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17 May, 01:53

A heat engine receives heat from a source at 1100 K at a rate of 400 kJ/s, and it rejects the waste heat to a medium at 320 K. The measured power output of the heat engine is 120 kW, and the environment temperature is 25∘C. Determine

(a) the reversible power,

(b) the rate of irreversibility and

(c) the second-law efficiency of this heat engine

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  1. 17 May, 03:55
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    part (a) 283.6 KW

    part (b) 163.6 KW

    part (c) 42.3%

    Explanation:

    The reversible power is the power produced by a reversible heat engine operating between the specified temperature limits.

    part (a)

    η_th (max) = η_th (rev) = 1-T_l/T_h

    we replace the values in the equation

    η_th (max) = 1 - (320 K/1100 K)

    = 0.7091

    therefore

    W_rev (out) = η_th (rev) Q_in

    we replace the values in the equation

    W_rev (out) = (0.7091) * (400 KJ/s)

    W_rev (out) = 283.6 KW

    The irreversibly rate is the difference between the reversible power and the actual power output.

    part (b)

    I=W_rev (out) - W_u (out)

    we replace the values in the equation

    I=163.6 KW

    The second law efficiency is determined from its definition.

    part (c)

    η_2=W_u (out) / W_rev (out)

    we replace the values in the equation

    η_2=42.3%
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