494 Solutions Manual • Fluid Mechanics, Fifth Edition

Solution: Use Prandtl’s correlation for the left hand side of Eq. (7.32) in the text:

2 1/4 2 2 2w

1/4 1/4 5/4 1/4

d d 70.0225 U ( /U ) U U , cancel U and rearrange:

dx dx 72

4d 0.2314( /U) dx, Integrate: 0.2314( /U) x

5

θτ ρ ν δ ρ ρ δ ρ

δ δ ν δ ν

� �≈ = ≈ � �� �

= =

Take the (5/4)th root of both sides and rearrange for the final thickness result:

1/5 4/50.37( /U) x , or: (a)Ans.δ ν≈1/5x

0.37x Re

δ ≈

1/5

w f 1/4

2(0.0225)Substitute (x) into : C , or (b)

Ux(0.37)Ans.

νδ τ � �� �� �

≈ ≈f 1/5x

0.0577C

Re

1

D f f0

x 5Finally, C C d C (at x L) (c)

L 4Ans.

� �= = = ≈� �� �

� 1/5L

0.072

Re

7.34 A thin equilateral-triangle plate is immersed parallel to a 12 m/s stream of water 20°C, as in Fig. P7.34. Assuming Retr = 5 × 105, estimate the drag of this plate.

Solution: Use a strip dx long and (L − x) wide parallel to the leading edge of the plate, as shown in the figure. Let the side length of the triangle be a:

Fig. P7.34

Strip dA = 2(L − x) tan 30° dx, where L = a sin 60° and a = 2 m = side length. 1/2

3/2

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

: 0.332 2( ) tan30 (2 )

20 : 1.328( ) tan 30 2

3

lam w

crit lam crit crit

Laminar part dF dA U L x dx sidesx

Integrate from to x F U Lx x

ρµτ

ρµ

� �= = − °� �� �

� �= ° −� �� �

( ) ( )

1/72

1/7 13/7 13/7 6/7 13/7 13/7

: 0.027 2( ) tan 30 (2 )2

:

7 70.054 tan30

6 13

turb w

crit

turb crit crit

UTurbulent part dF dA L x dx sides

Ux

Integrate from x to L

F U L Lx L x

ρ ντ

ρν

� � � �= = − °� �� � � �� �

� �= ° − − − � �

Chapter 7 • Flow Past Immersed Bodies 495

The total force is, of course, Flam + Fturb. For the numerical values given, L = 1.732 m. For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Evaluate xcrit and F:

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

1/713/7 13/7

998(12)5 5 , or:

0.001

21.328[998(0.001)] (12) tan30 2(1.732)(0.042) (0.042) 22 N

3

0.001 70.054(998) (12) tan30 {(1.732) 1.732(0.

998 6

critcrit crit

lam

turb

xUxE x

F

F

ρµ

= = = =

� �= ° − =� �� �

� �= ° − � �

Re 0.042 m

6/7042) }���

13/7 13/77{(1.732) (0.042) } 703 N; 22 703 .

13 totalF Ans�− − = ∴ = + =��

725 N

7.35 Repeat Problem 7.26 for turbulent flow. Explain your results.

Solution: The turbulent formula 1/7D LC 0.031/Re= means that CD ∝ L−1/7. Thus:

11/71

(a) (4 ) . (a)(2 )

aconst

F A AnsL

= = 13.62F

11/71

(b) (4 ) . (b)(4 )

bconst

F A AnsL

= = 13.28F

The trailing areas have slightly less shear stress, hence we are nearly quadrupling drag.

7.36 A ship is 125 m long and has a wetted area of 3500 m2. Its propellers can deliver a maximum power of 1.1 MW to seawater at 20°C. If all drag is due to friction, estimate the maximum ship speed, in kn.

Solution: For seawater at 20°C, take ρ = 1025 kg/m3 and µ = 0.00107 kg/m⋅s. Evaluate

L D 1/7 1/7L

2 20/71/7

UL 1025V(125) 0.031 0.00217Re (surely turbulent), C

0.00107 Re V

0.0217 1025Power FV V (3500) V 1.1E6 watts, or V 282.0

2V

ρµ

= =

� �� �= = = ≈� � � �� �

= =

Solve for V 7.2 m/s . Ans.= ≈ 14 knots