Thermodynamics

512 | ThermodynamicsT, K130030021q inP = const.w compr p = 8P = const.FIGURE 9–35T-s diagram for the Brayton cyclediscussed in Example 9–5.3w turb4q outsgas temperature at the exits of the compressor and the turbine, (b) the backwork ratio, and (c) the thermal efficiency.Solution A power plant operating on the ideal Brayton cycle is considered.The compressor and turbine exit temperatures, back work ratio, and the thermalefficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptionsare applicable. 3 Kinetic and potential energy changes are negligible.4 The variation of specific heats with temperature is to be considered.Analysis The T-s diagram of the ideal Brayton cycle described is shown inFig. 9–35. We note that the components involved in the Brayton cycle aresteady-flow devices.(a) The air temperatures at the compressor and turbine exits are determinedfrom isentropic relations:Process 1-2 (isentropic compression of an ideal gas):T 1 300 K S h 1 300.19 kJ>kgP r1 1.386P r2 P 2P 1P r1 18211.3862 11.09 S T 2 540 K1at compressor exit2Process 3-4 (isentropic expansion of an ideal gas):T 3 1300 K S h 3 1395.97 kJ>kgP r4 P 4P 3P r3 a 1 8 b1330.92 41.36 S T 4 770 K1at turbine exit2(b) To find the back work ratio, we need to find the work input to the compressorand the work output of the turbine:Thus,That is, 40.3 percent of the turbine work output is used just to drive thecompressor.(c) The thermal efficiency of the cycle is the ratio of the net power output tothe total heat input:Thus,P r3 330.9w comp,in h 2 h 1 544.35 300.19 244.16 kJ>kgw turb,out h 3 h 4 1395.97 789.37 606.60 kJ>kgr bw w comp,in 244.16 kJ>kgw turb,out 606.60 kJ>kg 0.403q in h 3 h 2 1395.97 544.35 851.62 kJ>kgw net w out w in 606.60 244.16 362.4 kJ>kgh th w netq inh 2 544.35 kJ>kgh 4 789.37 kJ>kg362.4 kJ>kg 0.426 or 42.6%851.62 kJ>kg