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DEPARTMENT OF CHEMICAL ENGINEERING, UNIVERSITY OF ILORIN, ILORIN. LECTURE GUIDE FUNDAMENTALS OF FLUID MECHANICS CHE 241 - 3 Credits 2010/2011 SESSION COURSE: CHE CHE 241 – FUNDAMENTALS OF FLUID MECHANICS (3 credits /Compulsory). Lecturer: OGUNLEYE,Oladipupo Olaosebikan B.Tech. Chemical Eng.(Ogbomoso),M.Sc. Industrial Eng. (Ibadan),Ph.D. Industrial Eng.(Ibadan). Reg. Engr.(COREN), MAiCHE,MSIAM. Department of Chemical Engineering, Faculty of Engineering and Technology, University of Ilorin, Ilorin, Kwara State, Nigeria.. [email protected] E-mail: Office Location: – Room 5, Ground Floor, Department of Chemical Engineering Building. Consultation Hours: 11.00-1.00pm Mondays and Wednesdays. Course Content: Properties of fluids, Fluids Statics, Basic conservation laws, friction effect and losses in laminar and turbulent follows in ducts and copies. Dimensional analysis and dynamic similitude, principles of construction and operation of selected hydraulic machinery. Hydropower systems. 45h (T); C Course Description: The course is designed to introduce students in the Faculty of Enginnering and Technology especially to lower level course in process engineering fluid mechanics, which emphasizes the systematic application of fundamental principles (e.g., macroscopic mass, energy, and momentum balances and economics) to the analysis of a variety of fluid problems of a practical nature. The scope of coverage includes internal flows of Newtonian and non- Newtonian incompressible fluids, adiabatic and isothermal compressible flows .Applications include dimensional analysis and scale-up, piping systems with fittings for Newtonian and non-Newtonian fluids, compressible pipe flows, flow measurement, and control, pumps, compressors. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 2 Course Justification: Nigeria being an oil-rich nation transports crude oil and its allied products through pipe line where so many phenomena occur due to the nature of the fluid being transported, the channel of transportation and the forces behind the transportation. Fluid mechanics gives clear insight into the understanding of such systems as this. The economics of crude oil processing will not be complete without an application of fluid mechanics to imrove pump efficiency in the course of crude oil transportation thereby minimizing the cost and maximizing profit. This principle is what many developed nations had applied to the management of their pipeline networks whose integrity has been secured through Artificial Intelligent Systems. Course Objectives: The objectives of this course as an integral part for the award of B. Eng. are : • For the student to know how the fundamental principles underlying the behavior of fluids can be applied in an organized and systematic manner to the solution of practical engineering problems. • To provide a ready reference and basic methods for the analysis of a variety of problems encountered in the movement of fluids through pipes , pumps and all kinds of process equipment. Course Requirements: This is a compulsory course for all students studying Engineering In view of this, students are expected to participate in all the course activities and have minimum of 75% attendance to be able to write the final examination. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 3 Methods of grading: No 1. 2. 3. Item Assignment/ Quiz / Monthly Test Mid Semester Test Examination Total Score % 10 20 70 100 Course Delivery Strategies: The lecture will be delivered through face-to-face method, lecture guide (lecture note) will be provided during lectures. Students will be encouraged and required to read around the topics. The delivery strategies will also be supported by tutorial sessions. LECTURES Week 1 - 2: General Introduction of Fluid Mechanics Objective: The students will be able to explain the basic concept of fluid mechanics. Description: The course outline will be introduced with emphasis on the objectives and delivery strategies, definitions, the continuum concept , types of fluids, units of measurements, properties of fluids, Viscosity, Compressibility of fluids, surface tension and capillarity, vapour pressure. Week 3: Dimensional Analysis and Dynamic Similitude Objective: Students will gain the necessary knowledge of how independent variables are expressed in terms of their dependent variables without experimentation. Description: Dimensional Analysis, Techniques of Dimensional Analysis, Interpretation of Dimensionless Numbers and Dynamic Similitude. Week 4: Fluids Statics. Objective: Student will be able to understand the forces which keep the body of fluid in static equilibrium. Description: Concept of stress and pressure, pressure transmission, pressure measurement, Forces on submerged surfaces,Buoyancy and Static Forces on Solid Boundaries. Student will also be assessed on the topics covered so far through a short Monthly Quiz Week 5 - 6: Basic Conservation Laws. 6: Laws Objective: The students will be exposed to the mathematical laws guiding the movement of mass and energy. Description: The concept of system, Conservation of Mass, Conservation of Energy, Irreversible Effects and Conservation of Momentum. Week 7 - 8 : Pipe Flows Objective: Students will be able to understand various flow regimes and their effects on flow velocities in pipes. Description: Flow Regimes, Pressure Drop in Laminar Flow, Pressure Drop in Turbulent Flow, Velocity Profiles in Turbulent Flow, Velocity Distribution, Pressure Losses in Pipe Fittings and Pipeline Problems. Student will also be assessed on the topics covered so far through a short Monthly Quiz. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 4 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 Week 9-10: Compressible Flows 10: Objective: Students will understand the effect of pressure and temperature on fluid density. Description: Gas Properties, Energy equation for compressible flow, Isentropic flow, Compressible flow in pipes. Week 11 11: Pumps and Compressors Objective: Students will understand the working principles of pumps and compressors. Description: Pumps, Pump Characteristics, Pumping Requirements and Pump Selection, Cavitation and Net Positive Suction Head (NPSH) , Compressors. Week 12: 12: Mid Semester Test Objective: To assess students’ mastery of the course. Description: Students will be tested on all the topics taken so far. This is to serve as a preparation for the final examination at the end of the semster Week 13 13 -14: 14: Flow Measurement and Control. Objective: Students will have the understanding of the construction and operation of the hydraulic machineries. Description: The Pitot Tube, The Venturi and Nozzle, The Orifice Meter, Loss Coefficient, Orifice Problems and Control Valves. Week 15: Revision/ Tutorial Exercises Objective: Student will have opportunity to ask questions on all the topics covered in the course. Description: A general overview of the course will be made. Students are expected to seek explanation on any difficult concept or topic treated during the course. LIST OF BOOKS FOR FURTHER READING: 1 Ron Darby (2001) .Chemical Engineering Fluid Mechanics. Second Edition. Marcel Dekker, Inc. New York. 2 McCabe,W. L., Smith,J.C. and Harriott, P.(1993) .Unit Operations of Chemical Engineering. Fifth Edition. McGraw Hill Co. Singapore. 3 Ogboja ,O. (2005). Fluid Mechanics .UNESCO, Nairobi Kenya. 4 Coulson, J.M. and Richardson, J.F., Backhurst,J.R. and Harker, J.H. (2001 ) Chemical Engineering Volume 1, Sixth Edition, Butterworth Heinmann Inc. Oxford. 5 Douglas,J.F. and Gasiorek, J. M. (1997) Fluid Mechanics. Third Edition,Longman Press, Singapore. 6 Rajput, R.K. (1998) . A Textbook of Fluid Mechanics in SI Unit. First Edition. S. Chand & Company Ltd. New Delhi. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 5 MODULE 1 1.0 1.1 General Introduction of Fluid Mechanics Definitions Fluid mechanics is the study of the behaviour of fluids under the influence of forces and it can be studied under two broad topics: fluid statics and fluid dynamics. Fluid statics is the study of the forces that keep fluids in static equilibrium while fluid dynamics deals with the motion of fluids and the forces that keep them in motion. A fluid is a substance that undergoes continuous deformation when subjected to a shear stress. Liquids and gases are called fluids. An attempt to change the shape of a mass of fluid results in layers of fluid sliding over one another until a new shape is attained. During the change in shape, shear stresses exist, the magnitude of which depends upon the viscosity of the fluid and the rate of sliding, but when a final shape has been reached, all shear stresses will disappear. A fluid in equilibrium is free from shear stresses. It is easier to define these materials in terms of how they respond (i.e. deform or flow) when subjected to an applied force in a specific situation such as the simple shear situation illustrated in Figure 1. A fluid is bounded by two large parallel plates, of area A, separated by a small distance H. The bottom plate is held fixed. Application of a force F to the upper plate causes it to move at a Velocity V. The fluid continues to deform as long as the force is applied, unlike a solid, which would undergo only a finite deformation. The material is assumed to adhere to the plates, and its properties can be classified by the way the top plate responds when the force is applied. Figure 1.1 Deformation of a fluid subjected to a shear stress. 1.2 The Continuum Concept Although gases and liquids consist of molecules, it is possible in most cases to treat them as continuous media for the purposes of fluid flow calculations. On a length scale comparable to the mean free path between collisions, large rapid fluctuations of properties such as the velocity and density occur. However, fluid flow is concerned with the macroscopic scale: the typical length scale of the equipment is many orders of magnitude greater than the mean free path. Even when an instrument is placed in the fluid to measure some property such as the pressure, the measurement is not made at a point-rather, the instrument is sensitive to the properties of a small volume of fluid around its measuring element. Although this measurement volume may be minute compared with the volume of fluid in the equipment, it will generally contain millions of CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 6 molecules and consequently the instrument measures an average value of the property. In almost all fluid flow problems it is possible to select a measurement volume that is very small compared with the flow field yet contains so many molecules that the properties of individual molecules are averaged out. It follows from the above facts that fluids can be treated as continuous media with continuous distributions of properties such as the pressure, density, temperature and velocity. Not only does this imply that it is unnecessary to consider the molecular nature of the fluid but also that meaning can be attached to spatial derivatives, such as the pressure gradient dP/dx, allowing the standard tools of mathematical analysis to be used in solving fluid flow problems. Two examples where the continuum hypothesis may be invalid are low pressure gas flow in which the mean free path may be comparable to a linear dimension of the equipment, and high speed gas flow when large changes of properties occur across a (very thin) shock wave. 1.3 Types of Fluids So many researchers have classified fluids based of their physical and reohological properties. Fluids may be classified as ideal and real or non-ideal fluids. The ideal fluids does not exhibit viscous properties and cannot sustain frictional and shear stresses when in motion. Its motion is being purely sustained by pressure forces and so, it cannot dissipate mechanical energy into heat. The real fluid possesses viscous properties, sustains frictional and shear stresses and dissipates mechanical energy into heat. In practice, the ideal fluid does not exist, but the flow of many real fluids can be analyzed by assuming that they are ideal especially if their viscosities are low. 1.3.1 Classification of Fluids Fluids may also be classified in two different ways: 1 According to their behaviour under the action of externally applied pressure. Incompressible fluids are those whose volume of the element is independent of its pressure and temperature, but if its volume changes, it is said to be Compressible. 