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DEN233 Low Speed Aerodynamics:
Basic Concepts and Elementary
FlowsSergey Karabasov
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Part1.1: Basics of inviscid flow
Introductory lectures: descriptive (0.5-1 week) :
Aerofoil characteristics, dimensionless
coefficients, brief introduction on the effects of
Reynolds number & Mach number, boundary
layers, separation, stall. Types of aerofoil forspecific applications. Lift/drag ratio. Maximum lift
coefficient; high lift devices. Control surfaces.
Finite wings: aspect ratio, wing sweep, slender
wings.
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Part1.2: Basic concepts and
elementary flowsElementary flows: (2-2.5 weeks) Fundamentals ofinviscid, incompressible flow: Continuity andBernoulli equations. Definitions of circulation andvorticity. Stokes' theorem. Irrotational flow.Definitions of stream function and velocity
potential with formulation as Laplace's equation.Kelvin's theorem. Elementary flows: uniform flow,source/sink, doublet and vortex. Complex flows bysuperposition including Rankine oval and circularcylinder with and without circulation. Comparisonwith real flow around a circular cylinder. Kutta-Joukowski lift theorem, aerofoil starting vortex.
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Literature for Part 2
Low Speed AerodynamicsLectures by H.P. Horton & R.C.Raichura (available on QMplus)
Fluid DynamicsLectures by S.Nazarenko, Warwick University, 2003
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Concepts used in fluid visualisation
Eulerian (time line)
Lagrangian (path line)
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Streaklines
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Streamlines
For steady flows: Streamline = Streakline
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Stream functions
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Streamlines, Contd
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Velocity components and stream
function
Streamline slope:
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Dynamics of Ideal
FluidsFluid is called Ideal if:
= const
(fluid can be incompressible but is not constant. E.g. stratifiedocean water due to variation intemperature or salinity)
viscosity = 0.
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- Navier-Stokes eqn
Special case for viscosity =0:
-Euler Equation
(ideal flow equation)
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When can viscosity be ignored?
Ratio of the last term on the RHS to the 2nd LHS term:
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Example of ideal flow,
UL/v = Re>>1.
Water: v= 10-6 m2/sec
Air: v= 1.5 10-5 m2/sec
U=1m/sec (walking)
L=1.5m (human)
Re=105 >>1
BUT! High Re flow often
unstable/turbulent reductionof L => less ideal
Re=15,000
IDEAL Laminar
flow with large L
NON-IDEAL
turbulence small L
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Aerodynamic bodies: more
idealattached flow
Non-aerodynamicAerodynamic
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Ideal flow: derivation of continuity eqn
s u = 0.
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Ideal flow: derivation of the Momentum
equation
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Material derivative
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Pressure force and Gravity
Thus we derived the Euler equations for an ideal flow
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Properties of Ideal Flow
VorticityBernoulli Theorem
Potential
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Vorticity: fluid rotation
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Angular velocity in 2D
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Relation with vorticity
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Other form of the Euler equation
s H(x,y,z,t)
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3D Vorticity equations
Take curl of both sides:
Euler equation:
Note:p does not enter intovorticity eqn
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Two-dimensional flow
Vorticity is conserved along fluid paths
E.g. vorticity is a Lagrangian invariant
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2D vortex dynamics
Vorticity moves with the fluid
Different colors correspond to
different values of vorticity
In these examples there are
only 4 values of vorticity at all
times
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Bernoullis thm for steady
irrotationalflow
Definition. The flow is irrotationalif thevorticity =0 in this flow.
