Question 1 of 6: Two-Column Benzene–Toluene–Xylene Distillation Train

Question A1: Two-Column Benzene–Toluene–Xylene Distillation Train (Part A — equal value)

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. Feed (basis $F=100$ mol): 30 mol benzene (Bz), 25 mol toluene (Tol), 45 mol xylene (Xy). Column 1 bottoms $W_1$: 98 mol% Xy, zero Bz, and captures 96% of the feed xylene. Column 1 overhead $D_1$ feeds Column 2. Column 2 overhead $D_2$: 94 mol% Bz, remainder Tol (no Xy), containing 97% of the benzene fed to Column 2. Column 2 bottoms $=W_2$.

Find. (a) A labelled flowchart, a degree-of-freedom (DOF) proof that the system is fully determined, and the ordered solution sequence; (b) the % of feed benzene reporting to $D_2$; (c) the % of feed toluene reporting to $W_2$.

Approach. This is a mixed part: a DOF argument first, then two short recovery calculations. Representing every stream by its component molar flows, count unknowns against independent balances and specifications to show DOF = 0; because all Column 1 specifications reference only the (known) feed, the streams solve sequentially with no tear (simultaneous) equations.

(a) Flowchart, DOF analysis, and solution order

Describe each stream by its three component molar flows, so the unknowns are the component flows of $W_1$, $D_1$, $D_2$ and $W_2$ — $4\times3=12$ unknowns (the feed is fixed by the basis). The independent equations are:

Material balances (6): a component balance on each species about each column, $F=D_1+W_1$ and $D_1=D_2+W_2$ (3 species × 2 columns).

Specifications (6): (i) $n_{Bz,W_1}=0$; (ii) 96% xylene recovery, $n_{Xy,W_1}=0.96\,n_{Xy,F}$; (iii) $W_1$ purity, $n_{Xy,W_1}=0.98\,W_1$; (iv) 97% benzene recovery in Column 2, $n_{Bz,D_2}=0.97\,n_{Bz,D_1}$; (v) $D_2$ purity, $n_{Bz,D_2}=0.94\,D_2$; (vi) no xylene overhead, $n_{Xy,D_2}=0$.

Hence $\text{DOF}=12-(6+6)=\boxed{0}$: the process is exactly determined for the assumed basis. The equations are solved in the following order, each yielding the circled unknown(s):

1. Fix the basis. Take $\boxed{F=100\ \text{mol}}$, so $n_{Bz,F}=30,\ n_{Tol,F}=25,\ n_{Xy,F}=45$.
2. Xylene to $W_1$ (recovery spec). Solve for $\big(n_{Xy,W_1}\big)$: $\quad n_{Xy,W_1}=0.96\,n_{Xy,F}.$
3. Total bottoms $W_1$ (purity spec). Solve for $\big(W_1\big)$: $\quad n_{Xy,W_1}=0.98\,W_1.$
4. Toluene in $W_1$. With $n_{Bz,W_1}=0$, solve for $\big(n_{Tol,W_1}\big)=W_1-n_{Xy,W_1}$.
5. Overhead $D_1$ (three Column 1 component balances). Solve for $\big(n_{Bz,D_1},n_{Tol,D_1},n_{Xy,D_1}\big)=n_{i,F}-n_{i,W_1}$.
6. Benzene to $D_2$ (recovery spec). Solve for $\big(n_{Bz,D_2}\big)=0.97\,n_{Bz,D_1}$.
7. Total overhead $D_2$ and its toluene (purity spec). Solve for $\big(D_2\big)$ from $n_{Bz,D_2}=0.94\,D_2$, then $\big(n_{Tol,D_2}\big)=0.06\,D_2,\ n_{Xy,D_2}=0$.
8. Bottoms $W_2$ (three Column 2 component balances). Solve for $\big(n_{i,W_2}\big)=n_{i,D_1}-n_{i,D_2}$, completing every stream.

No step requires a previously-unknown quantity from a later step, confirming the train is sequential (no simultaneous solve). The numeric results of this sequence — needed for parts (b) and (c) — are shown on Figure A1.

(b) Benzene recovered overhead in Column 2

All feed benzene rises in Column 1 (the bottoms $W_1$ contains no benzene), so the benzene entering Column 2 equals the feed benzene, $n_{Bz,D_1}=n_{Bz,F}=30$ mol. Specification (iv) then gives directly:

$$\%\,\text{Bz to }D_2=\frac{n_{Bz,D_2}}{n_{Bz,F}}\times100=\frac{0.97\,(30)}{30}\times100=\boxed{97.0\%}.$$

The 97% column-recovery equals the feed-basis percentage precisely because no benzene is lost to $W_1$.

(c) Toluene recovered in the Column 2 bottoms

Working the ordered sequence numerically: $n_{Xy,W_1}=0.96(45)=43.2$, so $W_1=43.2/0.98=44.08$ mol and $n_{Tol,W_1}=44.08-43.2=0.882$ mol. Toluene up to Column 2 is $n_{Tol,D_1}=25-0.882=24.118$ mol. In Column 2, $D_2=0.97(30)/0.94=30.96$ mol carries $n_{Tol,D_2}=0.06(30.96)=1.857$ mol of toluene, so the toluene dropping to $W_2$ is

$$n_{Tol,W_2}=n_{Tol,D_1}-n_{Tol,D_2}=24.118-1.857=22.26\ \text{mol},$$

$$\%\,\text{Tol to }W_2=\frac{22.26}{25}\times100=\boxed{89.0\%}.$$