Question 5 of 6: Part B: Differential (Rayleigh) Distillation of Acetic Acid–Water
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: open-book, three-hour exam; Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6); all problems 25 points, at least two attempted per part. Every problem is worked in full below.
Reference texts: J. M. Coulson & J. F. Richardson, Chemical Engineering Vols. 1&2 (Coulson & Richardson) — boiling curves, tube-bank convection, gas absorption and column sizing; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free-convection correlations and the gas-property tables reproduced in the paper; J. R. Welty, C. E. Wicks & R. E. Wilson, Fundamentals of Momentum, Heat and Mass Transfer — the laminar vertical-plate local Nusselt relation; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — stagnant-film diffusion, differential (Rayleigh) distillation and packed-tower flooding; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 5 — Part B: Differential (Rayleigh) Distillation of Acetic Acid–Water (25 points)
Given. Basis $F=100$ mol charge, $x_F=0.25$ mole-fraction acetic acid. 60 % distilled, so residue $W=40$ mol and distillate $D=60$ mol. Acetic acid is the less volatile component ($y<x$ throughout the table), so the residue enriches in acid. Equilibrium ($x$ = liquid, $y$ = vapour mole fraction of acetic acid):
$x$
0.07
0.15
0.27
0.50
0.62
0.72
0.82
0.90
1.00
$y$
0.05
0.11
0.20
0.38
0.49
0.60
0.73
0.84
1.00
Find. The residue composition $x_W$ and the (average) distillate composition $y_D$.
Figure 5 — Batch (differential) distillation: vapour in instantaneous equilibrium with the pot liquid is continuously withdrawn. Acetic acid is heavy, so it concentrates in the pot; the Rayleigh integral tracks the falling — here rising in acid — pot composition.
Approach. Apply the Rayleigh equation $\ln(F/W)=\int dx/(x-y^{*})$ over the pot composition path, integrating numerically against the interpolated equilibrium curve to find $x_W$, then close an overall acid balance for the average distillate $y_D$.
Rayleigh equation. For differential distillation the pot mole number $L$ and composition $x$ obey
$$\ln\frac{F}{W}=\int_{x_W}^{x_F}\frac{dx}{x-y^{*}(x)},$$
where $y^{*}(x)$ is the equilibrium vapour. Here $60\%$ distilled gives $W=40$, so the left side is
$$\ln\frac{100}{40}=\ln 2.5=0.9163.$$
Set up the integral. Acetic acid is heavy ($y^{*}<x$), so $x-y^{*}>0$ and the pot enriches in acid: the residue composition $x_W$ lies above $x_F=0.25$. We integrate $1/(x-y^{*})$ from $x_F$ upward to the $x_W$ that makes the area equal 0.9163.
Evaluate numerically. Interpolating the table, sample $1/(x-y^{*})$ between $x=0.25$ and trial upper limits and accumulate the area (trapezoidal). The driving gap $x-y^{*}$ runs $\approx0.05\to0.13$ over the range, so the integrand is $\sim8$–$20$; the area reaches 0.9163 at
$$\boxed{x_W\approx 0.316\ \text{(acetic acid in the residue).}}$$
Distillate by overall balance. An acid balance $F x_F=W x_W+D\,y_D$ gives the average distillate composition:
$$y_D=\frac{F x_F-W x_W}{D}=\frac{(100)(0.25)-(40)(0.316)}{60}=\frac{25-12.6}{60}=\boxed{0.206.}$$
As expected the distillate is water-rich ($y_D<x_F$) and the residue acid-rich ($x_W>x_F$), consistent with acetic acid being the heavier component.
Quantity
Result
$\ln(F/W)$ (Rayleigh target)
0.916
Residue composition $x_W$
≈ 0.316 mol frac. acid
Average distillate $y_D$
≈ 0.206 mol frac. acid
Check
The residue composition depends only mildly on the numerical scheme: a coarser graphical evaluation of the Rayleigh integral typically returns $x_W$ in the range 0.31–0.32 and $y_D\approx0.20$–0.21, so the reported values are robust to the integration method.
