Surface temperature of a natural-convection heating panel: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q7

Question 7 — Surface temperature of a natural-convection heating panel (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. A thin vertical panel, height $H=0.75$ m, length $L_w=1.5$ m, dissipating 690 W by natural convection from both faces into still air at 20 °C. Find. the panel surface temperature $T_s$.

Quantity	Value
Panel height (characteristic length), $H$	0.75 m
Panel length	1.5 m
Total area (both faces)	2 × (0.75 × 1.5) = 2.25 m²
Heat dissipated, $\dot Q$	690 W
Air temperature, $T_\infty$	20 °C

Figure 7 — Thin vertical panel losing heat by natural convection from both faces; the boundary layers rise along the 0.75 m height, which is the characteristic length for the Rayleigh number.

Approach. Balance the dissipated power against natural-convection loss from both faces, using the Churchill–Chu vertical-plate correlation with air properties at the film temperature, and iterate on $T_s$.

Energy balance. $\dot Q = h\,A_\text{tot}(T_s-T_\infty)$ with $A_\text{tot}=2.25\ \text{m}^2$, so the required flux is $\dot Q/A_\text{tot}=306.7\ \text{W/m}^2$ and $h(T_s-20)=306.7$.
Rayleigh number (film properties). At the converged film temperature $T_f\approx 48.8$ °C, air has $\nu\approx1.79\times10^{-5}$, $\alpha\approx2.54\times10^{-5}\ \text{m}^2/\text{s}$, $k\approx0.0280$, $Pr\approx0.71$. With $\beta=1/T_f$ and $\Delta T\approx57.5$ °C, $$Ra_H = \frac{g\beta\Delta T\,H^3}{\nu\alpha} \approx 1.64\times10^{9}$$
Churchill–Chu Nusselt number (all-$Ra$ vertical plate). $$Nu = \left\{0.825 + \frac{0.387\,Ra_H^{1/6}}{\left[1+(0.492/Pr)^{9/16}\right]^{8/27}}\right\}^2 \approx 143$$ $$h = Nu\,k/H \approx 5.33\ \text{W/m}^2\text{}\cdot\text{°C}$$
Solve the balance. Iterating $T_s$ until $h(T_s-20)=306.7$ converges gives $$T_s - 20 = \frac{306.7}{5.33} \approx 57.5\ \text{°C}$$ $T_s \approx 77.5\ ^\circ\text{C}$

Check — radiation neglected.

Only natural convection is counted, as the question states. If the panel also radiated ($\varepsilon\approx0.9$) it would shed extra heat and the true surface temperature would be somewhat lower; the convection-only figure is the conservative (higher) surface temperature the question asks for.

Quantity	Result
Required convective flux	306.7 W/m²
Rayleigh number $Ra_H$	≈ 1.64 × 10⁹
Nusselt number / $h$	≈ 143 / 5.33 W/m²·°C
Surface temperature $T_s$	≈ 77.5 °C

Concept & Theory

Topic. Free (natural) convection from a vertical plate; iterative surface-temperature solution because the coefficient depends on the unknown $\Delta T$.

Key relations. $\dot Q = h A_\text{tot}\Delta T$; $Ra_H = g\beta\Delta T H^3/(\nu\alpha)$; Churchill–Chu $Nu$ for a vertical plate; $h=Nu\,k/H$; ideal-gas $\beta=1/T_f$.

Why this works. Because the buoyant coefficient itself depends on the wall-to-air temperature difference, the balance is implicit and must be iterated: guess $T_s$, evaluate properties at the film temperature, compute $Ra$, $Nu$ and $h$, and update until the heat balance closes. The height is the correct characteristic length for a vertical plate, and both faces share the load.

Common pitfall. Using the panel length (1.5 m) instead of the vertical height as the characteristic length, forgetting the factor of two for two faces, or evaluating properties at $T_\infty$ rather than the film temperature.

Source. Çengel & Ghajar, Heat and Mass Transfer §9 (natural convection; Churchill–Chu vertical-plate correlation).

Question 7 of 8: Surface temperature of a natural-convection heating panel

Question 7 — Surface temperature of a natural-convection heating panel (Part B, equal value)