I Explain Bernoulli at the molecular level?

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Static pressure remains equal to atmospheric pressure regardless of vehicle speed, as demonstrated by static ports in pitot tubes. The discussion highlights that Bernoulli's principle does not apply in this context, as static pressure does not decrease with increased airflow speed. Instead, the focus shifts to understanding how static pressure is transmitted at the molecular level and the role of pressure gradients in accelerated flow. The conversation also emphasizes that pressure is frame invariant, while speed is frame dependent, complicating the relationship between the two. Overall, the dialogue seeks to clarify the conditions under which Bernoulli's principle is valid and the physical implications of static pressure in various scenarios.
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This is not homework question, I use numbers to easier explain my question.

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?

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

I think a) is correct answer, Bernulli cant apply here(faster speed = lower static pressure)

Is this correct and why here pressure dont drop with speed(as Bernoulli state)?
 
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russ_watters said:
A. Is correct. Bernoulli says this: p is just p.
So it isnt less bouncing force from molecules in normal direction at higher speeds?
 
russ_watters said:
No. Unless you have something else in mind that you haven't stated you seem to be describing this:

https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/pitot-static-tube-speedometer/
Yes static port at pitot tube show same static pressure for every speeds, so it show that static pressure dont depend on airflow speed.
I am try to find out why different speeds have same static pressure and why accelerated flow reduce static pressure, using explanation of pressure at molecular level("molecules hitting walls")?
 
user079622 said:
I am try to find out why different speeds have same static pressure and why accelerated flow reduce static pressure, using explanation of pressure at molecular level("molecules hitting walls")?
It is my firm belief that "molecules hitting walls" is a hindrance to understanding rather than an aid.
 
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user079622 said:
Yes static port at pitot tube show same static pressure for every speeds, so it show that static pressure dont depend on airflow speed. I am try to find out why...
That's the definition/goal of the device.
 
jbriggs444 said:
It is my firm belief that "molecules hitting walls" is a hindrance to understanding rather than an aid.
Then we have to answer the question of what static pressure actually is, how is it physically transmitted at the molecular level to the walls?
Bernoulli doesn't answer this, he just set correlation between something, that it turns out that we dont know what it is..
 
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russ_watters said:
That's the definition/goal of the device.
It is property of constant velocity, accelerated flow reduce pressure.
If you blow between two flat plates it will not comes together, but if you blow between two curved surface it will comes together because flow accelerated=reduce s. pressure.
 
  • #10
user079622 said:
I am try to find out why different speeds have same static pressure ...
Speed is frame dependent, while pressure differences are frame invariant. So it wound not make sense if the later was determined by the former.
user079622 said:
... and why accelerated flow reduce static pressure, using explanation of pressure at molecular level("molecules hitting walls")?
The more important part here is molecules hitting each other. In order to accelerate you need an a force imbalance. In order for bouncing molecules to accelerate in bulk you need an imbalance in momentum transferred to them over time. That imbalance in momentum transfer rate (force) is a pressure gradient.
 
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  • #11
user079622 said:
Then we have to answer the question of what static pressure actually is, how is it physically transmitted at the molecular level to the walls?
Bernoulli doesn't answer this, he just set correlation between something, that it turns out that we dont know what it is..
While I agree with @jbriggs444 that this is a potentially unnecessary and unhelpful complication, I'll give it a go:

If you have X number of molecules per square meter bouncing off a horizontal surface per second, that number doesn't change just because the surface is moving horizontally. That would be like the weight of a block changing just because you're pushing it horizontally along the ground. Obviously it doesn't.

user079622 said:
It is property of constant velocity, accelerated flow reduce pressure.
If you blow between two flat plates it will not comes together, but if you blow between two curved surface it will comes together because flow accelerated=reduce s. pressure.
OK?
 
