I Explain Bernoulli at the molecular level?

[erobz]

Your static port at post #85 will show higher pressure than atmospheric, pressure will be higher as speed increase. If you installed your static port at probe 3m in front of car, in clean air/freestream, it will show atmospheric pressure at any constant speed.

Ok, that aligns with what I think too. So you are after something different, so I'll stop and maybe ask this question myself in a separate thread. Feeling it and showing it to be the case seems to be challenging, and I don't think I'll be satisfied until I can make something to that effect pop out of the equations.

 

[erobz]

Also, if the pressure is constant along a streamline and is initially atmospheric, then it is atmospheric everywhere right?

That's kind of what I thought is the case in steady flow.

However, in streamlines that go around the care the pressure is not constant, nor atmospheric. Also, the velocity is not equal to $V_1$. Why do you think it is?

Well there is this whole business with turning I suppose. The $\frac{V^2}{r}$ "centripetal acceleration" should have some say about that. It gives me pause but then we need to consider rotational, irrotational flow. Am I missing something, things clearly turn in Bernoulli's too, so its confusing me.

Funnily enough this is what Newton used to predict lift forces of a wing, it is not correct. But conceptually it is correct that the static pressure increases at the red dot which is correlated with the angle of the surface at that location.

At least I'm in good company a few hundred years ago!

This is indeed conservation of mass and momentum. You can use those to derive Bernoulli. You don't need conservation of energy as well. But you can also use conservation of energy and mass to derive Bernoulli. Both will work.



Well that's not trivial at all. So there are some analytical solutions of the flow equations (with lots and lots of simplifications), like the Joukowsky transform. Those solutions will tell you what the static pressure and velocity is at a certain location. But that does not follow from Bernoulli.

I can't solve for what the force $F$ is equivalent to, without some invocation/simplification of the First Law - Bernoulli's? I used all three equations. I get this:

$$F = \rho V_1^2 A \sin \theta ( 1 + \cos \theta - \sin \theta ) $$

EDIT: I was trying to derive that in the ground frame before, I probably messed up. Also, I assumed hastily $A_1 = A_2$.

Just to be clear this was the simplification:

Summary of the solution ( assuming uniform velocity distributions remove integrations - steady flow - incompressible flow):

Mass Continuity:

$$ 0 = ( \boldsymbol {V_1} \cdot d \boldsymbol {A_1} ) + ( \boldsymbol {V_2} \cdot d \boldsymbol {A_2} )$$
$$ \implies A_2 = \frac{V_1}{V_2} \frac{A_1}{\sin \theta} \tag{1}$$

Momentum in the horizontal direction ( the frame is inertial - all velocities in equations now referenced from frame $cv$ traveling with cart ):

$$ \boldsymbol {F} = \rho \left[ \boldsymbol {V_1} ( \boldsymbol {V_1} \cdot d \boldsymbol {A_1} ) + \boldsymbol {V_2} ( \boldsymbol {V_2} \cdot d \boldsymbol {A_2} ) \right] $$

$$ F \boldsymbol {i} = \rho \left[ -V_1\boldsymbol {i} ( -V_1\boldsymbol {i} \cdot 1 \boldsymbol {i} ) + \langle -V_2 \cos \theta \boldsymbol {i} + V_2 \sin \theta \boldsymbol {j} \rangle ( \langle -V_2 \cos \theta \boldsymbol {i} + V_2 \sin \theta \boldsymbol {j} \rangle \cdot \langle 0 \boldsymbol {i} + 1 \boldsymbol {j} \rangle) \right] $$

$$F = \rho \left[ V_1^2 A_1 - V_2^2 \cos \theta \sin \theta A_2\right] \tag{2}$$

And Bernoulli's a reduction of the Energy Equation:

$$ \frac{P_1}{\rho g } + \frac{V_1^2}{2g} = \frac{P_2}{\rho g } + \frac{V_2^2}{2g} $$

$$ \implies V_1 = V_2 \tag{3}$$

Combining (1),(2), and (3) results in:

$$ F = \rho V_1^2 A_1 ( 1 - \cos \theta ) $$

so as a "Drag Coefficient" in $\frac{1}{2} \rho A C_d V^2$

$$C_d = 2 ( 1-\cos\theta )$$

So anyhow, I know its a gross simplification but text list the drag coefficient of a cube ( suspended in flow ) as 1.10. At a right angle $\theta$ this gives 2 since in the textbook flow is over entire surface. I'm sure its vastly to simple but at least it was fun trying.

