444
Solutions Manual x Fluid Mechanics, Fifth Edition
Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation
reduces to:
p1 V12
z
U g 2g 1
0 0 (L l )
For laminar flow, h f
p2 V22
z h
U g 2g 2 f
128P LQ
SUgd 4
Solve for 't
0 0 0 h f , or: h f | L l
and, for uniform draining, Q
128 P LX
SU gd 4 ( L l )
X
't
Ans. (a)
(b) Apply to 't 6 s. For water, take U 998 kg/m3 and P 0.001 kg/m˜s. Formula (a) predicts:
't
128(0.001 kg/m˜s)(0.12 m)(8E6 m3 )
,
S (998 kg/m3 )(9.81 m/s2 )d 4 (0.12 0.02 m)
6s
Solve for d | 0.0015 m
Ans. (b)
6.18 To determine the viscosity of a
liquid of specific gravity 0.95, you fill, to a
depth of 12 cm, a large container which
drains through a 30-cm-long vertical tube
attached to the bottom. The tube diameter
is 2 mm, and the rate of draining is found
to be 1.9 cm3s. What is your estimate of
the fluid viscosity? Is the tube flow laminar?
Fig. P6.18
Solution: The known flow rate and diameter enable us to find the velocity in the tube:
Q
A
V
Evaluate U liquid 0.95(998)
and the tube exit:
0.605
m
s
948 kgm3. Write the energy equation between the top surface
pa
Ug
or: 0.42
1.9 E6 m 3 /s
(S /4)(0.002 m)2
2
Vtop
2g
ztop
V 2 32P LV
2g
Ugd 2
pa V 2
0 hf ,
Ug 2 g
(0.605)2
32P (0.3)(0.605)
2(9.81) 948(9.81)(0.002)2