  • #12
russ_watters said:
While I agree with @jbriggs444 that this is a potentially unnecessary complication, I'll give it a go:

If you have X number of molecules per square meter bouncing off a horizontal surface per second, that number doesn't change just because the surface is moving horizontally. That would be like the weight of a block changing just because you're pushing it horizontally along the ground. Obviously it doesn't.
Yes that is good explanation. How can we use this analogy to explain pressure drop in accelerated flow?
russ_watters said:
OK?
Yes that is correct, I can post video from these experiments.
A.T. said:
Speed is frame dependent, while pressure differences are frame invariant. So it wound not make sense if the later was determined by the former.
P1+1/2pv1^2=P2+1/2pv2^2

If I enter here v1=500km/h and v2=200km/h, p=1.2kg/m3,
P2 would be not same as P1

Why here Bernoulli is not valid?
 
  • #13
user079622 said:
Yes that is good explanation. How can we use this analogy to explain pressure drop in accelerated flow?
We can't. Different situations are different.
 
  • #14
jbriggs444 said:
It is my firm belief that "molecules hitting walls" is a hindrance to understanding rather than an aid.
How would you explain physically, what is static pressure and why accelerated flow reduce pressure?
If we do F=ma for every molecule that hit a wall, will this correspond to pressure?
 
  • #15
user079622 said:
How would you explain physically, what is static pressure and why accelerated flow reduce pressure?
What is "accelerated flow"? What are you claiming here?
 
  • #16
russ_watters said:
What is "accelerated flow"? What are you claiming here?
Airflow that increase speed, for example over wing upper surface.
 
  • #17
user079622 said:
Airflow that increase speed, for example over wing upper surface.
A wing and venturi are not flat surfaces. Do you recognize that this question is completely different from what you were asking before?
 
  • #18
russ_watters said:
A wing and venturi are not flat surfaces. Do you recognize that this question is completely different from what you were asking before?
Yes, I want to find where can I use Bernoulli and where not
 
  • #19
user079622 said:
How would you explain physically, what is static pressure and why accelerated flow reduce pressure?
If the flow accelerates, there must be a pressure gradient consistent with the acceleration.

##F=ma## You cannot have acceleration without a net force.
 
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  • #20
user079622 said:
Yes...
So when I said "Unless you have something else in mind that you haven't stated...", this was it?
I want to find where can I use Bernoulli, why I cant use it...
What exactly is it that you don't understand? Bernoulli's principle obviously applies to both situations if you use it correctly.
 
  • #21
russ_watters said:
So when I said "Unless you have something else in mind that you haven't stated...", this was it?

What exactly is it that you don't understand? Bernoulli's principle obviously applies to both situations if you use it correctly.
This was what?
How would use it Bernoulli correctly for my first post?
 
  • #22
user079622 said:
This was what?
The different thing you wanted to discuss that you weren't saying.
user079622 said:
How would use it Bernoulli correctly for my first post?
It's in the link I gave you in post #4.
 
  • #23
@jbriggs444

1. So there is not good explanation with molecules, what happen with them when static pressure drop, with constant density and temperature?
(In case density reduce, than is easy, distance between molecules increase..)

2. If we sum all normal component of hits from molecules into the wall, is this correspond to static pressure at that location? Is this mathematically correct or this is pure explantion?

3. If pressure is caused by molecules bouncing the walls, why then in zero gravity, pressure is zero?





russ_watters said:
It's in the link I gave you in post #4.
There are lots of questions like my, that are confused by Bernoulli
https://www.pprune.org/tech-log/88839-why-isn-t-static-pressure-speed-dependent.html
 
  • #24
user079622 said:
3. If pressure is caused by molecules bouncing the walls, why then in zero gravity, pressure is zero?
You will need to specify the scenario better than than just "zero gravity".

A tank full of compressed gas has pressure in zero gravity.

The same mass of gas allowed to spread through an essentially unlimited volume in interstellar space has essentially zero pressure.
 
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  • #26
jbriggs444 said:
You will need to specify the scenario better than than just "zero gravity".