The factors involving $\theta$ at the end acts like a drag coefficient, with some respectability (as far as I can tell)?

But anyhow you can turn the attention back to what the OP is actually after, which is not what I thought. I appreciate the responses.

 

[user079622]

No, this explanation also works for incompressible flow. I think it is tightly related to the Kinetic Theory of Gases, in which the molecules are far enough apart that the only interactions are through collisions but close enough that statistics makes sense. So gas at around atmospheric pressure or less I think (I'm not sure what exactly the limitations of this theory are)

You can describe a bulk of molecules as having some bulk velocity (just the averaged velocity vector) and some random component added on top of that (which is every molecules velocity vector minus that averaged velocity vector). That random component has some statistical properties. So things like average space between molecules, amount of collisions per unit time, average kinetic energy, things like that. So this is the average kinetic energy of the random motions, not to be mistaken with the kinetic energy of a fluid parcel! This kinetic energy and amount of collisions amounts to a pressure (roughly, I'm no expert in the theory). When entering the Venturi, the properties of the random part of the velocity change so that the average kinetic energy and/or the amount of collisions become less. This amounts to lower pressure. Something like that.

What is difference for molecules if they travel at const. 200km/h over static port of aircraft during flight and molecules that travel const. 200km/h over static port in Bernoulli narrow section. In both case streamlined are flat, not curved.
Why at plane not reduce pressure and in narrow section does, that is what make people confused.

Is main difference that streamlines in narrow section has tighter spacing compare to aircraft case(that must have"big" freestream spacing)?

 

[Arjan82]

That's kind of what I thought is the case in steady flow.

Euh, but that would mean that for steady flow there cannot be any other than atmospheric pressure anywhere? Flow over a car or a wing is steady but the pressure is not constant at all.

Well there is this whole business with turning I suppose. The $\frac{V^2}{r}$ "centripetal acceleration" should have some say about that. It gives me pause but then we need to consider rotational, irrotational flow. Am I missing something, things clearly turn in Bernoulli's too, so its confusing me.

That was my point in my fist post of this thread (#69). If you have curvature in the flow then there must be a pressure gradient (and thus a non-constant pressure). Note that rotating flow can be irrotational... irrotational flow means no vorticity anywhere. No vorticity means that the integration of the tangential velocities around a fluid parcel should equal zero. So a fluid parcel (a point in flow) should not be rotating. But a fluid parcel can move in a circle around something without rotating itself. Notice that even a vortex is mostly irrotational flow, up to the very core of that vortex!

Also, with only Bernoulli you simply cannot solve for the flow around anything.

At least I'm in good company a few hundred years ago!

True, cheers for that

I can't solve for what the force $$F$$ is equivalent to, without some invocation/simplification of the First Law - Bernoulli's? I used all three equations. I get this:

$$F = \rho V_1^2 A \sin \theta ( 1 + \cos \theta - \sin \theta ) $$

If you can accurately compute $F$ using any equation, give me a call. Also, there is a Nobel Prize waiting for you .

 

[erobz]

Euh, but that would mean that for steady flow there cannot be any other than atmospheric pressure anywhere? Flow over a car or a wing is steady but the pressure is not constant at all.

I thought it was to form drag/ shear stress...a fundamentally viscous effect?

If you can accurately compute $F$ using any equation, give me a call. Also, there is a Nobel Prize waiting for you .

