aSSSBSBBBSSSaaSSSSSSSSSSSBSSS (Qacttell ltmoer0tty Sitbtart| 3ti(aca, ^tm ^ntk BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. SAGE 1891 Cornell University Librai^ TJ 265.C33 1922 Notes and exam nples on the theoi^ of heat 3 1924 003 955 881 The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924003955881 NOTES AND EXAMPLES ON THE THEORY OF HEAT AND HEAT ENGINES LONDON AGENTS: SIMPKIN, MARSHALL, HAMILTON, KENT & Co. Ltd. NOTES AND EXAMPLES ON THE THEORY OF HEAT & HEAT ENGINES BY JOHN CASE, MA. Cambridge W. HEFFER AND SONS Ltd. 1922 Second Edition, Revised and Enlarged, 1922 First Issued in 1913 under the Title: "A Synopsis of the Elementary Theory of Heat and Heat Engines " I ALL rights reserved] PREFACE. TTHE present volume is an enlargement of my "Synopsis of the Elementary Theory of Heat and Heat Engines," which has been out of print for some time. The most important new feature is the addition of a large number of examples worked out in the text, with many others for the student to work out himself. 1 hope that these additions will considerably increase the value of the book. The book is not intended as an exhaustive text-book, as sufficient ejtcellent treatises exist already, but rather as a companion to lectures, to help the student tp see at a glance the important points of the subject,' and to^ assist him with his revision for examinations. I hope also that the practical engineer, who has to deal with the elementary thermodynamics of heat engines will find the book of value, and to this end all the more important formulae are printed in heavy type. I am greatly indebted to Mr. C. Barclay-Smith for his labour of proof reading. JOHN CASE. Bristol, 1922. CONTENTS. PAGES General Principles I - - i Perfect Gases I 8 General Principles II - i8 Perfect Gases II - 29 Steam and other Vapours - 36 The Steam Engine 50 Surface Condensers 63 Engine Trials - - 65 Orifices and Nozzles - 72 Theory of Injectors - 76 Refrigeration - - - 88 Air Compressors - 94 Combustion - 102 Internal Combustion Engines 109 Miscellaneous Examples 118 GENERAL PRINCIPLES L 1. If heat be supplied to a system, the system may perform work ; if work be done on a system, heat may be produced. Therefore heat is a form of energy. 2. The effects of supplying a system with heat are in general (i.) to change the physical state* of the system; (iiO to make the system do work. The energy of a system depends only on the physical state of the system. Changing the physical state of a system changes its store of energy. In general we are here only concerned with its store of thermal energy. All the energy which is not kinetic is called internal energy. 3. The conservation of energy gives : Total heat increase of work done increase of energy = kinetic + by the + internal absorbed by energy, if system energy a system any whenever heat is supplied to a system. Hence (i) if there be no increase of kinetic energy and no work done we can define internal energy by saying : 4. When a system absorbs heat without doing external work or increasing its kinetic energy, it is said to change its internal energy by an amount equal to the heat energy absorbed. (ii.) If there be no change of kinetic or internal energy, all the heat energy absorbed is converted into work energy. Thus we arrive at 5. The First Law of Thermodynamics : When work is performed by a system at the expense of heat energy, for every unit of heat that goes out of existence one * The term ' physical state ' is meant to include such things as position, velocity, etc., besides the thermal, electrical, etc. state. unit of work is performed, and conversely, provided of course they are both measured in the same units. But we generally measure heat energy on one scale and mechanical energy on another, and a correcting factor is necessar)- ; this we call J. 6. The Unit of Heat that we shall employ is the quantity of heat necessary to raise the temperature of one pound of water 1° C. This is called a Standard Thermal Unit, and is equal to 1400 ft.-lbs. Hence on these scales J = 1400. 1 Th. Unit = 1400 ft.-lbs. 1 Ft.-lb. = 1/1400 Th. Unit.* Sometimes this unit of heat is called a pound-calorie. 7. The Specific Heat of a substance is the ratio of the heat required to raise a mass of the substance through a certain temperature rise to that required to raise b\' the same amount the temperature of an equal mass of water. By §6 the Sp. Ht. is numerically equal to heat required to raise the temperature of lib. of the substance 1° C. 8. Let dQ = heat absorbed by a system from without in any change. „ ciW = the work done by the system. „ liE = the increase of internal energy. Let there be no change of kinetic energy. Then, by § 3 : dQ = dW + dE. 9. An adiabatic change is one in which no heat is absorbed or given up, and then d\V + dE = 0. An isothermal change is a change at constant temperature. • Sometimes A is used for l/J. 10. A cycle fs a series of clianges such that at the end of the series the state of the substance undergoing the change is the same as before the changes commenced. 11. Absolute temperature. The absolute zero is the temperature at which a perfect gas would have zero volume. It is 273° below the freezing point of water on the centigrade scale. Hence, ii T = the absolute temperature, and t the centigrade temperature, T = t + 273. EXAMPLES. 1. If water is forced through a porous plug under a pressure of 1000 lbs./in.^ find the approximate rise of temperature on the supposition that all the heat produced remains in the water (l cubic foot of water weighs 62i lbs.) (Special Exam. Cambridge, 1910.) Suppose the area of the section of the porous plug is one square foot, and that the velocity with which the water passes through is x' ft./sec. Then the work done = 1000 X 144 v ft. Ibs./sec. 144,000 ,_, _,, „ .^ , = ^ , '.^n^ 1' = 103 V Th. Units/sec. 1400 The volume of water dealt with = v ft.'Vsec. ,, weight „ „ ,, „ = 62*5 v Ibs./sec. Let t°C. be the rise in temperature, then 62-5 V X 1 X ^ = 103 V. ' = el = - 2. In turning a steel shaft, 4 lbs. of metal are removed per minute. The cutting speed is 200 ft. per minute, and the pressure on the point of the tool in the direction of cut is 1700 lbs. Assuming that all the work done by the tool goes to heating the turnings, find their increase of temperature as they leave the tool. The specific heat of steel may be taken as 0"12. (Special Exam. Cambridge, 1912.) The work done per minute =1700X200=340,000 ft. lbs. 340,000 ,_^ = l400 =-"^3Th.Units. Let ^°C. = the rise of temperature of the turnings. Then 4X0-12X^=the heat given to the turnings = 243 Th. Units. 2+3 t — „ .„ ^ 500°C. approximately. 3. 1000 gallons of water are pumped per minute to a height of 80 ft. The pumps have an efficiency of 60%, and are driven by a steam engine. Find the H.p. of the engine. The kinetic energy of the water may be neglected. If the engine and boiler use 18,000 Th. Units per h.p. per hour, find the over-all efficiency of the whole plant. A gallon of water weighs 10 lbs. The work done by the pumps = 10,000 X 80 = 800,000 ft. lbs. per minute. The work done by the engine ^ 800,000 X 100 60 = 1,333,000 ft. lbs. per minute. = 40-6 H.p. The heat supplied to the boiler , ^ 40-6 X 18,000 60 = 12,200 Th. Units per minute. In work units this is 12,200 X 1400 = 17,080,000 ft. lbs. per minute, the over-all efficiency ^ work done ^ 800, 000 "work supplied 17,080,000 = -0468 = 4'7 % appro.x. 4. An iron bullet is fired obliquely at a hard steel plate with a velocity of 1600 ft. /sec. The bullet is deformed but not broken, and its velocity after striking is 800 ft./sec. The plate is unaffected. If the temperature of the bullet before striking is 20°C., shew that its temperature after striking is about 190°C. The average sp. ht. of iron is 0'12 for the range 20° to 100", and 0-13 for the range 100° to 200°C. (Mech. Sci. Trip. Cambridge, 1913.) The heat acquired by the bullet is equal to its loss of kinetic energy, which is -^ (1600' - 80tf) =29,800 Wit. lbs. 64-4 where W is the weight of the bullet. In heating the bullet from 20° to 100° the heat absorbed = W X 0-12 X 80 = 9-6 W Th. Units. In heating it from 100° to fC, the heat absorbed is W X 0-13 X (t - 100) = W iO-Ut - 13). 29800 Hence H^(0-13i- 13) + 9-6 W = ^^ W = 21-3 0-13f = 21-3 + 13 - 9-6 = 24-7 t = 190°C. 5. An engine develops 500 H.P., the power being absorbed by a brake. The brake is cooled by a stream of water which enters the brake drum at 20°C., and leaves it at 80°C. How much water is used per minute? (Special Exam. Cambridge, 1912.) 6. A shell has an average specific heat of 0"1. It hits a target with a velocity of 2000 ft./sec, and is brought to rest by it. If the heat is at the moment all in the shell, find the rise of temperature that ensues. (Special Exam. Cam- bridge, 1913.) 7. Water, flowing in a pipe, with velocity Vi ft./sec, passes abruptly into a pipe of larger section where its velocity is Va ft./sec. The loss of head due to the sudden enlargement is (vi—Vi) /2g feet. If all the energy dissipated is expended in heating the water, find the rise of temperature when 40 cubic feet of water pass per second from a pipe of sectional area 1 sq. ft. to one of area 4 sq. ft. (Intercoll. Exam. Cambrid,^e, 1908.) 8. An iron wire is suddenly loaded to a stress of 40,000 lbs. /in.^ and stretches under this load by ^o of its length. Shew that the temperature rises about 4°C. The sp. ht. of the wire is 0"11, and its density 480 lbs. per cubic foot. (Mech. Sc. Trip., 1914.) PERFECT GASES. I. 12. The Characteristic Equation of a perfect gas is pv = RT = R(273 + t) ( i> = the pressure in lbs. per ft.' V = volume of 1 lb. in cubic ft. where < t = temperature °C. T = absolute temperature. >i? = a constant. In calculations we must take 144pv = R{273 + t) when p is in lbs. per sq. inch. For air at 0'^ C, 760 mm., R = 96, v = 124, using lbs. and feet as units. 13. The Specific Heat of a gas has different values according as heat is absorbed at constant pressure or constant volume. kp = sp. ht. at constant pressure kv = sp. ht. at constant volume. For air kp = "237, *,, = -169, both in heat units. 14. The Internal Energy of a Gas: — (i.) Joules Experiment. Joule took two vessels and filled one with air under pressure, while the other had a vacuum, the two being connected by a pipe with a tap in it. The tap was opened and the air expanded into the empty vessel ; no change of temperature occurred, after the air had come to rest. Now (a) no work was done ; no heat was absorbed or given out ; there was no final change of kinetic energy. .'. , by the energy equation (B), the internal energy had not changed. But (b) the pressure and volume had changed, the temperature did not change, and the internal energy had not changed. .'■ Conclusion : the internal energy of a gas depends only on the temperature. (ii,) Take the case of gas absorbing heat at constant volume and so not doing any work. By §7 we have change of internal energy = heat taken in by gas = ^w X mass X change of temperature. .'• for one lb. of gas : increase of internal energy is given by E2 - Ei = Kv (T2 - Ti) Th.U. This is true in whatever manner the gas takes in heat. 15. Work done by gas in Expanding at constant pressure. Suppose we have a pound of gas shut up in a cylinder behind a piston which fits the cylinder closely. Let S = section of cylinder (sq. it), li = length of cylinder occupied by gas before expansion (ft.), /j = ditto after expansion (ft.), p = pressure of gas (lbs. /ft".) Supply heat to gas in such a way that p is constant. Total force on piston = pS, distance moved by piston = / -~ h, .'. work done = pS{h — h) = p{Sh - Sh) = Piv., - V,) = R{T., - Ti) Ft.-lbs. 10 where v^ and I'l are the volumes of the gas before and after expansion. In a small change, when vi = v, and t\. = v + dx', the work done is pdv, or, expressed in thermal units, pdv/J. 16. To prove that kp - kv = R, let one pound of gas expand, behind a piston, at constant pressure. Then Q = heat received = k/, (T-i — Ti). Th. Units, W = work done = R {T^ - Tj Ft. lbs. [by 515. = A{T,-T,) Th. Units; ••• , by §8, E = change of internal energy, = Q- W = kp{T., -T,) -^ (^*-'^'). -U-I-) 'T,~ rj; also E = k,. {T, - Ti) hy Hi, " kp ~ J ~ kv, or kj, - ^v = R, if R be supposed already expressed in Th. Units. 1 7. Adiabatic Equation for a gas : to prove pv^ = constant, where y — k±^ kv Take a small change in which (p remains constant, j V changes to v + dv, ^T „ „T + dT; ( dQ = since change is adiabatic, then j dW = pdv (ft.-lbs.) [by §17 (i.)] = ^^ Th. Units, ^ dE = kvdTiTh. U.)(by §14); 11 .-. ^+k^dT = Q (by §8) (i.) but pv = RT, •• P (ii.) V and logj^ + log v = log T + const (iii.) From (i.) and (ii.) : — RT dv , ^ — + kvdT = 0, J V ^^- T ° •'■ -y log u + kv log T = constant ; or (kp — kv) log V + kv log T = constant. •'• {kp — kv) log V + kv (log p + log v) = constant, by (iii.) .'■ kp log V + kv log p = constant, ■7 log V + log p = constant, kv .'. pv '" = constant, ■y or pv = constant, where V = — = ^''^^^ ^°^ S'i'^- kv 18. Isothermal Expansion oS a Gas. (r = constant). (i.) Work done. Suppose that, while the volume under- goes a small change from v to v + dv, the pressure remains constant. Wehavej^v = RT = constant since T is constant. 12 Work done in small change = pdv. : „ „ „ whole „ from {pivd to (^»vj. 2 = W = J pdv, 1 /.fa r »i i?r loge -', I.e. W = PjV, log. ^-RT, log.;^ (ii.) Change of Internal Energy. T is constant, .•. there is no change of internal energy. (iii.) Heat supplied = work done + change of E. Q = i?T, loge -^ In isothermal compression W is the work done on the gas, and Q is the heat given out. 19. Expansion oi a gas in general. The general law of expansion is pv" = C, where n and C are constants. (i.) Work done = J pdv = J C^ V, - c(v;--v,'-") or ^^ 1 - « ' Now C = p,v," = p,v,". . ^ _ PaVa - PiV, p,V, - PsjV, 1 - n n - 1 _H(T,-T,). n- 1 Not including the case of unresisted expansion 13 (ii.) Change of Temperature. P^v^ = piv", (i.) and M»=i? = lm, ,.(ii.) J 2 i 1 .*. v"~ Ta = v" Ti by division of (i.) by (ii.) n— 1 •■■-^'-(sr-^-d;)"-^- (iii.) Change of Internal Energy. E = kv{T^ - Ti). (iv.) Heat Supplied = W + E, = ^^^'~,^^^ + KiT, - T.), M — 1 20. Adiabatic Expansion of a Gas. Put n = V in above : _ i)i Pi - j'aPa _ J?(ri - Ta) y-1 y-1 ' Q = 0. A B = - W, r-i -■=&:)"'-- (I) ^^.- 14 EX-VMPLES. 1. One pound of gas at 15^C.. JO lbs. per sq. inch absolute pressure is enclosed in a cj-linder with a moveable piston. The gas occupies 9'65 ft.' J4 Th. Units of heat are supplied to the gas, and the temperature rises to 9J°C., the pressure being kept constant. Find the external work done, and the increase of internal energy. Find also the specific heat at constant volume. 