2 According to the effect produced by the action of a shear stress. The most important physical properties affecting the stress distribution within the fluid is its viscosity. In many problem involving the flow of gas or liquid, the viscous stress are important and give rise to velocity gradients within the fluid, and dissipation of energy occurs as a result of the frictional forces set up. In gases and most pure liquids where the ratio of the shear stress to the rate of shear is constant and equal to the viscosity of the fluid, such fluids are said to be Newtonian in their behaviour. However, in some liquids, particularly those containing a second phase in suspension, the ratio is not constant and the apparent viscosity of the fluid is a function of the rate of shear. The fluid is said to be Non-Newtonian and to exhibit rheological properties. Fluids that exhibit a nonlinear relationship between stress and strain rate are termed non-Newtonian fluids. Many common fluids that we see everyday are non-Newtonian. Paint, peanut butter, and toothpaste are good examples. High viscosity does not always imply non-Newton behavior. Honey is viscous and Newtonian while 5W30 motor oil is not very viscous, but it is non-Newtonian. There are several types of non-Newtonian fluids. Figure 1.2 shows several of the more common types. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 7 Figure 1.2 : Types of Non- Newtonian Fluid Non-Newtonian fluids include those for which a finite stress τy is required before continuous deformation occurs; these are called yield-stress materials. 1 The Bingham plastic fluid is the simplest yield-stress material; its rheogram has a constant slope µ∞, called the infinite shear viscosity. Highly concentrated suspensions of fine solid particles frequently exhibit Bingham plastic behavior. 1.1 2 Shear-thinning fluids are those for which the slope of the rheogram decreases with increasing shear rate. These fluids have also been called pseudoplastic, but this terminology is outdated and discouraged. Many polymer melts and solutions, as well as some solids suspensions, are shear-thinning. Shear-thinning fluids without yield stresses typically obey a power law model over a range of shear rates. 1.2 The apparent viscosity is 1.3 The factor K is the consistency index or power law coefficient, and n is the power law exponent. The exponent n is dimensionless, while K is in units of kg/(m .s2 − n). For shearthinning fluids, n < 1. The power law model typically provides a good fit to data over a range of one to two orders of magnitude in shear rate; behavior at very low and very high shear rates is often Newtonian. Shear-thinning power law fluids with yield stresses are sometimes called Herschel-Bulkley fluids. Numerous other rheological model equations for shear-thinning fluids are in common use. 3 Dilatant, or shear-thickening, fluids show increasing viscosity with increasing shear rate. Over a limited range of shear rate, the power law model with n > 1 may describe them. Dilatancy is rare, observed only in certain concentration ranges in some particle suspensions. 4 Time Dependent Fluids :Time dependent fluids are fluids that increase or decrease in viscosity over time at a constant shear rate. The construction industry uses a cement slurry that thins with time at a constant pump rate because it is easier to pump and fills forms easily. There are two types of time dependent fluids: Rheopectic and Thixotropic. Rheopectic Fluids CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 8 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 Rheopectic fluids increase in viscosity with time at constant shear rate. There are few examples of rheopectic behavior. In the British literature, rheopectic behavior is called anti-thixotropic behavior. Rheopectic behavior has been observed in bentonite sols, vanadium pentoxide sols, and gypsum suspensions in water as well as in some polyester solutions. Thixotropic Fluids These fluids lose viscosity over time at a constant shear rate. Dispersing agents for cements tend to make them thixotropic. Viscosity is recovered after the cessation of shear. In some systems, the time it takes to recover is so short that a sheared sample cannot be poured out of a container before it gets too thick to flow. Examples of thixotropic fluid include mayonnaise, clay suspensions used as drilling muds, and some paints and inks, that show decreasing shear stress with time at constant shear rate. Viscoelastic Fluids These fluids exhibit elastic recovery from deformation when stress is removed. Polymeric liquids comprise the largest group of fluids in this class. A property of viscoelastic fluids is the relaxation time, which is a measure of the time required for elastic effects to decay. Viscoelastic effects may be important with sudden changes in rates of deformation, as in flow startup and stop, rapidly oscillating flows, or as a fluid passes through sudden expansions or contractions, where accelerations occur. In many fully developed flows where such effects are absent, viscoelastic fluids behave as if they were purely viscous. In viscoelastic flows, normal stresses perpendicular to the direction of shear are different from those in the parallel direction. These give rise to such behaviors as the Weissenberg effect, in which fluid climbs up a shaft rotating in the fluid, and die swell, where a stream of fluid issuing from a tube may expand to two or more times the tube diameter. A parameter indicating whether viscoelastic effects are important is the Deborah number, which is the ratio of the characteristic relaxation time of the fluid to the characteristic time scale of the flow. For small Deborah numbers, the relaxation is fast compared to the characteristic time of the flow, and the fluid behavior is purely viscous. For very large Deborah numbers, the behavior closely resembles that of an elastic solid. 1.4 Units of Measurements Fluid motion and behaviour can be described in units of mass, length and time. These are the basic units from which a number of derived units can be obtained as listed in Table 1.1. Several different sets of units are used in both scientific and engineering systems of dimensions. These can be classified as either metric - Système International d’Unites (SI and cgs) or English – foot-pound-seconds (fps). Although the internationally accepted standard is the SI scientific system, English engineering units are still very common and will probably remain so for some time. Therefore, it is necessary for students master these two systems and become adept at converting them from one to the other. Typical conversion factors are given in the Table 1.2 below. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 9 Table 1.1 Derived Units CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 10 Table 1.2: Units Conversion Factor Source: Chemical Engineering Fluid Mechanics by Ron Darby. Marcel Dekker, Inc. New York,2nd Edition, 2001 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 11 1.5 1.5.1 Properties of Fluids Density Density is an extremely important property of matter. The density of a material can be considered continuous except at the molecular level. Density can also be thought of as the constant that relates mass to volume. It is defined as mass per unit volume. 1.4 1.5.2 Specific Weight The specific weight is a quantity that is used frequently in fluid mechanics. In the American Engineering Series (AES) of units, it is numerically equal to the density. The units are lbf rather than lbm. It is defined as weight per unit volume. w = 1.5 1.5.3 Specific Gravity Specific gravity is used instead of density to tabulate data for different materials. Using the specific gravity, the density in any set of units may be found by picking the reference density in the desired units. Note: When the reference density is expressed in g/cm3 , the density and specific gravity have the same numerical value. 1.6 1.6 Viscosity It is a measure of the fluids resistance to flow because of the cohesive forces of particles. All real fluids resist any force tending to cause one layer to move over another. Consequently, adjacent layers experience shearing stresses when the fluid is flowing. The shearing stresses are similar to the frictional forces experienced by two solids sliding on each other. This resistance to the movement of one layer of the fluid over an adjacent one is called the viscosity of the fluid. It is the ratio of shear stress to shear rate is the viscosity. The SI units of viscosity are kg/(m.s) or Pa.s (pascal second). The cgs unit for viscosity is the poise; 1 Pa.s equals 10 poise or 1000 centipoise (cP) or 0.672 lbm/(ft.s). Consider Figure 1.2 where fluid flows over a solid boundary. Due to the tendency of fluid particles to cling to the surface of the solid, the particle closest to the boundary are brought to rest such that the fluid velocity increases gradually from zero on the wall to the free stream velocity U far away from the wall. The velocity distribution is as shown. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 12 Figure 1.2 : The Velocity Distribution Source: Fluid Mechanics by Ogboja ,Olu. UNESCO, Nairobi Kenya,2005. Consider adjacent two thin layer of fluid moving together at a distance of dy apart and a velocity difference of dV. As result of difference in their velocities, friction or shear stress will develop between the two layers. The shear stress between the two layers can be obtained as follows. Assuming the velocity profile given in Figure 1.3 below,the flow is effected by subjecting the top plate to a force F which consequently moves with velocity U. The fluid layer adjacent to the plate will adhere to it and move with velocity U while the lower layer will move with velocity lower than U. The opposing action between the two layers where the upper accelerates and the lower retards results in shearing of the fluid. The velocity gradually reduces towards the lower plate where it finally assume zero. Figure 1.3 : Shearing of Fluid. Source: Fluid Mechanics by Ogboja , Olu. UNESCO, Nairobi Kenya, 2005. