Thin disc with high thermal conductivity

  • #1
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Homework Statement:
A very thin disc of radius R, made of a material with extremely high thermal conductivity is placed inside a large, homogeneous medium with thermal conductivity κ. In the plane of the disk, concentrically with it, there is a thin circular loop of radius R√2 made of conductive wire and connected to a battery. The heating power generated by this loop per unit length is uniform, equal P/(2πR√2) and does not change with time. The effects of the wires connecting the element to the battery can be neglected. Suppose the steady state is reached, which means that the temperature of each point in space does not change with time. Determine the difference ΔT between the temperature of the disk and the temperature at a point of the medium very far from the disk. Neglect thermal radiation.
Relevant Equations:
Fourier's heat equation, Laplace's equation
I've tried to explicitly solve the Fourier's equation in cylindrical coordinates but I'm getting some messy integrals which cannot be solved analytically. Additionally my instructor said that there's a neat trick for this problem and it's possible to obtain the answer in a rather elementary way.Could someone point out to me this neat trick?
 

Answers and Replies

  • #3
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On the surface of the disc heat flux density vector must be perpendicular to that surface, so I'd assume very near the disc that is also the case. There is a problem on the edge, obviously, but since the disc is very thin, I think we are supposed to neglect that.
 
  • #4
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Oh, by the heat flux density vector i mean vector $$\mathbf{Q}=-\nabla T$$.
 
  • #5
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I haven't figured out how to do this yet, but here are some thoughts:

If the disk were not present, the isotherms would be perpendicular to the disc, and the heat flow vector would be parallel to the disk.

The temperature of the disc is going to match the medium temperature at the edge of the disc. So, in the plane of the disk, heat is going to flow into the disk at its edge, and then out of the disk at its flat surfaces. So the disk is going to be giving off heat as if the entire disk were at its edge temperature.
 
  • #6
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Have you considered using the Green's function method for solving this problem? The Green's function for the heat equation satisfy the equation,
$$
\nabla^2 G-\frac{1}{\alpha}\frac{\partial G}{\partial t}=\frac{1}{r}\delta (r-\rho)\delta (z-\zeta)\delta (\theta - \phi)\delta (t-\tau)
$$
where
$$
\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{\partial z^2}
$$
and ##\alpha## is the diffusion constant. In this particular case you could try to find two Green's functions, one for the region of the disk and one for the region outside the disk. You would have to find the boundary conditions but once you find the Green's function for the region outside the disk you can evaluate the solution by integrating it against your forcing function. For a somewhat similar example see https://www.researchgate.net/public...onduction_problems_in_multi-layered_cylinders
 
  • #7
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Honestly I'm not very familiar with Green's functions. So far we have only studied the method of separation of variables and I don't think we are supposed to use such a heavy mathematical machinery like in the article you linked. As I mentioned, my instructor said that there's an elementary apporach which requiered only some basic integrals. I'd love to see that kind of solution (however any solution would be helpful) because I've been thinking about this problem for a few days and still don't how to solve it.
 
  • #8
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Has your professor provided you with solutions to some standard cases, like the loop without the disc or the disc at constant temperature without the loop?
 
  • #9
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There was another HW to prove that when the disc without the loop is kept at constant temperature ΔT relative to temperature at a point of the medium very far from the disc then the total amount of heat transfered per unit time through any closed surfuce surrounding the disc is $$P=8\kappa R\Delta T$$. However, I can't prove that either.
 
  • #10
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There was another HW to prove that when the disc without the loop is kept at constant temperature ΔT relative to temperature at a point of the medium very far from the disc then the total amount of heat transfered per unit time through any closed surfuce surrounding the disc is $$P=8\kappa R\Delta T$$. However, I can't prove that either.
Is it that you don't understand the solution for this case?

Regarding the pure loop without the disc, would you know how to solve this, say using superposition of point sources.
 
  • #11
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I don't know the solution. It was another homework.

