23-Chem-A2 Unit Operations and Separation Processes · December 2014
Question 2 of 6: Plate-and-Frame Filter Press — Constant-Pressure Filtration
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2014. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, crystallisation); Perry's Chemical Engineers' Handbook (9th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.).
Find. (a) equivalent filtrate volume $V_e$; (b) specific cake resistance $\alpha$ and porosity $\varepsilon$; (c) specific surface area $S_0$.
Figure A2 — Constant-pressure filtration linearises as $t/V = \dfrac{\mu\alpha c}{2A^2\Delta P}V + \dfrac{\mu R_m}{A\Delta P}$; slope gives $\alpha$, intercept gives the medium resistance.
Approach. Convert the cake data to porosity and to the mass of solids deposited per unit filtrate ($c$); regress $t/V$ versus $V$ to get slope and intercept; then read $\alpha$, $R_m$ (hence $V_e$) from those, and finally invert the Kozeny–Carman relation for the specific surface.
Cake porosity and solids loading $c$. From the dried-cake density, $$\varepsilon = 1-\frac{\rho_{\mathrm{dry}}}{\rho_s}=1-\frac{1603}{2830}=0.434.$$ The wet-to-dry cake mass ratio is $m = 1+\dfrac{\varepsilon\rho_f}{(1-\varepsilon)\rho_s}=1.270$, and the solids deposited per unit filtrate volume is $$c=\frac{\rho_f s}{1-m\,s}=\frac{998.3(0.0723)}{1-1.270(0.0723)}=79.5\ \mathrm{kg/m^3}.$$
Linear regression of $t/V$ vs $V$. A least-squares fit of the 14 points (with $V$ in $\mathrm{m^3}$, $t$ in s) gives $$\text{slope}=8.74\times10^{6}\ \tfrac{\mathrm{s}}{\mathrm{m^6}},\qquad \text{intercept}=6.98\times10^{3}\ \tfrac{\mathrm{s}}{\mathrm{m^3}}.$$ Writing $K_p = 2\times\text{slope} = 1.748\times10^{7}$.
(a) Equivalent filtrate volume. The medium+piping resistance is expressed as the filtrate volume that would build a cake of equal resistance, $V_e = \text{intercept}/K_p$: $$V_e=\frac{6.98\times10^{3}}{1.748\times10^{7}}=3.99\times10^{-4}\ \mathrm{m^3}=0.399\ \mathrm{L}.$$ $V_e \approx 0.40\ \mathrm{L}$ (equivalent medium resistance)
(b) Specific cake resistance and porosity. From the slope, $$\alpha=\frac{K_p A^2\Delta P}{\mu\,c}=\frac{(1.748\times10^{7})(0.0263)^2(2.76\times10^{5})}{(1.03\times10^{-3})(79.5)}=4.08\times10^{10}\ \mathrm{m/kg},$$ and the medium resistance from the intercept is $R_m=\dfrac{\text{intercept}\cdot A\,\Delta P}{\mu}=4.92\times10^{10}\ \mathrm{m^{-1}}$. The porosity is that from Step 1. $\alpha = 4.1\times10^{10}\ \mathrm{m/kg}$, $\varepsilon = 0.43$
(c) Specific surface area of the cake. Inverting the Kozeny–Carman form $\alpha=\dfrac{5\,S_0^{2}(1-\varepsilon)}{\varepsilon^{3}\rho_s}$ for the surface area per unit solid volume, $$S_0=\sqrt{\frac{\alpha\,\varepsilon^{3}\rho_s}{5(1-\varepsilon)}}=1.82\times10^{6}\ \mathrm{m^2/m^3},$$ equivalent to a mean sphere diameter $d = 6/S_0 = 3.3\ \mu\mathrm{m}$. $S_0 = 1.8\times10^{6}\ \mathrm{m^2/m^3}$ ($d\approx3.3\ \mu\mathrm{m}$)
