# Heat equation for a cylinder in cylindrical coordinates

So I have a description of a Partial differential equation given here.

$\text{Consider a long pipe of inner radius a}=2.5cm \text{ and outer radius b}=3 cm, \text{made of copper for which the thermal conductivity is K=400 W/(mK).}\\ \text{The pipe carries water at a surface temperature of }20 ^{\circ} \text{and lies half buried} \\ \text{on the surface of the ground in the desert. The outside surface has a} \\ \text{temperature of } (50+10\cos(\theta))^{\circ}C \text{ while the inside surface has a temperature of } {20^{\circ}C}, \text{where } r,\theta,z \\ \text{are the cylindrical coordinates. Using separation of variables, solve for the steady state temperature} \\ {\text{distribution in the pipe and the heat power transferred to the water per metre length of pipe.} }$

I ended up reaching the point where you have:

$\frac{\partial }{\partial r}\left(kr \frac{\partial T}{\partial r} \right)+ \frac{1}{r} \frac{\partial }{\partial \theta}\left(k \frac{\partial T}{\partial \theta} \right)= 0$

There is no forcing term, the pipe is at steady state and it is a long pipe.

I am a little bit unsure how to proceed with separation of variables.

I assume a solution of the form $T(r,\theta)=R(r)\psi(\theta)$. If I plug it back in, I get:

$\frac{r}{R}\frac{\partial }{\partial r}\left(kr \frac{\partial T}{\partial r} \right)+ \frac{k}{\psi} \frac{\partial^2 \psi }{\partial \theta^2}= 0$

However, this can only be true for some separation constant $-\lambda$. So essentially, we have:

$\frac{k}{\psi}\frac{\partial^2 \psi}{\partial \theta^2}=-\lambda$ or $\frac{\partial^2 \psi}{\partial \theta^2} = \frac{-\psi \lambda}{k}$.

I'm stuck on where to go from here....

EDIT: So I ended with my final summation as:

$T(r,\theta)=\sum_{n=0}^{\infty} r^n \left(A_{n}\cos(n\theta)\right)+B_{n}\sin(n\theta))$

I'm still not really sure how to use the boundary conditions on this though. If I plug in $a$ , I'll just get $T(a,\theta)=20=\sum_{n=0}^{\infty} 20^n \left(A_{n}\cos(n\theta)\right)+B_{n}\sin(n\theta))$ . I'm not really sure how that's going to help me.

• Fine but the problem is not homogeneous though right? There's a forcing term added to it. In addition, the pipe is n steady state so shouldn't the right hand side vanish anyways? – Future Math person Feb 1 '18 at 7:25
• Well this is the most general form of the heat equation. I was just looking at which terms cancelled to simplify the equation slightly. I don't even know if I am approaching this correctly. – Future Math person Feb 1 '18 at 7:37
• If you were given this as an assignment, shouldn't you know how to solve it? As far as I see there should be no source term and no time derivative. So you can apply separation of variables and then look to satisfy boundary conditions – Yuriy S Feb 1 '18 at 7:40
• I made some progress. See my edit. I'm just stuck on what boundary conditions I should use now. – Future Math person Feb 1 '18 at 8:27

The temperature is a function of $r,\theta$. The boundary conditions are given for $r=a,b$, so:

$$T(a,\theta)=20 \\ T(b,\theta)=50+10 \cos \theta$$

The angle variable is called $\theta$, not $\phi$ in the problem statement, if case you were confused.

$K$ the thermal conductivity is constant, which means you should be able to write your steady state equation as Laplace equation:

$$\Delta T =\frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial T}{\partial r} \right)+ \frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} = 0$$

Turns out you don't actually need the value of thermal conductivity to find the temperature distribution, only to find the heat current later.

Now use separation of variables as intended and the boundary conditions stated above.

Since you appear to have a problem with applying separation of variables to the equation, here's how you do it:

$$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r} \frac{\partial T}{\partial r}+ \frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} = 0$$

Substitute $T=f(r)g(\theta)$ into the equation:

$$g \cdot f''+\frac{1}{r} g \cdot f'+ \frac{1}{r^2} f \cdot g'' = 0$$

Divide everything by $f \cdot g$ and multiply by $r^2$:

$$r^2 \frac{f''}{f}+r \frac{f'}{f}+\frac{g''}{g} = 0$$

Now proceed with the separation constant (the sign is chosen for convenience, considering the boundary conditions):

$$r^2 \frac{f''}{f}+r \frac{f'}{f} = \lambda^2$$

$$\frac{g''}{g} = -\lambda^2$$

The above are just ordinary differential equations, the first one gives Bessel functions, the second one trigonometric functions, I hope you know how to write the general solutions and apply the boundary conditions.

• I made some progress on my separation of variables in my edit. I'm a little unsure how I would deal with the different $\lambda$ cases here since there is a mix of both $\psi$ and $k$ instead of just $\lambda$. – Future Math person Feb 1 '18 at 9:31
• There's no mix. Look at my edit. You made a mistake when applying the separation of variables to the equation – Yuriy S Feb 1 '18 at 9:32
• Okay that helped a lot. So I ended with my final summation as: $T(r,\theta)=\sum_{n=0}^{\infty} r^n \left(A_{n}\cos(n\theta)\right)+B_{n}\sin(n\theta))$ I'm still not really sure how to use the boundary conditions on this though. If I plug in $a$ , I'll just get $T(a,\theta)=20=\sum_{n=0}^{\infty} 20^n \left(A_{n}\cos(n\theta)\right)+B_{n}\sin(n\theta))$ . I'm not really sure how that's going to help me. – Future Math person Feb 1 '18 at 10:02
• Look at your boundary conditions. Also, your solution for $r$ should be in terms of Bessel functions, not $r^n$ – Yuriy S Feb 1 '18 at 10:10
• We never covered Bessel functions in the course yet so I am assuming there is another way to solve this. – Future Math person Feb 1 '18 at 10:14

Apologies for this incredible breach of protocol... But if you just want the numbers and not the fancy equations...

you can use the logarithmic average of the tube's thickness - and then calculate it directly as if it was a regular prism with dimensions:

• surface = area of a tube, length l, @ logarithmic average
• thickness = logarithmic average

From this, the conductivity of Cu and the ${\Delta}$T you can get the power transferred per whatever unit you want (l) directly. When you know the power transmitted, you can calculate backwards again to find the distribution, assuming symmetry...

At least for future references, if someone wants to actually know the temperatures with the minimum amount of work and find this question...

• That's nice as far as practical advice goes, but I think this one was an assignment question – Yuriy S Feb 3 '18 at 13:07
• @Yuriy Yes, I think so too. I meant the answer for the someone else who comes here, searching for "heat conduction in a cylinder" having an actual problem, looking for the numbers. You seem to be a regular here; if you disagree, if you mean that is beyond the scope of this stack, I'll delete it np. – Stian Yttervik Feb 3 '18 at 16:29
• @ Stian Yttervik, I do not have anything against this answer staying up, I agree that it's useful in the general context – Yuriy S Feb 3 '18 at 17:08