FLUID MECHANICS FUNDAMENTALS AND APPLICATIONS E-BOOK OF NORSHAHRIZAN BIN RASIP SUYANI BINTI ARIFIN KEJURUTERAAN MEKANIKAL
i E-BOOK OF FLUID MECHANICS FUNDAMENTALS AND APPLICATIONS NORSHAHRIZAN BIN RASIP SUYANI BINTI ARIFIN
ii WRITER Norshahrizan Bin Rasip Suyani Binti Arifin EDITOR Dr. Mohd Shahril Bin Mohd Hassan @ Abdul Ghani Khairil Bin Che Mat DESIGNER Norshahrizan Bin Rasip Suyani Binti Arifin Terbitan Edisi 2022 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Publisher. PUBLISHED BY: Politeknik Muadzam Shah, Lebuhraya Tun Abdul Razak, 26700 Muadzam Shah, Pahang
iii PREFACE This e-book is inscribed to learn the fundamentals and applications of Fluid Mechanics for polytechnic students especially in Department of Mechanical Engineering, Politeknik Muadzam Shah as a reference in their study. This e-book emphasizes fundamentals and applications of fluid mechanics principles related to the fluid properties and behavior in static and dynamic situations. The contents of this book are based on the Polytechnic syllabus and total of five chapter are included. Hopefully the content of this ebook will help the students for guidance of their studies.
iv ACKNOWLEDGEMENT I am really grateful for the completion of this Fluid Mechanics ebook. This book could not have been accomplished without the support of my family and friends. Finally, I would like to express my sincere gratitude to Allah for letting me through all the difficulties.
v 1.0 INTRODUCTION OF FLUID 1.1 Introduction 1.2 Fluid characteristics 1.3 Physical properties of fluid 1.4 Classification of Pressure Problems 1 2 5 10 17 2.0 FLUID STATICS 2.1 Pascal’s Law and hydraulic jack 2.2 Pressure measurement devices 2.3 Buoyancy Problems 19 24 35 37 3.0 FLUID DYNAMICS 3.1 Classification of flow 3.2 Volume flow rate and mass flow rate 3.3 Continuity and energy equation 3.4 Bernoulli Theorem Problems 40 42 44 47 62 4.0 ENERGY LOSS IN PIPELINE 4.1 Velocity profile in the round pipe system 4.2 Types of head loss 4.3 Head loss in pipeline systems Problems 64 65 71 76 5.0 REFERENCES 77
1 INTRODUCTION OF FLUID
CHAPTER 1 INRODUCTION OF FLUID 2 1.1 INTRODUCTION Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid statics). Both liquids and gases are classified as fluids. Fluid mechanics has a various application in mechanical and chemical engineering, in biological systems, and in astrophysics. The handling of liquids is much simpler, cheaper, and less troublesome than handling solids. Figure 1.1: Application of fluid mechanics in daily life 1.2 FLUID CHARACTERISTICS In everyday life, we recognize three states of matter: solid, liquid and gas. Although different in many respects, liquids and gases have a common characteristic in which they differ from solids. Both are fluids, but lack the ability of solids to offer a permanent resistance to a deforming force.
