Fluid Mechanics and Hydrology Fundamentals

Ministry of Higher Education and Scientific research
Erbil Polytechnic University
Technical Engineering College
Highway Engineering Department
Fluid Mechanics and Hydrology
Prepared by:
Dr. Rawaz Kurda
2022-2023

Topics for study
Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda
• Physical Introduction (FluidProperties)
• Dimensional Analysis
• Fluid Static
• Accelerated Fluid
• Fluid Dynamics
• Fluid Viscosity and Turbulences
• Fluid Resistance
• Flow in closed conduits(pipes)
• Flow in open conduits (openchannels)
• Pumps and Turbines
• Introduction to hydrology
• Domain and objective of Engineering hydrology
• Practical applications of hydrology
• Water cycle
• Evaporation
• Evapotranspiration
• Precipitation
• Infiltration
• Measurement techniques of each Hydrologic cycle

Course Book
Rawaz M. S. Kurda
Highway Engineering/Erbil Technical Engineering College
Rawaz.kurda@epu.edu.iq
Three hours - Theory
1. Course name
2. Lecturer in charge
3. Department/College
4. Contact
5.Time (in hours)per
week
6. Office hours
7. Coursecode
8. Teacher'sacademic
profile
2 days a week depending on the classes & appointments
HE206
PhD in Civil Engineering at University of Lisbon (2017).
https://academicstaff.epu.edu.iq/faculty/rawaz.kurda
https://scholar.google.com/citations?user=KesSqb4AAAAJ&hl=ar&oi=sra
9. Keywords
10. Course overview:
The course focuses on analyzing different engineering problems and examining various aspects of
the problem through focusing on each component through a holistic approach that recombines the
different elements of the problem back into an integrated problem. Fundamentals of Fluid
mechanics are learned through real life problem solving and they are set as basis for solving open
ended questions. Additionally, this course is to understand fundamentals of engineering hydrology.
It is the study of water in all its forms (rain, snow and water on the earth’s surface), and from its
origins to all its destinations on the earth. The study of hydrology that concerned mainly with
engineering applications is known as applied hydrology. Engineering Hydrology deals with
estimation of water resources, the study of processes such as runoff, precipitation and their
interaction, the study of problems such as Floods, Droughts and strategies to overcome them.
• 11. Course objective:
• To familiarize the student basic principles and equations of fluid mechanics and show their
application real life engineering example.
• To give the students the correct intuition when comes to Fluids and their application
• The student uses fundamentals of math needed for solving complicated problems
• Brain storming on open ended problems and using learned method to get educated
approximation for solving them
• The course will focus on explaining the background of Applied hydrology,
• The application of hydrology in different engineering structures.
13. Forms of teaching
• Students are provided with handouts for each chapter to be available with them during the
lectures.
The handout includes explanations, examples, problems and homework.
• Notes and questions are explained on white board
• Since there is not any laboratory experimental work, multiple videos will be shown during the
class.
14. Assessmentscheme
• 4%Quiz
• 40% Activities (14%homework, 2% Class activity, 8% Report, 8% Seminar, 8% Project)
• 16% Mid. Term Theory exam
• 40% Final Theory Exam
15. Studentlearning outcome:
After successful completion of the course, students are expected to:
• understand the basic concepts of Fluid Mechanics (Recognize the various types of fluid flow
problems encountered in practice.)
• understand how the main concept of the Fluid Mechanics and Hydrology is
used in the Civil Engineering (Water, Drainage and Sewerage)
• model engineering problems and solve them in a systematic manner.
• a working knowledge of accuracy, precision, and significant digits, and recognize the
importance of dimensional homogeneity in engineering calculations.
16. Course Reading List and References:
i. Fluid Mechanics Fundamental and application/ Yunus A Cengel /3th edition
ii. Fundamental of Fluid Mechanics / Bruce R.Munson/ 7th edition
iii. Fluid Mechanics / Frank M. White/ 7th edition

17. The Topics:
Fluid Mechanics _ properties of fluids
Introduction
Properties of fluids
Density and volume
Viscosity
Thermodynamic properties
Compressibility and bulk Modulus
Surface Tension and capillarity
Vapour pressure and cavitation
01.
1.1
1.2
1.2.1
1.2.2
1.2.3
1.2.4
1.2.5
1.2.6
2weeks
Pressure and its measurement (Fluid Statics)
Fluid pressure at a point
Pascal’slaw
Pressure variation a fluid at rest
Measurement of pressure
Simple nanometers
Differential Manometers
02.
2.1
2.2
2.3
2.4
2.5
2.6
2weeks
Buoyancy and floatation
Introduction
Bouncy
Center of Bounce
Meta-centricheight
Analytical Method for meta-center Height
ConditionsofEquilibriumof a floatingand sub-mergedbody
ExperimentalmethodofdeterminationofMeta-centricheight
03.
3.1
3.2
3.3
3.4
3.5
3.6
3.7
2weeks
Kinematics of flow and ideal flow
Introduction
Methods of describingfluid motion
Type of fluid flow
Rate of flow ordischarge
Continuity Equation
Continuity equation in three dimensions
Velocity and acceleration
Velocity potentialfunction and stream function
04.
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
1week
Dynamics of fluid flow
Introduction
Equation ofMotion
Euler’s equation of motion
Bernoulli’s equation form Euler’s equation
Assumptions
Bernoulli’s Equation for real fluid
Practicalapplications of Bernoulli’s equation
The Momentum Equation
Kinetic energy correction factor
05.
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
1week
06. Fundamentals of engineeringhydrology
6.1 Introduction to hydrology
6.2 Domain and objective of Engineering hydrology
6.3 Practicalapplications of hydrology
I. Hydraulic structures
II. Water supply
III. Wastewater treatment and disposal
IV. Irrigation
V. Drainage
VI. Hydropower generation
VII. Floodcontrol
VIII. Navigation
IX. Erosion and sediment control
X. Salinity control
XI. Pollution abatement
XII. Recreational use of water
XIII. Fish and wildlife protection
6.4 Hydrologic cycle
6.5. Steps of the HydrologicCycle
6.6 Measurementtechniques
6.7 Estimated world water quantities
6.8 Hydrologic budget in details
1 week
6.9 Test for consistency of record
6.10 Analysis of Precipitation Records
I. Precipitation depth
II. Precipitation intensity
III. Computation of Average RainfallDepth over a Basin
IV. Arithmetic Average Method
V. Thiessen Polygon Method 1 week
6.11 Analysis of Evaporation losses
I. Factorsaffectingevaporation
I. Meteorological Data:
II. Type of Surface:
III. Water Quality:
II. Measurement or estimation of evaporation
I. Water budget method
II. Empirical formulae
III. Energy budget method
IV. Mass transfer method
V. Actual observations
VI. Pan observations
2weeks
18. Practical Topics (If there is any)
19. Examinations:
Solving Problems
Such as solve,derive, find, determine, ... etc.
Explanations and graphing
Such as explain,why, show that, prove that ..., etc.
Number of Questions: 3-6
Number of Assignments:2-4
Recommendations for Students at Exams
• Read the questions carefully and at least twice.
•Think about the answers and don't hurry.
Answer the questions with the easiest first
At the end, review the answers.
20. Extra notes
For the above time schedule, 12 weeks of teaching is
considered, hence, the completion of the program is
dependent on the available number of weeks. However,
some changes might happen to optimize the available time.

Objectives
❑ Understand the basic
concepts of Fluid Mechanics
(Recognize the various types
of fluid flow problems
encountered in practice.)
❑ Show how it is used in the
Civil Engineering (Water,
Drainage and Sewerage)
• Model engineering problems
and solve them in a
systematic manner.
• Have a working knowledge
of accuracy, precision, and
significant digits, and
recognize the importance of
dimensional homogeneity in
engineering calculations.
5
U-tube manometer

Chapter one
Fluid Mechanics _ properties of fluids

Why should we know this?

1.1 Introduction
Fluid Mechanics
What are fluids, and what is Fluid mechanics?
Fluids:
With exception to solids, any other matters can be categorized as fluid, such as water, blood,
milk, oil and gases.
Fluid Mechanics:
It is a physical science concerned with the behaviour of fluid (liquid and gaseous state) at rest,
motions and the force acting on them.
Fluid statics Fluid kinematics Fluid dynamics
The study of fluids at rest
matters
The study of fluids in
motion
The study of the effect of
forces on fluid motion

Fluid Mechanics: may be divided in three parts
❑ Statics
• Hydrostatic: the study of incompressible fluid
under static condition.
• Aerostatic: the study of compressible gases under
static condition.
❑ Kinematics
it deals with the velocity, accelerations and
patterns of flow and not the forces or energy
causing them.
❑ Dynamics
It deals with the relations between
velocities, acceleration of fluid with the
forces or energy causing them.

Fluid can be classified as:
▪ Liquid: it is a fluid which possesses a definite volume, which varies only slightly with temperature
& pressure
▪ Gas: it possesses not definite volume and it is compressible.
▪ Vapour: it is a gas whose temperature and pressure are such that it is very near to gas state (e.g.
steam).
Solid
()
Gas
(water vapor)
Liquid
(water)
10

Shear stress and matters
A solid can resist an applied shear stress by
deforming.
A fluid deforms continuously under the influence
of a shear stress, no matter how small.
In solids, stress is proportional to strain, but in
fluids, stress is proportional to strain rate.
When a constant shear force is applied, a
solid eventually stops deforming at some
fixed strain angle, whereas a fluid never
stops deforming and approaches a constant
rate of strain. Deformation of a rubber block
placed between two parallel plates
under the influence of a shear force.
The shear stress shown is that on
the rubber—an equal but opposite
shear stress acts on the upper plate.

