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Example: Consider a steam power plant operating on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at
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given deta Aty 10 4-1 mointeinauce = 491, 81 t(0896) C2392-1) 3 the piessure at ohich stcam can be ReheateAIS puusmp

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steam power plant operating on the ideal reheat Rankine cycle is considered. For a specified moisture content at the turbine exit, the reheat pressure and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig. 10–13. We note that the power plant operates on the ideal reheat Rankine cycle. Therefore, the pump and the turbines are isentropic, there are no pressure drops in the boiler and condenser, and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure. (a) The reheat pressure is determined from the requirement that the entropies at states 5 and 6 be the same: State 6: P6 5 10 kPa x6 5 0.896 (sat. mixture) s6 5 sf 1 x6sfg 5 0.6492 1 0.896(7.4996) 5 7.3688 kJ/kg·K Also, h6 5 hf 1 x6hfg 5 191.81 1 0.896(2392.1) 5 2335.1 kJ/kg s T,°C Reheating 600 4 5 1 6 3 2 10 kPa 15 MPa 15 MPa 10 kPa 15 MPa Low-P turbine 6 Boiler High-P turbine Reheater P4 = P5 = Preheat Condenser Pump 3 2 5 4 10 kPa 1 FIGURE 10–13 Schematic and T-s diagram for Example 10–4. cen98179_ch10_553-606.indd 567 11/28/13 6:51 PM 568 VAPOR AND COMBINED POWER CYCLES Thus, State 5: T5 5 6008C s5 5 s6 f P5 5 4.0 MPa h5 5 3674.9 kJ/kg Therefore, steam should be reheated at a pressure of 4 MPa or lower to prevent a moisture content above 10.4 percent. (b) To determine the thermal efficiency, we need to know the enthalpies at all other states: State 1: P1 5 10 kPa Sat. liquid f h1 5 hf @ 10 kPa 5 191.81 kJ/kg v1 5 vf @ 10 kPa 5 0.00101 m3 /kg State 2: P2 5 15 MPa s2 5 s1 wpump,in 5 v1(P2 2 P1) 5 (0.00101 m3 /kg) 3 [(15,000 2 10)kPa]a 1 kJ 1 kPa·m3 b 5 15.14 kJ/kg h2 5 h1 1 wpump,in 5 (191.81 1 15.14) kJ/kg 5 206.95 kJ/kg State 3: P3 5 15 MPa T3 5 6008C f h3 5 3583.1 kJ/kg s3 5 6.6796 kJ/kg·K State 4: P4 5 4 MPa s4 5 s3 f h4 5 3155.0 kJ/kg (T4 5 375.58C) Thus qin 5 (h3 2 h2) 1 (h5 2 h4) 5 (3583.1 2 206.95) kJ/kg 1 (3674.9 2 3155.0) kJ/kg 5 3896.1 kJ/kg qout 5 h6 2 h1 5 (2335.1 2 191.81) kJ/kg 5 2143.3 kJ/kg and hth 5 1 2 qout qin 5 1 2 2143.3 kJ/kg 3896.1 kJ/kg 5 0.450 or 45.0%

answered by: Usman Khan
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