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- University of Tennessee - Knoxville
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- Mechanical Engineering 331
- Bond
- Thermo Ch 9 Solutions Manual

Billy D.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-1 Chapter 9 GAS POWER CYCLES Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions 9-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot be approximated using the hardware of actual power producing devices. 9-2C It is less than the thermal efficiency of a Carnot cycle. 9-3C It represents the net work on both diagrams. 9-4C The cold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature. 9-5C Under the air standard assumptions, the combustion process is modeled as a heat addition process, and the exhaust process as a heat rejection process. 9-6C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state. 9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center. 9-8C It is the ratio of the maximum to minimum volumes in the cylinder. 9-9C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. 9-10C Yes. 9-11C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remain the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car gets older as a result of wear and tear. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-2 9-12C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines. 9-13C Stroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the minimum volume formed in the cylinder. 9-14 The temperatures of the energy reservoirs of an ideal gas power cycle are given. It is to be determined if this cycle can have a thermal efficiency greater than 55 percent. Analysis The maximum efficiency any engine using the specified reservoirs can have is 0.678= + + ?=?= K 273)(627 K 273)(17 11 Carnotth, H L T T ? Therefore, an efficiency of 55 percent is possible. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-3 9-15 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) The properties of air at various states are () () () kJ/kg 828.149.101310 kPa 2668 kPa 100 kPa 2668kPa 800 K 539.8 K 1800 1310 kJ/kg 1487.2 K 1800 K 539.8 kJ/kg 389.22 088.111.386 kPa 100 kPa 800 386.1 kJ/kg 300.19 K300 4 3 4 2 2 3 3 2 22 3 33 3 3 2 2 1 2 1 1 34 3 12 1 =???=== ===???= = = ???= = = ???=== = = ???= hP P P P P T T P T P T P P u T T u P P P P P h T rr r rr r vv From energy balances, kJ/kg570.1 9.5270.1098 kJ/kg 527.919.3001.828 kJ/kg 1098.02.3892.1487 outinoutnet, 14out 23in =?=?= =?=?= =?=?= qqw hhq uuq (c) Then the thermal efficiency becomes 51.9%=== kJ/kg1098.0 kJ/kg570.1 in outnet, th q w ? v P 1 2 4 3 q in q out s T 1 2 4 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-4 9-16 EES Problem 9-15 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] P[2] = 800 [kPa] T[3]=1800 [K] P[4] = 100 [kPa] "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2" q_12 -w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3" q_23 -w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=0.287*T[4]} "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant pressure heat rejection" {P[4]*v[4]/T[4]=P[1]*v[1]/T[1]} "Conservation of energy for process 4 to 1" q_41 -w_41 = DELTAu_41 w_41 =P[1]*(v[1]-v[4]) "constant pressure process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-5 T 3 [K] ? th q in,total [kJ/kg] W net [kJ/kg] 1500 50.91 815.4 415.1 1700 51.58 1002 516.8 1900 52.17 1192 621.7 2100 52.69 1384 729.2 2300 53.16 1579 839.1 2500 53.58 1775 951.2 5.0 5.3 5.5 5.8 6.0 6.3 6.5 6.8 7.0 7.3 7.5 200 400 600 800 1000 1200 1400 1600 1800 2000 s [kJ/kg-K] T [K] 100 kPa 800 kPa Air 1 2 3 4 10 -2 10 -1 10 0 10 1 10 2 10 1 10 2 10 3 4x10 3 v [m 3 /kg] P [k P a ] 300 K 1800 K Air 1 2 3 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-6 1500 1700 1900 2100 2300 2500 50.5 51 51.5 52 52.5 53 53.5 54 T[3] [K] ? th 1500 1700 1900 2100 2300 2500 800 1000 1200 1400 1600 1800 T[3] [K] q in,to t a l [kJ/kg] 1500 1700 1900 2100 2300 2500 400 500 600 700 800 900 1000 T[3] [K] w ne t [k J/kg] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-7 9-17 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (b) From the ideal gas isentropic relations and energy balance, () () K 579.2 kPa 100 kPa 1000 K 300 0.4/1.4/1 1 2 12 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT () ()() K 3360==?????= ?=?= 3max3 2323in 579.2KkJ/kg 1.005kJ/kg 2800 TTT TTchhq p (c) ()K 336K 3360 kPa 1000 kPa 100 3 3 4 4 4 44 3 33 ===???= T P P T T P T P vv ()() ()() ()() 21.0%=?=?= = ??+??= ?+?= ?+?=+= kJ/kg 2800 kJ/kg 2212 11 kJ/kg 2212 K300336KkJ/kg 1.005K3363360KkJ/kg 0.718 in out th 1443 1443out41,out34,out q q TTcTTc hhuuqqq p ? v Discussion The assumption of constant specific heats at room temperature is not realistic in this case the temperature changes involved are too large. v P 1 2 4 3 q 34 q 41 q in s T 1 2 4 3 q in q 41 q 34 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-8 9-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (b) The properties of air at various states are Btu/lbm 129.06 Btu/lbm, 92.04R 540 111 ==???= huT () () Btu/lbm 593.22317.01242 psia 57.6 psia 14.7 1242 Btu/lbm 48.849 R 3200 psia 57.6psia 14.7 R 540 R 2116 Btu/lbm 1.537,R2116 Btu/lbm 04.39230004.92 4 3 4 3 3 1 1 2 2 1 11 2 22 22 in,1212 12in,12 34 3 =???=== = = ???= ===???= == =+=+= ????= hP P P P P h T P T T P T P T P hT quu uuq rr r vv From energy balance, Btu/lbm 464.1606.12922.593 38.312300 Btu/lbm 312.381.53748.849 14out in23,in12,in 23in23, =?=?= =+=+= =?=?= hhq qqq hhq Btu/lbm612.38 (c) Then the thermal efficiency becomes 24.2%=?=?= Btu/lbm612.38 Btu/lbm464.16 11 in out th q q ? v P 3 4 2 1 q 12 q 23 q out s T 3 4 2 1 q out q 23 q 12 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-9 9-19E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 0.240 Btu/lbm.R, c v = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis (b) () ()() () ()( )( ) Btu/lbm 217.4R22943200RBtu/lbm 0.24 psia 62.46psia 14.7 R 540 R 2294 R 2294 R540Btu/lbm.R 0.171Btu/lbm 300 2323in,23 1 1 2 2 1 11 2 22 2 2 1212in,12 =??=?=?= ===???= = ?= ?=?= TTchhq P T T P T P T P T T TTcuuq P vv v Process 3-4 is isentropic: () () ()( )( ) Btu/lbm 378.55402117Btu/lbm.R 0.240 4.217300 R 2117 psia 62.46 psia 14.7 R 3200 1414out in,23in,12in 0.4/1.4/1 3 4 34 =?=?=?= =+=+= = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? TTchhq qqq P P TT p kk Btu/lbm517.4 (c) 26.8%=?=?= Btu/lbm 517.4 Btu/lbm 378.5 11 in out th q q ? v P 3 4 2 1 q 12 q 23 q out s T 3 4 2 1 q out q 23 q 12 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-10 9-20 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the heat rejected and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (b) () () K 579.2 kPa 100 kPa 1000 K 300 0.4/1.4/1 1 2 12 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT ()() ()( )( ) K 12662.579KkJ/kg 1.005kg 0.004kJ 2.76 33 2323in =?????= ?=?= TT TTmchhmQ p Process 3-1 is a straight line on the P-v diagram, thus the w 31 is simply the area under the process curve, () () kJ/kg 7.273 KkJ/kg 0.287 kPa 1000 K 1266 kPa 100 K 300 2 kPa 1001000 22 area 3 3 1 113 31 13 31 = ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? + = ? ? ? ? ? ? ? ? ? + =? + == P RT P RTPPPP w vv Energy balance for process 3-1 gives ()[] ()( )( )[]kJ 1.679=?+?= ?+?=???= ?=??????=? K1266-300KkJ/kg 0.718273.7kg 0.004 )( )( 31out31,31out31,out31, 31out31,out31,systemoutin TTcwmTTmcmwQ uumWQEEE vv (c) 39.2%=?=?= kJ 2.76 kJ 1.679 11 in out th Q Q ? v P 32 1 q in q out s T 3 2 1 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-11 9-21 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work per cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) The properties of air at various states are () () ()( )( ) ()( )( ) kJ0.422 651.1073.2 kJ 1.651kJ/kg290.16840.38kg 0.003 kJ 2.073kJ/kg206.91897.91kg 0.003 kJ/kg 840.3851.8207.2 kPa 380 kPa 95 2.207kJ/kg, 897.91 K 1160K 290 kPa 95 kPa 380 kJ/kg 290.16 kJ/kg 206.91 K 029 outinoutnet, 13out 12in 3 2 3 2 1 1 2 2 1 11 2 22 1 1 1 23 2 =?=?= =?=?= =?=?= =???=== ==??? ===???= = = ???= QQW hhmQ uumQ hP P P P Pu T P P T T P T P h u T rr r vv (c) 20.4%=== kJ 2.073 kJ 0.422 in outnet, th Q W ? v P 3 2 1 q in q out s T 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-12 9-22 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work per cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (b) From the isentropic relations and energy balance, () () () ()() ()( )( ) ()() ()( )( ) kJ0.39 48.187.1 kJ 1.48K290780.6KkJ/kg 1.005kg 0.003 kJ 1.87K2901160KkJ/kg 0.718kg 0.003 K 780.6 kPa 380 kPa 95 K 1160 K 1160K 290 kPa 95 kPa 380 outinoutnet, 1313out 1212in 0.4/1.4/1 2 3 23 1 1 2 2 1 11 2 22 =?=?= =??= ?=?= =??= ?=?= = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ===???= ? QQW TTmchhmQ TTmcuumQ P P TT T P P T T P T P p kk v vv (c) 20.9%=== kJ1.87 kJ0.39 in net th Q W ? v P 3 2 1 q in q out s T 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-13 9-23 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2). Analysis The minimum pressure in the cycle is P 3 and the maximum pressure is P 1 . Then, or () () () kPa 3.935 K 300 K 900 kPa 20 1.4/0.41/ 3 2 32 /1 3 2 3 2 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? kk kk T T PP P P T T The heat input is determined from Then, () ()( )()( ) ()( ) kJ 0.393=== =?=?= =?=?= ?=??=?=? kJ 0.58890.667 .7%66 K 900 K 300 11 kJ 0.5889KkJ/kg 0.2181K 900kg 0.003 KkJ/kg 0.2181 kPa 2000 kPa 935.3 lnKkJ/kg 0.287lnln inthoutnet, th 12in 1 2 0 1 2 12 QW T T ssmTQ P P R T T css H L H p ? ? ? s T 3 2 q in q out 4 1 900 300 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-14 9-24 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second- law efficiency of an actual cycle operating between the same temperature limits are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperatures are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2). Analysis (a) The minimum temperature is determined from ()( ) ( )( ) K 350K750KkJ/kg 0.25kJ/kg 100 12net =?????=?????= LLLH TTTTssw The pressure at state 4 is determined from or () () kPa 1.110 K 350 K 750 kPa 800 4 1.4/0.4 4 1/ 4 1 41 /1 4 1 4 1 =??? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? PP T T PP P P T T kk kk The minimum pressure in the cycle is determined from () kPa 46.1=?????=?? ?=??=? 3 3 3 4 0 3 4 3412 kPa 110.1 lnKkJ/kg 0.287KkJ/kg 25.0 lnln P P P P R T T css p ? (b) The heat rejection from the cycle is kgkJ/ 87.5==?= kJ/kg.K) K)(0.25 350( 12out sTq L (c) The thermal efficiency is determined from 0.533=?=?= K 750 K 350 11 th H L T T ? (d) The power output for the Carnot cycle is kW 9000kJ/kg) kg/s)(100 90( netCarnot === wmW & & Then, the second-law efficiency of the actual cycle becomes 0.578=== kW 9000 kW 5200 Carnot actual II W W & & ? s T 3 2 q in 4 1 750 K w net =100 kJ/kg q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-15 9-25 An ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per cycle are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The temperature of the low-temperature reservoir can be found by applying the isentropic expansion process relation K 481.1=? ? ? ? ? ? += ? ? ? ? ? ? ? ? = ? ? 14.1 1 1 2 21 12 1 K) 2731027( k TT v v Since the Carnot engine is completely reversible, its efficiency is 0.630= + ?=?= K 273)(1027 K 481.1 11 Carnotth, H L T T ? The work output per cycle is kJ/cycle 20 min 1 s 60 cycle/min 1500 kJ/s 500 net net =? ? ? ? ? ? == n W W & & According to the definition of the cycle efficiency, kJ/cycle 31.75===???= 0.63 kJ/cycle 20 Carnotth, net in in net Carnotth, ? ? W Q Q W 9-26E The temperatures of the energy reservoirs of an ideal gas Carnot cycle are given. The heat supplied and the work produced per cycle are to be determined. Analysis According to the thermodynamic definition of temperature, Btu/cycle 340= + + == R 460)(40 R 460)(1240 Btu) (100 L H LH T T QQ Applying the first law to the cycle gives Btu/cycle 240=?=?= 100340 net LH QQW s T 3 2 q in q out 4 11300 K s T 3 2 q in q out 4 1 1700 R 500 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-16 Otto Cycle 9-27C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat addition, (3) isentropic expansion, and (4) v = constant heat rejection. 9-28C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency. 9-29C For actual four-stroke engines, the rpm is twice the number of thermodynamic cycles; for two- stroke engines, it is equal to the number of thermodynamic cycles. 9-30C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes. 9-31C It increases with both of them. 9-32C Because high compression ratios cause engine knock. 9-33C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667. 9-34C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines. 9-35 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The definition of cycle thermal efficiency reduces to 0.630=?=?= ?? 11.41 th 12 1 1 1 1 k r ? The rate of heat addition is then kW 318=== 0.630 kW 200 th net in ? W Q & & v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-17 9-36 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The definition of cycle thermal efficiency reduces to 0.602=?=?= ?? 11.41 th 10 1 1 1 1 k r ? The rate of heat addition is then kW 332=== 0.602 kW 200 th net in ? W Q & & v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-18 9-37E An Otto cycle with non-isentropic compression and expansion processes is considered. The thermal efficiency, the heat addition, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then () R 11958R) 520( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k s rTTT v v With the isentropic compression efficiency, the actual temperature at the end of the compression is R 1314 0.85 R 520)(1195 R) 520( 12 12 12 12 = ? += ? +=??? ? ? = ? ? TT TT TT TT ss Similarly for the expansion, R 1201 8 1 R) 4602300( 1 14.11 3 1 4 3 34 =? ? ? ? ? ? +=? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k k s r TTT v v R 1279R )1201(2760)95.0(R) 2760()( 4334 43 43 =??=??=??? ? ? = s s TTTT TT TT ?? The specific heat addition is that of process 2-3, Btu/lbm 247.3=??=?= R)13142760)(RBtu/lbm 171.0()( 23in TTcq v The net work production is the difference between the work produced by the expansion and that used by the compression, Btu/lbm 5.117 R)5201314)(RBtu/lbm 171.0(R)12792760)(RBtu/lbm 171.0( )()( 1243net = ?????= ???= TTcTTcw vv The thermal efficiency of this cycle is then 0.475=== Btu/lbm 247.3 Btu/lbm 117.5 in net th q w ? At the beginning of compression, the maximum specific volume of this cycle is /lbmft 82.14 psia 13 R) 520)(R/lbmftpsia 3704.0( 3 3 1 1 1 = ?? == P RT v while the minimum specific volume of the cycle occurs at the end of the compression /lbmft 852.1 8 /lbmft 82.14 3 3 1 2 === r v v The engine?s mean effective pressure is then psia 49.0= ? ? ? ? ? ? ? ? ? ? = ? = Btu 1 ftpsia 404.5 /lbmft )852.182.14( Btu/lbm 5.117 MEP 3 3 21 net vv w v P 4 1 3 2 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-19 9-38 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression. ()() () ()kPa 2338kPa 100 K 308 K 757.9 9.5 K 757.99.5K 308 1 1 2 2 1 2 1 11 2 22 0.4 1 2 1 12 = ? ? ? ? ? ? ? ? ==???= == ? ? ? ? ? ? ? ? = ? P T T P T P T P TT k v vvv v v Process 3-4: isentropic expansion. ()() K 1969== ? ? ? ? ? ? ? ? = ? 0.4 1 3 4 43 9.5K 800 k TT v v Process 2-3: v = constant heat addition. ()kPa 6072= ? ? ? ? ? ? ? ? ==???= kPa 2338 K 757.9 K 1969 2 2 3 3 2 22 3 33 P T T P T P T P vv (b) ()( ) () kg10788.6 K 308K/kgmkPa 0.287 m 0.0006kPa 100 4 3 3 1 11 ? ×= ?? == RT P m V ()()( )( )( ) kJ 0.590=??×=?=?= ? K757.91969KkJ/kg 0.718kg106.788 4 2323in TTmcuumQ v (c) Process 4-1: v = constant heat rejection. ()( )( )( ) kJ0.2 40K308800KkJ/kg 0.718kg106.788)( 4 1414out =??×?=?=?= ? TTmcuumQ v kJ 0.350240.0590.0 outinnet =?=?= QQW 59.4%=== kJ 0.590 kJ 0.350 in outnet, th Q W ? (d) ()() kPa 652= ? ? ? ? ? ? ? ? ? ? = ? = ? = == kJ mkPa 1/9.51m 0.0006 kJ 0.350 )/11( MEP 3 3 1 outnet, 21 outnet, max 2min r WW r VVV V VV v P 4 1 3 2 Q in Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-20 9-39 An Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression. ()() () ()kPa 2338kPa 100 K 308 K 757.9 9.5 K 757.99.5K 308 1 1 2 2 1 2 1 11 2 22 0.4 1 2 1 12 = ? ? ? ? ? ? ? ? ==???= == ? ? ? ? ? ? ? ? = ? P T T P T P T P TT k v vvv v v Process 3-4: polytropic expansion. ()( ) () kg10788.6 K 308K/kgmkPa 0.287 m 0.0006kPa 100 4 3 3 1 11 ? ×= ?? == RT P m V ()() ()()()() kJ 0.5338 1.351 K1759800KkJ/kg 0.287106.788 1 9.5K 800 4 34 34 35.0 1 3 4 43 = ? ??× = ? ? = == ? ? ? ? ? ? ? ? = ? ? n TTmR W TT n K 1759 v v Then energy balance for process 3-4 gives () () () ()() kJ 0.0664kJ 0.5338K1759800KkJ/kg 0.718kg106.788 4 in34, out34,34out34,34in34, 34out34,in34, outin =+??×= +?=+?= ?=? ?=? ? Q WTTmcWuumQ uumWQ EEE system v That is, 0.066 kJ of heat is added to the air during the expansion process (This is not realistic, and probably is due to assuming constant specific heats at room temperature). (b) Process 2-3: v = constant heat addition. ()kPa 5426= ? ? ? ? ? ? ? ? ==???= kPa 2338 K 757.9 K 1759 2 2 3 3 2 22 3 33 P T T P T P T P vv ()() ()()kJ 0.4879K757.91759KkJ/kg 0.718kg106.788 4 in23, 2323in23, =??×= ?=?= ? Q TTmcuumQ v Therefore, kJ 0.5543=+=+= 0664.04879.0 in34,in23,in QQQ (c) Process 4-1: v = constant heat rejection. ()()( )( )( ) kJ 0.2398K308800KkJ/kg 0.718kg 106.788 4 1414out =??×=?=?= ? TTmcuumQ v kJ 0.31452398.05543.0 outinoutnet, =?=?= QQW 56.7%=== kJ 0.5543 kJ 0.3145 in outnet, th Q W ? (d) ()() kPa 586= ? ? ? ? ? ? ? ? ? ? = ? = ? = == kJ mkPa 1/9.51m 0.0006 kJ 0.3145 )/11( MEP 3 3 1 outnet, 21 outnet, max 2min r WW r VVV V VV v P 4 1 3 2 Q in Q out Polytropic 800 K 308 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-21 9-40E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression. 32.144 Btu/lbm92.04 R540 1 1 1 = = ???= r u T v () Btu/lbm 11.28204.1832.144 8 11 2 1 2 222 =???==== u r rrr vv v v v Process 2-3: v = constant heat addition. Btu/lbm 241.42=?=?= = = ???= 28.21170.452 419.2 Btu/lbm 452.70 R2400 23 3 3 3 uuq u T in r v (b) Process 3-4: isentropic expansion. ()( ) Btu/lbm 205.5435.19419.28 4 3 4 334 =???==== ur rrr vv v v v Process 4-1: v = constant heat rejection. Btu/lbm 50.11304.9254.205 14out =?=?= uuq 53.0%=?=?= Btu/lbm 241.42 Btu/lbm 113.50 11 in out th q q ? (c) 77.5%=?=?= R 2400 R 540 11 Cth, H L T T ? v P 4 1 3 2 q in q out 2400 R 540 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-22 9-41E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable with argon as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon are c p = 0.1253 Btu/lbm.R, c v = 0.0756 Btu/lbm.R, and k = 1.667 (Table A-2E). Analysis (a) Process 1-2: isentropic compression. ()() R 21618R 540 0.667 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT v v Process 2-3: v = constant heat addition. () ()() Btu/lbm 18.07= ?= ?=?= R 21612400Btu/lbm.R 0.0756 2323in TTcuuq v (b) Process 3-4: isentropic expansion. () R 600 8 1 R 2400 0.667 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?k TT v v Process 4-1: v = constant heat rejection. ()( )( ) Btu/lbm 4.536R540600Btu/lbm.R 0.0756 1414out =?=?=?= TTcuuq v 74.9%=?=?= Btu/lbm 18.07 Btu/lbm 4.536 11 in out th q q ? (c) 77.5%=?=?= R 2400 R 540 11 Cth, H L T T ? v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-23 9-42 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are c p = 1.110 kJ/kg·K, c v = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) Process 1-2: polytropic compression ()() ()() kPa 199510kPa 100 K 4.66410K 333 1.3 2 1 12 1-1.3 1 2 1 12 == ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? n n PP TT v v v v Process 2-3: constant volume heat addition () K 2664 kPa 1995 kPa 8000 K 664.4 2 3 23 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT () ()()kJ/kg 1646K664.42664KkJ/kg 0.823 2323in =??= ?=?= TTcuuq v Process 3-4: polytropic expansion. () K 1335=? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? 1-1.3 1 4 3 34 10 1 K 2664 n TT v v () kPa 9.400 10 1 kPa 8000 1.3 1 2 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = n PP v v Process 4-1: constant voume heat rejection. ()( )( ) kJ/kg 8.824K3331335KkJ/kg 0.823 1414out =??=?=?= TTcuuq v (b) The net work output and the thermal efficiency are kJ/kg 820.9=?=?