAMICS where y= pounds of steam per hour -5- 1000; load (in this case in K.W.)-^1000; pounds of steam per K.W. hr. P Y P A similar equation fits fairly well the curve of Fig. 115, representing th 7000 H.P. piston engines of the Interborough Railway, Fifty-ninth Stm station, as well as the combined piston engine and low-pressure steam turbia 7000 9000 1000 P IF Kilowatts Fia. 114.— Performance of a 10000-K.W. Steam Turbine. taking its exhaust steam, in the same station, but with different numeric constants, as below: Piston engine, -p = ' + .6-|- 1 .85P, Y 89.4 Combined piston engine and turbine, p = ~p — 2.90+.713P. A third case of smaller size is shown in Fig. 116, representing the jut formance of a 1000-K.W. Corliss piston engine driving a generator for whi the equation is Y 9.8 WORK OF PISTON ENGINES 385 10000 11060 12000 laooo 14000 15000 laooo Kilowatts - 1000 P Fia. 115. — Performance of a 7000-H.P. Piston Engine alone and with a Low-pressure Steam Turbine, 26 24 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 1000-K.W. Steam Turbine. 386 ENGINEERING THERMODYNAMICS These illustrations could be multiplied indefinitely, but those given will suffice to establish the fact that the two following equations are fundamental over ihe working range of any steam engine of whatever type: Water rate line, ^=^-+B+CP (597 Water per hour, Willans line, Y=A+BP+CI^, (598 in which Y is the weight of steam per hour and P the engine load whether expressed in indicated or brake horse-power, or in kilowatts. At the most economical load the water rate is a minimum, so that i(?)=<'-^(p+«+'^'') dP whence the most economical load is i^' = >J|. (599) Where the Willans line is straight, C=0, and the most economical load is the greatest load. Two engines carrying the same load must divide it and some one pro- portion may be best. To find out, consider first any number of similar engiius. that is, engines that have the same constants A, B, and C, denoting each ca^e by subscripts. Then Let P= total load; '' Pu P2, Pd) etc. = individual engine loads; *' Y = total water per hour; '* Yi, Y2y F3 = water per hour for each engine. Then r=Fi+F2+F3+. . .+r„ = nA+B{Pi+P2 + P3 + . . . Pn)+C(Pl^ + P2' + P3'+. . .+Pn') ==nA+BP+C{Pi^+P2^+P3^+. . .+Pn2). Only the last term is variable and this is a minimum when Pl=P2 = P3 = P»/ Therefore for similar engines^ ihe best division of load is an equal division. WORK OF PISTON ENGINES 387 When the engines are dissimilar it is convenient to first consider the case of straight Willans lines for which C=0. Then for two such engines Y=Ai+A2+BiPi+B2P2 = (Ai+A2) + Bi(P-P2)+B2P2 = {Ai+A2)+BiP+{B2-Bi)P2. At any ^ven load P the first two terms together will be constant, and the ivater per hour will be least when the last term is least. As neither factor jan be zero, this will occur when P2 is least. Therefore for two dissimilar engines the best division of load is thai which Tuts the greatest possible share on the one with the smaller value of B, in its equation, yrmded each has a straight WiUans line. Two dissimilar engines of whatever characteristics yield the equation, Y==Ai+A2+BiPi+B2P2+CiPi^ + C2P2^ = (Ai+A2+B,P+CiP2) + (B2-2PCi-Bi)P2 + {Cl + C2)P2^. DiflFerentiation with respect to P2, and solving for P2, the load for the second (ngine that makes the whole steam consumption least, gives, = constant+constant XP. Therefore, fJie load division must be linear and Eq. (600) gives the numerical alue, when any two engines share a given load. This sort of analysis can be carried much further by those interested, but pace forbids any extension here. It is proper to point out, however, that y means of it the proper switch-in points for each imit in a large power station an be accurately found, to give most economical operation on an increasing tation load. 26. Graphical Solution of Problems on Horse-power and Cylinder Sizes . Tie diagram for mean effective pressures in terms of initial and back ressure, clearance, compression and cut-off, Fig. 117, facilitates the solution f Eq. (262) in Section 5. The mean effective pressure is the difference etween mean forward and mean back pressure. The former is dependent XK)n clearance, cut-off and initial pressure. In the example shown on the gure by letters and dotted lines, clearance is assumed 5 per cent, shown at L. Project horizontally to the point P, on the contour line for the assumed ut-off, 12 per cent. Project downward to the logarithmic scale for " mean 388 ENGINEERING THERMODYNAMICS 0) 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.) =49.5-3.2=46.3 lbs. Fig. 118 is arranged to show what conditions must be fulfilled in order to obtain equal work with complete expansion in both cylinders in a compound engine, finite receiver, logarithmic law, no clearance, Cycle VII, when low- pressure admission and high-pressure exhaust are not simultaneous. This is discussed in Section 11, and the diagram represents graphically the conditions expressed in Eqs. (376), (377), (378), (379). To illustrate its use assume that in an engine operating on such a cycle, the volume of receiver is 1.5 times the high-pressure displacement, 1,5 = y, then 1 ~= .667. Locate the point A on the scale at bottom of Fig. 118, corresponding y to this value. Project upward to the curve marked ''ratio of cut-offs " and at the side, C, read ratio of cut-offs l^=.672. Next extending the line 45 to its intersection D, with the curve (7/f, the point D is found. From D project horizontally to the contour line representing the given ratio of initial to back pressure. In this case, initial pressure is assumed ten times back pressure. Thus the point E is located. Directly above E at the top of the sheet is read the cylinder ratio, at /^, D L fic=n-=2.4. If cylinder ratio and initial and final pressures are the fimdamental data of the problem, the ratio of cut-offs and ratio of high-pressure displacements to receiver volume may be found by reversing the order. 390 ENaiNEERING THERMODYNAMICS to to m JO ORBH WORK OF PISTON ENGINES 391 GENERAL PROBLEMS ON CHAPTER IH. Prob. 1, How much steam will be required to run a 14 xl8-in. ^double-acting engine with no clearance at a speed of 200 R.P.M. when initial pressure is 100 lbs. per square in. gage, back pressure 28 ins. of mercury (barometer reading 30 ins ), and cut-off is i? What will be horse-power imder these conditions? Note: 8 for 100 lbs. =.26, for 28 ins. =.0029. Prob. 2. Draw the indicator cards and combined diagram for a compound steam engine without receiver, and with 3 per cent clearance in low pressure and 5 per cent in high, when initial pressure is 100 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high-pressure cut-off J, high-pressure compression iV, and low-pressure compression i, Prob. 3. A simple double-acting engine, 18x24 ins., is running at 100 R.P.M. on compressed air, the gage pressure of which is 80 lbs. The exhaust is to atmosphere. If the clearance is 6 per cent and cut-off f , and compression 10 per cent, what horse- power is being developed, the expansion being adiabatic, and how long can engine be run at rated load on 1000 cu.ft. of the compressed air? Prob. 4. Will the work be equally distributed in a 12xl8x24-in. engine with infinite receiver and no clearance when cut-off is J in high pressure cyhnder, and f in low, expansion being logarithmic, initial pressure 150 lbs. per square inch absolute and back pressure atmosphere? What will be work in each cylinder? Prob. 6. The receiver of a 15X20x22 in. engine is 4 times aa large as high- pressure cylinder. What will be the horse-power, steam used per hour, and variation in receiver pressure for this engine, if clearance be considered, zero and initial pressure is 125 lbs. per square inch gage, back pressure 5 lbs. per square inch absolute, cut-offs J and I in high- and low-pressure cylinders respectively, and piston speed is 550 ft. per minute? Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. Prob. 6. At no load an engine having 7 per cent clearance, cuts off at 4 per cent of its stroke, while at full load it cuts off at 65 per cent of its stroke. At no load, compression is 40 per cent and at fuU load 5 per cent. What percentage of full-load horse-power is required to overcome friction, and what percentage of steam used at full load, is used on friction load, if initial pressure is constant at 100 lbs. per square inch gage, back pressure constant at 5 lbs. per square inch absolute, and expansion is logarithmic? Note: 8 for 100 lbs. =.262, for 5 lbs. =.014. • Prob. 7. The initial pressure on which engine is to run is 115 lbs. per square inch gage, and steam is superheated and known to give a value of s = 1.3. For an engine in which clearance may be neglected, work is to be equal, and expansion complete in both cylinders, when back pressure is 10 lbs. per square inch absolute. What must be the cut-offs and cylinder ratio to accomplish this when receiver is 3i times high pressure cylinder volume? Prob. 8. A 12-in. and 18x24 ins. double-acting engine with zero clearance and infinite receiver operates on an initial pressure of 150 lbs. per square inch gage, and 392 ENGINEERING THERMODYNAMICS a back pressure of 5 lbs. per square inch absolute. What will be the release and receiver pressures, horse-power, and steam consumption when speed is 150 R.P M., expansion logarithmic, and cut-off i in each cylinder? Note: 8 for 150 lbs. =.367, for 5 lbs. =.014. Prob. 9. If a third cylinder 24 ins. in diameter were added to engine of Prob. 8 and cut-off in this made i, how would horse-power, steam consumption, receiver and release pressures change? Prob. 10. What would have to be size of a single cylinder to give iSame horse-power at same revolutions and piston speed as that of engine of Prob. 8 under same conditions of pressure and cut-off? Prob. 11. With the higli-pressure cut-off at f, and low and intermediate cut-offs at A, what will be the horse-power, water rate and receiver pressures of a 30 X 48 X 77 X 72- in. engiae running at 102 R.P.M. on an initial pressure of 175 lbs. per square inch gage and a back pressure of 26 fins, of mercury (barometer reading 30 ins.), if the receiver be considered infinite and expansion logarithmic, clearance zero? What change in intermediate and low-pressure cut-offs would be required to give equal work distribu- tion? Note: 8 for 175 lbs. =.419, for 26 ins. =.0058. Prob. 12. If it had been intended to have all the cut-offs of the engine of Prob. 11, equal to i, what should have been the size of the intermediate and low-pre^ure cylinders to give equal work for same pressure range and same high-pressure cylinder? • Prob. 13. To attain complete expansion in all cylinders of the engine of Prob. 11, with the initial and back pressures as there given, what cut-offs would be required and what receiver pressures would result? Prob. 14. A compoimd locomotive has no receiver, the high- pressure clearance is 8 per cent, and low-pressure clearance 5 per cent. The cylinders are 22 and 33x48 ins., high-pressure cut-off |, high- and low-pressure compression each 10 per cent, initial pressure 175 lbs. per square inch gage, back pressure one atmosphere, and expansion and compression logarithmic. What will be the horse-power at a speed of 40 miles per hour, the engine having 7-ft. driving wheels? At this speed, how long will a tank capacity of 45,000 gallons last? Note: 5 for 175 lbs. =.419, for 15 lbs. =.038. Prob. 16. A superheater has been installed on engine of Prob. 14 and expansion and compression, now follow the law PV ^c, when s = 1.2. What effect will this have on the horse-power and steam consumption? Prob. 16. What will be the maximiun receiver pressure work done in each cylinder and total work for a cross-compoimd engine 36 and 66x48 ins., running at 100 R.P.M. on compressed air of 100 lbs. per square inch gage pressure, exhausting to atmosphere if the high pressure cut-off is J, clearance 6 per cent, compression 20 per cent, low- pressure cut-off is f , clearance 4 per cent, compression 15 per cent, and receiver volume is 105 cu.ft.? Prob. 17. A manufacturer gives the horse-power of a 42x64x60-in. engine as 2020, when run at 70 R.P.M. on an initial pressure of 110 lbs. per square inch gage, atmospheric back pressure, and .4 cut-off in high-pressure cylinder. How does this value compare with that found on assumption of 5 per cent clearance in high, 4 per cent in low, and complete expansion and compression is each cj^linder? Prob. 18. A mine hoisting engine is operated on compressed air with a pressure of 150 lbs. per square inch absolute and exhausts to atmosphere. The cylinder sizes are 26X48x36 ins., and clearance is 5 per cent in each. At the start the high-pressure WORK OF PISTON ENGINES 393 cut-off 18 1 and low pressure i, while normally both cut-offs are J. The exhaust from high-pressure cylinder is into a large receiver which may be considered infinite. The compression is zero at all times. Considering the exponent of expansion to be 1.4, what will be the horse-power imder the two conditions of cut-off given, for a speed of 100 R.P.M.? Prob. 19. What must be ratio of cylinders in the case of a compound engine with infinite receiver, to give equal work distribution complete expansion and com- pression if the least clearance which may be attained is 5 per cent in the high- pressure cylinder, and 3 per cent in the low-pressure. The engine is to run non- condensing on an initial pressure of 125 lbs. per square inch gage, with expansion exponent equal to 1.3? What must be the cut-offs and compressions to satisfy these conditions? Prob. 20. Assuming 7 per cent clearance in high-pressure cylinder and 5 per cent in low, infinite receiver, and no compression, how will the manufacturer's rating of 2100 H.P. check, for a 36X6x48-in. engine running at 85 R.P.M. on an initial pressure of 110 lbs. per square inch gage, and a back pressure of 26 ins. vacuum, with .3 cut-off in high pressure cylinder? Prob. 21. For a 25X40x36-in. engine, with 5 per cent clearance, i cut-off and 20 per cent compression in each cylinder, what will be horse-power for an initial pressure of 100 lbs. per square inch gage, and a back pressure of 17.5 lbs. per square inch absolute, with logarithmic expansion and compression? Prob. 22. What must be the cylinder ratio and cut-off to give complete expansion in a no-clearance, 14 and 22 x24-in. engine with no receiver and logarithmic expansion, when initial pressure is 100 lbs. per square inch gage, and back pressure 10 lbs. per square inch absolute? What will be the horse-power and steam used for these conditions at a speed of 150 R.P.M.? Note: 8 for 100 lbs. =.262, for 10 lbs. =.026. Prob. 23. A 24X20x24 in. engine with no receiver or clearance, runs on com- pressed air of 120 lbs. per square inch gage pressure, and exhausts to atmosphere. When running at a speed of 125 R.P.M., with high-pressure cut-off f, what horse- power will be developed and how many cubic feet of compressed air per minute will be required to run the engine, the expansion being adiabatic? WiU the work be equally divided between the two cylinders? Prob. 24. It is desired to run the above engine as economically as possible. What change in cut-off wiU be required, and will this cause a decrease or increase in horse- power and how much? How will the quantity of air needed be affected? Prob. 26. A mill operates a cross-compound engine with a receiver 3 times as large as high-pressure cylinder, on an initial pressure of 125 lbs. per square inch gage, and a back pressure of 10 lbs. per square inch absolute. The engine may be considered as without clearance, and the expansion as logarithmic. As normally run the cut-off in high-pressure cylinder is | and in low, \, It has been found that steam is worth 25 cents a thousand pounds. What must be charged per horse-power day (10 hours) to pay for steam if the missing water follows Heck's formula? Note 8 for 125 = .315, for 10 = .026. Prob. 26. By installing a superheater the value of 8 in Prob. 25 could be changed to 1.3. The cost of st am would then be 30 cents a thousand pounds. From the effect on value of s alone would the installation of the superheater pay? Prob. 27. When a 26X48x36-in. cross-compound engine with a receiver volume of 35 cu.ft. and zero clearance, is being operated on ste^frn of 125 lbs. per square inch 394 ENGINEERING THERMODYNAMICS gage initial pressure, and atmospheric exhaust, is the work distribution equal, when high-pressure cut-off is i and low-pressure cut-off J? For these cut-offs what i- fluctuation in receiver pressure and what steam will be used per horse-power hour? Note; 5 for 125 =.315, for 15 lbs. =.038. Prob. 28. To operate engine of Prob. 27 under most economical conditions, what values must be given to the cut-offs, and what values will result for receiver pressims, horse-power, and eteam used per hour? Prob. 29. What will be the horse-power and steam used by a 20x30x36-m. engine with infinite receiver and no clearance, if expansion be such, that 8 = li5, high-pressure cut-off i, low-pressure cut-off i, initial pressure 100 lbs. per square IlcIi gage, back pressure 3 lbs. per square inch absolute, and speed 100 R.P.M. Note: 8 for 100 lbs. =.262, for 3 lbs. =.0085. Prob. 30. The following engine with infinite receiver and no-clearance, runs od steam which expands according to the logarithmic law. Cylinders 9, and 13X1S ins., initial pressure 125 lbs. per square inch gage, back pressure 5 lbs. per square incli absolute, high-pressure cut-off f, low-pressure |, speed 150 R.P.M. What will be horse-power and steam consumption hypothetical and probable? Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. Prob. 31. By graphical means find the (m.e.p.), of a 15X22x30-in. cto^ compound engine, with 5 per cent clearance in each cylinder, if the receiver volume is 8 cu.ft., initial pressure 125 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high-pressure cut-off J, low-pressure A, high-pressure compres- sion 40 per cent, low-pressure 20 per cent, high-pressure crank following 90°, logarithmic expansion. Prob. 32. Show by a series of curves, assuming necessary data, the effect on (m.e.p.) of cut-off, back pressure, clearance, and compression. Prob. 33. Show by curves, how the indicated, and actual water rate, of an 18 X24- in. engine with 5 per cent clearance, and running at 125 R.P.M. on an initial pressure of 125 lbs. per square inch gage, and a back pressure of 10 lbs. per square inch absolute, may be expected to vary with cut-off from A to }. TABLES 395 Table XIII PISTON POSITIONS FOR ANY CRANK ANGLE ^ROM BilGINNINO OF STROKE AWAY FROM CrANK ShAFT TO FiND PiSTON POSITION FROM Dead-Center Multiply Stroke by Tabular Quantity Crank Angle. -'-4 r -^-4.6 r r -=5.5 r r -?=7 r ^=8 r r 5 .0014 .0015 .0015 .0016 .0016 .0016 .0017 .0019 10 .0057 .0059 .0061 .0062 .0063 .0065 .0067 .0076 15 .0128 .0133 .0137 .0140 .0142 .0146 .0149 .0170 23 .0228 .0237 .0243 .0248 .0253 .0260 .0266 .0302 25 .0357 .0368 .0379 .0388 .0394 .0405 .0413 .0468 30 ! .0513 .0531 .0545 .0556 .0566 .0581 .0592 .0670 35 .0698 .0721 .0740 .0754 .0767 .0787 .0801 .0904 40 .0910 .0939 .0962 .0981 .0997 .1022 .1041 .1170 45 .1152 .1187 .1215 .1237 .1256 .1286 .1308 .1468 50 .1416 .1458 .1491 .1518 .1541 .1576 .1607 .1786 55 .1713 .1769 .1828 .1827 .1863 .1892 .1922 .2132 60 .2026 .2079 .2122 .2167 .2186 .2231 .2296 .2500 65 .2374 .2431 .2477 .2514 .2546 .2594 .2630 .2886 70 .2730 .2794 .2844 .2885 .2929 .2973 .3013 .3290 75 .3123 .3187 .3239 .3282 .3317 .3372 .3414 • .3706 80 .3516 .3586 .3642 .3687 .3725 .3784 .3828 .4132 85 .3944 .4013 .4068 .4113 .4151 .4210 .4264 .4664 90 .4365 .4437 .4495 .4547 .4680 .4641 .4686 .5000 95 .4816 .4885 .4940 .4985 .5022 .6081 .5126 .6436 100 .5253 .5323 .5378 .5424 .5461 .5520 .6564 .5868 105 .5711 .5775 .5828 .6870 .5905 .5961 .6002 .6294 110 .6150 .6214 .6265 .6306 .6340 .6393 .6630 .6710 115 .6600 .6657 .6703 .6740 .6771 .6820 .6866 .7113 120 .7026 .7080 .7122 .7167 .7186 .7231 .7265 .7500 125 .7449 .7495 .7533 .7563 .7588 .7628 .7668 .7868 130 .7844 .7885 .7920 .7947 .7969 .8004 .8030 .8214 135 .8223 .8258 .8286 .8308 .8327 .8367 .8379 .8535 140 .8570 .8600 .8623 .8642 .8668 .8682 .8703 .8830 145 .8889 .8913 .8931 .8946 .8968 .8978 .8993 .9096 150 .9173 .9191 .9204 .9216 .9226 .9241 .9262 .9330 155 .9420 .9432 .9452 .9451 .9457 .9468 .9476 .9631 160 .9625 .9633 .9640 .9645 .9650 .9656 .9661 .9698 165 1 .9787 .9792 .9796 .9799 .9802 .9806 .9809 .9829 170 .9905 .9908 .9909 .9911 .9912 .9913 .9916 .9924 175 .9976 .9977 .9977 .9977 .9978 .9978 .9979 .9981 180 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 396 ENGINEERING THERMODYNAMICS Table XIV VALUES OF z FOR USE IN HECK'S FORMULA FOR MISSING WATER Absolute Absolute Absolute Steam Preaeure. X Steam Pressure. X Steam Pressure. % 170 70 297.5 165 393 1 175 75 304 170 397 2 179 80 310 180 405 3 183 85 316 185 409 4 186 90 321.5 190 413 6 191 95 327 irs 4165 8 196 100 332.5 200 420 10 200 105 338 210 427 15 210 110 343 220 431 20 220 115 348 230 441 25 229 120 353 240 447.5 30 238 125 358 250 454 35 246 130 362.5 260 460.5 40 254 135 367 270 467 45 262 140 371.5 280 473 50 269.5 145 376 290 479 55 277 150 380.5 300 485 60 284 155 385 65 291 1 160 389 • Table XV SOME ACTUAL ENGINE DIMENSIONS Simple 7X9 7iX15 16 X18 151X24 24 X36 8X9 8iXl5 16JX18 16 X24 26 X36 9X9 12 X15 17 X18 18 X24 26iX38 5iX10 13 X15 17iX18 20 X24 28 X36 6iX10 14 X15 18 X18 22 X24 14 X42 8 XIO 14iX15 19 X18 24 X24 15 X42 9 XIO 15 X15 20 X18 16iX27 16 X42 10 XIO 16 X15 29 X19 17iX27 18 X42 11 XIO 17JX15 12 X20 10 X30 20 X42 OiXlOJ 11 X16 14 X20 12 X30 22 X^2 lOiXlOJ 12 X16 18 X20 16 X30 24 X42 7fX12 13 X16 19 X20 18 X30 26 X42 8 X12 14iXl6 28 X20 18iX30 28 y.