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During the derivation of velocity of efflux
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Bernoulli's principle is used,

$$P_A + \rho gH + \frac{1}{2}\rho V_A^2 = P_B + \rho gh+ \frac{1}{2}\rho V_B^2$$

My question is why $P_A=P_B=P_0$ = atmospheric pressure? Definitely the pressure of fluid at points $A$ and $B$ are different. Also, during derivation of Bernoulli's theorem, $P$ was taken as pressure applied by the fluid to do work, so $P$ is definitely pressure applied by fluid not any other pressure and it should be different at $A$ and $B$.

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    $\begingroup$ Bernoulli is the name. $\endgroup$
    – my2cts
    Commented Feb 16 at 19:03

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The picture is oversimplified. In fact, pressure decreases with distance along the flow as long as the flow velocity changes with that distance. This is happening outside the pipe, where the liquid beam discharges and starts falling down, because the liquid beam gets thinner with increasing distance from the pipe end.

If $B$ is still inside the container or close to the wall but at such a point of the liquid flow where the flow velocity isn't the final velocity of discharge(such as inside the pipe), then pressure there isn't atmospheric. But if you place the point B far enough from the pipe end, there is a point where the horizontal flow does not get thinner anymore, thus does not change horizontal velocity with distance anymore, and thus the liquid pressure is very close to atmospheric. The derivation of the final horizontal velocity assumes the point B is far enough from the pipe's end so the liquid pressure is atmospheric, but the picture is misleading, since it puts B inside the pipe where that is not the case.

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  • $\begingroup$ "But if you place the point B far enough from the pipe end, there is a point where the horizontal flow does not get thinner anymore, thus does not change horizontal velocity with distance anymore, and thus the liquid pressure is very close to atmospheric" why is that? wouldn't the weight due to (H-h) column of liquid add extra pressure to point B? $\endgroup$
    – SHINU_MADE
    Commented Feb 17 at 14:07
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    $\begingroup$ No. It would if the liquid was at rest, but the liquid is accelerating, and for acceleration, net force is needed. This force is due to pressure acting from the left is higher than pressure acting from the right. Pressure from the right falls with distance and eventually reaches atmospheric pressure. $\endgroup$
    – Ján Lalinský
    Commented Feb 17 at 14:20
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Inside the container, as the fluid converges toward the exit hole, the velocity of the fluid is speeding up until, at B, it is approximately equal to the velocity coming out of the pipe. This speeding up all happen within a few hole diameters approaching the exit hole (inside the container). So, at B, the pressure is approximately the same aw in the pipe exit.

Draw a schematic and show what you visualize as the flow streamlines approaching the exit hole

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