NATIONAL UNIVERSITY OF SINGAPORE Department of Mechanical Engineering

ME2135 Fluid Mechanics II

Part 2 External Incompressible Viscous Flow

Solution to Tutorial 2

1. The Blasius solution for laminar boundary layer flow on a flat plate is given as numerical results in Table 3.1 of Prof Winoto’s Lecture Notes p. 45:

a) Evaluate the distribution of shear stress by plotting dimensionless τ/τw versus y/δ. b) Evaluate the vertical component of velocity by plotting v/U versus y/δ for Rex = 105. c) Obtain an algebraic expression for the x component of the acceleration (ax) of a fluid

particle in the laminar boundary layer. Plot ax versus η to determine the maximum x component of acceleration at a given x.

y/δ Plot of τ/τw versus y/δ

( )

12

'

12

1 12

0

) Similarity variable defined as: =

From Eq. (12): u = Uf

shear stress "( )

wall shear stress "( ) 0.332

y

w

Ua yx

u UUfy x

U UUf Ux x

η

η

ηn

η

τ µ µ ηn

τ µ η µn n

∂∂

=

∂ = = ∂

= =

2

in Blasius

"( )

solutionFrom Bl

asius solution, u/U = 0.99 whe

0.332

wheren = 5; that

"( ) is given for various y = at 5

is:

w

f

f η

τ ητ

ηηη δ

=

=

Or 5

y ηδ

=

τ/τw 1

0

0.5

1

1.5

2

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000

y/δ v/U f f ′′ X

[ ]

[ ]

12

b) From Eq.(13) : v f ( ) f( ) 4

v 1 f ( ) f( )2 Rex

Ux

U

n η η η

η η η

′= −

′= −

( ) ( ) [ ] ( )

x

1 12 2'

x

2

c) From Eq. (9), acceleration a = v

Subst. u, v from Eqns (12) & (13) and , from p. 47:

a = Uf f '' f ( ) f( ) f ''2 4

= f(2

u uux y

u ux y

U U UUx x x

Ux

η nη η η η η ηn

η

∂ ∂+

∂ ∂∂ ∂∂ ∂

′− + −

−

( )

( ) [ ]2

x max

) f ''

From graph max f( ) f '' 0.23 at 3; and a = 0.115 Ux

η

η η η= ≈ −

2

00.20.40.60.8

11.21.41.61.8

0.0000000.0005000.0010000.0015000.0020000.0025000.003000

00.05

0.10.15

0.20.25

0 2 4 6 8 10

Numerical result of shear stress, vertical velocity, and particle acceleration

η f f' f'' y/δ τ/τw v/U f f '' 0 0 0 0.3321 0 1.0003 0.000000 0.0000

0.5 0.0415 0.1659 0.3309 0.1 0.9967 0.000066 0.0137 1 0.1656 0.3298 0.323 0.2 0.9729 0.000260 0.0535

1.5 0.3701 0.4868 0.3026 0.3 0.9114 0.000569 0.1120 2 0.65 0.6298 0.2668 0.4 0.8036 0.000964 0.1734

2.5 0.9963 0.7513 0.2174 0.5 0.6548 0.001394 0.2166 3 1.3968 0.846 0.1614 0.6 0.4861 0.001804 0.2254

3.5 1.8377 0.913 0.1078 0.7 0.3247 0.002147 0.1981 4 2.3057 0.9555 0.0642 0.8 0.1934 0.002397 0.1480

4.5 2.7901 0.9795 0.034 0.9 0.1024 0.002558 0.0949 5 3.2833 0.9915 0.0159 1 0.0479 0.002647 0.0522

5.5 3.7806 0.9969 0.0066 1.1 0.0199 0.002692 0.0250 6 4.2796 0.999 0.0024 1.2 0.0072 0.002711 0.0103

6.5 4.7793 0.9997 0.0008 1.3 0.0024 0.002718 0.0038 7 5.2792 0.9999 0.0002 1.4 0.0006 0.002720 0.0011

7.5 5.7792 1 0.0001 1.5 0.0003 0.002721 0.0006 8 6.2792 1 0 1.6 0.0000 0.002721 0.0000

2. A thin flat plate, L = 0.25 m long and b = 1 m wide, is installed in a water tunnel as a splitter.

The freestream speed is U = 1.75 m/s, and the velocity profile in the boundary layer is given by the Blasius solution. The kinematic viscosity of water is 10-6 m2/s. Evaluate the momentum thickness θL and the total drag on both sides.

