Note – throughout figures the boundary layer thickness is greatly exaggerated!

Note – throughout figures theboundary layer thickness is

greatly exaggerated!

CHAPTER 9: EXTERNAL INCOMPRESSIBLE

VISCOUS FLOWS

•Can’t have fully developed flow•Velocity profile evolves,

flow is accelerating• Generally boundary layers(Rex > 104) are very thin.

Laminar Flow/x ~ 5.0/Rex

1/2

THEORY

Turbulent Flow (Rex > 106)/x ~ 0.16/Rex

1/7

EXPERIMENTAL

“At these Rex numbersbdy layers so thin that displacement effect onouter inviscid layer is small”

Blasius showed theoretically that /x = 5/Rex (Rex = Ux/)

BOUNDARY LAYER THICKNESS: is y where u(x,y) = 0.99 U

This definition for is completely arbitrary, why not 98%, 95%, etc.

Because of the velocity deficit, U-u, within the bdy layer, the mass flux through b-b is less than a-a. However if we displace the plate a distance *, the mass flux along each section will be identical.

DISPLACEMENT THICKNESS:

* = 0 (1 – u/U)dy

u

U[(u)([U-u])]

[Flux of momentum deficit]

U2 = total flux of momentum

deficit

The momentum thickness, , is defined as the thickness of a layer of fluid, with velocity Ue, for which the momentum flux is equal to the deficit of momentum flux through the boundary layer.

Ue2

= 0 u(Ue – u)dy = 0 �[u/Ue] (1 – u/Ue)dy

Want to relate momentum thickness, , with drag, D, on plate Ue is constant so p/dx = 0; Re < 100,000 so laminar,

flow is steady, is small (so * is small) so p/y ~ 0

Conservation of Mass: -Uehw + 0

huwdy + 0Lvwdx = 0

Ignoring the fact that because of *, Ue is not parallel to plate

Jason Batin, what are forces on control volume?

X-component of Momentum Equation

p/dx = 0p/dy = 0

Ue/y = 0

-D = - Ue2hw + 0

hu2wdy + 0L Uevwdx

v is bringing Ue out of control volume

a-b c-d b-c

X-component of Momentum Equation

p/dx = 0Ue/y = 0

-D = - Ue2hw + 0

hu2wdy + 0L Uevwdx

Conservation of Mass: -Uehw + 0

huwdy + 0Lvwdx = 0

-Ue2hw + 0

huUewdy + 0LvUewdx = 0

X-component of Momentum Equation

p/dx = 0Ue/y = 0

-D = - Ue2hw+0

hu2wdy +Ue2hw-0

huUewdy -D = 0

hu2wdy +0huUewdy

D/(Ue2w) = 0

h (-u2/Ue2) dy + 0

h (u/Ue)wdyD/(Ue

2w) = 0h (u/Ue)[1-u/Ue] dy

~ 0 (u/Ue)[1-u/Ue] dy =

p/dx = 0Ue/y = 0

D/(Ue2w) =

D = Ue2w

dD/dx = Ue2w(d/dx)

D = 0L wallwdx (all skin friction)

dD/dx = wallw =Ue2w(d/dx)

p/dx = 0Ue/y = 0

D = Ue2w

dD/dx = wallw =Ue2w(d/dx)

• knowing u(x,y) then can calculate and from can calculate drag and wall

• the change in drag along x occurs at the expense on an increase in which represents a decrease in the momentum of the fluid

SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS

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

Moreover: u(x,y)/U vs y/ is self similarand wall shear stress w(x).

Dimensionless velocity profile for a laminar boundary layer:comparison with experiments by Liepmann, NACA Rept. 890,1943. Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991

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 andevery particle within the boundary layer.

WHERE WE WANT TO GET…

Deriving MOMENTUM INTEGLAL EQso can calculate (x), w.

u(x,y)

Surface Mass Flux Through Side ab

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;Fsx will be composed of shear force on boundaryand pressure forces on other sides of c.v.

