Radial heat flow in a cylinder

**dwsmith** · February 1st 2011, 11:11 AM

A case of heat flow which is virtually one-dimensional arises when the conducting medium is a circular cylinder and the temperature function u depends only on the time t and the distance r from the axis of the cylinder, $u=(r,t)$ . For example, imagine a cylindrical pipe filled with a hot fluid and suppose that one wishes to study the loss of heat through the sides of the pipe.

Let c, p, k, K denote the thermal constants of the cylinder. By considering the heat energy contained in a section of pipe of length H and lying between the radii r = a, r = b (Why are the radii different? Shouldn't they be the same?) show that

$\displaystyle 2\pi H\int_a^b\left[c\cdot p\cdot r\frac{\partial u}{\partial t}-\frac{\partial}{\partial r}\left(k\cdot r\frac{\partial u}{\partial r}\right)\right]dr=\text{source term}$

Hence, obtain the equation for the source-free radial heat flow in a cylinder:

$\displaystyle\frac{\partial u}{\partial r}=\frac{K}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)=K\left[u_{rr}+\frac{1}{r}u_r\right]$ .

How do I start obtaining the equation?

**dwsmith** · February 1st 2011, 12:48 PM

Source Term:

$\displaystyle\int_a^b\left[c\cdot p\frac{\partial u}{\partial t}(x,t)-\frac{\partial}{\partial x}\left(k\frac{\partial u}{\partial x}(x,t)\right)\right]Adx$

$A=2\pi Hr$

$\displaystyle 2\pi Hr\int_a^b\left[c\cdot p\frac{\partial u}{\partial t}(r,t)-\frac{\partial}{\partial r}\left(k\frac{\partial u}{\partial r}(r,t)\right)\right]dr$

$\displaystyle 2\pi H\int_a^b\left[c\cdot p\cdot r\frac{\partial u}{\partial t}(r,t)-r\cdot\frac{\partial}{\partial r}\left(k\frac{\partial u}{\partial r}(r,t)\right)\right]dr$

$\displaystyle 2\pi H\int_a^b\left[c\cdot p\cdot r\frac{\partial u}{\partial t}(r,t)-\frac{\partial}{\partial r}\left(k\cdot r\frac{\partial u}{\partial r}(r,t)\right)\right]dr$

Correct?

It seems logical.

**dwsmith** · February 1st 2011, 04:38 PM

For the source-free radial heat flow, I believe this what needs to be done, but I don't know how to manipulate it into the correct format.

$\displaystyle\left[c\cdot p\cdot r\frac{\partial u}{\partial t}-\frac{\partial}{\partial r}\left(k\cdot r\frac{\partial u}{\partial r}\right)\right]=0\Rightarrow c\cdot p\cdot r\frac{\partial u}{\partial t}=\frac{\partial}{\partial r}\left(k\cdot r\frac{\partial u}{\partial r}\right)$

**dwsmith** · February 2nd 2011, 02:30 PM

Here is the finished product in case anyone wants it.

$\displaystyle u_t=\frac{k}{c\cdot r\cdot p}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$

$\displaystyle K=\frac{k}{c\cdot p}$

$\displaystyle u_t=\frac{K}{r}\left(1\cdot u_r+ru_{rr}\right)\Rightarrow u_t=K\left(\frac{1}{r}u_r+u_{rr}\right)$

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