Lecture 17 Final07

7/28/2019 Lecture 17 Final07

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Lecture 17Final Version

Contents Lift on an airfoil

Dimensional Analysis

Dimensional Homogeneity Drag on a Sphere / Stokes Law

Self Similarity

Dimensionless Drag / Drag

Coefficient


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Recall : Cylinder with Circulation in a Uniform Flow

Without performing calculation, can see that a uniform flow around a fix cylinder

gives no net lift or drag on cylinder since pressure distribution on surface is

symmetric aboutx- andy-axis..

2 KpG =

Note that this does not violate the flow around cylinder: line vortex produces a u

component of velocity only. Hence, we are still adhering to condition that flowcannot pass through cylinder boundary.

Working from S.F. for cylinder in uniform flow additional inclusion of line vortex

gives:

CrKr

rUr lnsin

sin,

originatDoublet

flowUniform

originatvortexLine

constantArbitrary

Use result that radius

of resulting cylinder is : And set :

UR

RKC ln

(1)

(1) R

rK

r

RrU lnsin

2

Velocity

Components

2

2

1cos1

r

RU

rur

2

2

R Ku U sin 1

r r rq

yq

= - = - + +

In orderto generate lift need tobreak symmetry. Achieved by introducing

line vortex of strength,K, at origin which introduces circulation .


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Continued...

So, on surface (r=R), velocity components are:

0ruR

KUu sin2

Surface Stagnation points also need: 0u

UR

K

2sin

Note: By setting vortex strength zero (K=0), recover flow over cylinder in

uniform flow with stagnation points at ,0

Plotting, Choose value forK, Now first get value of S.F. forr=R,... then

set S.F. equal to that value, then compile table r vs. angle This gives

particular streamline through stagnation points.

Then choose any other point in flow field not on stagnation streamline,

determine value of S.F. for this point, set S.F. equal to that value, then

compile table r vs. angle This gives streamline through the chosen particular

points Then choose another point in flow field etc (compare flow chart

from beginning of lecture). For various values ofKthe following, flow fields

emerge...

0K 1K

2K 3K


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Pressure Distribution Around the Cylinder

To evaluate press. on cyl. surface use Bernoulli Eq. along S.L. that

originates far upstream where flow is undisturbed. Ignoring grav.

forces:22

2

1

2

1SS UpUp

flowdundisturbe

Upstream

surfacecylinderOn

Substituting for surface flow speed : with

Ku 2U sin

Rq

q

= - +

Ru 0 ,=

222 sin

4sin41

2

1

UR

K

UR

KUppS

difference in pressures between surface and undisturbed free stream

(1)

In particular for non-rotating cylinder

where K=0:

22 sin4121 UppS

(2)

2

2

sin41

2

1

U

ppC Sp

f.: Pressure Coefficient Only top half of cyl. shown.

2 2 2

S rU u uq

= +


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Continued...

2

2

2sin

4sin41

2

1 UR

K

UR

K

U

ppC Sp

Qualitative behaviour of

for various values of .RUK

Best way of interpreting above graphs is to think of flow velocity and radius being constant

while vortex strength is increasing from one plot to next.

RUK When plotting graphs I did not explicitly specify velocity or radius! I simply used different

numeric values for in order to illustrate behaviour of graph. I have not considered

if any of these cases may not be realizable in reality or not!.

,cyl.ofTop:57.12,cyl.ofRear:0( )cyl.ofBott.:71.423,cyl.ofFront:14.3


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Continued...

Equation (1)

can be used to calculate net lift and drag acting on cylinder!

2

22 sin4

sin412

1

UR

K

UR

KUppS

Sketch (A) Sketch (B)

In Sketch (b) ...

sinsin pppL S coscos pppD S

Hence, integration around cylindersurface yields total L and D ...

