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### Ch.10

1. 1. 10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLESCarnot Vapor Cycle10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisturecontent allowed is about 10%.10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heattransfer processes to two-phase systems to maintain isothermal conditions severely limits the maximumtemperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisturecontent which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the quality at the end of the heat rejection process, and the net work output are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We note that TH = Tsat @180 psia = 373.1°F = 833.1 R T TL = Tsat @14.7 psia = 212.0°F = 672.0 Rand 1 180 psia 2 T 672.0 R ηth,C = 1− L = 1− = 19.3% qin TH 833.1 R(b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R, 14.7 psia 4 3 s4 − s f 0.53274 − 0.31215 x4 = = = 0.153 s s fg 1.44441(c) The enthalpies before and after the heat addition process are h1 = h f @ 180 psia = 346.14 Btu/lbm h2 = h f + x 2 h fg = 346.14 + (0.90)(851.16) = 1112.2 Btu/lbmThus, q in = h2 − h1 = 1112.2 − 346.14 = 766.0 Btu/lbmand, wnet = η th q in = (0.1934)(766.0 Btu/lbm) = 148.1 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
2. 2. 10-210-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermalefficiency becomes TL 333.1 K η th,C = 1 − = 1− = 0.3632 = 36.3% T TH 523 K(b) The heat supplied during this cycle is simply theenthalpy of vaporization, 2 1 250°C q in = h fg @ 250oC = 1715.3 kJ/kg qinThus, ⎛ 333.1 K ⎞ 20 kPa ⎜ 523 K ⎟(1715.3 kJ/kg ) = 1092.3 kJ/kg TL q out = q L = q in = ⎜ ⎟ 4 3 TH ⎝ ⎠ qout s(c) The net work output of this cycle is wnet = η th q in = (0.3632 )(1715.3 kJ/kg ) = 623.0 kJ/kg10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermalefficiency becomes TL 318.8 K η th, C = 1 − =1− = 39.04% T TH 523 K(b) The heat supplied during this cycle is simply theenthalpy of vaporization, 1 2 250°C q in = h fg @ 250°C = 1715.3 kJ/kg qinThus, ⎛ 318.8 K ⎞ 10 kPa ⎜ 523 K ⎟(1715.3 kJ/kg ) = 1045.6 kJ/kg TL q out = q L = q in = ⎜ ⎟ 4 3 TH ⎝ ⎠ qout s(c) The net work output of this cycle is wnet = η th q in = (0.3904)(1715.3 kJ/kg ) = 669.7 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
3. 3. 10-310-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The thermal efficiency is determined from TL 60 + 273 K T η th, C = 1 − = 1− = 46.5% TH 350 + 273 K(b) Note that 1 2 s2 = s3 = sf + x3sfg 350°C = 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·KThus, 60°C 4 3 T2 = 350°C ⎫ s ⎬ P2 ≅ 1.40 MPa (Table A-6) s 2 = 7.1368 kJ/kg ⋅ K ⎭(c) The net work can be determined by calculating the enclosed area on the T-s diagram, s 4 = s f + x 4 s fg = 0.8313 + (0.1)(7.0769) = 1.5390 kJ/kg ⋅ KThus, wnet = Area = (TH − TL )(s 3 − s 4 ) = (350 − 60)(7.1368 − 1.5390) = 1623 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
4. 4. 10-4The Simple Rankine Cycle10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump,(2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heatrejection in a condenser.10-8C Heat rejected decreases; everything else increases.10-9C Heat rejected decreases; everything else increases.10-10C The pump work remains the same, the moisture content decreases, everything else increases.10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involvefriction and pressure drops in various components and the piping, and heat loss to the surrounding mediumfrom these components and piping.10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) thesame as the boiler inlet pressure in ideal cycles.10-13C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam willadversely affect the turbine work output.10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher thanthe temperature of the cooling water.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
5. 5. 10-510-15E A simple ideal Rankine cycle with water as the working fluid operates between the specifiedpressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 6 psia = 138.02 Btu/lbm v 1 = v f @ 6 psia = 0.01645 ft 3 /lbm T wp,in = v 1 ( P2 − P1 ) 3 ⎛ ⎞ 500 psia 1 Btu = (0.01645 ft 3 /lbm)(500 − 6)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ = 1.50 Btu/lbm ⎝ ⎠ 2 qin h2 = h1 + wp,in = 138.02 + 1.50 = 139.52 Btu/lbm 6 psia 1 4 P3 = 500 psia ⎫ h3 = 1630.0 Btu/lbm qout ⎬ T3 = 1200°F ⎭ s 3 = 1.8075 Btu/lbm ⋅ R s s 4 − s f 1.8075 − 0.24739 P4 = 6 psia ⎫ x 4 = = = 0.9864 ⎬ s fg 1.58155 s 4 = s3 ⎭ h = h + x h = 138.02 + (0.9864)(995.88) = 1120.4 Btu/lbm 4 f 4 fgKnowing the power output from the turbine the mass flow rate of steam in the cycle is determined from & & WT,out 500 kJ/s ⎛ 0.94782 Btu ⎞ WT,out = m(h3 − h4 ) ⎯ & ⎯→ m = & = ⎜ ⎟ = 0.9300 lbm/s h3 − h4 (1630.0 − 1120.4)Btu/lbm ⎝ 1 kJ ⎠The rates of heat addition and rejection are & Qin = m(h3 − h2 ) = (0.9300 lbm/s)(1630.0 − 139.52)Btu/lbm = 1386 Btu/s & & = m(h − h ) = (0.9300 lbm/s)(1120.4 − 138.02)Btu/lbm = 913.6 Btu/s Qout & 4 1and the thermal efficiency of the cycle is & Qout 913.6 η th = 1 − = 1− = 0.341 Q& 1386 inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
