Potential FlowChapter Three

Dr. Hamdy A. Kandil

PART TWO

Stokes’s Theorem The transformation from a line integral to a surface integral in

three-dimensional space is governed by Stokes’s theorem:

where n dA is a vector normal to the surface, positive when pointing outward from the enclosed volume, and equal in magnitude to the incremental surface area.

In words, the integral of the normal component of the curl of the velocity vector over any surface A is equal to the line integral of the tangential component of the velocity around the curve C which bounds A.

Stokes’s theorem is valid when A represents a simply connected region in which V is continuously differentiable.

Thus, it is not valid if the area A contains regions where the velocity is infinite.

Potential Vortex The curl of the velocity vector for the potential vortex can be

found using the definition for the curl of V in cylindrical coordinates

But and

which simplifies in two dimensions to:

Although the flow is irrotational ( = 0), we must remember that the velocity is infinite at the origin (i.e., when r = 0 ).

In fact, the flow field at the origin is rotational and vorticity exists there.

We will now calculate the circulation around a closed curve C1which encloses the origin. We can choose a circle of radius r1

The circulation is

Recall that Stokes’s theorem, is not valid if the region contains points where the velocity is infinite, which is true for vortex flow at the origin.

However, if we calculate the circulation around a closed curve C2, which does not enclose the origin, such as that shown in Fig. b, we find that

Therefore, the circulation around a closed curve not containing the origin is zero.

Paths for the calculation of the circulation for a potential vortex: (a) closed curve C1, which encloses origin; (b) closed curve C2, which does

not enclose the origin.

Shape of the free surface

2

2

2

p Vgz const

2 21 2

2 2

V Vz

g g

at the free surface p=0:

2

2 28z

r g

Bernolli’s equation

&

V1 = 0&

Elementary Planar Irrotational FlowsLine Vortex

If vortex is moved to (x,y) = (a,b)

Source and Sink Consider a source of strength K at (-a, 0) and a sink of K at (a, 0) For a point P with polar coordinate of (r, ). If the polar coordinate

from (-a,0) to P is (r2, 2) and from (a, 0) to P is (r1, 1), Then the stream function and potential function obtained by

superposition are given by:

12

12

lnln2

, 2

rrK

K

Source and Sink

Hence,

Since

We have

We have

12

1212 tantan1

tantantan

2tan

K

22

sin22tan

ar

ar

K

ar

r

ar

r

cos

sintan

cos

sintan 12 and

22

1- sin2tan

2 ar

arK

)

Source and Sink

We have

Therefore,

The velocity component are:

cos2cossin

cos2cossin22222

1

222222

arararrr

arararrr

cos2

cos2ln

2 22

22

arar

ararK

=

sin2

sin

sin2

sin

2

cos2

cos

cos2

cos

2

2222

2222

arar

r

arar

rKv

arar

ar

arar

arKvr

Doublet The doublet occurs when a source and a sink of the same

strength are collocated the same location, say at the origin. This can be obtained by placing a source at (-a,0) and a sink of

equal strength at (a,0) and then letting a 0, and K , with Ka kept constant, say aK/2=B

For source of K at (-a,0) and sink of K at (a,0)

and sin2

tan2 22

1-

ar

arK

cos2

cos2ln

2 22

22

arar

ararK

Under these limiting conditions of a 0, K , we have

cos2

cos2

cos2lnlim

sin2sin2tanlim

22

22

0a

221-

0a

r

a

arar

arar

r

a

ar

ar

Doublet (Summary)

Adding 1 and 2 together, performing some algebra, Therefore, as a0 and K with aK/2=B then:

B is the doublet strengthThe velocity components for a doublet may be found the same way we found them for the source

&

Examples of Irrotational Flows Formed by SuperpositionSuperposition of sink and vortex : bathtub vortex

Superposition of sink and vortex : bathtub vortex

Sink Vortex

Superposition of Source and Uniform Flow Assuming the uniform flow U is in x-direction and the source of K

stregth at(0,0), the potential and stream functions of the superposed potential flow become:

&

&

Source in Uniform Stream The velocity components are:

A stagnation point (vr=v=0) occurs at

Therefore, the streamline passing through the stagnation point when

The maximum height of the curve is

sin2

cos

Ur

vr

KU

rvr and

22

KUr

U

Kr ss

and

UrK

ss 2

2

Ks

and as rU

Krh 0

2sin

Source in Uniform Stream

2m

ψ 2m

ψ

0ψStag. point

Superposition of basic flows Streamlines created by

injecting dye in steadily flowing water show a uniform flow.

Source flow is created by injecting water through a small hole.

It is observed that for this combination the streamline passing through the stagnation point could be replaced by a solid boundary which resembles a streamlined body in a uniform flow.

