# Homework Help: Heat equation problem

1. Apr 17, 2014

### joshmccraney

1. The problem statement, all variables and given/known data
A nuclear fuel of thickness $2L$ has a steel slab to the left and right, each slab of thickness $b$. Heat generates within the rod at a rate $\dot{q}$ and is removed by a fluid at $T_{\infty}$ (the question doesn't say, but I believe $T_{\infty}$ is temperature of the fluid), which is to the right of the rightmost slab of steel and is convecting by a coefficient $h$. The other surface (to the left of the leftmost steel slab) is well insulated, and the fuel and steel have thermal conductivities of $k_f$ and $k_s$, respectively.

Obtain an equation for the temperature distribution $T(x)$ in the nuclear fuel. Express answers in the above variables.

2. Relevant equations
the heat equation: $$\frac{\partial T}{\partial t} = \alpha \frac{\partial ^2 T}{\partial x^2} + \dot{q}$$

3. The attempt at a solution
nothing is said about time, thus we may assume $\frac{\partial T}{\partial t} = 0$. Additionally, we are only transferring heat in 1-dimension.

Before trying to solve anything, i believe i will have to use the heat equation for each of the 3 materials (the leftmost steel slab $T_l$, the nuclear fuel in the middle $T_m$, and the rightmost slab of steel $T_r$).

First, i'll try prescribing boundary conditions, assuming the center of the fuel is $x=0$, the end of the far right steel slab is $x=L+b$ and the end of the far left steel slab is $x=-L-b$.
$$T_l '(-L-b) = 0$$
I'm just not sure how to deal with the other boundary conditions? My professor said something about, at $x=L+b$ we have that $-k_s T_r'(x) = hA(T_r-T_{\infty})$ where I think $A=b$ since we are in 1-dimension. Can someone explain this relation?

Also, what about $x=L,-L$? Can't we prescribe some kind of flux similarity, such as $T_l'(-L) = T_m'(-L)$ and $T_m'(L) = T_r'(L)$?

Similarly, can't we also say $T_l(-L) = T_m(-L)$ and $T_m(L) = T_r(L)$?

Please help! I've been working very hard and think I just need a little push in the correct direction.

Thanks everyone!

2. Apr 17, 2014

### AlephZero

OK so far!

If you have a steady temperature distribution in the system, the heat flux out of the slab must be the same as the internal heat generation. You correctly said there is no heat lost at the $x=-L-b$ boundary.

$hA(T_r-T_{\infty})$ looks like Newton's law of cooling, and it is the heat flux out of the steel at $x = L+b$, where $A$ is the area of the free surface. The question seems to imply this is a "large" slab of material and you can ignore the edge effects, so you might as well consider a piece of the slab with unit area.

You are right to think about the heat flux either side of the boundary between the fuel and the steel, but the flux depends on the temperature gradient and the thermal conductivity of the two materials.

There is no heat generation in the steel plates. What does that tell you about the heat flux through the steel, and the "shape" of the temperature distribution through the steel, as a function of $x$?

When you have got a clear idea what happens in the steel plates (and you don't really need differential equations to solve that part of the question) you will have the boundary conditions for the fuel.

You will need to write a differential equation for the temperature in the fuel, and solve it.

Note: I'm not sure what $\dot q$ means in the question. It could be the heat generated per unit volume of the fuel, or the total heat generated in the slab. It doesn't make much difference to the thought process in the question either way.

Last edited: Apr 17, 2014
3. Apr 17, 2014

### Staff: Mentor

joshmccraney,

You did a nice job of analyzing the problem, and AlephZero correctly pointed out that you needed to include the thermal conductivities in matching the heat fluxes at the interfaces.

Regarding the equation $-k_s T_r'(x) = hA(T_r-T_{\infty})$ :

There shouldn't be an A on the right hand side of this equation. The units on both sides of the equation have to match, and they don't with the A present.

Chet

4. Apr 19, 2014

### joshmccraney

I'm not too sure what to do here. are you referring to fourier's law of heat transfer $\vec{q} = -k \nabla T$, or in our 1-dimensional case, $q = -k \frac{dT}{dx}$? If so, it seems that at each boundary of $\pm L$ we have that $-k_f \frac{dT_m}{dx} = -k_s \frac{dT_i}{dx}$ for the $i^{th}$ steel slab?

ahh, you're being clever, if i understand you correct. it would seem as though $\frac{dT_l}{dx}=\frac{dT_m}{dx}=0$ when $x=-L$ since we are in steady state. thus, no heat will transfer out, so $T$ is constant along $[-L-b,-L]$. is this correct? As for the shape, it seems the temperature will decrease from left to right. do you agree?

so, if the above reasoning is correct, our boundary conditions are: $$T'(-L)=0$$ $$k_f T_m'(L) = k_s T_r'(L)$$ $$-k_s T_r'(L+b) = h(T_r - T_{\infty})$$ but dont we need a B.C. for the temperature profile (not only derivatives)?

the equation we will use is first $$-T_r''(x) = \dot{q}$$ and next is $$-T_m''(x) = \dot{q}$$ and it seems we actually don't need the $T_l$ profile (if the above logic is correct)?

5. Apr 19, 2014

### joshmccraney

good call! of course, what are the units associate with $h$? i know $A$ has m^2 and $T$ has kelvins. i'm just confused why area is present at all? is it because the rate of cooling is proportional to surface area and the temperature gradient, which in this case is ambient temp minus surface temp? thus, $hA(T_r-T_{\infty})$ takes units Watts per square meter?

thanks! please let me know if this is correct.

6. Apr 19, 2014

### joshmccraney

thinking about this question more (after alephzero's and chet's comments) do you two think the heat is being generated throughout the entire fluid? if so, isn't temperature constant from $[-L-b,L]$ since the generation is taking place continuously, constantly, and everywhere through $[-L,L]$?

7. Apr 19, 2014

### Staff: Mentor

You are correct to say that the temperature from $[-L-b,L]$ is constant, because no heat is flowing through this region at steady state. This is going to be at the highest temperature in your system. So you might as well take the insulated boundary to be at -L; it will make no difference in the calculated temperature profiles to the right of -L.

Chet

8. Apr 22, 2014

### joshmccraney

thanks for the help guys! i think i have a solution, but can one of you comment on this brief analysis: $\dot{q} + k_f T''(x) = 0$ implies $\dot{q}$ takes units Watts per cubic meter. However, for my energy balance I have that $\dot{q} = \frac{k_s}{b}A(T_m-T_r)$. how do i get rid of the area term? my professor said we have $\dot{q}2L = \frac{k_s}{b}(T_m-T_r)$ but i don't see how these units work out. how can we divide by area and multiply the left side by length?

9. Apr 22, 2014

### joshmccraney

nevermind, i see that the q term is my flux. thanks to you both!!!!