6. 6. 10-610-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressurelimits. The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis For maximum thermal efficiency, the quality at state 4 would be at its minimum of 85% (mostclosely approaches the Carnot cycle), and the properties at state 4 would be (Table A-5) P4 = 30 kPa ⎫ h4 = h f + x 4 h fg = 289.27 + (0.85)(2335.3) = 2274.3 kJ/kg ⎬ x 4 = 0.85 ⎭ s 4 = s f + x 4 s fg = 0.9441 + (0.85)(6.8234) = 6.7440 kJ/kg ⋅ KSince the expansion in the turbine is isentropic, P3 = 3000 kPa ⎫ ⎬ h3 = 3115.5 kJ/kg T s 3 = s 4 = 6.7440 kJ/kg ⋅ K ⎭Other properties are obtained as follows (Tables A-4, A-5, and A-6), 3 3 MPa h1 = h f @ 30 kPa = 289.27 kJ/kg 2 qin v 1 = v f @ 30 kPa = 0.001022 m 3 /kg 30 kPa wp,in = v 1 ( P2 − P1 ) 1 4 ⎛ 1 kJ ⎞ qout = (0.001022 m 3 /kg )(3000 − 30)kPa ⎜ ⎟ = 3.04 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ s h2 = h1 + wp,in = 289.27 + 3.04 = 292.31 kJ/kgThus, q in = h3 − h2 = 3115.5 − 292.31 = 2823.2 kJ/kg q out = h4 − h1 = 2274.3 − 289.27 = 1985.0 kJ/kgand the thermal efficiency of the cycle is q out 1985.0 η th = 1 − = 1− = 0.297 q in 2823.2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
7. 7. 10-710-17 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressurelimits. The power produced by the turbine and consumed by the pump are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg v 1 = v f @ 20 kPa = 0.001017 m 3 /kg T 3 wp,in = v 1 ( P2 − P1 ) 4 MPa ⎛ 1 kJ ⎞ = (0.001017 m 3 /kg)(4000 − 20)kPa ⎜ ⎟ = 4.05 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ 2 qin h2 = h1 + wp,in = 251.42 + 4.05 = 255.47 kJ/kg 20 kPa 1 qout 4 P3 = 4000 kPa ⎫ h3 = 3906.3 kJ/kg ⎬ T3 = 700°C ⎭ s 3 = 7.6214 kJ/kg ⋅ K s s4 − s f 7.6214 − 0.8320 P4 = 20 kPa ⎫ x 4 = = = 0.9596 ⎬ s fg 7.0752 s 4 = s3 ⎭ h = h + x h = 251.42 + (0.9596)(2357.5) = 2513.7 kJ/kg 4 f 4 fgThe power produced by the turbine and consumed by the pump are & WT,out = m(h3 − h4 ) = (50 kg/s)(3906.3 − 2513.7)kJ/kg = 69,630 kW & & W P,in = mwP,in = (50 kg/s)(4.05 kJ/kg) = 203 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
8. 8. 10-810-18E A simple ideal Rankine cycle with water as the working fluid operates between the specifiedpressure limits. The turbine inlet temperature and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2Kinetic and potential energy changes are negligible. TAnalysis From the steam tables (Tables A-4E, A-5E,and A-6E), 3 2500 psia h1 = h f @ 5 psia = 130.18 Btu/lbm v 1 = v f @ 5 psia = 0.01641 ft 3 /lbm 2 qinwp,in = v 1 ( P2 − P1 ) 5 psia ⎛ 1 Btu ⎞ = (0.01641 ft /lbm)(2500 − 5)psia ⎜ 3 ⎟ 1 qout 4 ⎜ 5.404 psia ⋅ ft 3 ⎟ = 7.58 Btu/lbm ⎝ ⎠ h2 = h1 + wp,in = 130.18 + 7.58 = 137.76 Btu/lbm s P4 = 5 psia ⎫ h4 = h f + x 4 h fg = 130.18 + (0.80)(1000.5) = 930.58 Btu/lbm ⎬ x 4 = 0.80 ⎭ s 4 = s f + x 4 s fg = 0.23488 + (0.80)(1.60894) = 1.52203 Btu/lbm ⋅ R P3 = 2500 psia ⎫ h3 = 1450.8 Btu/lbm ⎬ s 3 = s 4 = 1.52203 Btu/lbm ⋅ R ⎭ T3 = 989.2 °FThus, q in = h3 − h2 = 1450.8 − 137.76 = 1313.0 Btu/lbm q out = h4 − h1 = 930.58 − 130.18 = 800.4 Btu/lbmThe thermal efficiency of the cycle is q out 800.4 η th = 1 − = 1− = 0.390 q in 1313.0PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
9. 9. 10-910-19 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressurelimits. The power produced by the turbine, the heat added in the boiler, and the thermal efficiency of thecycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg T v 1 = v f @ 100 kPa = 0.001043 m 3 /kg w p,in = v 1 ( P2 − P1 ) 15 MPa ⎛ 1 kJ ⎞ 3 = (0.001043 m 3 /kg )(15,000 − 100)kPa ⎜ ⎟ 2 qin = 15.54 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ 100 kPa h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg 1 qout 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K s s4 − s f 5.3108 − 1.3028 P4 = 100 kPa ⎫ x 4 = = = 0.6618 ⎬ s fg 6.0562 s 4 = s3 ⎭ h = h + x h = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg 4 f 4 fgThus, wT,out = h3 − h4 = 2610.8 − 1911.5 = 699.3 kJ/kg q in = h3 − h2 = 2610.8 − 433.05 = 2177.8 kJ/kg q out = h4 − h1 = 1911.5 − 417.51 = 1494.0 kJ/kgThe thermal efficiency of the cycle is q out 1494.0 η th = 1 − = 1− = 0.314 q in 2177.8PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