275 -^ 9 ' The new volume = ^^^^ X 9'65 = 1J"3 ft." The work done = JO X l-ff (12-3 - 9-65) = 7660 ft. lbs. = 5-47 Th. Units. Increase of internal energy = heatsupplied — work done. = J4 - 5-47 = lS-53Th. Units. This also = h. (92 - 15) = 77*v. • ^„=l^;53=o-J41. -*^ 2. For a certain gas, which may be assumed perfect, the weight of a cubic foot, at 0°C., 14-7 lbs. per sq. in. pressure, is found to be 0-0S93 lbs., and the sp. ht. at constant volume is 0'155. Find the \-alue of y. (Intercoll. Exam. Cam- bridge. 1911.) We have r = the volume of one pound =^-^i — = ll-Tft* 0-0S95 ^^-"• P = 14-7 X 144 lb. ft.' r = J73. From §IJ, we have p _j>f _ 147 X 144 X IIJ _,r „ ^ ~ Y ^73 ^ ^^'^ ^*- ^^- ^its. = 0-062 Th. Units. From §16, kp = R + /v = 0J17 > = |^ = ^!; = i.4o. 15 3. A cylinder, provided with a piston, contains 1 lb. of dry air at 273°C. and 367'5 Ibs./in.^ absolute. The air expands adiabatically to five times its original volume ; it is then compressed isothermally to its original volume, and the cycle completed by supplying heat at constant volume until the initial conditions are obtained. Find (l) the work done in foot-pounds, (2) the heat taken in and rejected. (Mech. Sc. Trip., 1910.) We shall use the equations of § 20 for dealing with the first part of the process. We have Vi 1 Ti = 546 and — = V Vi 5 .■. taking y = 1-406, ir )-406 Ta = ( ^ ) X 546 = 284. The work done by the gas No heat is taken in or rejected since the process is adiabatic. For the second stage we use § 19. We now have Tx = 284. The work done by the gas = RTt loge - = 96 X 284 X loge \ = - 27,300 X 1-61 = - 44,000 ft. lbs. i.e. the work done on the gas = + 44,000 ft. lbs. There is no increase of internal energy, and the thermal equivalent of this (= 44,000 -^ 1400 = 31-4 Th. Units) is rejected. In the last stage, no work is done since the volume remains constant. The heat supplied is k^ (546 - 284) = 0-169 X 262 = 44-3 Th. Units. 16 Summing up we have 1st 2nd 3rd Stage. Stage. Stage. Heat taken in (Th. Units) -314 44-3 Total. 12-9 Work done (ft. lbs.) ... 62,000 - 44,000 18,000 (= 12-9 Th. Units.) The last column serves as a check on the work; for always : heat received — heat rejected = work done. 4. Unit mass of a perfect gas is made to expand in such a way that its volume is at any instant proportional to its pressure, the initial values being Vi and pi respectively. If the pressure be changed from ^ to ^ + dp, find the values of dE and dQ in terms of p and dp. (Intercoll. Exam. Cambridge, 1909.) We have (§14) dE = kvdT (i.) Also, since pv = RT, we have dp dv _dT P "*" V ~ T ^rr._rr ( ^P , dv\ PV/dP dv\ But, by the conditions of the problem, ^ ^and^ =^ V p Vi p dT = 2 pv dp _ ,7 vi pdp R p pi R Hence, from (i.) Again, dE = 2^' -^^ -pdp (ii.) dQ=dE+ dW K pi P\ J -!;■*■*• (4'+j) <"'•) (ii.) and (iii.) are the expressions required. 17 5. Two lbs. of air at 17°C. are compressed suddenly from an absolute pressure of 151bs./in.^ to an absolute pressure of 45 Ibs./in.'' Taking R = 95-7 and 7 = 14, find the volume before compression, and the volume and tempera- ture immediately after. (Special Exam. Cambridge, 1911.) 6. Twenty cubic feet of gas at 15°C. and atmospheric pressure are suddenly compressed (the law being ^u^'^ = const.) to one-fourth the volume. Find the pressure and temperature attained. The gas is then allowed to cool to its original temperature : find the final pressure and the work, in foot pounds, lost in compression. (Special Exam. Cambridge, 1913.) 7. One pound of air, at a pressure of 151bs./in.^ and temperature 15-5°C., is compressed adiabatically to a pressure of 301bs./in.^, and at this pressure is cooled to its initial temperature. It is then further compressed adiabatically to a pressure of 60 Ibs./in.^, and again cooled at this pressure to its initial temperature. Determine the temperature and volume at the end of each compression, and sketch the pv diagram. (Intercoll. Exam. Cambridgev 1909.) 8. A cartridge containing 4 lbs. of air at lOOOlbs./in.^ by gauge and 15°C. is placed in the chamber of a gun behind a light frictionless piston fitting the bore of the gun. The cartridge is perforated and the piston just reaches the muzzle of the gun. Calculate the mean temperature of the air and the volume of the gun, on the assumption that the air absorbs no heat from the walls of the gun. Atmospheric pressure = 14-71bs./in.' (Mech. Sc. Trip. 1912.) 9. Air is compressed adiabatically into a vessel of V cubic ft. capacity to m times the atmospheric density. Show that, if p be the atmospheric pressure, the work expended is foot-pounds. (Trin. Coll., Cambridge, Schol. Exam. 1905.) IS GENERAL PRINCIPLES II. 21. In the theory of heat engines we deal with processes in which a substance or system changes its state. These processes may be reversible or irreversible. A rcVCrsiblc process is one in which all the parts of the acting system can be brought back to their original condition without leaving any change in other bodies or systems. All other processes are irreversible. That a process should be irreversible it is not enough that it should not be directly re\-ersible ; it must be impossible, even with the assistance of all the agents in nature, to restore everywhere the exact initial conditions, leaving no changes in the agents used to effect the reversal. Examples of reversible processes : The motion of a projectile in an unresisting medium ; adiabatic and isothermal expansion. Examples of irreversible processes : The motion of a projectile in a resisting medium ; the unresisted expansion of a gas ; generation of heat by friction ; transference of heat by conduction. 22. Second Law of Thermodynamics : Heat cannot pass from one body to a wanner body without some compensating transformation taking place (without work being done). 23. Carnot's Cycle. We require a standard to indicate the maximum amount of work that we can hope to get out of an ideal engine using a gi^■en quantity of heat. We have (l) A "source" of heat consisting of an infinite body at temperature Ti absolute. (2) A " cooler " in the form of an infinite body at temperature Ta, and T, < Ti. (3) An engine using any material for its working substance. 19 Since the source and the cooler are infinite, their temperatures will be unaltered, however much heat we take from the one or give to the other. We perform the following series of operations which together constitute Carnot's cycle : (1) Let the working substance be at Ti and take heat Q' from the source. This is isothermal. (2) Take away the source and let the substance pass adiabatically to the lower temperature Ta. (3) Apply the cooler and let the working substance give out Q2 to it at Tj. This again is isothermal. (4) Take away the cooler and make the working substance pass adiabatically back to Ti, so that it is in the same state as when it started, and so has gone through a cycle. Now the engine has received Qi, given back Qa, and has none left. Hence an amount of heat Qi — Q^ has gone out of existence, and therefore an equal amount of work (measured in heat units) has been done, and we have W = Qi - Q2 heat units. ., re ■ energy obtained in work the emciency = ^^ energy supplied in heat a Qi' 24. Reversibility of Carnot's Cycle. The following things have happened: The source has lost Qi, the cooler has gained Q2, the working substance has done work = 01 — 02 through the engine. Now let the cooler give up the heat 02 to the substance at temperature T^, and let the engine do work 0i — 02 on the substance raising its temperature to Ti, so that it now has heat 0i, which it can give up to the source at Tu Thus everything is left as it started, and no changes have occurred outside the system. Hence the cycle is reversible. 20 Note that here all the reception and rejection of heat takes place at the highest and lowest temperatures respectively, and this is an essential condition for the reversibility. If at any time, durii^ the supply of heat, there were a difference of temperature between the source and the working substance, the transference of heat from the former to the latter would involve conduction which is an irreversible process. Similarly with the rejection of the heat. 25. Is this the best we can do ? To prove that any reversible engine is more efficient than any irreversible one, and hence that all engines working on any reversible cycle have the same efficiency. Take a reversible engine A , and an irreversible one B. If possible let B be more efficient than A. Let B work direct and drive .4. reversed. Let B receive heat Oi at the high temperature Ti and produce work W. Let A receive work W and produce heat Os also at Ti. If A worked direct it would do work W when supplied with Qs, whereas B requires Qi. But, B is more efficient than A. •■• Qi < Qs or Qs > 0,. The nett effect of the combination is that no external work is done, the source gives up Qi and receives Qa which is > Qi, which is impossible by the second law. .". B cannot be more efficient than A. This argument is in no wise upset if we substitute ' reversible ' for ' irreversible ' in the case of B. Our result then is that no one reversible engine B can be more efficient than any other reversible engine A. Therefore all reversible engines have the same efficiency. 21 26. To prove that the efliciency of any T reversible engine is 1 - — ', where Ti is the tempera- -1 1 ture at which the heat is suppUed, and Ta is the lowest available temperature, i.e. the temperature of the cooler. Since all reversible engines have the same efficiency it suffices to find the efficiency of any one. Take one which works with a perfect gas as its working substance. Carnot's Cycle is then represented graphically as in Fig. 1, where, along : — 1—2, heat is supplied at constant temperature Ti. 2 — 3 is adiabatic expansion. 3 — 4, heat is rejected at tem- perature Ti. 4 — 1 is adiabatic compression to temperature Ti. Qi = heat received along 1 — 2 Fig. 1. = i?T. log, ^' by §18. Qi = heat rejected along 3 — 4- = i?r.ioge-. No heat is received or rejected along 2 — 3 or 4 — 1. Now piVi = RTi = piVi, piV^ Vi y _ pivjv^ piV^ P'V^V^ Vi' V^~'^ piVf^ Va _Vi V4 Vi' Vj Vl w f/ Vi l-> Vi 1-y RT, loge :: Hence RTi loge - O T .: the efficiency (§23) = 1 - ^' = 1 - -'.which applies to any engine working on any reversible cycle. 27. Available Energy. By §§23 — 26 the maximum amount of work obtainable from a quantity of heat Ql drawn from a source at temperature Tj when the temperature of the cooler is Ta is v^ = Q.-e. = Q.(i-|)=Q:(i-f;). In other words, Ql (1 ~ 7^ I is available energy, of the heat Qi \ „ the remainder, Qi — ^ = Qs, is iinavailahle. \ i 1 If we had another cooler capable of receiving heat con- tinuously at a lower temperature To, we could get out of Qs an amount of work or available energy -o-('-f:> the unavailable energy then — Ui r^ — Ui ^ ;ir — Ui r^T — loTi; — Qo say. ^9 -I 1 J 2 -1 J i 1 Hence the unavailable energy associated with a given quantity of heat is (i) proportional to the lowest absolute temperature available in the cooler, (ii) inversely proportional to the temperature of the body which the heat is entering. Dei : The Available Energy of a system sub- ject to given external conditions is the maximum amount of 23 mechanical work theoretically obtainable from the system without violating the given conditions.* Hence %, may be regarded as a measure of the unavail- ability or the factor which only has to be multiplied by any assumed auxiliary temperature To in order to give the quantity of unavailable energy relative to that temperature. This factor is called entropy. 28. Hence, immediately, we have two definitions of entropy : (i.) If a system at temperature T receive a small quantity of heat dQ the quotient -^ is called the increase of entropy of the system arising from this cause. If A and B denote two different states of a system which are capable of being connected together by a continuous series of reversible transformations the change of entropy of the system in passing from A to S is defined as f^ dQ_ J T 2 'a where the S extends to all parts of the system and the / is taken along the reversible series of transformations referred to. (ii.) If the unavailable energy of a system with reference to an auxiliary medium of temperature To undergo any *Note the difference between total energy measured above a given zero and available energy when the same zero point is taken, e.g. the total potential energy of a mass of 10 pounds 10ft. above the earth's surface is 100 ft.-lbs., the earth's surface being taken as position of zero energy ; now interpose a table underneath the weight and 5ft. high, this does not alter the potential energy of the weight but only 10 X 5 = 50 ft.- lbs. are now available since the weight can only fall as far as the table. In the above the whole heat energy measured above 0° C is Qi, but ™ly Qi (1-To/Ti) is available. 24 positive or negative increase, and if this increase be divided by To the quotient is called the increase of entropy of the system. \^Tien (i.) is given it must be given as precisely as above. (i.) is inapplicable to most irreversible phenomena. (ii.) makes no restrictions as to the nature of the trans- formations which take place, and holds for irreversible changes. 2Sa. To prove that for maixiinum efficiency an engine must taike in all its heat supply at the highest temperature, and give out all its rejected heat at the lowest temperature. If a quantity of heat Q be received at temperature T, when the lowest available temperature is To, the unavailable energj- is "^To. Hence, if an engine receive part of its heat at temperatures below T, this unavailable energy is increased, and we cannot get so much work out of the total heat supply. Similarly the available energy is reduced if some of the heat rejection take place at temperatures above the lowest a^-ailable. Hence an engine can only have its maximum eflBciency when the whole of the heat supply is taken in at the highest temperature, and all its rejected heat is given out at the lowest temperature. 29. To prove that there is no total change of entropy when a substance or system is taken through a reversible cycle, i.e. that rdQ / T =» for the whole cycle. The unavailable energy cannot change imless the physical state changes. •"•) by Def. (ii.) of IJS, the entropy cannot change unless the physical state changes. 