邊界層理論1).pdf · 2018. 10. 26. · 2-D Flow on Positive Curved Surface . Effect of Pressure...

邊界層理論〈Theory of Incompressible Boundary Layers〉

方富民教授編著〈Dr. Fuh-Min Fang〉

Department of Civil Engineering

National Chung Hsing University

September 2018

Outline Textbook： BOUNDARY LAYER THEORY – by Hermann Schlichting Subjects:

1. INTRODUCTION

Mathematical Example Physical Example Simple Successive Approximation

2. BOUNDARY LAYER EQUATION

Order Analysis Falkaner–Skan Equation (Flat-Plate Boundary Layer) 2-D Flow on Positive Curved Surface Effect of Pressure Gradient on the Fluid Motion Flow Past A Wedge Free Shear Flows – Jet, Mixing Layer, Wake

3.Turbulent Boundary Layer

Karman–Polhausen Method Separation Depth – Averaged in Open Channel Flows Smooth / Rough Flow Inner – Outer layer Joint Condition Law of the Wake Power Law Developed Boundary-Layer Flows in Open Channel

*** Initiation & Participation in discussions are highly encouraged.

A Mathematical Example

Given a 2nd-order linear ordinary differential equation (O.D.E.) as

0yxdyd

=++ε , where ε > 0 and ε << O[1]

with boundary conditions (B.C.’s) as 0)0(y = and 1)1(y = It is plausible to approximate the O.D.E. as

0yxdyd

As the solution is in a form as xey λ= , substitution yields

0ee xx =+λ λλ

or 0e)1( x =+λ λ

Since 0e x ≠λ , 1−=λ . The general solution is x-eAy = .

B.C. @ 1=)1(y x-1ey eA =⇒=∴

B.C. @ 0=)0(y (N.G.) 0ey 0-1 ≠=

actual soluation x

approximate solution

Region that approximate solution does not work (B.L.)

Comments:

* Sometimes, even ε is small, a negligence might cause a serious problem (near x = 0) because of a loss of information.

* In some flow problems, viscosity may be neglected so as to simplify flow analyses. However, if the viscosity is neglected recklessly, one might lose the essence of the problem.

Physical Example

In a one-dimensional (1-D) flow with a heated fluid in a pipe (flow moves slowly, only molecular problem is of interest. That is, the flow is laminar.)

Within △, 2

ε is important. (Although △ is small, gradient is very large.)

Beyond △, 2

ε can be neglected.

x=00θ=θ x x+dx

dx x=L01 2

1θ=θ=θ

Temp is free to vary in the region

t∂∂

(heat in box)＝(heat inflow rate)－(heat outflow rate)＋(rate of internal production)

Flux ＝ rate of transfer across boundaries /time/ boundary area

＝ (diffusive flux)＋(convective flux) ………….……………(1) � Diffusive flux:

xkFD ∂

θ∂−=

k：molecular diffusivity of heat (joules / m‧sec) θ：absolute temperature x：spatial coordinate

� Convective flux＝ volumequantity

‧(velocity along direction of interest)

Define specific heat (at constant pressure) ＝ Cp (joule / kg‧°K)

3oo3p mjoule � K K kg

joule mkg � C θρ

uCF pC θρ=

Therefore, flux x

kuCFFF pDC ∂θ∂

−θρ=+= ……………………..(2)

* Assumption for heat produced by reaction

( )0-timevolume

heatθθβ=

⋅ ( in

K secmjoules

o3 ⋅ )

From (1), therefore, one gets

( ) dxA-FA-FAdxA]C[t 0dxxxp θθβ+=θρ

∂∂

+ ( ∀= ddxA )

For a constant cross-sectional area (A) of the pipe

( ) ( )0dxxx

p -xdFF

θθβ+−

=θρ∂∂ +

Assume Cp is not a function of time and for constant density (ρ),

( )0p xF

tC θ−θβ+

∂∂

−=∂θ∂

Substitution of F gives (equation 2 on pp. 3)

tC θ−θβ+

∂θ∂

+∂θ∂

ρ−=∂θ∂

Assume steady ( )0t=

∂∂ , setting Uu −= leads to

( ) 0x

=θ−θβ+∂θ∂

ρ+∂θ∂

Normalization:

Let Lxx = and

θ−θθ−θ

=θ ( )( )θθ−θ+θ=θ 010

Noted that as 0x = , 0x = , 0θ=θ , 0=θ

Lx = , 1x = , 1θ=θ , 1=θ

By substitution,

( )( ) ( )( ) ( )01010p0102

θ−θθβ+θθ−θ+θρ+θθ−θ+θ

θ−θρ×

01pUCL gives 0

=θρβ

ρ …(3)

Chain rule gives ( ) xdd

( ) 2222Lx2

∂∂

Define kinematic heat diffusivity as

ρ=α ( ) ν~ sec

which is independent of the substance transfer.

Hence ULLUC

ρ (similar to

For simplicity, set 1* =β . Equation (3) on pp. 4 becomes

=θ+θ

with boundary conditions as 0)0( =θ and 1)1( =θ .

For large ULα , 1<<ε . As it is compared to the mathematical example as

0y'y"y =++ε 1<<ε

@ ( ) 10y = ( ) 11y =

⇒ One gets a similar problem.

Simple Successive Approximation (Perturbation Method)

Purpose: to overcome the result of unsatisfactory boundary condition.

To get the zeroth-order solution, set 0=ε then solve

0y'y 00 =+ ( ) 11y0 = x10 ey −=→

To get the 1st-order solution of 0y'y"y 110 =++ε

"yy'y 011 ε−=+ with B.C. as ( ) 11y1 =

To get the 2nd-order solution of 0y'y"y 221 =++ε

………………………………

Formalize in terms of a regular (outer) asymptotic expansion, assume

∑ ε=∞

+ε+ε+= 2020100 yyy

By substitution, it yields

0..)yyy(..)'y'y'y(..)"y"y"y( 022

0100022

0100 =+ε+ε+++ε+ε+++ε+ε+ε

Re-arranging according to the exponential of ε leads to

( ) ( ) ( ) ( ) 0O''yy'y''yy'yy'y 3010202

20001010000 =ε+++ε+++ε++

For any 1<<ε , which makes the above exist, all in ( ) should be equal to zero.

Accordingly, for ( ) ( ) ( ) ( ) 00y0y0y0y 022

0100 =+ε+ε+=

( ) 00y00 =∴ ( ) 00y01 =

Proof :

If 0A0i

ii =∑ ε∞

= for all 1* <<ε<ε , then 0AAA 210 ====

0AAA 22

10 =+ε+ε+⇒

As 0→ε , 0A0 = then ( ) 0AA 21 =+ε+ε

0AA0 21 =+ε+→≠ε

As 0→ε , 0A1 =

Similarly, one can show that ∞A,A,A 32 are all zero.

( ) 00y = ( ) 11y =

O[0] 0y'y 0000 =+ ( ) 00y00 = ( ) 11y00 =

O[1] ''yy'y 000101 −=+ ( ) 00y01 = ( ) 01y01 =

O[2] ''yy'y 010202 −=+ ( ) 00y02 = ( ) 01y02 =

Supposed one uses the boundary condition at 0x =

0y'y 0000 =+

Let x00 ey λ= , substitution gives ( ) 0e1 x =+λ λ

The general solution is x00 eCy −= . For 0e x ≠λ 1−=λ∴ .

@ ( ) 00y00 = then 0C = ( ) 0xy00 =∴

For 0y'y 0101 =+ , the general solution is x01 eDy −=

@ ( ) 0D0y01 == thus ( ) x101 ex1y −−=

Similarly,

Now, use the boundary condition at 1x =

0y'y 0000 =+ @ ( ) 10y00 = x100 ey −=⇒

Then x100 e'y −−= x1

00 e''y −=

x10101 ey'y −−=+ @ ( ) x1

01 e1y −= Homogeneous solution ( ) x

ey h01

Particular solution ( ) x1xey p01

( ) ( ) ( ) ( ) 11y1y1y1y 022

0100 =+ε+ε+=

( )[ ] ( ) ( ) 01y1y11y 022

01000 =+ε+ε+−ε∴

( ) 11y00 =∴ ( ) 01y01 = ( ) 01y02 =

⇒ x101 e)x1(y −−=

( ) 0xy01 =

( ) 0xy02 = ⇒ trivial solutions

( ) 0xy03 =

Now, go back to the beginning

])(O)x1(1[eyyyy 2x102

20100 ε+−ε+=+ε+ε+= −

⇒ Now, look at this small region using a “microscope”.

Define ε

0 1ε x

inner outer

Check : x1x

01 execy −− −=

x1x1x01 exeec'y −−− +−−=

x10101 ey'y −−=+∴ (O.K.)

The solution obtained by using perturbation method is even worse!

Why? ⇒ due to bias at the start.

Substituting into 0y'y"y =++ε leads to

2 =+ζε

or 0yd

=ε+ζ

< diffusion > < convection > < reaction >

The boundary condition applies ( ) 00y ==ζ .

