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Special 10-77 Topic: Binary Vapor Cycles 10-82C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency. 10-83C Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottoming cycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potential energies, and assuming the heat exchanger is well-insulated, the steady-flow energy balance relation yields & & & E in - E out = E system & & m h = m h e e 0 (steady) =0 & & E in = E out & & & & & & m A h2 + m B h4 = m A h1 + m B h3 or m A (h2 - h1 ) = m B (h3 - h4 ) i i Thus, & m A h3 - h4 = & mB h2 - h1 10-84C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low. 10-85C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization. 10-86C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle. 10-78 Review Problems 10-87 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant cc can be expressed as cc = g + s - gs where g = Wg / Qin and s = Ws / Qg,out are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as Wtotal Q = 1 - out Qin Qin Wg Qin = 1- Qg,out Qin cc = g = s = Ws Q = 1 - out Qg,out Qg,out where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle. Using the relations above, the expression g + s - g s can be expressed as g,out + 1 - out g + s - g s = 1 - Qin Qg,out Q Q = 1- = 1- = cc Qg,out Qin Qout Qin +1- Qout Qg,out Qg,out Q - 1 - 1 - out Q Qin g,out Qg,out Q Q -1+ + out - out Qin Qg,out Qin Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combined cycle is determined to be cc = g + s - g s = 0.4 + 0.30 - 0.40 0.30 = 0.58 10-88 The thermal efficiency of a combined gas-steam power plant cc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as cc = g + s - gs . It is to be shown that the value of cc is greater than either of g or s . Analysis By factoring out terms, the relation cc = g + s - gs can be expressed as cc = g + s - g s = g + s (1 - g ) > g 1 24 4 3 Positive since g <1 or cc = g + s - g s = s + g (1 - s ) > s 1 24 4 3 Positive since s <1 Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbine cycles alone. 10-79 10-89 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6), P6 = 10 kPa h6 = h f + x6 h fg = 191.81 + (0.92 )(2392.1) = 2392.5 kJ/kg x6 = 0.92 s6 = s f + x6 s fg = 0.6492 + (0.92 )(7.4996) = 7.5488 kJ/kg K T5 = 600 C P5 = 2780 kPa s5 = s6 T 600 C SINGLE 25 MPa 2 10 kPa 1 6 s 1 3 5 T 600 C DOUBL 25 MPa 2 10 kPa 8 s 3 5 7 (b) Double Reheat : P3 = 25 MPa s = 6.3637 kJ/kg K T3 = 600 C 3 P4 = Px P = Px and 5 T5 = 600 C s 4 = s3 4 4 6 Any pressure Px selected between the limits of 25 MPa and 2.78 MPa will satisfy the requirements, and can be used for the double reheat pressure. 