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1. 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

2. 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

3. Course Book Fluid Mechanics and Hydrology 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda

4. 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) Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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.

5. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 5 U-tube manometer

6. Chapter one Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda Fluid Mechanics _ properties of fluids

7. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda Why should we know this?

8. 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 and Hydrology _ Dr. Rawaz Kurda

9. 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.

10. 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

11. 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.

12. 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

13. 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

14. 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.

15. 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

16. 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

17. 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

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

19. 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

20. 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

21. 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

22. 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.

23. 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 𝜌 = Mass of fluid 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

25. 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

26. 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.

27. 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

28. General notes: Differentiation rules (Table A1) 30

29. General notes: Integration rules (Table A2) 31

30. 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

31. 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

32. 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

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

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

35. 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

36. 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

37. 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

38. Example 5: 𝑑𝑦 𝜏 = 𝜇 𝑑𝑢  𝜇 = 𝜏 𝑑y 𝑑u 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 40 Given data Upper – lower Already given the difference

39. Example 6: 𝜏 = 𝜇 𝑑𝑢 𝑑𝑦 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 𝑓𝑜𝑟𝑐 𝜇 = 𝑎𝑟𝑒𝑎 𝑑 𝑢 𝑑 𝑦 41 Given data 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠(𝜏) = 𝑓𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠(𝜏) = 𝑓𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎

40. Example 7: cos 600 degrees is½ 𝜏 = 𝜇 𝑑𝑢 𝑑𝑦 42 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 𝑓𝑜𝑟𝑐 𝜇 = 𝑎𝑟𝑒𝑎 𝑑 𝑢 𝑑 𝑦 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠(𝜏) = 𝑓𝑜𝑟𝑐𝑒 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

41. Excel → COS pi()/4

42. Example 8: 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 44 Given data

43. 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 𝜏 = 𝜇 𝑑𝑢 𝑑𝑦 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 𝑓𝑜𝑟𝑐 𝜇 = 𝑎𝑟𝑒𝑎 𝑑 𝑢 𝑑 𝑦 Given data (Only upper given) (2.5-0 = 2.5) (98.1-0 = 98.1) 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠(𝜏) = 𝑓𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎

44. Example 10: 46 S for Liquids = Weight density (or density) of liquid Weight density (or density) of water The density of a liquid = S x weight density of water = S x 1000 kg/m3 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 (𝜇) Kinematic Viscosity (v) = density (𝜌) 𝜏 = 𝜇 𝑑𝑢 𝑑𝑦 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 𝑓𝑜𝑟𝑐 𝜇 = 𝑎𝑟𝑒𝑎 𝑑 𝑢 𝑑 𝑦 Given data

45. Example 11: 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) Given data

46. Example 12: 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) Given data

47. Example 13: 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 49

48. Example 14: 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) minutes tosecond 𝜏 = 𝜇 𝑑𝑢 𝑑𝑦 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑡) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑢) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑦) 𝑓𝑜𝑟𝑐 𝜇 = 𝑎𝑟𝑒𝑎 𝑑 𝑢 𝑑 𝑦

49. 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.

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

52. 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

53. 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 Absolute pressure of a gas (p) x 1 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 (𝜌) = Gas constant (R) x Absolute temperature (T) 𝑅 𝑥 1 = 𝑝 𝜌 𝑅 = 𝑝 𝜌 𝑅 = 𝑝 𝜌 𝑘

54. Gas constant(R): (R4) Universal Gas constant (in order to be used for cases): = n x M 56 Absolute pressure of a gas (p) x 1 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 (𝜌) = Gas constant (R) x Absolute temperature (T)

55. Example 14: (Celsius to Kelvin) Mass density = Specific weight (𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦) Acceleration due to gravity 57 Absolute pressure of a gas (p) x 1 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 (𝜌) = Gas constant (R) x Absolute temperature (T)

56. 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 Absolute pressure of a gas (p) x 1 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑠 (𝜌) = Gas constant (R) x Absolute temperature (T)

57. 59 Two volumes due to compression

58. Example 15: 60 Two volumes

59. 1.2.4 Compressibility and Bulk Modulus 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝) 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 −𝑑∀ ∀ 61

