KULIAH X
EXTERNAL
INCOMPRESSIBLE
VISCOUS FLOW
Nazaruddin Sinaga
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Page 1
Main Topics
•
•
•
•
•
The Boundary-Layer Concept
Boundary-Layer Thickness
Laminar Flat-Plate Boundary Layer: Exact Solution
Momentum Integral Equation
Use of the Momentum Equation for Flow with Zero
Pressure Gradient
• Pressure Gradients in Boundary-Layer Flow
• Drag
• Lift
The Boundary-Layer Concept
The Boundary-Layer Concept
Boundary Layer Thickness
Boundary Layer Thickness
• Disturbance Thickness, d where
Displacement Thickness, d*
Momentum Thickness, q
Boundary Layer Laws
1. The velocity is zero at the wall (u = 0 at y = 0)
2. The velocity is a maximum at the top of the layer (u =
um at = d )
3. The gradient of BL is zero at the top of the layer (du/dy
= 0 at y = d )
4. The gradient is constant at the wall (du/dy = C at y = 0)
5. Following from (4): (d2u/dy2 = 0 at y = 0)
Navier-Stokes Equation
Cartesian Coordinates
Continuity
X-momentum
Y-momentum
Z-momentum
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Governing Equations
• For
incompresible
steady 2D
cases:
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Boundary Conditions
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Equations are Coupled, Nonlinear, Partial Differential
Equations
• Blassius Solution:
– Transform to single, higher-order, nonlinear, ordinary
differential equation
Boundary Layer Procedure
• Before defining and d* and q are there analytical
solutions to the BL equations?
– Unfortunately, NO
• Blasius Similarity Solution boundary layer on a flat
plate, constant edge velocity, zero external pressure
gradient
Blasius Similarity Solution
• Blasius introduced similarity variables
• This reduces the BLE to
• This ODE can be solved using RungeKutta technique
• Result is a BL profile which holds at
every station along the flat plate
Blasius Similarity Solution
Blasius Similarity Solution
• Boundary layer thickness can be computed by
assuming that d corresponds to point where U/Ue =
0.990. At this point,  = 4.91, therefore
Recall
• Wall shear stress w and friction coefficient Cf,x can
be directly related to Blasius solution
Displacement Thickness
• Displacement thickness d* is the imaginary
increase in thickness of the wall (or body), as
seen by the outer flow, and is due to the
effect of a growing BL.
• Expression for d* is based upon control
volume analysis of conservation of mass
• Blasius profile for laminar BL can be
integrated to give
(1/3 of d)
Momentum Thickness
• Momentum thickness q is another
measure of boundary layer
thickness.
• Defined as the loss of momentum
flux per unit width divided by U2
due to the presence of the growing
BL.
• Derived using CV analysis.
q for Blasius solution,
identical to Cf,x
Turbulent Boundary Layer
Black lines: instantaneous
Pink line: time-averaged
Illustration of unsteadiness of a
turbulent BL
Comparison of laminar and
turbulent BL profiles
Turbulent Boundary Layer
• All BL variables [U(y), d, d*, q] are determined
empirically.
• One common empirical approximation for the
time-averaged velocity profile is the oneseventh-power law
• Results of Numerical Analysis
Momentum Integral Equation
• Provides Approximate Alternative to
Exact (Blassius) Solution
Momentum Integral Equation
Equation is used to estimate the boundary-layer
thickness as a function of x:
1. Obtain a first approximation to the freestream velocity
distribution, U(x). The pressure in the boundary layer is
related to the freestream velocity, U(x), using the
Bernoulli equation
2. Assume a reasonable velocity-profile shape inside the
boundary layer
3. Derive an expression for w using the results obtained
from item 2
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Simplify Momentum Integral Equation
(Item 1)
