Z A I R I N I B I N T I M O H A M M A D M e c h a n i c a l E n g i n e e r i n g D e p a r t m e n t P O L I M A S
Zairini Binti Mohammad Mechanical Engineering Department Editor Neza Nurulhuda binti Nekmat Azijan bin Murad Pereka Bentuk Siti Ameerah Diyana Binti Jamal Ilustrator Siti Ameerah Nadiah Binti Jamal Bandar Darulaman 06000 Jitra Published by Azijan Bin Murad
FLUID MECHANICS 2022 Zairini binti Mohammad, Published by : Politeknik Sultan Abdul Halim Mu’adzam Shah Address : Politeknik Sultan Abdul Halim Mu’adzam Shah, Bandar Darulaman, 06000, Jitra, Kedah. Copyright : All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without any the prior written permission of the publisher.
PREFACE ______________________________________________ This book was written as a textbook or guidebook on fluid mechanics for polytechnic students or technicians who are studying mechanical engineering. The recent progress in the science of visualization and computational fluid dynamics is seen to be astounding. In this book, effort has been made to introduce students to fluid mechanics by making explanations easy to understand. Fluid Mechanics provides students with a strong understanding of the fundamentals of fluid mechanics principles related to the fluid properties and behavior in statics and dynamic situations. Though every care has been taken to check the proofs, yet claiming perfection is not prudent. We shall be grateful to the readers for pointing the mistakes that might have occurred. Suggestions for improvement are most welcome and would be incorporated in the next edition. ZAIRINI BINTI MOHAMMAD 2022
CONTENTS PAGES 1.0 CHAPTER 1 : INTRODUCTION OF FLUID Objectives ………………………………………………………… Introduction ………………………………………………………. 1 1 1.1 Fluid Characteristics …………………………………………….. 2 1.2 Pressure ………………………………………………………….. 3 1.3 Pressure In Fluid ………………………………………………… 4 Exercise 1.1 ……………………………………………………… 5 2.0 CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID Objectives ………………………………………………………… Introduction ……………………………………………………….. 8 8 2.1 Physical Properties Of Fluid ……………………………………. 9 Exercise 2.1 ………………………………………………………. 11 3.0 CHAPTER 3 : FLUID STATICS Objectives ………………………………………………………… Introduction ………………………………………………………. 13 13 3.1 Pascal Law And Hydraulic Jack ………………………………... 14 Exercise 3.1 ………………………………………………………. 17 3.2 Piezometer, Barometer And Manometer ……………………… 19 Exercise 3.2 ………………………………………………………. 25 3.3 Bourdon Gauge ………………………………………………….. 32 3.4 Buoyancy …………………………………………………………. 32 Exercise 3.3 ………………………………………………………. 33 4.0 CHAPTER 4 : FLUID DYNAMICS Objectives ………………………………………………………… Introduction ……………………………………………………….. 35 35 4.1 Types of Flow …………………………………………………….. 36 4.2 Flow Rate …………………………………………………………. 38 4.3 Continuity Equation Law ………………………………………… 39 Exercise 4.1 ………………………………………………………. 40 4.4 Bernoulli Theorem ……………………………………………….. 43 Exercise 4.2 ………………………………………………………. 52 5.0 CHAPTER 5 : ENERGY LOSS IN PIPELINE Objectives ………………………………………………………… Introduction ……………………………………………………….. 59 59 5.1 The Round Pipe System ………………………………………… 60 5.2 Head Loss ………………………………………………………… 60 Exercise 5.1 ………………………………………………………. 63
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 1 CHAPTER 1 : INTRODUCTION OF FLUID OBJECTIVES · Define the nature of a fluid. · Show where fluid mechanics concepts are common with those of solid mechanics and indicate · Introduce viscosity and show what are Newtonian and non-Newtonian fluids · Define the appropriate physical properties and show how these allow differentiation between solids and fluids as well as between liquids and gases. INTRODUCTION What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids - both liquids and gases. The analysis of the behavior of fluids is based on the fundamental laws of mechanics that relate continuity of mass and energy with force and momentum together with the familiar solid mechanic’s properties. Fluid mechanics is a division in applied mechanics related to the behaviour of liquid or gas, which is either in rest or in motion. The study related to a fluid in rest or stationary is referred to fluid static, otherwise it is referred to as fluid dynamic.
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 2 Figure 1.1 : Arrangement of molecules 1.1 FLUID CHARACTERISTICS 1.1.1 FLUID Fluid can be defined as a substance that can deform continuously when being subjected to shear stress at any magnitude. In other words, it can flow continuously because of shearing action. This includes any liquid or gas. We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids, they lack the ability to resist deformation. Because a fluid cannot resist the deformation force, it moves, it flow under the action of the force. Its shape will change continuously as long as the force is applied. A solid can resist a deformation force while at rest, this force may cause some displacement but the solid does not continue to move indefinitely. 1.1.2 CHARACTERISTIC OF FLUID The major states or phases of matter are solid, liquid, and gas. Since liquids and gases have some common characteristics that are different from solids, they have been classified as fluids. One common characteristic of fluids is that they do not have a fixed shape and are easily deformed. In space, liquids tend to take the form of a sphere, while gases seem shapeless. Under the influence of gravity, liquids take on the shape of the container, while gases often flow out of any container. Both liquids and gases can be made to flow. 1.1.3 Difference Between Solid, Liquid and Gas. Solid Liquid Gas Arrangement of molecules : close order and in order patterns. Intermolecular forces : the cohesive force and repulsive force are strong. The movement of molecules : only molecules vibrate at a fixed position. The shape and volume : the volume and shape remain. Arrangement of molecules : less closely than solid. Intermolecular forces : cohesive force is weak but repulsive force is strong. The movement of molecules : molecules move freely and can slip. The shape and volume : the volume is fixed but not in shape. Arrangement of molecules : irregular and far from each other. Intermolecular forces : cohesive forces and repulsive forces are weak. The movement of molecules : free and randomly filling the space. The shape and volume : do not have a fixed volume and shape.
