Question

Air enters the compressor of a regenerative air-standard Brayton cycle with a volumetric flow rate of 20 m3/s at 0.8 bar, 280 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 1950 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. For a regenerator effectiveness of 86%, determine: (a) the net power developed, in MW. (b) the rate of heat addition in the combustor, in MW. (c) the percent thermal efficiency of the cycle. Part A Determine the net power developed, in MW. W cycle = MW

Answer 1

This question is based on the Brayton Cycle with Regeneration arrangement.

Brayton cycle is an air standard power cycle used mostly in aircrafts and prime movers. A regenerator is used in the cycle to increase the overall thermal efficiency of the cycle by increasing the mean temperature of heat addition.

The step by step analysis of the Brayton Cycle with the calculations is given below:-

A Brayton cycle with Regeneration as WT 2 processes in 1 -2= Compressor 2-15 Regenerator, 5-3= Combuster 3-4= Turbine. 4-62. Regenerator, 6-1 = Heat Exchanges Waha T-S Diagram Given, Air volumetric flow rate, at înlet; VE 20 m/s. P = 0.8 bar 280 K. pressure ratio, P2 = 20 P3 Pa. max. cycle temp. T= 1950k. isentropic efficieny of compersson = 0.92. and. turbine, n 2 0.95. € =0-95. E and regenerateppectivevest Assuming Ideal Air conditions, Co=1.005 KJ 1 kg R = 0.287 kJ 1kg K. 7 = 1.4

I deal 0.287 X 280 gas equation, P, W, = MRT, (0.8x100) x 20 RT, T 19.910 kgls. This is the value of air mase flow rate in the Yele supplied. for is entropic process. 1-2, Tz (P2 T2 1.4-1 4. Yt (207.4. 280 To = 658.993k. and for process, 3-4 similarly, P27 FIN 1950 Th (20) T2 = 828.537 K. I sentropic power input to the compressor We = ń (2 h). = ŕ Sp CT-T.). W = 19.910 X 1.005* (658.993-280) w 7583.479 kW. h = 7.583 MW.

and. turbine power output (isentropich. ni (h₂ ha) = rig (Tg - Tad W = 19.920 x 1.005x (1950-828.537) > T = 22.44 MW.

Compressor efficieney, y Isentropic power Actual power > 0.92 = 7.583 M Low (We) Act. (Wa) Act 8. 242 MW. and, Turbine efficiency Actual power Isentropic power. T 0,95 (WT) Act. >> 22.44 MW (wit) 21. 318 Act. = MW. (9) the Net. Power developed, Whet Wrach - (Wihad. = 21.318 – 8,242 Wrnet = 13.076 MW. Ans. € = Tsaka Regenerator Effectiveness, is given by Actual temperature rise maximum temperature rise. Tarta! 0.96 Ts - II Ta'-Ta >

where, T2 and Ta' are actual temperaturas at the compressor and turbine exit. Chelact. = 8,242.mw. inco CT-T,) = 8.242.8103 kw. T2 = 691.903 k. > and, (WT) an 21.318 Mw. > in cp (Tz - Th ) = 21.318 x 103 Kw. Th=884.609 K. y putting values into the expression 0.86 ts- 691.903 884.609-691.903 Ts - 857.53016. K. so (6) The rate of feat Added in the Combustor, r Cp (T3-Ts) aina 19.910X1-005X (1950 ein = 21.858 MW, 857.630) Ans.

(C) The thermal efficiency of the cycle, Whet n o. 5982 th 13.076 21.858. - ) 59.82%