Natural Convection from a Vertical Oil-Filled Heating Panel: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2017, Q7

Question 7: Natural Convection from a Vertical Oil-Filled Heating Panel

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 rectangular panel losing heat by natural (free) convection from BOTH faces into still air, with a known total dissipation and ambient temperature.

Given data
Quantity	Symbol	Value
Panel height (characteristic length)	$H$	0.75 m
Panel width	$W$	1.5 m
Heat dissipated	$\dot Q$	690 W
Ambient (quiescent) air temperature	$T_\infty$	$20\,{}^{\circ}\text{C}$

Find. The panel surface temperature $T_s$.

Thin vertical panel losing heat by natural convection from both faces into quiescent air.

Approach. Both faces convect into the same still air, so by symmetry each removes half the total load; the characteristic length is the panel height $H$ (vertical-plate natural convection). Because the convection coefficient itself depends on the unknown $T_s$ (through the Rayleigh number and the film-temperature-evaluated air properties), solve $\dot Q=hA_{tot}(T_s-T_\infty)$ iteratively with the Churchill–Chu correlation, which is valid across the whole laminar-to-turbulent range.

Total convecting area. $$A_{tot}=2HW=2\times0.75\times1.5=2.25\text{ m}^2$$ (both faces, since the panel is thin and loses heat from front and back alike).
Iterative solve for $T_s$. At each trial $T_s$, evaluate air properties at the film temperature $T_f=(T_s+T_\infty)/2$, form $Ra_H=g\beta(T_s-T_\infty)H^3Pr/\nu^2$, and $$Nu_H=\left\{0.825+\frac{0.387\,Ra_H^{1/6}}{\left[1+(0.492/Pr)^{9/16}\right]^{8/27}}\right\}^2, \qquad h=\frac{Nu_H\,k}{H}$$ Root-finding $hA_{tot}(T_s-T_\infty)=\dot Q$ converges to $$\boxed{T_s=77.5\,{}^{\circ}\text{C}}$$ At this state: film temperature $48.8\,{}^{\circ}\text{C}$, $Ra_H=1.64\times10^9$ (into the turbulent range, $Ra\gt10^9$), $Nu_H=142.9$, $h=5.33\text{ W/m}^2\cdot\text{K}$.
Check. $$\dot Q=hA_{tot}(T_s-T_\infty)=5.33\times2.25\times(77.5-20)=690\text{ W}\ \checkmark$$

Question 7 — results
Quantity	Value
Total convecting area $A_{tot}$	2.25 m²
Rayleigh number $Ra_H$	$1.64\times10^9$
Convection coefficient $h$	5.33 W/m²·K
Surface temperature $T_s$	$77.5\,{}^{\circ}\text{C}$

Topic. Natural (free) convection from a vertical isothermal plate — an implicit, iteratively-solved surface temperature.

Key relations. Churchill–Chu correlation $Nu_H=\{0.825+0.387Ra_H^{1/6}/[1+(0.492/Pr)^{9/16}]^{8/27}\}^2$ (valid all $Ra$); $Ra_H=g\beta(T_s-T_\infty)H^3Pr/\nu^2$; $\dot Q=hA(T_s-T_\infty)$.

Why this works. $h$ depends on $T_s$ through both $Ra_H$'s $\Delta T$ term and the film-temperature-evaluated properties, so the surface energy balance is implicit in $T_s$; the problem is solved by guessing/root-finding $T_s$ until the convected heat from both faces matches the panel's known 690 W dissipation. Both faces see the same $T_\infty$ and $H$, so by symmetry each carries exactly half the load.

Common pitfall. Using only one face's area (halving the true $A_{tot}$, which would roughly double the apparent $T_s$); or evaluating air properties at $T_\infty$ instead of the film temperature $(T_s+T_\infty)/2$ that the correlation requires.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 9 (free convection, vertical-plate correlations).

Question 7 of 8: Natural Convection from a Vertical Oil-Filled Heating Panel

Question 7: Natural Convection from a Vertical Oil-Filled Heating Panel