Question

The steam turbine for a power plant in Sydney is operating at 873.15K at 5000kPa. The steam at the exhaust exits the turbine at 323.15K with 5% by mass, of the steam that is condensed to liquid. Neglect pump work, kinetic and potential energy changes. What is the thermal efficiency of the cycle and the turbine efficiency? State assumptions.

Answer 1

$92 At state I (P=5Mp9, T = 873.15k) TO room AH & pressure fasce , P=SMRA, Tsat = 263.941° c = 536.94k Tr> Tsat ; Hence, state I lies in superheated state. Now, from superheated table At D - 5M na T=873.156 = 600°c h = 3666.8 W si = 7.2605 1 108 Topse At state 2 (T= 323.15k, x=0.95) Since At the exit of tossine 5% of steam Condensed into vquid me = 5% my 95% A New Tas diagram my=950) 1 = 0.95 5%+95%. Mytme 25 2 ha= hiloc Sisoc Esk WT-Wp from temperature table, T=323.156 = 50°c hea Psap = 0.012352 mm Turline work (WA) & xxhts/sooc Wy= (he - H2) = (366618 - 2472.15) = 1194.65 lossing - 209.34 +0.95% 238.9= 2472.15 h S2 - & xx sssloc Heat supplied (op) = 0.70381 Qs = Chichu) = (36668- 214.39) 0.95% 7.3710=7.70626 lg = 3452.41 lglig for Isoenfropic tustine, S=525 7.2605 lg lig4.65 si-Szs 7.2605 as 3452.41 of our Stefsoc - 7.2605 = 0.346 on 34.6%. - 0.70381 + x 7.3710 = 7.2605 hi ha 3666.8 - 2472.15 = xy = 0.889 or 89.23% it Max haf 209.34f 0.889x2351.9 = 2328.09 slug At state & Mcycle Salou ชาๆ haa hofre 3666.8 - 2328.09 - 0.8923 At state y hg = holoof 209.34 14 * 15-0.012352)X10 a) hy = 209.34 +000101215x (5-01 012352x10? - 214.39 penly hy-hz uslove