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.