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SOLUTION FOR THE BOUNDARY LAYER ON A FLAT PLATEConsider 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!U

U

y

x late

u

H1

A STEADY RECTILINEAR POTENTIAL FLOW HAS ZERO PRESSURE GRADIENT EVERYWHEREA steady, rectilinear potential flow in the x direction is described by the relations xJ xJ J ! Ux , u ! !U , v ! !0 xx xy According to Bernoullis equation for potential flows, the dynamic pressure of the potential flow ppd is related to the velocity field as

ppd

V(u v ) ! const

Between the above two equations, then, for this flow

xppd xx

!

xppd xy

!0

U

U late

u

H2

BOUNDARY LAYER EQUATIONS FOR A FLAT PLATEFor the case of a steady, aminar boundary ayer on a f at p ate at 0 ang e of incidence, with vanishing imposed pressure gradient, the boundary ayer equations and boundary conditions become (see S ide 15 of BoundaryLayerApprox.ppt with dppds/dx = 0)xu xu 1 dppds x 2u ! u v R 2 xy xx dx xy xu xv !0 xx xy

x 2u xu xu !R 2 u v xx xy xy xu xv !0 xx xy

u y !0 ! 0

,

v y !0 ! 0

, u y !g ! U

Tangentia and norma ve ocities vanish at boundary: tangentia ve ocity = free stream ve ocity far from p ateU

U

y

x p ate

u

H 3

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

u( x, y) y !H ! 0.99 UU

u = 0.99 U

u U x H

U

y

x plate

u

H

y

The choice 0.99 is arbitrary; we could have chosen 0.98 or 0.995 or whatever we find reasonable.4

STREAMWISE VARIATION OF BOUNDARY LAYER THICKNESSConsider a plate of length L. Based on the estimate of Slide 11 of BoundaryLayerApprox.ppt, we can estimate H asH ~ (Re) L/2

L , Re ! R RL H ! C U1/ 2

or thusH RL U1/ 2

or

where C is a constant. By the same arguments, the nominal boundary thickness up to any point x e L on the plate should be given asH Rx U1/ 2

or

Rx H ! C U

1/ 2

U

U

x p ate L

u

H5

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:

xu u xx

xu x 2u v !R 2 xy xy

,

xu xx

xv !0 xy

u y !0 ! 0 ,

v y !0 ! 0 , u y ! g ! U

6

SIMILARITY SOLUTIONSuppose the solution has the property that when u/U is plotted against y/H (where H(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/H = 0 and u/U = 0.99 at y/H = 1, no matter what the value of x. Similarity is satisfied if a plot of u/U versus y/H defines exactly the same function regardless of the value of x.U U

U

y

x

u x1 late

u x2

H1 fil t x1

1 fil s t x1 x2

y/H

Simil rity s tisfi

y/H

Simil rity ot s tisfi0 0 u/U profil t x2 17

0 0 u/U 1

SIMILARITY SOLUTION contd.So for a solution obeying similarity in the velocity profile we must haveu y ! f1 U H(x )

where g1 is a universal function, independent of x (position along the plate). Since we have reason to believe that Rx Rx or H ! C H~ U U where C is a constant (Slide 5), we can rewrite any such similarity form as1/ 2 1/ 2

u y U !y ! f (L) , L ! 1/ 2 U Rx Rx U

Note that L 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).

8

BUT DOES THE PROBLEM ADMIT A SIMILARITY SOLUTION?Maybe, maybe not, you never know until you try. The problem is:

xu xu x 2u u v !R 2 xx xy xy

,

xu xv !0 xx xy

u y !0 ! 0 ,

v y !0 ! 0

, u y!g ! U

This problem can be reduced with the streamfunction (u = x]/xy, v = x]/xx) to:

x 3] x] x] x] x] !R 3 2 xy xy xxxy xx xyx] !0 xy y ! 0 , x] !0 , xx y ! 0 x] !U xy y ! g

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

9

SOLUTION BY THE METHOD OF GUESSINGWe want our streamfunction to give us a velocity u = x]/xy satisfying the similarity formu ! f (L) , U U L!y Rx

So we could start off by guessing

] ! F(L)where f is another similarity function. But this does not work. Using the prime to denote ordinary differentiation with respect to L, if ] = f(L) then

x] xL d) u! ! F (L xy xyBut

xL ! xy

U Rx

so that

u 1 ! Fd ) (L U UR x

10

SOLUTION BY THE METHOD OF GUESSING contd.So if we assume

] ! F(L)then we obtainnot OK OK

u 1 F (L) ! U UR xThis form does not satisfy the condition that u/U should be a function of L alone. If F is a function of L alone then its first derivative F(L) is also a function of L alone, but note the extra (and unwanted) functionality in x via the term (URx)-1/2! So our first try failed because of the term (URx)-1/2. Lets not give up! Instead, lets learn from our mistakes!11

