A long steam pipe of outer radius 1 cm is surface-held at 120^circC. It is lagged by two coaxial cylindrical layers of different materials in perfect…

For steady radial conduction the heat current per unit length is the same through every cylindrical surface, $H=\dfrac{2\pi k\,\Delta T}{\ln(r_2/r_1)}$, so each layer acts as a thermal resistance $\mathcal{R}=\dfrac{\ln(r_2/r_1)}{2\pi k_{layer}}$ and the layers are in series. Inner layer: $\mathcal R_1=\dfrac{\ln 2}{2\pi(3k)}$. Outer layer: $\mathcal R_2=\dfrac{\ln(4/2)}{2\pi k}=\dfrac{\ln2}{2\pi k}$. The temperature drop is proportional to resistance: $\dfrac{T_i-20}{120-20}=\dfrac{\mathcal R_2}{\mathcal R_1+\mathcal R_2}=\dfrac{1}{1/3+1}=\dfrac34$. Therefore $T_i=20+75=95^\circ\mathrm{C}$. Using the plane-slab formula (resistance $\propto$ thickness$/k$) would instead give a wrong interface temperature, because the area grows with radius.