boundary layers thickness-displacement-momentum

Boundary Layers

As a fluid flows over a body, the no-slip condition ensures that the fluid next to the boundary is subject to large shear. A pipe is enclosed, so the fluid is fully bounded, but in an open flow at what distance away from the boundary can we begin to ignore this shear?

There are three main definitions of boundary layer thickness:1. 99% thickness2. Displacement thickness3. Momentum thickness

99% ThicknessU

U is the free-stream velocity

(x)

x

y

(x) is the boundary layer thickness when u(y) ==0.99U

Displacement thickness

There is a reduction in the flow rate due to the presence of the boundary layer

This is equivalent to having a theoretical boundary layer with zero flow

y

u

y

uU

U

d

Displacement thicknessThe areas under each curve are defined as being equal:

0

dyuUq and Uδq d

0d dy

Uu1δ

Equating these gives the equation for the displacementthickness:

Momentum thicknessIn the boundary layer, the fluid loses momentum, so imagining an equivalent layer of lost momentum:

0

dyuUρum and m2δρUm

0m dy

Uu1

Uuδ

Equating these gives the equation for the momentumthickness:

Laminar boundary layer growth + d

dy

x

y

Boundary layer => Inertia is of the same magnitude as Viscosity

a) Inertia Force: a particle entering the b.l. will be slowed from a velocity U to near zero in time, t. giving force FI U/t. But u=x/t => t l/U where U is the characteristic velocity and l the characteristic length in the x direction.

Hence FI U2/l

b) Viscous force: F /y 2u/y2 U/2

since U is the characteristic velocity and the characteristic length in the y direction

(x)

Laminar boundary layer growthComparing these gives:

ρUμδ l

So the boundary layer grows according to l

Alternatively, dividing through by l, the non-dimensionalisedboundary layer growth is given by:

lRl1δ

Note the new Reynolds numbercharacteristic velocity and characteristic length

υU

μρU llRl

)(ρUμ5δ Blasiusl

U2/l U/2

Boundary layer growth

Length Reynolds Number

μρUlRl

l

U

Flow at a pipe entry

l

Ud

δ

If the b.l. meet while the flow is still laminar the flow in the pipe will be laminar

If the b.l. goes turbulent before they meet, then the flow in the pipe will be turbulent

Length Reynolds number and Pipe Reynolds number

The critical Reynolds number for flow along a surface is Rl=3.2*105

In a pipe, the Reynolds number is given by

dumRe

Considering a pipe as two boundary layers meeting, d=2a=2and from above

ρUμ5δ l

lRll 10μ

ρU10ρUμ10.

μρURe

The mean velocity in the pipe, um, is comparable to the free-stream velocity, U

If Rl=3.2*105 then Re=5657

Boundary layer equations for laminar flow

These may be derived by solving the Navier-Stokes equationsin 2d.

0

yv

xu

dtdu

yu

xu

ρμ

xp

ρ1

2

2

2

2

Continuity MomentumU

Assume:1. The b.l. is very thin compared to the length2. Steady state

Boundary layer equations for laminar flow

yuv

xuu

yu

ρμ

xp

ρ1

2

2

This gives Prandtl’s b.l. equation:

rate of change of u with x is small compared to y

Blasius produced a perfect solution of these equations validfor 0<x<3.2*105, and demonstrated the shape of the boundarylayer profile

Blasius Solution

0

5

0 1

u/U

y'

y' f' (or u/U) f'' 0 0 0.3321 0.330 0.3322 0.630 0.3233 0.846 0.2674 0.956 0.1615 0.992 0.0646 0.999 0.0027 1.000 0.000

lμρUy'y

Laminar skin frictionThe shear stress at the surface can be found by evaluatingthe velocity gradient at the surface

00 y

uμτ

The friction drag force along the surface is then found byintegrating over the length

dxyubμF

0y0f

l

where b is the breadth of the surface

Laminar skin frictionFrom the Balsius solution, the gradient of the velocityprofile at y=0 yields the result:

