Prove that if you had a refrigerator whose COP was better than the ideal value (4.9), you could hook it up to an ordinary Carnot engine to make an engine that produces no waste heat.

Short Answer

Expert verified

It is verified that a refrigerator with COP better than the ideal value can be hooked up to a Carnot engine to make an engine not producing waste heat.

Step by step solution

01

Concept Introduction

Let us write the expression of the ideal value of COP for Carnot engine

COP=TcoldTholTcoll

Here, Tcoldis temperature of the cold reservoir and Thotis temperature of hot reservoir.

02

Explanation

A refrigerator that has a coefficient of performance greater than the maximum Carnot value of COP requires an amount of work that is less than the work done in a Carnot engine.

A Carnot engine extracts heat from the hot reservoir and transfers it to the cold reservoir. The refrigerator, when hooked up to the Carnot engine, causes the composite system to yield a net surplus of work.

The compound system does not produce any waste heat because the refrigerator extracts the same amount of heat that the Carnot engine transfers.

03

Conclusion

Thus, it is verified that a refrigerator with COP better than the ideal value can be hooked up to a Carnot engine to make an engine not producing waste heat.

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Most popular questions from this chapter

A common (but imprecise) way of stating the third law of thermodynamics is "You can't reach absolute zero." Discuss how the third law, as stated in Section 3.2, puts limits on how low a temperature can be attained by various refrigeration techniques.

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

A power plant produces1GWof electricity, at an efficiency of 40%(typical of today's coal-fired plants).

(a) At what rate does this plant expel waste heat into its environment?

(b) Assume first that the cold reservoir for this plant is a river whose flow rate is 100m3/s.By how much will the temperature of the river increase?

(c) To avoid this "thermal pollution" of the river, the plant could instead be cooled by evaporation of river water. (This is more expensive, but in some areas it is environmentally preferable.) At what rate must the water evaporate? What fraction of the river must be evaporated?

Liquid HFC-134a at its boiling point at 12 bars pressure is throttled to 1 bar pressure. What is the final temperature? What fraction of the liquid vaporizes?

Table 4.3. Properties of the refrigerant HFC-134a under saturated conditions (at its boiling point for each pressure). All values are for 1kgof fluid, and are measured relative to an arbitrarily chosen reference state, the saturated liquid at -40°c. Excerpted from Moran and Shapiro (1995).

Consider an ideal Hampson-Linde cycle in which no heat is lost to the environment.

(a) Argue that the combination of the throttling valve and the heat exchanger is a constant-enthalpy device, so that the total enthalpy of the fluid coming out of this combination is the same as the enthalpy of the fluid going in.

(b) Let xbe the fraction of the fluid that liquefies on each pass through the cycle. Show that

x=Hout-HinHout-Hliq,

where Hinis the enthalpy of each mole of compressed gas that goes into the heat exchanger, Houtis the enthalpy of each mole of low-pressure gas that comes out of the heat exchanger, and Hliqis the enthalpy of each mole of liquid produced.

(c) Use the data in Table 4.5to calculate the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle operating between 1 bar and 100 bars, with an input temperature of 300K. Assume that the heat exchanger works perfectly, so the temperature of the low-pressure gas coming out of it is the same as the temperature of the high-pressure gas going in. Repeat the calculation for an input temperature of 200K.

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