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- 1. 10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content 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. The thermal efficiency, the quality at the end of the heat rejection process, 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) We note that TH = Tsat @180 psia = 373.1°F = 833.1 R T TL = Tsat @14.7 psia = 212.0°F = 672.0 R and ηth,C 672.0 R T = 1− L = 1− = 19.3% 833.1 R TH 1 180 psia 2 qin 14.7 psia (b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R, x4 = s4 − s f s fg = 0.53274 − 0.31215 = 0.153 1.44441 4 3 s (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/lbm Thus, q in = h2 − h1 = 1112.2 − 346.14 = 766.0 Btu/lbm and, wnet = η th q in = (0.1934)(766.0 Btu/lbm) = 148.1 Btu/lbm 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.
- 2. 10-2 10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes η th,C = 1 − TL 333.1 K = 1− = 0.3632 = 36.3% 523 K TH (b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250oC = 1715.3 kJ/kg T 250°C 2 1 qin Thus, q out = q L = 20 kPa ⎛ 333.1 K ⎞ TL q in = ⎜ ⎟ ⎜ 523 K ⎟(1715.3 kJ/kg ) = 1092.3 kJ/kg TH ⎠ ⎝ 4 qout 3 s (c) The net work output of this cycle is wnet = η th q in = (0.3632 )(1715.3 kJ/kg ) = 623.0 kJ/kg 10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes η th, C = 1 − TL 318.8 K =1− = 39.04% 523 K TH (b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250°C = 1715.3 kJ/kg T 250°C 2 1 qin Thus, q out = q L = ⎛ 318.8 K ⎞ TL q in = ⎜ ⎜ 523 K ⎟(1715.3 kJ/kg ) = 1045.6 kJ/kg ⎟ TH ⎝ ⎠ 10 kPa 4 qout 3 (c) The net work output of this cycle is wnet = η th q in = (0.3904)(1715.3 kJ/kg ) = 669.7 kJ/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. s
- 3. 10-3 10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 η th, C = 1 − TL 60 + 273 K = 1− = 46.5% 350 + 273 K TH T (b) Note that s2 = s3 = sf + x3sfg 350°C 1 2 4 3 = 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K Thus, 60°C T2 = 350°C ⎫ ⎬ P2 ≅ 1.40 MPa (Table A-6) s 2 = 7.1368 kJ/kg ⋅ K ⎭ s (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 ⋅ K Thus, wnet = Area = (TH − TL )(s 3 − s 4 ) = (350 − 60)(7.1368 − 1.5390) = 1623 kJ/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.
- 4. 10-4 The Simple Rankine Cycle 10-7C 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 heat rejection 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 involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping. 10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same 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 will adversely 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 than the temperature of the cooling water. 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.
- 5. 10-5 10-15E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, 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), 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 ) ⎛ 1 Btu = (0.01645 ft 3 /lbm)(500 − 6)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 1.50 Btu/lbm h2 = h1 + wp,in = 138.02 + 1.50 = 139.52 Btu/lbm ⎞ ⎟ ⎟ ⎠ 500 psia 2 3 qin 6 psia 1 qout P3 = 500 psia ⎫ h3 = 1630.0 Btu/lbm ⎬ T3 = 1200°F ⎭ s 3 = 1.8075 Btu/lbm ⋅ R 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 fg 4 s Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from & & & WT,out = m(h3 − h4 ) ⎯ ⎯→ m = & WT,out h3 − h4 = 500 kJ/s ⎛ 0.94782 Btu ⎞ ⎜ ⎟ = 0.9300 lbm/s (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 1 and the thermal efficiency of the cycle is η th = 1 − & Qout 913.6 = 1− = 0.341 & 1386 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.
- 6. 10-6 10-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to be determined. 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% (most closely 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 ⋅ K Since the expansion in the turbine is isentropic, P3 = 3000 kPa ⎫ ⎬ h3 = 3115.5 kJ/kg s 3 = s 4 = 6.7440 kJ/kg ⋅ K ⎭ T 3 Other properties are obtained as follows (Tables A-4, A-5, and A-6), h1 = h f @ 30 kPa = 289.27 kJ/kg v 1 = v f @ 30 kPa = 0.001022 m 3 /kg wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001022 m 3 /kg )(3000 − 30)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 3.04 kJ/kg h2 = h1 + wp,in = 289.27 + 3.04 = 292.31 kJ/kg 3 MPa 2 qin 30 kPa 1 qout 4 Thus, q in = h3 − h2 = 3115.5 − 292.31 = 2823.2 kJ/kg q out = h4 − h1 = 2274.3 − 289.27 = 1985.0 kJ/kg and the thermal efficiency of the cycle is η th = 1 − q out 1985.0 = 1− = 0.297 2823.2 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. s
- 7. 10-7 10-17 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. 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 wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001017 m 3 /kg)(4000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 4.05 kJ/kg h2 = h1 + wp,in = 251.42 + 4.05 = 255.47 kJ/kg 4 MPa 2 3 qin 20 kPa 1 qout P3 = 4000 kPa ⎫ h3 = 3906.3 kJ/kg ⎬ T3 = 700°C ⎭ s 3 = 7.6214 kJ/kg ⋅ K 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 fg 4 s The 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 10-8 10-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The turbine inlet temperature 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. T Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), 2500 psia h1 = h f @ 5 psia = 130.18 Btu/lbm v 1 = v f @ 5 psia = 0.01641 ft 3 /lbm wp,in = v 1 ( P2 − P1 ) ⎛ 1 Btu = (0.01641 ft /lbm)(2500 − 5)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.58 Btu/lbm 3 2 ⎞ ⎟ ⎟ ⎠ 3 qin 5 psia 1 qout 4 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 °F Thus, q in = h3 − h2 = 1450.8 − 137.76 = 1313.0 Btu/lbm q out = h4 − h1 = 930.58 − 130.18 = 800.4 Btu/lbm The thermal efficiency of the cycle is η th = 1 − q out 800.4 = 1− = 0.390 1313.0 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. 10-9 10-19 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine, the heat added 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 From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg v 1 = v f @ 100 kPa = 0.001043 m 3 /kg T 15 MPa w p,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001043 m 3 /kg )(15,000 − 100)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 15.54 kJ/kg h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg 2 qin 3 100 kPa 1 qout 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K s4 − s f 5.3108 − 1.3028 P4 = 100 kPa ⎫ x 4 = = = 0.6618 6.0562 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg 4 f 4 fg Thus, 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/kg The thermal efficiency of the cycle is η th = 1 − q out 1494.0 = 1− = 0.314 q in 2177.8 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. s
- 10. 10-10 10-20 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. 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 v 1 = v f @ 100 kPa = 0.001043 m 3 /kg T 15 MPa wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001043 m /kg )(15,000 − 100)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 15.54 kJ/kg h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg 3 2 qin 3 100 kPa 1 qout 4s 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K 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 f 4s 4 s fg s 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 ηT = h3 − h4 2610.8 − 1997.8 = = 0.877 h3 − h4 s 2610.8 − 1911.5 Thus, q in = h3 − h2 = 2610.8 − 433.05 = 2177.8 kJ/kg q out = h4 − h1 = 1997.8 − 417.51 = 1580.3 kJ/kg The thermal efficiency of the cycle is η th = 1 − q out 1580.3 = 1− = 0.274 q in 2177.8 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.
- 11. 10-11 10-21E 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 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), h1 = h f @ 1 psia = 69.72 Btu/lbm v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 ) ⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm T ⎞ ⎟ ⎟ ⎠ 2500 psia 2 qin 1 psia 1 qout P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R 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 ηT = 3 4s 4 h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h4 s Thus, 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/lbm The 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 η th = & W net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3718 & 2549 Btu/s ⎝ 1 kJ Qin ⎠ 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. s
- 12. 10-12 10-22E 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 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), h1 = h f @ 1 psia = 69.72 Btu/lbm T v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 ) ⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm ⎞ ⎟ ⎟ ⎠ 2500 psia 2 qin 1 psia 1 qout P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R 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 ηT = 3 4s 4 s h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h4 s The 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 η th = & W net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3779 & 2508 Btu/s ⎝ 1 kJ Qin ⎠ The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermal efficiency caused by neglecting the pump work is then Error = 0.3779 − 0.3718 × 100 = 1.64% 0.3718 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.