Since ε is small, the equation can be approximated as

then Cyd

The general solution is CeDy += ζ− (inner solution)

A ( ) CD00y +== , thus one has ( )1eDy −= ζ−

To find out the behavior at the “outer edge” of the boundary, take limit as ∞→ζ

( ) Dy −=∞→ζ

Matching the solution at the edge as 0x

lim→

(outer region) = ∞→ζ

lim(inner region)

One has De −=

outer approx

inner approx

combined solution

Chain rule: ζε

xdd , 2

Finally,

inner solution: ( )ζ−−= e1ey

outer solution: x1ey −=

Review

e∂∂

=∂∂

+∂∂

Gravitational force / mass of fluid:

3xgG −= (force potential, 22

gkGF −=∇= (downward)

3ii gF δ−=

Divergence:

eu ⋅∂∂

=⋅∇

jji ∂

∂∂

+∂∂

⇒∂∂

=∂∂

δ= (3-D)

Laplacian:

e∂∂

⋅∂∂

=∇⋅∇

jjjiji xxxx

∇⇒∂∂

∂∂

=∂∂

∂∂

Continuity:

0u =⋅∇

∂∂

j (valid for incompressible flows)

Momentum (Navier-Stokes) equation:

Guxp1uu

∇+∇ν+∂∂

ρ−=∇⋅+

∂∂

+∂∂

∂∂

ν+∂∂

ρ−=

∂∂

+∂∂ ( )3xgG −=

Scaling： t~ULt = , p~Up 2ρ= , G~UG 2=

∇=∇ ~L1 u~Uu =

Normalization yields

Continuity： 0u =⋅∇ 0u~~LU

=⋅∇ or 0u~~ =⋅∇

Momentum：(x-direction)

u~U~L1u~U

∇⋅+

∂∂ G~~

LUu~~U

∇+∇ν

+∇ρ

ρ−=

Dropping “~” and multiplying

L gives

Gupuutu 2 ∇+∇ε+−∇=∇⋅+∂∂

where LURe

1 ν==ε

airfoil

* When the Reynolds number (Re) is small, ε can be large.

* For a large Re, if u2∇ε is neglected, the solution will not be satisfactory within a thin region where u2∇ε is not small.

In the region where Re >> 1: ( )1<<ε

Droping u2∇ε terms yields 0=φ∇×∇ (irrotational flow, φ exists)

Also, as it can be proved that

( )uuu21uu

2×∇×−∇=∇⋅

Navier-Stokes (momentum) equation then becomes

( ) Gpuuu21

∇+−∇=×∇×−∇+∂∂

For a potential flow (incompressible and irrotational),

φ∇=u where φ is the potential function

∂φ∂

∇=φ∇∂∂

=∂∂

also ( )[ ] 0u)u(u ≡φ∇×∇×=×∇×

One gets

0Gpu21

∂φ∂

∇ …………………….(∆)

Integrating along a streamline yields

( )tFGpu21

=−++∂φ∂

(Bernoulli’s equation)

The continuity equation

0u =⋅∇

can become ( ) 02 =φ∇=φ∇⋅∇ (Laplace equation)

Two-Dimensional Boundary Layer on a Flat Plate

The momentum equations are (2-D laminar; gravity effect neglected)

x-direction:

∂∂

+∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

y-direction:

∂∂

+∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

The continuity equation is 0yv

=∂∂

+∂∂

As the Prandtl’s slender flow assumption is taken as 1L<<

=∂∂

+∂∂

0VOLUO =

V~LU or U

L~V δ or 1

x-momentum: (UL~t )

∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

δ UUL

ν << 2

× ) [ ]1O [ ]1O [ ]1O ? LUν 2

2LLU δν

ε << 2L

∆1 ∆2

Blasius Layer

Suppose one considers a limiting case as 0→ε and drops 1∆ and 2∆ , the

case returns to inviscid flows again since viscosity is no longer involved.

Accordingly, one must have

[ ]1O~L 2

or LRe

1ε~Lδ

= where νLUReL =

y-momentum:

ν=εε

; U~UL

∂∂

+∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

UU Lδ ( )

δ 22L U A∆ 2

LUδν 2

L B∆

δν L

ε ε ε ? 23ε ε

Therefore, for a large LRe , 0yp1→

∂∂

Then )x(pp ≅ . That is, p is not a function of y .

( ) ( )xpy,xp limp psolution

flow-potential0y=≅

∂∂

ν is negligible compared to 2

∂∂ .

* εδ ~L

Outside the boundary layer: ( )smallis u2∇ν

∂∂

ρ−=

∂∂

+∂∂

∂∂

ρ−=

∂∂

+∂∂

=∂∂

+∂∂

Within the boundary layer:

∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

=∂∂

+∂∂

* 1) The above are valid for incompressible flows.

* 2) Both forms are approximated outcomes.

21 QQ > , flow in section <2> will transport momentum upwards a little bit.

Generally, however, it will not significantly affect the free stream velocity since the thickness of boundary layer is small.

d 1Q 2Q

< 1 > < 2 >

some fluid are thrown out of b .l.

3 unknowns：u, v, p

3 equations

2 unknowns：u, v

2 equations

Euler equations

“entrainment”

Normalized form of Navier-Stokes equations:

∂∂

+∂∂

ε+∂∂

−=∂∂

+∂∂

LURe1 ν

∂∂

+∂∂

ε+∂∂

−=∂∂

+∂∂

Let ε

=ξy , vv ε=

then ξ∂

=∂ξ∂

ξ∂∂

=∂∂ v1

The continuity equation becomes 0v1xu

=ξ∂

+∂∂

or 0vxu

=ξ∂

∂∂

x-momentum equation:

ξ∂∂

+∂∂

ε+∂∂

−=ξ∂

⋅ε+∂∂

+∂∂

2uxpuv

ξ∂∂

+∂∂

−=ξ∂

∂∂

+∂∂ (neglect 2

∂∂ )

y-momentum equation:

2 vxvp1vv

ξ∂∂

ε+∂∂

εε+ξ∂

−=ξ∂

+∂∂

ε+∂∂

( )ε× 2

ξ∂∂

ε+∂∂

ε+ξ∂

∂−=

ξ∂∂

ε+∂∂

0 0 0 0 0

For smallRe1ε →= 0p

→ξ∂

Laminar Boundary Layer Equation for Blasius Layer (steady 0

∂∂ ; zero-pressure gradient 0

xdpd= )

The governing equations are

∂∂

ν=∂∂

+∂∂ ……………. (1)

=∂∂

+∂∂ …………….. (2)

with B.C.’s: 0)0,x(v)0,x(u == (no-slip, impermeable condition)

U),x(u =∞

Prandtl: 1) The larger the L, boundary layer thickness will grow.

2) The larger the U, the thinner the boundary layer thickness.

3) The larger the ν, the thicker the boundary layer thickness.

Due to dimensional homogeneity, one has

UL~ ν

δ or LU

Accordingly, one has xU

νδ or Ux~ ν

Question: Are all )x(

)y,x(uδ

the same? (δ is “some” thickness)

( )xδy

For a steady, laminar boundary layer with 0xdpd p = (U = constant), Prandtl

assumes: (similarity assumption)

Uu where g is a similarity function of

For Ux~ ν

δ , η=νδ xUy~y

Uy ν=

∂η∂ ,

21]xUy[

xx2/32/1 η

−=ν

−=ν∂

∂η∂ −−

One looks for a solution of the form (steady, 0xdpd= for Blasius layer)

and tries to make it satisfy the continuity equation.

Let’s define a stream function, ψ , which satisfies

yyu ψ−=

∂ψ∂

−= (Some defines y

u∂ψ∂

v∂ψ∂

−= )

and xxv ψ=

∂ψ∂

Choose ( )ηfxUνψ −=

then 'fUx

u =νη

ν=∂ψ∂

−= or g'fUu

Suppose one chooses ( )η−=ψ f

( )ην

=∂η∂

∂ψ∂

−= 'fx

then ( )ην

= 'fxU

1Uu (N.G.)

since 'f and LHS are dimensionless and xUν is dimensional.

yxx xuu ψ−=

∂∂

=⇒ xyy yvv ψ=

∂∂

yyy yuu ψ−=

∂∂

= yyyyyu ψ−=

Substituting into equations (1) and (2) on pp. 17 gives

0xyyx ≡ψ+ψ− (O.K.)

yyyyyxyxy ψν−=ψψ−ψψ+ ………………….. ( )∆

For ( )ην−=ψ fxU where x

∂η∂

3 η−=

∂η∂ −

( )ην∂∂

−=∂ψ∂

=ψ fxUyyy yd

fdxU∂η∂

ην−= 'fU

xνU'fxUν −=−=

( )'fUxx

yyx −

∂∂

=∂ψ∂

=ψ "fx2

"fU η+=

∂η∂

( )]fUx[xxx ην

∂∂

−=∂ψ∂

'fxUfxU

∂η∂

ν−ν

η−ν−

ν−=

x2'fxUf

21 ( )f'f

−ην

UUyy ν−=ψ

UUyyy ν−=ψ

* 'fUu y =ψ−= ,

Substitution into equation )(∆ on pp. 19 leads to

( ) '''fx

UUf'f"fxU

νν=−η

or ( ) '''fxUf'f"f

2U 222

=−η+η

Finally, 0f"f'''f2 =+ → 3rd order non-linear equation

with 3 boundary conditions as

( ) 00f0v0y

==η⇒==

( ) 00'f0u0y

==η⇒==

( ) 1η'fUuy

=∞=⇒=∞=

The solution techniques：Frohenius Method

Finite difference (numerical) Galerkin-finite element

Noted that for 0.5≅η %99Uu

and ( )ηf is linear as 0.5>η (free-stream portion)

( ) )(f)5(d'fd'fd'fd'ff 505

500 η=−η+∫ η=∫ η+η∫=η∫=η ηη

Nominal boundary layer thickness (Some chooses 95δ ):

