# Power Output of an Engine

1. Feb 5, 2015

### Abid Rizvi

1. The problem statement, all variables and given/known data
During the power stroke in a four-stroke automobile engine, the piston is forced down as the mixture of combustion products and air undergoes an adiabatic expansion. Assume (1) the engine is running at 3400 cycles/min; (2) the gauge pressure immediately before the expansion is 20.0 atm; (3) the volumes of the mixture immediately before and after the expansion are 50.0 cm3 and 400 cm3, respectively; (4) the time interval for the expansion is one-fourth that of the total cycle; and (5) the mixture behaves like an ideal gas with specific heat ratio 1.40. Find the average power generated during the power stroke

2. Relevant equations
W = integral of PdV
PV^(1.4) = constant

3. The attempt at a solution
So first I found out that it takes approximately .005s for the power stroke.
I found the constant to be approximately (using pascals and m^3) 1.9289. I then said P = 1.9289/V^1.4.
I put that into the integral so integral from .00005 to .0004 of 1.9289/V^1.4 dV. I get about 143 J. After doing the conversions to kw, I am wrong. I found (i know the answer) that I should be getting 150 J to turn into kW, so I don't understand what is wrong with my integral. The other solution that gives 150 J uses dE = nCv dT instead of PdV, but isnt' it the same thing as PdV? Thanks in advance!

2. Feb 5, 2015

### Simon Bridge

What did the model answer use for Cv considering that neither the volume nor the pressure is constant?

The method sounds OK - to see the problem we'd have to see the details.
i.e. how does the pressure being quoted as gauge pressure affect the calculation?
how does using more exact values affect the calculation (you are not far off the model answer.)

3. Feb 5, 2015

### Abid Rizvi

The model answer used 5/2 R for Cv. The model for how to get the right answer is on this link: <http://www.cpp.edu/~skboddeker/132/assign/ch21h.htm>. Sorry, but which details are you looking for? All of my work or was it all of the work done in the link above?

Edit: I just went through the calculation done in the link above, and I realized that although he says he got 150, his equation gives 136 ...

Last edited: Feb 5, 2015
4. Feb 5, 2015

### Simon Bridge

Oh I see ... the model approach is less than ideal since it makes assumptions about information not given.
Compare:

$$W=P_1V_1^\gamma \frac{V_2^{1-\gamma}-V_1^{1-\gamma}}{1-\gamma}$$

Maybe the model answer is in error?
The "details" would be step-by-step including the reasoning rather than "I did A and got B".
I gave an example in the i.e.

5. Mar 13, 2015

### Abid Rizvi

Sorry this is way late. In the end I figured out the issue. You were right to suspect looking at gauge pressure. I simply needed to add that extra atm... After that everything works! Thanks

6. Mar 13, 2015

### Simon Bridge

No worries and well done.