# Application of Bernoullis identity for problem concerning hole in a container

Consider the following problem:

We have a container of water with a hole on the side, and we want to determine the velocity of the stream coming out of the hole (at $S_2$):

If we suppose $S_1>>S_2$, we can suppose that the problem is stationary, so we can apply Bernoullis identity, i.e. $\frac{\rho u^2}{2}+p+\varphi$ is constant (on a line of flow). Now according to my professor, the pressure at $S_1$ as well as at $S_2$ is equal to the atmospheric pressure, and thus: $$\frac{\rho u_1^2}{2}+p_{atm}+\rho gh=\frac{\rho u_2^2}{2}+p_{atm}$$ But why isn't the pressure at $S_2$ equal to $p_2=p_{atm}+\rho gh$? As pressure is isotropic and uniform on the same water level, this seems right to me. I see that this would lead to $u_1=u_2$ which would't make sense, but I can't explain why $p_2=p_{atm}+\rho gh$ is false.

• I think you meant "hole", not "whole" – SJuan76 Jun 16 '16 at 22:22

The Bernoulli equation inherently takes into account the fact that the flow approaching the exit hole is converging toward the exit hole and thereby accelerating. So the pressure in close proximity to the exit hole (within just a few exit hole diameters away) is decreasing while the flow velocity is increasing. This is how the pressure decreases from values approaching $p_{atm}+\rho gh$ at horizontal locations just a few hole diameters away from the hole to $p_{atm}$ at the hole. And the flow velocity speeds up to the exit hole velocity in this region.
the water is no longer constant at the same level because it moves. It is likely very close to $p_{atm}+\rho g h$ an the far left side, but not on left side where the hole is, it should be less because it is moving. Luckily the details are not important because you know that the hole is open to the atmosphere, so the pressure there must be $1 atm$, why would it be different?
Your professor is correct. The hole $S_2$ is exposed to the atmosphere and the pressure at that point is indeed $p_{atm}$.
The $\rho gh$ term you worry about has in fact already been 'accounted' for in the Bernoulli equation (third term on the LHS).