What is the Work Done by a Heat Engine Operating on a Given Cycle?

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The discussion focuses on calculating the work done by a heat engine operating on a specific cycle, with an initial incorrect calculation of work at 36,477 J. Participants emphasize the importance of accurately determining the height from the graph, suggesting conversions from atm to Pa for precise calculations. The correct approach involves using the area of the parallelogram formed by the cycle on the graph, with attention to the direction of the engine's flow affecting the sign of the work. Despite attempts to clarify the calculations, participants continue to encounter errors, indicating a need for careful analysis of the graph and the work formula. The conversation highlights the complexities involved in thermodynamic calculations and the significance of unit conversions.
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Homework Statement



A heat engine operates on the cycle shown below. How much work is done by the engine (J) per cycle, if Pmax = 0.03 atm?

http://i995.photobucket.com/albums/af79/huybinhs/workprb.gif

Homework Equations



W = P * Delta V

The Attempt at a Solution



Delta V which is area on the graph is 8 - 2 = 6 * 2 = 12 m^2

W = 12 * 3039.75 N/m^2 = 36477 J = INCORRECT.

Any ideas? Thanks!
 
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Work will be the area of the box the system creates.

You can approximate the area just by looking at the graph, and using formula for area of a parallelogram, which this appears to be.
 
jagged06 said:
Work will be the area of the box the system creates.

You can approximate the area just by looking at the graph, and using formula for area of a parallelogram, which this appears to be.

Yes, I looked on the graph:

Base = 6 , h= 2 so Area = 6*2 = 12, correct?
 
huybinhs said:
Yes, I looked on the graph:

Base = 6 , h= 2 so Area = 6*2 = 12, correct?

It looks to me like h is wrong. It looks like it is 1/3 of Pmax
 
Ok, so 6 * 0.01 = 0.06 then * 1013.25 = 60.795 = answer correct?

just submited it and INCORRECT :(
 
huybinhs said:
Ok, so 6 * 0.01 = 0.06 then * 1013.25 = 60.795 = answer correct?

just submited it and INCORRECT :(

My suggestion is to convert .01 atm to Pa FIRST.

as in .01 atm = 1013.25 Pa <-- that is now your height

there is no need to convert the Volume so

W = 1013.25 * 6

Answer will be in Joules

OH THE CONVERSIONS!
 
jagged06 said:
My suggestion is to convert .01 atm to Pa FIRST.

as in .01 atm = 1013.25 Pa <-- that is now your height

there is no need to convert the Volume so

W = 1013.25 * 6

Answer will be in Joules

OH THE CONVERSIONS!

6079.5 J is still INCORRECT! How come?
 
Oh I'm very sorry, did you try negative?

How many chance do you get?
 
I got 2 left. How is it negative then?
 
  • #10
Because of the flow of the engine.

Let's take it in sections:

starting at the top left corner, the system moves down and to the right, and everything below this part of the graph would be positive. Delta V in this case would be positive.

Once it goes back up and back to the left Delta V in this case, would be negative, and the area below it is larger.

Direction matters here, I'm just not finding a good example. I wish I had one.
 
  • #11
jagged06 said:
Because of the flow of the engine.

Let's take it in sections:

starting at the top left corner, the system moves down and to the right, and everything below this part of the graph would be positive. Delta V in this case would be positive.

Once it goes back up and back to the left Delta V in this case, would be negative, and the area below it is larger.

Direction matters here, I'm just not finding a good example. I wish I had one.

Perfect. Could u help me with this one?

https://www.physicsforums.com/showthread.php?t=402433
 
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