Vertical acceleration of the water bubbles in a kettle?

[mastermechanic]

Question is simple, as we all know water boils at the bottom surface and it forms tiny bubbles. These bubbles grow up and rise in the water until they detach. What is the acceleration of these bubbles compared to gravitational acceleration?

- Is it constant velocity?
- Is it approximately equal to gravitational acceleration?
- Or is it significantly larger than gravitational acceleration?

 

[Arjan82]

By far the most part of the rise of the bubble is on it's terminal velocity, so the acceleration part is very short but is indeed driven by gravity, so it is definitely not higher than the gravitational acceleration.

I would say that only at the instant that the bubble breaks free from the bottom the acceleration is about equal to gravitational acceleration. But it almost immediately drops off to a constant velocity. This constant velocity is dependent on the bubble size among other things.

see e.g.:
https://www.sciencedirect.com/science/article/pii/S1738573316303667

 

[anorlunda]

This constant velocity is dependent on the bubble size among other things.

Yes, but the bubble size increases as it rises because of reduced hydrostatic pressure. I've always been curious about the speed of the bubbles as a function of elevation. I don't know the answer.

 

[berkeman]

Yes, but the bubble size increases as it rises because of reduced hydrostatic pressure. I've always been curious about the speed of the bubbles as a function of elevation. I don't know the answer.

Anecdotally, I don't think the vertical velocity of the bubbles changes much with depth. When ascending while scuba diving, it is often taught to ascend at the same rate as your bubbles. That usually doesn't involve any speeding up (kicking) or slowing down (flaring) that I've been aware of. Just a nice constant ascent...

 

[Arjan82]

Anecdotally, I don't think the vertical velocity of the bubbles changes much with depth. When ascending while scuba diving, it is often taught to ascend at the same rate as your bubbles. That usually doesn't involve any speeding up (kicking) or slowing down (flaring) that I've been aware of. Just a nice constant ascent...

As from the paper I linked to, it seems that the terminal velocity doesn't change much over the range of diameters from 1 to 20mm (and possibly a bit beyond):

 

[tech99]

By far the most part of the rise of the bubble is on it's terminal velocity, so the acceleration part is very short but is indeed driven by gravity, so it is definitely not higher than the gravitational acceleration.

I would say that only at the instant that the bubble breaks free from the bottom the acceleration is about equal to gravitational acceleration. But it almost immediately drops off to a constant velocity. This constant velocity is dependent on the bubble size among other things.

see e.g.:
https://www.sciencedirect.com/science/article/pii/S1738573316303667

As far as I can see, no bouyant body can accelerate upwards faster then -g. This is because for there to be upward hydrostatic pressure on it, there must be water beneath it, and as the object rises, water has to fall at acceleration g to fill the void beneath.

 

[erobz]

We can try to make a model with some basic (narrowly applicable) assumptions:

1. Bubble is Ideal Gas
2. Isothermal expansion of the bubble as it rises
3. The bubble remains spherical
4. Quadratic Drag
5. Pressure inside the bubble is equivalent to outside the bubble
6. Neglecting pressure gradient across bubble as it expands

Newtons Second Law:

$$ F_b - W - F_D = m \ddot h \tag{Eq 1}$$

For the buoyant force:

$$ F_b = \rho g V$$

Using the Ideal Gas Law (assumption 1) (isothermal expansion) to find the volume of the bubble:

$$ P_{abs} = \frac{mRT}{V} $$

Equate that to the absolute hydrostatic pressure in the fluid at depth $h_o -h$ (assumption 5):

$$ \frac{mRT}{V} = \rho g (h_o - h) + P_{atm} \implies V = \frac{mRT}{\rho g (h_o - h) + P_{atm} }$$

Substitute:

$$ F_b = \rho g V = \rho g \frac{mRT}{\rho g (h_o - h) + P_{atm} }$$

Force of weight:

$$W = mg$$

Force of Drag:

$$F_D = \frac{1}{2} C \rho \pi r^2 \dot h ^2$$

The cross-sectional area of the bubble will ultimately be a function of the depth, through the volume of the bubble $V$ and spherical bubble (assumption 3):

$$\frac{4}{3}\pi r^3 = V = \frac{mRT}{\rho g (h_o - h) + P_{atm} }$$

This implies that:

$$ r^2 = \left( \frac{3mRT}{4 \pi} \right)^{2/3} \left( \rho g (h_o - h) + P_{atm} \right)^{-2/3}$$