^1 8iX12 15 X16 21 X20 20 X30 18 X48 9 X12 151X16 22 X20 24 X30 20 X48 10 X12 16 X16 12 X21 22 X33 22 X48 11 X12 17 X16 13 X21 24 X33 24 X48 lliXl2 18 X16 18iX21 10 X36 26 X48 12 X12 18iX17 20 X21 12 X36 28 X48 12iXl2 23 X17 20 X22 14 X36 24 X54 13 X12 26 X17 18 X24 16 X36 26 X54 14 X12 10 X18 10 X24 18 X36 28 X54 10 X14 11 X18 12 X24 20 X36 28 X60 11 X14 15 X18 14i X24 22 X36 TABLES 397 Table XV. — Cordinued Compound NoTSs: 1 to run condenBins or 2 to run condenmng or 3 to run condensing or 4 to run condensing or 5 to run condensing or 6 to run condensing or 7 to run condensing or non-condensing non-condensing non-condensing non-condensing non-condensing non-condensing non-condensing on initial on initial on initial on initial on initial on initial on initial pressure pressure pressure pressure pressure pressure pressure of 100-160. of 100. of 126. of 90-100. of 110-130. of 140-160. of 126 4i- 8 X 6 6 -10 X 6 7 -13 X 8 6 -12 XIO 7 -12 X 10 8 -12 XIO 8i-15jX10 7 -14 XIO 8 -14 XIO 9i-15 Xll 7H3iXl2 9 -15JX12 19 -14 X 12 10 -16 XX2 10-18 X12 U -16 X12 9 -18 X14 10 -18 X14 10 -17iX14 11 -19 X14 11 -18 X14 12 -18 X14 12 -20 X14 1 1 1 1 1 2 3 3 1 1 1 1 1 1 3 3 1 1 1 13 -18 13 -20 7M3i 9 -15J 11 -19 13 -19 7i-13i 9 -15J 10 -17i 11 -19 11 -22 12 -21 13 -22 13 -22J 14 -22 14J-25 15 -22 15i-26J 16 -25 13 -23 15 -26 16i-29 9 -15J X14 1 X14 1.2 X15 3 ' X15 3 1 X15 4 ! X15 5 X16 3 X16 3 X16 3 X16 3 ! X16 1 1 X16 3 X16 1 X16 3 • X16 2 ' X16 3 X16 1 X16 3 ; X16 2 1 X17 4,6 X17 4,6 X17 4 ; X18 3 11 -19 X18 12 -21 X18 13 -22JX18 14 -24 X18 15i-26JX18 16 -24 X18 16 -26 X18 161-28iX18 8 -12 X20 9 -14 X20 16 -28 X20 17 -30 X20 18 -28 X20 19 -30 X20 19 -30 X22 9 -15JX21 12 -21 X21 13 -22JX21 14i-26iX21 15i-28JX21 18i-32i X21 20 -36 X21 13 -23 X22 3 3 1 3 1 3 7 7 1 1 1 1 1 3 3 3 3 3 3 3 5 14i-26 X22 18 -32 X22 10 -17JX24 11 -19 X24 12 -18 X24 13 -20 X24 14 -22 X24 16J-28JX24 17i-30JX24 22 -38 X24 24 -A2 X24 12 -21 X27 13 -22 J X 27 16J-28i X27 17J-30iX27 14i-25 X30 15l-26iX30 18i-32iX30 20 -36 X30 28i-50 X30 30 -54 X30 16i-28JX33 17iX30iX33 4 5 3 3 7 7 7 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 22 -38 X33 24 -42 X33 181-32 J X 36 20 -36 X36 26J-46 X36 281-50 X36 14i-25 X42 15i-26iX42 18J-32JX42 20 -36 X42 16 J-28i X48 17i-30iX48 22 -38 X48 24 -A2 X48 18J-321X54 20 -36 X64 26i-46 X54 28i-50 X54 22 -38 X60 24 -42 X60 30 -54 X60 32i-57 X60 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 34 -%Q X60j 3 Triple Note: All condensing and to run of initial pressure as given. Size. V Size. P Size. V 10 -151-26X15 200 27 -43. (41x39 180 30-50-82X48 180 11 -18 -30X20 250 25-41-68X48 190 12 -20 -34X24 180 18 -281-48X40 180 27-45-75X54 190 12 -19 -32X24 190 22 -37 -63X42 180 28-45-72X54 185 175 22 -38 -64X42 185 1 28-46-75X54 180 12^22 -36X24 180 321-53 -{5[x48 265 ' 29-47-83X54 160 14 -23 -28X26 190 1 32-52-92X50 200 18 -29 -47X30 161-24 -41X30 18 -30 -50X30 16 -25M3X30 200 180 200 190 35-67-/^X48 36 -57 -{76X48 265 295 ' 34-56-100X60 35-58 |^qX60 34-57-104X63 200 190 200 161-24 -41 X30 180 28 -45 -72X48 180 CHAPTER IV HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS BETWEEN THE HEAT CONTENT OF SUBSTANCES AND THEIR PHYSICAL- CHEMICAL STATE, 1. Substances and Heat Effects Important in Engineering. It has been shown in preceding chapters concerned with work in general and with the deter- mination of quantity of work that may be done in the cyUnders by, or on expansive fluids that (a) Fluids originally at low may be put in a high-pressure condition by the expenditure of work; (b) Fluids under high pressure may do work in losing that pressure. That work may be done, fluids under pressure are necessary and thai the greatest amount of work mxiy be done per unit of fluid the fluid itself must be expansive, that is, it mv^t be a gOrS or a vapor. Gases or vapors under pressure are, therefore, prerequisites to the economical use of fluids for the doing of work, and that this work may be done at the expense of heat or derived from heat, it is only necessary that the heat be used to create the necessary primary con- dition of high pressure in vapors and gases. There are two general ways of accomplishing this purpose — first, to apply the heat to a boiler supplied with liquid and discharging its vapor at any pressure as high as desired or as high as may be convenient to manage; second, to apply the heat to a gas confined in a chamber, raising its pressure if the chamber be kept at a fixed volume, which is an intermittent process, or increasing the fluid volume if the size of the chamber be allowed to increase, the fluid pressure being kept constant or not, and this latter process may be intermittently or continuously carried out. These two processes are fundamental to th^ steam and gas engines that are the characteristic prime movers or power generators of engineering practice, utilizihg heat energy, and with the exception of water-wheels the sole commer- cially useful sources of power of the industrial world. Thus, the heating of gases and the evaporation of liquids are two most important thermal processes to be examined together with their inverse, cooling and condensation, and necessarily associated in practical apparatus with the heating and cooling of solid containers or associated Uquids. From the power standpoint, the effects of heat on solids, liquids, gases and vapors, both without change of state and with change of state are fundamental, and the substances to be studied as heat carriers do not include the whole, known chemical world, but only those that are cheap enough to be used in engineering practice or otherwise essential thereto. These substances of supreme importance are, of course, air and water, with all their 398 HEAT AND MATTER 399 physical and chemical variations, next the fuels, coal, wood, oil, alcohol and ombustible gases, together with the chemical elements entering into them ind the chemical compounds which mixed together may constitute them. Probably next in importance from the standpoint of engineering practice ^re the substances and thermal processes entering into mechanical refriger- ation and ice making. There are but three substances of commercial importance lere — ammonia, pure and in dilute aqueous solution, carbonic acid and air. The process of heating or cooling solids, liquids, gases and vapors, together vith solidification of water into ice, evaporation and condensation, fundamental power problems, are also of equal importance here, but there is added an idditional process of absorption of ammonia vapor in water and its discharge rom the aqueous solution. Many are the practical applications of heat transfer or transmission, some ){ which call into play other substances than those named. In the heating )f buildings there is first combustion with transfer of heat to water in boilers, iow of the hot water or steam produced to radiators and then a transfer of heat .0 the air of the room; in feed-water heaters, heat of exhaust steam warms ^ater on its way to the boilers; in economizers, heat of hot flue gases is trans- 'erred to boiler feed water; in steam superheaters, heat of hot flue gases is trans- ferred to steam previously made, to raise its temperature, steam pipes, boiler surfaces and engine cylinders transfer heat of steam to the air which is opposed by covering and lagging, in steam engine condensers heat of exhaust steam is transferred to circulation water; in cooling cold storage rooms and making ice, 1 solution of calcium or sodium chloride in water is circulated through pipes and tanks and is itself kept cool in brine coolers in which the brine transfers the heat absorbed in the rooms and tanks, to the primary substance ammonia 3r carbonic acid and evaporates it. While evaporation and condensation as processes are fundamental to the machinery and apparatus of both power and refrigeration, they also are of importance in certain other industrial fields. In the concentration of s:olutions to promote crystallization such, for example, as sugar, evapora- tion of the solution and condensation of the distillate are primary processes as also is the case in making gasolene and kerosene from crude oil, in the making of alcohol from a mash, and many other cases found principally in chemical manufacture. These are examples of evaporation and condensation in which little or no gases are present with the vapor but there are other cases in which a gas is present in large proportion, the thermal characteristics of which are different as will be seen later. Among these processes are: the humidification or moistening of air with water in houses and factories to prevent excessive- skin evaporation of persons breathing the air, excessive shrinkage of wood-work and to facilitate the manufacturing processes like tobacco working and thread spinning. Conversely, air may be too moist for the purpose, in which case it is dried by cooling it and precipitating its moisture as rain or freezing it out as ice, and this is practiced in the Gayley process of operating blast furnaces, where excess of moisture will on dissociating absorb heat of coke combustion and reduce 400 ENGINEERING THERMODYNAMICS the iron output per ton of coke, and in the factories where, for example, collodion is worked, as in tb'^ manufacture of photographic films, with which moisture seriously interferes*. Of course, himiidification of air by water is accomplished only by evaporation of water, and evaporation of water is only to be accomplished by the absorption of heat, so that humidification of air by blowing it over water or spraying water into it must of necessity cool the water, and this is the prin- ciple of the cooling tower or cooling pond for keeping down the temperature of condenser circulating water, and likewise the principle of the evaporative condenser, in which water cooler and steam condenser are combined in one. The same process then, may serve to cool water if that is what is wanted, or to moisten air, when dry air is harmful, and may also serve to remove moisture from solids like sand, crystals, fabrics, vegetable or animal matter to be reduced to a dryer or a pulverized state. There are some important examples of humidification in which the substances are not air and water, and one of these is the humidification of air by gasolene or alcohol vapor to secure* explosive mixtures for operating gas engines. Here the air vaporizes enough of the fuel, humidifjring or carburetting itself to serve the purpose, sometimes without heat being specifically added and sometimes with assistance from the hot exhaust. A somewhat similar action takes place in the manufacture of carburetted water gas when the water gas having no illuminating value is led to a hot brick checkerwork chamber supplied with a hydrocarbon oil, the vapor of which humidifies the gas, the heat of vaporization being supplied by the hot walls and regularly renewed as the process is inter- mittent. Of course, in this case some of the vapors may really decompose into fixed gases, a peculiar property of the hydrocarbon fuels, both liquid and gaseous, and frequently leaving residues of tar, or soot, or both. Finally, among the important processes there is to be noted that of gasifica- tion of solid and liquid fuels in gas producers and vaporizers, a process also carried on in blast furnaces in which it is only an accidental accompaniment and not the primary process. Some of the actions taking place in gas producers are also common to the manufacture of coal gas, and coke, in retorts, beehive and by-product ovens. From what has been said it should be apparent that engineers are concerned not with any speculations concerning the nature of heat but only with the kind and quantity of effect that heat addition to, or abstraction from, substances may be able to produce and not for all substances either. While this interest is more or less closely related to philosophic inquiry, having for its object the development of all embracing generalizations or laws of nature, and to the relation of heat to the chemical and physical constitution of matter, subject matter of physical chemistry, the differences are marked, and a clearly defined field of application of laws to the solution of nurtierical problems dealing with identical processes constitutes the field of engineering thermodynamics. It is not possible or desirable to take up and separately treat every single engineering problem that may rise, but on the contrary to employ the scientific methods of grouping thermal processes or substance effects into types. HEAT AND MATTER 401 Prob. 1. Water is forced by a pump through a feed-water heater and economizer to a boiler where it is changed to steam, which in turn passes through a superheater to a cylinder from which it is exhausted to a condenser. Which pieces of apparatus have to do with heat effects and which with work? Point out similarities and differences of process. Prob. 2. Air is passed over gasolene in a carburetter; the mixture is compressed, burned and allowed to expand in a gas engine cylinder. Which of the above steps have to do with heat effects and which with work effects? Prob. 3. In certain types of ice machines liquid ammonia is allowed to evaporate, the vapor which is formed being compressed and condensed agmn to liquid. Which of these steps is a work phase and which a heat phase? Compare with Problem 1. Prob. 4. When a gun is fired what is the heat phase and what is the work phase? Are they separate or coincident? Prob. 5. Air is compressed in one cylinder, then it is cooled and compressed to higher pressure and forced into a tank. The air in the tank cools down by giving up heat to the atmosphere. From the tank it passes through a pipe hne to a heater and then to an engine from which it is exhausted to the atmosphere. Which steps in the cycle may be regarded as heat and which as work phases? Compare with Problem 2. 2. Classification of Heating Processes. Heat Addition and Abstraction withy or without Temperature Change. Qualitative Relations. That heat will pass from a hot to a less hot body if it gets a chance is axiomatic, so that a body acquiring heat may be within range of a hotter one, the connection between them being, either inmiediate, that is they touch each other, or another body may connect them acting as a heat carrier, or they may be remote with no more provable connection than the hypothetic ether as is the case with the sun and earth. A body may gaiii heat in other ways than by transfer from a hotter body, for example, the passage of electrical current through a conductor will heat it, the rubbing of two solids together will heat both or perhaps melt one, the churning of a liquid will heat it, the mixing of water and sulphuric acid will produce a hotter liquid than either of the components before mixture, the absorp- tion by water of ammonia gas will heat the liquid. All these and many other similar examples that might be cited have been proved by careful investigation partly experimental, and partly by calculation based on various hypotheses to be examples of transformation of energy, mechanical, electrical, chemical, into the heat form. While, therefore, bodies may acquire heat in a great many different concrete ways they all fall under two useful divisions: (a) By transfer from a hotter body; (6) By transformation into heat of some other energy manifestation. One body may be said to be hotter than another when it feels so to the sense of touch, provided neither is too hot or too cold for injury to the tissues, or more generally, when by contact one takes heat from the other. Thus, ideas of heat can scarcely be divorced from conceptions of temperature and the definition of one will involve the other. As a matter of fact temperature as indicated by any instrument is merely an arbitrary number located by some- body on a scale, which is attached to a substance on which heat has some visible 402 ENGINEERING THERMODYNAMICS effect. Temperature is then a purely arbitrary, though generally accepted, number indicating some heat content condition on a scale, two points of which have been fixed at some other conditions of heat content, and the scale space between, divided as convenient. Examination of heat effects qualitatively will show how thermometers might be made or heat measured in terms of any handy effect, and will also indicate what is hkely to happen to any substaHce when it receives or loses heat. Some of the more common heat effects of various degrees of importance in engineering work are given below: Expansion of Free Solids. Addition of heat to free solids will cause them to expand, increasing lengths and volumes. Railroad rails and bridges are longer in siunmer than winter and the sunny side of a building becomes a little higher than the shady side. Steam pipes are longer and boilers bigger hot, than cold, and the inner shell of brick chinmeys must be free from the outer to permit it to grow when hot without cracking the outer or main supporting stack body. Shafts running hot through lack of lubrication or overloading in comparatively cool bearing boxes may be gripped tight enough to twist off the shaft or merely score the bearing. Stressing of Restrained Solids. A solid being heated may be restrained in its tendency to expand, in which case there will be set up stresses in the mate- rial which may cause rupture. Just as with mechanically applied loads, bodies deform in proportion to stress up to elastic limit, as stated by Hooke's law, so if when being heated the tendency to expand be restrained the amount of deformation that has been prevented determines the stress. A steam pipe rigidly fixed at two points when cold will act as a long column in compression and buckle when hot, the budding probably causing a leak or rupture. If fixed hot, it will tend to shorten on cooUng and being restrained will break something. Cylinders of gas engines and air compressors are generally jacketed with water and becoming hot inside, remaining cold outside, the inner skin of the metal tends to expand while the outer skin does not. One part is, therefore, in tension and the other in compression, often causing cracks when care in designing is not taken and sometimes in spite of care in large gas engines. Expa^^^n of Free Liquids. Heating of liquids will cause them to expand just as do solids, increasing their volume. Thus, alcohol or mercury in glass tubes will expand and as these liquids expand more than the glass, a tube which was originally full will overflow when hot, or a tube of very small bore attached to a bulb of cold liquid will on heating receive some liquid; the movement of liquid in the tube if proportional to the heat received will serve as a thermometer. If the solid containing the Uquid, expanded to the same degree as the liquid there would be no movement. Two parts of the same liquid mass may be unequally hot and the hotter having expanded will weigh less per cubic foot, that is, be of less density. Because of freedom of movement in liquids the lighter hot parts will rise and the cooler heavy parts fall, thus setting up a circulation, the principle of which is used in hot water heating systems, the hot water from the furnace rising to the top of the house through one pipe and cooling on its downward path through radiators and return pipe. In general then, liquids HEAT AND MATTER 403 crease in density on heating and increase in density on cooling, but a most iportant exception is water, which has a point of maximum density just ove the freezing-point, and if cooled below this becomes not heavier but lighter. )nsequently, water to be cooled most rapidly should be cooled at first from e top and after reaching this point of maximum density, from the bottom, if it to h6 frozen. Rise of Pressure in Confined Liquids. When liquids are restrained from panding under heating they suffer a rise of pressure which may burst the ntaining vessel. For this reason, hot water heating systems have at their gbest point, open tanks, called expansion tanks, which contain more water len the system is hot than when cold, all pipes, radiators and furnaces being nstantly full of water. Should this tank be shut off when the water is cold mething would burst, or joints leak, before it became very hot. Expansion of Free Gases. Just as solids and liquids when free expand under mating, so also do gases and on this principle chinmeys and house ventilation stems are designed. The hot gases in a chimney weigh less per cubic foot lan cooler atmospheric air; they, therefore, float as does a ship on water, le superior density of the water or cold gas causing it to flow under and t the ship or hot gas, respectively. Similarly, hot-air house furnaces and ^ntilating systems having vertical flues supplied with hot air can send it upward f simply allowing cold air to flow in below and in turn being heated flow up id be replaced. Rise of Pressure in Confined Gases. Gases when restrained from expanding ider heat reception will increase in pressure just as do liquids, only over greater inges, and as does the internal stress increase in solids when heated under straint. It is just this principle which lies at the root of the operation of ins and gas engines. Confined gases are rapidly heated by explosive combus- on and the pressure is thus raised sufficiently to drive projectiles or pistons I their cylinders. Melting of Solids. It has been stated that solids on being heated expand at it should be noted that this action cannot proceed indefinitely. Continued ^ting at proper temperatures will cause any solid to melt or fuse, and the pre- iously rising temperature will become constant during this change of state, hus, melting or fusion is a process involving a change of state from soUd ) liquid and takes place at constant temperature. The tanks or cans of ice- laking plants containing ice and water in all prc^ortions retain the same ?mperature until all the water becomes ice, provided there is a stirring or cir- ilation so that one part communicates freely with the rest and provided also le water is pure and contains no salt in solution. Impure substances, such as quid solutions, may suffer a change of temperature at fusion or solidification, or