Question 5 of 8: Critical Radius of Pipe Insulation

Question 5: Critical Radius of Pipe Insulation (Part B, 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bare pipe diameter $d_i=1.27$ cm ($r_i=0.635$ cm), pipe surface uniformly at $T_i=82.5^{\circ}\text{C}$; insulation $k=0.156$ W/m·K; ambient $T_\infty=20^{\circ}\text{C}$, outer-surface $h=8.5$ W/m²·K.

Find. The insulation thickness(es) for which heat loss EXCEEDS the bare-pipe loss.

Approach. This is a small-diameter pipe, so it is worth checking the critical radius of insulation $r_{cr}=k/h$ before assuming insulation helps at all; heat loss per unit length as a function of outer radius $r_o$ is then solved against the bare-pipe baseline.

1. Bare-pipe heat loss (baseline). $$q'_{bare}=2\pi r_i h(T_i-T_\infty)=2\pi(0.00635)(8.5)(62.5)=\boxed{21.20\ \text{W/m}}.$$
2. Critical radius of insulation. $$r_{cr}=\frac{k}{h}=\frac{0.156}{8.5}=\boxed{1.835\ \text{cm}},$$ which is nearly THREE TIMES the bare pipe's own radius ($r_i=0.635$ cm). Because $r_{cr}>r_i$, adding insulation initially INCREASES total heat loss (the added convective surface area outpaces the added conduction resistance) — the classic small-pipe critical-radius effect. Heat loss per unit length as a function of outer radius, wall conduction neglected per the problem statement, is $$q'(r_o)=\frac{T_i-T_\infty}{\dfrac{\ln(r_o/r_i)}{2\pi k}+\dfrac{1}{2\pi r_o h}}.$$ At $r_o=r_{cr}$, $q'=29.72$ W/m — a $40\%$ INCREASE over the bare pipe, the worst case.
3. Crossover radius (root-find $q'(r_o)=q'_{bare}$, $r_o\neq r_i$). $q'(r_o)$ rises from $q'_{bare}$ at $r_o=r_i$, peaks at $r_{cr}$, then falls back through $q'_{bare}$ again at a larger radius. Solving numerically (bisection) for that second crossing, $$r^{*}=\boxed{9.40\ \text{cm}}\quad\Rightarrow\quad t^{*}=r^{*}-r_i=9.40-0.635=\boxed{8.77\ \text{cm}}.$$
4. Answer. ANY insulation thickness in the open interval $$0 \lt t \lt 8.77\ \text{cm}$$ gives MORE heat loss than the bare pipe (worst at $t=r_{cr}-r_i=1.20$ cm, $+40\%$). Only past $t\approx8.77$ cm does adding more insulation finally reduce the loss below the bare-pipe value — a substantial, impractically thick jacket for a 1.27 cm pipe, which is itself the point of the problem: on pipes this small, "a little insulation" is counter-productive.