# CL: 1-D Conduction through a Pipe

Recall that we have proven to ourselves, at least conceptually (for now), that the heat flow may not vary spatially in a steady one dimensional conduction problem.

Let's consider a more difficult example...

A pipe with inside diameter (ID) of 2mm has an outside diameter (OD) of 2.5mm. If the inside wall of the pipe is held at Ti = 300K and the outside is kept at To = 350K. What is the rate of heat flow per length of pipe if the thermal conductivity of the pipe is k = 10 W/mK?

$\displaystyle{\frac{q}{A} = -k\frac{dT}{dr}}$

Our first difference between the slab and the cylinder is that we need to write out the correct area (because it depends on r!):

$\displaystyle{\frac{q}{2\pi rL} = -k\frac{dt}{dr}}$

The second issue is when we write the integrals...what can we pull out of the left hand side?!

$\displaystyle{ \frac{q}{2\pi L}\int_{R_i}^{R_o} \frac{dr}{r} = -k \int_{T_1}^{T_2}dT}$

Integration gives:

$\displaystyle{\frac{q}{2\pi L} \ln{\left(\frac{R_o}{R_i}\right )} = -k(T_2-T_1)}$

which can be rearranged to:

$\displaystyle{q = \frac{2\pi L k(T_1-T_2)}{ \ln{\left(\frac{R_o}{R_i}\right )}}}$

Writing this in terms of a resistor gives

$\displaystyle{q = \frac{\Delta T}{R}}$

where R is given as

$\displaystyle{R_{cyl} = \frac{ \ln{\left(\frac{R_o}{R_i}\right )}}{2\pi kL}}$

##### OUTCOME:

Calculate the thermal resistance and magnitude of conductive heat flow/flux through a cylindrical shell