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I Fluid mechanics in a water container, reference from Feynman

  1. Mar 7, 2017 #1
    Hi all,

    I have a little problem while reading Feynman lecture note volume 2, chapter 40, section 40-3.

    In that section Mr Feynman talked about Bernoulli equation and a example of a filled water container.

    In high school we learned that if we drill a hole near bottom of container, we can get the water outflow speed from Bernoulli equation. However Feynman mentioned that it only considered the conservation of energy (energy conservation is equivalent to Bernoulli equation of fluid) but not the conservation of momentum.

    There is a net outward momentum carried by the outflow water. Thus in the most simplest case the static pressure at the opposite position of the hole is not the normal static water pressure in nearby environment.

    Mr Feynman wrotes " Then the static pressure at any point on the side of the tank must be matched by an equal pressure at the point on the opposite wall, except at the points on the wall opposite the charge tube. If we calculate the momentum poured out through the jet by this pressure".

    I understand how Feynman argued that the static pressure at the *opposite-hole point* is different from the momentum argument. However I don't understand HOW BERNOULLI EQUATION FAILED in giving the right momentum if I compare the static water surface and the static *opposite-hole point*.

    I would like to know why the Bernoulli equation failed in this situation or is there any possible way to modify the Bernoulli equation to make it work again?

    I am always fascinated in reading Feynman's materials.
    Thanks for any help in this interesting question.
  2. jcsd
  3. Mar 7, 2017 #2
    Are you talking about horizontal flow out of a hole in the side wall?
  4. Mar 7, 2017 #3
    Yes I am talking about the horizontal flow out of a hole on the side wall.
    As water pressure near that hole is higher than the outer atmosphere pressure water will come out through the hole.
  5. Mar 7, 2017 #4
    What do you think the flow streamlines look like inside the tank in the region approaching the hole? What do you think the pressure distribution looks like inside the tank in the region approaching the hole?
  6. Mar 7, 2017 #5
    Thanks for you reply.
    I (intuitively) think that the streamline should be flowing from some upper region up in the container to the hole which is at some lower level region.
    The pressure along the streamline should be gradually decreasing as the further away water tends to be more still and the closer water is at some speed being sucked to that hole.
    However I am focusing on the point opposite to the hole. If the water tank is large in radius the opposite location should be nearly still.
  7. Mar 7, 2017 #6
    Actually, the flow velocities within most of the tank are very slow. Only in very close proximity to the hole (say 2 or 3 hole diameters away from the exit hole) does the flow converge strongly toward the hole. In this region, all the streamlines are pointing directly toward the hole. And the flow velocity is speeding up as the fluid approaches the hole. So, immediately at the hole, the pressure in the tank matches the outside pressure, but a few diameters away from the hole, you have a higher pressure (close to the static pressure head at that depth).

    My next question is what is the horizontal velocity component of the fluid in the tank at a horizontal distance several diameters away from the hole?

    With regard to the Bernoulli equation and the Euler equation, the Bernoulli equation is obtained from the Euler equation by dotting the Euler equation with the local fluid velocity along a given streamline. As such, the Bernoulli equation contains less information than the Euler equation regarding the details of the flow. This is why, although the Bernoulli equation is very useful at describing the relationship between the pressure changes, elevation changes, and velocity magnitude changes along a streamline, it cannot tell us everything. And, one of the things that it can't tell us is details of the momentum balance contained in the Euler equation. To determine the horizontal force that the tank is exerting on the fluid (and, in turn, the horizontal force that the fluid is exerting on the tank), one must perform a macroscopic momentum balance, which is essentially and integrated version of the Euler equation (but not dotted with the velocity vector).
  8. Mar 8, 2017 #7
    Responding to your question raised, sorry I cannot do a well analysis due to my limited skills but I can roughly estimate the speed near the hole as normal
    physics students would do.

    I will claim the streamline near the hole to be isotropic (which is obviously an approximation as the gravity will break the asymmetry). In order to satisfy the continuity of fluid, volume entering the hole's neighborhood is filling the volume of water flowing out of the hole. That gives us
    4 π r2 v(r) = π d2 v(hole) , d is diameter of hole and r is distance from the hole
    v(r) = (1/4π) (d/r)2 v(hole)

    I am not sure how well it works at a larger length scale where isotropic symmetry will break.
    However I am still worrying about the pressure at the OPPOSITE POINT to the hole, which is supposed to be far away from the hole.
    Can you give me more hint on finding that point's pressure?
  9. Mar 8, 2017 #8
    This is very nicely done. And, since the velocity is speeding up as the flow approaches the hole, the pressure is decreasing. This is a very local effect near the hole.

