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7/30/2019 Aerodynamics (Chapter 1)
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Aerodynamics
)(
3
rd
year Mechanical Engineering
2012/2011
..
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Syllabus )(
:/683
::5
::2
:1
:1
Subject Number: ME\683
Subject: AerodynamicsUnits: 5
Weekly Hours: Theoretical: 2
Experimental: 1
Tutorial : 1ContentsWeek
1
2
3
4






NavierStokes equations
 Introduction
 Derivation
 Laminar flow between parallel plates
 Couette flow Hydrodynamic lubrication
 Sliding bearing
 Laminar flow between coaxial
rotating cylinders
1
2
3
4
5
6
7
8
9
10








Boundary layer theory
 Introduction
 Displacement, Momentum, and
Energy thicknesses
 Momentum equation for the
boundary layer
 Laminar boundary layer
 Turbulent boundary layer
 Transition from laminar to turbulent
flow
 Effect of pressure gradient
 Separation and pressure drag
5
6
7
8
9
10
11
)(


Potential flow theory
(Ideal fluid)
 Introduction Continuity equation
 Vorticity equation
11
12



Basic concepts in potential flow
 Stream function
 Potential function
 Circulation
12
13




Basic flow patterns
 Uniform flow
 Source , Sink
 Doublet
 Free vortex
13
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14
15




Combination of basic flows Flow past a half body
 Flow past a Rankine oval
 Flow past a cylinder
 Flow past a cylinder with circulation
14
15
16
17




Incompressible flow over airfoils
 Introduction
 The Kutta condition
 Kelvins circulation theorem
 Thin airfoil theory
16
17
18




Airfoil characteristics
 Wind tunnel tests
 Estimation of aerodynamic coefficients
from pressure distribution
 Compressibility effects
 Reynolds number effects
18
19



Airfoil maximum lift characteristics
 Geometric factors effects
 Effect of Reynolds number
 Effect of leading and trailing edges
devices
19
20
21



Incompressible flow over wings
 Introduction
 Circulation, downwash, lift and
induced drag
 Finite wing theory
20
21
22



Wing stall
 Stall characteristics
 Effect of planform and twist
 Stall control devices
22
23


Lift control devices
 High lift devices
 Spoilers
23
24


Flow control devices
 Boundary layer control
 Reduction of drag24
25
26
27
28




Propellers Momentum theory
 Simple blade element theory
 Combined blade element theory and
momentum theory
 Propeller performance
25
26
27
28
29
30


Computational methods
 Introduction to panel methods
for airfoils
 Introduction to panel methods
for wings
29
30
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References
1 Fluid Mechanics By Streeter & Wylie
2 Introduction to Fluid Mechanics
By Fox & McDonald
3 Aerodynamics for Engineering Students
By Houghton & Carpenter
4 Airplane Aerodynamics and Performance
By Roskam & Edward Lan
:
.
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Chapter One
Navier Stokes EquationsContents
1 NavierStokes equations.
2Steady laminar flow between parallel flat plates.
3 Hydrodynamic lubrication.4 Laminar flow between concentric rotating cylinders.5 Example.
6 Problems; sheet No. 1
1 NavierStokes equations:
The general equations of motion for viscous incompressible, Newtonian fluids may
be written in the following form:
x direction:
(1)2
2
2
2
2
2
+
+
+
=
+
+
+
z
u
y
u
x
u
x
pg
z
uw
y
uv
x
uu
t
ux
y direction:
(2)2
2
2
2
2
2
+
+
+
=
+
+
+
z
v
y
v
x
v
y
pg
z
vw
y
vv
x
vu
t
vy
Equations (1) and (2) are called: Navier Stokes equations.
2 Steady laminar flow between parallel flat plates:
The fluid moves in the x direction without acceleration.
v= 0 , w= 0 , 0=t
the NavierStokes equation in the x direction (eq. 1) reduces to:
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(3)2
2
dy
ud
dx
dpg
x =+
dx
dhgggx == sin.
eq. 3 will be
( )(4)
12
2
dx
hpd
dy
ud
+=
Integration of eq. 4:
( )Ay
dx
hpd
dy
du+
+=
1
( )(5)
2
1 2BAyy
dx
hpdu ++
+=
B.C (Two fixed parallel plates)B
( )dx
hpdaAuay
uy
+===
===
20
000
eq. 5 will be
( ) ( ) (6)2
1 2ayy
dx
hpdu
+=
B.C (One plate is fixed and the other plate moves with a constant
velocity U) (Couette flow)
( )dx
hpda
a
UAUuay
Buy
+===
===
2
000
eq. 5 will be
( ) ( ) (7)a
Uy21 2 ++= ayy
dxhpdu
For the case of horizontal parallel plates:
0=dx
dh
eq. 7 will be
( ) (8)aUy212 += ayydx
dpu
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The location of maximum velocity umax may be found by evaluatingdy
duand setting it to
zero.
The volume flow rate is
=
a
dyuwQ0
(9).
3 Hydrodynamic lubrication:
Sliding bearing
Large forces are developed in small clearance when the surfaces are slightly inclined
and one is in motion so that fluid is wedged into the decreasing space. Usually the oils
employed for lubrication are highly viscous and the flow is of laminar nature.
Assumptions:
The acceleration is zero.
The body force is small and can be neglected.
Also2
2
2
2
2
2
2
2
and
z
u
y
u
x
u
y
u
The NavierStokes equation in the xdirection (eq. 1) reduces to:
dx
dp
dy
ud
12
2
=
Integration:
BAyydx
dpu ++
= 22
1
B.C
x
xx
h
U
dx
dphAuhy
UBUuy
===
===
20
0
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( )
+=
x
xh
yUyhy
dx
dpu 1
2
1 2
The volume flow rate in every section will be constant.
1wassume.
0
== xh
dyuwQ
dx
dphUhQ xx
122
3
= (*)
** For a constant taper bearing:
l
hh 21 =
( )xhhx = 1
Sub in eq.(*) and solving for
dx
dpproduces:
( ) ( )312
1
126
xh
Q
xh
U
dx
dp
=
Integration gives:
( ) ( )C
xh
Q
xh
Uxp +
=
2
11
66)(
(**)
B.C
0
00
===
===
o
o
pplx
ppx
( )2121
21 6andhh
UC
hh
hUhQ
+
=+
=
With these values inserted in eq.(**) we obtain the pressure distribution inside the bearing.( )
( )212
26)(
hhh
hhUxxp
x
x
+
=
The load that the bearing will support per unit width is:
=l
dxxpF0
).(
( )
+
=
1
)1(2ln
62
21
2
k
kk
hh
UlF
where
2
1
h
hk =
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4 Laminar flow between concentric rotating cylinders:
Consider the purely circulatory flow of a fluid contained between two long
concentric rotating cylinders of radius R1 and R2 at angular velocities 1 and 2.
In this case the NavierStokes equations in cylindrical coordinates are used.
r direction:
rrrrrrrr
rr g
z
uu
r
u
rr
u
r
ur
rrr
p
z
uw
r
uu
r
u
r
uu
t
u+
+
+
+
=
+
+
+
2
2
22
2
22
22111
 direction:
g
z
uu
r
u
rr
u
r
ur
rr
p
rz
uw
r
uuu
r
u
r
uu
t
urr
r +
+
+
+
+
=
++