# Boundary Layer Theory_P2

• View
28

3

Embed Size (px)

DESCRIPTION

Chapter presentation on Boundary Layer Theorems

Transcript

• CVEN 311 Fluid Dynamics 1

Further the equation can be re-written as

MOMENTUM INTEGRAL EQUATION

( ) 00 0

h hd dU dUu u U dy udy h U

dx dx dx

=

( ) ( ) 00 0

h hd dUu U u dy U u dy

dx dx

+ =

or

( ) ( )2 * 0d dUU Udx dx + =

For a flat plate at zero incidence (i.e. no imposed

pressure, dp/dx = 0, then

Hence,

( ) ( )2 * 0d dUU Udx dx + =

MOMENTUM INTEGRAL EQUATIONMOMENTUM INTEGRAL EQUATION of

the boundary layer which forms the

basis for approximate methods of

solving boundary layer problems

0=dxdUU

02

dU dx

=

• CVEN 311 Fluid Dynamics 2

In order to use this equation to estimate the boundarylayer thickness as a function of x, we must first: Obtain a first approximation to the freestream velocity

distribution, U(x). This is determined from inviscid flow theory(the velocity that would exist in the absence of boundarylayer) and depends on body shape.

Assume a reasonable velocity-profile shape inside theboundary layer

Derive an expression for 0 using the result obtained from theprevious item.

Reviewing the assumptions made in the derivation, it can be seen thatthe equation is:

Restricted to steady, incompressible, two-dimensional flow with nobody forces parallel to the surface.

Valid for either a laminar or turbulent boundary layer flow.

Step 1. Assume a suitable velocity profile for u(y)

inside the BL

For example, assume a linear velocity

distribution inside a BL

Determine constants A and B with the boundary

conditions as

u(0) = 0 B = 0

u() = U A = U/

VON KARMANS MOMENTUM INTEGRAL APPROACH

BAyyu +=)(

y

Uu

=

• CVEN 311 Fluid Dynamics 3

Thus U

yu

y

=

=

=00

VON KARMANS MOMENTUM INTEGRAL APPROACH

*

0 0

1 12

u ydy dyU

= = =

0 0

1 16

u u y ydy dyU U

= = =

dxd

U

=20 1

6ddx U U

= =

2( ) 12d

dx U

=

Integrating

Local drag coefficient

Skin friction drag

VON KARMANS MOMENTUM INTEGRAL APPROACH

2 12 x CU = +

At x=0, = 0, hence C = 0

2 12 xU =

2

212 12

Rexx Ux

= =12 3.464Re Rex xx

= =

02

2 0.5774Re

2

fx

cU U

= = =

3 2 1 2 1 20

0

0.5774L

DF Bdx U L B= =

• CVEN 311 Fluid Dynamics 4

VON KARMANS MOMENTUM INTEGRAL APPROACH Average drag coeffcient

Displacement thickness

Momentum thickness

21.1548

2 ReD

fL

F BLCU

= =

where UL

L =Re

* 1.7322 Re x

x = =

0.57746 Re x

x = =

This problem dealt with linear velocity profile as an approximate solution. The results

obtained are rough. However the exercise illustrates the use of the momentum

integral method. Practice this method with other types of approximated velocity

profile, such as parabolic, sinusoidal, etc.

One key point to remember

Be careful not to confuse the calculation for cf and Cf.

cf is a local calculation at a particular x location

(including x=L) and can only be used to calculate local

shear stress, NOT drag force. Cf is an integrated

average over a specified length (including any x L)

and can only be used to calculate average shear

stress and the integrated force over the length

• CVEN 311 Fluid Dynamics 5

EXAMPLE PROBLEM

Water at 15C flows over a flat plate at a speed of 1 m/s. Theplate is 0.4 m long and 1 m wide. The boundary layer on eachsurface of the plate is laminar. Assume that the velocity profilemay be approximated as linear. Determine the drag force on theplate.

Given

Working fluid is water at T = 15 C = 999 kg/m3 & = 1.14 10-3Ns/m2

U = 1 m/s

L = 0.4 m

W = 1 m

The boundary layer on each surface of the plate is laminar

Velocity profile is linear (assuming approximately)

Assumptions

Incompressible fluid flow

Laminar boundary layer

System diagramU = 1 m/s

L = 0.4 m

• CVEN 311 Fluid Dynamics 6

Governing Equations

Skin friction coefficient definition:

Reynolds number definition for a flat plate:

2

21 U

C wf

=

Ux

x =Re

LAMINAR BOUNDARY LAYER ON A FLAT PLATE:

APPROXIMATE SOLUTION USING PARABOLIC

VELOCITY PROFILE

Consider two-dimensional laminar boundary

layer flow along a flat plate. Assume the

boundary layer as parabolic.

