Homework Help: Heat flux through a composite wall

1. Nov 8, 2015

plata_o_plomo

1. State the problem

A plane wall consisting of three different materials, each of constant thermal conductivity k. Assume steady-state temperature distribution.

a) Comment on relative magnitudes of q(dot)''2 and q(dot)''3 and of q(dot)''3 and q(dot)''4.

b) Comment on the relative magnitudes of kA and kB and of kB and kC.

c) Sketch the heat flux as a function of x.

2. Relevant equations

q'' = -k dT/dX
d2T/(dx)2 = 0 for A and B
d2T/(dx)2 = -Qgen/k for c

3. The attempt at a solution

a)

I will assume that q(dot)'' = q'' (not sure why they wrote q(dot)'', but in the graph it's just q''). Steady state and no heat generation in A and B means that q''2=q''3. In C, the parabola indicates that there's some heat generation occurring, thus q''3<q''4.

b)

Since q''2=q''3 ⇒ kA / kB = Δx12⋅ΔT23 / Δx23⋅ΔT12

kB / kC = ???

d)

Horizontal line through A and B somewhere below the x-axis. Through C I'm not certain. It should be 0 at T4 and the highest at T3.

Last edited: Nov 8, 2015
2. Nov 8, 2015

BvU

Heat flux is a flow of heat. Hence the dot in $\dot q$.
If there is a steady state, there is a constant heat flux (otherwise the temperature would not be constant in time).
Check in your textbook or here to understand the meaning of your relevant equation. The double quote is a derivative wrt x.
Do try to be consistent in your notation ! Better $\ {d^2T\over dx^2} \$ than $\ {d^2t\over dx^2} \$.
Your expression is correct, though !

which way is the heat flux ? So is it heat generation or does the wall give off heat ?

3. Nov 8, 2015

plata_o_plomo

Fixed the notation, thanks!

So, q'' = dq/dx? I thought it was dq/dt.

Heat flux is in the direction of the arrows. Not sure about your other question.

4. Nov 8, 2015

Staff: Mentor

Your conclusion about the flux line for A and B is correct. For C, it is a straight line, running from the A/B line at 3 to zero at 4.

Chet

5. Nov 8, 2015

plata_o_plomo

Thanks, Chet!

Does this mean that

a) Heat flux in C can never be equal to heat flux in A or B.
b) Heat flux in C is always greater than heat flux in B.

Furthermore, shouldn't heat flux be the same in all three materials according to conservation of energy?

6. Nov 9, 2015

plata_o_plomo

Solved. You may close the thread.

7. Nov 9, 2015

BvU

One of my colleagues has a signature "Don't tell us that you have found the solution; tell us the solution you have found" !

8. Nov 9, 2015

plata_o_plomo

Sorry, BvU. Didn't expect anyone to care about my solution.

My solution for a) and c) below.

The easy part: Temperature profile in A and B is linear, thus heat flux through A and B is the same.

The harder part: Heat flux through C?

According to the heat equation, for a steady state problem with internal heat generation:

(q'' should be q(dot)'')

$$\ {d^2T\over dx^2} \ = -{Q_g\over k}$$

where $$Q_g$$ is internal heat generation.

Fourier's law

$$\frac{-q''}{k} = \frac{dT}{dx}$$

thus if we use that in our heat equation

$$\frac{d}{dx}[q''(x)] = Q_g$$

Ergo, heat flux through C

$$q''(x) = Q_g\cdot x + c$$

This is a first degree polynomial. This confirms what Chet wrote earlier. If we picture the graph (horizontal line from 1-3 below the axis, and then a diagonal line to y=0 from 3-4), we see that the magnitude of heat flux in B is always gte than the magnitude of heat flux C.

9. Nov 9, 2015

BvU

Thanks for posting !
I think I get it. The parabola in C confused me (not an expert, just interested and curious). So this C is some kind of wall heating device and then the rest follows. Most of the heat generated should flow to the right if all is well.

10. Nov 9, 2015

Staff: Mentor

Actually, all the heat generated in C flows to the left. The very right boundary of C is insulated (zero heat flux).

Chet

11. Nov 9, 2015

BvU

hence the ${dT\over dx} = 0$, I take it ?

12. Nov 9, 2015

Staff: Mentor

Yes. You can see that in the figure.

Chet

13. Nov 9, 2015

plata_o_plomo

Exactly, BvU!

Here's a derivation of the heat diffusion equation: