heat diffusion

[logistic_guy]

here is the question

Derive the solution of the two dimensional steady-state heat diffusion equation using the method of separation of variables with the boundary conditions in figure 1.16. Use this solution to find the temperature at the midpoint \(\displaystyle T(1,0.5)\) with the boundary temperatures in figure 1.17. Show that the temperature at the midpoint is \(\displaystyle 94.5^{\circ}C\) by considering the first five nonzero terms of the infinite series that must be evaluated.

Hint: Use the transformation \(\displaystyle \theta \equiv \frac{T - T_1}{T_2 - T_1}\) and solve \(\displaystyle \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} = 0.\)

figure 1.16


figure 1.17



my attemb

this is the solution i must get



i think they want me to use the first part

\(\displaystyle \theta(x,y) = \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1} + 1}{n}\)

which look like infinite series of calculus

 

[mario99]

Read this thread, it will help you a lot to derive the solution.

Laplace PDE that will Blow your Mind: partial^2-u/partial-x^2 + partial^2-u/partial-y^2 = 0,...

\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 0 < x < 1, \ \ \ \ \ 0 < y < 2 u(x,0) = x, \ \ \ \ \ u(x,2) = x^2 u(0,y) = y, \ \ \ \ \ u(1,y) = y^2 This is a Laplace partial differential equation. I don't know how to solve this problem. I looked for similar...

www.freemathhelp.com

 

[logistic_guy]

Read this thread, it will help you a lot to derive the solution.

Laplace PDE that will Blow your Mind: partial^2-u/partial-x^2 + partial^2-u/partial-y^2 = 0,...

\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 0 < x < 1, \ \ \ \ \ 0 < y < 2 u(x,0) = x, \ \ \ \ \ u(x,2) = x^2 u(0,y) = y, \ \ \ \ \ u(1,y) = y^2 This is a Laplace partial differential equation. I don't know how to solve this problem. I looked for similar...

www.freemathhelp.com

thank

it look different and it don't explain any steps

 

[mario99]

thank

it look different and it don't explain any steps

Did you check the links given by professor Khan?

 

[khansaheb]

Derive the solution of the two dimensional steady-state heat diffusion equation

Heat is "transported" by (i) conduction, (ii) convection and (3) radiation. What does "heat diffusion" include?

 

[logistic_guy]

Did you check the links given by professor Khan?

no

Heat is "transported" by (i) conduction, (ii) convection and (3) radiation. What does "heat diffusion" include?

i don't understad heat diffusion but i think the question is about the two dimensional steady state conduction

 

[mario99]

no

Open the link in post #3. Example 2 is exactly the same as the partial differential equation in your OP. Just follow the steps. If you got stuck anywhere, tell us.

 

[logistic_guy]

Open the link in post #3. Example 2 is exactly the same as the partial differential equation in your OP. Just follow the steps. If you got stuck anywhere, tell us.

i open link, example 2 is very different

 

[mario99]

i open link, example 2 is very different

It is the same, it just uses different variables. I will show you.

This is your partial differential equation:

[imath]\displaystyle \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} = 0[/imath]

[imath]\displaystyle \theta(0,y) = 0[/imath]
[imath]\displaystyle \theta(L,y) = 0[/imath]
[imath]\displaystyle \theta(x,0) = 0[/imath]
[imath]\displaystyle \theta(x,W) = 1[/imath]

This is the partial differential equation in Example 2:

[imath]\displaystyle \frac{\partial^2 u_3}{\partial x^2} + \frac{\partial^2 u_3}{\partial y^2} = 0[/imath]

[imath]\displaystyle u_3(0,y) = 0[/imath]
[imath]\displaystyle u_3(L,y) = 0[/imath]
[imath]\displaystyle u_3(x,0) = 0[/imath]
[imath]\displaystyle u_3(x,H) = f_2(x)[/imath]

Compare between them and you will notice only this difference:

[imath]\displaystyle u_3 = \theta[/imath]
[imath]\displaystyle H = W[/imath]
[imath]\displaystyle f_2(x) = 1[/imath]

Read the solution of Example 2, and anywhere you see [imath]\displaystyle u_3, H, \ \text{or} \ f_2(x),[/imath] just replace it with [imath]\displaystyle \theta, W, \ \text{or} \ 1,[/imath] respectively.