In a liquid, groups of molecules can move relative to each other, but
the volume remains relatively constant because of the strong
cohesive forces between the molecules. As a result, a liquid takes the
shape of the container it is in, and it forms a free surface in a larger
container in a gravitational field.
A gas expands until it encounters the walls of the container and fills
the entire available space. This is because the gas molecules are
widely spaced, and the cohesive forces between them are very small.
Unlike liquids, a gas in an open containercannot form a free surface.
Fluid behavior
Unlike a liquid, a gas does not form
a free surface, and it expands to fill
the entire available space
Why? See next slide

The arrangement of atoms in different phases: (a) molecules are at relatively fixed positions in a solid, (b) groups of
molecules move about each other in the liquid phase, and (c) individual molecules move about at random in the gas phase.
❑ Solid: The molecules in a solid are arranged in a pattern that is repeated throughout.
❑ Liquid: In liquids molecules can rotate and translate freely.
❑ Gas: In the gas phase, the molecules are far apart from each other, and molecular
ordering is nonexistent.
Intermolecular bonds
Strongest Weakest

Dimensions and units
Units needed to properly express a physical quantity
Systems to be used:
1- S.I. (International System of Units)
● Adopted in 1960
● Used by nearly in every major country, except the U.S.
● Likely to be adopted by the U.S. in the near future
2- B.G. (British Gravitational system)
● Used in the technical literature for years
● Preferred system in the U.S.

General note:
One kilogram is equal to 9.81 Newtons. To convert Newtons to kilograms, divide by 9.81.
For instance, 20 Newtons would be equivalent to 20/9.81 or 2.04 kilograms

Density and
volume
Specific Volume
Specific Gravity:
Specific mass
Specific weight
Dynamic Viscosity
Kinematic Viscosity
Newton’s Law of Viscosity
Non Newtonian Fluid
Cohesion
Adhesion
Viscosity
Variation of viscosity with
temperature
Classification of fluids
Surface
tensionand
capillarity
Compressibility
and bulk
modulus
Vapour
pressure and
cavitation
Compressible
Modulus of Elasticity
Incompressible
1.2 Properties of fluids
18

Specific mass (mass density)
Density or specific mass density of a fluid is defined as the ratio of the mass of a fluid to its volume. Thus mass per
unit volume of a fluid is called density. It is denoted the symbol 𝜌 (rho). The unit of mass density in SI unit is kg per
cubic meter, i.e., kg/m3. The density of liquids may be considered as constant while that of gases changes with the
variation of pressure and temperature. Mathematically, mass density is written as
𝜌 = Mass of fluid
Volume of fluid
Note: the value of density of water is 1 gm /cm3 or 1000 kg /m3.
Specific Weight (weight density)
Specific weight or weight density of a fluid is the ratio between the weight of a fluid to
its volume. Thus weight per unit volume of a fluid is called weight density and it is
denoted by the symbol w.Thus, mathematically,
Mass of fluid × Acceleration due to gravity
Volume of fluid
= 𝜌 𝑥 𝑔
w =
Weight of fluid
Volume of fluid
=
Note: the value of specific weight or weight density (w) of water is 9.81 × 1000 Newton / m³ in SI units.
1 m3
1 m
1 m
1 m
Mass density =
Specific weight
Acceleration due to gravity
Newton
1.2.1 Density and volume
18

Specific Volume
Specific volume (ɏ) of a fluid is defined as the volume of a fluid occupied by a unit mass or volume
per unit mass of a fluid is called specific volume. Mathematically, it is expressed as
Specific volume = ɏ =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
Thus specific volume is the reciprocal of mass density. It is expressed as m3/ kg. It is commonly applied to gases.
Specific Gravity (S):
Specific gravity is defined as the ratio of the weight density (or density) of a fluid to the weight density
(or density) of a standard fluid. For liquids, the standard fluid is taken water and for gases., The
standard fluid is taken air. Specific gravity is also called relative density. It is dimensionless quantity and
is denoted by the symbol S.
Weight density of a liquid =
S x weight density of water = S x 1000 x 9.81
𝑆 𝑓𝑜𝑟 𝑙𝑖𝑞𝑢𝑖𝑑𝑠 =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Note (i): If the specific gravity of a fluid is known, then the density of the fluid will be equal to specific
gravity of fluid multiplied by the density of water. For example, the specific gravity of mercury is 13.6,
hence density of mercury = 13.6 x 1000 = 13600 kg/m.
Note (ii): According to the International Standard Atmosphere (ISA) values—15° C at sea level—the
density of dry air is at: In Metric units: 1.225 kg/m3.
𝑆 𝑓𝑜𝑟 𝑔𝑎𝑠𝑒𝑠 =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎𝑖𝑟
The density of a liquid =
S x density of water = S x 1000

Example 1:
Calculate the (1) specific weight, (2) density and (3) specific gravity of one liter of a liquid
which weighs 7 N.
Solution:
Volume = 1 liter =
1
1000
𝑚3 (1 liter = 1/1000 m3 = 1000 cm3)
Weight = 7 N (note that the value given in N. Thus, it means that the value is
Mass of fluid × Acceleration due to gravity)
𝑣𝑜𝑙𝑢𝑚𝑒
𝑊𝑒𝑖𝑔ℎ𝑡
(1) Specific weight (w) = = = 7000 N/m3
(2) Density (𝜌) = 𝑤
= 7000
= 713.5
𝑔 9.81
kg/m3 (weight density)
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 713.5
(3) Specific gravity = 𝐷𝑒𝑠𝑛𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
= 1000
= 0.7135
General note
Specific Weight = Mass of fluid x g
Volume of fluid
mass density
Densityand
volume
Specific Volume
Specific Gravity:
Specific mass
Specific weight
7 𝑁
1
1000

Example 2:
Calculate the density, specific weight and weight of one liter of petrol, when its specific gravity equal to 0.70.
Solution:
22
S for Liquids =
Weight density (density) of liquid
Weight density (density) of water

Solution:
Example 3:
10 m3
Weight = 136 x 104 N
Filled
with
Mercury
Specific Weight = Mass of fluid x g
Volume of fluid
mass density
23

Homework 1:
Calculate the density, specific weight
and weight of one liter of Bitumen,
when its specific gravity is equal to
1.03.
In case, you have VL km of road (10 m
width) and required to cover with 3
mm of Bitumen. Show how much
bitumen we need in weight.
Bitumen Emulsion road
Note:
Value from the list (VL) is the constant number given in the list where every student has different number. Thus, the final
results of the homework for each students will bedifferent.

Quiz 1:
Q1:
Calculate the density, specific weight and weight of one
liter of asphalt, when its specific gravity is equal to
1.03. In case, you have VL km of road (10 m width) and
required to cover with 3 mm of Bitumen. Show how
much bitumen we need in weight.
Bitumen Emulsion road
Volume of fluid
Mass of fluid × Acceleration due to gravity
Volume of fluid
w =
Weight of fluid
Volume of fluid
=
𝑆 𝑓𝑜𝑟 𝑙𝑖𝑞𝑢𝑖𝑑𝑠 =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑
𝑊𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Q2:
Define the following terms of the Fluid Mechanics
1. Statics
2. Kinematics
3. Dynamics

Ministry of Higher Education and Scientific research
Erbil Polytechnic University
Technical Engineering College
Highway Engineering Department
Prepared by:
Dr. Rawaz Kurda
2021-2022

Dynamic Viscosity
The viscosity of a fluid is a measure of its “resistance to deformation.” In other
words, viscosity is due to the internal frictional force that develops between
different layers of fluids as they are forced to move relative to each other.
Thus, viscosity can be also defined as the property of a fluid which offers
resistance to the movement of one layer of fluid over another adjacent layer of
the fluid. When two layers of a fluid, a distance “dy” apart, move one over the
other at different velocities, say “u” and “u + du” (right Figure).
As shown in the Figures, the viscosity together with relative velocity causes
shear stress acting between the fluid layers. The top layer causes a shear stress
on the adjacent lower layer while the layer stress on the adjacent top layer.
The mentioned shear stress is proportional to the rate of change of velocity (𝜏 ∝
𝑑𝑢
𝑑𝑦
)
with respect to y. It is denoted by symbol 𝜏 called Tau.
𝑑𝑦
Mathematically, 𝜏 = 𝜇 𝑑𝑢
→ 𝜇 =𝜏
dy
𝑑𝑢
27
Where 𝜇 (called mu) is the constant of proportionality and is known as the co-
efficient of dynamic viscosity or only viscosity.
𝑑𝑢
𝑑𝑦
represents the rate of shear
strain or rate of shear deformation or velocity gradient.
𝜏 called Tau
𝜇 called mu
1.2.2 Viscosity
Explained in the next slides

viscosity can be also defined as the property of a fluid
which offers resistance to the movement of one layer
of fluid over another adjacent layer of the fluid.

