Numerical Methods in Incompressible Fluid mechanics

8/4/2019 Numerical Methods in Incompressible Fluid mechanics

1/52

- ,:; p 2.0_- .,'. . . . . . . . . . ._.. .

13.024 - Numerical Methods in Incompressible Fluid Mechanics

Lecture Notes - Version 3

By Jerome H. Milgram

Spring 2003


2/52

INCOMPRESSIBLE FLUID MECHANICS BACKGROUND

Conservation of Mass, Continuity Equation

divV = V V = 0

o u+ o v + ow=0o x o y o z

Newtonian Dynamics, Navier-Stokes Equations

DV oV _, _ , 1 2 _,- =- + (V V)V = --VP+ vV VDt at p

P is the dynamic pressure. The total pressure, P T , is the sum of thedynamic pressure and the hydrostatic pressure, -p 9 z, where z is positive

< upwards. PT =

P - pg z.

1


3/52

PARTICLE IMAGE VELOCIMETRY

Camera

L a s e r

LightS h e e t

~~-------+~~

FluidFlowRegion

L a s e r

2


4/52

PIV Example

u and v can be measured, so ~~and ~~are known.

O W a u in:----oz ax oy

If PIV is done on multiple planes inside a fluid domain, then ~~ is knownover the whole domain. At a rigid boundary, w = 0 and, in principle, wcan be found anywhere by:

1z ow

W= -dzboundary O Z

Example

In a domain bounded by 0 < z < 4, u = (3e Z - zez - 3) sinx and v = 0over a range of x and in 0 < z < 2. In this sub-domain,

a uax = 3e Z - ze" - 3) cos(x)

aw= -(3e Z _ ze" - 3) cos(x)O Z

w = rzowdz =- [3e Z - 3 - ze' + eZ - 1 - 3z] cos(x)io O Z

w = (Z oWdz =- [4e Z - 4 - ze' - 3z] cos( x)io O Z

3


5/52

Velo city F ie ld a t z = 0. 412

2

: ; : : : . . :::;::.- e : : : : EC - :>- --: : :z;. : ; : : : . . EC . . : : : : : - >- --: ; : : : . . :;::.. e : : : : ;:::.. e : : : : - -- ;:::.. - - -=-~ >-


6/52

60

50

40

30

20

10

0

-10

-20

-300 0 .5 1

dw/dz and w(z) at x =1

1 .5 2 2 .5 3 .5 4

z

5


7/52

A More Interesting PIV Example

Consider the following flow for z > 0 in a range of x and y. Of course, inan experiment you would not know the mathematical formulation. Rather

you would just measure u and v over a set of (x,y,z) points.

u(x, y, z) = ( 3 e O .1Z - 3 cos z) sin y cos x

v(x, y, z) = ( 3 e O .1Z - 3 cos z) cosy

The x and y derivatives of the velocities can be computed numerically

from the measurements. If the experiment were done well, they would

have values according to the following formulae:

~~ = - ( 3 e O .1z - 3 cos z ) sin y sin x

8v- =- (3eO.1Z - 3 cos z) sin yBy

OwThen, the continuity equation is: a z

a u Bv----ax By

The experimentally determined values would have the values given by:

8woz = ( 3 e O .1Z - 3 cos z ) sin y(sin x + 1)

Integrating ~~ from 0 to z at a prescribed value of (x,y) would give w(z)there. The values obtained would obey:

w= (30 eO .

1z-

30 - 3 sin z ) sin y (sin x + 1)

6


8/52

F low in P la ne a t z= t

7

6

>- 5

4

3

2

1

r r r f \ \ \ \ \ \f [{{{H\ \ \ \'\ '\ \ t1///11\\,\,\'\ t r r

o c______--- -L-.J_-'-- -'----'-~____j___J_J..____L_l__...L___l.___L_~~_____l-2 0 2 4 6 8 10 12

x

7


9/52

dw/dz and w(z) at x=t, y=190~---,~---.~---.-----.-----.-----.-----.-----,-----.----,

50

40

30

20

10

00 1 2 3 4 5 6 7 8 9 10

z

70

80

60

8


10/52

AVERAGED NAVIER-STOKES EQUATIONS

a (V+V i) [R J -) ]( -= - -) 1 ( -= - -)at +" v + v' . \7 V + v' = - P\7 ( p + p ') + v\72 V + v'

Take Average of above equation:

ail [(.... -) ] (- -) 1 - 2 -= -at + v + v' . \7 V + v' =- P\7 P + v \7 V

Thus, the Reynolds-Averaged Equation is:

ail -= - -= - - - 1 - -= -- + (V . \7)V + (v' \7)v' = --"\7 P + v\72Vat p

The" Reynolds Stress Term" is:

- - -( au' au' aU')(v' \7)v' = i u'- + v'- + w'- +ax ay az-:( ,av' ,av' ,aV')Ju-+v-+w- +ax ay az

~(aW' aw' aW')k u'-+v'-+w'-ax ay az

9


11/52

THE PRESSURE EQUATION FOR AN INCOMPRESSIBLEFLUID

Start with the Navier-Stokes Equation

8Y ... . . . . . .1 2""- + (V \7)V = --\7P + v\7 Va t p

Take its divergence.