At 99y δ= U99.0u = (see Table on pp. 21)

x99 ν=

δ → Laminar (Blasius) boundary layer

( )0x0y

=η=−η

ν=ψ=

1'f = when η > 5

Falkner-Skan Equation

Table of the function )(f η for the boundary layer along a flat plate at zero incidence after L. Howarth

f Uuf =′

f ′′

0 0.2 0.4 0.6 0.8 1.0

1.2 1.4 1.6 1.8 2.0

2.2 2.4 2.6 2.8 3.0

3.2 3.4 3.6 3.8 4.0

4.2 4.4 4.6 4.8 5.0

5.2 5.4 5.6 5.8 6.0

6.2 6.4 6.6 6.8 7.0

7.2 7.4 7.6 7.8 8.0

8.2 8.4 8.6 8.8

0 0.00664 0.02656 0.05974 0.10611 0.16557

0.23795 0.32298 0.42032 0.52952 0.65003

0.78120 0.92230 1.07252 1.23099 1.39682

1.56911 1.74696 1.92954 2.11605 2.30576

2.49806 2.69238 2.88826 3.08534 3.28329

3.48189 3.68094 3.88031 4.07990 4.27964

4.47948 4.67938 4.87931 5.07928 5.27926

5.47925 5.67924 5.87924 6.07923 6.27923

6.47923 6.67923 6.87923 7.07923

0 0.06641 0.13277 0.19894 0.26471 0.32979

0.39378 0.45627 0.51676 0.57477 0.62977

0.68132 0.72899 0.77246 0.81152 0.84605

0.87609 0.90177 0.92333 0.94112 0.95552

0.96696 0.97587 0.98269 0.98779 0.99155

0.99425 0.99616 0.99748 0.99838 0.99898

0.99937 0.99961 0.99977 0.99987 0.99992

0.99996 0.99998 0.99999 1.00000 1.00000

1.00000 1.00000 1.00000 1.00000

0.33206 0.33199 0.33147 0.33008 0.32739 0.32301

0.31659 0.30787 0.29667 0.28293 0.26675

0.24835 0.22809 0.20646 0.18401 0.16136

0.13913 0.11788 0.09809 0.08013 0.06424

0.05052 0.03897 0.02948 0.02187 0.01591

0.01134 0.00793 0.00543 0.00365 0.00240

0.00155 0.00098 0.00061 0.00037 0.00022

0.00013 0.00007 0.00004 0.00002 0.00001

0.00001 0.00000 0.00000 0.00000

For 0.5>η

( )ηf )ttancons0.5( −−η= ]ηd)η('fηd1[η5

0∫ ∫−−= ( )∫ η−−η=5

area of deficit ∆

( )η= 'fUu

=η , dyx

( ) ( ) η∫ η=η η d'ff 0

( )η∫

η= η d

( )dyyuU1

0∫ν=

Then ( )ην= fxUq

Summary: (for Blasius layer)

1) Laminar boundary layer

2) Zero-pressure gradient; UUp = , 0xdpd p =

3) Similarity profile: ( )η= 'fUu where

xUyyν

5=η , ( ) 99.0'fUu

≅η=

dyuq y0∫= ( ) ηη

ν= ∫

( ) ( )ην=ηην= ∫η

fxUd'fxU0

] [0] → [Lt

Entrainment (transverse) velocity at the edge of boundary layer:

( ) ( )∞ψ=∞ xv ( )f'flimUx

−ην

=∞→η

(see # on pp. 20)

As ∞→η , 1'f = ( )[ ]∫ η−η−+⋅ην

⇒∞→η

0d)'f1(1lim

( )∫ η−ν

area of deficit

[ ])028.3()05(xU

−−−ν

= xU86.0 ν

Or ( )xRe

86.0xU

See ∆ on pp. 22

Generally, 0v y ⇒∞→

Since we are looking at the problem in a “micro” scale, the error of x-velocity will happen far away from the area of interest.

Drag on a Flat Plate

In a 2-D case, ( )0or0yw xv

∂∂

+∂∂

For ( )η= 'fUu ,

( )y'fU

∂η∂

=∂∂

yd'fdU

∂η∂

thus ( )0η

w xνUη"fUμτ

= ( )x

U0"fUν

UU332.0ν

( )↑v

Local friction coefficient: (normalized form of wall shear stress)

1332.02U

wf νρ

τ= ρ

Another definition:

xU664.0 ν

= ⇒ 'C21C f

xRe1664.0=

Re1332.0

Total drag on one side of plate:

dxbτDL

0 w∫ ⋅= where 0y

w yuμτ

=∂∂

∫ −⋅ν

0dxxbUU)332.0(D 2

1 ( ) LbUU)664.0(ν

bU664.0 2 νρ=

0τ singularity (solution is not satisfactory )

Hopefully, for small sx , dxτsx

0 w∫ →small << dxτL

x ws∫

∼ 2/1x −

Global Friction Factor: (flat-plate, Length ＝ L ; ν

=LUReL )

( )Lb2UD2C 2

21f ρ

LULbU)664.0(2

328.1= ()

Displacement Thickness: ( )1δ

Total discharge deficit:

Udy)uU( 10δ≡−∫

therefore dyUu1

01 ∫∞

−=δ

Recall that ( )η= 'fUu where

=η , ην

= dUxdy

For Blasius layer

[ ] ηη−ν

=δ ∫∞

d)('f1Ux

01 ( ))(flimUx

η−ην

=∞→η U

x72.1 ν=

Re72.1

δ⇒ , where

( )uU −

Some use *δ .

e.g. pick up 5>η 72.1)08.78.8(8.8 =−⇒=η

Momentum thickness: ( )2δ

uρ → smmkg

3 ⋅ → volumemomentumx −

Flux of momentum deficit )uU(u −ρ⋅=

Discharge of momentum deficit dy1)uU(u0

⋅⋅−ρ= ∫∞

22U δρ≡

Thus dyUu1

02 ∫∞

−=δ

For Blasius layer [ ] ηη−ην

= ∫∞

d)('f1)('fUx

Re664.0

Integral Mass Balance

For incompressible flows, the continuity equation is

=∂∂

+∂∂

If one tries to calculate ∫∞

∂∂

xu , it leads to an infinite number. However, if

one replaces xu

∂∂ by )uU(

∂∂

∫∞

xU , the result is

finite.

Noted that U is neither a function of x nor y for Blasius layer.

some uses θ

numerical result＝0.664

dyyvdy)uU(

x 00 ∫∫∞∞

∂∂

+−∂∂

( ) ∞∞ +∫ −∂∂

−= 00 vdyuUx

[ ( ) 00yv == → non-penetrable ]

( ) 0vxd

Ud 1 ⇒+δ

−= ∞ U is a spatial constant.

Entrainment velocity x~1δ

( )↑δ∞v~

vd<∞ (decreases with x)

* The discharge deficit will result in upward entrainment.

Since area of velocity deficit increases as x increases, ( )1Uxd

dδ is positive.

Therefore, ∞v is also positive ( )↑ . Accordingly, within the Blasius layer, the

growth of velocity deficit (due to long distance of friction effect) tends to produce upward entrainment.

Shape Factor:

1Hδδ

For Blasius Layer (laminar)

==664.072.1H 2.59

Noted that the shape factor of a turbulent boundary layer is about 1.3 .

Summary: for Blasius Layer

(Laminar boundary layer with 0xdpd p = or U = constant)

xUy ν

x99. ν=

)('fUu

)(fxUq ην=

xνUUν332.0τw =

xU664.0'Cf

Re72.1

Re664.0

xRe860.0

59.2H =

Two-Dimensional Flow on a Convex Curved Surface

The momentum equation in the normal direction (y-direction) is

+∇ν+∂∂

ρ−=−

∂∂

+∂∂

+∂∂ v

(body force terms)

UU Lδ

δ 2L U

O ⇒ For large LRe , A○1 E

A, A○2 E

A, A○3 E

If [ ]1OR

<<δ , 0

ydpd→ or p＝p(x only)

same as that in the flat plate case

⇒ Boundary layer approximations still holds for large LRe and small Rδ .

~Uv δ ,

UL~t ,

2U~p ρ

Boundary Layer Equation on a 2-D Body with Outer Potential Flow

with 0xdpd p ≠

( or )x(UU = ) and R <<δ

=∂∂

+∂∂

∂∂

−=∂∂

+∂∂

In the potential flow, along a streamline (boundary layer edge), Bernoulli’s equation yields

=ρ++ρ zgpU2 p

2p constant

By neglecting z effect (assume on a horizontal plane)

p2p pU

2 constant

Then 0xdpd

U ppp =+ρ or

Uxdpd1 p

where )x(Up is the potential-flow (tangential) velocity (inviscid)

extrapolated to y = 0 (surface).

x：arc axis

y：axis normal to x (dependent on x)

（I） 0xdpd<

→↑

smalleargl ppU

0xdUd> earglsmall UU → → Favorable (negative) pressure gradient

（II） 0dx

→↓

earglsmall ppU

0xdUd< smalleargl UU → → Adverse (positive) pressure gradient

Recall：Potential-flow solution of flow around a circular cylinder

u=0u=0

Rxsin2

AdverseFavorableAdverse Favorable

> 0dxdp

< 0dxdp

> 0dxdp

possible region of separation

Von Karman’s Momentum Integral Equation For incompressible boundary layer can be expanded to TBL

The boundary layer equations are：(unsteady form)

=∂∂

+∂∂ …………………….. (1)

∂∂

−=∂∂

+∂∂

+∂∂ …………………….. (2)

where pp is the pressure at the boundary layer edge.

