What psi can be achieved from a 60mph wind

[Cyclonus]

If I put vents on the hood of my car and drove at 60mph what kind of psi could be achieved? I don't care about what kind of technique is used just curious about what kind psi could be created. I apologize if I'm not asking this correctly but I have no background in fluid dynamics.

 

[russ_watters]

Bernoulli's equation can be used to calculate the answer. A shortcut using common units can be found here:
http://www.eng.utoledo.edu/~akumar/IV/Workbook/IVWB_GeneralPrinciples.htm

60 mph equates to a stagnation pressure of about 1.7" W.G., or about 0.06 psi.

 

[Baluncore]

The wind pressure is proportional to the air density multiplied by the speed squared.
My very old engineering text gives pressure due to wind as; p = 0.0032 * v^2
where pressure p is in pounds per square foot and v is miles per hour.
For pressure in psi divide by 144 to get; p = .00002222 * v^2
So if v = 60 mph, then p = 0.08 psi.
This estimate agrees reasonably well with russ_watters result of 0.06 psi.

 

[russ_watters]

The rolled-up constant I used was 4005, which inverted is .000250. Not sure where the discrepancy comes from. But not a big deal.

 

[Baluncore]

I based my computation on an old formula to confirm the order of magnitude of russ_watters result. We agreed.

The coefficient should really be 0.00255646, not the 0.0032 as obtained from my 1938 engineering text that used a coefficient recommended by a paper published in 1911. I believe the result discrepancy comes about because Engineers are conservative and so overestimate the effect of wind on their structural designs. Standards have also been redefined during the last 102 years.

Further examination of the “rolled up” constant k based on;
dynamic pressure = half * density * velocity^2
Assuming air at 15°C and sea level, the density is 1.225 kg/m3
and knowing that 1 psi = 6.8948*10^3 Pa

k = (0.5 * 1.225 * 1609.344^2) / (6894.8 * 3600^2)
So k = 17.7532e-6
And 1 / k = 56327.87
Then psi = mph^2 * 17.7532e-6
Or psi = mph^2 / 56327.87

At 60 mph the theoretical dynamic pressure will be 0.0639115 psi

What the computation does confirm is that at the speed of road vehicles, the force of the wind on a large exposed surface can be very great, but the pressure per unit area is small when compared to atmospheric pressure. This limits the utility of ram air intakes and explains why a turbo-charger or super-charger must be used to get a significant charge boost. Fundamentally, a turbo-charger gives a greater pressure boost than a ram air intake because the blades of the compressor can move significantly faster than the vehicle.