Thermodynamics 1 Problem: A refrigerator freezer compartment is maintained at a

Question

Thermodynamics 1 Problem:

A refrigerator freezer compartment is maintained at a temperature of -2 C by removing heat at a rate of 4500 nr. The refrigerator is located in a room with a temperature of 24°C The power input required to run the refrigerator is 1200 kJlhr. (a) Calculate the COP B for the refrigeration cycle. (b) Calculate the rate of heat transfer from the refrigerant to the room. (c) Calculate the power to run the cycle and the heat transfer rate from the refrigerant to the room if the refrigeration cycle was reversible. Assume that the heat transfer rate from the freezer compartment to the refrigerant is the same 4500 Btu/hr as for the actual cycle

Explanation / Answer

>> T2 = - 26 C = 247 K

>> T1 = 24 C = 297 K

>> Q2 = Heat Removed From Cold Body = 4500 KJ/hr

>> W = Power Input required to run the refrigerator = 1200 KJ/hr

>> Now, as COP = Coefficient of Performance = Q2/W = 4500/1200 = 3.75

Part (b).

>> Q1 = Rate of Heat Transfer to the room = Q2 + W = 5700 KJ/hr

Part (c).

>> Now, Refrigeration Cycle is reversible, that means, it behaves like as a Carnot Cycle

So, Q1/Q2 = T1/T2 = 297/247 = 1.202

>> As, Q2 = 4500 Btu/hr

=> Q1 = 1.202*4500 = 5410.93 Btu/hr .....Rate of Heat Transfer to the room.........

So, Power Required to run the refrigerator, W = Q1 - Q2 = 5410.93 - 4500 = 910.93 Btu/hr

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