Noted that, in the outer (potential flow) region, local pressure along the edge of boundary layer is related to inviscid flow velocity along the surface (y = 0)

based on the Navier-Stokes equation

∂∂

ν∂∂

ν dropped y

y ρ−=

∂∂

+∂∂

at boundary layer edge xdpd p⇒

(2)⇒ 2

∂∂

ν+∂∂

+∂∂

=∂∂

+∂∂

+∂∂ ……….. (2a)

where U is the edge velocity. Re-arranging equation (2a) gives

xuu)Uu(

t ∂∂

ν+∂∂

=∂∂

+∂∂

+−∂∂ ………………. (2b)

1∆ 0 (for incompressible flows)

∂∂

+∂∂

=∆yv

xuu1 )vu(

∂∂

+∂∂

= )vu(y

∂∂

+∂∂

[ ] )Uv(y

)Uu(vy2 ∂

∂+−

∂∂

[ ]yUv

yvU)Uu(v

y ∂∂

+∂∂

+−∂∂

= since U is function of x only

∂∂

−−∂∂

=xuU)Uu(v

xUu)Uu(

x)Uu(v

y ∂∂

+∂∂

−−∂∂

Equation (2b) then becomes

[ ] [ ] 2

yu)Uu(v

yxU)Uu()Uu(u

t ∂∂

ν=−∂∂

+∂∂

−+−∂∂

+−∂∂ … (2c)

Integrating equation (2c) with respect to y from 0 to ∞ (or δ ) gives

[ ] [ ]∞∞

∞∞∞

∂∂

ν=−+∂∂

−+−∂∂

+−∂∂

∫∫∫00

000 yu)Uu(vdy

xU)Uu(dy)Uu(u

xdy)Uu(

0 yu00dy1

∞∞∞

∂∂

ν−=+

∂∂

∫∫∫ Changing sign yields

=δ∂∂

+δ∂∂

+δ∂∂ w

1 xUU)U(

von Karman’s integral equation Suppose velocity profiles are of a similar type (not always)

( )η= GUu , ( )x

=η , 10 ≤η≤

* 0)Uu( =− ∞ and 0v0 → (impermeable boundary)

∂∂

∫∞

−=δ

01 dyUu1

−δ=

[ ] ηη−δ= ∫ d)(G11

∫∞

−=δ

02 dyUu1

−δ=

[ ] ηη−ηδ= ∫ d)(G1)(G1

Example: (a steady case)

≤η≤η>η

=η=10 ,

1 ,1)(f

and U ＝ constant (independent of x & t ) The von Karman’s integrate equation is

=δ∂∂

+δ∂∂

+δ∂∂ w

1 xUUU

One has ρτ

=δ∂∂ w

Substitution leads to U6

xdd ν

δ and further to get U

d 2 ν=δ

B.C. Supposed at the leading edge (x = 0), 0=δ , then

Ux122 ν

=δ ⇒ Ux12 ν=δ ⇒

xRe46.3

For 21δ

η−δ=δ ∫ d)f1(f1

02 η−δ=δ ∫ d)f1(1

ηη−ηδ= ∫ d)1(1

0 ηη−δ= ∫ d)1(

32 δ=

η−ηδ=

Also, δν

ρτ UUw

1 =δδ

= ( Blasius layer: 2.59 , pp. 29)

Re73.1

δ (Blasius layer: 1.72)

Re576.0

δ (Blasius layer: 0.664) → more sensitive

For δν

=ρτ Uw

121 3ν

the local friction coefficient is

'C =τ

= ρ (Blasius Layer: x

f Re664.0'C = )

Homework:

Assume

1) 22Uu

η−η= , δ

⋅η=2

Redo the calculation. Tabulate the results and compare to the Blasius solution.

Effect of Pressure Gradient on the Fluid Motion (Approximated Analysis)

For a small fluid parcel with 1-D

motion (xdpd

= constant)

0xdpd< 21 pp > (favorable)

0xdpd> 21 pp < (adverse)

pΔpp 12 +=

Newton’s second law gives

tdudmFx =∑ xA∆ρ

then ( )tdudxΔAρ

tdudmΔApp 21 ==−

and xdpd1

ρ−= λ−=

Assume for a small t∆ , tΔUxΔ = , where U is the local velocity independent of x (actually not quite) then tdUxd =

One gets ( ) λ−=U

xdud or

Uxdud λ

Then cxU

−= with I.C. as 0x = ( ) 0U0u =

By substitution, one gets xU

Uu 0λ

−= xxdpd

11Ap 22Ap

cross-sectionalarea=A

)(xΔxdpd

（I） 0xdpd< (favorable), and

xdpd = constant (neither a function of x nor y)

** Profile becomes fuller as x increases.

Initial velocity profile (x=0)

U1Uu 0

（II）xdpd = constant > 0 (adverse pressure gradient)

* Separation would occur near the bottom since this area contains the least

momentum.

0xx = y

1xx = y

U1Uu 0

ρ−=

Steady Flow Past A Wedge

The potential-flow (analytical) solution at 0y = is (for 02 >β> )

m1 xu)x(U = (Note: 0x = 0U = → stagnation point)

where 1u is “some” velocity and β−

(** For 02 >β> , ↑β , ↑m )

(i) 0m = ( )0=β → flow along a flat plate

(ii) 1m = ( )1=β → stagnation-point flow

(iii) 0m < ( )0<β

For xd

)x(Ud)x(Uxdpd1 p =

ρ− ( ) 1m22

1 xumxum)xu( −− == ,

the boundary layer equations (steady) become

=∂∂

+∂∂

1 yuxum

∂∂

ν+=∂∂

+∂∂ − ……………………… ( )∆

*1uU =

(singularity , separation maybe occur )

(Imagined) → physically does not exist

To get a similarity solution,

Let xν

1myη += so as to get a simpler form of O.D.E.

For m1 xu)x(U =

1my 1−

=η 21m

d 1−

Introduce )(f1m

2)x(U η+

ν−=ψ

)(fxu1m

1 ην+

(Compared to that on pp. 18 .)

yu ψ−=y∂ψ∂

−= 21m

1m)('fxu1m

)('fxu m1 η= )('fU η=

Or )('fUu

By substituting into ( )∆ on pp. 40, one gets

0)'f1("ff'''f 2 =−β++ where 1m

=β (see pp. 20)

B.C.’s: (3)

0)0(f = 0v 0y =← = (same reason as # on pp. 20)

0)0('f = 0u 0y =← = (no-slip)

1)('f =∞ )x(Uu y =← δ=

** For a flat plate case ( )0m = , xν

Uy21η =

which is a little bit different from the previous one on pp. 18.

As 091.0m −< ( )199.0−<β , separation occurs.

When separation occurs, ( )( )

p velocity

A 0dxdpp <

Favorable (negative)pressure gradient

p velocity

A 0dxdpp >

Adverse (positive)pressure gradient

=∂∂=τ

* If 0=β (flat-plate case), one gets

0f"f'''f =+ → (Previously on pp. 20, the equation is 0f"f'''f2 =+ . The difference is due to the inconsistency of η definition.)

Blasius

Incipient separation occurs when τw = 0 (for 2-D boundary layers)

3-D B.L. separation criterion is different.

For wedge flows,

∂∂

µ=τ )('fUu

η= , x

0yydud

=∂η∂

ηµ= m

1 xuU =

1m)0(''fU 1−

For the flow around a circular cylinder: (according to the inviscid solution)

Favorable Adverse pressure gradient

u=0 u=0

If separation occurs, if must be within this region

(where or )0τ 0yu→

∂∂

Nominal boundary layer thickness (δ) is evaluated at )('f99.0Uu

η== .

?99 =η

For wedge flows, 21m

21myη 1

21mδη 1

−+=∴

If separation occurs, it must be within this region.

Since 99η , 1u , ν are constant, 21m

x~−−δ .

Growth rate of boundary layer along the wedge surface:

d +−−−

δ , 23m

d 2+−−

⇒ 02 ≥β> 02

m ≥β−

m1 xu)x(U =

1m1 xum

xdUd −=

d +−−−

2+−−δ

< I > m > 0,

0xdUd> → Favorable

(i) m > 1

δ decreases with x .

Strength of favorable pressure gradient overcomes natural growth of the boundary layer.

U=0βπ

U=0 (stagnation point)

Reasons of increase of δ: 1) natural growth 2) adverse pressure gradient

* Favorable pressure gradient tends to make boundary layer thinner.

(ii) 1 > m > 0

δ increases with x

Strength of favorable pressure gradient is not strong enough to overcome the natural growth rate of boundary layer.

Also, since 0xd

<δ , the growth rate of

xdd δ decreases with x .

(iii) m = 1 ( 1=β , angle =π ) → stagnation-point flow

< II > m < 0 (Imagined, 0<β ) → Stagnation point does not exist.

0xdUd< → Adverse pressure gradient

δ , δ grows with x .

At large x, b.l. thickness asymptotically equals to d

xuU 1=

1uxdUd=

δ , δ does not grow.

Free Shear Flows (for which boundary layer approximation holds)

Free stream velocity is constant, 0xdpd=

(1) Wake

(2) Jet

(3) Mixing Layer

velocity deficit

Laminar Wake Far Downstream of A Flat Plate ( 0xdpd= )

* The momentum deficit is due to the action of viscous drag on the plate.

Continuity equation:

=∂∂

+∂∂

Integration yields 0dyyv

∂∂

+∂∂

∫∞

(upper half)

Then 0)0(vVdyuxd

≡−+∫∞

∫∞

∞ −=∴0

dV ………..……………. (1)

Momentum equation: (boundary-layer equation)

∂∂

ν=∂∂

+∂∂

Integrating with respect to y from 0 to ∞ leads to (upper half)

width = ”b ” δ

impermeable (x < L) and symmetric (x > L)

For yuv

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

∞∞∞

∂∂

ν=+∫0

yuvudyu

∂∂

−∂∂

ν=−+∞

yu)vuVU(dyu

xdd ……............... (2)

At plate, u = v = 0 (no-slip, impermeable) After plate, v = 0 (symmetric)

Substituting (1) into (2) gives

−=− ∫∫∞∞ 000

2 dyuxd

dUdyuxd

or ρτ

=−∫∞ 00

dy)uU(uxd

* ])x(δU[xd

dLHS 22=

Integrating with respect to x from 0 to x ( > L) leads to

( ) ∫∫ τρ

0 22 dx)x(bdx)x(U

xddb ( 0τ0 = for x > L)

( ) dx)L(Uxd

dbLHSL

0 22∫ δ=∴

Let 22 )L( δ=δ (at the trailing edge of the plate) ⇒ bUD 22 δρ=

One tries to find a similarity solution at far wake ( Lx >> , Uu1 << ).