2

0 sin dRbppL S

2

0 cos dRbppD S

where b is width (into paper) of cylinder. Substituting for pressure using

Eq. (1), and integrating (most terms drop out), leads to following results:

bKU

RbUR

KUL

2

4

2

1 2

0D

Or, lift per unit width:

UKUb

L 2

Thus, drag zero

a remarkable result!

TheoremLift

JoukowskiKutta

Paradox

sAlembert'd'


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Continued...

Net lift is indicated in sketch below. ... Note that if a line vortex is used which

rotates in mathematically positive sense (anti-clockwise) then resulting lift is

negative, i.e. downwards.

Ub

L U

L

Final notes:How is lift generated? ... From sketch above and from

pressure profiles plotted earlier it is evident how this is physically

achievedBreaking of the flow symmetry inx-axis means that flow

round lower part of cylinder is faster than round top - this means that

pressure is lower round bottom and so a net downward force results.

Notice that symmetry in y-axis is retained symmetry of pressure

on left-hand and right-hand faces is retained and so there is no netdrag force. Keep in mind that our analysis was for an ideal fluid (i.e.

there is no viscosity). In a real flow would fore-aft symmetry be

retained?

Lastly, since lift is proportional to circulation, we wish to make

circulation large to generate a large lifting force. In applications ofabove flow this is achieved by spinning cylinder to produce large

vorticity but is there a limit to how much circulation we should

produce?

------------------------End of Recall--------------------


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Circulation and Lift for Aerofoil Applications

If a thin symmetric aerofoil is placed at zero incidence in an inviscid,

irrotational and imcompressible uniform flow, the flow pattern shown

in Fig. (1) below ensues. There is no circulation and the aerofoil does

not generate lift. (This case is analogous to the cylinder with )0

Fig. (1)

In case of cylinder, can generate vorticity by spinning cylinder. For

airfoil section this can be achieved by setting it at incidence or by

using a non-symmetric shape (which shape to get lift? and to get

negative lift?). Placed at incidence, flow past a thin symmetric

aerofoil is shown in Fig. (2).

Fig. (2)

Clearly airfoil experiences upward force - compare flow speeds on

upper and lower surfaces. We have seen that this type offlow speed

differential can be modelled by using line vortices which yield

circulation and hence lift. (In Fig. (2) line vortices would have

negativeKto give clockwise velocity contribution.)


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Continued...

If we wish to calculate lift per unit span on aerofoil section using

Kutta-Joukowski Lift Theorem ...

Ub

L

then need to know value of circulation for a given aerofoil at a

specific flow speed and for particular angle of incidence. Key to

finding unique value of circulation lies in modelling flow at trailing

edge of aerofoil. ConsiderFig. (3a-c) below ...

nCirculatio:SpeedFlow:AirDensity: U

Evidently the correct valueKutta

Fig. (3)

The KUTTA CONDITION ...

has been used in Fig. (3c).

states that flow from upper and lower surfaces must leave trailing

edge with same speed. The Kutta condition thus determines correct

value of circulation when performing a calculation of flow around a

lifting aerofoil.


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Continued...

Increasing either

SpeedFlow:orIncidenceofAngle: U

increases Kutta and hence the lift.

Is there a limit to how large one can make angle of incidence and

hence ?Kutta

For a flat plate with the lift experienced isincidence

UcUb

L

where c is length of plate.Non-dimensional lift coefficient given by

2

2

1 2

cbU

LCL

This result (as with calculations for aerofoils) is achieved by using a

line of vortices - a vortex sheet - within aerofoil section to generate

circulation, rather than a single line vortex as used for cylinders in our

earlier considerations.


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Continued...

Qualitative comparison of pressure coefficient for NACA 0012 airfoil at incidence

with one of the earlier graphs for pressure coefficient of a rotating cylinder.

Upper wing surface

Lower wing surface

Lower cyl. hal

Upper cyl. hal

Note: In order to get negative pressure coefficient on top half of

cylinder and (i.e. upward lift) need to reverse sense of rotation of line

vortex used in example for flow around rotating cylinder.

