I Explain Bernoulli at the molecular level?

[user079622]

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.

Focus only for their molecular explanation for reduction in pressure, not Bernoulli law.

quote from video from post #61 :
"Atoms inside narrow section have smaller component of perpendicular velocity to the pipe and larger component of their parallel velocity to the pipe. Hence the atoms in narrow section will exert smaller pressure on the walls , but will have larger flow velocity."

If you apply this rule to this case below: constant cross section, P1=200Pa, flow velocity= 10km/h and then just increase velocity to 50km/h and ask how much is P2.(where Bernoulli is not valid, because flow is unsteady) you will automatically conclude that P2 will be smaller then P1, because "atoms have smaller component of perpendicular velocity and larger component of their parallel velocity over surface of static pressure manometer".
If their explanation for reduction in pressure is really physically correct, then P2 must be smaller then P1. But of course this is not a case.
Correct answer is P1=P2, and this is proof that their "components velocity" explanation is not "reason" for reduction in pressure, it is not physically correct.

 

[Arjan82]

Focus only for their molecular explanation for reduction in pressure, not Bernoulli law.

But this is exactly what you are not allowed to do with this explanation. The explanation is about Bernoulli's law, not about pressure in general...

quote from video from post #61 :
"Atoms inside narrow section have smaller component of perpendicular velocity to the pipe and larger component of their parallel velocity to the pipe. Hence the atoms in narrow section will exert smaller pressure on the walls , but will have larger flow velocity."

If you read "Atoms inside the narrow section have a smaller component of perpendicular velocity" than you have to ask yourself: "smaller than what?" The answer is of course "smaller than in the wide section". This means it is a comparison between two locations of the same flow. You cannot extend this explanation to two unrelated cases...

If you blindly apply this rule to this case below:

If you do things blindly you are bound to hurt yourself. You need to keep on using those brain cells...

constant cross section, P1=200Pa, flow velocity= 10km/h and then just increase velocity to 50km/h and ask how much is P2.(where Bernoulli is not valid, because flow is unsteady)

If Bernoulli is not valid, then this explanation in the video from post #61 is not valid. The video explains Bernoulli, so if you apply it to a case for which Bernoulli is not valid, then the explanation is not valid...

 

[user079622]

But this is exactly what you are not allowed to do with this explanation. The explanation is about Bernoulli's law, not about pressure in general...

But they explanation is at fundamental level, atoms, physical explanation of pressure at atoms level must be general and usable in all cases. Must be consistent in all cases
If fail at even one case, it is wrong. That's my guiding principle.

If you read "Atoms inside the narrow section have a smaller component of perpendicular velocity" than you have to ask yourself: "smaller than what?" The answer is of course "smaller than in the wide section". This means it is a comparison between two locations of the same flow. You cannot extend this explanation to two unrelated cases...

Of course you can't extend this explanation to two unrelated cases , because explanation is incorrect, so it doesn't "work" in general.


Am I the only one who finds this explanation inconsistent?

 

[erobz]

Am I the only one who finds this explanation inconsistent?

I think it should be like merging traffic. If there is a lane closure on the sides, the molecules should have to "wait for an opening" to merge into the open lanes. I would expect more molecules to pass though the unobstructed lanes than obstructed ones per unit time. An ideal traffic model seems like what we should be using as an analog here.

 

[Lnewqban]

Focus only for their molecular explanation for reduction in pressure, not Bernoulli law.

If their explanation for reduction in pressure is really physically correct, then P2 must be smaller then P1. But of course this is not a case.
Correct answer is P1=P2, and this is proof that their "components velocity" explanation is not "reason" for reduction in pressure, it is not physically correct.


View attachment 360937

 

[Swamp Thing]

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.

Is it correct to say that sometimes, if we have only one boundary, then the Coanda effect could provide another "virtual boundary" so that the flow behaves as if it is subject to additional constraints? Alternately, maybe one could say that rather than a container, a single boundary surface plus the flow attachment effect is sufficient to create interesting Bernouilli effects without actually needing a "container"?

 

[erobz]

Is it correct to say that sometimes, if we have only one boundary, then the Coanda effect could provide another "virtual boundary" so that the flow behaves as if it is subject to additional constraints? Alternately, maybe one could say that rather than a container, a single boundary surface plus the flow attachment effect is sufficient to create interesting Bernouilli effects without actually needing a "container"?