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 3 Figure 1.2 : Pressure diagram 1.2 PRESSURE Pressure is another important characteristic of fluids. This is the force exerted over a given area. In a liquid, the pressure comes from the weight of the fluid and the weight of the air above it, which we call the atmosphere. You feel this pressure when you swim under water. The deeper you go, the more pressure you feel. This is because there is more weight pressing down on you from above as you increase your depth. The same principle is true for atmospheric gases. The lower in elevation you are, the more weight and pressure you experience. However, even though these gases are constantly putting pressure on you, you do not feel it because your body is the same pressure as the surrounding air. Pressure is an effect that occurs when a force is applied on a surface. Pressure is the amount of force acting on a unit area. The symbol of pressure is P. It is usually more convenient to use pressure rather than force to describe the influences upon fluid behavior. Pressure cannot be measured directly. Normally, pressure difference measuring device is considered with the surrounding air. It is known as Pressure Gauge including Hydraulic Jack, U-Tube Manometers (Simple U-tube Manometer, Differential U-tube Manometer and Inverted U-Tube Manometer), Bourdon Gauge, Piezo Tube, Barometer and many others. There are many different units that can be used to express pressure, like pounds per square inch (psi), millimeters of mercury (mmHg), bar, Pascal and Newton per meter squared (N/m2 ). It all depends on the force and the area you are measuring. Where, 1 bar = 105 N/m2 1 bar = 105 Pa 1 Pa = 1 N/m2 1.2.1 TYPES OF PRESSURE i. Atmospheric Pressure (PO) - the air pressure that considered as reference / datum pressure. Often they do not have a fixed value that depends on several factors of weather conditions and altitude. ii. Gauge Pressure (PG) - the difference in pressure between the fluid and the ambient air measured by the gauge (Pressure Gauge). iii. Absolute Pressure (PA) - gauge pressure + atmospheric pressure [PA = PG + PO] Gauge Pressure Absolute Pressure Atmospheric Pressure iv. Vacuum pressure (PV) - space that has a value of zero pressure.
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 4 Figure 1.3 : Pressure and depth 1.3 PRESSURE IN FLUID 1.3.1 RELATIONSHIP BETWEEN PRESSURE AND DEPTH For equilibrium: force acting = liquid weight Where: Pressure, A F P = F = PA --------------- (1) Specific weight, Ah W V W ω = = W = ωAh --------------- (2) Because of: Force = Weight F = W PA = ωAh → from (1) & (2) P = ωh P = ρgh 1.3.2 EQUATION BASED ON RELATIONSHIP BETWEEN PRESSURE AND DEPTH P = ρgh P = h 1.3.3 HEAD OF PRESSURE, H METER Head of Pressure at a point in a fluid column can be expressed in the fluid causing the pressure. If is the specific weight of a fluid, the pressure, P = h where h is the height of the column or head pressure of the fluid at that point. 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 is the weight of the fluid above it. The more dense the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid. specific weight of liquid, ω h A Pressure, P
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 5 EXERCISE 1.1 : 1. Define the following terms : a. Pressure (P) b. Atmospheric Pressure (Patm) c. Gauge Pressure (PG) d. Absolute Pressure (PA) e. Vacuum Pressure (Pv ) [refer to the notes for the answer!!!] Answer : 2. What is the pressure gauge of air in the cylinder if the atmospheric pressure is 101.3 kN/m2 and absolute pressure is 460 kN/m2 . [358.7 kN/m2] Answer : 3. A Bourdon pressure gauge attached to a boiler located at sea level shows a reading pressure of 7 bar. If atmospheric pressure is 1.013 bar, what is the absolute pressure in that boiler (in kN/m2 ) ? [801.3 kN/m2] Answer :
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 6 4. Assume the density of water to be 1000 kg/m3 at atmospheric pressure 101 kN/m2 . What will be: a. the gauge pressure of water at depth 12 m from sea level [117.72 kN/m2] b. the absolute pressure of water at a depth of 2000 m below the free surface? [19721 kN/m2] Answer : 5. Determine in Newton per square metre, the increase in pressure intensity per metre depth in fresh water. The mass density of fresh water is 1000 kg/m3 . [9.81 kN/m2] Answer : 6. A diver is working at a depth of 20 m below the surface of the sea. How much greater is the pressure intensity at this depth than at the surface? Take into consideration specific weight of water is 10000 N/m3 . [P = 200 kN/m2 ] Answer :
CHAPTER 1 : INTRODUCTION OF FLUID FLUID MECHANICS 7 7. Find the height of a water column which is equivalent to the pressure of 2 N/m2 . [h = 2.04 x 10-4 m] Answer : 8. Find the head, h of water corresponding to an intensity of pressure, P of 340 000 N/m2 . [h = 34.7 m] Answer :
CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID FLUID MECHANICS 8 CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID OBJECTIVES · Introduce the concept of physical properties of fluid; · Have a working knowledge of the basic properties of fluids and understand the continuum approximation · Have a working knowledge of viscosity and the consequences of the frictional effects it causes in fluid flow · Calculate the capillary rises and drops due to the surface tension effect INTRODUCTION One property that all fluids do share is that they have density. This is simply the amount of matter in a given space for that substance. Another way of saying this is that the density is the amount of matter per unit volume, or in equation form: density = mass/volume.
CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID FLUID MECHANICS 9 2.1 PHYSICAL PROPERTIES OF FLUID There are certain properties that fluids share, though the specifics of these may be slightly different for each type of fluid. The main difference between the two fluids mentioned here is that gas particles are much farther apart than the particles of a liquid. Both will spread out to fill their container, but a liquid only does so beneath its surface. This important difference helps us understand that a gas is compressible, which means its volume can easily be increased or decreased, while a liquid is incompressible, meaning its volume cannot easily be changed. In other words, you can more easily press gas particles together than you can the particles of a liquid. This is because there is more space between those gas particles, while the liquid particles are already about as close together as they can get. One property that all fluids do share is that they have density. This is simply the amount of matter in a given space for that substance. Another way of saying this is that the density is the amount of matter per unit volume, or in equation form: density = mass/volume. a. Mass Density The mass density or density of a material is defined as its mass per unit volume. The symbol most often used for density is ρ (rho). = V m (kg/m3 ) b. Relative Density / Specific Gravity Relative density or specific gravity is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. Specific gravity usually means relative density with respect to water. The term "relative density" is often preferred in modern scientific usage. The symbol most often used for density is S. Ssubstance = water substance ω ω = water substance ρ ρ (no unit) c. Specific Weight The specific weight (also known as the unit weight) is the weight per unit volume of a material. The symbol of specific weight is (omega). = v W = v mg = g (N/m3 ) d. Specific Volume Specific volume is the volume of a substance divided by mass. It is the reciprocal of density. The symbol of specific volume is (upsilon). = m V = ρ 1 (m3 /kg)
CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID FLUID MECHANICS 10 e. Fluid Compressibility Compressibility (K) is a measure of the relative volume change of a fluid or solid as a response to a pressure (or mean stress) change. Compressible flow describes the behavior of fluids that experience significant variations in density. For flows in which the density does not vary significantly, the analysis of the behavior of such flows may be simplified greatly by assuming a constant density. This is an idealization, which leads to the theory of incompressible flow. However, in the many cases dealing with gases (especially at higher velocities) and those cases dealing with liquids with large pressure changes, the significant variations in density can occur, and the flow should be analyzed as a compressible flow if accurate results are to be obtained. f. Viscosity Viscosity is a measure of the resistance of a fluid that is being deformed by either shear stress or tensile stress. Viscosity is an internal friction of a fluid. Viscosity of fluid effects can be seen in the pipe flow, blood flow and others. Fluid viscosity depends on the density of the fluid. Viscosity has a direct proportion to fluid density. In simpler word, fluid with a higher density will have a higher in viscosity and vice versa. Viscosity is an important fluid property when analyzing liquid behaviour and fluid motion near solid boundaries. Viscosity of a fluid is a measure of its resistance to gradual deformation by shear stress or tensile stress. The shear resistance in a fluid is caused by inter molecular friction exerted when layers of fluid attempt to slide by one another. • viscosity is the measure of a fluid's resistance to flow • molasses is highly viscous • water is medium viscous • gas is low viscous There are two related measures of fluid viscosity i) Dynamic (absolute) Viscosity Absolute or dynamic viscosity - coefficient of absolute viscosity is a measure of internal resistance. Dynamic (absolute) viscosity is the tangential / shear force per unit area required to move one horizontal plane with respect to another plane at a unit velocity when maintaining a unit distance apart in the fluid. The symbol of specific volume is (eta) in Ns/m2 or kg/ms. = (Ns /m2 ) ; = A F ii) Kinematic Viscosity Kinematic viscosity - is the ratio of absolute (or dynamic) viscosity to density. A quantity in which no force is involved. Kinematic viscosity can be obtained by dividing the absolute viscosity of a fluid with the fluid mass density. The symbol of specific volume is (nu) in m2 /s. = ρ (m2 /s)
CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID FLUID MECHANICS 11 Quantity Symbol Formula Unit Mass Density ρ (rho) ρ = V m kg/m3 Specific Gravity / Relative Density s S = water substance ω ω = water substance ρ ρ no unit Specific Weight (omega) = v W = v mg = g N/m3 Specific Volume (upsilon) = m V = ρ 1 m3 /kg Dynamic Viscosity (eta) = Ns/m2 @ kg/ms @ Pa s (pressure X time) Kinematic Viscosity (nu) = ρ m2 /s @ Nm/kg @ Js/kg (specific energy X time) Table 1.1 : Summary of the quantities of fluid EXERCISE 2.1 : 1. What is the mass density, ρ of fluid (in kg/m3 ) if mass is 450 g and the volume is 9 cm3 ? [50 x 103 kg/m3] Answer : 2. What is the specific weight, of fluid (in kN/m3 ) if the weight of fluid is 10N and the volume is 500 cm3 . Also calculate the specific gravity and the specific volume of the fluid? [20 kN/m3, 2.039, 4.905 x 10-4 m3/kg] Answer :
CHAPTER 2 : PHYSICAL PROPERTIES OF FLUID FLUID MECHANICS 12 3. Given specific weight of fluid is 6.54 kN/m3 and its mass is 8.3 kg, calculate the following: a) volume of fluid [0.0124 m3] b) specific volume of fluid [0.0015 m3/kg] c) density of fluid [669.355 kg/m3] d) specific gravity of fluid [0.6667] Answer : 4. Given oil specific gravity is 0.89, find : a) density of oil [0.89 x 103 kg/m3] b) specific weight of oil [8730.9 N/m3] c) specific volume of oil [0.00112 m3/kg] Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 13 CHAPTER 3 : FLUID STATICS OBJECTIVES · Introduce the concept of pressure; · Prove it has a unique value at any particular elevation; · Show how it varies with depth according to the hydrostatic equation and · Show how pressure can be expressed in terms of head of fluid. INTRODUCTION Fluid statics (also called hydrostatics) is the science of fluids at rest, and is a sub-field within fluid mechanics. The term usually refers to the mathematical treatment of the subject. It embraces the study of the conditions under which fluids are at rest in stable equilibrium. The use of fluid to do work is called hydraulics.
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 14 Figure 3.1 : Pascal Law Figure 3.2 : Hidraulic Jack 3.1 PASCAL LAW AND HYDRAULIC JACK 3.1.1 PASCAL LAW Pascal's Law states that pressure acts equally in all directions; the force acts at right angles to any surface in contact with the fluid. It applies to fluids at rest. P1 P8 P2 P7 P3 P1 = P2 = P3 = P4 = P5 = P6 = P7 = P8 (Pressure at any point is the same in all directions) P6 P4 P5 3.1.2 HYDRAULIC JACK A mechanical jack is a device, which lifts heavy equipment. Hydraulic jacks are typically used for shop work, rather than as an emergency jack to be carried with the vehicle. A hydraulic jack uses a fluid, which is incompressible, that is forced into a cylinder. Pascal's Principle allows a hydraulic lift to generate large amounts of force from the application of a small force. Basically, the principle states that the pressure in a closed container is the same at all points.
