23-Chem-A2 Unit Operations and Separation Processes · Undated paper
Question 1 of 6: Rate of Wash-Water Flow through a Batch Centrifugal Filter
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 A1: Rate of Wash-Water Flow through a Batch Centrifugal Filter (25 marks)
Final cake thickness $\Rightarrow$ inner cake face $r_c=r_2-0.15$
0.225 m
Inner liquid-surface radius (wash) $r_1$
0.200 m
Cake porosity $\varepsilon$ / dry-solid density $\rho_s$
0.435 / 2000 kg/m³
Specific cake resistance $\alpha$
$9.5\times10^{11}$ m/kg
Medium resistance $R_m$
$2.5\times10^{10}$ m−1
Speed $N$
2000 rpm
Water $\mu$ / $\rho$
$9\times10^{-4}$ Pa·s / 998 kg/m³
Find. The volumetric rate of wash water (= final filtrate rate) in m³/h.
Radial section: wash water sits inside $r_1=0.20$ m, drives outward through the annular cake ($r_c\!\to\!r_2$) and the medium at $r_2$.
Approach. Compute the dry-cake mass from the cake geometry, form the log-mean and arithmetic-mean cake areas, then apply Geankoplis’s centrifugal-filtration flow equation with the centrifugal driving pressure $\rho\omega^2(r_2^2-r_1^2)/2$.
Dry-cake mass. The cake fills the annulus $r_c\!\to\!r_2$: $$V_\text{cake}=\pi\big(r_2^2-r_c^2\big)b=\pi\big(0.375^2-0.225^2\big)(0.45)=0.1272\ \mathrm{m^3},$$ $$m_c=V_\text{cake}(1-\varepsilon)\rho_s=(0.1272)(0.565)(2000)=143.8\ \mathrm{kg}.$$ (The 50 g/L feed concentration is not needed — the cake is specified directly by its thickness.)
Centrifugal driving pressure (liquid from $r_1$ to $r_2$): $$\Delta P=\frac{\rho\,\omega^2\big(r_2^2-r_1^2\big)}{2}=\frac{(998)(209.4)^2\big(0.375^2-0.200^2\big)}{2}=2.20\times10^{6}\ \mathrm{Pa}.$$
Filtrate (= wash) flow rate. Geankoplis’s centrifugal-filter equation puts the cake resistance on the mean-area product and the medium resistance on $A_2$: $$Q=\frac{\Delta P}{\mu\!\left(\dfrac{\alpha\,m_c}{A_a A_{lm}}+\dfrac{R_m}{A_2}\right)}.$$ The cake term $\alpha m_c/(A_aA_{lm})=1.94\times10^{14}\ \mathrm{m^{-3}}$ dwarfs the medium term $R_m/A_2=2.4\times10^{10}$, so the cake controls. Substituting, $$Q=\frac{2.20\times10^{6}}{(9\times10^{-4})(1.94\times10^{14})}=1.26\times10^{-5}\ \mathrm{m^3/s}.$$
Convert to per-hour. $$Q=1.26\times10^{-5}\times3600=0.0454\ \mathrm{m^3/h}.$$ Rate of washing $\approx 0.045\ \mathrm{m^3/h}$ ($\approx 45\ \mathrm{L/h}$).