(Comparison included here to highlight where corresponding points / regions are

and to practice how to read such graphs.)

NACA

airfoil

Rotating

cylinder


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Dimensional Analysis and Model Testing

Introduction to Dimensional Analysis

Consider drag D of sphere .

On what quantities does it depend?

Flow Speed,

Fluid Density,

Fluid Viscosity,

Vd

,,,VdFD

Diameter,

Write

What does the above mean in terms of the

measurements we have to carry out to collect

data for all possible spheres in all types of fluids?

(1)

Note: Eq.(1) reads Drag, D, is a Function of ...


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Continued ...

WE NEED ...

1 page for Drag as

function of 2 variables

(e.g. velocity and

diameter)

dincreases

from curve

to curve

1 book for Drag as

function of 3 variables

(e.g. velocity, diameter,

density)

1 page for

each valueof

Shelf of books for Drag as afunction of 4 variables

(velocity, diameter, density,

viscosity)

If we want 10 data points per curve, at 10 each

experiment, this will cost...

000,1001010101010

THERE MUST BE

A BETTER WAY !?!?


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Continued...

GOAL IS TO COMPRESS SHELF

OF BOOKS INTO ONE SINGLE

GRAPH...

DRAG

4 Independent

Experimental Parameters

How Could We Possibly

Achieve This?


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Dimensional Analysis for Re


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Continued...

So, we are restricted to flow with..

HIGH RE NUMBERT

U

R

B

U

L

E

N

T

LOW RE NUMBERL

A

M

I

N

A

R

Restrictions exclude ...


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Continued...

,,VdFD The expression Eq.(2) ...

(2) - repeated

represents a VERY general statement!!!

CRUCIAL NEXT STEP:

Ensure that function Fhas such a form that

one ends up with same dimensions

on both sides of equal sign.

Hence, we may NOT choose a function that produces anon-sense statement where units are for instance ...

132 VdD

Units:2s

mkgUnits:

4

4

3

32

s

mkg

ms

kg

s

mm

QUESTION:

How Do I Have To Choose Exponents

Such That Units AreThe Same on

Both Sides Of Equation?


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Continued...

Answer question by determining conditions for

exponents under which one gets same units on both

sides of equation ...

,,VdFD (2) - repeated

Units:

Dimensions:

2

s

mkgN m

s

m

sm

kg

2TLM L1TL 11 TLM

Mass:M Length:L Time:T

WANT !!!

dVD

,,

such that

gives

2TLM 2TLM

(3a-c)

(4)

Find by subst. Eq.(3a-c) into Eq.(4) ...


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Continued...

dVD (4) - repeated

1TL L11 TML2TML

Collect corresponding terms ...

TLMTLM 2

By comparing exponents ...

ofMon left and right hand side of Eq.(5)

(5)

1 (6a)

ofL on left and right hand side of Eq.(5)

1 (6b)

ofTon left and right hand side of Eq.(5)

2 (6c)

Substitut Eq.(6a) into Eq.(6c) ...

12 1

(6d)

Substitut Eq.(6a) and Eq.(6d) into Eq.(6b) ...

111 1


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Continued...

In summary we get...

dVD (4) - repeated where ...

1 1 1

This is the ONLY possible solution for the three

simultaneous linear equations Eqs.(6a-c)!

It is the ONLY possible solution that ensures ...

DIMENSIONAL HOMOGENEITY

This solution ...

dVconstD

3 for sphere. Must be obtainedfrom experiments or theory

STOKES LAWRecall, that it is only valid for low Re!

is the ...


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Continued...

While we assumed Re


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A note on .

Self Similarity

A common view is that scaling or power-law relations arenothing more than the simplest approximations to the

available experimental data, having no special advantagesover other approximations. ...

Recall that we used a function of type...

dVDfor the dimensional analysis. This is called a power-law

relation.

IT IS NOT SO !

Scaling laws give evidence of a very deep property of the

phenomena under consideration their ...

