Question 2 of 7: Fall time of an acetone / dibutyl-phthalate level
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC 04-CHEM-A3 Mass Transfer Operations, May 2016 — three-hour open-book exam. Seven questions in three parts: Part A (Q1–2, molecular diffusion), Part B (Q3–4, film mass transfer), Part C (Q5–7, staged and equilibrium separations). The candidate answers one of Q1–2, one of Q3–4 and two of Q5–7 (four questions of equal value). All seven are worked in full below.
Reference texts. J.M. Coulson & J.F. Richardson, Chemical Engineering Vol. 1 (Fluid Flow, Heat and Mass Transfer) and Vol. 2 (Particle Technology & Separation Processes) — the source of all seven problems; C.J. Geankoplis, Transport Processes and Separation Process Principles; R.E. Treybal, Mass-Transfer Operations; Bird, Stewart & Lightfoot, Transport Phenomena; Perry’s Chemical Engineers’ Handbook.
Question 2 — Fall time of an acetone / dibutyl-phthalate level (25 marks)
Given. Tube $d=6\text{ mm}$ ($A=2.83\times10^{-5}\text{ m}^2$); starting path $L_0=11.5\text{ mm}$; acetone (A) evaporates through stagnant air while DBP (non-volatile) remains. As A leaves, both the path $L$ lengthens and the acetone mole fraction $x_A$ (hence surface pressure $p_A=x_Ap_A^{*}$) falls.
Find. The time for the surface to drop from 11.5 mm to 50 mm below the top.
Property
Acetone (A)
DBP
Initial volume
2 cm³
2 cm³
Density
764 kg/m³
1048 kg/m³
Molar mass
58 g/mol
279 g/mol
Initial moles
0.02634 mol
0.00751 mol
Only acetone leaves; the surface recedes as its volume shrinks, lengthening the diffusion path while $x_A$ (and $p_A=x_Ap^{*}_A$) declines.
Approach. Because bulk flow is neglected, use the simple Fick flux across the path; express the path length and the Raoult surface pressure as functions of the remaining acetone moles, then integrate the rate equation numerically from the final to the initial amount.
Rate of acetone loss (Fick, no bulk flow). With $p_A(\text{top})=0$,
$$-\frac{dn_A}{dt}=N_A A=\frac{D_{AB}A}{RT\,L}\,p_A=\frac{D_{AB}A}{RT\,L}\,x_A\,p_A^{*}.$$
Geometry: path length vs. remaining acetone. DBP volume is fixed, so the surface set by the acetone volume gives
$$L(n_A)=L_0+\frac{(n_{A,0}-n_A)\,M_A/\rho_A}{A},\qquad x_A=\frac{n_A}{n_A+n_{\text{DBP}}}.$$
End states. Initially $x_{A,0}=\dfrac{0.02634}{0.02634+0.00751}=0.778$. The level reaches 50 mm when the acetone volume has fallen enough to recede the surface by $38.5$ mm, i.e. $n_{A,f}=0.01201$ mol and $x_{A,f}=0.615$.
Integrate the fall time. Separating variables,
$$t=\int_{n_{A,f}}^{n_{A,0}}\frac{RT\,L(n_A)\,(n_A+n_{\text{DBP}})}{D_{AB}A\,p_A^{*}\,n_A}\;dn_A .$$
Numerical (trapezoidal) evaluation gives
$$\boxed{t\approx7.9\times10^{4}\ \text{s}=22.0\ \text{hours}.}$$
The integral is dominated by the late stages, where the path is longest and $x_A$ (so the driving pressure) is smallest — both slow the evaporation.