For irrotational flow:sH=0
I.e. His the same constant at any
streamline
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Applications of Bernoullis thm
Pipe is horizontal:z1 = z2Same flux (Au=const) => u1 < u2Bernoulli: H=p/ + u2/2 =const =>
p1 > p2
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Flow down a slope
nSlope: z1 > z2
nFree surface =>p1 = p2= patmBernoulli: H=p/ + u2/2 +z =const =>
nu2 > u1
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Lift on an aerofoil
nShape: l1 > l2=> u1 > u2
nBernoulli: H=p/ + u2/2 =const =>
np2 > p1 => Net force is UP (lift)
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Velocity potential
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and
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Relation between velocity and
potential
So:
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Vorticity via potential
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Equipotential lines
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Relation between potential and
streamfunction
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Geometrical interpretation
Wasnt that obvious? Recall
the definition of potential as
a contour integral across the
flow
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Both potential and streamfunction
satisfy Laplace equation
Irrotational flow condition
As we saw previously, from the incompressibility condition:
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Combination of several flows
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Finding the potential and velocity of
irrotational flow
Laplace equation
incompressibility
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Example of Bernoulli for a time
dependent irrotational flow
This is Bernoullis thm for a time dependent irrotational flow
It is useful for findingp via given u, but not for finding u
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Conservation laws for Ideal fluids (rotational &
irrotational)Euler equation:
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Why it is only the kinetic energy and
not the sum of the kinetic and
potentialenergies that is conserved?
Potential energy is conserved too, - due to incompressibility
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Circulation
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Circulation, Contd
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Stokes theorem
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Example
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Stokes theorem example, Contd
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Kelvin circulation theorem
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Proof
Use
Integrating Euler eqn, , we have:
But this change =0 becausep and are single-valued functions.
Thus we proved the Theorem.
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Remarks about Kelvin theorem
1. C denotes a dyedcontour, composed of the same fluid
particles at any time. The result would not be true for afixed in space C.
2. Constant density is not essential: Kelvin established his
result for compressible fluids too.
3. KT does not require Cto be simply connected, i.e. it does
not require Cto be spannable by a surface S lying wholly inthe fluid (see the vortex shedding example).
4. Invicid eqns were used on Conly. If viscosity happens to be
important elsewhere in the flow away from Cthen KT still
holds
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Example: vortex shedding
Viscous only: near aerofoil surface, thinwake, rolled-up starting vortex
Choose C away from these regions, C
remains 0
C = A + B , B = 0A = 0
Cauchy Lagrange thm persistence
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Cauchy-Lagrange thm: persistence
of irrotational motion
Under conditions of KT (ideal flow,conservative forces):
If a portion of the fluid is initially in theirrotational motion, it will remainirrotational at any time
U=const:irrotational
irrotational
irrotational
Not irrotational
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Proof of CL theorem
Prove by contradiction: suppose theflow is irrotational initially but not at alater time for the same portion of
fluid. Then, because of the Gauss f-la
one can find contour Cs.t. circulation is
not 0, but that violates KT
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Remarks about Cauchy-Lagrange
theorem
CLT is obvious in 2Dbecause of the vorticityconservation along the
fluid paths
It is not obvious in 3D
because of the vortex
stretching
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2D ideal flow
Consider the 2D vorticityequation
Define a streamfunction y
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Elementary flows
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Elementary flows
(a)
(b)
(c)
(d)
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Uniform flow (u,0)
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Point sink or source
Mass flow rate through a closed surface = constant (+ve = source, -ve=sink)
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Point source, Contd
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Contd
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Point source/sink not at the origin
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Irrotational vortex
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Irrotational vortex, contd
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Circulation along the path including
the centreStokes thm:
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For case 5b the circulation is zero
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Strength of the vortex
0
For several vortices:
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Streamfunction and velocity potential
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Vortex: contd
R ki t
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Rankine vortex
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Point vortex
Irrotational flow everywhereexcept for r=0
V t di l
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Vortex dipole
Vorticity moves withthe fluid
Each vortex is movedby velocity inducedby another vortex atits location
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Dipole example: wingtip vortices
RAF Tornado
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A gasof point vortices
Each vortex is moved by the Vwhich is a vector sum of Vs
produced by all other vortices atits location
Bio-Savart
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Example: Karman vortex street
Cloud pattern behind an island (satelliteimage)
Laboratory experiment: wakebehind an obstacle
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Source-sink pair
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Contd
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Calculating the velocity
and
There is no stagnation point in this flow!
