Mylan Nejyar

1. 1
University of Duhok
College of Engineering
Water Resources Eng.Department
Fluid Mechanics II
Course II: Fluid Dynamics
Second Year Students
(2015- 2016)

2. 2
Subjects
Chapter one: Fluid dynamics, Types of flow (Steady flow, Unsteady flow,
Uniform flow, Non- uniform flow), Definitions ( One, two and three
dimensional flow), stream line, Separation zone, Continuity equation,
Control volume and control surface, Bernoulli equation, solved
problems.
Chapter two: Application of Bernoulli equation, Flow over weir,
Rectangular weir, Rectangular contracted weir, Triangular weir (or V-notch
weir) , Trapezoidal weir, Flow under sluice gate, Venturi meter, Orifice
meter, Pitot tube, Flow through orifice, Free orifice, Submerged orifice,
Flow in a siphon, Cavitation, A free liquid jet, Unsteady Flow, Time of
emptying a tank through an orifice, solved problems.
Chapter three: Momentum equation (or Dynamic Forces), Examples of
momentum forces,(jet, elbow pipe, tee pipe, pipe transition) solved problems
Chapter four: Viscous Flow in Pipes, Types of fluid, (Ideal fluid, Real
fluid), Laminar flow Turbulent flow, Reynolds Number (Re), Friction head
loss in pipes, Darcy Weisbach equation, Moody diagram, Minor head loss in
pipes, Piping systems, Single pipe., Pipes in series, Pipes in parallel,
Combination of pipes in series and in parallel, Pump selection, solved
problems.
References
1-Frank M. white, (2011), Fluid Mechanics, seventh edition,
McGraw-Hill, Inc.
2-Yunus A.Gengel and Joun M. Cimbala, (2006), Fluid Mechanics
Fundamentals and Applications, first edition, McGraw- Hill series.
3-Clayton T. crowe, Donald F. Elger, Barbara C. Williams, John A.
Roberson, (2011), Engineering Fluid Mechanics, ninth edition,
publisher: Wiley
4-Edward J. Shaughnessy, Jr.,Ira M. Katz, James P. Schaffer, (2005),
Introduction to Fluid Mechanics, Published by Oxford University
Press, Inc.
5- Bruce R. Munson and Donald F. Young, (2002), Fundamentals
of Fluid Mechanics, fourth edition, John Wiley & Sons, Inc.
6-Jack B. Evett and Cheng Liu,(1988), 2500 solved problems in Fluid
s solved problems series, revised first,Mechanics and Hydraulics, Schaum
edition, McGraw-Hill, Inc.

3. 3
Chapter One
Fluid dynamics: Is a study of fluids in motion. Three significant equations
are used for solving the problem of fluids flow.
1- Continuity equation (conservation of mass).
2- Bernoulli equation (conservation of energy).
3- Momentum equation (or Dynamic forces).
Types of flow:
1-Steady flow: The velocity at a point remains constant with respect to time.
𝐝𝐯
𝐝𝐭
= 𝟎
2-Unsteady flow: The velocity changes either in magnitude or in direction
with respectto time.
𝐝𝐯
𝐝𝐭
≠ 𝟎
3-Uniform flow: The velocity remains constant with respectto distance.
𝐝𝐯
𝐝𝐬
= 𝟎

4. 4
4-Non- uniform flow: The velocity changes either in magnitude or in
direction with respectto distance.
𝐝𝐯
𝐝𝐬
≠ 𝟎
Changed diameter pipe elbow pipe
Definitions:
1- One, Two and three dimensional flow
One dimensional flow two dimensional flow
2-stream line: Is an imaginary line with in the flow for which the tangent at
any point represent the direction of motion at that point.
Pip Orifice Weir

5. 5
Separation zone
Separa
tion
zone
3-Separation zone: In regions where boundaries turn away from the flow so
as to cause the stream lines to diverge, the flow usually separates from the
boundary and recirculation pattern is generated in the region.
Expansion pipe Elbow pipe
Continuity equation:
The continuity equation is based on the conservation of mass as it applies
to the fluids flow. It is applicable to:
1- Steady flow
2- One dimensional flow.
3- Incompressible fluid.
Control volume and control surface:
CV = control volume
CS=control surface

6. 6
The controlvolume is a region in spacethat one established to aid in the
solution of flow problems, and control surface is the surface surrounding
the control volume.
In most problems, part of the control surface will coincide with the wall
of pipe.
Flow into and out of controlvolume:
Rate of flow past a given area A is,
Q = VA
Rate of change of flow in CV = rate of flow out – rate of flow in
dQ = V2 A2 - V 1A1
Multiplying the equation by ρ
d ρ Q = ρ V2 A2 - ρ V 1A1
2 2 1 1
Vol mass
Q
t t
dmass
V A V A
dt
 
 
 
 
From the principle of conservation of mass,
0
dmass
dt


7. 7
Example (1-1): Water flows through a pipes shown in fig., if VA = 2m/sec,
VD = 4m/sec, calculate VB and VC?
Solution:
 