Because \7 . Y = 0, the only non-zero terms are:..... .... 1 2

div (V \7)V = --\7 Pp

Working out the details of the LHS and interchanging the LHS and the

RHS results in:

In2p _ (8U )2 (8V )2 (8w )2 2 8 v8 u 8w 8 u 8w a v--v - -

+ _ + - + --+2--+2--p a x a y a z a x a y a x a z 8 y a z

n 2 p _ { (8 u )2 (B v .)2 (B w )2 2 a va u B u : 8 u 2 awy - -p - + _ + - + _- +2-- + --8 x a y B z a x 8 y 8 x 8 z 8 y8z

The pressure, P, satisfies Poisson's Equation driven by products of the

spatial derivatives of the velocity. This is different than the commonBernoulli Equations because here the flow can be unsteady and rotational.

10


12/52

The Vorticity equation

vorticity = w - curl iT -\7 x V

Start with the Navier Stokes equation:

a v - . - . 1 2 _,- + (V. \7)V = --\7P + 1/\7 Va t p

Take the curl of this equation, term by term:

a w _ , _ ,a t + (V . \7)w + (w . \7)V = 1/\7 2w

Dsx: _, 2- = -(w. \7)V + 1/\7 wa t

The first term on the right hand side is the rotation and stretching of thevorticity by the non-uniform velocity field.

12


13/52

Inviscid Fluid Mechanics, Euler's Equation

Set the viscosity, J 1 and the kinematic viscosity, v to zero. Apply these"settings" to the Navier Stokes Equation.

DV a v -+ -+ 1- = -+ V V)V = --VPDt at p

au au au au 1 aP- + u- + v- + w- = ---at ax ay az p ax

av av av av 1 aP-+u-+v-+w-- =---at ax ay az p ay

aw aw aw aw lap- + u- + v- + w- = ---at ax ay az paz

13


14/52

Bernoulli Theorems for Inviscid Flow

Theorem 1 - Irrotational Flow

Vecto r Id en tity

H = ~(IVI)2 + P2 p

For Irrotational Flow, V = Vand Vx V = 0

Let

a a -V+ VH = V(- + H) = 0a t a t

~~+ H = f(t)

The functio n f(t) can be absorbed into by letting = '+ fLf(t)dt and

V = V'. 8 . 8/8t = at + f(t) so, 8' H = 0a t +

Finally rename 'as .

a H=Oa t +Remember that the total pressure,P T = P - pgz.

8+ ~(V)2 + PT + gz = 0 .a t

2 p

15


15/52

Theorem 2 - Steady Flow

B V _ , -> 1

- + (V \7)V=

--\7P{)t p

B _ ,{)t \7>+ \7 H - V X W = 0

Thus, for steady flow:_,

Vxw=\7H

Streamlines and vortex lines are perpendicular to \7 H.Along either a streamline or a vortex line, H is a constant. So on anyoneof these lines,

1 _, 2 PI_, 2 PTH = -(V) + - = -(V) + - + gz = constant

2 p 2 p .

If the flowis both steady and irrotational, H is the same everywherebecause \7 H = .

16


16/52

Vorticity Dynamics and Kelvin's Circulation Theorem

Circulation = r = f V .dr = s ta- ds

....DV 1 2 ....- = --\7P+v\7 VDt P

Identity for an incompressible fluid: \72V = - -c \7X W....

DV 1-=--\7P-v\7xwDt p

Following a closed material curve,

d rd t

....

f ~~.dr= - f >P .dr - v j(V X IV ) .dr-v f(\7 x tv) .dr

Kelvin's Circulation Theorem

In an inviscid fluid, ~ = 0

pathlines

Corollary: In an inviscid fluid with no circulation (such as starting from

rest) the circulation remains zero.

17


17/52

Practical Implication

In a high Reynolds number streaming flow, fluid which has not passed close

to a boundary or a free surface has negligible vorticity.

Therefore, in high Reynolds number streaming flows, vorticity is limited

to boundary layers, separated zones, and wakes.

