1

SOLUTION FOR THE BOUNDARY LAYER ON A FLAT PLATE

Consider the following scenario.

1. A steady potential flow has constant velocity U in the x direction.

2. An infinitely thin flat plate is placed into this flow so that the plate is parallel to the potential flow (0 angle of incidence).

Viscosity should retard the flow, thus creating a boundary layer on either side of the plate. Here only the boundary layer on one side of the plate is considered. The flow is assumed to be laminar.

Boundary layer theory allows us to calculate the drag on the plate!

xy d

U

U

u

plate

2

A STEADY RECTILINEAR POTENTIAL FLOW HAS ZERO PRESSURE GRADIENT EVERYWHERE

xy d

U

U

u

plate

A steady, rectilinear potential flow in the x direction is described by the relations

According to Bernoulli’s equation for potential flows, the dynamic pressure of the potential flow ppd is related to the velocity field as

Between the above two equations, then, for this flow

0y

v,Ux

u,Ux

const)vu(2

1p 22

pd

0y

p

x

p pdpd

3

BOUNDARY LAYER EQUATIONS FOR A FLAT PLATE

xy d

U

U

u

plate

For the case of a steady, laminar boundary layer on a flat plate at 0 angle of incidence, with vanishing imposed pressure gradient, the boundary layer equations and boundary conditions become (see Slide 15 of BoundaryLayerApprox.ppt with dppds/dx = 0)

0y

v

x

u

y

u

y

uv

x

uu

2

2

Uu,0v,0uy0y0y

Tangential and normal velocities vanish at boundary: tangential velocity = free stream velocity far from plate

0y

v

x

u

y

u

dx

dp1

y

uv

x

uu

2

2pds

4

NOMINAL BOUNDARY LAYER THICKNESS

xy d

U

U

u

plate

Until now we have not given a precise definition for boundary layer thickness. Here we use d to denote nominal boundary thickness, which is defined to be the value of y at which u = 0.99 U, i.e.

U99.0)y,x(uy

d

x

y

u

U

u = 0.99 U

d

The choice 0.99 is arbitrary; we could have chosen 0.98 or 0.995 or whatever we find reasonable.

5

STREAMWISE VARIATION OF BOUNDARY LAYER THICKNESS

Consider a plate of length L. Based on the estimate of Slide 11 of BoundaryLayerApprox.ppt, we can estimate d as

or thus

where C is a constant. By the same arguments, the nominal boundary thickness up to any point x L on the plate should be given as

d UL,)(~

L2/1 ReRe

2/12/1

U

LCor

U

L~

d

d

2/12/1

U

xCor

U

x~

d

d

xy d

U

U

u

plateL

6

SIMILARITYOne triangle is similar to another triangle if it can be mapped onto the other triangle by means of a uniform stretching.

The red triangles are similar to the blue triangle.

The red triangles are not similar to the blue triangle.

Perhaps the same idea can be applied to the solution of our problem:

0y

v

x

u,

y

u

y

uv

x

uu

2

2

Uu,0v,0uy0y0y

7

SIMILARITY SOLUTIONSuppose the solution has the property that when u/U is plotted against y/d (where d(x) is the previously-defined nominal boundary layer thickness) a universal function is obtained, with no further dependence on x. Such a solution is called a similarity solution. To see why, consider the sketches below. Note that by definition u/U = 0 at y/d = 0 and u/U = 0.99 at y/d = 1, no matter what the value of x. Similarity is satisfied if a plot of u/U versus y/d defines exactly the same function regardless of the value of x.

plate

xy d

U

U

uu

U

x1 x2

00

1

1u/U

y/dprofiles at x1 and x2

00

1

1u/U

y/dprofile at x1

profile at x2

Similarity satisfied

Similarity not satisfied

8

SIMILARITY SOLUTION contd.So for a solution obeying similarity in the velocity profile we must have

where g1 is a universal function, independent of x (position along the plate). Since we have reason to believe that

where C is a constant (Slide 5), we can rewrite any such similarity form as

Note that is a dimensionless variable.

If you are wondering about the constant C, note the following. If y is a function of x alone, e.g. y = f1(x) = x2 + ex, then y is a function of p = 3x alone, i.e. y = f(p) = (p/3)2 + e(p/3).

d

)x(

yf

U

u1

2/12/1

U

xCor

U

x~

d

d

x

Uy

Ux

y,)(f

U

u2/1

9

BUT DOES THE PROBLEM ADMIT A SIMILARITY SOLUTION?

