3
$\begingroup$

Why the pressure at point $y=0$ in the given example (see image) is atmospheric pressure $p_o$. How can it be atmospheric pressure? The pressure at that point should be the pressure exerted by the fluid flowing through the hole but the author of my book "Fundamentals of Physics by Halliday/Resnick/Walker" has substituted atmospheric pressure for the pressure at hole. enter image description here

$\endgroup$
0

2 Answers 2

5
$\begingroup$

Within the tank, the streamlines of the flow converge to the outlet hole, and, as a result, the flowing fluid speeds up as it approaches the outlet hole. This all happens mainly within just a few hole diameters of the exit. As the flow speeds up, due to Bernoulli, its pressure decreases, and, at the very outlet, it reaches atmospheric pressure. The atmospheric pressure is imposed on the flow at the outlet hole by the surrounding atmosphere. Don't forget that, according to Pascal's law, pressure at a given location in a fluid acts equally in all directions. So the radial pressure from the atmosphere at the hole exit is equal to the axial pressure of the fluid.

$\endgroup$
8
  • $\begingroup$ If the pressure of fluid is the atmospheric pressure then how does the fluid flow outwards? I think there should be a pressure gradient for flow $\endgroup$
    – Babu
    Oct 6, 2020 at 11:38
  • $\begingroup$ @Buraian the pressure at the bottom of the tank is different than atmospheric pressure. $\endgroup$ Oct 6, 2020 at 11:40
  • $\begingroup$ How did you deduce that @BioPhysicist? $\endgroup$
    – Babu
    Oct 6, 2020 at 11:41
  • $\begingroup$ @Buraian Because there is a tank's worth of fluid above the bottom of the tank. $\endgroup$ Oct 6, 2020 at 11:42
  • 1
    $\begingroup$ @Buraian As I explained in my answer, the pressure is not discontinuous around the hole. It varies within the tank in the region approaching the hole, as a result of the fluid flow speeding up as it approaches the hole. At the very exit, the pressure is atmospheric. So, inside the tank, there is a pressure gradient in the immediate region approaching the hole. physics.stackexchange.com/questions/248094/… $\endgroup$ Oct 6, 2020 at 12:30
3
$\begingroup$

The pressure at that point should be the pressure exerted by the fluid flowing through the hole

This is a typical misunderstanding of Bernoulli's principle. The pressure term is not the pressure the fluid would exert on something if it were to collide with the fluid. The pressure term is the pressure pushing on the fluid itself (which could be due to other parts of the fluid, or other external effects).

This is evident if you look at typical derivations of Bernoulli's principle. You are considering the work being done on the fluid by the net force exerted on it from pressure on either side of the fluid. Therefore, at the hole you need to be considering the pressure pushing on the fluid at the outside of the hole. Since this hole exits to the atmosphere, this is atmospheric pressure. i.e., the air is pushing on the fluid outside of the hole.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.