Concept of Vortex Lines

vor tex lines

This lin e rep res en ts th e vorticity inth is area wh ich is th e circu lationaround itsboundary

E ach lin e rep resen ts circu lation aroun d an area which is the sameas th e vorticity in sid e th e area.

-+ -+

ti: = curl V = 7 x V

Therefore: div zo = \7 . to = 0

The vortex field is solenoidal. Vortex lines are continuous. They can have

curves and turns,but they cannot have ends in the fluid.

13


18/52

v

W /A . /G -

r

r

19


19/52

za : a :w wI- 0

en 0:Ja :

20


20/52

r r: 9 f

Potential Flows and Mostly Potential Flows

For an irrotational fluid \7 x V = 0

This means that there exists a velocity potential, , such that,

For an incompressible fluid \7 . V = 0

Thus, \7 . (\7) = 0

For a completely potential flow, the velocity potential satisfies Laplace's

equation.

For an incompressible flow that is "nearly" irrotational except in bound-

ary layers and wakes, the flow outside these boundary layers and wakes

is approximately described by a velocity potential that satisfies Laplace's

equation.

21


21/52

Green Functions, Green's Theorem and Boundary Integral Equations

The following development is for three-dimensional flows. The development issimilar for two-dimensional flows except that two dimensional source functionsare involved and the dimensionality of some integrals and associated constantsare different.

Green's Theorem

If and ' l /Jboth satisfy Laplace's equation ( \J2 = 0, \ J2 ' l / J= 0), then:

Green Functions A three-dimensional (~, 1] , ( ) space is considered with a "sink" atlocation (x , y, z) . The "sink" has the velocity potential ' l / Js ,

1 1~= =-

J(x - ~)2 + (y - 1])2+ (z - ()2 r

where: r = J(x - ~)2 + (y - ' r / )2+ (z - ()2

Importantly, \J2' l / Js= 0 both for the differentiations done in (~, 'r /, ( ) space as wellas for the differentiations done in (x , y, z) space. The followingdevelopment canbe formulated either way and we will choose to differentiate over (~, 'r /, ( ).

A Green Function, Gis:

G = ' l / J s (x ,u,z,~,'r /, ( ) + ' l / J r (x ,y, z,~,n.()

where, \ J 2 r = 0 in the fluid domain, and ( x , y, z) is called the point P.

,::::LUI 0

22


22/52

If P is outside the fluid domain, the bracketed terms on the right hand side inGreen's Theorem are zero. However, if P is inside the fluid domain, ' \ J21 j Ji- 0 atP. Then, if P is enclosed by a small sphere of radius e, which is excluded from thefluid domain, Green's Theorem applies in the modified fluid domain. However,

now the integrals include an integral about the small sphere.

FLUID

1 r [Be - eB ]ds - 0is+sphere an an ~,"7,( -

1 1ec -1

1>-8 ds = -[P(x, y, z)]-2 47rf = 47r1>[P(x,y, x)]sphere n f

j Is[4 >~ ~ ~G~~] d s~,",(= ~47r4>[P (x , y,z )]If P(X, y, z) is on the boundary, the integral is not defined. However, if we replacethe real boundary by one which has an infinitesmal hemisphere syrrounding P,

the Green Function integral is zero because the functions have no singularities in

the revised fluid domain.

1f [ oc a]1>- - G- ds = 0S+hemisphere an an ~,"7,(fA oc

-1. -a ds = -[P(x, y, z)]-2 27rf = 27r[P(x,y, x)]

hemisphere n f

j Is[4 > ~ ~ ~G~~] d s~,",(= ~47r4>[P(x,y,z)1

23


23/52

Putting the preceding parts together, if a closed fluid domain of surface S isconsidered with ii being the outward normal vector (out of the fluid) and 'IjJis

taken as G with proper exclusion of the singular point of G when (x, y, z) is insidethe domain or on its boundary,

[8G 8] { 0 (x , y, z) outside SJisa - Ga dS = -2Jr(x, y, x) (x, y, z) on Sn n -47r(x, y, x) (x, y, z) inside S

The integral is over the closed area in (~, 'T I , ( ) .When the singular point is on thesurface, an infinitesimally small circle surrounding the singular point is excluded

from the integral.

24


24/52

Example of method of solution

Generate integral equation on surface of an object in a uniform flow.

< u

Suppose uniform flow onto an object is known

~ is known.

q,=-Ux+

Boundary condition: a~ = 0a n '

U~ - aA . 0- ln+~=a n '

d~

J fs [~~ - c tn n ]= -27r

J fs ~~ a s+ 27r = JfsC u i. s a sRight hand side is known in integral equation for on boundary.

