23-Chem-A2 Unit Operations and Separation Processes · May 2017
Question 6 of 6: Packed-Tower Height for Air Drying with Caustic
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 2017. 3 hours, open book (one text). Six problems — Part A Unit Operations (A1–A3) and Part B Separation Processes (B1–B3), each worth 25 marks; the rubric asks for at least two problems from each part (only the first two per part are marked). All six are worked below for completeness as a study resource.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, filtration, crystallization, drying, absorption); Perry's Chemical Engineers' Handbook (9th ed.); Badger & Banchero, Introduction to Chemical Engineering.
Notes on chart / table reads. A1 uses the ISO nozzle discharge-coefficient chart ($C_n$ vs $Re$, $\beta$); B1's isotherm concentrations are printed as “g/cm³” but are read as g/L (= kg/m³) so the feed concentration (0.25 kg/m³) sits on the same scale as the data — a units slip in the source; B3's equilibrium $Y$ column carries an implied ×10−4 multiplier (the printed integers 0–202 cannot be literal mol/mol, and the inlet gas $Y_1=0.0177$ coincides exactly with the $X=10$ entry of 177, which fixes the scale). Engineering choices are flagged in Check callouts.
Part A — Unit Operations
Question B3: Packed-Tower Height for Air Drying with Caustic (25 marks)
Figure B3 — Equilibrium curve (points, $Y\times10^{4}$) and the operating line at $1.25\,L_\text{min}$. The S-shaped equilibrium makes the minimum-liquid pinch an internal tangent near $X\approx6$, not the bottom intersection.
Approach. Convert the humidities to molar ratios; fix the entering-liquid composition from the 50% caustic; locate the minimum-liquid pinch (here an internal tangent because the equilibrium curve is concave), scale the slope by 1.25, then integrate $N_{OG}=\int dY/(Y-Y^{*})$ along the operating line and multiply by $H_{OG}$.
Gas molar ratios. Using $M_\text{air}/M_{H_2O}=28.97/18.0$, $$Y_1=\frac{0.011}{18}\Big/\frac{1}{28.97}=0.0177,\qquad Y_2=0.00322\ \ (\text{i.e. }177\text{ and }32.2\times10^{-4}).$$ Note $Y_1$ coincides with the $X=10$ table entry (177) — this fixes the $\times10^{-4}$ scale.
Entering liquid. A 50 wt% NaOH solution has $$X_2=\frac{50/18}{50/40}=2.22\ \text{mol H}_2\text{O/mol NaOH}.$$
Minimum liquid — internal tangent pinch. The equilibrium curve is S-shaped, so a straight operating line drawn from the top point $(X_2,Y_2)$ to the bottom would cut below equilibrium. The binding constraint is the steepest tangent from $(X_2,Y_2)$ to the curve, which occurs near $X\approx6$: $$\left(\frac{L}{G}\right)_\text{min}=24.8\ (\text{in }Y\times10^{4}\text{ units}).$$
Operating slope and bottom composition. Scaling up by 25%, $$\frac{L}{G}=1.25\times24.8=31.0,\qquad X_1=X_2+\frac{Y_1-Y_2}{L/G}=2.22+\frac{177-32.2}{31.0}=6.89.$$ The operating line $(2.22,32.2)\rightarrow(6.89,177)$ clears the equilibrium everywhere (positive driving force).
Number of transfer units. Integrating along the operating line with $Y^{*}$ read from the equilibrium curve, $$N_{OG}=\int_{Y_2}^{Y_1}\frac{dY}{Y-Y^{*}}=6.4.$$ (Numerical quadrature over the operating range.)
Check: the printed equilibrium $Y$ column has no multiplier; it is taken as $\times10^{-4}$ mol/mol because (i) the integers 0–202 are impossible as literal mol/mol, and (ii) the inlet $Y_1=0.0177$ matches the $X=10$ entry (177) exactly, so minimum liquid would otherwise pin at the bottom — but the concave curve makes the internal tangent the true minimum-liquid constraint, which governs the design.
