b'\n\n\n\xe2\x96\xa0f.-.f \xe2\x80\xa2\xe2\x96\xa0\' \'--.\'.\':\xe2\x80\xa2 \'-\'^\'\xe2\x80\xa2y>;t^\'-y<f-\'^P\'nCta;-y^>e4v-*\'\'. \xe2\x96\xa0 \n\n\n\n\nGop)iight }J? \n\n\n\nCOP^\'RIGHT DEPOSIT. \n\n\n\nENGINEERING \nTHERMODYNAMICS \n\n\n\nBY \n\nJAMES AMBROSE MOYER, S. B., A. M. \n\nProfessor of Mechanical Engineermg in The \nFenfisylvania State College \n\n. AND \n\nJAMES PARK CALDERWOOD, M. E. \n\nAssociate Professor of Mechanical Engineering in The \nPennsylvania State College \n\n\n\nFIRST EDITION \n\nFIRST THOUSAND \n\n\n\nNEW YORK \n\nJOHN WILEY & SONS, Inc. \n\nLondon: CHAPMAN & HALL, Limited \n\n1915 \n\n\n\n\n\n\nCopyright, 191 5, \n\nBY \n\nJ. A. MOYER AND J. P. CALDERWOOD \n\n\n\n/i^^^-^ \n\n\n\nfc) \n\n\n\nstanhope iprcss \n\nF. H. GILSON COMPANY \nBOSTON, U.S.A. \n\n\n\nNOV I 1915 \n\n\xc2\xa9CI.A414330 \n\n\n\nPREFACE \n\n\n\nFor years there has been an important demand for a text- \nbook on thermodynamics which would be brief and concise, but \nat the same time so clearly written as regards explanation that \nstudents of average ability in our large technical schools could \nread it without difhculty. A professor of thermodynamics wrote \nrecently as follows: \'\'I like the idea of making the text largely \nself-explanatory. Too many books require the reading of several \nlines between every two Hnes of the text." This book has been \nprepared to meet this demand and in writing it the authors have \nkept in mind these requirements. Further, it has been the idea \nof the authors to make this book particularly suitable for use in \nthe larger technical schools where it is possible to give special \ncourses on the subjects of steam turbines, pneumatic machinery, \ninternal combustion engines, refrigeration and pumping ma- \nchinery. Usually in the courses on these subjects the advanced \nand special theory of thermodynamics as it relates to each of \nthem is taken up with completeness. At present there is too \nmuch duplication of subject matter in the large volumes on \nthermodynamics now available, which are made to include the \ndescriptive matter and the applications of these special subjects. \n\nThe authors are particularly indebted to Professor Roy B. Fehr \nof The Pennsylvania State College for invaluable assistance and \ncriticisms in the preparation of this work. Acknowledgments \nare due also to President Ira N. Hollis of Worcester Polytechnic \nInstitute; Professor Lionel S. Marks of Harvard University and \nMassachusetts Institute of Technology; Professor H. C. Ander- \nson of the University of Michigan; Dr. S. A. Moss of the \nGeneral Electric Company; Dr. William Kent of Montclair, \nN. J.; Professor A. M. Greene of Rensselaer Polytechnic In- \n\n\n\nIV PREFACE \n\nstitute; Professor A. L. Westcott of University of Missouri; \nProfessor A. A. Atkinson of Ohio University; and Professors \nJ. A. Bursley and C. H. Fessenden of the University of Michigan, \nfor assistance in various ways. \n\nA book on applied thermodynamics published privately by \nthe late Professor H. W. Spangler and with the preparation of \nwhich one of the authors was intimately associated has been \nconsulted freely in the preparation of the last chapters. This \nbook, as a whole, includes many of Professor Spangler\'s ideas as \nregards subject matter to be included in a book of this kind. \n\nAcknowledgment of the services of Messrs. W. M. Sides and \nH. J. Hartranft of State College, Pa., is due for assistance in \npreparation and proof-reading. \n\nJ. A. MOYER \n\nJ. P. Calderwood \n\nState College, Pa., \nAugust, 1915. \n\n\n\nCONTENTS \n\n\n\nPage \nChapter I. \xe2\x80\x94 Introduction to the Theory or Heat Engines i \n\nHistorical; Definition, Scope, and Object of Thermodynamics; First and \nSecond Laws of Thermodynamics; Units Employed: British Thermal \nUnits, Density, Specific Volume, Mechanical Equivalent of Heat; Ther- \nmal Efficiency; Working Substances. \n\nChapter II. \xe2\x80\x94 Properties of Perfect Gases 9 \n\nExamples of Perfect Gases and Vapors; Laws of Perfect Gases: Boyle\'s \nLaw, Charles\' Law, Combination Law (PV = MRT); Absolute Tempera- \nture and Pressure; Gas Thermometers; Heat and Its Effects : Heat Added \n= Increase of Internal Energy + Work Done; Internal or Intrinsic \nEnergy; External Work; Specific Heats : Cj, and C^; Joule\'s Law; Rela- \ntion between Cp and Cv\', Ratio of Specific Heats; Low Temperature Re- \nsearches. \n\nChapter III. \xe2\x80\x94 Expansion and Compression of Gases 29 \n\nGraphical Diagrams; Indicator Diagrams; Isothermal Expansion and \nCompression; Adiabatic Expansion and Compression; Change of Internal \nEnergy during Adiabatic Processes. \n\nChapter IV. \xe2\x80\x94 Cycles of Heat Engines 44 \n\nCamot\'s Cycle; Heat Engines and Refrigerating Machines; Efficiency of \nCamot\'s Cycle; Reversible Cycles; Camot\'s Principle; Perfection in a \nHeat Engine; Reversed Camot\'s Cycle; Regenerative Air Engines. \n\nChapter V. \xe2\x80\x94 Properties of Steam 60 \n\nSteam in Heat Engines; Process of Steam Formation; Relation of Tem- \nperature, Pressure and Volume in Saturated Steam; Heat of the Liquid; \nLatent Heat of Vaporization; External Work of Evaporation; Total Heat \nof Steam; Intemal Energy of Evaporation and of Steam; Steam Formed \nat Constant Volume; Wet and Dry Saturated Steam; Superheated Steam; \nDetermination of Moisture in Steam; Throtthng or Superheating Calo- \nrimeters; Separating Calorimeters; Condensing or Barrel Calorimeters; \nEquivalent Evaporation; Factor of Evaporation, \n\nChapter VI. \xe2\x80\x94 Practical x\\pplications of Thermodynamics 88 \n\nRefrigerating Machines or Heat Pumps; Systems in Use; Coefficients of \nPerformance; Compressed Air; Indicator Diagrams from Compressors; \nComparison of Actual and Ideal Compression Curves; Multi-stage Com- \npressors. \n\n\n\nVI CONTENTS \n\nPage \nChapter VII. \xe2\x80\x94 Entropy loi \n\nPressure- volume Diagrams and their Uses; Heat Diagrams; Explanation \nof Entropy; Camot\'s Cycle on a Temperature-entropy Diagram; Tem- \nperature-entropy Diagrams for Steam; Isothermal and Adiabatic Lines \nfor Steam; Saturation Line. \n\nChapter VIII. \xe2\x80\x94 Practical Steam Expansions and Cycles 113 \n\nSteam Engine Efficiency With and Without Expansion of the Steam; \nQuality of Steam during Adiabatic Expansion; MoUier Diagram; Graphical \nDetermination of Quality of Steam by Throttling Calorimeter; Available \nEnergy from Adiabatic Expansion; Rankine Cycle; EfiQciency of Various \nPrime Movers; Combined Indicator Diagrams; Hirn\'s Analysis of the \nSteam Engine. \n\nChapter IX. \xe2\x80\x94 Flow of Fluids 163 \n\nFlow of Air; Thermodynamic Formulas; Fliegner\'s Formulas; Experi- \nmental Data; Flow of Steam; Grashof\'s Formula; Napier\'s Formula; \nFlow through Nozzles; Restriction of Flow Due to Over or Under Expan- \nsion; Correction for Flow in Poorly Designed Nozzles. \n\n\n\nSYMBOLS \n\n\n\nA = area in square feet. \nB.t.u. = British thermal units (=778 ft. lbs.). \n\nCy = specific heat at constant pressure in B.t.u. per pound per degree. \nCv = specific heat at constant volume in B.t.u. per pound per degree. \nC = a general constant in equations of perfect gases. \nD = degrees of superheat. \nE = external work in B.t.u. per pound; also sometimes used to express efficiency, \n\nusually as a decimal. \nEa = available energy in B.t.u. per pound. \nF = force in pounds. \nH = heat per pound in B.t.u.* \n^sup = total heat of superheated steam, B.t.u. per pound. \nIh = total internal energy of steam (above 32\xc2\xb0 F.) in B.t.u. per pound. \nIl \xe2\x80\x94 internal energy of evaporation of steam in B.t.u. per pound. \n/ = reciprocal of mechanical equivalent of heat = yf g (use becoming obsolete) . \nK = specific heat in foot-pound units. \nL = latent heat of evaporation in B.t.u. per pound. \nM = mass (pounds). \n\nP = pressure in general or pressure in pounds per square foot. \nQ = quantity of heat in B.t.u. \nR = thermodynamic constant for gases; for air it is 53.3 (in foot-pound units \n\nper pound.) \nT = absolute temperature, in Fahr. degrees = 460 -|- t. \n\nV = volume in cubic feet, also specific volume and velocity in feet per second. \nW. E. = Warme Einheit = kilogram calorie. \nW = work done in foot-pounds. \na = area in square inches. \nc = constant of integration. \nd = distance in feet. \n\ne = subscript to represent base of natural logarithms. \ng = acceleration due to gravity = 32.2 feet per second per second. \nh = heat of the hquid per pound in B.t.u. (above 32\xc2\xb0 F.). \n- k = a. constant, \nlog = logarithm to base 10. \nloge = logarithm to natural base e (Naperian) . \n\nn = general exponent for V (volume) in equations of perfect gases, also some- \ntimes used for entropy of the Uquid in B.t.u. per degree of absolute tem- \nperature. \n\n* In steam tables it is total heat above 32\xc2\xb0 F. \nvii \n\n\n\nvm SYMBOLS \n\np = pressure in pounds per square inch. \n\nq = sometimes used for heat of the liquid in B.t.u. per pound (above 32\xc2\xb0 F.). \n\nr = ratio of expansion (see page $3) > also sometimes used for latent heat of \n\nevaporation in B.t.u. per pound. \n/ = temperature in ordinary Fahr. degrees. \n\nV = specific volume, in cubic feet per pound (in some steam tables). \nw = weight per cubic foot = density. \nX = quality of steam expressed as a decimal. \n\nA = differential symbol. \n\n(^ \ny = ratio of specific heats -pr" \n\ndO \n\n<i> = total entropy -^\xe2\x80\xa2 \n\n= entropy of the liquid in B.t.u. per pound per degree of absolute temperature. \n\n\n\nEngineering Thermodynamics \n\nCHAPTER I \nINTRODUCTION TO THE THEORY OF HEAT ENGINES \n\nThe beginning of our knowledge of the theory of heat on a \nscientific basis is probably Black\'s doctrine of latent heat.* The \nimportant relations as to the equivalence of heat and work were \nnot established until about the middle of the nineteenth century, \nlong after the heat engines invented by Newcomen and Watt \nhad found general use. The greatest development in the theory \nof heat as regards its use in engines started with the pubhca- \ntions of Carnot in 1824, when he showed that work is performed \nby a heat engine in the same proportion as the absolute tem- \nperature of the working fluid is changed from a higher to a lower \ndegree. The next important development in heat theory was \nmade by Joule who showed by his experiments in 1843 ^^^-t for \nevery unit of heat there was an exact equivalent in mechanical \nwork. Although these experiments and the conclusions from \nthem were epoch-making, there was Httle attempt at accurate \ncalculations for designing steam and other engines until Reg- \nnault in 1847 published his classical data on the properties of \nsteam. In the next few years important contributions to our \nknowledge of what the ideal engine should be, particularly as \nregards its efficiency in the way of heat conversion, were made \nby Rankine, Clausius, and Lord Kelvin. It was undoubtedly \nRankine\'s philosophical treatment of the subject, as presented \nin various books and papers of which he was the author, that \ngave steam engineers scientific methods for the development of \nnew and improved designs. \n\nThermodynamics is that branch of engineering science which \ndeals with the interconversion of heat and work.f The object \n\n* Encyclopedia Britannica, iithed., vol. 4, page 18. \n\nt In general, thermodynamics may be defined as that branch of physical science \nwhich treats of the effects produced by heat. \n\n\n\n2 ENGINEERING THERMODYNAMICS \n\nof the study of thermodynamics is to consider in the light of the \nmost recent investigations, questions relating to the influence \non the efficiency of heat engines of increased pressure, of higher \nvacuums and expansions, of higher superheats (meaning also a \ngreater range of expansion), of jacketing cylinders, of using the \nworking medium in several cylinders one after the other; that \nis, compounding instead of expanding it only in a single cylinder, \nof putting receivers between the cylinders, and of reheating the \nworking medium as it passes through them from one cylinder \nto the next. All these subjects require the most careful con- \nsideration by the engineer, even though theoretical considera- \ntions are not always directly applicable, and the final results \nwhich determine our engineering practice are largely determined \nby the theoretical analysis of carefully conducted experiments. \nThe student should, therefore, keep in mind that although the \nactual conditions existing as regards the operation of our heat \nengines are relatively complex, the exact theory of their action is \nof great practical value, and he should, for this reason, give the \nstudy of the theory of heat engines his best attention. Because, \nalso, of this complexity in the action of engines the study of \nthe theory underlying their operation becomes all the more \nessential, as it must serve as a guide in deciding what conditions \nare most important for securing the highest efficiency. \n\nAnother important service which the study of thermodynamics \nrenders is that of showing us what maximum efficiency is attain- \nable for any engine operating under a given set of conditions. \nIt often happens that the enthusiastic inventor presents data \nshowing results which indicate an efficiency very much better \nthan we can obtain with any of our present types of engines. \nIn such cases it requires usually only a very little calculation, \nstarting from the fundamental theory of thermodynamics, to \nshow conclusively that the results claimed are absolutely im- \npossible. \n\nIn many cases this study is also useful to the engineer in check- \ning up his own work. For example, in calculating the results of \na test made on a gas engine of the ordinary types, if it is found \n\n\n\nINTRODUCTION TO THE THEORY OF HEAT ENGINES 3 \n\nthat the efficiency as computed and reported is higher than \n50 per cent, then obviously our knowledge of thermodynamics, \nand of the maximum possible efficiency of such engines operat- \ning under ordinary conditions, shows that the results are impos- \nsible and absurd. The ability to interpret correctly the results \nof experiments performed on all kinds of heat engines and simi- \nlar apparatus requires a thorough knowledge of the basic prin- \nciples of thermodynamics. \n\nLaws of Thermodynamics. The heat engine in its simplest \nform does work by utilizing heat developed in a furnace or \ngenerated by the combustion of fuel within the engine itself. A \npart of the heat thus suppHed is spent in doing mechanical work \nso that it no longer exists in the form of heat energy, while the \nremainder is rejected by the engine, still in the form of heat. \nThe heat action thus described depends on two fundamental \nlaws of thermodynamics called generally the First and Second. \n\nFirst Law of Thermodynamics. The statement of the first \nof these is that the amount of heat which disappears in the \naction of the heat engine is proportional to the amount of me- \nchanical work performed by the engine. In other words this \nlaw is nothing more than a simple statement of the conserva- \ntion of energy as regards the equivalence of mechanical work \nand heat. With more exactness this first law may be stated \nas follows: \n\nA definite quantity of heat goes out of existence for every \nunit of mechanical work that is performed, and, conversely, \nwhen heat is developed by the performance of mechanical work, \na definite quantity of heat comes into existence for every unit \nof work. \n\nSecond Law of Thermodynamics. A self-acting machine can- \nnot transmit heat from one body at a lower temperature to \nanother body at a higher temperature unless aided by some ex- \nternal agency; that is, " heat cannot pass from a cold body to \na hot body by a purely self-acting process" (Clausius). This \nlaw really states as regards heat engines the limits to their pos- \nsible performance, which would be otherwise unlimited, if only \n\n\n\n4 ENGINEERING THERMODYNAMICS \n\nthe "first law" is considered. It states further, as will be \ndiscovered in the discussion that follows, that no heat engine \nconverts or can convert into work all of the heat supplied to \nit. A very large part of the heat suppHed is necessarily rejected \nby the engine in the form of unused heat. \n\nENGINEERING UNITS \n\nEngineers in English-speaking countries use as the standard \nof measurements the units generally known as the foot-pound- \nsecond-system. In this system the unit of length is the foot, \nthe unit of mass is the pound (equivalent to 0.4535 kilogram), \nand the unit of time is the second. On the same basis we use \nas derived units the square foot as the unit of area in all theo- \nretical calculations, unless it is expressly stated that the area \nis given in some other units, as for example in square inches. \nPractically all equations and formulas are stated with the volume \ngiven in cubic feet. Unit pressure, or what we call briefly \n** pressure," is the total applied force in pounds divided by the \ntotal area in square feet over which it is exerted, and is then \nexpressed in terms of pounds per square foot. \n\nSpecific Volume is the term applied to the volume in cubic \nfeet of a pound of a substance. \n\nDensity is the mass in pounds, of a cubic foot of a substance. \nIt is, therefore, the reciprocal of the specific volume. \n\nFor engineering purposes the ordinary unit of work is the \nfoot-pound, which is the quantity of work performed by a force \nof one pound in moving through a distance of one foot. Quan- \ntities of heat are usually expressed in terms of the British ther- \nmal unit (B.t.u.). This unit is the quantity of heat required to \nraise the temperature of one pound of water one degree on the \nFahrenheit scale. The corresponding unit of heat on the Centi- \ngrade scale, used almost universally by physicists and chemists, \nis called the calorie (French and English), and Warme Einheit \n(German).* It is the quantity of heat required to raise the \n\n* Temperatures in Centigrade degrees are converted into Fahrenheit by mul- \ntiplying by I and adding 32. Kilogram-calories multiplied by 3.968 give the \n\n\n\nINTRODUCTION TO THE THEORY OF HEAT ENGINES 5 \n\ntemperature of one gram of water one degree on the Centigrade \nscale. \n\nTo make the definition of the B.t.u. accurate it is necessary \nto state at what temperature the rise of 1\xc2\xb0 F. as stated is to occur, \nbecause the specific heat of water is sHghtly variable. This is \nalso one of the reasons for some of the variations in the tables \nof the properties of steam and other vapors that we shall observe \nin our calculations. Some tables are based on the assumption \nthat the B.t.u. is the amount of heat required to raise the tem- \nperature of water 1\xc2\xb0 F. at the condition of maximum density of \nwater, that is, between 39\xc2\xb0 and 40\xc2\xb0 F. Other tables are based \non the amount of heat required to raise the temperature 1\xc2\xb0 F. \nfrom 60 to 61 degrees. Still another table, which is the latest \nand is generally considered to be the most accurate, uses for the \nB.t.u. one one-hundred-and-eightieth (yjo) ^^ ^^^ amount of heat \nrequired to raise the temperature of water from 32\xc2\xb0 to 2i2\xc2\xb0F. \nIn other words, according to this last definition the B.t.u. is \nthe average value of the amount of heat required to raise the \ntemperature of one pound of water one degree between the con- \nditions of freezing and boiling at atmospheric pressure. \n\nMechanical Equivalent of Heat. It has already been stated \nthat for every heat unit there is a corresponding exact equivalent \nwhich can be expressed in the mechanical units of work. The \nnumber of work units corresponding to i B.t.u. is called the \nmechanical equivalent of heat. This quantity was necessarily \ndetermined by experiments. The most reliable results show \nthat the average value of the mechanical equivalent of heat \nis 778 foot-pounds. This means, in other words, that a given \nnumber of B.t.u. multipHed by 778 gives the number of foot- \npounds of work corresponding to the amount of heat as ex- \npressed by the thermal units. \n\nThis value of the mechanical equivalent of heat is the one \ngiven by Rowland, Griffith, Osborne Reynolds and Morby. \n\nequivalent British thermal units (.B.t.u.), and kilogram-calories per kilogram \ntimes 1.8 give B.t.u. per pound. A "small" calorie, or gram-calorie, is one- \nthousandth as large as a kilogram-calorie. \n\n\n\n6 ENGINEERING THERMODYNAMICS \n\nThe first determination of this factor was made by Joule. In \n1843 h^ obtained the value of 772 foot-pounds. In later experi- \nments he found the value to lie between 774 and 775. A great \nmany modern physicists believe that the most exact value is \nsomewhere between 779 and 780 foot-pounds. The value stated \nabove as given by Rowland and others, that is, 778 foot-pounds, \nis, however, the value most generally accepted and is invariably \nused in engineering calculations. \n\nTHERMAL EFFICIENCY OF A HEAT ENGINE \n\nHeat converted into work \n\n\n\nThermal efficiency = \n\n\n\nHeat suppHed to the engine \n\nSince only a part of the heat supplied to an engine can be con- \nverted into work, the above ratio is a fraction always less than \nunity. \n\nSince heat is not a form of matter, it must be transmitted, in \na physical sense, by the use of some medium which is called the \nworking substance. It is this working substance that takes in \n\nSpring \nDetent \n\n\n\n\nFig. I. \xe2\x80\x94 "Ratchet and Pawl" Type of Heat Engine. \n\nand rejects heat. It may be in the form of a gas, a liquid or a \nsoHd. In general, there is a considerable change of volume in \na suitable working substance for heat engines as that substance \ndoes work in overcoming resistance. \n\n\n\nINTRODUCTION TO THE THEORY OF HEAT ENGINES 7 \n\nAs the simplest imaginable example of a heat engine in which \nthe working substance is neither a gas nor a liquid, but a solid \nmaterial, we may think of the working substance of the engine \nas being a long rod of brass arranged to operate as the pawl \nof a ratchet wheel (Fig. i) with teeth relatively close together. \nNow if the pawl is heated, it will elongate sufficiently to drive the \ntooth of the ratchet wheel with which it is in contact far enough \nforward to have the wheel held in its last position by a click or \ndetent. When the pawl cools, as, for example, by pouring on \ncold water, it will contract and fall into the position where it \nwill engage the next succeeding tooth on the wheel, which upon \napplication of heat will again be driven forward and held in \nposition, while the contracting of the pawl is again repeated. \nIn fact, if the ratchet wheel referred to above is fastened at its \ncenter to a round shaft, its movement can be readily made to do \nwork by raising the weight attached to a cord winding on the \nshaft. This heat engine has heat supplied to it at a high tem- \nperature. A small part of this heat is transformed into mechan- \nical work in moving the ratchet wheel and with it the weight to \nbe lifted, but by far the greatest part of the heat supplied is \nrejected in being absorbed by the water used for cooling and \nrequired to bring about the necessary contraction of the brass \nrod which is the working substance. This action as described \nhere for the metaUic working substance is typical of the action \nof all heat engines. They must take in heat at a relatively \nhigh temperature and reject it at a comparatively low tempera- \nture, and in the process they convert a small amount of heat \nsupplied into mechanical work. This process is also somewhat \nsimilar to the performance of a water wheel which does work \nin bringing water from a higher to a lower level. In the case \nof the water wheel we say that the water has lost potential \nenergy in dropping from a higher to a lower level. Similarly \nthe working substance in heat engines loses heat energy, which \nis in fact (lost) potential energy as it drops from a higher to a \nlower temperature. \n\n\n\n8 ENGINEERING THERMODYNAMICS \n\nPROBLEMS \n\n1. One pound of fuel has a heating value of 14,500 B.t.u. How manj \nfoot-pounds of work is it capable of producing, if all this heat could b( \nconverted into work ? Ans. 11,281,000 ft. -lbs. \n\n2. An engine developed 15,560 ft. -lbs. of work. How much heat was \nrequired? Ans. 20 B.t.u. \n\n3. A stone weighing 2 lbs. is let fall from a height of 389 ft. into i lb, \nof water. How many degrees is the water raised? Ans. 1\xc2\xb0 F. \n\n4. A heat engine receives 100,000 B.t.u. of heat in the form of fuel \nand during the same period 30,000 B.t.u. are converted into work. What \nis the thermal efificiency of the engine? Ans. 30 per cent. \n\n5. A gas engine receives 20,000 B.t.u. of heat in the form of fuel, and \nduring the same period 3,112,000 ft. -lbs. of work are developed. What is \nthe thermal efhciency of the engine? Ans. 20 per cent. \n\n6. Show that kilogram-calories per kilogram X 1.8 give B.t.u. per pound. \n\n7. Convert \xe2\x80\x9440\xc2\xb0 C. into degrees Fahrenheit. Ans. \xe2\x80\x9440\xc2\xb0 F. \n\n\n\nCHAPTER II \n\n\n\nPROPERTIES OF PERFECT GASES \n\nIn the study of thermodynamics we shall have to deal mostly \nwith fluids in the condition of a perfect gas, or a vapor. When \nthe word ^\' gas " is used it refers to what is more properly \ncalled a perfect gas, which is a fluid remaining in the gaseous \ncondition even when subjected to moderately high pressures \nand low temperatures. Oxygen, hydrogen, nitrogen, air, and \ncarbon dioxide are examples of what are called perfect gases. \nThey are fluids which require a very great reduction in tempera- \nture and increase in pressure to bring them to the Hquid state. \nVapors, on the other hand, are fluids which are readily trans- \nformed into Hquids by a very moderate reduction in tempera- \n\n\n\n.Connection fo itir Pump \n\n\n\n\nture or increase in pressure. Com- \nmon examples of vapors with which \nengineers have to deal are steam and \nammonia. \n\nFor the present all our studies \n\nwill be confined to the consideration \n\nof the properties of the perfect gases. \n\nRelation between Pressure and \n\nFig. 2. \xe2\x80\x94 Constant Temperature Volume of a Perfect Gas. In prac- \n\nApparatus for Demonstrating tically all heat engines, work is done \n\nRelation between Pressure and ^ ^j^ ^f ^^j^^^ ^f ^ ^^.^ ^^^ \n\nVolume of a Gas. ^ r \n\nthe amount of work performed de- \npends only on the relation of pressure to volume during such \nchange and not at all on the form of the vessel containing this \nfluid. Since a good understanding of pressure and volume \nrelations is most important, an illustration with a practical \nexample will not be out of place. \n\nFig. 2 shows a vessel filled with a perfect gas and surrounded \n\n\n\nlO ENGINEERING THERMODYNAMICS \n\nby a jacket filled with cracked ice. Its temperature will, \ntherefore, be at 32\xc2\xb0 F. This vessel has a tightly fitting piston \nP of which the lower flat side has an area of one square foot. \nIn the position shown the piston is two feet from the bottom of \nthe vessel so that the volume between the piston and the bottom \nof the vessel is two cubic feet. The pressure on the gas is that \ndue to the. piston and the weights shown. Assume this total \nweight is 100 pounds and that the air pump connected to the \ntop of the vessel maintains a practically perfect vacuum above \nthe piston. Then we say the pressure on the gas below the \npiston is 100 pounds per square foot. If now the weights are \nincreased to make the pressure on the gas 200 pounds per \nsquare foot the piston will sink down until it is only one foot \nfrom the bottom of the vessel, provided the ice keeps a con- \nstant temperature.* \n\nSimilarly, if the weight on the gas were 50 pounds and the \nvessel were made high enough, the lower side of the piston \nwould be four feet from the bottom of the vessel. Examina- \ntion of these figures shows that for all cases the product of \npressure and volume is constant, and in this particular case is \nalways 200 foot-pounds. Similarly, if the volume is expressed by \nV in cubic feet per pound and the pressure by P in pounds per \nsquare foot the product is a constant quantity, which for air at \n32\xc2\xb0 F. is 26,220 foot-pounds. These facts are expressed by \nBoyle\'s Law, which states that the volume of a given mass of \ngas varies inversely with the pressure, provided the tempera- \nture is kept constant; that is, V varies inversely as P or \n\nPF = a constant, (i) \n\nand also, if Pi and Vi represent some initial condition of pres- \nsure and volume and P2 and V2 are corresponding final con- \nditions, then \n\nPiVi = P2F2 = a constant. (la) \n\n* If the temperature is not maintained constant, because of the tendency of \ngases to expand with increase in temperature, it will be necessary to apply a \ntotal weight greater than 200 pounds to reduce the volume to one cubic foot. \n\n\n\nPROPERTIES OF PERFECT GASES II \n\nExperiment shows that a permanent gas when heated in- \ncreases in pressure if the volume is kept constant, and likewise \nincreases in volume if the pressure is kept constant. If the \npressure of a gas is kept constant and the temperature is raised \n1\xc2\xb0 F. the gas will expand 4^2 * ^f its volume at 32\xc2\xb0 F.; in \nother words, a volume of 492 cubic inches of gas will expand at. \nconstant pressure to 493 cubic inches when the temperature is. \nraised from 32\xc2\xb0 to 33\xc2\xb0 F. \n\nSimilarly, this same volume will be doubled if the gas is \nheated 492 degrees (i.e., to 524\xc2\xb0 F.) at a constant pressure; and \non the other hand, if the volume remains constant and the \npressure is allowed to vary, the same increase in temperature \n(492 degrees) will double the pressure which the gas had ini- \ntially. \n\nConversely, if the same law could be held to apply at very \nlow temperatures, at 492\xc2\xb0 below 32\xc2\xb0 F. or \xe2\x80\x94460\xc2\xb0 F. ( \xe2\x80\x94 273\xc2\xb0 C.)t \nthe volume would be zero. \n\nAbsolute Zero of Temperature. The examples above show \nthat if temperatures are calculated, not from the ordinary zero \nbut from 460 degrees below the ordinary Fahrenheit zero, the \nvolume of the gas, if kept at constant pressure, will be propor- \ntional to the temperature reckoned from that zero point; and \nsimilarly, if the volume is kept constant, the pressure will be pro- \nportional to the temperature above that zero, which is called the \nabsolute zero. Temperatures measured from the absolute zero \nas a basis are called absolute temperatures. \n\nAbsolute Temperature Conversion. If we represent the tem- \nperatures on the ordinary Fahrenheit scale by t and the corre- \nsponding absolute temperatures by T then \n\nT = t-\\- 460. (2) \n\n* This value is probably not exact. Various authorities give the following \n\nvalues: (i) \xe2\x80\x94 , (2)-^, (3)^\xe2\x80\x94, (4) \xe2\x80\x94 , (5) ^-, (6) \xe2\x80\x94 ,etc. Thevalue \n\n492\' 491-6\' "" 491-4 493 492-7 49i \n\ngiven above is probably the best average value and is certainly close enough for \nall engineering calculations. \n\nt See end of chapter for a brief description of the latest low temperature re- \nsearch. \n\n\n\n12 ENGINEERING THERMODYNAMICS \n\nIf /\' is the temperature on the Centigrade scale and V the cor- \nresponding absolute temperature then \n\nr = t\' + 273. \n\nThe laws of thermodynamics illustrated by the foregoing ex- \namples, dealing with volume and pressure changes corresponding \nto temperature variations, may be stated in two short paragraphs \nas follows : \n\n(i) Under constant pressure the volume of a given mass of gas \nvaries directly as the absolute temperature. \n\n(2) Under constant volume the absolute pressure of a given \nmass of gas varies directly as the absolute temperature. \n\nThese fundamental principles, often called Charles\' Laws, may \nalso be stated thus : \n\nV T \nWith pressure constant, \xe2\x80\x94 ^ = -7, (3) \n\nK 2 i 2 \n\nP T \n\nWith volume constant, \xe2\x80\x94 ^ = -7, (4) \n\nP2 T2 \n\nwhere Vi and V2 are respectively the initial and final volimies, \n\nPi and P2 are the initial and final absolute pressures, and Ti \n\nand T2 are the absolute temperatures corresponding to the \n\npressures and volumes of the same subscripts. \n\nThe following problem shows applications of Charles\' laws: \n\nA gas has a volume of 2 cubic\' feet, a pressure of 14.7 pounds \n\nper square inch absolute and a temperature of 60\xc2\xb0 F. \n\n(a) What will be the volume of this gas if the temperature is \nincreased to 120\xc2\xb0 F., the pressure remaining constant? \n\n(b) What will be the pressure if the temperature is increased \nas in (a) but the volume remain constant? \n\nSolution, (a) Since pressure remains constant and the sub- \nstance is a gas, the volume varies directly as the absolute tem- \nperature. \n\nLetting Vi and Ti be the initial conditions and Vo and T2 be \nthe final conditions, then \n\nYl \xe2\x80\x94 Tl _^ _ 60 + 460 \n\nV2 T2 Fo 120 + 460 \n\nF2 = 2.33 cubic feet. \n\n\n\nPROPERTIES OF PERFECT GASES 13 \n\n(b) Since the volume remains constant, \n\n?1 ^Xl 147 _ 60 + 460 ^ \n\nP2 T2 P2 120 + 460 \n\nPi = 16.39 pounds per square inch absolute. \n\nAbsolute Pressure.* The ordinary types of pressure gages \nused in engineering work do not measure the pressure from a \ntrue basis for comparison. Such instruments measure only the \nexcess of pressure above that of the atmosphere. The pressure \nindicated by the instrument is called gage pressure, and may \nbe expressed as a \'\' true " or absolute pressure by adding the \nactual pressure of the atmosphere as obtained from the reading \nof a barometer. In all thermodynamic calculations in this \nbook, unless it is expressly stated that gage pressures are meant, \nit is to be understood that absolute pressures are to be used. \nBoyle\'s and Charles\' laws hold only for cases where the pressure \nP is in absolute terms. Furthermore, the same statement ap- \nplies to all rational formulas in thermodynamics because all \nhave their fundamental basis on these laws. \n\nGas Thermometers. If temperatures are measured by ther- \nmometers filled with a permanent gas like oxygen, hydrogen, air, \nor any other permanent gas used as the expansive medium, the \nincrements of expansion are proportional to the temperature \n\n* It is frequently necessary to reduce the pressures in inches of mercury or of \nwater to the equivalent in pounds per square inch. Since the weight of a cubic \ninch of mercury at 70\xc2\xb0 F. is 0.4906 pound and of water at the same temperature \nis 0.0360 pound, pressures in inches of mercury at the usual "room" temperatures \ncan be reduced to pounds per square inch by multiplying by 0.491 or by dividing \nby 2.035; a^iid similarly, inches of water can be converted to pounds per square \ninch by multiplying by 0.0360 or by dividing by 27.78. Centimeters of mer- \ncury are reduced to pounds per square inch by multiplying by 0.1903. \n\nKilograms per square centimeter are reduced to pounds per square inch by \nmultiplying the kilograms per square centimeter by 14.223 or by dividing by \n0.0703. Grams divided by 28.35 s-^e ounces avoirdupois, or one gram is approxi- \nmately 3^0 ounce. \n\nA cubic foot of water at 70\xc2\xb0 F. weighs 62.3 pounds and at 32\xc2\xb0 F. 62.4 pounds. \nAt the ordinary room temperature the pressure due to 2.31 feet of water is equiv- \nalent to one pound per square inch. \n\n\n\n14 ENGINEERING THERMODYNAMICS \n\nmeasured from the absolute zero. The air thermometer, some- \ntimes used by physicists, is one having the bulb and tube filled \nwith air in such a way as to exert pressure upon a mercury col- \numn in a manometer or " U-tube " attached to the end of the \nair tube. Expansions and contractions in volume of the air \nin the bulb will change the relative levels of the mercury in the \nmanometer, which can be graduated to indicate temperatures. \n\nCombination of Boyle\'s and Charles* Laws. Equations (i), \n(3) and (4) cannot often be used in actual engineering problems \nas they stand, because it does not often happen that any one of \nthe three variables {P, V and T) remains constant. A more gen- \neral law must be developed, therefore, allowing for variations \nin all of the terms P, V and T. This is accomplished by com- \nbining equations (la), (3) and (4). It will be assumed that we \nare dealing with a pound of gas of which the initial conditions \nof pressure, volume and temperature are represented by Pi, Vi \nand Ti, while the corresponding final conditions, given by P2, \nV2 and T2, are arrived at in two steps. The first step is in chang- \ning the volume from Vi to V2 and the pressure from Pi to \nsome intervening pressure P2 while the temperature Ti remains \nconstant. This change can be expressed by Boyle\'s law (equa- \ntion la). \n\nWith constant temperature (Pi), \n\n(5) \n\nfrom which, by solving, P2\' = " ^\' ^ , (6) \n\nwhere P2 is the resulting pressure of the gas when its volume \nis changed from Vi to V2, with the temperature remaining \nconstant at Pi. \n\nThe second step is in the change in pressure from P2\' to P2 \nand the temperature from Pi to P2, while the volume remains \nconstant at V2. This step is expressed by equation (4) as \nfollows: \n\n\n\n\n\nP.\' \n\n\nP/ \n\n\nPiFi \nF. \' \n\n\n\nPROPERTIES OF PERFECT GASES \n\n\n\n15 \n\n\n\nWith constant volume (F2), \n\nP2 Ti \n\n\n\nwhich may be written \n\n\n\nP2 T,\' \nP2T2 \n\n\n\nP, = \n\n\n\nTi \n\n\n\nSubstituting now the value of P2 from (6) in (8), we have \n\nwhich may be arranged to read, \n\nPiVi ^ P^ \n\n\n\n(7) \n(8) \n\n(9) \n\n\n\n(10) \n\n\n\nThese steps are shown diagrammatically in Figs. 3 and 4. \nIn Fig. 3 a surface is shown in which the lines indicating these \nchanges lie. \n\n\n\n\nFig. 3. \xe2\x80\x94 "Surface" Diagram Illustrating Derivation of "Combination" Law \n\nof Gases. \n\nThe following problem shows the cippHcation of equation 10: \nA quantity of air at atmospheric pressure has a volume of \n2000 cubic feet when \'the barometer reads 28.80 inches of mer- \ncury and the temperature is 40\xc2\xb0 C. What will be the volume \n\n\n\ni6 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nof this air at a temperature of o\xc2\xb0 C. when the barometer reads \n29.96 inches of mercury? \n\n\n\n\nLetting \n\n\n\nFig. 4. \xe2\x80\x94 Isometric Drawing Illustrating Derivation of "Combination" Law of \n\nGases. \n\nSolution. Volume, pressure and temperature vary in this \ncase as in the following equation, \n\nPiVi _ P2V2 \n\nPi, Fi, Ti = initial conditions, \nP2, F2, r2 = final conditions, \n\n28.80 X 2000 _ 29.96 X V2 \n40 + 273 o + 273 \' \n\nF2 = 1676.42 cubic feet. \n\nNow, since P2, V2 and Ti in equation (10) are any other \nsimultaneous conditions of the gas, we may also write the follow- \ning more general relations: \n\nPiFi P2F2 PzVz \n\n\n\nthen \n\n\n\nTi \n\n\n\nand, therefore, \n\nwhere R is the " gas constant." \n\n\n\nT2 Ts \n\nPV = RT \n\n\n\n= a constant, \n\n\n\n(10\') \n(11) \n\n\n\nPROPERTIES OF PERFECT GASES 1 7 \n\nThis constant is a most important quantity in thermody- \nnamic calculations, its value varying only with the kind of gas \ndealt with. It must be remembered that equation (ii) was \nderived for one pound of gas, and therefore, in order to make \nit applicable for any mass of gas, it is only necessary to mul- \ntiply the constant R by the mass M, the volume V in the \nequation being the volume corresponding to the mass M. We \nthus obtain the final form of the equation representing the \n" combination law " of perfect gases: \n\nPV = MRT, (iiO \n\nwhere P = absolute pressure in pounds per square foot, \nV = actual volume in cubic feet, \nM = mass of gas in pounds, \n\nR = the "gas constant\'\' for one pound of gas in^ foot- \npound units, \nT = the absolute temperature in Fahrenheit degrees. \n\nThis equation is applicable to any perfect gas within the \nlimits of pressure and temperature employed in common en- \ngineering practice. The \'\'thermodynamic" state of a gas is \nknown when its pressure, volume, temperature, mass and com- \nposition * are known, and when any four of these quantities \nare known the fifth can be found by equation (iiO- \n\nValues of the constant R in the table on page 23 have been \ncalculated from experimental data of the specific volumes of the \nvarious gases mentioned. Thus, the specific volume of air is \ngiven as 12.38 cubic feet per pound at 32\xc2\xb0 F. (492 degrees \nabsolute) and at atmospheric pressure (14.7 X 144= 21 17 pounds \nper square foot). Substituting these data in equation (iiQ, we \nhave \n\n2117 X 12.38 = iXR X492, \nR = 53-3 (nearly). t \n\nIt is also useful to observe that the value of R for any gas \n\n* The gas constant R depends upon the composition of the gas. \nt The tables on page 23 were calculated from slightly more exact data, but the \nagreement shown is good enough for all engineering calculations. \n\n\n\nl8 ENGINEERING THERMODYNAMICS \n\ncan be approximately calculated by dividing the number 1544 \nby its molecular weight. For example, the value of R for acet- \nylene gas (C2H2), having a molecular weight of 24 + 2 or 26, is \n1544 -^ 26 or 59.4, which is the value commonly given for this \ngas. \n\nThis latter method gives a very rapid and sufficiently exact \nmeans for determining the value of R of any gas for which the \nchemical formula is known. \n\nThe following problem shows the application of equation 11\': \n\nWhat is the volume of a tank that will hold a mass of 5 \npounds of air when the pressure is 200 pounds per square inch \nabsolute and the temperature 40\xc2\xb0 C? \n\nSolution. Three unknowns are given and the fourth is to be \nfound by the equation, \n\nPiVi = MiRiTi. \n\nChanging values given to the proper units, \n200 pounds per square inch absolute X 144 \n\n= 28,800 pounds per square foot absolute. \n\n40\xc2\xb0 C. X I + 32 = 104\xc2\xb0 F. \nThen substituting in equation 11\', \n\n28,800 X Fi = 5 X 53.3 (104 + 460); \nVi = 5.22 cubic feet. \n\nHeat and Its Effect on Expansion. It is most important \nto discuss here, before going on to the more difficult prob- \nlems having to do with the expansions of gases, the most \nessential considerations involved in this study. In general the \neffect of adding heat to a gas is to raise its temperature. If \nheat is added so that at the same time the gas expands and \ndoes work its temperature may either rise or fall according as \nthe amount of heat added is greater or less than the heat equiva- \nlent of the external work done. The following concise statement \nalways appHes, and will be found very useful: \nHeat added = increase in internal energy -f- external work (12) \n\nSpecific Heat. The amount of heat required to raise the \ntemperature of a unit mass of a substance one degree is called its \n\n\n\nPROPERTIES OF PERFECT GASES 19 \n\nspecific heat. In the English system the specific heat is the \nnumber of British thermal units (B.t.u.) required to raise the \ntemperature of a pound of the substance 1\xc2\xb0 F. \n\nThe specific heat of gases and vapors changes considerably \nin value according to the conditions under which the heat is \napplied. If heat is applied to a gas or a vapor held in a closed \nvessel, with no chance for expansion, no work is performed, and, \ntherefore, all the heat added is used to increase the temperature. \nThis is the condition in a boiler, for example, when no steam is \nbeing drawn off. In this case we use the symbol C^ to represent \nthe specific heat during its appHcation at constant volume. If, \non the other hand, the heating is done while the pressure is kept \nconstant and the volume is allowed to change to permit expan- \nsion and the performance of work,* we use the symbol Cp to \nrepresent the specific heat during its application at constant \npressure. Heat application at constant pressure is the condition \nthat is most interesting to the engineer. When his engines are \nrunning his boilers are making steam at constant pressure. \nThe heat energy absorbed by a pound of a substance for raising \nonly the temperature must certainly be the same regardless of \nthe conditions of pressure and volume. Since, however, for \nconstant pressure conditions some external work is always done, \nrequiring a correspondingly larger amount of heat energy than \nfor the case when the volume is constant, it follows that Cp is \nalways greater than Cy. In other words, Cp is equal to Cy plus \nthe heat equivalent of the work done by one pound of the sub- \nstance in expanding at constant pressure, while the tempera- \nture is raised 1\xc2\xb0 F., if Cp and C^ are in B.t.u. \n\nSpecific heat can also be expressed in units of work (foot- \npounds). When Cp and Cv are expressed in B.t.u. then the \ncorresponding values in foot-pounds will be 778 Cp and \n778 Q. \n\nExternal Work. The external work or the work done by a \ngas in its expansion is represented graphically by Fig. 5. This \n\n* Work performed as the result of expansion of a gas or vapor is always done \nat the expense of an equivalent amount of heat energy. \n\n\n\n20 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nW//////////////////y/////(\'/. \n\n\n\ngE= \n\n\n\nPi \xe2\x80\x94 \n\n\n\n///<><\'/<\'/// . ,\'<\'/<\'<\'{<\'//<\'/ \n\n\n\nis the simplest sort of figure to show that the area under the \n\nexpansion line BC is proportional to the work done in the expan- \nsion. Let us represent the initial \ncondition of the gas at B as regards \npressure and volume by Pi and Vi \nand the final condition at C by Pi \nand V2 (expansion being at con- \nstant pressure), then elementary \nmechanics teaches that the force \nP moved through a distance rep- \nresented by the difference of ab- \nscissas (F2 \xe2\x80\x94 Fi) is a measure of \nthe work done. Obviously the area \nunder the line BC divided by the \n\nhorizontal length (F2 \xe2\x80\x94 Fi) is the average value of the force P. \n\nIf, further, and in general, we represent the area under BC by \n\nthe symbol A^ then we can write, \nA \n\n\n\nV: \n\n\n\nFig. 5.- \n\n\n\nVolume \n\nExternal Work by Expan- \nsion. \n\n\n\nF2-F1 \n\n\n\n= average value of P, \n\n\n\nwhether or not P is constant. And also, \n\nA \n\n\n\nWork done = \n\n\n\nF2-F1 \n\n\n\nX (F2 - V,) = A. \n\n\n\nThe same principle applies whether the Kne BC is a straight \nline as shown or a very irregular curve, as will be shown later. \n\nInternal Energy, The heat energy possessed by a gas or vapor, \nor, we may say, the heat energy which is in a gas or vapor in \na form similar to " potential " energy, is called its internal \nenergy.