10. 10. 10-1010-20 A simple Rankine cycle with water as the working fluid operates between the specified pressurelimits. The isentropic efficiency of the turbine, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg T v 1 = v f @ 100 kPa = 0.001043 m 3 /kg wp,in = v 1 ( P2 − P1 ) 15 MPa 3 ⎛ 1 kJ ⎞ 3 = (0.001043 m /kg )(15,000 − 100)kPa ⎜ ⎟ 2 qin = 15.54 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg 100 kPa 1 qout 4s 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K s s4 − s f 5.3108 − 1.3028 P4 = 100 kPa ⎫ x 4 s = = = 0.6618 ⎬ s fg 6.0562 s 4 = s3 ⎭ h = h + x h = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg 4s f 4 s fg P4 = 100 kPa ⎫ ⎬ h4 = h f + x 4 h fg = 417.51 + (0.70)(2257.5) = 1997.8 kJ/kg x 4 = 0.70 ⎭The isentropic efficiency of the turbine is h3 − h4 2610.8 − 1997.8 ηT = = = 0.877 h3 − h4 s 2610.8 − 1911.5Thus, q in = h3 − h2 = 2610.8 − 433.05 = 2177.8 kJ/kg q out = h4 − h1 = 1997.8 − 417.51 = 1580.3 kJ/kgThe thermal efficiency of the cycle is q out 1580.3 η th = 1 − = 1− = 0.274 q in 2177.8PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11. 11. 10-1110-21E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate,the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm T wp,in = v 1 ( P2 − P1 ) 3 2500 psia ⎛ 1 Btu ⎞ = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ = 7.46 Btu/lbm ⎝ ⎠ 2 qin h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm 1 psia 1 4s 4 P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm qout ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 ⎬ s fg 1.84495 s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg h3 − h4 ηT = ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg ⎯ h3 − h4 sThus, q in = h3 − h2 = 1302.0 − 77.18 = 1224.8 Btu/lbm q out = h4 − h1 = 839.13 − 69.72 = 769.41 Btu/lbm wnet = q in − q out = 1224.8 − 769.41 = 455.39 Btu/lbmThe mass flow rate of steam in the cycle is determined from W& 1000 kJ/s ⎛ 0.94782 Btu ⎞ & & ⎯→ m = net = W net = mwnet ⎯ & ⎜ ⎟ = 2.081 lbm/s wnet 455.39 Btu/lbm ⎝ 1 kJ ⎠The power output from the turbine and the rate of heat addition are & ⎛ 1 kJ ⎞ WT,out = m(h3 − h4 ) = (2.081 lbm/s)(1302.0 − 839.13)Btu/lbm⎜ & ⎟ = 1016 kW ⎝ 0.94782 Btu ⎠ & Qin = mq in = (2.081 lbm/s)(1224.8 Btu/lbm) = 2549 Btu/s &and the thermal efficiency of the cycle is & W net 1000 kJ/s ⎛ 0.94782 Btu ⎞ η th = = ⎜ ⎟ = 0.3718 & Qin 2549 Btu/s ⎝ 1 kJ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
12. 12. 10-1210-22E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate,the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm T 3 wp,in = v 1 ( P2 − P1 ) 2500 psia ⎛ 1 Btu ⎞ = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ 2 qin = 7.46 Btu/lbm ⎝ ⎠ h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm 1 psia 1 qout 4s 4 P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 ⎬ s fg 1.84495 s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg h3 − h4 ηT = ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg ⎯ h3 − h4 sThe mass flow rate of steam in the cycle is determined from & W net 1000 kJ/s ⎛ 0.94782 Btu ⎞ & W net = m(h3 − h4 ) ⎯ & ⎯→ m = & = ⎜ ⎟ = 2.048 lbm/s h3 − h4 (1302.0 − 839.13) Btu/lbm ⎝ 1 kJ ⎠The rate of heat addition is & ⎛ 1 kJ ⎞ Qin = m(h3 − h2 ) = (2.048 lbm/s)(1302.0 − 77.18)Btu/lbm⎜ & ⎟ = 2508 Btu/s ⎝ 0.94782 Btu ⎠and the thermal efficiency of the cycle is & W net 1000 kJ/s ⎛ 0.94782 Btu ⎞ η th = = ⎜ ⎟ = 0.3779 & Qin 2508 Btu/s ⎝ 1 kJ ⎠The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermalefficiency caused by neglecting the pump work is then 0.3779 − 0.3718 Error = × 100 = 1.64% 0.3718PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
13. 13. 10-1310-23 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between thespecified pressure limits. The overall plant efficiency and the required rate of the coal supply are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 25 kPa = 271.96 kJ/kg v 1 = v f @ 25 kPa = 0.001020 m 3 /kg T w p ,in = v 1 (P2 − P1 ) 3 ( 3 ) = 0.00102 m /kg (5000 − 25 kPa )⎜ ⎛ 1 kJ ⎜ 1 kPa ⋅ m 3 ⎞ ⎟ ⎟ 5 MPa · = 5.07 kJ/kg ⎝ ⎠ Qin 2 h2 = h1 + w p ,in = 271.96 + 5.07 = 277.03 kJ/kg 25 kPa P3 = 5 MPa ⎫ h3 = 3317.2 kJ/kg 1 · 4 Qout ⎬ T3 = 450°C ⎭ s 3 = 6.8210 kJ/kg ⋅ K s P4 = 25 kPa ⎫ s 4 − s f 6.8210 − 0.8932 ⎬ x4 = = = 0.8545 s 4 = s3 ⎭ s fg 6.9370 h4 = h f + x 4 h fg = 271.96 + (0.8545)(2345.5) = 2276.2 kJ/kgThe thermal efficiency is determined from qin = h3 − h2 = 3317.2 − 277.03 = 3040.2 kJ/kg qout = h4 − h1 = 2276.2 − 271.96 = 2004.2 kJ/kgand q out 2004.2 η th = 1 − = 1− = 0.3407 q in 3040.2Thus, η overall = η th ×η comb ×η gen = (0.3407 )(0.75)(0.96 ) = 24.5%(b) Then the required rate of coal supply becomes & W net 300,000 kJ/s & Qin = = = 1,222,992 kJ/s η overall 0.2453and & Qin 1,222,992 kJ/s ⎛ 1 ton ⎞ m coal = & = ⎜ ⎟ = 0.04174 tons/s = 150.3 tons/h C coal 29,300 kJ/kg ⎜ 1000 kg ⎟ ⎝ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
