AerodynamicsB, AE2115 I, Chapter III

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

2
Overview
3
Stationary, 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

4
Stationary, inviscid, incompressible flow in the
direction of a streamline

• Similar manipulations can also be performed in
  the y- and z-component of the momentum equation
• Adding these three relations together gives

5
Stationary, inviscid, incompressible flow in the
direction of a streamline

• For an incompressible flow the density is
  constant so if we integrate these infinitisimal
  change between two points on the same streamline
  we get

6
Bernouillis 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

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

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

9
Flow through a converging-diverging nozzle

• Assumption We assume that the flow is quasi 1
  dimensional, therefore
• Continuity
• Bernoulli
• Eliminate

10
The pressure coefficient for incompressible flow

• Bernoulli

11
Inviscid, incompressible, irrotational flows

• Continuity
• Irrotational
• Laplace equation (insert potential equation in
  continuity equation)

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

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

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

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

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!)

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

18
Uniform parallel flow

• Consider a uniform parallel flow given by

• Satisfies the continuity condition
• Irrotational, therefore a potential flow
• Stream function

19
Uniform parallel flow
a
C
D

• Circulation

B
A

Uniform parallel flow, under an angle a

20
Source flow

• I 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

• The volume flow through a circle with radius R
  is equal to

So the velocity components are given by
21
Source flow (cont.)

• If Qgt0 the flow is called a source flow (fluid is
  emanating from the origin) and when Qlt0 the flow
  is called a sink flow (fluid disappears at the
  origin)

The stream function belonging to this flow can be
found by solving

Note that
22
Uniform flow source flow

• Since 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

Uniform flow
Source

• What does this flow look like??
• Calculate the flow field
• Determine stagnation points
• Special streamlines

23
Uniform flow source flow

• Stagnation points
• Streamlines

A

B
C
24
Uniform flow source flow

• The 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

25
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

26
Uniform parallel flow source (Q) sink (-Q)

• Velocity components
• Stagnation points

27
Uniform parallel flow source (Q) sink (-Q)

• 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!

28
Doublet flow

• Consider the flow of a source and a sink placed
  at a distance l at either side of the origin.

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

29
Doublet flow

• 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

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

31
Uniform flow over a circular cylinder

• Note that for rR 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

32
Flow over a circular cylinder

• Streamline , then either
• The velocity at the cylinder (rR) are given by
• The pressure coefficient now provides the
  pressure over the cylinder

Was this to be expected??
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

Compare in 2D

Velocity components
34
Source in 3D

• The volume flow

35
Doublet (dipole) in 3D

• 3D-source
• Now let with

36
Flow over a Sphere

• Uniform parallel flow
• Addition of the doublet

Stagnation points
37
Flow over a Sphere

• Stagnation points (R,0) and (R,p)
• Note For rR
• Materialize region (sphere)
• Parallel flow doublet incompressible,
  inviscid, irrotational, steady flow over a sphere
• For rR

38
Comparison between the flow over a cylinder and a
sphere

• Cylinder
  Sphere

Note the different definitions of q
39
Vortex flow

• Consider a 2 dimensional potential function
• in order to satisfy the Laplace equation
• Velocity components

Circulation
??????
Note this flow irrotational! (Why?)
40
Vortex flow

• Note that unless c0 the circulation will be
  non-zero, that means the flow field cannot be
  irrotational according to Stokes.
• However, if c0 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.

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.

42
Vortex flow

• Calculation of the stream function

43
Elementary Potential Flows
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

Just a constant
Uniform flow
Doublet
Vortex
45
Flow over a cylinder with circulation

• Velocity field
• Velocity at the cylinder
• Stagnation points

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 G, the resulting
  flow will be the flow around a cylinder, so the
  potential solution allows for infinitely many
  cylinder flows.

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

48
Calculation of the Drag

• Drag Only the pressure contributes to the total
  drag, since body and viscous forces have been
  neglected.

Independent of G. Use
49
Calculation of the Lift over the Cylinder
Using the definition of the lift coefficient, we
get
The Kutta-Joukowski Theorem
Use
50
The Kutta-Joukowski Theorem
Contour B
Contour A

• 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.

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.

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 is a parameter along the contour and l is the
source strength per unit length which can be
positive (source) or negative (sink)
53
Source panel method (cont.)

• An infinitisimal elemenent ds has a source
  strength of ldsand 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 l(s)
  such that the desired profile is approximated?

54
Source panel method (cont.)

• Approximate the contour by a finite number of
  straight line segments and assume that l is
  constant within each segment, so we have a finite
  number, N, of source strengths li 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 li .
  Imposing the boundary conditions at all segments,
  gives N equations for N unknowns, which in
  general to a unique solution.

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

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

57
Comparison with real flows

• This finalizes chapter 3 (and 6) on potential
  flows.
• I wish you good luck with the two remaining
  chapters.