10-80 10-90E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as the working fluid is considered. The exit temperature of the geothermal water from the vaporizer, the rate of heat rejection from the working fluid in the condenser, the mass flow rate of geothermal water at the preheater, and the thermal efficiency of the Level I cycle of this plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steadyflow energy balance on the geothermal water (brine), - 22,790,000 Btu/h = (384,286 lbm/h )(1.03 Btu/lbm F)(T2 - 325 F) T2 = 267.4 F & & Q brine = m brine c p (T2 - T1 ) (b) The rate of heat rejection from the working fluid to the air in the condenser is determined from the steady-flow energy balance on air, & & Qair = m air c p (T9 - T8 ) = (4,195,100 lbm/h )(0.24 Btu/lbm F)(84.5 - 55 F) = 29.7 MBtu/h (c) The mass flow rate of geothermal water at the preheater is determined from the steady-flow energy balance on the geothermal water, & - 11,140,000 Btu/h = m geo (1.03 Btu/lbm F)(154.0 - 211.8 F) & m geo = 187,120 lbm/h & & Qgeo = m geo c p (Tout - Tin ) (d) The rate of heat input is & & & Qin = Qvaporizer + Qreheater = 22,790,000 + 11140,000 , and = 33,930,000 Btu / h & Wnet = 1271 - 200 = 1071 kW Then, th = & 3412.14 Btu W net 1071 kW = 10.8% = & 33,930,000 Btu/h 1 kWh Qin 10-81 10-91 A steam power plant operates on the simple ideal Rankine cycle. The turbine inlet temperature, the net power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required 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 @ 7.5 kPa = 168.75 kJ/kg v 1 = v f @ 7.5 kPa = 0.001008 m 3 /kg T1 = Tsat @ 7.5 kPa = 40.29 C wp,in = v 1 (P2 - P1 ) 1 kJ = 0.001008 m 3 /kg (6000 - 7.5 kPa ) 1 kPa m 3 = 6.04 kJ/kg T 3 ( ) 6 MPa 2 1 qin 7.5 kPa qout 4 s h2 = h1 + wp,in = 168.75 + 6.04 = 174.79 kJ/kg h4 = h g @ 7.5 kPa = 2574.0 kJ/kg s 4 = s g @ 7.5 kPa = 8.2501 kJ/kg P3 = 6 MPa s3 = s4 h3 = 4852.2 kJ/kg T = 1089.2 C 3 (b) qin = h3 - h2 = 4852.2 - 174.79 = 4677.4 kJ/kg qout = h4 - h1 = 2574.0 - 168.75 = 2405.3 kJ/kg wnet = qin - qout = 4677.4 - 2405.3 = 2272.1 kJ/kg and th = Thus, wnet 2272.1 kJ/kg = = 48.6% q in 4677.4 kJ/kg & & W net = th Qin = (0.4857 )(40,000 kJ/s ) = 19,428 kJ/s (c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the condenser, which is 40.29 C, & & & Qout = Qin - Wnet = 40,000- 19,428 = 20,572 kJ/s & mcool = & 20,572 kJ/s Qout = = 194.6 kg/s (4.18 kJ/kg C)(40.29 - 15 C) cT 10-82 10-92 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. The thermal efficiency of the cycle and the mass flow rate of the steam 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 @ 5 kPa = 137.75 kJ/kg v1 = v f @ 5 kPa = 0.001005 m3/kg wp,in = v1 (P2 - P ) 1 1 kJ = 0.001005 m3/kg (15,000 - 5 kPa ) 1 kPa m3 = 15.07 kJ/kg 15 MPa 2 5 kPa T 5 MPa 1 MPa 3 4 5 6 7 ( ) h2 = h1 + wp,in = 137.75 + 15.07 = 152.82 kJ/kg P3 = 15 MPa h3 = 3310.8 kJ/kg T3 = 500 C s3 = 6.3480 kJ/kg K P4 = 5 MPa h4 = 3007.4 kJ/kg s4 = s3 P5 = 5 MPa h5 = 3434.7 kJ/kg T5 = 500 C s5 = 6.9781 kJ/kg K P6 = 1 MPa h6 = 2971.3 kJ/kg s6 = s5 1 8 s P7 = 1 MPa h7 = 3479.1 kJ/kg T7 = 500 C s7 = 7.7642 kJ/kg K s8 - s f 7.7642 - 0.4762 = = 0.9204 P8 = 5 kPa x8 = s fg 7.9176 s8 = s7 h8 = h f + x8h fg = 137.75 + (0.9204 )(2423.0 ) = 2367.9 kJ/kg Then, qin = (h3 - h2 ) + (h5 - h4 ) + (h7 - h6 ) = 3310.8 - 152.82 + 3434.7 - 3007.4 + 3479.1 - 2971.3 = 4093.1 kJ/kg qout = h8 - h1 = 2367.9 - 137.75 = 2230.2 kJ/kg wnet = qin - qout = 4093.1 - 2230.2 = 1862.9 kJ/kg Thus, th = (b) & m= wnet 1862.9 kJ/kg = = 45.5% 4093.1 kJ/kg q in & Wnet 120,000 kJ/s = = 64.4 kg/s wnet 1862.9 kJ/kg 10-83 10-93 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis 5 Boiler 6 y 4 Open fwh P II 3 Turbine 1-y 7 Condenser 2 4 T qin 10 MPa y 6 6s 7s 1-y 7 s 5 3 0.5 MPa 10 kPa 1 qout 2 PI 1 (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = hf @ 10 kPa = 191.81 kJ/kg v1 = v f @ 10 kPa = 0.00101 m3/kg wpI,in = v1(P - P ) / p 2 1 1 kJ = 0.00101m3/kg (500-10 kPa) 1 kPa m3 /(0.95) = 0.52 kJ/kg ( ) h2 = h1 + wpI,in = 191.81+ 0.52 = 192.33 kJ/kg P = 0.5 MPa h3 = hf @ 0.5 MPa = 640.09 kJ/kg 3 v = v 3 sat.liquid f @ 0.5 MPa = 0.001093 m /kg 3 wpII,in = v 3 (P4 - P3 ) / p 1 kJ = 0.001093 m3/kg (10,000 - 500 kPa ) 1 kPa m3 / (0.95) = 10.93 kJ/kg ( ) h4 = h3 + wpII,in = 640.09 + 10.93 = 651.02 kJ/kg P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500 C s 5 = 6.5995 kJ/kg K 6.5995 - 1.8604 = = 0.9554 4.9603 s fg P6 s = 0.5 MPa h6 s = h f + x 6 s h fg = 640.09 + (0.9554)(2108.0) s 6s = s5 = 2654.1 kJ/kg x6s = s6s - s f T = h5 - h6 h6 = h5 - T (h5 - h6 s ) h5 - h6 s = 3375.1 - (0.80)(3375.1 - 2654.1) = 2798.3 kJ/kg 10-84 6.5995 - 0.6492 = = 0.7934 7.4996 s fg P7 s = 10 kPa h7 s = h f + x 7 s h fg = 191.81 + (0.7934 )(2392.1) s7s = s5 = 2089.7 kJ/kg x7s = s7s - s f T = h5 - h7 h7 = h5 - T (h5 - h7 s ) h5 - h7 s = 3375.1 - (0.80)(3375.1 - 2089.7 ) = 2346.8 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 ke pe 0 , & & & E in - E out = E system & & m h = m h i i 0 (steady) =0 & & Ein = E out e e & & & m6 h6 + m2 h2 = m3 h3 yh6 + (1 - y )h2 = 1(h3 ) & & where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, y= h3 - h2 640.09 - 192.33 = = 0.1718 h6 - h2 2798.3 - 192.33 Then, qout = (1 - y )(h7 - h1 ) = (1 - 0.1718)(2346.8 - 191.81) = 1784.7 kJ/kg wnet = qin - qout = 2724.1 - 1784.7 = 939.4 