Question

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind speed of 1 m/s causes a pressure drop of 0.645 Pa, what pressure drop is predicted by Bernoul's equation for a wind speed of 5 m/s? Multiple Choice 3.325 Pa 129 Pa 194 Pa 16 125 Pa

Answer 1

Answer:

16.125 Pa

Explanation:

The Bernoulli equation despising the height changes is:

$\frac{P_{2}-P_{1}}{pg}+(\frac{v_{2}^{2}-v_{1}^{2}}{2g})=0$

The gravity constant can be cancelled.

Applying the equation to the first situation, $v_{1}=0$ , $v_{2}=1$, $P_{2}-P_{1}=-0.645$

The density 'p' may be calculated because it is the only unknown.

$p=\frac{-2(P_{2}-P_{1})}{v_{2}^{2}}=\frac{-2*-0.645}{1^{2}}=1.29\frac{kg}{m^{3} }$

Applying the equation to the second situation, where the only unknown is the pressure drop ($P_{2}-P_{1}$):

$P_{2}-P_{1}=-p*(\frac{v_{2}^{2}-v_{1}^{2}}{2})$

$P_{2}-P_{1}=-1.29*(\frac{5_{2}^{2}}{2})=-16.125$

In both cases the assumption is  $v_{1}=0$ because its supposed that the air is stored.