Question

01: Steam enters the first turbine at 15.0 MPa and 600°C. The pressure in the condenser is 20.0 kPa. While some steam is extracted from the high-pressure turbine at 5 MPa and sent to the closed feedwater heater, the remaining steam is reheated to 600°C. The extracted steam is condensed as saturated liquid at 5.0 MPa and trapped to the open feedwater heater. Some steam is extracted from the lower-pressure turbine at 1.0 MPa and sent to the open feedwater heater out 10 5 MPa 6000 • • Determine the steam fraction extracted from the turbines. Determine the thermal efficiency of this cycle.

Answer 1

Determining properties of working fluid at all points: state 6: 15 Mpa, 60oC from Superheated steam ho= 3582.92 kJ lkg table, S6 = 6.679 kJ/kgK State H: 5 MPa. Expansion process is isentropic. S6=51 = 6.679 kJ k gk from super heated steam tab h7= 3220 Kolkg. State 8: 5 mpa, 6ool hg= 3666.48 kJ/kg S8 = 7.26 kaligk Now S8= 5g = S10 = 7.260 kilkg K

IN state g: 1 MPa, from superheated hg=3130 kilkg 7.26 kJ/kgk. steam table, 7.26 kJ ligK. state 10: 20 kPa At 20 lapa, hf = 251.4 kolkg Sf = 0.832 KollcgK , hg=2608.95 kJ 1kg Sg = 7.907 kolicgK 4-0.001 m lkg. Sg = 5,0 = Sitx, &g-Sf) > 7.26 = 0.832 +40 7.907-0-832) = 240 - 0.9085 hio = hf+40(hg-ht) = 251.4+ 0.9085 (2608.95-251.4) = 2393-35 KJ/kg State Ii 20 kpa, sat. liquid h = 251.4 kolkg 5, = 0.832 kilkak state 2: compressed water at 1 MPa Wola 4 (1₂-1) = 0.001C1000-20) wpi= hz-hi ) 2 = 252.38 kilkg 0.98 kulka state 3: sat. liauid, I MPa z=ht= 762.68 kJ/kg

State 4: compressed water is MPa Wp2 = 4 (Pu-P3) = 0.001 (15-1) x 10 = Wp2 = hu-h3 => hu=776.68 Kolig 14 kilky state s: saturated water, 15 MPa - h5= 1610.15 KJ/kg I state II: saturated water, 5 MPa h = 1154.5 kJ/kg derice. Hence, a state 12 : Toup is throttling - 112= h = 1154.5 KJ/kg I Now, 4 kg steam is extracted from the per kg mass of water flowing through ALSO I kg Steam is extracted from per kg mass of water flowing through HP Turbine the boiler. the LP turbine the boiler Applying energy balance for dosed FWH:

4 kg y Cha-hi) = (hg-hu) 1 => y = hs-hu hy-hu I kg 9 19 11 1610.15-776.68 3220-1154.5 0.4035 y = Applying Energy balance for open FWH, g 12kg h3(1) = 2(hg) + ych 12) +(1-y-7) hz -y-2) kgt Ikg 17 762.68 = 2(3130)+4 (1154S) +(1-4-2) 252-38 24 kg - 762.68=31302 + 4 65.841 + 252.38 - 101.835 - Z (252-38) Z= 0.05084

Thermal efficiency: WT = 1 (ho-h7) + (1-4) (hg-hg) + (1-4-2)(hg-ho) - 1( 3582.92 - 3220) + (1 -0.4035) (3666.48-3130) + (1-o-1035-0.05084) (3130 - 2393.35) Wp = 1084.89 kJiky Net work Done : → Wher = wy - Wpi-WP2 = logh.89 -0.98-14 = 1069.9) Kulkg Heat input ! Qin = 1 Cho-hs) t 1-4) (hg-ha) _ . 103582.92 - 1610.15) + (-0.4035) (3666-48-3220) = 1972.77 +266.33 = 2239.095 kJ lig Thermal efficiency: Na What a 1069,9) 47.78 % 2239.095