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

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

4. 4

5. 5 Refrigerators, air conditioners and heat pumps

6. 6 Refrigerators/‘air conditioners’

7. 7 Refrigerator

8. 8 Heat Pump

9. 9 Coefficient of Performance

10. 10

11. 11

12. 12

13. 13 TEAMPLAY

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

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

17. 17

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

19. 19

20. 20

21. 21 Analytical form of KP Statement: Conservation of Energy for a cycle says ?E = 0 = Qcycle - Wcycle, or Qcycle = Wcycle We have not limited the number of heat reservoirs (or work interactions, for that matter). Qcycle could be QH - QC, for example.

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

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.

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 Qcycle = Wcycle.

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

28. 28 Carnot’s first corollary Each engine receives identical amounts of heat QH and produces WR or WI. 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

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

31. 31

32. 32 Carnot’s first corollary

33. 33

34. 34 Carnot’s first corollary So, WI ? WR, and

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

36. 36

37. 37 Carnot’s second corollary Both engines receive QH, and Qcycle = 0 and Wcycle= 0 for both engines with one reversed because they are both reversible. Now, with engine 1 reversed. Wcycle = 0 = WR,1-WR,2 and WR,1 = WR,2

38. 38 Carnot’s second corollary And so