I didn't say it was accurate, it was a gross simplification of an incompressible, inviscid flow. uniform velocity distribution at angle $\theta$ over the outlet. The best I could do ( at least for now). I meant it seems behaved in some way in the limits of $\theta = 0$, and $\theta = 90$. Thats somewhat satisfying to me as a stay at home dad with a physics hobby. But I still want more...

 

[Arjan82]

I thought it was to form drag/ shear stress...a fundamentally viscous effect?

form drag is actually also pressure drag. I think most of the drag is pressure drag, but viscous drag is also important.

I didn't say it was accurate, it was a gross simplification of an incompressible, inviscid flow. unifor velocity distribution at angle $\theta$ over the outlet. The best I could do ( at least for now). I meant it seems behaved in some way in the limits of $\theta = 0$, and $\theta = 90$.

I also didn't say it was accurate. I said that if you could find and accurate solution you should give me a call. In other words: this is a hopeless exercise. It will not work for any but the simplest of cases.

 

[erobz]

I also didn't say it was accurate. I said that if you could find and accurate solution you should give me a call. In other words: this is a hopeless exercise. It will not work for any but the simplest of cases.

Rest assured that is not my goal. I just want to figure out these equations as practice for a game I'll never play. Really, to occupy some time. I think I got them at least tentatively explored for seemingly "simple" cases and then something comes along and poof, my confidence with them vanishes into thin air. The problem with people like me (low education level) is I don't know what can't be done.

 

[Arjan82]

What is difference for molecules if they travel at const. 200km/h over static port of aircraft during flight and molecules that travel const. 200km/h over static port in Bernoulli narrow section. In both case streamlined are flat, not curved.
Why at plane not reduce pressure and in narrow section does, that is what make people confused.

There is no difference. The question is what pressure you are comparing to. You can definitely design a Venturi where at the constriction the pressure is atmospheric. It can even be higher than atmospheric. The only thing Bernoulli says is that the pressure at the constriction is lower than before the constriction. Before the constriction velocity is lower and thus pressure is higher.

In front of the air plane the static pressure is atmospheric (let's say it flies really really low...) and the velocity is equal to the airplane's velocity, really really far in front of the air plane the pressure is still atmospheric and the velocity is still equal to the airplane's velocity (because you are, and should be, still in the frame of reference of the moving airplane).

Is main difference that streamlines in narrow section has tighter spacing compare to aircraft case(that must have"big" freestream spacing)?

No, the space between streamlines is a choice. But for a given choice they constrict if the velocity increases and widen if the velocity decreases.

 

[erobz]

@Arjan82

You talk about the pressure changing over a wing in atmosphere. Lets just talk about any obstruction, and slow flows, please. So what is mathematically responsible for these points along a streamline to vary in static pressure. We are ignoring elevation head over the obstruction (tiny in air). Which just leaves pressure and velocity to specify what the various static pressures are along a streamline. under constrained. There must be some "container" that is forcing the shape of the flow? If it could just flow over the obstruction and not be forced to change shape by interacting with static/dynamic layers of atmosphere above it, it wouldn't?

 

[user079622]

There is no difference. The question is what pressure you are comparing to.

So overall molecules hits / exerted "force" at the surface of ports is same for both case?

 

[russ_watters]

So overall molecules hits / exerted "force" at the surface of ports is same for both case?

Maybe, maybe not. You haven't fully specified the Venturi scenario. As said: "The question is what pressure you are comparing to."

 

[Arjan82]

So overall molecules hits / exerted "force" at the surface of ports is same for both case?

They can be the same, if the absolute static pressure and temperature is the same.

 

[Arjan82]

@Arjan82

You talk about the pressure changing over a wing in atmosphere. Lets just talk about any obstruction, and slow flows, please. So what is mathematically responsible for these points along a streamline to vary in static pressure. We are ignoring elevation head over the obstruction (tiny in air). Which just leaves pressure and velocity to specify what the various static pressures are along a streamline. under constrained. There must be some "container" that is forcing the shape of the flow? If it could just flow over the obstruction and not be forced to change shape by interacting with static/dynamic layers of atmosphere above it, it wouldn't?

