Calculating wind pressure against a rigid wall

[SW VandeCarr]

I would check your math again. It appears to me as though you added a zero. 1.2*100/2=60 You prob multiplied by 12 instead of 1.2

Yes? No?

Of course you're correct. This misplaced decimal was picked up and corrected during the thread. If you read through the thread, several of the posters got the same answer using several different approaches (including Bernoulli's equation). (All are essentially the same, including yours.)

 

[AIR&SPACE]

Of course you're correct. This misplaced decimal was picked up and corrected during the thread. If you read through the thread, several of the posters got the same answer using several different approaches (including Bernoulli's equation). (All are essentially the same, including yours.)

MAN, I'm horrible at this. Well good then. All's well that end's well.

 

[Dadface]

Everybody here seems to be calculating a pressure of 60 Pa but I still calculate 120Pa so if I am wrong I would appreciate it If somebody could show me where I go wrong.With my method I use P= change of momentum per second per unit area and this seems to be at odds with the answer given by using Bernoullis equation where the pressure is given by P=0.5*air density*V^2.I think that this equation is good for finding the pressure at the instant just before the air hits the wall but not when it hits the wall when there is a sudden change of V and therefore pressure.

 

[AIR&SPACE]

Everybody here seems to be calculating a pressure of 60 Pa but I still calculate 120Pa so if I am wrong I would appreciate it If somebody could show me where I go wrong.With my method I use P= change of momentum per second per unit area and this seems to be at odds with the answer given by using Bernoullis equation where the pressure is given by P=0.5*air density*V^2.I think that this equation is good for finding the pressure at the instant just before the air hits the wall but not when it hits the wall when there is a sudden change of V and therefore pressure.

I think I have found the reason.

I went through your process and have attempted to think as you might have.
{
All right we've got 1.2 kg of air, traveling at 10m/s.
Since the wall is 1 square meter in area this means that our volume of air has dimensions 1x1x1.
So if the cube is 1 meter in depth, it will take 0.1 seconds for the last particles to reach the wall. (am I inline with your thinking so far?)
So then,
\frac{1.2kg*10\frac{m}{s}}{0.1s*1m^2}=120\frac{N}{m^2}
}

I believe the problem you face has to do with essentially creating a solid. I think the root of your math is saying that the air is not a fluid but actually constrained like a solid and that the wall has to impart the change of momentum across the whole solid, during the entire tenth of a second.
During that last segment of time your math is still assuming the entire mass of the air, when in reality, at anyone point in time the wall really only has to stop the mass that is found in the control volume Area\cdot du

The best example I can give is a video. It's kind of imperfect but it's kind of a good mental model.
http://www.youtube.com/watch?v=oxbFrGFNVO0&feature=related




Does this help?

 

[Dadface]

Everybody here seems to be calculating a pressure of 60 Pa but I still calculate 120Pa so if I am wrong I would appreciate it If somebody could show me where I go wrong.With my method I use P= change of momentum per second per unit area and this seems to be at odds with the answer given by using Bernoullis equation where the pressure is given by P=0.5*air density*V^2.I think that this equation is good for finding the pressure at the instant just before the air hits the wall but not when it hits the wall when there is a sudden change of V and therefore pressure.

I think I have found the reason.

I went through your process and have attempted to think as you might have.
{
All right we've got 1.2 kg of air, traveling at 10m/s.
Since the wall is 1 square meter in area this means that our volume of air has dimensions 1x1x1.
So if the cube is 1 meter in depth, it will take 0.1 seconds for the last particles to reach the wall. (am I inline with your thinking so far?)
So then,
\frac{1.2kg*10\frac{m}{s}}{0.1s*1m^2}=120\frac{N}{m^2}
}

I believe the problem you face has to do with essentially creating a solid. I think the root of your math is saying that the air is not a fluid but actually constrained like a solid and that the wall has to impart the change of momentum across the whole solid, during the entire tenth of a second.
During that last segment of time your math is still assuming the entire mass of the air, when in reality, at anyone point in time the wall really only has to stop the mass that is found in the control volume Area\cdot du

The best example I can give is a video. It's kind of imperfect but it's kind of a good mental model.
http://www.youtube.com/watch?v=oxbFrGFNVO0&feature=related




Does this help?

Thank you very much AIR&SPACE but like SW VandeCaar I was making simplifying assumptions and ignoring details like compressibility.I think the real answer is very complicated and depends upon so many factors that it can best be estimated by say wind tunnel testing.

 

[AIR&SPACE]

Thank you very much AIR&SPACE but like SW VandeCaar I was making simplifying assumptions and ignoring details like compressibility.I think the real answer is very complicated and depends upon so many factors that it can best be estimated by say wind tunnel testing.

In the context of this question it is very easy. We wouldn't be able to test this particular question in a wind tunnel because the OP specifically said he didn't want to take aerodynamics (such as turbulence) into consideration.
So essentially he was asking, without realizing, that he was asking for Dynamic Pressure.

I agree though, that if we truly wanted a real life answer that this would be pretty complicated. There is a reason why empirical tests are done to find coefficients of drag.

