# How long does it take for oil to coalesce in water?

I was studying the process of coalescence in emulsions. We considered $$N$$ bubbles of liquid 1 floating in liquid 2. The result we derived, is that if there are some dissipative forces (diffusion) the most stable configuration is for the $$N$$ bubbles to coalesce/fusion and become 1 bubble (to reduce surface tension).

So I tried to make the experiment at home to see how long would it take. I emulsified oil in water. After one day, it has not changed much.

I tried to estimate the time it would take using Stokes-Einstein equation: $$\begin{equation} D=\frac{k_{\rm B}T}{6\pi \eta r} \end{equation}$$
where $$\eta$$ is the viscosity of (olive) oil (?) and $$r$$ the radius of the bubbles (about 1 mm).

And then using the result of Brownian motion for 2D: $$\overline{x^2}=4Dt$$ I replaced $$\overline{x^2}$$ by the surface of my container, about $$R=$$ 4 cm of radius so $$\overline{x^2}=\pi R^2$$. Mixing both equations, I can derive the time it takes (for room temperature). But I get an enormous duration of $$10^5$$ years.

Is this the true value or I am doing something wrong?

Update: Day five, the smallest bubbles ($$r<1$$ mm) are gone but still $$N>100$$.

Update2: Day six. No coalescence. I stopped the experiment.

Update3: Obvious problem of leaving the experiment for so long is that the water starts to evaporate.

• are you sure you want to use the Stokes-Einstein equation, since it is specifically for low Reynold's number? – N. Steinle Dec 18 '18 at 13:51
• A few points : 1. olive oil is not pure, so don't expect to see a real phase separation. 2. In your calculation, you should replace $x$ not by $R$, but by the average distance $d$ between the bubbles. 3. Gravity plays also a role since oil is less dense. Personnally, I would rather write $t=(R\rho v /gv\Delta \rho)^{1/2}$ that womes out from $ma=F$ with $v$ the volume of the small bubbles and $\rho$ their density, and $\Delta\rho$ the difference of density with water. – J.A Dec 18 '18 at 14:01