Question 1 of 6: Process Flowsheet Diagram

Question 1: Process Flowsheet Diagram (20 marks)

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. A single-feed, two-product separation: reactor effluent (C₄–C₁₅ paraffins, feed rate 650,000 m³/yr) is split into an overhead motor alkylate (light, predominantly C₄–C₉, the automotive blendstock) and a bottom heavy alkylate (predominantly C₁₀–C₁₅, furnace fuel). Feed and product mole fractions are tabulated above. The stem locates the cut "between C₈ and C₁₀ fractions"; the tabulated product specifications place it more precisely, since C₈ and C₉ both report almost entirely overhead (recoveries of about 99 % and 88 % of the feed species) while only about 21 % of the C₁₀ does. The operating key pair is therefore C₉ (light key) / C₁₀ (heavy key), and C₉ is the distributed component.

Find. A process flow sheet for the alkylate-splitter module (distillation column and its ancillaries), plus the overall split of feed to distillate versus bottoms that the flow sheet must deliver.

Approach. A single sharp split between adjacent carbon numbers is a textbook single-column distillation; identify the column and its essential ancillaries (feed preheat, condenser, reflux drum, reboiler), then close a quick overall mole balance on the key components to fix the distillate-to-feed ratio the design must achieve.

(a) Overall split — how much goes overhead?

Let $\beta = D/F$ be the fraction of feed leaving as distillate (motor alkylate). An overall component balance on each species reads $z_i = \beta\,x_i + (1-\beta)\,w_i$, where $x_i$ and $w_i$ are the distillate and bottoms mole fractions. Because the printed product specifications are nominal (they do not close every species exactly), $\beta$ is obtained as the best fit over all twelve components:

1. Set up the least-squares estimate of the split. Writing $a_i = x_i-w_i$ and $b_i = z_i-w_i$, the balance $b_i=\beta a_i$ is over-determined, so $$\beta = \frac{\sum_i a_i b_i}{\sum_i a_i^{2}} = 0.913$$ The C₈ key alone gives the same picture: $0.730 = \beta(0.79)+(1-\beta)(0.11)\Rightarrow\beta = 0.91$, and the C₁₀ key $0.043 = \beta(0.01)+(1-\beta)(0.41)\Rightarrow\beta = 0.92$.
2. Interpret the result. About $\boxed{91\%}$ of the feed (mole basis) leaves overhead as motor alkylate and $\approx 9\%$ as heavy-alkylate bottoms — consistent with the feed being 73 mol% C₈ and dominated by species lighter than the C₉/C₁₀ cut point.

(b) Process flow sheet

The physical separation is a single distillation column (the alkylate splitter). The essential flow sheet is the column plus the four ancillaries that make a distillation practical: a feed surge drum, a feed preheater (or feed/bottoms exchanger) to bring the feed near its bubble point, an overhead condenser and reflux (accumulator) drum returning liquid reflux and drawing off distillate, and a reboiler supplying the boil-up. Motor alkylate is drawn from the reflux drum; heavy alkylate is drawn from the column bottoms.

Reading the flow sheet as a sequence of steps: (1) reactor effluent enters a feed surge drum that dampens upstream flow swings; (2) a feed preheater (economically, a feed/hot-bottoms exchanger) raises the feed to near its bubble point so it enters the column partially vaporised; (3) in the alkylate splitter the light C₄–C₉ paraffins rise and the heavy C₁₀+ fall, the split set between the C₉ light key and C₁₀ heavy key; (4) overhead vapour is totally condensed and collected in the reflux drum, from which reflux is returned to the top tray and net distillate is drawn as motor alkylate; (5) a reboiler vaporises part of the bottoms to provide boil-up, and the net liquid bottoms leave as heavy alkylate to furnace-fuel storage.