The fridge could be made to operate in the reverse mode, extracting 80 kJ of heat every second from a hot environment (say steam at 100 0C ), converting some of this energy to useful work (e.g. electrical energy) and then dumping the rest as heat to the surrounding air at 30 0C. e.  What is the minimum amount of heat per second that has to be dumped to the air? f.  Using your answer to part (e), show that the maximum useful work that we can hope to obtain from this operation is around 15 kW (i.e. show that the efficiency of this machine is only 15/80 = 18.7 %). g.  Why can’t the efficiency of this machine ever be 100 % (i.e. why can’t we hope to design an engine that converts all of the extracted 80 kJ/s of heat completely to work)?

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
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Chapter6: Forced Convection Over Exterior Surfaces
Section: Chapter Questions
Problem 6.51P
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*Only need parts E, F and G*

A large fridge freezer is to extract 80 kJ of energy every second from inside of the fridge, maintained at temperature of -3 0C, and to dump it to the air outside which is at 30 0 To do so, the fridge consumes 7 kW of electrical energy which is also converted and dissipated as heat to the surrounding air.

a. Calculate the entropy change per second in the surrounding air and that inside the fridge.

b. Show that such a fridge freezer can never be realised in practice.

c. What is the minimal electrical power necessary to achieve the operation of this fridge?

d. If the electrical power used is to remain at 7kJ/s, but we still require to extract 80 kW of heat from inside, what is the lowest temperature in the fridge that we could theoretically hope to achieve?

The fridge could be made to operate in the reverse mode, extracting 80 kJ of heat every second from a hot environment (say steam at 100 0C ), converting some of this energy to useful work (e.g. electrical energy) and then dumping the rest as heat to the surrounding air at 30 0C.

e.  What is the minimum amount of heat per second that has to be dumped to the air?

f.  Using your answer to part (e), show that the maximum useful work that we can hope to obtain from this operation is around 15 kW (i.e. show that the efficiency of this machine is only 15/80 = 18.7 %).

g.  Why can’t the efficiency of this machine ever be 100 % (i.e. why can’t we hope to design an engine that converts all of the extracted 80 kJ/s of heat completely to work)?

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