Question 1 of 7: Diffusivity by the falling-level (Winkelmann) method
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 1 — Diffusivity by the falling-level (Winkelmann) method (25 marks)
Given. CCl4 evaporating through stagnant air in a vertical tube; $T=321\text{ K}$, $P=101.3\text{ kN/m}^2$, saturation pressure $p_A^{*}=37.6\text{ kN/m}^2$, $\rho_L=1540\text{ kg/m}^3$, $M=153.8\text{ g/mol}$. Air passes over the mouth so $p_A\approx0$ there. The measured level-drop $x$ grows with time $t$ (10 data points).
Find. The gas-phase diffusivity $D_{AB}$ of CCl4 vapour in air.
$t$ (s)
1600
11 000
27 400
80 200
117 500
168 600
199 700
289 300
383 100
$x$ (cm)
0.25
1.29
2.32
4.39
5.47
6.70
7.38
9.03
10.48
Winkelmann tube: CCl4 diffuses up the growing air column of length $L(t)$ to the swept mouth where $p_A\approx0$.
Approach. Write the steady Stefan flux of A through stagnant B over the instantaneous path $L$, equate it to the rate the liquid surface recedes, integrate, and recognise a straight line of $t/x$ against $x$ whose slope yields $D_{AB}$ independently of the unknown initial path length.
Stefan flux of A through stagnant B. With B (air) not transferring, the molar flux across a path of length $L$ is
$$N_A=\frac{D_{AB}\,C_T}{L}\,\ln\frac{C_{B2}}{C_{B1}}=\frac{D_{AB}\,C_T}{L}\,\ln\frac{P}{P-p_A^{*}},$$
where $C_T=P/RT$ is the total molar concentration and subscripts 1, 2 denote the liquid surface and the mouth.
Couple the flux to the receding surface. Each mole leaving lowers the liquid by $M/\rho_L$ per unit area, so $N_A=\dfrac{\rho_L}{M}\dfrac{dL}{dt}$. Equating and separating variables,
$$\frac{\rho_L}{M}\,L\,dL=C_T D_{AB}\ln\!\frac{P}{P-p_A^{*}}\;dt .$$
Integrate over the drop $x=L-L_0$. Integrating from $L_0$ to $L_0+x$ gives a quadratic in $x$ which rearranges to a straight line:
$$\frac{t}{x}=\underbrace{\frac{\rho_L L_0}{M\,K}}_{\text{intercept}}+\underbrace{\frac{\rho_L}{2M\,K}}_{\text{slope}}\,x,\qquad K=C_T D_{AB}\ln\frac{P}{P-p_A^{*}} .$$
Plotting $t/x$ against $x$ removes the unknown $L_0$ entirely — only the slope carries $D_{AB}$.
Evaluate the constants. $C_T=\dfrac{101\,325}{8.314\times321}=37.97\text{ mol/m}^3$, $\ \ln\dfrac{P}{P-p_A^{*}}=\ln\dfrac{101.3}{63.7}=0.4638$, $\ \dfrac{\rho_L}{M}=\dfrac{1540}{0.1538}=1.001\times10^{4}\text{ mol/m}^3.$
Least-squares slope and diffusivity. The regression of the nine points (figure) gives slope $=2.99\times10^{7}\text{ s/m}^2$. Hence
$$D_{AB}=\frac{\rho_L/M}{2\,C_T\ln\!\frac{P}{P-p_A^{*}}\;\text{slope}}=\frac{1.001\times10^{4}}{2(37.97)(0.4638)(2.99\times10^{7})}.$$
$$\boxed{D_{AB}\approx9.5\times10^{-6}\ \text{m}^2/\text{s}}$$
The intercept implies an initial path $L_0\approx8.5$ mm, a sensible tube geometry that confirms the fit.
The data fall on a straight line $t/x$ vs. $x$; the slope fixes $D_{AB}=9.5\times10^{-6}\text{ m}^2/\text{s}$, close to the literature value for CCl4–air.