= 8.8241646 outinoutnet, qqw 0.499=== kJ/kg 1646 kJ/kg 820.9 in outnet, th q w ? (c) The mean effective pressure is determined as follows ( )( ) ()() kPa 954.3= ? ? ? ? ? ? ? ? ? ? = ? = ? = == == ?? == kJ mkPa 1/101/kgm 0.9557 kJ/kg 820.9 )/11( MEP /kgm 0.9557 kPa 100 K 333K/kgmkPa 0.287 3 3 1 outnet, 21 outnet, max 2min max 3 3 1 1 1 r ww r P RT vvv v vv vv 1 Q in 2 3 4 P V Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-24 (d) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are 3 3 m 0002444.0 m 0022.0 10 =??? + =??? + = c c c c dc r V V V V VV 3 1 m 002444.00022.00002444.0 =+=+= dc VVV The total mass contained in the cylinder is ()() kg 0.002558 K 333K/kgmkPa 0.287 )m 444kPa)/0.002 100( 3 3 1 11 = ?? == RT VP m t The engine speed for a net power output of 70 kW is rev/min 4001=? ? ? ? ? ? ? == min 1 s 60 cycle)kJ/kg kg)(820.9 002558.0( kJ/s 70 rev/cycle) 2(2 net net wm W n t & & Note that there are two revolutions in one cycle in four-stroke engines. (e) The mass of fuel burned during one cycle is kg 0001505.0 kg) 002558.0( 16AF =??? ? =??? ? == f f f f ft f a m m m m mm m m Finally, the specific fuel consumption is g/kWh 258.0=? ? ? ? ? ? ? ? ? ? ? ? ? ? == kWh 1 kJ 3600 kg 1 g 1000 kJ/kg) kg)(820.9 002558.0( kg 0001505.0 sfc net wm m t f PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-25 9-43E The properties at various states of an ideal Otto cycle are given. The mean effective pressure is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis At the end of the compression, the temperature is () R 12529R) 520( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v while the air temperature at the end of the expansion is R 9.813 9 1 R) 1960( 1 14.11 3 1 4 3 34 =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k k r TTT v v Application of the first law to the compression and expansion processes gives Btu/lbm 81.70 R)5201252)(RBtu/lbm 171.0(R)9.8131960)(RBtu/lbm 171.0( )()( 1243net = ?????= ???= TTcTTcw vv At the beginning of the compression, the specific volume is /lbmft 76.13 psia 14 R) 520)(R/lbmftpsia 3704.0( 3 3 1 1 1 = ?? == P RT v while the specific volume at the end of the compression is /lbmft 529.1 9 /lbmft 76.13 3 3 1 2 === r v v The engine?s mean effective pressure is then psia 31.3= ? ? ? ? ? ? ? ? ? ? = ? = Btu 1 ftpsia 404.5 /lbmft )529.176.13( Btu/lbm 81.70 MEP 3 3 21 net vv w v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-26 9-44E The power produced by an ideal Otto cycle is given. The rate of heat addition and rejection are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis The thermal efficiency of the cycle is 5848.0 9 1 1 1 1 11.41 th =?=?= ??k r ? According to the definition of the thermal efficiency, the rate of heat addition to this cycle is Btu/h 609,100= ? ? ? ? ? ? ? ? == hp 1 Btu/h 2544.5 0.5848 hp 140 th net in ? W Q & & The rate of heat rejection is then Btu/h 252,900=×?=?= Btu/h) 5.2544140(100,609 netinout WQQ &&& 9-45 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The temperature at the end of the compression varies with the compression ratio as 1 1 1 2 1 12 ? ? = ? ? ? ? ? ? ? ? = k k rTTT v v since T 1 is fixed. The temperature rise during the combustion remains constant since the amount of heat addition is fixed. Then, the maximum cycle temperature is given by 1 1in2in3 ? +=+= k rTqTqT The smallest gas specific volume during the cycle is r 1 3 v v = When this is combined with the maximum temperature, the maximum pressure is given by () 1 1in 13 3 3 ? +== k rTq RrRT P vv v P 4 1 3 2 q out q in v P 4 1 3 2 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-27 9-46 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis During a polytropic process, constant constant /)1( = = ? nn n TP Pv and for an isentropic process, constant constant /)1( = = ? kk k TP Pv If heat is lost during the expansion of the gas, s TT 44 > where T 4s is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur when kn ? v P 4 1 3 2 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-28 9-47 An ideal Otto cycle is considered. The heat rejection, the net work production, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The mass in this system is kg 004181.0 K) 300)(K/kgmkPa 287.0( )m kPa)(0.004 90( 3 3 1 11 = ?? == RT P m V The two unknown temperatures are () K 4.6537K) 300( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v K 8.642 7 1 K) 1400( 1 14.11 3 1 4 3 34 =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k k r TTT v v Application of the first law to four cycle processes gives kJ 061.1K)3004.653)(KkJ/kg 718.0(kg) 004181.0()( 1221 =??=?= ? TTmcW v kJ 241.2K)4.6531400)(KkJ/kg 718.0(kg) 004181.0()( 2332 =??=?= ? TTmcQ v kJ 273.2K)8.6421400)(KkJ/kg 718.0(kg) 004181.0()( 4343 =??=?= ? TTmcW v kJ 1.029=??=?= ? K)3008.642)(KkJ/kg 718.0(kg) 004181.0()( 1414 TTmcQ v The net work is kJ 1.212=?=?= ?? 061.1273.2 2143net WWW The thermal efficiency is then 0.541=== kJ 2.241 kJ 1.212 in net th Q W ? The minimum volume of the cycle occurs at the end of the compression 3 3 1 2 m 0005714.0 7 m 004.0 === r V V The engine?s mean effective pressure is then kPa 354= ? ? ? ? ? ? ? ? ? ? = ? = kJ 1 mkPa 1 m )0005714.0004.0( kJ 212.1 MEP 3 3 21 net VV W v P 4 1 3 2 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-29 9-48 The power produced by an ideal Otto cycle is given. The rate of heat addition is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m 3 /kg.K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The compression ratio is 667.6 150 1 1 2 1 === v v v v . r and the thermal efficiency is 5318.0 6.667 1 1 1 1 11.41 th =?=?= ??k r ? The rate at which heat must be added to this engine is then kW 126.2= ? ? ? ? ? ? ? ? == hp 1 kW 0.7457 0.5318 hp 90 th net in ? W Q & & v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-30 Diesel Cycle 9-49C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine. 9-50C The Diesel cycle differs from the Otto cycle in the heat addition process only; it takes place at constant volume in the Otto cycle, but at constant pressure in the Diesel cycle. 9-51C The gasoline engine. 9-52C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem. 9-53C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As the cutoff ratio decreases, the efficiency of the diesel cycle increases. 9-54 An expression for cutoff ratio of an ideal diesel cycle is to be developed. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis Employing the isentropic process equations, 1 12 ? = k rTT while the ideal gas law gives 1 1 23 TrrrTT k cc ? == When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is )()( 1 1 1 1 23in TrTrrcTTcq kk cpp ?? ?=?= When this is solved for cutoff ratio, the result is 1 1 in 1 Trrc q r k cp c ? += v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-31 9-55 An ideal diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m 3 /kg?K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by using the process types to fix the temperatures of the states. K 5.921K)(18) 290( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v K 1382=== ? ? ? ? ? ? ? ? = K)(1.5) 5.921( 2 2 3 23 c rTTT v v Combining the first law as applied to the various processes with the process equations gives 6565.0 )15.1(4.1 15.1 18 1 1 )1( 11 1 4.1 11.41 th = ? ? ?= ? ? ?= ?? c k c k rk r r ? According to the definition of the thermal efficiency, kW 227.2= ? ? ? ? ? ? ? ? == hp 1 kW 0.7457 0.6565 hp 200 th net in ? W Q & & v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-32 9-56 A Diesel cycle with non-isentropic compression and expansion processes is considered. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m 3 /kg?K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then K 5.921K)(18) 290( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k s rTTT v v With the isentropic compression efficiency, the actual temperature at the end of the compression is K 7.991 0.90 K )290(921.5 K) 290( 12 12 12 12 = ? += ? +=??? ? ? = ? ? TT TT TT TT ss The maximum temperature is K 1488=== ? ? ? ? ? ? ? ? = K)(1.5) 7.991( 2 2 3 23 c rTTT v v For the isentropic expansion process, K 7.550 18 1.5 K) 1488( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k s r r TTT v v since 4 3 24 23 2 4 2 3 / / v v vv vv v v v v == ? ? ? ? ? ? ? = = r r r r c c The actual temperature at the end of expansion process is then K 6.597K )7.550(1488)95.0(K) 1488()( 4334 43 43 =??=??=??? ? ? = s s TTTT TT TT ?? The net work production is the difference between the work produced by the expansion and that used by the compression, kJ/kg 5.135 K)2907.991)(KkJ/kg 718.0(K)6.5971488)(KkJ/kg 718.0( )()( 1243net = ?????= ???= TTcTTcw vv The heat addition occurs during process 2-3, kJ/kg 8.498K)7.9911488)(KkJ/kg 005.1()( 23in =??=?= TTcq p The thermal efficiency of this cycle is then 2717.0 kJ/kg 498.8 kJ/kg 135.5 in net th === q w ? According to the definition of the thermal efficiency, kW 548.9= ? ? ? ? ? ? ? ? == hp 1 kW 0.7457 0.2717 hp 200 th net in ? W Q & & v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-33 9-57 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The specific volume of the air at the start of the compression is /kgm 8701.0 kPa 95 K) 288)(K/kgmkPa 287.0( 3 3 1 1 1 = ?? == P RT v The total air mass taken by all 8 cylinders when they are charged is kg 008665.0 /kgm 8701.0 m)/4 12.0(m) 10.0( )8( 4/ 3 2 1 2 cyl 1 cyl === ? = ?? vv V SB NNm The rate at which air is processed by the engine is determined from kg/s 1155.0 rev/cycle 2 rev/s) 1600/60kg/cycle)( (0.008665 rev === N nm m & & since there are two revolutions per cycle in a four-stroke engine. The compression ratio is 20 05.0 1 ==r At the end of the compression, the air temperature is () K 6.95420K) 288( 14.11 12 === ??k rTT Application of the first law and work integral to the constant pressure heat addition gives kJ/kg 1325K)6.9542273)(KkJ/kg 005.1()( 23in =??=?= TTcq p while the thermal efficiency is 6867.0 )12.1(4.1 12.1 20 1 1 )1( 11 1 4.1 11.41 th = ? ? ?= ? ? ?= ?? c k c k rk r r ? The power produced by this engine is then kW 105.1= = == kJ/kg) 67)(1325kg/s)(0.68 (0.1155 inthnetnet qmwmW ?&& & v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-34 9-58E An ideal dual cycle has a compression ratio of 20 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are () R 175720R) 530( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v () psia 92820psia) 14( 4.1 1 2 1 12 === ? ? ? ? ? ? ? ? = k k rPPP v v psia 1114==== psia) 928)(2.1( 23 PrPP px R 2109 psia 928 psia 1114 R) 1757( 2 2 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT x x R 2742=== ? ? ? ? ? ? ? ? = R)(1.3) 2109( 3 3 cx x x rTTT v v R 8.918 20 1.3 R) 2742( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k r r TTT v v Applying the first law and work expression to the heat addition processes gives Btu/lbm 212.1= ??+??= ?+?= R)21092742)(RBtu/lbm 240.0(R)17572109)(RBtu/lbm 171.0( )()( 32in xpx TTcTTcq v The heat rejected is Btu/lbm 48.66R)5308.918)(RBtu/lbm 171.0()( 14out =??=?= TTcq v Then, 0.687=?=?= Btu/lbm 212.1 Btu/lbm 66.48 11 in out th q q ? v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-35 9-59E An ideal dual cycle has a compression ratio of 12 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are () R 143212R) 530( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v () psia 9.45312psia) 14( 4.1 1 2 1 12 === ? ? ? ? ? ? ? ? = k k rPPP v v psia 544.7==== psia) 9.453)(2.1( 23 PrPP px R 1718 psia 453.9 psia 544.7 R) 1432( 2 2 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT x x R 2233=== ? ? ? ? ? ? ? ? = R)(1.3) 1718( 3 3 cx x x rTTT v v R 9.917 12 1.3 R) 2233( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k r r TTT v v Applying the first law and work expression to the heat addition processes gives Btu/lbm 172.5= ??+??= ?+?= R)17182233)(RBtu/lbm 240.0(R)14321718)(RBtu/lbm 171.0( )()( 32in xpx TTcTTcq v The heat rejected is Btu/lbm 33.66R)5309.917)(RBtu/lbm 171.0()( 14out =??=?= TTcq v Then, 0.615=?=?= Btu/lbm 172.5 Btu/lbm 66.33 11 in out th q q ? v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-36 9-60E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression. 32.144 Btu/lbm 40.92 R 540 1 1 1 = = ???= r u T v () Btu/lbm 402.05 R 1623.6 93.732.144 2.18 11 2 2 1 2 112 = = ???==== h T r rrr vv v v v Process 2-3: P = constant heat addition. 1.848===???= R 1623.6 R 3000 2 3 2 3 2 22 3 33 T T T P T P v vvv (b) Btu/lbm 388.6305.40268.790 180.1 Btu/lbm 790.68 R 3000 23in 3 3 3 =?=?= = = ???= hhq h T r v Process 3-4: isentropic expansion. () Btu/lbm 91.250621.11180.1 848.1 2.18 848.1848.1 4 2 4 3 4 3334 =???===== u r rrrr vv v v v v v v Process 4-1: v = constant heat rejection. (c) 59.1% Btu/lbm 158.87 =?=?= =?=?= Btu/lbm 388.63 Btu/lbm 158.87 11 04.9291.250 in out th 14out q q uuq ? v P 4 1 2 3 q in q out 3000 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-37 9-61E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 0.240 Btu/lbm.R, c v = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis (a) Process 1-2: isentropic compression. ()() R172418.2R540 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT v v Process 2-3: P = constant heat addition. 1.741===???= R 1724 R 3000 2 3 2 3 2 22 3 33 T T T P T P v vvv (b) ()( )( ) Btu/lbm 306R17243000Btu/lbm.R 0.240 2323in =?=?=?= TTchhq p Process 3-4: isentropic expansion. () R 1173 18.2 1.741 R 3000 741.1 0.4 1 4 2 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? kk v TTT v v v Process 4-1: v = constant heat rejection. (c) () ()() 64.6% Btu/lbm 108 =?=?= =?= ?=?= Btu/lbm 306 Btu/lbm 108 11 R0541173Btu/lbm.R 0.171 in out th 1414out q q TTcuuq ? v v P 4 1 2 3 q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-38 9-62 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression. ()() K 971.120K 293 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. 2.265 K971.1 K2200 2 3 2 3 2 22 3 33 ===???= T T T P T P V VVV Process 3-4: isentropic expansion. () ()( )( ) ()( )( ) 63.5%=== =?=?= =??=?=?= =??=?=?= =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ?? kJ/kg 1235 kJ/kg 784.4 kJ/kg 784.46.4501235 kJ/kg 450.6K293920.6KkJ/kg 0.718 kJ/kg 1235K971.12200KkJ/kg 1.005 K 920.6 20 2.265 K 2200 265.2265.2 in outnet, th outinoutnet, 1414out 2323in 0.41 3 1 4 2 3 1 4 3 34 q w qqw TTcuuq TTchhq r TTTT p k kk ? v V V V V (b) ( )( ) () ()() kPa933 kJ mkPa 1/201/kgm 0.885 kJ/kg 784.4 /11 MEP /kgm 0.885 kPa 95 K 293K/kgmkPa 0.287 3 3 1 outnet, 21 outnet, max 2min max 3 3 1 1 1 = ? ? ? ? ? ? ? ? ? ? = ? = ? = == == ?? == r ww r P RT vvv v vv vv v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-39 9-63 A diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression. ()() K 971.120K 293 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. 2.265 K 971.1 K 2200 2 3 2 3 2 22 3 33 ===???= T T T P T P V VVV Process 3-4: polytropic expansion. () ()( )( ) ()( )( ) kJ/kg 526.3K 2931026KkJ/kg 0.718 kJ/kg 1235K 971.12200KkJ/kg 1.005 K 1026 20 2.265 K 2200 r 2.2652.265 1414out 2323in 0.351n 3 1n 4 2 3 1 4 3 34 =??=?=?= =??=?=?= =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ?? TTcuuq TTchhq TTTT p n v V V V V Note that q out in this case does not represent the entire heat rejected since some heat is also rejected during the polytropic process, which is determined from an energy balance on process 3-4: ()( )( ) () () kJ/kg 120.1 K 22001026KkJ/kg 0.718kJ/kg 963 kJ/kg 963 1.351 K 22001026KkJ/kg 0.287 1 34out34,in34,34out34,in34, systemoutin 34 out34, = ??+= ?+=????=? ?=? = ? ?? = ? ? = TTcwquuwq EEE n TTR w v which means that 120.1 kJ/kg of heat is transferred to the combustion gases during the expansion process. This is unrealistic since the gas is at a much higher temperature than the surroundings, and a hot gas loses heat during polytropic expansion. The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the air table, we would obtain () kJ/kg 1.128)4.18723.781(963 34out34,in34, ?=?+=?+= uuwq which is a heat loss as expected. Then q out becomes kJ/kg 654.43.5261.128 out41,out34,out =+=+= qqq and 47.0%=== =?=?= kJ/kg 1235 kJ/kg 580.6 kJ/kg 580.64.6541235 in outnet, th outinoutnet, q w qqw ? (b) ( )( ) () ()() kPa 691= ? ? ? ? ? ? ? ? ? ? = ? = ? = == == ?? == kJ mkPa 1 1/201/kgm 0.885 kJ/kg 580.6 /11 MEP /kgm 0.885 kPa 95 K 293K/kgmkPa 0.287 3 3 1 outnet, 21 outnet, max 2min max 3 3 1 1 1 r ww r P RT vvv v vv vv v P 4 1 2 3 q in q out Polytropic PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-40 9-64 EES Problem 9-63 is reconsidered. The effect of the compression ratio on the net work output, mean effective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: Procedure QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) q_in_total = 0 q_out_total = 0 IF (q_12 > 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12 If q_23 > 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23 If q_34 > 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34 If q_41 > 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41 END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K] n=1.35 {r_comp = 20} "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant pressure heat addition" P[3]=P[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =P[2]*(V[3] - V[2])"constant pressure process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is polytropic expansion" P[3]/P[4] =(V[4]/V[3])^n s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 - w_34 = DELTAu_34 "q_34 is not 0 for the ploytropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) P[3]*V[3]^n = Const w_34=(P[4]*V[4]-P[3]*V[3])/(1-n) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent" "The mean effective pressure is:" MEP = w_net/(V[1]-V[2]) r comp ? th MEP [kPa] w net [kJ/kg] 14 47.69 970.8 797.9 16 50.14 985 817.4 18 52.16 992.6 829.8 20 53.85 995.4 837.0 22 55.29 994.9 840.6 24 56.54 992 841.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 s [kJ/kg-K] T [K] 9 5 k P a 3 40. 1 k Pa 5 920 k Pa 0 . 0 4 4 0 . 1 0 . 8 8 m 3 / k g Air 2 1 3 4 10 -2 10 -1 10 0 10 1 10 2 10 1 10 2 10 3 10 4 10 1 10 2 10 3 10 4 v [m 3 /kg] P [k Pa] 293 K 1049 K 2200 K 5 . 6 9 6 . 7 4 k J / k g - K Air PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-42 14 16 18 20 22 24 790 800 810 820 830 840 850 r comp w ne t [k J/kg] 14 16 18 20 22 24 47 49 51 53 55 57 r comp ? th 14 16 18 20 22 24 970 975 980 985 990 995 1000 r comp ME P [k Pa] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-43 9-65 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined. Assumptions 1 The cold air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Process 1-2: isentropic compression. ()() K 101917K 328 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. ()( ) K 2241K 10192.22.2 22 2 3 3 2 22 3 33 ====???= TTT T P T P v vvv Process 3-4: isentropic expansion. () ()() ()() ()() ()()kW 46.6=== =?=?= =??×= ?=?= =??×= ?=?= ×= ?? == =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ?? kJ/rev 1.864rev/s 1500/60 kJ/rev 864.1174.1038.3 kJ 174.1K328989.2KkJ/kg 0.718kg10473.2 kJ 3.038K)10192241)(KkJ/kg 1.005)(kg10.4732( kg10473.2 )K 328)(K/kgmkPa 0.287( )m 0.0024)(kPa 97( K 989.2 17 2.2 K 2241 2.22.2 outnet,outnet, outinoutnet, 3 1414out 3 2323in 3 3 3 1 11 4.01 3 1 4 2 3 1 4 3 34 WnW QQW TTmcuumQ TTmchhmQ RT P m r TTTT v p k kk & & V V V V V Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions). v P 4 1 2 3 Q in Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-44 9-66 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined. Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are c p = 1.039 kJ/kg·K, c v = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Process 1-2: isentropic compression. ()() K 101917K 328 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. ()( ) K 2241K 10192.22.2 22 2 3 3 2 22 3 33 ====???= TTT T P T P v vvv Process 3-4: isentropic expansion. () ()() () ()() ()() ()() ()() ()()kW 46.6=== =?=?= =??×= ?=?= =??×= ?=?