Lect14 - Le cture14 Heat Engines(2 Pumping Heat Heat pumps...

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Lecture 14, p 1 Lecture 14 Heat Engines (2) Pumping Heat: Heat pumps and Refrigerators Available Work and Free Energy Work from Hot and Cold Bricks Reading for this Lecture: Elements Ch 10 Reading for Lecture 16: Elements Ch 11

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Lecture 14, p 2 Run the Engine in Reverse The Carnot cycle is reversible (each step is reversible) : When the engine runs in reverse: Heat is transferred from cold to hot by action of work on the engine . Note that heat never flows directly from cold to hot; the cold gas is being heated by adiabatic compression (process 3). Q c / Q h = T c / T h is still true. (Note: Q h , Q c , and W are still positive!) Hot reservoir at T h Cold reservoir at T c Q h Q c W Engine V T c T h V b V a V c V d p 1 2 3 4 Reverse all the arrows Q h Q c
Lecture 14, p 3 Refrigerators and Heat Pumps Hot reservoir at T h Cold reservoir at T c Q h Q c W Engine Refrigerators and heat pumps are heat engines running in reverse. How do we measure their performance? It depends on what you want to accomplish. Refrigerator: We want to keep the food cold (Q c ). We pay for W (the electric motor in the fridge). So, the coefficient of performance , K is: Heat pump: We want to keep the house warm (Q h ). We pay for W (the electric motor in the garden). The coefficient of performance , K is: c c c h c h c Q Q T K W Q Q T T = = - - h h h h c h c Q Q T K W Q Q T T = = - - It’s not called “efficiency”.

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Lecture 14, p 4 Example: Refrigerator There is a 70 W heat leak (the insulation is not perfect) from a room at temperature 22° C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10° C? Assume Carnot performance. Hint: Q c must exactly compensate for the heat leak. Hot reservoir at T h Cold reservoir at T c Q h Q c W Engine Heat leak
Lecture 14, p 5 Solution There is a 70 W heat leak (the insulation is not perfect) from a room at temperature 22° C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10° C? Assume Carnot performance. Hint: Q c must exactly compensate for the heat leak. Hot reservoir at T h Cold reservoir at T c Q h Q c W Fridge Heat leak 8.2 c h c T K T T = - The coefficient of performance is: We need Q c = 70 J each second. Therefore we need The motor power is 8.5 Watts. Watt = Joule/second. 70 J 8.5 J 8.2 c Q W K = = This result illustrates an unintuitive property of refrigerators and heat pumps: When T h - T c is small, they pump more heat than the work you pay for, because K > 1.

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Lecture 14, p 6 Refrigerators work less well as T h - T c becomes large. The colder you try to go, the less efficient the refrigerator gets. The limit as T C goes to zero is zero efficiency ! Since heat leaks will not disappear as the object is cooled, you
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This note was uploaded on 02/21/2011 for the course PHYS 213 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Lect14 - Le cture14 Heat Engines(2 Pumping Heat Heat pumps...

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