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

61. Relationship between Bulk Modulus (K) and Pressure for a Gas 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝) 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 −𝑑∀ ∀ 63

62. 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝) 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 −𝑑∀ ∀ Example 17: 64 (0.15 have been decreased for each 100)

63. 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝) 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 −𝑑∀ ∀ Example 18: 65

64. 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

65. 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

66. 68

67. Example 19: 69

68. Example 20: 70

69. 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

70. 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

71. Capillary 73

72. Capillary 75

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

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

75. Example 21: 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 The density of Liquid = S x density of water 78

76. Example 22: 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 79

79. 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

80. 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: 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.

82. 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)

83. 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) 𝐴𝑟𝑒𝑎 (𝐴) 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 forces or energy causing them. ❑ Kinematics It deals with the velocity, accelerations and patterns of flow and not the forces or energy causing them. This chapter 86

84. 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

85. 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)

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

87. 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.

88. 91 https://www.youtube.com/watch? v=iVddJWB9rXs Video 7

89. 92

90. Example 25: 93 𝑝 = 𝑓𝑜𝑟𝑐𝑒 (𝐹) 𝐴𝑟𝑒𝑎 (𝐴) 𝑝 = 𝑓𝑜𝑟𝑐𝑒 (𝐹) 𝐴𝑟𝑒𝑎 (𝐴)

91. Example 26: p1 = 𝐴𝑟𝑒𝑎 (𝐴) 𝑓𝑜𝑟𝑐𝑒 (𝐹) = 𝑝2 = 𝑓𝑜𝑟𝑐𝑒 (𝐹) 𝐴𝑟𝑒𝑎 (𝐴) p1 p2 PASCAL'S LAW 94

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

94. Example 29: Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H) Mass density = S x density of water 97 PASCAL'S LAW

96. 2.4 Absolute, gauge, Atmospheric and vacuum pressures Absolute pressure = Atmospheric pressure + Gauge pressure 99

97. 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

98. 2.6 Simple Manometers Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H) 101

99. Absolute pressure = Atmospheric pressure + Gauge pressure Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H) 𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2 102 𝑝 + 𝜌1 ∗ 𝑔 ∗ ℎ1 + 𝜌2 ∗ 𝑔 ∗ ℎ2 = 0

100. Example 31: Absolute pressure = Atmospheric pressure + Gauge pressure 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 = 0 103 Consider the sp. Gr. Of mercury is 13.6.

101. Example 32: Absolute pressure = Atmospheric pressure + Gauge pressure 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 = 0 104

102. Example 33: 105 Pressure (p) = mass density (𝜌) x Acceleration due to gravity (g) x Height (H) 𝜌1 ∗ 𝑔 ∗ ℎ1 = 𝜌2 ∗ 𝑔 ∗ ℎ2

103. 106

104. Example 34: 107

105. 108

106. 109

107. Example 35: 110

108. 111

109. Example 36: 112

110. Example 37: 113

111. Example 38: 114

112. 115

113. Example 39: 116

114. Example 40: 110

115. Example 41: 118

116. Example 42: 119

117. 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: 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. Absolute pressure = Atmospheric pressure + Gauge pressure 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

118. 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. Absolute pressure = Atmospheric pressure + Gauge pressure 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 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑑𝑝) 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 −𝑑∀ ∀

120. 3 Buoyancy and floatation 3.1 Introduction Submarine 3.2 Bouncy 123

121. 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

122. 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?

123. Example 43: Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity(g) Volume of displaced liquid = Weight of displaced liquid Weight density Position of center ofBuoyancy: Volume of water displaced = volume of wooden block in water Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V 12 6 1.5 m 2.5 m 6.0 m

124. 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 Volume of displaced liquid = Weight of displaced liquid Weight density 127

125. Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g) FB = Specific weight of fluid * Volume displaced = γ .V Volume of displaced liquid = Weight of displaced liquid Weight density 120

126. 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

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

128. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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.

