Fluid Mechanics and Fluid Power (4020410) Diploma in Mechanical Engineering Notes PDF

DEPARTMENT OF MECHANICAL ENGINEERING Course Name : 1020-Diploma in Mechanical Engineering Subject Code : 4020410 Semester : IV Subject Title : Fluid Mechanics and Fluid Power NOTES OF LESSON Prepared by S Prakash, HOD/MECH S Di...

DEPARTMENT OF MECHANICAL ENGINEERING Course Name : 1020-Diploma in Mechanical Engineering Subject Code : 4020410 Semester : IV Subject Title : Fluid Mechanics and Fluid Power NOTES OF LESSON Prepared by S Prakash, HOD/MECH S Divakaran, Lect /MECH  STATE BOARD OF TECHNICAL EDUCATION &TRAINING, TAMILNADU DIPLOMA IN ENGINEERING / TECHNOLOGY SYLLABUS N - SCHEME (To be implemented for the students admitted from the year 2020 - 2021 onwards) Course Name : 1020 Diploma in Mechanical Engineering Subject Code : 4020410 Semester : IV Subject Title : Fluid Mechanics and Fluid Power TEACHING AND SCHEME OF EXAMINATION No of weeks per semester: 16 weeks Subject Instructions Examination 4020410 Marks Hours / Hours / Fluid Internal Board Duration Week Semester Total Mechanics Assessment Examinations and Fluid 5 80 25 100* 100 3 Hrs. Power * Examinations will be conducted for 100 marks and it will be reduced to 75 marks for result. Topics and Allocation of Hours Unit No Topics Hours I Properties of Fluids & Fluid Pressure 12 II Fluid Flow, Flow Through Pipes & Impact of Jet 17 Hydraulic Turbines, Centrifugal Pumps & Reciprocating 16 III Pumps IV Hydraulic Systems 16 V Pneumatic Systems 12 Test and Model Exam 7 Total 80 83 RATIONALE: The purpose of this subject is to teach the students the fundamentals of engineering fluid mechanics in a very general manner so that they can understand the way that forces are produced and transmitted by fluids that are, first, essentially at rest and, second, in motion. This will allow them to apply the physical principles behind some of the most common applications of fluid mechanics in engineering. OBJECTIVES:  To study the basic fluid properties and types of flow;  To understand the transmission of pressure in liquids and its application to hydraulics;  To calculate hydrostatic forces on plane and curved submerged surfaces;  To employ the concept of continuity of flow and use Bernoulli's equation to measure flow rate and velocity;  To apply the momentum principle to liquids in jets and pipes.  To understand the working of hydraulic machines like, turbines, pumps.  To identify the various components of a Hydraulic & Pneumatic systems and select them for design of hydraulic and pneumatic circuits for Engineering applications. 4020410 FLUID MECHANICS AND FLUID POWER DETAILED SYLLABUS Contents: Theory Unit Name of the Topics Hours I PROPERTIES OF FLUIDS & FLUID PRESSURE Chapter: 1.1: Properties of Fluids 4 Fluid – definition-classification. Properties – density, specific gravity, specific weight, specific volume, dynamic viscosity, kinematic viscosity, surface tension, capillarity, vapour pressure and compressibility – Problems Chapter: 1.2: Fluid Pressure & Its Measurement 8 Fluid Pressure – Hydrostatic law - Pressure head, Pascal’s Law – proof - 84  applications - Hydraulic press- Hydraulic jack. Concepts of absolute, vacuum, gauge and atmospheric pressures. Pressure measurements – Simple U tube manometers and differential manometers and their types – Problems - Bourdon tube pressure gauge. Pressure sensor technologies - classification only. Total Pressure, Centre of pressure on immersed bodies (flat vertical., flat vertical) – Problems. II FLUID FLOW, FLOW THROUGH PIPES & IMPACT OF JET Chapter: 2.1: Fluid Flow 6 Types of Fluid flow - Laminar, turbulent, steady, unsteady, uniform, non-uniform, rotational, irrotational. Continuity equation, Bernoulli’s theorem - assumptions- derivation - applications and limitations - Problems. Venturimeter – Construction - working principle, coefficient of discharge - derivation for discharge. Orificemeter - Construction working principle, coefficient of discharge- derivation for discharge. Problems. Pitots Tube – Construction and working principle only. 6 Chapter: 2.2: Flow through Pipes Laws of fluid friction for Laminar and turbulent flow- Darcy’s equation and Chezy’s equation for frictional losses – Problems. Minor Losses - description. Hydraulic gradient line and Total energy line. Hydraulic Power transmission through pipes – problems. 5 Chapter: 2.3: Impact of Jet Impact of jet on fixed vertical plate - Impact of jet on moving vertical flat plates in the direction of jet - Impact of jet on a series of moving plates or vanes - Problems on work done and efficiency. III HYDRAULIC TURBINES, CENTRIFUGAL PUMPS & RECIPROCATING PUMPS Chapter: 3.1: Hydraulic Turbines 4 Classification of hydraulic turbines and their applications. Construction and working principle of Pelton wheel, Francis and Kaplan turbine. Draft tubes – types and construction, Concept of cavitation in turbines, Surge tank and its need. 85  Chapter: 3.2: Centrifugal Pumps 6 Construction - Principle of working. Types of casings and impellers. Concepts of multistage. Priming and its methods. Manometric head, work done, manometric, mechanical and overall efficiencies - problems Chapter: 3.3: Reciprocating Pumps 6 Construction, working principle and applications of single and double acting reciprocating pumps. Discharge - Theoretical power required coefficient of discharge – Problems Concepts of slip – negative slip. Cavitation and separation. Use of air vessel. Indicator diagram with effect of acceleration head and friction head. IV HYDRAULIC SYSTEMS Chapter: 4.1: Introduction to Fluid power systems 4 Fluid power systems - general layout - components of hydraulic & Pneumatic systems. Practical applications of Fluid power systems. Comparison - Advantages and limitations. Chapter: 4.2: Components of Hydraulic systems 8 Types, construction, working Principle and symbol of the following components. Pump – vane, gear and piston pumps. Valves: Pressure Control valves – pressure relief. valve, pressure reducing valve, pressure unloading valve. Direction control valve – poppet valve, spool valve, 3/2, 4/2 & 4/3 DC valves, sequencing valve.Flow control valve – pressure compensated – non pressure compensated.Actuators – Linear actuactors – single acting & double acting – rotory actuators – hydraulic motors. Accessories – Intensifiers and Accumulators. Chapter: 4.3: Hydraulic Circuits 4 Double acting cylinder with Meter in, Meter out circuits, Pump unloading cut, Bleed off circuit, sequencing circuit. Hydraulic circuits for milling machine, shaping machine. Motion synchronisation circuit. V PNEUMATIC SYSTEMS Chapter: 5.1: Components of pneumatic systems 8 Types, construction, working Principle and symbol of the following components. Compressor – Reciprocating & Rotary Compressors. 86  Valves. Pressure Control valves – pressure relief valve, pressure regulating valves.Direction control valves – 3/2, 5/2 & 5/3 DC valves, sequencing valve.Flow control valve – throttle valves – shuttle valves- quick exhaust valves. Actuators – Linear actuactors – single acting & double acting – rotary actuators – air motors.Accessories.- FRL unit. Chapter: 5.2: Pneumatic Circuits Double acting cylinder with Meter in, Meter out circuits, speed control 4 circuit and sequencing circuit Reference Books: 1. A Textbook of Fluid Mechanics and Hydraulic Machines, R. K. Bansal, Laxmi Publications (P).,Ltd, New Delhi, 2010 2. Hydraulics and Fluid Mechanics, Modi P.N. and Seth, S.M. Standard Book House, New Delhi, 2013. 3. Fluid Power with Applications, Anthony Esposito,, Pearson Education 2005. 4. A Textbook of Fluid Mechanics, R. K Rajput, S.Chand & Co, New Delhi, 2019 5. Engineering Fluid Mechanics, Kumar K. L., Eurasia Publishing House (P) Ltd., New Delhi, 2016. 6. Oil Hydraulics Systems- Principles and Maintenance”, Majumdar S.R., Tata McGraw- Hill, 2001. 7. Hydraulic and Pneumatic Controls, Shanmugasundaram.K, Chand & Co, 2006 87  DEPARTMENT OF MECHANICAL ENGINEERING Course Name : 1020-Diploma in Mechanical Engineering Subject Code : 4020410 Semester : IV Subject Title : Fluid Mechanics and Fluid Power UNIT –I PROPERTIES OF FLUIDS AND PRESSURE MEASUREMENTS UNIT –I PROPERTIES OF FLUIDS AND PRESSURE MEASUREMENTS Objectives:  Study about Static fluid and dynamic fluid.  Explain the classification of fluid.  Discuss the liquid and gaseous fluid.  Define fluid pressure, atmospheric pressure, gauge pressure, vacuum pressure and absolute pressure and derive the relation between them.  Solve problems in Properties of a fluid and pressure relations.  Derive Pascal’s law and notify its applications  Explain the construction and working principle of Hydraulic press and Hydraulic jack.  Explain the type of manometers and solve problems in manometers  Discuss about Mechanical Gauges  Explain the construction and working principle of different types of Mechanical Gauges. 