Question 11.11: Figure 11.13 shows a crude hurricane model in which the flow ...
Figure 11.13 shows a crude hurricane model in which the flow is circular, and the wind speed increases linearly with radius from 0 at the center of the eye to 150 km/h at R = 50 km. Estimate the pressure difference from the eye of the hurricane to the indicated location R.
Circular flow implies circular streamlines with r pointing in the n direction. A linear θ velocity profile fitted to the data is given by
v_{\theta }(r)=\alpha r=\left(\frac{150\ km/h}{50\ km} \right)r=\left(3\frac{ km/h}{ km} \right)r
thus α = 3(km/h)/km = 3/h. To determine the pressure distribution, we will apply Eq. 11.20b, using V = vθ∂/∂n = ∂/∂r and ℜ = r to write the equation as
\frac{\partial p}{\partial n}=\rho \left(\frac{V^2}{ℜ} \right) (11.20b)
\frac{dp}{dr}=\rho \frac{V^2}{r}=\rho \frac{(\alpha r)^2}{r}=\rho \alpha ^2r
Integrating to find the pressure difference, we have \int_{p_0}^{p_R}{dp} =\int_{0}^{R} \rho \alpha ^2r\ dr, which yields
p_R − p_0=\rho \frac{\alpha ^2r^2}{2}\left.\right|_0^R=\rho \frac{\alpha ^2R^2}{2}
Inserting the data, we find that the pressure in the eye of the hurricane is lower than that at R = 50 km by the amount
p_R − p_0=1.225\ kg/m^3\frac{(3/h)\left(\frac{1\ h}{3600\ s} \right)^2[(50\ km)(10^3\ m/1\ km)]^2}{2} =1063\ N/m^2
In reality, hurricanes are complex 3D phenomena; therefore, this result of approximately 10 millibars somewhat underestimates recorded pressure differences for hurricanes of this strength. Can you confirm this result using Eq. 11-17a?
\rho\left(\frac{\partial v_r}{\partial t}+v_r \frac{\partial v_r}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_r}{\partial \theta}+v_z \frac{\partial v_r}{\partial z}-\frac{v_\theta^2}{r}\right) =\rho f_r-\frac{\partial p}{\partial r} (11.17a)