Problem 9.61

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Problem 9.61 [Difficulty: 3]
Given: Turbulent boundary layer flow with 1/6 power velocity profile: u
U
y
δ




1
6
 η
1
6

The given or available data (Table A.9) is U 1
m
s
 L 1 m ν 1.00 10
6

m
2
s
 ρ 999
kg
m
3

Find: Expressions for δ/x and Cf using the momentum integral equation; evaluate drag for the conditions given
Solution: We will apply the momentum integral equation
Governing
Equations:
τw
ρ x
U
2
θ d
d
δdisp U x
Ud
d






 (Momentum integral equation)
Cf
τw
1
2
ρ U
2

 (Skin friction coefficient)
Assumptions: (1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
(3) Incompressible flow
(4) Wall shear stress is: τw 0.0233 ρ U
2

ν
U δ




0.25

Applying the assumptions to the momentum integral equation yields: τw ρ U
2

x
θ
d
d






 ρ U
2

x
δ
0
1
η
u
U
1
u
U









d








d
d









Substituting for the velocity profile: τw ρ U
2

x
δ
0
1
ηη
1
6
η
2
6











d










d
d










 ρ U
2

6
56

x
δ
d
d






 Setting our two τw's equal:
0.0233 ρ U
2

ν
U δ




0.25
 ρ U
2

6
56

x
δ
d
d






 Simplifying and separating variables: δ
1
4
dδ 0.0233
56
6

ν
U




1
4
 dx
Integrating both sides:
4
5
δ
5
4
 0.0233
56
6

ν
U




1
4
 x C but C = 0 since δ = 0 at x = 0. Therefore: δ
5
4
0.0233
56
6

ν
U




1
4
 x






4
5

In terms of the Reynolds number:
δ
x
0.353
Rex
1
5

For the skin friction factor:
Cf
τw
1
2
ρ U
2


0.0233 ρ U
2

ν
U δ




1
4

1
2
ρ U
2

 0.0466
ν
U x




1
4

x
δ




1
4
 0.0466 Rex
1
4


Rex
1
5
0.353








1
4
 Upon simplification:
Cf
0.0605
Rex
1
5

The drag force is: FD
0
L
xτw b



d
0
L
x0.0605
1
2
 ρ U
2

ν
U




1
5
 x
1
5

 b






d
0.0605
2
ρ U
2

ν
U




1
5
 b
0
L
xx
1
5





d
Evaluating the integral: FD
0.0605
2
ρ U
2

ν
U




1
5
 b
5
4
 L
4
5
 In terms of the Reynolds number: FD
0.0378 ρ U
2
 b L
ReL
1
5

For the given conditions and assuming that b = 1 m: ReL 1.0 10
6
 and therefore:
FD 2.38 N