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Ron19932017

- 32

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I have a little problem while reading Feynman lecture note volume 2, chapter 40, section 40-3.

In that section Mr Feynman talked about Bernoulli equation and a example of a filled water container.

In high school we learned that if we drill a hole near bottom of container, we can get the water outflow speed from Bernoulli equation. However Feynman mentioned that it only considered the conservation of energy (energy conservation is equivalent to Bernoulli equation of fluid) but not the conservation of momentum.

There is a net outward momentum carried by the outflow water. Thus in the most simplest case the static pressure at the opposite position of the hole is not the normal static water pressure in nearby environment.

Mr Feynman wrotes " Then the static pressure at any point on the side of the tank must be matched by an equal pressure at the point on the opposite wall,

*except*at the points on the wall opposite the charge tube. If we calculate the momentum poured out through the jet by this pressure".

I understand how Feynman argued that the static pressure at the *opposite-hole point* is different from the momentum argument. However I don't understand HOW BERNOULLI EQUATION FAILED in giving the right momentum if I compare the static water surface and the static *opposite-hole point*.

I would like to know why the Bernoulli equation failed in this situation or is there any possible way to modify the Bernoulli equation to make it work again?

I am always fascinated in reading Feynman's materials.

Thanks for any help in this interesting question.