= ×= ?? == =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ?? kJ/rev 1.863rev/s 1500/60 kJ/rev .8631175.1037.3 kJ 1.175K328989.2KkJ/kg 0.743kg102.391 kJ 3.037K10192241KkJ/kg 1.039kg102.391 kg10391.2 K 328K/kgmkPa 0.2968 m 0.0024kPa 97 K 989.2 17 2.2 K 2241 2.22.2 outnet,outnet, outinoutnet, 3 1414out 3 2323in 3 3 3 1 11 0.41 3 1 4 2 3 1 4 3 34 WnW QQW TTmcuumQ TTmchhmQ RT P m r TTTT p k kk & & v V V V V V Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions). v P 4 1 2 3 Q in Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-45 9-67 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by fixing the temperatures at all states. () K 7.92418K) 291( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v () kPa 514818kPa) 90( 4.1 1 2 1 12 === ? ? ? ? ? ? ? ? = k k rPPP v v kPa 5663kPa) 5148)(1.1( 23 ==== PrPP px K 1017 kPa 5148 kPa 5663 K) 7.924( 2 2 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT x x K 1119K) 1017)(1.1( 3 === xc TrT K 8.365 18 1.1 K) 1119( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k r r TTT v v Applying the first law to each of the processes gives kJ/kg 0.455K)2917.924)(KkJ/kg 718.0()( 1221 =??=?= ? TTcw v kJ/kg 5.102K)10171119)(KkJ/kg 005.1()( 33 =??=?= ? xpx TTcq kJ/kg 26.29K)10171119)(KkJ/kg 718.0(5.102)( 333 =???=??= ?? xxx TTcqw v kJ/kg 8.540K)8.3651119)(KkJ/kg 718.0()( 4343 =??=?= ? TTcw v The net work of the cycle is kJ/kg 1.1150.45526.298.540 21343net =?+=?+= ??? wwww x The mass in the device is given by kg 003233.0 K) 291)(K/kgmkPa 287.0( )m kPa)(0.003 90( 3 3 1 11 = ?? == RT P m V The net power produced by this engine is then kW 24.8=== cycle/s) 0/60kJ/kg)(400 115.1kg/cycle)( 003233.0( netnet nmwW & & v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-46 9-68 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by fixing the temperatures at all states. () K 7.92418K) 291( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k s rTTT v v K 1037 0.85 K )291(924.7 K) 291( 12 12 12 12 = ? += ? +=??? ? ? = ? ? TT TT TT TT ss () kPa 514818kPa) 90( 4.1 1 2 1 12 === ? ? ? ? ? ? ? ? = k k rPPP v v kPa 5663kPa) 5148)(1.1( 23 ==== PrPP px K 1141 kPa 5148 kPa 5663 K) 1037( 2 2 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT x x K 1255K) 1141)(1.1( 3 === xc TrT K 3.410 18 1.1 K) 1255( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k s r r TTT v v K 8.494K )3.410(1255)90.0(K) 1255()( 4334 43 43 =??=??=??? ? ? = s s TTTT TT TT ?? Applying the first law to each of the processes gives kJ/kg 6.535K)2911037)(KkJ/kg 718.0()( 1221 =??=?= ? TTcw v kJ/kg 6.114K)11411255)(KkJ/kg 005.1()( 33 =??=?= ? xpx TTcq kJ/kg 75.32K)11411255)(KkJ/kg 718.0(6.114)( 333 =???=??= ?? xxx TTcqw v kJ/kg 8.545K)8.4941255)(KkJ/kg 718.0()( 4343 =??=?= ? TTcw v The net work of the cycle is kJ/kg 95.426.53575.328.545 21343net =?+=?+= ??? wwww x The mass in the device is given by kg 003233.0 K) 291)(K/kgmkPa 287.0( )m kPa)(0.003 90( 3 3 1 11 = ?? == RT P m V The net power produced by this engine is then kW 9.26=== cycle/s) 0/60kJ/kg)(400 42.95kg/cycle)( 003233.0( netnet nmwW & & v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-47 9-69E An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are () R 158015R) 535( 14.11 1 1 2 1 12 === ? ? ? ? ? ? ? ? = ?? ? k k rTTT v v () psia 2.62915psia) 2.14( 4.1 1 2 1 12 === ? ? ? ? ? ? ? ? = k k rPPP v v psia 1.692psia) 2.629)(1.1( 23 ==== PrPP px R 1738 psia 629.2 psia 692.1 R) 1580( 2 2 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT x x R 2433R)(1.4) 1738( 3 3 === ? ? ? ? ? ? ? ? = cx x x rTTT v v R 2.942 15 1.4 R) 2433( 14.11 3 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?? ? k c k r r TTT v v Applying the first law to each of the processes gives Btu/lbm 7.178R)5351580)(RBtu/lbm 171.0()( 1221 =??=?= ? TTcw v Btu/lbm 02.27R)15801738)(RBtu/lbm 171.0()( 22 =??=?= ? TTcq xx v Btu/lbm 8.166R)17382433)(RBtu/lbm 240.0()( 33 =??=?= ? xpx TTcq Btu/lbm 96.47R)17382433)(RBtu/lbm 171.0(Btu/lbm 8.166)( 333 =???=??= ?? xxx TTcqw v Btu/lbm 9.254R)2.9422433)(RBtu/lbm 171.0()( 4343 =??=?= ? TTcw v The net work of the cycle is Btu/lbm 124.2=?+=?+= ??? 7.17896.479.254 21343net wwww x and the net heat addition is Btu/lbm 193.8=+=+= ?? 8.16602.27 32in xx qqq Hence, the thermal efficiency is 0.641=== Btu/lbm 193.8 Btu/lbm 124.2 in net th q w ? v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-48 9-70 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2) Analysis The thermal efficiency of a dual cycle may be expressed as )()( )( 11 32 14 in out th xpx TTcTTc TTc q q ?+? ? ?=?= v v ? By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result ? ? ? ? ? ? ? ? ?+? ? ?= ? 1)1( 1 1 1 1 th pcp k cp k rrkr rr r ? where 2 P P r x p = and x c r v v 3 = When r c = r p , we obtain ? ? ? ? ? ? ? ? ?+? ? ?= + ? 1)( 1 1 1 2 1 1 th ppp k p k rrrk r r ? For the case r = 20 and r p = 2, 0.660= ? ? ? ? ? ? ? ? ?+? ? ?= + ? 12)22(4.1 12 20 1 1 2 14.1 14.1 th ? v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-49 9-71 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The thermal efficiency of a dual cycle may be expressed as )()( )( 11 32 14 in out th xpx TTcTTc TTc q q ?+? ? ?=?= v v ? By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result ? ? ? ? ? ? ? ? ?+? ? ?= ? 1)1( 1 1 1 1 th pcp k cp k rrkr rr r ? where 2 P P r x p = and x c r v v 3 = When r c = r p , we obtain ? ? ? ? ? ? ? ? ?+? ? ?= + ? 1)( 1 1 1 2 1 1 th ppp k p k rrrk r r ? Rearrangement of this result gives 1 th 2 1 )1( 1)( 1 ? + ?= ?+? ? k ppp k p r rrrk r ? v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-50 9-72 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are c p = 1.110 kJ/kg·K, c v = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) Process 1-2: Isentropic compression ()() ()() kPa 434117kPa 95 K 7.88117K 328 1.349 2 1 12 1-1.349 1 2 1 12 == ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? k k PP TT v v v v The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are 3 3 m 0002813.0 m 0045.0 17 = + =??? + = c c c c dc r V V V V VV 3 1 m 004781.00045.00002813.0 =+=+= dc VVV The total mass contained in the cylinder is ()() kg 0.004825 K 328K/kgmkPa 0.287 )m 781kPa)(0.004 95( 3 3 1 11 = ?? == RT P m V The mass of fuel burned during one cycle is kg 000193.0 kg) 004825.0( 24 =??? ? =??? ? == f f f f f f a m m m m mm m m AF Process 2-3: constant pressure heat addition kJ 039.88)kJ/kg)(0.9 kg)(42,500 000193.0( HVin === cf qmQ ? K 2383=????=????= 3323in K)7.881(kJ/kg.K) kg)(0.823 004825.0(kJ 039.8)( TTTTmcQ v The cutoff ratio is 2.7=== K 881.7 K 2383 2 3 T T ? (b) 3 3 1 2 m 0002813.0 17 m 0.004781 === r V V 23 14 33 23 m 00076.0)m 0002813.0)(70.2( PP = = === VV VV ? 1 Q in 2 3 4 Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-51 Process 3-4: isentropic expansion. () () kPa 2.363 m 0.004781 m 0.00076 kPa 4341 K 1254 m 0.004781 m 0.00076 K 2383 1.349 3 3 4 3 34 1-1.349 3 3 1 4 3 34 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? k k PP TT V V V V Process 4-1: constant voume heat rejection. () ( )( ) kJ 677.3K3281254KkJ/kg 0.823kg) 004825.0( 14out =??=?= TTmcQ v The net work output and the thermal efficiency are kJ 4.361=?=?= 677.3039.8 outinoutnet, QQW 0.543=== kJ 8.039 kJ 4.361 in outnet, th Q W ? (c) The mean effective pressure is determined to be kPa 969.2= ? ? ? ? ? ? ? ? ? ? = ? = kJ mkPa m)0002813.0004781.0( kJ 4.361 MEP 3 3 21 outnet, VV W (d) The power for engine speed of 2000 rpm is kW 72.7=? ? ? ? ? ? == s 60 min 1 rev/cycle) 2( (rev/min) 2000 kJ/cycle) (4.361 2 netnet n WW & & Note that there are two revolutions in one cycle in four-stroke engines. (e) Finally, the specific fuel consumption is g/kWh 159.3=? ? ? ? ? ? ? ? ? ? ? ? ? ? == kWh 1 kJ 3600 kg 1 g 1000 kJ/kg 4.361 kg 000193.0 sfc net W m f PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-52 Stirling and Ericsson Cycles 9-73C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less. 9-74C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less. 9-75C The Stirling cycle. 9-76C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle. 9-77 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis (a) The entropy change during process 3-4 is KkJ/kg 0.5 K 300 kJ/kg 150 0 out34, 34 ??=?=?=? T q ss and () KkJ/kg 0.5 kPa 120 P lnKkJ/kg 0.287 lnln 4 3 4 0 3 4 34 ??=??= ?=? P P R T T css p ? It yields P 4 = 685.2 kPa (b) For reversible cycles, ()kJ/kg 600kJ/kg 150 K 300 K 1200 outin in out ===???= q T T q T T q q L H H L Thus, kJ/kg 450=?=?= 150600 outinoutnet, qqw (c) The thermal efficiency of this totally reversible cycle is determined from 75.0%=?=?= K1200 K300 11 th H L T T ? s T 3 2 q in q out 4 1 1200 K 300 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-53 9-78 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The net work produced per cycle and the thermal efficiency of the cycle are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis Since the specific volume is constant during process 2-3, kPa 7.266 K 300 K 800 kPa) 100( 3 2 32 =? ? ? ? ? ? == T T PP Heat is only added to the system during reversible process 1-2. Then, kJ/kg .6462)KkJ/kg 0.5782)(K 800()( KkJ/kg 5782.0 kPa 2000 kPa 266.7 ln)KkJ/kg 0.287(0 lnln 121in 1 2 0 1 2 12 =?=?= ?= ? ? ? ? ? ? ??= ?=? ssTq P P R T T css ? v The thermal efficiency of this totally reversible cycle is determined from 0.625=?=?= K 800 K 300 11 th H L T T ? Then, kJ 289.1=== kJ/kg) kg)(462.6 (0.625)(1 inthnet mqW ? s T 3 2 q in q out 4 1 800 K 300 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-54 9-79 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The power produced and the rate of heat input are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis Since the specific volume is constant during process 2-3, kPa 7.266 K 300 K 800 kPa) 100( 3 2 32 =? ? ? ? ? ? == T T PP Heat is only added to the system during reversible process 1-2. Then, kJ/kg .6462)KkJ/kg 0.5782)(K 800()( KkJ/kg 5782.0 kPa 2000 kPa 266.7 ln)KkJ/kg 0.287(0 lnln 121in 1 2 0 1 2 12 =?=?= ?= ? ? ? ? ? ? ??= ?=? ssTq P P R T T css ? v The thermal efficiency of this totally reversible cycle is determined from 0.625 K 800 K 300 11 th =?=?= H L T T ? Then, kJ 289.1kJ/kg) kg)(462.6 (0.625)(1 inthnet === mqW ? The rate at which heat is added to this engine is kW 3855=== cycle/s) /60kJ/kg)(500 462.6kg/cycle)( 1( inin nmqQ & & while the power produced by the engine is kW 2409=== cycle/s) 500/60kJ/cycle)( 1.289() netnet nWW & & s T 3 2 q in q out 4 1 800 K 300 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-55 9-80E An ideal Stirling engine with hydrogen as the working fluid operates between the specified temperature limits. The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator per cycle are to be determined. Assumptions Hydrogen is an ideal gas with constant specific heats. Properties The properties of hydrogen at room temperature are R = 5.3224 psia·ft 3 /lbm.R = 0.9851 Btu/lbm·R, c p = 3.43 Btu/lbm·R, c v = 2.44 Btu/lbm·R, and k = 1.404 (Table A-2Ea). Analysis The mass of the air contained in this engine is lbm 006832.0 R) 1100)(R/lbmftpsia 3224.5( )ft psia)(0.1 400( 3 3 1 11 = ?? == RT P m V At the end of the compression, the pressure will be psia 40 ft 1 ft 0.1 psia) 400( 3 3 2 1 12 = ? ? ? ? ? ? ? ? == V V PP The entropy change is RBtu/lbm 268.2 psia 400 psia 40 ln)RBtu/lbm 0.9851(0 lnln 1 2 0 1 2 4312 ?= ? ? ? ? ? ? ? ? ??= ?=?=? P P R T T cssss ? v Since the processes are reversible, Btu 17.0=?=?= )RBtu/lbm 268.2)(R 0011(lbm) 0.006832()( 121in ssmTQ Btu 7.75=?=?= )RBtu/lbm 268.2)(R 005(lbm) 0.006832()( 343out ssmTQ Applying the first law to the process where the gas passes through the regenerator gives Btu 10.0=??=?= R)5001100)(RBtu/lbm 44.2(lbm) 0.006832()( 41regen TTmcQ v s T 3 2 q in q out 4 1 1100R 500 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-56 9-81E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to and rejected by this cycle, and the net work produced by the cycle are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R = 0.06855 Btu/lbm·R, c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 3-4 gives psia 100psia)(10) 10( 4 3 34 === v v PP Since process 4-1 is a constant volume process,, R 3360 psia 100 psia 600 R) 560( 4 1 41 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT According to first law and work integral, Btu/lbm 530.3=?=== ? R)ln(10) 3360)(RBtu/lbm 0.06855(ln 1 2 121in v v RTwq and Btu/lbm 88.4=? ? ? ? ? ? ?=== ? 10 1 R)ln 560)(RBtu/lbm 0.06855(ln 3 4 343out v v RTwq The net work is then Btu/lbm 441.9=?=?= 4.883.530 outinnet qqw 9-82E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3 /lbm.R, c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 1-2 gives psia 60 10 1 psia) 600( 2 1 12 =? ? ? ? ? ? == v v PP Since process 2-3 is a constant-volume process, R 3360 psia 10 psia 60 R) 560( 3 2 32 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = P P TT Application of the first law to process 2-3 gives Btu/lbm 478.8=??=?= R)5603360)(RBtu/lbm 171.0()( 32regen TTcq v s T 3 2 q in q out 4 1 560 R s T 3 2 q in q out 4 1 560 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-57 9-83 An ideal Ericsson cycle operates between the specified temperature limits. The rate of heat addition is to be determined. Analysis The thermal efficiency of this totally reversible cycle is determined from 0.6889 K 900 K 280 11 th =?=?= H L T T ? According to the general definition of the thermal efficiency, the rate of heat addition is kW 726=== 0.6889 kW 500 th net in ? W Q & & 9-84 An ideal Ericsson cycle operates between the specified temperature limits. The power produced by the cycle is to be determined. Analysis The power output is 500 kW when the cycle is repeated 2000 times per minute. Then the work per cycle is kJ/cycle 15 cycle/s (2000/60) kJ/s 500 net net === n W W & & When the cycle is repeated 3000 times per minute, the power output will be kW 750=== kJ/cycle) 5cycle/s)(1 000/603( netnet WnW & & s T 3 2 q in q out 4 1 900 K 280 K s T 3 2 q in q out 4 1 900 K 280 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-58 Ideal and Actual Gas-Turbine (Brayton) Cycles 9-85C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume. 9-86C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection. 9-87C For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum, and then decreases. 9-88C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines. 9-89C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-59 9-90E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Noting that process 1-2 is isentropic, T h P r 1 1 1 12147 =??? = = 520 R 124.27 Btu / lbm . ()( ) Btu/lbm 240.11 147.122147.110 2 2 1 2 12 = = ???=== h T P P P P rr R996.5 (b) Process 3-4 is isentropic, and thus () Btu/lbm 38.88283.26571.504 Btu/lbm 115.8427.12411.240 Btu/lbm 265.834.170.174 10 1 0.174 Btu/lbm 504.71 R 2000 43outT, 12inC, 4 3 4 3 3 34 3 =?=?= =?=?= =???=? ? ? ? ? ? == = = ???= hhw hhw hP P P P P h T rr r Then the back-work ratio becomes 48.5%=== Btu/lbm 238.88 Btu/lbm 115.84 outT, inC, bw w w r (c) 46.5%=== =?=?= =?=?= Btu/lbm 264.60 Btu/lbm 123.04 Btu/lbm 123.0484.11588.238 Btu/lbm 264.6011.24071.504 in outnet, th inC,outT,outnet, 23in q w www hhq ? s T 1 2 4 3 q in q out 2000 520 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-60 9-91 [Also solved by EES on enclosed CD] A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Noting that process 1-2s is isentropic, T h P r 1 1 1 1 5546 =??? = = 310 K 310.24 kJ / kg . ()( ) () () ()( ) kJ/kg 89.167 19.69292.123082.092.1230 K 3.680 and kJ/kg 692.1990.252.207 8 1 2.207 kJ/kg 1230.92 K 1160 kJ/kg 646.7 75.0 24.31058.562 24.310 K 557.25 and kJ/kg 562.5844.125546.18 4334 43 43 44 3 4 3 3 12 12 12 12 22 1 2 34 3 12 = ??= ??=??? ? ? = ==???=? ? ? ? ? ? == = = ???= = ? += ? +=??? ? ? = ==???=== sT s T ssrr r C ss C ssrr hhhh hh hh ThP P P P P h T hh hh hh hh ThP P P P ?? ? ? Thus, T 4 = 770.1 K (b) kJ/kg 105.3=?=?= =?=?= =?=?= 92.4782.584 kJ/kg 478.9224.31016.789 kJ/kg 584.27.64692.1230 outinoutnet, 14out 23in qqw hhq hhq (c) 18.0%=== kJ/kg 584.2 kJ/kg 105.3 in outnet, th q w ? s T 1 2s 4s 3 q in q out 1160 K 310 K 2 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-61 9-92 EES Problem 9-91 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the diagram window method for supplying data to EES is to be developed. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 310 [K] P[1]= 100 [kPa] T[3] = 1160 [K] m_dot = 20 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-62 Bwr ? P ratio W c [kW] W net [kW] W t [kW] Q in [kW] 0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700 0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241 5.0 5.5 6.0 6.5 7.0 7.5 0 500 1000 1500 s [kJ/kg-K] T [K] 100 kPa 800 kPa 1 2 s 2 3 4 4 s Air Standard Brayton Cycle Pressure ratio = 8 and T max = 1160K 2 4 6 8 10 12 14 16 18 20 0.00 0.05 0.10 0.15 0.20 0.25 0 500 1000 1500 2000 2500 P ratio Cycle efficiency, W ne t [kW] ? W net T max =1160 K Note P ratio for maximum work and ? ? c = 0.75 ? t = 0.82 ? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-63 9-93 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the compressor and turbine efficiency relations, () ()() () () () () () () () ()( ) K733.9 4.640116082.01160 K 645.3 75.0 3105.561 310 K 640.4 8 1 K 1160 K 561.58K 310 4334 43 43 43 43 12 12 12 12 12 12 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 = ??= ??=??? ? ? = ? ? = = ? += ? +=??? ? ? = ? ? = =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? sT sp p s T C s p sp s C kk s kk s TTTT TTc TTc hh hh TT TT TTc TTc hh hh P P TT P P TT ?? ? ? (b) ()( )( ) ()( )( ) kJ/kg91.3 0.4263.517 kJ/kg 426.0K310733.9KkJ/kg 1.005 kJ/kg 517.3K645.31160KkJ/kg 1.005 outinoutnet, 1414out 2323in =?=?= =??=?=?= =??=?=?= qqw TTchhq TTchhq p p (c) 17.6%=== kJ/kg 517.3 kJ/kg 91.3 in outnet, th q w ? s T 1 2s 4s 3 q in q out 1160 K 310 K 2 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-64 9-94 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Using the isentropic relations for an ideal gas, K .1706 kPa 100 kPa 2000 K) 300( 0.4/1.4 /)1( 1 2 12 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT Similarly, K 9.424 kPa 2000 kPa 100 K) 1000( 0.4/1.4 /)1( 3 4 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT Applying the first law to the constant-pressure heat addition process 2-3 produces kJ/kg 4.295K)1.7061000)(KkJ/kg 1.005()( 2323in =??=?=?= TTchhq p Similarly, kJ/kg 5.125K)3009.424)(KkJ/kg 1.005()( 1414out =??=?=?= TTchhq p The net work production is then kJ/kg 169.9=?=?= 5.1254.295 outinnet qqw and the thermal efficiency of this cycle is 0.575=== kJ/kg 295.4 kJ/kg 169.9 in net th q w ? s T 1 2 4 3 q in q out 1000 K 300 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-65 9-95 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Using the isentropic relations for an ideal gas, K .1706 kPa 100 kPa 2000 K) 300( 0.4/1.4 /)1( 1 2 12 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT For the expansion process, K 9.424 kPa 2000 kPa 100 K) 1000( 0.4/1.4 /)1( 3 4 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 4.482 )9.4241000)(90.0(1000 )( )( )( 4334 43 43 43 43 = ??= ??=??? ? ? = ? ? = sT sp p s T TTTT TTc TTc hh hh ?? Applying the first law to the constant-pressure heat addition process 2-3 produces kJ 4.295K)1.7061000)(KkJ/kg 1.005)(kg 1()()( 2323in =??=?=?= TTmchhmQ p Similarly, kJ 3.183K)3004.482)(KkJ/kg 1.005)(kg 1()()( 1414out =??=?=?= TTmchhmQ p The net work production is then kJ 112.1=?=?= 3.1834.295 outinnet QQW and the thermal efficiency of this cycle is 0.379=== kJ 295.4 kJ 112.1 in net th Q W ? 4s s T 1 2 4 3 q in q out 1000 K 300 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-66 9-96 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the compression process, K .1706 kPa 100 kPa 2000 K) 300( 0.4/1.4 /)1( 1 2 12 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 6.807 80.0 3001.706 300 )( )( 12 12 12 12 12 12 = ? += ? +=??? ? ? = ? ? = C s p sp s C TT TT TTc TTc hh hh ? ? For the expansion process, K 9.424 kPa 2000 kPa 100 K) 1000( 0.4/1.4 /)1( 3 4 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 4.482 )9.4241000)(90.0(1000 )( )( )( 4334 43 43 43 43 = ??= ??=??? ? ? = ? ? = sT sp p s T TTTT TTc TTc hh hh ?? Applying the first law to the constant-pressure heat addition process 2-3 produces kJ 4.193K)6.8071000)(KkJ/kg 1.005)(kg 1()()( 2323in =??=?=?= TTmchhmQ p Similarly, kJ 3.183K)3004.482)(KkJ/kg 1.005)(kg 1()()( 1414out =??=?=?= TTmchhmQ p The net work production is then kJ 10.1=?=?= 3.1834.193 outinnet QQW and the thermal efficiency of this cycle is 0.0522=== kJ 193.4 kJ 10.1 in net th Q W ? 