# Heat Equation for Cylinder Wire Problem

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#### member 428835

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!

heat produced
Q=RI 2 πr 2 i L
Care to tell us how you came up with this expression?

@Chestermiller , @Orodruin

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

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

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

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But what are the units of $k\nabla^2T$ (k should be the thermal diffusivity)?
Watts per cubic meter, right?

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)?

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

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

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

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?

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

Chet

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.

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.
OK. 0 to rT' evaluated at ri.

Chet

OK. 0 to rT' evaluated at ri.

Chet
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?

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

Chet

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?

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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?

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?

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.

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

Chet

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

Chet
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.

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

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.

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

The first equation is correct only at r = ri. The second equation is correct at all radial locations.
Can you elaborate here. I'm wondering what I have done wrong in the flux balance. It really looked right to me.

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

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? Chet
##\pi r^2 L / (\pi r_i^2 L) = (r/r_i)^2##
What is the rate of heat generation within the wire per unit volume?
Chet
##R I^2 / (\pi r_i^2 L)##
What is the rate of heat generation between r = 0 and arbitrary radial position r?
Chet
$$\int_0^L \int_0^{2 \pi} \int_0^r \frac{R I^2}{ \pi r_i^2 L} (r dr d \theta d z) = R I^2 \left(\frac{r}{ r_i}\right)^2$$

Am I missing something though? How does this relate to flux (if we are doing the flux balance)?

Am I missing something though? How does this relate to flux (if we are doing the flux balance)?
OK. Now go back to that equation I indicated and multiply both sides by 2πrL. Show us what you get. Then see if you can interpret what each side of the equation represents physically.

Chet

The r.h.s. is (obviously) the heat generation at some arbitrary distance r, as you've already said. and i agree that the left hand side has the same units, but it's difficult for me to see this without the r.h.s (i understand that it is heat generation, but i don't think it's obvious that it's total heat generation from 0 to r).

but the relation is obvious now! thanks! although what did i do wrong in trying to make the flux balance with an arbitrary r? as you've said, it's only correct when ##r=r_i##.

The r.h.s. is (obviously) the heat generation at some arbitrary distance r, as you've already said. and i agree that the left hand side has the same units, but it's difficult for me to see this without the r.h.s (i understand that it is heat generation, but i don't think it's obvious that it's total heat generation from 0 to r).

but the relation is obvious now! thanks! although what did i do wrong in trying to make the flux balance with an arbitrary r? as you've said, it's only correct when ##r=r_i##.
The flux balance is wrong because it implicitly assumes that all the heat generation takes place between r = 0 and radial location r, and none of the heat is generated between r and ri.

Chet

The flux balance is wrong because it implicitly assumes that all the heat generation takes place between r = 0 and radial location r, and none of the heat is generated between r and ri.

Chet
ahh yes, this makes sense! so, if we were to look at the flux at some ##r > r_i## would we be able to use the flux argument?

ahh yes, this makes sense! so, if we were to look at the flux at some ##r > r_i## would we be able to use the flux argument?
r > r1 is outside the wire. We don't know what's happening out there, do we?

Chet

r > r1 is outside the wire. We don't know what's happening out there, do we?

Chet
sorry, I'm speaking in hypotheticals. and yea, if it was the same material but no heat generation.

sorry, I'm speaking in hypotheticals. and yea, if it was the same material but no heat generation.
Then it would be OK.

Chet

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Then it would be OK.

Chet
Thanks chet!