Question

Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5160 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.

Answer 1

Answer:

$COP_R=1.433\\Q_H=2.433kW$

Explanation:

Hello,

In this case, the coefficient of performance of this refrigerator is defined in terms of the removed heat and the work input as:

$COP_R=\frac{Q_{L}}{W_{in}} =\frac{5160\frac{kJ}{h}*\frac{1h}{3600s} }{1kJ/s} =1.433$

Moreover, the rate of heat rejection to the outside air tuns out:

$W_{in}=Q_H-Q_L\\Q_H=W_{in}+Q_L=1kW+5160\frac{kJ}{h}*\frac{1h}{3600s} =2.433kW$

Best regards.

Answer 2

Answer:

Explanation:

Given:

Ql = 5160 kJ/h

Win = 1 kW

The coefficient of performance or COP of a refrigerator is defined as the ratio of useful heating or cooling provided, Ql to the work required, Win.

COP = Ql/Win

= 5160 kJ/h/1 kW

= 5160 kJ/h × 1 h/3600 s × 1/1 kW

= 1.43

B.

Energy balance of the rate of heat gained and lost in the refrigerator system

Rate of Heat rejection, Q = Ql + Win

= 1.43 kW + 1 kW

= 2.43 kW

= 2.43 × 3600

= 8760 kJ/h