25 But a necessary condition for a cycle is that the physical state should be same at the end as at the beginning. •'• There can be no change of entropy. •'• the sum of all the changes of entropy, i.e. the sum of all T the -^s, is zero, ■/ — — 0, taken round the cycle. dQ T 30. To prove that the value of f 7^ is, 1 the same for all reversible paths connecting the states 1 and 2. Suppose a reversible cycie repre- sented by a curve 1 A 2 B 1 (Fig. 2). Then by §29 dQ T I = 0, dQ ■■■ /f + /f = o. B Fig. 2. 1A2 2B1 • f ^= _ f 4G= f dQ " J T J T J T' 1A2 2B1 1B2 But 1A2 and 1B2 are any two reversible paths connecting 1 and 2, fdQ. is constant for all reversible paths connecting 1 and 2, i.e. the change of entropy in passing from one state to another depends only on the two states, and not on the mode of passage. Thus the entropy of a substance under given conditions depends only on the state of the substance and not at all on its past history. We are not concerned here with the entropy of irrever- sible changes, but in these cases there is a loss of availability, and consequently a gain of unavailability of energy, i.e. an increase of entropy. 26 EXAMPLES. 1. Find the maximum amount of work which can theoretically be obtained from 1 lb. of water at 100°C. if all the heat wasted may be given out at 15°C. Assume the sp. ht. of water to be constant and equal to unity. (Intercoll- Exam. Cambridge, 1911.) The heat contained in the water = 100 Th. Units. The portion which must be wasted is . 100 X 11 = 77-2 ."■ The work which could theoretically be obtained is 22-8 X 1400 = 32,000 ft. lbs. 2. A heat engine is used to drive a reversed heat engine as a refrigerator. The heat engine takes in its heat at absolute temperature Ti and the refrigerator at an absolute temperature Ta. Both engines reject heat at To. Shew that the maximum amount of heat which can be extracted by the refrigerator per unit of heat taken in by the heat engine is r.(r, - To) T,(ro - T,) (Intercoll. Exam. Cambridge, 1913.) If the engine receive heat Qi it will do an amount of work Wi = Qi(l-^). If the refrigerator, working as an engine, receives heat Q^ at temperature To and reject it at Ta, it will do work T and reject heat = Qo ^zr ~ To Conversely, working as a refrigerator, if it receive work Wa T and heat Qo ~ it will give out heat = Qo. -I 27 If Wi comes from the first engine, it must = Wi. . _ To (r. - To) The heat which the refrigerator has extracted at temperature Ta is Qo :=r. which = -r-j^ -A Qi. The amount per unit of heat supplied to the first engine is T,{T^ - To) TxiTo - T,) 3. The specific heat of a substance at an absolute temperature T may be taken to be given by a -\- bT, where a and h are constants. Shew that, if the heat from 1 lb. of the substance be used as efficiently as possible in doing work, the quantity of heat wasted will be equal to ro|6(T,- ro) + aioge^}. Tl being the initial temperature of the substance, and To the lowest available temperature. Of a quantity of heat, dQ, taken in at temperature T, an amount must be wasted equal to ^dQ = '^^{a + bT)dT. The body can go on giving out heat until its temperature is To. Hence the total waste will be Tl f{a+bT)dT Tl A To Tl To + b)dT = Toia log« :^+b{Ti Jo ro)}. 28 4. A heat engine takes in 500 Th. Units at a temperature of 150°C. and rejects its heat to a body whose temperature rises 1°C. for every 5 Th. Units of heat absorbed. If the temperature of the body be initially 15°C. shew that the maximum work which the engine can do is 115 Th. Units, and that the final temperature of the body will then !be 92°C. (Mech. So. Trip., 1913.) 29 PERFECT GASES II. 31. We use ^ to denote entropy. By definition (i.) § 28, also dQ = dE + dW by §8. pdv T = dE + ^ by §15. for a small change. .'. dE = Td4> - -pdv, J .'■ , for a reversible cycle, since E must be same at the end as at the beginning, or O = J dE = J Td(f> - jj pdv, JJTd = J pdv. We can represent a cycle by plotting T and ^ or by plotting p and v, and this equation shows that the area (expressed in ft. -lb.) of the temperature-entropy diagram equals the area of the pressure-volume diagram, and therefore represents the work done in the cycle. 32. In the T-^ diagram, an adiabatic is represented by ^ = constant, since dQ = 0, i.e. by a line parallel to the T axis, and an isothermal is T = constant and so parallel to the axis. 30 e.g. A Caraot cycle consists of two adiabatics and two isothermals, and so is drawn as in Fig. 3, where the numbers Fig. 3 9 correspond in each diagram. 33. To find the change of entropy of a perfect gas in changing from (piViTi) to {piViT^. For small changes, dW =- _ pdv T by §15 dE = k^dT by §14.ii. ,_, dO dE + dW ,dT 1 pdv '" T J T ' but pv = RT and R V , dT R dv also log i) + log V = log i? + log T dp . dv _ dT P (i.) (ii.) By (i.) and (ii.) we can express

== kJ^ + ^) + ^ ^' p ^ P ^ "i' V ■'■ 2-l = k, loge ^ + kp log.^ (A) (b) Eliminates: d

i~ (j>i = ^>loge^ - {kp - ^Jloge-^ (B) Ji pl (c) From (i) i>2— (pl = kvloge-Tj^ + {kp — ^,;)loge— (C) 1 1 Vi 34. To calculate the entropy of a gas in a given state {p v T). Entropy can only be measured above some arbitrary state of zero entropy. We take 0° C (273° abs.), 14-7 lbs/in' as state of zero entropy. Hence Ti = 273, pi = 147 lbs/in'. Find the corresponding value of Vi ; put - r,) - kpjTi - Ti ) kpiTs — Tz) Ef&ciency = 1 - Ts - T,' Y-i = '-?; = '- (i;)~='.-(sr where r = — 37, The Otto Cyclei (Two constant-volume lines and two adiabatics.) Fig. 6. We have by §19 4 — 1. Adiabatic compression. No heat received. 1 — 2. 01 = heat received = k, (Tj-Ti) per lb. Fig. 6. 34 2 — 3. Adiabatic expansion. No heat received or rejected. 3 — +• Qi = heat rejected = k,, (T, - Tj per lb. W = work done = Qi — Qa. efficiency Qi T,-T, r, U/ = 1 0) 35 EXAMPLES. 1. Calculate the change of entropy when the state of 1 lb. of air is changed from 15 Ibs./in.', 15°C., to 2250 Ibs./in'., 10°C. We shall use equation (B) of § 33. A J. z. 1 283 ,. , . , 2250 = volume, in cubic ft., of lib. of liquid at pres- sure p. t — temperature, °C. (1) To force the liquid into the cylinder. Originally the piston is at the bottom, and volume underneath is zero. When the liquid has been pushed in the volume underneath is to ft' ; the cross section is 1 ft". Therefore the piston has been raised oo ft. •■• work done = po) ft. -lbs. 37 (2) To raise the temperature of the liquid to t° C. Taking the sp. ht, of the liquid as constant and = c Heat required = mass X sp. ht. X t. = a-t. Work done = 0, if we neglect expansion of the liquid. .*. Iw = total heat of liquid at t° C, pressure p. = (Tt +^. (3) To evaporate the liquid. When the temperature reaches the temperature of evaporation, the liquid begins to evaporate, and the piston is raised. Thus there is an increase of internal energy and work is done. At any instant let there be q lbs. of vapour and (l — q) lbs. of liquid. Then q is called the dryness fraction. Let L = the latent heat of the vapour, = heat required to evaporate lib. = work done + increase of internal energy. Then to evaporate q lbs. will require heat qL. The volume of vapour will he qV : an increase of q{V - ->). ■'• the work done = qp{V — Iw = total heat of liquid = + p(V- to) /J f Is = total heat of dry saturated vapour = I„+L=E,+ ^. E = total internal energy of vapour p + o, the entropy in the initial state, as zero, and To = 273, T 'Pw = entropy of liquid = cloge 07^* Hence the ^ — T diagram for heating water is a logarithmic curve, as shown by O A in Fig. 7. (2) When the liquid reaches the temperature of evaporation for the given pressure, evaporation commences, the temperature remaining con- stant. Therefore in the a horizontal line A B. To evaporate to dryness q requires heat qL. T' T diagram we go along the increase of entropy .'. at any instant during evaporation where T = saturation temperature. 40 = O- lOge HS=5. + ^• When evaporation is complete we have ^s = entropy of dry saturated vapour 273 ^ T' We thus arrive at point B (Fig. 7). For the wet vapour we can write, then, (3) In superheating, to increase the temperature by dT requires heat dQ = Kp dT. dT .'■ total increase of in superheating to a temperature t' (T' absolute) rTl =/ Kp-^ = Kploge—, if Kp be supposed constant. •"• ^'s = entropy of superheated vapour On the — T diagram we go up another logarithmic curve BC (Fig. 7). 42. For every different pressure there is a different temperature of evaporation, giving a point A on the line OA, and a corresponding point B. If we plot the points A and B for different pressures we get two curves AA\ BB\ (Fig. 8). 43. In §41 we saw that to evaporate to drj'ness q we go along A B a distance — , and for corn- Fig. 8. plete evaporation we go a distance -. Hence if we have 41 steam in any condition, and know its entropy and temperature, i.e. its " state point" Ci say, we can find its dryness by drawing the line of evaporation, Ai Bi, through Ci and then Ai C, <^ — (l>„i q = A, B, '#'»i - wi' 44. Specific Heat of Superheated Vapours. We have taken Kp as constant, but this is not correct. Hence to calculate the entropy of a superheated vapour we must find the mean value of Kp over the range t to t' , thus : We have I's = Kp{t' - t) + I, ^ t -t and the correct value of I' s (calculated by other means) is given in the book of tables. 45. Adiabatic Expansion of Vapours. To find the adiabatic equation. Let a vapour, in the state denoted by suffix 1, expand adiabatically to the state denoted by 2. The condition of adiabatic expansion is ^ = constant. •■• 's = q = 639-9 442-4 ^ = 4-72:3 = O-^^^- 4. Two boilers, whose volumes are in the ratio of 2 to 3, are under steam at pressures 200 and 150 Ibs./in." abs. respectively. The first boiler is \ full of water, and the second is half full. A connection is opened between them, and the resulting pressure is 1801bs./in.^ Assuming that no steam has escaped, find the total quantity of heat, per unit of total volume, rejected from, or supplied to, the whole system during the equalization of pressures. (Mech. Sc. Trip., 1913.) Let the volumes be 2 V and 3 V cubic feet. Then the volumes of water are 0-5 V and 1-5 V and „ „ steam „ 1-5 V „ 1-5 V weights of water „ 31-25 V „ 93-751^ lbs. and „ steam (from tables) „-^;jY^ 1/ „ ^;;^„ „ total weights are 31-91^ and 9424 V respectively, or 126-1 F for the two boilers. 47 The total value is 5 V. Let xV he the volume of water, the weight of which will be 62-5 xV. The volume of steam at 180 Ibs./in.'' will be (5 — x) V. The pressure is 180 lbs. /in.^ and from the tables we find that the volume of 1 lb. of steam is 2'558 ft.' {5 — x)V Therefore the weight of steam is — tttt^ — a' J JO Thus the total weight is '2-5^- + ^7558- But this must be the same as the initial weight. ^2-5. + 1^; = 126-1 which gives x = 2 nearly. To find the heat supplied or lost, we apply the energy equation of § 8 to the whole system. No external work has been done, and therefore the heat supplied = the increase of internal energy. For the first boiler the initial internal energy is (from the tables) y(31-25 X 197-2 + 0-65 X 622-1) = 6555V', for the second it is K(93-75 X 183-6 + 0-494 X 619-8) = 17,500 V, making a total of 24,055 V. In the final stage the weight of water = 125 V, and the weight of steam = 1-17 V lbs. The internal energy is V(125 X 192 + 1-17 X 621-2)= 24,730 V. Thus there is an increase of internal energy equal to 675 V. .-. The heat supplied per unit of total volume = ^|P^ = 135 Th. Units. 48 5. In order to warm the feed-water of a boiler, saturated steam at ISOlbs./in." abs. is blown into it. How much steam must be used per pound of water in order to raise the temperature from 20°C. to 60°C. ? (Special Exam. Cambridge, 1912.) 6. A cylinder fitted with a piston contains 1 lb. of steam and water at 120°C. Find how much steam and how much water is present when the volume is 10 cub. ft. (Special Exam. Cambridge, 1912.) 7. A boiler evaporates 2,000 lbs. of water per hour, the temperature of the feed being 50°C., and the pressure in the boiler 1801bs./in.^ abs. The efficiency of the boiler is 70%, and the calorific value of the coal used is 8000 C. Th. Units per lb. How much coal is used per hour ? 8. A pound of steam expands from lOOlbs./in. abs. to 20 Ibs./in.^ abs., being kept just dry all the time. Plot a curve to show how the volume varies with the pressure and find the work done. (Special Exam. Cambridge, 1913-) 9. Steam at a pressure of 250rDS./in. abs., dryness fraction 0"96, expands adiabatically to a pressure of 20 Ibs./in.^ abs. What will be the dryness fraction ? (IntercoU. Exam. Cambridge, 1904.) 10. Steam at 150 lbs./in^ abs., dryness fraction 0*9, expands {a} at constant volume, (b) adiabatically, (c) at constant total heat, {d) at constant dryness. The final pressure is 40 Ibs./in.^ Plot the

