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25 May, 14:14

An air-conditioning system is used to maintain a house at 70°F when the temperature outside is 100°F. The house is gaining heat through the walls and the windows at a rate of 800 Btu/min, and the heat generation rate within the house from people, lights, and appliances amounts to 100 Btu/min. Determine the minimum power input required for this air - conditioning system.

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  1. 25 May, 14:21
    0
    the minimum power input required for this air - conditioning system is 1.20 hp.

    Explanation:

    Let W' be power input required for this air - conditioning system

    Conversion factor;

    Fahrenheit (°F) to Rankine (R)

    1 °F = 460R

    Let the source temperature of the air-conditioner be Th = 100°F

    Th = 100 + 460 = 560R

    Let the sink temperature of the air-conditioner be TL = 70°F

    TL 70 + 460 = 530R

    Q' = heat gain from the wall + heat generation rate.

    Q' = (800 + 100) Btu/min

    COP = (Th/TL) - 1

    By combining the general and maximum COP, W' can be found as;

    W' = Q' * { (Th/TL) - 1}

    W' = (800+100) * 60 * 0.000393 { (560/530) - 1}

    W' = 1.20 hp
  2. 25 May, 14:42
    0
    Minimum power output required = 1.1977 hp

    Explanation:

    Given dа ta:

    Temperature outside = 100°F.

    House temperature = 70°F

    Rate of heat gain (Qw) = 800 Btu/min

    Generation rate within (Ql) = 100 Btu/min.

    Converting the outside temperature 100°F from fahrenheit to ranking, we have;

    1°F = 460R

    Therefore,

    100°F = 460+100

    To = 560 R

    Converting the house temperature 70°F from fahrenheit to ranking, we have;

    1°F = 460R

    Therefore,

    70°F = 460+70

    Th = 530 R

    Consider the equation for coefficient of performance (COP) of refrigerator in terms of temperature;

    COP = Th / (To-Th)

    = 530 / (560-530)

    = 530/30

    = 17.66

    Consider the equation for coefficient of performance (COP) of refrigerator;

    COP = Desired output/required input

    = Q/Wnet

    = Ql + Qw / Wnet

    Substituting into the formula, we have;

    17.667 = (100 + 800) / Wnet

    17.667 = 900/Wnet

    Wnet = 900/17.667

    = 50.94 Btu/min.

    Converting from Btu/min. to hp, we have;

    1 hp = 42.53 Btu/min.

    Therefore,

    50.94 Btu/min = 50.94 / 42.53

    = 1.1977 hp =

    Therefore, minimum power output required = 1.1977 hp
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