A cyclone is assumed to have the following velocity distribu

A cyclone is assumed to have the following velocity distribution: u_theta(r) = {omega r r lessthanorequalto r_0 omegar^2 _0/r r > r_0 u_r = 0 r_0 = 10m omega = 10 1/s H = 100 m rho = 1.25 kg/m^3 Sketch u_theta(r) Determine the circulation for a circle around the axis of the cyclone for r r_0 Show that for r > r_0 the flow is irrotational. How large is the kinetic energy in a cylinder of radius R = 2r_0 and height H?

Solution

solution:

1)here it is given in cylindrical coordinate that velocity are

Uh=0,Ur=0

Uk=r*w for r<ro

Uk=ro^2*w/r for r>ro

hence sketch of Uk is straight inclined line for r< ro with positive slope w and again decreases linearly with slope ro^2*w

2)here rotation of fluid element about verticle axis h is given by

wh=1/r[d/dr(r.Uk)-dUr/dk]

Ur=0

wh=1/r[d/dr(r.Uk)]

Uk=rw ,on putting and differentiating we get

here rotation for r< ro

wh=w

hence circulation is

circulation=2*rotation about perpendicular axis*area

circulation=2*pi*r^2*w

for r=ro

rotation is

wh=w

circultion is

circulation=2*pi*ro^2*w

for r>ro

wh=1/r[d/dr(r.Uk)]

Uk=ro^2*w/r

on putting and differentiang we get that

wh=0

hence circulation is zero

circulation=0

4)as here outside the cyclone r>ro

Uh=Ur=0

hence rotation vector along

wk=wr=0

and as we get for h direction again

wh=0

as rotation is given by

w=wr.r^+wk.k^+wh.h^

w=0

hence flow is irritational outside the cyclone

5)kinetic energy inside cylinder of radius R=2ro and height H is given by integrating kinetic energy over volume and we get

elementla KE=.5*m*Uk^2

total energy as on integrating over volume within limit for r from 0>r> ro + ro >r> 2ro and K from 0>k>2*pi and height 0>h>H,so we get final relation as

E=4*density*pi^2*ro^6*w^2*H*(.25+ln2)

on putting alue we get that

E=total energy s

E=4.65*10^11 J

E=