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An air-conditioning system is used to maintain a house at 70^\circ{} ∘ F when the temperature outside is 100^\circ{} ∘ 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. Answer: 1.20 hp

User Katsuya
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Answer:

the minimum power input required for this air-conditioning system is 1.20 hp

Step-by-step explanation:

Given the data in the question;

Temperature inside ( Sink ) T
_L = 70°F = ( 70 + 460 )R = 530 R

temperature outside ( source )T
_H = 100°F = ( 100 + 460 )R = 560 R


Q_W = 800 Btu/min


Q_L = 100 Btu/min

Now, from the equation of coefficient of performance of refrigerator;


COP_{Ref = Desired output / Required input


COP_{Ref =
Q_{out /
W_{net --------- let this be equation 1


COP_{Ref = (
Q_L + Q_W ) /
W_{net

where
Q_W is the rate heat gained through the wall


Q_L is the heat generation from people and lights and appliances.

Now, lets consider the equation coefficient of performance of refrigerator in terms of temperatures;


COP_{Ref =
T_L / (
T_H -
T_L )

we substitute


COP_{Ref = 530 / ( 560 - 530 )


COP_{Ref = 530 / 30


COP_{Ref = 17.667

so we substitute into equation 1;


COP_{Ref =
Q_{out /
W_{net ---------


COP_{Ref = (
Q_L + Q_W ) /
W_{net

17.667 = ( 100 + 800 ) /
W_{net

17.667 = 900 /
W_{net


W_{net = 900 / 17.667


W_{net = 50.94 Btu/min


W_{net = ( 50.94 / 42.53 ) hp


W_{net = 1.198 hp ≈ 1.20 hp

Therefore, the minimum power input required for this air-conditioning system is 1.20 hp

User Shevski
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