# Homework: Ideal Flows

date post

17-Feb-2016Category

## Documents

view

279download

13

Embed Size (px)

description

### Transcript of Homework: Ideal Flows

THE UNIVERSITY OF TEXAS AT DALLAS, MECHANICAL ENGINEERING DEPARTMENT

Incompresible Fluid Mechanics, MECH-6370,HW 3

Edgardo Javier Garca Cartagena

October 21, 2015

PROBLEM 1

The streamfunction for flow over a circular cylinder is =Ur sin(1 r

20

r 2

)Evaluate the pres-

sure distribution on the cylinder.

vr = 1r

=U(

1 r20

r 2

)cos()

v =

r=U

(1+ r

20

r 2

)sin()

On the surface of the cylinder (r = r0) vr = 0 and v =2Usin. Using the Bernoulli equationwe obtain:

P0+ 12U2 = Ps + 1

2v2s

Ps = P0+ 12U2 1

2

(U(

1+ r20

r 2

)sin()

)2

Ps = P0+ 12U2 1

2U24sin2

Ps = P0+ 12U2(14sin2)

1

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

3.0

2.5

2.0

1.5

1.0

0.5

0.0

0.5

1.0

PsP

01 2U

2

Figure: Pressure distribution on the surface of a circular cylinder

2

PROBLEM 2

Find the streamfunction for the aymptotic suction pforile, u(y)=U0[1exp(yV0/)], whichoccurs when a streaming flow with outer velocity,U0, flows above a porous wall with suctionvelocity, V0. Use Matlab to plot streamlines with equal increments of .

v =x

V0 =x

(x, y)=V0x+ f (y)

u = y

U0[1exp(yV0/)]= y

(x, y)=U0y +U0 V0

eyV0 + f (x)

(x, y)=V0x+U0y +U0 V0

eyV0

0.0 0.2 0.4 0.6 0.8 1.0

X

0.0

0.2

0.4

0.6

0.8

1.0

Y

1.500

3.000

4.500

6.000

7.500

9.000

10.500

(a)

0.0 0.2 0.4 0.6 0.8 1.0

X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (a) Streamfunction, (b) Velocity Vector Field

3

PROBLEM 3

The velocity potential for a spiral vortex flow is given by: (r,) = (/2pi) (m/2pi)log (r ), where and m are constants. Show that the angle, , between the velocity vector, u =ur ir +Ui and the radial direction, r , remains constant throughout the flow field.

ur = r = m2piru = 1r

= 2pir

tan()= uur

= /2pirm/2pir

= m

= tan1(m

)= constant

PROBLEM 4

Velocity potential for flow impinging on a flat plate is described by the streamfunction, =Axy . Using MatLab (or some other application), draw streamlines in equal increments of for this flow. The addition of a source of strength, m, at the flow origin, O, creates thepresence of a bump of height, h, protruding in the wall-normal direction. By conservation(i.e. 2 = 0), streamlines will flow around this bump. Add the source to your code, andexplore the relationship between h, A, and m (5 different combinations will be adequate).

= Axy + m2pi

= A2r 2sin(2)+ m

2pi

Stagnation point is at x = 0, y = h ( = pi2 ,r = h) where h is the height of the bump.

vr = 1r

= Arcos(2)+ m

2pir

v =

r= Ar sin(2)

At this stagnation point the velocity is zero meaning ur = 0 and v = 0, therefore from previ-ous equations the following relation is obtained:

0= Ahcos(pi)+ m2pih

or

h2 = m2piA

4

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

-0.5

00-0.

250

0.000

0.000

0.25

0

0.250

0.500 0.750

1.000

(a)

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (A = 1,m = 1,h = 1/2pi) (a) Stremfunction, (b) Velocity Vector Field

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

-0.7

50

-0.5

00

-0.2

50

0.000

0.250

0.500

0.750

(a)

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (A = 1,m = 0,h = 0) (a) Stremfunction, (b) Velocity Vector Field

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

0.000

0.000

0.250

0.250

0.50

00.

500

0.750

0.750

1.000

(a)

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (A = 1,m = 2,h = (1/pi)1/2) (a) Stremfunction, (b) Velocity Vector Field

5

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

-1.2

00

-0.6

00

0.000

0.00

0

0.600

1.200

1.800

(a)

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (A = 2,m = 1,h = (1/(4pi))1/2 (a) Stremfunction, (b) Velocity Vector Field

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

0.080

0.16

0

0.24

0

0.320

0.400

0.480

(a)

1.0 0.5 0.0 0.5 1.0X

0.0

0.2

0.4

0.6

0.8

1.0

Y

(b)

Figure: (A = 0,m = 1,h =) (a) Stremfunction, (b) Velocity Vector Field

6

PROBLEM 5

Show that the volume of an arbitrary region is given by

dVARdt

=niwidS

the volume of an arbitrary region can be wrriten by the integral

VAR =R(t )