 The Momentum Integral Equation becomes
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Laminar Flow
– Example: Assume a Polynomial Velocity Profile (Item 2)
•
The wall shear stress w is then (Item 3)
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Laminar Flow Results
(Polynomial Velocity Profile)
Compare to Exact (Blassius) results!
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
• Turbulent Flow
– Example: 1/7-Power Law Profile (Item 2)
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Turbulent Flow Results
(1/7-Power Law Profile)
Pressure Gradients in Boundary-Layer Flow
DRAG AND LIFT
• Fluid dynamic forces are
due to pressure and viscous
forces acting on the body
surface.
• Drag: component parallel
to flow direction.
• Lift: component normal to
flow direction.
Drag and Lift
• Lift and drag forces can be found by
integrating pressure and wall-shear stress.
Drag and Lift
• In addition to geometry, lift FL and drag FD forces are
a function of density  and velocity V.
• Dimensional analysis gives 2 dimensionless
parameters: lift and drag coefficients.
• Area A can be frontal area (drag applications),
planform area (wing aerodynamics), or wettedsurface area (ship hydrodynamics).
Drag
• Drag Coefficient
with
or
Drag
• Pure Friction Drag: Flat Plate Parallel to
the Flow
• Pure Pressure Drag:
Flat
Plate
Perpendicular to the Flow
• Friction and Pressure Drag: Flow over a
Sphere and Cylinder
• Streamlining
Drag
• Flow over a Flat Plate Parallel to the Flow: Friction
Drag
Boundary Layer can be 100% laminar,
partly laminar and partly turbulent, or
essentially 100% turbulent; hence
several different drag coefficients are
available
Drag
• Flow over a Flat Plate Parallel to the Flow: Friction
Drag (Continued)
Laminar BL:
Turbulent BL:
… plus others for transitional flow
Drag Coefficient
0.140
0.120
0.100
CD Laminar
0.080
CD Turbulen
0.060
0.040
0.020
0.000
1.E+00
1.E+01
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
1.E+08
1.E+09
Drag
• Flow over a Flat Plate Perpendicular to the
Flow: Pressure Drag
Drag coefficients are usually obtained empirically
Drag
• Flow over a Flat Plate Perpendicular to the
Flow: Pressure Drag (Continued)
Drag
• Flow over a Sphere : Friction and Pressure Drag
Drag
• Flow over a Cylinder: Friction and Pressure
Drag
Streamlining
• Used to Reduce Wake and Pressure Drag
Lift
• Mostly applies to Airfoils
Note: Based on planform area Ap
Lift
• Examples: NACA 23015; NACA 662-215
Lift
• Induced Drag
Lift
• Induced Drag (Continued)
Reduction in Effective Angle of Attack:
Finite Wing Drag Coefficient:
Lift
• Induced Drag (Continued)
Fluid Dynamic Forces and
Moments
Ships in waves present one of the most
difficult 6DOF problems.
Airplane in level steady flight: drag =
thrust and lift = weight.
Example: Automobile Drag
Scion XB
CD = 1.0, A = 25 ft2, CDA = 25 ft2
Porsche 911
CD = 0.28, A = 10 ft2, CDA = 2.8 ft2
• Drag force FD=1/2V2(CDA) will be ~ 10 times larger for Scion XB
• Source is large CD and large projected area
• Power consumption P = FDV =1/2V3(CDA) for both scales with V3!
Drag and Lift
• For applications such as tapered wings, CL and
CD may be a function of span location. For
these applications, a local CL,x and CD,x are
introduced and the total lift and drag is
determined by integration over the span L
Friction and Pressure Drag
Friction drag
• Fluid dynamic forces are comprised
of pressure and friction effects.
• Often useful to decompose,
– FD = FD,friction + FD,pressure
– CD = CD,friction + CD,pressure
Pressure drag
Friction & pressure drag
• This forms the basis of ship model
testing where it is assumed that
– CD,pressure = f(Fr)
– CD,friction = f(Re)
Streamlining
• Streamlining reduces drag by
reducing FD,pressure, at the
cost of increasing wetted
surface area and FD,friction.
• Goal is to eliminate flow
