How does the standard derivation of Bernoulli's Equation work? So, the derivation itself (the work energy theorem one) is pretty simple, but the work-energy theorem is supposed to be applied to particles, not systems of particles. And even if I did that, shouldn't we include the potential energy related to the arrangement of different particles in the liquid also be included? Or is there any reason it remains unchanged? Why do we only use the change in kinetic energy and gravitational potential energy?

Another thing that bugs me is how the net work done is calculated $$p_1A_1v_1\cdot dt - p_2A_2v_2\cdot dt$$ In the end, we're just multiplying the force on each end of the flow tube with the distance traveled by the ends of the flow tube, not the displacement of the system, so it technically isn't the work done on the system (at least that's how I see it), the work done should be the product of the force and the displacement of the center of mass, I believe, which this clearly is not. Can someone explain to me how this actually is true then?

  • $\begingroup$ > "the work-energy theorem is supposed to be applied to particles, not systems of particles." Why do you think that? The work-energy theorem I know applies to any chosen set of particles. In the usual derivation of the Bernoulli equation, this chosen set consists of fluid particles that form a small body in shape of cylinder or rectangular parallelepiped and move together along its long axis. $\endgroup$ – Ján Lalinský Aug 26 '17 at 22:39
  • $\begingroup$ The calculation of the work done on the body in the derivation is application of the standard definition of external work - it is the sum of works of external forces. The relevant displacement to be used in calculating work is always displacement of the point where the force is acting, not necessarily displacement of center of mass of the body! In this case, those external forces are pressure forces and they are acting on the faces of the body, so it is their displacements that determine the work. $\endgroup$ – Ján Lalinský Aug 26 '17 at 22:45

What happens is that the standard derivation of Bernoulli's equation often involves some hand waving, omitting justifications for the (sometimes implicit) assumptions. So it may indeed be unconvincing.

In this sense, I strongly recommend checking other derivations, like the ones in this highly informative thread in PhysicsForuns, whose starting point is either Euler's (see also this question) or Navier-Stokes equations.

But now to your questions:

the work-energy theorem is supposed to be applied to particles, not systems of particles

You can also apply it, for example, to a rigid body. The fluid of course isn't a rigid body, but we are considering it's restricted to a streamtube and any relevant change in fluid element shape is reflected by a change of speed and accounted for by the kinetic term in Bernoulli's equation.

That's also why you can apply $W=F\Delta x$ at the ends of the streamtube: what's happening microscopically is complicated, but you still have a force acting over some distance on an incompressible mass.

shouldn't [...] the potential energy related to the arrangement of different particles in the liquid also be included?

When relevant it can be done. One example is when particles can be pushed closer together - when we must consider compressibility. Bernoulli's eq. can then be generalized to compressible flows by substituting its pressure term by an appropriate pressure potential.

What brings us to OP's last question:

Why do we only use the change in kinetic energy and gravitational potential energy?

Because it's simple and in many cases enough. But when it's not enough, all kinds of things can be lumped together into the term pressure, since "pressure is the capacity for a closed system to do work when it changes volume". As for the gravitational potential term, $gz$, it can be substituted by any other relevant bulk potential $\Psi$.


I am not sure what you mean by "particle" here. If by that you mean "molecule of fluid", then recall that Bernoulli equation is founded upon continuum model of fluid (for a derivation see for e.g. Fluid dynamics by Kundu & Cohen) and not on the molecular picture. So any question framed in terms of particles of fluid is irrelevant to Bernoulli equation. The relevant question then becomes "why does Bernoulli equation derived for continuum model work for a real fluid made of particles?" and the answer to it is simply the justification for the applicability of the continuum model to systems made of (large number of) closely spaced particles, in this case fluids.

But I suppose you meant "particle" in the sense in which it used in continuum mechanics. The "arrangement of different particles in the liquid" is indeed accounted for. Different particles of fluid possess different gravitational potential energy by virtue of their vertical separation and this term does appear in Bernoulli equation. If you are thinking in terms of energy of interaction between molecules (I already said this would be an irrelevant question in the context of a continuum, but then...) these are short range forces which are modeled as stresses in the continuum fluid. Bernoulli equation accounts for presence of normal stresses (pressure) but is limited to flows in which there is no shear stress.

Now coming to your second question which asks why work term should equal $p_1A_1v_1~dt-p_2A_2v_2~dt$. This equation is written for a stream tube of finite length for which pressure and velocity may be assumed to be uniform over its inlet and outlet cross-section. In this derivation stream tube is treated as a black-box which exerts pressure on its surrounding fluid. Now pushing fluid into or out of the stream tube through its inlet as well as its outlet, against the pressure prevalent there, requires work, and this rate of work is what is expressed by the $p_1A_1v_1~dt-p_2A_2v_2~dt$ term. Center-of-mass of the stream tube never changes by virtue of our assumption of incompressible flow (there can be no net mass inflow/outflow from a stream tube of fixed geometry).

  • $\begingroup$ OK, the first part satisfied me, but for the last part, it's fine for it to be true, all I want is the proof that that really is the case, since the way the work done is derived feels quite wonky to me $\endgroup$ – Shahe Ansar Aug 27 '17 at 9:16
  • $\begingroup$ @ShaheAnsar If you are imagining marking the fluid in the stream tube at one instant and are talking about displacement of its center of mass of this fluid then this is accounted for in the potential energy term. $\endgroup$ – Deep Aug 27 '17 at 10:13
  • $\begingroup$ No, my problem is the way we derive the work done, since this isn't exactly a rigidbody, all I want is a justification for why pAdt works as a measure of work done. $\endgroup$ – Shahe Ansar Aug 27 '17 at 17:30

Bernouilli's Equation is only an approximation. It ignores many local features of real fluid flows, even in the incompressible and inviscid case.

If you think about the simplest situation, i.e. all the fluid "particles" have identical mass and volume, their volume does not change (because the fluid is incompressible,) and the flow is steady, you should be able to see that

  1. The total potential energy summed over all the particles entering or leaving the control volume are as given in the derivation, even though in any small time interval $dt$ the individual particles at the inlet and outlet are different ones.

  2. Similarly, you can calculate the net work done on any particular particle as it flows through the control volume and then sum over the whole volume. Since the work done is constant over time for a steady flow, again it doesn't matter that in an interval $dt$ the individual particles involved are all at different positions in the flow - the sum over all the particles in the volume is the same.

If the fluid is homogeneous, the same arguments work for a mixture of different particle masses and sizes (e.g. the mixture of gases in air) because in a statistical sense the distribution of the particles doesn't change over time.

In the more general case - for example flow through a helical pipe which acts similarly to a centrifuge - these assumptions may not be true, and strictly speaking Bernoulli's equation doesn't apply, though it may still be a useful approximation in practical situations.

  • 1
    $\begingroup$ I do not think you answered the questions. Regarding the helical pipe - why would the Bernoulli equation not apply? If the flow is stationary and friction is negligible, it should apply just as it does in a straight pipe. $\endgroup$ – Ján Lalinský Aug 26 '17 at 22:34

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