1d

using Leibnitz theorem

dVARdt

= ddt

R(t )

1d=1

td+

niwi 1dS

1

td= 0

dVARdt

=niwi 1dS

7

PROBLEM 6

Prob. 6.1 Panton Stokes flow over a sphere has velocity components

vr =Ucos()[

1+ 12

(r0r

)3 3

2

(r0r

)]

v =Usin()[1+ 1

4

(r0r

)3+ 3

2

(r0r

)]Compute all components of the viscous stress tensor in r,, coordinates.The viscous stress tensor is defined by

=23v+2S

if incompressible= 2S

r r = 2vrr

= 2Ucos()[3

2

r 30r 4+ 3

2

r0r 2

]

= 2[

1

r

v

+ vrr

]= 2

[Ucos()

r

[1+ 1

4

(r0r

)3+ 3

2

(r0r

)]+Ucos()

r

[1+ 1

2

(r0r

)3 3

2

(r0r

)]]= 3

2

Ucos()

r

(r0r

)3 = 2

[1

r sin()

v

+ vrr+ vcot ()

r

]= 2

[vrr+ vcot ()

r

]= 2

[Ucos()

r

[1+ 1

2

(r0r

)3 3

2

(r0r

)]+Ucos()

r

[1+ 1

4

(r0r

)3+ 3

2

(r0r

)]]= 3

2

Ucos()

r

(r0r

)3

8

r = r =[r

r

(vr

)+ 1r

vr

]=[r

r

[Usin()

[1r+ 1

4

r 30r 4+ 3

2

r0r 2

]]+ 1r

[Ucos()

[1+ 1

2

(r0r

)3 3

2

(r0r

)]]]

=[Usin()

(1

r r

30

r 4 6

2

r0r 2

)Usin()

(1

r+ 1

2

r 30r 4 3

2

r0r 2

)]

=Usin()[3

2

r 30r 4+ 3

2

r0r 2

]

= =[sin()

r

(v

sin()

)+ 1r sin()

v

]= 0

r = r =[

1

r sin()

vr

+ r r

(vr

)]= 0

9

PROBLEM 7

Prob. 6.3 Panton An ideal "inviscid" flow over a cylinder has the velocity components givenin problem 5.1. Compute all components of the viscous stress tensor. Compute . Why isthis flow called inviscid?

vr =Ucos()[

1(r0r

)2]v =Usin()

[1+(r0r

)2]p = 1

2U2

[2(r0r

)2(12sin2())

(r0r

)4]the viscous stress tensor is defined as

=23(v)+2S

first the divergence of the velocity is computed in cylindrical coordinates

v= 1r

r(r vr )+ 1

r

v

+ vzz

expanding and setting vz = 0

v= 1r

[rvrr+ vr]+ 1r

v

vrr

= 2Ucos()r20

r 3

v

=Ucos()(

1+ r20

r 2

)

substituting back into the divergence equation of the velocity

v= 1r

[2Ucos()

r 20r 3Ucos()

[1(r0r

)2]+Ucos()

(1+ r

20

r 2

)]

= 1rUcos()

[2r

20

r 21+1+ r

20

r 2+ r0r 2

]= 0

the viscous stress tensor is reduced to

10

= 2S

r r = 2vrr

=4Ucos()r20

r 3

= 2(

1

r

v

+ vrr

)= 2

(1

r

(Ucos()

[1+ r

20

r 2

]) 1r

(Ucos()

[1 r0

r 2

]))

= 2rUcos()

[1+ r

20

r 21+ r

20

r 2

]

= 4Ucos()r20

r 3

zz = 0

r = r =[r

r

(vr

)+ 1r

vr

]=[vr

vr+ 1r

vr

]=[2Usin()r

20

r 3Usin()

r

(1+ r

20

r 2

)+Usin()

r

(1 r

20

r 2

)]

=Usin()[2r

20

r 3 1r r

20

r 3+ 1r r

20

r 3

]

=4Usin()r20

r 3

zr = r z = 0z = z = 0

[]r = 1r

r(rr r )+ 1

r

r

r

= 1r

8Ucos()r 20r 3 1r

4Ucos()r 20r 34Ucos r

20

r 4

= 0

[] =1

r 2

r(r 2r)+

1

r

= 1r 2uUsin()

r 20r 2+ 1r

4U (sin())r20

r 3

= 0

11

[]z = 0

= 0since this is the term that accounts for viscous forces in the momentum equation , we can saythat the flow is "inviscid"

12

PROBLEM 8

Prob. 6.7 Panton For a Newtonian fluid, show that the dissipation is given by

:v=23(v)2+2S : S (0.1)

the viscous tress tensor is

=23v+2S (0.2)

:v=(2

3v+2S

):v

=(2

3i j

vkxk