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# Heat Engine

Heat Engine. A heat engine is a device that absorbs heat ( Q ) and uses it to do useful work ( W ) on the surroundings when operating in a cycle. Sources of heat include the combustion of coal, petroleum or carbohydrates and nuclear reactions. Télécharger la présentation ## Heat Engine

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1. Heat Engine • A heat engine is a device that absorbs heat (Q) and uses it to do useful work (W) on the surroundings when operating in a cycle. • Sources of heat include the combustion of coal, petroleum or carbohydrates and nuclear reactions. • Working substance: the matter inside the heat engine that undergoes addition or rejection of heat and that does work on the surroundings. Examples include air and water vapour (steam). • In a cycle, the working substance is in the same thermodynamic state at the end as at the start.

2. Hot Body (source of heat) Q1 W Q2 Cold Body (absorbs heat) Heat Engine E

3. Example of a Heat Engine Open system

4. Work per cycle = Area inside Q=0 Q=0 Comparison of Otto and Diesel Cycles combustion

5. Nuclear Power Plant: A Very Large Heat Engine

6. Hot Body (source of heat) Q1 W Q2 Cold Body (absorbs heat) Heat Engine E

7. Substituting: Efficiency of a Heat Engine Efficiency,h = Work out/Heat in: Apply First Law to the working substance: DU = Q1 – Q2 – W But in a cycle,DU= 0 Thus, W = Q1 – Q2. Lesson:his maximum when Q2 is minimum.

8. The Stirling Engine See: http://www.animatedengines.com/ltdstirling.shtml • Closed system • Operates between two bodies with (small) different temperatures. • Can use “stray” heat

9. isothermal Heat in Heat out isothermal The Stirling Cycle TH >TC (TH - TC ) is proportional to the amount of work that is done in a cycle. = air temp =hot water 2

10. Hot Reservoir T1 Q1 W Q2 Cold Reservoir T2 Carnot Cycle C

11. a • Q1 b • T1 Q=0 Q=0 • d • T2 Q2 c Carnot Cycle Pressure Volume

12. a • Q1 b • T1 W Q=0 Q=0 • d • T2 Q2 c Carnot Cycle Pressure Volume

13. From b to c: adiabatic, Q = 0, so that TVg-1 is constant. Thus, T1Vbg-1 = T2Vcg-1 or Similarly, d to a: adiabatic, Q = 0, so that TVg-1 is constant. Thus, T2Vdg-1 = T1Vag-1 or Carnot Cycle From a to b: isothermal, so that DU = 0 and Q = - W Thus, Q1 = +nRT1ln(Vb/Va)(+ve quantity) Similarly, from c to d: isothermal, so thatDU= 0 and Q = - W Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve)

14. Which means that But as the volume ratios are equal: Carnot Cycle We see that: Now also: This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot Cycle.

15. What’s Special about a Carnot Cycle? (1) Heat is transferred to/from only two reservoirs at fixed temperatures, T1 and T2 - not at a variety of temperatures. (2) Heat transfer is the most efficient possible because the temperature of the working substance equals the temperature of the reservoirs. No heat is wasted in flowing from hot to cold. (3) The cycle uses an adiabatic process to raise and lower the temperature of the working substance. No heat is wasted in heating up the working substance. (4) Carnot cycles are reversible. Not all cycles are!

16. • Recall that for any cycle, the efficiency of a heat engine is given as: • For an engine using a Carnot cycle, the efficiency is also equal to: What’s Special about a Carnot Cycle? (5) The Carnot theorem states that the Carnot cycle (or any reversible cycle) is the most efficient cycle possible. The Carnot cycle defines an upper limit to the efficiency of a cycle. • Where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively, in degrees Kelvin. As T2 > 0,hcis always <1.

17. Efficiency of a Stirling Engine Question: What is the maximum possible efficiency of a Stirling engine operating between room temperature (25 °C) and boiling water (100 °C)? Maximum efficiency would be achieved by a Carnot cycle operating between reservoirs at T1 = 373 K and T2 = 298 K. Question: What is the maximum possible efficiency of a Stirling engine operating between room temperature (25 °C) and ice (0 °C)?

18. Kelvin-Planck Statement of the Second Law of Thermodynamics “It is impossible to construct a device that - operating in a cycle - will produce no other effect than the extraction of heat from a single body and the performance of an equivalent amount of work” Or…A cyclical engine cannot convert heat from a single body completely into work. Some heat must be rejected at a lower temperature. Thus,efficiency, h < 1!

19. Hot Body (source of heat) Q1 Q2 = 0 Heat Engine E W= -Q1 Cold Body (absorbs heat)

20. Q2 = W Heat Engine Hot Body (source of heat) Q1= 0 Examples: friction creating heat; isothermal compression of ideal gas E W POSSIBLE! Cold Body (absorbs heat)

21. Refrigerator: A heat engine operating in reverse Hot Body Q1 Refrigerator Efficiency: E W Q2 Cold Body

22. For a Carnot refrigerator, the efficiency is: Refrigerator Efficiency First Law tells us that Q2 + W -Q1 = 0. Thus, W = Q1 – Q2 Efficiency is usually >1! The smaller the T difference, the more efficient is the refrigerator.

23. Clausius Statement of the Second Law of Thermodynamics (applies to refrigerators) “It is impossible to construct a device that - operating in a cycle - will produce no other effect than heat transfer from a colder body to hotter body.” “Or…Heat cannot flow from a cold body to a hotter body by itself. Work has to be done in the process.”

24. The efficiency is defined as the amount of heat pumped in to the hot body per the amount of work done: The First Law tells us that W = Q1-Q2 So, substituting, we find: Efficiency of a Heat Pump The purpose of a heat pump is to extract heat from a cold body (such as the River Thames) and “pump” it to a hot body (such as an office building). hhpis always > 1! For maximumh, T2 should be T1 (just slightly less).

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