Note: I took the boundary conditions of your partial differential equation from figure 1.16.

 

[logistic_guy]

It is the same, it just uses different variables. I will show you.

This is your partial differential equation:

[imath]\displaystyle \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} = 0[/imath]

[imath]\displaystyle \theta(0,y) = 0[/imath]
[imath]\displaystyle \theta(L,y) = 0[/imath]
[imath]\displaystyle \theta(x,0) = 0[/imath]
[imath]\displaystyle \theta(x,W) = 1[/imath]

This is the partial differential equation in Example 2:

[imath]\displaystyle \frac{\partial^2 u_3}{\partial x^2} + \frac{\partial^2 u_3}{\partial y^2} = 0[/imath]

[imath]\displaystyle u_3(0,y) = 0[/imath]
[imath]\displaystyle u_3(L,y) = 0[/imath]
[imath]\displaystyle u_3(x,0) = 0[/imath]
[imath]\displaystyle u_3(x,H) = f_2(x)[/imath]

Compare between them and you will notice only this difference:

[imath]\displaystyle u_3 = \theta[/imath]
[imath]\displaystyle H = W[/imath]
[imath]\displaystyle f_2(x) = 1[/imath]

Read the solution of Example 2, and anywhere you see [imath]\displaystyle u_3, H, \ \text{or} \ f_2(x),[/imath] just replace it with [imath]\displaystyle \theta, W, \ \text{or} \ 1,[/imath] respectively.

Note: I took the boundary conditions of your partial differential equation from figure 1.16.

give me some time to read this

 

[logistic_guy]

i'm back

you don't tell me about this variable

\(\displaystyle u_3(x,y) = h(x)\varphi(y)\)

what is \(\displaystyle h(x)\)?
what is \(\displaystyle \varphi(x)\)?

 

[khansaheb]

what is φ(x)\displaystyle \varphi(x)φ(x)?

There is no Φ(x) - it is Φ(y)

This is separable variable method of solving PDE.

If you are not familiar with it - do a google search with "separable variable method of solving PDE" as key words.

Please let us know what you find

 

[logistic_guy]

There is no Φ(x) - it is Φ(y)

This is separable variable method of solving PDE.

If you are not familiar with it - do a google search with "separable variable method of solving PDE" as key words.

Please let us know what you find

i don't understand

i'm using this site

Differential Equations - Laplace's Equation

In this section we discuss solving Laplace’s equation. As we will see this is exactly the equation we would need to solve if we were looking to find the equilibrium solution (i.e. time independent) for the two dimensional heat equation with no sources. We will also convert Laplace’s equation...

tutorial.math.lamar.edu

 

[mario99]

i'm back

you don't tell me about this variable

\(\displaystyle u_3(x,y) = h(x)\varphi(y)\)

what is \(\displaystyle h(x)\)?
what is \(\displaystyle \varphi(x)\)?

This is the main step in any PDE when solved by the method of separation of variables. You can choose any two variables you want, not necessarily [imath]h(x) \ \text{and} \ \varphi(y)[/imath].

If I was the one who solved that PDE in that website, I would choose these two variables:

[imath]u_3(x,y) = X(x)Y(y)[/imath]

I advise you to do the same, so start with this:

[imath]\theta(x,y) = X(x)Y(y)[/imath]

 

[logistic_guy]

This is the main step in any PDE when solved by the method of separation of variables. You can choose any two variables you want, not necessarily [imath]h(x) \ \text{and} \ \varphi(y)[/imath].