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 13 According to Newton, the force F is directly proportional to the product of the area of the moving plate, A ,and the velocity gradient : U FαA 1.7 d U F = µA 1.8 d Where µ is a constant of proportionality known as coefficient of dynamic viscosity or simply viscosity .The equation can be rearranged as F U τ= =µ 1.9 A d τ is the shear stress at the plate/liquid interface. This equation is valid if d is small and the velocity profile is assumed to be linear .If d is large and the velocity profile is therefore parabolic, the differential form of the equation should be used: dV 1.10 τ=µ x dy The above equation is valid when the origin of the coordinate, which is traversed to the direction of flow, is chosen such that the velocity increases as y increases. Otherwise, it is reversed as: dV 1.11 τ = −µ x dy Viscosity of a fluid can be determined through: (i) Cup-and-Bob (Couette) Viscometer and (ii) (ii) Tube Flow (Poiseuille) Viscometer. Kinematic viscosity is sometimes used and it is the ratio of dynamic viscosity to the density: µ υ= 1.12 ρ The unit of kinematic viscosity is m2/s. 1.7 Compressibility of Fluids The nature of particles of fluids does not allow their handling like that of solid that can be compressed or stretched .The volume of a given mass of fluid can be changed by increasing or decreasing the pressure on it. If the fluid is gas ,its volume can also be altered by changing its temperature. The new state of the perfect gas can be predicted using the perfect gas law. For many practical purposes, the liquid can be regarded as incompressible. When subjected to large pressure, the compressibility of the liquid cannot be ignored. If a liquid of volume V is subjected to a sudden increase in pressure, dp with a corresponding decrease in volum of dV .The compressibility of the liquid is given as : dp κ=− 1.13 dV V Where κ is the bulk modulus of elasticity .The negative sign takes care of the fact that dV is negative since κ cannot be negative. In terms of density, if m = ρV is differentiated then , we have dm = ρdV + Vdρ 1.14 Since m is constant, dm =0 and CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 14 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 − dV dρ = V ρ 1.15 dρ dV for in the initial equation yields ρ V dp κ= 1.16 dρ ρ κ varies with temperature and at any particular temperature, the higher the compression pressure ,the higher the resistance to compression and the higher the value of κ .The variation of κ with temperature are available in literature for different liquids. Substitute 1.8 1.8.1 Surface Tension and Capillarity Surface Tension Surface tension, ( σ ), is a property of a liquid surface. It describes the strength of the surface interactions. The units on surface tension are Force/length . Surface tension is the driving force for water beading on a waxy surface and free droplets of liquid assuming a spherical shape. Lowering of surface tension can be accomplished by adding species that tend collect at the surface. This breaks up the interactions between the molecules of the liquid and reduces the strength of the surface. Surface tension varies with temperature. Water has a surface tension at 200 is 0.074N/m and at 1000C it is 0.059N/m. Surface tension can be reduced by adding surface-active agents (surfactants) to the liquid. A relationship can be derived between surface tension, σ ,and the excess pressure above atmospheric pressure, p, by considering the force balance on the fluid under consideration. For instance, a cylinder whose length is l, the surface tension force and the excess pressure force are 2 σ l and 2prl respectively. The force balance equation is : 2rl (p j − p a ) = 2σl 1.17 And so, σ p = (p j − p a ) = 1.18 r Capillary rise in liquids with adhesive properties and capillary depression in liquids with cohesive properties are caused by surface tension. Cohesion is the intermolecular attraction between molecules of the same liquid while adhesion is the attraction between molecules of a liquid and the molecules of solid bounded surface in contact with the liquid. If the adhesive force between molecules of a particular liquid and a particular solid is greater than the cohesive force between the liquid molecules, the liquid tends to stick to the solid and the area of contact between them tend to increase. This explains why water wets glass but mercury does not and water will not also wet wax or greasy surface. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 15 1.8.2 Capillarity Figure 1.3: Capillary rise and capillary depression When a glass tube with small diameter is partially immersed in water and held vertically, the surface tension will cause the liquid level in the tube to rise above the level of the liquid in the container. The liquid surface in the tube will form a fairly parabolic meniscus. If the liquid wets glass, the meniscus is concave. The surface tension force is in the direction which is inclined to the tube at an angle, θ ,the angle between the tangent to the meniscus and the tube wall. θ is called contact angle. The capillary rise can be calculated as follows: 2σ cos θ h= 1.19 ρgr A parameter for determining when surface tension is important in the flow is call capillary constant, a. Surface tension effect cannot be ignored when capillary rise is of the order of a. 2σ cos θ a = rh = 1.20 ρg 1.9 Vapour Pressure Evaporation occurs when a molecule on the surface of liquid gathers enough energy to break away from the liquid into the space above it. As molecules escape into the space and random movement continues, pressure is exerted on the liquid surface. This pressure is called vapour pressure. When the rate of evaporation is equal to the rate of reabsorption, equilibrium is said to be reached and vapour pressure at that point in time is called saturation vapour pressure. Evaporation of is a molecular activity which is promoted by temperature .Therefore, vapour pressure and saturation vapour pressure depends on the liquid temperature and increase with it. When the pressure above a liquid at any given temperature is equal to the vapour pressure of the liquid, the liquid boils and the temperature is called the boiling point. Boiling can occur at any temperature as long as the pressure above the liquid can be made equal the vapour pressure. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 16 1.10 1 2 3 4 5 6 Problems. Convert a force of 17 dynes to Newton, poundals and lbf. A gas whose molecular weight is 28 is kept under a pressure of 200kN/m2 and a temperature of 600C .Calculate the density of the gas. The viscosity of an organic liquid is 15cP. Calculate the corresponding values on the fps and SI system. Two plates with area 10m2 are 2mm apart. The lower plate is fixed while the upper plate is moved at a constant velocity of 1m/s by a force of 50N. Determine the viscosity of the liquid between the plates. A bubble of a liquid whose surface tension is 0.0289N/m has a radius of 1mm.Calculate the pressure within the bubble if the atmospheric pressure is 101kN/m2. Two glass tubes diameter 2 and 4 mm respectively, are attached to the side of a water tank to measure the level inside the tank,( σ = 0.074N / m). Use this information to express the capillary rise in the tube in the form h= mr + c where m and c are constants and r is the tube radius and hence determine the ideal tube diameter. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 17 MODULE 2 2.0 Dimensional Analysis and Dynamic Similitude 2.1 Dimensional Analysis In building a complex and giant devices in engineering, the usual practice is to start by designing and building a scale model of the device. The scale model is geometrically exact replica of the prototype but it may be smaller of bigger in size depending on the convenience. The model is studied fro performance and efficiency, making alterations until the performance parameters are satisfactory. The dimensions and performance of the prototype are then deduced from those of the model by using dimensional anlaysis and similarity criteria which require that the model and prototype be similar geometrically, kinematically and dynamically. Dimensional analysis is used to determine which groupings of the flow properties affect the performance of the model and prototype. This approach saves energy and cost, and tremendously increases the chance of producing a near perfect design of the prototype at first attempt. The procedure for obtaining the dimensionless parameters is called Dimensional Analysis . Dimensional analysis helps to reduce the number of variables by grouping them into dimensionless parameters. It is then easier to determine the functional relationship between the parameters by experiment. Where there are many parameters, the relative influence of each parameter can be determined from experiment; less influential parameters can be dropped while the more influential parameters can be related empirically using appropriate coefficients and exponents. Many approaches have been proposed for converting a functional relationship in dimensional variables to a functional relationship in nondimensional or dimensionless variables. Among them are , the step-by-step method, Buckingham II Theorem. 2.1.1 Units and Dimensions The dimensions of a quantity identify the physical character of that quantity, e.g., force (F), mass (M), length (L), time (t), temperature (T), electric charge (e), etc. On the other hand, Units identify the reference scale by which the magnitude of the respective physical quantity is measured. Many different reference scales (units) can be defined for a given dimension; for example, the dimension of length can be measured in units of miles, centimeters, inches, meters, yards, angstroms, furlongs, light years, kilometers, etc. Dimensions can be classified as either fundamental or derived. Fundamental dimensions cannot be expressed in terms of other dimensions and include length (L), time (t), temperature (T), mass (M), and/or force (F) (depending upon the system of dimensions used). Derived dimensions can be expressed in terms of fundamental dimensions, for example, area ([A] = L2), volume ([V] =L3), energy ([E]=FL=ML2/t2), power ([HP]=FL/t=ML2/t3),viscosity ([µ]=Ft/L2=M/Lt) e.t.c. 2.1.2 Conservation of Dimensions It states that the terms in an equation which is used to express a relationship between physical quantities must be dimensionally homogenous. For any equation to be valid, every term in the equation must have the same physical character, i.e., the same net dimensions (and consequently the same units in any consistent system of units).. This principle was enunciated by Fourier in 1922. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 18 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 The principle simply implies that if : a = b+c 2.1 a, b and c must have the same dimensions of if: d=ef 2.2 then the dimensions of d must equal to the product of those of e and f. Note that what gives the homogeneity of dimension here is the combined dimensions of e and f and not individual dimensions. Take for instance, the equation which expresses the distance moved by a body falling under gravity: 1 s = gt 2 2.3 2 Since g is a constant, one can rewrite the equation as : s = kt 2 2.4 Where k is a constant that must be dimensional constant having the same dimension as g since s is not equal in dimension to t2. 