Using superposition of point sources to find heat flux density vector generated by the loop itself at any point would be rather hard imo. Maybe for some specific points (e.g. points on the axis) it is feasible.
 
  • #12
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here is a guide for solving the heat equation on a disc using the method of separation of variables.
 
  • #13
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I don't think these problems are the same. Here the problem is 3D with azimuthal symmetry so that the Poisson's eq. in cylindrical coordinates (r,φ,z) for temperature function Θ that I'd need to solve is $$\frac{\partial^2 \Theta}{\partial r^2}+\frac{1}{r}\frac{\partial \Theta}{\partial r}+\frac{\partial^2\Theta}{\partial z^2}=-\frac{\rho(\mathbf{r})}{\kappa}$$, with the distribution of heat sources given by $$\rho(\mathbf{r})=\frac{P}{2\pi r}\delta\left(r-R\sqrt{2}\right)\delta(z)$$ and boundary conditions are: I) Θ is constant on the the disc, II) heat transferred per unit time through any closed surface containing the disc but not the loop is 0, III) partial derivative of Θ with respect to z is 0 at z=0. The paper that you send is rather for a case of infinitely long cylinder when Θ is the same for all planes which cut the cylinder perependicular to its axis and the problem is actually 2D even though it 'lives' in 3D space.
 
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  • #14
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However, as I said, I don't know how to solve this Poisson's eq. with such boundary conditions. I'm almost sure it's not what we are supposed to do here since the problem doesn't ask to explicitly find Θ(r,z) but 'just' to find the difference ΔT.
 
  • #15
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Here is the solution to the problem of a loop without a disk. For a point heat source at the origin of power P at the origin, the heat conduction equation in spherical coordinates reads: $$P=-4\pi r^2k\frac{dT}{dr}$$
The solution to this equation for the temperature distribution is:
$$T-T_{\infty}=\frac{P}{4\pi r k}$$
From this it follows that for a loop of total power P and radius ##R_0## situated in the x-y plane, the temperature distribution is $$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\theta}{\sqrt{(x-R_0\cos{\theta})^2+(y-R_0\sin{\theta})^2+z^2}}}$$where ##\theta## is the polar angle about the z axis, and x,y,z are the Cartesian coordinates of the point. Switching to polar coordinates with ##x=r\cos{\phi}## and ##y=r\sin{\phi}##, we obtain:$$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\theta}{\sqrt{R_0^2+r^2-2R_0r\cos{(\theta-\phi)}+z^2}}}$$or$$T-T_{\infty}
=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\alpha}{\sqrt{R_0^2+r^2-2R_0r\cos{\alpha}+z^2}}}$$In the x-y plane, this reduces to:$$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\alpha}{\sqrt{R_0^2+r^2-2R_0r\cos{\alpha}}}}$$In our problem, $$R_0=R\sqrt{2}$$
 
  • #16
Steve4Physics
Homework Helper
Gold Member
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No claim to mathematical rigour, but here's an approach mainly based on physical arguments...

From the question’s wording and the instructor's hint, I'd guess that we are meant to consider a distance sufficiently far from the disc, such that we can assume the flux and the temperature-distribution are effectively spherically symmetric.

Since the disc has an an ‘extremely high’ (and unspecified) thermal conductivity, also assume the disc is at a uniform temperature, ##T_D##. The position of the heating-loop inside the disc is not important.

The disc’s surface area is ##2\pi R^2## (two faces). In the context of the question we can replace the disc with a sphere of radius ##\frac {R} {\sqrt 2}## (so it's surface area is ##2\pi R^2##) and surface temperature ##T_D##. A leap of faith, but only a small one.

The system can now be approximated to one with complete spherical symmetry. This is then easily solved with Fourier’s law, requiring only separation of variables and simple integration.

I don’t really see how the system significantly differs from the one described in the question quoted in @Rlwe's Post# 9. Using the above approach for the Post #9 question, I get ##P= 2\pi \sqrt2 \kappa R\Delta T##. This is tantalisingly similar to ##P=8 \kappa R\Delta T## (the answer given in Post #9) which suggests the above approach not unreasonable, at least as an approximate method.
 
