Chapter 1 bounadary layer theory_13_14

CE CE C312 HYDRAULICS ENGINEERINGC312 HYDRAULICS ENGINEERING 11BY Prof. AJIT PRATAP SINGHBY Prof. AJIT PRATAP SINGH

Boundary Layer TheoryBoundary Layer Theory

By

Prof. Ajit Pratap Singh

Civil Engineering Department

� When a real fluid past a solid boundary, a

layer of fluid which comes in contact with

the boundary surface adheres to it on

account of viscosity. Since this layer of

fluid can not slip away from the boundary

surface it attains the same velocity as that of

the boundary.

� This is called no slip condition.

� When a real fluid past a solid boundary, a

layer of fluid which comes in contact with

the boundary surface adheres to it on

account of viscosity. Since this layer of

fluid can not slip away from the boundary

surface it attains the same velocity as that of

the boundary.

� This is called no slip condition.

Boundary LayersBoundary 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% thickness: Nominal Thickness

2. Displacement thickness

3. Momentum thickness

Boundary Layer FormationBoundary Layer Formation

Thickness of Boundary LayerThickness of Boundary Layer

� Velocity within the boundary layer

increases from zero at the boundary surface

to the velocity of the main stream

asymptotically

� That distance from the boundary surface in

which the velocity reaches 99% of the

velocity of the main stream.

� Velocity within the boundary layer

increases from zero at the boundary surface

to the velocity of the main stream

asymptotically

� That distance from the boundary surface in

which the velocity reaches 99% of the

velocity of the main stream.


99% Thickness99% Thickness

V

U is the free-stream velocity

δ(x)

x

y

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

Displacement Thickness Displacement Thickness

� The distance by which the boundary surface

would have to be displaced outwards so that

the actual discharge deficit would be same

as that of an ideal (or frictionless) fluid past

the displaced boundary.

� The distance by which the boundary surface

would have to be displaced outwards so that

the actual discharge deficit would be same

as that of an ideal (or frictionless) fluid past

the displaced boundary.

Displacement thicknessDisplacement 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

v

y

vV

V

δd

Displacement Thickness…Displacement Thickness…

Displacement thicknessDisplacement thickness

The areas under each curve are defined as being equal:

( )∫∞

−=

0

dyvVq and Vδq *=

∫∞

−=

0

*dy

V

v1δ

Equating these gives the equation for the displacement

thickness:

Momentum Thickness Momentum Thickness

� The distance from the actual boundary

surface such that the momentum flux

corresponding to the main stream velocity

V through this distance is equal to the

deficiency or loss in momentum due to the

boundary layer formation.

� The distance from the actual boundary

surface such that the momentum flux

corresponding to the main stream velocity

V through this distance is equal to the

deficiency or loss in momentum due to the

boundary layer formation.


Momentum thicknessMomentum thickness

In the boundary layer, the fluid loses momentum, so

imagining an equivalent layer of lost momentum:

( )∫∞

•

−==

0

DX dyvVρvmF and θρVm 2=•

∫∞

−=

0

dyV

v1

V

v θ

Equating these gives the equation for the momentum

thickness:

Boundary Layer ConceptsBoundary Layer Concepts

� Two flow regimes

� Laminar boundary layer

� Turbulent boundary layer

� with laminar sub-layer

� Calculations of

� boundary layer thickness

� Shear (as a function of location on the surface)

� Drag (by integrating the shear over the entire surface)





� Calculations of

� boundary layer thickness



Flat Plate: Parallel to FlowFlat Plate: Parallel to Flow

Ux

y

V V V

δ

τ

Why is shear maximum at the leading edge of

the plate?

boundary

layer

thickness

shear

du

dyis maximum

Factors Affecting the Thickness of

BL

Factors Affecting the Thickness of

BL

� BLT increases as the distance from the leading edgeincreases

� BLT decreases with the increase in the velocity of flow ofapproaching stream fluid

� Greater is the kinematic viscosity, greater is the BLT

� Considerably affected by the pressure gradient in thedirection of flow

� If pressure gradient is –ve in the case of converging flow,boundary layer growth is retarded because the resultingpressure force acts in the direction of flow and itaccelerates the retarded flow in the BL.

