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1. a Steam Power Plant Operates on a Simple Ideal Rankine Cycle Between the Specified Pressure Limits

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1. a Steam Power Plant Operates on a Simple Ideal Rankine Cycle Between the Specified Pressure Limits 1. 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 3 v1= v f @ 10 kPa = 0.00101 m /kg w p,in=v 1()PP 2 − 1 3 ⎛ 1 kJ ⎞ = ()0.00101 m3 /kg() 7,000− 10 kPa ⎜ ⎟ 7 MPa ⎜1 kPa⋅ m 3 ⎟ ⎝ ⎠ qin = 7.06 kJ/kg 2 10 kPa h2= h 1 + w p ,in =191.81 + 7.06 = 198.87 kJ/kg 1 4 qout P3 = 7 MPa ⎫h3 = 3411.4 kJ/kg ⎬ s T3 =500 ° C ⎭ s3 = 6.8000 kJ/kg⋅ K P4 =10 kPa ⎫ s4 − s f 6.8000− 0.6492 ⎬ x4 = = = 0.8201 s4= s 3 ⎭ s fg 7.4996 h4 = hf + x4 h fg =191.81 + ()() 0.8201 2392.1= 2153.6 kJ/kg Thus, qin= h 3 − h 2 =3411.4 − 198.87 = 3212.5 kJ/kg qout= h 4 − h 1 =2153.6 − 191.81 = 1961.8 kJ/kg wnet= q in − q out =3212.5 − 1961.8 = 1250.7 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& = = = 36./ 0 kg s wnet 1250.7 kJ/kg (c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = mq& out = ()()35.98 kg/s 1961.8 kJ/kg= 70,586 kJ/s Q& out 70,586 kJ/s ΔTcoolingwater = = =8.4 ° C ()mc& coolingwater ()()2000 kg/s 4.18 kJ/kg⋅° C 2. 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 T 3 v1= v f @ 20 kPa = 0.001017 m /kg 3 5 w=v () P − P p,in 1 2 1 ⎛ 1 kJ ⎞ = ()0.001017 m3 /kg() 6000− 20 kPa ⎜ ⎟ 6 MPa ⎜1 kPa⋅ m3 ⎟ = 6.08 kJ/kg ⎝ ⎠ 4 h2= h 1 + w p ,in =251.42 + 6.08 = 257.50 kJ/kg 2 P3 = 6 MPa ⎫ h3 = 3178.3 kJ/kg 20 kPa ⎬ T3 =400 ° C ⎭ s3 = 6.5432 kJ/kg⋅ K 1 6 s P4 = 2 MPa ⎫ ⎬ h4 = 2901.0 kJ/kg s4= s 3 ⎭ P5 = 2 MPa ⎫ h5 = 3248.4 kJ/kg ⎬ T5 =400 ° C ⎭ s5 = 7.1292 kJ/kg⋅ K s6 − s f 7.1292− 0.8320 P6 = 20 kPa ⎫ x6 = = = 0.8900 ⎬ s fg 7.0752 s6= s 5 ⎭ h6 = hf + x6 h fg =251.42 + ()() 0.8900 2357.5= 2349.7 kJ/kg The turbine work output and the thermal efficiency are determined from wT,out=()() h 3 − h 4 + h 5 − h 6 =3178.3 − 2901.0 + 3248.4 − 2349.7 = 1176 kJ/kg and qin=()() h 3 − h 2 + h 5 − h 4 =3178.3 − 257.50 + 3248.4 − 2901.0 = 3268 kJ/kg wnet= wT , out − w p,in =1176 − 6.08= 1170 kJ/kg Thus, wnet 1170 kJ/kg η th = = =0.358 = 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 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= h sat@ 10 kPa = 191.81 kJ/kg T 3 v1= v sat@ 10 kPa = 0.00101 m /kg 3 5 w p,in=v 1()PP 2 − 1 15 ⎛ 1 kJ ⎞ = 0.00101 m3 /kg() 15,000− 10 kPa ⎜ ⎟ 4 () ⎜ 3 ⎟ ⎝1 kPa⋅ m ⎠ = 15.14 kJ/kg 2 h= h + w =191.81 + 15.14 = 206.95 kJ/kg 2 1p ,in 10 kPa 1 P3 = 15 MPa ⎫ h3 = 3310.8 kJ/kg 6 ⎬ s T3 =500 ° C ⎭ s3 = 6.3480 kJ/kg⋅ K P6 = 10 kPa ⎫ h6 = hf + x6 h fg =191.81 + ()() 0.90 2392.1= 2344.7 kJ/kg ⎬ s6= s 5 ⎭ s6 = sf + x6 s fg =0.6492 + ()() 0.90 7.4996= 7.3988 kJ/kg⋅ K T5 =500 ° C ⎫ P5 = 2150 kPa ()the reheat pressure ⎬ s5= s 6 ⎭ h5 = 3466.61 kJ/kg P4 = 2.15 