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Derivation of Bernoulli's Equation via conservation of E

  1. May 23, 2014 #1
    So I'm not OK with how some people derive this equation.

    These people consider a pipe whose endings have cross-sectional areas and heights which are different. They then use the conservation of energy principle by saying dW = dK + dU (Where W is work, K is kinetic energy, and U is potential energy).

    For this they consider that the work done on the system would be due to external pressure forces exerted on the whole system of water along the pipe. And here comes the part where I disagree: they use this Work (this value) calculated on the entire system of fluid to calculate the change in Potential and Kinetic energy for an infinitesimally small slab of water within the whole system. This is completely invalid isn't it? I mean you would have to consider the entire system, I think.

    My way of interpreting the derivation is if you consider just one slab the whole time. Is this a valid way of thinking?

    How would you derive this equation via conservation of E?


    edit: In fact, in one video I saw, the person just says "the middle chunk of water stays the same the whole time, so we can just ignore it".
  2. jcsd
  3. May 23, 2014 #2
    Yes, you apply the energy conservation to the whole system but most of the system remains unchanged so any changes to the energy must be accounted by the single slab that does change, as you described. In essence, the statement that you added by edit "the middle chunk of water stays the same the whole time, so we can just ignore it" is correct.
  4. May 23, 2014 #3
    And why does it remain unchanged? I just can't see this being true :(

    edit: To me, every slab of water inside the whole system gained energy. I don't see why we're only considering the slabs at both extremes of the pipe...
    Last edited: May 23, 2014
  5. May 23, 2014 #4
    You are studying a steady state flow situation. The flow at any give point in space is constant over time. So the potential energy and kinetic energy of the middle section is constant since the flow configuration doesn't change. But there is a change due to the fact that one slab entered the pipe section we're considering at one end while another slab left at the other end. That must be taken into consideration.
  6. May 23, 2014 #5
    Think about it that way. If you have a stack of bricks and lift them just enough that the bottom brick replaces the next one over and that one replaces the third brick and so one, you might compute the energy increase for each brick and add it all up or you might just remember that all the bricks are identical and pretend you removed the bottom brick and placed it at the top of the stack. The final configuration is the same but only one brick moved from the very bottom of the stack to the very top and all the other bricks remained where they were. All you have to do is to compute the energy change for that one single brick which is much easier than what you did before, but you get the same result.
    Last edited: May 23, 2014
  7. May 23, 2014 #6
    If the flow is going against gravity (it has some sort of inclination), then as one slab enters and another one exists, every fluid particle in the middle chunk gains PE. So PE didn't remain constant, isn't this so?

    edit: When dealing with KE, it makes sense to me if I consider just the last slab as dt has passed and what was happening to that same slab before dt had passed.
  8. May 23, 2014 #7
    Did you see post #5?
  9. May 23, 2014 #8
    Sorry, for some reason I couldn't see it before. This analogy was extremely useful, but does it hold for KE too?

    edit: My guess is it doesn't, but I can consider a case where the only change in KE was done on the last slab of water that passes through a narrower pipe at the end of time interval dt
  10. May 23, 2014 #9
    Why wouldn't it work for kinetic energy? The same logic applies.
  11. May 23, 2014 #10
    Well I don't get how the change for KE of the whole system, will be equal to the 1/2 * mass of bottom block * (v2-v1)^2 where v2 is the velocity of the top block and v1 is the velocity of the bottom block.

    Or put in fluid terms: how the change in KE in the whole fluid is just equal to 1/2 * mass of slab * (v2-v1)^2 where v2 is the speed of the last slab, and v1 is the speed of the first slab.

    Mathematically if I try this out for PE I get m(3*dy) when I'm considering a system of three blocks which is lifted in the vertical direction. And for KE I get (1/2)m(3dv^2).

    Now for potential energy I can see how it means I just took out the bottom block and inserted it on the top, and that equals the same amount as if I were to add dPE for each individual block. But I can't figure out what the thing I did for KE entails. :(

    edit: I don't see how the same logic applies mathematically speaking. Because the bernoulli equation states that (1/2)ρdV(Vf-Vi)^2 where Vi is the speed of the first slab, and Vf is the speed of the last slab, and I can't get to that conclusion mathematically via the analogy/post#5.
    Last edited: May 23, 2014
  12. May 23, 2014 #11
    Bernoulli's equation doesn't have the term (1/2)ρdV(Vf-Vi)2. The correct therm is (1/2)ρdV(Vf2-Vi2)
  13. May 24, 2014 #12
    Sorry I got messed up. But I'm still confused about how you arrive at that term via the analogy. :/... I mean I get how the analogy is useful for PE, and to some extent for KE as well, but I wouldn't get that mathematical term.
  14. May 24, 2014 #13