Define the velocity deficit as uUu1 −= , or 1uUu −=

Continuity:

x)uU( 1 =

∂∂

+∂−∂ or 0

∂∂

+∂∂

Note：For x > L, ρ

=τρ ∫∫

Ddx)x(bdx)x(b L

Momentum:

v)uU(x

)uU( 12

111 −∂∂

ν=−∂∂

+−∂∂

∂∂

ν=∂∂

+∂∂

−∂∂

A A○B E

A A○C E

For 1Uu1 << ( far wake ), A○AE

A >> A○B E

A , A○AE

A >> A○C E

Since 21

∂∂

ν cannot be neglected (for viscous flows)

ν LRe

⇒ For a large LRe , the approximate boundary layer equation (wake) is

∂∂

ν=∂∂ ………………………… (3)

B.C.’s:

A○1 E

A For 00y=τ

1 =∂−∂

or 0yu

1 =∂∂

at x > L

A○2 E

A 0uy1 =

∞→

Note: ( ) *)t(uuLHS 1

∂∂

→∂∂

⇒ . The form is similar to a heat equation.

(symmetric)

Since 1u must decay with x due to diffusion,

Let Uu1 ~ m)

Lx( − , where m ＞ 0

Guess )(g)Lx(C

Uu m1 η= − → similarity solution

where x

At far wake Uu1 << 1 ,

Then D≒ ∫∞

12 dyUuUb ∫ η

νρ= ∞

UxUb ……….…. (4)

Or D≒ ∫∞− ηη

m2 d)(g)Lx(C

Since RHS becomes independent of x as x＞L, thus 21m = .

⇒ )(g)Lx(CUu 2

1 η= − , where x

∂η∂

η−=

∂η∂

Examine equation (3) on pp. 49

∂∂

ν=∂∂

Recalled that (pp. 48)

22UbD δρ= → dose not change with x as x＞L

∫ =−ρ ∞0

112 Ddy)Uu1(

[ ]'ggx1)

x2)((g)

Lx(CU)(gLxCU

231 η+−=

η−η′+η−=

∂∂ −−−−

xU)(g)

νη′=

∂∂ −

xU)(g)

ν′′=

ννη′′=

∂∂ −−

Substitution gives

η′′ν=

′η+− −−

xU)(g)

Lx(UC)gg(

Or g)gg(21 ′′=′η+− then g)g(

21 ′′=η

Thus 0)g(dd

(O. D. E.)

Or 0)g(21

Integration leads to =η+η

dgd C1

A At 0y = , 0yu1 =∂∂ ( x ＞ L; far wake; symmetric)

Or for 0=η , 0d

η thus C1 = 0

ηη−= d21

ggd then

eCgη−∗=

⇒ Take 1C =∗ (this will be absorbed into C )

Noted that from a mathematic table, π=η∫ η∞ d)(g0

From (4) on pp.50

η∫ ην

ρ= ∞− d)(g)Lx(C

UxUbD 0

ρ= CLU

Since in the flat plate case, LU

LUb664.0D 2 νρ= (see pp. 25)

Comparison leads to π

=664.0C

Finally, the solution is

41exp)

Lx(664.0

Laminar Plane Jet (Ambient fluid is still)

* Mostly, jets are turbulent due to high Reynolds number (ν

bURe > 40)

Continuity:

=∂∂

+∂∂

Integrating with respect to y from ∞− to ∞+ leads to

0vdyuxd

d≡+ ∞

∞−

∞−∫

Define entrainment velocity eV (positive downwards)

then eee V2)V(Vv =−−=∞∞−

Momentum: (boundary-layer equation)

∂∂

ν=∂∂

+∂∂

Integration gives ∞∞−

∞∞−

∞∞−∫

∂∂

ν=+yuvudyu

0-0 0-0

Thus 0dyuxd

d 2 =∫∞∞− or =∫

∞∞− dyu2 constant = 0

20 bU = K ….(A)

Similarity:

Note: For the flat plate case, the entrainment is positive outwards.

At 0x = , 0Uu = , width 0b=

αx~uc )0( <α ; βx~b )0( >β

Assume a similarity profile ))x(b

y(fun~)x(u

then )xy(funx~u β

α ⋅

Try a stream function )xy(fx~ q

pψ , ( qxy

=η , qx1

∂η∂ , 1qx

yqx +−=

∂η∂ )

For qqp

xy(fx~

yu ′

∂ψ∂

−= , then qp −=α , q=β .

Therefore, )xy(fx~u q

qp ′− [ b(x) 〜 xq ]

As qxdyd =η or η= dxdy q , substituting into (A) on pp. 53 yields

=η∫ η′⋅ ∞∞−

− d)(fxx 2q)qp(2 constant independent of x

∴ 0qp2 =− ………………………………… (B)

Now, examine the momentum equation as

∂∂

ν=∂∂

+∂∂

For )(fx~u qp η′−

qp1qp x~)x

yq(fxfx)qp(~xu −−

+−−− −′′+′−

∂∂

)("fxy

)(''fx~yu q2pqp η=

∂η∂

η∂∂ −− )('''fx~

yu q3p2

η∂∂ −

∂∂

ν=∂∂

+∂∂

)1qp()qp(x −−+− q3px −

After comparisons, one gets q3p1q2p2 −=−− ……………….. (C)

From (B) and(C), therefore, one gets 31p = ,

x~x~uq

−−

Now, go back to the stream function.

Recall that )xy(gx~)

xy(gx~

qp−ψ ,

32x~b )x~

Take )(g)bx(c 3

01 η−=ψ + where

There are 2 unknowns to be determined ( 1c & 2c ) and the condition that has

to be satisfied is

2 bUconstdyuK ∫ === ∞∞− (pp. 53)

The solutions are

)tanh1()bx()(Re4543.0

ξ−= −

[ ]ξ−ξ−ξ= −− tanh)tanh1(2)bx()(Re5503.0

where ν

bURe and 3

)()(Re2752.0

Noted that 32

bx()(Re5503.0

−−∞

∞→−==

( Cu ∼ u )

Potential Core of Jets

( )xuc

31x − (Laminar)

The analysis is valid at x >> pcx

The equations are suggested to work by starting x from the end of the potential core.

Laminar Boundary Layer with A Non-Zero Pressure Gradient (steady case)

Assume a similarity solution as 0xdpd≠ , )x(UU =

432 dcba)x(U)y,x(u

η+η+η+η+α= where δ

=ηy , )1

η∂∂

∂∂

Similar to the derivation of von Karman integral equation, the profile is an approximation.

Matching conditions:

○1 E

A 0u 0y ==

A○2 E

A Uu y =δ=

A○3 E

A boundary layer equation along wall (y = 0)

xuu =∂

∂ν+=

∂∂

+∂∂

or xdUdU

−=∂∂

A○4 E

y =∂∂

A○5 E

=∂∂

δ= (smooth matching)

A○1 E

A → 0Uu

0==η , 0=α

A○2 E

A → 1Uu

1==η , 1dcba =+++

A○3 E

A → xdUdU

2 ν−=

ηδ =η

xdUdU)d12c6b2(U

2 ν−=η+η+

δ =η

Λ−=νδ

−=21

δ=Λ xd

normalized velocity gradient

A○4 E

A → 0dd

η =η , 0d4c3b2a =+++

A○5 E

A → 0dd

=η =η , 0d12c6b2 =++

Finally , 0=α Λ+=612a Λ−=

21b Λ+−=

212c Λ−=

And 432 )611()

)x(U)y,x(u

ηΛ−+ηΛ+−+ηΛ−ηΛ+=

)22( 43243 η−η+η−ηΛ

+η+η−η= )(G)(F ηΛ+η=

A○AE

A: Approximated profile without pressure gradient

A○B E

A: Due to non-zero pressure gradient 3)1(6

η−ηΛ

Favorable pressure gradient ( 0xdpd< , 0

xdUd> ) , 0>Λ

Adverse pressure gradient ( 0xdpd> , 0

xdUd< ) , 0<Λ

Zero pressure gradient 0=Λ

○A ○B

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

-0.2 0 0.2 0.4 0.6 0.8 1 1.20

300.7052-6 0

At the point of separation, 0yu

0y =∂∂

= ( valid for 2-D cases only)

For 0612a

d)U/u(d

0 =Λ+==η =η , then 12

−=νδ

∴ It happens when 0xdUd< (or 0

xdpd> , adverse pressure gradient)

Displacement thickness:

∫ η−δ=δ 101 d)

Uu1( )

120103( Λ−δ=

Momentum thickness:

η∫ −δ=δ d)Uu1(

02 )144155

2Λ−

Wall shear stress:

0y0 yu

=∂∂

µ=τ 0d)U/u(dU

=ηηδµ

2(U Λ+

= or 6

0 Λ+=

µδτ

Define Λδ

ν′δ

2 )(Uk 2 = ΛΛ

− 22

2 )144155

37()631( (still dimensionless)

Shape factor:

14415537(

)12010

3(6312

Also, δδ

µδτ

=µδτ 2020

14415537()

2Λ−

Λ+= )k(f2=

Let νδ

2Z (dimensional) , xd

d2xdZd 22 δ

then UZUk2

2 ′=ν

′δ=

Substitution into von Karman integral equation (steady form) as (see pp. 34)

=δ+δ+δ∂∂ 0

1 xdUdUU

=′δ+δ+δ′ 012

22 UU'UUU2

U'UU2 20202122

µδτ

=νρδτ

=νδδ′

+δδν

+νδ′

A○1 E

A A○2 E

A A○3 E

For A○2 E

A xdZdU

= and A○1 E

A+ A○3 E

A )2(k2

⇒ [ ] )k(f)k(f2kxdZdU

21 =++

Let [ ])k(f2k2)k(f2)k(F 12 +−= , then U

)k(FxdZd=

Karman-Polhausen Method →Predict ( )x2δ growth under action of pressure gradient by extrapolation.

Based on ( )UkF

xdZd= , ( ) x

∆+=∆+

Typically, the calculation starts from the stagnation point where ( ) 0y,xU =

However, problem occurs since locally ∞→==

)k(FxdZd

Thus, )k(F is forced to be zero to assure not being blown out.

That is 00

U)k(Flim

xdZdlim

0x0x→=

→→

The limit can be achieved by the application of the I’Hopital’s rule.

