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3 June, 20:36

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

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  1. 3 June, 23:47
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    (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.
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