* Thus, an amount of heat added to a substance when \nno work is performed is all added to the internal energy of that \nsubstance. On the other hand, when heat is added while work \nis being performed, the internal energy is increased only by the \ndifference between the heat added and the work done. \n\nInternal energy may also be defined as the energy which a \n\n* It is also sometimes called intrinsic energy, since it may be said to "reside" \nwithin the substance and has not been transferred to any other substance. \n\n\n\nPROPERTIES OF PERFECT GASES 21 \n\ngas or vapor possesses by virtue of its temperature, and may \nbe expressed as follows: \n\nInternal energy = C^T (in B.t.u.), \n\nwhere T is the absolute temperature and Cv the specific heat \nat constant volume. From the paragraph on specific heat it \nwill be remembered that Cv takes into account only that heat \nrequired to raise the temperature, since under constant volume \nconditions no external work is done; and therefore, in dealing \nwith internal energy, since we do not care anything about the \nexternal work that may have been done, Cv is always used. \n\nIt is with the change in internal energy that we are generally \nconcerned, and this change is, obviously. \n\nIncrease in internal energy \n\n= Cv {T2 \xe2\x80\x94 Ti) in B.t.u. (for one pound of gas). (13) \n\nThis formula will always apply, for in whatever way the \ntemperature of a substance is changed from Ti to T2 the change \nin its internal energy is the same since, by definition, it depends \nonly upon the tempera ture^ \n\nJoule\'s Law. In the case of ideally perfect gases such as \nthese thermodynamic equations must deal with, it is assumed, \nwhen a gas expands * without doing external work and with- \nout taking in or giving out heat (and, therefore, without \nchanging its stock of internal energy) that its temperature does \nnot change. \n\nRelation of Specific Heats and Gas Constant. In order to \nderive this relation we must consider certain fundamental equa- \ntions in thermodynamics and in general physics. \n\nA formula in physics for the total heat added, H, is as follows: \n\nH = mass X specific heat X difference in temperature. \n\n* It was for a long time supposed that when a gas expanded without doing \nwork, and without taking in or giving out heat, that its temperature did not change. \nThis fact was based on the famous experiments of Joule. Later investigations \nby Lord Kelvin and Linde have shown that this statement is not exactly correct \nas all known gases show a change in temperature under these conditions. This \nchange in temperature is known as the "Joule-Thomson " effect. \n\n\n\n22 ENGINEERING THERMODYNAMICS \n\nIf the heat is added at constant pressure, then obviously, in \nB.t.u., \n\nH = MC^{T,-T,). (14) \n\nAlso, by equation (13), the increase in internal energy when \nheat is added = MC^{T2-Ti). (15) \n\nAnother formula in physics is stated as follows: \n\nWork (external) = force X distance = Fd \n\n= PAd, where F= force in pounds, \nd = distance moved in feet, \nP = unit pressure in pounds per square foot, \nand A = area in square feet. \n\nCombining the factors A and d we obtain the change of \nvolume, \n\nAd = Av, where A^; is the change in volume. \n\nTherefore, External work = PAv. \n\nWhen the pressure remains constant, equation (15) may be \nexpressed as follows: \n\nExternal work = P (V2 \xe2\x80\x94 Vi), foot-pounds \n\nRemember the fundamental equation (12), which is as follows: \nHeat added = increase in internal energy + external work. \nSubstituting in the above equation (14), (15), and (16), we \nhave \n\nMC, {r, - Td = MQ (T, - T,) + ^^\\~^\'^ \' (17) \n\n778 \n\nBy equation (iiO> \n\nPV2 = MRT2 and PVi = MRTi. \nSubstituting these values ii^i (17), \n\nMC, (T, - ro = Mc. (r. - r,) + ^^^J^"^\'^ \xe2\x80\xa2 (is) \n\n778 \n\nSimplifying, Cp = Cv + \xe2\x80\x94 r- (19) \n\n778 \n\n\n\nPROPERTIES OF PERFECT GASES \n\n\n\n23 \n\n\n\nRatio of Specific Heats (7). The ratio of the specific heat at \nconstant pressure to the specific heat at constant volume enters \ninto many thermodynamic equations and is of so much im- \nportance in engineering calculations that it is commonly repre- \nsented by the symbol 7; thus \n\n\n\n\n\n\n(20) \n\n\n\nFrom the previous discussion of the relative values of Cp and C\xe2\x80\x9e \nit is evident that 7 must always have a value greater than unity. \nBy means of equations (19) and (20), the relation between \nCv, 7 and R may be obtained: \n\nR \n\n\n\nC/p \xe2\x80\x94 Ov I \n\n\n\nFrom (b), \nSubstituting in (a) \n\n\n\n\n\n\nTherefore, \n\n\n\n778 \n\n\n\nCv (7 - i) = \n\n\n\na \n\n\n\n778 \n\nR \n\n\n\n778 \n\n\n\nR \n\n\n\n778 (7 - i) \n\n\n\n(a) \n(b) \n\n\n\n(21) \n\n\n\nDATA FOR VARIOUS GASES \n\n\n\nName of gas. \n\n\n\nAir (pure) \n\nCarbon dioxide (CO2) .... \nCarbon monoxide (CO) . . . \n\nHydrogen (H2) \n\nMarsh gas (CH4) \n\nNitrogen (N2) \n\nOxygen (O2) \n\nSulphur dioxide (SO2) .... \n\n\n\nMolecu- \nlar \nweight. \n\n\n\n29 \n\n44 \n28 \n2 \n16 \n28 \n32 \n64 \n\n\n\nSpe- \ncific * \nvolume, \ncu. ft, \nper lb. \n\n\n\n12.38 \n\n8.17 \n12.80 \n\n179-65 \n22.45 \n12.81 \nII .22 \n\n5-61 \n\n\n\nDen- \nsity,* \nlbs. per \ncu. ft. \n\n\n\n0.0807 \n0.1224 \n0.0779 \n0.0056 \n0.0445 \n0.0779 \n0.0891 \n0.1781 \n\n\n\nValue \nof R per \nlb. (ft.- \n\nIb. \nunits). \n\n\n\nS2> \n35 \n55. \n766, \n96, \n\n55 \n48, \n\n24 \n\n\n\n0.237 \n0.200 \no . 243 \n3-4IO \n\n0-593 \n0.244 \n0.217 \n0.154 \n\n\n\n.169 \n\xe2\x80\xa2154 \n\xe2\x80\xa2173 \n\xe2\x80\xa2435 \n.470 \n\xe2\x80\xa2173 \n\xe2\x80\xa2155 \n.123 \n\n\n\n* Under standard conditions, Le., 32\xc2\xb0 F. and 14.7 lbs. per sq. in. absolute pressure. \n\n\n\n24 ENGINEERING THERMODYNAMICS \n\nLOW TEMPERATURE RESEARCHES \n\nIn Harper^ s Monthly Magazine"^ is described the remarkable \nachievement of Professor Kamerlingh Onnes, of Leyden, in ap- \nproaching the absolute zero of temperature, viz., \xe2\x80\x94273\xc2\xb0 C, \nwithin 1.8 degrees, when using helium. This is 93 degrees colder \nthan the temperature of liquid air. Professor Onnes in 1913 \nwon the Nobel Prize for researches in the science of physics, \nand in 19 14 cooled a coil of lead wire to a temperature where \nthe electrical resistance becomes zero, the current, once started, \ncontinuing to flow without supply so long as its temperature is \nkept at about 270 degrees below 0\xc2\xb0 C. \n\nAt this low temperature many of the properties of matter at \nordinary temperatures were expected to disappear, or at least \nto become greatly modified. The temperature of liquid air is \n\xe2\x80\x94 180\xc2\xb0 C, and by its evaporation it is possible to reach \xe2\x80\x94205\xc2\xb0 C. \nAt these temperatures ordinary soft lead becomes hard and \nbrittle like cast iron, rubber becomes fragile like glass, and \nalcohol can be frozen to a white solid and a candle can be \nmade from it. The next lowest depression in temperature was \nmade by Sir James Dewar, who, in liquefying hydrogen, attained \na temperature of 253 degrees below 0\xc2\xb0 C, only 20 degrees from \nabsolute zero of temperature, but no startHng effects were ob- \nserved there. \n\nA reduction in temperature by evaporating a liquid with a \nvery low boiling point results from the consumption of the \nheat within the liquid, making it necessary to prevent the \nentrance of heat from the outside. This is accomplished, firstly, \nby evaporating the liquid in a vacuum-jacketed glass vessel \nhaving the heat-insulating property of a \'\' thermos " bottle; sec- \nondly, by surrounding the vessel with a Kquid of low temperature. \nUntil recently the lowest temperature produced, about \xe2\x80\x94260\xc2\xb0 C, \nwas obtained by evaporating liquid hydrogen under reduced \npressure in a vessel surrounded by liquid air. This tempera- \nture was thought to be the lowest limit attainable, liquid hydro- \n\n* Vol. 129 (1914), pages 783-9. \n\n\n\nPROPERTIES OF PERFECT GASES 25 \n\ngen having a lower boiling point than any other substance then \nknown. \n\nHowever, when the peculiar and extremely rare gas known as \nhelium was discovered, which is formed by the spontaneous de- \ncomposition of the element radium, the attainment of the ab- \nsolute zero seemed feasible, provided a sufficient supply of \nhelium could be obtained. This gas proved to be even more \ndifficult to Hquefy than hydrogen, its boiling point being lower. \nHowever, it requires about 1000 cubic feet of gas to form one \ncubic foot of hquid. Professor Onnes succeeded in building an \napparatus which enables him now to produce about half a pint \nof liquid heHum in a few hours. The glass vessel in which the \nhelium is liquefied must be surrounded by a vessel containing \nliquid hydrogen, boihng under reduced pressure, which in turn \nmust be surrounded by liquid air. The problem is comparable \nin difficulty to the construction of an apparatus for making \nice inside of a furnace. Helium sells for about $50 a quart, \nbut by the generosity of an American private commercial source, \nProfessor Onnes obtained an enormous quantity after searching \nEuropean markets in vain. \n\nLiquefaction of the gas helium is accomplished by means of \nan apparatus similar to that employed for the liquefaction of \nair, except that the compression pump, pipes and receiver must \nbe absolutely gas-tight. Professor Onnes employs a pump with \nmercury pistons; every joint and valve in the system is im- \nmersed in a bath of oil, so that leaks will manifest themselves \nby small bubbles rising through the Hquid. The apparatus is \nsufficiently perfect to permit the compression of heliimi to 3000 \npounds per square inch. The compressed fluid is passed through \npipes immersed in liquid air or hydrogen, then through the \ncoiled tube of the Kquefier, whereby a portion becomes liquid \nand drops into a glass vessel, while the remainder passes off \nthrough another pipe and returns to the compressor \xe2\x80\x94 the proc- \ness repeating itself over and over again without any loss of gas. \n\nAt atmospheric pressi^re Hquid heHum boils at \xe2\x80\x94 268.6\xc2\xb0 C. or \n4.4 degrees absolute; but by reducing the pressure in the tube \n\n\n\n26 ENGINEERING THERMODYNAMICS \n\nby means of an air pump, Professor Onnes has succeeded in \nreaching an actual measured temperature of only 1.8 degrees \nabsolute. Such temperatures can be measured only by means \nof a gas thermometer filled with helium under very low pressure, \ninstead of the usual hydrogen-filled thermometer. \n\nPROBLEMS \n\n1. The pressure in a gas tank is 90 lbs. per sq. in. by the gage. If \natmospheric pressure is 15 lbs. per sq. in. absolute, what is the absolute \npressure in the tank ? \n\nAns. 105 lbs. per sq. in. \n\n2. What is the absolute pressure within the above gas tank in pounds \nper square foot? Ans. 15,120 lbs. per sq. ft. \n\n3. The gas in Problem i is at 60\xc2\xb0 F. temperature. What is the absolute \nFahrenheit temperature of this gas ? Ans. 520 degrees. \n\n4. Air is at a temperature of 40\xc2\xb0 C. What is the absolute Fahrenheit \ntemperature? Ans. 564\xc2\xb0 absolute F. \n\n5. Air at constant pressure has an initial volume of 2 cu. ft. and \ntemperature of 60\xc2\xb0 F. ; it is heated until the volume is doubled. What is \nthe resulting temperature in degrees Fahrenheit ? \n\n. J 1040\xc2\xb0 absolute F. \n\nJxnS . \xe2\x96\xa0{ n n ^\xe2\x80\xa2 t-\xc2\xab \n\nI 580 ordmary F. \n\n6. Air is cooled under constant volume. The initial condition is pres- \nsure of 30 lbs. per sq. in. absolute and temperature of 101\xc2\xb0 F. The final \ncondition has a temperature of 50\xc2\xb0 F. What is the final pressure ? \n\nAns. 27.27 lbs. per sq. in. absolute. \n\n7. One pound of hydrogen is cooled under constant pressure from a \nvolume of i cu. ft. and temperature of 300\xc2\xb0 F. to a temperature of 60\xc2\xb0 F. \nWhat is the resulting volume? Ans. 0.684 cu. ft. \n\n8. A tank whose voliune is 50 cu. ft. contains air at 105 lbs. per sq. in. \nabsolute pressure and temperature of 80\xc2\xb0 F. How many pounds of air does \nthe tank contain ? Ans. 26.26 lbs. \n\n9. An automobile tire has a mean diameter of 34 in. and 4 in. width. \nIt is pumped to 80 lbs. per sq. in. gage pressure at a temperature of 60\xc2\xb0 F.; \natmospheric pressure 15 lbs. per sq. in. absolute. \n\n(a) How many pounds of air does the tire contain ? \n\nAns. 0.38 lb. \n\n(b) Assuming no change of volume, what would be the gage pres- \n\nsure of the tire if placed in the sun at 100\xc2\xb0 F. ? \n\nAns. 87 lbs. per sq. in. \n\n\n\nPROPERTIES OF PERFECT GASES 27 \n\n10. An acetylene gas tank is to be made to hold 0.25 lb. of this substance \nwhen the pressure is 250 lbs. per sq. in. gage, atmospheric pressure 15 lbs. \nper sq. in. absolute, and the temperature of the gas 70\xc2\xb0 F. What will be \nthe volume in cu. ft.? Ans, 0.206 cu. ft. \n\n11. A quantity of air at a temperature of 70\xc2\xb0 F. and a pressure of 15 lbs. \nper sq. in. absolute has a volume of 5 cu. ft. What is the volimie of the \nsame air when the pressure is changed at constant temperature to 60 lbs. \nper sq. in. absolute? Ans. 1.25 cu. ft. \n\n12. How many pounds of air are present in Problem 11? \n\nAns. 0.383 lb. \n\n13. The volume of a quantity of air is 10 cu. ft. at a temperature of \n60\xc2\xb0 F. when the pressure is 15 lbs. per sq. in. absolute. What is the pres- \nsure of this air when the volume becomes 60 cu. ft. and the temperature \n60\xc2\xb0 F.? Ans. 2.5 lbs. per sq. in. absolute. \n\n14. How many pounds of air are present in Problem 13 ? \n\nAns. 0.779 lb. \n\n15. A tank contains 200 cu. ft. of air at a temperature of 60\xc2\xb0 F. and \nunder a pressure of 200 lbs. per sq. in. absolute. \n\n(a) What weight of air is present? Ans. 207.5 lbs. \n\n(b) How many cubic feet will this air occupy at 15 lbs. per sq. \n\nin. absolute and temperature of 100\xc2\xb0 F. ? \n\nAns. 2880 cu. ft. \n\n16. The volume of a quantity of air at 70\xc2\xb0 F. under a pressure of 15 lbs. \nper sq. in. absolute is 20 cu. ft. What is the temperature of this air when \nthe volume becomes 5 cu. ft. and the pressure 80 lbs. per sq. in. absolute? \n\nAns. 707\xc2\xb0 F. absolute. \n\n17. Air at constant pressure has a specific heat of 0.237 B.t.u. How \nmany B.t.u. are required to raise 2 lbs. from 60\xc2\xb0 to 100\xc2\xb0 F. ? \n\nAns. 18.96 B.t.u. \n\n18. Three pounds of a substance has 75 B.t.u. suppHed to it to change \nthe temperature 100\xc2\xb0 F. What is its specific heat? Ans. 0.25 B.t.u. \n\n19. If the specific heat of air under constant pressure is 0.237 and the \nvalue of R is 53.3, find the value of the specific heat under constant volume. \n\nAns. 0.169 B.t.u. \n\n20. From the data in Problem 19 find the value of 7 for air. \n\nAns. 1.41. \n\n21. An auto tire has a volume of 0.66 cu. ft. and is to be pumped to a \npressure of 85 lbs. per sq. in. gage at a temperature of 50\xc2\xb0 F. What vol- \nume tank will be required to inflate four such tires if the air can be stored \nin this tank at 250 lbs. per sq. in. gage pressure and temperature of 70\xc2\xb0 F.? \n\nSuggestion. Assume the tires are filled with air at atmospheric pressure \nbefore starting to pump, and also when all tires have been inflated that \nthe tank will contain its volume of air at the final pressure in the tires. \n\n\n\n28 ENGINEERING THERMODYNAMICS \n\nLet Pa = atmospheric pressure; Pt = gage pressure in tires; Pk = gage \npressure in tank;* T = absolute temperature in tires or tank; Vt = vol- \nume of tires ; Vk = volume of tank ; Ma = mass (pounds) of air in tires \nwhen at atmospheric pressure; Mb = total mass (pounds) of air in tires \nwhen at the tire pressure; Mc = total mass (poimds) of air in tank at final \ntire pressure; Md = total mass (pounds) of air in tank at the final tank \npressure before beginning to fill the tires. \n\nThen \n\n(Mb \xe2\x80\x94 Ma) = Mass suppUed tires = Md \xe2\x80\x94 Mc \niPt-hPa)Vt = MbXRxTt \n\nPaXVt = MaXRXTt \n\nPtXVt= (Mb - Ma) RTt \nSolving this for (Mb \xe2\x80\x94 Ma) : \n\niPjc + Pa)Vk = MdXRXTk \n(Pt + Pa)Vk-McXRxTk \n(Pk-Pt)Vk=(Md-Mo)RTu. \n\nSubstitute the value of (Mb \xe2\x80\x94 M,^ found above for (Md \xe2\x80\x94 Mc) and then \nsolve for Vk- Ans. Vk = i-4i cu. ft. \n\n22. How many B.t.u. are required to double the volume of i lb. of air \nat constant pressure from 50\xc2\xb0 F.; specific heat is 0.237 B.t.u.? \n\nAns, 121 B.t.u. \n\n23. A tank filled with 200 cu. ft. of air at 15 lbs. per sq. in. absolute and \n60" F. is heated to 150\xc2\xb0 F. \n\n(a) What will be the resulting air pressure in the tank ? \n\nAns. 17.6 lbs. per sq. in. absolute. \n\n(b) How many B.t.u. will be required to heat the air? \n\nAns. 240 B.t.u. \n\n24. A tank contains 200 cu. ft. of air at 60\xc2\xb0 F. and 40 lbs. per sq. in. \nabsolute. If 500 B.t.u. of heat are added to it, what will be the resulting \npressure and temperature? . 145-5 lbs. per sq. in. absolute. \n\n"\'\'"^\' 1 131.3\xc2\xb0 F. \n\n* All pressures are in pounds per square foot. \n\n\n\nCHAPTER III \nEXPANSION AND COMPRESSION OF GASES \n\nThe general equation in the form PV = MRT"^ for the \nexpansion or compression of gases has three related variables, \n(i) pressure, (2) volume and (3) temperature. For a given \nmass of gas with any two of these variables given, obviously, the \nthird is fixed. As regards the analysis of the action of heat \nengines, the pressure and volume relations are most important, \nand graphical diagrams, called pressure- volume or P-V dia- \ngrams, are frequently needed to assist in the analysis. The \n\n\n\n\xc2\xa9100 \n\nft \n\n\n\nPV3 \n\n\nPVi \n\n\nS^^ \n\n\nV \nI 1 \n\n\nj \n\n1 ! 1 \n\n\n\xe2\x80\x94 (p \n\n1 L.. 1 \n\n\n\nP^ 2 4 6 8 10 \n\nVolume, eu. ft. \n\nFig. 6. \xe2\x80\x94 Diagram of Expansion and Compression at Constant Pressure. \n\nsimplest sort of diagram of this kind is shown in Fig. 6, in \nwhich the vertical scale of coordinates represents pressures and \nthe horizontal, volumes. \n\nAssume that the pressure and volume of a pound of a given \ngas are given by the coordinates P and Fi, which are plotted in \nthe middle of the diagram. It will be assumed further that the \npressure remains constant in the changes to be indicated. Now \nif the gas is expanded until its volume becomes V2, then its con- \ndition as regards pressure and volume would be represented at \nPV2. If, on the other hand, the gas had been compressed while \na constant pressure was maintained, its final condition would \n\n* Equation (iiO, P- i7- \n29 \n\n\n\n30 ENGINEERING THERMODYNAMICS \n\nbe represented by the point PF3 to the left of PFi. Similarly, \nany line whether straight or curved extending from the initial \ncondition of the gas at PVi will represent an expansion when \ndrawn in the direction away from the zero of volumes and will \nrepresent a compression when tending toward the same zero. \n\nIt is readily shown that areas on such diagrams represent the \nproduct of pressure and volume, and, therefore, work or energy \n(see page 20). Thus in Fig. 6 the area under the curve PF3 to \nPV2 represents on the scales given 100 (pounds per square foot) \nX (9 \xe2\x80\x94 i) cubic feet or 800 foot-pounds irrespective of whether \nit is an expansion or a compression from the initial condition. \nWhen a series of curves are joined together to form a closed \nfigure which shows the varying conditions of pressure and vol- \nume of a gas, the figure is called an indicator diagram. \n\nMost of the lines to be studied in heat engine diagrams are \neither straight or else they can be exactly or approximately \nrepresented by an equation in the form \n\nPF" = a constant, (22) \n\nwhere the index n, as experimentally determined, has varying \nnumerical values, but is almost invariably constant for any \none curve. When the lines of the diagram are straight the \nareas of simple rectangles and triangles need only be calculated \nto find the work done, and for these cases no discussion is here \nnecessary. The two most common forms of curves to be dealt \nwith in expansions are (i) when there is expansion with addi- \ntion of heat at such a rate as to maintain the temperature of \nthe gas constant throughout the expansion. Such an expansion \nis called isothermal. The other important kind of expansion \n(2) occurs when work is done by the gas without the addition \nor abstraction of heat. To do this work some of the internal \nheat energy contained in the gas must be transformed in propor- \ntion to the amount of work done. Such an expansion is called \nadiabatic. \n\nThe following problems show the application of the foregoing \ntheory to straight line expansion: \n\n\n\nEXPANSION AND COMPRESSION OF GASES 31 \n\nCase I. One pound of air having an initial temperature of \n60\xc2\xb0 F. is expanded to 100\xc2\xb0 F. under constant pressure. Find \n\n(a) External work during expansion; \n\n(b) Heat required to produce the expansion. \n\nSolution. From the fundamental theory, the heat added \nequals the increase in internal energy plus the external work \ndone. In solving then for the heat added or required during \nany expansion it is only necessary to find the external work \n(which is equal to the area under the expansion curve) and add to \nit the heat needed to increase the internal energy. \n\nThe external work W = Pi (V2 \xe2\x80\x94 Vi) or its equivalent \nMR{T2 -Ti)= 1 X 53.3 (100 + 460) - (60 + 460) = 2132 ft.-lbs. \n\nThe increase in internal energy \n\n= Ma {T2 - Ti) \n\n= I X 0.169 (100 + 460) \xe2\x80\x94 (60 + 460) \n= 6.75 B.t.u. \n\n21^2 \nHeat required = 6.75 + \xe2\x80\x94 ^ = 9.50 B.t.u. \n\n778 \n\nAs part of the data Cp is known. Then for this case of con- \nstant pressure expansion, the heat required equals by another \nmethod, \n\nMCp {T2 - Ti) = I X 0.237 (100 + 460) - (60 + 460) \n= 9.5 B.t.u. approximately. \nCase 2. One pound of air having an initial temperature of \n60\xc2\xb0 F. is heated at constant volume until the final temperature \nis 120\xc2\xb0 F. Find \n\n(a) External work; \n\n(b) Heat required. \n\nSolution. Applying the same theory as above, \nHeat added = increase in internal energy + external work. \nExternal work = o, since there is no area under the curve. \nThen \n\nHeat added = increase in infernal energy + o \n= MC. (r2 - Ti) + o \n= I X o\'i69 (100 + 460) \xe2\x80\x94 (60 + 460) + o \n= 6.76 B.t.u. \n\n\n\n32 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nIsothermal Expansion and Compression. In an isothermal \nexpansion or compression the temperature of the working sub- \nstance is kept constant throughout the process. When the \ntemperature of the gas is kept constant, while the pressure and \nvolume change, Boyle\'s Law (page lo) applies and we have \nsimply \n\nPV = C = a constant. (23) \n\nThis is the equation of a curve which is known in analytic \ngeometry as a rectangular hyperbola. It is the special case of \nthe general equation FV"" = constant (22), in which the index \nn = 1. If various values are substituted for V in equation (23) \nand the values found together with the corresponding calculated \n\n\n\n\nVolume \nFig. 7. \xe2\x80\x94 -Work done by Isothermal Expansion and Compression. \n\nvalues of P are plotted, a curve Hke the one in Fig. 7 is obtained. \nAn equation for the work done in the isothermal expansion of a \ngas will now be determined, starting with the initial condition \nof pressure and volume represented by Pi and Vi. The external \nwork performed is shown graphically by the shaded area under \nthe curve between A and B. Two vertical lines close together \nin the figure are the limits of a narrow closely shaded area and \nindicate an infinitesimal volume change dV, so small that the \npressure may be assumed constant for the interval. Work \ndone during this small change of volume is, then, \n\ndW = P dV, \n\n\n\nEXPANSION AND COMPRESSION OF GASES 33 \n\nand for a finite change of volume of any size as from Vi to V2 \nthe work done, W (foot-pounds), is \n\n\n\n= J PdV. (24) \n\n\n\nw \n\nFor integration of this form it is necessary to substitute P \nin terms of V. Assume that P and V are values of pressure \nand volume for any point on the curve of expansion of a gas of \nwhich the equation is \n\nPV = C (see equation 23). \nThen ^ " F* \n\nSubstituting this value of P in equation (24), we have \n\nTF = C(logeF2-log. Fi). (25) \n\nSince the initial conditions of the gas are Pi and Vi, we have \nPV = C = PiVi, \nand substituting this value of C in equation (25), we obtain \n\nW = PiVi (loge V2 - loge Fi) \n\nor IF = Pi Fi loge\xe2\x80\x94 ^ (in foot-pounds). (26) \n\nVi \n\nThe above equation is quite general in application and can be \nused for any values of volume represented by Vi and F2. \n\nUnits of mass do not enter. For the work done by a certain \nmass of gas under the same conditions we could write, since \n\n\xe2\x96\xa0 PiVi = MRT (in foot-pounds) and ]^ = \xc2\xa7, \n\nVi P2 \n\nW = MRT loge ^ = MRT log, ^ (in foot-pounds) . (27) \nyi P2 \n\nV2 / \nOften the ratio -f is called the ratio of expansion and is rep- \n\n\n\n34 ENGINEERING THERMODYNAMICS \n\nresented by r. Making this substitution we have, in foot- \npounds, \n\nW = MRT loge r. (28) \n\nThese equations refer to an expansion from PiFi to FiVi. \nIf on the other hand, we wanted the work done in a compression \nfrom P\\V\\ to P3F3 the curve of compression would be from A \nto C and the area under it would be its graphical representa- \ntion. Equations (26), (27) and (28) would represent the work \ndone the same as for expansion except that the expression \nwould have a negative value; that is, work is to be done upon \nthe gas to decrease its volume. \n\nThe isothermal expansion or compression of a " perfect " gas \ncauses no change in its stock of internal energy since T is constant \n(see page 20). During such an expansion the gas must take in \nan amount of heat just equal to the work it does, and conversely \nduring such a compression it must reject an amount of heat \njust equal to the work spent upon it. This quantity of heat R \n(in B.t.u.) is, from equation (27), \n\nH = \xe2\x80\x94 \xe2\x80\x94 - loge \xe2\x80\x94 (m B.t.u.). (29) \n\nThe following problem shows the application of the foregoing \ntheory to isothermal expansions: \n\nAir having a pressure of 100 pounds per square inch absolute \nand a volume of i cubic foot expands isothermally to a volume \nof 4 cubic feet. Find \n\n(a) External work of the expansion; \n\n(b) Heat required to produce the expansion; \n\n(c) Pressure at end of expansion. \nSolution, (a) Since expansion is isothermal. \n\nExternal work, W = PiVi loge* ^ \n\nVi \n\n= 100 X 144 X I X 1.3848 \n\n= 19,941 foot-pounds. \n\n* 2.3 X log base 10 = log base e. Tables of natural logarithms are given on \npages 195-196. \n\n\n\nEXPANSION AND COMPRESSION OF GASES 35 \n\n(b) Since the heat added equals increase in internal energy \nplus external work, and since the temperature remains constant \n(requiring therefore no heat to increase the internal energy), the \ninternal energy equals zero and the heat added equals the work \ndone. \n\nThen Heat added = external work = ^\'^\'7 = 2t;.6 B.t.u. \n\n778 \n\n(c) Since PiFi = P2F2, \nthen 100 X I = ^2 X 4, \n\nP2 = 25 pounds per square inch absolute. \n\nIf a gas expands and does external work without receiving a \nsupply of heat from an external source, it must derive the amount \nof heat needed to do the work from its own stock of internal \nenergy. This process is then necessarily accompanied by a low- \nering of temperature and the expansion obviously is not iso- \nthermal. \n\nAdiabatic Expansion and Compression. Another most im- \nportant mode of expansion and compression from the viewpoint \nof the engineer is that in which the working substance neither \nreceives nor rejects heat as it expands or is compressed and is \ncalled adiabatic. A curve which shows the relation of pressures \nto volumes in such a process is called an adiabatic line (see \nFig. 8). In any adiabatic process the substance is neither \ngaining nor losing heat by conduction or radiation or internal \nchemical action. Hence the work which a gas does in such an \nexpansion is all done at the expense of its stock of internal \nenergy, and the work which is done upon a gas in such a com- \npression all goes to increase its stock of internal energy. We \ncould secure ideally adiabatic action if we had a gas expanding, \nor being compressed, in a cylinder which in all parts was a per- \nfect non-conductor of heat. The compression of gas in a cylin- \nder is approximately adiabatic when the process is very rapidly \nperformed, but when done so slowly that the heat has time to \nbe dissipated by conduction the compression is more nearly \nisothermal. Fig. 8 shows on a pressure- volume diagram the \n\n\n\n36 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\ndifference between an isothermal and an adiabatic for expansion \nor compression from an initial condition PiVi at A to final \nconditions at B and C for expansions, and at D and E for com- \npressions. \n\n\n\n\nVolume \nFig. 8. \xe2\x80\x94 Isothermal and Adiabatic Expansion Lines. \n\nDerivation of Equation for Adiabatic Expansion (or Com- \npression). In order to derive the pressure- volume relation for \na gas expanding adiabatically, consider the fundamental equa- \ntion (page 1 8), \n\nHeat added = increase in internal energy + external work or \n\nH = MK. {T2 -Ti)+P dV (foot-pounds), (30) \n\nwhere K^, is the specific heat in foot-pound units, i.e., 778 C^. \nNow in adiabatic expansion no heat is added or taken away \nfrom the gas by conduction or radiation, and, therefore, the \nleft hand member of the above equation becomes zero. Fur- \nthermore, since the combination law (page 17) of Boyle and \nCharles can always be applied to perfect gases, the following \nsimultaneous equations may be written: \n\no = MK,dT + PdV, (31) \n\nPV = MRT. (32) \n\nWhen P, V and T vary, as they do in adiabatic expansion, \nequation (32) may be written as follows: \n\nPdV +VdP = MRdT, (33) \n\nand \n\nPdV-\\-VdP \n\n\n\ndT = \n\n\n\nMR \n\n\n\nEXPANSION AND COMPRESSION OF GASES 37 \n\nSubstituting the value of dT in (31), we have \n\n^ ,,, , -,,^ PdV +VdP f . \n\nPdV + MK. -^ = o (34) \n\nRPdV + K,PdV + K.VdP \xe2\x96\xa0\xe2\x96\xa0= o. \nTo separate the variables divide hy PV: \n\n^dV ^ ^ dV ^ ^ dP , . \n\nCollecting terms, \n\nIntegrating, \n\n{R + K,) log F + i:, log P = a constant = c. (36) \n\n^ + ^nog F + log P = c. \n\n\n\nR + K \n\n\n\nlogPV ^^ =c. (37) \n\nSince from equation (19), \n\nR ~f" Kv = ivp, \nwhere Kp and Kv are respectively the specific heats at constant \npressure and at constant volume in foot-pound units. \n\nEquation (37) becomes then \n\nKp \n\nlog PV^^ = c, \nand from equation (20), \n\n\xe2\x80\x94 ^ = 7 and, therefore, \xe2\x80\x94 ^ = 7, \n\nwe may write \n\nPV = a constant. (38) \n\nFollowing the method used for obtaining an expression for \n\nthe work done in isothermal expansion (equation 26), we can \n\nwrite again for the work done, W (in foot-pounds), for a change \n\nof volume from Vi to F2, \n\n\xe2\x96\xa0W^TpdV. (39) \n\n\n\n38 ENGINEERING THERMODYNAMICS \n\nAgain we shall substitute, for purposes of integration, P in \nterms of V as outlined below. In the general expression PV"" \n= c, a, constant (see equation 22), where P and V are values of \npressure and volume for any point on the curve of expansion \nof a gas of which the initial condition is given by the symbols \nPi and Vi, we can then write, \n\n^=fn- (4\xc2\xb0) \n\nAnd substituting (40) in (39), \n\nL\xe2\x80\x94n + iJfi \n\n[ V^l-n _ 7 l-n -| \n\nSince PV"" = c = PiFi" = P2F2", we can substitute for c in \n(42) the values corresponding to the subscripts of V as follows: \n\nn \xe2\x80\x94 n \n\nl\xe2\x80\x94n \n\nor W = ^"^^ ~ ^\'^^ (foot-pounds). (43) \n\nn \xe2\x80\x94 I \n\nSince PV = MRT, \n\nW = ^^(^1-^^) (foot-pounds). (44) \n\nft \xe2\x80\x94 I \n\nEquations (43) and (44) apply to any gas undergoing expan- \nsion or compression according to PV^ = sl constant. In the \ncase of adiabatic expansion of a perfect gas n = y (see equation \n\n38). \n\nChange of Internal Energy During Adiabatic Processes. \n\nSince in adiabatic expansion no heat is conducted to or away \nfrom the gas, the work is done at the expense of the internal \nenergy and, therefore, the latter decreases by an amount equiva- \n\n\n\nEXPANSION AND COMPRESSION OF GASES 39 \n\nlent to the amount of work performed. This loss in internal \nenergy is readily computed by equation (43) or (44). The \nresult must be divided by 778 in order to be in B.t.u. \n\nDuring adiabatic compression the reverse occurs, i.e., there \nis a gain in internal energy and the same formulas apply, the \nresult coming out negative, because work has been done on the \ngas. \n\nRelation between Volume, Pressure and Temperattxre in \nAdiabatic Expansion of a Perfect Gas. Since P, V and T vary \nduring adiabatic expansion, it will be necessary to develop for- \nmulas for obtaining these various quantities. It will be remem- \nbered that equation (10), page 15, applies to perfect gases at \nall times. Therefore, in the case of adiabatic expansion or com- \npression we can write the two simultaneous equations: \n\n^ = f^^- (A) \n\nBy means of these two equations we can find the final condi- \ntions of pressure, volume and temperature, having given two \ninitial conditions and one final condition. \n\nFor instance, having given F1F2 and Ti, to find T2, divide (A) \nby (B)j member for member. Then \n\nFi V2 \n\n\n\nTiFi^ T2V2\'\' \n\n\n\nor \n\n\n\n\n\n\nIn like manner the following formulas can be obtained: \n\n\n\n-\xc2\xa9 \n\n\n\n(46) \n\n\n\n40 ENGINEERING THERMODYNAMICS \n\n(47) \n\n\n\n-\xe2\x96\xa0(PJ \n\n\n\n(48) \n\n1 \n\n^2 = Fi(g)\', (49) \n\n\n\n= "\'\xc2\xa9\'"\xe2\x96\xa0 \n\n\n\n(50) \n\n\n\nIt should be noted that the above formulas can be used for \nany expansion of a perfect gas following PF" = a constant, \nprovided 7 in the formulas is replaced by n. \n\nIt is also to be noted that these equations can be used for \nany system of units so long as the same system of units is \nemployed throughout an equation. \n\nThe application of these equations to a practical example \n\nwill now be shown by way of illustration. Take the case of a \n\nquantity of pure air in a cylinder at a temperature of 60\xc2\xb0 F. \n\n(Ti = 460 + 60 = 520 degrees absolute) which is suddenly (adi- \n\nFi 2 \nabatically) compressed to half its original volume. Then \xe2\x80\x94 - = -\xc2\xbb \n\nV2 I \n\nand taking 7 from the table on page 23 as 1.40, the temper- \nature immediately after compression is completed, T2 is calcu- \nlated by equation (45) as follows: \n\nT2 = Ti (pj = 520 i^-J = 520 X 20-40 = 688\xc2\xb0 absolute, \n\nor ^2 in ordinary Fahrenheit is 688 \xe2\x80\x94 460 or 228 degrees. \n\nThe work done in adiabatic compression of one pound of this \nair is calculated by equation (44) : \n\n^^. _ MR (T, - T,) _ 5.V3 (.^20 - 688) _ .=;.V3 ( - 168) _ \n\n7 \xe2\x80\x94 1 1.40 \xe2\x80\x94 I 0.40 \n\nfoot-pounds per pound of air compressed. The negative sign \nmeans that work has been done on the gas. If the sign had \n\n\n\nEXPANSION AND COMPRESSION OF GASES 41 \n\nbeen positive it would have indicated an expansion. As the \nresult of this compression the internal energy of the gas has \n\nbeen increased by \xe2\x80\x94 \'-"^ B.t.u., but if the cy Under is a con- \n\n778 \n\nductor of heat, as in practice it always is, the whole of this heat \nwill become dissipated in time by conduction to surrounding \nair and other bodies, and the internal energy will gradually \nreturn to its original value as the temperature of the gas comes \nback to the initial temperature of 60\xc2\xb0 F. \n\nDuring the compression the pressure rises according to equa- \ntion (38): \n\nPV^ = constant, \n\nand just at the end the value is greater than the original pressure \nby the ratio of r^ to i,* or 2^-^^ to i, or 2.65 to i. If, as before, \nwe now assume the temperature drops gradually to the initial \ntemperature before compression (60\xc2\xb0 F.) without changing the \nvolume, the pressure will fall with the temperature until it \nhas at 60 degrees a value only twice as great as the original \npressure. In other words, after cooling the pressure becomes \ninversely proportional to the change in volume produced by \nthe compression. \n\nThere are many cases of expansions which are neither adiabatic \nnor isothermal and which are not straight lines on P-V diagrams. \nIt will, be observed from the equations in the discussion of the \ninternal work done by an expanding gas and for the change of \ninternal energy, that if in the general equation PF"* = a constant \nthe exponent or index n is less than 7, the work done is greater \nthan the loss in internal energy. In other words for such a \ncase, the expansion lies between an adiabatic and isothermal and \nthe. gas must be taking in heat as it expands. On the other \nhand, if n is greater than 7 the work done is less than the loss \nof internal energy. \n\n* Remember r is called the ratio of expansion and is f/- (See equation (28), \npage 34). \n\n\n\n42 ENGINEERING THERMODYNAMICS \n\nPROBLEMS \n\n1. How many foot-pounds of work are done by 2 lbs. of air in expand- \ning to double its volume at a constant temperature of 100\xc2\xb0 F. ? \n\nAns. 41,400 ft.-lbs. \n\n2. Three pounds of air are to be compressed from a volume of 2 to i cu. ft. \nat a constant temperature of 60\xc2\xb0 F. How many B.t.u. of heat must be \nrejected from the air? Ans. 74.2 B.t.u. \n\n3. An air compressor has a cylinder voliune of 2 cu. ft. If it takes air \nat 15 lbs. per sq. in. absolute and 70\xc2\xb0 F. and compresses it isothermally to \n100 lbs. per sq. in. absolute, find \n\n(a) Pounds of air in cylinder at beginning of compression stroke. \n\nAns. 0.15 lb. \n\n(b) The final volume of the compressed air. Ans. 0.30 cu. ft. \n\n(c) The foot-pounds of work done upon the gas during compression. \n\nAns. 8200 ft.-lbs. \n\n(d) The B.t.u. absorbed by the air in increasing the internal energy. \n\nAns. o. \n\n(e) The B.t.u. to be abstracted from the cylinder. \n\nAns. 10.46 B.t.u. \n\n4. Air at 100 lbs. per sq. in. absolute and a volume of 2 cu. ft. expands \nalong ann = 1 curve to 25 lbs. per sq. in. absolute pressure. Find \n\n(a) Work done by the expansion. Ans. 39,900 ft.-lbs. \n\n(b) Heat to be supplied. Ans. 51.3 B.t.u. \n\n5. A quantity of air at 100 lbs. per sq. in. absolute pressure has a tem- \nperature of 80\xc2\xb0 F. It expands isothermally to a pressure of 25 lbs. per \nsq. in. absolute when it has a volume of 4 cu. ft. Find (i) the mass of air \npresent, (2) Work of the expansion in foot-pounds, (3) Heat required in \nB.t.u. Ans. (i) 0.50 lb. \n\n(2) 19,950 ft.-lbs. \n\n(3) 25.6 B.t.u. \n\n6. Air at 100 lbs. per sq. in. absolute pressure and 2 cu. ft. expands to \n25 lbs. per sq. in. absolute adiabaticaUy. What is the final volume? \n\nAns. 5.4 cu. ft. \n\n7. One cubic foot of air at 60\xc2\xb0 F. and a pressure of 15 lbs. per sq. in. \nabsolute is compressed without loss or addition of heat to 100 lbs. per sq. \nin. absolute pressure. Find the final temperature and volume. \n\nAns. 891\xc2\xb0 F. absolute; 0.257 cu. ft. \n\n8. Two pounds of air are expanded from a temperature of 300\xc2\xb0 F. to \n200\xc2\xb0 F. adiabaticaUy. How many foot-pounds of work are developed ? \n\nAns. 26,300 ft.-lbs. \n\n9. A quantity of air having a volume of i cu. ft. at 60\xc2\xb0 F. under a pres- \nsure of 100 lbs. per sq. in. absolute is expanded to 5 cu. ft. adiabaticaUy. \n\n\n\nEXPANSION AND COMPRESSION OF GASES 43 \n\nFind the pounds of air present, the final temperature of the air and how \nmuch work will be done during this expansion. Ans. 0.52 lb. \n\n273\xc2\xb0 F. absolute. \n16,900 ft. -lbs. \n10. Data same as Problem 3 but the compression is to be adiabatic. Find \n\n(a) The final volume of the compressed air. \n\nAns. 0.516 cu. ft. \n\n(b) The final temperature of the compressed air. \n\nAns. 911\xc2\xb0 F. absolute. \n\n(c) The foot-pounds of work to compress this air. \n\nAns. 7510 ft.-lbs. \n\n(d) The B.t.u. absorbed by the air in increasing the internal \n\nenergy. Ans. 9.65 B.t.u. \n\n(e) The B.t.u. to be abstracted from the gas. Ans. o B.t.u. \nII. A pound of air at 32\xc2\xb0 F. under atmospheric pressure is compressed \n\nto 4 atmospheres (absolute). What will be the final volume and the work \nof compression if the compression is (a) isothermal, (b) adiabatic ? \n\nAns. (a) V2 = 3.09 cu. ft.; work = 33,800 ft.-lbs. \n(b) V2 = 4.60 cu. ft.; work = 31,500 ft.-lbs. \n\n\n\nCHAPTER IV \nCYCLES OF HEAT ENGINES \n\nWhen a gas or vapor undergoes a series of processes in which \nthere is an interchange of heat quantities and is finally brought \nback to the condition, as regards its physical properties, which \nit had initially, the gas is said to have gone through or per- \nformed a cycle.* The most important cycle with which we have \nto deal is the Carnot cycle, because it is typical of the maximum \nefficiency obtainable. \n\nCarnot Cycle. Very important conclusions regarding theo- \nretically perfect heat engines are now to be drawn from the \nconsideration of the action of an ideal engine in which the \nworking substance is a perfect gas which is made to go through \na cycle of changes involving both isothermal and adiabatic \nexpansions and compressions. This ideal cycle of operations \nwas invented and first explained in 1824 by Carnot, a French \nengineer, and gave us really the first theoretical basis for com- \nparing heat engines with an ideally perfect engine. For ex- \nplaining this Carnot cycle assume a piston and cylinder as \nshown in Fig. 9, composed of perfectly non-conducting material, \nexcept the cylinder-head (left-hand end of the cylinder) which is \na good conductor of heat. The space in the cyhnder between \nthe piston and the cylinder head is occupied by the working \nsubstance, which we shall assume to be a perfect gas. There \nis provided a hot body H of unlimited heat capacity, always \nkept at a temperature Ti, also a perfectly non-conducting cover \nN and a refrigerating or cold body R of unlimited heat receiving \ncapacity, which is kept at a constant temperature T2 (lower \n\n* A thermodynamic machine performing a cycle in which heat is changed into \nwork is called a heat engine, and one performing a cycle in which heat is trans- \nferred from a medium at a low temperature to one at a higher temperature is \ncalled a refrigerating machine. \n\n44 \n\n\n\nCYCLES OF HEAT ENGINES \n\n\n\n45 \n\n\n\nthan Ti). It is arranged that H, N or R can be apphed, as \nrequired, to the cyHnder head. Assume that there is a charge \nof one pound of gas in the cylinder between the piston and the \ncylinder head, which at the beginning of the cycle, with the piston \nin the position shown, is at the temperature T\\, has a volume \nVa and has a pressure Pa> The subscripts attached to the \n\n\n\n\nFig. 9. \xe2\x80\x94 Apparatus and Diagram Illustrating a Reversible (Camot) Cycle. \n\nletters V and P refer to points on the pressure- volume diagram \nshown in the figure. This diagram shows, by curves connecting \nthe points a, b, c and d, the four steps in the cycle. \n\nThe operation of this cycle will be described in four parts as \nfollows: (i) Apply the hot body or heater H to the cylinder \nhead * at the left-hand side of the figure. The addition of heat \n\n* It will be remembered that the head of this cylinder is a perfect conductor of \nheat. \n\n\n\n46 ENGINEERING THERMODYNAMICS \n\nto the gas will cause it to expand isothermally because the tem- \nperature will be maintained constant during the process at Ti. \nThe pressure drops sHghtly to Ph when the volume becomes F?,. \nDuring this expansion external work has been done in advanc- \ning the piston and the heat equivalent of this work has been \nobtained from the hot body H. \n\n(2) Take away the hot body H and at the same time attach \nto the cylinder head the non-conducting cover N. During this \ntime the piston has continued to advance toward the right, \ndoing work without receiving any heat from an external source \nso that the expansion of the gas in this step has been done at \nthe expense of the stock of internal energy in the gas. The \ntemperature has continued to drop * in proportion to the loss of \nheat to the value T^. Pressure is then Pc and the volume is Vc. \n\n(3) Take away the non-conductor N and apply the refriger- \nator R. Then force the piston back into the cylinder. The \ngas will be compressed isothermally at the temperature T2. \nIn this compression, work is being done on the gas, and heat is \ndeveloped, but all of it goes into the refrigerator R, in which the \ntemperature is always maintained constant at T^. This com- \npression is continued up to a point d in the diagram, so selected \nthat a further compression (adiabatic) in the next (fourth) \nstage will cause the volume, pressure and temperature to reach \ntheir initial values as at the beginning of the cycle.f \n\n(4) Take away the refrigerator R and apply the non-conduct- \ning cover N. Then continue the compression of the gas without \nthe addition of any heat. It will be adiabatic. The pressure \nand the temperature will rise and, if the point d has been prop- \nerly selected, when the pressure has been brought back to its \ninitial value Pa the temperature will also have risen to its initial \nvalue Ti. The cycle is thus finished and the gas is ready for a \n\n* A pressure- volume diagram of a perfect gas does not show, graphically, changes \nof temperature. \n\nt Briefly the third stage of the cycle must be stopped when a point d is reached, \nso -located that an adiabatic curve {PV^ = constant) drawn from it will pass \nthrough the "initial" point a. \n\n\n\nCYCLES OF HEAT ENGINES 47 \n\nrepetition of the same series of processes comprising the \ncycle. \n\nTo define the Carnot cycle completely we must determine \nhow to locate algebraically the proper place to stop the third \nstep (the location of d). During the second step (adiabatic \nexpansion from b to c) we can write, by applying equation (45), \nthe following temperature and volume relations: \n\n\n\nTo \n\n\n\n=[f:r\' \n\n\n\nalso for the adiabatic compression in the fourth step we can \nstate similarly, \n\nII \nT2 \n\n\n\nHence, \n\n\n\n[W\'-m"- \n\n\n\nSimplifying and transposing, we have \n\nObserve that \xe2\x80\x94 is the ratio of expansion r (page 33) for the \n\n^ a \n\nisothermal expansion in the first step of the cycle. This has been \n\nshown to be equal to -f in the isothermal compression in the third \n\nyd \n\nstep in order that the adiabatic compression occurring in the \nfourth step shall complete the cycle. \n\nA summary of the heat changes to and from the working \ngas (per pound) in the four steps of the Carnot cycle is as fol- \nlows: \n\n(ab). Heat taken in from hot body = RTi ^^ge\xe2\x80\x94 (by \n\n* a \n\nequation (2*8), in foot-pounds). (52) \n\n(be) . No heat taken in or rejected. \n\n\n\n48 ENGINEERING THERMODYNAMICS \n\n(cd). Heat rejected to refrigerator = i^r2 log* 7^ \n\ny c \n\nor \xe2\x80\x94 RT2 loge~ (by equation (28), in foot- \ned \n\npounds). (53) \n\n(da). No heat taken in or rejected. \n\nHence, the net amount of work done, W, by the gas in this \ncycle, being the mechanical equivalent (foot-pounds), of the \nexcess of heat taken in over that rejected, is the algebraic \'sum \n\nof (52) and (53): \n\nW = r(t, log. |-^ - T, log.|-j = R(T^- T,) loge^- (54) \n\nIf the curves in Fig. 9 are accurately plotted to scales of pressure \nand volume, then the work in foot-pounds as calculated, from \nthe measured area included in the P-V diagram will be found \nto agree exactly with the result given by equation (54) above. \n\nEfficiency of Carnot\'s Cycle. The thermal or heat efficiency \nof a cycle is usually defined as the ratio of \n\nHeat equivalent of work done \nHeat taken in \n\nThe heat equivalent of work done is, by equation (54), \n\nV a \n\nand the heat taken in is, by equation (52), \n\n* a \n\nThe ratio above representing the efficiency E is \n\n\n\nj?