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2-D heat transfer problem

  • Thread starter leoflc
  • Start date
  • #1
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Homework Statement


a rectangular plate (15 x 30) with the following boundary conditions:

u_y(x,0) = 0
u_y(x,30) = 0
u_x(0,y) = 0
u_x(15,y) = y(20-y)

the derivative B.Cs describe the heat flux through the boundaries.

solve the the steady-state temp u(x,t)


Homework Equations


steady-state: [tex]\nabla[/tex]^2 * u =0


The Attempt at a Solution



I set u(x,y)=F(x)G(y)

with the BCs, I got

G'(0)=0
G'(30)=0
F'(0)=0
F'(15)=y(20-y)

let F=A*cos(px) + B*sin(px)

F'(0)=0; so B=0
F'(15)=-A*sin(15*p)=y(20-y)


this is where I got stock.
How can I solve for 'p' or A with the B.C. that has 'y' in there?

Am I doing the right thing?
Thank you very much!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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First you are also going to have to solve for the general form for G(y)!
Since the condition for F'(15) is not itself a sin or cosine, you will need to write this as a Fourier cosine series. Then expand y(20-y) (a constant with respect to x) as a Fourier cosine series.
 
  • #3
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The standard way in solving this is indeed by considering that the solution can be written as a product of two functions each depending only on one variable. This is thus:

[tex]u(x,y)=X(x)\cdot Y(y)[/tex]

By substituting this into the partial differential equation you end up with two ordinary differential equations both equal to a constant. Remember that this constant can be smaller, equal to or larger than zero. Consider all three cases separately, one will give you the zero solution which is to be discarded, one will give you a constant and the last one will give you a fundamental solution. The total solution will be the sum of the constant and all functions of the fundamental form. After this you can apply the remaining condition to search for the unknown constants remaining. It will not be possible to find one constant because you only have derivatives on the boundaries. This means that your solution is determined up to a constant which is physically interpretable by stating that the solution can differ in temperature level.

To make it more easy on yourself you could consider to change the size of the rectangle to B and L (width and height) and the last boundary condition to y(a-y). You will avoid making calculation errors and the obtained solution is more generally valid for different rectangles. In case of problems, just ask.
 
  • #4
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Thank you for the responses!
For F'(15)
I expand y(20-y) as a Fourier cosine series, but I got

F'(15) = -A*sin(15*p) = FourierCosine series in terms of 'y'

I still don't know what to do with the y term. Do I just set x=y(20-y) and solve for y, then put it back to the Fourier Consine series?
Also for G'(y), I got

G'(y) = A*p*sinh(p*y)

with B.C (A is a const)

G'(30) = A*p*sinh(30p) = 0

which will make p=0, and I don't think that should be it.

Thank you again for the help!
 
  • #5
279
0
Thank you for the responses!
For F'(15)
I expand y(20-y) as a Fourier cosine series, but I got

F'(15) = -A*sin(15*p) = FourierCosine series in terms of 'y'

I still don't know what to do with the y term. Do I just set x=y(20-y) and solve for y, then put it back to the Fourier Consine series?
Also for G'(y), I got

G'(y) = A*p*sinh(p*y)

with B.C (A is a const)

G'(30) = A*p*sinh(30p) = 0

which will make p=0, and I don't think that should be it.

Thank you again for the help!
Mmm, I think you are doing this incorrect. You don't mention a series here at all. OK, the first thing to do is to assume that the solution can be written as a product of two functions, each depending on only one of the two independent variables. This means:

[tex]u(x,y)=X(x)\cdot Y(y)[/tex]

You need to substitute this in the original partial differential equation. This gives you (leaving out the variable naming):

[tex]X''Y+XY''=0[/tex]

Dividing this with the function itself gives:

[tex]\frac{X''Y+XY''}{XY}=0[/tex]

Or:

[tex]\frac{X''}{X}=-\frac{Y''}{Y}[/tex]

Because the left hand side and the right hand side both are depending on one of the variables and not the other they both have to be equal to a constant. You have now:

[tex]\frac{X''}{X}=-\frac{Y''}{Y}=\alpha[/tex]

Now you have obtained two ordinary differential equations. The constant can be smaller, equal to or larger than 0. Use these three cases to find the solutions to the ordinary differential equations.

What I've just written is the basic handling of boundary value problems using Fourier series. If you don't understand this at all, you should consult a book or get some more explanation of the teacher.

Come back with what you have obtained and we'll go to the next step.
 

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