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23-Chem-A2 Unit Operations and Separation Processes · May 2016

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. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, May 2016. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section (only the first two per section are marked). All six are worked below for completeness.

Reference texts. Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, sedimentation, fluidization, filtration, crystallization, evaporation) and Vol. 1 (heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on saturation data. Sections B2 and B3 need water saturation temperatures and latent heats that the exam expects from an open-book steam table. These are taken from standard tables (and, for reproducibility). The B1 latent heat of evaporation is not printed in the exam and is taken as 2370 kJ/kg at the cooling range; this and other engineering choices are flagged in Check callouts.

Question A3: Constant-Pressure Plate-and-Frame Filtration (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. Constant-pressure filtration data pairs $(t,V)$; press area $A=4.287\times10^{-2}$ m²; $\Delta P=68.9$ kPa; slurry solids fraction $s=0.00495$; cake solids fraction $0.2937$; $\mu=10^{-3}$ Pa·s; $\rho_{liq}=1000$ kg/m³.

Find. The filter-medium resistance $R_m$ and the average specific cake resistance $\alpha$.

Linearized filtration: t/V vs V V (m³) t/V (s/m³) slope = μαc/(2A²ΔP) intercept = μR_m/(AΔP) intercept ≈ 6217
Figure A3 — Plotting $t/V$ against $V$ linearizes the constant-pressure filtration equation; the slope gives $\alpha$ and the intercept gives $R_m$.

Approach. Integrate the constant-pressure filtration equation to the linear form $t/V=K_p V+B$, least-squares-fit the eleven data points, then back out $\alpha$ from the slope and $R_m$ from the intercept after computing the mass of cake solids per unit filtrate volume, $c$.

  1. Solids deposited per unit filtrate, $c$. With wet/dry cake ratio $m=1/0.2937=3.405$, $$c=\frac{\rho\,s}{1-m\,s}=\frac{1000(0.00495)}{1-3.405(0.00495)}=5.04\ \mathrm{kg/m^3}.$$ The moist-cake correction ($m s$) is small here, so $c$ is close to $\rho s$.
  2. Linearized filtration equation. Integrating $\dfrac{dt}{dV}=\dfrac{\mu\alpha c}{A^2\Delta P}V+\dfrac{\mu R_m}{A\Delta P}$ at constant $\Delta P$ gives $$\frac{t}{V}=\underbrace{\frac{\mu\alpha c}{2A^2\Delta P}}_{\text{slope}}V+\underbrace{\frac{\mu R_m}{A\Delta P}}_{\text{intercept}}.$$
  3. Least-squares fit of $t/V$ vs. $V$. Regressing the eleven $(V,\,t/V)$ pairs: $$\text{slope}=1.616\times10^{6}\ \mathrm{s/m^6},\qquad \text{intercept}=6.22\times10^{3}\ \mathrm{s/m^3}.$$
  4. Average specific cake resistance. Solving the slope relation for $\alpha$: $$\alpha=\frac{2A^2\Delta P\,(\text{slope})}{\mu c}=\frac{2(4.287\times10^{-2})^2(6.89\times10^{4})(1.616\times10^{6})}{(10^{-3})(5.04)}=\;\boxed{8.13\times10^{10}\ \mathrm{m/kg}}.$$
  5. Filter-medium resistance. From the intercept, $$R_m=\frac{(\text{intercept})A\Delta P}{\mu}=\frac{(6.22\times10^{3})(4.287\times10^{-2})(6.89\times10^{4})}{10^{-3}}=1.84\times10^{10}\ \mathrm{m^{-1}}.$$ $\alpha\approx8.1\times10^{10}$ m/kg, $R_m\approx1.8\times10^{10}$ m⁻¹
QuantityResult
Cake solids per filtrate volume $c$5.04 kg/m³
Slope $\mu\alpha c/2A^2\Delta P$$1.62\times10^{6}$ s/m⁶
Intercept $\mu R_m/A\Delta P$$6.22\times10^{3}$ s/m³
Specific cake resistance $\alpha$$8.1\times10^{10}$ m/kg
Medium resistance $R_m$$1.8\times10^{10}$ m⁻¹