General notes:
Integration
Derivative ‫اوە‬ ‫ر‬
‫داتاش‬
𝑑𝑦
𝑑𝑥
‫تەواوکارى‬
⇄
𝑑𝑥
𝑑𝑦
= 𝑓 (𝑥)
𝑦 = ‫׬‬ 𝑓 𝑥 𝑑 𝑥 +𝐶
𝑑𝑦 = 𝑓 (𝑥) dx න 𝑑 𝑦 = න 𝑓 𝑥 𝑑 𝑥
𝑑
𝑑𝑥
= 3𝑥2 +10𝑥
𝑦 = (𝑥𝑛 + 𝐶) 𝑑
𝑑𝑦
= n𝑥𝑛−1+0
Example 2:
y = 𝑥3 + 5𝑥2 -2
The derivative of y at x=a:
∆𝑥→0
lim 𝑦 𝑎+∆𝑥 −𝑦(𝑎)
∆𝑥 𝑑𝑥
𝑑𝑦
Notation: y`(a) or (a)
න
𝑑𝑦
𝑑𝑥
‫داتاشراوە‬ ‫پێچەوانەى‬ ‫کردارى‬ ‫تەواوکارى‬

General notes: Differentiation rules (Table A1)
30

General notes: Integration rules (Table A2)
31

Units of Viscosity
𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦(𝜇) =
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝜏)
𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢)
𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦)
=
𝑓𝑜𝑟𝑐𝑒
𝑎𝑟𝑒𝑎
𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 1
𝑇𝑖𝑚𝑒 𝐿𝑒𝑛𝑔𝑡ℎ
=
𝐹𝑜𝑟𝑐𝑒 𝑥 𝑡𝑖𝑚𝑒
(𝑙𝑒𝑛𝑔𝑡ℎ)2
- SI units N . s /m2 → pascal x s (pascal = N/m2)
- MKS units kgf . s /m2 (Meter-Kilogram-Second units)
N or kgf
Second
Meter
32
Notes :
• 𝜇 is often expressed in Poise (P) = gm/cm.s (CGS: Centimetre–gram–second system)
• I poise = 1 gm/cm.s = kg/(10 m.s) ; 1 centipoise = (1/100) poise
10 poise = 1 N.s/m2
• 𝜇 of water at 15 °C is 1.14 x 10-3 N.s/m2

Kinematic Viscosity
It is defined as the ratio between the dynamic viscosity (μ) and density (ρ) of fluid. It is donated by Greek symbol (v)
called “nu”. In addition, Kinematic Viscosity is also called "momentum diffusivity“.
In the MKS and SI, the unit of the Kinematic Viscosity is meter2/sec or m2/sec, while in CGS unites, it is written as
cm2/s. In CGS units, Kinematic Viscosity is also known as a stoke.
33
Length
of
drop
Sticky, thick, viscous (stronger bonds)
difference
V =
µ
𝜌

Types of Fluid
i. Newtonian Fluid
A fluid which obey the above relation known as Newtonian fluids, which its viscosity constant with change in shear
force, Newtonian fluids have constant values of µ, such as water, Kerosene, air….etc.
Ƭ: shear stress
µ :viscosity
du :change in velocity and
dy :change in distance
ii. Non-Newtonian Fluid
Some fluids do not have constant µ. They do not obey Newton’s Law of
viscosity, they are called Non-Newtonian fluids, such as blood, gels and
ketchup.
Which A, B and n are constants
𝜏 = 𝜇 𝑑𝑢
𝑑𝑦
𝜏 = 𝐴+ 𝐵 (
𝑑𝑦
)
𝑑𝑢 𝑛
Velocity gradient
34

Types of Fluid https://www.youtube.com
/watch?v=JJfppydyGHw
Can YouWalk on Water Non Newtonian Fluid Pool?
Video 1
35

Types of Fluid
https://www.youtube.com
/watch?v=2mYHGn_Pd5M
Can YouWalk on Water Non Newtonian Fluid Pool?
Video 2
36

Types of Fluid
iii. Plastic Fluid
It is a non Newtonian fluid with an initial yield stress is to be exceeded to cause a continuous deformation. This
substance are represented by straight line intersecting the vertical axis at the yield stress, there are two types.
a- Ideal plastic fluid (Bingham plastic) has a definite yield stress and a
constant linear relationship.
𝜏 = 𝑐 + 𝜇
𝑑𝑢
b- Thixotropic fluid has a definite yield stress and a non linear
relationship
𝜏 = 𝑐 + 𝜇 (
𝑑𝑦
)
𝑑𝑢 𝑛
Shear
stress
Velocity gradient
iv. Ideal Fluid
A fluid, which is incompressible and is having no viscosity. is known as an ideal fluid. Ideal fluid is only an
imaginary fluid as all the fluids, which exist, have some viscosity. In other words, no such fluid is exist in the
nature. This assumption helps in simplifying the mathematical analysis.
τ = 0
37
𝑑𝑦

Viscosity relation to pressure and temperature
❑ Viscosity is partially dependent of pressure. It
can be said that the viscosity of fluid under
ordinary condition is not considerably
affected by the change in pressure. However,
the viscosity of some oils has been found to
increase with increase in pressure.
❑ Viscosity depends mostly upon temperature,
increase of temperature reduces viscosity. Viscosity and Density of water
Determination of Viscosity:
38
• Capillary Tube
• Sphere Resistance
• Rotating Cylinder
• Viscometer

Example 4:
If the velocity distribution over
2
a plate is given by u = 𝑦 − 𝑦2
3
in which u is the velocity in
meter per second at a distance
y meter above
determine the shear
the plate,
stress
when y = 0 and 0.15 m.
Consider that dynamic viscosity
of fluid as 8.63 poises.
Solution: Derivative
Ƭ: shear stress
µ :viscosity
du :change in velocity and
dy :change in distance
39
𝝉 = 𝝁
𝒅𝒖
𝒅𝒚

Example 5:
𝑑𝑦
 𝜇 = 𝜏
𝑑y
𝑑u
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇
40
Given
data
Upper – lower
Already given the difference

Example 6:
𝑑𝑦
𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 =
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡)
𝑓𝑜𝑟𝑐
𝜇 = 𝑎𝑟𝑒𝑎
𝑑 𝑢
𝑑 𝑦
41
Given
data
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠(𝜏) =
𝐴𝑟𝑒𝑎
𝐴𝑟𝑒𝑎

Example 7:
cos 600 degrees is½
𝑑𝑦
42
𝑓𝑜𝑟𝑐
𝑑 𝑢
𝑑 𝑦
𝐴𝑟𝑒𝑎
Calculate the dynamic viscosity of an oil, which is used for lubrication between a square plate of size 0.8
m x 0.8 m and an inclined plane with angle of inclination 30° as shown in the Fig The weight of the
square plate is 300 N and it slides down the inclined plane with a uniform velocity of 0.3 m/s. The
thickness of oil film is 1.5 mm.
τ = μ
du
dy

Example 8:
44
Given
data

Example 9:
S for Liquids = Weight density (or density) of liquid
Weight density (or density) of water
The density of a liquid = S x weight density ofwater
= S x 1000 kg/m3
𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 (𝜇)
Kinematic Viscosity (v) = density (𝜌)
45
𝑑𝑦
𝑓𝑜𝑟𝑐
𝑑 𝑢
𝑑 𝑦
Given
data
(Only upper given)
(2.5-0 = 2.5)
(98.1-0 = 98.1)
𝐴𝑟𝑒𝑎

Example 10:
46
The density of a liquid = S x weight density of water
= S x 1000 kg/m3
𝑑𝑦
𝑓𝑜𝑟𝑐
𝑑 𝑢
𝑑 𝑦
Given
data

Example 11:
= S x 1000 kg/m3
Given
data

Example 12:
= S x 1000 kg/m3
Given
data

Example 13:
49

Example 14:
minutes tosecond
𝑑𝑦
𝑓𝑜𝑟𝑐
𝑑 𝑢
𝑑 𝑦

Homework 2 :
The gap between two square flat parallel boards (each side of the board is VL cm) is full of bitumen. The thickness of the
bitumen film was 13 mm. The upper board moves 3 m/s and needs a force of 98.1 N to obtain the target speed. Find (i)
the dynamic viscosity of the bitumen in poise, and (ii) the kinematic viscosity of the bitumen in stokes if the specific
gravity of the bitumen is 1.03.
Note:
Value from the list (VL) is the constant number given in the list where every student has different number. Thus,
the final results of the homework for each students will bedifferent.

Homework 3 :
Find practical application on the “fluid mechanics and hydrology” or “reliable sources (references)”.

1.2.3 Thermodynamic Properties
Fluids consist of liquids or gases. Gases are compressible fluids and hence thermodynamic properties play an
important role. With the change of pressure and temperature, the gases undergo large variation in density. The
relationship between pressure (absolute), specific volume and temperature (absolute) of a gas is given by the
equation of state as
𝑝
∀
=
𝑅
𝑇 or
𝑝
𝜌
= 𝑅𝑇
Where:
p = Absolute pressure of a gas in N/m2
∀= Specific volume
R = Gas constant
T = Absolute temperature in 0K
𝜌 = Density of a gas
Gas constant (R):
(R1)Regular case :
The gas constant, R, depends on the particular gas. The dimension of R can be obtained according the
following equation.
𝑅 =
𝑝
54
𝜌𝑇
Absolute pressure of a gas (p) x
1
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 (𝜌)
=
Gas constant (R) x Absolute temperature (T)
p x ∀ = R x T
p x ∀ = m x R x T
m = n x M

Gas constant (R):
(R2) Isothermal Process:
If the changes in density occurs at constant temperature, then the process is called isothermal and
relationship between pressure (p) and density (P) can be directly considered as a gas constant.
(R3) Adiabatic Process:
If the change in density occurs with no heat exchange from the gas, the process is called adiabatic. And if no
heat is generated within the gas due to friction, the relationship between pressure and density can be as the
following equation.
Where
k = Ratio of specific heat of a gas at constant pressure and constant volume.
= 1.4 for air
55
1
=
𝑅 𝑥 1 =
𝑝
𝜌
𝑅 =
𝑝
𝜌
𝑅 =
𝑝
𝜌
𝑘

Gas constant(R):
(R4) Universal Gas constant (in order to be used for cases):
= n x M
56
1
= Gas constant (R) x Absolute temperature (T)

Example 14:
(Celsius to Kelvin)
Mass density =
Specific weight (𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦)
Acceleration due to gravity
57
1
=

Example 16:
𝑅 =
𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑔𝑎𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
p x ∀ = n x M x R x T
p x ∀ = m x R x T
= n x M
58
m
1
=

59
Two volumes due to compression

1.2.4 Compressibility and Bulk Modulus
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝)
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛
−𝑑∀
∀
61

Gases are easiest to compress, solids most difficult
Compressibility in gas, liquid andsolid
https://www.youtube.com/
watch?v=WrM5SQrRTMM
Video 3
62

