09 09 19sol

Pagina 1

Problem 1 Data cogen. Steam sat. temperature 110 °C T5 = 120 °C ma = 10 kg/s Tamb = 15 °C pamb = 1.013 bar beta,c = 6 TIT = 800 °C cp,a = 1.05 kJ/kg/K Dp,i = 1 % MMa = 28.8 kg/kmole eta,is = 80 % eta,me = 94 % U WHR = 45 W/m2K ma' = 6 kg/s UWHR' = 20 W/m2K b) cv,a = 0.761 kJ/kg/K gamma = 1.379 p1 = 1.003 T1 = T0 = 15.0 °C = 288.15 K p2 = p1*beta,c = 6.017 bar T2is = T1 * beta,c^((gamma-1)/gamma) = 471.6 K = 198.4 °C T2 = T1 + (T2is-T1)/eta,is = 244.3 °C = 517.4 K T3 = 800.0 °C = 1073.15 K p3 = p2*(1-Dp,i) = 5.957 bar p5 = pamb = 1.013 bar p4 = p5/(1-Dp,i) = 1.023 bar beta,t = p3/p4 = 5.822 T4is = T3 * beta,t^((1-gamma)/gamma) = 661.2 K = 388.0 °C T4 = T3 - (T3-T4,is)*eta,is = 470.4 °C = 743.6 K Qin = ma * cp,a * (T3-T2) = 5834.9 kW Pc = ma * cp,a * (T2-T1) = 2407.6 kW Pt = ma * cp,a * (T3-T4) = 3460.5 kW Pnet = (Pt - Pc)*eta,me = 989.8 kW eta,e = Pnet/Qin = 16.96 % QWHR = ma * cp,a * (T4-T5) 3679.5 kW Qin ~ 1 5 4 32 0 waste heat recuperator Qin steam evaporator c) p3' = p3*ma'/ma = 3.574 bar from definition of non-dim mass flow rate p4' = p4 = 1.023 bar beta,t' = 3.493 p2' = p3'/(1-Dpi) = 3.610 bar p1' = p1 = 1.003 bar beta,c' = 3.600 T2is' = T1* beta,c'^((gamma-1)/gamma) = 409.8 K = 136.6 °C T2' = T1 + (T2is'-T1)/eta,is = 167.1 °C = 440.2 K T4is' = T3 *beta,t'^((1-gamma)/gamma) = 760.9 K = 487.7 °C T4' = T3 - (T3-T4,is')*eta,is = 550.2 °C = 823.3 K d) DTm = (T4-TSAT)-(T5-TSAT)/LN(T4-TSA T 97.756 A = QWHR /DTmU 836.430 Cmin' 6300.000 NTU' = U'A/Cmin' 2.655 effectiveness' = 1-exp(-NTU') 0.930 QWHR' = ma' * cp,a * (T4'-TS)*effectivene s 2578.3 kW Problem 2 green background = input data m steam, kg/s 18.00 p steam turbine in, bar 60.00 T steam turbine in, °C 480.00 p steam turbine out, bar 18.00 Turbine isoentropic efficiency 0.85 Turbine mechanical efficiency 0.98…