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23 September, 14:46

Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by hot air. The hot air enters at 300°F and is to leave at 200°F. Calculate the total lb mol air/h needed. The mean heat capacity of the oil is 0.45 btu/lbm · °F.

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  1. 23 September, 16:12
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    2062 lbm/h

    Explanation:

    The air will lose heat and the oil will gain heat.

    These heats will be equal in magnitude.

    qo = - qa

    They will be of different signs because one is entering iits system and the other is exiting.

    The heat exchanged by oil is:

    qo = Gp * Cpo * (tof - toi)

    The heat exchanged by air is:

    qa = Ga * Cpa * (taf - tai)

    The specific heat capacity of air at constant pressure is:

    Cpa = 0.24 BTU / (lbm*F)

    Therefore:

    Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

    Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

    Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
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