� BLT increases as the distance from the leading edgeincreases

� BLT decreases with the increase in the velocity of flow ofapproaching stream fluid

� Greater is the kinematic viscosity, greater is the BLT

� Considerably affected by the pressure gradient in thedirection of flow

� If pressure gradient is –ve in the case of converging flow,boundary layer growth is retarded because the resultingpressure force acts in the direction of flow and itaccelerates the retarded flow in the BL.

Boundary Layer ConceptsBoundary Layer Concepts





� Calculations of

� Boundary layer thickness







� Calculations of

� Boundary layer thickness



Conservation of MassConservation of Mass

� Basic Law for a System� Basic Law for a System



� Rectangular Coordinate System� Rectangular Coordinate System





“Continuity Equation”“Continuity Equation”



“Del” Operator“Del” Operator





Incompressible Fluid:Incompressible Fluid:

Steady Flow:Steady Flow:


Momentum EquationMomentum Equation

� Newton’s Second Law� Newton’s Second Law


� Forces Acting on a Fluid Particle� Forces Acting on a Fluid Particle


� Forces Acting on a Fluid Particle� Forces Acting on a Fluid Particle


� Differential Momentum Equation� Differential Momentum Equation


� Newtonian Fluid: Navier-Stokes Equations� Newtonian Fluid: Navier-Stokes Equations


� Special Case: Euler’s Equation� Special Case: Euler’s Equation


Navier-Stokes equationsNavier-Stokes equations

Looking back to Euler’s equation with unsteadiness, the Looking back to Euler’s equation with unsteadiness, the

gravity term is simply the component of gravity, gravity term is simply the component of gravity, ggss..

Introducing viscosity as well gives 3 similar equations:Introducing viscosity as well gives 3 similar equations:

dt

duρ

z

u

y

u

x

uµ

x

pρg

2

2

2

2

2

2

x =

∂

∂+

∂

∂+

∂

∂+

∂

∂−

dt

dvρ

z

v

y

v

x

vµ

y

pρg

2

2

2

2

2

2

y =

∂

∂+

∂

∂+

∂

∂+

∂

∂−

dt

dwρ

z

w

y

w

x

wµ

z

pρg

2

2

2

2

2

2

z =

∂

∂+

∂

∂+

∂

∂+

∂

∂−

Navier-Stokes equationsNavier-Stokes equations

� There is no general solution to the N-S

equations

� Some analytical solutions may be obtained

by simplification

� The equations may be written in vector

(div/grad) notation:

� There is no general solution to the N-S

equations

� Some analytical solutions may be obtained

by simplification

� The equations may be written in vector

(div/grad) notation:

dt

udρuµpgρ 2 =∇+∇−

Force and momentum in fluid

mechanics - refresher

Force and momentum in fluid

mechanics - refresher

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

The von Karman Integral

Equation (VKI)

A

B

C

D

Flow enters on AB and BC, and leaves on CD

δ1 δ2

δ2 - δ1

∆x

U

u1(y)u2(y)

VKIVKI

The momentum change between entering and leaving the control volume

is equal to the shear force on the surface:

( )12

2

δ

0

2

1

δ

0

2

20 δδρUdyρudyρuxτ12

−−−=∆− ∫∫(CD) (AB) (BC)

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

also leave, therefore

∫∫ −=−12

0

1

0

212 )(

δδ

δδ dyudyuU

( ) ( )∫∫ −−−=∆−12 δ

0

1

2

1

δ

0

2

2

20 dyUuuρdyUuuρxτ

Force on

fluid

VKIVKI

dx

dθρVτ

20 =

( )

−=− ∫

δ

0

2

0 dyUuuρddxτ

As ∆x →0, the two integrals on the right become closer and

the equation may be written as a differential:

−=⇒ ∫

δ

0

2

0 dyU

u1

U

u

dx

dρUτ

The integral is the definition of the momentum thickness, so

if V(x)dx

dVVρδ*+


Laminar boundary layer growthLaminar boundary layer growth

τ + dτ

τ

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 ∝ ρV/t. But u=x/t => t ∝ l/V

where V is the characteristic velocity and l the characteristic length in the

x direction.

