Tuesday, September 30, 2008

TE 303 - HW #4, P3 - COMPUTER ANALYSIS: P-V Data - 20 pts

Measured data for pressure versus volume during the expansion of gases within the cylinder of an internal combustion engine are given in the table below:

Data Point V (cm^3) P (bar)
1 300 15
2 361 12
3 459 9
4 644 6
5 903 4
6 1608 2

Using data from the table and using EXCEL, complete the following:
(a) Determine the value of d such that the data are fit by an equation of the form PVd = constant. (HINT: take the log of both sides and rearrange into a linear equation.) You are expected to extract the value of d from either doing a curve fit with something like least squares, or from the slope or intercept function in Excel for the appropriate graph.

(b) Evaluate analytically the work done by the gases, in kJ, using the appropriate equation from class along with the result from (a).

(c) Using numerical integration of the data, evaluate the work done by the gases, in kJ (NOTE: Be careful about converting your units!). A common numerical integration method is the Trapezoidal Rule for Unevenly Spaced Data. Using this method, you can approximate the integral in the following way:



where Ai is the area of each rectangular interval, and n is the number of rectangular intervals.

For more information on the Trapezoidal Rule, please see the following website: http://oregonstate.edu/~haggertr/487/integrate.htm

For the "analysis" section of your Engineering Model, be sure to compare the different methods for estimating the work used in parts (b) and (c).

12 comments:

Anonymous said...

ANSWERS:
a) Delta about 1.1996 (depending on how you solve for it)

b) W = +/- 0.643 kJ (YOU determine the sign)

c) Your work value should be in the same range as b).

CORRECTION: In the trapezoid equation, n is the number of data points, not intervals.

HINTS: See problem description.

Anonymous said...

I was wondering if the units mattered for part c. I used logV in m^3 and logP in bar. I was not sure what the answer comes out as when you use logs.

My answer seems right when I do this but I just want to make sure that the answer you get from this does not need to be converted.

Anonymous said...

Nevermind...ignore the last comment. I am not supposed to use logs...

Anonymous said...

When doing this my sign for A is negative but I got the right answer am I looking at something wrong?

Anonymous said...

I got it no negative work

Anonymous said...

Dear anonymous posters,

I think that you answered your own questions; if not, please post again!

özlem said...

can you place the solution here? I couldn't understand the trapezoidal rule..especially intervals?? thanks

Dr. B said...

Hello Ozlem,
It is not easy to post solutions on this blog. The answers are here and here is a link to a great explanation of the trapezoidal rule.

http://en.wikipedia.org/wiki/Trapezoidal_rule

Best of luck to you! Write again if you need some help after looking at the Wiki.

Anonymous said...

I got the answer as 0.658 kJ..is it exactly right your answer? 0.643 kJ? did I have a mistake? thank you very much..I used the trapezoidal equation that you gived for explanation of answer..

Dr. B said...

Anonymous 12:04,

I cannot say for certain whether you solved the problem correctly from the information you provided. I can say that 0.658 kJ is not all that close to 0.643 kJ. So, I suspect you made a minor mistake.
Best of luck to you!

Anonymous said...

Excuse ne..How to find the Delta itself??

Dr. B said...

Hi Anon 12:04AM,
The HINT is the key. PV^d = constant. Take the LN of this eqn. Keep in mind that LN(V^d) = d*LN(V). PLot LN(P) vs LN(V) and you should get something nearly linear.
Best of luck,
Dr. B