CHAPTER 1 INRODUCTION OF FLUID 3 1.2.1 CHARACTERISTIC OF LIQUID, GAS AND SOLID Table 1.1 Characteristic of liquid, gas and solid LIQUID GAS SOLID SHAPE No Shape No Shape Definite Shape VOLUME Definite Volume No Definite Volume Definite Volume COMPRESSIBILITY No Yes No DENSITY Medium Low High DEFORM CONTINUOSLY Yes Yes No KINETIC ENERGY BETWEEN PARTICLE Medium High Low FLUID DEFINITION A fluid is a substance that continually flows or deforms when we subject it to shear stress or external force. (Vedantu, 2022) Substance that can flow , has no fixed shape, and offers little resistance to an external stress. (Collins, 2014)
CHAPTER 1 INRODUCTION OF FLUID 4 DISTANCE BETWEEN PARTICLE Medium Large Small a) Example of Liquid and Gas Figure 1.2: Examples of liquid and gas Examples 1.1 Compare the characteristics between liquid, gas and solid in terms of shape and volume. Answer Liquid - No definite shape (takes the shape of its container) - Has definite volume Gas - No definite shape (takes the shape of its container)
CHAPTER 1 INRODUCTION OF FLUID 5 1.3 PHYSICAL PROPERTIES OF FLUID Fluid properties are intimately related to fluid behaviour. It is obvious that different fluids can have grossly different characteristics. For example, gases are light and compressible, whereas liquids are heavy and relatively incompressible. To quantify the fluid behaviour differences certain fluid properties are used. The fluid properties are mass density, relative density @ specific gravity, specific weight, specific volume, fluid compressibility and viscosity. The relative change in fluid volume related to a unit change in pressure. Gas has higher compressibility than liquid (oil or water). a) Mass Density, Mass density is defined as the mass per unit volume. The symbol is “rho” denoted by the Greek symbol, ρ. Density is amount of matter within a certain volume. (SI units, kg/m3 ) PHYSICAL PROPERTIES OF FLUID MASS DENSITY SPECIFIC WEIGHT SPECIFIC GRAVITY VISCOSITY COMPRESS IBILITY SPECIFIC VOLUME
CHAPTER 1 INRODUCTION OF FLUID 6 b) Specific Weight, Specific weight, (omega) is defined as the weight per unit volume of a material. The SI unit for specific weight is N/m3 . Or Specific weight equal to the force exerted by gravity on a unit volume of fluid c) Specific Volume, Specific volume, v is defined as the reciprocal of mass density. It is used to mean volume per unit mass. (SI units, m3 /kg). Or = Weight,W Volume,V = mg V = = specific weight = density g = gravity = Volume,V (3) Mass (kg) = 1
CHAPTER 1 INRODUCTION OF FLUID 7 d) Specific Gravity, Specific gravity can be defined in two ways which are: i. Ratio of density of a substance to the density of water at 4°C. ii. Ratio of specific weight of a substance to the specific weight of water at 4°C. Or e) Compressibility i. Compressibility is the property of being reduced to a smaller space by pressure. This property is a consequence of porosity, and the change of bulk comes from the particles being brought closer together by the pressure. ii. The compressibility of fluid is basically a measure of the change in density that will be produced in the fluid by a specified change in pressure. Gases are, in general, highly compressible whereas most liquids have very low compressibility. iii. The fact that compressibility effects can have a large influence on fluid flow. The study of these flows in which the changes in density and temperature are important is basically what is known as compressible fluid flow or gas dynamics. f) Viscosity A fluid at rest cannot resist shearing forces but once it is in motion, shearing forces are set up between layers of fluid moving at different velocities. The viscosity of the fluid determines the ability of the fluid in resisting these shearing stresses. Kinematic viscosity, v is defined as the ratio of dynamic viscosity to mass density (SI unit: m2 /s) = ρ ρ = ω ω = μ