Hence FI ∝ ρV2/l

b) Viscous force: Fµ ∝ ∂τ/∂y ∝ µ∂2u/∂y2 ∝ µV/δ2

since V is the characteristic velocity and δ the characteristic length in the y

direction

δ(x)

Laminar boundary layer growthLaminar boundary layer growth

Comparing these gives:

ρV

µδ

l∝

So the boundary layer grows according to l

Alternatively, dividing through by l, the non-dimensionalised

boundary layer growth is given by:

lRl

1δ∝

Note the new Reynolds number

characteristic velocity and

characteristic lengthυ

U

µ

ρV llRl ==

)(ρV

µ5δ Blasius

l=ρV2/l ∝ VU/δ2

Boundary layer growthBoundary layer growth Length Reynolds NumberLength Reynolds Number

µ

ρUlRl =

l

U

Flow at a pipe entryFlow 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

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 byµ

ρ dum=Re

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

ρU

µ5δ

l=

lRll

10µ

ρU10

ρU

µ10.

µ

ρURe ===

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


laminar flow

These may be derived by solving the Navier-Stokes equations

in 2d.

0=∂

∂+

∂

∂

y

v

x

u

dt

du

y

u

x

u

ρ

µ

x

p

ρ

12

2

2

2

=

∂

∂+

∂

∂+

∂

∂−

Continuity MomentumU

Assume:

1. The b.l. is very thin compared to the length

2. Steady state


laminar flow


laminar flow

y

uv

x

uu

y

u

ρ

µ

x

p

ρ

12

2

∂

∂+

∂

∂=

∂

∂+

∂

∂−

This gives Prandtl’s boundary layer

equation:

rate of change of u with

x is small compared to y

Blasius produced a perfect solution of these equations valid

for 0<x<3.2*105, and demonstrated the shape of the boundary

layer profile

0y

p

ρ

1

0

y

u

ρ

µ

xy

uv

x

uu

OR

y

u

ρ

µ

x

p

ρ

1

y

uv

x

uu

2

2

2

2

=∂

∂−

=∂

∂+

∂

∂

∂

∂+

∂

∂=

∂

∂+

∂

∂

∂

∂+

∂

∂−=

∂

∂+

∂

∂

y

v

x

u

VV

Blasius SolutionBlasius Solution

0

5

0 1

u/U

y'

y' f' (or u/U) f''

0 0 0.332

1 0.330 0.332

2 0.630 0.323

3 0.846 0.267

4 0.956 0.161

5 0.992 0.064

6 0.999 0.002

7 1.000 0.000

lµ

ρUy'y =

Laminar skin frictionLaminar skin friction

The shear stress at the surface can be found by evaluating

the velocity gradient at the surface

0

0y

uµτ

∂

∂=

The friction drag force along the surface is then found by

integrating over the length

dxy

ubµF

0y0

f

=

∫

∂

∂=

l

where b is the breadth of the surface

Laminar skin frictionLaminar skin friction

From the Balsius solution, the gradient of the velocity

profile at y=0 yields the result:

0.5

x0 Rx

U0.332µτ

=

The shear force can be obtained by integration along the surface

0.5

0

0f R0.664UbdxτbF l

l

µ== ∫

The frictional drag coefficient can then be calculated

21

R33.1ρAU

FC

2

21

ff

−== l

Coefficients used for values of Boundary Layer

thickness and coefficient of drag for different

Velocity Distribution

Coefficients used for values of Boundary Layer

thickness and coefficient of drag for different

Velocity Distribution

v/V δ CD cD δ* ɵ

3.464 1.1548 0.5774 1.732 0.5774

5.48 1.46 0.73

4.64 1.292 0.646

5.84 1.37 0.686

4.8 1.31 0.654

Blacius

Solution

4.91 ≈5 1.328 0.664 1.729 0.664

2y

- y

2 V

vParabolic

=

δδ

3y

2

1-

y

2

3

V

vCubic

=

δδ

43y

2- y

2 V

v

+

=

δδδ

y

y

2Sin

V

vSin wave

=

δ

π

y

V

vLinear

=

δ

eL

D

ex R

1.1548 C andx

R

3.464δ ==


Solution By BlaciusSolution By Blacius

x

x

*

2

0

2/12/12/3

x

Re

0.664

x

Re

1.729

x

δ

328.1

2

/

664.0

664.0

0.5Re

5.0

x

δ

=

=

==

==

=

==

∫

θ

ρ

υρτ

τ

υ

EL

Df

L

oD

ex

o

RV

BLFC

BLVBdxF

R

V

x

Turbulent boundary layersTurbulent boundary layers

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 is steeper in a turbulent boundary layer

Comparison of dimensionless laminar and turbulent

flat-plate

velocity profiles.

Comparison of dimensionless laminar and turbulent

flat-plate

velocity profiles.

CE CE F312: Hydraulic Engineering F312: Hydraulic Engineering by by Prof. Prof. A. P. SinghA. P. Singh

Reynolds Experiment in Pipe

Flow

Reynolds Experiment in Pipe

Flow

� Reynolds Number

� Laminar flow: Fluid moves in

smooth streamlines

� Turbulent flow: Violent mixing,

fluid velocity at a point varies

randomly with time

� Reynolds Number

� Laminar flow: Fluid moves in

smooth streamlines

� Turbulent flow: Violent mixing,

fluid velocity at a point varies

randomly with time

∝>

−

∝<

=2lowfTurbulent4000

flowTransition40002000

flowLaminar2000

Re

Vh

VhVD

f

f

µ

ρ

LaminarLaminar TurbulentTurbulent


The assumption is made that the flat plate approximates to

the behaviour in a pipe. The free stream velocity, V,

corresponds to the velocity at the centre, and the boundary

layer thickness, δ, corresponds to the radius, R.

1/7 Power Law

From experiments, one possibility for the shape of the

boundary layer profile is 71

δ

y

V

v

=

and measurements of the shear profile give

41

Vδ

υ0.0225τ 2

0

= ρV



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

equations for displacement and momentum thickness

δ72

7,

8

δδd == θ

δ=99%

δd

δm


becomesdx

dδρV

72

7τ 2

0 =

Equating this to the experimental value of shear stress:

41

Vδ

υ0.0225

d

d

72

7

=

x

δ

Integrating gives:

51

υ

Vx0.370xδ

−

=

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

laminar boundary layer.

dx

dθρVτ 2

0 =


Momentum thickness51

υ

Vx0.036xδ

72

7θ

−

==

To find the total force, first find the shear stressdx

dθρVτ 2

0 =

then integrate over the plate length

θρVdxdx

dθρVdxτF

2

0

2

0

0 === ∫∫ll

f

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

51

υ

Vb0.036F

2

−

=

llρVf

51

0.074RC−

= elf)10R10*5( 75 << l

Logarithmic boundary layerLogarithmic boundary layer

From the mixing length hypothesis it can be shown that

the profile is logarithmic, but the experimental values

are different from those in a pipe

υ

yVln85.556.5

V

u *

*+=

and the friction coefficient ( )( ) ll

fR

A

Rlog

455.0C

58.2−=

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

)10R0( 9<< l

( )( ) rit

rit

58.2

rit

RR

1.328

Rlog

455.0c

cc

A

−=

Problem 1Problem 1

� A plate 0.50m X 0.20m has been placedlongitudinal in a stream of crude oil which flowswith undisturbed velocity of 6.0 m/sec. Given thatoil has a specific gravity of 0.9 and kinematicviscosity of 1 stoke, calculate the boundary layerthickness and shear stress at the middle of plate.Also calculate friction drag on one side of theplate.

� A plate 0.50m X 0.20m has been placedlongitudinal in a stream of crude oil which flowswith undisturbed velocity of 6.0 m/sec. Given thatoil has a specific gravity of 0.9 and kinematicviscosity of 1 stoke, calculate the boundary layerthickness and shear stress at the middle of plate.Also calculate friction drag on one side of theplate.

