Calculating wind pressure against a rigid wall

[SW VandeCarr]

Consider a 10 m/sec wind blowing against a perfectly rigid wall. I want to calculate the pressure the wind produces on the wall. I'm ignoring the details of the actual physics of compression and turbulence. I'm reducing the problem to one cubic meter of (incompressible) air with a mass of 1.2 kg striking one square meter of a perfectly rigid wall surface.

An analytic solution doesn't seem to work since the pressure goes to infinity as the time approaches zero. However it turns out that the dimensions of pressure (P=M(L^-1)(T^-2)) are equivalent to the dimensions of energy density, although the former is a vector while the latter is a scalar. Assuming they are equivalent, the kinetic energy density is (1.2)(100)/2=600 joules/m^3 which which would be equivalent to 600 Newtons/m^2.

I know you need to be careful with units of measurement when dealing with a combination of vector and scalar quantities. Is this a valid way to go about the problem or have I missed something obvious?

 

[tiny-tim]

Consider a 10 m/sec wind blowing against a perfectly rigid wall. I want to calculate the pressure the wind produces on the wall.

Hi SW VandeCarr!

Just go back to basics …

force = pressure/area = rate of change of momentum …

so how fast is the wall destroying momentum?

 

[SW VandeCarr]

Hi SW VandeCarr!

Just go back to basics …

force = pressure/area = rate of change of momentum …

so how fast is the wall destroying momentum?

If the mass is incompressible and the wall perfectly rigid, then deceleration is infinite, and momentum is destroyed in zero time.

EDIT: The part of the question you quoted completely ignores the stated specifications. Treat the cubic meter air packet as a mass which cannot decompress or change direction. I did fail to specify that the force vector is perpendicular to the wall. If air is considered incompressible, treat it like a solid incompressible mass hitting perfectly rigid immovable wall. Show me how you can get an analytic solution.
Also, if you have perfect elastic recoil, no momentum is destroyed. I'm not allowing any recoil.

By the way, is your answer suggests momentum can be destroyed. In the real world, the assumption is it can't. However, I'm ignoring all the complicated measurements and calculations that would be necessary to obtain the actual transfer of momentum and assuming that I can get a fairly good estimate by the method I described.

 

[dave_baksh]

If we went a little further back to basics, we might realize that force = pressure TIMES area

 

[russ_watters]

Use Bernoulli's equation instead and find the velocity pressure of air movign at 10m/s.

 

[Bob S]

Another approach is to calculate the power in the wind.

P = (1/2)ρAv³

where ρ is air density, A = area, and v = wind velocity.
Since power is force times velocity, the force is

F = P/v = (1/2)ρAv²

This assumes the air velocity is zero after hitting the wall, which obviously cannot be true because there would be a big localized increase in density and pressure. So there has to be a factor like the Betz factor β for wind turbines. So the force is now

F = P/v = (1/2)βρAv²

So for a density of 1.2 kg/m³, A = 1 m², and v= 10 m/sec,

F = 60β Kg m/sec² = 60β Newtons on a 1 m² area

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

 

[SW VandeCarr]

Use Bernoulli's equation instead and find the velocity pressure of air movign at 10m/s.

Thanks, but I further clarified the problem in an edit to post 3. Bernoulli's equations assume lateral flow and a stagnation point. My example is assumes an unrealistic model for to what is a complicated real world problem involving involving a lot of parameters. By assuming I can substitute energy density for pressure and that all of the energy is dissipated instantly into the wall (with no wall motion) within the one square meter area, I can get a quick and dirty answer which is reasonable. The answer I get is a reasonable value for a 10 m/sec wind. The basic question is, can I assume that the dimensional equivalence of pressure and energy density allows me to substitute one for the other in this particular problem?

 

[Bob S]

... the kinetic energy density is (1.2)(100)/2=600 joules/m^3 ...

This looks more like 60 joules per m³

 

[SW VandeCarr]

This looks more like 60 joules per m³

Yep. It sure does. Then it agrees with your calculation if \beta=1. I don't know if this validates my approach since I have no idea how to derive \beta. 60 Newtons/m^2 seems too low.

EDIT: On second thought, it may not be too low. In more familiar units 36 km/hr is a fresh breeze (as defined by NOAA) that scatters leaves but rarely causes damage.

 

[russ_watters]

Another approach is to calculate the power in the wind.

P = (1/2)ρAv³

where ρ is air density, A = area, and v = wind velocity.
Since power is force times velocity, the force is

Note that that's just Bernoulli's equation with area in it and V^3 instead of V^2...

 

[SW VandeCarr]

Note that that's just Bernoulli's equation with area in it and V^3 instead of V^2...

It's also the same as using energy density.

 

[tiny-tim]

(just got up … :zzz:)

If we went a little further back to basics, we might realize that force = pressure TIMES area

oops!

thanks, dave!

