Answer:
[tex]COP_R=1.433\\Q_H=2.433kW[/tex]
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:
[tex]COP_R=\frac{Q_{L}}{W_{in}} =\frac{5160\frac{kJ}{h}*\frac{1h}{3600s} }{1kJ/s} =1.433[/tex]
Moreover, the rate of heat rejection to the outside air tuns out:
[tex]W_{in}=Q_H-Q_L\\Q_H=W_{in}+Q_L=1kW+5160\frac{kJ}{h}*\frac{1h}{3600s} =2.433kW[/tex]
Best regards.
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