5L 6

1.75 0.25Renolds number Re 4.38 10

which justify assumption that the boundary layer is lami10

0.664Momentum thickness, from Eq. (17):

nar f

= Re

or the entire length

x

UL

xq

n −

×= = = ×

3

5

2 3 2 3

= = 0.251x10 4.38 10

From Eq. (23): drag 10 1 1.75 0.251x10 0.766

Total drag on both sides is 1

0.664 0.

.

5

5 N

2

3

D bU Nr

q

q

−

−

×

×= = × × =

3

3. Assume laminar boundary-layer flow to estimate the drag on the

plate shown when it is placed parallel to a 7.5m/s air flow, with kinematic viscosity n = 1.5 x 10-5 m2/s. The shape of the plate is given by x = y2/25, where x and y are in cm.

( )2

4

3/

2

5

/ 25

1.5 1

shape of plate is

25 / 2length is L= cm=6.25 cm =0.0625 m

257.5 0.0625Reynolds number Re 3.125 10

which justify assumption of laminar

From Eq. (18), wall shear

0

stress: 0.332

L

w

x y

xUL x

U

n

τ

−

×= = =

=

=

[ ]

2

0

3/2

0

3/20

Drag D = ( )

where plate width is found fron plate shape: w(x)= W

Drag D = 0.332 W dx

1 =0.332 0.332

L

w

area

L

L

x

w x dx

xL

xUx L

U W x WL

rµ

τ

rµ

r n r n× =

∫

∫

3

3 3

3

50.332 1.2 0.25 0.0625 7.5 1.98 10

Total drag for both sides = 3.96

1.5

x10

10

N

LU

x Nx − −

−

= × × × × =

4

4. Assume laminar boundary-layer flow to estimate the drag (both sides) on four square plates (each 7.5 cm x 7.5 cm) placed parallel to a 1 m/s water flow, for the two configurations shown. Before calculating, which configuration do you expect to experience the lower drag? Assume that the plates attached with string are far enough apart for wake effects to be negligible and that the kinematic viscosity of water is 10-6 m2/s.

Drag is much lower on composite plates compared to separate plates. This is because τw is largest near leading edges and falls off rapidly. In this problem, the separate plates experience leading edges four times! 5

L 6

L

1 0.075For separate plates, Reynolds number Re 0.75 10

which justify assumption that the boundary layer is laminar for the entire length

1 4 0.07For composite plates, Reynolds number e

10

R

UL

xUL

n

n

−

×= = = ×

×= =

( )6

5

2

53 10

which is still laminar on all 4 plates

From Eq. (23) drag is related momentum thickness:

10

0.664 = Re

For separate pl

where from Eq.

ates, on eac

(17):

h plate is

x

L L

D bU

x

r qq

q q

− = ×

=

-3

5

2 3 2 -3 3

3

= 0.075=0.182x10 m0.75 10

Drag on each plate (one side) is: = 10 0.075 1 0.182x10 13.65 10Total drag for 4 plates and both sides is: 8 13.65

0.664 0.664= LRe

For compo

10 0.109

site

L

D bU x x Nx N N

r q −

−

×

= =

× =

( )5

-3

0.664 0.664plates, on each plate is = L 4 Re

= 0.0754 0.75 10

= 2 0.182x10 m

Drag

compositon plate (one sie de

L LL

q q ×× ×

×

( )

( )

2 3 2 -3 3

3

) is: = 10 0.075 1 2 0.182x10 2 13.65 10

Total drag (both sides) is: 2 2 13.65 10 0.054

D bU x x N

x N N

r q −

−

= × = ×

× × =

5