I Explain Bernoulli at the molecular level?

[Frabjous]

Maybe this discussion will help
https://www1.grc.nasa.gov/beginners...ppears in many,a constant throughout the flow.

 

[user079622]

?? The particles are kilometers apart.

If they are on same streamline, why does is matter?

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.

Yes that is correct, I never think about it.

Maybe this discussion will help
https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/bernoullis-equation/#:~:text=The equation appears in many,a constant throughout the flow.

Quote from your source:
"If a gas is at rest, all of the motion of the molecules is random and the pressure that we detect is the total pressure of the gas. If the gas is set in motion or flows, some of the random components of velocity are changed in favor of the directed motion. The directed motion is called “ordered,” as opposed to the disordered random motion.

We can associate a “pressure” with the momentum of the ordered motion of the gas. We call this pressure the dynamic pressure. The remaining random motion of the molecules still produces a pressure called the static pressure."

I don't agree with this view, because it suggest if airflow travel at higher speed, random components of velocity is reduced so static pressure is reduced. We know that is not case, for reduction in static pressure we must have acceleration of flow, not constant velocity. Otherwise aircraft static port will show increase in altitude as he fly at higher speeds.

 

[A.T.]

Quote from your source:
"If a gas is at rest, all of the motion of the molecules is random and the pressure that we detect is the total pressure of the gas. If the gas is set in motion or flows, some of the random components of velocity are changed in favor of the directed motion. The directed motion is called “ordered,” as opposed to the disordered random motion.

We can associate a “pressure” with the momentum of the ordered motion of the gas. We call this pressure the dynamic pressure. The remaining random motion of the molecules still produces a pressure called the static pressure."

I don't agree with this view, because it suggest if airflow travel at higher speed, random components of velocity is reduced so static pressure is reduced. We know that is not case, for reduction in static pressure we must have acceleration of flow, not constant velocity. Otherwise aircraft static port will show increase in altitude as he fly at higher speeds.

It would be clearer, if they stated the condition as:... if the gas accelerates relative to an inertial frame of reference.

Then you need a real non-zero net force on the air packets, and thus a pressure gradient.

 

[user079622]

It would be clearer, if they stated the condition as:... if the gas accelerates relative to an inertial frame of reference.

Then you need a real non-zero net force on the air packets, and thus a pressure gradient.

If we have case where air blower blow air in atmosphere. If I increase power of blower in time interval A to B, from 10% to 100%, airflow is accelerated in this time interval, so there is positive pressure gradient in flow direction.
(Here we can't use Bernoulli, because it is unsteady flow.)
If I put pitot-static tube in front of air nozzle, what will static port show in this time interval?

 

[Frabjous]

I don't agree with this view, because it suggest if airflow travel at higher speed, random components of velocity is reduced so static pressure is reduced. We know that is not case, for reduction in static pressure we must have acceleration of flow, not constant velocity. Otherwise aircraft static port will show increase in altitude as he fly at higher speeds.

For an ideal gas, P is proportional to density and temperature. If P drops, T must also drop for low speed (negligible density change, <.3 sound speed) flows.

 

[A.T.]

If I put pitot-static tube in front of air nozzle, what will static port show in this time interval?
View attachment 360838

I don’t understand your scenario and what interval you mean. You need a steady flow in an inertial reference frame.

 

[user079622]

I don’t understand your scenario and what interval you mean. You need a steady flow in an inertial reference frame.

Yes I know, but I do unsteady flow deliberately , here airflow speed increase during measurement. Will static port read lower pressure that atmospheric pressure?

 

[erobz]

Yes I know, but I do unsteady flow deliberately , here airflow speed increase during measurement. Will static port read lower pressure that atmospheric pressure?

My intuition is that the static pressure would be higher than atmospheric in the impact zone(front of the vehicle). What is your opinion, or anyone?

If the car is sitting still in still air, atmospheric pressure registers on the gauge. If it starts forward motion a normal force develops on the car, that normal force is also acting on the flow via Newtons Third Law. I expect that normal force distributes over the front to the wedge in addition to atmospheric pressure.

$$ P_{gauge} \approx P_{atm} + \frac{N \sin \theta }{A_1} $$

 

[A.T.]

Yes I know, but I do unsteady flow deliberately

Why? It's more complicated to solve and requires more specifications.

 

[user079622]

Why? It's more complicated to solve and requires more specifications.

To see how probe react to accelerated flow.

