# Why do steam bubbles increase in size as they rise?

In the following video (a customer's review of a glass kettle), we can observe water boiling: http://youtu.be/jByY5I7Xk7w?t=2m55s

As the kettle starts to boil at around 2:55, we can see large steam bubbles being formed at the bottom, where the heating element is, and these bubbles shrink as they rise. Presumably this is because they are coming into contact with cooler water. Then we get a crazy convection current for a bit before the element switches off again.

After the chaotic motion has died down (and the fluid is presumably very well mixed) we see small steam bubbles being forming at the bottom, which grow as they rise. I can think of two possible explanations for this, and I'm curious as to which is correct:

1. The water is superheated. Nucleation sites exist on the bottom of the kettle, so that's where steam bubbles form. Steam is produced at the interface between steam and water, which causes the bubbles to grow as they rise.

2. The pressure at the bottom is slightly higher than at the top. Assuming a depth of 15cm, the boiling point at the bottom of the water is about $100.3^\circ \mathrm{C}$, compared to $100.0^\circ \mathrm{C}$ at the top. Bubbles form at the bottom because the heating element is still slightly hotter than $100.3^\circ \mathrm{C}$, and as they rise they drag hot water up into the slightly lower-pressure area, where it turns to steam because its boiling point lowers, and this increases the size of the bubble.

In particular I'm interested in whether the second of these explanations plays a role. If it doesn't happen in a boiling kettle, is there any situation in which it does?

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Why did you turn down the simpler explanation - that as the bubbles rise the surrounding pressure decreases, and they simply expand ($V=\frac{Nk_bT}{P}$) without transfer of mass or energy? – yohBS Mar 5 '12 at 19:23
@yohBS purely because the change in pressure is only about 1.5 Pa, compared to the ambient pressure of 100 Pa, so the volume change due to that effect could only be a couple of percent at most - the amount of expansion visible in the video is much more than that. – Nathaniel Mar 5 '12 at 19:30
There are times where bubbles rising in boiling water would shrink. Can you think why? – John Berryman Mar 6 '12 at 4:41
@JohnBerryman I mentioned shrinking bubbles in the question - in this case I'm fairly sure it's because before it gets mixed, the water at the top is colder. (The same thing happens in the bubble tubes in old juke boxes, which contain water at low pressure, with a heating element at the bottom.) Or are you thinking of some other situation? – Nathaniel Mar 6 '12 at 8:53

The discussion of nucleation sites is very much to the point. Water at atmospheric pressure without nucleation sites will theoretically boil only at $320.7 {}^o C$. The bubbles act as nucleation sites that reduce the energy required for evaporation. In the case of a bubble, the effective contact angle between the superheated liquid and the bubble surface is $180 {}^o$ which reduces the superheat needed to evaporate the water to $0$. Interestingly, there is actually an impediment to bubble growth caused by the reduced temperature of the vapor inside the bubble and a corresponding lower superheat boundary layer of liquid surrounding it.

FYI: My information is based on Collier and Thome pages 138 and 549.

In that text, an equation is given for the rate of bubble growth as:

$$R = \sqrt{\frac{12\alpha_f}{\pi}} \frac{\rho_f c_{pf} \Delta T}{\rho_g i_{fg}}\mbox{Sn} t^{1/2}$$

where

$$\mbox{Sn} = \left[ 1-(y-x)\sqrt{\frac{\alpha_f}{D}}\left(\frac{c_{pf}}{i_{fg}}\right)\left(\frac{\partial T}{\partial x}\right)_p\right]^{-1}$$

and the variables are:

$R$ - rate of bubble growth

$\alpha_f$ - thermal diffusivity of liquid

$\rho_f$ - density of liquid

$c_{pf}$ - specific heat of liquid phase

$\Delta T$ - temperature difference

$\rho_g$ - gas density

$i_{fg}$ - latent heat of vaporization

$t$ - time

$D$ - molar diffusion coefficient

It's been a while since I looked at this in depth, but I think the $x$ and $y$ variables refer to position relative to a uniformly heated tube coaxial to the $y$ axis. Honestly, I don't expect you to actually use this formula, but hopefully it will impress that there are people who have spent a great deal of time on this subject. If you find it interesting, you might have a promising career in power plant boiler engineering in general or nuclear power plant engineering in particular.

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Huh, I never knew the threshold for boiling in the absence of nucleation sites was so high - I always thought of superheated water as being a rather unusual state, rather than the normal state for hot water in a kettle. So, in order to observe the second phenomenon I mentioned, I would need to heat a very tall column of dirty water (i.e. with plenty of impurities to form nucleation sites), right? – Nathaniel Mar 6 '12 at 9:02
It is quite astounding how much superheat is required to cause homogenious boiling (boiling without nucleation sites) but note that I said "theoretically;" even in closely controlled experiments no one has ever obtained that much superheat. If they did, it would be a very unstable situation. The Mythbusters had a bit on "exploding water" that is relevant. As to the dirty water, it depends on which phenomenon you are interested in. The dirt would provide nucleation sites, but would not much alter bubble growth. Now we're talking about Mentos in soda. :) – AdamRedwine Mar 6 '12 at 13:20
What I meant was, if I heat a very tall column of dirty water, I should be able to see bubbles growing because of (possibly bubble-induced) convection carrying hot water into lower pressure regions, so that its boiling point lowers, rather than because the water is superheated. It sounds like a rather hard experiment to try, however. (I'll check out the Mythbusters episode as soon as I have time :) ) – Nathaniel Mar 6 '12 at 15:48
It seems to me that dependance of bubble growth rate from liquid pressure would be minimal. There are a number of other effects that alter growth rate; I'll add some of the math to my answer. – AdamRedwine Mar 6 '12 at 16:32