Question

1.) A power operates between 2 surfaces. One surface temperature is 300K and the other surface temperature is 285K. During a steady state cycle, 14,000 kJ/min is rejected by heat transfer while the cycle developed a power output of 10 kW. Det ermine the thermal efficiency of the power cycle and the maximum thermal efficien cy for any such power cycle.

Answer 1

Solution:

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solution: Ti = 300k, T2 = 285k, W= lokw, Q2 = 14000 kJ / min here, heat rejection, Q2 = 14000 kJ /min Q2 = 14000 233.33 kJ/sec 60 92 = 233.33 kW = 0.05 n = Ti-Tz 300 (9) The Thermal Efficiency of the power cycle is, 300-285 51 Ti & other Way to determine thermal Efficiency is, n = w/a, here ai= heut Input W = 9 - Q2 9= wta2 QI = 10 + 233.33 , : 243.33 kW There Fore Efficiency, n = wa n = 10/243.33 no = 0.041 n = 4.14 These are the Thermal Efficiencies of the Cycle. (b) The Muximum Thermal Efficiency of the power cycle is the efficiency which is obtained by the temperature i. e. 5% muximum Thermal Efficiency, n = 5% Thank you.