    I'm having trouble understanding what these questions are about. Is this related to the objective of determining the horizontal reaction force that the fluid exerts on the tank, and the corresponding horizontal force that the tank exerts on the fluid.
  10. Mar 8, 2017 #9
    Maybe I did not make the question clear before.

    Yes you are exactly right. I am concerning the reaction force exert by fluid on the wall at the position opposite to the hole.
    Lets say the hole is at the right bottom of the tank so there is a water jet pointing to right.
    By conservation of momentum there should be a leftward force exerting on the left bottom wall of the tank (I guess the position is just opposite to the hole).

    All my concern is can we estimate that pressure exert on the wall opposite to the hole, preferentially using Bernoulli equation. (Although you pointed out that Bernoulli equation don't contain the "information" about momentum, I think that opposite point is still satisfying some Bernoulli equation)
  11. Mar 8, 2017 #10


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    There are forces by the fluid on the walls everywhere, but they cancel if there is an opposite wall, with an equal pressure on it. The wall at the point opposite to the hole doesn't have an opposite wall, so in total the horizontal forces by the fluid on the container are not balanced.

    But note that aside from hydrostatic pressure, there are also dynamic forces. As the container empties, the water flows down, and then turns to the right. That requires a force on the water, and increased pressure on the left wall and bottom. To visualize this better consider an extreme case: a container that is a vertical pipe bending 90° to the right at the bottom.
    Last edited: Mar 8, 2017
  12. Mar 8, 2017 #11
    A.T. thanks for your reply.
    I can follow 100% of your statement and I think this is exactly the physical picture I want.
    However can you furthermore explain how to get the dynamics pressure term using Bernoulli equation?
    I am not sure if it is possible either.
  13. Mar 8, 2017 #12


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    The hole in a tank case is a bit complicated for a first attempt at understanding forces associated with fluid flows .

    Consider to begin with forces generated by fluid flows in curved ducts of smoothly varying flow area . There are many worked examples online and in undergraduate text books .
    Last edited: Mar 8, 2017
  14. Mar 8, 2017 #13


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    The pressure forces generating the reaction force on the tank do not just act on one small area opposite the nozzle .

    With the nozzle closed there is a simple hydrostatic distribution of pressure in the tank . Opening the nozzle and allowing fluid to flow modifies the pressure distribution in the tank . This modified pressure distribution is biased so that there is a net pressure force on the tank acting in opposite direction to outflow direction .

    You can look at this another way . Since the fluid starts off flowing vertically downwards and then bends round to go out of the nozzle there must be pressure gradients in the fluid in the tank causing this to happen .
    Last edited: Mar 8, 2017
  15. Mar 8, 2017 #14
    If the objective is to determine the horizontal frictional force that the ground must exert on the bottom of the tank to hold it in place against the force of the fluid jet through the hole, here is how it is done using Macroscopic Momentum balances.

    The horizontal momentum balance on the fluid is as follows: $$F-p_{atm}A=\rho A v^2$$
    where F is the horizontal force that the walls of the tank exert on the fluid in the +x direction, v is the jet velocity, ##\rho## is the fluid density, A is the area of the exit hole, and ##p_{atm}## is the atmospheric pressure at the exit hole. The term on the right hand side is the rate of horizontal momentum exiting the tank (there is no horizontal momentum entering).

    The horizontal momentum balance on the tank is as follows: $$-F+G+p_{atm}A=0$$ where G is the frictional force exerted by the ground on the tank in the +x direction. The term ##p_{atm}A## is present in this equation because the air outside the tank is acting on all exposed surfaces outside the tank, except for the exit hole (where the exit hole has been drilled out). There is pressure acting on the exit hole, but this pressure is on the fluid, not the tank metal. So, there is a net forward horizontal force of ##p_{atm}A## exerted by the outside air on the outside surface of the tank. This might be what Feynman was referring to.

    If we combine the above two equations, we obtain the frictional force: $$G=\rho Av^2=2\rho ghA$$

    This equation could never have been derived solely by use of the Bernoulli equation.
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