kJ/kg qin = h5 - h4 = 3375.1 - 651.02 = 2724.1 kJ/kg and & m= & Wnet 150,000 kJ/s = = 159.7 kg/s wnet 939.4 kJ/kg q out 1784.7 kJ/kg = 1- = 34.5% q in 2724.1 kJ/kg (b) The thermal efficiency is determined from th = 1 - Also, P6 = 0.5 MPa s6 = 6.9453 kJ/kg K h6 = 2798.3 kJ/kg s3 = s f @ 0.5 MPa = 1.8604 kJ/kg K s2 = s1 = s f @ 10 kPa = 0.6492 kJ/kg K Then the irreversibility (or exergy destruction) associated with this regeneration process is 0 q iregen = T0 sgen = T0 me se - mi si + surr = T0 [s3 - ys6 - (1 - y )s2 ] TL = (303 K )[1.8604 - (0.1718)(6.9453) - (1 - 0.1718)(0.6492 )] = 39.25 kJ/kg 10-85 10-94 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes negligible. are Analysis 5 Boiler 6 y 4 Open fwh P II 3 Turbine 1-y 7 Condenser 2 T qin 4 10 MPa y 6 1-y qout 7 s 5 3 0.5 MPa 10 kPa 2 PI 1 1 (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa @ 10 kPa = 191.81 kJ/kg = 0.00101 m 3 /kg v1 = v f wpI,in = v 1 (P2 - P1 ) 1 kJ = 0.00101 m 3 /kg (500 - 10 kPa ) 1 kPa m 3 h2 = h1 + wpI,in = 191.81 + 0.50 = 192.30 kJ/kg ( ) = 0.50 kJ/kg P3 = 0.5 MPa sat.liquid h3 = h f @ 0.5 MPa = 640.09 kJ/kg v = v 3 f @ 0.5 MPa = 0.001093 m /kg 3 wpII,in = v 3 (P4 - P3 ) 1 kJ = 0.001093 m3/kg (10,000 - 500 kPa ) 1 kPa m3 = 10.38 kJ/kg h4 = h3 + wpII,in = 640.09 + 10.38 = 650.47 kJ/kg ( ) P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500 C s5 = 6.5995 kJ/kg K s6 - s f 6.5995 - 1.8604 = = 0.9554 P6 = 0.5 MPa x6 = s fg 4.9603 s6 = s5 h6 = h f + x6 h fg = 640.09 + (0.9554)(2108.0 ) = 2654.1 kJ/kg s7 - s f 6.5995 - 0.6492 = = 0.7934 P7 = 10 kPa x7 = s fg 7.4996 s7 = s5 h7 = h f + x7 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg 10-86 The fraction of steam extracted is determined from the steady-flow energy equation applied to the & & feedwater heaters. Noting that Q W ke pe 0 , & & & E in - E out = E system & & m h = m h i i e e 0 (steady) & & = 0 E in = E out & & & m6 h6 + m2 h2 = m3 h3 yh6 + (1 - y )h2 = 1(h3 ) & & where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, y= h3 - h2 640.09 - 192.31 = = 0.1819 h6 - h2 2654.1 - 192.31 Then, qout = (1 - y )(h7 - h1 ) = (1 - 0.1819)(2089.7 - 191.81) = 1552.7 kJ/kg wnet = qin - qout = 2724.6 - 1552.7 = 1172.0 kJ/kg & Wnet 150,000 kJ/s = = 128.0 kg/s wnet 1171.9 kJ/kg qin = h5 - h4 = 3375.1 - 650.47 = 2724.6 kJ/kg and & m= (b) The thermal efficiency is determined from th = 1 - Also, q out 1552.7 kJ/kg = 1- = 43.0% q in 2724.7 kJ/kg s 6 = s 5 = 6.5995 kJ/kg K s3 = s f @ 0.5 MPa = 1.8604 kJ/kg K = 0.6492 kJ/kg K s 2 = s1 = s f @ 10 kPa Then the irreversibility (or exergy destruction) associated with this regeneration process is 0 q iregen = T0 sgen = T0 