 

[tiny-tim]

… With my method I use P= change of momentum per second per unit area and this seems to be at odds with the answer given by using Bernoullis equation where the pressure is given by P=0.5*air density*V^2.I think that this equation is good for finding the pressure at the instant just before the air hits the wall but not when it hits the wall when there is a sudden change of V and therefore pressure.

I think Dadface is right …

imagine water flowing along a horizontal pipe of constant size going through two right-angles, East then North then East again (with rounded corners, so that the streamlines don't suffer "geodesic deviation") …

it's incompressible, so continuity means that the speed is constant, and so Bernoulli's equation says that the pressure is constant …

but obviously there's a force on the side of the pipe to make the water change direction, and that doesn't come from Bernoulli's equation.

Bernoulli's equation applies along a streamline, and so has nothing to say about forces perpendicular to streamlines.

I think the fundamental difference is between energy and momentum: Bernoulli's equation is an energy equation, and has no application where energy does not change (or, to be precise, where energy-per-mass does not change).

When a flow changes direction, there is no reason to believe that energy will be lost (why would energy be lost by air flowing past a wall … would it warm the wall up, or make a noise? ), and so Bernoulli's equation says nothing, and momentum must be considered rather than energy.

I think.

 

[AIR&SPACE]

I understand where you're coming from, it just that in the way the question is defined we essentially have a stagnation point. For example, there are two ways to set this up.

1. Assume a 1 m^2 square pipe that is closed on one end. Assume there is a vacuum inside this pipe. At one instant 1.2 kg of air in magically inserted into vent and given initial velocity 10m/s. A mechanism ensures consant density of the air
What is the force the end of the pipe?

or

2. Assume an infinite wall (in both x,z) in a velocity field of 10m/s. The constituent of this field is air @ a density of 1.2kg/m^3. Determine the amount of force that a 1m^2 imparts.


The thing is that if one wants to ignore all flow characteristics then (i feel) WE HAVE to assume that we are essentially dealing with a stagnation point.

 

[tiny-tim]

online calculator

I think the fundamental difference is between energy and momentum: Bernoulli's equation is an energy equation, and has no application where energy does not change (or, to be precise, where energy-per-mass does not change).

When a flow changes direction, there is no reason to believe that energy will be lost (why would energy be lost by air flowing past a wall … would it warm the wall up, or make a noise? ), and so Bernoulli's equation says nothing, and momentum must be considered rather than energy.

In "Pipe bends and thrust blocks forces" at http://www.engineeringtoolbox.com/forces-pipe-bends-d_968.html (link supplied by physical1 in another thread), there's an online Pipe Bend Resulting Force Calculator

 

[SW VandeCarr]

I posed the question in terms of an incompressible cubic meter of air moving, not in a vacuum, but in a pseudo-continuum of incompressible air of the same density. The wind flow is perpendicular to, not parallel to the wall. The model is therefore stagnation. The imagined block of air does not decompress over a finite time. tiny-tim is imagining the block of air as continuing to move as it makes contact with wall over a period of 0.1 sec. The resulting flows cannot be calculated in any systematic way. They must be measured. I was looking for a way to calculate the pressure against a perfectly rigid wall in a simplified model. The above reference of flow through a bent pipe is not analogous to the situation I described. Since the incompressible block of air is imagined to have stopped dead in zero time, the kinetic energy of the block does change. The energy is transferred to the rigid wall, presumably as heat. As I said, it's an idealized model. The real situation would a very difficult or intractable calculation without direct measurements. Perfectly rigid walls and incompressible air do not exist.

 

[rcgldr]

The problem with an infinitely large wall is that the situation is similar to have a pipe with one end closed and fluid flowing into the open end of the pipe.

I assume the rate of accumulation is related to the speed of sound. If the fluid is incompressable, the speed of sound is infinite, and it would not be possible to have any fluid moving into the open end of the pipe. If the fluid is compressable, then a higher pressure at the open end of the pipe would gradually increase the overall pressure inside the pipe until the pressure throughout the pipe equalized.

For a pitot tube moving through still air, the air inside a pitot tube is stagnant relative to the pitot tube chamber. In front of the pitot is a inverted cone type wedge of stagnant air that interacts with the still air via viscosity at the edges of the cone, accelerating that affected air in the direction of the pitot tube movment through the air, coexistant that affected air exerting a force in the opposite direction, increasing the pressure of the air inside the chamber of the pitot tube. Only a very tiny amount of air is accelerated from zero (speed of still air) to the speed of the pitot tube moving through the air directly in front of the center of the stagnant air wedge. The rest of the air in the stagnant wedge is just that, stagnant, and at the edges of the wedge, a vicous interaction accelerates the surrouding air to some fraction of the pitot speed.

The stagnation pressure for a compressable flow such as air is explained here:

wiki_stagnation_pressure_compressible_flow.htm

and here:

wiki_Dynamic_Pressure#Alternative_form.htm

wiki's pitot link, which refers stagnation pressure link above:

http://en.wikipedia.org/wiki/Pitot

 

[Cyrus]

See the drag force equation for high (turbulent; R_e> ~ 3000) velocity wind at
http://en.wikipedia.org/wiki/Drag_(physics ).