(rx-r.)iog.p \n\n* Log i^=- logy-- \n\n\n\n(ss) \n\n\n\nCYCLES OF HEAT ENGINES 49 \n\nThis efficiency represents the proportion of the total amount of \nheat given to the gas employed in a Carnot cycle, which the \nengine converts into work. At the temperature of the hot body \nH, the engine takes in an amount of heat proportional to the \nabsolute temperature of this body or Ti, and in the course of \nthe cycle rejects to the refrigerator R an amount of heat pro- \nportional to its absolute temperature 7*2 . The range of temper- \nature for the cycle is, therefore, between Ti and T2. In the \nlowering of temperature corresponding to this range the engine \nconverts into work that part of the heat taken in that is repre- \nsented by equation (55). Observe the conditions affecting the \nmaximum efficiency of this cycle. For a given heat supply \nproportional to Ti the only way to increase efficiency is by \nreducing 7^2, so that the smaller T2 is the greater the tempera- \nture range (T1 \xe2\x80\x94 T2) becomes and the higher the efficiency will be. \n\nReversible Cycles. A heat engine which is capable of dis- \ncharging to the "source of heat" when running in the reverse \ndirection from that of its normal cycle the same quantity of \nheat that it would take from this source when it is running \ndirect and doing work is said to operate with its cycle reversed, \nor, in other words, the engine is reversible. A reversible heat \nengine then is one which, if made to follow its indicator diagram \nin the reverse direction, will require the same horse power to drive \nit as a refrigerating machine as the engine will deliver when \nrunning direct, assuming that the quantity of heat used is the \nsame in the two cases. An engine following Carnot\'s cycle is, \nfor example, a reversible engine. The thermodynamic idea of \nreversibility in engines is of very great serviceableness because \nit will be shown that no heat engine can be more efficient than a \nreversible engine when both work between the same limits of \ntemperature; that is, when both engines take in the same amount \nof heat at the same higher temperature and reject the same \namount at the same lower temperature. \n\nCarnot\'s Principle. It was first proved conclusively by Car- \nnot that no other heat engine can be more efficient than a re- \nversible engine when both work between the same temperature \n\n\n\n50 ENGINEERING THERMODYNAMICS \n\nlimits. To illustrate this principle, assume that there are two \nengines A and B. Of these let us say A is reversible and B is \nnot. In their operation both take heat from a hot body or \nheater H and reject heat to a refrigerator or cold body R. Let \nQh be the quantity of heat which the reversible engine A takes \nin from the hot body H for each unit of work performed, and \nlet Qr be the quantity of heat per unit of work which it dis- \ncharges to the refrigerator R. \n\nFor the purpose of this discussion, assume that the non- \nreversible engine B is more efficient than the reversible engine A. \nUnder these circumstances it is obvious that the engine B will take \nin less heat than A and it will reject correspondingly less heat \nto R per unit of work performed. The heat taken in by the \nnon-reversible engine B from the hot body H we shall designate \nthen by a quantity less than & or Q^ \xe2\x80\x94 X and the heat rejected \nby B to the refrigerator R by Q^j. \xe2\x80\x94 X. Now if the non-reversible \nengine B is working direct (when converting heat into work) \nand is made to drive the reversible engine A according to its \nreverse cycle (when converting work into heat), then for every \nunit of work done by the engine B in driving the reversible \nengine A, the quantity of heat mentioned above, that is, Qu \xe2\x80\x94 X, \nwould be taken from the hot body H by the non-reversible \nengine B and, similarly, the quantity of heat represented by Qu \nwould be returned to the hot body H by the reverse action of \nthe cycle of operations performed by A. This follows because \nthe engine A is reversible and it returns, therefore, to H, when \noperating on the reverse cycle, the same amount of heat as it \nwould take in from H when working on its direct cycle. By \nthis arrangement the hot body H would be continually receiving \nheat, in the amount represented by X for each unit of work \nperformed. At the same time the non-reversible engine B dis- \ncharges to the refrigerator R a quantity of heat represented by \nQji \xe2\x80\x94 X, while the reversible engine A removes from the refriger- \nator R a quantity represented by Qr. As a result of this last \noperation the cold body will be losing continually per unit of work \nperformed a quantity of heat equal to X. The combined per- \n\n\n\nCYCLES OF HEAT ENGINES 5 1 \n\nformances of the two engines, one working direct as a normal heat \nengine and the other, according to its reverse cycle, as a com- \npressor or what might be called a "heat pump," gives a constant \nremoval of heat from the refrigerator R to the hot body H, \nand as a result a degree of infinite coldness must be finally pro- \nduced in the refrigerator. \n\nIf we assume that there is no mechanical friction, this \ncombined machine, consisting of a normal heat engine and \ncompressor, will require no power from outside the system. \nFor this reason the assumption that the non-reversible engine B \ncan be more efficient than the reversible engine A has brought \nus to a result which is impossible from the standpoint of expe- \nrience as embodied in the statement of the " Second Law of \nThermodynamics" (see page 3); that is, it is impossible to \nhave a self-acting engine capable of transferring heat, infinite \nin quantity, from a cold body to a hot body. We should, \ntherefore, conclude that no non-reversible engine, as B for \nexample, can be more efficient than a reversible engine A when \nboth engines operate between the same temperature limits. \nMore briefly, when the source of heat and the cold receiver are \nthe same for both a reversible heat engine and any other engine, \nthen the reversible engine must have a higher possible efficiency; \nand if both engines are reversible it follows that neither can be \nmore efficient than the other. \n\nPerfection in a Heat Engine. A reversible engine is perfect \nfrom the viewpoint of efficiency; that is, its efficiency is the best \nobtainable. No other engine than a reversible engine which \ntakes in and discharges heat at identical temperatures will \ntransform into work a greater part of the heat which it takes in. \nFinally, it should be stated as regards this efficiency that the \nnature of the substance being expanded or compressed has \nabsolutely no relation to the thermal efficiency as outlined above. \n\nReversed Carnot\'s Cycle. If an engine operating on Carnot\'s \ncycle is reversed in its action so that the same indicator diagram \nshown in Fig. 9 would Be traced in the opposite direction, the \nreversed cycle, when beginning as before at a with a per- \n\n\n\n52 ENGINEERING THERMODYNAMICS \n\nfeet gas at the temperature Ti, will consist of the following \nstages : \n\n(i) When the non-conductor N is applied and the piston is \nadvanced toward the right by the source of power performing \nthe reversed cycle, the gas will expand, tracing the adiabatic \ncurve ad, with constant lowering of temperature which at the \npoint d will be T2. \n\n(2) When the non-conductor N is now removed, the refrig- \nerator R is applied, and the piston continues on its outward stroke. \nThe gas will expand isothermally at the constant temperature Tij \ntracing the curve dc. During this stage the gas is taking heat \nfrom the refrigerator R. \n\n(3) When the refrigerator R is removed and the non-conductor \nN is again applied, which will be on the back stroke of the engine, \nthe gas will be compressed, and on the indicator diagram another \nadiabatic curve cb will be traced. At the point b the temper- \nature will be obviously Ti. \n\n(4) When the non-conductor N is removed and the hot body \nH is again applied, with the compression continuing along the \nisothermal curve ba, heat will be discharged to the hot body H, \nwhile the temperature is maintained constant at Ti. The cycle \nhas now been traced in a reverse direction from the beginning \nback to the starting point at a, and is now complete. During \nthis process no work has been done, but on the contrary an \namount of work represented by the area of the indicator dia- \ngram, equivalent in foot-pounds to \n\ni^logeT^ (ri - To) (see equation (54), page 48), \n\nhas been converted into heat. First, heat was taken from the \nrefrigerator R, represented in amount by \n\nRT2\\oge-zf^ \n\ny d \n\nand second, heat was rejected to the hot body H in the amount \ni^Tilog,^^ or -RTiloge \xe2\x80\x94 \n\nVb Va \n\n\n\nCYCLES OF HEAT ENGINES 53 \n\nAs in direct operation of Carnot\'s cycle no heat is given or lost \n\nin the first and third stages outlined above. The algebraic sum \n\nVb Vc \nof these two quantities, remembering that \xe2\x80\x94 = \xe2\x80\x94 , gives the \n\n\'a ^ d \n\nnet amount of work done, W, on the gas, and, therefore, the \nnet amount of heat (foot-pound units) transferred from the cold \nbody R to the hot body H or, \n\nW=RT,\\oge~-RT,\\oge^= -R\\oge^{T,-T,). (56) \n\n\'a \'a \'a \n\nSince the result is the same as given by equation (54), although \nopposite in sign on account of being work of compression, it \nwill be observed that in the reverse cycle the same amount \nof heat is given to the hot body H as was taken from it in the \ndirect operation of the same cycle, and that the same amount \nof heat is now taken from the refrigerator R as was in the other \ncase given to it. \n\nConclusions from the Above Discussion. In the explanation \nthat has preceded of the performance of heat engines a perfect \ngas took in heat at the temperature of the source of supply of \nheat and discharged heat at the temperature of the refrigerator, \nor receiver of heat, the changes of temperature occurring as the \nresult of adiabatic expansion or adiabatic compression. For a \nperfect gas we have determined, as in equation (55), that the \nefficiency of the heat engine was {Ti \xe2\x80\x94 T2) -^ T\\. It has also \nbeen shown that Carnot\'s cycle was reversible, and that accord- \ning to the "Second Law of Thermodynamics" no heat engine \ncan have a higher efficiency than such a reversible engine when \ntaking in and discharging heat at the same two temperatures \nTi and T2. Finally, we are brought to the important conclusion \nthat all reversible heat engines receiving and discharging heat \nat the same temperatures Ti and T2 are of equal efficiency, and \nthat this efficiency, \n\n\xe2\x96\xa0\xc2\xa3 = ^^^ (57) \n\n\n\n54 ENGINEERING THERMODYNAMICS \n\nhaving been determined for one reversible engine is also the \nefficiency of any other reversible engine, and is the maximum \nefficiency attainable with any engine. \n\nIt should be observed in connection with the statement of an \nequation of efficiency like (57) above that it is impossible to \nutiKze the whole of any supply of heat for conversion into work \nbecause it is impossible to reach the absolute zero of tempera- \nture, or in other words to make T2 in this equation practically \nzero. Considering only practical conditions, therefore, we may \nsay that with given limits of temperatures Ti and T2 it is neces- \nsary for the attainment of the greatest efficiency that no heat \nshall be taken in by an engine except during the isothermal at \nthe highest temperature and that no heat shall be rejected ex- \ncept at the isothermal at the lowest temperature T2. \n\nThe following problem shows the application of the theory of \nCarnot\'s cycle: \n\nIn Carnot\'s cycle the gas has an initial condition of 100 pounds \nper square inch absolute pressure, volume of i cubic foot and \ntemperature of 300\xc2\xb0 F. The volume at the end of isothermal \nexpansion is 2 cubic feet. Exhaust temperature is 60\xc2\xb0 F. Find \n\n(a) Heat supplied to the cycle; \n\n(b) Efficiency of the cycle; \n\n(c) Net work of the cycle. \nSolution. The heat supplied equals \n\n2 \n100 X 144 X I X loge- = 9970 foot-pounds or 12.8 B.t.u. \n\nThe efficiency is \n\n(^00 + 460) \xe2\x80\x94 (60 -f- 460) ^ ^ 4. \n\n^ \xe2\x80\x94^ \xe2\x80\x94 ^ -^ \xe2\x80\x94 - = 0.316 or 31.6 per cent. \n\n300 -{- 460 \n\nThe net work of the cycle is \n\n9970 foot-pounds X 0-3^6 = 3150 foot-pounds. \n\nRegenerative Air Engines. The previous discussion has dealt \nentirely with ideal engines following what is known as Carnot\'s \ncycle. Such an engine has never been built. Another engine \nhaving theoretically a reversible cycle has, therefore, the same \n\n\n\n(CYCLES OF HEAT ENGINES 55 \n\npractical application and will be next described. This engine, \nknown as Stirling\'s, consists of two cylinders side by side, one \nof which is heated and the other cooled. It performs its cycle \naccording to the following four stages: \n\n(i) Air which has been previously heated in a regenerator * \nto a temperature Ti is expanded isothermally from a volume \nrepresented by Vi to a volume F2. During this stage heat is \nbeing taken in from the furnace and the piston is raised as the \nexpansion proceeds. The heat taken in during this stage in \nfoot-pounds per pound of air is \n\ni^Ji loge\xe2\x80\x94 (see equation 52). \n\n\n\nV \n\n\n\n(2) During the next stage this air which is now highly heated \nis made to pass through the regenerator placed between the hot \nand the cold cylinders, and in its passage through the regenera- \ntor it gives up heat and has its temperature reduced to T2, with- \nout a change of volume. The heat absorbed by the regenerator \nis Cv (Ti \xe2\x80\x94 T2) . There is, of course, a drop in pressure corre- \nsponding to the reduction in temperature, although there is no \nchange of volume. \n\n(3) In the cooled cylinder the air is compressed isothermally \nat the temperature T2 to its original volume. The heat dis- \ncharged in this stage is \n\nRT2log^^= -RT2 log^- \n\n(4) The air is now passed back through the regenerator from \nthe cooled cylinder to the heated cylinder, absorbing heat from \nthe regenerator on the way and having its temperature raised \nto that at the beginning of the cycle or Ti. Heat taken in from \nthe regenerator in this stage is Cv (Ti \xe2\x80\x94 T\'^. \n\n* The regenerator consisting of a series of iron plates serves as a heat accumu- \nlator. When hot gases are passed through these plates they become heated. On \nthe other hand, when cold gases are passed through the plates the heat previously- \nabsorbed would be given up to the gas. \n\n\n\n56 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nThe thermal efficiency of the whole cycle is then \nj^ _ Heat taken in \xe2\x80\x94 Heat discharged \nHeat taken in \nThis would be expressed by the symbols already used as \n\n\n\nE = \n\n\n\nRTAogey - RT,\\oge^ \n\n\n\nRTi loge \n\n\n\nVi \n\n\n\nCanceling out the common terms, we have more simply \n\n\n\nE = \n\n\n\nTi \n\n\n\nA theoretical indicator diagram from Stirling\'s engine is shown \nin Fig. 10. \n\n\n\n\nVolume \nFig. io. \xe2\x80\x94 Indicator Diagram of Stirling\'s Air Engine. \n\nThis engine, although not commercially a success, is, however, \nimportant because it represents the only type besides Carnot\'s \nthat is reversible. The application of regenerators in heat \nengines is very limited and is confined almost exclusively to \nengines using hot air as the working substance. \n\nAnother method of using air as the working substance in heat \nengines was developed by Ericsson, who used a cycle consisting \nof constant pressure and isothermal lines. The expansion took \nplace at constant pressure while the air was passing through \nthe regenerator. \n\nThe following problem shows the application of the theory to \ncycles other than the Carnot. (Stirling engine.) \n\n\n\nCYCLES OF HEAT ENGINES 57 \n\nAssume a cylinder of i cubic foot volume which contains air \nat 60\xc2\xb0 F. and 15 pounds per square inch absolute pressure and \nthe cycle of operation to be performed as follows: \n\n(i) The gas is compressed adiabatically until the pressure \nequals 100 pounds per square inch absolute. \n\n(2) Heat is then supplied without change of volume of the gas \nand raises the temperature to 100\xc2\xb0 F. \n\n(3) The gas then expands adiabatically to a volume of i \ncubic foot. \n\n(4) Heat is then rejected without change of volume until the \ntemperature is lowered to 60\xc2\xb0 F. The mass of air present is \n\nM= 15X144X1 3^^^ \n\n53.3 (60 + 460) \nThe temperature at the end of adiabatic compression is \n\n7^2 = (60 + 460) j^^ = 895\xc2\xb0 F. absolute. \n\nThe temperature after addition of the heat is \n\nTz = 895 + 100 = 995\xc2\xb0 F. absolute. \n\nThe volume after adiabatic compression is \n\nT/ 0.078 X 53.3 X 895 o -. \n\nV2 = \xe2\x80\x94 ^^-^ ^ = 0.258 cu. ft. \n\n100 X 144 \n\nThe temperature after adiabatic expansion equals \nT, = 995 i\xc2\xb0-^^^y-\'-\' = 578\xc2\xb0 F. absolute. \n\nThe heat supplied to the cycle is \n\n0.078 X 0.169 X 100 = 1.32 B.t.u. \n\nThe heat exhausted from the cycle is \n\n0.078 X 0.169 X (578 \xe2\x80\x94 520) = 0.76 B.t.u. \n\nThe net work of the cycle is \n\n1.32 \xe2\x80\x94 0.76 = 0.56 B.t.u. or 436 foot-pounds. \n\nThe efficiency of the cycle is \n\n0-56 * \n\n\xe2\x80\x94 "^ = 0.42 or 42 per cent. \n\n1.32 \n\n\n\n58 ENGINEERING THERMODYNAMICS \n\nPROBLEMS \n\n1. A Carnot engine containing lo lbs. of air has at the beginning of the \nexpansion stroke a volume of lo cu. ft. and a pressure of 200 lbs. per sq. in. \nabsolute. The exhaust temperature is 0\xc2\xb0 F. If 10 B.t.u. of heat is added \nto the cycle, find \n\n(a) Efficiency of the cycle. Ans. 15 per cent. \n\n(b) Work of the cycle. Ans. 1160 ft.-lbs. \n\n2. A Carnot cycle has at the beginning of the expansion stroke a pressure \nof 75 lbs. per sq. in. absolute, a volume of 2 cu. ft. and a temperature of \n200\xc2\xb0 F. The volume at the end of isothermal expansion is 4 cu. ft. The \nexhaust temperature is 30\xc2\xb0 F. Find \n\n(a) Heat added to cycle. Ans. 19.3 B.t.u. \n\n(b) Efficiency of cycle. Ans. 25.8 per cent. \n\n(c) Work of cycle. Ans. 3860 ft.-lbs. \n\n3. A cycle made up of two isothermal and two adiabatic curves has a \npressure of 100 lbs. per sq. in. absolute and a volume of i cu. ft. at the \nbeginning of isothermal expansion. At the end of adiabatic expansion the \npressure is 10 lbs. per sq. in. absolute and the volume is 8 cu. ft. Find \n\n(a) Efficiency of cycle.* Ans. 20 per cent. \n\n(b) Heat added to cycle. Ans. 28.5 B.t.u. \n\n(c) Net work of cycle. Ans. 4360 ft.-lbs. \n\n4. In a steam power plant the steam is generated at 400\xc2\xb0 F. and is ex- \nhausted at 216\xc2\xb0 F. If the heat in the steam could be transformed accord- \ning to Carnot\'s cycle, what would be the efiiciency of the plant ? \n\nAns. 21.4 per cent. \n\n5. In a Carnot cycle the heat is added at a temperature of 400\xc2\xb0 F. and \nrejected at 70\xc2\xb0 F. The working substance is i lb. of air which has a volimie \nof 2 cu. ft. at the beginning and a volume of 4 cu. ft. at the end of isothermal \nexpansion. Find \n\n(a) Volume at end of isothermal compression. Ans. 6.70 cu. ft. \n\n(b) Heat added to the cycle. Ans. 40.7 B.t.u. \n\n(c) Heat rejected from cycle. Ans. 25.2 B.t.u. \n\n(d) Net work of the cycle. Ans. 12,150 ft.-lbs. \n\n6. Air at a pressure of 100 lbs. per sq. in. absolute, having a volume of \nI cu. ft. and a temperature of 200\xc2\xb0 F., passes through the following opera- \ntions: \n\n15^. Heat is suppHed to the gas while expansion takes place under \nconstant pressure untO the volume equals 2 cu. ft. \n\n2nd, It then expands adiabatically to 15 lbs. per sq. in. absolute \npressure. \n\n* ^^., PiVi - P2V2 _ Ti-Ti \xe2\x80\xa2 \n\n\n\nEfiiciency = 5-^7 or \n\n\n\nCYCLES OF HEAT ENGINES 59 \n\n;^d. Heat is then rejected while compression takes place under con- \nstant pressure. \n\n4th. The gas is then compressed adiabatically to its original volume \nof I cu. ft. \nFind (a) Pounds of air used. Ans. 0.41 lb. \n\n(b) Temperature at end of constant pressure expansion. \n\nAns. 1320\xc2\xb0 F. absolute. \n\n(c) Heat added to the cycle. Ans. 64.3 B.t.u. \n\n(d) Net work of cycle. Ans. 21,160 ft.-lbs. \n\n(e) Efficiency of cycle. Ans. 42.2 per cent. \n\n\n\nCHAPTER V \nPROPERTIES OF STEAM \n\nSteam in Heat Engines. The discussion on the preceding \npages has had to do largely with the action of perfect gases in \nheat engines. Now we shall consider in this chapter a more \nlimited field, confining ourselves exclusively to the action of \nwater vapor or steam in such engines. The unusual physical \nproperties of steam must, therefore, be explained here in con- \nsiderable detail as well as the use and appHcation of tables of \nsteam properties, called, for short, steam tables. Nearly all our \ndealings with steam in an engineering way will have to do with \nits formation at constant pressure. This is the condition of \nsteam formation in a power plant boiler when the engines are \nat work. \n\nTo make perfectly clear the process of steam formation in a \nboiler, assume that steam is to be made in a cylinder having a \ncross-sectional area of one square foot, closed at one end on \nwhich it stands. This cylinder is fitted with a frictionless piston \n(Fig. ii),* which is loaded so that it will exert a constant pres- \nsure of I GO pounds per square foot on the fluid in the cylinder \nbelow the piston. To begin the explanation of the properties \nof water vapor or steam, assume that there is in the bottom of \nthe cylinder a quantity of water, say one pound, at just 32\xc2\xb0 F. \n(not ice) . If, now, heat is applied to the bottom of the cylinder, \nit will pass through the walls of the cylinder and will enter the \nwater where it will produce the following changes in three \nstages: \n\n(i) As the water takes in heat its temperature rises until a \ncertain temperature t is reached, at which steam begins to form. \nThe value of t depends on the particular pressure which the \n\n* Compare with Fig. 2, page 9. \n60 \n\n\n\nPROPERTIES OF STEAM \n\n\n\n6l \n\n\n\npiston and its load exerts. Until the temperature t is reached \nthere is nothing but water below the piston. \n\n(2) After heating the water to the temperature t corresponding \nto the particular pressure exerted, and then adding more heat, \nthere is no further rise in temperature, but water vapor (steam)\' \nbegins to form. With this formation of steam there is a rapid \nincrease in volume, and the frictionless piston which is sup- \n\n\n\n\n\n\n\n\n\nPerfect \nVacuum \n\n\n\\ \n\n\nA \n\n\n\n\nI \n\nConnection \nto Air Pump \n\n\n\n\n\n\nWeight ^ \n\n\n\n\nm \nM \n\n\nWMM \n\n\n\n\n\n\nWater \n\n\n\nFig. II. \n\n\n\n\xe2\x80\xa2Simple Apparatus to Illustrate Pressure and Volume Relations of \nSteam. \n\n\n\nposed to exert a constant pressure will be raised. The forma- \ntion of steam at constant temperature and constant pressure \ncontinues throughout this stage until all the water is converted \ninto steam. All the steam which is formed during this stage is \nsaid to be saturated. \n\n(3) If, after all the water has been evaporated into steam, \nstill more heat is added and taken in, the volume will be still \nfurther increased and there will also be an increase in tempera- \nture. The steam in this last stage is then said to be super- \nheated. To make clear this distinction between saturated and \nsuperheated steam, it should be added that if at a given pres- \nsure steam exists at a temperature higher than the temperature \n(t) " of saturation," it is said to be superheated. The difference \nbetween saturated and superheated steam may also be expressed \nby saying that if water (at the temperature of the steam) be \n\n\n\n62 ENGINEERING THERMODYNAMICS \n\nmixed with the steam some of this water will be gradually evapo- \nrated if the steam is superheated, but not if the steam is satu- \nrated.* \n\nThe properties of saturated steam differ very much from those \nX)i the perfect gases which we have been stud3dng, but when the \nsteam is superheated to a high degree its properties approach \nvery closely those of an ideally perfect gas. It is most impor- \ntant to remember that saturated steam at any particular pres- \nsure is always at the same temperature, while, on the other hand, \nsuperheated steam can have any temperature higher than that \ncorresponding to saturated steam at the same pressure. \n\nRelation of Temperature, Pressure and Volixme in Saturated \nSteam. The important relations of temperature, pressure and \nvolume were first determined in a remarkable series of experi- \nments conducted by a French engineer named Regnault, and \nit has been on the basis of his data, first pubHshed in 1847, \nthat even our most modern steam tables are based. Later ex- \nperimenters have found, however, that these data were some- \nwhat in error, especially for values near the dry saturated \ncondition. These errors resulted because it was difficult in the \noriginal apparatus to obtain steam entirely free from moisture. \n\nThe pressure of saturated steam increases very rapidly as the \ntemperature increases in the upper limits of the temperature \nscale. It is very interesting to examine a table of the proper- \nties of steam to observe how much more rapidly the pressure \nmust be increased in the higher limits for a given range of tem- \nperature. \n\nIt should be observed that in most tables the pressure is almost \ninvariably given in terms of pounds per square inch, while in \nnearly all our thermodynamic calculations the pressure must \nbe used in pounds per square foot. \n\nHeat in the Liquid (Water) (h) . The essentials of the process \nof making steam have been above described in a general way. \n\n* It is not unusual at all to find in practice hot water existing indefinitely in \nthe presence of superheated steam. See Moyer\'s Steam Turbines, page 262, and \nJPower Plant Testing by the same author, page 316. \n\n\n\nPROPERTIES OF STEAM * 63 \n\nThe relation of this process to the amount of heat required will \nnow be explained. If a pound of water which is initially at \nsome temperature to is heated at a constant pressure P (pounds \nper square foot) to the boiling point corresponding to this pres- \nsure and then converted into steam, heat will first be absorbed \nin raising the temperature of the water from to to t, and then \nin producing vaporization. During the first stage, while the \ntemperature is rising, the amount of heat taken in is approxi- \nmately (t\xe2\x80\x94 to) heat units, that is, British thermal units (B.t.u.), \nbecause the specific heat of water is approximately unity and \npractically constant. This number of B.t.u. multiplied by 778 \ngives the equivalent number of foot-pounds of work. For the \npurpose of stating in steam tables the amount of heat required for \nthis heating of water, the initial temperature to must be taken at \nsome definite value; for convenience in numerical calculations \nand also because of long usage, the temperature 32\xc2\xb0 F. is invari- \nably used as an arbitrary starting point for calculating the \namount of heat " \'taken in." The symbol h (or sometimes q) is \nused to designate the heat required to raise one pound of water \nfrom 32\xc2\xb0 F. to the temperature at which it is vaporized into \nsteam. In other words, " the heat of the liquid " (h) is the \namount of heat in B.t.u. required to raise one pound of water \nfrom 32\xc2\xb0 F. to the boiling point. \n\nIt is obvious, therefore, that we can write the value of the \nheat absorbed by water in being raised to the steaming tem- \nperature (h) in B.t.u., approximately, by the formula \n\nh = t - 32 (in B.t.u.). (58) \n\nMore accurate values of h, taking into consideration the varia- \ntion in the specific heat of water, will be found in the usual \nsteam tables. During this first stage, before any steaming has \noccurred, practically all the heat applied is used to increase the \nstock of internal energy. The amount of external work done \nby the expansion of water as a liquid is practically negligible. \n\nLatent Heat of Evaporation (L). In the second stage of the \nformation of steam as described, the water at the temperature t, \n\n\n\n64 ENGINEERING THERMODYNAMICS \n\ncorresponding to the pressure, is changed into steam at that \ntemperature. Although there is no rise in temperature, very \nmuch heat is nevertheless required to produce this evaporation \nor vaporization. The heat taken in during this stage is the \nlatent heat of steam. In other words, the latent heat of steam \nmay be defined as the amount of heat which is taken in by a \npound of water while it is changed into steam at constant pres- \nsure, the water having been previously heated up to the tem- \nperature at which steam forms. The symbol L (also sometimes \nr) is used to designate this latent heat of steam. Its value \nvaries with the particular pressure at which steaming occurs, \nbeing somewhat smaller in value at high pressures than at low. \n\nExternal Work of Evaporation (E). A part of the heat taken \nin during the \'\' steaming " process is spent in doing external \nwork.* Only a small part of the heat taken in is represented \nby the external work done in making the steam in the boiler, \nand the remainder of the latent heat (L) goes to increase the \ninternal energy of the steam. The amount of heat that goes \ninto the performing of external work is obviously equal to P (the \npressure in pounds per square foot) times the change of volume \noccurring when the water is changed into steam. \n\nExample. At the usual temperatures of the working fluid in \nsteam engines the volume of a pound of water is about g^ of a \ncubic foot. The external work, W, done in making one pound of \nsteam having finally a volume of V (cubic feet) at a constant pres- \nsure P (pounds per square foot) may be written in foot-pounds: \n\nW = External work = P(V- q\\). (59) \n\nThis last equation can be expressed in British thermal units \n(B.t.u.) by dividing by 778. It is apparent also from this equa- \n\n* The external work done in the formation of steam at constant pressure would \nbe illustrated by the apparatus used for explaining the three stages of steam \nformation on page 61. It should be observed that while heat is being added \nmerely to raise the temperature of the water there is practically no movement \nof the piston as there is scarcely any change in volume, but when steam is being \nmade the volume increases rapidly and the piston will rise high above its initial \nposition. \n\n\n\nPROPERTIES OF STEAM 65 \n\ntion that the external work done in making steam is less at low \npressure than at high,* because there is less resistance to over- \ncome, or, in other words, P in equation (59) is less. The heat \nequivalent of the external work is, therefore, a smaller propor- \ntion of the heat added at low temperature than at high. \n\nTotal Heat of Steam (H). The heat added during the process \nrepresented by the first and second stages in the formation of a \npound of steam, as already described, is called the total heat of \nsaturated steam or, for short, total heat of steam, and is repre- \nsented by the symbol H. Using the symbols already defined, \nwe can write, per pound of steam, \n\nH = h + L(B.t.u.). (60) \n\nIn other words, this total heat of steam is the amount of heat \nrequired to raise one pound of water from 32\xc2\xb0 F. to the tempera- \nture of vaporization and to vaporize it at that temperature \nunder a constant pressure. \n\nRemembering that h for water is approximately equal to the \ntemperature less 32 degrees corresponding to the pressure at \nwhich the steam is formed (t) , we can also write approximately, \n\nH = (t - 32) + L. (61) \n\nTo illustrate that equation (61) is approximately correct, take \nthe case of steam being formed in a boiler at an absolute pressure \nof 115 pounds per square inch. \n\nFrom the steam tables we find that the temperature / of the \nsteam at this pressure is 338\xc2\xb0 F., the latent heat of vaporiza- \ntion is 880 B.t.u. per pound and the total heat of the steam is \n1 189 B.t.u. per pound. To check these values with equation \n(59)5 we have, by substituting values of L and t, \n\n^ = (338 - 32) + 880 = 1 186 B.t.u. per pound. \n\n* Although at the lower pressure the volum\xc2\xabj of a given weight of steam is \ngreater than at a higher pressure, the change of pressure is relatively so much greater \nin the process of steam formq,tion that the product of pressure and change of \nvolume, P (F2 \xe2\x80\x94 Fi), which represents the external work done, is less for low \npressure steam than for high. \n\n\n\n66 ENGINEERING THERMODYNAMICS \n\nWhen steam is condensed under constant pressure, obvi- \nously the process which we have called the " second stage " is \nreversed and the amount of heat equal to the latent heat of \nevaporation (L) is given up during the change that occurs in the \ntransformation from steam to water. \n\nInternal Energy of Evaporation and of Steam. It was ex- \nplained in a preceding paragraph that when steam is forming \nnot all of the heat added goes into the internal or ** intrinsic " \nenergy of the steam, but that a part of it was spent in per- \nforming external work. If, then, we represent the internal \nenergy of evaporation by the symbol 1^, we can write, similarly \nto equation (59), in B.t.u. per pound of steam, \n\nI^ = ^_p(K^_M*. (62) \n\n778 \n\nThis equation represents the increase in internal energy which \ntakes place in the changing of a pound of water at the tem- \nperature t into steam at the same temperature. In all the \nformulas dealing with steam that we have used we adopt the \nstate of water at 32\xc2\xb0 F. as the arbitrary starting point from \nwhich the taking in of heat was calculated. This same arbitrary \nstarting point is used also in expressing the amount of internal \nenergy in the steam. This is the excess of the heat taken in \nover the external work done in the process. The total internal \nenergy (I^) of a pound of saturated steam at a pressure P in \npounds per square foot is equal to the total heat (H) less the \nheat equivalent of the external work done; thus, \n\nI^ = H-P^^^^^- (63) \n\n778 \n\nSuch reference is made here to the internal energy of steam \nbecause it is very useful in calculating the heat taken in and \nrejected by steam during any stage of its expansion or com- \n\n* It must be remembered that whenever there is a product of pressure and \nvolume the result is in foot-pounds. To combine such a result with other terms \nin B.t.u. we must divide by 778. \n\n\n\nPROPERTIES OF STEAM 67 \n\npression. It is well to recall the following brief and simple \nstatement (equation 12, page 18) : \n\nHeat taken in = increase of internal energy + external work \ndone. \n\nWhen we are deaHng with a compression instead of an expan- \nsion then the last term above (external work) will be a negative \nvalue to indicate that work is done upon the steam instead of \nthe steam doing work by expansion. \n\nThe following problem shows the calculation of internal energy \nand external work: \n\nExample. A boiler is evaporating water into dry and satu- \nrated steam at a pressure of 300 pounds per square inch abso- \nlute. The feed water enters the boiler at a temperature of 145\xc2\xb0 F. \n\nThe internal energy of evaporation per pound of steam is \n\n7^ _ 2, _ -P (^ - bV) _ o^j.^^ 300 X 144 (1.551 - fiV) \n778 \' 778 \n\n= 811.3 -85.3 = 726.0 B.t.u., \n\nor taken directly from saturated tables equals 726.8 B.t.u. \n\nThe total internal energy suppHed above 32\xc2\xb0 F. per pound of \nsteam is \n\nIh =H- ^ ^^ ~ ^^\'^^ = 1204.1 - 85.3 = 1118.8 B.t.u., \n778 \n\nor taken directly from saturated tables equals 1118.5 B.t.u. \n\nExternal work done above 32\xc2\xb0 F. per pound of steam as calcu- \nlated from steam tables is \n\nH \xe2\x80\x94 Ih = 1204. 1 \xe2\x80\x94 1118.5 = 85.6 B.t.u. \n\nExternal work of evaporation per pound of steam as calcu- \nlated from the steam tables is \n\nL - Il = 811.3 - 726.8 = 84.5 B.t.u. \n\nThe external work done in raising the temperature of a pound \nof water from 32\xc2\xb0 F. to the boiling point is \n\n85.6 - 84.5 = I.I B.t.u. \n\n\n\n68 ENGINEERING THERMODYNAMICS \n\nThe external work done in raising the temperature of a pound \nof water from 32\xc2\xb0 F. to 145\xc2\xb0 F. is \n\n300 X 144 (0.0163 - 0.01602*) ^ ooj g 1-^ \n\n778 \n\nThe external work done in forming the steam from water at \n145\xc2\xb0 F. is, then, \n\n84.5 + 1. 10 \xe2\x80\x94 o.oi = 85.59 B.t.u. \nas compared with 85.3 B.t.u. as calculated from \n\n778 \n\nIt is to be noticed that the external work done during the addi- \ntion of the heat of the liquid is small as compared with other \nvalues, and for most engineering work it is customary to assume \nthat no external work is done during the addition of heat to the \nwater. With this assumption \n\nIu = h + L- ^^^~J^^ =H-{L-Il). \n778 \n\nSteam Formed at Constant Volume. When saturated steam \nis made in a boiler at constant volume, as, for example, when the \npiping connections from the boiler to the engines are closed, then \nno external work is done, and all the heat taken in is converted \ninto and appears as internal energy Ih of the steam. This \nquantity is less than the total heat H of steam, representing its \nformation at constant pressure, by quantity P {V \xe2\x80\x94 -^-q) -^ 778, \nwhere P represents the absolute pressure at which the steam is \nformed in pounds per square foot and V is the volume of a \npound of steam at this pressure in cubic feet. \n\nWet Steam. In all expansions studied thus far, deaHng with \nsaturated steam, it has been assumed that the steaming process \nwas complete and that the water had been completely converted \ninto steam. Now in actual engineering practice it is not at all \nunusual to have steam leaving boilers which is not perfectly \n\n* See Table 6 in Marks and Davis\' Steam Tables and Diagrams for volumes of \nwater. \n\n\n\nPROPERTIES OF STEAM 69 \n\nand completely vaporized; in other words, the boilers are sup- \nplying to the engines a sort of mixture of steam and water. \nThis mixture we call wet steam. It is steam which carries \nactually in suspension minute particles of water, which remain \nthus in suspension almost indefinitely. The temperature of this \nwet steam is always the same as that of completely saturated \nsteam as given in the steam tables so long as any steam remains \nuncondensed. The ratio of the weight of moisture or water in \na pound of wet steam to a pound of completely saturated steam \nis called the degree of wetness ; and when this ratio is expressed \nas a per cent, we have then what we call percentage of moisture or \n** per cent wet"; thus, if in a pound of wet steam there is 0.04 \npound of water in suspension or entrained, we speak of the steam \nas being four per cent wet. Another term, called the quality of \nsteam, which is usually expressed by the symbol x, is also fre- \nquently used to represent the condition of wet steam. Quality \nof steam may be defined as the proportion of the amount of \ndry or completely evaporated steam in a pound of wet steam. \nTo illustrate with the example above, if there is 0.04 pound \nof water in a pound of wet steam; the quality in this case \nwould be I \xe2\x80\x94 0.04 or 0.96. In this case we say then that the \nquaHty of this steam is ninety-six one-hundredths. \n\nWith this understanding of the nature of wet steam it is obvi- \nous that latent heat of a pound of wet steam is xL. Similarly, \nthe total heat of a pound of wet steam is h -{- xL, and the \nvolume of a pound of wet steam is xV -h g^ (i \xe2\x80\x94 x) or approxi- \nmately equal to xV, because the term gV (^ ~ ^) i^ negligibly \nsmall except in cases where the steam is so wet as to consist \nmostly of water. Similarly, the internal energy in a pound of \nwet steam is \n\nl^ = h + x\'^L-^^^^^]^- (64) \n\nSuperheated Steam. When the temperature of steam is higher \nthan that corresponding to saturation as taken from the steam \ntables, and is, therefore; higher than the standard temperature \ncorresponding to the pressure, the steam is said to be super- \n\n\n\n70 ENGINEERING THERMODYNAMICS \n\nheated. In this condition steam begins to depart and differ \nfrom its properties in the saturated condition, and when super- \nheated to a very high degree it begins to behave somewhat \nlike a perfect gas. \n\nThere are tables of the properties of superheated steam just \nas there are tables of saturated steam. Tables of superheated \nsteam are very much larger and cover many more pages than \nthose of saturated steam, for the reason that for every pressure \nthere are innumerable values for temperature and also for vol- \nume. When dealing with saturated steam there is always only \none possible temperature and only one specific volume to be \nconsidered. With superheated steam, on the other hand, for a \ngiven pressure we may have any temperature above that of \nsaturated steam, and corresponding to each temperature there \nwill be, of course, definite values for specific volume and total \nheat. Like a perfect gas the specific volume or the cubic feet \nper pound increases with the increase in temperature. \n\nTotal heat of superheated steam is obviously greater than \nthe total heat of saturated steam which is not wet by the amount \nof heat that must be added to dry * saturated steam to produce \nthe required degree of superheat. Thus, since the total heat \nof dry saturated steam is, as before, h -\\- L, the total heat of \nsuperheated steam with D degrees of superheat is \n\nKs = h + L-\\-CpXD; \nor if we call the temperature of the superheated steam /gup and \n/sat is the temperature of saturated steam corresponding to the \npressure, we can write, similarly, \n\n"Bis = h -\\- L -\\- Cp (/gup \xe2\x80\x94 /gat ) \xe2\x80\xa2 . \n\nThe formulas above for superheated steam are for the total \nheat of steam at constant pressure as shown by the use of Cp. \nIn practically all engineering calculations it is only the condition \nof total heat at constant pressure that interests us. \n\n* In practice we speak of steam as being dry saturated when it is exactly sat- \nurated and has no moisture. It is the condition known simply as saturated steam \nas regards the properties given in the ordinary steam tables. The dry saturated \ncondition is the boundary between wet steam and superheated steam. \n\n\n\nPROPERTIES OF STEAM 7 1 \n\nIt should be carefully observed that the total heat of super- \nheated steam is the amount of heat required to produce a pound \nof steam with the required degrees of superheat from water at \n32\xc2\xb0 F. \n\nTo obtain the amount of internal energy of a pound of steam \n(superheated) corresponding to this total heat as stated above, \nthe external work expended in the steaming process must be \nsubtracted; that is, \n\nIh= h + L - ^^^-\\~^\'^^ +C,feup - 4at) - ^^^^^% ^"^^ \n778 778 \n\n= h ^L+C, feup -4at) - ^ ^^^^^ 7 \'^ ^ = Hs - -^ (^-up - /o) . \n\n778 778 \n\nThe term ^ can be neglected in the equations above for prac- \ntically all engineering calculations as the maximum error from \nthis is not likely to be more than one in one thousand or ^0 per \ncent.* The accuracy of our steam tables for values of latent and \ntotal heat is not established to any greater accuracy. Making \nthis approximation, the equation above becomes \n\nlH = h + L + C, (4up - U - (^^^^) = H. - 1^<^ . \n\nV 778 / 778 \n\nTo make these matters clearer, examine, for example, the prop- \nerties of superheated steam as given in Marks and Davis\' Steam \nTables f for superheated steam at 165 pounds per square inch \nabsolute pressure and 150\xc2\xb0 F. superheat. We read as follows: \n\nPress, lbs. per Deg. of Sup, \n\nsq. in. abs. F. \n\n165 / 516.0 \n\ny 3-43 \n\nHs or hX 1277.6 \n\n* The error in the value of total internal energy due to neglecting the term -^q \nin the exercise worked out on page 67 is 0.85 B.t.u. per pound (one per cent of \nthe external work), or an error of 0.76 per cent in the final result. \n\nt Page 48, line 13 (First Edition). \n\nt In their tables Marks and Davis represent the total heat of superheated steam \nby h. In this book the symbol Hs is used for greater clearness and to be con- \nsistent with the symbols used in the preceding formulas. They use also v for \nspecific volume in place of V as above. \n\n\n\n72 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nThe values of h and L for saturated steam at this pressure are \nrespectively 338.2 and 856.8. From the curves given, Fig. 12,* \nwe find that the specific heat of superheated steam at constant \npressure (Cp) is 0.552. From these data we could obtain, then, \n^a = ^-f L + Cp Xi5o= 338.2 +856.8+ 0.552 X 150= 1277.7 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nf ^ \n\n\nr \n\n\n\\ ^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nPC \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n, \n\n\n\n\n\\ \\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n\\^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.W \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nV \' \n\n\nV ^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\\ \n\n\n\\> \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x80\xa2W \n\n\n\n\n\n\n\n\n\n\n\n\n\n\nV \n\n\n\\ \n\n\n> \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\nv^ \n\n\n\\\\ \n\n\n\\, \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x80\xa2UU \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\\ \n\n\n\n\n\\ \n\n\n\\ \n\n\ns \n\n\ns \n\n\ny \n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.58 \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\nN \n\n\n^ \n\n\nV, \n\n\ns\\ \n\n\n\\ \n\n\n\'v \n\n\ns \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\\ \n\n\n\\ \n\n\nV \n\n\nS \n\n\n\\ \n\n\n\\ \n\n\n\\ \n\n\n^ \n\n\nV \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nrp \n\n\n\n\n\n\n\n\n\n\n\n\nN \n\n\n\n\ns \n\n\n\n\ns \' \n\n\ns \n\n\n\\ \n\n\nSJ \n\n\n.\'^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\ni \n\n\n\n\nN \n\n\n\n\n^\\ \n\n\n\\ \n\n\n\n\n\\ \n\n\nN \n\n\n\n\n"V \n\n\n^ \n\n\n\n\n\n\n\n\n\n\n4.4 \n\n\nLb \n\n\ns.Abi \n\n\n\n\n\n\n\n\n\n\n\n\n/s \n\n\n\n\n\n\nV \n\n\n\n\nN \n\n\ns. \n\n\nN \n\n\n\\ \n\n\n\n\n-^ \n\n\n\n\n\n\nJ \n\n\ng \n\n\nJii \n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\nH \n\n\n\n\n\n\n^s \n\n\n\n\n\\ \n\n\n\n\n"^ \n\n\n\xe2\x96\xa0^ \n\n\n\n\n\n\n\n\n\xe2\x96\xa022 \n\n\n7.2 \n).0- \n0.4 \n2.2 \n3.6 \n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\xe2\x96\xa0^ \n\n\n\n\n\n\n\n\n> \n\n\n\n\n\n\n\xe2\x96\xa0->. \n\n\n^ \n\n\n\n\n\n\n\xe2\x80\x94 \xe2\x80\xa2 \n\n\n\xe2\x80\x94 \n\n\n-17 \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x80\xa2*~. \n\n\n^ \n\n\n-J \n\n\n^ \n\n\n\n\n\n\n\n\n_ \n\n\n-11 \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x96\xa0^ \n\n\n\n\n- \n\n\n~U \n\n\n\n\n\n\n\n\n\n_ \n\n\n.8i2 \n\n\n\xc2\xbb \n\n\n\n\n\n\n\n\n\n\nJ_ \n\n\n\n\n\n\n\n\n\xe2\x96\xa0"" \n\n\nr \xe2\x80\x94 \' \n\n\n\n\n\n\n\n\n_ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.5^.8 \n\n\n\xc2\xbb \n\n\n\n\n^8 \n\n\n\n\n\nV \n\n\nr \n\n\n\n\n\n\n\n\n\n\n\n_ \n\n\n\n\n\n\n\n_ \n\n\n- \n\n\nJ \n\n\n\n\n\n_ \n\n\n\n\n\n\n_ \n\n\n-1 \n\n\n=5 \n\n\nc \n\n\nJL \n\n\n\n\n\n\n\n7 \n\n\n</. \n\n\n- \n\n\n\n\n\n\n\n\n\xe2\x80\x94 \n\n\n- \n\n\n\n\n\n\n- \n\n\n\n\n:z \n\n\n\n\n^ \n\n\n\n\n\n\n-= \n\n\nn \n\n\n- \n\n\n\n\n\n\n\n\n/.4G \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n" \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.42 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nFig. 12. \n\n\n\n200 250 300 350 400 450 500 550 600 050 700 750 \nTemperature \xc2\xb0 F \n\n\xe2\x96\xa0Mean Values of Cp Calculated by Integration from Knoblauch \nand Jakob\'s Data. \n\n\n\n(nearly). Observe that this value agrees with that given in \nthe tables as indicated above. The specific volume (F) is cal- \nculated from the empirical formula derived from experimental \nresults and is expressed as follows: \n\nV = [0.5962 T-p{.+ 0.0014 p) [ \'\'\xc2\xb0\'\'^\'\xc2\xb0\xc2\xb0\xc2\xb0 - 0.0833)] ^, \n\n* The curves for values of Cp, as given in Fig. 12, are for average and not for \ninstantaneous values such as are- given in Fig. 13 and also in Fig. 5, page 97, of \nMarks and Davis\' Tables and Diagrams. Great caution must be observed in the \nuse of curves of this kind. Those giving instantaneous values can only be used \nfor the value of Cp in the formulas given for the total heat of superheated steam \nafter the average value has been found by integrating the curve representing \nthese values for a given pressure. \n\n\n\nPROPERTIES OF STEAM \n\n\n\n73 \n\n\n\nwhere p is in pounds per square inch, V is in cubic feet per pound \nand r = / + 460 is the absolute temperature on the Fahrenheit \nscale.