Full Elementary Aerodynamics Course by MIT

7/28/2019 Full Elementary Aerodynamics Course by MIT

1/158

Sensitivity Analysis

Often in engineering analysis, we are not only interested in predicting theperformance of a vehicle, product, etc, but we are also concerned with the

sensitivity of the predicted performance to changes in the design and/or errors inthe analysis. The quantification of the sensitivity to these sources of variability iscalled sensitivity analysis.

Lets make this more concrete using an example. Suppose we are interested inpredicting the take off distance of an aircraft. From Andersons, Intro. To Flight,an estimate for take-off distance is given by Eqn (6.104):

7f max

244.1

L LO SC g

W S

U

To make the example concrete, lets consider the jet aircraft CJ-1 described by Anderson. In that case, the conditions were:

0.1

19815

7300

/2.32

@/002377.0

318

max

2

3

2

f

LC

lbsW

lbsT

s ft g

level sea ft slugs

ft S

U

Thus, under these nominal conditions:

73000.1318002377.02.32

1981544.1 2 LOS

ft S LO 3182 @ nominal conditions

Suppose that we were not confident of the value and suspected that itmight be 0.1 from nominal.

max LC

That is,

0.9 1.1max L

C


2/158


Also, lets suppose that the weight of the aircraft may need to be increased for ahigher load. Specifically, lets consider a 10% weight increase:

1) Linear sensitivity analysis

2) Nonlinear sensitivity analysis (i.e. re-evaluation)They both have their own advantage and disadvantages. The choice is oftenmade based on the problem and the tools available. Well look at both options.

Linear Sensitivity Analysis

Linear sensitivity based on Taylor series approximations. Suppose we wereinterested in the variation of with w & LOS max LC Then:

W W

S C

C S

C S

W W C C S

LO L

L

LO L LO

L L LO

'w

w'

ww

#''

max

max

max

maxmax

)(

),(

That is, the change in is: LOS

W W

S C

C S

S LO L L

LO LO 'w

w'

ww

{'max

max

The derivativesW

S C S LO

L

LO

ww

ww

&max

are the linear sensitivities of to changes in

& , respectively.

LOS

max LC jW

Returning to the example:

maxmaxmax

2

244.1

L

LO

L L

LO

C S

T SC g

W C S

ww

f U

W S

T SC g W

W S LO

L

LO 244.12

maxw

w

f U

These can often be more information by looking at percent or fractional changes:

W W

W S

S W

C

C

C S

S

C

W W S

S C C S

S S S

LO

LO L

L

L

LO

LO

L

LO

LO L

L

LO

LO LO

LO

'w

w'

ww

'ww

'ww

#'

max

max

max

max

max

max

11

Fractional sensitivities

16.100 2


3/158


For this problem:

W W

S S

W S

S W

C

C

S S

C

S

S

C

LO

LO LO

LO

L

L

LO

LO

L

L

LO

L

'|

'w

w

'|

'w

w

0.20.2

10.1max

max

max

maxmax

L LO

LO'

|'

max

max

LO

LO '|'

.

Thus, a small fraction change in will have an equal but opposite effect onthe take-off distance.

max LC

The weight change will result in a charge of which is twice as large and inthe same direction.

LOS

Thus, is more sensitive to W than changes at least according to linear

analysis.

LOS max LC

Example

We were interested in varying 0.1 which according to linear analysis

would produce variation in take-off distance:max L

C

LOS 1.0P

ft S S C

ft S S C

LO LO L

LO LO L

3181.01.1

3181.09.0

max

max

|'o

|'o

For a weight increase of 10% we find

ft

S S lbW LO LO636

1.02198151.1

|

|'o

Nonlinear Sensitivity Analysis

For a nonlinear analysis, we simply re-evaluate the take off distance at thedesired condition (including the perturbations). So, to assess the impact of the

variations we find:max L

C

ft C S

C S

L LO

L LO

3535)9.0(

)7300)(9.0)(318)(002377.0)(2.32()19815(44.1

)9.0(

max

max

2

ft C S L LO 6.353)1.0( max''

which agrees well with linear result

t S S C

t S C

L

L L

3181.01.1

3181.090

max

max

'o

'o

16.100 3


4/158


Similarly,

ft C S L LO 28921.1max

ft C S L LO 2901.0max'

Finally, a 10% W increase to 21796lbs gives:

ft W W S

ft lbW S

LO

LO

6681.0

385021796

''

16.100 4


5/158

Kinematics of a Fluid Element

Convection: uK

Rotation rate :

1 1

2 2

i j k

u x y z u v w

vorticity

= =

=

KK K

K K

K

1

2

w v u w v ui j k

y z z x x y

= + +

KK K

Normal strain rates :

x

xx

x

dLudt

L x

= =

y yy

dL vdt z

= =

z ZZ

dL wdt z

= =

Shear strain rates :

Angle between edge1 1

along and along2 2

jiij ji

j i

uu d

i j x x dt

= + = =

Strain rate tensor :

xx xy xz

yx yy yz

zx zy zz

Convection Rotation Compression/Dilation(Normal strains) Shear Strain

Lx

Ly


6/158

Kinematics of a Fluid Element

16.100 2002 2

Divergence

( )/d Volumeu v wu Volume x y z dt

= + + =

K

Substantial or Total Derivative

u

Du v w

Dt t x y z

= + + +

K

=rate of change (derivative) as element move through space

Cylindrical Coordinates

x x r r u u e u e u e = + +K K K K

1 x r r xx rr

u u u u x r r r

= = = +

1 1

2r

r

u ur

r r r

= +

1

2r x

rx

u u x r

= +

1 1

2 x

xu u

r x

= +

( )1 1

r x

uu ru e

r r r

=

K K1

xr

u ue

r x

+ K r xu u e

x r

+ K

( )1 1r x ruu uu x r r r

= + +

K


7/158

Stress-Strain Relationship for a Newtonian Fluid

First, the notation for the viscous stresses are:

x

y

z

xx

xy

xz

zx

zy

zz

yx

yy

yz

ij = stress acting on the fluid element with a face whose normal is in + xidirection and the stress is in + x j direction.

Common assumption is that the net moment created by the viscous stresses arezero.

ij = ji


8/158

Stress-Strain Relationship for a Newtonian FluidLets look at this in 2-D:

dx

dy x

y

xx

xy

yx

yy

yy

yx

xy

xx

Mz

dx dy xy 2

dy yx dx 2 = Net moment about center= z

dx + xy dy yxdy

dx 2 2

M = ( xy )dxdy

z yx 2

Thus, for M = 0, xy = z yx

Assumptions for Newtonian fluid stress-strain:

1) ij is at most a linear function of ij .

2) The fluid is isotropic, thus its properties are independent of direction stress-strain relationship cannot depend on choice of coordinate axes.

3) When the strain rates are zero, the viscous stresses must be zero.

16.100 2002 2


9/158

Stress-Strain Relationship for a Newtonian Fluid

To complete the derivation, we consider the stress-strain relationship in theprincipal strain axes (i.e. where ij = 0 i for j ).

Thus,

11 = C 11 11 + C 12 22 + C 13 33

22= C 21 + C 22 22 + C 23 33

11

33= C 31 + C 32 22 + C 33 3311

But, to maintain an isotropic relationship:

C

C 11 = C 22 = C 3312

= C 21 = C 31 = C = C 23 = C 3213

which leaves only two unknown coefficients.

