CHAPTER 9: EXTERNAL* INCOMPRESSIBLE VISCOUS FLOWS *(unbounded)

CHAPTER 9: EXTERNAL INCOMPRESSIBLE VISCOUS FLOWS adverse pressure gradient leads to separation difficult to use theory thicker Re = U  x/ ; Re = U  c/ ; … laminar and turbulent boundary layers displaced inviscid outer flow adverse pressure gradient and separation = separated bdy layer

Boundary Layer Provides Missing Link Between Theory and Practice Boundary layer, , where viscous stresses (i.e. velocity gradient) are important we’ll define as where u(x,y) = 0 to 0.99 U  above boundary.

In August of 1904 Ludwig Prandtl, a 29-year old professor presented a remarkable paper at the 3 rd International Mathematical Congress in Heidelberg. Although initially largely ignored, by the 1920s and 1930s the powerful ideas of that paper helped create modern fluid dynamics out of ancient hydraulics and 19 th -century hydrodynamics. (only 8 pages long, but arguably one of the most important fluid-dynamics papers ever written)

Prandtl assumed no slip condition Prandtl assumed thin boundary layer region where shear forces are important because of large velocity gradient Prandtl assumed inviscid external flow Prandtl assumed boundary so thin that within it  p/  y  0; v << u and  /  x <<  /  y Prandtl outer flow drives boundary layer boundary layer can greatly effect outer outer “inviscid” flow if separates

Theodore von Karman Prandtl Fluid Motion with Very Small Friction 2-D boundary layer equations Blasius The Boundary Layers in Fluids with Little Friction Solution for laminar, 0-pressure gradient flow von Karman Integral form of boundary layer equations Sir Horace Lamb Hydrodynamics ~ one paragraph on bdy layers Sir Horace Lamb Hydrodynamics ~ entire section on bdy layers BOUNDARY LAYER HISTORY

INTERNALEXTERNAL FULLY DEVELOPED? CAN BENEVER WAKE?NEVERUSUALLY - PLATE IS EXCEPTION THEORY LAMINAR PIPES, DUCTS,..FLAT PLATE & ZERO PRESSURE GRADIENT GROWING BOUNDARY LAYER? NOT WHEN FULLY DEVELOPED ALWAYS ADVERSE PRESSURE GRADIENT PIPE/DUCT=N0 DIFFUSER=YES PLATE=MAYBE BODIES=USUALLY TURBULENT EXPERIMENT PIPE (EXAMPLE) u(r)/U c/l = (y/R) 1/n PLATE (EXAMPLE) u(y)/U o = (y/  ) 1/n

Note – throughout figures the boundary layer thickness*, , is greatly exaggerated! (disturbance layer*) Airline industry had to develop flat face rivets.

Re = 20,000 Angle of attack = 6 o Symmetric Airfoil 16% thick

Flat Plate (no pressure gradient) ~ what is velocity profile? ~ wall shear stress/drag? ~ displacement of free stream? ~ laminar vs turbulent flow? Immersed Bodies ~ wall shear stress/drag? ~ lift? ~ minimize wake

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FLAT PLATE – ZERO PRESSURE GRADIENT

Laminar Flow  /x ~ 5.0/Re x 1/2 THEORY Turbulent Flow Re xtransition > 500,000 u(y)/U  = (y/  ) 1/7  /x ~ 0.382/Re x 1/5 EXPERIMENTAL

“At these Re x numbers bdy layers so thin that displacement effect on outer inviscid layer is small” No simple theory for Re < 1000; (can’t assume  is thin)

Re L = 10,000 Visualization is by air bubbles see that boundary + layer, , is thin and that outer free stream is displaced,  *, very little. FLAT PLATE – ZERO PRESSURE GRADIENT + Disturbance Thickness,  (x) (pg 412); boundary layer thickness,  (x) (pg 415) outside  (x), U is constant so P is constant u(x,y) is not constant,  (x) is thin so assume P inside  (x) is impressed from the outside

Re L = 10,000 Visualization is by air bubbles see that boundary + layer, , is thin and that outer free stream is displaced,  *, very little. FLAT PLATE – ZERO PRESSURE GRADIENT + Disturbance Thickness,  (x) (pg 412); boundary layer thickness,  (x) (pg 415) Re x = U  x/ Assume Re xtransition ~ 500,000 x Re L = U  x/ L

SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS

Development of laminar boundary layer (0.01% salt water, free stream velocity 0.6 cm/s, thickness of the plate 0.5 mm, hydrogen bubble method). * * * * * Re x  1000

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FLAT PLATE – ZERO PRESSURE GRADIENT:  (x) BOUNDARY OR DISTURBANCE LAYER

 (x)  *  BOUNDARY OR DISTURBANCE LAYER

Definition: u(x,  ) = 0.99 of U=U  =U e (within 1 % of U  ) Boundary + Layer Thickness  (x) + Disturbance

 is at y location where u(x,y) = 0.99 U  Because the change in u in the boundary layer takes place asymptotically, there is some indefiniteness in determining  exactly.

NOTE: boundary layer is much thicker in turbulent flow. Blasius showed theoretically for laminar flow that  /x = 5/(Re x ) 1/2 (Re x =  U  x/  )   x 1/2 Experimentally found* for turbulent flow that   x 4/5

NOTE: velocity gradient at wall (  w =  du/dy) is significantly greater. At same x: U/  L > U /  T At same x:  wL <  wT

Note, boundary layer is not a streamline! From theory (Blasius 1908, student of Prandtl):  = 5x/(Re x 1/2 ) = 5x/(U/[ x]) 1/2 = 5 1/2 x 1/2 /U 1/2 d  /dx = 5 ( /U) 1/2 (½) x -1/2 = 2.5/(Re x ) 1/2 V/U = dy/dx  streamline = 0.84/(Re x 1/2 ) dy/dx  streamline  d  /dx so  not streamline

Behavior of a fluid particle traveling along a streamline through a boundary layer along a flat plate. U  = U = U e

ASIDE ~ LAMINAR TO TURBULENT TRANSITION

LAMINAR TO TURBULENT TRANSITION

Emmons spot ~ Re x = 200,000 Spots grow approximately linearly downstream at downstream speed that is a fraction of the free stream velocity. NOTE: Turbulence is not initiated at Re tr all along the width of the plate

Emmons spot, Re x = 400,000 smoke in wind tunnel

Transition not fixed but usually around Re x ~ 500,000 (2x x10 6, MYO) For air at standard conditions and U = 30 m/s, x tr ~ 0.24 m x=0 Turbulent boundary layer is thicker and grows faster.

Nevertheless: Treat transition as it happens all along Re x = 500,000 Experimentally transition occurs around Re x ~ 5 x 10 5 Water moving around 4 m/s past a ship, transitions after about 0.14 m from the bow, representing only about 0.1 % of total length of 143 m long ship.

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FLAT PLATE – ZERO PRESSURE GRADIENT:  * (x) DISPLACEMENT THICKNESS

  * (x)  DISPLACEMENT THICKNESS

Definition:  * = 0  (1 – u/U)dy Displacement Thickness  * (x)   * is displacement of outer streamlines due to boundary layer

By definition, no flow passes through streamline, so mass through 0 to h at x = 0 is the same as through 0 to h +  * at x = L. x = L Displacement thickness  *

 Uh = 0  h+  *  udy = 0  h+  *  (U + u - U)dy Uh = 0  h+  * Udy + 0  h+  * (u - U)dy Uh = U(h +  *) + 0  h+  * (u - U)dy -U  * = 0  h+  * (u -U)dy  * = 0  h+  * (– u/U + 1)dy  0   (1 – u/U)dy

 *  0   (1 – u/U)dy  0  (1 – u/U)dy displacement of outer streamlines due to  (x) function of x! 

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Definition:  * = 0   (1 – u/U)dy  U  * w = 0    (U – u)dyw  0    (U – u)dyw the deficit in mass flux through area  w due to the presence of the boundary layer. = mass flux passing through an area [  * w] in the absence of a boundary layer Displacement Thickness  *

Definition:  * = 0   (1 – u/U)dy  U  * = 0    (U – u)dy  0    (U – u[y])dy Displacement Thickness  *

 * problem

Displacement Thickness,  *, Problem Suppose given velocity profile: u = 0 from y = 0 to  u = U e for y > . Show that  * =  = 