Problem 2Problem 2

� A plate 3.0 m X 1.5 m is held in water moving at

1.25 m/sec parallel to its length. If the flow in the

boundary layer is laminar at the leading edge of

the plate, determine the distance from the leading

edge where the boundary layer flow changes from

laminar to turbulent flow. The thickness of the

boundary layer at this section, and The frictional

drag on the plate considering both its sides.

� A plate 3.0 m X 1.5 m is held in water moving at

1.25 m/sec parallel to its length. If the flow in the

boundary layer is laminar at the leading edge of

the plate, determine the distance from the leading

edge where the boundary layer flow changes from

laminar to turbulent flow. The thickness of the

boundary layer at this section, and The frictional

drag on the plate considering both its sides.


Problem 3Problem 3

A rectangular plate of length ‘a’ and width ‘b’ is

towed lengthwise through water with velocity Va

and subsequently widthwise with velocity Vb. The

boundary layer is laminar and the plate experiences

equal drag in both the cases. Determine the ratio of

velocities Va and Vb in terms of dimensions of the

plate.

A rectangular plate of length ‘a’ and width ‘b’ is

towed lengthwise through water with velocity Va

and subsequently widthwise with velocity Vb. The

boundary layer is laminar and the plate experiences

equal drag in both the cases. Determine the ratio of

velocities Va and Vb in terms of dimensions of the

plate.

Problem 4Problem 4

� A smooth rectangular plate 1.25 m wide x 25 mlong moves through water in the direction of itslength. The drag force on the two sides of the plateis estimated to be 8500 N. Workout (i) velocitywith which the plate moves, (ii) boundary layer atthe trailing edge of the plate and (iii) whetherboundary layer changes from laminar to turbulent?If so determine distance at which the laminarboundary layer existing at the leading edgetransforms into turbulent boundary layer. AssumeCf = 0.0018

� A smooth rectangular plate 1.25 m wide x 25 mlong moves through water in the direction of itslength. The drag force on the two sides of the plateis estimated to be 8500 N. Workout (i) velocitywith which the plate moves, (ii) boundary layer atthe trailing edge of the plate and (iii) whetherboundary layer changes from laminar to turbulent?If so determine distance at which the laminarboundary layer existing at the leading edgetransforms into turbulent boundary layer. AssumeCf = 0.0018

Problem 5Problem 5

� Water flows down a smooth wide concrete apron intoa river. Assuming that a turbulent boundary layerforms, estimate shear stress and the boundary layerthickness 50 m downstream of the entrance to theapron. Use following data:

� Main stream velocity V = 7.0 m/sec

� Dynamic viscosity

� Velocity distribution is given by

� Shear stress at the wall

� Water flows down a smooth wide concrete apron intoa river. Assuming that a turbulent boundary layerforms, estimate shear stress and the boundary layerthickness 50 m downstream of the entrance to theapron. Use following data:

� Main stream velocity V = 7.0 m/sec

� Dynamic viscosity

� Velocity distribution is given by

� Shear stress at the wall

kg/m.sec 101.14 µ -3×=

1/7

δ

y

V

v

=

( )1/4δ

2

0Re

ρV 0.0225τ =

Problem 6Problem 6

� Air moves over a flat plate with a uniform free

stream velocity of 10 m/sec. At a position 15 cm

away from the front edge of the plate, what is the

boundary layer thickness? Use a parabolic profile

in the boundary layer. For air, take kinematic

viscosity 1.5 x 10-5 m2/sec and density of air 1.23

kg/m3.

� Air moves over a flat plate with a uniform free

stream velocity of 10 m/sec. At a position 15 cm

away from the front edge of the plate, what is the

boundary layer thickness? Use a parabolic profile

in the boundary layer. For air, take kinematic

viscosity 1.5 x 10-5 m2/sec and density of air 1.23

kg/m3.

