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[tex]_L[/tex] = 70°F = ( 70 + 460 )R = 530 R
temperature outside ( source )T[tex]_H[/tex] = 100°F = ( 100 + 460 )R = 560 R
[tex]Q_W[/tex] = 800 Btu/min
[tex]Q_L[/tex] = 100 Btu/min
Now, from the equation of coefficient of performance of refrigerator;
[tex]COP_{Ref[/tex] = Desired output / Required input
[tex]COP_{Ref[/tex] = [tex]Q_{out[/tex] / [tex]W_{net[/tex] --------- let this be equation 1
[tex]COP_{Ref[/tex] = ( [tex]Q_L + Q_W[/tex] ) / [tex]W_{net[/tex]
where [tex]Q_W[/tex] is the rate heat gained through the wall
[tex]Q_L[/tex] is the heat generation from people and lights and appliances.
Now, lets consider the equation coefficient of performance of refrigerator in terms of temperatures;
[tex]COP_{Ref[/tex] = [tex]T_L[/tex] / ( [tex]T_H[/tex] - [tex]T_L[/tex] )
we substitute
[tex]COP_{Ref[/tex] = 530 / ( 560 - 530 )
[tex]COP_{Ref[/tex] = 530 / 30
[tex]COP_{Ref[/tex] = 17.667
so we substitute into equation 1;
[tex]COP_{Ref[/tex] = [tex]Q_{out[/tex] / [tex]W_{net[/tex] ---------
[tex]COP_{Ref[/tex] = ( [tex]Q_L + Q_W[/tex] ) / [tex]W_{net[/tex]
17.667 = ( 100 + 800 ) / [tex]W_{net[/tex]
17.667 = 900 / [tex]W_{net[/tex]
[tex]W_{net[/tex] = 900 / 17.667
[tex]W_{net[/tex] = 50.94 Btu/min
[tex]W_{net[/tex] = ( 50.94 / 42.53 ) hp
[tex]W_{net[/tex] = 1.198 hp ≈ 1.20 hp
Therefore, the minimum power input required for this air-conditioning system is 1.20 hp
