
#1
Jun3009, 05:07 PM

P: 2,490

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? 



#2
Jun3009, 06:00 PM

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P: 26,167

Just go back to basics … force = pressure/area = rate of change of momentum … so how fast is the wall destroying momentum? 



#3
Jun3009, 07:46 PM

P: 2,490

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. 



#4
Jun3009, 08:04 PM

P: 21

Calculating wind pressure against a rigid wall
If we went a little further back to basics, we might realise that force = pressure TIMES area




#5
Jun3009, 08:13 PM

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P: 22,008

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




#6
Jun3009, 08:47 PM

P: 4,664

Another approach is to calculate the power in the wind.
P = (1/2)ρAv^{3} 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)ρAv^{2} 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)βρAv^{2} So for a density of 1.2 kg/m^{3}, A = 1 m^{2}, and v= 10 m/sec, F = 60β Kg m/sec^{2} = 60β Newtons on a 1 m^{2} 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 



#7
Jun3009, 08:51 PM

P: 2,490





#8
Jun3009, 08:51 PM

P: 4,664





#9
Jun3009, 09:06 PM

P: 2,490

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. 



#10
Jun3009, 09:14 PM

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P: 22,008





#11
Jun3009, 09:30 PM

P: 2,490





#12
Jul109, 03:01 AM

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P: 26,167

(just got up … )
thanks, dave! (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
Jul109, 01:31 PM

P: 2,490





#14
Jul109, 04:11 PM

P: 4,664

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



#15
Jul109, 05:18 PM

P: 2,490

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



#16
Jul109, 07:20 PM

P: 4,664

Air (wind) drag coefficients can be found at
http://www.engineeringtoolbox.com/dr...entd_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. 



#17
Jul109, 09:07 PM

P: 261





#18
Jul109, 09:38 PM

P: 2,490

[QUOTE=Cantab Morgan;2257099]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. 


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