Question

is 6 kPa. Stel vapor enters the Nuid in an iden the steam is ezt expands through bine at 10 mpang 0.7 MPa. The remeted and diverted to an elastage turbine to 0. Problem #4 125 Points Rankine Cycle-Reneration - Clased Feed Water Heater Water is the working fluid in an ideal penerative Rankine Cycle Superheated vapor enters the turbine at 10 MP. ANC, and the condent preure is 6 kPa. Steam expands through the first stage turbine to 0.7 MPa were women the steam is extracted and diverted to a closed feed water heater operating at 0.7 MPa. The remaining expands through the second stage turbine to the condenser pressure of 6 kPa. Condensate drains from the feed water heater as a saturated liquid at 700 kPa and is trapped back into the condenser. The feed water leaves the heater at 10 MPa and a temperature equal to the saturation temperaturen 700 kPa. Schematic: 151 Process Diagram: 151 Assume: 151 hı = 3321.4 kJ/kg h2 = 2684.8 kJ/kg h3 = 2009.8 kJ/kg h4 = 151.53 kJ/kg v4 = 1.0064 x 10 m/kg hs kJ/kg Wp = hs - 697.22 kJ/kg kJ/kg hs kJ/kg ha= 697.22 kJ/kg kJ/kg WLP WHPt=_ kJ/kg QsG/m1 = kJ/kg y= Wp/mi_ kJ/kg WTou'mi = kJ/kg kj/kg Q Conde

Answer 1

Ikals 10000KR 70015. Fa - б кла hi= 3321.4 kJ/kg hu= 151.53 kJ/kg h2=26848 k Jlky h3= 2009.8 kolly hs= hyt wp (Pump work) Wp = V4 (P2-13) = 1.0064x10-3(700-6) = 0.698413/159 hs= 152.02844 KJ /kg ho = 697-22 KJ/kg ho=y hz + (ty Jhz & Subst hos hasta we get - y = 0.22 kgs- Wypt = hi-h2= 33214 2684.8 = 636.6.15 Jika Wert = (1-9) (hz-hy) = 526-5kJ / kg W Total = 636.6+ 526.5 = 1163.1KJ/kg W pumpa = h7-b6=0 kJ llig Whet = WTotal-Wp = 1162.4 kJ/kg Ocord 1 = 0.226h2-hod = 437.2676 kJ/ing Blond 2 = 0.18Ch3-64) = 1449. 45 kJ/kg.