Terminal Velocity and Matched Fluidized-Bed Densities: 23-Chem-A2 Unit Operations and Separation Processes, December 2019, Q2

Question A2: Terminal Velocity and Matched Fluidized-Bed Densities (25 points: 12.5 + 12.5)

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.

Particle	Diameter $d$	Density $\rho_p$
Glass	12 mm = 0.012 m	2500 kg/m³
Metal	1.5 mm = 0.0015 m	7500 kg/m³
Water	—	1000 kg/m³
Drag group $R'/\rho_f u^2 = 0.22$ (Newton's-law regime); $(u_c/u_0)=e^{2.3}$

Find. (a) The terminal velocities of both particles; (b) the single water velocity $u_c$ at which the two fluidized beds have equal bulk (bed) density.

Figure A2 — At terminal velocity, drag balances submerged weight. In fluidization the looser (higher-voidage) bed is the lighter one; equating bulk densities $\rho_b=\rho_p(1-e)+\rho_f e$ fixes the single superficial velocity that satisfies both particle systems.

Approach. (a) Balance submerged weight against Newton-regime drag with $R'=0.22\rho_f u^2$ to solve for $u_0$. (b) Express each bed's voidage from $e=(u_c/u_0)^{1/2.3}$, write $\rho_b=\rho_p(1-e)+\rho_f e$ for both, and find the $u_c$ that makes the two bed densities equal.

Terminal-velocity force balance. Submerged weight $=\tfrac{\pi}{6}d^3(\rho_p-\rho_f)g$ equals drag $=R'\cdot\tfrac{\pi}{4}d^2$ with $R'=0.22\rho_f u_0^2$. Cancelling and solving, $$u_0=\sqrt{\frac{d(\rho_p-\rho_f)g/6}{0.22\,\rho_f/4}}.$$
Glass and metal terminal velocities. Substituting each particle: $$u_{0,\text{glass}}=\sqrt{\frac{0.012(1500)(9.81)/6}{0.22(1000)/4}}=\boxed{0.732\ \text{m/s}},\qquad u_{0,\text{metal}}=\sqrt{\frac{0.0015(6500)(9.81)/6}{0.22(1000)/4}}=\boxed{0.538\ \text{m/s}}.$$ The small dense metal particle settles slower than the large lighter glass sphere.
Voidage from the fluidization law. For any common water velocity $u_c$, $e=(u_c/u_0)^{1/2.3}$, so a given $u_c$ gives a looser bed for the particle with the lower $u_0$ (metal). The bed bulk density is $\rho_b=\rho_p(1-e)+\rho_f\,e$.
Match the bed densities. Setting $\rho_{b,\text{glass}}=\rho_{b,\text{metal}}$ and solving numerically gives $$\boxed{u_c=0.495\ \text{m/s}},\quad e_\text{glass}=0.844,\ e_\text{metal}=0.964,$$ at which both beds have $\rho_b\approx1235\ \text{kg/m}^3$. Note $u_c=0.495

Quantity	Glass	Metal
Terminal velocity $u_0$	0.732 m/s	0.538 m/s
Voidage $e$ at match	0.844	0.964
Matched water velocity $u_c$	0.495 m/s
Common bed density $\rho_b$	≈ 1235 kg/m³

Concept & Theory

Topic. Newton-regime terminal velocity and Richardson–Zaki-type bed expansion.

Key relations. Force balance $\tfrac{\pi}{6}d^3(\rho_p-\rho_f)g=0.22\rho_f u_0^2\tfrac{\pi}{4}d^2$; expansion law $u_c/u_0=e^{n}$ (here $n=2.3$); bed density $\rho_b=\rho_p(1-e)+\rho_f e$.

Why this works. A constant drag coefficient ($C_D\approx0.44$, i.e. $R'/\rho_f u^2=0.22$) is Newton's-law regime, valid for large $Re_p$; it makes $u_0\propto\sqrt{d(\rho_p-\rho_f)}$ so the two particles have comparable terminal velocities despite very different sizes. In fluidization, a fixed superficial velocity fixes each bed's voidage through the expansion law, and because bulk density falls as voidage rises, a unique $u_c$ equalises the two.

Common pitfall. Forgetting buoyancy (using $\rho_p$ instead of $\rho_p-\rho_f$), or checking that the chosen $u_c$ stays below the smaller terminal velocity so both beds remain fluidized rather than transported.

Source. Coulson & Richardson Vol. 2 ch. 3 (motion of particles) & ch. 6 (fluidization); McCabe ch. 7.

Question 2 of 6: Terminal Velocity and Matched Fluidized-Bed Densities

Question A2: Terminal Velocity and Matched Fluidized-Bed Densities (25 points: 12.5 + 12.5)