Blasius Solution and Integral Parameters

2004 Mecânica dos Fluidos II Prof. António Sarmento - DEM/IST

Boundary layer Equations Contents:

– Boundary Layer Equations;– Boundary Layer Separation;– Effect of londitudinal pressure gradient on boundary layer

evolution– Blasius Solution– Integral parameters: Displacement thickness and momentum

thickness


Laminar Thin Boundary Layer Equations (<<x) over flat plate

Steady flow, constant and . Streamlines slightly divergent0 yp

dxdpxp e

2

2

2

21yu

xu

xp

yuv

xuu

2D Navier-Stokes Equations along x direction:

Compared with 2

2

yu

dxdpe


Laminar Thin Boundary Layer Equations (<<x) over flat plate

2

21yu

dxdp

yuv

xuu e

Laminar thin boundary layer equations (<<x) for flat plates

pe external pressure, can be calculated with Bernoulli’s Equation as there are no viscous effects outside the Boundary Layer

Note 1. The plate is considered flat if is lower then the local curvature radius

Note 2. At the separation point, the BD grows a lot and is no longer thin


zwu

yvu

xuu

yu

dxdp

yuv

xuu e

2

21

Turbulent Thin Boundary Layer Equations (<<x) over flat plate 2D Thin Turbulent Boundary Layer Equation

(<<x) to flat plates:

Resulting from Reynolds Tensions (note the w term)

0 0


Boundary Layer Separation Boundary Layer Separation: reversal of the flow by

the action of an adverse pressure gradient (pressure increases in flow’s direction) + viscous effects

mfm: BL / Separation / Flow over edges and blunt bodies


Boundary Layer Separation Boundary layer separation: reversal of the flow by the

action of an adverse pressure gradient (pressure increases in flow’s direction) + viscous effects


Boundary Layer Separation Bidimensional (2D) Thin Boundary Layer (<<x)

Equations to flat plates:

2

21yu

dxdp

yuv

xuu e

Close to the wall (y=0) u=v=0 :

dxdp

yu e

y 1

02

2

Similar results to turbulent boundary layer - close to the wall there is laminar/linear sub-layer region.


Boundary Layer Separation Outside Boundary layer: 02

2

yu

The external pressure gradient can be:o dpe/dx=0 <–> U0 constant (Paralell outer streamlines):

o dpe/dx>0 <–> U0 decreases (Divergent outer streamlines):o dpe/dx<0 <–> U0 increases (Convergent outer streamlines):

Close to the wall (y=0) u=v=0 :

dxdp

yu e

y1

02

2

Same sign


Zero pressure gradient:dpe/dx=0 <–> U0 constant (Paralell outer streamlines):

yu

Inflection point at the wall

No separation of boundary layer

02

2

yyu

00

2

2

yyu

Boundary Layer Separation

Curvature of velocity profile is constant


Favourable pressure gradient:dpe/dx<0 <–> U0 increases (Convergent outer streamlines):

02

2

yyu y

00

2

2

yyu

Curvature of velocity profile remains constant

No boundary layer separation



Adverse pressure gradient:dpe/dx>0 <–> U0 decreases (Divergent outer streamlines):

02

2

yyu

00

2

2

yyu Curvature of velocity

profile can change

Boundary layer Separation can occur

y

P.I.


Separated Boundary Layer


Sum of viscous forces:2

2

yu

Become zero with velocity

Can not cause by itself the fluid stagnation (and the separation of Boundary Layer)



Effect of longitudinal pressure gradient:

0dxdpe (Convergent outer

streamlines)0

dxdpe (Divergent outer

streamlines)

Viscous effects retarded Viscous effects reinforced

Fuller velocity profiles

Less full velocity profiles

...11

dxdp

uxu e

Decreases BL growth Increases BL growths



Effect of longitudinal pressure gradient:

Fuller velocity profiles

Less full velocity profiles

...11

dxdp

uxu e

Decreases BL growth Increases BL growthsFuller velocity profiles – more resistant

to adverse pressure gradientsTurbulent flows (fuller profiles)- more resistant to

adverse pressure gradients



Boundary Layer Sepaation

Longitudinal and intense adverse pressure gradient does not cause separation

=> there’s not viscous forces


Blasius Solution to Laminar Boundary Layer Equation over a flat plate with dpe/dx=0