1.1 Introduction Fluid Mechanics Fluid Mechanics is the branch of engineering which deals with the properties and behavior of fluids at rest and in motion. Hydraulics Hydraulics is the branch of engineering deals with the properties and behavior of water. 1.2. Definition of Fluid Fluid can be defined as the substance which can flow with or without the aid of force A fluid may be in three form like as liquid (or) a vapour (or) a gas 1.3. Types of Fluid Fluids are classified as follows. 1. Ideal (or) perfect fluid A fluid having density only as property is called Ideal fluid. Ideal fluid one which has no viscosity, surface tension, cohesion and adhesion etc. 5  Ex. Imaginary fluid 2. Real fluid (or) Practical fluid A fluid having viscosity, surface tension, cohesion, adhesion and density is called Real Fluid. Ex; water, air, lubricating oil 3. Newtonian Fluid A fluid which obeys Newton’s Law of viscosity is called Newtonian fluid. Ex; Water, Lubricating oil etc. 4. Non – Newtonian Fluid A fluid which does not obey Newton’s Law of viscosity is called Non- Newtonian fluid. Ex; Paints, Plastics etc. 1.4. Properties of Fluid 1. Density It is defined as the mass per unit volume. Density = Unit is kg/m3 Density of water is 1000 kg/m3. 2. Specific weight (weight density) It is defined as the weight per unit volume. Specific weight = Unit is kN/m3 Relation between the Specific weight and density is w = xg 3. Specific volume It is defined as the volume per unit mass. Specific volume = (or) 6  Unit is m3/kg. Relation between the Specific volume and density is v = 4. Relative density (or) Specific gravity It is defined as the ratio between the density of any liquid to the density of water. Density of fluid Relative density = Density of water Specific weight of liquid Relative density = Specific weight of water Specific gravity of water = 1 Specific gravity of mercury = 13.6 5. Compressibility It is the change in volume due to change in pressure of fluid. As the change in volume of a liquid under pressure is so small , a liquid considered to be an incompressible fluid. Air considered to be a compressible fluid. 6. Cohesion The inter molecular attraction between the molecules of same liquid is known as cohesion. 7. Adhesion It is the attraction between molecules of different substances. Ex. Water molecules stick in the container wall. In this way water has both Adhesion and Cohesion properties. But mercury has no Adhesion but it has Cohesion property. 8. Viscosity It is one of the important of fluid property which resist the flow of fluid. It is denoted by “µ”. Unit is Ns/m2. 7 Ex. Oil has high Viscosity and water has less Viscosity 9. Kinematic viscosity It is defined as the ratio between absolute viscosity and density of fluid Absolute viscosity Kinematic viscosity = Density of liquid Unit =m2/sec 10. Surface Tension The surface tension of liquid is defined as the tangential force per unit length acting at right angles on either side of the surface. Ex: Falling drops of rain water become sphere. 11. Capillarity When a smaller diameter capillary tube with open ends is dipped in a liquid, the liquid surface inside the tube rises (or) falls relative to the adjacent general level of liquid. It is known as capillarity. It can be capillary rise(water) (or) capillary fall(mercuryr) as shown in fig.1.1. Fig.1.1 12. Vapour pressure When liquid in the closed vessel evaporates, the vapour exerts a pressure, on the liquid surface. This pressure is known as vapour pressure. 8 1.5. Worked Examples 1. One Litre of petrol weighs 7 Newton. Calculate its Specific weight, Density, Specific Volume and Relative density. Given Data Volume of petrol = 1 litre = 1 x 10-3 m3 Weight of petrol = 7 N Solution 1. Specific weight = = =7000 N 2. Density = Mass = = = kg Density = = 713.557kg/m3 3. Specific Volume = = = 9  = 1.40 x 10-3 m3/kg 4. Relative Density = = = 0.7135 No unit 2. One cubic meter of crude oil weights 9.44 KN. Calculate the density, specific weight, specific volume and relative density. Given Data Volume = 1 m Weight of petrol = 9.44 k N = 9440 N Mass = W/g = 9440/9.81 = 962.28 kg Solution 1. Density = = = = 962.28 kg/m3 2. Specific weight = = = 9440 N/m3 3. Specific Volume = = = 0.00104 m3/kg 4. Relative Density = = = 0.962 No unit 10  3. If the density of liquid is 837 kg/m3. Find its specific weight , Specific volume and Relative density Given Data Density = 837 kg/m3 To find 1.Specific weight 2.Specific volume 3.Relative density Solution 1. Specific weight = (0r) Density(ρ) x g = 873 x 9.81 N/m3 = 8210.97 N/m3 2. Specific Volume = = = = = 1.19 x 10-3 m3/kg 3. Relative Density = = = 0.837 No unit 1.6. Important Formula 1.Weight W= m x g m-Mass g-acceleration due to gravity 2. Density(ρ) = = 3. Specific volume = = 4. Specific weight = = 5. Relative Density = (or) = 11  6. Specific gravity of water = 1 7. Specific gravity of mercury = 13.6 8. Density of water = 1000 kg/m3 9. Density of mercury = 1000x13.6 kg/m3 10. Specific weight water = 9810 N/m3 EXERCISE A. Theoretical Questions 1. Define fluid. 2. Define density. 3. Define specific volume. 4. Define specific weight. 5. Define relative density. 6. What is viscosity? 7. What is the unit of surface tension? 8. How fluids are classified? 9. Which fluid is called as Newtonian fluid? 10. Give the example of real fluid. 11. Distinguish between absolute viscosity and kinematic viscosity. 12. State the relation between absolute viscosity and kinematic viscosity. 13. What is the difference between ideal and real fluid? B. Numerical Problems 1. If a liquid weighs 200 N and occupies 2.5 m3. Find its specific weight, mass density and relative density. 2. One liter of fuel weighs 8.02 N. Calculate its,  Specific weight.  Density.  Specific volume.  Relative density. 1.7.PRESSURE OF A FLUIDS Pressure of Fluid The normal force acting on unit area is known as “pressure” (or) “intensity of pressure”. 12  Intensity of pressure (P) = Unit of pressure is N/m2 (or) kN/m2. 1 Pascal = 1 N/m2 1 bar = 105 N/m2 = 105pascal. 1 Torr (Torricelli) = 1 mm of Hg. 1 pieza = 1 KPa. 1.8.Law of pressure The intensity of pressure of a liquid at rest is always acts normal to the surface. The intensity of pressure of a fluid at rest, is acting equally in all directions. This is known as pascal’s law. The intensity of pressure depends only upon the vertical height of the liquid and not upon the size and shape of the vessel. If a vessel contains a fluid, a slight in cress in the intensity of pressure at any point will be immediately transmitted to all other points in the fluid. When a pressure gauge is connected to the points A, B, C, &D respectively. It will indicate the same reading as shown in fig. Pressure at A = pressure at B = pressure at C Pressure at D = pressure at B Fig 1.2 13  pressure at a point in a liquid pressure at a point in a liquid is, due to the weight of liquid column above the point. consider a small area “a” at a depth of H below the liquid surface. Force acting on the area a is due to weight of liquid prism standing over is as shown in fig.1.3 Fig.1.3 Pressure(p)=weight of liquid prism/area (p) =volume of the prism x sp.weight of liquid/area P = where P = wH. H- Pressure head w- specific weight pressure head of a liquid The vertical height of the liquid corresponding to particular pressure is known as pressure head. Pressure head = Atmospheric Pressure  Atmospheric Pressure is due to the weight of air column acting on unit area, at sea level.  Barometer is used to measure the atmospheric pressure.  Atmospheric pressure is 101.325 kN/m2 (or) 760 mm mercury (or) 10.33 m of water. Gauge pressure If the pressure to be measured is more than the atmospheric pressure, then the pressure indicated by a gauge is known as Gauge pressure. Ex :Boiler steam pressure, Air compressor pressure. Vacuum pressure If the pressure to be measured is less than the atmospheric pressure , than the pressure indicated by a gauge is known as “vacuum pressure”. 14  The relationship between the absolute pressure gauge pressure and vacuum pressure are as shown in fig. Mathematically 1.Absolute pressure = Atmospheric pressure+ gauge pressure 2.vacuum pressure =atmospheric pressure – absolute pressure Fig1.4 1.9.Pascal’s law Pascal’s law states that the intensity of pressure at any point in a fluid at rest, is acting equally in all direction. Fig.1.4 proof Px – Intensity of pressure acting on AB, Py – Intensity of pressure acting on BC PZ – Intensity of pressure acting on AC, θ- Angle between AC & BC 15 W-weight of element Consider unit length of element in a direction perpendicular to the direction of paper Force on the face AB = Px x AB Force on the face BC = Py x BC Force on the face AC = Pz x AC Self weight of the element W Resolving the forces horizontally Fx = Fz sin θ Px x AB = Pz x AC x sin θ From triangle ABC sin θ = AB = AC sin θ put AB value in the previous equation Px x AC sin θ = Pz x AC x sin θ Px = P z -----------------------------(1) Resolving the forces vertically Fz cos θ+W = Fy Pz AC cos θ+W = Py BC W is very small. So it is negligible Pz AC cos θ = Py BC From triangle ABC cos θ = BC = AC cos θ Pz AC cos θ = Py AC cos θ Pz = Py ---------------------------------------(2) From equation (1) & (2) Px = Py= Pz hence Pascal’s law proved. 