My intuition is that the static pressure would be higher than atmospheric in the impact zone(front of the vehicle). What is your opinion, or anyone?

In my test there is no impact zone.

In your test from post #45, yes some distance in front of car is higher pressure than atmospheric. The highest pressure is at stagnation point.

 

[Swamp Thing]

@user079622 .. Maybe this video will help ...



(Other members may care to comment if there are issues with the explanation given there).

 

[Demystifier]

Bernulli cant apply here(faster speed = lower static pressure)

Is this correct and why here pressure dont drop with speed(as Bernoulli state)?

The Bernoulli principle says that, in the absence of external forces on the fluid,
$$\frac{v^2}{2}+\frac{p}{\rho}={\rm constant}$$
along a streamline. But the "constant" is not a universal constant, it depends on the streamline.

What happens with this equation when you go from Galilean frame of reference to another? The $\rho$ and $p$ do not change, they are invariant. The $v$ changes. This is consistent with the Bernoulli principle because the constant changes as well. In other words, the Bernoulli principle still applies, but in this case faster speed implies larger constant, not lower pressure.

This change of $v$ due to change of the Galilean frame of reference should be distinguished from the more common change of $v$ due to change of geometry (such as width of the pipe or shape of the wing). In the latter case the constant is the same along the streamline, so the change of $v$ is accompanied with the change of $p$.

 

[A.T.]

To see how probe react to accelerated flow.

You can have accelerated steady flow, where Bernoulli still applies. So it's not clear why you introduce unsteady flow, in a thread about Bernoulli.

 

[user079622]

@user079622 .. Maybe this video will help ...



(Other members may care to comment if there are issues with the explanation given there).

This explanation is wrong, as I said before, if that is true, as speed increase aircraft will read lower and lower static pressure, because molecules will hit walls of sensor "with less and less force". That is not happening.

 

[erobz]

@user079622 .. Maybe this video will help ...



(Other members may care to comment if there are issues with the explanation given there).

They seem to be showing a highly compressible flow, very sparsely packed molecules. I think it more or less agrees with what I expect, but with far less space for molecules to bounce around with random velocity in "incompressible" flow I feel it needs some more explanation... careful thought.

I would think that in a tightly packed liquid flow this explanation would lead to a "selection" of flow from near the entrance of the narrow section. This seems to be what is observed, but I would have like that explained.

 

[russ_watters]

You can have accelerated steady flow, where Bernoulli still applies. So it's not clear why you introduce unsteady flow, in a thread about Bernoulli.

Did you mean "can't"?

 

[erobz]

Did you mean "can't"?

No, I think they are correct. You can have accelerated steady flow. The molecules entering a convergent section are accelerating, its non uniform flow, but still steady ( not time varying, but spatially varying ).

 

[user079622]

Did you mean "can't"?

I think he mean acceleration of flow in narrow section as shown in every Bernoulli example, not acceleration that is product of increasing power of my blower/source,(because then is not steady flow).

@erobz
Bernoulli works only for steady flow, so water level(source ) in this tank must be constant.

 

[Arjan82]

So, Bernoulli is in my opinion one of the most confusion inducing equations out there. So, I didn't read everything thoroughly, so forgive me if I re-state some things. I'll probably also do not adres all earlier examples.

Bernoulli is an energy conservation statement. This means that it only applies as long as you do not add energy to the flow!

Bernoulli equates to a constant (the total pressure usually), but that constant changes with reference frame. So you cannot directly compare a 5km/h car with a 300km/h car with Bernoulli. Going from 5 to 300 adds energy somewhere, so Bernoulli does not apply.

If you compress air in a confined space (e.g. a syringe like), you add energy -> Bernoulli does not apply.

If you heat or cool air you add energy (provided that this heating or cooling is not due to the adiabatic compression or expansion of the air)-> Bernoulli does not apply

If you accelerate flow by some fan or propeller -> Bernoulli does not apply

There are Bernoulli variants for unsteady flow or compressible flow, they look different however.

So the statement "accelerating flow leads to lower pressure" is misleading and certainly not true in general. Take a fire hose which is open at one end (no nozzle). The pressure at the end (where the water flows into the open atmosphere) is very close to atmospheric pressure. Ok, now add a nozzle at the end. This constricts the outflow area and usually greatly increases the flow velocity. Is the pressure at the nozzle exit now lower? Nope... it is still about atmospheric pressure. But you need energy to accelerate the flow, what happens? It is actually the pressure of the flow inside the hose that increases! Of course, the pump that supplies this pressure must be able to do so, though.

