Sublimation Rate from a Falling Iodine Sphere: 23-Chem-A3 Heat and Mass Transfer, May 2014, Q2

Question 2: Sublimation Rate from a Falling Iodine Sphere (equal value)

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. Convective sublimation from a falling sphere; the supplied correlation gives the mass-transfer coefficient, and the iodine vapour pressure fixes the surface concentration.

Quantity	Value
$T$, $P$	348.15 K, 1.0 atm
Diameter $d_p$	1.5 mm; terminal $v=1.47$ m/s
$D_{AB}$ (I₂–air)	0.108 cm²/s $=1.08\times10^{-5}$ m²/s
Vapour pressure I₂	11.2 mm Hg $=1.49$ kPa
$M_{I_2}$, $M_{air}$	253.8, 28.8 g/mol
$\mu_{air}$	$2.07\times10^{-5}$ Pa·s

Find. The rate of mass loss (mg/s) at $d_p=1.5$ mm.

Figure 2 — A solid iodine sphere falls through air at its terminal velocity; iodine sublimes from the whole surface into the surrounding air, whose bulk iodine concentration is essentially zero.

Approach. Evaluate $Re$ and $Sc$ for the sphere, apply the supplied correlation to get $k^{\prime}_c$, take the surface concentration from the vapour pressure (bulk $\approx 0$), and multiply the flux by the sphere area and molar mass.

Air density. $\rho = PM_{air}/RT = (101{,}325)(0.0288)/(8.314\times348.15) = 1.008\ \text{kg/m}^3.$
Reynolds and Schmidt numbers. $Re = \dfrac{\rho v d_p}{\mu} = \dfrac{(1.008)(1.47)(1.5\times10^{-3})}{2.07\times10^{-5}} = 107.4$; $\quad Sc = \dfrac{\mu}{\rho D_{AB}} = \dfrac{2.07\times10^{-5}}{(1.008)(1.08\times10^{-5})} = 1.90.$
Sherwood number and coefficient. $Sh = 0.768[Re\,Sc^{0.5}]^{0.62} = 0.768[107.4\sqrt{1.90}]^{0.62} = 17.0$, so $$k^{\prime}_c = \frac{Sh\,D_{AB}}{d_p} = \frac{17.0(1.08\times10^{-5})}{1.5\times10^{-3}} = \boxed{0.123\ \text{m/s}}.$$
Surface concentration of I₂ vapour. $c_{As} = p^{vap}/RT = 1493/(8.314\times348.15) = 0.516\ \text{mol/m}^3$ (bulk air $c_{A\infty}\approx0$).
Mass-loss rate. With sphere area $A=\pi d_p^2 = 7.07\times10^{-6}\ \text{m}^2$, $$\dot m = k^{\prime}_c\,c_{As}\,(\pi d_p^2)\,M_{I_2} = (0.123)(0.516)(7.07\times10^{-6})(253.8) = \boxed{0.113\ \text{mg/s}}.$$

Quantity	Result
$k^{\prime}_c$	0.123 m/s
Surface conc. $c_{As}$	0.516 mol/m³
Mass-loss rate	≈ 1.13×10⁻⁴ g/s (0.113 mg/s)

Concept & Theory

Topic. Convective mass transfer (sublimation) from a sphere in forced flow.

Key relations. $Sh=k^{\prime}_c d_p/D_{AB}=0.768[Re\,Sc^{0.5}]^{0.62}$; $c_{As}=p^{vap}/RT$; $\dot m=k^{\prime}_c c_{As}\pi d_p^2 M$.

Why this works. Iodine leaves as a dilute vapour into a large air stream, so the far-field concentration is zero and the surface concentration is set by equilibrium (the vapour pressure). The empirical Sherwood correlation converts the flow field into a film coefficient, and the flux times the sphere area and molar mass gives the mass-loss rate.

Common pitfall. Forgetting to convert the vapour pressure (mm Hg) to Pa before dividing by $RT$, or leaving the coefficient in cm/s; also, $c_{A\infty}$ is not exactly zero only if the air is recirculated.

Source. Welty et al., Momentum, Heat and Mass Transfer §30 (sphere correlations); Geankoplis §7.3.

Question 2 of 7: Sublimation Rate from a Falling Iodine Sphere

Question 2: Sublimation Rate from a Falling Iodine Sphere (equal value)