MPa ⎫ ⎬ h4 = 2817.2 kJ/kg s4= s 3 ⎭ (b) The rate of heat supply is Q& in = m& []()() h3 − h 2 + h 5 − h 4 = ()(12 kg/s 3310.8− 206.95 + 3466.61 − 2817.2) kJ/kg = 45,039 kW (c) The thermal efficiency is determined from Q& out = m(& h6 − h 1 ()() = 12 kJ/s 2344.7− 191.81 ) kJ/kg= 25,835 kJ/s Thus, Q& out 25,834 kJ/s η th =1 − = 1 − = 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), h1= h f @ 5 kPa = 137.75 kJ/kg 3 v1= v f @ 5 kPa = 0.001005 m /kg ⎛ 1 kJ ⎞ w =v ()PP − = 0.001005 m3 /kg() 200− 5 kPa ⎜ ⎟ = 0.20 kJ/kg pI ,in 1 2 1 ()⎜ 3 ⎟ ⎝1 kPa⋅ m ⎠ h2= h 1 + w pI ,in =137.75 + 0.20 = 137.95 kJ/kg h= h = 504.71 kJ/kg P3 = 0.2 MPa ⎫ 3f @ 0.2 MPa ⎬ 3 sat.liquid ⎭v3= v f @ 0.2 MPa = 0.001061 m /kg ⎛ ⎞ 3 ⎜ 1 kJ ⎟ w pII ,in=v 3()PP 4 − 3 = ()0.001061 m /kg() 600− 200 kPa ⎜1 kPa⋅ m 3 ⎟ = 0.42 kJ/kg ⎝ ⎠ h4= h 3 + w pII ,in =504.71 + 0.42 = 505.13 kJ/kg h= h = 670.38 kJ/kg P5 = 0.6 MPa ⎫ 5f @ 0.6 MPa ⎬ 3 sat.liquid ⎭v5= v f @ 0.6 MPa = 0.001101 m /kg ⎛ ⎞ 3 ⎜ 1 kJ ⎟ w pIII ,in=v 5()PP 6 − 5 = ()0.001101 m /kg() 10,000 − 600 kPa ⎜1 kPa⋅ m 3 ⎟ = 10.35 kJ/kg ⎝ ⎠ h6= h 5 + w pIII,in =670.38 + 10.35 = 680.73 kJ/kg P7 = 10 MPa ⎫ h7 = 3625.8 kJ/kg ⎬ T7 =600 ° C ⎭ s7 = 6.9045 kJ/kg⋅ K P8 = 0.6 MPa ⎫ ⎬ h8 = 2821.8 kJ/kg s8= s 7 ⎭ s9 − s f 6.9045− 1.5302 P9 = 0.2 MPa ⎫ x9 = = = 0.9602 ⎬ s fg 5.5968 s9= s 7 ⎭ h9 = hf + x9 h fg =504.71 + ()() 0.9602 2201.6= 2618.7 kJ/kg 5. 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), h1= h f @ 20 kPa = 251.42 kJ/kg 3 v1= v f @ 20 kPa = 0.001017 m /kg 4 Turbine wp,in=v 1() P 2 − P 1 3 ⎛ 1 kJ ⎞ = (0.001017 m /kg)(3000− 20)kPa ⎜ ⎟ 5 ⎝1 kPa⋅ m 3 ⎠ = 3.03 kJ/kg Boiler 6 h2= h 1 + w p,in =251.42 + 3.03 = 254.45 kJ/kg P4 = 3000 kPa ⎫ h4 = 3116.1 kJ/kg Condenser ⎬ 7 8 T4 =350 ° C ⎭ s4 = 6.7450 kJ/kg⋅ K Closed P = 1000 kPa ⎫ 1 5 h = 2851.9 kJ/kg fwh Pump ⎬ 5 3 2 s5= s 4 ⎭ s6 − s f 6.7450− 0.8320 P6 = 20 kPa ⎫ x6 = = = 0.8357 ⎬ s fg 7.0752 s6= s 4 ⎭ h6 = hf + x6 h fg =251.42 + (0.8357)(2357.5)= 2221.7 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 = 762.51 kJ/kg 4 7 7 ⎬ 3 MPa x7 = 0 ⎭ T7 =179.9 ° C qin h8= h 7 = 762.51 kJ/kg 3 1 MPa y P3 = 3000 kPa ⎫ ⎬ h3 = 763.53 kJ/kg 2 7 5 TT3= 7 =209.9 ° C ⎭ 1-y An energy balance on the heat exchanger gives the 8 20 kPa 1 6 fraction of steam extracted from the turbine ( = m&&5/ m 4 ) qout for closed feedwater heater: s ∑m&&i h i = ∑ me h e m&&&&5 h 5+ m 2 h 2 = m 3 h 3 + m 7 h 7 yh5+1 h 2 = 1 h 3 + yh 7 Rearranging, h− h 763.53− 254.45 y = 3 2 = = 0.2437 h5− h 7 2851.9− 762.51 Then, wT,out= h 4 − h 5 +(1 − y )( h5 − h 6 ) = 3116.1− 2851.9+ (1− 0.2437)(2851.9− 2221.7) = 740.9 kJ/kg wP,in = 3.03 kJ/kg qin= h 4 − h 3 =3116.1 − 763.53 = 2353 kJ/kg Also, wnet= w T,out − w P,in =740.9 − 3.03= 737.8 kJ/kg wnet 737.8 η th = = = 0.3136 qin 2353.
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