- 13. 10-13 10-23 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply 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 @ 25 kPa = 271.96 kJ/kg v 1 = v f @ 25 kPa = 0.001020 m 3 /kg w p ,in = v 1 (P2 − P1 ) ( ) ⎛ 1 kJ = 0.00102 m /kg (5000 − 25 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 5.07 kJ/kg h2 = h1 + w p ,in = 271.96 + 5.07 = 277.03 kJ/kg 3 T ⎞ ⎟ ⎟ ⎠ P3 = 5 MPa ⎫ h3 = 3317.2 kJ/kg ⎬ T3 = 450°C ⎭ s 3 = 6.8210 kJ/kg ⋅ K 3 2 1 5 MPa · Qin 25 kPa · Qout 4 s 4 − s f 6.8210 − 0.8932 P4 = 25 kPa ⎫ = = 0.8545 ⎬ x4 = s 4 = s3 s fg 6.9370 ⎭ h4 = h f + x 4 h fg = 271.96 + (0.8545)(2345.5) = 2276.2 kJ/kg The 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/kg and η th = 1 − Thus, q out 2004.2 = 1− = 0.3407 q in 3040.2 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 and & W net 300,000 kJ/s & = = 1,222,992 kJ/s Qin = 0.2453 η overall & m coal = & Qin 1,222,992 kJ/s ⎛ 1 ton ⎞ ⎜ ⎟ = 0.04174 tons/s = 150.3 tons/h = 29,300 kJ/kg ⎜ 1000 kg ⎟ C coal ⎝ ⎠ 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. s
- 14. 10-14 10-24 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be determined. 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 ⎝ = 0.58 kJ/kg ⎞ ⎟ ⎟ ⎠ 2 h2 = h1 + w p ,in = 88.82 + 0.58 = 89.40 kJ/kg P3 = 1.4 MPa ⎫ h3 = h g @ 1.4 MPa = 276.12 kJ/kg ⎬ sat.vapor ⎭ s 3 = s g @ 1.4 MPa = 0.9105 kJ/kg ⋅ K 1.4 MPa qin R-134a 3 0.7 MPa 1 qout 4 s s 4 − s f 0.9105 − 0.33230 P4 = 0.7 MPa ⎫ = = 0.9839 ⎬ x4 = s 4 = s3 s fg 0.58763 ⎭ h4 = h f + x 4 h fg = 88.82 + (0.9839)(176.21) = 262.20 kJ/kg Thus , 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/kg and η th = (b) wnet 13.34 kJ/kg = = 7.1% q in 186.72 kJ/kg & & 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 10-15 10-25 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 cooling 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 v 1 = v f @ 10 kPa = 0.00101 m 3 /kg T w p ,in = v 1 (P2 − P1 ) ( ) ⎛ 1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 7.06 kJ/kg ⎞ ⎟ ⎟ ⎠ h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg 3 7 MPa qin 2 10 kPa 1 qout 4 P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = s 4 = s3 s fg 7.4996 ⎭ h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg Thus, 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/kg and η th = (b) & m= wnet 1250.7 kJ/kg = = 38.9% q in 3212.5 kJ/kg & Wnet 45,000 kJ/s = = 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission. s
- 16. 10-16 10-26 A steam power plant operates on a simple nonideal 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 cooling 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 ( ) ⎛ 1 kJ = 0.00101 m /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 8.11 kJ/kg 3 ⎞ ⎟ / (0.87 ) ⎟ ⎠ h2 = h1 + w p ,in = 191.81 + 8.11 = 199.92 kJ/kg P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K 2 2 7 MPa qin 3 10 kPa 1 qout 4 4 s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = 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 ηT = h3 − h4 ⎯ ⎯→ h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 3411.4 − (0.87 )(3411.4 − 2153.6) = 2317.1 kJ/kg Thus, 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/kg and η th = (b) & m= wnet 1086.2 kJ/kg = = 33.8% q in 3211.5 kJ/kg & 45,000 kJ/s Wnet = = 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 ΔTcooling water = & Qout & (mc) cooling water = 88,051 kJ/s = 10.5°C (2000 kg/s )(4.18 kJ/kg ⋅ °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. s
- 17. 10-17 10-27 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle 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 w p ,in = v1 (P2 − P ) 1 ( Rankine cycle T ) ⎛ 1 kJ ⎞ ⎟ = 0.001017 m3 /kg (10,000 − 20 ) kPa ⎜ ⎜ 1 kPa ⋅ m 3 ⎟ ⎝ ⎠ = 10.15 kJ/kg h2 = h1 + w p ,in = 251.42 + 10.15 = 261.57 kJ/kg P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.6159 kJ/kg ⋅ K 3 2 4 1 s s4 − s f P4 = 20 kPa ⎫ 5.6159 − 0.8320 = = 0.6761 ⎬ x4 = s 4 = s3 7.0752 s fg ⎭ 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 η th = 1 − q out 1594.0 = 1− = 0.353 2463.9 q in (b) Carnot Cycle analysis: P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg ⎬ x3 = 1 ⎭ T3 = 311.0 °C T2 = T3 = 311.0 °C ⎫ h2 = 1407.8 kJ/kg ⎬ x2 = 0 ⎭ s 2 = 3.3603 kJ/kg ⋅ K x1 = s1 − s f = P1 = 20 kPa ⎫ s fg ⎬h = h +x h s1 = s 2 f 1 fg ⎭ 1 3.3603 − 0.8320 = 0.3574 7.0752 T Carnot cycle 2 3 1 4 s = 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 η th = 1 − q out 751.4 = 1− = 0.430 1317.7 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.
- 18. 10-18 10-28 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the 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 for geothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ h2 − h f P2 = 500 kPa ⎫ ⎬x2 = h2 = h1 = 990.14 kJ/kg ⎭ h fg 990.14 − 640.09 = 2108 2 = 0.1661 The mass flow rate of steam through the turbine is separator 4 condenser Flash chamber = (0.1661)(230 kg/s) (b) Turbine: steam turbine 6 & & m3 = x 2 m1 = 38.20 kg/s 3 production well 1 5 reinjection well 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 ⎭ ηT = h3 − h4 2748.1 − 2344.7 = = 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 η th = & WT,out 15,410 = = 0.0757 = 7.6% & 203,622 E 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.
- 19. 10-19 10-29 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second 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 for geothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg ⎭ 3 & & m3 = x2 m1 = (0.1661)(230 kg/s) = 38.20 kg/s & & & m6 = m1 − m3 = 230 − 0.1661 = 191.80 kg/s P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg x3 = 1 ⎭ P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭ steam turbine 8 4 separator 2 P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭ P7 = 150 kPa ⎫ T7 = 111.35 °C ⎬ h7 = h6 ⎭ x 7 = 0.0777 P8 = 150 kPa ⎫ ⎬h8 = 2693.1 kJ/kg x8 = 1 ⎭ 6 Flash chamber 7 separator Flash chamber 1 production well condenser 5 9 reinjection well (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 η th = & WT, out 15,410 + 5193 = = 0.101 = 10.1% & 203,621 E 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.
- 20. 10-20 10-30 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined 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/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = 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 x3 = 1 ⎭ 3 separator steam turbine P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭ condenser 4 6 P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭ 9 1 isobutane turbine 2 T7 = 90°C ⎫ ⎬ h7 = 377.04 kJ/kg x7 = 0 ⎭ BINARY CYCLE 8 The isobutane properties are obtained from EES: pump heat exchanger flash chamber P8 = 3250 kPa ⎫ ⎬ h8 = 755.05 kJ/kg T8 = 145°C ⎭ 1 1 7 production well P9 = 400 kPa ⎫ ⎬ h9 = 691.01 kJ/kg T9 = 80°C ⎭ 5 air-cooled condenser reinjection well P10 = 400 kPa ⎫ h10 = 270.83 kJ/kg ⎬ 3 x10 = 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/kg An 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 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 η th,binary = & W net, binary 6139 = = 0.122 = 12.2% & 50,454 Qin, binary 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 η th, plant = & & WT,steam + Wnet, binary 15,410 + 6139 = = 0.106 = 10.6% & 203,622 E 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.
