Steady-State Temperature of a Charcoal-Lighter Element: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2013, Q7

Question 7: Steady-State Temperature of a Charcoal-Lighter Element (Part B, 20 marks)

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 cylindrical element, $D=1$ cm, $L=70$ cm, electrical input $P=425$ W; still ambient air at $T_\infty=28^{\circ}\text{C}$; $\varepsilon=0.30$; no forced convection (negligible wind).

Find. Steady-state surface temperature $T_s$.

Approach. At steady state, electrical input equals natural convection plus radiation loss from the cylinder's lateral surface; since the natural-convection coefficient itself depends on the (unknown) surface temperature through the Rayleigh number, solve the coupled energy balance iteratively (Churchill–Chu correlation for a horizontal cylinder).

Energy balance to be solved. With $A=\pi DL=0.02199$ m², $$P=h(T_s)\,A\,(T_s-T_\infty)+\varepsilon\sigma A\,(T_s^4-T_\infty^4),$$ where $h(T_s)$ comes from natural convection off a horizontal cylinder, evaluated at the film temperature $T_f=(T_s+T_\infty)/2$.
Churchill–Chu natural-convection correlation. $$Ra_D=\frac{g\beta(T_s-T_\infty)D^3}{\nu\alpha},\qquad Nu_D=\left\{0.60+\frac{0.387\,Ra_D^{1/6}}{\left[1+(0.559/Pr)^{9/16}\right]^{8/27}}\right\}^{2}, \qquad h=\frac{Nu_D\,k}{D}.$$
Iterative (bisection) solution. Sweeping $T_s$ until the energy balance closes converges to $$T_s=\boxed{893.0\ \text{K}=619.8^{\circ}\text{C}},$$ at which $Ra_D=2.54\times10^{3}$, $Nu_D=3.19$, $h=14.6$ W/m²K, giving $Q_{conv}=190.2$ W and $Q_{rad}=234.8$ W (sum $=425.0$ W $\checkmark$, radiation slightly dominant at this high a temperature).

Final results — Question 7
Quantity	Value
Natural-convection coefficient $h$	$14.6$ W/m²K
Convective loss	$190.2$ W
Radiative loss	$234.8$ W
Steady-state surface temperature	$\boxed{619.8^{\circ}\text{C}}$ ($893$ K)

Topic: Coupled natural-convection + radiation steady-state surface temperature of a bare heated cylinder.

Key relations: Churchill–Chu horizontal-cylinder correlation; Stefan–Boltzmann radiation $Q_{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)$; energy balance $P=Q_{conv}(T_s)+Q_{rad}(T_s)$.

Why this works: because $h$ itself is a (weak, roughly $\Delta T^{1/3}$ for turbulent-regime, or here a mixed $Ra_D^{1/6}$-inside-a-square dependence) function of the unknown $T_s$, the energy balance is implicit and must be solved by iteration/root-finding rather than algebraically; at the ~600°C surface temperature found here, radiation (a $T^4$ effect) has overtaken natural convection even though it started smaller at low $\Delta T$.

Common pitfall: using a constant "typical" $h$ (e.g. $10$ W/m²K) instead of solving self-consistently, or neglecting radiation entirely on what looks like a "convection problem" — here radiation is the LARGER of the two loss mechanisms.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 9 (free convection, Churchill–Chu correlation), Ch. 13 (radiation exchange).

Question 7 of 8: Steady-State Temperature of a Charcoal-Lighter Element

Question 7: Steady-State Temperature of a Charcoal-Lighter Element (Part B, 20 marks)