I Work - Energy Principle Application to Fluid Flow

[vanhees71]

That's a more general derivation similar to the one I've given above for the special case of ideal fluids. It's a great book!

 

[Dale]

I thought that too, but I think this is not the case because the way derivation is carried out is like with control volume approach. Nowhere is it written that this approach is used, but if you are familiar with that type of derivation, you can see that is actually the case.

I am familiar with control volume type derivations. But, if you are going to reject what a derivation actually explicitly says and replace it with your own version and criticize your own version then if the resulting derivation is wrong you really have only yourself to blame. It is at that point no longer the Wikipedia derivation but your own personal strawman derivation. To criticize a derivation you must not reinterpret what they say, but accept what they actually do say.

Here they explicitly consider a fluid element. This element moves and changes KE. That is legitimate. Then, since the KE of any given location is constant, they can remove the portion of the fluid element which is shared by both the initial and final element and attribute the change in KE of the whole element to a difference in KE between the unshared portions.

Again, the derivation is sloppy because it does not clearly relate the COM of the work energy theorem to the KE of these unshared portions. But it is not a wrong approach.

 

[Dario56]

I am familiar with control volume type derivations. But, if you are going to reject what a derivation actually explicitly says and replace it with your own version and criticize your own version then if the resulting derivation is wrong you really have only yourself to blame. It is at that point no longer the Wikipedia derivation but your own personal strawman derivation. To criticize a derivation you must not reinterpret what they say, but accept what they actually do say.

Here they explicitly consider a fluid element. This element moves and changes KE. That is legitimate. Then, since the KE of any given location is constant, they can remove the portion of the fluid element which is shared by both the initial and final element and attribute the change in KE of the whole element to a difference in KE between the unshared portions.

Again, the derivation is sloppy because it does not clearly relate the COM of the work energy theorem to the KE of these unshared portions. But it is not a wrong approach.

I didn't reinterpret what they said. Problem arises because I disagree that moving fluid element is used, but control volume actually. Nowhere it is stated that fluid volume element moves nor is it tracked in time as it moves, but masses of fluid that enter and exit control volume are in fact tracked which is exactly control volume approach. In derivation it is stated that in time $\Delta t$ masses $\Delta m$ ENTER AND EXIT the system carrying kinetic and potential energy with it. It didn't say that in time $\Delta t$ fluid element moves as a whole and than tracks its COM .In my opinion, bad derivation which shouldn't be tought in this way.

 

[Dale]

I didn't reinterpret what they said. Problem arises because I disagree that moving fluid element is used, but control volume actually. Nowhere it is stated that fluid volume element moves (emphasis added)

They said “The system consists of the volume of fluid, initially between the cross-sections A1 and A2. In the time interval Δt fluid elements initially at the inflow cross-section A1 move over a distance s1 = v1 Δt” (emphasis added)

I am not sure how you can be unclear. They are clearly talking about the motion of the fluid elements. The use of the work energy theorem is legitimate here.

If you wish to criticize a derivation, then do so, criticize it on its own ground. But don’t make your own strawman and criticize that strawman. I don’t like this derivation either, but your specific criticism of it is a strawman fallacy.

 

[Dario56]

They said “The system consists of the volume of fluid, initially between the cross-sections A1 and A2. In the time interval Δt fluid elements initially at the inflow cross-section A1 move over a distance s1 = v1 Δt” (emphasis added)

I am not sure how you can be unclear. They are clearly talking about the motion of the fluid elements. The use of the work energy theorem is legitimate here.

If you wish to criticize a derivation, then do so, criticize it on its own ground. But don’t make your own strawman and criticize that strawman. I don’t like this derivation either, but your specific criticism of it is a strawman fallacy.

Yes, but this a mistake of this derivation. They say first what you quoted, there is a fluid element between cross sections and than procced by deriving equation like there is a control volume. They mixed two approaches because they say one thing and than two the other. We agreed that if you use fluid element approach, you must compute how COM moves which clearly isn't done and derivation progresses with control volume approach even though it didn't start with that.

 

[Dale]

They mixed two approaches

There is nothing wrong with mixing approaches.

 

[Dario56]

There is nothing wrong with mixing approaches.

I disagree. If you pick fluid element you should compute how its COM moves in space and time not do energy balance on control volume. Approaches can't be mixed.

Imagine a professor in mechanics class saying to class that he will derive equation of motion of some body via Newton's laws (Newtonian mechanics) and than starting to derive it via Lagrangian mechanics. This is similar to what was done in this derivation. They said one thing and than made the other, confusing people.