Bidimensional (2D) Thin Boundary Layer (<<x) Equations to flat plates:

2

2

yu

yuv

xuu

0

yv

xu

Boundary Condition: y=0 u=v=0y=∞ u=U


Blasius hypothesis: with fUu

The introdution of η corresponds to recognize that the nondimension velocity profile is stabilized.

nxAy

A and n are unknowns


Remark: eyx

Ay n

xny

xnA

x n

1



Procedure: oUsing current function:

yu

x

v

o Remark:

yydd

AxUf

Axu

nn

F

dfAxU

n

o Replace u/U=f(η) e at the boundary layer equation, choose n such that the resulting equation does not depend on x and A in order to simplify the equation.

.

xv



o FUy

u

results:

Fx

nUxu o

o Fx

UAyu

n

o F

xUA

yu

n

2

2

2

2

o FxnFnxAU

xv nn

11

FAxU

n

From:



We will obtain:

02

12

FFA

UnxFn

2

2

yu

yuv

xuu

00, xu

00, xv

Uxu ,

Boundary Conditions:

o Making n=1/2 and the equation comes:UA

02 FFF xUy

with

00 FU 00 F 00 FF 00 F

UFU 1F


Graphical Solution:


0 0 0,33211 0,3298 0,3232 0,6298 0,26683 0,8461 0,16144 0,9555 0,06425 0,9916 0,00596 0,999 0,00247 0,999 0,00028 1 0,0001

xUy

FUu F

0

0,4

0,8

1,2

0 2 4 6 8 10

xUy

FUu

F


Solution:


0 0 0,33211 0,3298 0,3232 0,6298 0,26683 0,8461 0,16144 0,9555 0,06425 0,9916 0,00596 0,999 0,00247 0,999 0,00028 1 0,0001

xUy

FUu F

0Fx

UU

x

FUx Re

664,002

00

yy

u

oShear stress at the wall

2

0

21 U

fc

o Friction coefficent


Solution:


0 0 0,33211 0,3298 0,3232 0,6298 0,26683 0,8461 0,16144 0,9555 0,06425 0,9916 0,00596 0,999 0,00247 0,999 0,00028 1 0,0001

xUy

FUu F

dxDL

o 0

o Drag

LD

LU

DCRe328,1

21 2

o Drag Coefficent

UL

L Re

021 F

LUU


Solution :


Uyu 99,0

o Boundary layer thickness

0 0 0,33211 0,3298 0,3232 0,6298 0,26683 0,8461 0,16144 0,9555 0,06425 0,9916 0,00596 0,999 0,00247 0,999 0,00028 1 0,0001

xUy

FUu F

xUxx Re55

η=5

%8,105

0

FF

o Shear stress at y=


Displacement thickness:


0

* 1 dyuUUd

0

1 dyuUUd

0

udyUU d

U

0

dyuU

Ideal Fluid flow rate

Real Flow rate

Déficit of flow rate due to velocity reduction at BD


Displacement thickness :


0

1 dyuUUd

0

1 dyuUUd

0

udyUU d

Ideal Fluid flow rate

Real Flow rate

Déficit of flow rate due to velocity reduction at BD

dUq



Displacement thickness :

0

1 dyuUUd

0

1 dyuUUd

0

1 udyUd

Initial deviation of BD

δ

Deviation of outer streamlines

Section where the streamline become part of boundary layer

δdq/U LC



Blasius Solution for displacement thickness:

x

d

x Re72,1

δ dq/U LC

Ux

x Recom

344,0dou



Momentum thickness:

0

2

1 udyuUUm

0

2

1 udyuUUm

dU mdUdyu

2

0

2

mUudyUdyu

2

00

2

0

2

0

2 dyuudyUU m



Momentum flow rate through a section of BD:

mdqm UUUdyuqx

222

0

2

Momentum flow rate of uniform

profile

UU

Reduction due to deficit of flow

rate

dUU

Reduction due to deficit

momentum flow rate at BD

mUU



Longitudinal momentum balance between the leading edge and a cross section at x:

xxqmxqm xx

qqD

0

dU 2 mU d 2mU 2

δ d-d LCx



Blasius Solution to momentum thickness:

x

m

x Re664,0

Ux

x Rewith

133.0mor


Laminar Boundary Layer Equations Contents:

