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Heat engine

  • Thread starter roam
  • Start date
  • #1
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Homework Statement


I need some help with this problem:

A model Stirling engine uses n = 7.44 × 10–3 mol of gas (assumed to be ideal) as a working substance. It operates between a high temperature reservoir at TH= 95.0°C and a low temperature reservoir at Tc = 24.0°C. The volume of its working substance doubles during each expansion stroke. It runs at a rate of 0.6 cycles per second. Assume the engine is ideal.

How much Work does the engine do per cycle (include the sign)?

Homework Equations



[tex]W=nRT ln \frac{V_i}{V_f}[/tex]

The Attempt at a Solution



Using the formula above I get

(7.44 × 10-3)(8.314) T ln (vi/2vi)

But I don't know how to continue since I don't know the value for the initial volume or temprature and I don't know how to find them... I mean for the temprature at least, I'm given two different tempratures TH and Tc and I don't see which one to use!
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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The volume doubles, so Vi/Vf = 1/2 for the expansion processes and Vi/Vf=2 for the compression process.

Take a look at this graph: http://upload.wikimedia.org/wikipedia/commons/2/25/Stirling_Cycle.svg

You have to calculate the work done for each leg and add them together to calculate total work. The expansion/compression legs are isothermal, so there's only one T.
 
  • #3
1,266
11
The volume doubles, so Vi/Vf = 1/2 for the expansion processes and Vi/Vf=2 for the compression process.

Take a look at this graph: http://upload.wikimedia.org/wikipedia/commons/2/25/Stirling_Cycle.svg

You have to calculate the work done for each leg and add them together to calculate total work. The expansion/compression legs are isothermal, so there's only one T.
THANK YOU! It makes perfect sense now. :smile: But I have another question; what if they ask 'what is the Power of the engine?'

Should we just divide the total work done (that we just calculated) by the number of cycles per second? I tried that but it didn't give me the correct answer...
 
  • #4
1,434
2
You should multiply the work with the number of cycles per second, cause P = Q/t and Q = W(one cycle) * (cycles in a second) if t = 1 sec.
 

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