CHAPTER 1 INRODUCTION OF FLUID 8 Figure 1.3: Viscosity of fluid Examples 1.2 Given a fluid weight 32 N and volume 750 cm3 . Calculate specific weight, (in kN/m3 ) and specific gravity, s. Solution: Given; weight = 32 N volume = 750 cm3 = 750 x (102 ) 3 = 750 x 10-6 m3 = = 32 750 10−6 = 42666.7 N/m3 = 42.67 kN/m3 s = = 42.67 103 1000(9.81) = 4.35
CHAPTER 1 INRODUCTION OF FLUID 9 Examples 1.3 A glass bottle with a volume of 50 cm3 full with fluid has relative density of 1.3. if the total mass is 180 g and the mass density of glass bottle is 2000 kg/m3 , determine: i. Glass bottle mass ii. Glass bottle volume Solution: Given: Vglass bottle = 50 cm3 , sfluid = 1.3 , mtotal = 0.18 kg , glass bottle = 2000 kg/m3 Volume = 50 cm3 x 1 3 1 106 = 5 x 10-5 m3 fluid = 1.3 x 1000 = 1300 kg/m3 mfluid = fluid x volume = 1300 x (5 x 10-5 ) = 0.065 kg i) Glass bottle mass, m = mtotal - mfluid = 0.18 – 0.065 = 0.115 kg ii) Glass bottle volume, V = = 0.115 2000 = 5.75 x 10-5 m3
CHAPTER 1 INRODUCTION OF FLUID 10 1.4 CALSSIFICATION OF PRESSURE a) Atmospheric Pressure, Patm The earth is surrounded by an atmosphere many miles high. Atmospheric pressure can be defining as the pressure due to this atmosphere at the surface of the earth depends upon the head of the air above the surface. The air is compressible; therefore, the density is different at different height. Due to the weight of atmosphere or air above the surface of earth, it is difficult to calculate the atmospheric pressure. So, atmospheric pressure is measured by the height of column of liquid that it can support. Atmospheric pressure at sea level is about 101.325 kN/m2 , which is equivalent to a head of 10.35 m of water or 760 mm of mercury approximately, and it decreases with altitude. Figure 1.4: Atmospheric pressure
CHAPTER 1 INRODUCTION OF FLUID 11 b) Gauge Pressure, PG It is the pressure, measured with the help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum which is the atmospheric pressure at the gauge scale is marked zero. The gauge pressure can be either positive or negative depending on whether the pressure is above atmospheric pressure (a positive value) or below atmospheric pressure (a negative value). Figure 1.4: Device to measure gauge pressure c) Absolute Pressure, Pabs It is the pressure equal to the algebraic sum of the atmospheric and gauge pressures. d) Vacuum Pressure, Pv In a perfect vacuum which is a completely empty space, the pressure is zero. Atmospheric Pressure Patm Pressure Gauge PG Absolute Pressure Pabs
CHAPTER 1 INRODUCTION OF FLUID 12 1.4.1 RELATIONSHIP BETWEEN PA, PG, PATM & PV Figure 1.5: Relationship between PA, PG, Patm and PV Examples 1.4 A bourdon gauge is attached to a boiler located at sea level with pressure of 5 bar. If the atmospheric pressure is 1.013 bar, calculate the absolute pressure in the boiler. Your answer must be in kN/m2 . Solution: Given PG = 5 bar, Patm = 1.013 bar Pabs = PG + Patm = 5 bar + 1.013 bar = 6.013 bar Convert to kN/m2 Pabs = 6.013 x 105 N/m2 = 601.3 kN/m2 Notes: 1 bar = 1 x 105 N/m2 1 Pascal (Pa) = 1 N/m2
CHAPTER 1 INRODUCTION OF FLUID 13 Examples 1.5 A bourdon gauge is attached to a boiler located at sea level with pressure of 5 bar. If the Calculate pressure gauge of water in the cylinder. Given atmospheric pressure is 101.3 kN/m2 and absolute pressure is 1.75 bar. Solution: PG = Pabs – Patm = (1.75 x 105 ) – (101.3 x 103 ) = 73700 N/m2 @ 73.7 kN/m2 1.5 PRESSURE AND DEPTH IN FLUID The pressure in a liquid is different at different depths. Pressure increases as the depth increases. The pressure in a liquid is due to the weight of the column of water above. Since the particles in a liquid are tightly packed, this pressure acts in all directions. For example, the pressure acting on a dam at the bottom of a reservoir is greater than the pressure acting near the top. This is why dam walls are usually wedge-shaped. (BBC Bitesize, 2022). Figure Figure 1.6: Relationship between pressure and depth in Dam The pressure P acting on a fluid is the force exerted perpendicularly per unit of the fluid’s surface area.
