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6 June, 23:04

Air is compressed from 100 kPa and 150C to a pressure of 1000 kPa while being cooled at a rate of 20 kJ/kg by circulating water through the compressor casing. The volume flow rate of the air at the inlet conditions is 140 m3/min, and the power input to the compressor is 520 kW Determine: (a) The mass flow rate of the air, and (b) The temperature at the compressor exit.

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  1. 6 June, 23:24
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    a) 115.7 kg/min

    b) 240 C

    Explanation:

    To determine the mass flow rate we need the specific volume of the air. We can use the gas state equation for that:

    p * v = R (air) * T (no mass term because we want specific volume)

    v = R * T / p

    The specific R for air is R = 287 J / (kg * K)

    Also 150 C = 423 K

    Then

    v = 287 * 423 / 100e3 = 1.21 m^3/kg

    Them the mass flow is

    G / v = 140 / 1.21 = 115.7 kg/min

    The flow rate can also be expressed as 115.7 / 60 = 1.93 kg/s

    Then we find that the compressor applies a work of

    L = 520 / 1.93 = 269.4 kJ/kg to the air

    By enthalpy:

    H2 = H1 + Q + L

    H2 = cv * T2 + p2 * v2

    H1 = cv * T1 + p1 * v1

    Then

    cv * T2 + p2 * v2 = cv * T1 + p1 * v1 + Q + L

    Also

    (p1 * v1) / T1 = (p2 * v2) / T2

    p2 * v2 = (p1 * v1 * T2) / T1

    Therefore:

    cv * T2 + (p1 * v1 * T2) / T1 = cv * T1 + p1 * v1 + Q + L

    T2 * (cv + (p1 * v1) / T1 = cv * T1 + p1 * v1 + Q + L

    T2 = (cv * T1 + p1 * v1 + Q + L) / ((cv + (p1 * v1) / T1)

    The cv of air is cv = 0.72 kJ / (kg*K)

    T2 = (0.72 * 423 + 100e3 * 1.21 - 20 + 269.4) / ((0.72 + (100e3 * 1.21) / 423) = 513 K = 240 C
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