A tank full of compressed gas has pressure in zero gravity.

The same mass of gas allowed to spread through an essentially unlimited volume in interstellar space has essentially zero pressure.
question 1. and 2.?
russ_watters said:
Ok, and?
just say
 
  • #27
user079622 said:
just say
Say what? I answered your question. You don't seem to like the answer but you haven't said why or where we go from here. Do you understand the answer/that link? Are we finished with the question you asked in the OP/"How would use it Bernoulli correctly for my first post?"?
 
  • #28
Imagine you are compressing fluid in a piston cylinder assembly with the end closed . You push on the plunger; the water is pushing in any way it can to relive the elastic potential being stored in it. It cant move much(compress) itself, and the pipe is trying to contain it (so it is expanding). If the nozzle opens under this force of the plunger a flow begins to accelerate (from rest) as that elastic potential in the pipe walls and fluid is converted to kinetic energy of the flow. So the two things are coupled. Accelerating flow is the outcome of internal energy change to kinetic energy. In that scenario to ##t=0## the pressure will start high and decline as the flow accelerates out of the end of the tube. What I'm talking about is a time varying Bernoulli's to get a handle on how pressure is generally converted to kinetic energy, its easier to get a handle on this way. It's not "Classic Bernoulli" you encounter in steady flow (flow not varying in time) but it's an extension that I think helps clarify. Why this translates to steady incompressible flows that experience a change in their container shape is because of mass conservation. So for steady flow is spatially varying pressure, mass is entering across some surface at some rate and unless it's being stored or temporarily compressed it's got to exit. So it's using that internal pressure energy to accelerate the flow to satisfy the mass conservation constraint. So it seems that that mass constraint "decides" what must happen, and it takes from the available energy reserve it can alter.

$$ 0 = \frac{ d}{dt} \int_{cv} \rho d V\llap{-} + \int_{cs} \rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) $$

Bernoulli's is a special case of the First Law of Thermodynamics for a control volume:

$$ \dot Q - \dot W_s = \frac{d}{dt} \int_{cv} \rho \left( \frac{V^2}{2} + gz + u \right) d V\llap{-} + \int_{cs} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} $$

It's not hard to be "deceived" by it, there is much to study. I've been trying to chip away at it for many years myself.
 
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  • #29
user079622 said:
1. So there is not good explanation with molecules, what happen with them when static pressure drop, with constant density and temperature?
Depends on what is considered a 'good explanation'
user079622 said:
2. If we sum all normal component of hits from molecules into the wall, is this correspond to static pressure at that location?
That's the total absolute pressure.
user079622 said:
3. If pressure is caused by molecules bouncing the walls, why then in zero gravity, pressure is zero?
Pressure doesn't have to be zero in zero gravity.
 
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  • #30
russ_watters said:
Say what? I answered your question. You don't seem to like the answer but you haven't said why or where we go from here. Do you understand the answer/that link? Are we finished with the question you asked in the OP/"How would use it Bernoulli correctly for my first post?"?
I didnt tell that I dont agree with Bernoulli, I just say that Bernoulli tell nothing about how pressure is exerted on walls, at molecular level(real physics behind pressure).
Yes I understand that total pressure = dynamic p. + static p. , Do I know how and when to use Bernoulli correctly? No

A.T. said:
Depends on what is considered a 'good explanation'

Can you write explanation?
 
  • #31
user079622 said:
.... Bernoulli tell nothing about how pressure is exerted on walls, at molecular level(real physics behind pressure).
True. That's not what it is for.
user079622 said:
Yes I understand that total pressure = dynamic p. + static p. , Do I know how and when to use Bernoulli correctly? No
What exactly is it that you don't understand about how to apply Bernouuli's principle/equation to the situation described in the link in Post #4?
 
  • #32
russ_watters said:
True. That's not what it is for.

What exactly is it that you don't understand about how to apply Bernouuli's principle/equation to the situation described in the link in Post #4?
Bernoulli said......Total p.=static p. +dynamic p.

