Flash vs. Differential Distillation of Benzene–Toluene: 23-Chem-A2 Unit Operations and Separation Processes, December 2018, Q4

Question B1: Flash vs. Differential Distillation of Benzene–Toluene (25 points)

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.

Quantity	Value
Feed $F$ / composition $z$	100 mol / $z=0.70$ benzene
Fraction vaporized $f=V/F$	$1/3$ (so $W/F=2/3$)
Relative volatility $\alpha$	2.5 (benzene/toluene)
Pressure	1 atm

Find. Distillate (vapour) and bottoms (liquid) benzene mole fractions for (a) an equilibrium flash and (b) a Rayleigh (differential) distillation, both removing one-third of the feed as vapour.

Figure B1 — Flash: the operating line $z=f\,y+(1-f)x$ intersects the equilibrium curve at a single point. Differential (Rayleigh) removes vapour continuously, so the still liquid follows the curve and the average distillate is richer.

Approach. Both use the constant-$\alpha$ equilibrium $y=\dfrac{\alpha x}{1+(\alpha-1)x}$. The flash couples that with a single overall mole balance; the differential case integrates the Rayleigh equation (which has a closed form for constant $\alpha$) and recovers the average distillate by mole balance.

Equilibrium relation. With $\alpha=2.5$, $$y=\frac{2.5\,x}{1+1.5\,x}.$$
(a) Flash — operating balance. A component balance on benzene with $f=V/F=1/3$ gives $z=f\,y+(1-f)x$. Substituting $y(x)$ and $z=0.70$: $$0.70=\tfrac13\cdot\frac{2.5x}{1+1.5x}+\tfrac23\,x.$$ Solving this single equation, $$x_W=\boxed{0.641},\qquad y_D=\frac{2.5(0.641)}{1+1.5(0.641)}=\boxed{0.817}.$$ Check: $\tfrac13(0.817)+\tfrac23(0.641)=0.700$ ✓.
(b) Differential — Rayleigh integral (closed form). For constant $\alpha$ the Rayleigh equation $\ln\dfrac{F}{W}=\displaystyle\int_{x_W}^{z}\frac{dx}{y-x}$ integrates to $$\ln\frac{F}{W}=\frac{1}{\alpha-1}\ln\!\frac{z(1-x_W)}{x_W(1-z)}+\ln\frac{1-x_W}{1-z}.$$ With $F/W=100/66.7=1.5$, $\ln1.5=0.405$; solving for $x_W$ gives $$x_W=\boxed{0.633}.$$
Average distillate composition. An overall benzene balance closes the differential case: $$y_{D,\text{avg}}=\frac{Fz-Wx_W}{F-W}=\frac{100(0.70)-66.7(0.633)}{33.3}=\boxed{0.835}.$$ Because the vapour is withdrawn continuously (always in equilibrium with an ever-leaner still), the differential distillate is slightly richer (0.835) and the bottoms slightly leaner (0.633) than the single-stage flash (0.817 / 0.641) for the same cut.

Quantity	Flash (a)	Differential (b)
Bottoms $x_W$ (benzene)	0.641	0.633
Distillate $y_D$ (benzene)	0.817	0.835 (avg)
Cut $V/F$	1/3

Concept & Theory

Topic. Single-stage flash vs. Rayleigh (differential/batch) distillation with constant relative volatility.

Key relations. Equilibrium $y=\alpha x/[1+(\alpha-1)x]$; flash balance $z=(V/F)y+(1-V/F)x$; Rayleigh $\ln(F/W)=\int_{x_W}^{x_F}dx/(y-x)$, with the constant-$\alpha$ closed form above; average distillate from $Fz=Wx_W+Dy_{D,\text{avg}}$.

Why this works. In a flash the vapour and liquid leave in mutual equilibrium at one composition, so a single balance suffices. In differential distillation the vapour is removed the instant it forms, so the still liquid moves continuously down the equilibrium curve; integrating the instantaneous balance over the batch (Rayleigh) tracks that path, and an overall balance recovers the time-averaged distillate. The differential mode gives marginally sharper separation than one flash stage for the same cut.

Common pitfall. Applying the Rayleigh average distillate as if it were an equilibrium vapour, or using an arithmetic-mean $\alpha$ where composition varies widely; also forgetting that $W/F=1-V/F$ fixes the integration limit.

Source. Treybal ch. 9 (distillation); McCabe ch. 21; Geankoplis §11.3–11.4 (flash & simple distillation).

Question 4 of 6: Flash vs. Differential Distillation of Benzene–Toluene

Question B1: Flash vs. Differential Distillation of Benzene–Toluene (25 points)