Question

A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is estimated to be 3,300 kJ/h and the heat rejection in the condenser is 4,800 kJ/h. If the refrigeration cycle was as efficient as the Carnot cycle, how much power (in kW) would be required to remove 3,300 kJ/h heat from the cooled space?

Answer 1

Answer: P = 0.416 kW

Step-by-step explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1