» tt), = (T,- TJ (l + ll) - r. loge I + ^-«- Of this work, the part (^i — pi)<^/J is used in pumping the water from the condenser to the boiler. Hence the external work L,N .(ii.) = (T,-T,)(l + ^|)-T,logeT| - = I. - 1., if we neglect the w term, which is very small indeed. If proper values be given to /i and /a to suit the case, the formula W = Ii — h applies whether the steam be supplied wet, dry, or superheated, but (ii.) only applies when the steam starts dry. 53 50. Efficiency of Rankine Cycle. One lb. of water enters the boiler at ^a and is turned into steam. The water contains heat Iwi when it enters the boiler, and the heat of the steam formed is /i. •"• Heat required per lb. = A — /«,2. efficiency = — = -z = — . By substituting the above value of W, we have (ri-T.)(i + |i)-r,ioge^ ~- . \ ii/ in eihciency = _ . 51. To find the number of pounds of steam used by a Rankine Engine per I.H.P. hour. Let H = H.P. output. 1 I.H.P. hour = 33000 X 60 ft. lbs. Work done per lb. of steam = W = h — h. 33000 X 60H, steam required /i - /, -lbs. per hour. 52. Mollier's Diagram of Entropy and Total Heat. This affords quite the simplest method of dealing with questions of the adiabatic ex- pansion of steam, and other problems. The diagram is drawn with the horizontal axis as axis of and the ver- tical axis as axis of I. A sketch is shown in Fig. 13, and diagrams drawn to scale are obtainable.* On the Fig. 13. ' From the Publishers, price Is. nett. 54 diagram are drawn lines of constant pressure such as pi p%, etc., lines of constant dryness, c?i q^, etc., in the wet region, and lines of constant. temperature, h U, etc., in the superheat region. In adiabatic processes ^ is constant, and so the adiabatics are vertical, while lines of constant / are horizontal. 53. To use MoUicr's Diagram. (1) For Rankine's Cycle. Take the pressure line corresponding to the initial pressure, and follow it up until the proper tempera- ture line is reached, at A say, if the steam be superheated, or the proper dryness line, at C, if it be wet. This gives the starting point ; read off the value of 7i. Now follow a vertical line downwards till the line of condenser pressure is reached at B or D. At this point read I2, and the dryness can also be read if desired. Then the vertical distance between the two points gives at once the work done, measured in heat units, by the engine per lb. of steam used, since C D or A B, according to conditions, is equal to the heat drop, /i — /a. (2) For steam passing through a throttling valve or other small orifice. By §46 we have I2 = h. Hence, to use the diagram to find the condition of the steam after it has passed through the orifice : find the point corresponding to the initial state of the steam, e.g. C, and go along a line of constant /, i.e. a horizontal, until the line of low pressure is reached, at E say, then the condition of the steam can be read off at once. The diagram is very useful, and saves a lot of time, and the student is advised to obtain one and familiarize himself with its use. 54. To draw the Saturation Curve on the Indicator Diagram. The object of this is to examine the state of the steam throughout the stroke. 55 (i.) In the test of the engine the amount of steam used per hour or per minute will be measured. From this the steam used per stroke can be calculated when the r.p.m. are known. (ii.) Calculate the cushion steam : — the piston does not travel to the end of the cylinder, and the volume between the end of the cylinder and the piston at the same end of its stroke is called the clearance volume. In this space a certain amount of steam is enclosed, which is called the cushion steam. To find its Fig. 14. amount: — MUrk the point A, on the indicator card, where compression begins (where the line begins to go round the corner). The whole stroke is LN ; hence the volume of the steam shut up in the cylinder = v = clearance vol. (supposed known) + y-r X vol. swept out by piston in its stroke. Take the steam as dry, and read its pressure OB. In the tables find the volume of one lb. of steam at this pressure, = V say. Then V wt. of cushion steam = — lbs. (iii.) Then the total wt. of steam present during expansion = w = wt. used per stroke + wt. of cushion steam. (iv. For a series of points x, y, z, . . . on the expansion line calculate the actual volume of the steam as 56 shown for finding volume of cushion steam. Suppose, at X, vol. = Vi. Find the volume of one lb. at the pressure at x from the tables, = Vi say. Then the volume of w lbs. (wt. actually in the cyclinder) = wVi. This is the volume the steam would have if it were dry; the actual volume is Vi. Hence the dryness = ffi = —7-. The real volume is represented by DX. Hence, the volume it would have if dry is repre- DX sented by DF, where :p-j- = qi. In this way find the points FGH . . . , and the curve through them is the saturation curve. 55. To show the exchange of heat between steam and cylinder walls on the ^ — T diagram. (Fig. 15.) From the saturation curve drawn on the indicator diagram (§54), find the dryness of the steam at different points X, y, z (Fig. 14). Draw on the i = 0-95 X 1-578 + 0-05 X 0-515 = 1-526 02 = l-94ga + 0-158 (1 -q^) = 0-158 + 1-78232. Also 4> t S W ' S ' w at 150 Ibs./in.' 666-7 183-9 1-578 0-515 181-2°C. at 31bs./in.' 622-9 61 1-885 0-202 60-9°C. (i.) Efficiency in the first case. To find the dryness at the end of expansion, we have 1-885 q + {l - q) 0-202 = 1-578 whence q = 0-815 Then h = 0-815 X 622-9 + 0-185 X 61 = 507 J .1- ra ■ 666-7 - 507 -^,0, and the efficiency = -rrri, tt- = 26-47o. bob'/ — oi 70 (ii.) Let ^' = the temperature, °C., to which the steam is superheated. Then 273 -I- ^1 ^.rrrrrr,-^ — ^ ^ lto2. .^^^'"''^''''''''^^^^^^ = 4-i- per lb. ^''■''• Work done on steam entering (l) by the steam behind it = PiVi. Work done by steam leaving (2) on steam in front of it = P2V2. Loss of internal energy = Ei — E^. Hence 2 2 "' „ ' =E,- E^ + P,V, - P2K2, 2g or 2g — » 2 U; If f 1 be negligible or 0, V, 2 2^ = '"-'^ (2) The heat drop, h — h, is most easily found from Mollier's 0—1 diagram draw a vertical line from the , starting point on the Pi line to the Ps line. By §49, h -h=J VdP, Pi taken along. the adiabatic. 2 _ ^. a /-Pi 2g" -'--^'^f V.dP (3) P3 Suppose PK" = C. Then ^■=-5.^(P..,_p.^J=-^(,_(S)"-}p.^.,4, 73 Hence, taking Vi = 0, the velocity of discharge is given by 62. Mass of Discharge from Nozzle. At the smallest section let the pressure volume be Po and Vo, then Mass discharged per unit area of section, then ^-V.-v\pJ -^ n~lVS\¥j ~\pj I- 63. To find the value of p^which gives the maximum discharge. Let "" = X, then, for maximum, -^ = 0. Pi ax '' 2gn Pi / I ?s±l ^ V« - 1 Fi ^* -* For maximum -y «" - ;e " must be maximum. 1 2 n+ 1 _1 ' « + 1 «\ / » » \ « „ 2:«» ' - («+l) = 0, Hence for maximum discharge l:=("-f')" Pn Taking n = 1-135, this gives— = 0'S8. ri 74 64. Maximum Discharge. Inserting this value of Po/Pi in the expression for Q, gives per unit area of section. If n = ri35 and g = 32'2, Q=3-Vw 65. If the flow is into a region where the pressure is less than given above (about 0"58 Pi) the steam must expand further, thereby gaining velocity, provided the jet is allowed to take up its natural shape. To design a nozzle for a gjveh discharge the area of the throat can be found by the above formula (§ 65), and the area at the discharge end can be found from the formula of (§ 63) substituting Pj for Po, neglecting friction. EXAMPLES. 1. Dry saturated steam is to expand through a nozzle from 2001bs./in.^ abs. to 5 Ihs./in.^ abs. Assuming n = 1'135, find the diameters of the nozzle at the throat and at the dis- charge end for a flow of 50 lbs. of steam per minute. From the tables Vi =2*316 ft.'/lb. The flow per unit area 3.C a/200 X 144 , = 3-6 V ^.3^^ = 402 lbs. per sec. .'. area required 50 = go ^ 4Q2 ^ 0-002075 sq. ft. = 0-299 sq. ins. .'. diameter of throat = 0-618". 75 At the discharge end, the flow per unit area / m 2'135 / 64-4 X 1-135 200X144 // A V"'_/ A V''" '^ 0-135 ' 2-316 ■lUoO/ V200/ / 1762 r88 = V 541 X 12430 {(0-025) -(0-025) } = 58-8 lbs. per sec. , . J 50 X 144 ^„. area required = = 2-04 sq. uis. 60 X 58-0 diameter must = 1-613". 2. Taking the same data as above, find the diameter at exit, assuming that 10% of the heat drop is lost in friction in the nozzle. If the flow were frictionless the velocity of discharge would be given by v' = 2g {I, - I,) actually it will be given by v' = 0-9 X 2g (/, - h) we find in the usual way 52 = 0-82, h — h = 142 Th. Units v" = 0-9 X 64-4 X 142 X 1400 = 11,500,000 V = 3400 ft./sec. The volume of the steam = 0-82 X 73-39 = 60-2 ft' per lb. . ,■ , • 3400 ,,,,^ . . discharge per unit area = = 5-65 lbs. per sec. 60-2 . 50 X 144 ■ , area required = = 2 12 in. 60 X 36-5 .'. diameter required = 1-643". 3. Find the principal dimensions of a nozzle to expand steam from 180 lbs. /in. abs. to 2 lbs. /in. abs., with a flow of 75 lbs. per minute, (a) neglecting friction, (6) assuming that 12% of the kinetic energy is lost in friction. 4. Steam at 2001bs./in.^ abs., superheated to 350°C., expands through a nozzle to 5 lbs./ in. ^ Design a nozzle to deliver 60 lbs. of steam per minute, assuming adiabatic flow, and that 10% of the heat drop is lost in the nozzle. 76 THEORY OF INJECTORS. 66. Let suffix (l) refer to the steam as supplied to injector, suffix (2) refer to water entering injector, (3) refer to discharge. Dlschdrj e^- — j-> To find the weight oi water lifted per lb. of steam. Let, each pound of steam entering A draw x lbs of water from B, so that 1 + x lbs. are delivered at C. Heat energy per lb. of steam at A = qiLi + /«,! Th. Units. Heat „ „ ,, water entering = 1^2 „ Kinetic „ „ „ „ « = ^ Ft.-lbs. Heat ,, „ „ ,, delivered = /u.8 Th. Units. Kmetic „ „ „ „ „ = — Ft.-lbs. Where Vs = velocity of water where it mixes with the steam, and V., = ditto in smallest section of delivery pipe. Then, by conservation of energy, J{q^L^ + /«,.) + x(j . /„„ + ^') = (l + x) (/ . /,„«+ ^J (l) The Vi' and V,^ terms can generally be neglected, then : Ii + Xlw» = (1 +X)I„3 (1) Hence x can be found. 77 Let Vi = velocity of issuing steam. Tlien the conserva- tion of momenturn gives f X — = (1 + X) — , & & g or Vi + X V2 = (1 + x) V3 (2) Equations (1) (in either form) and (2) are the equations to be used for solving injector problems. If the injector lifts the water a term must be introduced into right hand side of (l) for the work done in Hfting the water. 78 EXAMPLES. 1. An injector is working on a boiler with lOOlbs./in . pressure : the feed water enters at 10°C. and is delivered to the boiler at 80°C. Find the weight of feed water supplied per lb. of boiler steam, assumed dry. (Mech. Sc. Trip., 1916.) Let X = the weight of feed water per lb. of steam. The heat in the feed water =10 Th. Units per lb. „ „ steam supplied = 662 Th. Units per lb. ,, ,, water when it^ enters the boiler = 80-3 Th. Units per lb. Neglecting the work done we have 10;t:+-662 =-(\+x) 80-3 661 „,,, whence x = zrr^ = 9"41bs. 2. A steam ejector fitted to a destroyer will discharge 40 tons of water per hour, from the bilge, through a lift of 12 ft. The ejector is supplied with steam of a dryness fraction 0"9 at 250 lbs. /in. ^ by gauge. The temperature of the bilge water is lO'C, and that of the water discharged from the ejector 29°C. Estimate approximately the quantity of steam" used per ton of water pumped from the bilge, and the efficiency of this ejector as a pump. (R.N.C., Greenwich, 1912.) 79 67. Classification of Turbines. The principle of all steam turbines is the conversion of the thermal energy of the steam into kinetic energy, by letting the steam dis- charge, through nozzles, against blades or vanes fixed to rotating wheels. Turbines may be broadly divided into'(l) Impulse Turbines, in which the total fall of pressure takes place in the nozzles, only the velocity changing in the passage through the blades; (2) Reaction Turbines, in which the pressure of the steam is reduced, as well as the velocity, by passing through the vanes. TURBINE TYPES. Type Name Description Single Stage. De Laval. Impulse. 1 set of fixed nozzles and 1 set of moving blades. Multi-Stage. Rateau 1 Each stage consists of a set of Zoelly I nozzles and 1 ring of blades mounted on a wheel fixed to shaft. Each is an impulse turbine ; Rateau has 20 to 30 stages, Zoelly about 10. Parsons. Calle ' reaction, but really partly reaction and partly impulse. Each stage has 1 set of fixed blades and 1 set of moving blades. Curtis. One or more stages, each con- sisting of one set of nozzles, followed by alternate rings of moving and fixed blades. 68. Velocity Diagram for Impulse Turbines. In Fig. 19, OA = ui = velocity » , of steam leaving nozzle, and AB = u = velocity of blades. Then OB = Vri = velocity of steam relative to blades, which gives the proper angle for the blades at entry. The steam comes out of the channel between the blades with the same relative velqcity, D E = Vs- Then, if FD = «, FE = Vi = absolute velocity of steam at exit, which must be as small as possible, as the kinetic energy is wasted, and FE should be as nearly as possible perpen- dicular to FD. 80 69. The Velocity Diagram for a Parsons' Turbine is shown in Fig. 20, for one stage constructed in Fig. 20 the same way as above. In the diagram, u = velocity of moving blades. Vo = velocity of steam entering fixed blades. Vi = velocity of steam leaving fixed blades and enter- ing moving blades. Vs = relative velocity of same. Vs = relati\'e velocity of steam leaving moving blades to enter the next set of fixed blades. f4 = absolute velocity of same. 70. Output o{ Single- Stage Impulse Tur- bines (De Laval) (see Fig. 19). The work done per lb. of steam = change of k.e. 2g ' The energy supplied = — per lb. The erhciency = 5 — . Vi 81 If /3„_ /3i (which is usual), Fig. 19 can be drawn as in Fig 21 , since v^ = i% where OK is drawn = Vi = OC, and at same angle as OC. / / \ \ d^K ,' ^'^ ^'• u C K It 6 ii A ^^^ = OK' = OA' + AK' - 2 OA . AK cos a. = Vi +4 11 — 4 u Vi cos Oi, .'. Vi" — v,^ = 4-11 {v, cos tti — m) • CC ■ Vij-vl 4« / ;, \ . . efficiency = i? = ^ — = — I cos oi ). 2u * For maximum efficiency cos ai = — , Vi then max. efficiency ~- cos' a,. The process to be followed when fii^P^, and va^^Vi will be seen from the example below. 71 Output of Reaction (Parsons, etc.) Turbines (Fig. 20). For one stage : v^ — v ' Change of K.E. in fixed blades = = \V, 2 2 .. moving „ = '^~^'-- — M\ 2g K.E. carried away = -- 2g' work done per lb. of steam is 2 2 2 9 2 ■ 2g 2g 2g W and Efficiency = . Hence, to find the efficiency, find the v's, substitute their values in the expressions for W, \Vi, and W-^, and hence find the efficiency. * Differentiate i? witli respect tow. 82 EXAMPLES. 1. In a De Laval turbine, in which the full pressure drop is used up in the nozzles, and the blades have their outlet and inlet angles equal, steam is supplied dry and saturated at 180 lbs./in.^ and the exhaust pressure is 2'5 lbs. /in. The peripheral speed of the blades is 1250 ft. /sec. and the nozzles make an angle of 20° with the direction of motion of the blades. Assuming adiabatic flow in the nozzles and neglecting friction, find the velocity of discharge from the nozzles, the inlet and outlet angles of the blades, and the work done per pound of steam. (Mech. Sc. Trip., 1914.) The velocity of discharge from the nozzles is given by (§ 61) 2g ^' ^'- h — h is found as for a Rankine engine (p. ) or from a Mollier diagram. We find /, — /.^ = 158 lbs. calories. v' = 158 X 1400 X 64-4 = 14,200,000 .'. fi = 3770ft./sec. Let /i = the blade angles. „ Q 3770 X sin 20° „^,, Tan p = --— --0 — — = 0-564 3770 cos 20 — 1250 ■ /3= 29-4°. Fig. 22. From the figure : — r/ = i\ = 3770' + 1250' - 2 . 3770 . 1250 . cos 20° = 6,910,000 83 v., = ua = 2630 ft./sec. Again : v/ = 1250' + v/ - 2v, . 1250 . cos B = 2,740,000 V, = 1655 ft./sec. The work done per lb. of steam 2 S ^Vi — Vi ' 2g (For max. efficiency we should have cos a ^2500 3770 178,000 ft. lbs. hould have cos •663, or a = 48-5° instead of 20°.' 2. A De Laval turbine has blade angles of 30° at inlet and 38*5 at outlet. The nozzle axis makes 20° with the plane of the disc. The steam enters the blades without shock and leaves them with a relative speed which is 80% of the relative speed at entry. The steam is dry and saturated at entry and expands from 100 lbs. /in. to 1 Ib./in. absolute in the nozzle. Determine the thermal efficiency of the turbine, neglecting losses in the nozzle, the temperature of the boiler feed being 38°C. (Mech. Sc. Trip., 1912.) T ^o' -3»— u ^^ Fig. 23. 