1. A plant operates on a simple ideal Rankine cycle between the specified limits. The of the cycle, the mass flow rate of the steam, and the rise of the cooling are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

= hh f @1 10 kPa = .81191 kJ/kg 3 = vv f @1 10 kPa = 00101.0 /kgm

w p v ()−= PP 121in, 3 ⎛ 1 kJ ⎞ = ()0.00101 3 ()−107,000/kgm kPa ⎜ ⎟ 7 MPa ⎜1 ⋅mkPa 3 ⎟ ⎝ ⎠ qin = 7.06 kJ/kg 2 10 kPa whh p in,12 =+=+= 198.8706.781.191 kJ/kg 1 4 qout P3 = 7 MPa ⎫h3 = 411.43 kJ/kg ⎬ s T3 °= C500 ⎭ s3 = 8000.6 ⋅ KkJ/kg

P4 =10 kPa ⎫ 4 − ss f − 6492.08000.6 ⎬ x4 = = = 8201.0 = ss 34 ⎭ s fg 4996.7

4 f 4hxhh fg +=+= ()()= 3.62151.23928201.081.191 kJ/kg Thus,

hhq 23in =−=−= 5.321287.1984.3411 kJ/kg

hhq 14out =−=−= 8.196181.1916.2153 kJ/kg

qqw outinnet =−=−= 7.12508.19615.3212 kJ/kg and

wnet 1250.7 kJ/kg η th == = 38.9% qin 3212.5 kJ/kg

W&net 45,000 kJ/s (b) m& == = 036 /. skg wnet 1250.7 kJ/kg (c) The rate of rejection to the cooling water and its temperature rise are

& out & qmQ out == ()()35.98 1961.8kg/s = ,58670kJ/kg kJ/s

Q& out 70,586 kJ/s ΔTcoolingwater = = °= C8.4 & cm )( coolingwater ()()2000 4.18kg/s °⋅ CkJ/kg

2. A steam power plant that operates on the ideal reheat Rankine cycle is considered. The 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),

= hh f @1 20 kPa = 42.251 kJ/kg T 3 = vv f @1 20 kPa = 700101.0 /kgm 3 5 v ()−= PPw p 121in, ⎛ 1 kJ ⎞ = ()001017.0 3 ()− 206000/kgm kPa ⎜ ⎟ 6 MPa ⎜1 ⋅ mkPa 3 ⎟ = .086 kJ/kg ⎝ ⎠ 4

whh p in,12 =+=+= .5025708.642.251 kJ/kg 2

P3 = 6 MPa ⎫ h3 = 3.3178 kJ/kg 20 kPa ⎬ T3 °= C400 ⎭ s3 = 5432.6 ⋅ KkJ/kg 1 6 s P4 = 2 MPa ⎫ ⎬ h4 = 0.2901 kJ/kg = ss 34 ⎭

P5 = 2 MPa ⎫ h5 = 4.3248 kJ/kg ⎬ T5 °= C400 ⎭ s5 = 1292.7 ⋅ KkJ/kg

6 − ss f − 8320.01292.7 P6 = 20 kPa ⎫ x6 = = = 8900.0 ⎬ s fg 0752.7 = ss 56 ⎭ 6 f 6hxhh fg +=+= ()()= 7.23495.23578900.042.251 kJ/kg The turbine work output and the thermal efficiency are determined from

()()hhhhw 6543outT, 7.23494.32480.29013.3178 =−+−=−+−= 1176 kJ/kg and

()()hhhhq 4523in =−+−=−+−= 32680.29014.324850.2573.3178 kJ/kg

,net www poutT in, −=−= =117008.61176 kJ/kg Thus,

wnet 1170 kJ/kg η th == 358.0 == 35.8% qin 3268 kJ/kg 3. 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 , 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),

= hh @sat1 10 kPa = 81.191 kJ/kg T

3 = vv @sat1 10 kPa = 00101.0 /kgm 3 5

w p v ()−= PP 121in, 15 ⎛ 1 kJ ⎞ = 00101.0 3 ()−10000,15/kgm kPa ⎜ ⎟ 4 () ⎜ 3 ⎟ ⎝1 ⋅mkPa ⎠ = .1415 kJ/kg 2 whh =+=+= 95.20614.1581.191 kJ/kg p in,12 10 kPa 1 P3 = 15 MPa ⎫ h3 = .83310 kJ/kg 6 ⎬ s T3 °= C500 ⎭ s3 = 3480.6 ⋅ KkJ/kg

P6 = 10 kPa ⎫ 6 f 6 hxhh fg +=+= ()()= 7.23441.239290.081.191 kJ/kg ⎬ = ss 56 ⎭ 6 f 6 sxss fg +=+= ()()= 3988.74996.790.06492.0 ⋅ KkJ/kg

T5 °= C500 ⎫ P5 = 2150 kPa ()the reheat pressure ⎬ = ss 65 ⎭ h5 = 61.3466 kJ/kg

P4 = 15.2 MPa ⎫ ⎬ h4 = 2.2817 kJ/kg = ss 34 ⎭ (b) The rate of heat supply is

& in & []()()−+−= hhhhmQ 4523 = ()(12 −+− )kJ/kg2.281761.346695.2068.3310kg/s = 45,039 kW (c) The thermal efficiency is determined from

