Question 6 of 6: Differential (Rayleigh) Distillation of Methanol–Water
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistance networks, critical radius, convection correlations, radiation exchange between surfaces; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — heat-exchanger sizing and diffusion through stagnant films; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential (Rayleigh) distillation and penetration-theory absorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6). The rubric asks for at least two problems from each part; every problem is solved here for completeness. Open-book, three-hour paper.
Part A — Heat Transfer
Question 6: Differential (Rayleigh) Distillation of Methanol–Water (25 points)
Given. Feed $x_F=0.40$ mole-fraction methanol (basis $F=1$ mol); 60% of the charge is distilled, so the residue is $W/F=0.40$ and distillate $D/F=0.60$. Equilibrium ($x,y^{*}$): (0.05, 0.27), (0.10, 0.42), (0.20, 0.57), (0.30, 0.66), (0.40, 0.73), (0.50, 0.78).
$x$ (liquid)
$y^{*}$ (vapour)
0.05
0.27
0.10
0.42
0.20
0.57
0.30
0.66
0.40
0.73
0.50
0.78
Find. The residue composition $x_W$ and the average distillate composition $y_{D,\text{avg}}$.
Figure 6 — Equilibrium $x$–$y$ diagram for methanol–water. The Rayleigh integral $\int_{x_W}^{x_F}dx/(y^{*}-x)$ is evaluated between the feed composition and the (lower) residue composition; the accumulated distillate $y_{D,\text{avg}}$ follows from an overall balance.
Approach. Apply the Rayleigh equation for a differential still, integrate the tabulated equilibrium numerically to locate the residue composition that consumes the required fraction of charge, then close the overall mole balance for the average distillate.
Rayleigh equation. For a differential (batch) distillation,$$\ln\frac{F}{W}=\int_{x_W}^{x_F}\frac{dx}{y^{*}-x}.$$With $W/F=0.40$, the left side is $\ln(1/0.40)=0.916$.
Integrate the equilibrium data. Interpolating the $x$–$y^{*}$ table and integrating $1/(y^{*}-x)$ downward from $x_F=0.40$, the lower limit is adjusted (by bisection) until the integral equals $0.916$. This gives$$x_W=\boxed{0.081}\ \text{mole fraction methanol}.$$
Average distillate by overall balance. On the basis $F=1$, $Fx_F=Wx_W+Dy_{D,\text{avg}}$:$$y_{D,\text{avg}}=\frac{x_F-(W/F)x_W}{D/F}=\frac{0.40-0.40(0.081)}{0.60}=\boxed{0.612}.$$
Sense check. The residue (0.081) is much leaner and the accumulated distillate (0.612) much richer than the feed (0.40), as expected when the more-volatile methanol is preferentially boiled off; the two compositions bracket the feed and balance it in the ratio $D:W=0.6:0.4$.