separation and minimize
total drag FD
• Also improves structural
acoustics since separation
and vortex shedding can
excite structural modes.
Streamlining
Streamlining via Active Flow
Control
• Pneumatic controls
for blowing air from
slots: reduces drag,
improves fuel
economy for heavy
trucks (Dr. Robert
Englar, Georgia Tech
Research Institute).
CD of Common Geometries
• For many geometries, total drag CD is
constant for Re > 104
• CD can be very dependent upon
orientation of body.
• As a crude approximation,
superposition can be used to add CD
from various components of a system
to obtain overall drag. However, there
is no mathematical reason (e.g., linear
PDE's) for the success of doing this.
CD of Common Geometries
CD of Common Geometries
CD of Common Geometries
Flat Plate Drag
• Drag on flat plate is solely due to friction created by
laminar, transitional, and turbulent boundary layers.
Flat Plate Drag
• Local friction coefficient
– Laminar:
– Turbulent:
• Average friction coefficient
– Laminar:
– Turbulent:
For some cases, plate is long enough for turbulent flow,
but not long enough to neglect laminar portion
Effect of Roughness
• Similar to Moody Chart
for pipe flow
• Laminar flow unaffected
by roughness
• Turbulent flow
significantly affected: Cf
can increase by 7x for a
given Re
Cylinder and Sphere Drag
Cylinder and Sphere Drag
• Flow is strong function of
Re.
• Wake narrows for turbulent
flow since TBL (turbulent
boundary layer) is more
resistant to separation due
to adverse pressure
gradient.
• qsep,lam ≈ 80º
• qsep,turb ≈ 140º
Effect of Surface Roughness
Lift
• Lift is the net force (due
to pressure and viscous
forces) perpendicular to
flow direction.
• Lift coefficient
• A=bc is the planform
area
Computing Lift
• Potential-flow approximation gives accurate
CL for angles of attack below stall: boundary
layer can be neglected.
• Thin-foil theory: superposition of uniform
stream and vortices on mean camber line.
• Java-applet panel codes available online:
http://www.aa.nps.navy.mil/~jones/online_t
ools/panel2/
• Kutta condition required at trailing edge:
fixes stagnation pt at TE.
Effect of Angle of Attack
• Thin-foil theory shows that
CL≈2 for  < stall
• Therefore, lift increases linearly
with 
• Objective for most applications
is to achieve maximum CL/CD
ratio.
• CD determined from wind-tunnel
or CFD (BLE or NSE).
• CL/CD increases (up to order 100)
until stall.
Effect of Foil Shape
• Thickness and camber
influences pressure
distribution (and load
distribution) and
location of flow
separation.
• Foil database compiled
by Selig (UIUC)
http://www.aae.uiuc.ed
u/m-selig/ads.html
Effect of Foil Shape
• Figures from NPS airfoil java
applet.
• Color contours of
pressure field
• Streamlines through
velocity field
• Plot of surface pressure
• Camber and thickness shown
to have large impact on flow
field.
End Effects of Wing Tips
• Tip vortex created by
leakage of flow from highpressure side to lowpressure side of wing.
• Tip vortices from heavy
aircraft persist far
downstream and pose
danger to light aircraft. Also
sets takeoff and landing
separation at busy airports.
End Effects of Wing Tips
• Tip effects can be
reduced by attaching
endplates or winglets.
• Trade-off between
reducing induced drag
and increasing friction
drag.
• Wing-tip feathers on
some birds serve the
same function.
Lift Generated by Spinning
Superposition of Uniform stream + Doublet + Vortex
Lift Generated by Spinning
• CL strongly depends on rate of
rotation.
• The effect of rate of rotation
on CD is small.
• Baseball, golf, soccer, tennis
players utilize spin.
• Lift generated by rotation is
called The Magnus Effect.
The End
Terima kasih
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Derivation of the boundary layer equations II
*
U 1
*
x1
*