Relationship between Bulk Modulus (K) and Pressure for a Gas
−𝑑∀
∀
63

−𝑑∀
∀
Example 17:
64
(0.15 have been decreased for each 100)

−𝑑∀
∀
Example 18:
65

1.2.5 Surface tension and Capillarity
Surface tension
Surface tension is a property that results from the attractive forces between
molecules, as such manifests itself only in liquids at an interface usually a
liquid -gas interface.
Surface tension is expressed as force per unit length.
The forces between molecules in the bulk of a liquid are equal in all
directions and as a result no net force is exerted on the molecules.
However at an interface the molecules exert a force that has a resultant in
the interface layer.
Surface tension (σ) = force / length = F/L = N/m
(i) Cohesion:
Cohesion means intermolecular attraction between molecules
of the same liquid. It enables a liquid to resist small amount of
tensile stresses. Cohesion is a tendency of the liquid to remain
as one assemblage of particles.
(ii) Adhesion:
Adhesion means attraction between the molecules of a liquid
and the molecules of a solid boundary surface in contact with
the liquid. The property enables a liquid to stick to another
body.
66

The adhesion of liquid also depends on the solid boundary surface in contact with the liquid.
Video 4
Hydrophobic Surface
67
https://www.youtube.com/watch?v=MU_tPRIrclE
https://www.youtube.com/watch?v=e1b_kP4h_jw

Capillary
Capillary is a phenomenon by which a liquid (depending on its specific gravity) rises into a thin glass tube above or
below its general level.
This phenomenon is due to the combined effects of cohesion and adhesion of liquid particles.
Upward surface tension force (lifting force) = weight of the water column in the tube (gravity force)
71

Capillary
𝜃
𝜃
p
p
D
ℎ =
4 𝛔 cos 𝜃
𝜌 𝑔 𝑑
h = hight of the liquid in the tube
𝛔 = surface tension of liquid
𝜃 = Angle of contact between liquid and glass tube
𝜌 = Density of liquid
d = glass tube diameter
g = acceleration due to gravity
Note:
72

Capillary Geosynthetics with enhanced lateral drainage capabilities in roadway systems (Zornberg et al. 2017)
76

Capillary Geosynthetics with enhanced lateral drainage capabilities in roadway systems (Zornberg et al. 2017)
77

Example 21:
ℎ =
4 𝛔 cos 𝜃
𝜌 𝑔 𝑑
The density of Liquid = S x density of water
78

Example 22:
ℎ =
4 𝛔 cos 𝜃
𝜌 𝑔 𝑑
79

Example 23:
ℎ =
4 𝛔 cos 𝜃
𝜌 𝑔 𝑑
80

Example 24:
ℎ =
4 𝛔 cos 𝜃
𝜌 𝑔 𝑑
81

1.2.6 VAPOUR PRESSURE AND CAVITATION
A change from the liquid State to the gaseous state is known as vaporization. The vaporization (which depends upon
the prevailing pressure and temperature condition) occurs because of continuous escaping of the molecules through
the free liquid surface.
Consider a liquid (say water) which is confined in a closed vessel. Let the temperature Of liquid is and pressure is
atmospheric. This liquid will vaporise at 1000C. When vaporization takes place, the molecules escapes from the free
surface of the liquid. These vapour molecules get accumulated in the space between the tree liquid surface and top of
the vessel. These accumulated vapours exert a pressure on the liquid surface. This pressure is known as vapour
pressure of the liquid. Or this is the pressure at which the liquid is converted into vapours.
Again consider the same liquid at 200C at atmospheric pressure in the closed vessel. If the pressure above the liquid
surface is reduced by some means, the boiling temperature will also reduce. If the pressure is reduced to such an
extent that it becomes equal to or less than the vapour pressure, the boiling of the liquid will start, though the
temperature of the liquid is 200C. Thus a liquid may boil even at ordinary temperature. if the pressure above the liquid
surface is reduced so as to be equal or less than the vapour pressure of the liquid at that temperature.
Now consider a flowing liquid in a system. If the pressure at any point in this flowing liquid becomes equal to or less
than the vapour pressure, the vaporization of the liquid starts. The bubbles of these vapours are carried by the flowing
liquid into the region of high pressure where they collapse. giving rise to high impact pressure. The pressure developed
by the collapsing bubbles is so high that the material from the adjoining boundaries gets eroded and cavities are
formed on them. This phenomenon is known as cavitation.
Hence the cavitation is the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the
pressure of the liquid falls below the vapour pressure and sudden collapsing of these vapour bubbles in a region of
higher pressure. When the vapour bubbles collapse, a very high pressure is created. The metallic surfaces. above which
the liquid is flowing. is subjected to these high pressures. which cause pitting action on the surface. Thus cavities are
formed on the metallic surface and hence the name is cavitation. 82

Homework 3 :
83
Find the height of the liquid in the tube with VL mm diameter, when submerged in (a) mercury, and (b) water. the
values of the surface tension of water in contact with air is 0.08 N/m, and it is 0.60 for mercury when their
temperature are 20oC. In addition, the angle of contact for mercury is 20.
Note:

2 Pressure and its measurement (Fluid Statics)
2.1 Fluid pressure at a point
Consider a small area in large mass of fluid. If the fluid is stationary, then the force exerted by the surrounding
fluid on the area dA will always be perpendicular to the surface dA. Let dF is the force acting on the area dA in the
normal direction. Then, the ratio of is known as the intensity of pressure or simply pressure and this ratio is
represented by p. Hence, mathematically the pressure at a point in a fluid at restis
If the force (F) is uniformly distributed over the area (A),
then pressure at any point is givenby
The pressure of a fluid on a surface will always
act normal to the surface. Pressure unit =
N/m2 or Pa orbar
1 kPa = 1 kN/m2 = 1000 Pa = 1000 N/m2
1 bar = 100 kPa
Concrete dam
85
𝑝 =
𝑑𝐹
𝑑𝐴
𝑝 =
𝑓𝑜𝑢𝑟𝑐𝑒 (𝐹)
𝐴𝑟𝑒𝑎 (𝐴)

2 Pressure and its measurement (Fluid Statics)
2.1 Fluid pressure at a point
Consider a small area in large mass of fluid. If the fluid is stationary, then the force exerted by
the surrounding fluid on the area dA will always be perpendicular to the surface dA. Let dF is
the force acting on the area dA in the normal direction. Then, the ratio of is known as the
intensity of pressure or simply pressure and this ratio is represented by p. Hence,
mathematically the pressure at a point in a fluid at rest is
𝑝 = 𝑑𝐹
𝑑
𝑝 = 𝑓𝑜𝑢𝑟𝑐𝑒 (𝐹)
If the force (F) is uniformly distributed over the area (A), then pressure at any point is givenby
The pressure of a fluid on a surface will always act normal to the surface. Pressure unit = N/m2 or Pa or bar
1 kPa = 1 kN/m2 = 1000 Pa = 1000 N/m2 1 bar = 100 kPa
❑ Statics
• Hydrostatic: the study of incompressible fluid under static condition.
• Aerostatic: the study of compressible gases under static condition.
❑ Dynamics
It deals with the relations between velocities, acceleration of fluid with the
❑ Kinematics
It deals with the velocity, accelerations and patterns of flow and not the
This chapter
86

2.2 PASCAL'SLAW
It states that pressure or intensity pressure at a pint in a static fluid is equal in all directions. This is proved as: The
fluid element is of very small dimensions i.e., dx, dy and dz.
The intensity of pressure at any point in a liquid at rest, is the same in all directions.
Proof : Consider very small wedge shaped element of liquid (LMN)
px=intensity of horizontal pressure on the element of the
liquid py=intensity of vertical pressure on the element of
the liquid
pz=intensity of pressure on the diagonal of the right angled triangular element
α=Angle of the element of the liquid
Px=Total pressure on the vertical side LN = px *LN
Py=Total pressure on the horizontal side MN = py
*MN Pz=Total pressure on the diagonal LM = pz
*LM
87

2.3 Pressure variation a fluid at rest
A liquid is subjected to pressure due to its own weight, this pressure increase as the depth of
liquid increases.
Specific weight (w) =
mass density (𝜌) x Acceleration due to gravity (g)
Z
88
Z * w
Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (Z)
pressure
w
Z or H (the same)

Video 5
89
https://www.youtube.com/watch?v=K5g6P8-GmBg

https://www.youtube.com/ watch?v=hgjAgz1iswo
Video 6
90
Pressure of liquid does not
depend on the size, shape
and surface area of
container as long as the
depth f water the same.

91
https://www.youtube.com/watch?
v=iVddJWB9rXs
Video 7

Example 25:
93
𝑝 =
𝑓𝑜𝑟𝑐𝑒 (𝐹)
𝑝 =

Example 26:
p1 =
= 𝑝2 = 𝑓𝑜𝑟𝑐𝑒 (𝐹)
p1
p2
PASCAL'S LAW
94

Example 27:
Pressure (p) =
mass density (𝜌) x Acceleration due to gravity (g) x Height (H)
Mass density = S x density of water
95

Example 28:
Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H)
96

Example 29:
97
PASCAL'S LAW

Example 30:
Mass density = S x density of water 98

2.4 Absolute, gauge, Atmospheric and vacuum pressures
Absolute pressure = Atmospheric pressure + Gauge pressure
99

2.5 Measurement of pressure
Manometers Mechanical gauges
Simple Manometers
Differential Manometers
Diaphragm pressure gauge
Dead-weight pressure gauge
Bourdon tube pressure gauge
Bellows pressure gauge
Measurement of pressure
Piezometer
U-tube Manometer
Single column Manometer
U-tube Differential Manometer
Inverted U-tube Differential Manometer
100

2.6 Simple Manometers
101

𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2
102
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0

Example 31:
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0
103
Consider the sp. Gr. Of mercury is 13.6.

