Gornell University Library Sthara, Nem York THE ALEXANDER GRAY MEMORIAL LIBRARY ELECTRICAL ENGINEERING THE GIFT OF Engineering _ BASEMENT STORAGE Cornell Univ: ersity Ir tabi iii engr STEAM CHARTS STEAM CHARTS A Table of Theoretical Jet Velocities and The Corrections of Mercury Columns WITH FIFTY ILLUSTRATIVE PROBLEMS BY F. O. ELLENWOOD Professor of Heat Power Engineering, Sibley School of Mechanical Engineering, Cornell University; Member of the American Society of Mechanical Engineers NEW YCRK JOHN WILEY & SONS, Inc. Lonpon: CHAPMAN & HALL, Limrtrep } Copyright, 1914, by F. 0. ELLENWOOD Copyrighted in Great Britain PRESS OF BRAUNWORTH & COs 8/24 BOOK MANUFACTURERS BROOKLYN, N. Y- PREFACE Turis little book is intended to be of assistance to engineers and students when making calculations involving wet or superheated steam. The chief aim of the author has been to prepare a set of steam charts which shall be accurate and comprehensive, and at the same time convenient to handle, and easy to read. An attempt has also been made to give, concisely, the corrections to be applied to the readings of mercury columns, and to prepare a table of velocities, which it is hoped may prove useful. In order to illustrate some of the uses of the charts and tables, and also to aid those who may desire it, a number of problems, with their solutions, have been added. To make these of more assistance, they have been indexed. For the further aid of those who may desire a brief review of the thermodynamics of steam, and in order to make clear the meaning of all terms used, the few pages of Fundamental Principles were written. For the main chart, total heat and specific volume were chosen as coordinates because of the fact that upon these two values could be plotted lines of constant pressure, entropy and quality (or superheat), so that each pair of the five sets of lines will make clear intersections. The total heat entropy chart does not permit this. To complete the set of values ordinarily needed, the curve was added, showing the heat of the liquid and temperature of vapori- zation. The supplementary chart, Plate 8, enables one to read the external work, and therefore obtain easily the intrinsic heat. The index chart for Plates 1 to 6 was made to give a general idea of the relative position and shape of each set of lines, to show quickly the limiting values for each of the six sections, and to assist in determining the particular plate needed. The range of pressures, qualities, and superheats is intended to be more than sufficient for present practice. For the wet region the inch of mercury was used as the main unit to represent pressures iii lv PREFACE less than one pound absolute, as it is believed that this is the more convenient one for practical work. Special endeavor has been made to prevent confusion of these two units by using broken lines to represent pressures in inches of mercury, and by putting the proper units with each numeral representing pressure in this region. The book form of chart was chosen because the author believes that it will be of greater convenience and easier to read than a large folded chart made to the same scales. By making the plates small the eye has to travel only a short distance to read the scales, and this may also be done without requiring any desk space whatever. The book form also has the advantages of better protecting the chart, permitting a quicker reference, and wasting less space in the corners, than does the same chart when in the form of a large folded sheet. To Prof. Lionel 8. Marks and to Dr. Harvey N. Davis, and to their publishers, Longmans, Green & Co., the author desires to express his thanks for permission to use their steam tables in pre- paring these charts. He also wishes to acknowledge his indebted- ness to Prof. Albert W. Smith, Director of Sibley College, and to Prof. William N. Barnard, for their many helpful criticisms; and to Mr. C. H. Berry and Mr. E. T. Jones, instructors in Sibley College, for their able assistance in preparing the charts and problems. F. O. E. ItHaca, New York, August, 1914. CONTENTS PAGE INTRODUCTION Fundamental Principles. . . . re ‘ oor ee a oe OL Preparation and Use of the Steam Charts and Tables . .. . 14 Atmospheric Pressure and ‘Barometric Corrections . . . . .18 CHARTS Index Chart .. . . eo ce ee ew & w « 23 Plates 1 to 7, the Total Heat-Volume Chart. . . . . . . . 24 Plates 8a and 8b, the External Work-Volume Chart . . . . . 38 Plate 9a, Correction of Mercury Column Due to Temperature . . 40 Plate 9b, Correction of Barometric Readings Due to Change in Elevation 2 « « 6 «© % «© = 6 © # «© # «© = « » « 40 TABLES 1. Correction of Barometric Readings to 45° Latitude . . . . 41 2. Correction of the Barometer for Capillarity . . . .. . .41 3. Density of Mercury . . .. ...... 2... « . Al 4. Theoretical Velocities of Steam . ......2.2. 2. ~« «42 PROBLEMS: a). Gyo -8> aay Che chy Woe ay ne coe eS ths ee, La ei, RO ENDER se 5. GS op RE ae: ap Se Bw He es Gh SS es ae): ABE INTRODUCTION FUNDAMENTAL PRINCIPLES Pressure-Volume and Temperature-Entropy Diagrams.—In the study of thermodynamics of vapors the pressure-volume and the temperature-entropy diagrams are of very great importance, for the reason that by their aid the areas representing work and heat, respectively, may usually be shown. Since the engineer is con- cerned with the transformation of heat into work, any diagrams which will assist him to understand how this is accomplished will always be very useful. Other diagrams may be of more assistance in obtaining numerical results, but these two will always be fore- most in analyzing thermodynamic processes. Pressure-Volume Diagram.—Referring to Fig. la, where the absolute pressure, P, in pounds per square foot, is represented by the ordinates, and the volume, V, in cubic feet by abscisse, the area abhg underneath curve ab represents the work done in foot- pounds by the substance in passing along the constant-pressure line from a to b. This might be expressed in this manner: b Work | = f. Vo Pdv = P [Vp — Val = area abhg. a a In general, it may be stated that if a substance expands or is compressed in such a manner that its pressure-volume history is definite for the entire change, as from the point a to some point c along the path adc, Fig. 1a, the following equation is true: Work |. = J. s PdV = area adchg. This would represent work done by the substance upon some other body. In case the substance had been compressed along the 1 2 FUNDAMENTAL PRINCIPLES path cda, then the area cdagh represents the work done upon this substance in order to compress it, oy, _ Vv. or Work | = Te Pav = Te PdV = — [area adchg] = area cdagh. Temperature-Entropy Diagram.—If, by the addition of heat, a substance may be made to change its state point a to some other b f 1 1 ' | a | 5 2 | ' 2 1 | = { ! 3 ¢ { g g d i 5 ' o { ‘ 3 ! & c ! i é : Pa \ 3 : | ( 1 ' 9 i 1 1 | ee i \ | 2 ! 1 | \ I < l | 1 1 1 | 1 1 1 1 1 'g ‘h inn n "e Volume Entropy (a) (b) Fic. 1. condition 6, in such a manner that its temperature and pressure are uniform* throughout the entire mass of the substance during this change, then the heat which has been added to this substance is equal to the area mabn, Fig. 1b. This is true by reason of the definition of entropy. In other words, for such processes as this, if H = Heat added to the substance in B. t. u. T = Absolute temperature (= 460 + t° F.) ¢ = Entropy then the change of entropy from a to 6 is defined by the equation dH d¢ T b bp = Tdg= bn or HI die = area abnm * It is particularly important not to confuse this word uniform with the word constant. FUNDAMENTAL PRINCIPLES 3 For the condition given above, if the heat had been added at constant temperature, as from 6 to f, Fig. 1b, then, Hy} = = ia de =T[ oy — 4 | = area bflen or, of — $5 = ly 2 = distance bf or nk. For both of these cases the addition of heat to.the substance caused an increase of entropy. Had the heat been abstracted, the entropy would have been decreased. Note that it is always the change in entropy that engineers are interested in, rather than its absolute values. Adiabatics——The term adiabatic means no transfer of heat. Hence an adiabatic expansion or compression means one in which no heat is added or abstracted during the process. Reversible and Irreversible Adiabatics—The two general classes into which all adiabatics may be divided, are called reversible and irreversible adiabatics. An adiabatic expansion or compres- sion, which is frictionless and which takes place in such a manner that the substance passes through a continuous series of equilib- rium states, would be called a reversible adiabatic. A reversible adia- batic is also called a constant entropy line or isentropic, because a vertical line on the temperature-entropy chart is the only one which permits the area underneath it to become equal to zero, thus satis- fying the definitions of adiabatics and entropy. It is impossible to have these conditions fulfilled in actual cases, but it is useful to study them and compare them with the actual conditions which may be made to approach very close to the ideal. Thus, a substance may expand in a single cylinder, with but very little friction, and with but very little transfer of heat to or from the cylinder walls, and slowly enough so that all of the substance in this cylinder is almost exactly in the same state. Such cases may properly be treated as reversible adiabatics or isentropics. On the other hand, sudden or ‘‘free’” expansions are sure to set up eddies whose kinetic energy will soon reappear im the form of heat, thus causing an increase of entropy. Friction will have a 4 FUNDAMENTAL PRINCIPLES similar effect. Such expansions are called irreversible, and if they have taken place without any heat being transferred to or from another substance, they would be called irreversible adiabatics. In the case of adiabatic expansion of gases or steam through a properly-formed nozzle in which there are no eddies formed, and in which there is no friction, the transformation of the available heat energy into velocity is complete and may therefore be considered as a reversible adiabatic. In actual steam turbine nozzles the loss due to friction and eddies is extremely small, so that in turbine design the expansion in the nozzle is justly assumed to be a reversible adiabatic. The main losses in the turbine occur in the endeavor to transfer the energy of the jet to the turbine blades. These losses cause considerable reheating, so that in any actual case there is considerable increase in the entropy of the steam in passing through the turbine. As an illustration, see problems 41 and 42. The term adiabatic will be used hereafter to mean reversible adiabatic, unless specifically stated otherwise. Work and Heat Depend upon Path.—From the consideration of the PV and T¢ charts it will be evident that it is not sufficient to give merely the initial and final states of a substance in order to find the work done and the heat required to go from one state to another. It is also necessary to give the exact path to be followed. Thus, in Figs. la and 1b, heat might be added and abstracted in such a manner that the state c is finally reached by means of the constant pressure line ab, and the constant volume line bc, or this same state might have been reached more directly by some such line as ade. By observing the areas it will be seen that the work done in the first case is abhg, while for the path adc the work done is represented by the area adchg. Also from the T¢ chart it may be seen that in going from a to b an amount of heat equal to the area abnm was supplied to the substance, and in going along the constant volume line, bc, an amount of heat equal to the area bemn was abstracted from the substance. The substance might also have reached the state c by expanding adiabatically along the line adc; but note the difference in areas representing the work and heat. FUNDAMENTAL PRINCIPLES 5 Cycles.—If a substance should be made to follow a series of paths such as ab, bc, and ca, so that it is returned to its initial state, the substance would then have completed a cycle, and the substance used would have been called the working substance. Had the cycle been completed in the order of the letters, abc, an amount of work in foot-pounds, equal to the area, abc, Fig. 1a, would have been done upon some external mechanism. The heat equivalent in B. t. u. of this work, would be represented by the area abc, Fig. 1b. This heat would have been supplied to the working substance from some external source. Properties which are Independent of the Path.—After a cycle has been completed, the substance is in exactly the same condition as at the beginning, so its pressure, volume, temperature, entropy, and intrinsic heat must all be the same as originally. Likewise, if a substance is in a certain state, such as c, it merely means that all of the above properties have some definite value and may be determined regardless of the manner in which this substance may have reached this state c. Intrinsic Heat or Intrinsic Energy.