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AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
1
Fundamentals of Inviscid, Incompressible Flow
Dr.ir. M.I. Gerritsma & Dr.ir. B.W. van Oudheusden
Delft University of Technology
Department of Aerospace Engineering
Section Aerodynamics
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
2
Overview
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
3Stationary, inviscid, incompressible flow in the direction of a streamline
The x-component of the momentum equation for a stationary, invisvid, incompressible flow reads
Multiplying this equation by dx gives
Now bear in mind that along a streamline we have
Using this in the momentum equation gives
1u u u pu v w
x y z x
1u u u pu dx v dx w dx dx
x y z x
0
0
u dz wdx
v dx u dy
1u u u pu dx dy dz u du dx
x y z x
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
4
Similar manipulations can also be performed in the y- and z-component of the momentum equation:
Adding these three relations together gives
Stationary, inviscid, incompressible flow in the direction of a streamline
21 1
2
pu du du dx
x
2
2
1 1
2
1 1
2
pdv dy
y
pdw dz
z
2 2 2 2
2
1 1 1 1
2 2
1 1
2
p p pdV d u v w dx dy dz dp
x y z
dV dp
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
5
For an incompressible flow the density is constant so if we integrate these infinitisimal change between two points on the same streamline we get
Stationary, inviscid, incompressible flow in the direction of a streamline
2 2
1 1
2 22 1
2 1
2 21 1 2 2
2 2
1 1
2 2
p V
p V
dp V dV
V Vp p
p V p V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
6
Bernouilli’s Equation
So, although the momentum equation consists of 3 partial differential equations, the equation along a streamline reduces to a single algebraic relation.
The use of ‘Bernoulli’ is limited. Necessary requirements are
• stationary flow
• inviscid, no body forces
• incompressible
• only valid along a streamline!
‘Bernoulli’ is valid for rotational flows also. However, when the flow is irrotational ‘Bernoulli’ is valid everywhere in the flow field and there is no restriction that it is only satisfied along streamlines
21.
2p V const
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
7
Proof Bernoulli for irrotational flows
Home work:
Show that Bernoulli holds everywhere in the flow field (I.e. the constant is the same everywhere) if the flow is irrotational.
In order to prove this either use
• The definition of a irrotational flow
• The following vector identity
a b a b b a b a a b
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
8
Application of Bernoulli
1. Airfoil at se level conditions: Consider an airfoil in a flow at sea level conditions with a freestream velocity of 50 m/s. At a given point on the airfoil, the pressure is equal to . Calculate the velocity at this point.
Solution:
At Standard see level conditions and . Hence
5 20.9 10 N/m
31.23 kg/m 5 21.01 10 N/mp
2 2
522
1 1
2 2
2 2 1.01 0.9 1050
1.23
142.8 m/s
p V p V
p pV V
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
9
Flow through a converging-diverging nozzle
Assumption: We assume that the flow is quasi 1 dimensional, therefore:
Continuity
Bernoulli
Eliminate
, , etc.p p x V V x
1 1 1 2 2 2 1 2 1 1 2 2V A V A V A V A
2 21 1 2 2
1 1
2 2p x V x p x V x
2 122 1 2
1
2
2
1
p pV V
AA
1 1 1, ,A p V 2 2 2, ,A p V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
10
The pressure coefficient for incompressible flow
Bernoulli:
2
:12
p
p p p pC
q V
2
2 2
2
1 11
12 22
p
p p Vp V p V C
VV
2
1p
VC
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
11
Inviscid, incompressible, irrotational flows
• Continuity
• Irrotational
• Laplace equation (insert potential equation in continuity equation)
0 0u v w
Vx y z
0
, , .
w v
y z
u wV
z xv u
x y
V u v wx y z
2 2 2
2 2 20 0
x y z
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
12
Potential Equation
• This equation is also applicable to unsteady flows in which
• By introducing the potential, the irrotationality requirement is identically satisfied
• Every inviscid, incompressible, irrotational flow is described by a potential which satisfies the above potential equation.
• Conversely, every solution of the Laplace equation generates a valid inviscid, incompressible, irrotational flow.
