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If I reduce the surface tension to half of the original value, what would be the effect on the size of an air bubble in the fluid?

What would be the effect on buoyancy and drag forces?

Also, if I have two bubbles one large and one small in a fluid, which will rise faster?

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The effect of surface tension on bubble size

The Young-Laplace law describes the relation of the pressure difference $\Delta p$ on the curvature $C=2/R$ of a spherical bubble: $$\Delta p = 2\frac{\sigma}{R}$$ where proportionality constant is known as the surface tension $\sigma$.

Now, if the surface tension were reduced to half its original value $\sigma/2$, at the same pressure difference, this bubble would exist at a larger radius $2R$. You can understand this as the surface tension being the force which resist an increase in surface area (i.e. radius $R$), and lowering this resistance allows the bubble to expand under its own over-pressure $\Delta p$. An experiment you can do at home is to add some dishsoap to a glass of bubbly soda and see it foam over by the increase in bubble area.

However, most likely what will happen for the case of halving the surface tension, is that the surface tension is unable to support such a sudden increase in radius and the bubble will simply burst. If you were to reduce the surface tension gradually to half the original value, you may be able to see the bubble increase twice in size.

Bouyancy and drag forces on spherical bubbles at low speeds

Assuming the bubble exist with radius $2R$ due to the decreased surface tension, let's look at the bouyancy and drag force for slowly moving spherical droplets. The bouyancy force $F_b=\rho_agV_b$ is the upward force resulting from the bubble occupying a volume $V_b$ in water. As this force is proportional to $V_b$, and the volume is proportional to $R^3$, the bouyancy force will increase by factor 8. The drag force $F_d=6\pi\mu Rv$ is given by Stokes' law and is proportional to $R$, however the speed $v$ may also depend on $R$ so this needs to be investigated first.

A steady-state force balance on the bubble involving gravity, bouyancy and drag forces gives: $$0 = \left(\rho_w-\rho_a\right)gV_b - 6\pi\mu R v$$ solving for the terminal velocity of a spherical bubble: $$v = \frac{\left(\rho_w-\rho_a\right)gV_b}{6\pi\mu R} \propto R^2$$

so the terminal velocity of a bubble is proportional to $R^2$, this makes the drag force proportional to $R^3$. Like the bouyancy, the drag force will increase by a factor 8 and the terminal velocity will increase by factor 4.

This analysis is only valid for spherical bubbles and at low speeds such that $\text{Re}\ll1$ (see below); at all times the results need to be checked against these assumptions.

A more general treatment

The shape a bubble will take depends on which regime it is in as shown in this figure. The regimes are characterized by three dimensionless numbers: $$\text{Re}=\frac{\rho_wvD}{\mu_w}\quad\text{Eo}=\frac{\Delta\rho_wgD^2}{\sigma}\quad\text{Mo}=\frac{g\mu_w^4\Delta\rho}{\rho_w^2\sigma^3}$$ The Reynolds number $\text{Re}$ describes the relative importance of viscous to inertial forces and the Eotvos number $\text{Eo}$ describes the relative importance of bouyancy to surface tension forces. The Morton number $\text{Mo}$ characterizes the type of phases as it contains mostly material properties.

For bubbles at $\text{Re}\ll1$ we clearly see that the bubble shape is spherical inline with the treatment above for slow moving spherical bubbles. While the bubble regimes are presented on a log-log-log scale in which small variations have little impact on the regime the bubble is in, however, for higher Eotvos and Morton numbers we clearly see regime transitions as function of these dimensionless numbers. Now, by halving the surface tension and doubling the diameter we find:

$$\bar{\text{Eo}}=\frac{\Delta\rho_wg(2D)^2}{\sigma/2}=8\text{Eo}\quad\bar{\text{Mo}}=\frac{g\mu_w^4\Delta\rho}{\rho_w^2(\sigma/2)^3}=8\text{Mo}$$

Eotvos and Morton have almost increased by an order of magnitude. Depending on the original regime, this may have cause the bubble shape to shift into a new regime. As this will result in a separate Reynolds number which also depends on the new diameter, the terminal velocity of the bubble as well as the shape may have changed.

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    $\begingroup$ Presumably, it is the number of air molecules inside the bubble that stays constant, not the pressure difference. $\endgroup$ – Mark Eichenlaub Jul 8 '16 at 21:53
  • $\begingroup$ @MarkEichenlaub I would think this as well. I would say that if $\sigma / R$ is constant and $\sigma$ halves, then $R$ must halve as well. What actually happens is that when the surface tension is reduced, then the surface tension can no longer balance the gas's overpressure, so the bubble expands reducing its density until the new gas pressure balances the new overpressure. I get from a naive estimate that the new radius is $\sqrt{2}$ of the old one, since overpressure has one power of the radius, but gas pressure has three. $\endgroup$ – Brian Moths Jul 8 '16 at 22:10

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