Question 8 of 9: Tundish Inclusion Removal — Stokes Flotation of Alumina and Argon Micro-Bubbles

Question 8: Tundish Inclusion Removal — Stokes Flotation of Alumina and Argon Micro-Bubbles (20 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. $\mu=7\times10^{-3}$ Pa s, $\rho_{steel}=7000\ \text{kg/m}^3$ (consistent with Q7’s liquid-steel data), $\rho_{Al_2O_3}=3000\ \text{kg/m}^3$, $\rho_{Ar}=1\ \text{kg/m}^3$, depth $=500$ mm, tundish residence time $=8$ min $=480$ s, inclusion diameter $25\ \mu\text{m}$, bubble diameter $600\ \mu\text{m}$.

Find. Stokes rise time for the bare inclusion and for an argon-bubble-carried inclusion; assess whether either clears the tundish inside the 8-minute residence time.

Approach. Balance buoyancy against Stokes drag to get terminal rise velocity for a sphere much less dense than the surrounding steel (both alumina and argon qualify), then divide the rise distance by that velocity; compare each rise time against the stated residence time.

1. Stokes terminal velocity. Force balance (buoyancy = drag) for a sphere of radius $r$: $$\dfrac{4}{3}\pi r^3(\rho_f-\rho_p)g=6\pi\mu rU_\infty\ \Rightarrow\ U_\infty=\dfrac{2}{9}\dfrac{(\rho_f-\rho_p)g\,r^2}{\mu}$$
2. Bare alumina inclusion (25 μm dia., $r=12.5\ \mu\text{m}$). $$U_\infty=\dfrac{2}{9}\dfrac{(7000-3000)(9.81)(12.5\times10^{-6})^2}{7\times10^{-3}}$$ $$\boxed{U_\infty=0.195\ \text{mm/s}}\ \Rightarrow\ t=\dfrac{0.5\ \text{m}}{0.195\times10^{-3}\ \text{m/s}}=\boxed{2569\ \text{s}\approx42.8\ \text{min}}$$ Reynolds check: $Re=\rho U_\infty(2r)/\mu=0.0049\ll1$ — Stokes’ law is fully valid here.
3. Argon micro-bubble (600 μm dia., $r=300\ \mu\text{m}$), carrying the inclusion attached. $$U_\infty=\dfrac{2}{9}\dfrac{(7000-1)(9.81)(300\times10^{-6})^2}{7\times10^{-3}}$$ $$\boxed{U_\infty=196\ \text{mm/s}}\ \Rightarrow\ t=\dfrac{0.5}{0.196}=\boxed{2.55\ \text{s}}$$ Reynolds check: $Re=\rho U_\infty(2r)/\mu\approx118$ — this is well outside the Stokes ($Re\ll1$) regime, so the true bubble rise velocity is somewhat lower than the Stokes estimate (form drag becomes significant); even a substantial correction, however, leaves the float-out time orders of magnitude below the residence time.
4. Compare against the 8-minute (480 s) residence time. $$t_{inclusion}=2569\ \text{s} \gg 480\ \text{s}\ \Rightarrow\ \text{does NOT clear the tundish unaided}$$ $$t_{bubble}=2.55\ \text{s} \ll 480\ \text{s}\ \Rightarrow\ \text{clears easily, with large margin}$$

A bare 25 μm alumina inclusion has essentially no chance of floating out on its own — its 43-minute Stokes rise time is more than five times the tundish residence time, so most such inclusions report directly to the mould unless removed by another mechanism. Attaching it to a 600 μm argon micro-bubble changes the outcome completely: the vastly larger buoyancy-to-drag ratio (bubble radius squared, and a much larger density deficit) cuts the rise time to a few seconds, well inside the residence window, which is exactly why gas-bubble flotation (and, more practically, well-designed flow-control/impact-pad geometry that maximises quiescent surface-directed flow and minimises short-circuiting) is the standard tool for improving inclusion removal in tundish design, alongside dams, weirs and impact pads that lengthen the effective path length and residence time available for inclusions like the bare 25 μm particle to have any chance at all.