View attachment 360859

Essentially the geometry you put in the flow is responsible for the variation in static pressure. The point is that flow cannot go through an object so the normal velocity at the wall of an object needs to be zero^*). But air parcels will only change their velocity if there is a pressure gradient. So there needs to be a pressure gradient (and thus by extension different static pressures at different locations) to avoid any air parcel to flow through the object.

I guess you can start reasoning from the object's surface. An air parcel hitting that surface will cause the pressure on the surface to rise as much as is necessary to avoid the parcel to go through the surface. That means a high pressure at the surface and an atmospheric pressure far away. There is your pressure gradient. How this gradient is exactly taking shape is dependent on the details of the solution to the Navier-Stokes equation.

Maybe one other thing to realize is that the Navier-Stokes equations themselves are not enough to describe the flow around an object. You also need boundary conditions. Zero velocity on the object's wall is a boundary condition, but also a certain constant speed and pressure far away from the object is a boundary condition. So, the equations themselves actually do not give rise to a change in pressure. It is the boundary conditions that enforce that.

*) As it turns out, the tangential velocity also needs to be zero, but that is not necessary to get different static pressures. An important flow model called 'potential flow' actually cannot prescribe zero tangential flow, but it predicts the pressures around an airfoil just fine.

 

[user079622]

Maybe, maybe not. You haven't fully specified the Venturi scenario. As said: "The question is what pressure you are comparing to."

They can be the same, if the absolute static pressure and temperature is the same.

Yes you can set pressure in venturi what ever you want.

quote:
"As the flow is accelerated by the airfoil, the acceleration of molecules is now biased in the direction of the flow (b)
This acceleration on the molecule mass creates a force in the direction of the flow, which integrates as the pressure exerted by air. This force is reduced in all directions, but the direction of the flow in which it is increased."

They use acceleration for explanation for reduction in pressure, in our youtube video(post #61) use velocity, everyone is pushing their own theory.

 

[Arjan82]

They use acceleration for explanation for reduction in pressure, in our youtube video(post #61) use velocity, everyone is pushing their own theory.

If you only blame your surroundings for not explaining things well enough instead of trying to own your misunderstanding and do something about it, then your surroundings will always fail you.

There is just one theory, we are all saying the same things in a little bit different way and from a little bit of a different angle. This is so we enlighten all puzzle pieces from all angles, but you really have to put them together yourself.

 

[erobz]

Maybe one other thing to realize is that the Navier-Stokes equations themselves are not enough to describe the flow around an object. You also need boundary conditions. Zero velocity on the object's wall is a boundary condition, but also a certain constant speed and pressure far away from the object is a boundary condition. So, the equations themselves actually do not give rise to a change in pressure. It is the boundary conditions that enforce that.

Forgive me if I misrepresent, but in enthusiastic layman's terms it does indeed seem to me like you are agreeing there needs to be a "container" (like I suggest) i.e. a boundary or boundary conditions.

 

[russ_watters]

They use acceleration for explanation for reduction in pressure, in our youtube video(post #61) use velocity, everyone is pushing their own theory.

View attachment 360870

Acceleration changes velocity. It's just slightly different ways of saying the same thing.

You've gotten a ton of good answers from multiple people on many different aspects of the issue. At this point it just seems like you're choosing not to believe them.

 

[user079622]

Acceleration changes velocity. It's just slightly different ways of saying the same thing.

You've gotten a ton of good answers from multiple people on many different aspects of the issue. At this point it just seems like you're choosing not to believe them.

If you only blame your surroundings for not explaining things well enough instead of trying to own your misunderstanding and do something about it, then your surroundings will always fail you.

There is just one theory, we are all saying the same things in a little bit different way and from a little bit of a different angle. This is so we enlighten all puzzle pieces from all angles, but you really have to put them together yourself.

I will explain where I enter in "contradiction loop."


We have two surface with same area(A) over witch water travel at v=50km/h, in one case static pressure of water is P1=1 bar, in second case is P2=2bar.
density=const.(water is incompressible)
temperature=const.

