Question

A proposed steam power plant design consists of an ideal Rankine cycle with reheat and regeneration. Steam enters Turbine 1 at Pl and T1 at the rate of mi and exits at P2. A fraction () of the steam exiting Turbine 1 is diverted to an open feedwater heater while the remainder is reheated to T3 before entering Turbine 2. The condenser operates at P4. Saturated liquid exits the condenser and is fed to Pump 1. The outlet of Pump 1 is fed into the open feedwater heater. Saturated liquid exits the feedwater heater and is fed to Pump 2. All turbines and pumps are isentropic. --Given Values-- mi (kg 3) = 35 P1 (Bar) = 100 T1 (C) = 520 P2 (Bar) = 5 T3 (C) = 530 P4 (Bar) = 0.08 a) Determine the specific enthalpy (kJ/kg) at the inlet of turbine 1. Your Answer=3425.1 Correct! Exact Answer=3425.1 + - 1.16E-00 b) Determine the specific enthalpy (kJ/kg) at the exit of turbine 1. Your Answer = c) Determine the specific enthalpy (kJ/kg) at the inlet of turbine 2. Your Answer=1 d) Determine the specific enthalpy (kJ/kg) at the exit of turbine 2. Your Answer = e) Determine the specific enthalpy (kJ/kg) at the condenser exit. Your Answer = f) Determine the specific enthalpy (kJ/kg) at the exit of the low pressure pump. Your Answer g) Determine the specific enthalpy (kJ/kg) at the exit of the feedwater heater. Your Answer = h) Determine the specific enthalpy (kJ/kg) at the exit of the high pressure pump. Your Answer = i) Determine the fraction (y') of flow diverted to the open feedwater heater. Your Answer = 1) Determine the power (MW) produced by turbine 1. Your Answer = k) Determine the power (MW) produced by turbine 2. Your Answer 1) Determine the power (kW) required (a positive number) by the low pressure pump. Your Answer=L m) Determine the power (kW) required a positive number) by the high pressure pump Your Answer n) Determine the total rate of heat transfer (MW) supplied to the boiler. Your Answer = o) Determine the thermal efficiency (%) of the power plant.

Answer 1

4 LO Pq = 0.08bas: se 20.5925: 59 28, 2274 kJ/K9k. P, = 100 bag : 7 =5206: From superheated tables @ P, FTI. can h = 3426.3 kJ/kg S,2616647 kJ/xgK. (6) using sies, [@ P₂ = 5bay S9, 2618205 kJ/kakes, & 2' is in mixed] 21-542+42 [Sg2+2); 6:6647 = 1.8605 +*2*[6.8205–118605) > >2=0.968 ha a hfzt ha Chg h = 640 18+ 01968 [2748:1-640118] = 2680.6 kJ/169) (C) Prom Supeas heated tables @p=5 bary: Tz=530°C hz = 3549,1 KT/kg $z=9:1712 IT/kgk. (d) using Sza54 [@pq = 0.08bas: se = 0.5925: 594 3.20 8.712 20.5925+*[8.2274-015925) 719=0.9926. - 173 854 792 C331 2–1939 s )=2ss8 skyga 1 ce) (his theolog bas = 173.95 kJ/159) Usato? m?kg 1 ton (f) using to 3 [5-0.08]x 100 = hehs hghs = 0.492 T [h6 = 174.3 kJ/K9 & 4 eg) (hz ehf@ 5bar = 640.18 kJ/kg) Chi 103 [100-5]*100 = hq hz hg hz 9.5 h 9 = 649.68 kJ/kg S yiha +(35–43 h6 = 35xha. => [9= 6.50 kg/s? WT, = 35x(h, he] =26. 1 mW) (k) we = 135-6.5) (hghan) - 28.2 MW] CL) Wep =(35-6.5) (ho-hs) = 16102 kW) cm) Wap= 35x Chg hz)=(332.5 kW) en] H'S = 35xCh, the) +(35-y) (ba-b27= 97.181+24.752 </21.93 MW) (26:1+28.2) - (0.332 +0.014) -49,24). CO o 12 — 121.93