# Heat EQ

1. Jan 7, 2015

### joshmccraney

hi pf!

i'm wondering if you can help me with the heat eq for a basic cylinder wire problem. namely, we have a wire with radius $r_i$ and length $L$and resistance is $R$ and current is $I$. Thus heat produced $$Q = R I^2 \pi r_i^2 L$$. When using the heat eq, we assume time rate of change is negligable. flux is governed by fouriers law, and the divergence theorem gives us the following: $$\int_V k \nabla^2 T dv + \int_V \frac{Q}{\pi r^2 L}dv = 0$$.is this right though? namely, is $Q$ divided by an arbitrary $r$ or the radius $r_i$?

thanks so much!

2. Jan 8, 2015

### Bystander

Care to tell us how you came up with this expression?

@Chestermiller , @Orodruin

Last edited: Jan 8, 2015
3. Jan 8, 2015

### joshmccraney

I knew units for heat generation in 3-D need to be watts per cubic meter. So I simply tracked units. This Q multiplied by dv gives us watts, which is the unit we're after.

4. Jan 8, 2015

### bigfooted

But what are the units of $k\nabla^2T$ (k should be the thermal diffusivity)?

[edit: I mean k=thermal conductivity of course]

Last edited: Jan 8, 2015
5. Jan 8, 2015

### joshmccraney

Watts per cubic meter, right?

6. Jan 8, 2015

### bigfooted

and of the term $\frac{Q}{\pi r^2 L} = I^2 R \frac{\pi r_i^2 L}{\pi r^2 L}$ (with $I^2R$ the electric power)?

7. Jan 8, 2015

### joshmccraney

watts? am i missing something here? seems like you are eluding to something.

8. Jan 8, 2015

### joshmccraney

oh shoooot! i should have defined $Q : Q = I^2 R / \pi r_i ^2 L$ right? but is it $r_i$ or $r$?

Last edited: Jan 8, 2015
9. Jan 8, 2015

### Staff: Mentor

ri. The rate of heat generation per unit volume in the wire is constant. ri should also be what appears in the equation with the integrals.

Chet

10. Jan 9, 2015

### joshmccraney

Hi chet!

ok, so what i should have modeled from the start is $$\int_v k \nabla ^2T dv = \int_v Q dv : Q = I^2 R / \pi r_i^2 L$$ do you all agree? if so, solving would be (using 1-D radial flow in polar coordinates) $$-k\frac{1}{r}\frac{d}{dr} ( r T') = Q \implies \\ -k d(r T') = rQdr \implies \\ \int_?^{??} -k d(r T') = \int_0^{r_i}rQdr$$
but what are my bounds for integration? any ideas?

Last edited: Jan 9, 2015
11. Jan 9, 2015

### Staff: Mentor

Q is a constant, so it comes out of the integral. You integrate both sides from 0 to ri.

Chet

12. Jan 9, 2015

### joshmccraney

Are you sure? I'm thinking if we had a differential equation over some interval of time $[0,T]$, say, $$\frac{dy}{dt} = k \implies \\ \int_0^T \frac{dy}{dt} dt = \int_0^T k dt \implies \\ \int_{y(0)}^{y(T)} dy = \int_0^T k dt$$ but notice we do not have $[0,T]$ on both sides.

13. Jan 9, 2015

### Staff: Mentor

OK. 0 to rT' evaluated at ri.

Chet

14. Jan 9, 2015

### joshmccraney

But in this case we wouldn't have a function of $r$. Would we instead just integrate from $0,r$ generally so we can still have a profile rather than a number?

15. Jan 9, 2015

### Staff: Mentor

That's fine, but you seemed to be applying the equation over the entire volume. Integrating out to R is just fine.

Chet

16. Jan 9, 2015

### joshmccraney

So if I'm understanding this correctly we would have $$-k\int_{0*T'(0)}^{r*T'(r)} d(r T') = Q\int_0^r r dr \implies \\ -k r T'(r) = Qr^2/2 \implies \\ -kT'(r) = Qr/2 \implies \\ -kT'(r) = \frac{R I^2}{2 \pi r_i^2 L r}$$ is this right so far?

Last edited: Jan 9, 2015
17. Jan 9, 2015

### joshmccraney

But then if I made a flux balance we could write $q = R I^2 / (2 \pi r L)$ watts/sq. meter. Fourier's law implies also $q = -k T'(r)$ (1-D radial flow). Thus, $$-kT'(r) = \frac{R I^2} { 2 \pi r L}$$ which doesn't agree with the above. Can you help me with what I'm doing wrong?

18. Jan 9, 2015

### Staff: Mentor

No. Check your algebra.

19. Jan 9, 2015

### joshmccraney

Sorry, we would have $$-kT'(r) = \frac{RI^2 r}{2 \pi r_i^2 L}$$ right? But this still doesn't agree with the flux balance.

20. Jan 9, 2015

### Staff: Mentor

Who says? Multiply both sides by 2πrL and see what you get.

Chet

21. Jan 9, 2015

### joshmccraney

I must be missing something. The flux balance states $-kT'(r) = R I^2 / (2 \pi r L)$ yet the heat eq method states $-kT'(r) = R I^2 r / (2 \pi r_i^2 L)$. These two are different. I must have made a mistake but I'm not seeing it.

Thanks so much for your help (and please continue)!

22. Jan 9, 2015

### joshmccraney

If the $r_i^2$ was simply $r^2$ then we would have agreeing equations.

23. Jan 9, 2015

### Staff: Mentor

The first equation is correct only at r = ri. The second equation is correct at all radial locations.

24. Jan 9, 2015

### joshmccraney

Can you elaborate here. I'm wondering what I have done wrong in the flux balance. It really looked right to me.

25. Jan 9, 2015

### Staff: Mentor

If the rate of heat generation within the wire is spatially uniform, what fraction of the heat is generated between r = 0 and arbitrary radial position r? What is the rate of heat generation within the wire per unit volume? What is the rate of heat generation between r = 0 and arbitrary radial position r?

Chet

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