Maybe, maybe not, you never know until you try. The problem is:

This problem can be reduced with the streamfunction (u = /y, v = - /x) to:

Note that the stream function satisfies continuity identically. We are not using a potential function here because boundary layer flows are not potential flows.

0y

v

x

u,

y

u

y

uv

x

uu

2

2

Uu,0v,0uy0y0y

3

3

2 yyxyxy

Uy

,0x

,0y

y0y0y

10

SOLUTION BY THE METHOD OF GUESSING

We want our streamfunction to give us a velocity u = /y satisfying the similarity form

So we could start off by guessing

where f is another similarity function.

But this does not work. Using the prime to denote ordinary differentiation with respect to , if = f() then

But

x

Uy,)(f

U

u

)(F

y)(F

yu

x

U

y

so that )(FxU

1

U

u

11

SOLUTION BY THE METHOD OF GUESSING contd.

So if we assume

then we obtain

This form does not satisfy the condition that u/U should be a function of alone. If F is a function of alone then its first derivative F’() is also a function of alone, but note the extra (and unwanted) functionality in x via the term (Ux)-1/2!

So our first try failed because of the term (Ux)-1/2.

Let’s not give up! Instead, let’s learn from our mistakes!

)(F

)(FxU

1

U

u

not OK OK

12

ANOTHER TRY

This time we assume

Now remembering that x and y are independent of each other and recalling the evaluation of /y of Slide 10,

or thus

Thus we have found a form of that satisfies similarity in velocity!

But this does not mean that we are done. We have to solve for the function F().

)(FxU

)(FUx

U)(FxU

y)(FxU

yu

)(F)(f,)(fU

u

13

REDUCTION FROM PARTIAL TO ORDINARY DIFFERENTIAL EQUATION

Our goal is to reduce the partial differential equation for and boundary conditions on , i.e.

to an ordinary differential equation for and boundary conditions on f(), where

To do this we will need the following forms:

3

3

2 yyxyxy

Uy

,0x

,0y

y0y0y

x

Uy,)(FxU

x

U

y

x2

1x

Uy

2

1

x2/3

14

The next steps involve a lot of hard number crunching. To evaluate the terms in the equation below,

we need to know /y, 2/y2, 3/y3, /x and 2/yx, where

Now we have already worked out /y; from Slide 12:

Thus

REDUCTION contd.

3

3

2 yyxyxy

x

Uy,)(FxU

x2

1

x,

x

U

y

)(Fx

UU

y)(FU

y2

2

)(FUy

)(Fx

UU

y)(F

x

UU

y3

3

15

Again using

we now work out the two remaining derivatives:

REDUCTION contd.

x

Uy,)(FxU

x2

1

x,

x

U

y

)(F)(Fx

U

2

1

x)(FxU)(F

x

U

2

1)(FxU

xx

)y(Fx

U

2

1

y)(F)(F)(F

x

U

2

1

)(F)(Fx

U

2

1

yxyyx

2

16

Summarizing,

REDUCTION contd.

x

Uy,)(FxU

x2

1

x,

x

U

y

)(F)(Fx

U

2

1

x

)y(Fx

U

2

1

yx

2

)(FUy

)(Fx

UU

y2

2

)(Fx

UU

y3

3

17

Now substituting

into

yields

or thus

Similarity works! It has cleaned up the mess into a simple (albeit nonlinear) ordinary differential equation!

REDUCTION contd.

)(F)(Fx

U

2

1

x

)y(Fx

U

2

1

yx

2

)(FUy

)(Fx

UU

y2

2

)(F

x

UU

y3

3

3

3

2 yyxyxy

Fx

UF)FF(

x

U

2

1FF

x

U

2

1 222

0FFF2

18

From Slide 9, the boundary conditions are

But we already showed that

Now noting that = 0 when y = 0, the boundary conditions reduce to

Thus we have three boundary conditions for the 3rd-order differential equation

BOUNDARY CONDITIONS

)(F)(Fx

U

2

1

x

)(FUy

0FFF2

Uy

,0x

,0y

y0y0y

x

Uy

1)(F,0)0(F,0)0(F

19

There are a number of ways in which the problem

can be solved. It is beyond the scope of this course to illustrate numerical methods for doing this. A plot of the solution is given below.

SOLUTION

0FFF2 1)(F,0)0(F,0)0(F

Blasius Solution, Laminar Boundary Layer

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

f, f', f''

f()

f'()

f''() F

FF

F,F,F

20

To access the numbers, double-click on the Excel spreadsheet below. Recall that

SOLUTION contd.