CHAPTER 1 INRODUCTION OF FLUID 14 Unit of pressure is the N/m2 or Pascal; 1 N/m2 = 1 Pa (Pascal). Atmospheric pressure at sea level is 1 atmosphere (atm) = 1.013 x 105 Pa. If a fluid is within a container, then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because of the weight of the fluid above it. The denser the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid. Figure 1.7: Relationship between pressure and depth
CHAPTER 1 INRODUCTION OF FLUID 15 The formula that gives the pressure, p on an object submerged in a fluid is: Where, (rho) is the density of the fluid, g is the acceleration of gravity h is the height of the fluid above the object Examples 1.5 Calculate the pressure at a point of 500 m depth in sea water. Give your answer in Pascal and bar unit. (Density sea water is 1025 kg/m3 ) Solution: h = 500 m g = 9.81 m/s2 = 1025 kg/m3 P = g h = 1025 x 9.81 x 500 OR = 5027625 N/m2 @ 5027625 Pa In Bar unit: 5027625 N/m2 / 105 = 50.276 bar = = 1025 x 9.81 = 10055.25 N/m3 P = ℎ = 10055.25 x 500 = 5027625 N/m2 @ 5027625 Pa
CHAPTER 1 INRODUCTION OF FLUID 16 Examples 1.5 Calculate the height of water column which is equivalent to the pressure of 4.35 kN/m2 . ( water = 1000 kg/m3 x 9.81m/s2 ) Solution: P = h 4.35 x 103 = 9810h h = 4.35 ×103 9810 = 0.443 m
CHAPTER 1 INRODUCTION OF FLUID 17 PROBLEMS 1. If the volume and weight of certain oil are 9 m3 and 4500 N respectively, calculate the: a. Mass Density b. Specific weight c. Specific gravity d. Specific volume 2. A tank of oil contains a full capacity of 5000 liter and weights 50000 N. Calculate: a. Mass density of oil, b. Specific weight of oil, c. Specific gravity of oil, s 3. A metal cube with 50 mm side is inserted in to a container filled with fluid. The mass of spilled fluid is 50 kg. Calculate: a. Mass density of fluid b. Specific weight of fluid c. Specific volume of fluid 4. State the difference between gauge pressure, atmospheric pressure and absolute pressure. 5. The gauge pressure in a cylinder is 78 kN/m2 . Determine: i. Pressure in terms of height for oil with specific gravity of 0.89. ii. The absolute pressure in the cylinder when the atmospheric pressure is 101.3 kN/m2 . 6. There is the gauge pressure attach to a boiler which is located at sea level and its reading 9 bar. If atmospheric pressure is 1.013 bar, determine: i. The absolute pressure in kN/m2 . ii. The pressure head of water, h. 7. State in kN/m2 the local atmospheric pressure at Mount Tahan if a pressure gauge used at that mountain indicates 0.45 bar and absolute pressure is 1.35 bar. 8. Given A bourdon pressure gauge is attached to a boiler located at sea level shows a reading pressure of 20 bar. If atmospheric pressure is 101.3 kPa, determine: i. The absolute pressure in kN/m2 ii. The pressure head of water, h
CHAPTER 1 INRODUCTION OF FLUID 18 2 FLUID STATICS
CHAPTER 2 FLUID STATICS 19 2.1 PASCAL LAW AND HYDRAULIC JACK 2.1.1 PASCAL LAW a) Pascal’s Law at a point b) Pascal’s Law at any point The Pascal’s law states that “the pressure or intensity of pressure at a point in a static fluid is equal in all direction”. SO, P1 = P2 = P3 = P4 = P5 = P6 = P7 = P8 Pascal’s law states that when there is an increase in pressure at any point in confined fluid, there is an equal increase at every other point in the container.
CHAPTER 2 FLUID STATICS 20 2.1.2 HYDRAULIC JACK What does Hydraulic Jack mean? A hydraulic jack is a hydraulically powered jacking device consisting of a hydrostatic press with its pump and reservoir containing a supply of liquid, as oil, used for lifting, moving, or forcing objects or loads into place. A hydraulic jack is a device that uses force to lift heavy loads. The primary mechanism by which force is applied varies, depending on the specific type of jack, but is typically a screw thread or a hydraulic cylinder. Figure 2.1: Floor hydraulic jack How do hydraulic jack works? The pressure applied to a fluid stored in a container will be distributed equally in all directions. The important components of a hydraulic jack are cylinders, a pumping system, and hydraulic fluid (oil is used commonly). The hydraulic jack fluid is selected by considering certain fluid properties like viscosity, thermal stability, filterability, hydrolytic stability, and more. If a compatible hydraulic fluid is selected, it will offer maximum performance, self-lubrication, and smooth operation. The hydraulic jack design that will contain two cylinders (one small and another big) connected to each other using piping. Both cylinders are filled partially using hydraulic fluids. While applying a small pressure on the smaller cylinder, the pressure will be transmitted equally to the larger cylinder through the incompressible fluid. Now, the larger cylinder will experience a force multiplication effect. The force exerted on all points of both cylinders will be the same. But, the force produced by the larger cylinder will be higher and it is directly proportional to the surface area. Other than cylinders, a hydraulic jack will contain a