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 15 P1 = P2 P1 = F/A1 _________ (1) P2 = W/A2 ________ (2) 2 A W 1 A F = W F P1 P2 A1 A2 P1 = P2 ; if A1 = A2 2 A W 1 A F = F = 2 A W x A1 W P1 P2 A1 A2 F fluid fluid Figure 3.2.1 Figure 3.2.2 Situation I if the small piston and the large piston are at the same level Situation I I if both piston are in the same size and at the same level F = W F = 2 A W x A1
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 16 F W A1 A2 P1 P2 h fluid W A2 F A1 P1 P2 h fluid Figure 3.2.3 Figure 3.2.4 Situation I I I if the small piston and the large piston are not in the same level (i) P1 = 1 A F ; P2 = 2 A W P2 = P1 + fluid h 2 A W = 1 A F + fluid h (ii) P1 = 1 A F ; P2 = 2 A W P1 = P2 + fluid h 1 A F = 2 A W + fluid h F = ( 2 A W - fluid h ) A1 F = ( 2 A W + fluid h ) A1
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 17 EXERCISE 3.1 : 1. A force of 500 N is applied to the smaller cylinder of a hydraulic jack. The area of a small piston is 20 cm2 while the area of a larger piston is 200 cm2 . What mass can be lifted on the larger piston? [m = 509.684 kg] Answer : 2. A force of 650 N is applied to the smaller cylinder of a hydraulic jack. The area of a small piston is 15 cm2 and the area of a larger piston is 150 cm2 . What load, W can be lifted on the larger piston if : a. the pistons are at the same level ? [W = 6500 N] b. the large piston is 0.65 m below the smaller piston ? [W = 6.5956 kN] c. the small piston is 0.40 m below the larger piston? [W = 6.441 kN] Answer : 3. In a hydraulic jack, a force is applied to a small piston that lifts the load on the large piston. If the diameter of the small piston is 15 mm and that of the large piston is 180 mm, calculate the value of force required to lift 1000 kg. [F = 68.2 N] Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 18 4. A pipe containing water connects two cylinders with pistons. Their diameters are 75 mm and 600 mm respectively and the face of the smaller piston is 6 m above the larger. What force on the smaller piston is required to maintain a load of 3500 kg on the larger piston? [F = 276.5 N] Answer : 5. A pipe filled with oil connects two cylinders. Small and large piston with diameter cylinder is 82 mm and 710 mm respectively. The small piston height is 0.8 m higher than the larger piston. Force of 160 N is applied on the small piston to lift 10000 kg load by the larger piston. Determine the specific gravity of the oil. [ωoil = 271865.69 N/m2 ; soil = 27.71] Answer : 6. A hydraulic jack used to lift a vehicle of mass 1200 kg. Small piston diameter is 120 mm. Maximum force that may be imposed by the small piston is 120 N. What is the diameter of the large piston hydraulic jack to lift the vehicle? [d2 = 1.1886 m] Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 19 3.2 PIEZOMETER, BAROMETER AND MANOMETER Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure pressure are called pressure gauges. Types of pressure measuring instrument : • Anemometer (used to determine wind speed) • Barometer used to measure the atmospheric pressure. • Manometer see pressure measurement • Pitot tube (used to determine speed) • Tire-pressure gauge in industry and mobility PIEZOMETER (Pressure Tube) Figure 3.3 : Piezometer inside a pipe A Piezometer is used for measuring pressure inside a vessel or pipe in which liquid is there. A tube may be attached to the walls of the container (or pipe) in which the liquid resides so that liquid can rise in the tube. By determining the height to which liquid rises and using the relation p1 = ρgh, gauge pressure of the liquid can be determined. It is important that the opening of the device is to be tangential to any fluid motion; otherwise, an erroneous reading will result. Although the Piezometer tube is a very simple and accurate pressure measuring device, it has several disadvantages. It is only suitable if the pressure in the container (pipe or vessel) is greater than the atmospheric pressure (otherwise air would be sucked into system), and the pressure to be measured must be relatively small so that the required height of column is reasonable. In addition, the fluid in the container in which the pressure is to be measured must be a liquid rather than a gas. BAROMETERS Figure 3.4 : Mercury Barometer A Barometer is a device used for measuring atmospheric pressure. A simple Barometer consists of a tube of more than 30 inch (760 mm) long inserted into an open container of mercury with a closed and evacuated end at the top and open tube end at the bottom and with mercury extending from the container up into the tube. Strictly, the space above the liquid cannot be a true vacuum. It contains mercury vapour at its saturated vapour pressure, but this is extremely small at room temperatures (e.g. 0.173 Pa at 20oC). The atmospheric pressure is calculated from the relation patm = ρgh where ρ is the density of fluid in the barometer. There are two types of Barometer; Mercury Barometer and Aneroid Barometer.
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 20 Figure 3.5 : Simple U-Tube Manometer MANOMETER A manometer usually limited to measuring pressures near to atmospheric. The term manometer is often used to refer specifically to liquid column hydrostatic instruments. A manometer consisting of a U-shaped glass tube partly filled with a liquid of known specific gravity; when the legs of the manometer are connected to separate sources of pressure, the liquid rises in one leg and drops in the other; the difference between the levels is proportional to the difference in pressures and inversely proportional to the liquid's specific gravity. Also known as liquid-column gauge. One of the most common manometers is the water filled u-tube manometer used to measure pressure difference in pitot or orifices located in the airflow in air handling or ventilation system. The advantage of this tube is it can measure a pressure in positive and negative value. There are several types of U-Tube Manometer : a. Simple U-Tube Manometer The simplest manometer is a tube bent in U-shape. One end of which is attached to the gauge point and the other is open at the top, which is attached to the top of a vessel containing liquid at a pressure (higher than atmospheric) to be measured. An example can be seen in the figure below. As the tube is open to the atmosphere, the pressure measured is relative to atmospheric so is gauge pressure. The liquid used in the bent tube or simple manometer is generally mercury, which is 13.6 times heavier than water. Hence, it is also suitable for measuring high pressure. Now consider a simple manometer connected to a pipe containing a light liquid under high pressure. The high pressure in the pipe will force the heavy liquid, in the left-hand limb of the Utube, to move downward. This downward movement of the heavy liquid in the left-hand limb will cause a corresponding rise of the heavy liquid in the right-hand limb. 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, as shown in figure. PB = PC ; PD = 0 PB = PA + water h1 PC = PD + Hg h2 PA + water h1 = PD + Hg h2 PA = PD + Hg h2 - water h1 PA = PD + Hg h2 - water h1 PA = Hg h2 - water h1 water mercury A B