But, Bernoulli is still valid in the sense that the pressure just before the nozzle is higher than halfway of the nozzle, and this pressure is again higher than the one at the exit. This is because this pressure is converted to speed (i.e. kinetic energy) by the nozzle. Now Bernoulli states that the sum of pressure and kinetic energy remains constant, so if the kinetic energy increases the pressure must decrease. This is all fine, but you cannot directly compare the case with nozzle to the case without nozzle...

So, now take external flow. I.e. the flow around a car or airplane or whatever. As long as it is not somehow confined. Also, take a case for which Bernoulli applies and remember what I said: adding energy to the flow causes Bernoulli to not be valid anymore. How do you accelerate external flow without adding energy to it?!? Think about it...

A Venturi doesn't work, that's internal flow. Accelerating it in, say, X-direction would require energy right? So how it that possible? Well, if you think about it, you'll notice that just accelerating the flow in only one direction simply will not be possible without adding energy. So, the kinetic energy that an air parcel picks up in one direction, it must lose in some other direction... That means the air parcel must be going around a corner! The airflow is bent somehow. This air bending requires pressure to do, and that's what changes the pressure around e.g. a wing. The shape of the wing makes the air take a turn, but the only thing that changes the direction or velocity of an air parcel is pressure, so the pressure has to change to make the flow go around a wing (or a car, or whatever).

This change of pressure is described by Bernoulli, but only if you know the velocity. So how do you know the velocity? For that you must solve some other equation (the flow potential, or the Navier-Stokes equation, or whatever), and this actually solves both velocity and pressure (and density, and temperature, if you add it). In other words Bernoulli is adhered to by Navier-Stokes (you can derive Bernoulli from Navier-Stokes), and this is also true for the velocity potential equation.

So, now to add the last bit of complexity:
First take some air parcel that goes around a circular arc at constant velocity for whatever reason. The kinetic energy stays equal (since energy has no direction), and thus the pressure must stay equal. However, an air parcel must feel some pressure gradient (directed to the center of the circle) to be willing to go around a corner (you need to 'pull'/'push' the mass of the air parcel around the corner somehow).

So what does that mean, say the pressure of that parcel is everywhere exactly atmospheric. Then the air parcel that is right next to this parcel at the inside corner must be at a lower pressure, otherwise there would not be the necessary pressure gradient to let both parcels go around the corner. But if it is at a lower pressure, it must also be at a higher velocity due to Bernoulli. And exactly the opposite is true for the air parcel at the other side of this atmospheric parcel, this one needs to decelerate and be at a higher pressure for the gradient to exist. So, for air to go around a corner, at least some of the flow must be accelerated or decelerated somehow. But acceleration of an air-parcel in one direction without deceleration in another direction is not possible without adding energy.

I hope this clarifies some things, and hopefully does not add too much confusion . Bernoulli can never give the full explanation for why air has a certain pressure somewhere. So you cannot use just Bernoulli to compute the airflow around a wing, you'll need other equations as well. Bernoulli is in the end just an energy conservation statement...

 

[erobz]

On a pathline we have:

$$ \boldsymbol{a} = \frac{d \boldsymbol{V} }{dt} = \frac{d}{dt} V(s,t) \boldsymbol{u_t} + V(s,t) \frac{d}{dt}\boldsymbol{u_t} $$

$$ \boldsymbol{a} = \frac{d \boldsymbol{V} }{dt} = \left( \frac{\partial V}{\partial s} \frac{ds}{dt} + \frac{\partial V}{\partial t} \right)\boldsymbol{u_t} + \frac{V^2}{r}\boldsymbol{u_n} $$

In the Bernoulli derivation along a streamline the tangential component is the relevant bit, and the flow is assumed steady.

This is Euler Equation which arises from a force balance:

$$ -\frac{\partial }{ \partial s} \left( p + \rho g z \right) = \rho V \frac{\partial V }{\partial s} $$

They can all become normal derivatives, since everything just depends on $s$:

with $V \frac{dV}{ds} = \rho \frac{d}{ds}\left( \frac{V^2}{2}\right)$

you get:

$$ \frac{d}{ds} \left( p + \rho g z + \rho \frac{V^2}{2} \right) = 0 $$

$$ \implies p + \rho g z + \frac{V^2}{2} = C $$

So in Bernoulli's we apply in instances without temporal variation at a location, rotation, viscous, and compressibility effects.