14. 14. 10-1410-24 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as theworking fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13), h1 = h f @ 0.7 MPa = 88.82 kJ/kg v 1 = v f @ 0.7 MPa = 0.0008331 m 3 /kg T w p ,in = v 1 (P2 − P1 ) ( ) ⎛ 1 kJ = 0.0008331 m 3 /kg (1400 − 700 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎞ ⎟ ⎟ 1.4 MPa 3 ⎝ ⎠ qin = 0.58 kJ/kg 2 R-134a h2 = h1 + w p ,in = 88.82 + 0.58 = 89.40 kJ/kg 0.7 MPa 1 qout 4 P3 = 1.4 MPa ⎫ h3 = h g @ 1.4 MPa = 276.12 kJ/kg s ⎬ sat.vapor ⎭ s 3 = s g @ 1.4 MPa = 0.9105 kJ/kg ⋅ K P4 = 0.7 MPa ⎫ s 4 − s f 0.9105 − 0.33230 ⎬ x4 = = = 0.9839 s 4 = s3 ⎭ s fg 0.58763 h4 = h f + x 4 h fg = 88.82 + (0.9839)(176.21) = 262.20 kJ/kgThus , q in = h3 − h2 = 276.12 − 89.40 = 186.72 kJ/kg q out = h4 − h1 = 262.20 − 88.82 = 173.38 kJ/kg wnet = q in − q out = 186.72 − 173.38 = 13.34 kJ/kgand wnet 13.34 kJ/kg η th = = = 7.1% q in 186.72 kJ/kg(b) Wnet = mwnet = (3 kg/s )(13.34 kJ/kg ) = 40.02 kW & &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
15. 15. 10-1510-25 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits.The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the coolingwater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg v 1 = v f @ 10 kPa = 0.00101 m 3 /kg T w p ,in = v 1 (P2 − P1 ) 3 ( ) ⎛ 1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎞ ⎟ ⎟ 7 MPa ⎝ ⎠ qin 2 = 7.06 kJ/kg 10 kPa h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg 1 4 qout P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg s ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K P4 = 10 kPa ⎫ s 4 − s f 6.8000 − 0.6492 ⎬ x4 = = = 0.8201 s 4 = s3 ⎭ s fg 7.4996 h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kgThus, q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kgand wnet 1250.7 kJ/kg η th = = = 38.9% q in 3212.5 kJ/kg & Wnet 45,000 kJ/s(b) m= & = = 36.0 kg/s wnet 1250.7 kJ/kg(c) The rate of heat rejection to the cooling water and its temperature rise are Qout = mq out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s & & & Qout 70,586 kJ/s ΔTcooling water = = = 8.4°C (mc) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C ) &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
16. 16. 10-1610-26 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressurelimits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of thecooling water are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg T v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 ) / η p 3 ( 3 ) ⎛ 1 kJ = 0.00101 m /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎞ ⎟ / (0.87 ) ⎟ 2 7 MPa qin ⎝ ⎠ 2 = 8.11 kJ/kg h2 = h1 + w p ,in = 191.81 + 8.11 = 199.92 kJ/kg 10 kPa 1 qout 4 4 P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ s T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K P4 = 10 kPa ⎫ s 4 − s f 6.8000 − 0.6492 ⎬ x4 = = = 0.8201 s 4 = s3 ⎭ s fg 7.4996 h4 s = h f + x 4 h fg = 191.81 + (0.820)(2392.1) = 2153.6 kJ/kg h3 − h4 ηT = ⎯→ h4 = h3 − ηT (h3 − h4 s ) ⎯ h3 − h4 s = 3411.4 − (0.87 )(3411.4 − 2153.6) = 2317.1 kJ/kgThus, qin = h3 − h2 = 3411.4 − 199.92 = 3211.5 kJ/kg qout = h4 − h1 = 2317.1 − 191.81 = 2125.3 kJ/kg wnet = qin − qout = 3211.5 − 2125.3 = 1086.2 kJ/kgand wnet 1086.2 kJ/kg η th = = = 33.8% q in 3211.5 kJ/kg & Wnet 45,000 kJ/s(b) m= & = = 41.43 kg/s wnet 1086.2 kJ/kg(c) The rate of heat rejection to the cooling water and its temperature rise are Qout = mq out = (41.43 kg/s )(2125.3 kJ/kg ) = 88,051 kJ/s & & & Qout 88,051 kJ/s ΔTcooling water = = = 10.5°C (mc) cooling water & (2000 kg/s )(4.18 kJ/kg ⋅ °C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
17. 17. 10-1710-27 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankinecycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg v1 = v f @ 20 kPa = 0.001017 m 3 /kg T Rankine cycle w p ,in = v1 (P2 − P ) 1 ( ) ⎛ 1 kJ ⎞ = 0.001017 m3 /kg (10,000 − 20 ) kPa ⎜ ⎟ ⎜ 1 kPa ⋅ m 3 ⎟ 3 = 10.15 kJ/kg ⎝ ⎠ h2 = h1 + w p ,in = 251.42 + 10.15 = 261.57 kJ/kg 2 P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg 1 4 ⎬ x3 = 1 ⎭ s 3 = 5.6159 kJ/kg ⋅ K s P4 = 20 kPa ⎫ s4 − s f 5.6159 − 0.8320 ⎬ x4 = = = 0.6761 s 4 = s3 ⎭ s fg 7.0752 h4 = h f + x 4 h fg = 251.42 + (0.6761)(2357.5) = 1845.3 kJ/kg q in = h3 − h2 = 2725.5 − 261.57 = 2463.9 kJ/kg q out = h4 − h1 = 1845.3 − 251.42 = 1594.0 kJ/kg wnet = q in − q out = 2463.9 − 1594.0 = 869.9 kJ/kg q out 1594.0 η th = 1 − = 1− = 0.353 q in 2463.9(b) Carnot Cycle analysis: Carnot T P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg cycle ⎬ x3 = 1 ⎭ T3 = 311.0 °C T2 = T3 = 311.0 °C ⎫ h2 = 1407.8 kJ/kg 2 3 ⎬ x2 = 0 ⎭ s 2 = 3.3603 kJ/kg ⋅ K s1 − s f 3.3603 − 0.8320 x1 = = = 0.3574 1 4 P1 = 20 kPa ⎫ s fg 7.0752 ⎬h = h +x h s s1 = s 2 ⎭ 1 f 1 fg = 251.42 + (0.3574)(2357.5) = 1093.9 kJ/kg q in = h3 − h2 = 2725.5 − 1407.8 = 1317.7 kJ/kg q out = h4 − h1 = 1845.3 − 1093.9 = 751.4 kJ/kg wnet = q in − q out = 1317.7 − 752.3 = 565.4 kJ/kg q out 751.4 η th = 1 − = 1− = 0.430 q in 1317.7PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
