Reduction of heat loss from a steam pipe by insulation: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2018, Q5

Question 5 — Reduction of heat loss from a steam pipe by insulation (Part B, 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. Steel tube, inside radius $r_1=0.075$ m, outside radius $r_2=0.085$ m (10 mm wall). Wet steam inside at $T_i=200$ °C; ambient air $T_\infty=27$ °C. Bare loss $q'_\text{bare}=2000$ W/m. Insulation 50 mm thick ($r_3=0.135$ m), $k_\text{ins}=0.35$, $k_\text{pipe}=45\ \text{W/m°C}$; $h_\text{metal}=1.40\,h_\text{ins}$.

Find. the heat loss per metre after insulating.

Figure 5 — The insulated pipe as a series thermal network (per metre of length): steam-side wall ≈ 200 °C, then conduction through the steel and the insulation, then convection to the 27 °C air through the reduced coefficient $h_\text{ins}=h_\text{metal}/1.4$.

Approach. First back out $h_\text{metal}$ from the bare-pipe loss (steel wall resistance is negligible, so the outer wall sits at ≈200 °C); reduce it to $h_\text{ins}$; then sum the series resistances of the insulated pipe and divide the 173 °C driving temperature by the total.

Metal-to-air coefficient from the bare pipe. The steel wall drops less than 1 °C at 2000 W/m, so the bare outer surface is essentially at the steam temperature. With outer area $2\pi r_2=0.534\ \text{m}^2/\text{m}$: $$h_\text{metal}=\frac{q'_\text{bare}}{2\pi r_2(T_i-T_\infty)}=\frac{2000}{0.534(200-27)}=21.6\ \text{W/m}^2\text{°C}$$
Insulation-to-air coefficient. $$h_\text{ins}=\frac{h_\text{metal}}{1.40}=\frac{21.6}{1.40}=15.5\ \text{W/m}^2\text{°C}$$
Series resistances (per metre). $$R'_\text{pipe}=\frac{\ln(r_2/r_1)}{2\pi k_\text{pipe}}=\frac{\ln(0.085/0.075)}{2\pi(45)}=4.4\times10^{-4}$$ $$R'_\text{ins}=\frac{\ln(r_3/r_2)}{2\pi k_\text{ins}}=\frac{\ln(0.135/0.085)}{2\pi(0.35)}=0.2104,\qquad R'_\text{conv}=\frac{1}{h_\text{ins}\,2\pi r_3}=\frac{1}{15.5(0.848)}=0.0763$$ All in units of °C·m/W; the total is $R'_\text{tot}=0.2871$ °C·m/W.
Insulated heat loss. $$q'_\text{ins}=\frac{T_i-T_\infty}{R'_\text{tot}}=\frac{200-27}{0.2871}=603\ \text{W/m}$$ $q'_\text{ins}\approx 603\ \text{W/m}$ — a reduction of about 70 % from the bare 2000 W/m

Quantity	Result
$h_\text{metal}$ (from bare loss)	21.6 W/m²°C
$h_\text{ins}=h_\text{metal}/1.4$	15.5 W/m²°C
Insulated loss $q'_\text{ins}$	≈ 603 W/m
Reduction from bare pipe	≈ 70 %

Concept & Theory

Topic: steady radial conduction through a composite cylinder with external convection; thermal-resistance network.

Key relations: cylindrical conduction resistance $R'=\ln(r_o/r_i)/2\pi k$ (per unit length); convection $R'=1/(h\,2\pi r)$; series total $q'=\Delta T/\sum R'$.

Why this works: the bare pipe loss is dominated by the outside film (the steel is a near-perfect conductor), which lets us extract the metal-to-air coefficient directly. Because a $r_3=0.135$ m outer radius is far above the critical radius $r_c=k_\text{ins}/h_\text{ins}=0.35/15.5=0.023$ m, adding insulation unambiguously reduces the loss; the insulation conduction resistance (0.21) then dominates the network.

Common pitfall: forgetting that the two convection coefficients differ (the problem deliberately links them), or neglecting to check the critical radius before assuming insulation helps. Here it clearly does, but for a thin bare pipe with a low $h$ it might not.

Source: Incropera et al., Fundamentals of Heat and Mass Transfer 8th ed., §3.3 (radial systems), §3.4 (critical radius).

Question 5 of 8: Reduction of heat loss from a steam pipe by insulation

Question 5 — Reduction of heat loss from a steam pipe by insulation (Part B, equal value)