Question

A power cycle operating at steady state receives energy by heat transfer at a rate Q˙ H at TH = 1500 K and rejects energy by heat transfer to a cold reservoir at a rate Q˙ C at TC = 500 K. For each of the following cases, determine whether the cycle operates reversibly, irreversibly, or is impossible. If it is impossible, state why it is impossible. (a) Q˙ H = 550 kW, Q˙ C = 100 kW (b) Q˙ H = 500 kW, W˙ cyc = 200 kW, Q˙ C = 200 kW (c) W˙ cyc = 300 kW, Q˙ C = 150 kW (d) Q˙ H = 500 kW, Q˙ C = 200 kW

Answer 1

(a) The cycle is not possible.

(b) The cycle operates irreversibly.

(c) The cycle operates irreversibly.

(d) The cycle is not possible.

Explanation:

Given -

Temperature of hot reservoir, Th = 1500K

Temperature of cold reservoir, Tc = 500K

(a)

Heat transfer at hot reservoir, Qh = 550 kW

Heat transfer at cold reservoir, Qc = 100 kW

Maximum net work -

Wcycle = Qh - Qc

Wcycle = 550 - 100 = 450kW

Maximum efficiency -

ηmax = 1 - Tc/Th

ηmax = 1 - 500/1500 =  0.667

ηmax = 66.7%

η(actual) = Wcycle/ Qh

η(actual) = 450 kW/ 550 = 0.8

η(actual) = 80%

Since the actual efficiency is higher than the maximum efficiency, the cycle is not possible.

(b)

Heat transfer at hot reservoir, Qh = 500 kW

Heat transfer at cold reservoir, Qc = 200 kW

Maximum net work -

Wcycle = Qh - Qc

Wcycle = 500 - 200 = 300kW

Maximum efficiency -

ηmax = 1 - Tc/Th

ηmax = 1 - 500/1500 =  0.667

ηmax = 66.7%

η(actual) = Wcycle/ Qh

η(actual) = 300 kW/ 500 = 0.6

η(actual) = 60%

Since the actual efficiency is lower than the maximum efficiency, the cycle operates irreversibly.

(c)

Heat transfer at hot reservoir, Qh = 300 kW

Heat transfer at cold reservoir, Qc = 150 kW

Maximum net work -

Wcycle = Qh - Qc

Wcycle = 300 - 150 = 150 kW

Maximum efficiency -

ηmax = 1 - Tc/Th

ηmax = 1 - 500/1500 =  0.667

ηmax = 66.7%

η(actual) = Wcycle/ Qh

η(actual) = 150 kW/ 300 = 0.5

η(actual) = 50%

Since the actual efficiency is lower than the maximum efficiency, the cycle operates irreversibly.

(d)

Heat transfer at hot reservoir, Qh = 500 kW

Heat transfer at cold reservoir, Qc = 100 kW

Maximum net work -

Wcycle = Qh - Qc

Wcycle = 500 - 100 = 400kW

Maximum efficiency -

ηmax = 1 - Tc/Th

ηmax = 1 - 500/1500 =  0.667

ηmax = 66.7%

η(actual) = Wcycle/ Qh

η(actual) = 400 kW/ 500 = 0.8

η(actual) = 80%

Since the actual efficiency is higher than the maximum efficiency, the cycle is not possible.