Question 5.1: A tornado has winds that essentially move along horizontal c......
A tornado has winds that essentially move along horizontal circular streamlines, Fig. 5–3. Within the eye, 0 ≤ r ≤ r_0, 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 r = r_0 the pressure is p = p_0.
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).
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
\frac{dp}{dr} – 0 = \frac{\rho(ωr)^2}{r}dp = \rho ω^2 r dr
Notice that the pressure increases, +dp, as we move farther away +dr from the center. Since p = p_0 at r = r_0, then
\int _p ^{p_o} dp = \rho ω^2 \int _r ^{r_0} r drp = p_0 – \frac{\rho ω^2}{2}(r_0^2 – r^2)
We will extend this analysis in Example 7.9.