Homework Set #3 of Course AerE 541

1. Consider the incompressible, irrotational flow where the potential function is:

22ln YXK ,

where K is an arbitrary constant.

(a) What is the velocity field for this flow?

What is the magnitude and direction of

the velocity at ( 0,2 ), at ( 2,2 ) and

at( 2,0 )

(b) Is the flow physically possible? Is the

flow field irrotational?

(c) What is the stream function for this flow?

Sketch the stream line pattern.

(d) Sketch the lines of constant potential.

How do the iso-potential lines related to

the streamlines?

(e) For the region shown in the figure, evaluate ?)( AdV

and ?sdV

to

demonstrate that Stokes’s theorem is valid.

(1,1)

(1,0) (0,0)

X

Y

Solution to question (e)

2. In an ideal, 2-D incompressible irrotational flow field, the fluid is flowing past a wall

with a sink of strength K per unit length at the origin as shown in the Figure. The potential

function of a 2-D sink is 22ln2

YXK

. At infinity the flow is parallel to wall and of

uniform velocity U .

(a) Determine the location of the stagnation point 0X at the wall in terms of U and K.

(b) Find the pressure distribution along the wall as a function of X. Taking the free stream

statitc pressure at infinity to be p , express the pressure coefficient as a function of 0/ XX .

(c) Sketch the resulting pressure distribution.

wall Sink of strength K at X=0

X

U

p Y

3. Superposition of a uniform stream ( V ) and a point source (strength q ) located at the origin

produces the flow over a smooth blunt-nosed body. The blunt-nosed body is usually called

Rankine nose. The radius of the

Rankine nose at Z is

V

qr

(f) To express r

Z as a function

of .

(g) To express the radius of the

Rankine nose (r

r) as a

function of , and plot the

curve of r

rchanging with

r

Z .

(h) Derived the flow velocity (

2

2

V

V) on the surface of the Rankine nose as a function of

, and plot the curve of V

Vchanging with

r

Z .

(i) Derived the pressure coefficient (2

2

1

V

ppC p

) on the surface of the Rankine nose as

a function of , and plot the curve of 2

2

1

V

ppC p

changing with r

Z.

Solution:

The stream function for the combined flow is:

cos4

sin2

22

qRV

The corresponding velocity field is:

sin

4cos

2

VV

R

qVVR

At stagnation point 0;0 VVR

Z

Y

rr

X

i.e., 04

cos2

R

qVVR

and 0sin VV

0sin RVV or0

0

04

cos2

R

qVVR

V

qRstag

4

2 It is impossible

0

4cos

2R

qVVR

V

qRstag

4 Possible solution

Stream line passing the stagnation point:

4cos

4sin

2

22 qqRV

stagnation

Therefore, the equation of the streamline passing the stagnation point is given by:

2cos

2

)1(cos

2

)1(cossin

2

)1(cos

sinsin2

)1(cos

sin

)1(cos

2

sin

)1(cos

2sin

2

)1(cos4

0)1(cos4

sin2

22

22

2

22

r

r

r

R

r

V

q

V

qR

V

q

V

q

R

qRV

Since )2/sin(

cos

2

1)2/cos(

)2/cos()2/sin(2

cos

2

)1(cos

sin

costan/

r

ZrZ

Therefore:

(a). )2/sin(

cos

2

1

r

Z

(b). 2

cos2

)1(cos

r

r

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

0 30 60 90 120 150 180

r/r

Z/r

, deg.

r/r

, Z

/r

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

-2 0 2 4 6 8 10 12

Z/r

r/r

The stream line passing the stagnation point can be treated as the surface of a solid body since no

flow will pass the stream line according to the definition of streamline.

On the surface of Rankin nose:

2sin

)1(cos

2

V

qR

Thus,

sin

)]2/(sin[cos)1(cos2

sincos

4cos 2

2

2

VV

VV

VR

qVVR

Therefore:

)2/(sin)2/(sincos21)2/(sin)2/(sincos2cossin)( 4242222 V

V

)2/(sin)2/(sincos21)( 422 V

V

)2/(sin)2/(sincos2

)]2/(sin)2/(sincos21[1

)(1

2

1

42

42

2

2

V

V

V

ppC p

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

-2 0 2 4 6 8 10 12

Cp

U2/V

2

Z/r

CP

,U2/V

2

4. Consider air flowing past a hemisphere resting on a flat surface, as shown in the Figure.

Neglecting the effects of viscosity, if the internal pressure is ip , find an expression for the

pressure force on the hemisphere. At

what angle location should a hole be

cut in the surface of the hemisphere

so that the net pressure force acting

on the hemisphere will be zero.

Solution:

On the surface of the sphere, the

local pressure is:

)sin4

91()sin

4

91(

2

1 222 qPVPP ;

2

2

1 Vq

The area of a small element on the sphere can be expressed as:

RddRdA sin

The x-component of the force (i.e., along x-direct) acting on the

small element at the outer surface:

ddRqP

ddRPRddRPdAPdFx

cossin)]sin4

91([

cossinsincossincossin

222

22

Therefore, the lift force the due to the external pressure acting on the

surface of the hemisphere is

)16

11()

16

271((]

8

3

4

9

2)[(2

])32

4sin

4

2sin

8

3(

4

9)

4

2sin

2)([(2

]sin4

9sin)[(2

sin)]sin4

91([2

cossin)]sin4

91([

cossin)]sin4

91([

222

00

2

0

4

0

22

0

222

0

2

2

222

0

2

2

222

qPRqPRqqPR

qqPR

dqdqPR

dqPR

ddqPR

ddRqPLoutside

If the pressure inside the semi-sphere shell is insideP , then, the total lift force (i.e., along x-

component) acting on the semi-sphere due to the inner pressure will be insideinside PRL 2

If the net lift acting on the semi-sphere shell is to be zero:

qPPqPRPRLL insideinsideoutsideinside16

11)

16

11(22

On the surface of the sphere, the local pressure is: )sin4

91( 2 qPP

Therefore, at the location where the hole is made will have:

O

O

qPqPP

120

60

2

3sin

4

3sin

16

27sin

4

9

16

11)sin

4

91(

16

11)sin

4

91(

2

22

2