CHAPTER 2 FLUID STATICS 21 pumping system to push fluid into a cylinder through a one-way valve. This valve will restrict the backflow of hydraulic fluid from the cylinder. a) Piston at the same level P1 = P2 HYDRAULIC JACK If small and large cylinder/ piston at SAME LEVEL If SMALL cylinder/piston BELOW large cylinder/piston If LARGE cylinder/piston BELOW small cylinder
CHAPTER 2 FLUID STATICS 22 P1 = P2 F A1 = 2 b) The small piston is below the larger piston P1 = P2 + ℎ 1 = 2 + ℎ c) The large piston is below the smaller piston P1 + ℎ = P2 1 + ℎ = 2 F W P1 P2 Area 1, A1 Area 2, A2 h P1 P2 h W
CHAPTER 2 FLUID STATICS 23 Examples 2.1 A force of 800 N is applied to smaller cylinder of a hydraulic jack. The area A1 of the small piston is 20 cm2 and area A2 of larger piston is 120 cm2 . Calculate the load W if: ( = 1000 kg/m3 ) a) The piston is the same level b) The large piston is 0.5 m below the small piston c) The small piston is 0.4 m below the larger piston Solution: A1 = 20 cm2 = 0.002 m2 ; A2 = 120 cm2 = 0.012 m2 a) P1= P2 1 = 2 800 0.002 = 0.012 W = 4800 N b) P1 + gh = P2 1 + gh = 2 800 0.002 + (1000 9.81 0.5) = 0.012 W = 4858.86 N c) P1 = P2 + gh 1 = 2 + gh 800 0.002 = 0.012 + (1000 9.81 0.4) W = 4752.9 N
CHAPTER 2 FLUID STATICS 24 2.2 PRESSURE MEASUREMENT DEVICES 2.2.1 PIEZOMETER TUBE Figure 2.2: Piezometer Piezometers are used in geotechnical, environmental, and hydrological applications. They can be installed in boreholes, placed in fill materials or open wells to measure water levels or pore water pressures to enable engineers to verify design assumptions and control placement of fill. Other application of piezometer is clinical applications like measuring blood pressure and in physiotherapy. Piezometers are used to measure the pressure in pipes where the liquid is in motion. A manometer is one of the earliest and simplest devices used for measurement of gauge pressure and differential pressures.
CHAPTER 2 FLUID STATICS 25 Examples 2.2 A pressure tube is used to measure the pressure of oil (mass density 640 kg/m3 ) in a pipeline. If the oil rises to a height 1.45 m above the centre of a pipe, what is the gauge pressure in N/m2 at the point? (gravity = 9.81 m/s) Solution: P = gh = 640 x 9.81 x 1.45 m = 9103.68 N/m2 2.2.2 BAROMETER A barometer is a device that measures atmospheric pressure. The word “barometer” comes from the Greek words for “weight” and “measure.” Changes in atmospheric pressure recorded by barometers are most often used in meteorology for forecasting weather. Barometers and pressure altimeters (the most basic and common type of altimeter) are essentially the same instrument, but used for different purposes. An altimeter is intended to be used at different levels matching the corresponding atmospheric pressure to the altitude, while a barometer is kept at the same level and measures subtle pressure changes caused by weather and elements of weather. The average atmospheric pressure on the earth’s surface varies between 940 and 1040 hPa (mbar). The average atmospheric pressure at sea level is 1013 hPa (mbar). Figure 2.3: Barometer
CHAPTER 2 FLUID STATICS 26 Examples 2.3 What is the atmospheric pressure in N/m2 if the lever of mercury in barometer tube is 760 mm above the level of mercury in the bowl? Given the specific gravity of mercury is 13.6 and specific weight of water is 9.81 x 103 N/m3 . Solution: P = mgh = 13600 x 9.81 x 0.76 m = 101.3 kN/m2 2.2.3 BOURDON GAUGE Bourdon Gauge is used to measure pressure differences that are more than 1.2 bar. Pressure applied to the tube tends to cause the tube to straighten out, and the deflection of the end of the tube is communicated through a system of levers to a recording needle. This motion is transmitted via linkage to the pointer, which would directly indicate on the calibrated scale or dial on the gauge pressure. Figure 2.4: Bourdon gauge water = water/9.81 = 1000 kg/m3 m = s x water = 13.6 x 1000 = 13600 kg/m3
CHAPTER 2 FLUID STATICS 27 Examples 2.4 A pressure tube is used to measure the pressure of oil (mass density = 640 kg/m3 ) in a pipeline. If the oil rises to a height of 1.3 above the centre of the pipe, what is the gauge pressure in at that point? (gravity = 9.81 m/s2 ) Solution: Pgauge = gh = 640 x 9.81 x 1.3 m = 8.162 kN/m2 2.2.4 MANOMETERS a) Simple Manometer for Gauge Pressure The horizontal surface, at which the heavy and light liquid meet in the left-hand limb is known as a common surface or datum line. Let B-C be the datum line. Pressure at A, PA = Pressure in pipe h1 = height of the liquid in the left-hand limb above the common surface in m. h2 = height of the liquid in the right-hand limb above the common surface in m.
