Lecture 16Final Version

• Combinations of Solutions: Solid Bodies in a Potential Flow (Rankine Oval etc.)

• Cylinder in Uniform Flow• Cylinder with Circulation in a

Uniform Flow• Pressure Distribution Around the

Cylinder• Kutta-Joukowski Lift Theorem• Circulation and Lift for Aerofoil

Applications

Contents

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also via pigeon holes)

COMBINATIONS OF SOLUTIONS: SOLID BODIES IN A POTENTIAL FLOW

• Recall: Can use PRINCIPLE OF SUPERPOSITIONPRINCIPLE OF SUPERPOSITION for velocity potential.

Cartesian Coordinates:

Uniform Flow and Source: THE RANKINE BODY

What happens if we combine...

sourceflowuniform

Polar Coordinates:

xymyUyx 1tan,

mrUr

sin,

• (1)/(2) represent complete descriptions of flow field. But what does it look like?...

(1)

(2)

•To graph lines of constant , first look for STAGNATION POINTS.

There ... 0V

Thus, both velocity components must be zero… Differentiate to get expressions for velocity components ...

• In addition, have shown that for incompressible, irrotational flow, stream function also satisfies Laplace Eq. So can similarly construct flow solutions by combining S.F. associated with uniform flow, source/sink flow and line-vortex flow.

• In fact, we will almost exclusively use stream function here because we are interested in pattern of streamlines; once we find stream function, we can use fact that it is constant along streamlines to plot out streamlines.

?

Continued...

• Cartesian Coordinates:

xymyUyx 1tan,

u

222

cosr

rmUyx

xmUy

cosrmU

v sinrm

222

sinr

rmyx

ymx

(3)

(4)

• For v=0 Eq. (4) requires ...

xmUu

0

u

Umx...with this Eq. (3) gives...

0y

STAGNATION POINT at 0, Um

• Polar Coordinates: mrUr sin,

ru u

r1

r

rmU cos sin Uand

need0For u ...,3,2,0,

Since m, r positive choose... to get a solution for 0ru

rmU

rmUur cos

0

ru

Umr

STAGNATION POINT at ,Um

Continued...

• In both cases same location for Stagnation Point ...

Cartesian Coordinates

Polar Coordinates

0, Um

,Um

We repeated ourselves to demonstrate that either coordinate system can be used. In general choose the one that makes the analysis easiest.

• Now use for distance between origin and stagnation point:

Uma

• Find S.L. that arrives at stagnation point and divides there. Using ...

.const

along this S.L., use a known point - the stagnation point - to evaluate constant. With Eq (2) from above ...

mrUr sin,

mUmU sin

where suffix s denotes ‘along particular streamline through stagnation point’.

• Streamline found by equating Eq (2) to the constant and rearranging...

s m

mrUm sin

sinsin

aUmr

PLOT

(5)

Continued...

sin

arPlotting stagnation streamline:

r 2 2a

23 2a 0 w i t h r

ar sin

e t c . . . .

• Now only have stagnation/surface. To get other S.L.’s, choose point, determine constant for S.L. through point and then sketch particular S.L. through this point by compiling a table as above.

• Stagnation streamline defines shape of (imaginary) solid half-body which may be fitted inside streamline boundary; remember flow does not cross streamline … or solid boundary. Call this special S.L. SURFACE STREAMLINE.

• Body shape named after Scottish engineer W.J.M. Rankine (1820-1872).

Continued...

How Does Flow Speed Vary Along Surface Streamline?

cosrmUu sin

rmv

• Recall velocity components in cartesian coord. from Eqs. (3)/(4) :

• Flow speed is: 222 vuvU

22sincos

rm

rmU 2

2

22

2

22 sincoscos2

rm

rm

rmUU

222

22 cossincos2

r

mr

mUU 2

22 cos2

rm

rmUU

Uma NOW RECALL:

22

22 cos21

Urm

UrmU

2U

cos21

2

222

ra

raUU

• To find flow speed on body surface, VS , evaluate Eq. (6) subject to

(6)

sin

ar (5) for surface streamline. This gives ...