Example 32:
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0
104

Example 33:
105
𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2

Homework 4 :
120
The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the
left limb is connected to a pipe in which a fluid of sp. gr.
1.0 is flowing. The center of the pipe is VL cm below the level of mercury in the right limb. Find the
pressure of fluid in the pipe if the difference of mercury level in the two limbs is 30 cm.
Note:
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0

Quiz
Q1: Define the following three branched of Fluid Mechanics
Statics, Kinematics and Dynamics
Q2: what is the difference between Kinematic Viscosity and dynamic Viscosity?
Q4:
Q3:
The right limb of a simple U-tube manometer containing mercury is open to the
atmosphere while the left limb is connected to a pipe in which a fluid of sp. gr. 1.0 is
flowing. The center of the pipe is VL cm below the level of mercury in the right limb. Find
the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 25 cm.
Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H) 𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2
𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0
25 cm
−𝑑∀
∀

3 Buoyancy and floatation
3.1 Introduction
Submarine
3.2 Bouncy
123

FB = buoyancy force = WABCKD –WABCHD
The buoyancy force on a submerged body is equal to the weight of the fluid it displaces
W > FB
γs . V > γw . V
γs > γw
W = FB
γs . V = γw . V
γs = γw
W < FB
γs . V < γw . V
γs < γw
W < FB
γs . V < γw . V
γs < γw
Specific weight (w) = mass density (𝜌) x Acceleration due to gravity (g)
FB = Specific weight of fluid * Volume displaced = γ . V
124

3.3 Centre of Bounce
Specific weight or weight density (w) =
mass density (𝜌) x Acceleration due to gravity(g)
Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) =
Specific weight of fluid * Volume displaced = γ . V
Volume of displaced liquid=
Weight of displaced liquid
Weight density
Position of center of Buoyancy:
Volume of water displaced = volume of object inwater
Centre of Bounce
What is the difference between
center of mass and center of volume?
2 m
2 m
4 m
1 m
125
What if it is Solid or Hollow?

Example 43:
Volume of displaced liquid = Weight of displaced liquid
Weight density
Position of center ofBuoyancy:
Volume of water displaced = volume of wooden block in water
12
6
1.5 m
2.5 m
6.0 m

Example 44:
Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g)
FB = Specific weight of fluid * Volume displaced = γ .V
Weight density
127

Weight density
120

Application of the Buoyancy and floatation
The longest rail ‘n’ road bridge (in Europe)
The Oresund Bridge stretches from Copenhagen in Denmark to Malmö in Sweden, allowing cars
and trains to cross the 16km stretch of water which separates one country from the other. And if
the image above looks like an optical illusion it’s because the 2-track railway and 4-lane highway
abruptly descend into the sea as the bridge becomes a tunnel for the last stretch. Is it a tunnel or
a bridge (a brunnel)? Either way it’s the continent’s longest road and rail bridge by a mile!
129

World's First Floating Tunnel Project In Norway:
130
https://www.youtube.
com/watch?v=Gt1MD
QHLjF8

Practical application of the Fluid Mechanics and Hydrology
‫بكات‬ ‫دروست‬ ‫لواسراو‬‫هە‬ ‫پردى‬ ‫كە‬ ‫بوو‬‫نە‬ ‫ى‬‫وه‬‫ئە‬ ‫ى‬‫پاره‬ ‫ت‬‫حكومە‬ ‫الم‬‫بە‬ ‫دروستكرا‬ ‫يۆنان‬ ‫لە‬ ‫جسره‬ ‫و‬‫ئە‬
(
‫بريدج‬ ‫سپينشن‬‫سە‬
)
‫بوو‬ ‫خراپ‬ ‫زۆر‬ ‫شى‬‫كە‬‫گلە‬ ‫بوو‬ ‫قول‬ ‫زۆر‬ ‫ش‬‫كە‬‫ئاوه‬ ‫و‬‫ئە‬ ‫وه‬
,
‫بە‬
‫جۆرى‬ ‫پردى‬ ‫و‬ ‫ن‬‫بكە‬ ‫كۆنكريت‬ ‫كە‬‫ئاوه‬ ‫ناو‬‫لە‬ ‫توانرا‬‫ده‬ ‫نە‬ ‫ك‬‫يە‬‫شێوه‬ ‫هيج‬
(
‫يد‬‫ستە‬ ‫يبل‬‫كە‬
)
‫ن‬‫بكە‬
.
‫كرديان‬ ‫دواتر‬ ‫الم‬‫بە‬
,
‫بوو‬ ‫جۆره‬ ‫م‬‫بە‬ ‫كە‬‫فكره‬ ‫كورتى‬ ‫بە‬ ‫چۆن؟‬
:
‫خر‬ ‫بۆشى‬ ‫ناو‬ ‫كي‬‫كۆنكريتيە‬ ‫كوتلە‬ ‫كە‬‫ئاوه‬ ‫رۆخى‬ ‫لە‬
90
‫ب‬ ‫واتا‬ ‫وه‬‫بێتە‬ ‫رز‬‫بە‬ ‫ى‬‫وه‬‫ئە‬ ‫بۆ‬ ‫كە‬‫بۆشە‬ ‫ناو‬ ‫كۆنكريتيە‬ ‫كوتلە‬ ‫رى‬‫وروبە‬‫ده‬ ‫خستە‬ ‫ئاويان‬ ‫پاشان‬ ‫كرد‬ ‫دروست‬ ‫ريان‬‫ميتە‬‫دايە‬ ‫تر‬‫مە‬
‫نا‬
‫ى‬‫غە‬
‫م‬‫لە‬‫بە‬ ‫بە‬ ‫پاشان‬ ‫و‬ ‫وێت‬‫بكە‬ ‫ئاو‬ ‫ر‬‫سە‬‫بە‬ ‫ى‬‫وه‬‫ئە‬ ‫بۆ‬ ‫كرد‬ ‫دروستيان‬ ‫مێك‬‫لە‬‫بە‬ ‫ى‬‫شێوه‬ ‫لە‬ ‫كە‬‫جسره‬
(
‫پاپۆر‬
)
‫ست‬‫بە‬‫مە‬ ‫شوێنى‬ ‫بۆ‬ ‫وه‬‫گواستيانە‬
....
‫ت‬ ‫كانى‬‫كاره‬ ‫بوو‬‫هە‬ ‫پيلۆنت‬ ‫كە‬ ‫جسريشا‬ ‫لە‬ ‫ئيتر‬
‫ئاسان‬ ‫ر‬
‫بێت‬‫ئە‬
.
RION-ANTIRION
BRIDGE(Greece)
After transported
precast foundation to
exact location then
they continued work
on rest of structure;
Each steps increased
on weight of structure
and it effected to sink
down step by step.

Example 45:
mass density (𝜌) x Acceleration due to gravity (g)
392.4
196.2
Weight of water displaced =
weight of object in air – weight of object inwater
One kilogram is equal to 9.81 Newtons
Volume of fluid
Sp. gr. = Density of object
Density ofwater
13
2
Note: This example is useful to find volume and
weight of concrete cubs in laboratory

Example 46:
Weight density
They are the same
Weight in water=
weight of object in air - weight of object displaced
One kilogram is equal to 9.81 Newtons
133
1.5 m
1.0 m
2.0 m

Example 47:
Water
Mercury
Object
13
4

Example 48:
Weight density
135

Weight density
136

4.4. Meta-centre
137
G – center of gravity
B – center of buoyancy
M – Meta center

4.4. Meta-centric hight
The metacentric height (GM) is a measurement of the initial static stability of a floating body. It is calculated
as the distance between the center of gravity of an object (e.g. ship) and its metacenter.
130
M – Meta center

∀
140
𝐺𝑀 =
𝐼
− 𝐵𝐺
𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵
M – Meta center

3.6
141
Conditions of Equilibrium of a floating and sub-merged body

-For stable equilibrium, the position of metacenter (M) remains
higher than center of gravity of the body (G).
-For unstable equilibrium, the position of metacenter (M)
remains lower than (G).
-For neutral equilibrium, the position of metacenter (M)
coincides with (G)
Metacenter height (GM) = BM ± BG
+ ve sign : when G is lower than B
- ve sign : when G is higher than B
Briefly, metacenter height is the distance between the
center of gravity of a floating body and the metacenter.
142
M – Meta center

Example 49:
5 m
1.2m
3 m
∀
145
𝐺𝑀 =
𝐼
− 𝐵𝐺
(1.2/2)
(+) For stable equilibrium, the
position of metacenter (M)
remains higher than center of
gravity of the body (G).

Example 50:
∀
𝐺𝑀 =
I
− 𝐵𝐺 𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵
3 m
2 m Water level
1 m 0.8 m
147

Example 51:
∀
𝐺𝑀 =
𝐼
− 𝐵𝐺
Specific weight of fluid * Volume displaced = γ . V Wood
Specific weight of fluid * Volume displaced = γ . V Water
= ?
14
9
Considered it all

Example 52:
4 m
4 m
∀
𝐺𝑀 =
𝐼
−𝐵𝐺
150
Main
differences
A solid cylinder of diameter 4.0 m has a height of 4.0 m. Find the meta-centric height of the cylinder if the specific
gravity of the material of cylinder = 0.6 and it is floating in water with its axis vertical. State whether the equilibrium is
stable or unstable.

