aerodynamics (chapter 1)

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

-

-

-

--

-

-

Navier-Stokes 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- Navier-Stokes equations.

2-Steady laminar flow between parallel flat plates.

3- Hydrodynamic lubrication.4- Laminar flow between concentric rotating cylinders.5- Example.

6- Problems; sheet No. 1

1- Navier-Stokes 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 Navier-Stokes 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


7/14

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 Navier-Stokes equation in the x-direction (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 Navier-Stokes 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 +

+

+

+

+

=

++

+

+

2

2

22

2

22

2111

In the above equations:ur= 0

w = 0

0,0,0 =

=

=

pu

t

body force = 0

The equation in - direction reduces to:

02

2

=

+

r

u

dr

d

dr

ud


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Integration:

( ) Arudr

d

r=

1

r

BAru += ---------(i)

B.C

( )

( )1221

2

2

2

2

2

1

122

1

2

2

2

21

222

111

=

+=

== ==

RR

RRB

RR

RA

RuRrRuRr

Sub. in eq.(i) yields:

( ) (

= 1222

212

11

2

222

1

2

2

1

r

RRrRR

RRu ) -------(ii)

The shear stress may be evaluated by the equation:

=r

u

dr

dr

By using eq.(ii):

( )1222

2

2

12

1

2

2

2 = rRR

RR

5- Example:

1- Using the Navier-Stokes equation in the flow direction, calculate the power required to

pull (1m 1m) flat plate at speed (1 m/s) over an inclined surface. The oil between the

surfaces has ( = 900 kg/m3 , = 0.06 Pa.s).The pressure difference between points 1 and

2 is (100 kN/m2) .

Solution:


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The Navier-Stokes equation in x- direction

+

+

+

=

+

+

+

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

uv

x

uu

t

ux

We have: Acceleration =0 , v=0 , w=0 ,2

2

2

2

,0z

u

x

u

=

The equation reduces to:

xgdx

dp

dy

ud

=

12

2

Integration

BAyygydx

dpu

Aygydx

dp

dy

du

x

x

++=

+=

22

22

1

1

B.C (b=10 mm)B

xg

b

dx

dpb

b

UAUuby

uy

22

000

+

===

===

xx gb

dx

dpb

b

Uygy

dx

dp

dy

du

22

1 +=

The shearing force on the moving plate:

areadyduF

areaF

by

o

=

=

=

area=1 m2

xgb

dx

dpb

b

UF

22+=

We havel

p

dx

dpggx

== ,sin

(Ans)5281528

528

30sin81.9900201.0

110100

201.0

01.0106.0

3

WPower

UFPower

NF

F

==

=

=

=


12/14

University of Technology Sheet No. 1

Mechanical Engineering Dep. Navier-Stokes Equations

Aerodynamics (3 rd year) 2011/2012

1- Using the Navier-Stokes equations, determine the pressure gradient along flow, the

average velocity, and the discharge for an oil of viscosity 0.02 N.s/m2

flowing betweentwo stationary parallel plates 1 m wide maintained 10 mm apart. The velocity midway

between the plates is 2 m/s. [-3200 N/m2 per m ; 1.33 m/s ; 0.0133 m3/s]

2- An incompressible, viscous fluid is placed between horizontal, infinite, parallel plates

as shown in figure. The two plates move in opposite directions with constant velocities U 1

and U2. The pressure gradient in the x-direction is zero. Use the Navier-Stokes equationsto derive expression for the velocity distribution between the plates. Assume laminar flow.

[ ( ) 221 UUUby

u += ]

3- Two parallel plates are spaced 2 mm apart, and oil ( = 0.1 N.s/m2 , S = 0.8) flows at arate of 2410

-4m

3/s per m of width between the plates. What is the pressure gradient in the

direction of flow if the plates are inclined at 60o with the horizontal and if the flow isdownward between the plates? [-353.2 kPa/m]

4- Using the Navier-Stokes equations, find the velocity profile for fully developed flow of

water ( = 1.1410-3 Pa.s) between parallel plates with the upper plate moving as shown infigure. Assume the volume flow rate per unit depth for zero pressure gradient between the

plates is 3.7510-3

m3/s. Determine:

a- the velocity of the moving plate.

b- the shear stress on the lower plate.

c- the pressure gradient that will give zero shear stress at y = 0.25b. (b = 2.5 mm)

d- the adverse pressure gradient that will give zero volume flow rate between the plates.

[3 m/s ; 1.37 N/m2

; 2.19 kN/m2

per m ; -3.28 kN/m2

per m]

5- A vertical shaft passes through a bearing and is lubricated with an oil ( = 0.2 Pa.s) asshown in figure. Estimate the torque required to overcome viscous resistance when the

shaft is turning at 80 rpm. (Hint: The flow between the shaft and bearing can be treated as

laminar flow between two flat plates with zero pressure gradient). [0.355 N.m]

6- Determine the force on the piston of the figure due to shear, and the leakage from the

pressure chamber for U = 0. [295.1 N ; 1.63610-8

m3/s]


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7- A layer of viscous liquid of thickness b flows steadily down an inclined plane. Show

that, by using the Navier-Stokes equations that velocity distribution is:

( )

sin2

2

2ybyu = and that the discharge per unit width is:

sin

3

3bQ =

8- A wide moving belt passes through a container of a viscous liquid. The belt movingvertically upward with a constant velocity Vo, as illustrated in figure. Because of viscous

forces the belt picks up a film of fluid of thickness h. Gravity tends to make the fluid drain

down the belt. Use the Navier-Stokes equations to determine an expression for the average

velocity vav of the fluid film as it is dragged up the belt. Assume the flow is laminar,

steady, and uniform. [

3

2h

Voav=v ]

9- Determine the formulas for shear stress on each plate and for the velocity distribution

for flow in the figure when an adverse pressure gradient exists such that Q = 0.

[b

yU

b

yUu

b

U

b

Uby 23;

4;

22

2

0 ==

= ==

y ]

10- A plate 2 mm thick and 1 m wide is pulled between the walls shown in figure at speed

of 0.4 m/s. The space over and below the plate is filled with glycerin ( = 0.62 N.s/m2).The plate is positioned midway between the walls. Using the Navier-Stokes equations,

determine the force required to pull the plate at the speed given for zero pressure gradient;

and the pressure gradient that will give zero volume flow rate.[496 N ; 372 kN/m

2.m]

11- A slider plate 0.5 m wide constitutes a bearing as shown in figure. Estimate:

a- the load carrying capacity.

b- the drag.

c- the power lost in the bearing.

d- the maximum pressure in the oil and its location.

[739.6 kN ; 348.6 N ; 348.6 W ; 12500 kN/m2

; 150 mm]

12- Consider a shaft that turns inside a stationary cylinder, with a lubricating fluid in the

annular region. Using the Navier-Stokes equation in -direction, show that the torque per

unit length acting on the shaft is given by:

1

42

2

1

2

1

=

R

R

RT

Where: = angular velocity of the shaft.R1 = radius of the shaft.R2 = radius of the cylinder.


14/14

Problem No. 2 Problem No. 4




aerodynamics (chapter 1)

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