Pressure and Lift around a Wing

[fog37]

Hello everyone,

I am pondering on the airflow around a wing:

When air flows over a wing, part of the air goes below the wing and part over the wing. Even when the air is moving, the pressure at a point in space is always isotropic. The pressure on top of the wing is slightly lower than than the free stream pressure. This pressure distribution produces a net force $L _{top}$ on top of the wing directed downward. The pressure at the bottom of the wing is slightly larger than the free stream pressure farther away from the wing and produces a net lift force $L _{bottom}$. The difference between these two forces, one pointing down and one pointing up, produces the overall lift force $L_{total}= L _{top} - L_{bottom}$ directed upward.

Does the static air located far above the wing rush toward the top of wing since it is at a higher pressure than the air closer to the wing? Also, the higher pressure air below the wing should push both on the bottom surface of the wing and also on the air farther below the wing. Is that what happens?

Thanks!

 

[FactChecker]

Does the static air located far above the wing rush toward the top of wing since it is at a higher pressure than the air closer to the wing?

The airflow over the wind does get drawn downward. The net result is that the airflow gets diverted downward. This can be looked at as essential for the lift force on the wing. They are equal and opposite forces and reactions -- the airflow is diverted down and the wing is pushed up.

Also, the higher pressure air below the wing should push both on the bottom surface of the wing and also on the air farther below the wing. Is that what happens?

Yes.

 

[fog37]

Thanks a lot! This is how I envision it:

It is clear, from Bernoulli's equation, that fast moving air has a lower isotropic pressure than slower moving air. That said, I am not clear why from a molecular standpoint? Does anyone have a clear interpretation?

 

[FactChecker]

There is a great danger of oversimplifying this subject and only looking at one aspect while ignoring others. A good insight article was written by @boneh3ad that is worth reading: https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/

 

[fog37]

Thanks. I read that. I will read again.

 

[FactChecker]

Thanks. I read that. I will read again.

Good. You should be aware that any simple answer to your question is over-simplified and will have to be discarded if you get deeper into the subject. The only real way to calculate lift is to apply Computational Fluid Dynamics (CFD), which traces tiny bits of air around the wing while calculating all the pressures and forces on it. Even that is very limited. Wind tunnel tests are done to verify and modify the CFD results to get an aerodynamic model. The flight tests are done in a careful sequence to make sure that flight results and aerodynamic model agree before more dangerous flights are attempted.

 

[fog37]

You are right. I don't want to look to much into it. I am just trying to grasp some qualitative understanding of the situation.

Thanks.

 

[rootone]

I am just trying to grasp some qualitative understanding of the situation..

The shape of the wing forces air to go 'down', so therefore the wing obtains a force going 'up' (lift).

 

[russ_watters]

[Mod Note:]
A severely wrong post and several responses were deleted.

Also, let's please keep this focused on the OP's questions and perhaps most importantly respect the OP's request to keep this as basic as possible. Not every discussion about how a car is propelled (for example) needs to get into the thermodynamics of the Otto cycle.

Thread re-opened.

 

[sophiecentaur]

The shape of the wing forces air to go 'down', so therefore the wing obtains a force going 'up' (lift).

It amazes me that this inescapable fact seems to be accepted in discussions about a helicopter rotor but not accepted for a wing traveling in a straight line. The details of how the downward motion of air is achieved is very complicated and, as has been pointed out, there are many different levels for the analysis. You don't need an aerofoil to produce lift but it's just a good design which involves relatively little drag.

 

[FactChecker]

It amazes me that this inescapable fact seems to be accepted in discussions about a helicopter rotor but not accepted for a wing traveling in a straight line.

That's an interesting point. I suspect that the helicopter version is accepted in an over-simplified form as though its blades operate like a screw digging into the air rather than as wings with a profile like an airplane wing. (Many early attempts at flight made that mistake.)

