1

Lecture 17Final Version

Contents• Lift on an airfoil• Dimensional Analysis• Dimensional Homogeneity• Drag on a Sphere / Stokes Law• Self Similarity• Dimensionless Drag / Drag

Coefficient

2

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 about x- and y-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 flow cannot pass through cylinder boundary.

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

CrKr

rUr lnsin

sin,

originat Doublet

flowUniform

originat vortexLine

constantArbitrary

Use result that radius of resulting cylinder is : And set :

U

R

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¥

æ ö¶ ÷ç ÷= - = - + +ç ÷ç ÷ç¶ è ø

• In order to generate lift need to break symmetry. Achieved by introducing

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

3

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 for K,… Now first get value of S.F. for r=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 of K the 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:

222

1

2

1SS UpUp

flowdundisturbe

Upstream

surfacecylinderOn

• Substituting for surface flow speed : with

Ku 2U sin

Rq q¥= - +

Ru 0 ,=

2

22 sin4

sin412

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 sin412

1 UppS

(2)

2

2sin41

21

U

ppC S

p

Def.: Pressure Coefficient

Only top half of cyl. shown.

2 2 2S rU u uq= +

5

Continued...

2

2

2sin

4sin41

21 UR

K

UR

K

U

ppC S

p

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

6

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 cylinder surface yields total L and D ...

2

0sin dRbppL S

2

0cos 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!

TheoremLiftJoukowskiKutta

ParadoxsAlembert'd'

7

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 achieved… Breaking of the flow symmetry in x-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 net drag 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 of above 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 of flow speed differential can be modelled by using line vortices which yield circulation and hence lift. (In Fig. (2) line vortices would have negative K to 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. Consider Fig. (3a-c) below ...

nCirculatio:SpeedFlow:AirDensity: U

Evidently the correct value Kutta

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.

10

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 is incidence

UcUb

L

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

2

21 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.

11

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. half

Upper cyl. half

• 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 ...

13

Continued ...

WE NEED ...

1 page for Drag as

function of 2 variables

(e.g. velocity and diameter)

d increases from curve

to curve

1 book for Drag as

function of 3 variables

(e.g. velocity, diameter, density)

1 page for each value of

Shelf of books for Drag as a function of 4 variables

(velocity, diameter, density, viscosity)

If we want 10 data points per curve, at £10 each experiment, this will cost...

000,100£10£10101010

THERE MUST BE

A BETTER WAY !?!?

14

Continued...

GOAL IS TO COMPRESS SHELF OF BOOKS INTO ONE SINGLE

GRAPH...

DR

AG

4 Independent Experimental Parameters

How Could We Possibly

Achieve This?

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•Dimensional Analysis for Re<<1

1Re

(Note: Later we will relax this restriction and look at larger Re.)

What does imposed restriction mean?

1Force Viscous

Force Inertia..

Force Viscous

Momentum of Change of Rate

ei

Forces ViscousForces Inertia

Viscous Forces are the dominant forces!

•Inertia Forces are associated with density of Fluid

•Consequently, if Inertia Forces << Visc. Forces then, to a good approximation DRAG DOES NOT DEPEND ON DENSITY of FLUID!

Thus, Eq.(1)...

,,,VdFD

,,VdFD

reduces to ...

(2)

(1) - repeated

16

Continued...

So, we are restricted to flow with..

HIGH RE NUMBERTURBULENT

LOW RE NUMBERLAMINAR

Restrictions exclude ...

17

Continued...

,,VdFD •The expression Eq.(2) ...

(2) - repeated

… represents a VERY general statement!!!

•CRUCIAL NEXT STEP:

Ensure that function F has 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 a non-sense statement where units are for instance ...

132 VdD

Units:2s

m kg Units: 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:

2s

m kgN m

s

m

sm

kg

2TLM L 1TL 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) ...

19

Continued...

dVD (4) - repeated

1TL L 11 TML 2TML

Collect corresponding terms ...

TLMTLM 2

By comparing exponents ...

•… of M on left and right hand side of Eq.(5)

(5)

1 (6a)

•… of L on left and right hand side of Eq.(5)

1 (6b)

•… of T on 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

20

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 obtained

from experiments or theory

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

is the ...

21

Continued...

•While we assumed Re<<1 experiments show that Stokes law is, in fact, valid for Re<2.

•For flow regime where Stokes law is valid drag is proportional to velocity. Hence, doubling velocity results in double drag. We will later see that this is not the case for higher Reynolds numbers.

•The constant in Stokes law can, in principle be obtained from one single experiment.

•Think about all this an ‘let it sink in’… We have determined formula for drag forces acting on sphere without knowing anything about the physics of the flow. The only thing we did was ensuring dimensional homogeneity! Of course the whole strategy can only yield correct results if we have identified all parameters involved in problem.

dVconstD

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

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A note on ….

Self Similarity

•A common view is that scaling or power-law relations are nothing more than the simplest approximations to the available experimental data, having no special advantages over other approximations. ...

•Recall that we used a function of type...

dVD

for 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 intermediate asymptotics. Cambridge University Press, 1996

23

Continued...

‘Reproducing themselves’ means ...

2a

… that wake behind an inclined flat plate looks the same as flow ...

… the wake of a grounded tankship.

Power laws are ‘Magnifying Glasses’…Example:

Going from

to 22a

gives24a

(I)

(II)

Apart from scale factor 4 Eq.(II)

is the same as Eq.(I)

24

Continued...

Some Background Info:Classic example illustrating how powerful dimensional analysis can be ...

Explosion of Atomic Bomb

• By measuring radius r as a function of time, t, G.I. Taylor was able to deduce energy released when bomb explodes by means of dimensional analysis alone from analyzing freely available cine films of explosions.

• The figure was considered ‘Top Secret’ back in the 1940s

• Taylor’s result caused ‘much embarrassment in US government circles.

G. I. Taylor

tr

Ground Ground

100 m

25

Continued...

TIP:•Use requirement for dimensional homogeneity as a quick

check for correctness of unfamiliar equations!

Example•Someone claims that drag force, D, acting on sphere with diameter

d 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 [ ] 22

MLN kg . m/ s

Té ù= =ê úë û

ViscosityDiameter Velocity 2

LT

´ ( )2L ´ M

LT

2

2

MLT

=

Different dimensions on both sides of the equation. Hence, formula cannot be right!

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

26

Continued...

GOAL WAS…

TO COMPRESS SHELF OF BOOKS WITH DRAG DATA INTO ONE SINGLE GRAPH...

DR

AG

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

Stagnation

Point

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 to pressure 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 to CROSS-SECTIONAL AREA

2

2

d

A (5)Diameter:d

•Since PRESSURE= DRAG/AREA

2

2Drag

dpps

With Eq. (1) 22

22Drag

dV

•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 of the drag force

29

Continued...

Notes:•Main assumption was to neglected viscosity. This means we are

dealing 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

22

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.