—Either of these terms may be used to represent all of the heat energy contained within a substance measured above some convenient standard. It is to be carefully noted that this is a different quantity in general from the heat required to bring the substance to a particular state from the standard condition. This is on account of the fact that, when heat is added to a substance, in which no heat is lost by radiation or otherwise, in which there is no change in electrical or chemical energy, or in which there is no change in the kinetic energy of the substance due to its mass velocity, then in going from a to b, a general equation may be written, thus: Heat 7° x Gain in ey External on added, heat i done ii The external work done means that work is done by the sub- stance upon some external mechanism. In case this term is nega- tive, it means that the work is done by the external mechanism upon the substance. Pressure 6 FUNDAMENTAL PRINCIPLES Steam Formed at Constant Pressure.—For nearly all commercial purposes steam is generated in a steam boiler at a pressure which is maintained nearly constant. It is therefore very important to consider carefully the formation of steam at constant pressure. It has been found by experiment that during the change of state of any substance from a liquid to a vapor, that if the pressure be kept constant the temperature of the vapor in contact with its liquid will also remain constant until the vaporization has been completed. Q Absolute Temperature eS as ee ots fs ae eee ied Oy. pen hh thy m my v ming Volume Entropy (a) (b) Fic. 2. Suppose that a pound of water at a temperature of melting ice, 32° F., is contained in a metal cylinder with a tight-fitting piston resting upon the surface of the water. The area of this piston and the total weight resting upon the water being known, the abso- ‘lute pressure on the water may be determined. Let this value in pounds per square inch be denoted by p, and in pounds per square foot by P. The volume of this pound of water will be very small, only .016 cubic feet. We may represent this starting point by the letter a in Figs. 2a and 2b. (We locate the point a on the zero entropy line merely as a matter of convenience, as we shall be con- cerned only with the heat measured above 32° F.) FUNDAMENTAL PRINCIPLES 7 Let heat be applied to the cylinder now, and it will be found that at first the piston moves up only a little, due to the slight increase in the volume of the water. The temperature, however, is ob- served to rise rapidly until it reaches some definite point dependent entirely upon the pressure imposed by the piston. Let this state be represented by 6. The substance is now all in the form of liquid, but at the particular temperature known as the temperature of vaporization for this given pressure. Since the volume has changed only very slightly, the points a and b on the PV diagram will almost exactly coincide. Even though this water had been heated to a temperature of 400° F., as it might have been if under a pressure of 250 pounds or more, its volume would still only be .0187 cubic foot. This extremely small change in volume means that very little work, shown by the area abkk’, Fig. 2a, has been done on the piston, and consequently virtually all of the heat that has been applied has been used to increase the sensible heat of the water. The Heat of the Liquid is the term applied to this heat, which is used to heat a unit weight of water from 32° F. to the temperature of vaporization. The area representing the heat of the liquid is abmm, Fig. 2b. With further addition of heat it will now be found that the volume is increased very much and that the temper- ature remains constant until some state c is reached. The water has now all been evaporated, and exists in the form of dry saturated steam. Any further addition of heat will cause it to become super- heated, and any abstraction of heat will cause it to be partially condensed. The heat which was necessary to vaporize this pound of water under this pressure is represented by the area be n m, and is called the latent heat of vaporization, or latent heat of steam. The Total Heat of Dry Saturated Steam is equal to the heat of the liquid plus the latent heat of vaporization and is equal to the area abcnm, Fig. 2b. The Saturation Curve on any diagram is the curve which shows the two coordinate values of dry saturated steam for the entire range of scales on that diagram. Thus in Fig. 2a the specific volume of dry saturated steam may be determined for any pressure by 8 FUNDAMENTAL PRINCIPLES means of the saturation curve, cc: ¢. The saturation curve may also be seen for other coordinates as in Fig. 2b, Plate 1b, or Plate 8a. Wet Steam and Quality During the formation of a pound of dry saturated steam in the manner just outlined, it might have been possible to stop the addition of heat before all the water was evaporated and the resultant mixture of water and steam would be what is known as wet steam. That portion by weight of this mix- ture which is dry vapor would be known as the Quality of the steam. Thus if x represents quality and if nine-tenths of the pound of water have been vaporized, this fact would be expressed by the equation x = 90%. In Fig. 2b the quality at ¢ is unity and at b the quality is zero; or xe = 100% and x» =0. It is im- portant to note that this interpretation of quality is the most useful one in using the T¢ chart. In this chart it is understood that a unit weight of substance is being considered unless it be defi- nitely stated otherwise, and it is sometimes necessary to carry this unit weight of substance throughout the extreme range of quality in order to analyze a cycle or to understand many important thermodynamic relations. On the other hand, steam coming from a boiler often carries with it a considerable amount of moisture. If a sample of this mixture be obtained, then the weight of the dry steam present divided by the weight of this entire mixture would be the quality of the steam coming from the boiler, the weight of water still re- maining in the boiler not being considered at all in this determi- nation of quality. The Total Heat of Wet Steam is equal to the heat of the liquid plus the quality multiplied by the latent heat of vaporization, or h + xL, where h represents the heat of the liquid and L the latent heat. If c’, Fig. 2b, represents this state of the steam, then the area abe'n'm is equal to the total heat at c’. Superheated Steam.—If heat is added to dry saturated steam, the pressure remaining constant as before, the volume and temper- ature will be found to increase; and the steam is said to be super- heated. Superheated steam may therefore be defined as any steam, regardless of how formed, having a temperature higher than FUNDAMENTAL PRINCIPLES 9 the temperature of saturated steam of the same pressure. The difference between these two temperatures is called the Degrees of superheat, and is nearly always represented by the letter D. For the state d, Fig. 2b, D = Tag —T,. If the specific heat of super- heated steam at constant pressure be represented by ¢,, then ¢, D is equal to the area cdn,n and is called by either of the terms, heat of superheat, heat of superheating or superheat. The value of the specific heat of superheat steam is variable, and depends upon both the pressure and the degrees of superheat. The Total Heat of Superheated Steam is equal to the total heat of dry saturated steam plus the heat of superheat. Thus for the state point d, Fig. 2b, the total heat at dis equal to the area abcdnim. Total Heat may now most conveniently be defined in the manner in which it is used by engineers, that is, a general definition which will hold for wet, dry, or superheated steam. Total Heat of Steam in any Given State is the amount of heat required to heat at constant pressure a unit weight of water from the temperature of melting ice to the state under consideration. Thus referring to Fig. 2b: Total Heat | area abe'n’m = h + x,’ L c Total Heat | = area abenm=h+L c Total Heat | area abednim = h + L + epD. In the United States the units are almost always B. t. u. per pound. All values of total heat which are likely to be needed will be found on Plates 1 to 7 inclusive. External Work at Constant Pressure.—During the period of vaporization the volume of a pound of water is changed to the very much larger volume of dry saturated steam, as from Vp to Ve, Fig. 2a. If this volume is represented in cubic feet by V and the pressure 10 FUNDAMENTAL PRINCIPLES in pounds per square foot by P, the work done during vaporization is c Work ] = P [ Ve — Vo | ft-lbs. = areadchh, or = AP (V¢_ — Vp) B. t. u. A being taken as the reciprocal of the mechanical equivalent of 1 1 heat, or A = 773 0 775 AP (V- — Vb) is commonly known as the ezternal latent heat of vaporization. During superheating the external work done is d Work il; =P [ va — Vz | ft.-lbs. = area cdhih or = AP (Va — V-~) B. t. u. For the entire process starting with water at 32° when its specific volume would accordingly be .016 cubic feet per pound, the external work done in order to reach some state such as c’ or d would be Work |” = P[ Ve — Vo ] ft-bs.= AP [Ve —.016 ] B. 6. u. Work |’ =P [Va —Vo ] ft-lbs. = AP[Va— 016] B. +. u. This is what is called the constant pressure external work, and is given for all conditions of steam by Plates 8a and 8b. This work has been done upon the piston, so that after some state such as d has been reached, the piston and all weights resting upon it have more potential energy by the amount of AP [Va —Val B. t. u. than they had at the beginning of application of heat to the water in condition a. This increase in potential energy of the piston and its weights has come from the heat which was neces- sary to form the pound of steam along the constant pressure path abcd. Now, if from the total amount of heat which has been added to a substance in order to reach a certain state by going along a certain path, all of the heat which has been used to do external work be subtracted, the amount left would be the gain in intrinsic energy of the substance, provided there were no losses FUNDAMENTAL PRINCIPLES 11 and that no energy had been used to impart velocity to the sub- stance. The Intrinsic Energy, Internal Energy, or Intrinsic Heat of steam, then, merely means the heat energy contained within the steam above 32°, and may always be found by subtracting the constant pressure external work from the total heat. The total heat may be obtained directly from Plates 1 to 7, while the external work may be obtained from Plates 8a and 8b. Since we can not reduce the temperature of any substance to absolute zero, it is not possible to reach the state in which the intrinsic heat is zero. For work with steam we need only be concerned with the intrinsic heat above 32° F. Specific Volume of steam means the volume of one pound of the steam in its given state. Unless some quality or superheat is given, it is understood to refer to dry saturated steam. The specific volume of wet steam may be determined by means of the quality in the following manner: Let Vsat. = volume of 1 lb. of dry saturated steam of given pres- sure, Vw = volume of 1 lb. of water at the temperature of vapori- zation, = .016 cu. ft. at 32° F., .017 cu. ft. at 250° F., .018 cu. ft. at 350° F., .019 cu. ft. at 420° F. Let u = change of volume during vaporization, = Vsat. a Vw. Then for some point such as c’ having a quality x,’, the specific volume is Vo = Vw + X'U, = Vw + Xe’ (Vsat. — Vu), = (1 — Xe’) Vw + Xe’ Vsat. This becomes equal to x,’ Vsa. almost exactly, except when the quality is very low or the pressure is higher than atmospheric. 12 FUNDAMENTAL PRINCIPLES The specific volume may be read directly from Plates 1 to 8 inclu- sive, Entropies.—For convenience, entropy of water at 32° is taken as zero. Hence the entropy of the liquid per pound of water, from Fig. 2b, is 6’ dH Ts ¢, dT ae tp = i TE Where cp = the specific heat of water at constant pressure. This value depends upon the temperature, so the integration is not often made, but instead the entropy of the liquid is usually obtained from the steam tables. The Entropy of Vaporization is, from Fig. 2b, ecdH L _ Latent heat oc — ob = i 7h T Ts abs. Temp. The Entropy of Superheat or the change of entropy due to superheating is, from Fig. 2b, sa-ge= f B= f™ of where cy, represents the constant pressure specific heat of super- heated steam. The Entropy of Steam in any state means the total entropy up to that state, measured above the assumed zero of entropy. For Wet Steam the Total Entropy is equal to 7 Entropy of the liquid + (Quality) (Entropy of vaporization) For Superheated Steam the Total Entropy is equal to Entropy of the liquid + Entropy of vaporization + Entropy of Superheat The Total Entropies are plotted on Plates 1 to 7 inclusive and these should serve nearly every purpose for which the engineer has need of their numerical values. Rankine and Clausius Cycles.