I remember seeing an experiment where a toy helicoptor inside a box begins to hover within the box. The weight of the box remains identical to the weight when the helicoptor just sat on the floor with no blade rotation. So obviously the weight of the hovering helicoptor was transferred by the air downdraft to the floor. I wonder if an equivalent experiment can be done with a model airplane. I think that it would be much more difficult to set up in such a convincing way.

 

[sophiecentaur]

That's an interesting point. I suspect that the helicopter version is accepted in an over-simplified form as though its blades operate like a screw digging into the air rather than as wings with a profile like an airplane wing. (Many early attempts at flight made that mistake.)

I remember seeing an experiment where a toy helicoptor inside a box begins to hover within the box. The weight of the box remains identical to the weight when the helicoptor just sat on the floor with no blade rotation. So obviously the weight of the hovering helicoptor was transferred by the air downdraft to the floor. I wonder if an equivalent experiment can be done with a model airplane. I think that it would be much more difficult to set up in such a convincing way.

I would be surprised (very) at an experiment that could show Newton's Laws of motion don't apply.

 

[FactChecker]

I would be surprised (very) at an experiment that could show Newton's Laws of motion don't apply.

Sorry for the confusion. (I believe in Newton's laws as much as you do.) The example of a hovering helecopter, only supported by air is relatively simple. But it would be hard to keep a wing in place in a wind tunnel without some supports that complicate the logic of the conclusion.

 

[sophiecentaur]

Sorry for the confusion. (I believe in Newton's laws as much as you do.) The example of a hovering helecopter, only supported by air is relatively simple. But it would be hard to keep a wing in place in a wind tunnel without some supports that complicate the logic of the conclusion.

I was not disagreeing with you. What you describe is supporting Newton.
A wind tunnel would need to be designed from scratch with facilities for measuring its weight. A lot of trouble to prove what can be proved in other, easier ways.

 

[olivermsun]

Sorry for the confusion. (I believe in Newton's laws as much as you do.) The example of a hovering helecopter, only supported by air is relatively simple. But it would be hard to keep a wing in place in a wind tunnel without some supports that complicate the logic of the conclusion.

For the purposes of your demonstration, what would be wrong with using a 2-d wing section supported by the sides of the wind tunnel?

 

[FactChecker]

For the purposes of your demonstration, what would be wrong with using a 2-d wing section supported by the sides of the wind tunnel?

The point is to show that the force of the air on the bottom of the box is exactly the same as the weight of the plane that is floating above. Any support and forces from it can be all accounted for, but it is not as clear a demonstration. The hovering helicopter which requires no attachments is ideal.

 

[olivermsun]

I am confused as to the point of this demonstration.
The plane is floating, so it follows that the downward force exerted by the plane is equal to its weight.
Why is it also necessary to demonstrate that the box enclosing the plane weighs the same whether the plane is floating or parked?

 

[FactChecker]

I am confused as to the point of this demonstration.
The plane is floating, so it follows that the downward force exerted by the plane is equal to its weight.
Why is it also necessary to demonstrate that the box enclosing the plane weighs the same whether the plane is floating or parked?

I'm afraid that I have diverted this thread too much. I just thought that the helicopter demonstration was such a good demonstration of "equal and opposite" in the aerodynamics context that it impressed me at the time. It wouldn't bother me if a monitor cut this out.

 

[sophiecentaur]

For the purposes of your demonstration, what would be wrong with using a 2-d wing section supported by the sides of the wind tunnel?

A narrow wind tunnel would be cheaper but there would be an issue with the interaction of the air flow and the sides / corners. The way N3 applies to these things is easily observed when you see the result of turning a heavy boat with a rudder. There is a visible disturbance of the water, way into the inside of the curve, with eddies moving inwards by a significant distance. That 'Momentum' is very visible, unlike the air that trails behind an aircraft.

 

[olivermsun]

I didn't say a "narrow" wind tunnel, I said a 2-d wing section — you have a relatively wide section and tunnel where the wing does not vary across the flow, and then you try to observe a 2-d flow away from the edges of the tunnel.