Solve for values of on boundary (panel methods).

Then and ~ are known on boundary.

Green's Theorem then gives in all space.

25


25/52

. rrn h 00 f'" .I - (J J ", j

Interpretation of Boundary Integral Equationin terms of source and Dipole Layers

J r [2_8 G _ 2_ 8G] a s = { 0 (x , u ,z) outside Sis 47f8n 47f 8n (x,y,x) (x,y,z) inside 3"Inside S" means inside the fluid and outside "3" means outside the fluid.G is the potential of a "unit sink" and -8g / im is the potential of a unitdipole.

The Green's Theorem integrals are integrals of sink distributions per unit

area of (1/47f)8/8n over the object and of dipole distributions of strengthper unit area of /47f over the object.

The sinks

Consider the effect of a unit sink.

1V n=2"'r

1 2influx = -47fT = 47 fr 2

Now, look at the small patch of area A on the surface:

pqtc~

B is the effect of the integrals on the remainder of the object. Call thesink strength per unit area (5 . Total sink Strength on the patch is (5A.

Net influx, based on the velocities is 2AVs.

2A Vs= 47f(5A 2Vs= 47f(5

26


26/52

2V sis the jump in normal velocity. This must equal the normal velocity,8/8n in the fluid at the boundary since V = 0 inside the object.

a- = 47r(Ja n

1 a(J =---

47ron

Now consider an infinitesmal dipole patch of strength fL

Inside the infinistemally thin dipole layer of thickness f,

47rfLV=--f in fluid - inside object = 47rfL

since: inside object = 0,

1fL = 47rin fluid

27


27/52

Kelvin-Neumann Problem

zu

y

x

28


28/52

The Kelvin-Neumann Problem

is the perturbation potential (does not include -U x).

The integral over 81, which is the part of the ship hull below the waterlineis of the same form as a Green's theorem or "panel method" integral forany finite size body, except here the top is open.

and C decay with distance from the ship fast enough for the integralover 8 3 to vanish.

The integral over 8 2, which is the free surface external to the ship ISspecial. We consider it here and call it 2.

1 1 1 [O O C ]2= - C- - - dx dy4~ ~ O n on

Since n is a unit vector in the z direction on the mean free surface,

1 If [O O C ]2 = - C- - - dx dy4~ ~ o z o z

On the mean free surface, ~ = - U ;~ :Wand we choose G (the Kelvin-Neumann Green function) such that it satisfies the same boundary con-dition 8C = _ u 2 8 2C

, 8z g 8x2

Then, applying these boundary conditions, 2 becomes,

1 U2 [0 2 0 2C ]2 =- 4~ 9 is2G ox 2 - ox 2 dx dy

1 U2 I! 0 [O O C ]2=--- - C- - - dxdy

4~ g S2 oX ox ox

29


29/52

Now, the integral over x can be done. The contributions at x = oo vanishso the result is:

(P 2= _ !_U2{ [ca _ aC] dy __ !_U

2{ [Ca _ aG] dy

471"g Jfore ax ax 4 7 1 "g Jaft . ax ax

The curve of the waterline is called C, with the part forward of the max-imum beam called C] and the part aft of this is called Ca. Consider theintegrals taken along C in the counterclockwise direction. Then, dy ispositive on the forebody (C j) and negative on the afterbody (C a).

2 = _ !_U2

( [Ca _ aC] dy471"g Jc ax ax

If v is the vector in the horizontal plane that is perpendicular to the wa-

terline and pointed out of the fluid into the ship, and dl! is the differentialof arc length along C, along the waterline dy = vxd., so,

1 Jf [ .o ac] 1 U2 [a ac] =- .G- - - d5 - -- C-- - vxdl!471" s, an an 4 7 1 "g c ax ax

Now, consider a potential function, ' defined in the region bounded by 51and 54. In other words, ' is some function of space such that \12' = 0in the region where it is defined outside the actual fluid. The boundary

condition we impose on ' on 54 is the same as the one we impose on on52. On Sl we impose ' = c p . z= - ~2 ~x{ 0 f) SfIn the fluid region,

n = -n pointed into the fluid on 51 and n' is a unit vector in the zdirection on 54.

Call the contribution to ' from the integral on 54 by ~.

