Question

A regenerative gas turbine power plant has two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 12. The air enters each stage of compressor at 300 K and each stage of turbine at 1,180 K. Assume ideal Brayton cycle and effectiveness of regenerator as 100%. The compression or expansion ratios across each compression or expansion stage are also identical. The heating value of the fuel is 39,000 kJ/kg. The minimum fuel consumption rate (kg/s) needed to develop a net power output of 100 MW is:

Answer 1

overall pressure ratto Comp) = 12 The a'r enters each t'Stage of compression = 3001 (TT] and each stage of turbine (Tol Too) = 1180 Effectiveness of regenerator s = 100%, - Regenerator + 5) Repeater toy (4) I (6) Conyol Comp.2 Tussinet Fursine 2 + 22 + 1 Intercaler 114

- Assumptions & I o Air is treated as ideal gas with constant specific 1. heat Cp = locos KT regu. & Ake and ope is neglected in each case. E o mas> ing so we don't involve ing for calculation in SFEE. Since pressure ratio in each comp. and turbine is identical in each case WW hehet (given) WWW overall pressure ratio = P6 =12 w c 12 & since 2-3 is inter cooling in constant 2 pressure Process P2 = P2 ) =12 la = 12 - 3:404) Theu fe y - Py = lo a po 3-464 * Www

Process (1-2) ( sentropic compression in a k = ratio of specific heat = 104 1.4 +(3-464.Iz 2 со T = 427084 4 e Then it passes to intercooler where it rejects hert at constant pressure (P2 = P₂) and enters second stage of compressor at T3 = 3 cole. since it is also isentropic for some pressure ratio 3.464 1. Then Ts = T2 = 4270844 k) WW Then it enters to regenertor where it heat up by enhaust gas of second stage of tussine, The maximum Teme which B can be obtained for compressed air is (Ts) which is equal to tempo of exhaust gas TTS = To enters to twosine-t Then it enters to coco and then at To = 1180k Pentropic expansion in tussinet 1471 => (3-404) 1982 150 Tt TIS WWWWWWWWEES

T = 827-402 k - Then it enters to reheater where it get heat up to Tg = 1180 le and enters to gas tweine sime it is isentropic with some pressure ratio Po/po = 30464 so Tg = Ty = 8270402 So Temp at all imp. Polat Ti = T3 = 3ook Tz = Ty = 427-8446 1. T6 = To = 1180 l Ts - Tz - Tg = 827-402 l Net wore output of cycle . Whet = at co o wg. worie produced by turbines. - wir auch-ha) tina Chochg) - wymalo [To-Tz + T8 - Tg) 80-827.402 + 1/80-827.402) consumed W7 = mig (705-196) —>0 wc - worre popelitten by compressor - we = ma (hz-hi) tuna Chu-ha) -D wo = maco ( Tz - T + Ty - P3)

wc = mg Cp (427.844- 300+ 42.7. 844 – 300) wc = rng (255.686) Weet = wir so > 100x103 (kW) = ma (705-196) - mc (255685) 100 x103 - mig (705.196- 255.688) max 1.605 love e 1449.50g) ke Iwx103 - lows x 4 49.Sog - me - X ma - 221.358 leys mia = 221355 let mass flow rate of air in cycle Heat input to cycle & Cin = 64-5 + 627 1 Coco Reheater Cin = max (ho-hs) tina (46-47) Gin = max(ho-hst hq-hy) Cin = maxco (To - Ts + T8 - Ty) Gin 2 221•355 J x 1.205 (liter) (1180-827.402 + 1180 - 82704031 Gain = 156881-74 kW

fuel This heat is provided by burning of heat released by fuel = Gin =) mex cou. - 156881.79 kw > p (143) x 39000 Gen - 156861.74 Tmp = 4:0 2260 le mass of fuel consumed during process,