The streamlines are typical made more visible by injecting smoke or other tracers.

 

[sophiecentaur]

I didn't say a "narrow" wind tunnel, I said a 2-d wing section — you have a relatively wide section and tunnel where the wing does not vary across the flow, and then you try to observe a 2-d flow away from the edges of the tunnel.

The streamlines are typical made more visible by injecting smoke or other tracers.

Oh yes. A misunderstanding. I expect most basic wing design and optimisation can be done with 2D sections. What happens at the ends can be very relevant though - as demonstrated by all these fancy new wings on passenger jets with the clever bits on the tips. (You can tell I am speaking not as an aerodynamicist)
But, of course, the streamlines would need a massive space to develop fully and to reveal a downward motion that is actually providing the N3 lift.

 

[rcgldr]

Does the static air located far above the wing rush toward the top of wing since it is at a higher pressure than the air closer to the wing? Also, the higher pressure air below the wing should push both on the bottom surface of the wing and also on the air farther below the wing. Is that what happens?

Correct. Using the air as a frame of reference (instead of the wing), the air accelerates downwards and it's pressure decreases as it approaches the plane swept out by a wing as a wing passes through the air. As the air passes through the plane swept out by a wing, there's an increase in pressure with little or no increase in speed. This pressure jump corresponds to the energy added to the air by a wing passing through a volume of air. With this increase in pressure, the air continues to accelerate downwards as it's pressure returns to ambient. The speed of the air when it's pressure returns to ambient is called the "exit velocity". Nasa has an article on propellers that explains this, although the diagrams ignore the viscous interaction with the surrounding air. Note that from the air's frame of reference, Bernoulli is violated as the air flows through the plane swept out by the wing, because work is performed by the wing by increasing it's pressure. The Nasa article mentions this.

https://www.grc.nasa.gov/www/k-12/airplane/propanl.html

 

[fog37]

So, inside a fluid is moving in the direction $x$ at uniform speed v, a fluid particle (small blob of fluid composed of many molecules) will feel an isotropic pressure $p_0$ from all the other fluid particles around it. Neglecting depth effect on pressure, if the fluid particle is brought to rest (at a stagnation point), the new pressure the particle would experience is increased to $p_0+\rho \frac {v^2} {2}$. But without bringing the fluid parcel to rest, the pressure will only be and remain $p_0$ and be isotropic and the fact that the fluid is moving at speed $v$ in the x direction will not show it effect.

Does that mean that the pressure $p_0$ would be the same as when the fluid is at rest even if the fluid is actually moving unless we bring it to a stop?

 

[boneh3ad]

Your nomenclature is somewhat nonstandard. Let me try rewriting this. Let's assume for a moment that a given flow is steady, inviscid, and adiabatic. A given fluid element moving with velocity $V$ in an incompressible flow will have total (or stagnation) pressure
p_0 = p + \dfrac{1}{2}\rho V^2.
We typically call $p_0$ (sometimes written $p_t$) the total or stagnation pressure, which is the pressure that would be felt if the flow was isentropically (i.e. no heat added, not energy dissipated) brought to rest. $p$ is the static pressure and is what is felt by a surface immersed in said fluid whether it is moving or not. The $V^2$ term is dynamic pressure and isn't truly a pressure but a kinetic energy term. The whole statement really amounts to a conservation of energy statement. In an isentropic flow, $p_0$ is a constant, so if you bring the velocity to zero, $p = p_0$ and the surface at the stagnation point feels that pressure. Anywhere else in the flow, and the pressure felt by the surface is $p$, which is some fraction of $p_0$ based on the value of the dynamic pressure.

The static pressure (sometimes called thermodynamic pressure) is always isotropic. It acts equally in all directions. This is universally true of static/thermodynamic pressure. Now, if you are talking about the actual stress tensor in the fluid, that is not necessarily isotropic in all cases, but the pressure itself is.

 

[fog37]

Thanks boneh3ad.