– Thin Boundary Layer Equations with Zero Pressure Gradient;

– Boundary Layer Separation;– Effect of longitudinal pressure gradient on the evolution of

Boundary Layer– Blasius Solution– Local Reynolds Number and Global Reynolds Number– Integral Parameters: displacement thickness and momentum

thickness


Recommended study elements:– Sabersky – Fluid Flow: 8.3, 8.4– White – Fluid Mechanics: 7.4 (sem método de Thwaites)



=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

Exercise

L=2m

U=2m/s 0dxdpe

Large plate with neglectable thickness, lenght L=2m. Parallel and non-disturbed air flow. (=1,2 kg/m3, =1,810-5 Pa.s) with U=2 m/s. Zero pressure gradient over the flat plate. Transition to turbulent at Rex=106.


a) Find boundary layer thickness at sections S1 and S2, at distance x1=0,75 m and x2=1,5 m of the leading edge

Exercise

mU

x cxc 5,72

2,1108,110Re5

6

Find xc:

xx Re5

Laminar Boundary layer at x1 and x2 – We can apply Blasius Solution

=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

L=2m

U=2m/s 0dxdpe


Exercise

xx Re5

Laminar Boundary layer at x1 and x2 – We can apply Blasius Solution

mx 75,01 5

51 10105,1

75,02Re

x m0119,010

575,051

mx 5,12 52 102Re x

m0168,02

a) Find boundary layer thickness at sections S1 and S2, at distance x1=0,75 m and x2=1,5 m of the leading edge


Exercise

b) Check that it is a thin boundary layer.

=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

L=2m

U=2m/s 0dxdpe

A: Thin Blayer if /x<<1: 0159,075,0

0119,0

1

x

0112,05,1

0168,0

2

xWhy/x at 2 is lower than

/x at 1?

y=(x)


Exercise

d) Find the value of y1 at x1 of the streamline passing through the coordinates x2=1,5 and y2=.

=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

L=2m

U=2m/s 0dxdpe

Streamline

x2=1,5mx1=0,75m

y1=?

A: We have the same flow rate between the streamline and the plate at both cross sectionsFlow rate through a cross section of BD: dUq

Flow rate through section 2: 22 dUq

Flow rate through section 1: 11 yU 1dU 1

1

11

001

yy

udyudyudyq

1221 ddy


Exercise

d) Find the value of y1 at x1 of the streamline passing through the coordinates x2=1,5 and y2=.

=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

L=2m

U=2m/s 0dxdpe

Linha de corrente

x2=1,5mx1=0,75m

y1=?

A: We have the same flow rate between the streamline and the plate at both cross sections

1221 ddy Laminar BD: 344,0d

0,0168m 0,0058m 0,0041m y1=0,0151m


Exercise

e) Find the force per unit leght between sections S1 and S2.

L=2m

U=2m/s 0dxdpe=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

xmx UD 2,0

A: There are no other forces applied except that imposed by the resistance (Drag) of plate:

The applied force between the leading edge and the cross section at x is:

Laminar BD: 133,0m

Drag force to section 2: D0,2=0,0107N/mmm 00223,0133,0 22 mm 00158,0133,0 11 Drag force to section 1: D0,1=0,0076N/m

Drag force between 1 and 2: D1,2=D0,2-D0,1=0,0031N/m


Exercise

f) True or False?: ”Under the conditions of the problem, if the plate was sufficiently long (L ), the boundary layer would eventually separate?

L=2m

U=2m/s 0dxdpe=1,2 kg/m3, =1,810-5 Pa.s(Rex)c =106.

False: The BD will separate only with adverse pressure gradient. The drag forces will decrease with the velocity

over the plate. The drga forces are not able to stop the fluid flow.

Documents

Blasius Solution and Integral Parameters