How is the power of water flowing through a pipe directly proportional to $v^3$ and not $v^2$? (Where $v$ is the velocity of water)

I have been trying to write power in terms of $v^3$ but I'm unable to figure it out. Though according to me, it should have been $v^2$ because power is the work done divided by time.


Consider a kilogram of water flowing past a particular spot in the piping.

If you double the velocity of the water, what happens to the kinetic energy of that kilogram?

And what happens to the time it takes for that kilo to pass the given point?

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  • $\begingroup$ How can a mass of water flow past a point? $\endgroup$ – Lamar Latrell Oct 27 '18 at 7:47
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    $\begingroup$ @LamarLatrell I assume DJohnM meant a cross-section. $\endgroup$ – J.G. Oct 27 '18 at 8:06

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