& out & ()()(hhmQ 16 =−= 12 − )= ,83525kJ/kg81.1917.2344kJ/s kJ/s Thus,

Q& out 25,834 kJ/s η th 11 −=−= = 42.6% Q& in 45,039 kJ/s 4. 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

7

Turbine 7 Boiler 10 MPa 8 6 10 5 0.6 MPa 8 6 4 9 y 1 - y 0.2 MPa 3 9 fwh II fwh I Condenser 2 1 - y - z 5 kPa 4 2 1 1 10 3 P III 5 P II P I s

(a) From the steam tables (Tables A-4, A-5, and A-6),

= hh f @1 5 kPa = 75.137 kJ/kg 3 = vv f @1 5 kPa = 001005.0 /kgm ⎛ 1 kJ ⎞ w v ()PP =−= 0.001005 3 ()− 5200/kgm kPa ⎜ ⎟ = 02.0 kJ/kg pI 121in, ()⎜ 3 ⎟ ⎝1 ⋅ mkPa ⎠ whh pI in,12 =+=+= 95.13720.075.137 kJ/kg = hh = 71.504 kJ/kg P3 = 2.0 MPa ⎫ f @3 2.0 MPa ⎬ 3 liquidsat. ⎭ = vv f @3 2.0 MPa = 001061.0 /kgm ⎛ ⎞ 3 ⎜ 1 kJ ⎟ w pII v ()PP 343in, =−= ()0.001061 ()− 200600/kgm kPa ⎜1 ⋅ mkPa 3 ⎟ = 0.42 kJ/kg ⎝ ⎠

whh pII in,34 =+=+= 13.50542.071.504 kJ/kg = hh = 38.670 kJ/kg P5 = 6.0 MPa ⎫ f @5 6.0 MPa ⎬ 3 liquidsat. ⎭ = vv f @5 6.0 MPa = 001101.0 /kgm ⎛ ⎞ 3 ⎜ 1 kJ ⎟ w pIII v ()PP 565in, =−= ()0.001101 ()10,000/kgm − 600 kPa ⎜1 ⋅mkPa 3 ⎟ = 10.35 kJ/kg ⎝ ⎠

56 whh pIII in, =+=+= 73.68035.1038.670 kJ/kg

P7 = 10 MPa ⎫ h7 = 8.3625 kJ/kg ⎬ T7 °= C600 ⎭ s7 = 9045.6 ⋅ KkJ/kg

P8 = 6.0 MPa ⎫ ⎬ h8 = 8.2821 kJ/kg = ss 78 ⎭

9 − ss f − 5302.19045.6 P9 = 2.0 MPa ⎫ x9 = = = 9602.0 ⎬ s fg 5968.5 = ss 79 ⎭ 9 f 9 hxhh fg +=+= ()()= 7.26186.22019602.071.504 kJ/kg 5. An ideal regenerative Rankine cycle with a closed is considered. The work produced by the turbine, the work consumed by the , 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),

= hh f @1 20 kPa = 42.251 kJ/kg 3 = vv f @1 20 kPa = 001017.0 /kgm 4 Turbine v −= PPw 121inp, )( 3 ⎛ 1 kJ ⎞ = 001017.0( − kPa)203000)(/kgm ⎜ ⎟ 5 ⎝1 ⋅mkPa 3 ⎠ = 03.3 kJ/kg Boiler 6 whh inp,12 =+=+= 45.25403.342.251 kJ/kg

P4 = 3000 kPa ⎫ h4 = 1.3116 kJ/kg Condenser ⎬ 7 8 T4 °= C350 ⎭ s4 = 7450.6 ⋅ KkJ/kg Closed P = 1000 kPa ⎫ 1 5 h = 9.2851 kJ/kg fwh ⎬ 5 3 2 = ss 45 ⎭

6 − ss f − 8320.07450.6 P6 = 20 kPa ⎫ x6 = = = 8357.0 ⎬ s fg 0752.7 = ss 46 ⎭ 6 f 6 hxhh fg +=+= = 7.2221)5.2357)(8357.0(42.251 kJ/kg 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 T extraction pressure. P = 1000 kPa ⎫ h = 51.762 kJ/kg 4 7 7 ⎬ 3 MPa x7 = 0 ⎭ T7 °= C9.179 qin hh 78 == 51.762 kJ/kg 3 1 MPa y P3 = 3000 kPa ⎫ kJ/kg 53.763 ⎬ h3 = 53.763 kJ/kg 2 7 5 TT 73 °== C9.209 ⎭ 1-y An energy balance on the gives the 8 20 kPa 1 6 fraction of steam extracted from the turbine ( = / mm && 45 ) qout for closed feedwater heater: s ii = ∑∑ && hmhm ee

+=+ &&&& hmhmhmhm 77332255

11 +=+ yhhhyh 7325 Rearranging, − hh − 45.25453.763 y = 23 = = 2437.0 − hh 75 − 51.7629.2851 Then,

54outT, hhyhhw 65 =−−+−= − + − − )7.22219.2851)(2437.01(9.28511.3116))(1( = 740.9 kJ/kg

w inP, = 3.03 kJ/kg

hhq 34in 53.7631.3116 =−=−= 2353 kJ/kg

Also, www inP,outT,net −=−= = 8.73703.39.740 kJ/kg

wnet 8.737 η th === 3136.0 qin 2353