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 21 b. Differential U-Tube Manometer It is a device used for measuring the difference of pressures, between two points in a pipe, or in two different pipes. A differential manometer consists of a U-tube, containing a heavy liquid with two ends connected to two different points. We are required to find the difference of pressure at these two points, as shown in figure. A differential manometer is connected to two different points A and B. A little consideration will show that the greater pressure at A will force the heavy liquid in the U-tube to move downwards. This downward movement of the heavy liquid, in the left-hand limb, will cause a corresponding rise of the heavy liquid in the right-hand limb as shown in figure. The horizontal surface C-D, at which the heavy liquid meet in the left-hand limb, is the datum line. As we know that the pressures in the left-hand limb and right-hand limb, above the datum line are equal. PX = PY PX = PA + water h1 PY = PB + water h3 + Hg h2 PA + water h1 = PB + water h3 + Hg h2 PA = PB + water h3 + Hg h2 - water h1 PA - PB = water (h3 - h1) + Hg h2 X Y mercury H2O Let : h = Height of the light liquid in the left-hand limb above the datum line. h1 = Height of the heavy liquid in the right-hand limb above the datum line h2 = Height of the light liquid in the right-hand limb above the datum line pA = Pressure in the pipe A, expressed in term of head of the liquid in cm pB = Pressure in the pipe B, expressed in term of head of the liquid in cm ωP = Specific weight of the light liquid ωQ = Specific weight of the heavy liquid Figure 3.6 : Differential U-Tube Manometer Figure 3.6.1
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 22 Figure 3.7 : Inverted Differential U-Tube Manometer Figure 3.7.1 Figure 3.7.2 c. Inverted Differential U-Tube Manometer Inverted U-tube manometer is used for measuring pressure differences in liquids. The space above the liquid in the manometer is filled with air, which can be admitted or expelled through the tap on the top, in order to adjust the level of the liquid in the manometer. It is a particular type of differential manometer, in which an inverted U-tube is used. 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. The two ends of the U-tube are connected to the points where the difference of pressure is to be found out as shown in figure below. Now consider an inverted differential manometer whose two ends are connected to two different points A and B. Let us assume that the pressure at point A is more than that at point B, a greater pressure at A will force the light liquid in the inverted U-tube to move upwards. This upward movement of liquid in the left limb will cause a corresponding fall of the light liquid in the right limb as shown in figure. Let us take C-D as the datum line in this case. We know that pressures in the left limb and right limb below the datum line are equal. (i) (ii) ωP ωQ Let : h = Height of the heavy liquid in the left-hand limb below the datum line, h1= Height of the light liquid in the left-hand limb below the datum line , h2= Height of the light liquid in the right-hand limb below the datum line, ωP = Specific weight of the light liquid ωQ = Specific weight of the heavy liquid PC = PD PA = PC + Q h + P h1 PB = PD + P h2 PA - PB = (PC + Q h + P h1) - (PD + P h2) = Q h + P h1 - P h2 PA - PB = Q h + P (h1 - h2) ωQ ωP PA = PC + Q h1 PB = PD + P h2 PA - PB = (PC + Q h1) - (PD + P h2) PA - PB = Q h1 - P h2 ωP ωQ air / gas C
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 23 Figure 3.8 : Combined U-Tube Manometer d. Combined / Compound Differential U-Tube Manometer Compound Manometer is a 'U' Tube manometer having 'T' joints at appropriate equal elevation in the columns of U tube manometer where the impulse lines are to be connected to measure Differential Pressure. T joint allows the use two liquids simultaneously in the manometer. I.e. the liquid whose differential pressure is to be measured being lighter it will gauge at the top and a heavier liquid that does not dissolve with the lighter fluid will gauge at the bottom. Elevation difference in lighter liquid as well as heavier liquids gives their respective differential pressure. Both the differential pressure will be equal in terms of Pressure Units. Since two liquids are used simultaneously, this type of U tube manometer is called Compound Manometer. A compound U-tube manometer consists of more than one U-tube in series between one system and another. Example 3.1 : (one of the method for the solution) (i) P1 = 0 N/m2 (atm pressure) 1→2 : P2 = P1 + (water x 0.9) 2→3 : P3 = P2 + (Hg x 0.6) P3 = [P1 + (water x 0.9)] + [(Hg x 0.6)] 3→4 : P3 = P4 + oil (1.5 + 0.3) P1 + (water x 0.9) + (Hg x 0.6) = P4 + (oil x 1.8) P4 = P1 + (water x 0.9) + (Hg x 0.6) - (oil x 1.8) 4→A : PA = P4 + (water x 1.5) = P1 + (water x 0.9) + (Hg x 0.6) - (oil x 1.8) + (water x 1.5) = P1 + (water x 0.9 + 1.5) + (Hg x 0.6) - (oil x 1.8) = (9810 x 2.4) + (133416 x 0.6) - (7848 x 1.8) = (23544) + (80049.6) - (14126.4) = 89467.2 N/m2 PA = 89.4672 kN/m2 ════════════ Water = 9810 N/m3 Hg = 13.6 x 9810 = 133416 N/m3 Oil = 0.8 x 9810 = 7848 N/m3 1 2 3 4 A
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 24 ii) A→1 : P1 = PA + water (0.6 + 0.5) = PA + water (1.1) = PA + 9800 (1.1) P1 = PA + 10780 N/m2 1→2 : P1 = P2 + (fluid x 0.6) = P2 + (2.6 x 9800 x 0.6) P1 = P2 + 15288 P2 = P1 – 15288 N/m2 2→B : PB = P2 + water (1.3 - 0.5) PB = (P1 – 15288) + 9800 (0.8) = P1 – 15288 + 7840 PB = P1 – 7448 = (PA + 10780) – 7448 PB = PA + 3332 = - 3332 N/m2 PA – PB = - 3.332 kN/m2 ═══════════════ 50 cm 60 cm 130 cm A B Sfluid = 2.6 H2O PA - PB = ? water = 9800 N/m3 50 cm 60 cm 130 cm A B Sfluid = 2.6 H2O 2 1 60 cm 50 cm 50 cm 60 cm
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 25 EXERCISE 3.2 : 1. A U-tube manometer similar to that shown in figure below is used to measure the gauge pressure of water. What will be the gauge pressure at A if h1 is 45 cm and h2 is 25 cm above BC. [PA = 88.98 x 103 N/m2 ] Answer : 2. Figure shown a U-tube manometer that used to measure the gauge pressure of water. Find the value of ‘h’ if the gauge pressure at A is 175 kN/m2 and the height ‘k’ is 1.8 m. Given that, the specific gravity of raw oil is 0.8. [h = 24.55 m] Answer : ωwater ωmercury Raw oil water • k h
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 26 3. A U-tube manometer similar to that shown in figure is used to measure the gauge pressure of a fluid P. If fluid P is water and liquid Q mercury, what will be the gauge pressure at A. Take into consideration patm = 101.3 kN/m2 . [PA = 66.474 kN/m2 ] Answer : 4. What is the gauge pressure of the water at A if h1 = 60 cm and the mercury in the right hand limb, h2 = 90 cm as shown in the figure below? [PA = 114.19 kN/m2 ] Answer : 150mm 250mm Liquid Q water mercury