F F' F''0 0 0 0.33206

0.1 0.00166 0.033206 0.3320510.2 0.006641 0.066408 0.3319870.3 0.014942 0.099599 0.3318120.4 0.02656 0.132765 0.3314730.5 0.041493 0.165887 0.3309140.6 0.059735 0.198939 0.3300820.7 0.081278 0.231892 0.3289250.8 0.106109 0.264711 0.3273920.9 0.134214 0.297356 0.325435

1 0.165573 0.329783 0.323011.1 0.200162 0.361941 0.3200741.2 0.237951 0.393779 0.3165921.3 0.278905 0.42524 0.3125311.4 0.322984 0.456265 0.3078681.5 0.370142 0.486793 0.3025831.6 0.420324 0.516761 0.2966661.7 0.473473 0.546105 0.2901141.8 0.529522 0.574763 0.2829331.9 0.5884 0.602671 0.275138

2 0.65003 0.62977 0.2667532.1 0.714326 0.656003 0.257811

By interpolating on the table, it is seen that u/U = F’ = 0.99 when = 4.91.

)(FU

u

21

Recall that the nominal boundary thickness d is defined such that u = 0.99 U when y = d. Since u = 0.99 U when = 4.91 and = y[U/(x)]1/2, it follows that the relation for nominal boundary layer thickness is

Or

In this way the constant C of Slide 5 is evaluated.

NOMINAL BOUNDARY LAYER THICKNESS

91.4x

U

d

91.4C,U

xC

2/1

d

22

Let the flat plate have length L and width b out of the page:

The shear stress o (drag force per unit area) acting on one side of the plate is given as

Since the flow is assumed to be uniform out of the page, the total drag force FD acting on (one side of) the plate is given as

The term u/y = 2/y2 is given from (the top of) Slide 17 as

DRAG FORCE ON THE FLAT PLATE

L

b

0y0y

o y

u

y

u

L

0 ooD dybdAF

)(Fx

UU

yy

u2

2

23

The shear stress o(x) on the flat plate is then given as

But from the table of Slide 20, f’’(0) = 0.332, so that boundary shear stress is given as

Thus the boundary shear stress varies as x-1/2. A sample case is illustrated on the next slide for the case U = 10 m/s, = 1x10-6 m2/s, L = 10 m and = 1000 kg/m3 (water).

DRAG FORCE ON THE FLAT PLATE contd.

)0(Fx

UU)0(F

x

UUo

Ux

,)(332.0U x

2/1x2

o ReRe

24

Boundary Shear Stress

0

0.0001

0.0002

0.0003

0 0.02 0.04 0.06 0.08 0.1

x (m)

o (

Pa)

Sample distribution of shear stress o(x) on a flat plate:

DRAG FORCE ON THE FLAT PLATE contd.

x

UU332.0o

Note that o = at x = 0.Does this mean that the drag force FD is also infinite?

U = 0.04 m/sL = 0.1 m = 1.5x10-5 m2/s = 1.2 kg/m3

(air)

25

No it does not: the drag force converges to a finite value!

And here is our drag law for a flat plate!

We can express this same relation in dimensionless terms. Defining a diimensionless drag coefficient cD as

it follows that

For the values of U, L, and of the last slide, and the value b = 0.05 m, it is found that ReL = 267, cD = 0.0407 and FD = 3.90x10-7 Pa.

DRAG FORCE ON THE FLAT PLATE contd.

2/1D

2/1L

0

2/1

L

0

2/1L

0 oD

bLUU664.0F

L2dxx

dxxUbU332.0dxbF

bLU

Fc

2D

D

UL

,)(664.0c 2/1D ReRe

26

The relation

is plotted below. Notice that the plot is carried only over the range 30 ReL 300. Within this range 1/ReL is sufficiently small to justify the boundary layer approximations. For ReL > about 300, however, the boundary layer is no longer laminar, and the effect of turbulence must be included.

DRAG FORCE ON THE FLAT PLATE contd.

UL

,)(664.0c 2/1D ReRe

Blasius Drag Law for Laminar Flow over Flat Plate

0.01

0.1

1

10 100 1000

ReL

c D

27

The solution presented here is the Blasius-Prandtl solution for a boundary layer on a flat plate. More details can be found in:

Schlichting, H., 1968, Boundary Layer Theory, McGraw Hill, New York, 748 p.

REFERENCE