~ = !_J {[Ga, - ,OG] dx dy= -_ !_ U2 J {[Ca2' - ,a2C] dx dy471" J S4 az az 4 7 1 "g J S4 ax2 ax2

30


30/52

Carrying out the integral over x gives,

' =_ !_U2 r [GOI _ IOG] dy+ _ !_U2 r [GOI _ IOG ] dy

4 47 r 9 1fore ox ax 41 f 9 laft ax ax

1 [a1 OG ] 1 u2 [OI aG ]0=-/ r G-+- dS+-- r G--- vxdf41 f lSI an' an 41 f 9 1c ax ax

To this, we add the equation for derived before:

The sum is:

1 [a a,] 1 U2 [o a/] =- / r G - + - dS - -- r G - - - vxdf41 f lSI on on' 41 f 9 1c ox ox

4~ [~ +~] is the source (actually it is a sink) strength CT .

The normal derivative of 4~ jumps at the interface by the source strength,CT . The tangential derivative of is continuous across the interface because is continuous. ~ jumps across the interface by the jump in the normalderivative times n x. Therefore,

31


31/52

The Kelvin-Neumann Green Function

The Kelvin Neumann Green Function, Gk(x, y, z) is the velocity potentialfor a source located at (a , b , c) and moving at speed U and which satisfiesthe linearized free surface boundary condition:

G~x(x,u ,0) + vG~ = 0, 9v = U2

This function is:

1 1--+-+r rl

4v (7r/2 de 1 00 ek(z+c) cos[k(x - a) cos e ] cos[k(y - b )sin e Jdk +7r io fo k cos? e - v4v fo7r/2 eV(z+c) sec

2B sin[v(x - a) sec e] cos[v(y - b)sin e sec 2 e Jsec 2 ede

where:

32


32/52

Source Only and Dipole Only Distributions

(X I~I~)

REALFLUIDDOMAIN

HYPOTHETICALFLUIDDOMAIN

t / J '

nl

n=-naan

aan'

is the velocity potential in the fluid.

' is a function that satisfies \l2' = 0 in the region inside the object.

For a field point in the fluid domain, the following equations apply:

= _ !_J JG ( a + a , )dB - _ !_J J( - ') a Gds41r a n a n ' 41r a n

33


33/52

Suppose / is chosen as the harmonic function whose values on S are thesame as . The hypothetical interior flowwould have the same tangentialvelocity on the object as the real outer flow. Then:

>= _2_I IG(D + a ,) a s41r a n a n 'This is a representation for >in terms of surface sources only.

This representation does not apply to lifting flows since they have wakesacross which the potential jumps.

Now consider the case for which ' is chosen such that on B, 1 n=_~.The normal velocity is continuous across the surface for this case. Then:

>=_ 2_I I( > _') a Gds41r a nThis is a distribution of dipoles on the object surface S.

34


34/52

Green's Theorem in Two Dimensions

For two dimensional flow, the source potential is In r and the Green functionbecomes:

C(x, y, e ,' f ])= In r + 'lfJr(x,y, e ,'T ] )

FLUtO

The analysis proceeds exactly the same as in the 3D case. When a pointP = (x, y), which is the local origin for In r, is outside the fluid domain,

f [ac _ Ca ]ds = 0is a n a nWhen the point P is inside the fluid domain, Green's Theorem is valid in adomain in which the point P is excluded by a small circle, circle f surroundingit since '\72 = I-0 at P.

1 [ oc a ] t [8 C a ]Then: -. - - C ds + - - C ds = 0circle a r a r s a n a n

1 [ oc a ]Here: -. -a -G ads = -27r(P)circle r roc a ] {0 (x , y)outside S

Therefore: Is[a - G a ds = 7r(x,y) (x,y) on Sn n 27r(x,y) (x, y) inside S

35


35/52

Sometimes, the two-dimensional Green function is taken as:

G(x, u.t, 17)= -In r + 'l/Jr(x,y,~, 1 7 )

Then,

{

0 (x , y ) outside St [oG - G O ]ds = -7r(x, y ) (x, y ) on SS on an -27r(x, y ) (x, y) inside S

36


36/52

Force on a Vortex

-:/

/

\

\

\

lI

/

\

n = -icosB - ksinB

u; = -~sinB Wv = ~cosB2~T 2~T

P { ( r .)2 (r )2 2 }=-- U--smB + -cosB -U2 2~T 27fT

p { ru J ( r )2 }=-- --sInB+ -2 7fT 27fT

F P c Pn ds-- : r (-USinO + 4~ r )( - ic o s o -k S i nO )rdO~t27 r pur . 2 ~ purk ---SIn BrdB = -k--

o 27fT 2

Momentum influx = Min Uin = iu . n = -U cos B

~Ftotal = Fe + FM = kpUr

37

FM6 V J


37/52

Lift on a Vortex in a Cylinder

When a vortex is in a uniform stream, to determine the lift force both the pressureforce and the momentum influx into a circular cylinder must be considered.