- 22. 10-22 The Reheat Rankine Cycle 10-31C 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 in Figure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the lowpressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%. SteamIAPWS 900 800 700 T [K ] 1 600 3 6 2 6000 kPa 4000 kPa 500 5 500 kPa 400 0.2 300 0.4 20 kPa 0.6 0.8 4 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 average temperature 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 10-23 10-34 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine cycle 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 v1 = v f @ 20 kPa = 0.001017 m3 /kg w p ,in = v1 (P2 − P ) 1 ( T 3 ) ⎛ 1 kJ ⎞ ⎟ = 0.001017 m 3 /kg (8000 − 20 kPa )⎜ ⎜ 1 kPa ⋅ m3 ⎟ ⎝ ⎠ = 8.12 kJ/kg h2 = h1 + w p ,in = 251.42 + 8.12 = 259.54 kJ/kg P3 = 8 MPa ⎫ h3 = 3399.5 kJ/kg ⎬ T3 = 500°C ⎭ s3 = 6.7266 kJ/kg ⋅ K 5 8 MPa 4 2 20 kPa 1 6 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 = = 0.9051 P6 = 20 kPa ⎫ x6 = s fg 7.0752 ⎬ s6 = s5 ⎭ h6 = h f + x6 h fg = 251.42 + (0.9051)(2357.5) = 2385.2 kJ/kg The 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/kg and 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/kg Thus, η th = wnet 1358.3 kJ/kg = = 38.9% 3492.5 kJ/kg 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.
- 24. 10-24 10-35 EES Problem 10-34 is reconsidered. The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at 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)' end Fluid$='Steam_IAPWS' P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,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$,s=s_s[6],P=P[6]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 10-25 Ts[6]=temperature(Fluid$,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_p Eff=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 T [C] 4 00 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 ] SOLUTION Eff=0.389 Fluid$='Steam_IAPWS' h[3]=3400 [kJ/kg] h[6]=2385 [kJ/kg] P[1]=20 [kPa] P[4]=3000 [kPa] Q_in=3493 [kJ/kg] s[2]=0.8321 [kJ/kg-K] s[5]=7.236 [kJ/kg-K] s_s[6]=7.236 [kJ/kg-K] T[3]=500 [C] T[6]=60.06 [C] v[1]=0.001017 [m^3/kg] v[4]=0.08968 [m^3/kg] W_p=8.117 [kJ/kg] W_t_lp=1072 [kJ/kg] x[6]=0.9051 Eta_p=1 h[1]=251.4 [kJ/kg] h[4]=3105 [kJ/kg] hs[4]=3105 [kJ/kg] P[2]=8000 [kPa] P[5]=3000 [kPa] Q_out=2134 [kJ/kg] s[3]=6.727 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[4]=345.2 [C] Ts[4]=345.2 [C] v[2]=0.001014 [m^3/kg] vs[6]=6.922 [m^3/kg] W_p_s=8.117 [kJ/kg] x6s$='' Eta_t=1 h[2]=259.5 [kJ/kg] h[5]=3457 [kJ/kg] hs[6]=2385 [kJ/kg] P[3]=8000 [kPa] P[6]=20 [kPa] s[1]=0.832 [kJ/kg-K] s[4]=6.727 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K] T[2]=60.4 [C] T[5]=500 [C] Ts[6]=60.06 [C] v[3]=0.04177 [m^3/kg] W_net=1359 [kJ/kg] W_t_hp=294.8 [kJ/kg] x[1]=0 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.
- 26. 10-26 10-36E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, 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 ) ⎛ 1 Btu = (0.01659 ft 3 /lbm)(600 − 10)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 1.81 Btu/lbm h2 = h1 + wp,in = 161.25 + 1.81 = 163.06 Btu/lbm 3 ⎞ ⎟ ⎟ ⎠ 5 600 psia 2 200 psia 10 psia P3 = 600 psia ⎫ h3 = 1289.9 Btu/lbm ⎬ 1 T3 = 600°F ⎭ s 3 = 1.5325 Btu/lbm ⋅ R s 4 − s f 1.5325 − 0.54379 P4 = 200 psia ⎫ x 4 = = = 0.9865 1.00219 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 355.46 + (0.9865)(843.33) = 1187.5 Btu/lbm 4 f 4 fg 4 6 s 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 1.50391 s fg ⎬ s6 = s5 ⎭ h = h + x h = 161.25 + (0.9266)(981.82) = 1071.0 Btu/lbm 6 f 6 fg Thus, 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/lbm The mass flow rate of steam in the cycle is determined from & W 5000 kJ/s ⎛ 0.94782 Btu ⎞ & & & W net = mwnet ⎯ ⎯→ m = net = ⎜ ⎟ = 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 η th = & W net 5000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.2790 & 16,990 Btu/s ⎝ 1 kJ Qin ⎠ 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.
- 27. 10-27 10-37E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, 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 ) ⎛ 1 Btu = (0.01659 ft 3 /lbm)(600 − 10)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 1.81 Btu/lbm h2 = h1 + wp,in = 161.25 + 1.81 = 163.06 Btu/lbm 3 5 600 psia ⎞ ⎟ ⎟ ⎠ 2 100 psia 10 psia P3 = 600 psia ⎫ h3 = 1289.9 Btu/lbm 1 ⎬ T3 = 600°F ⎭ s 3 = 1.5325 Btu/lbm ⋅ R s 4 − s f 1.5325 − 0.47427 P4 = 100 psia ⎫ x 4 = = = 0.9374 1.12888 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 298.51 + (0.9374)(888.99) = 1131.9 Btu/lbm 4 f 4 fg 4 6 s 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 1.50391 s fg ⎬ s6 = s5 ⎭ h = h + x h = 161.25 + (0.9808)(981.82) = 1124.2 Btu/lbm 6 f 6 fg Thus, 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/lbm The mass flow rate of steam in the cycle is determined from & W 5000 kJ/s ⎛ 0.94782 Btu ⎞ & & & W net = mwnet ⎯ ⎯→ m = net = ⎜ ⎟ = 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 η th = & W net 5000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.2729 & 17,360 Btu/s ⎝ 1 kJ Qin ⎠ Discussion The thermal efficiency for 200 psia reheat pressure was determined in the previous problem to be 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 10-28 10-38 An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inlet of both turbines, 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. T Analysis From the steam tables (Tables A-4, A-5, and A-6), 3 h1 = h f @ 10 kPa = 191.81 kJ/kg v 1 = v f @ 10 kPa = 0.001010 m 3 /kg w p,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001010 m 3 /kg )(4000 − 10)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 4.03 kJ/kg h2 = h1 + wp,in = 191.81 + 4.03 = 195.84 kJ/kg 5 4 MPa 2 500 kPa 10 kPa 1 4 6 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 °C Thus, 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/kg and η th = 1 − q out 2152.9 = 1− = 0.335 3235.4 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.
- 29. 10-29 10-39 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiency of 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 ) ⎛ 1 kJ ⎞ = (0.001030 m 3 /kg )(17500 − 50)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 17.97 kJ/kg h2 = h1 + wp,in = 340.54 + 17.97 = 358.51 kJ/kg P3 = 17,500 kPa ⎫ h3 = 3423.6 kJ/kg ⎬ T3 = 550°C ⎭ s 3 = 6.4266 kJ/kg ⋅ K P4 = 2000 kPa ⎫ ⎬ h4 = 2841.5 kJ/kg s 4 = s3 ⎭ 3 17.5MPa 5 2 MPa 4 2 50 kPa 1 6 s 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 6.5019 s fg ⎬ s6 = s5 ⎭ h = h + x h = 340.54 + (0.8732)(2304.7) = 2352.9 kJ/kg 6 f 6 fg Thus, 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/kg and η th = 1 − q out 2012.4 = 1− = 0.380 3247.8 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.
