# Why does water falling slowly from a tap bend inwards?

This is something anyone could easily verify. When we open a tap slowly, water bends inwards (towards the axis) while maintaining its laminar flow. After a certain height below the opening, the flow becomes turbulent. I've approximately illustrated the shape of water near the top portion in the following diagram:

I tried to explain the above phenomenon based on my knowledge on fluid dynamics. Let us consider the following diagram:

Here, $$A_1$$ and $$A_2$$ are the areas of cross-section and $$v_1$$ and $$v_2$$ are the speeds of water molecules at two different heights (indicated by dotted red lines).

Since, the shape of water remains fairly constant and the flow is laminar, in a time interval $$\Delta t$$, the volume of water passing through level 1 must be equal to the volume of water passing through level 2. Mathematically, we can say:

$$A_1v_1\Delta t=A_2v_2\Delta t$$ $$A_1v_1=A_2v_2$$

Or in other words, the product of the area of cross section and the speed remains the same at all heights and this is known as the equation of continuity. Since water molecules are under the gravitation force of attraction, they are accelerated downwards. So, $$v_1. As the product of area of cross section and the speed must be a constant, $$A_1>A_2$$. This explains why water bends towards the axis while falling slowly from a tap.

But the above explanation fails at much lower heights above the fluctuating flow zone (where flow fluctuates from laminar to turbulent). Let us consider another diagram:

The area of cross-section remains almost constant at the intermediate heights above the red zone. It doesn't decrease in accordance to the equation of continuity. Further, my method of explanation involves a lot of assumptions and I've also neglected surface tension, viscosity etc. I'm unable to imagine how these forces would affect our results.

Is this a correct reason for "Why does water falling slowly from a tap bend inwards?" or is there any better explanation for this phenomenon?

Image Courtesy: My own work :)

• After a certain height below the opening, the flow becomes turbulent. How do you know this? – Gert Jan 19 '20 at 14:35
• @Gert: You asked "After a certain height below the opening, the flow becomes turbulent. How do you know this?". I've seen this myself. I can also explain it based on Reynold's number which is directly proportional to the speed. As the speed increases so does the Reynold's number. This explains why the flow is initially laminar, then fluctuating between laminar and turbulent and then finally turbulent. If you wish, you may also verify it yourselves :) – Guru Vishnu Jan 19 '20 at 14:38
• It might be that your observation is correct but it is my observation that after falling for a particular height the water falling from the tap forms droplets. – Kenzo Tenma Jan 19 '20 at 14:48
• it's just surface tension and the work of gravity but if you placed your finger under the water you will see some ripples... that's more fun i guess ;D – user6760 Jan 19 '20 at 15:20
• @user6760: That is not turbulence; that is simply a standing wave, and if you open the tap just a tiny bit you can easily observe such standing waves especially the nearer you put your finger to the top (where there should be less turbulence). – user21820 Jan 20 '20 at 4:02

You can actually predict the shape of the profile precisely using the arguments you mention above, which are by and large correct. To do so, you can make the following assumptions:

• Neglect viscosity (not a great assumption, but it's a start).
• The pressure is the same everywhere in the fluid—the edges are free surfaces, so this is reasonable.
• The flow is axially symmetric (i.e. the top-down cross section is always circular).

If you do this, and take the location of the faucet as the origin, you can then state the relationship between the gravitational potential energy and the flow speed using Bernoulli's equation as:

$$\rho g h + \rho \frac{1}{2}v^2 = \rho \frac{1}{2}v_0^2$$

where $$v$$ is the speed of the fluid as a function of height $$h$$, $$\rho$$ is density, and $$v_0$$ is the speed at which the water leaves the faucet.

Solving for $$v$$, you'll find that:

$$v = \sqrt{v_0^2 - 2gh}$$

As the fluid moves further down (i.e. as $$h$$ becomes further negative), the speed increases as you'd expect.

