23-Chem-A2 Unit Operations and Separation Processes · December 2014
Question 3 of 6: Terminal Settling Velocity of Solid Particles
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 2014. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, crystallisation); Perry's Chemical Engineers' Handbook (9th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.).
Question A3: Terminal Settling Velocity of Solid Particles (25 marks)
Given. Gravity settling of dense particles in water; the drag chart is used because the Reynolds number falls in the intermediate (transition) regime where Stokes' law does not apply.
Quantity
Value
Particle diameter $d$
$1.5\times10^{-4}\ \mathrm{m}$
Particle density $\rho_p$
$2800\ \mathrm{kg/m^3}$
Water viscosity $\mu$
$8\times10^{-4}\ \mathrm{kg/m\,s}$
Water density $\rho$
$996\ \mathrm{kg/m^3}$
Separator acceleration (b)
$390\ \mathrm{m/s^2}$
Find. (a) terminal velocity under gravity; (b) settling velocity under $390\ \mathrm{m/s^2}$.
Figure A3 — At terminal velocity the net body force equals drag. The group $C_D\mathrm{Re}^2$ is independent of $v$, so the chart is entered without iterating on velocity.
Approach. Form the velocity-free group $C_D\mathrm{Re}^2$ from the force balance; read (or fit) the drag curve to get $\mathrm{Re}$, then back out $v$. The identical procedure applies in the separator with $g$ replaced by the imposed acceleration.
Velocity-independent drag group. At terminal velocity, weight−buoyancy = drag gives $C_D\mathrm{Re}^2$ free of $v$: $$C_D\mathrm{Re}^2=\frac{4\,d^{3}\rho(\rho_p-\rho)g}{3\mu^{2}}=\frac{4(1.5\times10^{-4})^3(996)(1804)(9.81)}{3(8\times10^{-4})^2}=123.9.$$
Read the chart via the intermediate-law fit. In the transition regime the sphere curve of Fig. A3 is well represented by $C_D = 18.5\,\mathrm{Re}^{-0.6}$, so $C_D\mathrm{Re}^2=18.5\,\mathrm{Re}^{1.4}$: $$\mathrm{Re}=\left(\frac{123.9}{18.5}\right)^{1/1.4}=3.89.$$
(a) Terminal velocity. Inverting the Reynolds number, $$v_t=\frac{\mathrm{Re}\,\mu}{\rho\,d}=\frac{3.89(8\times10^{-4})}{996(1.5\times10^{-4})}=0.0208\ \mathrm{m/s}.$$ Since $\mathrm{Re}=3.9>1$, Stokes' law would have over-predicted $v_t$; the chart value governs. $v_t \approx 0.021\ \mathrm{m/s} = 2.1\ \mathrm{cm/s}$
(b) Settling in the separator. With $g\to a=390\ \mathrm{m/s^2}$ the group scales linearly, $C_D\mathrm{Re}^2 = 123.9\times\dfrac{390}{9.81}=4926$, so $$\mathrm{Re}=\left(\frac{4926}{18.5}\right)^{1/1.4}=54.0\;\Rightarrow\; v=\frac{54.0(8\times10^{-4})}{996(1.5\times10^{-4})}=0.289\ \mathrm{m/s}.$$ The $\sim$40-fold acceleration raises the settling velocity about 14-fold (sub-linear, because drag stiffens as $\mathrm{Re}$ rises). Chart-reading check: the $18.5\,\mathrm{Re}^{-0.6}$ fit lies slightly above the plotted sphere curve near $\mathrm{Re}\approx50$; reading the full standard sphere curve (Schiller–Naumann, $C_D=\tfrac{24}{\mathrm{Re}}(1+0.15\,\mathrm{Re}^{0.687})$) gives $\mathrm{Re}\approx59$ and $v\approx0.32\ \mathrm{m/s}$ here (and $v_t\approx0.020\ \mathrm{m/s}$ in (a), within 3.5 %), so a careful chart read lands in the band 0.29–0.32 m/s. $v \approx 0.29\ \mathrm{m/s}$ at $390\ \mathrm{m/s^2}$
