Terminal Velocity and Equal-Bed-Density Fluidization: 23-Chem-A2 Unit Operations and Separation Processes, May 2016, Q2

Question A2: Terminal Velocity and Equal-Bed-Density Fluidization (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. Two spheres settling / fluidizing in water at $\rho=1000$ kg/m³; Newton's-law drag with a constant friction factor $R'/\rho u^2=0.22$; Richardson–Zaki index $n=2.3$.

Particle	Diameter $d$	Density $\rho_s$
Glass	12 mm	2500 kg/m³
Metal	1.5 mm	7500 kg/m³

Find. (a) The terminal velocities $u_0$ of each; (b) the single water velocity at which the two fluidized beds carry the same bed density.

Figure A2 — (a) At terminal velocity, net gravity equals Newton-regime drag. (b) Both beds are fluidized by the same upward water velocity; we seek the velocity that makes their voidage-weighted bed densities equal.

Approach. For part (a) balance net gravity against constant-friction-factor (Newton-law) drag to get $u_0$. For part (b) express each bed's density as $\rho_b=\rho_s(1-\varepsilon)+\rho\varepsilon$ with $\varepsilon=(u_c/u_0)^{1/2.3}$, and solve for the common velocity $u_c$ that equalizes them.

Terminal-velocity force balance (Newton regime). Equating buoyant weight to drag with $R'/\rho u^2=0.22$ acting on the projected area gives $\tfrac{\pi}{6}d^3(\rho_s-\rho)g=0.22\,\rho u^2\cdot\tfrac{\pi}{4}d^2$, i.e. $$u_0=\sqrt{\frac{2\,d\,(\rho_s-\rho)g}{3(0.22)\rho}}.$$
Glass terminal velocity. $$u_{0,\text{glass}}=\sqrt{\frac{2(0.012)(1500)(9.81)}{3(0.22)(1000)}}=\sqrt{0.535}=\;\boxed{0.731\ \mathrm{m/s}}.$$
Metal terminal velocity. $$u_{0,\text{metal}}=\sqrt{\frac{2(0.0015)(6500)(9.81)}{3(0.22)(1000)}}=\sqrt{0.290}=0.538\ \mathrm{m/s}.$$ The dense metal is smaller, so despite its higher density it settles slightly slower than the large glass sphere.
Bed density as a function of voidage. A fluidized bed of solids density $\rho_s$ at voidage $\varepsilon$ has bulk density $\rho_b=\rho_s(1-\varepsilon)+\rho\,\varepsilon$. Richardson–Zaki gives the voidage at velocity $u_c$: $\varepsilon=(u_c/u_0)^{1/2.3}$.
Equal-bed-density condition. Setting $\rho_{b,\text{glass}}(u_c)=\rho_{b,\text{metal}}(u_c)$ and substituting $\varepsilon_{\text{glass}}=(u_c/0.731)^{1/2.3}$, $\varepsilon_{\text{metal}}=(u_c/0.538)^{1/2.3}$ yields one equation in $u_c$. Solving numerically (bisection): $$u_c=0.495\ \mathrm{m/s}.$$ At that velocity $\varepsilon_{\text{glass}}=0.844$ and $\varepsilon_{\text{metal}}=0.964$, and both beds share $$\rho_b=\;\boxed{\approx 1235\ \mathrm{kg/m^3}}.$$ Physically the looser (higher-voidage) metal bed offsets its denser solid so the two suspensions weigh the same.

Quantity	Glass	Metal
Terminal velocity $u_0$	0.731 m/s	0.538 m/s
Voidage at $u_c$	0.844	0.964
Common fluidizing velocity $u_c$	0.495 m/s
Common bed density $\rho_b$	≈ 1235 kg/m³

Check (assumptions): the "constant friction factor 0.22" is read as the Coulson&Richardson group $R'/\rho u^2$ (Newton's-law regime), giving a drag coefficient $C_D=8(0.22)/3\approx0.59$; particles are treated as monosized smooth spheres ($\phi_s=1$) and the Richardson–Zaki index is fixed at 2.3 as stated, independent of Reynolds number.

Concept & Theory

Topic. Hindered settling and liquid fluidization — terminal velocity plus Richardson–Zaki expansion.

Key relations. Force balance $\tfrac{\pi}{6}d^3(\rho_s-\rho)g=\tfrac12 C_D\rho u^2\tfrac{\pi}{4}d^2$; Newton regime $C_D\approx$ const; Richardson–Zaki $u_c/u_0=\varepsilon^{n}$ ($n\approx2.3$ turbulent); bed density $\rho_b=\rho_s(1-\varepsilon)+\rho\varepsilon$.

Why this works. A single particle reaches terminal velocity when net weight equals drag; in a fluidized bed each velocity fixes a voidage through Richardson–Zaki, and the voidage sets how much solid is suspended per unit volume. Because glass and metal have different $u_0$, the same velocity produces different voidages, and there is exactly one velocity where the two bulk densities cross.

Common pitfall. Using Stokes' law (valid only for $Re<1$) when the problem specifies a constant Newton-regime friction factor; and confusing bed bulk density with solids density — the voidage term is essential.

Source. Coulson & Richardson, Chemical Engineering Vol. 2 ch. 3 (sedimentation) and ch. 6 (fluidization).

Question 2 of 6: Terminal Velocity and Equal-Bed-Density Fluidization

Question A2: Terminal Velocity and Equal-Bed-Density Fluidization (25 marks)