Max Temp of Water Around Heated Square Rod

[skrat]

Homework Statement

A long squared rod with sides $l$, thermal conductivity $\lambda$, specific heat $c_p$ and density $\rho$ is cooled with water with constant temperature. Inside our squared rod we have a long square heater with sides $l/2$ and is completely centered inside the rod. Heat sources per unit volume in this is equal to $q [W/m^3]$. (When calculating the temperature, you can use only the biggest three terms of the sum)
What is the maximum possible temperature of the surrounding water , before the heater starts to melt down? Melting temperature is $T_c$.

Homework Equations

The Attempt at a Solution

I am really not sure this is correct, but at least it is something:
So we have two squares, one with sides $l$ and the other one centered in the first one with sides $l/2$. I have put the origin of my coordinate system in the left bottom corner of the bigger square. This may not be a really good idea but there is a good reason why I chose this option.

And the reason is, that if the origin is there, than the function that solves $\nabla ^2T(x,y)=0$ is $$T(x,y)=\sum_{n,m}A_{n,m}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)$$ Of course the equation I am solving is not $\nabla ^2T=0$ instead I have to consider the heater! Therefore $\nabla ^2 T(x,y)=-\frac{q}{\rho c_p D}$.

Meaning I should write $\frac{q}{\rho c_p D}$ in terms of this sum $T(x,y)=\sum_{n,m}C_{n,m}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)$, so let's do that: $$\frac{q}{\rho c_p D}\int_{l/4}^{3l/4}\int _{l/4}^{3l/4}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)dxdy=C_{n,m}(\frac l 2)^2$$ Which brings me to $$C=\frac{8q}{\rho c_p D\pi ^2}\frac{1}{mn}$$ but only for (both) odd $m$ and $n$.
So finally $$\frac{q}{\rho c_p D}=\sum_{m,n}\frac{8q}{\rho c_p D\pi ^2}\frac{1}{mn}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)$$

Now I can start solving $\nabla ^2 T(x,y)=-\frac{q}{\rho c_p D}$ $$\sum _{m,n}A_{m,n}\frac{\pi ^2}{l^2}(m^2+n^2)sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)=-\sum_{m,n}\frac{8q}{\rho c_p D\pi ^2}\frac{1}{mn}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)$$ Which brings me to
$$A_{m,n}=\frac{8qL^2}{\rho c_pD\pi^4}\frac{1}{mn(m^2+n^2)}$$ Therefore the solution should be $$ T(x,y)=\sum_{\text{odd} m,n}\frac{8qL^2}{\rho c_pD\pi^4}\frac{1}{mn(m^2+n^2)}sin(\frac{n\pi}{l}x)sin(\frac{n\pi}{l}y)$$

Now to calculate the temperature I think I should do this, but I am really not so sure about this part: $$P=\lambda S\frac{\Delta T}{L/4}$$ where $P=\int qdV$ so $$\int qdV=\lambda S\frac{T_c-T_x}{L/4}$$ where I used notation $T_x$ for the temperature that I hope the problem is asking after...

Hmm. What do you think?

 

[BvU]

This is not a contribution, just some questions:
I imagine the rod is solid and the inner rod, same material properties, is the heat source. I would expect from symmetry that it's enough to consider one eigth, e.g. (with the origin at the axis of the rod) from x-axis to the line y=x. x-axis points to the middle of a side, y=x to an edge.
Is that the right idea ?

Heat is generated from x=0 to x=l/4 and the boundary conditions are: T gradient $\perp$ the lines mentioned and T = T_water at x =l/2 .
Sound reasonable ?

 

[skrat]

This is not a contribution, just some questions:
I imagine the rod is solid and the inner rod, same material properties, is the heat source. I would expect from symmetry that it's enough to consider one eigth, e.g. (with the origin at the axis of the rod) from x-axis to the line y=x. x-axis points to the middle of a side, y=x to an edge.
Is that the right idea ?

I can't see any problems with it. By the way, that is an interesting idea.

Heat is generated from x=0 to x=l/4 and the boundary conditions are: T gradient $\perp$ the lines mentioned and T = T_water at x =l/2 .
Sound reasonable ?

Sounds very reasonable! Hopefully I will have some time left to check if the results match with the original post.