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Chapter 1 : Boundary Layer

INTRODUCTION

The condition of zero fluid velocity at the solid

surface is referred to as ‘no slip’ and the layer of

fluid between the surface and the free stream fluid is

termed BOUNDARY LAYER.

1


2


Shear stress, τ

yu∂∂

= µτ

Reynolds number

The criterion which determines whether flow is

laminar or turbulent.

Reynolds number along a smooth flat plate:

υµρ xUxU ss

x ==Re

Re < 5 x 105 : Laminar

Re ≈ 5 x 105 : Transition (Engineering critical

Reynolds number)

Re > 5 x 105 : Turbulent

3


4


Turbulent layer usually has a greater

velocity gradient at the surface, which cause greater

shear stress.

From a turbulent layer, there is a more ready

interchange of particles with the main flow, and this

explains the more rapid increase in thickness of a

turbulent layer.

The thickness of a laminar boundary layer

increases as x0.5 (when pressure is uniform), a

turbulent layer thickens approximately as x0.8.

5


BOUNDARY LAYER IN PIPE

Laminar flow; Re < 2000

120 pipe diameters (=120D)

Turbulent flow; Re > 2000

60 pipe diameters (=60D)

6


DEFINITION

z Boundary layer thickness, δ

z Displacement thickness, δ*

z Momentum thickness, θ

BOUNDARY LAYER THICKNESS, δ

Boundary layer thickness is defined as that distance

from the surface where the local velocity equals

99% of the free stream velocity.

)99.0( sUuy ==δ

7


DISPLACEMENT THICKNESS, δ*

The displacement thickness for the boundary layer is

defined as the distance the surface would have to

move in the y-direction to reduce the flow passing

by a volume equivalent to the real effect of the

boundary layer.

∫=δ

δ0

* 1( − )dyUu

s

8


MOMENTUM THICKNESS θ

Momentum thickness is the distance that, when

multiplied by the square of the free stream velocity,

equals the integral of the momentum defect.

Alternatively, the total loss of momentum flux is

equivalent to the removal of momentum through a

distance θ. It is a theoretical length scale to quantify

the effects of fluid viscosity near a physical

boundary.

dyUu

Uu

ss∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

δθ

01

9


FLOW OVER IMMERSED BODY

When the object is completely surrounded by the fluid

and the flows are termed external flows.

Lift and drag is also called the fluid force.

Theoretical (analytical and numerical techniques) and

experimental approached are used to obtain information

on the fluid forces developed by external flows

One of the method to obtain flow data is by wind tunnel

testing works.

1


(a) Full scale wind tunnel test

(b) Model scale wind tunnel test

2


GENERAL EXTERNAL FLOW CHARACTERISTIC

Such as a airplane flying through still air, the fluid far

from the body is stationary and the body moves through

the fluid with velocity U.

Such as the wind blowing past a building, the body is

stationary and the fluid flows past the body with velocity

U.

To simplify the evaluation, we treat the situation as fluid

flowing past a stationary body with velocity U, called the

upstream velocity.

The velocity is assumed a uniform and constant velocity.

3


Three general categories of bodies are shown below.

1. Two-dimensional objects (infinitely long and of

constant cross-sectional size and shape)

2. Axisymmetric bodies (formed by rotating their

cross-sectional shape about the axis of symmetry)

3. Three-dimension bodies that may or may not possess a

line or plane of symmetry.

4


LIFT AND DRAG CONCEPTS

The resultant force in the direction of the upstream

velocity is termed the drag, D.

The resultant force normal to the upstream velocity is

termed the lift, L.

5


The resultant of the shear stress and pressure

distributions can be obtained by integral the effect of

these two quantities on the body surface as shown below.

The x and y components are;

θτθ sin)(cos)( dApdAdF wx += θτθ cos)(sin)( dApdAdF wy +−=

The drag, D is;

∫∫∫ +== dAdApdFD wx θτθ sincos

The lift, L is;

∫∫∫ +−== dAdApdFL wy θτθ cossin

6


The widely used alternative is to define dimensionless

lift and drag coefficients.