Displacement Thickness  * Suppose u = 0 from y = 0 to  and u = U e for y > . Show that  * =   * = 0   (1 – u/U e )dy = 0   (1 – 0/U e )dy +    (1 – U e /U e )dy  * = 0   dy =  = 

 * = Distance that an equivalent inviscid flow is displaced from a solid boundary as a consequence of slow moving fluid in the boundary layer.  *(x) ~ 1 / 3  (x)  Blasius * = 1.721x/(Re x ) 1/2 Laminar flow on flat plate in uniform free stream  Blasius = 5x/(Re x 1/2 )

Displacement Thickness  * Definition:  * = 0   (1 – u/U e )dy From the point of view of the flow outside the boundary layer,  * can be interpreted as the distance that the presence of the boundary layer appears to “displace” the flow outward (hence its name). To the external flow, this streamline displacement also looks like a slight thickening of the body shape.  and  * are greatly magnified  * (x)  (x)

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  * (x)  EXAMPLE

2-D DUCT PROBLEM: Find U e as a function of U , D and  * U e (x) U e (x) = ?

Continuity equation (per W):  {U  D = -D2  D/2 udy + -D2  D/2 U e dy - -D2  D/2 U e dy} U  D = -D2  D/2 U e dy – { -D2  D/2 (U e -u)dy} u(y) U   f(y) U e (x)

Continuity equation (per W): U  D = -D2  D/2 U e dy – { -D2  D/2 ( U e -u)dy} U  D = -D2  D/2 U e dy–{ (-D/2)  (-D/2+  ) (U e -u)dy+ (D/2-  )  D/2 ( U e -u)dy} (used approximation that outside boundary layer u =U e ) (note that U e = function of x but not y) INLET OF DUCT

Continuity equation (per W): U  D = -D2  D/2 U e dy – { (-D/2)  (-D/2+  ) (U e -u)dy + (D/2-  )  D/2 ( U e -u)dy} U  D = -D2  D/2 U e dy – U e { (-D/2)  (-D/2+  ) (1-u/U e )dy} + U e { (D/2-  )  D/2 (1-u/U e )dy} INLET OF DUCT ** **  * = 0   (1 –u/U e )dy  0   (1 – u/U e )dy

INLET OF DUCT ** ** Continuity equation (per W): U  D = -D2  D/2 U e dy–U e { (-D/2)  (-D/2+  ) (1-u/U e )dy+ (D/2-  )  D/2 (1-u/U e )dy} -{  * +  * } U  D= U e D – 2U e  * = U e [D – 2  * ] U e (x) = U  D/[D – 2  *(x)] QED

from Continuity Equation U  D = U e [D – 2  * ] We see that the free stream velocity in the duct is given by the effective decrease in the cross sectional area due to the growth of the boundary layers, and this decrease in area is measured by the displacement thickness!!!

FLAT PLATE – ZERO PRESSURE GRADIENT:  (x) MOMENTUM THICKNESS

  *  (x) Momentum Thickness MOMENTUM THICKNESS

Definition:  = 0  u/U(1 – u/U)dy Momentum Thickness  (x)  How define drag due to boundary layer?

Momentum Thickness  Definition:  = 0   u/U e (1 – u/U e )dy  U e 2  w = 0    u(U e – u)dyw The momentum thickness, , is defined as the thickness of a layer of fluid, with velocity U e, for which the momentum flux is equal to the deficit of momentum flux through the boundary layer. (Mass flux through  = 0    udyw)

u U  [(u)(  [U-u])]xdy [Flux of momentum deficit] U  U w 

 = 0.99U (within 1 %)  * = 0   (1 – u/U e )dy   0  (1 – u/U e )dy  = 0   u/U e (1 – u/U e )dy  0  u/U e (1 – u/U e )dy  * &  Easier to calculate form data and more physical significance but “can be” dependent on    Summary:

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Control volume analysis of drag force on a flat plate due to boundary shear Relate  to Drag

PRESSURE IS UNIFORM - NO NET PRESSURE FORCE! - NO DU/DX!

