Calculating wind pressure against a rigid wall

[Cyrus]

Thanks for the link. My intent was to ignore mass flows completely since I had no way to model them. Why was I able to get the "right" answer? Two posters got 120 Pa, but these had to involve mass flows. Are they correct? Those that got 60 Pa assumed stagnation along the entire wall front which is how I set up the problem.

My main concern is that you have a slug of mass with an initial momentum, mv, and then you said it comes to a dead stop. Well, where did the rest of the momentum go?

If you use one of the drag coefficients in the link I gave you, then the pressure is simply:

P=\frac{F}{A}=\frac{1}{2}\rho V^2 C_D​

How simpler does it get? Based on the bluff body values in that link, you're looking at a C_D \~= 1.1-1.3. Keep in mind, this is an averaged pressure over the entire face, with edges that can have turbulence. (Which is probably more realistic than your "infinite" wall anyways)Note: The only difference between the formula above and what you will get with Bernoulli is a factor of C_D. So, because C_D=1.1-1.3 the value you got earlier is off by 10-30%. (The first value was an under prediction).

 

[SW VandeCarr]

My main concern is that you have a slug of mass with an initial momentum, mv, and then you said it comes to a dead stop. Well, where did the rest of the momentum go?

If you use one of the drag coefficients in the link I gave you, then the pressure is simply:

P=\frac{F}{A}=\frac{1}{2}\rho V^2 C_D​

How simpler does it get? Based on the bluff body values in that link, you're looking at a C_D \~= 1.1-1.3. Keep in mind, this is an averaged pressure over the entire face, with edges that can have turbulence. (Which is probably more realistic than your "infinite" wall anyways)

OK. Without C_d I get 60 Pa, with C_d I get 66-78 Pa. Not too bad for government work. Thanks.

 

[Cyrus]

OK. Without CD I get 60 Pa, with Cd I get 66-78 Pa. Not too bad for government work. Thanks.

Ok, now I want you to recite to me all the assumptions made so I know you're not blindly using these equations and numbers. I don't care if your numbers are close or not, I want you to be able to understand what it is you are doing.

Now, I never told you all the assumptions (only some), so you'll have to think carefully about the equations before you reply.

 

[rcgldr]

Why don't you look at the ppt slides
BluffBodyAero.ppt

If a flate plate with tiny hole type static port was oriented so that it was a flat plate bluff body pitot port, would the pressure inside the chamber (same as the pressure at the hole in the middle of the flat plate) correspond to a C_d of 1.0 or 1.1? I assume that a real pitot tube is designed to have the equivalent of a C_d of 1.0?

 

[Cyrus]

If a flate plate with tiny hole type static port was oriented so that it was a flat plate bluff body pitot port, would the pressure inside the chamber (same as the pressure at the hole in the middle of the flat plate) correspond to a C_d of 1.0 or 1.1? I assume that a real pitot tube is designed to have the equivalent of a C_d of 1.0?

Ah, I see what you're asking. Well, the problem would be how do you measure the Pitot pressure. It's certainly not directly perpendicular to the wall face because the freestream air is forced to make a 90 degree bend as it stagnates and flows outward to the edges of the wall. So the flow along the face of the wall perpendicular to the flow will have a component of velocity perpendicular to the freestream.

Hope this helps clear my description:

http://arch.ced.berkeley.edu/kap/images/windbldg.gif​

[/URL]

What direction do you point the Pitot probe on that wall, and at what point on the wall? You have to take the probe and turn it until the pressure reads a maximum value. Then you know the direction of stagnation pressure.

So the first question is, where do you want the tube, and what direction do you point it?

The second question is, does C_D really have meaning in the context of what were describing? If you look at Bernoullis equation, C_D doesn't appear anywhere. What you're really doing is multiplying the pressure by some factor (C_D) which is where you are accounting for losses in the streamline. But instead of saying you have P = P_{stag} + losses, you are lumping into the pressure as P = C_D*P_{stag}.

So C_D is (in the context of how you used it here) nothing more than C_D= \left(1+\frac{loss}{P_{stag}} \right)

But this C_D is not the same C_D for the wall. If you did use the value of C_D of the wall, the value of "losses" would adjust so the equality holds. It's basically a tradeoff. You're taking the losses and moving it around to different terms, some of it in C_D, and the rest in "losses".

 

[SW VandeCarr]

My main concern is that you have a slug of mass with an initial momentum, mv, and then you said it comes to a dead stop. Well, where did the rest of the momentum go?

I'm imagining that a stagnant pool of air along the wall front could result from the cancellation of the outgoing momentum vector (rebound) by the incoming momentum vector (wind) although this might not be consistent with a "dead stop". The real answer is I don't know. It's not a physical model. It's an extrapolation to a limit value (zero time).