—The most useful cycle in power- plant work is the one represented by abfe, Figs. 5 and 6, Problem 29. This cycle is sometimes known as the Clausius Cycle and some- times as the Rankine Cycle, but since the analysis of this cycle was FUNDAMENTAL PRINCIPLES 13 first published by each of these men at the same time, there will probably always be a difference of opinion as to which name should be used. If, instead of allowing the adiabatic expansion to continue until the back pressure is reached, as at f, the expansion had been stopped at the point c, thus cutting off the ‘toe’ in each diagram, there would be another cycle, abcde, which is preferred by many engineers as a basis of comparison for the performance of recipro- cating engines. This latter cycle is oftentimes called the Rankine and the other the Clausius, thus making a convenient distinction. They are also spoken of as the ‘‘complete expansion”? and the ‘incomplete expansion”’ cycles. These latter terms will be used in the problems of this book in order that there may be no possible confusion of the two cycles by using the terms Rankine and Clausius. Available Energy is a term that may have many meanings, but when used in connection with steam cycles it means the energy of the steam which would be converted into work by an ideal mech- anism that would carry out exactly the theoretical cycle. Thus, in using Table IV, the engineer would consider the term to represent the net area of the complete expansion cycle, by assuming that all of this energy is available for the production of velocity in the ideal nozzle. Then with an ideal turbine all of this velocity energy could be transformed into work. When designing a multi-stage turbine, however, it is necessary to find the energy available for each stage. This amount of energy is affected by the reheating in each of the preceding stages, as well as the drop in pressure in the stage being considered. For any steam prime mover, the available energy is a small part of the heat supplied to it, since a very large part of this heat must necessarily be given up to the exhaust. PREPARATION AND USE OF THE STEAM CHARTS AND TABLE OF VELOCITIES The general idea of the main diagram, which consists of Plates 1 to 6, inclusive, may be readily obtained from the index chart. It is seen to be divided into twelve equal parts, the top and bottom halves of each section being indicated by the subscripts a and 8, respectively. These two halves will be found facing each other, so that wherever this chart is opened, a complete section may be seen. The same total heat scale is used throughout, but the volume scale for each section is changed so that the general relation of each family of curves to one another will remain about the same. Having established the scale of volumes and total heats, con- stant pressure lines were then plotted from the values given by the steam tables of Marks and Davis. In the superheated region these lines are slightly curved, but in the wet region they are straight. For those pressures not given in the steam tables—all fractional pressures and those given in inches of mercury—the volumes and total heats were determined in two ways. For the wet region, special auxiliary curves were drawn by computing for certain qualities the total heats and volumes for those pressures given by the steam tables. From such auxiliary curves the desired values could then be determined, and the curves plotted in their correct relative positions. In the same manner the corresponding values were found for all fractional pressures above one pound in the superheated region. For those pressures less than one pound the volumes in the superheated field were found from Linde’s equation and the total heats were determined by adding cpD to the total heat of the dry saturated steam. The values of the specific heat for these low pressures were determined by assuming as correct the specific heats used by Marks and Davis for the pressures from one to four pounds. A curve of specific heats was then constructed through these points and extended into the region of lower pres- sures. From such a curve the specific heats were read. This was the best method by which the lines in the superheated region of 14 STEAM CHARTS AND TABLE OF VELOCITIES 15 Plate 6a could be drawn so that they would all continue as smooth curves. For the region of very low pressures, such as those given on Plate 6a, it was found that constant temperature lines would almost coincide with the lines of constant total heat when in the super- heated region. It was, therefore, considered worth while to put on the scale of approximate temperatures as given on the upper right-hand corner of this sheet. This will be found to be of service when it is desired to determine the pressure corresponding to a certain temperature and specific volume. For a large part of this superheated region this temperature scale agrees with the temper- ature as obtained from the degrees of superheat and the temper- ature of vaporization. It is not intended, however, to be used as an accurate means of obtaining the temperature, as the error in using it may easily be one degree. Entropies.—To draw accurately the lines of constant entropy it was necessary to construct a large total heat-entropy diagram from which could be obtained the values of total heat for the various pressure lines and any entropy line. In the superheated region this was done for each entropy line, but for the wet region cnly every fifth line was obtained in this manner, as the others could be put in by dividers. Qualities.—The lines of constant quality were obtained by com- puting the total heats for such qualities as 70, 80, and 90 per cent., and after these were plotted, the intermediate ones were obtained by dividers. The Heat of the Liquid Curve.—It was desired to have the heat of the liquid for all pressures, if it could be obtained without inter- fering with the other lines of the chart. The lower left-hand portion of each section was the only space available, and it was found by trial that this curve would fit there very conveniently. Since it is important to be able to read the heat of the liquid just as accu- rately as the total heat, these values should naturally be plotted on scales of equal magnitude. This has been done by supplying the numbers in parentheses, thus establishing the heat of the liquid scale. 16 STEAM CHARTS AND TABLE OF VELOCITIES Temperature of Vaporization—Since the temperature of dry saturated steam is the same as that of wet steam having the same pressure, it is possible to construct a scale on any line intersecting the constant pressure lines in the wet region, so that such a scale will represent the temperatures of vaporization. Inasmuch as the heat of the liquid is often desired for some definite temperature, it is natural to try to place these two curves as close together as pos- sible. They were therefore combined by merely graduating the heat of the liquid curve, already drawn, so that it would also give the temperature of vaporization for each pressure. In order to find the temperature of the superheated steam, it is merely necessary to add the degrees of superheat to the temper- ature of vaporization. Plate No. 7 is a special addition to the main heat-volume chart in order to enable one to work with unusually high superheats for pressures ranging from 10 to 45 pounds per square inch absolute, such as are used when reheating steam to superheats of 500 or 600°, as is done in the Ferranti turbine. It was not thought worth while to show this plate on the Index Chart. Plate No. 8 gives the external work in B. t. u. done by a pound of steam during its formation at constant pressure from water at 32° F., until it reaches the state under consideration. For Plate 8a there are two volume scales and for 8b four are used. The small drawing in the upper right-hand corner is intended to give the external work for the low pressures when the quality is relatively high, so that the volumes become too great for the main part of 8b. For pressures below one pound absolute, there is a region for which no values are given, but it will seldom, if ever, happen that such values will be desired. Table of Velocities—Any body having a weight of one pound and a velocity of v feet per second, will have kinetic energy due to this velocity equal to v? v? 29" 64.34 vy? 64.34 X 777.5 ft.-Ibs. which is equal to B. t. u. STEAM CHARTS AND TABLE OF VELOCITIES 17 Calling this kinetic energy in B. t. u., E, the equation becomes v = V64.34 X 777.5 E = 223.7 VE Then, for a nozzle which transforms the available heat energy into velocity without any losses, the velocity is Vv = 223.7 Vavailable energy, B. t. u. Table IV was prepared by using this equation. The available energy in a steam nozzle is usually taken as the total heat at entrance to the nozzle minus the total heat for the nozzle back pressure, and the same entropy as at entrance. This neglects the small difference between P,; V,, and P2 V,, where P represents pressure in pounds per sq. ft. and V,, the volume of the water. This difference is usually far too small to be considered in actual cases. Thus, taking .017 as the average volume in cubic feet of a pound of hot water, this difference is equal to O17 X 144 778 For a drop of 100 pounds per square inch in passing through the nozzle, this difference would, therefore, become only 0.3 B. t. u. per pound. (pi — pe) = .00315 (pi — pz) B. t. u. per lb. Using the Charts.—Whenever any two of the properties of super- heated steam are given, that is sufficient to locate the state point on the chart from which all of the other values may be determined at once. This is also true for wet steam, except for the case in which the two values given are pressure and temperature. These two ‘alone are not sufficient to determine the state point, since the tem- perature of wet steam is independent of its quality. The problems will supply many illustrations of the use of the charts and tables. ATMOSPHERIC PRESSURE AND BAROMETRIC CORRECTIONS The average pressure of the atmosphere at sea level is about 14.7 pounds per square inch, and for many purposes this value is all that is needed. On the other hand, it is often necessary to determine carefully the atmospheric pressure at the time and place desired. This is usually done by measuring the height of a column of mercury which is just balanced by the atmosphere. Pressures less than atmospheric are also often determined by means of mercury col- umns. All such measurements may be accurately made, if care is exercised in obtaining the readings and in applying the proper correc- tions. The Standard Atmospheric Pressure * is equivalent to the height of a column of mercury 760 millimetres high, at a temper- ture of 0° C., at sea level and 45° latitude. Reducing this to the inch basis the standard atmospheric pressure becomes 760 X .03937 = 29.9212 inches at 32°, at sea level.and latitude of 45°. Since the engineer is not concerned with any readings of pressure closer than the thousandth of aninch of mercury, and very often only to the nearest hundredth, this is equivalent to the value usually given, viz., 29.921 inches at 32° F. The Thirty-Inch Barometer—When measuring the pressure in a condenser, engineers often speak of the vacuum referred to a thirty-inch barometer. The meaning of this expression may not always be the same, as it may be interpreted differently. However, a logical meaning and one quite generally used } is that 30 inches * The exact value adopted for international use by the Third General Conference on Weights and Measures is that of a column of mercury 760 millimetres high, the mercury being at a temperature of 0° C. and the acceleration of gravity being 980.665 centimetres per second per second. See Vol. XII, p. 66, of the ‘Travaux et Mémoirs du Bureau International des Poids et Mesures.” + See Genhardt’s ‘‘Steam Power Plant Engineering,” p. 463, 4th edition. Also “Steam Tables for Condenser Work” by the Wheeler Condenser and Engineering Company, page 5. 18 ATMOSPHERIC PRESSURE AND BAROMETRIC CORRECTIONS 19 of mercury at some definite temperature will mean the same pressure as do the 29.921 inches at 32°F. To find this temperature the equation 30.00 = 20.921 [1 +a (t — 32) | may be used where t = the required temperature, and a = .000101 = coefficient of cubical expansion of mercury per degree F. From this equation : t= Se. i, Had the standard pressure been taken as 29.92 inches instead of 29.921 inches, this temperature would, from the above equation, become 58.4° F. In making the above correction for temperature, it was assumed that the scale by which the mercury is measured is correct at this temperature. This is often the case, as when very short scales or scales made of non-expansive materials are used. However, full- length brass scales, or their equivalent, are sometimes used. If such a brass scale be correct at 62° F., as is usually intended, the temperature at which 30 such inches represent the standard pressure may be found from the equation: 30.00 [1 +b t— 62) |= 29.921[ 1 +a (b— 32) | where a= .000101 as above, and b = .00001 = coefficient of expansion of brass. The solution of this equation will give, t = 57.7° F. The difference in the above three values for this temperature is not of great importance. In all the steam charts used in this book the temperature of the mercury has been taken as 58.1° F., as that would seem to serve a more general use than does the value 57.6° derived on the assumption of a full-length brass scale having a definite coefficient of expansion. Correction Due to Temperature.