- #1

member 428835

## Homework Statement

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.

## Homework Equations

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

## 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!