The body is open at the downstream end and is thus called a halfbody.

Rankine Ovals The 2D Rankine ovals are the results of the superposition of equal

strength (K) sink and source at x=a and –a with a uniform flow in x-direction.

Rankine Ovals Equivalently,

The velocity components are given by:

The stagnation points occur atwhere V = 0 with corresponding s = 0

221

22

22

sin2tan

2sin

cos2

cos2ln

2cos

ar

raKrU

raar

raarKrU

sin2

sin

sin2

sin

2

cos2

cos

cos2

cos

2

2222

2222

arar

r

raar

rK

rv

arar

ar

raar

arK

rvr

0

12

1

2

1

2

s

ss

y

aU

K

a

xa

U

Kax

i.e., ,

Rankine Ovals The maximum height of the Rankine oval is located at

when = s = 0 ,i.e.,

which can only be solved numerically.

20

,r

a

r

K

aU

a

r

a

r

ar

arKrU

o0

2tan1

2

1

02

tan2

2

0

220

010

or

Flow around a Cylinder: Steady Cylinder Flow around a steady circular cylinder is the limiting case of a

Rankine oval when a0. This becomes the superposition of a uniform parallel flow with a

doublet in x-direction. Under this limit and with B = a.K /2 =constant, the radius of

the cylinder is:.2

1

U

BrR s

The stream function and velocity potential become:

The radial and circumferential velocities are:

sin1sin

sin

cos1cos

cos

2

2

2

2

r

RrU

r

BrU

r

RrU

r

BrU

and

sin1 cos12

2

2

2

r

RU

rrv

r

RU

rrvr and

Flow around a Cylinder: Steady Cylinder

Steady Cylinder

On the cylinder surface (r = R)

Normal velocity (vr) is zero, Tangential velocity (v) is non-zero slip condition.

sin2 0 Uvvr and

Pressure Distribution on a Circular Cylinder Using the irrotational flow approximation, we can calculate and plot

the non-dimensional static pressure distribution on the surface of a circular cylinder of radius R in a uniform stream of speed U .

The pressure far away from the cylinder is p Pressure coefficient:

Since the flow in the region of interest is irrotational, we use the Bernoulli equation to calculate the pressure anywhere in the flow field. Ignoring the effects of gravity

Bernoulli’s equation:

Rearranging Cp Eq. , we get

2

21

U

ppCp

2tan

2

22

Uptcons

Vp

2

2

21

21

U

V

U

ppCp

Pressure Distribution on a Circular Cylinder

We substitute our expression for tangential velocity on the cylinder surface, since along the surface V2 = v2

; the Eq. becomes

In terms of angle , defined from the front of the body, we use the transformation = - to obtain Cp in terms of angle :

We plot the pressure coefficient on the top half of the cylinder as a function of angle , solid blue curve.

22

22

sin41)sin2(

1

U

UCp

2sin41pC

Pressure distribution on a fish Somewhere between the front stagnation point and the

aerodynamic shoulder is a point on the body surface where the speed just above the body is equal to V, the pressure P is equal to P , and Cp = 0. This point is called the zero pressure point

At this point, the pressure acting normal to the body surface is the same (P = P), regardless of how fast the body moves

through the fluid. This fact is a factor in the location of fish eyes .

If a fish’s eye were located closer to its nose, the eye would experience an increase in water pressure as the fish swims—the faster it would swim, the higher the water pressure on its eye would be. This would cause the soft eyeball to distort, affecting the fish’s vision. Likewise, if the eye were located farther back, near the aerodynamic shoulder, the eye would experience a relative suction pressure when the fish would swim, again distorting its eyeball and blurring its vision.

Experiments have revealed that the fish’s eye is instead located very close to the zero-pressure point where P = P , and the fish can swim at any speed without distorting its vision.

Incidentally, the back of the gills is located near the aerodynamic shoulder so that the suction pressure there helps the fish to “exhale.”

The heart is also located near this lowest pressure point to increase the heart’s stroke volume during rapid swimming.

Pressure distribution on a fish

Rotating Cylinder The potential flows for a rotating cylinder is the free vortex flow. Therefore, the potential flow of a uniform parallel flow past a

rotating cylinder at high Reynolds number is the superposition of a uniform parallel flow, a doublet and free vortex.