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 27 5. In the figure below, fluid at A is water and fluid B is mercury (sHg = 13.6). What will be the difference in level h if the pressure at X is 140 kN/m2 and a =150 cm? [h = 1.16 m] Answer : 6. Assuming that the atmospheric pressure is 101.3 kN/m2 find the absolute pressure at A in the figure below when a. fluid P is water, fluid Q is mercury (sHg = 13.6), a = 1 m and h = 0.4 m. [PA = 38.124 kN/m2 ] b. fluid P is oil (s = 0.82), fluid Q is brine (sbrine = 1.10), a = 20 cm and h = 55 cm. [PA = 93.751 kN/m2 ] Answer : X D
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 28 7. A U-tube manometer measures the pressure difference between two points A and B in a liquid. The U-tube contains mercury. Calculate the difference in pressure if h =150 cm, h2 = 75 cm and h1 = 50 cm. The liquid at A and B is water (ω = 9.81 × 103 N/m3 ) and the specific gravity of mercury is 13.6. [PA – PB = 54.45 kN/m2 ] Answer : 8. In the figure below, fluid P is water and fluid Q is mercury (specific gravity = 13.6). If the pressure difference between A and B is 35 kN/m2 , a = 100 cm and b = 30 cm, what is the difference in level h? [h = 0.3069 m] Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 29 9. According to the figure in question 8, fluid P is oil (specific gravity = 0.85) and fluid Q is water. If a = 120 cm, b = 60 cm and h = 45 cm, what is the difference in pressure in kN/m2 between A and B? [PA – PB = - 4.34 kN/m2 ] Answer : 10. In the figure below, fluid Q is water and fluid P is oil (specific gravity = 0.9). If h = 69 cm and z = 23 cm, what is the difference in pressure in kN/m2 between A and B? [PB – PA = 1.58 kN/m2 ] Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 30 [Pgas = 82.361 N/m2 ] 11. In question 10, fluid Q is water and fluid P is air. Assuming that the specific weight of air is negligible, what is the pressure difference in kN/m2 between A and B? [PB – PA = 4.51 N/m2 ] Answer : 12. Base on the diagram below, calculate Pgas ? Answer :
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 31 13. In the figure below, given SFW = 1, and SSW = 1.03. What is the difference in pressure in kN/m2 between Q and R? Answer : R Q
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 32 Figure 3.8 : Combined U-Tube Manometer 3.3 BOURDON GAUGE / BOURDON TUBE The Bourdon Tube is a non liquid pressure measurement device. It is widely used in applications where inexpensive static pressure measurements are needed. A typical Bourdon tube contains a curved tube that is open to external pressure input on one end and is coupled mechanically to an indicating needle on the other end, as shown schematically below. Typical Bourdon Tube Pressure Gages Internal linkages have been simplified. The external pressure is guided into the tube and causes it to flex, resulting in a change in curvature of the tube. These curvature changes are linked to the dial indicator for a number readout. Alternatively, a strain gage circuit can be attached on the tube to convert the pressure-induced deflections into electric voltage signals. These signals can then be output electronically, rather than mechanically with the dial indicator. 3.4 BUOYANCY The mathematician Archimedes discovered much of how buoyancy works almost 2000 years ago. In his research, Archimedes discovered that an object is buoyed up by a force equal to the weight of the water displaced by the object. In other words, a inflatable boat that displaces 100 pounds (45 kilograms) of water is buoyed up by that same weight of support. An object that floats in the water is known as being positively buoyant. An object that sinks to the bottom is negatively buoyant, while an object that hovers at the same level in the water is neutrally-buoyant. This same idea helps to determine what will float in water and what will sink. If an object weighs more than the weight of the water it displaces, it will sink. If the object weighs less, it will float. This helps explain why a heavy ship can easily float in the water, while a much smaller and lighter brick will sink quickly. It isn't the size or shape of an object that primarily determines buoyancy, but the relation between an object's weights compared to the weight of the water the object displaces. Buoyancy is the ability of an object to float in a liquid, such as water. This concept helps to explain why some things float while other objects sink. Buoyancy is an important factor in the design of many objects and in a number of water-based activities, such as boating or scuba diving.
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 33 EXERCISE 3.3 : 1. A rectangular pontoon has a width B of 6 m, a length, l of 12 m, and a draught D of 1.5 m in fresh water (density 1000 kg/m3 ). Calculate : a. the weight of the pontoon b. its draught in sea water (density 1025 kg/m3 ) c. the load (in kN) that can be supported by the pontoon in fresh water if the maximum draught permissible is 2 m. Solution When the pontoon is floating in an unloaded condition, Up trust on immersed volume = weight of pontoon Since the up trust is equal to weight of the fluid displaced, Weight of pontoon = weight of fluid displaced, So, W = g Bl D a) In fresh water, 3 =1000 kg/ m and D = 1.5 m ; Therefore, Weight of pontoon, W = 1000 9.81 6 121.5 N W = 1059.5 kN b) In the sea water, 3 =1025 kg/ m ; therefore, Draught in sea water, g B l W D = = 1.46 m c) For maximum draught of 2 m in fresh water, Total up trust = weight of water displaced = g Bl D = 10009.81612 2 N = 1412.6 kN Load which can be supported = Up thrust – weight of pontoon = 1412.6 −1059.5 = 353.1 kN
CHAPTER 3 : FLUID STATICS FLUID MECHANICS 34 2. A rectangular pontoon 5.4 m wide by 12 m long, has a draught of 1.5 m in fresh water. Calculate: a. the mass of the pontoon, [m = 97000 kg] b. its draught in the sea water (density 1025 kg/m3 ) [1.47 m] Answer : 3. A ship floating in sea water displaces 115 m3 . Find : a. the weight of the ship if sea water has a density of 1025 kg/m3 , [118000 kg] b. the volume of fresh water (density 1000 kg/m3 ) which the ship would displace. [m2= 97000 kg] Answer :