4s s T 1 2s 4 3 q in q out 1000 K 300 K 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-67 9-97 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the compression process, K .1706 kPa 100 kPa 2000 K) 300( 0.4/1.4 /)1( 1 2 12 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 6.807 80.0 3001.706 300 )( )( 12 12 12 12 12 12 = ? += ? +=??? ? ? = ? ? = C s p sp s C TT TT TTc TTc hh hh ? ? For the expansion process, K 0.428 kPa 1950 kPa 100 K) 1000( 0.4/1.4 /)1( 3 4 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 2.485 )0.4281000)(90.0(1000 )( )( )( 4334 43 43 43 43 = ??= ??=??? ? ? = ? ? = sT sp p s T TTTT TTc TTc hh hh ?? Applying the first law to the constant-pressure heat addition process 2-3 produces kJ 4.193K)6.8071000)(KkJ/kg 1.005)(kg 1()()( 2323in =??=?=?= TTmchhmQ p Similarly, kJ 1.186K)3002.485)(KkJ/kg 1.005)(kg 1()()( 1414out =??=?=?= TTmchhmQ p The net work production is then kJ 7.3=?=?= 1.1864.193 outinnet QQW and the thermal efficiency of this cycle is 0.0377=== kJ 193.4 kJ 7.3 in net th Q W ? 4s s T 1 2s 4 3 q in q out 1000 K 300 K 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-68 9-98 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the isentropic relations, () ()() () () ()( )( ) ()( )( ) kg/s 352=== =?=?= =??=?=?= =??=?=?= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? kJ/kg 199.1 kJ/s 70,000 kJ/kg 99.1175.31184.510 kJ/kg 510.84K491.71000KkJ/kg 1.005 kJ/kg 311.75K300610.2KkJ/kg 1.005 K 491.7 12 1 K 1000 K 610.212K 300 outnet,s, outnet, inC,s,outT,s,outnet,s, 4343outT,s, 1212inC,s, 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 w W m www TTchhw TTchhw P P TT P P TT s sps sps kk s kk s & & (b) The net work output is determined to be ()( ) kg/s 1037=== =?= ?=?= kJ/kg 67.5 kJ/s 70,000 kJ/kg 67.50.85311.7584.51085.0 / outnet,a, outnet, inC,s,outT,s,inC,a,outT,a,outnet,a, w W m wwwww a CT & & ?? s T 1 2s 4s 3 1000 K 300 K 2 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-69 9-99 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas. Analysis (a) Assuming constant specific heats, () ()() () () () () ()( ) kW 15,680=== = ? ? ?= ? ? ?= ? ? ?=?= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? kW 35,0000.448 448.0 3.5251100 2902.607 1111 K 607.2 8 1 K 1100 K 525.38K 290 inthoutnet, 23 14 23 14 in out th 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 QW TT TT TTc TTc q q P P TT P P TT p p kk s kk s && ? ? (b) Assuming variable specific heats (Table A-17), ()( ) () ()( ) kW 15,085=== = ? ? ?= ? ? ?==?= =???=? ? ? ? ? ? == = = ???= =???=== = = ???= kW 35,0000.431 431.0 11.52607.1161 16.29037.651 111 kJ/kg 651.3789.201.167 8 1 1.167 kJ/kg 1161.07 K 1100 kJ/kg 526.128488.92311.18 2311.1 kJ/kg 290.16 K 290 inoutnet, 23 14 in out th 4 3 4 3 3 2 1 2 1 1 34 3 12 1 QW hh hh q q hP P P P P h T hP P P P P h T T rr r rr r && ? ? s T 1 2 4 3 q in q out 1100 K 290 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-70 9-100 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Using the isentropic relations, Th 11 22 =???= =???= 300 K 300.19 kJ / kg 580 K 586.04 kJ / kg () ()()( )kJ/kg 542.0905.831536.0486.0 kJ/kg 285.8519.30004.586 kJ/kg 05.83973.6711.474 7 1 11.474 kJ/kg1536.0404.586950 7 100 700 43outT, 12inC, 4 3 4 323in 1 2 34 3 =?=?= =?=?= =???=? ? ? ? ? ? == =? =+=????= === sT srr r p hhw hhw hP P P P P hhhq P P r ? Thus, 52.7%=== kJ/kg 542.0 kJ/kg 285.85 outT, inC, bw w w r (b) 27.0%=== =?=?= kJ/kg 950 kJ/kg 256.15 kJ/kg 256.15285.85.0542 in outnet, th inC,outT,net.out q w www ? s T 1 2s 4s 3 950 kJ/kg 580 K 300 K 2 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-71 9-101 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air- standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using constant specific heats, () ()( ) () () ()( )( ) ()()()( )( ) kJ/kg 562.2K874.81525.3KkJ/kg 1.00586.0 kJ/kg 281.4K300580KkJ/kg1.005 K 874.8 7 1 K 1525.3 K 1525.3 KkJ/kg 1.005/kJ/kg 950K 580 / 7 100 700 4343outT, 1212inC, 0.4/1.4 /k1k 3 4 34s in232323in 1 2 =??=?=?= =??=?=?= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ?+= +=????=?= === ? spTsT p pp p TTchhw TTchhw P P TT cqTTTTchhq P P r ?? Thus, 50.1%=== kJ/kg 562.2 kJ/kg 281.4 outT, inC, bw w w r (b) 29.6%=== =?=?= kJ/kg 950 kJ/kg 280.8 kJ/kg 280.8281.42.562 in outnet, th inC,outT,outnet, q w www ? s T 1 2s 4s 3 950 kJ/kg 580 K 300 K 2 4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-72 9-102E A simple ideal Brayton cycle with argon as the working fluid operates between the specified temperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are R = 0.2686 psia?ft 3 /lbm·R (Table A-1E), c p = 0.1253 Btu/lbm·R and k = 1.667 (Table A-2Ea). Analysis At the compressor inlet, /lbmft 670.9 psia 15 R) 540)(R/lbmftpsia 2686.0( 3 3 1 1 1 = ?? == P RT v lbm/s 05.62 /lbmft 670.9 ft/s) )(200ft 3( 3 2 1 11 === v VA m& According to the isentropic process expressions for an ideal gas, R 1357 psia 15 psia 150 R) 540( 70.667/1.66/)1( 1 2 12 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT R 7.660 psia 150 psia 15 R) 1660( 70.667/1.66/)1( 3 4 34 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk P P TT Applying the first law to the constant-pressure heat addition process 2-3 gives Btu/s 2356=??=?= R)13571660)(RBtu/lbm 1253.0(lbm/s) 05.62()( 23in TTcmQ p & & The net power output is Btu/s 1417= ?+??= ?+?= R)13575407.6601660)(RBtu/lbm 1253.0(lbm/s) 05.62( )( 2143net TTTTcmW p & & The thermal efficiency of this cycle is then 0.601=== Btu/s 2356 Btu/s 1417 in net th Q W & & ? s T 1 2 4 3 q in q out 1660 R 540 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-73 9-103 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process, K .1527K)(10) 273( 0.4/1.4/)1( 12 === ? kk p rTT The heat addition is kJ/kg 500 kg/s 1 kW 500 in in === m Q q & & Applying the first law to the heat addition process, K 1025 KkJ/kg 1.005 kJ/kg 500 K 1.527 )( in 23 23in = ? +=+= ?= p p c q TT TTcq The temperature at the exit of the turbine is K 9.530 10 1 K) 1025( 1 0.4/1.4 /)1( 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT Applying the first law to the adiabatic turbine and the compressor produce kJ/kg 6.496K)9.5301025)(KkJ/kg 1.005()( 43T =??=?= TTcw p kJ/kg 4.255K)2731.527)(KkJ/kg 1.005()( 12C =??=?= TTcw p The net power produced by the engine is then kW 241.2=?=?= kJ/kg)4.2556kg/s)(496. 1()( CTnet wwmW & & Finally the thermal efficiency is 0.482=== kW 500 kW 241.2 in net th Q W & & ? s T 1 2 4 3 q in q out 273 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-74 9-104 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process, K .8591K)(15) 273( 0.4/1.4/)1( 12 === ? kk p rTT The heat addition is kJ/kg 500 kg/s 1 kW 500 in in === m Q q & & Applying the first law to the heat addition process, K 1089 KkJ/kg 1.005 kJ/kg 500 K 8.591 )( in 23 23in = ? +=+= ?= p p c q TT TTcq The temperature at the exit of the turbine is K 3.502 15 1 K) 1089( 1 0.4/1.4 /)1( 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT Applying the first law to the adiabatic turbine and the compressor produce kJ/kg 6.589K)3.5021089)(KkJ/kg 1.005()( 43T =??=?= TTcw p kJ/kg 4.320K)2738.591)(KkJ/kg 1.005()( 12C =??=?= TTcw p The net power produced by the engine is then kW 269.2=?=?= kJ/kg)4.3206kg/s)(589. 1()( CTnet wwmW & & Finally the thermal efficiency is 0.538=== kW 500 kW 269.2 in net th Q W & & ? s T 1 2 4 3 q in q out 273 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-75 9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Compression KkJ/kg 7159.5 kPa 100 C30 kJ/kg 60.303C30 1 1 1 11 ?= ? ? ? = °= =???°= s P T hT kJ/kg 37.617 kJ/kg.K 7159.5 kPa 1200 2 12 2 = ? ? ? == = s h ss P kJ/kg 24.686 60.303 60.30337.617 82.0 2 212 12 C =??? ? ? =??? ? ? = h hhh hh s ? Process 3-4: Expansion kJ/kg 62.792C500 44 =???°= hT ss hh h hh hh 43 3 43 43 T 62.792 88.0 ? ? =??? ? ? =? We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h 3 = 1404.7 kJ/kg, T 3 = 1034ºC, s 3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3) The mass flow rate is determined from ()() kg/s 875.2 K 27330K/kgmkPa 0.287 )s/m 0kPa)(150/6 100( 3 3 1 11 = +?? == RT P m V & & The net power output is kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, =?=?= hhmW & & kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, =?=?= hhmW & & kW 659=?=?= 11001759 inC,outT,net WWW &&& (b) The back work ratio is 0.625=== kW 1759 kW 1100 outT, inC, bw W W r & & (c) The rate of heat input and the thermal efficiency are kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in =?=?= hhmQ & & 0.319=== kW 2065 kW 659 in net Q W th & & ? 1 Combustion chamber Turbine 2 3 4 Compress. 100 kPa 30°C 500°C 1.2 MPa PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-76 Brayton Cycle with Regeneration 9-106C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber. 9-107C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease. 9-108C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ?, and is defined as ? = q regen, act /q regen, max . 9-109C (b) turbine exit. 9-110C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since in / QW=? and W increases while Q in remains constant. Steam can be obtained by utilizing the hot exhaust gases. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-77 9-111 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas, K 8.530K)(8) 293( 0.4/1.4/)1( 12 === ? kk p rTT K 3.592 8 1 K) 1073( 1 0.4/1.4 /)1( 45 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT When the first law is applied to the heat exchanger, the result is 6523 TTTT ?=? while the regenerator temperature specification gives K 3.582103.59210 53 =?=?=TT The simultaneous solution of these two results gives K 8.540)8.5303.582(3.592)( 2356 =??=??= TTTT Application of the first law to the turbine and compressor gives kJ/kg 244.1 K )2938.530)(KkJ/kg 005.1(K )3.5921073)(KkJ/kg 005.1( )()( 1254net = ?????= ???= TTcTTcw pp Then, kg/s 6145.0 kJ/kg 244.1 kW 150 net net === w W m & & Applying the first law to the combustion chamber produces kW 303.0=??=?= K)3.5821073)(KkJ/kg 005.1(kg/s) 6145.0()( 34in TTcmQ p & & Similarly, kW 153.0=??=?= K)2938.540)(KkJ/kg 005.1(kg/s) 6145.0()( 16out TTcmQ p & & s T 1 2 5 4 q in 1073 K 293 K 3 6 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-78 9-112 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis For the compression and expansion processes we have K 8.530K)(8) 293( 0.4/1.4/)1( 12 === ? kk ps rTT K 3.566 87.0 2938.530 293 )( )( 12 12 12 12 = ? += ? +=??? ? ? = C s p sp C TT TT TTc TTc ? ? K 3.592 8 1 K) 1073( 1 0.4/1.4 /)1( 45 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p s r TT K 9.625 )3.5921073)(93.0(1073 )( )( )( 5445 54 54 = ??= ??=??? ? ? = sT sp p T TTTT TTc TTc ?? When the first law is applied to the heat exchanger, the result is 6523 TTTT ?=? while the regenerator temperature specification gives K 9.615109.62510 53 =?=?=TT The simultaneous solution of these two results gives K 3.576)3.5669.615(9.625)( 2356 =??=??= TTTT Application of the first law to the turbine and compressor gives kJ/kg 7.174 K )2933.566)(KkJ/kg 005.1(K )9.6251073)(KkJ/kg 005.1( )()( 1254net = ?????= ???= TTcTTcw pp Then, kg/s 8586.0 kJ/kg 174.7 kW 150 net net === w W m & & Applying the first law to the combustion chamber produces kW 394.4=??=?= K)9.6151073)(KkJ/kg 005.1(kg/s) 8586.0()( 34in TTcmQ p & & Similarly, kW 244.5=??=?= K)2933.576)(KkJ/kg 005.1(kg/s) 8586.0()( 16out TTcmQ p & & s T 1 2 5s 4 q in 1073 K 293 K 3 6 q out 5 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-79 9-113 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel- flow and counter-flow arrangements of the regenerator are to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas, K 9.510K)(7) 293( 0.4/1.4/)1( 12 === ? kk p rTT K 5.573 7 1 K) 1000( 1 0.4/1.4 /)1( 45 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT When the first law is applied to the heat exchanger as originally arranged, the result is 6523 TTTT ?=? while the regenerator temperature specification gives K 5.56765.5736 53 =?=?=TT The simultaneous solution of these two results gives K 9.5169.5105.5675.573 2356 =+?=+?= TTTT The thermal efficiency of the cycle is then 0.482= ? ? ?= ? ? ?=?= 5.5671000 293516.9 111 34 16 in out th TT TT q q ? For the rearranged version of this cycle, 6 63 ?=TT An energy balance on the heat exchanger gives 6523 TTTT ?=? The solution of these two equations is K 2.545 K 2.539 6 3 = = T T The thermal efficiency of the cycle is then 0.453= ? ? ?= ? ? ?=?= 2.5391000 293545.2 111 34 16 in out th TT TT q q ? s T 1 2 5 4 q in 1000 K 293 K 3 6 q out s T 1 2 5 4 q in 1000 K 293 K 3 6 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-80 9-114E An ideal Brayton cycle with regeneration has a pressure ratio of 8. The thermal efficiency of the cycle is to be determined with and without regenerator cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis According to the isentropic process expressions for an ideal gas, R 8.923R)(8) 510( 0.4/1.4/)1( 12 === ? kk p rTT R 1082 8 1 R) 1960( 1 0.4/1.4 /)1( 45 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT The regenerator is ideal (i.e., the effectiveness is 100%) and thus, R 8.923 R 1082 26 53 == == TT TT The thermal efficiency of the cycle is then 0.529= ? ? ?= ? ? ?=?= 10821960 510923.8 111 34 16 in out th TT TT q q ? The solution without a regenerator is as follows: R 8.923R)(8) 510( 0.4/1.4/)1( 12 === ? kk p rTT R 1082 8 1 R) 1960( 1 0.4/1.4 /)1( 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT 0.448= ? ? ?= ? ? ?=?= 8.9231960 5101082 111 23 14 in out th TT TT q q ? s T 1 2 5 4 q in 1960 R 510 R 3 6 q out s T 1 2 4 3 q in q out 1960 R 510 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-81 9-115 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Analysis The expressions for the isentropic compression and expansion processes are kk p rTT /)1( 12 ? = kk p r TT /)1( 34 1 ? ? ? ? ? ? ? ? ? = For an ideal regenerator, 26 45 TT TT = = The thermal efficiency of the cycle is kk p kk p kk p r T T r r T T TT TT T T TT TT T T TT TT q q /)1( 3 1 /)1( /)1( 3 1 34 12 3 1 35 16 3 1 53 16 in out th 1 1 1 1 )/(1 1)/( 1 )/(1 1)/( 111 ? ?? ? ?= ? ? ?= ? ? ?= ? ? ?= ? ? ?=?=? s T 1 2 4 3 q in 5 6 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-82 9-116E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air. Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows: ()( ) () Btu/lbm 372.253.5712.230 4 1 12.230 Btu/lbm 549.35 R 2160 Btu/lbm 0.192544.5386.14 386.1 Btu/lbm 129.06 R 054 4 3 4 3 3 2 1 2 1 1 34 3 12 1 =???=? ? ? ? ? ? == = = ???= =???=== = = ???= srr r srr r hP P P P P h T hP P P P P h T and Btu/lbm 63.407 2.37235.549 35.549 0.80 Btu/lbm 74.207 06.129 06.1290.192 0.80 4 4 43 43 turb 2 212 12 comp =? ? ? =? ? ? = =? ? ? =? ? ? = h h hh hh h hhh hh s s ? ? Then the thermal efficiency of the gas turbine cycle becomes Btu/lbm 179.9)74.20763.407(9.0)( 24regen =?=?= hhq ? Btu/lbm 63.0)06.12974.207()63.40735.549()()( Btu/lbm 161.7=9.179)74.20735.549()( 1243inC,outT,outnet, regen23in =???=???=?= ??=??= hhhhwww qhhq 0.39 Btu/lbm 161.7 Btu/lbm 63.0 in outnet, th %39==== q w ? Finally, the mass flow rate of air through the turbine becomes lbm/s 1.07= ? ? ? ? ? ? ? ? == hp 1 Btu/s 7068.0 Btu/lbm 63.0 hp 95 net net air w W m & & s T 1 2s 4s 3 q in 2160 R 540 R 5 4 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-83 9-117 [Also solved by EES on enclosed CD] The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are ()( ) () kJ/kg 23.82547.485.712 7.14 1 5.712 kJ/kg 1710.0 K 1561C1288 kJ/kg 3.643765.182765.17.14 2765.1 kJ/kg 293.2 K 293=C20 4 3 4 3 3 2 1 2 1 1 34 3 12 1 =???=? ? ? ? ? ? == = = ???=°= =???=== = = ???°= srr r srr r hP P P P P h T hP P P P P h T The net work output and the heat input per unit mass are C650 kJ/kg 54.9582.29334.665 kJ/kg 34.66566.3720.1038 kJ/kg 0.67210381710 kJ/kg 0.1038 0.359 kJ/kg 372.66 kJ/kg 66.372 h 1 s 3600 kg/h 1,536,000 kW 159,000 41out414out netinout 3223in th net in net net °=?=+=+=??= =?=?= =?=?=??= === =? ? ? ? ? ? == Thqhhhq wqq qhhhhq w q m W w in ? & & Then the compressor and turbine efficiencies become 0.924 0.849 = ? ? = ? ? = = ? ? = ? ? = 2.293672 2.2933.643 23.8251710 54.9581710 12 12 43 43 hh hh hh hh s C s T ? ? When a regenerator is added, the new heat input and the thermal efficiency become 0.496=== ?=?= ?= kJ/kg 751.46 kJ/kg 372.66 kJ/kg 751.46=54.2861038 kJ/kg 286.54=672.0)-.54(0.80)(958=)( newin, net newth, regeninnewin, 24regen q w qqq hhq ? ? Discussion Note an 80% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 49.6%. s T 1 2s 4s 3 q in 1561 K 293 K 5 4 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-84 9-118 EES Problem 9-117 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 20 [C] P[1]= 100 [kPa] {T[4]=589 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.80 Eta_c = 0.892 "Compressor isentorpic efficiency" Eta_t = 0.926 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-85 m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-86 ? t ? c ? th,noreg ? th,withreg Qinnoreg [kW] Q inwithreg [kW] W net [kW] 0.7 0.892 0.2309 0.3405 442063 299766 102.1 0.75 0.892 0.2736 0.3841 442063 314863 120.9 0.8 0.892 0.3163 0.4237 442063 329960 139.8 0.85 0.892 0.359 0.4599 442063 345056 158.7 0.9 0.892 0.4016 0.493 442063 360153 177.6 0.95 0.892 0.4443 0.5234 442063 375250 196.4 1 0.892 0.487 0.5515 442063 390346 215.3 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 -100 100 300 500 700 900 1100 1300 1500 1700 s [kJ/kg-K] T [C] 100 kPa 1470 kPa T-s Diagram for Gas Turbine with Regeneration 1 2 s 2 5 3 4 s 4 6 0.7 0.75 0.8 0.85 0.9 0.95 1 100 120 140 160 180 200 220 ? t W ne t [ k W] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-87 0.7 0.75 0.8 0.85 0.9 0.95 1 275000 310000 345000 380000 415000 450000 ? t Q d o t,in no regeneration with regeneration 0.7 0.75 0.8 0.85 0.9 0.95 1 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 ? t Et a th with regeneration no regeneration PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-88 9-119 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The properties of air at various states are ()( ) () ( )() () () ()( ) kJ/kg 803.14711.801219.250.821219.25 kJ/kg 711.8059.2815.200 7 1 15.200 kJ/kg 1219.25 K 1501 kJ/kg 618.260.75/310.24541.2610.243/ kJ/kg 541.2688.105546.17 5546.1 kJ/kg 310.24 K 310 4334 43 43 4 3 4 3 3 1212 12 12 2 1 2 1 1 34 3 12 1 =??=??=??? ? ? = =???=? ? ? ? ? ? == = = ???= =?+=?+=??? ? ? = =???=== = = ???= sT s T srr r Cs s C srr r hhhh hh hh hP P P P P h T hhhh hh hh hP P P P P h T ?? ?? Thus, T 4 = 782.8 K (b) ()( ) ()( kJ/kg 108.09= ???= ???=?= 24.31026.61814.80325.1219 1243inC,outT,net hhhhwww (c) () ()( ) kJ/kg 738.43 618.26803.140.6526.618 2425 24 25 = ?+= ?+=??? ? ? = hhhh hh hh ?? Then, 22.5%=== =?=?= kJ/kg 480.82 kJ/kg 108.09 kJ/kg 480.82738.4319.2512 in net th 53in q w hhq ? s T 1 2s 4s 3 q in 1150 K 310 K 5 6 4 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-89 9-120 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible. Properties When assuming constant specific heats, the properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17. Analysis (a) Assuming constant specific heats, () ()() () () () () ()( ) kW 39,188=== = ? ? ?= ? ? ?= ? ? ?=?= ====???= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? kW 75,0000.5225 5225.0 2.6071100 2903.525 1111 K 525.3andK 607.2%100 K 607.2 8 1 K 1100 K 525.38K 290 innet 53 16 53 16 in out th 2645 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 QW TT TT TTc TTc q q TTTT P P TT P P TT T p p kk kk && ? ? ? (b) Assuming variable specific heats, ()( ) () ()( ) kW 40,283=== = ? ? ?= ? ? ?=?= ====???= =???=? ? ? ? ? ? == = = ???= =???=== = = ???= kW 75,0000.5371 5371.0 37.65107.1161 16.29012.526 111 kJ/kg 526.12andkJ/kg 651.37%100 kJ/kg 651.3789.201.167 8 1 1.167 kJ/kg 1161.07 K1100 kJ/kg 526.128488.92311.18 2311.1 kJ/kg 90.162 K029 innet 53 16 in out th 2645 4 3 4 3 3 2 1 2 1 1 34 3 12 1 QW hh hh q q hhhh hP P P P P h T hP P P P P h T T rr r rr r && ? ? ? s T 1 2 4 3 75,000 kW 1100 K 290 K 5 q out 6 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-90 9-121 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The properties at various states are () () ()( ) ()()( ) kJ/kg152.5 04.58688.79772.0 kJ/kg 797.88 75.71979.127786.079.1277 kJ/kg 719.7575.290.238 8 1 238.0 kJ/kg 1277.79K 2001 kJ/kg 86.045K 580 kJ/kg 00.193K 300 8100/800/ 24regen 4334 43 43 4 3 4 33 22 11 12 34 3 =?