Dot Physics

Speed of a rising drop of oil

The oil spill is still in the news (sadly). One thing that keeps coming up is the speed that the oil bubbles rise to the surface. This is important in different oil-capture methods. The common statement is that smaller bubbles of oil can take quite a long time to reach the surface and larger bubbles can take about 2 days.

This is one of those cases where things do not scale quite the same. Suppose there is a spherical oil bubble rising at a constant speed. Here is a force diagram for such a bubble:

i-0da9d0612a249170ab9739baa17c2a50-2010-06-25_untitled_4.jpg

If this drop is going at a constant speed, then all these forces have to add up to the zero vector. That is fine, but here is the interesting part. Let me describe these three forces:

Gravitational Force

Close to the surface of the Earth, I can just say this force has a magnitude of mg where m is the mass of the drop and g is the gravitational field (9.8 N/kg). The mass is the interesting part. If I assume an oil density of ρoil and a radius of r, then the mass would be:

i-26ec81517e483ba48759a6ec28ab4439-2010-06-25_la_te_xi_t_1_4.jpg

The main point here is that the weight is proportional to r3.

Buoyancy Force

I am not going to go into the details of the buoyancy force (but here are some posts on that topic). Let me just say that the buoyancy force depends on the volume of the oil. So it also has a dependency on r3.

Drag Force

Is this drag force proportional to the velocity or the velocity squared? You know what? It doesn’t matter. What matters is that it depends on the cross sectional area of the oil drop. The bigger the drop, the bigger the drag force. Suppose that this drag force is proportional to the velocity, then I can write the magnitude as:

i-78296fde95d7ffb18210a13cbfdcbfb0-2010-06-25_la_te_xi_t_1_18.jpg

Maybe you can already see the point. This force depends on the radius squared. If I put all of these forces together and solve for the velocity, I get (these are just the y-components of the forces):

i-b7291deec0d70da74c360de82a5eda47-2010-06-25_la_te_xi_t_1_19.jpg

There you have it. Since the buoyancy and the weight essentially depend on the volume (r3), but the drag depends on the area (r2) the r-dependency doesn’t go away. Instead you have a terminal speed that depends on the size of the drop.

Our common intuition says that if you make a bigger drop, all things should be bigger to make the same effect. However, this doesn’t always work. If you double the radius, the volume is 8 times bigger, but the cross sectional area is only 4 times bigger.