me se - mi si + surr = T0 [s3 - ys6 - (1 - y )s2 ] TL = (303 K )[1.8604 - (0.1819 )(6.5995) - (1 - 0.1819 )(0.6492)] = 39.0 kJ/kg 10-87 10-95 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration 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 f @ 15 kPa = 225.94 kJ/kg v1 = v f @ 15 kPa = 0.001014 m 3 /kg 5 wpI,in = v 1 (P2 - P1 ) 1 kJ = 0.001014 m 3 /kg (600 - 15 kPa ) 1 kPa m 3 = 0.59 kJ/kg ( ) Boiler 6 7 Turbine 1-y 9 Condens. 2 PI 1 8 y h2 = h1 + wpI,in = 225.94 + 0.59 = 226.53 kJ/kg P3 = 0.6 MPa sat. liquid h3 = h f @ 0.6 MPa = 670.38 kJ/kg v = v 3 f @ 0.6 MPa = 0.001101 m /kg 3 4 Open fwh P II 3 wpII,in = v 3 (P4 - P3 ) 1 kJ = 0.001101 m3/kg (10,000 - 600 kPa ) 1 kPa m3 = 10.35 kJ/kg T ( ) h4 = h3 + wpII,in = 670.38 + 10.35 = 680.73 kJ/kg P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500 C s5 = 6.5995 kJ/kg K P6 = 1.0 MPa h6 = 2783.8 kJ/kg s6 = s5 P7 = 1.0 MPa h7 = 3479.1 kJ/kg T7 = 500 C s7 = 7.7642 kJ/kg K P8 = 0.6 MPa h8 = 3310.2 kJ/kg s8 = s7 1 MPa 7 5 10 MPa 4 3 2 15 kPa 1 9 s 0.6 MPa 6 8 s9 - s f 7.7642 - 0.7549 = = 0.9665 P9 = 15 kPa x9 = s fg 7.2522 s9 = s7 h9 = h f + x9 h fg = 225.94 + (0.9665)(2372.3) = 2518.8 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 ke pe 0 , & & & E in - E out = E system & & m h = m h i i e e 0 (steady) & & = 0 Ein = E out & & & m8 h8 + m 2 h2 = m3 h3 yh8 + (1 - y )h2 = 1(h3 ) & & where y is the fraction of steam extracted from the turbine ( = m8 / m3 ). Solving for y, y= h3 - h2 670.38 - 226.53 = = 0.144 h8 - h2 3310.2 - 226.53 (b) The thermal efficiency is determined from q in = (h5 - h4 ) + (h7 - h6 ) = (3375.1 - 680.73) + (3479.1 - 2783.8) = 3389.7 kJ/kg q out = (1 - y )(h9 - h1 ) = (1 - 0.1440)(2518.8 - 225.94 ) = 1962.7 kJ/kg q out 1962.7 kJ/kg = 1- = 42.1% 3389.7 kJ/kg q in and th = 1 - 10-88 10-96 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration 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 5 Boiler 6 7 Turbine 1-y 9 Condenser 2 T 5 4 3 7 8 y 6 6s 8s 8 y 1-y 4 Open fwh P II 3 2 PI 1 1 9s 9 s (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg v 1 = v f @ 15 kPa = 0.001014 m 3 /kg w pI ,in = v 1 (P2 - P1 ) 1 kJ = 0.001014 m 3 /kg (600 - 15 kPa ) 1 kPa m 3 = 0.59 kJ/kg h3 = h f @ 0.6 MPa = 670.38 kJ/kg v = v 3 f @ 0.6 MPa = 0.001101 m /kg 3 ( ) h2 = h1 + w pI ,in = 225.94 + 0.59 = 226.54 kJ/kg P3 = 0.6 MPa sat. liquid wpII,in = v 3 (P4 - P3 ) 1 kJ = 0.001101 m3/kg (10,000 - 600 kPa ) 1 kPa m3 = 10.35 kJ/kg ( ) h4 = h3 + wpII,in = 670.38 + 10.35 = 680.73 kJ/kg P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500 C s5 = 6.5995 kJ/kg K P6 s = 1.0 MPa h6 s = 2783.8 kJ/kg s6 s = s5 T = h5 - h6 h6 = h5 - T (h5 - h6 s ) h5 - h6 s = 3375.1 - (0.84 )(3375.1 - 2783.8) = 2878.4 kJ/kg 10-89 P7 = 1.0 MPa h7 = 3479.1 kJ/kg T7 = 500 C s7 = 7.7642 kJ/kg K P8 s = 0.6 MPa h8 s = 3310.2 kJ/kg s8 s = s7 T = h7 - h8 h8 = h7 - T (h7 - h8 s ) = 3479.1 - (0.84 )(3479.1 - 3310.2 ) h7 - h8 s = 3337.2 kJ/kg s 9 s - s f 7.7642 - 0.7549 = = 0.9665 P9 s = 15 kPa x9 s = s fg 7.2522 s9s = s7 h9 s = h f + x9 s h fg = 225.94 + (0.9665)(2372.3) = 2518.8 kJ/kg T = h7 - h9 h9 = h7 - T (h7 - h9 s ) = 3479.1 - (0.84)(3479.1 - 2518.8) h7 - h9 s = 2672.5 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 ke pe 0 , & & & E in - E out = E system & & m h = m h i i 0 (steady) =0 & & E in = E out e e & & & m8 h8 + m2 h2 = m3 h3 yh8 + (1 - y )h2 = 1(h3 ) & & where y is the fraction of steam extracted from the turbine ( = m8 / m3 ). Solving for y, y= h3 - h2 670.38 - 226.53 = = 0.1427 h8 - h2 3335.3 - 226.53 (b) The thermal efficiency is determined from qin = (h5 - h4 ) + (h7 - h6 ) = (3375.1 - 680.73) + (3479.1 - 2878.4 ) = 3295.1 kJ/kg qout = (1 - y )(h9 - h1 ) = (1 - 0.1427 )(2672.5 - 225.94 ) = 2097.2 kJ/kg and th = 1 - q out 2097.2 kJ/kg = 1- = 36.4% q in 3295.1 kJ/kg 10-90 10-97 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for the open feedwater heater, the thermal efficiency of the cycle, and the net power output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis 8 Boiler 9 10 11 y 5 Closed fwh 4 6 P II 3 Open fwh I 2 PI 12 z High-P Turbine Low-P Turbine 13 1-y-z 2 T 15 MPa 6 0.6 MPa 9 1 MPa 8 5 4 3 10 y 11 Condenser 1 1 z 12 0.2 MPa 7 1-y-z 5 kPa 13 s 7 (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 5 kPa @ 5 kPa = 137.75 kJ/kg = 0.001005 m 3 /kg v1 = v f wpI,in = v 1 (P2 - P1 ) 1 kJ = 0.001005 m 3 /kg (200 - 5 kPa ) 1 kPa m 3 = 0.20 kJ/kg ( ) h2 = h1 + wpI,in = 137.75 + 0.20 = 137.95 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 wpII,in = v 3 (P4 - P3 ) 1 kJ = 0.001061 m3/kg (15,000 - 200 kPa ) 1 kPa m3 = 15.70 kJ/kg h4 = h3 + wpII,in = 504.71 + 15.70 = 520.41 kJ/kg ( ) P6 = 0.6 MPa h6 = h7 = h f @ 0.6 MPa = 670.38 kJ/kg 3 sat.liquid v 6 = v f @ 0.6 MPa = 0.001101 m /kg T6 = T5 h5 = h6 + v 6 (P5 - P6 ) 1 kJ = 670.38 + 0.001101 m 3 /kg (15,000 - 600 kPa ) 1 kPa m 3 = 686.23 kJ/kg ( ) P8 = 15 MPa T8 = 600 C h8 = 3583.1 kJ/kg s = 6.6796 kJ/kg K 8 P9 = 1.0 MPa h9 = 2820.8 kJ/kg s 9 = s8 P10 = 1.0 MPa h10 = 3479.1 kJ/kg T10 = 500 C s10 = 7.7642 kJ/kg K 10-91 P = 0.6 MPa 11 h11 = 3310.2 kJ/kg s11 = s10 P = 0.2 MPa 12 h12 = 3000.9 kJ/kg s12 = s10 x13 = s13 - s f s fg = 7.7642 - 0.4762 7.9176 = 0.9205 P = 5 kPa 13 h13 = h f + x13h fg s13 = s10 = 137.75 + (0.9205)(2423.0 ) = 2368.1 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 