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Data for a regenerative vapor power cycle using an open and a closed feedwater heater are provided in the table below. Steam enters the turbine at 14 MPa, 560C,560^{\circ} \mathrm{C}, state 1, and expands isentropically in three stages to a condenser pressure of 80 kPa, state 4. Saturated liquid exiting the condenser at state 5 is pumped isentropically to state 6 and

Statep(kPa)T(C)h (kJ/kg) s(kJ/kgK)x114,0005603486.06.5941 -21,0002781.66.5941 -32002497.06.5941 0.9048 48093.52357.66.5941 0.8645 58093.5391.661.2329 06200391.701.2329 -7200504.701.5301 0814,000504.711.5301 -914,000170719.212.0419 -101,000762.812.1387 011200762.812.1861 0.1172 \begin{matrix} \text{State} & \text{p(kPa)} & T(^\circ C) & \text{h (kJ/kg) } & s(kJ/kg \cdot K) & \text{x}\\ \text{1} & \text{14,000} & \text{560} & \text{3486.0} & \text{6.5941 } & \text{-}\\ \text{2} & \text{1,000} & \quad & \text{2781.6} & \text{6.5941 } & \text{-}\\ \text{3} & \text{200} & \quad & \text{2497.0} & \text{6.5941 } & \text{0.9048 }\\ \text{4} & \text{80} & \text{93.5} & \text{2357.6} & \text{6.5941 } & \text{0.8645 }\\ \text{5} & \text{80} & \text{93.5} & \text{391.66} & \text{1.2329 } & \text{0}\\ \text{6} & \text{200} & \quad & \text{391.70} & \text{1.2329 } & \text{-}\\ \text{7} & \text{200} & \quad & \text{504.70} & \text{1.5301 } & \text{0}\\ \text{8} & \text{14,000} & \quad & \text{504.71} & \text{1.5301 } & \text{-}\\ \text{9} & \text{14,000} & \text{170} & \text{719.21} & \text{2.0419 } & \text{-}\\ \text{10} & \text{1,000} & \quad & \text{762.81} & \text{2.1387 } & \text{0}\\ \text{11} & \text{200} & \quad & \text{762.81} & \text{2.1861 } & \text{0.1172 }\\ \end{matrix}

enters the open feedwater heater. Between the first and second turbine stages, some steam is extracted at 1 MPa, state 2, and diverted to the closed feedwater heater. The diverted steam leaves the closed feedwater heater as saturated liquid at 1 MPa, state 10, undergoes a throttling process to 0.2 MPa, state 11, and enters the open feedwater heater. Steam is also extracted between the second and third turbine stages at 0.2 MPa, state 3, and diverted to the open feedwater heater. Saturated liquid at 0.2 MPa exiting the open feedwater heater at state 7 is pumped isentropically to state 8 and enters the closed feedwater heater. Feedwater exits the closed feed water heater at 14 MPa, 170C,170^\circ C, state 9, and then enters the steam generator. If the net power developed by the cycle is 300 MW, determine a. the cycle thermal efficiency. b. the mass flow rate into the first turbine stage, in kg/s. c. the rate of heat transfer from the working fluid as it passes through the condenser, in MW.

Question

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480C,480^{\circ} \mathrm{C}, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440C440^{\circ} \mathrm{C} and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210C,210^{\circ} \mathrm{C}, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle a. the heat transfer to the working fluid passing through the steam generator, in kJ per kg of steam entering the first-stage turbine. b. the thermal efficiency. c. the heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam entering the first-stage turbine.

Solution

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Given data:

p1=12MPa=120barp_1=12\mathrm{MPa}=120\mathrm{bar}

T1=480CT_1=480\mathrm{^{\circ}C}

p1=2MPa=20barp_{1'}=2\mathrm{MPa}=20\mathrm{bar}

T1=440CT_{1'}=440\mathrm{^{\circ}C}

p2=2MPa=20barp_2=2\mathrm{MPa}=20\mathrm{bar}

p3=0.3MPa=3barp_3=0.3\mathrm{MPa}=3\mathrm{bar}

pc=p4=6kPa=0.06barp_c=p_4=6\mathrm{kPa}=0.06\mathrm{bar}

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