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
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