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 win…

Question

asked Jul 23, 2022 80.1k views

1 Answer

Shevski · Answer 1 · 2022-07-27T23:54:19+0000

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
= 70°F = ( 70 + 460 )R = 530 R

temperature outside ( source )T
= 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