Continued...

22

21

21

SS VpUp

• Note: Since we know surface flow speed, we can evaluate static pressure at any point on surface. Using Bernoulli equation along central streamline that divides into surface streamline and, as usual, ignoring gravitational term...

flowdundisturbe

Upstream

surfaceOn

Hence we get the non-dimensional pressure coefficient :

2

2

21

21

U

V

U

ppC Ssp

2221

21

SS VppU

2

2

22121

211

U

V

U

pp SS

2

2

221

U

V

U

pp SS

2

2

21

21

U

V

U

pp SS

Continued...

Uniform Flow and Sink(instead of Source)

xymyUyx 1tan,

mrUr sin,

Source (previous case)

Cart. Coord.

Pol. Coord.

Sink

xymyUyx 1tan,

mrUr sin,

Stream Function

• Only difference: plus sign(s) changed into minus sign(s)!

Hence, for sink expect a very similar analysis as above for source.

SinkSource

• Note: In real world (inviscid) flow pattern for sink would not be observed! Flow would initially follow body contour but (due to viscosity) detach at separation points indicated by S.P. in sketch for sink. The phenomenon of SEPARATION will be covered later. At this stage learn that...

Potential flows do NOT model ALL features of a real flow!!!

This has lead to potential flow often being termed IDEAL FLOW.

S.P.

•Obvious question now is what happens for ...

Uniform Flow + Source + SinkWe consider ‘symmetric’ case where:

Source

Sink

Strength Location m

m

0,c

0,c

We cannot have both at origin now! What would happen if both at origin?

• By considering the two sketches on previous slide we can anticipate shape of surface streamline and resulting body...

… an oval.

• Using superposition, can readily write stream function for this flow:

cxym

cxymyUyx 11 tantan,

flow

Uniform

)0,(atSource c )0,(atSink c

• Second and third terms can be combined using:

1tantantan 111

To give a more concise form for stream function

222

1 2tan,cyx

ycmyUyx

(1)

Continued...

• Now find stagnation points, where u=v=0. From Eq. (3) one sees that when y=0 then v=0.

2222 ycxcx

ycxcxmU

yu

222211

ycxycxym

xv

• 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 for x are:

LUcmcx

21

21

• Hence, stagnation points at:

0,Land 0,L

•Now determine value of S.F. for surface streamline from Eq (1).

cxym

cxymyUyx 11 tantan, (1) - repeated

It can be seen that this is trivial and that 0S

Continued...

• Rankine Oval then looks like ...

• We already determined value of L. 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...

21

21

cUm

cL

cUm

ahch

2cot

22max

121

chcUm

Uu

cUmd

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 circular ‘cylinder’. 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

Overall Strategy for Plotting Streamlines from Stream Function was...

Write down stream function for flow by appropriately combining individual solutions for source, sink and line vortex as a sum:

yxyxyxyx ,,,, 321

Calculate expressions for vel. components u, v from

xv

yu

,

Note: Huge choice as far as selction of parameters is concerned! Souce strength, vortex direction 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 point x,y

From previous it should be obvious how one can find stream function for a cylinder (circle) in a uniform flow ...

Cylinder in a Uniform Flow

•Turn Rankine oval into circle by allowing ... cUm

• Achieved by moving source and sink closer to origin ... 0c

Limit (c=0) would ultimately ‘cancel’ pair!

0c with .constmc

222

1 2tan,cyx

ycmyUyx• Recall that Rankine oval had S.F. ...

0let c

• Noting that smallfortan 1

22,yx

yyUyx

Note: Merging of source and sink as above produces structure known as DOUBLET.

cm2

flowUniform

originat DOUBLET

• Ensure their influence remains by allowing m to increase in size. Necessary limit is...

• Now then the argument of tan-1 goes to zero...

gives

2222,

cyxycmyUyx

• Define DOUBLET STRENGTH:

Stream Function for Cylinder Flow

(1)

Continued...