  • #17
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No claim to mathematical rigour, but here's an approach mainly based on physical arguments...

From the question’s wording and the instructor's hint, I'd guess that we are meant to consider a distance sufficiently far from the disc, such that we can assume the flux and the temperature-distribution are effectively spherically symmetric.

Since the disc has an an ‘extremely high’ (and unspecified) thermal conductivity, also assume the disc is at a uniform temperature, ##T_D##. The position of the heating-loop inside the disc is not important.

The disc’s surface area is ##2\pi R^2## (two faces). In the context of the question we can replace the disc with a sphere of radius ##\frac {R} {\sqrt 2}## (so it's surface area is ##2\pi R^2##) and surface temperature ##T_D##. A leap of faith, but only a small one.

The system can now be approximated to one with complete spherical symmetry. This is then easily solved with Fourier’s law, requiring only separation of variables and simple integration.

I don’t really see how the system significantly differs from the one described in the question quoted in @Rlwe's Post# 9. Using the above approach for the Post #9 question, I get ##P= 2\pi \sqrt2 \kappa R\Delta T##. This is tantalisingly similar to ##P=8 \kappa R\Delta T## (the answer given in Post #9) which suggests the above approach not unreasonable, at least as an approximate method.
The heating loop is outside the disc, not inside.
 
  • #18
Steve4Physics
Homework Helper
Gold Member
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Aha, I misread ##R\sqrt 2## as ##R/\sqrt 2##. (Should have gone to Specsavers.) That totally screw's up my argument for the size of the "equivalent" spherical heat source.

That's life. I was just hunting for an approach to "obtain the answer in a rather elementary way".
 
  • #19
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It's rather obvious from the dimensional analysis that the solution must have the form c*P/κR, where c is some numerical prefactor so I don't think we are supposed to make such simplifications. Extremely high thermal conductivity in my opinion just means that we should assume that for any sort of heat source the surface of the disc is an isothermal surface. 'Very far from the disc' just means a point where the influence of the considered system on the temperature at such point can be neglected (i.e. point at infinity). I don't think that you can replace the disc with a sphere since the problem specifically asks to find the temperature of the disc (some 'normalised' temperature Θ=T-T(∞)) and therefore changing the shape would definitely affect the answer. Additionaly there is no word 'approximate' in the problem's wording so I'd assume that there is a fully analytical answer.

Here is the solution to the problem of a loop without a disk. For a point heat source at the origin of power P at the origin, the heat conduction equation in spherical coordinates reads: $$P=-4\pi r^2k\frac{dT}{dr}$$
The solution to this equation for the temperature distribution is:
$$T-T_{\infty}=\frac{P}{4\pi r k}$$
From this it follows that for a loop of total power P and radius ##R_0## situated in the x-y plane, the temperature distribution is $$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\theta}{\sqrt{(x-R_0\cos{\theta})^2+(y-R_0\sin{\theta})^2+z^2}}}$$where ##\theta## is the polar angle about the z axis, and x,y,z are the Cartesian coordinates of the point. Switching to polar coordinates with ##x=r\cos{\phi}## and ##y=r\sin{\phi}##, we obtain:$$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\theta}{\sqrt{R_0^2+r^2-2R_0r\cos{(\theta-\phi)}+z^2}}}$$or$$T-T_{\infty}
=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\alpha}{\sqrt{R_0^2+r^2-2R_0r\cos{\alpha}+z^2}}}$$In the x-y plane, this reduces to:$$T-T_{\infty}=\frac{P}{8\pi^2k}\int_0^{2\pi}{\frac{d\alpha}{\sqrt{R_0^2+r^2-2R_0r\cos{\alpha}}}}$$In our problem, $$R_0=R\sqrt{2}$$

This is what I meant when writing that it would be rather hard for an arbitrary point since to my knowledge this integral cannot be represented in elementary terms. I also don't think that it does solve the problem since the temperature field in the original problem won't be simply a superposition of the solutions for loop without the disc and disc without the loop because then the boundary conditions would not be satisfied i.e. the surface of the disc would not be isothermal.
 
