Determine heat flux from temperature profile

Question

The heating element and the insulator are of equal thickness L. Heat transfer in the air film adjacent to the heater is assumed negligible.

I've noticed that I find these type of problem the hardest to understand, so before I ask my question, does anyone know a good video or website which explains in detail how you should interpret temperature/concentration profiles?

My answer to (a) would be that both of the points should have the same heat flux. I know that one of the relationships used for layered wall is that q=constant, meaning that it is the same through every layer, however is that the correct way of thinking for this problem? I don't think so and don't understand how I should interpret the profile.

Another way I thought about it was that the heat flux for the layers should be:

$$q_1=(k_1/L)\cdot(T_\text{rear} - T_{12})$$
$$q_3=(k_3/L)\cdot(T_{23} - T_\text{front})$$

and that:

$$k_3>k_1$$

$$(T_\text{rear} - T_{12})\gt(T_{23} - T_\text{front})$$

If I just for simplicity assume that $L=1\ \mathrm m$. Then I would get that

$$q_1=(\text{something small})\cdot(\text{something big})$$
$$q_3=(\text{something big})\cdot(\text{something small})$$

Hence another reason why I think that the heat fluxes are the same.

Answer 1

Your first line of thinking to part a) is correct. For conductive heat transfer through rectangular layers, the heat flux needs to be constant and equivalent through all layers for steady state, or you would have a 'build up' of heat at between one layer. This would locally increase the temperature, and we wouldn't be at steady state.

For b) your second approach to a) is useful. You know the relative sizes of the temperature gradients from the sketch, and know from a) that the heat flux must be the same for each slab. Combine these two and you will find a requirement on the relative magnitudes of the conductivities.

Answer 2

Starting from the following equation (the one dimensional Fourier law) for heat flux $q$: $$q=-k\frac{\mathrm{d}T}{\mathrm{d}x}$$ where $k$ is the material's thermal conductivity, I obtained the following relations:

$$|q_{2\to1}|=q_{1}=k_1\frac{\Delta{T_1}}{L}$$

$$|q_{2\to3}|=q_{3}=k_3\frac{\Delta{T_3}}{L}$$

$$|q_{2\to1}|=k_2\frac{\Delta{T_{12}}}{L_{12}}$$

$$|q_{2\to3}|=k_2\frac{\Delta{T_{23}}}{L_{23}}$$

where $$\Delta{T_1} = T_{12}-T_{rear}$$ $$\Delta{T_3} = T_{23}-T_{front}$$ $$\Delta{T_{12}} = T_{max}-T_{12}$$ $$\Delta{T_{23}} = T_{max}-T_{23}$$

For instance, $q_{2\to1}$ is the heat flux within layer 2 in the direction of 1 (away from the maximum).

Now if I assume that $L_{23}=L_{12}$ (that the T maximum in layer 2 is positioned halfway in the layer), then $$\frac{|q_{2\to1}|}{|q_{2\to3}|}=\frac{\Delta{T_{12}}}{\Delta{T_{23}}}$$ Since $$\Delta{T_{23}}>\Delta{T_{12}}$$ I conclude that $$|q_{2\to3}|>|q_{2\to1}|$$ This answers (a). But the accuracy of this comes down to the assumption $L_{23}=L_{12}$ and I suspect there may be another way to solve this part.

The result of (a) can now be used, together with

$$\frac{|q_{2\to1}|}{|q_{2\to3}|}=\frac{|q_1|}{|q_3|}$$ and expressions above, to show that $$k_3\Delta{T_3}>k_1\Delta{T_1}$$ $$\frac{k_3}{k_1}>\frac{\Delta{T_1}}{\Delta{T_3}}$$ Now since $$1<\frac{\Delta{T_1}}{\Delta{T_3}}$$ we have that $$\frac{k_3}{k_1}>1$$ which answers part (b).