1.10.Hydraulic press Hydraulic press is used to lift heavy weights by the applications of a much smaller force. it is working under Pascal’s law. Elements Hydraulic press 1. Reservoir 2.plunger 3. Ram 4.check valves 5. Release valve 6.Handle 16  Fig.1.5 Construction Hydraulic press consists of two cylinders. One is larger then the other. larger cylinder contains a ram and smaller cylinder contains a plunger. These two cylinders are connected by a pipe. Cylinders and pipe contain a liquid through which pressure is transmitted. one plate is fixed and other plate is attached to the ram. A release valve is fitted at the bottom of the ram side. Fig.1.6. Hydraulic press 17 Working principle When the handle moves upward then upward movement of the plunger take place. Partial vacuum is created in the bottom of the plunger. The liquid from the reservoir is drawn through the check valve “v2” and enters into the small cylinders. Now the delivery valve will be in closed position. The plunger increases the liquid pressure at its bottom. At same time, the check valve “V1” closes and check valve “V2” opens. The liquid is forced to the ram cylinder through the check valve “V2”. The high pressure liquid acting at the bottom of the ram moves the ram up. Thus the load on the ram is lifted up and then the object is pressed between the two plates. To lower the load the release valve is opened to allow the liquid the reservoir. This caused the ram to move down. 1.11.Hydraulic jack The hydraulic jack is used to lift heavy loads by the application of a much smaller force. It is working under Pascal’s law. The diagram of hydraulic jack is shown in fig.1.7 Elements of Hydraulic Jack 1.Ram 2.plunger with handle 3.Suction&delivery valve. 4 Reservoirs Working principle The plunger will be moved up and down by actuating the handle. During the upward movement of the plunger, partial vacuum is produced in the plunger side. Now liquid flows to the plunger side from the reservoir by opening the suction valve. Now the delivery valve will be in closed position. During the downward movement of the plunger the liquid moves at the bottom of ram and the ram be moves up. The heavy load at the top of the ram is lifted. There is lowering screw at the bottom of ram side. It is unscrewed to allow the liquid to the reservoir. Hence the ram will be moved downward to lower the load. 18  Fig. 1.7.Hydraulic jack 1.12.PRESSURE MEASUREMENTS The pressure of a fluid is measured by the following devices. 1. Manometer 2. Mechanical gauges 1.12.1. Manometers Manometer are the pressure measuring instruments based on the principle of balancing the liquid columns. Uses of manometer Used for accurate measurements of *Low pressure *Vacuum pressure 19  *Difference of pressure Advantages *simple construction *precise measurement *-ve pressure can be measured. Types of Manometers  Barometer  Piezometer tube  U tube manometer  Differential manometer  Inverted u tube manometer  Micro manometer  Inclined manometer 1.12.2. Barometer Barometer is used to measure the atmospheric pressure. Types of Barometer 1. Mercury Barometer 2. Aneroid Barometer Mercury barometer Fig. 1.8. Mercury barometer Mercury barometer is used to measure atmospheric pressure. It consists of a vertical glass tube having one end closed and other end opens to atmosphere. Glass 20  tube is graduated in “m.m”. To measure the atmospheric pressure the glass tube is completely filled with mercury and then it is immersed upside down into a vessel containing mercury as shown in fig.1.8. It is seen that the mercury stands in the tube to a height of 760 mm above the mercury level in the vessel. Atmospheric pressure = 760mm of hg 1.12.3. Piezometer tube Piezometer tube is used to measure the low pressure. It consists of a vertical graduated glass tube with open ends. The length of the tube is so selected that the liquid will rise in the tube freely with our overflow. If the pressure of water flowing in a pipe is to be measured then the piezometer tube is connected to the pipe the water will rise in the tube corresponding to the pressure of water available in the pipe as shown in fig.1.9 Fig.1.9 Pressure in the pipe = Rise of water in the tube is” h” mm P = ” h” mm of water. 1.12.4. Simple U tube manometer This is used to measure the pressure at a point in a static or dynamic fluid. Gauge pressure and vacuum pressure can be measured. 21  Fig.1.10 It consists of a glass tube bent in the form of “u” us shown in fig.1.10. This glass tube is open at both ends. General the tube contains mercury. Mercury is 13.6 times heavier than water. It is suitable for measuring high pressure. One end of the “U” tube is connected to the pipe whose pressure is measured and the other end is open to atmosphere. The pressure of water in the pipe forces the mercury level in the left arm to go down and a corresponding amount will rise in the right arm. Let Take Datum line is A-A h1= height of light liquid above the datum line. H2=height of heavy liquid above the datum line. S1=specific gravity of light liquid. 22  S2=specific gravity of heavy liquid. Total pressure head in the left limp = Total pressure head in the right limp h + h1s1 = h2s2 h=h2s2-h1s1 Measuring negative (vacuum) (or) suction pressure To measure the suction pressure connect one of the limbs of the u-tube to the pipe and other limb is open to atmosphere. The level of the mercury in the manometer will be shown in fig.1.11. Let Take Datum line is A-A. Fig.1.11 h=vacuum pressure in pipe line. h1=height of liquid in the left limb above mercury level. h2=height of mercury in the left limb above AA. S1- Specific gravity of liquid in pipe line. S2=specific gravity of mercury. Total pressure head in the left limp = Total pressure head in the right limp h+h1 s1+h2s2=atmospheric pressure pr.head h=-(h1s1+h2s2) Since atmospheric pressure is neglected _ ve sign indicate the vacuum pressure. 1.12.15. Differential U tube manometer. Differential manometer is used to measure the difference of pressures between two points in the same pipe line or in two different pipes. 23  It consists of a simple u tube and containing a heavy liquid whose two ends are connected to the points whose difference of pressure is to be measured. Let the two points A and B are at different level and also contains liquids of different specific gravity. Fig.1.12 These two points are connected to the u tube differential manometer as shown in fig.1.12. Let hA=pressure of liquid in the pipe A. hB=pressure of liquid in the pipe B. h1=height of liquid of pipe A in the left limb above AA. h2=height of mercury in the right limb above AA. h3=height of liquid of pipe B in the left limb above mercury level. S1=specific gravity of liquid in pipe “A” S2=specific gravity of mercury in pipe “A”. S3=specific gravity of liquid in pipe “B”. 24 Total pressure head in the left limp = Total pressure head in the right limp ha+h1s1=hb+h2s2+h3s3 ha - hb =h2s2+h3s3 - h1s1 m of water. Let the two points A & B are same level and contains liquid of different specific gravity as shown in fig.1.13.then Fig.1.13. Total pressure head in the left limp = Total pressure head in the right limp hA + h1s1 = hB + h2s2 + h3s3 hA - hB = h2 s2+ h3s3-h1s1 = h2 s2+ h3s3-(h2 + h3)s1 hA - hB = h2(s2-s1) + h3(s3-s1) 1.12.16. Inverted Differential U-tube manometer This manometer is used to measure the difference of low pressure between the two points. Fig.1.14 25  It consists of an inverted “U” tube. It containing a light liquid. The two ends of the tube are connected to the point whose difference of pressure is to be measured. Fig.1.14. shows an inverted U-tube differential manometer connected to the two points A and B. The pressure at “A” is more than the pressure at “B”. Let h1=height of liquid in left limb below the datum line AB. h2 = height of liquid in right limb. h=difference of light liquid. S1=specific gravity of liquid at “A”. S2=specific gravity of liquid at “B”. S3=specific gravity of light liquid. ha=pressure of liquid in the pipe “A”. hb=pressure of liquid in the pipe “A”. Taking A-A as datum line. Pressure in the left lime below AA = pressure in the right limb below “AA”. ha- h1s1=hb-h2s2- hs3 (ha -hb)=h1s1-h2s2-hs3 1.12.17. Micro manometer Micro manometer is used for measuring low pressure with high degree of accuracy. Micro manometer is a modified form of manometer. In this manometer cross sectional area of one of limb (left limb) is made much larger than (about 100 times) that of the other limb. Micro manometer is shown in fig.1.15 Types of micrometer 1. Vertical tube micro manometer 2. Inclined tube micro manometer 1. Vertical tube micro manometer Now consider a vertical tube micro manometer connected to a pipe containing light liquid under a very high pressure. The pressure in the pipe will force the light