Engineering Fluid Mechanics
Solution Manual
Prof. T.T. Al-Shemmeri
,Engineering Fluid Mechanics Solution Manual Contents
Contents
Book Description: 5
Author Details: 5
1 Chapter One Tutorial Problems 6
2 Chapter Two Tutorial Problems 13
3 Chapter Three Tutorial Problems 21
4 Chapter Four Tutorial Problems 28
5 Chapter Five Tutorial Problems 31
Sample Examination Paper 33
Class Test – Fluid Mechanics 34
Formulae Sheet 44
4
,Engineering Fluid Mechanics Solution Manual Chapter One Tutorial Problems
1 Chapter One Tutorial Problems
1.1 Show that the kinematic viscosity has the primary dimensions of L2T-1.
Solution:
The kinematic viscosity is defined as the ratio of the dynamic viscosity by the density of the fluid.
The density has units of mass (kg) divided by volume (m3); whereas the dynamic viscosity has the units
of mass (kg) per meter (m) per time (s).
Hence:
ML−1T −1 L2T −1
= = =
ML−3
1.2 In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid
has absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient and
shear stress at the boundary assuming a linear velocity distribution? Determine its kinematic
viscosity.
[Ans: 15 s-1, 0.72Pa.s; 5.257x10-5 m2/s]
Solution:
, Engineering Fluid Mechanics Solution Manual Chapter One Tutorial Problems
dV 1.125
Gradient = = = 15 s −1
dy 0.075
dV
= = 0.048x15 = 0.720 Pa.s
dy
0.048
= = = 5.257x10−5 m 2 / s
913
1.3 A dead-weight tester is used to calibrate a pressure transducer by the use of known weights
placed on a piston hence pressurizing the hydraulic oil contained. If the diameter of the piston
is 10 mm, determine the required weight to create a pressure of 2 bars.
[Ans: 1.6 kg]
Solution:
(a) P = F/A
A= D2 = x0.0102 = 7.854x10−5 m2
4 4
Hence the weight =PxA/g
= 2 x 105 x7.854x10-.81
= 1.60 kg
Solution Manual
Prof. T.T. Al-Shemmeri
,Engineering Fluid Mechanics Solution Manual Contents
Contents
Book Description: 5
Author Details: 5
1 Chapter One Tutorial Problems 6
2 Chapter Two Tutorial Problems 13
3 Chapter Three Tutorial Problems 21
4 Chapter Four Tutorial Problems 28
5 Chapter Five Tutorial Problems 31
Sample Examination Paper 33
Class Test – Fluid Mechanics 34
Formulae Sheet 44
4
,Engineering Fluid Mechanics Solution Manual Chapter One Tutorial Problems
1 Chapter One Tutorial Problems
1.1 Show that the kinematic viscosity has the primary dimensions of L2T-1.
Solution:
The kinematic viscosity is defined as the ratio of the dynamic viscosity by the density of the fluid.
The density has units of mass (kg) divided by volume (m3); whereas the dynamic viscosity has the units
of mass (kg) per meter (m) per time (s).
Hence:
ML−1T −1 L2T −1
= = =
ML−3
1.2 In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid
has absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient and
shear stress at the boundary assuming a linear velocity distribution? Determine its kinematic
viscosity.
[Ans: 15 s-1, 0.72Pa.s; 5.257x10-5 m2/s]
Solution:
, Engineering Fluid Mechanics Solution Manual Chapter One Tutorial Problems
dV 1.125
Gradient = = = 15 s −1
dy 0.075
dV
= = 0.048x15 = 0.720 Pa.s
dy
0.048
= = = 5.257x10−5 m 2 / s
913
1.3 A dead-weight tester is used to calibrate a pressure transducer by the use of known weights
placed on a piston hence pressurizing the hydraulic oil contained. If the diameter of the piston
is 10 mm, determine the required weight to create a pressure of 2 bars.
[Ans: 1.6 kg]
Solution:
(a) P = F/A
A= D2 = x0.0102 = 7.854x10−5 m2
4 4
Hence the weight =PxA/g
= 2 x 105 x7.854x10-.81
= 1.60 kg
