# Trouble with Bernoulli's principle

• Erland
In summary, the first derivation of Bernoulli's principle uses Newton's second law to claim that the effective force on a parcel of fluid is ##-Adp##, but the pressure should actually be ##-Adp-pdA##. The second derivation uses conservation of energy to assume that the only forces acting on a volume of fluid between two cross sections doing net work are the forces caused by the pressures at the cross sections at the ends, and gravity. However, my objection is that these assumptions do not hold in many cases, and that Bernoulli's equation only applies to a steady flow.
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.

Erland said:
But how do we know that these are the only forces acting on the volume doing net work?
It is an assumption.

Erland said:
Why can't there be inner forces in the volume doing net work?
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.

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?

Erland said:
Ok, but on what grounds can we saý that this assumption holds?
Usually you just measure the pressure and velocity and verify experimentally that it holds.

Erland said:
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?
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.

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.

Dale said:
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.
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.

I like Serena said:
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.
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.

Erland said:
you wrote in post #2 that such forces can do net work.
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.

Erland said:
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.
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.

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Dale said:
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.
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.

Erland said:
the total (infintesimal) work the particles do on each other
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

Erland said:
Of course the total energy is constant, but this doesn't mean that net work isn't done by these internal forces.
Yes, that is exactly what it means.

But still, work is done by the internal forces.

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).

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Erland said:
But still, work is done by the internal forces.

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.

Erland said:
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.

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.

Erland said:
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.

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.

I like Serena said:
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.
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.

Erland said:
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.
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.

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Greg Bernhardt
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.

Erland said:
Of course the total energy is constant,
Note, this fact is the important fact for the derivation.

Erland said:
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.
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##.

Jano L. said:
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##.
But in such a situation, the pressure is also the same.

According to the second Wikipedia derivation for incompressible fluids

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

the change of the kinetic energy of the system equals the net work done on the system.

Why only kinetic energy? Why can't the work done on the system also change the potential energy? (Note: I don't mean gravitational potential energy - here, we only consider the case where the elevation is the same for the whole system.)

I'm thinking (correct me if I'm wrong) that in a region with higher pressure, there is an internal potential energy which is greater than in a region with low pressure. It seems logical that if that the harder the molecules in the fluid press against each other, the more work can be done on a neighbouring part of the fluid.

If it wasn't so, how could work be done at all on the system by the pressure of the neighbouring fluid at the ends of the system? Where does the energy doing the work come from?

And if there is such an internal potential energy, why doesn't it change when work is done on the system? Why does only the kinetic energy change?

Bernoulli'sdoesn't equation has three terms in it's classical form: pressure, dynamic pressure, and hydrostatic pressure.

Dynamic pressure (##\rho V^2/2##) is the bulk kinetic energy per unit volume of the fluid in motion.

The hydrostatic pressure term (##\rho g h#\$) is effectively gravitational potential energy per unit volume.

The pressure (##p##) is also an energy per unit volume term. You can think of it one of two ways: it is either a stored potential energy sort of like a spring, or else it is a continuum representation of the kinetic energy due to the motion of all of the individual fluid molecules. It behaves the same way either way.

All of these vary (or potentially vary) when work is done on the system, but their sum remains constant. in that sense, Bernoulli'sdoesn't equation is a conservation of energy statement.

Erland said:
But in such a situation, the pressure is also the same.
Only as a special case. In general, pressure is a function of position and usually changes in downstream direction.

Erland said:
According to the second Wikipedia derivation for incompressible fluids

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

the change of the kinetic energy of the system equals the net work done on the system.

Why only kinetic energy? Why can't the work done on the system also change the potential energy? (Note: I don't mean gravitational potential energy - here, we only consider the case where the elevation is the same for the whole system.)
Internal energy of the fluid can change as well. This can only happen if the fluid gets compressed or if internal friction in the fluid is significant (large gradients of velocity). When that is so, the assumption of incompressibility that the Bernoulli equation derivation is based on is no longer applicable and things get more complicated. For example, flow in thin tubes such as blood flow in capillaries manifests lots of internal friction in the fluid and the Bernoulli equation does not apply. In the simplest case the friction can be taken into account - see the Hagen-Poiseuille equation.

https://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

But there are many cases where the fluid does not compress much and where the internal friction can be neglected, if only to get a rough idea of the flow. Then the internal energy of the fluid is constant, can be ignored and the Bernoulli equation is applicable.
I'm thinking (correct me if I'm wrong) that in a region with higher pressure, there is an internal potential energy which is greater than in a region with low pressure. It seems logical that if that the harder the molecules in the fluid press against each other, the more work can be done on a neighbouring part of the fluid.
That is true, but usually*, if the fluid does not experience extreme pressures, the change of internal energy is very minuscule. In liquids, to change internal energy appreciably, extreme pressures would be needed since the work pressure forces can do is constrained by quite a small volume changes allowed by the liquid state. In common cases of flow of water, changes in internal energy are minuscule and can be very well ignored.

* This is not true in gas flows that experience significant changes in pressure, such as air flows around fast moving bodies, or expansion of gas from pressurized tank ,or liquid flows near sharp propeller blades where bubbles form. There, the Bernoulli equation may not be even roughly correct description, if the fluid changes temperature and volume too much.

If it wasn't so, how could work be done at all on the system by the pressure of the neighbouring fluid at the ends of the system? Where does the energy doing the work come from?
Internal energy is not needed to explain the work of pressure forces. It is not needed to explain mechanical work on incompressible solids and similarly it is not needed for mechanical work on incompressible fluids.

The work on any fluid element is done by contact pressure forces of the surrounding fluid. The energy comes from other energies in the body pushing and moving that surrounding liquid, maintaining pressure and moving the fluid so it can do work. For example, work done on a hydraulic lever is done by a motor that pushes a piston in contact with the hydraulic liquid.

## 1. What is Bernoulli's principle?

Bernoulli's principle is a scientific law that states that as the speed of a fluid increases, the pressure within the fluid decreases. This principle is based on the conservation of energy and is often used to explain various phenomena in fluid dynamics.

## 2. What is the trouble with Bernoulli's principle?

The main trouble with Bernoulli's principle is that it is often misunderstood or misapplied. Many people mistakenly believe that it can be used to explain all fluid-related phenomena, when in fact it only applies to certain situations. Additionally, there are some exceptions to the principle that can make it seem inconsistent or inaccurate.

## 3. Can Bernoulli's principle be violated?

No, Bernoulli's principle is a fundamental law of physics and cannot be violated. However, there are cases where it may not accurately predict the behavior of a fluid due to certain factors such as viscosity, compressibility, and turbulence.

## 4. How is Bernoulli's principle used in real-life applications?

Bernoulli's principle has a wide range of applications in various fields such as aviation, meteorology, and engineering. It is used to explain lift in airplane wings, the behavior of weather systems, and the design of pipes and ventilation systems.

## 5. How can I better understand Bernoulli's principle?

To better understand Bernoulli's principle, it is important to have a strong foundation in physics, specifically in fluid dynamics. It is also helpful to study real-life examples and applications of the principle and to understand its limitations and exceptions. Experimentation and hands-on activities can also aid in understanding the concept more deeply.

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