Posted by **Catie** on Monday, August 15, 2011 at 12:11am.

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?

- Math -
**MathMate**, Monday, August 15, 2011 at 9:58am
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:

(Broken Link Removed)

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.

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