Heat Loss from a Pipe Before and After Adding Insulation: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q5

Question 5: Heat Loss from a Pipe Before and After Adding Insulation (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.

Radius	Value (m)
Pipe outer $r_0$	0.150
A\|B interface $r_1$	0.175
B outer $r_2$ (= B\|C interface)	0.215
C outer $r_3$	0.235
Insulation C conductivity	$k_C=0.2$ W/m·°C

Figure 5 — Radial layers pipe→A→B→C. The known conductivity of C fixes its resistance and hence the "after" heat rate; that pins the combined A+B resistance, which (being unchanged) then gives the "before" heat rate from the original 360 °C drop.

Approach. Insulation C is the only layer with a known conductivity, so its measured 150 °C drop gives the "after" heat rate; steady-state continuity then fixes the combined A+B resistance, which is a fixed property of the (unchanged) inner layers and delivers the "before" heat rate.

Resistance of layer C. $R'_C=\dfrac{\ln(r_3/r_2)}{2\pi k_C}=\dfrac{\ln(0.235/0.215)}{2\pi(0.2)}=0.0708$ m·°C/W.
Heat rate after adding C. The 180→30 °C drop is entirely across C: $$q'_\text{after}=\frac{180-30}{R'_C}=\frac{150}{0.0708}$$ ==**$q'_\text{after}\approx2119$ W/m.**==
Combined A+B resistance. In the "after" state the same $q'$ crosses A and B under a 500→180 °C drop: $$R'_{A+B}=\frac{500-180}{q'_\text{after}}=\frac{320}{2119}=0.1510\ \text{m}\cdot\text{°C/W}$$
Heat rate before adding C. A and B are physically unchanged, so with the original 400→40 °C drop, $$q'_\text{before}=\frac{400-40}{R'_{A+B}}=\frac{360}{0.1510}$$ ==**$q'_\text{before}\approx2384$ W/m.**==

Check

The method needs no individual $k_A$, $k_B$: the two operating states share the same fixed $R'_{A+B}$, so one measured state calibrates it for the other. Adding C lowers the loss (2384 → 2119 W/m) even though the pipe now runs hotter (500 vs 400 °C) — the extra resistance more than offsets the larger driving temperature. The conductivities are assumed temperature-independent between the two states.

Quantity	Result
Resistance of layer C	0.0708 m·°C/W
Combined A+B resistance	0.151 m·°C/W
Heat loss before adding C	≈ 2384 W/m
Heat loss after adding C	≈ 2119 W/m

Concept & Theory

Topic. Steady radial conduction through composite cylindrical insulation with resistances in series.

Key relations. Cylindrical conduction resistance per unit length $R'=\ln(r_o/r_i)/(2\pi k)$; series law $q'=\Delta T/\sum R'$; continuity of $q'$ through all layers at steady state.

Why this works. At steady state the same radial heat rate passes through every layer, so a temperature drop measured across a layer of known conductivity yields $q'$ directly. The inner layers' combined resistance is a geometric/material constant, independent of the boundary temperatures, so it transfers unchanged between the two operating conditions and lets the un-instrumented "before" case be solved.

Common pitfall. Using the plane-wall form $L/k$ instead of the logarithmic cylindrical resistance, or assuming the heat rate is the same in both states (it is not — only $R'_{A+B}$ is shared).

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §3.3 (the cylinder — radial conduction through composite walls).

Question 5 of 8: Heat Loss from a Pipe Before and After Adding Insulation

Question 5: Heat Loss from a Pipe Before and After Adding Insulation (equal value)