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23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2017

Question 3 of 6: Two-Column Distillation Train (BTX)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 16-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar volumes at STP, gas constant, Rankine conversion) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — recycle balances, humidity and adiabatic-flame (energy) balances; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — multi-unit separation balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — cubic equations of state, residual properties, reaction/phase equilibrium and gas-cycle work; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 3: Two-Column Distillation Train (BTX) (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. Basis 100 kg feed: benzene B = 40, toluene T = 30, xylene X = 30 (wt). Column 1 distillate $D_1$ = 99.5% B, 0.5% T (no X). Column 1 residue $R_1$ feeds column 2. Column 2 distillate $D_2$ = 2% B, 97% T, 1% X. Column 2 residue $R_2$ = 5% T, 95% X (no B).

Find. (a) mass and composition of the three products $D_1,D_2,R_2$; (b) mass and composition of the intermediate $R_1$.

Column 1Column 2Feed 100 kg40% B / 30% T / 30% XD1: 99.5% B, 0.5% TR1 (intermediate)D2: 97% T, 2% B, 1% XR2: 95% X, 5% T
Figure 2 — Feed enters column 1; overhead D1 is nearly pure benzene. The column-1 bottoms R1 (the intermediate stream) feeds column 2, which splits it into a toluene-rich distillate D2 and a xylene-rich bottoms R2.

Approach. Write overall component balances for B, T, X across the whole train (three equations, three unknowns $D_1,D_2,R_2$); then a column-2 balance gives the intermediate $R_1=D_2+R_2$.

  1. Overall benzene balance. Benzene leaves only in $D_1$ (99.5%) and $D_2$ (2%): $$0.995\,D_1+0.02\,D_2=40.$$
  2. Overall xylene balance. Xylene leaves only in $D_2$ (1%) and $R_2$ (95%): $$0.01\,D_2+0.95\,R_2=30.$$
  3. Overall toluene balance. Toluene appears in all three products: $$0.005\,D_1+0.97\,D_2+0.05\,R_2=30.$$
  4. Solve the 3×3 system (part a). Solving simultaneously: $$D_1=\boxed{39.62\ \text{kg}},\qquad D_2=\boxed{29.11\ \text{kg}},\qquad R_2=\boxed{31.27\ \text{kg}},$$ which sum to 100.0 kg, closing the overall mass balance. Their compositions are as specified: $D_1$ = 99.5% B / 0.5% T; $D_2$ = 2% B / 97% T / 1% X; $R_2$ = 5% T / 95% X.
  5. Intermediate stream $R_1$ (part b). A balance around column 2 gives $R_1=D_2+R_2=29.11+31.27=60.38$ kg. Its components are the sum of what leaves in $D_2$ and $R_2$: $$\text{B}=0.02(29.11)=0.58,\quad \text{T}=0.97(29.11)+0.05(31.27)=29.81,\quad \text{X}=0.01(29.11)+0.95(31.27)=30.00\ \text{kg}.$$ As mass fractions: $\boxed{R_1 = 0.96\%\ \text{B},\ 49.4\%\ \text{T},\ 49.7\%\ \text{X}\ \ (60.38\ \text{kg})}.$

Sanity check on the overall toluene split: $0.5\%$ of 39.62 (=0.20) $+\,97\%$ of 29.11 (=28.24) $+\,5\%$ of 31.27 (=1.56) $=30.0$ kg — matches the 30 kg fed.

StreamMass & composition
(a) $D_1$ benzene product39.62 kg — 99.5% B, 0.5% T
(a) $D_2$ toluene product29.11 kg — 2% B, 97% T, 1% X
(a) $R_2$ xylene product31.27 kg — 5% T, 95% X
(b) $R_1$ intermediate60.38 kg — 0.96% B, 49.4% T, 49.7% X