2.2 Techniques of Dimensional Analysis 2.2.1 Step-by- Step Method In this procedure, dimensions of variables of functional relationship are rendered dimensionless in mass, length and time by step by step. At each step, one of the variables is taken and combined with it to eliminate it and with others to render them dimensionless in whatever dimensions one desire. It may be necessary to use multiples of the variables in rendering some of the others dimensionless. For example, the pressure drop ∆p experienced by a fluid in turbulent motion over a length l of a smooth pipe has been found to be a function of the fluid velocity V, the pipe diameter D, the fluid density ρ, and the fluid viscosity µ. Use the step-by-step method to reduce the functional relationship to that of dimensionless terms The relationship can be expressed as follows ∆p = f (V , D, ρ, µ ) l The dimensions of the variables are: M ∆p = l L2 T 2 [V ] = L T [D] = L 2.5 [ρ] = M3 L [µ ] = M LT We can render the variables dimensionless in mass (M) by dividing them through by ρ to obtain ∆p ρ µ = f1 V , D, , 2.6 lρ ρ ρ And since ρ/ρ = 1, CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 19 ∆p µ = f1 V , D, ρ lρ The dimensions of the terms in this equation are: ∆p L lρ = 2 T [V ] = L T [D] = L 2.7 2.8 µ L2 ρ = T Next ,we render the equation dimensionless in length L by dividing ∆p/lρ ,V, and D by D and µ/ρ by D2 to obtain V µ ∆p = f 2 , 2 2.9 lρD D ρD With the dimensions ∆p 1 lρD = 2 T V 1 2.10 D = T µ 1 2= ρD T And finally, we render this equation dimensionless in time (T) by dividing the left term by (µ/ρD2)2 and the right hand terms by just µ/ρD2 to obtain ρVD ∆pρD 3 2.11 = f 3 2 lµ µ Inspection will show that both terms of this equation are dimensionless. Note that in the last step, we used µ/ρD2 to render the equation dimensionless in T; we could have used V/D and obtain µ ∆pD 2 2.12 = f 4 lρDV 2 ρVD Note that the final results are not the same, only as a result of the choice of the variable we have used in nondimensionalising in T. Both approaches are correct and experiment can be employed to prove this point. In obtaining the dimensionless parameter in the above example, we adopted the order of M,L and T. Do note that there is no rigidity as to which order to adopt. We ended up with two parameter having carried out exercise on five variables employing three basic dimensions. 2.2.2 Buckingham Π Theorem The theorem proves the conclusion we reached at the end of the last section that is, if a functional relationship contains m variables with a total of n basic dimensions, on the application of dimensional analysis, the relationship will contain m-n groups of dimensionless groups . In the step –by-step CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 20 procedure, the dimensionless groups are determined together whereas in this procedure ,the groups are determined one by one and each is designated as a Π .The procedure is as follows as illustrated in the following example. ∆p Example: Solve the problem = f (V, D, ρ, µ, g ) using Buckingham’s procedure. l Solution: The problem can be stated as ∆p f , V, D, ρ, µ, g = 0 2.13 1 l There are 6 variables and 3 basic dimensions M, L and T. Therefore, the solution should yield 3 Π terms. In selecting the repeating variables, it has been found to be helpful to the interpretation of the functional relationship to select one to reflect the rate of flow of the fluid, another to reflect mass of the fluid and the last one to reflect the characteristic dimension of the conduit. Consequently, for this problem ,V, ρ and D are selected. The Π terms are therefore : ∆p Π 1 = V a1ρ a 2 D a3 l b1 b 2 b 3 Π2 = V ρ D µ 2.14 Π 3 = V c1ρ c 2 D c3 g Substituting dimensions for the variables, we have, a1 a2 L M M L T = 3 T L 0 0 b1 L M M L T = 3 T L 0 0 (L )a3 M L T2 0 b2 0 L M 0 L0 T 0 = T c1 M 3 L c2 2 (L )b3 M L T 2.15 (L )c3 L T2 Equating the exponents for M, L and T in the first equation above gives: M: a2 +1 =0 L: a1 -3a2 +a3 -2 =0 T: -a1 -2=0 Solving these three equations gives : a1= -2 ,a2 = -1 and a3=1 Therefore, ∆p D Π1 = l ρV 2 Equating the exponents for M, L and T in the second equation above gives: M: b2+1 =0 L: b1-3b2+b3-1 =0 T: - b1-1=0 Solving these three equations gives : b1 =-1, b2 =-1 and b3 = -1 Therefore CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 2.16 21 µ ρVD Equating the exponents for M, L and T in the third equation above gives: M: c2 = 0 L: c1-3c2+c3+1 = 0 T: - c1 - 2 = 0 Solving these three equations gives :c1 = 2; c2 = 0 and c3 =1 And consequently, gD Π3 = 2 V Finally, the resulting dimensionless equation is ∆p D µ gD f 2 , , 2 = 0 2 l ρV ρVD V Π2 = 2.3 2.17 2.18 2.19 Interpretation of Dimensionless Numbers Table 2.1 lists some dimensionless groups that are commonly encountered in fluid mechanics problems. The name of the group, and its symbol, definition, significance, and most common area of application are given in the table. Wherever feasible, it is desirable to express basic relations (either theoretical or empirical) in dimensionless form, with the variables being dimensionless groups, because this represents the most general way of presenting results and is independent of scale or specific system properties. Table 2.1 Dimensionless Groups in Fluid Mechanics CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 22 Source: Chemical Engineering Fluid Mechanics by Ron Darby. Marcel Dekker, Inc. New York,2nd Edition, 2001 2.4 Similitude Similitude is the science of predicting the behaviour of a large object from the behaviour of the smaller object which is geometrically similar. The former is called prototype and the later model. The model ids built and subjected to various forces that the prototype would encounter. The information thereby obtained is used to improve the design. Model studies can be carried out on components to further improve the performance of the prototype. In order to achieve total similarity between the prototype and the model, there must be kinematic and dynamic similarity in addition to geometrical similarity. Kinematic similarity implies that the ratio of velocity and the ratios of quantities derived from velocity at corresponding points of the model and prototype must be constant while the dynamic similarity implies that the ratio of dynamic forces at corresponding points of the prototype and model must be constant. The dynamic similarity implies that the dimensionless numbers in Table 2.1 must be the same for corresponding points of the model and the prototype. 2.5 Application of Similarity Theory 2.5.1 Submerged Bodies The behaviour of submerged bodies are studied in water and wind tunnels; water being preferred where very high velocities are required. Tunnel studies are used to determine drag characteristics and compressibility characteristics where very high air velocities are involved. The Reynolds model is used to achieve dynamic similarity between prototype and model flows. Example: The drag characteristics of a new brand of motor car is to be studied in a wind tunnel .The speed of the car in still air is to be limited to 150km/hr. A scale model CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 23 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 of 1:10 is to be used. What is the maximum speed that should be employed in the tunnel? Solution : The flow characteristics as depicted by Reynolds number should be used in the study. Therefore Re should be the same in both model and prototype : Rem = Rep ρ m Vm L m ρ p Vp L p = µm µp Vm = Vp Lp ρpµ m Lm ρmµ p = 150 x 10 x 1 x1 = 1,500km/hr. 2.5.2 Pipe Flow A typical flow of incompressible fluids in closed conduit is the flow in pipes. For such flows, gravity, surface tension and compressibility forces have no effect and the flow is controlled solely by pressure and viscosity. Therefore, the Reynolds model is applicable for the analysis of the flow. The flow velocity varies across the pipe, being zero at the wall and maximum at the centre. Depending on the mean velocity through the pipe, the mode of flow can be laminar or turbulent. For Re ≤ 2000, the flow is laminar and viscous forces predominate. When Re > 2000, inertia forces predominate. Complete similarity is achieved between the model and the prototype for laminar flow under all conditions. However, for turbulent flow, there has to be geometric similarity between the roughness of the pipes and their laminar sub layer before dynamic similarity can be attained between the model and the prototype flows. Example : A venture meter is to be used to measure the flow through a pipe of 1m diameter. The throat of the prototype is 0.1m in diameter and the fluid velocity is 5m/s.Determine the discharge through the model for a scale of 1:10. Solution ρVD ρVD = The equation µ m µ p Reduces to (VD )m = (VD )p If the same liquid is used in both model and prototype. Substituting the given values yields Vm x 0.01=0.1 x 5 Vm = 50m/s The discharge through the model is πd 2 Qm = Vm 4 π = x0.012 x50 4 = 0.004m3/s 2.6 Problems 1 An ethylene storage tank in your plant explodes. The distance that the blast wave travels from the blast site (R) depends upon the energy released in the blast (E), the density of the air ( ρ ), and CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 24 2 3 4 5 6 time (t). Use dimensional analysis to determine the dimensionless group(s) that can be used to describe the relationship between the variables in the problem Determine the number of the dimensionless parameters for the following functional relationship: ∆h = f 6 (V , D, ρ, µ, g ) using step –by step method l In general, the pressure drop ∆p of a fluid in motion in a conduit depends on the length of the pipe l ,its diameter D, the fluid velocity V, density ρ, viscosity µ, surface tension σ, bulk modulus k , gravitational acceleration g and pipe roughness e. Determine the Π term for the functional relationship Water is used to simulate the flow characteristics of oil in an industrial pipeline. The diameter of the test pipeline is 1/10th of that of the industrial pipeline. The viscosities of water and oil are 3.0 x10-4 and 2.9 x 10-2 N.s/m2 while their densities are 1000 and 1720 kg/m3 respectively. If the water velocity in the test pipeline is 10m/s, what is the corresponding oil velocity in the industrial pipeline? The head on a spillway is 1.5m and the discharge is 100m3/min. What is the head on the scale model of 1:80. Determine the model discharge. Water discharge through a pipe of length 50m and diameter 0.5m at 3m3/s. For this flow, f = 0.025 . Determine the head loss through the model pipe for a scale ratio of 1:10. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 25 MODULE 3 3.0 Fluids Statics. 