# If a free jet of liquid is exiting from a tube into the air, why is the pressure of the liquid exiting atmospheric when not submerged?

The second question is the closest to my question, but my flow is not governed solely by gravity. In my case, my flow is powered by a pump, so I can't just use the Bernoulli equation; I should use the full energy and total head loss equations (5-77 and 8-58 from "Fluid Mechanics Fundamentals and Applications" by Y. A. Cengel, and J. M. Cimbala). I'm assuming

• incompressible flow
• subsonic flow
• exit is open to the atmosphere
• there is a single pump and no turbine
• the major and minor losses are NOT negligible
• and the tubing has a constant diameter $$D$$.

$$\frac{P_1}{\rho g}+\alpha_1\frac{V_1^2}{2g}+z_1+h_{pump,u}=\frac{P_2}{\rho g}+\alpha_2\frac{V_2^2}{2g}+z_2+h_L$$ $$h_L=\sum_{i}^{}f_i\frac{L_i}{D_i}\frac{V_i^2}{2g}+\sum_{j}^{}K_{L,j}\frac{V_j^2}{2g}$$

Here is a schematic. The fluid is pulled up at $$z_1$$, and exits at $$z_2$$. Why is the pressure in the jet and at the outlet atmospheric? I understand that it's surrounded by atmospheric pressure, and any hydrostatic pressure within the jet is minimal. It seems that there would be a sharp discontinuity in pressure right at the outlet because within the tubing, it certainly is not atmospheric.

Later, when the tubing is submerged $$z_{2b}$$, the pressure is hydrostatic, so $$P_{atm}+\rho g z_{2b}$$ based on where I placed my datum $$z=0$$. It seems there would be a similar discontinuity in this case as well.

Can someone explain why this discontinuity isn't the case?