Counter Intuitive Result in Heat Transfer Analysis

Factual Questions
1

To anyone that is already familiar with heat transfer problems I’m hoping my question seems simple. It’s based off a hw problem from an introductory heat / momentum transfer class.

I’m working on a hypothetical system where heat is transferred through a plane wall. We approximate the direction of heat flux as if it were moving in only one dimension, in this case the x-direction. We know the thickness of the wall to be 2L, where it is decided for convenience to analyze the wall about position = 0, where the left side ranges from -L : 0 and the right side from 0 : L; the ambient temperature to the left of the wall is to be v (dimensions K), the convective heat flow coefficient is s (W/ m^2 - K); on the right side the environment temperature is given as w (K), while the convective heat coefficient is z (W/ m^2 - K).

The conductive coefficient of the wall is also given as k (W/m-K), which does not vary as a function of temperature or position. Within the wall there is also a certain amount of heat generation (incidentally due to electron flow through a wire), which we denote as q. q is the final variable of interest, which I’d like to represent as a function of a single temperature variable (besides q, all the values so far given are to be taken as constants).

Now the differential equation that seems appropriate to model this system is T’’ = -q/k, where T is a function of position. The environmental temperatures that are known are not directly linked to this equation. Given Initial condition specifications (T(-L) = x, T(L) = y) we find temperature as a function q, where variables x and y represent the temperatures at the surface of the wall (recall that these temperatures are not provided). These temperatures, x and y, may be linked to the environmental temperature by stipulating that the heat flux at -L and L are to be equal to the convective heat flux from / towards the wall. This reduces to

T ’ (- L) = (s/k)(x - v)
T ’ (+L) = (z/k)(y - w)

These equations may be related to the general solution of T’’ = -q/k by implementing the boundary conditions so far stipulated. The final system of equations obtained is thus:

(x - y)/(2L) = -(s/k)(x - v),
-(qL/k) + (x - y)/(2L) = -(z/k)*(y - w)

Where x and y are the variables of interest. Given these I may readily represent q as a function of temperature. However, I found solving for x and y to be too laborious by hand, so I asked Wolfram|Alpha to solve it for me. Wolfram readily provides a pair of solutions, but it appears that the it represents y and x as not being a function of w. This seems counter intuitive to me as this would imply that the temperature distribution of the system would be somehow independent of the environmental heat to the right of the wall (here represented as w). A link to the solution may be found here:

For the life of me I can’t see the issue in my approach. Are there any experienced dopers out there which feel generous enough to spare me some snarky wisdom?

2

I’m confused by this:
T(-L) = x, T(L) = y
Isn’t x your coordinate?

In any event, if I’m understanding the question correctly, you should see a symmetry in all of your equations between v and w, and between T(-L) and T(L) (whatever you call those last two). It doesn’t look like you have that symmetry in your input, so it’s not surprising that it doesn’t show up in the output.

At a guess, I’d say the problem is probably this:

If you’re somehow enforcing that the heat flux moves only left-to-right, then it’s not surprising that the solution is independent of what’s on the right.

3

I should have clarified this earlier, but the x-direction coordinate is never explicitly referred to in my equations (I write T’’ instead of T’’(x) ). This is because x was arbitrarily chosen to represent the temperature of the left surface of the plane wall, while implicitly representing the position coordinate for which I would measure Temperature. Selection of variable names were meant only to be consistent with my Wolfram|Alpha input (using subscripts tends to take too long or cause mistakes). So in this case x may be replaced with another other variable, I’ll say x → T_s_1, which represents the temperature on the left side of the wall. In a similar sense y → T_s_2, which is the right side.

But I actually would expect the system to lack symmetry. The temperatures on either side of the wall are different (v =/= w), as are the convective heat flow coefficients (s =/= z).

The temperature distribution is obtained from T’’(x) = -* q/k*; this result is assumed to apply in this situation; here q is the heat generated within the wall, which is an unknown. Using my new notation, my boundary conditions become T(-L) = T_s_1, T(L) = T_s_2 (i.e., the temperatures at either surface are set equal to the general solution of T(x) at the boundaries of the wall). Also, the temperatures of the environment, which is the temperature relevant to convective heat flows, were originally represented as v and w. So to be more explicit v → T_air_1, w → T_air_2. These represent the temperatures of the air on either side of the wall.

This results in the equation for temperature as a function of x ( x as in direction) and the surface temperatures on either side of the wall:
T(x) = [(qL^2)/(2k)](1 - (x/L)^2) + (T_s_2 - T_s_1)(x/2L) + (T_s_2 - T_s_1)(1/2)

Now as stated in the OP, I think the convective heat flows at these boundary conditions are to be set equal to the conductive heat flows evaluated at these boundaries. Fourier’s Law being (q_conduction)/A = -kT’(x).
Conduction heat flow at x = L: q_con/A = -kT’(L)
Convective heat flow at the boundary: q_convection = (h_1)*[(T_s_1) - (T_air_1)]
Solving for T ’ (L) and T ’ (-L) results in the set of equations I typed earlier (note the prime’s on T’(L)).

These equations are basically saying that the heat flow from convection on the left side must equal to the conduction of heat evaluated at the boundary L and - L. Each side of the wall has different conditions of temperature / convective heat coefficients. Heat generation is assumed to occur uniformly throughout the wall, and not depend on the position of the wall within the range of -L:L.

Sorry for all the confusion.

4

I didn’t say that the system was symmetric, just that the equations should be. That is, if you take any of your equations, and swap v with w, x with y, and s with z, then you should get one of your equations (possibly the same one) back again. The physics is symmetric, even if the particular situation isn’t.

5

I understand, thanks. I’m currently reworking the problem; the equations actually are symmetric, but the ones that I posted were not.