3.7 Experimental method of determination of Meta-centricheight
152

VL m
2 m
5 m
0.9m
Water level
Homework 5 :
Note:
A rectangular pontoon of size VL m long x 5 m wide x 2 m deep floats in water to support a bridge build on
river. (i) What is the weight of the body if depth of immersion is 0.9 m ? (ii) In order to confirm the stability of
the bridge, determine the meta-centric height. (iii) In addition, confirm whether the bridge is stable or not,
and explain the reason.
∀
𝐺𝑀 =
1
− 𝐵𝐺
Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 =
Buoyancy force (FB) = Specific weight of
fluid * Volume displaced = γ . V
155

Chapter Four
04.
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Fluid Mechanics and Hydrology _ Dr.
Rawaz Kurda
15
6
Kinematics of flow and Fluid Resistance
Introduction
Methods of describing fluid motion
Type of fluid flow
Rate of flow or discharge
Continuity Equation
Continuity equation in three dimensions
Velocity and acceleration
Velocity potential function and stream function

Kinematics of flow and Fluid Resistance
5.1 INTRODUCTION
Kinematics defined as that branch of science which deals with motion of particles without
considering the forces causing the motion. The velocity at any point in a field at any time is
studied in this branch of fluid mechanics. Once the velocity is known, then the pressure
distribution and hence forces acting on the fluid can determined. In this chapter, the
methods of determining velocity and acceleration are discussed.
❑ Kinematics
It deals with the velocity, accelerations and patterns of
flow and not the forces or energy causing them.
❑ Statics
• Hydrostatic: the study of incompressible fluid under
static condition.
• Aerostatic: the study of compressible gases under
static condition.
❑ Dynamics
It deals with the relations between velocities,
acceleration of fluid with the
This
chapter
157

5.2 Methods of Describing fluid motion
150
The fluid motion is described by two methods. They are (i) Lagrangian method, and (ii) Eulerian Method.
In the Lagrangian method, a single fluid practice is followed during its motion and velocity, acceleration,
density, etc. are described.
In case of Eulerian method, the velocity, acceleration, pressure, density etc. are described at a point in
flow field. The Eulerian method is commonly used in fluid mechanics.
3. Types of fluid flow
Fluid flow is classified as:
1. Steady and unsteady flows
2. Uniform and non-uniform flows
3. Laminar and turbulent flows
4. Compressible and incompressible flows
5. Rotational and irrotational flows
6. One, two and three-dimensional flows.

5.3.1 Steady and Unsteady Flows
• Steady flow is defined as that type of flow in which
the fluid characteristics like velocity, pressure,
density. etc. at a point do not change with time. Thus
for steady flow, mathematically. we have
• Unsteady flow is that type of flow, in which the velocity,
pressure, density at a point changes with respect to
time. Thus, mathematically, for unsteady flow
where (xo, yo, Zo) is a fixed point in fluid field.
15
9

5.3.2 Uniform and Non-uniform Flows
Uniform flow is defined as that type of now in which the velocity at any given time does not change with
respect to space (i.e. , length of direction of the flow). Mathematically, for uniform flow as
where 𝞉𝑣 = Change of velocity
𝞉𝑠 = Length of flow in the direction S.
Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 160
Non-uniform flow is that type of flow in which
the velocity at any given time changes with
respect to space. Thus, mathematically, for
non-uniform flow

5.3.3 Laminar and TurbulentFlows
• Laminar flow
A flow is said to be laminar if the fluid particles move in layers such that one
layer of the fluid slides smoothly over an adjacent layer. The viscosity property
of the fluid plays a significant role in the development of a laminar flow. The
flow pattern exhibited by a highly viscous fluid may in general be treated as
laminar flow.
• Turbulent flow
If the velocity of flow increases beyond a certain value, the flow becomes
turbulent. The movement of fluid particles in a turbulent flow will be random.
This mixing action of the colliding fluid particles generates turbulence, thereby
resulting in more resistance to fluid flow and hence greater energy losses as
compared to laminar flow.
a zig-zagway
161
For a pipe flow, the type of flow is determined by a non-dimensional number
𝑉𝑑
𝑣
called the Reynold number.
Where:
D = Diameter of pipe
V Mean velocity of flow in pipe
v = Kinematic viscosity of fluid.
If the Reynold number is less than 2000, the flow is called laminar. If the Reynold number is more
than 4000, it is called turbulent flow. If the Reynold number lies between 2000 and 4000, the flow may be laminar or
turbulent.

5.3.4 Compressible and Incompressible Flows
• Incompressible
In case of in compressible fluid flow, the density of the fluid remains
constant during the flow. (i. e. 𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡). Practically, all liquids are
treated as incompressible.
• Compressible
The flow in which density of the fluid varies during the flow is called
compressible fluid flow. (i. e. 𝜌 ≠ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡). This is applicable in gas flow.
It can be seen that the control volume remains constant for a flow that is
incompressible and control volume is squeezed for compressible flow.
Bernoulli's equation is applicable only when flow is assumed to be incompressible. In
case of compressible flow, Bernoulli's equation becomes invalid since the very basic
assumption for Bernoulli's equation is density r is constant
2
For compressible flow, p + 1
𝜌𝑉2

5.3.5 Rotational and IrrotationalFlows
Rotational flow is that type of flow in which the fluid particles while flowing along stream-lines, also rotate about
their own axis. And if the fluid particles while flowing along stream-lines, do not rotate about their own axis that
type of now is called irrotational flow.
Video - Whirlpool

5.3.6 One, Two and Three-Dimensional Flows
One-dimensional flow
One-dimensional flow is that type of flow in which the flow parameter such as velocity is a function of time and one space co-ordinate only, say x. For a
steady one-dimensional flow, the velocity is a function of one-space-co-ordinate only. The variation of velocities in other two mutually perpendicular
directions is assumed negligible.
Hence mathematically, for one-dimensional flow
u = f(x), v = 0 and w=0 where u, v
and w are velocity components in x, y and z directions respectively.
Examples: flow in pipe where average flow parameters considered for analysis.
Two-dimensional flow
Two-dimensional flow is that type of now in which the velocity is a function of time and two rectangular space co-ordinates say x and y. For a steady two-
dimensional flow the velocity is a function of two space co-ordinates only. The variation of velocity in the third direction is negligible. Thus, mathematically
for two-dimensional flow.
u = f1(x,y), v = f2(x,y) and w=0
Examples: (i) flow between parallel plates of infinite extent.
(ii) flow in the main stream of a wide river.
Three-dimensionalflow
Three-dimensional flow is that type of flow in which the velocity is a function of time and three mutually perpendicular directions. But for a steady three-
dimensional flow the fluid parameters are functions Of three space co-ordinates (x. y and z) only. Thus, mathematically. for three- dimensional flow.
u = f1(x,y,z), v = f2(x,y,z) and w= f3(x,y,z)
Examples: (i) flow in a converging or diverging pipe or channel
(ii) flow in a prismatic open channel in which the width and the water depth are of the same order of magnitude.
164

5.4 RATE OF FLOW OR DISCHARGE (Q)
It is defined as the quantity of a fluid flowing per second through a section of a pipe or a channel.
For an incomrressible fluid (or liquid) the rate of flow or discharge is expressed as the volume of fluid flowing across
the section per second.
For compressible fluids, the rate of now is usually expressed as the weight of fluid flowing across the section. Thus,
(i) For liquids the units of Q are m3/s or liters/s
(ii) For gases the units Of Q is kgf/s or Newton/s. Consider a liquid flowing
through a pipe in which A = Cross-sectional area of pipe and V = Average velocity
of fluid across the section
Then discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V)
V1 𝜌1 A1 = V2 𝜌2 A2
The above Equation is applicable to the compressible as well as incompressible fluids and is called Continuity
Equation. If the fluid is incompressible, then 𝜌1 = 𝜌2 and continuity equation reduces topipe.
V1 A1 = V2 A2
Direction of flow
Fluid flowing through apipe
165
5.5 CONTINUITY EQUATION
The equation based on the principle of conservation of mass is called continuity
equation. Thus for a fluid flowing through the pipe at all the cross- section, the
quantity of fluid per second is constant.
Consider two cross-sections of a pipe as shown in the Fig. Let
V1 = Average velocity at cross-section 1-1
𝜌1 = Density at section 1-1
A1 = Area of pipe at section1-1
and V2, 𝜌2, A2 are corresponding valves at section, 2-2. Then rate of flow at section 1-1 = V1 𝜌1 A1 Rate of flow at
section 2-2 = V2 𝜌2 A2
According to law of conservation of mass, Rate of flow at section 1-1 = Rate of flow at section 2-2

DISCHARGE (Q)
V= Average velocity
𝜌= Density at section
A = Area of pipe

Example 56:
167
V1 𝜌1 A1 = V2 𝜌2 A2
Discharge (Q) = Cross-sectional area of pipe (A) *
Average velocity of fluid across the section (V)

Example 57:
V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3
160
A 30 cm diameter pipe, conveying water, branches into two pipes of diameters20 cm and 15 cm
respectively. If the average velocity in the 30 cm diameter pipe is 2.5 m/s, find the discharge in this
pipe. Also determine the velocity in 15 cm pipe if the average velocity in 20 cm.

Example 58:
169
V1 𝜌1 A1 = V2 𝜌2 A2

170
Source
Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V)
V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3
10.3 Bernoulli’s equation
Public
water
system
-
Dead
end
Fluid Mechanics and Hydrology_ Dr. Rawaz Kurda
/12

171
Source 1
V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3
10.3 Bernoulli’s equation
Public
water
system
-
Dead
end
/12
Source 2

172
Source
Average velocity of fluid across the section (V) =
Discharge (Q)
Cross−sectional area of pipe (A)
Public
water
system
-
Dead
end
/12

173
Source
V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3
Public
water
system
-
continues
/12

4.6 Continuity equation in threedimensions
174

Example 59:

4.7 Velocity and acceleration

Local acceleration
The local acceleration captures the change rate of velocity of a certain particle with respect to
time and vanishes if its flow is steady.
Convective acceleration
The convective acceleration, on the other hand, captures the change of velocity flow in the spatial
space and, therefore, it increases when particles move through the region of spatially varying
velocity.
In this case, one can say that the local acceleration characterizes the particle velocity field in the
temporal domain, while the convective acceleration represents the velocity change due to the
spatial variation of the flow particle along its trajectory.

