An industrial refrigerator rejects heat at a rate of 24,750 kJ/min to the surroundings. If the refrigeration cycle has a COP of b = 3.3, determine Q_{C} and W_{cycle}, each in kJ/min. 





Read : 
This one is a
straightforward application of the definition of the the 1st
Law and the COP of a refrigeration cycle. 
Diagram : 













Given: 
COP = b 
3.3 



Find: 
Q_{C} 
??? 
kJ/min 


Q_{H} 
24,750 
kJ/min 



W_{cycle} 
??? 
kJ/min 












Assumption: 
 The cycle only exchanges heat with the two thermal reservoirs. 












Equations
/ Data / Solve: 









1st
Law applied to the refrigerator: 


Eqn 1 













Definition of COP for a refrigerator : 



Eqn 2 













Degree
of freedom analysis: 2
eqns in 2 unknowns: Q_{C} and W_{cycle}. 














Solve Eqn 2 for Q_{C} and use the result to eliminate Q_{C} from Eqn 1 : 















Eqn 3 



Eqn 4 













Next, solve Eqn 4 for W_{cycle} in terms of the known quantities Q_{H} and b. 





















Eqn 5 













Plug numbers into Eqn 5 : 



W_{cycle} 
5755.8 
kJ/min 













Now, use this value
for W_{cycle} and the given value of b in Eqn 3
to evaluate Q_{C} : 




















Q_{C} 
18994.2 
kJ/min 












Verify: 
The only assumption
cannot be verified. 

















Answers : 
Q_{C}
= 
19000 
kJ/min 



W_{cycle} = 
5760 
kJ/min 

































