Question 2 of 6: Free vs. Forced Convection from a Vertical Plate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A (Q1–Q3) Heat Transfer and Part B (Q1–Q3) Mass Transfer; at least two questions must be attempted from each part and only the first two in each part are marked, so four questions (each 25 points) constitute a complete paper. All six questions are solved below for completeness. Property values not printed on the paper (molar masses, water latent heat, the dimensionless free-convection peak velocity $f'_{max}$, benzene/toluene physical properties) are stated explicitly in each Given block as open-book look-ups.
Reference texts: Coulson & Richardson (Backhurst, Harker & Richardson), Chemical Engineering, Vol. 1 — Fluid Flow, Heat Transfer and Mass Transfer (6th ed., Butterworth-Heinemann) — the source family for the crystalliser, tube-condenser, Stefan-tube and distillation problems; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (free- and forced-convection correlations, the Ostrach similarity solution); Treybal, Mass-Transfer Operations (3rd ed.) and McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.) — Stefan diffusion, Chilton–Colburn analogy, McCabe–Thiele; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.
The obscured verb in Part B Q1 is taken as “calculate”, and the Part B Q2 water viscosity at 27 °C is $8.76\times10^{-4}$ Pa·s. One column of the Part A Q2 air-property table, the buoyancy group $g\beta/\nu^{2}$, is internally inconsistent: its printed values are about 40% below what $\rho$, $\mu$ and $\beta=1/T_f$ give. That column is not used; the Grashof number is built from the $\rho$ and $\mu$ columns and $\beta=1/T_f$.
Question 2: Free vs. Forced Convection from a Vertical Plate (Part A — 25 pts)
Given. A $0.15\times0.15$ m isothermal plate at $T_s=40$ °C in still air at $T_\infty=20$ °C. Air properties are taken at the film temperature $T_f=30$ °C $=303$ K, interpolated from the paper’s table.
Air property at $T_f=303$ K
Value
Density $\rho$
1.166 kg/m³
Viscosity $\mu$
1.867×10⁻⁵ Pa·s
Conductivity $k$
0.02636 W/m·K
Kinematic $\nu=\mu/\rho$
1.602×10⁻⁵ m²/s
Thermal diffusivity $\alpha$
2.248×10⁻⁵ m²/s
Prandtl $Pr=\nu/\alpha$
0.713
Find. (i) the free-convection heat-transfer rate; (ii) the maximum velocity in the free-convection boundary layer; (iii) the forced-convection rate at twice that velocity, and the ratio of the two.
Figure 2 — Buoyancy drives a rising film up the hot plate. The velocity profile $u(y)$ climbs from zero at the wall to a peak $u_{max}$ inside the boundary layer, then decays to the still air; the forced-convection case blows air across the same plate at $2\,u_{max}$.
Approach. Build the Grashof and Rayleigh numbers from the reliable property columns, get the free-convection $h$ from a laminar vertical-plate correlation; extract $u_{max}$ from the Ostrach similarity solution; then treat the plate as a flat plate in parallel flow at $V=2u_{max}$ and compare.
Grashof and Rayleigh numbers. With $\beta=1/T_f$ and $L=0.15$ m, $\Delta T=20$ K:$$Gr_L=\frac{g\beta\Delta T L^{3}}{\nu^{2}}=\frac{(9.81)(1/303)(20)(0.15)^3}{(1.602\times10^{-5})^2}=8.52\times10^{6},\quad Ra_L=Gr_L\,Pr=\boxed{6.07\times10^{6}}.$$Since $Ra_L<10^{9}$ the film is laminar.
Free-convection coefficient. Using the laminar vertical-plate relation $\overline{Nu}=0.59\,Ra_L^{1/4}$:$$\overline{Nu}=0.59(6.07\times10^{6})^{1/4}=29.3,\quad h_{free}=\frac{\overline{Nu}\,k}{L}=\frac{29.3(0.02636)}{0.15}=\boxed{5.15\ \text{W/m}^2\text{K}}.$$The area is $A=0.15^2=0.0225$ m², so $Q_{free}=h_{free}A\Delta T=5.15(0.0225)(20)=\boxed{2.32\ \text{W}}.$
Maximum velocity in the buoyant film. Ostrach’s similarity solution gives $u=(2\nu/x)\,Gr_x^{1/2}\,f'(\eta)$, whose dimensionless profile $f'(\eta)$ peaks at $f'_{max}\approx0.275$ for $Pr\approx0.7$. Evaluated at the top of the plate ($x=L$, largest $Gr_x$):$$u_{max}=\frac{2\nu}{L}Gr_L^{1/2}f'_{max}=\frac{2(1.602\times10^{-5})}{0.15}(8.52\times10^{6})^{1/2}(0.275)=\boxed{0.171\ \text{m/s}}.$$The forced-convection test velocity is therefore $V=2u_{max}=0.343$ m/s.
Forced convection at $V=0.343$ m/s. The Reynolds number over the 0.15 m plate is$$Re_L=\frac{VL}{\nu}=\frac{0.343(0.15)}{1.602\times10^{-5}}=3.21\times10^{3}\;(<5\times10^{5},\ \text{laminar}),$$so the average $\overline{Nu}=0.664\,Re_L^{1/2}Pr^{1/3}=0.664(3210)^{1/2}(0.713)^{1/3}=33.6$, giving$$h_{forced}=\frac{33.6(0.02636)}{0.15}=\boxed{5.91\ \text{W/m}^2\text{K}},\qquad Q_{forced}=5.91(0.0225)(20)=\boxed{2.66\ \text{W}}.$$
Comparison. The ratio is$$\frac{Q_{forced}}{Q_{free}}=\frac{5.91}{5.15}=\boxed{1.15}.$$Even at twice the natural peak velocity, forced convection beats free convection by only ~15% — buoyancy is already doing most of the work at this small $\Delta T$.
Check (which data are trusted): the paper’s $g\beta/\nu^{2}$ column is discarded (it is ~40% low and internally inconsistent, as flagged above); $Gr$ is built from $\rho,\mu$ and $\beta=1/T_f$ instead. Using the Churchill–Chu correlation instead of $0.59\,Ra^{1/4}$ lowers $h_{free}$ by ~10% but leaves the forced/free ratio near unity — the qualitative comparison is robust.