Find expressions for:

The rate of growth of as a function of x.

The displacement thickness, *, as a function of x.

The total friction force on a plate of length L and width b.

• CVEN 311 Fluid Dynamics 7

In 1908 Blasius, a student of Prandtl, obtained an exact solution of thefollowing BL equations for a flat plate and demonstrated the shape ofthe boundary layer profile.

With the following boundary conditions u(y = 0) = 0

v(y = 0) = 0

u U as y

Blasius exact solution is valid only for laminar BL flow with no pressuregradient.

LAMINAR BOUNDARY LAYER (BLASIUS EQN.)

Parallel flow along a

plate with zero

LAMINAR BOUNDARY LAYER Inside the boundary layer since the viscous forces

are predominant Reasonable to assume: inertial and viscous forces are of

the same order of magnitude in a laminar boundary layer

Inertial forces/unit volumex

Ux

uu

2

For a flat plate

Viscous forces/unit volume 22

2

U

yu

yu

yy

=

=

If these forces are proportional to each other, then

x

kUx

kx Re2

2

=

=

2

2

Uk

x

U=

x

Uk

= A non-dimensional parameter

Rex

kkx Ux

= =

• CVEN 311 Fluid Dynamics 8

RESULTS OF BLASIUSEXACT SOLUTION

u approaches to 99 % of U at k = 5.

In other words when k becomes 5, y becomes 5.

Therefore using the definition of k, BL thickness atany x becomes

Using local Reynolds number definition in theabove equation we get

Ux 5=

grows with

Uxx =Re

x

x

Re5

=

Shear stress

Since in the boundary layer

Replacing in the above equation

RESULTS OF BLASIUSEXACT SOLUTION

U

yuU

yu

yy

=

== 00

0

;

x

U

3

0 constant =

xx

f Uxc

U Re0.664

Reconstantconstant

2

20

====

Blasiusexact analytical solution

cf = local drag coefficient

• CVEN 311 Fluid Dynamics 9

Total horizontal force (or skin friction drag)

Average drag coeffcient

Displacement thickness

Momentum thickness

RESULTS OF BLASIUSEXACT SOLUTION

BLUBdxFL

D21

0

21230 664.0 ==

L

Df U

BLFCRe328.1

22==

where UL

L =Re

x

x

Re729.1*

=

x

x

Re664.0

=

VON KARMANS MOMENTUM INTEGRAL APPROACH

Blasiusexact solution

laminar BL

over a flat plate

Momentum Integral Approach (MIA)

both laminar and turbulent BLs

over flat and curved surfaces

for any known U(x) and poutside (x) distributions

• CVEN 311 Fluid Dynamics 10

COMPARISON OF DIFFERENT SOLUTIONS

Source: Munson, Yong, Okiishi, Fundamentals of Fluid Mechanics, 3rd ed. Willey, 1998

PROBLEM

Air flows over a sharp edged flat plate 1 m long,

3 m width with a velocity of 2 m/s. For one side

of the plate, determine at the end of the plate,

0 at the middle of the plate, FD. [ = 1.23 kg/m3;

=1.46 10-5 m2/s]

• CVEN 311 Fluid Dynamics 11

TURBULENT BOUNDARY LAYER

Turbulent Boundary layers are usually thicker

than laminar ones.

Velocity distribution in a turbulent boundary

layer is much more uniform than that in a

laminar boundary layer

Large velocity change occur in a relatively small

vertical distance

Velocity gradient (dv/dy) is steeper in a turbulent

boundary layer than in laminar boundary layer

• CVEN 311 Fluid Dynamics 12

From experiments,

Velocity distribution in a turbulent boundary profilefollows 1/7th power law i.e.

Satisfactorily describes velocity distribution for mostof the region of turbulent boundary layer but givesinfinite slope at the wall,

Therefore it can not be used to predict 0

TURBULENT BOUNDARY LAYER

17u y

U

=

( ) 617 7u 1 7 U at y = 0yy

= =

Instead experimentally obtained measurements of the

shear profile are used such as

Putting the expression for the 1/7 power law into the

equations for displacement and momentum thickness

14

20 0.0225 U U

=

* 7,

8 72 = =

=99%

dm

TURBULENT BOUNDARY LAYER

• CVEN 311 Fluid Dynamics 13

Using momentum thickness obtained and the above 0relation in the integral momentum equation we can

obtain

Equating this to the experimental value of shear stress:

20

772

dUdx

=

147 0.0225

72ddx U

=

Integrating gives:

The turbulent boundary grows as x4/5, faster than the lamin