ANOTHER TRYThis time we assume

] ! UR x F(L)Now remembering that x and y are independent of each other and recalling the evaluation of xL/xy of Slide 10,

x] xL U u! ! UR x F (L) ! UR x F (L) ! UF (L) xy xy Rxor thus

u ! f (L) , U

f (L) | F (L)

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(L).12

REDUCTION FROM PARTIAL TO ORDINARY DIFFERENTIAL EQUATIONOur goal is to reduce the partial differential equation for and boundary conditions on ], i.e.

x] x] x] x] x 3] !R 3 2 xy xxxy xx xy xyx] !0 xy y ! 0 , x] !0 , xx y ! 0 x] ! xy y ! g

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

U ] ! UR x F(L) , L ! y RxTo do this we will need the following forms:

xL ! xy

U Rx

1L 1 U 3 / 2 xL x ! y ! 2x 2 xx R

13

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

x] x] x] x] x 3] !R 3 2 xy xxxy xx xy xywe need to know x]/xy, x2]/xy2, x3]/xy3, x]/xx and x2]/xyxx, whereU ] ! UR x F(L) , L ! y Rx

xL ! xy

U Rx

,

xL 1L ! xx 2x

Now we have already worked out x]/xy; from Slide 12:

x] ! UFd ) (L xyThus

U x 2] xL d d (L Fd ) (L ! UFd ) !U 2 xy xy Rx

U U x 3] xL d d d d Fd ) (L (L !U ! U Fd ) 3 xy Rx xy Rx 14

REDUCTION contd.Again usingU ] ! UR x F(L) , L ! y Rx

xL ! xy

U Rx

,

1L xL ! xx 2x

we now work out the two remaining derivatives:

1 UR x] x xL UR x F(L) ! F(L) UR x Fd ) (L ! ! 2 x xx xx xx 1 UR ?F(L) LFd )A (L 2 x

1 x 2] x x] x UR ?F(L) LFd )A ! (L ! ! 2 xxxy xy xx xy x UR xL 1 d d ?Fd ) Fd ) LFd )A ! 1 U LFd ) (L (L (L (y xy 2 2x x

15

REDUCTION contd.Summarizing,U ] ! UR x F(L) , L ! y Rx

xL ! xy

U Rx

,

1L xL ! xx 2x

x] ! UFd ) (L xyU x 2] d Fd ) (L !U 2 xy Rx U x 3] d d ( ! U FdL) 3 xy Rx

x] 1 UR ?F(L) LFd )A ! (L xx 2 xx 2] 1U d ! LFd ) (y xxxy 2x16

REDUCTION contd.Now substitutingx] ! UFd ) (L xy x 2] U d !U Fd ) (L 2 xy Rxx 3] U d d ! U FdL) ( 3 xy Rx

x] 1 UR d ?F(L) LFd )A ! (L xx 2 x

1U x 2] d (y ! LFd ) 2x xxxy

into

x] x] x] x] x 3] !R 3 2 xy xxxy xx xy xy

yields

U2 1 U2 1 U2 d d LFdd ! Fd (F LFd d )F d Fd 2 x 2 x xor thus

d d d 2Fd FFd 0 !Similarity works! It has cleaned up the mess into a simple (albeit nonlinear) ordinary differential equation!17

BOUNDARY CONDITIONSFrom Slide 9, the boundary conditions are

x] !0 xy y ! 0x] ! UFd ) (L xy

,

x] !0 , xx y ! 0

x] !U xy y! gL! y

But we already showed thatx] 1 UR ?F(L) LFd )A ! (L xx 2 x

U Rx

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

Fd ) ! 0 , F(0) ! 0 , Fd ) ! 1 (0 (gThus we have three boundary conditions for the 3rd-order differential equation

d d d 2Fd FFd 0 !18

SOLUTIONThere are a number of ways in which the problem

d d d 2Fd FFd 0 !

Fd ) ! 0 , F(0) ! 0 , Fd ) ! 1 (0 (g

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.Blasius Solution, Laminar Boundary Layer6

5

4

f'( d FL) d f''(L) Fd

f(L)

L

3

F

2

1

0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

Ff, Fd d , f', ,f'' d F

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