Surface Temperature of an Electric Charcoal-Lighter Element: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013, Q7

Question 7: Surface Temperature of an Electric Charcoal-Lighter Element (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. Horizontal cylinder $D=0.01$ m, $L=0.70$ m ($A=\pi DL=0.0220\ \text{m}^2$); power 425 W; ambient 28 °C = 301 K; emissivity 0.30; still air (natural convection only).

Find. The steady-state surface temperature $T_s$.

Figure 7 — Horizontal heated element: the 425 W dissipated is balanced by natural convection plus radiation to the 28 °C surroundings; both depend on the (unknown) surface temperature, so the balance is solved iteratively.

Approach. Write a steady-state surface energy balance equating the electrical dissipation to convection (Churchill–Chu natural convection on a horizontal cylinder) plus radiation, and iterate on $T_s$.

Energy balance. At steady state,$$\dot Q=hA(T_s-T_\infty)+\varepsilon\sigma A\,(T_s^4-T_\infty^4)=425\ \text{W},$$ with $\sigma=5.67\times10^{-8}$ and $h$ from natural convection.
Convection correlation. For a horizontal cylinder, $Nu=\Big\{0.60+\dfrac{0.387\,Ra^{1/6}}{[1+(0.559/Pr)^{9/16}]^{8/27}}\Big\}^2$ with $Ra=\dfrac{g\beta(T_s-T_\infty)D^3}{\nu\alpha}$, air properties evaluated at the film temperature. At convergence $Ra\approx2.6\times10^3$, $Nu\approx3.2$, giving $h\approx14.7\ \text{W/m}^2\text{}\cdot\text{K}$.
Iterate to balance. Solving the balance (radiation carries the majority of the load at this flux), the convection contributes ≈192 W and radiation ≈233 W, summing to 425 W at$$\boxed{T_s\approx619\ ^{\circ}\text{C}\;(\approx892\ \text{K})}.$$

Quantity	Result
Surface area	0.0220 m²
Natural-convection coefficient	≈ 14.7 W/m²·K
Convection / radiation split	≈ 192 W / 233 W
Surface temperature	≈ 619 °C (892 K)

Concept & Theory

Topic. Coupled natural-convection and radiation surface energy balance.

Key relations. $\dot Q=hA\Delta T+\varepsilon\sigma A(T_s^4-T_\infty^4)$; Churchill–Chu horizontal-cylinder correlation; Rayleigh number $Ra=g\beta\Delta T D^3/(\nu\alpha)$.

Why this works. With no forced flow, the element sheds heat by buoyant convection and by radiation, both increasing with $T_s$; the surface temperature is whatever makes their sum equal the 425 W input. Because radiation scales as $T_s^4$, it dominates at the high temperatures reached here, so ignoring it would badly under-predict the loss and over-predict $T_s$.

Common pitfall. Neglecting radiation (a large error at ~900 K) or evaluating air properties at ambient rather than the film temperature.

Source. Çengel & Ghajar, Heat and Mass Transfer §9 (natural convection, Churchill–Chu) with §12 radiation exchange.

Question 7 of 8: Surface Temperature of an Electric Charcoal-Lighter Element

Question 7: Surface Temperature of an Electric Charcoal-Lighter Element (equal value)