Example 60:
180

Example 61:
170

Example 62:
171

_ Dr. Rawaz Kurda 172

4.8 Velocity potential function and stream function
Velocity potential function and stream function are two scalar functions that help study whether the given fluid
flow is rotational or irrotational. Both the functions provide a specific Laplace equation. The fluid flow can be
rotational or irrotational flow based on whether it satisfies the Laplace equation or not.
Relation between Stream Function and Velocity Potential Function
Rawaz Kurda
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5

Homework 6 :
Note:
A 45 cm diameter pipe, conveying water, branches into THREE pipes Of diameters 20 cm and 15 cm, 10 cm,
respectively. If the average velocity in the 45 cm diameter pipe is VL m/s, find the discharge in this pipe. Also
determine the velocity in 10 cm pipe if the average velocity in 20 cm and 15 cm pipes are 4 m/s and 6 m/s,
respectively.
V3 = 6
m/s D3 =
15 cm
V4 = ?
m/s D4 =
10 cm
V1 = VL
m/s D1 =
45cm
V2 = 4 m/s
D2 = 20 cm

Example 57:
V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3
160

Homework 6 :
Note:
A 45 cm diameter pipe, conveying water, branches into THREE pipes Of diameters 20 cm and 15 cm, 10 cm,
respectively. If the average velocity in the 45 cm diameter pipe is VL m/s, find the discharge in this pipe. Also
determine the velocity in 10 cm pipe if the average velocity in 20 cm and 15 cm pipes are 4 m/s and 6 m/s,
respectively.
V4 = 25 m/s
D4 = 15 cm
V5 = ? m/s
D5 = 10 cm
V1 = 5 m/s
D1 = 40cm
V3 = 20 m/s
D3 = 20 cm
V2 = 7 m/s
D2 = 35cm

What are the four different types of water supply
distribution system?
The aim of a distribution network is to supply a
community with the appropriate quantity and quality of
water. There are four network types: dead end, gridiron,
circular and radial systems

‫بێت؟‬ ‫داخيل‬ ‫کۆتا‬ ‫محازەرەى‬ ‫دوو‬
‫يان‬
‫بێت؟‬ ‫داخيا‬ ‫هەمووى‬

Contents:
05.
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
194
Dynamics of fluid flow
Introduction Equation of Motion
Euler’s equation of motion
Bernoulli’s equation form Euler’s equation
Assumptions
Bernoulli’s Equation for real fluid
Practical applications of Bernoulli’s equation
The Momentum Equation
Kinetic energy correction factor

5. Dynamics of fluidflow
5.1 Introduction
❑ Dynamics
It deals with the relations between velocities, acceleration
of fluid with the forces or energy causing them.
❑ Kinematics
It deals with the velocity, accelerations and patterns of flow and
not the forces or energy causing them.
❑ Statics
• Hydrostatic: the study of incompressible fluid under static condition.
• Aerostatic: the study of compressible gases under static condition.
This chapter
Rawaz Kurda
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5

5.2 Equation of Motion
Bernoulli’s equation

5.3 Euler’s equation of motion
Derivative
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7

5.4 Bernoulli’s equation form Euler’s equation
5.5 Assumptions
Total head (energy) = Pressure head + Kinetic head + datum head
=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
Integration Euler’s equation Derivative
Bernoulli’s equation
Rawaz Kurda
19
8

=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
potential head or datumhead
𝑣2
𝑝
From Bernoulli's principle, the total energy at a
given point in a fluid is the energy associated
with the movement of the fluid (2𝑔
), plus
energy from static pressure in the fluid (𝜌𝑔
),
plus energy from the height of the fluid relative
to an arbitrary datum (constant).
Rawaz Kurda
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Rawaz Kurda
20
0

Example 63:
=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
Rawaz Kurda
20
1

Example 64:
Total head (energy) = Pressure head +
𝑝
𝜌
= +
Kinetic head
𝑣2
2𝑔
+ datum head
+ constant
D1 = 20cm
V1 = 40m/sec
D2 = 10cm
?
Rawaz Kurda
20
2

Example 65:
Total head (energy) = Pressure head + Kinetic head + datumhead
=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
?
Rawaz Kurda
20
3

Example 65:
Total head (energy) = Pressure head + Kinetic head + datumhead
=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
Rawaz Kurda
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4

Example63:
=
𝑝
𝜌
+
𝑣2
2𝑔
Rawaz Kurda
20
5
+ constant

=
𝑝
𝜌
+
𝑣2
2𝑔
+ constant
Rawaz Kurda
20
6

=
𝑝1
+ 𝑣 2
1
𝜌𝑔 2𝑔 1 𝐴
+ 𝑧 + ℎ =
𝑝2
+
𝑣2
2
𝜌𝑔 2𝑔 2
+ 𝑧 + ℎ𝐿
5.6 Bernoulli’s Equation for real fluid
Bernoulli’s equation earlier derived was
based on the assumption that the fluid is
non-viscous and therefore friction-less .
Practically all fluid are real (and not
ideal) and therefore are viscous and such
there are always some losses in fluid
flows. This losses have, therefore , to be
taken into consideration in the
application of Bernoulli's equation which
gets modified (between section 1 and 2)
for real fluid as follows:
Where
hA = energy added between section 1 and 2, and = considered as zero (insignificant)
hL = loss of energy between section 1 and 2
Rawaz Kurda
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7

Example 64:
𝑝1
= +
𝑣1
2
+ 𝑧 + ℎ =
𝑝2
+
Rawaz Kurda
20
8
𝑣2
2
𝜌 𝑔 2𝑔 2
+ 𝑧 + ℎ𝐿

Example 64:
𝑝1
= +
𝑣1
2
+ 𝑧 + ℎ =
𝑝2
+
Rawaz Kurda
190
𝑣2
2
𝜌𝑔 2𝑔 2
+ 𝑧 + ℎ𝐿

5.7 Practical applications of Bernoulli’s equation
Rawaz Kurda
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0

Rawaz Kurda
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1

Rawaz Kurda
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2

Rawaz Kurda
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3

Rawaz Kurda
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4

Rawaz Kurda
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5

Rawaz Kurda
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6

Rawaz Kurda
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7

Project
Rawaz Kurda
21
8
Relationship Between Highway Applications and Fluid Mechanics and Hydrology
Report
Design of multiple highway projects on/in water with the use of Fluid Mechanics
and Hydrology
Introduction
Methodology
Results and discussion
Conclusion

Contents:
220
Fundamentals of Engineering Hydrology
1. Introduction to hydrology
2. Domain and objective of Engineering hydrology
3. Practical applications of hydrology
4. Hydrologic cycle
5. Steps of the Hydrologic Cycle
6. Measurement techniques
7. Estimated world water quantities
8. Hydrologic budget in details
9. Test for consistency of record
10. Analysis of Precipitation Records
11. Analysis of Evaporation losses

6.1 Introduction to hydrology
Study of the hydrologic cycle;
occurrence, distribution,
movement, physical and chemical
properties of waters of the earth
and their environmental
relationships.
6.2 Domain and objective
of Engineering hydrology
The role of applied hydrology is to
help analyze the problems involved
in these tasks shown in the figure
( → ) and to provide guidance for
the planning and management of
water resources.

6.3 Practical applications of hydrology
Practical applications of hydrology are found in such tasks as the design and operation of (I-XIII):
1. Hydraulic structures
2. Water supply
3. Wastewater treatment and disposal
4. Irrigation
5. Drainage
6. Hydropower generation
7. Flood control
8. Navigation
9. Erosion and sedimentcontrol
10. Salinity control
11. Pollution abatement
12. Recreational use of water
13. Fish and wildlife protection
Hydraulic structures

➢ Water supply

➢ Wastewater treatment and disposal
Rawaz Kurda
22
4

➢ Irrigation
➢ Hydropower generation
USA
➢ Recreational use of water and fish and
wildlife protection
Kuriqi et al.(2019)
225

➢ Flood control
https://www.youtube.com/watch?v=eUKd5K4E6mM

➢ Drainage
➢ Navigation
227

Porous Asphalt Demonstration https://www.youtube.com/watch?v=I16WGau3jxE

https://www.youtube.com/watch?v=5rsODZXPyCY
Permeable Paving in Action
https://www.youtube.com/watch?
v=JOIM2yR_B6k

➢ Drainage

➢ Erosion and sediment control
➢ Pollution abatement
231

6.4 Hydrologic cycle
The hydrologic cycle is known as the Global
water cycle or the H2O cycle describes the
storage and circulation of water between the
biosphere, atmosphere, lithosphere, and the
hydrosphere. In other words, it is the continuous
movement of water on, above and below the
surface of the Earth in different forms.
6.5. Steps of the Hydrologic Cycle
1. Evaporation (E) + Transpiration (T) = Evapotranspiration (ET)
2. Precipitation (P)
3. Infiltration (F)
4. Runoff (R)
Precipitation = Evapotranspiration + Infiltration + Runoff
P = ET + F + R

1.Evaporation is the change of state of
water (a liquid) to water vapor (a gas). On
average, about 47 inches (120 cm) is
evaporated into the atmosphere from the
ocean each year.
2.Transpiration is evaporation of liquid
water from plants and trees into the
atmosphere. Nearly all (99%) of all water
that enters the roots transpires into the
atmosphere.
3.Sublimation is the process where ice
and snow (a solid) changes into water
vapor (a gas) without moving throughthe
liquid phase.
4.Condensation is the process where
water vapor (a gas) changes into water
droplets (a liquid). This is when webegin
to see clouds.
5.Transportation is the movement of
solid, liquid and gaseous water through
the atmosphere. Without thismovement,
the water evaporated over the ocean
would not precipitate over land.
6.Precipitation is water that falls to the
earth. Most precipitation falls as rain but
includes snow, sleet, drizzle, and hail. On
average, about 39 inches (980 mm) of
rain, snow and sleet fall each year around
the world.
١
.
‫ئاسايى‬ ‫بوونى‬ ‫هەلم‬ ‫بە‬
_________
٢
.
‫دارەوە‬ ‫رێگەى‬ ‫لە‬ ‫بوون‬ ‫هەلم‬ ‫بە‬
___
٣
.
‫هەلم‬ ‫بۆ‬ ‫سەهۆل‬ ‫گۆرانى‬
,
٤
.
‫چربوونەوە‬
(
‫ئاو‬ ‫بۆ‬ ‫هلەم‬ ‫گۆرانى‬
)
_
٥
.
‫هەور‬ ‫جوالنى‬
_____________
٦
.
‫بارين‬ ‫باران‬

