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If we had a flow in an open vertical pipe kept in atmosphere so that external pressure is equal at both ends and if we try to apply Bernoulli's equation to it would stand to reason that due to pipe being vertical, potential energy is smaller at the lower end so, kinetic energy must be bigger, which is sensible because gravity did positive work and increased fluid's kinetic energy.

But, if pipe's diameter is the same on both ends how can we agree this with continuity principle? From this principle, velocity should have been the same. Is Bernoulli's continuity equation applicable here in its simplest form ? Although I am not quite sure what is missing here.

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    $\begingroup$ Below the top of the pipe, the fluid loses contact with the surface of the pipe (so that the diameter of the fluid stream decreases with position downward) $\endgroup$ Apr 11, 2021 at 0:41
  • $\begingroup$ Please see this post. $\endgroup$
    – kbakshi314
    Apr 11, 2021 at 3:40
  • $\begingroup$ @Chet Miller. Interesting, can you point to a reference in which I can see this? $\endgroup$ Apr 11, 2021 at 8:52
  • $\begingroup$ @kb314 Thank you. Questioner there had a similiar misunderstanding, but not quite the same and I know how to answer his question. He tried to apply Bernoulli's equation at the time where flow was non - steady which is what confused him. Here question is different since flow is vertical. $\endgroup$ Apr 11, 2021 at 8:55
  • $\begingroup$ The continuity equation is a consequence of conservation of mass. Also, note that a constant velocity on both ends of the vertical pipe only occurs for an incompressible fluid (e.g., water). $\endgroup$ Apr 11, 2021 at 15:08

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Consider an open-top parallel-sided vertical pipe filled with an inviscid fluid and plugged at the bottom. At $t=0$ the plug is removed. As the fluid is being acted on by gravity and is now unconstrained both above and below, it starts to fall. It starts form rest and accelerates uniformly downwards so that $v=-gt$. This velocity is the same for all vertical heights $z$ in the pipe, so the fluid not have to separate from the walls.

Let us compare this motion with Bernoulli. Bernoulli's equation for non-steady flow states that
$$ \frac{\partial \phi}{\partial t} +\frac 12 v^2 + \frac P \rho +gz $$ is independent of $z$. Here $\phi$ is the velocity potential defined by $v=\partial_z \phi$, so for $v= -gt$ we have $\phi=- gzt$ and $$ \frac{\partial \phi}{\partial t}=-gz $$ Bernoulli now claims that $$ -gz+\frac 12 g^2t^2 + \frac P \rho +gz $$ is independent of $z$. This means that he claims that $P$ is independent of $z$. There is thus no problem with the pressure being atmospheric at both ends of the pipe. Indeed, as the fluid is in free fall, it effectivly sees no gravity, and so $ P= P_0-m g_{\rm effective}z$ with $g_{\rm effective}=0$.

I think that the discussion in the comments shows that people are not very familar with the non-steady version of of Bernoulli's equation, so here is a dervation from Euler's equation $$ \rho\left(\frac{\partial v}{\partial t}+ v\partial_z v\right)=-\partial_z P- \rho g $$ for one-dimensional incompressible flow in the $z$ direction. We set $v=\partial_x \phi$ and note that $v\partial_z v= \partial_z(v^2/2)$, so Euler becomes $$ \rho \frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial t}+ \frac 1 2 v^2+ \frac P \rho - gz\right)=0. $$

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Bernoulli's equation goes like this: $$p_1 + h_1\rho_1 g + \frac12\rho_1 v_1^2 = p_2 + h_2\rho_2 g + \frac12\rho_2 v_2^2$$

Continuity principle states: $$A_1V_1 = A_2V_2$$

When the cross-section is same the velocity at both topmost point and bottom most point gets cancelled out from Bernoulli's equation and the kinetic energy term vanishes.So we are left with the pressure term and the $\rho gh$ term. So the difference in potential energy (represented by the $\rho gh$ term) is balanced by the pressure difference between the two points i.e $$p_2 - p_1 = h_1\rho _1g - h_2 \rho_2g$$

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  • $\begingroup$ Problem here is that if pipe is open on both ends pressure should be atmospheric on both ends due to Pascal's Law. I need to take a closer look on Bernoulli's equation derivation. $\endgroup$ Apr 11, 2021 at 8:29
  • $\begingroup$ Pressure difference of hpg will be there since one point is h height above the other point.The hpg present in Bernoulli's equation is often confused as pressure difference but it actually is difference in Potential energy.So hpg pressure difference is included in the p1-p2 term of Bernoulli's equation. Let me know if u wanna know more about it , I will edit my answer accordingly. $\endgroup$
    – Möbius
    Apr 11, 2021 at 10:31
  • $\begingroup$ Yes, that term is gravitational potential energy, but I am not sure if static pressure (p) is bigger on the bottom since container is opened on both sides to the atmosphere and thus pressure should be atmospheric. In any problem in fluid mechanics where you need to apply Bernoulli's equation, wherever you have cross section opened to the atmosphere, pressure is atmospheric. Like when you want to derive Torricelli's law for fluid flowing through orifice in vertical container. $\endgroup$ Apr 11, 2021 at 11:58
  • $\begingroup$ If its a non compressible fluid gravity will try to stretch it (something like when u get stretched like sphagetti when u cross the event horizon of a black hole but on a much smaller scale) which will give rise to reverse tensile forces inside the fluid .Those forces might create a pressure difference between the two points that we haven't accounted for in our equation.Also Bernoulli's equation holds only for streamlined flow but I don't think flow under gravity will be streamlined since particles at a point will have different velocities depending on their initial height above that point. $\endgroup$
    – Möbius
    Apr 11, 2021 at 13:48
  • $\begingroup$ Yes, thank you. I will need to take a better look on Bernoulli's equation derivation to create better understanding. $\endgroup$ Apr 11, 2021 at 15:24
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Continuity Theorem is always applicable. When thrown a stream vertically upwards [not in a pipe] it widens its area, and when it flows vertically downwards, it narrows its area.

When fluid gains Kinetic energy in significant amount then, to obey Rate Flow of Mass it narrows the stream-flow area. Just imagine the vertical pipe to be very long, considering that atmospheric pressure is same at both ends, when made to flow a fluid, it will become nearly impossible to get a stream which exits with an area equal to the area of the ends of the pipe.

It is reasonably difficult to clarify this mathematically but needs just some practical examples to justify ourselves. If someone happens to know any mathematical formulation then do share it.

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  • $\begingroup$ So, you say that stream of fluid being accelerated by gravity would become narrower than pipe's cross section at the other end? If so, this makes sense as it is compatible with continuity equation and Bernoulli's equation. $\endgroup$ Apr 11, 2021 at 15:21
  • $\begingroup$ Yes, As it starts to gain velocity, it will tend to narrow down, to maintain the Mass per unit time which is flowing. $\endgroup$ Apr 12, 2021 at 0:56
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The fluid in a vertical pipe which is open at both ends will be in free fall, unless you consider adhesion and viscosity.

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