I was trying to solve the following problem, but my results seem illogical when I graph my solution.

Assuming we have a bubble at the end of a coke glass, I want to solve for the speed of the bubble as a function of its position in the glass. enter image description here

Let's say the height of the glass is $y = H$, and we're looking for the speed of the bubble at $y=h$. What I did was that I tried to find the amount of work done on the bubble. There are two forces exerted on the bubble. The buoyant force is $\rho \times V \times g$, in which $\rho$ is the density of water and $V$ is the volume of the bubble at that height. The other force is the gravitational force which is equal to $n \times g \times M$, in which $n$ is the number of gas moles in the bubble, $M$ is the molar mass, and $g$ is the gravitational constant.

So the overall force is: $(-nM \times g) + (\rho \times V \times g) = (-nM \times g) + (\rho \times \dfrac{nRT}{P} \times g)$

$P(y) = P_0 + \rho \times g \times (H - y)$

Since we have the overall force, we can calculate the work done on the bubble. If we assume the bubble goes straight up(without deviation), we can do so by integrating the overall force with respect to time. The result is:

$-nM \times g \times h + nRT \times ln(\dfrac{P_0 + \rho \times g \times H}{P_0 + \rho \times g \times (H - h)})$

Now using $W = 1/2 \times m \times v^2$, we get:

$v = \sqrt{\dfrac{2RT}{M}ln(\dfrac{P_0 + \rho gH}{P_0 + \rho g(H - h)}) - 2gh}$

When I graphed my results, my answers didn't make sense. What am I doing wrong?

  • 2
    $\begingroup$ The problem is that you're not including fluid drag. I would guess, due to its low mass and high surface area, that a bubble would reach terminal velocity very quickly. $\endgroup$ – probably_someone Jan 28 '18 at 18:45
  • $\begingroup$ In fact, watch the streams of bubbles from small imperfections in the glass. You can see the acceleration as increased bubble distance, and the onset of terminal velocity when it becomes constant. $\endgroup$ – Anders Sandberg Jan 28 '18 at 18:54
  • $\begingroup$ I'm also curious what specifically you mean by "my answers didn't make sense." $\endgroup$ – probably_someone Jan 28 '18 at 18:59
  • $\begingroup$ As well, do you have to include the work done in expanding the volume of the bubble? Or at least take into account the changing volume of the bubble. Finally, shouldn't you have $\rho_{gas}$ somewhere? $\endgroup$ – jim Jan 28 '18 at 19:20
  • $\begingroup$ Why do you think this is an ideal gas? $\endgroup$ – QuIcKmAtHs Jan 28 '18 at 21:50

The kinetic energy of a bubble rising through liquid has hardly anything to do with the mass of the gas, which is negligible, but almost everything to do with the fluid flowing around the bubble. The rate of ascent is governed by the balance between buoyancy and drag, with a tiny correction for acceleration as the bubble grows. Since the bubbles are very small, drag is largely due to viscosity, and the force should scale as the product of viscosity, velocity, and radius.

How fast does the bubble grow? Unless the glass is very tall indeed, ambient hydrostatic pressure changes very little. The bubble grows due to outgassing of carbonation. You might estimate the rate of outgassing as follows: Find the rate that carbon dioxide molecules impinge on the surface of the bubble, given their concentration in the beverage and thermal velocity distribution. The molecules will then have to surmount an energy step, so divide by an appropriate Boltzmann factor. I would use the ratio of the equilibrium concentration in solution at 1 atm to the concentration in the bubble at the same pressure.

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  • $\begingroup$ Basically a totally different solution. ... $\endgroup$ – Alchimista Jan 29 '18 at 12:09

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