129. Example 45: Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g) FB = Specific weight of fluid * Volume displaced = γ .V 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 𝜌 = Mass of fluid 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

130. Example 46: Volume of displaced liquid = Weight of displaced liquid Weight density They are the same Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V 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

131. Example 47: Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g) FB = Specific weight of fluid * Volume displaced = γ .V FB = Specific weight of fluid * Volume displaced = γ .V FB = Specific weight of fluid * Volume displaced = γ .V Water Mercury Object 13 4

132. Example 48: Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g) FB = Specific weight of fluid * Volume displaced = γ .V Volume of displaced liquid = Weight of displaced liquid Weight density 135

133. Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity (g) FB = Specific weight of fluid * Volume displaced = γ .V Volume of displaced liquid = Weight of displaced liquid Weight density 136

134. 4.4. Meta-centre 137 G – center of gravity B – center of buoyancy M – Meta center

135. 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 G – center of gravity B – center of buoyancy M – Meta center

136. 139

137. ∀ 140 𝐺𝑀 = 𝐼 − 𝐵𝐺 𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵 G – center of gravity B – center of buoyancy M – Meta center

138. 3.6 141 Conditions of Equilibrium of a floating and sub-merged body

139. -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 G – center of gravity B – center of buoyancy M – Meta center

140. 143

141. 144

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

143. 146

144. Example 50: ∀ 𝐺𝑀 = I − 𝐵𝐺 𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵 Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V 3 m 2 m Water level 1 m 0.8 m 147

145. 140

146. Example 51: ∀ 𝐺𝑀 = 𝐼 − 𝐵𝐺 𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵 Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V Wood Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V Water = ? 14 9 Considered it all

147. Example 52: 4 m 4 m ∀ 𝐺𝑀 = 𝐼 −𝐵𝐺 Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V Specific weight or weight density (w) = mass density (𝜌) x Acceleration due to gravity(g) 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.

148. Example 53: 15 1

149. 3.7 Experimental method of determination of Meta-centricheight 152

150. Example 54: 153

151. Example 55: 154

152. VL m 2 m 5 m 0.9m Water level Homework 5 : 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. 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 𝐵𝐺 = 𝐴𝐺 − 𝐴𝐵 Weight of displaced liquid 𝑎𝑠 𝑎 𝑓𝑜𝑟𝑐𝑒 = Buoyancy force (FB) = Specific weight of fluid * Volume displaced = γ . V 155

153. 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

154. 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 forces or energy causing them. This chapter 157

155. 5.2 Methods of Describing fluid motion Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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.

156. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 15 9

157. 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

158. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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.

159. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 162 For compressible flow, p + 1 𝜌𝑉2

160. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 163

161. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 164

162. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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

163. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 166 DISCHARGE (Q) V= Average velocity 𝜌= Density at section A = Area of pipe

164. Example 56: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 167 V1 𝜌1 A1 = V2 𝜌2 A2 Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V)

165. Example 57: V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 160 Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V) 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.

166. Example 58: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 169 V1 𝜌1 A1 = V2 𝜌2 A2 Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V)

167. 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

168. 171 Source 1 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 Source 2

169. 172 Source Average velocity of fluid across the section (V) = Discharge (Q) Cross−sectional area of pipe (A) Public water system - Dead end Fluid Mechanics and Hydrology_ Dr. Rawaz Kurda /12

170. 173 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 Public water system - continues Fluid Mechanics and Hydrology_ Dr. Rawaz Kurda /12

171. 4.6 Continuity equation in threedimensions Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 174

172. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 175

174. Example 59: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 177

175. 4.7 Velocity and acceleration Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 178

176. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 179

182. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 18 5

183. Homework 6 : 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. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 186 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

184. Example 57: V1 𝜌1 A1 = V2 𝜌2 A2 + V3 𝜌3 A3 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 160 Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V)

185. Homework 6 : 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. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 188 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

186. 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

187. ‫بێت؟‬ ‫داخيل‬ ‫کۆتا‬ ‫محازەرەى‬ ‫دوو‬ ‫يان‬ ‫بێت؟‬ ‫داخيا‬ ‫هەمووى‬

189. 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

190. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 19 5

191. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 177 5.2 Equation of Motion Bernoulli’s equation

192. 5.3 Euler’s equation of motion Derivative Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 19 7

193. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 19 8

194. Total head (energy) = Pressure head + Kinetic head + datum head = 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). Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 180

195. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 0

196. Example 63: Total head (energy) = Pressure head + Kinetic head + datum head = 𝑝 𝜌 + 𝑣2 2𝑔 + constant Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 1

197. Example 64: Total head (energy) = Pressure head + 𝑝 𝜌 = + Kinetic head 𝑣2 2𝑔 + datum head + constant D1 = 20cm V1 = 40m/sec D2 = 10cm Discharge (Q) = Cross-sectional area of pipe (A) * Average velocity of fluid across the section (V) ? Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 2

198. Example 65: Total head (energy) = Pressure head + Kinetic head + datumhead = 𝑝 𝜌 + 𝑣2 2𝑔 + constant ? Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 3

199. Example 65: Total head (energy) = Pressure head + Kinetic head + datumhead = 𝑝 𝜌 + 𝑣2 2𝑔 + constant Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 4

200. Example63: Total head (energy) = Pressure head + Kinetic head + datum head = 𝑝 𝜌 + 𝑣2 2𝑔 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 5 + constant

201. Total head (energy) = Pressure head + Kinetic head + datum head = 𝑝 𝜌 + 𝑣2 2𝑔 + constant Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 6

202. = 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 7

203. Example 64: 𝑝1 = + 𝑣1 2 𝜌𝑔 2𝑔 1 𝐴 + 𝑧 + ℎ = 𝑝2 + Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 20 8 𝑣2 2 𝜌 𝑔 2𝑔 2 + 𝑧 + ℎ𝐿

204. Example 64: 𝑝1 = + 𝑣1 2 𝜌𝑔 2𝑔 1 𝐴 + 𝑧 + ℎ = 𝑝2 + Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 190 𝑣2 2 𝜌𝑔 2𝑔 2 + 𝑧 + ℎ𝐿

205. 5.7 Practical applications of Bernoulli’s equation Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 21 0

213. Project Fluid Mechanics and Hydrology _ Dr. 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

215. 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

216. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 221

217. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 222

218. ➢ Water supply Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 223

219. ➢ Wastewater treatment and disposal Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 22 4

220. ➢ Irrigation ➢ Hydropower generation USA ➢ Recreational use of water and fish and wildlife protection Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda Kuriqi et al.(2019) 225

221. ➢ Flood control Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 226 https://www.youtube.com/watch?v=eUKd5K4E6mM

222. ➢ Drainage ➢ Navigation Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 227

223. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 228 Porous Asphalt Demonstration https://www.youtube.com/watch?v=I16WGau3jxE

224. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 229 https://www.youtube.com/watch?v=5rsODZXPyCY Permeable Paving in Action https://www.youtube.com/watch? v=JOIM2yR_B6k

225. ➢ Drainage Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 230

226. ➢ Erosion and sediment control ➢ Pollution abatement Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 231

227. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 210 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

228. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 233 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

229. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 234

230. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 213 6.6 Measurement techniques Precipitation Evapotranspiration ➔ discharge ➔ flow depth (stage) Surface waters (velocity) Note: This methods will be discussed in the next slides →

231. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 236

232. 6.8 Hydrologic budget in details Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 237

233. 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) = Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 238 50 cm Wetted with rain Area = 37500 Rain

234. Example 66: 38880000000000 3000000000000 Solution: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 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

235. Example 67: April May June July August September October November January February March 60 m3 ?? m3 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 24 0 Total storage (∆S) = Inflow - outflow

236. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 24 1

239. Solution: Continuous Point of start changing ~ Similar Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 244 Suspicious data

240. Continuous Cumulative value + Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 24 5

241. Continuous Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 246 Point of start changing

242. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 247 1 or 1.89

243. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 226 180 160 140 120 100 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

244. 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

245. 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

246. Homework 7: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 251

247. 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 Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 252 Fig. B

248. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 253

249. 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. Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 230

250. EXAMPLE 69: Calculate the average precipitation depth for the area given below with four stations and rainfall depths given in the Table. Solution: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 255

251. EXAMPLE 70: Calculate the average precipitation depth for the area given below with four stations and rainfall depths given in the Figure. Solution: Fluid Mechanics and Hydrology _ Dr. Rawaz Kurda 25 6

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