—In order to facilitate the reduction of readings of a mercury column at any temperature to the temperature of 58.1°, Plate 9a has been prepared It js correct 20 ATMOSPHERIC PRESSURE AND BAROMETRIC CORRECTIONS for those cases in which a non-expansive scale is used, and may also, of course, be used indirectly to correct to any other temperature desired, say 32°. When a Full-Length Brass Scale is used to measure the height of a column of mercury of some considerable magnitude, such as a barometer, and the temperature of this scale is not close to 62°, the correction should be made to the scale as well as to the mercury. Thus, for barometers having such scales, it may be desired to reduce the reading to 32° for the mercury and 62° for the brass scale. The following values * give this correction for a column 30 inches high. Hence, for any other column, say h, it is only necessary to multiply by (3) Temp: 8, | Commperign miivorss |) emp, n, | ComnaETQN sy Ince 100 : Subtract .193 40 Subtract .031 95 .180 90 166 35 .017 , 32 .009 85 .153 30 .004 13) . io 28.5 Add .000 : 25 .010 70 .112 20 .024 65 .099 62 091 15 .037 : 10 -051 60 085 5 .065 58.1 .080 5 noe 55 .072 : — 5 .092 50 .058 —10 .106 .045 = —15 .119 Variation of Atmospheric Pressure with Altitude.—It may hap- pen that it is not convenient or practicable to have a barometer at the same place at which it is desired to measure the pressure of the atmosphere. It does not require much of a difference in elevation in * From Table 1, “Circular F, U. S. Weather Bureau No, 472.” ATMOSPHERIC PRESSURE AND BAROMETRIC CORRECTIONS 21 order to make quite a variation in atmospheric pressure. To obtain the true pressure at some elevation which is not extremely different from that at which a barometer is read, it is only necessary to observe the temperature of the atmosphere and obtain the alti- tude of each place. Then, by means of Plate 9b, this correction may be readily obtained. This plate* was intended to serve only in those cases in which the change in altitude might be 100 feet or less. But after completion it was found to agree with the correc- tions obtained from the above table for variations in altitude of as much as 1,000 feet. The average altitude is to be used when ob- taining the correction from this plate. Reduction to Standard Gravity——When it is desired to make a true comparison of pressures determined by heights of mercury columns, it is necessary to reduce them to a common standard of gravity. The standard usually adopted for this is that at sea level and a latitude of 45°. Table 1} will give these corrections for the various latitudes. The variation of gravity due to change in altitude is usually much too small to require correction of mercury columns. This correc- tion is proportional to the height of the mercury and for a 30-inch column amounts to about .0018-inch for each 1,000 feet above sea-level. It is, of course, to be subtracted. It is important not to confuse this correction with the variation of the pressure of the atmosphere itself, due to change in altitude, as has already been given. Correction Due to Capillarity—The capillary action between mercury and glass will cause a depression of the column if the pressure is constant so that the mercury becomes stationary. The correction to be made for this action may be very large in case the tube or the mercury is dirty or if a very small tube is used. The height of the meniscus and the diameter of the tube are the two most important factors required to determine the amount of this correction. Table II will give these corrections for nearly all cases. * Prepared from Table 21, Vol. II, ‘Report of the Chief of the Weather Bureau, 1900-1901.” + Abridged from Table 101 of the ‘Smithsonian Tables.” 22 ATMOSPHERIC PRESSURE AND BAROMETPIC CORRECTIONS Reduction of Mercury Column to Pounds per Square Inch.—For the most important work in connection with the measurement of steam pressures by means of mercury columns, it is not usually necessary to convert the reading to any other unit. Thus the absolute pressure in a condenser being measured by the difference in the heights of two mercury columns, it is convenient and proper te use the inch of mercury as the unit to express such pressures. Density of Mercury.—When it is necessary to convert inches of mercury to pounds per square inch, Table III may be used. This table was prepared by using the following equation: One cubic inch at 32° F. is equal to 1 + (.000101) (t — 32) cu. in. at t° F- The density, 0.49117 pounds per cubic inch at 32° F., was obtained from the equivalent, 13.59545 grams per c.c. at 0° C.* The den- sities for the other temperatures were then computed and were afterward found to agree with Table 76 of the Smithsonian Tables. * This is the value given by Table 19 of “Landolt and Bornstein Phystkalisch-Chemische Tabellen.” (23) "WND OFET 1145 B.t.u. WUD 08 é r : 5 E i 3 w a 8 < ay /S ee = &g Oo ve WENO eel pIrS ie) A 3 g : : e fa a 3 “IFND BRE 3 pz 3 ; a ~ Sg> © 3 = ep en “ g 3 lacysuy a 8 “ a} "SOUry a . © Rg LP ND 601 2 oo SE a ° ‘3 3 8 »/ 13 ey ol Is 2 IVa = TT - 3 hsND 6'€ 3 a & = M35 OTT 70H 1830.L Specific. Volume (24) PLATE 1A Specific Volume, Cu. Ft. per Lb. 1.6 1.8 2.0 2.2 24 2.6 2.8 3.0 3.2 i 3.6 3.8 1470 1450 ~ ea wo €> B.t.u. per Lb. 8 Total Heat, B.t.u. per Lb. Total Heat 1310 1.2 14 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 Specific Volume, Cu. Ft. per Lb. PLAte ip 1.2 14 1.6 1.8 2.0 ‘1270 1240 8 S Total Heat, B.t.u. per Lb. 1170 (380) (360 Liquid in parentheses Heat of (330) 1110 (320) 1100 tH! Specific Volume, Cu. Ft. per Lb. 2.2 2.4 2.6 2.8 3.0 22 240°. 28 Specific Volume, Cu. Ft. per Lb. : =e a © 3.2 32° 3.4 3.4 3.6 3.8 (25) 1270 1260 1240 1230 1220 1200 1190 1180 1170 1160 1150 1140 1130 1110 Total Heat, B.t.u. per Lb. (26) PLATE 2A Specific Volume, Cu. Ft. per Lb. 7 8 1440 1440 1430 1420 1410 1410 1400 1380 1370 g 1330 Total Heat, B.t.u. per Lb. 1310 1300 1250 1240 4 5 6 7 8 9 10 Specific Volume, Cu. Ft. per Lb. Total Heat,B.t.u. per Lb, PLatE 2B (27) Specific Volume, Cu. ft. per Ib. 1230 1220 1220 1210 1210 1200 1200 1190 1190 1180 1180 1170 1170 1160 1160 Total Heat B.t.u. per lb. 150 1150 a _= iI o 1140 1140 3 < 130 1130 a eo q ‘Buz 1120 ° & 1110 1110 1100 1100 - (300) 090 1090 (290) @ 3 & 1080 1080 S (280) q 2 a = 1070 1070 = (270) 3 a 1060 1060 . (260) a os Son 2 1050 1050 © (50) x 1040 ts 1030 30) 7 8 Specific Volume, Cu. ft. per Ib. (28) PLATE 34 Specific Volume, Cu. Ft. per Lb. 38 1350 1350 340 1340 11330 1330 1320 1320 1310 1310 1300 1300 1290 1290 1280 1280 m 1270 v7 = 9 S a a 3 ° S 1260 1260 8 —Q aa} s s S 1250 1250 3 q x vl —_ i i 1240 1240 e f 1230 1230 1220 1220 1210 1210 1200 1200 1190 1190 1180 1180 1170 4170 1160 1160 fe 1150 1150 22 24 30 32 34 36 38 12 Fe) 16 18 20 2% 28 Specific Volume, Cu. Ft. per Lb. PLATE 33 12 14 16 18 1140 1120 1110 1100 1090 1080 1070 1060 B.t.u. per Lb. 1040 Total Heat 1020 \ 1010 (220) 1000 (210) S Bo se so se S ae os "5 (180) 960 170) Heat of the Liquid in parenthese. 12 14 16 18 Specific Volume, Cu. Ft. per Lb. 20 22 24 28 30 x oa Or Hor OF v; Pon tig ° > re °. ais th; 4g cups e On 2t leg, 20 2 24 26 28 30 Specific Volume, Cu. Ft. per Lb. (29) 32 34 38 40 1130 1110 1100 1090 Total Heat B.t.u. per Lb. 1000 990 980 970 960 950 940 32 34 36 38 1280 1270 1260 1250 1240 Total Heat, B.t.u. per Lb. = = 1130 41120 1110 1100 1090 1080 (30) 40 50 50 60 60 Specific Volume, Cu. Ft. per Lb. 70 80 90 100 70 80 90 100 Specific Volume, Cu. Ft. per Lb. 110 110 120 120 PLATE 4A SO 96 130 ev 130 1280 1270 1200 Total Heat, B.t.u. per Lb. Piate 42 (31) Specific Volume, Cu. ft. Ib. 40 60 1070 1060 1050 105¢ 1040 1030 1020 1010 1000 1000 990 990 = 2 o e) 980 980 & 3 3 + s 979 970 64 a i x Vv — 960 960 a 8 —_— : i & 950 950 * 940 940 930 930 920 920 n oO Dn a 910 q o a 7m i} z 3 90 890 co) s 3 = 880 880 x 870 i 870 100 110 120 130 40 50 60 70 80 90 Specific Volume, Cu. ft. per lb. 1210 1200 1190 1180 1170 1l 1150 1140 1130 1120 .t.u. per Lb, B. 4 1110 Total Heat 1080 1060 1050 1040 1030 1020 (32) 150 200 Specific Volume, Cu. Ft. per Lb. 350 250 300 350 Specific Volume, Cu. Ft. per Lb. 400 PuaTEe 5A 450 Total Heat,B.t.u. per Lb. (33) PuatTE 5B 300 350 400. 450 250 Specific Volume, Cu. Ft. per Lb. 1000 990 & R 980 980 970 970 960 960 950 940 940 930 930 ‘qT Jod -n'3°g vax [vI0.L a o Ss eS Ss Ss an a S B S > e e > a — Ss a a n n Nn a} ao *g] s9d ‘n'}"g Qvoy [e10., 870 870 860 860 850 850 840 830 820 810 & a on oO sosoyjudsed UT pInbry oy} jo waxy 800 250 300 350 400 450 Specific Volume, Cu. Ft. per Lb, 200 150 (34) PLATE 6A Specific Volume, Cu. Ft. per Lb. 500 600 700 800 900 1000 1100 1200 1300 1490 1600 1800 2000 2200 1140 1130 1120 1110 1100 1090 1030 1070 1060 t.u. per Lb. e oS 5 1040 1030 Total Heat, B.t.u. per Lb, Total Heat, — o iy So 1010 1000 900 1000 1100 «1200-1360 :1400 1800 Specific Volume,-Cu. Ft. per Lb. 500 600 709 800 Total Heat, B.t.u. per Lb. Heat of the Liquid in parentheses 930 920 910 900 890 870 oo nn o> oo ag So o a So oo R = 810 800 790 780 (50) an 2a SS: y as S we Ss x wn Ss y Ss = co s PLATE 6B 500 500 600 600 700 700 800 800 Specific Volume, Cu. Ft. per Lb. 900 1000 1100 1200 1300 11u0 1200 1300 Specific Volume, Cu. Ft. per Lb. 1400 1600 1600 1800 1800 2000 2000 (35) 2200 2200 930 920 910 900 890 880 870 860 g & Total Heat, B.t.u. per Lb. oo eo Ss oO yD Ss 810 800 790 780 770 760 750 740 730 Total Heat, B.t.u. per Lb. (36) 16 1440 1430 1420 1410 1400 1390 8 oS 1370 1360 1350 1340 1330 1320 1310 1300 1290 16 18 18 PLATE 7 Specific Volume, Cu. Ft. per Lb. 20 2 24 26 28 30 40 50 60 1440 1430 1420 1410 1400 1380 Total Heat, B.t.u. per Lb. 22 26 28 30 40 Specific Volume, Cu. Ft. per Lb. THIS PLATE IS TO SUPPLEMENT PLATES 3A AND 4A FOR THOSE EXCEPTIONAL CASES IN WHICH EXTREMELY HIGH SUPERHEAT IS USED FOR COMPARATIVELY LOW PRESSURES. THE TWO FOLLOWING PLATES GIVE THE EXTERNAL WORK DONE DURING THE CONSTANT PRESSURE FORMATION OF STEAM FROM WATER AT 32° F. PLATE 8A (38) (aInssorg }URISUOD) *q7y Jod *n'}°g ‘HIOAA JeusoyXW = s & B R 3 3 & 30 20 10 0 0 8 8 6 6 Specifie Volume, Cu. Ft. per Lb. 5 Specific Volume, Cu. Ft. per Lb. =< eS S = S o 2 5 3 sS z a a = Ss a = 3 » re = - m= -~ _ = (ginsserg }UBISUOD) “qT sed °n'}'g SYIOAA [eusIa} XY (39) PLATE 8B & (ainssarg yuRSUOD) "qT N'Y ‘HIOAA [BOUIN XY Specific Volume, Cu. Ft. per Lb. (aanssazg yuejsu0D) -q’'] Jod *n'}°g ‘YIOAA [eUIOIX| Ss = > Ss 3s = 3 an oo wn Ss R & & 3 (2ansserq yuejsu0D) *q’] Jod “n'y "gq ‘YIOAA [BUIDIX| 50 10 160 100 140 70 Specific Volume, Cu. Ft. per Lb. 20 Correction in Inches of Mercury | (40) PLATE 9A Showing correction of Mercury Column due to Temperature, when scale is correct at the observed temperature Observed reading in Inches of Mercury 0 15 20 25 30 e & a 2 2 ercu rature of 02 -00 -02 04 - 06 08 -10 2 ald 16 10 15 20 25 30 Observed reading in Inches of Mercury PLATE 9B Showing correction of Barometric Reading due to change in Elevation 2000 Altitude in Feet 3000 4000 12 ll 10 09 +08 Correction, in. of Hg., per 100 ft. . —J s 3000 Altitude in Feet 4000 _ nN S Ss > ao ‘Correction, in. of Hg., per 100 ft. .20 -18 16 4 12 .10 08 7) 06S 04S rcury 02 8 oy 00 028 a 04 § 06 2 .08 5 LO 2 i 18 20 (41) TABLE I SHOWING CORRECTIONS TO REDUCE BAROMETRIC READINGS TO 45° LATITUDE * For These Lati- Barometer Reading in Inches of Mercury For These tudes the Cor- Latitudes the rection is to be’ Correction is Subtracted 22 23 24 25 26 27 28 29 30 to be Added 1 0° .059 | .061 | .064 | .067 | .069 | .072 | .074 | .077 | .080 90° 10° .055 | .058 | .060 | .063 | .065 | .068 | .070 | .073 | .075 80° 20° .045 | .047 | .049 | .051 | .053 | .055 | .057 | .059 | .061 70° 30° .029 | .031 | .032 | .033 | .035 | .036 | .037 | .039 | .040 60° 40° .010 | .O11 | .011 | .012 | .012 | .012 | .013 | .013 | .014 50° 45° .000 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | .000 45° * Abridged and rearranged from Table 101 of the Smithsonian Tables, TABLE II CORRECTION OF THE BAROMETER FOR CAPILLARITY f Correction to be added in inches Diameter Height of Meniscus in Inches Tube in Inches OL +02 203 -04 205 -06 -O7 -08 15 024 . 047 . 069 . 092 -116 alesse .20 O11 . 022 . 033 045 . 059 .079 dates .25 . 006 .012 .019 .028 .037 047 . 059 Batya .30 004 . 008 .013 -018 . 023 .029 035 . 042 .385 phe’ 005 . 008 .012 .015 .019 .022 .027 .40 ee . 004 . 006 . 008 .010 .012 .014 .016 .45 cod sets . 003 . 005 . 007 . 008 .010 .012 .50 Sisto SfGee . 002 . 004 005 . 006 . 006 . 007 .55 bestexs Siti . 001 . 002 003 . 004 005 005 + From Table 103 Smithsonian Tables, modified by giving the correction only to the nearest thousandth of an inch. TABLE III DENSITY OF MERCURY f Temperature Pounds per Temperature Pounds per oF, Cubic Inch oF, Cubic Inch 0 .4928 58.1 .4899 10 . 4923 60 .4898 20 .4918 70 -4893 30 -4913 80 .4888 32 -4912 90 . 4883 40 .4907 109 .4878 50 .4903 110 4873 tSee page 22. (42) TABLE IV SHOWING THE THEORETICAL VELOCITIES ATTAINED BY STEAM EXPANDING ADIABATICALLY IN A FRICTIONLESS NOZZLE The velocities are given in feet per second for each B. t. u. up to 599. Available Energy B. t. u. per Ib. 