- 30. 10-30 10-40 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiency of 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 ) ⎛ 1 kJ ⎞ = (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 3 P3 = 17,500 kPa ⎫ h3 = 3423.6 kJ/kg ⎬ T3 = 550°C ⎭ s 3 = 6.4266 kJ/kg ⋅ K P4 = 2000 kPa ⎫ ⎬ h4 = 2841.5 kJ/kg s 4 = s3 ⎭ 5 17.5MP 3 2 2 MPa 50 kPa 1 4 6 s 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 6.5019 s fg ⎬ s6 = s5 ⎭ h = h + x h = 340.54 + (0.9968)(2304.7) = 2638.0 kJ/kg 6 f 6 fg Thus, 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/kg and η th = 1 − q out 2297.4 = 1− = 0.396 3802.6 q in The thermal efficiency was determined to be 0.380 when the temperature at the inlet of low-pressure turbine 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 and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 10-31 10-41 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits 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 w p ,in = v 1 (P2 − P1 ) ( ) ⎛ 1 kJ = 0.00101 m 3 /kg (15,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 15.14 kJ/kg 3 ⎞ ⎟ ⎟ ⎠ 5 15 4 2 h2 = h1 + w p ,in = 191.81 + 15.14 = 206.95 kJ/kg P3 = 15 MPa ⎫ h3 = 3310.8 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.3480 kJ/kg ⋅ K 10 kPa 1 6 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 Thus, & & Qout = m(h6 − h1 ) = (12 kJ/s )(2344.7 − 191.81)kJ/kg = 25,835 kJ/s η th = 1 − & Qout 25,834 kJ/s = 1− = 42.6% & 45,039 kJ/s 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. s
- 32. 10-32 10-42 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 3 ⎭ s 3 = 6.6317 kJ/kg ⋅ K Turbine Boiler P4 = 2 MPa ⎫ 4 ⎬ h4 s = 2948.1 kJ/kg s 4s = s3 ⎭ 6 h3 − h4 ηT = 5 Condenser h3 − h4 s Pump 2 →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 s ηT = 5 6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) h5 − h6 s = 3358.2 − (0.85)(3358.2 − 2948.1) = 3027.3 kJ/kg The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from 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 )⎜ 3⎟ ⎜ 1 kPa ⋅ m / (0.90 ) ⎝ ⎠ = 14.02 kJ/kg h2 = h1 + w p ,in = 189.57 + 14.02 = 203.59 kJ/kg Cycle 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% 3603.8 kJ/kg 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.
- 33. 10-33 Regenerative Rankine Cycle 10-43C Moisture content remains the same, everything else decreases. 10-44C This is a smart idea because we waste little work potential but we save a lot from the heat input. The extracted steam has little work potential left, and most of its energy would be part of the heat rejected anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work output. 10-45C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing. 10-46C Both cycles would have the same efficiency. 10-47C To have the same thermal efficiency as the Carnot cycle, the cycle must receive and reject heat isothermally. Thus the liquid should be brought to the saturated liquid state at the boiler pressure isothermally, and the steam must be a saturated vapor at the turbine inlet. This will require an infinite number of heat exchangers (feedwater heaters), as shown on the T-s diagram. T Boiler inlet qin Boiler exit qout 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.
- 34. 10-34 10-48 Feedwater is heated by steam in a feedwater heater of a regenerative The required mass flow rate of the steam is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4 through A-6 or EES), h1 ≅ hf @ 70°C = 293.07 kJ/kg P2 = 200 kPa ⎫ ⎬ h2 = 2789.7 kJ/kg T2 = 160°C ⎭ h3 = hf @ 200 kPa = 504.71 kJ/kg Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: & & & min − m out = Δmsystem 0 (steady) Water 70°C 200 kPa 10 kg/s =0 & & min = m out & & & m1 + m 2 = m3 Energy balance: & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 4 Rate of change in internal, kinetic, potential, etc. energies =0 1 Steam 200 kPa 160°C 3 2 Water 200 kPa sat. liq. & & E in = E out & & & & & m1 h1 + m 2 h2 = m3 h3 (since Q = W = Δke ≅ Δpe ≅ 0) & & & & m1 h1 + m 2 h2 = (m1 + m 2 )h3 & Solving for m 2 , and substituting gives & & m 2 = m1 h1 − h3 (293.07 − 504.71) kJ/kg = 0.926 kg/s = (10 kg/s) h3 − h2 (504.71 − 2789.7) kJ/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.
- 35. 10-35 10-49E In a regenerative Rankine cycle, the closed feedwater heater with a pump as shown in the figure is arranged so that the water at state 5 is mixed with the water at state 2 to form a feedwater which is a saturated liquid. The mass flow rate of bleed steam required to operate this unit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Steam Properties From the steam tables (Tables A-4E through A-6E), 160 psia 3 400°F P1 = 200 psia ⎫ ⎬ h1 ≅ h f @ 350° F = 321.73 Btu/lbm 2 6 T1 = 350°F ⎭ 1 P3 = 160 psia ⎫ ⎬ h3 = 1218.0 Btu/lbm Feedwater 200 psia T3 = 400°F ⎭ 200 psia sat. liq. 4 5 P6 = 200 psia ⎫ 350°F ⎬ h6 = h f @ 200 psia = 355.46 Btu/lbm 2 lbm/s x6 = 0 ⎭ Analysis We take the entire unit as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 Rate of net energy transfer by heat, work, and mass = & ΔE system 0 (steady) 1442443 4 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & m1 h1 + m3 h3 + m3 wP,in = m 6 h6 & & & & & m1 h1 + m3 h3 + m3 wP,in = (m1 + m3 )h6 & Solving this for m3 , & & m3 = m1 h6 − h1 355.46 − 321.73 = (2 lbm/s) = 0.0782 lbm/s (h3 − h6 ) + wP,in 1218.0 − 355.46 + 0.1344 where wP,in = v 4 ( P5 − P4 ) = v f @ 160 psia ( P5 − P4 ) ⎛ 1 Btu = (0.01815 ft 3 /lbm)(200 − 160) psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ ⎞ ⎟ = 0.1344 Btu/lbm ⎟ ⎠ 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.
- 36. 10-36 10-50E The closed feedwater heater of a regenerative Rankine cycle is to heat feedwater to a saturated liquid. The required mass flow rate of bleed steam is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties From the steam tables (Tables A-4E through A-6E), Steam 300 psia 400°F P1 = 400 psia ⎫ ⎬ h1 ≅ h f @ 370° F = 342.88 Btu/lbm T1 = 370°F ⎭ P2 = 400 psia ⎫ ⎬ h2 = h f @ 400 psia = 424.13 Btu/lbm x2 = 0 ⎭ P3 = 300 psia ⎫ ⎬ h = 1257.9 Btu/lbm T3 = 500°F ⎭ 3 P4 = 300 psia ⎫ ⎬ h4 = h f @ 300 psia = 393.94 Btu/lbm x4 = 0 ⎭ 400 psia Sat. liq. Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as 300 psia Sat. liq. Feedwater 400 psia 370°F Mass balance (for each fluid stream): & & & m in − m out = Δm system 0 (steady) & & & & & & & & = 0 → m in = m out → m1 = m 2 = m fw and m 3 = m 4 = m s Energy balance (for the heat exchanger): & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & & & m1 h1 + m3 h3 = m 2 h2 + m 4 h4 (since Q = W = Δke ≅ Δpe ≅ 0) Combining the two, & & m fw (h2 − h1 ) = m s (h3 − h4 ) & Solving for m s : & ms = h2 − h1 & m fw h3 − h4 Substituting, & ms = 424.13 − 342.88 (1 lbm/s) = 0.0940 lbm/s 1257.9 − 393.94 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.
- 37. 10-37 10-51 The closed feedwater heater of a regenerative Rankine cycle is to heat feedwater to a saturated liquid. The required mass flow rate of bleed steam is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties From the steam tables (Tables A-4 through A-6), P1 = 4000 kPa ⎫ ⎬ h1 ≅ h f @ 200°C = 852.26 kJ/kg T1 = 200°C ⎭ P2 = 4000 kPa ⎫ ⎬ h2 ≅ h f @ 245°C = 1061.5 kJ/kg T2 = 245°C ⎭ Steam 3 MPa x = 0.90 4 MPa 245°C P3 = 3000 kPa ⎫ ⎬ h3 = h f + x 3 h fg x 3 = 0.90 ⎭ = 1008.3 + (0.9)(1794.9) = 2623.7 kJ/kg P4 = 3000 kPa ⎫ ⎬ h4 = h f @ 3000 kPa = 1008.3 kJ/kg x4 = 0 ⎭ 3 MPa Sat. liq. Feedwater 4 MPa 200°C Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance (for each fluid stream): & & & min − m out = Δmsystem 0 (steady) & & & & & & & & = 0 → min = m out → m1 = m 2 = m fw and m3 = m 4 = m s Energy balance (for the heat exchanger): & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & & & m1 h1 + m3 h3 = m 2 h2 + m 4 h4 (since Q = W = Δke ≅ Δpe ≅ 0) Combining the two, & & m fw (h2 − h1 ) = m s (h3 − h4 ) & Solving for m s : & ms = h2 − h1 & m fw h3 − h4 Substituting, & ms = 1061.5 − 852.26 (6 kg/s) = 0.777 kg/s 2623.7 − 1008.3 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.