pure substances, melting and freezing, or fusion and solidification, are onstant temperature heat effects, involving changes from solid to liquid, or quid to solid states. Boiling of Liquids. Ebuilition. Continued heating of solids causes fusion, nd similarly continued heating of liquids causes boiling, or change of state from 404 ENGINEERING THERMODYNAMICS liquid to vapor, another constant temperature process — ^just what temj>erati will depend on the pressure at the time. So constant and convenient is tl temperature pressure relation, that the altitude of high mountains can be foi from the temperature at which water boils. The abstraction of heat from vapor will not cool it, but on the contrary cause condensation. Steam boi and ammonia refrigerating coils and coolers are examples of evaporating apps tus, and house heating radiators and steam and ammonia condensers of c«3 densing apparatus. Evaporation of Liquids; Humidification of Gases, When dry winds b!oi over water they take up moisture in the vapor form by evaporation at temperature. This sort of evaporation then must be distinguished from ebi lition and is really a heat effect, for without heat being added, liquid c{ change into vapor; some of the necessary heat may be supplied by the watt and some by the air. This process is general between gases and liquids and the active principle of cooling towers, carburetters, driers of solids like woa kilns. The chilling of gases that carry vapors causes these to condense in part As a matter of fact it is not necessary for a gas to come into contact to produa this sort of evaporation from a liquid, for if the Uquid be placed in a vacuui some will evaporate, and the pressure finally attained which depends on the t^m perature, is the vapor pressure or vapor tension of the substance, and the amoun that will so evaporate is measured by this pressure and by the rate of remova of that which formed previously. Evaporation of Solids. Sublimation. Evaporation, it has been shown, mai take place from a liquid at any temperature, but it may also take place directli from the solid, as ice will evaporate directly to vapor either in the presena of a gas or alone. Ice placed in a vacuum will evaporate until the vapor tensioi is reached, and it is interesting to note that the pressure of vapors above thei solids is not necessarily the same as above their liquids at the same temperature though they merge at the freezing-point. This is the case with ice-wata" water vapor. Change of Viscosity. Heating of liquids may have another effect measured by their tendency to flow, or their viscosity. Thus, a thick oil will flow eaae when heated, and so also will any liquid. If, therefore, the time for a giva quantity to flow through a standard orifice under a given head or pressure U measured, this time, which is the measure of viscosity, will be less for any Mquk hot, than cold, for the same liquid. Viscosity then decreases with heat additioi and temperature rise. Dissociation of Gases. When gases ndi simple are heated and the heatinf continued to very high temperatures, they will split up into their elements oi perhaps into other compound gases. This may be called decomposition or, bett€f, dissociation, and is another heat effect. Thus, the hydrocarbon C2H4 will split up with solid carbon soot C and the other hydrocarbon CH4 and steam H2O into hydrogen and oxygen. This is not a constant temperature process, but the per cent dissociated increases as the temperature rises. Dissociation of Liquids. Similar to the dissociation of gases receiving heat al HEAT AND MATTER 405 ^ temperature is the decomposition of some liquids in the liquid state, notably ^ fuel and lubricating oils, or hydrocarbons which are compounds of H and C various proportions, each having different properties. Sometimes these wges of H and C groupings from the old to the new compounds under the luenee of heating will be at constant and at other times at varying tempera- ■es; sometimes the resulting substances remain liquid and sometimes soot C separates out, and this is one of the causes for the dark color of some inder oils. Absorption of Gases in Liquids. Liquids will absorb some gases quite freely; IS, water will absorb very large quantities of ammonia, forming aqua anmionia. idition of heat will drive off this gas so that another heat effect is the expul- n of gases in solution. Use is made of this industrially in the absorption {tem of ammonia refrigeration. SolvbilUy of Solids in Liquids, The heating of liquids will also affect their ubility for solid salts; thus, a saturated solution of brine will deposit crystals heat abstraction and take them back into solution on heat addition, rtain scale-forming compounds are thrown down on heating the water in- ided for boilers, a fact that is made use of in feed-water heating purifiers; ' these salts increase of temperature reduces solubility. In general then heat dition affects the solubility of liquids for solid salts. Chemical Reaction, Combustion. If oxygen and hydrogen, or oxygen and rbon, be heated in contact, they will in time attain an ignition temperature at lich a chemical reaction will take place with heat liberation called combus- •n, and which is an exothermic or heat-freeing reaction. Another and ferent sort of reaction will take place if CO2 and carbon be heated together, • these will together form a combustible gas, CO, under a continuation of heat leption. This is an endothermic or heat-absorbing reaction. Neither of ese will take place until by heat addition the reaction temperature, called lition temperature for combustion, has been reached. Electrical and Magnetic Effects. Two metals joined together at two separate ints, one of which is kept cool and the other heated, will be found to carry electric current or constitute a thermo-electric couple. Any conductor nrjing an electric current will on changing temperature suffer a change of flstance so that with constant voltage more or less current will flow; this is second electrical heat effect and like the former is useful only in instru- jnts indicating temperature condition. A fixed magnet will suffer a change magnetism on heating so that heat may cause magnetic as well as electric ects. These heat effects on substances as well as some others of not so great engi- ering importance may be classified or grouped for further study in a variety ways, each serving some more or less useful purpose. Reversible and Non-reversible Processes, There may be reversible and in-reversible thermal processes, when the process may or may not be con- iered constantly in a state of equilibrium. For example, as heat is applied boiling water there is a continuous generation of vapor in proportion to the r" 406 ENGINEERING THERMODYNAMICS heat received; if at any instant the heat application be stopped the evap^ ration will cease and if the flow of heat be reversed by abstraction, condens* tion will take place, indicating a state of thermal equilibrium in which lh effect of the process follows constantly the direction of heat flow and is co> stantly proportional to the amoimt of heat numerically, and in sign, of directitjs, As an example of non-reversible processes none is better than combustion, which the chemical substances receive heat with proportional temperature until chemical reaction sets in, at which time the reception of heat has no fur- ther relation to the temperatures, because of the liberation of heat by coai- bustion which proceeds of itself and which cannot be reversed by h abstraction. Even though a vigorous heat abstraction at a rate greater th it is freed by combustion may stop combustion or put the fire out, no amo of heat abstraction or cooling will cause the combined substances to chaoa back into the original ones as they existed before combustion. The effect d heat in such cases as this is, therefore, non-reversible. Constant and Variable Volume or Density. When gitses, liquids or soKii are heated they expand except when prevented forcibly from so doing, and ai a consequence they suffer a reduction of density with the increase of volume this is, of course, also true of changing liquids to their vapors. It should noted that all such changes of volume against any resistance whatever, oc with corresponding performance of some work, so that some thermal proce* may directly result in the doing of work. Heating accompanied by no vol change and during which restraints are applied to keep the volume invariable cannot do any work or suffer any change of density, but always results in chan^ of pressure in liquids, gases and vapors and in a corresponding change of intemi stress in solids. Constant and Variable Temperature Processes, Another useful division, aa that most valuable in the calculation of relations between heat effect and hei quantity, recognizes that some of the heating processes and, of course, coolki^ occur at constant temperature and others with changing temperature- Fi example, the changes of state from liquid to solid, and solid to liquid, or freeziu and fusion, are constant temperature processes in which, no matter how mud heat is supplied or abstracted, the temperature of the substance changing st.at4? i not affected, and the same is true of ebullition and condensation, or the changin of state from liquid to vapor, and vapor to liquid. These latter constant temperature processes must not be confused with evaporation, which m*] proceed from either the solid or liquid state at any temperature whether constan or not. Prob. 1. From the time a fire is lighted under a cold boiler to the time steaa first comes off, what heat effects take place? Prob. 2. What heat effects take place when a piece of ice, the temperature o which is 20*^ F., is thrown onto a piece of red-hot iron? Prob. 8. What heat effects must occur before a drop of water may be evaporate from the ocean, and fed back into it as snow? Prob. 4. What heat changes take place when soot 13 formed from coal or oil? HEAT AND MATTER 407 Prob. 6. In a gas producer, coal is burned to COi, which is then reduced to CO. Steam is also fed to the producer, and H and formed from it. Give all the heat effects which occur. Prob. 6. By means of what heat effects have you measured temperature changes, or have known them to be measured? Prob. 7. When the temperature changes from 40° F. to 20° F., give a list of all heat effects you know that commonly occur for several common substances. Do the same for a change in the reverse direction. Prob. 8. If a closed cyUnder be filled with water it will burst if the temperature be lowered or raised sufficiently. What thermal steps occur in each case? Prob. 9. If salt water be lowered sufficiently in temperature, a cake of fresh ice and a rich salt solution will be formed. State the steps or heat effects which occur during the process. 3. Thennometiy Based on Temperature Change Heat Effects. Ther- mometer and Absolute Temperature Scales. Those thermal proceeses in which heat addition or abstraction is followed as a result by a corresponding and more or less proportional temperature change, are quite numerous and important both in engineering practice and as furnishing a means for thermometer-mak- ing, and temperature definition and measurement. . According to Sir William Thomson " every kind of thermoscope must be foimded on some property of matter continuously varying with the temperature " and he gives the fol- lowing: (a) Density of fluid under constant pressure. (6) Pressure of a fluid under a constant volume envelope. (c) Volume of the liquid contained in a solid holder (ordinary mercury or spirit thermometer). (d) Vapor pressure of a solid or liquid. (e) Shape or size of an elastic solid under constant stress. (/) Stress of an elastic solid restrained to constant size. ig) Density of an elastic solid under constant stress. {h) Viscosity of a fluid, (t) Electric current in a thermo-couple. (j) Electric resistance of a conductor. {k) Magnetic moment of a fixed magnet. Any, or all of these — ^pressure, volume, shape, size, density, rate of flow, magnetic or electrical effects, may be measured, and their measure constitutes a measure of temperature indirectly, so that instruments incorporating these temperature effects to be measured, are also thermometers. Any temperature-indicating device may be called a thermometer, though those in use for high temperatures are generally called pyrometers, which indicates the somewhat important fact that no thermometer is equally useful for all ranges of temperature. Practically all thermometers in use for tempera- tures short of a red heat, depend on certain essential relations between the density or volume, the pressure and temperature of a fluid, though metals are used in some little-used forms in which change of size is measured, or change of shape 408 ENGINEERING THERMODYMAMICS of a double metallic bar, often brass and iron, consisting of a piece of each fastened to the other to form a continuous strip. The two metals are expanded by the temperature different amounts causing the strip to bend under heating. There are also in use electric forms for all temperatures, and these are the only reliable ones for high temperatures, both of the couple and resistance types except one dependent on the color of a high temperature body, black when cold. That most useful and common class involving the interde- pendence of pressure and temperature, or volume and temperature, of a fluid is generally found in the form of a glass bulb or its equivalent, to which is attached a long, narrow glass tube or stem which may be open or closed at the end; open when the changes of fluid voliune at constant pressure are to be observed and closed when changes of contained fluid pressure at constant restrained volume are to be measured as the effect of temperature changes. For the fluid there is used most commonly a liquid alone such as mercury, or a gas alone such as air; .though a gas may be introduced above mercury and there may be used a liquid with its vapor above. When the fluid is a liquid, such as mercury, in the common thermometer, the stem is closed at the end so that the mercury is enclosed in a constant-volume container or as nearly so a? the expansion or deformation of the glass will permit, which is not filled with mercury, but in which a space in the stem is left at a vacuum or filled with a gas under pressiu'e, such as nitrogen, to resist evaporation of the mercur}-^ at high temperatures. Gas-fill6d mercury thermometers, as the last form is called, are so designed that for the whole range of mercury expansion the pressure of the gas opposing it does not rise enough to offer material resistance to the expansion of mercury or to unduly stress the glass container. It should be noted that mercury thermometers do not measure the expansion of mercun- alone, but the difference between the voliune of mercury and the glass envelope, but this is of no consequence so long as this difference is in proportion to the expansion of the mercury itself, which it is substantially, with proi)er glass composition, when the range is not too great. Such thermometers indicate temperature changes by the rise and fall of merciu'y in the stem, and any numeri- J cal value that may be convenient can be given to any position of the mercur}^ or any change of position. Common acceptance of certain locations of the scale number, however, must be recognized as rendering other possible ones unneces- sary and so undesirable. Two such scales are recognized, one in use with metric units, the centigrade, and the other with measurements in English units, the Fahrenheit, both of which must be known and familiar, because of the frequent necessity of transformation of numerical values and heat data from one sj'stem to the other. To permit of the making of a scale, at least two points must be fixed with a definite nmnber of divisions between them, each called one degree. The two fixed points are first, the position of the mercury when the thermometer is in the vapor of boiling pure water at sea level, or under the standard atmos- pheric pressure of 29.92" = 760 mm. of mercury absolute pressure, and second, the position of the mercury when the thermometer is surrounded by melting ice at the same pressure. These are equivalent to the boiling- or con- HEAT AND MATTER 409 densation, and melting- or freezing-points, of pure water at one atmosphere pressure. The two accepted thermometer scales have the following character- istics with respect to these fixed points and division between them: THERMOMETER SCALES Pure Water Freeiiag-poiDt. at one aim. pr. Pure Water Boiling-point, at one atm. pr. Number of Equal Divisions Between Freesing and BoUing. Centigrade scale 32 100 212 100 Fahrenheit scale 180 From this it appears that a degree of temperature change is on the centigrade scale, ^l^ of the linear distance between the position of the mercury surface at the freezing- and boiling-^ints of water, and on the Fahrenheit scale, yf^ of the same distance. From this the relation between a degree temperature change for the two scales can be given. One degree temperature change centigrade 180^9 100 5 of one degree temperature change Fahrenheit; or One degree temperature change Fahrenheit of one degree temperatv/re change centigrade. It is also possible to set down the relation between scale readings, for when the temperature is 0^ C, it is 32° F., and when it is 100° C. it is (180+32) =212° F., so that 9 Temperature Fahrenheit =32+— (Temperature centigrade), or Temperature centigrade = q (Temperature Fahrenheit— 32). For convenience of numerical work tables are commonly used to transform temperatures from one scale to the other and such a transformation is shown in a curve. Fig. 119, and in Table XXIX at end of the Chapter. By reason of the lack of absolute proportionality between temperature and effect, other fixed points are necessary, especially at high temperatures, and the following of Table XVI have been adopted by the U. S. Bureau of Stand- ards and are considered correct to within 5° C, at 1200° C. ENGINEEHINa THERMODYNAMICS Dcsrcea CoutlBnule FiQ. 119. — Graphical Relation between Centigrade and Fahrenheit Thermometer Scalea. HEAT AND MATTER 411 TABLE XVI FIXED TEMPERATURES U. S. BUREAU OP STANDARDS Temperftture, •c. Temponture, Determined by the Point at which 232 449 Liquid tin solidifies 327 621 Liquid lead solidifies 419.4 787 Liquid zinc solidifies 444.7 832.6 Liquid sulphur boils 630.5 1167 Liquid antimony solidifies ^ 658 1216 Liquid aluminum, 97.7% pm'e, solicUfies 1064 1947 Solid gold melts 1084 1983 Liquid copper solidifies 1435 2615 Solid nickel melts 1546 2815 Solid palladium melts 1753 3187 Solid platinum melts Thermometers in which a liquid and its vapor exist together, depend on a property to be noted in detail later, the relation of vapor pressure to tempera- ture and its independence of the volume of vapor. So long as any vapor exists above the liquid the temperature will depend only on the pressure of that vapor so that such thermometers will indicate temperature by the pressure measure- ment, after experimental determination of this pressure-temperature relation of vapors. Conversely, temperature measurements of vapors by mercury ther- mometers will lead to pressure values, and at the present time some steam plants are introducing mercury thermometers on the boilers and pipe lines, in place of the proverbially inaccurate pressure gages. Gas thermometer, is the name generally applied to the class in which the fluid is a gas, whether air, hydrogen, nitrogen or any other, and whether the pressure is measured for a fixed contsdned volume, or the volume measured when acted on by a constant pressure. These gas thermometers are so bulky as to be practically useless in ordinary engineering work and are only employed as standards for comparison and for tests of extraordinary delicacy in investi- gation work. They give much larger indications than mercury thermometers because the changes of gas volume under constant pressure are far greater than for mercury or any other liquid. Regnault was the first to thoroughly investigate air thermometers and reported that the second form, that of constant gas volume with measurement of pressure, was most useful. Using the centigrade scale, fixing freezing point at 0^ C, and making the corresponding pressure po, atmospheric at this point, and reading at 100** C. another pressure pioo, he found experimentally a relation between these two pressiures and the temperature corresponding to any other pressure p, as given by the empiric formula. <=100-P=PO-. Pioo—po (601) 412 ENGINEERING THERMODYNAMICS He also determined the pressure at the boiling-point to be related to the pressure at the freezing-point, by Pioo= 1.3665 po, which on substitution gives This constant, 272.85, the reciprocal of which is .003665, is, of course, the pressure increase factor per degree C. rise of temperature for a gas held at constant volume, received extended investigation and it vxxs found that it had about the same value applied to the other type of thermometer in which gas volumes are measured ai constant pressure. This was true even when the pressure used was anything from 44 to 149 cm. of mercury, though it is reported that for 44 cm. pressure the value 272.98, and for 149 cm. pressure, the value 272.7, seemed closer. For hydrogen it was found that the constant was sub- stantially the same as for air, while for carbonic acid it was 270.64, and while the hydrogen thermometers agreed with the air over the whole scale, showing proportional efifects, this was hardly true of carbonic acid. Such uncertainty in the behavior of these thermometers and in the fixing of the constants was traced to the glass in some cases, but there still remained differences charge- able only to the gases themselves. Comparison of the air with mercury ther- mometers showed that there was not a proportional change