0.5x0 R

xU0.332μτ

The shear force can be obtained by integration along the surface

0.5

00f R0.664UbdxτbF l

l

μ The frictional drag coefficient can then be calculated

21

R33.1ρAUFC 2

21

ff

l

Force and momentum in fluid mechanics - refresher

Newton’s laws still apply. Consider a stream tube:

u1,A1

q1=u1A1

u2,A2

q2=u2A2

mass entering in time, δt, is ρu1A1δt

momentum entering in time, δt, is m1 = (ρu1A1δt)u1

momentum leaving in time, δt, is m2 = (ρu2A2δt)u2

Impulse = momentum change, F = (m2 – m1)/ δt = ρ(u22A2-u1

2A1)

The von Karman Integral Equation (VKI)

Boundary Layer

A

B

C

D

Flow enters on AB and BC, and leaves on CD

1 2

2 - 1

x

U

u1(y)u2(y)

VKIThe momentum change between entering and leaving the control volume is equal to the shear force on the surface:

122

δ

0

21

δ

0

220 δδρUdyρudyρuxτ

12

(CD) (AB) (BC)

By conservation of fluid mass, any fluid entering the control volume must also leave, therefore

12

01

0212 )(

dyudyuU

12 δ

01

21

δ

02

220 dyUuuρdyUuuρxτ

Force on fluid

VKI

δ

0

20 dyUuuρddxτ

As x 0, the two integrals on the right become closer andthe equation may be written as a differential:

δ

0

20 dy

Uu1

Uu

dxdρUτ

The integral is the definition of the momentum thickness, so

dxdδρUτ m2

0 dxdUUρδd if U(x)

Turbulent boundary layersThe assumption is made that the flat plate approximates to the behaviour in a pipe. The free stream velocity, U, corresponds to the velocity at the centre, and the boundary layer thickness, , corresponds to the radius, R.

1/7 Power LawFrom experiments, one possibility for the shape of the boundary layer profile is

71

δy

Uu

and measurements of the shear profile give41

UδυU0.0225τ 2

0

ρ

Turbulent boundary layersPutting the expression for the 1/7 power law into the equations for displacement and momentum thickness

δ727δ,

8δδ md

=99%

d

m

Turbulent boundary layers

dxdδρUτ m2

0 becomesdxdδρU

727τ 2

0

Equating this to the experimental value of shear stress:41

Uδυ0.0225

dd

727

x

Integrating gives:51

υUx0.37xδ

The turbulent boundary grows as x4/5, faster than the laminar boundary layer.

Turbulent boundary layersMomentum thickness

51

υUx0.036xδ

727δm

To find the total force, first find the shear stressdx

dδρUτ m20

then integrate over the plate length

m2

0

m2

00 δρUdx

dxdδρUdxτF

ll

f

For a plate of length, l, and width b, 51

υUbU0.036F 2

llρf51

0.074RC lf )10R10*5( 75 l

Logarithmic boundary layerFrom the mixing length hypothesis it can be shown thatthe profile is logarithmic, but the experimental valuesare different from those in a pipe

υyVln85.556.5

Vu *

*

and the friction coefficient llf R

ARlog455.0C 58.2

(A is a correction constant if part of the b.l. is laminar)

)10R0( 9 l

ritrit

58.2rit

RR

1.328Rlog455.0

ccc

A

Quadratic

0

1

0 1

u/U

y/

Blasius (exact)

Quadratic approximation to the laminar boundary layer

Quadratic approximation to the laminar boundary layer

2

δy

δy2

Uu

Remember - boundary layer theory is only applicable insidethe boundary layer.

This is sometimes written with =y/ and F()=u/U as

2η2ηηF

It provides a good approximation to the shape of the laminar boundary layer and to the shear stress at the surface

Turbulent Boundary Layer

Laminar Sub-Layer