23-Chem-A2 Unit Operations and Separation Processes · December 2019
Question 3 of 6: Constant-Pressure Plate-and-Frame Filtration
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — December 2019, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 pump-power from a full mechanical-energy balance, A2 terminal velocity & matched fluidized-bed densities, A3 constant-pressure plate-and-frame filtration; Part B (Separation Processes) — B1 countercurrent rotary-drier air rate, B2 Swenson–Walker crystallizer sizing, B3 double-effect forward-feed evaporator capacity. The rubric asks for two problems per part; all six are worked in full below.
Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (5th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §11 heat transfer, §12 drying, §18 solid–liquid separation).
Find. (a) The equivalent filtrate volume $V_e$; (b) the specific cake resistance $\alpha$ and porosity $\varepsilon$; (c) the specific surface area $S$ of the cake.
Figure A3 — The integrated constant-pressure filtration law $t/V=\tfrac{K_c}{2}V+K_cV_e$ plots as a straight line; the slope gives the cake resistance and the intercept (with the slope) gives the medium-equivalent volume $V_e$.
Approach. Linearise the constant-pressure law by plotting $t/V$ against $V$; the intercept locates $V_e$, the slope gives the specific cake resistance $\alpha$ (after finding the solids concentration $c$), the dry/solid density ratio gives porosity $\varepsilon$, and Kozeny–Carman converts $\alpha$ into the specific surface area.
Linear regression of $t/V$ vs $V$. The integrated law is $\dfrac{t}{V}=\dfrac{K_c}{2}V+K_cV_e$. Least-squares fit of the 14 points gives $$\text{slope}=\tfrac{K_c}{2}=8.74\ \text{s/L}^2,\qquad\text{intercept}=K_cV_e=6.979\ \text{s/L}.$$
(a) Equivalent filtrate volume. Since intercept $=K_cV_e$ and slope $=K_c/2$, $$V_e=\frac{\text{intercept}}{2\times\text{slope}}=\frac{6.979}{2(8.74)}=\boxed{0.399\ \text{L}}\ (\approx0.40\ \text{L}).$$ This is the filtrate volume that would build a cake whose resistance equals that of the medium plus piping.
Cake porosity. From the dry-cake and solid densities, $$\varepsilon=1-\frac{\rho_\text{dry}}{\rho_s}=1-\frac{1603}{2830}=\boxed{0.434}.$$
Solids concentration in the filtrate. Converting mole fraction to mass fraction, $s=\dfrac{xM_{\text{CaCO}_3}}{xM_{\text{CaCO}_3}+(1-x)M_w}=0.0725$; accounting for liquid retained in the cake ($m=1+\varepsilon\rho_w/[(1-\varepsilon)\rho_s]$), the mass of solids deposited per m³ of filtrate is $$c=\frac{\rho_w s}{1-ms}=79.8\ \text{kg/m}^3.$$
(b) Specific cake resistance. With slope in SI units ($8.74\ \text{s/L}^2=8.74\times10^{6}\ \text{s/m}^6$) and $\text{slope}=\dfrac{\mu\alpha c}{2A^2\Delta P}$, $$\alpha=\frac{2A^2\Delta P\,(\text{slope})}{\mu c}=\frac{2(0.0263)^2(275{,}600)(8.74\times10^{6})}{(1.03\times10^{-3})(79.8)}=\boxed{4.05\times10^{10}\ \text{m/kg}}.$$
(c) Specific surface area. Kozeny–Carman relates $\alpha$ to the volume-specific surface $S$: $\alpha=\dfrac{5S^2(1-\varepsilon)}{\varepsilon^3\rho_s}$, so $$S=\sqrt{\frac{\alpha\,\varepsilon^3\rho_s}{5(1-\varepsilon)}}=\sqrt{\frac{(4.05\times10^{10})(0.434)^3(2830)}{5(0.566)}}=\boxed{1.82\times10^{6}\ \text{m}^2/\text{m}^3}.$$