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 0 0 707 1001 1226 1415 1582 1732 1872 2000 2122 2237 2346 2450 2550 2647 2740 2830 2917 3001 3083 3164 3241 3318 3392 3465 35387 3607 3676 3743 3810 1 224 742 1026 1246 1433 1598 1747 1885 2013 2134 2248 2356 2460 2560 2657 2749 2839 2925 3010 3091 3172 3249 3325 3400 3473 3544 3614 3683 3750 3817 2 316 775 1050 1266 1450 1613 1761 1898 2026 2146 2259 2367 2470 2570 2666 2758 2848 2934 3018 3100 3180 3257 3332 3407 3480 3551 3620 3689 3757 3823 3 387 806 1073 1285 1467 1628 1775 1911 2038 2158 2270 2378 2480 2580 2675 2767 2857 2942 3026 3108 3188 3265 3340 3414 3487 3558 3627 3696 3763 3829 4 4A7 837 1097 1304 1484 1643 1789 1924 2050 2169 2281 2389 2490 2590 2684 2776 2866 2951 3034 3116 3196 3273 3348 3422 3494 3565 3634 3703 3770 3835 5 500 866 1120 1323 1501 1658 1803 1937 2062 2180 2292 2399 2500 2600 2694 2785 2874 2960 3042 3124 3204 3280 3355 3430 3501 3572 3641 3710 3777 3842 6 548 895 1141 1342 1517 1673 1817 1950 2074 2191 2303 2409 Q511 2609 2703 2794 2882 2968 3050 3132 3211 3288 3363 3437 3508 3579 3648 3717 3783 3849 592 922 1163 1361 1533 1688 1831 1963 2086 2202 2314 2419 2521 2619 2712 2803 2891 2976 3059 3140 3219 3296 3370 34.44 3516 3586 3655 3723 3790 3855 8 633 949 1184 1379 1550 1703 1844 1976 2098 2214 2325 2430 2531 2628 2721. 2812 2900 2984 3067 3148 3227 3303 3377 3451 3523 3593 3662 3730 3796 3861 671 975 1205 1397 1566 1718 1858 1988 2110 2226 2336 2440 2540 2637 2730 2821 2908 2993 3075 3156 3234 3310 3384 3458 3530 3600 3669 3737 3803 3868 TABLE IV SHOWING THE THEORETICAL VELOCITIES ATTAINED BY STEAM EXPANDING ADIABATICALLY IN A FRICTIONLESS NOZZLE (43) The velocities are given in feet per second for each B. t. u. up to 599. Available Energy B. t. a. per Ib. 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 0 3874 3939 4002 4063 4125 4185 4245 4302 4361 4418 4473 4530 4585 4639 4693 AT45 4797 4849 4901 4952 5002 5052 5101 5150 5198 5246 5294 5341 5387 5434 1 3881 3946 4008 4070 4131 4191 4251 4308 4367 4424 4479 4536 4590 4644 4698 . 4750 4802 4854: 4906 4957 5007 5057 5106 5155 5203 5251 5299 5346 5392 5439 2 3888 3952 4014 4076 4137 4197 4257 4314 4372 4430 4485 4541 4596 4650 4703 4755 4808 4859 4911 4962 5012 5062 5111 5160 5208 5256 5303 5350 5397 5443 3 3894 3958 4020 4082 4143 4203 4263 4320 4378 4435 4490 4547 4601 4655 4709 4761 4813 4865 4917 4967 5017 5067 5116 5164 5212 5260 5308 5355 5401 5448 4 3900 3964 4027 4088 4149 4209 4268 4326 4383 4440 44.96 4552 4607 4661 4714 4766 4818 4871 4922 4972 5022 5072 5121 5169 5217 5265 5312 5359 5406 5452 5 3907 3970 4033 4094 4155 4215 42°74 4332 4389 4446 4502 4558 4612 4666 4719 A771 4823 4875 4927 4977 5027 5077 5126 5174 5222 5270 5317 5364 5411 5457 6 3913 3976 4039 4100 4161 4221 4280 4338 4395 4451 4508 4563 4617 4671 4724, 4776 4828 4880 4932 4982 5032 5082 5131 5179 5227 5275 5322 5369 5416 5461 7 3920 3982 4045 4107 4167 4227 4286 43.44 4401 4457 4513 4569 4623 4677 4729 4781 4833 4885 4937 4987 5037 5087 5136 5184 5232 5280 5327 5373 5420 5466 8 3926 3989 4051 4113 4173 4233 4291 4350 4407 4.462 4519 4574 4628 4682 4735 4787 4839 4891 4942 4992 5042 5091 5140 5188 5236 5284 5331 5378 5425 5470 9 3932 3995 4058 4119 4179 4239 4297 4356 4413 4468 4524 4580 4634 4688 4740 4792 4844 4896 4947 4997 5047 5096 5145 5193 5241 5289 5336 5382 5429 5475 (43 a) TABLE V THEORETICAL VELOCITIES IN FEET PER SECOND FOR EACH TENTH OF A B.T.U. UP TO 80. B.T.U. Ree 30 al. ae ak OB 0 0 71 100-122 141 158 173 187 =—.200 212 0 1 224 237 248 259 269 278 287 295 303 310 1 2 316 325 333 340 347 354 360 367 374 381 2 3 387 392 400 406 412 418 424 430 436 442 3 4 447 453 458 463 469 AT 4 480 486 491 496 4 5 500 505 510 515 520 525 530 535 540 544 5 6 548 553 557 562 567 572 576 580 584 588 6 7 592 596 600 604 608 613 617 621 625 629 7 8 633 637 641 645 649 653 656 659 663 667 8 9 671 675 678 682 685 689 692 696 700 704 9 10 707 26710 714 718721725728 78Q_s7BH_C“‘ySD 10 11 742 746 749 753 757 760 763 766 769 772 11 12 775 779 782 785 788 791 794 797 800 803 12 13 806 810 813 816 819 821 824 827 830 833 13 14 836 839 842 845 848 851 854 857 860 863 14 15 866 869 872 875 878 881 884 887 890 893 15 16 895 898 901 904 907 909 912 915 918 921 16 17 923 925 928 931 934 936 939 942 945 947 17 18 949 952 955 957 960 962 965 967 970 972 18 19 975 977 980 982 985 987 990 993 995 998 19 20 1001 1003 1006 1008 1011 1013 1015 1018 1020 1023 20 21 1026 «©1028 §=©1031 «1033 «1036 61038) =61041 = 1043) 1046 =: 1048 21 22 1050 1053 «1055 1057 1060 1062 1064 1067 1070 1072 22 23 1074 1076 1079 1081 1083 1085 1088 1090 1093 1095 23 24 1097 «©1099 1101-01104 =. 1106S 1108) 1111S 1113) 1115-1118 24 25 1120-1122 1124) 1126-1128) 1180) 1133) 11385) -1137 1139 25 26 1141-1143) 11450-1148) 11150): 1152) 1154 = 1156) =: 1159s: 1161 26 27 1163) «1165 1167) -1169) 11711174 = 1176): 1178 =: 1180-1182 a7 28 «=««1184 «1186-1188 1190-1192 1195-1197 1199 1201. «12038 29 1205 1207 1209 1211 1213 1216 1218 1220 1222 1224 29 30 1226 =1228) =—-1230) 12382) 1234 = 1236) = 1238) 1240) 1242) 1244 30 31 1246 1248 1250 1252 1254 1256 1258 1260 1262 1264 31 32 1266 «©1268 «61270 «1272 1274 = 1276 1278) 1280): 1281 = 1283 32 33 1285 1287 1289 1290 1292 1294 1296 1298 1300 1302 33 34 1304 1306 1308 1310 1312 1314 1316 1317 1319 1321 34 35 1323 1325-1327) :1329 1331): 1333) 1334 =: 1336) 1338 «1340 35 36 1342 1344 «1346 «©1348 §=1350 «1352 1353 1355 1357 1359 36 37 1361 1363 «1365 «61366 =—:1368)— 1370-13872) 1374 = 13875) 1877 37 38 1379 §=1381 1383 1384 1386 §=61388 =—:1390) Ss -1392—S «1393 :1395 38 39 1397 1399 1401 1402 1404 1406 1408 1410 1411 1413 39 40 1415 1417 1419 1420 1422 1424 1426 1428 1429 1431 40 TABLE V First Srx Lines CoRREcTED (43 a) TABLE V THEORETICAL VELOCITIES IN FEET PER SECOND FOR EACH TENTH OF A B.T.U. UP TO 80. B.T.U. ere oC 2d 2 2 & & £ F ss «Ser 0 0 71 100 =. 122 141-158 173 187 200 212 0 1 224 235 Q45 Q55 265 Q74 283 292 300 308 1 2 316 324 332 339 847 354 360 367 374 381 2 3 387 = 304 400 406 412 418 424 430 436 442 3 4 447 453 458 463 469 ATA 480 486 491 496 4 5 500-505 ssB10.—té‘é ‘dU SSC“‘ gg = 35.8%. PROBLEMS 49° 10. Steam having an absolute pressure of 80 pounds per square inch and a quality of 98% is passed through a superheater on its: way to an engine. If the steam leaving the superheater has a tem-. perature of 450° F., find: (a) Degrees of superheat, (b) Heat added per pound by superheater, (c) Specific volume of the superheated steam, (d) What per cent. of the heat added by this superheater was: required to complete the vaporization. Solution: For steam having an absolute pressure of 80 pounds per square inch, we find, from Plate 2b: Temperature of vaporization = 312° F. Hence (a) Degrees of superheat = 450 — 312 = 138° F. Also from this same plate, for the given pressure and quality, we observe: Initial total heat = 1164 B. t. u. Then, by running along the 80-pound pressure line until it inter- sects the 138° superheat line, we find, from Plate 2a: Final total heat = 1253 B. t. u. Hence (b) Heat added by superheater = 1253 — 1164 = 89 B. t. u. Also from Plate 2a, we may read: (c) Specific volume = 6.65 cu. ft. per pound. For the given pressure we may read, from Plate 2b: (d) Total heat of dry saturated steam = 1182.3 B. t. u. Hence the heat required to complete the vaporization of the wet steam is 1182.3 — 1164 = 18.3 B.t. u. Therefore the portion of the heat which was added by the super= heater in order to complete the vaporization is 18.3 i 20.6% 50 PROBLEMS 11. If one pound of steam expands adiabatically from an ab- solute pressure of 100 pounds per square inch and 145° superheat, to an absolute pressure of 1.4 inches of mercury, find: (a) Final quality. (b) Final volume. (c) Ratio of expansion. (d) Work done. Solution: From Plate 2a, for the 100 pound line and 145° superheat, we read: Initial total heat = 1262.3 B. t. u. Initial entropy = 1.69 Initial volume = 5.45 cu. ft. Then turning to Plate 5b, and running along the entropy line 1.69 until it intersects the pressure line 1.4 inches of mercury, we read: Final total heat = 925 B. t. u. (a) Final quality = 83.4% (b) Final volume = 397 cu. ft. From these results we have: (c) Ratio of expansion = BINA SOUe gg Be = 72.8 In order to obtain the work done during this expansion, it is necessary to obtain the intrinsic heats for the initial and final states, since the work done during adiabatic expansion is equal to the loss of intrinsic heat. The intrinsic heat for any state may be obtained by subtracting the external work, as obtained from Plate 8, from the total heat. For the initial pressure of 100 pounds and superheat of 145°, from Plate 8a, we read: The constant pressure external work = 100.3 B. t. u. Hence Initial intrinsic heat = 1262.83—100.3 = 1162 B. t. u. PROBLEMS 51 For the final pressure of 1.4’’ Hg. and a quality of 83.4%, we read from the upper right-hand corner of Plate 8b: The constant pressure external work = 50 B. t. u. Hence Final intrinsic heat = 925 — 50 = 875 B. t. u. (d) Work done during adiabatic expansion = 1162 — 875 = 287 B. t. u. or 223300 ft. Ibs. 12. Plot to scale, on a pressure volume diagram, the line rep- resenting the adiabatic expansion for the previous problem. 100 Pressure Volume Diagram 90 For a pound of Steam expanding Adiabatically froma 80 Pressure of 100 #/3” and 145° Superheat to a final Pressure of 1.4'Hg. 70 60 50 40 30 20 Absolute Pressure, Ibs. per sq. in. 10 0 % 50 75 100 125 150 175 200 225 250 275 300 325 350 375 Volume, cu. ft. Fia. 3. Solution: This curve is easily obtained by reading the volumes determined by the constant entropy line, 1.69, cutting the various pressure lines, assumed as desired. Thus: Plate used Pressure Volume 2a, 100 5.45 2b 60 8.00 2b 40 10.85 52 PROBLEMS Plate used Pressure Volume 3b 20 19.5 3b 10 35.9 4b 5 66.4 4b 2.6 119.3 5b 1.2 237. Bb 1.4” He. 397. 13. A closed metallic tank having a cubical content of 30 cubic feet contains 3 pounds of steam having an absolute pressure of 75 pounds per square inch. If heat is now abstracted from this tank by pouring cold water over it until the pressure of the steam within has fallen to 40 pounds per square inch absolute, find: (a) The initial superheat. (b) The final quality. (c) The heat abstracted. Solution: Specific Volume = 2 = 10 cu. ft. per lb. Hence by turning to Plate 2a, and running along the 75-pound line until it intersects the 10 cubic feet specific-volume line, we find: (a) Initial superheat = 490° F. Initial total heat = 1421 B. t. u. per lb. (b) This must be a constant volume abstraction of heat since the steam is confined in a closed tank. Hence for a specific volume of 10 cubic feet per pound and an absolute pressure of 40 pounds per square inch we may read from Plate 2b: Final quality = 95.3% Final total heat = 1126 B. t. u. per lb. (c) The heat abstracted during any constant-volume process is equal to the loss of intrinsic heat, since no work is done during such a process. The intrinsic heat for any state may be obtained by subtracting the external work, as obtained from Plate 8, from the total heat. PROBLEMS 53. At the intersection of the 10 cubic feet specific-volume line and the 490° superheat line on Plate 8a, we read: The constant pressure external work = 138.5 B. t. u. per lb. Hence Initial intrinsic heat = 1421—138.5 = 1282.5 B. t. u. per lb. From the same plate, at the intersection of the 40-pound pressure line and the 10 cubic feet specific-volume line, we read: The constant pressure external work = 73.8 B. t. u. per lb. Hence final intrinsic energy must be 1126 — 73.8 = 1052.2 B. t. u. per lb. Therefore for the 3 pounds of steam cooled at constant volume, we have: Heat abstracted = 3 [1282.5—1052.2] 3 [230.3] = 690.9 B. t. u. ll 14. A pound of steam having a pressure of 62 pounds per square inch absolute and a temperature of 340° F. is heated isothermally until its pressure becomes 20 pounds per square inch absolute. Find: (a) Initial and final superheats. (b) Initial and final entropies. (c) Initial and final total heats. (d) Initial and final volumes. (e) Initial and final intrinsic heats. (f) Heat required to effect the change. (g) Work done during this expansion. Solution: From Plates 2b and 3b respectively, we may read: Temperature of vaporization for 62 Ibs. = 295° F. Temperature of vaporization for 20 lbs. = 228° F. Therefore (a) Initial superheat = 340 — 295 = 45°F. Final superheat = 340 — 228 = 112°F. Then from Plate 2b, for the pressure of 62 pounds and 45° superheat we may read: 54 PROBLEMS (b:) Initial entropy 1.67 (c1) Initial total heat = 1201 B. t. u. (di) Initial volume = 7.46 cu. ft. Also from Plate 3a, for the pressure of 20 pounds and 112° superheat we may read: ll (b2) Final entropy 1.803 (c,) Final total heat = 1209.3 B. t. u. (dz) Final volume = 23.7 cu. ft. From Plate 8a, for the pressure of 62 pounds and the superheat of 45° we read: The constant pressure external work = 85 B. t. u. From Plate 8b, for the pressure of 20 pounds and the super- heat of 112° we read: The constant pressure external work = 87.3 B. t. u. Then we may obtain: (e) Initial intrinsic heat = 1201 -— 85 = 1116B.t.u. Final intrinsic heat = 1209.3 — 87.3 = 1122 B.t.u. The heat required to go along a constant temperature line is equal to the change in entropy multiplied by the absolute tem- perature. Hence for this case (f) Heat required = (1.803 — 1.67) (840 + 460) = 106.4 B. t. u. given to the steam. Representing the initial and final states by the subscripts 1 and 2, we then have, from the fundamental equation of thermodynamics: Work]? ~—_— Heat 2 Gain in 2 sgl ameill telnead done |: supplied |, intrinsic heat |, = 106.4 — [1122 — 1116] = 106.4 —6 = 100.4 B. t. u. Notre.