 

[erobz]

So, now take external flow. I.e. the flow around a car or airplane or whatever. As long as it is not somehow confined. Also, take a case for which Bernoulli applies and remember what I said: adding energy to the flow causes Bernoulli to not be valid anymore. How do you accelerate external flow without adding energy to it?!? Think about it...

So in the case where there is still air and a force pushing a car through it at constant velocity, energy is being added to generate the flow. Is it fundamentally different if that force were instead a restraining force and a flow was passing over the still cart? I'm asking in reference to post 45, and post 58. Where I'm trying to analyze the OP's problem.

 

[user079622]

So the statement "accelerating flow leads to lower pressure" is misleading and certainly not true in general. Take a fire hose which is open at one end (no nozzle). The pressure at the end (where the water flows into the open atmosphere) is very close to atmospheric pressure. Ok, now add a nozzle at the end. This constricts the outflow area and usually greatly increases the flow velocity. Is the pressure at the nozzle exit now lower? Nope... it is still about atmospheric pressure.

Yes pressure is atmospheric but water velocity is still constant, just higher then with no nozzle.

Do you have any explanation for reduction in pressure over wing with const. density/temperature using molecular level?

So, now take external flow. I.e. the flow around a car or airplane or whatever. As long as it is not somehow confined. Also, take a case for which Bernoulli applies and remember what I said: adding energy to the flow causes Bernoulli to not be valid anymore. How do you accelerate external flow without adding energy to it?!? Think about it...

You can have external energy, but you must keep constant.
In case for plane engine give energy to accelerate flow, but you cant increase engine power, because Bernoulli is not valid any more.

 

[erobz]

Yes pressure is atmospheric but water velocity is still constant, just higher then with no nozzle.

Do you have any explanation for reduction in pressure over wing with const. density/temperature using molecular level?

The firehose example has a pump that obscures things in this example. The pressure inside the nozzle would be greater.

If you applied Bernoullis to a straight section of pipe, the pressure inside the pipe is still atmospheric, it neglects viscous loss. If you apply it across a convergent nozzle its interior pressure is $P = \frac{1}{2} \rho \left( V_a^2 - V_b^2 \right)$

 

[erobz]

As long pump work at constant power, Bernulli is valid.
you mean pressure at exit of nozzle?

Pressure at nozzle exit is constant (approximately atmospheric) in both cases. But pressure in a few inches inside of a straight pipe (actually the whole pipe if straight) is still atmospheric there "according to Bernoulli". In the nozzle entrance the pressure is elevated ( and the rest of the pipe if straight) " according to Bernoulli"

 

[user079622]

Pressure at nozzle exit is constant (approximately atmospheric) in both cases. But pressure in a few inches inside of a straight pipe (actually the whole pipe if straight) is still atmospheric there "according to Bernoulli". In the nozzle entrance the pressure is elevated ( and the rest of the pipe if straight) " according to Bernoulli"

Pressure in pipe(before nozzle) is higher than atmospheric

 

[erobz]

Pressure in pipe(before nozzle) is higher than atmospheric

We are comparing two separate cases, correct? A straight section of pipe discharging to atmosphere, vs a convergent nozzle discharging to atmosphere. At the nozzle entrance Pressure will be higher than the pressure at the entrance to the straight section of pipe.

 

[russ_watters]

No, I think they are correct. You can have accelerated steady flow. The molecules entering a convergent section are accelerating, its non uniform flow, but still steady ( not time varying, but spatially varying ).

Yeah, I needed to wait longer after waking up for that, thanks. I think the constantly changing scenarios are partly to blame though: OP's recent examples are unsteady flow. The air isn't accelerating through a Venturi, they're measuring from an accelerating car or fan.

 

[russ_watters]

If they are on same streamline, why does is matter?

The common definition I see implies but I suppose does not require that you're following a single particle. Sure, a different particle may follow the same streamline. But in your example the flow is unsteady because the car is accelerating, so it's not the same streamline.

If we have case where air blower blow air in atmosphere. If I increase power of blower in time interval A to B, from 10% to 100%, airflow is accelerated in this time interval, so there is positive pressure gradient in flow direction.
(Here we can't use Bernoulli, because it is unsteady flow.)
If I put pitot-static tube in front of air nozzle, what will static port show in this time interval?

I agree with the concerns of others regarding this scenario, so how about this: Why don't you tell us the answer based on all you've learned about fluid dynamics over the course of this thread.

