# Calculating wind pressure against a rigid wall

1. ### 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?

Last edited: Jun 30, 2009
2. ### tiny-tim

26,016
Hi SW VandeCarr!

Just go back to basics …

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

so how fast is the wall destroying momentum?

3. ### SW VandeCarr

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.

Last edited: Jun 30, 2009
4. ### dave_baksh

21
If we went a little further back to basics, we might realise that force = pressure TIMES area

### Staff: Mentor

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

6. ### 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

Last edited: Jun 30, 2009
7. ### SW VandeCarr

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?

Last edited: Jun 30, 2009
8. ### Bob S

This looks more like 60 joules per m3

9. ### SW VandeCarr

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.

Last edited: Jul 1, 2009

### Staff: Mentor

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

11. ### SW VandeCarr

It's also the same as using energy density.

12. ### tiny-tim

26,016
(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 … ?

13. ### SW VandeCarr

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?

Last edited: Jul 1, 2009

15. ### SW VandeCarr

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.

16. ### 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.
 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.

Last edited: Jul 1, 2009
17. ### Cantab Morgan

261
Wouldn't that be a solid?

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.)

18. ### SW VandeCarr

Last edited: Jul 1, 2009
19. ### tiny-tim

26,016

Nooo … it's a cube

a good cube take time!