Airplane wings -- How do they work and why do they change shape?

  • #76
sophiecentaur
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Sure; when gauge pressure is negative, that's the surface pulling on the air. When positive, that's pushing.
That's a convention I would need to get used to but it's hardly rocket Science.
 
  • #77
boneh3ad
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Rather than reply to the last few posts individually, I will try to address these last few issues all at once since they are related.

It is correct to say that differential pressure (or more correctly, gauge pressure, since it is generally referenced to atmosphere) is used extremely frequently when dealing with the pressure distribution over an airfoil. It is definitely the most common method of doing things, typically in the form of the dimensionless pressure coefficient, ##C_p##. This is done because using gauge pressure is more mathematically compact and lends itself to collapsing the data into relationships that work over a range of free-stream conditions.

That said, gauge pressure is also more physically misleading. If the person trying to learn about the process in the first place does not understand the distinction, then he or she will note that the upper surface shows a negative pressure, which implies that the air should "pull" on the wing on that side. While this still all works out mathematically, it leads to physically incorrect conclusions in the hands of a novice. So, circling back to the concerns about beginners, I still maintain that doing this in terms of gauge pressure right off the bat is misleading and the wrong approach.

On top of all of that, the fundamental concept of the top "contributing more" is still wrong. It is a given that the gauge pressure on the top is going to be negative. It is true that in some cases, the integrated force from that upper gauge pressure would be greater than on the lower surface. That is not a universally true fact, however. This is where gauge pressures can again be misleading. Consider the pressure coefficient, which varies across an airfoil surface:
[tex]C_p = \dfrac{p-p_{\infty}}{\frac{1}{2}\rho V_{\infty}^2} = \dfrac{p_{gauge}}{\frac{1}{2}\rho V_{\infty}^2}.[/tex]
The value of ##C_p## is effectively constant for any velocity so long as the flow is incompressible[1]. So, let's rewrite this equation
[tex]p_{gauge} = \dfrac{1}{2}C_p\rho V_{\infty}^2.[/tex]
Since ##C_p## and ##\rho## are constants, then increasing the velocity is going to increase ##p_{gauge}## on the lower surface (where ##C_p## is positive) and decrease the gauge pressure on the upper surface (where ##C_p##) is negative. Here's the kicker, though: while gauge pressure can become arbitrarily high, it cannot be arbitrarily low (after all, you can't have a negative amount of gas in a volume). So, as you increase speed, eventually the magnitude of the gauge pressure on the bottom must become larger than that on the top surface, at which point your argument about the top surface "contributing more" becomes invalid.

[1] Anderson Jr, J. D. (2016). Fundamentals of aerodynamics. Chapter 3
 
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  • #78
A.T.
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Sure; when gauge pressure is negative, that's the surface pulling on the air.
So by your definition, a surface exerting an upwards force on the air above it, is actually "pulling down on the air"?
 
  • #79
jbriggs444
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So by your definition, a surface exerting an upwards force on the air above it, is actually "pulling down on the air"?
It's a negative gauge pressure and might be visualized as a pull. Similar to the way a vacuum cleaner "pulls" dirt from a rug.
 
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  • #80
russ_watters
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So by your definition, a surface exerting an upwards force on the air above it, is actually "pulling down on the air"?
By this convention, that is what it says, yes. That's what it means when the pressure and thus the force is negative.
 
  • #81
anorlunda
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@doglover9754 ,

I'm sorry. Sometimes adults get so excited about the best way to say something that they can't answer a middle school student's simple question. It's not like this all the time on PF. Usually, simple questions get simple answers.

Thread closed.
 
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  • #82
russ_watters
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It is correct to say that differential pressure (or more correctly, gauge pressure, since it is generally referenced to atmosphere) is used extremely frequently when dealing with the pressure distribution over an airfoil. It is definitely the most common method of doing things, typically in the form of the dimensionless pressure coefficient, ##C_p##. This is done because using gauge pressure is more mathematically compact and lends itself to collapsing the data into relationships that work over a range of free-stream conditions.
Thank you!
That said, gauge pressure is also more physically misleading.

If the person trying to learn about the process in the first place does not understand the distinction, then he or she will note that the upper surface shows a negative pressure, which implies that the air should "pull" on the wing on that side. While this still all works out mathematically, it leads to physically incorrect conclusions in the hands of a novice.
Fine. So in the future, i would hope that you will recognize the convention without it needing to be described to you, so we can avoid repeating much of this.
So, circling back to the concerns about beginners, I still maintain that doing this in terms of gauge pressure right off the bat is misleading and the wrong approach.
Conventions are personal choices and cannot be right or wrong if both work when applied correctly. You prefer your convention, and I prefer mine and both work and that's fine. Please be ok with that. Or at least pretend to be for the sake of the OP!
On top of all of that, the fundamental concept of the top "contributing more" is still wrong. It is a given that the gauge pressure on the top is going to be negative. It is true that in some cases, the integrated force from that upper gauge pressure would be greater than on the lower surface.
So again: something true cannot be wrong just because you prefer a different convention or verbiage. The word "contribution" is used in academic settings. You don't like that. That doesn't make it wrong, but you don't have to use it. But more importantly - again - please try to recognize differences in conventions/wording and not to be so quick to argue over them. It is not helpful to the OP.
That is not a universally true fact, however.
Agreed/understood. But it *is* true for the example first given to the OP in post #2, which is a common example because it is common in real life and simple for the very first thing someone learns about lift. This isn't something that should create a disagreement, but rather a stepping stone to the next, deeper point.
 
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