—This problem is of much more importance as a drill in thermodynamics than as a practical question; because it is very difficult to arrange the necessary apparatus to permit steam to expand isother- mally while in the superheated field. Those who think that the heat required to effect this change should be equal to the difference in total heats are referred to the meaning of that term as given in the introduction. PROBLEMS 55 15. The pressure in a steam pipe is 105 pounds per square inch absolute, and in a throttling calorimeter, which is connected to this pipe, the pressure is 15 pounds per square inch absolute. If the temperature in the calorimeter is 238° F., find the quality of the steam. Solution: From Plate 3b, for a pressure of 15 pounds we read: Temperature of vaporization = 213° F. Hence Superheat in calorimeter = 238 — 213 = 25°F. Then from the opposite page, Plate 3a, for a pressure of 15 pounds and 25° superheat we read: Total heat = 1163 B. t. u. The throttling in a calorimeter takes place without any ap- preciable loss of heat by conduction or radiation, so the total heat remains constant. Hence by running along this total heat line of 1163 B. t. u. until we intersect the 105-pound pressure line, we may read, from Plate 2b: Quality = 97.38% 16. If the temperature in the calorimeter of the previous prob- lem had been 276° F., and the pressure in the calorimeter had been: 16 pounds per square inch absolute, find the quality of the steam in the main having a pressure of 187 pounds per square inch ab- solute. Solution: Proceeding as before, from Plate 3b we find the superheat in the calorimeter to be 276 — 216.3 = 59.7°F. Then, from Plate 3a, for the 16 pound line and 59.7° superheat we read: Total heat = 1180 B. t. u. From Plate 1b, at the intersection of the 1180-total-heat line 56 PROBLEMS and the 187-pound-pressure line, which is readily found by the eye between the 185- and 190-pound lines, we read: Quality = 98% 17. The pressure in the seventh stage of a twelve-stage turbine is 10.5 pounds per square inch absolute, and its quality is estimated to be 94%. Supposing that a fair sample of the steam in this stage might be obtained, what pressure would have to be main- tained in a throttling calorimeter in order that it might be used to determine this quality, provided that there shall be at least 10° superheat in the calorimeter? Solution: From Plate 3b, for an absolute pressure of 10.5 pounds and a quality of 94% we read: Total heat = 1085 B. t. u. Then, running along this total-heat line until we intersect the 10° superheat line, we read from Plate 6a: Pressure in calorimeter = 0.16 lbs. per sq. in. abs. Note.—This extremely low pressure required in the calorimeter is very difficult to obtain and the scheme is not therefore a very practical one. 18. If steam having a quality of 99% is generated in an auto- mobile boiler at a pressure of 400 pounds per square inch absolute, and is then throttled down to an absolute pressure of 100 pounds per square inch, find: (a) Final superheat. (b) Drop in temperature due to throttling. (c) Increase in volume due to throttling. Solution: From Plate 1b, for the pressure of 400 pounds and the quality of 99% we read: Initial total heat = 1200 B. t. u. per pound Initial temperature = 445° F. (almost) Initial specific volume = 1.16 cubic feet per pound Then from Plate 2b, running along the 1200-total-heat line until we intersect the 100-pound line, we read: PROBLEMS 57 (a) Final superheat = 24° F. Final specific volume = 4.6 cubic feet per pound Temperature of vaporization for 100 pounds = 328° (almost) Therefore final temperature = 328 + 24 = 352° F. (b) The drop in temperature due to throttling then is 445 — 352 = 93° F. (c) The increase in specific volume due to throttling is 4.6 — 1.16 = 3.44 cu. ft. per lb. 19. The test on a boiler gave the following data: Absolute pressure, pounds per square inch................. 185 Buber heal Ol Bia oarwaltabnccaoyxen = 7000 pounds per hr. Hence Wg = 5000 pounds per hr. = 12000 | PROBLEMS 61 25. The exhaust steam from an engine is delivered to the heating system of a building. If the pressure in this system is maintained constant at 17 pounds per square inch absolute, and if the quality of the steam exhausted by the engine is 88%, find the amount of heat that would be given up by this system when the engine is using 3,000 pounds of steam per hour. The water in the return pipes is delivered to the boiler room at a temperature of 210° F. Solution: From Plate 3b, for the pressure of 17 pounds and quality of 88% we read: Total heat = 1037 B. t. u. The heat returned to the boiler room is equal to Heat of the liquid = 210 — 32 = 178 B.t. u. The heat given up by the total weight of exhaust steam therefore is 3000 (1037 — 178) = 2577000 B. t. u. per hr. 26. A 12,000-horse-power steam turbine required 9.6 pounds of steam per horse-power hour when supplied with steam having a pressure of 175 pounds per square inch absolute and 132 degrees of superheat. If the exhaust pressure were 2 inches of mercury absolute, find (a) Heat supplied per h.p. hr. (b) Delivered thermal efficiency. (c) Cycle efficiency. (d) Theoretical water rate. (e) Efficiency ratio. Solution: Let abcde, Fig. 4, represent the cycle upon which the ideal turbine operates. Heat supplied per pound of steam = area mabcdn = Total heat], — Heat of the liquid], = 1270 —69 = 1201 B. t. u. This value of total heat at the point d is read directly from Plate 1b for the pressure of 175 pounds and superheat of 132°. 62 PROBLEMS The heat of the liquid for the point a is read from Plate 5b, at the intersection of the heat of the liquid curve and the 2 inches of mercury pressure line. 2 Absolute Temperture Sp ooo oop cee es 2 UNO Entropy Fia, 4. Then, since the turbine requires 9.6 pounds of steam per h.p. hour, (a) Heat supplied per h.p. hr. = 9.6 X 1201 = 11529.6 B.t.u. One horse-power hour in B.t.u. Heat supplied per h.p. hr. 2545 (b) Delivered thermal efficiency = = Tisu0.6 = te : Area abcde (c) Cycle efficiency = Area, mabedn _ Total heat]a — Total heat]e | 1270 — 915 = 1201 ~ 1201 355 Ion = 29.55% This value of total heat for the point e is obtained from Plate 5b, by reading the total heat at the intersection of the pressure line PROBLEMS 63 for 2 inches of mercury and the 1.64 entropy line. This entropy is obtained from Plate 1b, for the state d. 2545 Net work per lb. of steam = -” = 7.17 pounds per h.p. hr. (d) Theoretical water rate = Actual thermal efficiency Cycle efficiency .221 = "9055 ~ (P17 Theoretical water rate Actual water rate rey aR = 74.7% 27. Suppose that all the conditions of operation remain the same as in the previous problem, except the water rate and the vacuum. If, by reducing the exhaust pressure to 1 inch of mercury absolute, the water rate was actually reduced to 9 pounds per h.p. hour, find (a) (b) (c) (d) and (e) as before. Solution: Referring to the same figure modified by the line a’e’ and proceeding in the same manner, we have: From Plate 5b, Heat of the liquid] = 47 B. t. u. (a) Heat supplied per hr. = 9 [1270 — 47] = 11007 B. t. u. (b) Delivered thermal efficiency = a = 23.1% From Plate 6b Total heat]e = 882 B. t. u. *, Net work per cycle = 1270 — 882 = 388 B. t. u. Hence (e) Efficiency ratio = or Efficiency ratio = g 388 (c) Cycle efficiency = a = 31.7% : 2545 (d) Theoretical water rate = 333 = 6.56 pounds per h.p. hr. .231 (e) Efficiency ratio = 37 = 72.8% 64 PROBLEMS 28. A steam turbine receives steam at a pressure of 80 pounds per square’inch absolute and a superheat of 77° F. If this steam expands in a single set of nozzles to a condenser pressure of 1.4 lbs. per square inch absolute, and if we assume no nozzle losses, find, for the steam leaving nozzles: (a) Quality. (b) Specific volume. (c) Velocity. Solution: From Plate 2b, for the initial state we read: Total heat = 1223 B. t. u. Entropy = 1.67 From Plate 5b, for the final pressure and the entropy 1.67, we read: Total heat = 950 B. t. u. (a) Quality 84.4% (b) Specific volume = 204 cu. ft. per lb. The energy which has been transformed into velocity is equal to the difference in total heats. Initial total heat — Final total heat = 1223 — 950 = 273 B.t.u. per lb. Then, from Table IV, we read: (c) Velocity = 3696 feet per second. 29. The theoretical steam engine operates on the incomplete expansion cycle, as shown by abcde, Figs. 5and 6. The pres- sure at the throttle is 165 pounds per square inch absolute and the superheat is 143° F. If the back pressure is 2.8 pounds per square inch absolute and the expansion ratio is 6, find, for the ideal cycle: (a) Pressure at release. (b) Quality at release. (c) Net work of cycle in B. t. u. (d) Cycle efficiency. (e) Water rate of an ideal engine working on this cycle. PROBLEMS 65 DpBqBiPV Pressure Volume Fie. 5. Absolute Temperture gp ecu eke ettee Entropy Fie. 6. Solution: From Plate 1b, for the pressure of 165 pounds and the superheat of 143° we read: Initial total heat 1274 B. t. u. per lb. Initial specific volume = 3.4 cu. ft. per lb. Initial entropy 1.65 66 PROBLEMS Since the expansion ratio is 6, the specific volume at release must be Ve = 6 Vs = 6 X 3.4 = 20.4 cu. ft. per lb. For the ideal cycle, the expansion line is an adiabatic. Hence by following the 1.65 entropy line until it intersects the 20.4 specific-volume line, we may read from Plate 3b: (a) Release pressure = 18.4 lbs. per sq. in. abs. (b) Release quality = 93.7% Total heat], = 1094.5 B. t. u. In order to find the net work of the cycle we may proceed in several ways, one being to subtract the heat rejected from the heat supplied. Referring to the temperature entropy diagram, Fig. 6, we have: Heat supplied = area meabn = Total heat], — Heat of the liquid], 1274 — 107 = 1167 B. t. u. per lb. The heat of the liquid at the point e is obtained from Plate 4b for the 2.8 pounds back-pressure line. Heat rejected = area cdemn. _ oon i aauad oa ~~ heat 4 heat z done _|, = [1094.5 — 69] — 107 + 11 = 929.5 B. t. u. per lb. This intrinsic heat at c is obtained by subtracting from the total heat for this point the constant pressure external work as ob- tained from Plate 8b. We know the specific volume at c to be 20.4 cubic feet per pound, and the quality at c has already been found to be 93.7%. Therefore we may read from this plate the value of this external work, 69 B. t. u., as above given. The intrinsic heat at e is merely the heat of the liquid for the exhaust pressure and has already been found to be 107 B. t. u. The work done in going from c to e is the same as that done in going from d to e since the work done from c to d is zero, the volume being constant. But the work from d to e may be obtained directly from Plate 8b. Thus by running down the constant specific-volume PROBLEMS 67 line, 20.4, until it intersects the 2.8 pound back-pressure line, we may read: Work done |" = 11 B. t. u. per lb. Then from the above values, we have: (c) Net work of cycle = Heat supplied — Heat rejected = 1167 — 929.5 = 237.5 B. t. u. per lb. _ Net work of cycle — 287.5 _ ~ Heat supplied ~ 1167 ~ oh 8073 _ 2545 2545 ~ Net work per lb. of steam ~ 237.5 = 10.71 lbs. per h.p. hr. Note.—For the cycle abcde, Fig. 5 would commonly be drawn with a different ratio of scales so that the distance representing the admission pressure would be nearly the same as the distance representing the volume at d. In this case, however, it was desired to draw to scale the entire diagram, including the ‘‘toe’’ cfd, so that is why the volume at d may seem to be entirely too small for any real engine having a ratio of expansion equal to six. Since a pound of water occupies a space of only about .016 cubic feet, it is evidently impossible in Fig. 5 to represent such a volume with any line other than the zero volume line ea. Even with a scale of volumes as large as that used for Fig. 7 with Problem 34, the volume of the water can scarcely be shown. (d) Cycle efficiency (e) Ideal water rate 30. Find the net work of cycle in the previous problem by the method of combining the two cycles abce’ and e’cde, Figs. 5 and 6. Solution: ae a = Total _ Total y ; heat _]b heat _Je abce = 1274 — 1094.5 = 179.5 B. t. u. per lb. These values are obtained from Plates 1b and 4b as before. Net work of e 2 cycle a Work | 2 Work | e’cde # d = 69 — 11 = 58 B.t. u. per lb. These values may be obtained from Plate 8b as before. Hence the net work of the cycle abcde is 179.5 + 58 = 237.5 B. t. u. per 1b. 68 PROBLEMS 31. For the previous problem find the amount of work lost due to incomplete expansion and thereby check the result of preceding problem. Solution: The work lost due to incomplete expansion is rep- resented by cfd in Fig. 5 or 6. By inspection of these figures it will be seen that cfd = e’cfe — e’cde Work Total Total e e lost oe cheat A 7 Work lee Work Jt = 1094.5 — 977.3 — 69 4+ 11 = 659.2 B. t. u. per lb. The total heat at c is obtained from Plate 3b, as before. The total heat at f is found from Plate 4b, for the given back pressure of 2.8 pounds and the 1.65 entropy line. Work |’ is found from Plate 8b as before. Work | is found from Plate 8b as before. To check the net work of the incomplete expansion cycle we may subtract the “toe” from the complete expansion cycle. Thus abcde = abfe — cfd Total Total heat |p heat _|y = 1274 — 977.3 = 296.7 B. t. u. per lb. Hence the cycle abcde is equal to 296.7 — 59.2 = 2387.5 B. t. u. per lb. This checks the results of 29 and 30. But the cycle abfe = 32. Find the ratio of volumes of the cylinders necessary for the complete and incomplete expansion cycles of problem 29, and also find the per cent. reduction in the net work due to the incomplete expansion. PROBLEMS 69 Solution: From problem 31 we have: Per cent. reduction in net work is 59.2 506.7 = 19.9% From problem 29 the specific volume at e was 20.4 cubic feet per pound. From Plate 4b, for the given back pressure of 2.8 pounds and the entropy 1.65, we obtain Specific volume at f = 108.2 cu. ft. per lb. Then, since the cylinder volumes for any case would be propor- tional to the specific volumes as thus found for the ideal cycle, we have. Volume for complete expansion _ 108.2 Volume for incomplete expansion 20.4 = 5.3 33. Find the per cent. increase in the net work of the cycle of problem 30 for a back pressure of 1 pound per square inch absolute instead of 2.8 pounds. Solution: By referring to the solution of problem 30, we may see that by changing the back pressure only, the solution remains the same except the value of the work done during exhaust. From Plate 8b, running down the specific volume line 20.4, which is the specific volume at release, until we intersect the 1 pound back- pressure line, we obtain Work |" = 3.8 B. t. u. per lb. In problem 30 the work done against the exhaust pressure was 11 B. t. u. per pound. This is an increase of 7.2 B. t. u. per pound. Hence Per cent. increase in the net work of the cycle is 7.2 937.5 — 3-08% Nore.