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.

Question

Azul Casey

Engineering

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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Answers (2)

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Devin Villanueva · Answer 1

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

Taliyah · Answer 2

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