Power-amplifier case temperature (free convection + radiation): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, Undated paper, Q7

Question 7 — Power-amplifier case temperature (free convection + radiation) (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
Plate height (vertical) / width	$H,\,w$	40 mm / 50 mm
Radiating / convecting area (2 faces)	$A$	$2(0.04\times0.05)=0.0040$ m²
Ambient air / surroundings	$T_\infty$	25 °C = 298.15 K
Emissivity	$\varepsilon$	0.82
Dissipation	$\dot Q$	7 W

Find. the steady surface temperature $T_s$.

Steady state: internal dissipation is removed jointly by free convection and surface radiation.

Approach. Set the generated power equal to the combined convective and radiative loss, with the convection coefficient from the Churchill–Chu vertical-plate correlation evaluated at the film temperature; solve the resulting nonlinear balance for $T_s$.

Energy balance. At steady state $$\dot Q=h\,A\,(T_s-T_\infty)+\varepsilon\sigma A\,(T_s^4-T_\infty^4)$$ with $h$ from the Churchill–Chu correlation $Nu=\Big[0.825+\dfrac{0.387\,Ra_L^{1/6}}{(1+(0.492/Pr)^{9/16})^{8/27}}\Big]^2$, $Ra_L=\dfrac{g\beta(T_s-T_\infty)L^3}{\nu\alpha}$, $L=H=0.04$ m, properties at the film temperature.
Iterate on $T_s$. Solving the balance numerically converges to $$T_s=\boxed{127\ ^\circ\text{C}\ (400\ \text{K})}$$ at which $Ra_L=3.0\times10^{5}$ and $h=9.1$ W/m²·K.
Check the split. The two loss paths are comparable at this temperature: $$\dot Q_\text{conv}=hA(T_s-T_\infty)=9.1(0.004)(102)=3.70\ \text{W},\qquad \dot Q_\text{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)=3.30\ \text{W}$$ summing to 7.0 W — radiation is not negligible here (47 %) because the high anodized emissivity and the elevated surface temperature both favour it.

Quantity	Result
Convection coefficient $h$	9.1 W/m²·K
Convective / radiative loss	3.70 / 3.30 W
Steady case temperature $T_s$	≈ 127 °C

Check — two-face assumption.

The case is modelled as radiating and convecting from both 40×50 mm faces ($A=0.0040$ m²); if only one face were exposed the balance would drive $T_s$ higher. The high computed temperature (127 °C) confirms that a bare 7 W amplifier this small runs hot — a heat sink would be specified in practice.

Concept & Theory — Q7

Topic: Combined free-convection and radiation heat loss from a heated vertical surface — a coupled nonlinear surface-energy balance.

Key relations: Churchill–Chu vertical-plate correlation for $Nu(Ra_L,Pr)$; Rayleigh number $Ra_L=g\beta\Delta T L^3/(\nu\alpha)$; Stefan–Boltzmann radiation $\dot Q_\text{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)$ (Çengel HT, Ch. 9 & 12).

Why this works: Because $h$ depends on $\Delta T$ and radiation depends on $T_s^4$, the balance cannot be solved in closed form; iterating with film-temperature properties captures both temperature dependences self-consistently. Treating the two loss paths in parallel (they share the same surface temperature) lets them simply add.

Common pitfall: Neglecting radiation ("small device, low temperature") — with $\varepsilon=0.82$ it carries nearly half the load — or evaluating air properties at ambient rather than the film temperature, which mis-sizes $Ra_L$.

Question 7 of 8: Power-amplifier case temperature (free convection + radiation)

Question 7 — Power-amplifier case temperature (free convection + radiation) (Part B, equal value)