Chapter 10, Problem 45P

To determine

The air pressure inside a given category hurricane is to be calculated.

Answer to Problem 45P

The air pressure inside a given category hurricane is $96876Pa$ .

Explanation of Solution

Given info:

The wind’s speed, $v=300Km/h=297\times 1000\times \frac{1}{3600}=82.5m/s$

Formula used:

• The dynamic pressure is given as,

  $P_d=\frac{1}{2}\rho v^2$

Where, $P_d$ is dynamic pressure, $\rho$ is the density if the fluid and $v$ is the velocity of the dynamic fluid.

• The Bernoulli’s equation is,

  $P_1+g{h}_1\rho +\frac{1}{2}\rho v_1{}^2=P_2+g{h}_2\rho +\frac{1}{2}\rho v_2{}^2$

Calculation:

Considered a point outside the hurricane at the same height of that point inside the hurricane which means that $h_1=h_2$ and also the first speed is zero because its air is still. So, the Bernoulli’s equation becomes,

  $\begin{array}{l}{P}_1+g{h}_1\rho +\frac{1}{2}\rho v_1{}^2=P_2+g{h}_2\rho +\frac{1}{2}\rho v_2{}^2\\ P_1+\frac{1}{2}\rho v_1{}^2=P_2+\frac{1}{2}\rho v_2{}^2\\ P_1=P_2+\frac{1}{2}\rho v_2{}^2\left(use{v}_1=0\right)\\ P_2=P_1-\frac{1}{2}\rho v_2{}^2\end{array}$

Now the standard atmospheric pressure is $P_1=101,300Pa$ . So, the above equation becomes

  $P_2=101,300Pa-\frac{1}{2}\rho v_2{}^2$

Now use the given values in this reduced equation and then the resulted outcome is

  $\begin{array}{l}{P}_2=101,300Pa-\frac{1}{2}\left(1.3kg/m^3\right){\left(82.5m/s\right)}^2\\ P_2=101,300Pa-4424=96876Pa\end{array}$

Conclusion:

The air pressure is $96876Pa$ .