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23-Chem-A2 Unit Operations and Separation Processes · May 2014

Question 3 of 6: Rotary-Drum Vacuum Filter Throughput

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, May 2014 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems are worth 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below as a study resource.

Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction and loss coefficients, agitator power correlations (Fig. 9.13, Table 9.3), and filtration theory; de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance, fitting equivalent lengths, and the appended Fanning chart; Coulson & Richardson, Chemical Engineering Vol. 2 — rotary-drum filtration; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Lienhard, A Heat Transfer Textbook — composite-wall resistance networks, the Colburn (Chilton–Colburn) analogy, and LMTD/ε–NTU cross-flow exchangers.

Property note. Water properties are as printed in each question; the appended Table A.2 loss coefficients ($k$) and Table A3 turbine constants ($K_T$) are used directly, and pipe-friction factors are taken from Colebrook (identical to the appended Fanning chart, Fig. A1) so the numbers are reproducible without reading a graph.

Question A3: Rotary-Drum Vacuum Filter Throughput (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. Incompressible-cake rotary-drum filtration with a finite cloth resistance.

QuantityValue
Slurry solids / $\rho_s$20 wt% / $2000\ \mathrm{kg/m^3}$
Drum area $A$ / submergence$3\ \mathrm{m^2}$ / 30%
Driving pressure $-\Delta P$30 kPa
Filtrate $\rho_f$ / $\mu$$1000\ \mathrm{kg/m^3}$ / $1\times10^{-3}\ \mathrm{Pa\,s}$
Voidage $\varepsilon$ / resistance $r$0.4 / $2\times10^{12}\ \mathrm{m^{-2}}$
Drum speed / cloth $L$0.5 rpm / 1 mm cake-equivalent

Find. the filtrate production rate (m$^3$/s and m$^3$/h).

slurry (20 wt%) 30% submerged 0.5 rpm vacuum 30 kPa filtrate out
Figure A3 — Rotary-drum vacuum filter. Each element cakes only while submerged (30% of a revolution); the internal vacuum draws filtrate through the growing cake and the cloth.

Approach. First fix $v$ (cake volume per filtrate volume) from a slurry mass basis; then note each drum element filters only during the submerged fraction of one revolution, so integrate the given rate over that time to a quadratic in $V$ per revolution, and divide by the cycle time for the throughput.

  1. Cake-to-filtrate ratio $v$. Basis 100 kg slurry: 20 kg solids, 80 kg water. Solids volume $=20/2000=0.0100\ \mathrm{m^3}$; with voidage $\varepsilon=0.4$ the cake volume is $$V_{\text{cake}}=\frac{0.0100}{1-0.4}=0.01667\ \mathrm{m^3}.$$ Liquid retained in the cake $=\varepsilon V_{\text{cake}}\rho_f=0.4(0.01667)(1000)=6.67\ \mathrm{kg}$, so filtrate $=80-6.67=73.3\ \mathrm{kg}\Rightarrow 0.0733\ \mathrm{m^3}$, giving $$v=\frac{V_{\text{cake}}}{V_{\text{filtrate}}}=\frac{0.01667}{0.0733}=0.227.$$
  2. Filtration and cycle time. One revolution takes $t_{\text{cyc}}=60/0.5=120\ \mathrm{s}$; each element cakes only while submerged, so $$t_{\text{filt}}=0.30\,t_{\text{cyc}}=0.30(120)=36\ \mathrm{s}.$$
  3. Integrate the rate equation. Separating $\dfrac{dV}{dt}=\dfrac{A^2(-\Delta P)}{r\mu v[V+LA/v]}$ and integrating from 0 to $V$ over $0\to t_{\text{filt}}$ gives the standard quadratic $$\frac{r\mu v}{2}V^2+r\mu L A\,V=A^2(-\Delta P)\,t_{\text{filt}}.$$ Substituting $r\mu v/2=2.27\times10^{8}$, $r\mu LA=6.0\times10^{6}$, and $A^2(-\Delta P)t_{\text{filt}}=3^2(30{,}000)(36)=9.72\times10^{6}$: $$2.27\times10^{8}\,V^2+6.0\times10^{6}\,V-9.72\times10^{6}=0.$$
  4. Solve for filtrate per revolution. The positive root is $$V_{\text{rev}}=\frac{-6.0\times10^{6}+\sqrt{(6.0\times10^{6})^2+4(2.27\times10^{8})(9.72\times10^{6})}}{2(2.27\times10^{8})}=0.194\ \mathrm{m^3}.$$
  5. Throughput. Dividing the filtrate collected each revolution by the cycle time, $$\dot V=\frac{V_{\text{rev}}}{t_{\text{cyc}}}=\frac{0.194}{120}=1.62\times10^{-3}\ \mathrm{m^3/s}=5.82\ \mathrm{m^3/h}.$$ $\dot V\approx 5.82\ \mathrm{m^3/h}$ of filtrate
QuantityResult
Cake-to-filtrate ratio $v$0.227
Cycle / filtration time120 s / 36 s
Filtrate per revolution $V_{\text{rev}}$0.194 m$^3$
Filtrate production rate$1.62\times10^{-3}\ \mathrm{m^3/s}=5.82\ \mathrm{m^3/h}$
Assumption — meaning of "internal pressure of 30 kPa" A rotary drum filter pulls vacuum inside the drum, so the printed 30 kPa is taken as the pressure difference across the cake and cloth, $-\Delta P=30\ \mathrm{kPa}$ (i.e. a 30 kPa vacuum). If it were instead read as an absolute internal pressure, $-\Delta P=101.3-30=71.3\ \mathrm{kPa}$, and the same quadratic gives $V_{\text{rev}}=0.306\ \mathrm{m^3}$ and $2.55\times10^{-3}\ \mathrm{m^3/s}\approx9.2\ \mathrm{m^3/h}$. Only the pressure reading changes; the method is identical.