1. Lec 21: Isentropic efficiencies, air standard cycle, Carnot cycle, Otto cycle

2. For next time:
Read: § 8-6 to 8-7
HW11 due Wednesday, November 12, 2003
Outline:
Isentropic efficiency
Air standard cycle
Otto cycle
Important points:
Realize that we already know how to analyze all these new cycles, we just need to define what the cycle steps are
Know the difference between the air standard cycle and the cold air approximations
Know how to solve cycles using variable specific heats and constant specific heats

3. Isentropic Efficiencies
We can use the isentropic process as an ideal by which to compare real processes in different engineering devices.
COMPRESSORS
Actual compressors take more work than isentropic compressors. The efficiency will vary between zero and one.

4. Compressor Isentropic Efficiency
For a steady-state, adiabatic compressor
If q, KE, and PE are all zero, then: 2 1 h w - =
General expression s 2 1 h w - =
Isentropic compressor a 2 1 h w - =
Actual compressor

5. Compressor Isentropic Efficiency
The compressor efficiency is then:
For an ideal gas with constant specific heats, ) T ( c h w s 2 1 p - = ) T (T c h w 2a 1 p a - =
Thus:

6. Compressor Isentropic Efficiency
Note that the work is directly proportional to T with constant specific heats. Real gases will also have dependence on P.
The work can then be represented by a change on the T-axis of a Ts diagram.

7. Compressor Isentropic Efficiencies2a 2s
T1–T2a
T1–T2s 1 S

8. Isentropic Efficiencies
TURBINES
With turbines, we’re interested in the work/power output not input. An isentropic turbine will produce the maximum output. Efficiency is given by:
Again, isentropic efficiency will vary between zero and one.

9. Isentropic Efficiencies1 2a
Actual path will vary depending on amount of irreversibilities... 2a 2s 2a S

10. Isentropic Efficiencies
For nozzles, the isentropic efficiency is given by

11. TEAMPLAY
Work problem 7-89

12. Chapter 8, Gas Cycles
Carnot cycle is the most efficient cycle that can be executed between a heat source and a heat sink.
However, isothermal heat transfer is difficult to obtain in reality--requires large heat exchangers and a lot of time.

13. Gas Cycles
Therefore, the very important (reversible) Carnot cycle, composed of two reversible isothermal processes and two reversible adiabatic processes, is never realized as a practical matter.
Its real value is as a standard of comparison for all other cycles.

14. Gas Cycles Assumptions of air standard cycle
Analyze two cycles in detail
Otto
Brayton

15. Assumptions of air standard cycle
Working fluid is air
Air is ideal gas
Combustion process is replaced by heat addition process
Heat rejection is used to restore the fluid to its initial state and complete the cycle
All processes are internally reversible
Constant or variable specific heats can be used

16. Gas cycles have many engineering applications
Internal combustion engine
Otto cycle
Diesel cycle
Gas turbines
Brayton cycle
Refrigeration
Reversed Brayton cycle

17. Some nomenclature before starting internal combustion engine cycles

18. More terminology

19. Terminology Bore = d Stroke = s Displacement volume =DV =
Clearance volume = CV
Compression ratio = r

20. Mean Effective Pressure
Mean Effective Pressure (MEP) is a fictitious pressure, such that if it acted on the piston during the entire power stroke, it would produce the same amount of net work.

21. The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume

22. Real Otto cycle

23. Real and Idealized Cycle

24. Idealized Otto cycle

25. Idealized Otto cycle 1-2 - ADIABATIC COMPRESSION (ISENTROPIC)
CONSTANT VOLUME HEAT ADDITION
ADIABATIC EXPANSION (ISENTROPIC)
CONSTANT VOLUME HEAT REJECTION

26. Performance of cycle
Efficiency:
Let’s start by getting heat input:

27. Cycle Performance Get net work from energy balance of cycle:
Substituting for qin and qout:
Efficiency is then:

28. Cycle Performance Substituting for net work and heat input:
We can simplify the above expression:

29. Teamplay
Problem 8-36

30. Cold air standard cycle
cp, cv, and k are constant at ambient temperature ( 70 °F) values.
Assumption will allow us to get a quick “first cut”approximation of performance of cycle.

31. Cycle performance with cold air cycle assumptions
If we assume constant specific heats:

32. Cycle performance with cold air cycle assumptions
Because we’ve got two isentropic processes in the cycle, T1 can be related to T2, and T3 can be related to T4 with our ideal gas isentropic relationships….
Details are in the book!
Thus

33. Cycle performance with cold air cycle assumptions
This looks like the Carnot efficiency, but it is not! T1 and T2 are not constant.
What are the limitations for this expression?

34. Differences between Otto and Carnot cycles

35. Effect of compression ratio on Otto cycle efficiency

36. Sample Problem
The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22C and the cylinder volume is 5600 cm3. The compression ratio is 9 and 8.6 kJ are added during the heat addition process. Calculate:
(a) the temperature and pressure after the compression and heat addition process
(b) the thermal efficiency of the cycle
Use cold air cycle assumptions.

37. Draw cycle and label points
r = V1 /V2 = V4 /V3 = 9
Q23 = 8.6 kJ
T1 = 299 K
P1 = 95 kPa

38. Major assumptions Kinetic and potential energies are zero
Closed system
1 is start of compression
Ideal cycle: 1-2 isentropic compression, 2-3 const. volume heat addition, etc.
Cold cycle const. properties

39. Carry through with solution
Calculate mass of air:
Compression occurs from 1 to 2:
But we need T3!

40. Get T3 with first law:
Solve for T3:

41. Thermal Efficiency

42. Let’s take a look at the Diesel cycle.

43. Idealized Diesel cycle
ADIABATIC COMPRESSION (ISENTROPIC)
CONSTANT PRESSURE HEAT ADDITION
ADIABATIC EXPANSION (ISENTROPIC)
CONSTANT VOLUME HEAT REJECTION

44. Performance of cycle Efficiency:
Heat input occurs from 2 to 3 in constant pressure process:
Why enthalpies?

45. TEAMPLAY
Work problem 8-16