﻿ Tangential Velocities Around Vortices: The Explanation of Velocities in Fluid Drainage from One Potential Energy Level to Another
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley,
& the Gateway
to the Rockies
USA

Tangential Velocities Around Vortices:
The Explanation of Velocities in Fluid Drainage
from One Potential Energy Level to Another

## Introduction

The ultimate purpose of this analysis is to explain the wind velocities in tropiical cyclones (hurricanes, typhoons, etc.) and tornadoes. It is however convenient for the analysis to start with water draining down drains.

## Energy Changes in Drainage

When water flows from a vertical level of z1 to vertical level of z2 it loses potential energy per unit mass equal to gΔz where Δz equals (z1−z2]) and g is the acceleration due to gravity. That loss of potential energy must go into kinetic energy. The gain in kinetic energy per unit mass in the fluid is Δ(½v(z)²) where v(z) is fluid velocity at vertical level z. Consider the changes over a time interval Δt and the limits as that interval goes to zero.

Thus a balance of energy requires that

#### g(dz/dt) = d(½v(z)²)/dt and hence gv(z) = v(z)(dv(z)/dt) and further (dv(z)/dt) = g

This just says that acceleration is equal to g.

But this can be expressed as

#### d(v(z)/dz)(dz/dt) = g or, equivalently v(z)d(v(z)/dz) = g or, also d(½v(z)²./dz) = g which, upon integration gives ½v(z)² = C + gz

where C is a constant.

The quantity z is just the distance which the water has fallen. If z=0 then v(z)=0 so the contant C is equal to zero. Therefore

## The Components of Fluid Velocity

The fluid velocity v(z) is made up of two components: The vertical velocity vz and the tangential velocity vθ, where

#### v(z)² = vz² + vθ²

Assume the drain is circular with the radius R being a decreasing function of vertical level; i.e., R(z). Assume for now that the vertical velocity is the same at throughout a vertical level. The mass per unit time M that passes through the circular area at level z is given by

#### M = μπR(z)²vz

where μ is the density of water.

Thus

Therefore

#### 2gz = (M/(μπ))²/R(z)4 + vθ(z)²

But the tangential velocity is proportional to the distance r from the center of the drain; i.e.;

#### vθ(z, r) = r(z)ω(z)

where ω(z) is the angular rate of twisting of the water stream at level z. Its value is give by the condition

#### 2gz = (M/(μπ))²/R(z)4 + R(z)ω(z) and hence ω(z) = 2gz/R(z) − (M/(μπ))²/R(z)5

(To be continued.)