Question

Can you please help me with a detailed solution to this problem?

An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 250 K and 25 kPa. The compressor pressure ratio is 10. The turbine inlet temperature is 1800 K. For the cycle: (a) the heat addition and work done in each process, in kJ/kg, (b) the thermal efficiency (c) the back work ratio

Answer 1

2 V = (rp) + Vu=2.064x (10) ty [va = 10.691 m/kg The heat addition & work done in each process in kalkg. The heat addition & work done for each process is calculated as, 8-1 0 Process 1-2- CIGerstropic process) work supply to compressor is Wi-2= Pivi - PzV2 = (2.87X 25-0.554x250) x 103 1.4-1 Wi-2--166-875 *1031/79 Waz= -166.875 kJ 1kg) – Cove sign means work supplies to compressor) Process 2-3 ( constant pressure process) - Heat Supplig ot copstast pressure. H.G. = mcpa (T3-T2 ) = 1.005% (1800-482-99) H. S= 1, 323.6 kJ/kg Process 34 - (Isentropic process) - Turbine work W3-4 = (P3V3 - P4 Vy) (250% 2.064- 25*10-691) 1.4-1 7-1 W3-4 = 621.8125 kJ/kg Process 44 - C constant pressure process) - Heat rejected at costat pressure H.R - mcpa (Ta-TI) = 1.oosl931.7 - 250) H.R. = 685.11 kg/kg

3 The thermal efficiency of cycle chth) th=1- tas - Ett Turbine Whet (WT-WK) His th 1.4 - -0.4824 lo 1.4 hth = 48.24% © The back work ratio (hback (BWR) : BWR React Wcomp WTurb 166.875 621-8125 BWR=0 2684