The lift coefficient CL is defined as;

AULCL 2

21 ρ

=

The drag coefficient CD is defined as;

AUDCD 2

21 ρ

=

A is a characteristic area of the object.

ρ is the density of flowing fluid

U is the upstream velocity

7


Typically, A is taken to be frontal area - the projected

area seen by a person looking toward the object from a

direction parallel to the upstream velocity U.

In other situations A is taken to be the platform area – the

projected area seen by an observer looking toward the

object from a direction normal to the upstream velocity.

8


LAMINAR BOUNDARY LAYER

For flat plate with zero pressure gradient.

Shear stress at wall ;

0=⎟⎟⎠

⎞⎜⎜⎝

⎛=

yo dy

duµτ

10


BOUNDARY LAYER EQUATION

0=∂∂

+∂∂

yv

xu

x-axis;

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

2

2

2

2

yu

xu

xp

yuv

xuu µρ

y-axis;

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

2

2

2

2

yv

xv

xp

yvv

xvu µρ

y-axis component could be ignored.

( v << u ) ( )0/ =∂∂ yp

( )(xpp = ) ( )0/ 22 ⇒∂∂ xu

11


Navier-Stokes equation;

0=∂∂

+∂∂

yv

xu

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

2

2

yu

xp

yuv

xuu µρ

=== ( )0/ ⇒∂∂ xp could be ignored ===

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

2

2

yu

yuv

xuu µρ

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∂∂

+∂∂

2

2

yu

yuv

xuu υ

Where υ is kinematics viscosity

12


boundary conditions;

y = 0 , u = 0

y = δ , u = U , ( )0/ =dydu

assumption from Blasius;

)(ηgUu= δ

η y∝

Base on Stokes’ idea, Blasius also said;

Uxυδ ∝ x

Uyυ

η =

From flow function ψ ;

yu

∂∂

=ψ

xv

∂∂

−=ψ

From above assumption, Navier-Stokes equation

could be written as;

13


3

3

2

22

yyxyxy ∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂∂ ψυψψψψ

Flow function was assumed as;

( )xU

fυψη =

Component velocity could be written as;

ηψ

ddfU

yu =

∂∂

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

∂∂

−= fddf

xU

xv

ηηυψ

21

differentiate ( )ηddfUu /= to x and y ;

14


2

2

2 ηη

η dfd

xUf

xU

xu

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

2

2

ηυη dfd

xUUf

yU

yu

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

3

32

2

2

ηυ dfd

xU

yu

−=∂∂

Replace above equation into Navier-Stokes equation,

and we found;

3

3

2

2

2

2

21

2 ηυυ

ηυηηυ

ηη

η dfd

xU

dfd

xUUf

ddf

xU

dfd

xU

ddfU =⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

It could be simplified as;

02 2

2

3

3

=+∂ ηη d

fdffd

15


Or;

0 2 =′′+′′′ fff

Or;

0 21

=′′+′′′ fff

Boundary conditions;

0=η 0==ηd

dff

∞=η 1==ηd

dff

16


Solution of the Blasius laminar flat plate boundary

layer in similarity variables. Function )(ηf is

solved using the Runge-Kutta numerical technique.

xUy υη = f U

uf =′ f ′′

3.0 1.3968 0.8460 0.1614

3.5 1.8377 0.9130 0.1078

4.0 2.3057 0.9555 0.0642

4.5 2.7901 0.9795 0.0340

5.0 3.2833 0.9915 0.0159

5.5 3.7806 0.9969 0.0066

6.0 4.2796 0.9990 0.0024

6.5 4.7793 0.9997 0.0008

7.0 5.2792 0.9999 0.0002

7.5 5.7792 1.0000 0.0001

17

BLASIUS SOLUTION WITH RUNGE-KUTTA METHOD.