Its mechanism is fundamentally viscous (my belief is there is no Coanda effect without viscosity). I feel like that is already in murky theoretical water as Bernoulli's is non viscous. Do we know what would happen in an inviscid open atmospheric flow where there is no clear boundary on "one side"? I suspect the "virtual boundary" is a decay phenomenon of energy as it spreads out due to shear stress work (heat is generated).

Just thinking out loud though.

 

[russ_watters]

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

Are you saying there's no such thing as a static force in physics? I'm not sure I've ever heard that before.

 

[jbriggs444]

Are you saying there's no such thing as a static force in physics? I'm not sure I've ever heard that before.

No. That is not what @A.T. is saying.

A force is indeed a rate of momentum transfer. But it need not be the only momentum transfer into a particular body.

A book on the table has upward momentum continuously transferred from the table via contact. And downward momentum continuously transferred from the Earth via gravity. Yet the book remains in place.

 

[russ_watters]

The mean free path of liquid water is less than a nm.

I thought it was zero in a liquid by definition? The molecules in a liquid are chemically bound, aren't they?

 

[russ_watters]

No. That is not what @A.T. is saying.

A force is indeed a rate of momentum transfer. But it need not be the only momentum transfer into a particular body.

A book on the table has upward momentum continuously transferred from the table via contact. And downward momentum continuously transferred from the Earth via gravity. Yet the book remains in place.

Momentum transfer without motion? I guess I didn't realize physicists always thought of force in terms of momentum transfer, as opposed to sometimes just...force.

Still, then, is there a way to differentiate when you're talking about momentum transfer involving motion and momentum transfer not involving motion? Because that's what I was trying to get it: If you exclude molecular motion, the "momentum transfer" is just opposing static forces. The reason I think this is significant is that the addition of molecular motion doesn't add anything of value to the scenario/understanding how pressure arises in a liquid.

What I was trying to convey is if we take the rubber/steel balls model and assume they aren't vibrating, the force is just the weight of the balls over the area. If we assume they are vibrating vertically a little, the bottom sheet of balls is still mostly static force, plus a time-varying force due to the vibrating such that the average total force is the same as if they weren't vibrating. If they are vibrating so much they lose contact with the bottom, then the average force is all due to the time-varying force of the bouncing -- no more static force. The vibrating affects the nuts and bolts of how the force is applied to the floor, but it's not directly relevant to what the force is and how it works macroscopically.

My point was a reference back to your first point: "It is my firm belief that "molecules hitting walls" is a hindrance to understanding rather than an aid."

 

[jbriggs444]

Momentum transfer without motion? I guess I didn't realize physicists always thought of force in terms of momentum transfer, as opposed to sometimes just...force.

I do not know about "always". But if you have a force, you have a rate of momentum transfer.

Still, then, is there a way to differentiate when you're talking about momentum transfer involving motion and momentum transfer not involving motion? Because that's what I was trying to get it: If you exclude molecular motion, the "momentum transfer" is just opposing static forces. The reason I think this is significant is that the addition of molecular motion doesn't add anything of value to the scenario/understanding how pressure arises in a liquid.

So you might have a bunch of impulsive interactions transferring momentum in discrete chunks. Or you might have a continuous interaction. Either way, the [average] rate of momentum transfer is the same. I would consider the collection of impulsive interactions to amount to a [average] force.

Yes, I agree that there is no value in differentiating between the two.

 

[Swamp Thing]

the addition of molecular motion doesn't add anything of value to the scenario/understanding how pressure arises in a liquid.

Not in the sense of advancing human knowledge, maybe not. But if an individual student is at the stage where they need to make an effort to add an alternate picture to their overall understanding, then that effort is often, um, worth the effort, so to speak.

 

[user079622]

Speed does not produce low pressure, increasing it does!

 

[Frabjous]

Speed does not produce low pressure, increasing it does!

For steady flow along a streamline.

 

[user079622]

For steady flow along a streamline.

Exactly.

Case where Bernoulli is not valid:
Do we know how will static port react to horizontal pressure gradient when aircraft accelerate horizontally?

 

[russ_watters]

Case where Bernoulli is not valid:
Do we know how will static port react to horizontal pressure gradient when aircraft accelerate horizontally?

