Heat transfer: Temperature distribution inside a sphere submerged in a fluid

[happyparticle]

Homework Statement

What is the temperature distribution inside a sphere and fluid

Relevant Equations

$$\kappa \nabla^2 T = 0$$

$$T(r,\theta) = \sum_{m=0}^{\infty} = (A_m r^m + \frac{B_m}{r^{m+1}}) P_m(cos \theta)$$

Consider a sphere of thermal diffusivity $\kappa_2$ is submerged in an incompressible and stationary fluid of thermal diffusivity $\kappa_1$.

The fluid is held between 2 large plates ( at T_0 for the top plate and T_1 for the bottom plate).

What is the stationary temperature distribution inside the sphere and in the fluid ?

Since $\frac{\partial}{\partial t} = 0$ and we have symmetry in $\phi$ coordinates, the heat conduction equation becomes:


$\kappa \nabla^2 T = 0$

With the following solution in spherical coordinates :

$$T(r,\theta) = \sum_{m=0}^{\infty} = (A_m r^m + \frac{B_m}{r^{m+1}}) P_m(cos \theta)$$.

Also, since $T \rightarrow \infty$ at $r \rightarrow 0$

B must be 0.

This is as far as I can go. I'm not sure what are the boundary conditions to find the value of A and m.

 

[Chestermiller]

Is the temperature distribution axisymmetric (a function of r only)? Please provide a diagram of the situation. From your description, I am unable to visualize the system.

 

[pasmith]

You need one solution inside the sphere and a separate solution outside the sphere. Since $T$ is specified on each of the plates, you must start by finding the outer solution. Having done so, the inner solution is found from the requirement that $\kappa\frac{\partial T}{\partial r}$ must be continuous on the surface of the sphere.

For the outer solution, if the plates are at $z = r \cos \theta = \pm h$, then we have the conditions $$\begin{split} \sum_{m=0}^\infty B_m \frac{1}{r^{m+1}}P_m\left(\frac{h}{r}\right) &= T_0 \\ \sum_{m=0}^\infty B_m \frac{1}{r^{m+1}}P_m\left(-\frac{h}{r}\right) &= T_1.\end{split}$$ Exploiting the parity of the $P_m$, we can add and subtract these to get one condition for even $m$
and one condition for odd $m$: $$\begin{split} \sum_{m=0}^\infty \frac{B_{2m}}{r^{2m+1}} P_{2m}\left( \frac hr \right) = \frac{T_0 + T_1}{2} \\ \sum_{m=0}^\infty \frac{B_{2m+1}}{r^{2m+2}} P_{2m+1}\left( \frac hr \right) = \frac{T_0 - T_1}{2} \end{split}$$ The left hand sides are power series in $r^{-1}$, and by comparing coefficients of powers of $r^{-1}$ it should be possible to determine the $B_m$.

EDIT: I' not sure if it's $T$ or $\kappa \frac{\partial T}{\partial r}$ which is continuous across the surface of the sphere. Also it may be necessary to include $P_0$ and $rP_1$ in the outer solution, since far from the sphere the temperature should approximate $C + Dr\cos \theta$.

 

[PhDeezNutz]

In order to relate $A$ to $B$ and solve the system we really need to know about the boundary conditions across the sphere. Both in terms of $T$ and $\frac{\partial T}{\partial R}$. I have no domain experience for this type of thermodynamics problem BUT it is very similar to electrostatics (Laplace's equation shows up everywhere doesn't it?).

I'm going to assume $T$ is continuous but $\frac{\partial T}{\partial r}$ is not continuous and it's related to $\kappa$ of each medium across the boundary.

I don't know what that relationship is but if we did this problem would be a lot easier to solve.

 

[happyparticle]

Thank you for the answers. Sorry, I was really busy lately. I took the last couple days to get back on this problem.
@pasmith I'm not sure how to find m and $B_m$, should I find them before finding T(r) inside the sphere ?

As far as I know both $T$ and $\kappa \frac{\partial T}{\partial r}$ are continuous at $r = R$.

I made a diagram, it might help.

 

[pasmith]

Yes, you need to find the outer solution first. Setting $T_1 = T_0 + \Delta T$, we should have as $r \to \infty$ that $T$ tends to $T_0 + \frac{\Delta T}2 - \frac{\Delta T}{2h} r \cos \theta$; for that reason I would write $T = T_0 + \frac{\Delta T}2 - \frac{\Delta T}{2h} r \cos \theta + u$ where $u = 0$ on $r \cos \theta = \pm h$. We can then seek a solution for $u$ which is even in $\cos \theta$, so that $$u(r, \theta) = \begin{cases} \sum_{n=0}^\infty A_n r^{2n} P_{2n}(\cos \theta) & r < R \\ \sum_{n=0}^\infty B_n r^{-2n-1} P_{2n}(\cos \theta) & r > R \end{cases}$$ where the condition or $r = h\sec \theta$ is $$\sum_{n=0}^\infty \frac{B_{n}}{h^{2n+1}} (\cos^{2n+1} \theta)P_{2n}(\cos \theta) = 0.$$ ($B_n \equiv 0$ is a solution.) I do not think that you can require continuity of both $T$ and $\kappa \partial T/\partial r$ on the sphere: you have a second-order operator, and you are already imposing the condition that $T$ is finite at the origin.

 

[PhDeezNutz]

If I insist that both $T$ and $\frac{\partial T}{\partial R}$ are continuous at $r=R$ then I get contradicting statements.

With the continuity of $T$ I get

$B_m = A_m R^{2m+1}$

with the continuity of $\frac{\partial T}{\partial R}$ I get

$B_m = - \frac{m}{m+1} A_m R^{2m+1}$

There has to be (in my opinion) a discontinuity condition for the derivative.

hmm...I see @pasmith beat me to it.

