Torricelli's law and pressure at the drain

[fog37]

Hello Forum,

The speed of water exiting from a side narrow hole at depth h in a large bucket is given by $v=\sqrt{2gh}$.
This result is obtained applying Bernoulli's equation along a streamline that goes from the top free surface to the drain on the side of the bucket at depth h.

• The water speed at the free surface is practically zero.
• The pressure at the free surface is ##p_{atm}. This pressure pushes on the liquid in the downward direction.
• The liquid pressure at depth h since the liquid, at the same depth as the drain, is $p = p_{atm} + \rho g h$. But in the application of Bernoulli's equation we take the pressure at the hole to be the atmospheric pressure $p_{atm}$. Why? Is it because, whenever we apply Bernoulli's equation, we should use the pressure (of the air) on the water pushing towards the inside of the drain instead of the liquid pressure exerted towards the outside of the drain by the water on the air?

The term ($p_{atm} + \rho g h$ - $p_{atm}$ )$A_{drain}$should be the net force pushing the water out of the drain and making it produce projectile motion...is that correct?

Thanks!

 

[BvU]

Yes on both counts. Bernoulli is an energy balance -- you deal with a difference between free top surface and exiting liquid.

 

[fog37]

Ok, but to pick the atmospheric pressure to be the pressure at the orifice, we need to consider a point outside the orifice. For points to the left of the orifice, the pressure is not atmospheric but larger. Right at the orifice, the pressure distribution must be complicated and depend on the orifice shape. Once the water stream is out, considering a thin stream, the pressure on it and on every point inside the stream is the atmospheric pressure.

Am I on the right track?