Turn on your faucet. Go on, I’ll wait. Make sure you keep the speed of the water low enough that the water still appears clear (i.e. laminar). Now consider the diameter of the stream near the faucet and then closer to the basin. You should notice that the stream thins as it flows down into the sink. How strange!

Stream thinning is not just a quirk of your sink! It is, in fact, a consequence of the very fundamental law of conservation of mass and energy density in the flow.

To see how this works, let’s consider the equation for the Conservation of Energy Density

P_1 + 1/2\rho v^2_1 + \rho g h_1 = P_2 + 1/2 \rho v^2_2 + \rho gh_2

Here P is pressure, \rho is density of the fluid, v is the velocity of the fluid, g is the gravitational acceleration acting on the fluid, and h is the height of the flow defining its potential energy. This is just our friendly neighborhood Conservation of Energy Equation taken over a unit volume with an additional pressure term.

\frac{KE}{Vol} = \frac{1/2mv^2}{Vol} = 1/2 \rho v^2 \\

\\ \frac{PE}{Vol} = \frac{mgh}{Vol} = \rho gh

From the conservation of energy density we can also define the continuity of mass flow given by

A_1v_1 = A_2v_2

where A is the cross-sectional area of the flow.

Now let’s consider your sink!

From conservation of energy density we have

\cancel{P_1} + 1/2\rho v^2_1 + \cancel{\rho g h_1} = \cancel{P_2} + 1/2 \rho v^2_2 + \rho gh_2

The pressure is the same at both point 1 and point 2 (i.e. atmospheric pressure). We can also arbitrarily assign point 1 to be the point of zero potential energy such that the flow gains potential energy as it moves down towards the basin.

Now let’s solve for the velocity as the flow just leaves your sink.

1/2\rho v^2_1  = 1/2 \rho v^2_2 + \rho gh_2\\

v_1 = \sqrt{v^2_2 - 2gh_2}

This can be rewritten in a clever way to give us a better intuition into the behavior of the flow at different heights.

v_1 = v_2 \sqrt{1 - \frac{2g h_2}{v_2^2}}

Which can be rewritten for the velocity at the base of the flow as

v_2 = v_1 \sqrt{1 + \frac{2gh_2}{v_1^2}}

So as the height of the flow (distance from the faucet to the basin) increases, so will the striking velocity.

Now consider how this affects the continuity of mass.

A_1v_1 = A_2v_2 \\

\pi r_1^2 v_1 = \pi r_2^2 v_2 \\

\\ v_2 = (\frac{r_1}{r_2})^2v_1

So combining both equations we get,

(\frac{r_1}{r_2})^2v_1 = v_1 \sqrt{1 + \frac{2gh_2}{v_1^2}} \\

r_2 = r_1(1 + \frac{2gh_2}{v_1^2})^{-1/4}

We can clean this up through by removing the dimensionality of the equation.

\chi = \frac{v_1^2}{2g}

which is the characteristic distance of the stream. Also, let h_2 = z such that

\frac{r_2}{r_1} = (1 + \frac{z}{\chi})^{-1/4}

Now we can let r^* = r_2/r_1 and z^* = z/\chi such that

r^* = (1 + z^*)^{-1/4}

In conclusion, the stream thins as it moves down because it has been accelerated by gravity and must thin to maintain mass flow and energy!