84 As before, v, = \ 2g (A - I-^ J = 3850 ft./sec. From the diagram : Vi sin 20° Vi = = 2630 ft./sec. sin 30 11 = Vi cos 20° — Vi cos 30° = 1340 ft./sec. va = 0-8 vj = 2100 ft./sec. i>/ = v/ + m' - 2 jws cos 38-5° = 1,800,000 Vi = 1340 ft./sec. The work done per lb. of steam '2g "' ""' = 202,000 ft. lbs. = 144 lbs. calories. 144 ^ 144 The thermal efficiency = r" — r~. = 0-231. /i — Iw2 624 3. A Curtis (impulse) turbine has two stages, each with one set of nozzles, three rotating and two fixed sets of blades, as shown below diagrammatically. The nozzles are inchned Moyi/\/G Moving Fig. 24. 85 at 20° to the plane of rotation, and the peripheral velocity of the blades is 400 ft. /sec. The moving blades all have the same angles at entry and exit ; draw a velocity diagram for one stage and determine the angles of the moving and stationary blades. The velocity of the steam leaving the nozzles is 2800 ft. /sec. Neglect friction losses. Fig. 25, In the diagram, AB = velocity of steam leaving nozzle, and CB = velocity of blades. Then AC = relative velocity entering first row of moving blades, and ACD (=23"1 ) is the blade angle at entry and exit. This is to be the same for all the rows of moving blades. Make EC = volocity of blades and CF = CA = relative velocity of steam. Then EF= absolute velocity of steam 86 leaving first moving blades and entering first fixed blades, so that the apgle EFG (= 27'6°) is the entry angle of fixed blades. Since the angle of entry for the second set of moving blades is to be 23" 1", the direction of the relative velocity of the steam must make this angle with FG ; make FK = velocity of blades, and draw KH at 23'1° to FG; then draw an arc with centre F radius FE cutting KH in H, so that FH = FE. Complete the parallelogram FKHL ; then the angle GFH (= 19°) gives exit angle for these fixed blades. Repeat the process until the last set of moving blades has been dealt with. MO and RT represent the absolute velocity of the steam leaving the second and third rows of moving blades. The angles of entry and exit for the second row of fixed blades are 30"1° and 17"1° respectively. 4. In a single stage impulse turbine compare the efficiency found by formula, neglecting friction, with the efficiency actually obtained under the following provisions : — Velocity of exit from nozzles = 1800ft./sec. ,, ,, wheel periphery = 600 ,, Angle at which the inlet and outlet of the vanes are inclined to the direction of motion = 30°. Assume a loss of 8% due to friction in the moving vanes. (R.N.C. Greenwich, 1910.) 5. A 500 kW steam turbine using dry steam at a pressure of 140 Ibs./in." abs. consumes 22-6 lbs. of steam per kilowatt hour and condenses at a pressure of 1 lb. per. sq. in. abs. The condensing water measures 5620 lbs. per minute and its rise of temperature is 18-3°C. Determine the dryness of the steam as it leaves the turbine and find the loss of work in the cylinder of the turbine due to the expansion not following the adiabatic line on the entropy-temperature diagram. (Mech. Sc. Trip., 1912.) 87 6. Draw to scale a velocity diagram for one complete constant pressure stage of a Curtis turbine, consisting of a set of nozzles, two sets of fixed and three sets of rotating blades ; find the absolute speed of discharge and the speed of axial steam flow, from the last row of moving blades, having given that the blade velocity is lOOft./sec, the speed of the steam leaving the nozzles 1200 ft. /sec, and the angle of nozzles to the plane of rotation is 20°. (R.N.C. Greenwich, 1912.) 7. ■ Steam issues from a nozzle, inclined at 17 to the circumference of a turbine wheel, at 3600 ft./sec. The peripheral speed of the blades is 1200 ft./sec. The inlet and outlet angles of the blades are equal. Find this angle, the velocity of discharge of the steam, the indicated H.P. perlb. of steam, and the efficiency of the blades neglecting all losses.' REFRIGERATION. 72. In the heat engine working direct we supply heat and obtain work ; in the refrigerator we do work to extract heat from a substance which we wish to cool : we ' lift " heat from one body to another of higher temperature at the expense of work. 73_ Heat 'lifted' (extracted) ^^ ^^j,^j ^^^ coefficient work done oi performance. 74. Refrigerator working on reversed Joule cycle : Bell-Coleman Refrigerator (Fig 26). Cold Chamber Ti| Fig. 26. Used for cooling cold-chambers in ships. Air is drawn ra by B from (Y2 in pv diagram) chamber A, and is com- pressed along 2 3 until the pressure equals the pressure in C, which raises the temperature above that of C. The com- pressed air is delivered (along 3X) into C. At the same time D draws (along X4) an equal quantity of air from C and expands it along 4 1, the temperature falling below that of A. The cold air is discharged into A on return stroke of D. 89, 75. To find the coefficient of Performance of the Bell-Coleman Refrigerator (see Fig. 26). Along 1 2, heat taken in from cold chamber = kf, {Ti — Ti) = Q]2 Along 3 ff, heat given to cooler = kp (Ta - Tj = Qa The work done = the area 12 3 4 = Qu - Qn = kp {Ts '- T, - T, - Ti) ; •'■ the coefficient of performance The heat extracted work supplied kp {T,-T,) 1 T,+ T,) ' ' 1 76. Vapour Compression Refrigerators. The chief substances used are NHs, COj, SO2. In the forward 1^ 1, t 4' 4 /: A \ : \ /Ti \ ifr z 3' |3 3 ' :m denser I Pi B I I i ■h- A Comp? Cylr Refrigerator Fig. 27. stroke of the pistoa, vapour is drawn from A (2 3), it is then compressed (3 4) and driven (4 l) into the- condenser. An equal quantity of substance expands through the expansion 90 valve from B to A. This operation is irreversible. The (ji — T diagram is shown to the left of Fig. 27, the com- pression ending with the vapour just dry, but the compression line might equally well be 3' 4' or 3" 4". 1 — -2 is adiabatic passage through the valve, 3 — 4 is adiabatic compression in the cylinder. 77. Equations for Vapour Refrigerators. From the (p — T diagram : Heat taken up = q^L^ — ql loge-^ = Wi = piXh = piVi -- u'i by the air on the suction side of piston = ^it'i ==^ w.i W = Total work done = xc, -t- iv^ — xv„- pii\ loge ' Pi 95 (ii.) If compression follow pv" = constant. — PlVi Wi = n - 1 Wi = paVi, Wi ^ piVi, • • W — + piVi - piv„ n — 1 — T piVi I 7 — M — 1 \plVi Work done 0. 1^-1(1:)"-^ (1) The rise oi temperature is given by (see §19, ii.) T,= Pi .(2) The heat taken up by the air = W, - kv{T, - Ti) /?(r2 - Ti) 71-1 — kviTi — Ti) = (— - - /fe.)(T,^-r,) \n — 1 ' = (;r^i-''>^'TM(|;)""'-'} (3) 81. Two-stage Com- ^ pressor Si Air is compressed along 1 2 to pa, cooled to initial Pj temperature, compressed along 3 4 and again cooled. Thus the ^^ area 2 3 4 4' is saved. Fig. 31. Area B 21 A =-^..«,if!;)"-4 by § 80- 96 Area C 4 3 B = n-l n n - 1 and piVs = piVi. ^-^-Kft) " -il ''^"''■ ..H' = B.IA + C43B.,;^«r,{(J;)n(f:)^--2t. 82. To find the relation between the pressures so that W shall be a minimum. (?:)"+(£)" "-^' a minimum. Let n-l Then A-2 , ^3 3; = — H Xi Xt must be a minimum, dy _ \ ^3 dXi Xy Xi = 0, , ■ 2 ^'2 "'" XiX'i / • i>2 = pips. which is the relation sought. 83. To find the Cylinder ratios. If the isothermal is pv = C, and the above condition is satisfied, C" _ ^ C 2 — , '1, Vi pi t\ .". l\ — V, \/~. If the stroke of both cylinders be the same, the ratio of the diameters is given by pi • ^' - */^ 97 84. Three-stage Compressors (Fig. 32). W can be found as before, and we must have for minimum work p2 = pipit P^^ = Pip-i _._ P± = ^ = Pi pa pi Pl' and similarly for any number of stages ; i.e. the pressures must be in G.P. Similarly we can shew Fig. 32. \pj di \PJ 85. The effect of Clearance. In Fig. 33, CL = stroke volume = V, FL = clearance „ =cV. At the beginning of suction stroke, the air left behind ex- pands, and the inlet valves do not open until pi is reached at H. K A Dl B \ P, \c L H » Fig. 33. Since air does not accumulate in the cylinder, fi(F.g.33)^f-|(Fig.34) (where there is no clearance), for AH and BC follow pv" = const. Then Vi=V + cV=(l+c)V, and x'2 = c F. .-. the area KBCF Fig. 34. «-l Pl . Vil + c) if)"~'l 98 2 also FH= I r)"cV, And the volume sucked in per stroke instead of V. 99 EXAMPLES. 1. An air compressor takes in air at a constant pressure po, and, after compressing it adiabatically, delivers it to a receiver at constant pressure pi. Show that the work done by the compressor is given by /? — h, where /i and U are the values of / at the beginning and end of compression. If the initial pressure po is 15 lbs. /in. , and the initial temperature is 15°C., tind the temperature at which the air is delivered, the higher pressure ^i being 500 lbs. /in." Find also the work done by the compressor per pound of air. (Intercoll. Exam. Cambridge, 1914.) The proof of the first part of the question follows the lines of § 49. The final temperature [§ 80,(2)] is 0J4 /500N1* ^'^(15) >^ 288 = 785 or ti = 5)2 C. The work done per pound of air [§ 80, (l)] is ;-:|x,ax.s»[(f)"-.] = 167,000 ft. lbs. 2. What is the ratio of the efficiencies of two air compressors, compressing air from 15 to 150 lbs. 'in. abs., one in a single stage, the other in two stages of equal pressure ratios ? The compression follows the law pv^ ^ = const. The air is cooled to its initial temperature between the stages in the second case. Neglect clearance and mechanical losses. (Mech. Sc. Trip., 1913.) In the single stage compressor the work done per lb. of air is r — "I — X 95 X Ti lO''^ -1 = 3-&4 X 96Ti. 0-3 L J 100 If the compression were isothermal, no work would be wasted in heating the air, and the work done would be 9671 loge 10 = 2-303 X 967,. Taking this as unity the efficiency of the single stage 2'303 compressor = ^ „. = 0'758, or 75' 8%. 3-0+ For the two stage compressor we must have pr ^p2 jh. As or ^2 = \ pip^ = V 2250 = 47-5. The work done (§81) per pound of air = 2-64 X 96 Ti 2-30^ The efficiency = - ^f = 0-874, or 87-4%. 2-54 The ratio of the efficiences is 2 stage ,^ 874 1 stage ^ 75-8 = 1-15. 3i A single stage compressor compresses air from 15 to 150 Ibs./in.^ abs., the clearance volume being 1 % of the piston displacement volume. At what point of the suction stroke will the inlet valve open if the spring on it has negligible tension ? (Mech. Sc. Trip., 1912.) Referring to Fig. 33, we have C = 0-01. _i_ ■ I FH = lo'" X 0-01 V = 0-05176y FL = 0-01V LH =0-04176 r LC = V The inlet valve will open at H, z'.e. at 0-04176 of the stroke. 4. Calculate the H.p. required to drive a single stage compressor which takes in 250 ft." of air per minute and 101 compresses it adiabatically from 15 lbs. /in. to 150 Ibs./in.^ The clearance is 0"5 % of the stroke volume, and the mechanical efficiency of the compressor 85 %■ 5. A three stage compressor is used to charge a receiver of 20 ft." capacity from atmospheric pressure to 2000 lbs. /in.* Find the total work done while the pressure in the receiver rises to the final first and second stage pressures (« = TS). Assume that the compressor has been designed for maximum efficiency when working against a steady back pressure of ,20001bs./in.' (R.N.C. Greenwich, 1912.) 6. The clearance volume in a single stage compressor is 0"5 % of the volume swept out by the piston, which is 2 ft.* Calculate the number of strokes of the piston required to deliver 300 ft.' at 100 Ibs./in.^ the volume being measured at the higher pressure after cooling to the initial temperature of 15-5C°. (R.N.C. Greenwich, 1912.) 7. In a system of power transmission by compressed air, air at 15°C. is compressed adiabatically to lOOlbs./in.* abs. It passes to the motor through pipes in which the temperature falls to the original atmospheric temperature, the pressure remaining constant. The air then expands adiabatically in the motor cylinder, the expansion ratio being 3-5:1. Find the efficiency of the transmission neglecting friction losses. (Trin. Coll. Cambridge, 1913.) 102 COMBUSTION. 86. Data : Element:— H O N C S Atomic wt.:— 1 16 14 12 32 Gas:— H2O CO CO2 C2H4 NH, CH, Molecularwt.:- 18 28 44 28 17 16 Heat given out when lib. of C burns in O to form CO,, = 8060 Th. Units. „ C „ „ O „ CO = 2450 „ H combines with O „ H2O = 34500 r . . O 23 , The composition of air is r~ = ;^ by weight, ^ _ 23 °'' Air" 100 At 0° C, 760 mm. pressure, 1 cub. ft. of air = 0-0807 lbs. 87. Fundamental Combinations. (i.) C + O2 = CO2 To obtain the proportions by weight in which C and O combine, substitute the atomic weights of the elements : — 12 + 32 = 44 i.e. 12 lbs. of C combine with 32 lbs. of O to give 44 lbs. of CO2 or 1 lb» of C requires f lbs. of oxygen for complete combustion. Hence for complete combustion 2'56 lbs. of oxygen, or 11-5 lbs. of air = 142-5 ft.' at 0° C. 760 mm. V = 150 ft." at 15° C. 760 min. (ii.) C + O = CO 12 16 28 Whence, as above, 1 lb. of C requires 5-77 lbs. of air. (iii.) 2Hj + 02 = 2H,0 4 32 36 1 8 9 i.e. 1 lb. of hydrogen requires 8 lbs. of oxygen for complete com- bustion, and forms 9 lbs. of water, giving out 34,500 Th. Units. lib. of C requires 103 88. Draught in a, Chimney. Let V = velocity of gases ; ha — height of column of air corresponding with difference of pressure (draught) above and below grate. Then V = s'Zgha . Let h = draught in inches of water (l inch of water is equal to 5"2 lbs. per ft.^) Then 5'2h ha = j^TT^Tn: = SA'Sh at standard temp, and pressure 0'0807 V = '/U9gh = \ 4150/1 ft./sec. Let H = height of chimney above grate (feet) Ti = temperature inside chimney (abs.) Ta = „ outside „ A = area of section of chimney (ft. ) in = lbs. af air per lb. of fuel. Vol. of chimney = AH. .'. wt. of this vol. of external air , „ 0-0807 X 273 = AH = J 2 .". wt. of this vol. of chimney gases „ 0-0807 X 273 m + l = AH. ;= • Ti III Difference between these is equivalent to draught, and so Ah(^ -^^^-ti.^\ 0-0307 X 273 =5-2 Ah \Ti in TiJ Hi^--'^^^-^\=Q^2Z6h. 104 89. Specific Heat of Gas Mixtures. The sp. ht. of a mixture of uh lbs. of a gas of sp. ht. k^, 111-2 lbs. of a gas of sp. ht. k^, etc., is iiiik, + ni-iki + k = till + iii.j-\- The following may be taken as the mean values of kp for the gases named : — COa 0-216 CO 0-245 O., 0-218 N., 0-244 H2O 0-480 1-05 EXAMPLES. 1. — A certain coal has the following composition by weight: C = 797; H = 4-9 ; O = lO'S, %; the rest is ash, nitrogen, etc. Find its gross and nett calorific values, and the' air supply required per lb. of coal. (Mech. Sc. Trip., 1910.) Consider 100 lb. of coal: What O there is exists in the form of moisture. .". , by ?83 (ii.), there is i as much H as O in. corribina- tion with the O, 10-3 .