C2 — Evaporation time of a suspended toluene drop: 23-Chem-B1 Transport Phenomena, December 2014, Q6

Question 6: C2 — Evaporation time of a suspended toluene drop

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. Toluene drop evaporating into still air; quasi-steady Stefan diffusion of $A$ through stagnant $B$.

Symbol	Value
Temperature $T$	$25.9\ \text{°C}=299.05\ \text{K}$
Initial radius $r_1$; total $P$	$2\times10^{-3}\ \text{m}$; $101{,}325\ \text{Pa}$
$P_{A1}$ (surface); $P_{A2}$ (far)	$3840\ \text{Pa}$; $\approx0$
$\rho_A,\ M_A,\ D_{AB}$	$866\ \text{kg/m}^3,\ 0.09214\ \text{kg/mol},\ 8.6\times10^{-6}\ \text{m}^2/\text{s}$

Find. (a) derive the closed-form $t_f$; (b) evaluate it numerically.

Fig. C2: A toluene drop on a wire evaporates by Stefan diffusion of vapour $A$ radially outward through stagnant air $B$. Quasi-steady transport plus a shrinking-drop mass balance yields the $r_1^2$ (“$d^2$”) evaporation law.

Approach. Treat the vapour field as quasi-steady (the drop shrinks slowly), write the spherical Stefan flux of $A$ through stagnant $B$, then equate the surface molar flow to the rate the liquid drop loses moles and integrate the radius from $r_1$ to zero.

Quasi-steady Stefan flux. For diffusion of $A$ through stagnant $B$ radially outward from the drop, the molar flow $W_A=4\pi r^2 N_A$ is constant; integrating from the surface ($r_1$) into the still air gives the surface flux $$N_{A1}=\frac{D_{AB}\,P}{R\,T\,r_1\,P_{BM}}\,(P_{A1}-P_{A2}),\qquad P_{BM}=\frac{P_{B2}-P_{B1}}{\ln(P_{B2}/P_{B1})},$$ where $P_B=P-P_A$ is the stagnant-air partial pressure and $P_{BM}$ its log-mean.
Liquid mass balance. The drop loses moles as its radius shrinks: $$4\pi r_1^2\,N_{A1}=-\frac{\rho_A}{M_A}\frac{d}{dt}\!\left(\tfrac{4}{3}\pi r_1^3\right)=-\frac{\rho_A}{M_A}\,4\pi r_1^2\,\frac{dr_1}{dt}\ \Rightarrow\ N_{A1}=-\frac{\rho_A}{M_A}\frac{dr_1}{dt}.$$
Separate and integrate. Equating the two flux expressions and separating variables, $$r_1\,dr_1=-\frac{M_A D_{AB} P\,(P_{A1}-P_{A2})}{\rho_A R T\,P_{BM}}\,dt .$$ Integrating $r_1:\,r_1\!\to\!0$ over $t:\,0\!\to\!t_f$ gives $\tfrac12 r_1^2$ on the left and delivers $$\boxed{\,t_f=\frac{\rho_A\,r_1^2\,R\,T\,P_{BM}}{2\,M_A\,D_{AB}\,P\,(P_{A1}-P_{A2})}\,}\qquad(\text{part a}).$$
Evaluate the log-mean (part b). $P_{B1}=P-P_{A1}=97{,}485\ \text{Pa}$, $P_{B2}=P=101{,}325\ \text{Pa}$: $$P_{BM}=\frac{101{,}325-97{,}485}{\ln(101{,}325/97{,}485)}=99{,}393\ \text{Pa}.$$
Compute the time. With $P_{A2}=0$, $$t_f=\frac{(866)(2\times10^{-3})^2(8.314)(299.05)(99{,}393)}{2(0.09214)(8.6\times10^{-6})(101{,}325)(3840)}=\boxed{1388\ \text{s}\approx23.1\ \text{min}} .$$

Quantity	Value
Log-mean air pressure $P_{BM}$	$99{,}393\ \text{Pa}$
Evaporation time $t_f$	$1388\ \text{s}\approx23.1\ \text{min}$

Concept & Theory

Topic. Quasi-steady Stefan diffusion and the $d^2$ (or $r_1^2$) evaporation law.

Key relations. Spherical Stefan flux $N_{A1}=D_{AB}P(P_{A1}-P_{A2})/(RTr_1P_{BM})$; drop balance $N_{A1}=-(\rho_A/M_A)\,dr_1/dt$; $P_{BM}$ = log-mean of the stagnant-gas partial pressure.

Why this works. Because the vapour field relaxes far faster than the drop shrinks, the concentration profile is effectively steady at each instant (quasi-steady). Stefan flow (the induced bulk motion of air away from the surface) is captured by the $P_{BM}$ factor; integrating the resulting $r_1\,dr_1\propto dt$ gives the square-radius law — the mass-transfer twin of the droplet-combustion $d^2$ law.

Common pitfall. Dropping $P_{BM}$ (i.e. using the dilute Fick flux) — valid only when $P_{A1}\ll P$; here $P_{BM}/P=0.981$, so the correction is small but should be shown. Keep $R T$ in SI ($8.314\ \text{J/mol}\cdot\text{K}$, $T$ in kelvin) and pressures in pascals.

Source. Geankoplis, Ch. 6 (diffusion of $A$ through stagnant $B$, evaporating drops); Bird, Stewart & Lightfoot, §18.2 (Stefan diffusion).

Question 6 of 6: C2 — Evaporation time of a suspended toluene drop

Question 6: C2 — Evaporation time of a suspended toluene drop