First slide

Fluid dynamics

Question

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 $m^{2}$ . Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be
$(ρ_{air} = 1.2 kg / m^{3})$

Moderate

A

$2.4 \times 10^{5} N, upwards$
B

$2.4 \times 10^{5} N, downwards$
C

$4.8 \times 10^{5} N, downwards$
D

$4.8 \times 10^{5} N, upwards$

Solution

By Bernoulli's theorem,
$\underset{inside}{\underset{⏟}{P_{1} + \frac{1}{2} ρ v_{1}^{2}}} = \underset{outside}{\underset{⏟}{P_{2} + \frac{1}{2} ρ v_{2}^{2}}}$
Assuming that the roof width is very small Pressure difference,
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Here, ρ = 1.2 kg m^{- 3}, v_{2} = 40 m s^{- 1}, v_{1} = 0$
$A = 250 m^{2}$
$P_{1} - P_{2} = \frac{1}{2} \times 1.2 (40^{2} - 0^{2})$
$= \frac{1}{2} \times 1.2 \times 1600 = 960 N m^{- 2}$
$Force acting on the roof F = (P_{1} - P_{2}) \times A = 960 \times 250$
$= 2.4 \times 10^{5} N upwards$

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