with the temperature and that temperatures on the two, consistently departed. Examination of Eq. (602), giving the relation between two temperatures and the corresponding gas pressures, will show a most important relation. If in Eq. (602), the pressure be supposed to drop to zero and it is assumed that the relations between pressure and temperature hold, then when p=0, i=— 272.85. This temperaiure has received the name of the absolute zero, and may be defined as the temperature at which pressure disappears or becomes zero at constant volume, and correspondingly, at which the volume also dis- appears, since it was foimd that similar relations existed between volume and temperature at constant pressure. Calling temperature on a new scale begin- ning 272.85** below the centigrade zero by the name absolute temperatures, then Absolute temperature 1 ^.o^o or_i f Scale temperature centigrade J ' [ centigrade As this constant or absolute temperature of the centigrade scale zero, is an experimental value, it is quite natural to find other values presented by differ- ent investigators, some of them using totally different methods. One of these methods is based on the temperature change of a gas losing pressure without doing work, generally described as the porous plug experiment, and the results HEAT AND MATTER 413 as the Joule-Thomson effect, and another is based on the coefficient of expansion of gases being heated. Some of these results agreed exactly with Regnault's value for hydrogen between 0° C. and 100° C. for which he gave —273° C. = —491.4° F. Still other investigations continued down to the last few years yielded results that tend to change the value slightly to between — 491.6°F., and —491.7° F., and as yet there is no absolute agreement as to the exact value.. In engineering problems, however, it is seldom desirable or possible to work to such degrees of accuracy as to make the uncertainty of the absolute zero a matter of material importance, and for practical purposes the following values may be used with sufficient confidence for all but exceptional cases which are to be recognized only by experience. I Centigrade =273] Absolute Temperature (T) j Fahrenheit =460 +Scale Temperature (0 When great accuracy is important it is not possible at present to get a better Fahrenheit value than 459.65, the mean of the two known limits of 459.6 and 459.7, though Marks and Davis in their Steam Tables have adopted 459.64, which is very close to the value of 459.63 adopted by Buckingham in his excellent Bulletin of U. S. Bureau of Standards and corresponding to 273.13 on the centigrade scale. These experiments with the gas thermometers, leading to a determination of temperature as a function of the pressure change of the gas held at constant volume, or its voliune change when held at constant pressure, really supply a definition of temperature which before meant no more than an arbitrary number, and furnished a most valuable addition to the generalization of relations between heat content of a body and its temperature or physical state. A lack of proportionality between thermometer indication and temperature, has already been pointed out, and it is by reason of this that two identical ther- mometers, or as nearly so as can be made, with absolute agreement between water boiling- and freezing-points, will not agree at all points between, nor will the best constructed and calibrated mercury thermometers agree with a similarly good gas thermometer. The temperature scale now almost universally adopted as standard is that of the constant volume hydrogen gas thermometer, on which the degree F. is one one-himdred-and-eightieth part of the change in pressure of a fixed voliune of hydrogen between melting pure ice, and steam above boiling pure water, the initial pressure of the gas at 32° being 100 cm. = 39.37 ins. Hg. A mercury in glass thermometer indication is, of course, a measure of the proper- ties of the mercury and glass used, and its F. degree of temperature is defined in parallel with the above as one one-hundred-and-eightieth part of the volume of the stem between its indications at the same two fixed points. A comparison of the hydrogen thermometer and two different glasses incorporated in mercury thermometers is given below, Table XVII, from the Bulletin of the U. S. Bureau of Standards, v. 2, No. 3, by H. C. Dickinson, quoting Mahlke, but it must be 414 ENGINEERING THERMODYNAMICS remembered that other glasses will give different results ana even different thermometers of the same glass when not similarly treated. Table XVII FAHRENHEIT TEMPERATURES BY HYDROGEN AND MERCURY THERMOMETERS Temporsture by Hydrogen Thermometer. 32 212 302 392 428 464 500 536 672 Difference in Reading bv Mercury m Jena 50'' Glass. + 1.3 Difference in Reading by Mercury in 18" Jena Glass. - .18 + .072 + .39 -h .83 +1.79 +2.4 +3.53 Temperature by Hydrogen Thermometer. 617 662 707 752 797 842 887 932 Difference in Reading bv Mercury in Jena 59" Glass. +10.6 + 16.6 +18.7 +24.6 +28.2 +38.3 +41.4 +50.0 Difference in Reading by Mercury in 16" Jena Glass. Useful comparisons of air, hydrogen, nitrogen, and other gases, with alcohol and mercury, in various kinds of glass, are given in the Landolt-Bomstein- Meyerhoffer, and in the Smithsonian Physical Tables, but are seldom needed for engineering work. One sort of correction that is often necessary in mercury thermometer work is that for stem immersion. Thermometers are calibrated as a rule with the whole stem immersed in the melting ice or the steam, but are ordinarily used with part of the stem exposed and not touching the substance whose tem- perature is indicated. For this condition, the following correction is recom- mended by the same Bureau of Standards Bulletin: When n t h Stem correction = .000088 n{t-tiyF- number of degrees exposed ; temperature indicated Fahrenheit degrees; =mean temperature of emergent stem itself, which must necessarily be estimated and most simply by another thermometer next to it, and entirely free from the bath. Prob. 1. What will be the centigrade scale and absolute temperatures, for the following Fahrenheit readings? -25^, 25^, llO*', 140°, 220*^, 263° scale, and 300°, 460°, 540°, 710°, 2000° absolute. Prob. 2. What will be the Fahrenheit scale and absolute temperatures, for the follow- ing centigrade readings? -20°, 10°, 45°, 80°, 400°, 610° scale, and 200°, 410°, 650°, 810°, 2500° absolute. Prob. 8. By the addition of a certain amount of heat the temperature of a quantity of water was raised 160° F. How many degrees C. was it raised? HEAT AND MATTER 415 Prob. 4. To bring water from 0° C. to its boiling-point under a certain pressure equired a temperature rise of 150** C. What was the rise in Fahrenheit degrees? Prob. 6. For each degree rise Fahrenheit, an iron bar will increase .00000648 of is length. How much longer will a bar be at 150** C. than at 0** C? At 910** C. bsolute than at 250** C. absolute? Prob. 6. The increase in pressure for SO2 for a rise of 100° C. is given as .3845 at onstant volimie. What would have been absolute zero found by Regnault had he sed SOs rather than air? Prob. 7. A thermometer with a scale from 40° F. to 700° F. is placed in a thermome- sr well so that the 200° mark is just visible. The temperature as given by the [lerraometer is 450°. If the surrounding temperature is 100° F., what is true tempera- are in the well? 4. Calorimetiy Based on Proportionality of Heat E£fects to Heat Quantity, rnits of Heat and Mechanical Equivalent. Though it is generally recognized rom philosophic investigations extending over many years, that heat is one lanifestation of energy capable of being transformed into other forms such s mechanical work, electricity or molecular arrangement, and derivable from hem through transformations, measurements of quantities of heat can be made rithout such knowledge, and were made even when heat was regarded as a ubstance. It was early recognized that equivalence of heat effects proved fleets proportional to quantity; thus, the melting of one pound of ice can cool » poimd of hot water through a definite range of temperature, and can cool wo pounds through half as many degrees, and so on. The condensation of , pound of steam can warm a definite weight of water a definite number of legrees, or perform a certain niimber of pound-degrees heating effect in water, lo that taking the pound-degree of water as a basis the ratio of the heat liberated >y steam condensation to that absorbed by ice melting can be found. Other ubstances such as iron or oil may suffer a certain number of poimd degree hanges and affect water by another number of pound-degrees. The unit if heat quantity might be taken as that which is liberated by the condensation ►f a pound of steam, that absorbed by the freezing of a pound of water, that to aise a poimd of iron any number of degrees or any other quantity of heat iffoct. The heat unit generally accepted is, in metric measure, the calorie, »r the amount to raise one kilogramme of pure water one degree centigrade, )T in English units, the British thermal unit, that necessary to raise one pound if water one degree Fahrenheit. Thus, the calorie is the kilogramme degree lentigrade, and the British thermal unit the poimd degree Fahrenheit, and the atter is used in engineering, usually abbreviated to B.T.U. There is also iccasionally used a sort of cross unit called the centigrade heat unit, which is he pound degree centigrade. The relation between these is given quantitatively by the conversion table it the end of this Chapter, Table XXX. All the heat measurements are, therefore, made in terms of equivalent ^ater heating effects in pound degrees, but it must be understood that a water )ound degree is not quite constant. Careful observation will show that the 416 ENGINEERING THERMODYNAMICS melting of a pound of ice will not cool the same weight of water from 200** F, to 180° F., as it will from 60° F. to 40° F., which indicates that the heat capacity of water or the B.T.U. per pound-degree is not constant. It is, thereforr. necessary to further limit the definition of the heat unit, by fixing on som* water temperature and temperature change, as the standard, in addition to the selection of water as the substance, and the pound and degree as units of capacity. Here there has not been as good an agreement as is desirable, some using 4° C. = 39.4° F. as the standard temperature and the range one-half degree both sides; this is the point of maximum water density. Others have used 15' C. = 59° F. as the temperature and the range one-half degree both sides; still others, one degree rise from freezing point 0° C. or 32° F. There are gtx reasons, however, for the most common present-day practice which will prol)- ably become universal, for taking as the range and temperatures, freezing- point to boiling-point, and dividing by the niunber of degrees. The heat unit so defined is properly named the mean calorie or mean British thermal unit; therefore, Mean calorie =t7v7j (amount of heat to raise 1 Kg. water from 0° C. to 100° C). Mean B.T.U. =— (amount of heat to raise 1 lb. water from 32° F. to 212° F.i. In terms of the heat unit thus defined, the amount of heat per degree tem perature change is variable over the scale, but only in work of the most accurate character is this difference observed in engineering calculations, but in accurate work this difference must not be neglected and care must be exercised in using other physical constants in heat units reported by different observers, to be sure of the unit they used in reporting them. It is only by experience that judgment can be cultivated in the selection of values of constants in heat units reported for various standards, or in ignoring differences in standards entirely. The great bulk of engineering work involves uncertainties greater than these differ- ences and they may, therefore, be ignored generally. By various experimental methods, all scientifically carried out and exetnding over sixty years, a measured amount of work has been done and entirely eon- verted into heat, originally by friction of solids and of liquids, for the deter- mination of the foot-pounds of work equivalent to one B. T. U., when the conversion is complete, that is, when all the work energy has been converted mto heat. This thermo-physical constant is the mechanical equivalent of kat Later, indirect methods have been employed for its determination by cakula- tion from other constants to which it is related. All of these experiments have led to large number of values, so that it is not surprising to find doubt as to the correct value and different values are used even by recognized authori- ties. The experiments used include: HEAT AND MATTER 417 (a) Compression and expansion of air; Joule. (6) Steam engine experiments, comparing heat in supplied and exhausted steam; Him. (c) Expansion and contraction of metals; Edlund and Haga. (d) Specific volume of vapor; Perot. (e) Boring of metals; Rumford and Him. (/) Friction of water; Joule and Rowland. (g) Friction of mercury; Joule. (A) Friction of metals; Him, Puluj, Sahulka. (i) Crushing of metals; Him. 0) Heating of magneto electric currents; Joule. (k) Heating of disk between magnetic poles; Violle. (I) Flow of liquids (water and mercury) under pressure; Him, Bartholi. (m) Heat developed by wire of known absolute resistance; Quintus Icilius, Weber, Lenz, Joule, Webster, Dieterici. (n) Diminishing the heat contained in a battery when the current produces work; Joule, Favre. (o) Heat developed in, and voltage of Daniell cells; Weber, Boscha, Favre and Silberman, Joule, (p) Combination of electrical heating and mechanical action by stirring water; Griffiths. {q) Physical constants of gases. The results of all of these were studied by Rowland in 1880, who himself experimented also, and he concluded that the mechanical equivalent of heat was nearly 778.6 ft.-lbs. = 1 B.T.U., at latitude of Baltimore, or 774.5 ft.-lbs. = 1 B.T.U., at latitude of Manchester. with the following corrections to be added for other latitudes. Latitude 0** 10*» 20'' 30'' 40'' 50** eO** 70** 80* 90* Ft.-lbs 1.62 1.50 1.15 .62 .15 -.75 -1.41 -1.93 -2.30 -2.43 Since that time other determinations have been made by Re3molds and Morby, using mechanical, and Griffiths, Schuste and Gannon, Callendar and Barnes, using electrical transformation into heat. Giving these latter deter- minations equal weight with those of Joule and Rowland, the average is 1 small calorie at 20® C. (nitrogen thermometer) =4.181X10^ ergs. ^ 418 ENGINEERING THERMODYNAMICS On the discussion of these results by Smith, Marks and Davis accept and use the mean of the results of Reynolds, and Morby and Barnes, which is 1 mean calorie = 4. 1834 X 10^ ergs, = 3.9683 B.T.U. 1 mean B.T.U. = 777.52 ft.-lbs., when the gravitational constant is 980.665 cm. sec^, which corresponds to 32.174 lbs., and is the value for latitude between 45*^ and 46**. For many years it has been most common to use in engineering calculations, the round number 778, and for most problems this round number is still the best available figure, but where special acciu'acy is needed it is likely that no closer value can be relied upon than anything between 777.5 and 777.6 for the above latitude. ^ Example. To heat a gallon of water from 60^ F. to 200^ F. requires the heat equivalent of how many foot-pounds? 1 gallon =8.33 lbs.,' 200^ F. -60** F. =140^ F. rise, 8.33 X 140 = 1665 pound-degrees, = 1665 B.T.U. =778X1665, ft.-lbs. -90,800 ft.-lb8. Prob. 1. A feed-water heater is heating 5000 gallons of water per hour from 40* F. to 200** F. What would be the equivalent energj'^ in horse-power units? Prob. 2. A pound of each of the following fuels has the heating values as given. Change them to foot-pounds. Average bituminous coal, 14,000 B.T.U. per lb., Average kerosene, 18,000 B.T.U. per lb. " small anthracite, 12,000 B.T.U. per lb., " alcohol, 10,000 B.T.U. per lb. Average gasolene, 20,000 B.T.U. per lb. Prob. 8. A cubic foot of each of the following gases yields on combusture, the number of heat units shown. Change them to foot-pounds. Natural gas (average), 880 B.T.U. per cu.ft., Carburetted water gas, 700 B.T.U. per cu.ft. Coal gas, 730 B.T.U. per cu.ft., Mond gas, 150 B.T.U. per cu.ft. Blast fimiace gas, 100 B.T.U. per cu.ft. Prob. 4. A pool contains 20,000 cu.ft. of water and must be warmed from 40** F. to 70** F. How much work might be done with the equivalent energy? HEAT AND MATTER 419 Prob. 6. How many calories and how many centigrade heat units would be required in Prob. 4? Prob. 6. In the course of a test a man weighing 200 lbs. goes up a ladder 25 ft. high, every 15 minutes. If the test lasted 12 hours how many B.T.U. did he expend? Prob. 7. A reservoir contains 300 billion gallons of water which are heated each year from 39® F. to 70° F. What is the number of foot-poimds of work equivalent? Prob. 8. A pound of water moving at the rate of 450 ft. per second is brought to rest, so that all of its energy is turned into heat. What will be the temperature rise? Prob. 9. For driving an automobile 30 horse-power is being used. How long will a gallon of average gasolene, sp.gr. = .7, last, if 10% of its energy is converted into work? Prob. 10. Power is being absorbed by a brake on the flywheel of an engine. If the engine is developing 50 horse-power how many B.T.U. per minute must be carried off to prevent burning of the brake? 6. Temperature Change Relation to Amount of Heat, for Solids, Liquids, Gases, and Vapors, not Changing State. Specific Heats. Provided gases do not decompose, vapors condense, liquids freeze or evaporate, and solids melt, under addition or abstraction of heat, there will always be the same sort of relation between the quantity of heat gained or lost and the temperature change for all, differing only in degree. As the reception of heat in each case causes a temperature rise proportional to it and to the weight of the sub- stances, this constant of proportionality once determined will give numerical relations between any temperature change and the corresponding amount of heat. Making the weight of the substance unity, which is equivalent to the consideration of one pound of substance, the constant of proportionality may be defined as the quantity of heat per degree rise, and as thus defined is the specific heat of the substance. Accordingly, the quantity of heat for these cases is equal to the product of specific heat, temperature rise and weight of substance heated. The heat, as ahready explained, may be added in two characteristic ways: (o) at constant volume or density, or (6) at constant pressure. It might be expected that by reason of the increase of volume and performance of work under constant pressure heating, more heat must be added to raise the tempera- ture of one pound, one degree, than in the other case where no such work is done, and both experimental and thermodynamic investigations confirm this view. There are, therefore, two specific heats for all substances, capable of definition: (a) The specific heat at constant volume, and (6) The specific heat at constant pressure. These two specific heats are quite different both for gases and for vapors, which suffer considerable expansion imder constant pressure heating, but for solids and liquids, which expand very little, the difference is very small and is to be neglected here. As a matter of fact, there are no cases of common engineering practice involving the specific heat of liquids and solids under constant volume, and values for the specific heats of liquids and solids are always without further definition to be understood as the constant pressure values. 420 ENGINEERING THERMODYNAMICS Let C, be the specific heat of solids and liquids suffering no change of state. C,, be the specific heat of gases and vapors at constant pressure and suffering no change of state. C„ be the specific heat of gases and vapors at constant volume and suf- fering no change of state. <2 and^i, be the maximum and min i mum temperatures for the process. w, be the weight in pounds. Then will the heat added, be given by .the following equation, if the tempera- ture rise is exactly proportional to the quantity of heat, or in other words, ij the specific heat is constant. Q=Cw(t2^ti)y for solids and liquids (603) Q=sC.io(b— <i), for gases and vapors (not near condensation) when voliune is constant (6(H) Q=Cpti?