1.From (0,0) to (0,h): Vn = -U o = -U e ; V = U e 2.From (0,h) to (L,  ): streamline, Vn = 0 (no shear) 3.From (L,  ) to (L,0): Vn = u(x=L,y) 4.From (L, 0) to (0,0): Vn = 0 (shear)  F x = -D on CV = (d/dt)  udV ol +  u  (Vn)dA D 0 (NO BODY FORCES) (STEADY)

 F x = -D =  u  (Vn)dA  u  (Vn)dy(w) =   0 h U o (-U o )wdy +   0  u(L,y) u(L,y)wdy 1.From (0,0) to (0,h): Vn = -U o = -U e ; V = U e 2.From (0,h) to (L,  ): no shear, Vn = 0 3.From (L,  ) to (L,0): Vn = u(x=L,y); V = u(x,y) 4.From (L, 0) to (0,0): Vn = 0

-D = -  U o 2 wh +   0 u(L,y) 2 wdy Not so useful because don’t know h as a function of  (x) 

-D = -  U o 2 wh +   0 u(x,y) 2 wdy Get rid of h by using conservation of mass.    (Vn)dA= 0 -   0 U o wdy +   0 u(x,y) wdy = 0 U o h =  0 u(x,y) dy -D = -  U o w  0 u(x,y) dy +   0 u(x,y) 2 wdy     h 

-D = -  U o w  0  (x) u(y)dy +   0  (x) u(x,y) 2 wdy D =  w [  0  (x) U o u(x,y)dy - u(x,y) 2 dy] D =  w [  0  (x) u(x,y) [U o - u(x,y)]dy D   w [  0  u(x,y) [U o - u(x,y)]dy  U o 2  w [  0  [u(x,y)/U o ][1 o – [u(x,y)/U o ]]dy D(x) = U o 2  w  (x) (First derived by Von Karman in 1921)

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D(x) = U o 2  w  (x) D(x) =  0 x  w wdx = U o 2  w  (x)  w (x) =  d/dx {U o 2  (x)} No pressure gradient; dU/dx = 0  w (x)/  = U o 2 d/dx {  (x)} The shear stress at the wall is proportional to the rate of change of momentum thickness with distance. (C f =  w (x)/[( 1 / 2 )  U o 2 )])

Book ~ derivation includes pressure gradient case  w (x)/  = d/dx { U o (x) 2  (x)}  *U o (x)dU o (x)/dx NOTE Class ~ derivation for constant pressure case  w (x)/  = U o 2 d/dx {  (x)}

 w (x)/  = d/dx { U o (x) 2  (x)}  *U o (x)dU o (x)/dx book me U o = constant; dP/dx = 0 U o  constant dP/dx  0

breath Estimating  (x),  * (x),  (x), Drag For laminar / turbulent / and pressure gradient flows

 w (x)/  = U o 2 d/dx {  (x)}  w (x)/  = d/dx { U e (x) 2  (x)}  *U e (x)dU e (x)/dx book me

 w (x)/  = U o 2 d/dx {  (x)} No pressure gradient; dU/dx = 0  w (x)/  = U o 2 d/dx { 0  u/U e (1 – u/U e )dy  u(x)/U is assumed to be similar for all x and is normally specified as a function of y/  (x) Defining  = y/   w (x)/  =U o 2 d/dx{  0  1 u/U e (1– u/U e )d  }  w (x)/  =U o 2 d  /dx{ 0  1 u/U e (1– u/U e )d  }

Dimensionless velocity profile for a laminar boundary layer: comparison with experiments by Liepmann, NACA Rept. 890, Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991  y/ 

Turbulent bdy layer* ~ u/U e = (y/  ) 1/7 ; U e = 1000 cm/s y/  (x) u(x)/U * du/dy| wall = 

Blasius developed an exact solution (but numerical integration was necessary) for laminar flow with no pressure variation. Blasius could theoretically predict boundary layer thickness  (x), velocity profile u(x,y)/U  vs y/ , and wall shear stress  w (x). Von Karman and Poulhausen derived momentum integral equation (approximation) which can be used for both laminar (with and without pressure gradient) and turbulent flow

Von Karman and Polhausen method (MOMENTUM INTEGRAL EQ. Section 9-4) devised a simplified method by satisfying only the boundary conditions of the boundary layer flow rather than satisfying Prandtl’s differential equations for each and every particle within the boundary layer.