• The Laplace equation is linear, therefore we can use the principle of superposition. So if and are solutions of the Laplace equation, so is . So complicated flow paterns can be obtained by a suitable combination of elementary flows. (Although it is not known in advance how and which elementary flow patterns to combine)
• Once the equation for has been solved the velocity components are obtained from
2 2 2
2 2 20
x y z
1 2 1 1 2 2
, , ,x y z t
, , .u v w
x y z
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
13
Potential Equation
• If the potential flow is stationary (why?) the pressure coefficient is given by
• The Laplace equation in Cylindrical coordinates
• The Laplace equation in Spherical coordinates
22 2
2
2 21 1p
x y zVC
V V
2 2
2 2 2
1 1r
r r r r z
22
1 1sin sin
sin sinr
r r r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
14
Stream function for incompressible flow, 2D
• Stream function
• The velocity components obtained from a stream function automatically satisfy the incompressibility constraint:
•For 2D irrotational flow we have
• Inserting the velocity components obtained from the stream function gives the 2D Laplace equation
, , ,x y t u vy x
0x y y x
0v u
x y
0x x y y
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
15
Boundary conditions• Note: flows in all kinds of different geometries (a sphere, airfoil, cone) are governed by the same equation
• Question: How can one equation generate solutions for so many different flow problems?
• Answer: The difference between the various geometries and flows is the domain in which the Laplace equation has to be solved and the the boundary conditions that are imposed at the boundary of the domain.
• Boundary conditions at infinity: We assume that any disturbances caused by a object placed in the flow have vanished at “infinity”, so we set:
0
, 0 ,
, 0.
x u V v
y u V v
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
16
Boundary conditions at a solid wall• At a solid wall we assume that the flow cannot enter the object, nor will fluid emerge from the object. Since we assume that the flow is inviscid, the fluid is allowed to slide along the solid. If viscous effects are taken into account the friction will prevent the fluid from sliding along the surface of the object (the so-called no-slip condition). In the latter case a so-called boundary layer will develop. However, these viscous phenomena cannot be described by potential equations (why?).
• So at a solid interface the velocity component perpendicular to the surface must be set to zero, i.e.
•In terms of the potential function this condition can be written as
•In terms of the stream function this can be written as (the wall is a streamline!)
0V n
0nn
0 . along streamlineconsts
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
17
Inviscid, incompressible, irrotational flow
Solution Strategy:
• Solve the Laplace equation for or which satisfy the appropriate boundary conditions.
• Determine the velocity components using
• Determine the pressure distribution using Bernoulli
visc
2 2 2
2 2 2
0, ., 0,
0
F const V
x y z
or ,V u vy x
2 21 1
2 2p V p V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
18
Uniform parallel flow
Consider a uniform parallel flow given by
, 0
.
u V v
V const
• Satisfies the continuity condition
• Irrotational, therefore a potential flow
• Stream function
, .0
u Vx
x y V x constv
y
, 0 ,u V v x y V yy x
y
x
.const
.const
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
19
Uniform parallel flowCirculation:
A B
CD
a
0 0 0
0
B C D A
A B C D
V d s
V a V a
Uniform parallel flow, under an angle
V
V
x
y cos
sin
u Vx y
v Vy x
cos sin , cos sinV x y V y x
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
20
Source flowI am looking for a potential flow solution which only depends on r (in polar coordinates), so
inserting this in the two dimensional Laplace equation gives
r
2
2 2
1 10r
r r r r
1 2lnr c r c
So the velocity components are given by:
,
10.
r
cu
r r
ur
y
x
r
ruu
.const
.const
• The volume flow through a circle with radius R is equal to
2 2
is called the source strength and 2
rQ R u R c
QQ c
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
21
Source flow (cont.)If Q>0 the flow is called a source flow (fluid is emanating from the origin) and when Q<0 the flow is called a sink flow (fluid disappears at the origin)
ln2 2
, 02r
Q Qc r r
Qu u
r
y
x
r
ruu
.const
.const
The stream function belonging to this flow can be found by solving
1,
2 2
0 ,2
r
Q Qu r f r
r rQ
u rr
Note that:
. rays .
. circles .
const const
const r const
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
22
Uniform flow + source flowSince the Laplace equation is linear, we are able to add elemenatary flows together. So one may ask what the resulting flow would be if we define a source at the origin in a uniform parallel flow along the x-axis.
Converting y to polar coordinates gives
,2
Qr V y
, sin2
Qr V r
Source Uniform flow
What does this flow look like??
