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

Question 3 of 6: Rotary-Drum Filter Speed and Throughput

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 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, drying) and Vol. 1 (fluid flow, heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Perry's Chemical Engineers' Handbook (9th ed.); Kern, Process Heat Transfer.

Note on chart and property data. A1 uses a Fanning friction-factor chart, B1 a KNO3 solubility curve (mole-fraction basis), and B2/B3 standard steam-table / transport-property values. Engineering choices are flagged in Check callouts.

Question A3: Rotary-Drum Filter Speed and 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.

QuantityValue
Drum diameter × length2 m × 2 m; 40% immersed
Leaf test area200 cm² at $\Delta P=17$ kN/m²
Filtrate collected300 cm³ @ 60 s; +140 cm³ @ 120 s (total 440)
Slurry / dry-cake / filtrate density40% solids; 1500 kg/m³ cake; 1000 kg/m³ filtrate
Minimum removable cake5 mm

Find. (a) the drum rotational speed giving maximum throughput; (b) the corresponding slurry mass fed per unit time.

Rotary vacuum drum filter — 40% submerged slurry trough (40% immersion) rotation n cake forms (submerged) cake scraped off (5 mm min)
Figure A3 — Each element of the drum surface filters only while submerged (fraction 0.40 of a revolution), then dewaters and is scraped. Maximum throughput is set by the fastest speed that still builds the 5 mm minimum removable cake.

Approach. Fit the constant-pressure filtration law $t=C_1V^2+C_2V$ to the two leaf-test points to get area-independent constants; relate cake thickness to filtrate volume through the slurry loading; set the cake to the 5 mm minimum (the speed limit for a removable cake, which is where throughput peaks) to get the submerged time, hence the cycle time and speed; finally scale filtrate to the whole drum for the slurry throughput.

  1. Constant-pressure filtration constants from the leaf test. With cumulative $(V,t)=(300\ \mathrm{cm^3},60\ \mathrm{s})$ and $(440,120)$ in $t=C_1V^2+C_2V$: $$C_1=5.195\times10^{-4}\ \mathrm{s/cm^6},\qquad C_2=0.04416\ \mathrm{s/cm^3}.$$ Both leaf-test points reproduce to $\pm0.3$ s.
  2. Recast per unit filter area. Writing the depth of filtrate collected per unit area as $v_f=V/A$, the law becomes $t=a\,v_f^2+b\,v_f$ with $a=C_1A^2=20.78$ and $b=C_2A=8.83$ (leaf $A=200$ cm²) — now applicable to the drum.
  3. Cake volume per unit filtrate (slurry loading). Per m³ of filtrate (1000 kg water = 60% of the slurry), solids $c=\rho_f\,s/(1-s)=1000(0.40/0.60)=667\ \mathrm{kg}$, giving cake volume $v_c=c/\rho_\text{cake}=667/1500=0.444$ m³ per m³ filtrate.
  4. Filtrate depth that gives a 5 mm cake. Cake thickness $=v_c\,v_f$, so $$v_f=\frac{0.005}{0.444}=0.01125\ \mathrm{m}=1.125\ \mathrm{cm}.$$
  5. Submerged filtration time. $$t_f=a\,v_f^2+b\,v_f=20.78(1.125)^2+8.83(1.125)=36.2\ \mathrm{s}.$$
  6. Cycle time and drum speed. Filtration occupies the 40% submerged arc, so the full revolution takes $t_\text{cyc}=t_f/0.40=90.6$ s: $$n=\frac{60}{t_\text{cyc}}=\;\boxed{0.66\ \mathrm{rev/min}}\quad(\text{one turn per }\approx91\ \mathrm{s}).$$
  7. Maximum throughput. Every element of the $A_\text{drum}=\pi DL=12.57$ m² surface delivers $v_f$ once per revolution, so filtrate rate $=A_\text{drum}\,v_f\,n=1.56\times10^{-3}$ m³/s. The slurry fed is the filtrate (liquid) plus deposited solids: $$\dot m_\text{slurry}=\frac{\rho_f\,\dot V_f}{1-s}=\frac{1000(1.56\times10^{-3})}{0.60}=\;\boxed{2.60\ \mathrm{kg/s}\ (9.4\ \mathrm{t/h})}.$$
QuantityValue
Filtration constants $C_1,C_2$$5.20\times10^{-4}$, $4.42\times10^{-2}$ (cm, s)
Filtrate depth for 5 mm cake1.125 cm
Submerged time / cycle time36.2 s / 90.6 s
(a) Drum speed0.66 rev/min
Filtrate rate1.56 L/s
(b) Slurry throughput2.60 kg/s ≈ 9.4 t/h
Check (cake-moisture assumption): the solids loading $c$ is computed neglecting liquid retained in the cake (all slurry liquid reports to filtrate). Accounting for the moisture held in a 5 mm cake would raise $c$ and modestly lower the throughput; the exam data (dry-cake bulk density, filtrate density) support the simplified balance used here.