-. H present as H,0 = -^ -= 1-29 lbs. O and the O „ „ = 10-3. .-. total weight of H,0 -■= 10-3 + 1-29= 11-59 lbs., and nett weight of free H -= 4-9 - 1-29= 3-61 lbs. To find the calorific value : — Heat produced by combination of 79'7 lbs. of C = 79-7 X 8060 = 642,000 Th.U. ditto, 3-61 lbs. of H= 3-61 x 34,500= 124,400 766,400 .'. calorific value of 1 lb. = 7664 Th. Units. This is called the gross calorific value. A lot of this heat goes up the chimney, but we charge this to the boiler, not the coal. The coal is charged with the latent heat of the stsam carried away, which is subtracted from the gross calorific value to give the nett calorific value I— Total wt. of H-jO originally present = 11'59 lbs. 106 The combustion of the 3-61 lbs. of H with the oxygen of the air gives 3-61 X 9 = 3249 lbs. .■- total wt. of HaO going up the chimney = 44-08 lbs. Tlie latent heat of this = 44-08 X 539 = 23700 Th.U. or 237 per lb. of coal burnt. .•. nett calorific value = 7664 - 237 = 7427 Th. Units per pound. To find the atnount of air required for combustion : — By (i.) of §83, for 79-7 lbs. of carbon we require 79-7 X ft = 212-5 lbs. of O By (iii.) of §83, for 3-61 lbs. of H we require 3-61 X 8 = 28-88 lbs. of O. .■. each 100 lbs. of coal requires 212-5 + 28-88 = 241-38 lbs. of O = 241-38 X VV" lbs. of air = 1049 lbs., or, air req. per lb. of coal = ■10-49 lbs. 2. — The flue gases from a boiler are found to contain, by volume, COs = 12% CO =1-3%, 02= 6-5%. F'ind the actual supply of air to the boiler. (Mech. Sc. Trip., 1910.) The volumes present are COu : CO : Oj = 12 : 13 : 5-5 The relative densities are = 44 : 28 : 32 .'. The weights present are CO2 : CO : O2 = 12 X 44 : 1-3 X 28 : 6-5 X 32 = 528 : 36-4 : 208 1 lb. of CO2 requires tt lbs. of C and ri lbs. of O2, ••• 528 „ „ 143-7 „ C „ 384 „ „ O2 ; 1 lb. of CO requires hi lbs. of C and if lbs. of O2, ••• 36-4 lbs. „ „ 15-6 „ C „ 20-8 „ „ O2. 107 Hence we can say, that 1437 + 15-6 = 169-3 lbs. of carbon were burnt, producing 528 lbs. of CO2 and 36-4 lbs. of CO, and that 384 + 20'8 = 404-8 lbs. of O2 were supplied for combustion, and 208 lbs. besides, or 612-8 lbs. of O2 altogether. Thus each lb. of C had TTqT^ "^ 3-62 lbs. of O2 supplied, 3-62 X Vf = 15-7 lbs. of air. 3. — Find the specific heat, at constant pressure, of the flue gases whose composition by weight is CO2 17-5 CO 1*2 O2 6-9 N2 74-4 lOO'O The sp. ht. is 17'5 X 0'216 + V2 X 0-245 + 6'9 X 0-218 + 74'4 X 0*244 100 =0'^37. 108 4. The gas from a suction gas plant gives the following analysis by volume : — CH... .... 0-65 COa 6-57 H, 18-73 CO 25-07 N. 48-98 100-00 Find the volume of air required for the complete com- bustion of 100 ft." of the gas, and the lower calorific value of the gas per standard cubic foot. (Mech. Sc. Trip., 1907.) 5. A sample of petrol consists of 80% of CeHia, 18% of C7H16 and 2% of CeHia. Fipd the weight of air required for the complete combustion of 1 lb. of this petrol and the' lower calorific value. (Trin. Coll. Cambridge, 1912.) 6. A sample of coal contains 80% by weight of carbon and 5% of hydrogen, some of which is combined in the form of moisture. ' When the sample is tested the lower calorific value is found to be 7500 Th. Units per pound ; find the percentage of moisture originally present. (Mech. Sc. Trip., 1916.) 7. The boiler room of a battleship contains six Yarrow boilers; the dimensions of each of the grates being 7' X 8'6". Forty pounds of coal are burned per square foot of grate per hour, the composition being: C, 90%; H, 4%; O, 2%. The fan for the forced draught produces a pressure in the suction duct of 0'04" of water below atmospheric pressure. If the supply of air by the fan is 100% in excess of the minimum required for complete combustion, calculate the section of the air ducks, the temperature of the air being 15 C, the specific volume at that temperature and atmos- pheric pressure being 13 ft." per lb. (Mech. Sc. Trip., 1912.) 109 INTERNAL COMBUSTION ENGINES. Fig. 35. 90. Cycles. ^ We shall work out the efficiencies of internal combustion engines, working on different cycles. (1) Fig. 35. 0—1 is suction. 1 — 2 is explosion. 2 — 3 is adiabatic expansion. The only supply of heat is along 1 — 2, and is instantaneous. Heat supplied = Qi=kv{T% — Ti), „ rejected = Qa = *^ (Ts - Ti), efficiency=l--=l-^-^r^^^- T, Fig. 36. (2) Fig. 36. Two cylinders -used, a compressing pump cylinder and a motor cylinder. The action is : — — 1 = suction ) t pump. 1 — 2 = adiabatic compression ' 2 — A = passage into receiver. A23 = passage into motor cylinder, and supply of heat 3C B = adiabatic expansion. no With complete expansion, we have, since this amounts to a Joule cycle (see §36), the efficiency = 1 — ( — ) With incomplete expansion, assume cooling at constant V, as CD, and then at constant pressure, Dl. Then Q2 = kv {Tc - Td) + kp {Td - Ti) 1 and efficiency =1 (Tc— Td) + [Td - T.) (3) The Otto Cycle (Fig. 37). The compression takes place in the motor cylinder — 1 = suction (1st stroke) of ex- plosive mixture. 1 — 2 = adiabatic compression (2nd stroke). 2 — 3 = explosion = heat supply at constant volume. 3 — 4 = adiabatic expansion (3rd stroke). 4 — 1 = cooling at constant volume. 1 — = exhaust (4th stroke). As in §37 we have. Fig. 37 efficiency 91. The Diesel Engine. The distinctive feature is the adiabatic compression of 3.ir to such a high temperature that it ignites the liquid fuel injected into the cylinder (Fig. 38). In Fig. 38, the action is as follows : 1 = suction of air into the cylinder (1st stroke). 12 = adiabatic compression of air (2nd stroke). Ill 2 3 = injection of fuel and burning 3 4 = adiabatic expansion 4 1= cooling at constant volume. 10 = exhaust (4th stroke). j (3rd stroke). Fig. 38. Assume that the burning (2 — 3) is isothermal. To find the efficiency : Qi — heat supplied = RT^ loge Qi — heat rejected = kv{Ti — Ti). 7-1 efficiency — 1 — -*"^' {{v,y '-''1. RT^ log. 112 If, instead of assuming that the burning along 2 3 is isothermal, we assume that it takes place at constant pressure, we shall have Q, = k^ {T, - Ti) Q. = ki, (Ta - rj and the efficiency = 1 - ki,(T,- T,y 92. Air Standard. Since the working substance always consists mainly of air, it is usual to suppose that it is all air, and in the above expressions give k^, kp, and A the values they have for air. The efficiency so calculated is called the Air Standard Efficiency. 113 93. Tests. (i.) B.H.P. (see §59). (ii-) i.H.P. (see § 56). (iii.) Heat supplied by combustion of fuel in given time. = quantity of fuel used X lower calorific value. (iv.) Heat carried in by jacket water = wt. of water used X its initial temperature. (v.) Heat carried away by jacket water = wt. of water used X its temperature leaving jacket. (vi.) Heat supplied to engine by air taken in. Let V = vol. of air used. V = specific volume. w = wt. of water vapour in the air. t = temperature of air, °C. Heat supplied by air = — . ks, t. V „ „ „ water vapour = wlw. To find w : let t^ = the dew point, °C. Find pressure of steam at temperature <' from tables ; subtract this from the barometric pressure : this gives partial pressure of dry air (j>, say). Then vol. of 1 lb. of dry air = R{t + 273)/p. This will also be vol. of water vapour present at teinperature t^, and hence the weight of water vapour can be found. (vii.) Heat carried away by exhaust gases. = wt. of dry gases X kp X .temperature °C. + wt. of steam X 7 of steam and pressure and temp, of exhaust. If an exhaust calorimeter is used : heat carried from engine in exhaust gases. = heat carried away from calorimeter by gases + heat given to calorimeter cooling water. 114 EXAMPLES. 1. A gas engine developing 10 i.h.p. consumes 180 ft.' of gas per hour, the calorific value of which is 320 Th. Units per ft." The piston displacement is 1*2 it." and the clearance volume 0*24 ft." Determine the thermal efficiency of the engine, and compare this with the efficiency of the standard engine. (Intercoll. Exam. Cambridge, 1911.) Heat supplied = 180 X 320 Th. Units per hour. = 960 Th. Units per minute. Ti, 1 A 10 X 33,000 ^__ ^^ . The work done = rr^rx = 236 Th. Units per min. 1400 thermal efficiency = r— = 0-246, or 24-6%. 960 Efficiency of standard engine 0-4 /0-24\ - [y^J - 0'512, or 51-2%. 2. A Diesel engine is supplied with oil whose calorific value is 8000 Th. Units per pound. On test it was found that the I.H.P. = 44-6, b.h.p. = 30-2, oil used per hour = 14" libs., circulating water 870 lbs. per hour with a tem- perature rise of 40"5°C. Find the thermal efficiency, the mechanical efficiency, and estimate the heat lost in exhaust and radiation. (Mech. Sc. Trip., 1916.) Heat supplied = — Th. Units per mmute. = 1880 Th. Units per minute. Indicated work 44-6 X 33,000 1400 Useful work ^ 30-2 X 33,000 1400 = 1050 Th. Units per minute. = 710 Th. Units per minute. Hence the thermal efficiency = -— - = 0'56 115 710 Hence the mechanical efficiency = -^-^ = 0*675 Heat carried away by cooling water per minute 870 = ^TV X "l-O-S = 588 Th. Units. We have then : Th. Units per minute. Heat supplied = 1880 1880 Useful work = 710 Mechanical losses = 1050 -710 = 340 Carried away by cooling water = 588 Exhaust and radiation = 242 1880 3. Describe how you would carry out a trial of a gas engine in order to obtain data for a heat balance sheet. In a trial of an engine rated at 50 b.h.p. the following data were obtained : Duration of trial, 60 minutes. Indicated horse-power, 55"9. Brake horse-power, 48" 1. Total gas used at standard temperature and pressure, 880 cubic feet. Lower calorific value of gas in T.U. per cubic foot, 310. Jacket water per hour, 980 pounds. Temperature of jacket water at entry, 7*2 C. Temperature of jacket water at exit, 70 C. Cooling water to exhaust calorimeter, 4300 pounds. Temperature of water at entry to exhaust calorimeter, 6-6°C. Temperature of water at outlet of exhaust calorimeter, 33-3°C. Loss due to radiation in terms of gross heat supply, 5 %• 116 From these figures make out a balance sheet shewing the distribution of the heat supply to the engine. (Mech. Sc. Trip. 1910.) Heat supplied by fuel = 880 X 310 = 273,000 Th. Units. 33 000 Indicated work = 55*9 X -773 r X 60 = 79,000 Th. Units. 1400 Mechanical work 33 000 = 48'1 X ^777^ X 60 = 68,000 Th. Units. 1400 Heat carried in by cooling water = 980 X 7"2 = 7050 Heat carried away by cooling water = 980 X 70 = 68,500 Th. Units. Heat given up by exhaust gases = 4300 (33-3 - 6-6) = 115,000 Th. Units. Loss due to radiation = 0-05 X 273,000 = 13,650 Th. Units. Hence we have : Th. Units per hour, reckoned from 0°C. Heat supplied by fuel= 273,000 Heat in jacket water 7,050 280,050 Useful work 68,000 Mechanical losses 11,000 Radiation loss 13,650 Heat carried away by jacket water 68,500 Heat from exhaust gases 115,000 Heat unaccounted for 3,900 280,050 4. An engine working on the Otto cycle has a bore and stroke of lO" and 15" ; the compression ratio is 1 : 5 and the engine takes in mixture at 15°C., one cubic foot of the gas weighing 0'07 lb. The maximum temperature reached during explosion is 1800°C. ; y = 1-4 and ife„ = 0-18. Find the indicated work per cycle. (Trin. Coll. Cambridge, 1914.) 117 Referring to Fig. 37 : Ti = 288 Ta= (~^ Ti = 5°'" X 288 = 548 Ts = 2073 T.= 2^3 = 1090. Heat received ^ = kviTs - 7,)= 0-18 X 1525 .= 275 Th. Units per lb. Heat rejected = kv {Ti - Ti) = 0-18 X 802 = 144 ■'. Work done = 131 „ „ The stroke volume =~x~ = 0-681 ft.' 144 4 This is the volume of gas sucked in per stroke. .'. wt of gas used per stroke = -681 X 0-07 = 0-0477 lbs- Hence work done per stroke = 131 X 0-0477 X 1400 = 8750 ft. lbs. 118 MISCELLANEOUS EXAMPLES. 1. 18,000 gallons of water are being pumped per hour to a height of 70 ft. by a pump driven by an oil engine. The pump has an efficiency of 65%. Determine the I.H.P. of the oil engine if its mechanical efficiency is 75%. The kinetic energy of the water may be neglected. If the oil is of sp. gr. 0'8, and the engine requires O'l gallon of oil per l.H.p. hour, what is the efficiency of the whole plant if the calorific value of the oil is 10,000 Th. Units per pound ? (Intercoll. Exam. Cambridge, 1914.) 2. In the experiments of Osborne Reynolds to deter- mine the mechanical equivalent of heat, a paddle was fixed to the shaft of an engine and rotated in water within a closed hollow vessel, freely mounted on the shaft, and prevented from turning round by weights attached to its side. At 300 R.P;M. of the engine 470 lbs. of water were found to have a rise of 99'5°C. in 62 minutes. Determine the mechanical equivalent of heat if the load on the side of the vessel was 141 lbs. at a leverage of 4 ft. (Intercoll. Exam. Cambridge, 1913.) 3. A bar of steel 3 sq. ins. section is quickly stretched in a testing machine by 1/2000 of its length. What change would be required in the balancing weight as the temperature recovers its original value if the stretch is maintained con- stant ? coeff. of expansion 0-000013/°C. ; sp. ht. =0-1098; sp. gr. = 7-55. E. = 30-10' Ibs./in.') (R.N.C. Greenwich, 1912.) 4. The thrust bearing of a liner is water cooled. The water used is 5750 lbs. per hour, and the temperature rise is 9 C. What horse-power is wasted ? 5. In the brake-horse-power test of a certain gas engine the nett load on the brake is 21 lbs., the diameter of the wheel is 5'6", and the speed 130 R.P.M. How much heat is gene- rated per minute ? 119 6. In the trial of a gas engine the following observations were made : Duration of trial 60 minutes, i.H.P. 60, b.h.P. 52, total gas used 950 cubic ft., total air used 8000 cubic ft. both being at standard temperature and pressure ; higher calorific value of the gas 360 Th. Units per standard cubic foot, jacket water per hour 1060 lbs., temperature of jacket water at entry 12°C., at exit 75° C, cooling water to exhaust calorimeter 5000 lbs., temperature of water at entry to exhaust calorimeter 12°C., at exit 36°C., temperature of gas in the exhaust pipe 50°C. The specific heat of the gases is 0'24 and the temperature of the gas and air supply is 15°C. Make out a complete balance-sheet shewing the distribution of the heat supply to the engine. (IntercoU. Exam. Cambridge, 1913). 