Solve Heat Conduction for 3D Rectangular Solid

[danai_pa]

Given the 3-D rectangular solid with sides of length a, b and c in the x, y and z direction respectively.
Find T(x,y,z) in the interior of the solid when laplace T = 0
Boundary condition are following conditions:
1) x=0, T=0
2) x=a, dT/dx=0
3) y=0, dT/dy=0
4) y=b, dT/dy=0
5) z=0, T=0
6) z=c, T=f(x,y)
please suggest me, How to solve it?

 

[Astronuc]

Well, start with the general solution of the Laplace equation, and apply the boundary conditions. This is a steady-state heat conduction problem in 3 dimensions.

 

[quicknote]

To solve this problem, we first need to understand the concept of heat conduction. Heat conduction is the transfer of heat energy through a material due to a temperature gradient. In this case, we are looking at a 3D rectangular solid, where heat is being transferred through the material in all three dimensions.

The equation we are given, laplace T = 0, is known as the Laplace equation and it describes the behavior of heat conduction in a steady state. This means that the temperature distribution within the solid does not change with time.

To solve this problem, we need to use the method of separation of variables. This method involves assuming a solution of the form T(x,y,z) = X(x)Y(y)Z(z) and then solving for each variable separately.

First, we will solve for the x-direction. Using the boundary conditions given, we can see that at x=0, T=0 and at x=a, dT/dx=0. This means that the temperature at the left and right boundaries of the solid are fixed at 0, and there is no heat flux at these boundaries. This leads to the solution X(x) = sin(nπx/a), where n is a positive integer.

Next, we will solve for the y-direction. Using the boundary conditions given, we can see that at y=0 and y=b, dT/dy=0. This means that the temperature at the top and bottom boundaries of the solid are also fixed at 0, and there is no heat flux at these boundaries. This leads to the solution Y(y) = sin(mπy/b), where m is a positive integer.

Finally, we will solve for the z-direction. Using the boundary conditions given, we can see that at z=0, T=0 and at z=c, T=f(x,y). This means that the temperature at the front and back boundaries of the solid are fixed at 0 and f(x,y), respectively. This leads to the solution Z(z) = sinh(λz) + C cosh(λz), where λ is a constant and C is a constant of integration.

Now, we can combine these solutions to get the final solution for T(x,y,z) = Σ(A_nm sin(nπx/a)sin(mπy/b)(sinh(λz) + C cosh(λz)), where A_nm is a constant to be determined.

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