wind velocity

Ry122

If I know the force that a wind is exerting on an object, is it be possible to determine what velocity the wind is moving at? I also know the air density.

Danger

That's pretty much how anemometers work, so it's definitely possible.
I'm afraid that I can't supply any formulae, though.

Daiquiri

The aerodynamic force acting on a solid body can be expressed by a general equation:

F = [tex]1/2 \rho V^2 S C[/tex]

Where:
C is a force coefficient (either lift Cl, drag Cd or Cx, or whatever you're looking for)
S is a reference area (either frontal area, wing area for airplanes etc.)

The problem lies in the coefficient C. It depends on some adimensional numbers (Mach number and Reynolds number in most cases) and is usualy measured in a wind tunnel on a scaled model of the object.

If you were able to measure the force acting on the solid body, and if you knew the value of the aerodynamic coefficient C, then you wuld be also able to calculate the velocity from the formula above.

If you only knew the values of C vs. V (through Reynolds number) then it would necessarily be an iterative process, since you would have to estimate an initial V, then you would calculate C for that V, then you would recalculate V with that value of C, and then recalculate C with the new value of V, and so on and so on and so on... till the convergence of the result.

On the other hand, if you had data (from real-scale or wind-tunnel measurements) which relate directly F to V (usually for incompressible flows and near-standard temperatures) then the process is straightforward: measure F --> read V from the F-V curve.

russ_watters

In practice, that method is difficult, but what you could do is use a pito-static tube.

RadeK

Quote by Daiquiri

The aerodynamic force acting on a solid body can be expressed by a general equation:

F = [tex]1/2 \rho V^2 S C[/tex]
...

Can you add units I shall use to calculate force?

I found that Cx for flat wall is 1.5

R

Redbelly98

C has no units. The other quantities on the right-hand-side of the equation combine to give force units.

As long as you use a consistent set of units for the different quantities, the result will be a force. For example:

[tex]\rho[/tex] in kg/m^3
V in m/s
S in m^2

Resulting units are
(kg / m^3) * (m/s)^2 * m^2
= (kg / m^3) * (m^2 / s^2) * m^2
= (kg / m^3) * m^4 / s^2
= kg * m / s^2
= Newtons

RadeK

Thank you

rcgldr

As mentiond the pitot-static tube measures wind velocity, and does so via comparason of the pressure (force per unit area) in 2chambers connected to tubes, one that face into the wind and and one perpendicular to the wind.

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Ry122	wind velocity Tweet If I know the force that a wind is exerting on an object, is it be possible to determine what velocity the wind is moving at? I also know the air density.

Danger Recognitions: Gold Member	That's pretty much how anemometers work, so it's definitely possible. I'm afraid that I can't supply any formulae, though.

russ_watters Mentor	wind velocity In practice, that method is difficult, but what you could do is use a pito-static tube.

Redbelly98 Mentor Blog Entries: 10	C has no units. The other quantities on the right-hand-side of the equation combine to give force units. As long as you use a consistent set of units for the different quantities, the result will be a force. For example: [tex]\rho[/tex] in kg/m^3 V in m/s S in m^2 Resulting units are (kg / m^3) * (m/s)^2 * m^2 = (kg / m^3) * (m^2 / s^2) * m^2 = (kg / m^3) * m^4 / s^2 = kg * m / s^2 = Newtons

rcgldr Recognitions: Homework Help	As mentiond the pitot-static tube measures wind velocity, and does so via comparason of the pressure (force per unit area) in 2chambers connected to tubes, one that face into the wind and and one perpendicular to the wind.