A reversible power cycle receives energy QH by heat transfer

A reversible power cycle receives energy Q_H by heat transfer from a hot reservoir at T_H and rejects energy Q_c by heat transfer to a cold reservoir at T_c. If T_H = 1077 K and T_c = 431 K, what is the thermal efficiency? If T_H = 592 degree C, T_c = 24 degree C, and W_cycle = 1062 kJ, what are Q_H and Q_c, each in kJ? If n = 68% and T_c = 42 degree F, what is T_H, in degree F? If eta = 55 % and T_H = 747 degree C, what is T_c, in degree C?

Solution

Remember:

W = Qh - Qc

n = efficiency = W/Qh = (Qh - Qc)/Qh = 1 - (Qc/Qh)

Qh/Qc = Th/Tc

n = 1 - (Tc/Th)

A.

Tc = 431 K

Th = 1077 K

n = 1 - (431/1077) = 0.599 = 0.6

B.

Tc = 24 C = 297 K

Th = 592 C = 865 K

Wcycle = 1062 kJ

W = Qh - Qc = 1062 kJ

Qc/Qh = Tc/Th = 297/865 = 0.343

Qc = 0.343*Qh

W = Qh - 0.343*Qh = 0.657Qh

W = 1062 kJ = 0.657*Qh

Qh = 1062/0.657 = 1616.44 kJ

Qc = 0.343*1616.44 = 554.44 kJ

C.

n = 68% = 0.68

Tc = 42 F = 278.71 K

0.68 = 1 - 278.71/Th

278.71/Th = 0.32

Th = 278.71/0.32 = 870.97 K = 1108.08 F

D.

n = 0.55

Th = 747 K = 1020 K

n = 0.55 = 1 - Tc/1020 K

Tc = 0.45*1020 = 459 K

Tc = 186 C