Edit: If I had to guess from my exposure to E&M and dielectrics

$\kappa_2 \frac{\partial T}{\partial r} = \kappa_1 \frac{\partial T}{\partial r}$ at $r=R$

Edit 2: What I just said doesn't contradict OP at all never mind.

 

[Orodruin]

There is a discontinuity condition for the derivative. The continuity statement is for the current, which is the derivative multiplied by $\kappa_i$. Since the $\kappa_i$ are different, you obtain a discontinuity in the derivative.

you have a second-order operator, and you are already imposing the condition that T is finite at the origin.

The required boundary conditions are coordinate independent and there is nothing special about the origin in Cartesian coordinates. Instead, the function behaving well at the origin is not so much a boundary comdition as it is a requirement for the Laplace equation to hold. This fact is skimmed over and not really covered in many texts.

As for the continuity actoss the surface, compare to the one-dimensional case where you wish to solve $\kappa(x) f’’(x) = 0$ with $\kappa(x)$ being a step function with the step at zero*. The solution is linear on both sides and any boundary condition away from $x=0$ may fix only a single constant out of two for the solution on the corresponding side. In order to fix the two remaining constants you will need two boundary conditions at the interface: one on the temperature and one on the current. (In the 1D case, the current is not only continuous, but constant)

* With current being constant and proportional to $\kappa(x) f’(x)$, the DE actually comes from requiring the current to have zero divergence. The DE is therefore actually $\kappa’(x) f’(x) + \kappa(x) f’’(x) = 0$. The $\kappa’(x)$ term is a delta distribution at $x=0$.

 

[Chestermiller]

The thermal diffusivities are not the correct parameters to use in this steady state problem. Thermal diffusivities only come into play in transient problems. You should be using thermal conductivities in your analysis.

I would approach this problem differently. To a first approximation, the temperature in both regions should approach a linear gradient in z. So the temperature can be written as the linear temperature profile plus a perturbation. In the perturbation problem, the boundary conditions would be zero at the two boundaries, and a specified normal heat flux at the surface of the sphere (the latter being determined from the first approximation linear temperature profile in the two regions).

 

[happyparticle]

Isn't the inner solution $T(r,\theta) = \sum_{m=0}^{\infty} (C_m r^m P_m(cos \theta)$, since $D_m = 0$ to prevent $\lim_{T \to \infty}, \lim_{r \to 0}$ ? $C_m$ can be found by studying the conditions at the boundary between the fluid and the sphere. Knowing that I have 2 equations for the boundary conditions, I should have 2 coefficients. However, despite your help, I'm struggling with the outer solution.

Yet, my starting point is $T(r,\theta) = \sum_{m=0}^{\infty} (A_m r^m + \frac{B_m}{r^{m+1}}) P_m(cos \theta)$. In both post 3 and 6, one coefficient is missing and I don't know why. Also, I have no idea what value to choose for m. I'm wondering if m = 0. Since if m = 0, the right hand side will not go to infinity.

 

[happyparticle]

I forgot something...

Since $T \neq \infty$ when $r \to \infty$ and $r \to 0$

I have $T(r,\theta) = \sum_{m=0}^{\infty} ( \frac{B_m}{r^{m+1}}) P_m(cos \theta)$, $r > R$

$T(r,\theta) = \sum_{m=0}^{\infty} (C_m r^m) P_m(cos \theta)$, $r < R$

Then, I can find the coefficients using the fact that $T$ and $\frac{\partial T}{\partial R}$ are continuous at $r = R$.

I only need to find the value of $m$.

 

[Orodruin]

Since $T \neq \infty$ when $r \to \infty$ and $r \to 0$

I have $T(r,\theta) = \sum_{m=0}^{\infty} ( \frac{B_m}{r^{m+1}}) P_m(cos \theta)$, $r > R$

Not quite. You cannot allow arbitrary $r$ while not constraining $\theta$. Therefore, as $r\to \infty$ you must also have $\cos\theta \to 0$. This will allow more terms.

I only need to find the value of m.

No. The summation variable $m$ is not something you ”find”. You need to determine the constants for all $m$. There will not be a single $m$ value.

 

[happyparticle]

No. The summation variable $m$ is not something you ”find”. You need to determine the constants for all $m$. There will not be a single $m$ value.

I'm really don't know how to solve this problem. For instance, I had a problem where I had to find the velocity around a cylinder. I did find a solution for the Laplace's equation with a specific value for m by analyzing the conditions far from the cylinder. Here is an example, this is pretty much the approach I'm trying to use for this problem.

Here, I really have no idea what to do.

Edit:
Also, why $\cos\theta \to 0$ when $r \to \infty$. r is independent of $\theta$. Otherwise, how @pasmith found his solution in post #3.

 

[Orodruin]

I'm really don't know how to solve this problem. For instance, I had a problem where I had to find the velocity around a cylinder. I did find a solution for the Laplace's equation with a specific value for m by analyzing the conditions far from the cylinder. Here is an example, this is pretty much the approach I'm trying to use for this problem.

The boundary conditions may be such that only one $m$ contributes and you may be lucky that this is the case - for example when the boundary condition itself is expressable as a single eigenfunction as in the example.

Edit:
Also, why $\cos\theta \to 0$ when $r \to \infty$. r is independent of $\theta$. Otherwise, how @pasmith found his solution in post #3.

Because of the boundaries of your region. Your region is given by $|z| < h$, ie $r|\cos\theta| < h$ which can be reshuffled to $|\cos\theta| < h/r$. You can make $r$ as large as you want, but that means the angle $\theta$ is restricted. The region where the cosine is larger is simply not in your solution’s domain.