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12 June, 01:16

At steady state, a reversible heat pump cycle discharges energy at the rate to a hot reservoir at temperature TH, while receiving energy at the rate from a cold reservoir at temperature TC. (a) If TH = 16°C and TC = 2°C, determine the coefficient of performance. (b) If 10.5 kW, 8.75 kW, and TC = 0°C, determine TH, in °C. (c) If the coefficient of performance is 12 and TH = 27°C, determine TC, in °C.

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  1. 12 June, 02:40
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    A) β_max = 20.64

    B) TH = 68.25°C

    C) TC = 54.27°C

    Explanation:

    A) We are given;

    TH = 16°C = 16 + 273K = 289K

    TC = 2°C = 2 + 273K = 275K

    Formula for maximum cycle coefficient of performance is given as;

    β_max = TH / (TH - TC)

    β_max = 289 / (288 - 275)

    β_max = 20.64

    B) We are given;

    Heat rejected to system at hot reservoir; Q_H = 10.5 KW

    Heat provided to system at cold reservoir; Q_C = 8.75 KW

    Cold reservoir temperature; TC = 0°C = 0 + 273K = 273K

    Formula for actual cycle COP is given as;

    β_actual = Q_C/W_cycle

    Where W_cycle is the work done and is given by;

    W_cycle = Q_H - Q_C

    W_cycle = 10.5 - 8.75 = 1.75 KW

    Thus,

    β_actual = 8.75/1.75

    β_actual = 5

    Actual cycle COP is defined as;

    β_actual = TH / (TH - TC)

    And we are looking for TH.

    Thus,

    TH = TC / (1 - (1/β_actual))

    TH = 273 / (1 - 1/5)

    TH = 273 / (4/5)

    TH = 341.25K = 341.25 - 273°C = 68.25°C

    C) We are given;

    TH = 27°C = 27 + 273 = 300°C

    β_max = 12

    Thus, from,

    β_max = TH / (TH - TC)

    TC = TH (1 - (1/β_max))

    TC = 300 / (1 - 1/12)

    TC = 327.27K = 327.27 - 273 °C = 54.27°C
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