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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
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Sensitivity Analysis
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
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Sensitivity Analysis
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
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Sensitivity Analysis
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
''
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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
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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
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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
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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.
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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
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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
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Coordination Transformations for Strain & Stress Rates
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 = +
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Coordination Transformations for Strain & Stress Rates
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'
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Coordination Transformations for Strain & Stress Rates
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 = + +
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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
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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
=
+ = +
=
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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?
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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
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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
= + + +
= + +
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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
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Equations of Aircraft Motion
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
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Equations of Aircraft Motion
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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:
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Equations of Aircraft Motion
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
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Equations of Aircraft Motion
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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
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Equations of Aircraft Motion
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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= +
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Equations of Aircraft Motion
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.
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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
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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 =
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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~
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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 .
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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 =
.
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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
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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
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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
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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>
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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
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Quick Visit to Bernoulli Land
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)
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Quick Visit to Bernoulli Land
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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)
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Quick Visit to Bernoulli Land
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 ,
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Quick Visit to Bernoulli Land
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
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Quick Visit to Bernoulli Land
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
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Quick Visit to Bernoulli Land
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
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Quick Visit to Bernoulli Land
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
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Quick Visit to Bernoulli Land
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 += + = + +
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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
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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.
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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.
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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)
0=nu at control pt #2:
Now, following the same procedure for control pt #2:
=10 miles
V = 100 mph 8
1
2
3 x
y
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Solution
16.100 2002 2
2
2 32 2
1 32 2
02
n i j
V u n
= +
= + =
2 3
2V
=
(2)
0=nu at control pt #3:
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
=
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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
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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
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Thin Airfoil Theory Summary
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 =
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Thin Airfoil Theory Summary
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
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Thin Airfoil Theory Summary
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.
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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 =
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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
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Prandtls Lifting Line Introduction
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
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Prandtls Lifting Line Introduction
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)
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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
= = =
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Force Calculations for Lifting Line
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
=
=
= =
= +
=
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Force Calculations for Lifting Line
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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
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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
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Trefftz Plane Analysis of Induced Drag
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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
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Trefftz Plane Analysis of Induced Drag
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)!
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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
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Problem #1
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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 = +
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Problem #1
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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 .
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Problem #1
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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
=
= =
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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
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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 & .
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Viscous Flow Stress-Strain Relationship
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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= .
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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
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Integral Boundary Layer Equations
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
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Integral Boundary Layer Equations
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)
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Integral Boundary Layer Equations
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
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Integral Boundary Layer Equations
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
= + +
=
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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
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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
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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
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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
===
=
=
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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
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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.
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Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging
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
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Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging
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
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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
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Poiseuille Flow Through a Duct in 2-D
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
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Poiseuille Flow Through a Duct in 2-D
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:
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Poiseuille Flow Through a Duct in 2-D
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
=
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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
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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
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Lecture 28
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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
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Lecture 28
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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 =
=
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Lecture 28
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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
=
7/28/2019 Full Elementary Aerodynamics Course by MIT