We define these two coefficients by:

(

= 2 + + 22 + )33ii ii 11K u

dynamic viscosity coefficient 2nd or bulk viscosity coefficient

For general axes (i.e. not in principal axes):

ij = 2 ij + + 22 + )ij ( 11 33

or, ij =

xui

j

+ u

xi

j

+ uK

ij

where

0 i j. ij =

1 i=

j

16.100 2002 3


10/158

Coordination Transformations for Strain & Stress Rates

To keep the presentation as simple as possible, we will look at purely two-dimensionalstress-strain rates. Given an original coordinate system ( x, y ) and a rotated system( y x, ) as shown below:

x

x'yy'

Recall that the strain rates in the x-y coordinate system are:

u 1 u v v = = = xx xy yy x 2 y

+ x y

Or, in index notation:

= ij

1

2

xu i

j

+ u

xi

j

Also, we note that the unit vectors for the rotated axes are:

KiK = cos i

K + sin jK K K

j = sin i + cos j

Thus, the location of a point in ( y x, ) is:

cos sin x= y y sin cos

Similarly, the velocity components are related by:

cos sin u u= v v sin cos

For differential changes, we also have

cos sin dx dx= dy dy sin cos


11/158


16.100 2002 2

Thus, defining T as the rotation matrix, we note that:

cos sin sin cos

x x x y

T y y x y

= =

Inverting this:

1

1 cos sin cos sin

sin cos sin cos

x x x y

T y y x y

= = =

Thus to find xu

in terms of vu , and their derivatives:

cos sin

u u x u y u u x x x y x x y

= + = +

Then, substituting sincos vuu += :

2 2

2 2

2

cos sin ( cos sin )

cos cos sin cos sin sin

cos 2cos sin sin

sin cos ( sin cos )

sin sin cos sin cos

xx xx xy yy

uu v

x x y

u v u v x x y y

vu v

y x y

u v x x

= + +

= + + +

= + +

= + + =

2

2 2

cos

sin 2sin cos cos

1 1 sin cos ( cos sin ) cos sin ( sin cos ) 2 2

yy xx xy yy

u v y y

u v u v v y x x y x y

+

= + +

+ = + + + + + 2 2

sin cos ( ) (cos sin ) xy yy xx xy = +


12/158


16.100 2002 3

If y x are the principal strain directions, then y y y x x x xy &,0 and= are

)

22

22

(cossin

cossin

sincos

xx yy y x

yy xx y y

yy xx x x

=

+=

+=

if 0 xy =

The next step is to relate the stresses in ( ) ( ) y xto y x ,, . Consider a differential surfacewith y normal:

The resultant stress is given as the vector K

and the force on the surface is .ds K

Decomposing the stress vector into the coordinate axes gives:

( )( ) ( )

yx yy

xx yx yy xy

dx i j dx

dy dx i dx dy j

= +

= + +

K KK

K K

Note that:

cossin

cos sin sin cos

dx dx

dy dx

i i j j i j

=

=

=

= +

K KKK KK

dx

dy

yx

yy

xx

xy

yy

yx

d x

x

x'yy'


13/158


16.100 2002 4

Thus, the second line becomes:

( sin cos )(cos sin )

( cos sin )(sin cos ) ( )

xx yx

yy xy yx yy

i j dx

i j dx i j dx

+

+ + = +

K K

K K K K

So collecting all the ji & terms (and enforcing yx xy = ) gives:

2 2

2 2

( )sin cos (cos sin )

sin cos 2 sin cos

yx yy xx yx

yy xx yy yx

= +

= +

For the principal strain axes,

( )0

22

=

++=

++=

xy

yy xx yy yy

yy xx xx xx

Plugging this into y y x y and gives

( )

( ) ( )

2 2

2 sin cos

2 sin cos

xy

xx yy yy

yx yy xx

yy xx yy xx yy

+

=

= + + +

Thus, we arrive at the known result:

( )

2

2

yx xy

yy yy xx yy

=

= + +

A similar derivation would give:

( ) 2 xx xx xx yy = + +


14/158

CompressibleVi

Compressibleiscid

IncompressibleVi

Compressible

CompressiblePotential

IncompressibleInvicid

Incompressible

Incompressible

Potential

c

scous

Inv scous Boundary Layer

Boundary Layer

Small Disturbance

Comp. Potential

Small DisturbanceInc. PotentialSubsonic Transoni Supersonic

*Each arrow indicates an additional assumption has been applied*More than one path to a given approximation


15/158

Compressible Viscous Equations

Also known as the compressible Navier-Stokes equations:

Mass: ( ) 0=

+

j

j

v xt

Momentum:( ) ( ) ( ), 1,2,3i i j ij

j i j

pi

t x x x

+ = + =

Energy: 2 21 1

( ) ( )2 2 j j

e v e v vt x

+ + + ( ) ( ) ( ) j ij i j j j j

p q x x x

= + +

( ), , ( )

ji k ijij

j i k

j

x x x

q k e e p state relationship ideal gas x

= + + = =

Incompressible Viscous Equations

In this case, we assume .const =

Mass: 0=

j

j

x

Momentum: ( )

+

+

=

+

i

j

j

i

ji ji

j

i

x x x x p

xt

Usually, ( )= . Often, when temperature variations are small, =const. is assumed.

0 j

i j

j ji i

j j i j j j i

x x

i

j j

x x x x x x

x x

=

+ = +

=


16/158

16.100 2

Usual form of momentum for incompressible flow:

j j

i

i ji

j

i

x x

v

x

pvv

xt

v

+

=

+

)(

In this case, the energy equation is not needed to find pvi & .

Incompressible Inviscid Equations

In this case we assume that the effects of viscous stresses are small compared toacceleration and pressure forces:

Mass: 0=

j

j

x

Momentum: ( )i i j j i

pt x x

+ =

These are known as the incompressible Euler equations.

Incompressible Potential Flow

In potential flow, we assume the flow is irrotational (i.e. 0V =K

). This allows thevelocity to be written as the gradient of a scalar potential:

,ii x

=

= Potential

Mass: 0 j j j x x

=

Also written out as: 022

2

2

2

2

=

+

+

z y x

or 2 20, Laplacian =

This is a single equation for a single unknown . It is the same for steady and unsteadyflows.

What happens to momentum?


17/158

16.100 1

Compressible Equations

Conservation of mass

( )

( )

0

0

0

0,

volume surface

d dv u ndS

dt

ut

Du

Dt u incompressible

+ =

+ =

+ =

=

K K

K

K

K

Conservation of Momentum

( ) ( ) ( )( ) ( )

Net viscous force in

:

volume surface surface surface

xx yx zx xy yy zy xz yz zz

xx xx zx

d udv uu ndS pn idS idS

dt

Noteds dx dy dz i dx dy dz j dx dy dz k

u puu

t x x y z

+ = +

= + + + + + + + +

+ = + + +

K KK K K K

KK KK

K

x

yx xx zx Du p Dt x x y z

= + + +

Recall:

2( ), 3

jiij ij

j i

uuu

x x

T

= + + = =

K

Incompressible Equations in Cartsian Coordinates

2

2

2

0, i.e. 0

1

1

1

u v wu

x y z

Du pu

Dt x Dv p

v Dt y

Dw pw

Dt z

+ + = =

= +

= + = +

K


18/158

Compressible Equations

16.100 2002 2

Where

2 2 22

2 2 2

, kinematic viscosity

Du v w

Dt t x y z

x y z

= + + +

= + +

Incompressible Equations for Cylindrical Coordinates

( ) ( ) ( )

( )

2 22 2

22 2

2

1 10

1 1 2

1 2

1

r x

r r r

r r

x x

ru u ur r r x Du p u u

u u Dt r r r r

Du u u p u uu

Dt r r r r

Du pu

Dt x

+ + =

= +

+ = + + = +

Where

2 2

22 2 21 1

, kinematic viscosity

r z

D uu u

Dt t r r z

r r r r r z

= + + +

= + +


19/158

Equations of Aircraft Motion

Force Diagram Conventions

Definitions

V flight speed angle between horizontal & flight path angle of attack (angle between flight path and chord line)W aircraft weight

L lift, force normal to flight path generated by air acting on aircraft D drag, force along flight path generated by air acting on aircraft pitching moment

T

propulsive force supplied by aircraft engine/propeller T

angle between thrust and flight path

To derive the equations of motion, we apply

= am F KK

(1)

Note: we will not be including the potential for a yaw force.

Applying (1) in flight path direction:dV F ma mdt

= = & &

and examining the force diagramcos sin

T F T D W = &

cos sinT

dV T D W m

dt = (2)

C h o r d l i n e

Fligh t pa th

T

M

W

D

T

Horizontal


20/158


16.100 2002 2

Now applying (1) in direction to flight path

2

c

V F ma m

r = =

where cr radius of curvature of flight path

sin cosT F L T W = +

2

sin cosT c

V L T W m

r + = (3)

Equations (2) & (3) give the equations of motion for an aircraft (neglecting yawingmotions) and are quite general. One important specific case of these equationsis level, steady flight with the thrust aligned w/ the flight path.

0, , 0, 0c T dV

r dt

= = =

Level, steady flightT D

L W

=

=

Moment definitionsThe pitching moment must be defined relative to a specific location. The twotypical locations are:

leading edge

1 ,4c quarter of mean chord


21/158


16.100 2002 3

Force & Moment CoefficientsTypically, aerodynamicists use non-dimensional force & Moment coefficients.