If the vortex is in a flow whose streamlines form a cylinder around it, there is no

momentum influx so the pressure force is the complete force.

A closed circle in a stream can be represented by a dipole.

o>

The velocity potential of a 2D dipole is < P d= A 2+x 2.X z

For the flow to make a circle of radius equal to 1 in a stream of speed U, A = U.the x- and z-directed speeds on the circle of radius 1 due to the dipole are:

Wd = -2Uzx = -2UsinOcosO = -Usin20

The speeds obn the circle due to the vortex are:

. rU v = --sinO

27r

rWv = -cosO

27r

The pressure on the circle is:

p = -~ [(u-Ucos2{}- ~S i n { } r+ (-USin2{} + ~COS { } ) '-U']- f!_[U 2 + ( i_)2 _ 2U 2 cos 20 _ ur sin 0 + ur sin 0 cos 20 _ ur cos 0 sin 20j

2 27r 7r 7r 7r

38


38/52

The vertical force, Fw is:

{27r A (27rFw = 1 0 Pn kdi) = 1 0 - P sin ()d()

Fw p~ :o27r (- sirr' () + sin2 () cos 2() - sin ()cos ()sin 2()) aopur 1027r ( 2 1 2 1. 2 )_- - sin () - - cos 2() - - sm 2() d()2 IT 0 2 2pur (-IT _ IT _ IT )= -pur2 IT 2 2

39


39/52

Example: Design of 2D Airfoil Mean Line using Dipoles and Vortices

121 2t~ " ,. ...._= 2PU - 2P (U + u) ~ -pUu

~ -.............. J]-~~x-------~------~\~\------

~. ~= 0 ~= 1

e = x/c x = c e 'f]=y/c

Design Condition: P top = -1.0pU2e (1 - e)

Ut = 1.0U~(1 - ~)

(~ 2 ~ 3 )[Dipole StrengthJioil = I- l= 2.0U c 2 - 3 1.0J.Lwake= T Uc

C = In r = In [ (x - xo)2 + (y - Yo)2 J 1/2 = ~ In [ (x - xo )2+ (y - Yo)2 ]

(8C) 8C- --8n t 8yo (

8C) 8C8n b = 8yo

40


40/52

Consider the upper surface:

(ac) oc y - Yoan t = - ayo = (x - xo)2 + (y - Yo)21 ro o Y - Yo

(x, y) = 7r 10 p,(xo) (x - xo )2+ (y _ Yo)2dxo

Bay

The above analysis has an incorrect non-integrable singularity at x = Xo be-cause a careful limiting analysis requiring \l2 = 0 was not done.

However, another, and simpler, approach exists.

A dipole represents a jump in the potential. Another way to achieve a potentialjump is a vortex distribution.

GCCCCGCCIn length dx.; vortex strength =l'(xo)d xo. l'(xo) is vorticity/unit-length.

l'(x)U t ( x )= U + -2-

41


41/52

Now, we can solve for the mean line shape of the airfoil

For an arbitrarily defined pressure distribution, the integral for the slope canbe done numerically. Here, for the particular pressure distribution given, wewill solve for the slope analytically.

Then the shape is found by integrating the slope, s(~). This will be donenumerically.

42


42/52

C:\MATLAB6pl \work\ fo i l td .mJuly 11, 2002

Page9:24:51 A

format compactx = 0 : 0.01 : 1. 0;fac = -1.0/pi;s = fac.*(O.S - (LO-x) .*( x.*log1.0-x+eps) ./(x+eps))+1.0));h(l) = 0.0;for i = 2:101

h(i) = h(i-1) + (x(i)-x(i-1))*0.5*(s(i)+s(i-1));end;fid = fopen('ht.dat', 'w');for m = 1:101fprintf(fid, '%6.2f %7.4f %7.4f\n' ,x(m) I s(m), h(m));end;fclose (fid);plot(x,s)ylabel (,Slope' )xlabel (,\xi ')pause

plot(x,h)ylabel ( 'Camber' )xlabel (,\xi ')title('Camber vs \xi')axis([O 1 0 0.066])grid

43


43/52

j. . f ; J , oPt- IfE:~I6--ftT0.00 0.1592 0.00000.01 0.1705 0.00160.02 0.1771 0.00340.03 0.1818 0.00520.04 0.1853 0.00700.05 0.1878 0.00890.06 0.1895 0.0108

