# thermodynamics

posted by
**jody** on
.

3.4 A heat engine operates between two heat reservoirs with temperatures of 560 and 280 K

a) The heat engine itself has a temperature of 560 K during heat addition from the 560 K

reservoir and a temperature of 280 K during heat rejection to the 280 K reservoir. The

thermal efficiency of the actual unit is 0.4, and the heat transfer rate from the 560 K

reservoir is 1000 kJ/min. Determine the entropy generation rate (kJ/min-K) for (1) the

heat engine itself, (2) the low temperature heat transfer process, and (3) the high

temperature heat transfer process. (.3572, 0, 0 kJ/K-min)

b) Now, for the same reservoir temperatures, heat transfer rate, and engine thermal

efficiency the engine itself receives heat at 540 K and rejects heat at 300 K. As before,

determine the entropy generation rate (kJ/min-K) for (1) the heat engine itself, (2) the

low temperature heat transfer process, and (3) the high temperature heat transfer process.

Which process has the largest irreversibility? (.148, .143, .0662 kJ/K-min)

c) Finally, for the same reservoir temperatures, heat transfer rate, and engine temperatures cited in part b) the thermal efficiency is increased to 0.42. For these conditions, determine the entropy generation rate for (1) the heat engine itself, (2) the low

temperature heat transfer process, and (3) the high temperature heat transfer process.

Which process now has the largest irreversibility? (.0814, .138, .0662 kJ/K-min)

We will be happy to critique your thinking or work on this. The question seems rather straighforward to me.

for part a I used thermal eff = Wout/Qh to find Wout. After that I used Qnet =Wnet.... Qin+Qout=Wout+Wbdry, where Wbdry is 0. and i solved for Q out. I then used Sgen = -((Qin/Thot)-(Qout/Tlow)) and I got .44 but the answer is .3572. Any words of wisdom?

I get .3572 for the a1, same numbers you did.

1000/560 - 400/280

i must be smoking crack..... thank you.i think i got the wrong answer about 5 times. I will try part b now.

For the second part of A. The only thing going on in the low temperature part is the rejection into the resivoir. there is no work being done. which then gives thermal eff = work /heat = o/600 = o. does that sound right or am i looking at this incorrectly?