(fe— <i), for gases and vapors (not near condensation) when pressure is constant (605) When, however, the specific heat is variable, as is the case for many sub- stances, probably for all, the above equation cannot be used except when the specific heat average value, or mean specific heat is used. If the variation is irregular this can be found only graphically, but for some substances the variation is regular and integration will give the mean value. It has been the custom to relate the specific heat to the temperature above the freez- ing-point of water, expressing it as the siun of the value at 32^ F., and some fraction of the temperatm^ above this point to the first and second powers, as in Eq. (606). Specific heat at temperature (0 =a+b(t-Z2)+c{t-32)^ . (606) In this equation a is the specific heat at 32^, while b and c are constants, different for different substances, c being generally zero for liquids. When this is true, the heat added is related to the temperature above 32^ by a differential expression which can be integrated between limits Wt-32 Q= I [a+6(i-32)+C(i-32)2](ft Jtt -32 = a[(fe-32)-(<i-32)]+|[(fe-32)2-(ii-32)2]+|[(fe-32)3-(<i-32)3]. (607) Usually the heats are calculated above 32® so that the heats between any two temperatures will be the difference between the heats from 32® to those two temperatures. In this case /i=32®, and, t2 = t, whence B.T.U. per lb., from 32® to <,= ra+|(i-32)+|(^-32)2l(/-23). . . (608) HEAT AND MATTER 421 ''or this range, of temperature 32** to t, the quantity of heat may be ex- tressed as the product of a mean specific heat and the temperature range Heat from 32^ to t = (mean sp. heat from 32° to r) X (<-32). . (609) ::omparing Eq. (608) with Eq. (609) it follows that Mean specific heat from 32° P. tor P. } =a+|(<-32)+|(<-32)2 (610) The coefficient of (1—32) in the mean specific heat expression, is half that in he expression for specific heat at (, and the coefficient of (i— 32)^, is one-third, rhis makes it easy to change from specific heat at a given temperature ibove 32°, to the mean specific heat from 32° to the temperature in question. The specific heats of some substances are directly measured, but for some others, notably the gases, this is too difficult or rather more difficult than cal- culation of values from other physical constants to which they are related. • It often happens that in engineering work the solution of a practical problem requires a specific heat for which no value is available, in which case the general law of specific heats, known as the law of Dulong and Petit, for definite compounds may be used as given in Eq. (611). (Specific heat of solids) X (atomic weight) = 6.4. . . . (611) This is equivalent to saying that all atoms have the same capacity for heat, and while it is known to be not slrictly true, it is a useful relation in the absence of direct determinations. Some values, experimentally determined for the specific heats of solids, are given in Table XXXI at the end of this Chap- ter, together with values calculated from the atomic weights to show the degree of agreement. The- atomic weights used are those of the International Com- mittee on Atomic Weights (Jour. Am. Chem. Soc, 1910). When the specific heat of a solid varies with temperature and several determinations are avail- able, only the maximum and minimum are given with the corresponding tem- peratures, as these usually suffice for engineering work. To illustrate this variability of specific heat of solids, the values deter- mined for two samples of iron are given in Figs. 120 and 121, the former showing the variation of the mean specific heat as determined by OberhofiFer and Harker from 500** F. up, and the latter the amount of heat per pound of iron at any temperature above the heat content at 500® F., which is gen- erally called its total heat above the base temperature, here 500** F. It is extremely probable that the specific heats of liquids all vary irregularly with temperature so that the constant values given in Table XXXII at the end of the Chapter must be used with caution. This is certainly the case for water, and is the cause of the difficulty in fixing the unit of heat, which is best solved by the method of means. In Fig. 122 are shown in curve form the values for the 422 INGINEERINQ THERMODYNAMICS specific heats of water at temperatures from 20** F. to 600® JF., as accepted by Marks and Davis after a critical study of the experimental results of •18 .'16 X |.14 OQ S .12 .10 ■ (a) (6) Oberi Hark loffer jr / V *^^m^^^^ — — — >^^ aia^M IS) f 1/ y X ^.. (6) y} '/ y -A 'y -;? ■:>^ y 1000 aooo ^Temperature in Degrees Fahr. Fig. 120. — Mean Specific Heat of Iron above 500** F., Illustrating Irregular Variations not Yielding to Algebraic Expression. Barnes and Dieterici and adjustment of the differences. The integral curv^e is plotted in Fig. 123 which, therefore, gives the heat of water from 32®F. to any >'30O > < paoo o a § glOO I (6) From Oberh t ffer Data Harker ^^ y fiOO 1000 ./^ y l^-'- ^^ :'(6' ^Td) 2300 laoo 2000 Temperature la Degrees Fabr. Fio 121. — ^Total Heat of Iron above 500" F., Illustrating its Approximation to a Straight Line Relation in Spite of Wide Variation in Specific Heat Given in Fig. 120. temperature up to the highest used in steam practice and which is designated in steam tables, summarizing all the properties of water and steam, as the HEAT AND MATTER 423 > > / > A /^ ^ X X" ^ ^ •^ > ^ ^ . It » » N) » X) 400 600 1.15 I B-LIO «3 I rni^OS LOO Temperature Degrees F&hr. Fig. 122. — Specific Heat of Water at Various Temperatures. iOO 400 Temperature In Degrees Fabr. Fig. 123.— Total Heat of Water from 32*^ F., to any Temperature, tfie Heal of the Liquid at that Temperature above 32** F. 424 ENGINEERING THERMODYNAMICS heat of the liquid. For the purpose of comparison, the mean specific heat of water is given in Fig. 124 from 32** F. to any temperature which is obtained from the heat of the liquid above 32® F. by dividing it by the temperature above 32** F. In the table of specific heats of liquids there is a column giving the value calculated from the atomic weights to show at a glance the degree with which liquids satisfy the Dulong and Petit law. Variability of specific heat is especially noticeable in liquids that are solu- tions with different amounts of dissolved substance, in which case the specific heat varies with the density and temperature. Problems of refrigeration involve four cases of this kind: (a), calcium, and (b), sodium chloride, XO 400 Temperature in Degrees Fahrenheit Fig. 124. — Mean Specific Heat of Water from 32** to any Temperature. brines, the densities of which vary considerably but which are used with but little temperature range, .seldom over 20® F. and often not over 5® F., (c), anhydrous ammonia and (d)y carbonic acid. As the density of brines is often reported on the Baum6 scale and liquid fuels always so, a comparison of this with specific gravities is given in Table XXXIII in connection with the specific heat tables at the end of this Chapter to facilitate calculation. One of the best-known solutions so far as accuracy of direct experimental data is concerned, is calcium brine, results for which, from 35® C. to 20® C. given below, are from U. S. Bureau of Standards BuUetui by Dickinson, Mueller and George, for densities from 1.175 to 1.250. • For chemically pure HEAT AND MATTER 425 calcium chloride in water, it was found that the following relation be- tween density D, and specific heat (7, at 0® C, Z)=2.8821-3.6272C+1.7794C2, (612) and these results plotted in Fig. 125 show the specific heat variation with temperature to follow the straight line law very nearly. This being the case the mean specific heat for a given temperature range is closely enough the arithmetical mean of the specific heat at the two limiting temperatures. To the figure are also added dotted, the specific heats for some commercial brines, not pure calciiun chloride, but carrying magnesium and sodium chloride of density 1.2. It might be conveniently noted here that the relation between freezing- point and density for pure calciiun chloride by the same bulletin is given in Table XVIII below: Table XVIII FREEZING-POINT OF CALCIUM CHLORIDE . U. S. Bureau of Standards Density of Solution. Per cent CaClt by Wt. Freesing-point, Freeaing^point, 1.12 14.88 - 9 16.8 1.14 16.97 -13 8.6 1.16 19.07 -16 3.2 1.18 21.13 -20 - 4.0 1.20 23.03 -24 -11.2 1.22 24.89 -29 -20.2 1.24 26.77 - -34 -29.2 1.26 28.55 -40 -40.0 Other values for the specific heats of brines as commonly used are given m Table XIX, the accuracy of which is seriously in doubt and whigh may be checked by more authoritative values at different points where deter- minations have been made. Anhydrous ammonia liquid, has a variable specific heat with temperature, but the exi)erimental values are too few to make its value and law quite certain. Several formulas have been proposed, however, that tend to give an impression of accuracy not warranted by the facts though quite convenient in preparing tables. Specific heat of NH« liquid at t^ F. 1.0135+.00468 (<-32) (o) ) 1.118 +.001156 (t-32) (6) Authority Zeuner Dieterici Wood Ledoux 1.1352+.00438 (^-32) (c) 1.0057+.00203 (t-32) (d) (613) 426 ENGINEERING THEBMODYNAMIO0 Temperature In Degrees Fahr. Fig. 125. — Specific Heat of Calcium Chloride Brine of Various Densities D at Temperatures -10*' F. to +70^ F. HUAT AND MATTER 427 Table XIX SPECIFIC HEAT OF SODIUM CHLORIDE BRINE Density, B6 8p.gr. 1 Per cent NaCl byWi. Sp. Heat. Temp. F. Authority. 1 1.007 1 1.6 4.9 5.0 10.0 10.3 10.3 11.5 12.3 15.0 18.8 18.8 20.0 24.3 24.5 25 .992 .978 .995 .960 .892 .892 .912 .887 .871 .892 .841 .854 .829 .7916 .791 .783 -0 64.4 66-115 -0 -0 59-120 59-194 61-126 64.4 -0 63-125 68-192 64-68 64 Common Thomsen WinVelmftnn 5 10 1.037 1.073 Common Common Teudt Teudt Marignac Winlf«'n)ftnn 15 1.115 Common Teudt Teudt 19 1.150 Common WinkAlrnann Thomsen '23 " 1.191 Common From these expressions the mean specific heat follows by halving the coefficient of (<— 32) F., and these were determined and plotted to scale^ together with some direct experimental values of Drewes, in Fig. 126. Giving greatest weight to Drewes and Dieterici, a mean curve shown by the solid line is located as the best probability of the value for liquid anhydrous and it has the Eq. (614). trSSfL'm^^^^^^ ! =1.07+.00056(.-32).. (614) From this value the heat of liquid ammonia above 32° F. has been determined and is presented graphically in Fig. 127 from which, and the equation, the tabular values at the end of the Chapter were determined. Anmionia dissolved in water, giving an aqueous solution as used in the absorp- tion refrigerating sjrstem, has a nearly constant specific heat so closely approxi- mating unity as shown by Thomsen, who gives 3 per cent NH3 in water solution, sp.ht., 1.8 per cent NH3 in water solution, sp.ht., .9 per cent NH3 in water solution, sp.ht.. .997, at 66° F. .999, at 66° F. .999, at 66° F., that it is customary in these calculations to ignore any departure from unity, the value for water. Liquid carbonic acid, another important substance in engineering, especially in mechanical refrigeration, is less known as to its specific heat than is ammonia, and that is much too uncertain. There is probably nothing better available at present for the necessary range than the results of Amagat and Mollier, reported by Zeuner for the heat of the liquid, which are reproduced in Fig. 128, and used in the table at the end of this Chapter. 428 ENGINEERING THERMODYNAMICS It is, however, with gases that the most complex situation exists with respect to specific heats. As iias already been pointed out, gases may be heated at Li tz; I I L2 1.0 99 .8 B -60 AA-Zeuner BB-Ledouz CO -Wood DD-Dieterioi E£ Drewes FF - Mean Used in Book Temperature in Degrees Fah^ 150 Fig. 126. — Mean Specific Heat of Liquid Anhydrous Ammonia from —50*" F. to 150"* .F GO 100 Temperature in Degrees Fahr . Fig. 127. — Heat of Liquid Anhydrous Ammonia above — 50** F. constant volume, doing no external work while being heated, or at constant pressure, in which latter case work is done by expansion of the gas against the resisting constant pressure. Therefore, there must be two difiFerent specific HEAT AND MATTER 429 heats for each gas, one Cp at constant pressure and the other C« at constant volume, the difference between them representing the heat equivalent of the ■ 1 50 / 1 / » / • fa' o < / / a* / o / / w / / / / / / / / / / y / -a / / 26 s s * 6 % Temperature in Degreea Fahr. Fig. 128.— Heat of Liquid Carbonic Acid above 32** F. work of expansion done during the rise of temperature. Most experimental determinations of the specific heats of gases have been made at constant pressure 430 ENGINEERING THERMODYNAMICB and the constant volume value found from established relations between it and other physical constants. These relations most commonly used are two, Eq. (615) connecting the difiFerence with a constant R and the other Eq. (616) 777.52(Cp-C.) = fi, (615) §f-T (616) connecting their ratio to a constant y. These constants have each a special significance that may be noted here and proved later, thus R is the ratio of the PV product of a pound of gas to the absolute temperature, and t the particular value taken by the general exponent s in PV'^Cj when the expansion represented takes place with no heat addition or abstraction, i.e., adiabatic, it is also a function of the velocity of sound in gases. Table XXXIV at the end of this Chapter gives some authentic values, with those adopted here designated by heavy type. YanabiUty of specific heats of gases and vapors is most marked and of some engineering importance, because so many problems of practice involve highly heated gases and vapors, the most common being superheated steam and the active gases of combustion in furnaces, gas producers and explosive gas engines. In f act^ with regard to the latter it may be regarded as quite impossible with even a fair degree of accuracy to predict the temperature that will result in the gaseous products from the liberation of a given amount of heat of combustion. The first fairly creditable results on the variability of the specific heats of gases of combustion at high temperatures were announced by Mallard and LeChatelier, Vieille and Berthelot, all of whom agree that the specific heat rises, but who do not agree as to the amount. A general law was proposed by LeChatelier, giving the specific heat as a function of temperature by an equation of the following form: Specificheatatr F., (F=C), = C,=a+b(«-32), (617) in which a = specific heat at constant volume at 32** F. This yields, Mean-specific heat from 32.^F.,tor F., (F=C), ^=C'. = a+|(i-32) (619) The specific heat at constant pressure is obtained by adding a constant to the value for constant volume according to HEAT AND MATTEB 431 whence Specific heat at t" F., (P=C), = C,=a4 R 777.52 f6(<-32), (621) B.T.U. per lb. from 32°F.,to«''F., (P=(7), =Q 32 to t '[' f ^^+|(<-32)](<-32) . . (622) [Mean specific heat from \ ^r' —n\ ^ \^ff Q9^ rao'Vi i32^F., to <^F.,(P-0, J ''"■^777.52+2^^"^^^ ^^^"^^ The values of these constants have been determmed by LeChatelier, Clerk, Callender, and Holbom and Austm, from which the following values are selected. Table XX SPECIFIC HEAT CONSTANTS, GASES, Gm. o . R *'"*"777.52 h b 2 Authority. CO, .1477 .1944 .000097 .0000484 LeChatelier CO, .2010 .0000824 .0000412 Holborn and Austin N, .170 .2404 .0000484 .0000242 LeChatelier N, .2350 .000021 .0000106 Holborn and Austin to 2606* F. N, .2350 .0000208 .0000104 Callender 1644* F. to 2440* F. 0, .1488 .2126 .0000424 .0000212 LeChatelier H,0 .3211 .000122 .000061 LeChatelier Air .2431 .000135 .0000675 CaUcnder (1544* F. to 2440* F.) For purposes of comparison the following curves are plotted, showing all these results of specific heat at constant volume, at temperature t^ F., the total heat above 32^ F. per poxmd of gas, and the mean specific heat from 32^ F. to r F. in Fig. 129. Probably there is now more known of the specific heat of superheated steam than of any conunon gaseous substance, and it is likely that other substances will be found in time to have somewhat similar characteristics. Pure computa- tion from the laws of perfect gases indicates that the specific heat of gases or superheated vapors must be either a constant, or a function of temperature only, and this is what prompted the form of the LeChatelier formula. Bold experimentation on steam, disregarding the law, or rather appreciating that superheated steam is far from a perfect gas, principally by Knobloch and Jacob and by Thomas, showed its specific heat to be a function of both pres- sure and temperature. Results were obtained that permitted the direct solu- tion of problems of heat of superheat, or the heat per pound of vapor at any temperature above that at which it was produced, or could exist in contact with the liquid from which it came. Critical study of various results by Marks and Davis led them to adopt the values of Knobloch and Jacob with slight modifi- 432 ENGINEERING THERMODYNAMICS ■«n"X"a ^] 88 ©AOQB SBO JO punoi jad ^toh 3 a o "S o CO a 2 O o CO s o «5 o to c § O eS eg o ■ HEAT AND- MATTER 433 •gfl'X'arnonBJTHiig aAoqB niBois JO pnno^i Jaj ^bsh a «^ ts s :i| § 3 i - » g OS eg 1 T s. S '^. s 434 ENGINEERING THERMODYNAMICS cationS; for which evidence was in existence, raising the specific heats at low pressures and temperatures, and their conclusions are adopted in this work. In Fig. 130 is shown (A) the Marks and Davis modification of the Cp curve of Knobloch and Jacobs, the integral of which (C) gives the heat of superheat from any temperature of steam generation to actual steam temperature, while (B) shows the values for the mean specific heat above the temperature of satura- tion for the particular pressure in question. When substances of the same class are mixed so that toi, W2, wz^ etc., lbs. of the diflFerent substances having specific heats Ci, C2, C3, etc., or Cpi, C,2, C^f etc., or Cfi, C92, C,3, etc., then the specific heat of the mixture is given by wi+W2+wz+eic. ' ^ ^ C9iWi+C^ W2+C^W3 +etc. ,^25A wi+v>2+wz+eiQ, ' ^ • C piwi + Cp2W2 + Cpzws +et c. ,^^^. tp= ; : j — 7 [oZb) Example. If 5 lbs. of olive oil at a temperature of 100° F., 10 lbs. of petroleum at a temperature of 150° F., and 50 lbs. of water at 50° F. are mixed together, what will be the resultant temperature and how much heat will be required to heat the miA- ture 100° above this temperature? Sp. ht. of olive oil « .4, Sp. ht. petroleum » .511, Sp. ht. water =1.000. Let x-the final temp. The heat given up by the substances falling in tempera- ture is equal to that gained by those rising, hence 50(x-50) XI -5(100-0:) X.4+10(150-x) X.511, 50x -2500 -200 -2a: +766 -5.11a:, 57.11a: =3466, or, a: =60.7° F., Sp.ht. of nuxture- ; , from Eq. (611), 5X.4-H0x.511-f50Xl 57.11 ._. ^ z z~z z~^ ~" A_ ^ .o7oO« 5-1-10-1-50 65 whence the heat required will be 65X.8786 = 57 B.T.U. Prob. 1. To change a pound of water at 32° F. to steam at 212° F. requires 1150.4 B.T.U's. If the same amount of heat be given to a cubic foot of each of the following subsfahces at 32° F., what will be final temperature in each case? (a) cop- per; (6) iron; (c) mercury; (d) clay; (e) stone. HEAT AND MATTER 435 Prob, 2. How many pounds of the following substances could be wanned 10® F. )y the heat required to raise 100 lbs. of water from 40° F. to 200** F.? (a) Ethyl alcohol from 100** F.; (6) Sea water from 60° F., (density = 1.045); (c) Glycerine from 60° F; (d) Tin from 480° F. Prob. 3. If 150 lbs. of water at 200° F. are added to a tank containing 200 lbs )f petroleum at 70° F., what will be the resultant temperature, neglecting any heat ibsorbed or given up by the tank itself? Prob. 4. To melt 1 lb. of ice requires 144 B.T.U. How much would this lower :he temperature of 1 lb. of the following substances (1) at constant pressure; (2) at »>nstant volume; (a) air; (6) oxygen; (c) ammonia; (d) hydrogen; (e) nitrogen? Prob. 6. What would be the specific heats of the following mixture? Hydrogen ) lbs., oxygen 1 lb., nitrogen 7 lbs., carbon dioxide 20 lbs., carbon monoxide 10 lbs.? Prob. 6. Air is approximately 77 per cent N2, and 23 per cent O2 by weight. By means of the specific heats of the components, find its specific heats at constant pres- sure, and at constant volume. Prob. 7. By means of the specific heats, find the values of R and y most correct at atmospheric temperature (60° F.) for, hydrogen, air, carbon dioxide, carbon monoxide and nitrogen. Prob. 8, How much water could be heated from 40° F. to 60° F. by the heat needed to superheat 10 lbs. of steam at 200 lbs. per square inch absolute to 700° F.? Prob. 9. A building containing 250,000 cu. ft. of space is heated by a hot-water system. Considering the air to change eight times per hour, how many pounds of water per hour must be circulated if the drop in temperature of the water is from 200° to 100° and the temperature of the outside air is 30° F. while that of the room is 60° F. neglecting wall conducted heat? Prob. 10. How much heat would be required to warm a pound of liquid COi from zero to 80° F.? Compare with water and ammonia. 