Find the rate of heat flow through a cross-section of the rod shown in figure (θ2> θ1). Thermal conductivity of the material of the rod is K.
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Let’s redraw the diagram

From the above diagram we can say that ΔABE is similar to ΔACD.
By the property of similar triangles-
= ![]()
⇒ x = r1 + (r2 – r1 ) ![]()
Lets assume-
a = ![]()
⇒ r = ax + r1 (1)
Thermal resistance is given by –
dR = ![]()
Now area ![]()
dR = ![]()
dR = ![]()
= ![]()
Solving above integral
R =![]()
⇒ R = ![]()
–
Rate of heat flow =![]()
q =
Kπr1r2