xU

yυ

η = f Uu

f =′ f ′′

0 0 0 0.3321 0.5 0.0415 0.1659 0.3309 1.0 0.1656 0.3298 0.3230 1.5 0.3701 0.4868 0.3026 2.0 0.6500 0.6298 0.2668 2.5 0.9963 0.7513 0.2174 3.0 1.3968 0.8460 0.1614 3.5 1.8377 0.9130 0.1078 4.0 2.3057 0.9555 0.0642 4.5 2.7901 0.9795 0.0340 5.0 3.2833 0.9915 0.0159 5.5 3.7806 0.9969 0.0066 6.0 4.2796 0.9990 0.0024 6.5 4.7793 0.9997 0.0008 7.0 5.2792 0.9999 0.0002 7.5 5.7792 1.000 0.0001 8.0 6.2792 1.000 0.0000


Refer to above table, we found that;

5=η 09915.0=Uu

Value of Uu is suitable for boundary layer

thickness δ . ( )99.0=Uu

From;

xUyυ

η = with 5=η , δ=y

xUυ

δ=5

x

xRe5

=δ

18


Displacement thickness *δ ;

∫ ⎟⎠⎞

⎜⎝⎛ −=

δδ

01* dy

Uu

Uxddy υη=

x

xRe729.1* =δ

Momentum thickness θ ;

x

xdyUu

Uu

Re664.01

0=⎟

⎠⎞

⎜⎝⎛ −= ∫

δθ

Shear stress oτ ;

xy

o uU

dydu Re)3321.0(

0

µµτ =⎥⎦

⎤⎢⎣

⎡=

=

19


Local skin friction Cf ;

x

of U

CRe664.0

221

==ρτ

Average skin friction CF ;

x

AF

F UC

D

Re328.1

221

==ρ

20

VON KÁRMÁN INTEGRAL EQUATION

Control volume ABCD

Flow in : AB and BC

Flow out : CD

Assumption : incompressible flow, a unit width

surface, zero pressure gradient

21

Continuity equation ;

∫ ⋅∂∂

=−=δρ

0dxudy

xmmm inoutBC &&&

From 2nd law of Newton ;

∑ −−= BCinoutx MMMF &&&

( )

∫ ∫

∫∫∫∫

⎟⎠⎞

⎜⎝⎛∂∂

−∂∂

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

∂∂

+=

−+⎟⎠⎞

⎜⎝⎛

∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

++

δ δ

δδδδ

ρρ

ρρρρ

τδδδδ

0 0

2

0

2

00

22

0

21

Uudydxx

dydxux

Uudydxx

dyudydxux

dyu

dxddxxppddx

xppp o

22

Simplify ;

∫ ∫ ⎟⎠⎞

⎜⎝⎛∂∂

−∂∂

=−∂∂

−δ δ

ρρτδ0 0

2 Uudydxx

dydxux

dxdxxp

o

Divide by (-dx) ;

∫ ∫∂∂

−∂∂

=∂∂

+δ δ

ρρδτ0 0

22 dyux

dyux

Uxp

o

Simplify ;

∫ ∫−=+δ δ

ρρδτ0 0

22 dyudxddyu

dxdU

dxdp

o

(Von Kármán equation)

Pressure gradient ; (from Bernoulli equation)

CUP =+ 221 ρ (Neglect the potential energy)

dxdUU

dxdp

dxdUU

dxdp

ρ

ρ

−=

=+ 0

23

Von Kármán equation becomes ;

( )dxdUUU

dxdo *2 δθ

ρτ

+=

if ( ) 0=dxdp , and 0=dx

dU

Von Kármán equation becomes ;

dxdUoθρτ 2=

if δη y= and ηδddy =

∫ ⎟⎠⎞

⎜⎝⎛ −=

1

0

2 1 ηδρτ dUu

Uu

dxdUo

if ∫ ⎟⎠⎞

⎜⎝⎛ −=

1

01 ηα d

Uu

Uu

dxdUoδαρτ 2=

24

Use idea of Blasius ;

Shear stress ;

0=⎟⎟⎠

⎞⎜⎜⎝

⎛=

yo dy

duµτ

from Karman equation ;

0

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

yo dy

dudxdU µδαρτ

which ∫ ⎟⎠⎞

⎜⎝⎛ −=

1

01 ηα d

Uu

Uu

Assumption ;

