Problem 1.1
1.1
[3]
A number of common substances are
Tar
Sand
Silly Putty
Jello
Modeling clay
Toothpaste
Wax
Shaving cream
Some of these materials exhibit characteristics of both solid and fluid behavior under different conditions. Explain
and give examp
Problem 13.1
Given:
Air extracted from a large tank
Find:
[2]
Mass flow rate
Solution:
Basic equations:
mrate = V A
h1 +
V1
2
V2
2
p
= h2 +
2
2
k
= const
The mass flow rate is given by
T0 = ( 70 + 273) K
p0 = 101 kPa
J
cp = 1004
kg K
k = 1.4
mrate = A V
A
Problem 12.1
Given:
Air flow through a filter
Find:
[2]
Change in p, T and
Solution:
Basic equations:
(
h2 h1 = cp T2 T1
)
p = R T
Assumptions: 1) Ideal gas 2) Throttling process
In a throttling process enthalpy is constant. Hence
h2 h1 = 0
so
T2 T1 = 0
Problem 11.1
Given:
Rectangular channel flow
Find:
[1]
Discharge
Solution:
2
Basic equation:
1
1
3
2
Q = A R S0
n
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width Bw = 2 m and depth y = 1.5 m we fin
Problem 10.1
[2]
Problem 10.2
Given:
Geometry of centrifugal pump
Find:
[2]
Estimate discharge for axial entry; Head
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
= 1.94
slug
ft
3
= 1500 rpm
r1 = 4 in
r2 = 7.5 in
1 =
Problem 9.1
Given:
Model of riverboat
Find:
[2]
Distance at which transition occurs
Solution:
U x
U x
=
Basic equation
Rex =
For water at 10oC
= 1.30 10
Hence
xp =
For the model
xm =
2
6 m
Rex
U
xp
18
s
and transition occurs at about Rex = 5 10
5
(Tabl
Problem 8.1
Given:
Air entering duct
Find:
[1]
Flow rate for turbulence; Entrance length
Solution:
The governing equations are Re =
V D
2
D V
4
Q=
D = 6 in
From Table A.9
= 1.62 10
Llaminar = 0.06 Recrit D
The given data is
Recrit = 2300
or, for turbulen
Problem 7.1
[2]
Problem 7.2
[2]
Problem 7.3
Given:
Equation for beam
Find:
[2]
Dimensionless groups
Solution:
Denoting nondimensional quantities by an asterisk
A* =
Hence
A
L2
A = L2 A *
Substituting into the governing equation
The final dimensionless equ
Problem 6.1
[2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
NOTE: Units of B are s-1 not ft-1s-1
Basic equations
(2
ax = u
x
u + v
) B x
2
u ( x , y) = A y x
For this flow
v ( x , y) = 2 A x y + B y
(
)
(
Problem 5.1
[1]
Problem 5.2
Given:
Velocity fields
Find:
[2]
Which are 3D incompressible
Solution:
Basic equation:
x
u+
y
v+
z
w=0
Assumption: Incompressible flow
2
2
u ( x , y , z , t) = y + 2 x z
v ( x , y , z , t ) = 2 y z + x y z
w ( x , y , z , t) =
Problem 4.1
Given:
Data on mass and spring
Find:
[1]
Maximum spring compression
Solution:
The given data is
M = 3 kg
h = 5 m
k = 400
N
m
Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (the spring has gravitiona
Problem 2.1
Given:
Velocity fields
Find:
[1]
Whether flows are 1, 2 or 3D, steady or unsteady.
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
V
V
V
V
V
V
V
V
=V
=V
=V
=V
=V
=V
=V
=V
( y)
1D
( x)
1D
( x , y)
2D
( x , y)
2D
( x)
1D
( x , y , z)
3D
( x , y)
2D
(
Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald
Answers to Selected Problems, Chapter 1
1.5
1.7
1.19
1.21
1.23
1.25
1.27
M = 5913 kg
L = 27.25 in.
W2
y = 2.05 2
gt
d = 0.109 mm
y = 0.922 mm
a) Nm/s, lbfft/s
f) Ns, lbfs
a) 6.89 kPa
a