– Calculate the flow field
– Determine stagnation points
– Special streamlines
1cos
2
sin
r
Qu V
r r
u Vr
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
23
* Stagnation points:
* Streamlines
Uniform flow + source flow
0ru u
0, 0cos 0 22
sin 0,
2
r
QQ r Qu V Vr
u V Qr
V
0
A
B
C2
Q
2
Q
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
24
Uniform flow + source flowThe streamline ABC gives the contour of a semi-infinite body. The streamline passes through the stagnation point B, so
So the streamline passing through ABC is given by
The half-width of the semi-infinite body tends to (prove this).
for
: ,2 2
Q QB r
V
1sin .
2 2 2 sin
Q Q QV r r
V
x 2
Q
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
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Uniform parallel flow + source (+Q) + sink (-Q)• Consider a uniform parallel flow in the x-direction
• and a source and a sink placed at a distance 2b from eachother in the x-direction.
Rankine oval
1 2sin2 2
Q QV r
21
bb
1
sintan
cos
r
b r
2
sintan
cos
r
r b
,P r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
26
* Velocity components:
* Stagnation points:
Uniform parallel flow + source (+Q) + sink (-Q)
1 2
1 2
1cos
2
sin2
r
Qu V
r
Qu V
r r r
1 2sin2 2
Q QV r
2
2
: ,
: 0,
QbA r b
V
QbB r b
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
27
Remarks: Because the total strength of the source and the sink (+Q-Q) is equal to zero, a closed streamline through the stagnation points A and B will appear.
All the mass created by the source is consumed by the sink
Since the flow is assumed to be inviscid, the closed streamline can be cosidered as the shape of the Rankine oval placed in a uniform flow. Materializing the inner domain does not effect the outer flow.
Note that the Rankine oval is not an ellips!
Uniform parallel flow + source (+Q) + sink (-Q)
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
28
Doublet flow
Consider the flow of a source and a sink placed at a distance l at either side of the origin.
P
Q Q
12
r
l
1 2,2 2
Q Qr
Now let the distance l shrink to zero and let the source strength Q grow to infinity, such that the product . This will result in a doublet.
Since it follows that
Ql const
doublet 0 0
fixed fixed
lim lim2 2l l
Q
l
0
sinliml l r
doublet
sin
2 r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
29
Using: we find that
• Streamlines:
So the streamlines consist of circles with midpoint (0,1/2c) and radius 2c
• The doublet is oriented in the direction of the x-axis. The tilted doublet may be obtained by rotating the frame of reference. This yields:
Doublet flow1
1
rur r
ur r
doublet
cos
2 r
2 2
2 22
sin
1 1
2 2
const c y c x yr
x yc c
doublet
sin
2 r
doublet
cos
2 r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
30
Uniform flow over a circular cylinder
Now we add together a uniform parallel flow in the x-direction and a doublet at the origin oriented in the x-direction.
Set gives
2
2
coscos cos 1
2 2
sinsin sin 1
2 2
V r V rr V r
V r V rr V r
2
2R
V
2
2
2
2
cos 1
sin 1
RV r
r
RV r
r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
31
Note that for r=R the stream function vanishes identically, therefore the circle with radius R is a streamline. If we ‘materialize’ the region inside the cylinder, we obtain the potential flow solution over a circular cylinder.
• The velocity field
• Stagnation points
Uniform flow over a circular cylinder2
2sin 1
RV r
r
2
2
2
2
1cos 1
sin 1
r
Ru V
r r
Ru V
r r
0
: 0,
: ,
0
r
A B
u u
A r R
B r R
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
32
Flow over a circular cylinder
Streamline , then either
The velocity at the cylinder (r=R) are given by
The pressure coefficient now provides the pressure over the cylinder
0 0 : the positive x-axis
: the circular cylinder
= : the negative x-axis
r R
2
2
2
2
1cos 1 0
sin 1 2 sin
r
Ru V
r r
Ru V V
r r
Was this to be expected??
