Question

Problem B. Analyze a Rankine cycle with single-reheat (so include a HPT and LPT) with cycle mass flowrate of 500 kg/s. The main steam-generator produces super-heated vapor at 320 bar, 600 °C. The reheater operates at constant pressure of 20 bar and produces super-heated vapor at 540 °C. The condenser operates at 0.06 bar. The isentropic efficiency of both turbines is 90%. The boiler-feed-pump (BFP) isentropic efficiency is 75%. Neglect external heat transfer from the pump and turbines. Neglect the small frictional flow resistances in the (constant pressure) heat transfer components. Changes in kinetic and potential energy are insignificant. Complete the cycle analysis to determine component energy transfer rates: 6Q 1,5G = _ _ MW, 20 3,RH = __ MW, W 2.HPT = __ MW, 3W 4,LPT=__ MW, 5W 6BFP = __ MW Also calculate the cycle thermal efficiency = __ %

Answer 1

Draw the property diagram for Rankine cycle with single reheat. TA [email protected]°C 4@540°C 320 bar 20 bar 0.06 bar 5S 5 Find the properties of the steam at 320 bar and 600°C. h = 3423.8 kJ/kg S2 =6.185 kJ/kg .K Find the properties at state2 at 15 bar. Szs = S2 = 6.379 kJ/kg .K ho = 2723.6 kJ/kg The fluid is in two phase region. Calculate the actual specific enthalpy at state 3. h = h, - [nopt (ha – h)] h = 3423.8-[0.9x(3423.8 – 2723.6)] h = 2793.62 kJ/kg.K Find the properties at 20 bar and 540°C.

ha = 3556.2 kJ/kg S4 = 7.543 kJ/kg .K Find the properties at state 5 and 0.06 bar. Sigg = S4 = 7.543 kJ/kg · K<($). Therefore, the inlet of the condenser will be in the saturation region. @0.1 bar Calculate the dryness fraction at state 5s. $-(st) @0.06 bar X5, T58)@o.oo bar 7.572 – 0.521 Xgs = 7.807 X53 = 0.903 Find the specific enthalpy at state 5s. hos =(hy) acol bar + x3, (hi) 2.00 bar his = 151.5+(0.903 x 2415) his = 2332.24 kJ/kg Calculate the actual specific enthalpy at state 5. h; = h- [Nupt (ha – hşs)] h = 3556.2 -[0.9x (3556.2 – 2332.24)] ho = 2454.636 kJ/kg <(h), (still, in the saturation region) g) @ 0.1 bar Find the properties of saturated liquid at 0.06 bar. ho = (hq) cool bar = 151.5 kJ/kg So = (st) = 0.521 kJ/kg.K 0.06 bar

Find the properties at s, and 320 bar. hs = 183.5 kJ/kg Calculate the actual specific enthalpy at state 1. nho-hotho 1 ПВЕР 183.5–151.5 -151.5 m = 0.75 h = 194.16 kJ/kg Calculate the heat addition to the cycle. (9 total )in = 91-2 +93-4 (4 total)in = (ha - h) +(ha – hz) (total) = (3423.8–194.16) – (3556.2 – 2793.62) (qtotal) in = 2467.06 kJ/kg (Qtotal ) = 2467.06x 500 kW (Ciotal); n = 1233.53 MW Calculate the power output from both the turbines. (Wout rotal = WHPT + WLPT (Wout ) rotal = (h – hz)+(ha – ha) (Wout Data = (3423.8–2793.62)+(3556.2 – 2454.636) (Wout) total = 1731.74 kJ/kg (Wour)otal = 1731.74x 500 kW (Wout me = 865.87 MW out) total Calculate the work consumed by the boiler feed pump.

(win )BFP = (h – ho) (Win) Ep = 194.16-151.5 (Win )pfp = 42.66 kJ/kg (W. BFP = 42.66x 500 kW (Winerp = 21.33 MW cycle Calculate the cycle work output. (Wout Seycle = ( w out ) total – (Win) BFP (wout hari = 1731.74 – 42.66 (Wout cycle = 1689.08 kJ/kg (Wout )exote = 1689.08x 500 kW (Wout) = 844.54 MW Calculate the thermal efficiency of the cycle. (Wout)cycle Ncycle TQin total 844.54 Ncycle – 1233.53 Ncycle = 0.6846 %n cycle = 68.46%