As far as my overall thinking I'll get back to you. It involves the dimensional equivalence of pressure and energy density which I discuss in the first post. As far as assumptions made, I'm not sure you're talking about the assumptions I made (no mass flows) or the assumptions I perhaps I should have made (buff body dynamics). I frankly had no knowledge of the latter, but it was mentioned by others in the early posts of this long thread.

Right now it's 12:40 am and I'm going to bed.

 

[Cyrus]

I'm imagining that a stagnant pool of air along the wall front could result from the cancellation of the outgoing momentum vector (rebound) by the incoming momentum vector (wind) although this might not be consistent with a "dead stop". The real answer is I don't know. It's not a physical model. It's an extrapolation to a limit value (zero time).

As far as my overall thinking I'll get back to you. It involves the dimensional equivalence of pressure and energy density which I discuss in the first post..

Without the equations, I'm having trouble understanding what you're doing exactly and what you have described is clearly not right. I don't think you should make up a simulation with a nonphysical model when you have the answer in one simple equation.

 

[rcgldr]

How do you measure the Pitot pressure.

Compare to a chamber connected to a static port that is at ambient pressure, or have a really good pressure gauge.

What direction do you point the Pitot probe on that wall, and at what point on the wall?

The pitot probe is embedded into the center of the wall, so that it's just a flush mounted hole in the center of the wall. Basically it's just a static port mounted to the nose of an aircraft so it acts as a flat plate:

staticports.html

The hole is the end of a pipe connected to a pressure measuring chamber imbeddeded within the "wall".

The second question is, does C_D really have meaning in the context of what were describing?

Not directly, but instead as a coefficient for "losses" in the streamline as you mentioned before.

The real question would be if I mount a static port on the nose of an aircraft so it's essentially a flat plate, and compare it's pressure reading versus that of a conventional pitot tube (both would have pipes feeding internal chambers as usual), will the sensed pressure be different, and if so, by some approximate ratio? I assume there's some reason that static ports are flush mounted and pitot ports are extended tubes and not flush mounted.

I know that static ports need to be flush mounted because the end of a tube perpendicular to air flow experiences a vortice that reduces pressure greatly, enough to draw fluid up through a nozzle for the purpose of a spray pump. One of my pet peaves are the web sites or articles that use the end of straws in a cross flow to measure the "lower pressure" of air blowing across the end of the straw by drawing water up the straw to demonstrate Bernoulli. Stick the end of a straw in a spool of sewing thread to make a crude static port and the results are quite different. Drill a hole in a board and stick the end of the straw into the board so it's flush mounted or receded a bit, to make a better static port, and compare to having the end of the straw extended into the wind.

 

[SW VandeCarr]

Without the equations, I'm having trouble understanding what you're doing exactly and what you have described is clearly not right. I don't think you should make up a simulation with a nonphysical model when you have the answer in one simple equation.

P=\frac{mv^2}{2V} = (1.2 kg)(10 m/s)^2/2(1 m^3)=60 Pa This is the equation and calculation I used in post 1 (with the error corrected).

This assumes no mass flows, just stagnation. The result some others got (120 Pa) seems to represent all mass flows, no stagnation. It seems the correct answer is between these calculated limits, and is determined by experimentally determined parameters such as C_D.

 

[Cyrus]

P=\frac{mv^2}{2V} = (1.2 kg)(10 m/s)^2/2(1 m^3)=60 Pa This is the equation and calculation I used in post 1 (with the error corrected).

This assumes no mass flows, just stagnation. The result some others got (120 Pa) seems to represent all mass flows, no stagnation. It seems the correct answer is between these calculated limits, and is determined by experimentally determined parameters such as C_D.

What you did here is calculate the dynamic pressure, Q. Not the pressure on the face of the wall, P. The dynamic pressure, Q, is related to the actual pressure P, by the factor C_D.

I don't know what others did to get 120Pa and that number is not correct. I would add that the correct answer is not between these 'limits' because they are not limits. It's simply the right answer and the wrong answer, there is no limiting process going on here.

 

[Cyrus]

Compare to a chamber connected to a static port that is at ambient pressure, or have a really good pressure gauge. The pitot probe is embedded into the center of the wall, so that it's just a flush mounted hole in the center of the wall. Basically it's just a static port mounted to the nose of an aircraft so it acts as a flat plate:

I think you missed what I was saying. That setup won't measure the total stagnation pressure because there is flow perpendicular to the wall face. Just look at the picture I posted and you can see the air flowing transverse to the free stream.

staticports.html

The hole is the end of a pipe connected to a pressure measuring chamber imbeddeded within the "wall".

You would just be measuring the static pressure at the wall face.

