NATIONAL UNIVERSITY OF SINGAPORE Department of Mechanical Engineering

ME2135 Fluid Mechanics II

Part 2 External Incompressible Viscous Flow

Solution to Tutorial 3

1. An airplane flies at a speed of 640 km/h at an altitude of 3,000 m. If the boundary layers on the wing surfaces behave as those on a flat plate, estimate the extent of laminar boundary layer flow along the wing. Assume a transitional Reynolds number of Rex,cr = 5 x 105. If the airplane maintains its 640 km/h speed but descends to sea-level elevation, will the portion of the wing covered by a laminar boundary layer increase or decrease compared with its value at 3,000 m? The kinematic viscosity of air is 1.5 x 10-5 m2/s at sea level. Note that at different heights, the dynamic viscosities are approximately the same but the densities are different.

1.

5

5,

2

5 5 dynamic viscosity 1.2 =1.8

At sea level, kinematic viscosity =1.5 10 m /s

= 1.5 10 10 Pa 640 10

critical Reynolds number Re

s

5 10 crx cr

x

x x

Ux

nr

rn

n

=

=

= = =

3

3300

5

50

5 2

1.5 10

At 3000 m, air density

1.8 10 kinematic viscosity

3600

x 0.042

0.83 kg/m

critica

= =2.17 10 m /s0.83

l Rey nolds

cr

cr

x

x

x x

mr

nr

=

=

=

3

5, 5

640 103600number Re 5 10

2.17

= 0.061 m10

crcr

x cr

cr

x

xx

Uxn

= = =

1

2. Water flows at U = 1 m/s past a flat plate with L = 1 m in the flow direction. The boundary layer is tripped so it becomes turbulent at the leading edge. Evaluate the boundary layer thickness, , displacement thickness, *, and wall shear stress, w, at x = L. Assume a 1/6-power turbulent velocity profile. Develop an algebraic expression for the variation of wall shear stress with distance along the surface. Integrate to obtain an algebraic expression for the total skin friction drag on the surface. Evaluate the drag for the given conditions.

0

1/6 16

1 116 6

017 8

6 6

0

Momentum thickness (1 )

yVelocity profile = = where Y =

= (1 )

= 7 / 6 8 / 6

u u dyU U

u yYU

Y Y dY

Y Y

d d

d

d

=

2

142

142 2

3 = 28

3 28

Momentum integral equation:

Empiric

0.0243

0.0243

al shear stress formula:

Combine:

Substitute

w

w

dUdx

UU

dU UU dx

t r

n

d

t rd

n rd

d

r

=

=

=

=

142 2

1144

5144

00

0.0243

328 :

0.227

0.227

54

x

dU U

U dx

dx dU

xU

d

nr rd

n d

n d

d

d

=

=

=

2

45

1

66

*

0

5

1 1/55

1

1/5

0.365 0.365 0.365 Re

1 1At x=1 m, Re =

0.365 Re

1010

0.023

Displacement thickness (1 )

x

x m

x

xU x U

mu

x

dyU

x

nd

d

d

d

d

n

=

= = =

=

=

=

=

17

1 616

0

0

*

1 = (1 ) = = 7 / 6 7

1At x=1m = 0.023 0.0033 7

Empirical shear stress formu 0.l 0a: 2 w

YY dY Y

m

d d d

t

d

=

=

45

142

16 4

3 21

14

22

1 1/55

0.023

0.0329At any x:

43

10At x=1 m, shear stress is 0.0243 10 1 1.97 Pa1

0.0243

0.365

Find friction drag from

=

w x m

w

UU

UUUx

U xU

nrd

t

n rt rn

n

=

= =

=

1

01/5

12

0.

1/512 4/5

0

1/5623

integration of shear stress : D

Subst. shear stress: D

1In

10

tegrate : D 0.0314 / 5

1Subst. limits : D 1 0.031 11 4 /

1

0. 31

0

0

f w

f

f

f

b dx

b U dxUx

b U xU

t

nr

nr

=

=

=

=

( )4/5 4/51 05

D 5 N2.4f

=

3

3. A three-bladed helicopter blade rotates at 200 rpm. If each blade is 3.6 m long and 0.5 m wide, estimate the torque needed to overcome the friction on the blades if they act as flat plates. For simplicity, assume laminar boundary layer throughout but check the assumption.

top

2Df

Df 1/2

2 3/2 1/2 1/21/2

3/2

Torque from drag on elemental area dA is dM=(D +D )y

1=2 U C2

where U= y1.328C =

1 1.328dM=2 U ( y) 1.3282

2 3.1421.2(200 ) 1.60

bottom

dA y

U

dy y ydyU

r

n

r r n

n

=

=

3.65 1/2 1/2 5/2

03,63.6 7/2

5/2

0 07/2

328(1.5 10 ) 0.5

0.418 0.4187 / 2

3.63 0.4187 / 2

Torque will be greater if boundary layer was turbulent2 3.142At tip, y=3.6 m. y=200 3.6 74.88m/s

6

31

0

Re

.7

y dy

yy dy

M Nm

= =

= =

=

65

74.88 0.5 2.56 10 which is greater than critical value1.5 10

Thus boundary layer is laminar at hub and turbulent at tip

Un

= = =

4