Heat Equation for Cylinder Wire Problem

AI Thread Summary
The discussion focuses on deriving the heat equation for a cylindrical wire with given parameters such as radius, length, resistance, and current. The heat generated in the wire is expressed as Q = RI^2πr_i^2L, and participants debate the correct application of Fourier's law and the divergence theorem in modeling heat flow. There is confusion regarding whether to use the inner radius r_i or an arbitrary radius r in calculations, with clarification that r_i should be used for the rate of heat generation per unit volume. The conversation concludes with the realization that the flux balance must account for heat generation occurring within the entire volume of the wire, not just between the center and an arbitrary radius. Understanding these principles is crucial for accurately modeling thermal behavior in cylindrical conductors.
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!
 
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joshmccraney said:
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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Bystander said:
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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bigfooted said:
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)?
 
bigfooted said:
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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joshmccraney said:
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
 
  • #10
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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  • #11
Q is a constant, so it comes out of the integral. You integrate both sides from 0 to ri.

Chet
 
  • #12
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
joshmccraney said:
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
 
  • #14
Chestermiller said:
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?
 
  • #15
joshmccraney said:
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
 
  • #16
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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  • #17
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
joshmccraney said:
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?
No. Check your algebra.
 
  • #19
Chestermiller said:
No. Check your algebra.
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
joshmccraney said:
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
 
  • #21
Chestermiller said:
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.

Thanks so much for your help (and please continue)!
 
  • #22
If the ##r_i^2## was simply ##r^2## then we would have agreeing equations.
 
  • #23
joshmccraney said:
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)!
The first equation is correct only at r = ri. The second equation is correct at all radial locations.
 
  • #24
Chestermiller said:
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.
 
  • #25
joshmccraney said:
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
 
  • #26
Chestermiller said:
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##
Chestermiller said:
What is the rate of heat generation within the wire per unit volume?
Chet
##R I^2 / (\pi r_i^2 L)##
Chestermiller said:
What is the rate of heat generation between r = 0 and arbitrary radial position r?
Chet
$$\int_0^L \int_0^{2 \pi} \int_0^r \frac{R I^2}{ \pi r_i^2 L} (r dr d \theta d z) = R I^2 \left(\frac{r}{ r_i}\right)^2$$
 
  • #27
Am I missing something though? How does this relate to flux (if we are doing the flux balance)?
 
  • #28
joshmccraney said:
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
 
  • #29
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##.
 
  • #30
joshmccraney said:
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
 
  • #31
Chestermiller said:
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?
 
  • #32
joshmccraney said:
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
 
  • #33
Chestermiller said:
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.
 
  • #34
joshmccraney said:
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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  • #35
Chestermiller said:
Then it would be OK.

Chet
Thanks chet!
 
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