2

2

12 3D Drag/Lift coefficients

12

L

D

LC V S

DC

V S

where

is freestream density

V is freestream velocity (flight speed)S is a reference area (problem dependent)

21 Freestream dynamic pressure2

q V

The moment coefficient requires another length scale:

212

M ref

M C

V S

A

ref A reference length scale (problem dependent)

For 2-D problems, such as an airfoil, the forces are actually forces/length. So, for example

3D force 2D force/length

L L

D D

Similarly, M . The non-dimensional coefficients for 2-D are defined:


22/158


16.100 2002 4

2

2

2 2

1

2

1

2

1

2

l

ref

d

ref

m

ref

LC

V c

DC

V c M

C V c

where ref c is a reference length such as the chord of an airfoil.

Forces on Airfoils

The forces & moments on airfoils are normalized by the chord length. So,

2,

1

2

l L

C V c

2 2 2,

1 1

2 2

d m D M

C C V c V c

Force coefficients data is generally plotted in 2 forms:

Lift curve

Cl max

Cl


23/158


16.100 2002 5

Drag polar

Forces on Wings

)( yc chord distribution b wing span

S planform area =

2

2

b

b

cdy

A aspect ratioS

b 2

We can think of the 3-D or total lift on the wing as being the sum (i.e. integral) of the 2-D lift acting on the wing.

( )2

2

b

b

L L y dy

= where ( ) L y = lift distribution

Cl

Cd

Cd min

Cl

Cd

Cd min

x

y

b

c(y)

Wing planform


24/158


16.100 2002 6

The average 2-D lift on the wing L can be defined:

2

2

1b

b

L L L dy

b b =

Plugging that into LC :

2

2

2 2 21 1 1

2 2 2

b

b

L

L dy

L L bC

V S V S V S

= =

But, the average chord or mean chord can be defined as:

2

2

1b

b

S c cdy

b b =

2 21 1

2 2

L L L

C V S V c

=

In other words, we can think of the 3-D lift coefficient as the mean value of the 2-D lift coefficient on the wing. The same is true for drag and moment:

2 2

2 2 2

1 1

2 2

1 1

2 2

D

M

ref

D DC

V S V c

M M C

V S V c

=

=

A

where ref c=A is used.

A Closer Look at Drag

The drag coefficient can be broken into 2 parts:

NN

2

, L

D D e

parasiteinduced drag drag

C C C

eA= +


25/158


16.100 2002 7

where e = span efficiency factor (more on this when we get to lifting line).

The parasite drag contains everything except for induced drag including: skin friction drag wave drag pressure drag (due to separation)

It is a function of , thus, we can also think of , D eC as being a function of LC .The parasitic drag can be well-approximated by:

0

2, D e D L

C C rC = +

where0

DC drag at 0 LC = , r = empirically determined constant.

0

21 D D L

C C r C eA

= + +

Finally, we can re-define e to include r :

0

21 D D L

C C C eA

= +

where

1

ee

r eA

+ .

This re-defined e is known as the Oswald efficiency factor.

We will refer to0

DC as the parasite drag coefficient from now on.


26/158

16.100 2002 1

Brequet Range Equation

f

i

t

t

Range Vdt = In level flight at const speed:

L W = , Lift = WeightT D= , Thrust = Drag

The aircraft weight changes during flight due to use of fuel. Relate weightchange to time change:

dW dW dt

dt fuel weight

dt time

fuel weight T dt

time T

=

=

=

The quantity:

1 fuel weight time T

is known as the specific fuel consumption or sfc. It has units of:

sfc units =( ) fuel lb force

or lb hr force time

1 f

i

W

W

dW sfc T dt

V Range dW

sfc T

=

=

i

But since T D= and L W = we have:

f

i

W

W

V L dW Range

sfc D W =

Range

V

t=t iW=Wi

t=t f W=Wf


27/158

This is the general form of the range equation. The Breguet range equation is

found by assuming that D L

sfcV

is constant for the entire flight. In that case:

f f

i i

W W

W W

V L dW V L dW Range

sfc D W sfc D W = =

Breguet range equation:

logi

f

V L W Range

sfc D W =


28/158

Aerodynamic Center 1

Suppose we have the forces and moments specified about some referencelocation for the aircraft, and we want to restate them about some new origin.

ref M = Pitching moment about ref

new M = Pitching moment about new x

ref x = Original reference location

new x = New origin N = Normal force L for small

= A Axial force D for small

Assuming there is no change in the z location of the two points:

( )ref new ref new M x x L M = +

Or, in coefficient form:

N

. .

ref new

new ref m L M

meana c

x xC C C

c

= +

The Aerodynamic Center is defined as that location ac x about which the pitchingmoment doesnt change with angle of attack.

How do we find it?

1 Anderson 1.6 & 4.3

x

z

x ref x new

Mnew

Mref

A A

N ~ L~N ~ L~


29/158

Aerodynamic Center

16.100 2002 2

Let new ac x=

Using above:( )

ref ac

ac ref

L M

x x

C C C c

= +

Differential with respect to :

ref ac M M ac ref LC C x x C

c

= +

By definition:

0ac M C

=

Solving for the above

,ref

ref

M ac ref L

M ac ref ref

L

C x x C or

c

C x x x

c c C

=

=

Example:

Consider our AVL calculations for the F-16C 0ref x = - Moment given about LE

Compute ref M L

C

C

for small range of angle of attack by numerical

differences. I picked 0 03 to 3 = = .

This gave 2.89ac xc

.

Plotting M C vs about ac x

c

shows 0 M C

.

Note that according to the AVL predictions, not only is 0 @ 2.89 M acC

x

= , but

also that 0 M C = . The location about which 0 M C = is called the center of pressure .


30/158

Aerodynamic Center

16.100 2002 3

Center of pressure is that location where the resultant forces act and about whichthe aerodynamic moment is zero.

Changing new to cp and ref to NOSE to correspond to AVL:

NOSE

NOSE cp

NOSE

cp M L M

M cp

L

x xC C C c

C x

c C

= +

=

So if:

NOSE NOSE

L

M M

L

C C

C C

=

,

this will be true. This means that

0 at 0 M LC C =

.


31/158

C m

v s

A l p h a

f o r

F 1 6 C f r o m

A V L ( M =

0 )

- 3 . 5 - 3

- 2 . 5 - 2

- 1 . 5 - 1

- 0 . 5 0

0 . 5 1

1 . 5

- 1 0

- 5

0

5

1 0

1 5

2 0

A l p h a

C m

X r e

f =

0

X r e

f =

X a c =

2 . 8

9


32/158

W i n d T u n n e l

T e s

t A e r o

d y n a m

i c C e n

t e r

C h a r a c

t e r i s t

i c s

f o r

W a s

h o u

t a n

d R i g i d W i n g s

( A l t i t u

d e

= 1 0

, 0 0 0 f t

. )

1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 00 .

5

0 . 6

0 . 7

0 . 8

0 . 9

1

1 . 1

1 . 2

1 . 3

M a c h

N u m

b e r

A . C . ( % M A C )

R i g i d W i n g

W a s

h o u t

W i n g


33/158

Notes on CQ #1

( )

( )1

2

1 1

2 2

2

Re

Re

( )

d

d

d

D q cC

D q cC

Drag scales withV C V

=

=

=

=

==

=

V C V

V C V

qV pq

V pq

22

11

12

22

211

Re

Re

42

121

NN

12

2

12

1

2

1 1337 72

(Re)

From data, Re Laminar flow behavior

Re

, . . Turbulent Re ,

d

d

V V

d

D q c C

C

D q c

D V c f C D V

D withV

=

Note dependence on chord6172 , c.f. Turbulent D c D c

Drag Polar

N N

or L l ref ref

usually usuallywing chord planform

L LC C

q S q

A

C L

C D

C D min C D 0

0

C L min increasing

stall


34/158

Notes on CQ

16.100 2

For many aircraft,

( )min min2

min D D L LC C k C C +

Also, sincemin min

& 0o D D L

C C C 2

O D D LC C kC +

The first option will be slightly more accurate, but both are reasonableapproximations.