CHAPTER 2 FLUID STATICS 28 Examples 2.5 A U-tube manometer similar to that shows in figure is used to measure the gauge pressure of water (mass density = 1000 kg/m3 ). If the density of mercury is 13.6 x 103 kg/m3 , what will be the gauge pressure at A if h1 = 0.55 m and D is 0.8 m above BC. Solution: Left-hand limb PB = Pressure A at A + Pressure to depth, h1 of water PB = PA + watergh1 = PA + (1000 x 9.81 x 0.55) = PA + 5395.5 Right-hand limb Pc = Pressure D at D + Pressure to depth, h2 of mercury PD = 0, PC = 0 + mercurygh2 = 13600 x 9.81 x 0.8 = 106732.8 N/m2 Since PB = PC PA + 5395.5 = 106732.8 PA = 101337.3 N/m2 PA = 101.34 kN/m2
CHAPTER 2 FLUID STATICS 29 Examples 2.6 A U-tube manometer below is used to measure the gauge pressure of water (Swater = 1). What will be the gauge pressure at A if h1 = 0.2 m and D is 0.3 m above BC. (Smercury = 13.6) (Patm = 101.3 kN/m2 ) Solution: Left Limb Right Limb PB = PA + Qgh1 + Pgh2 PC = Patm Since, PB = PC PA + Qgh1 + Pgh2 = Patm PA = Patm – Qgh1 – Pgh2 PA = 101.3 kN/m3 – (1000 x 9.81 x 0.2) – (13600 x 9.81 x 0.3) PA = 59313.2 N/m2 PA = 59.31 kN/m2
CHAPTER 2 FLUID STATICS 30 b) Differential manometer A differential manometer consists of a U-tube, containing a heavy liquid with two ends connected to two different points.
CHAPTER 2 FLUID STATICS 31 Examples 2.7 A U-tube manometer measures the pressure difference between two points A and B in a liquid. The U tube contains mercury and the liquid at A and B is oil. Calculate the difference in pressure if h = 1.5 m, h1 = 0.5 m and h2 = 0.7 m. (Soil = 0.9, Smercury = 13.6) Solution: Left limb PC = Pressure at A + Pressure to depth, h of oil = PA + oilh = PA + 8829(1.5) = PA + 13243.5 N/m2 Right limb PD = Pressure at B + Pressure to depth, h1 of mercury + Pressure to depth, h2 of oil = PB + Hgh1 + oilh2 = PB + 133416(0.5) + 8829(0.7) = PB + 72888.3 N/m2 Since PC = PD PA + 13243.5 = PB + 72888.3 PA – PB = 59644.8 N/m2
CHAPTER 2 FLUID STATICS 32 c) Inverted Differential Manometer An inverted differential manometer is used for measuring the difference of low pressure, where accuracy is the prime consideration. It consists of an inverted U-tube, containing a light liquid. Fluid Q Fluid P
CHAPTER 2 FLUID STATICS 33 Examples 2.8 An inverted U-tube manometer filled with mercury in a u-tube and water in pipe A and B. Calculate the difference in pressure if h = 70 cm, h1 = 75 cm and h2 = 150 cm. (SHg = 13.6) Solution: Left limb PC = Pressure at A – Pressure to depth, h of mercury – Pressure to depth, h1 of water = PA – Hggh – watergh1 = PA – (13600 x 9.81 x 0.7) – (1000 x 9.81 x 0.75) = PA – 86033.7 N/m2 Right limb PD = Pressure at B + Pressure to depth, h2 of water = PB – watergh2 = PB – (1000 x 9.81 x 1.5) = PB – 14715 N/m2 Since PC = PD PA – 86033.7 = PB – 14715 PA – PB = 71318.7 N/m2
CHAPTER 2 FLUID STATICS 34 d) Combined/Multiple U-Tube Manometer Starting from 1 to 2, PA = PE + waterh1 – oilh2 + waterh3 PE = Patm (the tube is open to atmosphere) So, PA = PATM + waterh1 – oilh2 + waterh3 Examples 2.9 A multiple U-tube manometer is fitted to a pipe with a centre at A as shown in figure below. Determine the pressure at A if h1 = 0.9 m, h2 =0.45 m and h3 =0.5 m. (Soil = 0.9) Solution: Given; ℎ1 = 0.9 ℎ2 = 0.45 ℎ3 = 0.5 = 0.9 = = + ℎ1 − ℎ2 + ℎ3 = + ℎ1 − ℎ2 + ℎ3 = + (9810)(09) − (0.9 9810)(0.45) + (9810)(0.3) = + .