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    What is used for the derivation is the KE per unit mass, not the KE of the entire mass of fluid in the system. The other term in the derivation process is the potential energy per unit mass, which you already correctly specified as a potential term (as opposed to potential energy). Since Bernoulli's equation uses terms based on energy per unit volume, instead of energy per unit mass, it includes a density (mass per unit volume) factor in the equation.
  15. May 24, 2014 #14
    Yeah I can see that, but at first (in the derivation) it's not about energy per unit volume. In my textbook they proceed from dW = dK + dU, and then treat dK and dU for just one small slab; I can see how post#5 relates to dU, but not KE (mathematically). Like how is it that dK of the entire system will be equal to just (1/2)ρdV(vf2-vi2) where vi is the velocity of the slab at the first ending, and vf is the velocity of the slab at the last ending.
  16. May 24, 2014 #15
    dauto says the same logic from post#5 applies, but I want to know why it does. I can prove it mathematically for PE, but not for KE.
  17. May 24, 2014 #16
    Or in general, why is the analogy from post#5 true? If you can answer me mathematically, that would be better.
  18. May 24, 2014 #17
    Slice the fluid into slabs of equal mass. Call them slab 1, slab 2, ... , slab N. The total energy of the slab i is Ei = Ki + Ui. the total energy of the fluid is E = E1 + E2 + ... + EN. Now wait until the fluid moves enough that slab 1 replaces slab 2, slab 2 replaces slab 3, ... , slab N replaces slab N+1. The total energy is E' = E2 + E3 + ... + EN + EN+1. the energy change is ΔE = E' - E = (E2 + E3 + ... + EN + EN+1) - (E1 + E2 + ... + EN) = EN+1 - E1. Notice how energy of all the intermediary slabs cancelled out leaving only the first and the last.
  19. May 24, 2014 #18
    Wow, thanks a lot. I guess that solves all my confusions... One last thing though: I keep hearing "faster moving fluid causes lower pressure", technically it's the other way around, right? Namely: "lower pressure has caused the fluid to move faster".
  20. May 24, 2014 #19


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    It's not the entire system. Note that density x dV = dm, (where m is mass) so dK = (1/2)dm(vf^2-vi^2), the energy change per dm (delta mass).

    The cause is that fluid accelerates from higher pressure regions to lower pressure regions, and absent external forces involved in the transition (no friction with the pipe, no viscosity issues), except for gravity if there is a gravitational potential energy per unit volume term, then Bernoulli's equation relates the change in pressure versus the change in speed^2 and/or height. This ignores the fact that some external energy source has to maintain the pressure differentials at both ends of the observed system.
    Last edited: May 24, 2014
  21. May 24, 2014 #20
    The dW was done on the entire system. The reason it's expressed as dK = (1/2)dm(vf^2-vi^2) is because when you calculate Efinal-Einitial all the slabs cancel except the one that went in and the one that went out.

    By the way, does a lower pressure cause higher velocity, or higher velocity cause lower pressure? They keep saying that the higher velocity has caused lower pressure on an airplane wing, but I don't get this... Bernoulli's equation is talking about pressures outside the system.
  22. May 24, 2014 #21


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    The dm component in the formula means the formula is working with a component of mass not the entire mass of the fluid in the system. It makes more sense determine the power which is the work done on the entire system per unit time: you'd need to divide both sides by dt (delta time), then integrate to determine work done per unit time (power) for a given mass flow (dm/dt, which is a constant), and the initial and final velocities in to and out of the system.

    Note that to get dU in terms of mass instead of volume, dU = (Pi - Pf) / ρ

    It's not the cause. It would better to state that higher velocities and lower pressures coexist at any point along a wing. This ignores the fact that interaction with the wing adds some energy to the air, if the air is used as a frame of reference. Say you have an ideal wing that effectively diverts the relative flow near the wing downwards by some small angle θ with no change in air speed if using the wing as a frame of reference. Switching to using the air as a frame of reference, the air ends up with a downwards (lift) and somewhat forwards (drag) velocity at the moment the affected air's pressure returns to ambient, so the affected air's speed has increased from zero to non-zero, which means the wing has performed some work on the air, which violates the assumptions made for Bernoulli principle. If a wing is efficient, such as a wing with a very long wing span, then the increase in energy is small, and Bernoulli's equation applied to the averaged effect on the affected air is a close approximation. An alternative approach to show that energy is added to the air is to note as a wing passes through a parcel of air, it exerts a downwards and somewhat forwards onto the air for some period of time, the force x time is the impulse exerted onto the air, and the change in momentum of that parcel of air equals the impulse, changing from zero velocity to non-zero velocity (using the air as a frame of reference).
    Last edited: May 24, 2014
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