1. 1 Lec 16: Refrigerators, heat pumps, and the Carnot cycle

2. 2 For next time: –Read: § 6-5 to 6-6 and 6-8 to 6-9 Outline: –Refrigerators –Heat pumps –Carnot cycle Important points: –Understand the different performance measures for cyclic devices. –Realize that COPHP and COPR are different –Start learning to recognize systems that violate the 2nd Law of Thermodynamics

3. 3 Second Law of Thermodynamics

4. 4 Review--Heat Engine or Cycle Efficiency

5. 5 Refrigerators, air conditioners and heat pumps Hot reservoir at T H Cold reservoir at T L System

6. 6 Refrigerators/‘air conditioners’ Remember: the purpose of a refrigerator or ‘air conditioner’ is to remove heat Q L from a cold region at T L.

7. 7 Refrigerator Basic components and typical operating conditions

8. 8 Heat Pump Remember: the purpose of a heat pump is to add heat Q H to a warm region at T H.

9. 9 Coefficient of Performance Refrigerators/Air conditioners

10. 10 Coefficient of Performance Refrigerators/Air conditioners

11. 11 Coefficient of Performance for Heat Pumps

12. 12 Coefficient of Performance for Heat Pumps

13. 13 TEAMPLAY Problem 6-52

14. 14 Perpetual Motion Machines (PMM) PMM1--A perpetual motion machine of the first kind violates the first law or the law of conservation of energy. An example would be an adiabatic system that supplies work with no change in internal energy, kinetic energy or potential energy.

15. 15 Perpetual Motion Machines (PMM)--Teamplay PMM2--A perpetual motion machine of the second kind violates the second law of thermodynamics. Your book has Figure 6-34 and goes into a correct explanation of why it violates the first law. The contraption in Figure 6-34 also violates the second law, as does the machine in Figure 6-35. Why?

16. 16 Carnot Cycle Composed of four internally reversible processes. –Two isothermal processes –Two adiabatic processes

17. 17 Carnot cycle for a gas. T L =const

18. 18 The Carnot cycle for a gas might occur as visualized below. TLTL QLQL

19. 19 This is a Carnot cycle involving two phases--it is still two adiabatic processes and two isothermal processes. It is always reversible--a Carnot cycle is reversible by definition. TLTL TLTL TLTL

20. 20 Carnot refrigeration cycle for a gas.

21. 21 Analytical form of KP Statement: Conservation of Energy for a cycle says E = 0 = Q cycle - W cycle, or Q cycle = W cycle We have not limited the number of heat reservoirs (or work interactions, for that matter). Q cycle could be Q H - Q C, for example.

22. 22 Analytical form of KP statement. Let us limit ourselves to the special case of one TER (thermal energy reservoir): TER HE W Q

23. 23 TEAMPLAY Can the system on the previous slide do work while operating in a cycle? If not, what does it violate?

24. 24 Analytical form of the KP statement. However, it would not violate the KP statement if work were done on the system during the cycle, or if work were zero. These are analytical forms of the KP statement.

25. 25 Analytical forms of the KP statement. Both the equations may be regarded as analytical forms of the KP statement. It can be shown that the equality applies to reversible processes and that the inequality applies to irreversible processes. Consider a cycle for which the equality applies, that is Q cycle = W cycle.

26. 26 Carnot’s first corollary The thermal efficiency of an irreversible power cycle is always less than the thermal efficiency of a reversible power cycle when each operates between the same two reservoirs.

27. 27 R I QHQH QHQH WIWI WRWR Q C =Q H -W R Hot reservoir Cold reservoir

28. 28 Carnot’s first corollary Each engine receives identical amounts of heat Q H and produces W R or W I. Each discharges an amount of heat Q to the cold reservoir equal to the difference between the heat it receives and the work it produces.

29. 29 R I QHQH QHQH QHQH WIWI WRWR WRWR QCQC Q C =Q H -W R Hot reservoir Cold reservoir

30. 30 Carnot’s first corollary. Taken together, Now reverse the reversible engine.

31. 31 R I QHQH QHQH QHQH WIWI WRWR QCQC Q C =Q H -W R Cold reservoir

32. 32 Carnot’s first corollary If W I  W R, the system puts out net work and exchanges heat with one reservoir. This violates KP. So, W I cannot be  W R.

33. 33 Carnot’s first corollary If W I = W R, Q C = Q′ C And the irreversible engine is identical to the reversible engine, i.e., it is just the reversible engine.

34. 34 Carnot’s first corollary So, W I  W R, and So  th,I   th,R

35. 35 Carnot’s second corollary All reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiencies.

36. 36 R1R1 R2R2 QHQH QHQH QHQH W R,2 W R,1 QCQC QCQC Hot reservoir Cold reservoir QCQC

37. 37 Carnot’s second corollary Both engines receive Q H, and Q cycle = 0 and W cycle = 0 for both engines with one reversed because they are both reversible. Now, with engine 1 reversed. W cycle = 0 = W R,1 -W R,2 and W R,1 = W R,2

38. 38 Carnot’s second corollary And so