Auxiliary function for the approximate calculation of laminar boundary layers after Holstein and Bohlen (1940)

Λ k ( )kF ( ) 122

11 Hkf =

= ( )U

kf 022 µ

15 0.0884 -0.00658 2.279 0.346

14 0.0928 -0.0885 2.262 0.351

13 0.0941 -0.00914 2.253 0.354

12 0.0948 -0.00948 2.25 0.356

11 0.0941 -0.00912 2.253 0.355

10 0.0919 -0.008 2.26 0.351

9 0.0882 -0.00608 2.273 0.347

8 0.0831 -0.00338 2.289 0.34

7.8 0.0819 -0.00271 2.293 0.338

7.6 0.0807 -0.00203 2.297 0.337

7.4 0.0794 -0.00132 2.301 0.335

7.2 0.0781 -0.00051 2.305 0.333

7.052 0.077 0 2.308 0.332

7 0.0767 0.0021 2.309 0.331

6.8 0.0752 0.0102 2.314 0.33

6.6 0.0737 0.0186 2.318 0.328

6.4 0.0721 0.0274 2.323 0.326

6.2 0.0706 0.0363 2.328 0.324

6 0.0689 0.0459 2.333 0.321

5 0.0599 0.0979 2.361 0.31

4 0.0497 0.1579 2.392 0.297

3 0.0385 0.2255 2.427 0.283

2 0.0264 0.3004 2.466 0.268

1 0.0135 0.382 2.508 0.252

0 0 0.4698 2.554 0.235

-1 -0.014 0.5633 2.604 0.217

-2 -0.0284 0.6609 2.647 0.199

-3 -0.0429 0.764 2.716 0.179

-4 -0.0575 0.8698 2.779 0.16

-5 -0.072 0.978 2.847 0.14

-6 -0.0862 1.0877 2.921 0.12

-7 -0.0999 1.1981 2.999 0.1

-8 -0.113 1.308 3.085 0.079

-9 -0.1254 1.4167 3.176 0.059

-10 -0.1369 1.529 3.276 0.039

-11 -0.1474 1.6257 3.383 0.019

-12 -0.1567 1.7241 3.5 0

-13 -0.1648 1.8169 3.627 -0.019

-14 -0.1715 1.9033 3.765 -0.037

-15 -0.1767 1.982 3.916 -0.054

Example:. ( )

1x2lim

limxxlim

→→→

* Note:

At the stagnation point,

052.7=Λ

077.0k =

( ) 0kF =

> favorable,0

At a location with known velocity profile (to get 2δ ) and free-stream velocity distribution along wall ( ))x('U ,)x(U , the procedure is as follows:

Firstly, start from the stagnation point ( )0x =

0U = 0)k(F = , 077.0k =

( )UkFlim

= where 'UZk =

I’Hopital Rule dxdUdxdF

0xlim→

⇒( )

→ UkF

'U''UK

dkdFlim 20x

( )20x0x 'U''U

kdFdklim

UkFlim

→→=

2077.0kdK

dFdKdF

0x0x 'U

UkFlim

0x2'U''U0652.0

−= at stagnation point

Procedure：

(1) Grid generation (determine ∆x)

(2) Must know U , 'U as functions of x, also ''U and 2δ at 0x =

(3) At next location ( )x0x ∆+= , x'U''U0652.0ZZ

∆⋅−=∆+

then for ( )kF'UZk →= ( ) xUkFZZ

∆⋅+=∆+

( )kf1→

Separation would occur at 12−=Λ or 1567.0k −= ( 00 =τ )

( )kf2

( )'U'Z''ZUdkdF

U)k(F''U

* At the stagnation point, 'U

077.0Z 2 =νδ

''U0652.0xdZd

Improved Method: (Dimensionless form of Karman-Polhausen method)

Let 0U and L be the velocity and length scales

=ξ ( ) ( )ξ== wU

and the normalized “Z” as p

= where ν

=LUR 0

Now, one has ξν

UxdUdk 0

20 wh =

By substitution, ( )U

whFxdZd =

Then ( ) ( )wUwhF

)wh(FL

LUdd 2

Thus w

)wh(Fd

hd⇒=

* For a stagnation-point flow

xu α=yv α−=

const=α(potential flow)

'U077.0Z

→ calculate 2δ , ν

At x ＝ 0 (stagnation point) 0'U''U0652.0

== 0xdxx

ZZZ ==+

⇒ Nothing would change with x (δ , H, 0τ , U)

→ Boundary layer remains the same shape throughout.

At the wall ( )0y = :

( ) xxU α= ( )0v =

0''U =

Homework:

Use Karman-Polhausen method to calculate the growth of 2δ on a 2-D cylinder in a uniform velocity field. Find out the location )(θ of separation.

RxsinU2xU 0

Also, plot

Turbulent Boundary Layer

i =∂∂

∂∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

Time Average:

'dt)'t(uT1u

2/Tt ii ∫+

where T is a sufficiently long period.

* Note： ii uu = , 0'ui = (white noise)

Reynold’s Averaging Conditions: ( 'fff += , 'ggg += )

(1) gfgf +=+ (f and g are stationary process.)

'gg'ff)'gg()'ff( +++=+++

(2) fafa = where “a” is a constant independent of t .

(3) sf

∂∂

=∂∂ where s can be a function of x, y, z and t .

( )sf'f

∂∂

=∂∂

+∂∂

=∂∂

+∂∂

=+∂∂

=∂∂

(4) gfgf =

O(x, y, z)

)t('uuu iii +=

)t('ppp +=

ensemble average fluctuation

* Note: )'gg()'ff(gf ++= 'g'fg'f'gfgf +++= 'g'fg'f'gfgf +++=

'g'fgf +=

Continuity:

0)'uu(x ii

=+∂∂ or 0

i =∂∂

+∂∂

Taking a time average gives

∂∂

+∂∂

=∂∂

+∂∂ ⇒ 0

i =∂∂ …..…………. (1)

Momentum:

∂∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

* Note: j

∂−

∂∂

Taking a time average gives

∂∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

Since 'u'uuuuu jijiji +=

'u'uxx

∂∂

+∂∂

=∂∂

Assume µ is a constant independent of jx Reynold’s Averaged equation

∂∂

µ∂∂

∂∂

ρ−=

∂∂

+∂∂ 'u'u

1∆ 2∆

1∆ ：laminar shear stress ; 2∆ ：turbulent shear stress (Reynold’s stress)

(continuity)

O(x, y, z)Lam.Trans.

500U0 ≅νδ (transition)

* Mostly, 22 'v~'u~'v'u

○1 E

A: If laminar stress term is important,

A○2 E

A If turbulent stress term is important,

Generally, for a 2-D turbulent boundary layer

=∂∂

+∂∂

∂τ∂

ρ−=

∂∂

+∂∂ where 'v'u

ρ−∂∂

* It can be shown that 0yp=

∂∂ (same as before).

For 2-D steady (stationary) flows

=∂∂

+∂∂

~V δ∴ ( )UV <<

∂∂

−∂∂

∂∂

+∂∂

ν+∂∂

ρ−=

∂∂

+∂∂

δν<<ν

22 'uL'u

L× ) 1 1 1

2LULUL

δ<< 2

Two-Dimensional Turbulent Jet → ambient fluid is still

Momentum: All variables denote averaged quantities.

∂∂

=∂∂

+∂∂ ……………………………. A○1 E

where 'v'uyu

ρ−∂∂

Continuity:

=∂∂

+∂∂ ……………………………………. A○2 E

For turbulent jets, integrate A○1 E

A with respect to y from 0 to ∞ gives

∂τ∂

∂∂

+∂∂∞

or ∞∞∞ τρ

=+∫ 0002 1vudyu

0y = , 0)0(v = , 0)0( =τ ( 0'v'u → , 0yu

=∂∂ due to symmetry )

∞=y , 0)(u =∞ , 0)( =∞τ ( 0'v'u → in free stream, 0yu

=∂∂

Therefore, 0dyuxd

2 =∫∞

or kdyu0

2 =∫∞

a constant independent of x (see also pp. 53)

Integration of A○2 E

A leads to 0vdyuxd

d00 =+∫∞∞

( )0Ve −− (due to symmetry)

Thus ∫= ∞0e dyu

xddV (see also pp. 53 for a laminar jet)

negative (upwards)

Experimentally, similarity is satisfied well in the turbulent cases.

( )( ) ( )

xuy,xu

Experimental results show that both Ve and u c decrease with respect to x . Accordingly, guess the entrainment coefficient (dilution factor) as

e α= (about 0.07 experimentally) or ce u'V α=

Then c0u'dyu

α=∫∞

………….. (∆)

Recall that kdyu0

2 =∫∞

(pp. 69)

By taking )x(b

y=η , η= dbdy

Substitution gives kd)(f)x(b)x(u0

22c =ηη∫

Then =)x(b)x(u2c constant independent of x

Also, from (∆) c0c u'αηd)η(fbuxd

Then [ ] =)x(b)x(uxd

constant independent of x

=bu2c constant

( ) =buxd

constant

Assume αx~uc , βx~b

=β+α+−α−=β+α

or 01=−β

One gets 21x~uc

Experimental results for plane turbulent jests are as follows:

= (3.2 ~ 3.6)

00 bx10.0

= (0.09 ~ 0.115) * x starts from pcxx =

053.0UV

e = (0.05 ~ 0.07)

Recall that for laminar plane jets (pp. 55),

31x~uc

− , 32x~b

Turbulent Plane Channel Flows (Fully-Developed)

When the flow is fully developed, )y(u , 0τ and other variables (except p)

are constant (independent of x ).

Choose a velocity scale as *0 u=ρτ (friction/shear velocity)

Related length scales are δ , ks and *uν

ν≅δ

thickness of viscous sublayer

If 1ks

L >>δ , the turbulent boundary layer roughness elements are lubricated

(smooth turbulent boundary layer).

