Terminal velocity of a gas bubble in a liquid

[Gilded]

As I was watching CO2 bubbles accelerating upwards in my glass of coke and wondered how fast they'll eventually go in a very tall glass, in other words what would be a bubble's terminal velocity, I realized that although it would be interesting to calculate an estimate I needed hydrodynamics, which is something I hate.

 

I've been brushing up my rather non-existant previous hydrodynamics knowledge but I'm still a bit confused. Are small bubbles (let's say 0.001m in diameter) practically rigid spheres? That would certainly make things easier, as I could just use the "normal" terminal velocity formula, right?

[Bignose]

Treating a bubble as a sphere would be a decent 1st approximation. But, a bubble, being a fluid, actually will form circulation zones inside the bubble, breaking that assumption for rigidity.

 

I have some books that give appropriate drag coefficients for bubbles. I'll try to look them up for you later today or tomorrow.

[Mr Skeptic]

Hate to make things even more complicated, but would not the bubble expand significantly as it rises if the glass was very tall?

[Gilded]

Hate to make things even more complicated, but would not the bubble expand significantly as it rises if the glass was very tall?

 

Yeah it sounds logical that it would happen. What also worries me is the possibility that the bubbles would act as nucleation sites themselves which would expand them even more. Although I'm not sure if this happens to a significant extent with small bubbles in an idealized mixture of just water and CO2.

[D H]

Hate to make things even more complicated, but would not the bubble expand significantly as it rises if the glass was very tall?

Conceptually, yes. In a typical drinking glass, no. The ratio of the pressure at the bottom of a 6 inch tall glass to the pressure at the top is only 1.0015. Those tiny bubbles expand by a lot more than 1.5% as they rise from the bottom of the glass to the top.

 

What also worries me is the possibility that the bubbles would act as nucleation sites themselves which would expand them even more. Although I'm not sure if this happens to a significant extent with small bubbles in an idealized mixture of just water and CO2.

The best vehicle for performing such experiments is a crystal champagne glass, filled with champagne of course. A regular drinking glass has too many large nucleation sites. The crystal is smoother, so the bubbles that form are tiny. Ahh, those tiny hypnotizing bubbles ...

 

Anyhow, those bubbles grow considerably more than the 1.5% the can be explained through pressure as they rise from the bottom of the glass to the top. It must be something else, and that something else is just what you said: The bubbles themselves act as nucleation sites.

[swansont]

The drag is proportional to surface area, but the buoyancy force is proportional to volume, so as the bubble gets bigger the latter should grow faster. There shouldn't be a terminal velocity in a carbonated beverage.

 

In regular water or other fluid, you should get a terminal velocity, and bigger bubbles should rise faster than small ones. All of this gets confounded if the bubble deviates from being a sphere, of course.

 

edit: the Reynolds number of a bubble in water is apparently 660

http://sciencelinks.jp/j-east/article/200702/000020070207A0005118.php

 

 

Bubble hydrodynamics:

http://www.bubbleology.com/Hydrodynamics.html

[CaptainPanic]

A decent approximation is Stokes' Law:

 

[math]V_s = \frac{2}{9}\frac{\left(\rho_p - \rho_f\right)}{\mu} g\, R^2 [/math]

 

It does assume that you have a single perfect spherical bubble.

 

However, bubbles are not spherical. The fact that they rise up with many friends joining them influences the terminal velocity significantly (much more than the change in pressure).

The bubble in fact does grow a lot (which is obviously not taken into account in Stokes' Law), but not because of the change in pressure, which is minimal as D H already said. It's due to more CO2 coming out of solution and into the bubble.

 

[edit] Here's a link to some bubble shapes. It's a bit of a weird link: the page won't open (you gotta pay) but Google images does find the pictures. I guess linking to Google pictures cannot be copyright infringement.

Edited August 25, 2008 by CaptainPanic

[booker]

A decent approximation is Stokes' Law:

 

[math]V_s = \frac{2}{9}\frac{\left(\rho_p - \rho_f\right)}{\mu} g\, R^2 [/math]

 

It does assume that you have a single perfect spherical bubble.

 

However, bubbles are not spherical. The fact that they rise up with many friends joining them influences the terminal velocity significantly (much more than the change in pressure).

The bubble in fact does grow a lot (which is obviously not taken into account in Stokes' Law), but not because of the change in pressure, which is minimal as D H already said. It's due to more CO2 coming out of solution and into the bubble.

 

CP,

More than that, the gas in the bubble itself can flow. So I wounder how this effects the boundry layer, if it can be called such.

 

In the case where the viscosity of the gas is taken to the zero limit, the profile of the boundry layer doesn't seem to drop to zero at the interface as it would with a solid sphere but remain at a value determined by Stoke's law of ideal fluids. This would seem to imply that the boundry layer would remain attached from nose to stern.

 

With the velocity of the fluid undiminished from nose and stern, the profile drag should be zero.

 

In addition, would there is no self-sustaining oscillation in the wake, the induced drag would be zero, where the ideal case of zero viscosity gas is still in force.

 

It's hard to imaging where any drag would come from to slow the bubble down at all!

 

Am I missing some crucial factor?

Edited August 26, 2008 by booker

[CaptainPanic]

I'm not sure I understood your point / question...

 

The viscosity term in the formula above is that of the liquid, not of the gas.

 

[math]V_s = \frac{2}{9}\frac{\left(\rho_p - \rho_f\right)}{\mu} g\, R^2[/math]

 

[math]V_s[/math] = terminal velocity of bubble in m/s

[math]\rho_p[/math] = density of particle (in this case that's a bubble!) in kg/m3

[math]\rho_f[/math] = density of fluid (the soda, so take that of water) in kg/m3

[math]{\mu}[/math] = viscosity of fluid (water) in Pa s

[math]g[/math] = gravitational acceleration (m/s2)

[math]R[/math] = radius of particle (radius of bubble, half of the bubble diameter) in m

 

It is true that the gas in the bubble will move, this is induced by the outside of the bubble. (Simplified: the liquid outside of the bubble is stationary, and the gas tries to move upwards through this). There is indeed a boundary layer, both in the gas phase and in the liquid phase.

 

Movement will reduce the thickness of this layer, which enhances mass transport (and perhaps heat transport). So the bubble moving through the liquid already reduces the boundary on the liquid side. Additionally, the turbulence that might be created increases mixing. Inside the bubble, the movement of gas will increase mixing. But in the case of soda bubbles, this is just 1 single molecule (CO2) so mixing in the gas phase is irrelevant for mass transport.

 

I don't understand how you see the equations. But I think you are right: if the liquid viscosity would be zero, there would be no drag. It's kinda theoretical though.

[brian334]

Things that can be tested should be. Things that can not be tested should be guessed about.

[dirtyamerica]

I'm guessing that the terminal velocity would be reached quite rapidly.

 

The viscosity of the liquid should be taken into account too.

[CaptainPanic]

I'm guessing that the terminal velocity would be reached quite rapidly.

 

The viscosity of the liquid should be taken into account too.

 

You're right, and you're right. Check the formula:

[math]{\mu}[/math] = viscosity of fluid (water) in Pa s