Boundary Layer Correction

Air enters a two-dimensional wind tunnel as shown. Because of the presence of the boundary layer, fluid will be displaced away from the surface. In order to maintain a constant velocity inside the tunnel, it is necessary to increase the cross-sectional size of the tunnel. (a) Determine the channel height, H(x), as a function of the distance measured from the inlet of the tunnel, x. The tunnel velocity is 10 m/s and the tunnel inlet height is 5 m. Assume the boundary layer has a profile u(y)=Usin(ybWhat is the momentum thickness (x) of the boundary layer flow.

U=10 m/s

5 m

x

Sphere, 0.1 m diameter

Boundary Layer Correction (cont.)

Solution: In order to maintain a constant velocity inside the tunnel, the tunnel wall

has to be displaced outward in order to accomodate the growth of the boundary layer.

The mass flow displaced by t

0 0

he presence of the boundary layer can be related to U *

by the definition of the displacement thickness. Therefore, the tunnel height has to be

H(t)=5+2 *(x).

u y 2*= (1- ) [1-sin( )]

U 2dy dy

0

3

ycos ( )

2

2 2(0 ) (1 ) ( ) 0.363 ( )

( ) 5 0.726 ( )

( ) ( ) 5 2 * ( ) 5 1.158 10 ( )

x x

H x x

a H x x x m

Boundary Layer Correction (cont.)

2w

w 0

0

0 0

2

0 0

( ) From the momentum thickness equation:

u uwhere = (1- ) and | .

U U

y y[sin ( )] cos ( )

2 2 2 2

y y y ysin ( )(1-sin ( )) [sin ( )- sin ( )]

2 2 2 2

2co

y

w y y

db U

dx

udy

y

y

dy dy

0 0

0

2

2

y 1 1s ( ) [ cos( )]

2 2 2

2 2 2 1sin( ) ( )

2 2 2 2

2 1 2 1( ) ( )

42 2 2 ( 1)

ydy

y

d d dU dx d

dx dx dx U

Boundary Layer Correction (cont.)

2 23

22

3 4

2Integrate: , ( ) 1.595 10 ( )

4 2 (4 )( 1)

(b) The momentum thickness

2 1 2 1(x)=( ) ( )(1.595 10 ) 2.18 10 ( )

2 2

x x x x mUU

x x m

Boundary Layer Correction (cont.)

c) If a sphere has a diameter of 0.1 m is placed in the center of the wind tunnel, what is the drag force exerted on the sphere. air=1.2 kg/m3, =1.5x10-5 m2/s. Use the following graph for CD verse Re data.

CD

CD=24/Re, when Re<1

Re=VD/

0.5

400 300,000

CD=24/Re0.646, when 1<Re<400

5

2 2

2 2

(c) For the sphere:

U (10)(0.1)Re= 66,667,

1.5 10300 Re 300,000 0.5

,1 ( )2 4

(0.5)(1/ 2)(1.2)(10) ( / 4)(0.1)

0.235( ) drag force

D

DD

D

D

C

FC

U D

F

N