Notes on CQ #2

(1) First, we note that &O D

C k will almost certainly depend on the Reynoldsnumber. But, this dependence is probably weak since the b.l. flow will beturbulent. So, we assume &

O DC k remain constant to good approximation.

Also important is that for a general aviation aircraft, we expect no wave dragsince the flight is subsonic.

(2)

( )

0

2 2

2

2 2

2

2 22

12

1 112 22

1 1122

O

O

O

L

D L

D

D

D

D

D V S C kC

W V SC k V S

V S

k D SC V W

V S

= +

= +

= +

So we see that 20 21

& L D V D V

(3) Often at cruise, 0C C L D D for prop-aircraft.

0 02C C C C L D D D D= + =

At approach:

0 0 0

1 14 4

4 4C C C A D D D D= + =

Note: C D D>


35/158

Quick Visit to Bernoulli Land

Although we have seen the Bernoulli equation and seen it derived before, thisnext note shows its derivation for an uncompressible & inviscid flow. Thederivation follows that of Kuethe &Chow most closely (I like it better than

Anderson).1

Start from inviscid, incompressible momentum equation

1uu u p

t

+ =

There is a vector calculus identity:

( )2

,

12

vorticity

u u u u u

=

21 12

uu p u

t

+ + =

From here, we can make the final re-arrangement:

212

u p u u

t

+ =

Two common applications:1. Steady irrotational flow

0 0 Irrotational

Steady

ut

= =

2

2

10

2

1.

2

p u

p u const for entire flow

+ =

+ =

1 Kuethe and Chow, 5 th Ed. Sec 3.3-3.5


36/158


16.100 2002 2

2. Steady but rotational flow

0 0 Rotational

Steady

ut

=

212

p u u + =

This is a vector equation. If we dot product this into the streamwisedirection:

uu

s streamwise direction

( )( )

2

0,

2

2

12

1 02

1.

2

u u

s p u s u

d p uds

p u const along streamline

=

+ =

+ =

+ =

Vortex Panel Methods 2

Step#1: Replace airfoil surface with panels

Step #2: Distribute singularities on each panel with unknown strengths

In our case we will use vortices distributed such that their strength varies linearly

from node to node:

Recall a point vortex at the origin is:

1tan2 2

y x

= =

2 Kuethe and Chow, 5 th Ed. Sec. 5.10

1234

m m+1

i-1i

i+1

Original airfoil m-panels (m+1 nodes)


37/158


16.100 2002 3

A point vortex at y x , is:

1 tan2

y y x

=

Next, consider an arbitrary panel:

At any j s , we will place a vortex with strength ( ) j s ds :

( ) 1( )

, tan 2 j j

j

s ds y yd x y x x

=

where

( )

1

1

( )

j j j j j

j

j j j j j

j

s x x x x

S

s y y y y

S

+

+

+

+

Thus, the potential at any ( , ) x y due to the entire panel j is:

( ) ( ) 10

, tan2

jS

j j j

j

s y y y ds x x

We will assume linear varying on each panel:

( ) ( )1 j j j j j j

s s

S += +

s j S j (s j)

ds j

Vortex of strength (s j)ds

j+1

j

(x j,y j)

(x j+1,y j+1)

expandedview

Midpoints willbe (x j,y j)


38/158


16.100 2002 4

With this type of panel, we have m+1 unknowns = 1,,1...3,2,1 + mmm , so weneed m+1 equations.

Step#3: Enforce Flow Tangency at Panel Midpoints

The next step is to enforce some approximation of the boundary conditions at theairfoil surface. To do this, we will enforce flow tangency at the midpoint of eachpanel.

Panel method lingo: control point is a location where onu = is enforced.

To do this, we need to find the potential and the velocity at each control point.

The potential has the following form:

#

panels

freestream individual panel

potential potential

= +

Suppose freestream has angle :

( ) ( ) ( ) 11 0

, cos sin tan

2

jS m j j

j j

s y y x y V x y ds

x x

=

= +

The required boundary condition is ( ), 0 1i ii

x y for all i mn = =

So, lets carry this out a little further:

( ) ( ) 1,0 ,component of freestream normal

to surface of panel inormal velocity due to panel jat control poi

( )cos sin tan

2

j

i i

S j j

i i i j i j x y

s y y x y V i j n ds

n n x x

= +

0

nt of panel i

0m

j=

=

And recall ( ) ( ) j

j j j j j S

s s += +1 .

We can re-write these integrals in a compact notation:

( )1

11 2 1

0 , Influence of panel j due tonode j on control point of panel i

tan

2

j

ij ij

i i n ij

S j j

n j n ji j x y C

s y yds C C

n x x

+

=

= +

i.e. 1ijn jC =normal velocity from panel j due to node j on control point of panel i .

ii y x ,


39/158


16.100 2002 5

2ijnC = Influence of panel j due to node 1 j + on control point at panel i

Total normal velocity at control point of panel i due to panel j 1 2 1ij ijn j n jC C += + So, lets look at the control point normal velocity

So, for panel i , flow tangency looks like:

( ) ( )1 2 11

cos sin , for all 1ij ij

i

m

n j n j i j

V n

C C V i j n i m

+ =

+ = + = i

We can write this as a set of m equations for m+1 unknowns.

Question: What can we do for one more equation?

Step#4: Apply Kutta condition

We need to relate Kutta condition to the unknown vortex strengths j . To do this,consider a portion of a vortex panel.

ds


40/158


16.100 2002 6

Put a contour about differential element ds

Recall: [ ]

( ) ( )2 2 1 1

1 2 1 2

u ds

ds V dn U ds V dn U ds

V V dn U U ds

=

= +

=

Now let & 0dn ds :( )1 2

2 1

,

, or

, in generaltop bottom

dn O ds U U ds

U U

U U

= = =

So, since the Kutta condition requires top bottomU U = at TE:

. . 0, Kutta conditiont e =

For the vortex panel method, this means:

1 1 0m ++ =

Step#5: Set-up System of Equations & Solve

ds V2V

1

U2

U1

ds

dn


41/158


16.100 2002 7

1

2

8

11 12 19 1

21 22 2

31 32 33 3

41

51

61

71

881

9

0@ 1

0@ 2

.

. .

. .

.

0@

1 0 0 0 1 0

n

n

n

I

V I I I u n iV I I u n i

I I I

I

I I

I V u n i m I

Kutta

= = = = =

= =

i

i

i

nV

Where =ij I total influence of node j at control point i

For example:3637 2137 nn

C C I +=

The problem thus reduces to =Ax b , or, using our notation

nI = V ,

which we solve to find the vector of s!

1234

6

58

7

9Cn137Cn236

NodesControl points


42/158


16.100 2002 8

Step #6: Post-processing

The final step is to post-process the results to find the pressures and the liftacting on the airfoil.

airfoil L V V ds = = So, for our method, this reduces to:

( )11

1

2 1

12

12

m

i i ii

mi i i

l i

L V S

L S C

V V cV c

+=

+

=

= +

= = +

Vortex Panel Method Summary

In practice, the vortex panel method used for airfoil flows is a little different thanthe strategy used in the windy city problem. Heres a summary:

Step #1: Replace airfoil surface with panels

Note: the trailing edge is double-numbered 1 points, panelsm m + .

Step #2: Distribute vortex singularities with linear strength variables on eachpanel

( )1( ) j j j j j j

s sS

+= +

1234

m m+1

ds

S j

j

j+1

(j)

(j+1)

(s j)

s j ds


43/158


16.100 2002 9

This means we have m+1 unknowns:

1,............,3,2,1 +mm

Step #3: Enforce flow tangency at panel midpoints

0u n = at midpoint of every panel

equationsm

Step#4: Apply Kutta condition

Kutta condition becomes:

. . 1 10 01 equations & 1 unknowns

t e m

m m += + = + +


44/158

Kutta Condition

Thought Experiment 1

Suppose we model the flow around an airfoil using a potential flow approach.

We know the following: L V = u =

0 D = 0 = 0u n =

i

Bernoulli applies

Question: How many potential flow solutions are possible? Answer: Infinitely many!

For example:

Both of these flows have circulation which are not all equal

1 2 1 2 L L

1 Anderson, Sec. 4.5

V 8

p 8

L'

Flow #1 Flow #2


45/158

Kutta Condition

16.100 2002 2

Another difference can be observed at the trailing edge:

As a result of this and the physical evidence, Kutta hypothesized:

In a physical flow (i.e. having viscous effects), the flow will smoothly leave asharp trailing edge. - Kutta Condition

Flow #1 is physically correct!

Lets look at Flow #1 a little more closely:

Finite angle T.E. )0(>

te

Flow leaves t.e. smoothly.Velocity is not infinite.

Flow turning around a sharpcorner has infinite velocityat corner for potential flow.

Flow #1

Flow #2

t.e.

Vlower

Vupper T.E.

Upper and lower surfacevelocities must still betangent to theirrespective surfaces.

This implies 2 differentvelocities at TE.


46/158

Kutta Condition

16.100 2002 3

Only realistic option:

0 for finite angle T.E.lower upper V V = =

Note: from Bernoulli, this implies

2 2. . . .

0

1 12 2t e t e

p p V V =

= +

2. .

. .

12

is a stagnation point

with total pressure

t e

t e

p p V

TE

p

= +

Cusped TE )0( =te

In order for the pressure at the TE to be unique:

upper lower V V =

Vlower

Vupper

T.E.

In this case, velocitiesfrom upper and lowersurface are aligned.


47/158

Solution

0=nu at control pt #1:

The velocity at control pt #1 is the sum of the freestream + 3 point vorticesvelocities at that point:

1 2 31

2 2 22 2 2

u V i i i j

= + +

The normal at control pt #1 is:

1

1 21 1 0

2 22 2

n i

u n V

=

= + =

Rearranging:

1 2 V

=

(1)


Now, following the same procedure for control pt #2:

=10 miles

V = 100 mph 8

1

2

3 x

y


48/158

Solution

16.100 2002 2

2

2 32 2

1 32 2

02

n i j

V u n

= +

= + =

2 3

2V

=

(2)


3

1 33 3

1 32 2

02

n i j

V u n

=

= + =

1 3

2V

+ =

(3)

Final System of Equations

Combine the numbered equations:

1

2

3

vortexstrengths(unknowns)

Influence matrix

1 10

1 10

21 1

02

iV n

V

V

V

=

i

The problem with these equations is that they have infinitely many solutions.One clue is that the determinant of the matrix is zero. In particular we can add aconstant strength to any solution because:

0

0

0

influence 0matrix

=


49/158

Solution

16.100 2002 3

Given a solution0

0

0

, then

1 0

2 0

3 0

+

is also a solution where 0 is arbitrary.

So, how do we resolve this? Answer: the Kutta condition!

2 2. . . .

1 12 2t e upper t e lower

p V p V + = +

0upper lower V V =

Whats the Kutta condition for the windy city problem:

Kutta: 3 0 = no flow around node 3!

So, we can now solve our system of equations starting with 03 =

1

2

3

2

20

V

V

=

= =

V 8

1

2

3


50/158

Thin Airfoil Theory Summary

Replace airfoil with camber line (assume smallc

)

Distribute vortices of strength )( x along chord line for 0 x c .

Determine )( x by satisfying flow tangency on camber line.

0

( )0

2 ( )

cdZ d V

dx x

=

The pressure coefficient can be simplified using Bernoulli & assuming smallperturbation:

x

z

c

(x) = thickness

z(x) = camber line

x

z

c

z(x) = camber line

x

z

c

(x)dx


51/158


16.100 2002 2

{ }

2

2 2 2

2 2

22

2 2 2

2

2 2

2

higher order

12

1 1( )

2 2( )

112

21

2

p

p pc

V

p V u V p V

p p V u V V V

V V u u V V

u u V V V

=

+ + + = +

+ + =

+ + +=

+=

2 pu

C V =

It can also be shown that

( ) ( ) ( )

2( )

lower upper

upper lower

p p p upper lower

x u x u x

C C C u uV

=

= =

( )( ) 2 p

xC x

V

=

Symmetric Airfoil Solution

For a symmetric airfoil (i.e. 0dz dx

= ), the vortex strength is:

sin

cos12)(

+= V

But, recall:

(1 cos )2c

x =


52/158


16.100 2002 3

2

2

cos 1 2

sin 1 cos

1 1 2

sin 2 (1 )

xc

c

x xc c

=

=

=

=

1( ) 2

(1 )

c x V x xc c

=

1( ) 2 c x V

c

=

Thus,1

4 pcC

c

= .

Some things to notice: At trailing edge 0= pC . Kutta condition is enforced which requires upper lower p p=

At leading edge, pC ! Suction peak required to turn flow aroundleading edge which is infinitely thin.

The instance of a suction peak exists on true airfoils (i.e. not infinitely thin)though pC is finite (but large).

Suction peaks should be avoided as they can result in1. leading edge separation2. low (very low) pressure at leading edge which must rise towards trailing edge adverse pressure gradients boundary layer separation.

xc

Cp


53/158


16.100 2002 4

Cambered Airfoil Solutions

For a cambered airfoil, we can use a Fourier serieslike approach for the vortexstrength distribution:

1

flat plate camberedcontributions contributions

1 cos( ) 2 sin

sino nnV A A n

=

+

= +

Plugging this into the flow tangency condition for the camber line gives (after some work):

0 0

0

0 00

1

2cosn

dz A d

dxdz

A n d dx

=

=

After finding the n A s, the following relationships can be used to find , acmC C A , etc.

42 1

0 0 0 10

2 ( )

( )4

1 1(cos 1)2

c ac

LO

m m

LO

C

C C A A

dz d A Adx

=

= =

= =

A

Note: in thin airfoil theory, the aerodynamic center is always at the quarter-chord( 4

c ), regardless of the airfoil shape or angle of attack.


54/158

Important Concepts in Thin Airfoil Theory

1. This airfoil theory can be viewed as a panel method with vortex solutionstaking the limits of infinite number of panels & zero thickness & zero camber

{ }0

1

00 1 ( )

12

2

lim lim vortex panel thin airfoil theoryc

N

j ij i j

thickness N camber d dz

V K V n x dx

=

= =

= i

2. 2 ( )l LOC =

0

1(1 cos )

(1 cos )2

LO o o

o

dz d

dx

c x

=

=

0= LC for 0dz dx= {i.e. symmetric airfoils}

thickness does not affect l C to 1st order

3. Moment at4c

is constant with respect to according to thin airfoil theory

4c

= aerodynamic center

4c only depends on camber!

4

0c M = for symmetric airfoil

4. Thin airfoil theory assumes:

2-dimensions Inviscid* Incompressible* Irrotational* Small Small max c Small max z c

* ' 0 D =


55/158

Prandtls Lifting Line Introduction

Assumptions: 3-D steady potential flow (inviscid & irrotational) Incompressible

High aspect ratio wing Low sweep Small crossflow (along span)

flow looks like 2-D flow locally with adjusted

Outputs: Total lift and induced drag estimates Rolling moment Lift distribution along span Basic scaling of

i DC with LC & geometry

2

ei L

D

C C

A = dominated by A , where

2b A S =

2

2

22

2

1

i

i

Lq S Dbq S eS

L L D

b q e q e b

=

= =

21

i L

Dq e b

=

In steady level flight where W L = , we have the following options for reducing i D :

Raise q (i.e. raise cruise velocity) Friction increases

Decrease span loading, W b

bW & coupled due to structures

Improve wing efficiency, e Can be difficult


56/158


16.100 2002 2

Geometry & Basic Definitions for Lifting Line

Chord is variable, )( ycc = Angle of attack is variable, )( y = and is a sum of two pieces

N N N

( ) ( ) g local freestream geometric

twist

y y

= +

Effective angle of attack is modified by downwash from trailing vortices

N( ) ( ) ( )eff i

induced

y y y

=

Note: 0i > for downwash

Local lift coefficient linear with eff

( )[ ( ) ( )]l o eff LOC a y y y =

Fundamental Lifting Line Equation

Basic model results by:

Assuming Kutta-Joukowsky locally gives L :

( ) ( )

( ) 2 ( )( )

( ) ( )l

L y V y

L y yC y

q c y V c y

=

= =

yc(y)

Cl( eff ,y)

x

y=-b/2 y=b/2


57/158


16.100 2002 3

Distributing infinitely many horseshoe vortices along wing 4c -line to model

induced flow:

[ ]

2 ( )( ) ( ) ( )

( )

2 ( )( ) ( ) ( ) ( )

( )

l o eff LO

o i LO

yC a y y y

V c y

ya y y y y

V c y

= =

= =

2 ( )( ) ( ) ( )

( ) ( ) LO io

y y y y

a y V c y

= + +

where2

2

( )( ) 1

( )4

bo o

ib o

d y dyw y dy yV V y y

= =

Note: only unknown is )( y !

