Settling Velocity to Minimum-Fluidization Velocity Ratio: 23-Chem-A2 Unit Operations and Separation Processes, May 2015, Q2

Question A2: Settling Velocity to Minimum-Fluidization Velocity Ratio (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. Uniform spheres fluidized by a liquid; the ratio requested is the free terminal (settling) velocity $u_t$ over the minimum-fluidization superficial velocity $u_{mf}$.

Quantity	Value
Void fraction at $u_{mf}$, $\varepsilon_{mf}$	0.48
Particle diameter $d_p$	3 mm = 0.003 m
Particle density $\rho_p$	4200 kg/m³
Liquid viscosity $\mu$	$1\times10^{-3}$ N·s/m²
Liquid density $\rho$	1100 kg/m³

Find. $u_t/u_{mf}$. (The bed height $l$ cancels because both velocities are superficial.)

Figure A2 — Minimum fluidization (bed of voidage 0.48 on the point of suspension) versus free terminal settling of a single sphere; the ratio of the two velocities characterises the operable fluidization window.

Approach. Get $u_{mf}$ from the full Ergun-equation force balance (bed weight = pressure drop) written as a quadratic in $\mathrm{Re}_{mf}$, and $u_t$ from Newton's law for the coarse, high-Reynolds settling regime; both are superficial velocities so $l$ drops out.

Archimedes number. This dimensionless group sets both correlations: $$\mathrm{Ar}=\frac{d_p^3\rho(\rho_p-\rho)g}{\mu^2}=\frac{(0.003)^3(1100)(3100)(9.81)}{(10^{-3})^2}=9.03\times10^{5}.$$
Minimum-fluidization velocity — full Ergun quadratic. At incipient fluidization the pressure drop equals the buoyed bed weight; substituting the Ergun equation gives $$\frac{1.75}{\varepsilon_{mf}^3}\mathrm{Re}_{mf}^2+\frac{150(1-\varepsilon_{mf})}{\varepsilon_{mf}^3}\mathrm{Re}_{mf}-\mathrm{Ar}=0.$$ With $\varepsilon_{mf}=0.48$: $15.82\,\mathrm{Re}_{mf}^2+705.3\,\mathrm{Re}_{mf}-9.03\times10^5=0$, so $\mathrm{Re}_{mf}=217.7$ and $$u_{mf}=\frac{\mathrm{Re}_{mf}\,\mu}{\rho\,d_p}=\frac{217.7(10^{-3})}{1100(0.003)}=0.066\ \mathrm{m/s}.$$ $u_{mf}=0.066\ \mathrm{m/s}$ Because $\mathrm{Re}_{mf}\approx218$ (well outside the laminar range), the inertial Ergun term is essential — a laminar-only estimate would be invalid.
Terminal settling velocity — Newton's law. For a large sphere in the Newton (constant-$C_D\!\approx\!0.44$) regime, $$u_t=1.74\sqrt{\frac{g\,d_p(\rho_p-\rho)}{\rho}}=1.74\sqrt{\frac{9.81(0.003)(3100)}{1100}}=0.501\ \mathrm{m/s}.$$ $u_t=0.501\ \mathrm{m/s}$ Check: $\mathrm{Re}_t=\rho u_t d_p/\mu = 1650$, which lies in Newton's-law range ($10^3<\mathrm{Re}<2\times10^5$), so the assumption is self-consistent.
Form the ratio. $$\frac{u_t}{u_{mf}}=\frac{0.501}{0.066}=7.6.$$ $u_t/u_{mf} \approx 7.6$ The bed height $l$ appears in neither velocity, so it cancels identically. Drag-law sensitivity: $C_D=0.44$ is slightly high at $\mathrm{Re}_t\approx1650$; the standard sphere drag curve (Coulson & Richardson Galileo-number correlation) gives $u_t\approx0.53\ \mathrm{m/s}$ and a ratio of about 8.1, so expect $u_t/u_{mf}\approx7.6$–8.1.

Quantity	Result
Archimedes number	$9.03\times10^5$
$\mathrm{Re}_{mf}$ (full Ergun)	217.7
Minimum fluidization velocity $u_{mf}$	0.066 m/s
Terminal velocity $u_t$ (Newton)	0.501 m/s ($\mathrm{Re}_t\approx1650$)
Ratio $u_t/u_{mf}$	7.6

Check (assumptions): spheres are uniform and monosized; $\varepsilon_{mf}=0.48$ is taken as the packed-bed voidage at incipient fluidization; particle sphericity $\phi_s=1$; the liquid is Newtonian; and the terminal Reynolds number confirms the Newton-law (rather than Stokes or intermediate) settling regime.

Concept & Theory

Topic. Liquid–solid fluidization: minimum-fluidization vs. terminal-settling velocity and the operable bed-expansion window.

Key relations. Ergun equation; incipient-fluidization force balance $\Delta P=(1-\varepsilon)(\rho_p-\rho)gL$; Newton's law $u_t=1.74\sqrt{gd(\rho_p-\rho)/\rho}$ ($C_D\approx0.44$); Archimedes number $\mathrm{Ar}=d^3\rho\Delta\rho g/\mu^2$ (MSH ch. 7; Coulson & Richardson Vol. 2 ch. 6).

Why this works. $u_{mf}$ marks where the upward drag first balances the buoyed particle weight, so equating the Ergun pressure drop to the bed weight gives a single quadratic in $\mathrm{Re}_{mf}$; $u_t$ is the velocity at which drag balances gravity on one isolated particle. Their ratio ($\approx u_t/u_{mf}$) sets how far the superficial velocity can be raised before particles wash out — larger for coarse, high-$\mathrm{Re}$ solids.

Common pitfall. Using only the laminar (Kozeny–Carman) term for $u_{mf}$ when $\mathrm{Re}_{mf}\gtrsim20$ — here it would give a wildly high, invalid velocity; and applying Stokes' law for $u_t$ when the true $\mathrm{Re}_t\approx1650$ demands Newton's law.

Source. McCabe, Smith & Harriott, Unit Operations ch. 7 (fluidization); Coulson & Richardson Vol. 2, ch. 6.

Question 2 of 6: Settling Velocity to Minimum-Fluidization Velocity Ratio

Question A2: Settling Velocity to Minimum-Fluidization Velocity Ratio (25 marks)