Yes, of course the behavior of static pressure ports is fully understood by physicists/engineers, as discussed at length here.

 

[Lnewqban]

Speed does not produce low pressure, increasing it does!

See post 125.
Airplanes fly at constant speed while their wings induce lower and higher than atmospheric static pressure on certain areas next to their skin (even when reducing speed for landing).

 

[rcgldr]

Somehow, no one has mentioned that a pitot static tube changes the relative air velocity to zero with respect to an aircraft. If there is a port perpendicular to the air flow placed at a point where the relative air velocity across the port matches the surrounding relative air velocity, then it will sense a decrease in pressure as relative air speed increases, and this is the basis for an air speed indicator.

Assuming total energy is not changed, then at the molecular level, the Brownian motion becomes more organized as air speed increases, and the pressure exerted onto walls parallel to the airflow decreases as air speed increases. There is a image in the Wikipedia article that shows this:

https://en.wikipedia.org/wiki/Venturi_effect

Note - lost an edit here, corrected now.

 

[russ_watters]

Somehow, no one has mentioned that a pitot static tube changes the relative air velocity to zero with respect to an aircraft. If there is a port perpendicular to the air flow placed at a point where the relative air velocity across the port matches the surrounding relative air velocity, then it will sense a decrease in pressure as relative air speed increases, and this is the basis for an air speed indicator.

I'm not understanding the way you are using the term "relative air velocity" or how what you are saying jives with how a pito-static system works. Here is a diagram of a typical system for a light aircraft:

Can you describe it again in terms of what each of the two ports is sensing/seeing? I don't think I've heard it described as a pressure going down anywhere, as airspeed increases.

 

[Arjan82]

Somehow, no one has mentioned that a pitot static tube changes the relative air velocity to zero with respect to an aircraft. If there is a port perpendicular to the air flow placed at a point where the relative air velocity across the port matches the surrounding relative air velocity, then it will sense a decrease in pressure as relative air speed increases, and this is the basis for an air speed indicator.

That's not true. The static pressure as measured perpendicular to the airflow (or the static port in @russ_watters figure) stays equal to the pressure of the surroundings.

A pitot tube also measures the stagnation pressure (or total pressure). This is the pressure of the air that enters the hole in the front where it's velocity is decreased to zero with respect to the airplane. This pressure increases with speed. The difference between these pressures is a measure for the speed.

 

[user079622]

Somehow, no one has mentioned that a pitot static tube changes the relative air velocity to zero with respect to an aircraft

Yes airflow at static port is zero, but pressure don't change in boundary layer.

Quote Fundamentals of Aerodynamics by JD Anderson:
"It can be shown experimentally and theoretically that the pressure through the boundary layer in a direction perpendicular to the surface is constant. That is, letting pa and pb be the pressures at points a and b, respectively, in Figure 1.47, where the y-axis is perpendicular to the body at point a, then pa = pb."

If there is a port perpendicular to the air flow placed at a point where the relative air velocity across the port matches the surrounding relative air velocity, then it will sense a decrease in pressure as relative air speed increases, and this is the basis for an air speed indicator.

No.
If static port is located at location where airflow velocity is same as freestream(as it should be), then it will show atmospheric pressure of that altitude.


@Arjan82 , @Dale
I am not sure what will static port show, when is exposed to pressure gradient in flow direction when missile accelerate horizontally in subsonic regime?

@russ_watters
Here is nice animation how static-pitot instruments work

 

[Arjan82]

I am not sure what will static port show, when is exposed to pressure gradient in flow direction when missile accelerate horizontally in subsonic regime?

It will still show the pressure equal to the surroundings. The stagnation pressure however does feel the acceleration. It will indicate a somewhat higher pressure, until the velocity is constant again.

[Mod edit: fixed quotes]

 

[user079622]

It will still show the pressure equal to the surroundings.

Yes that seems logic to me. Is it mathematically justified?

 

[Arjan82]

I think the confusion comes from the fact that when you are accelerating the rocket and look at it from the rocket's frame of reference, that frame is not inertial anymore. The unsteady potential flow equation you show is derived in an inertial frame of reference. The equation changes when you apply it in an accelerating frame of reference. This change exactly cancels the change in ambient pressure you would otherwise see.