 

[russ_watters]

This explanation is wrong, as I said before, if that is true, as speed increase aircraft will read lower and lower static pressure, because molecules will hit walls of sensor "with less and less force". That is not happening.

No, what you are saying is wrong: an aircraft increasing its speed is not in steady flow, or from the opposite direction, what they are trying to measure is the static pressure of the air independent of what the plane is doing (the freestream pressure/ when the air isn't moving). Nothing happened to the air to change its static pressure. The accelerating plane and Venturi are different scenarios and you are mis-applying the principle. You seem to have all the information you need here, but you aren't using it/don't believe it.

 

[Arjan82]

Yes pressure is atmospheric but water velocity is still constant, just higher then with no nozzle.

Water velocity is constant compared to what? Velocity is constant in time at a specific location, that is correct. But the water accelerates through the nozzle, so it is not constant in space.

Do you have any explanation for reduction in pressure over wing with const. density/temperature using molecular level?

The concept of pressure is hard to combine with the concept of molecules. Pressure at the molecular level is indeed roughly the bouncing of molecules against a surface (or at each other). But it is more complex than that. For pressure to be defined you need the continuum hypothesis. Pressure makes no sense in very thin air, like near 'the edge' of the atmosphere and space. In other words, Navier-Stokes equations do not work for re-entrant vehicles for the first part of the re-entry.

You can have external energy, but you must keep constant.

But if the external energy is constant everywhere (which is the necessary condition for Bernoulli to be valid), you might as wel take it to be zero. This is because the amount of energy is relative anyway in these problems.

In case for plane engine give energy to accelerate flow, but you cant increase engine power, because Bernoulli is not valid any more.

That is not true. What you are referring to is unsteady flow. This is flow that changes its velocity and pressure in time at a specific location. But you can also have steady flow for which Bernoulli does not apply. Steady flow is flow which does not change its velocity or its pressure in time at any specific location. However, velocity and pressure can change from location to location and still be called steady.

If you take a streamline that enters and exits the engine, the energy of the flow over that streamline has increased. This means that over this streamline Bernoulli cannot be applied. But the flow through an engine which does not change power, is steady flow.

 

[Arjan82]

So in the case where there is still air and a force pushing a car through it at constant velocity, energy is being added to generate the flow.

That depends on your frame of reference. You can say there is no energy added if you attach your frame of reference to the car. So from the car's perspective there is no energy added and Bernoulli can be applied.

If you fix your reference frame to the road (to earth) the car indeed adds energy to the system and you cannot use (unsteady) Bernoulli.

Is it fundamentally different if that force were instead a restraining force and a flow was passing over the still cart? I'm asking in reference to post 45, and post 58. Where I'm trying to analyze the OP's problem.

Energy is frame dependent, yes.

 

[user079622]

No, what you are saying is wrong: an aircraft increasing its speed is not in steady flow

What does it matter steady flow with their explanation how molecules exert less force on the walls when they travel faster? Steady or unsteady is irrelevant for their molecular explanation.

 

[russ_watters]

What does it matter steady flow with their explanation how molecules exert less force on the walls when they travel faster? Steady or unsteady is irrelevant for their molecular explanation.

No, its' not irrelevant. Their explanation is relevant to the situation they are describing, which is a steady flow situation. I assure you they would not try to apply that explanation to your different scenario.

And the "why" is that when you accelerate the car or plane or whatever, you are now dealing with a different set of flow conditions than you were before; A higher energy flow because you put energy into accelerating the plane against the air.

 

[Arjan82]

I think he mean acceleration of flow in narrow section as shown in every Bernoulli example, not acceleration that is product of increasing power of my blower/source,(because then is not steady flow).

You are mistaken here. You can still have acceleration in steady flow. You are talking about acceleration at a specific point that changes in time, that indeed results in unsteady flow.

However if you have a flow on a streamline that has some velocity V1 at location x1, and velocity V2 at location x2 on that same streamline, than the flow is also accelerated.

Steady flow does not mean every air parcel stays at the same location during the entire time. Steady flow means just that the time derivative in the Eulerian flow description is equal to zero. In the Lagrangian flow description of that same flow the time derivative is not necessarily zero.

Eulerian flow description:
You look from an inertial reference frame to a region in space where flow passes in and out of this region

Lagrangian flow description:
You ride on the back of a fluid parcel and follow a streamline.