—In an actual engine operating under these conditions much less gain than this would be realized on account of greater loss due to cylinder condensation, more work required for vacuum pump, and more heat required to heat the feed water. 34. A direct-acting steam pump operates on a non-expansive cycle as shown by abcd, Fig. 7. 70 PROBLEMS The engine is supplied with steam having a pressure of 120 pounds per square inch absolute and a quality of 97%. If the ex- haust pressure is 18 pounds per square inch absolute, find the net work of the “ b cycle, and the ideal water rate. Solution: From Plate 8a, for the in- itial pressure of 120 pounds and quality of 97% we find: Pressure b Work ] = 80B. t. u. per lb. i. F Specific volume at b = 3.6 cu. ft. per lb. Volume Then, running down this constant vol- ume line until we intersect the back pres- sure of 18 pounds, we have: Fic. 7. d Work J. = 12 B. t. u. per lb. Then the net work of the cycle becomes 80 — 12 = 68 B. t. u. per lb. and the ideal water rate is therefore a» = 37.4 lbs. per h.p. hr. 68 35. In a uni-flow locomotive the clearance is 16% and com- pression takes place during 90% of the stroke. If the back pressure is 16 pounds per square inch absolute and the quality of the steam at the beginning of compression is 96%, find the condition of the steam at the end of the stroke, assuming the compression to be adiabatic. Solution: From Plate 3b, for the pressure of 16 pounds and quality of 96% we may read: Entropy = 1.6925 Initial specific volume = 23.8 cu. ft. per lb. PROBLEMS 71 We must next find the specific volume at the end of compression, which is equal to the specific volume at the beginning of com- pression divided by the ratio of compression. Ratio of compression = WF = 6.63 Hence 23.5 6.63 Then by going along the entropy line 1.6925, until we intersect this specific volume line 3.59, we read from Plate la: Final specific volume = = 3.59 cu. ft. per lb. Final pressure = 175 lbs. per sq. in. abs. Final superheat = 240° F. 36. An engine having a steam jacket has a clearance of 15% of the piston displacement which is 3.17 cubic feet. It is found from test results that the weight of steam inside the cylinder dur- ing expansion is 0.163 pounds. If release occurs at 95% of the stroke at a pressure of 19 pounds per square inch absolute, find the condition of the steam at release. ~~ Solution: Actual volume of the steam in the cylinder is equal to the clearance volume plus 95% of the piston displacement. But since the clearance is given in terms of the piston displacement, we nave: Actual volume = 1.10 X 3.17 = 8.49 cu. ft. Specific volume of the steam in the cylinder is Actual volume 3.49 Weight = 0.163 = 21.4 cu. ft. per lb. Then from Plate 3a, for the pressure of 19 pounds and a specific volume of 21.4 we find: Superheat = 9° F. 37. The temperature in a condenser is 87° F. and the absolute pressure is equal to 1.43 inches of mercury. Find the weight of air 72 PROBLEMS which must be removed per pound of exhaust steam if the steam entering the condenser has a quality of 85%. Solution: From the curve giving temperature of vaporization, Plate 5b, for 87°, we find: Pressure of wet steam = 1.29” Hg. and for this pressure and a quality of 85% Specific volume = 442. cu. ft. per lb. The air pressure is equal to the total pressure less the pressure of the steam, that is, 1.43 — 1.29 = 0.14” Hg. 1 inch of mercury at 58.1° being equal to 0.49 pound per square inch, the air pressure in the condenser is P= .14 X .49 X 144 Ib. per sq. ft. The absolute temperature in the condenser is T = 460 + 87 = 547°. Hence the weight of air existing in the same space as each pound of steam is 38. If the temperature of the atmosphere is 68.5° F. and the relative humidity is 70%, find the pressure due to this moisture and the weight of moisture in each 1,000 cubic feet of the atmosphere. Solution: From the temperature of vaporization curve of Plate 6b, for the temperature of 68.5° we may read: Saturation pressure = 0.7” Hg. Since the relative humidity is almost exactly equal to the Actual vapor pressure Saturation pressure » it follows that PROBLEMS 73 Actual vapor pressure = .70 X 0.7 = 0.49” Hg. = .49 X .49 = 0.24 lb. per sq. in. Then from Plate 6a, for this pressure, 0.24 pound and the temperature 68.5° given by the temperature scale on the right-hand side of the sheet, we may read: Specific volume = 1,310 cu. ft. per lb. Superheat = 10° F. It is interesting to observe that this means the vapor in the air in this condition is merely superheated steam having a pressure of 0.24 pound per square inch absolute or 0.49 inch of mercury and a superheat of 10°. Hence the weight of this steam contained in 1,000 cubic feet of space would be 1000 1310 = 0.764 lb. Then by Dalton’s law of partial pressures, for the given vapor pressure and temperature, this same weight will be in this space when it is also full of air. Notre.—This scale of approximate temperatures, placed on the upper right-hand corner of Plate 6a, is not intended to be sufficiently accurate to solve all hygrometric problems. See page 15 for the variation of this scale from the true temperatures. 39. Find the diameter of pipe to furnish steam for a 9,000 kw. turbine, if it requires 12.8 pounds per kw.-hr. when running at full load with a pressure of 200 pounds per square inch absolute and 145° superheat at the throttle. Allow a velocity of 8,000 feet per minute. Solution: From Plate 1b, for the steam at the throttle with a pressure of 200 pounds and superheat of 145° we find: Specific volume = 2.84 cu. ft. per lb. The area of the pipe in square feet is equal to Volume in cu. ft. per min. _ lbs. per kw.-hr. X kw. X sp. vol. Velocity in ft. per min. 60 X velocity 74 PROBLEMS But if the diameter of the pipe in inches is d, then the area in square zd? 4x 144 '4 xX 144 X water rate X load X sp. vol. \ 7 X 60 X velocity feet is also equal to Hence d = 1.748 ae rate X load X sp. vol. velocity in ft. per min, 12.8 X 9,000 XK 2.84 = 1.748 8,000 = 11.16”. This means, of course, that a 11-inch pipe would be used. 40. The Parson’s turbine at the Fiske Street Station of the Commonwealth Edison Co., Chicago, has an exhaust opening to the condenser of 252 square feet. If the water rate for a back pressure of 1 inch of mercury and a load of 25,000 kw. is 11.65 pounds per kw.-hr., find the velocity through this opening, assuming the steam has a quality of 80%. Solution: From Plate 6b, for the pressure of 1 inch of mercury and quality of 80% we find: Specific volume of exhaust steam = 523 cu. ft. per lb. Then Volume — 11.65 X 25,000 X 523 Velocity = an = 60 x 252 = 10,080 ft. per min, 41. If, in the second stage of a Curtis turbine, the pressure at the entrance to the nozzle is 56 pounds per square inch absolute, the superheat 64°, and the pressure in the next stage 36.8 pounds per square inch absolute, find the cross-sectional area of each nozzle for this stage in order that 112 such nozzles will give a flow of 96,000 pounds of steam per hour. Assume the coefficient of velocity for this condition to be 96%. PROBLEMS 75 Solution: From Plate 2b, for a pressure of 56 pounds and super- heat of 64° we find: Total heat Entropy 1,208 B. t. u. 1.69 Then, by following this entropy line until we intersect the 36.8- pound Iine on Plate 3a, we may read: Total heat = 1,173 B. t. u. Specific volume = 11.6 cu. ft. per lb. Then the theoretically available energy to produce velocity in this stage is 1,208 ~— 1,173 = 35 B. t. u. per lb. and from Table IV this velocity is 1,323 feet per second. For the given nozzle coefficient the velocity would therefore be .96 X 1,323 = 1,270 ft. per sec. Let A = area in sq. in. of each nozzle. Let N = number of nozzles. Let v = velocity in ft. per sec. Let V = specific volume of steam as it passes the throat. Then the flow in pounds per hour is p _3000NAv _ 25NAv = aaaye oe _ FV _ 96,000 X 11.6 _ “25 Nv 25 X 112 X 1270 ~ For this and the following problem it will be observed that. the ratio of pressures between stages is considerably over 57%, which means that the maximum velocity occurs at the throat of the nozzle and that the nozzle will therefore have no divergent. part. aes 0.313 sq. in. 42. Find the size of nozzles and the pressures in the third and fourth stages for the turbine of the previous problem if in each of these stages there are to be 82 nozzles and the theoretically avail- able energy is to be 35 B. t. u. per stage. Assume that the nozzle, bucket, and rotational losses amount to 25% of this energy. Neglect. 76 PROBLEMS radiation loss and leakage of steam and assume nozzle coefficient to be 96% as before. Solution: The losses in each stage go to reheat the steam, and this reheating is assumed to take place at constant pressure after the expansion in the nozzle, since the nozzle loss is very small. Hence, from the previous problem, the steam enters the third-stage nozzle in the following condition: Pressure = 36.8 lbs. Total heat = 1,173 + .25 X 85 = 1,181.8 B.t.u. Then, from Plate 3a, for these two values we obtain: Entropy = 1.70+ Superheat 27° Since it is desired to give up 35 B. t. u. in this stage, we now follow down this entropy line until the total heat becomes 1,181.8 — 35 = 1,146.8 B.t.u. and from Plate 3b we find for this total heat and entropy of 1.70+ Pressure = 23.5 |bs. Specific volume = 17.1 cu. ft. The velocity corresponding to 35 B. t. u. is from Table IV equal to 1,323 feet per second. Hence, using the same notation as before, the area of the throat of the nozzle for the third stage is - ff V. .. 96,000 X 17.1 >" 25Nv 25 X 82 X (.96 X 1,323) A = 0.632 sq. in. Then the steam enters the nozzle of the fourth stage in the following condition: Pressure = 23.5 lbs. Total heat = 1,146.8+ .25 X 35 = 1,155.6 From these two values and Plate 3a we obtain Entropy = 1.713 Quality = 99.5% PROBLEMS 77 and by following this entropy line until the total heat becomes 1,155.6 —35 = 1,120.6B.t.u. we have from Plate 3b Pressure = 14.6 lbs. Specific volume = 26.2 cu. ft. Hence the area of the throat of the nozzle for the fourth stage is FV 96,000 X 26.2 Ai = 55 Nv — 35 x82 X (.96 X 1,323) = 0.967 sq. in. 43. Steam expands through a properly shaped divergent nozzle from an initial pressure of 125 pounds per square inch absolute and 220° of superheat to a final pressure of 1 pound per square inch absolute. Find the proper cross-sectional areas of this nozzle to permit a flow of one pound per second, assuming adiabatic expansion and no friction. Solution: From Plate 2a, for the pressure of 125 pounds and superheat of 220° we may read: Total heat = 1,305 B. t. u. Entropy = 1.71 and from Plate 5b for this entropy and a pressure of 1 pound we find; Total heat = 955 B. t. u. Specific volume = 285 cu. ft. Then the total available energy is 1,305 — 955 = 350 B. t. u. per lb. Let the nozzle be divided into sections, so that in each one the expansion will be sufficient to use equal parts of this total available energy. If we compute the velocity for 10 such sections we shall have 35 B. t. u. liberated in each one. By the charts we may now follow the entropy line 1.71 until it intersects the desired total heat line, and then read the corresponding pressure and specific 78 PROBLEMS volume. By Table IV the velocity is obtained, and the area of the nozzle at any section may then be found from the equation (Cu. ft. flowing per sec.) 144 Velocity in ft. per sec. ~ Area in sq. in. = Hence the following values may be at once tabulated: Available Pressure Specific * section | Entropy | Eneey | Tie | Lbs per | Volume | Yeu | Neate (EH Seri. B. t. ue ‘Abs. per Lb. Second Sa. In. Entrance| 1.71 0 1,305 125. 