# I Trouble with Bernoulli's principle

1. Oct 19, 2017

### Erland

I have some trouble with the derivation of Bernoulli's principle. The Wikipedia gives two derivations, for an incompressible fluid, and I have trouble with both of them:

https://en.wikipedia.org/wiki/Bernoulli's_principle#Derivations_of_the_Bernoulli_equation

In the first derivation, using Newton's second law, it is claimed that the effective force on a parcel of the fluid is $-Adp$, but in my opinion, it should be $-Adp-pdA$. I see no the reason why the cross-sectional area should vary less than the pressure.

In the second derivation, using conservation of energy, which seems to be more common in texts, it is assumed that the only forces acting on a volume of the fluid between two cross sections doing net work are the forces caused by the pressures at the cross sections at the ends, and gravity.
But how do we know that these are the only forces acting on the volume doing net work? Why can't there be inner forces in the volume doing net work?

Actually, it seems to me that my two objections are related. Clearly, if the cross-sectional area is constant, then the inner forces pressing on parcels of the fluid cancel each other out, by Newton's third law, together with the fact that all parcels have the same velocity, so these forces do no net work.
But if the cross-sectional area is constant, then so is the pressure, making Bernoulli's equation meaningless in this case.

2. Oct 19, 2017

### Staff: Mentor

It is an assumption.

There can, and then Bernoulli principle does not hold. It isn't a fundamental law of nature. It is a derived principle that holds when the assumptions are valid.

3. Oct 19, 2017

### Erland

Ok, but on what grounds can we saý that this assumption holds?

For example, in a rigid body, we can say the internal forces do no work, since the relative motions of the particles are perpendicular to their relative positions, which means that the internal forces do no work, assuming that these forces are central forces. Can we say something similar in this case?

4. Oct 19, 2017

### Staff: Mentor

Usually you just measure the pressure and velocity and verify experimentally that it holds.

Internal forces never do work on a system. Work is a transfer of energy and an internal force cannot transfer energy to or from a system. It has nothing to do with a rigid body, just the fact that the force is internal.

5. Oct 19, 2017

### I like Serena

We only look at the external forces, and we only look at the net result when applying those forces during an infinitesimal time interval.
Just like with rigid bodies, all internal forces cancel out when we have a steady flow.
Note that Bernoulli's equation only applies to a steady flow.

So we have some system with changing cross section A.
For our model, without loss of generality, we put a small extension on both sides with constant cross section.
The external forces are applied on those extensions.
The pressure p in those extensions does change due to internal forces, even though the cross sections of those extensions have been modeled to be constant.

6. Oct 20, 2017

### Erland

Then you must mean something else by "internal forces" than I do. I just mean forces by which particles of a system act on each other, and you wrote in post #2 that such forces can do net work. As another example, take a system of celestial bodies acting on each other by gravitation. In this system, there is a trade-off between potential and kinetic energy which means that work is being done.

7. Oct 20, 2017

### Erland

But I don't understand how we can assume that the cross section is constant and not the pressure. I my understanding, the cross section and the pressure vary together, if one of these quantities changes, so does the other one.
If we have a thin parcel with width $dx$, with an effective force $pA$ acting on its left side and an effective force $-(p + dp)(A+dA)$ acting on its right side, then the net effective force on the parcel is $-p\,dA-A\,dp$ (the term $-dp\,dA$ is negligible). I see no reason that the term $-p\,dA$ can be discarded if the term $-A\,dp$ cannot.

8. Oct 20, 2017

### Staff: Mentor

Oops, my apologies. That is wrong. Internal forces can never do work. There are other assumptions that can be violated to invalidate the Bernoulli equations, but that isn't one.

Not if they are internal to the system. No work is being done on the system. The energy of the system is constant, so no work is being done.

Last edited: Oct 20, 2017
9. Oct 20, 2017

### Erland

But if we have a system with two particles with the same mass falling towards each other by their gravitation along a straight line with equal speeds, with no external forces acting upon them, then they act on each other by equal but opposite forces, $F$ and $-F$, say, and with equal but opposite infinitesimal displacements, $dr$ and $-dr$, say. Then the total (infintesimal) work the particles do on each other is $Fdr+(-F)(-dr)=2Fdr$ which is nonzero (although infinitesimal). This work represents potential energy being converted to kinetic energy. Of course the total energy is constant, but this doesn't mean that net work isn't done by these internal forces.

10. Oct 20, 2017

### Staff: Mentor

At this point in your analysis you are subtly changing the system from the one system composed of both particles to two systems each composed of one particle.

The whole system of the two particles has a potential energy and a kinetic energy. The internal forces can change the system's energy from potential to kinetic and vice versa. But the total energy of the system does not change precisely because the forces are internal

Yes, that is exactly what it means.

11. Oct 20, 2017

### Erland

But still, work is done by the internal forces.

12. Oct 20, 2017

### Staff: Mentor

Not on the two particle system. It is only work if you break it up and look at each particles as separate systems.

The problem in your analysis is the changing of the systems. Think about this, if an internal force does work then is the work done on the system (increasing its energy) or by the system (decreasing its energy).

Last edited: Oct 20, 2017
13. Oct 20, 2017

### Mister T

Sure, but it doesn't change the system's internal energy; you need work done by external forces (or heat transferred to or from the system) to accomplish a change in internal energy.

14. Oct 21, 2017

### I like Serena

Taking your example of celestial bodies, yes, if they approach each other they do positive work on each other.
And when they move away from each other again, they do negative work on each other.
When the system is back in its original state, the net work is 0.

You're looking at an infinitesimal element somewhere inside the system, but that's not intended.
Instead we create artificial extensions on the outside of the system with constant cross section.
That way we can tell how much work an external force applies to the system, which is how much the internal energy increases.

15. Oct 21, 2017

### Erland

Could you please draw a figure which explains what you mean with those artificial extensions.
I know this could be quite much to ask for, I am very poor myself at drawing figures and upload them to the web, but I would very much appreciate if you could do this.

16. Oct 21, 2017

### I like Serena

It's on the wiki page where the proof is:

In the middle we have the actual system with a changing cross section.
And to the left and right we have artificial extensions with constant cross section intended to say something meaningful about how much energy we add/remove.

17. Oct 21, 2017

### Erland

Thanks, but in each of the extensions, both area and pressure are constant. On the other hand, comparing the extensions, both area and pressure are unequal.

18. Oct 21, 2017

### Staff: Mentor

Note, this fact is the important fact for the derivation.

19. Nov 5, 2017

### Jano L.

The element of the fluid is conveniently chosen so its two lids are both perpendicular to velocity and have the same area. Only then total force on the element in the direction of flow is $Adp$.

20. Nov 5, 2017

### Erland

But in such a situation, the pressure is also the same.