Cumulative jets

When a drop of water falls into a pool of water, one can easily notice that it produces a little fountain. Two principal stages of that process are sketched below. First of all, afer striking the surface, the drop makes a cone in water and then this cone collapses. I will show below that during this collapse an upward cumulative jet is unavoidably produced, making thus the fountain.

Since the process is absolutely left-right symmetrical, I will pay attention only to the left side, imagining that this half cone collapses near a wall. Besides, for convenience, I will draw the picture horizontally (now the Z axis points to the right):

So, this is our geometry: the surface of the cone moves with speed u, angle theta is the half of the cone's vertex angle, the point A is the contact point which moves along axis z with speed:
v_A = u / (Sin theta)

The next step is to come to the reference frame of point A. In this frame of reference, the point A does not move at all, and the liquid just flows near it. This can be see on the next figure:

Note that here I have simplified the problem a bit, replacing the infinite water medium by size-limited flows. This is not crutial for the jet formation. Finally, I will treat only 2D problem, replacing half-cone by simple angle. It does not change the physics, but it simplifies geometry.

So, let see what we have. The incoming flow with width h₀ moves at speed v₀ and strikes the wall at the angle theta. The value of v₀ is related to u and theta by

v₀ = u * cot theta

Then suppose that this flow breaks into two jets - with parameters v₁, h₁ and v₂, h₂ respectively. We will find these quantities by means of three hydrodynamics laws.

Flux conservation

Just see, how much water came in and came out in a small time interval dt. h₀ v₀ dt = h₁ v₁ dt + h₂ v₂ dt which results in

h0 v0 = h1 v1 + h2 v2

(1)

Energy conservation

We consider our water uncompressable. In this limit, kinetic energy is conserved and is not converted into heat. Again, write down kinetic energy brought in and brought out. The common factors (pho dt)/2 in both rhs and lhs of the equation can be leaved out, which yields:

h0 v03 = h1 v13 + h2 v23

(2)

Equations (1) and (2) can simultaneously be true, if only

v0 = v1 = v2

(3)

h0 = h1 + h2

(4)

Momentum conservation along z axis

Now the equation reads: h₀ v₀² Cos theta = h₁ v₁² - h₂ v₂² which can be reduced to:

h0 Cos theta = h1 - h2

(5)

Solving (4) and (5) together, one gets

h1 = h0 Cos2 theta/2

(6.1)

h2 = h0 Sin2 theta/2

(6.2)

That's is it. We established that there must be two jets coming out in opposite directions. So, there exists the jet which goes ahead of point A at fig.3 Now - coming back to our problem (drop in the water), we have to come back to the pool's frame of reference. Which means that our cumulative jet becomes even more fast. Now its velocity is almost doubled:

vjet = u (1 + Cos theta)/(Sin theta) ~ 2 u /(Sin theta)

(7)

So, thiis is the jet that makes the fountain. Thanks.


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Spark