Now just make the following substitutions:

$$ \lambda = \rho g (h_o - h) + P_{atm}$$

$$ \implies \dot h^2 = \left( \frac{1}{\rho g} \right)^2 \dot \lambda^2$$

$$ \implies \ddot h =\frac{-1}{\rho g} \ddot \lambda$$

Making all the substitutions into $( \rm{Eq 1})$:

$$ \frac{m}{\rho g} \ddot \lambda + \rho g m R T \lambda^{-1} -mg - \frac{1}{2} C \frac{\pi}{\rho g^2} \left( \frac{3mRT}{4 \pi} \right)^{2/3} \lambda^{-2/3} \dot \lambda^2 = 0$$That’s at least half the battle. Then just decide on some parameters, initial conditions and solve (numerically) the non-linear, second order ODE.

 

[berkeman]

We can try to make a model with some basic (narrowly applicable) assumptions:

That's not what my bubbles look like...

https://www.google.com/url?sa=i&url...ved=0CAwQjRxqFwoTCJjZhOmYlfsCFQAAAAAdAAAAABAD

 

[erobz]

View attachment 316729

That's not what my bubbles look like...

View attachment 316730
https://www.google.com/url?sa=i&url=https://www.shutterstock.com/image-photo/scuba-diver-bubbles-1028169967&psig=AOvVaw0vMQQ6uWTWMWsXH0edQDUk&ust=1667674199000000&source=images&cd=vfe&ved=0CAwQjRxqFwoTCJjZhOmYlfsCFQAAAAAdAAAAABAD

Yeah I know! But tell that to the spherical cow.

 

[erobz]

In all seriousness, the big ones in that chaos appear to be approximately hemispherical…the small ones are probably close to spherical?

 

[berkeman]

In all seriousness, the big ones in that chaos appear to be approximately hemispherical…the small ones are probably close to spherical?

I tried to find a good picture of the way I remember my bubbles, and that's the closest I found with a quick Google Images search. It's been a while since I went scuba diving (I'm mostly a free diver now), but I remember the flatter ~hemispherical shape a lot more in my bubbles as I was rising among them. They also fluttered a lot as they rose, which I would think would also make them rise a bit slower compared to a smoothly rising bubble.

It seems like the bubbles in little aquarium tanks are a lot more spherical -- I wonder if it's because they are smaller at the source or something. May @BillTre has some thoughts based on all of the aquariums he's worked with...

https://www.google.com/url?sa=i&url...ved=0CAwQjRxqFwoTCOjBkoarlfsCFQAAAAAdAAAAABAL

 

[BillTre]

Just looking at the bubbles pictured in this thread makes me think it involves a balance of two tendencies:

1. The bubble's surface tension will want to draw the bubble into a sphere.
2. the resistance of the water to the bubble raising will tend to flatten it out as the lower end of the raising bubble goes faster that the front.

This flattens out the bubble and increases its resistance to raising (more horizontal surface area). At this point the bubble may break into smaller bubbles seeking their own lower resistance paths up.
This effect should be affected by the viscosity of the water (reynolds number) and the strength of the surface tension between the water and the air in the bubble.

Aquarists usually like small bubbles because they have more surface area for gas exchange. Alternatively air bubbles can be used in air lift tubes where water is driven up through a tube by introducing air at the bottom. This reduces the density of the water causing it to raise wrt the surrounding unaerated water. it also drives the water up with the force of the raising bubbles. In this case, bigger bubbles can drive the water better. Air lifts are very energy efficient but do not make very high pressures.

There are other ways to make bubbles besides running air through a porous surface, like an airstone. You can also make bubbles with a venturi valve )draws air into a flow of water by the Bernoulli effect).

WRT to original posts questions, the rate of raise will vary with bubble size. It can be very slow (vs. expectations).