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Author Topic: Do small bubbles rise at a different speed than large bubbles?  (Read 4333 times)

Offline AndroidNeox

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Today, a friend commented that small bubbles rise faster than large ones. Is that true?

I haven't noticed that about bubbles. In scuba diving, they tell you to rise no faster than the bubbles. It always seemed to me they rose at pretty much the same speed.


 

Offline Lmnre

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Assume a spherical bubble (I always wanted to say that) of radius r.

Its force of buoyancy is a function of its volume (ie, ~r, and its force of drag is a function of its cross-sectional area and the square of its speed (ie, ~rv). Terminal velocity (ie, constant velocity) is reached when force of buoyancy equals force of drag, or r = rv.

So, using this simplified math, for a bubble with a radius r = 1,

1 = 1v, or 1 = v

and so v = 1.

For a larger bubble of radius r = 2,

2 = 2v, or 8 = 4v, or 2 = v

and so v = 2, or v = √2 = 1.414...

So, actually, larger bubbles should rise faster than smaller ones, the speed being proportional to the square root of its radius. Assuming spherical bubbles, of course.

You can also simplify the original equation, r = rv, down to r = v, or v = √r, which shows the relationship more clearly.

Someone check my assumptions and math please.
« Last Edit: 15/04/2013 17:47:53 by Lmnre »
 

Offline imatfaal

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Assume a spherical bubble (I always wanted to say that) of radius r.

Its force of buoyancy is a function of its volume (ie, ~r, and its force of drag is a function of its cross-sectional area and the square of its speed (ie, ~rv). Terminal velocity (ie, constant velocity) is reached when force of buoyancy equals force of drag, or r = rv.

So, using this simplified math, for a bubble with a radius r = 1,

1 = 1v, or 1 = v

and so v = 1.

For a larger bubble of radius r = 2,

2 = 2v, or 8 = 4v, or 2 = v

and so v = 2, or v = √2 = 1.414...

So, actually, larger bubbles should rise faster than smaller ones, the speed being proportional to the square root of its radius. Assuming spherical bubbles, of course.

You can also simplify the original equation, r = rv, down to r = v, or v = √r, which shows the relationship more clearly.

Someone check my assumptions and math please.


looks fairly sound to me.  Only possible snaglet (not as big as a snag) is that larger bubbles might be more susceptible to deformation - flattening in direction of motion (increasing CS Area without increasing Buoyancy) - than small ones.
 

Offline AndroidNeox

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Thanks for the analysis... nicely done. I seem to recall that the larger bubbles did tend to be flattened horizontally.
 

Offline dlorde

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I saw a fascinating recent TV prog about bubbles, where a researcher into champagne bubbles (how lucky is that?) mentioned that deeper glasses tend to produce bigger bubbles (they expand as they rise due to more gas infiltrating and lowering liquid pressure) which rise faster than smaller bubbles. This significantly changes the way they burst at the surface and the amount of scent molecules they expose and throw into the air (the scent molecules stick to the surface of the bubbles and are released in a plume at the surface as the bubble bursts). This actually changes the taste of the champagne; drinking the same champagne from a narrow, deep glass tastes different from a wider, shallow glass.
 

Offline AndroidNeox

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This actually changes the taste of the champagne; drinking the same champagne from a narrow, deep glass tastes different from a wider, shallow glass.

Brilliant! Thanks. That makes total sense.
 

Offline evan_au

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That agrees with my recollection of playing with rising bubbles in a viscous fluid at the museum.
You can pump a number of small bubbles, which rise slowly through the liquid column, and then make a big bubble, which rises faster and gobbles them up (like Pac-Man!).
 

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