Vertical Power-Amplifier Case — Combined Natural Convection and Radiation: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, Undated paper, Q7

Question 7: Vertical Power-Amplifier Case — Combined Natural Convection and Radiation

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. A small vertical plate dissipates a fixed heat load to still air by natural convection AND radiation acting together (in parallel, same $\Delta T$); both faces of the thin case are taken as active heat-transfer area.

Given data
Quantity	Symbol	Value
Plate height (vertical, characteristic length)	$H$	40 mm
Plate width	$W$	50 mm
Ambient / surroundings temperature	$T_\infty$	$25\,{}^{\circ}\text{C}$
Surface emissivity	$\varepsilon$	0.82
Heat dissipated	$\dot Q$	7 W

Find. The steady surface temperature $T_s$.

Approach. Both faces of the case ($A=2HW$) lose heat by natural convection (Churchill–Chu correlation, valid for all Rayleigh numbers) and by radiation to the same $25\,{}^{\circ}\text{C}$ surroundings; because $h$ itself depends on $T_s$ through the film properties, solve $h(T_s)A(T_s-T_\infty)+\varepsilon\sigma A(T_s^4-T_\infty^4)=\dot Q$ iteratively.

Set up the coupled convection+radiation balance and iterate. With $A=2\times0.04\times0.05=0.004$ m² and air properties evaluated at the film temperature $T_f=(T_s+T_\infty)/2$ on each pass: $$Ra_H=\frac{g\beta(T_s-T_\infty)H^3}{\nu\alpha},\qquad Nu=\left\{0.825+\frac{0.387\,Ra_H^{1/6}}{\left[1+(0.492/Pr)^{9/16}\right]^{8/27}}\right\}^2,\qquad h=\frac{Nu\,k}{H}$$ $$h(T_s)A(T_s-T_\infty)+\varepsilon\sigma A(T_s^4-T_\infty^4)=\dot Q$$ converges (Newton iteration on $T_s$) to $$\boxed{T_s=127.0\,{}^{\circ}\text{C}\ (400.1\text{ K})}$$ with $Ra_H=3.03\times10^5$, $Nu=12.1$, $h=9.08$ W/m²·K.
Split the 7 W between the two mechanisms as a check. $$q_{conv}=hA(T_s-T_\infty)=9.08\times0.004\times(127.0-25.0)$$ $$\boxed{q_{conv}=3.70\text{ W}}$$ $$q_{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)=0.82\times5.67\times10^{-8}\times0.004\times(400.1^4-298.15^4)$$ $$\boxed{q_{rad}=3.30\text{ W}}$$ $$q_{conv}+q_{rad}=3.70+3.30=7.00\text{ W}\ \checkmark$$

Question 7 — results
Quantity	Value
Rayleigh number $Ra_H$	$3.03\times10^5$
Convection coefficient $h$	9.08 W/m²·K
Surface temperature $T_s$	$127.0\,{}^{\circ}\text{C}$
Convective / radiative split	3.70 W / 3.30 W

Topic. Combined free (natural) convection and surface radiation from a small vertical plate to quiescent air.

Key relations. Churchill–Chu correlation for a vertical plate (all $Ra$); Rayleigh number $Ra_H=g\beta\Delta TH^3/(\nu\alpha)$; radiation $q_{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)$.

Why this works. Neither mechanism alone can be solved in closed form because $h$ depends on the unknown $T_s$ through the film-temperature properties and $Ra$; iterating the combined energy balance (rather than guessing how the 7 W splits) is what makes the two checks in step 2 come out to EXACTLY 7.00 W, confirming convergence.

Common pitfall. Using only ONE face's area for a "thin plate" that is physically exposed to ambient air on both sides, and evaluating air properties at $T_\infty$ instead of the (much hotter) film temperature, which would under-predict $h$ and over-predict $T_s$.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 9 (free convection, Churchill–Chu correlation) & Ch. 13 (surface radiation exchange).

Question 7 of 8: Vertical Power-Amplifier Case — Combined Natural Convection and Radiation

Question 7: Vertical Power-Amplifier Case — Combined Natural Convection and Radiation