liquid to push the heavy liquid in the basin downwards. Due to larger area of the basin, the fall of heavy liquid level will be very small. This downward movement of the heavy liquid in the basin will cause a considerable rise of the heavy liquid in the right limb. 26  Let us consider our datum line Z-Z corresponding to heavy liquid level before the experiment. Fig 1.15 Vertical Micro manometer ᵟh=fall of heavy liquid level in the basin in cm. h1=height of liquid above the datum line in cm. h2=height of heavy liquid (after experiment) in the right limb above this datum line in cm. h=pressure in the pipe, expressed in terms of head of water in cm. A=cross sectional area of the basin in cm2. a=cross sectional area of the tube in cm2. S1=specific gravity of the light liquid and S2=specific gravity of the heavy liquid. We known that the fall of heavy liquid level, in the basin, will cause a corresponding rise of heavy liquid level, A ᵟh =ah2 or ᵟh = h2 Now let us take horizontal surface in the basic, at which the heavy and light liquid meet, as datum line. We also known that the pressures in the left limb and right limb above the datum line are equal. Pressure in the left limb above the datum line =h+s1h1+s1ᵟh cm of water and pressure in the right limb above the datum line. 27  =s2h2+s2ᵟh cm of water equating these two pressure, h+s1h1+s1 ᵟh = s2h2+s2ᵟh or h=s2h2+s2ᵟh -s2h1-s1ᵟh h=s2h2-s1h1+ h2(s2-s1) 2. Inclined tube micromanometer Fig. 1.16 Sometimes, the vertical tube of the micromanometer is made inclined as shown in fig.1.16 This type of inclined micromanometer is more sensitive than the vertical tube.Due to inclination the distances moved by the heavy liquid, in the narrow tube, will be comparatively more and thus it can give a higher reading for then given pressure. From he geometry of figure, we find h2 =sin α h2=1.sin α By substituting the value of h2 in the micro manometer equation, we can find out the required pressure in the pipe. 1.12.18. Problem 1) Determine the pressure in a pipe containing a liquid of specific gravity 0.8. A micro manometer was used a s shown in fig.1.17. The ratio of area of the basin to that of the limp is 50. 28  Fig. 1.17 Given Sp. Gravity of liquid in the pipe S = 0.8 Ratio of area of the basin to that of the limb (A/a) = 50 Height of liquid in the left Limb h1 = 10 cm = 0.1 m Height of liquid in the right Limb h2 = 20 cm = 0.2 m Sp.Gravity of Mercury S2 = 13.6 Solution Pressure in the pipe h =s2h2-s1h1+ h2(s2-s1) m of water = (13.6x20) – (0.8x10) + x 20(13.6-0.8) cm of water = 269.12 cm of water h = 2.6912 m of water Pressure in the pipe p = W h = 9810 x 2.6912 = 26.40 x 103 N/m2 Note: W – Specific weight of water = 9810 N/m3 1.13. MECHANICAL GAUGES Introduction Mechanical gauges are one of the direct pressure measuring instruments. Example the pressure gauges are known as mechanical gauges. Types of mechanical pressure gauges  Bourdons tube pressure gauge.  Diaphragm pressure gauge.  Dead weight pressure gauge.  Bellows pressure gauge. 29 Advantages of mechanical gauges  Durable  Portable  Giving direct reading  Long life 1.13.1. Bourdons pressure gauge Fig.1.18 30  Fig.1.19 This gauge used for measuring high pressure. The principle used in this gauge is an elastic deformation of a metallic tube which is proportional to the fluid pressure as shown in fig.1.18. It consists of the following parts. Parts 1.Elliptical tube 2.Toothed sector 3.Links 4.Pinion 5.Pointer Function When the gauge is connected to the pressure enters into an elliptical spring tube. The fluid under the pressure of fluid, the tube tends to deform. As a result of increased pressure of fluid the tube tends to deform. One end of elliptical tube is fixed. The free end moves out. This elastic deformation of the bourdons tube rotates the pointer through the links. This pointer moves over a calibrated dial which shows the pressure reading directly. 31 1.14. Important Formula 1. Pressure p = W x H 2. Absolute Pressure = Atmospheric pressure + Gauge pressure 3. Absolute Pressure = Atmospheric pressure - Vacuum pressure 4. Atmospheric pressure = 760 mm of Hg = 1.01325x105 Pascal = 1.01325x105 N/m2 5. Simple U tube Manometer h = h2S2 – h1S1 6. Differential Manometer ha-hb = h2S2+h3S3-h1s1 7. Inverted Differential Manometer ha-hb = h1s1- h2S2-hS3 1.15. Simple u tube manometer problems 35 Fig. 1.23 Fig 1.23 36 Problems 1. Determine the pressure at a depth of 20 m in an oil of relative density 0.8. Given data Depth of Oil h = 20 m Relative density of oil = 0.8. Solution Relative Density = 0.8 = Specific weight of oil = 0.8 x 9810 Pressure of oil = Specific weight of oil x h = 0.8 x 9810 x 20 Pressure of oil =156.96x103 kN/m2 2. Express the pressure intensity in absolute pressure in i)kN/m2 ii)in meter of water. if gauge pressure is 0.7356 N/mm2 Given Data Gauge Pressure = 0.7356 N/mm2 = 0.7356x 106 N/m2 37 Solution i) absolute pressure = Atmospheric pressure +Gauge pressure = 1.01325 x 105 + 0.7356x 106 absolute pressure = 8.37 x 105 N/m2 ii) Atmospheric pressure = 10.33 m of water Gauge Pressure = 0.7356x 106 N/m2 = [P=wH] = 74.985 m of water absolute pressure = Atmospheric pressure +Gauge pressure = 10.33 + 74.985 absolute pressure = 85.315 m of water 3. A gauge fitted to a gas cylinder records a pressure of 16.27 kN/m2. Compute the corresponding absolute pressure in i)kN/m2 ii)in meter of water. The Atmospheric pressure is 700 mm of Hg Given Data Atmospheric pressure = 1.01315 x 105 N/m2 = 101.325 KN/m2 Note: The given pressure is lower than the Atmospheric pressure. Hence the pressure is taken as vacuum pressure. vacuum pressure = 16.27 kN/m2 given atmospheric pressure = 700 mm of Hg = 0.7 m of Hg = 0.7x13.6 m of water = 0.7x13.6x9.810 KN/m2 = 93.39 KN/m2 Solution i) absolute pressure = atmospheric pressure - vacuum pressure = 93.39 - 16.27 Pabs=77.12 KN/m2 ii) absolute pressure in m of water P=wH H=P/w = 77.12/9.810 38  H = 7.86 m of water 4. A gauge records a pressure of 24.52 kN/m2 vacuum. Compute the corresponding absolute pressure in i) kN/m2 ii) in meter of water. The local atmospheric pressure is 0.75 m of Hg. Specific gravity of mercury is 13.6. Given Data vacuum pressure = 24.52 kN/m2 local atmospheric pressure(h) = 0.75m of Hg = 0.75x13.6 m of water = 0.75x13.6x9.810 KN/m2 [p=wh] p= 100.062 KN/m2 w-specific weight of water W=9810 N/m2 = 9.810 KN/m2 Solution i) absolute pressure = atmospheric pressure - vacuum pressure = 100.062 - 24.52 Pabs=75.542 KN/m2 ii) absolute pressure in m of water P=wH H=P/w = 75.542 /9.810 H = 7.7 m of water 5. The pressure of water in a pipe line was measured by means of a Simple manometer containing mercury. The mercury level in the open tube is 150 mm higher than that of the left tube. The height of water level in the left tube is 40 mm. Determine the pressure in the pipe i) in m of water ii) kN/m2 Given data 39  h1= 40 mm h2= 150 mm S1(Specific gravity of water) = 1 S2(Specific gravity of mercury) = 13.6 Solution total pr.head in left limp = total pr.head in right limp hA + h1s1 = h2s2 i) = (150x13.6)-(40x1) ha = 2000 mm = 2 m of water ii) pressure P = w ha = 9.810x2 p = 19.62 KN/m2 6. The left limb of a U tube manometer containing mercury is open to the atmosphere and the right limb is connected to the pipe line carrying water under pressure. The centre of pipe is at the face surface of mercury. Find the difference in level of mercury in limbs if the absolute pressure of water in the pipe is 12.5 m of water. Given Data absolute pressure = 12.5 m Gauge pressure (ha) = absolute pressure – atmospheric pressure = 12.5-10.33 atmospheric pressure = 10.33 m ha = 2.17 m of water solution total pr.head in left limp = total pr.head in right limp 40  h1s1 = hA + h2S2 here h1 = h2 let take h1 = h2 = h h (S2-S1) = hA h = hA/(S2-S1) h = 2.17 / (13.6-1) Difference in Hg level h = 0.172 m 7. A differential manometer connected to two point A & B in a pipe line containing an oil of relative density 0.8as shown in fig 1.26. Difference in mercury level is 100mm. Determine the difference in pressure between the two points in terms of i) m of water ii) kN/m2 abs Given data Relative density (or) Specific gravity of oil (S1)= 0.8 Specific gravity of mercury (S2)= 13.6 Difference in mercury level h2=100mm To find i) ha-hb in m of water ii) Pa-Pb in kN/m2 Solution Note Both points connected in the same pipe and same liquid i) ha-hb = h2(S2-S1) = 100(13.6-0.8) = 1280 mm of water = 1.280 m of water ii) Pa-Pb = w(ha-hb) = 9.810x1.280 = 12.556 kN/m2 Pr difference (gauge) Pa-Pb = 12.556 kN/m2 Pr difference (abs) Pa-Pb = Atm Pr + Gauge Pr = 101.325 + 12.556 41  Pr difference (abs) Pa-Pb = 113.856 kN/m2 8 A vacuum pressure in a pipe line carrying water was measured by U tube manometer, the deflection of mercury between the lines was 0.05 m and the free surface of mercury in the open limb was 0.1 m below the centre line of the pipe. Find the pressure in the pipe in absolute unit in terms of m of water Given data h1 - height of liquid in the left