Question

Air streams horizontally past an air plane .The speed over the top surface is 60 m/s and that under the bottom surface is 45 m/s .the density of air is 1.293 kg /m cube .then he diference in pressure is ????????????????????????????????????

Answer 1

Air streams horizontally past an air plane. The speed over the top surface is 60 m/s and that under the bottom surface is 45 m/s. The density of air is 1.293 kg/m cube. Then he difference in pressure is?

Bernoulli's principle can be expressed as a mathematical equation:

$$\frac {v ^ {2}}{2} + g h + \frac {p}{\rho} = c o n s t$$

where $v$ is the air streams speed, $g$ is the acceleration due to gravity,

$h$ is the height, $p$ is the pressure, and $\rho$ is the density of the air.

In our case:

$$\frac {v _ {t} ^ {2}}{2} + \frac {p _ {t}}{\rho} = \frac {v _ {b} ^ {2}}{2} + \frac {p _ {b}}{\rho}$$

$v_{t}, v_{b}$ - speeds over the top and that under the bottom surfaces, $p_{t}, p_{b}$ - pressures over the top and that under the bottom surfaces.

$$p _ {b} - p _ {t} = \frac {\rho}{2} (v _ {t} ^ {2} - v _ {b} ^ {2}) = \frac {1 . 2 9 3 \frac {k g}{m ^ {3}}}{2} (6 0 ^ {2} - 4 5 ^ {2}) \frac {m ^ {2}}{s ^ {2}} = 1 0 1 8 P a$$

Answer: 1018 Pa