=?= = ??= ??=??? ? ? = =???=? ? ? ? ? ? == = =???= =???= =???= === hhq hhhh hh hh hP P P P P hT hT hT PPr sT s T srr r p ? ?? (b) ()( ) ()( () ( ) 36.0%=== =??=??= =???= ???=?= kJ/kg 539.23 kJ/kg 194.06 kJ/kg 539.23152.52586.041277.79 kJ/kg 194.06300.19586.0488.79779.1277 in net th regen23in 1243inC,outT,net q w qhhq hhhhwww ? s T 1 2s 4s 3 q in 1200 K 300 K 5 6 4 580 K 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-91 9-122 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) Using the isentropic relations and turbine efficiency, () () () () () ()( ) ()()()( )( ) kJ/kg 114.2=??=?=?= = ??= ??=??? ? ? = ? ? = =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = === ? K580737.8KkJ/kg 1.0050.72 K 737.8 5.662120086.01200 K 662.5 8 1 K 1200 8100/800/ 2424regen 4334 43 43 43 43 4.1/4.0 /1 3 4 34 12 TTchhq TTTT TTc TTc hh hh P P TT PPr p sT sp p s T kk s p ?? ?? (b) ()( ) ()()[] () () ( 36.0%=== =???= ??=??= =????= ???=?= kJ/kg 508.9 kJ/kg 183.1 kJ/kg 508.9114.2K5801200KkJ/kg 1.005 kJ/kg 183.1K300580737.81200KkJ/kg 1.005 in net th regen23regen23in 1243inC,outT,net q w qTTcqhhq TTcTTcwww p pp ? s T 1 2s 4s 3 q in 1200 K 300 K 5 6 4 580 K 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-92 9-123 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The properties of air at various states are () () ()( ) ()()( ) kJ/kg148.3 04.58688.79770.0 kJ/kg 797.8875.71979.127786.079.1277 kJ/kg 719.7575.290.238 8 1 238.0 kJ/kg1277.79K2001 kJ/kg86.045K580 kJ/kg00.193K300 8100/800/ 23regen 4334 43 43 4 3 4 33 22 11 12 34 3 =?=?= =??=??=??? ? ? = =???=? ? ? ? ? ? == = =???= =???= =???= === hhq hhhh hh hh hP P P P P hT hT hT PPr sT s T srr r p ? ?? (b) ()( ) ( ) ( ) () ( ) 35.7%=== =??=??= =???=???=?= kJ/kg 5543. kJ/kg 194.06 kJ/kg 543.5148.3586.041277.79 kJ/kg 194.06300.19586.0488.79779.1277 in net th regen23in 1243inC,outT,net q w qhhq hhhhwww ? s T 1 2s 4s 3 q in 1200 K 300 K 5 6 4 580 K 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-93 Brayton Cycle with Intercooling, Reheating, and Regeneration 9-124C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle. 9-125C (a) decrease, (b) decrease, and (c) decrease. 9-126C (a) increase, (b) decrease, and (c) decrease. 9-127C (a) increase, (b) decrease, (c) decrease, and (d) increase. 9-128C (a) increase, (b) decrease, (c) increase, and (d) decrease. 9-129C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output. 9-130C (c) The Carnot (or Ericsson) cycle efficiency. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-94 9-131 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then, ()( ) () ()( ) ()( ) kJ/kg 62.86636.94679.127722 kJ/kg 2.142219.30026.41122 kJ/kg 946.3633.79238 3 1 238 kJ/kg 77.7912 K 2001 kJ/kg 411.26158.4386.13 386.1 kJ/kg 300.19 K 300 65outT, 12inC, 86 5 6 75 5 42 1 2 1 1 56 5 12 1 =?=?= =?=?= ==???=? ? ? ? ? ? == = == ???= ==???=== = = ???= hhw hhw hhP P P P P hh T hhP P P P P h T rr r rr r Thus, 33.5%=== kJ/kg 662.86 kJ/kg 222.14 outT, inC, bw w w r ()()( )( ) 36.8%=== =?=?= =?+?=?+?= kJ/kg 1197.96 kJ/kg 440.72 kJ/kg 440.72222.1486.662 kJ/kg 1197.9636.94679.127726.41179.1277 in net th inC,outT,net 6745in q w www hhhhq ? (b) When a regenerator is used, r bw remains the same. The thermal efficiency in this case becomes ()()( ) 55.3%=== =?=?= =?=?= kJ/kg 796.63 kJ/kg 440.72 kJ/kg 796.6333.40196.1197 kJ/kg 33.40126.41136.94675.0 in net th regenoldin,in 48regen q w qqq hhq ? ? s T 3 4 1 5 q in 1200 K 300 K 86 7 10 9 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-95 9-132 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then, ()( ) () () () () ()( ) ()( ) ()( ) kJ/kg 63.44507.99679.127722 kJ/kg 77.68219.30003.43922 kJ/kg 96.079 36.94679.127785.079.1277 kJ/kg 946.3633.79238 3 1 238 kJ/kg 1277.79K 1200 kJ/kg 9.0343 80.0/19.30026.41119.300 / kJ/kg 411.26158.4386.13 386.1 kJ/kg 300.19K 300 65outT, 12inC, 65586 65 65 86 5 6 755 12142 12 12 42 1 2 11 56 5 12 1 =?=?= =?=?= = ??= ??==??? ? ? = ==???=? ? ? ? ? ? == = ==???= = ?+= ?+==??? ? ? = ==???=== = =???= hhw hhw hhhhh hh hh hhP P P P P hhT hhhhh hh hh hhP P P P P hT sT s T rr r Cs s C ssrr r ?? ?? Thus, 49.3%=== kJ/kg 563.44 kJ/kg 277.68 outT, inC, bw w w r ()()( )( ) 25.5%=== =?=?= =?+?=?+?= kJ/kg 1120.48 kJ/kg 285.76 kJ/kg 5.762868.27744.563 kJ/kg 1120.48996.071277.79439.031277.79 in net th inC,outT,net 6745in q w www hhhhq ? (b) When a regenerator is used, r bw remains the same. The thermal efficiency in this case becomes ()()( ) 40.7%=== =?=?= =?=?= kJ/kg 702.70 kJ/kg 285.76 kJ/kg .7070278.41748.1120 kJ/kg 17.78403.43907.99675.0 in net th regenoldin,in 48regen q w qqq hhq ? ? s T 3 4 1 5 q in 86 7 1 9 2 86s 4 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-96 9-133E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis According to the isentropic process expressions for an ideal gas, R 7.711R)(3) 520( 0.4/1.4/)1( 142 ==== ? kk p rTTT R 1023 3 1 R) 1400( 1 0.4/1.4 /)1( 697 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? == ? kk p r TTT The regenerator is ideal (i.e., the effectiveness is 100%) and thus, R 7.711 R 1023 210 75 == == TT TT The net work output is determined as follows Btu/lbm 94.8802.9296.180 Btu/lbm 96.180R )1023R)(1400Btu/lbm 2(0.24)(2 Btu/lbm 02.92R 0)52R)(711.7Btu/lbm 2(0.24)(2 inC,outT,net 76outT, 12inC, =?=?= =??=?= =??=?= www TTcw TTcw p p The mass flow rate is then lbm/s 7.947= ? ? ? ? ? ? ? ? == hp 1 Btu/s 0.7068 Btu/lbm 94.88 hp 1000 net net w W m & & Applying the first law to the heat addition processes gives Btu/s 1438= ?+??= ?+?= R )102314001023R)(1400Btu/lbm 4lbm/s)(0.2 947.7( )()( 7856in TTcmTTcmQ pp && & Similarly, Btu/s 731= ?+??= ?+?= R )5207.711520R)(711.7Btu/lbm 4lbm/s)(0.2 947.7( )()( 32110out TTcmTTcmQ pp && & s T 3 4 1 6 1400 R 520 R 9 7 8 2 5 1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-97 9-134E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis For the compression and expansion processes, we have R 7.711R)(3) 520( 0.4/1.4/)1( 142 ==== ? kk pss rTTT R 8.737 88.0 5207.711 520 )( )( 12 142 12 12 = ? += ? +==??? ? ? = C s p sp C TT TTT TTc TTc ? ? R 1023 3 1 R) 1400( 1 0.4/1.4 /)1( 697 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? == ? kk p ss r TTT R 1049)10231400)(93.0(1400 )( )( )( 76697 76 76 =??= ??==??? ? ? = sT sp p T TTTTT TTc TTc ?? The regenerator is ideal (i.e., the effectiveness is 100%) and thus, R 8.737 R 1049 210 75 == == TT TT The net work output is determined as follows Btu/lbm 94.6354.10448.168 Btu/lbm 48.168R )1049R)(1400Btu/lbm 2(0.24)(2 Btu/lbm 54.104R 0)52R)(737.8Btu/lbm 2(0.24)(2 inC,outT,net 76outT, 12inC, =?=?= =??=?= =??=?= www TTcw TTcw p p The mass flow rate is then lbm/s 11.05 hp 1 Btu/s 0.7068 Btu/lbm 94.63 hp 1000 net net = ? ? ? ? ? ? ? ? == w W m & & The rate of heat addition is then Btu/s 1862= ?+??= ?+?= R )104914001049R)(1400Btu/lbm 4lbm/s)(0.2 05.11( )()( 7856in TTcmTTcmQ pp && & s T 3 4s 1 6 1400 R 520 R 9 7 8 2 5 1 4 2s 7s 9s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-98 9-135 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows K 9.430K)(4) 290( 0.4/1.4/)1( 142 ==== ? kk p rTTT K 9.450209.43020 45 =+=+=TT K 4.749 KkJ/kg 1.005 kJ/kg 300 K 9.450 )( in 56 56in = ? +=+= ?= p p c q TT TTcq K 3.504 4 1 K) 4.749( 1 0.4/1.4 /)1( 67 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT K 8.802 KkJ/kg 1.005 kJ/kg 300 K 3.504 in 78 = ? +=+= p c q TT K 2.540 4 1 K) 8.802( 1 0.4/1.4 /)1( 89 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT K 2.520202.54020 910 =?=?=TT The heat input is kJ/kg 600300300 in =+=q The heat rejected is kJ/kg 373.0 R )2909.430290K)(520.2kJ/kg (1.005 )()( 32110out = ?+??= ?+?= TTcTTcq pp The thermal efficiency of the cycle is then 0.378=?=?= 600 0.373 11 in out th q q ? s T 3 4 1 6 290 K 9 7 8 2 5 1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-99 9-136 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows K 9.430K)(4) 290( 0.4/1.4/)1( 1642 ===== ? kk p rTTTT K 9.450209.43020 67 =+=+=TT K 4.749 KkJ/kg 1.005 kJ/kg 300 K 9.450 )( in 78 78in = ? +=+= ?= p p c q TT TTcq K 3.504 4 1 K) 4.749( 1 0.4/1.4 /)1( 89 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT K 8.802 KkJ/kg 1.005 kJ/kg 300 K 3.504 in 910 = ? +=+= p c q TT K 2.540 4 1 K) 8.802( 1 0.4/1.4 /)1( 1011 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT K 7.838 KkJ/kg 1.005 kJ/kg 300 K 2.540 in 1112 = ? +=+= p c q TT K 4.564 4 1 K) 7.838( 1 0.4/1.4 /)1( 1213 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT K 4.544204.56420 1314 =?=?=TT The heat input is kJ/kg 900300300300 in =++=q The heat rejected is kJ/kg .9538 R )2909.4302909.430290K)(544.4kJ/kg (1.005 )()()( 5432114out = ?+?+??= ?+?+?= TTcTTcTTcq ppp The thermal efficiency of the cycle is then 0.401=?=?= 900 9.538 11 in out th q q ? s T 3 4 1 290 K 9 8 2 5 1 11 6 7 12 13 14 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-100 9-137 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis Since all compressors share the same compression ratio and begin at the same temperature, K 9.430K)(4) 290( 0.4/1.4/)1( 1642 ===== ? kk p rTTTT From the problem statement, 40 137 ?=TT The relations for heat input and expansion processes are p p c q TTTTcq in 7878in )( +=????= kk p r TT /)1( 89 1 ? ? ? ? ? ? ? ? ? = p c q TT in 910 += , kk p r TT /)1( 1011 1 ? ? ? ? ? ? ? ? ? = p c q TT in 1112 += , kk p r TT /)1( 1213 1 ? ? ? ? ? ? ? ? ? = The simultaneous solution of above equations using EES software gives the following results K 7.596 K, 7.886 K, 2.588 K, 0.874 K 5.575 K, 2.855 K, 7.556 13121110 987 ==== === TTTT TTT From am energy balance on the regenerator, K 9.470409.43040)40( 6141413613 141367 =+=+=????=?? ?=? TTTTTT TTTT The heat input is kJ/kg 900300300300 in =++=q The heat rejected is kJ/kg 0.465 R )2909.4302909.430290K)(470.9kJ/kg (1.005 )()()( 5432114out = ?+?+??= ?+?+?= TTcTTcTTcq ppp The thermal efficiency of the cycle is then 0.483=?=?= 900 0.465 11 in out th q q ? s T 3 4 1 290 K 9 8 2 5 1 11 6 7 12 13 14 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-101 Jet-Propulsion Cycles 9-138C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity. 9-139C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio. 9-140C It reduces the exit velocity, and thus the thrust. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-102 9-141E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia?ft 3 /lbm?R (Table A-1E), c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields R 8.868R)(10) 450( 0.4/1.4/)1( 12 === ? kk p rTT R 1.725 10 1 R) 1400( 1 0.4/1.4 /)1( 35 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives R 2.981)4508.868(1400)( 1234 =??=??= TTTT The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, 2 )( 2 inlet 2 exit 54 VV mTTcm ppe ? =? && which when solved for the velocity at which the air leaves the propeller gives ft/s 9.716 )ft/s 600( Btu/lbm 1 /sft 25,037 R)1.7252.981)(RBtu/lbm 24.0( 20 1 2 )(2 2/1 2 22 2/1 2 inlet54exit = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? +?= VTTc m m V p p e & & The mass flow rate through the propeller is lbm/s 2261 /lbmft 84.20 ft/s 600 4 ft) 10( 4 /lbmft 84.20 psia 8 R) 450()ftpsia 3704.0( 3 2 1 1 2 1 1 3 3 1 ==== = ? == ?? vv v VDAV m P RT p & The thrust force generated by this propeller is then lbf 8215= ? ? ? ? ? ? ? ? ? ?=?= 2 inletexit ft/slbm 32.174 lbf 1 600)ft/s.9lbm/s)(716 (2261)( VVmF p & s T 1 2 4 3 q in 5 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-103 9-142E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia?ft 3 /lbm?R (Table A-1E), c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields R 8.868R)(10) 450( 0.4/1.4/)1( 12 === ? kk p rTT R 1.725 10 1 R) 1400( 1 0.4/1.4 /)1( 35 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives R 2.981)4508.868(1400)( 1234 =??=??= TTTT The mass flow rate through the propeller is lbm/s 1447 /lbmft 84.20 ft/s 600 4 ft) 8( 4 /lbmft 84.20 psia 8 R) 450()ftpsia 3704.0( 3 2 1 1 2 1 1 3 3 1 ==== = ? == ?? vv v VDAV m P RT p & According to the previous problem, lbm/s 1.113 20 lbm/s 2261 20 === p e m m & & The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, 2 )( 2 inlet 2 exit 54 VV mTTcm ppe ? =? && which when solved for the velocity at which the air leaves the propeller gives ft/s 0.775 )ft/s 600( Btu/lbm 1 /sft 25,037 R)1.7252.981)(RBtu/lbm 24.0( lbm/s 1447 lbm/s 1.113 2 )(2 2/1 2 22 2/1 2 inlet54exit = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? +?= VTTc m m V p p e & & The thrust force generated by this propeller is then lbf 7870= ? ? ? ? ? ? ? ? ? ?=?= 2 inletexit ft/slbm 32.174 lbf 1 600)ft/slbm/s)(775 (1447)( VVmF p & s T 1 2 4 3 q in 5 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-104 9-143 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa?m 3 /kg?K, c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis The total mass flow rate is kg/s 1.676 /kgm 452.1 m/s 200 4 m) 5.2( 4 /kgm 452.1 kPa 50 K) 253()mkPa 287.0( 3 2 1 1 2 1 1 3 3 1 ==== = ? == ?? vv v VDAV m P RT & Now, kg/s 51.84 8 kg/s 1.676 8 === m m e & & The mass flow rate through the fan is kg/s 6.59151.841.676 =?=?= ef mmm &&& In order to produce the specified thrust force, the velocity at the fan exit will be m/s 284.5 N 1 m/skg 1 kg/s 591.6 N 50,000 m/s) (200 )( 2 inletexit inletexit = ? ? ? ? ? ? ? ? ? +=+= ?= f f m F VV VVmF & & An energy balance on the stream passing through the fan gives K 232.6= ? ? ? ? ? ? ? ? ?= ? ?= ? =? 22 22 2 inlet 2 exit 45 2 inlet 2 exit 54 /sm 1000 kJ/kg 1 )KkJ/kg 005.1(2 )m/s 200()m/s 5.284( K 253 2 2 )( p p c VV TT VV TTc s T 1 2 4 3 q in 5 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-105 9-144 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa?m 3 /kg?K, c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis Working across the two isentropic processes of the cycle yields K 1.527K)(10) 273( 0.4/1.4/)1( 12 === ? kk p rTT K 5.374 10 1 K) 723( 1 0.4/1.4 /)1( 35 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives K 9.468)2731.527(723)( 1234 =??=??= TTTT The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, 2 )( 2 inlet 2 exit 54 VV TTc p ? =? which when solved for the velocity at which the air leaves the propeller gives [] m/s 528.9= ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ??= +?= 2/1 2 22 2/1 2 inlet54exit )m/s 300( kJ/kg 1 /sm 1000 K)5.3749.468)(KkJ/kg 005.1(2 )(2 VTTcV p The mass flow rate through the engine is kg/s 7.721 /kgm 306.1 m/s 300 4 m) 2( 4 /kgm 306.1 kPa 60 K) 273()mkPa 287.0( 3 2 1 1 2 1 1 3 3 1 ==== = ? == ?? vv v VDAV m P RT & The thrust force generated is then N 165,200= ? ? ? ? ? ? ? ? ? ?=?= 2 inletexit m/skg 1 N 1 00)m/s39kg/s)(528. (721.7)( VVmF & s T 1 2 4 3 q in 5 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-106 9-145 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ? 0). Diffuser: () () ()( ) () () kPa 62.6 K 241 K 291.9 kPa 32 K 291.9 /sm 1000 kJ/kg 1 KkJ/kg 1.0052 m/s 320 K 241 2 2/0 2 02/2/ 1.4/0.41/ 1 2 12 22 22 1 12 2 112 2 1 0 2 2 12 2 22 2 11 outin (steady) 0 systemoutin = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? ? +=+= ??= ? +?=???+=+ =????=? ?kk p p T T PP c V TT VTTc VV hhVhVh EEEEE ? ? &&&&& Compressor: ( )()()( ) () ()() K 593.712K 291.9 kPa 751.2kPa 62.612 0.4/1.4 /1 2 3 23 243 == ? ? ? ? ? ? ? ? = ==== ? kk p P P TT PrPP Turbine: ( ) ( ) 54235423outturb,incomp, TTcTTchhhhww pp ?=?????=????= or K1098.2291.9593.71400 2345 =+?=+?= TTTT Nozzle: () () ()2/0 2 0 2/2/ K 568.2 kPa 751.2 kPa 32 K 1400 2 656 0 2 5 2 6 56 2 66 2 55 outin (steady) 0 systemoutin 0.4/1.4/1 4 6 46 VTTc VV hh VhVh EEEEE P P TT p kk +?=??? ? +?= +=+ =????=? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? &&&&& or, s T 1 2 4 3 Q i 5 6 · PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-107 ()( )( ) m/s1032 kJ/kg 1 /sm 1000 K568.21098.2KkJ/kg 1.0052 22 6 = ? ? ? ? ? ? ? ? ??=V (b) ()()()() kW 13,670= ? ? ? ? ? ? ? ? ?=?= 22 aircraftinletexit /sm 1000 kJ/kg 1 m/s 320m/s3201032kg/s 60VVVmW p & & (c) ()()( )( )( ) kg/s 1.14=== =??=?=?= kJ/kg 42,700 kJ/s 48,620 HV kJ/s 48,620K593.71400KkJ/kg 1.005kg/s 60 in fuel 3434in Q m TTcmhhmQ p & & && & PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-108 9-146 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ? 0). Diffuser: () () ()( ) () () kPa 62.6 K 241 K 291.9 kPa 32 K 291.9 /sm 1000 kJ/kg 1 KkJ/kg 1.0052 m/s 320 K 241 2 2/0 2 0 2/2/ 1.4/0.41/ 1 2 12 22 22 1 12 2 112 2 1 0 2 2 12 2 22 2 11 outin (steady) 0 systemoutin = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? ? +=+= ??= ? +?= +=+ = ?=? ?kk p p T T PP c V TT VTTc VV hh VhVh EE EEE ? ? && &&& Compressor: ( )()()( ) () ()() K 593.712K 291.9 kPa 751.2kPa 62.612 0.4/1.4 /1 2 3 23 243 == ? ? ? ? ? ? ? ? = ==== ? kk s p P P TT PrPP () () () ( )() K 669.20.80/291.9593.71.929/ 2323 23 23 23 23 =?+=?+= ? ? = ? ? = Cs p sp s C TTTT TTc TTc hh hh ? ? Turbine: or, ( ) ( ) K 1022.7291.9669.21400 2345 54235423outturb,incomp, =+?=+?= ?=?????=????= TTTT TTcTTchhhhww pp () () () ( ) () () kPa 197.7 K 1400 K 956.1 kPa 751.2 K 956.185.0/1022.714001400/ 1.4/0.41/ 4 5 45 5445 54 54 54 54 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = =??=??= ? ? = ? ? = ?kk s Ts sp p s T T T PP TTTT TTc TTc hh hh ? ? s T 1 2 4 3 Q in 5s 6 · 5 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-109 Nozzle: () () ()2/0 2 0 2/2/ K 607.8 kPa 197.7 kPa 32 K 1022.7 2 656 0 2 5 2 6 56 2 66 2 55 outin (steady) 0 systemoutin 0.4/1.4/1 5 6 56 VTTc VV hh VhVh EE EEE P P TT p kk +?= ? +?= +=+ = ?=? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? && &&& or, ()( )( ) m/s 913.2= ? ? ? ? ? ? ? ? ??= kJ/kg 1 /sm 1000 K607.81022.7KkJ/kg 1.0052 22 6 V (b) () ()( )() kW 11,390= ? ? ? ? ? ? ? ? ?= ?= 22 aircraftinletexit /sm 1000 kJ/kg 1 m/s 320m/s320913.2kg/s 60 VVVmW p & & (c) ()()( )( )( ) kg/s 1.03=== =??=?=?= kJ/kg 42,700 kJ/s 44,067 HV kJ/s 44,067K669.21400KkJ/kg 1.005kg/s 60 in fuel 3434in Q m TTcmhhmQ p & & && & PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-110 9-147 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A17. Analysis (a) Using variable specific heats for air, Compressor: 386.1 kJ/kg 300.19K 300 1 11 = =???= r P hT ()( ) ()( ) 27.396 kJ/kg 1464.65854610.65 kJ/kg 854 kg/s 10 kJ/s 8540 kJ/s 8540kJ/kg 42,700kg/s 0.2HV kJ/kg 610.6563.16386.112 3 12 2323in in in fuelin 2 1 2 =??? =+=+=????= === ==×= =???=== r in rr P qhhhhq m Q q mQ hP P P P & & & & Turbine: or, kJ/kg 741.17300.19610.65464.651 1234 4312outturb,incomp, =+?=+?= ?=????= hhhh hhhhww Nozzle: () 2 0 2/2/ kJ/kg 41.79702.33 12 1 27.396 0 2 4 2 5 45 2 55 2 44 outin (steady) 0 systemoutin 5 3 5 35 ? ? VV hh VhVh EE EEE h P P PP rr ? +?= +=+ = ?=? =???=? ? ? ? ? ? = ? ? ? ? ? ? ? ? = && &&& or, ()()( ) m/s 908.9 kJ/kg 1 /sm 1000 kJ/kg741.171154.1922 22 545 = ? ? ? ? ? ? ? ? ?=?= hhV Brake force = Thrust = ()()() N 9089= ? ? ? ? ? ? ? ? ? ?=? 2 inletexit m/skg 1 N 1 m/s0908.9kg/s 10VVm& s T 1 3 2 q in 4 5 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-111 9-148 EES Problem 9-147 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-112 "Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2]) Brake Force [N] m [kg/s] T 3 [K] T 1 [C] 9971 11.86 1164 -20 9764 11.41 1206 -10 9568 10.99 1247 0 9383 10.6 1289 10 9207 10.24 1330 20 9040 9.9 1371 30 4.5 5.0 5.5 6.0 6.5 7.0 7.5 200 400 600 800 1000 1200 1400 1600 s [kJ/kg-K] T [ K ] 9 5 k P a 1 14 0 k P a Air 1 2s 3 4s 5s -20 -10 0 10 20 30 9000 9200 9400 9600 9800 10000 T 1 [C] B r ake F o r ce [ N ] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-113 9-149 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield Th 11 22 280 28013 700 713 27 =???= =???= KkJ/kg. . () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? ? +?= +=++ = ?=? 22 22 2 2 1 2 2 12in 2 22 2 11in outin (steady) 0 systemoutin /sm 1000 kJ/kg 1 2 m/s 300 13.28027.713kg/s 16kJ/s 000,15 2 )2/()2/( V VV hhmQ VhmVhmQ EE EEE & & && & && &&& ? It gives V 2 = 1048 m/s Thus, ()()( ) N 11,968=?=?= m/s3001048kg/s 16 12 VVmF p & 15,000 kJ/s 1 7°C 300 m/s 16 kg/s 2 427°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-114 Second-Law Analysis of Gas Power Cycles 9-150E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-60 and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-60E, q out = 158.9 Btu/lbm, T 1 = 540 R, T 4 = 1420.6 R, and v 4 = v 1 . At T avg = (T 4 + T 1 )/2 = (1420.6 + 540)/2 = 980.3 R, we have c v,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is () RBtu/lbm 0.1741 R 1420.6 R 540 lnRBtu/lbm 0.180lnln 0 4 1 4 1 41 ??