6. Volume or Density Variation with Temperature of Solids, Liquids, Gases and Vapors, Not Changing State. Coefficients of Expansion. Coefficients of Pressure Change for Gases and Vapors. Solids increase in length or in any linear dimension, a certain fraction of their original length for each degree temperature rise and the expansion is usually assumed to be in proportion to temperature rise. The relation between original and final length can be set down in an equation involving the coefficient of expansion. Let a = coefficient of linear expansion = fractional increase in length per degree. '^ h and <i= original length or any other linear dimension and the cor- responding temperature; '' I2 and <2= length which h becomes after heating and the corresponding temperature. Then Increase in length = Z2—ii = aZi(te—<i), (627) New length Z2 = ii+aZi(fa — ii), = li[l+a{t2-ti)] (628) 436 ENGINEERING THERMODYNAMICS Solids, of course, expand eubicaJly and the new volume will be to the old as the cubes of the linear dimension. Let a = coefl5eient of volumetric expansion; t;i= original volume; t;2= final volume after heating. Then when the temperature rises one degree, ^=(|y = (l+a)3 = l+3a+3a2+a3 = l+a . . . . (629) If a is small, and it is generally less than j^j then the square and cube eao be neglected in comparison with the first power, whence l+a = l+3a and a = 3a. so that the coefficient of volumetric expansion may be taken as sensibly equal to three times the coefficient of linear expansion, and similarly, the coefficient of surface expansion as twice the coefficient of linear expansion. Liquids, by reason of the fact that they must always be held in solid containers, may be said to have no linear expansion, and therefore, although the expansion may be one direction only, the amount is due to the total change of volume rather than the change of length along the direction of freedom to expand. The same is true of gases, so that for gases and liquids only coefficients of volumetric expansion are of value and these are given in the Tables XXXV, XXXVI, XXXVII, and XXXVIII at the end of this Chapter. With liquids and gases it is usual to take the volume at 0® C. or 32° F. and 29.92 ins. Hg pressure as a standard, and the coefficient gives the increase sls a fraction of this, per degree departure from the freezing-point. This is the universal practice with gases. It appears that the coefficients of expansion for solids are quite different from one another, ranging from over 15X10"* for wax, to.085XlO~* for Jena normal glass, a range of over two hundred and sixty times. Detennina- tions of the value at various temperatures for any one substance indicate a variation with temperature, which proves that proportionality of increase of dimensions to temperature rise, does not hold true, a fact which has led to formulas of the form I2^lill+x{t2-ti)+y(t2-ti)^, the value of which is dependent on the determination of the constant and veri- fication of correctness of form, which has not by any means been conclusively done. For most engineering work the constant values nearest the temperature range will suffice except for certain liquids, vapors, and gases. A more marked tendency to follow such a law of variation with temperature is found with liquids and coefficients for some are given in the standard physical tables. HEAT AND MATTER 437 The two important liquids, mercury and water^ have been separately studied in greater detail and the latter exhibits a most important exception to the rule. For mercury, according to Broch t;2 = vi(l+.00O455e+54Xl0-i2i2+602Xl0-i*^), . . . (630) which exhibits a refinement of value only in instnunent work such as barometers and thermometers. Water, as already mentioned, has its maximum density at 39.1^ F. and expands with both fall and rise of temperature. Its expansion is given by a similar formula by Scheel, as follows: t;2=t;i(l-.03655Xl0"^<+2.625Xl0^®^2_ii61^). . . (631) Most commonly the expansion of water is not considered in this^way, but by comparing densities at varying temperatures, and all sets of physical tables contain values which in this work are significant only as affecting the change of volume in turning water to steam and siLch values as are needed are incorporated in the steam tables later. The study of the expansion of gases and vapors at constant pressure, and rise of pressure at constant volume, per degree has perhaps been fairly com- plete and is of greatest significance, because from it most of the important laws of thermodynamics have been derived. This work may be said to have started with the Regnault air and gas thermometer work, already described. Some of the authentic values collected in the Landolt, Bomstein, Myerhoflfer, and Smithsonian Physical Tables, are given at the end of this Chapter, where oe, is the coeflicient of pressure change at constant volume, and a, the coefficient of expansion, or volume change at constant pressure. The remarkable thing about the coefficients for these gases and vapors is the approach to constancy for most of the gases, not only of the coefficients of expansion for P = c nor the similar constancy of the coefficients of pressure rise for F=c, but more remarkable than either of these is the similarity of the two constant coeffir- dents. These facts permit of the generalizing of effect when P^c, and when F=c, and of the announcement of a law by means of which ail such problems can be solved instead of applying separate coefficients for every substance and every different temperature necessary for solids and liquids where, for example, the maximum coefficient was over 260 times as great as the least. The average coefficient for all gases, applying both to pressures and volumes, is the same as enters into the gas thermometer work and its best value is found to be a = VqYko ^ -^2034, per degree F. a = 27^13 = 003661 , per degree C. • , .... (632) 438 ENGINEERING THERMODYNAMICS and approximately a = 2^2 = .00203, per degree F. a = 070 == .00366, per degree C. . (633) These are the same as the reciprocals of the absoliUe temperature of i}te ice-meUing point, and are but expressions of conditions for reduction of the volume and pressure at the ice-melting temperature to zero by constant pressure and constant volume abstraction of heat respectively, and by stating the amount of reduction per degree give by implication the nimibcr of degrees for complete reduction, j Example. The rails on a stretch of railroad are laid so that they just touch when the temperature is 120** F. How much total space will there be between the rails per mile of track at 0** F.? For wrought iron a will be nearly the same for Bessemer steel = .00000648. Hence the linear reduction in 5280 ft. for a change of 120° F. will be 5280 X 120 X. 00000648*4.1 ft. Prob. 1. A steam pipe is 700 ft. long when cold (60** F.), and is anchored at one end. How much will the other end move, if steam at a temperature of 560** F. is turned into the pipe? Prob. 2. A copper sphere is one foot in diameter at 50** F. What must be the diameter of a ring through which it will pass at a temperature of 1000° F.? Prob. 3. A hollow glass sphere is completely filled with mercury at 0° F. What per cent of the mercury will be forced out if the temperature rises to 300° F.? Prob. 4. A room 100 ft.XoO ft.XlO ft. is at a temperature of 40° F. The tem- perature rises to 70° F. How many cubic feet of air have been forced from the room? Prob. 6. The air in a pneumatic tire is at a pressure of 90 lbs. per square inch gage and at a temperature of 50° F. Due to friction of the tire on the ground in running, the temperat re rises to 110° F. What will be the pressure? Prob. 6. A brick lighthouse is approximate y 200 ft. high. Should it be exactly this at 0° F., what would it be at 100° F.? Prob. 7. Show that if a glass tube is rigidly held at each end by brackets attached to an iron tank it will break if the tank is warmed. Prob. 8. From Eq. (618) find the density of water at 60° F., 100° F., 212° F., and compare with the values in the steam tables. Prob. 9. A drum containing COs gas at a pressure of 250 lbs. per square inch gage is raised 100° F. above its original temperature. What will be the new pressure? 7. Pressure, Volume and Temperature Relations for Gases. Perfect and Real Gases. Formulating the relations between the pressure change at constant volume and the volume change at constant pressure, Let P and V be the simultaneous pressure and volume of gas; *' < be its scale temperature at the same time, F.; " r be its absolute temperature at the same time, F. ^ HEAT AND MATTER 439 Then at constant volume the pressure reached at condition (a) after heating from 32** F. is given by Pa — -P32* = 409^32 X (4i — 32) . Pa = -P32* L ^ 492 I Similaxly for another temperature U, the pressure will be Pt=Pz2' L^ 492 J- Whence or Similarly 1 4.??r_32 Pa^ 49 2 ^492--32+<a^<a+460 P* ti,-S2 492-32+fe fo+460' ■*■ 492 §5=^, for 7 constant, (634) ^b -lb V T 7~=^, for P constant (635) Vb lb Both EJqs. (634) and (635) are true, for no gas all the time, but very nearly true for all, under any range of change, and a hypothetical gas is created for which it is exactly true all the time^ known as a perfect gaSy about which calculations can be made as would be impossible for real gases and yet the results of which are so close to what would be the result with real gases, as to be good enough for engineering practice. Therefore, with a mental reservation as a guard against too great confidence in the work, all real gases will be assumed perfect and to follow Eqs. (634) and (635) except when experience shows the results are too far wrong to be useful. These laws, known by the names both of Charles and Gay-Lussac, are closely associated with another also doubly named as Boyle's or Mariotte's and like- wise an idealization of experimental observations known to be nearly true for all gases. This is to the effect that so long as temperatures are kept constant the pressures of gases vary inversely as their volume, or that. Pa Vb -^^T7, and, PaVa'^PbVb^^Qon&i^iiiy for T constant . (636) i^b y a Study of the PY product, for various gases has revealed a good deal on the general properties of matter, especially as to the transition from one state to another. This is most clearly shown by curves which may be plotted in two 440 ENQINEEBINa THERMODYNAMICS 1 • «« «=>. I ^1 a i W/l /llll /in "5 / /m / f J jJX- r 6 y X • u ^ ^ ^ "5 — — ..^ "^ A 9 — — =S ^2 ^ • > i r 1 r^ Ml3 9om^V U|g Bdj 1S6^ *d r ~ _ U • ^ ^ X 1^ d ^ ^ ^ i j^ ^ u c • ■ I - 1 ^ s ^ ^ \ \ L t3 ^ •1 g ^ ^ \ V \ \ L »: -^ ^ « 1^ i ^ \ Sr- \ \ A \ — (MM . .^ — — K 218 \ V \ \ \v II • ^ •^ V \ > V^ s.! ^ N \ Ov - -5 ^ ■^ 53 N ^^N ' 1 ^ ^ V. V x> ^ g ^ s s — ^ O -^ - < > ^ ■^ a. *^ ^ ■ — ' — ' ■"■^ m ^^ Job HEAT AND MATTER 441 ways. To coordinates of pressure and volume a family of equilateral hyper- bolas one for each temperature, would represent the true PF=C or isoiAermaZ relation and any variation in the constancy of the product would be shown by its departure from the hyperbola. Still more clearly, however, will the depart- ure appear when the product PV is plotted against pressures, constancy of product would require all lines to be straight and inconstancy appear by departures from straight lines. To illustrate, the data from Young for car- bon dioxide are plotted both ways in Fig. 131, from 32° F. to 496° F., the values of PV at 32° and 1 atm. are taken as unity on one scale. It appears th^t up to the temperature of 88° F, known as the critical temperature, each isothermal plotted to P and PV coordinates consists of three distinct parts: (a) a curved line sloping to the right and upwards; (6) a straight line nearly or exactly horizontal; (c) a nearly straight line sloping upward rapidly and to the left. In this region then the isothermals are discontinuous, and this is caused by the liquification or condensation of the gas, during which increase of pressure, produces no change of volume, provided the temperature is low enough. It also appears that each PV line has a minimum point and these minima joined result in a parabola. At the end of this Chapter are given in Table XXXIX the values of PV at three different temperatures and various pressures for oxygen, hydrogen, carbon dioxide and ammonia, in terms of the values at 32° and 1 atm. for further comparison and use. Further study along these hues is not profitable here and the topic while extremely interesting must be dropped with the observation, that except near the point of condensation or liquefaction, gases or vapors, which are the same thing except as to nearness to the critical state, follow the Boyle law closely enough for engineering purposes. None of these approximate laws, Eqs. (634), (635) and (636) can be con- sidered as general, because each assumes one of the variables to be constant, but a general law inclusive of both of these follows from further investigation of a fixed mass of gas suffering all sorts of pressure volume and temperature changes, such as occur in the cylinders of compressors and gas engines. A table of simultaneous experimental values of pressure, volume, and temperature, for any gas will reveal the still more general relation inclusive of the preceding three as follows: Paya _ Pbyb _ Py _f^ /AQTN m "~ m ~" /T7 ~^^> woi; in which Cg is approximately constant for any one gas and assumed constant for perfect gases in all calculations. For twice the weight of gas at the same pressures and temperatures Cg would be twice as large, so that taking a constant R for one pound, and generally known as the ** gas constant," and introducing a weight factor u;, the general characteristic equation for the perfect gas is, PV^wRT (638) 442 ENGINEERING THERMODYNAMICS This general law may be derived from the three primary laws by imagining in Fig. 132, two points, A and B, in any position and representing any two states of the gas. Such points can always be joined by three lines, one constant A X \ .^ Y B pv Y Diaamim to derive Law ^=Cq Fig. 132. — Curve of ContinuouB Relation between P, F, and T for Gases. pressure A to X, one constant temperature X to F, and the other constant volume F to B. For these the following relations hold, passing from A to B But and whence or Passing to B, y w" y xm • la F,= r,=r, Vf Va- _ y P» To " ^'Pa Ty' PaVa _P,V, - Ty ■ P,= .pTy HEAT AND MATTER 443 But T " Ta Ty Ty Th ' PV -jp- = constant. = wR or in general when the weight of gas is w lbs. For numerical work, the values of R must be fixed experimentally by direct measurement ojf simultaneous pressure, volume, and temperature, of a known weight of gas or computed from other constants through established relations. One such relation already mentioned but not proved is fl = 777.52(Cp-C.) (639) It is extremely imlikely that the values of R f oimd in both ways by a multi- tude of observers under all sorts of conditions should agree, and they do not, but it is necessary for computation work that a reasonable consistency be attained and that judgment in use be cultivated in applying inconsistent data. In the latter connection the general rule is to use that value which was determined by measurement of quantities most closely related to the one being dealt with. Thus, if jB is to be used to find the state of. a gas as to pressure, volume, and tem- perature, that value of R determined from the first method should be selected, but the second when specific heats or Joule's equivalent are involved. Of coiu^e, a consistency could be incorporated for a perfect gas, but engineers deal with real gases and must be on guard against false results obtained by too many hypoth- eses or generalizations contrary to the facts. Accordingly, two values of R are given in Table XL, at the end of this chapter, one obtained from measure- ments of specific heats at constant pressure and determinations of the ratio of specific heats unfortunately not always at the same temperature and gen- erally by different people, and the other by direct measure of gas volume at standard 32° F. temperature and 1 atm. pressure. These measurements are separately reported in Sections (5) and (8), respectively. If a gas in condition A, Fig. 133, expand in any way to condition B, then it has been shown that in which « has any value and which becomes numerically fixed only when the process and substance are more definitely defined. Comparing the temperatures at any two points A and jB, it follows that 1 ^E^ and T.=^*Z? « wR' ^"^ ^' wR' 4AA X. X X whence ENGINEERING THERMODYNAMICS n p.n But and 1 a -to ^ a whence Ta~\Vj ' (640) A • I v \ V • « -^ B V Fio. 133. — Expansion or Compression of Gas between A and B, Causing a Change of Tem- peratiure. and (641) Eqs. (640) and (641) give the relation between temperatures and volumes But Vg _ TgPo TbPg which, substituted in above, gives or ©■-©■"■• HEAT AND MATTER 445 and rArf <««> or k-iW'"- ■ • • <'^> Eqs. (629) and (630), give the relation between pressures and temperatures. It is convenient to set down the volume and pressure relations again to complete the set of three pairs of most important ga& equations. Pa \Vj (645) These are perfectly general for any expansion or compression of any gas, but are of value in calculations only when s is fixed either by the gas itsel/ or by the thermal process as will be seen later. Example. A pound of air has a volume of 7.064 cu.ft. at a pressure of two atmos-. pheres and a temperature of 100** F. Find the value of R for air from the data; also the final volume and temperature if expansion bccurs so that «*«1.4 until the pressure becomes i an atmosphere. PV^wRT, or 2116X2X7.064 = 1 Xi2x560, or fi«63.38, »-l .4 ^(g) •-<«"-'•«• .-. T, = ^1 -^1.49 = ^7^ =352 abs. = -108** F. 1.49 HW-^^- h^[^y =2.7, or 7, =2.7 7i =19.1 cu.ft. Prob. 1. A perfect gas is heated in such a way that the pressure is held constant. If the original volume was 10 cu.ft. and the temperature rose from 100^ F. to 400** F., what was the new volume? Prob. 2. The above gas was under a pressure of 100 lbs. per square inch gage at the beginning of the heating. If the volume had been held constant what would have been the pressure rise? Prob. 3. A quantity of air, 5 lbs. in weight, was found to have a volume of 50 cu.ft. and a temperature of 60** F. What was the pressure? Prob. 4. A cylinder holding 12 cu.ft. has a pressure of 250 lbs. per square inch gage, and the temperature is 50*^ F. What would be the weight of its contents were it filled with (a) CO,; (6) NH,; (c) Oxygen; (d) Hydrogen? Cp. C 3.409 2.412 .217 .1535 .2175 .1551 .2438 .1727 446 ENGINEERING THERMODYNAMICS Prob. 6. At a pressure of 14.696 lbs. per square inch and a temperature of melting ice, one pound of air has a volume of 12.387 cu.ft. From the data find the value of R for air. The specific heats of air are given by one authority as Cp = .2375 and C» = .1685. Find R from the data and see how the two values obtained compare. Prob. 6. At the same pressure and temperature as in Prob. 5, a pound of the following substances has a volume as shown. From the data and the valu^ of specific heats, find R by the two methods. Substance. Cu.ft. per lb. Hydrogen 178.93 Carbon dioxide ..... 8.15 Oxygen 11.21 Nitrogen 12.77 Prob. 7. 5 cu.ft. of gas at a pressure of 3 atmospheres absolute and a tempera- ture of 50** F. expand to atmospheric pressure. What will be the final volume and temperature, if s = 1.35? Prob. 8. 1000 cu.ft. of gas at atmospheric pressure and 60** F. are compressed into a tank of 100 cu.ft. capacity. What will be the pressure in the tank and the temperature of the gas at the end of the process, if the gas is COj and the com- pression adiabatic? Prob. 9. What will be the final volume, pressure and temperature, if a pound of air at atmospheric pressure (14.7 lbs. per square inch) and a temperature of 60® F. be compressed adiabatically until its absolute temperature is six times its original value? 8. Gas Density and Specific Volume and its Relation to Molecular Weight and Gas Constant. The density of a gas is best stated for engineering purposes as the weight of a cubic foot, but as this becomes less on rise of temperature or decrease of pressure it is necessary to fix a standard condition for reporting this important physical constant. It is best to take one atmosphere 760 mm. or 29.92 ins. of mercury as the pressure, and 0® C. = 32°F. as the standard temperature, though it is in some places customary in dealing with commercial gases, such for example as those used for illumination, to take the temperature at 60^ F. and illuminating gas at this condition is often known among gas men as standard gas. In this work, however, the freezing-point and standard atmosphere will be tmderstood where not specifically mentioned, as the conditions for reporting gas density and its reciprocal, the specific volume of gases or the cubic feet per pound. The chart, Fig. 134, shows the relation of volume and density at any pressure and temperature to the volume and density under standard conditions. These constants have been pretty accurately determined by many investi- gators, whose figures, to be sure, do not agree absolutely, as is always the case in experimental work, but the disagreement is found only in the last significant figures. Some selected values of reliable origin are reported at the end of this Chapter in Table XLI for the important gases and these may be used in computation work. HEAT AND MATTER 447 It often happens in dealing with gases and especially superheated vapors that a value is needed for which no determination is available, so that general Pressure in Founds Per Sq. In. Abs. U 13 la 11 4) 1 1 1 1 M» 1 1 1 1 1 1 u 1.1 1 Temperature « uegreeB Fahr. I I I 1 1 1-90 I I I I 1^ I I I I I t I I I I I .7 I I I I I II I '.A I 1^8 I I I I ^ ^„, ^^.. Density at 82 A 29.W^^ _ _ , „ ^, Volume at 32"^ 29.98'' Upper Soale =Batlo Density at any T » P I^^er Sc^le=Ratio v,.imi»^ atanv T^p Inner Scale - Ratio Volu5^aJ_321R ^^^^ ^^^ ^ ^^^.^ Den«lt_y at^2^ Volume at any T Density at any T Fig. 134. — Equivalent Gas Densities At Different Pressures and Temperatures. laws of density or specific volumes of substances are necessary to permit the needed constant to be estimated. These relations may be applied to vapors 448 ENGINEERING THERMODYNAMICS as well as to gases even though the standard conditions are those for the liquid state, on the assumption that all gases and vapors will expand under temperature, or contract imder pressure rise, to the same degree, retaining the same relative relations between all substances as exist at the standard atmosphere and freezing-point. A vapor thus reported below its point of condensation and assumed to have reached that condition from one of higher temperature at which it exists as vapor is often called steam gas, or alcohol gas, for example in the case of water and alcohol. Such general relations between the densities of gases as are so desirable and useful in practical work have been found by studying the manner in which gases chemically combine with respect to the volume relations before and after the reaction. Following several experimenters, who reported observed rela- tions, Gay-Lussac stated a general law, as follows: When two or more gaseous substances combine to form a compound, the vol- umes of the combining gases bear a simple raiio to each other and also to thai of the compound when it is also a gas. He also attempted to derive some relation between this law and Dalton's atomic combining law, which states that, in combining chemically, a simple numerical relation exists between the number of atoms of different elements which unite to form a compound. This was not successful, but Avagadro later foim^d the expected relation by assuming that it is a particle, or a number of atoms, or a molecule, that is important in combining, and the law stated is as follows: Equal volumes of different gases Tneasured at the same pressure arid tempera- ture contain the same number of molecules. It is possible by analysis of these two laws to get a relation between the volumes of gases and the weights of their molecules because the molecular relation of Avagadro, combines with the combining law of Gay-Lussac to define the rela- tion between the number of combining molecules. At the same time the weight relations in chemical reactions, based on atomic weights, may be put into a similar molecular form, since the weight of any one substance entering is the product of the number of its molecules present and the weight of the molecule. Applying the relation between the number of molecules derived previously, there is fixed a significance for the weight of the molecule which for simple gases like hydrogen and oxygen is twice the atomic weight and for compound gases, like methane and carbon dioxide, is equal to the atomic weight. Appl\dng this to the Avagadro law, the weights of equal volumes of different gases must be proportional to their molecular weights, as equal volumes of all contain the same number of molecules. Putting this in symbolic form and comparing any gas with hydrogen, as to its density, because it is the lightest gas of all and has well determined charac- HEAT AND MATTER 449 teristics, requires the following symbols, denoting hydrogen values by the subscript h. Let m= molecular weight of a gas, 8 = density in lbs. per cu.ft. =— , then 81 Vfll 8.