18. 18. 10-1810-28 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Themass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output fromthe turbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water forgeothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ 3P2 = 500 kPa ⎫ h2 − h f ⎬x2 = steamh2 = h1 = 990.14 kJ/kg ⎭ h fg turbine 990.14 − 640.09 separator 4 = 2108 = 0.1661 2 condenserThe mass flow rate of steam 6through the turbine is 5 Flash m3 = x 2 m1 & & chamber = (0.1661)(230 kg/s) = 38.20 kg/s production 1 reinjection well well(b) Turbine: P3 = 500 kPa ⎫ h3 = 2748.1 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 6.8207 kJ/kg ⋅ K P4 = 10 kPa ⎫ ⎬h4 s = 2160.3 kJ/kg s 4 = s3 ⎭ P4 = 10 kPa ⎫ ⎬h4 = h f + x 4 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg x 4 = 0.90 ⎭ h3 − h4 2748.1 − 2344.7 ηT = = = 0.686 h3 − h4 s 2748.1 − 2160.3(c) The power output from the turbine is & WT,out = m3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW &(d) We use saturated liquid state at the standard temperature for dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ & E in = m1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW & & WT,out 15,410 η th = = = 0.0757 = 7.6% & E in 203,622PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
19. 19. 10-1910-29 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Thetemperature of the steam at the exit of the second flash chamber, the power produced from the secondturbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kgx1 = 0 ⎭P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661h2 = h1 = 990.14 kJ/kg ⎭ 3m3 = x2 m1 = (0.1661)(230 kg/s) = 38.20 kg/s& &m6 = m1 − m3 = 230 − 0.1661 = 191.80 kg/s& & & steam 8 turbineP3 = 500 kPa ⎫ separator 4 ⎬ h3 = 2748.1 kJ/kgx3 = 1 ⎭ 2P4 = 10 kPa ⎫ 7 6 condenser ⎬h4 = 2344.7 kJ/kg separatorx 4 = 0.90 ⎭ Flash 5P6 = 500 kPa ⎫ Flash 9 ⎬ h6 = 640.09 kJ/kg chamber chamberx6 = 0 ⎭P7 = 150 kPa ⎫ T7 = 111.35 °C ⎬ 1 reinjectionh7 = h6 ⎭ x 7 = 0.0777 production wellP8 = 150 kPa ⎫ well ⎬h8 = 2693.1 kJ/kgx8 = 1 ⎭(b) The mass flow rate at the lower stage of the turbine is m8 = x7 m6 = (0.0777)(191.80 kg/s) = 14.90 kg/s & &The power outputs from the high and low pressure stages of the turbine are & WT1, out = m3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW & & WT2, out = m8 (h8 − h4 ) = (14.90 kJ/kg)(2693.1 − 2344.7)kJ/kg = 5191 kW &(c) We use saturated liquid state at the standard temperature for the dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ & Ein = m1 (h1 − h0 ) = (230 kg/s)(990.14 − 104.83)kJ/kg = 203,621 kW & & WT, out 15,410 + 5193 η th = = = 0.101 = 10.1% E& 203,621 inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
20. 20. 10-2010-30 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heatsource. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbineand the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kgx1 = 0 ⎭P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661h2 = h1 = 990.14 kJ/kg ⎭m3 = x2 m1 = (0.1661)(230 kg/s) = 38.20 kg/s& &m6 = m1 − m3 = 230 − 38.20 = 191.80 kg/s& & &P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg 3 steamx3 = 1 ⎭ separator turbineP4 = 10 kPa ⎫ 4 condenser ⎬h4 = 2344.7 kJ/kgx 4 = 0.90 ⎭ air-cooled 5 condenserP6 = 500 kPa ⎫ 6 ⎬ h6 = 640.09 kJ/kg 9x6 = 0 ⎭ isobutane 1 2 turbineT7 = 90°C ⎫ BINARY ⎬ h7 = 377.04 kJ/kg CYCLEx7 = 0 ⎭ 8 pumpThe isobutane propertiesare obtained from EES: heat exchanger 1 flash chamberP8 = 3250 kPa ⎫ 7 ⎬ h8 = 755.05 kJ/kg 1T8 = 145°C ⎭ production reinjectionP9 = 400 kPa ⎫ ⎬ h9 = 691.01 kJ/kg well wellT9 = 80°C ⎭P10 = 400 kPa ⎫ h10 = 270.83 kJ/kg ⎬ 3x10 = 0 ⎭ v 10 = 0.001839 m /kg w p ,in = v10 (P − P ) / η p 11 10 ( ) ⎛ 1 kJ ⎞ = 0.001819 m 3/kg (3250 − 400 ) kPa ⎜ ⎟ ⎜ 1 kPa ⋅ m3 ⎟ / 0.90 = 5.82 kJ/kg. ⎝ ⎠ h11 = h10 + w p ,in = 270.83 + 5.82 = 276.65 kJ/kgAn energy balance on the heat exchanger gives m 6 (h6 − h7 ) = miso (h8 − h11 ) & & (191.81 kg/s)(640.09 - 377.04)kJ/kg = m iso (755.05 - 276.65)kJ/kg ⎯ & ⎯→ miso = 105.46 kg/s &(b) The power outputs from the steam turbine and the binary cycle are & WT,steam = m3 (h3 − h4 ) = (38.19 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
21. 21. 10-21 & WT,iso = miso (h8 − h9 ) = (105.46 kJ/kg)(755.05 − 691.01)kJ/kg = 6753 kW & & & W net,binary = WT,iso − miso w p ,in = 6753 − (105.46 kg/s)(5.82 kJ/kg ) = 6139 kW &(c) The thermal efficiencies of the binary cycle and the combined plant are & Qin,binary = miso (h8 − h11 ) = (105.46 kJ/kg)(755.05 − 276.65)kJ/kg = 50,454 kW & & W net, binary 6139 η th,binary = = = 0.122 = 12.2% & Qin, binary 50,454 T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ & E in = m1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW & & & WT,steam + Wnet, binary 15,410 + 6139 η th, plant = = = 0.106 = 10.6% & E 203,622 inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
