1. Unit 61: Engineering Thermodynamics Lesson 9: Carnot Engine Cycles

2. Objective The purpose of this lesson is to examine the Carnot engine cycle.

3. The Carnot Engine The heat engine that operates the most efficiently between a high-temperature reservoir and a low temperature reservoir is the Carnot engine. It is an ideal engine that uses reversible processes to form its cycle of operation. – thus it is also called a reversible engine.

4. The Carnot Engine The Carnot engine is very useful since its efficiency establishes the maximum possible efficiency of any real engine. If the efficiency of a real engine is significantly lower than the efficiency of a Carnot engine operating between the same limits, then additional improvements may be possible.

5. The Carnot Cycle THTH Insulator TLTL W 1-2 W 2-3 W 3-4 W 4-1 1 2 2 3 3 4 4 1 QHQH QLQL

6. The Carnot Cycle 3 2 1 4 v p T = constant Q = 0

7. The Carnot Engine Considering an ideal gas… 1  2: An isothermal expansion – Heat is transferred reversibly from high- temperature reservoir at constant temperature T H. The piston in the cylinder is withdrawn and the volume increases.

8. The Carnot Engine 2  3: An adiabatic reversible expansion – The cylinder is completely insulated so that no heat transfer occurs during this reversible process. The piston continues to be withdrawn with the volume increasing. 3  4: An isothermal compression – Heat is transferred reversibly to the low- temperature reservoir at constant temperature T L. The piston compresses the working substance with the volume decreasing.

9. The Carnot Engine 4  1: An adiabatic reversible compression – The completely insulating cylinder allows no heat transfer during reversible process. The piston continues to compress the working substance until the original volume, temperature and pressure are reached thereby completing the cycle.

10. The Carnot Engine Applying the first law to the cycle gives… Q H – Q L = W net Thus thermal efficiency… η = Q H – Q L = 1 - Q L Q H Q H

11. The Carnot Engine Postulate 1: – It is impossible to construct an engine, operating between two temperature reservoirs, that is more efficient than the Carnot engine. – To show this assume that an engine exists, operating between two reservoirs, that has greater efficiency than that of a Carnot engine; also assume that a Carnot engine operates as a refrigerator between the same two reservoirs.

12. The Carnot Engine THTH TLTL QHQH W’ Q’ L QHQH W QLQL Engine Carnot Refrigerator W’ = Q H – Q’ L W = Q H - Q L Q’ L < Q L Thus W’ > W

13. The Carnot Engine TLTL QHQH Q’ L QLQL Engine Carnot Refrigerator W’ - W System boundary

14. The Carnot Engine Let the heat transferred from the high-temperature reservoir to the engine be equal to the heat rejected by the refrigerator; then the work produced by the engine will be greater than the work required by the refrigerator (that is Q’ L < Q L ) since the efficiency of the engine is greater than that of the Carnot engine. The second diagram thus show that the engine drives the refrigerator using the rejected heat from the refrigerator with the net work (W’-W) leaving the system.

15. The Carnot Engine The net result is the conservation of energy from single reservoir, a violation of the second law. Thus the Carnot engine is the most efficient engine operating between two particular reservoirs.

16. The Carnot Engine Postulate 2: – The efficiency of a Carnot engine is not dependent on the working substance used or any particular design feature of the engine – Suppose that a Carnot engine drives a Carnot refrigerator

17. The Carnot Engine THTH TLTL QHQH W QLQL Q’ H W’ QLQL Carnot Engine Carnot Refrigerator Q H – Q’ L = W Q’ H – Q L = W’ Q H > Q’ H Thus W > W’

18. The Carnot Engine QHQH Q’ H QLQL Carnot Engine Carnot Refrigerator W – W’ System boundary W’ THTH

19. The Carnot Engine Let the heat rejected by the engine be equal to the heat required by the refrigerator. Suppose the working fluid in the engine results in Q H being greater than Q’ H : then W would be greater than W’ ( a consequence of the 1 st law). The net result is a transfer of heat (Q H – Q’ H ) from a single reservoir and the production of work, a clear violation of the second law. Thus the efficiency of the Carnot engine is not dependent upon the working substance

20. Carnot Efficiency The Carnot engine is only dependent upon the two reservoir temperatures. Assume the working substance to be an ideal gas. Then the heat transfer for each of the four processes is… 1  2 : Q H = W 1-2 = PdV = mRT H ln(V 2 /V 1 ) 2  3: Q 2-3 = 0 V2V2 V1V1

21. Carnot Efficiency 3  4 : Q L = -W 3-4 = - PdV = - mRT L ln(V 4 /V 3 ) 4  1: Q 4-1 = 0 We want Q L to be positive, as in any thermal efficiency relationship hence the negative sign. V4V4 V3V3

22. Carnot Efficiency The thermal efficiency is thus… η = 1 – Q L /Q H = 1 + [T L ln(V 4 /V 3 )]/ T H ln(V 2 /V 1 )] During the reversible adiabatic processes 2  3 and 4  1, we know that… T L /T H = (V 2 /V 3 ) k-1 and that (T L /T H ) (V 1 /V 4 ) k-1 Thus (V 3 /V 2 ) = (V 4 /V 1 ) or (V 4 /V 3 ) = (V 1 /V 2 ) Thus… η = 1 – (T L /T H )

23. Carnot Efficiency We have simply replaced Q L /Q H with T L /T H. We can do this for all reversible engines or refrigerators. Thus the thermal efficiency of a Carnot engine is dependent only on the high and low absolute temperatures of the reservoirs. The fact that we used an ideal gas to perform the calculations is not important as the Carnot efficiency is independent of the working substance. The Carnot engine when operated in reverse becomes a heat pump or refrigerator depending upon the desired heat transfer

24. Carnot Efficiency The Coefficient of Performance of a heat pump becomes… COP HP = Q H /W net = [Q H /(Q H -Q L )] = 1/[1-(T L /T H )] The coefficient of performance of a refrigerator becomes… COP R = Q L /W net = [Q L /(Q H -Q L )] = 1/[(T H /T L ) - 1] Rather than list the COP of refrigerators or air- conditioners, manufacturers often list the EER (energy- efficiency ratio): EER = 3.412 x COP

25. Carnot Efficiency A Carnot engine operates between two temperature reservoirs maintained at 200 o C and 20 o C, respectively. If the desired output of the engine is 15kW, determine the heat transfer from the igh- temperature reservoir and the heat transfer to the low-temperature reservoir.

26. Carnot Efficiency T H = 200 o C T L = 20 o C QHQH QLQL W = 15kW

27. Carnot Efficiency The efficiency of a Carnot engine is… η = W/Q H = 1 – T L /T H Thus… Q H = W/[1-(T L /T H )] = 15/(1 – 293/473) = 38.42 KW Using the 1 st law we have… Q L = Q H – W = 39.42 – 15 = 24.42 kW