* \n\n\n\n0.90 \n0.85 \n0.80 \n\n0.75 \n\n0.70 \n\nc \n\nI \n|o.65 \n\nI \n\nf^O.60 \n0.55 \n\'^ \n\n\n\n\n\n\n20 \n\n\n\n\n\n\n\xe2\x80\xa2 \n\n\n\n\n\n\n1 \n\n\nH \n\n\n\n\n\n\n\n\n\n\n\n\nIG \n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n14/ \n\n\n1 \n\n\n\n\n\n\n\n\n\n\n12^ \n\n\n\\l \n\n\n\n\n\n\n\n\n\n\n10/ \\ \n\n\nWW \n\n\n\n\n\n\n\n\n\n\nc/\\ \n\n\n\\vl \n\n\nPressu \nsq. cm. a \n\n\nres in Kg \nbsolute. \n\n\nm. per \n\n\n/ \n\n\nK^ \n\n\n^ \n\n\n^ \n\n\n\n\n\n\n\n\nA \n\n\n\n\n"~^ \n\n\n\n\n::r \xe2\x80\x94 -^ \n\n\n\n\nI_\xe2\x80\x94 \xe2\x80\x94 ^ \n\n\n^^^^^ \n\n\ne.45 \n\n\n\xe2\x80\x94 \n\n\n-^;;r^ \n\n\n^^^^^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n100 150 200 250 ^300 350 400 \n\nTemporature C. \n\nFig. 13. \xe2\x80\x94 Values of the "True " Specific Heat of Superheated Steam. \n\nDrying of Steam by Throttling or Wire-drawing. When steam \nexpands by passing through a very small opening, as, for example, \nthrough a valve only partly open in a steam line, the pressure \n\n\n\n* The value of 7 or -^ of superheated steam used in ordinary engineering calcu- \nlations is 1.3. \n\n\n\n74 ENGINEERING THERMODYNAMICS \n\nis considerably reduced. Steam engineers usually call this \neffect throttling or wire-drawing. The result of expansion of \nthis kind when the pressure is reduced and no work is done \nis that if the steam is initially wet it will be drier and if it is \ninitially dry or superheated the degree of superheat ^dll be in- \ncreased. The reason for this is that the total heat required to \nform a pound of dry saturated steam {H) is considerably less at \nlow pressure than at high, but obviously the total quantity of \nheat in a pound of steam must be the same after wire-drawing \nas it was before, neglecting radiation. Now if steam is initially \nwet and the quahty is represented by Xi, then the total heat in \nthe steam is represented by hi plus XiLi, in which hi and Li repre- \nsent the heat of the Kquid and the latent heat of the steam at \nthe initial pressure. If, also, the quality, heat of liquid and the \nlatent heat of the steam after wire-drawing are represented \nrespectively hy oc^, h^ and L2, then hi + XiLi = h. -\\- x^Li. It \nhappens in many engineering calculations that all of the terms \nin the last equation are known except x^ and this can be solved \nas shown in the following equation: \n\n_ XiLi -\\- hi \xe2\x80\x94 h. \n\nx^ \xe2\x80\x94 \n\n\n\nThis drying action of steam in passing through a small open- \ning or an orifice is very well illustrated by steam discharging \nfrom a small leak in a high pressure boiler into the atmosphere. \nIt will be observed that no moisture is visible in the steam a \nfew inches from the leak but farther off it becomes condensed \nby loss of heat, becomes clouded and plainly visible. An im- \nportant application is also to be found in the throttling calo- \nrimeter (page 75). \n\nDetermination of the Moisture in Steam. Unless the steam \nused in the power plant is superheated it is said to be either dry \nor wet, depending on whether or not it contains water in sus- \npension. The general types of steam calorimeters used to de- \ntermine the amount of moisture in the steam may be classified \nunder three heads: \n\n\n\nPROPERTIES OF STEAM 75 \n\n1. Throttling or superheating calorimeters. \n\n2. Separating calorimeters. \n\n3. Condensing calorimeters. \n\nThrottling or Superheating Calorimeters. The type of steam \ncalorimeter used most in engineering practice operates by passing \na sample of the steam through a very small orifice, in which it \nis superheated by throttling. A very satisfactory calorimeter of \nthis kind can be made of pipe fittings as illustrated in Fig. 14. \nIt consists of an orifice O discharging into a chamber C, inta \nwhich a thermometer T is inserted, and a mercury manometer \nis usually attached to the cock V3 for observing the pressure in \nthe calorimeter. \n\nIt is most important that all parts of calorimeters of this \ntype, as well as the connections leading to the main steam pipe, \nshould be very thoroughly lagged by a covering of good insu- \nlating material. One of the best materials for this use is hair \nfelt, and it is particularly well suited for covering the more or \nless temporary pipe fittings, valves and nipples through which \nsteam is brought to the calorimeter. Very many throttling \ncalorimeters have been declared useless by engineers and put \ninto the scrap heap merely because the small pipes leading to \nthe calorimeters were not properly lagged, so that there was too \nmuch radiation, producing, of course, condensation, so that the \ncalorimeter did not get a true sample. It is obvious that if the \nentering steam contains too much moisture the drying action \ndue to the throttling in the orifice may not be sufiicient to super- \nheat. It may be stated in general that unless there is about \n5\xc2\xb0 to 10\xc2\xb0 F. of superheat in the calorimeter, or, in other words, \nunless the temperature on the low pressure side of the orifice is \nat least about 5\xc2\xb0 to 10\xc2\xb0 F. higher than that corresponding to the \npressure in the calorimeter, there may be some doubt as to the \naccuracy of results.* The working limits of throtthng calo- \n\n* The same general statement may be made as regards determinations of \nsuperheat in engine and turbin^ tests. Experience has shown that tests made \nwith from o to 10 degrees Fahrenheit superheat are not reliable, and that the \nsteam consumption in many cases is not consistent when compared with results \n\n\n\n76 ENGINEERING THERMODYNAMICS \n\nrimeters vary with the initial pressure of the steam. For 35 \npounds per square inch absolute pressure the calorimeter ceases \nto superheat when the percentage of moisture exceeds about \n2 per cent; for 150 pounds absolute pressure when the moisture \nexceeds about 5 per cent; and for 250 pounds absolute pressure \nwhen it is in excess of about 7 per cent. For any given pressure \nthe exact limit varies slightly, however, with the pressure in the \ncalorimeter. \n\nIn connection with a report on the standardizing of engine \ntests, the American Society of Mechanical Engineers* published \nthe following instructions regarding the method to be used for \nobtaining a fair sample of the steam from the main pipes. It is \nrecommended in this report that the calorimeter shall be con- \nnected with as short intermediate piping as possible with a \nso-called calorimeter nipple made of |-inch pipe and long enough \nto extend into the steam pipe to within | inch of the opposite \nwall. The end of this nipple is to be plugged so that the steam \nmust enter through not less than twenty J-inch holes drilled \naround and along its length. None of these holes shall be less \nthan I inch from the inner side of the steam pipe. The sample \nof steam should always be taken from a vertical pipe as near as \npossible to the engine, turbine or boiler being tested. A good \nexample of a calorimeter nipple is illustrated in Fig. 15. \n\nNever close and do not usually attempt to adjust the discharge \nvalve V2 without first closing the gage cock V3. Unless this \nprecaution is taken, the pressure may be suddenly increased in \nchamber C, so that if a manometer is used the mercury will be \nblown out of it; and if, on the other hand, a low-pressure steam \ngage is used it may be ruined by exposing it to a pressure much \nbeyond its scale. \n\nUsually it is a safe rule to begin to take observations of tem- \nperature in calorimeters after the thermometer has indicated a \n\nobtained with wet or more highly superheated steam. The errors mentioned \nwhen they occur are probably due to the fact that in steam, indicating less than \n10 degrees Fahrenheit superheat, water in the hquid state may be taken up in \n*\' slugs" and carried along without being entirely evaporated. \n\n* Proceedings American Society of Mechanical Engineers, vol. XXI. \n\n\n\nPROPERTIES OF STEAM 77 \n\nmaximum value and has again slightly receded from it. The \nquahty or relative dryness of wet steam is easily calculated by \nthe following method and symbols : \n\npi = steam pressure in main, pounds per square inch \n\nabsolute, \np2 = steam pressure in calorimeter, pounds per square \n\ninch absolute, \ntc = temperature in calorimeter, degrees Fahrenheit, \nLi and hi = heat of vaporization and heat of liquid corre- \nsponding to pressure ^1, B.t.u., \n^2 and /2 = total heat (B.t.u.) and temperature (degrees \nFahrenheit) corresponding to pressure ^2, \nCp = specific heat of superheated steam. Assume 0.47 \n\nfor low pressures existing in calorimeters, \nX\\ = initial quality of steam. \n\nTotal heat in a pound of wet steam flowing into orifice is \n\nXiLi + ^1, \n\nand after expansion, assuming all the moisture is evaporated, \nthe total heat of the same weight of steam is \n\nH2 + Cp [tc \xe2\x80\x94 h). \n\nThen assuming no heat losses and putting for Cp its value 0.47 we \nhave \n\nociLi + hi = H2 + 0.47 {tc - k) \n\nX ^ H2 + 0.47 {ic -h) -hi .^ s \n\nLi \n\nThe following example shows the calculations for finding the \nquality of steam from the observations taken with a throttling \ncalorimeter : \n\nExample. Steam at a pressure of loc pounds per square inch \nabsolute passes through a throttling calorimeter. In the calo- \nrimeter the temperature Of the steam becomes 243\xc2\xb0 F. and the \npressure 15 pounds per square inch absolute. Find the quahty. \n\n\n\n78 ENGINEERING THERMODYNAMICS \n\nSolution. By taking values directly from tables of properties \nof superheated steam,* the total heat of the steam in the calorim- \neter at 15 pounds per square inch absolute pressure and 243\xc2\xb0 F. \nis 1 164.8 B.t.u. per pound or can be calculated as follows: \n\nH + Cp {to- k) = 1150.7 + 0.47 (243 - 213) \n= 1 164.8 B.t.u. per pound. \n\n(Note, tc = temperature in calorimeter and ^2 = temperature \ncorresponding to calorimeter pressure.) \n\nThe total heat of the steam before entering the calorimeter \nis h -\\- xL. At 100 pounds per square inch absolute pressure, \nthis is 298.3 + 888.0 X in B.t.u. per pound. Since the heat in \nthe steam per pound in the calorimeter is obviously the same as \nbefore it entered the instrument, we can equate as follows: \n\n298.3 + 888.0 X = 1 164.8 \nX = 0.976 \n\nor the steam is 2.4 per cent wet. \n\nBarrus Throttling Calorimeter. This is an important varia- \ntion from the type of throttling calorimeter shown in Fig. 14 and \nhas been quite widely introduced by Mr. George H. Barrus. In \nthis apparatus the temperature of the steam admitted to the \ncalorimeter is observed instead of the pressure and a very free \nexhaust is provided so that the pressure in the calorimeter is \natmospheric. This arrangement simplifies very much the obser- \nvations to be taken, as the quahty of the steam Xi can be calcu- \nlated by equation (65) by observing only the two temperatures \nti and tc, taken respectively on the high and low pressure sides of \nthe orifice in the calorimeter. This calorimeter is illustrated in \nFig. 15. The two thermometers required are shown in the \nfigure. Arrows indicate the path of the steam. f \n\nThe orifice in such calorimeters is usually made about -^^ i^^h \nin diameter, and for this size of orifice the weight of steam { \n\n* Marks and Davis\' Steam Tables and Diagrams (ist ed.), page 24. \nt Transactions of American Society of Mechanical Engineers, vol. XI, page 790. \nX In boiler tests corrections should be made for the steam discharged from the \nsteam calorimeters. \n\n\n\nPROPERTIES OF STEAM \n\n\n\n79 \n\n\n\nConnection for \nManometer \n\n\n\n\nFig. 14. \xe2\x80\x94 Simple Throttling Steam Calorimeter. \n\n\n\ni k I \n\n//, /\' ooooo<>oo6ooo o lidrii \n\n\n\n\n\nFig. 15. \xe2\x80\x94 Barrus\' Throttling Steam Calorimeter. \n\n\n\ndischarged per hour at 175 pounds per square inch absolute \npressure is about 60 pounds. It is important that the orifice \nshould always be kept clean, because if it becomes obstructed \nthere will be a reduced quantity of steam passing through the \n\n\n\n8o ENGINEERING THERMODYNAMICS \n\ninstrument, making the error due to radiation relatively more \nimportant. \n\nIn order to free the orifice from dirt or other obstructions the \nconnecting pipe to be used for attaching the calorimeter to the \nmain steam pipe should be blown out thoroughly with steam \nbefore the calorimeter is put in place. The connecting pipe \nand valve should be covered with hair felting not less than \nI inch thick. It is desirable also that there should be no leak \nat any point about the apparatus, either in the stuf&ng-box of \nthe supply valve, the pipe joints or the union. \n\nWith the help of a diagram* giving the quahty of steam directly \nthe Barrus calorimeter is particularly well suited for use in \npower plants, where the quality of the steam is entered regu- \nlarly on the log sheets. The percentage of moisture is obtained \nimmediately from two observations without any calculations. \n\nSeparating Calorimeters. It was explained on page 76 that \nthrottling calorimeters cannot be used for the determination of \nthe quality of steam when for comparatively low pressures the \nmoisture is in excess of 2 per cent, and when for average boiler \npressures in modern engineering practice it exceeds 5 per cent. \nFor higher percentages of moisture than these low limits sepa- \nrating calorimeters are most generally used. In these instru- \nments the water is removed from the sample of steam by mechan- \nical separation just as it is done in the ordinary steam separator \ninstalled in the steam mains of a power plant. There is pro- \nvided, of course, a device for determining, while the calorimeter \nis in operation, usually by means of a calibrating gage glass, the \namount of moisture collected. This mechanical separation de- \npends for its action on changing very abruptly the direction of \nflow and reducing the velocity of the wet steam. Then, since \nthe moisture (water) is nearly 300 times as heavy as steam at \nthe usual pressures delivered to the engine, the moisture will be \ndeposited because of its greater inertia. \n\nWhen a U-tube manometer is used to determine the pressure \nin a calorimeter of the type illustrated in Fig. 14, this pressure \n* See page 117, and also Moyer\'s Power Plant Testing, 2d edition, pages 58-60. \n\n\n\nPROPERTIES OF STEAM \n\n\n\n8l \n\n\n\ncan be obtained very accurately, and an excellent means is pro- \nvided for calibrating the thermometer in the calorimeter just as \nit is to be used. The calibration would be made, of course, by \nthe method of comparing with the temperature corresponding to \nknown pressures. In order to avoid having superheated steam \nin the calorimeter for this calibration the felt or similar material \nusually needed for covering the valves and nipples between the \nmain steam pipe and the calorimeter \nshould be kept saturated with cold \nwater. \n\nFig. 1 6 illustrates a form of \nseparating calorimeter having a \nsteam jacketing space which re- \nceives Hve steam at the same tem- \nperature as the sample. Steam is \nsupplied through a pipe A, dis- \ncharging into a cup B. Here the \ndirection of the flow is changed \nthrough nearly i8o degrees, causing \nthe moisture to be thrown outward \nthrough the meshes in the cup into \nthe vessel V. The dry steam passes \nupward through the spaces between \nthe webs W, into the top of the- \noutside jacketing chamber J, and \n\nis finally discharged from the Fig. i 6. \xe2\x80\x94 Separating Calorimeter. \n\nbottom of this steam jacket through \n\nthe nozzle N. This nozzle is considerably smaller than \nany other section through which the steam flows, so that there \nis no appreciable difference between the pressures in the calo- \nrimeter proper and the jacket. The scale opposite the gage \nglass G is graduated to show in hundredths of a pound at the \ntemperature corresponding to steam at ordinary working pres- \nsures, the variation of the level of the water accumulating. A \nsteam pressure gage P indicates the pressure in the jacket J, \nand since the flow of steam through the nozzle N is roughly \n\n\n\n\n82 ENGINEERING THERMODYNAMICS \n\nproportional to the pressure, another scale in addition to the \none reading pressures is provided at the outer edge of the dial. \nA petcock C is used for draining the water from the instru- \nment, and by weighing the water collected corresponding to a \ngiven difference in the level in the gage G the graduated scale \ncan be readily calibrated. Too much reliance should not be \nplaced on the readings for the flow of steam as indicated by the \ngage P unless it is frequently calibrated. Usually it is very \nlittle trouble to connect a tube to the nozzle N and condense \nthe steam discharged in a large pail nearly filled with water. \nWhen a test for quality is to be made by this method the pail \nnearly filled with cold water is carefully weighed, and then at \nthe moment when the level of the water in the water gage G has \nbeen observed the tube attached to the nozzle N is immedi- \nately placed under the surface of the water in the pail. The \ntest should be stopped before the water gets so hot that some \nweight is lost by \'\' steaming." The gage P is generally cali- \nbrated to read pounds of steam flowing in ten minutes. For the \nbest accuracy it is desirable to use a pail with a tightly fitting \ncover into which a hole just the size of the tube has been cut. \n\nIf W is the weight of dry steam flowing through the orifice N \nand w is the weight of moisture separated, the quality of the \nsteam is \n\n(66) \n\n\n\nW -\\-w \n\n\n\nCondensing or Barrel Calorimeter. For steam having a large \npercentage of moisture (over 5 per cent) the condensing or \nbarrel calorimeter will give fairly good results if properly used. \nIn its simplest form it consists of a barrel placed on a platform \nscale and containing a known weight of cold water. The steam \nis introduced by a pipe reaching nearly to the bottom of the \nbarrel. The condensation of the steam raises the temperature \nof the water, the loss of heat by the wet steam being equal to \nthe gain of heat by the cold water. It will, therefore, be neces- \nsary to observe the initial and final weights, the initial and \n\n\n\nPROPERTIES OF STEAM 83 \n\nfinal temperatures of the water in the barrel and the temperature \nof the steam. \n\nLet W = original weight of cold water, pounds. \nw = weight of wet steam introduced, pounds. \nti = temperature of cold water, degrees Fahrenheit. \n4 = temperature of water after introducing steam. \nts = temperature of steam. \nL = latent heat of steam at temperature ts. \nX = quahty of steam. \nThen, \n\nHeat lost by wet steam = heat gained by water. \n\nwxL + w{ts-h) =W {h- k) \n\n^ ^ W{k-k) -w{ts-h) ^^^^ \n\nThe accuracy of the results depends obviously upon the accu- \nracy of the observation, upon thorough stirring of the water so \nthat a uniform temperature is obtained and upon the length of \ntime required. The time should be just long enough to obtain \naccurate differences in weights and temperatures; otherwise, \nlosses by radiation will make the results much too low. \n\nEquivalent Evaporation and Factor of Evaporation. For the \ncomparison of the total amounts of heat used for generating \nsteam (saturated or superheated) under unlike conditions it is \nnecessary to take into account the temperature k at which the \nwater is put into the boiler as well as also the pressure P at \nwhich the steam is formed.* These data are of much impor- \ntance in comparing the results of steam boiler tests. The basis \nof this comparison is the condition of water initially at the boil- \ning point for \'\'atmospheric" pressure or at 14.7 pounds per \nsquare inch; that is, at 212\xc2\xb0 F. and with steaming taking place \nat the same temperature. For this scandard condition, then, \n\n* As the pressure P increases the total heat of the steam also increases; but \nas the initial temperature of the water ("feed temperature") increases the value \nof the heat of the hquid decreases. \n\n\n\n84 ENGINEERING THERMODYNAMICS \n\nh = o and H = L = 970.4 B.t.u. per pound. Evaporation \nunder these conditions is described as, \n\n" from (a feed- water temperature of) and at (a pressure corre- \nsponding to the temperature of) 212\xc2\xb0 F." \n\nTo illustrate the application of a comparison with this stand- \nard condition let it be required to compare it with the amount \nof heat required to generate steam at a pressure of 200 pounds \nper square inch absolute with the temperature of the water sup- \nplied (feed water) at 190\xc2\xb0 F. \n\nFor P = 200 pounds per square inch absolute the heat of the \nliquid h = 354.9 B.t.u. per pound and the heat of evaporation \n(L) is 843.2 B.t.u. per pound. For t = 190\xc2\xb0 F. the heat of the \nliquid (ho) is 157.9 B.t.u. per pound. The total heat actually- \nrequired in generating steam at these conditions is, therefore, \n\n843.2 + (354.9 \xe2\x80\x94 157.9) = 1040.2 B.t.u. per pound. \n\nThe ratio of the total heat actually used for evaporation to \nthat necessary for the condition defined by \'\' from and at 212\xc2\xb0 F.\'^ \nis called the factor of evaporation. In this case it is the value \n\n1040.2 -7- 970.4 = 1.07. \n\nIf we write F for factor of evaporation, h and L respectively \nthe heats of the liquid and of evaporation corresponding to the \nsteam pressure and ho the heat of the liquid corresponding to the \ntemperature of feed water, then \n\np ^ L + Qi-h) \n970.4 \n\nThe actual evaporation of a boiler (expressed usually in pounds \nof steam per hour) multiplied by the factor of evaporation is \ncalled the equivalent evaporation. \n\nPROBLEMS \n\n1. Dry and saturated steam has a pressure of 100 lbs. per sq. in. abso- \nlute. What is the temperature of the steam ? ^wj. 327.8\xc2\xb0 F. \n\n2. What is the volume per pound of this steam ? ^;z5. 4.429 cu. ft. \n\n3. What is the heat of the liquid per pound of this steam? \n\nAns. 298.3 B.t.u. \n\n\n\nPROPERTIES OF STEAM 85 \n\n4. What is the latent heat per pound of this steam ? \n\nAns. 888.0 B.t.u. \n\n5. What is the total heat above 32\xc2\xb0 F. per pound of this steam? \n\nAns. 1 186.3 B.t.u \n\n6. Dry and saturated steam has a temperature of 300\xc2\xb0 F. What is \nits pressure? Ans. 67.0 lbs. per sq. in. absolute. \n\n7. How many British thermal units would be required to raise the tem- \nperature of I lb. of water from 32 degrees to the boiling point as stated in \nproblem 6? Ans. 269.6 B.t.u. \n\n8. How many British thermal units are required to evaporate i lb. of \nthis water into dry and saturated steam under the conditions of problem 6 ? \n\nAns. 909.5 B.t.u. \n\n9. How many British thermal units are required to generate i lb. of \ndry and saturated steam from water at 32\xc2\xb0 F. under the conditions of \nproblem 6? Ans. 11 79.1 B.t.u. \n\n10. A closed tank contains 9 cu. ft. of dry and saturated steam at a \npressure of 150 lbs. per sq. in. absolute. \n\n(a) What is its temperature? Ans. 358.5** F. \n\n(b) How many pounds of steam does the tank contain? \n\nAns. 2.988 lbs. \n\n11. A boiler generates dry and saturated steam under a pressure of 200 \nlbs. per sq. in. absolute. The feed water enters the boiler at 60\xc2\xb0 F. \n\n(a) What is the temperature of the steam? Ans. 381.9\xc2\xb0 F. \n\n(b) How many British thermal units are required to generate \n\nI lb. of this steam if this feed water is admitted at 32\xc2\xb0 F.? \n\nAns. 1198.1 B.t.u. \n\n(c) How many British thermal units are required to raise the \n\ntemperature of i lb. of water from 32\xc2\xb0 to 60\xc2\xb0 F.? \n\nAns. 28.08 B.t.u. \n\n(d) How many British thermal units are required to generate \n\nI lb. of this steam from feed water at 60 degrees into the \nsteam at the pressure stated at the beginning of this prob- \nlem? Ans. 1170.02 B.t.u. \n\n12. One pound of dry and saturated steam is under a pressure of 250 \nlbs. per sq. in. absolute. \n\n(a) What is its internal energy of evaporation ? \n\nAns. 742.0 B.t.u. \n\n(b) What is its total internal energy above 32\xc2\xb0 F. ? \n\nAns. 1116.4 B.t.u. \n\n(c) How much external work was done during its formation from \n\n32\xc2\xb0 F.? * Ans. 85.1 B.t.u. \n\n(d) How much external work was done during the evaporation? \n\nAns. 84.3 B.t.u. \n\n\n\n86 ENGINEERING THERMODYNAMICS \n\n(e) How much external work was done during the change in \ntemperature of the water from 32 degrees to the boiling \npoint corresponding to the pressure? Ans. 0.8 B.t.u. \n\n13. Dry and saturated steam is generated in a boiler and has a tempera- \nture of 400\xc2\xb0 F. The feed water enters the boiler at 200\xc2\xb0 F. \n\n(a) What pressure is carried in the boiler ? \n\nAns. 247.1 lbs. per sq. in. absolute. \n\n(b) What is the total heat supplied to generate i lb. of this steam? \n\nA7ZS. 1033.36 B.t.u. \n\n(c) How much external work was done during its formation? \n\nAns. 84.97 B.t.u. \n\n(d) How much heat was used in increasing the internal energy? \n\nAns. 948.39 B.t.u. \nCheck this by (^2 \xe2\x80\x94 ^1 + II) noting that this assumes no \nexternal work done in the heating of the liquid. \n\nAns. 949.16 B.t.u. \n\n14. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has \na quality of 90 per cent dry. What is its temperature? Ans. 327.8\xc2\xb0 F. \n\n15. How many British thermal units would be required to raise the \npound of steam in the above problem from 32\xc2\xb0 F. to the boiling point corre- \nsponding to the pressure stated? Ans. 298.3 B.t.u. \n\n16. What would be the volume of a pound of steam for the conditions \nstated in problem 14? Ans. 3.99 cu. ft. \n\n17. How many heat units (latent heat) are required to evaporate the \nsteam in problem 14? Ans. 799.2 B.t.u. \n\n18. What is the amount of the total heat (above 32\xc2\xb0 F.) of the steam in \nproblem 14? Ans. 1097.5 B.t.u. \n\n19. What would be the external work of evaporation of the steam in \nproblem 14? Ans. 73.26 B.t.u. \n\n20. How much external work (above 32\xc2\xb0 F.) is done in making steam as \nin problem 14? Ans. 73.56 B.t.u. \n\n21. What is the internal energy of evaporation of the steam in prob- \nlem 14? Ans. 725.9 B.t.u. \n\n22. What is the total internal energy of the steam in problem 14? \n\nAns. 1023.94 B.t.u. \n\n23. A tank contains 9 cu. ft. of steam at 100 lbs. per sq. in. absolute \npressure which has a quality of 95 per cent. How many pounds of steam \ndoes the tank contain ? \' Ans. 2.14 lbs. \n\n24. Two pounds of steam have a volume of 8 cu. ft. at a pressure of 100 \nlbs. per sq. in. absolute. What is the quality? Ans. 90.5 per cent dry. \n\nCalculate its total heat above 32\xc2\xb0 F. Ans. 2003.8 B.t.u. \n\n25. One pound of steam having a quahty of 0.95 has a temperature of \n325\xc2\xb0 F. What is the pressure? Ans. 96.15 lbs. per sq. in. absolute. \n\n\n\nPROPERTIES OF STEAM 87 \n\n26. One pound of steam at a pressure 225 lbs. per sq. in. absolute has a \ntemperature of 441.9\xc2\xb0 F. Is it superheated or saturated? How many \ndegrees of superheat has it? Ans. 50\xc2\xb0 F. sup. \n\nWhat is the total heat required to generate such steam from water at \n32\xc2\xb0 F.? Ans. 1232.7 B.t.u. \n\nWhat is its volume? Ans. 2.23 cu. ft. \n\nHow much external work was done (above 32\xc2\xb0 F.) in generating it? \n\nAns. 92.2 B.t.u. \nHow much internal energ>^ above 32\xc2\xb0 F. does it contain? \n\nAns. 1 140.5 B.t.u. \n\n27. One pound of steam at a pressure of 300 lbs. per sq. in. absolute has \na volume of 1.80 cu. ft. \n\nIs it saturated or superheated? \n\nWhat is its temperature? Ans. 507.5\xc2\xb0 F. \n\nHow much superheat has it? Ans. 90\xc2\xb0 F. \n\nHow much is its total heat above 32\xc2\xb0 F.? -Ans. 1262.8 B.t.u. \n\nWhat is its total internal energy ? Ans. 1163.8 B.t.u. \n\n28. Steam in a steam pipe has pressure of 110.3 lbs. per sq. in. by the \ngage. A thermometer in the steam registers 385\xc2\xb0 F. Atmospheric pres- \nsure is 14.7 lbs. per sq. in. absolute. Is the steam superheated, and if \nsuperheated how many degrees ? \n\n29. Steam at a pressure of 200 lbs. per sq. in. absolute passes through \na throttling calorimeter. After expansion into the calorimeter the tem- \nperature of this steam is 250\xc2\xb0 F. and the pressure 15 lbs. per sq. in. abso- \nlute. What is the quality ? Ans. 0.965. \n\n30. Steam at a temperature of 325\xc2\xb0 F. passes through a throttling \ncalorimeter. In the calorimeter the steam has a pressure of 16 lbs. per \nsq. in. absolute and a temperature of 236.3\xc2\xb0 F. What is the quality? \n\nAns. 0.973. \n\n31. Steam at 150 lbs. per sq. in. absolute pressure passes through a \nthrottling calorimeter. Assuming that the lowest conditions in the calorim- \neter for measuring the quality is 10\xc2\xb0 F. superheat and the pressure in the \ncalorimeter is 15 lbs. per sq. in. absolute, what is the largest percentage of \nwetness the calorimeter is capable of measuring under the above conditions ? \n\nAns. 4.3 per cent wet or a quality of 0.957. \n\n32. In a ten-minute test of a separating calorimeter the quantity of dry \nsteam passing through the orifice is 9 lbs. The quantity of water sepa- \nrated was I lb. What was the quality? Ans. 0.90. \n\n33. A barrel contains 400 lbs. of water at a temperature of 50\xc2\xb0 F. Into \nthis water steam at a pressure of 125 lbs. per sq. in. absolute is admitted \nuntil the temperature of the .water and condensed steam in the barrel \nreaches a temperature of 100\xc2\xb0 F. The weight of the water in the barrel \nwas then 418.5 lbs. What was the quality? Ans. 0.958. \n\n\n\nCHAPTER VI \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS TO \nTHERMAL MACHINERY \n\nRefrigerating Machines or Heat Pumps. By a refrigerating \nmachine or heat pump is meant a machine which will carry heat \nfrom a cold to a hotter body.* This, as the second law of \nthermodynamics asserts, cannot be done by a self-acting proc- \ness, but it can be done by the expenditure of mechanical work. \nAny heat engine will serve as a heat pump if it be forced to \ntrace its indicator diagram backward, so that the area of the \n\n\n\n\nVolume \nFig. 17. \xe2\x80\x94 Pressure- volume Diagram of Camot Cycle. \n\ndiagram represents work spent on, instead of done by, the \nworking substance. Heat is then taken in from the cold body \nand heat is rejected to the hot body. \n\nTake, for instance, the Carnot cycle, using air as working sub- \nstance (Fig. 17), and let the cycle be performed in the order \ndcba, so that the area of the diagram is negative, and repre- \nsents work spent upon the machine. In the stage dc, which is \n\n* This statement is not at variance with our knowledge that heat does not \nflow of itself from a cold body to a hotter body. \n\n88 \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS 89 \n\nisothermal expansion in contact with the cold body R (as in Fig. \n9, page 45), the air takes in a quantity of heat from R equal to \nMRT2 loge r (equation (29)), and in stage ba it gives out to the \nhot body H a quantity of heat equal to MRTi loge r. There is \nno transfer of heat in stages cb and ad. Thus R, the cold body, \nis constantly being drawn upon for heat and can therefore be \nmaintained at a temperature lower than its surroundings. In an \nactual refrigerating machine operating with air, of the kind that \nmight be used for making ice, the cold body R consists of a coil \nof pipe through which brine circulates while " working " air is \nbrought into contact with the outside of the pipe. The brine is \nkept, by the action of the machine, at a temperature below 32\xc2\xb0 F. \nand is used in its turn to extract heat by conduction from the \nwater which is to be frozen to make ice. The \'\'cooler" H, which \nis the relatively hot body, is kept at as low a temperature as \npossible by means of circulating water, which absorbs the heat re- \njected to H by the " working " air. \n\nThe use of a regenerator, as in Stirling\'s engine (page 55), \nmay be resorted to in place of the two adiabatic stages in the \nCarnot cycle just explained with .the advantage of making the \nmachine much less bulky. Refrigerating machines of this kind \nusing air as working substance, with a regenerator, were intro- \nduced by Dr. A. C. Kirk and have been widely used.* The \nworking air is completely enclosed, which allows it to be in a \ncompressed state throughout, so that even its lowest pressure \nis much above that of the atmosphere. This makes a greater \nmass of air pass through the cycle in each revolution of the \nmachine, and hence increases the performance of a machine of \ngiven size. \n\nIn another class of refrigerating machines the working sub- \nstance, instead of being air, consists of a Hquid and its vapor, \nand the action proceeds by alternate evaporation under a low \npressure and condensation under a relatively high pressure. A \n\n* See Kirk, On the Mechanical Production of Cold, Proc. Inst, of C. E., vol. \nXXXVII, 1874. Also lectures on Heat and its Mechanical Applications, in the \nsame proceedings for 1884. \n\n\n\ngo ENGINEERING THERMODYNAMICS \n\nliquid must be chosen which evaporates at the lower extreme of \ntemperature under a pressure which is not so low as to make \nthe bulk of the engine excessive. Ammonia, ether, sulphurous \nacid, and other volatile liquids have been used. Ether machines \nare inconveniently bulky and cannot be used to produce intense \ncold, for the pressure of that vapor is only about 1.3 pounds per \nsquare inch at 4\xc2\xb0 F., and to make it evaporate at any tempera- \nture nearly as low as this would require the cylinder to be exces- \nsively large in proportion to the performance. This would not \nonly make the machine clumsy and costly, but would involve \nmuch waste of power in mechanical friction. The tendency of \nthe air outside to leak into the machine is another practical ob- \njection to the use of so low a pressure. With ammonia a dis- \ntinctly lower limit of temperature is practicable: the pressures \nare rather high and the apparatus is compact. \n\nThe standard systems of mechanical refrigeration are : * \n\n(A) The dense-air system, so-called because the air which is \nthe medium is never allowed to fall to atmospheric pressure, so \nas to reduce the size of the cylinders and pipes through which a \ngiven weight is circulating. \n\n(B) The compression system, using ammonia, carbon dioxide \nor sulphur dioxide, and so-called to distinguish it from the third \nsystem, because a compressor is used to raise the pressure of the \nvapor and deliver it to the condenser after removing it from the \nevaporator. \n\n(C) The absorption system, using ammonia, and so-called be- \ncause a weak water solution removes vapor from the evaporator \nby absorption, the richer aqua ammonia so formed being pumped \ninto a high-pressure chamber called a generator in communica- \ntion with the condenser, where the ammonia is discharged from \nthe liquid solution to the condenser by heating the generator, \nto which the solution is delivered by the pump. \n\nNo matter what system is used, circulating water is employed \nto receive the heat, the temperature of which Kmits the highest \n\n* Lucke\'s Engineering Thermodynamics, page 1148. \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS g\'l \n\ntemperature allowable in the system and indirectly the highest \n\npressure. \n\nThe dense or closed air system is illustrated in Fig. i8, in \n\n\n\nA \n\n\n\n\n\n\n\n\n\n1 \n\n\nCirculatin \n\n\n^ T \n\n\n\n\n\n\n\n\n\n\n1 Water 1 \n1 1 \n\n\n\n\nA \n\n\n\n\n\n\nWater Cooler \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x96\xa0^ \n\n\n\n\nCool Com- \n\n\n\n\n\n\n\n\nHot Com- \n\n\npressed Air \n\n\n\n\npressed Air \n\n\n\\f j\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^ \n\n\n\n\n\n\n\n\n\n\nExpansion _ \nCylinder \n\n\n\n\nCompressor \nCylinder \n\n\n\n\nE \nEngine \nCylinder \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^ \n\n\nV \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n\n\n\n\nCold Low \n\n\n\n\nWarmed Low \n\n\nPressure Air \n\n\n\n\n\n\n\n\nPressure Air \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\'-\' \n\n\n\xe2\x96\xa0 ^ \n\n\n\\ \n\n\n\n\n^ \n\n\n\n\n\n\n\n\n\n\n1 ^ B . \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 " \n\n\n\n\n\n\nBrine Tank \'\' \n\n\n\n\n\n\n1 "^ \n\n\n\n\n\n\n1 \n\n\n-^ \n\n\n\n\n\njCirculating^ Brmej \nWarm i Pipes \\!/Cold \n\nBrine Brine \n\nI : \n\nFig. 1 8. \xe2\x80\x94 Dense Air System of Refrigeration. \n\nwhich air previously dried of moisture is continuously circulated. \nThe engine cylinder, E, furnishes power * to drive the compressor \n\n* Since the work done by the expansion of the cool-compressed air is less than \nthat necessary for the compressing of the air taken from the brine coils through \nthe same pressure conditions, a means must be employed to make up for the dif- \nference, and for this purpose the engine cylinder is used. \n\n\n\nc \n\n\n^ M \n\n\nB \n\n\n\n\n\n\n\n\n\xe2\x96\xa01 \n\n\n\n\nk \n\n\n\n\nD \n\n\n\n\n\n\nw \n\n\n\\A \n\n\n\n\n\n\n\n\nN \n\n\n\n\n\n92 ENGINEERING THERMODYNAMICS \n\ncylinder, F. This cylinder delivers hot-compressed air into a \ncooler, A, where it is cooled, and then passed on to the expan- \nsion cyHnder, G (tandem-connected to both, the compressor, \nF, and to the engine cyHnder, E), which in turn sends cold low- \npressure air first through the refrigerating coils in the brine \ntank, B, and then back to the compressor cyHnder, F; thus \nthe air cycle is completed. The courses of the circulating water \nand also of the brine are shown by the \ndotted Hues. \n\nThe dense-air cycle in a pressure- \nvolume diagram is represented in Fig. 19, \nin which BC is the deHvered volume of \nhot-compressed air; CM is the volume of \n\nVolume , T . 1.1-1 \n\nFig. 19.\xe2\x80\x94 Pressure- volume cooled air admitted in the expansion \nDiagram of Dense Air cyHnder; MB the reduction in volume \nyceo engera ion. ^^^ ^^ ^^ water cooler; MN, the expan- \nsion; NA the refrigeration or heating of the air by the brine, \nand AB the compression. This operation is but a reproduction \nof that previously described. \n\nThe compression system for ammonia or similar condensable \nvapors is shown in Fig. 20. The figure only illustrates the \nessential members of a complete compression refrigerating sys- \ntem. B represents the direct-expansion coil in which the work- \ning medium is evaporated; F, the compressor or pump for \nincreasing the pressure of the gasified ammonia; E, the engine \ncyHnder, \xe2\x80\x94 the source of power; W, the condenser for cooling \nand Hquefying the gasified ammonia; and V a throttHng valve \nby which the flow of liquefied ammonia under the condenser \npressure is controlled as it flows from the receiver R to the ex- \npansion coils; B (the brine tank), in which a materiaUy lower \npressure is maintained by the pump or compressor in order \nthat the working medium may boil at a sufficiently low tem- \nperature to take heat from and consequently refrigerate the \nbrine which is already cooled. \n\nThese descriptions of the refrigeration systems will serve as a \nfoundation for a general understanding of refrigerating. \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS \n\n\n\n93 \n\n\n\nCOEFFICIENT OF PERFORMANCE OF REFRIGERATING MACHINES \n\nHeat extracted from the cold body \nWork expended \n\nThis ratio may be taken as a coefficient of performance in esti- \nmating the merits of a refrigerating machine from the thermo- \n\nI i \n\nI Circulating I \n\nY Water A \n\n! I \n\n\n\nHigh Pres-^ \' \nsure Liquid \n\n\n\nW \nCondenser \n\n\n\nR \n\nAmmonia \nReceiver \n\n\n\nHigh Pres- \n> ^sure Vapor \n\n\n\nCompreBBor \nCylinder \n\n\n\nE \nEngine \nCylinder \n\n\n\n^El \n\n\n\nThrottling or \nExpansion Valve \n\nLow Pressure Liquid \n\n\n\na \n\n\n\nBrine ^ Tank \n\n\n\nLow Pres- \nsure Vapor \n\n\n\nl\xc2\xab \xe2\x80\x94 Circulating Brine \xe2\x80\x94 s-l \n\nI Pipes i \n\nWarm | I Cold \n\nBrine I | Brine \n\nFig. 20. \xe2\x80\x94 Compression System of Refrigeration. \n\ndynamic point of view. When the limits of temperature Ti \nand T2 are assigned it is very easy to show by a sHght variation \nof the argument used in Chapter IV that no refrigerating machine \n\n\n\n-94 ENGINEERING THERMODYNAMICS \n\ncan have a higher coefficient of performance than one which is \nreversible according to the Carnot method. For let a refriger- \nating machine S be driven by another R which is reversible \nand is used as a heat-engine in driving S. Then if S had a higher \ncoefficient of performance than R it would take from the cold \nbody more heat than R (working reversed) rejects to the cold \nbody, and hence the double machine, although purely self-actingj \nwould go on extracting heat from the cold body in violation of \nthe Second Law (page 3). Reversibility, then, is the test of \nperfection in a refrigerating machine just as it is in a heat-engine. \nWhen a reversible refrigerating machine takes in all its heat, \nnamely Qc at T2 and rejects all, namely Qa at Ti and if we repre- \nsent the heat equivalent of the work done by PT = Q^ ~ Qc, then \nthe coefficient of performance is as already defined, \n\nQc _ Qc T^ \n\nW Qa-Qc T,-T2 \n\nHence \xe2\x80\x94 and the inference is highly important in practice \xe2\x80\x94 \nthe smaller the range of temperature, the better is the perform- \nance. To cool a large mass of any substance through a few \ndegrees will require much less expenditure of energy than to \ncool one-fifth of the mass through five times as many degrees, \nalthough the amount of heat extracted is the same in both cases. \nIf we wish to cool a large quantity, say of water 1 or of air, it is \nbetter to do it by the direct action of a refrigerating engine \nworking through the desired range of temperature, than to cool \na portion through a wider range and then let this mix with the \nrest. This is only another instance of a wide, general principle, \nof which we have had examples before, that any mixture or \ncontact of substances at different temperatures is thermody- \nnamically wasteful because the interchange of heat between \nthem is irreversible. Ah ice-making machine, for example, \nshould have its lower limit of temperature only so much lower \nthan 32\xc2\xb0 F. as will allow heat to be conducted to the working \nfluid with sufficient rapidity from the water that is to be \nfrozen. \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS 95 \n\nCOMPRESSED AIR \n\nAir when compressed may be used as the working medium in \nan engine, in exactly the same way as steam. Furthermore, it \nis an agent for the transmission of power and can be distributed \nvery easily from a central station for the purpose of driving \nengines, operating quarry drills and various other pneumatic \ntools. The type of machine used for the compression of air is \nthat known as a piston-compressor, and consists of a cylinder \nprovided with valves and within which there is a reciprocating \npiston. \n\nInasmuch as the work performed in the air-cylinder of a com- \npressor depends on so many variable conditions, it can only be \nstudied successfully from an indicator diagram. Imagine in such \na compressor, that the compression is performed very slowly \nin a conducting cyhnder, so that the air within may lose heat \nby conduction to the atmosphere as fast as heat is generated by \ncompression; the process will in that case be isothermal, at the \ntemperature of the atmosphere. Imagine further that the com- \npressed air is distributed to be used in compressed air motors * \nor engines without a change of temperature, and that the \nprocess of expansion in the compressed air motors or engines is \nalso indefinitely slow and consequently isothermal. In that \ncase (if we neglect the losses caused by friction in pipes) there \ncould be no waste of power in the whole process of transmission. \nThe indicator diagram would then be the same per pound of air \nin the compressor as in the air motor or engine, although the \ncourse of the cycle would be the reverse \xe2\x80\x94 that is, it would re- \ntrace itself. \n\nImagine, on the other hand, that compression and expansion \nare both adiabatic \xe2\x80\x94 a state of things which would be approxi- \nmately true, if expansion and compression were performed very \nquickly. The diagram of the compression, Fig. 21, is FCBE \nand that of the air engine as in Fig. 22, is EADF, and, therefore, \n\n* Compressed air motors are really engines just like steam engines but use \ncompressed air instead of steam. \n\n\n\n96 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nCB and AD are both adiabatic lines. The change of volume \nof the compressed air from that of EB to EA occurs through its \ncooling in the distributing pipes, from the temperature pro- \nduced by adiabatic compression down to the temperature of \nthe atmosphere. Suppose both diagrams of compressor and of \n\n\n\n\nVolume \nFig. 21. \xe2\x80\x94 Diagram of Compressor. \n\n\n\nVolume \nFig. 2 2. \xe2\x80\x94 Diagram of Air Engine. \n\n\n\nair engine be superimposed as in Fig. 23, and then sketch an \nimaginary isothermal line between the points A and C, both of \nwhich are at atmospheric conditions as regards temperature. \n\nThis simple sketch shows that the use of adiabatic compres- \nsion causes a waste of power which is measured by the area \n\n\n\n\nVolume \nFig. 23. \xe2\x80\x94 Superimposed Diagrams of Figs. 21 and 22. \n\nABC, while the use of the adiabatic expansion in the air engine \ninvolves a further waste, shown by the area ACD. \n\nIn practice the compression cannot be made strictly isother- \nmal for want of time, \xe2\x80\x94 \xe2\x96\xa0 the operation of the piston would be \ntoo slow for practice. The difference between isothermal and \nadiabatic compression (and expansion) can be very clearly shown \ngraphically as in Figs. 24 and 25. In this illustration the termi- \nnal points are correctly placed for a certain ratio for both com- \npression and expansion. Note that in the compressing diagram \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS \n\n\n\n97 \n\n\n\n(Fig. 24), the area between the two curves ABC represents the \nwork lost in compressing due to heating, and the area between \nthe two curves, ACMNF (in Fig. 25), shows the work lost by \ncooling during the expansion. The isothermal curve AC will \nbe the same for both cases. Illustrations of this sort show the \n\n\n\n\n\nVolume \nFig. 24. \xe2\x80\x94 Compression Diagram. \n\n\n\nVolume \nFig. 25. \xe2\x80\x94 Expansion Diagram. \n\n\n\neffect of reheating before expansion, cooling before compression, \nheating during expansion, etc. \n\nThe temperature of the air is prevented as far as possible \nfrom rising during the compression by injecting water into the \ncompressing cylinder, and in this way both the isothermal and \nadiabatic curves will change. The curves which would have \nbeen PV = a. constant, if isothermal and PV^"^ = a constant, if \nadiabatic will be very much modified. In perfectly adiabatic \nconditions the exponent ^\'w" = 1.40 for air, but in practice the \ncompressor cylinders are water-jacketed, and thereby part of \nthe heat of compression is conducted away, so that ^\'w" becomes \nless than 1.40. This value of "w" varies with conditions; gen- \nerally the value is taken as 1.2. \n\nThe problem of economy, obviously, becomes one of abstract- \ning the heat generated in the air during the process of compres- \nsion. As previously mentioned, this is partially accomplished by \nwater-jacketing the cylinders, and also by water injection. \nNevertheless, owing to the short interval within which the com- \npression takes place, and the comparatively small volume of \nair actually in contact with the cylinder walls, very little \nreally occurs. The practical impossibility of proper cooling to \nprevent waste of energy\' leads to the alternative of discharging \nair from one cylinder after partial compression has been effected, \n\n\n\n98 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\ninto a so-called inter-cooler, intended for removing the heat \ngenerated during the first compression, and then compressing \nthe air to the final pressure in another cyHnder. This opera- \ntion is termed " two-stage " compression and when repeated \none or more times for high pressures, the term " multi-stage " \ncompression applies. \n\nReferring to Fig. 26 and assuming the compression in a two- \nstage compressor to be adiabatic for each cylinder, the compres- \n\n\n\n\nVolume \nFig. 26. \xe2\x80\x94 Indicator Diagram of Two-stage Air Compressor. \n\nsion curve is represented by the broken line ABDE ; the compres- \nsion proceeds adiabatically in the first or low-pressure cyHnder \nto B ; the air is then taken to a cooler and cooled under practi- \ncally constant pressure until its initial temperature is almost \nreached, and its volume reduced from HB to HD; it is then \nintroduced to the second or high-pressure cylinder and com- \npressed adiabatically along the line DE to the final pressure \ncondition that was desired. It is seen that the compression \ncurve approaches the isothermal line FA.* The isothermal con- \ndition is ob\\dously desired and, in consequence, air-machines \nare built to approach that condition as nearly as possible. \n\nNumerous devices may be applied by the engineer to make \nthe expansion curves in his air engines approximate more nearly \nto the isothermal Kne; that is, he may use a preheater or inject \nhot water; or use a compound engine, allowing the air time to \ntake up enough heat to restore it more or less nearly to atmos- \npheric temperature between one stage of expansion and the next. \nBy these means the efficiency of the transmitting system as a \nwhole (neglecting all losses due to friction in the distributing \n\n* The line FE represents further cooUng. \n\n\n\nPRACTICAL APPLICATIONS OF THERMODYNAMICS 99 \n\npipes, in the valves of the engines, etc.), may be considerably \nincreased. \n\nThere is, however, another point to be considered. If the \ntemperature be allowed to fall materially during expansion, the \nexpanding air tends to deposit dew or even snow. To prevent \nthis the practice is often followed of passing the compressed air \nthrough a stove or \'^preheater" in order to raise its temperature \njust before it is allowed to expand and so prevent the deposit of \nfrozen moisture. When preheaters are used the extra heat \nwhich they supply is, of course, itself partly converted into \nwork.