limb above the mercury level = 0.05 m h2 - height of liquid in the left limb above AA = 0.05 m S1- Specific gravity of water = 1 S2- Specific gravity of mercury = 13.6 Solution Formula h = h1S1+ h2S2 = 0.05x1+0.05x13.6 h= 0.73 m of water absolute pr= Atm Pr – vacuum Pr = 10.33-0.73 = 9.6 m of water Absolute Pr h = 9.6 m of water 9 A Differential manometer connected to two pipes A & B as shown in fig 1.28. The pipe A contains liquid of specific gravity 1.5 and the pipe B contains liquid of specific gravity 0.85 and the difference in pressure between the two pipes is 10 m of water. Find the difference in mercury level Given data S1 = 1.5 42  S2 = 13.6 S3 = 0.85 ha- hb = 10 m h3 = 3 m h 1 = h2 + 3 Solution ha + h1S1 = hb + h2S2 + h3S3 (ha -hb) + h1S1 = h2S2 + h3S3 10 +( h2 + 3)1.5 = h2 x13.6 + 3 x 0.85 after simplification h2 = 0.988 m difference in mercury level h2 = 0.988 m = 988 mm 10 A Differential manometer connected to two pipes A & B. The pipe A contains carbon tetra chloride having relative density 1.594 under a pressure of 118 kN/m2.The pipe B contains oil of specific gravity 0.8 under a pressure of 200 kN/m2.The pipe A lies 2.5 m above pipe B.The centre of pipe B is at the level of face surface of mercury in the pipe A. Find the difference in mercury level Given data PA = 118 kN/m2 PB= 200 kN/m2 S 1= 1.594 S 2= 13.6 S3= 0.8 ha=Pa/W = 118 / 9.810 = 12.03 m of water hb=Pb/W = 200 / 9.810 = 20.387 m of water Solution sum of pressure in the left limb = sum of pressure in the right limb ha + h1S1+h2S2 = hb +h3S3 12.03+2.5x1.594+13.6 h2 = 20.387 +h2x0.8 h2 = 0.3416 m Difference in mercury level h2 = 0.3416 m 43  11 An inverted U tube Differential manometer connected to two pipes A & B both are containing same liquid of specific gravity 1.4. Its two ends are at same horizontal line.The relative density of manometric liquid is 0.8. Find the difference in pressures between the two pipes. The manometer reading is 370 mm. Fig1.30 Given Data S1=S3=1.4 S2=0.8 h2=370 mm = 0.37 m h1=h2+h3 = 0.37 + h3 To find Difference in pressure Pa-Pb Solution ha- h1S1 = hb -h2S2-h3S3 ha – (0.37+ h3) x 1.4 = hb - 0.37 x 0.8 – h3 x 1.4 ha – hb = - 0.37 x 0.8 – h3x1.4+0.37x1.4 + h3x1.4 = 0.222+0 = 0.222 m of water Pa-Pb = w(ha-hb) = 9.810 x 0.222 Pa-Pb = 2.178 kN/m2 44 1.16. Numerical Problems Note: Pr – Pressure Atm – atmospheric abs-absolute Ans-answer respt-respectively 1. Convert a pressure head of 500 mm of mercury into oil of relative density o.75 Ans 9.066 m of Oil 2. A driver can sustain a pressure of 110 kN/m2 under sea water. Upto what depth under a sea water he can work. Ans 11.213 m 3. The pressure at a point which is 40m below the surface of sea is 439.49 kN/m 2. Find the density of sea water. 4. A gauge fitted to a compressor shows a reading of 30 kN/m 2. Compute the corresponding absolute pressure. Assume the local atm Pr is 101.325 kN/m 2. Ans 131.325 kN/m2 5. Measurement at the base and top of mountain are 720 and 550 mm of mercury respt. Compute the height of mountain when the air has a constant density of 11.97 kN/m2. Ans 1.89 m. 6. A siple manometer is used to measure the Pr of water flowing in a pipe line. Its right limb is open to atm and the left limb is connected to the pipe. The centre of the pipe is in level with that of mercury in the right limb. Determine the Pr in the pipe if the difference of mercury level is 125 mm Ans 1.575 m of water. 7. A simple U tube manometer is used to measure the pressure of oil of relative density 0.75 flowing in a pipe line. Its right limb is open to atm and the left limb is connected to the pipe. The centre of pipe is 0.1 m below the level of mercury in the right limb. If the difference of mercury level is 0.2m. Determine the Abs Pr of oil in the pipe 8. A differential manometer connected to two point A & B in a horizontal venturimeter containing an oil of relative density 0.8. If the deflection of mercury level is 0.8 m. Determine the difference in pressure between the two points in terms of kN/m2 Ans 100.54 kN/m2 9. A differential manometer connected to a vertical pipe. The oil of relative density 0.85 is flowing through the pipe. The distance between the two gauge points is 1 m. Difference of mercury level is 0.2m. Determine the difference in pressure between the two points in terms of kN/m2. Ans 70.87 kN/m2 10. An inverted U tube Differential manometer connected to two pipes A & B carrying liquid of specific gravity 1.6. Its two ends are at same horizontal line. The relative density of manometric liquid is 0.75. Find the difference in pressures between the two pipes. The manometer reading is 400 mm. Ans 0.34 m of water. 45  DEPARTMENT OF MECHANICAL ENGINEERING Course Name : 1020-Diploma in Mechanical Engineering Subject Code : 4020410 Semester : IV Subject Title : Fluid Mechanics and Fluid Power UNIT-II Fluid Flow, Flow Through Pipes & Impact of Jet UNIT-II FLOW OF FLUIDS AND FLOW THROUGH PIPES Objectives  Discuss the types of fluid  Explain the energies of fluid  State and prove Bernoulli’s theorem  Explain the applications Bernoulli’s theorem  Problems solve in venturi meter orifice meter  Discuss about hydraulic coefficients and their relations  Explain the methods of finding hydraulic coefficients experimentally.  Explain the major & minor losses of energies in flow through pipes  Derive Darcy’s & Chezy’s formula  Solve problems in power transmission through pipes 2.1. FLOW OF FLUIDS Introduction  Hydro statics deals with the study of liquid when it is at rest.  Hydro dynamics deals with the motion of liquid by taking into consideration the force (or) energy causing the flow.  Hydro kinematics deals with the study of velocity and acceleration of the liquid particles with out taking into consideration the force or energy causing the flow. Types of fluid flow The type of flow depends upon the conditions of fluid flow. 1. Laminar flow 5.uniform flow 2. Turbulent flow 6.Non-uniform flow 3. Steady flow 7.compressible flow. 4. Unsteady flow 8.In compressible flow. 46 2.1.1. Laminar flow Fig. 2.1.a Fig. 2.1.b  Laminar flow is a smooth and regular flow.  Laminar flow is a flow in which each liquid particle has a definite path, and the paths of individual particles do not across with each other.  This type of flow is called as stream line flow. Conditions for this type of flow  Low velocity of flow.  Highly viscous fluid.  Reynolds number is less than 2000.  Ex : high viscous oil moves slowly in a very small pipe. 2.1.2.Turbulent flow (eddy flow)  Turbulent flow is flow in which the velocity of a particle will very in magnitude and direction form point and time to time.  In this flow each liquid particles does not have a definite path and moves in and irregular way.  Turbulent flow is shown in fig.2.1 (b)  It is also called eddy flow. Conditions for this type of flow  High velocity of flow.  Reynolds number is more than 2000. Ex: flow of water in the pump. flow in river during floods, Waterfalls. 2.1.3.Steady flow  If at any section of flow the quantity of liquid flowing per seconds is constant, the flow is called steady flow. (or)  If the liquid properties such as velocity, pressure, density at any point of liquid do not change with respect to time, it is called steady flow. Ex: Flow through a tap, when the water level in a tank is constant. 47 2.1.4.Unsteady flow If at any section of flow, the quantity of liquid flowing per second is not constant the flow is called unsteady flow. Ex: flow through a tap when the head is not constant. 2.1.5.Uniform flow If the magnitude and direction of velocity at a point of liquid in motion do not vary with respect to time then the flow is called uniform flow. Ex: flow of fluids through a pipe line whose cross sectional area is constant. 2.1.6. Un – uniform flow If the magnitude and direction of velocity change from point to point in a liquid flow, it is called uniform flow. 2.1.7.Compressible flow Compressible flow in which density of fluid is same in all sections. Ex: liquid flow in pipe. 2.1.8. Incompressible flow In compressible flow in which density of fluid is not same in all section. Ex: Gas flow in pipe. 2.2. Type of lines in fluid flow 1. Path line 2. Stream line 2.2.1.Path line Fig 2.2. Path lines The path followed by a fluid particle in motion is called a path line. Thus the path line shows the direction of a particles, for a certain period of time. 2.2.2.Stream line Fig 2.3 stream line 48  The imaginary line drawn in the fluid in such way that the tangent to which at any point gives the direction of motion at that point is called stream line. Stream line shows the direction of motion of a number of particle at the same time. 2.3. Rate of flow (or) discharge Fig.2.4 Volume of liquid flowing per seconds is known as the rate of discharge or simply discharge. Consider a liquid flowing through the pipe as shown in fig.2.4 Let a=area of the pipe. V=velocity of flow Q=rate of discharge Discharge =volume of cylinder =area x velocity Discharge Q =a x v unit :m 3 / sec. 2.4.Continuity equation Fig 2.5a “If an incompressible liquid is continuously flowing through a pipe or a channel whose section may or may not be uniform the quantity of liquid passing per sections”. Consider a tapering pipe through which some liquid is flowing as shown in fig.2.5a Let a1 = area of pipe in section 1-1 v1 = velocity of liquid in section 1-1 Similarly a2v2 = corresponding values at section 2-2 Total quantity of fluid passing through section 1-1 Q1 = a1 v1 49  Total quantity of fluid passing through section 2-2 Q2 = a2 v2 Similarly Total quantity of fluid passing through section 3-3 Q3 = a3v3 Discharge is same through all sections. Hence Q= a1 v1= a2 v2= a3v3 This is called as continuity equation. 