The real question would be if I mount a static port on the nose of an aircraft so it's essentially a flat plate, and compare it's pressure reading versus that of a conventional pitot tube (both would have pipes feeding internal chambers as usual), will the sensed pressure be different, and if so, by some approximate ratio? I assume there's some reason that static ports are flush mounted and pitot ports are extended tubes and not flush mounted.

They will both be wrong due to installation errors. They will absolutely be different from each other because the upwash/downwash at the wing will change the reading of the probe at that station. The reason why Pitot tubes are not in the nose is because that is prime real estate on an aircraft. People put things like radar and antennae in the nose.

I know that static ports need to be flush mounted because the end of a tube perpendicular to air flow experiences a vortice that reduces pressure greatly, enough to draw fluid up through a nozzle for the purpose of a spray pump. One of my pet peaves are the web sites or articles that use the end of straws in a cross flow to measure the "lower pressure" of air blowing across the end of the straw by drawing water up the straw to demonstrate Bernoulli. Stick the end of a straw in a spool of sewing thread to make a crude static port and the results are quite different. Drill a hole in a board and stick the end of the straw into the board so it's flush mounted or receded a bit, to make a better static port, and compare to having the end of the straw extended into the wind.

That's right. You don't want any burr from the end of the static port extruding into (or outside of) the boundary layer.

 

[Cyrus]

You might like to read this Jeff:

http://img17.imageshack.us/img17/596/pg1q.jpg

http://img177.imageshack.us/img177/9561/pg2.jpg

http://img10.imageshack.us/img10/7754/pg3p.jpg

http://img41.imageshack.us/img41/8629/pg4g.jpg

 

[rcgldr]

hole in center of wall

That setup won't measure the total stagnation pressure because there is flow perpendicular to the wall face.

Perhaps the center of a flat plate where the perpendicular flow is minimal (or perhaps this isn't possibe).

static pitot at nose of aircraft

They will absolutely be different from each other because the upwash/downwash at the wing will change the reading of the probe at that station.

Wing? I mentioned nose of aircraft.

The reason why Pitot tubes are not in the nose

Yet that's where they are on the Stratoliner in the example shown.

So the issue with the forward facing static port flat plate is that horizontal flow across the hole would be an issue. Since the static port is flush mounted, there is horizontal flow, but the hole "hides" behind a boundary layer that transitions from fuselage speed to free stream speed through viscous layers, and with proper installation it senses the ambient pressure of the freestream flowing across it.

Why isn't perpendicular flow (even if just a small amount) an issue for pitot tubes? Essentially the pitot tube's opening contains a cross section of air that is the equivalent of a tiny flat plate.

 

[Cyrus]

Perhaps the center of a flat plate where the perpendicular flow is minimal (or perhaps this isn't possibe).

I don't know, but that's why I said you would have to vary the probe until you find a maximum reading of the dynamic pressure (if that's even possible).

Wing? I mentioned nose of aircraft.

I misread what you wrote, now I see what you mean.

Yet that's where they are on the Stratoliner in the example shown.

Well, it doesn't have radar in the nose: it's an old airplane. Ideally, you would like to put the Pitot probe on a boom extending out the nose, and many aircraft do this. My point is that sometimes things get in the way that take priority over the probe:

or this:

http://www.totalexperience.co.nz/img/Cessna%20nose.jpg

So the issue with the forward facing static port flat plate is that horizontal flow across the hole would be an issue. Since the static port is flush mounted, there is horizontal flow, but the hole "hides" behind a boundary layer that transitions from fuselage speed to free stream speed through viscous layers, and with proper installation it senses the ambient pressure of the freestream flowing across it.

Yes, that's right. The static port measures the static pressure inside the boundary layer because of the no slip condition.

Why isn't perpendicular flow (even if just a small amount) an issue for pitot tubes? Essentially the pitot tube's opening contains a cross section of air that is the equivalent of a tiny flat plate.

My estimation is that the component is so small it's ignored, but I'll think about this one some more. I want to say it's a matter of "Good enough". I know flows can be tracked using PVIs (http://en.wikipedia.org/wiki/Particle_image_velocimetry" ) if you want very accurate measurements of flow properties. The guys in some of our labs use this method for rotorcraft flow.

 

[rcgldr]

Why isn't perpendicular flow (even if just a small amount) an issue for pitot tubes? Essentially the pitot tube's opening contains a cross section of air that is the equivalent of a tiny flat plate.

My estimation is that the component is so small it's ignored, but I'll think about this one some more.

I was also thinking about the affect of AOA. An F16 fighter AOA can exceed 20 degrees (such as a 9 g turn). Wondering if the pitot tube is on a motorized mount to maintain it's orientation with the airstream or if it's just taken care via math in the electronics.