Minimum-Fluidization and Transport Mass Flux: 23-Chem-A2 Unit Operations and Separation Processes, December 2016, Q2

Question A2: Minimum-Fluidization and Transport Mass Flux (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. Upflow oil $\mu=3\times10^{-3}$ Pa·s, $\rho=900$ kg/m³ through spheres $d_p=0.1$ mm, $\rho_s=2600$ kg/m³, bed voidage $\varepsilon=0.48$.

Find. (a) the superficial mass flux $G_{mf}=\rho u_{mf}$ that just fluidizes the bed; (b) the mass flux $G_t=\rho u_t$ at which particles are carried out (transport = terminal velocity).

Figure A2 — As the superficial velocity rises, the bed passes from fixed, to fluidized at $u_{mf}$, to pneumatic/hydraulic transport once $u$ reaches the single-particle terminal velocity $u_t$.

Approach. For these very fine particles the flow around them is creeping (low $Re_p$), so minimum fluidization follows the laminar (Kozeny) limit of the Ergun equation and the transport velocity is the Stokes terminal velocity; convert each velocity to a mass flux $\rho u$ and confirm the Reynolds numbers justify the laminar forms.

Minimum-fluidization velocity (laminar Ergun limit). $$u_{mf}=\frac{\varepsilon^3}{1-\varepsilon}\,\frac{d_p^2(\rho_s-\rho)g}{150\,\mu}=\frac{0.48^3}{0.52}\,\frac{(10^{-4})^2(1700)(9.81)}{150(3\times10^{-3})}.$$ Evaluating, $u_{mf}=7.88\times10^{-5}$ m/s.
Check the regime. $Re_{mf}=\rho u_{mf}d_p/\mu=2.4\times10^{-3}\ll1$, so the creeping-flow (Kozeny) form is valid.
Mass flux to fluidize. $$G_{mf}=\rho\,u_{mf}=900(7.88\times10^{-5})=\;\boxed{0.071\ \mathrm{kg/m^2\,s}}.$$
Transport velocity = single-particle terminal (Stokes). $$u_t=\frac{d_p^2(\rho_s-\rho)g}{18\,\mu}=\frac{(10^{-4})^2(1700)(9.81)}{18(3\times10^{-3})}=3.09\times10^{-3}\ \mathrm{m/s}.$$ Prose connector: with $Re_t=\rho u_td_p/\mu=0.093<1$, Stokes' law is justified.
Mass flux for transport. $$G_t=\rho\,u_t=900(3.09\times10^{-3})=\;\boxed{2.78\ \mathrm{kg/m^2\,s}}.$$ The window between incipient fluidization and carry-over is a factor $u_t/u_{mf}\approx39$ — the wide operating band typical of fine particulate fluidization.

Condition	Superficial velocity	Mass flux $\rho u$
(a) Minimum fluidization	$7.9\times10^{-5}$ m/s	0.071 kg/m²s
(b) Transport (terminal)	$3.1\times10^{-3}$ m/s	2.78 kg/m²s

Concept & Theory

Topic. Fluidization of fine particles — incipient fluidization and elutriation/transport limits.

Key relations. Ergun laminar limit $u_{mf}=\tfrac{\varepsilon^3}{1-\varepsilon}\tfrac{d_p^2(\rho_s-\rho)g}{150\mu}$; Stokes terminal $u_t=\tfrac{d_p^2(\rho_s-\rho)g}{18\mu}$; mass flux $G=\rho u$.

Why this works. At minimum fluidization the pressure drop equals the buoyant bed weight; for $Re_p<1$ only the viscous Ergun term survives, giving the Kozeny expression. Transport begins when the upward drag on an isolated particle balances its net weight — exactly the terminal-settling condition — so the same drag law (Stokes here) sets the upper limit.

Common pitfall. Using a turbulent drag law (Newton's-law $C_D=0.44$) for sub-100 µm particles: the terminal Reynolds number is well below 1, so Stokes applies. Also forgetting to subtract the fluid density from the solid in the buoyant weight.

Source. Coulson & Richardson Vol. 2, ch. 6 (fluidization) and ch. 3 (motion of particles in a fluid); Perry's §17.

Question 2 of 6: Minimum-Fluidization and Transport Mass Flux

Question A2: Minimum-Fluidization and Transport Mass Flux (25 marks)