Question

A combined gas–steam power plant has been designed with a net power output of 450 MW. The pressure ratio of the gas-turbine cycle is 14. Air enters the compressor at 300 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 8 MPa to 400 C in a heat exchanger. The combustion gases leave the heat exchanger at 460 K. An open feedwater heater incorporated with the steam cycle operates at a pressure of 0.6 MPa. The condenser pressure is 20 kPa. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate ration of air to steam, (b) the required rate of heat input in the combustion chamber, and (c) thermal efficiency of the combined cycle.

Answer 1

(2) given T Wreta 450 mW YP = 14 5 Тg = 3 cok smpa 4 To= 1400k Ege 160K=Tiz mai О. 6м Ра 3 Mateam ZokPa 7 din = combustion chamber Nagcle=? S Now at state-i P = rokpa= 0.02mpa fat lie from steam tabl. h = 251.4 KJ lkg Vis 0.001017 m3lky at state-2 h₂= hit V, CP2-P) = 251.4+ 0.001019% (0.6-0.0237105 h₂ = 252 kilkg at staten 3 h3= 670.38 ks 1kg V3= 0.001101 m3lkg at p= 0.6mpa fatlis at statent ha= hit V3 x (Pq-P3) 670.38 +0.001101*(8-0.6) 678.53 killeg

at state-s Ps= 8MP4 hs=3138.0kJ lig (2) (turbine Inlet) T5= 400c Ss= 6.65 KJ 1kgK at state-G » Stz S5 2 = 16 So- Suq, ovo Gudlity at stage 6) 6.3658- 19308 Serop, sro 4.8285 716 oogig seropa So- Sf ats ho= hue hox nfg - 670.38+ 0.90 x 2085.8 no 2587.2 Kglig at state-2 es 17= St-Suq 200 Stig - 6.3658 0.832 7.3752 17=0.78 ht nię, 20+ haxa hfg = 251.4 to 7 8x 2357.5 ha 2095 kilkg at State (8) +8= 3ook, rp= Pg P8 - 14 Tg T8 (la lry Ag Tg = T8X (A) 637.6k at starter 10 - 7107 took

mair Caa (Tiia T12) = 141 at state - 11 =) Tio TU = (A = ( 141 14 (3) TO Tila - 1400 (1490484 (14)04/14 Til = 658.7k Now Wo= Cp ( Tg-T8) = ricos (637.6-300) Wc = 339.2880 KJ lkg Work done by turbine Wr= Cp ( Tlo-T11) =loos ( 1400- 658.7) = 745.0065 KTlk Clin to gas cycle > Oin= Cp ( Fio Tg) Elicos( 1400 - 637.6) =766.2120 kg 1 kg # vet work done by gas cycle Wired = WT-WC = 745icors-339.288 Wires = 405.77 kalky # ratio of mals determined from an energy balance on heat exchanger mair Chin-hi2) mlhs-hu) Rt

mair ns-ne 3138-678.51 (4) meteem CpaCTI-T2) loos(658. 7460) Mair 12.31 Answers m glean for onen feed water. y = har hz. norm 670.38- 252 2551.23-252 y = 0.179 fleam you Wy - Wp - WP (hs- b )+(-9) ha-ha )-(-3)(h, họ ) 4b4-b1) (3138 - 2587.2)+(-0.179) (2587-2- 2095) - (1-0:179) x(252-2514) - (678.53 - 670.38) 954.8-8.64 Writean = 946.25 Waira Wirean 405.77+9 46.25 nagu 3ler din 766-2120

Wtotal Want aans who Mar (5) total work Quithelt per unit (40 405.77 + leo tut 946.25 ) mass of oir 482.63 Whet 450000 ń How rate of air Wtot =932:38 kg ree 452.62 Oina m (@ingir) = 932.38X 766.21 F 714466.21 kw Anawen Then thermal efheney of cycle és w 4 Soouo n= ele = 0.62 714406.21 or 62% n=62% Anaces Thang like Pleast