- 38. 10-38 10-52 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net power output of the power plant 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 T 7 Turbine Boiler 8 10 6 4 9 fwh II fwh I 4 P III 7 6 10 MPa 5 Condenser P II 0.2 MPa 3 y 5 kPa 2 3 2 8 5 0.6 MPa 1 1 PI 1-y 9 1-y-z 10 s (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 5 kPa = 137.75 kJ/kg v 1 = v f @ 5 kPa = 0.001005 m 3 /kg ( ) ⎛ 1 kJ w pI ,in = v 1 (P2 − P1 ) = 0.001005 m 3 /kg (200 − 5 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ h =h +w = 137.75 + 0.20 = 137.95 kJ/kg 2 1 pI ,in ⎞ ⎟ = 0.20 kJ/kg ⎟ ⎠ P3 = 0.2 MPa ⎫ h3 = h f @ 0.2 MPa = 504.71 kJ/kg ⎬ 3 sat.liquid ⎭ v 3 = v f @ 0.2 MPa = 0.001061 m /kg ( ) ⎛ 1 kJ w pII ,in = v 3 (P4 − P3 ) = 0.001061 m 3 /kg (600 − 200 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.42 kJ/kg ⎞ ⎟ ⎟ ⎠ h4 = h3 + w pII ,in = 504.71 + 0.42 = 505.13 kJ/kg P5 = 0.6 MPa ⎫ h5 = h f @ 0.6 MPa = 670.38 kJ/kg ⎬ 3 sat.liquid ⎭ v 5 = v f @ 0.6 MPa = 0.001101 m /kg ( ) ⎛ 1 kJ w pIII ,in = v 5 (P6 − P5 ) = 0.001101 m 3 /kg (10,000 − 600 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 10.35 kJ/kg ⎞ ⎟ ⎟ ⎠ h6 = h5 + w pIII ,in = 670.38 + 10.35 = 680.73 kJ/kg P7 = 10 MPa ⎫ h7 = 3625.8 kJ/kg ⎬ T7 = 600°C ⎭ s 7 = 6.9045 kJ/kg ⋅ K P8 = 0.6 MPa ⎫ ⎬ h8 = 2821.8 kJ/kg s8 = s 7 ⎭ 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.
- 39. 10-39 s9 − s f 6.9045 − 1.5302 = = 0.9602 P9 = 0.2 MPa ⎫ x 9 = s fg 5.5968 ⎬ s9 = s7 ⎭ h9 = h f + x 9 h fg = 504.71 + (0.9602)(2201.6) = 2618.7 kJ/kg s10 − s f 6.9045 − 0.4762 = = 0.8119 P10 = 5 kPa ⎫ x10 = s fg 7.9176 ⎬ s10 = s 7 ⎭ h10 = h f + x10 h fg = 137.75 + (0.8119)(2423.0) = 2105.0 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the & & feedwater heaters. Noting that Q ≅ W ≅ Δke ≅ Δpe ≅ 0 , FWH-2: & & & E in − E out = ΔE system 0 (steady) =0 & & E in = E out & & ∑m h = ∑m h i i e e & & & ⎯ ⎯→ m8 h8 + m 4 h4 = m5 h5 ⎯ ⎯→ yh8 + (1 − y )h4 = 1(h5 ) & & where y is the fraction of steam extracted from the turbine ( = m8 / m5 ). Solving for y, y= h5 − h4 670.38 − 505.13 = = 0.07133 h8 − h4 2821.8 − 505.13 FWH-1: & & ∑m h = ∑m h i i e e & & & ⎯ ⎯→ m9 h9 + m 2 h2 = m3 h3 ⎯ ⎯→ zh9 + (1 − y − z )h2 = (1 − y )h3 & & where z is the fraction of steam extracted from the turbine ( = m9 / m5 ) at the second stage. Solving for z, z= h3 − h2 (1 − y ) = 504.71 − 137.95 (1 − 0.07136) = 0.1373 2618.7 − 137.95 h9 − h2 Then, qin = h7 − h6 = 3625.8 − 680.73 = 2945.0 kJ/kg qout = (1 − y − z )(h10 − h1 ) = (1 − 0.07133 − 0.1373)(2105.0 − 137.75) = 1556.8 kJ/kg wnet = qin − qout = 2945.0 − 1556.8 = 1388.2 kJ/kg and & & W net = mwnet = (22 kg/s )(1388.2 kJ/kg ) = 30,540 kW ≅ 30.5 MW (b) η th = 1 − q out 1556.8 kJ/kg = 1− = 47.1% q in 2945.0 kJ/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.
- 40. 10-40 10-53 [Also solved by EES on enclosed CD] A steam power plant operates on an ideal regenerative Rankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam through the boiler for a net power output of 250 MW 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), 8 h1 = h f @ 10 kPa = 191.81 kJ/kg v 1 = v f @ 10 kPa = 0.00101 m /kg ( ) ⎛ 1 kJ = 0.00101 m 3 /kg (300 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.29 kJ/kg 9 y 11 5 3 w pI ,in = v 1 (P2 − P1 ) Turbine Boiler 4 Closed fwh ⎞ ⎟ ⎟ ⎠ 10 3 Open fwh P II Condenser 2 6 1-y-z z 1 7 PI h2 = h1 + w pI ,in = 191.81 + 0.29 = 192.10 kJ/kg P3 = 0.3 MPa ⎫ h3 = h f @ 0.3 MPa = 561.43 kJ/kg ⎬ 3 sat. liquid ⎭ v 3 = v f @ 0.3 MPa = 0.001073 m /kg w pII ,in = v 3 (P4 − P3 ) ( T 8 12.5 MPa ) ⎛ 1 kJ ⎞ ⎟ = 0.001073 m 3 /kg (12,500 − 300 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎟ ⎝ ⎠ = 13.09 kJ/kg h4 = h3 + w pII ,in = 561.43 + 13.09 = 574.52 kJ/kg h6 = h7 = h f @ 0.8 MPa = 720.87 kJ/kg P6 = 0.8 MPa ⎫ 3 ⎬ v 6 = v f @ 0.8 MPa = 0.001115 m /kg sat. liquid ⎭ T6 = Tsat @ 0.8 MPa = 170.4°C 5 4 2 0.8 MPa 9 y 6 3 7 1 0.3 MPa 10 z 1-y-z 10 kPa 11 s T6 = T5 , P5 = 12.5 MPa → h5 = 727.83 kJ/kg P8 = 12.5 MPa ⎫ h8 = 3476.5 kJ/kg ⎬ T8 = 550°C ⎭ s 8 = 6.6317 kJ/kg ⋅ K s9 − s f 6.6317 − 2.0457 = = 0.9935 P9 = 0.8 MPa ⎫ x 9 = s fg 4.6160 ⎬ s 9 = s8 ⎭ h9 = h f + x 9 h fg = 720.87 + (0.9935)(2047.5) = 2755.0 kJ/kg s10 − s f 6.6317 − 1.6717 = = 0.9323 P = 0.3 MPa ⎫ x10 = 10 s fg 5.3200 ⎬ s10 = s8 ⎭ h10 = h f + x10 h fg = 561.43 + (0.9323)(2163.5) = 2578.5 kJ/kg s11 − s f 6.6317 − 0.6492 = = 0.7977 P = 10 kPa ⎫ x11 = 11 s fg 7.4996 ⎬ s11 = s8 ⎭ h11 = h f + x11h fg = 191.81 + (0.7977 )(2392.1) = 2100.0 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the & & feedwater heaters. Noting that Q ≅ W ≅ Δke ≅ Δpe ≅ 0 , 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.
