Consider a power cycle with a net work for the cycle of 1000

Consider a power cycle with a net work for the cycle of 10,000 kJ and thermal efficiency of 0.4. Find the amount of energy transferred into the cycle by heat (Q_in) and out of the cycle by heat (Q_out). (3b) Consider a power cycle with a energy transfer by heat into the cycle, Q_in, of 500 MJ. What is the net work developed, in MJ, if the cycle thermal efficiency is 30%? What is the value of Q_out, in MJ?

Solution

sol:

3a)

for the power cycle thermal efficiency = (Qin - Qout)/Qin

Given thermal efficiency = 0.4

therefore (Qin - Qout)/ Qin = 0.4

Qin -Qout = 0.4Qin

0.6Qin - Qout = 0---------(1)

Given net work = 10000 kJ

therefor Qin - Qout = 10000 kJ -------(2)

after solving (1) and (2) we get Qin = 16666.67 kJ and Qout = 6666.67 kJ

3b) Given Thermal efficiency = 0.30 and Qin = 500 MJ

for the power cycle thermal efficiency = (Qin - Qout)/Qin

0.3 = (500-Qout)/Qout

Qout = 384.61 MJ

Work output = Qout-Qin = 500-384.61 = 115.384 MJ