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.

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Accepted Answer

Circular ﬂow implies circular streamlines with r pointing in the n direction. A linear θ velocity proﬁle ﬁtted to the data is given by

[latex]v_{ heta }(r)=\alpha r=\left(\frac{150\ km/h}{50\ km} ight)r=\left(3\frac{ km/h}{ km} ight)r [/latex]

thus α = 3(km/h)/km = 3/h. To determine the pressure distribution, we will apply Eq. 11.20b, using V = v_θ∂/∂n = ∂/∂r and ℜ = r to write the equation as

[latex]\frac{\partial p}{\partial n}= ho \left(\frac{V^2}{ℜ} ight) [/latex] (11.20b)

[latex]\frac{dp}{dr}= ho \frac{V^2}{r}= ho \frac{(\alpha r)^2}{r}= ho \alpha ^2r[/latex]

Integrating to ﬁnd the pressure difference, we have [latex]\int_{p_0}^{p_R}{dp} =\int_{0}^{R} ho \alpha ^2r\ dr[/latex], which yields

[latex]p_R − p_0= ho \frac{\alpha ^2r^2}{2}\left. ight|_0^R= ho \frac{\alpha ^2R^2}{2}[/latex]

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

[latex]p_R − p_0=1.225\ kg/m^3\frac{(3/h)\left(\frac{1\ h}{3600\ s} ight)^2[(50\ km)(10^3\ m/1\ km)]^2}{2} =1063\ N/m^2[/latex]

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?

[latex] ho\left(\frac{\partial v_r}{\partial t}+v_r \frac{\partial v_r}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_r}{\partial heta}+v_z \frac{\partial v_r}{\partial z}-\frac{v_ heta^2}{r} ight) = ho f_r-\frac{\partial p}{\partial r}[/latex] (11.17a)

Question 11.11: Figure 11.13 shows a crude hurricane model in which the ﬂow ...

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