3.1 Concept of Stress and Pressure Fluid statics is the study of the forces that keep a body of fluid in static equilibrium. Static equilibrium implies that the resultant of all forces does not result in translation or rotation of the body. Consequently, the resultant moment must be zero. Linear and angular acceleration are therefore zero and shear stresses are also zero. Since there is no rotation, the body of fluid is in equilibrium under the influence of normal force. In the absence of mechanical, electrical ,thermal, magnetic and gravitational forces, the normal forces are hydrostatic forces which are directly proportional to the weight of the fluid above the body. The most important characteristics of static fluid is the pressure it exerts on the body floating on or immersed in it. The pressure on a given plane in a fluid is defined as the fluid force acting perpendicularly to the plane, that is F p= 3.1 A Where F is the force which is acting perpendicular ly on area A. This equation is based on the assumption that force F will be the same all over the points of the area A. If the force is not uniformly distributed, than expression is for the average pressure on A. The pressure at any point on A can be determined by noting the force ∆F on a small area ∆A surrounding the point. When ∆A tends to a point, the ration of ∆F and ∆A yields pressure at a point as expressed below: ∆F 3.2 ∆A Force is a vector quantity, which consists of normal surface force, and body force that is due to the weight of the element, the hydrostatic pressures in the fluid acts in the three Cartesian coordinates while the normal pressure may act in any direction normal to the point of action. Since the hydrostatic pressures in various directions are mutually perpendicular to each other but not to the normal pressure, it is then concluded that the three directions represent arbitrary directions and therefore the pressure at a point acts with equal magnitude in all directions. Conclusively, pressure is therefore taken as a scalar quantity. p = lim ∆A → 0 3.1.1 Pressure Head The pressure at a point in a fluid is usually expressed in terms of the height h of the fluid above the point. The height h is usually referred to as the head of the liquid above the point. The relationship between pressure at a point and the height of fluid above that point can be deduced as follows: Consider the column of fluid of arbitrary cross sectional area ∆A and height h standing perpendicularly on area ∆A in the fluid as shown in figure 3.1. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 26 h ∆A Figure 3.1 : Pressure Head If the fluid acting on the upper bounding area of the column is ignored, the force on ∆A is solely that due to the weight of the elemental column which is ρ(h∆A )g .Therefore ,the pressure due to this weight is : ρ(h∆A )g p= = ρhg 3.3 ∆A Since g is constant for a given location on the earth’s surface and ρ is constant for a given fluid, p can be computed easily once h is known and vice versa. 3.1.2 Pressure Variation with Depth Consider Figure 3.2 below, the effective force on the element are hydrostatic pressure forces in the z –direction and the weight of the element. Hydrostatic pressure acts inwards. Since the fluid is at rest, shear forces are zero and the forces on opposite side of the pipe in the direction x and y directions are equal and opposite and therefore cancel out. Since the element is in static equilibrium , the sum of the forces on it must be equal to zero. Figure 3.2 : Pressure Variation with Depth δp p∆x∆y − p + ∆z ∆x∆y − ρg∆x∆y∆z = 0 δz Which reduces to δp = − ρg δz 3.4 3.5 Since p is a function of z only, we can write CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 27 dp = −ρgdz 3.6 This is the hydrostatic equation and it is the general expression for pressure variation in static fluids. For fluids that are incompressible, ρ is constant and therefore the general equation can be integrated to give : p z = − ρgz + p 0 3.7 If the fluid is a liquid with a free surface whose surface is open to the atmosphere, p0 then represents air pressure on the free surface and pz the pressure at depth –z from the free surface. If h is substituted for –z and p for pz - p0 ,we have p = ρgh 3.8 Note that h is measured downwards while z is measured upwards. The above equation is the hydrostatic law of pressure variation in static fluids and it is used to determine the pressure difference between two points in a static fluids. For pressure variation in compressible fluid, the density varies with position. If we assume the fluid to be a perfect gas, we can make use of the equation of state to relate pressure and density: p = RT 3.9 ρ Combining this equation with the general hydrostatic law equation gives: dp g =− dz 3.10 p RT If we assume an isothermal case (T= constant) and integrate between limits: z = z 0 , p = p0 z=z,p=p where p0 is a place whose pressure is known such as earth’s surface, we obtain p dp g Z =− 3.11 ∫ ∫ dz RT Z0 p0 p which yields p g g (z − z 0 ) (z − z 0 )} ln = − or p = p 0 exp{− 3.12 RT RT p0 3.2 Pressure Transmission Compressible fluids decrease in volume when subjected to increased pressure while incompressible fluids do not change noticeably when subjected to increased pressure. When an incompressible fluid experiences increased pressure at any point, the increase in pressure is transmitted equally in all directions to the other points such that the difference in pressure between any tow points in the fluids is that due to hydrostatic head. Consider figure 3.3 in which a tank containing liquid is open to the atmosphere. The surface pressure is the atmospheric pressure pa .which is pressure at point A. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 28 Figure 3.3 Pressure Transmission The pressure at depth h below the surface is higher than the pressure at the surface by an amount due to the head of liquid of height h. The pressure at point B and C at the same depth h is given as : 3.13 p B = p C = p A + ρhg = p a + ρhg This is also applicable to figure 3.3 b for point B and P: p B = p P = p a + ρHg 3.14 This is applicable in determining the pressure transmitted by an hydraulic jack as shown on figure 3.4. Suppose pressure pA due to a weight of mass m is applied to the surface of the liquid at A to produce pressure pD at D to raise the bigger mass M. Figure 3.4 . The Hydraulic Jack From the earlier derivations, p B = p A + ρhg p C = p D + ρHg Since B and C are on the same level in the liquid, pB = pC and therefore CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 3.15 29 p A + ρhg = p D + ρHg 3.16 If the head terms are small or if h ≈H, then pA = pD 3.17 This equation can be used to calculate the effort required to raise a given load. 3.3 Pressure Measurement Pressure is a measure of the force of the bombardment of the fluid particles in a space on the wall containing the fluid particles. When the space is completely empty of particle, it is called a vacuum and its pressure is zero pressure. Pressure measured with reference to zero pressure is called absolute pressures. Atmospheric pressure is the pressure of the atmosphere measured with reference to zero pressure. In most practical situations, such pressures measured with respect to the atmospheric pressures are called gauge pressures. Many devices are available for measuring pressures 3.3.1 Barometers : These devices are used to measure atmospheric pressure. The simplest of them is the mercury barometer. Aneroid barometers are used to measure absolute pressure. The Bourdon gauges are used to measure gauge pressure. Figure 3.5 : The Bourdon Gauge 3.3.2 Manometers : are devices which employ change in elevation to determine pressure or pressure differences. Simple manometers measure gauge pressures while differential manometers measure pressure difference. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 30 Figure 3.6 : Simple Open-end manometer. Figure 3.7 : Differential manometer. 3.4 Forces on Submerged Surface Every point on a submerged surface experiences an amount of pressure, which is directly proportional to the head of fluid above it. The total fluid force on a submerged surface can be obtained by summing the forces on the elemental areas that make up the surface. The resultant force can be assumed to act through a point on the surface called the centre of pressure. 3.4.1 Horizontal Plane Surface Consider the figure 3.8 below where z- axis is assumed to be the fluid’s free surface. If the surface is at depth y below the surface of the fluid and the surface on z –axis is horizontal where all the elements of the surface experiences same pressure as discussed before, then the total or resultant force is therefore normal to the surface and is given as : F = ∫ pdA = pA A Where A is the area of the surface. The line of action passes through the centre of pressure which is located at the centre of the area or centroid. The distance of the line of action form y-axis can be obtained by taking moments about the y-axis. Fx = pAx = ∫ pxdA 3.18 A CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 31 And since p is constant x= 1 ∫ xdA AA 3.19 Figure 3.8 Fluid force on horizontal Surface 3.4.2 Vertical Surface Consider fluid force on a vertical surface AB which is perpendicular to x-axis as shown in figure 3.9. Points on the surface are at different depth and hence experiences different pressures. The coordinate axes have to be chosen such that the h axis is parallel to AB and both axes intersect at the liquid surface at O. The centre of area (centroid) is represented by C and the centre of pressure is P. The pressure at any point h is p = ρhg 3.20 And the force on the elemental area δA is δF = ρhgδA 3.12 The total forces on the surface is therefore F = ρ g ∫ h δA 3.22 A The term in integral is defined as the first moment of area about the x-axis which is usually written as hCA ,with hC being the depth of the centroid of the area and consequently, F = ρgh C A 3.23 But the expression for pressure at the centroid is p C = ρgh C 3.24 Therefore, F = p C A 3.25 The resultant hydrostatic force on the vertical surface is the product of the pressure at the centroid and the area of the surface CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 32 Figure 3.9 : Vertical Submerged Surface 3.4.3 Inclined Surface The pressure distribution on an inclined surface AB is as shown on figure 3.10.The force on an elemental area δA is δF = pδA = ρghδA = ρgy sin αδA 3.26 Where h is the depth of the element and y its distance from O. The total force on the surface after integration is given as : F = ρgAy C sin α 3.27 But from the notation in the figure, the depth of the centroid is: h C = y C sin α 3.28 Therefore , F = ρgh C A = p C A 3.29 Where pC is the pressure at the cenroid of the surface. This is the same as the result obtained for vertical surface. The y coordinate of the centre of pressure can be determined by equating the moments of the resultant force about x-axis to those of the distributed forces: y P F = ∫ ypdA 3.30 A On substituting ρgAy C sin α for F and ρgy sin α for p in the above equation 1 Ix 2 yp= 3.31 ∫ y dA = yCA A yCA Where Ix is the second moment of area of the plane about the x-axis. By the application of the parallel axes theorem, the equation becomes: I yp= C + yC 3.32 yCA CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 33 Figure 3.10: Force on an Inclined Surface 3.4.4 Curved Surface The analysis of the hydrostatic forces on a curved surface can be simplified by considering the horizontal and Vertical components of the forces separately. Consider the curved surface shown on figure 3.11a.The fluid force on the surface has horizontal and vertical components FH and FV respectively. It is convenient to consider FH as acting on the surface CD which is the projection of the curved surface on the vertical plane. Similarly, FV can be considered as acting on the plane ED which is the projection of the surface on the horizontal plane. The magnitudes and line of action of these components can be determined as has been done for plane surfaces and the magnitude and line of action of their resultant forces can then be determined from vector consideration. Figure 3.11: Force on Curved Surface Since we are considering a static situation, the surface must develop a reaction to the fluid forces whose components are RH and RV respectively. For equilibrium to occur, FH and RH must be equal ,opposite and have same line of action. This must also be true for FV and RV . FH = ρg x area of CD x depth of centroid of CD 3.33 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 34 FH acts through centre of pressure of the vertical projection of the curved surface while FV is equal to the weight of the fluid and acts through its centre of gravity. Figure 3.11b can be treated in similar manner. 