Turblent flow does not mean Re > 3000. This is typically stated for pipe flow. I've seen you generalize this statement to other flows in more than one thread.

 

[Cyrus]

I posed the question in terms of an incompressible cubic meter of air moving, not in a vacuum, but in a pseudo-continuum of incompressible air of the same density. The wind flow is perpendicular to, not parallel to the wall. The model is therefore stagnation. The imagined block of air does not decompress over a finite time. tiny-tim is imagining the block of air as continuing to move as it makes contact with wall over a period of 0.1 sec. The resulting flows cannot be calculated in any systematic way. They must be measured. I was looking for a way to calculate the pressure against a perfectly rigid wall in a simplified model. The above reference of flow through a bent pipe is not analogous to the situation I described. Since the incompressible block of air is imagined to have stopped dead in zero time, the kinetic energy of the block does change. The energy is transferred to the rigid wall, presumably as heat. As I said, it's an idealized model. The real situation would a very difficult or intractable calculation without direct measurements. Perfectly rigid walls and incompressible air do not exist.

I don't think you understand the basic premis of compressibility. Is your Ma > 0.3?

If not, why are we even talking about it?

I'm not worried about compressibility when I run wind tunnel tests at 160MPH+. Is your wind stronger than that? -no.

For the record, compressibility comes into play when the density varies due to Ma > 0.3 because the density changes. This is a fundamental concept about the physics of the problem you should think deeply about.

 

[rcgldr]

For the record, compressibility comes into play when the density varies due to Ma > 0.3 because the density changes.

Only because the 5% or so change in density at speeds due to Ma ~= .03 are ignored. 5% would be significant to some.

 

[Cyrus]

Only because the 5% or so change in density at speeds due to Ma ~= .03 are ignored. 5% would be significant to some.

Not the OP. Calculate the Ma and then tell me if you think its important.


Edit: I'll save you the trouble, its: Ma = 0.029.

Absolutely not. It's an order of magnitude lower than the value (Ma = 0.3) when it varies by only 5%.

 

[rcgldr]

Only because the 5% or so change in density at speeds due to Ma ~= .03 are ignored. 5% would be significant to some.

Not the OP.

But the OP was clarified to mean an infinitely large wall, essentially the same as a pipe closed at one end. If the air is incompressable, then there can be no flow at the open end of the pipe, just a change in pressure.

 

[Cyrus]

So if the pipe is closed on the end, why is there any flow in it at all (there isnt)...this is just a pitot tube problem. The answer is one equation, not 3 pages of posts.

Note: I'm not saying anything against what you posted. I'm curious as to why people are trying to calculate forces in pipe bends.

Your observation of being like a closed pipe is a good one.

 

[SW VandeCarr]

I don't think you understand the basic premis of compressibility. Is your Ma > 0.3?

If not, why are we even talking about it?

I ignored a lot physical properties in order to arrive at solution that turned out to be identical to the one given by the Bernoulli equation for incompressible fluids. That's the only reason I used the term. In fact, I imagined a cubic meter of a non-deformable solid with mass of 1.2 kg crashing into a perfectly rigid wall at 10 m/sec and stopping dead in zero time. This highly unrealistic model gave the same solution as several others got from the usual form(s) of the Bernoulli equation for the zone of stagnation. A dumb arithmetical error caused my first calculation to be off by a factor of 10, which was quickly pointed out. Most of the posters, using the Bernoulli equation got a solution of 60 Pa which was my solution after correcting the error. I really don't have much experience in fluid dynamics.

 

[Cyrus]

I ignored a lot physical properties to arrive at solution that turned out to be identical to the one given by the Bernoulli equation for incompressible fluids. That's the only reason I used the term. In fact, I imagined a cubic meter of a non-deformable solid with mass of 1.2 kg crashing into a perfectly rigid wall at 10 m/sec and stopping dead in zero time. This highly unrealistic model gave the same solution as several others got from the usual form(s) of the Bernoulli equation for the zone of stagnation. A dumb arithmetical error caused my first calculation to be off by a factor of 10, which was quickly pointed out. Most of the posters got a solution at or very close to 60 Pa.

I don't think that's the right way to model it. You should be looking into the Raleigh Transport Theorem with mass flow rates, not a slug of mass slamming into a wall. Your simulation seems to be nonphysical to me, and violates conservation of momentum.Why don't you look at the ppt slides on the first link here:

http://www.google.com/search?rlz=1C1CHNG_enUS329US330&sourceid=chrome&ie=UTF-8&q=bluff+body"

 

[SW VandeCarr]

I don't think that's the right way to model it. You should be looking into the Raleigh Transport Theorem with mass flow rates, not a slug of mass slamming into a wall. Your simulation seems to be nonphysical to me, and violates conservation of momentum.Why don't you look at the ppt slides on the first link here:

http://www.google.com/search?rlz=1C1CHNG_enUS329US330&sourceid=chrome&ie=UTF-8&q=bluff+body"

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.

 

[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.