=?=+=? ? v v R T T css v Thus, () Btu/lbm 64.9= ? ? ? ? ? ? ? ? +??= ? ? ? ? ? ? ? ? +?= R 540 Btu/lbm 158.9 RBtu/lbm 0.1741R540 41, 41041 destroyed, R R T q ssTx Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from ()()( ) 040040044 vv ?+???= PssTuu? where RBtu/lbm 0.1741 0 RBtu/lbm 9.581 1404 1404 out1404 ?=?=? =?=? ?==?=? ssss quuuu vvvv Thus, ()()( ) Btu/lbm 64.9=+??= 0RBtu/lbm 0.1741R540Btu/lbm 158.9 4 ? Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-115 9-151 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-91 is to be determined. Analysis From Prob. 9-91, q in = 584.62 kJ/kg, q out = 478.92 kJ/kg, and KkJ/kg2.67602kJ/kg789.16 KkJ/kg3.13916K1160 KkJ/kg2.47256kJ/kg6.364 KkJ/kg.734981K310 44 33 22 11 ?=???= ?=???= ?=???= ?=???= o o o o sh sT sh sT Thus, () () ( )() () () () ( )() () kJ/kg 220 kJ/kg 41.4 kJ/kg 93.4 kJ/kg 43.6 = ? ? ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +??= ? ? ? ? ? ? ? ? +?== =???= = ? ? ? ? ? ? ? ? ??=?== = ? ? ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ? +??= ? ? ? ? ? ? ? ? +?== =???= = ? ? ? ? ? ? ? ? ??=?== K 310 kJ/kg 478.92 2.676021.73498K 310 ln 1/8lnKkJ/kg 0.2873.139162.67602K 310 ln K 1600 kJ/kg 584.62 2.472563.13916K 310 ln 8lnKkJ/kg 0.2871.734982.47256K 310 ln out 0 4 1 410 41, 41041gen,041 destroyed, 3 4 34034034gen,034 destroyed, in 0 2 3 230 23, 23023gen,023 destroyed, 1 2 12012012gen,012 destroyed, LR R HR R T q P P RssT T q ssTsTx P P RssTssTsTx T q P P RssT T q ssTsTx P P RssTssTsTx ? ? oo oo oo oo PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-116 9-152E The exergy loss of an ideal dual cycle described in Prob. 9-58E is to be determined. Analysis From Prob. 9-58E, q out = 66.48 Btu/lbm, T 1 = 530 R, T 2 = 1757 R, T x = 2109 R, T 3 = 2742 R, and T 4 = 918.8 R. Also, Btu/lbm 9.151R)21092742)(RBtu/lbm 240.0()( Btu/lbm 19.60R)17572109)(RBtu/lbm 171.0()( 33,in 22,in =??=?= =??=?= ? ? xx xx TTcq TTcq v v kJ/kg 3.146K)2918.494)(KkJ/kg 718.0()( 14out =??=?= TTcq v The exergy destruction during a process of the cycle is ? ? ? ? ? ? ? ? +??== sink out source in 0gen0dest T q T q sTsTx Application of this equation for each process of the cycle gives 0 2-dest,1 =x (isentropic process) RBtu/lbm 03123.00 R 1757 R 2109 ln)RBtu/lbm 171.0( lnln 22 2 ?=+?= +=? v v v xx x R T T css Btu/lbm 4.917 R 2742 Btu/lbm 60.19 RBtu/lbm 0.03123R) 530( source -2 in, 20-2 dest, =? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= T q ssTx x xx RBtu/lbm 06299.00 R 2109 R 2742 ln)RBtu/lbm 240.0(lnln 33 3 ?=??=?=? xx px P P R T T css Btu/lbm 4.024 R 2742 Btu/lbm 151.9 RBtu/lbm 0.06299R) 530( source 3- in, 303- dest, =? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= T q ssTx x xx 0 4-dest,3 =x (isentropic process) RBtu/lbm 09408.00 R 8.918 R 530 ln)RBtu/lbm 171.0(lnln 4 1 4 1 41 ??=+?=+=? v v v R T T css Btu/lbm 16.62 R 530 Btu/lbm 66.48 RBtu/lbm 0.09408R) 530( sink out 4101-dest,4 =? ? ? ? ? ? +??= ? ? ? ? ? ? ? ? +?= T q ssTx The largest exergy destruction in the cycle occurs during the heat-rejection process s. The total exergy destruction in the cycle is Btu/lbm 25.6=++= 62.16024.4917.4 totaldest, x v P 4 1 2 3 q out x q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-117 9-153E The entropy generated by the Brayton cycle of Prob. 9-102E is to be determined. Analysis From Prob. 9-102E, R 540 R 1660 Btu/s 93914172356 Btu/s 2356 netinout in = = =?=?= = L H T T WQQ Q &&& & No entropy is generated by the working fluid since it always returns to its original state. Then, RBtu/s 0.320 ?=?=?= R 1660 Btu/s 2356 R 540 Btu/s 939 inout gen HL T Q T Q S && & 9-154 The exergy loss of each process for a regenerative Brayton cycle described in Prob. 9-112 is to be determined. Analysis From Prob. 9-112, T 1 = 293 K, T 2 = 566.3 K, T 3 = 1073 K, T 4 = 625.9 K, T 5 = 615.9 K, T 6 = 576.3 K, and r p = 8. Also, kJ/kg 7.284K)2933.576)(KkJ/kg 005.1()( kJ/kg 4.659K)9.6151073)(KkJ/kg 005.1()( 16out 53in =??=?= =??=?= TTcq TTcq p p The exergy destruction during a process of a stream from an inlet state to exit state is given by ? ? ? ? ? ? ? ? +??== sink out source in 0gen0dest T q T q ssTsTx ie Application of this equation for each process of the cycle gives kJ/kg 19.2= ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?= )8ln()287.0( 293 566.3 (1.005)ln)293(lnln 1 2 1 2 02-1 dest, P P R T T cTx p kJ/kg 38.0= ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= 1073 4.459 0 615.9 1073 (1.005)ln)293(lnln 5 3 5 3 03-5 dest, source in p T q P P R T T cTx kJ/kg 16.1= ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?= 8 1 ln)287.0( 1073 625.9 (1.005)lnK) 293(lnln 3 4 3 4 04-3 dest, P P R T T cTx p kJ/kg 85.5= ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +?= 293 7.284 0 576.3 293 (1.005)ln)293(lnln sin6 1 6 1 01-6 dest, k out p T q P P R T T cTx kJ/kg 0.41= ? ? ? ? ? ? += ? ? ? ? ? ? ? ? +=?+?= ?? 625.9 576.3 (1.005)ln 566.3 615.9 (1.005)ln)293( lnln( 4 6 2 5 064520regendest, T T c T T cTssTx pp s T 1 2 4s 3 q in 1073 K 293 K 5 6 q out 4 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-118 9-155 The total exergy destruction associated with the Brayton cycle described in Prob. 9-119 and the exergy at the exhaust gases at the turbine exit are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-119, q in = 480.82, q out = 372.73 kJ/kg, and KkJ/kg 08156.2kJ/kg 738.43 KkJ/kg 69407.2kJ/kg 14.803 KkJ/kg 12900.3K 1150 KkJ/kg 42763.2kJ/kg 26.618 KkJ/kg 1.73498K 310 55 44 33 22 11 ?=???= ?=???= ?=???= ?=???= ?=???= o o o o o sh sh sT sh sT and, from an energy balance on the heat exchanger, KkJ/kg 2.52861 kJ/kg 97.682)26.61843.738(14.803 6 66425 ?=??? =??=????=? o s hhhhh Thus, () () ( )() () () ( )() ()()[]()()[] ()( ) () () kJ/kg 126.7 kJ/kg 78.66 kJ/kg 4.67 kJ/kg 38.30 kJ/kg 41.59 = ? ? ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +??= ? ? ? ? ? ? ? ? +?== = ? ? ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ???= ? ? ? ? ? ? ? ? ??== =?+?= ?+?=?+?== =???= ? ? ? ? ? ? ? ? ??=?== =???= ? ? ? ? ? ? ? ? ??=?== K 310 kJ/kg 372.73 2.528611.73498K 310 ln K 1800 kJ/kg 480.82 2.608153.12900K 310 ln 2.694072.528612.427632.60815K 310 1/7lnKkJ/kg0.2873.129002.69407K 310 ln 7lnKkJ/kg 0.2871.734982.42763K 310 ln out 0 6 1 610 61, 61061gen,061 destroyed, 0 5 3 530 53, 53053gen,053 destroyed, 4625046250regengen,0regen destroyed, 3 4 34034034gen,034 destroyed, 1 2 12012012gen,012 destroyed, LR R H in R R T q P P RssT T q ssTsTx T q P P RssT T q ssTsTx ssssTssssTsTx P P RssTssTsTx P P RssTssTsTx ? ? oo oo oooo oo oo Noting that h 0 = h @ 310 K = 310.24 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from ()() 0 6 0 2 6 060066 2 ? ? gz V ssThh ++???=? where KkJ/kg 0.793631.734982.52861ln 0 1 6 161606 ?=?=??=?=? ? P P Rssssss oo Thus, ()( ) kJ/kg 126.7=???= KkJ/kg 0.79363K 310310.2497.682 6 ? s T 1 2s 4s 3 q in 1150 K 310 K 5 6 4 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-119 9-156 EES Prob. 9-155 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Given" T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 epsilon=0.65 T_H=1800 [K] T0=310 [K] P0=100 [kPa] "Analysis for Problem 9-156" q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical" "Analysis for Problem 9-119" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-120 "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in Ratio_P x total [kJ/kg] x6 [kJ/kg] 6 279.8 120.7 7 289.9 126.7 8 299.8 132.5 9 309.5 138 10 318.8 143.4 11 327.9 148.6 12 336.7 153.7 13 345.2 158.6 14 353.4 163.4 6 7 8 9 10 11 12 13 14 270 280 290 300 310 320 330 340 350 360 120 125 130 135 140 145 150 155 160 165 170 175 180 Ratio P x to t a l [ kJ/ kg ] x6 [ k J/kg] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-121 9-157E The exergy loss of each process for a reheat-regenerative Brayton cycle with intercooling described in Prob. 9-134E is to be determined. Analysis From Prob. 9-134E, T 1 = T 3 = 520 R, T 2 = T 4 = T 10 = 737.8 R, T 5 = T 7 = T 9 = 1049 K, T 6 = T 8 = 1400 R, and r p = 3. Also, Btu/lbm 27.52R)5208.737)(RBtu/lbm 240.0( )( Btu/lbm 24.84R)10491400)(RBtu/lbm 240.0( )( 1103-out,21-out,10 568-in,76-in,5 =??= ?== =??= ?== TTcqq TTcqq p p The exergy destruction during a process of a stream from an inlet state to exit state is given by ? ? ? ? ? ? ? ? +??== sink out source in 0gen0dest T q T q ssTsTx ie Application of this equation for each process of the cycle gives Btu/lbm 4.50)3ln()06855.0( 520 737.8 (0.24)ln)520(lnln 1 2 1 2 04-3 dest,2-1 dest, = ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?== P P R T T cTxx p Btu/lbm .734 1400 24.84 0 1049 1400 (0.24)ln)520(lnln source 6-in,5 5 6 5 6 08-7 dest,6-5 dest, = ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??== T q P P R T T cTxx p Btu/lbm 3.14 3 1 ln)06855.0( 1400 1049 (0.24)ln)520(lnln 6 7 6 7 09-8 dest,7-6 dest, = ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?== P P R T T cTxx p Btu/lbm 8.61 520 27.52 0 737.8 520 (0.24)ln)520(lnln sink out 10 1 10 1 03-2 dest,1-10 dest, = ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +?== T q P P R T T cTxx p Btu/lbm 0 1049 737.8 (0.24)ln 737.8 1049 (0.24)ln)520( lnln( 9 10 4 5 0109540regendest, = ? ? ? ? ? ? += ? ? ? ? ? ? ? ? +=?+?= ?? T T c T T cTssTx pp The greatest exergy destruction occurs during the heat rejection processes. s T 3 4s 1 6 1400 R 520 R 9 7 8 2 5 10 4 2s 7s 9s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-122 9-158 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-137 is to be determined. Analysis From Prob. 9-137, r p = 4, q in,7-8 = q in,9-10 = q in,11-12 = 300 kJ/kg, q out,14-1 = 181.8 kJ/kg, q out,2-3 = q out,4-5 = 141.6 kJ/kg, T 1 = T 3 = T 5 = 290 K , T 2 = T 4 = T 6 = 430.9 K K 9.470 K, 7.596 K, 7.886 K, 2.588 K, 0.874 K 5.575 K, 2.855 K, 7.556 1413 121110 987 == === === TT TTT TTT The exergy destruction during a process of a stream from an inlet state to exit state is given by ? ? ? ? ? ? ? ? +??== sink out source in 0gen0dest T q T q ssTsTx ie Application of this equation for each process of the cycle gives 0kJ/kg 0.03 ?= ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?=== )4ln()287.0( 290 430.9 (1.005)ln)290( lnln 1 2 1 2 06-5 dest,4-3 dest,2-1 dest, P P R T T cTxxx p kJ/kg 27.0= ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= 7.886 300 0 556.7 855.2 (1.005)ln)290(lnln source 8-in,7 7 8 7 8 08-7 dest, T q P P R T T cTx p kJ/kg 23.7= ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= 7.886 300 0 575.5 874.0 (1.005)ln)290(lnln source 10-in,9 7 8 9 10 010-9 dest, T q P P R T T cTx p kJ/kg 21.5= ? ? ? ? ? ? ??= ? ? ? ? ? ? ? ? ??= 7.886 300 0 588.2 886.7 (1.005)ln)290(lnln source 12-in,11 11 12 11 12 012-11 dest, T q P P R T T cTx p 0kJ/kg 0.06 ??= ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?= 4 1 ln)287.0( 855.2 575.5 (1.005)ln)290(lnln 8 9 8 9 09-8 dest, P P R T T cTx p 0kJ/kg 0.42 ?= ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?= 4 1 ln)287.0( 874.0 588.2 (1.005)ln)290(lnln 10 11 10 11 011-10 dest, P P R T T cTx p 0kJ/kg 0.05 ??= ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?= 4 1 ln)287.0( 886.7 596.7 (1.005)ln)290(lnln 12 13 12 13 013-12 dest, P P R T T cTx p kJ/kg 40.5= ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +?= 290 8.181 0 470.9 290 (1.005)ln)290(lnln sink 1-out,14 14 1 14 1 01-14 dest, T q P P R T T cTx p kJ/kg 26.2= ? ? ? ? ? ? +?= ? ? ? ? ? ? ? ? +?== 290 6.141 0 430.9 290 (1.005)ln)290(lnln sink 3-out,2 2 3 2 3 05-4 dest,3-2 dest, T q P P R T T cTxx p kJ/kg 5.65= ? ? ? ? ? ? += ? ? ? ? ? ? ? ? +=?+?= ?? 596.7 470.9 (1.005)ln 430.9 556.7 (1.005)ln)290( lnln)( 13 14 6 7 01413760regendest, T T c T T cTssTx pp s T 3 4 1 9 8 2 5 10 11 6 7 12 13 14 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-123 9-159 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 500ºC = 773 K are c p = 1.093 kJ/kg·K, c v = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A- 2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case () K 6.505 kPa 100 kPa 700 K 303 1)/1.357-(1.357 /)1( 1 2 12 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT 0.881= ? ? = ? ? = K)303533( K)3036.505( 12 12 TT TT s C ? (b) The total mass flowing through the turbine and the rate of heat input are kg/s 81.12kg/s 21.0kg/s 6.12 60 kg/s 6.12 kg/s 6.12 AF =+=+=+=+= a afat m mmmm & &&&& kW 85557)kJ/kg)(0.9 00kg/s)(42,0 21.0( HVin === cf qmQ ?& & The temperature at the exit of combustion chamber is K 1144)K533kJ/kg.K)( 3kg/s)(1.09 81.12(kJ/s 8555)( 3323in =????=????= TTTTcmQ p & & The temperature at the turbine exit is determined using isentropic efficiency relation () K 7.685 kPa 700 kPa 100 K 1144 1)/1.357-(1.357 /)1( 3 4 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk s P P TT K 4.754 K)7.6851144( K)1144( 85.0 4 4 43 43 =??? ? ? =??? ? ? = T T TT TT s T ? The net power and the back work ratio are kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12()( 12inC, =?=?= TTcmW pa & & kW 5455)K4.754144kJ/kg.K)(1 3kg/s)(1.09 81.12()( 43outT, =?=?= TTcmW p & & kW 2287=?=?= 31685455 inC,outT,net WWW &&& 0.581=== kW 5455 kW 3168 outT, inC, bw W W r & & (c) The thermal efficiency is 0.267=== kW 8555 kW 2287 in net th Q W & & ? The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is, 735.0 K 1144 K 303 11 3 1 max =?=?= T T ? and 0.364=== 735.0 267.0 max th II ? ? ? 1 Combustion chamber Turbine 2 3 4 Compress. 100 kPa 30°C Diesel fuel 700 kPa 260°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-124 9-160 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are c p = 1.110 kJ/kg·K, c v = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are xc c c c dc r VVV V V V VV ===??? + =??? + = 2 3 3 m 0002154.0 m 0028.0 14 4 3 1 m 003015.00028.00002154.0 VVVV ==+=+= dc Process 1-2: Isentropic compression ()() ()() kPa 334114kPa 95 K 9.82314K 328 1.349 2 1 12 1-1.349 1 2 1 12 == ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? k k PP TT v v v v Process 2-x and x-3: Constant-volume and constant pressure heat addition processes: K 2220 kPa 3341 kPa 9000 K) 9.823( 2 2 === P P TT x x kJ/kg 1149K)9.8232220(kJ/kg.K) (0.823)( 2-2 =?=?= TTcq xx v K 3254=????=????== ? 3333-2 K)2220(kJ/kg.K) (0.823kJ/kg 1149)( TTTTcqq xpxx (b) kJ/kg 229811491149 3-2in =+=+= ? xx qqq 333 3 m 0003158.0 K 2220 K 3254 )m 0002154.0( === x x T T VV Process 3-4: isentropic expansion. () () kPa 9.428 m 0.003015 m 0.0003158 kPa 9000 K 1481 m 0.003015 m 0.0003158 K 3254 1.349 3 3 4 3 34 1-1.349 3 3 1 4 3 34 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? k k PP TT V V V V Process 4-1: constant voume heat rejection. ()( )( ) kJ/kg 7.948K3281481KkJ/kg 0.823 14out =??=?= TTcq v The net work output and the thermal efficiency are kJ/kg 1349=?=?= 7.9482298 outinoutnet, qqw 1 Q in 2 3 4 P V Q out x PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-125 0.587=== kJ/kg 2298 kJ/kg 1349 in outnet, th q w ? (c) The mean effective pressure is determined to be ()() kg 0.003043 K 328K/kgmkPa 0.287 )m 015kPa)(0.003 95( 3 3 1 11 = ?? == RT P m V kPa 1466= ? ? ? ? ? ? ? ? ? ? = ? = kJ mkPa m)0002154.0003015.0( kJ/kg) kg)(1349 (0.003043 MEP 3 3 21 outnet, VV mw (d) The power for engine speed of 3500 rpm is kW 120=? ? ? ? ? ? == s 60 min 1 rev/cycle) 2( (rev/min) 3500 kJ/kg) kg)(1349 003043.0( 2 netnet n mwW & & Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa. 908.0 K 3254 K )27325( 11 3 0 max = + ?=?= T T ? and 0.646=== 908.0 587.0 max th II ? ? ? The rate of exergy of the exhaust gases is determined as follows () ()( ) kJ/kg 6.567 100 9.428 ln287.0 298 1481 kJ/kg.Kln 110.1()298(29814810.823 lnln)( 0 4 0 4 004040044 = ? ? ? ? ? ? ???= ? ? ? ? ? ? ???=???= P P R T T cTTTcssTuux pv kW 50.4=? ? ? ? ? ? == s 60 min 1 rev/cycle) 2( (rev/min) 3500 kJ/kg) kg)(567.6 003043.0( 2 44 n mxX & & PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-126 Review Problems 9-161 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have (a) cycle)mech kJ/cyl 8.16(= hp 1 Btu/min 42.41 rev/min) (1200cylinders) (16 hp 3500 cycles) mechanical of (No.cylinders) of (No. producedpower Total mechanical ??= ? ? ? ? ? ? ? ? = = cycle mechBtu/cyl 7.73 w (b) cycle) thermkJ/cyl 16.31(= hp 1 Btu/min 42.41 rev/min) (1200/2cylinders) (16 hp 3500 cycles) amic thermodynof (No.cylinders) of (No. producedpower Total micthermodyna ??= ? ? ? ? ? ? ? ? = = cycle thermBtu/cyl 15.46 w 9-162 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as () () () () ()kk p kk kk p kk r T P P TT rT P P TT /1 3 /1 3 4 34 /1 1 /1 1 2 12 1 ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? = Setting T 2 = T 4 and solving for r p gives () 16.7= ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? 1.4/0.812/ 1 3 K 300 K 1500 kk p T T r Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7. s T 1 2 4 3 q in q out T 3 T 1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-127 9-163 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) We treat air as an ideal gas with variable specific heats, () () kJ/kg 2377.7 kJ/kg 1775.3K 2100 K 300 kPa 100 kPa 700 kJ/kg 523.90702.9386.1 kPa 100 kPa 700 386.1 kJ/kg 214.07K 300 3 33 1 1 3 3max 1 11 3 33 2 1 2 11 12 1 = =???= = ? ? ? ? ? ? ? ? ===???= =???= ? ? ? ? ? ? ? ? == = =???= h uT T P P TT T P T P hP P P P P uT rr r K2100 vv (c) kJ/kg 1853.8 kJ/kg 1561.23 11 kJ/kg1561.2307.2143.1775 kJ/kg 1853.89.5237.2377 in out th 13out 23in 15.8%=?=?= =?=?= =?=?= q q uuq hhq ? v P 32 1 q in q out s T 3 2 1 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-128 9-164 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (b) We treat air as an ideal gas with constant specific heats. Process 1-2 is isentropic: ( ) () () K2100 K 300 kPa 100 kPa 700 K 523.1 kPa 100 kPa 700 K 300 1 1 3 3max 1 11 3 33 0.4/1.4/1 1 2 12 = ? ? ? ? ? ? ? ? ===???= = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? T P P TT T P T P P P TT kk vv (c) () ()() () ()() 18.5%=?=?= =??= ?=?= =??= ?=?= kJ/kg 1584.8 kJ/kg 1292.4 11 kJ/kg 1292.4K3002100KkJ/kg 0.718 kJ/kg 1584.8K523.12100KkJ/kg 1.005 in out th 1313out 2323in q q TTcuuq TTchhq v p ? v P 32 1 q in q out s T 3 2 1 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-129 9-165 [Also solved by EES on enclosed CD] A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression. () kJ/kg 475.11 51.841.676 8 11 1.676 kJ/kg 206.91K 029 2 1 2 11 112 1 =??? ==== = =???= u r uT rrr r vv v v v v Process 2-3: v = constant heat addition. ()() () ()()()kJ0.715 kJ/kg475.111487.2kg 107.065 kg 107.065 K 290K/kgmkPa 0.287 m 0.0006kPa 98 994.3 kJ/kg 1487.2K 1800 4 23in 4 3 3 1 11 33 3 =?×=?= ×= ?? == = =???= ? ? uumQ RT P m uT r V v (b) Process 3-4: isentropic expansion. ()( ) kJ/kg 693.2395.31994.38 4 3 4 334 =???==== ur rrr vv v v v Process 4-1: v = constant heat rejection. ()( )( ) 51.9%=== =?=?= =?×=?= kJ 0.715 kJ 0.371 kJ 0.371344.0715.0 kJ/kg206.91693.23kg 107.065 in net th outinnet -4 14out Q W QQW uumQ ? kJ0.344 (c) rpm 4852= ? ? ? ? ? ? ? ? × == min 1 s 60 kJ/cycle) 0.371(4 kJ/s 60 rev/cycle) 2(2 cylnet,cyl net Wn W n & & Note that for four-stroke cycles, there are two revolutions per cycle. v P 4 1 3 2 Q in Q out 1800 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-130 9-166 EES Problem 9-165 is reconsidered. The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated. Also, the T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=(17+273) [K] P[1]=98 [kPa] T[3]=1800 [K] V_cyl=0.6 [L]*Convert(L, m^3) r_v=8 "Compression ratio" W_dot_net = 60 [kW] N_cyl=4 "number of cyclinders" v[1]/v[2]=r_v "The first part of the solution is done per unit mass." "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat transfer (s=const.) with work input" w_in = DELTAu_12 DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zero for v=const, heat is added" q_in = DELTAu_23 DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is zero for v=const; heat is rejected" - q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent" "The mass contained in each cylinder is found from the volume of the cylinder:" V_cyl=m*v[1] "The net work done per cycle is:" W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-131 ? th [%] r v w net [kJ/kg] 42.81 5 467.1 46.39 6 492.5 49.26 7 509.8 51.63 8 521.7 53.63 9 529.8 55.35 10 535.2 56.85 11 538.5 10 -2 10 -1 10 0 10 1 10 2 50 100 1000 8000 v [m 3 /kg] P [ k P a ] 290 K 1800 K Air Otto Cycle P-v Diagram s = const 1 2 3 4 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 0 500 1000 1500 2000 2500 s [kJ/kg-K] T [ K ] 98 kPa 4866 kPa Air Otto Cycle T-s Diagram 1 2 3 4 v = const PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-132 5 6 7 8 9 10 11 460 470 480 490 500 510 520 530 540 r v w ne t [ kJ/ kg ] 5 6 7 8 9 10 11 42 44 46 48 50 52 54 56 58 r v ? th [ % ] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-133 9-167 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2a). Analysis From the definition of the thermal efficiency of a Carnot heat engine, K 576= ? + = ? =????