- 41. 10-41 & & & Ein − E out = ΔEsystem 0 (steady) =0 & & Ein = E out & & ∑m h = ∑m h i i e e & & ⎯ ⎯→ m9 (h9 − h6 ) = m5 (h5 − h4 ) ⎯ ⎯→ y (h9 − h6 ) = (h5 − h4 ) & & where y is the fraction of steam extracted from the turbine ( = m10 / m5 ). Solving for y, y= h5 − h4 727.83 − 574.52 = 0.0753 = h9 − h6 2755.0 − 720.87 For the open FWH, & & & E in − E out = ΔE system 0 (steady) =0 & & E in = E out & & ∑m h = ∑m h i i e e & & & & ⎯ ⎯→ m7 h7 + m 2 h2 + m10 h10 = m3 h3 ⎯ ⎯→ yh7 + (1 − y − z )h2 + zh10 = (1)h3 & & where z is the fraction of steam extracted from the turbine ( = m9 / m5 ) at the second stage. Solving for z, z= (h3 − h2 ) − y(h7 − h2 ) = 561.43 − 192.10 − (0.0753)(720.87 − 192.10) = 0.1381 h10 − h2 2578.5 − 192.10 Then, qin = h8 − h5 = 3476.5 − 727.36 = 2749.1 kJ/kg qout = (1 − y − z )(h11 − h1 ) = (1 − 0.0753 − 0.1381)(2100.0 − 191.81) = 1500.1 kJ/kg wnet = qin − qout = 2749.1 − 1500.1 = 1249 kJ/kg and & m= (b) & Wnet 250,000 kJ/s = = 200.2 kg/s wnet 1249 kJ/kg η th = 1 − q out 1500.1 kJ/kg = 1− = 45.4% q in 2749.1 kJ/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.
- 42. 10-42 10-54 EES Problem 10-53 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[8] = 12500 [kPa] T[8] = 550 [C] P[9] = 800 [kPa] "P_cfwh=300 [kPa]" P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=250 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages" Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages" Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages" Eta_pump = 100/100 "Pump isentropic efficiency" "Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[11] P[2]=P[10] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[5]=P[8] P[4] = P[5] P[3]=P[10] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Closed Feedwater Heater analysis" P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 43. 10-43 h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]" T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]" s[5]=entropy(Fluid$,P=P[6],h=h[5]) h[6]=enthalpy(Fluid$,P=P[6],x=0) T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]" s[6]=entropy(Fluid$,P=P[6],x=0) "Trap analysis" P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "Boiler analysis" q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler" h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) "Turbine analysis" ss[9]=s[8] hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9]) Ts[9]=temperature(Fluid$,s=ss[9],P=P[9]) h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9])"Definition of turbine efficiency for high pressure stages" T[9]=temperature(Fluid$,P=P[9],h=h[9]) s[9]=entropy(Fluid$,P=P[9],h=h[9]) ss[10]=s[8] hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10]) Ts[10]=temperature(Fluid$,s=ss[10],P=P[10]) h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10])"Definition of turbine efficiency for Intermediate pressure stages" T[10]=temperature(Fluid$,P=P[10],h=h[10]) s[10]=entropy(Fluid$,P=P[10],h=h[10]) ss[11]=s[8] hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11]) Ts[11]=temperature(Fluid$,s=ss[11],P=P[11]) h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11])"Definition of turbine efficiency for low pressure stages" T[11]=temperature(Fluid$,P=P[11],h=h[11]) s[11]=entropy(Fluid$,P=P[11],h=h[11]) h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine" "Condenser analysis" (1-y-z)*h[11]=q_out+(1-y-z)*h[1]"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net 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.
- 44. 10-44 ηturb 0.7 0.75 0.8 0.85 0.9 0.95 1 ηturb 0.7 0.75 0.8 0.85 0.9 0.95 1 ηth 0.3916 0.4045 0.4161 0.4267 0.4363 0.4452 0.4535 m [kg/s] 231.6 224.3 218 212.6 207.9 203.8 200.1 0.46 235 ηturb =ηpump 0.45 230 0.43 220 215 0.41 210 0.4 205 0.39 0.7 0.75 0.8 0.85 0.9 ηturb m [kg/s] 225 0.42 ηth 0.44 200 1 0.95 Steam 600 8 500 T [C] 400 300 200 12500 kPa 800 kPa 5,6 10 3,4 100 0 0 9 7 300 kPa 11 10 kPa 1,2 2 4 6 8 s [kJ/kg-K] 10 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.
- 45. 10-45 10-55 An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat added in the boiler 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), T 4 Turbine 4 3 MPa qin 5 6 Boiler 3 Condenser 2 3 2 y 5 1-y 7 7 Closed fwh 1 MPa 20 kPa Pump 1 1 qout 6 s h1 = h f @ 20 kPa = 251.42 kJ/kg v 1 = v f @ 20 kPa = 0.001017 m 3 /kg wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001017 m 3 /kg )(3000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 3.03 kJ/kg h2 = h1 + wp,in = 251.42 + 3.03 = 254.45 kJ/kg P4 = 3000 kPa ⎫ h4 = 3116.1 kJ/kg ⎬ T4 = 350°C ⎭ s 4 = 6.7450 kJ/kg ⋅ K P5 = 1000 kPa ⎫ ⎬ h5 = 2851.9 kJ/kg s5 = s 4 ⎭ s6 − s f 6.7450 − 0.8320 P6 = 20 kPa ⎫ x 6 = = = 0.8357 s fg 7.0752 ⎬ s6 = s 4 ⎭ h = h + x h = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg 6 f 6 fg For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. P7 = 1000 kPa ⎫ h7 = 762.51 kJ/kg ⎬ x7 = 0 ⎭ T7 = 179.9°C P3 = 3000 kPa ⎫ ⎬ h = 763.53 kJ/kg T3 = T7 = 209.9°C ⎭ 3 & & An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m5 / m 4 ) for closed feedwater heater: 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.
- 46. 10-46 & & ∑m h = ∑m h i i e e & & & & m5 h5 + m 2 h2 = m3 h3 + m 7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging, y= h3 − h2 763.53 − 254.45 = = 0.2437 h5 − h7 2851.9 − 762.51 Then, wT,out = h4 − h5 + (1 − y )(h5 − h6 ) = 3116.1 − 2851.9 + (1 − 0.2437)(2851.9 − 2221.7) = 740.9 kJ/kg wP,in = 3.03 kJ/kg q in = h4 − h3 = 3116.1 − 763.53 = 2353 kJ/kg Also, wnet = wT,out − wP,in = 740.9 − 3.03 = 737.8 kJ/kg η th = wnet 737.8 = = 0.3136 q in 2353 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.
- 47. 10-47 10-56 An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The change in thermal efficiency when the steam serving the closed feedwater heater is extracted at 600 kPa rather than 1000 kPa 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), T 4 Turbine 4 3 MPa qin 5 6 Boiler 3 Condenser 2 3 2 5 1-y 7 20 kPa 7 Closed fwh 0.6 MPa y Pump 1 1 qout 6 s h1 = h f @ 20 kPa = 251.42 kJ/kg v 1 = v f @ 20 kPa = 0.001017 m 3 /kg wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001017 m 3 /kg )(3000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 3.03 kJ/kg h2 = h1 + w p,in = 251.42 + 3.03 = 254.45 kJ/kg P4 = 3000 kPa ⎫ h4 = 3116.1 kJ/kg ⎬ T4 = 350°C ⎭ s 4 = 6.7450 kJ/kg ⋅ K s5 − s f 6.7450 − 1.9308 P5 = 600 kPa ⎫ x 5 = = = 0.9970 s fg 4.8285 ⎬ s5 = s 4 ⎭ h = h + x h = 670.38 + (0.9970)(2085.8) = 2750.0 kJ/kg 5 f 5 fg s6 − s f 6.7450 − 0.8320 P6 = 20 kPa ⎫ x 6 = = = 0.8357 s fg 7.0752 ⎬ s6 = s 4 ⎭ h = h + x h = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg 6 f 6 fg For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. P7 = 600 kPa ⎫ h7 = 670.38 kJ/kg ⎬ x7 = 0 ⎭ T7 = 158.8°C P3 = 3000 kPa ⎫ ⎬ h = 671.79 kJ/kg T3 = T7 = 158.8°C ⎭ 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.
