Question 5 of 7: Distillation of Ethanol–Water with a Liquid Side Stream
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.
Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 5: Distillation of Ethanol–Water with a Liquid Side Stream (Part C — equal value)
Given. Basis $F=100$ mol, $z_F=0.16$, $x_D=0.77$, $x_W=0.02$, side stream $x_S=0.50$ carrying 25% of the feed ethanol; feed a saturated liquid ($q=1$); reflux ratio $R=2$.
Find. the number of theoretical plates and the side-draw plate location.
Approach. Close the overall and component balances for $D$, $S$, $W$; write the three operating lines (rectifying above the draw, middle between draw and feed, stripping below feed); then step off stages against the equilibrium curve.
Product, side and bottoms flows. Feed ethanol $=16$ mol; side ethanol $=0.25(16)=4$ mol, so $S=4/0.50=8$ mol. Overall $D+W=92$ and ethanol $0.77D+0.02W=12$ give $$\boxed{D=13.55,\quad S=8.0,\quad W=78.45\ \text{mol}}.$$
Rectifying line (above the side draw). $L=RD=27.1$, $V=L+D=40.65$: $$y=\tfrac{L}{V}x+\tfrac{D x_D}{V}=0.667x+0.257.$$
Middle line (between side draw and feed). The liquid side stream lowers the internal liquid to $L'=L-S=19.1$ (vapour unchanged): $$y=\tfrac{L'}{V}x+\tfrac{Dx_D+Sx_S}{V}=0.470x+0.355.$$ It meets the rectifying line at $x=x_S=0.50$, so the draw is taken there.
Stripping line (below feed). The saturated-liquid feed adds to the liquid, $L''=L'+F=119.1$: $$y=\tfrac{L''}{V}x-\tfrac{Wx_W}{V}=2.93x-0.0386,$$ meeting the middle line on the vertical $q$-line $x=0.16$.
Step off stages. Starting from $(x_D,x_D)$ and alternating between the equilibrium curve and the appropriate operating line, the staircase needs about 9 equilibrium stages (8 plates + reboiler). The liquid composition passes 0.50 at the 4th plate from the top (side draw) and 0.16 at the 6th plate (feed).
Figure 5 — McCabe–Thiele construction: equilibrium curve (blue), the three operating lines and the stepped stages. The rectifying and middle lines cross at $x_S=0.50$ (side draw, ~plate 4); the middle and stripping lines cross on the $q$-line at $z_F=0.16$ (feed, ~plate 6).
