Sublimation of a UF₆ Cylinder in Cross-flow: 23-Chem-A3 Heat and Mass Transfer, May 2017, Q5

Question 5: Sublimation of a UF₆ Cylinder in Cross-flow (25 points)

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. Cylinder $D = 0.01$ m, $L = 0.10$ m; air velocity $u = 1$ m/s normal to the axis; surface $303$ K with $p^\*_{UF_6} = 27$ kPa; bulk air $325$ K, 1 atm, containing no UF₆. $D_{AB} = 9.04\times10^{-6}\ \text{m}^2/\text{s}$, $\mu = 2.29\times10^{-5}\ \text{kg/m}\cdot\text{s}$, $M_{UF_6} = 352.0\ \text{g/mol}$.

Quantity	Value
Cylinder $D$ / $L$	0.01 m / 0.10 m
Air velocity	1 m/s (normal)
Surface T / $p^\*$	303 K / 27 kPa
$D_{AB}$ / $\mu$	$9.04\times10^{-6}$ m²/s / $2.29\times10^{-5}$ kg/m·s

Find. The mass rate at which the cylinder sublimes.

Figure 5 — UF₆ cylinder in a normal airstream. UF₆ vapour at its surface partial pressure (27 kPa) diffuses into the essentially UF₆-free air; the given Sherwood correlation is the mass-transfer twin of the Hilpert heat-transfer relation.

Approach. Evaluate $Re$ and $Sc$ for the crossflow, get $Sh$ and the mass-transfer coefficient $k_c$ from the given correlation, then multiply the surface–bulk concentration difference by $k_c$ and the lateral area.

Gas density and dimensionless groups. Take air at the bulk 325 K: $\rho = PM/RT = (101325)(0.02897)/(8.314\times325) = 1.086\ \text{kg/m}^3$. Then $$Re = \frac{\rho u D}{\mu} = \frac{1.086(1)(0.01)}{2.29\times10^{-5}} = 474,\qquad Sc = \frac{\mu}{\rho D_{AB}} = \frac{2.29\times10^{-5}}{1.086(9.04\times10^{-6})} = 2.33.$$
Sherwood number and $k_c$. $$Sh = 0.43 + 0.532(474)^{0.5}(2.33)^{0.31} = 15.5,\qquad k_c = \frac{Sh\,D_{AB}}{D} = \frac{15.5(9.04\times10^{-6})}{0.01} = \boxed{0.0140\ \text{m/s}}.$$
Surface concentration driving force. The bulk air carries no UF₆, so $c_{A\infty}=0$; at the surface $$c_{As} = \frac{p^\*}{R T_s} = \frac{27000}{8.314(303)} = 10.7\ \text{mol/m}^3.$$
Molar and mass sublimation rate. Lateral area $A = \pi D L = \pi(0.01)(0.10) = 3.14\times10^{-3}\ \text{m}^2$: $$\dot N = k_c\,c_{As}\,A = 0.0140(10.7)(3.14\times10^{-3}) = 4.72\times10^{-4}\ \text{mol/s},$$ $$\dot m = \dot N\,M_{UF_6} = 4.72\times10^{-4}(352.0) = \boxed{0.166\ \text{g/s} \approx 0.60\ \text{kg/hr}}.$$

Quantity	Result
$Re$ / $Sc$ / $Sh$	474 / 2.33 / 15.5
Mass-transfer coefficient $k_c$	0.0140 m/s
Surface concentration $c_{As}$	10.7 mol/m³
Sublimation rate	$4.72\times10^{-4}$ mol/s = 0.166 g/s ≈ 0.60 kg/hr

Concept & Theory

Topic. Convective mass transfer from a cylinder in cross-flow (sublimation), via the heat–mass transfer analogy.

Key relations. $Sh = 0.43 + 0.532Re^{0.5}Sc^{0.31}$ (mass-transfer analog of the Hilpert $Nu$); $k_c = Sh\,D_{AB}/D$; flux $N_A = k_c(c_{As}-c_{A\infty})$; surface concentration $c_{As}=p^\*/RT_s$.

Why this works. The Chilton–Colburn analogy lets an empirical Nusselt correlation be reused for mass transfer by swapping $Nu\!\to\!Sh$ and $Pr\!\to\!Sc$. The vapour pressure fixes the surface concentration; the swept air keeps the bulk at zero UF₆, so the full $c_{As}$ is the driving force. Multiplying the flux by the wetted area gives the sublimation rate.

Common pitfall. Evaluating $c_{As}$ at the bulk temperature instead of the surface temperature, or forgetting to convert $D_{AB}$ from cm²/s to m²/s. At $p^\*/P = 0.27$ a rigorous treatment would add a drift-flux (log-mean) correction; the low-flux film form is the standard exam estimate.

Source. Incropera & DeWitt §7.4 and §14.1 (external-flow convection and the heat–mass analogy); Geankoplis §7.3.

Question 5 of 6: Sublimation of a UF₆ Cylinder in Cross-flow

Part A — Heat Transfer

Question 5: Sublimation of a UF₆ Cylinder in Cross-flow (25 points)