From F=PxA, we know that force exert on surface is higher in 2bar case.

How can we explain at molecular level why in 2bar case, overall impact(higher force) at surface is higher IF density=const(same number of molecules hits surface), temp.=const(at same speed molecules hit surface) and with same "normal component" hit the surface(caused by water flow, explained in video post #61)?

 

[Frabjous]

density=const.(water is incompressible)
temperature=const.

These are approximations. In reality, the EOS is P=f(density, internal energy or temperature). This approximation is biting you at the molecular level.

 

[user079622]

These are approximations. In reality, the EOS is P=f(density, internal energy or temperature). This approximation is biting you at the molecular level.

Yes something is missing in molecular view , so I end up in contradiction.
I think this tiny change in density(even it is water) must somehow increase intermolecular forces, so I look at two molecules like two magnets that are repel harder you press them, so by Newton 3 law they push back to the surface with same force..

 

[Frabjous]

Yes something is missing in molecular view , so I end up in contradiction.
I think this tiny change in density(even it is water) must somehow increase intermolecular forces, so I look at two molecules like two magnets that are repel harder you press them, so by Newton 3 law they push back to the surface with same force..

You do need to relax these approximations at the continuum level also. For example, the sound speed requires a dependence of pressure on density.

 

[russ_watters]

We have two surface with same area(A) over witch water travel...

How can we explain at molecular level why ...(caused by water flow, explained in video post #61)?

Water is a liquid and the video in Post 61 is about gasses. They are different, so you can't use that video for water(though water should be easier). And yet again you haven't provided all relevant details of the scenario.

[edit] Actually, to save time I'll just make up the details to answer it in a way that is easy:

The two surfaces are sitting at the bottom of a pool in a room with very low atmospheric pressure (so it's negligible). The reason the pressure at B is twice the pressure at A, at a molecular level, is A is in the shallow end of the pool and B is in the deep end, so there are twice as many water molecules on top of B as above A. Easy peasey.

 

[A.T.]

The two surfaces are sitting at the bottom of a pool in a room with very low atmospheric pressure (so it's negligible). The reason the pressure at B is twice the pressure at A, at a molecular level, is A is in the shallow end of the pool and B is in the deep end, so there are twice as many water molecules on top of B as above A. Easy peasey.

Easy peasey, because it ignores the actual question: How does the doubled pressure look on the level of water molecules hitting the pool bottom. How do they transfer double the momentum per second to the same area? Are there twice the number of collisions? Is the momentum transferred per collision doubled? A combination of those two?

That's what @user079622 means by explanation on molecular level. Not some trivial stuff like : Taller water column weights more because it contains more molecules.

 

[jbriggs444]

Easy peasey, because it ignores the actual question: How does the doubled pressure look on the level of water molecules hitting the pool bottom. How do they transfer double the momentum per second to the same area? Are there twice the number of collisions? Is the momentum transferred per collision doubled? A combination of those two?

One model that fits these facts (but which is not entirely correct) is to imagine the water as a collection of perfectly elastic and approximately rigid tiny balls that nearly fill the volume. Like little steel balls. They vibrate enough to keep, on average, a small vacant area around each ball.

At the bottom of the pool, pressure is increased. The vacant area around each ball is reduced. Squeezed out. Collisions between adjacent balls are more frequent since the balls are closer to each other. The result is increased pressure on the bottom of the pool.

If the volume of the balls is large compared to the volume of the vacant area then this increased pressure does not come with a significant increase in density.

 

[user079622]

@A.T.
@jbriggs444
@Frabjous

Explanation in video from post #61 tell that larger component of velocity parallel to pipe(flow ), reduce perpendicular component of velocity to pipe(static pressure).
This explanation leads to conclusion, that molecules that travel faster parallel to surface(flow),always has lower static pressure then molecules that travel slower or they are at rest.
We know that is not always the case, it is only case for some specific situation, this fact automatically refutes this explanation.

I consider this explanation wrong and misleading, it will only create additional confusion at velocity-pressure related problems and leads to hasty conclusions what I call the "velocity = low pressure misinterpretation" .
Do you agree with me?
If not, explain why.