( )ηfUu= , δ

η y=

Integral of ∫ ⎟⎠⎞

⎜⎝⎛ −=

1

01 ηα d

Uu

Uu

Will becomes constant, and assume that ;

25

( )0=

⎥⎦

⎤⎢⎣

⎡=

yddfηηβ

dxdUU δαρβ

δµ 2=

multiple with Uδ

and do integral;

∫∫ = δδαρµβ dUdx

pemalarUx +=2

2δαρµβ

at 0=δ and ; 0=x

( )21

21

21

Re

22

x

xU

x⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

αβ

αρµβδ

µρUx

x =Re is called local Reynolds number

21

22

Re2 ⎟⎟⎠

⎞⎜⎜⎝

⎛==

xo U

dxdU αβρδαρτ

26

Local skin friction coefficient ;

21

221 Re

2⎟⎟⎠

⎞⎜⎜⎝

⎛==

x

of U

C αβρτ

Drag force ;

( )21

32 LUFD αβµρ=

Average skin friction coefficient ;

( )( )2

1

21

Re

22

L

FC αβ=

27

conditions ;

0=u , and 0=y Uu = , δ=y

shear stress 0=τ at δ=y

( )yKdydu

−=⎟⎟⎠

⎞⎜⎜⎝

⎛= δµτ

Integral ;

cyyKu +⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

δµ

0=u , , 0=y 0=c

ηδ=y , and divided by Uµ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

cUK

Uu

2

22 ηηµδ

Uu = , 1=η , 22

=U

Kµδ

22)( ηηη −=f

28

From this ;

152

=α , 2=β

Boundary layer thickness ;

xx

xRe48.5

=δ

Displacement thickness ;

x

xRe826.1* =δ

Momentum thickness ;

x

xRe730.0

=θ

shear stress ;

xo

URe

365.0 2ρτ =

29

Local skin friction coefficient ;

xfC

Re73.0

=


LFC

Re46.1

=

Other assumptions ;

)sin(

32

byaudycybyau

=+++=

boundary conditions ;

(1) , 0=y 0=u

(2) , 0=y 02

2

=dy

ud

(3) δ=y , Uu =

(4) δ=y , 0=dydu

30

TURBULENT BOUNDARY LAYER ON FLAT PLATE

Shear stress in pipe ;

)2/(2 fUo ρτ =

R is radius of pipe ;

41

2

221

⎟⎠⎞

⎜⎝⎛=

RUUo

υρτ

If R is assumed as boundary layer thickness ;

41

2constant ⎟⎠⎞

⎜⎝⎛×=

δυρτ

UUo

From Von Karman equation ;

dxdd

Uu

UuU

UU δηρ

δυρ

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛× ∫

1

0

241

2 1constant

32

From Prandtl, constant = 0.0229

41

41 0229.0

⎟⎠⎞

⎜⎝⎛

=

υα

δδUdx

d

∫ ⎟⎠⎞

⎜⎝⎛ −=

1

01 ηα d

Uu

Uu

ONE-SEVENTH-POWER LAW

Velocity of flow ;

71

71

ηδ

=⎟⎠⎞

⎜⎝⎛=

yUu

1271

1

0=⎟

⎠⎞

⎜⎝⎛ −= ∫ ηα d

Uu

Uu

41

41

727

0229.0

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

=

υ

δδUdx

d

Integral ;

( )51

Re

376.0

x

xx

=δ

33

( )51

Re

047.0*x

x=δ

( )51

Re

037.0

x

x=θ

41

0229.0

⎟⎠⎞

⎜⎝⎛

=

υδα

δ

Udxd

Shear stress ;

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

−515

141

2

3769.010.0229 x

UUo

υρτ

Drag force ;

LUL

UFD

51

2036.0 ⎟⎠⎞

⎜⎝⎛=υρ

34


545

1

221

/ LuLU

AFCF ⎟⎠⎞

⎜⎝⎛=υ

ρ

35

FRICTION COEFFICIENT FOR MIXED

(TURBULENT + LAMINAR) FLOW

Assumption ;

Length of transition area is ignored.

