Insulation Thickness for a 95% Heat-Loss Reduction: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2016, Q5

Question 5: Insulation Thickness for a 95% Heat-Loss Reduction

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 copper tube OD 2.54 cm, wall thickness 0.54 cm, inner-surface (steam-side) temperature 100°C, room air 27°C, forced-convection coefficient 55 W/m²°C, insulation $k=0.0875$ W/m°C, target: 95% reduction in the heat-loss rate per unit length.

Check: copper's thermal conductivity ($k\approx 400$ W/m·K) is four orders of magnitude above the insulation's, so the copper wall's own conduction resistance is neglected — the outer surface of the bare tube is taken at the given inner-surface temperature $T_i=100\,{}^{\circ}\text{C}$, which is where the insulation layer begins.

Given data
Quantity	Symbol	Value
Tube outside diameter	$D_o$	2.54 cm ($r_1=1.27$ cm)
Inner-surface temperature	$T_i$	$100\,{}^{\circ}\text{C}$
Room air temperature	$T_\infty$	$27\,{}^{\circ}\text{C}$
Convection coefficient	$\bar{h}$	55 W/m²°C
Insulation conductivity	$k$	0.0875 W/m°C

Find. The insulation thickness $t$ that cuts the per-unit-length heat loss to 5% of the bare-tube value, and the resulting outside surface temperature $T_o$.

Approach. Model the bare tube as a single convective resistance to get the baseline loss $q'_{\text{bare}}$; model the insulated tube as a conduction-then-convection series resistance in cylindrical coordinates, set $q'_{\text{insulated}}=0.05\,q'_{\text{bare}}$, and solve implicitly for the outer insulation radius $r_2$.

Baseline (bare-tube) heat loss per unit length. $$q'_{\text{bare}}=\bar{h}(\pi D_o)(T_i-T_\infty)=55\,\pi(0.0254)(100-27)=320.4\text{ W/m}$$ Target: $q'_{\text{target}}=0.05(320.4)=16.02\text{ W/m}$.
Insulated-tube resistance network (per unit length). $$q'(r_2)=\frac{T_i-T_\infty}{\dfrac{\ln(r_2/r_1)}{2\pi k}+\dfrac{1}{\bar{h}\,2\pi r_2}}$$ with $r_1=0.0127$ m fixed. Solve $q'(r_2)=16.02$ W/m for $r_2$ (implicit — $r_2$ appears in both the log and the $1/r_2$ term; solved numerically): $$\boxed{r_2\approx 153.95\text{ mm}}$$
Insulation thickness. $$t=r_2-r_1=153.95-12.7=\boxed{141.3\text{ mm}}$$
Outside surface temperature under these conditions. With $q'=16.02$ W/m flowing through the final convective film at $r_2$: $$T_o=T_\infty+\frac{q'}{\bar{h}\,2\pi r_2}=27+\frac{16.02}{55\cdot2\pi(0.15395)}=27+0.30=\boxed{27.3\,{}^{\circ}\text{C}}$$

141 mm of insulation on a 25 mm tube is a large ratio, but it is the direct consequence of the GIVEN combination of a low insulation conductivity (0.0875 W/m°C) and a comparatively high convection coefficient (55 W/m²°C, forced convection): the outer convective film resistance shrinks as $1/r_2$ while the conductive resistance only grows as $\ln r_2$, so a very thick layer is needed before conduction dominates enough to choke the loss down to 5%.

Question 5 — results
Quantity	Value
Bare-tube heat loss, $q'_{\text{bare}}$	320.4 W/m
Target heat loss (5%), $q'_{\text{target}}$	16.0 W/m
Insulation outer radius, $r_2$	154.0 mm
Insulation thickness, $t$	141.3 mm
Outside surface temperature, $T_o$	$27.3\,{}^{\circ}\text{C}$

Topic. Steady 1-D radial conduction through a cylindrical insulation layer with a convective outer boundary.

Key relations. Cylindrical conduction resistance $R'_{\text{cond}}=\ln(r_2/r_1)/(2\pi k)$; convective resistance $R'_{\text{conv}}=1/(\bar{h}\,2\pi r)$; series resistances add; $q'=\Delta T/\sum R'$.

Why this works. Both resistances are expressed PER UNIT LENGTH so they can be combined directly regardless of tube length, and adding insulation always increases the conduction resistance while decreasing the outer convection resistance (larger outer radius, more convective area) — for a plane wall this trade-off doesn't exist, which is why "critical radius of insulation" is a distinctly cylindrical-geometry effect worth checking is not the binding constraint here (it is not, since the required $r_2$ is already far beyond the critical radius $k/\bar{h}=1.6$ mm).

Common pitfall. Solving for $r_2$ by isolating it from only the logarithmic term (ignoring that it also appears in the convective term) gives an inconsistent, non-physical result; the equation must be solved as one implicit relation in $r_2$.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 3 (conduction through a cylindrical wall with convective boundaries, critical radius of insulation).

Question 5 of 8: Insulation Thickness for a 95% Heat-Loss Reduction

Question 5: Insulation Thickness for a 95% Heat-Loss Reduction