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The doublet
St f ti
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=
Streamfunction
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Potential
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Overall Strategy for Plotting Streamlines from
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Stream Function
Write down stream function for flow by appropriately combining individual
solutions for source, sink and line vortex as a sum:
( ) ( ) ( ) ( ) KK+++= yxyxyxyx ,,,, 321 yyyy
Calculate expressions for vel. components u, vfrom
xv
yu
-=
=
yy,
Note: Huge choice as far as selction of parameters is concerned! Souce strength, vortexdirection of rotation, strength ...
Determine coord.of stagnation point(s) via u=0 , v=0.
Determine value of stream function passing through (stagnation) point by
substituting coordinates of (stagnation) point(s) into the stream function.
Set stream function equal to the value you have determined for point in
question.
Determine values of x, y (or r, ) that satisfy this expression and plot to
obtain streamline.
Choose new pointx,y
q
Obvious question now is what happens for ...
Uniform Flow + Source + Sink
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Uniform Flow + Source + Sink
We consider symmetric case where:
Source
Sink
Strength Location
m
m-
( )0,c-
( )0,cUsing superposition, can readily write stream function for this flow:
( )
--
++= --
cx
ym
cx
ymyUyx
11tantan,y
{
flowUniform
44 344 21
)0,(atSource c-44 344 21
)0,(atSink c
Second and third terms can be combined using:
( ) ( )
+
-=- ---
ba
baba
1tantantan 111
To give a more concise form for stream function
( )
-+-= - 222
1 2tan,cyx
ycmyUyxy
(1)
Continued...
From either of the two forms of S F on previous slide one can determine velocity components
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Now find stagnation points, where u=v=0. From Eq. (3) one sees that when y=0 then v=0.
( ) ( )
+-
--
++
++=
= 2222 ycx
cx
ycx
cxmU
yu
y
( ) ( )
+-
-
++
=
-=
2222
11
ycxycx
ym
x
v y
From either of the two forms of S.F. on previous slide, one can determine velocity components
(2)
(3)
Substitute y=0 into Eq. (2) and then find value of x which gives that u=0.
After some manipulation the solutions forxare:
LUcmcx =
+=
2
1
21
Hence, stagnation points at:
( )0,Land( )0,L-
Now determine value of S.F. for surface streamline from Eq (1).
( )
--
++= --
cx
ymcx
ymyUyx
11
tantan,y (1) - repeated
It can be seen that this is trivial and that 0=Sy
Continued...
Rankine Oval then looks like
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Rankine Oval then looks like ...
We already determined value ofL. Can
find points of maximum velocity and
minimum pressure at shoulders +/-h, of
oval using similar methods. All these
parameters are a function of the...
2
1
21
+=
cU
m
c
L ( )22max121
chcUm
Uu
++=
cU
md
=
In summary one obtains
As one increases dimensionless parameter d from
zero to large values, oval shape increases in size and
thickness from flat plate of length 2c to huge, nearly
circularcylinder. Here think of increase when
All Rankine ovals, except very thin ones, have large
adverse pressure gradient on leeward surface. Thus,
boundary-layer will separate in rear, broad wake
flow develops, inviscid pattern unrealistic in that
region.
constUandconstc == .
basic dimensionless parameter
=
cUm
ahch
2cot
=Uma
Incompressible, inviscid, two-
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dimensional flow over a cylinder:
uniform flow + doublet
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Contd
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Stagnation streamline
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Contd
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Pressure variation on the cylinder
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Contd
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Re=
Resolution of the DAlembert paradox
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Resolution of the D Alembert paradox
Drag on the bluff bodies is finitebecause the flow fails to be idealbehind them due to the flowseparation
No separation
minimal drag
aerodynamic shape
Turbulent re attachment
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Drag crisisTurbulent re-attachmentleads to drag reduction
Uniform flow + doublet +vortex (lifting
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cylinder)
d
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Contd
l d b
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Velocity distribution
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Case r=R
C d
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Contd
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Case: /2 3/2
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Case: =/2, 3/2
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C td
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Contd
P di t ib ti d f
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Pressure distribution and forces
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Symmetry Force calculation
Lift
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Lift
C td
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Contd
K tta J ko ski
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Kutta-Jukowski
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Zhukovski lift theorem
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