DEN233 Low Speed Aerodynamics-week2-3

5/20/2018 DEN233 Low Speed Aerodynamics-week2-3

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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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DEN233 Low Speed Aerodynamics-week2-3

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