# Conduction Heat Transfer

This problem is important in the process industries, but we do need to make the distinction between thick and thin walled pipes. In general thin walled pipes can be considered by the previous analysis – but assuming that the pipe wall is effectively unwrapped so that it looks like a flat plate, with the process fluid on one side and the ambient condition on the other. The situation for thick pipes is, however, more complex. [pic]

The figure shown above represents the condition in a thick walled pipe. The area for heat flow is proportional to the radius – as may be seen, the area at the outside wall of the pipe is much greater than the middle. As a result the temperature gradient is inversely proportional to the radius.

The heat flow ‘per unit length of pipe’ at any radius r, is

[pic]

cf. [pic]

Note: Area,[pic]

Note there is no length of pipe (l) in this equation as we choose to deal with loss per unit length of pipe instead – later we shall introduce the length again, to calculate the total heat loss.

Integration the above equation between r1 and r2 gives

[pic]

or

[pic]

Which, if we define rm as a logarithmic mean radius then

[pic]

[pic]

In Coulson and Richardson Vol 1 it is said that for thin walled pipes it is sufficient to use the arithmetic mean radius ra giving:

[pic](Used for thin cylinder)

Compare this equation with the previous statement about treating a thin walled pipe like a flat plate. Hint, consider a 1m length of a 1m id inside diameter pipe of 0.01m thickness. Use inside diameter as a flat plate

[pic]

[pic]

Conduction through the wall of a sphere

All arguments applied to the thin/thick walled pipe apply to this case, except we can go straight to the thin wall case because most spheres, for example those for storing liquid natural gas (LNG) have huge diameters relative the wall thickness. Note, one could also go straight away...

Please join StudyMode to read the full document