Carnot Efficiency of a heat engine


by AlaskanPow
Tags: carnot, efficiency, engine, heat
AlaskanPow
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#1
Nov8-12, 11:10 PM
P: 13
My problem only gives me joules to work with. Is it possible to convert from joules of energy to temperature (Kelvin)?? If so how?
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Basic_Physics
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#2
Nov9-12, 01:15 AM
P: 358
Best to state the problem. Maybe we can see a way around it.
AlaskanPow
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#3
Nov9-12, 01:29 AM
P: 13
An engine transfers 2.00x10^3 Joules of energy from a hot reservoir during a cycle and transfers 1.50x10^3 Joules as exhaust to a cold reservoir. Find the actual efficiency of the engine and then compare it to the carnot efficiency.

The actual efficiency is 75% I calculated. For carnot efficiency i need temperature, but I dont know how to get it.

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#4
Nov9-12, 01:45 AM
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Carnot Efficiency of a heat engine


Your calculation of the efficiency is wrong. Also, if you don't have the temperature of the reservoirs, you can't get the Carnot efficiency.
Basic_Physics
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#5
Nov9-12, 03:31 AM
P: 358
The actual efficiency of the engine

[itex]\epsilon=\frac{work \; done \; by \; the \; engine}{total \; energy \; used \; by \; it}[/itex]
AlaskanPow
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#6
Nov9-12, 03:43 AM
P: 13
My book says (efficiency=energy output/energy input)
Which would give me 75% I believe.
So there is no way to get the carnot efficiency?
Basic_Physics
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#7
Nov9-12, 03:57 AM
P: 358
The engine extracts energy between the input and output reservoirs of the engine, this it converts into work. Look in the section on the 2nd law of Thermodynamics how to convert the transferred heat to temperatures.
Basic_Physics
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#8
Nov9-12, 04:05 AM
P: 358
The 75% is the Carnot engine's efficiency.
AlaskanPow
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#9
Nov9-12, 04:16 AM
P: 13
Hmmm interesting. My book does not go in to depth very much, because im in a survey class, so I dont understand how you could calculate it without having your temperatrure. The only equation I got in my book for carnot efficiency is
carnot efficiency=(Thot-Tcold)/ Thot
Basic_Physics
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#10
Nov9-12, 04:24 AM
P: 358
For a Carnot engine it is assumed that all of the extracted heat, QH - QL (high, low), is converted into work by the engine, so that its efficiency is given by

[itex]\epsilon=\frac{Q_{H}-Q_{L}}{Q_{H}}[/itex]
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#11
Nov9-12, 05:34 AM
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Quote Quote by Basic_Physics View Post
For a Carnot engine it is assumed that all of the extracted heat, QH - QL (high, low), is converted into work by the engine, so that its efficiency is given by

[itex]\epsilon=\frac{Q_{H}-Q_{L}}{Q_{H}}[/itex]
W = Qh-Ql for any heat engine. So this is the definition of efficiency for any heat engine. To calculate the Carnot efficiency, the maximum efficiency of any possible heat engine operating between these two reservoirs, you would need to know the temperatures, Th and Tc.

AM
vela
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#12
Nov9-12, 05:57 AM
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Quote Quote by AlaskanPow View Post
My book says (efficiency=energy output/energy input)
Which would give me 75% I believe.
So there is no way to get the carnot efficiency?
The heat exhausted to the cold reservoir is not the energy output of the heat engine.


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