Then the possible length scales leave δ and *uν .

It is noted that ks*s

ku6.11

δ where ν

For smooth turbulent boundary layers, 1R k <<

In practice, =kR 3 70

smooth transitional rough

k s (roughness height)

Define a cross-sectional average velocity (bulk velocity) as

0dyu1U

(i) Plane (2-D) channel flows ( )H=δ

νδ turbulent

(ii) Circular pipe flows

2000DU>

or 1000U>

νδ turbulent

Hydraulic Radius: Hr = area wettedofperimeter

(1) Open channel (plane pipe) flows (2-D)

δ=⋅δ

(2) Circular pipe flows

=δπδπ

⇒ Pipe flows 1000rU H >ν

turbulent

Friction Reynolds number: (for a 2-D turbulent boundary layer)

6.11sublayer viscousof thicknesscknesslayer thiboundary u

** ×≅

Core region (outer region)ν not important

wall regionν important

viscous sublayer

In the core region: (ν is relatively not important.)

Flow is almost independent of ν (or *uν ). *u and δ are important.

In the wall region:

Flow is affected by ν ; *u and *uν are important.

(A) In the core region, use “vorticity” to represent the flow (2-D)

( )y , ,ufunctionydud

*z δ=−≈∂∂

−∂∂

Due to dimensional homogeneity, by Rayleigh’s method one has

=δ yg

Let δ

=ηy (outer variable), one has

)u/u(d * η=η

<< (for a boundary layer)

Integration yields 'd)'(guu

ηη= ∫η

( 'η is a dummy variable),

)η(F~

which leads to the Outer Law for turbulent boundary layers as

0 η=− [ At y = δ, u = u0 ]

)(F η is a universal function and is dependent on domain geometry and pressure gradient. (See pp. 81 for experimental results.)

(B) In the wall region,

uν ,ufun

Dimensional analysis yields let ν

yy (inner variable)

uu/ or )y(f~

yd)u/u(d * +

Integration yields )y(fu

= [ At y (or y+) = 0, u = 0 ]

which is the Inner (Wall) Law for turbulent boundary layers.

(C) In the viscous sublayer, ( )6.11y <+

δy is so small that local shear stress ≈ wall shear stress ＝ constant

Thus, for a very small y

ydud = 2

*u ( *u is a function of x only.)

Integration gives cyuu2* +ν

(Turbulent stress is nearly zero.)

B.C.: @ 0y = , 0u = (no-slip)

Then yuu2*

ν= or

= or ++ = yu .

Summary:

Outer Law: )(Fu

0 η=− where

Inner Law: )y(fu ++ = where *u

uu =+ and ν

=+ *uyy

Viscous sublayer: ++ = yu

2-D Turbulent Channel Flows

dxdp0−⇒ (const, developed)

Assume nothing varies with x except for po

≠= 0const

xd)x(pd 0

Continuity: (u, v and p denote time-averaged quantities)

=∂∂

+∂∂

Because 0xu=

∂∂ thus 0

∂∂ (v is independent of y)

Since at 0y = , 0v = therefore 0v ≡ (everywhere)

x-momentum:

xyu'v'u

∂∂

ν+∂∂

−∂∂

ν+∂∂

−∂∂

ρ−=

∂∂

+∂∂

y-momentum:

xv(ν'v

∂∂

−∂∂

+∂∂

−∂∂

−=∂∂

+∂∂

Integration with respect to y yields )x(F'vp1

2 =+ρ

→ independent of y

Taking derivative w.r.t. x yields )x(Fxp1

2=∂∂

ρ → independent of y and x

=∂∂

xp1 constant

xdpd1 0

(fully- developed)

( = constant; fully-developed )

0 0 fu l ly-developed

0 0 0 0 fu l ly-deve loped

Integrating the x-momentum equation with respect to y gives

cydud'v'uy

xdpd10 0 +ν+−

ρ−= ……………………. (1)

@ at 0y = , 0'v'u = (lower wall) then 2*

uydudc −=

−=ν−=

@ at h2y = (upper wall) 0'v'u = and 2*

−=ρτ

−=ν

Thus, 2*

0 u)u(0)h2(xdpd10 −−−

ρ−=

then 2*

0 uhxdpd1

− , or hu

xdpd1 2

*0 =ρ

Substituting into (1) yields

ydud'v'u

hyu0 −ν+−=

−=ν+−

ydud'v'u 2

* …………………..(#)

Normalized Form I ⇒ for outer (core) region, νδ

η1ηd

)u/u(dR1

u'v'u *

−=+− , where hy

=η , h=δ

Normalized Form II ⇒ for inner (wall) region

y1ydud

u'v'u +

−=+−

⇒ In the limit →νδ

uR large, where )hy(=η is finite (but →+y large)

Form I becomes (outer region)

η−=− 1u

'v'u2*

(valid in outer region)

⇒ In the limit →*R large, [ ]1Oy ≡+

Form II becomes (inner region)

=+− +

(valid in inner region)

For a smooth turbulent boundary layer ( )sL kδ >> ,

Inner layer: (wall region)

Since )y(fuu

+= (see pp. 77)

then )y(gu

'v'u2*

Note: η=ν

** Rhyhuyuy ++

ydfduu

ydud 2

Outer layer: (core region)

0 η=− (see pp. 77)→ Velocity Defect Law or Friction Law

Note: yd

Fduydud

hu* F is still unknown.

Experimentally, η=η ln5.2)(F in the core region.

Law of the Wall

Universal Velocity Distribution Laws for Smooth and Rough Pipes (δ = R: radius)

(Schlichting’s book, pp. 607)

Ryln5.2

0 −=−

Ry11(ln[1

0 −+−−κ

−=−

0 )Ry(08.5

(The case for a circular pipe is similar to the 2-flat-plate case; R → h or δ)

In the buffer zone or inertia sublayer (the transition part between the outer and inner regions), matching the vorticity (- du/dy) as (see pp. 80)

∞→+y (inner)

0→η (outer)

one has ηη

yd)y(fdu *

hy * where

Then κ1

ηd)η(Fdη

yd)y(fdy ⇒=+

+ where κ is Karman constant.

)y(fun + )(fun η

aln1)(F +ηκ

=η⇒

cylnκ1)y(f += ++

uuu −

valid in the inertia sublayer

Noted that

Outer: η−=− 1u

'v'u2*

As 0→η 2*u'v'u ≅−⇒

Inner: 1dy

)y(dfu

'v'u2*

=+− +

As ∞→+ y 0y1

yd)y(fd →

κ=⇒ ++

Therefore, in the inertia sublayer, 2*u'v'u ≅−

The profile：(for a pipe flow or flow between two flat plates)

ln1uu *

which is called the logarithmic law (or log law).

Also, since aln1u

uu)(F*

0 +ηκ

=η …...……...…………….

cyln1uu)y(f

== ++ .……....…….………….

As both relationship are valid within the inertial sublayer, － yields

( )acyln1uu

0 −+ηκ

( )acRln1* −+

*y* Ru/uyy

=η δ

(κ, c) = (0.4 , 5.0)

(0.41, 5.5) more popular

Coles Law of the Wake

Deviation of the velocity profile/maximum deviation at the boundary layer edge:

=−δ−−−

5.5ln5.2u5.5yln5.2u

Note: at δ=y , 2W =

Experiential result shows that

sin2yW 2 in the outer region

Then, the composite law is

κ= ++ yWcyln1u

where 55.0=π for flat plate case

)5.0(8.0 +β≈ , where β is a pressure parameter

Law of the Wake

pipe center

For a flow in a circular pipe：(stationary, fully-developed)

Momentum: (in an index form; averaged form)

'u'uxxx

u jijjj

ij ∂

∂−

∂∂∂

ν+∂∂

ρ−=

∂∂

For )y(uu = , 0wv ≡= (everywhere)

xuu ⇒

∂∂

+∂∂

x-momentum: xdpd1 0

uxp10 ji

∂∂

−∇ν+∂∂

ρ−= (see also pp. 78)

Converting into the cylindrical coordinate system：

Recall that

∂∂

=∂∂

+∂∂

=∇rur

ru)r(u 2

θ∂∂

+∂∂

=•∇ θAr1

r1A xr

where 'u'uA xrr = , 'u'uA xxx = , 'u'uA xθθ =

One gets

( )'u'urrd

xd)x(pd1

xro −

symmetry

Multiplying by r and integrating leads to

c'u'urrdudr

xd)x(pd1

+−ν=ρ

(at centerline r = 0, c* = 0)

or 'u'urdud

xr0 −ν=

At Rr = (wall), 0'u'u xr =

then one has Rr

ν=ρ 0yyd

ν−= 2*

0 u−=ρτ

Or R2u

xdpd1 2

*0 −=

C.V.x∆

p xdxdp0 ∆+

Then 'u'urdud

Rru xr

2* −ν=−

Changing variables: rRy −=

v'u'u yr −=−= 'u'ux =

'v'uydud

* +ν−=−

−⇒

−=ν+−

ydud'v'u 2

* (compare to (#) on pp. 79)

dydr −=

For a fully developed flow,

( )2o0 Rx

xdpdx)R2( π

∆−=∆πτ

then R2u

xdpd1 2

*o −=

Eddy Viscosity near the Wall

(I) Inertia sublayer (Logarithmic region)

ydudu'v'u *

tt2* κ

ν=ν==− (based on log-law; see pp. 82)

yu*t κ=ν∴

From Prandtl’s mixing-length concept

t =ν yuy

*2 κ=κ

= yκ=⇒

How does 'v'u− behave in the log-region (inertia sublayer)?

From 0y'v

=∂∂

+∂∂ dy

0∫ ∂∂

For a small y, expand 'u in power series as

+++= 2210 y)t,x(fy)t,x(f)t,x(f'u (note: 0y = , 0'u = , 0f0 = )

then dy)yxfy

221∫ +∂∂

+∂∂

−= −∂∂

−∂∂

−=xfy

21 2312

∂∂

−∂∂

−++= xfy

21)yfyf('v'u 23122

)y(Oxffy

∂∂

Note: 0f21

xxff 2

1 ⇒∂∂

=∂∂ ⇒ Fully developed, average is independent of x.

