Diffusion Equation on a plate - 2 dimensions

In summary, the problem is to determine the steady-state temperature distribution in a thin plate with the edges held at specific temperatures. The plate is lying along the x-y plane with one corner at the origin, a height of 1m along the y-axis, and a length of 2m along the x-axis. The edges along the y-axis and x-axis are held at 0 C, while the edges parallel to the x-axis and y-axis are held at 0 C and 50sin(pi*y) C respectively. The equation for this problem is a diffusion equation with two space variables, x and y. The solution involves separating the variables and solving for X(x), Y(y), and T(t). The solution process involves determining
  • #1
samee
60
0

Homework Statement



The edges of a thin plate are held at the temperature described below. Determine the steady-state temperature distribution in the plate. Assume the large flat surfaces to be insulated.

If the plate is lying along the x-y plane, then one corner would be at the origin. The height of the plate would be 1m along the y-axis and the length would be 2m along the x-axis. The edge along the y-axis is being held at 0 C. The edge along the x-axis is being held at 0 C. The edge parallel to the x-axis is being held at 0 C. The edge parallel to the y-axis is being held at 50sin(pi*y) C.

Homework Equations



So this question is actually just a diffusion equation. Alpha2uxx=ut
and
u(x,t)=X(x)T(t)=(C1coskx+C2sinkx)e-K2alpha2t+C3+C4x

The Attempt at a Solution



So the problem I'm having is previously everything was done on a rod. Now I have a plate. I know that for a rod, my boundary conditions would be the temperature at the ends, for example, if this were a rod I would say that u(0,t)=0 and u(2,t)=50sin(piy). BUT! How can I say that if I have corners and a second dimension to deal with?
 
Physics news on Phys.org
  • #2
samee said:

Homework Statement



The edges of a thin plate are held at the temperature described below. Determine the steady-state temperature distribution in the plate. Assume the large flat surfaces to be insulated.

If the plate is lying along the x-y plane, then one corner would be at the origin. The height of the plate would be 1m along the y-axis and the length would be 2m along the x-axis. The edge along the y-axis is being held at 0 C. The edge along the x-axis is being held at 0 C. The edge parallel to the x-axis is being held at 0 C. The edge parallel to the y-axis is being held at 50sin(pi*y) C.

Homework Equations



So this question is actually just a diffusion equation. Alpha2uxx=ut
and
u(x,t)=X(x)T(t)=(C1coskx+C2sinkx)e-K2alpha2t+C3+C4x
No, it isn't. There are two space variables, x and y. The equation is [tex]\alpha^2(u_{xx}+ u_{yy})= u_t[/tex].
(Corrected thanks to LCKurtz.)

The Attempt at a Solution



So the problem I'm having is previously everything was done on a rod. Now I have a plate. I know that for a rod, my boundary conditions would be the temperature at the ends, for example, if this were a rod I would say that u(0,t)=0 and u(2,t)=50sin(piy). BUT! How can I say that if I have corners and a second dimension to deal with?
 
Last edited by a moderator:
  • #3
Halls has a typo there, he means$$
\alpha^2(u_{xx} + u_{yy})=u_t$$
in case it isn't obvious.
 
  • #4
Okay, so I'm going to expand the formula from alpha2uxx to alpha2(uxx+uyy)=ut

Then, u(x,y,t)=(X(x)+Y(y))T(t)

and

X'"X+Y"/Y=T'/alpha2T=λ and λ=0

so, X(x)=ax+b, Y(y)=cy+d and T(t)=e

Then u(x,y,t)=c(ax+cy+b+d)

right?
 
  • #5
LCKurtz said:
Halls has a typo there, he means$$
\alpha^2(u_{xx} + u_{yy})=u_t$$
in case it isn't obvious.

samee said:
Okay, so I'm going to expand the formula from alpha2uxx to alpha2(uxx+uyy)=ut

Then, u(x,y,t)=(X(x)+Y(y))T(t)

and

X'"X+Y"/Y=T'/alpha2T=λ and λ=0

Assuming that X''' is a typo meaning X'', please show us how you got that equation with that substitution for u(x,y,t). Or are you just guessing?
 
  • #6
I was just guessing. Since u(x,t)=X(x)T(t), I assumed that U(x,y,t)=[X(x)+Y(y)]T(t) because x and y are vectors and T isn't. And yes, I meant X"/X, not X'''/X
 
  • #7
samee said:
I was just guessing. Since u(x,t)=X(x)T(t), I assumed that U(x,y,t)=[X(x)+Y(y)]T(t) because x and y are vectors and T isn't. And yes, I meant X"/X, not X'''/X

X, Y, and T are not vectors, they are ordinary functions. And it is one thing to guess what the answer might be but surely if you are doing mathematics you would check your guess to see if you got lucky. If you try it, you will not be able to verify your guess.