If I was the one who solved that PDE in that website, I would choose these two variables:

[imath]u_3(x,y) = X(x)Y(y)[/imath]

I advise you to do the same, so start with this:

[imath]\theta(x,y) = X(x)Y(y)[/imath]

thank

i reach this step

\(\displaystyle \theta_n(x,y) = B_n \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

the next step i don't understand

 

[mario99]

thank

i reach this step

\(\displaystyle \theta_n(x,y) = B_n \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

the next step i don't understand

This step:

[imath]\displaystyle \theta_n(x,y) = B_n \sinh\left(\frac{n\pi y}{L}\right)\sin\left(\frac{n\pi x}{L}\right), \ \ \ \ \ n = 1, 2, 3, \cdots[/imath]

tells you that you have infinite eigenfunctions. Each one of them is a solution.

And this step:

[imath]\displaystyle \theta(x,y) = \sum_{n = 1}^{\infty} B_n \sinh\left(\frac{n\pi y}{L}\right)\sin\left(\frac{n\pi x}{L}\right)[/imath]

tells you that the summation of all eigenfunctions is also a solution which is called the Superposition Principle.

The next step that comes after the Superposition Principle is very important. It will help you to find the constant [imath]B_n[/imath] by using the nonhomogeneous boundary condition, [imath]\theta(x,W) = 1[/imath].

Continue.

 

[logistic_guy]

\(\displaystyle B_n = \frac{2}{L\sinh(\frac{n\pi W}{H})}\int_{0}^{L}\sin(\frac{n\pi x}{L}) dx \ \ \ n = 1,2,3,...\)

it say this is \(\displaystyle B_n\) but i don't understand why

 

[mario99]

\(\displaystyle B_n = \frac{2}{L\sinh(\frac{n\pi W}{H})}\int_{0}^{L}\sin(\frac{n\pi x}{L}) dx \ \ \ n = 1,2,3,...\)

it say this is \(\displaystyle B_n\) but i don't understand why

For now, you don't need to know why [imath]B_n[/imath] is equal to that integral. Let your goal be just to arrive to the same solution in post #1. Later, we would fill the gaps that you did not understand about the solution.

Your solution still does not look like the given one. And your task now is to solve this integral:

[imath]\displaystyle \int_{0}^{L}\sin\left(\frac{n\pi x}{L}\right) \ dx = \ ?[/imath]

 

[logistic_guy]

For now, you don't need to know why [imath]B_n[/imath] is equal to that integral. Let your goal be just to arrive to the same solution in post #1. Later, we would fill the gaps that you did not understand about the solution.

Your solution still does not look like the given one. And your task now is to solve this integral:

[imath]\displaystyle \int_{0}^{L}\sin\left(\frac{n\pi x}{L}\right) \ dx = \ ?[/imath]

this easy

\(\displaystyle \int_{0}^{L}\sin(\frac{n\pi x}{L}) dx = \cos(\frac{n\pi x}{L}) + c\)

i'm not sure if i should change the constant \(\displaystyle c\) to \(\displaystyle L\). i think it make more sense to use the limit of integration, so the answer is \(\displaystyle \cos(\frac{n\pi x}{L}) + L\)

\(\displaystyle B_n = \frac{2}{L\sinh(\frac{n\pi W}{H})}\cos(\frac{n\pi x}{L}) + L \ \ \ n = 1,2,3,...\)

i don't know how to simplify it more

 

[mario99]

this easy

\(\displaystyle \int_{0}^{L}\sin(\frac{n\pi x}{L}) dx = \cos(\frac{n\pi x}{L}) + c\)

i'm not sure if i should change the constant \(\displaystyle c\) to \(\displaystyle L\). i think it make more sense to use the limit of integration, so the answer is \(\displaystyle \cos(\frac{n\pi x}{L}) + L\)

\(\displaystyle B_n = \frac{2}{L\sinh(\frac{n\pi W}{H})}\cos(\frac{n\pi x}{L}) + L \ \ \ n = 1,2,3,...\)

i don't know how to simplify it more

no comment



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[imath]\displaystyle \int_{0}^{L}\sin\left(\frac{n\pi x}{L}\right) \ dx = \frac{L(1 - \cos n\pi)}{n\pi}[/imath]

Now think about it. What happens if [imath]n[/imath] is odd and what happens if [imath]n[/imath] is even?