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

[huybinhs]

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!

 

[jagged06]

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.

 

[huybinhs]

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?

 

[jagged06]

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

 

[huybinhs]

Ok, so 6 * 0.01 = 0.06 then * 1013.25 = 60.795 = answer correct?

just submited it and INCORRECT :(

 

[jagged06]

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!

 

[huybinhs]

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?

 

[jagged06]

Oh I'm very sorry, did you try negative?

How many chance do you get?

 

[huybinhs]

I got 2 left. How is it negative then?

 

[jagged06]

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.

 

[huybinhs]

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