Differential Equation Help

  • Thread starter Wildcat04
  • Start date
  • #1
34
0

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 seperation 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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
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.
 
  • #3
34
0
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))
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
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.
 
  • #5
34
0
Sir,

I appriciate 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.
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
619
Sir,

I appriciate 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?
 
  • #7
34
0
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 appriciate your approval / laughter.

=)
 
  • #8
60
0
You are very close. I think you may have screwed up your initial integration (mine looks a lot different). This might help...
 

Attachments

  • Calculus_Cheat_Sheet_Integrals.pdf
    183.5 KB · Views: 102
  • #9
34
0
I like the cheat sheet! I am not quite sure that I follow you though.
 
  • #10
60
0
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...
 
  • #11
34
0
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.
 
  • #12
60
0
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?
 
  • #13
34
0
Hrmmm, just plotted it and it doesn't make sense. Back to the drawing board.
 
  • #14
60
0
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.
 
  • #16
60
0
did you get it?
 

Related Threads on Differential Equation Help

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
974
  • Last Post
Replies
1
Views
959
Top