# Efflux velocity vessel filled with air and fluid

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.

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.
• 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. Sep 14, 2021 at 16:37