DCDBDAD FFFF +−=

FDA : Friction force for turbulent from 0 to L

FDB : Friction force for turbulent from 0 to Xt

FDC : Friction force for turbulent from 0 to Xt

( ) ( ) ( )⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

LX

LX

LUF t

Xt

t

XtL

D

21

5158.2

102

21

Re

461.1

Re

073.0Relog455.0

ρ

Assumption ;

5105Re ×=Xt , L

Xtt

LX

ReRe

=

LLFC

Re1612

)Re(log4.0

58.210

−=

36

BOUNDARY LAYER THICKNESS FOR MIXED

(TURBULENT + LAMINAR) FLOW

Assume as Le

5105Re ×=Xt

⎟⎠⎞

⎜⎝⎛×=

×=

UUX Xt

tυυ 5105Re

distance from A to transition point;

xXLLe tturbulent ′+−=)(

with value of (Le)turbulent, we could defined displacement

thickness, momentum thickness and shear stress.

37


Question 1

Determine the δδ *

and δθ

for below mentioned velocity distribution.

1. 21

⎟⎠⎞

⎜⎝⎛=δy

Uu

2. 2

2 ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

δδyy

Uu

(parabolic profile)

3. ⎟⎠⎞

⎜⎝⎛=

δπ y

Uu

2sin (sinusoidal profile)

Question 2

If velocity distribution in a laminar boundary layer over a flat plate is

assumed to be given by second order polynomial, determine its form

using the necessary boundary conditions.


Question 3

In the boundary layer over the face of a high spillway, the velocity

distribution was observed to have the following form;

22.0

⎟⎠⎞

⎜⎝⎛=δy

Uu

The free stream velocity U at a certain section was observed to be 30m/s

and a boundary layer thickness of 60mm was estimated from the velocity

distribution measured at the section. The discharge passing over the

spillway was 6m3/s per meter length of spillway. Calculate ;

1. The displacement thickness

2. The energy thickness

3. The loss of energy up to the section under consideration


Question 4

1. Explain what you understand by boundary layer thickness and

displacement thickness.

2. Assume that in the laminar boundary layer the flow obeys the law,

shear stress dyduµτ = , where µ is the viscosity, which lead to the

velocity profile ( ) ( )2ykuU −=− δ , where U is the free stream velocity,

u is the velocity at a distance y above the plate and k is a constant.

Determine the displacement thickness.

3. The velocity distribution in the turbulent boundary layer is given by

71

⎟⎠⎞

⎜⎝⎛=δy

Uu

. Determine the displacement thickness.


SOALAN Sekeping plat rata, lebarnya b(m) dan panjangnya L(m), terendam di dalam satu aliran. Lapisan sempadan laminar terbentuk pada kedua-dua belah permukaan. Profil (taburan halaju) aliran lapisan sempadan adalah di dalam bentuk sinusoidal seperti berikut;

⎟⎠⎞

⎜⎝⎛=

δyBA

Uu sin

Dengan A dan B ialah pemalar dan U ialah halaju utama. Dapatkan nilai pemalar A dan B. Tunjukkan bahawa;

21

2327.0 ⎟⎠⎞

⎜⎝⎛=Ux

Uoυρτ

Dan

( )21

3308.1 LUbFD ρµ=

Halaju aliran minyak 3.0m/s melintasi plat rata, nipis dengan lebarnya 1.25(m) dan panjang 2.5(m). Berdasarkan taburan halaju di atas, tentukan; 1. Tebal lapisan sempadan. 2. Tegasan riceh pada pertengahan plat. 3. Jumlah daya geseran bagi kedua-dua belah plat. Diberi ρ = 850 kg/m3

υ = 10-5 m2/s


Q 1

A submarine can be assumed to have cylindrical shape with rounded nose. Assuming its length to be 50m

and diameter 5.0m. Determine the total power required to overcome boundary friction if it cruises at 8m/s

velocity in sea water at 20 degree Celsius.

( , ) 3/1030 mkg=ρ sm /101 26−×=υ

Q 2

A barge with a rectangular bottom surface 30m long times 10m wide is traveling down a river with a

velocity of 0.6m/s. a laminar boundary layer exist up to a Reynolds number equivalent to 5 x 105 and

subsequently abrupt transition occurs to turbulent boundary layer.