* \n\nPROBLEMS \n\n1. If 200 cu. ft. of free air per minute (sea-level) is compressed isother- \nmally and then delivered into a receiver, the internal pressure of which is \n102.9 lbs. per sq. in. absolute, find the theoretical horse-power required. \n\nAns. 24,93 h.p. \n\n2. What will be the net work in foot-pounds per stroke by an air com- \npressor displacing 3 cu. ft. per stroke, compressing air from an atmospheric \npressure of 15 lbs. per sq. in. absolute, to a gage pressure of 75 lbs.? (Iso- \nthermal.) Ans. 11,595. \n\n3. What horse-power will be needed to compress adiabatically 1500 \ncu. ft. of free air per minute to a gage pressure of 58.8 lbs., when n equals \n1.4? Ans. 197 h.p. \n\n4. A compressed-air motor without clearance takes air at a condition \nof 200 lbs. per sq. in. (gage) and operates under a cut-off at one-fourth \nstroke. What is the work in foot-pounds that can be obtained per cubic \nfoot of compressed air, assmning free air pressure of 14.5 lbs. and n equal \nto 1. 41? Ans. 54,936 ft.-lbs. \n\n5. Find the theoretical horse-power developed by 3 cu. ft. of air per \nminute having a pressure of 200 lbs. per sq. in. absolute, being admitted \nand expanded in an air engine with one-fourth cut-off. The value of n is \n1.2. (Neglect clearance.) Ans. 5.01 h.p. \n\n6. Compute the net saving in energy that is effected by compressing \nisothermally instead of adiabatically 50 cu. ft. of free air to a pressure of \n\n* On the subject of transmission of power by compressed air, reference should \nbe made to papers by Professor Richards in Bulletin No. 63 of Engineering Experi- \nment Station of Univ. of 111.; Weymouth on " Problems in Gas Engineering," \nTrans. American Society of Mechanical Engineers, vol. 34 (1912), pages 185-234; \nBaker on "Expansion and Temperature Drop of Compressed Air," Trans. \nA. S. M. E., vol. 2)2)1 pages 918-919. \n\n\n\nlOO ENGINEERING THERMODYNAMICS \n\n200 lbs. per sq. in. gage. Barometer = 14 lbs. per sq. in., and a tempera- \nture of 70\xc2\xb0 F. What is the increase in intrinsic energy during each kind \nof compression? How much heat is lost to the jacket-water during each \nkind of compression ? \n\nAns. 144,000 ft.-lbs.; \xe2\x80\x94o isothermal; 299.500 ft. -lbs. adiabatic; 275,500 \nft.-lbs. isothermal; o adiabatic. \n\n7. Let a volume of 12 cu. ft. of free air be adiabaticaUy compressed in \none stage from atmospheric pressure (15 lbs.) to 85 lbs. gage; the initial tem- \nperature of the air being 70\xc2\xb0 F. \n\n(a) What is the volume and temperature of the air after the com- \n\npression? Ans. 3.12 cu. ft.; 920\'\' F. absolute. \n\n(b) Suppose this heated and compressed air be cooled to an initial \n\ntemperature of 60\xc2\xb0 F., what is its pressure for that condi- \ntion? Ans. 56.65 lbs. absolute. \n\n(c) Now if the air occupies such a voliune as foimd in (a) and at \n\nan absolute pressure as in (b), at a temperature of 60\xc2\xb0 F., \nand is then allowed to expand adiabaticaUy down to atmos- \npheric pressure (15 lbs.), what is the temperature of the \nexpanded air in Fahrenheit degrees ? Its volume as well ? \n\nAns, \xe2\x80\x94106.72\xc2\xb0 F.; 8 cu. ft. \n\n\n\nCHAPTER VII \nENTROPY \n\nPressure-volume diagrams are useful for determining the \nwork (in foot-pounds), done during a cycle, but they are of very \nlimited use in analyzing the heat changes involved. It has, \ntherefore, been found desirable to make use of a diagram which \nshows directly by an area the number of heat units (instead of \nfoot-pounds) involved during the processes constituting a cycle. \nIn order that an area shall represent heat units instead of work \n\n\n\n\n\n\n\n\n\n\n\nTg \n\n\n^^ \n\n\nC^^ \n\n\n\n\n\n\nft \n\n\nTx \n\n\n\n\n\n\n\n\n-2 \n\n\n\n\ne \n\n\n\xe2\x96\xa0SH \n\n\n/ \n\n\n\ndH \n\n\n\nFig. 27. \xe2\x80\x94 Diagram of Entropy. \n\n\n\nEntropy- (p \nFig. 28. \xe2\x80\x94 Analysis of Entropy Diagram. \n\n\n\nunits the coordinates must be such that their product will give \nheat units. If the ordinates are in absolute temperature, the \nabscissas must be heat units per degree of absolute temperature, \n\n77 77 \n\nthat is, \xe2\x80\x94 ;, for then T X ~ = H, the amount of heat added \n\nduring the process from A to B (Fig. 27). \n\nThis simple ratio, heat added divided by the absolute tempera- \nture during addition, can be employed when the temperature re- \nmains constant; but when the temperature changes, a different \nform of expression must be developed. Suppose that the heat \nH is divided up into a number of small increments dH (Fig. 28), \n\n\n\nI02 ENGINEERING THERMODYNAMICS \n\nand that each small increment of heat is divided by the average \n\nabsolute temperature at which the heat change occurs. We \n\ndH \nwill then have a series of expressions \xe2\x80\x94 which, when summed \n\nup, will give the total change in the abscissas. This quantity \n\n/ \xe2\x80\x94 when multiplied by the average absolute temperature \n\nbetween C and D will give the total amount of heat added \nduring the process. Mathematically expressed, the change in \nthe abscissas is \n\nd<f>=\xe2\x80\x94 ^^ \'^^J\'Y ^^ \n\nand the heat change involved is \n\ndH = Tdcj> or H=CTdcf>, (69) \n\nThe quantity 4> in the equations is known as the increase in en- \ntropy of the substance, and may be defined as a quantity which, \nwhen multiplied by the average absolute temperature occurring \nduring a process, will give the number of heat units (in B.t.u.) \nadded or abstracted as heat during the process. The " increase \nin entropy " is employed rather than entropy itself, because we \nare concerned only with the differences in entropy, and further- \nmore we could not calculate the absolute entropy, because the \nspecific heat of a substance is not accurately known at low \ntemperatures. \n\nThis definition of entropy means that in a diagram such as \nFig. 28, where the ordinates are absolute temperatures, and the \nabscissas are entropies as calculated above some standard tem- \nperature, the area under any line CD gives the number of heat \nunits added to the substance in passing from a temperature Ti \nand entropy 0i = Oe, to a temperature T2 and entropy 02 = Of \n(or the number of heat units abstracted in passing from T2 to Ti) . \n\nLet us apply the above conceptions of entropy to the analysis \nof a Carnot cycle. A pound of the working substance is expanded \nisothermally. On a T-^ (temperature-entropy) diagram (Fig. \n\n\n\n\nENTROPY 103 \n\n29), this process would be represented by line AB, where the \ntemperature remains constant at Ti, and where the entropy in- \ncreases from Oe to Of, because of the addition of heat that is \nrequired to keep the temperature constant. The amount of \nheat that is added is given by the area ABfe. \n\nThe next process is adiabatic expansion from Ti to T^. Heat \nis neither added to nor abstracted from the substance during \nthis expansion. Hence the entropy \nremains constant, as indicated by BC. l^ \n\nThe substance is now isothermally | \ncompressed along CD, the temperature, | \nof course, remaining constant at T2 and 1 \nthe entropy decreasing because of the *^ \nabstraction of heat equal to the area Entropj\'-^ \n\nunder CD, i.e., CDef. Fig. 29. \xe2\x80\x94 Entropy Diagram of \n\nThe last process of the cycle is ^^^\'^ ^^ ^\' \n\nadiabatic compression from D to A, no heat being added or \nabstracted, and the entropy, therefore, remaining constant.* \n\nTo determine the amount of net work done during this cycle \nwe can employ the familiar relation \n(Heat equivalent of) work done = heat added \xe2\x80\x94 heat rejected. \n\n(70) \n\nApplying equation (70) to Fig. 29 we have \n\nHeat added = ABfe ; heat rejected = CDef. Therefore, \n\nWork done (in B.t.u.) = ABfe - CDef \n\n= ABCD. (71) \n\nAppl)dng the expression for efficiency, namely, \n^ work done \nheat added \nwe obtain then, \n\n\xe2\x80\x9e^ . ABCD \n\nEinciency = -^i \n\n^ . ABfe \n\nNow the question may arise as to why the entropy remains \n\nconstant during adiabatic expansion, as from B to C, when it is \n\n* Adiabatic lines are sometimes called isentropic lines (lines of equal entropy). \n\n\n\nI04 ENGINEERING THERMODYNAMICS \n\nknown that at C there is less intrinsic energy (i.e., less heat) in \nthe substance than at B, the decrease being MCv {Ti \xe2\x80\x94 T2). \nThe answer is that while heat disappeared from the substance \nduring the process, it disappeared as work and not as heat. \nThe case of constant entropy during adiabatic expansion is thus \nfound to be in accord with the explanation of entropy (as given \non page 102), which states that the heat involved in a change of \nentropy must be added or abstracted as heat, that is, must be \nconducted or radiated to or from the substance. \n\nFrom the foregoing discussion two important conclusions may \nbe drawn in regard to the use of the T-<f> diagram: \n\n1. If any Jieat process be represented by a curve on a T-cl> \ndiagram, the heat involved during the process is equal to the \narea under the curve, that is, between the curve and the axis of \nabsolute temperature. \n\n2. If a cycle of heat processes be represented on a T-<f) \ndiagram by a closed figure, the net work done is equal to the \nenclosed area, that is, the enclosed area measures the amount \nof heat that was converted into work. \n\nAs has been stated previously, T-(f) diagrams are very use- \nful for analyzing heat processes, and are often referred to as \n^\'heat diagrams." They find particularly useful apphcation in \nsteam engineering, as indicated by the following: \n\n1. Graphical analysis of heat transfers in a steam engine \ncylinder (pages 129-133). \n\n2. Determination of quality of steam during adiabatic ex- \npansion (page 116). \n\n3. Quick calculation of efficiency according to Rankine cycle \n(page 129). \n\n4. Steam turbine calculations (page 123). \n\nThe applications of entropy to heat engineering will be readily \nappreciated as the subject is developed. \n\nTemperature-Entropy Diagrams for Steam. Practically all \nthe diagrams that have been shown in the preceding chapters \nindicated the relations between pressure and volume. Such \ndiagrams show very well the action of steam in a reciprocating \n\n\n\nENTROPY \n\n\n\n105 \n\n\n\no800- \n\n3 \n\n\xe2\x96\xa0\xc2\xa7600- \n\nt^oo- \n\nH c \n,.\xe2\x80\xa2 200 \n\n\n\n\nsteam engine, where steam is admitted at the beginning of a \nstroke, expanded and exhausted or rejected from the cylinder \non the return stroke. Steam turbines operate differently, as \nthere is a continual flow of steam into the high pressure end and \na continual flow of low pressure steam from the exhaust. The \nresulting continual and uninterrupted drop in pressure and tem- \nperature cannot be satisfactorily represented with the pressure- \nvolume diagrams that have preceded. \n\nAnother kind of diagram is, therefore, universally used by \nsteam engineers. In this diagram, as has been stated, any \nsurface represents accurately to given scales, a definite quan- \ntity of heat. Absolute temperatures \n(r) are the ordinates, and the entro- \npies (</)) are the abscissas. \n\nFigure 30 shows a simple heat dia- \ngram laid out with absolute tempera- \nture and entropy for the coordinates. \nSteam at a certain condition of tem- \nperature and entropy is represented \nhere by the point A. Then if some \nheat is added, increasing both tem- \nperature and entropy, the final condition is represented by \nthe point B, and the area ABCD represents the heat added in \npassing from the condition at A to the condition at B. Such a \ndiagram is called a temperature-entropy diagram, although the \nname " heat diagram " is just as appropriate, since every area \nrepresents a definite amount of heat. \n\nAnother temperature-entropy diagram is shown in Fig. 31 rep- \nresenting by the various shaded areas the heat added to water \nat 32\xc2\xb0 F. to completely vaporize it at the pressure Pi. The \nunshaded area under the irregular curve AA\'B represents the \nheat in a pound of water at the freezing point, 32\xc2\xb0 F. or 492 de- \ngrees in absolute Fahrenheit temperature. A\'B represents the \nincrease in entropy due to the latent heat of fusion. The area \nOBCD is the heat added to the water to bring it to the tempera- \nture of vaporization, or in other words this last area represents \n\n\n\n1.0 \n\nEntropy (0) \n\n\n\n2.0 \n\n\n\nFig. 30. \xe2\x80\x94 Simple Temperature- \nentropy Diagram of Steam. \n\n\n\nio6 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nthe heat of the liquid {h) given in the steam-tables for the pres- \nsure Pi. Further heating after vaporization begins is at the \nconstant temperature Ti corresponding to the pressure Pi, and \n\n\n\n\nDry Steam or \nSaturation Line \n\n\n\n0.5 1.0 1.5 2.0 \n\nEntropy (0) \n\nFig. 31. \xe2\x80\x94 Temperature-entropy Diagram of Steam. \n\n\n\nOrd. \n\n\nAbs. \n\n\n/ \n\n\n\n\n\n\nEahr.^ \n328- \n\n\nSfJ^o- o/Ti s ek \n\n\n\n\n\n\n"1 \n\n\n139 \n32\xc2\xb0< \n\n\n5007T2 \n\n\nP \\m \n\n\n\n\n2 \n\n\n\n\n\n\n\n\n\\ \n\n\n\n\nTo \n\n\n\n\n1 g \n\n\n\n\n\n\n2 \n\n\n\n\n\n\n1 13 \n\n\n\n\n\n\n3 \n\n\n\n\n\n\n! - \n\n\n\n\n\n\n\n\n\n\n\n\n1 -t^ \n\n\n\n\n\n\nc3 \n\n\n\n\n\n\n1 \'^ \n\n\n\n\n\n\nj^ \n\n\n\n\n\n\n\xe2\x96\xa0^ \n\n\n\n\n\n\nK \n\n\n\n\n\n\n\n\n\n\n\n\na \n\n\n\n\n\n\n-< \n\n\n\n\n\n\na> \n\n\n\n\n\n\n\n\n\n\n\n\nH \n\n\n\n\n\n\n\n\n\n\nV \n\n\n\n\n\ntt \n\n\n6 fc / ^ ^ \n\n\ni \n\n\nEntropy- \n\n\n)ge T2\xe2\x80\x94 H \n\n\n\n\n\\OEe\\-^, \n\n\nFig. ^ \n\n\n52.- \n\n\n-Dia \n\n\ngram for Calculation of \n\n\nEntro \n\n\npy of Steam. \n\n\n\nis represented by an increasing area under line CE. When \n\'^ steaming " is complete, the latent heat, or the heat of vaporiza- \ntion (L), is the area DCEF. If, after all the water is vaporized, \nmore heat is added, the steam becomes superheated, and the \n\n\n\nENTROPY 107 \n\nadditional heat required would be represented by an area to \nthe right of E. \n\nCalculation of Entropy for Steam. In order to lay off the \nincrease in entropy as abscissas in the heat diagram for steam, \nit is necessary to develop formulas for calculating </>. For con- \nvenience 32\xc2\xb0 F. has been adopted as the arbitrary starting point \nfor calculating increase of entropy, as well as for the other \nthermal properties of steam. Referring to Fig. 32 the entropy \nof water is seen to be o at 32\xc2\xb0 F. In order to raise the tem- \nperature from To to Ti, h heat units are required, and by the \nequation (68) , the entropy of one pound of the liquid 6 will be \n\n\'^1 dh f V \n\n(72) \n\n\n\nJ To \n\n\n\nT \n\nor, assuming the specific heat of water to be unity we can write \n\'T,dT \n\n\n\ne \n\n\n\nCTidT \nJto T \n\n\n\n= lOge Ti - loge To = loge -=7\' (iS) \n\nIn order to evaporate the water into steam at the boihng \npoint Ti, the latent heat, L, must be added at constant tempera- \nture Ti. The increase of entropy during the "steaming" proc- \ness is represented by \xe2\x80\x94 and is obviously equal to \n\nEntropy of Evaporation = -rp-- (74) \n\nTi \n\nThe total entropy, <f>, of dry saturated steam above 32\xc2\xb0 F. at \ntemperature Ti is, then, \n\n= \xc2\xab + ^ = log.|l + fi, (75) \n\nI loll \n\nand for steam at temperature T2, we have \n\n* = log.J + ^- \nIq 1i \n\nValues of these entropies can be found in nearly all steam-tables. \n\n\n\nio8 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nIsothermal Lines of Steam. When the expansion of steam \noccurs at constant pressure as, for example, in the conversion \nof water into steam in a boiler when the engines are working, we \nhave isothermal expansion. It must be obvious from the pre- \nceding explanation that steam (or any other vapor) can be ex- \npanded or compressed isothermally only when wet. Isothermal \nlines for wet steam, which consists of a mixture of water and its \nvapor, are, therefore, straight lines of uniform pressure. On a \npressure-volume diagram an isothermal line is consequently \nrepresented by a horizontal line parallel to the axis of abscissas. \nThe horizontal parts of the indicator diagram as illustrated in \n\n\n\n\nVolume \n\nFig. 33. \xe2\x80\x94 Indicator Diagram of Ideal Cycle Using Steam. \n\nFig* 33 are Knes of constant pressure and, therefore, isothermal \nKnes of steam. On a T-(f) diagram, the isothermal line is repre- \nsented by a line of constant temperature, i.e., by a line parallel \nto the X-axis. \n\nAdiabatic Lines for Steam. Adiabatic lines will have differ- \nent curvature as they represent expansion or compression of \ndifferent substances. It will be remembered that the values of \n7 are different for the various gases discussed in preceding \nchapters, and therefore the adiabatic line for each of these gases \nwould have a different curvature. In the same way the curva- \nture of adiabatic lines of steam will vary with the relative amounts \nof steam and water in wet steam. It is worth mentioning here \nthat steam which is initially dry, if allowed to expand adiabati- \ncally, will become wet, the percentage of moisture which it will \ncontain depending on the extent to which the expansion is car- \n\n\n\nENTROPY 109 \n\nried. Also, on any r-0 diagram, an adiabatic (isentropic) line \nis represented by a line parallel to the Y-axis, i.e., by a line of \nconstant entropy. If steam is initially wet and is expanded adia- \nbatically, it becomes wetter as a rule.* \n\nIn general, in any sort of an expansion in order to keep steam \nat the same relative dryness as it was initially, while it is doing \nwork some heat must be suppHed and taken in. And if the ex- \npansion is adiabatic so that no heat is taken in, a part of the \nsteam will be condensed and will form very small particles of \nwater suspended in the steam, or it will be condensed as a sort \nof dew upon the surface of the enclosing vessel. \n\nThe relation between pressure and temperature as indicated \nby the steam-tables continues throughout an expansion, pro- \nvided the steam is initially dry and saturated or wet. \n\nAdiabatic Cixrve for Steam. Whether steam is initially dry \nand saturated or wet, the adiabatic curve may be represented \nby the formula: PV"" = constant. The value of the index n \ndepends on the initial dryness of the steam. Zeuner has deter- \nmined the following relation \n\noc \n\nn = LOS"; H \n\n^^ 10 \n\nSolving this when x = unity (dry and saturated steam) the \nvalue of n is 1.135, and when x is 0.75, n has the value i.ii.t \n\nExample. One pound of steam having a quality of 0.95 at a \npressure of 100 pounds per square inch absolute expands adia- \nbatically to 15 pounds per square inch absolute. \n\nWhat is the quality at the final condition? \n\n* When the percentage of water in wet steam is very great and the steam is \nexpanded adiabatically there is in many cases a tendency at the beginning of the \nexpansion for the steam to become drier. This is very evident from an inspection \nof diagrams like Fig. 32. \n\nt Rankine gave the value oi n = -\\\xc2\xb0-, which obviously from the results given \nis much too low if the steam is at all near the dry and saturated condition. His \nvalue would be about right for the condition when a; = 0.75. In an actual steam \nengine, the expansion of steam has, however, never a close approximation to the \nadiabatic condition, because there is always some heat being transferred to and \nfrom the steam and the metal of the cyUnder and piston. \n\n\n\nno ENGINEERING THERMODYNAMICS \n\nSolution. The total entropy at the initial condition equals \ne-Vx- = 0.4743 + 0.95 X 1. 1277 = 1.5456. \n\nThe total entropy at the end of the expansion equals \ne-\\-x\xe2\x80\x94 = 0.3133 + i.44i6:XJ. \n\nSince entropy is constant in adiabatic expansion \n0.3133 + 1.4416X = 1.5456, \nfrom which x = 0.854. \n\nHow much work is done during the expansion? \nSince there is no heat added the work done equals the loss in \ninternal energy. \nThe internal energy at the end of the expansion equals \nh + xiL = 181.0 + 0.854 X 896.8 = 946.9 B.t.u. \nThe internal energy at the initial condition equals \n\nh + xIl = 298.3 + 0.95 X 806.6 = 1064.6 B.t.u. \nThe work equals \n\n1064.6 \xe2\x80\x94 946.9 = 1 17.7 B.t.u. or 91,576 foot-pounds. \n\nExample. One pound of steam has a pressure of 100 pounds \nper square inch absolute and a quality of 0.95. It expands \nalong amn = I curve to 20 pounds per square inch absolute. \nWhat is the quality at the end of the expansion? \nSolution. The volume of the steam at the initial condition is \nxXV. \n\nX X V = 0.95 X 4.429 = 4.207 cubic feet. \nObviously, PiFi" = P2F2" and since n = i, \n\n100 X 4-?o7 = 20 X V2, \n\nV2 = 21.035 cubic feet. \n\nThe volume of dry saturated steam at the end of the expan- \nsion or at 20 pounds per square inch is 20.08. Therefore the \nsteam is superheated at end of expansion. How is this known? \n\n\n\nENTROPY III \n\nFrom the superheated steam-tables the amount of superheat cor- \nresponding to a specific volume of* 2 1.035 is found to be 29\xc2\xb0 F. \nWhat is the work done during the expansion? \n\nWork = PiFi loge^/ = 100 X 144 X 4-207 loge^^^^ > \nVi 4.207 \n\n= 100 X 144 X 4.207 X 1.6094 \n\n= 97,499 foot-pounds or 125.3 B.t.u. \n\nHow much heat must be added? \n\nThe energy of the steam at the end of the expansion equals, \n\nsince the pressure is 20 pounds per square inch absolute and \n\n29 degrees superheat: \n\n1169.9 ^^ ^ ^^ = 1169.9 \xe2\x80\x94 77.8 = 1092. 1 B.t.u. \n\n778 \n\nThe internal energy of the steam at the initial condition is \n\nh + xIl = 298.3 + 0.95 X 806.6 = 1064.6. \nThe internal energy of the steam was increased during the \nexpansion by 1092. i \xe2\x80\x94 1064.6 = 27.5. Therefore 27.5 B.t.u. of \nheat must be suppKed to compensate this increase as well as \nthe heat necessary to do the work. The total heat supplied to \nthe steam during the expansion would thus be \n\n27.5 + work of the expansion = 27.5 + 125.3 \n\n= 152.8 B.t.u. \n\nExample. One pound of steam at a pressure of 150 pounds \nper square inch absolute and a volume of 1.506 cubic feet expands \nunder constant pressure until it becomes dry and saturated. \n\nWhat is the quahty at the initial condition? \n\nSolution. The volume of dry and saturated steam at the \ngiven pressure is 3.012 cubic feet per pound. \n\nThe quality then is -^^ \xe2\x80\x94 = o. t;o. \n\n3.012 \n\nWhat is the volume of the steam at the final condition? \n\nThe volume a pound then is 3.012 cubic feet since the steam \nis dry and saturated. *What is the work done during the expan- \nsion? \n\n\n\n112 ENGINEERING THERMODYNAMICS \n\nWork Pi (F2-F1) = 150 X 144 (3-012 - 1.506) = 32,530 \n\nfoot-pounds. \nHow much heat is required? \nHeat added = H2 \xe2\x80\x94 {h + XiLi) \n\n= 11934 -.(330-2 +0.5 X 863.2) \n= 11934 \xe2\x80\x94 761.8 = 431.6 B.t.u, \n\nPROBLEMS \n\n1. One pound of water is raised in temperature from 60\xc2\xb0 to 90\xc2\xb0 F. \nWhat is the increase in entropy? Ans. 0.056. \n\n2. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute. \n{a) What is the entropy of the Uquid? Ans. 0.4743. \n\n(b) What is the entropy of evaporation ? Ans. 1.1277. \n\n(c) What is the total entropy of the steam ? Ans. 1.6020. \n\n3. Steam at 150 lbs. per sq. in. absolute has a quahty of 0.90. \n\n(a) What is the entropy of the liquid? Ans. 0.5142. \n\n(b) What is the entropy of evaporation? Ans. 0.9490. \n\n(c) What is the total entropy of the steam ? Ans. 1.4632. \n\n4. Steam having a temperature of 300\xc2\xb0 F. has an entropy of evapora- \ntion of 1. 1900. What is its quality? Ans. 0.994. \n\n5. Steam having a pressure of 200 lbs. per sq. in. absolute has a total \nentropy of 1.5400. \n\n{a) What is the total entropy of dry and saturated steam under the \ngiven pressure ? Ans. 1.5456. \n\n(b) Is the steam wet or dry ? \n\n(c) What is its quality? Ans. 0.994. \n\n6. Steam having a pressure of 125 lbs. per sq. in. absolute is super- \nheated 100\xc2\xb0 F. \n\n(a) What is its total entropy ? Ans. 1.6484. \n\n{b) What is the total entropy of dry and saturated steam under the \n\ngiven pressure ? Ans. 1.5839. \n\n(c) What is the entropy of the superheat ? Ans. 0.0645. \n\n7. Steam having a pressure of 150 lbs. per sq. in. absolute has a total \nentropy of 1.6043. \n\n(a) What is the total entropy of dry and saturated steam under the \n\ngiven pressure ? \' Ans. 1.5692. \n\n(b) Is the above steam saturated or superheated? How can you tell? \n\n(c) How much superheat has the steam ? Ans. 50\xc2\xb0 F. \n\n8. A boiler generates steam of 0.90 quality at a temperature of 350\xc2\xb0 F. \nwith the feed water admitted at 90\xc2\xb0 F. What is the increase in entropy? \n\nAns. 1.3594- \n\n\n\nCHAPTER VIII \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\nEfficiency of an Engine Using Steam Without Expansion. In \n\nthe early history of the steam engine, nothing was known about \nthe \'\'expansive" power of steam. Up to the time of Watt in \nall steam engines the steam was admitted at full boiler pressure \nat the beginning of every stroke and the steam at that pressure \ncarried the piston forward to the end of the stroke without any \ndiminution of pressure. Under these circumstances the volume \nof steam "used at each stroke at boiler pressure is equal to the \nvolume swept through by the piston. \n\n\n\n\n\xe2\x80\x94 ^ Volume \nFig. 34. \xe2\x80\x94 Indicator Diagram of an Engine using Steam without Expansion. \n\nAn indicator diagram representing the use of steam in an \nengine without expansion is shown in Fig. 34. This diagram \nrepresents steam being taken into the engine cyHnder at i at \nthe boiler pressure. It forces the piston out to the point 2 \nwhen the exhaust opens and the pressure drops rapidly from \n2 to 3. On the back stroke from 3 to 4 steam is forced out of \nthe cylinder into the exhaust pipe. At 4 the pressure rises \n\n113 \n\n\n\n114 ENGINEERING THERMODYNAMICS \n\nrapidly to that at i due to the rapid admission of fresh steam \ninto the cyHnder. In this case the thermal efficiency (E) is \nrepresented by \n\nJ, _ work done _ (Pi - F^) (V2 - Vi) ^ .^. \n\nheat taken in 778 (xiLi -\\- h \xe2\x80\x94 Ih) \n\nwhere the denominator represents the amount of heat taken in, \nwith the feed water at temperature of exhaust, ^. In actual \npractice the efficiency of engines using steam without expansion \nis about 0.06 to 0.07, when the temperature of condensation is \nabout 100\xc2\xb0 F. When steam is used in an engine without ex- \npansion and also without the use of a condenser the value of \nthis efficiency is still lower. It will be observed that under the \nmost favorable conditions obtainable the efficiency of an engine \nwithout expansion cannot be made under normal conditions to \nexceed about 7 per cent. \n\nIn the actual Newcomen steam engines efficiency was very \nmuch lower than any of the values given because at every stroke \nof the piston a very much larger amount of steam had to be \ntaken in than that corresponding to the volume swept through \nby the piston on account of a considerable quantity of steam \ncondensing on the walls of the cylinder. \n\nQuality of Steam During Adiabatic Expansion. A very im- \nportant equation, having wide application in steam engineering \nwill now be developed. This equation is used in finding the \nquality of steam after adiabatic expansion, and can easily be \nderived after a further study of the T-cf) diagram. \n\nReferring to Fig. 32 (page 106), the line TiC represents the in- \ncrease of entropy due to the latent heat added during the steam- \ning process. If this steaming process had stopped at some \npoint such that the steam was wet, having a quality x, this con- \ndition of the steam could be denoted by the point s, where \n\nTiC \n\nThis relation is obvious, for a distance along TiC represents \nthe entropy of steam, which is proportional to the latent heat \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 1 15 \n\nadded, which in turn is proportional to the amount of dry steam \nformed from one pound of water. In like manner, ~\xe2\x80\x94 is the \nquality of steam that has been formed along T2d at temperature \n\nIt is thus apparent that any point on a 7-0 diagram will give \nfull information in regard to the steam. The proportional dis- \ntances on a line drawn through the given point between the \nwater and the dry steam lines and parallel to the X-axis give \nthe quality of the steam as shown above. The ordinate of the \npoint gives the temperature and corresponding pressure, while \nits position relative to other lines such as constant volume and \nconstant total heat lines which can be drawn on the same dia- \ngram, will give further important data. Such a point (as m) \nis said to be the " state point " of the steam. \n\nFor the present we are particularly interested in the quality of \nsteam during adiabatic expansion. Since in adiabatic expansion \nno heat transfer takes place, the entropy remains constant, \nand, therefore, on a 7-0 diagram, this condition is represented by \na straight vertical line as cgf (Fig. 32) or smk. Equation 75, \n(page 107), must be modified for wet steam as follows: \n\nL T L \n\n<i> = e + x- = loge ;^ + xi -r, (77) \n\n-L i ^1 \n\nwhere x = the quality or dryness fraction of the steam, and \n4> = the total entropy of dry saturated steam, as before. \n\nSince in adiabatic expansion the entropy remains constant, \nthe following equation can be written \n\n01 = 02 (total entropies) \nor \n\nei + xi^ = e2 + x2^\' (78) \n\nKnowing the initial conditions of steam, the quality of the \nsteam at any time during adiabatic expansion can be readily \ndetermined. Thus, suf)pose the initial pressure of dry satu- \nrated steam to be 100 pounds per square inch absolute, and the \n\n\n\nIi6 ENGINEERING THERMODYNAMICS \n\nfinal pressure after adiabatic expansion 17 pounds per square \ninch. \n\nFrom the steam-tables we find that the total entropy 0, for \ndry steam at 100 pounds pressure is 1.6020; that is^ \n\n4)1 = 1.6020. \n\nThe entropies at 17 pounds pressure are also obtained from the \ntables, and we have, substituting in equation (78) \n\n1.6020 = 0.3229 + 0C2 1.4215 \nwhence, \n\n0C2 = 0.899. \n\nFor a rapid and convenient means of checking the above \nresult, the " Total Heat-entropy " diagram (see Marks and Davis\' \nSteam Tables, Diagram I), can be used. From the intersection \nof the 100-pound pressure line and that of unit quality ("satura- \ntion fine"), is dropped a vertical line (fine of constant entropy \n= 1.602) to the 17-pound pressure line. This latter intersec- \ntion is found to lie on the 0.90 quahty line. \n\nGraphical Determination of Quality of Steam by Throttling \nCalorimeter and Total Heat-entropy Diagram. It will be re- \nmembered that the throttling calorimeter (pages 74-77) depends \nfor its action upon the fact that the total heat of steam which ex- \npands without doing work remains the same, the heat in excess \nof that required to keep the steam dry and saturated going to \nsuperheat the steam. Suppose that steam enters the calorimeter \nat a pressure of 150 pounds per square inch absolute, and is throt- \ntled down to 17 pounds per square inch, the actual temperature \nbeing 240\xc2\xb0 F. Since the saturation temperature for steam at \n17 pounds pressure is 219.4, the steam in the calorimeter is \nsuperheated 240\xc2\xb0 \xe2\x80\x94 219.4\xc2\xb0 or 20.6 degrees. In order to find the \nquahty of the live steam refer to the " M oilier Diagram " (Total \nHeat-entropy Diagram, Fi\'g. 35) and find the intersection of the \n20.6 degrees superheat line with the 17-pound pressure Hne. \nFrom this point follow a horizontal line (line of constant total \nheat) to the left until it intersects the 150-pound pressure line. \nThis point of intersection is found to He on the 0.96 quaKty line. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\n117 \n\n\n\nFormula 65 (page 77) gives the following result in close \nagreement with the diagram: \n\n1153.1 + 0.47 (240 - 21Q.4) - 330-2 \n863.2 \n= 0.965. \n\n\n\nXi = \n\n\n\n\nEntropy \nFig. 35. \xe2\x80\x94 Mollier Diagram for Determining QuaKty of Steam. \n\nConstruction of Adiabatic Curve by Total Heat-entropy Dia- \ngram. By means of this " Mohier " diagram the curve for \nadiabatic expansion can be very readily drawn on a pressure- \nvolume diagram when the initial quahty of steam at " cut-off " \nis known. Assume that one pound of wet steam at the initial \ncondition of pressure and quahty as determined above is ad- \nmitted to the engine cyHnder per stroke, and that there is previ- \nously in the clearance space 0.2 cubic feet of steam (see Fig. \n36), at exactly the same\' condition. The volume of a pound of \ndry saturated steam at this initial pressure of 150 pounds is, from \n\n\n\nii8 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nthe steam-tables, 3.012 cubic feet. At 0.965 quality it will be \nI (0.965 X 3.012) = 2.905 cubic feet. On the scale of abscissas \nthis amount added to the 0.20 cubic feet in clearance gives \n\n\n\n\xe2\x80\xa2 Clearance Steam \n^Admission \n\n\n\nI Cut-ofE \n\n\n\n\n0.20 3.15 \n\nVolume of Steam in Cylinder (Cu. ft.) \n\nFig. 36. \xe2\x80\x94 Illustrative Indicator Diagram of Engine Using Steam with Expansion. \n\n\n\n400- \n\n\n\n\xe2\x80\xa253100 \n\nr \n\n03 \n\na \n\nt \no \n\n\n\n) \n\n\n/ V- \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\\ \n\n\nSupeiheaied \nW Steajn \n\n\n\n\n\n\n\n\nG \n\n\n\\ \n\n\ng\' \n\n\n/ \n\n\nr \n\n\n\n\n\n\n\n\n\\ \n\ns \n\n\n\n\nM \n\n\nD \n\n\nF \n\n\n\n\nf\' \n\n\n\n0.130 \n\n\n\n.566 \n\n\n\n1.528 \n\n\n\nFig. 37. \n\n\n\nEntropy \xe2\x80\x94 (p \n\xe2\x80\xa2Temperature-entropy Diagram of Steam Engine. \n\n\n\n3.105 cubic feet, the volume to be plotted at cut-off. Other \npoints in the adiabatic expansion curve can be readily plotted \nafter determining the quality by the method given on page 116. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 119 \n\nThe use of the temperature-entropy diagram in exhibiting \nthe behavior of steam in an adiabatic expansion and the various \nheat exchanges in the passage of steam through a steam cycle \nwill now be discussed and illustrated with a practical example. \n\nFig. 37 illustrates the heat process going on when feed water \nis received in the boilers of a power plant at 100\xc2\xb0 F., is heated \nand converted into steam at a temperature of 400\xc2\xb0 F., and then \nloses heat in doing work. When the feed water first enters \nthe boiler its temperature must be raised from 100\xc2\xb0 to 400\xc2\xb0 F. \nbefore any " steaming " begins. The heat added to the Hquid \nis the area MNCD. This area represents the difference between \nthe heats of the liquid (374 \xe2\x80\x94 68) or about 306 B.t.u. The \nhorizontal or entropy scale shows that the difference in entropy \nbetween water at 100\xc2\xb0 and 400\xc2\xb0 F. is about 0.437.* \n\nEvery reader should understand how such a diagram is con- \nstructed and especially how the curves are obtained. In this \ncase the curve NC is constructed by plotting from the steam- \ntables the values of the entropy of the liquid for a number of \ndifferent temperatures between 100\xc2\xb0 and 400\xc2\xb0 F. \n\nIf, now, water at 400\xc2\xb0 F. is converted into steam at that tem- \nperature, the curve representing the change is necessarily a \nconstant temperature line and therefore a horizontal, CE. Pro- \nvided the evaporation has been complete, the heat added in the \n"steaming" process is the latent heat or heat of evaporation \nof steam {L) at 400\xc2\xb0 F., which is approximately 827 B.t.u. \n\nThe change in entropy during evaporation is, then, the heat \nunits added (827) divided by the absolute temperature at which \nthe change occurs (400 + 460 = 860\xc2\xb0 F. absolute) or \nr 827 \n\nr = 865 = \xc2\xb0-9^^- \n\nThe total entropy of steam completely evaporated at 400\xc2\xb0 F. \nis, therefore, 0.566 -f 0.962, or 1.528.! To represent this final \n\n* As actually determined from Marks and Davis\' Steam Tables (pages 9 and \n15), the difference in entropy is 0.5663 \xe2\x80\x94 0.1295 or 0.4368. Practically it is im- \npossible to construct the scales in the figure very accurately. \n\nt Entropy like the total heat {H), and the heat of the liquid {h) is measured \nabove the condition of freezing water (32\xc2\xb0 F.). \n\n\n\nI20 ENGINEERING THERMODYNAMICS \n\ncondition of the steam, the point E is plotted where entropy \nmeasured on the horizontal scale is 1.528 as shown in the figure.* \nThe area MNCEF represents, then, the total heat added to a \npound of feed water at 100\xc2\xb0 F. to produce steam at 400\xc2\xb0 F., \nand then the area OBCEF represents, similarly, the total heat \n{H in the steam- tables) above 32 degrees required to form one \npound of steam at 400\xc2\xb0 F. \n\nAdiabatic Expansion and Available Energy. A practical ex- \nample as to how the temperature-entropy diagram can be used \nto show how much work can be obtained by a theoretically per- \nfect engine from the adiabatic expansion of a pound of steam \nwill now be given. When steam expands adiabatically \xe2\x80\x94 with- \nout a gain or loss of heat by conduction \xe2\x80\x94 its temperature falls. \nRemembering that areas in the temperature-entropy diagram \nrepresent quantities of heat and that in this expansion there is \nno exchange of heat, it is obvious that the area under a curve of \nadiabatic expansion must be zero ; this condition can be satisfied \nonly by a vertical line which is a line of constant entropy. \n\nThe work done during an adiabatic process, while it cannot be \nobtained from a \'\'heat diagram," can very readily be determined \nfrom the area under the adiabatic curve of a pressure-volume \ndiagram, or better still by the use of steam- tables as follows : In \nan adiabatic expansion the amount of work done is the mechani- \ncal equivalent of the loss in internal energy as explained in Chap- \nter III. Therefore, it is only necessary to determine the internal \nenergy of the steam at the beginning and end of the adiabatic \nexpansion. \n\nIiH = h-\\- ocJiL, \nI2H = h + 0C2 I2L\' \n\n* The point E is shown located on another curve RS, which is determined \nby plotting a series of points calculated the same as E, but for different pressures. \nIf more heat has been added than was required for evaporation, the area DCEF \nwould have been larger and E would have fallen to the right of RS, indicating by- \nits position that the steam had been superheated. The curve RS is therefore a \n"boundary-line" between the saturated and superheated conditions. This curve \ncan also be plotted from the values obtained from a table of the entropies of dry \nsaturated steam. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 121 \n\nWork during adiabatic expansion = loss in internal energy \n\n= Qii + xiliL \xe2\x80\x94 h. \xe2\x80\x94 0C2I2L) 77^? in foot-pounds. \n\nFor the case in Fig. 37 the adiabatic expansion curve will He \nalong the line EF, and if the temperature falls to 100\xc2\xb0 F., the ex- \npansion will be from E to G, and during this change some of the \nsteam has been condensed. If now heat is removed from this \nmixture of steam and water until all the steam is reduced to the \nliquid state, but without further lowering of the temperature, \nthe horizontal line GN * will represent the change in its condi- \ntion. The quantity of heat absorbed in this last process, techni- \ncally known as condensing the steam, is represented by the \narea MNGF. The difference between this heat rejected by the \nsteam at 100 degrees and the total heat added above 100\xc2\xb0 F. is \nthe \'* available energy " of the steam,t and is represented by \nthe area NCEG. By means of diagrams like those in the pre- \nceding figures, it will now be shown how the available energy \nof dry saturated steam for any given conditions can be readily \ncalculated from the data given in the steam- tables. \n\nFig. 38a is a temperature-entropy diagram representing dry \nsaturated steam which is expanded adiabatically from an ini- \ntial temperature Ti, corresponding to a pressure Pi, to a lower \nfinal temperature T2 corresponding to a pressure P2. The other \ninitial and final conditions of total heat (H) and entropy (0) \nare represented by the same subscripts i and 2. The available \nenergy or the work that can be done by a perfect engine under \n\n* That the steam might have been dry and saturated, the expansion would \n\nhave had to follow the curve ES and G would have appeared at G\'. The heat of \n\nthe liquid, h, of a pound of steam at 100\xc2\xb0 F. is represented by OBNM, and the \n\nheat of evaporation (L) is MNG\'F\', so that the total heat (h-\\-L or H) is OBNG\'F\'. \n\nThe total heat of wet steam is expressed hy h-\\- xL, where x is the quality or \n\nrelative dryness. In the case of this adiabatic expansion, then, h is as before \n\nOBNM and xL is MNGF. It is obvious also that the hnes NG and NG\' have \n\nthe same relation to each other as the areas under them, so that \n\nlineNG area MNGF xL NG \n\nT- \xe2\x80\x94 ^if7=r/ = ---.\xe2\x96\xa0-\xe2\x96\xa0-_.-,. = \xe2\x80\x94r, or .-\xe2\x96\xa0-\xe2\x96\xa0^ . = x (see pa^e 118) \n\nImeNG\' area MNG\'F\' L\' NG\' v fs j \n\nshowing that the quality of the steam at any point, G, on a constant temperature \nline is given by the ratio of NG to NG\'. \n\nt It is equal to the net work of the Rankine cycle, and depends upon the ini- \ntial and final conditions of the steam. \n\n\n\n122 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nthese conditions is the area NCEG. It is now desired to obtain \na simple equation expressing this available energy \xc2\xa3\xe2\x80\x9e in terms \n\n\n\n\n\n\n\nC/ Ti Pi \\ \n\n\ni,E \n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n\n\n/ \n\n\n^^^^^^^^^ \n\n\no\\ \n\n\np\' \n\n\nB \n\n\n\n\n\n\nN \n\n\n\n\n\n\n\nF \n\n\n\n\nF\'\' \n\n\n\nEntropy <p^ <}>\xe2\x80\x9e \n\nFig. 38a. \xe2\x80\x94 Temperature-entropy Diagram for Dry Saturated Steam \nExpanded Adiabatically. \n\nof total heat, absolute temperature and entropy. Explanations \nof the preceding figures should make it clear that \n\nHi = area OBNCEF, \n\nH2 = area OBNG\'F\', \n\nEa = area (OBNCEF + FGG\'FO - OBNG\'F\', \n\nEa = Hi-H2 + (ct>2- 0l) r2.* \xe2\x80\xa2 (79) \n\nAn appHcation of this equation will be made at once to deter- \nmine the heat energy available from the adiabatic expansion \nof a pound of dry saturated steam from an initial pressure of \n165 pounds per square inch absolute to a final pressure of 15 \npounds per square inch absolute. \n\nExample. Pi = 165 ri = 826 degrees, from steam- tables. \nP2= 15 T2= 673.0 \n\nHi = 1195.0 B.t.u. \nH2 = 1150.7 B.t.u. \n\' </)i = 1. 5615 \n\n<h = 1.7549 \n\n* It should be observed that this form is for the case where the steam is ini- \ntially dry and saturated. For the case of superheated steam a slightly different \nform is required which is given on page 125. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 123 \n\nSubstituting these values in equation (79), we have \n\nEa = II95-0 - 1150-7 +(i-7549 - i-S^is) 673 = i74-5 ^-t-u. \nper pound of steam. \n\nNow if in a suitable piece of apparatus like a steam turbine \nnozzle, all this energy that is theoretically available could be \nchanged into velocity, then we have by the well-known formula \nin mechanics, for unit mass,* \n\ny2 \n\n\xe2\x80\x94 = Ea (foot-pounds) = Ea (B.t.u.) X 778, \n\n2g __ \n\nV = V778 X2gEa= 223.8 VEa, (80) \n\nwhere V is the velocity of the jet and g is the acceleration due to \ngravity (32.2), both in feet per second. \n\nSolving then for the theoretical velocity obtainable from the \navailable energy we obtain the following: \n\n\n\nV = 223.8 V174.5 = 223.8 X 13.22 = 2956 feet per second. \n\nThe important condition assumed as the basis for determining \nequation (79), that the steam is initially dry and saturated, must \nnot be overlooked in its apph\'cation. There are, therefore, two \nother cases to be considered: \n\n(i) When the steam is initially wet, \n\n(2) When the steam is initially superheated. \n\nAvailable Energy of Wet Steam. The case of initially wet \nsteam is easily treated in the same way as dry and saturated \nsteam. Fig. 38b is an example of the case in hand. At the ini- \ntial pressure Pi the total heat of a pound of wet steam (hi -f XiLi) \nis represented in this diagram by the area OBNCE\'\'F\'\'. The \ninitial quality of the steam (xi) is represented by the ratio of the \n\nCE\'\' \n\nlines \xe2\x80\x94 \xe2\x80\x94 \xe2\x80\x94 . The available energy from adiabatic expansion from \nCE \n\nthe initial temperature Ti (corresponding to the pressure Pi) to \n\nthe final temperature T^ (corresponding to the pressure P2) is the \n\n* See Church\'s Mechanics of Engineering, page 672, or Jameson\'s Applied Me- \nchanics and Mechanical Engineering, vol. I, page 47. \n\n\n\n124 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\narea NCE\'^G\'\'. If we call this available energy Eaw, we have by \n\nmanipulation of the areas, \n\nEa^ = area OBNCEF + FGG\'F\' - OBNG\'F\' - G"E\'\'EG, \n\nE.,, = Hi-H2 + (ch-ct>i)T2-{cl>i-<}>,){T,-T2) * (8i) \n\nor, \n\nEa^ = H^ - H2 + {cf>2 - <h) T2 - ^{i - x,){Ti - T2). (82) \n\n\n\n\n\nC/ Ti P, E \n\n\n"\\e \n\n\nN \n\n/ \n\n\n/ T, P, g" \n\n\n!g \\g\' \n\n\nf" \n\n\nIf y \n\n\n\nEntropy (p^ (p^ \n\n\n\n4>\xe2\x80\x9e \n\n\n\nFig. 38b. \xe2\x80\x94 Temperature-entropy Diagram of Wet Steam Expanded Adiabatically. \n\nThe velocity corresponding to this energy is found by substi- \ntution in equation (80) , just as for the case when the steam was \ninitially dry and saturated. \n\nExample. Calculations for the velocity resulting from adia- \nbatic expansion for the same conditions given in the preceding \nexample, except that the steam is initially 5 per cent wet, are \ngiven below. \n\nTi = 826 degrees from tables \nT2 = 673.0 degrees " \nHi = 1195.0 B.t.u. " \n\nH2 = 1150.7 " \n\' 01 = 1.5615 \n\n02 = 1.7549 \n\nLi = 856.8 B.t.u. \nXx = i.oo \xe2\x80\x94 0.05 = 0.95 \n\n<tl = ^ + ^1, 0x = Xl ^ + ^1, <^l \xe2\x80\x94 0a: = :p (l \xe2\x80\x94 \xc2\xbbl)- \nil i\\ J-\\ \n\n\n\nPi = 165 lbs. absolute. \nP2 = 15 lbs. absolute. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 125 \n\nEav, = (1195-0 - 11507) + (1-7549 - 1-5615) 673 - -f^ \n\nX (i - 0.95) (826 - 673) = 44-3 + 130.2 - 7-93 \nEaw = 166.5 B.t.u. per pound of wet steam. \n\nV = 223.8 VeZ = 223.8 X V166.5 \n\n= 223.8 X 12.9 = 2886 feet per second. \n\nThis result can be checked very quickly by the \'\'total heat- \nentropy" or \'\'MoUier" diagram included in Marks and Davis\' \nand some other steam-tables. The intersection of the 0.95 \nquality line and 165 pounds pressure line is found to lie on the \n1 152 B.t.u. total heat line. Since the expansion is adiabatic, \nthe entropy remains constant. Therefore, following the verti- \ncal or constant entropy Hne (entropy = about 1.507) down to \nits intersection with the 15 pounds pressure Kne, we find that \nthe total heat at the end of adiabatic expansion is 985 B.t.u. and \n1152 \xe2\x80\x94 985 = 167 B.t.u. available energy, as above. \n\nThe velocity can be readily checked by the scale at the left of \nthe diagram. \n\nIf the steam were superheated to begin with, the available \nenergy during adiabatic expansion could be obtained in the \nsame way by means of the diagram. \n\nAvailable Energy of Superheated Steam. The amount of \nenergy that becomes available in the adiabatic expansion of \nsuperheated steam is very easily expressed with the help of Fig. \n39. Two conditions after expansion must be considered: \n\n(i) When the steam in the final condition is superheated, \n\n(2) When the steam in the final condition is wet (or dry \nsaturated) . \n\nUsing Fig. 39 with the notation as before except E^s is the avail- \nable energy from the adiabatic expansion of steam initially super- \nheated in B.t.u. per pound, (f>s and Hs are respectively the total \nentropy and the total heat of the superheated steam at the \ninitial condition, then obviously from the diagram, when the \nsteam is wet at the final* condition, \n\nE\xe2\x80\x9e, = H,-H2 + (<A2 - 0.) T2. (83) \n\n\n\n126 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nWhen the steam is superheated at the final condition, \n\nEaa = Hs \xe2\x80\x94 H2 \xe2\x80\x94 (03 \xe2\x80\x94 02 ) 2^2 . (84) \n\nIt will be observed that these equations (83) and (84) are the \nsame in form as (79), and that equation (83) differs only in hav- \ning the terms Ha and <f>s in the place of Hi and 0i. In other \n\n\n\nLHs \n\n\n\n\n\nTi Hi \n\n\n/ \n\n\n) \n\n\n\n\n/ T^ \n\n\n\\ \n\n\ni\' \n\n\n\n\n\n\n/ T2 \n\n\n\n\n\\ \n\n\n\\.H. \n\n\n\n\n*. \n\n\n*: \n\n\ni \n\n\n% \n\n\n\n\n\nEntropy- (^ \nFig. 39. \xe2\x80\x94 Temperature-entropy Diagram for Superheated Steam. \n\nwords equation (79) can be used for superheated steam if the \ntotal heat and entropy are read from the steam tables for the \nrequired degrees of initial superheat. \n\nThe following examples illustrate the simplicity of calcula- \ntions with these equations: \n\nExample i. Steam at 150 pounds per square inch absolute \npressure and 300\xc2\xb0 F. superheat is expanded adiabatically to i \npound per square inch absolute pressure. How much energy in \nE.t.u. per pound is made available for doing work ? \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 127 \n\nSolution. Hs = 1348.8 B.t.u. per pound, \nH2 = 1103.6 " " " \n<^ = 1.980, \n\n<f>s= 1.732, \n\nr2 = 559-6" F., \n\nEas = 1348.8 - IIO3.6 + (1.980 - 1.732) 559.6 \n\n= 383.9 B.t.u. per pound. \n\nThe result above may be checked with the total heat-entropy \nchart in Marks and Davis\' Steam Tables and Diagrams (Dia- \ngram I), and obtain thus (1349 \xe2\x80\x94 967) or 382 B.t.u. per pound. \n\nExample 2. Data same as in preceding example except that \nthe final pressure is now 35 pounds per square inch absolute. \n(Final condition of steam is superheated.) Calculate E^i. \n\nSolution. Hs = 1348.8 B.t.u. per pound, \nH2\' = 1166.8 " " \n\n<t>s= 1.732, \n\n<^\' = 1.6868, \nr/ = 718.9\xc2\xb0 F., \n\nEas = 1348.8 - 1166.8 - (1.7320 - 1.6868) 718.9 \n= 149.5 B.t.u. per pound. \n\nThe Rankine Cycle.* In Chapter IV it was shown that the \nCarnot cycle gave the maximum efficiency obtainable for a heat \nengine operating between given units of temperature. In order \nthat a steam engine may work on a Carnot cycle, the steam \nmust be evaporated in the cylinder instead of in a separate \nboiler, and condensed in the cylinder, instead of being rejected \nto the air or to a separate condenser. Such conditions are \nobviously impracticable, and it has, therefore, been found nec- \nessary to adopt some other cycle which conforms more with \npractical conditions. The most efficient practical steam cycle, \nand the one which has, therefore, been adopted as the standard \nwith which the efficiency of all steam engines may be compared, \nis the Rankine Cycle. The pressure-volume diagram of this cycle \n\n* Also known as the Clausius Cycle, having been published simultaneously \nand independently by Clausius. \n\n\n\n128 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nis shown in Fig. 40. Steam is admitted at constant pressure and \ntemperature along ab. At b cut-off occurs, and the steam ex- \npands adiabatically from b to c, some of it condensing during the \nprocess. The steam is then discharged at constant pressure \nand temperature along the back pressure line cd. Line da \nrepresents the rise in temperature and pressure at constant \nvolume when the inlet or admission valve opens. \n\n\n\n\nVolume \n\nFig. 40. \xe2\x80\x94 Indicator Diagram of Ideal Rankine Cycle. \n\n\n\nThe four stages of the Rankine cycle may also be stated as \nfollows: \n\n(i) Feed water raised from temperature of exhaust to tem^ \nperature of admission steam. (Line da.) \n\n(2) Evaporation at constant admission temperature. (Line \nab.) \n\n(3) Adiabatic expansion down to back pressure. (Line be.) \n\n(4) Rejection of steam at the constant temperature corre- \nsponding to the back pressure. (Line cd.) \n\nFollowing the usual method for calculating the net work done \nin a cycle, we have, assuming one pound of steam: \n\nWab = YT 8" (^i^i^i) = external work of evaporation, Ei \n\n(in B.t.u.), \nWbc = loss in internal energy \n\n= hi + xJli - (^2 + 0C2IL2) B.t.u., (85) \n\nWdc = - 7T8 (PcVc) = - yys (^2F2-\'^2) = external work of \n\nevaporation at temperature of exhaust (B.t.u.), \nW,, = o. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 129 \n\nAdding, \n\nNet work of cycle, W = jjg ^i^^i^i + ^h + xJu \xe2\x80\x94 h \xe2\x80\x94 ^^m \n\nbut \n\nyy-g- PiFi + Il\\ = Li (equation (62), page 66). \n\nTherefore, \n\nW = hi-\\- XiLi -h- 0C2L2 (B.t.u.). (86) \n\nThis means that the net work of the Rankine cycle is equal to \nthe difference between the total heat of the steam admitted and \nthe total heat of the steam exhausted. This statement applies \nwhether the steam is initially wet, dry or superheated, and, there- \nfore, it becomes a very simple matter to determine the work \ndone in a Rankine cycle by referring to the total heat-entropy \ndiagram. \n\nIt should be noted that this net work of the Rankine cycle is \nthe so-called ** available energy " of steam as defined and dis- \ncussed in the preceding section, and that equation (82) will give \nthe same result as equation (79). (The student should check \nthis statement.) \n\nFig. 41 is the r-(/) diagram for a Rankine cycle using dry \nsaturated steam to begin with. The letters ahcd refer to the \ncorresponding points in the pressure- volume diagram. The net \nwork of the cycle is B -)- C, which is the difference between the \ntotal heats at admission and exhaust. The heat added per cycle \nis A -f B -1- C + D and the \n\nThermal efficiency = ^ j^ ^ j^ q + D ^^^^ \n\n_ h + ociLi \xe2\x80\x94 Jh \xe2\x80\x94 OC2L2 \nhi -f- XiLi + h \n\nIt should be carefully observed that /^ in the denominator \nmust always be subtracted from hi + XiLi (the total heat above \n32\xc2\xb0 F.), in order to give. the total heat above the temperature of \n\n* The final quality can be determined by equation (78). \n\n\n\nI30 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nfeed water, which in engine tests is always assumed for the \npurpose of comparison to be the same as the exhaust temperature. \n\n\n\n\nEntropy\xe2\x80\x94 \nFig. 41 . \xe2\x80\x94 Temperature-entropy Diagram of Rankine Cycle. \n\nOne pound of steam at a pressure of 160 lbs. per sq. in. abso- \nlute and quality of 0.95 performs a Rankine cycle exhausting at 5 \nlbs. per sq. in. absolute. \n\nWhat is the quality of the exhaust ? \n\nSolution. The total entropy at the initial condition \n\n= .5208 + .95 X 1.0431- \n\nThe total enthropy at the exhaust \n\n= .2348 + 1.6084X. \n\nThen .5208 + .95 X 1.043 1 = -2348 + i.6o84:t:. \n\nFrom which rr = 0.794. \n\nWhat is the net work of the cycle ? \nWork = Hi-H2 = 335.6 + .95 X 858.8 - 130.1 - .794 \nX 1000.3 ~ 227.2 B.t.u. \n\nWhat is the efficiency of the cycle ? \n\n2 27.2 \n\n\n\nEfficiency = \n\n\n\n- = .222 or 22.2 per cent. \n\n\n\n335.6 +.95X858.8-130.1 \n\nThe Practical or Actual Steam Engine Cycle. In the steam \nengine designed for practical operation it is impossible to expand \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\n131 \n\n\n\nthe steam down to the back-pressure Hne; and, furthermore, it \nis evident that some mechanical clearance must be provided. \nThe result is that in the indicator diagram from an actual steam \nengine, we have to deal with a clearance volume, and both incom- \nplete expansion and incomplete compression as shown in Fig. \n42. In order to calculate the theoretical efficiency of this prac- \n\n\n\n\nVolurae \nFig. 42. \xe2\x80\x94 Indicator Diagram of Practical Engine Cycle. \n\ntical cycle, it is necessary to assume that the expansion Hne cd \nand the compression Hne fa are adiabatic. Knowing then the \ncyHnder feed of steam per stroke and the pressure and volume \nrelations as determined from the indicator diagram, one can \ncalculate the theoretical thermal efficiency by obtaining the net \narea of the diagram (expressed in B.t.u.) and dividing by the \nheat suppHed per cycle. In order to obtain the net area of the \ndiagram, the latter may be divided up into several simple parts \nas follows: \n\ngcdi \xe2\x80\x94 a Rankine cycle, \n\nidej \xe2\x80\x94 a rectangle, \n\nhafj \xe2\x80\x94 a Rankine cycle (negative), \ngbah \xe2\x80\x94 a rectangle (negative) . \n\nThe above areas can be evaluated in B.t.u. by methods pre- \nviously explained (see pages 113 and 129), and thus the net work \nof the cycle can be determined. \n\n\n\n132 ENGINEERING THERMODYNAMICS \n\nThe heat added per cycle is equal to \nTotal heat of cylinder feed (b to c) \xe2\x80\x94 cylinder feed X heat of \nHquid of feed water at the temperature of the exhaust. \nExercise. Calculate the thermal efficiency for the \'\'practi- \ncal " indicator diagram shown in Fig. 42, having given the fol- \nlowing data: \n\nCylinder feed = i cu. ft., \nInitial steam pressure = 165 lbs. per sq. in. absolute (dry \nsaturated), \nExhaust steam pressure = 2 lbs. per sq. in. absolute. \n\nPROBLEMS \n\n1. What is the entropy of the liquid of steam of 92 per cent quality at a \npressure of 15 lbs. per sq. in. absolute? Ans. 0.313. \n\n2. With a quality of 0.90, what is the entropy of evaporation of steam \nat a pressure of 25 lbs. per sq. in. absolute? Ans. 1.224. \n\n3. If there is 1.2 lbs. of water in 8 lbs. of wet steam, what is its quality? \n\nAns. 0.85. \n\n4. What is the total entropy of steam of 94 per cent quahty at a pres- \nsure of 100 lbs. per sq. in. absolute? Ans. 1.534. \n\n5. It was found that an engine operating under a pressure of 155 lbs. \nper sq. in. absolute was using steam at a temperature of 561\xc2\xb0 F. What was \nthe condition of the steam? Ans. Superheat 200\xc2\xb0 F. \n\n6. If steam at 200 lbs. per sq. in. absolute, 95 per cent quality, is caused \nto expand adiabatically to 228\xc2\xb0 F., what are the properties at the lower \npoint ? (That is, final total entropy, entropy of evaporation, quality and \nvolume.) Ans. 1.495; i-i6oo; 0.831; 16.7 cu. ft. \n\n7. One pound of steam at a pressure of 100 lbs. per sq. in. absolute \nand a quality of 50 per cent is expanded isothermally until it is dry and \nsaturated. Find the heat supplied and the work done. \n\nAns. 444.0 B.t.u. and 31,890 ft. -lbs. \n\n8. What will be the final total heat of dry saturated steam that is ex- \npanded adiabatically from 150 lbs. per sq. in. absolute down to 10 lbs. \nper sq. in. absolute? Ans. 999.7 B.t.u. \n\n9. Steam having a quality of 20 per cent is compressed along an adia- \nbatic curve from a pressure of \'20 lbs. per sq. in. absolute to a pressure whose \ntemperature is 293\xc2\xb0 F. What is the final quality? Ans. 15.4 per cent. \n\n10. Determine the final quality of the steam and find the quantity of \nwork performed by 2 lbs. of steam in expanding adiabatically from 250 lbs. \nper sq. in. absolute pressure to 100 lbs. per sq. in. absolute, the steam \nbeing initially dry and saturated. Ans. 93.4 per cent; 12948 B.t.u. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 133 \n\n11. Assume that i lb. of steam of a pressure of 160 lbs. per sq. in. abso- \nlute and a quality of 95 per cent performs a Rankine cycle, being exhausted \nat a pressure of 5 lbs. per sq. in. absolute. Compute the quality of the steam \nexhausted and the efficiency of the cycle. Find the final volume. \n\nAns. 79.4 per cent; 22.2 per cent; 58.2 cu. ft. \n\n12. What is the work of a Rankine cycle if the steam initially at \n200 lbs. per sq. in. absolute pressure, superheated 200\xc2\xb0 F., goes through such \na cycle with a back pressure of i lb. per sq. in. absolute? If, instead of ex- \npansion in such a cycle, the steam (for same conditions) were to expand in \na turbine nozzle, what would be the velocity of the steam leaving ? \n\nAns. 385 B.t.u.; 4390 ft. per sec. \n\n13. One pound of the steam at a pressure of 100 lbs. per sq. in. absolute \nwith a quality of 0.90 performs a Rankine cycle exhausting at a back pres- \nsure of 2 lbs. per sq. in. absolute. \n\nWhat is the net work of the cycle? Ans. 235 B.t.u. \n\nWhat is the efficiency of the cycle? Ans. 23.4 per cent. \n\n14. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute \nand a volume of 8.34 cu. ft. performs a Rankine cycle. The exhaust pres- \nsure is 25 lbs. per sq. in. absolute. \n\nWhat is the net work of the cycle? Ans. 260 B.t.u. \n\nWhat is the efficiency of the cycle? Ans. 12.5 per cent. \n\n\n\nApplication of Temperature-entropy Diagram to Analysis of \nSteam Engine. The working conditions of a steam engine, as \nstated before, can be shown not only by the indicator card, but \nalso by the empIo3anent of what is known as a \'\'temperature- \nentropy" diagram. These diagrams represent graphically the \namount of heat actually transformed into work, and in addition, \nthe distribution of losses in the steam engine. \n\nFor illustration, a card was taken from a Corliss steam engine \nhaving a cylinder volume of 1.325 cubic feet, with a clearance \nvolume of 7.74 per cent, or 0.103 cubic feet; the weight of steam \nin pounds per stroke (cyHnder feed plus clearance) was 0.14664 \npounds. Barometer registered atmospheric condition as 14.5 \npounds per square inch. The scale of the indicator spring used \nin getting the card was 80 pounds to the inch. Steam, chest \npressure was taken as 153 pounds per square inch (absolute), \nand a calorimeter determination showed the steam to be practi- \ncally dry and saturated. \n\n\n\n134 ENGINEERING THERMODYNAMICS \n\nThe preliminary work in transferring the indicator card to \na T-(j) diagram, consists first in preparing the indicator card. \nIt was divided into horizontal strips at pressure intervals of lo \npounds with the absolute zero line taken as a reference; this \nline was laid off 14.5 pounds below atmospheric conditions. \n(See Fig. 43a.) For reference, the saturation curve was drawn. \nHaving known the weight of steam consumed per stroke and \nspecific volume of steam (from the Steam Tables), for various \npressures taken from the card, the corresponding actual vol- \numes could be obtained; this operation is, merely, weight of \nsteam per stroke multiplied by specific volume for some pressure \n(0.14664 X column 5 in the table below), the resulting value be- \ning the volume in cubic feet for that condition. These pressures \nand volumes were plotted on the card and the points joined, re- \nsulting in the saturation curve, 2" -d" -%"-()" . \n\nThe next step consisted in constructing a " transformation \'* \ntable with the columns headed as shown. All the condensing and \nevaporation processes are assumed to take place in the cyhnder \nand the T-<^ diagram is then worked up for a total weight of \none pound of steam as is customary. Column i shows the re- \nspective point numbers that were noted on the card; column 2, \nthe absolute pressures for such points; column 3, the corre- \nsponding temperatures for such pressures; column 4, the vol- \nume in cubic feet up to the particular point measured from the \nreference Hne of volumes; column 5, the specific volume of a \npound of dry and saturated steam at the particular pressure \n(Steam Tables); column 6, the volume of actual steam per \npound, obtained by dividing the volumes in column 4 by 0.14664 \npound (total weight of steam in cyHnder per stroke); column 7, \nthe dryness factor \'^x," found by dividing column 6 by column 5; \n\ncolumn 8 is the entropy of evaporationf\xe2\x80\x94 jfor particular condi- \ntions (Steam Tables) ; column 9 is the product of column 7 and \ncolumn 8; column 10 is the entropy of the liquid at various \nconditions as found in the Steam Tables; column 11 is the sum \nof column 9 and column 10, giving the total entropy. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\n135 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nENTROPY \n\n\nDIAGRAM \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n. \n\n\n\n\n\n\n\n\n_A \n\n\n/ \n/ \n\n\nB \n\n\xe2\x80\x94r \n\n\n1\' \n\n\n__ \n\n\n._. \n\n\n\n\n._. \n\n\n5\' \n"h- \n\n\n.._. \n\n\n4^ \n\n\n! \n\n\n\n\n\n\n\n\n\n\nson \n\n\n: \n\n\n\n\n\n\n\n\n/ \n/ \n\n\n1 \n\n\nP22 \n\n\n23 \n\n\n\n\n\n\n2 \n\n\n\n\n31 \n\n6 \ns/ \n\n\nt \n\n\n\n\n1 \n1 \n1 \n\n\n\\ \n\n\n\n\n\n\n\n\n\n\nocn \n\n\n- \n\n\n\n\n\n\nrl \n\n\n/ \n/ \n\n\nf20 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n10 \nIT. \n\n\ni \n\n\n^ \n\n\n1 \n1 \n1 \n\n1 \n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n^ 200 \n\n\n- \n\n\n\n\n\n\n/ \n/ \n/ \n/ \n\n\nI \n\n\n\n\n\n\n^ \n\n\n^ \n\n\n^ \n\n\n< \n\n\n\n\n\n\n1 \n\n1 \n\n1 \n\n\n\n\n1 \n1 \n1 \n\n1 \n\n1 \n\n\n\n\nV \n\n\n\n\n\n\n\n\nft \xe2\x80\x94 \n150^ \n\n\n- \n\n\n\n\n1 \n\n\n/ \n\n\n\n\nQ^ \n\n\nG \n\n\n^^ \n\n\n\n\n\n\n\n\nL_. \n\n\n\xe2\x80\x94 \n\n\n\n\n\n1 \n.1 \xe2\x80\x94 \n\nf\' \n\n\n\xe2\x80\x94 \n\n\n1 \n1 \n\n\n\n\nV \n\n\n\\ \n\nV, \n\n\n\\ \n\\ \n\n\n\n\nt \n\n\nI8|l7 \n\n\n16 15 \n\n\n14 \n\n\n\n\n: \n100- \n\n\n/ \n\n\n1 \n\n1 \n1 \n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\nr \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\\' \n\n\nKn \n\n\n"r( \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nAt \n\n\nL? \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n- \n\n\n- \n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nM \n\n\nM \n\n\n\n\n5 \n\n\n\n\n\n\n\' \n\n\n1 \n1. \n\n\n1 \n\n\n\n1 \n\n\n\n\n\n\n1. \n\n\n5N \n\n\n\n\n\n\n\n\n2. \n\n\nJ \n\n\n\nEntropy \nFig. 43a. \xe2\x80\x94 Temperature-entropy Diagram of Actual Steam Engine Indicator \n\nDiagram. \n\n\n\n136 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nTRANSFORMATION TABLE \n\n\n\nI \n\n\n2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n9 \n\n\n10 \n\n\nII \n\n\na \n\n\n\n\nIS \n\n\nS.I \n\ni \n\n>8 \n\n\nIII \n\n\n2^ S ^ \n\n\nC.2 \nQ \n\n\n\n\n\n\n12 \n\n0. \n\nS \n+-> \n\n\nIs \n\n\nI \n\n\n145 \n\n\n356 \n\n\n0.1050 \n\n\n3-II \n\n\n0.716 \n\n\n0.2302 \n\n\nI .0612 \n\n\n0.2440 \n\n\n0.5107 \n\n\n0.7547 \n\n\n2 \n\n\n140 \n\n\n353 \n\n\n0.2250 \n\n\n3 \n\n\n22 \n\n\n1-536 \n\n\n0.4770 \n\n\n1.0675 \n\n\n. 5090 \n\n\n0.5072 \n\n\nI. 0162 \n\n\n5 \n\n\n120 \n\n\n341 \n\n\n0.4230 \n\n\n3 \n\n\n73 \n\n\n2.890 \n\n\n0.7740 \n\n\n1-0954 \n\n\n0.8475 \n\n\n0.4919 \n\n\n1-3394 \n\n\n9 \n\n\n48 \n\n\n279 \n\n\n. 8500 \n\n\n8 \n\n\n84 \n\n\n5.810 \n\n\n0.6580 \n\n\n1-2536 \n\n\n0.8250 \n\n\n0.4077 \n\n\n1.2327 \n\n\n12 \n\n\n20 \n\n\n228 \n\n\n1.4225 \n\n\n20 \n\n\n08 \n\n\n9.700 \n\n\n0.4830 \n\n\n1-3965 \n\n\n0.6740 \n\n\n0.3355 \n\n\n1.0095 \n\n\n15 \n\n\n8 \n\n\n183 \n\n\n0.9758 \n\n\n47 \n\n\n27 \n\n\n6.660 \n\n\n0.1410 \n\n\n1-5380 \n\n\n0.2168 \n\n\n0.2673 \n\n\n0.4841 \n\n\n18 \n\n\n8 \n\n\n183 \n\n\n0.3500 \n\n\n47 \n\n\n27 \n\n\n2.390 \n\n\n0.0506 \n\n\n1.5380 \n\n\n0.0778 \n\n\n0.2637 \n\n\n0.3451 \n\n\n20 \n\n\n30 \n\n\n250 \n\n\n0.1825 \n\n\n13 \n\n\n74 \n\n\n1-245 \n\n\n0.0907 \n\n\nI-33II \n\n\n0.1208 \n\n\n0.3680 \n\n\n0.4888 \n\n\n\nAbove table is employed for transferring the P-V diagram to the T-(f> \ndiagram. \n\nAfter this table was completed, columns 3 and 1 1 were plotted. \nConvenient scales were selected, the ordinates as temperatures \nand the abscissas as entropies. The various points, properly \ndesignated, were connected as shown on the T-(j) diagram, the \nclosed diagram resulting. This area shows the amount of heat \nactually transformed into work. This diagram is the actual \ntemperature-entropy diagram for the card taken and may be \nsuperimposed upon the Rankine cycle diagram in order to de- \ntermine the amount and distribution of heat losses. \n\nThe water line, A-A\\ and the dry steam line, C-C , were \ndrawn directly by the aid of Steam Table data, i.e., the entropy \nof the Kquid and the entropy of the steam taken at various \ntemperatures, and plotted accordingly. \n\nBefore the figure could be studied to any extent, the theoreti- \ncal (Rankine) diagram had to be plotted, assuming that the steam \nreaches cut-off under steam-chest conditions; that it then ex- \npands adiabatically down to back pressure and finally exhausts \nat constant pressure to the end of the stroke without compres- \nsion. This diagram is marked, A-C-E-H, on the T-(j> plane. \nThe steam-chest pressure of 153 pounds per square inch absolute \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 137 \n\nfixes the point, C, when the temperature line cuts the steam line, \nC-C. The rest of the cycle is self-evident. \n\nReferring to the T-^ diagram, Fig. 43a, H-A-C-E is the Ran- \nkine operation with no clearance for one pound of working fluid. \nThe amount of heat supplied is shown by the area, Mi-H-A-C-N, \nand of this quantity, the area Mi-H-E-N * would be lost in \nthe exhaust while the remainder, H-A-C-E, would go into \nwork. This is theoretical, but in practice there are losses, and \nfor that reason, the Rankine cycle is used merely for comparison \nwith the actual card as taken from a test. The enclosed irregu- \nlar area, 1-2-3 \xe2\x80\xa2 \xe2\x80\xa2 \xe2\x80\xa2 22-23, is the amount of heat going into \nactual work. By observation, it is evident that a big area re- \nmains; this must represent losses of some sort or other. That \nquantity of work represented by the area, 1-5-5\'-!\', is lost on \naccount of wire-drawing; the area $\'-C-D-F, shows a loss due \nto initial condensation; the loss due to early release is shown \nby the area F-12-14-F\' for the real card, and by D-G-E for the \nmodified Rankine cycle (such a loss, in other words, is due to \nincomplete expansion); that quantity represented by 2 2-^-1 \'-i \nis lost on account of incomplete compression, and H-A-B-iS is \nthe loss due to clearance. The expansion line from 5 on to 9 in- \ndicates that there is a loss of heat to the cylinder walls, causing \na loss of entropy; from 9 on to 10, re-evaporation is taking place \n(showing a gain of entropy). \n\nAll of the heat losses are not necessarily due to the transfer \nof heat to or from the steam, as there may be some loss of steam \ndue to leakage. In general, however, the T-cf) diagram is satis- \nfactory in showing heat losses. \n\nFig. 43a was constructed for the purpose of showing how the \nactual thermal efficiency and the theoretical thermal efficiency \n(based on the Rankine cycle) can be obtained from the T-<l) \ndiagram. The letters in Fig. 43a refer to the same points as \nin Fig. 43b, the only difference between the two diagrams \n\n* The areas MiHACN and MiHEN should have added to them, the rectangu- \nlar area between Mi \xe2\x80\x94 N, and the absolute zero line (Fig. 216), which is not shown \non the diagram (Fig. 43a). \n\n\n\n138 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\n800 \n\n\n\n700 \n\n\n\n500 \n\na\' \n\n\n\n\n^ 400 \n\n\n\nMl \n\n\n\nFahr. Line \n\n\n\n300 \n\n\n\n200 \n\n\n\n100 \n\n\n\nThermal Eff . = \n\n\n\n_ W \n\n\n\nM\'iHaJcn\' \n\n\n\nliiLLM \n\n\n\nN \n\n\n\n1.6 \n\n\n\n2.0 \n\n\n\n.4 .8 1.2 \n\nEntropy \n\nFig. 43b. \xe2\x80\x94 Temperature (absolute) -entropy Diagram of Actual Steam Engine \n\nIndicator Diagram. \n\n\n\n.PRACTICAL STEAM EXPANSIONS AND CYCLES 139 \n\nbeing the addition of the absolute zero temperature line to \nFig. 43b. \n\nWork done \n\n\n\nThermal efficiency = \nActual thermal efficiency = \n\n\n\nHeat added \nShaded area W \n\n\n\nMi\'H-A\'C-N\' \n\n^ ^5 \xe2\x80\x94 1 \xe2\x80\x94 _i_ (pianimeter) \n26.00 sq. m. \n\n= 0.096, or 9.6 per cent.* \n\nRankine cycle efficiency (theoretical thermal efficiency) \n\nE-A-C-E \n\nMi\'-H-A-C-N\'\' \n\n= ^f^ - (Planimeter) \n26.00 \n\n= 0.203, or 20.3 per cent. \n\nDiscussion of Heat Engine Efficiencies. The action of all \nheat engines is, in general, to produce motion against a resistance \nor to perform work, that is to say, engines or machines by whose \nagency one form of energy is converted into an energy of another \nkind. The heat engine may be a motive power engine, in which \nenergy in the form of heat is converted into energy in the form \nof mechanical work; or it may be one in which mechanical \nenergy is transformed into the potential energy of a gas. \n\nAll these engines depend for their action on the variation in \nvolume, temperature and pressure of some gas or vapor, or upon \nthe explosion of some gaseous mixture. The gas or vapor which \nundergoes these changes in a heat engine is called the working \nfluid. In an air compressor, the working fluid is air. In the gas \nengine, work is done by the expansion of the products of com- \n\n* This value may be regarded as the actual thermal efficiency for one stroke, \n\ninasmuch as in the succeeding strokes the "cushion" steam will be used over and \n\nover again and hence will not constitute a heat loss. The actual thermal efficiency \n\nW \nof a steam engine under runnirtg conditions would then be given by \xe2\x80\x9e, _ p^,w \n\nM 18 nCiv \n\nwhere 18 BCE is a Rankine cycle with clearance and complete compression. \n\n\n\nI40 ENGINEERING THERMODYNAMICS \n\nibustion of a mixture of air and combustible gas. In the oil en- \ngine, the action is the same except that oil is used. The steam \nengine, the engine with which we are mostly concerned, uses the \nT-apor of water for its working fluid. \n\nIn all these heat engines the working fluid undergoes a regular \nseries of changes, the same change occurring in the same order, \nover and over again. So, we can say, the working fluid goes \nthrough a circuit or ** cycle " of operation. Beginning at a par- \nticular condition it passes through a series of successive states \nof pressure, volume and temperature and returns to the initial \ncondition. Thus, the engine has started from a certain point, \nhas gone through a regular series of movements in a fixed order, \nand has returned to the place from which it started. Such a \nseries of regular changes is called a cycle. \n\nThe simplest form of reversible cycle is the Camot wherein \nthe working fluid undergoes four reversible processes between \nthe temperatures Ti and T2 (page 45) . \n\nAll heat engines work between some range of temperature, as \nin this case, Ti to T2, and in view of this fact, the available \nrelation of heat is, as explained in Chapter III, \n\nEx=^^^ (89) \n\nThe heat represented by the temperature range between Ti \nand T2 is the only heat which under any consideration can be \nturned into work. It represents the ** maximum possible effi- \nciency " (El), In fact, an ideal steam engine is impossible. \nSuch an engine with this cycle cannot be actually constructed \nbecause of the various thermodynamic losses. \n\nThe above expression, however, represents the efficiency of \nthe Carnot cycle and merely signifies that it is the highest value \nthat is obtainable for such a cycle working between such tem- \nperature limits. \n\nThe amount of work performed by the working fluid upon the \npiston of an engine is obviously equal to the mechanical equiva- \nlent of the difference between the heat supplied (Hi) and the \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 141 \n\nheat abstracted (^^2); and the theoretical thermal efficiency \n(E2), of the cycle is therefore given by the expression, \n\nE. = ^^- (90) \n\nWe know that Ei is the highest possible efficiency that is \nattainable. \xc2\xa32, the theoretical thermal efficiency of the cycle, \nmust then approach Ei as its Kmit. So the ratio of the one to \nthe other gives what is known as the " type efficiency " (E3). \n\nE,=f. . (91) \n\nThe closer the type efficiency approaches unity, the better \nthe cycle should be. \n\nIt is a known fact that steam engines have very large heat \nlosses, \xe2\x80\x94 losses that cannot be entirely removed. If for a cer- \ntain amount of work (W) actually accomplished (equivalent \nheat units), the amount of heat used in the production of that \nwork is Ha, then the " actual thermal efficiency " (E4) is stated \nas \n\nE.= ^. (9.) \n\nTheory implies that there is an efficiency that the actual \nthermal efficiency must approach as a limit. This relation be- \ntween the actual thermal efficiency (\xc2\xa34) and the theoretical \nthermal efficiency (\xc2\xa32) is known as the practical efficiency (E5) \nand shows how nearly the theoretical efficiency is approached, by \nthe actual efficiency, or \n\nE. = |. (93) \n\nMost important of all efficiencies is that for the " mechanical \nefficiency," meaning the comparison of the useful work per- \nformed with the amount hi work theoretically possible to obtain \nwith a perfect machine. In other words, in a heat engine the \n\n\n\n142 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nmechanical efficiency (Ee), is the ratio of the brake horse \npower to the indicated horse power, or \n\nb.h.p. \ni.h.p. \n\n\n\nEfi \n\n\n\n(94) \n\n\n\nAnother efficiency often applied is what is known as the \n"overall efficiency" (E7), and is the relation of the output, or \nthe useful work performed, divided by the heat supplied. Such \nan efficiency can be expressed in various ways, as \n\nE7 = El X E3 X E5 X Ee, \n= E2 X E5 X Ee, \n= E4 X Ee. \n\nIt is noted that the "overall efficiency" is the product of \nseveral efficiencies, and in view of that fact, the increasing of \nany of the other six respective efficiencies means a relative in- \ncrease in the \'\'overall efficiency." \n\n\n\n\n\na \n\n\nh \n\n\n\n\n2 \n\n\n\\ \n\n\n\n\n\n\n3 \n\n\ni \n\n\n\n\n\n\n^ \n\n\nK \n\n\n\n\n\n\n\n\n\n\n\\d \n\n\n^""^^--^ c \n\n\n\n\n1 \ni \n\n\n1 \n\n\n\n\n\na h \n\n\nS \n\n\na\'; \n\n\n\n\n\n\n\n\n1 \n\n\n\n\n* \n\n\n\n\n1 \n\n\n\n\n\n\nS \n\n\n1 \n\n\n\n\n\n\n\n\n\nJ \n\n\n\n\n1 \n\n\np- \n\n\nI \n\n\n\n\n\\ \n\n\na \n\n\nId \n\n\n\n\n\n\n\n\n\n\n\n\nH \n\n\n1 1 \n/ 1 \n\n/ 1 \n\n\n\\ \n\n\n\nm \n\n\n\nP \n\n\n\nVolume \n\nFig. 44a. \xe2\x80\x94 Pressure- volume Diagram of \nCarnot Cycle of Steam Engine. \n\n\n\no m n \n\nEntropy\xe2\x80\x94 \nFig. 44b. \xe2\x80\x94 Temperature-entropy \nDiagram of Carnot Cycle of \nSteam Engine. \n\n\n\nReferring to the Carnot cycle for a steam engine (Fig. 44a), \nthe amount of work (W) actually done is shown by the area \nabed on the pressure- volume diagram and also on the tempera- \nture-entropy diagram (Fig. 44b). Of course the area mabcn (Fig. \n44a) represents the maximum work possible with the heat (^"1) \nthat was supplied, while madcn represents the amount of heat \n{H-^ that was abstracted from the cycle. This abstraction of \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 143 \n\nheat is a loss, so that the net work done is the difference be- \ntween mabcn and mdcn, or area abed. The same is expressed \nby symbols of heat units as Hi \xe2\x80\x94 H2. \n\nTaking the Carnot cycle on the temperature-entropy dia- \ngram (Fig. 44b), the efficiency (Ei) is \n\n\xe2\x80\x9e abed , . , , Hi \xe2\x80\x94 n2 \n\nEl = \xe2\x80\x94 ; \xe2\x80\x94 , which equals \xe2\x80\x94 \xe2\x80\x94 \n\nraabn Hi \n\nIf in this cycle heat was not added along the admission line \nab but along some other line such as a\'b, the heat added would \nbe decreased by the triangular area a\'ab, and consequently the \nwork done would be correspondingly reduced. Hence in this \ncase, the efficiency of the modified cycle (Ei\') is \n\nabed \xe2\x80\x94 a\'ab \n\n\n\nEl\' \n\n\n\nmabn \xe2\x80\x94 a\'ab \n\n\n\nSuch a deerease in the heat supplied means an actual de- \nerease in effieiency. From this analysis, one can say that for \nthe obtaining of the maximum efficiency (Ei), heat must either \nbe added or subtracted along a constant temperature line, such \nas ab and not a\'b, and Ti and T2 which govern the range of \ntemperatures should be separated as widely as possible. This \nstatement m_ust not be interpreted, however, to mean that when \nheat is added with a varying temperature (finally, attaining a \nhigher value), that it may not be more efficient than one in \nwhich heat is added at a constant temperature of lower value. \nCycles of such kinds are numerous. From an inspection of \nFig. 45 it can be seen that abed is more efficient than mnop. \nHence the greatest efficiency is obtained when all the heat is \nadded at the highest possible temperature and all the heat re- \nmoved at the lowest possible. This is true for all cycles. \n\nThus in Fig. 46, with saturated steam, the efficiency of the \nsteam engine having such a temperature-entropy diagram, would \n\nbe ; and with superheated steam, a slight increase in \n\nfdabg \n\nefficiency is observed, as expressed by , . The relations, \n\nf dabb h \n\n\n\n144 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\n\xe2\x80\x94 \xe2\x80\x94 \xe2\x80\x94 and \xe2\x80\x94 \xe2\x80\x94 - , are practically equal. Now, if the little area \nfdabg fdaeh j -i i \n\nheW is added to both numerator and denominator of the last ex- \npression the efficiency for the superheated cycle is found. This \nadding of the triangle beb\' has the same relative effect as the \n\n\n\n\nEntropy \xe2\x80\xa2 \nFig. 45. \xe2\x80\x94 Temperature-entropy \nDiagram Showing Heat Added \nat Increasing Temperature. \n\n\n\n\nEatropy-0 \nFig. 46. \xe2\x80\x94 Temperature-entropy Dia- \ngram for Engine Using Superheated \nSteam. \n\n\n\nsubtracting of the triangle a\'ab in the Camot cycle (Fig. 44b). \nIn the latter case the efficiency was increased while in the former \nit was diminished. Hence we can say that superheated steam \nwill increase the theoretical thermal efficiency of the engine. \n\nVarious explanatory solutions for efficiencies of thermal ma- \nchinery, such as steam engines, steam turbines, " locomobiles " \nunits, gas producers and engines, blast-furnace gas engines, oil \nengines and pumping engines will now be given. \n\nI. Simple Non-condensing Steam Engine. An engine devel- \noping 135 i.h.p. and using dry saturated steam at 125 pounds \nper square inch absolute, running with a back pressure of 0.2 \npound per square inch gage and with a barometer pressure \nequal to 14.8 pounds per square inch, uses 32 pounds of steam \nper i.h.p. per hour. At end of expansion, the pressure is 20 \npounds per square inch gage. The brake horse power is 125. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 145 \n\nFind the various efficiencies (\xc2\xa31, \xc2\xa32, \xc2\xa33, \xc2\xa34, ^5, Ee and \xc2\xa37) and \nalso the number of thermal units consumed per i.h.p. per \nminute. \n\n(a) Maximum Possible Efficiency (Ei). \n\nEl = \xe2\x80\x94 ^-^; \xe2\x80\x94 -, where Ti is the absolute temperature of the \n\nsteam at the initial pressure condition, and T2 is the absolute \ntemperature of the steam at atmospheric. \n\n(344.4 + 460) - (213 + 460) \n\n^1 = . \xe2\x80\x94 7 \n\n344.4 + 460 \n\n= ^ = 0.1014 or 16.14 per cent. \n804.4 \n\n\n\n(b) Theoretical Thermal Efficiency (E2). \n\xc2\xa32 = \n\n\n\nHi \xe2\x80\x94 H2 \n\n\n\nHi \n\n_ (hi + XiLi) -{h + X2L2) + IM (P4 - P5) X :^F4 ^ \n(hi + XiLi) - h \n\nThe quantity (h + iCiLi), the total heat, is read from the \n"Mollier Diagram,"* the steam being at an initial condition \nof 125 pounds per square inch \nabsolute with a quality Xi of \nunity. From this point the ex- \npansion takes place adiabatically \nto the end of expansion at 34.8 \npounds per square inch absolute, \nwhere the quality X2 is found to volume \n\nbe 0.9225 and the total heat ^^^- ^^\' ^\'^^^^^^ ^\'"^^ ^""^^ \n(^2 + oc^L^ is 1093 B.t.u.; P4 is \n\nthe pressure at end of expansion, P5 is absolute back pressure, \nF4 is the volume corresponding to the pressure P4 (from Steam \nTables) and fe, the heat of the liquid at the back pressure \n(14.8 + 0.2) or 15 pounds per square inch absolute, then \n\n* Diagram I, Marks and Davis\' Steam Tables and Diagrams. \n\n\n\n\n146 ENGINEERING THERMODYNAMICS \n\nP> ^ 1 189 - 1093 + yH X (29.8 - 20) X 0.9225 X 11.89 \n^ I189 - 181 \n\n_ 1 189 - 1093 + 19-85 \nI189 - 181 \' \n\n115-85 \n\n= \xe2\x80\x94 "^\xe2\x80\x94f- = O.I 15 or 1 1.5 per cent. \n1008. \n\n(c) Type Efficiency (E3). \n\n= , = 0.713 or 71.3 per cent. \n16.14 \n\n(d) Actual Thermal Efficiency (\xc2\xa34). \n\nwhere (W) is the amount of work actually accomplished per hour \nand (Ha) is the amount of heat used in developing that work, then \n\n\xc2\xa3, = ms^ , \n\n32(1189-181) \n\n^545 0.0788 or 7.88 per cent. \n\n\n\n32 X 1008 \n(e) Practical Efficiency (\xc2\xa35). \n\xc2\xa34 \n\n\n\n^-S\' \n\n\n\n= -^ \xe2\x80\x94 = 0.686 or 68.6 per cent. \nII. 5 \n\n(f) Mechanical Efficiency (Eq). \n\nJ. b.h.p. \n\nl.h.p. \n\n= \xe2\x80\x94 - = 0.926 or 92.6 per cent. \n135 \n\n* Both numerator and denominator will be expressed in terms of B.t.u. per hour. \nThe numerator is the heat equivalent of one horse-power-hour or (33,000 X 60) -j- \n778 = 2545 B.t.u. per hour. The denominator is the net heat used at the rate of \n32 pounds of steam per hour. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 147 \n\n(g) OveraU Efficiency (\xc2\xa37). \n\nEt = E4 X Eq \n\n= 0.0788 X 0.926 \n\n= 0.073 or 7.3 per cent. \n\nFor a check, by using another formula, we have \n\nEt = E2XE^X Ee \n\n= 0.115 X 0.686 X 0.926 = 0.073 or 7.3 per cent. \n\n(h) Heat units per i.h.p. per minute. \n\nWe know that one i.h.p. represents 33,000 foot-pounds of \nwork, and furthermore that one B.t.u. equals 778 foot-pounds. \nFrom this relation we find that the number of B.t.u. per i.h.p. \n\nper mmute (no losses) is^^^^^ \xe2\x80\x94 - \xe2\x80\x94 , or 42.42. \n\n778 \n\nBut the actual thermal efficiency (\xc2\xa34) is but 7.88 per cent, so \n\nthe actual number of B.t.u. necessary per minute will equal \n\nt^,\' \xc2\xb0^ 538 B.tu. \n\n0.0705 \n\nII. Compound High-speed Non-condensing Steam Engine. \n\nAssume that this engine works under the same pressure condi- \ntions as the above engine with the steam quality at unity. \nFind all efficiencies with the number of B.t.u. required per i.h.p. \nper minute when it operates at 130 i.h.p. and uses 25 pounds of \nsteam per hour. Prony brake test gives no b.h.p. \n\nNote. The values for the efficiencies, Ei, Ei and E3 will be the same for this \nengine as for the case of the simple engine as just calculated. \n\n(a) Maximum Possible Efficiency (\xc2\xa31) = 16.14 per cent. \n\n(b) Theoretical Thermal Efficiency (\xc2\xa32) = n-S per cent. \n\n(c) Type Efficiency (\xc2\xa33) = 71.3 per cent. \n\n(d) Actual Thermal Efficiency (E4). \n\n\xc2\xa3, = , ^545 * ^ \n\n25 (1189 \xe2\x80\x94 181) ^ \n\n= "^-^ \xe2\x80\x94 - = o.ioi or 10. 1 per cent. \n\n25^ X 1008 ^ \n\n* See page 146 (foot-note). \n\n\n\n148 ENGINEERING THERMODYNAMICS \n\n(e) Mechanical Efficiency (Eg). \n\ni.h.p. \n= x\xc2\xa57 = 0-847 or 84.7 per cent. \n\n(f ) Practical Efficiency (\xc2\xa35) . \n\n^\' = ^ \n\n= \xe2\x80\x94 \xe2\x80\x94 = 0.878 or 87.8 per cent. \n\n\' o \n\n(g) Overall Efficiency (\xc2\xa37). \n\nE7 = E^XE6 \n\n= o.ioi X 0.847 = 0.0854 or 8.54 per cent. \n\n(h) Heat per i.h.p. per minute. \n\nRefer to the former problem for the method of obtaining the \nvalue of 42.42 B.t.u. per i.h.p. per minute (no losses). \n\nB.t.u. per i.h.p. per minute = ? \n\nO.IOI \n\n= 420 B.t.u. \n\nIII. Steam Tixrbine. A turbine using steam at 200 pounds \nper square inch absolute pressure at 160\xc2\xb0 F. superheat and a \nback pressure of 0.5 pound per square inch absolute with a \nbarometer of 29.92 inches (of mercury), consumes 12 pounds of \nsteam per kilowatt-hour. Find maximum possible efficiency, \nthe theoretical thermal efficiency, the type efficiency, actual \nthermal efficiency and the practical efficiency, for such condi- \ntions. Calculate the heat consumption per kilowatt-minute and \nalso per electrical horse-power-minute. \n\n(a) Maximum Possible Efficiency (Ei) . \n\n\n\nEi = \n\n\n\n^ (541 >9 + 460) - (80 4- 460) \n541.9 -h 460 \n\n= -^ \xe2\x80\x94 \xe2\x80\x94 = 0.461 or 46.1 per cent. \n1001.9 \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 149 \n\n(b) Theoretical Thermal Efficiency (\xc2\xa32), \n\n1288 - 886 \n\n1288 - 80 \nwhere 1288 is the total heat in B.t.u. per pound at 200 pounds per \nsquare inch absolute, from which point the expansion takes place \nadiabatically down to 0.5 pound per square inch absolute. The \ntotal heat at the lower pressure is 886 B.t.u. per pound. The \nheat of the Hquid at 0.5 pound per square inch absolute \n(back pressure) is 80 B.t.u. per pound, or, \n\n^ = "^^ = 0-333 or 33.3 per cent. \n1200 \n\n(c) T3^e Efficiency, \xc2\xa33 = tt * \n\n-Ss = \'"^^ \xe2\x96\xa0 = 0.722 or 72.2 per cent. \n0.461 \n\nW \n\n(d) Actual Thermal Efficiency, E^ = -\xe2\x80\x94- \n\nThen, E ^-^^ \n\n\n\n12 (1288 -80) X 0.746 \n\n\n\nNote. Denominator is multiplied by the value 0.746 because the number of \nhorse power multiplied by this coefficient gives the equivalent power in kilowatts. \n\nP - 2545 \n\n12 X 1208 X 0.746 \n= 0.226 or 22.6 per cent. \n\nE\xc2\xb1 \n\n(e) Practical Efficiency, \xc2\xa35 = \xe2\x80\x94 * \n\nE2 \n\n(f) Heat per kilowatt-minute, \n\n\xc2\xa35 = \xe2\x80\x94 ^ = 0.679 or 67.9 per cent. \n33-3 \n\nWe know that there are 42.42 B.t.u. per horse power per \nminute (no losses) and that one horse power equals 0.746 kilo- \nwatt. Then, similarly for no losses, we have \n\n\\. = \'^6.6 B.t.u. per kilowatt per minute. \n0.746 ^ ^ ^ ^ \n\n* See page 146 (foot-note). \n\n\n\nI50 ENGINEERING THERMODYNAMICS \n\nBut the test shows an actual thermal efficiency of 22.6 per \ncent and the actual heat unit consumption must be established \non this basis, thus, \n\nB.t.u. per kilowatt per minute = -^-~ = 2\';o B.t.u. \n^ ^ 0.226 ^ \n\nThe heat per electrical horse power per minute can readily be \nfound by multiplying the B.t.u. per kilowatt-minute by 0.746 \n(which is the kilowatt equivalent of a horse power), or 250 X \n0.746 = 187 B.t.u. per electrical horse power per minute. \n\nIV. Unit Consisting of Steam Boiler, Steam Engine and \nElectric Generator. A \'^ locomobile " engine * gives the follow- \ning test results: i.h.p. = 185; kilowatts of generator =115; \nsteam consumption per i.h.p. per hour = 10 pounds; steam pres- \nsure = 210.3 pounds per square inch gage; superheat = 260\xc2\xb0 F.; \nbarometer, 14.7 pounds per square inch; vacuum = 12.7 pounds \nper square inch (nearly 26 inches of mercury) ; feed water tem- \nperature = 142\xc2\xb0 F. \n\nFind the various efficiencies, Ei, E2, \xc2\xa33, E^, E5 and Eq and de- \ntermine the number of B.t.u. per i.h.p. per minute. The coal \nthat was used in the boiler was found to have 14,350 B.t.u. per \npound. With an evaporating capacity of 9 pounds of steam per \npound of coal, what is also the boiler efficiency? \n(a) Maximum Possible Efficiency (\xc2\xa31), \n\nTi - T2 (Ti + ^i) - T2 \n\'" Ti " T^ + h \' \nwhere h is the degrees Fahrenheit of superheat; f other values \nare the same as in previous problems, then, \n\nJ. (391.9 H- 260 -{- 460) - (126 -f- 460) \n\n391.9 -f- 260 + 460 \n^ _ 1111.9- 586 \nmi. 9 \n525-9 \n\n\n\n0.473, or 47.3 per cent. \nI III. 9 \n\n* A "locomobile" is a combination of steam boiler and engine, arranged with \nthe engine placed on top of the boiler. \n\nt The 126\xc2\xb0 F. corresponds to (14.7 \xe2\x80\x94 12.7), or 2 lbs. per sq. in. pressure. (See \nSteam Tables.) \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 15 1 \n\n(b) Theoretical Thermal Efficiency (\xc2\xa32), \n\nj^ Hi \xe2\x80\x94 H2 \n\n^^ 1325-1005 ^ \n^3^5 - 94 \n\nwhere 1325 B.t.u. is the heat read from the "MoUier Diagram " \nat 225 pounds per square inch absolute and 260\xc2\xb0 F. superheat. \nFrom this point the steam expands adiabatically to the lower \ncondition of 2 pounds per square inch absolute (14.7 \xe2\x80\x94 12.7) \nwhere the total heat is 1005 B.t.u. per pound; the value 94 is \nthe heat of the liquid at 2 pounds per square inch absolute \npressure. Then, \n\nT^ ^20 . , \n\nIL2 = \xe2\x80\x94 \xe2\x96\xa0 = 0.26, or 26 per cent. \n1231 \n\n(c) Type Efficiency (\xc2\xa33). \n\n\xc2\xa33 = \xe2\x80\xa2 = 0.55, or 55 per cent. \n\n47-3 \n\n(d) Actual Thermal Efficiency (\xc2\xa34). \n\n\n\nE,= \n\n\n\n2545 \n\n\n\n10 X (1325 - 94) \n\n\n\n= ^ = 0.2666, or 20.66 per cent. \n\n10 X 1231 \n\n(e) Practical Efficiency (E5). \n\nj^ 20.66 , \n\n\xc2\xa35 == \xe2\x80\x94 \xe2\x80\x94 = o-794j or 79.4 per cent. \n26.00 \n\n(f) Mechanical Efficiency (Eq). \n\n^ b.h.p. kilowatt output \ni.h.p. indicated kilowatts \' \n\nwhere ^\'kilowatt output " of the power developed by the gener- \nator and the " indicated kilowatts " are the number of kilowatts \ncorresponding to i.h.p. Then, \n\n^\' = 185 X 0.746 = \xc2\xb0-^33\' "\' ^3-3 per cent. \n\n\n\n152 ENGINEERING THERMODYNAMICS \n\n(g) B.t.u. per i.h.p. per minute. \n\nB.t.u. per i.h.p. per minute = = 2ot:. \n\n^ ^ 0.2066 ^ \n\n(h) Efficiency of the Boiler. \n\nr^rr Lbs, steam evap. per lb. coal X B.t.u. avail, per lb. of steam.* \n\nB.t.u. per lb. of coal \n\n^ 9 [1325 - (142 -32)] \n14,350 \n\no X 1215 ^ ^ \n\n= ~ == 0.7625, or 76.25 per cent. \n\n14,350 \n\nV. Gas-producer and Engine. If the mechanical efficiency \nof a producer-gas engine is known to be 85 per cent, what is the \ni.h.p., provided the machine is running under Prony brake test \nat 658 r.p.m. with a load of 400 pounds (net) at an 8-foot radius? \nSuppose the producer has an efficiency of 60 per cent, find the \nB.t.u. per i.h.p. per minute and also the B.t.u. per b.h.p. per \nminute if the coal consumed per i.h.p. per hour is i pound. \nHeating value of the coal is 14,000 B.t.u. per pound. \n\n(a) Calculation of Indicated Horse Power. \n\nMechanical Efficiency (Ee) = .\' \' \xe2\x96\xa0 , or i.h.p. = \'"\' > \n\nI.h.p. Ae \n\n, , 2 X3141 XrXN XW \n\nD.n.p. = J \n\n33,000 \n\nwhich is the equation for b.h.p., where r is the radius of the brake \nin feet, N is the number of revolutions per minute of the engine, \nand W is the net load on the brake arm at the measured radius, \nthen, \n\n= 2X3.141X8X658X400 ^ ^^^. \n33,000 \n\n\n\nalso, \n\n\n\n. , 400 \n\n* B.t.u. per pound above the temperature of feed water. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 153 \n\n(b) B.t.u. per i.h.p. per minute. \n\nThis value is dependent on the actual thermal efficiency (\xc2\xa34), \nthus, since \n\nE,=- 2545 ., \n\n0.00 X 1. 00 X 1400 \n= 0.303, or 30.3 per cent. \n\n.*. B.t.u. per i.h.p. per minute = \xe2\x80\x94 \'\xe2\x96\xa0 \xe2\x80\x94 = 140. \n\n0-303 \n\n(c) B.t.u. per b.h.p. per minute. \n\nSince the mechanical efficiency is 85 per cent, B.t.u. per b.h.p. \n\nper minute = \xe2\x80\x94^ , or 164.8. \n0.85 \n\nVI. Blast-furnace Gas Engine. The results of a test of a \nblast-furnace gas engine show that 100 cubic feet of gas is con- \nsumed per i.h.p. per hour. Heating value of gas was 120 B.t.u. \nper cubic foot; the i.h.p. developed 800 and b.h.p. = 580. Find \nthe mechanical efficiency {E^ of the machine. Find the actual \nthermal efficiency (\xc2\xa34). What are the heat equivalents B.t.u. \nper i.h.p. per minute, and also per b.h.p.? \n\n(a) Mechanical Efficiency {E^. \n\nEq = ^ = 0.725, or 72.5 per cent. \n000 \n\n\n\n(b) Actual Thermal Efficiency (E4). \n\n\xc2\xa34 = \'^^^ = \n120 X 100 \n\n(c) B.t.u. per i.h.p. per minute. \n\n\n\n\xc2\xa34 = \xe2\x96\xa0 = 0.212, or 21.2 per cent. \n\n120 X 100 \n\n\n\nB.t.u. per i.h.p. per minute = = 200. \n\n0.212 \n\n(d) B.t.u. per b.h.p. per minute. \n\nSince the mechanical efficiency was found to be 72.5 per cent, \n\nB.t.u. per b.h.p. per minute = , or 276. \n\n0.725 \n\nVn. Oil Engine. In tests of an oil engine, the switchboard \nreadings showed a generator output of 375 kilowatts, and the in- \n\n\n\n154 ENGINEERING THERMODYNAMICS \n\ndicator diagrams 560 i.h.p. If the machine consumed 0.32 pounds \nof oil per i.h.p. per hour, what are the efficiencies, \xc2\xa34 and E^} \nHow many B.t.u. were required per b.h.p. and i.h.p. per minute? \nHeat value of the oil was 19,000 B.t.u. per pound. \n\n(a) Mechanical Efhciency {E^. \n\nThe output of the machine is 375 kilowatts which is equal to \n\n\xe2\x96\xa0 ^\'^ , or 502 \'^electrical " horse power. From this, we can say, \n0.746 \n\nC02 \nEe = ^ = 0.8975, or 89.75 per cent. \n560 \n\n(b) Actual Thermal Efficiency (E4). \n\nE4 = ^ = 0.418, or 41.8 per cent. \n\n0.32 X 19,000 \n\n(c) B.t.u. required per i.h.p. per minute. \n\n-r\xc2\xbb \xe2\x80\xa2 1 \xe2\x80\xa2 42.42 \n\nB.t.u. per I.h.p. per mmute = \xe2\x80\x94 ^ = 101.4. \n\n0.418 \n\n(d) B.t.u. required per b.h.p. per minute. \n\nB.t.u. per b.h.p. per minute = = 113.1. \n\n0.8975 \n\nVin. Pumping Engine. A pumping engine uses 12 pounds \nof steam per i.h.p. per hour under the following conditions: \nsteam pressure = 190 pounds per square inch absolute; quality \nof steam = 0.98; barometer = 14.8 pounds per square inch; \npressure at end of expansion = 6 pounds per square inch; back \npressure = 2.2 pounds per square inch absolute. What are the \nvarious efficiencies if the i.h.p. is 850 and the \'\' delivery " horse \npower 825? Find the number of B.t.u. per i.h.p. per minute, \nand also the " duty " per million B.t.u. \n\n(a) Maximum Possible Efficiency {Ei). \n\n^ ^ (377-6 + 460) - (130 + 460) \n\n377.6 + 460 \n\n247.6 \n= ^ = 0.2952, or 29.52 per cent. \n\no / \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES 1 55 \n\n(b) Theoretical Thermal Efficiency (\xc2\xa32). \nUsing the \'\'Mollier Diagram," \n\n1181 - 946 + 4M X (6 - 2.2) X 0.811 X 61.80 \n\nJtL2 = \n\n1181 \xe2\x80\x94 97 \n\n_ 1181 - 946 + 35-3 \n\n1181 \xe2\x80\x94 97 \n\n270.