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

Step-by-step 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

Answer 2

`COP_R=1.433\\Q_H=2.433kW`

Step-by-step 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=(Q_(L))/(W_(in)) =(5160(kJ)/(h)*(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(kJ)/(h)*(1h)/(3600s) =2.433kW`

Best regards.