Question

A common problem involves the radial heat flow through a material between two concentric cylinders, say through the insulation between an inner pipe and its outer jacket. Consider an inner cylinder of radius r1 at temperature T1 and an outer cylinder of radius r2 at temperature T2. Show that the radial rate of heat flow per unit length, L, is given by:

(1/L) dQ/dt = 2p k(T1 – T2) / ln(r2 /r1).

Assume the thermal conductivity of the material, k, is constant. Note that unlike the case dealt with in your text, wherein the cross-sectional area between the points at which the temperature is measured remains constant, here it varies with the distance between the concentric cylinders. In the steady state, in either case, by conservation of energy, dQ/dt is the same for all cross-sections. However, as the cross-sectional area varies in the present case, dT/dr, is no longer the same for all cross-sections.

Answer 1

consider an inner cylinder of radius r1, temperature T1
outer cylinder radius r2, temperature T2
length of cylinder = L
hence
radial rat eof flow of heat = q'
at radius r, consuder thickness dr
r1 < r < r2
then

q' = k*2*pi*r*l*dT/dr
hence
dr/r = 2*pi*l*k*dT/q' ( where k is thermal conductivity of the material of the cylinder)
hence
integrating
ln(r2/r1) = 2*pi*l*k(T2 - T1)/q'
hence
rate of heat transfer per unit length
q'/L = 2*pi*k(T2 - T1)/ln(r2/r1)