• More convenient to work in polar coordinates! S.F. can be written ...

rU

rUr

rUr sinsinsin,

• From Eq. (1) can get velocity components in usual way...

(1)

Continued...

2

21cos1

rRU

rur

2

21sin

rRU

ru

Wherewe

used...

UR

• CHECK that this flow really does represent a cylinder in uniform flow.

Stagnation points:

(2)

(3)

Eq. (3) : 00 u0 u

• Substitute these angles into Eq.(2) ...get to0setand ru

RrrR

rRU

2

2

2

2110cos0

RrrR

rRU

2

2

2

211cos0

Hence,... Stagnation points: ,and0, RR

• Surface S.L. VALUE by substituting one stag. point into Eq. (1)...

00sin0,

RU

RUR

• Now get equation for Surface S.L. by equating Eq. (1) to zero ...

Ur 2 Rr or

As required surface S.L. is circle with radius UR PLOT

:0For

:For

Ur

Continued...

• Velocity Components on cylinder surface are obtained, by setting r=R, from...

2

21cos

rRUur

2

2

1sinrRUu

Eq. (2)

Eq. (3)

0ru

sin2 Uu

Might have expected to find that radial flow component is zero on surface - flow cannot pass through (solid) cylinder wall!

• Note also that maximum flow speed occurs at

where it is UU 2and22

3and2

respectively.

Means in both cases (top and bottom half of cyl.) flow is from left to right! On top negative value as velocity points in clockwise (negative angle) direction. On bottom in anti-clockwise (positive angle) direction.

• Finally, note symmetry of flow about both the x- and y- axes. What does this tell you about the pressure distribution on the cylinder surface… remember the Bernoulli Equation!

Uniform Flow + Doublet = Flow over a Cylinder

Continued...

• Since we can now mathematically describe …

...we can, in principle, also describe flow through an arbitrary array of cylinders as, for instance, the flow shown in the photo below. We simply need to put several doublets in our uniform flow.

Cylinder with Circulation in a Uniform Flow• Without performing calculation, can see in preceding flow no net lift or drag on

cylinder since pressure distribution on surface symmetric about x- and y-axis..

2 KpG=

• Note that this does not violate the flow around cylinder: line vortex produces a ucomponent 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 lnsinsin,

originat Doublet

flowUniform

originat vortexLine

constantArbitrary

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

U

R RKC ln

(1)

(1) RKrKrU

rUr lnlnsin,

RrKrU

rU lnln1sin

RrK

rRrU lnsin

2

Velocity Components

2

21cos1

rRU

rur

2

2R Ku U sin 1r r rq

y q¥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 .

Continued...

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

0ruRKUu sin2

• Surface Stagnation points also need: 0u

UR

K2

sin

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

Continued...

• Can now also describe flow through an arbitrary array of cylinders when each of them is rotating! (Note: In photo below cylinders are not rotating)

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:

2221

21

SS UpUp

flowdundisturbe

Upstream

surfacecylinderOn

• Re-arranging...

2

22 1

21

UUUpp S

S

• Substituting for flow speed gives...)sin2RKUu ,0( Ru

2

22

2

222 1

211

21

UuU

UuuUpp R

S

222 /sin2121

URKUU

222 sin4sin41

21

URK

URKUppS

… difference in pressures between surface and undisturbed free stream

(1)

In particular for non-rotating cylinder

where K=0:

22 sin4121

UppS

(2)

2

2sin41

21

U

ppC Sp

Def.: Pressure Coefficient

Only top half of cyl. shown.

Continued...

2

2

2sin4sin41

21 UR

KURK

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

Continued...

Equation (1) …

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

222 sin4sin41

21

URK

URKUppS

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

RbURKUL

2

421 2

0D Or, lift per unit width:

UKUbL 2 Thus, drag zero…

a remarkable result!

TheoremLiftJoukowskiKutta

ParadoxsAlembert'd'

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.

UbL

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?