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I want to calculate the effelux velocity for this open vessel with area $A_1$ with a small outlet with area $A_2$ as a function of time. The atmospheric pressure is $p_0$.

The vessel has height $h = h_{air} + h_{fluid}$ and is filled with a fluid. I want to take into account the velocity and potential head losses of the air. I am not interested in other losses.

An idea is to calculate the velocity starting from this equation.

$$(p_0-\rho_{air}g(h_{air}+h_{air} - z) -\frac{1}{2}\rho_{air}v^2)+\frac{1}{2}\rho_{fluid}v^2+\rho_{fluid}gz = p_0 + \rho_{fluid}\frac{A_1^2}{2A_2^2} v^2$$

I am in doubt as I don't know how to deal with the equilibrium at the interface between the fluid and the air. I know that in principle Bernouilli is not valid for a flow with two different fluids.

Vessel

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1 Answer 1

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I think the point of the question may be that the air contribution is completely negligible. You are simply converting the potential energy of the still water at height $h_\text{fluid}$ to kinetic energy at the bottom.

If you did want to account for the air head anyway, simply assume $P_0$ at the top of the water is $P_0+\rho g h_\text{air}$ instead. This would also be the same if it were a closed tank, with pressurized gas, what's called Ullage Pressure.

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  • $\begingroup$ What about the effect of the air compressibility for the closed tank case? $\endgroup$ Sep 14, 2021 at 12:15
  • $\begingroup$ Not sure what you mean. $\endgroup$
    – RC_23
    Sep 14, 2021 at 13:57
  • $\begingroup$ In the closed tank situation, the gas pressure decreases as the volume of gas increases, as described by the combination of ideal gas law and barotropic equation. $\endgroup$ Sep 14, 2021 at 14:01
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    $\begingroup$ Correct. That is why in a water cooler/dispenser, you constantly get slugs of air bubbling up as you drain it, to replace the water that goes out. Otherwise the air pressure would decrease. In a rocket fuel tank, there is a system that draws hot gases from the engines and delivers it all the way up to the ullage of the tank, to add gas and maintain the pressure draining the liquid fuel. Otherwise the fuel would not drain fast enough and you'd also get those slugs of gas like the water cooler. $\endgroup$
    – RC_23
    Sep 14, 2021 at 14:17
  • $\begingroup$ It is an open vessel. The atmospheric pressure of air in the neigbourhood that is not moving is $p_{0}$. The air in the vessel is moving. This decreases the pressure. That is why I expect something like $p0-\rho_{air} v^2/2$. But correct me if I am wrong. $\endgroup$ Sep 14, 2021 at 16:37

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