Question

A hurricane wind blows across a 6.00 m times 15.0 m flat roof at a speed of 190 km/hr.
What is the pressure difference? Use 1.28 kg/m^3 for the density of air.

How much force is exerted on the roof?

If the roof cannot withstand this much force, will it "blow in" or "blow out"?

Help?

Answer 1

Use Bernoulli's equation:
P1+ρgh1+(1/2)ρv1² = P2+ρgh2+(1/2)ρv2²

where
P1=exterior pressure
P2=interior pressure
v1=exterior wind velocity
v2=interior wind velocity = 0
h1-h2 is the negligible height differences
ρ=density of air (1.293 kg-m^-3)

which by neglecting ρg(h2-h1), and substituting v1=0 we get
P2-P1=(1/2)ρv1²
v1=190km/h=52.778m/s
ρ=1.293kg/m³=12.7N/m³

Suction force (blows out):
=area*pressure difference
=6m*15m*(1/2)*1.293kg/m³*(52.8m/s)²
=162,073 N

The above calculation of pressure assumes:
1. the pressure factor is -1 (which is not far for a flat roof).
See:
http://www.nrc-cnrc.gc.ca/eng/ibp/irc/cbd/building-digest-68.html

2. the STP value of ρ=1.293kg/m³ instead of 1.28 as stipulated. You will need to make adjustments to get the correct answer.