# Thermodynamics problem

1. May 11, 2010

### huybinhs

1. The problem statement, all variables and given/known data

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

2. Relevant equations

W = P * Delta V

3. 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!

2. May 11, 2010

### 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.

3. May 11, 2010

### huybinhs

Yes, I looked on the graph:

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

4. May 11, 2010

### jagged06

It looks to me like h is wrong. It looks like it is 1/3 of Pmax

5. May 11, 2010

### huybinhs

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

just submited it and INCORRECT :(

6. May 11, 2010

### jagged06

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

OH THE CONVERSIONS!!!

7. May 11, 2010

### huybinhs

6079.5 J is still INCORRECT!!! How come????

8. May 11, 2010

### jagged06

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

How many chance do you get?

9. May 11, 2010

### huybinhs

I got 2 left. How is it negative then?

10. May 11, 2010

### 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.

11. May 11, 2010

### huybinhs

Perfect. Could u help me with this one?