Question

1. A steam power plant steam power plant operates with a high pressure of 5 MPa and has a oller exit temperature of 600°C receiving heat from 700°C source. The Dient at 20°C provide cooling for the condenser so it can maintain 45°C side. All components are ideal except for the turbine, which has an exit state with a quality of 97%. i) Find the work and heat transfer in all components per kg of water and the turbine isentropic efficiency, ii) find the rate of entropy generation per kg of water in the boiler, turbine, pump.

Answer 1

2s State 1 Turbine inlet 9, = 5MPa, T, = 600c from supex heated steam table, his 3666-48 kilka , Sia 7.26 kilkg State 2: Turbine exit T2 = 45 c, xz = 0.97 At Coc Prat = 0.01 MPG, FR hf = 187.81 Koleg Sf = 0.637 kilkg hg = 2582.19 kilky Sg = 8.166 kilkg - 103 m 1kg - Now, ha = het 2 Chg-hif) - 2510.355 kilkg similarly, S2 = 7.94 kolkok

State 3 condensor outlet, Sat llouid at usc hz=h = 187.81 KT kg. Sf=5370-637 KolkgK. State u Boiler inlet. Pump work wp=Vf de = 103( 5-0.01) x103 wpa4.99 kulkg) Wp = hahz hu- 192.8 kJ/kg - Assuring isentropic expansion of steam in strobine then I - - = S=525 Si= Sft Xrs (Sg-ff) 7.26 = 0.637 + das (8.166 -0.637) x2 = 87.97 % has = hfts(hg- hf) pln 187.81+ 0-879762582-19-187-81) 2 2294-15 Kolka

Entropy Generation : ms = 1 kgls. asg in boiler : (ASO) B = riscs-su) - 1(7-26-0-637) = 6.623 kwk. b) oso in Trobine: I (DSg) = ms (G-S7) = 17.94-7.26) = 0.68 kwik Dsg in Pump: AS Given Pumping in question, all processes are ideal. Hence, process is taken as isentropic. (Asap = 0 Ink