7. The following data were obtained in a test of a Diesel engine : Duration of test, 60 minutes. I.H.P. = 12-95. ; B.H.P. = 9-^4. Oil used 5"441bs. ; lower calorific value 93401b. cal. per lb. Cooling water used 700 lbs. ; rise of temperature, 16"2°C. Calculate the thermal efficiency relative to I.H.P. and to B.H.P., and find what percentage of the heat supplied is not accounted for in the above observations. What becomes of the heat which is not accounted for ? The air compressor is driven direct from the engine shaft and takes I'l H.P., of which 75% may be taken as usefully returned to the engine cylinder in the air supply. If this is taken into account how does it affect the values of the thermal efficiency obtained above? (Mech. Sc. Trip., 1914.) 8. A gas engine has a stroke volume of 1450 ins.° and a clearance volume of 250 ins.", and uses 15 ft'. of gas per I.H.P. hour, the, calorific value of the gas being 310 Th. Units 120 per ft.' Find the thermal efficiency of the engine and com- pare this with the efficiency of the standard engine. (Inter- coll. Exam. Cambridge, 1914.) 9. In a trial of a Diesel oil engine the oil used had a calorific value of 10,700 thermal units per pound, and 18"5 lbs. of oil were used per hour. The b.H.p. was 40"4 and the i.h.p. 52'6. Cooling water passed through the jackets at the rate of 30'2 lbs. per minute and its rise of temperature was 29°C. It was calculated that the exhaust gases carried off 940 thermal units per minute. Determine the rnechanical efficiency, the heat used per I.H.P. minute, and draw up a heat account, calculating the percentage of heat unaccounted for. (Inter- coll. Exam. Cambridge, 1908). 10. A balloon of 5000 cubic ft. capacity is to be so far filled with hydrogen at 30" of mercury and 15°C., that after ascending td a height where the pressure is 20" of mercury and the temperature 0°C., the silk envelope is then fully extended, no gas having escaped. Calculate the mass of hydrogen required and its original volume. The density of hydrogen = 0'0056 lbs. per ft." at standard temperature and pressure. (Mech. Sc. Trip., 1912.), 11. A cubic ft. of air weighing' Jib. expands from a pressure of 100 ibs./in.^ to a volume of 10 ft.' at 20 ibs./in.^ the pressure falling uiiif ormly. Find the heat taken in by the air. (Mech. Sc. Trip., 1916.) 12. One pound of air at 0°C. is expanded at constant pressure to three times its initial volume. Calculate the final temperature, the work done, and the heat supplied. 13. One pound of water at Ti° abs. is mixed with one pound of water at Ti° abs. Show that there is a gain of entropy equal to , (Ti+T,)' log, -^pf^- (Intercoll. Exam. Cambridge, 1912.) 121 14. Prove that if ever a method is found for reversing the conduction of heat, a perpetual motion with enormous power behind it will have been discovered. (Mech. Sc. Trip., 1912.) 15. One pound of air, initially at 10°C., 15 lbs. /in. ^ is heated at constant volume to a temperature of 350°C. ; it is then expanded adiabatically and brought back to its original condition by isothermal compression. Draw the T-^ diagram and calculate the efficiency of the cycle. 16. Ten cubic feet of gas at 100 lbs. /in.^ 150°C. are expanded adiabatically to a pressure of 20 lbs. /in. , and then compressed to the original volume according to the law pv'"^= const., and finally heated at constant volume to the original pressure. Determine the work dope, and the efficiency of the cycle. (Intercoll. Exam. Cambridge, 1913.) 17. A deflated gas bag of small capacity is charged from a steel bottle containing air under high pressure, and at 10°C. When the connection- between the bag and the bottle is shut, the pressure in the bag is 15"5 lbs/in.^ abs. The barometer stands at 29'2 , there is no loss of heat and no work done in stretching the bag. Find the temperature of the air in the bag. 18. A quantity of gas of volume V and at atmospheric pressure 11 is confined behind a piston of area A. A spring bears on the other side of the piston and is compressed when the piston moves out. If the force necessary to compress the spring a unit distance be X, show that the quantity of heat, in work units, that must be given to the gas so that it slowly expands and compresses the spring a distance x is : V±i. ^+^^(yIlA + ^ y- 1 2 ^ y-\V A (Intercoll. Exam. Cambridge, 1912.) 122 19. Steam initially at 200°C., and dryness 0'8 expands adiabatically to a temperature of 90°C. Find the dryness fraction at the end of expansion. If the expstnsion curve is to be represented by pv" = constant., find n. 20. Steam at 210 Ibs./in.^ abs. and dryness 0"8 is throttled down to a pressure of 40 lbs/in." abs. Find the final state of the steam. 21. How much heat must be given to 10 lbs. of water at 10°C. in order to convert it into steam at 220 lbs. per sq. in., with 100°C. of superheat. 22. On a temperature- entropy diagram for steam draw the constant volume line from 60 Ibs./in.^ to 20 Ibs./in.^, taking the volume as that of 1 lb. of dry steam at 60 Ibs./in.^ 23. A boiler of 200 cubic feet capacity contains equal volumes of water and steam when the pressure is 30 Ibs./in.'' Find the quantity of heat which must be given to the boiler to raise the pressure to 120 lbs/in.^ abs. If the efficiency of the boiler is 69%, and the calorific value of the coal ' 8500 Th. Units per lb., how much coal will be consumed? (IntercoU. Exam. Cambridge, 1911.) 24. The average total pressure of a slide valve on its seat is 2 tons, and the coefficient of friction is O'OS. The travel of the valve is 3'5", and the speed of the engine 200 R.P.M. If half the heat developed by friction goes to evaporate water in the steam passing into the cylinder, find the water evaporated per lb. of steam when there are 42 lbs. of steam used per minute, the steam chest pressure being 1501bs./in.^ (Intercoll. Exam. Cambridge, 1913.} 25. 10 lbs. of water at 15°C. are introduced into an exhausted vessel of 20 cubic feet capacity. How much water will be evaporated, and by how much will thg temperature fall ? The vessel is then heated until the pressure rises to 100 lbs. per sq. in. absolute : how much heat is taken in, and what is then the internal energy ? 123 Steam is then blown off until the pressure falls to 50 lbs. per sq. in. absolute : how much steam is blown off ? (Trin. Coll. Cambridge, 1914.) 26. The expansion curve of an indicator diagram is found to have the equation pv^'^ = const. ; determine how much heat has been taken in from the cylinder walls, per pound of steam, as the steam expands from the pressure 100 lbs. /in." abs. to 201bs./in.^ abs., the steam being dry at the higher pressure. (Mech. Sc. Trip., 1912.) 27. If the dryness fraction of a vapour remain constant during adiabatic expansion, show that the relation between the latent heat and the temperature must be of the form T = ae 'a-r where a is a constant, and o- the specific heat of the liquid, (Mech. Sc. Trip., 1913.) 28. A steam main 6 inches in external diameter conveys 120 lbs. of steam per minute from a boiler working at 200 lbs. pressure per square inch absolute to an engine 90 feet away. The pressure at the engine stop valve is 150 lbs. absolute and the dryness fraction of the steam is 0*93. Calculate the loss per minute in T.U. per square foot of pipe due to conduction and radiation, if the effective length of the pipe is 100 feet The pipe is afterwards covered by lagging and the steam then reaches the engine at a pressure of 190Jbs. per square inch absolute and dryness fraction 0"97. Calculate the saving in T.U. per minute effected by the lagging. (Mech. Sc. Trip. 1910.) 29. A boiler contains 3000 lbs. of steam and water at 1201bs./in.^ abs., half the total volume being water. Find the amount of steam which must be blown off to reduce the pres- sure to 50 Ibs./in.^ abs. 30. The indicator diagram shewn (Fig. 39) is taken from a single-acting engine using 1200 lbs. of steam per hour and 124 making 90 revolutions per minute. The clearance volume is 10 per cent, of the stroke volume. The stroke of the engine is 2 feet and the diameter of the cylinder is 18 inches. Deter- mine the I.H.P. of the engine and find the dryness of the steam at the point marked A in the diagram. Atmospheric pressure = 15 lbs. per sq. in. (Intercoll. Exam. Cambridge, 1913.) Spring 40 Fig. 39. 31. Two points on the expansion line of an indicator card gave pi — 75 lbs. per sq. in. , ^a = 15 lbs. per sq. in. Vi = 1'3 cub. feet v^ = 5"1 cub. feet. the pressures being measured from the atmospheric line and the volumes from the beginning of the stroke. Atmospheric pressure 15 lbs. per sq. in. The total mass of steam expanding in the cylinder was estimated at 0'48 lbs. and the volume of the clearance space d'46 cub. feet. Assuming the expansion line to be represented by the equation PV" = constant, find the work done by the steam between the points (l) and (2) and the interchange of heat between the steam and the cylinder. (Intercoll. Exam. Cambridge, 1914.) 32. The indicator diagram taken from the end of a single acting cylinder is as shewn in Fig. 40. The cylinder is 1 foot 3 inches in diameter and the stroke is 1 foot 9 inches. Determine the indicated horse-power of the engine. 125 The clearance volume at either end of the cylinder is 10 per cent, of the piston displacement and the air pump dis- charge is 40 lbs. per min. If the atmospheric pressure is 14"7 lbs. per sq. inch, determine the dryness of the steam at the points A and B and shew on the diagram supplied to you the points on the saturation curve corresponding to A and B. (Mech. Sc. Trip., 1912.) Spring -^ 200 Revs, per- min. Atmospheric line Fig. 40. 33. A steam engine consumes 3000 lbs. of steam per hour and develops 320 1.H.P. The steam is supplied at 1501bs./in.^ abs. and 400°C. The pressure in the condenser is l"251bs./in.^ abs. What is the thermal efficiency, and ■vvhat would it have been if the engine had followed the Ran- kine cycle ? 34. A boiler supplies steam at a pressure of 200 lbs. per sq. in. abs., the dryness fraction being 0'9. After passing through a reducing valve, with a reduction of pressure to 160 lbs. per sq. in. the steam enters the high-pressure cylinder of a compound engine where it expands adiabatically to 40 lbs. per sq. in. It is exhausted from this cylinder at 22 lbs. per sq. in. entering the low-pressure cylinder, where it expands adiabatically to 8 lbs. per sq. in., exhaust taking place at 2 lbs. per sq. in. Estimate the volume of the steam per pound at cut-ofF in each cylinder and at the beginning of exhaust in the low-pressure cylinder. (Mech. Sc. Trip., 1910.) 35. Dry steam is supplied to a single-cylinder steam engine at 102 lbs. per sq. in. absolute pressure and the engine 126 uses 19 lbs. per i.h.P. per hour. The steam-jacket condenses 0*95 lb. of steam per I.H.P. per hour at the temperature of supply. The engine has a jet condenser to which water is supplied at 10°C. Neglecting loss of heat by radiation, find how many pounds of water must be supplied to the condenser per pound of steam supplied to the engine, in order, that the air pump discharge may have a temperature of 35 C. (Mech. Sc. Trip., 1906.) 36. Dry steam is admitted to an engine cylinder at a pressure of a 100 lbs. per sq. in. and 20% of it is condensed during admission, without fall of pressure. The steam expands down to a pressure of 20 lbs. per sq. in. and during expansion one half of the heat absorbed by the walls is returned to the steam at a uniform rate as the temperature falls. Find the dryness fraction of the steam at the end of expansion. Find also the amount of work obtainable in the cylinder per pound of steam from the returned heat. Shew that if the steam expand adiabatically to the same final volume the final pressure will be a little below 18 lbs. per sq. in. (Mech. Sc. Trip., 1914.) 37. Calculate the work available per pound of steam in a Rankine Cycle," the exhaust being at 2 Ibs./in." abs : (a) when the supply pressure is 150 Ibs./in.^ abs. and the steam is dry and saturated ; (b) when dry steam- at 150 Ibs./in.'' abs. is throttled just before admission to 100 lbs. /in." abs. (R.N.C. Greenwich, 1912.) 38. The temperature inside a surface condenser is 46*6°C. and the pressure 2 lbs. per sq. in. The steam space is of 40 cubic feet capacity. Shew that the weight of air present is 0-093 lb., and is 53 per cent, of the weight of steam present. The engine with this condenser is of 50 l.H.p. and con- sumes 16 lbs. of steam per I.H.P. hour. Assuming that the exhaust steam entering the condenser is free of air, determine the proper volume of the air pump cylinder, making 120 127 strokes per minute necessary to overcome a leakage of air into the condenser at the rate of 1 lb. per hour. (Intercoll. Exam. Cambridge, 1914.) 39. Dry hot air is passed into a cooling tower containing tiles which are kept wetted by a supply of water at 15 C. The air emerges saturated with water-vapour at a temperature of 60°C. Shew that the quantity of water evaporated is about 0-15 pound per pound of air, and that the temperature of the entering air is about 440°C. The density of dry air may be taken as '08 pound per standard cubic feet. (Mech. Sc. Trip., 1914.) 40. A fluid expands adiabatically from 120 lbs. absolute to 15 lbs. absolute in an expanding nozzle according to the law pv^ " = constant. If the flow per second is 0'2 lb., find the area of the minimum section and the area of the outlet section of the nozzle. The specific volume at the higher pressure is 3"7 cubic feet per pound. (Mech. Sc. Trip., 1910.) 41. Steam is discharged through a nozzle from an initial pressure of 160 lbs/in. abs. into the casing of a De Laval turbine where the pressure is 2"5 Ibs./in.^ abs. Neglecting esses, find the velocity of discharge. 42. Steam initially dry and at rest under 180 lbs. pressure flows down a nozzle with no external communication of heat. It is estimated that at the section where the pressure is 40 lbs. per. sq. in. the kinetic energy of the steam is 20% less than it would be at that pressure in the case of frictionless flow. Find the percentage increase of section, at this pressure, as compared with frictionless flow, in order that the discharge may be the same in both cases. Also if the pv curve during the expansion down the nozzle is of the form ^v" = a constant, find the value of n, and determine how much heat has been communicated, to each pound of steam, between the given pressures. (Mech. Sc. Trip., 1910, B.) 128 43. An ejector is supplied with dry steam at 245 lbs/in.^ abs., and lifts water from a tank at the rate of 48 tons per hour, the lift being 12 feet. The water in the tank is at 5°C., and the discharged water at 28°C'. Find the steam used per ton of water, and the efficiency of the ejector considered as a pump. 44. In a Curtis turbine, with an initial pressure of 250 Ibs./in.^ abs. and final pressure 1 Ib./in.