 

[erobz]

That depends on your frame of reference. You can say there is no energy added if you attach your frame of reference to the car. So from the car's perspective there is no energy added and Bernoulli can be applied.

In 45 and 58 the frame is the cart, and its inertial. The air is still in the ground frame, the cart is moving with constant velocity $v=V_1$ to the right. Bernoulli's applied for the velocity of the flow yields $V_1 = V_2$ along any of those streamlines. Pressure is also constant along a streamline, and is initially atmospheric on all the streamlines. Therein lies the problem, because I think the pressure at a streamline near the "hood" is actually more like:

$$ P_{gauge} \approx P_{atm} + \frac{N \sin \theta}{ A_1} $$

If I do the problem in the way I was "taught" using conservation of mass, momentum, and energy and some uniformly distributed flow velocities with the equations:

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

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

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

I don't seem to find that result? This is I think what the OP is after (as far as I can tell) "What a static gauge reads on the front face of the cart". The apparent contradiction gives me pause in my understanding of a typical simplified analysis of this problem.

 

[user079622]

You are mistaken here. You can still have acceleration in steady flow. You are talking about acceleration at a specific point that changes in time, that indeed results in unsteady flow.

However if you have a flow on a streamline that has some velocity V1 at location x1, and velocity V2 at location x2 on that same streamline, than the flow is also accelerated.

Steady flow does not mean every air parcel stays at the same location during the entire time. Steady flow means just that the time derivative in the Eulerian flow description is equal to zero. In the Lagrangian flow description of that same flow the time derivative is not necessarily zero.

Eulerian flow description:
You look from an inertial reference frame to a region in space where flow passes in and out of this region

Lagrangian flow description:
You ride on the back of a fluid parcel and follow a streamline.

I agree with everything you write here, I know that acceleration is in steady flow. Indeed every Bernoulli example where pipe decrease profile, has acceleration, because velocity can't not increase without acceleration.
Dont know what part you think I made mistake.. You are mixing my answer for different cases

 

[Arjan82]

They seem to be showing a highly compressible flow, very sparsely packed molecules. I think it more or less agrees with what I expect, but with far less space for molecules to bounce around with random velocity in "incompressible" flow I feel it needs some more explanation... careful thought.

I would think that in a tightly packed liquid flow this explanation would lead to a "selection" of flow from near the entrance of the narrow section. This seems to be what is observed, but I would have like that explained.

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.

 

[erobz]

@user079622

Are you trying to figure out what the gauge I have shown ( post 85) reads when the cart is traveling to the right with constant speed $V_1$ as a function of $V_1$ like I think, or am I mistaken and you are really concerned with what it reads while the cart is explicitly accelerating? If it's the latter I'll tap out, because I think what it reads when the cart is just moving with constant velocity is challenging as it is (and would be enlightening regardless), and a place to try and make some forward progress with structured analysis.

 

[user079622]

Are you trying to figure out what the gauge I have shown ( post 85) reads when the cart is traveling to the right with constant speed $V_1$ as a function of $V_1$ like I think, or am I mistaken?

No. Topic is explanation of reduction in static pressure (const. density/temp.) at molecular level. Physicall reason. I should have named title differently, because topic turn into Bernoulli.

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.

I am not sure what will static port(probe 3m in front) read, when car accelerate, because now we have pressure gradient in direction of flow.

 

[Arjan82]

View attachment 360858

In 45 and 58 the frame is the cart, and its inertial. The air is still in the ground frame, the cart is moving with constant velocity $v=V_1$ to the right. Bernoulli's applied for the velocity of the flow yields $V_1 = V_2$ along any of those streamlines. Pressure is also constant along a streamline, and is initially atmospheric on all the streamlines. Therein lies the problem, because I think the pressure at a streamline near the "hood" is actually more like:

If by $V_1$ you mean 'the free and undisturbed flow in front of the car' and with $V_2$ 'the free and undisturbed flow downstream of the car' than you are correct to say $V_1 = V_2$, but anywhere between this does not need to be true..

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

$$ P_{gauge} \approx P_{atm} + \frac{N \sin \theta}{ A_1} $$

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.

If I do the problem in the way I was "taught" using conservation of mass, momentum, and energy and some uniformly distributed flow velocities with the equations:

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

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

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.

I don't seem to find that result? This is I think what the OP is after (as far as I can tell) "What a static gauge reads on the front face of the cart". The apparent contradiction gives me pause in my understanding of a typical simplified analysis of this problem.

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.

 

[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?