- 48. 10-48 & & An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m5 / m 4 ) for closed feedwater heater: & & ∑m h = ∑m h i i e e & & & & m5 h5 + m 2 h2 = m3 h3 + m 7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging, y= h3 − h2 671.79 − 254.45 = = 0.2007 h5 − h7 2750.0 − 670.38 Then, wT,out = h4 − h5 + (1 − y )(h5 − h6 ) = 3116.1 − 2750.0 + (1 − 0.2007)(2750.0 − 2221.7) = 788.4 kJ/kg wP,in = 3.03 kJ/kg q in = h4 − h3 = 3116.1 − 671.79 = 2444 kJ/kg Also, w net = wT,out − wP,in = 788.4 − 3.03 = 785.4 kJ/kg η th = wnet 785.4 = = 0.3213 2444 q in When the steam serving the closed feedwater heater is extracted at 600 kPa rather than 1000 kPa, the thermal efficiency increases from 0.3136 to 0.3213. This is an increase of 2.5%. 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.
- 49. 10-49 10-57 EES The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES. Analysis The EES program used to solve this problem as well as the solutions are given below. "Given" P[4]=3000 [kPa] T[4]=350 [C] P[5]=600 [kPa] P[6]=20 [kPa] P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/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.
- 50. 10-50 ηth 0.32380 0.32424 0.32460 0.32490 0.32514 0.32534 0.32550 0.32563 0.32573 0.32580 0.32585 0.32588 0.32590 0.32589 0.32588 0.32585 0.32581 0.32576 0.32570 0.32563 0.326 0.3258 0.3255 0.3253 η th P 6 [kPa] 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 0.325 0.3248 0.3245 0.3243 0.324 0.3238 100 140 180 220 260 300 Bleed pressure [kPa] 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.
- 51. 10-51 10-58 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency 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), T 4 Turbine 4 3 MPa qin 5 6 Boiler Condenser 3 2 7 3 2 Pump 1 y 5s 1 5 1-y 20 kPa 7 Closed fwh 1 MPa qout 6s 6 s h1 = h f @ 20 kPa = 251.42 kJ/kg v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w p,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001017 m 3 /kg )(3000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 3.03 kJ/kg h2 = h1 + w p,in = 251.42 + 3.03 = 254.45 kJ/kg P4 = 3000 kPa ⎫ h4 = 3116.1 kJ/kg ⎬ T4 = 350°C ⎭ s 4 = 6.7450 kJ/kg ⋅ K P5 = 1000 kPa ⎫ ⎬ h5 s = 2851.9 kJ/kg s 5s = s 4 ⎭ s6s − s f 6.7450 − 0.8320 P6 = 20 kPa ⎫ x 6 s = = = 0.8357 7.0752 s fg ⎬ s 6s = s 4 ⎭ h = h + x h = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg 6s f 6 s fg ηT = h4 − h5 ⎯ ⎯→ h5 = h4 − η T (h4 − h5s ) = 3116.1 − (0.90)(3116.1 − 2851.9) = 2878.3 kJ/kg h4 − h5 s ηT = h 4 − h6 ⎯ ⎯→ h6 = h4 − η T (h4 − h6s ) = 3116.1 − (0.90)(3116.1 − 2221.7) = 2311.1 kJ/kg h 4 − h6 s For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. 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.
- 52. 10-52 P7 = 1000 kPa ⎫ h7 = 762.51 kJ/kg ⎬ x7 = 0 ⎭ T7 = 179.9°C P3 = 3000 kPa ⎫ ⎬ h = 763.53 kJ/kg T3 = T7 = 209.9°C ⎭ 3 & & An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m5 / m 4 ) for closed feedwater heater: & & ∑m h = ∑m h i i e e & & & & m5 h5 + m 2 h2 = m3 h3 + m 7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging, y= h3 − h2 763.53 − 254.45 = = 0.2406 h5 − h7 2878.3 − 762.51 Then, wT,out = h4 − h5 + (1 − y )(h5 − h6 ) = 3116.1 − 2878.3 + (1 − 0.2406)(2878.3 − 2311.1) = 668.5 kJ/kg wP,in = 3.03 kJ/kg q in = h4 − h3 = 3116.1 − 763.53 = 2353 kJ/kg Also, wnet = wT,out − wP,in = 668.5 − 3.03 = 665.5 kJ/kg η th = wnet 665.5 = = 0.2829 2353 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.
- 53. 10-53 10-59 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency 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), T 4 Turbine 4 3 MPa qin 5 6 Boiler 3 Condenser 2 7 7 3 Closed fwh 2 1 MPa y 5s 1-y 20 kPa Pump 1 1 5 qout 6s 6 s When the liquid enters the pump 10°C cooler than a saturated liquid at the condenser pressure, the enthalpies become P1 = 20 kPa T1 = Tsat @ 20 kPa ⎫ h1 ≅ h f @ 50°C = 209.34 kJ/kg ⎬ − 10 = 60.06 − 10 ≅ 50°C ⎭ v 1 ≅ v f @ 50°C = 0.001012 m 3 /kg wp,in = v 1 ( P2 − P1 ) ⎛ 1 kJ ⎞ = (0.001012 m 3 /kg )(3000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 3.02 kJ/kg h2 = h1 + wp,in = 209.34 + 3.02 = 212.36 kJ/kg P4 = 3000 kPa ⎫ h4 = 3116.1 kJ/kg ⎬ T4 = 350°C ⎭ s 4 = 6.7450 kJ/kg ⋅ K P5 = 1000 kPa ⎫ ⎬ h5 s = 2851.9 kJ/kg s 5s = s 4 ⎭ s6s − s f 6.7450 − 0.8320 P6 = 20 kPa ⎫ x 6 s = = = 0.8357 s fg 7.0752 ⎬ s 6s = s 4 ⎭ h = h + x h = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg 6s f 6 s fg ηT = h4 − h5 ⎯ ⎯→ h5 = h4 − η T (h4 − h5s ) = 3116.1 − (0.90)(3116.1 − 2851.9) = 2878.3 kJ/kg h4 − h5 s ηT = h 4 − h6 ⎯ ⎯→ h6 = h4 − η T (h4 − h6s ) = 3116.1 − (0.90)(3116.1 − 2221.7) = 2311.1 kJ/kg h 4 − h6 s For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. 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.
- 54. 10-54 P7 = 1000 kPa ⎫ h7 = 762.51 kJ/kg ⎬ x7 = 0 ⎭ T7 = 179.9°C P3 = 3000 kPa ⎫ ⎬ h = 763.53 kJ/kg T3 = T7 = 209.9°C ⎭ 3 & & An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m5 / m 4 ) for closed feedwater heater: & & ∑m h = ∑m h i i e e & & & & m5 h5 + m 2 h2 = m3 h3 + m 7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging, y= h3 − h2 763.53 − 212.36 = 0.2605 = h5 − h7 2878.3 − 762.51 Then, wT,out = h4 − h5 + (1 − y )(h5 − h6 ) = 3116.1 − 2878.3 + (1 − 0.2605)(2878.3 − 2311.1) = 657.2 kJ/kg w P,in = 3.03 kJ/kg q in = h4 − h3 = 3116.1 − 763.53 = 2353 kJ/kg Also, wnet = wT,out − wP,in = 657.2 − 3.03 = 654.2 kJ/kg η th = wnet 654.2 = = 0.2781 q in 2353 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.