2
21 1 4sinp
VC
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
33
Source in 3D
In order to find the source in 3D we insert a potential function, which only depends on the radius into the Laplace equation in sperical coordinates
22
1 1sin sin 0
sin sinr
r r r
121 2
12
sinsin
,sin
0
cr c
r r rc
r cr
cr
r
Compare in 2D: lnr c r
Velocity components: 2
1 1, 0 , 0 .
sinr
cu u u
r r r r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
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Source in 3D
The volume flow: 2
2
4 44
and4 4
r
r
R u R c c
ur r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
35
Doublet (dipole) in 3D
3D-source:
Now let with
P1r
rl
x
z
4 r
1
1 1
1 1
4 4
r r
r r rr
r
z
x
y
0l .l const
1
21
cos cos
4
r r l
r r r
3
3
cos
21 sin
41
0sin
rur r
ur r
ur
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
36
Flow over a Sphere
Uniform parallel flow
Addition of the doublet
zV V e cos
sin
0
ru V
u V
u
3 3
3 3
0
coscos cos
2 2
sinsin sin
4 4
0
ru V Vr r
u V Vr r
u
Stagnation points:
3 3
0 0,
02r
u
u r RV
3
2R
V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
37
Flow over a SphereStagnation points: (R,0) and (R,)
Note: For r=R :
Materialize region (sphere)
Parallel flow + doublet = incompressible, inviscid, irrotational, steady flow over a sphere
For r=R :
0ru
r R
3
3
21
3sin sin
4 2V
R
u V VR
2 2
2
31 1 sin
2
91 sin
4
p
p
VC
V
C
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
38Comparison between the flow over a cylinder and a sphere
Cylinder Sphere
2
2
2
2
2
2
1 cos
1 sin
1 4sin
r
p
RV
Ru V
r
Ru V
r
C
3
3
3
3
3
2
2
1 cos
1 sin
91 sin
4
r
p
RV
Ru V
r
Ru V
r
C
Note the different definitions of
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
39
Vortex flowConsider a 2 dimensional potential function
in order to satisfy the Laplace equation
Velocity components:
2
2 2
1 10r
r r r r
1 2c c c
0
1
rur
cu
r r
ruu
Note this flow irrotational! (Why?)
Circulation
2
0
2
V d s
crd c
r
??????
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
40
Vortex flow
Note that unless c=0 the circulation will be non-zero, that means the flow field cannot be irrotational according to Stokes.
However, if c=0 then there will be no flow at all!
How do we resolve this problem??
If we calculate the vorticity, we will find that the vorticity is zero everywhere in the flow field, except at the origin where the
vorticity is infinitely large.
So all contours which enclose will the origin will have a non-zero circulation. In complex function theory where similar phenomena
occur the plane is usually cut to prevent contours around the origin.
2 c
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
41
Vortex flow
Instead of the integration constant c we usually express the strength of the vortex in terms of its circulation.
Note that the velocity is constant along the streamlines and therefore the pressure is constant along the streamlines.
2 c
c
2
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
42
Vortex flow
Calculation of the stream function1
0ln
2
2
rur r
ur r
. rays .
. circles .
const const
const r const
ru
u
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
43
Elementary Potential Flows
Potential Stream function
Uniform flow in
the x-direction
Source flow ln2 2
cos sinDoublet flow
2 2
Vortex flow ln2 2
V x V y
Q Qr
r r
r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
44
Flow over a cylinder with circulation
Consider again the flow over a cylinder, but now we add a vortex at the origin. The stream function for this flow is given by
This can be succinctly written as
sinsin ln ln
2 2 2V r r R
r
Uniform flow Doublet VortexJust a constant
2
2sin 1 ln
2
R rV r
r R
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
45
Flow over a cylinder with circulation
* Velocity field:
* Velocity at the cylinder:
* Stagnation points:
2
2sin 1 ln
2
R rV r
r R
2
2
2
2
1cos 1
sin 12
r
Ru V
r r
Ru V
r r r
10
2 sin2
rur
u Vr R
, sin4
r RRV
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
46
Flow over a cylinder with circulation
* Stagnation points (cont.)
• : Two stagnation points,
• Two stagnation points underneath the cylinder
• One stagnation point at
• Two stagnation points, one outside the cylinder, one inside the cylinder,
Note that for every value of , the resulting flow will be the flow around a cylinder, so
the potential solution allows for infinitely many cylinder flows.
, arcsin4
r RRV
0 0 ,
0 4 RV
4 RV 3,
2r R
4 RV
2
23,
2 4 4r R
V V
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
47
Flow over a cylinder with circulation
The velocity field at the cylinder was found to be
So the pressure distribution over the cylinder is given by
10
2 sin2
rur
u Vr R
22 22
2
2 sin1 1 4sin
2r
p
u uC
V RV RV
2
21 2 sin1 4sin
2 2p p V
RV RV
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
48
Calculation of the Drag
Drag: Only the pressure contributes to the total drag, since body and viscous forces have been neglected.