 

[vanhees71]

I'm not sure what you are after, but of course you must distinguish two approaches in the very beginning of formulating fluid dynamics.

(a) Lagrangian formulation

This is a formulation which is most closely related to point-particle mechanics, where you have a set of $N$ bodies, described by trajectories $\vec{x}_j(t)$, i.e., the position of the $j$th particle as a function of time.

In continuum mechanics you don't deal with a discrete set of point particles but you use a continuum description. In the Lagrangian approach you start by describing the fluid in some standard configuration, filling in this standard configuration a volume $V_0$ (which can also be $\mathbb{R}^3$ if you idealize a fluid where boundary conditions don't play a role). Then each individual fluid element is labelled with a vector $\vec{\xi} \in V_0$ and you can describe it as a function $\vec{x}(\vec{\xi},t)$. For a fluid (liquid or gas) you can take $\vec{\xi}$ as the position of the fluid element at some initial time $t=0$.

(b) Eulerian formulation

That's however not so convenient. We are rather used to looking at a fluid at a certain place $\vec{x}$ and describing it by the flow field $\vec{v}(t,\vec{x})$, which is the velocity of the fluid element being at time $t$ at the position $\vec{x}$. Both descriptions are of course equivalent.

To follow trajectories of fluid elements, the direct way is of course to use the Lagrangian formulation, but nothing prevents you from using the Eulerian description to solve for the flow field and then calculate the pathlines via
$$\dot{\vec{x}}=\vec{v}(t,\vec{x}), \quad \vec{x}(0)=\vec{\xi}.$$

 

[Dale]

I disagree. If you pick fluid element you should compute how its COM moves in space and time not do energy balance on control volume. Approaches can't be mixed

Approaches absolutely can be mixed. In fact, typically one approach is derived from the other. If they are derived from one another then there is no way to claim that they cannot be mixed. In this case, since they explicitly started with the fluid element approach if you believe that implies the control volume approach, then doesn’t that in itself demonstrate that the approaches are compatible? One implies the other here, by your interpretation.

Often it is convenient to use one approach to provide an input to another approach. As long as the final result satisfies both approaches then it is completely legitimate to combine approaches.

You have recognized that the work energy theorem is legitimate in a fluid element approach. The derivation in question explicitly uses a fluid element approach. Anything that might be implied certainly does not make what was explicitly stated magically disappear. The work energy theorem is explicitly applicable in this derivation.

 

[Dario56]

Approaches absolutely can be mixed. In fact, typically one approach is derived from the other. If they are derived from one another then there is no way to claim that they cannot be mixed. In this case, since they explicitly started with the fluid element approach if you believe that implies the control volume approach, then doesn’t that in itself demonstrate that the approaches are compatible? One implies the other here, by your interpretation.

Often it is convenient to use one approach to provide an input to another approach. As long as the final result satisfies both approaches then it is completely legitimate to combine approaches.

You have recognized that the work energy theorem is legitimate in a fluid element approach. The derivation in question explicitly uses a fluid element approach. Anything that might be implied certainly does not make what was explicitly stated magically disappear. The work energy theorem is explicitly applicable in this derivation.

In my opinion, the fact they are combined implies not that they are compatible, but that this derivation isn't correct in sense that is says one thing and than does the other. I used analogy to describe it.

Derivation says that it is going to use fluid element in the beggining and than actually uses control volume approach and than falsely applies work - energy theorem.

 

[Dale]

Derivation says that it is going to use fluid element in the beggining and than actually uses control volume approach and than falsely applies work - energy theorem.

If I say I am going to build a barn using lumber and I do use lumber and also nails, did my use of the nails somehow make the lumber disappear or did it make my statement that I would use lumber false? Obviously not.

I think that the derivation is a flawed derivation, but I think your complaint about it is not the problem at all.

Your complaint is not even valid because you are trying to pretend that they didn’t do something that they explicitly did. I even quoted the relevant sentence for you. Your claim that they didn’t do what they said they were going to do is demonstrably false.

 

[Dario56]

If I say I am going to build a barn using lumber and I do use lumber and also nails, did my use of the nails somehow make the lumber disappear or did it make my statement that I would use lumber false? Obviously not.

I think that the derivation is a flawed derivation, but I think your complaint about it is not the problem at all.

Your complaint is not even valid because you are trying to pretend that they didn’t do something that they explicitly did. I even quoted the relevant sentence for you. Your claim that they didn’t do what they said they were going to do is demonstrably false.

I don't agree with the analogy.