ke pe 0 , & & & E in - E out = E system & & m h = m h i i 0 (steady) =0 & & Ein = E out e e & & m11 (h11 - h6 ) = m5 (h5 - h4 ) y (h11 - h6 ) = (h5 - h4 ) & & where y is the fraction of steam extracted from the turbine ( = m11 / m5 ). Solving for y, y= h5 - h4 686.23 - 520.41 = = 0.06287 h11 - h6 3310.2 - 670.38 & & & E in - E out = E system & & m h = m h i i 0 (steady) For the open FWH, =0 & & E in = E out e e & & & & m 7 h7 + m 2 h2 + m12 h12 = m3 h3 yh7 + (1 - y - z )h2 + zh12 = (1)h3 & & where z is the fraction of steam extracted from the turbine ( = m12 / m5 ) at the second stage. Solving for z, z= (h3 - h2 ) - y(h7 - h2 ) h12 - h2 = 504.71 - 137.95 - (0.06287 )(670.38 - 137.95) = 0.1165 3000.0 - 137.95 (b) q in = (h8 - h5 ) + (h10 - h9 ) = (3583.1 - 686.23) + (3479.1 - 2820.8) = 3555.3 kJ/kg q out = (1 - y - z )(h13 - h1 ) = (1 - 0.06287 - 0.1165)(2368.0 - 137.75) = 1830.4 kJ/kg wnet = q in - q out = 3555.3 - 1830.4 = 1724.9 kJ/kg and th = 1 - (c) q out 1830.4 kJ/kg = 1- = 48.5% 3555.3 kJ/kg q in & & Wnet = mwnet = (42 kg/s )(1724.9 kJ/kg ) = 72,447 kW 10-92 10-98 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies 7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating 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), 6 Turbine h1 = h f @ 15 kPa = 225.94 kJ/kg Boiler 7 h2 h1 h3 = h f @ 120 C = 503.81 kJ/kg P6 = 8 MPa h6 = 3399.5 kJ/kg T6 = 500 C s6 = 6.7266 kJ/kg K P7 = 1 MPa h7 = 2843.7 kJ/kg s7 = s6 8 5 Process heater 3 P II 4 2 PI 9 Condenser 1 P8 = 1 MPa h8 = 3479.1 kJ/kg T8 = 500 C s8 = 7.7642 kJ/kg K T s9 - s f 7.7642 - 0.7549 = = 0.9665 x9 = s fg 7.2522 P9 = 15 kPa h9 = h f + x9 h fg = 225.94 + (0.9665)(2372.3) s9 = s8 = 2518.8 kJ/kg 6 8 MPa 5 1 MPa 4 3 7 8 The mass flow rate through the process heater is & Qprocess 7,000 kJ/s & = = 2.992 kg/s m3 = h7 - h3 (2843.7 - 503.81) kJ/kg Also, or, 2 15 kPa 1 9 s & & & & & WT = m6 (h6 - h7 ) + m9 (h8 - h9 ) = m6 (h6 - h7 ) + (m6 - 2.993)(h8 - h9 ) & & 3000 kJ/s = m6 (3399.5 - 2843.7 ) + (m6 - 2.992 )(3479.1 - 2518.8) & m6 = 3.873 kg/s & & & m9 = m6 - m3 = 3.873 - 2.992 = 0.881 kg/s in out It yields and Mixing chamber: & & E -E & = E system 0 (steady) =0 & & & m4 h4 = m2 h2 + m3h3 & & m h = m h i i & & E in = E out e e or, Then, h4 h5 = & & m2 h2 + m3h3 (0.881)(225.94 ) + (2.992)(503.81) = = 440.60 kJ/kg & m4 3.873 & & & Qin = m6 (h6 - h5 ) + m8 (h8 - h7 ) = (3.873 kg/s )(3399.5 - 440.60 kJ/kg ) + (0.881 kg/s )(3479.1 - 2843.7 kJ/kg ) = 12,020 kW (b) The fraction of steam extracted for process heating is & m3 2.992 kg/s = = 77.3% y= & m total 3.873 kg/s

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