If the mass is incompressible and the wall perfectly rigid, then deceleration is infinite, and momentum is destroyed in zero time.

EDIT: … treat it like a solid incompressible mass hitting perfectly rigid immovable wall. Show me how you can get an analytic solution.
Also, if you have perfect elastic recoil, no momentum is destroyed. I'm not allowing any recoil.

Hi SW VandeCarr!

(ignore Bernoulli's equation … you don't need it )

In time t, a block of air of volume Avt stops dead.

So the change of momentum is … ?

 

[SW VandeCarr]

(just got up … :zzz:)oops!

thanks, dave! Hi SW VandeCarr!

(ignore Bernoulli's equation … you don't need it )

In time t, a block of air of volume Avt stops dead.

So the change of momentum is … ?

The change in momentum is -(1.2)(10)= -12 kg m/s in zero time. So? To get from momentum to pressure dimensionally you need to multiply by (L^-2)(T^-1). What's that?

 

[Bob S]

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

 

[SW VandeCarr]

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

Thanks, but this equation requires a drag co-efficient. (By the way, the link failed, so I searched on 'drag force equation'). I guess we could apply this equation to this problem by treating the wall area as a moving object through still air.

I edited post 9. The solution we got does seem reasonable for a 36 km/hr breeze, assuming beta equal to one.

 

[Bob S]

Air (wind) drag coefficients can be found at
http://www.engineeringtoolbox.com/drag-coefficient-d_627.html
For a cube, it is about 0.8. For a square flat plate, it is 1.18.
For wind in HAWTs (horizontal axis wind turbines) the Betz (beta) coefficient is about 0.593. This is the percentage of wind energy that can be theoretically extracted by a wind machine. It accounts for air stagnation problems.
[Edit] The Betz (beta) factor is the theoretical maximum fraction of the incident wind energy that can be extracted by a HAWT (horizontal axis wind turbine). The rotor blades are airfoils that create minimum turbulence. The rotor blade tip speeds are about 6 x the wind velocity. Drag coefficients for flat plates, rigid walls, and cubes are not airfoils, and create lots of turbulence (which heats the air rather than extracts energy). So drag coefficients and the Betz factor probably do not belong in the same equation.

 

[Cantab Morgan]

Treat the cubic meter air packet as a mass which cannot decompress or change direction.

Wouldn't that be a solid?

Also, if you have perfect elastic recoil, no momentum is destroyed. I'm not allowing any recoil.

Elastic recoil is not required to conserve momentum, it's required to conserve mechanical energy. If a wad of putty collides with something and sticks, momentum is still conserved. (But it gets warmer.)

 

[SW VandeCarr]

Wouldn't that be a solid?

As I said, this is a simple model for complex set of parameters (compressibility, turbulence, directional flows, etc) I'm thinking of this as a collision between two perfectly rigid bodies which cannot be deformed. As it turns out, this quick and dirty approach gives an answer which is identical to the Bernoulli equation result if you take beta to be one (see previous posts). That's because my approach seems to be equivalent to the Bernoulli equation.

 

[tiny-tim]

add one cube …

… I'm reducing the problem to one cubic meter of (incompressible) air with a mass of 1.2 kg striking one square meter of a perfectly rigid wall surface.

The change in momentum is -(1.2)(10)= -12 kg m/s in zero time.

Nooo … it's a cube …

a good cube take time! ​

 

[Dadface]

To find the pressure find the mass of air hitting unit area of the wall per second and multiply by the change of velocity.Minimum velocity change =10-0=10 m/s(if the wind stops on impact).Maximum velocity change=10--10=20m/s(if the wind bounces back elastically with the same speed).The real answer will lie between this minimum and maximum value and you can estimate it.

 

[rcgldr]

Why not just use the formula for a pitot static tube to get the total pressure then multiply this pressure by the effective wall area? (This is not a realistic case, because the air is either accumlating in front of a massive wall, or it's flowing around a the wall and only the stagnation zone of the wall fits the original post question).

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

http://www.grc.nasa.gov/WWW/K-12/airplane/pitot.html

You can use this pitot-static simulator to calculate the pressure versus speed

http://www.luizmonteiro.com/Learning_Pitot_Sim.aspx

1 m/s = 1.9438444924406 knots, so for 10 m/s use 19.438444924406 as the speed in knots. The dynamic pressure is only about 0.06% (.6/1013.2 or .2/29.92). The static pressure at sea level is 101325 Pa (Pa = N/m²) , and the net pressure on the wall would be .06% of this or 61 N/m².

 

[SW VandeCarr]

Why not just use the formula for a pitot static tube to get the total pressure then multiply this pressure by the effective wall area? (This is not a realistic case, because the air is either accumlating in front of a massive wall, or it's flowing around a the wall and only the stagnation zone of the wall fits the original post question).