-m,' (^6> and • ^=^ (647) But as the molecular weight of hydrogen is for engineering purposes equal to 2 closely enough and hydrogen weighs .00562 lb. per cu.ft. = Bjsr, at 32® F and 29.92 ins. Hg, Lbs. percu.ft. = 8i = .00281mi (648) To permit of evaluation of Eq. (648) it is necessary that there be available a table of molecular weights of gases and the atomic weights of elements from which they are derived, and the values given at the end of this Chapter in Table XLII are derived from the international table. As atomic weights are purely relative they may be worked out on the basis of any one as imity, and originally chemists used hydrogen as imity, but for good reasons that are of no importance here, the custom has changed to ^ the value for oxygen as unity. These atomic weights are not whole numbers but nearly so, therefore, for con- venience and suflScient accuracy the nearest whole number will be used in this work and hydrogen be taken as imity except where experience shows it to be undesirable. The reciprocal expression to Eq. (648) can be set down, giving the specific volume of a gas or its cubic feet per pound at 32® F. and 29.92 ins. Hg., as follows: Cu.ft.perlb.=^ = ;^l28r"i=^ (649) This is a most important and useful conclusion as applied to gases and vapors for which no better values are available, and in words it may be stated as follows: The cubic feet per pound of any gas or vapor at 32^ and 29,92 ins, Hg, is equal to 855,87 divided by its molecular weight, or The molecular weight of any gas or vapor in pounds j will occupy a volume of 355,87 cu,ft. at 32"* and 29,92 ins, Hg, The approach to truth of these general laws is measured by the values given for specific volume and density at the end of this Chapter (a) experiment- ally derived and, (b) as derived from the hydrogen value by the law. 450 ENGINEERING THEHMODYNAMICS Another and very useful relation of a similar nature is derivable from what has been established, connecting the gas constant R with molecular weights. The general law PV^wRT when put in the density form by making 8=- becomes J = ftr (650 Whence, comparing gases with each other and w^th hydrogen at the same pressure and temperature ll = h=^^ (651, P2 §1 R2 T2 — = ^, and oi=-p— (652) which indicate that the densities of gases are irwersely proportional to (hi ga^ cansiarUs, or the density of any gas is equal to the density of hydrogen times the gas constant for hydrogen divided by its own. Inserting the values of density at 32^ and 29.92 ins. Hg and of the gas con- stant for hydrogen, it follows that for any gas Lbs. per cu.ft. = Bi=-^5 — , (653) the reciprocal of which gives the specific volume at 32** F. and 29.92 ins. Hg, or Cu.ft. perlb. = Fi=^^ (654) Example. 1. Explanation and use of Chart, Fig. 134. This diagram is for the purpose of finding the cubic feet per pound, or pounds per cubic foot, of a gas at 32*^ F. and a pressiue of 29.92 ins. of Hg, if its volume or weight per cubic foot be known at any pressiire and temperature. The curves depend upon the fact that the pounds per cubic foot (8) vary directly as the pressure and inversely as the temperature. That is „ ^ T 29.92 032 , 29.92 = °5rp;^ "T"' The line of least slope is so drawn that for any temperature on the horizontal scale its value when divided by 492 may be read on the vertical scale. The group of lines with the greater slope is so drawn that for any value on the vertical scale this quantity 29 92 times — '- — may be used on the horizontal scale. That is, the vertical scale gives the HEAT AND MATTER 451 ratio of densities as affected by temperature for constant pressure, while horizontal scale gives the ratio as affected by both temperature and pressure. A reciprocal scale is given in each case for volume calculations. To find the pounds per cubic foot of gas at 32** F. and 29.92 ins. of mercury when its value is known for 90^ and 13 lbs. per sq.in. On the temperature scale, pass vertically until the temperature line is reached, then horizontally imtil the curve for 13 lbs. absolute is reached. The value on the scale below is found to be 1.265, so that the density imder the standard conditions is 1.265 of the value under known conditions. Had it been required to find the cubic feet per pound the process would be precisely the same, the value being taken from the lower scale, which for the example reads .79, or, the cubic feet per pound imder standard conditions is 79 per cent of the value under conditions assumed. Example 2. By means of the molecular weight find the density of nitrogen at 32^ F. and 29.92 ins. Hg, and the cubic feet per pound for these conditions. From Eq. (646) li mi ^ 28 X. 00562 — =— , or 8i"= . Oh f^H J Hence 8 for nitrogen = .07868 pounds per cu.ft. and, 1 1 8 .07868 12.709. cu.ft. perlb. Prob. 1. Taking the density of air from the table, find the value of R for air, by means of Eq. (653) and compare its value with that found in Section 7. Prob. 2. Compare the density of carbon monoxide when referred to 32° F. and 60° F. as the standard temperature, as foimd both ways. Prob. 3. By means of their molecular weights find the density of oxygen, nitrogen and carbon dioxide at 32° F. and 29.92 ins. Hg. Prob. 4. What are the cubic feet per pound of acetylene, methane and ammonia at 32° F. and 29.92 ins. Hg? Prob. 6. An authority gives the following values for R. Compare the densities found by this means with the densities for the same substance found by the use of the molecular weights. Oxygen 48.1 Hydrogen 764.0 Carbon monoxide 55.0 Prob. 6. What will be the volume and density under standard conditions, of a gas which contains 12 cu.ft. per pound at a temperature of 70° F. and a pressure of 16 lbs. per square inch absolute? Prob. 7. What will be the difference in volume and density of a gas when con- sidered at 60° and 29.92 ins. of Hg, and at 32° F. and 29.92 ins. of Hg? 9. Pressure and Temperature Relations for Vapor of Liquids or Solids. Vaporization^ Sublimation and Fusion Curves. Boiling- and Freezing-points for Pure Liquids and Dilute Solutions. Saturated and Superheated Vapors. Substances may exist in one of three states, solid, liquid or gas, the latter being generally called vapor when, at ordinary temperatures the conmion state is that 452 ENGINEEBING THERMODYNAMICS of liquid or solid, or when the substance examined is near the point of lique- faction or condensation, and just which state shall prevail at any time dep>end5 on thermal conditions. Within the same space the substance may exist in two of these three states or even all three at the same time under certain special condi- tioas. These conditions may be such as to gradually or rapidly make that part in one state, turn in to another state, or may be such as to maintain the relative amounts of the substance in each state constant; conditions of the latter sort are known as conditions of equilibrium. These are experimental conclusions, but as in other cases they have been concentrated into general laws of which they are but special cases. The study of the conditions of equilibrium, whether of physical state or chemical constitution, is the principal function of ph^^ical chemistry, in the pursuit of which the Gibbs phase rule is a controlling prin- ciple. According to this rule each possible state is called a phase, and the number of variables that determine which phase shall prevail or how many phases may exist at the same time in equilibrium for one chemical substance like water, is given by the following relation, which is but one of the conclusions of this general principle of equilibrium. Number of undefined variables =3— number of phases. Now it is experimentally known that if water be introduced into a vacuum chamber some of it will evaporate to vapor and that, therefore, water and its vapor may coexist or the number of phases is two, but this does not state how or when equilibrium will be attained. The rule above, however, indicates that for this case there can be but one undefined or independent variable and, of course, since the pressure rises more when the temperature is high than when low, the two variables are pressure and temperature, of which accordingly only one is free or independent, so that fixing one fixes the other. In other words when a vapor and its liquid are together the former will condense or the latter evaporate until either pressure or temperature is fixed, and fixing the one the other cannot change, so that the conditions of equilibrium are indicated by a curve to coordinates P and T, on one side of which is the vapor state and on the other that of liquid. Such a curve is the vapor pressure-temperature curve of the substance, sometimes called its vapor tension curve, and much experi- mental information exists on this physical property of substances, all obtained by direct measurement of simultaneous pressiu'es and temperatures of a vapor above its liquid, carefully controlled so that the pressiu'e or the temperature is at any time uniform throughout. The conditions of equilibrium between vapor and liquid, defined by the vapor tension curve extend for each substance over a considerable range of pressure and temperature, but not indefinitely, nor is the range the same for each. At the high-pressure and temperature end a peculiar interruption takes place due to the expansive eflfect of the temperature on the Uquid and the compressive eflfect of the pressure on the vapor, the former making Uquid less dense and the latter making vapor more dense, the two densities become equal at some pressure and temperature. The point at which this occurs is the " critical point " at which the equilibrium between liquid and vapor that previously existed, HEAT AND MATTER 453 ends and there is no longer any difference between vapor and liquid. This point is a most important one in any discussion of the properties of matter, and while difficult to exactly locate, has received much experimental attention, and some of the best values are given below in Table XXI for the pressure, density, and temperature defining it, for the substances important in engineering Table XXI THE CRITICAL POINT Substance. Hydrogen Ozysen Nitrogen Amntonia. Ammonia Carbon dioxide. . . Carbon dioxide. . . Water Water Water Water Water Water Symbol. Ht Ot Nt NHs NH; CO« COa HiO HsO HsO HsO HsO HsO Critical Temp. 0«C. -243.5 -118.1 -146. » +130.0 +131.0 + 31.35 + 30.021 +358.1 +364.3 +365.0 +374. +374.6 +374.5 O'^F. -390.1 -180.4 -232.8 266. 267.8 88.43 87.67 676.4 687.7 680. 705.2 706.3 706.1 Critical Pres- sures. Atm. 20 601 35.1 115. 113. 72.9 77.1 194.61 200.5 Lbs. per Sq.in. 294 735 515 1690 1660 1070 1130 2859 2944 3200 3200 Critical Density Water at 4«»C-1. ■ • • • 65* 44S 464 45> .429 Authority. Olssewski 1 Wroblewski *Dewar 1 Olssewski > Wroblewski Dewar Vincent and Chappuis Amagat 1 Andrews * Cailletet And Mathias Nadejdini Batteli Cailletet and Colardeau Traube and Teichner Holborn and Baumann Marks Criti- cal vol. Cu.ft. per Lb. 26.8 13. Authority. Nadejdini Batteli To illustrate this discussion there is presented the vapor tension curves of water, ammonia and carbon dioxide to a large scale in chart form derived from the tabular data both at the end of this Chapter, while a small scale dia- gram for water is given in Fig. 136. These data are partly direct experimental determinations and partly corrections obtained by passing a smooth curve representing an empiric equation of relation between pressure and tem- perature, through the major part of the more reliable experimental points. These pressure-temperature points are very accurately located for water, the first good determinations having been made by Renault in 1862 and the last by Holborn and Bamnann of the German Bureau of Standards in the last yoar. The data presented are those of Regnault corrected by various investigations by means of curve plotting, and empiric equations by Wiebe, ThieGsen and Schule, and those of various later observers, including Battelli, Holborn, Hen- ning, Baumann, Ramsay and Young, Cailletet and Colardeau, somes eparately, but all together as xinified by Marks and Davis in their most excellent steam 454 ENGINEERING THERMODYNAMICS tables, and later by Marks alone for the highest temperatures 400** F. to the critical point, which he accepts as being located at 706.1® F. and 3200 lbs. square 2800 S400 i a 2000 QQ f leoo I I 1200 800 400 / / / , / / / / / / ' / / / / \ / • • / • . / i i . 1 J w ate I* • / V ape r . i t / / / i / / / f y / [^ ^^^ u K) » X) U iO « iT Temperature in Degrees Ffthr. Fig. 136. — ^Vapor of Water, Pressure-temperature Curve over Liquid (Water). inch. In calculations the values of Marks and Davis, and Marks, will be accepted and used. HEAT AND MATTER 455 Carbon dioxide and ammonia are by no means as well known as steam, md the original data plotted, while representing the best values obtainable, must ye accepted with some uncertainty. A smooth curve Figs. (139) and (140) las been drawn for each through the points at locations that seem most fair, or both these substances and the values obtained from it are to be used in calculations; these curves have been located by the same method as used by Marks in his recent paper and described herein later. The equalized values ire given in the separate table at the end of the Chapter with others for latent 1 1 / / • cr CO / • OD ;3.05 a •pi* 00 2 • Ice 1 / / CO / A. ipor / • y .^ -^ '^ -e -36 -1( ) +15 440 Temperature in Degrees Fahr. Fig. 136. — ^Vapor of Water, Pressure-temperature Curve over Solid (Ice). heats and volumes, but while consistent each with the other are probably but little more correct than values reported by others which are inconsistent. The curves and the equivalent tabular data are most useful in practical work, as they indicate the temperature at which the vapor exists for a given pressure, either as formed during evaporation or as disappearing during con- densation, or the other way round, they indicate the pressure which must be maintained to evaporate or condense at a given temperature. Just as the vapor-Uquid curves indicate the conditions of equilibrium between 456 ENGINEERING THERMODYNAMICS vapor and its liquid, dividing the two states and fixing the transition pressure or temperature from one to the other, so also does a similar situation exist with respect to the vapor-solid relations. In this case the curve is that of " siA- limadon " and indicates the pressure that will be developed above the solid by direct vaporization at a given temperature in a closed chamber. In Fig. • 136 is plotted a curve of sublimation of vapor-ice, based on Juhlin's data, Table XXII, which indicates that the line divides the states of ice from that of vapor so that at a .constant pressure, decrease of temperature will caus* vapor to pass directly to ice and at constant temperature a lowering of pres- sure will cause ice to pass directly to vapor. Table XXII JUHLIN'S DATA ON VAPOR PRESSURE OF ICE Temperature. • Pressure. C. F. Min. Hg. Lbs. sq.in. • -60 -58. .050 .001 -40 -40. .121 .0023 -30 -22. .312 .006 -20 - 4. .806 .0156 -15 5. 1.279 .0247 -10 14. 1.999 .0386 - 8 17.6 2.379 .0459 - 6 21.2 2.821 .0544 - 4 24.8 3.334 .0643 - 2 28.4 3.925 .0758 - 32. 4.602 .0888 Likewise the liquid, water, may pass to the solid, ice, by lowering temperature at a fixed pressure as indicated in Fig. 137, plotted from data by Tamman, Table XXIII, and which becomes then the curve of " fusion.^' Table XXIII TAMMAN^S DATA ON FUSION PRESSURE AND TEMPERATURE OF WATER-ICE Temperature. Pressure. C. F. Kg. sq.cm. Lbs. sq.in. 32. 1 1423 - 2.5 27.5 336 4779. - 5. 23. 615 8747.4 - 7.5 18.5 890 13658.8 -10.0 14. 1155 16428. -12.5 9.5 1410 20055. -16. 5. 1625 23113. -17.6 .5 1835 26100. -20. - 4. 2042 27044. -22.1 - 7.8 2200 31291. HEAT AND MATTER 457 80000 aeooo 22000 S1800Q m o Suooo 10000 eooo 2000 Temperature in Degrees Fahrenheit Fig, 137. — Water — Ice, Pressure-temperature Curve. 458 ENGINEERING THERMODYNAMICS These three curves plotted to the same scale meet at a point located at a pressure of 4.6 mm. Hg. = .18 ins. Hg., and temperature +.0076*' C. =32.01^ F.,ordi. narily taken at 32^ F., which point is named the triple point, as indicated in Fig. 138. The fact that the vapor pressure for water extends below freezing- point and parallels more or less that of ice indicates the condition of supercooled .2 «0 < u o CQ '^ 1 9 O I u Pi4 Ice Triple Point 50 Fig. 138.— Water Vapoi Temperature Degrees Fahr. -Water — loe, Combined Curves of Prebsure-temperature Rela- tion. The THpU Paint. water, one of unstable equilibrium instantly dispelled by the introduction of a little ice at the proper stable state for this temperature. Ordinary en^neering work is not concerned with the entire range indicated in Fig. 138 for any substance, but with the higher temperature ranges for some and the low for others, with transition from solid to liquid state for metals and similar solids and the transition from liquid to vapor for a great many, of which water comes first in importance, then the refrigerating fluids, ammonia HEAT AND MATTER 459 and carbon dioxide^ and last certain fuels like alcohol and the petroleum oils with their distillates and derivatives. MeUing-paintSy or the fusion temperature of such solids as are important, are usually given for only one pressure, the standard atmosphere, as in ordinary practice these substances are melted only at atmosphere pressure, and some such values are given at the end of the Chapter in Table XLIII. This is not the case, however, for boHing-pointa, which must be defined a little more closely before discussion. The vapor pressure curves indicate that as the temperature of a liquid rises, the pressure rises also if the substance is enclosed, but if the pressiu'e were relieved by opening the chamber to a region of lower pressure and kept constant, then the temperature would no longer rise and boiling or ebullition would take place. The boiling-point then is the highest temperature to which the liquid and its vapor could rise under the existing pressure. When not otherwise defined the term boilmg-point must be taken to mean the temperature of ebullition for atmospheric pressure of 29.92 ins. Hg, and values for several substances are given at the end of this Chapter in Table XLIV. Vapor having the temperatiu'e required by the pressure of the pressure- temperature curve is known as aaturcUed vapor, and this may be defined as vapor having the lowest temperature at which it could exist as vapor, under the given pressure. Vapors may, however, be superheated, that is, have higher temperatures than satiu'ated vapors at the same pressure, but cannot so exist for long in the presence of liquid. Superheating of vapors, therefore, implies isolation from the liquid, and the amount of superheat is the number of degrees excess of temperature possessed by the vapor over the saturation temperature for the pressure. In steam power plant work, especially with turbines, it is now customary to use steam with from 75® F. to 150® F. of superheat, and it might be noted that all so-called gases like oxygen and nitrogen are but superheated vapors with a great amount of superheat. It has abeady been mentioned that the saturated vapor pressure-temperature curve of direct experiment is seldom accurate as foimd, but must be corrected by empiric equations or smooth average curves, and many investigators have sought algebraic expressions for them. These equations are quite useful also in another way, since they permit of more exact evaluation of the rate of change of pressure with temperature, which in the form of a differential coefficient is found to be a factor in other physical constants. One of these formulas for steam as adopted by Marks and Davis in the calculation of their tables is given in Eq. (655), the form of which, was suggested by Thiessen: (i-l-459.6) Iog^ = 5.409(i-212®)-3.71X10-I0[(689-0*-477^1, • (655) in which i= temperature F.; and p= pressure lbs. sq.in. This represents the truth to within a small fraction of one per cent up to 400® F., but having been found inaccurate above that point Professor Marks has 460 ENGINEERING THERMODYNAMICS very recently developed a new one, based on Holbom and Baumann's high temperature measurements, which fits the entire range, its agreement with the new data being one-tenth of 1 per cent, and with the old below 100® F., about one-fifth of 1 per cent, maximum mean error. It appears to be the best ever found and in developing it the methods of the physical chemists have been followed, according to which a pressure is expressed as a fraction of the critical pressure and a temperature a fraction of the critical temperature. This gives a relation between reduced pressures and temperatures and makes use of the principle of corresponding states according to which bodies having the same reduced pressure and temperature, or existing at the same fraction of their critical are said to be in equivalent states. The new Marks formula is given in Eqs. (656) and (657), the former containing symbols for the critical 1 X m i_ r and the latter giving to them their numerical values, temperature I e abs. J ^ « in pressure pounds per square inch, and temperature absolute F. log 2l = 3.006854 (y-l) Tn-. 