Problem 7.87
CD
For drag we can use
Model:
L=
=
For water
=
D
1
V
2
2
As a suitable scaling area for A we use L
2
3
[Difficulty: 3]
CD
A
D
1
V 2 L2
2
ft
1.94
slug/ft3
2.10E-05
lbfs/ft2
Wave Drag
The data is:
V (ft/s)
D Wave (lbf)
D Friction (lbf)
Fr
Re

Problem 7.88 (In Excel)
[Difficulty: 4]
Given: Data on centrifugal water pump
Find: groups; plot pressure head vs flow rate for range of speeds
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of

Problem 7.90
[Difficulty: 3]
Given:
Data on model propeller
Find:
Speed, thrust and torque on prototype
Solution:
We will use the Buckingham Pi-theorem to find the functional relationships between these variables. Neglecting the
effects of viscosity:
1
F

Problem 7.89
Given:
Model of water pump
Find:
[Difficulty: 3]
Model head, flow rate and diameter
Solution:
From Buckingham
h
2
2
Q D2
D3
f
D
Neglecting viscous effects
Qm
3
m Dm
Hence if
p Dp
3
hm
then
3
2
3
m
3
hp
2
p Dp
Pm
and
2
3
5
m Dm
Pp
3
p Dp

Problem 7.91
[Difficulty: 3]
Given:
For a marine propeller (Problem 7.40) the thrust force is: FT FT( D V g p )
For ship size propellers viscous and pressure effects can be neglected. Assume that power and torque depend on
the same parameters as thrust.
F

Problem 7.92
Given:
Water drop mechanism
Find:
[Difficulty: 2]
Difference between small and large scale drops
Solution:
3
Given relation
5
d D ( We)
V2 D
D
For dynamic similarity
V 2 D
m m
Dm
dm
dp
V 2 D
p p
Dp
2
Hence
dm
dp
5
1 5
20
1
3
5

Problem 7.93
Given:
Find:
Solution:
[Difficulty: 2]
Kinetic energy ratio for a wind tunnel is the ratio of the kinetic energy flux in the test section to the drive power
Kinetic energy ratio for the 40 ft x 80 ft tunnel at NASA-Ames
nmi 6080 ft
hr
ft
From

Problem 7.95
Given:
Flapping flag on a flagpole
Find:
[Difficulty: 4]
Explanation of the flapping
Solution:
Discussion: The natural wind contains significant fluctuations in air speed and direction. These fluctuations tend to disturb the flag
from an init

Problem 7.96
[Difficulty: 3]
Given:
A 1:16 scale model of a bus (152 mm x 200 mm x 762 mm) is tested in a wind tunnel at 26.5 m/s. Drag force is 6.09
N. The axial pressure gradient is -11.8 N/m2/m.
Find:
(a) Horizontal buoyancy correction
(b) Drag coeffic

Problem 7.97
[Difficulty: 5]
Discussion: The equation given in Problem 7.2 contains three terms. The first term contains surface tension and gives a
speed inversely proportional to wavelength. These terms will be important when small wavelengths are
consi

Problem 8.4
[Difficulty: 2]
Given:
That transition to turbulence occurs at about Re = 2 300
Find:
Plots of average velocity and volume and mass flow rates for turbulence for air and water
Solution:
The basic equations are
From Tables A.8 and A.10
For the

Problem 8.12
Given:
Piston-cylinder assembly
Find:
[Difficulty: 3]
Mass supported by piston
Solution:
Basic equation
Available data
Q
l
3
=
a p
This is the equation for pressure-driven flow between parallel plates; for a small gap a,
the flow between the

Problem 8.17
Given:
Navier-Stokes Equations
Find:
[Difficulty: 2]
Derivation of Eq. 8.5
Solution:
The Navier-Stokes equations are
4
3
u v w
+
+
=0
x y z
1
4
5
3
(5.1c)
6
4
3
u u u
u
u
u
u
p
+u
+v
+ w = g x
+ 2 + 2 + 2
x
x
y
z
x
y
z
t
2
1
4
5
3
2

Problem 8.21
[Difficulty: 3]
Given:
Laminar velocity profile of power-law fluid flow between parallel plates
Find:
Expression for flow rate; from data determine the type of fluid
Solution:
n
h p
n h
u=
1
k L n + 1
n+ 1
1
The velocity profile is
Q = w

Problem 8.24
[Difficulty: 3]
Given:
Properties of two fluids flowing between parallel plates; applied pressure gradient
Find:
Velocity at the interface; maximum velocity; plot velocity distribution
Solution:
dp
Given data
=k
dx
k = 50
kPa
m
h = 5 mm
1 = 0

Problem 8.26
[Difficulty: 2]
Given:
Computer disk drive
Find:
Flow Reynolds number; Shear stress; Power required
Solution:
For a distance R from the center of a disk spinning at speed
V = R
The gap Reynolds number is
Re =
V = 25 mm
V a
Re = 22.3
m
s
1

Problem 8.27
[Difficulty: 2]
Given:
Velocity profile between parallel plates
Find:
Pressure gradients for zero stress at upper/lower plates; plot
Solution:
U y
y 2
p
2 x a
2
y
u=
The shear stress is
y
1
yx =
= +
p 2
2a
2 x
dy
a
a
a
+
a
From Eq. 8

Problem 8.30
[Difficulty: 3]
Given:
Data on flow of liquids down an incline
Find:
Velocity at interface; velocity at free surface; plot
Solution:
2
Given data
h = 10 mm
= 60 deg
1
2 =
5
m
1 = 0.01
s
2 = 2 10
2
3m
s
(The lower fluid is designated fluid 1,

Problem 8.31
[Difficulty: 2]
Given:
Velocity distribution on incline
Find:
Expression for shear stress; Maximum shear; volume flow rate/mm width; Reynolds number
Solution:
u(y) =
From Example 5.9
du
y
g sin( )
2
h y
2
For the shear stress
=
is a max