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WoPhO 2012, Теория — Задача 3.2. The shape of tornado

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Всички материали от WoPhO 2012

Theoretical 3: Tornado

Внимание: Бележка към темата

The official solutions (Ver. 1.1, June 2012) label their sub-parts differently from the problem sheet: under [Q.2], (a) answers question 2(c), (b) answers 2(b), (c) answers 2(d) and (d) answers 2(e); under [Q.3], (a) answers question 2(a), and (b)–(e) answer 3(a)–3(d). The solutions are reproduced with their printed labels.

Условие

Inside the tornado’s core the water vapor condenses into liquid droplets as the air spirals into the core forming condensation funnel. The water vapor condenses when the temperature drops below certain point called dew point. This temperature drop is caused by pressure drop. Thus the region where the water vapor starts to condense marks a boundary of equal pressure called isobar boundary layer that stretches from the base of the cumulonimbus cloud down to the base of tornado (shown as red boundary in Figure 2). This is the boundary between region I and II.

Now only consider region I. Consider a reference point G (Fig. 2) very close to the ground ( $z \approx 0$ ) located at radius $r_G$ . The speed $v_G$ can be treated as the ground rotation speed of the tornado.

We further assume: (i) The tornado is stationary (only has rotation and no translation); (ii) The wind radial velocity is negligible. Velocity $v$ is only tangential (along the circle), not radial. (iii) The wind velocity $v$ is independent of altitude $z$ , it only depends on the radial position $r$ . (iv) We ignore turbulence very close to the ground. (v) We assume air mass density ( $\rho_{\mathrm{air}}$ ) is constant.

(a) Show that in both region I and II along $r$ :

\frac{\partial P}{\partial r} = \rho_{\mathrm{air}}\frac{v^2}{r}

[0,4 т.]

(b) In region I calculate the tangential wind velocity $v$ as a function of $r$ and in terms of $v_C$ and $r_C$ (velocity and radius at the core boundary)at any given altitude ( $z$ )! [0,5 т.]

(c) Estimate the air speed $v_G$ at the base of tornado at point G! [0,5 т.]

(d) Derive the shape of the condensation funnel or the tornado core i.e. the function $r_C(z)$ , express them in terms of $r_G$ and $v_G$ and altitude $z$ ! Plot or sketch this tornado shape in dimensionless quantities $z/h$ vs. $r/r_G$ , where $h$ is the height defined in Fig. 2! [2 т.]

(e) Most tornadoes look like funnel (the radius is larger at higher altitude) while some is more uniform in diameter like a pipe. Given everything the same, which one do you think has the higher ground rotation speed $v_G$ ? [0,5 т.]

Отговори

Решение

(a) Because point G is in the isobar boundary layer we have: $P_G = P_B$ . Using Bernoulli’s equation at point A and G, and assuming constant air mass density :

P_A + \cancel{\tfrac{1}{2}\rho v_A^2} + \cancel{\rho g z_A} = P_G + \tfrac{1}{2}\rho v_G^2 + \cancel{\rho g z_G} \qquad (8)

v_G = \sqrt{\frac{2(P_A - P_B)}{\rho}} = \sqrt{\frac{2(1e5 - 0.887e5)}{1.2}} = 137\ \mathrm{m/s} \qquad (9)

which is within the range of speed for a deadly tornado ( $v \sim 100$ m/s).

(b) Due to conservation of angular momentum we have (using point G as a reference or any point P at altitude $z$ ) :

m v_C r_C = m v_G r_G = m v r \qquad (10)

Thus:

v(r) = \frac{v_G r_G}{r} \qquad (11)

Cross-section diagram: CUMULONIMBUS CLOUD band at the top with a dashed base line, two blue funnel curves, REGION II between them and REGION I outside, a vertical z axis and horizontal r axis from the origin on the GROUND, point G near the ground on the right funnel curve and point P higher on the same curve, with a horizontal segment r_C(z) from the axis to P. — Figure 2. Tornado cross section diagram

(c) Consider point P along the boundary between region I and II as shown above. This boundary layer is also an isobar boundary layer thus $P_P = P_G$ . We have:

\cancel{P_G} + \tfrac{1}{2}\rho v_G^2 + \cancel{\rho g z_G} = \cancel{P_P} + \tfrac{1}{2}\rho v_P^2 + \rho g z \qquad (12)

v_G^2 = v_P^2 + 2 g z \qquad (13)

From part (b) we have: $v(r) = v_G r_G / r$ . Note that $v$ is only a function of $r$ not altitude $z$ as stated in the problem. Thus we have:

z = \frac{v_G^2}{2g}\left[1 - \left(\frac{r_G}{r}\right)^2\right] \qquad (14)

Alternatively we can express the tornado shape as a function of radius versus altitude below and plot it in Figure 3 below.

\frac{r}{r_G} = \frac{1}{\sqrt{1 - \frac{2 g z}{v_G^2}}} \qquad (15)

As we can see, in comparison with a real life tornado picture, this equation describe the general feature of tornado shape remarkably well despite a very simple model.

(a) Plot titled Tornado shape function: horizontal axis Radius (r/r_G) from 8 through 0 to 8, right vertical axis Altitude (z/h) from 0.0 to 1.0; a red solid pair of curves labelled v_G = 137 m/s flaring out near the top, a black dashed narrower pair labelled v_G = 160 m/s, shaded region between, label h = 1km. (b) Black-and-white photograph of a wide tornado touching the ground. — Figure 3. (a) The tornado shape function for two speeds with h = 1 km. The higher the speed the more pipe-like it looks like. (b) A real life tornado for comparison (Valparaiso, Nebraska, US, 2001).

(d) Higher speed will make the term $2 g z / v_G^2$ in Eq. (15) smaller, thus $r$ becomes a weaker function of $z$ , in other words the tornado shape becomes more uniform or pipe-like. This is also illustrated in Figure 3(a).

Оригинал в Архива: WoPhO_2012_Q13.pdf · официални решения: WoPhO_2012_S13.pdf