Question

Design a small gas turbine engine to produce 130kW of net power. Use an air-standard analysis (const. gas prop.) and assume air enters the compressor at 100kPa and $20^\circ C$ . the compressor pressure ratio is 10, the maximum cycle temperature is $750^\circ C$ , and the cold air stream leaves the regenerator $15^\circ C$ 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 95%. Determine the rates of heat addition and rejection required to produce 130kW. cp=1.005 kJ/kg

Answer 1

TA 7500 26_3 2 1-2 - Compression 2-3 → Gnst.prepuye heat add 3-4 = Sentropic Capanion 4-1 = const Pressure A 200 PEC 7QR heat rejection Fig. T's diogram of Gas turbine Given: Net work done (W), 130 KW Wart = Wp - We Where Wy Turbine work We compresor work Pressure Ratio P2 10 les B2 Plepi Proces (1-2); Adiabatic 21 P2 l (6) T2 TI Given: Ti = 2ỐC = 293 %). = 293%. , ? for Air -14 CS ScannedTgat SÓC = 102315 CamScanner

1:-) та е (10) 1, 293 Т. 565-696 (+ в ка» 4] . 14-) (lb) гч o 23. 4 Т - . 529. 86 ), Gon: Lлитерс ара сте се мрти - 8*lb - е - 11 T, -Т, - 8+* - -Т 0-31 = 565 69 — 293 care T, 242 1, 606-435 Bentopic eff: Of tuibine = 95%) = int. 1 - 0-43= 5-7 T3-19. 1023 - D-93- Ib23- 529-83 ту - 564. 3+қ. CS Scanned with CamScanner

Given What Wp - c = 130kW Where Wp = mcp [T3 - Ty Wes Mice [T- Whene mos mais facio rate 130 = rcp ( 13 - Tu - 72 +T BD. = "X 1.005[1023 – 564.34 606-437 243 Bi M = 0.89 kgls. ... Heat Sopplied (05) = 0.89'x 1p05[1023- 606-43] Mcp [T3 - 77 Qs = 372 96 kW Heat Supplied Cake CPLTG - T D81x10056437-243] Pa = 242.72190 CS Scanned with CamScanner