Question

Design a small gas turbine engine to produce 150 kW of net power. Use an air-standard analysis (constant gas properties) and assume air enters the compressor at 100kPa and 20°C. The compressor pressure ratio is 8, the maximum cycle temperature is 800°C, and the cold air stream leaves the regenerator 10°C cooler than the hot air stream at the inlet of the regenerator. Assume a compressor isentropic efficiency of 87% and a turbine isentropic efficiency of 93%. Determine the rates of heat addition and rejection required to produce 150 kW. Assume constant specific heats at room temperature. Helpful equations, parameters: Isentropic relations: (k-1)/K T, ET Gas Properties: k= 1.4 P. P₂ Cp = 1.005 kJ/kg R=0.287 kJ/kg-K

Answer 1

Sohn Gas Turbine with Regeneration Missing In formation We cannot find inlet temperature of Regenerator with given data in question. For that am assuming Regenerator effectiveness { = 0.91 90% efficient 4 3 T(K) 5 2 4 2. S kJ/kg k For 1-2 Process To K-1 к K=1-4 TI P2 P r = 8 P Same for compressor assuming a and turbine T, 1.4-1 = Ti (e) 1.4 Ti = 20c or 293K = 293 0.2857 X (8) T = 530.738K Mc Tz - Ti Mc = 87.7. T, Ti 0.87 - Tz - Ti T. - Ti

т, - т, Т. - Т. о. 8 т 530. 38 993 + 293 О. 87 T 566. 262 К For 3 - ч Process К - | K T3 Tz = 800°c 9 То) 3к 0.2 851 Къ (к) Ty 1o 3 Ty (8) -2 8 57 КА 59 2. 36 К от та - ти Т. - Та т аз :). о - 93 (Io13 - 542 . 30) Т. Tú ту - (To73 - 446- 99 52) Ty 626. 00 Ч8 К

150 = mx 174.6018 m = 0.859 Kg s Heat addition as = m (T3 - Ts) - 0.859 x 1.005 (1073 - 620.03) Qs = 391.09 KW Heat Rejection OR = m (p (To-T.) OR = 0.859 X 1.005 1.005 ( 337.03 - 293) Q R = 38.01kW R Efficiency MGT Net Power Heat addition 150 kW 391.09 kW 38.35°). GT