Windy day wind pressures.

As the UK experiences Storm Amy It’s windy. Trees are swaying when we step out the door to walk our Black Labrador. This sparks thoughts about what generates motion. I think back to my younger years learning Newton’s Laws of motion. How simple they appear yet how powerful they are. I notice the wind speed has doubled since the previous walk, but the force on me feels much more than double. I muse about what is happening …

The trees and pedestrians feel the wind by the impact of the mass of the air as it hits them. The air is moving. This movement is generated by a difference in pressure from place to place on our planet. Pressure is higher where the wind has come from and lower where it is going to. I think about all those particles of air, with mass, that are travelling on their way. When they hit me, I temporarily bring their journey to a halt.

Let’s give our wind mass $m$ and assume that a small area $da$ stops the wind. There is a length of this wind tube upstream that we will call $dx$ . The wind is travelling in this direction. We will assume when the wind stops it suffers a negative acceleration $a$ to impart a force $F$ over the area $da$ to generate a pressure $dp$ . Newton’s Second Law of motion states $F= ma$ . Newton’s Third Law states that to every action there is an equal and opposite reaction. The static equilibrium condition, as the wind stops momentary, is given by:

(1) $\begin{equation*}m \frac{dv}{dt} = (p+\frac{dp}{dx}dx)da - pda\end{equation*}$

If the mass density of the volume $dx \; da$ remains constant then $m=\rho \; dx \; da$ with $\rho$ being the density of air. Place this in Equation 1:

(2) $\begin{equation*}\rho \; dx \; da \frac{dv}{dt} = (p+\frac{dp}{dx}dx)da - pda \implies \rho \frac{dv}{dt} = \frac{dp}{dx} \implies \rho v dv = dp \end{equation*}$

This differential equation can be solved by simple integration using the limits of zero velocity when it stops and a velocity of $v$ upstream:

(3) $\begin{equation*}\rho \int v \; dv = \int dp \implies p = \rho \frac{v^2}{2}\end{equation*}$

The pressure exerted by the wind is a function of the velocity squared! If we double the wind speed we feel an increase in force of 4 x. The motion on our walk now feels fully justified for a wind speed that is only double that of a breezy day in the UK.

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