Problem 7Problem 7

A wind tunnel has a cross section at its inlet of 1.0

m by 1.0 m and a length of 10.0 m. Wind at

uniform velocity of 15 m/sec enters the tunnel at

20 oC. Determine the cross sectional dimensions at

the end of the test section which will yield zero

pressure gradient along its length. Assume

velocity distribution in turbulent boundary layer to

follow the law and

A wind tunnel has a cross section at its inlet of 1.0

m by 1.0 m and a length of 10.0 m. Wind at

uniform velocity of 15 m/sec enters the tunnel at

20 oC. Determine the cross sectional dimensions at

the end of the test section which will yield zero

pressure gradient along its length. Assume

velocity distribution in turbulent boundary layer to

follow the law and51

y

V

v

=

δ51053.1 −×=υ m2/sec

Problem 8Problem 8

� A plate 3.0 m X 1.5 m is held in water moving at1.25 m/sec parallel to its length. If the flow in theboundary layer is laminar at the leading edge ofthe plate, determine

� Whether the flow changes from laminar toturbulent? If so, calculate the distance from theleading edge where the boundary layer flowchanges from laminar to turbulent flow.

� The thickness of the boundary layer at this section,and The frictional drag on the plate consideringboth its sides. (Assume that constant A is notknown).

� A plate 3.0 m X 1.5 m is held in water moving at1.25 m/sec parallel to its length. If the flow in theboundary layer is laminar at the leading edge ofthe plate, determine

� Whether the flow changes from laminar toturbulent? If so, calculate the distance from theleading edge where the boundary layer flowchanges from laminar to turbulent flow.

� The thickness of the boundary layer at this section,and The frictional drag on the plate consideringboth its sides. (Assume that constant A is notknown).


Problem 9Problem 9

A small low-speed wind tunnel is being designed for calibrationof hot wires. The air is at 190C. The test section of the windtunnel is 30 cm in diameter and 30 cm in length. The flowthrough the test section must be as uniform as possible. Thewind tunnel speed ranges from 1 to 8 m/s, and the design is tobe optimized for an air speed of V = 4.0 m/s through the testsection. (a) For the case of nearly uniform flow at 4.0 m/s at thetest section inlet, by how much will the centerline air speedaccelerate by the end of the test section? (b) Recommend adesign that will lead to a more uniform test section flow.

A small low-speed wind tunnel is being designed for calibrationof hot wires. The air is at 190C. The test section of the windtunnel is 30 cm in diameter and 30 cm in length. The flowthrough the test section must be as uniform as possible. Thewind tunnel speed ranges from 1 to 8 m/s, and the design is tobe optimized for an air speed of V = 4.0 m/s through the testsection. (a) For the case of nearly uniform flow at 4.0 m/s at thetest section inlet, by how much will the centerline air speedaccelerate by the end of the test section? (b) Recommend adesign that will lead to a more uniform test section flow.

ProblemProblem

� Below are given the wind tunnel measurement of velocity at different elevations from the boundary. What is the free stream velocity? Determine the thickness of nominal boundary. Also determine the exponent n in the equation.

� Below are given the wind tunnel measurement of velocity at different elevations from the boundary. What is the free stream velocity? Determine the thickness of nominal boundary. Also determine the exponent n in the equation.

y,

mm

1 2 5 7 10 15 20 25 50 75 100 150 200

v,

m/s

6.1 6.8 7.98 8.4 8.95 9.58 10.0

5

10.2

5

10.4 10.3 10.6 10.2 10.4

2

Problem 8Problem 8

� If the velocity in laminar boundary layer over aflat plate is assumed to be given by the secondorder polynomial, determine its form using thenecessary boundary conditions. Hence calculatedisplacement and momentum thickness.

� If the velocity in laminar boundary layer over aflat plate is assumed to be given by the secondorder polynomial, determine its form using thenecessary boundary conditions. Hence calculatedisplacement and momentum thickness.

Quadratic

0

1

0 1

u/U

y/ δδ δδ

Blasius (exact)

Quadratic approximation to the

laminar boundary layer







2

δ

y

δ

y2

U

u

−=

Remember - boundary layer theory is only applicable inside

the boundary layer.

This is sometimes written with η=y/δ and F(η)=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 LayerTurbulent Boundary Layer


Laminar Sub-LayerLaminar Sub-Layer

Thank YouThank You

Chapter 1 bounadary layer theory_13_14

Engineering

Transcript of Chapter 1 bounadary layer theory_13_14