CHAPTER 2 FLUID STATICS 35 2.3 BUOYANCY Principle of Archimedes: Up thrust on body = weight of fluid displaced by the body Up thrust on upper part, R1 = 1 gv1 acting through G1, the centroid of v1, Up thrust on lower part, R2 = 2 gv2 acting through G2, the centroid of v2, Total up thrust = 1 gv1 + 2 gv2 Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. If the weight of the water displaced less than the weight of the object, the object will sink Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.
CHAPTER 2 FLUID STATICS 36 Examples 2.10 A wooden block has mass of 0.5 kg and density of 350 kg/m3 floated in water. ( =1000kg/m3 ). Calculate the buoyant force acting on the wooden block. Solution: Given; = 0.5 = 350/3 = = = 0.5 350 = . − = = (1000)(9.81)(1.429 10−3 ) = .
CHAPTER 2 FLUID STATICS 37 PROBLEMS 1. Describe working principle of hydraulic Jack. 2. A hydraulic jack filled with oil having a large piston, 650 mm in diameter lifts a metal block, with mass, m = 4000 kg under the action of an 80 mm piston. Given specific weight, ω, of oil is 8.89 kN/m3 , calculate force required on the small piston to lift the weight if: a) Both piston at same level. b) The small piston is 7 m above the larger piston? 3. A force, P of 650 N is applied to a small cylinder of the hydraulic jack. A hydraulic jack has are ration between the two piston of 10:1. The area of the small piston is 25 cm2 . What is the load of W that can be lifted at a large piston if the small piston is 1.2 m below the large piston? (Mass density fluid is 103 kg/m3 ). 4. A hydraulic jack has diameter ratio between two pistons of 10: 1. The diameter of the larger system is 485 mm and it is required to support mass of 3455 kg. The hydraulic jack is filled with a hydraulic fluid which has specific gravity 0.85. Fill in kN the force required on the smaller piston when the smaller piston is 1.26 m below the larger piston. 5. Figure 2.6 below shows a U-tube manometer, used to measure the fluid pressure difference between pipe A and pipe B. The U-tube is containing the mercury. Calculate the pressure between the pipe A and pipe B for given h1 = 120 cm, h2 = 30 cm and h3 = 50 cm. Fluid A and B (water = 1000 kg/m3) and specific gravity of mercury is 13.6. Figure 2.6 6. U-tube manometer as shown in figure 2.7 below. Calculate PA – PB, if h1 = 45 cm, h2 = 25 cm and h3 = 75 cm. The liquid W1 and W2 is water ( = 1000kg/m3 ), while Wg is mercury with specific gravity 13.6.
