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Boundary Layers

Hydromechanics VVR090

Boundar Layer on a Flat Plate

Boundary layer: the zone in which the velocity profile is governed by frictional action

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Boundary layer characterized by a Reynolds number:

3900 laminar flow3900 turbulent flow

δ

δ

δ

δ=

ν

<>

oVR

RR

Viscous sublayer exist close to the surface.

Mathematical implications of boundary layer:

, ,

0,

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

∂ ∂→ ≈ ≈

∂ ∂

u u v vy x x y

u v

p p dpy x dx

von Karman momentum integral equation

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Von Karman momentum integral equation

Apply momentum equation for control volume ABCD.

Height of control volume extends beyond the edge of the boundary layer (to the outer flow).

At edge of the boundary layer: po(x), Vo(x) (known).

Small curvature along body.

Conservation of mass:

[ ]0

0=

⎛ ⎞ρ + ρ =⎜ ⎟

⎝ ⎠∫h

y h

d udy vdx

Conservation of momentum:

[ ] [ ]2

0= =

⎛ ⎞− − τ + τ = ρ + ρ⎜ ⎟

⎝ ⎠∫h

oo y h y h

dp dh u dy uvdx dx

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Develop conservation of mass equation:

0

⎛ ⎞ρ = − ρ⎜ ⎟

⎝ ⎠∫h

o odv udydx

Develop conservation of momentum equation:

2

0

⎛ ⎞− − τ = ρ +ρ⎜ ⎟

⎝ ⎠∫h

oo o o o

dp dh u dy V vdx dx

Combine conservation of mass and momentum:

2

0 0

⎛ ⎞ ⎛ ⎞− − τ = ρ − ρ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫h h

oo o

dp d dh u dy V udydx dx dx

Euler eqution for the outer flow:

0=+ρ

= −ρ

oo o

o

o oo o

dp V dV

dp dVVdx dx

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2

0 0

⎛ ⎞ ⎛ ⎞−τ = ρ − ρ −ρ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫h h

oo o o o

dVd du dy V udy V hdx dx dx

Combining:

Define displacement thickness:

( )10

ρ δ = ρ −ρ∫h

o o o oV V u dy

Flow rate with and without boundary layer

0

ρ < ρ δ∫h

o oudy V

10

1⎛ ⎞ρ

δ = −⎜ ⎟ρ⎝ ⎠∫h

o o

u dyV

In the same manner, define momentum thickness:

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1⎛ ⎞ρ

δ = −⎜ ⎟ρ ⎝ ⎠∫h

o o o

u u dyV V

Substitute in d1 and d2 in momentum equation and express τo in terms of a local friction coefficient cf:

2

2δ

=fc ddx

(constant density, no pressure gradients)

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Laminar Boundary Layer along a Flat Plate

Assume parabolic velocity profile:

( )2

2

2δ −=

δoV y y

u

2

2

2

215

22 15

2215

fo

o

oo o

c dV dx

VdVdx

δ = δ

τ δ= =

ρ

δτ = ρ = μ

δ

2

2

2215

152

30

oo

o

x

VdVdx

xV

x R

δρ = μ

δ

δ μ=ρ

δ=

Integrate the equation:

2 82 15

oo

x

VR

ρτ =

Substitute in dd/dx in shear stress expression:

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Local friction coefficient:

815f

x

cR

≈

Mean friction coefficient along the plate:

0

1 3215

x

f fx

C c dxx R

= =∫

Relationship between Rx and Rd:

2

30xRR δ=

Rd = 3900 Rx = 500,000

(transition between laminar and turbulent conditions in the boundary layer)

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Drag Coefficient for Smooth, Flat Plates

212f f oD C V A= ρ

A: surface area of plate