# A tornado may be modeled as the circulating flow shown below, with

## Question:

A tornado may be modeled as the circulating flow shown below, with {eq}v_r = v_2 =0 {/eq} and

{eq}v_{\theta} = \left\{\begin{matrix} \omega r \ r\leq R \\\frac{\omega R^2}{r} \end{matrix}\right . {/eq}{eq}r > R {/eq}

a) Using cylindrical coordinates, determine if the inner region is rotational.

b) Using cylindrical coordinates, determine if the outer region is rotational.

Moving away from the center, {eq}r\rightarrow \infty {/eq}, we recover the free stream pressure p\approx p_\infty .

From the {eq}\widehat{r} {/eq}- momentum equation: {eq}\frac{dp}{dr} = \frac{\rho v^2 \theta}{r} {/eq}

Using appropriate boundary conditions

c) Find the pressure distribution {eq}p(r) {/eq} for {eq}r > R. {/eq}

d) Find the pressure distribution {eq}p(r) {/eq} for {eq}r \leq R {/eq}.

e) Find the minimum pressure.

## Momentum:

The momentum is the quantity that could be evaluated by finding the product of mass with the velocity of the moving particle. It is represented in terms of kilogram meter per seconds.

## Answer and Explanation:

**Given data**

The velocity condition is,

{eq}{v_\theta } = \left\{ {\begin{align*} {\omega R,\;r \le R}\\ {\dfrac{{\omega {R^2}}}{r},\;r > R} \end{align*}} \right. {/eq}

**(a)**

The expression for the rotational component for inner region {eq}r \le R {/eq} is,

{eq}{\Omega _{in}} = \dfrac{1}{r}\dfrac{{d\left( {r{v_\theta }} \right)}}{{dr}} {/eq}

Substitute values.

{eq}\begin{align*} {\Omega _{in}} &= \dfrac{1}{r}\dfrac{{d\left( {r\omega r} \right)}}{{dr}}\\ {\Omega _{in}} &= 2\omega \\ {\Omega _{in}} &\ne 0 \end{align*} {/eq}

Thus the inner region is **rotational**.

**(b)**

The expression for the rotational component for outer region {eq}r > R {/eq} is,

{eq}{\Omega _{out}} = \dfrac{1}{r}\dfrac{{d\left( {r{v_\theta }} \right)}}{{dr}} {/eq}

Substitute values.

{eq}\begin{align*} {\Omega _{out}} &= \dfrac{1}{r}\dfrac{{d\left( {r\dfrac{{\omega {R^2}}}{r}} \right)}}{{dr}}\\ {\Omega _{out}} &= \dfrac{1}{r}\dfrac{{d\left( {\omega {R^2}} \right)}}{{dr}}\\ {\Omega _{out}} &= 0 \end{align*} {/eq}

Thus the outer region is **irrotational**.

**(c)**

The {eq}\hat r {/eq} momentum equation is,

{eq}\dfrac{{dp}}{{dr}} = \dfrac{{\rho v_\theta ^2}}{r} {/eq}

Integrate above equation.

{eq}p = \int {\dfrac{\rho }{r}{{\left( {{v_\theta }} \right)}^2}dr} ??(1) {/eq}

Substitute values for outer region.

{eq}\begin{align*} {p_{outer}} &= \int {\dfrac{\rho }{r}{{\left( {\dfrac{{\omega {R^2}}}{r}} \right)}^2}dr} \\ {p_{outer}} &= \left( {\rho {\omega ^2}{R^4}} \right)\int {\dfrac{1}{{{r^3}}}dr} \\ {p_{outer}} &= \dfrac{{ - \rho {\omega ^2}{R^4}}}{{2{r^2}}} + c \end{align*} {/eq}

Here, **c** is the constant.

Apply the boundary condition; at {eq}r = \infty {/eq}, {eq}p = {p_\infty } {/eq} in above equation.

{eq}\begin{align*} {p_\infty } &= \dfrac{{ - \rho {\omega ^2}{R^4}}}{{2{{\left( \infty \right)}^2}}} + c\\ c &= {p_\infty } \end{align*} {/eq}

The pressure at outer region is,

{eq}{p_{outer}} = \dfrac{{ - \rho {\omega ^2}{R^4}}}{{2{r^2}}} + {p_\infty } {/eq}

The pressure distribution at r = R is,

{eq}\begin{align*} {p_{R = r}} &= \dfrac{{ - \rho {\omega ^2}{R^4}}}{{2{R^2}}} + {p_\infty }\\ {p_{R = r}} &= \dfrac{{ - \rho {\omega ^2}{R^2}}}{2} + {p_\infty } \end{align*}??(2) {/eq}

**(d)**

Substitute values for inner region in equation (1).

{eq}\begin{align*} {p_{inner}} &= \int {\dfrac{\rho }{r}{{\left( {\omega {r^2}} \right)}^2}dr} \\ {p_{inner}} &= \dfrac{{\rho {\omega ^2}{r^2}}}{2} + c' \end{align*}??(3) {/eq}

Here {eq}c' {/eq} is constant.

The pressure distribution at {eq}R = r {/eq} is,

{eq}{p_{R = r}} = \dfrac{{\rho {\omega ^2}{R^2}}}{2} + c'??(4) {/eq}

Equate equation (2) and equation (4).

{eq}\begin{align*} \dfrac{{ - \rho {\omega ^2}{R^2}}}{2} + {p_\infty } &= \dfrac{{\rho {\omega ^2}{R^2}}}{2} + c'\\ c' &= - \rho {\omega ^2}{R^2} + {p_\infty } \end{align*} {/eq}

Substitute values in equation (3).

{eq}{p_{inner}} = \dfrac{{\rho {\omega ^2}{r^2}}}{2} - \rho {\omega ^2}{R^2} + {p_\infty } {/eq}

**(e)**

The minimum pressure obtain at the origin.

Substitute r = 0 in equation (5).

{eq}\begin{align*} {p_{\min }} &= \dfrac{{\rho {\omega ^2}{{\left( 0 \right)}^2}}}{2} - \rho {\omega ^2}{R^2} + {p_\infty }\\ {p_{\min }} &= - \rho {\omega ^2}{R^2} + {p_\infty } \end{align*} {/eq}

#### Learn more about this topic:

from UExcel Physics: Study Guide & Test Prep

Chapter 7 / Lesson 2