I "think" we are on the same page. There is one and only one pressure which is always isotropic, from the standpoint of a fluid parcel. In a fluid at rest (hydrostatics), the pressure would be $p$ and only vary with depth. If the fluid was in motion, the fluid parcel would still feel the same pressure $p$ unless we brought the flow to a stop. At the stopping point (stagnation point) the pressure, still isotropic, would become $p_0$ and be larger since the momentum of the flow is now converted in extra pressure. What has always confused me was the fact that, either at rest or in motion, the fluid parcel would apparently feel the same pressure, the thermodynamic pressure, unless it was brought to rest.

Bernoulli's equation is often read as the local isotropic pressure as being different for different flow speeds. that does not seem to agree with what I say above. If the fluid is moving fast or slow, the pressure a parcel exerts on other parcels and the pressure the parcels exert on is just $p_0$ since the pressure increase only derives from converting the momentum into force.

That said, I see how in a tube narrowing, based on Bernoulli's equation, there must be a gradient of pressure to accelerate the fluid (faster in the narrowing). So, in that case, the pressure $p$ seems to be different for different speeds...

 

[rcgldr]

That said, I see how in a tube narrowing, based on Bernoulli's equation, there must be a gradient of pressure to accelerate the fluid (faster in the narrowing). So, in that case, the pressure $p$ seems to be different for different speeds...

Correct. The assumption is no external forces that could change the total energy within the tube, so total energy: pressure x volume + kinetic energy is constant. For the transition into a narrower tube, some of the pressure energy is converted into kinetic energy, resulting in a lower pressure $p$ in the faster moving fluid.

Since the title of this thread is about wings and lift, the situation changes depending on the frame of reference. From the wings frame of reference, assuming an idealized wing, the flow is diverted with no change in speed (no change in total kinetic energy). From the air's frame of reference, as a wing travels through a volume of air, it accelerates the air downwards, increasing it's energy. As posted earlier, most of this increase in energy is a pressure jump that occurs as the air crosses through the path swept out by a wing passing through a volume of air.

 

[fog37]

Thanks rcgldr. I fully agree with your comments. But think of these scenarios: two cylindrical pipes of the same diameter $D$ and uniform cross-section (no narrowings or expansions). The same fluid flows at speed $V_1$ inside the pipe. The other fluid flows at $V_2 > V_1$. Without bringing the fluid to rest and creating a stagnation point, wouldn't the isotropic pressure felt by a fluid parcel be the same in both cases even if the fluid speeds are the different? the speed difference will only become apparent if the fluids are stopped. Is there any truth in what I am saying?

 

[rcgldr]

Thanks rcgldr. I fully agree with your comments. But think of these scenarios: two cylindrical pipes of the same diameter $D$ and uniform cross-section (no narrowings or expansions). The same fluid flows at speed $V_1$ inside the pipe. The other fluid flows at $V_2 > V_1$. Without bringing the fluid to rest and creating a stagnation point, wouldn't the isotropic pressure felt by a fluid parcel be the same in both cases even if the fluid speeds are the different? the speed difference will only become apparent if the fluids are stopped. Is there any truth in what I am saying?

I don't understand your point here. You have two independent flows, each of which can have a different static pressure and different dynamic pressure. You can define the situation so that the two flows have the same static pressure. The velocities are relative to some frame of reference. Consider the case of two planes moving at different speeds and the pressures sensed inside each planes pitot tube.

 

[fog37]

rcgldr, Sorry if I am not clear. Let me try one more last time.

Consider two vertical pipes with fluid flowing from left to right at uniform but different speeds $V_1$ and $V_2$. I think these are possible and feasible scenarios.
Even if the two blue fluid parcels travel to the right but at different speeds, I am arguing that the isotropic pressure at those two identical points in the fluid is the same regardless of the difference in speed. Of course, different fluid parcels will occupy that position as time goes on. Is that incorrect?

If we brought the bottom parcel traveling at speed $V_2$ to rest, the pressure at the stagnation point would higher compared to the case above with speed $V_1$.