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  • #20
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It's rather obvious from the dimensional analysis that the solution must have the form c*P/κR, where c is some numerical prefactor so I don't think we are supposed to make such simplifications. Extremely high thermal conductivity in my opinion just means that we should assume that for any sort of heat source the surface of the disc is an isothermal surface. 'Very far from the disc' just means a point where the influence of the considered system on the temperature at such point can be neglected (i.e. point at infinity). I don't think that you can replace the disc with a sphere since the problem specifically asks to find the temperature of the disc (some 'normalised' temperature Θ=T-T(∞)) and therefore changing the shape would definitely affect the answer. Additionaly there is no word 'approximate' in the problem's wording so I'd assume that there is a fully analytical answer.



This is what I meant when writing that it would be rather hard for an arbitrary point since to my knowledge this integral cannot be represented in elementary terms. I also don't think that it does solve the problem since the temperature field in the original problem won't be simply a superposition of the solutions for loop without the disc and disc without the loop because then the boundary conditions would not be satisfied i.e. the surface of the disc would not be isothermal.
In my judgment, this is definitely going to be part of the solution. Just because you can't think of an analytic form for the integral doesn't mean that one doesn't exist. I would take this solution and see how the temperature distribution predicted for the case of loop without disc would vary over the region where the disc would be present. It probably wouldn't come out to be constant, but the average would probably be an excellent approximation to the actual answer. Plus, if we superimposed sources and sinks on the disk (i.e., added the negative of the loop variation relative to the average temperature at the surface of the disc), they would almost certainly cancel out at large distances from the disc.
 
  • #21
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Okay, so with a help of WolframAlpha this integral can be written as $$\Theta_\text{loop}(r)=\frac{P}{2\pi^2\kappa}\frac{1}{|R_0-r|}K\left\{\frac{-4rR_0}{(r-R_0)^2}\right\}\,,$$ where K(u) is the complete elliptic integral of the first kind. I was also able to prove that for a case of disc kept at constant temperature Θ_0 the temperature at any point in the plane of the disc is given by $$\Theta_\text{disc}(r)=\frac{\Theta_0}{\pi^2}\int_{0}^{2\pi}\int_0^R\frac{r'\,dr'\,d\phi}{\sqrt{R^2-r'^2}\sqrt{r^2+r'^2-2rr'\cos\phi}}\,,$$ where R is the radius of the disc. What should I do know in order to find the solution for the case loop+disc?
 
  • #22
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OK. my next step would be to evaluate the variation of this loop solution with respect to r, and plot it up vs r/R. I would also determine the average value of this, integrated over the area.
 
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  • #23
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Okay, so with a help of WolframAlpha this integral can be written as $$\Theta_\text{loop}(r)=\frac{P}{2\pi^2\kappa}\frac{1}{|R_0-r|}K\left\{\frac{-4rR_0}{(r-R_0)^2}\right\}\,,$$ where K(u) is the complete elliptic integral of the first kind.
At r = 0, this solution should reduce to $$\Theta_{loop}(0)=\frac{P}{4\pi \kappa \sqrt{2}R}$$
Does it?
 
  • #24
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Yes, it does. According to Wolfram K(0)=π/2.
 
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  • #25
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For the case of a loop only, I integrated the equations numerically to get the radial temperature distribution and the average temperature over the region where the disk would be present. I defined the dimensionless temperature as $$\frac{(T-T_{\infty})4\pi k R}{P}$$The minimum value of this dimensionless temperature was ##1/\sqrt{2}=0.707## at r = 0, and the maximum value was 0.916 at r = R. The average value over the region where the highly conductive disc would be present was 0.786. This is the value that the conductive disk would exhibit over its entire area.
 
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