Hollow cylinder conduction and critical radius

(a) Hollow cylinder

For steady radial flow without generation the conduction equation reduces to

$$\frac{d}{dr}\left(r\frac{dT}{dr}\right)=0$$

Integrating twice: $T = C_1\ln r + C_2$. Boundary conditions: $T = T_1$ at $r = r_1$ and $T = T_2$ at $r = r_2$. Hence

$$C_1 = -\frac{T_1 - T_2}{\ln(r_2/r_1)},\qquad \frac{T - T_1}{T_2 - T_1} = \frac{\ln(r/r_1)}{\ln(r_2/r_1)}$$

The temperature varies logarithmically. Heat flow by Fourier's law with $A = 2\pi r L$:

$$Q = -k(2\pi rL)\frac{dT}{dr} = \frac{2\pi kL\,(T_1-T_2)}{\ln(r_2/r_1)}$$

The thermal resistance is $R = \dfrac{\ln(r_2/r_1)}{2\pi kL}$.

(b) Critical radius

When a cylinder is covered with insulation, its outer radius increases. The conduction resistance of the insulation rises, but the outer surface area grows, so the convection resistance falls. For an insulation outer radius $r_o$, with fluid at $T_\infty$ and coefficient $h_o$:

$$R_{total} = \frac{\ln(r_o/r_1)}{2\pi k L} + \frac{1}{h_o\,2\pi r_o L}$$

Heat loss is maximum where $R_{total}$ is minimum:

$$\frac{dR}{dr_o} = \frac{1}{2\pi kL\,r_o} - \frac{1}{2\pi h_o L\,r_o^2} = 0 \;\Rightarrow\; r_c = \frac{k}{h_o}$$

The second derivative is positive at $r_c$, so $R$ is minimum and $Q$ is maximum there.

Discussion

 Q ^        .---.
   |      /       \
   |    /           ` ----
   |  /
   +--------|--------------> r_o
           r_c

• If the bare pipe radius $r_1 < r_c$, adding insulation first increases heat loss until $r_o = r_c$, and further insulation reduces it. This applies to small wires and thin pipes (for example electric cables, where the extra heat loss is desirable).
• If $r_1 \ge r_c$, any insulation reduces the heat loss. This applies to steam pipes of normal size.
• For a sphere the critical radius is $r_c = 2k/h_o$.