 

[boneh3ad]

What has always confused me was the fact that, either at rest or in motion, the fluid parcel would apparently feel the same pressure, the thermodynamic pressure, unless it was brought to rest.

Bernoulli's equation is often read as the local isotropic pressure as being different for different flow speeds. that does not seem to agree with what I say above.

That's because what you wrote above is incorrect. The static pressure absolutely does change when velocity changes. The total pressure represents the total energy per unit volume in the flow. If the fkownlocally accelerates, it's kinetic energy per unit volume (dynamic pressure) increases and its static pressure has a corresponding decrease.

From the wings frame of reference, assuming an idealized wing, the flow is diverted with no change in speed (no change in total kinetic energy). From the air's

Well that's not true. There is energy dissipated by the forces causing drag on the wing, meaning the total pool of energy absolutely decreases in either frame of reference.

Thanks rcgldr. I fully agree with your comments. But think of these scenarios: two cylindrical pipes of the same diameter $D$ and uniform cross-section (no narrowings or expansions). The same fluid flows at speed $V_1$ inside the pipe. The other fluid flows at $V_2 > V_1$. Without bringing the fluid to rest and creating a stagnation point, wouldn't the isotropic pressure felt by a fluid parcel be the same in both cases even if the fluid speeds are the different? the speed difference will only become apparent if the fluids are stopped. Is there any truth in what I am saying?

You are neglecting the fact that each pipe can have a different toal pressure since they are in no way related other than being in the same problem. There's nothing we can say about the pressure in either pipe other than the fact that it will be isotropic.

 

[rcgldr]

From the wings frame of reference, assuming an idealized wing, the flow is diverted with no change in speed (no change in total kinetic energy).

There is energy dissipated by the forces causing drag on the wing, meaning the total pool of energy absolutely decreases in either frame of reference.

For my concept of an idealized wing, there is no drag. In the air's frame of reference, and using a glider in a steady descent to simplify things, then the energy of the air is increased, and the gravitational potential energy of the glider is decreased.

 

[sophiecentaur]

For my concept of an idealized wing, there is no drag.

Can drag be eliminated, ever?

 

[russ_watters]

For my concept of an idealized wing, there is no drag.

Can drag be eliminated, ever?

I don't think it can. Even if skin friction is ignored, the same pressure forces that produce lift produce induced drag.

https://en.m.wikipedia.org/wiki/Lift-induced_drag

 

[rcgldr]

There is energy dissipated by the forces causing drag on the wing, meaning the total pool of energy absolutely decreases in either frame of reference.

For my concept of an idealized wing, there is no drag. In the air's frame of reference, and using a glider in a steady descent to simplify things, then the energy of the air is increased, and the gravitational potential energy of the glider is decreased.

Can drag be eliminated, ever?

Not in a real world situation, which was the point of using an idealized wing to focus on just the diversion of flow. High end cross country gliders, like the Nimbus-4 come close. At 1500 pounds (including pilot), it has about a 60 to 1 glide ratio at somewhat over 60 mph. Assuming it is 60 to 1 at 60 mph, that's 60 mph forward speed with a 1 mph descent rate, which translates into 4 hp.

I don't think it can. Even if skin friction is ignored, the same pressure forces that produce lift produce induced drag.

boneh3ad's post mentions the fact that drag dissipates energy (it converts the energy into heat). This is different than induced drag which is related to diversion of flow. For an "idealized" wing, the flow is diverted, but the speed is not changed. The components of velocity are changed, the horizontal component is decreased, and the vertical component is increased (from zero to some non-zero "exit" velocity). The aerodynamic force vector points upwards (lift) and a bit backwards (induced drag). The point here is that lift induced drag is related to the diversion of flow, not dissipation of mechanical energy being converted into heat, such as friction or viscous related drag.

 

[sophiecentaur]

For an "idealized" wing, the flow is diverted, but the speed is not changed.