22. 22. 10-22The Reheat Rankine Cycle10-31C The pump work remains the same, the moisture content decreases, everything else increases.10-32C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown inFigure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressureturbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressureturbine to state 7. Increasing the pressure for reheating increases the average temperature for heat additionmakes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus thereheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as thereheating pressure increases, the amount of condensation increases during the expansion process in the low-pressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisturecontent of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds toquality in the range of 85 to 90%. SteamIAPWS 900 800 700 1 3 6 600 2 T [K] 6000 kPa 4000 kPa 500 500 kPa 5 400 20 kPa 0.2 0.4 0.6 0.8 4 300 7 200 0 20 40 60 80 100 120 140 160 180 s [kJ/kmol-K]10-33C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the averagetemperature at which heat is added will be higher in this case.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
23. 23. 10-2310-34 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankinecycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg T v1 = v f @ 20 kPa = 0.001017 m3 /kg 3 5 w p ,in = v1 (P2 − P ) 1 ( ) ⎛ 1 kJ ⎞ = 0.001017 m 3 /kg (8000 − 20 kPa )⎜ ⎟ ⎜ 1 kPa ⋅ m3 ⎟ 8 MPa = 8.12 kJ/kg ⎝ ⎠ 4 h2 = h1 + w p ,in = 251.42 + 8.12 = 259.54 kJ/kg 2 P3 = 8 MPa ⎫ h3 = 3399.5 kJ/kg 20 kPa ⎬ 1 6 T3 = 500°C ⎭ s3 = 6.7266 kJ/kg ⋅ K s P4 = 3 MPa ⎫ ⎬ h4 = 3105.1 kJ/kg s4 = s3 ⎭ P5 = 3 MPa ⎫ h5 = 3457.2 kJ/kg ⎬ T5 = 500°C ⎭ s5 = 7.2359 kJ/kg ⋅ K s6 − s f 7.2359 − 0.8320 P6 = 20 kPa ⎫ x6 = = = 0.9051 ⎬ s fg 7.0752 s6 = s5 ⎭ h6 = h f + x6 h fg = 251.42 + (0.9051)(2357.5) = 2385.2 kJ/kgThe turbine work output and the thermal efficiency are determined from wT,out = (h3 − h4 ) + (h5 − h6 ) = 3399.5 − 3105.1 + 3457.2 − 2385.2 = 1366.4 kJ/kgand q in = (h3 − h2 ) + (h5 − h4 ) = 3399.5 − 259.54 + 3457.2 − 3105.1 = 3492.0 kJ/kg wnet = wT ,out − w p ,in = 1366.4 − 8.12 = 1358.3 kJ/kgThus, wnet 1358.3 kJ/kg η th = = = 38.9% q in 3492.5 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
24. 24. 10-2410-35 EES Problem 10-34 is reconsidered. The problem is to be solved by the diagram window data entryfeature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam qualityat the low-pressure turbine exit Also, the T-s diagram is to be plotted.Analysis The problem is solved using EES, and the solution is given below."Input Data - from diagram window"{P[6] = 20 [kPa]P[3] = 8000 [kPa]T[3] = 500 [C]P[4] = 3000 [kPa]T[5] = 500 [C]Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"}"Pump analysis"function x6\$(x6) "this function returns a string to indicate the state of steam at point 6" x6\$= if (x6>1) then x6\$=(superheated) if (x6<0) then x6\$=(subcooled)endFluid\$=Steam_IAPWSP[1] = P[6]P[2]=P[3]x[1]=0 "Satd liquid"h[1]=enthalpy(Fluid\$,P=P[1],x=x[1])v[1]=volume(Fluid\$,P=P[1],x=x[1])s[1]=entropy(Fluid\$,P=P[1],x=x[1])T[1]=temperature(Fluid\$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid\$,P=P[2],h=h[2])s[2]=entropy(Fluid\$,P=P[2],h=h[2])T[2]=temperature(Fluid\$,P=P[2],h=h[2])"High Pressure Turbine analysis"h[3]=enthalpy(Fluid\$,T=T[3],P=P[3])s[3]=entropy(Fluid\$,T=T[3],P=P[3])v[3]=volume(Fluid\$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid\$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid\$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid\$,P=P[4],h=h[4])s[4]=entropy(Fluid\$,T=T[4],P=P[4])v[4]=volume(Fluid\$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine""Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid\$,T=T[5],P=P[5])h[5]=enthalpy(Fluid\$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid\$,s=s_s[6],P=P[6])PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
25. 25. 10-25Ts[6]=temperature(Fluid\$,s=s_s[6],P=P[6])vs[6]=volume(Fluid\$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=QUALITY(Fluid\$,h=h[6],P=P[6])"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler""Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid\$,h=h[6],P=P[6])s[6]=entropy(Fluid\$,h=h[6],P=P[6])x6s\$=x6\$(x[6])"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in 7 00 Id e a l R a n k in e c yc le w ith re h e a t 6 00 5 00 3 5 4 00 T [C] 4 3 00 8 0 0 0 kP a 3 0 0 0 kP a 2 00 1 00 1 ,2 2 0 kP a 6 0 0.0 1 .1 2 .2 3.3 4 .4 5 .5 6.6 7 .7 8.8 9.9 1 1.0 s [k J /k g -K ]SOLUTIONEff=0.389 Eta_p=1 Eta_t=1Fluid\$=Steam_IAPWS h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg]h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg]h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg]P[1]=20 [kPa] P[2]=8000 [kPa] P[3]=8000 [kPa]P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa]Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K]s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K]s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K]s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C]T[3]=500 [C] T[4]=345.2 [C] T[5]=500 [C]T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C]v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg]v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg]W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg]W_t_lp=1072 [kJ/kg] x6s\$= x[1]=0x[6]=0.9051PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
