384
Solutions Manual x Fluid Mechanics, Fifth Edition
(b) If we swing the pendulum on the moon at the same 20q, we may use similarity:
1/2
§g ·
T1 ¨ 1 ¸
© L1 ¹
1/2
§ 9.81 m/s2 ·
(2.04 s) ¨
¸
© 1.0 m ¹
or: T2
1/2
6.39
2.75 s
§ 1.62 m/s2 ·
T2 ¨
¸ ,
0.3
m
©
¹
Ans. (b)
5.27 In studying sand transport by ocean waves, A. Shields in 1936 postulated that the bottom
shear stress W required to move particles depends upon gravity g, particle size d and density Up, and
water density U and viscosity P. Rewrite this in terms of dimensionless groups (which led to the
Shields Diagram in 1936).
Solution: There are six variables (W, g, d, Up, U, P) and three dimensions (M, L, T), hence
we expect n j 6 3 3 Pi groups. The author used (U, g, d) as repeating variables:
W
U gd
§ U g1/2 d 3/2 Up ·
, ¸
fcn ¨
P
U¹
©
Ans.
The shear parameter used by Shields himself was based on net weight: W /[(Up U)gd].
5.28 A simply supported beam of diameter D, length L, and modulus of elasticity E is subjected to
a fluid crossflow of velocity V, density U, and viscosity P. Its center deflection G is assumed to be a
function of all these variables. (a) Rewrite this proposed function in dimensionless form. (b) Suppose
it is known that G is independent of P, inversely proportional to E, and dependent only upon UV2, not U
and V separately. Simplify the dimensionless function accordingly.
Solution: Establish the variables and their dimensions:
G
{L}
fcn(
U
, D , L ,
E
,
V ,
P )
{M/L3} {L} {L} {M/LT2} {L/T} {M/LT}
Then n 7 and j 3, hence we expect n j
and selected by myself, as follows (a):
Well-posed final result:
73
G
L
4 Pi groups, capable of various arrangements
§ L U VD E ·
fcn ¨ ,
,
Ans. (a)
2 ¸
D
P
V
U
©
¹
(b) If P is unimportant, then the Reynolds number (UVD/P) drops out, and we have already cleverly
combined E with UV2, which we can now slip out upside down:
385
Chapter 5 x Dimensional Analysis and Similarity2
If P drops out and G v
GE
U V2L
or:
1
G
, then
E
L
§L·
fcn ¨ ¸
©D¹
UV
§L·
fcn ¨ ¸ ,
E
©D¹
Ans. (b)
5.29 When fluid in a pipe is accelerated linearly from rest, it begins as laminar flow and then
undergoes transition to turbulence at a time ttr which depends upon the pipe diameter D, fluid
acceleration a, density U, and viscosity P. Arrange this into a dimensionless relation between ttr and
D.
Solution: Establish the variables and their dimensions:
ttr
{T}
U , D ,
fcn(
a
,
P )
{M/L3} {L} {L/T2} {M/LT}
Then n 5 and j 3, hence we expect n j 5 3 2 Pi groups, capable of various arrangements
and selected by myself, as required, to isolate ttr versus D:
§
U a2
t tr ¨
¨ P
©
1/3
·
¸
¸
¹
ª § U 2 a ·1/3 º
¸ »
« © P2 ¹ »
¬
¼
fcn « D ¨
Ans.
5.30 The wall shear stress Ww for flow in a narrow annular gap between a fixed and a rotating
cylinder is a function of density U, viscosity P, angular velocity :, outer radius R, and gap width 'r.
Using (U, :, R) as repeating variables, rewrite this relation in dimensionless form.
Solution: The relevant dimensions are {Ww} {ML1T2}, {U} {ML3}, {P} {ML1T1},
{:} {T1}, {R} {L}, and {'r} {L}. With n 6 and j 3, we expect n j k 3 pi
groups. They are found, as specified, using (U, :, R) as repeating variables:
U a : b R cW w
31
U : RP
3
U a : b R c 'r
b
c
­ M ½ a ­ 1 ½b
c­ M ½
® 3 ¾ ® ¾ {L} ®
¾
¯ L ¿ ¯T ¿
¯ LT ¿
1
32
a
­ M ½ a ­ 1 ½b
c­ M ½
® 3 ¾ ® ¾ {L} ® 2 ¾
¯ L ¿ ¯T ¿
¯ LT ¿
1, b
M 0 L0T 0 , solve a
1
M 0 L0T 0 , solve a 1, b 1, c
­ M ½ a ­ 1 ½b
c
0 0 0
® 3 ¾ ® ¾ {L} {L} M L T , solve a
¯ L ¿ ¯T ¿
0, b
0, c
1
The final dimensionless function has the form:
31
fcn(3 2 , 33 ), or:
2, c
W wall
U: 2 R 2
§ U:R2 'r ·
fcn ¨
, ¸
R¹
© P
Ans.
2
2