Natural-Convection Heat Loss From a Steel Plate: 21-Mat-A2 Materials Transport Phenomena, December 2019, Q2

Question 2: Natural-Convection Heat Loss From a Steel Plate (25 marks: (a) 15, (b) 10)

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 side length	$a$	$2.5\ \text{m}$ (square, $A_s=6.25\ \text{m}^2$)
Plate thickness	—	$2.5\ \text{mm}$ (edge area negligible)
Initial surface temperature	$T_s$	$430\ \text{K}$
Ambient temperature	$T_\infty$	$295\ \text{K}$
Air properties (Table 1 in the exam appendix)	—	interpolated at the film temperature $T_f$

Find. The initial rate of heat loss $\dot Q$ (W), from both faces of the plate, for (a) the plate hung horizontally and (b) the plate hung vertically.

Approach. At the instant the plate leaves the oven its whole (thin) cross-section is still at $T_s=430\ \text{K}$, so the "initial" heat loss is a pure surface free-convection problem: evaluate air properties at the film temperature from the exam's own Table 1, then apply the standard horizontal-plate (top and bottom separately, since a hung plate loses heat from both faces) and vertical-plate natural-convection correlations.

Film temperature and air properties. $T_f=(T_s+T_\infty)/2=(430+295)/2=362.5\ \text{K}$, which falls $25\%$ of the way from the Table 1 row at 360 K to the row at 370 K. Interpolating linearly: $$k=0.03056\ \text{W/m}\cdot\text{K},\qquad \frac{g\beta}{\nu\alpha}=39.8\times10^{6}\ \text{m}^{-3}\text{K}^{-1}\quad(\text{read directly from Table 1's own column})$$ With $\Delta T=T_s-T_\infty=135\ \text{K}$, the Rayleigh number for any characteristic length $L_c$ is $Ra_{L_c}=\dfrac{g\beta}{\nu\alpha}\,\Delta T\,L_c^3$.
(a) Horizontal plate — both faces. Characteristic length $L_c=A_s/P=6.25/10=0.625\ \text{m}$, giving $Ra_{L_c}=1.31\times10^{9}$ (turbulent range for a horizontal plate). Top face (hot surface facing up, unstable/enhanced plume detachment) and bottom face (hot surface facing down, stable/suppressed) use different constants: $$\begin{aligned}Nu_{top}&=0.15\,Ra_{L_c}^{1/3}=164.2\\ h_{top}&=\frac{Nu_{top}\,k}{L_c}=8.03\ \text{W/m}^2\text{K}\\ \dot q_{top}&=h_{top}A_s\Delta T=6774\ \text{W}\end{aligned}$$ $$\begin{aligned}Nu_{bot}&=0.27\,Ra_{L_c}^{1/4}=51.4\\ h_{bot}&=\frac{Nu_{bot}\,k}{L_c}=2.51\ \text{W/m}^2\text{K}\\ \dot q_{bot}&=h_{bot}A_s\Delta T=2120\ \text{W}\end{aligned}$$ $$\boxed{\dot Q_{horizontal}=\dot q_{top}+\dot q_{bot}=6774+2120=8894\ \text{W}}$$
(b) Vertical plate — both faces. Characteristic length is now the plate height, $L_c=2.5\ \text{m}$, giving $Ra_{L_c}=8.40\times10^{10}$ (turbulent, $Ra>10^9$): $$\begin{aligned}Nu&=0.10\,Ra_{L_c}^{1/3}=438\\ h&=\frac{Nu\,k}{L_c}=5.35\ \text{W/m}^2\text{K}\\ \dot q_{face}&=hA_s\Delta T=4516\ \text{W}\end{aligned}$$ Since the hung plate presents an identical vertical face on each side: $$\boxed{\dot Q_{vertical}=2\,\dot q_{face}=2(4516)=9032\ \text{W}}$$

Quantity	Value
Film temperature, $T_f$	362.5 K
Horizontal: $q_{top}$	6774 W
Horizontal: $q_{bottom}$	2120 W
Horizontal: total initial heat loss	8894 W
Vertical: $q$ per face	4516 W
Vertical: total initial heat loss (2 faces)	9032 W

Check

"Initial" heat loss is taken at the instant the plate is removed, so the whole 2.5 mm section is still uniformly at 430 K (Biot number for a thin steel plate is small, so the lumped/uniform-temperature assumption is reasonable only at $t=0$, before any internal gradient develops). Edge losses (thickness 2.5 mm against a 2.5 m face) are neglected as negligible surface area (<0.5% of a face). The answer is the convective loss only: the exam supplies air properties (Table 1) and nothing else, so with no emissivity given the radiative component cannot be evaluated and is excluded by assumption — worth stating, because a bare steel surface at 430 K into 295 K surroundings would radiate on the same order as it convects, so 8.9 kW and 9.0 kW are lower bounds on the true initial heat loss rather than estimates of it.

Concept & Theory

Topic. Free (natural) convection from horizontal and vertical flat plates.

Key relations. $Ra_L=\dfrac{g\beta}{\nu\alpha}\Delta T L^3$; horizontal plate, hot face up: $Nu=0.54Ra^{1/4}$ ($10^4$–$10^7$) or $0.15Ra^{1/3}$ ($10^7$–$10^{11}$); hot face down: $Nu=0.27Ra^{1/4}$ ($10^5$–$10^{10}$); vertical plate: $Nu=0.59Ra^{1/4}$ (laminar, $Ra<10^9$) or $0.10Ra^{1/3}$ (turbulent, $Ra>10^9$).

Why this works. Buoyancy-driven flow detaches readily from a hot surface facing up (plumes rise straight off the plate, thinning the boundary layer and enhancing $h$) but is suppressed under a hot surface facing down (the buoyant fluid is trapped against the plate, thickening the boundary layer and reducing $h$) — the same physical distinction is why an "up" and a "down" face need different empirical constants even though $Ra$ is otherwise identical.

Common pitfall. Using the plate's shorter side (or diameter-style length) as $L_c$ for a horizontal plate instead of the correct $A_s/P$; forgetting that a plate hung in open air loses heat from both faces, not one.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer, Ch. 9 (free convection, horizontal and vertical plate correlations).

Question 2 of 5: Natural-Convection Heat Loss From a Steel Plate

Question 2: Natural-Convection Heat Loss From a Steel Plate (25 marks: (a) 15, (b) 10)