0.07 0.1905 0.01270.08 0.1909 0.01460.09 0.1908 0.01650.10 0.1903 0.01840.11 0.1893 0.02030.12 0.1879 0.02220.13 0.1862 0.02400.14 0.1842 0.02590.15 0.1818 0.02770.16 0.1792 0.02950.17 0.1763 0.03130.18 0.1731 0.03310.19 0.1697 0.03480.20 0.1661 0.03640.21 0.1623 0.03810.22 0.1583 0.03970.23 0.1541 0.04130.24 0.1497 0.04280.25 0.1451 0.04420.26 0.1405 0.04570.27 0.1356 0.04710.28 0.1306 0.04840.29 0.1255 0.04970.30 0.1203 0.05090.31 0.1150 0.05210.32 0.1095 0.05320.33 0.1040 0.0543

0.34 0.0983 0.05530.35 0.0926 0.05620.36 0.0868 0.05710.37 0.0809 0.05800.38 0.0749 0.05870.39 0.0689 0.05950.40 0.0628 0.06010.41 0.0567 0.06070.42 0.0505 0.06120.43 0.0443 0.06170.44 0.0380 0.06210.45 0.0317 0.06250.46 0.0254 0.06280.47 0.0191 0.06300.48 0.0127 0.06320.49 0.0064 0.06320.50 0.0000 0.06330.51 -0.0064 0.06320.52 -0.0127 0.06320.53 -0.0191 0.06300.54 -0.0254 0.06280.55 -0.0317 0.06250.56 -0.0380 0.0621

44


44/52

0.57 -0.0443 0.06170.58 -0.0505 0.06120.59 -0.0567 0.06070.60 -0.0628 0.06010.61 -0.0689 0.05950.62 -0.0749 0.05870.63 -0.0809 0.0580

0.64 -0.0868 0.05710.65 -0.0926 0.05620.66 -0.0983 0.05530.67 -0.1040 0.05430.68 -0.1095 0.05320.69 -0.1150 0.05210.70 -0.1203 0.05090.71 -0.1255 0.04970.72 -0.1306 0.04840.73 -0.1356 0.04710.74 -0.1405 0.04570.75 -0.1451 0.04420.76 -0.1497 0.04280.77 -0.1541 0.04130.78 -0.1583 0.03970.79 -0.1623 0.03810.80 -0.1661 0.03640.81 -0.1697 0.03480.82 -0.1731 0.03310.83 -0.1763 0.03130.84 -0.1792 0.02950.85 -0.1818 0.02770.86 -0.1842 0.02590.87 -0.1862 0.02400.88 -0.1879 0.02220.89 -0.1893 0.02030.90 -0.1903 0.0184

0.91 -0.1908 0.01650.92 -0.1909 0.01460.93 -0.1905 0.01270.94 -0.1895 0.01080.95 -0.1878 0.00890.96 -0.1853 0.00700.97 -0.1818 0.00520.98 -0.1771 0.00340.99 -0.1705 0.00161. 00 -0.1592 -0.0000

45


45/52

0.15

0.1

0.05

< ll0..

0U 5

-0.05

-0.1

-0.15

0.1 0.2 0.3 0.4 0.5~

0 . 6 0.7 0 . 8 0 . 9 1

46


46/52

Cambervs S

0.04

0.1 0.2 0.3 0.4 0.5

S0.6 0.7 0.8 0.9 1

0.06 .

0.05

....Q).. cEeno 0.03

0.02

0.01

47


47/52

0.2

w 0.1c,0 0. . . . . Jen -0.1

-0.20 0.2 0.4 0.6 0.8 1.0

0.08. . . . .s: 0.060>_Q)

0.04:.

~0.02

0.2 0.4 0.6 0.8 1.0

48


48/52

foiltdforma t c ompa c tx = 0 : 0 . 01 : 1 . 0 ;fa c = -1. O /pi ;s = fa c. *( 0. 5 - ( 1. 0- x) .* ( x .* log (( 1. 0- x+ eps ). /( x+ eps )) +1 .0 )) ;h(l) = 0 .0 ;

for i = 2:101h(i) = h ( i- 1) + ( x( i) -x ( i- 1) ) *0 . 5* (s ( i) +s ( i- 1) ) ;end;fid = fopen( 'ht .dat ' , 'w');for m = 1:101fprin tf( fid, '%6 .2 f %7 .4 f %7 .4 f\n ' , x (m) , sCm) , hem) ) ;end;fc los e ( fid) ;plot(x,s)y l abe l C 'S lope ' )x la b e l C' \x i ' )pause

plot(x,h)y labe l ( ' camber ' )x la be l C' \x i ' )title ( 'c amber vs \x i' )ax is ( [O 1 0 0 .066 J )grid

