Efficiency of an engine

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Homework Statement



http://img291.imageshack.us/img291/7786/effx.jpg [Broken]


Homework Equations



e=1-1/(V1/V2)^(gamma-1)

The Attempt at a Solution



This is a practice test where he gives the solutions, as you can see the first thing he asks for is the efficiency, and lists the answer as 0.53. How did he get this? T1*V1^gamma-1=T2*V2^gamma-1 with t1=20 and T2=350, what am I doing wrong? (I assume this, but what am I supposed to do instead?)
 
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Answers and Replies

  • #2
Delphi51
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I don't recognize that formula for efficiency with the V1, V2 and gamma, so perhaps I am off on the wrong track! The usual way to find the efficiency of a heat engine is given here
http://en.wikipedia.org/wiki/Heat_engine#Efficiency
and is 1 minus the square root of (Tc/Th) where Tc is the temperature of the exhaust and Th the temperature of the hot input gas. The temperatures MUST be in KELVIN degrees, so change your 20 to 293.
 
  • #4
Delphi51
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Sorry, beyond me! Hope someone else will help.
Do try your equation with Kelvin temperatures.
 
  • #5
Andrew Mason
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Using the equation you linked gave .31, which isnt as listed. The TV^gamma minus one is because its an adiabatic system.

http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_case.29

I think what you linked to is a bit more advanced, we are just doing carnot engines and such. Anything?
You do not need to apply the adiabatic condition here. All you have to know, as Delphi51 has correctly pointed out, is the relationship between efficiency of a Carnot engine and operating temperatures.

Efficiency = Output/Input = W/Qh = (Qh-Qc)/Qh = 1 - Qc/Qh.

In a Carnot engine, [itex]\Delta S = 0[/itex]. Since [itex]\Delta S = \Delta S_c + \Delta S_h = Q_c/T_c - Q_h/T_h [/itex] it follows that: [itex]Q_c/Q_h = T_c/T_h[/itex]. So the efficiency of a Carnot engine is always:

[tex]\eta = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_c}{T_h} [/tex]

Apply that to the problem.

AM


[NOTE: A Carnot engine uses adiabatic expansion and compression, as well as isothermal expansion and compression. You could actually calculate the heat flows and work done in each stage of the cycle to determine the efficiency, which is what you seem to be trying to do. But that is quite unnecessary.]
 

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