SELF SIMILARITY

Such phenomena reproduce themselves, so to speak, in time

and space.

From: G.I. Barenblatt, Scaling, self-similarity, and intermediateasymptotics. Cambridge University Press, 1996


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Continued...

Reproducingthemselves means ...

2a

that wake

behind an inclined

flat plate looks thesame as flow ...

the wake of agrounded tankship.

Power laws are MagnifyingGlasses

Example:

Going from

to 2

2a

gives24a

(I)

(II)

Apart from scale factor 4 Eq.(II)

is the same as Eq.(I)


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Continued...

Some Background Info:Classic example illustrating how powerful dimensional

analysis can be ...

Explosion of Atomic Bomb

By measuring radius r as a function oftime, t, G.I. Taylor was

able to deduce energy released when bomb explodes by means

ofdimensional analysis alone from analyzing freely available

cine films of explosions.

The figure was considered TopSecretback in the 1940s

Taylors result caused much embarrassment in US government

circles.

G. I. Taylor

tr

Ground Ground100 m


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Continued...

TIP:

Use requirement for dimensional homogeneity as a quickcheck for correctness of unfamiliar equations!

Example

Someone claims that drag force, D, acting on sphere with diameterd moving with velocity V through a fluid of viscosity is given

by ...

23 dVD (Formula is wrong!)

Use dimensional arguments to show that this formula cannot be

right!

Left-hand side of equation is:

Solution

Right-hand side of equation is:

D is a force, hence, dimensions are [ ] 2 2ML

N kg . m/sT

= =

ViscosityDiameterVelocity 2

L

T ( )

2

L

M

LT

2

2

ML

T=

Different dimensions on both sides of the

equation. Hence, formula cannot be right!

([..] tells that you take the dimension of the unit)


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Continued...

GOAL WAS

TO COMPRESS SHELF OF BOOKS WITH DRAG

DATA INTO ONE SINGLE GRAPH...

DRA

G

4 Independent

Experimental Parameters

Briefly recall where we were coming from and where

we are heading for.

We are not there yet but

we are getting closer...


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Dimensionless Drag / Drag Coefficient

Flow

VV

StagnationPoint

pp

0, Vps

Apply Bernoulli along streamline to stagnation point ...

spVp 2

2

(1)2

2 Vpps

(2)

Pressure in wake must be approximately equal topressure in free stream

ppw

If one neglects viscosity then drag arises only because of

different pressures on front and back of sphere.

ws ppp

With Eq. (2) ppp s

(3)

(4)


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Continued...

Pressure forces act on area approximately equal toCROSS-SECTIONAL AREA

2

2

d

A

(5)Diameter:d

Since PRESSURE= DRAG/AREA

2

2

Drag

d

pps

With Eq. (1) 22

22Drag

d

V

Divide through by right hand side and DEFINE the

(6)

DRAG COEFFICIENT CD

22

22

Drag

d

V

CD

CD is a non-dimensional number

CD is a non-dimensional representation ofthe drag force


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Continued...

Notes:

Main assumption was to neglected viscosity. This means we aredealing with Reynolds numbers for which Stokes law is NOT

applicable. Hence, we have considered high Reynolds numbers, i.e.

Re>>1. Only for these an extended wake exists.As right-hand and left-hand side approximately equal in Eq.(6) the

drag coefficient must be of order 1 under the assumptions (Re>>1)

we made.

On previous page defined drag coefficient for a sphere. This

definition can be extended to include bodies of arbitrary shape...General Definition of Drag (and Lift) Coefficient

Coefficient of

Drag

Lift

DC

LCArea

2

2

V

Drag or Lift

Notes:

Carefully check exact definitions of quantities such as CD, CL or Re

Before using data found in literature! Definitions may vary!Usually one uses projected area (cross-sectional), i.e. area one sees

when looking towards body from upstream for CD. But for CL for

airfoils one uses one uses the planform area.

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Lecture 17 Final07