For ease of analysis we usually attach the frame of reference to the object. We also do this trick when we do measurements in windtunnels. In that case you are comparing:
1) a stationary rocket sitting in a windtunnel that is generating a certain airspeed, say V,
2) a rocket moving at that same speed V through stationary air.
This works fine because in both cases the frame of reference is inertial.

If you now add acceleration, in the first case, in the windtunnel the air actually needs to be accelerated, generating a pressure gradient. But in the second case, a rocket accelerating through the air, it is not the air that is accelerating but your frame of reference (and the rocket of course). For this case no pressure gradient needs to exist in the air to facilitate the acceleration. This is because the air is not accelerating with respect to an inertial frame of reference.

 

[A.T.]

What is that diagram supposed to show? Pressure gradient (which is frame invariant) is associated with frame invariant proper acceleration of the fluid, not with its frame dependent coordinate acceleration in some non-inertial frame (like the rest frame of the rocket).

 

[rcgldr]

I'm not understanding the way you are using the term "relative air velocity" ...

The static pressure as measured perpendicular to the airflow ...

Yes airflow at static port is zero, but pressure don't change in boundary layer. ...

I lost an edit in my prior post. It's fixed now. I recall that static port has to be located so that the (air velocity) shear boundary layer just outside of the port results in the static port chamber being at the same pressure as the surrounding ambient air.

I had mixed up what happens in a Venturi tube, with the air flow sensed by an aircraft. It's not clear to me if the OP was referring to what happens in a Venturi tube.

 

[erobz]

@russ_watters
Here is nice animation how static-pitot instruments work

Excellent videos!

 

[erobz]

But in the second case, a rocket accelerating through the air, it is not the air that is accelerating but your frame of reference (and the rocket of course). For this case no pressure gradient needs to exist in the air to facilitate the acceleration. This is because the air is not accelerating with respect to an inertial frame of reference.

So if the cart here is accelerating to the right in still air, the force from the "airflow" on the cart at a particular instant is not different than the force from the air on a cart as if it were moving at a constant velocity? Here the purple arrows are area Normals, not velocity.

I don't know what a simplified velocity $V_2$ looks like in this frame... can it actually be parallel to the slope in the cart frame?

For me its challenging not to see it as this in the cart frame:

And infer it must be this in the inertial frame $G$:

The problem is that I get different forces ( from the airflow - opposing $F$) for the two scenarios, and I think it's because I'm cheating (inadvertently) too much with the Energy Equation, or everything I'm doing is a blunder.

 

[rcgldr]

Static pressure is caused by bouncing molecules in normal direction at walls, the faster air travel, the less force at normal direction(static pressure) and more force in direction of flow(dynamic pressure). Is this correct?

Only if the total energy of the air, static + kinetic does not increase. An example of this is Venturi effect.

https://en.wikipedia.org/wiki/Venturi_effect

Atmospheric pressure is 101325 Pa, car drive at two speeds, first at 10km/ and then at 300km/h at flat road, at same altitude.
Is static pressure:
a) for both speeds same as atmospheric pressure
b) for both speeds lower than atmospheric pressure, and lower at 300km/h than at 10km/h

As posted by others, a) is the correct answer, the static pressure | energy remains constant, while kinetic energy increases. This would be easier to visualize in a wind tunnel, where the car is placed at a point in the tunnel where static pressure | energy is constant. Assuming a propeller is driving the air, more kinetic energy is added when accelerating the air to 300 km/h than when accelerating air to 10 km/h, so the air at 300 km/h has more kinetic energy than the air at 10 km/h, but the static pressure | energy remains constant. The total energy is the sum of kinetic and static related energies, so the air has more energy at 300 km/h than at 10 km/h, while static pressure | energy remains constant.

In case you're wondering about wings, Bernoulli is violated since a wing increases the energy of the air from the air's frame of reference or it decreases the energy of the air from the wing's frame of reference.

Using the air as a frame of reference is easier to visualize with a propeller (assume zero initial air velocity, a propeller that is spinning, but not advancing through the air). The work down by a propeller mostly occurs as a pressure jump as the air flows through the disc swept out by the propeller. The velocity of the affected air when its pressure returns to ambient is called the exit velocity, and reflects the work performed by the propeller.