0505476^-1- .629547 (-^-.7875^1, . (656) log p = 10.515354 - 4873.71 r- 1 - .00405096^-1- .000001392964T2. . (657) As the method used in arriving at this formula is so rational and scientific, it has been adopted for a new determination, from old data to be sure, of the relations between p and T for ammonia and carbon dioxide, so important as substances in refrigeration, especially the former. According to this method if pc and Tc are the critical pressures and temperatures, both absolute, and p and T those corresponding to any other point, then according to Van der Waals, Iog^=/(^'-lj (658) Accordingly, the logarithm of the critical divided by any other pressure, is to be plotted against the quantity [(critical temperatur edivided by the tem- perature corresponding to the pressure) — 1], and the form of curve p>ennits of the determination of the fimction, after which the values of the critical point are inserted. This has been done for NH3 and CO2 with the result for NH3 log £^ = . 045+2.75^1?- 1) +.325^1?- 1^, .... (659) which on inserting the critical constants, Pc=114 atm. = 1675.8 lbs. per square inch rc== 727.4^ F. absolute which are the Vincent and Chap- puis values, HEAT AND MATTER 461 becomes, logp=5.60422-1527.54r"i-17196ir-2 .... (660) For CO2 it was found that log2?=.038+2.65^p-l) + 1.8(~^-iy, .... (661) which on inserting the critical constants, Pe=77 atm. = 1131.9 lbs. per square inch ' which are Andrews' values, rc=547.27^F. abs. becomes, log p = 7.46581-4405.765r"i+1617501.366r"2-257086165.8706r"3. (662) Curves showing the relation of reduced and actual temperatures and pressures are given in Fig. 139 for ammonia and in Fig. 140 for carbon doixide. For the past half century far more time and effort have been devoted to making other formulas of relation of p to T for saturated vapor not only for steam, but also for other vapors, than would have sufficed for accurate exper- imental determination, and as these help not at all they are omitted here. Equa- tions of physical relations can be no better than the data on which they are based, and for the substances ammonia and carbonic acid the charts or formulas must be used with a good deal of suspicion. In all engineering calculations requiring one of these constants even for steam no one is justified in using a formula; the nearest tabular or chart value must be employed and it will be as accm*ate as the work requires. Time is at least as important as accuracy, if not more so, for if too much time is required to make a calculation in commercial work, it will not be made because of the cost, indirect and approximate methods being substituted. It is sometimes useful in checking the boiling-point of some substance little known, to employ a relation between boiling-points of different substances at the same pressure appUed to a substance well-known. Let Ta and T^ be absolute temperatures of boiling for substances A and B under same pressure; Ta and Th be absolute temperature of boiling for substances A and B under some other pressure. Then, ^! = f;+c(n'-n) (663) Such equations as this are useful in finding the saturation curve of other sub- stances from that for water, which is now so well established, when enough points are known for the other substances to establish, the constant c. Also Ta the ratio =^ plotted against the temperature difference Tb—Ti, should give TjT 462 ENGINEKRlNa THERMODYNAMICS .. I / • > / > /^ / /^ |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 ■5-4 .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 Pressure-temperature Relations, for Saturated Vapor. HEAT AND MATTER 463 .2 .3 .4 .5 (Critical Temperatare Divided by any otlier Temperature)-! —100 50 50 lemperatureB in Degrees Fahr. FiQ. 140. — Carbon Dioxide Pres6ure-temperature Relatipng for Saturaled Vapor. 464 ENGINEERING THERMODYNAMICS a straight line, and if the line is not straight the experimental values may k wrong or the law untrue. This procedure has been followed in Fig. 141, in checking the curves for CO2 and NH3 against those for water, but it is impos- sible to say whether the discrepancies for CO2 are due to a failure of the law or bad experimental values, probably both, as the law holds poorly for water itself. 250 200 10 CO ^150 0) I 100 50 o O .5 .6 Values of Fig. 141.— Curves for CO, and NHi to Check the Linear Relation Eq. (663). All of the preceding refers, of course, to pure substances, but in practid work there are frequently encountered problems on solutions where large differences may exist compared to the pure liquids. Thus, for salts in water, it is well known that addition of a salt lowers the freezing-point, that more salt lowers it more, and it was first thought that the depression was in proportion to the amount dissolved. This being found to be untrue, recourse was had HEAT AND MATTER 465 again to molecular relations by Raoult, who announced the general law that the molecular depression of the freeztng-point is a constant Atm Molecular lowering of freezing-point J?'= = const., . . (664) w in which Af = depression of freezing-point in degrees F.; ti7= weight dissolved in 100 parts of solvent; m= molecular weight of substance dissolved. From Eq. (664) the freezing-point for brines may be found as follows: w Freezing-point of aqueous solutions = 32°— (const.) X — . . . (665) m As examples of the degree of constancy of the " constant " the following values Table XXIV, taken from Smithsonian Tables are given: Table XXIV LOWERING OF FREEZING POINTS Salt. g. Mol. 1000 g. H«0* Molecular Lowering. AuthoritioB. NaCl .004 .01 3.7 3.67 ■ .022 3.55 Jones .049 3.51 Loomis .108 .232 .429 3.48 3.42 3.37 Abegg Roozeboon .7 3.43 NH4CI .01 3 6 .02 3.56 .035 .1 3.5 3.43 Loomis .2 3.4 .4 3.39 • CaCli .01 .05 5.1 4.85 .1 4.79 .508 5.33 • .946 2.432 3.469 3.829 .048 .153 5.3 8.2 11.5 14.4 5.2 4.91 Arrhenius Joneff^letman Jones-Chambers Loomis Roozeboon .331 5.15 .612 5.47 • .788 6 34 466 ENGINEERING THERMODYNAMICS Just as the pressure of dissolved substances in liquids lowers the freezing- point, so also does it lower the vapor pressure at a given temperature or raii^e the boiling-point at a given pressure. Investigation shows that a similar formula expresses the general relation: Mm Molecular rise of boiling-point = S = = constant = 5.2, . (666) when water is the solvent. From Eq. (666) the rise of the boiling-point is found to be Rise of boiling-point =5.2 — (667) When liquids are mixed, such as is the case with all fuel oils and ^ith denatured alcohol, the situation is different than with salts in solution, and these cases fall into two separate classes: (a) liquids infinitely miscible like alcohol and water or like the various distillates of petroleum with each other, and (&) those not miscible, like gasolene and water. The vapor pressure for miscible liquid mixtures is a function of the pressure of each separately and of the molecular per cent of one in the other when there are two. This rule, which can be symbolized, is no use in engineering work, because in those cases where such mixtures must be dealt with there will be generally more than two liquids, the vapor pressure characteristic and molec- ular per cent of each, or at least some of which will be imknown. When, however, the two liquids in contact or in fact any number are non-miscible they behave in a Very simple manner with respect to each other, in fact are quite independent in action. Each liquid will evaporate imtil it.< own vapor pressure is established for the temperature, as if the other were not there, and the vapor pressure for the mixture will be the sum of all the separate ones. On the other hand the boiling-point will be the temperature at which all the vapor pressures together make up the pressure of say the atmosphere, and this is necessarily lower than the highest and may be lower than the lowest value for a single constituent. This action plays a part in vaporizers and carburettors using alcohol and petroleum products. To permit of some approximations, however, a few vapor tension curves for hydrocarbons and alco- hols are given later in the Section on vapor-gas mixtures, and data on the vapor pressure and temperature relations of ammonia-water solution are given in the section on the solution of gases in liquids. Example 1. Through how many degrees has ammonia vapor at a pressure of 50 lbs. per square inch absolute been superheated, when it is at the temperature at which steam is formed under a pressure of 100 lbs. per square inch absolute? From the curve of pressure and temperature of steam the temperature is 328* F. for the pressure of 100 lbs. From the similar curve for anmionia vaporization occurs under a pressure of 50 lbs. at a temperature of 22^ F. Hence, superheat «=32S- 22«306^F, HEAT AND MATTER 467 Prob. 1. Three tanks contain the foUowing liquids together: water, ammonia, and carbon dioxide respectively, and at a temperature of 30^ F. What pressure exists in each tank? If the temperature rises to 70^ F. how much will the pressure rise in each? Prob. 2. The pressure exerted by water vapor in the atmosphere when saturated, is that due to the temperature and is independent of the pressure of the air. The total pressure read by a barometer is the sum of the air pressure and the water vapor pressure. What is the pressure due to each under a saturated condition for tem- peratures of 50** F., 100** F., 150** F., and 200* F., the barometer m each case being 29.92 inches of Hg? Prob. 3. In order to secure a sufficiently high rate of heat transfer the steam in a radiator must be at a much higher temperature than the room to be warmed. If it is to btf 150** above room temperature what must be its pressure for room temperatures of 50** F., 60** F., 70** F., 80** F., and 125** F.? Prob. 4. In one tj^e of ice machine ammonia gas is condensed at a high pressure and evaporated at a low pressure. What is the least pressure at which gas may be condensed with cooling water of 70° F., and what is the highest pressure which may be carried in the evaporating coils to maintain a temperature in them of 0** F.? Prob. 5. Should carbon dioxide be substituted in the above machine what pressures would there be in the condensing coils, and in the evaporating coils? Prob. 6. How many degrees of superheat have the vapors of water, anunonia and carbon dioxide at a pressure of 15 atmospheres and a temperature of 500** F.? Prob. 7. Change the following pressures in pounds per square inch absolute xo reduced pressures for water, ammonia, and carbon dioxide, 15 lbs., 50 lbs., 100 lbs., 500 lbs. Prob. 8. At the temperature of melting ice what will be the vapor pressure of ammonia and carbon dioxide? At the temperature of melting tin what will be the pressure of water vapor? At this same temperature how many degrees of superheat would ammonia vapor under 100 lbs. pressure have, and how many degrees superheat would carbon dioxide vapor have under 1000 lbs. pressure? Prob. 9. If 10 lbs. of common salt, NaCl, be dissolved in 100 lbs. of water, what will be the boiling point of the solution at atmospheric pressure, what the freezing-point? m 10. Change of State with Amount of Heat at Constant Temperature. Latent Heats of Fusion and Vaporization. Total Heats of Vapors. Relation of Spe- cific Volume of Liquid and of Vapor to the Latent Heat As previously explained, a liquid boils or is converted into a vapor at constant temperature when the pressure on the surface is constant. Then during the change of state the amount of heat added is indirect proportion to the amount of vapor formed. The amount of vapor to convert a pound of liquid into vapor at any one steady tempera- ture, is the latent heat of vaporization some values for which are given at the end of this chapter in Table XLV, and it must be understood that this latent heat is also the amount given up by the condensation of a pound of vapor. Latent heat is not the same for different pressures or temperatures of vapori- sation but is intimately associated with the volume change in the transition From the liquid to the vapor state. That this should be so, is clear on purely rational grounds because there is necessarily external mechanical work done 468 ENGINEERING THERMODYNAMICS in converting the liquid to the vapor, since this is accompanied by a change of volume against the resisting pressure at which the conversion takes place. Thus, if Vv= specific volume of the vapor in cubic feet per pound; Vl = specific volume of the liquid in cubic feet per pound ; P= pressure of vaporization lbs. per sq.ft. absolute. Then Mechanical external work done dur- ] „,-, rr \ ^i. ii_ iooc. ... r 1 lu f =F(7v— yj it.lbs (^8 mg vaponzation of 1 lb. J Of course, at high temperatures the volume of a poimd of liquid is greater than at low because of its expansion with temperature rise, and imder the cor- responding higher pressures the volimie of a pound of vapor is less, because of the compressional effect of the pressure, than at low pressures, so that as pressures and temperatures rise the difference Vv—Vl becomes less and dis- appears at the critical point where it is zero. The latent heat being thus asso- ciated with a factor that becomes less in the higher ranges of temperature and pressure may be expected, likewise to become less unless some other factor tends to increase. All the energy of vaporization making up the latent hea: may be said to be used up in (a) doing external work as above, or (6) overcom- ing attraction of the* molecules for each other. As at the critical point there is no molecular change and no external work, the latent heat becomes zero at this point. This relation between latent heat and volume change was formulated by Clausius and Clapeyron, but Eq. (669) is generally known as ClapeyronV equation: Let L = latent heat ; " J = mechanical equivalent of heat = 778, or better 777.52, in such cases as this; " T = absolute temperature of vaporization ; dP " -7^= rate of increase of vapor pressure per degree change of cone sponding temperature. Then ^=77?52S(^-^^) (^■' This formula is used to calculate latent heat from the specific volumes of vapor and liquid and from the curvature of the saturation curve when they are known, but as these volumes are especially diflBcult to measure, direct experimental determination of the latent heat should be depended upon to get numerical values wherever possible. The formula will then be useful for the inverse process of calculating specific volumes from latent heats or as a means of HEAT AND MATTER 469 checking experimental values of both, one against the other. It is, however, dP just as useful to calculate latent heats from the specific volumes, and -— of dT the vapor curve, when the latent heats are less positively determined than the volumes or densities. Another simpler relation of a similar general character exists and is useful in estimating latent heats approximately for some little known substances like, for example, the liquid fuels, and in the use of which accurate physical data are badly needed. Despretz announced that Vv-Vi is nearly constant for all substances, and this was simplified by Ramsay and Trouton on the assumption, first, that the voliune of the liquid is very small at ordinary temperatures and may be neglected, in comparison with the volume of the vapor, and second, that the volume of the vapor is inversely proportional to the molecular weight m and directly proportinal to absolute temperatures so that (Trouton's law) m-jf = constant = C (670) or m the constant c is given the following values by Young: CO2 c = 21.3 NH3 c = 23.6 Hydrocarbons c = 20.21 Water and alcohols c = 260 For such substances as water and steam, the properties of which must be accurately known, general laws like the above are of no value compared with, direct experimental determination except as checks on its results, and even these checks are less accurate than others that are known. These experimental data are quite numerous for water, but as generally made include the heat of liquid water from some lower temperature to the boiling-point. The amount of heat necessary to warm a pound of liquid from temperature 32° F. to some boiling-point, and to there convert it entirely into vapor is designated as the total heat of the dry saturated vapor above the origi- nal temperature. This is, of course, also equal to the heat given up by the con- densation of a pound of dry saturated vapor at its temperature of existence and by the subsequent cooling of the water to some base temperature taken univer- sally now as 32° F. in engineering calculations. 470 ENGINEERING THERMODYNAMICS From observations by Regnault and formulated by him in 1863 the present knowledge of the total heat of water may be said to date. He gave the expression, Eq. (671), in which the first term is the latent heat at 32^ and one atmospheric pressure: Total heat per pound dry saturated steam^H = 1091.7+.305(<-32). (671) This was long used as the basis of steam calculations, but is now to be discarded in the light of more recent experimental data, the best of it based on indirect measurements by Grindley, Griessmann, Peake, who observed the behavior of steam issuing from an orifice, together with the results of Knobloch and Jacob and Thomas on specific heats of superheated steam, and in addition on direct measurements by Dieterici, Smith, GriflSths, Henning, Joly. All this work has been recently reviewed and analyzed by Davis, who accepts 1150.3 B.T.U. as the most probable value of the total heat under the standard atmos- phere and the following formula as representing total heats from 212^ up to 400° F. i3r=1150.3+.3746(<~212)-.000550(«-212)2 . . . (672) The Davis curve containing all the important experimental points ^nd the accepted line, extended dotted from 212° to 32°, is presented in Fig. 142. From the total heats given by this formula the latent heat is obtainable by subtraction, according to the relation. Latent heat (L)= total heat of vapor above 32° F. (If)— heat of liquid from 32° F. to boiling point (A), (673) in which the heat of the liquid is computed from a mean curve between Dieterici's and Regnault's values, having the equation A = .9983-. 0000288 («- 32) + .0002133(^—32)2. This is the basis of the values for latent and total heats in the Marks and Davis steam tables referred to, and accepted as the best obtain- able to-day. From these tables a pair of charts for latent heat and total heat of dry saturated steam are given at the end of this Chapter. The specific volume and density of dry saturated steam, given in the charts and table are calculated, as this seems to promise more exact results than direct experiment, the method of calculation involving three steps:* dp (a) From the pressure-temperature equation the ratio of — is foimd by differentiation as follows: log p = 10.516354 -4873.71 r~i-. 004050967+ .000001392964r2, whence ^= f^^%^ -.00405096+. 0000027859287) p. dT \ T^ HEAT AND MATTER 471 (b) From the latent heats the diflFerence between specific volume of vapor and liquid, {Vv—Vl) is calculated by substituting (a) in Clapeyron's equation. (c) From the Landolt, Bomstein, Myerhoffer tables for density of water the volume Vl is taken, whence by addition the volume of the vapor Vv is found, For ammonia and carbonic acid there are no data available on total heats by either direct measure or by the orifice expansion properties, and very few a8° 50° eS*' 86^ 104° 122° 140** 158° 176'' IW" 212" 230° ti»° 266° 284'* JJ08° 820" 838" 356° 874° 888' Temperature In. Degrees K Fig. 142.— Total Heat of Dry Saturated Steam above 32° F. (Davis). determinations of the latent heat itself, so that the process that has proved so satisfactory with steam cannot be directly followed with these substances. Accordingly, a process of adjustment has been used, working from both ends, beginning with the pressure temperature relations on the one hand and specific volumes of liquid and vapor on the other, the latent heat is determined by Clapeyron's equation and where this does not agree with authentic values an adjustment of both latent heat and specific volume is made. 472 ENGINEERING THERMODYNAMICS This process is materially assisted by the so-called Cailletet and Mathia> law of mean diameter of the curves of density of liquid and vapor, which are given in Figs. 143, 144 and 145, for water, ammonia and carbon dioxide, ol which the points are marked to indicate the source of information. On each of these curves the line BD is the line of mean density, its abscissa being given by the following general equation, 8 = i{^+yij=(i+bt+cfi (6: Of course, this mean density line passes through the critical volume B. For these three cases this Ekj. (674) is found to have the form. For water s = 28.7 -.015(<- 300) -.000015(<- 300)2. (a) For ammonia s = 20 - .022(< - 30) . (6) For carbonic acid.. s = 33.1-.0219(/+20)-.00016(<-20)2. (c) (675 - A more exact equation for water has been determined by Marks and Da%Ts in their steam tables and is s = 28.424 -.01650(^-320) -.0000132(^-320)2. . . . (676) From the smooth curve, which has the above equation, the volumes and densi- ties of liquid and vapor that are accepted have been derived, and are presented in chart form on a large scale and in tabular form at the end of the Chapter, the values for water being those of Marks and Davis. dp From these volume differences and the tz, relation the latent heats have dT been calculated and the newly calculated points are compared