We use a Fourier series to solve this.

=

= N

n

n n AbV 1

sin2 where cos

2

b y =

Thus, the governing equation is:

1 1

( )

4 sin( ) sin ( )

( ) ( ) sini

N N

n LO nn no

b n A n nA

a c

= == + +

y

x

(x)


58/158

Force Calculations for Lifting Line

Recall:

cos2

sin2)()(1

b y

n AbV y N

nn

=

== =

The local two-dimensional lift distribution is given by Kutta-Joukowsky:

)()( yV y L =

=

= N

nn n AV b L

1

2 sin2)(

To calculate the total wing lift, we integrate L :

2

2

2

1

( ) sin2

2 sin sin2

b

b

N

nno

b L L y dy dy d

bb V A n d

=

= =

=

But:0,

sin sin,

2o

m k m k d

m k

==

In this case, nm = and 1=k . So, the only non-zero term is for 1=n .

2(2 )2 2n

b L b V A

=

122

2 AV b L =

2

11

212

L L b A

C AS V S

= = =


59/158


16.100 2002 2

The induced drag is similar. In this case:

)()( y yV D ii =

From previous lecture,

=

= N

nni

nnA

1 sinsin

)(

1 1

sin2 sin

sin

N N

i n nn m

n D V nA bV A m

= =

=

Integrating along the wing:

( )

2

2

2 2

1 10

2 2

1 10

only 0 for

( )

sinsin sin

sin

sin sin

b

i ib

N N

n mn m

N N

n mn m

n m

D D y dy

nb V nA A m d

b V nA n A m d

= =

= =

=

=

= =

2 2 2

1

2 2 2

2

2

N

n

n

i n

b V nA

D b V nA

=

=

=

2

2 1

2

2

2 1

12

or (1 ),

where

i

i

N i

D nn

L D

N

n

n

DC nA

V S

C C

An A

=

=

= =

= +

=


60/158


16.100 2002 3

Lift Distributions

The lift distributions due to each of the n A terms can be plotted as well:

root

Elliptic Lift Distribution

Recall that minimum induced drag is achieved when 0=n A for 1>n . In thiscase:

sin2)(

sin2)(

12

2

AV b L

n AV b L n

=

=

but:

2

2sin 1 cos 12

yb

= =

2

21( ) 2 1

2

y L y b V A

b

=

Elliptic lift

)( L

20

tip at2b tip at

2b


61/158

Trefftz Plane Analysis of Induced Drag

Consider an inviscid, incompressible potential flow around a body (say a wing).We define a control volume surrounding the body as follows

Upstream flow is V and is in direction. Thus, drag is the force in x direction. Apply integral momentum in x to find induced drag.

++

=S S S S bodybody

dS n pdS nuu KKKK

First, on the body 0=nu KK , so:

+

=S S S body

dS n pdS nuu KKKK

Next, also on the body,

=bodyS

dS n p K force of body acting on fluid

We are interested in the exact opposite, i.e. the force acting on the body. In ,this is the drag, in z this is the lift, and in y this is a yaw or side force:

x, V 8

yz

Trefftz planeS T (part of S )

body

Sbody

S 8


62/158


16.100 2002 2

k L jY i DdS n pbodyS

KKKK =

S S

Di Yj Lk pndS u u ndS

+ + = KK K K K K K Now, lets pull out the drag:

=S S

dS nuudS in p D KKKKK

The next piece is to apply Bernoulli to eliminate the pressure:

)(21

21 2222 wvuV p p +++=

+++= S S

dS nuudS inwvuV p D KKKK )(21

21 2222

But, 2 2

0 for a closedsurface

1 1( ) ( )

2 2S S p V n idS p V n idS

=

+ = + K KK K

++=S S

dS nuudS inwvu D KKKK

)(21 222

Next, we divide the velocity into a freestream and a perturbation:

ww

vv

uV u

==

+=

where , ,u v w are perturbation velocities (not necessarily small).

Substitution gives:

2 2 2 21 ( 2 ) ( )2

S S

D V V u u v w n idS V u u ndS

= + + + + + KK K K

But, we note that

== S S

dS nuV dS nuV 0KKKK from conservation of mass


63/158


16.100 2002 3

+++= S S S

dS nuudS inwvudS inuV D KKKKKK )(

21 222

If we take the control volume boundary far away from the wing, then the velocityperturbations go to zero except downstream. Downstream the presence of trailing vortices will create non-zero perturbations (more on this in a bit).

So, , , 0u v w except on T S .

T S

D V udS = 2 2 21 ( ) (2T S

u v w dS u V + + + )T S

u dS +

2 2 21 ( )2 T S

D v w u dS = +

The final step is to note that far downstream the velocity perturbation must dieaway (in inviscid flow). The reason is that the trailing vortices, which far downstream must be in the x direction, cannot induce an component of velocity.

So, this brings us to the final answer

+=T S

dS wv D )(

2

1 22

In other words, the induced drag is the kinetic energy which is transferred into thecrossflow (i.e. the trailing vortices)!


64/158

Problem #1

Assume: Incompressible

2-D flow 0, 0 z V z

= =

Steady 0=t

Parallel 0=r V

a) Conservation of mass for a 2-D flow is:

1( r r V r r

N

0

1) ( ) 0

( ) 0 does not depend on

( )

V r

V V

V V r

=

+ =

= =

b) -mometum equation is:

V t

N

( ) r

steady

V V V V

r

+ +K

N

22

0

1 2(

r

r

V

p V V

r r

=

= + +

N

2

0

)

r V

V r

=

In cylindrical coordinates:

( ) r V V =K

N

0

1V

r r =

+

Thus,

1( )

V V V V

r

+

K

N

0 fromcontinuity

0

=

=

Also,

22

2 2

1 1V V V r

r r r r

= +

0=

r0

r1

1

0


65/158

Problem #1

16.100 2002 2

Combining all of these results gives:

2

this side is independent of

1 1 p V V r

r r r r r

=

Since the right-hand-side (RHS) is independent of , this requires that

constant p

= for fixed r . But as varies from 20

, it must be equal at

2&0 , that is )2()0( === p p . If not, the solution would be discontinuous.

Thus, 0 p

= constant must be zero!

The differential equation for V is:

0)(1

2=

r V

r V

r r r

A little rearranging gives:

0)(1 =

rV dr d

r dr d

Integrating once gives:

1)(1 C rV dr d

r =

Integrating again gives:

22

121

C r C rV +=

2112

C V C r

r = +


66/158

Problem #1

16.100 2002 3

Next, we must apply the no-slip boundary conditions to find V . Specifically,

at ooo r V r r == ,

at 111 , r V r r ==

because flow velocity equals wall velocity in a viscous flow.

So, apply 1& r r r r o == bcs:

11

12 1

1 1 0

0 12

1 1 1 1 1 121

1 0

0 1

1122

1( )

2

oo

o

o o oo

o o

r r r r C C r C r r r

r r r

C r C r r r

C r r r r r

== +

= + =

Or, rearranged a little gives:

1

11 1

1 1

1 1

o

oo o

o o

o o

r r r r r r r r

V r r r r r r

r r r r

= +

c) The radial momentum equation is:

2 22 2

1 1 2( )r r r r

V p V V V V V V

t r r r r

+ = + K

But 0&0 =

=

V V r so this reduces to:

r V

r p 2 =

Since2

0V r

always, then clearly 0

r p

.

Thus, pressure increases with r .


67/158

Problem #1

16.100 2002 4

d) On the inner cylinder, the moment is a result of the skin friction due to the fluidshear stress. For this flow in which only 0 V and is only a function of r , theonly non-zero shear stress is r and has the following form:

, the onlynon-zero strain

r

r V V V

r r r r r

=

= =


68/158

Problem #1

16.100 2002 5

Rotating Cylinders

For the problem you studied in the homework:1. What direction is the fluid element acceleration?