CHAPTER 2 FLUID STATICS 38 Figure 2.7 7. A U-tube manometer as shown in figure 2.8 below is used to measure pressure different between two points in a pipe. If the pressure difference between A and B is 45.3 kN/m2 , h=38 cm and h2=95 cm, calculate h1. Figure 2.8
CHAPTER 1 INRODUCTION OF FLUID 39 FLUID DYNAMICS 3 2
CHAPTER 3 FLUID DYNAMICS 40 3.1 CLASSIFICATION OF FLOW a) Uniform flow The cross-sectional area and velocity of the stream of fluid are the same at each successive cross-section. Example: flow through a pipe of uniform bore running completely full. Flow velocity is the same magnitude and direction at every point in the fluid. Figure 3.1 shown the uniform flow in the pipelines system. Figure 3.1: Uniform flow in pipelines b) Steady flow The cross-sectional area and velocity of the stream may vary from cross-section, but for each cross-section they do not change with time. Example: a wave travelling along a channel. If at any point in the fluid, the conditions change with time, the flow is described as unsteady flow as figure 3.2. Figure 3.2: Steady flow in pipelines
CHAPTER 3 FLUID DYNAMICS 41 c) Laminar flow Also known as streamline or viscous flow, in which the particles of the fluid move in an orderly manner and retain the same relative positions in successive cross-sections. In laminar flow, fluid particles flow in an orderly manner along path lines, and momentum and energy are transferred across streamlines by molecular diffusion. In fully developed laminar flow, each fluid particle moves at a constant axial velocity along a streamline and the velocity profile remains unchanged in the flow direction. There is no motion in the radial direction, and thus the velocity component in the direction normal to flow is everywhere zero. There is no acceleration since the flow is steady and fully developed. The flow is laminar when Reynolds number, Re is less than 2000. Figure 3.3: Laminar flow in pipelines d) Transition flow Transitional flow is a mixture of laminar and turbulent flow, with turbulence flow in the center of pipe and laminar flow near the edges pipe as figure 3.4. The flow is transition when Reynolds number, Re is in range 2000 to 4000. Figure 3.4: Transition flow in pipelines e) Turbulent flow Turbulent flow is a non-steady flow in which the particles of fluid move in a disorderly manner, occupying different relative positions in successive cross-sections. Turbulent flow is characterized by random and rapid fluctuations of swirling regions of fluid, called eddies, throughout the flow. These fluctuations provide an additional mechanism for momentum
CHAPTER 3 FLUID DYNAMICS 42 and energy transfer. In turbulent flow, the swirling eddies transport mass, momentum, and energy to other regions of flow much more rapidly than molecular diffusion, greatly enhancing mass, momentum, and heat transfer. The flow is turbulent when Reynolds number, Re is more than 4000. Figure 3.5: Turbulent flow in pipelines 3.2 VOLUME FLOW RATE AND MASS FLOW RATE 3.2.1 VOLUME FLOW RATE, Q The volume of liquid passing through a given cross-section in unit time is called the discharge. It is measured in cubic meter per second, or similar units and denoted by Q. Q = mass flow rate in pipelines (m3 /s) A = cross-sectional area through in the pipelines (m2 ) ν = fluid mean velocity through in the pipelines (m/s) Q = A x ν
CHAPTER 3 FLUID DYNAMICS 43 Examples 3.1 If the diameter d = 25 cm and the mean velocity, = 1.5 m/s, calculate the actual discharge in the pipe. Solution: , = 25 = 0.25 = 1.5 / = = 2 4 = (0.25) 2 4 1.5 = . / 3.2.2 MASS FLOW RATE, ṁ The mass of fluid passing through a given cross section in unit time is called the mass flow rate. It is measured in kilogram per second, or similar units and denoted by ṁ. ṁ = mass flow rate in pipelines (kg/s) ρ = density fluid flow in the pipelines (m3 /kg) A = cross-sectional area through in the pipelines (m2 ) ν = fluid mean velocity through in the pipelines (m/s) ṁ= ρ x A x ν
CHAPTER 3 FLUID DYNAMICS 44 Examples 3.2 Oil flows through a pipe at a velocity of 2 m/s. The diameter of the pipe is 10 cm. Calculate discharge and mass flowrate of oil. Take into consideration soil = 0.9. Solution: , = 10 = 0.10 = 2 / = 0.9 Discharge: = = 2 4 = (0.1) 2 4 2 = . / Mass flowrate: ṁ = ṁ = ṁ = 900 0.016 ṁ = . / = = = 0.9 1000 = / 3.3 CONTINUITY AND ENERGY EQUATION For continuity of flow in any system of fluid flow, the total amount of fluid entering the system must equal the amount leaving the system. This occurs in the case of uniform flow and steady flow.