I will give you a hint: Try u(x,y,t) = X(x)Y(y)T(t). No guessing from here. Plug it in and see what you get. Work it out.
 
  • #8
I'm confused because this was supposed to be solved with separation of variables. I don't know how to do that with 3 variables.
 
  • #9
alpha2(X"/X+Y"/Y)=T'/T

X"/X+Y"/Y=T'/(Talpha2)=λ=0

If I were just doing X"/X=T'/(Talpha2), I would use separation of variables method to solve and get that X(x)=ax+b and T(t)=c

Since this is not just X"/X, but X"/X+Y"/Y, I don't think I know how to solve the problem.
 
  • #10
samee said:
alpha2(X"/X+Y"/Y)=T'/T

X"/X+Y"/Y=T'/(Talpha2)=λ=0

If I were just doing X"/X=T'/(Talpha2), I would use separation of variables method to solve and get that X(x)=ax+b and T(t)=c

Since this is not just X"/X, but X"/X+Y"/Y, I don't think I know how to solve the problem.

There is no reason to assume ##\lambda=0##. Since each fraction uses a different variable you can assume each fraction is constant. So you might start with$$
\frac{X''} X = \lambda,\ \frac{Y''} Y = \mu,\ \frac{T''}{\alpha^2 T}=\lambda +\mu$$
Start with the ##Y## equation. Apply u=XYT to the boundary conditions to see what the boundary conditions for Y must be. Then solve the Y eigenvalue problem just like you did for a 1D rod. After that, do the X equation, followed by the T equation.

I have to leave for the rest of the afternoon but that should get you started.
 
  • #11
Okay, I'm looking more carefully through my book, and I found a problem with a temperature difference on a thin disk. They simplify the problem and explain, "since u does not vary with θ, we dropped the uθθ term and solved the reduced equation. If you have doubts about this step, observe that equation 79 does satisfy the boundary conditions."

So can I drop the uyy term since the plate have a temperature gradient along x and not y? That would make the whole problem much simpler.
 
  • #12
May I suggest, and I know LCKurtz is doing a fine job helping you, that you find "Basic Partial Differential Equations" by Bleecker and Csordas in the Library and just work through the example? And the reason I suggest this is because you seem to be having trouble with just the concepts, for example separation of variables and it's always nice to first just work through an example in the book, or two or three, then attempt to work on a new one yourself.
 
  • #13
samee said:
... Determine the steady-state temperature distribution in the plate...

Steady-state means [itex]u_t=0[/itex] so the equation to solve is [itex]u_{xx}+u_{yy}=0[/itex] and [itex]u=u(x,y)[/itex]
 

What is the diffusion equation on a plate in 2 dimensions?

The diffusion equation on a plate in 2 dimensions is a mathematical model used to describe the process of diffusion, where particles move from an area of high concentration to an area of low concentration. It is commonly used in fields such as physics, chemistry, and engineering to study the diffusion of heat, mass, and other quantities on a flat surface.

What are the variables and parameters in the diffusion equation on a plate in 2 dimensions?

The diffusion equation on a plate in 2 dimensions includes variables such as time, position, and the concentration of the diffusing particles. It also includes parameters such as the diffusion coefficient, which describes how quickly particles diffuse, and the boundary conditions, which specify the behavior of the diffusing particles at the edges of the plate.

How is the diffusion equation on a plate in 2 dimensions solved?

The diffusion equation on a plate in 2 dimensions can be solved using various numerical methods, such as the finite difference method or the finite element method. These methods involve discretizing the plate into a grid and solving the equation at each grid point using iterative techniques. Analytical solutions can also be obtained for simple cases with specific boundary conditions.

What are some real-world applications of the diffusion equation on a plate in 2 dimensions?

The diffusion equation on a plate in 2 dimensions has many practical applications, such as in the design of heat exchangers, chemical reactors, and diffusion-based separation processes. It is also used in the study of atmospheric and oceanic processes, as well as in medical imaging techniques such as MRI.

How does the diffusion equation on a plate in 2 dimensions differ from the 1-dimensional diffusion equation?

The main difference between the 1-dimensional and 2-dimensional diffusion equations lies in the number of dimensions that are considered. The 1-dimensional equation only accounts for diffusion along a single axis, while the 2-dimensional equation takes into account diffusion in both the x and y directions. This makes the 2-dimensional equation more versatile and applicable to a wider range of problems.

Similar threads

Replies
0
Views
456
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
12
Views
987
Replies
3
Views
1K
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
467
  • Calculus and Beyond Homework Help
Replies
2
Views
185
  • Differential Equations
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top