Question

3. (10 pts) A simple Rankine cycle operates with water as the working fluid between the pressures of 4 MPa and 20 kPa. The fluid leaves the boiler at 550°C, the turbine efficiency is 89% and the fluid exits the condenser as saturated liquid. The flow rate is 81.5 kg/s and the pump efficiency is 52%. Determine: a. The net power output, in kW. b. The rate of heat input in the boiler, in kW. C. The rate of heat rejection in the condenser, in kW. d. The overall efficiency of the Rankine cycle engine.

Answer 1

Given that, 1-2 Igentropic compression 2-3 const Preplete Heat abitinn 3-4 Isentropie expansion uri const pre Bule Heat T1 s > P2 = P₂ = umpa, 3=5500 - Pi = Py = 20kPa., M. 81.5kg/ / nr = 89% pump=521. at P = Pu = 20kPa, hf = h, = 251.40 kJ/kg ;hg = 2609.7 k5/4g sf = 0.8320 kJ/kg-16, sg=sey =7-9085 #3/kg-k. , a1.0172 xíó? m?leg. at P2=4mpa, 3 = 5500 h3 = 3559.26 43/49, 5z = 7.2398 kJ/E8-k. ha= hit cumpe. » Wrump = hachi sp, 10172x10° 4000 - 20) = 7.18 43/44 Wpump = 8 AP = 10172x103 [1000 = 20) npump. 0.52 Warmp = 7-78 KS/K9 -7 ha= hit Wpump = 251.040+ 7.78 [h2 = 259.18 kJ/kg) 33=54 => 72328= 0.9320 + xy [4.9085-0-9320] pu= 0.9d hur het dy [hg-hif] hu = 251.40 70.9022609.7-251.46] - The = 23 73-8713563

. actual enthalpy drop -ha-hul Murbine - Ideal enthalpy drop ha-hy 3559.26 - hal 0-89= => 3559.26-hu' = 1054.99 3559.26-2373.87 k) - 250 4.2 + 15/1| Wfurbine = hathu = 3559.26 -2584.27 whorbine = 1054.99 kJ/kg/ Heat addition, Qin = ha ha = 3559.26-259.18 Din = 3300.08 kJ/ky] a) nat power out where in (w when he in 8.5x11054.99-7.78] - W = 81.5X1047-21 = 85347.61 kW =85347-61kw/ “xan = 8 = 81.5X3300.08268956.52 kw lain = 268956-52 kw c) dorm Chis-hi] =81.5x (2504.27-/251.40] = 183608.90kw = 183608.90 kw 85347. 6) 268956:52 = 0.3173 - lover in loverall = din [ Moverall = 31.73 %)