Question 4 of 4: Natural Convection Cooling of a Tank Wall
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A3 Heat Engineering, National Exams
May 2013 — a three-hour open-book examination; any non-communicating
calculator is permitted. The cover page states that four (4) questions constitute a
complete exam paper and that only the first four as they appear in the answer book are
marked, that each question is of equal value, and that all questions require calculation.
All four printed problems are worked here.
Reference texts. Y.A. Çengel & A.J. Ghajar, Heat and Mass
Transfer: Fundamentals and Applications, 5th ed. (conduction, flat-plate solar
collectors); J.P. Holman, Heat Transfer, 10th ed. (natural-convection
correlations and boundary-layer thickness); R.F. Barron, Cryogenic Heat Transfer,
2nd ed. (radiation shields for cryogenic lines).
Problem 4: Natural Convection Cooling of a Tank Wall (25 points)
Find. The average convection coefficient h, the average heat flux q, and
the thermal boundary-layer thickness δ at the top of the wall (x = L).
Approach. Because the internal coefficient is very large, the outer wall
surface sits essentially at the tank fluid temperature (40°C), so this is an external
natural-convection problem on a vertical isothermal plate. Form the Grashof and Rayleigh
numbers at the given film-temperature properties, apply the laminar vertical-plate
correlation for the average Nusselt number, then use the companion integral-solution result
for the boundary-layer thickness at the top of the plate.
Natural-convection boundary layer on the tank's vertical side: warm air
rises from the bottom leading edge, thickening to δ ≈ 17.1 mm by the top
(x = L = 0.4 m).
Film temperature and driving ΔT. The wall runs at the tank fluid
temperature (internal h very large), so $T_s = 40\,{}^{\circ}\text{C}$,
$T_\infty=14\,{}^{\circ}\text{C}$, $\Delta T = 26\,\text{K}$; the given properties are
already evaluated at the correct film temperature,
$T_f=(40+14)/2=27\,{}^{\circ}\text{C}$.
Grashof and Rayleigh numbers. Using the full height as the
characteristic length ($L=0.4\ \text{m}$),
$$Gr_L=\frac{g\beta\Delta T L^3}{\nu^2}=\frac{9.81(0.00348)(26)(0.4)^3}{(1.556\times10^{-5})^2}
= 2.35\times10^{8}$$
$$Ra_L = Gr_L\,Pr = 2.35\times10^{8}\times0.711 = \boxed{1.67\times10^{8}}$$
This falls inside the laminar range for a vertical plate, $10^4 \lt Ra_L \lt 10^9$.
Average Nusselt number and convection coefficient. With the
Rayleigh number in the laminar range, McAdams' correlation applies,
$$\overline{Nu}_L = 0.59\,Ra_L^{1/4} = 0.59(1.67\times10^{8})^{1/4} = 67.1$$
$$h = \frac{\overline{Nu}_L\,k}{L}$$
Air's thermal conductivity at 300 K (27°C), consistent with the given ν and
α, is $k \approx 0.0262\ \text{W/m}\cdot\text{K}$ (standard air-property table), so
$$h = \frac{67.1\times0.0262}{0.4} = \boxed{4.40\ \text{W/m}^2\cdot\text{K}}$$
Average heat flux.
$$q = h\,\Delta T = 4.40\times26 = \boxed{114.4\ \text{W/m}^2}$$
Thermal boundary-layer thickness at the top (x = L). The companion
integral-solution result for the laminar natural-convection boundary layer on a vertical
isothermal plate gives
$$\frac{\delta}{x} = 3.93\,Pr^{-1/2}(0.952+Pr)^{1/4}Gr_x^{-1/4}$$
Evaluated at $x=L=0.4\ \text{m}$ (using $Gr_L$ found above),
$$\delta = 0.4\times3.93\times(0.711)^{-1/2}\times(0.952+0.711)^{1/4}\times
(2.35\times10^{8})^{-1/4} = \boxed{0.0171\ \text{m} = 17.1\ \text{mm}}$$
Quantity
Result
Rayleigh number, RaL
1.67×10⁸
Average convection coefficient, h
4.40 W/m²·K
Average heat flux, q
114.4 W/m²
Boundary-layer thickness at top, δ
17.1 mm
Check: air's thermal conductivity k is not printed in the source data
(only ν, α and Pr are given). k ≈ 0.0262 W/m·K is the standard
air-property-table value at 300 K, consistent with the given ν and α (their ratio
reproduces the stated Pr to within rounding) — this is the value an open-book
candidate would read from the same reference table.