7.Deposition is the reverse of
sublimation (3). Water vapor (a gas)
changes into ice (a solid) without
going through the liquid phase. This
is most often seen on clear, cold
nights when frost forms on the
ground.
8.Infiltration is the movement of
water into the ground from the
surface. Percolation is movement of
water past the soil going deep into
the groundwater.
9.Surface flow is the river, lake, and
stream transport of water to the
oceans. Groundwater is the flow of
water under- ground in aquifers.
The water may return to the surface
in springs or eventually seep into
the oceans.
10.Plant uptake is water taken
from the groundwater flow and soil
moisture. Only 1% of water the
plant draws up is used by the plant.
The remaining 99% is passed back
into the atmosphere.
234

6.6 Measurement techniques
Precipitation Evapotranspiration ➔ discharge ➔ flow depth (stage)
Surface waters (velocity)
Note: This methods will be discussed in the next slides →

6.7 Estimated world water quantities
Item
Area
(106
km2)
Volume (km3)
Percent
of total
water
Percent
of fresh
water
Ocean 361.3 1338000000 96.5
Groundwater
Fresh 134.8 10530000 0.76 30.1
Saline 134.8 12870000 0.93
Soil Moisture 82.0 16500 0.0012 0.05
Polar ice 16.0 24023500 1.7 68.6
Other ice and snow 0.3 340600 0.025 1.0
Lakes
Fresh 1.2 91000 0.007 0.26
Saline 0.8 85400 0.006
Marshes 2.7 11470 0.0008 0.03
Rivers 148.8 2120 0.0002 0.006
Biological water 510.0 1120 0.0001 0.003
Atmospheric water 510.0 12900 0.001 0.04
Total water 510.0 1385984610 100
Fresh water 148.8 35029210 2.5 100
Largest sphere represents all
of Earth's water
“Tiny" bubble represents fresh
water in all the lakes and rivers

6.8 Hydrologic budget in details

Example 65:
Evapotranspiration
cubic meter per second
𝑅𝑢𝑛𝑜𝑓𝑓 = 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑖𝑛 𝑑𝑒𝑝𝑡ℎ =
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
𝑎𝑟𝑒𝑎
Precipitation (p) – Evapotranspiration (ET) – Infiltration (I) – Runoff (R) = storage change (∆S)
Precipitation (p) – Evapotranspiration (ET) – Infiltration (I) – Runoff (R) = 0 (over long term)
𝑑𝑆 (𝑔𝑟𝑜𝑢𝑛𝑑 𝑤𝑎𝑡𝑒𝑟,𝑎𝑞𝑢𝑖𝑓𝑒𝑟)
𝑑𝑡
= Inflow (precipitation) – outflow (evaporation,transpiration)
Total storage (∆S) =
238
50 cm
Wetted
with rain
Area = 37500
Rain

Example 66:
38880000000000
3000000000000
Solution:
239
𝑅𝑢𝑛𝑜𝑓𝑓 = 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑖𝑛 𝑑𝑒𝑝𝑡ℎ =
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
𝑎𝑟𝑒𝑎
Precipitation (p) – Evapotranspiration (ET) – Infiltration (I) – Runoff (R) = storage change (∆S)
Precipitation (p) – Evapotranspiration (ET) – Infiltration (I) – Runoff (R) = 0 (over long term)
𝑑𝑆 (𝑔𝑟𝑜𝑢𝑛𝑑 𝑤𝑎𝑡𝑒𝑟,𝑎𝑞𝑢𝑖𝑓𝑒𝑟)
𝑑𝑡
= Inflow (precipitation) – outflow (evaporation,transpiration)
Total storage (∆S) =
Rain: 4 cm
precipitation
Area = 300
15 cms
inflow
13 cms
outflow
160 km2cm
storage

Example 67:
April
May
June
July
August
September
October
November
January
February
March
60 m3
?? m3
Rawaz Kurda
24
0
Total storage (∆S) = Inflow - outflow

6.9 Test for consistency of record
Double – Mass curveanalysis
180
160
140
120
100
80
60
40
20
0
0 5 10 15
Data
Months
Average data
Rawaz Kurda
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1

Example 68:

Solution:
Continuous
Point of
start
changing
~ Similar
Suspicious
data

Continuous
Cumulative value
+
Rawaz Kurda
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5

Continuous
Point of
start
changing

1
or
1.89

180
160
140
120
100
80
60
40
20
0
0 5 10 15
Data
Months
Before correction
D
ABC
200
180
160
140
120
100
80
60
40
20
0
0 5 10 15
Data
Months
After correction
D
ABC
𝑴𝑪
Correction factor = 𝑴𝑨
= 𝟏.𝟖𝟗
𝑴𝑨
Correction factor = 𝑴𝑪
= 𝟎. 𝟓𝟑
0
20
40
60
80
100
120
0 5 10 15
Data
Months
After correction
D
ABC

Example 69:
Test the consistency of the 22 years of data of the annual precipitation measured at station A. Rainfall data for
station A as well as the average annual rainfall measured for a group of eight neighboring stations located in a
meteorologically homogeneous region are given below as follows:
Sl. No. Year Annual Rainfall of Station A Average Annual Rainfall (AAR) of 8 Station Group
(mm) (mm)
1 1946 177 143
2 1947 144 132
3 1948 178 146
4 1949 162 147
5 1950 194 161
6 1951 168 155
7 1952 196 152
8 1953 144 117
9 1954 150 130
10 1955 191.9 190
11 1956 171.7 170
12 1957 191.9 190
13 1958 167.66 166
14 1959 151.5 150
15 1960 126.25 125
16 1961 141.4 140
17 1962 164.63 163
18 1963 146.45 145
19 1964 144.43 143
20 1965 136.35 135
21 1966 151.5 150
22 1967 166.65 165

Cumulative Station A rainfall
Cumulative of 8 Station
A.A.R.
(mm) (mm)
177 143 1336 1.17 153
321 275 1140 124
499 421 =(1.01/1.17) = 0.86 153
661 568 140
855 729 167
1023 884 145
1219 1036 169
1363 1153 124
1513 1283 129
1704.9 1473 1860.42 1.01 192
1876.6 1643 1842 172
2068.5 1833 192
2236.16 1999 168
2387.66 2149 152
2513.91 2274 126
2655.31 2414 141
2819.94 2577 165
2966.39 2722 146
3110.82 2865 144
3247.17 3000 136
3398.67 3150 152
3565.32 3315 167
153

Homework 7:

6.10.1 Precipitation depth
A recording gauge provides a record of the precipitation depth as a function of time.
Precipitation depth – time curve is called the mass curve (Fig. A).
6.10.2 Precipitation intensity
Precipitation depth in unit time is called precipitation intensity.
The curve showing the variation of precipitation intensity with time is called
hyetograph and is usually drawn in steps (Fig. B). The time interval Δt is chosen with
respect to the size of the region and usually in the range 1-6 hours.
Classification of Precipitation intensity
• Less than2.5 mm/hr is called light precipitation
• 2.5-7.5 mm/hr as medium
• More than 7.5mm/hr as heavy rain.
Usually average intensity reduces as the duration increases.
6.10 Analysis of Precipitation Records
Fig. A
Fig. B

6.10.3 Computation of Average Rainfall Depth over a Basin
In order to compute the average rainfall over a basin or catchment area, the rainfall is measured at a number of
rain-gauge stations suitably located in the area.
A network should be planned as to have a representative picture of the areal distribution of rainfall.
World Meteorological Organization (WMO) recommends the optimum density of one gauge per 600-900 km2 in
plains, and one gauge per 100-250 km2 in mountain regions, where the elevation difference must be less than 500
m.
If a basin or catchment area contains more than one gauge, the computation of average precipitation or rainfall
depth may be done by the following methods.
6.10.3.1 Arithmetic Average Method
If the rainfall is uniformly distributed on its areal pattern, the simplest method of estimating average rainfall is
to compute the average of the recorded rainfall values at various stations.
This method can be used in regions smaller than 500 km2 when the gauges are rather uniformly distributed.

6.10.3.2 Thiessen PolygonMethod
Thiessen method is a more common method of weighing the rain-gauge observation with respect to the area. This method is more accurate
than the arithmetic average method. The procedure to be followed in computing the average rainfall depth is;
i) Join the adjacent rain-gauge stations A, B, C, D,….. By straight lines.
ii) Draw the perpendicular bisectors of each of these lines.
iii) A Thiessen Polygon is thus constructed. The polygon formed by the perpendicular bisectors around a station encloses an area which is
everywhere closer to that station than any other station. Find the area of each of these polygons.
iv) Multiply the area of each Thiessen polygon by the rainfall value of the enclosed station.
v) Find the total area (ΣA) of the basin.
vi) Compute the average precipitation depth from the equation;
Thiessen polygon does not change in time, and is drawn only once.
The method can be used in regions 500-5000 km2 size. It considers
the non-uniformity of the areal distribution of gauges.
230

EXAMPLE 69: Calculate the average precipitation depth for the area given below with four stations and
rainfall depths given in the Table.
Solution:
255

EXAMPLE 70: Calculate the average precipitation depth for the area given below with four stations and
rainfall depths given in the Figure.
Solution:
25
6

Fluid Mechanics and Hydrology Fundamentals

Fluid Mechanics and Hydrology Fundamentals

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