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23-Chem-A2 Unit Operations and Separation Processes · Undated paper

Question 5 of 6: Absorber Operating Line in Solute-Free Coordinates

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

Notes on this paper

Paper format. National Exams — 16-CHEM-A2 Unit Operations and Separation Processes, May 2019 (this is the “undated” PDF; the cover sheet is a faint template reading MAY, while every interior page reads May 2019). 3 hours, open book. Six problems: Section A (Momentum-transfer / mechanical operations) A1–A3, and Section B (Separation processes) B1–B3, each 25 marks. Candidates attempt at least two from each section and only the first two per section are marked; all six are worked below for completeness.

Reference texts. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — filtration, packed beds, absorption, adsorption, packed-tower distillation; McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 2; Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on the data. The B1 feed is described as acetone–water but the equilibrium table supplied is explicitly acetone–ethanol — B1 is solved with the data actually provided (acetone–ethanol) and the discrepancy is flagged.

Question B2: Absorber Operating Line in Solute-Free Coordinates (25 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.

QuantityValue
Entering gas $G$ / $y_\text{in}$500 kmol/h / 0.30
Leaving gas $y_\text{out}$0.001
Equilibrium $y=mx$$m=2.8$
Solvent entering (pure) $X_\text{in}$0
Carrier gas Binsoluble in solvent

Find. Carrier flow $G_s$, terminal ratios, the minimum solvent rate, and the equilibrium + operating lines on solute-free ($X$,$Y$) axes.

Solute-free coordinates: equilibrium curve + operating line $X = x/(1-x)$ (liquid, mole ratio) $Y = y/(1-y)$ (gas) equilibrium $Y=\dfrac{2.8X}{1-1.8X}$ top (0, 0.001) bottom pinch (0.12, 0.4286) operating line, slope $L_s/G_s$ (minimum) 0 0.12
Minimum-solvent operating line: it runs from the clean-solvent top $(0,\,Y_\text{out})$ to the bottom, where the leaving liquid is in equilibrium with the entering gas — the line just touches the equilibrium curve (a pinch). Any real column uses a steeper line (more solvent).

Approach. Convert both terminal gas compositions to mole ratios, get the constant inert-gas flow $G_s$, find the richest liquid the entering gas could ever reach (equilibrium at the bottom) to fix the minimum solvent, then plot the (slightly curved) equilibrium line and the straight operating line.

  1. Carrier-gas flow (constant through the tower). $$G_s=G(1-y_\text{in})=500(1-0.30)=350\ \mathrm{kmol/h}.$$
  2. Terminal gas ratios. $$Y_\text{in}=\frac{0.30}{0.70}=0.4286,\qquad Y_\text{out}=\frac{0.001}{0.999}=0.0010.$$ A absorbed $=G_s(Y_\text{in}-Y_\text{out})=350(0.4276)=149.6\ \mathrm{kmol/h}$ — a 99.7% recovery.
  3. Minimum solvent (bottom pinch). The richest possible exit liquid is in equilibrium with the entering gas: $x_1^{*}=y_\text{in}/m=0.30/2.8=0.1071$, i.e. $X_1^{*}=0.1071/(1-0.1071)=0.120$. With clean solvent ($X_\text{in}=0$), $$\left(\frac{L_s}{G_s}\right)_\text{min}=\frac{Y_\text{in}-Y_\text{out}}{X_1^{*}-0}=\frac{0.4276}{0.120}=3.56,$$ $$L_{s,\text{min}}=3.56\times350=\boxed{1247\ \mathrm{kmol/h}}.$$
  4. Design operating line. Practical columns run at $L_s\approx1.2$–$2\times$ minimum; at $1.5\times$, $L_s=1871\ \mathrm{kmol/h}$ and the exit liquid drops to $X_1=G_s(Y_\text{in}-Y_\text{out})/L_s=0.080$, moving the operating line off the equilibrium curve (finite column). Minimum solvent 1247 kmol/h; recommended $\sim$1871 kmol/h at 1.5×.
  5. Lines to plot (solute-free). Equilibrium (curved): $Y=\dfrac{mX}{1-(m-1)X}=\dfrac{2.8X}{1-1.8X}$. Operating (straight): from top $(X_\text{in},Y_\text{out})=(0,0.001)$ to bottom $(X_1,Y_\text{in})$ with slope $L_s/G_s$. The minimum-solvent line touches the curve at the bottom; the design line is steeper.
QuantityValue
Carrier gas $G_s$350 kmol/h
$Y_\text{in}$ / $Y_\text{out}$0.4286 / 0.0010
A absorbed149.6 kmol/h (99.7% recovery)
Richest exit liquid $X_1^{*}$ (equilibrium)0.120
Minimum solvent $L_{s,\text{min}}$1247 kmol/h ($L_s/G_s=3.56$)
Design solvent (1.5×) / exit $X_1$1871 kmol/h / 0.080
Check. The problem statement asks to “determine and plot” the lines but does not print an operating solvent rate, so the uniquely-determined quantity is the minimum solvent (bottom pinch); the 1.5× design case is a standard engineering choice shown for context. If the exam expected a specified $L_s$, substitute it for the operating-line slope.