# If a hole is drilled at the bottom of a vessel, why is the pressure of the liquid leaving the vessel equal to atmospheric pressure?

My question is similar to this one : Should the liquid come out of the tank if a hole is drilled in the vertical wall? , but I still don't understand why the pressure within a free jet of liquid/fluid leaving an open vessel is equal to atmospheric pressure.

Let's take the same example as in the previous post.

Indeed, hydrostatic law says that pressure within the same fluid at the same height (horizontal level) doesn't change.

The argument, in the previous post, was that point $B$ is in the open air, another fluid, so $p_B$ (the pressure at point $B$) is equal to atmospheric pressure and hydrostatic law doesn't apply to that point. I'm ok with this explanation.

However, if we check the pressure $p_D$ inside of the liquid, at point $D$, we should have the same pressure as $p_A$, at point $A$. The previous argument doesn't hold : we are in the same fluid and we are at the same height. So why is $p_D$ (roughly) equal to atmospheric pressure ?

• Because if there is a pressure at D greater than the surrounding atmosphere, the liquid is free to move laterally to reduce that pressure. – Jon Custer Apr 7 '16 at 20:14
• The fluid is accelerated as it leaves the hole. The only thing that can accelerate a fluid is a difference in pressure. Inside the tank, there is pressure due to the height of fluid above the hole (+ atmospheric pressure). Outside the tank, the pressure is just atmospheric. That is a pressure difference. It makes the fluid accelerate. Whenever fluid accelerates, you will find a pressure difference. (That's Bernoulli, by the way.) – Mike Dunlavey Apr 8 '16 at 0:31
• @MikeDunlavey The fluid is not accelerated as it leaves the hole. It is accelerated as it approaches the hole (inside the tank). There is a pressure gradient approaching the hole inside the tank. See my submitted answer. – Chet Miller Apr 8 '16 at 11:30