So if dynamic(speed) goes up, then static must go down to keep total p. constant. So people automatically remember if speed goes up, static goes down, and vice versa.
And than people wonder why static port in pitot tube, dont show drop in pressure as plane increase speed!! :)
 
  • #33
user079622 said:
Bernoulli said......Total p.=static p. +dynamic p.

So if dynamic(speed) goes up, then static must go down to keep total p. constant. So people automatically remember if speed goes up, static goes down, and vice versa.
And than people wonder why static port in pitot tube, dont show drop in pressure as plane increase speed!! :)
Bernoulli's principle states: "The total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant."
[insert standard assumptions here]

https://byjus.com/physics/bernoullis-principle/

That's for one packet of fluid traveling from one place to another (aka, along a streamline). That's what the principle/equation are about. When you're trying to use it to describe two separate situations (a car at rest vs a car in motion), that's not what it's about/for.

And that should be fairly obvious from looking at the equation: there's one term of static pressure and one term for velocity/velocity pressure, unless you are comparing two points along the same flow in which case there's up to two of each.
 
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  • #34
russ_watters said:
That's for one packet of fluid traveling from one place to another (aka, along a streamline).
If I choose two arbitrary points above the wing, how will I know if they lie on the same streamline?
 
  • #37
@jbriggs444

If I somehow "capture" in theoretical box flow over the wing, inside the box I will have same number of molecules compare to box that I capture a freestream air? Of course density and temperature is constant.
 
  • #38
user079622 said:
If I somehow "capture" in theoretical box flow over the wing, inside the box I will have same number of molecules compare to box that I capture a freestream air? Of course density and temperature is constant.
If you are holding density constant then of course a box of a fixed volume will always capture the same number of molecules.

What are you really trying to ask?
 
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  • #39
jbriggs444 said:
If you are holding density constant then of course a box of a fixed volume will always capture the same number of molecules.

What are you really trying to ask?
If there is same numbers of molecules that hitting the wall, how then explain why pressure is reduced?
I just want to tell that something with this pressure= "molecules bouncing the wall" is not logical.
 
  • #40
user079622 said:
Pocket of air/water is stretched in flow direction and contracted perpendicular to flow when pressure drop?
When velocity rises.
 
  • #41
russ_watters said:
When velocity rises.
Stretched in flow direction and contracted perpendicular to flow in the way, that volume of "control volume" keep the same?
 
  • #42
user079622 said:
Stretched in flow direction and contracted perpendicular to flow in the way, that volume of "control volume" keep the same?
Yes.
 
  • #43
user079622 said:
If there is same numbers of molecules that hitting the wall, how then explain why pressure is reduced?
I just want to tell that something with this pressure= "molecules bouncing the wall" is not logical.
You're mixing and matching from different scenarios now. A gas in a closed, rigid box is not moving relative to the box. Bernoulli's equation/principle does not apply inside the closed box.

The derivation of Bernoulli's equation is in the link I gave you in Post #33.

Note also: incompressible flow is a simplifying assumption used in the derivation that works in low speed (low compressibility) flow.
 
  • #44
russ_watters said:
That's for one packet of fluid traveling from one place to another (aka, along a streamline). That's what the principle/equation are about. When you're trying to use it to describe two separate situations (a car at rest vs a car in motion), that's not what it's about/for.

And that should be fairly obvious from looking at the equation: there's one term of static pressure and one term for velocity/velocity pressure, unless you are comparing two points along the same flow in which case there's up to two of each.
in 10:05h car drive 10km/h in 10:07h car drive 300km/h, at front is probe with static port in clean air.
Why are these two separate situations, why this case is not same as particle flow in pipe from big diameter to small diameter? (except this is atmosphere, not constrain by pipe walls)


Unted.webp
 
  • #45
@user079622

I think it would be helpful if we tried to work off of this problem type I cooked up that attempt to examine the things you ask about in the OP.