2.5.Mean velocity of fluid flow (Vavg): Fig.2.5b The fluid velocity in a pipe changes from zero at the surface because of the no-slip condition to a maximum at the pipe center. In fluid flow, it is convenient to work with an Average velocity Vavg or mean velocity , which remains constant in incompressible flow when the cross-sectional area of the pipe is constant as shown in fig.2.5b. The regime of flow when velocity is lower than "critical" is called laminar flow (or viscous or streamline flow). At laminar regime of flow the velocity is highest on the pipe axis, and on the wall the velocity is equal to zero. When the velocity is greater than "critical", the regime of flow is turbulent 2.6.Energy of a liquid in motion Energy is defined as the capacity to do word. A fluid in motion is said to possess only the following three forms of energies. 1.potential energy 2.kinetic energy 3.pressure energy 2.6.1.potential energy of a liquid in motion Potential energy is the energy possessed by a liquid particle, by virtue of its position. If a liquid particle is “Z” meters above the horizontal datum- line. Potential energy = WZ unit N-m Where, W = Weight of liquid (N) Z=height of liquid above the datum line (m). 50 2.6.2.Kinetic energy Kinetic energy is the energy possessed by a liquid particle, by virtue of its moment (or) velocity. Kinetic energy = ½ m v2 =1/2 (w/g) v2 Where, m=mass of liquid m=wg W=weight of liquid (N) V=velocity of the liquid (m/sec.). 2.6.3.pressure energy Pressure energy is the energy possessed by virtue of its existing pressure. If a liquid particle is under a pressure of “P” (kg/m2) then the pressure energy of the particle will be p/w mk /kg of the liquid. Where w = specific weight of the liquid. Total Energy = Potential Energy + Kinetic Energy + Pressure Energy = WZ + W +W Total Energy = W[Z + + ] Datum head Datum head is defined as the height of the liquid above the datum line. It is also defined as the potential energy per unit weight of a liquid. Potential energy = W x Z Datum head = Z velocity head  Velocity head is defined as the height of liquid. Corresponding to a particular velocity of the liquid.  It is also defined as the kinetic energy of unit weight of a liquid.  Kinetic energy = W v2/2g  Velocity head = v2/2g Pressure head The height of liquid corresponding to a particle pressure is known as pressure head. Pressure head, H=P/W Total head of liquid Total head of a liquid is defined as the sum of datum head pressure head & velocity head. Total head = datum head+ pressure head + velocity head. 51  Total Head = [Z + + ] 2.7. Bernoulli’s theorem Bernoulli’s theorem states “for a perfect in compressible liquid, flowing in a continuous stream, the total energy of a particle remains same while the particle moves from one point to another point”. Total Energy = [Z + + ]= constant Fig 2.6 Bernoulli’s theorem Let take two sections AA and BB of the pipe. Let Z1 = Height of AA above the datum P1 = Pressure at AA V1 = Velocity at AA a1 = area of the pipe at AA Z2, P2, V2, a2 corresponding values at BB  Let the liquid move from AA to A’A’ and from BB to B’B’ through the small length dl 1 and dl2  Quantity of liquid movements is constant  Hence the shaded volumes are equal  Let “W” be the weight of the liquid between AA and A’A’ W=wa1dl1 = wa2dl2 a1dl1 = W/w--------------(1) a2dl2 = W/w ----------(2) W-Total weight of liquid a1dl1 = a2dl2 ----------(3) w- specific weight of liquid Work done by the pressure at AA in moving the liquid to A’A’= Force x Distance = P1a1dl1 52 Similarly Work done by the pressure at BB in moving the liquid to B’B’= -P2a2dl2 Note (-ve) sign is taken Pressure P2 is opposite to P1 Total workdone by the pressure = P1a1dl1 - P2 a2dl2 Here a1dl1 =a2dl2 = P1a1dl1 - P2 a1dl1 = a1dl1(P1-P2) = [P1-P2] from (1) Loss of potential energy = W[Z1-Z2]----------------------(4) Gain in kinetic energy = W[ - ]----------------(5) = [v22-v12] -------------------(6) Loss of potential energy + Work done by the pressure = Gain in kinetic energy W[Z1-Z2] + [P1-P2] = [v22-v12] [Z1-Z2] + [ - ] = z1 + + = z2 + + ------------(7) Hence Bernoulli’s equation is proved Assumptions in Bernoulli’s equation  Flow is incompressible and homogeneous.  Flow is steady.  Flow is continuous.  Flow is ideal.  Flow is one – dimensional.  Flow is along a stream line.  No energy transfer to the flow (or) from the flow. Limitations of Bernoulli’s theorem 1.velocity of flow is constant. 2.frictional force is neglected. 3.no loss of energy is assumed. Application of Bernoulli’s theorem 1. Venturimeter 2. Orifice meter 53  3. Pitot tube 4. Nozzle meter or flow Nozzle. 2.8.Verturimeter Verturimeter is an instrument used for measuring the discharge quantity of liquid flowing in a pipe Tthe diagram is shown in fig.2.7. It consists of 1. Convergent cone 2. Throat 3. Divergent cone. Fig: 2.7.Venturimeter Convergent cone It is a short pipe which converges from a diameter (d1) to a smaller diameter (d2). Convergent cone is also known as inlet of the venturimeter. Throat It is a small portion of circular pipe, in which diameter d2 is kept constant. Divergent cone It is a pipe which diverges from a diameter (d2) to a large diameter (d1). Divergent cone is also known as outlet of the discharge through a venturimeter. The pressure head difference between the enlarged and throat is measured by a differential manometer. Consider a horizontal venturimeter through which some fluid is flowing as shown in fig 2.8 Applying Bernoulli’s equation at sections 1-1 and 2-2 z1 + + = z2 + + For horizontal venturimeter Z1 = Z2 then the above equation becomes 54  + = + -------------------(1) - = - ------------------(2) Hence the discharge at section 1-1 and 2-2 are constant Q = a1v1 = a2v2 v1 = v12 = --------------------------------(3) Pressure head difference - =h = - = [v22-v12] From equation (3) h = [v22- ] a12-a22 After simplification v2 = ] --------------(4) From Equation (4) v2 value can be found Discharge through venturimeter (Qt) = a1v1 = a2v2 Coefficient of discharge Cd = Actual discharge Qa = Cd Qt = Cd a2 ] ------------------(5) Venturimeter constant C = a2 ] Put equation C value in (5) then (5) becomes Actual discharge Qa = CdC 2.9. Orifice meter Orifice meter is used to measure the discharge of the liquid flowing in a pipe. It consist of sharp edged circular hole plate and is fixed inside the pipe. The working principle of Orifice meter is similar to Venturimeter. A differential manometer is used to measure the pressure difference between the two sides of Orifice meter. 55  Fig.2.8.a Fig.2.8.b Fig 2.9 Orifice meter h =reading of differential manometer P1= Pressure at inlet V1 = velocity at inlet A1 = area of the pipe at inlet P2 v2 & a2 = corresponding values at throat. Now apply Bernoulli’s equation z1 + + = z2 + + For horizontal Orificemeter Z1 = Z2 then the above equation becomes + = + -------------------(1) 56  - = - ------------------(2) Hence the discharge at section 1-1 and 2-2 are constant Q = a1v1 = a2v2 v1 = v12 = --------------------------------(3) Pressure head difference - =h = - = [v22-v12] From equation (3) h = [v22- ] a12-a22 After simplification v2 = ] --------------(4) From Equation (4) v2 value can be found Discharge through venturimeter (Qt) = a1v1 = a2v2 Coefficient of discharge Cd = Actual discharge Qa = Cd Qt = Cd a2 ] ------------------(5) Orificemeter constant C= ] Put equation C value in (5), then (5) becomes Actual discharge of Orificemeter Q a = Cd C Differences between venturimeter & Orificemeter Si.No Venturimeter Orificemeter 1 Used in large pipes Used in small pipes 2 Cd is high Cd is less 3 Require more space Require less space 4 Losses are less Losses are high 5 High pressure recovery is attainable pressure recovery is poor 57 2.10.Pitot Tube Pitot tube is an important instrument used to measure the velocity of flow in a river or in open channel. It is a glass tube, both ends are open and bent through 90 deg as shown in fig 2.10. The Liquid flow in to the tube and rises in the tube until all its kinetic energy converted into the potential energy. Fig 2.10 pitot tube  By measuring the rises of liquid in the tube, the velocity of liquid can be determined. Let h = Height of liquid in the pitot tube above the water surface H = Depth of the tube in the