There have been a few questions already asked that are similar to what I am asking.

  1. Why is the pressure of a jet, as it exits, is always the ambient pressure? [closed]
  2. 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?

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

  • steady flow
  • 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?



1 Answer 1


There isn't a discontinuity in pressure of the liquid at the opening. The pressure decreases quickly with distance after the liquid leaves the pipe/tubing, and the liquid jet gets narrower (vena contracta). After some short distance is traveled, the jet does not shrink appreciably, it just approaches asymptotic diameter corresponding the final velocity and final pressure - the environmental pressure.

See vena contracta, for example here


  • $\begingroup$ I'll have to look more into this. I only know about the vena contracta that occurs when there is a sharp-edged inlet (i.e. going into a tube not exiting) not much about an outlet. $\endgroup$
    – WnGatRC456
    Apr 22, 2021 at 22:51
  • $\begingroup$ But if I understand correctly, the conclusion is the vena contracta results in the outlet pressure asymptoting to the atmospheric pressure. However, this distance is relatively a short number of diameters regardless of whether the flow is governed by gravity or a pump? $\endgroup$
    – WnGatRC456
    Apr 22, 2021 at 22:59
  • 1
    $\begingroup$ There isn't a distance where the jet contraction stops, but there is minimum possible diameter (due to finite final liquid speed) so after some distance from the opening it becomes hard to observe further contraction. Usually most of it is done after few diameters from the outlet. I don't know if this is always necessarily true. Maybe there can be a jet that contracts along a distance of many orifice diameters. $\endgroup$ Apr 22, 2021 at 23:13

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