3
$\begingroup$

I was in a friends garden and saw this:

enter image description here

My question is: what causes the water to flow towards the center? My first thought was that maybe the water in the center falls faster and thus creates a sort of force inwards, but because gravity doesn't care about weight I don't think thats correct...

I also noticed that the width of the water is everywhere the same, exept for the edges. The water which flows towards the center forms a small tube there.

What causes this effect? Or is it simply because of the design of this fountain? (I don't think so, I've seen this before on other designs)

$\endgroup$
  • 3
    $\begingroup$ Probably because of water's surface tension? $\endgroup$ – Kunal Pawar May 21 '17 at 16:53
  • $\begingroup$ @KunalPawar I believe other fluids behave like that as well.. But I don't really know :D $\endgroup$ – Sebastian Schneider May 21 '17 at 16:54
  • 2
    $\begingroup$ Yes, all fluids have the tendency to minimise their surface area. It takes energy to have a bigger surfav area. $\endgroup$ – Kunal Pawar May 21 '17 at 16:55
  • 1
    $\begingroup$ @innisfree it's difficult to make that out from the picture but it looks like it isn't being thrown towards the centre. $\endgroup$ – Kunal Pawar May 21 '17 at 16:57
  • 3
    $\begingroup$ I think that it's simply due to surface tension and the fact that the water moves faster as it falls. Clearly, the water stream can't have a constant cross-sectional area as it falls and gathers increasing speed because the volume of water passing by any point in the stream has to remain constant. So the cross-sectional area has to decrease. In this case of a stream with a long, thin, horizontal cross-section with a large surface area, surface tension causes the stream to contract horizontally as it falls. $\endgroup$ – Samuel Weir May 21 '17 at 17:15
5
$\begingroup$

My guess is conservation of mass flow rate. Let's imagine to cut the water flow with an imaginary surface: since mass must be conserved, the quantity of mass passing through this surface per unit time must be a constant:

$$\dot m = \rho \ u \ A = const$$

where $\rho$ is density, $u$ is the velocity of water and $A$ is the cross section of the water flow.

The density is constant in this situation (1). Moreover, let's assume that the thickness of the water flow is almost constant (it looks like it is, up to a good approximation). We will then have

$$A = const \cdot l$$

where $l$ is the width of the water flow. Therefore

$$u \cdot l = const \rightarrow l = \frac{const}{u}$$

Let's take a $z$ axis to start at the top of the fountain (when the water starts to fall) and to finish when the flow enters in the water; it is then easy to see from conservation of energy that

$$u (z) = \sqrt{2 g z}$$

so that

$$l(z) = \frac{const}{\sqrt z}$$

(yes, we are assuming that $g$ is also constant: quite reasonable in this case!)

Of course, this formula cannot be completely right, because it would give you a divergence in $z=0$. We are probably neglecting some other kind of energy (I am betting on surface tension), and the "true" form must probably be something like

$$l(z) = \frac{1}{\sqrt {a z + b}} \ \text{cm}$$

where $a$ and $b$ are constants. You can already see by plotting the function $1/\sqrt{z+1}$ that the shape looks similar to the one in the picture.

So, in conclusion, my opinion is that the shape of the water flow is not that a triangle, but rather that it behaves as $\sim 1/\sqrt z$.

(1) As a matter of fact, the density of water is almost always constant, since water is with very good approximation an incompressible fluid.

$\endgroup$
  • $\begingroup$ Thank you for you're answer. I think that's quite logical $\endgroup$ – Sebastian Schneider Jun 1 '17 at 23:24
  • 2
    $\begingroup$ Dang, when I read this question I was pretty sure it had something to do with conservation of mass/flow rate. I never thought to try and derive the stream width as a function of height. Great answer and great catch on the inverse square root relationship. $\endgroup$ – JMac Jun 1 '17 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.