= 0.501 K 273)(15 1 1 Carnotth, Carnotth, ? ? L H H L T T T T An isentropic process for an ideal gas is one in which Pv k remains constant. Then, the pressure ratio is 5.65=? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? )1667.1/(667.1 )1/( 1 2 1 2 K 288 K 576 kk T T P P Based on the process equation, the compression ratio is 2.83== ? ? ? ? ? ? ? ? = 667.1/1 /1 1 2 2 1 )65.5( k P P v v 9-168E An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy-source reservoir are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2Ea). Analysis From the definition of the thermal efficiency of a Carnot heat engine, R 1300= ? + = ? =????= 0.601 R )460(60 1 1 Carnotth, Carnotth, ? ? L H H L T T T T An isentropic process for an ideal gas is one in which Pv k remains constant. Then, the pressure ratio is 9.88=? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? )1667.1/(667.1 )1/( 1 2 1 2 R 520 R 1300 kk T T P P Based on the process equation, the compression ratio is 3.95== ? ? ? ? ? ? ? ? = 667.1/1 /1 1 2 2 1 )88.9( k P P v v s T 3 2 q in q out 4 1 T H 288 K s T 3 2 q in q out 4 1 T H 520 R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-134 9-169 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. 4 The combustion efficiency is 100 percent. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis The heat input to the cycle for 0.043 grams of fuel consumption is kJ 806.1kJ/kg) kg)(42,000 10043.0( 3 HVfuelin =×== ? qmQ The thermal efficiency is then 5537.0 kJ 1.806 kJ 1 in net th === Q W ? From the definition of thermal efficiency, we obtain the required compression ratio to be 7.52= ? = ? =????= ??? )14.1/(1)1/(1 th 1 th )5537.01( 1 )1( 11 1 kk r r ? ? 9-170 An equation is to be developed for )/( 1 1in ?k rTcq v in terms of k, r c and r p . Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The temperatures at various points of the dual cycle are given by 1 12 ? = k rTT 1 12 2 2 ? == ? ? ? ? ? ? ? ? = k pp x x rTrrT P P TT 1 1 3 3 ? == ? ? ? ? ? ? ? ? = k cpcx x x rTrrrTTT v v Application of the first law to the two heat addition processes gives )()( )()( 1 1 1 1 1 1 1 1 32in ???? ?+?= ?+?= k p k cpp kk p xpx rTrrTrrcrTrTrc TTcTTcq v v or upon rearrangement )1()1( 1 1 in ?+?= ? cpp k rkrr rTc q v v P 4 1 2 3 q out x q in v P 4 1 3 2 q out q in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-135 9-171 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression. () () () kPa 2129kPa 98 K 300 K 708.3 9.2 kJ/kg 9.518 K 3.70852.672.621 2.9 11 2.621 kJ/kg 214.07K 300 1 1 2 2 1 2 1 11 2 22 2 2 1 2 11 112 1 = ? ? ? ? ? ? ? ? ==???= = =???==== = =???= P T T P T P T P u T r uT rrr r v vvv vv v v v v Process 2-3: v = constant heat addition. ()( ) kJ/kg609.8 9.5187.1128 593.8 kJ/kg 1128.7K 1416.63.70822 23 322 2 3 3 2 22 3 33 3 =?=?= = =???====???= uuq uTT P P T T P T P in r v vv (b) Process 3-4: isentropic expansion. ()( ) kJ/kg 487.7506.79593.82.9 4 3 4 334 =???==== ur rrr vv v v v Process 4-1: v = constant heat rejection. kJ/kg336.1 7.2738.609 kJ/kg 273.707.21475.487 outinnet 14out =?=?= =?=?= qqw uuq (c) 55.1%=== kJ/kg 609.8 kJ/kg 336.1 in net th q w ? (d) ( )( ) () ()() kPa429 kJ 1 mkPa 1 1/9.21/kgm 0.879 kJ/kg 336.1 /11 MEP /kgm 0.879 kPa 98 K 300K/kgmkPa 0.287 3 3 1 net 21 net max 2min 3 3 1 1 1max = ? ? ? ? ? ? ? ? ? ? = ? = ? = == = ?? === r ww r P RT vvv v vv vv v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-136 9-172 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2 is isentropic compression: ()() () () kPa 2190kPa 98 K 300 K 728.8 9.2 K 728.89.2K 300 1 1 2 2 1 2 1 11 2 22 0.4 1 2 1 12 = ? ? ? ? ? ? ? ? ==???= == ? ? ? ? ? ? ? ? = ? P T T P T P T P TT k v vvv v v Process 2-3: v = constant heat addition. ()( ) ()( )( ) kJ/kg523.3 K728.81457.6KkJ/kg 0.718 K 457.618.72822 2323 22 2 3 3 2 22 3 33 =??=?=?= ====???= TTcuuq TT P P T T P T P in v vv (b) Process 3-4: isentropic expansion. () K 600.0 9.2 1 K 1457.6 0.4 1 4 3 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ?k TT v v Process 4-1: v = constant heat rejection. ()( )( ) kJ/kg307.9 4.2153.523 kJ/kg 215.4K300600KkJ/kg 0.718 outinnet 1414out =?=?= =??=?=?= qqw TTcuuq v (c) 58.8%=== kJ/kg 523.3 kJ/kg 307.9 in net th q w ? (d) ( )( ) () ()() kPa393 kJ 1 mkPa 1 1/9.21/kgm 0.879 kJ/kg 307.9 /11 MEP /kgm 0.879 kPa 98 K 300K/kgmkPa 0.287 3 3 1 net 21 net max 2min 3 3 1 1 1max = ? ? ? ? ? ? ? ? ? ? = ? = ? = == = ?? === r ww r P RT vvv v vv vv v P 4 1 3 2 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-137 9-173 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) The compression and the cutoff ratios are 2 cm 75 cm 150 16 cm 75 cm 1200 3 3 2 3 3 3 2 1 ====== V V V V c rr Process 1-2: isentropic compression. () kJ/kg 863.03 K 837.3256.421.676 16 11 1.676 kJ/kg 206.91K 029 2 2 1 2 11 112 1 = =???==== = =???= h T r uT rrr r vv v v v v Process 2-3: P = constant heat addition. ()( ) 002.5 kJ/kg 1848.9 K 1674.63.83722 3 3 22 2 3 3 2 22 3 33 = =??? ====???= r h TTT T P T P v v vvv Process 3-4: isentropic expansion. () kJ/kg 636.00 K 853.4016.40002.5 2 16 22 4 4 2 4 3 4 3334 = =???=? ? ? ? ? ? ==== u T r rrrr vv v v v v v v Process 4-1: v = constant heat rejection. () kPa 294.3= ? ? ? ? ? ? ? ? ==???= kPa 100 K 290 K 853.4 1 1 4 4 1 11 4 44 P T T P T P T P vv (b) ()( ) () kg 101.442 K 290K/kgmkPa 0.287 m 0.0012kPa 100 3 3 3 1 11 ? ×= ?? == RT P m V ()( )( ) ()()() kJ 0.803=?=?= =?×=?= =?×=?= 619.0422.1 kJ 0.619kJ/kg206.91636.00kg101.442 kJ 1.422863.081848.9kg101.442 outinnet 3- 14out -3 23in QQW uumQ hhmQ (c) () ()() kPa 714= ? ? ? ? ? ? ? ? ? ? = ? = ? = kJ 1 mkPa 1 1/161m0.0012 kJ 0.803 /11 MEP 3 3 1 net 21 net r WW VVV v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-138 9-174 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are c p = 0.5203 kJ/kg.K, c v = 0.3122 kJ/kg·K, R = 0.2081 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are 2 cm 75 cm 150 16 cm 75 cm 1200 3 3 2 3 3 3 2 1 ====== V V V V c rr Process 1-2: isentropic compression. ()() K 184316K 290 0.667 1 1 2 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. ()( ) K 3686184322 22 2 3 3 2 22 3 33 ====???= TTT T P T P v vvv Process 3-4: isentropic expansion. () K 920.9 16 2 K 3686 22 0.6671 3 1 4 2 3 1 4 3 34 =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ?? k kk r TTTT V V V V Process 4-1: v = constant heat rejection. () kPa 317.6= ? ? ? ? ? ? ? ? ==???= kPa 100 K 290 K 920.9 1 1 4 4 1 11 4 44 P T T P T P T P vv (b) ()( ) () kg101.988 K 290K/kgmkPa 0.2081 m 0.0012kPa 100 3 3 3 1 11 ? ×= ?? == RT P m V ()()( )( )( ) ()()()()() kJ 1.514=?=?= =??×=?=?= =??×=?=?= 392.0906.1 kJ 0.392K290920.9KkJ/kg 0.3122kg101.988 kJ 1.906K18433686KkJ/kg 0.5203kg101.988 outinnet 3- 1414out -3 2323in QQW TTmcuumQ TTmchhmQ p v (c) () ()() kPa 1346= ? ? ? ? ? ? ? ? ? ? = ? = ? = kJ 1 mkPa 1 1/161m 0.0012 kJ 1.514 /11 MEP 3 3 1 net 21 net r WW VVV v P 4 1 2 3 Q in Q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-139 9-175E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 0.240 Btu/lbm.R, c v = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis The mass of air is ()( ) () lbm103.132 R 550R/lbmftpsia 0.3704 ft 75/1728psia 14.7 3 3 3 1 11 ? ×= ?? == RT P m V Process 1-2: isentropic compression. ()() R 148612R 550 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-x: v = constant heat addition, ()() ()() R 2046R1486RBtu/lbm 0.171lbm103.132Btu 0.3 x 3 22in,2 =?????×= ?=?= ? ? x xxx TT TTmcuumQ v Process x-3: P = constant heat addition. ()() ()() 1.715 R 2046 R 3509 R 3509R2046RBtu/lbm 0.240lbm103.132Btu 1.1 33 3 33 33 3 33in,3 ====???= =?????×= ?=?= ? ? xx c x xx xpxx T T r T P T P TT TTmchhmQ V VVV Process 3-4: isentropic expansion. () R 1611 12 1.715 R 3509 715.1715.1 0.41 3 1 4 1 3 1 4 3 34 =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ?? k kk r TTTT V V V V Process 4-1: v = constant heat rejection. ()() ()() 59.4%=?=?= =??×= ?=?= ? Btu 1.4 Btu 0.568 11 Btu 0.568R5501611RBtu/lbm 0.171lbm 103.132 in out th 3 1414out Q Q TTmcuumQ ? v v P 4 1 2 3 1.1 Btu Q out x 0.3 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-140 9-176 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) The entropy change during process 1-2 is KkJ/kg 0.5 K 1800 kJ/kg 900 12 12 ?===? H T q ss and () ()() kPa 5873= ? ? ? ? ? ? ? ? ===???= =????=????+=? K 350 K 1800 5.710kPa 200 710.5lnKkJ/kg 0.287KkJ/kg 0.5lnln 3 1 1 2 3 3 1 1 3 31 1 11 3 33 1 2 1 2 1 2 0 1 2 12 T T P T T PP T P T P R T T css v v v vvv v v v v v v v ? (b) ()kJ/kg 725= ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?== kJ/kg 900 K 1800 K 350 11 ininthnet q T T qw H L ? s T 3 2 q in = 900 kJ/kg q out 4 1 1800 K 350 K PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-141 9-177 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis The properties at various states are Th P Th P r r 11 33 1 3 1386 330 9 =???= = =???= = 300 K 300.19 kJ / kg 1300 K 1395.97 kJ / kg . . For r p = 6, ()( ) () %9.37 kJ/kg 894.57 kJ/kg 339.46 kJ/kg 339.4611.55557.894 kJ/kg 555.1119.3003.855 kJ/kg 894.5740.50197.1395 kJ/kg 855.315.559.330 6 1 kJ/kg 501.40316.8386.16 in net th outinnet 14out 23in 4 3 4 2 1 2 34 12 === =?=?= =?=?= =?=?= =???=? ? ? ? ? ? == =???=== q w qqw hhq hhq hP P P P hP P P P rr rr ? For r p = 12, ()( ) () %5.48 kJ/kg 785.37 kJ/kg 380.96 kJ/kg 380.9641.40437.785 kJ/kg 404.4119.3006.704 kJ/kg 785.3760.61097.1395 kJ/kg 704.658.279.330 12 1 kJ/kg 610.663.16386.112 in net th outinnet 14out 23in 4 3 4 2 1 2 34 12 === =?=?= =?=?= =?=?= =???=? ? ? ? ? ? == =???=== q w qqw hhq hhq hP P P P hP P P P rr rr ? Thus, (a) ( )increase46.33996.380 net kJ/kg 41.5=?=?w (b) ( )increase%9.37%5.48 th 10.6%=?=?? s T 1 2 4 3 q in q out 2? 3? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-142 9-178 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For r p = 6, () ()() () () () ()() () ()() %1.40 kJ/kg 803.4 kJ/kg 321.9 kJ/kg 321.95.4814.803 kJ/kg 481.5K300779.1KkJ/kg 1.005 kJ/kg 803.4K500.61300KkJ/kg 1.005 K 779.1 6 1 K 1300 K 500.66K 300 in net th outinnet 1414out 2323in 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 === =?=?= =??= ?=?= =??= ?=?= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? q w qqw TTchhq TTchhq P P TT P P TT p p kk kk ? For r p = 12, () ()() () () () ()() () ()() %8.50 kJ/kg 693.2 kJ/kg 352.3 kJ/kg 352.39.3402.693 kJ/kg 340.9K300639.2KkJ/kg 1.005 kJ/kg 693.2K610.21300KkJ/kg 1.005 K 639.2 12 1 K 1300 K 610.212K 300 in net th outinnet 1414out 2323in 0.4/1.4 /1 3 4 34 0.4/1.4 /1 1 2 12 === =?=?= =??= ?=?= =??= ?=?= =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = == ? ? ? ? ? ? ? ? = ? ? q w qqw TTchhq TTchhq P P TT P P TT p p kk kk ? Thus, (a) ( )increase9.3213.352 net kJ/kg 30.4=?=?w (b) ( )increase%1.40%8.50 th 10.7%=?=?? s T 1 2 4 3 q in q out 2 3? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-143 9-179 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. () ()() () () () () () () () () () ()( ) () () () ()( ) ()()( )( ) ()()( )( ) kJ/kg 624.1K889.51200KkJ/kg 1.005222 kJ/kg 332.7K300465.5KkJ/kg 1.005222 K 770.8 5.4655.88972.05.465 K 889.5 9.838120086.01200 K 838.9 3.5 1 K 1200 K 465.5 78.0/3001.429300 / K 429.13.5K 300 7676outT, 1212inC, 4945 49 45 49 45 76679 76 76 76 76 0.4/1.4 /1 6 7 679 12124 12 12 12 12 0.4/1.4 /1 1 2 124 =??=?=?= =??=?=?= = ?+= ?+=??? ? ? = ? ? = = ??= ??==??? ? ? = ? ? = =? ? ? ? ? ? = ? ? ? ? ? ? ? ? == = ?+= ?+==??? ? ? = ? ? = == ? ? ? ? ? ? ? ? == ? ? TTchhw TTchhw TTTT TTc TTc hh hh TTTTT TTc TTc hh hh P P TTT TTTTT TTc TTc hh hh P P TTT p p p p sT sp p s T kk ss Cs p sp s C kk ss ?? ?? ?? Thus, 53.3%=== kJ/kg 624.1 kJ/kg 332.7 outT, inC, bw w w r ()()( ) ( )[ ] ([] 39.2%=== =?=?= =?+??= ?+?=?+?= kJ/kg 743.4 kJ/kg 291.4 kJ/kg 291.47.3321.624 kJ/kg 743.4K889.51200770.81200KkJ/kg 1.005 in net th inC,outT,net 78567856in q w www TTTTchhhhq p ? s T 3 4 1 6 9 7 8 5 2 97s 4 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-144 9-180 EES Problem 9-179 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[6] = 1200 [K] T[8] = T[6] Pratio = 3.5 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.72 "Regenerator effectiveness" Eta_c =0.78 "Compressor isentorpic efficiency" Eta_t =0.86 "Turbien isentropic efficiency" "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp_LP = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the HP compressor" P[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual compressor analysis:" h[3] + w_comp_HP = h[4] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-145 h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Intercooling heat loss:" h[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[4] + q_in_noreg = h[6] h[6]=ENTHALPY(Air,T=T[6]) P[6]=P[4]"process 4-6 is SSSF constant pressure" "HP Turbine analysis" s[6]=ENTROPY(Air,T=T[6],P=P[6]) s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio s_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit" Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7]) "Actual Turbine analysis:" h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7]) s[7]=ENTROPY(Air,T=T[7], P=P[7]) "Reheat Q_in:" h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8]) "HL Turbine analysis" P[8]=P[7] s[8]=ENTROPY(Air,T=T[8],P=P[8]) s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio s_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9]) "Actual Turbine analysis:" h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9]) s[9]=ENTROPY(Air,T=T[9], P=P[9]) "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat Eta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-146 Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio" "With the regenerator, the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[4] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:" h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10]) P[10]=P[9] "Cycle thermal efficiency with regenerator" q_in_total_withreg=q_in_withreg+q_in_reheat Eta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] s_s[10]=s[10] T_s[10]=T[10] ? C ? reg ? t ? th,noreg [%] ? th,withreg [%] q in,total,noreg [kJ/kg] q in,total,withreg [kJ/kg] w net [kJ/kg] 0.78 0.6 0.86 27.03 36.59 1130 834.6 305.4 0.78 0.65 0.86 27.03 37.7 1130 810 305.4 0.78 0.7 0.86 27.03 38.88 1130 785.4 305.4 0.78 0.75 0.86 27.03 40.14 1130 760.8 305.4 0.78 0.8 0.86 27.03 41.48 1130 736.2 305.4 0.78 0.85 0.86 27.03 42.92 1130 711.6 305.4 0.78 0.9 0.86 27.03 44.45 1130 687 305.4 0.78 0.95 0.86 27.03 46.11 1130 662.4 305.4 0.78 1 0.86 27.03 47.88 1130 637.8 305.4 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-147 4.5 5.0 5.5 6.0 6.5 7.0 7.5 200 400 600 800 1000 1200 1400 1600 s [kJ/kg-K] T [ K] 1 0 0 k P a 3 5 0 kP a 1 2 2 5 k P a Air 1 2 3 4 5 6 7 8 9 10 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 10 15 20 25 30 35 40 45 50 ? t ? th ? reg = 0.72 ? c = 0.78 With regeneration No regeneration 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 600 700 800 900 1000 1100 1200 ? t q in ,to t a l ? reg = 0.72 ? c = 0.78 No regeneration With regeneration PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-148 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 100 150 200 250 300 350 400 450 ? t w ne t [ kJ/ k g ] ? reg = 0.72 ? c = 0.78 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 25 30 35 40 45 50 ? reg ? th ? c = 0.78 ? t = 0.86 With regenertion No regeneration 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 15 20 25 30 35 40 45 50 ? c ? th ? reg = 0.72 ? t = 0.86 With regeneration No regeneration PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-149 9-181 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are c p = 5.1926 kJ/kg.K and k = 1.667 (Table A-2). Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. () ()() () () () () () () () () () ()( ) () () () ()( ) ()()( )( ) ()()( )( ) kJ/kg 4225.7K793.11200KkJ/kg 5.1926222 kJ/kg 2599.4K300550.3KkJ/kg 5.1926222 K 725.1 3.5501.79372.03.550 K 793.1 9.726120086.01200 K 726.9 3.5 1 K 1200 K 550.3 78.0/3002.495300 / K 495.23.5K 300 7676outT, 1212inC, 4945 49 45 49 45 76679 76 76 76 76 70.667/1.66 /1 6 7 679 12124 12 12 12 12 70.667/1.66 /1 1 2 124 =??=?=?= =??=?=?= = ?+= ?+=??? ? ? = ? ? = = ??= ??==??? ? ? = ? ? = =? ? ? ? ? ? = ? ? ? ? ? ? ? ? == = ?+= ?+==??? ? ? = ? ? = == ? ? ? ? ? ? ? ? == ? ? TTchhw TTchhw TTTT TTc TTc hh hh TTTTT TTc TTc hh hh P P TTT TTTTT TTc TTc hh hh P P TTT p p p p sT sp p s T kk ss Cs p sp s C kk ss ?? ?? ?? Thus, 61.5%=== kJ/kg 4225.7 kJ/kg 2599.4 outT, inC, bw w w r ()()( ) ( )[ ] ([] 35.5%=== =?=?= =?+??= ?+?=?+?= kJ/kg 4578.8 kJ/kg 1626.3 kJ/kg 1626.34.25997.4225 kJ/kg 4578.8K793.11200725.11200KkJ/kg 5.1926 in net th inC,outT,net 78567856in q w www TTTTchhhhq p ? s T 3 4 1 6 q in 97 8 5 2 97s 4 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-150 9-182 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle. Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is r p , which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes ? r p . Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as () () () () () () () () () () () ()kk p kk p kk p kk p kk p kk kk p kk p kk kk p kk rTrrTrT r T P P TT rT r T P P TTT rT P P TTT 2/1 1 2/1/1 1 2/1 2 /1 5 /1 5 6 56 2/1 3 /1 3 /1 3 4 347 /1 1 /1 1 2 125 1 1 ???? ? ? ? ? ? ? ? === ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? == = ? ? ? ? ? ? ? ? == Then, () ( ) ( ) () () 1 1 2/1 11616out 2/1 37373in ?=?=?= ?=?=?= ? ? kk ppp kk ppp rTcTTchhq rTcTTchhq and thus () ( ) () () kk pp kk pp rTc rTc q q 2/1 3 2/1 1 in out th 1 1 11 ? ? ? ? ?=?=? which simplifies to ()kk p r T T 2/1 3 1 th 1 ? ?=? The thermal efficiency of the single stage ideal regenerative cycle is given as ()kk p r T T /1 3 1 th 1 ? ?=? Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio r p . s T 1 2 7 q in 5 6 4 3 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-151 9-183 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K. Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Optimum intercooling and reheating pressure is kPa 4.346)1200)(100( 412 === PPP Process 1-2, 3-4: Compression KkJ/kg 7054.5 kPa 100 K 300 kJ/kg 43.300K 300 1 1 1 11 ?= ? ? ? = = =???= s P T hT kJ/kg 79.428 kJ/kg.K 7054.5 kPa 4.346 2 12 2 = ? ? ? == = s h ss P kJ/kg 88.460 43.300 43.30079.428 80.0 2 212 12 C =??? ? ? =??? ? ? = h hhh hh s ? KkJ/kg 5040.5 kPa 4.346 K 350 kJ/kg 78.350K 350 3 3 3 33 ?= ? ? ? = = =???= s P T hT kJ/kg 42.500 kJ/kg.K 5040.5 kPa 1200 4 34 4 = ? ? ? == = s h ss P kJ/kg 83.537 78.350 78.35042.500 80.0 4 434 34 C =??? ? ? =??? ? ? = h hhh hh s ? Process 6-7, 8-9: Expansion KkJ/kg 6514.6 kPa 1200 K 1400 kJ/kg 9.1514K 1400 6 6 6 66 ?= ? ? ? = = =???= s P T hT kJ/kg 9.1083 kJ/kg.K 6514.6 kPa 4.346 7 67 7 = ? ? ? == = s h ss P kJ/kg 1.1170 9.10839.1514 9.1514 80.0 7 7 76 76 T =??? ? ? =??? ? ? = h h hh hh s ? KkJ/kg 9196.6 kPa 4.346 K 1300 kJ/kg 6.1395K 1300 8 8 8 88 ?= ? ? ? = = =???= s P T hT s T 3 4 1 5 q in 8 6 7 10 9 2 8s 6s 4s 2s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-152 kJ/kg 00.996 kJ/kg.K 9196.6 kPa 100 9 89 9 = ? ? ? == = s h ss P kJ/kg 9.1075 00.9966.1395 6.1395 80.0 9 9 98 98 T =??? ? ? =??? ? ? = h h hh hh s ? Cycle analysis: kJ/kg 50.34778.35083.53743.30088.460 3412inC, =?+?=?+?= hhhhw kJ/kg 50.6649.10756.13951.11709.1514 9876outT, =?+?=?+?= hhhhw 0.523=== 50.664 50.347 outT, inC, bw w w r kJ/kg 317.0=?=?= 50.34750.664 inC,outT,net www Regenerator analysis: kJ/kg 36.672 83.5379.1075 9.1075 75.0 10 10 49 109 regen =??? ? ? =??? ? ? = h h hh hh ? KkJ/kg 5157.6 kPa 100 K 36.672 10 10 10 ?= ? ? ? = = s P h kJ/kg 40.94183.53736.6729.1075 5545109regen =????=?????=?= hhhhhhq (b) kJ/kg 54.57340.9419.1514 56in =?=?