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

http://www.grc.nasa.gov/WWW/K-12/airplane/pitot.html

You can use this pitot-static simulator to calculate the pressure versus speed

http://www.luizmonteiro.com/Learning_Pitot_Sim.aspx

1 m/s = 1.9438444924406 knots, so for 10 m/s use 19.438444924406 as the speed in knots. The dynamic pressure is only about 0.06% (.6/1013.2 or .2/29.92). The static pressure at sea level is 101325 Pa (Pa = N/m²) , and the net pressure on the wall would be .06% of this or 61 N/m².

Good. Note that you get virtually the same answer I got (60 N/m²), and Bob S got from the Bernoulli equation.

 

[Dadface]

The answer I get is twice what others have calculated i.e. 120Pa(this is the minimum pressure calculated by assuming that the wind stops on impact and by making other simplifying assumptions)I think that the 60Pa calculated from Bernoulli etc is the static pressure and not the pressure that would be exerted if the wind were brought to rest.

 

[SW VandeCarr]

The answer I get is twice what others have calculated i.e. 120Pa(this is the minimum pressure calculated by assuming that the wind stops on impact and by making other simplifying assumptions)I think that the 60Pa calculated from Bernoulli etc is the static pressure and not the pressure that would be exerted if the wind were brought to rest.

We calculated kinetic energy which assumes that all the energy goes into the meter square wall surface. Although no work is done assuming a perfectly rigid wall, I think the kinetic energy could be thought of as dissipated as heat.

 

[AIR&SPACE]

There's this thing called "Dynamic Pressure." This is the pressure that a control volume of a fluid would have if you could bring it to rest.

Airplanes have Pitot tubes for this purpose. They take air flowing into them and then measure the pressure of the air after it's brought to rest (with respect to the aircraft). Then they use this to calculate the velocity of the plane.

Dynamic Pressure = q_inf
q_inf = (1/2)*density*(velocity^2)

Since you say you don't want to take into consideration turbulence or other factors, we don't have to take the drag coefficient into consideration.
We can just say, F=P*A = (q_inf)*A = (1/2)*density*(velocity^2)*Area

What this essentially finds is the maximum possible drag force on the wall. Given your neglection of other fluid properties, I believe this is what you're looking for.

 

[AIR&SPACE]

There's this thing called "Dynamic Pressure." This is the pressure that a control volume of a fluid would have if you could bring it to rest.

Airplanes have Pitot tubes for this purpose. They take air flowing into them and then measure the pressure of the air after it's brought to rest (with respect to the aircraft). Then they use this to calculate the velocity of the plane.

I should clarify a bit. The pitot tube actually measures static pressure + dynamic pressure.
Then the static pressure, which is measured by another instrument called the static port, is subtracted from the pitot tube's measurement - thus giving us the dynamic pressure. Then this value is doubled and divided by the density of the air and this gives us the square of the velocity. Then you take the sqrt of that to find the velocity.

Just thought I'd clear that up.

 

[SW VandeCarr]

There's this thing called "Dynamic Pressure." This is the pressure that a control volume of a fluid would have if you could bring it to rest.



Dynamic Pressure = q_inf
q_inf = (1/2)*density*(velocity^2)

Since you say you don't want to take into consideration turbulence or other factors, we don't have to take the drag coefficient into consideration.
We can just say, F=P*A = (q_inf)*A = (1/2)*density*(velocity^2)*Area

What this essentially finds is the maximum possible drag force on the wall. Given your neglection of other fluid properties, I believe this is what you're looking for.

Yes. This is the equation I used in the first post where density and velocity were given and area was unity in SI units.

 

[AIR&SPACE]

Yes. This is the equation I used in the first post where density and velocity were given and area was unity in SI units.

Ahh. Good job then, sire. I must have skimmed over it. Nothing like clubbing dead seals!

 

[SW VandeCarr]

Ahh. Good job then, sire. I must have skimmed over it. Nothing like clubbing dead seals!

No problem. Thanks for your interest.

 

[AIR&SPACE]

An analytic solution doesn't seem to work since the pressure goes to infinity as the time approaches zero. However it turns out that the dimensions of pressure (P=M(L^-1)(T^-2)) are equivalent to the dimensions of energy density, although the former is a vector while the latter is a scalar. Assuming they are equivalent, the kinetic energy density is (1.2)(100)/2=600 joules/m^3 which which would be equivalent to 600 Newtons/m^2.
?

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

I did it as well to find the dynamic pressure instead of energy density (I'm an aero student using what I'm familiar with, sue me )

\frac{1.2\frac{kg}{m^3}*\left(10\frac{m}{s}\right)^2}{2}

=0.6\frac{kg}{m^3}*100\frac{m^2}{s^2}

=60\frac{kg*m}{s^2}*\frac{1}{m^2}

=60 \frac{Newtons}{m^2}

Yes? No?

 

[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.