26. 26. 10-2610-36E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition andrejection, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E or EES), h1 = h f @ 10 psia = 161.25 Btu/lbm T v 1 = v f @ 10 psia = 0.01659 ft 3 /lbm w p,in = v 1 ( P2 − P1 ) 3 5 ⎛ 1 Btu ⎞ 600 psia = (0.01659 ft 3 /lbm)(600 − 10)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ = 1.81 Btu/lbm ⎝ ⎠ 200 psia h2 = h1 + wp,in = 161.25 + 1.81 = 163.06 Btu/lbm 2 4 P3 = 600 psia ⎫ h3 = 1289.9 Btu/lbm 10 psia ⎬ 1 6 T3 = 600°F ⎭ s 3 = 1.5325 Btu/lbm ⋅ R s 4 − s f 1.5325 − 0.54379 s P4 = 200 psia ⎫ x 4 = = = 0.9865 ⎬ s fg 1.00219 s 4 = s3 ⎭ h = h + x h = 355.46 + (0.9865)(843.33) = 1187.5 Btu/lbm 4 f 4 fg P5 = 200 psia ⎫ h5 = 1322.3 Btu/lbm ⎬ T5 = 600°F ⎭ s 5 = 1.6771 Btu/lbm ⋅ R s 4 − s f 1.6771 − 0.28362 P6 = 10 psia ⎫ x 6 = = = 0.9266 ⎬ s fg 1.50391 s6 = s5 ⎭ h = h + x h = 161.25 + (0.9266)(981.82) = 1071.0 Btu/lbm 6 f 6 fgThus, q in = (h3 − h2 ) + (h5 − h4 ) = 1289.9 − 163.06 + 1322.3 − 1187.5 = 1261.7 Btu/lbm q out = h6 − h1 = 1071.0 − 161.25 = 909.7 Btu/lbm w net = q in − q out = 1261.7 − 909.8 = 352.0 Btu/lbmThe mass flow rate of steam in the cycle is determined from W& 5000 kJ/s ⎛ 0.94782 Btu ⎞ & & ⎯→ m = net = W net = mwnet ⎯ & ⎜ ⎟ = 13.47 lbm/s wnet 352.0 Btu/lbm ⎝ 1 kJ ⎠The rates of heat addition and rejection are & Qin = mq in = (13.47 lbm/s)(1261.7 Btu/lbm) = 16,995 Btu/s & & Qout = mq out = (13.47 lbm/s)(909.7 Btu/lbm) = 12,250 Btu/s &and the thermal efficiency of the cycle is & W net 5000 kJ/s ⎛ 0.94782 Btu ⎞ η th = = ⎜ ⎟ = 0.2790 & Qin 16,990 Btu/s ⎝ 1 kJ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
27. 27. 10-2710-37E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition andrejection, and the thermal efficiency of the cycle are to be determined for a reheat pressure of 100 psia.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E or EES), h1 = h f @ 10 psia = 161.25 Btu/lbm T v 1 = v f @ 6 psia = 0.01659 ft 3 /lbm w p,in = v 1 ( P2 − P1 ) 3 5 ⎛ 1 Btu ⎞ 600 psia = (0.01659 ft 3 /lbm)(600 − 10)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ = 1.81 Btu/lbm ⎝ ⎠ 100 psia h2 = h1 + wp,in = 161.25 + 1.81 = 163.06 Btu/lbm 2 4 10 psia P3 = 600 psia ⎫ h3 = 1289.9 Btu/lbm ⎬ 1 6 T3 = 600°F ⎭ s 3 = 1.5325 Btu/lbm ⋅ R s s 4 − s f 1.5325 − 0.47427 P4 = 100 psia ⎫ x 4 = = = 0.9374 ⎬ s fg 1.12888 s 4 = s3 ⎭ h = h + x h = 298.51 + (0.9374)(888.99) = 1131.9 Btu/lbm 4 f 4 fg P5 = 100 psia ⎫ h5 = 1329.4 Btu/lbm ⎬ T5 = 600°F ⎭ s 5 = 1.7586 Btu/lbm ⋅ R s 6 − s f 1.7586 − 0.28362 P6 = 10 psia ⎫ x 6 = = = 0.9808 ⎬ s fg 1.50391 s6 = s5 ⎭ h = h + x h = 161.25 + (0.9808)(981.82) = 1124.2 Btu/lbm 6 f 6 fgThus, q in = (h3 − h2 ) + (h5 − h4 ) = 1289.9 − 163.07 + 1329.4 − 1131.9 = 1324.4 Btu/lbm q out = h6 − h1 = 1124.2 − 161.25 = 962.9 Btu/lbm w net = q in − q out = 1324.4 − 962.9 = 361.5 Btu/lbmThe mass flow rate of steam in the cycle is determined from W& 5000 kJ/s ⎛ 0.94782 Btu ⎞ & & ⎯→ m = net = W net = mwnet ⎯ & ⎜ ⎟ = 13.11 lbm/s wnet 361.5 Btu/lbm ⎝ 1 kJ ⎠The rates of heat addition and rejection are & Qin = mq in = (13.11 lbm/s)(1324.4 Btu/lbm) = 17,360 Btu/s & & Qout = mq out = (13.11 lbm/s)(962.9 Btu/lbm) = 12,620 Btu/s &and the thermal efficiency of the cycle is & W net 5000 kJ/s ⎛ 0.94782 Btu ⎞ η th = = ⎜ ⎟ = 0.2729 & Qin 17,360 Btu/s ⎝ 1 kJ ⎠Discussion The thermal efficiency for 200 psia reheat pressure was determined in the previous problem tobe 0.2790. Thus, operating the reheater at 100 psia causes a slight decrease in the thermal efficiency.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
28. 28. 10-2810-38 An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inletof both turbines, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kineticand potential energy changes are negligible. TAnalysis From the steam tables (Tables A-4, A-5, and A-6), 3 5 h1 = h f @ 10 kPa = 191.81 kJ/kg 4 MPa v 1 = v f @ 10 kPa = 0.001010 m 3 /kg 500 kPa w p,in = v 1 ( P2 − P1 ) 2 ⎛ 1 kJ ⎞ 4 = (0.001010 m 3 /kg )(4000 − 10)kPa ⎜ ⎟ 10 kPa ⎝ 1 kPa ⋅ m 3 ⎠ = 4.03 kJ/kg 1 6 h2 = h1 + wp,in = 191.81 + 4.03 = 195.84 kJ/kg s P4 = 500 kPa ⎫ h4 = h f + x 4 h fg = 640.09 + (0.90)(2108.0) = 2537.3 kJ/kg ⎬ x 4 = 0.90 ⎭ s 4 = s f + x 4 s fg = 1.8604 + (0.90)(4.9603) = 6.3247 kJ/kg ⋅ K P3 = 4000 kPa ⎫ h3 = 2939.4 kJ/kg ⎬ s3 = s 4 ⎭ T3 = 292.2 °C P6 = 10 kPa ⎫ h6 = h f + x 6 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg ⎬ x 6 = 0.90 ⎭ s 6 = s f + x 6 s fg = 0.6492 + (0.90)(7.4996) = 7.3989 kJ/kg ⋅ K P5 = 500 kPa ⎫ h5 = 3029.2 kJ/kg ⎬ s5 = s6 ⎭ T5 = 282.9 °CThus, q in = ( h3 − h2 ) + ( h5 − h4 ) = 2939.4 − 195.84 + 3029.2 − 2537.3 = 3235.4 kJ/kg q out = h6 − h1 = 2344.7 − 191.81 = 2152.9 kJ/kgand q out 2152.9 η th = 1 − = 1− = 0.335 q in 3235.4PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
