Question 3 of 6: Constant-Pressure Filtration of a CaCO₃ Slurry

Question A3: Constant-Pressure Filtration of a CaCO₃ Slurry (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Run data ($t$ in s, filtrate weight in lb): (26, 5), (98, 10), (211, 15), (361, 20), (555, 25), (788, 30), (1083, 35).

Find. (a) fill volume and time; (b) $\alpha$; (c) $\varepsilon$; (d) $R_m$.

Approach. Convert filtrate weights to volumes and fit the constant-pressure law $t/V=K_c V+ (1/q_0)$; the slope yields $\alpha$ and the intercept $R_m$. Porosity comes from the dry-bulk vs true-solid densities, and the fill point from the frame volume.

1. Solids loading per unit filtrate. The mass of dry cake deposited per unit volume of filtrate is $$c=\frac{\rho\,s}{1-m\,s}=\frac{(62.2)(0.139)}{1-(1.473)(0.139)}=\boxed{10.9\ \mathrm{lb_m/ft^3}}.$$
2. Linearise the constant-pressure data. With $V=W/\rho$, plotting $t/V$ against $V$ gives a straight line $$\frac{t}{V}=\underbrace{\frac{\mu\,\alpha\,c}{2A^{2}\,\Delta p}}_{\text{slope}=3.27\times10^{3}\,\mathrm{s/ft^6}}\,V+\underbrace{\frac{\mu\,R_m}{A\,\Delta p}}_{\text{int}=73\,\mathrm{s/ft^3}}.$$ (Least-squares fit of the seven points.)
3. (b) Mean specific cake resistance. Solving the slope for $\alpha$ with $\Delta p=30\times144$ lb_f/ft² and $A=1$ ft² (units reconciled through $g_c$): $$\alpha=\frac{2A^{2}\Delta p\,(\text{slope})}{\mu\,c}=\boxed{1.45\times10^{11}\ \mathrm{ft/lb_m}}\;(\approx9.7\times10^{10}\ \mathrm{m/kg}).$$
4. (d) Filter-medium resistance. From the intercept, $$R_m=\frac{A\,\Delta p\,(\text{int})}{\mu}=\boxed{1.77\times10^{10}\ \mathrm{ft^{-1}}}\;(\approx5.8\times10^{10}\ \mathrm{m^{-1}}).$$ The medium term is two orders below the cake term, so the cake controls once a layer builds.
5. (c) Mean porosity. The cake solids occupy a fraction $\rho_\text{bulk}/\rho_\text{solid}$ of the volume, so $$\varepsilon=1-\frac{\rho_\text{bulk}}{\rho_\text{solid}}=1-\frac{73.8}{164}=\boxed{0.55}.$$ Cross-check via the wet/dry water content: each lb of dry cake holds 0.473 lb of water, so one ft³ of cake holds $(73.8)(0.473)/62.2=0.56$ ft³ of liquid, i.e. $\varepsilon\approx0.56$. The two routes agree within 0.01, so the mean porosity is about 0.55–0.56.
6. (a) Filtrate volume and time to fill the frames. The two 2-in frames fill from both faces, so the cake volume equals the total face area times the half-thickness: $V_\text{cake}=1\ \mathrm{ft^2}\times\tfrac{1}{12}\ \mathrm{ft}=0.083\ \mathrm{ft^3}$. Filtrate to reach this, $$V=\frac{\rho_\text{bulk}V_\text{cake}}{c}=\frac{(73.8)(0.083)}{10.9}=0.566\ \mathrm{ft^3}\;(35.2\ \mathrm{lb}),$$ and from the fitted law $t=\text{(slope)}V^2+\text{(int)}V\approx\boxed{1.09\times10^{3}\ \mathrm{s}\ (\approx18\ \mathrm{min})}.$ This lands right at the last data point — the run was carried to the fill point.