Imagine you have a cart like this in still air. It is moving with a constant velocity to the right being propelled by a force ##F##. The control volume is the dashed boundary encasing the cart and the induced flow in the "vicinity" of the cart, where things are obviously changing. The frame labeled ##cv## is inertial. What questions can we help to answer on a model like this (please excuse the drawing)?

1746735037415.webp



P.S. at Mods. What is going on with the spell checking here. Its doesn't even try to make suggestions lately? It is killing me trying to remember how to spell things!
 
  • #46
user079622 said:
in 10:05h car drive 10km/h in 10:07h car drive 300km/h, at front is probe with static port in clean air.
Why are these two separate situations, why this case is not same as particle flow in pipe from big diameter to small diameter? (except this is atmosphere, not constrain by pipe walls)


View attachment 360813
It is "like" pipe flow in the sense that the molecules bouncing off the cart are constrained by other molecules, which are constrained by more molecules etc... This is what I think anyhow. I don't know how a physicist reconciles this...as I think the pressure is independent of particle collision in an ideal gas. However, in a fluid model it makes perfect sense. So I am also intrigued to explore/discuss there as well.
 
  • #47
user079622 said:
in 10:05h car drive 10km/h in 10:07h car drive 300km/h, at front is probe with static port in clean air.
Why are these two separate situations, why this case is not same as particle flow in pipe from big diameter to small diameter? (except this is atmosphere, not constrain by pipe walls)
?? The particles are kilometers apart.

But another answer is that this example is unsteady flow, and Bernoulli's Priinciple assumes steady flow. It's like if you have a fan powered Venturi tube and you are turning up the fan speed while reading different points in the tube. The readings will not satisfy Bernoulli's Principle/ conservation of energy because the mass flow rate at the first point will be higher. than at the second point.
 
  • #48
erobz said:
It is "like" pipe flow in the sense that the molecules bouncing off the cart are constrained by other molecules, which are constrained by more molecules etc... This is what I think anyhow. I don't know how a physicist reconciles this...as I think the pressure is independent of particle collision in an ideal gas. However, in a fluid model it makes perfect sense. So I am also intrigued to explore/discuss there as well.
I'm pretty sure the detail here is that air is a gas and the devil is that the basic form of Bernoulli's Principle/equation assumes incompressible flow*. So you can't use a particle based gas model to explore the basic form of Bernoulli's. The basic form is technically wrong to apply to a gas, it's just not wrong enough to matter at low speed.

OP may prefer to jump straight to the compressible flow version to reconcile how a pressure can change without a change in volume, density or temperature, but then they haven't actually learned Bernoulli's Principle itself. That's not what it's about. As said in Post #6, it's just getting in the way of learning Bernoulli's Principle.

*Indeed, it was discovered through experiments with flowing water.
 
  • #49
russ_watters said:
I'm pretty sure the detail here is that air is a gas and the devil is that the basic form of Bernoulli's Principle/equation assumes incompressible flow*. So you can't use a particle based gas model to explore the basic form of Bernoulli's. The basic form is technically wrong to apply to a gas, it's just not wrong enough to matter at low speed.

OP may prefer to jump straight to the compressible flow version to reconcile how a pressure can change without a change in volume, density or temperature, but then they haven't actually learned Bernoulli's Principle itself. That's not what it's about. As said in Post #6, it's just getting in the way of learning Bernoulli's Principle.

*Indeed, it was discovered through experiments with flowing water.
Well, I feel for the OP (and myself). Problems that show (relatively) simple relationships as examples inspire confidence that isn't really grounded in reality. They make you feel like you should be able to say something, but then you find so many blind spots the problem becomes intractable in just a shallow dig.

So the problem I contrived in post 45 seems to be "steady flow" to me. How do you feel about that?
 
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  • #50
erobz said:
So the problem I contrived in post 45 seems to be "steady flow" to me. How do you feel about that?
Looks good to me.
 
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