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17 May, 21:33

Air enters a heat exchanger at a rate of 5000 cubic feet per minute at a temperature of 55 °F and pressure of 14.7 psia. The air is heated by hot water flowing in the same exchanger at a rate of 11,200 pounds per hour with a decrease in temperature of 10 °F. At what temperature does the air leave the heat exchanger?

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  1. 17 May, 23:02
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    75 °F

    Explanation:

    Air has a specific heat at constant pressure of:

    Cpa = 0.24 BTU / (lbm*F)

    The specific heat of water is:

    Cpw = 1 BTU / (lbm*F)

    The first law of thermodynamics:

    Q = L + ΔU

    The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.

    Then:

    Q = 0.

    We split the heat into the heat transferred by the air and the heat trnasferred by the water:

    Qa + Qw = 0

    Qa = - Qw

    The heat exchanged by the air is

    Qa = Ga * Cpa * (tfin - ti)

    And the heat exchanged by the water is:

    Qw = Gw * Cpw * Δt

    Replacing:

    Ga * Cpa * (tfin - ti) = - Gw * Cpw * Δt

    tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)

    tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti

    The G terms are mass flows, however we have volume flow of air.

    With the gas state equation we calculate the mass:

    p * V = m * R * T

    m = (p * V) / (R * T)

    55 °F = 515 °R

    The gas constant for air is R = 53.35 (ft*lb) / (lbm * °R)

    14.7 psi = 2117 lb/ft^2

    m = (2117 * 5000) / (53.35 * 515) = 385 lbm

    The mass flow is that much amount per minute

    The mass flow of water is

    11200 lbm/h = 186.7 lbm/min

    Then:

    tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F
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