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Chapter 4: 4.12 (page 129)

Explain why a rectangular P V cycle, as considered in Problems 1.34 and 4.1, cannot be used (in reverse) for refrigeration.


Short Answer

Expert verified

The temperature range at which working substance absorbs heat must be less than the temperature range at which working substance rejects heat. As the rectangular PV cycle doesn't show this property, therefore it cannot be used for refrigeration.

Step by step solution

01

Given Information

The rectangular PV cycle.

Explain why it cannot be used (in reverse) for refrigeration.

02

; Explanation

A square PV cycle is shown as below.

The PV diagram is a closed loop. The area of the diagram represents the amount of work done during the process.

The temperature range at which working substance absorbs heat must be less than the temperature range at which working substance rejects heat.

As the rectangular PV cycle doesn't show this property, therefore it cannot be used for refrigeration.

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Most popular questions from this chapter

The amount of work done by each stroke of an automobile engine is controlled by the amount of fuel injected into the cylinder: the more fuel, the higher the temperature and pressure at points 3 and 4 in the cycle. But according to equation 4.10, the efficiency of the cycle depends only on the compression ratio (which is always the same for any particular engine), not on the amount of fuel consumed. Do you think this conclusion still holds when various other effects such as friction are taken into account? Would you expect a real engine to be most efficient when operating at high power or at low power? Explain.

Recall Problem 1.34, which concerned an ideal diatomic gas taken around a rectangular cycle on a PVdiagram. Suppose now that this system is used as a heat engine, to convert the heat added into mechanical work.

(a) Evaluate the efficiency of this engine for the case V2=3V1,P2=2P1.

(b) Calculate the efficiency of an "ideal" engine operating between the same temperature extremes.

Under many conditions, the rate at which heat enters an air conditioned building on a hot summer day is proportional to the difference in temperature between inside and outside, Th-Tc. (If the heat enters entirely by conduction, this statement will certainly be true. Radiation from direct sunlight would be an exception.) Show that, under these conditions, the cost of air conditioning should be roughly proportional to the square of the temperature difference. Discuss the implications, giving a numerical example.

A heat pump is an electrical device that heats a building by pumping heat in from the cold outside. In other words, it's the same as a refrigerator, but its purpose is to warm the hot reservoir rather than to cool the cold reservoir (even though it does both). Let us define the following standard symbols, all taken to be positive by convention:
Th=temperature inside buildingTc=temperature outsideQh=heat pumped into building in1dayQc=heat taken from outdoors in1dayW=electrical energy used by heat pump in1day
(a) Explain why the "coefficient of performance" (COP) for a heat pump should be defined as Qh / W.
(b) What relation among Qh , Qc, and W is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?
(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Th and Tc alone.
(d) Explain why a heat pump is better than an electric furnace, which simply converts electrical work directly into heat. (Include some numerical estimates.)

In a real turbine, the entropy of the steam will increase somewhat. How will this affect the percentages of liquid and gas at point4in the cycle? How will the efficiency be affected?
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