4.82 Sehte s oll. ate 2a 1 1.71 35 1,270 89.8 6.18 1,323 0.674 2a 2 1.71 70 1,235 62.8 8.10 1,872 0.624 2a 3 1.71 105 1,200 42.4 10.9 2,292 0.685 2b 4 1.71 140 1,165 27.6 15.1 2,647 0.822 32 5 1.71 175 1,130 17.1 22.7 2,960 1.101 3b 6 1.71 210 1,095 10.3 35.4 3,241 1.572 3b 7 1.71 245 1,060 6.10 56.5 3,501 2.32 4b 8 1.71 280 1,025 3.44 94.0 3,743 3.62 4b 9 1.71 315 990 1.90 | 160. 3,970 5.81 5b 10 1.71 350 955 1.00 | 285. 4,185 9.82 5b If these sections are now laid off at equal intervals so that the total length becomes equal to the desired length of the nozzle the form will be such that it will give uniform acceleration of the steam. This is theoretically desirable, and is due to the fact that the sections have been chosen at such points that the energy transformed into velocity in each one is constant. Such a nozzle will have curved elements for its divergent part and will therefore be expensive to make. It has been found by experiment that the divergent part of the nozzle can be made of straight-line elements without any great loss in efficiency. Since there is a large amount of hand work re- quired in making nozzles, they are therefore nearly always made with the divergent part as the frustum of a pyramid or cone. The throat of the nozzle must be approached by a gradually decreasing cross-sectional area in order to prevent loss due to the sudden con- vergence of the stream lines. . No general rule can be given to determine the best length of nozzle for all conditions. The length of the rounded entrance to the PROBLEMS 79 throat is a very small portion of the total length of the nozzle, and it is common practice for some designers to have the divergent part of the nozzle taper about one in twelve, while others make it only about one in twenty. 44, For the expansion of the previous problem find the value of the exponent n which satisfies the equation povo” = p v" where p represents the pressure and »v the specific volume of the steam at any time during its passage through the nozzle and py and refer to the pressure and specific volume at the entrance to nozzle. Solution: Rewriting the equation we have UO ae a» Go Then by the aid of a log log slide rule, or logarithms, and the results of the previous problem the following table may at once be made: n ; Po er Quality or Superheat Section P v . = n ns Plate 0 125. 4.82 1.000 1 000 dah 220° 2a 1 89.8 6.18 1.392 1.283 1 325 | 166° 2a, 2 62.8 8.10 1.990; 1.680 1.825 | 112° 2a 3 42.4 10.9 2.950 | 2.265 1.325 60° 2b 4 27.6 15.1 4.535 3.131 1.325 5° 3a 5 17.1 22.7 7.32 4.71 1.285 97.6% 3b 6 10.3 35.4 12.14 7.35 1.250 95.1% 3b 7 6.10 56.5 20.50 11.73 1.227 92.6% 4b 8 3.44 94.0 36.38 19.50 1.210 90.2% 4b 9 1.90 | 160. 65.80 | 33.2 1.195 87.9% 5b 10 1.00 | 285. 125.00 | 59.1 1.185 85.6% 5b This table shows that so long as the steam was superheated the value of m was constant, but decreased rapidly after reaching the wet region. From Plate 3a, it may be seen that the 1.71 entropy line crosses the saturation curve at a pressure of a little more than 26 pounds. 80 PROBLEMS 45. With the value of n = 1.325 as obtained from problem 44, find the pressure at the throat of this nozzle by the following equation :* Pe _ Fat = (- =e ) n—1 where pP_ = pressure at the throat of the nozzle. Po = pressure at entrance to nozzle. Solution: le)” “ee 125 (4 5) =) 4.075 625 = 495 (V5 aS) = 67.75 Ibs. = 54.2% of po. 46. With the throat pressure, as determined from the equation given in the previous problem, determine the area of the throat. and then find by trial whether this is the minimum section. Solution: Taking pressures each side of the throat pressure as: found above, and obtaining the values needed as in problem 43, we may construct the following table: Total Available Section Entropy Pressure Heat Energy B. t. u. B. t. u. Velocity Specific Area Ft. per Sec.| Volume Sq. In. 0 1.71 | 125. 1,305. 0 fabs one Cie a ea! 70. 1,245.8} 59.2 | 1,721 | 7.46 | 0.624 b 1.71 | 68. 1,242.9] 62.1 | 1,762 | 7.62 | 0.623 Throat 1.71 | 67.75 | 1,242.5] 62.5 | 1,767 | 7.64 | 0.622 ¢ 17 |) ey, 1,241.5] 63.5 | 1,782 | 7.70 | 0.622 d 1.71 | 66. 1,240. | 65. 1,803 | 7.80.) 0.623 * For the derivation of this equation see any good book on turbine nozzles, PROBLEMS 81 From the above results, it is seen that the value of the throat pressure, as found from the equation of problem 45, is checked as well as can be desired. The computations from the charts and slide rule indicate that the pressure of 67 pounds would answer equally well. 47. Supposing that the steam is compressed, adiabatically, from the final condition given in problem 43, to the pressure of 125 pounds, find the value of m to satisfy the equation pio vig = pv™ The subscript 10 refers to the state of the steam when in the section 10 of problem 43. Solution: The equation may be put in the more convenient form De (a) Pio Vv Pp Vis Quality or Section Pp v = — m Superheat Pio v _ from 44 10 1.00 285. Ls Me APY eaters 85.6% 9 1.90 160. 1.9 1.783 1.110 87.9% 8 3.44 94. 3.44 3.03 1.115 90.2% 7 6.10 56.5 6.10 5.05 1.117 92.6% 6 10.3 35.4 10.3 8.05 1.118 95. % 5 17.1 22.7 17.1 12.55 1.122 96.6% 4 27.6 15.1 27.6 18.88 1.128 5° 3 42.4 10.9 42.4 26.12 1.148 60° 2 62.8 8.10 62.8 35.2 1.163 112° 1 89.8 6.18 89.8 46.2 1.175 166° 0 125. 4.82 125. 59.1 1.185 220° From an inspection of these exponents and those as given by the results of 44, it will be seen that the value of the exponent for adiabatic compression or expansion depends upon the state of the steam and a simple general expression to determine its value cannot be given. 82 PROBLEMS When any adiabatic expansion or compression of steam is plotted on the P-V diagram the resultant curve will always be smooth, even though the variation in the exponent may be con- siderable as in the above cases. This means that the exponents change gradually from one to the other even when crossing the saturation curve. The reason that the above tabulations do not show this more fully is due to the fact that the desire to make a short table necessitated the selection of points which are some distance apart. 48. In making a turbine test, the barometer was read as 29.55 inches at a temperature of 50° F. The diameter of the barometer tube was 0.25 inch and the height of the meniscus was 0.02 inch. The vacuum was determined by reading the height of mercury in a glass tube 0.2 inch in diameter, the bottom end of which was resting in a vessel of mercury. The height of this mer- cury column was 28.85 inches at a temperature of 77° F. and the meniscus was 0.04 inch. The elevation of the condenser was 500 feet above sea level and the barometer was in another building 50 feet higher than this. The temperature of the atmosphere was 40° F. Find the pressure in the condenser in inches of mercury. Solution: The barometer corrections are: (1) Due to capillarity from Table IT.................. + .012 in. (2) Due to temperature from Plate 9a............... + .024 “ (3) Due to change in elevation from Plate 9b......... + .055 “ HG alco sateks une neereeaee ae cle ates + .091 in. This third correction is obtained by running along the 40° tem- perature line until we intersect the line representing the average altitude, which for this case would be 525 feet, where the correction for 100 feet is seen to be.110. Then for 50-foot change in elevation the correction would be one-half of this, or .055, as above. The corrections to the mercury column attached to condenser are: PROBLEMS 83 (1) Due to capillarity from Table II.................. + .045 in. (2) Due to temperature from Plate 9a................ — .053 ‘ POM es wept Ga sg et ere eden cate eee ene — .008 in. Hence by using correction to nearest hundredth The barometer is 29.55 + .09 = 29.64 in. at 58.1° and the vacuum is 28.85 — .01 = 28.84 in. at 58.1° and absolute pressure in condenser is 29.64 — 28.84 = 0.8 in. He. Note.—Without making any of the above corrections the pressure in the condenser would appear to be 29.55 — 28.85 = 0.7 in. Hg. and from Plate 6b the difference in total heats between a pressure of 0.7 in. and 0.8 in. for some common entropy line is seen to be about 6 or 7B. t.u. This amount of heat might easily mean 2% of the total available energy or as much as 15 or 20% of the energy available in the last stage. 49. Absolute Temperature Entropy Fia. 8. Let the area abcdefka, Fig. 8, represent the net work in B. t. u. of the theoretical cycle for the Ferranti turbine.* * See Power, December 30, 1913, page 908, for a description of this turbine. 84 PROBLEMS If the steam is received at a pressure of 145 pounds per square inch absolute and 360 degrees of superheat, then expands, adia- batically, to 25 pounds per square inch absolute, and, after that, is reheated at constant pressure to the initial temperature and then finally expands, adiabatically, to the back pressure of 1.5 inches of mercury, find (a) Net work of the cycle. (b) Heat supplied per pound of steam. (c) Cycle efficiency. (d) Theoretical water rate. Solution: From Plate 2a, for the pressure of 145 pounds and superheat of 360° we find Total heat | d H, = 1,877 B. t. u. Entropy | = 1.76 and from Plate 3a for this entropy and a pressure of 25 pounds we have Total heat | =H, = 1195 B: tu, From Plates 1b and 3b respectively, we also find: Temperature of vaporization for 145 lbs. = t, = 356° F. Temperature of vaporization for 25 lbs. = t, = 240° F. Hence tg = 356 + 360 = 716° F. and since the temperature after reheating is to be the same as this, the superheat at f becomes t, — ty = 716 — 240 = 476° Then, from Plate 7, for the pressure of 25 pounds and 476° of superheat we have Total heat | =H, = 1986 Bit. w Sf Entropy | = 1.959+ or1.96— PROBLEMS 85 For this entropy and the condenser pressure of 1.5 inches of mercury, we may then obtain from Plate 5a Total heat | = H, = 1077 ask and from Plate 5b the heat of the liquid corresponding to the pressure of 1.5 inches of mercury is h, = 60 B. t. u. Then by inspection of Fig. 8 we have: (a) Net work of cycle = area b bedee + be'fka = H, —H, + H; — H, = 1377 — 1195 + 1386 — 1077 = 491 B. t. u. per lb. of steam. (b) Heat supplied per lb. of steam = area labcdefn = H, — H, + H, — h, = 1377 — 1195 + 1386 — 60 = 1508 B. t. u. (c) Cycle effici ee ee) c) Cycle efficiency = prog = 32.67% : . 2545 (d) Theoretical water rate is 0 5.19 lbs. per h.p. hr. ea = 6.94 lbs. per kw. hr. 50. Supposing the ordinary steam turbine cycle abcdj, Fig. 8, had been followed, find (a) (b) (c) (d) as before. Solution: The only additional numerical value needed is the total heat at the point 7. From Plate 5b, for the entropy 1.76 and the pressure of 1.5 inches of mercury we find: Total heat | = H, = 967.3 B. t. u. J Hence (a) Net work of cycle = H, — H, = 1377 — 967.3 = 409.7B. t. u. (b) Heat supplied per pound = H, — h, = 1377—60=1317 B. t. u. ‘ 409.7 (c) Cycle efficiency = Bl7 = 31.15% 86 PROBLEMS (d) Theoretical water rate = ae, = 6.22 Ibs. per h.p. hr. 412 or a = 8.32 lbs. per kw. hr. From the above results it is seen that for the theoretical cycles the decrease in water rate due to the reheating is 8.32 — 6.94 eo 19.9% and the increase in heat supplied is 1508 — 1317 191 iy iy 8% or from the cycle efficiencies the gain due to reheating is .3826 — .38115 3115 =onibe ge In an actual turbine built to operate on this cycle there would be the disadvantages of higher first cost, more complications, and greater radiation loss. On the other hand, by having the steam superheated for the entire passage through the turbine, or nearly so, the leakage, friction, and rotational losses are very much reduced. Whether the reduction of these losses and the slight increase in the cycle efficiency will be sufficient to make this type of turbine superior to others cannot yet be told. INDEX PAGE A, reciprocal of the mechanical equivalent of heat . . . . ..... =.=. .10 Absolute temperature, °F., constant for, prob.37 . . 2... 2 we ee 71 Abstraction of heat at constant volume, prob.18 . . . . ...... . . 52 Addition of heat at constant pressure, prob.9 . . . . . . ww ee ee 48 Adiabatic compression, in steam engine, prob. 35 . F Gere ea ie of steam, exponent for, prob. 47... sig) adds! Glee cael SS Adiabatic expansion of steam, exponent for, be 44 and 4s pte bah, . 79, 80 P-V diagram for, probh12 . . 2. 1. 2. ww, so & ad & G1 work during, prob. dd 2. 2 1. we ee en eo a wp ow Ga 2 BO Adiabatics, definition of. 2. 2. 2. 2... ee ee ee ee ee 8 Air in condenser, prob. 37 oS oe! anor g we Bec Ble cap ce OL Altitude, variation of atmospheric pressure with aa 4g : Rie diel pity at- a7 20) Area representing heat, general discussion . . . . ges eo) ve 780, O supplied turbine, prob. 26 . . . at. Gin Si akon Ge ae Jem ae Satna. bb) a ca Area representing work, general Riccueson ao Ga Bins Gs bugs . . 1,2, 4,5, 7,10 Atmosphere, relative humidity of, prob.38 . . 2. . 2. 1 we ee eee 72 Atmospheric pressure, the standard. . . . ........... . .18 variation of, with altitude . . . 2. . 2... 1 ee ee ee ee. 20 Available energy, definition of ee. shy ongacs “empian, Sar Joy AB Sits cil, cy auey ee en eS insteam nozzle . ... fr. iis Ten We CART RES Teel OG.