Radiant-heated evacuated tube: radial conduction & surface convection: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2019, Q5

Question 5 — Radiant-heated evacuated tube: radial conduction & surface convection (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.

Quantity	Symbol	Value
Inner / outer radius	$r_1,\,r_2$	0.010 m, 0.025 m
Inner-surface heat flux	$q''_i$	10 kW/m² (see the check note)
Outer-surface temperature	$T_2$	633 K
Cooling-fluid temperature	$T_\infty$	300 K
Tube conductivity	$k$	2.2 W/m·°C

Find. (a) $q'$ (W/m); (b) outer-surface $h$; (c) inner-surface temperature $T_1$.

Radial heat path: radiant flux enters at $r_1$, conducts outward through the tube wall, and is convected to the fluid at the outer surface. In steady state the same $q'$ crosses every radius.

Approach. The line rate $q'$ comes from the inner flux times the inner circumference; steady state passes that same $q'$ to the fluid, giving $h$; radial conduction through the wall then fixes the inner-surface temperature.

Heat transfer per unit length. All the flux entering at $r_1$ crosses each radius: $$q'=q''_i\,(2\pi r_1)=10\,000\,(2\pi)(0.010)=\boxed{628\ \text{W/m}}$$
Outer-surface heat-transfer coefficient. Newton's law at $r_2$, $q'=h\,(2\pi r_2)(T_2-T_\infty)$: $$h=\frac{q'}{2\pi r_2 (T_2-T_\infty)}=\frac{628}{2\pi(0.025)(633-300)}=\boxed{12.0\ \text{W/m}^2\text{K}}$$
Inner-surface temperature (cylindrical conduction). $T_1-T_2=\dfrac{q'\ln(r_2/r_1)}{2\pi k}$: $$T_1=633+\frac{628\,\ln(2.5)}{2\pi(2.2)}=633+41.6=\boxed{674.6\ \text{K}}$$

Check — inner-surface flux magnitude.

The paper prints "10 W/m²", which would give an absurd $q'=0.63$ W/m and $h=0.012$ W/m²K. For a high-temperature radiant heater driving a 633 K outer surface, the engineering-consistent value is 10 kW/m² (= 10⁴ W/m²), which yields a physically sensible $h\approx12$ W/m²K (low-pressure gas convection). The three answers all scale linearly with the assumed flux; the method is exact regardless.

Quantity	Result
(a) Heat rate per length	628 W/m
(b) Outer-surface $h$	12.0 W/m²K
(c) Inner-surface temperature	674.6 K (≈ 401 °C)

Concept & Theory — Q5

Topic: One-dimensional steady radial conduction through a cylindrical wall with a specified inner flux and convective outer boundary.

Key relations: Line rate $q'=q''\,2\pi r$; cylindrical conduction resistance $R'_\text{cond}=\ln(r_2/r_1)/(2\pi k)$; convective resistance $R'_\text{conv}=1/(h\,2\pi r_2)$; Newton's law of cooling.

Why this works: With no generation in the wall and steady state, energy conservation forces the same total rate $q'$ through every cylindrical shell, so the inner-surface flux immediately fixes $q'$. That single rate then drives two independent resistances in series — conduction across the wall and convection at the outer face — which can be solved one at a time: the outer convection gives $h$, and the conduction drop gives the inner-surface temperature.

Common pitfall: Using the outer radius (or an arithmetic-mean radius) in $q'=q''2\pi r$ — the flux is specified at $r_1$, so $r_1$ is correct. Another is dropping the logarithmic (cylindrical) form and using a slab $\Delta x/k$.

Source: Çengel & Ghajar, Heat and Mass Transfer 6th ed., §3-2 (conduction through cylindrical layers); Incropera 8th ed., §3.3.

Question 5 of 8: Radiant-heated evacuated tube: radial conduction & surface convection

Question 5 — Radiant-heated evacuated tube: radial conduction & surface convection (Part B, equal value)