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 35 CHAPTER 4 : FLUID DYNAMICS OBJECTIVES · Introduce concepts necessary to analyze fluids in motion · Identify differences between Steady/unsteady uniform/non-uniform compressible / incompressible flow · Demonstrate streamlines and stream tubes · Introduce the Continuity principle through conservation of mass and control volumes · Derive the Bernoulli (energy) equation · Demonstrate practical uses of the Bernoulli and continuity equation in the analysis of flow · Introduce the momentum equation for a fluid · Demonstrate how the momentum equation and principle of conservation of momentum is used to predict forces induced by flowing fluids INTRODUCTION This section discusses the analysis of fluid in motion - fluid dynamics. The motion of fluids can be predicted in the same way as the motion of solids are predicted using the fundamental laws of physics together with the physical properties of the fluid. It is not difficult to envisage a very complex fluid flow. Spray behind a car; waves on beaches; hurricanes and tornadoes or any other atmospheric phenomenon are all example of highly complex fluid flows which can be analyzed with varying degrees of success (in some cases hardly at all!). There are many common situations which are easily analyzed. In physics, fluid dynamics is a sub-discipline of fluid mechanics that deals with fluid flow—the natural science of fluids (liquids and gases) in motion. It has several sub-disciplines itself, including aerodynamics (the study of air and other gases in motion) and hydrodynamics (the study of liquids in motion). Fluid dynamics has a wide range of applications, including calculating forces and moments on aircraft, determining the mass flow rate of petroleum through pipelines, predicting weather patterns, understanding nebulae in interstellar space and reportedly modeling fission weapon detonation. Some of its principles are even used in traffic engineering, where traffic is treated as a continuous fluid. Fluid dynamics offers a systematic structure that underlies these practical disciplines, that embraces empirical and semi-empirical laws derived from flow measurement and used to solve practical problems. The solution to a fluid dynamics problem typically involves calculating various properties of the fluid, such as velocity, pressure, density, and temperature, as functions of space and time. Fluid mechanics assumes that every fluid obeys the following: • Conservation of mass • Conservation of energy • Conservation of momentum • The continuum hypothesis, which considers fluids to be continuous • A fluid is incompressible – that is, the density of the fluid does not change. • The viscosity of the fluid is zero (the fluid is inviscid)
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 36 4.1 TYPES OF FLOW a. Uniform flow The cross-sectional area and velocity of the stream of fluid are the same at each successive crosssection. Example: flow through a pipe of uniform bore running completely full. a1 = a2 v1 = v2 v1 v2 (1) (2) b. Steady Flow The cross-sectional area and velocity of the stream are different at each successive cross-section, but for each cross-section they do not change with time. Example: flow through a tapering pipe. a1 > a2 v1 v2 v1 < v2 (1) (2) 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. Laminar flow generally happens when dealing with small pipes and low flow velocities. Laminar flow can be regarded as a series of liquid cylinders in the pipe, where the innermost parts flow the fastest, and the cylinder touching the pipe isn't moving at all. Shear stress depends almost only on the viscosity - μ - and is independent of density - ρ. d. Transitional Flow Transitional flow is a mixture of laminar and turbulent flow, with turbulence in the center of the pipe, and laminar flow near the edges. Each of these flows behave in different manners in terms of their frictional energy loss while flowing, and have different equations that predict their behavior. 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. In turbulent flow vortices, eddies and wakes make the flow unpredictable. Turbulent flow happens in general at high flow rates and with larger pipes. Shear stress for turbulent flow is a function of the density - ρ.
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 37 f. Non-Uniform Flow If at a given instant, the velocity is not the same at every point the flow is non-uniform. (In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform, as the fluid at the boundary must take the speed of the boundary, usually zero. However, if the size and shape of the cross-section of the stream of fluid is constant the flow is considered uniform.) g. Non-Steady / Unsteady Flow If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practice, there are always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady). OR The cross-sectional area and velocity of the stream at any cross-section vary with time. Example: a wave travelling along a channel. The dimensionless Reynolds Number determines Turbulent or laminar flow. Combining the above, we can classify any flow in to one of four type: a. Steady uniform flow. Conditions do not change with position in the stream or with time. An example is the flow of water in a pipe of constant diameter at constant velocity. b. Steady non-uniform flow. Conditions change from point to point in the stream but do not change with time. An example is flow in a tapering pipe with constant velocity at the inlet - velocity will change as you move along the length of the pipe toward the exit. c. Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with time. An example is a pipe of constant diameter connected to a pump pumping at a constant rate, which is then switched off. d. Unsteady non-uniform flow. Every condition of the flow may change from point to point and with time at every point. For example, waves in a channel. Reynolds Number The Reynolds number is important in analyzing any type of flow when there is substantial velocity gradient (i.e. shear.) It indicates the relative significance of the viscous effect compared to the inertia effect. The Reynolds number is proportional to inertial force divided by viscous force. The flow is : • laminar when Re < 2300 • transient when 2300 < Re < 4000 • turbulent when 4000 < Re Re = η ρνd
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 38 4.2 FLOW RATE / DISCHARGE Flow rate or rate of fluid flow is the volume of fluid, which passes through a given surface per unit time. Volumetric Flow rate 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 usually represented by the symbol Q. So cubic meters per second [m3 s -1 ] in SI units. Q = A.v Mass Flow rate 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 • m . So kilogram per second [kg s-1 ] in SI units. • = × × = × • • m1 = m2 1 1 1 2 2 2 A v = A v A very simple way to measure the rate at which water is flowing along the pipe is by catching all the water that is coming out of the pipe in a bucket over a fixed time period. We can obtain the rate of accumulation of mass by measuring the weight of the water in the bucket and dividing this by the time taken to collect this water. This is known as the mass flow rate. A1 A2 i in out
