A tornado has winds that essentially move along horizontal circular streamlines, Fig. 5–3. Within the eye, 0 ≤ r ≤ r0, the wind velocity is V = ωr, which represents a forced vortex, that is, flow rotating at a constant angular rate ω as described in Sec. 2.14. Determine the pressure distribution

Question

A tornado has winds that essentially move along horizontal circular streamlines, Fig. 5–3. Within the eye, 0 ≤ r ≤ r₀, the wind velocity is V = ωr, which represents a forced vortex, that is, flow rotating at a constant angular rate ω as described in Sec. 2.14. Determine the pressure distribution within the eye of the tornado as a function of r, if at [latex] r=r_0 [/latex] the pressure is [latex] p=p_0 [/latex].

Accepted Answer

Fluid Description. We have steady flow, and we will assume the air is an ideal fluid, that is, it is inviscid and has a constant density ρ.

Analysis. The streamline for a fluid particle having a radius r is shown in Fig. 5–3. To find the pressure distribution as a function of r (positive outward), we must apply Euler’s equation in the n direction (positive inward).

[latex] -\frac{d p}{d n}- ho g \frac{d z}{d n}=\frac{ ho V^2}{R} [/latex]

Since the path is horizontal, dz = 0. Also, for an arbitrarily chosen streamline, R = r and dn = -dr. Since the velocity of the particle is V = ωr, the above equation becomes

[latex] \begin{aligned} \frac{d p}{d r}-0 & =\frac{ ho(\omega r)^2}{r} \ d p & = ho \omega^2 r d r \end{aligned} [/latex]

Notice that the pressure increases, +dp, as we move farther away +dr from the center. Since [latex] p=p_0 ext { at } r=r_0 [/latex], then

[latex] \begin{aligned} \int_p^{p_0} d p & = ho \omega^2 \int_r^{r_0} r d r \ p & =p_0-\frac{ ho \omega^2}{2}\left(r_0^2-r^2 ight) \end{aligned} [/latex]

We will extend this analysis in Example 7.9.

Question 5.1: A tornado has winds that essentially move along horizontal c...

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