 

[russ_watters]

Easy peasey, because it ignores the actual question: How does the doubled pressure look on the level of water molecules hitting the pool bottom. How do they transfer double the momentum per second to the same area? Are there twice the number of collisions? Is the momentum transferred per collision doubled? A combination of those two?

In a liquid, I thought there are no collissions/momentum transfer? Aren't they molecules always in contact with the surface and each other (to the extent that molecules can be in contact) except as moved via Brownian motion? In other words, more like squeezed rubber balls than bouncing ones?

 

[Frabjous]

First let me caveat that I have very little experience with molecular flow, but here are a couple of thoughts.

The pressure difference as a force is a continuum explanation so the video’s dismissal is wrong.
The mean free path of liquid water is less than a nm. The mean free path of air is less than a micron. Given this, the video’s mechanism argument is lacking.

That being said, I think that I interpret the video’s argument different than you. The video is arguing that the velocity distribution is changed by the change in pipe size. This means that in the frame where the mean velocity is zero (for either the thick or thin section) a maxwell-boltzmann distribution for the speeds can be determined. Since the distributions are different for the thick and thin sections, the pressures are also different.

 

[A.T.]

In a liquid, I thought there are no collissions/momentum transfer?

Momentum transfer is always there, if there is a force. Force is the rate of momentum transfer.

Aren't they molecules always in contact with the surface and each other (to the extent that molecules can be in contact) except as moved via Brownian motion? In other words, more like squeezed rubber balls than bouncing ones?

You still have Brownian motion, and the centers of mass of molecules 'in contact' with the bottom are accelating up and down. So instead of talking about frequency of collisions, you can talk about frequency of those accelerations.

 

[256bits]

At the bottom of the pool, pressure is increased. The vacant area around each ball is reduced. Squeezed out. Collisions between adjacent balls are more frequent since the balls are closer to each other. The result is increased pressure on the bottom of the pool.

If the volume of the balls is large compared to the volume of the vacant area then this increased pressure does not come with a significant increase in density.

I don't follow.
'... balls are closer to each other....'
'... not come with a significant increase in density.'
I would say this is the Pauli Exclusion principle and degenerative pressure description for a lay person.

 

[Arjan82]

Explanation in video from post #61 tell that larger component of velocity parallel to pipe(flow ), reduce perpendicular component of velocity to pipe(static pressure).
This explanation leads to conclusion, that molecules that travel faster parallel to surface(flow), always has lower static pressure then molecules that travel slower or they are at rest.
We know that is not always the case, it is only case for some specific situation, this fact automatically refutes this explanation.

I've underlined the error in your reasoning. Nowhere in the video it is said this is generally true, it is specifically explaining a Venturi nozzle. In other words, you are generalizing their statements up to a point that these statements are not valid anymore. This is your generalization, not theirs.

I consider this explanation wrong and misleading, it will only create additional confusion at velocity-pressure related problems and leads to hasty conclusions what I call the "velocity = low pressure misinterpretation" .
Do you agree with me?
If not, explain why.

Already in my first post (#69) I explained why there is no direct relation between velocity and absolute static pressure (what is 'low pressure'? Low compared to what?). Only a velocity difference that leads to a pressure difference. This tells you nothing about the absolute static pressure.

I think the reasoning in the video is based on the kinetic theory of gases, which is strictly speaking not valid for liquids (because it assumes essentially a large mean free path length compared to the particle size), but I think the reasoning holds well enough, conceptually at least. In this condition you essentially have the random motion of particles (responsible for static pressure, and also temperature actually) and their bulk motion (the velocity of the flow). In a Venturi nozzle essentially the random motion is converted to bulk motion. I may have cut some corners here, but I think this is the essence of it.

But note that in general it is perfectly well possible to have different bulk motions (different flow speeds) with equal random motion (equal pressure / temperature) of the particles. Actually, all combinations of bulk motion and random motion is possible (within reason, of course...).