^ \n= \xe2\x80\xa2 \' = 0.249, or 24.9 per cent. \n\n(c) Mechanical Efficiency (Ee). \n\nE^ = -\xe2\x80\x94^ = 971, or 97.1 per cent. \n850 \n\n\n\n(d) Type Efficiency (\xc2\xa33). \n\n\n\n212 \nEz = \xe2\x80\x94 \'\xe2\x80\x94 = 0.851, or 85.1 per cent. \n24.9 \n\n\n\n(e) Actual Thermal Efficiency (E^). \nE. \'-^ \n\n\n\n12 X (1181 - 97) \n= 0.1959, or 19.59 per cent. \n\n(f) Practical Efficiency (\xc2\xa35). \n\n\xc2\xa35 = = 0.786, or 78.6 per cent. \n\n24.9 \n\n(g) B.t.u. per i.h.p. per minute. \n\nB.t.u, per i.h.p. per minute = - = 216.4. \n\n0.1959 \n\n(h) Duty per Million B.t.u. \n\nPumping engines are usually reported as being capable of a \n\ncertain *\' duty," which is the amount of useful work done in \n\nfoot-pounds per million B.t.u. supplied. Duty per milHon B.t.u. \n\n= 778 X 1,000,000 X Et, where \xc2\xa37 is the overall efficiency. \n\nThen, \n\n\xc2\xa37 = \xc2\xa34 X Eq, and in this case \n\n= 0.1959 X 0.971 = 0.19, or 19 per cent. \n\nDuty of the pump pei" million B.t.u. = 778 X 1,000,000 X \n0.19, or 147,900,000 foot-pounds. \n\n\n\n156 ENGINEERING THERMODYNAMICS \n\n\n\nPROBLEMS \n\n1. Steam at a pressure of 100 lbs. per sq. in. absolute having a quality \nof 0.50 expands adiabatically to 15 lbs. per sq. in. absolute. What is the \nquality at the end of the expansion? Ans. 0.50. \n\n2. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has \na volume of 4 cu. ft. and expands adiabaticaUy to 15 lbs. per sq. in. absolute. \n\nWhat is the quaHty at the initial and final conditions ? \n\nAns. 0.905; 0.817. \nWhat is the work done during the expansion? Ans. 114 B.t.u. \n\n3. Two pounds of steam having a temperature of 330\xc2\xb0 F. and quaHty \nof 0.90 expand adiabatically to 230\xc2\xb0 F. \n\nWhat is the quality at end of expansion? Ans. 0,826. \n\nWhat is the work of the expansion? Ans. 193 B.t.u. \n\n4. One pound of steam having a pressure of 125 lbs. per sq. in. abso- \nlute and volume of 4.17 cu. ft. expands adiabatically to 25 lbs. per sq. in. \nabsolute. \n\nWhat is the quality at the initial and final conditions ? \n\nAns. 100 degrees Sup.; 0.953. \nWhat is the work of expansion? Ans. 95.6 B.t.u. \n\n5. Given the steam as stated in problem 4 but expansion complete at \n100 lbs. per sq. in. absolute. What is the quality at 100 lbs. pressure ? \n\nAns. 70\' F. Sup. \n\n6. What would the pressure be if the steam in problem 4 were expanded \nadiabatically until it became dry and saturated ? \n\nAns. 56,6 lbs. per sq. in. absolute. \n\n7. Two pounds of steam having a pressure of 100 lbs. per sq. in. abso- \nlute and a temperature of 377.8\xc2\xb0 F. expand adiabatically to 15 lbs. per sq. \nin. absolute. \n\nWhat is the quality at the initial and final conditions? \n\nAns. 50 degrees F. Sup.; 0.917. \nWhat is the work of the expansion? Ans. 241 B.t.u. \n\n8. One pound of dry and saturated steam has a pressure of 100 lbs. per \nsq. in. absolute, and expands to 20 lbs. per sq. in. absolute along a.n n = i \ncurve. \n\nWhat is the quality at end of the expansion? \n\nAns. 55\xc2\xb0 F. Sup. \nWhat is the work of the expansion? Ans. 102,750 ft.-lbs. \n\nWhat heat is required? Ans. 127.6 B.t.u. \n\n9. Two pounds of saturated steam at a temperature of 300\xc2\xb0 F. have a \ntotal volume of 12 cu. ft., and expand to a pressure of 15 lbs. per sq. in. \nabsolute along an w = i curve. \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\n157 \n\n\n\nWhat is the quahty at the initial and final conditions ? \n\nAns. 0.928; 12\xc2\xb0 F. Sup. \nWhat is the work of the expansion ? Ans. 173,000 ft. -lbs. \n\nWhat heat is required? Ans. 309 B.t.u. \n\n10. One pound of steam at a temperature of 360\xc2\xb0 F. has a quaUty of \n0.50, and expands under constant pressure to a volume of 3.4 cu. ft. \nWhat is the quality at the final condition ? \n\nAns. 90 degrees F. Sup\xe2\x80\x9e \nWhat is the work of the expansion? Ans. 9750 ft.-lbs. \n\nWhat heat is required? Ans. 482.1 B.t.u. \n\nII. Two pounds of steam at a pressure of 100 lbs. per sq. in. absolute \nhave a volume of 4 cu. ft., and expand under constant temperature to a \nvolume of 8 cu. ft. \n\nWhat is the quality at the initial and final conditions ? \n\nAns. 0.452; 0.904. \nWhat is the work of the expansion ? Ans. 57,600 ft.-lbs. \n\nHow much heat is required? Ans. 401 B.t.u. \n\n\n\n1^5.61* \n\n\n\n\n\n145.61^ , ^ \n\n/\' \xe2\x80\x94 \' ! 1 \n\n\n\n\nj/ 1 H.P. I \n\n\n^^ \n\n\n\xe2\x96\xa0^I [ C.E.I \n\n\n43.36* ,___+.\xe2\x80\x94 ^1 "1 I \n\n\n1 i \n\n\n^TaT\'Atm.iLine \' ! ! " ! \n\n\ni I ]39.36^{ \n\n\nsv.iFl 11! ! : ! \n\n\n! . 1 1 \n\n\n\nZero Line \n\n\n\n27. 96 1*\' \n\n\n\nAtm. Line j^^"^ \n\n\ny \n\n\nC.E.I \xc2\xab \n\n\n.9.76* .^.^.i\xe2\x80\x94 \xe2\x80\x94 ^-\'^ I ! \n\n\n\n\n\xe2\x96\xa0 J\' \n\n\n^"H \xe2\x80\x94 J ! i ! \n\n\n\\J\\ \n\n\ni6.76#i ; j ! i ; \n\n\n\n\n!5.96*; \n\n\n\nZero Line \n\nFig. 48. \xe2\x80\x94 High and Low Pressure Indicator Diagram of Compound Steam \n\nEngine. \n\n\n\nCombined Indicator Card of Compound Engine. The method \nof constructing indicator diagrams to a common scale of vol- \nume and pressure shows where the losses peculiar to a compound \nsteam engine occur, and, to the same scale, the relative work areas. \n\nAs the first step divide the length of the original indicator dia- \ngrams into any number \'of equal parts (Fig. 48), erecting per- \npendiculars at the points of division. \n\n\n\n158 ENGINEERING THERMODYNAMICS \n\nIn constructing a combined card, select a scale of absolute \npressure for the ordinates and a scale of volumes in cubic feet \nfor the abscissas. To the scale adopted draw in the atmos- \npheric pressure (" atm.") line, see Fig. 49. \n\nLay off the low-pressure clearance volume on the ii:-axis to \nthe scale selected. In like manner lay off the piston displace- \nment of the low-pressure cylinder and divide this length into the \nsame number of equal parts as the original indicator diagrams \nwere divided. From the original low-pressure card (Fig. 48), de- \ntermine the pressures at the points of intersection of the perpen- \ndiculars erected above the line of zero pressure, taking care \nthat the proper indicator-spring scale is used. Lay off these \npressures along the ordinates (Fig. 49), connect the points and \nthe result will be the low-pressure diagram transferred to the \nnew volume and pressure scales. \n\nThe high-pressure diagram is transferred to the new volume \nand pressure scale by exactly the same means as described for \nthe low-pressure diagram. \n\nThe saturation curve is next drawn. This curve represents \nthe curve of expansion which would be obtained if all the steam \nin the cylinder was dry and saturated. It is very probable \nthat these curves for each cylinder will not be continuous since \nthe weight of the cushion steam in the low-pressure cylinder is \nusually not the same as that in the high-pressure cylinder. The \nsaturation curve would be continuous for the two cards only \nwhen the weight of cushion steam in the high-pressure and low- \npressure cylinders is the same (assuming no leakage or other \nlosses) . \n\nOn the assumption that the steam caught in the clearance \nspaces at the beginning of compression is dry and saturated, the \nweight of the cushion steam can be calculated from the pressure \nat the beginning or end of compression, the corresponding cylin- \nder volumes and the specific volumes corresponding to the pres- \nsure (as obtained from steam-tables). \n\nThe total weight of steam in the cylinder is the weight of steam \ntaken into the engine per stroke plus the weight of steam caught \n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\n159 \n\n\n\n155 \n\n\n;i?^ \n\n\n...P. \n\n\nQua \n\n\nlity ( \n\n\n^urve 100 \n\n\ni \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\ns \n\n\n\\ 1 \n\n\n^ \n\n\n\n\n\n\ni..\' \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n175% \n\n\n\n\n\n\n\\. \n\n\nt5.6 \n\n\nf \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n. \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\niH \n\n\n.P.S \n\n\natur \n\n\ndtioi \n\n\niCu \n\n\nrve \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\na \n\n\n& \\c \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n! \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\nj \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n. 7 \n\n\n0.61^ \n\n\n\\ \n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\ni \n\n\nLP.C \n\n\n\\ \n\n^ard \n\n\n\\ \n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n,1( \n\n\noi \n\n\n\n\n/ \n\n\nLV.MlE.P \n=49.5 \n\n\n\\\' \n\n\n\\ \n\n\n!l. \n\n\nP.Qi \n\n\nlalit.^ \n\n\nCu] \n\n\n\xe2\x80\xa2ve \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x96\xa0 \n\n\n\n\n.i \n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\\ \n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 .\'i \n\n\n\xe2\x80\x9e* \n\n\n\n\nii \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n% \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n] \n\n\n^ \n\n\n\n\n\n\n\n\nIs: \n\n\n36^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^~- \n\n\n\xe2\x96\xa0 \n\n\n\n\n\n\nJ37. \n\n\n11*^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n27. \n\nr \n\n\n6^ \n\n\n\n\n\n\n^. \n\n\n36* \n\n\n\n\n^^, \n\n\n"--.-^ \n\n\nL.P \n\n\nSati \n\n\nirati \n\n\none \n\n\nurve \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nL.P \n\n\nCai \n\n\nd \n\n\n\n\n\\ \n\n\ns^ \n\n\n\n\n\xe2\x96\xa0-> \n\n\n^-^. \n\n\nk \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nAv. \n\n\nVI.E \n\n\nP.= \n\n\n11.22 \n\n\n\n\nX \n\n\n\\ \n\n\n^ \n\n\ni \n\n\n\n\nLin \n\n\nr-^- \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nL \n\n\n.76* \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n__,^ \n\n\n\n\n"~"- \n\n\n\xe2\x80\x94- \n\n\n~9k \n\n\n# \n\n\n\n\n\n\n\\ \n\n\n^^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nT^ \n\n\n? \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.5 1 1.5 2 2.5 3 3.5 i 4.5 \n\n\xc2\xbb Volume- Cu. Ft. \n\nFig. 49. \xe2\x80\x94 Combined Indicator Diagrams for Compound Engine. \n\n\n\nl6o ENGINEERING THERMODYNAMICS \n\nin the clearance space (cushion steam). The saturation curves \nmay now be drawn by plotting the volumes which the total \nweight of steam will occupy at different pressures, assuming it \nto be dry throughout the stroke. \n\nThe quality curve (Fig. 49) shows the condition of the steam \nas the expansion goes on. At any given absolute pressure, the \nvolume up to the expansion line shows the volume of the wet \nvapor, while the volume up to the saturation curve shows the \nvolume that the weight of the wet vapor would have if it were \ndry. Thus at any given absolute pressure, the ratio of the vol- \nume of the wet vapor (as given by the expansion line of the \nindicator card) to the total volume of the dry vapor (as obtained \nfrom the saturation curve) is the measure of the quality of the \nsteam. \n\nShowing the quality by the use of the figure, we have Vol. \nA \xe2\x80\x94B -r- Vol. A \xe2\x80\x94 C = quality. By laying off this ratio from \na horizontal Hne to any scale desired as shown, the quality curve \nmay be constructed. \n\nHirn\'s Analysis. By the use of an indicator diagram, it is \npossible to determine exactly the heat added to or lost from the \n\nsteam. Hirn\'s analysis shows \non a theoretical basis these \ngains and losses for each condi- \ntion of the steam. Investiga- \ntions have shown, however, \nthat the method is of no great \nvalue except to designers. \nFig. 50 is an indicator card, \n\nFig. "Jo. \xe2\x80\x94 Indicator Diagram to Illustrate _ \xe2\x80\xa2 i. \xe2\x80\xa2 \xe2\x80\xa2 . \n\nHim\'s Analysis. compression beginning at c. \n\nIf the quality of steam at this \npoint were known, the weight of and the heat in the steam can \nobviously be calculated. \n\nAssuming the correct quality at c is known, the heat added to \nor lost from the steam from c to d and for each point from c \nto d can be exactly determined. Laying off, for each point, \nthose quantities above the zero line (Fig. 51), the shape of the \n\n\n\n\nPRACTICAL STEAM EXPANSIONS AND CYCLES \n\n\n\nl6l \n\n\n\narea cdd\' is found which represents the amount of heat added \nto the steam. \n\nFrom d to a steam is taken in from the boiler. Determin- \ning this weight from the boiler or from the condenser, the \ntotal weight as well as the volume at a is known, thus the en- \ntropy at a can be calculated. By comparing the conditions of \n\n\n\n\n\n\n\n\n\nb" \n\n\n\n\n\n\nC\' A \n\n\nd\' \n\n\na ^- \n\n\n\xe2\x80\x94 \xe2\x80\x94 ^b\' \n\n\n\\L \n\n\nd\' \n\nA \n\n\nc \n\n\nd \nd" \n\n\n/ \n\n\nh \n\n\nc \n\n\nd \n\n\n\nAreas above line represent heat added to steam \n" below " " " taken from steam \n\nFig. si.\xe2\x80\x94 Diagram of Him\'s Analysis. \n\n\n\nthe steam when entering and at a, the loss to the steam due to \nthe heat stored up in the cylinder walls is shown by the area \ndd\'\'a\'a. The shape of this part of the diagram is unknown \nalthough its area is known if the original assumption at c is \ncorrect. A reasonable guess of the shape of the area may be \nmade. The greater part of the initial condensation probably \ntakes place at the beginning of admission and falls off very fast \nas the point of cut-off is approached, thus dd" would be very \nlarge and a\'a very small. \n\nFrom a to b the cylinder walls give up heat to the steam which \nin turn loses its heat because of the work done. The condi- \ntion of the steam at every point along ab can be determined, and \nthus the shape of the area is also determined. \n\nThe exhaust opens at b, the cylinder walls give up their heat \nto the steam at a greater rate than before because the steam \nis being discharged. \n\nThe area dd^\'a\'a is the quantity of heat exchanged between \nthe cylinder walls and the steam during the time steam is taken \ninto the cylinder and the results practically mean that during \nadmission so much heat must be stored up somewhere. The heat \ninterchanged during expansion is worked out on the basis that \n\n\n\nl62 ENGINEERING THERMODYNAMICS \n\nif the weight and pressure of a given weight of steam are known \nat the beginning and end of a certain period, the difference in \nheat is accounted for by the work done and the heat added to \nor lost from the cylinder walls. \n\n\n\n\nThroat or smallest \nsection of nozzle \n\n\n\nCHAPTER IX \nFLOW OF FLUIDS \n\nFlow through a Nozzle or Orifice. Thermodynamic prob- \nlems embrace the measurement of the flow of air or of a mix- \nture of a liquid and vapor through a nozzle or orifice. In the \nnozzle shown in Fig. 52, let A and B be two sections through \nwhich the substance passes. At A let \na pressure of P\\ be maintained and at \nB a pressure of P2. To maintain the \nconstant pressure at A of Pi let more \nsubstance be added, while at B allow \nenough of the substance to be dis- \ncharged so that the constant pressure Fig. 52. \xe2\x80\x94Typical Nozzle for \nof P2 is maintained. The quantity of Expanding Gases and Vapors. \n\nenergy and the mass passing into the section A must be accounted \nfor at section B and the relation of these quantities will determine \nthe change of the velocity of the substance. \n\nAfter uniform conditions have been estabhshed in the nozzle, \nthe same mass entering at A must be discharged at B during \nthe same time. Thus any mass may be considered as a work- \ning basis, but as a rule one pound of the substance is used. All \nformulas refer to one pound, unless another mass is definitely \nstated. \n\nThe same quantity of energy discharged at B must enter at \nA unless heat is added to or taken from the substance between \nthe sections A and B. Thus the general formula is derived: \n\nEnergy carried by substance at B = energy carried by sub- \nstance at A + energy added between the sections A and B. \n\nThe energy carried by the substance at the entering or dis- \ncharge end of the nozzle is made up of three quantities: (i) the \namoimt of work necessary to maintain constant pressure at each \n\n163 \n\n\n\n1 64 ENGINEERING THERMODYNAMICS \n\nend of the nozzle; (2) the internal energy of the substance, \n(3) the kinetic energy stored in the substance because of the \nvelocity which it has when passing the section. \n\nThe amount of work necessary to maintain a constant pres- \nsure (pounds per square foot) of Pi at A or of P2 at B is -^ or \n\n778 \n\n\xe2\x80\x94^, where Vi and v^ are the volumes of one pound of the sub- \n\n778 \n\nstance at A and B respectively. \n\nThe internal or " intrinsic " energy in B.t.u. per pound {Ih) \nof the substance passing A or B, calculating from 32\xc2\xb0 F. is for \n\nA- \xe2\x80\xa2 -1 j?x(r-4Q2)* . . \n\nAir or similar gases, ^ , (95) \n\n778 X 0.40 \n\nLiquid, h, (96) \n\nLiquid and Vapor, h -\\- xlL -\\ (97) \n\n^ 2^X778/ \n\nSuperheated Vapor, h -\\-Il \xe2\x80\x94 -j \n\nPfaup-^>sat) \n778 \n\n\n\n+Cp(rs\xe2\x80\x9ep-rsat) - "^^T./"^^ , (98) \n\n\n\n72 \nwhere is the kinetic energy in B.t.u. per pound of the \n\n2^X778 \n\nsubstance as it passes a section, V = velocity in feet per second \nand g \xe2\x80\x94 32.2 feet. \n\nIf Q represents in B.t.u. per pound the heat units added to the \nsubstance between the sections A and B (Fig. 52), the energy \nequation for air and similar gases can be found by equating the \ntotal heat energy put in at A plus the energy added between \nA and B to the energy discharged at B. This general formula \nreduces to \n\n\n\nFo^ - F\'2 \n\n\n\ng[^(Pi2;i-P2%) + 778e] \n\n\n\n2 \xe2\x80\x94 yi \n\n4 \n\nor = 2 g X 778 [Cp (Ji - T2) + Ql (99) \n\n* See equation (44) , page 38. \n\n\n\nFLOW OF FLUIDS 165 \n\nThis thermodynamic equation is the usual form for the flow \nof air or similar gases. \n\nThe energy equation for superheated steam can be derived \nby the use of the same general formula stated above, on the \nassumption that the substance is superheated at A and B. This \nformula reduces to \ny 2 \xe2\x80\x94 y 2 \n\n\xe2\x80\x94 = [hi + Li + Cp (Tsnp \xe2\x80\x94 Tsat)l] \n\n2^X775 \n\n-[h2+L2 + Cp(rgup - rsat)2] + Q, (100) \n\nor F2^ \xe2\x80\x94 Fi^ = 2 g X 778 [Tabular heat contenh^ \xe2\x80\x94 Tabu- \n\nlar heat content2 + Q]. (loi) \n\nFrom a thermodynamic standpoint, the relation between the \ninitial and final condition is that of adiabatic expansion when all \nthe heat which disappears as such is used in changing the veloc- \nity, providing the nozzle is properly shaped and Q is zero. The \ndiagram in Fig. 53 represents this condition of affairs on a tem- \nperature-entropy diagram for air and similar substances. The \narea acdf is Cp (T2 \xe2\x80\x94 492), and the area abef is Cp (Ti \xe2\x80\x94 492). \nThe quantity of heat energy changed into kinetic energy is there- \nfore the area bcde and is the difference between the internal \nenergy in the substance at the beginning and end of the opera- \ntion, together with the excess of work done to maintain the pres- \nsure of Pi at A over the pressure of P2 at B. The line cd would \nincline to the right if heat were added in the nozzle, since the \neffect would be to increase the velocity or increase the area cdeb. \nThe line cd would incline to the left from c if heat were lost \nsince the area cdeb would decrease. \n\nIn Fig. 54 the diagram represents the conditions for super- \nheated vapor. The area aa\'cdf represents the heat required to \nraise the substance from a Hquid at 32\xc2\xb0 F. to superheated vapor \nat the temperature of Tisup- The area ab\'bdf is the heat con- \ntent at b, the final condition. The area a\'cbb\' represents, there- \n\n* Tabular heat content means the total heat of superheated steam as read \nfrom tables of the properties of superheated steam as, for example, Marks and Davis\' \nTables, pages 22-65. \n\n\n\ni66 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nfore, the heat available for increasing the velocity. The areas \nrepresenting the heat available for increasing the velocity in Fig. \n\n\n\n\nFig. 53. \xe2\x80\x94 Temperature-entropy \nDiagram of Heat Available in \nAir. \n\n\n\nTiSup. \n\n\n\n\n1 \n\n\n\n\nTi sat. \n\n\na\' \n\n\n/ \n\n\n\n\nT2SUP. / \\ \n\n\nb \n\n\nT2sat. / \\ \n\n\n\n\n/"\' \n\n\n\n\n\\ \n\n\nu \n\n\nT.-Ent.-Diagram \n\n\n\n\n\\ \n\n\nf \n\n\n\n\n\n\nd \n\n\n\nFig. 54. \xe2\x80\x94 Temperature-entropy Diagram \nof \'the Heat Available in Superheated \nSteam for Increasing Velocity. \n\n\n\n53 and Fig. 54 are shown by the cross-hatched area in Fig. 55 \nand are really the representation of the work done (theoretically) \n\nin an engine giving such an in- \ndicator diagram. \n\nEvidently, the greater the \ndrop in pressure, the greater \nwill be the cross-hatched area \nin Fig. 55 and the greater will \nbe the velocity, regardless of \nthe substance. The line ab in \nFig. 56 represents the velocity \n\n\n\n\nFig. \n\n\n\n55. \xe2\x80\x94 Heat (Work) Available \nIncreasing Velocity. \n\n\n\ncurve with V as ordinates and \xe2\x80\x94 ^ as abscissas, but since with \n\nPi \n\n\n\nFLOW OF FLUIDS \n\n\n\n167 \n\n\n\nany substance expanding the weight of a cubic foot decreases as \nthe pressure drops, the line cd will represent to some scale not \nhere determined the weight of a cubic foot at any discharge \npressure. \n\nSince the product of the area (square feet) through which the \ndischarge takes place, the velocity in feet per second at the area \nand the weight (pounds) per cubic foot of the substance is equiv- \n\n\n\na \n\\ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\\ \n\n\n\n\n\n\n\n\ne \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n/ \n\n\nw \n\n\n;^:in.p \n\n\n-^ \n\n\n\\, \n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\nr \n\n\n\n\n\n\n\n\n\\ \n\n\n\\ \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\\ \n\n\n,K. \n\n\n\n\n\n\n\n\n\\ \n\n\n\n\nd \n\n\n\n\n/ \n\n\n/ \n\n\n\n\n^ \n\n\n\n\n\n\n^ \n\n\n^ ^ \n\n\n\\ \n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n< \n\n\n\n\n\n\n\\ \n\n\n\n\nj \n\n\n/ \n\n\n\n\n.f> \n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n\n\n1 \n\n\n\n\ni^A \n\n\nV \n\n\n\n\n\n\n\n\n\n\nN \n\n\n\n\n\n\n/ \n\n\n/ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\ \n\n\nh \n\n\n\n.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 \n\nFig. 56. \xe2\x80\x94 Illustrative Curves of Weight Discharge and Velodty. \n\n\n\nalent to the weight in pounds of the substance discharged per \nsecond. The product of the ordinates at any point of the curves \nab and cd is proportional to the weight discharged from a pres- \nsure of Pi to a condition where the pressure is P2. The line ceb \nrepresents this product. Evidently there is some low pressure \ninto which the weight discharged per square foot will be a maxi- \nmum, and this will be the pressure corresponding to the high \npoint e on the curve. \n\nWeight per Cubic Foot., From the formula for adiabatic ex- \npansion the weight per cubic foot can be obtained if the sub- \n\n\n\n1 68 ENGINEERING THERMODYNAMICS \n\nstance is similar to air.* The general formula (applied to air) is \nPiz^ii-^ = P22^2\'-^ (102) \n\nwhich can be reduced to \n\nI P2^XPl I Pi0.286p 0.714 \n\n- = \xe2\x80\x94 I \' or -=: \xe2\x80\x94 , (103) \n\nPi\'-\' X RTi \n\nwhich is the weight in pounds per cubic foot of discharge. Pres- \nsures are, of course, in pounds per square foot. If the supply to \nthe nozzle is from a large reservoir so that Vi can be taken as \nzero then the discharge velocity is \n\nV2 = V 2g X ^ (PiVi - P2V2) \n\' 0.4 \n\nF. = 109.6 v/ri[i-(^J\'\'^]- (104) \n\nAll quantities on the right-hand side of this equation must be \nobtained from the data of tests. Weight in pounds discharged \nthrough the area A (in square feet) is \n\n\n\np 0.286 p 0.714 / r /p \\0. 286-1 \n\n^ = ^x RT, X 109.6 v/ri[i-(^) ]\xe2\x80\xa2 (105) \n\nMaximum Discharge. This weight is a maximum when \n\n-\xe2\x80\x94 \xe2\x96\xa0 = o or when P2 = 0.525 Pi. The maximum quantity of air \ndP2 \n\nwill be discharged when the low pressure is 52.5 per cent of the \n\nhigh pressure. \n\nShape of Nozzle. See page 183, on the flow of steam. \n\nFlow of Air through an Orifice. Air under comparatively high \n\n* The exponent in the formula is the ratio of the specific heats (of air in this \ncase). \n\n\n\nFLOW OF FLUIDS 169 \n\npressure is usually measured in practice by means of pressure \nand temperature observations made on the two sides of a sharp- \nedged orifice in a diaphragm. The method requires the use of \ntwo pressure gages on opposite sides of the orifice and a ther- \nmometer for obtaining the temperature ti at the initial or higher \npressure pi. The flow of air w, in pounds per second, may then \nbe calculated by Fliegner\'s formulas: \n\nw = 0.530 X f X a -^zr when pi is greater than 2 p2, (106) \n\nVTi \n\nw = 1.060 X f X a y^^-\xe2\x80\x94^ \xe2\x80\x94 \xe2\x80\x94 when pi is less than 2 p2, (107) \n\xe2\x96\xbc Ti \n\nwhere a is the area of the orifice in square inches, f is a coeffi- \ncient, Ti is the absolute initial temperature in degrees Fahren- \nheit at the absolute pressure pi in the " reservoir or high-pressure \nside " and p2 is the absolute discharge pressure, both in pounds \nper square inch. When the discharge from the orifice is directly \ninto the atmosphere, p2 is obviously barometric pressure. \n\nWestcott\'s and Weisbach\'s experiments show that the values of f \nare about 0.925 for equation (106) and about 0.63 for equation \n(107). \n\nFor small pressures it is often desirable to substitute manom- \neters for pressure gages. One leg of a U-tube manometer can be \nconnected to the high-pressure side of the orifice and the other leg \nto the low-pressure side. Many engineers insert valves or cocks \nbetween the manometer and the pipe in which the pressure is to \nbe observed for the purpose of \'^ dampening" oscillations. This \npractice is not to be recommended as there is always the possi- \nbility that the pressure is being throttled.* A better method is \nto use a U-tube made with a restricted area at the bend between \nthe two legs. This will reduce oscillations and not affect the \naccuracy of the observations. \n\nDischarge from compressors and the air supply for gas engines \nare frequently obtained by orifice methods. \n\n* Report of Power Test Committee, Journal A.S.M.E., Nov., 1912, page 1695. \n\n\n\nlyo ENGINEERING THERMODYNAMICS \n\nWhen pi \xe2\x80\x94 p2 is small compared with pi, the simple law of \ndischarge * of fluids can be used as follows: \n\nfa \n\n\n\nw = \xe2\x80\x94 V2 g X 144 (Pi - P2) s, (108) \n\n144 \n\nwhere f is a coefficient from experiments, g is the acceleration \ndue to gravity (32.2), and s is the unit weight of the gas meas- \nured, in pounds per cubic foot, for the average of the initial and \nfinal conditions of temperature and pressure. If the difference \nin pressure is measured in inches of water h with a manometer \nthen \n\n144 (pi \xe2\x80\x94 P2) = \xe2\x80\x94 ^ X h (expressed in terms of pounds per square \n12 \n\nfoot), \n\nw = \xe2\x80\x94 1/2 ghs X \xe2\x80\x94 ^ (pounds per second), (109) \n\nwhere 62.4 is the weight of a cubic foot of water (density) at \nusual " room ^^ temperatures. \n\nThis equation can also be transformed so that a table of the \nweight of air is not needed, since by elementary thermodynamics \n144 pv = 53.3 T, where v is the volume in cubic feet of one \n\n\n\nIf the density is fairly constant, \n\n\n\n144 Pl , Vl^ 144 P2 , Vo^ \n\n\xe2\x80\x94 -j \xe2\x80\x94 -\\ , \n\nS 2g S 2g \n\nwhere Vi is the velocity in feet per second in the "approach" to the orifice, and Vo is \nthe velocity in the orifice itself. Since Vi should be very small compared with Vo, \n\nVo^ ^ 144 (pi - P2) \n2g S \' \n\n\n\n2 g X 144 (pl - P2) \n\n\n\nVo=\\/ \n\nw = favos = fas ^ ^ g X i44(Pi - P2 ) , \n\nor w = fa V2 g X 144 (pl \xe2\x80\x94 P2) s. \n\nProfessor A. H. Westcott has computed from accurate experiments that the \nvalue of the coefi&cient f in these equations is approximately 0.60. \n\n\n\nFLOW OF FLUIDS 171 \n\npound and T is the absolute temperature in Fahrenheit. Since \nV is the reciprocal of s, then \n\ns = 144 p-^ 53-3 T, \nand w = 0.209 fa \\/-^\' (no) \n\nHere p and T should be the values obtained by averaging the \ninitial and final pressures and temperatures. Great care should \nbe exercised in obtaining correct temperatures. When equa- \ntion (47) is used, for accurate work, corrections of s for humidity \nmust be made.* \n\nFor measurements made with orifices with a well-rounded \nentrance and a smooth bore so that there is practically no con- \ntraction of the jet the coefficient f in equations (47) and (48) \nmay be taken as 0.98. In the rounding portion of the entrance \nto such a nozzle the largest diameter must be at least twice the \ndiameter of the smallest section. For circular orifices with \nsharp corners Professor Dalby f in reporting very recent experi- \nments stated that the coefficient for his sharp-edged orifices in a \nthin plate of various sizes from i inch to 5 inches in diameter \nwas in all cases approximately 0.60; and these data agree very \nwell with those pubhshed by Durley.f \n\nWhen p2 -^ pi = 0.99 the values obtained with this coefficient \nare in error less than | per cent; and when p2 -^ pi = 0.93 the \nerror is less than 2 per cent. \n\nReceiver Method of Measuring Air. None of the preceding \nmethods are adaptable for measuring the volume of air at high \npressures as in the case of measuring the discharge in tests of \nair compressors. Pumping air into a suitably strong receiver \nis a method often used. The compressor is made to pump \n\n* Tables of the weight of air are given on page 181 and tables of humidity on \npage 368 in Moyer\'s Power Plant Testing (2d Edition). \n\nt Engineering (London), Sept. 9, 19 10, page 380, and Ashcroft in Proc. Insti- \ntution of Civil Engineers, vol. 173, page 289. \n\n% Transactions American Society of Mechanical Engineers, vol. 27 (1905), \npage 193. \n\n\n\n172 ENGINEERING THERMODYNAMICS \n\nagainst any desired pressure which is kept constant by a regu- \nlating valve on the discharge pipe: \n\nPi and P2 = absolute initial and final pressures for the test, \npounds per square inch. \n\nTi and T2 = mean absolute initial and final temperatures, de- \ngrees Fahrenheit. \n\nWi and W2 = initial and final weight of air in the receiver, \npounds. \n\nV = volume of receiver, cubic feet. \n\nPiV = WRTi, and P2V = W2RT2, where R is the constant \n53.3 for air, then weight of air pumped \n\nIn accurate laboratory tests the humidity of the air entering \nthe compressor should be measured in order to reduce this \nweight of air to the corresponding equivalent volume at atmos- \npheric pressure and temperature. \n\nPrincipal errors in this method are due to difficulty in measur- \ning the average temperature in the receiver. Whenever practi- \ncable the final pressure should be maintained in the receiver at \nthe end of the test until the final temperature is fairly constant. \n\nPROBLEMS \n\n1. Air at a temperature of 100\xc2\xb0 F. and pressure of 100 lbs. per sq. in. ab- \nsolute flows through a nozzle against a back pressure of 20 lbs. per sq. in. \nabsolute. Assuming the initial velocity to be zero, what will be the ve- \nlocity of discharge? Ans. 1570 ft. per sec. \n\n2. If the area at the mouth of the above nozzle is 0.0025 sq. ft. and the \ncoefficient of discharge is unity, how many pounds of air will be discharged \nper minute? Ans. 59 pounds. \n\n3. What will be the theoretical kinetic energy per minute of the above \njet assuming no frictional losses? Ans. 2901 B.t.u. \n\nFlow of Steam. In Fig. 52 suppose the sections A and B \nare so proportioned that the velocity of the substance passing \nsection A is the same as that at section B. Such a condition \nmight arise in a calorimeter or in the expansion of ammonia \n\n\n\nFLOW OF FLUIDS \n\n\n\n173 \n\n\n\nthrough a throttling or expansion valve (Fig. 20), as in an ice \n\nmachine. The pressure at A will be Pi which is greater than the \n\npressure at B of P2. Fig. 57 represents the entropy diagram \n\nfor such a condition. As the pressure falls from Pi to P2 the \n\nmaximum heat available to produce \n\nvelocity through the nozzle is the area \n\nacde. The value of the quaHty, \n\nrepresented by the symbol x for the \n\nsubstance after leaving the nozzle \n\ncorresponds to that of point c and the \n\narea acde is the excess kinetic energy \n\nrepresented by the increased velocity. \n\nThis excess kinetic energy is destroyed \n\nby coming into contact with the more \n\nslowly moving particles at B and with \n\nthe sides of the vessel. The area acde \n\nis equal to {ha + XaL^ \xe2\x80\x94 {he + Xc Lc) \n\nand the relation of Xa to Xc is obviously adiabatic. The area \n\nohdbf equals area oheag; thus the heat content at b is the \n\nsame as at a. The location of b can be found as follows: \n\n\n\n\n\n/e p^ a \n\n\n\n\n\n\n\n\nId \n\n\nc \n\n\nh \n\n\n/ \n\n\nI P2 \n\n\n\n\n\n\n/ \n\nh \n\n\nT-Ent.-Diagram \n\n\n\n\n\n\n\n\n\ng \n\n\n\n\nf \n\n\n\nFig. 57. \xe2\x80\x94 Diagram for no \nVelocity Change. \n\n\n\nXb \n\n\n\nha + XaLa \xe2\x80\x94 h \n\n\n\n(112) \n\n\n\nThe curve shown in Fig. 58 represents the discharge of a mixture \nof steam and water {x = 0.6 at 100 pounds per square inch abso- \nlute pressure) into a vessel having the pressures shown. The \npoints on this curve cannot be determined by entropy tables. \nAt 100 pounds per square inch pressure the total heat of the wet \nsteam is h -\\- 0.6 L or 298.5 + 0.6 X 887.\'6 = 831. i B.t.u. per \npound. \n\nAt 60 pounds per square inch absolute pressure the total heat \nmay be found by the use of the entropy diagram shown in \nFig. 59. The entropy values are taken directly from steam- \ntables. The entropy for the initial point is, then, \n\n\n\nab = 0.4748 + 0.6 X 1. 1273 = 1.1512, \n\n\n\n174 ENGINEERING THERMODYNAMICS \n\nThe distance \n\nde = ab \xe2\x80\x94 0.4279 = 0.7233, \n\nand X at 60 pounds \n\nde 0.72SS \n\n= = \xe2\x96\xa0 = 0.595. \n\n1. 2154 1. 2154 \n\n2.00 \n\n\n\n1.75 \n\n\n\no \n\nd \n\no \n\n:S 1.25 \n\no \n\n\n\n1.00 \n\n\n\n.75 \n\n\n\n\xc2\xbb .50 \n\n\n\n^ \n\n\n\n.25 \n\n\n\n\n\n\n\n\n\n\xe2\x96\xa0^^ \n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\\ \n\n\n\n\n\n\n\n\nto/ \n\n7 \n\n\n\n\n\n\n\n\n\n\ny \n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\\ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\' \n\n\n\n\n\n\n\n\n\n\n\n\n\n10 20 30 40 50 60 70 80 90 100 \nDischarge Pressure in lbs. per sq. in. \n\nFig. 58. \xe2\x80\x94 Discharge of Steam Under Various Pressures. \n\nThe total heat at 60 pounds pressure is \n\nh + 0.595 L = 262.4 + 0.595 X 914.3 = 805.4. \nThe velocity of flow is \nV2 X 32.2 X 778 (83 1. 1 \xe2\x80\x94 805.4) = 1135 feet per second. \nThe volume of one pound is \n\n0.016 (i.o \xe2\x80\x94 0.595) + 7-i66 X 0.595 = 4-27 cubic feet, \nand the weight per cubic foot is \n\n\n\n4.27 \n\n\n\n= 0.2342 pound. \n\n\n\nFLOW OF FLUIDS \n\nThe weight discharged per square inch per second is \n\n"35 X 0.2342 ^ ^ g ^^^^^ \n144 \n\n\n\n175 \n\n\n\n\n\ne h\\ \n\n\n\nT.- Ent.- Diagram \n\n\n\nFig. 59. \xe2\x80\x94 Temperature-entropy \nDiagram for Calculating the \nWeight of Discharge of Steam. \n\n\n\nFig. 60. \xe2\x80\x94 Diagram Illustrating \nRadiation Loss in Nozzle. \n\n\n\nVelocity of Flow as Affected by Radiation. Fig. 60 shows the \nradiation losses. The condition at entrance k represented at a \nand the area acde represents the quantity of heat lost by radi- \nation. Area aefg represents the velocity change while the point e \nrepresents the condition of the moving substance. \n\nIf, after passing through the nozzle, the velocity is reduced \nto that of entrance, a point located as at b will represent the \ncondition of the substance. This point would be so located that \n\n\n\neb \n\n\n\narea aefg \n\n\n\n(113) \n\n\n\nFriction Loss in a Nozzle. Fig. 61 shows the friction loss. \nThe energy converted into heat by friction varies with the square \nof the velocity. In this figure, a is the initial condition and \naefg is the energy available for change in velocity providing \nthere is no friction loss.^ The ratio of the areas acde to aefg is \nthe proportional loss by friction. The point c represents the \n\n\n\n176 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\ncondition of the substance at the return of the friction heat to \nthe substance. The heat is returned in exactly the same way as \nif it came from an outside source. The distance ch is the area \nac(ie.-^ Lc. The area edfg represents the energy expended in \nthe velocity change and the point h represents the state of the \nsubstance at the point of discharge. \n\nThis condition is found existing in the fixed nozzle of most \nturbines. Point a represents the condition on the high-pressure \nside of the nozzle and point h, the low-pressure side. The abso- \nlute velocity of discharge is really caused by the energy repre- \nsented by the area edfg. \n\nImpulse Nozzles. Suppose that the substance is discharged \nwith an absolute velocity corresponding to the area edfg (Fig. \n\n\n\n\nT.- Ent,- Diagram \n\n\n\n\nn I d \n\n\n\nh \n\n\n\nT.- Ent.- Diagram \n\n\n\nFig. 61. \xe2\x80\x94 Diagram Illustrating \nFriction Loss in Nozzle. \n\n\n\nFig. 62. ^Diagram of Heat Losses \nin a Steam Nozzle (Impulse). \n\n\n\n61), and that it passes into a moving nozzle, having the same \npressure on the discharge as on the intake side. The energy \nrepresented by the area edfg would be used up in the following \nways: (i) by the friction in moving nozzle; (2) residual abso- \nlute velocity; (3) and in driving the moving nozzle against the \nresistance. \n\nFig. 62 shows the quantities used up by each. Point a rep- \n\n\n\nFLOW OF FLUIDS 177 \n\nresents the condition of the substance before passing into the \nfixed nozzle while point h shows its condition leaving the fixed \nnozzle, the velocity corresponding to the area edfg. Area klde \nrepresents the energy used up in friction in the moving nozzle; \narea klnm residual velocity after leaving moving nozzle and \narea mnfg represents useful work used in moving the nozzle \nagainst its resistance. The condition of the substance leaving \nthe nozzle is shown at q and not at h, the distance h \xe2\x80\x94 q being \nthe area edlk divided by Lh. The substance leaves the moving \nnozzle with a velocity corresponding to the area klnm and it \nwill have done work corresponding to the area mnfg. \n\n\n\ne V \nFig. 63. \xe2\x80\x94 Impulse Nozzle and Velocity Diagrams. \n\nTurbine Losses. Fig. 63 is a simple velocity diagram show- \ning, for an impulse nozzle such as occurs in many turbines, the \nrelative value of those various losses. A is a stationary nozzle \ndischarging against the movable blades B. The path of the steam \nis shown by the dotted line. The line db marked v represents \nthe velocity of discharge of the stationary nozzle, which makes \nan angle a with the direction of motion of the moving blades. \nCall ezj the velocity of the moving blades, then h is the amount \nand direction of the relative velocity of the steam over the \nsurface of the moving blades. It loses a portion of this velocity \nas it passes over the surface of the blades and lh becomes the \nactual relative velocity of discharge. The direction of lh is de- \ntermined by the discharge edge of the moving blades, the angles \n\n\n\n178 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\na and jS being as shown. The residual absolute velocity is re- \npresented by r. \n\nThe total energy equivalent of the velocity developed \n\nc,2 \n\n\n\nin B.t.u. per pound = \n\n\n\n2g \n\n\n\nThe residual energy per pound = \n\n\n\n^g \n\n\n\n(114) \n(115) \n\n\n\nReaction Nozzles. When the substance leaving the station- \nary nozzle passes into a moving nozzle having the pressure at \nthe intake greater than at the discharge, the conditions differ \nfrom those just discussed. The velocity in this case is thermo- \ndynamically changed in passing through the moving nozzle. \nIn the equations given in the previous discussion it was assumed \nthat the moving nozzles were entirely filled with the substance, \n\nand when partly filled in the expand- \ning portion, coefficients of correction \nwere appHed, but in this case the \nnozzles should be so designed that \nthe substance entirely fills them, as \nthe corrections are unknown. \n\nIn Fig. 64 the lines of Fig. 62 are \nreproduced together with those relat- \ning directly to the reaction nozzle. \nPoint a corresponds to the condition \non entering the stationary nozzle, \npoint h the condition on leaving it \nwith a velocity corresponding to the \nDiagramof Heat Losses area cdfg. Point k represents the \npressure at the discharge end of the \nmoving nozzle, and if no friction losses or impact loss occur \nin moving nozzle, point k would represent the condition of the \ndischarged substance and the area egmkhd would be accounted \nfor as useful work done and residual velocity. But since friction \nlosses and impact loss do occur a portion of this area edhkno can \nbe set aside to represent these losses, a portion noqp represents \n\n\n\na, \n\n\n\n\n1 q\\ o\\ e \n\n\na \n\n\nf 1 ! \n\n\nh \n\n\n/ 1 1 ^ \n\n\n\n\nw./ \' \' \n\n\nI \n\n\n7 p n \n\n\nk \n\n\n. T.- Ent.- Diagram \n\n\n\n\n\nFig. 64 \nin a Steam Nozzle (Reaction) \n\n\n\nFLOW OF FLUIDS 1 79 \n\nthe residual velocity, while the remaining area pqgm represents \n\nthe useful work done. \n\nThe condition of the substance leaving the moving nozzle is \n\n, , ,, ,. , 1,1- area edhkno . \ngiven by k, the distance kl being . Area onpq rep- \n\nresents the residual velocity. \n\nCoefficient of Flow. Few experiments have been carried on \nfor determining the flow of steam in nozzles proportional for \nmaximum discharge or through nozzles exactly designed for an \nexact pressure. For nozzles having well rounded entrance and \nparallel portion of least diameter from 0.25 to 1.5 times the length \nof the converging entrance the coefficient of discharge is about \n1.05. For properly shaped entrances and for areas of orifices \nbetween 0.125 square inch and 0.75 square inch the coefficient of \ndischarge varies from 0.94, the two pressures being nearly alike, \nto unity, the ratio of the pressures being 0.57. For an orifice \nthrough a thin plate the coefficient is about 0.82, the ratio of the \npressures being 0.57. \n\nInjectors. In an injector, steam enters at A in Fig. 65 at \nthe pressure of the supply. The quantity of water entering at \nC, the cross-section of the pipe, and \nthe pressure of the water determine \nthe pressure at ^. At Z) the pres- \nsure should be zero (atmospheric) Fig. 65.\xe2\x80\x94 Essential Parts of an \nor equal to the pressure in the water Injector, \n\nsupply pipe to which D may be connected. The total hydrauKc \nhead should exist as velocity head at this point. At E the pres- \nsure should be sufficient to raise the check valve into the boiler \nand the velocity sufficient to carry the intended supply into the \ndischarge pipe. \n\nThe shape of the nozzle from ^ to ^ should be such as to con- \nvert the energy in the steam at A into velocity d.t B. At B the \nwater and steam meet, condensing the steam, heating the water \nand giving to the water a velocity sufficient to carry it through \nthe nozzle B-D. \n\nAll the energy accounted for at A and C must be accounted for \n\n\n\n\nl8o ENGINEERING THERMODYNAMICS \n\nat E. The heat lost by radiation may be neglected. The veloc- \nity at any section of the nozzle equals \n\nvolume passing in cubic feet \narea of section in square feet \n\nor F^= \xe2\x80\x94 , where a = area at any section corresponding to \n\nvelocity V and a a = area at section A . \n\nWeight of Feed Water per Pound of Steam. Assuming the \nsteam supply to be dry and reckoning from 32\xc2\xb0 F. the heat units \ncontained in the steam and feed water per pound and the heat \nin the mixture of steam and feed water per pound may be easily \ncalculated. \n\nKnowing the rise of temperature of the water passing through \nthe injector and neglecting radiation losses, the pounds of feed \nwater suppHed per pound of steam used by the injector may be \nobtained. Thus \n\nHeat units lost by steam = Kinetic energy of jet + Heat \nunits gained by feed water. \n\nThe term expression \'\'kinetic energy of jet " may be neglected \nsince it is very small, then, \n\nH -hf = W(hm-hf) \n\nwhere W = the weight of feed water lifted per pound of steam. \nhm = heat of liquid of mixture of condensed steam and \n\nfeed water. \nhf = heat of liquid of entering feed water. \n\nThermal Efficiency of Injector. The thermal efficiency of an \ninjector neglecting radiaition losses is unity. All the heat ex- \npended is restored either as work done or in heat returned to the \nboiler. \n\nMechanical Efficiency. The mechanical work performed by \nthe injector consists in lifting the weight of feed water and deliver- \n\n\n\nFLOW OF FLUIDS \n\n\n\nl8l \n\n\n\ning it into the boiler against the internal pressure. The effi- \nciency, considering the injector as a pump, is \n\nWork done \n\n\n\nor \n\n\n\nB.t.u. given up by steam to perform the work \n\n\n\nU \n\n\n\nH -h/ \nwhere U = { Wh -\\- {W + i) h\\ -^ 778 (in heat units). (117) \n\nIp = pressure head corresponding to boiler gage pressure, \n\nin feet. \nIs = suction head in feet. \nW = pounds of water deHvered per pound of steam. \n\nOrifice Measurements of the flow of steam are particularly \nrecommended by some engineers for ascertaining the steam con- \nsumption of the \'\' auxiliaries " in a power plant. This method \ncommends itself particularly because of its simplicity and accu- \nracy. It is best applied by inserting a plate | inch thick with \nan orifice one inch in diameter, with square edges, at its center, \nbetween the two halves of a pair of flanges on the pipe through \nwhich the steam passes. Accurately calibrated steam gages are \nrequired on each side of the orifice to determine the loss of pres- \nsure. The weight of steam for the various differences of pres- \nsure may be determined by arranging the apparatus so that the \nsteam passing through the orifice will be discharged into a tank \nof water placed on a platform scales. The flow through this \norifice in pounds of dry saturated steam per hour when the dis- \ncharge pressure at the orifice is 100 pounds by the gage is given \nby the following table: \n\n\n\nPressure drop, \nlbs. per sq. in. \n\n\nFlow of dry steam \nper hour, lbs. \n\n\nPressure drop, \nlbs. per sq. in. \n\n\nFlow of dry steam \nper hour, lbs. \n\n\nI \n2 \n\n3 \n4 \n\n\n430 \n\n615 \n\n930 \n\n1200 \n\n1400 \n\n\n5 \n10 \n\nIS \n\n20 \n\n\n1560 \n2180 \n2640 \n3050 \n\n\n\n\n\n\n\n1 82 ENGINEERING THERMODYNAMICS \n\nFlow of Steam through Nozzles. The weight of steam dis- \ncharged through any well-designed nozzle with a rounded inlet, \nsimilar to those illustrated in Figs. 66 and 67, depends only on \nthe initial absolute pressure (Pi), if the pressure against which \nthe nozzle discharges (P2) does not exceed 0.58 of the initial \npressure. This important statement is well illustrated by the \nfollowing example. If steam at an initial pressure (Pi) of 100 \npounds per square inch absolute is discharged from a nozzle, the \nweight of steam flowing in a given time is practically the same \nfor all values of the pressure against which the steam is dis- \ncharged (P2) which are equal to or less than 58 pounds per square \ninch absolute. \n\nIf, however, the final pressure is more than 0.58 of the initial, \nthe weight of steam discharged will be less, nearly in proportion \nas the difference between the initial and final pressures is re- \nduced. \n\nThe most satisfactory and accurate formula for the " constant \nflow " condition, meaning when the final pressure is 0.58 of the \ninitial pressure or less, is the following, due to Grashof ,* where w \nis the flow of steam f (initially dry saturated) in pounds per \nsecond, Aq is the area of the smallest section of the nozzle in \n\n* Grashof, Theoretische Maschinenlehre, vol. i, iii; Hiitte Taschenbuch, vol. i, \npage ^^:^. Grashof states the formula, \n\n\xe2\x96\xa0w = 0.01654 ^o-Pi,-^^> \n\nbut |the formula given in equation (118) is accurate enough for all practical \nuses. \n\nt Napier\'s formula is very commonly used by engineers and is accurate enough \nfor most calculations. It is usually stated in the form \n\nAoPi \n\nw = 5 \n\n70 \n\nwhere w, Pi, and Ao have the same significance as in Grashof s formula. The \nfollowing formula is given by Rateau, who has done some very good theoretical \nand practical work on steam turbines, but this formula is too compHcated for \nconvenient use: \n\nw = o.ooi AoPi [15.26 \xe2\x80\x94 0.96 (log Pi + log 0.0703)]. \nCommon or base 10 logarithms are to be used in this formula. \n\n\n\nFLOW OF FLUIDS \n\n\n\n183 \n\n\n\n\nFig. 66. \xe2\x80\x94 Example of a Well-designed Nozzle. \n\n\n\n\n\n\nDeLaval Type. \n\n\n\n\nNozzle \n\nDiaphragm \n\n\n\nCurtis Type. \n\nFig. 67. \xe2\x80\x94 Examples of Standard Designs of Nozzles. \n\n\n\n184 ENGINEERING THERMODYNAMICS \n\nsquare inches, and Pi is the initial absolute pressure of the steam \nin pounds per square inch, \n\nw=^^, (118) \n\n60 . ^ ^ \n\n\n\n\n\n\nor, in terms of the area, \n\n60 w \nPi- \n\nThese formulas are for the flow of steam initially dry and \nsaturated. An illustration of their applications is given by the \nfollowing practical example. \n\nExample. The area of the smallest section (^0) of a suitably \ndesigned nozzle is 0.54 square inch. What is the weight of the \nflow (w) of dry saturated steam per second from this nozzle when \nthe initial pressure (Pi) is 135 pounds per square inch absolute \nand the discharge pressure (P2) is 15 pounds per square inch ab- \nsolute? \n\nHere P2 is less than 0.58 Pi and Grashof \'s formula is applicable, \n\nor, w = ^^ ) ^^^ y \n\n60 \n\n0.54 X 116. 