4y~'v'u⇒ in the logarithmic region

(II) In the viscous sublayer ( ++ = yu )

1)/uy(d

)u/u(du

+− (see pp. 82)

As 0y → (at wall) , 0'v'u →

One has 1)/uy(d

)u/u(d*

For ++ = yu or ν

uyuu , one has yuu

Form Prandtl’s mixing-length concept,

ydud'v'u

one has 42 y~ or 2y~

Van Driest Correction

From Van Driest “On the turbulent flow near a wall”, Journal of Aeronautical Science, 23, 1007(1956)

For an equilibrium turbulent boundary layer

−κ=

+− A

where A = 26 (originally, 25) to give 1.5yln1uu

Define ν

Traditional profile for an equilibrium (smooth) turbulent boundary layer:

Power Law (An approximation valid in most regions of smooth turbulent boundary layers)

Experimentally, 7

uy74.8uu

ν= for 510Re ≈

Since at δ=y , 7

0 u74.8uu

= ……………………(*)

Thus 7

= (power-law)

Also, define a (local) friction coefficient (normalized wall shear stress) as 2

From (*), 7

)74.8(2C

u045.0C−

= for flat plate

= for circular pipe ry −δ= , R=δ

where n depends on Reynolds numbers.

Turbulent Boundary Layer Development on a Smooth Flat Plate 0u

x x’(virtual origin of f.b.l.)

500u≅

TransL

From the von Karman integral equation

=δ+δ w0012

udu)u(xd

d (see pp. 34)

w220 u

xddu =

For 2/u

= , 20

one has 20

f220 u

xddu =δ …………………………... (1)

Take the (1/7) power law assumption as 7

= (pp. 89)

−δ=δ ∫ d

2 , where δ

∫ ηη−ηδ=1

0d)1( 7

1δ=δ

−= 0972.0

u045.0C−

= (pp. 89), substituting into (1) gives

0u0225.0xd

d0972.0−

dxu0972.00225.0d

=δδ Cxu0972.00225.0

45 0 +

=δ−

xu371.0x

At virtual origin (x = 0), δ = 0,

thus C = 0

Turbulent boundary layer : 54x~δ

Laminar boundary layer : 21x~δ

( 0xdpd= )

500uo ≈νδ

(virtual origin of turbulent boundary layer)

Turbulent drag on the plate (2 sides)

∫ τ=L

0 w dxb2D

∫ρ=L

0 f20 dxC)u

dxu045.0ub L0

ρ=−

(pp. 89)

xu0.045ub L0

−−

4/15/10 xu0371.0

−−

ν (pp. 90)

xu0371.0xu045.0C 0of

ν= −

− 51

xu0577.0 0−

* For large L, Total Drag ≒ Turbulent Drag

5/4020

0 L)45(u)0577.0(ubdxxu)0577.0(ubD

1 −−

ρ= ∫

Global friction coefficient: (normalized drag)

Lu0721.0Lbu

DC 020

ρ= with

LuRe 0L

which is valid for 7L

5 101Re105 ×<<× . (see also pp. 26 )

Blasius Transition Fully Developed Turbulent

510U•

νδ∞

Growing wave

rough transition smooth

( ) 51 102.3xU×=

( ) 500U2 =νδ∞

Blasius layer (laminar):

x~δ 21

−≅δ where

ν= ∞ xURex

x1 Re72.1

x2 Re664.0

δ 6.2H2

1 ≅δδ

Turbulent boundary layer: ( for 66x 1018~ 101.7 Re ××= )

71 1315.0

10 δy737.0

= from Nikuradge

This (power) law is valid in the log-region. However, it is not valid to predict the shear stress at wall (tends to ∞).

≅δ δ≅δ727

2 30.1H ≅

139.0xf Re02296.0C −=

Example: Air-steam ( 4108.1 −×=ν ft2/s) at 50 ft/s passes a smooth flat

plate. It is assumed that as 50 102.3xu×=

ν the boundary layer changes

suddenly from laminar to turbulent. Calculate δ at x = 5 ft .

【Solution】

Formulae： TBL：5

xRe371.0

xuRe 0x = ( 0u = 50 ft/s)

LBL：xRe

δ Transition occurs at 5x 102.3Re ×=

(A) Check the nature of boundary layer at x = 5 ft

( ) 564x 102.31039.1

108.1550Re ×>×=

×= − (turbulent)

(B) Find the location of transition

For 4tr5

108.1x50102.3 −×

=× 152.1x tr = ft

(C) Calculate boundary layer thickness (laminar) at xtr

For xRe

δ ( )( )( ) 01018.0102.30.5152.1 215 =×=δ

− ft

(D) Find the virtual origin of turbulent boundary layer

371.0x

5108.1

50 x371.0

x01018.0

4−×

371.001018.0

108.150x 5

−×= , x = 0.256 ft

(E) Calculate turbulent boundary layer thickness at x = 5 ft

104.4256.0)152.15(x =+−= ft,

64x 1014.1

108.1)104.4(50Re ×=

×= −

in123.1ft0936.0)1014.1)(371.0)(104.4( 516 ==×=δ −

Power Law of Rough Flows

The profile 5.8kyln1

y30ln1 (see also pp. 82)

can be approximated as

s* ky34.9

= for 2006 <<

Power Friction Law:

dyky34.91uV

= δ where V：average of u within δ

δ= for 200

< (error is within 3%)......()

Rough Turbulent Boundary Layer:

In case that 70kuu/

, define sk

yy =+ and δ

For 1ks

>>δ , the log-law is wake correction

yW5.8kyln1

…….……… (1)

For channel flows, π is small ( )3.0< and

sin2yW 2

(see pp. 83)

@ as δ=y )2(5.8k

κ= ….…..…….. (2)

(2)－(1) leads to

−κπ

−=− yW2yln1

Recall the Darcy-Weisbach Equation (valid for fully developed pipe flows)

p p∆+(-)

cross sectional area A

Average vel=V

ρ=∆ 2V

Force balance yields ( )LDD4

p w2 π⋅τ=

π⋅∆

Therefore, 22

LDD ρ

2 uV8f

ρ⇒ρ=

One gets the Darcy friction coefficient 2

Since ∫δ

= δ0 dyu1V ∫

δ≅ δ

sk dyu1 ( )sk>>δ

δδκ

ssk (neglect wake correction term)

= ∫δ

δκη

κη−ηη

( ) 5.8k

ln1101

κ−⋅

11ln1or

g2Vfph

L =g∆

< Stanton-Moody Diagram >

70νku s* ≅

ReD = V D / ν

Fully rough

Laminar

64/ReD

Fully Developed Boundary Layer Flows in Open Channel

∂∂

At the free surface ( )δ=y , since 0yu→

∂∂ , 0→τ

δ−δρ=τ∴

At the bottom ( )0y = , since 2*w uSg ρ=δρ=τ

δ= Sgu*

δ= (from on pp. 94)_

Then *s

δ= ( V is the average velocity )

δ= Sg

skSg8 δ

n1δ= (from Manning’s formula)

The Manning’s coefficient 61

1n = 61

sk04.0≅ (in SI unit system)

The momentum equation is (fully developed)

∂τ∂

∂∂

ρ−=

∂∂

+∂∂

Or 0Sgy

∂τ∂

Integration gives ySgc ρ−=τ

General Turbulent Boundary Layer (2-D)

The law of the wall (for an equilibrium smooth turbulent boundary layer) is

cuyln1uyFunuu **

where ( )ρ

)x(xu w*

Noted that before separation, 0w ≥τ

after separation, 0w <τ . u* becomes meaningless.

Velocity defect law is generally in the form

=− x,

u)x(Uu

e due to non-equilibrium nature

In order to be consistent with law of the wall

)x(ayln1x,yF +δκ

Noted that )x(a , c, )x(Ue and )x(u* are related by the friction law

cauln1u

)x(U *

e +−νδ

∞→∞→

u)x(U ,u

Von Karman Integral Equation (Steady, Turbulent B.L.)

e ∂ρτ∂

+∂∂

ν+∂∂

=∂∂

+∂∂ where 'v'ut ρ−=τ

)Uu(yv)Uu(v

x)Uu(uLHS ee

ee −∂∂

−−∂∂

+∂∂

+∂−∂

e−∂∂

∂∂

+∂∂

ν=−∂∂

+−+−∂∂ t

ee yyu)Uu(v

yxdUd)Uu()Uu(u

Integrating with respect to y from 0 to ∞ leads to ( 2*

w0yy u

yuν −=−=

∂∂

−∂∂

=∞=)

0dy)Uu(xd

Uddy)Uu(uxd

e0 e +−+− ∫∫

∞∞∞

+−=0

or 2*1e

=δ+δ

Clauser (1984) introduced

∫∞ −

uU ( )1e* Uu δ=∆

δ= where ∆ is comparable to δ

Noted that ∫∞

−0 e dy)Uu(u

∫∫∞∞

−+−=0

2e0 ee dy)Uu(dy)Uu(U

( )∫ −+∆−= ∞0

2e*e dyUuuU

(see pp.33 for laminar cases)

Estimate

ν+⋅∆≈−∫

∞ 2e

uOudy)Uu(

−∞small

e2* dy

uUuu ∫ −

ν small

dy)Uu(u

+∆= 2*

Since 2

cauln1uu

κ∆ν

( ) 0RlnR1~ *R2

→ ∞→

For large *R , ∆−≅−∫∞

*e0 e uUdy)Uu(u

The approximated form is

e*e* u

xdUdu)Uu(

=∆+∆

邊界層理論1).pdf · 2018. 10. 26. · 2-D Flow on Positive Curved Surface . Effect of Pressure...

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