Question

1. Steam is the working fluid in the (less than ideal) Ranking vapor power cycle with superheat, shown in the figure with the operational data. Boiler Turbine i W 2 T2 = 560°C Ts=38°C 20 MP2 Cooling water Condenser out Pump Ts = 20°C mew = 70.7 kg's Р x h s (kJ/kg.K) (kJ/kg) 338.14 State TCI 20.1 MPa 560 20 MPa 0.5 bar 0.4 bar 2442.3 0 a. Identify states 1 through 4 on the power plant schematic. b. Fill the table for the missing information. Cells with "*" are not required. c. Determine the net work per unit mass flowing through the cycle; d. Determine the cycle's thermal efficiency; e. Determine the turbine's isentropic efficiency: f. (BONUS) The cooling water enters the condenser at 20°C with a mass flow rate of 70.7 kg/s and exits the condenser at 38°C. The cooling water can be considered an incompressible substance with specific heat c=4.211 kJ/kg.K. Determine the mass flow rate of steam in the power cycle.

Answer 1

Turbine Wt 6 To = 380 pum Rout AN & Garderders Ts=200 Meu Foutrols from given properties, & Tas diagram, inlet. Process 2-3 is expansion process. State 2 in steam is turbine turbine.

1 of State 3 is Condensor inlet as condensor wone Orosor works under vacuum. Process 3-4 is the process of conden Steam. Process ut is is Pumping pumping process. so, state pump u is process. inet & State 3 is pump outlet This Seed water enter into boiler where it convere into ctecom at state & & this steam is worth drive the steam turbine. State TCC) Pupala h kil19) Scratica 338.14 560 81.82 76 2001 MPa 20 MPa x 0.5 bar x ooh bar a 6.374 3425-134 2442-3 34.566 se026 Stafe 2: Steam P = 20 MPa. table, To sboc - from from a super heated h = 3925-134 Kolkg , 5226.374 kolica

State 31 P=0.5 bart so lepa, h=244 2.3 kJ/kg From steam table hg = 340.48 kilig Sf = log1 kilkg K. hg=2645.213 kilkg , T = 356-47K Sg = 7.593 kuficgk T=81-47c Szo sf + x₂ sfg & ha= hit X3 hfg h3= ht t3 Chg- hf) 2942,3 = 30.48 + 03( 2645-213-340.48) Z Xz = 2442.3-340048 2.645- 213-240.48 → 23 = 0.912 S3 = - 1.091 0.912 (7593-10091) 7.02 kJgK

Xu=0 state 4: P 4 = 0.4 bar = 40 kPa from property table of steam hf= 317.56 Elkg Of = 0.001 m3lkg Sf = 1.026 kJ lige To 349 K = 76 C hf = ha , SE S4 hu= 317.566 Kulkg 365 1.026 Koligk. Net work per unit mass. We = har hz I = 3425.134 - 2442.3. 982-83 kJlKg. Wp = hirnya = 388-16 -3 17.566 = 20.57 kJ kg. Whet = Wy-Wp What a 962-26 Klicg

thermal efficiency! Qs = hachi 3425-139-338-14 = 3086.99 kalig Whet 962.26 3086.99 31-17 % Isentropil efficiency of Turbine! Wactaal ni Weisen Wisen Sq=3 for finding isentropic finding hz by taking work done by tiabine. 6.374 - 1.091 of X, (7.593-1081) 3) X3 = 0.8125 ha= hft X₃ Cha-hs). - B400187 0.812512645.213 340.48) - 2213.08 kilka

Nya ha ha he ha 982-83 (3425.134-2213.08) Actua) bort . done 81.0g Twi= 20°C Two = 38°C Mows 70.7 lcals cpw = 4.211 kilkak mass flow rate of steam rs = (?) 2 For Condensor energy balance ear as below Heat Gain by cooling water - heat lost by steam

m (Two-Twi) Cw= ms (ha-hu) => 70.7x4-211* ( 38-20) = ņs (2442-3-319-5662 img= 2.522 kals