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Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of 220 m2 ? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89×104 N/m2 . (Bernoulli's principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)
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Anonymous
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Sahil Khan
1

Ignoring turbulence, we can use Bernoulli's equation:


P_1\ +\ \frac{1}{2}\ \rho v_1^2\ +\ \rho gh_1\ =\ P_2\ +\ \frac{1}{2}\ \rho v_2^2\ +\ \rho hg_2\


where the heights are the same : h_{1} = h_{2}, because we are concerned about above and below a thin roof. The velocity inside the house is zero, so v_{1} = 0.0 m/s, while the speed outside the house is v_{2} = 45.0 m/s.

The difference in pressures, P_{1}-P_{2} can then be found: P_{1}-P_{2} = \frac{1}{2}\rho v_{2}^{2}. Now we can relate the change in pressure to the force on the roof, using the equation

F = (P_{1} - P_{2}) A, because we know the area of the roof (A = 200 m^{2})

F\ =\ \left(P_1-P_2\right)A\ =\ \frac{1}{2}\ \rho\left(v_2^2\ -v_1^2\right)\ A and substituting in the values gives:

[math]F\ =\ \frac{1}{2}\left(1.14\ \frac{kg}{m^3}\right)\left[\left(45\ \frac{m}{s}\right)^{^2}-\ \left(0.0\ \frac{m}{s}\right)^{^2}\right]\left(220\ m^2\right)\ =\ 2.54\ \times10^5N[/math]