2
0
2 2
0 0
cos
1 1cos cos
2 1 2 2D p
D pRd
D pC d C d
q R q
0DC Independent of .
Use:2 2 2
2
0 0 0
cos 0, sin cos 0 , sin cos 0d d d
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
49
Calculation of the Lift over the Cylinder
x
y
R
p
Rd
2 2
0 0
22
0
2 1
1sin sin
2
1 2sin
2
L
L p
L
LC
q R
L pRd C C d
C dR V RV
Using the definition of the lift coefficient, we get
L V The Kutta-Joukowski Theorem
Use:2 2 2
3 2
0 0 0
sin 0, sin 0, sind d d
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
50
The Kutta-Joukowski Theorem
This expression is generally valid for 2D shapes in an icompressible, inviscid and irrotational flow. (Proof by means of complex potentials)
In order to describe the flow around an airfoil, one usually employes not one vortex, but a vortex distribution. The sum of the circulation induced by all these individual vortices appears in the KJ-Theorem.
Contour A
Contour B
L,V
A
L V V d s
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
51
Potential Flow around Bodies (overview)
• Combination of elementary flows
• Basic idea: Replace streamlines by solid wall
• Example 1.: Uniform flow + source
• Example 2.: Uniform flow + source + sink (Rankine oval)
• Example 3.: Uniform flow + doublet (flow around cylinder)
• Example 4.: Uniform flow + doublet + vortex (flow over cylinder with lift)
Extension: The approximation of arbitrary shapes by distributed sources on the body of the contour. The panel method.
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
52
Source panel method
• Numerical method for an approximate determination of the flow around bodies of arbitrary shape
• Idea: Distribute sources (and sinks) with a yet undetermined strength along the boundary of the object. Use the boundary condition at the wall of the object to determine the strength of the sources and sinks. Finally, determine the flow of the source distribution in a uniform parallel flow.
•In order to do this we have to introduce the concept of a source sheet, which is a continuous distribution of sources along a contour.
s
ds
sa
b
s is a parameter along the contour and is the source strength per unit length which can be positive (source) or negative (sink)
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
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Source panel method (cont.)
An infinitisimal elemenent ds has a source strength of dsand this induces a potential at the point P equal to
So the total potential induced by the coutour at the point P is given by
Problem: How do we determine the strength (s) such that the desired profile is approximated?
s
ds
sa
b
r ,P x y
ln2
dsd r
ln2
b
a
dsP r
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
54
Source panel method (cont.)Approximate the contour by a finite number of straight line segments and assume that is constant within each segment, so we have a finite number, N, of source strengths i to determine.
Calculate the contribution at an arbitrary point P to the total potential due to one segment.
Add all N contribution to obtain the total potential at a point P.
Place the point P at the midpoint of an arbitrary panel and set the derivative of the potential in the direction of the normal of the panel equal to zero (boundary condition), i.e.
This gives one equation for the N unknowns i . Imposing the boundary conditions at all segments, gives N equations for N unknowns, which in general to a unique solution.
0n
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
55
Source panel method (cont.)
The equation to be solved have the following form:
Remarks: The influence coefficients Iij do not depend on the flow, but only of the geometry of the profile.
Of course, increasing the number of panels, will improve the approximation (higher accuracy).
Modern panel techniques employ curved panels and a non-constant source distribution.
Once the source strengths have been obtained we can calculate the velocity along the panels, using
1
cos ln 02 2
ij
Nji
i ij jj ijj i
I
V r dsn
panel
1
sin ln2
Ni
i ij jij jj i
V V r dss s
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
56
Source panel method (cont.)
Once we have the velocity along the panel I, we can use Bernoulli to obtain the pressure acting on panel I.
For closed contour we should have
panel
1
sin ln2
Ni
i ij jij jj i
V V r dss s
2
panel , 1 i
p i
VC
V
1
0N
i ii
s
AdAerodynamics-B, AE2-115 I, Chapter IIIGerritsma & Van OudheusdenTUD
57
Comparison with real flows
This finalizes chapter 3 (and 6) on potential flows.
I wish you good luck with the two remaining chapters.