Calculate ;

i. The maximum distance from the leading edge up to which laminar boundary layer thickness

persists and the maximum boundary layer thickness at that point.

ii. The total drag force on the flat bottom surface of the barge.

iii. The power required to push the bottom surface through water at the given velocity.

( , ) 3/998 mkg=ρ sm /101 26−×=υ

Q 3

Sekeping plat rata nipis berbentuk segitiga sama terjunam ke dalam air dengan halaju seragam, U=3 m/s

seperti gambarajah. Tentukan ;

1. tempat berlaku aliran peralihan, diukur dari puncak plat

2. tebal lapisan sempadan pada sentroid plat

3. daya seret yang dialami plat

Q 4

Sebuah pelurus aliran terdiri daripada sebuah kotak dengan panjang L = 30 cm, luas keratan rentas A = 4 x

4 cm2 dipasang pada sebuah terowong angina. Jika 20 kotak serupa digunakan, tentukan daya seret yang

ditimbulkan oleh pelurus aliran tersebut. Halaju aliran yang memasuki kotak adalah seragam pada 10 m/s.


Question 1

Determine the δδ *

and δθ

for below mentioned velocity profile.

(a) ⎟⎠⎞

⎜⎝⎛=

δπ y

Uu

2sin

(b) 3

21

23

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

δδyy

Uu

Question 2

A laminar boundary layer velocity profile is given by; 4322)( ηηηη +−=f

where Uu

f =)(η and δ

η y=

Find the value of LDC Re where ReL is a Reynolds number for plate length L at trailing edge

and CD is drag coefficient (mean skin friction coefficient)

Question 3

Velocity distribution for a laminar boundary layer flow is given as

DyCByAu cossin +=

Determine Uu

using appropriate boundary layer conditions.

I_)

( - (

KE=-.-.

+ .% X X X Ub__aWT_l _TlXe g V aXfgg K

| UX ggYb Tg ba GYg X UbhaWL S

?

% X g| a X g g g T _ h b Y V _ T _ _ W ; - a g g X L T U _ X I _ K G O W T _ l l b e V T _ T _ T

L T U _ X I 5 e T g c _ T g X b X a _ Z 1 h _ g g Y b e i T a V f T f J A a X W U g g e T b j i X _ i c e G T _ X f

Hebgg ; TeTVgT ;e 7 ; 7

:J. f h f _ h b a

D a X T e

,76m%

HTeTUba V

+ ) 1 1 /

+ 023

,).-3

;

,)/1+

100 ,).,+

_K

V % X g X a X W _ X g b g T _ H b i X e g g g G c W h g W T + ) / = _ T a W _ X a Z W , , 0

" , + ' 3 a g l T e G Y V _ _ g _ X j F V b V g g D X

Y b j g b g b a g f g e Y = )

+'2

7- |

%-

K a jTiX V-

LT X ; 7 ' +S+2/7 q.( ,+++ ee

.'_k,+n '- f g

0 K

-

fE .-.

LeTaf gba % T _aTe gb h_Xag UbhaWT_l _TlXe Tbj TV __l bT_ef bNXe T

WVTZW, +YW_V 'W ag_ ; W_X iXg Heb WX TaW jT_ f XTe fgeXff

_T WXag Yb=aK)9 hfX Tcce Tg ba W afggba

gf g Tg g X TV XaghT_ W_ Xff bY g X UGhaWTi gTlXe e Tgaf VbaKgTag

9 f f T a _ _ g a b X e _ g _ _ a W _ a X f f ' T a W g _ Xb % Y G e

g bgg T cTZU+, _bVgg c gg_X gg T~_e2(Hb h_Xag

gl HebKD

,+ 1%

iX_b; gl ceG_; a T ggeUh_X_g UGhagg _TlXe baXa f TccebkB aTgXW Ul W_X

ecbj _TY XdhTg baS

=iT_hTgX 7r qYbeg X cbOXe_TjHeba_XS

,0 %

I-)x O Tg K T fgeXT ggaVg b)

