1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Misunderstanding of Pressure in the Bernoulli Equation

  1. Nov 11, 2012 #1
    First, this is is not a homework problem, per se, but it is a conceptual difficulty I am having with my physics 1 course, in which we are studying fluid mechanics (moderators please move this post if there is a more appropriate subforum).

    1. The problem statement, all variables and given/known data
    I was going over the derivation of Torricelli's law using Bernoulli's equation as in my text (Serway & Jewett) and online resources. Just a recap as I understand it: In the diagram below, let our origin be at the hole on the side of the container, so the height at the hole is 0 and the height at the surface is [itex]h[/itex]. Using Bernoulli's equation, with point 1 being at the surface of the water and point 2 being at the hole a distance [itex]h[/itex] down: [itex]P_1+\tfrac{1}{2}\rho v_1^2+\rho gh=P_2+\frac{1}{2}\rho v_2^2\Rightarrow v_2^2-v_1^2=2gh+\frac{2(P_1-P_2)}{\rho}[/itex].http://upload.wikimedia.org/wikipedia/commons/5/5b/TorricelliLaw.svg
    Now, since the area at the top is much larger than the area of the hole, we can simply set [itex]v_1=0[/itex]. We also set [itex]P_1=P_{atm}[/itex] the atmospheric pressure.

    So far, this makes sense to me. Where I have a hangup is in setting [itex]P_2=P_{atm}[/itex] as well. I think this has something to do with static pressure vs. dynamic pressure, but the definition of pressure as I understand it is a force divided by a cross sectional area. So at the interface of the hole (assuming the hole has area A) I understand that there is a force of magnitude [itex]F=P_{atm}A[/itex] pushing to the left on the water. But is there not also a force from the water pressure pushing to the right on a sliver of water at the interface of the hole? Shouldn't [itex]P_2[/itex] be the difference of these pressures?

    2. Relevant equations
    [itex]P_1+\frac{1}{2}\rho v_1^2+\rho g z_1=P_2+\frac{1}{2}\rho v_2^2+\rho g z_2[/itex]

    3. The attempt at a solution
    So, if I concede that [itex]P_2[/itex] should be the same as [itex]P_1[/itex], then it should apply to other problems whose solution is known without Bernoulli's equation. The simplest problem of this type I can think of is just a pipe, open at both ends, containing a column of water of height [itex]h[/itex]. I know from the first part of the semster, that the top of the water will fall at the same speed as the bottom of the water, so [itex]v_1=v_2[/itex]. Then our equation becomes [itex]v_2^2-v_1^2=2gh+\frac{2(P_1-P_2)}{\rho}\Rightarrow 0=2gh+\frac{2(0)}{\rho}\Rightarrow 2gh=0[/itex]. How should I explain this contradiction?
  2. jcsd
  3. Nov 12, 2012 #2
    Not following thee , but suppose that you are given a vessel filled with a non compressible and non viscous fluid , and there is a hole made in that vessel "h" metres below the free surface of fluid in the vessel , and the height of the vessel is "H" metres. Now you are given the task too calculate the velocity of "efflux" at the point just outside the hole. How would you apply Bernoulli's theorem then ?