I can see how the 'in vacuo' approach to this whole business leads to arguments and confusion. How is the paradox explained that air has to be deflected (along with its KE) in order to follow N3, yet no Energy is lost in the process

 

[rcgldr]

For an "idealized" wing, the flow is diverted, but the speed is not changed. The components of velocity are changed, the horizontal component is decreased, and the vertical component is increased (from zero to some non-zero "exit" velocity). The aerodynamic force vector points upwards (lift) and a bit backwards (induced drag). The point here is that lift induced drag is related to the diversion of flow, not dissipation of mechanical energy being converted into heat, such as friction or viscous related drag.

I can see how the 'in vacuo' approach to this whole business leads to arguments and confusion. How is the paradox explained that air has to be deflected (along with its KE) in order to follow N3, yet no Energy is lost in the process

I updated my last post (included in the quote in this post), that might help clarify my point. In an "ideal" circumstance, Newton laws would still apply, but without requiring any loss in energy, Bernoulli principle is an example of such an idealized case where it's assumed that there are no energy losses, where a flow is accelerated as it transitions into a narrower section of pipe and decelerated as it transitions into a wider section of pipe.

 

[boneh3ad]

You have to be careful when discussion an "ideal" flow around a wing, because you will predict zero drag. This is a concept called D'Alembert's paradox, so named because D'Alembert formulated the first mathematical theory of forces on an object moving through a fluid and it predicted zero drag despite there clearly being drag when observed in the real world. In order to predict a nonzero value of lift on an infinite ideal wing, one must consider viscosity, which leads to viscous drag and form drag due to boundary-layer separation. The other option is induced drag, which only affects finite wings. The induced drag on an infinite wing is zero (thus the reason gliders typically have very large wingspans).

 

[fog37]

boneh3ad,

For a steady flow, I see how the isotropic pressure on the surface of a fluid parcel moving at a constant speed inside a horizontal pipe decreases with an increase in the fluid parcel speed based on energy arguments: From a thermodynamic point of view, why does the external isotropic pressure on the surface of the fluid parcel decreases is moving at a higher speed?

 

[boneh3ad]

Conservation of energy is the First Law of Thermodynamics, so I am not sure where the disconnect here is. Bernoulli's equation is essentially a statement of the First Law of Thermodynamics and can be derived directly from the energy equation.

 

[rcgldr]

The induced drag on an infinite wing is zero (thus the reason gliders typically have very large wingspans).

I find conflicting articles about this. The main issue seems to be the case of an infinite wing producing a finite amount of lift, in which case the angle of the diversion of airflow approaches zero as the wing span approaches infinity. If instead the infinite wing produces some finite amount of lift per finite unit of length of the wing, then the angle of diversion of airflow and the corresponding effective angle of attack are non-zero, and there is lift induced drag since there is a non-zero angle of diversion.

 

[sophiecentaur]

D'Alembert formulated the first mathematical theory of forces on an object moving through a fluid and it predicted zero drag

That's fine as a basis for preliminary calculations but not for the sort of discussion in this thread which attempts to describe "what's really happening". Any deeper understanding of a phenomenon can't ignore the Energy considerations, can it?

 

[fog37]

Thanks boneh3ad

But doesn't Bernoulli's equation come from the Euler equation (which is NS with zero viscosity) when the flow is time steady and has zero vorticity?

 

[boneh3ad]

That's fine as a basis for preliminary calculations but not for the sort of discussion in this thread which attempts to describe "what's really happening". Any deeper understanding of a phenomenon can't ignore the Energy considerations, can it?

I was replying to @rcgldr and his use of an idealized flow. If the question is about lift, you can get a very, very accurate answer and understand all the fundamental ideas with that approach. It simply provides no insight into drag.

 

[boneh3ad]

Thanks boneh3ad

But doesn't Bernoulli's equation come from the Euler equation (which is NS with zero viscosity) when the flow is time steady and has zero vorticity?

You can derive it from the Euler equation. You can also derive it from the Reynolds transport theorem applied to energy.

