23-Chem-A2 Unit Operations and Separation Processes · May 2017
Question 4 of 6: Part B — Separation Processes
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
Part B — Separation Processes
Question B1: Adsorption Isotherm for Phenol on Activated Carbon (25 marks)
Figure B1 — Equilibrium data (points) with the fitted Langmuir isotherm (curve). The plateau near $C_{A,S}\!\approx\!0.14$ kg/kg marks the monolayer capacity $q_m$.
Approach. The data level off at high concentration, signalling a Langmuir (Type I) isotherm; linearise as $C_A/C_{A,S}$ vs $C_A$ to extract $q_m$ and $K$, then intersect the isotherm with the batch mass-balance (operating) line to find the equilibrium point.
(a),(b) Identify and fit the isotherm. The rise-then-plateau shape is characteristic of monolayer (Langmuir) adsorption. Linearising $$\frac{C_A}{C_{A,S}}=\frac{1}{q_m K}+\frac{C_A}{q_m}$$ and least-squares fitting the six points gives $$\boxed{q_m=0.141\ \mathrm{kg/kg},\quad K=60.5\ \mathrm{m^3/kg}}$$ with $R^2>0.99$. (A Freundlich fit, $C_{A,S}=0.296\,C_A^{0.39}$, does not capture the plateau as well.)
Batch operating line. A phenol balance on the batch links the carbon loading to the liquid depletion: $$C_{A,S}=\frac{V}{m}\,(C_0-C_A)=\frac{2.5}{3}\,(0.25-C_A)=0.833\,(0.25-C_A).$$
(c) Equilibrium point. Setting the operating line equal to the Langmuir isotherm and solving, $$0.833(0.25-C_A)=\frac{q_m K\,C_A}{1+K C_A}\ \Rightarrow\ \boxed{C_A=0.104\ \mathrm{kg/m^3},\ \ C_{A,S}=0.121\ \mathrm{kg/kg}}.$$
(d) Percentage recovered. The fraction of phenol captured by the carbon is $$\%=\frac{C_0-C_A}{C_0}\times100=\frac{0.25-0.104}{0.25}\times100=\boxed{58\%}.$$ Equivalently, $m\,C_{A,S}/(V C_0)=3(0.121)/(2.5\times0.25)=58\%$.
Quantity
Value
Isotherm type
Langmuir (Type I)
Monolayer capacity $q_m$
0.141 kg/kg
Langmuir constant $K$
60.5 m³/kg
Equilibrium $C_A$ / $C_{A,S}$
0.104 kg/m³ / 0.121 kg/kg
Phenol recovered
≈58%
Check: the source labels $C_A$ as “g/cm³.” Taken literally, the feed (0.25 kg/m³ = 2.5×10−4 g/cm³) would lie three orders below the smallest data point and no meaningful equilibrium exists. Reading the column as g/L (= kg/m³) places the feed just above the data and yields a physical 58% recovery, so that scale is used throughout.