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 39 4.3 CONTINUITY EQUATION LAW Continuity Equation For continuity of fluid flow system, the total amount of fluid entering the system is equal to the amount of fluid leaving the system. This occurs in the case of uniform flow and steady flow. Discharge at section P = Discharge at section R QP = QR AP vP = AR vR Application We can apply the principle of continuity to pipes with cross sections that have changes along their length. Consider the diagram below of a pipe with a contraction. A liquid is flowing from left to right and the pipe is narrowing in the same direction. By the continuity principle, the discharge must be the same at each section. The mass going into the pipe is equal to the mass going out of the pipe. Discharge at section 1 = Discharge at section 2 Q1 = Q2 1 1 2 2 A v = A v 2 1 3 P R SYSTEM P R QP QR QP =QR QP = discharge through cross-section P-P AP = cross-sectional area through P-P vp = fluid mean velocity through P-P QR = discharge through cross-section R-R AR = cross-sectional area through R-R vR = fluid mean velocity through R-R Section 1 Section 2 Q1 = A1 v1 Q2 = A2 v2 Q3 = A3 v3 Q1 = Q2 + Q3
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 40 EXCERSICE 4.1 : 1. If the diameter d = 15 cm and the mean velocity, v = 3 m/s, calculate the actual discharge in the pipe. [Q = 0.053 m3 /s] Answer : 2. Oil flows through a pipe at a velocity of 1.6 m/s. The diameter of the pipe is 8 cm. Calculate discharge and mass flow rate of oil. Take into consideration soil = 0.85. [Q = 8.042x10-3 m3 /s] [ • m = 6.836 kg/s] Answer : 3. The mass of an empty bucket is 2.0 kg. After 7 seconds of collecting water the mass of the bucket is 8.0 kg. Calculate the mass flow rate of the fluid. [ • m = 0.857 kg/s] Answer : 4. If the area A1 = 10 10-3 m2 and A2 = 3 10-3 m2 and the upstream mean velocity, v1=2.1 m/s, calculate the downstream mean velocity. [v2 = 7.0 m/s] Answer :
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 41 5. State the actual discharge equation for the following pipes. Answer : 6. Referring to the Figure the diameter at section 1 is d1 = 30 mm and at section 2 is d2=40 mm and the mean velocity at section 2 is v2 = 3.0 m/s. Calculate the velocity entering the diffuser. [v1 = 5.3 m/s] Answer : 7. A pipe PR of diameter 40 cm, in which the mean velocity is 2 m/s split into two at K and L of diameter 15 cm and 20 cm respectively. The velocity at RK pipe is twice compared to RL pipe. Calculate: a) discharge at RK and RL pipes [QRK = 0.1332 m3 /s, QRL = 0.118 m3 /s] b) velocity at RK and RL pipes [vRK = 7.524 m/s, vRL = 3.76 m/s] Answer : Section 1 Section Section Section 2 6 5 4 3 2 7 8 1 Ans : Q1 = Q2 + Q3 Q2 = Q4 + Q5 + Q6 Q7 = Q3 – Q8
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 42 8. A pipe is split into 2 pipes which are BC and BD as shown in the Figure. The following information is given: Answer : 9. Water flows through a pipe AB of diameter d1 = 50 mm, which is in series with a pipe BC of diameter d2 = 75 mm in which the mean velocity v2 = 2 m/s. At C the pipe forks and one branch CD is of diameter d3 such that the mean velocity v3 is 1.5 m/s. The other branch CE is of diameter d4 = 30 mm and conditions are such that the discharge Q2 from BC divides so that Q4 = ½ Q3. Calculate the values of Q1,v1,Q2,Q3,D3,Q4 and v4. Answer : A D C B diameter pipe AB at A = 0.45 m diameter pipe AB at B = 0.3 m diameter pipe BC = 0.2 m diameter pipe BD = 0.15 m Calculate: a) discharge at section A if vA = 2 m/s [QA = 0.318 m3 /s] b) velocity at section B and section D if velocity at section C = 4 m/s [vB = 4.5 m/s, vD = 10.86 m/s] [Q1 = 8.836 × 10-3 m3 /s, v1 = 4.50 m/s, Q3 = 5.891 × 10-3 m3 /s, Q4 = 2.945 × 10-3 m3 /s, d3 = 86.6 mm, v4 = 4.17 m/s] B E D C A
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 43 4.4 BERNOULLI THEOREM 4.4.1 BERNOULLI THEOREM Bernoulli’s Theorem states that the total energy of each particle of a body of fluid is the same if no energy enters or leaves the system at any point. Bernoulli's theorem is an expression of the conservation of the total energy; that is, the sum total of these energies in a fluid flow remains a constant along a streamline. Expressed concisely, the sum of the kinetic energy, pressure energy, and potential energy remains a constant. I. Potential energy If a liquid of weight W is at a height of z above datum line Potential energy = Wz Potential energy per unit weight = z The potential energy per unit weight has dimensions of Nm/N, is measured as a length or head z, and can be called the potential head. II. Pressure energy When a fluid flows in a continuous stream under pressure it can do work. If the area of crosssection of the stream of fluid is a, then force due to pressure p on cross-section is pa. If a weight W of liquid passes the cross-section Volume passing cross-section = ω W Distance moved by liquid = ωa W Work done = force distance = ωa W pa = ω p W pressure energy per unit weight = ω p = ρ g p Similarly, the pressure energy per unit weight p/W is equivalent to a head and is referred to as the pressure head. III. Kinetic energy If a weight W of liquid has a velocity v, Kinetic energy = 2 v g W 2 1 Kinetic energy per unit weight = 2g 2 v The kinetic energy per unit weight 2g 2 v is also measured as a length and referred to as the velocity head.
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 44 The division of this energy between potential, pressure and kinetic energy may vary, but the total remains constant. In symbols: constant 2g 2 v ω p H = z + + = By Bernoulli’s Theorem, Total energy per unit weight at section 1 = Total energy per unit weight at section 2 2g 2 2 v ω 2 p 2 z 2g 2 1 v ω 1 p 1 z + + = + + z = potential head p = pressure head g v 2 2 = velocity head H = Total head 4.4.2 LIMITS OF BERNOULLI THEOREM Bernoulli’s Equation is the most important and useful equation in fluid mechanics. It may be written, 2 2 2 2 1 2 1 1 2 2 p g v z p g v z + + = + + Bernoulli’s Equation has some restrictions in its applicability, they are : ▪ the flow is steady ▪ the density is constant (which also means the fluid is compressible) ▪ friction losses are negligible ▪ the equation relates the state at two points along a single streamline (not conditions on two different streamlines). Do you know : The Bernoulli equation is named in honour of Daniel Bernoulli (1700-1782). Many phenomena regarding the flow of liquids and gases can be analysed by simply using the Bernoulli equation.
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 45 4.4.3 APPLICATION OF BERNOULLI’S EQUATION Bernoulli’s equation commonly found applied in the following applications. It proven true in both horizontal and incline installation: a. Horizontal Normal Pipes b. Incline Normal Pipes c. Horizontal Venturi Meter d. Incline Venturi Meter e. Orifice Meter f. Pitot Tube 4.4.3.1 Normal Pipes Horizontal uniform diameter pipe H = Z + ρg P + 2g 2 v datum line Incline uniform diameter pipe 2g 2 2 V ρg 2 P 2 Z 2g 2 1 V ρg 1 P 1 Z + + = + + datum line Horizontal tapered pipe datum line Incline tapered pipe datum line Z Z1 Z2 Z Z1 Z2