# Is it possible to calcualte the yield-stress of a fluid by measuring the smalles bubble that rises through it?

Consider a yield-stress liquid in gravity. I assume that the buyouncy of a small gas-bubble will not be enough to overcome the yield stress (so the liquid doesn't behave liquid), thus leaving the bubble trapped. Is this so in theory, or is there a flaw in my thinking?

Sme argue that there are no yield-stress liquids in sense of one stress, below wich there is no yield, but rather a very high apparent viscosity. Would this imply that even miniscule bubbles will rise, but very slowly?

Lastly, have there been tries to measure viscosity or apparent viscosity or some measure for yield stress from rising bubbles?

-
AFAIK yield stress can be reasonably defined only for solid materials. How can you deform liquid plastically? –  Pygmalion May 29 '12 at 13:13
@Pygmalion: many liquids have a viscosity that rises with decreasing shear. Shampoo is a good example. A yield stress means the viscosity rises to infinity at a finite stress. There is some debate about whether this ever really happens: see my answer for details. –  John Rennie May 29 '12 at 14:01
@JohnRennie OK, I am aware that we have yield stress in terms of rheology, but yield stress is defined as a point when material starts to behave like liquid and not opposite. Or am I missing something? –  Pygmalion May 29 '12 at 14:49
Assuming there's no hysterisis (not always a safe assumption!) the yield stress is the same whether you define it as the lowest stress above which a solid behaves like a liquid or the highest stress below which a liquid behaves like a solid. Industrial rheometers normally start at the high stresses and ramp the stress down, so we think of the yield stress as the point at which the liquid stops flowing. Which is fine, but as I've said below for most non-Newtonian fluids the yield stress probably doesn't exist i.e. the liquid never behaves like a solid. –  John Rennie May 29 '12 at 15:09
So I guess I agree with you :-) –  John Rennie May 29 '12 at 15:09

Suppose your standard error in measuring the particle position is 1mm, then to measure the particle velocity to 95% accuracy you need it to move 3mm. Suppose you can't do the experiment for more than one day (e.g. because of Ostwald ripening) then the smallest velocity you can measure is 3mm/day or $3.5 \times 10^{-8}$m/sec. If you feed this velocity into Stokes' law you'll get the highest viscosity you can measure. If the viscosity is higher than this it will look as if you have a yield stress when actually the viscosity is still finite. –  John Rennie May 29 '12 at 14:34