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QUESTIONS

? In laminar flow along a plate,  (x),  (x),  * (x) and  w (x): (a)Continually decreases (b)Continually increases (c)Stays the same ? In turbulent flow along a plate,  (x),  (x),  * (x) and  w (x): (a)Continually decreases (b)Continually increases (c)Stays the same ?At transition from laminar to turbulent flow,  (x),  (x),  * (x) and  w (x): (a)Abruptly decreases (b)Abruptly increases (c)Stays the same

forced natural 6:1 ellipsoid  wall  Turbulent Laminar Turbulent Laminar

What is wrong with this figure?

Are Antarctic Icebergs Towable Arctic News Record – Summer 1984; 36 C f = 0.074/Re 1/5 (Fox:C f = /Re 1/5 ); Area = 1 km long x 0.5 km wide  sea water = 1030 kg/m 3 ; sea water = 1.5 x m 2 /sec Power available = 10 kW; Maximum speed = ?

P = UD 10 4 = U C f ½  U 2 A = U [ /5 /(U 1/5 L 1/5 )]( ½  U 2 A) 10 4 = U 14/5 [0.074(1.5x10 -6 ) 1/5 /1000 1/5 ] x [½ (1030)(1000)(500)] U 14/5 = U = m/s Re x ~ 3x10 8

80% of fresh water found in world – Antarctic ice Biggest problem not melting but is = ?

THE END

extra “stuff”

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Blasius developed an exact solution (but numerical integration was necessary) for laminar flow with no pressure variation. Blasius could theoretically predict boundary layer thickness  (x), velocity profile u(x,y)/U  vs y/ , and wall shear stress  w (x). Von Karman and Poulhausen derived momentum integral equation (approximation) which can be used for both laminar (with and without pressure gradient) and turbulent flow

MOMENTUM INTEGRAL EQUATION dP/dx is not a constant!

Deriving: MOMENTUM INTEGRAL EQ so can calculate  (x),  w.

u(x,y) Surface Mass Flux Through Side ab w

Surface Mass Flux Through Side cd

Surface Mass Flux Through Side bc

Assumption : (1) steady (3) no body forces Apply x-component of momentum eq. to differential control volume abcd u

mf represents x-component of momentum flux F sx will be composed of shear force on boundary and pressure forces on other sides of c.v.

X-momentum Flux =  cv u  V  dA Surface Momentum Flux Through Side ab u w

X-momentum Flux =  cv u  V  dA Surface Momentum Flux Through Side cd u

U=U e =U  X-momentum Flux =  cv u  V  dA Surface Momentum Flux Through Side bc u

a-bc-d b-c X-Momentum Flux Through Control Surface

IN SUMMARY X-Momentum Equation

X-Force on Control Surface w is unit width into page p(x) Surface x-Force On Side bc Note that p  f(y) inside bdy layer w

w is unit width into page p(x+dx) Surface x-Force On Side cd w

Note that since the velocity gradient goes to zero at the top of the boundary layer, then viscous shears go to zero there (bc). p + ½ (dp/dx)dx Surface x-Force On Side bc

Why w  and not w (bc)? pressure

Surface x-Force On Side bc p + ½ dp/dx along bc In x-direction: [p + ½ (dp/dx)] wd 

-(  w + ½ d  w /dx] x dx)wdx Surface x-Force On Side ad

F x = F ab + F cd + F bc + F ad F x = pw  -(p + [dp/dx] x dx) w(  + d  ) + (p + ½ [dp/dx] x dx)wd  - (  w + ½ (d  w /dx)  x dx)wdx F x = pw  -(p w  + p wd  + [dp/dx] x dx) w  + [dp/dx] x dx w d  ) + (p wd  + ½ [dp/dx] x dxwd  ) - (  w + ½ (d  w /dx)  x dx)wdx d  <<  = d  w <<  w p(x) **+ + # #

=

ad = 0 ab -cdbc U

Divide by wdx dp/dx = -  UdU/dx for inviscid flow outside bdy layer  =  from 0 to  of dy -  (dp/dx) = -(  0  dy) (-  UdU/dx) = (dU/dx)(  0   Udy)

dp/dx = -  UdU/dx for inviscid flow outside bdy layer  =  from 0 to  of dy p + ½  U 2 +  gz = constant outside boundary layer (“inviscid”) dp/dx +  UdU/dx = 0 outside bdy layer (dz = 0)

