An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rate of heat transfer into the dwelling through the walls and roof is 30,000 Btu/h and the net power input to the air conditioner compressor is 3.3 hp. Required:
a. Determine the coefficient of performance of the air conditioner.
b. Determine the power input required, in hp, and the coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures.

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature = 99° F = ( 99 + 460 )R = 559 R

Cooling Load = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

= Cooling Load  / power P

we substitute

= 30000 Btu/h / 8396.619 Btu/h

= 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

/ = / ( - )

we substitute

30000 Btu/h / = 530 R / ( 559 R - 530 R )

30000 Btu/h / = 530 R / 29 R

we solve for

 = ( 30000 Btu/h × 29 R ) / 530 R

 = ( 870000 Btu/h / 530 )

 = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

 = ( 1641.5094 / 2544.43 ) hp

 = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

= / ( - )

we substitute

= 530 R / ( 559 R - 530 R )

= 530 R / 29 R

= 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759