liquid V = velocity of the liquid Applying Bernoulli’s equation z1 + + = z2 + + H+ = H+h After simplification theoretical velocity (Vt) = Then actual velocity (Va) = Cv x 2.11. Important Formula 1. Rate of discharge Q = AV 2. Continuity equation a1v1 = a2v2= a3v3= constant 3. Potential energy of liquid in motion = WZ 4. Kinetic energy = W 58 5. Pressure energy =W 6. Total energy = WZ + W + W 7. Total head of liquid = Z + + 8. Bernoulli’s equation z1 + + = z2 + + 9. Venturimeter & Orificemeter Q a = Cd C Constant C = ] 10. Pitot tube actual velocity Va = Cv x 2.12.Solved problems 1. Determine the diameter of a pipe line which carries 100 lit/min of water with a velocity of 0.25 m/s Given Data Discharge(Q) = 100 lit/min = = 0.0017 m3/s Velocity (v) = 0.25 m/s To Find Diameter of pipe(d) Solution Q=axv = xV 0.0017 = x 0.25 After simplification d = 0.093 m = 93 mm 2. A pipe line tapers from 80 mm to 40 mm diameter. The discharge through the pipe is 0.2 m3/sec. Find the average velocities at the two sections Given Data d1 = 80 mm = 0.08 m 1 mm = (1/1000) m d2 = 40 mm = 0.04 m 59  Discharge Q = 0.2 m3/sec To find velocities V1 &V2 Solution Area of larger end a1 = = = 5.026 x 10-3 m2 Area of larger end a2 = = = m2 Discharge Q = a1v1 = a2v2 0.2 = 5.026 x 10-3 x V1 V1 = 39.789 m/sec Q = a2v2 0.2 = 1.2566 x 10-3 x v2 v2 = 159.15494 m/sec 3. A 0.3 m diameter pipe carrying water, branches into two pipes of diameter 0.2 m and 0.15 m. If the mean velocity of 0.3 m pipe is 2.5 m/s and that in the 0.2 m pipe is 2 m/s. Determine the discharge in the pipes and the velocity in the 0.15 m pipe. Given data D = 0.3 m d1 = 0.2 m d2 =0.15 m V = 2.5 m/s v1 = 2 m/sec v2 = ? 60  Solution Discharge in main pipe(Q) = Discharge in pipe-1 (Q1)+ Discharge in pipe-2 (Q2) Discharge in main pipe(Q) = A x V = xV = x 2.5 Discharge in main pipe(Q) = 0.1756 m3/sec Discharge in pipe -1 (Q1) = a1 x v1 = x v1 = x2 Discharge in main pipe(Q1)= 0.0628 m3/sec Q = Q1 + Q2 0.1756 = 0.0628 + Q2 Q2 = 0.1137 m3/sec Discharge in pipe -2 (Q2) = a 2 x v2 = x v2 0.1137 = x v2 After simplification V2 = 6.42 m/sec 4. A pipe is running full of water. At a point “A” in the pipe the velocity of flow is 1 m/sec and the pressure is 1.875 kN/m2. If “A” is 16.5 m above the datum, determine the total energy of 1 kg of water at point A. If 30 kg of water moving in the pipe then find the total energy in joules. Given Data V = 1 m/sec P = 1.875 kN/m2 Z = 16.5 m 61  Solution Total Energy = WZ + W + W here W = mg = 1x9.81 = 9.81 N = 9.81x16.5+9.81 x + 9.81 x = 161.865+1.875+0.5 Total Energy/kg = 164.24 Nm/kg (or) J/kg Total Energy for 30 kg of water = (Total Energy/kg) x 30 = 30x164.24 Total Energy for 30 kg of water = 4927.2 Joules 5. A pipe 300 m long has a slope of 1 in 100 and tapers from 1.5 m diameter at the higher end and 0.625 m diameter at the lower end. The discharge of water through the pipe is 100 lit/sec. The pressure at the higher end is 110 kN/m 2. Find the pressure at the lower end and neglecting the friction. Given Data Discharge(Q) = 100 lit/sec = 100/1000 = 0.1 m3/sec P2 = 110 kN/m2 d2 = 1.5 m d1 = 0.625 m length = 3 m slope = 1:100 To find Pressure at lower end P1 62 Solution slope = 1:100 Vertical height = x 300 = 3 m Hence Z1 = 0 & Z2= 3 m Discharge (Q) = a1v1 = a2v2 Q = a1v1 = x v1 0.1 = x v1 V1 = 0.32594 m/sec Q = a2v2 = x v2 0.1 = x v2 V2 = 0.05658 m/sec Apply Bernoulli’s equation z1 + + = z2 + + 0+ + =0+ + After simplification P1 = 139.378 k/Nm2 Pressure at lower end P1 = 139.378 k/Nm2 6. A pipe line is carrying full of water at a point ‘A’ in the pipe line the diameter is 600 mm and the pressure is 70 kN/m2 and velocity is 2.4 m/sec. At another point ‘B’ in the same pipe line the diameter is 300 mm and the pressure is 14 70 kN/m 2 and is 2 m higher than “A”. Determine the direction of flow. Given Data d1 = 600 mm = 0.6 m P1 = 70 kN/m2 V1 = 2.4 m/sec 63  d2 = 300 mm = 0.3m P2 = 14 kN/m2 To find Direction of flow Solution Find total head at point A & B and then find the Direction of flow Apply Bernoulli’s equation z1 + + = z2 + + -----(1) First find velocity at B V2 Continuity euation Q = a1v1 = a2v2 ------------------------------(2) a1 = = = 0.2827 m2 a2 = = = 0.07068 m2 put a1&a2 values in (2) 0.2827 x 2.4 = 0.07068 x V2 V2 = 9.6 m/sec Total head at point (1) = z1 + + =0+ + Total head at point (1) = 7.429 m Total head at point (2) = z2 + + =0+ + Total head at point (2) = 8.1243 m Hence Total head at point (2) > Total head at point (1) 64 Direction of flow is (2) to (1) 7. A tapered section pipe is running full of water. The diameter of pipe at inlet and outlet are1.0 m and 0.5 m respectively. The outlet is at a vertical height of 5 m above the inlet. The loss of head in the pipe is 1/10 th of velocity head at outlet. The pressure at the outlet section is 100 kN/m 2 and at inlet is 400 KN/m2. Calculate the rate of discharge through the pipe. Determine also velocities at inlet and outlet Given Data d2 = 0.5 m P1 = 400 kN/m2 V1 = 2.4 m/sec d1 =1 m Z1 =0m Z2 =5m Loss of head hc = 1/10 x velocity head at outlet = 1/10 x To find Discharge (Q), V1 & V2 Solution a1 = = = 0.78539 m2 a2 = = = 0.196349 m2 65  by continuity equation a1v1 = a2v2 0.78539 x 2.4 = 0.196349 x v2 v2 = 4 v1 Apply Bernoulli’s equation z1 + + = z2 + + + hc 0+ + = 5+ + + 0+ + = 5+ + + After simplification V1 = 5.498 m/sec V2 = 4 x V 1 = 4 x 5.498 V2 = 21.995 m/sec Discharge Q = a1v1 = 0.78539 x 5.498 Q = 4.3186 m3/sec 8. A venturimeter is to be fitted to a 250 mm diameter pipe in which the maximum flow is 120 lps and the pressure head 6 m. What is the minimum diameter at throat that there is no (-ve) head in it. Cd = 0.97. Given data Q = 120 lps(lit/sec) = 0.120 m3/sec d1 = 250 mm = 0.250 m Solution a1 = = = 0.0491 m2 Pressure head at inlet = H= 6 m Coefficient of discharge Cd = 0.97 no (-ve) head means , the pressure head at that point must be equal to zero hence Pr head at throat =0 Discharge Q = Cd.C 0.12 = 0.97 x Cx C = 0.0505 66  C= ] 0.0505 = ] a2 = 0.01172 m2 a2 = 0.01172 = Diameter at throat d2 = 0.112 m 9. An oil of specific gravity of 0.9 is flowing through a venturimeter having inlet diameter 20 cm and throat diameter 10 cm. The mercury manometer shows a reading 20 cm. Calculate the discharge of oil through the horizontal venturimeter. Take Cd = 0.98 Given Data d1 = 20 cm =0.2 m d2 = 10 cm= 0.1 m Specific gravity of oil (Soil) = 0.9 Cd = 0.98 Manometer reading = 20 cm = 0.2 m To find Discharge (Q) Solution Formula Venturi head H = h [ - 1] S1 – Specific gravity of venturimeter fluid = 0.9 S2 – Specific gravity of Manometric fluid = 13.6 Put all values in the above formula H = 0.2 [ - 1] H = 2.82 m Area of pipe at section 1 & 2 is a1 & a2 a2 = a1 = a2 = 0.00785 m2 & a1 = 0.0314 m2 67  C= ] = ] C = 0.03591 Discharge Q = Cd.C Discharge Q = 0.98x0.03591 Q = 0.059 m3/sec 10. Venturimeter has a diameter 0.15 m at the enlarged end and a diameter of 0.08 m at the throat. Oil of specificgravity0.8 is flowing at the rate of 4.2 m 3/min. If the coefficient of meter is 0.98. Determine the deflection of mercury gauge in mm which is used to find the pressure difference. Given Data d1 = 0.15 m d2 = 0.08 m Specific gravity of oil (Soil) = 0.8 Cd = 0.98 Q= 4.2 m3/min = 4.2/60 = 0.07 m3/min To find deflection of mercury(h) Solution Area at enlarged end a1 = = 0.01767 m2 Area at throat a2 = = 5.08 x 10-3 m2 C= ] = ] venturimeter constant C = 0.00232 no unit Discharge Q = Cd.C 68  0.07 = 0.98 x 0.0232x H = 9.4593 m Venturi head H = h [ - 1] 9.4593 =h[ - 1] Deflection of mercury gauge h = 0.5912 m Theoritical Questions Part A 1. Define uniform flow. 2. Define Turbulent flow 3. Give an examples of laminar flow. 4. Give an examples of Turbulent flow. 5. Write the equation of continuity flow 6. Define steady flow 7. Write any two applications of Bernoullis theorem. 8. Write any two assumptions of Bernoullis theorem. 9. Write any two limitations of Bernoullis theorem. 10. Write the formula to find the actual discharge of venturimeter. 11. Write the formula for the total energy of flow. 12. What is the use of pitot tube 13. List any two difference of venturimeter & Orificemeter. 2.13.Numerical Problems 1. Determine the size of the pipe line in which water is flowing at the rate of 3.5 m3/sec with a velocity of 2.5 m/sec. ans d = 1.35 m 2. Find the total head and total energy per kN of the flowing fluid at the section of pipe carrying oil of specific gravity 0.8. The pipe diameter is 0.35 m. Discharge 200 lit/sec pressure at the section is 400 kN/m2. The section is 3.5 m above the datum 3. 100 lit/min of glycerine flowing in 80 mm diameter pipe line. Calculate velocity of flow. Ans v=0.33157 m/sec 4. A pipe AB inclined at 60 deg to the horizontal. The diameter at the higher end “A” of the pipe is 0.5 m and diameter at the lower end “B” of the pipe is 0.15 m. The length of pipe is 30 m. The velocity of flow at B = 2.5 m/sec and the pressure at B is 450 kN/m 2. Compute the pressure at A. Ans Pr=198.2 kN/m2. 