Question

H1= 3756.1

I (30 pts.) A steam power plant with a two-stage turbine and a closed feedwater heater is used to generate power at a local power plant. A portion of the steam at the exit of the first stage of the turbine is used raise the temperature of the water before entering the boiler. The remainder is reheated and sent to the second stage of the turbine. Assume that the pump is isentropic and that the water entering the pump and steam trap is saturated liquid. Given the following, determine the unknowns below. m = 50kg/s T1=659°C, P1=12MPA P2=2.0MPA T3=600°C T1 T2 SG P4=0.1bar T7=210°C [4 kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg hi= h2= h3= h4= hs= h6- h7= hg= h9= CND FWH 6. 7. P5 9. FWH: y' =

S1 = 6.9665

H2= 3143.8

H3 = 3690.1

H4 = 2441.8

H5 = 191.85

H6 = 193.8

stuck on H7

Answer 1

from Steam.bababes 9 11859°c) 3 : 120 bar At 120 bar, 659°C 18 20 barn barwy Wy h = 3156 Kolkg $=6.96.65 Kulky- s oolbar Se At zo bar ,, 468 4 кајE) 1 , hea = 1888.4 Kolka Š; -2.4443 kilig-k $qs 3. 8 894 K5/1g- S, - $ +2 Spa x=1:1619 hy het alty 5.3102.48 kolky

At zobar, 6ooc h = 369011 kJ /kg S2 = 7.702 kJ / kg - K = S4 Atolbar he=191.9 kJ/kg = hs... hm = 2.393.0itling Szi Sp+a sig sf = 8.6494 kJ / kg-K . x=0.94018 Spg = 7.50 13 kJ/kg-k hy het dhe = 0.00 joi millig =244145 *otlig 65 hechy - 07 (120-01) X 100 = 12.1099 h = 204.0099 At If = 210 c Peut = 19,0 81 bar, hp = 897.719 kilka • 50001143 miling Qorx (tos y stry) to +ty = Fy

RAL h - 897.719 + (0.001143)(120-19.08N)X100 he = 909.5567 kJ / kg Apply Stedy Hlow Egery Ezuation for closed feed water heater hy (g) the = hely) thy y (hy-hy) = hy he y = h4 h6 = 0.32155 b-hg Y =3201557. Turbine cork w = (he-by) + (-y) (hgh) = 150011630 killiy W = 75.008 MW

Paul work we = (höchs) xmg = 0.6054 MW Heat supplied Q. = (h-hy) + (1-4) (hy ha) 3244.910 Kolly 2017 = 162.2.45 MW Thermal Efficiery half lop Olin 75.008-0.6054 20.45858 Uh 162.245 2 = 45.85%