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Suppose a situation where we are going to apply Bernoulli principle to find the velocity of a water stream leaving at the bottom of a large lake. When we apply the Bernoulli theory, we consider a stream line from the surface of the lake that runs through a hole located at the bottom of the dam, through which water is leaking. We consider a streamline and take the

pressure at the surface=atmospheric pressure

on the same streamline at the bottom, but inside the dam;

pressure =atmospheric pressure + hydrostatic pressure

so in this case the potential energy component at the top of the streamline has become the hydrostatic pressure component at the bottom of the streamline. in such a case should we neglect both hydrostatic pressure as well as potential energy component as they are going to be cancelled out?

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The pressure at the bottom in close proximity to the exit hole is not hydrostatic. In fact, it approaches atmospheric. All the flow streamlines converge toward the exit hole in this region, and the velocity increases while the pressure decreases. So even though, away from the exit hole, the pressure at the bottom is hydrostatic, in close proximity to the exit hole, the pressure rapidly decreases while the velocity rapidly increases.

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  • $\begingroup$ Thanks Chester. This helps. If we consider some point above the hole, hydrostatic pressure exists right? $\endgroup$ – Kosala Oct 1 '16 at 2:24
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    $\begingroup$ Hydrostatic pressure exists pretty much everywhere, except within a few hole diameters of the hole. This is where the flow most rapidly accelerates. $\endgroup$ – Chet Miller Oct 1 '16 at 12:01
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If you apply Bernoulli equation to streamline you mentioned, between surface and exit points, you get:

$p_{atm}=p_{exit}-\rho g h+\frac{1}{2}\rho v^2$

Now $p_{exit}=\rho g h$ only if fluid is static, i.e. if there is no flow. If there is flow then $p_{exit}$ is unknown, as is $v$, so the equation is not much help unless you make some other assumption. If flow through the hole is discharging into atmosphere, then you may take $p_{exit}=p_{atm}$, and thus you get the familiar $v=\sqrt{2gh}$ formula.

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