0 T %

9 gXT ggaVgUTZ iXa Ul ggX XkHUT i7- ) XgXe_a aX W_X iX_bV gl

T HVggHC.',%

0 %

9 fbheVX +-=,. f) TaW NGegXk j ,,. f gg ,+VTe W

Tgg X baggaXgVaa aX

D W_X ;T Ybe H;gXag gg TaW gXT TVg bD

_ S N X _ ; G H G a X T g g g P 7 A T l 7 + S 0 T a W

) f XgV W_X Tbj Y_X_W'

,0 TU%

I_ x

+

U%

%

9 hDLA C:MJML=J99F E=C9FgC9

m )S,,

MF N=JK'LA L= FG b?Y E9D9QKg9

e _ 5 K E = - . - . h e E = ; 9 E ; K

/ 9g_Zh_ bLY.+ ggh Xf'58 / + 4,

IhXfYba

T' O Tg W4 ggag UN~ ~ X 8

HeGiXggTg XibT_ JTa Eo Xa gXZeTe Xd a_ ggXa TfK TY

O Xe

m Yl7 -

5O_gg K XTefgeXff

c 5Jh W WX9f gl

v 7YeXV eXTaJ NX,4 gl

s 5 b g g V aXff

w 5' 5W fgTaVV

,+ ggY

+ ZheX I_ f bgf heggeT_ J )cTKg T aTg'_T S XgXe g X OT f XHe fgeXffUXggeXXa gg 7

0 L S o c T Y7g. p

H9W 0V nTgg J LT X°

Z 'aW e7 S.,k + Ff -Ybe TYYS

,+ eaT '

M7+)0 0

|

_? J=I__n

(-(

KE=-.-. K ..+.

+hXfg b _

9 _T aTe UbhaWTel _TlXe iX_bVceba_X f Z iXa Ul5

j XeX i f Th W T a fgeXT iX_bVggl' c lf VT_ g V aXff bY g X UbhaWTel_TlXe TaW f Th W iX_bV gl Tg iXeg VT_ W fgTaVX Yeb g X fb__W UbhaWTel)

XgX aX ;_';- TaW gg a g X Y+,Uj aZ Xdha

7d

7;-7

f T W fgTaVX bY UbhaWTel _TlXe Yeba_ _XTW aZ XWZX'' f W fH_TVX Xag

s f b Xagh g V_Vff TaW JXe K JXlab_Wf ah UXe Tg cb ag

%

_ %

_i VeX A

g V aXff'

,0 Te %

U%9 e Tg Tg bfc Xe V ceXffheX'abjf cTfgg X aTg c_TgX Tf f bgia a ZheX I_'haWXe

T aTe VbaW g ba Tf a IhXfg Ga_);T_;h_TgX 'r 'TaW s Tg g Tg W fgTaVX

D70+ Va_ Y 07,- f)9_fb VT_Vh_TgX T iX_bV gl TaW TUfb_hgX ceXffheX Tg cb ag

y j V f _bVTgXW - TUbiX g X c_TgX)

,+ Te %

l

HT bfc XeV

ZheX I_

70+ V

M

=AC := eggDAJ K K .;KS/

KggL- -

9 "TJ7 - T fT T%

9 5 q7,)--

) e 5 b 7, 0,+ f

? 7 3 f R

,

9 U ? 7 ? g g ; K f i X = D

%7- (- . /

2%7

gg iT_gg bYJ

l

27

/ = c_TgX

;H;X g:_XTa Kgga %T VVX X SXWe

K

;,1KK gg P k/+ f T

b - ) q g1 T c Tg - ? GY

' . U B 2 f g g f G Y 9 S Z g K

_ T l; e , 5 H e b _ X I _x

,+

U

g

,0d |

X c TV7 ZV

I-

.

-(

g K E = - . - .