69  Theoretical discharge Qt =1.390 x 10-3 m3/s Coefficient of Discharge Cd = = 0.00085/0.00139 Cd = 0.61 2.20. FLOW THROUGH PIPES Introduction Fig.2.22 78  Fig.2.23 2.21.Hydraulic gradient line The distance between the pipe centerline and the hydraulic grade line is the pressure head, or piezometric height, at the section. The line showing the pressure head, or piezometric height, at any point in a pipe. The slope of the hydraulic grade line is known as the hydraulic gradient. Fig.2.24 79 2.22.Total Energy line The Energy Line is a line that represent the total head available to the fluid and can be expressed as: EL = (p / w)+ (v2 / 2 g) + z= constant along a streamline where EL = Energy Line For a fluid flow without any losses due to friction (major losses) or components (minor losses) - the energy line would be at a constant level. In a practical world the energy line decreases along the flow due to losses. A turbine in the flow reduces the energy line and a pump or fan in the line increases the energy line Fig.2.25 80  Fig.2.26  A pipe is 2 closed conducted and it is used for carrying fluids or water under pressure.  General pipe is circular cross section  When the pipe is running full of flow is under pressure. 2.23.Laws of fluid friction Fig. 2.27 Whenever a liquid is flowing in a pipe, it loss of head or energy due to frictional resistance and other reasons. 1. major losses – due to frictional resistance. 2. Minor losses – due to sudden change in velocity of flow either in magnitude or direction 81  Fig. 2.28 The frictional resistance offered to the flow of a fluid depends on the type of flow laminar or turbulent. 2.23.1.Laws of fluid friction for laminar flow The frictional resistance in the laminar flow is  Proportional to the velocity of flow.  Proportional to the area of surface in contact.  Greatly decreased if the temperature of the fluid is increased  Independent of pressure.  Independent of pressure.  Independent of the nature of surface in contact. 2.23.2.Laws of fluid friction for turbulent flow The frictional resistance in the turbulent flow is  Proportional to the area of surface in contact.  Proportional to the square of velocity.  Proportional to the density of fluid.  Dependent on the nature of surface in contact.  Independent of pressure.  Slightly vary with temperature. 1.3.5 Reynolds number Reynolds number is a non-dimensional number. Reynolds number = where –density v-velocity - absolute viscosity 2.24. Critical velocity The velocity of flow changes from laminar to turbulent is called critical velocity. 2.25. Total energy line 82  If sum of pressure heads and velocity heads of a liquid flowing in a pipe be plotted as vertical ordinates on the center line of the pipe. 2.26.Hydraulic gradient line If the pressure head p/w of a liquid flowing in a pipe be plotted as vertical ordinates in the centre line of the pipe. The line joining the tops of such ordinates is known as hydraulic gradient line. Wetted perimeter(P) Wetted perimeter is the surface which is in contact with water. Consider a pipe of circular cross section in which a liquid as flowing in full. Wetted perimeter(P)=πd Hydraulic mean depth or Hydraulic radius Hydraulic mean depth = For circular pipe ,Hydraulic mean depth = = 2.27. Head loss due to friction Fig.2.29 Let P1- inlet pressure and P2-outlet pressure v-velocity of the flow L-length of the pipe and d-diameter of the pipe A- cross sectional area of pipe f’-frictional resistance per unit area(Froud’s constant) Consider section 1-1 & 2-2 of the pipe 83  P1A= P2A + Frictional resistance Frictional resistance = P1A-P2A Dividing both sides by w(specific weight of water) Frictional resistance = P1A-P2A w w Frictional resistance = P1-P2 = hf (loss of head due to friction) Aw w hf (loss of head due to friction) = Frictional resistance Aw 2 Frictional resistance = f’ x πdl x v then hf = hf = f’ = substitute f’ value in hf equation then Darcy’s formula for loss of head due to friction hf = where 4f – friction factor f – Darcy’s frictional coefficient. 2.28. Head loss due to friction in the pipe (Chezy’s formula) Fig.2.30 Let 84  P1- inlet pressure and P2-outlet pressure v-velocity of the flow L-length of the pipe and d-diameter of the pipe A- cross sectional area of pipe f’-frictional resistance per unit area(Froud’s constant) Consider section 1-1 & 2-2 of the pipe P1A= P2A + Frictional resistance Frictional resistance = P1A-P2A Dividing both sides by w(specific weight of water) Frictional resistance = P1A-P2A w w Frictional resistance = P1-P2 = hf (loss of head due to friction) Aw w hf (loss of head due to friction) = Frictional resistance Aw 2 Frictional resistance = f’ x πdl x v then hf = = where perimeter(p)= πd Hydraulic mean depth (m) = A/p hf = v2 = = xi where hf/l = i( slope of hydraulic gradient line) = C(chezy’s constant) v=C 2.29.Minor losses 1.Loss of head due to sudden Enlargement 2. Loss of head due to sudden contraction 3. Loss of head at entrance of pipe 4. Loss of head at exit of pipe 5. Loss of head due to an obstruction in a pipe 85 1. Loss of head due to sudden Enlargement Fig.2.30 Loss of head due to sudden Enlargement (hf) = V1-velocity at inlet V2-velocity at Exit 2.Loss of head due to sudden Contraction Fig.2.31 Loss of head due to sudden Contraction =0.375 v- velocity of flow at exit of pipe 3.Loss of head at Entrance of pipe Loss of head at Entrance of pipe = 0.5 4.Loss of head at Exit of pipe Loss of head at Exit of pipe = 5.Loss of head due to an obstruction in a pipe Loss of head due to an obstruction in a pipe = Where A- area of pipe a- Area of Obstruction v- velocity of liquid Cc- Coefficient of contraction Transmission of power through the pipe Total head available at outlet = Total head available at inlet – loss of head due to friction Total head available at outlet = H- hf 2.30.Power transmitted to outlet(p) P = w a v (H-hf) Where w- specific weight of fluid a- area of the pipe 86  v- velocity of flow H- total head at inlet hf- loss of head due to friction Efficiency of power Transmission 2.31.Efficiency = η = x 100 = x 100 2.32. Condition for Maximum power Transmission Power transmission = wav(H-hf) P =wav [H- ] For maximum power transmission differentiate the above equation with respect to V and then equate to zero (i.e) =0 =0 H - 3[ ]=0 H- 3hf = 0 hf = hence loss of head is 1/3 of the total head at inlet 2.33.Important Formula 1. Reynolds number = 2. Wetted perimeter(P)= πd 3. Hydraulic mean depth (m) = d/4 4. Head loss due to friction (Darcy’s formula) hf = 5. Chezy’s formula v = C 6. Loss of head due to sudden Enlargement (hf) = 87 7. Loss of head due to sudden Contraction =0.375 8. Loss of head at Entrance of pipe = 0.5 9. Loss of head at Exit of pipe = 10. Loss of head due to an obstruction in a pipe = 11.power transmitted in the pipe P =wav [H- ] 12. Efficiency = η = x 100 = x 100 13.condition for maximum power transmission hf = 2.34.Solved problem 1.Calculate the head loss due to friction in a pipe of 600 mm diameter and 1.5 km long. The velocity of flow of water is 2.5 m/sec and the friction factor is 0.02 Given: Diameter of pipe d = 600 mm = 0.6 m Length of pipe L = 1.5 km = 1500 m Velocity of flow v= 2.5 m/s Friction factor F or 4f = 0.02 Solution: Head loss due to friction (Darcy’s formula) hf = = = 15.92 m 2. Calculate the head loss due to friction in a pipe of 200 mm diameter and 500 m long. The velocity of flow of water is 5 m/sec and f is 0.008 Given: Diameter of pipe d = 200 mm = 0.2 m Length of pipe l = 500 m Velocity of flow v= 5 m/s Friction factor f = 0.008 88 Solution: Head loss due to friction (Darcy’s formula) hf = = = 101.9368 m 3. Two reservoirs are connected by a 50 mm diameter and 2 km long pipe line. The difference of water level between the two reservoirs is 20 m. calculate the discharge in lit/sec. take friction factor = 0.0248 Given: Diameter of the pipe d = 500 mm = 0.5 m Length of the pipe l = 2 km = 2000 m Difference of water level hf = 20 m Friction factor F = 4f = 0.0248 Solution: Head loss due to friction (Darcy’s formula) hf = 20 = V = 1.9889 m/s Discharge Q = av = (πd2/4)v = (π0.52/4)x1.9889 = 0.3905 m3/sec = 390.05 lit/sec 4. Compare the discharge of 150mm and 300 mm diameter pipes when the loss head due to friction in the pipe is the same. Consider both pipes having equal length and equal value of f. Given: d1 = 150 mm = 0.150 m d2 = 300 mm = 0.300 m Solution: the loss of head due to friction for pipe 1 = pipe 2 = 89 4f, l, 2g are cancelled = = 0.707 Discharge of pipe ratio = = x 0.707 = 0.17675 5. A pipe line 8 km long delivers a power of 40 kw at its outlet end. The pressure at inlet is 4000 kN/m2 and pressure drop per Km of pipe is 50 kN/m2. Take friction factor = 0.02. calculate the diameter of pipe and the efficiency. Given: Length of pipe l =8 km = 8000 m Power delivered p = 40 kw Pressure at inlet p1 = 4000 kN/m2 Pressure drop per km = 50 kN/m2 Total pressure drop for 8 km pipe = 8 x 50 = 400 kN/m2 Solution: Convert the pr at inlet and loss of pr into pr head at inlet and los