 
   
2 3
2
2 2
2 0.45 0.318 / sec
4
0.318 0.3
4
4.5 / sec
0.318 0.15 4 0.2
4 4
10.9 / sec
A A A
A
A A B B
B
B
A C D
C
C
Q V A
Q m
Q V A V A
V
V m
Q Q Q
V
V m


 
 
  
   
 

 
   


8. 8
Example(1-2): The river discharges in to the reservoir shown at a rate of
400000 ft3/sec, and the outflow rate from the reservoir through the flow
passages in the dam is 250000 ft3/sec. If the reservoir surface area is 40 mi2,
what is the rate of rise of water in the reservoir? (1 mi= 5280 ft)
Solution:
We first choosea controlvolume
   
 
1 2 3
3
4
2
400000 250000
150000 / sec
150000
1.34 10 / sec 60 60
40 5280
0.484 /
CS CSin out
rise
rise
rise
rise
VA VA
Q Q Q
Q
Q ft
V ft
V ft hr

  
 
 

    



9. 9
Example(1-3): From figure shown find the elevation of water
surface in the tank after 24 sec?
Solution:
Vrise1 =
Q
A1
=
0.392
π
4
12 = 0.5ft/sec
𝑉1 =
𝑑ℎ
𝑑𝑡
Vrise2 =
Q
A2
=
0.392
π
4
22
= 0.125ft/sec

10. 10
Bernoulli equation (or Energy equation):
Bernoulli equation is based on the conservation of energy, it is applicable in
steady flow, one dimension flow, and incompressible fluid.
Energy may be defined as the ability to do work and has the same units as
work (N.m), in pipe flow we have:
1- Pressure energy = 𝑊 × ℎ = 𝑊 ×
𝑃
𝛾
(where, h =depthof water)
2- Potential energy or elevation energy = 𝑾 × 𝒁 (where, Z= elevation)
3- Kinetic energy =
1
2
𝑀 𝑉2
=
1
2
𝑊
𝑔
𝑉2
Total energy = pressure energy + Potential energy + kinetic energy
(dimension of total energy = 𝐹 × 𝐿)
Total energy /unit weight = H =
𝑷
𝜸
+ 𝒁 +
𝑽 𝟐
𝟐𝒈
(dimension=L)
H = total energy head = (constant)
Bernoulli equation between two points for head loss h L = 0
Bernoulli equation between two points for head loss h L ≠ 0
2 2
1 1 2 2
1 2
2 2
L
p V p V
Z Z h
g g 
     
TEL= Total Energy Line=
𝑷
𝜸
+ 𝒁 +
𝑽 𝟐
𝟐𝒈
HGL=Hydraulic Grad Line or (Piezometric Line) = h =
𝑷
𝜸
+ 𝒁
H=h + V2/2g

11. 11
(B.E. between two points for hL=0)
(B.E. between two points for hL≠0

12. 12
Example (1-4): For a horizontal pipe shown in figure, find the direction of
flow between the two sections 1 and 2,the value of head loss, and draw TEL
and HGL if hL =0, (d1 = 7.5cm, P1 =3.5 N/cm2, d2 = 15cm, P2 = 7N/cm2 ).
Solution:
 
 
 
 
3
1 2
1
2
1
2
2
2
22
1 1
1
1
2
2
2
1 2
0.085 / sec
0.085
19.24 / sec
0.075
4
0.085
4.81 / sec
0.15
4
19.243.5 10000
3.57 18.87
2 9810 2 9.81
22.44
4.817 10000
7.14 1.18
9810 2 9.81
8.32
The flow wi
Q Q Q m
Q
V m
A
Q
V m
A
p V
H
g
H m
H
H m
H H



  
  
  

     



   



1 2
ll be from section toSection
22.44 8.32 14.12
L
L
h H H
h m
 
  
1 2

13. 13
Example (1-5): water flows steadily through the pipe shown in figure such
that the pressures at sections (1) and (2) are 300 Kpa and 100 kpa
respectively. Determine the diameter of the pipe at section (2) D2, if the
velocity at section (1) is 20m/sec.

14. 14
Solution:
Problems
1- Three pipes steadily deliver to a large exit pipe in figure. The velocity
V2=5m/sec, and the exit flow rate Q4=120m3/hr. Find V1,V3,V4 if it is known
that increasing Q3 by 20 percent would increase Q4 by 10 percent.
(ANS:V4=5.23m/sec)

15. 15
2-In figure shown water flows in the pipe from section A to section B at
Q = 13.2 ft3/sec, the pressure head at A is 22.1 ft, find the pressureat B and
draw the TEL and HGL if hL=0?
(ANS:PB=699lb/ft2
)
3-A small nozzle attached to a high pressure water pipe as shown in figure,
if the maximum pressure available in the water pipe is 100 psi , how high
will the water go ? At what velocity does the water exit the nozzle?
(ANS: Vexit = 121.9 ft/sec)