A power cycle operates between hot and cold reservoirs at 900 K and 300 K, respectively. At steady state the cycle develops a power output of 0.3 MW while receiving energy by heat transfer from the hot reservoir at the rate of 1 MW. Determine the thermal efficiency and the rate at which energy is rejected by heat transfer to the cold reservoir, in MW for the following. (a) The power cycle and conditions above. (b) A reversible power cycle operating between the reservoirs and receiving the same rate of heat transfer from the hot reservoir as in part (a).

A) η_actual = 0.3 or 30%

Q_c = 0.7 MW

B) η_max = 0.67

Q_c = 0.33 MW

Explanation:

We are given;

Cold reservoir temperature; T_c = 300K

Hot reservoir temperature; T_h = 900K

Heat added at hot reservoir; Q_H = 1 MW

Cycle work; W_cycle = 0.3 MW

A) For actual cycle, actual cycle efficiency is given as;

η_actual = W_cycle/Q_H

Thus, η_actual = 0.3/1

η_actual = 0.3 or 30%

the rate at which energy is rejected by heat transfer to the cold reservoir is given as;

Q_c = Q_H - W_cycle

Q_c = 1 - 0.3

Q_c = 0.7 MW

B) Maximum cycle efficiency is given as;

η_max = 1 - (T_c/T_h)

Thus; η_max = 1 - (300/900)

η_max = 1 - 0.33

η_max = 0.67

the rate at which energy is rejected by heat transfer to the cold reservoir for reversible power cycle operating between the reservoirs and receiving the same rate of heat transfer from the hot reservoir is gotten from;

η_max = W_cycle/Q_H

Where;

W_cycle = Q_H - Q_c

Thus;

η_max = (Q_H - Q_c) / Q_H

Making Q_c the subject, we have;

Q_c = Q_H (1 - η_max)

Q_c = 1 (1 - 0.67)

Q_c = 0.33MW