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Bernoulli's equation states

$P_1+{1\over2}\rho v_1^2+\rho g h_1 = P_2+{1\over2}\rho v_2^2+\rho g h_2$

In a classic "water tank with an open top and a leak" scenario, "point 1" is the surface water in the tank, and "point 2" is the leak. The equation could be rewritten for $v_2$ as

$v_2= \sqrt{2g\Delta h}$

This is simple, but suppose it involves a tank where the top is closed off. The above simplification will no longer yield the correct velocity.

How do I apply Bernoulli's equation for "water tank" scenarios in which the water tank is closed?

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For a tank that is either open at the top or has a (sufficiently large) venting hole in it above the fluid line, the outflow speed is indeed approximately:

$$v= \sqrt{2g\Delta h}$$

But with a closed tank pressure $p_0$ above the water and thus also hydrostatic pressure:

$$p=p_0+\rho g\Delta h,$$

drops and flow rate drops. This is because the air in the closed tank gets to occupy a larger volume and according to the Ideal Gas Law that means pressure must drop.

What really happens you can observe by overturning a half filled bottle of pop or such like. Very quickly the flow of liquid starts slowing down, then it becomes erratic but without actually stopping. The reason is that atmospheric pressure now pushes air through the bottom opening because atmospheric pressure has become higher than the pressure inside the bottle.

For that reason the flow never stops, rather it proceeds with stops and starts as liquid is trying to get out and air trying to get in, at the same time.

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It seems more complicated but I would just take the force which is pushing the water out (Gravity) minus the force by the difference in air pressure.

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A closed-off tank can be considered to be 0 Pa. Assuming the outside air is 1 atm in pressure, $P_1$ and $P_2$ would no longer cancel out. The simplified equation would not apply here.

(I'm doing homework, and Googled for multiple hours. I never found an explanation. I made an assumption it turned out to yield a correct answer, so I'm sharing my results with a self-answer.)

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If the leak is closed, the velocity goes to zero. So instead of the water down near the leak having extra flow velocity ($v_2 > v_1$), it has extra pressure ($P_2 > P_1$).

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Depends on your assumptions. In the extreme case, with vaccum on the top of the fluid you can set $P_1=0$. This yields $$v_2=\sqrt{2g\Delta h-\frac{p_{\infty}}{\rho}}$$ with $p_{\infty}$ as ambient pressure. Be aware that $v_2$ is only an approximation since in reality $\Delta h$ changes as fluid leaves the tank.

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