49


49/52

foi 1 tda .% vers ion of foiltd with one le s s loop for compu ting speed improvemen t

x = 0 : 0 .01 : 1 . 0 ;fa c = -1 .0 /p i ;s = fa c. *( O. 5 - ( 1. 0- x) .* ( x .* log (( 1. 0- x+ eps ). /( x+ eps )) +1 .0 )) ;

h(l) = 0 .0 ;for i = 2:101h( i ) = h (i- 1) + ( x (i) - x( i- 1) ) *0 . 5* ( s( i) +s ( i- 1 )) ;end;fid = fopen( 'hta .dat ' , IWl );q = [x ; s ;h ] ;fprin tf( fid, '%6 .2 f %7 .4 f %7 .4 f\n ' , q ) ;fc los e ( fid) ;plot(x,s)y l ab e l ( ' s lope I)x la b e l ( , \x i I)pause

plot(x,h)y l ab e l ( ' c ambe r I)x l a b el ( '\x i I)title ( ' c amber vs \x i' )a x is ( [O 1 0 0 . 066 ] )grid

50


50/52

foiltdb% This ve rS lon uses even more vec toriza tion and no " for" loops a t a ll.

% vers ion of foiltd with one les s loop for computing speed improvemen t

x = 0 : 0 .01 : 1 .0 ;

fac=

-1.0/pi ;5 = fa c. *(O. S - ( 1. 0- x) .* ( x .* log (( 1. 0- x+ eps ). /( x+ eps )) +1 .0 )) ;h(l) = 0.0 ;xd = [0 diff(x ) ] ; % This is [0 x(2) -x (1 ) x (3) -x (2 ) . . .]55 = [0 s (1 :end-1 ) + s (2 :end) ] % This is [0 5 (2 )+5(1) 5 (3 )+5(2)h = O.S*xd .* 55;h = cumsum(h) ; % Each elemen t is the sum of the ones bve fore it.

fid = fope n (' h tb . da t' , 'Wi) ;q = [x ; s ;h] ;fprin tf( fid, '%6.2f %7 .4 f %7.4f\n ' , q ) ;fc los e ( fid) ;plot(x,s)y la b e l ( ' S lope ' )

x la be l ( ,\x i ' )pause

plot(x,h)ylabel( 'camber ' )x la be l ( ,\x i ' )title ( ' camber vs \x i' )ax is ( [O 1 0 0 .066] )grid

51


51/52

C:\MATLAB6pl\work\foiltda.mJuly 11, 2002

Page9:23:45 AM

% Version of foiltd with one less loop for computing speed improvement

x = 0 : 0.01 : 1.0;fac = -1.0/pi;s = fac.*(0.5 - (1.0-x) .*( x.*log((1.0-x+eps) ./(x+eps))+1.0));h(l) = 0.0;

for i = 2:101h(i) = h(i-1) + (x(i)-x(i-1))*0.5*(s(i)+s(i-1));end;fid = fopen('hta.dat', 'w');q = [x;s;h];fprintf(fid, '%6.2f %7.4f %7.4f\n',q);fclose (fid);plot(x,s)ylabel (,Slope' )xlabel (,\xi ')pause

plot(x,h)ylabel ( 'Camber' )xlabel (,\xi ')title('Camber vs \xi')axis([O 100.066])grid

52


52/52

A:\foiltdb.mJuly 11, 2002

Page9:35:36 AM

% This version uses even more vectorization and no "for" loops at all.% version of foiltd with one less loop for computing speed improvement

x = 0 : 0.01 : 1.0;fac = -1.0/pi;s = fac.*(0.5 - (1.0-x) .*( x.*log1.0-x+eps) ./(x+eps))+1.0));

h(l) = 0.0;xd = [0 diff(x)]; % This is [0 x(2)-x(1) x(3)-x(2) ...]ss = [0 s(1:end-1) + s(2:end) ] % This is [0 s(2)+s(1) s(3)+s(2)h = 0.5*xd .* ss;h = cumsum(h); % Each element is the sum of the ones bvefore it.

fid = fopen('htb.dat', 'w');q = [x;s;h];fprintf(fid, '%6.2f %7.4f %7.4f\n',q);fclose (fid);plot(x,s)ylabel (,Slope' )

xlabel (,\xi ')pause

plot(x,h)ylabel ('Camber' )xlabel (,\xi ')title('Camber vs \xi')axis([O 1 0 0.066])grid

Numerical Methods in Incompressible Fluid mechanics

Documents

Transcript of Numerical Methods in Incompressible Fluid mechanics