 

[fog37]

Conservation of energy is the First Law of Thermodynamics, so I am not sure where the disconnect here is. Bernoulli's equation is essentially a statement of the First Law of Thermodynamics and can be derived directly from the energy equation.

Thanks for the patience. In hydrostatics, I get a clear picture of what is going on: at a certain depth inside water, the static pressure on a the walls of an infinitesimal fluid parcel located at point $P$ is isotropic and due to all the other fluid parcels surrounding the fluid parcel positioned at $P$.

When the fluid is in motion, instead, to the right at speed $V$, the isotropic pressure at the same point $P$ is lower compared to when the fluid is at rest. Of course, different fluid parcels will occupy point $P$ at different instants of time. But, without using energy arguments, I was wondering if I could get some insight from a more elementary point of view of the forces acting on the passing by fluid parcel at point $P$ and understand why the pressure would be reduced if the fluid is moving...

 

[jbriggs444]

When the fluid is in motion, instead, to the right at speed VVV, the isotropic pressure at the same point PPP is lower compared to when the fluid is at rest. Of course, different fluid parcels will occupy point PPP at different instants of time. But, without using energy arguments, I was wondering if I could get some insight from a more elementary point of view of the forces acting on the passing by fluid parcel at point PPP and understand why the pressure would be reduced if the fluid is moving...

There is no reason for the parcels in a moving fluid to be at a different pressures than in a stationary fluid. Whether the fluid is "moving" or "stationary" is, after all, decided when you pick a frame of reference. The choice of frame of reference cannot affect the pressure.

If you choose a single body of fluid which is moving in one region and stationary at another, now you have a real physical effect to look at. What does change pressure is the fact that a fluid parcel that is moving slowly as it passes one point and moving more rapidly as it passes another point must have accelerated to do so. That acceleration is the result of a pressure gradient.

 

[russ_watters]

There is no reason for the parcels in a moving fluid to be at a different pressures than in a stationary fluid. Whether the fluid is "moving" or "stationary" is, after all, decided when you pick a frame of reference. The choice of frame of reference cannot affect the pressure.

I'll add that for airplanes this should be particularly obvious. There are thousands of planes zooming around at different speeds at any given time, and the air may be considered moving with respect to any of them or stationary with respect to the surface of the Earth, and that doesn't affect the freestream static pressure.

...but it may be less obvious for flow in a wind tunnel or pipe, where there would seem to be a preferred reference frame (but doesn't have to be) and other factors affecting static pressure (such as loss through the pipe/duct, Venturi effects, etc.).

 

[olivermsun]

...without using energy arguments, I was wondering if I could get some insight from a more elementary point of view of the forces acting on the passing by fluid parcel at point $P$ and understand why the pressure would be reduced if the fluid is moving...

Consider Newton’s 2nd law along a stream line. The acceleration on a parcel of water is due to the net force on the parcel, here equal to the difference in pressure dp/dx, where x is along the streamline. If the flow is accelerating, then the net force must be positive in the direction of the flow, and hence the pressure gradient must be negative.

 

[boneh3ad]

@olivermsun gave the explanation I would have here. You can view Bernoulli from either a force persective or an energy perspective, and you can derive it from either direction as well. I suspect, perhaps, that it would be useful if I wrote another Insight article about Bernoulli's equation.

 

[fog37]

Thanks olivermsun and boneh3ad.

Given a fluid parcel moving along a straight streamline in the positive x-direction, I see that a pressure difference between two points (one before and one after point $P$) would cause a force $F$ on the fluid parcel which would accelerate. But what if the parcel is moving at constant speed? The force would zero and the pressure gradient would be zero. In that case, being pressure isotropic, the parcel experiences the same force from the back, from the front, from the top and from the bottom. This isotropic pressure is smaller compared to the static pressure at the same point $P$ when the fluid is not moving at all. I am not sure how to justify the fact that the speed makes the pressure smaller without using Bernoulli's equation.