Critical Radius of Insulation on a Hot-Water Pipe: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013, Q5

Question 5: Critical Radius of Insulation on a Hot-Water Pipe (equal value)

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 1.27 cm ($r_1=0.00635$ m); insulation $k=0.156\ \text{W/m}\cdot\text{K}$; pipe-surface temperature 82.5 °C; ambient 20 °C; outside coefficient $h=8.5\ \text{W/m}^2\text{}\cdot\text{K}$.

Find. The range of insulation thickness for which the insulated pipe loses more heat than the bare pipe (the critical-radius effect), and the boundary thickness where the loss returns to the bare value.

Figure 5 — Pipe cross-section: a thin small-diameter pipe gains surface area faster than it gains conduction resistance when first insulated, so heat loss rises until the critical radius r_c = k/h and does not return to the bare value until r* ≈ 9.4 cm.

Approach. Compare the bare-pipe convective loss with the insulated series (conduction + convection) resistance per unit length, and solve for the outer radius at which the two losses are equal again.

Bare-pipe loss. The bare pipe loses by convection directly:$$q_\text{bare}=h(2\pi r_1)(T_s-T_\infty)=8.5(2\pi\cdot0.00635)(62.5)=\boxed{21.2\ \text{W/m}}.$$
Critical radius. The insulated loss $q=\dfrac{2\pi(T_s-T_\infty)}{\dfrac{\ln(r_2/r_1)}{k}+\dfrac{1}{h r_2}}$ is maximised at $$r_c=\frac{k}{h}=\frac{0.156}{8.5}=0.0184\ \text{m}=1.84\ \text{cm},$$ where the loss peaks at 29.7 W/m — about 40 % above the bare pipe.
Thickness where loss returns to the bare value. Setting $q_\text{ins}=q_\text{bare}$ gives $\dfrac{1}{h r_1}=\dfrac{\ln(r_2/r_1)}{k}+\dfrac{1}{h r_2}$. Solving numerically for the root beyond $r_c$,$$r^\ast=0.0940\ \text{m}\;\Rightarrow\;t=r^\ast-r_1=\boxed{0.0877\ \text{m}\approx8.8\ \text{cm}}.$$ Any insulation thickness from 0 up to about 8.8 cm therefore increases the loss; only beyond 8.8 cm does insulation begin to reduce it.

Quantity	Result
Bare-pipe loss	21.2 W/m
Critical radius (max loss)	r_c = 1.84 cm (1.20 cm of insulation), 29.7 W/m
Outer radius where loss = bare	r* = 9.40 cm
Thickness bound (loss ≥ bare)	up to ≈ 8.8 cm

Concept & Theory

Topic. Critical radius of insulation on a cylinder.

Key relations. Cylindrical conduction resistance $R_\text{cond}=\ln(r_2/r_1)/(2\pi kL)$; convection resistance $1/(h\,2\pi r_2 L)$; critical radius $r_c=k/h$.

Why this works. Adding insulation raises the conduction resistance but also enlarges the outer surface, which lowers the convective resistance. For a small pipe with a low outside coefficient the second effect wins until $r_2=k/h$; the loss keeps exceeding the bare value until the growing conduction term finally overtakes it at $r^\ast$. Here $r_c=1.84$ cm is much larger than the 0.635 cm pipe radius, so thin insulation is counter-productive.

Common pitfall. Assuming insulation always reduces loss — true only once $r_2>r_c$; and confusing the peak radius $r_c$ with the break-even radius $r^\ast$.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §3.3 (critical radius of insulation).

Question 5 of 8: Critical Radius of Insulation on a Hot-Water Pipe

Question 5: Critical Radius of Insulation on a Hot-Water Pipe (equal value)