29. 29. 10-2910-39 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiencyof the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 50 kPa = 340.54 kJ/kg v 1 = v f @ 50 kPa = 0.001030 m 3 /kg T w p,in = v 1 ( P2 − P1 ) 3 ⎛ 1 kJ ⎞ = (0.001030 m 3 /kg )(17500 − 50)kPa ⎜ ⎟ 17.5MPa 5 ⎝ 1 kPa ⋅ m 3 ⎠ = 17.97 kJ/kg 2 MPa h2 = h1 + wp,in = 340.54 + 17.97 = 358.51 kJ/kg 4 2 P3 = 17,500 kPa ⎫ h3 = 3423.6 kJ/kg 50 kPa ⎬ 1 6 T3 = 550°C ⎭ s 3 = 6.4266 kJ/kg ⋅ K P4 = 2000 kPa ⎫ s ⎬ h4 = 2841.5 kJ/kg s 4 = s3 ⎭ P5 = 2000 kPa ⎫ h5 = 3024.2 kJ/kg ⎬ T5 = 300°C ⎭ s 5 = 6.7684 kJ/kg ⋅ K s6 − s f 6.7684 − 1.0912 P6 = 50 kPa ⎫ x 6 = = = 0.8732 ⎬ s fg 6.5019 s6 = s5 ⎭ h = h + x h = 340.54 + (0.8732)(2304.7) = 2352.9 kJ/kg 6 f 6 fgThus, q in = (h3 − h2 ) + ( h5 − h4 ) = 3423.6 − 358.51 + 3024.2 − 2841.5 = 3247.8 kJ/kg q out = h6 − h1 = 2352.9 − 340.54 = 2012.4 kJ/kgand q out 2012.4 η th = 1 − = 1− = 0.380 q in 3247.8PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
30. 30. 10-3010-40 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiencyof the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 50 kPa = 340.54 kJ/kg v 1 = v f @ 50 kPa = 0.001030 m 3 /kg T w p,in = v 1 ( P2 − P1 ) 3 5 3 ⎛ 1 kJ ⎞ 17.5MP = (0.001030 m /kg )(17500 − 50)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 17.97 kJ/kg h2 = h1 + wp,in = 340.54 + 17.97 = 358.52 kJ/kg 2 MPa 2 4 P3 = 17,500 kPa ⎫ h3 = 3423.6 kJ/kg 50 kPa ⎬ 1 6 T3 = 550°C ⎭ s 3 = 6.4266 kJ/kg ⋅ K P4 = 2000 kPa ⎫ s ⎬ h4 = 2841.5 kJ/kg s 4 = s3 ⎭ P5 = 2000 kPa ⎫ h5 = 3579.0 kJ/kg ⎬ T5 = 550°C ⎭ s 5 = 7.5725 kJ/kg ⋅ K s6 − s f 7.5725 − 1.0912 P6 = 50 kPa ⎫ x 6 = = = 0.9968 ⎬ s fg 6.5019 s6 = s5 ⎭ h = h + x h = 340.54 + (0.9968)(2304.7) = 2638.0 kJ/kg 6 f 6 fgThus, q in = (h3 − h2 ) + (h5 − h4 ) = 3423.6 − 358.52 + 3579.0 − 2841.5 = 3802.6 kJ/kg q out = h6 − h1 = 2638.0 − 340.54 = 2297.4 kJ/kgand q out 2297.4 η th = 1 − = 1− = 0.396 q in 3802.6The thermal efficiency was determined to be 0.380 when the temperature at the inlet of low-pressureturbine was 300°C. When this temperature is increased to 550°C, the thermal efficiency becomes 0.396.This corresponding to a percentage increase of 4.2% in thermal efficiency.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
31. 31. 10-3110-41 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressurelimits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler,and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = hsat @ 10 kPa = 191.81 kJ/kg T 3 v 1 = v sat @ 10 kPa = 0.00101 m /kg 3 5 w p ,in = v 1 (P2 − P1 ) 15 ( ) ⎛ 1 kJ = 0.00101 m 3 /kg (15,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎞ ⎟ ⎟ 4 = 15.14 kJ/kg ⎝ ⎠ 2 h2 = h1 + w p ,in = 191.81 + 15.14 = 206.95 kJ/kg 10 kPa P3 = 15 MPa ⎫ h3 = 3310.8 kJ/kg 1 6 ⎬ s T3 = 500°C ⎭ s 3 = 6.3480 kJ/kg ⋅ K P6 = 10 kPa ⎫ h6 = h f + x 6 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg ⎬ s6 = s5 ⎭ s 6 = s f + x 6 s fg = 0.6492 + (0.90)(7.4996 ) = 7.3988 kJ/kg ⋅ K T5 = 500°C ⎫ P5 = 2150 kPa (the reheat pressure) ⎬ s5 = s6 ⎭ h5 = 3466.61 kJ/kg P4 = 2.15 MPa ⎫ ⎬ h4 = 2817.2 kJ/kg s 4 = s3 ⎭(b) The rate of heat supply is Qin = m[(h3 − h2 ) + (h5 − h4 )] & & = (12 kg/s )(3310.8 − 206.95 + 3466.61 − 2817.2)kJ/kg = 45,039 kW(c) The thermal efficiency is determined from Qout = m(h6 − h1 ) = (12 kJ/s )(2344.7 − 191.81)kJ/kg = 25,835 kJ/s & &Thus, & Qout 25,834 kJ/s η th = 1 − = 1− = 42.6% Q& 45,039 kJ/s inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
32. 32. 10-3210-42 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure,the net power output, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), P3 = 12.5 MPa ⎫ h3 = 3476.5 kJ/kg ⎬ T3 = 550°C ⎭ s 3 = 6.6317 kJ/kg ⋅ K 3 Turbine P4 = 2 MPa ⎫ Boiler ⎬ h4 s = 2948.1 kJ/kg 4 s 4s = s3 ⎭ 6 h3 − h4 ηT = 5 h3 − h4 s Condenser 2 Pump →h4 = h3 − η T (h3 − h4 s ) 1 = 3476.5 − (0.85)(3476.5 − 2948.1) = 3027.3 kJ/kg T P5 = 2 MPa ⎫ h5 = 3358.2 kJ/kg ⎬ 3 5 T5 = 450°C ⎭ s 5 = 7.2815 kJ/kg ⋅ K 12.5 MPa P6 = ? ⎫ ⎬ h6 = 4s 4 x 6 = 0.95⎭ 2 2s P6 = ? ⎫ ⎬ h6 s = s6 = s5 ⎭ P=? 1 6s 6 h −h ⎯→ h6 = h5 − η T (h5 − h6 s ) s ηT = 5 6 ⎯ h5 − h6 s = 3358.2 − (0.85)(3358.2 − 2948.1) = 3027.3 kJ/kgThe pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EESfrom the above equations: P6 = 9.73 kPa, h6 = 2463.3 kJ/kg,(b) Then, h1 = h f @ 9.73 kPa = 189.57 kJ/kg v1 = v f @ 10 kPa = 0.001010 m3 /kg w p ,in = v1 (P2 − P ) / η p 1 ( ) ⎛ 1 kJ = 0.00101 m3 /kg (12,500 − 9.73 kPa )⎜ ⎞ ⎜ 1 kPa ⋅ m / (0.90 ) ⎟ 3⎟ = 14.02 kJ/kg ⎝ ⎠ h2 = h1 + w p ,in = 189.57 + 14.02 = 203.59 kJ/kgCycle analysis: q in = (h3 − h2 ) + (h5 − h4 ) = 3476.5 − 3027.3 + 3358.2 − 2463.3 = 3603.8 kJ/kg q out = h6 − h1 = 3027.3 − 189.57 = 2273.7 kJ/kg & W net = m(q in − q out ) = (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW &(c) The thermal efficiency is q 2273.7 kJ/kg η th = 1 − out = 1 − = 0.369 = 36.9% q in 3603.8 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.