Solving a Complicated Differential Equation

[Wildcat04]

Homework Statement

Well, I finally figured out the differential equation from a previous post (thank god) but know I am having some trouble getting the result I want as it has been 6 years since diff eq.

500dx = 20dT + 20T - 9.47(T-15)dx

Homework Equations

N/A

The Attempt at a Solution

Well, first shot I went for separation of variables which seemed logical to me:

500dx + 9.47(T-15)dx = 20dT + 20T

(500 + 9.47(T-15)dx = 20dT + 20T

dx = (20dT + 20T) / [500 + 9.47(T-15)]

And this seems unrealistic.

Someone kick me and tell me how wrong I am please. I have tried digging though my old books but I don't seem to have my diff eq or calc books anymore.

 

[Dick]

Either you aren't using parentheses right, or that's complete rubbish. 500dx = 20dT + 20T - 9.47(T-15)dx is crazy. Every term in that equation should have 'd' something in it if one does.

 

[Wildcat04]

Hrmmm.

Well, this was what I worked out with my professor last week and I have been trying to work on it for awhile.

I am not sure if the link to the other post I had worked or not, but basically it is a heat transfer problem. The control volume being from x to x+dx which corresponds to T and T+dT.

In the middle of typing this I think I may have figured out my error...I believe the first should have been T+dT.

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

Does that look better?

Continuing on that thought...

Integating both sides...

x = 2.11*ln(T+37.8) + c

x/2.11 = ln (T +37.8) + c

e^(x/2.11) = T + 37.8 + c

T = e^(x/2.11) - 37.8 + c

T(0) = 15 -> c = 51.8

T = e^(x/2.11) + 14

hrmmm...still along way from the books answer of T = 15 + 52.8(1-e^(-x/2.11))

 

[Dick]

Hrmmm.

Well, this was what I worked out with my professor last week and I have been trying to work on it for awhile.

I am not sure if the link to the other post I had worked or not, but basically it is a heat transfer problem. The control volume being from x to x+dx which corresponds to T and T+dT.

In the middle of typing this I think I may have figured out my error...I believe the first should have been T+dT.

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

Does that look better?

Looks better to me. At least the 'd's are balanced. I don't know the whole context of your problem, but at least it looks like a differential equation now.

 

[Wildcat04]

Sir,

I appreciate your help...but I think I am still missing some as you can see from my above answer. I am getting closer but not quite there yet. It may be that I am missing something in my equation still.

 

[Dick]

Sir,

I appreciate your help...but I think I am still missing some as you can see from my above answer. I am getting closer but not quite there yet. It may be that I am missing something in my equation still.

It doesn't look like anything is missing to me, except for actually solving the equation. Now you can integrate both sides, right?

 

[Wildcat04]

I believe so, I should have replied but I just edited the text. My solution is about 4 posts up. If you have time I would very much appreciate your approval / laughter.

=)

 

[staticd]

You are very close. I think you may have screwed up your initial integration (mine looks a lot different). This might help...

 

[Wildcat04]

I like the cheat sheet! I am not quite sure that I follow you though.

 

[staticd]

When I worked it out (already erased), my integration of both sides resulted in an x that was much different than yours. I used the integral of 1/at+b to get the x in terms of t. I also did u substitution and got the same answer...

...this is also very similar to a first order transient response in electric circuits...

 

[Wildcat04]

I have tried this problem several different ways and keep coming up with the answer that I originally had. I looked into some FO Transient Response problems and can't seem to get it to fit together.

 

[staticd]

Maybe the answer in the book is wrong... Have you tried plotting your results or running a simulation to see if your results make sense?

 

[Wildcat04]

Hrmmm, just plotted it and it doesn't make sense. Back to the drawing board.

 

[staticd]

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

This is wrong. You should have 20(t + dt) in the numerator. Based on the equation you first posted.

500dx = 20dT + 20T - 9.47(T-15)dx

Actually, I think you are going to have to use a transform to solve this.

 

[staticd]

Actually, this seems more plausible...

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

Since you have a diff-eq with two independent variables.

 

[staticd]

did you get it?