Heat Equation for Cylinder Wire Problem

In summary, heat is generated in a wire with radius r_i and length L due to resistance R and current I according to the equation Q = R I^2 pi r_i^2 L. The heat equation, which assumes a negligible time rate of change, is derived using Fourier's law and the divergence theorem, resulting in the equation ∫V k ∇^2T dv + ∫V Q/(pi r^2 L) dv = 0. There is some confusion about the correct form of the equation, but ultimately it should be -kT'(r) = RI^2r/(2pi r_i^2 L), which agrees with the flux balance equation.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
But what are the units of [itex]k\nabla^2T[/itex] (k should be the thermal diffusivity)?

[edit: I mean k=thermal conductivity of course]
 
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  • #5
bigfooted said:
But what are the units of [itex]k\nabla^2T[/itex] (k should be the thermal diffusivity)?
Watts per cubic meter, right?
 
  • #6
and of the term [itex]\frac{Q}{\pi r^2 L} = I^2 R \frac{\pi r_i^2 L}{\pi r^2 L}[/itex] (with [itex]I^2R[/itex] the electric power)?
 
  • #7
bigfooted said:
and of the term [itex]\frac{Q}{\pi r^2 L} = I^2 R \frac{\pi r_i^2 L}{\pi r^2 L}[/itex] (with [itex]I^2R[/itex] the electric power)?
watts? am i missing something here? seems like you are eluding to something.
 
  • #8
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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  • #9
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!
 

1. What is the heat equation for the cylinder wire problem?

The heat equation for the cylinder wire problem is a partial differential equation that describes the distribution of heat over time in a cylindrical object. It takes into account factors such as the material properties, initial temperature, and boundary conditions of the object.

2. How is the heat equation solved for the cylinder wire problem?

The heat equation for the cylinder wire problem is typically solved using separation of variables, where the solution is expressed as a product of functions of the spatial and temporal variables. This allows for the use of boundary conditions to solve for the coefficients in the solution.

3. What are the assumptions made in the heat equation for the cylinder wire problem?

The heat equation for the cylinder wire problem assumes that the material is homogeneous, isotropic, and has a constant thermal conductivity. It also assumes that there are no internal heat sources and that the temperature distribution is one-dimensional along the length of the cylinder.

4. How does the heat equation for the cylinder wire problem differ from the heat equation for a flat plate?

The main difference between the heat equation for the cylinder wire problem and a flat plate is the geometry of the object. The cylinder wire problem involves a cylindrical object, while the flat plate problem involves a two-dimensional object. This difference affects the boundary conditions and the methods used to solve the equations.

5. What are some applications of the heat equation for the cylinder wire problem?

The heat equation for the cylinder wire problem has various applications in fields such as engineering, physics, and materials science. It can be used to analyze the cooling of a heated wire, the heating of a wire in an electric circuit, and the thermal response of a cylindrical object to a changing environment.

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