They said they will use fluid element approach and than they switched to control volume. I won't repeat myself, there is no point. How explicitly was that said is a matter of debate, but we agreed that it was clear from what was written that they will use this approach. In this regard, I don't pretend they didn't say they will use this approach, but I am saying they didn't actually use it given arguments I said many times. Acta, non verba.

We didn't agree on every detail and that is okay. What we both agree upon is that this derivation is bad and sloppy as far as I understood you. Thanks for the discussion though, I enjoyed it. I also learned something new which is a point of making threads and discussing.

 

[anuttarasammyak]

I am sorry it is just step aside but when I consider your problem I have got primitive question on momentum as sketched. I should appreciate you would teach me. Thanks in advance.

[EDIT] I add one more primitive question: What kind of energy and how does it contained in fluids as "pressure energy".

Thanks again.

 

[Dale]

They said they will use fluid element approach … How explicitly was that said is a matter of debate

I directly quoted the sentence. They literally said the words “fluid elements” and described what the fluid elements did. It is hard to be more explicit than that.

I don't pretend they didn't say they will use this approach, but I am saying they didn't actually use it given arguments I said many times. Acta, non verba.

A proof is just words. There is no difference between saying and doing in a proof.

What we both agree upon is that this derivation is bad and sloppy as far as I understood you.

Yes, we agree on that. We don’t agree on “wrong” but we agree on “bad” and “sloppy”.

 

[Dale]

I am sorry it is just step aside but when I consider your problem I have got primitive question on momentum as sketched. I should appreciate you would teach me. Thanks in advance.

It is easier to consider energy. Those forces don’t move so they don’t do any work. Which means that, from an energy perspective, their complicated direction and unknown magnitude can be neglected.

In any case, the direct answer to your question is obviously that the forces to the right are greater than the forces to the left.

 

[vanhees71]

Approaches absolutely can be mixed. In fact, typically one approach is derived from the other. If they are derived from one another then there is no way to claim that they cannot be mixed. In this case, since they explicitly started with the fluid element approach if you believe that implies the control volume approach, then doesn’t that in itself demonstrate that the approaches are compatible? One implies the other here, by your interpretation.

Often it is convenient to use one approach to provide an input to another approach. As long as the final result satisfies both approaches then it is completely legitimate to combine approaches.

You have recognized that the work energy theorem is legitimate in a fluid element approach. The derivation in question explicitly uses a fluid element approach. Anything that might be implied certainly does not make what was explicitly stated magically disappear. The work energy theorem is explicitly applicable in this derivation.

A "mixed approach" (mixed beween Lagrange and Euler descriptions) is the approach I proposed in my longer posting above, using integrals over a material fluid volume applying Reynold's transport theorem, which is a special case of the Leibniz rules for deriving integrals wrt. an parameter with parameter-dependent boundaries (in continuum mechanics the parameter of course is time).

You describe everything in the Eulerian way, i.e., the kinematics with the flow field $\vec{v}(t,\vec{x})$ and mass (or particle) density and the dynamics with the energy-momentum balance equations. To close the system of equations you need also an equation of state. Then you derive the equations of motion for these Eulerian quantities (a field description of the fluid motion) using Reynold's transport theorem. The details are in my posting above.

 

[Dario56]

View attachment 291429
I am sorry it is just step aside but when I consider your problem I have got primitive question on momentum as sketched. I should appreciate you would teach me. Thanks in advance.

Forces (or pressures) you've sketched that have grip on the pipe wall do no work and so can't accelerate fluid. Reason is that these are normal forces of pipe surface and these are always perpendicular do direction of the flow (local fluid velocity) and so work of these forces is zero.

It is like work done by normal force when you pull some object on the horizontal surface or ground.

 

[anuttarasammyak]

Forces (or pressures) you've sketched that have grip on the pipe wall do no work and so can't accelerate fluid. Reason is that these are normal forces of pipe surface and these are always perpendicular do direction of the flow (local fluid velocity) and so work of these forces is zero.

Work would be zero but momentum change or velocity change might not be zero.
Say a ball bounce back from the floor, the ball speed changes sign by normal force from the floor which stays at rest.

 

[Dario56]

Work would be zero but momentum change or velocity change might not be zero.
Say a ball bounce back from the floor, the ball speed changes sign by normal force from the floor which stays at rest.

In your example, velocity or momentum does change, but not because of normal forces, but because of net work done by pressure forces on fluid.

Yes, if you bounce a ball perfectly elastically, velocity will have the same magnitude, but it will change direction. Since velocity is a vector, the change of direction of velocity vector requires a force due to Newton's 2nd law. This force is normal force.

In case of fluid flow, this force changes direction of local fluid velocity on pipe walls, but not its magnitude since work done is zero.