Then you can use conservation of mass for the rest. Assuming steady flow, you'll find that

$$A_1 v_1 = A_2 v_2$$

for any two cross-sections of the flow. Using the cross-sections at the faucet and another arbitrary cross-section, and declaring the faucet radius as $$r_0$$, you'll find:

$$\pi r_0^2 v_0 = \pi r^2 v$$ $$\pi r_0^2 v_0 = \pi r^2 \sqrt{v_0^2 - 2gh}$$

Solving for the radius $$r$$, you find up getting the following expression:

$$\boxed{r(h) = \frac{r_0 \sqrt{v_0}}{(v_0^2 - 2 g h)^{1/4}}}$$

This drop in the radius as the height decreases is consistent with your illustrations. For example, here is what I analytically determine as the flow profile when I use standard values for a bathroom sink faucet flow ($$r_0 = 1.5$$ centimeters, $$v_0 = 0.134$$ meters per second, and $$g = 9.81$$ meters per second squared):

Notice that the flow profile becomes effectively straight at distances observable in your common bathroom sink (4 inches or so). This is consistent with your observations.

After a certain point, the stream becomes so thin that surface tension effects along with shearing at the air-water interface begin to destabilize the shape and cause it to break up into droplets. In addition, the flow becomes turbulent after a certain distance from the faucet, so this prediction is only accurate for the early stages of such a flow (i.e. for "small" $$h$$).

• I did the same calculation as you, but only used different starting values. Thus, I'm not going to post it. However, I believe your $v_0 = 5.6um/s$ is not realistic, but you should use something like 1liter per 10sec, which yields $v_0 = 0.14m/s$. As this changes your graph considerably, would you mind updating your answer? – Semoi Jan 19 '20 at 17:48
• Ah, I caught a mistake in my calculations for the starting values—thank you for pointing it out. Will redo the graph with the correct values. – aghostinthefigures Jan 19 '20 at 18:03
• I think viscosity in this case is highly negligible. All water molecules are accelerated uniformly and hence there'll be no relative motion between different profiles. And thus the viscous force is negligible or the approximation is very close to reality. Am I right? – Guru Vishnu Jan 20 '20 at 3:44
• You should totally avoid speaking about molecules this way. You can talk about imaginary continuum particles if you want but if you are talking about continuum mechanics, you can't talk about molecules or at least not about their relative speeds. The molecules are not moving coherently, but in a random thermal movement. V1 and v2 are not molecules speeds but the average speed of many molecules in the vicinity. – Vladimir F Jan 20 '20 at 10:24
• Here's my take—you could try to approximate the viscous effects associated with this flow if you determine the flow field for this physical set-up using the calculations I did above and some vector calculus common sense (i.e. which way the flow is pointing being determined through symmetry/conservation of mass arguments). You could then calculate the velocity gradients associated with the flow and find where those gradients are biggest, as viscosity is proportional to velocity gradients—those are the outer parts of the flow and the initial bending region IMO. – aghostinthefigures Jan 20 '20 at 15:03

To enlarge slightly upon @aghostinthefigures' excellent exposition, for small gravity-driven jets the flow does not go turbulent- instead, it is subject to rayleigh instability when its cross-section gets small enough for surface tension forces to become dominant. At that point, any small perturbation of the jet will cause it to spontaneously break up into individual droplets before the flow in the jet has the opportunity to become turbulent.

• Right. The surface area associated with a given volume of water is increasing. That takes energy. Eventually, the path of least energy is to form droplets rather than a thinner stream. The ability to create a droplet stream of uniform size argues strongly against turbulent conditions. Both necking and drop formation treated here – Phil Sweet Jan 20 '20 at 2:00
• And the ripples thing:) The bizarre ripples that form in a stream of water – Phil Sweet Jan 20 '20 at 2:13
• One of my jobs in a former lifetime was to design and build a strobe microscope with which to photograph rayleigh instability in inkjet droplets leaving the nozzles of an inkjet printhead, this was very cool stuff. – niels nielsen Jan 20 '20 at 3:25
• @nielsnielsen I also use " ... in a former lifetime ..." -> it's quite fun to come upon it in contexts like this :-). In a previous lifetime I, well which one :-). China seemed to feature quite a lot. – Russell McMahon Jan 22 '20 at 8:45

From the continuity equation for steady flow, $$A_1v_1=A_2v_2$$ as the water leaves the tap opening which is at a height $$h$$ from the datum line then its velocity increases parabolically as height $$h$$ decreases according to third equation of motion $$v_2^2=v_1^2-2gh$$. The increase in velocity causes the cross-sectional area to decrease non linearly as the height $$h$$ decreases while maintaining laminar flow for certain fall in height. As a result the water falling from the tap becomes narrower & narrower in cross-section i.e. it bends towards its axis until the flowing water breaks into droplets (keeping volume constant but less surface area) due to Rayleigh Instability.