2. What direction are the net pressure forces on a fluid element?

3. What direction are the net viscous forces on a fluid element?

o

1


69/158

Viscous Flow: Stress Strain Relationship

Objective: Discuss assumptions which lead to the stress-strain relationship for a Newtonian, linear viscous fluid:

ji k ij ij

j i k

k

k

uu u x x

u u v wV

x x y z

= + + = + + = i

where =dynamic viscosity coefficient = bulk viscosity coefficient

Note:0,

1,ij

i j

i j

=

=

1shear strain rate in , plane

2 ji

ij i j j i

uu x x

x x

= +

Thus, written in terms of the strain rates, the stress tensor is:

( )viscous stress using indicial notationdue to shearing this isof a fluid element

viscous stress due to anoverall compression or expansion of the fluidelement's volume

2

kk

ij ij ij xx yy zz

= + + +

This stress-strain relationship can be derived by the following two assumptions:

1. The shear stress is independent of a rotation of the coordinate system2. The shear stress is at most a linear function of the strain rate tensor.

So, for example, xy :

zz yz yy xz xy xx xy aaaaaa 332322131211 +++++=

Clearly, assumption #2 gives 6 unknowns per shear, 1211 , aa , etc. Note: why do

zy zx yx &, not appear in this expression for xy ? The total number of unknownsfor all stresses are: 36. But, this can be eventually reduced by applying #1 to themost general linear form to the two unknowns & .


70/158

Viscous Flow Stress-Strain Relationship

16.100 2002 2

Stokes hypothesis

Stokes hypothesized that the total normal viscous stress on a fluid elementsurface,

))(32( zz yy xx zz yy xx +++=++

should be zero, i.e. 0=++ zz yy xx so that the normal force (stress) on asurface is only that due to pressure. This requires that

32

032 ==+

Comments

Experimental studies have indicated that 32 and in general is notnegative!

For incompressible flow, 0==++ V zz yy xx so 0=++ zz yy xx regardless of .

For most compressible flows V is small compared to shearing strains (i.e. yz xy , , etc.) so again, Stokes hypothesis has little impact. So, as a result,

common practice is to assume that 3

2= .


71/158

Integral Boundary Layer Equations

Displacement Thickness

The displacement thickness * is defined as:

*

0 0

1 1e e e

compress ible incompressible flow flow

u udy dy

u u

= =

The displacement thickness has at least two useful interpretations:

Interpretation #1

0

0

(1)

e

udy

u

dy

=

=

A

A + B

So, the difference is in area B .

* represents the decrease in mass flow due to viscous effects, i.e. lost* eevisc um =

y

u /ue

u /ue(y)

A

B


72/158


16.100 2002 2

Interpretation #2

Conservation of mass:

1 1

1 1

1

1

0 0

0 0

0

0

( )

1

y y y

e

y y

e e

y

e e

y

e

u dy udy

u dy udy yu

yu u u dy

u y dy

u

+

=

= +

=

=

Taking the limit of 1 y gives

*

0

1e

u y dy

u

= =

So, the external streamline is displaced by a distance * away from the bodydue to viscous effects.

Outer flow sees an effective body

s t r e am l in e

y

*(x)

y

y1u(y)

x

ue


73/158


16.100 2002 3

Karmans Integral Momentum Equation

This approach due to Karman leads to a useful approximate solution techniquefor boundary layer effects. It forms the basis of the boundary layer methodsutilized in Prof. Drelas XFOIL code.

Basic idea: integrate b.l. equations in y to reduce to an ODE in x .

Derivation:

Add )( u x continuity + momentum

2

2

( ) continuity momentum

ee

u x

u v u u du uu u v u

x y x y dx y

+ + + = +

2( )( ) ee

u du uuv u

y dx y y

+ = +

Now, we integrate from 0 to 1 y :

11 1

2

10 00

( ) y

y yee

u dudy uv u y

x dx

+ = +

Note:

1 11

100 0

( ) y y

y

e e ev u

uv u v y u dy u dy y x

= = =

So, the equation becomes:

x

y

*(x)ue(x)


74/158


16.100 2002 4

1 11

2

1 00 0

( ) y y

yee e

u u dudy u dy u y

x x dx

= +

After a little more manipulation this can be turned into (note we let 1 y also):

2 *( ) ew e ed du

u udx dx

= + (1)

where momentum thickness =0

1e e e e

u udy

u u

incompressible form = 1e eo

u udy

u u

Insight

Integrate (1) from stagnation point along airfoil & then down the wake

2 *

00 0

( ) ew e edu

dx u u dxdx

= +

But: 0=eu at stag. pt.

Bernoulli

( 0) & eedp du

x udx dx

= =

2 *

0 0drag (see AndersonSec 2.6 for proof)

e w xdpu dx dxdx

= +

N

*

0 0

friction form dragdrag

wdp

D dx dxdx

= +

x=0at stag. point

x

y

x

along wake

8


75/158


16.100 2002 5

Another common form of the integral momentum equation is derived below:

dxdu

uudxd e

eeeew*2 )( +=

2

*

(2 )

where

known as "shape parameter"

w e

e e e

d du H

u dx u dx

H

= + +

=


76/158

Correlation Methods for Integral Boundary Layers

We will look at one particularly well-known and easy method due to Thwaites in1949.

First, start by slightly re-writing the integral b.l. equation. We had:

dx

du

u H

dx

d

u

e

eee

w

)2(2 ++=

Multiply by vu

e :

)2(2

H dx

du

vdx

d

v

u

uee

e

w ++=

Then definedx

du

ve

2 = and this equation gives:

+= )2(2)( H

udx

dudx

d u

e

w

e

e

Thwaites then assumes a correlation exists which only depends on .Specifically:

)( H H = and )(

S

ue

w =

shape factor shear correlationcorrelation

[ ])(2()(2)( H S dx

dudx

d u

e

e +

now this is an approximation


77/158

Correlation Methods for Integral Boundary Layers

In a stroke of genius and/or luck, Thwaites looked at data from experiments andknown analytic solutions and found that

!!645.0(

dx

du

x

dx

d u

ee

This can actually be integrated to find:

dxuu

v x

o

e

e

= 562 45.0

where we have assumed 0)0( == x for this.

16.100 2002 2


78/158

Method of Assumed Profiles

Here are the basic steps:

1. Assume some basic boundary velocity profile for ),( y xu . For example, this is a

crude approach but illustrates the ideas:

,0 ( )( , )( )

( )1, ( )e

y y xu x y

xu x

y x

m then < 0dx

dp e favorable pressure gradient

if 0 0dxdp e adverse pressure gradient

These edge velocities result from the following inviscid flows:

m

m

+

1

2

y

)( xu e y

x

2

2

)( xu e

0= x

Flow around a corner (diffusion) Wedge flow02 20


83/158

Falkner-Skan Flows

Some important cases:

0,0 == m : flat plat (Blasius)1,1 == m : plane stagnant point

The boundary layer independent variable from the Blasius solution generalizesto:

vx xum

y e)(

21+

and )()(),( 1 m f xu y xu e=

An interesting case in 1,1 == m , i.e. stagnation point flow:

)( xu e

xe K u =

inviscid flow velocityincreases away fromstag. pt. at 0= x

y

x

x

v K

y2

11 +=

v

K y= is independent of x

Boundary layer at a stagnation point does notgrow with x !

The skin friction can be found from:

)(

)(

11

00

2

0

o f

yd f d

xu yu

ye

yw

===

=

=

16.100 2002 2


84/158

Falkner-Skan Flows

Sincevx

xum yvx

xum y ee

)(2

1)(2

1 +=

+=

)()(

2

1)( 11 o f

vx

xum xu eew

+= tabulated

The skin friction coefficient is normalized by )(21 2 xu e :

)()(2

12

)(21

)( 112

o f x xu

vm

xu xC

ee

w f

+=

v x xu

o f m

C

e x

x

f

)(Re

Re

)(2

12 11

+

=

Note: separation occurs when 0= f C which means . From the table,

this occurs for 0)(11 =o f

19884.0=

This is only an angle of 18 o!18 o

16.100 2002 3


85/158

Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging

In a turbulent flow, we can define the mean, steady flow as:

0

1( , , ) lim ( , , , )

T

T u x y z u x y z t dt

T =

This allows us to split the flow properties into a mean and a fluctuating part:

mean turbulent part fluctuating

part

( , , , ) ( , , ) ( , , , )

( , , , ) ( , , ) ( , , , )

( , , , ) ( , , ) ( , , , )

( , , , ) ( , , ) ( , , , )

u x y z t u x y z u x y z t

v x y z t v x y z v x y z t

w x y z t w x y z w x y z t

p x y z t p x y z p x y z t

= += +

= += +

Note: the mean of u is zero:

{ }0 0

1 1lim lim

T T

T T

u u u

u u dt u dt T T

=

=

0

1lim

T

T

u

udt u uT

=

N

0

u u u=

=

0u = mean of fluctuations is zero.

