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

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Homework Help Overview

The discussion revolves around calculating the work done by a heat engine operating on a specific cycle, with a focus on interpreting a pressure-volume graph. The original poster presents an initial calculation based on given parameters, including maximum pressure.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of calculating work as the area under the curve on the graph, with some suggesting the use of geometric formulas for area. There are questions about the accuracy of height measurements and conversions between units.

Discussion Status

Participants are actively exploring different interpretations of the graph and the calculations involved. Some have suggested conversion methods for pressure, while others are questioning the assumptions about the direction of work and the implications of negative values.

Contextual Notes

There is a noted confusion regarding unit conversions and the interpretation of the graph, particularly concerning the height used in the area calculation. The original poster has expressed frustration over incorrect submissions, indicating a need for clarity on the calculations.

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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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