    Hint : Apply Bernoulli's theorem for free surface of fluid and the point "just" outside the hole (at the orifice)....
  4. Nov 12, 2012 #3
    You can think of P2 as being the pressure just outside of the can, which is atmospheric pressure of cause.
  5. Nov 12, 2012 #4
    I know how to solve the original problem is to set [itex]P_2[/itex] to atmospheric pressure and then the solution is just [itex]v_2=\sqrt{2gh}[/itex]. What I am hung up on is why should [itex]P_2=P_1[/itex]?
    So if I put a small test swatch of area just outside the can, it will of course experience a force due to atmospheric pressure. But should it not also experience a force due to the water hitting it? Thus, the total force would be the vector sum of those two, and dividing its magnitude by the area, we would get a total pressure different from just [itex]P_{atm}[/itex].
  6. Nov 12, 2012 #5
    Neither of these answers seems to resolve the paradox I am getting. Here are diagrams of two other situations where the Bernoulli equation should have the same form as before: [itex]v_2^2-v_1^2=2gh+\frac{2(P_1-P_2)}{\rho}[/itex]. The first is simply a U-shaped tube with water in it at different heights (it's not in equillibrium so we expect the left level to move down as the right level moves up). The second is supposed to be a hollow cylinder with both ends open, so that atmospheric air pressure applies to both the top and bottom ends of the water. We expect the entire column of water to move downward.
    LXlB7.png oqlqR.png
    In either of these, suppose we set [itex]P_1=P_{atm}=P_2[/itex] as we did for Toricelli's theorem. Then we find that the speeds are different (namely, [itex]v_2^2-v_1^2=2gh[/itex]). We know from continuity, that since the cross-sectional area of the U-shaped tube is constant, that [itex]\lvert v_1\rvert=\lvert v_2\rvert[/itex]. We also know that in the second situation, the column of water will fall together as a whole so [itex]v_1=v_2[/itex] (this also follows from continuity).

    How should I resolve this inconsistency?
  7. Nov 12, 2012 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    In your examples, what are v1 and v2? As you noted, at any point in time all the fluid is traveling at the same speed. So the only meaning you can attribute to v1 and v2 are measurements of the speed of the fluid at two different points in time.

    In the first example, the velocity changes as Δh (the difference in height of the two columns) changes so the fluid oscillates back and forth. In the second example, the fluid is in freefall so the pressure at the bottom is the same as at a the top. The velocity of the whole fluid increases as the height of the centre of mass of the fluid (eg. the height above the ground) decreases. In both cases, if the velocity is 0 when Δh is maximum, [itex]v^2 = 2g\Delta h[/itex]

    Last edited: Nov 12, 2012
  8. Nov 12, 2012 #7
    So according to Bernoulli's equation as I understand it, [itex]v_1[/itex] is the instantaneous speed of the fluid at point 1, and likewise for [itex]v_2[/itex] at point 2.

    I understand what should happen if I just use Newton's laws and forget about fluid dynamics. As you said, in either example, the speed of the fluid at points 1 and 2 are the same (which is also what I said!). According to Bernoulli's equation, this either means that the pressures are different or that Bernoulli's equation doesn't apply for some reason.
  9. Nov 12, 2012 #8


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    These are good questions. I think that at least part of the resolution lies in the fact that Bernoulli's equation is derived under several assumptions. In particular, it is assumed that the flow is "steady". This means that at any point of the fluid that is fixed relative to the "lab frame", the velocity of the fluid is constant in time. This condition is not met in your examples.
  10. Nov 12, 2012 #9
    Okay, this is the first answer that seems to make sense to me (Bernoulli's principle doesn't apply since the speeds aren't constant). But I am still hung up on why in the original example, [itex]P_2=P_1[/itex].
  11. Nov 13, 2012 #10
    Bernoulli's equation uses the pressure while the fluid is in motion. Inserting something in the path of the water will change this condition. The pressure of the water changes abruptly as it exits the can, it can no longer sustain the pressure it exerted while in the can. The walls and the liquid pressing down on it creates this pressure but these conditions do not exist anymore once it exits the can, only atmospheric pressure determines the pressure once it exits.
  12. Nov 13, 2012 #11

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Bernoulli's equation deals with energy density of an incompressible fluid provided no energy is being added or lost. The total energy density does not change. This is due to the conservation of energy AND the incompressibility (ie. the fact that the volume does not change in any part of the flow) of the fluid.

    Pressure x volume represents a form of potential energy. So P represents the PV energy density. ρgh is the energy density of gravitational potential energy. And .5ρv2 is the kinetic energy density.

    Try applying that to your examples.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook