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$$P+\rho gh+\frac{\rho v^2}{2} = C$$

Does the $P$ here stand for the hydrostatic pressure exerted by the fluid on its surroundings or the pressure that’s exerted on a given portion of the fluid? I assumed it was the former but Torricelli’s Theorem made me think that I’m probably wrong.

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  • $\begingroup$ Why do you distinguish between "pressure exerted by the fluid on its surroundings" and "the pressure that’s exerted on a given portion of the fluid"? Pressure is just pressure, there aren't two kinds. $\endgroup$ – Deep Dec 2 '18 at 5:10
  • $\begingroup$ Is that not something you should do? I assumed so for some reason $\endgroup$ – Dahen Dec 2 '18 at 6:07
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A similar question has already been asked here. So basically P is the pressure energy. You can further view this answer What is Pressure Energy?

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  • $\begingroup$ As far as Pressure energy, it is a name used in some fields, but is not a widespread name for $P$ among physicists. $\endgroup$ – GiorgioP Dec 1 '18 at 23:46
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Pressure is a concept connected (but not coinciding) with the force exerted on a surface (not on a portion of fluid, which would be a volume).

In the formula you are referring to, $P$ stands for the local pressure in a point at height $h$ and where the local speed of the fluid is $v$. Calling it hydrostatic looks like a misname (since the fluid is moving), but the reason is that it is customary to call "dynamical pressure" the term $\rho v^2/2$.

I do not understand why you think that Torricelli's Theorem would prevent the identification.

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  • $\begingroup$ The confusion came from the fact that in the derivation, we cancel out the two $P$ terms from both sides of the equation since they are both equal to the atmospheric pressure exerted on the fluid surface at the top and at the opening at the bottom. I had thought that those two $P$ terms refer to the pressure that’s exerted by the fluid itself, regardless of other pressures that may be in the system. $\endgroup$ – Dahen Dec 2 '18 at 2:57

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