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23-Chem-A2 Unit Operations and Separation Processes · May 2018

Question 3 of 6: Velocity to Just Lift a Particle (Free Settling)

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

Notes on this paper

Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — May 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 filter-cake washing, A2 pipe flow & pump power, A3 free-settling of a particle; Part B (Separation Processes) — B1 packed-tower flooding/diameter, B2 adsorption isotherms, B3 multistage liquid–liquid extraction. The rubric asks for two problems per part; all six are worked in full below.

Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §14 packed towers, §15 extraction, §16 adsorption).

Question A3: Velocity to Just Lift a Particle (Free Settling) (25 points)

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
Particle diameter $d_p$$9.3\times10^{-3}$ in $=7.75\times10^{-4}$ ft
Particle density $\rho_p$$1.85\times62.43=115.5$ lbm/ft³
Air density / viscosity$7.52\times10^{-2}$ lbm/ft³ / $1.23\times10^{-5}$ lbm/ft·s
$g,\ g_c$32.2 ft/s², 32.2 lbm·ft/lbf·s²

Find. (a) lift (terminal) velocity $u_t$; (b) $Re_p$; (c) drag force $F_D$; (d) drag coefficient $C_D$.

Particle just suspended on the screen screen air up, u = uₜ dₚ F_D (drag) net weight
Figure A3 — "Just lifted" means upward drag balances the buoyant weight; the required air velocity equals the particle's terminal (free-settling) velocity.

Approach. Set drag equal to net weight to get $u_t^2=4gd_p(\rho_p-\rho_\text{air})/(3C_D\rho_\text{air})$; since $C_D$ depends on the unknown velocity, iterate with the intermediate-regime correlation, then evaluate the drag force from the (exactly balanced) net weight and back-check $C_D$.

  1. Force balance → velocity relation. Equating drag to buoyant weight for a sphere ($A_p=\tfrac{\pi}{4}d_p^2$, $V_p=\tfrac{\pi}{6}d_p^3$) and cancelling $g_c$: $$u_t^2=\frac{4\,g\,d_p(\rho_p-\rho_\text{air})}{3\,C_D\,\rho_\text{air}}=\frac{51.07}{C_D}.$$
  2. Identify the drag regime. A Stokes start ($C_D=24/Re$) gives $u_t\approx10$ ft/s at $Re\approx48$ — far above the Stokes limit ($Re<1$), so the flow is in the intermediate regime. Use $C_D=18.5/Re^{0.6}$ ($1
  3. Iterate $u_t\to Re\to C_D\to u_t$. Converging the coupled pair with $Re_p=\rho_\text{air}u_t d_p/\mu$: $$\boxed{u_t=4.02\ \text{ft/s}\ (1.23\ \text{m/s})},\qquad \boxed{Re_p=19.1},\qquad C_D=\frac{18.5}{19.1^{0.6}}=\boxed{3.16}.$$
  4. (c) Drag force = net weight at terminal velocity. The cleanest evaluation (with $g/g_c=1$): $$F_D=(\rho_p-\rho_\text{air})\tfrac{\pi}{6}d_p^3\tfrac{g}{g_c}=115.4\big(\tfrac{\pi}{6}\big)(7.75\times10^{-4})^3=\boxed{2.81\times10^{-8}\ \text{lbf}}.$$
  5. (d) Consistency check on $C_D$. Back-substituting into the drag definition confirms internal consistency: $$C_D=\frac{F_D}{A_p\,\rho_\text{air}u_t^2/2g_c}=3.16.\ \checkmark$$
QuantityValue
(a) Lift / terminal velocity $u_t$4.02 ft/s (1.23 m/s)
(b) Particle Reynolds number19.1 (intermediate regime)
(c) Drag force $F_D$2.81×10⁻⁸ lbf
(d) Drag coefficient $C_D$3.16
Check: the intermediate law $C_D=18.5/Re^{0.6}$ is a straight-line fit to the standard sphere-drag curve, and near $Re\approx20$ it sits about 20% above that curve. Iterating the full curve instead (Schiller–Naumann, $C_D=\tfrac{24}{Re}(1+0.15Re^{0.687})$, which tracks the Lapple–Shepherd chart) gives $u_t\approx4.52$ ft/s, $Re_p\approx21.4$ and $C_D\approx2.50$. A careful drag-chart solution therefore lands in the band $u_t=4.0$–$4.5$ ft/s. The drag force in (c) equals the net weight and does not depend on the drag law.