Hence, the stream function and the velocity potential are given by

The radial and circumferential velocities are given by

rr

RrU

r

RrU

ln2

sin1

2cos1

2

2

2

2

cos1

2

2

r

RU

rrvr

rr

RU

rrv

2sin1

2

2

Rotating Cylinder The stagnation points occur at

From

0 ssr vv

0cos1 2

2

s

sr

RU 0srv

0cos ss Rr :B Case OR :A Case

2

12

22

41 &

4sin

14

02

sin2 :

RURyRx

RURy

RU

RUvRr

ssss

sss

:A Case

when exits only Solution

Rotating Cylinder

sr real positivefor implies which

with sign :B Case

14

2

12

2

2

144

02

1 0

1sin0cos

RU

RURUR

r

Rr

RUv

s

s

ss

Rotating Cylinder The stagnation points occur at

Case 1:

Case 2:

Case 3:

14

RU

14

RU

14

RU

Rotating Cylinder

Case 1:

2

12

41

4

RUR

x

RUR

y ss

and

14

RU

Rotating Cylinder

Case 2:

The two stagnation points merge to one at cylinder surface where . Ryx ss ,0,

14

RU

Rotating Cylinder Case 3:

The stagnation point occurs outside the cylinder when where .

The condition of leads to

Therefore, as , we have

ss ry 0v

2

2

12

0

144

RUUrR

r

R

y ss

14

RU1

2

RUR

ys

14

RU

Rotating Cylinder Case 3:

14

RU

Stagnation points locations around a

cylinder

Lift Force The force per unit length of cylinder due to pressure on the

cylinder surface can be obtained by integrating the surface pressure around the cylinder.

The tangential velocity along the cylinder surface is obtained by letting r = R,

The surface pressure p0 as obtained from Bernoulli equation is

where p is the pressure at far away from the cylinder (free stream)

0

0 2sin2

RU

rv

Rr

2

22

sin2 2

2

0

Up

RU

p

Lift Force Hence,

The force due to pressure in x and y directions are then obtained by

222

22

2

04

sin2

sin412

URRU

Upp

] sin cos[ 000 jisjiF dRpdRpdpFFCCyx

URdpF

RdpFD

y

x

2

0 0

2

0 0

sin

0cos:arg

:Lift and

ji drd o sincos swhere

Lift Force The development of the lift on rotating bodies is called the

Magnus effect. It is clear that the lift force is due to circulation around the

body. An airfoil without rotation can develop a circulation around the

airfoil when Kutta condition is satisfied at the rear tip of the air foil.

Therefore, The tangential velocity along the cylinder surface is obtained by letting r = R

This forms the base of aerodynamic theory of airplane. Magnus Effect: The Magnus effect was first described by (and thusly named

after) Heinrich Magnus in 1852. Magnus discovered that a rotating cylinder experiences a force,

when held into a streaming fluid. The force is perpendicular to the direction of the streaming fluid and the axis of rotation.

Magnus Effect

Applications of Magnus effect

Funny soccer joke ~Reporter: How does understanding the laws of motion help with your game?Roberto: There are laws?

How Carlos scored an impossible goal?

The harder you kick a ball the more curve it will experience.

Curving ball is use to be trick goalies and to score amazing goals like this one from 35m out

48

APPLICATION: BASEBALL PITCH

49

EXAMPLES• Pitch: Overhand curveball

• Pitch: Split-Finger Fastball

– MLB Speed: 85-90 MPH

– 1300 RPM (10 Revolutions)

50

FLETTNER ROTOR SHIPLength: 100 ftDisplacement: 800 tonsRotors: 50 ft high, 9ft diameter

51

FLETTNER SHIP

Flettner rotor ship in NYC harbor, May 9, 1926 Since power to propel a ship varies as cube of its speed, 50 hp used

for this auxiliary propulsion system represented a large increase in fuel efficiency

FLETTNER ROTOR SHIP: EXAMPLE

Flettner Rotor Ship Data: Approximately 100 ft long, displaced 800 tons and wetted

area of 3,500 ft2

Two rotors each 50 ft tall and 9 ft diameter rotating at approximately 750 RPM

Measured ‘lift’ coefficient was 10 and measured ‘drag’ coefficient was 4

Water drag resistance coefficient of boat CD = 0.005

E-Ship 1 The E-Ship 1 is a RoLo cargo ship that made its first voyage with

cargo in August 2010. The ship is owned by the third-largest wind turbine

manufacturer, Germany's Enercon GmbH. It is used to transport wind turbine components.

Wind Power Station Utilizing Lift of a Rotating Cylinder

Experimental Wind Power Station Prototype from Russian was not possible to put it to practical

use. Spiral column was invented (national patent). April 2005 machine experiment of new Magnus Windmill 5 m

plant was completed - a real proof experiment begins for various data collections.

Wind Power Station Utilizing Lift of a Rotating Cylinder

Flettner Rotorflugzeug

Roman Fischer -Flettner Rotorflugzeug span: 150 cm , 4,3 kg, Electro Graduation project about the Magnus effecf

Gyrocopter

Gyrocopter

Fan Wing