I was wondering whether it was possible to apply Bernoulli's equation to determine what pressure your hand would experience if you were to hold it out the window of a driving car? If so, how could you use Bernoulli's equation to solve the problem?
I believe that you could, in a very basic sense, answer this question using Bernoulli's equation. Assuming no loss from the boundary layer of the vehicle, analyzing at a point in the center of the hand at which we assume the velocity of the air is approaching zero relative to the car.
P1 + ½ (ρ) v12 = P2 + ½ (ρ) v22
Ignoring the ρgh portions since they will cancel out.
Point 1 being a place adjacent to the vehicle traveling at (let's say) 88 ft/s.
Point 2 being a point at the center of the hand where the air speed is being forced to approach zero. ½ (ρ) v22 term then is estimated to be zero.
Now: P2 = P1 + ½ (ρ) v12
Assuming atmospheric pressure at 14.7 psi. Assuming density of air at 0.080713 lb/cubic foot. Assuming the vehicle travels at 88 ft/s.
½ × (0.080713) × 882 = 312.5 psf
312.5 psf / 144 = 2.17 psi
14.7 psi + 2.17 psi = 16.87 psi (or 2.17 additional psi)
Thoughts? That's just my first pass at it.
A lot of research has shown Bernoulli's equation doesn't work for lift calculation.
There are many articles that explain this and are easily searchable!
Very roughly if we could consider your hand as a wing,
- R= density of air 1.2 kgm3
- A= area of your hand 0.015 m2
- V= 20 m/s = 72 km/hr
Let.s assume CL is 1.0 for 12 degrees angle of attack, considering the geometry of the hand we are dealing with a wide fudge factor.
This will roughly give a lift of 3.8kg.
It is enough to lift your hand.