Question

2. Air enters the compressor of a regenerative gas turbine engine at 310 K and 100 kPa, where it is compressed to 900 kPa and 650 K. The regenerator has an effectiveness of 80%and the air enters the turbine at 1400 K. For a turbine isentropic efficiency of 90%, , then: (a) Sketch the T-s diagram of the cycle. (b) Determine the amount of heat transfer in the regenerator (c) Calculate the thermal efficiency of the cycle (d) Determine the net specific work output from the cycle in kJ/kg. (e) Determine the second law efficiency of the turbine (1) Calculate the exergy destructed in both the compressor and the turbine in kJ/kg.

Answer 1

энер тәЗоок P= lookPa ТА=69ок = op kra Та): 24 T = Hoo K 30. - с P, = доо ке ru= 0, = loo KPG 17 а) - -> (9) "чењә) =т, та ана. 286g K

ITH thoo-Tul 1400 - 7472868 * R = 8058 k 1 - 8125581 K amount of head to ensfes in the Degenerator 0= mcp {[31-? H = C, CT-0 (0.3) [817-588) – 650] =. Ty - Ecse 731 2 780.046 PS K g=mcp [780.046_650] Em6.005) (30.048)

Q - 13.6 67 m ma ei aina kg = = 130.697 ko regenerator heat transfer. In af cycle @ theomal officeny cp {3-12 o loos 21400-780.04648) Sing = 623.053 KILA EW = . Wy we = [] [] – SE-7] se { 1400 – 817:56e7] - <P[650-310] EW = 248.6791095 KF

EW n = Sin 248.679 623.057 Xloo 39912 * 23.39.az Yaylezo za r= 39.912 % kan net specific work output froney EW=Wg we - SP SETS FIA] - [5,-1] Tew=248,6791075 to na

è second low efficency of turbine. - Wact Wmase x 100 . X loo T-Ty 10o 1400 - 817.558) 1400 - 747.2862 - 89.99.9% = 0-899
as your guidlines we have solve first four sub parts thank thank u