K = ; e , + F 9 K g b a V b a g T a f g e X X d h X f g G a f ' a f j X e A O b G % d h X f g b a f

IggKLAGF _ b0 Tg%

NX_bV gl a Ybe T _gg ggTe UGhaWTel _TlXe TVj f Z iXa Tf

7RK ;Vbf

G XgX aX Tccebce gg UGhaWTelaW g

,+

G aW g X g V aXff bY UG ATlXe TaW g X f XTe fgeXff _)0 gg g X _XTW aZ

XWZX GYT c_TgX) c_TgX f - ,+aZ TaW _)3 j WX)f c_TVXW a jTggg j V f

b Z T iX_bV gl bY-0+(cXe fXVbaW)LT X tgge7b)bb_H

-

,0 aTe %

MfX XdhTg baf UX_bj eaXXWXW

X f -s7 7

f a-k76ge|K a_VGfA%;

IhXK ,

K bj g Tg g X UbhaWTgg _TlXe W fc_TVXg_aXag TaW b XaghAA_ g V aXff

s'TeX Z iXa Ul5

7 ,(5'm0 g=ADTe f%

1%NX_bV gg W Kgg Uhg ba bY _T aTe UbhaW _TlXe Tbj ba T Tg c_TgX f Z iXa

Ul'

7 -&' &V

XgX aX g X bbXTV Xagf bY qTaW u)

) 0 Te f%

G Mf aZg X iX_bV 3 W fge Uhg ba bUgT aXW a G'VT_VagX'

L X eTg b GYW ggTXgg g V aXff gb UbhaWT_l _TlXe g V aX

L X eTg bb Xagh ggV aX_b gg_TlXeg V aXff,+ ,aTe f%

IhXKg Ga -

L X gheUh_Xgg iX_bV g W fge U a f Z iXa Ul

7 5%!

aW g X VbXTV Xag bY WeTZ'; Tf T Vg) Y JXlab_W aJ_ UbD JX)LT X g X

XdhTg ba bYj f XTe fgeXff Tf

ej7+'+-80-6%!,+ Te f%

G ;T_Vh_TgX g X Tg + +YWeTZ YbeVX ba g X ggag TeTaW eXgg a_YbYg X Tg c_TgX

g_XaZg TaW j Wg a T ha YG))_ iX_A Yg X UGhaWTel _TlXe f gheUh_Xag

biXe g X j bgg c_TgX)

,+ ,,,Te f%

s7 m ,(g'

TURBULENTBOUNDARYLAYER

EMPIRICALFORM

The empirical method of predicting turbulent flow

quantitiesonaflatplatewithzeropressuregradient

isbasedentirelyondata.

Itismoreaccuratethanthepower-lawformbutalso

morecomplicated.

The time average turbulent velocity profile can be

divided into two regions, the inner region and the

outerregion.

Theinnerisdefinedas:

M

MN

= P

MNQ

R

MN=

ST

U

= shearvelocity

Theouterregionisdefinedas:

VW− M

MN

= P

Q

Y

VW− M = velocitydefect

Theequationsabove involve the shearvelocity,MN,

whichdependsonthewallshearstress,ST.Thereare

several such relationships used; one that gives

excellentresultsis:

\]= ^

_=

0.455

ln 0.06 ∙ efg

h

\]= localskinfriction

This local skin friction coefficient (local drag

coefficient)allowsustodetermineSTandthusM

Nat

anylocationofinterest.Thevelocityprofilescanbe

usedtocalculatedquantitiesof interestbutMNmust

beknown.

Assuming turbulent flow flow the leading edge, the

shearstresscanbeintegratedtoyieldthedrag.Then

the skin friction coefficient (Drag coefficient)

becomes:

]̂= ^

l=

0.523

ln 0.06 ∙ efo

h

Thisrelationisverygoodandcanbeusedupto

ef = 10qwithanerrorof2%orless.Evenat

ef = 10sTtheerrorisabout4%.

Finally,thisequationcanbesummarized:

VW

MN

= 2.44 ∙ ln

MN∙ Y

R

+ 7.4

this equation allows an easy calculation of Y by

knowingMN.

EXAMPLE1:

EXAMPLE2:

introduction - universiti teknologi malaysiasyahruls/resources/skmm2323/2-boundary-layer … ·...

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