5* , , \n\nw = ^- = 1.049 pounds per second. \n\n60 \n\nWhen steam passes through a series of nozzles one after the \nother as is the case in many types of turbines, the pressure is \nreduced and the steam is condensed in each nozzle so that it \nbecomes wetter and wetter each time. In the low-pressure noz- \nzles of a turbine, therefore, the steam may be very wet although \ninitially it was dry. Turbines are also sometimes designed to \noperate with steam which is initially wet, and this is usually the \ncase when low-pressure steam turbines are operated with the \nexhaust from non-condensing reciprocating engines \xe2\x80\x94 a practice \nwhich is daily becoming \'more common. In all these cases the \nnozzle area must be corrected for the wetness of the steam. \n\n* The flow (w) calculated by Napier\'s formula for this example isw = \xe2\x80\x94 \n\n1. 041 pounds per second. \n\n\n\nFLOW OF FLUIDS 185 \n\nF\'or a given nozzle the weight discharged is, of course, greater \nfor wet steam than for dry; but the percentage increase in the \ndischarge is not nearly in proportion to the percentage of \nmoisture as is often stated. The general equation for the theo- \nretic discharge (w) from a nozzle is in the form * \n\n\n\nw = K \n\n\n\n^f \n\n\n\nwhere Pi is the initial absolute pressure and vi is the specific vol- \nume (cubic feet in a pound of steam at the pressure Pi). Now, \nneglecting the volume of the water in wet steam, which is a usual \napproximation, the volume of a pound of steam is proportional \n\n* The general equation for the theoretic flow is \n\nwhere the symbols w, Aq, Pi, and g are used as in equations (117) and (118). A \nis the pressure at any section of the nozzle, vi is the volume of a pound of steam at \nthe pressure Pi, and ^ is a constant. The flow, w, has its maximum value when \n\n2 k+J. \n\nik /p\xe2\x80\x9e\\ k \n\n\n\n(\xc2\xbb)-(\xc2\xab \n\n\n\nis a maximum. Differentiating and equating the first differential to zero gives \n\n_k \nP2 \nPi \n\nP2 is now the pressure at the smallest section, and writing for clearness Po for \nP2, and substituting this last equation in the formula for flow (w) above, we haxe \n\n\n\n{^r \n\n\n\n-VrJ^O (S) \n\n\n\nNow regardless of what the final pressure may be, the pressure (Po) at the smallest \nsection of a nozzle (Ao) is always nearly 0.58 Pi for dry saturated steam. Making \nthen in the last equation Po = 0.58 Pi \'and putting for k Zeuner\'s value of 1.135 \nfor dry saturated steam, we may write in general terms the form stated above, \n\nT Vi \n\nwhere K is another constant. See Zeuner\'s Theorie der Turbinen, page 268 (Ed. of \n1899). \n\n\n\nl86 ENGINEERING THERMODYNAMICS \n\nto the quality (xi). For wet steam the equation above becomes \nthen \n\n\n\nw = K ^^ \n\nxiz;i \n\n\n\n\n\n\nThe equation shows, therefore, that the flow of wet steam is \ninversely proportional to the square root of the quality (xi). \nGrashof s equations can be stated then more generally as \n\n. 60 W Vxi f . \n\n^0 = -^^1 (121) \n\nThese equations become, of course, the same as (118) and \n(119) for the case where 0:1 = 1. \n\nFlow of Steam when the Final Pressure is more than 0.58 of \nthe Initial Pressure. For this case the discharge depends upon \nthe final pressure as well as upon the initial. No satisfactory \nformula can be given in simple terms, and the flow is most easily \ncalculated with the aid of the curve in Fig. 68 due to Rateau. \nThis curve is used by determining first the ratio of the final to \n\nthe initial pressure \xe2\x80\x94 ^, and reading from the curve the corre- \nP\\ \n\nsponding coefficient showing the ratio of the required discharge \nto that calculated for the given conditions by either of the equa- \ntions (118) or (120). The coefficient from the curve times the \nflow calculated from equations (118) or (120) is the required re- \nsult. Obviously the discharge for this condition is always less \nthan the discharge when the final pressure is equal to or less \nthan 0.58 of the initial. \n\nLength for Nozzles. Probably the best designers make the \nlength of the nozzle depend only on the initial pressure. In \nother words, the length of\' a nozzle for 150 pounds per square \ninch initial pressure is usually made the same for a given type \nregardless of the final pressure. And if it happens that there is \ncrowding for space, one or more of the nozzles is sometimes \nmade a little shorter than the others. \n\n\n\nFLOW OF FLUIDS \n\n\n\n187 \n\n\n\nDesigners of De Laval nozzles follow practically the same \n^\' elastic " method. The divergence of the walls of non-con- \ndensing nozzles is about 3 degrees from the axis of the nozzle, \nand condensing nozzles for high vacuums may have a divergence \n\n\n\n1.0 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n: \n\n\n\n\n\n\n\n\n\n\n"- \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^ \n\n\n\xe2\x80\x94 \n\n\n\xe2\x96\xa0"" \n\n\noT \n\n\n\n\n\n\n\n\n\n\n\n\n.9 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n^ \n\n\n^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nQ. \n\n\n\n\n\n\n\n\n\n\n\n\n.8 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\ny \n\n\n^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n[/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n/\' \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nft S * \n\n\n\n\n\n\n\n\n/ \n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nII .4 \n11-3 \n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n/ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-2 \n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nI \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n- \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\'. \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n.0 \n\n\n\n\n\n\n\n\n\xe2\x80\xa2 \n\n\n\n\n\n\n\nE{ \n\n\niftc \n\n\nof \n\n\nFu \n\n\nlal \n\n\nto \n\n\nIni \n\n\ntia \n\n\n7 \nLP \n\n\nresj \n\n\n3ur \n\n\n\' P \n\n\n2 \n1 \n\n\n) \n\n\n\n\n\n\n\n\n" \n\n\ni \n\n\n\n\n\nFig. 68. \xe2\x80\x94 Coefficients of the Discharge of Steam when the Final Pressure is \nGreater than 0.58 of the Initial Pressure. \n\n\n\nof as much as 6 degrees * for the normal rated pressures of the \nturbine. \n\nThe author has used successfully the following empirical \nformula to determine. a suitable length, L, of the nozzle between \nthe throat and the mouth (in inches) : \n\n\n\nwhere ^0 is the area at the throat in square inches. \n\n\n\n(122) \n\n\n\n* According to Dr. O. Recke, if the total divergence of a nozzle is more than \n6 degrees, eddies will hegin to, form in the jet. There is no doubt that a too rapid \ndivergence produces a velocity loss. \n\nt Moyer\'s Steam Turbines (2d Edition), pages 45-48. \n\n\n\ni88 \n\n\n\nENGINEERING THERMODYNAMICS \n\n\n\nEfficiency of Nozzles. Recent experience with nozzles of this \ntype does not bear out this statement, except in the case prob- \nably of square or rectangular nozzles with no rounding at the \nedges. An efficiency of 97 per cent is not unusual for properly \ndesigned square and rectangular shaped nozzles without any \n*\' square " edges; and circular nozzles have certainly never \ngiven 99 per cent efficiency. \n\nUnder- and Over-expansion. The best efficiency of a nozzle is \nobtained when the expansion required is that for which the nozzle \nwas designed, or when the expansion ratio for the condition of \nthe steam corresponds with the ratio of the areas of the mouth \nand throat of the nozzle. A little under-expansion is far better, \n\n\n\n10 \n\n\n\n\n\n\n"9 b A \n\n\n\n25 20 15 10 5 \npercentage Nozzle is too Small\' \naX ^outli (Under Expansion) \n\n\n\n\\- HH -W \n\n\n/ \n\n\n/ \n\n\n-I- ^ V \n\n\n/ \n\n\n-\\- 4- \xe2\x96\xa0 TTZ \n\n\nX \n\n\nA \n\n\n4^ Z_ \n\n\n\n\n^^^ ^^ \n\n\n\n5 10 15 20 25 30 \n\nPercentage Nozzle is too Large \n\nat Mouth (Over Elzpansion) \n\n\n\nFig. 69. \xe2\x80\x94 Curve of Nozzle Velocity Loss. \n\nhowever, than the same amount of over-expansion, meaning \nthat a nozzle that is too small for the required expansion is more \nefficient than one that is correspondingly too large.* Fig. 69 \nshows a curve representing average values of nozzle loss used by \nvarious American and European manufacturers \\ to determine \n\n* It is a very good method, and one often adopted, to design nozzles so that \nat the rated capacity the nozzles under-expand at least 10 per cent, and maybe \n20 per cent. The loss for these conditions is insignificant, and the nozzles can be \nrun for a large overload (with increased pressures) in nearly all types without \nimmediately reducing the efficiency very much. \n\nt C. P. Steinmetz, Proc. Am. Soc. Mech. Engineers, May, 1908, page 628; J. \nA. Moyer, Steam Turbines, page 50. \n\n\n\nFLOW OF FLUIDS 189 \n\ndischarge velocities from nozzles under the conditions of under- \nor over-expansion. \n\nNon-expanding Nozzles. All tiie nozzles of Rateau turbines \nand usually also those of the low-pressure stages of Curtis tur- \nbines are made non-expanding; meaning, that they have the \nsame area at the throat as at the mouth. For such conditions \nit has been suggested that instead of a series of separate nozzles \nin a row a single long nozzle might be used of which the sides \nwere arcs of circles corresponding to the inside and outside pitch \ndiameters of the blades. Advantages would be secured both on \naccount of cheapness of construction and because a large amount \nof friction against the sides of nozzles would be eliminated by \nomitting a number of nozzle walls. Such a construction has not \nproved desirable, because by this method no well-formed jets \nare secured and the loss from eddies is excessive. The general \nstatement may be made that the throat of a well-designed nozzle \nshould have a nearly symmetrical shape, as for example a circle, \na square, etc., rather than such shapes as ellipses and long rec- \ntangles. The shape of the mouth is not important. In Curtis \nturbines an approximately rectangular mouth is used because \nthe nozzles are placed close together (usually in a nozzle plate \nlike Fig. 67) in order to produce a continuous band of steam; \nand, of course, by using a section that is rectangular rather than \ncircular or elliptical, a band of steam of more nearly uniform \nvelocity and density is secured. \n\nFig. 70 shows a number of designs of non-expanding nozzles \nused by Professor Rateau. The length of such nozzles beyond \nthe throat is practically negligible. Curtis non-expanding noz- \nzles are usually made the same length as if expanding and the \nlength is determined by the throat area. \n\nMaterials for Nozzles. Nozzles for saturated or slightly super- \nheated steam are usually made of bronze. Gun metal, zinc \nalloys, and delta metal are also frequently used. All these \nmetals have unusual resistance for erosion or corrosion from_ the \nuse of wet steam. Because of this property as well as for the \nreason that they are easily worked with hand tools * they are \n\n* Nozzles of irregular shapes are usually filed by hand to the exact size. \n\n\n\n1 90 ENGINEERING THERMODYNAMICS \n\nvery suitable materials for the manufacture of steam turbine \nnozzles. Superheated steam, however, rapidly erodes all these \nalloys and also greatly reduces the tensile strength. For nozzles \nto be used with highly superheated steam, cast iron is generally \nused, and except that it corrodes so readily is a very satisfactory \n\n\n\n\nNon-expanding Nozzles. \n\n\n\nmaterial. Commercial copper (about 98 per cent) is said to \nhave been used with a fair degree of success with high super- \nheats; but for such conditions its tensile strength is very low. \nSteel and cupro-nickel (8 Cu + 2 Ni) are also suitable materials, \nand the latter has the advantage of being practically non-cor- \nrodible. \n\nThe most important part of the design of a nozzle is the deter- \nmination of the areas of the various sections \xe2\x80\x94 especially the \nsmallest section, if the nozzle is of an expanding or diverging \ntype. In order to calculate the areas of nozzles we must know \nhow to determine the quantity of steam (flow) per unit of \ntime passing through a unit area. It is very essential that the \nnozzle is well rounded on the *\' entrance " side and that sharp \nedges along the path of the steam are avoided. Otherwise it \nis not important whether the shape of the section is circular, \nelliptical, or rectangular with rounded corners. \n\nWhether the nozzle section is throughout circular, square, or \nrectangular (if these last sections have rounded corners) the \nefficiency as measured by the velocity will be about 96 to 97 per \ncent, corresponding to an equivalent energy efficiency of 92 to 94 \nper cent. Speaking commercially, therefore, it does not seem to \n\n\n\nFLOW OF FLUIDS 191 \n\nbe worth while to spend a great deal of time in the shops to make \nnozzles very exactly to some difficult shape. Simpler and more \nrapid methods of nozzle construction should be introduced. In \nsome shops the time of one man for two days is required for the \nhand labor alone on a single nozzle. \n\n\n\nAPPENDIX \n\n\n\nNAPERIAN LOGARITHMS \n\nc =2.7182818 log e = 0.4342945 = M \n\n\n\n1.0 \n\n1.1 \n1.2 \n1.3 \n\n1.4 \n1.5 \n1.6 \n\n1.7 \n1.8 \n1.9 \n\n2.0 \n\n2.1 \n2.2 \n2.3 \n\n2.4 \n2.5 \n2.6 \n\n2.7 \n2.8 \n2.9 \n\n3.0 \n\n3.1 \n3.2 \n3.3 \n\n3.4 \n3.5 \n3.6 \n\n3.7 \n3.8 \n3.9 \n\n4.0 \n\n4.1 \n4.2 \n4.3 \n\n4.4 \n4.5 \n4.6 \n\n4.7 \n4.8 \n4.9 \n\n5.0 \n\n5.1 \n5.2 \n5.3 \n\n5.4 \n5.5 \n5.6 \n\n\n\n0. 0000 \n\n0.09531 \n\n0.1823 \n\n0.2624 \n\n0.3365 \n0.4055 \n0.4700 \n\n0.5306 \n0.5878 \n0.6418 \n\n0.6931 \n\n0.7419 \n0.7884 \n0.8329 \n\n0. 8755 \n0.9163 \n0.9555 \n\n0.9933 \n1.0296 \n1.0647 \n\n1.0986 \n\n1.1314 \n1.1632 \n1.1939 \n\n1.2238 \n1.2528 \n1.2809 \n\n1.3083 \n1.3350 \n1.3610 \n\n1.3863 \n\n1.4110 \n1.4351 \n1.4586 \n\n1.4816 \n1.5041 \n1.5261 \n\n1.5476 \n1.5686 \n1.5892 \n\n1.6094 \n\n1.6292 \n1.6487 \n1.6677 \n\n1.6864 \n1.7047 \n1.7228 \n\n\n\n0.00995 \n\n0. 1044 \n0. 1906 \n0.2700 \n\n0.3436 \n0.4121 \n0.4762 \n\n0.5365 \n0.5933 \n0.6471 \n\n0.6981 \n\n0.7467 \n0.7930 \n0.8372 \n\n0.8796 \n0.9203 \n0.9594 \n\n0.9969 \n1.0332 \n1.0682 \n\n\n\n1.1019 \n\n1.1346 \n1.1663 \n1.1969 \n\n1.2267 \n1.2556 \n1.2837 \n\n1.3110 \n1.3376 \n1.3635 \n\n1.3888 \n\n1.4134 \n1.4375 \n1.4609 \n\n1.4839 \n1.5063 \n1.5282 \n\n\n\n5497 \n5707 \n5913 \n\n\n\n1.6114 \n\n1.6312 \n\n1.6506 \n\n6696 \n\n\n\n6882 \n\n7066 \n\n1.7246 \n\n\n\n0.01980 \n\n0.1133 \n0.1988 \n0.2776 \n\n0.3507 \n0.4187 \n0. 4824 \n\n0.5423 \n0. 5988 \n0.6523 \n\n0.7031 \n\n0.7514 \n0.7975 \n0.8416 \n\n0.8838 \n0.9243 \n0.9632 \n\n1.0006 \n1.0367 \n1.0716 \n\n\n\n1.1053 \n\n1.1378 \n1.1694 \n1.2000 \n\n1.2296 \n1.2585 \n1.2865 \n\n1.3137 \n1.3403 \n1.3661 \n\n1.3913 \n\n1.4159 \n1.4398 \n1.4633 \n\n1.4861 \n1.5085 \n1.5304 \n\n1.5518 \n1.5728 \n1.5933 \n\n1.6134 \n\n1.6332 \n1.6525 \n1.6715 \n\n1.6901 \n\n1.7884 \n1.7263 \n\n\n\n0.02956 \n\n0. 1222 \n0.2070 \n0.2852 \n\n0.3577 \n0.4253 \n0. 4886 \n\n0.5481 \n0. 6043 \n0.6575 \n\n0. 7080 \n\n0.7561 \n0.8020 \n0. 8459 \n\n0.8879 \n0.9282 \n0.9670 \n\n1.0043 \n1.0403 \n1.0750 \n\n\n\n4 \n\n\n\n1.1086 \n\n\n\n1.1410 \n1.1725 \n1.2030 \n\n\n\n1.2326 \n1.2613 \n1.2892 \n\n1.3164 \n1.3429 \n1.3686 \n\n1.3938 \n\n1.4183 \n1.4422 \n1.4656 \n\n1.4884 \n1.5107 \n1.5326 \n\n1.5539 \n1.5748 \n1.5953 \n\n1.6154 \n\n1.6351 \n1.6544 \n1.6734 \n\nL6919 \n1.7102 \n1.7281 \n\n\n\n0.03922 \n\n0.1310 \n0.2151 \n0.2927 \n\n0.3646 \n0.4318 \n0.4947 \n\n0.5539 \n0.6098 \n0.6627 \n\n0.7129 \n\n0. 7608 \n0. 8065 \n0. 8502 \n\n0.8920 \n0.9322 \n0.9708 \n\n1.0080 \n1.0438 \n1.0784 \n\n1.1119 \n\n1.1442 \n1.1756 \n1.2060 \n\n1.2355 \n1.2641 \n1.2920 \n\n1.3191 \n1.3455 \n1.3712 \n\n1.3962 \n\n1.4207 \n1.4446 \n1.4679 \n\n\n\n1.4907 \n1.5129 \n1.5347 \n\n1.5560 \n1.5769 \n1.5974 \n\n1.6174 \n\n1.6371 \n1.6563 \n1.6752 \n\n1.6938 \n1.7120 \n1.7299 \n\n\n\n6 \n\n\n\n0.04879 0.05827 0.06766 \n\n\n\n0.1398 \n0.2231 \n0.3001 \n\n0.3716 \n0.4382 \n0.5008 \n\n0.5596 \n0.6152 \n0. 6678 \n\n0.7178 \n\n0.7655 \n0.8109 \n0.8544 \n\n0. 8961 \n0.9361 \n0.9746 \n\n1.0116 \n1.0473 \n1.0818 \n\n1.1151 \n\n1.1474 \n1.1787 \n1.2090 \n\n1.2384 \n1.2669 \n1.2947 \n\n1.3218 \n1.3481 \n1.3737 \n\n1.3987 \n\n1.4231 \n1.4469 \n1.4702 \n\n\n\n1.4929 \n1.5151 \n1.5369 \n\n1.5581 \n1.5790 \n1.5994 \n\n1.6194 \n\n1.8390 \n1.6582 \n1.6771 \n\n1.6956 \n1.7138 \n1.7317 \n\n\n\n0. 1484 \n0.2311 \n0.3075 \n\n0.3784 \n0.4447 \n0.5068 \n\n0. 5653 \n0.6206 \n0.6729 \n\n0.7227 \n\n0.7701 \n0.8154 \n0. 8587 \n\n0.9002 \n0.9400 \n0.9783 \n\n1.0152 \n1.0508 \n1.0852 \n\n1.1184 \n\n1.1506 \n1.1817 \n1.2119 \n\n1.2413 \n1.2698 \n1.2975 \n\n1.3244 \n1.3507 \n1.3762 \n\n1.4012 \n\n1.4255 \n1.4493 \n1.4725 \n\n1.4951 \n\n.5173 \n\n1.5390 \n\n1.5602 \n1.5810 \n1.6014 \n\n1.6214 \n\n6409 \n6601 \n6790 \n\n6974 \n7156 \n7334 \n\n\n\n0. 07696 \n\n\n\n0.1570 0.1655 \n0.2390 0.2469 \n0.3148 0.3221 \n\n\n\n0.3853 \n0.4511 \n0.5128 \n\n0.5710 \n0.6259 \n0. 6780 \n\n0.7275 \n\n0. 7747 \n0.8198 \n0. 8629 \n\n0.9042 \n0.9439 \n0.9821 \n\n1.0188 \n1.0543 \n1.0886 \n\n1.1217 \n\n1.1537 \n1.1848 \n1.2149 \n\n1.2442 \n1.2726 \n1.3002 \n\n1.3271 \n1.3533 \n1.3788 \n\n1.4036 \n\n1.4279 \n1.4516 \n1.4748 \n\n1.4974 \n1.5195 \n1.5412 \n\n1.5623 \n1.5831 \n1.6034 \n\n1.6233 \n\n1.6429 \n\n6620 \n\n1.6808 \n\n\n\n6993 \n7174 \n7352 \n\n\n\n0.3920 \n0.4574 \n0.5188 \n\n0. 5766 \n0.6313 \n0.6831 \n\n0. 7324 \n\n0.7793 \n0. 8242 \n0.8671 \n\n0.9083 \n0.9478 \n0.9858 \n\n1.0225 \n1.0578 \n1.0919 \n\n1.1249 \n\n1.1569 \n1.1878 \n1.2179 \n\n1.2470 \n1.2754 \n1.3029 \n\n1.3297 \n1.3558 \n1.3813 \n\n1.4061 \n\n1.4303 \n1.4540 \n1.4770 \n\n1.4996 \n1.5217 \n1.5433 \n\n1.5644 \n1.5851 \n1.6054 \n\n1.6253 \n\n1.6448 \n1.6639 \n1.6827 \n\n1.7011 \n1.7192 \n1.7370 \n\n\n\n0.08618 \n\n0.1739 \n0.2546 \n0.3293 \n\n0.3988 \n0.4637 \n0.5247 \n\n0.5822 \n0.6366 \n0.6881 \n\n0.7372 \n\n0. 7839 \n0.8286 \n0.8713 \n\n0.9123 \n0.9517 \n0.9895 \n\n1.0260 \n1.0613 \n1.0953 \n\n1.1282 \n\n1.1600 \n1.1909 \n1.2208 \n\n1.2499 \n1.2782 \n1.3056 \n\n1.3324 \n\n3584 \n\n1.3838 \n\n1.4085 \n\n1.4327 \n\n1.4563 \n\n4793 \n\n1.5019 \n1.5239 \n1.5454 \n\n1.5665 \n\n1.5872 \n1.6074 \n\n1.6273 \n\n\n\n6467 \n\n6658 \n\n1.6845 \n\n\n\n7029 \n7210 \n7387 \n\n\n\n(195) \n\n\n\nNAPERIAN LOGARITHMS. \n\n\n\n\n\n\n\n\n1 \n\n\n2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n9 \n\n\n6.7 \n5.8 \n6.9 \n\n\n1.7405 \n1.7579 \n1.7750 \n\n\n1.7422 \n1.7596 \n1.7766 \n\n\n1.7440 \n1.7613 \n1.7783 \n\n\n1.7457 \n1.7630 \n1.7800 \n\n\n1.7475 \n1.7647 \n1.7817 \n\n\n1.7492 \n1.7664 \n1.7834 \n\n\n1.7509 \n1.7681 \n1.7851 \n\n\n1.7527 \n1.7699 \n1.7867 \n\n\n1.7544 \n1.7716 \n1.7884 \n\n\n1.7561 \n1.7733 \nL7901 \n\n\n6.0 \n\n\n1.7918 \n\n\n1.7934 \n\n\n1.7951 \n\n\n1.7967 \n\n\n1.7984 \n\n\n1.8001 \n\n\n1.8017 \n\n\n1.8034 \n\n\n1.8050 \n\n\n1.8066 \n\n\n6.1 \n6.2 \n6.3 \n\n\n1.8083 \n1.8245 \n1.8405 \n\n\n1.8099 \n1.8262 \n1.8421 \n\n\n1.8116 \n\n1.8278 \n1.8437 \n\n\n1.8132 \n1.8294 \n1.8453 \n\n\n1.8148 \n1.8310 \n1.8469 \n\n\n1.8165 \n1.8326 \n1.8485 \n\n\n1.8181 \n1.8342 \n1.8500 \n\n\n1.8197 \n1.8358 \n1.8516 \n\n\n1.8213 \n1.8374 \n1.8532 \n\n\n1.8229 \n1.8390 \n1.8547 \n\n\n6.4 \n6.5 \n6.6 \n\n\n1.8563 \n1.8718 \n1.8871 \n\n\n1.8579 \n1.8733 \n1.8886 \n\n\n1.8594 \n1.8749 \n1.8901 \n\n\n1.8610 \n\n1.8764 \n1.8916 \n\n\n1.8625 \n1.8779 \n1.8931 \n\n\n1.8641 \n1.8795 \n1.8946 \n\n\n1.8656 \n1.8810 \n1.8961 \n\n\n1.8672 \n1.8825 \n1.8976 \n\n\n1.8687 \n1.8840 \n1.8991 \n\n\n1.8703 \n1.8856 \n1.9006 \n\n\n6.7 \n6.8 \n6.9 \n\n\n1.9021 \n1.9169 \n1.9315 \n\n\n1.9036 \n1.9184 \n1.9330 \n\n\n1.9051 \n1.9199 \n1.9344 \n\n\n1.9066 \n1.9213 \n1.9359 \n\n\n1.9081 \n1.9228 \n1.9373 \n\n\n1.9095 \n1.9242 \n1.9387 \n\n\n1.9110 \n1.9257 \n1.9402 \n\n\n1.9125 \n1.9272 \n1.9416 \n\n\n1.9140 \n1.9286 \n1.9430 \n\n\n1.9155 \n1.9301 \n1.9445 \n\n\n7.0 \n\n\n1.&459 \n\n\n1.9473 \n\n\n1.9488 \n\n\n1.9502 \n\n\n1.9516 \n\n\n1.9530 \n\n\n1.9544 \n\n\n1.9559 \n\n\n1.9573 \n\n\n1.9587 \n\n\n7.1 \n7.2 \n7.3 \n\n\n1.9601 \n1.9741 \n1.9879 \n\n\n1.9615 \n1.9755 \n1.9892 \n\n\n1.9629 \n1.9769 \n1.9906 \n\n\n1.9643 \n1.9782 \n1.9920 \n\n\n1.9657 \n1.9796 \n1.9933 \n\n\n1.9671 \n1.9810 \n1.9947 \n\n\n1.9685 \n1.9824 \n1.9961 \n\n\n1.9699 \n1.9838 \n1.9974 \n\n\n1.9713 \n1.9851 \n1.9988 \n\n\n1.9727 \n1.9865 \n2.0001 \n\n\n7.4 \n7.5 \n7.6 \n\n\n2.0015 \n2.0149 \n2.0281 \n\n\n2.0028 \n2.0162 \n2.0295 \n\n\n2.0042 \n2.0176 \n2.0308 \n\n\n2. 0055 \n2.0189 \n2.0321 \n\n\n2.0069 \n2.0202 \n2.0334 \n\n\n2. 0082 \n2.0215 \n2.0347 \n\n\n2.0096 \n2. 0229 \n2.0360 \n\n\n2.0109 \n2.0242 \n2.0373 \n\n\n2.0122 \n2.0255 \n2.0386 \n\n\n2.0136 \n2.0268 \n2.0399 \n\n\n7.7 \n7.8 \n7.9 \n\n\n2.0412 \n2.0541 \n2.0668 \n\n\n2.0425 \n2.0554 \n2.0681 \n\n\n2.0438 \n2.0567 \n2.0694 \n\n\n2.0451 \n2.0580 \n2.0707 \n\n\n2.0464 \n2.0592 \n2.0719 \n\n\n2.0477 \n2.0605 \n2.0732 \n\n\n2.0490 \n2.0618 \n2.0744 \n\n\n2.0503 \n2.0631 \n2.0757 \n\n\n2.0516 \n2.0643 \n2.0769 \n\n\n2.0528 \n2.0656 \n2. 0782 \n\n\n8.0 \n\n\n2.0794 \n\n\n2.0807 \n\n\n2.0819 \n\n\n2.0832 \n\n\n2.0844 \n\n\n2.0857 \n\n\n2.0869 \n\n\n2.0881 \n\n\n2.0894 \n\n\n2.0906 \n\n\n8.1 \n8.2 \n8.3 \n\n\n2.0919 \n2.1041 \n2.1163 \n\n\n2.0931 \n2.1054 \n2.1175 \n\n\n2.0943 \n2.1066 \n2.1187 \n\n\n2.0956 \n2.1078 \n2.1199 \n\n\n2.0968 \n2.1090 \n2.1211 \n\n\n2.0980 \n2.1102 \n2.1223 \n\n\n2.0992 \n2.1114 \n2.1235 \n\n\n2. 1005 \n2.1126 \n2.1247 \n\n\n2.1017 \n2.1138 \n2.1258 \n\n\n2. 1029 \n2.1150 \n2. 1270 \n\n\n8.4 \n8.5 \n8.6 \n\n\n2. 1282 \n2.1401 \n2.1518 \n\n\n2. 1294 \n2.1412 \n2. 1529 \n\n\n2.1306 \n2. 1424 \n2.1541 \n\n\n2.1318 \n2.1436 \n2. 1552 \n\n\n2.1330 \n2. 1448 \n2. 1564 \n\n\n2.1342 \n2.1459 \n2.1576 \n\n\n2.1353 \n2.1471 \n2.1587 \n\n\n2.1365 \n2.1483 \n2.1599 \n\n\n2.1377 \n2.1494 \n2.1610 \n\n\n2. 1389 \n2.1506 \n2.1622 \n\n\n8.7 \n8.8 \n8.9 \n\n\n2.1633 \n2. 1748 \n2. 1861 \n\n\n2. 1645 \n2.1759 \n2.1872 \n\n\n2.1656 \n2.1770 \n2. 1883 \n\n\n2.1668 \n2.1782 \n2. 1894 \n\n\n2. 1679 \n2.1793 \n2. 1905 \n\n\n2.1691 \n\n2.1804 \n2.1917 \n\n\n2. 1702 \n\n2.1815 \n2.1928 \n\n\n2.1713 \n2.1827 \n2.1939 \n\n\n2.1725 \n2.1838 \n2. 1950 \n\n\n2.1736 \n2. 1849 \n2.1961 \n\n\n9.0 \n\n\n2. 1972 \n\n\n2.1983 \n\n\n2. 1994 \n\n\n2.2006 \n\n\n2.2017 \n\n\n2. 2028 \n\n\n2. 2039 \n\n\n2.2050 \n\n\n2.2061 \n\n\n2.2072 \n\n\n9.1 \n9.2 \n9.3 \n\n\n2.2083 \n2.2192 \n2. 2300 \n\n\n2.2094 \n2.2203 \n2. 2311 \n\n\n2.2105 \n\n2.2214 \n2. 2322 \n\n\n2.2116 \n2.2225 \n2.2332 \n\n\n2.2127 \n2.2235 \n2. 2343 \n\n\n2.2138 \n2.2246 \n2.2354 \n\n\n2.2148 \n2.2257 \n2.2364 \n\n\n2.2159 \n2.2268 \n2.2375 \n\n\n2.2170 \n2.2279 \n2. 2386 \n\n\n2.2181 \n2. 2289 \n2.2396 \n\n\n9.4 \n9.5 \n9.6 \n\n\n2. 2407 \n2.2513 \n2.2618 \n\n\n2.2418 \n2.2523 \n2.2628 \n\n\n2.2428 \n2.2534 \n2.2638 \n\n\n2.2439- \n\n2.2544 \n\n2.2649 \n\n\n2. 2450 \n2.2555 \n2.2659 \n\n\n2.2460 \n2.2565 \n2.2670 \n\n\n2.2471 \n2.2576 \n2.2680 \n\n\n2.2481 \n2.2586 \n2.2690 \n\n\n2.2492 \n2.2597 \n2.2701 \n\n\n2.2502 \n2.2607 \n2.2711 \n\n\n9.7 \n9.8 \n9.9 \n\n\n2.2721 \n2.2824 \n2. 2925 \n\n\n2.2732 \n2.2834 \n2.2935 \n\n\n2. 2742 \n2. 2844 \n2.2946 \n\n\n2.2752 \n2.2854 \n2.2956 \n\n\n2.2762 \n2.2865 \n2.2966 \n\n\n2.2773 \n\n2.2875 \n2.2976 \n\n\n2.2783 \n2.2885 \n2.2986 \n\n\n2.2793 \n2.2895 \n2.2996 \n\n\n2.2803 \n2.2905 \n2.3006 \n\n\n2.2814 \n2.2915 \n2.3016 \n\n\n10.0 \n\n\n2.3026 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(196) \n\n\n\nLOGARITHMS. \n\n\n\nNat. \nNos. \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nProportion \n\n\na Parts. 1 \n\n\n\n\n\n1 \n\n\n2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n9 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n10 \n\n\n0000 \n\n\n0043 \n\n\n0086 \n\n\n0128 \n\n\n0170 \n\n\n0212 \n\n\n0253 \n\n\n0294 \n\n\n0334 \n\n\n0374 \n\n\n4 8 \n\n\n12 \n\n\n17 21 \n\n\n25 \n\n\n29 \n\n\n33 37 \n\n\n11 \n\n\n0414 \n\n\n0453 \n\n\n0492 \n\n\n0531 \n\n\n0569 \n\n\n0607 \n\n\n0645 \n\n\n0682 \n\n\n0719 \n\n\n0755 \n\n\n4 8 \n\n\n11 \n\n\n15 \n\n\n19 \n\n\n23 \n\n\n26 \n\n\n30 34 \n\n\n12 \n\n\n0792 \n\n\n0828 \n\n\n0864 \n\n\n0899 \n\n\n0934 \n\n\n0969 \n\n\n1004 \n\n\n1038 \n\n\n1072 \n\n\n1106 \n\n\n3 7 \n\n\n10 \n\n\n14 \n\n\n17 21 \n\n\n24 \n\n\n28 31 \n\n\n13 \n\n\n1139 \n\n\n1173 \n\n\n1206 \n\n\n1239 \n\n\n1271 \n\n\n1303 \n\n\n1335 \n\n\n1367 \n\n\n1399 \n\n\n1430 \n\n\n3 6 \n\n\n10 \n\n\n13 \n\n\n16 \n\n\n19 \n\n\n23 \n\n\n26 29 \n\n\n14 \n\n\n1461 \n\n\n1492 \n\n\n1523 \n\n\n1553 \n\n\n1584 \n\n\n1614 \n\n\n1644 \n\n\n1673 \n\n\n1703 \n\n\n1732 \n\n\n3 6 \n\n\n9 \n\n\n12 \n\n\n15 \n\n\n18 \n\n\n21 \n\n\n24 27 \n\n\n15 \n\n\n1761 \n\n\n1790 \n\n\n1818 \n\n\n1847 \n\n\n1875 \n\n\n1903 \n\n\n1931 \n\n\n1959 \n\n\n1987 \n\n\n2014 \n\n\n3 6 \n\n\n8 \n\n\n11 \n\n\n14 \n\n\n17 \n\n\n20 \n\n\n22 25 \n\n\n16 \n\n\n2041 \n\n\n2068 \n\n\n2095 \n\n\n2122 \n\n\n2148 \n\n\n2175 \n\n\n2201 \n\n\n2227 \n\n\n2253 \n\n\n2279 \n\n\n3 5 \n\n\n8 \n\n\n11 \n\n\n13 \n\n\n16 \n\n\n18 \n\n\n21 24 \n\n\n17 \n\n\n2304 \n\n\n2330 \n\n\n2355 \n\n\n2380 \n\n\n2405 \n\n\n2430 \n\n\n2455 \n\n\n2480 \n\n\n2504 \n\n\n2529 \n\n\n2 5 \n\n\n7 \n\n\n10 \n\n\n12 \n\n\n15 \n\n\n17 \n\n\n20 22 \n\n\n18 \n\n\n2553 \n\n\n2577 \n\n\n2601 \n\n\n\'>625 \n\n\n2648 \n\n\n2672 \n\n\n2695 \n\n\n2718 \n\n\n2742 \n\n\n2765 \n\n\n2 5 \n\n\n7 \n\n\n9 \n\n\n12 \n\n\n14 \n\n\n16 \n\n\n19 21 \n\n\n19 \n\n\n2788 \n\n\n2810 \n\n\n2833 \n\n\n2856 \n\n\n2878 \n\n\n2900 \n\n\n2923 \n\n\n2945 \n\n\n2967 \n\n\n2989 \n\n\n2 4 \n\n\n7 \n\n\n9 \n\n\n11 \n\n\n13 \n\n\n16 \n\n\n18 20 \n\n\n20 \n\n\n3010 \n\n\n3032 \n\n\n3054 \n\n\n3075 \n\n\n3096 \n\n\n\n\n3118 \n\n\n3139 \n\n\n3160 \n\n\n3181 \n\n\n3201 \n\n\n2 4 \n\n\n6 \n\n\n8 \n\n\n11 \n\n\n13 \n\n\n15 \n\n\n17 19 \n\n\n21 \n\n\n3222 \n\n\n3243 \n\n\n3263 \n\n\n3284 \n\n\n3304 \n\n\n3324 \n\n\n3345 \n\n\n3365 \n\n\n3385 \n\n\n3404 \n\n\n2 4 \n\n\n6 \n\n\n8 \n\n\n10 \n\n\n12 \n\n\n14 \n\n\n16 18 \n\n\n22 \n\n\n3424 \n\n\n3444 \n\n\n3464 \n\n\n3483 \n\n\n3502 \n\n\n3522 \n\n\n3541 \n\n\n3560 \n\n\n3579 \n\n\n3598 \n\n\n2 4 \n\n\n6 \n\n\n8 \n\n\n10 \n\n\n12 \n\n\n14 \n\n\n15 17 \n\n\n23 \n\n\n3617 \n\n\n3636 \n\n\n3655 \n\n\n3674 \n\n\n3692 \n\n\n3711 \n\n\n3729 \n\n\n3747 \n\n\n3766 \n\n\n3784 \n\n\n2 4 \n\n\n6 \n\n\n7 \n\n\n9 \n\n\n11 \n\n\n13 \n\n\n15 17 \n\n\n24 \n\n\n3802 \n\n\n3820 \n\n\n3838 \n\n\n3856 \n\n\n3874 \n\n\n3892 \n\n\n3909 \n\n\n3927 \n\n\n3945 \n\n\n3962 \n\n\n2 4 \n\n\n5 \n\n\n7 \n\n\n9 \n\n\n11 \n\n\n12 \n\n\n14 16 \n\n\n25 \n\n\n3979 \n\n\n3997 \n\n\n4014 \n\n\n4031 \n\n\n4048 \n\n\n4065 \n\n\n4082 \n\n\n4099 \n\n\n4116 \n\n\n4133 \n\n\n2 3 \n\n\n5 \n\n\n7 \n\n\n9 \n\n\n10 \n\n\n12 \n\n\n14 15 \n\n\n26 \n\n\n4150 \n\n\n4166 4183 \n\n\n4200 \n\n\n4216 \n\n\n4232 \n\n\n4249 \n\n\n4265 \n\n\n4281 \n\n\n4298 \n\n\n2 3 \n\n\n5 \n\n\n7 \n\n\n8 \n\n\n10 \n\n\n11 \n\n\n13 15 \n\n\n27 \n\n\n4314 \n\n\n4330 \n\n\n4346 \n\n\n4362 \n\n\n4378 \n\n\n4393 \n\n\n4409 \n\n\n4425 \n\n\n4440 \n\n\n4456 \n\n\n2 3 \n\n\n5 \n\n\n6 \n\n\n8 \n\n\n9 \n\n\n11 \n\n\n13 14 \n\n\n28 \n\n\n4472 \n\n\n4487 \n\n\n4502 \n\n\n4518 \n\n\n4533 \n\n\n4548 \n\n\n4564 \n\n\n4579 \n\n\n4594 \n\n\n4609 \n\n\n2 3 \n\n\n5 \n\n\n6 \n\n\n8 \n\n\n9 \n\n\n11 \n\n\n12 14 \n\n\n29 \n\n\n4624 \n\n\n4639 \n\n\n4654 \n\n\n4669 \n\n\n4683 \n\n\n4698 \n\n\n4713 \n\n\n4728 \n\n\n4742 \n\n\n4757 \n\n\n1 3 \n\n\n4 \n\n\n6 \n\n\n7 \n\n\n9 \n\n\n10 \n\n\n12 13 \n\n\n30 \n\n\n4771 \n\n\n4786 \n\n\n4800 \n\n\n4814 \n\n\n4829 \n\n\n4843 \n\n\n4857 \n\n\n4871 \n\n\n4886 \n\n\n4900 \n\n\n1 3 \n\n\n4 \n\n\n6 \n\n\n7 \n\n\n9 \n\n\n10 \n\n\n11 13 \n\n\n31 \n\n\n4914 \n\n\n4928 \n\n\n4942 \n\n\n4955 \n\n\n4969 \n\n\n4983 \n\n\n4997 \n\n\n5011 \n\n\n5024 \n\n\n5038 \n\n\n1 3 \n\n\n4 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n10 \n\n\n11 12 \n\n\n32 \n\n\n5051 \n\n\n5065 \n\n\n5079 \n\n\n5092 \n\n\n5105 \n\n\n5119 \n\n\n5132 \n\n\n5145 \n\n\n5159 \n\n\n5172 \n\n\n1 3 \n\n\n4 \n\n\n5 \n\n\n7 \n\n\n8 \n\n\n9 \n\n\n11 12 \n\n\n33 \n\n\n5185 \n\n\n5198 \n\n\n5211 \n\n\n5224 \n\n\n5237 \n\n\n5250 \n\n\n5263 \n\n\n5276 \n\n\n5289 \n\n\n5302 \n\n\n1 3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n8 \n\n\n9 \n\n\n10 12 \n\n\n34 \n\n\n5315 \n\n\n5328 \n\n\n5340 \n\n\n5353 \n\n\n5366 \n\n\n5378 \n\n\n5391 \n\n\n5403 \n\n\n5416 \n\n\n5428 \n\n\n1 3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n8 \n\n\n9 \n\n\n10 11 \n\n\n35 \n\n\n5441 \n\n\n5453 \n\n\n5465 \n\n\n5478 \n\n\n5490 \n\n\n5502 \n\n\n5514 \n\n\n5527 \n\n\n5539 \n\n\n5551 \n\n\n1 2 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n9 \n\n\n10 11 \n\n\n36 \n\n\n5563 \n\n\n5575 \n\n\n5587 \n\n\n5599 \n\n\n5611 \n\n\n5623 \n\n\n5635 \n\n\n5647 \n\n\n5658 \n\n\n5670 \n\n\n1 2 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n10 11 \n\n\n37 \n\n\n5682 \n\n\n5694 \n\n\n5705 \n\n\n5717 \n\n\n5729 \n\n\n5740 \n\n\n5752 \n\n\n5763 \n\n\n5775 \n\n\n5786 \n\n\n1 2 \n\n\n3 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n9 10 \n\n\n38 \n\n\n5798 \n\n\n5809 \n\n\n5821 \n\n\n5832 \n\n\n5843 \n\n\n5855 \n\n\n5866 \n\n\n5877 \n\n\n5888 \n\n\n5899 \n\n\n1 2 \n\n\n3 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 \n\n\n9 10 \n\n\n39 \n\n\n5911 \n\n\n5922 \n\n\n5933 \n\n\n5944 \n\n\n5955 \n\n\n5966 \n\n\n5977 \n\n\n5988 \n\n\n5999 \n\n\n6010 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n7 \n\n\n8 \n\n\n9 10 \n\n\n40 \n\n\n6021 \n\n\n6031 \n\n\n6042 \n\n\n6053 \n\n\n6064 \n\n\n6075 \n\n\n6085 \n\n\n6096 \n\n\n6107 \n\n\n6117 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n8 \n\n\n9 10 \n\n\n41 \n\n\n6128 \n\n\n6138 \n\n\n6149 \n\n\n6160 \n\n\n6170 \n\n\n6180 \n\n\n6191 \n\n\n6201 \n\n\n6212 \n\n\n6222 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n42 \n\n\n6232 \n\n\n6243 \n\n\n6253 \n\n\n6263 \n\n\n6274 \n\n\n6284 \n\n\n6294 \n\n\n6304 \n\n\n6314 \n\n\n6325 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n43 \n\n\n6335 \n\n\n6345 \n\n\n6355 \n\n\n6365 \n\n\n6375 \n\n\n6385 \n\n\n6395 \n\n\n6405 \n\n\n6415 \n\n\n6425 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n44 \n\n\n6435 \n\n\n6444 \n\n\n6454 \n\n\n6464 \n\n\n6474 \n\n\n6484 \n\n\n6493 \n\n\n6503 \n\n\n6513 \n\n\n6522 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n45 \n\n\n6532 \n\n\n6542 \n\n\n6551 \n\n\n6561 \n\n\n6571 \n\n\n6580 \n\n\n6590 \n\n\n6599 \n\n\n6609 \n\n\n6618 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n8 9 \n\n\n46 \n\n\n6628 \n\n\n6637 \n\n\n6646 \n\n\n6656 \n\n\n6665 \n\n\n6675 \n\n\n6684 \n\n\n6693 \n\n\n6702 \n\n\n6712 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\n7 8 \n\n\n47 \n\n\n6721 \n\n\n6730 \n\n\n6739 \n\n\n6749 \n\n\n6758 \n\n\n6767 \n\n\n6776 \n\n\n6785 \n\n\n6794 \n\n\n6803 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n5 \n\n\n6 \n\n\n7 8 \n\n\n48 \n\n\n6812 \n\n\n6821 \n\n\n6830 \n\n\n6839 \n\n\n6848 \n\n\n6857 \n\n\n6866 \n\n\n6875 \n\n\n6884 \n\n\n6893 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n\n\n5 \n\n\n6 \n\n\n7 8 \n\n\n49 \n\n\n6902 \n\n\n6911 \n\n\n6920 \n\n\n6928 \n\n\n6937 \n\n\n6946 \n\n\n6955 \n\n\n6964 \n\n\n6972 \n\n\n6981 \n\n\n1 2 \n\n\n3 \n\n\n4 \n\n\n\n\n5 \n\n\n6 \n\n\n7 8 \n\n\n50 \n\n\n6990 \n\n\n6998 \n\n\n7007 \n\n\n7016 \n\n\n7024 \n\n\n7033 \n\n\n7042 \n\n\n7050 \n\n\n7059 \n\n\n7067 \n\n\n1 2 \n\n\n3 \n\n\n3 \n\n\n\n\n5 \n\n\n6 \n\n\n7 8 \n\n\n51 \n\n\n7076 \n\n\n7084 \n\n\n7093 \n\n\n7101 \n\n\n7U0 \n\n\n7118 \n\n\n7126 \n\n\n7135 \n\n\n7143 \n\n\n7152 \n\n\n1 2 \n\n\n3 \n\n\n3 \n\n\n\n\n5 \n\n\n6 \n\n\n7 8 \n\n\n52 \n\n\n7160 \n\n\n7168 \n\n\n7177 7185 1 \n\n\n7193 \n\n\n7202 \n\n\n7210 \n\n\n7218 \n\n\n7226 \n\n\n7235 \n\n\n1 2 \n\n\n2 \n\n\n3 \n\n\n\n\n5 \n\n\n6 \n\n\n7 7 \n\n\n53 \n\n\n7243 \n\n\n7251 \n\n\n7259 \n\n\n7267 \n\n\n7275 \n\n\n7284 \n\n\n7292 \n\n\n7300 \n\n\n7308 \n\n\n7316 \n\n\n1 2 \n\n\n2 \n\n\n3 \n\n\n\n\n5 \n\n\n6 \n\n\n6 7 \n\n\n54 \n\n\n7324 \n\n\n7332 \n\n\n7340 \n\n\n7348 \n\n\n7356 \n\n\n7364 \n\n\n7372 \n\n\n7380 \n\n\n7388 \n\n\n7396 \n\n\n1 2 \n\n\n^ \n\n\n3 \n\n\n\n\n5 \n\n\n6 \n\n\n6 7 \n\n\n\n(197) \n\n\n\nLOGARITHMS. \n\n\n\nNat. \nNos. \n\n\n\n\n\n1 \n\n\n2 \n\n\n3 \n\n\n4 \n\n\n5 \n\n\n6 \n\n\n7 \n\n\nQ \n\n\n9 \n\n\nProportional Parts. \n\n\n12 3 \n\n\n4 5 6 \n\n\n7 8 9 \n\n\n55 \n56 \n57 \n58 \n59 \n\n60 \n\n61 \n62 \n63 \n64 \n\n\n7404 \n\n7482 \n7559 \n7634 \n7709 \n\n7782 \n7853 \n7924 \n7993 \n8062 \n\n\n7412 \n7490 \n7566 \n7642 \n7716 \n\n7789 \n7860 \n7931 \n8000 \n8069 \n\n\n7419 \n7497 \n7574 \n7649 \n7723 \n\n7796 \n7868 \n7938 \n8007 \n8075 \n\n\n7427 \n7505 \n7582 \n7657 \n7731 \n\n7803 \n\n7875 \n7945 \n8014 \n8082 \n\n\n7435 \n7513 \n7589 \n7664 \n7738 \n\n7810 \n7882 \n7952 \n8021 \n8089 \n\n\n7443 \n7520 \n7597 \n7672 \n7745 \n\n7818 \n7889 \n7959 \n8028 \n8096 \n\n\n7451 \n\n7528 \n7604 \n7679 \n7752 \n\n7825 \n7896 \n7966 \n8035 \n8102 \n\n\n7459 \n7536 \n7612 \n7686 \n7760 \n\n7832 \n7903 \n7973 \n8041 \n8109 \n\n\n7466 \n7543 \n7619 \n7694 \n7767 \n\n7839 \n7910 \n7980 \n8048 \n8116 \n\n\n7474 \n7551 \n7627 \n7701 \n\n7774 \n\n7846 \n7917 \n7987 \n8055 \n8122 \n\n\n1 2 2 \n1 2 2 \n1 2 2 \n1 1 2 \n1 1 2 \n\n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n\n\n3 4 5 \n3 4 5 \n3 4 5 \n3 4 4 \n3 4 4 \n\n3 4 4 \n3 4 4 \n3 3 4 \n3 3 4 \n3 3 4 \n\n\n5 6 7 \n5 6 7 \n5 6 7 \n5 6 7 \n5 6 7 \n\n5 6 6 \n5 6 6 \n5 6 6 \n5 5 6 \n5 5 6 \n\n\n65 \n66 \n67 \n68 \n69 \n\n70 \n71 \n72 \n73 \n\n74 \n\n\n8129 \n8195 \n8261 \n8325 \n8388 \n\n8451 \n8513 \n8573 \n8G33 \n8692 \n\n\n8136 \n8202 \n8267 \n8331 \n8395 \n\n8457 \n8519 \n8579 \n8639 \n8698 \n\n\n8142 \n8209 \n8274 \n8338 \n8401 \n\n8463 \n8525 \n8585 \n8645 \n8704 \n\n\n8149 \n8215 \n8280 \n8344 \n8407 \n\n8470 \n8531 \n8591 \n8651 \n8710 \n\n\n8156 \n8222 \n8287 \n8351 \n8414 \n\n8476 \n8537 \n8597 \n8657 \n8716 \n\n\n8162 \n\n8228 \n8293 \n8357 \n8420 \n\n8482 \n8543 \n8603 \n8663 \n8722 \n\n\n8169 \n8235 \n8299 \n8363 \n8426 \n\n8488 \n8549 \n8609 \n8669 \n\n8727 \n\n\n8176 \n8241 \n8306 \n8370 \n8432 \n\n8494 \n8555 \n8615 \n8675 \n8733 \n\n\n8182 \n8248 \n8312 \n8376 \n8439 \n\n8500 \n8561 \n8621 \n8681 \n8739 \n\n\n8189 \n8254 \n8319 \n8382 \n8445 \n\n8506 \n8567 \n8627 \n8686 \n8745 \n\n\n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n\n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n\n1 1 2 \n\n\n3 3 4 \n3 3 4 \n3 3 4 \n3 3 4 \n2 3 4 \n\n2 3 4 \n2 3 4 \n2 3 4 \n2 3 4 \n2 3 4 \n\n\n5 5 6 \n5 5 6 \n5 5 6 \n4 5 6 \n4 5 6 \n\n4 5 6 \n4 5 5 \n4 5 5 \n4 5 5 \n4 5 5 \n\n\n75 \n76 \n77 \n78 \n79 \n\n80 \n81 \n82 \n83 \n\n84 \n\n\n8751 \n8808 \n8865 \n8921 \n8976 \n\n9031 \n9085 \n9138 \n9191 \n9243 \n\n\n8756 \n8814 \n8871 \n8927 \n8982 \n\n9036 \n9090 \n9143 \n9196 \n9248 \n\n\n8762 \n8820 \n8876 \n8932 \n8987 \n\n9042 \n9096 \n9149 \n9201 \n9253 \n\n\n8768 \n8825 \n8882 \n8938 \n8993 \n\n9047 \n9101 \n9154 \n9206 \n9258 \n\n\n8774 \n8831 \n8887 \n8943 \n8998 \n\n9053 \n9106 \n9159 \n9212 \n9263 \n\n\n8779 \n8837 \n8893 \n8949 \n9004 \n\n9058 \n9112 \n9165 \n9217 \n9269 \n\n\n8785 \n8842 \n8899 \n8954 \n9009 \n\n9063 \n9117 \n9170 \n9222 \n9274 \n\n\n8791 \n8848 \n8904 \n8960 \n9015 \n\n9069 \n9122 \n9175 \n9227 \n9279 \n\n\n8797 \n8854 \n8910 \n8965 \n9020 \n\n9074 \n9128 \n9180 \n9232 \n9284 \n\n\n8802 \n8859 \n8915 \n8971 \n9025 \n\n9079 \n9133 \n9186 \n9238 \n9289 \n\n\n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n\n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n1 1 2 \n\n\n2 3 3 \n2 3 3 \n2 3 3 \n2 3 3 \n2 3 3 \n\n2 3 3 \n2 3 3 \n2 3 3 \n2 3 3 \n2 3 3 \n\n\n4 5 5 \n4 5 5 \n4 4 5 \n4 4 5 \n4 4 5 \n\n4 4 5 \n4 4 5 \n4 4 5 \n4 4 5 \n4 4 5 \n\n\n85 \n86 \n87 \n88 \n89 \n\n90 \n91 \n92 \n93 \n94 \n\n\n9294 \n9345 \n9395 \n9445 \n9494 \n\n9542 \n9590 \n9638 \n9685 \n9731 \n\n\n9299 \n9350 \n9400 \n9450 \n9499 \n\n9547 \n9595 \n9643 \n9689 \n9736 \n\n\n9304 \n9355 \n9405 \n9455 \n9504 \n\n9552 \n9600 \n9647 \n9694 \n9741 \n\n\n9309 \n9360 \n9410 \n9460 \n9509 \n\n9557 \n9605 \n9652 \n9699 \n9745 \n\n\n9315 \n9365 \n9^_15 \n9465 \n9513 \n\n9562 \n9609 \n9657 \n9703 \n9750 \n\n\n9320 \n9370 \n9420 \n9469 \n9518 \n\n9566 \n9614 \n9661 \n9708 \n9754 \n\n\n9325 \n9375 \n9425 \n9474 \n9523 \n\n9571 \n9619 \n9666 \n9713 \n9759 \n\n\n9330 \n9380 \n9430 \n9479 \n9528 \n\n9576 \n9G24 \n9671 \n9717 \n9763 \n\n\n9335 \n9385 \n9435 \n9484 \n9533 \n\n9581 \n9628 \n9675 \n9722 \n9768 \n\n\n9340 \n9390 \n9440 \n9489 \n9538 \n\n9586 \n9633 \n9680 \n\n9727 \n9773 \n\n\n1 1 2 \n1 1 2 \n1 1 \n1 1 \n1 1 \n\n1 1 \n11 \n1 1 \n1 1 \n1 1 \n\n\n2 3 3 \n2 3 3 \n2 2 3 \n2 2 3 \n2 2 3 \n\n2 2 3 \n2 2 3 \n2 2 3 \n2 2 3 \n2 2 3 \n\n\n4 4 5 \n4 4 5 \n3 4 4 \n3 4 4 \n3 4 4 \n\n3 4 4 \n3 4 4 \n3 4 4 \n3 4 4 \n3 4 4 \n\n\n95 \n96 \n97 \n98 \n99 \n\n\n9777 \n9823 \n9868 \n9912 \n9956 \n\n\n9782 \n9827 \n9872 \n9917 \n9961 \n\n\n9786 \n9832 \n9877 \n9921 \n9965 \n\n\n9791 \n9836 \n9881 \n9926 \n9969 \n\n\n9795 \n9841 \n9886 \n9930 \n\n9974 \n\n\n9800 \n9845 \n9890 \n9934 \n9978 \n\n\n9805 \n9850 \n9894 \n9939 \n9983 \n\n\n9809 \n9854 \n9899 \n9943 \n9987 \n\n\n9814 \n9859 \n9903 \n9948 \n9991 \n\n\n9818 \n9863 \n9908 \n9952 \n9996 \n\n\n1 1 \n1 1 \n1 1 \n1 1 \n1 1 \n\n\n2 2 3 \n2 2 3 \n2 2 3 \n2 2 3 \n2 2 3 \n\n\n3 4 4 \n3 4 4 \n3 4 4 \n3 4 4 \n3 3 4 \n\n\n\n(198) \n\n\n\n\n\ni \n\n\n2 e. \n\n\n\xc2\xab ^ \n\n\nin o t> 00 \xc2\xbb -^ \n\n\n\n\nS <N \n\n\n\n\nM \n\n\n\n\n\xe2\x96\xa0* ift <o I- \n\n\n\xc2\xbb o> \n\n\nSt \n\n\n9 \n1 \n\n\n1 \n\n\n\n\n1 \n\n\n\n\ni^ii^^ \n\n\ni \n\n\n\xc2\xb1ife \n\n\n\n\nip--\' \'\' \' -\'-".^iiiiiiiiim \n\n\nHH \n\n\ni^ 1 :\xe2\x96\xa0\xe2\x96\xa0-\xe2\x96\xa0- \n\n\n\n\nffi- \n\n\n\n\nQ \n\n\n\n\n\n\n\n\nllt\xc2\xb1 \n\n\ni \n\n\n7-j: \n\n\nW^ \n\n\n_ _;Lritiy \n\n\n^ \n\n\n\xe2\x84\xa2 \n\n\n8 \n\n7 \n\n\n6 \n\n\n-\xe2\x96\xa0-\xe2\x96\xa0.-- \n\n\n\n\nn \n\n\n\xe2\x80\x94 \n\n\n\n\nw \n\n\n\n\n::::: \n\n\n::::: \n\n\n\n\n\n\n1 \n\n\n\n\n\n\n\n\n\n\n4 ^ \n\n\n: 6 \n\n\n\n\ni "-- \n\n\n--+\xe2\x80\x94 u-^~ \n\n\n:r.L:^^Pii \n\n\nd \n\n\ni^ --\xe2\x80\xa2--+- i \n\n\n""""^""""\xe2\x96\xa0""\xe2\x96\xa0""""\xe2\x96\xa0"^==si \n\n\ni \n\n\nBl\' \n\n\n3 \n\n\n:|Hp||ti \n\n\nEE \n\n\nitrtiJ \n\n\niR!-ll-4+ \n\n\nffl \n\n\nfH4 \n\n\ni+i+i- \n\n\nfl-H \n\n\n\n\n1 \n\n\n-^-+++++f \n\n\n-^r \n\n\nhas \n\n\nmtMMwS \n\niMfrnTTyrTTrr i \' i i ilh \n\n\ni \n\n1 \n\n\n1^3 \n\n\n2 \n\n\nI. + ..-1-1 \n\n\n11 \n\n\n:n;;::ii \n\n\n\xe2\x96\xa0Hl-rt+H \n\n\nm \n\n\nlit \n\n\n\n\n\n\n\n\n\xc2\xb1z \n\n\n\n\n15 \n\n\n!- \n\n\nll \n\n\n.::.::|i:^^ \n\n\nm \n\n\n\n\ni; \n\n\n\n\n\n\nfPffl \n\n\ni \n\n\n\n\n1 \n\n\n\n\n\\\\ \n\n\n^\xe2\x84\xa2P \n\n\n\n\niMii \n\n\n] \n\n\n\n\nIII! \n\n\n^ \n\n\nitK\' \n\n\nIf \n\xc2\xb1 \n\n\nifi|i|-]: \n\n\n\n\na-- \n\n\n\n\n8 \n1^ \n\n\n\n\n\n\nt,:-:: \n\n\n\xe2\x96\xa0j \n\n\n\n\n6 \n\n\n\n\n\n\n\n\nj \n\n\n\n\nffi \n\n\nWfef \n\n\n|j^^J|[. \n\n\n5 \n\n\n^ \n\n\nnil \n\n\nu \n\n\n\n\n\n\nEEE|!;i;::i::M: \n\n\nll \n\n\n^\xe2\x84\xa2 \n\n\npi \n\n\n-+~ \n\nnmP \n\n\n:: :::":::|::| : :: ; 5 \n::::::::: 4 \n\n\n\n\n\n\n|e \n\n\n\n\n\n\n\n\n:p \n\n\n|;;|^l|^li \n\n\nIp \n\n\n\n\n\n\n\n\n\n\n\n\n\n\nrt \n\n\n\n\neMM:;:;;; :- iiw - -;; 1 \n\n\n\xc2\xab> \n\n\n\n\nCI \n\n\n\n\n\n\n::::::::\xc2\xb1: iffl \n\n\nSffiE \n\n\n= :-:l-;5ij \n\n\nS \n\n\nH |i ilWL \n\n\n\n\n\n\n-- \n\n\n\n\n\n\n\n\n^: \n\n\n\n\nii:i \n\n\n!;;;;;;N; i|:|f|-i \n\n\n2 \n15 \n\n\n:::::::::^::\xc2\xb1::: \n\n\n\\\\ \n\n\n;;;B-;;; \n\n\n\n\nH \n\n\n1" \n1 \n\n\n:::i;;\xc2\xb1;:i+l \n\n\n=He \n\n\n\n\n^ \n\n\nt \n\n\n\n\ns \n\nvl \n\n\nWfl^ \n\n\n\n\ni \n\n\xe2\x96\xa0a \n\n\n\n\nM \n1 1 \n\n\n\n\n^1 \n\n\ni \n\n5 C \n\n\ns \n\n\ns \n\n\n:\xc2\xb1^ + ^ + j \n\n\n\n\nm \n\n\ni \n\n\n\n\n-iiitiiii\xc2\xb1:::::::::i: \n\n\n1 \n\n\n4 \n^1. \n\n\n\nFig. 71. \xe2\x80\x94 Logarithmic Cross-section Paper for Plotting Exponential Equations. \n\n\n\n(T99) \n\n\n\nINDEX \n\n\n\nAbsolute temperature, ii. \n\npressure, 13. \n\nzero, II, \nAbsorption system of refrigeration, 90. \nAdiabatic expansion, 30, 35, 39, 114. \n\ncompression, 35, 39, 46. \n\nlines of steam, 108. \nAir compressor, 95-99. \n\nengines, 54, 56. \n\nthermometer, 14. \nAir, table of properties of, 23. \nAmmonia machine, 92-94. \nAvailable energy of steam, 120-127, 129. \n\nBarrel calorimeter, 82. \nBarrus\' calorimeter, 78. \nBlack\'s doctrine of latent heat, i. \nBoyle\'s law, 10, 14. \nBritish thermal unit, 4, 5. \n\nCalorie, 4. \n\nCalorimeter, steam, 75-83. \n\nCamot, I, 44. \n\nCamot\'s cycle, 44. \n\ncycle, efficiency of, 48. \n\ncycle, reversed, 51, 88, \n\ncycle principle, 49. \nCentigrade, 4, 12. \nCharles\' law, 12, 14. \nClausius, I. \n\ncycle, 127. \nCoefficient of performance, 93-94. \nCombined diagrams, 96, 157. \nCombination law, 14, 36. \nCompressed air, 95-99. \nCompression, isothermal, 32, 95. \n\nof gases, 29. \nCompressor, air, 95-99. \nCondensing calorimeters, 82. \nConservation of energy, 3, \n\n\n\nConversion of pressures, 13 (foot-note). \n\nof temperatures, 4, 11. \nCross-section paper, logarithmic, 199. \nCycle, Carnot\'s, 44, \n\nClausius, 127. \n\ndefinition of, 44, 140. \n\nRankine, 127. \n\nDense air machine, 90-92. \n\nDensity, 4, 23. \n\nDiagram, temperature-entropy,io5-io7. \n\nindicator, 30, 157. \n\nMollier, 116, 117. \n\ntotal heat-entropy, 116, 117. \nDry saturated steam, 61. \nDrying of steam by throttling, 73. \n\nEfficiency, air engines, 56. \n\nCarnot cycle, 48, \n\nClausius cycle, 127, \n\nEricsson engine, 56. \n\nmechanical, 141. \n\nnon-expansive cycle, 144, 145. \n\nRankine cycle, 127, 129, 139. \n\nrefrigerating machine, 94, \n\nStirling engine, 56, \n\nthermal, 6, 56. \nEnergy, available, 120-127, 129. \n\ninternal, 20, 38, 66. \n\nintrinsic, 20, \nEngine, Ericsson, 56. \n\nhot air, 56. \n\nStirling, 56. \nEngineering units, 4. \nEntropy, 10 1. \n\ndiagram, 105-107. \n\no^ the evaporation, 107. \n\nof the liquid, 107. \nEquivalent evaporation, 8^. \nEricsson hot-air engine, 56. \n\n\n\n202 \n\n\n\nINDEX \n\n\n\nEvaporation, equivalent, 83. \n\nfactor of, 83. \n\ninternal energy of, 66. \n\nlatent heat of, 63. \nExpansion of gases, 29. \nExpansions, adiabatic, 30, 35, 39. \n\nisothermal, 30, 32, 102. \nExternal work, 19, 67. \n\nwork in steam formation, 61, 64. \n\nFactor of evaporation, 83. \nFirst law of thermodynamics, 3. \nFlow of air, 164-172. \n\nin nozzles, 163, 172. \n\nin orifices, 168-171, 181. \n\nof steam, 172-189. \nFoot-pound, 4. \nFoot, square, 4. \nFormation of steam, 60. \n\nGas constant (R), 16, 23. \n\nthermometers, 13. \nGas, perfect, 9. \nGram-calorie, 4. \n\nh (heat of liquid), 62. \n\nH (total heat of steam), 65. \n\nHeat engine, efficiencies, 139, 144. \n\nof liquid, 62. \n\nunits, 4. \nHim\'s analysis, 160. \nHistorical, i. \nHot-air engine, 54, 56. \nHyperbolic logarithms, 195, 196. \n\nIndicator diagram, 30. \nInternal energy, 20, 38, 66, 67. \n\nwork of evaporation, 66. \nIntrinsic energy, 20, 38, 66. \nIrreversible cycle, 50. \nIsentropic lines, 103 (f oot-n\'ote) . \nIsothermal expansion, 30, 32, 46. \n\ncompression, 32, 46. \n\nlines of steam, 108. \n\nJoule, I, 21. \nJoules\' law, 21. \n\nKelvin, i, 21. \n\n\n\nL (latent heat of steam), 6z. \nLatent heat of evaporation, 63. \nLaws of perfect gases, 10, 12, 14, 30, 37, \n39. 48. \n\nof thermodynamics, 3. \nLogarithmic cross-section paper, 199. \nLogarithms, natural, 34, 195, 196. \n\n"common" or ordinary, 34, 197, 198. \nLow temperature researches, 24. \n\nM (mass), 17. \n\nMass, 17. \n\nMean specific heat, 72. \n\nMechanical efficiency, 141. \n\nequivalent of heat, 5. \nMoisture in steam, 69, 74. \nMoUier diagram, 116, 117. \nMoyer\'s formula for nozzle losses, 188. \n\nNaperian logarithms, 195, 196. \nNatural logarithms, 34, 195, 196. \nNewcomen, i. \nNon-expansive cycle, 113. \nNon-reversible cycle, 50, 51. \nNozzle, flow through, 123. \nNozzle losses, 175, 188. \n\nOnnes, 24. \n\nOrifice, flow through, 163, 172. \n\nPerfect gas, 9, 39. \nPerfection of heat engine, 51. \nPorous plug experiment, 21. \nPound, 4. \n\nPressure-temperature relation, 12, 39, \n62. \n\nunits, 4. \nProperties of gases, 23. \n\nof steam, 61. \n\n9,63. \n\nQuaUty of steam, 69, 77, 114. \n\nR (thermodynamic or "gas" constant), \n\n16, 17, 21, 23. \nr, 64. \nRankine, i. \n\ncycle, 127, 129, 139. \n\n\n\nINDEX \n\n\n\n203 \n\n\n\nRatio of expansion (;-), 35, 41. \nRatio of specific heats, 23. \nRefrigerating machines, 44, 88-94. \nRefrigeration, appHcations of, 88-94. \nRegenerator, 55. \nRegnault, i, 62. \nReversed Carnot cycle, 51. \nReversibility, 49. \nReversible cycle, 49. \n\nSaturated steam, 61. \nSaturation curve, 138, 158. \n\nline of steam, 159. \nScales, thermometric, 4. \nSecond law of thermodynamics, 3, 51, \n\n53- \nSeparating calorimeter, 80. \nSmall calorie, 5 (foot-note). \nSpecific heat, 18. \n\nheat at constant volume, 19. \n\nheat, difference of, 19. \n\nheat, instantaneous values of, 73. \n\nheat, mean value of, 72. \n\nheat, ratio of, 23. \n\nheat, true, 72, 73. \n\nheat of gases, 23. \n\nheat of superheated steam, 72, 73. \n\nheat of water, 5. \n\nvolume of gases, 4, 23. \n\nvolume of saturated steam, 64. \n\nvolume of superheated steam, 72. \nSteam, dry, 61. \n\nengines, efficiencies of, 144, 147. \n\nformation of, 60. \n\nsaturated, 61. \n\nsuperheated, 61, 62. \n\ntables, 60. \n\ntotal heat of, 65, \n\nturbine, 123, 148. \n\nwet, 68. \nStirling\'s engine, 56. \nSuperheated steam, 61, 62, 69. \nSuperheating calorimeter, 75. \n\n\n\nTables, steam, 60. \nTemperature, absolute, 11. \n\n-entropy diagrams, 105-107. \n\nmeasurements, low, 24. \nThermal efficiency, 6, 56, 141. \n\nunit, British, 4, 5. \nThermodynamics, definition of, i. \nThermometer, air, 14. \n\ngas, 13. \nThermometric scales, 4. \nThrotthng calorimeter, 75, 78. \nTotal heat-entropy diagram, ji6, 117. \n\nheat of saturated steam, 65. \n\nheat of superheated steam, 70. \n\ninternal energy of steam, 66. \nTurbine, steam, 123. \n\nUnit of heat, i, 4, 5. \nUnits, engineering, 4. \n\nVaporization (see evaporation). \nVapors, examples of, 90. \nVelocity, 123. \nVolume, specific, 4, 23. \n\nWater, specific heat of, 5. \n\nweight of, 13 (foot-note). \nWatt, I. \n\nWeight, units of, 4, 13 (foot-note). \nWet steam, 68. \nWetness of steam, 69, 77. \nWire-drawing, 73. \nWork, external, 19, 67. \n\nof adiabatic expansion, 58. \n\nof Carnot cycle, 48. \n\nof Clausius cycle, 128. \n\nof isothermal expansion, 34. \n\nof Rankine cycle, 128. \nWorking substance, 6, 7. \n\nX (quality of steam), 69, 77. \n\nZero, absolute, 11. \n\n\n\n\xe2\x96\xa0if \'\' ^ \'\xe2\x80\xa2^y-\'i \n\n\n\n\n\n\n<-\xe2\x96\xa0" f\'-x \xe2\x80\xa2\'t.:^ \n\n\n\n\n\n\nr4k \n\n\n\n\n\n\n\xe2\x96\xa0 \n\nLIBRARY OF CONGRESS \n\n\n\n021 225 343 1 \n\n\n\n\'i Mv\xc2\xa5^ \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\'sf \n\n\n\n\n\n\n\n\n\n\n\n\n\xe2\x96\xa0m^mm^ \n\n\n\n'