- 55. 10-55 10-60 EES The effect of pressure drop and non-isentropic turbine on the rate of heat input is to be determined for a given power plant. Analysis The EES program used to solve this problem as well as the solutions are given below. "Given" P[3]=3000 [kPa] DELTAP_boiler=10 [kPa] P[4]=P[3]-DELTAP_boiler T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] eta_T=0.90 P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 56. 10-56 Solution with 10 kPa pressure drop in the boiler: DELTAP_boiler=10 [kPa] Eta_th=0.2827 P[3]=3000 [kPa] q_in=2352.8 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] y=0.2405 eta_T=0.9 Fluid$='steam_iapws' P[4]=2990 [kPa] w_net=665.1 [kJ/kg] w_T_out=668.1 [kJ/kg] Solution without any pressure drop in the boiler: DELTAP_boiler=0 [kPa] Eta_th=0.3136 P[3]=3000 [kPa] q_in=2352.5 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] y=0.2437 eta_T=1 Fluid$='steam_iapws' P[4]=3000 [kPa] w_net=737.8 [kJ/kg] w_T_out=740.9 [kJ/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.
- 57. 10-57 10-61E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of the plant, 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 High-P Turbine 7 T Boiler Low-P Turbine 8 9 6 Open fwh II P III 11 10 Open fwh I 2 P II 3 12 1-y-z 4 2 Condense z 5 4 6 1-y y 7 3 1 1 5 1500psia y 8 1 - y 10 250 psia z 11 140 psia 9 40 psia 1-y-z 1 psia 12 PI s (a) 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 @ 1 psia = 0.01614 ft 3 /lbm w pI ,in = v 1 (P2 − P1 ) ( ) ⎛ 1 Btu = 0.01614 ft 3 /lbm (40 − 1 psia )⎜ ⎜ 5.4039 psia ⋅ ft 3 ⎝ = 0.12 Btu/lbm ⎞ ⎟ ⎟ ⎠ h2 = h1 + w pI ,in = 69.72 + 0.12 = 69.84 Btu/lbm P3 = 40 psia ⎫ h3 = h f @ 40 psia = 236.14 Btu/lbm ⎬ 3 sat. liquid ⎭ v 3 = v f @ 40 psia = 0.01715 ft /lbm w pII ,in = v 3 (P4 − P3 ) ( ) ⎛ 1 Btu = 0.01715 ft 3 /lbm (250 − 40 psia )⎜ ⎜ 5.4039 psia ⋅ ft 3 ⎝ = 0.67 Btu/lbm ⎞ ⎟ ⎟ ⎠ h4 = h3 + w pII ,in = 236.14 + 0.67 = 236.81 Btu/lbm P5 = 250 psia ⎫ h5 = h f @ 250 psia = 376.09 Btu/lbm ⎬ 3 sat. liquid ⎭ v 5 = v f @ 250 psia = 0.01865 ft /lbm w pIII ,in = v 5 (P6 − P5 ) ( ) ⎛ 1 Btu = 0.01865 ft 3 /lbm (1500 − 250 psia )⎜ ⎜ 5.4039 psia ⋅ ft 3 ⎝ = 4.31 Btu/lbm ⎞ ⎟ ⎟ ⎠ h6 = h5 + w pIII ,in = 376.09 + 4.31 = 380.41 Btu/lbm P7 = 1500 psia ⎫ h7 = 1550.5 Btu/lbm ⎬ ⎭ s 7 = 1.6402 Btu/lbm ⋅ R T7 = 1100°F P8 = 250 psia ⎫ ⎬ h8 = 1308.5 Btu/lbm s8 = s 7 ⎭ 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.
- 58. 10-58 P9 = 140 psia ⎫ ⎬ h9 = 1248.8 Btu/lbm s9 = s7 ⎭ P10 = 140 psia ⎫ h10 = 1531.3 Btu/lbm ⎬ T10 = 1000°F ⎭ s10 = 1.8832 Btu/lbm ⋅ R P11 = 40 psia ⎫ ⎬ h11 = 1356.0 Btu/lbm s11 = s10 ⎭ x12 = s12 − s f = 1.8832 − 0.13262 = 0.9488 1.84495 s fg P12 = 1 psia ⎫ ⎬ h12 = h f + x12 h fg = 69.72 + (0.9488)(1035.7 ) s12 = s10 ⎭ = 1052.4 Btu/lbm The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the & & feedwater heaters. Noting that Q ≅ W ≅ Δke ≅ Δpe ≅ 0 , & & & Ein − E out = ΔEsystem 0 (steady) =0 & & Ein = E out FWH-2: & & ∑m h = ∑m h i i e e & & & ⎯ ⎯→ m8 h8 + m4 h4 = m5 h5 ⎯ ⎯→ yh8 + (1 − y )h4 = 1(h5 ) & & where y is the fraction of steam extracted from the turbine ( = m8 / m5 ). Solving for y, y= h5 − h4 376.09 − 236.81 = = 0.1300 h8 − h4 1308.5 − 236.81 & & & Ein − E out = ΔEsystem 0 (steady) =0 & & Ein = E out FWH-1 & & ∑m h = ∑m h i i e e & & & ⎯ ⎯→ m11h11 + m 2 h2 = m3 h3 ⎯ ⎯→ zh11 + (1 − y − z )h2 = (1 − y )h3 & & where z is the fraction of steam extracted from the turbine ( = m9 / m5 ) at the second stage. Solving for z, z= h3 − h2 (1 − y ) = 236.14 − 69.84 (1 − 0.1300) = 0.1125 h11 − h2 1356.0 − 69.84 Then, q in = h7 − h6 + (1 − y )(h10 − h9 ) = 1550.5 − 380.41 + (1 − 0.1300)(1531.3 − 1248.8) = 1415.8 Btu/lbm q out = (1 − y − z )(h12 − h1 ) = (1 − 0.1300 − 0.1125)(1052.4 − 69.72) = 744.4 Btu/lbm wnet = q in − q out = 1415.8 − 744.4 = 671.4 Btu/lbm and & m= & Qin 4 × 10 5 Btu/s = = 282.5 lbm/s q in 1415.8 Btu/lbm (b) ⎛ 1.055 kJ ⎞ & & Wnet = mwnet = (282.5 lbm/s)(671.4 Btu/lbm)⎜ ⎜ 1 Btu ⎟ = 200.1 MW ⎟ ⎝ ⎠ (c) η th = 1 − q out 744.4 Btu/lbm = 1− = 47.4% 1415.8 Btu/lbm 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.
- 59. 10-59 10-62 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, 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), h1 = h f @ 20 kPa = 251.42 kJ/kg v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w pI ,in = v 1 (P2 − P1 ) / η p = (0.001017 m 3 /kg )(12,500 − 20 kPa ) = 14.43 kJ/kg 1 0.88 High-P turbine 5 h2 = h1 + w pI ,in Low-P turbine Boiler = 251.42 + 14.43 6 = 265.85 kJ/kg P3 = 1 MPa ⎫ h3 = h f @ 1 MPa = 762.51 kJ/kg 4 ⎬ sat. liquid ⎭ v 3 = v f @ 1 MPa = 0.001127 m 3 /kg w pII ,in = v 3 (P11 − P3 ) / η p = (0.001127 m 3 /kg )(12,500 − 1000 kPa ) / 0.88 1-y 7 Mixing Cham. 10 9 8 Closed fwh Cond. 2 11 PII y 3 1 PI = 14.73 kJ/kg h11 = h3 + w pII ,in = 762.51 + 14.73 = 777.25 kJ/kg Also, h4 = h10 = h11 = 777.25 kJ/kg since the two fluid streams which are being mixed have the same enthalpy. P5 = 12.5 MPa ⎫ h5 = 3476.5 kJ/kg ⎬ T5 = 550°C ⎭ s 5 = 6.6317 kJ/kg ⋅ K P6 = 5 MPa ⎫ ⎬h6 s = 3185.6 kJ/kg ⎭ s6 = s5 ηT = h5 − h6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) h5 − h6 s = 3476.5 − (0.88)(3476.5 − 3185.6 ) = 3220.5 kJ/kg P7 = 5 MPa ⎫ h7 = 3550.9 kJ/kg ⎬ T7 = 550°C ⎭ s 7 = 7.1238 kJ/kg ⋅ K P8 = 1 MPa ⎫ ⎬h8 s = 3051.1 kJ/kg ⎭ s8 = s 7 ηT = h7 − h8 ⎯ ⎯→ h8 = h7 − η T (h7 − h8 s ) h7 − h8 s = 3550.9 − (0.88)(3550.9 − 3051.1) = 3111.1 kJ/kg P8 = 1 MPa ⎫ ⎬T8 = 328°C h8 = 3111.1 kJ/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.

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