Now, we will develop equations which govern the mean flow and try to developsome insight into how the fluctuations alter the mean flow equations.

Lets start with incompressible flow and look at the x momentum:

x momentum:

2 2 2

2 2 2

1u u u u p u u uu v w

t x y z x x y z

+ + + = + + +

Lets look at the averaging of this equation in time to develop an equation for the mean flow.


86/158


16.100 2002 2

[ ]time-average0of x-momentum

1lim or

T

T x mom dt x mom

T

The first term is:

0 0

1 1 ( )lim lim

T T

T T

u u udt dt

T t T t +=

But, 0=

t u

thus:

0 0

1 1lim lim

T T

T T

u u udt dt

T t T t t = =

Just as 0u = , we will assume 0=

t u

.

End result:

0=

t u

Lets skip over to the pressure term and look at its average:

p p p p p p x x x x x x

p p p x x x p p p x x x

= + = +

= +

= +

But, 0= p thus:

x p

x p

=

Similarly,

2

2

2

2

2

2

2

2

2

2

2

2

z u

yu

xu

z u

yu

xu

+

+

=

+

+

T


87/158


16.100 2002 3

Combining these into mom x , we now have:

2 2 2

2 2 2

must still work this out

1u u u p u u uu v w

y z x x y z

+ + = + + +

Lets work out the last term:

z u

w yu

v xu

u z u

w yu

v xu

u +

+

=

+

+

Thats the easy part. Now, consider xu

u

:

Question: What does xu

u

equal in terms of &u u only?

a) xu

u

b)u u

u u x

+

c) xu

u xu

u +

d) none of the above


88/158

Poiseuille Flow Through a Duct in 2-D

Assumptions:

Velocity is independent of 0, ==

xv

xu

x

Incompressible flow Constant viscosity, Steady Pressure gradient along length of pipe is non-zero, i.e. 0

x p

Boundary conditions:

No slip: ( ) 0 walls are not moving( ) 0

u y h

v y h

= =

= =

To be clear, we now will take the compressible, unsteady form of the N-Sequations and carefully derive the solution:

Conservation of mass:

0)( =+

V t

But 0=

t

because flow is steady and incompressible. Also, since =

constant, then V V = )(

0=

+

= y

v

x

uV

Finally, 0= xu

because of assumption #1 long pipe.

0=

yv

h y +=

h y =

y

x


89/158


16.100 2002 2

Now, integrate this:

C v ==constant

Apply boundary conditions: ( ) 0 ( ) 0v h v y = =

We expect this but it is good to see the math confirm it.

Now, lets look at momentum y .

Conservation of momentum y :

xy yy

xy yy

Dv p Dt y x y

v pV vt y x y

vt

= + +

+ = + +

0

steady

vu

x=

+

0

v

=

+0

xy yyv p y y x y

=

= + +

Now, what about yy xy &

xy

u v

y x

= +

0

2

xy

yy

u

y

v y

=

=

=

0v

u v x y

=

+ + 0

0 yy

V

=

=

i

So momentum y becomes:

0p u

y x y

= +

But 0u

x y =

because constant==

&0 xu


90/158


16.100 2002 3

0 (steady)

0 ( , , ) ( ) pt

p p x y t p x

y =

= =

Conservation of momentum x :

( )only

yx xx

p p x

Du dp Dt dx x y

=

= + +

ut

0

steady

uu

x=

+

0

v

=

+0

2u dp u y dx x x

=

= +

0

0

u y y

dp u

dx y y

=

+

= +

Now, we just need to solve this

u dp y y dx

=

so,&const )( yuu ==

2

2

only ( ) only ( )

1must be constant

f y f x

d u dpdy dx =

= const.dxdp

pressure can only be a linear function of !

Now, integrating twice in y gives:

21 0

1( )

2dp

u y y C y C dx

= + +

Finally, apply BCs:


91/158


16.100 2002 4

21 0

21 0

( ) 0

1( ) 0

2

1( ) 0

2

u h

dpu h h C h C

dx

dpu h h C h C

dx

=

+ = + + =

= + =

Solve for 0 1&C C gives:

20

1

12

0

dpC h

dx

C

=

=

2

2

1( ) 1

2

dp yu y

h dx h

=


92/158

Assume steady

0=

t

Assume 1>>h L

0V x

=

K

Assume 2-D

0,0 == z

w

Incompressible N-S equations:

1. 0=+

yv

xu

2.2 2

2 2

1u u u p u uu v

t x y x x y

+ + = + +

3.2 2

2 2

1v v v p v vu v

t x y y x y

+ + = + +

BCs

wuh xu

h xu

h xv

=+

==

),(

0),(

0),(

Turning the crank:

N

0

0 0 ( )

but 0 const

Apply bc's 0

x

u v vv v x

x y y

vv

x

v

=

+ = = =

= =

=

Now, momentum y : Since 0=v , we have:

0 ( , ) ( ) p

p x y p x y

= =

y

2hx

y=-h

y= h

p L pR

uw


93/158

Lecture 28

16.100 2002 2

Note: since the pressure does change from to L R p p over the length , ( ) L p p x= .

Finally momentum x :

N N

N

N

2 2

2 20

00

0

2

2

1

1,where

steady x

t

u u u dp u uu vt x y dx x y

u dp y dx

== = =

+ + = + +

=

Observe that )()( x g RHS y f LHS == and

L p p

dxdp

x y f

L R ==

==

const

const.)()(

For this problem, Ill just use the gradientdxdp

but realize this is specified by the

end pressures.

Next, integrate in y :

2

2

1

21

1

1

12 o

d u dpdy

dy dx

du dp y C dy

dy dx

dpu y C y C

dx

=

= +

= + +

Now, apply bcs:

wo

o

uC hC hdxdph yu

C hC hdxdp

h yu

=++=+=

=+==

12

12

21)(

021

)(

Solving for :& 1C C o


94/158

Lecture 28

16.100 2002 3

2

1

1 12

2

o w

w

dpC u h

dx

uC

h

= +

=

22

( ) 1 12 2

wh dp y u yu ydx h h

= + +

Suddenly started flat plate (Stokes 1 st Problem)

0IC: 0,

0

( ,0)BC: 0,

( ,0) 0

w

ut

v

u x ut

v x

==

==

>=

Assume infinite length, 0= x

Continuity:

N

0

0 ( )

but 0 so 0

u vv v x

x y

vv x

=

+ = =

= =

momentum y :

N N N

2 2

2 2

0 0 0 0

1

0 ( )

v v v p v vu v

t x y y x y

p p p x p

y

= = = =

+ + = + +

= = =

y p

u


95/158

Lecture 28

16.100 2002 4

momentum x :

N

N

N N

2 2

2 20

0 0

1

p p x x

u u u p u uu v

t x y y x y

= == =

+ + = + +

2

2

u ut y

=

This is the diffusion equation (also known as heat equation).

There are many ways to solve this equation Well use a similarity solution approach used in boundary layer theory.

Similarity Solution

Assume that ).,()(),( yt u yt u == where Reduce PDE to ODE. Usually, the assumption is made that:

y

b ybCt

y

t a

yaCt t

yCt

ba

ba

ba

==

==

=

1

1

d

dut

at d

dut u =

=


96/158

Lecture 28

16.100 2002 5

2

2

2 21

2

2 22

2

( )

( 1)

a b

a b

u u du d y y y y d dy

du b y d y

b du du b y y d d y y

b u dubCt y

y d y

b u dub b Ct y

y d

= = =

= +

= +

= +

d

du

ybb

d

ud

y

b

y

u22

22

2

2

)1()( +=

Thus:

2

2

2 2

2 2

becomes

( 1)

u uv

t y

a du b d u dub b

t d y d y d

=

= +

Re-arranging:

N

2 2

2 2

needs to be( ) only

1

f

b

d u a y b dud b t b d

yC

t

=

=

For simplicity, 11 and2

b C

= =

12

a =

yt

=

Documents

Full Elementary Aerodynamics Course by MIT