Integration by parts Multiply by U 2 /U 2 Multiply by U/U

QED

FLAT PLATE – ZERO PRESSURE GRADIENT LAMINAR FLOW Blasius Theoretical Solution for u(x,y),  (x),  * (x),  (x) Background

Blasius* (1908) developed an exact solution for laminar flow over a flat plate with no pressure variation. Blasius could theoretically predict:  (x),  *(x),  (x), velocity profile u(x,y)/U  vs y/ , and wall shear stress  w (x). *(first graduate student of Prandtl)

Blasius* (1908) developed an exact solution but required numerical integration, not good if there is a pressure gradient and not good for turbulent flow hence we develop the momentum integral equation.

Newton’s Law: F = ma u(  u/  x) + v(  u/  y) = -(1/  )  p/  x + (  2 u/  x 2 +  2 u/  y 2 ) u(  v/  x) + v(  v/  y) = -(1/  )  p/  y + (  2 v/  x 2 +  2 v/  y 2 ) Cons. of Mass  u/  x +  v/  y = 0 Boundary Conditions: u = v = 0 at y = 0 U e outside boundary layer 2-D Governing Equations Blasius Solution for Laminar, dP/dx=dU e /dx=0 Flow

v << u; d/dx << d/dy Conservation of Mass  u/  x +  v/  y = 0 Prandtl made some simplifying assumptions, based on the premise that the boundary layer is very thin (v and  small compared to U and L) so 2-D equations over flat plate more tractable: Blasius Solution for Laminar, dP/dx=dU e /dx=0 Flow

v << u; d/dx << d/dy Conservation of Y-Momentum u(  v/  x) + v(  v/  y) = -(1/  )  p/  y + (  2 v/  x 2 +  2 v/  y 2 )  p/  y  0 Prandtl made some simplifying assumptions, based on the premise that the boundary layer is very thin (v and  small compared to U and L) so 2-D equations over flat plate more tractable: Pressure can vary only along boundary layer not through it! Blasius Solution for Laminar, dP/dx=dU e /dx=0 Flow

v << u; d/dx << d/dy Conservation of X-Momentum u(  u/  x) + v(  u/  y) = -(1/  )  p/  x + (  2 u/  x 2 +  2 u/  y 2 )  p/  y  0 u(  u/  x) + v(  u/  y) = -(1/  ) dp/dx + (  2 u/  x 2 +  2 u/  y 2 ) u(  u/  x) + v(  u/  y)  UdU/dx + (1/  )(  /  y) Prandtl made some simplifying assumptions, based on the premise that the boundary layer is very thin (v and  small compared to U and L) so 2-D equations over flat plate more tractable: B.Eq. Blasius Solution for Laminar, dP/dx=dU e /dx=0 Flow

u(  u/  x) + v(  u/  y) = (  2 u/  y 2 )  u/  x +  v/  y = 0 u = v = 0 at y = 0 u = U e at y =  (outside boundary layer) With Prandtl’s simplifying assumptions plus no pressure gradient for laminar flow, left the 2-D mass and momentum equations over flat plate “ripe” for Blasius: BOUNDARY LAYER EQUATIONS Blasius Solution for Laminar, dP/dx=dU e /dx=0 Flow

By an ingenious coordinate transformation of the boundary layer equation Blasius showed that the dimensionless velocity distribution, u(x,y)/U e, was only a function of one composite dimensionless variable, [y/x][U e x/ ] 1/2. u(x,y)/U e = f([y/x][U e x/ ] 1/2 ) = f(  )  = y/  Dimensional analysis = u(x,y)/U e = f (U e x/, y/x)

Dimensionless velocity profile for a laminar boundary layer: comparison with experiments by Liepmann, NACA Rept. 890, Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991  y/  (x)

From Blasius solution it is found that:   = 5 ( x/U) 1/2 or  /x = 5(Re x ) -1/2  * = ( x/U) 1/2 or  * /x = 1.72(Re x ) -1/2  = ( x/U) 1/2 or  /x = 0.664(Re x ) -1/2  w = U 3/2 (  /x) u(x,y)/U e =1 at y(U e /[ x]) 1/2 =  u(x,y)/U e = 0.99 at y(U e /[ x]) 1/2  5.0

u(x,y)