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.


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


1