= hhq 0.553=== 54.573 0.317 in net th q w ? (c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is, 786.0 K 1400 K 300 11 6 1 max =?=?= T T ? and 0.704=== 786.0 553.0 max ? ? ? th II (d) The exergies at the combustion chamber exit and the regenerator exit are kJ/kg 930.7=???= ???= kJ/kg.K)7054.56514.6)(K 300(kJ/kg)43.3009.1514( )( 060066 ssThhx kJ/kg 128.8=???= ???= kJ/kg.K)7054.55157.6)(K 300(kJ/kg)43.30036.672( )( 010001010 ssThhx PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-153 9-184 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a three-stage gas turbine is to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg?K and k = 1.4 (Table A-2a). Analysis Two Stages: The pressure ratio across each stage is 416 == p r The temperatures at the end of compression and expansion are K 5.420K)(4) 283( 0.4/1.4/)1( min === ? kk pc rTT K 5.587 4 1 K) 873( 1 0.4/1.4 /)1( max =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p e r TT The heat input and heat output are kJ/kg 9.573K )5.587K)(873kJ/kg 2(1.005)(2 maxin =??=?= ep TTcq kJ/kg 4.276K )283K)(420.5kJ/kg 2(1.005)(2 minout =??=?= TTcq cp The thermal efficiency of the cycle is then 0.518=?=?= 9.573 4.276 11 in out th q q ? Three Stages: The pressure ratio across each stage is 520.216 3/1 == p r The temperatures at the end of compression and expansion are K 5.368K)(2.520) 283( 0.4/1.4/)1( min === ? kk pc rTT K 4.670 2.520 1 K) 873( 1 0.4/1.4 /)1( max =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p e r TT The heat input and heat output are kJ/kg 8.610K )4.670K)(873kJ/kg 3(1.005)(3 maxin =??=?= ep TTcq kJ/kg 8.257K )283K)(368.5kJ/kg (1.0053)(3 minout =??=?= TTcq cp The thermal efficiency of the cycle is then 0.578=?=?= 8.610 8.257 11 in out th q q ? s T 3 4 1 6 873 K 283 K 9 7 8 2 5 10 s T 3 4 1 9 8 2 5 10 11 6 7 12 13 14 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-154 9-185E A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia?ft 3 /lbm?R (Table A-1E), c p = 0.24 Btu/lbm?R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields R 0.918R)(9) 490( 0.4/1.4/)1( 12 === ? kk p rTT R 2.619 9 1 R) 1160( 1 0.4/1.4 /)1( 35 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT The work input to the compressor is Btu/lbm 7.102490)R-R)(918.0Btu/lbm 24.0()( 12C =?=?= TTcw p An energy balance gives the excess enthalpy to be Btu/lbm 09.27 Btu/lbm 102.7)R2.619R)(1160Btu/lbm 24.0( )( C53 = ???= ??=? wTTch p The velocity of the air at the engine exit is determined from 2 2 inlet 2 exit VV h ? =? Rearranging, () ft/s 1672 )ft/s 1200( Btu/lbm 1 /sft 25,037 Btu/lbm) 09.27(2 2 2/1 2 22 2/1 2 inletexit = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? = +?= VhV The specific impulse is then m/s 472=?=?= 12001672 inletexit VV m F & s T 1 2 4 3 q in 5 q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-155 9-186 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limit. The net work is to be determined using constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats: K 8.591K)(15) 273( 0.4/1.4/)1( 12 === ? kk p rTT K 7.402 15 1 K) 873( 1 0.4/1.4 /)1( 34 =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? kk p r TT kJ/kg 152.3= ?+??= ?+?= ???= ?= K)8.5912737.402873)(KkJ/kg 1.005( )( )()( 2143 1243 compturbnet TTTTc TTcTTc www p pp (b) Variable specific heats: (using air properties from Table A-17) 9980.0 kJ/kg 12.273 K 273 1 1 1 = = ???= r P h T kJ/kg 69.59297.14)9980.0)(15( 21 1 2 2 =???=== hP P P P rr kJ/kg 58.419461.4)92.66( 15 1 92.66 kJ/kg 76.902 K 873 43 3 4 4 3 3 3 =???=? ? ? ? ? ? == = = ???= hP P P P P h T rr r kJ/kg 163.6= ???= ???= ?= )12.27369.592()58.41976.902( )()( 1243 compturbnet hhhh www s T 1 2 4 3 q in q out 873 K 273 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-156 9-187 An Otto cycle with a compression ratio of 8 is considered. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are R = 0.287 kPa·m 3 /kg·K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats: 0.5647=?=?= ?? 11.41 th 8 1 1 1 1 k r ? (b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. 9.718 kJ/kg 9.201 K 283 1 1 1 = = ???= r u T v kJ/kg 76.46386.89)9.718( 8 11 222 1 2 2 =???==== u r rrr vv v v v Process 2-3: v = constant heat addition. kJ/kg 445.6376.46339.909 529.15 kJ/kg 39.909 K 1173 23 3 3 3 =?=?= = = ???= uuq u T in r v Process 3-4: isentropic expansion. kJ/kg 06.4082.124)529.15)(8( 433 3 4 4 =???==== ur rrr vv v v v Process 4-1: v = constant heat rejection. kJ/kg 16.2069.20106.408 14out =?=?= uuq 0.5374=?=?= kJ/kg 445.63 kJ/kg 206.16 11 in out th q q ? v P 4 1 3 2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-157 9-188 An ideal diesel engine with air as the working fluid has a compression ratio of 22. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats: Process 1-2: isentropic compression. K 7.991K)(22) 288( 0.4 1 2 1 12 == ? ? ? ? ? ? ? ? = ?k TT V V Process 2-3: P = constant heat addition. 485.1 K991.7 K 1473 2 3 2 3 2 22 3 33 ===???= T T T P T P V VVV Process 3-4: isentropic expansion. ()( )( ) ()( )( ) 0.698=== =?=?= =??=?=?= =??=?=?= =? ? ? ? ? ? =? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ?? kJ/kg 473.7 kJ/kg 330.7 kJ/kg 7.3300.1537.483 kJ/kg 0.153K288501.1KkJ/kg 0.718 kJ/kg 7.483K7.9911473KkJ/kg 1.005 K 1.501 22 1.485 K) 1473( 485.1485.1 in outnet, th outinoutnet, 1414out 2323in 0.41 3 1 4 2 3 1 4 3 34 q w qqw TTcuuq TTchhq r TTTT p k kk ? v V V V V (b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. 1.688 kJ/kg 48.205 K 288 1 1 1 = = ???= r u T v kJ/kg 73.965 K 2.929 28.31)1.688( 22 11 2 2 11 1 2 2 = = ???==== h T r rrr vv v v v Process 2-3: P = constant heat addition. 585.1 K 929.2 K 1473 2 3 2 3 2 22 3 33 ===???= T T T P T P v vvv kJ/kg 6.63773.96533.1603 585.7 kJ/kg 33.1603 K 1473 23in 3 3 3 =?=?= = = ???= hhq h T r v Process 3-4: isentropic expansion. kJ/kg 61.4353.105)585.7( 585.1 22 585.1585.1 433 2 4 3 3 4 4 =???===== u r rrrr vv v v v v v v Process 4-1: v = constant heat rejection. kJ/kg 230.1348.20561.435 14out =?=?= uuq Then 0.639=?=?= kJ/kg 637.6 kJ/kg 230.13 11 in out th q q ? v P 4 1 2 3 q in q out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-158 9-189 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Compression KkJ/kg 7159.5 kPa 100 C30 kJ/kg 60.303C30 1 1 1 11 ?= ? ? ? = °= =???°= s P T hT kJ/kg 37.617 kJ/kg.K 7159.5 kPa 1200 2 12 2 = ? ? ? == = s h ss P kJ/kg 24.686 60.303 60.30337.617 82.0 2 212 12 C =??? ? ? =??? ? ? = h hhh hh s ? Process 3-4: Expansion kJ/kg 62.792C500 44 =???°= hT ss hh h hh hh 43 3 43 43 T 62.792 82.0 ? ? =??? ? ? =? We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h 3 = 1404.7 kJ/kg, T 3 = 1034ºC, s 3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3) Also, kJ/kg 44.631C350 55 =???°= hT The inlet water is compressed liquid at 25ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature. 1 Combustion chamber Turbine 2 3 4 Compress. 100 kPa 30°C 500°C 1.2 MPa 350°C 25°C Sat. vap. 200°C Heat exchanger 5 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-159 kJ/kg 0.2792 1 C200 kJ/kg 27.106 kPa 1555 C25 2 2 2 1 1 1 = ? ? ? = °= = ? ? ? = °= w w w w h x T h P T The net work output is kJ/kg 03.61262.7927.1404 kJ/kg 64.38260.30324.686 43outT, 12inC, =?=?= =?=?= hhw hhw kJ/kg 39.22964.38203.612 inC,outT,net =?=?= www The mass flow rate of air is kg/s 3.487=== kJ/kg 39.229 kJ/s 800 net net w W m a & & (b) The back work ratio is 0.625=== 03.612 64.382 outT, inC, bw w w r The rate of heat input and the thermal efficiency are kW 2505)kJ/kg24.686.7kg/s)(1404 487.3()( 23in =?=?= hhmQ a & & 0.319=== kW 2505 kW 800 in net th Q W & & ? (c) An energy balance on the heat exchanger gives kg/s 0.2093=????=? ?=? ww wwwa mm hhmhhm && && )kJ/kg27.106(2792.0)kJ/kg44.63162kg/s)(792. 487.3( )()( 1254 (d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are kW 1.562)kJ/kg27.106.0kg/s)(2792 2093.0()( 12p =?=?= www hhmQ & & 0.544= + = + = kW 2505 1.562800 in pnet Q QW u & && ? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-160 9-190 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Diffuser, Process 1-2: kJ/kg 23.238C35 11 =???°?= hT kJ/kg 37.269 /sm 1000 kJ/kg 1 2 m/s) (15 /sm 1000 kJ/kg 1 2 m/s) (900/3.6 kJ/kg) 23.238( 22 2 22 2 2 22 2 2 2 2 2 1 1 =???? ? ? ? ? ? +=? ? ? ? ? ? + +=+ hh V h V h KkJ/kg 7951.5 kPa 50 kJ/kg 37.269 2 2 2 ?= ? ? ? = = s P h Compressor, Process 2-3: kJ/kg 19.505 kJ/kg.K 7951.5 kPa 450 3 23 3 = ? ? ? == = s h ss P kJ/kg 50.553 37.269 37.26919.505 83.0 3 323 23 C =??? ? ? =??? ? ? = h hhh hh s ? Turbine, Process 3-4: kJ/kg 8.1304C950 44 =???°= hT kJ/kg 6.10208.130437.26950.553 555423 =????=?????=? hhhhhh where the mass flow rates through the compressor and the turbine are assumed equal. kJ/kg 45.962 8.1304 6.10208.1304 83.0 5 554 54 T =??? ? ? =??? ? ? = s ss h hhh hh ? KkJ/kg 7725.6 kPa 450 C950 4 4 4 ?= ? ? ? = °= s P T kPa 147.4= ? ? ? ?== = 5 45 5 KkJ/kg 7725.6 kJ/kg 45.962 P ss h s (b) The mass flow rate of the air through the compressor is kg/s 1.760= ? = ? = kJ/kg )37.26950.553( kJ/s 500 23 C hh W m & & s T 1 2 4 3 q in 5 q out 6 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-161 (c) Nozzle, Process 5-6: KkJ/kg 8336.6 kPa 4.147 kJ/kg 6.1020 5 5 5 ?= ? ? ? = = s P h kJ/kg 66.709 kJ/kg.K 8336.6 kPa 40 6 56 6 = ? ? ? == = s h ss P kJ/kg 52.762 66.7096.1020 6.1020 83.0 6 6 65 65 N =??? ? ? =??? ? ? = h h hh hh s ? m/s 718.5=???? ? ? ? ? ? +=+ +=+ 6 22 2 6 2 6 6 2 5 5 /sm 1000 kJ/kg 1 2 kJ/kg 52.7620kJ/kg) 6.1020( 22 V V V h V h where the velocity at nozzle inlet is assumed zero. (d) The propulsive power and the propulsive efficiency are kW 206.1=? ? ? ? ? ? ?=?= 22 116 /sm 1000 kJ/kg 1 m/s) 250)(m/s 250m/s 5.718(kg/s) 76.1()( VVVmW p & & kW 1322kJ/kg)50.5538.1304(kg/s) 76.1()( 34in =?=?= hhmQ & & 0.156=== kW 1322 kW 1.206 in Q W p p & & ? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-162 9-191 EES The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 2000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-163 ? th [%] r comp w net [kJ/kg] 45.83 6 567.4 48.67 7 589.3 51.03 8 604.9 53.02 9 616.2 54.74 10 624.3 56.24 11 630 57.57 12 633.8 58.75 13 636.3 59.83 14 637.5 60.8 15 637.9 6 7 8 9 10 11 12 13 14 15 560 570 580 590 600 610 620 630 640 r comp w ne t [ k J/ kg ] 6 7 8 9 10 11 12 13 14 15 45 48.5 52 55.5 59 62.5 r comp ? th [ % ] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-164 9-192 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-165 Bwr ? P ratio W c [kW] W net [kW] W t [kW] Q in [kW] 0.254 0.3383 5 175.8 516.3 692.1 1526 0.2665 0.3689 6 201.2 553.7 754.9 1501 0.2776 0.3938 7 223.7 582.2 805.9 1478 0.2876 0.4146 8 244.1 604.5 848.5 1458 0.2968 0.4324 9 262.6 622.4 885 1439 0.3052 0.4478 10 279.7 637 916.7 1422 0.313 0.4615 11 295.7 649 944.7 1406 0.3203 0.4736 12 310.6 659.1 969.6 1392 0.3272 0.4846 13 324.6 667.5 992.1 1378 0.3337 0.4945 14 337.8 674.7 1013 1364 0.3398 0.5036 15 350.4 680.8 1031 1352 0.3457 0.512 16 362.4 685.9 1048 1340 0.3513 0.5197 17 373.9 690.3 1064 1328 0.3567 0.5269 18 384.8 694.1 1079 1317 0.3618 0.5336 19 395.4 697.3 1093 1307 0.3668 0.5399 20 405.5 700 1106 1297 0.3716 0.5458 21 415.3 702.3 1118 1287 0.3762 0.5513 22 424.7 704.3 1129 1277 0.3806 0.5566 23 433.8 705.9 1140 1268 0.385 0.5615 24 442.7 707.2 1150 1259 0.3892 0.5663 25 451.2 708.3 1160 1251 0.3932 0.5707 26 459.6 709.2 1169 1243 0.3972 0.575 27 467.7 709.8 1177 1234 0.401 0.5791 28 475.5 710.3 1186 1227 0.4048 0.583 29 483.2 710.6 1194 1219 0.4084 0.5867 30 490.7 710.7 1201 1211 0.412 0.5903 31 498 710.8 1209 1204 0.4155 0.5937 32 505.1 710.7 1216 1197 0.4189 0.597 33 512.1 710.4 1223 1190 0.4222 0.6002 34 518.9 710.1 1229 1183 5 10 15 20 25 30 35 500 525 550 575 600 625 650 675 700 725 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 P ratio W net [k W ] ? th PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-166 9-193 EES The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 85/100 Eta_t = 85/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-167 Bwr ? P ratio W c [kW] W net [kW] W t [kW] Q in [kW] 0.3515 0.2551 5 206.8 381.5 588.3 1495 0.3689 0.2764 6 236.7 405 641.7 1465 0.3843 0.2931 7 263.2 421.8 685 1439 0.3981 0.3068 8 287.1 434.1 721.3 1415 0.4107 0.3182 9 309 443.3 752.2 1393 0.4224 0.3278 10 329.1 450.1 779.2 1373 0.4332 0.3361 11 347.8 455.1 803 1354 0.4433 0.3432 12 365.4 458.8 824.2 1337 0.4528 0.3495 13 381.9 461.4 843.3 1320 0.4618 0.355 14 397.5 463.2 860.6 1305 0.4704 0.3599 15 412.3 464.2 876.5 1290 0.4785 0.3643 16 426.4 464.7 891.1 1276 0.4862 0.3682 17 439.8 464.7 904.6 1262 0.4937 0.3717 18 452.7 464.4 917.1 1249 0.5008 0.3748 19 465.1 463.6 928.8 1237 0.5077 0.3777 20 477.1 462.6 939.7 1225 0.5143 0.3802 21 488.6 461.4 950 1214 0.5207 0.3825 22 499.7 460 959.6 1202 0.5268 0.3846 23 510.4 458.4 968.8 1192 0.5328 0.3865 24 520.8 456.6 977.4 1181 5 9 13 17 21 25 380 390 400 410 420 430 440 450 460 470 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 P ratio W ne t [k W ] ? t h P ratio for W net,max W net ? th PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-168 9-194 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] {T[3] = 1000 [K]} r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-169 "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]" PerCentError [%] r comp ? th [%] ? th,ConstProp [%] ? th,easy [%] T 3 [K] 3.604 12 60.8 62.99 62.99 1000 6.681 12 59.04 62.99 62.99 1500 9.421 12 57.57 62.99 62.99 2000 11.64 12 56.42 62.99 62.99 2500 6 7 8 9 10 11 12 3.6 3.7 3.8 3.9 4 4.1 4.2 4.3 r comp P e rC e n tE r r o r [% ] Percent Error = |? th - ? th,ConstProp | / ? th T m ax = 1000 K 1000 1200 1400 1600 1800 2000 2200 2400 2600 4 6.2 8.4 10.6 12.8 15 T[3] [K] Pe r C e n t E r r o r [ % ] r comp = 6 =12 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-170 9-195 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-171 Bwr ? P ratio W c [kW] W net [kW] W t [kW] Q in [kW] 0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700 0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241 5.0 5.5 6.0 6.5 7.0 7.5 0 500 1000 1500 s [kJ/kg-K] T [ K ] 100 kPa 800 kPa 1 2 s 2 3 4 4 s Air Standard Brayton Cycle Pressure ratio = 8 and T max = 1160K 2 4 6 8 10 12 14 16 18 20 0.00 0.05 0.10 0.15 0.20 0.25 0 500 1000 1500 2000 2500 P ratio C y cl e ef f i ci en cy , W n e t [ kW ] ? W net T max =1160 K Note P ratio for maximum work and ? ? c = 0.75 ? t = 0.82 ? PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-172 9-196 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'} "Inlet conditions" h[1]=hFunc(WorkFluid$,T[1],P[1]) s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid$,T_s[2],P[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=hFunc(WorkFluid$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid$,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_s[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-173 Eta_th=W_dot_net/Q_dot_in"Cycle thermal efficiency" Call EtaCheck(Eta_th:EtaError$) Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) Bwr ? P ratio W c [kW] W net [kW] W t [kW] Q in [kW] 0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700 0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241 5.0 5.5 6.0 6.5 7.0 7.5 0 500 1000 1500 s [kJ/kg-K] T [ K ] 100 kPa 800 kPa 1 2 s 2 3 4 4 s Brayton Cycle P ressu re ratio = 8 and T m ax = 1160K 2 4 6 8 10 12 14 16 18 20 0.00 0.05 0.10 0.15 0.20 0.25 0 500 1000 1500 2000 2500 P ratio Cycl e ef f i c i ency, W ne t [k W ] ? W net T max =1160 K Note P ratio for maximum work and ? ? c = 0.75 ? t = 0.82 ? Brayton Cycle using Air m air = 20 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-174 9-197 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages" Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-175 "Cycle analysis" w_net=w_turb_total-w_comp_total "[kJ/kg]" Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]" ? th,Ericksson [%] ? th,Regenerative [%] Nstages 75 49.15 1 75 64.35 2 75 68.32 3 75 70.14 4 75 72.33 7 75 73.79 15 75 74.05 19 75 74.18 22 0 2 4 6 8 10 12 14 16 18 20 22 24 40 50 60 70 80 Nstages ? th [% ] Ericsson Ideal Regenerative Brayton PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-176 9-198 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages" {Nstages = 1} T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-177 "Cycle analysis" w_net=w_turb_total-w_comp_total Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]" ? th,Ericksson [%] ? th,Regenerative [%] Nstages 75 32.43 1 75 58.9 2 75 65.18 3 75 67.95 4 75 71.18 7 75 73.29 15 75 73.66 19 75 73.84 22 0 2 4 6 8 10 12 14 16 18 20 22 24 30 40 50 60 70 80 Nstages ? th [% ] Ericsson Ideal Regenerative Brayton PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-178 Fundamentals of Engineering (FE) Exam Problems 9-199 An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 24% (b) 43% (c) 52% (d) 57% (e) 75% Answer (d) 57% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). r=8.2 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value" 9-200 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is (a) Carnot (b) Stirling (c) Ericsson (d) Otto (e) All are the same Answer (d) Otto 9-201 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is (a) 0 (b) 0.300 kW/K (c) 0.353 kW/K (d) 0.261 kW/K (e) 2.0 kW/K Answer (c) 0.353 kW/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-179 9-202 Air in an ideal Diesel cycle is compressed from 3 L to 0.15 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 35% (b) 44% (c) 65% (d) 70% (e) 82% Answer (c) 65% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=3 "L" V2= 0.15 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency" 9-203 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is (a) 1790°C (b) 2060°C (c) 1240°C (d) 620°C (e) 820°C Answer (a) 1790°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-180 9-204 In an ideal Otto cycle, air is compressed from 1.20 kg/m 3 and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is (a) 612 kPa (b) 599 kPa (c) 528 kPa (d) 416 kPa (e) 367 kPa Answer (b) 599 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting" 9-205 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 800 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 46% (b) 54% (c) 57% (d) 39% (e) 61% Answer (a) 46% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=95 "kPa" P2=800 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-181 9-206 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is (a) 68 kJ/kg (b) 93 kJ/kg (c) 158 kJ/kg (d) 186 kJ/kg (e) 310 kJ/kg Answer (c) 158 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K" 9-207 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be (a) 74 kJ/kg (b) 95 kJ/kg (c) 109 kJ/kg (d) 128 kJ/kg (e) 177 kJ/kg Answer (b) 95 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-182 9-208 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 1200°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is (a) 490°C (b) 515°C (c) 622°C (d) 763°C (e) 895°C Answer (a) 490°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=1200+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-183 9-209 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is (a) 246°C (b) 846°C (c) 689°C (d) 368°C (e) 573°C Answer (e) 573°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-184 9-210 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is (a) 33% (b) 44% (c) 62% (d) 77% (e) 89% Answer (d) 77% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-185 9-211 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is (a) 38% (b) 46% (c) 62% (d) 58% (e) 97% Answer (c) 62% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K" Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)" 9-212 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is (a) 36% (b) 40% (c) 52% (d) 64% (e) 76% Answer (e) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 9-186 9-213 Air enters a turbojet engine at 260 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to the aircraft. The thrust developed by the engine is (a) 8 kN (b) 16 kN (c) 24 kN (d) 20 kN (e) 32 kN Answer (b) 16 kN Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel1=260 "m/s" Vel2=800 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation" 9-214 ··· 9-220 Design and Essay Problems. K J Erik J. Rose CHAPTER 14

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