X-momentumFlux =

cvuVdA

Surface Momentum Flux Through Side ab

u

X-momentumFlux =

cvuVdA

Surface Momentum Flux Through Side cd

u

U=Ue=U

X-momentumFlux =

cvuVdA

Surface Momentum Flux Through Side bc

u

a-b c-d

c-d b-c

X-Momentum Flux Through Control Surface

IN SUMMARYRHS X-Momentum Equation

X-Force on Control Surface

w is unit width into pagep(x)

Surface x-Force On Side ab

Note that p f(y)

w

w is unit width into pagep(x+dx)

Surface x-Force On Side cd

Surface x-Force On Side bc

p + ½ [dp/dx]dx is average pressure along bc

Force in x-direction: [p + ½ (dp/dx)] wd

w

Why and not (bc)?w

Surface x-Force On Side bc

psin in x-direction; (A)(psin) is force in x-direction

Asin = wSo force in x-direction = p w

psin

w

A

p

Note that since the velocity gradient goes to zero at the top of the boundary layer,

then viscous shears go to zero.

Surface x-Force On Side bc

-(w + ½ dw/dx]xdx)wdx

Surface x-Force On Side ad

Fx = Fab + Fcd + Fbc + Fad

Fx = pw -(p + [dp/dx]x dx) w( + d) + (p + ½ [dp/dx]xdx)wd - (w + ½ (dw/dx)xdx)wdx

Fx = pw -(p w + p wd + [dp/dx]x dx) w + [dp/dx]x dx w d) + (p wd + ½ [dp/dx]xdxwd) - (w + ½ (dw/dx)xdx)wdx

d <<

=

- ½ (dw/dxxdx)dx

dw << w

p(x)

** ++

#

#

=

ab -cd bc

U

Divide by wdx

dp/dx = -UdU/dx for inviscid flow outside bdy layer

= from 0 to of dy

Integration by parts

Multiply by U2/U2 Multiply by U/U

If flow at B did notequal flow at C then could connect and make perpetual motionmachine.

C

C

HARDEST PROBLEM – WORTH NO POINTS

…BUT MAYBE PEACE OF MIND

(plate is 2% thick, Rex=L = 10,000; air bubbles in water)

For flat plate with dP/dx = 0, dU/dx = 0

Realize (like Blasius) that u/U similar for all x when plotted as a function of y/ . Substitutions: = y/; so dy = d

= y/=0 when y=0=1 when y=

u/U

~ y/

= 0 u/Ue(1 – u/Ue)dy = y/; d = dy/

Strategy: obtain an expression for w as a function of , and solve for (x)

(0.133 for Blasius exact solution, laminar, dp/dx = 0)

Laminar Flow Over a Flat Plate, dp/dx = 0

Assume velocity profile: u = a + by + cy2

B.C. at y = 0u = 0 so a = 0 at y = u = U so U = b + c2

at y = u/y = 0 so 0 = b + 2c and b = -2c U = -2c2 + c2 = -c2 so c = -U/2

u = -2cy – (U/2) y2 = 2Uy/2 – (U/2) y2 u/U = 2(y/) – (y/)2 Let y/ =

u/U = 2 -2

Want to know w(x)

Strategy: obtain an expression for w as a function of , and solve for (x)

Laminar Flow Over a Flat Plate, dp/dx = 0

u/U = 2 -2

Strategy: obtain an expression for w as a function of , and solve for (x)

u/U = 2 -2

2 - 42 + 23 - 2 +23 - 4

Strategy: obtain an expression for w as a function of , and solve for (x)

2U/(U2) = (d/dx) (2 – (5/3)3 + 4 – (1/5)5)|01

2U/(U2) = (d/dx) (1 – 5/3 + 1 – 1/5) = (d/dx) (2/15)

Assuming = 0 at x = 0, then c = 0

2/2 = 15x/(U)

Strategy: obtain an expression for w as a function of , and solve for (x)

2/2 = 15x/(U)

2/x2 = 30/(Ux) = 30 Rex

/x = 5.5 (Rex)-1/2

x1/2

Strategy: obtain an expression for w as a function of , and solve for (x)

THE END

Illustration of strong interaction between viscid and inviscid regions in the rear of a blunt body.

Re = 15,000

Re = 17.9(separation at Re = 24)

Re = 20,000Angle of attack = 6o

Symmetric Airfoil16% thick

Laminar Flow Over a Flat Plate, dp/dx = 0

Assume velocity profile: u = a + by + cy2

B.C. at y = 0 u = 0 so a = 0 at y = u = U so U = b + c2

at y = u/y = 0 so 0 = b + 2c and b = -2c U = -2c2 + c2 = -c2 so c = -U/2

u = -2cy – (U/2) y2 = 2Uy/2 – (U/2) y2 u/U = 2(y/) – (y/)2 Let y/ = u/U = 2 -2