U
*
1
U1
*
x1

*

U 1
*
x 2
*

U
*
1
U2
*
x 2
0
2
*
 L   U1
 

 
*
Re L  d   x * 2
x1
2
0  
p *
2
1
p *
*
x 2
82
p
Blassius exact solution I
Boundary layer over a flat plate
= Ue1
Variable transformation:
,
X 1  x1
Wall boundary condition:
X 2  x2
Stream function definition:
,
  X 1, X 2  
U
 x1
U
 x1
G X 2 
U1 

x2
U2  

 x1
83
Blassius exact solution II
Boundary layer over a flat plate
Ordinary differential equation
2
G
Boundary conditions
3
d G
dX
2
2
2
d G
dX
3
2
0
G X 2  0  0
The analytical solution of the ordinary differential equation was obtained by Blasius
using series expansions
dG
dX
 0  0
2
dG
dX
X 2
X 2
 1  1
2
84
Blassius exact solution III
Boundary layer over a flat plate
Solution
Velocity along x1-direction:

U1 
dG
dX
2
Velocity along x2-direction:
1

U2 
Boundary layer thickness: U 

2 
Displacement thickness:
d

Re
x1
d


1
 0 . 8604
Re

 5 x1
x1
d

1
Re
Wall shear stress:

1
dG
X2


G


2
dX 2

x1
1

d d  1 . 7208 x 1
Re
 w  0 . 332  U 
2
x1
1
Re
x1
85
Blassius exact solution IV
Boundary layer over a flat plate
Solution
86
Von Karman integral momentum equation I
Momentum conservation along x1-direction
1


 p  x 1  dx 1  d d
2


1 dp
 pd d 
dx 1 d d
2 dx 1
F1 BC
pd
F1  AD
 pd
pd
F1 AD
 F  M 
1
1 AB
  w dx 1
(p 
p
x1
dx 1 ) d  d d

F1 CD
  p  x 1  dx 1 d  d d 
dp
dp
  p d  pd d 
d dx 1 
dx 1 d d
dx 1
dx 1
 BC  M 1 CD
 M 1
87
Von Karman integral momentum equation II
Solution
w 
Considerations:
dP
dx 1
d  U e1
 

 x1 
d
 

 U 1 dx 2  

0

x

1 


d
0

 U 1 dx 2 
2

p(x1) = pe(x1)
Ue
dU e

dx 1
1 dp e
0
 dx 1
U e  U e1
 w   U
2

 

 x1 

d
0
U1 
U
1  1
U  
U


 dx 2 



88
Approximate solutions I
Linear, quadratic, cubic and sinusoidal velocity profiles
1.
Assumption of a self-similar velocity profile
U1*= f (x2*)
2.
Specifications of the boundary conditions
3.
Resolution of the Von Karman integral momentum equation
 w   U
2

  
d
 
 x1 





U 1 1  U 1 dx 2 
0


1
89
Approximate solutions II
Example: quadratic velocity profiles
Velocity profile
General form
+
boundary conditions
=
velocity profile
along x1-dir
*
*
*
*
*
*
*
*
*2
U 1  2x2  x2
U 1 ( x 2  0)  0
*2
U 1  C1  C 2 x2  C 3 x2
U 1 ( x 2  1)  1
*
U 1
*
x 2
*
( x 2  1)  0
Von Karman integral momentum equation solution
 w   U
 d
*
d
2

 
d
* 

x1
*
x1
*
 15
*
    U 2 2 d  2  U 
(
2
x

x
)
(
1

2
x

x
)
dx

*
*
0

15  x 1
L d
1
Re
*
1
*
2


*
2
d
*
2
*
 5 . 477
*2
2
*
2
1
Re
*
x1
x1
boundary layer thickness
90
Approximate solutions III
Example: quadratic velocity profiles
Displacement thickness:
from the definition
dd 


0
1
U1
U e1
dx 2
=> *
*
d d  1 . 826 x 1
1
Re
x1
Velocity along x2-direction
from continuity equation
U 2  0 . 913 U 
1
Re
x1
Wall shear stress
from
w 
2
L

U
d
*
=>
U 
2
 w  0,365
Re
x1
91