Insulation Thickness to Cut a Steam-Tube Heat Loss: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2016, Q5

Question 5: Insulation Thickness to Cut a Steam-Tube Heat Loss (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.

Quantity	Value
Tube outer radius $r_2$ / inner $r_1$	0.0127 m / 0.0073 m
Inside surface temperature	100 °C
Room air temperature	27 °C
Insulation conductivity $k$	0.0875 W/m·°C
Convection coefficient $h$	55 W/m²·°C

Figure 5 — Radial resistance network: the (negligible) copper conduction, then insulation conduction $\ln(r_3/r_2)/2\pi k$, then the outer convective film $1/2\pi r_3 h$, all in series between the 100 °C inner surface and the 27 °C room.

Given. A copper tube (OD 2.54 cm, wall 0.54 cm) with a 100 °C inner surface in a 27 °C room, cooled by forced convection h = 55 W/m²·°C, to be lagged with k = 0.0875 W/m·°C insulation. Find. The insulation thickness that cuts the loss by 95 %, and the resulting outer-surface temperature.

Approach. Compute the bare-tube loss, target 5 % of it, then solve the series cylindrical-resistance equation for the outer insulation radius; the surface temperature follows from the convective leg.

Bare-tube heat loss (per metre). Copper is nearly isothermal, so the outer surface sits at ~100 °C and $$q'_\text{bare}=h(2\pi r_2)(T_i-T_\infty)=55(2\pi\cdot0.0127)(73)=320\ \text{W/m}.$$
Target loss. A 95 % reduction leaves $q'_\text{ins}=0.05(320)=16.0$ W/m.
Check the critical radius. $r_c=k/h=0.0875/55=1.6$ mm $
Solve for the outer radius. Setting $q'_\text{ins}=\dfrac{T_i-T_\infty}{\dfrac{\ln(r_3/r_2)}{2\pi k}+\dfrac{1}{2\pi r_3 h}}=16.0$ W/m and iterating gives $r_3=0.154$ m, so $$\text{thickness}=r_3-r_2=0.154-0.0127=\boxed{0.141\ \text{m}\approx14.1\ \text{cm}}.$$
Outside surface temperature. From the convective leg at $r_3$, $$T_o=T_\infty+\frac{q'_\text{ins}}{2\pi r_3 h}=27+\frac{16.0}{2\pi(0.154)(55)}=\boxed{27.3\,{}^\circ\text{C}}.$$

Check

The thick (14 cm) blanket is a direct consequence of the low-conductivity insulation and the demanding 95 % cut on a small pipe; with the surface now only 0.3 °C above ambient, essentially the entire 73 °C drop occurs across the insulation. The copper wall resistance ($\sim10^{-4}$ m·K/W) is four orders below the insulation resistance and is rightly neglected.

Quantity	Result
Bare-tube loss	≈ 320 W/m
Target (5 %) loss	≈ 16.0 W/m
Critical radius $k/h$	1.6 mm ( ≪ $r_2$ )
Required outer radius $r_3$	≈ 0.154 m
Insulation thickness	≈ 14.1 cm
Outside surface temperature $T_o$	≈ 27.3 °C

Concept & Theory

Topic. Radial (cylindrical) conduction with an outer convective film — the critical-radius-of-insulation problem.

Key relations. Cylindrical conduction $R'=\ln(r_3/r_2)/2\pi k$; convection $R'=1/2\pi r_3 h$; $q'=\Delta T/\sum R'$; critical radius $r_c=k/h$; surface temperature from the outer film.

Why this works. Heat flows radially through resistances in series driven by the fixed inner-to-ambient temperature difference; specifying the allowed loss fixes the total resistance, from which the single unknown $r_3$ is recovered. Because $r_2>r_c$, the added conduction resistance always outweighs the falling convective resistance, so more insulation monotonically lowers the loss.

Common pitfall. Forgetting to test the critical radius (on a very small pipe with low $h$, thin insulation could increase loss), or using the flat-wall $L/k$ resistance instead of the logarithmic cylindrical form.

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

Question 5 of 8: Insulation Thickness to Cut a Steam-Tube Heat Loss

Question 5: Insulation Thickness to Cut a Steam-Tube Heat Loss (equal value)