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Carnot Efficiency of a heat engine 
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#1
Nov812, 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?



#2
Nov912, 01:15 AM

P: 358

Best to state the problem. Maybe we can see a way around it.



#3
Nov912, 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. 


#4
Nov912, 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.



#5
Nov912, 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] 


#6
Nov912, 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? 


#7
Nov912, 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.



#8
Nov912, 04:05 AM

P: 358

The 75% is the Carnot engine's efficiency.



#9
Nov912, 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=(ThotTcold)/ Thot 


#10
Nov912, 04:24 AM

P: 358

For a Carnot engine it is assumed that all of the extracted heat, Q_{H}  Q_{L} (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] 


#11
Nov912, 05:34 AM

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#12
Nov912, 05:57 AM

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