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):
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 h_{0} moves at speed v_{0} and strikes the wall at the angle theta. The value of v_{0} is related to u and theta by
Then suppose that this flow breaks into two jets - with parameters v_{1}, h_{1} and v_{2}, h_{2} respectively. We will find these quantities by means of three hydrodynamics laws.
Flux conservation
h_{0} v_{0} = h_{1} v_{1} + h_{2} v_{2} | (1) |
Energy conservation
h_{0} v_{0}^{3} = h_{1} v_{1}^{3} + h_{2} v_{2}^{3} | (2) |
Equations (1) and (2) can simultaneously be true, if only
v_{0} = v_{1} = v_{2} | (3) |
h_{0} = h_{1} + h_{2} | (4) |
Momentum conservation along z axis
h_{0} Cos theta = h_{1} - h_{2} | (5) |
Solving (4) and (5) together, one gets
h_{1} = h_{0} Cos^{2} theta/2 | (6.1) |
h_{2} = h_{0} Sin^{2} 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:
v_{jet} = u (1 + Cos theta)/(Sin theta) ~ 2 u /(Sin theta) | (7) |
So, thiis is the jet that makes the fountain. Thanks.
Spark |
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