Vertical Power-Amplifier Plate — Convection and Radiation: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014, Q7

Question 7: Vertical Power-Amplifier Plate — Convection and Radiation (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. Vertical plate, height $L=0.040\ \text{m}$, width $0.050\ \text{m}$; both faces exposed, so surface area $A=2(0.040\times0.050)=4.0\times10^{-3}\ \text{m}^2$; $\varepsilon=0.82$; ambient/surroundings $T_\infty=25\ ^\circ\text{C}=298.15\ \text{K}$; dissipation $Q=7\ \text{W}$.

Find. The steady surface temperature $T_s$ with combined free convection and radiation.

Approach. Write the surface energy balance $Q=Q_\text{conv}+Q_\text{rad}$, with the free-convection coefficient from the Churchill–Chu vertical-plate correlation (evaluated at the film temperature) and radiation to large surroundings; solve iteratively for $T_s$.

Energy balance. $$Q=\bar h A(T_s-T_\infty)+\varepsilon\sigma A\!\left(T_s^4-T_\infty^4\right)=7\ \text{W}.$$
Free-convection coefficient (Churchill–Chu). With $L=0.04\ \text{m}$ and film properties, the Rayleigh number is $\text{Ra}_L=g\beta(T_s-T_\infty)L^3/(\nu\alpha)$, and$$\text{Nu}_L=\left\{0.68+\frac{0.670\,\text{Ra}_L^{1/4}}{\left[1+(0.492/\text{Pr})^{9/16}\right]^{4/9}}\right\},\quad \bar h=\frac{\text{Nu}_L k}{L}.$$
Iterate. Converging on $T_s\approx125\ ^\circ\text{C}$ (398 K): film ≈ 75 °C gives $\text{Ra}_L\approx2.9\times10^5$, $\text{Nu}_L\approx12.6$, $\bar h\approx9.4\ \text{W/m}^2\text{K}$.$$Q_\text{conv}=9.4(4.0\times10^{-3})(100)=3.8\ \text{W},\quad Q_\text{rad}=0.82(5.67\times10^{-8})(4.0\times10^{-3})(398^4-298^4)=3.2\ \text{W}.$$ Their sum is 7.0 W, so$$\boxed{T_s\approx125\ ^\circ\text{C}}.$$

Check

Both faces of the thin plate are assumed active for convection and radiation ($A=2LW=40\ \text{cm}^2$); if only one face dissipated, $T_s$ would be markedly higher. Convection (3.8 W) and radiation (3.2 W) carry comparable shares, so the emissivity matters — neglecting radiation would over-predict $T_s$ by tens of degrees. Laminar free convection ($\text{Ra}_L<10^9$) is confirmed.

Quantity	Result
Free-convection coefficient	≈ 9.4 W/m²K
Convective heat	≈ 3.8 W
Radiative heat	≈ 3.2 W
Surface temperature	≈ 125 °C

Concept & Theory

Topic. Combined natural convection and surface radiation from a vertical plate (an electronics-cooling energy balance).

Key relations. Churchill–Chu $\text{Nu}_L$; $\text{Ra}_L=g\beta\Delta T L^3/\nu\alpha$ with $\beta=1/T_f$; Stefan–Boltzmann $Q_\text{rad}=\varepsilon\sigma A(T_s^4-T_\infty^4)$; energy balance $Q=Q_\text{conv}+Q_\text{rad}$.

Why this works. The two loss mechanisms act in parallel from the same surface, so the dissipated power equals their sum at the (unknown) surface temperature. Because both depend on $T_s$ — convection roughly as $\Delta T^{5/4}$ and radiation as $T_s^4$ — the balance is nonlinear and solved by iteration; the anodized (high-emissivity) surface deliberately boosts the radiative share.

Common pitfall. Using $T_\infty$ instead of the film temperature for air properties and $\beta$, or omitting radiation (which here removes almost half the heat).

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §9.6 (free convection, vertical plate) and §1/§12 (radiation exchange with large surroundings).

Question 7 of 8: Vertical Power-Amplifier Plate — Convection and Radiation

Question 7: Vertical Power-Amplifier Plate — Convection and Radiation (equal value)