^ abs. the steam is expanded in 7 stages, in which the first develops i of the total powfer. If 20% of the available energy is lost in friction through the turbine, find the number of heat units converted into kinetic energy in the first stage. If 14% of the energy is lost in the nozzles, what is the velocity of exit from the nozzles of the first stage ? (R.N.C. Greenwich, 1910.) 45. The blades of a De I^aval turbine have angles of 35° at inlet and exit, and the nozzle makes an angle of 18° with the plane of rotation. The steam velocity at exit from the nozzle is 3600 ft. /sec. When the speed of the vanes is 1200 ft. /sec, will the steam impinge on the blades without shock ? Assuming that the relative velocity at exit is 80% of the relative velocity at inlet, calculate, at the above speed of the vanes, the work given to the vanes per pound of steam ~ and the internal efficiency of the turbine. (R.N.C. Green- wich, 1909.) 46. One stage of an impulse turbine of the Curtis type has three moving and two fixed rings. The blades on the moving rings have the angles of exit and of entry equal to one another,' and these are the same in all three rings. Steam issues from the nozzles with a velocity of 2650 ft./sec, and the nozzles make an angle of 20° with the direction of motion of the blades. The moving blades have a velocity of 530 ft./sec. Find the angles of entry and exit for both rings of fixed blades, and the common angle of exit and entry for the moving blades. 129 Find also the horse-power per pound of steam passing per second. All frictional effects are to be neglected (Mech. So. Trip., 1914, B.) 47. Steam issues from the nozzles of a Laval turbine at a velocity of 3325 ft. /sec. The axes of the nozzles are severally placed at angles of 18 degrees with the plane of revolution of the wheel. Find by measurement from a diagram. of velocities, drawn for maximum efficiency, (1) the angle of the vane at entry, (2) the absolute velocity of the steam at discharge from the wheel, assuming that the turbine- blades are symmetrical, (3) the efficiency of the turbine, (4) the horse-power of the turbine, assuming that 900 pounds of steam are used per hour. All frictional and eddy losses are to be neglected. (Mech. Sc. Trip., 1907.) 48. In an ammonia compression refrigerator the tem- perature in the evaporator is — , 10°C., and at the end of com- pression the pressure is 125 lbs. /in. The liquid passes to the expansion valve at 20°C. If the compression be adiabatic, what will be the coefficient of performance if the vapour is just dry (a) at the end, (b) at the beginning, of compression ? (R.N.C. Greenwich, 1909.) 49. An ammonia refrigerating machine is required to produce 3 tons of ice per' hour from water at 12°C. The temperatures of evaporation and condensation of the ammonia are — 10°C. and 25°C. respectively, and the condensed ammonia is driven through an expansion valve without pre- liminary cooling. Find the horse-power required in the compressor if the vapour ■ is compressed adiabatically and is just dry at the end of compression. The latent heat of water may be taken as 80. 130 If the vapour is just dry at the beginning of compression and is compressed adiabatically to the same final pressure, find the amount of superheat at the end of compression and the horse-power required. (Mech. Sc. Trip., 1914.) 50. Figure 41 shows a sketch of a portion of Mollier's (t>l diagram for carbonic acid. Explain the construction of the diagram and sketch on the figure one or two lines of constant pressure in the region of superheat. Describe the use of this diagram in refrigeration problems and make an estimate from it of the coefficient of performance of a CO2 cycle working between 20°C. and - 20°C., when the vapour is dry at the end of the com- pression. Check your estimate by making an evaluation of the coefficient from the data given in the CO2 table. (Mech. Sc. Trip., 1915, B.) ~~~ ■ -____ 1=70 ~~~-^--^-__ 60 ^^^ / ^'^~~ /~ — ~~~^ 50 ~~~^---------X/ 1 / / y '»«o5^ .0/ >7 V ^=orf5~-. — ^ \ 1 / / / "^-^ 20 ^^ //// / / 10 ^~^-^~^jO/ ^=^^05-^^^ ^F -— - -._ 1 = , Fig. 41. 51. Two refrigerating machines, one using SO2 and the other NHs, work betweeip the same limits of temperature, 131 viz. 10°C. and — 34"4°C. In each case the vapour is dry and saturated at the end of compression, and expansion takes place through a throttling valve. Determine in each case the refrigerating effect per pound of substance and the coefficient of performance. Also show that, if the machines are to have the same refrigerating effect per stroke, the volumes of their compressing cylinders must be approximately in the ratio 11:4. (Mech. Sc. Trip., 1908.) 52. A small reservoir for compressed air has a volume of 2 cubic feet, and is being charged from the atmosphere by means of a single-acting pump of diameter 6 inches and stroke 9 inches. At the beginning of one cycle of the pump, the reservoir pressure is 20 lbs. per sq. inch by gauge. Find the pressure after one cycle of the pump, neglecting clearance and assuming that the air remains at the temperature of the atmosphere, 15°C. The atmospheric pressure is 14'7 lbs. per sq. inch. Find also the work done by the pump and the mass of air pushed into the reservoir. (Intercoll. Exam. Cambridge, 1911.) 53. A single-acting reciprocating air engine, with a stroke of 6 inches and piston diameter 4 inches, is being driven from a reservior of compressed air of 10 cubic feet capacity ; the pressure of the reservoir falling as the air is used in the engine. At the beginning of a particular cycle of the engine, the reservoir pressure is 100 lbs. per sq. inch by gauge and the temperature 15 C. Find the work done in this cycle if the engine cuts off at i stroke and exhausts into the atmosphere. Neglect clearance and assume the process adiabatic. Find also what will be the temperature of the reservoir after the cycle. Pressure of atmosphere 14"71bs. per sq. inch. (Inter- coll. Exam. Cambridge, 1912.) 132 54. The piston of an air compressor displaces 8 cubic feet per stroke and makes 120 strokes per minute. It takes in atmospheric air at 15°C. and cortlpresses it, according to the law PV^'"* = constant, up to 75 lbs. gauge pressure ; finally delivering it at this pressure into a reservoir. Assuming no slip past the valves, no loss of head through them, and in the first place no clearance, calculate the work done upon the air in ' foot-pounds per minute, and the temperature at which it enters the reservoir. In the second place, if the clearance were 10 per cent, of the piston displacement, how would the work done per minute and the volumetric efficiency of the compressor be affected? Atmospheric pressure = 14"7 lbs. per sq. in. (Mech. Sc. Trip., 1906.) 55. The mean pressure during the suction stroke of a 4-cycle gas engine is 2 Ibs./in.^ below atmosphere. Shew that the temperature of the charge will be raised about lli°C. by the work done on it in drawing it in, the temperature of the outside air being 17°C. (Mech. Sc. Trip. B., 1910.) 56. A three-stage compressor, which was designed to give maximum efficiency when pumping against a constant back pressure ps from the initial pressure po, the interstage- pressures being pi and pi, is used to charge an air-vessel of volume V from atmospheric pressure po up to the pressure ps. Calculate the work done by the first-stage cylinder while the pressure in the air-vessel rises to pi ; also the work done in the first stage cylinder while the pressure in the air-vessel rises f rom^, to p^. (R.N.C., Greenwich, 1912.) 57. A gas-engine uses composition by volume producer gas of the following Per tent. CO 25 H COo 14 6 N 55 100-0 133 The exhaust-gases contain 15 per cent, of CO2. Shew that the ratio of the volumes of air to gas taken into the engine is 1-41. Air contains 21 per cent, of its volume of oxygen. What percentage of oxygen would you expect to find in the exhaust-gases ? (Mech. Sc. Trip., 1914.) 58. In the case of a land boiler fitted with a feed heater in the flue, the coal burnt per hour was 125"5 lbs. ; the feed per minute was 22"51bs., the temperature of the. products of combustion in the smoke-box and chimney bottom were 353°C, and 198°C. respectively, and the temperature of the feed water entering and leaving the feed heater were 54°C. and 109°C. respectively. If the specific heat of the products of combustion is 0"24 and the temperature of the boiler steam is 193°C., calculate the lbs. of air supplied per lb. of coal burnt ; and also (assuming the boiler steam dry) calculate the evaporation per lb. of coal from and at 100°C. (R.N.C., Greenwich, 1908.) 59. In the published results of a boiler test, the fuel and flue gas analyses are given as follows : Coal Analysis by Weight Flue Gas Analysis by Volume Carbon 83-44% CO, 11-7% H 3-99 O 3-85 O 2-88 CO 0-45 Ash 9-69 N 84-0 Are these figures consistent with reasonably complete combustion of the coal, and how could they be explained ? The atomic weight of carbon is- 12, that of oxygen is 16, and the ratio of the nitrogen to oxygen in the air is 376 by volume. (Mech. Sc. Trip., 1913.) 60. In the trial of a gas engine, the volume analysis of the gas supplied to the engine was CH4, 33% ; C2H4, 5% ; H, 42%; CO, 7%; N, 10%; CO,, 3%. The volume analysis of the exhaust gases was CO2, 5-8%;' O, 10-4%; N, 83-8%. Calculate the number of cubic feet of air 134 necessary for the complete combustion of one cubic foot of gas, and the excess air supplied to the engine per cubic foot of gas. Take the composition of air by volume as Oa 21% ; N, 79%. (R.N.C. Greenwich, 1911.) 61. A gas engine working the Otto cycle has a cylinder 10 inches diameter and a stroke of 20 inches. The com- pression space is 1 of the stroke volume. The inlet valve closes at the out-centre and the cylinder contents then consist of air having a temperature of 40°C. and a pressure of 14"7 lbs. per square inch. The pressure reached in compression is 160 lbs. per square inch absolute, the compression line of the indicator diagram following the curve ^t;'"'* = const. Find : — (1) The mean temperature of' the air at the end of compression. (2) The temperature reached at a point in the interior of the cylinder where there has been no loss of heat. (3) The quantity of heat lost to the cylinder walls in the course of compression. (Mech. Sc. Trip., 1907.) 62. The following are the results of the trial of a gas- engine working the Otto cycle, and governing by hit and miss : — Cylinder diameter, 10 inches. Stoke, 18 inches. Gas used at full load, -087 cub. ft. per explosion. The engine misses once in 10 cycks. Calorific value of gas, 308 C.T.u. per cubic foot at the temperature and pressure at which it is used. Brake Horse-power, 25. When the brakes are taken off the engine fires once in six cycles, the mean pressure of the explosion diagrams being 135 90 pounds per square inch. The speed is 180 revs, per minute whether loaded or unloaded. From these data estimate the indicated horse-power of the engine, when giving the load of 25 B.H.P., and the thermal efficiency (on the indicated horse-power.) (Mech. Sc. Trip., 1907.) 63. A gas engine works on an ideal cycle with adiabatic compression and expansion, receiving and rejecting heat only at constant volume. Obtain the expression for its efficiency. In such an engine the piston displacement per stroke is 1 cubic foot, the clearance volume 0*2 cubic foot, and at the beginning of compression the temperature of the cylinder contents is 333°C. abs., pressure being atmospheric. The engine receives 0'05 cubic foot of gas per cycle (calorific value 330C. Th. Units per cubic foot). Atmospheric pressure = 14"7 lbs. per sq. in. Find :— {a) Weight of cylinder contents. (b) Pressure and temperature at end of compression (take y = 1-38.) (c) Rise of temperature during explosion (neglect jacket loss and take ky = 0'18). (d) Pressure at end of explosion. (e) Temperature and pressure at end of expansion. if) Efficiency of the cycle. (g) Efficiency of an engine working on a Carnot cycle between the- same highest and lowest temperatures. (Mech. Sc. Trip., 1906.) 64. The following data were obtained in a complete test of an oil-engine : — Speed, 300. Barometer, 29*4 inches. B.H.P. 4J. Calorific value of oil used, 12,000 Th. Units per lb. 136 Oil entering engine per min., 0-7 lb. Air „ j, ,, i'y^ ,, Atmospheric temperature, 17° C. Dew point, 12 C. ( Quantity, 15 lb. per min. Jacket water ~ . / Inlet, 17 . 1 Temperature | q^^j^^^ 370^ Temperature of exhaust gases, 520°C. The chemical formula for petroleum may be taken as t^uH'so. Draw up a complete " heat balance sheet " for the engine, and deduce its thermal efficiency. (T.T., 1912.) INDEX The numbers refer to pages. Absolute temperature 3 adiabatic equation for gases 10 adiabatic equation for steam and vapours 41 adiabatic process 2 air compressors 94 ,, standard 112 available energy 22 B.H.P 66 Bell-Coleman refrigerator ... 83 Calorific value of coal 105 Carnot's cycle 18 characteristic equation of a gas 8 clearance, in steam engine ... ,, in air compressor combustion condensation conservation of energy cushion steam cycle 55 97 102 63 1 55, 59 ... ' 3 42 ' Dew point 113 Diesel engine 91 draught 103 dryness fraction 37, 44 Efficiency of reversible engine 20, 21 engine trials ^ 65, 67, 113 entropy, definition 23 ,, diagrams 30, 39, 51, 56 ,, of gas 30 ,, in reversible cycle 24 , , steam and vapours 39 entropy, water 39 of Rankine cycle ... 53 ,, nf steam engine ... 66 evaporation 37 exchange of heat between steam and cylinder 56, 61 expansion of gas at constant pressure 9 expansion of gas adiabatically 10, 13 ,, ,, isothermally 11 First Law of Thermo- dyramics 1 Gas engines ... tests ... 110 ... 113 Heat, balance sheet for steam engine 66 ,, for internal com- bustion engine 115 Heat, of formation 36 Heat, supply 65, 113 Heat, unit 2 I. H. P : 65 injectors 76 internal combustion engines 110 energy .. 1, of steam 37, 38 isothermal expansion ... 11 processes 2 Jets, steam 72 Joule's cycle 33, 88 ,, experiment 8 INDEX— continued . Latent heat ... 37 Stirling cycle ' ... . . ... 32 Mollier's Diagi am 53 superheated vapour surface condensers- ; ::: 40 63 Nozzles 72 Thermal unit 2 Otto cycle 33, 110 throttling 42 Rankine cycle refrigerators .. regenerators .. reversibility ... SO 88 32 18 total heat of steam vapours turbines Curtis de Laval and . 79, 79, 80 37 79 84 82 Saturation line Second Law dynamics of Thermo - 56 18 ,, Parsons Rateau .. Zoelly ... . 80 79 79 specific heat of gases gas mixtures 8 104 Vapour refrigerators . . 89 i> It superheated Work done, by gas ir ex- steam 41 pansion 9, 12 13 W. Heffer & Sons Ltd., Cambridgb, England.