Question

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

Answer 1

Answer:

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

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