3.5 Buoyancy When a body is immersed in a fluid, it experiences two fluid forces in each of the three coordinates directions. The two forces in each of the to horizontal directions are equal and opposite .There are therefore no net effects in these directions. The downward and upward forces in the vertical direction are not equal and therefore there is net upward force in this direction. The force is called Buoyant Force. Figure 3.12: Upthrust on a submerged body Considering figure 3.12 above, the two forces are opposite and the net upward fluid force is given by : F = (p 2 − p1 ) A 3.34 The net upward force F is the weight of the fluid corresponding to the volume of fluid displaced. This is Archimedes Principle. If the body is partially immersed the upthrust is equal to the volume of the fluid displaced. F = ρVg 3.35 Where V is the volume of the fluid displaced, ρV its mass and ρVg its weight. The upthrust acts upwards through the centroid of the displaced volume of fluid. The cenroid can be determined by taking moments about an axis normal to the vertical projection of the volume. Consider figure 3.13, the upthrust on the elemental volume whose cross sectional area is δA is given by: δF = (p 2 − p1 )δA = ρghδA = ρgδV 3.36 The upthrust on the whole volume is obtained by integrating the equation above: F = ∫ dF = ∫ ρgdV =ρg ∫ dV = ρgV 3.37 V V V We can now equate the sum of the moments due to all the elemental volumes which make up the object about the axis through point O to that of the total upthrust F: CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 35 Fx = ρg ∫ xdV 3.38 V Where x is the distance of the centroid from the selected axis. If we substitute for F and rearrange: 1 x = ∫ xdV 3.39 VV The centroid of the displaced fluid is called centre of buoyancy Figure 3.13: Determination of Centre of Buoyancy 3.5.1 Hydrometry The principle of buoyancy is applied to the subject of hydrometry for measuring the specific gravity of liquids. The device is call hydrometer and is as shown in figure 3.14. The stem of hydrometer cam be calibrated to measure specific gravity directly. The calibration equation is derived as follows . When immersed in pure water, the upthrust on the hydrometer is given as F = W = ρ w Vw g 3.40 When immersed in another liquid then, F = W = ρ l Vl g 3.41 Let the point on the hydrometer stem which coincide with the water surface be marked h=0. Suppose the second liquid is denser so that the hydrometer will sink less. Let h denote the distance of the new liquid level from the h=0 mark. If a is the cross sectional area of the stem, the displaced volume can be related as Vl = Vw − ah. 3.42 Combining the equations givens: ρ l (Vw − ah) = ρ w Vw 3.43 If we make h the subject of the equation and writes s=ρl/ρw , the specific gravity of the liquid is : V 1 V h=− w + w 3.44 a s a This equation is a straight line equation of the form y= mx+c. Therefore -Vw/a represents the gradient of the plot of h versus 1/s and Vw/a its intercept. This equation can be used to calibrate an hydrometer by measuring Vw and a and then determining h and s for various liquids. The specific CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 36 gravity values rater than h can then be marked on the stem of the hydrometer. This way the hydrometer can be used to measure specific gravities directly. Figure 3.14: The Hydrometer Figure 3.15 : Calibration graph for the Hydrometer 3.6 Static Forces on Solid Boundaries A floating object whether partially of fully immersed ia said to be stable if it tends to return to its original position after a small displacement. Such a body may be linearly or rotationally displaced. The upthrust on a fully immersed floating body is the weight of the fluid displaced by the volume of the body. Since the volume is constant, the upthrust will remain the same no matter the depth at which the body is. Displacing the body linearly does not change the equilibrium between the two forces since neither of the two is changed in magnitude by the displacement, but it only assumes new position. However, when a body is rotationally displaced, its ability to restore itself back to the original position after a small rotational displacement depends on the relative positions of the centre of CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 37 buoyancy and gravity. Figure 3.16 shows various positions of a loaded sphere to illustrated the effect of centre of buoyancy and gravity on its stability. (a) Stable (b) Neutral Figure 3.16: A fully –Immersed Floating body. (c) Unstable 3.6 Problems 1. If an egg can be crushed by pressure of 200kPa, at what depth will it burst in sean water of specific gravity 1.1? 2. Predict the atmospheric pressure at a height of 1000m above sea level on a day when the temperature is 300C.Assume that the air density at sea level is 1.15kg/m3 and take the gas constant as 287J/kg.K. State other assumptions you make. 3. If the effort and load arm areas of a hydraulic jack are in the ratio 1:25, determine the effort required to raise a load of one tonne mass. 4. A gate of a reservoir is hinged at a point and it is prevented from opening by the force in the spring. The gate is 3m high and 1m wide. The level of water is 2m above th top of the gate. Determine the force in the spring if the gate is just about to open. 5. The wall of a reservoir on which the slide gate is located is inclined at 300 to the horizontal. The gate is hinged at its upper end and a massive weight is pushed against its lower end to keep it closed. Determine the reaction of the weight and its direction if the gate has a circular cross section of diameter 3m. 6. A hydrometer whose mass is 25g is made up of a bulb of diameter 2cm and height 5cm and a stem of diameter 1cm and height 10cm. Determine the range of specific gravities that the hydrometer can be employed to measure. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 38 MODULE 4 4.0 BASIC CONSERVATION LAWS 4.1 THE SYSTEM The basic principles that apply to the analysis and solution of flow problems include the conservation of mass, energy, and momentum in addition to appropriate transport relations for these conserved quantities. For flow problems, these conservation laws are applied to a system, which is defined as any clearly specified region or volume of fluid with either macroscopic or microscopic dimensions (also referred to as a ‘‘control volume’’) as illustrated in Figure 4.1. Figure 4.1: A system with inputs and outputs. The general conservation law is : 4.1 where X is the conserved quantity, i.e., mass, energy, or momentum. In the case of momentum, because a ‘‘rate of momentum’’ is equivalent to a force (by Newton’s second law), the ‘‘rate in’’ term must also include any (net) forces acting on the system. It is emphasized that the system is not the ‘‘containing vessel’’ (e.g., a pipe, tank, or pump) but is the fluid contained within the designated boundary. The following sections will show how this generic expression is applied for each of these conserved quantities. 4.2 CONSERVATION OF MASS The continuity equation is a mathematical statement of the law of conservation of mass, which states that the mass of a system is constant. If the system is a open one, for example, a control volume into and from which fluid flows, then for conservation of mass, the rate of mass inflow from the control volume minus the rate of mass outflow from the control volume must equal the rate of increase of mass within the control volume. 4.2.1 Macroscopic Balance For a given system (e.g., Fig. 4.1), each entering stream (subscript i) will carry mass into the system & i ), and each exiting stream (subscript o) carries mass out of the system (at rate m & o ). Hence, (at rate m the conservation of mass, or continuity equation for the system is: 4.2 where ms is the mass of the system. For each stream, CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 39 4.3 that is, the total mass flow rate through a given area for any stream is the integrated value of the local mass flow rate over that area. Note that mass flow rate is a scalar, whereas velocity and area are vectors. Thus it is the scalar (or dot) product of the velocity and area vectors that is required. (The ‘‘direction’’ or orientation of the area is that of the unit vector that is normal to the area.). The corresponding definition of the average velocity through the conduit is : 4.4 r Where is the volumetric flow rate and the area A is the projected component of A that is r r r normal to V (i.e., the component of A whose normal is in the same direction as V ). For a system at steady state, Equation 4.1 reduces to or 4.5 4.2.2 Microscopic Balance The conservation of mass can be applied to an arbitrarily small fluid element to derive the ‘‘microscopic continuity’’ equation, which must be satisfied at all points within any continuous fluid. This can be done by considering an arbitrary (cubical) differential element of dimensions dx, dy, dz, with mass flow components into or out of each surface, e.g., 4.6 Dividing by the volume of the element (dx, dy, dz) and taking the limit as the size of the element shrinks to zero gives 4.7 This is the microscopic (local) continuity equation and must be satisfied at all points within any flowing fluid. For a steady state flow, the right hand side becomes zero. If the fluid is incompressible (i.e. constant ρ), Equation 4.7 reduces to 4.8 Example 4.1 : Water is flowing at a velocity of 7 ft/s in both 1 in. and 2 in. ID pipes, which are joined together and feed into a 3 in. ID pipe, as shown in Figure below. Determine the water velocity in the 3 in. pipe. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 40 Solution: Because the system is at steady state, Equation 4.5 applies: For constant density, this can be solved for V3: Example 4.2 : Water flows through a divergent conduit at the rate of 1000l/min. Determine the velocities at the ports if their diameters are 100mmand 150mm respectively. Solution 1 3 m /s 60 = 16.67 x10 −3 m 3 / s Q = 1000x10 − 3 x π x0.12 m 2 = 0.0079m 2 4 π A 2 = x0.152 m 2 = 0.0177m 2 4 From Equation 4.4 V1=Q/A1 and V2=Q/A2 ,therefore A1 = 16.67 x10 −3 = 2.11m / s 0.0079 16.67 x10 − 3 V2 = = 0.94m / s 0.0177 Example 4.3 The x- component of the velocity vector of a flow field is Vx= xy2-2y+x2. If the velocity vector satisfies the continuity equation for steady incompressible flow, determine the expression for Vy V1 = SOLUTION For 2 –dimensional incompressible steady flow, the continuity equation is CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 41 ∂Vx ∂Vy =0 + ∂x ∂y Therefore , Vy = ∫ ∂Vx ∂y + C ∂x Where C is constant of integration which may be a function of time and y. From the equation for Vx provided, ∂Vx = y 2 + 2x ∂x 1 Consequently, Vy = − y 3 + 2xy + C 3 If it is assumed that y =0, Vy = 0, then C = 0 and 1 Vy = − y 3 + 2xy 3 From which we obtain, ∂Vy ∂y ( = − y 2 + 2x ) Hence , ∂Vx ∂Vy =0 + ∂x ∂y 4.3 CONSERVATION OF ENERGY Energy can take a wide variety of forms, such as internal (thermal), mechanical, work, kinetic, potential, surface, electrostatic, electromagnetic, and nuclear energy. We will consider only internal (thermal), kinetic, potential (due to gravity), mechanical (work), and heat forms of energy. For the system illustrated in Figure 4.1,a unit mass of fluid in each inlet and outlet stream may contain a certain amount of internal energy (u) by virtue of its temperature, kinetic energy (V2/2) by virtue of its velocity, potential energy (gz) due to its position in a (gravitational) potential field, and ‘‘pressure’’ energy (P/ρ). The pressure energy is sometimes called the flow work, because it is associated with the amount of work or energy required to inject a unit mass of fluid into the system or eject it out of the system at the appropriate pressure. In addition, energy can cross the boundaries of the system other than with the flow streams, in the form of heat (Q) resulting from a temperature difference and shaft work (W). Shaft work is so named because it is normally associated with work transmitted to or from the system by a shaft, such as that of a pump, compressor, mixer, or turbine. The sign conventions for heat (Q) and work (W) are arbitrary and consequently vary from one authority to another. Heat is usually taken to be positive when it is added to the system, so it would seem to be consistent to use this same convention for work (which is the convention in most scientific references). However, engineers, being pragmatic, use a sign convention that is directly associated with value. That is, if work can be extracted from the system (e.g., to drive a turbine) then it is positive, because a positive asset can be sold to produce revenue. However, if work must be put into the system (such as from a pump), then it is negative, because it must be purchased (a negative asset). This CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 42 convention is also more consistent with the driving force interpretation of the terms in the energy balance, as will be shown later. The conservation of energy equation for any system is as follows: 4.9 Here, is the enthalpy per unit mass of fluid. Note that the inlet and exit streams include enthalpy (i.e., internal energy, u, and flow work, ), whereas the system energy includes only the internal energy but no flow work (for obvious reasons). If there are only one inlet stream and one exit stream and the system is at steady state, the energy balance becomes: 4.10 Where are the heat added to the system and work done by the system, respectively, per unit mass of fluid. This expression also applies to a system comprising the fluid between any two points along a streamline within a flow field. Specifically, if these two points are only an infinitesimal distance apart, the result is the differential form of the energy balance: 4.11 The d( ) notation represents a total or exact differential and applies to Where those quantities that are determined only by the state (T, P) of the system and are thus point properties. The δ( ) notation represents quantities that are inexact differentials and depend upon the path taken from one point to another. Note that the energy balance contains several different forms of energy, which may be generally classified as either mechanical energy, associated with motion or position, or thermal energy, associated with temperature. Mechanical energy is useful, in that it can be converted directly into useful work, and includes potential energy, kinetic energy, flow work and shaft work. The thermal energy terms, i.e., internal energy and heat, are not directly available to do useful work unless they are transformed into mechanical energy, in which case it is the mechanical energy that does the work. Let us take a closer look at the significance of enthalpy and internal energy, because these cannot be measured directly but are determined indirectly by measuring other properties such as temperature and pressure. 4.3.1 Internal Energy An infinitesimal change in internal energy is an exact differential and is a unique function of temperature and pressure (for a given composition). Since the density of a given material is also uniquely determined by temperature and pressure (e.g., by an equation of state for the material), the internal energy may be expressed as a function of any two of the three terms T, P and ρ Hence, 4.12 By making use of classical thermodynamic identities, this is found to be equivalent to CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 43 4.13 Where 4.14 is the specific heat at constant volume (e.g., constant density). For an ideal gas, 4.15 Thus Equation 4.13 reduces to 4.16 which shows that the internal energy for an ideal gas is a function of temperature only. For a non-ideal gas, Equation 4.15 is not valid, so 4.17 Consequently, the last term in Equation4.13 does not cancel as it did for the ideal gas, which means that 4.18 The form of the implied function, fn(T, P), may be analytical if the material is described by a non-ideal equation of state, or it could be empirical such as for steam, for which the properties are expressed as data tabulated in steam tables. For solids and liquids, ρ≈constant (or dυ=0), so 4.19 This shows that the internal energy depends upon temperature only ( just as for the ideal gas, but for an entirely different reason). 4.3.2 Enthalpy The enthalpy can be expressed as a function of temperature and pressure: 4.20 which, from thermodynamic identities, is equivalent to 4.21 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 44 Here, 4.22 is the specific heat of the material at constant pressure. Let us again consider some special cases. For an ideal gas 4.23 Thus Equation 4.21 for the enthalpy becomes 4.24 which shows that the enthalpy for an ideal gas is a function of temperature only (as is the internal energy). For a non ideal gas 4.25 which, like ∆u, may be either an analytical or an empirical function. All gases can be described as ideal gases under appropriate conditions (i.e., far enough from the critical point) and become more nonideal as the critical point is approached. That is, under conditions that are sufficiently far from the critical point that the enthalpy at constant temperature is essentially independent of pressure, the gas should be adequately described by the ideal gas law. For solid and liquids υ =1/ ρ ≈ constant, so that (∂υ/∂T)p= 0 and Therefore, 4.26 Or 4.27 This shows that for solids and liquids the enthalpy depends upon both temperature and pressure. This is in contrast to the internal energy, which depends upon temperature only. 4.3.3 Bernoulli Equation The Bernoulli equation states that the total energy of a flowing inviscid and frictionless fluid is constant p V2 + + z = cons tan t 4.28 ρ g 2g The energies of two points in the fluid can be related as p1 V1 2 p V 2 + + z1 = 2 + 2 + z 2 ρ g 2g ρg 2g Equation 4.29 is a form of energy equation 4.9 where: CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 4.29 45 • The fluid must not be receiving or delivering shaft work between the two points that are being considered. • There must be no heat transfer and, • The flow must be inviscid and frictionless. Equation 4.29 holds for ideal fluids. But real fluid experiences fluid friction which is represented by hL. Hence, the equivalent Bernoulli equation for real fluid is p1 V1 2 p V 2 + + z1 = 2 + 2 + z 2 + e f 4.30 ρ g 2g ρg 2g In other words, the Bernoulli equation for real fluids is the same as energy equation for case of no heat transfer and no work addition or rejection between the two points under consideration. 4.4 CONSERVATION OF MOMENTUM A macroscopic momentum balance for a flow system must include all equivalent forms of momentum. In addition to the rate of momentum converted into and out of the system by the entering and leaving streams, the sum of all the forces that act on the system (the system being defined as a specified volume of fluid ) must be included. This follows from Newton’s second law, which provides an equivalence between force and the rate of momentum. The resulting macroscopic conservation of momentum thus becomes 4.31 Note that because momentum is a vector, this equation represents three component equations, one for each direction in three-dimensional space. . If the system is also If there is only one entering and one leaving stream, then at steady state, the momentum balance becomes 4.32 4.4.1 One-Dimensional Flow in a Tube Consider the steady state momentum balance to a fluid in plug flow in a tube, as illustrated in Figure 4.2. (The ‘‘stream tube’’ may be bounded by either solid or imaginary boundaries; the only condition is that no fluid crosses the boundaries other that through the inlet and outlet planes.) The shape of the cross section does not have to be circular; it can be any shape. Figure 4.2 : Momentum balance on a slice in a stream tube. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 46 The fluid element in the slice of thickness dx is our system, and the momentum balance equation on this system is 4.33 The forces acting on the fluid result from pressure (dFP), gravity (dFg), wall drag (dFw), and external shaft work ( ): 4.34 is the stress exerted by the fluid on the wall (the reaction to the stress exerted on the fluid by Here, the wall), and Wp is the perimeter of the wall in the cross section that is wetted by the fluid (the ‘‘wetted perimeter’’). After substituting the expressions for the forces from Equation . 4.34 into the momentum balance equation, Equation 4.32, and dividing the result by where the result is 4.35 where is the work done per unit mass of fluid. Integrating this expression from the inlet (i) to the outlet (o) and assuming steady state gives 4.36 Comparing this with the Bernoulli equation shows that they are identical, provided 4.37 For steady flow in uniform conduit 4.38 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 47 4.39 is called the hydraulic diameter. Note that this result applies to a conduit of any cross-sectional shape. For a circular tube, for example, Dh is identical to the tube diameter D. 4.4.2 Conservation of Angular Momentum In addition to linear momentum, angular momentum (or the moment of momentum) may be conserved. For a fixed mass (m) moving in the x direction with a velocity Vx, the linear x-momentum (Mx) is mVx. Likewise, a mass m rotating counterclockwise about a center of rotation at an angular velocity has an angular momentum equal to where R is the distance from the center of rotation to m. Note that the angular momentum has dimensions of ‘‘length times momentum,’’ and is thus also referred to as the ‘‘moment of momentum.’’ If the mass is not a point but a rigid distributed mass (M) rotating at a uniform angular velocity, the total angular momentum is given by 4.40 where I is the moment of inertia of the body with respect to the center of rotation. For a fixed mass, the conservation of linear momentum is equivalent to Newton’s second law: 4.40 The corresponding expression for the conservation of angular momentum is 4.41 is the moment (torque) acting on the system and is the angular where acceleration. For a flow system, streams with curved streamlines may carry angular momentum into and/or out of the system by convection. To account for this, the general macroscopic angular momentum balance applies: 4.42 For a steady-state system with only one inlet and one outlet stream, this becomes 4.43 This is known as the Euler turbine equation, because it applies directly to turbines and all rotating fluid machinery. We will find it useful later in the analysis of the performance of centrifugal pumps. 4.4.3 Moving Boundary Systems and Relative Motion One can sometimes encounter a system that is in contact with a moving boundary, such that the fluid that composes the system is carried along with the boundary while streams carrying momentum and/or energy may flow into and/or out of the system. Examples of this include the flow impinging on CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 48 a turbine blade (with the system being the fluid in contact with the moving blade) and the flow of exhaust gases from a moving rocket motor. In such cases, we often have direct information concerning the velocity of the fluid relative to the moving boundary (i.e., relative to the system), Vr, and so we must also consider the velocity of the system, Vs, to determine the absolute velocity of the fluid that is required for the conservation equations. For example, consider a system that is moving in the x direction with a velocity Vs a fluid stream entering the system with a velocity in the x direction relative to the system of Vri, and a stream leaving the system with a velocity Vro relative to the system. The absolute stream velocity in the x direction Vx is related to the relative velocity Vrx and the system velocity Vsx by 4.44 The linear momentum balance equation becomes 4.45 4.5 Problems CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 49 MODULE 5 5.0 FLOW IN PIPES 5.1 FLOW REGIMES In 1883, Osborn Reynolds conducted a classical experiment, illustrated in Figure 5.1, in which he measured the pressure drop as a function of flow rate for water in a tube. He found that at low flow rates the pressure drop was directly proportional to the flow rate, but as the flow rate was increased a point was reached where the relation was no longer linear and the scatter in the data increased considerably. At still higher flow rates, the data became more reproducible, but the relationship between pressure drop and flow rate became almost quadratic instead of linear. To investigate this phenomenon further, Reynolds introduced a trace of dye into the flow to observe what was happening. At the low flow rates where the linear relationship was observed, the dye was seen to remain a coherent, rather smooth thread throughout most of the tube. However, where the data scatter occurred, the dye trace was seen to be rather unstable, and it broke up after a short distance. At still higher flow rates, where the quadratic relationship was observed, the dye dispersed almost immediately into a uniform cloud throughout the tube. The stable flow observed initially was termed laminar flow, because it was observed that the fluid elements moved in smooth layers relative to each other with no mixing. The unstable flow pattern, characterized by a high degree of mixing between the fluid elements, was termed turbulent flow. Although the transition from laminar to turbulent flow occurs rather abruptly, there is nevertheless a transition region where the flow is unstable but not thoroughly mixed. Careful study of various fluids in tubes of different sizes has indicated that laminar flow in a tube persists up to a point where the value of the Reynolds number (Re =DVρ ρ/µ µ) is about 2000, and turbulent flow occurs when Re is greater than about 4000, with a transition region in between. A good knowledge of the flow regimes in a given pipe is important for reasons of accurate flow analysis and pipe design. Laminar flows can be analysed mathematically whereas turbulent flows require both mathematical and experimental analysis because of their complexity. Figure 5.1 : Reynolds’ Experiment 5.2 PRESSURE DROP IN LAMINAR FLOW The pressure varies from point to point in the flowing fluid in a pipe. This variation is caused by the existence of shear stress in the fluid and consequently, the fluid pressure decreases in the direction of the flow. The expression for the mean velocity in laminar flow in a circular pipe is given as : CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 50 D2 d (p + ρgz ) − 5.1 32µ dx Where D is the pipe diameter,-dp is the pressure drop over a short length dx of the pipe and z is the vertical distance between the inlet and exit of length dx. If equation 5.1 is integrated between the two sections, an expression for the pressure drop can be obtained as follows: 32µVm p1 − p 2 = (z 2 − z 1 ) + L 5.2 ρg ρgD 2 Where L is the length of pipe (x2-x1). If equation 5.2 is rewritten in a way similar to energy equation then p1 p 5.3 + z1 = 2 + z 2 + h f ρg ρg 32µVm hf = L 5.4 ρgD 2 hf represent the frictional term. Vm = Example 5.1 Water flows in a pipe of diameter 25mm.What is the maximum mean velocity if the flow should always remain laminar? Calculate the head loss per metre length of the pipe. Take the viscosity of water as 1.2cP(1.2x10-3 N.s/m2) Solution For laminar flow, Re should not exceed 2000 and therefore the maximum mean velocity is given by ρVmax D = 2000 µ 2000x1.2x10 − 3 = 0.096m / s 1000x 25x10 − 3 The head loss in laminar flow can be obtained by substituting values into equation 5.4 32µVm 32x1.2x10 −3 x0.096 hf = L = = 6.01x10 − 4 m 2 3 −3 2 ρgD 10 x9.81x(25x10 ) Vmax = 5.3 PRESSURE DROP IN TURBULENT FLOW The pressure drop over a length L of pipe of diameter D and surface roughness ∈ can be expressed in functional terms as ∆p = f1(D,ρ,V,µ,L,∈) 5.5 Using dimensional Analysis and other mathematical manipulations: The frictional factor f is given by ∆p D 2g f= . . 5.6 ρg L V 2 The frictional head loss hf is given by L V2 hf = f. . 5.7 D 2g A chart of f versus Re with ∈ as a parameter has been developed by Moody as presented in Figure 5.2.For a pipe of given roughness ∈ and diameter D with flow regime Re, the frictional factor can be CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 51 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 read from the chart and the pressure head loss calculated from equation 5.7.For non circular pipes, D should be replaced by the equivalent diameter of hydraulic diameter De, which is defined as 4x flow area 5.8 De = wetted perimeter Figure 5.2 : Moody Diagram CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 52 Example 5.2 The relative roughness (∈/D) of a steel pipe is 6x10-5 and the pipe diameter is 0.82m.The pipe is carrying crude oil of specific gravity 0.86 and kinematic viscosity 6x10-5 m2/s.Determine the head loss per kilometer length of the pipe for discharge of 1m3/s. Solution π .0.82 2 = 0.528m 2 4 Q 1 Velocity : V = = = 1.89m / s A 0.528 VD 1.89x0.82 µ Re = = = 25,830 recall υ= −5 υ ρ 6x10 Using the values of Re and ∈/D given, f obtained from Moody’s chart is 0.025.Substituting into equation 5.7 gives L V2 1000 1.89 2 hf = f. . = 0.025. x = 5.6m D 2g 0.82 2x9.81 Pipe Area : A = 5.4 PRESSURE LOSSES IN PIPE FITTINGS Head loss in straight pipes can be calculated from equation 5.7. However, head losses in fittings like reducers, bends, valves and so on requires some similar means of measuring them. Some available methods of calculating losses in pipe fittings are presented as follows: 5.4.1 Sudden Expansion Fluids experiences sudden expansion when the diameter of the pipe in which it is flowing increases abruptly. For an incompressible flow, the head loss (hl) due to sudden expansion (enlargement) in the flow cross section is given as 2 V2 A V2 h l = 1 1 − 1 = K l 1 2g A 2 2g 5.9 2 A 5.10 Where V1 is the velocity in the smaller pipe and k l = 1 − 1 A2 5.4.2 Sudden Contraction A sudden reduction in the cross section of the conduit causes a reduction in the cross section of the flow area.The reduction continues until the smallest flow area called the vena contracta is achieved.The head loss due to contraction is given as V2 2 hc = 2g Where hc 2 2 A2 V2 2 1 V2 2 − 1 = − 1 = K c 2g C c 2g Ac denotes head loss due to sudden contraction, Cc =Ac /A 5.11 2 1 K c = − 1 5.12 C c Cc is described as coefficient of contraction,its values has been found experimentally to lie between 0.6 and 1.0. The values of Kc are also available in the literature. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 53 5.4.3 Pipe Entrance Loss The flow situation at the pipe entrance is similar to that of sudden contraction. The head loss is: V2 he = K e 5.13 2g Where Ke = 0.50. A rounded entrance prevents the formation of vena contracta and reduces the value of Ke to about 0.05.If the discharge pipe projects into the reservoir ,the entrance is called a re-entrant entrance .Such an arrangement increases the energy loss considerably. For projection of one pipe diameter or more, Ke is between 0.8 and 1.0. 5.4.4 Losses in Pipe Fittings In pipe networks, fluids losses pressure as a result of expansion or contraction through fittings such as couplings, bends, tees and reducers. Analytical expression for hf are only possible for a few of the fittings and therefore ,empirical approaches are commonly used. Two empirical approaches are available: • The equivalent length of the pipe fittings is determined from a chart and then hf is calculated from Le V 2 hf = f . . 5.14 D 2g Where Le denotes equivalent length of the pipe fittings • Here, hf is calculated from an equation of the form V2 hf = K 5.15 2g Where K is a loss coefficient .For both methods the values of Le/D and K are available in the literature based on the works of researchers. Solving for Le from equations 5.14 and 5.15 gives KD Le t = 5.16 f If there are n fittings in a line, the total equivalent length can be written as D n Le = ∑ K i 5.17 f i =1 The total length of the line is given by D n L t = L + ∑ K i = L + Le t 5.18 f i =1 The equation for total frictional loss can be written as 2 L + Le t V 2 L n V = f + ∑Ki 5.19 ∑ h f = f . D 2g D i =1 2g The equation for head loss for sudden enlargement uses the upstream velocity while contraction uses downstream. In order to avoid confusion when using equation 5.19, it is better to adopt V 2 L V2 n + ∑ K i fi 5.20 ∑ hf = f. . D 2g i =1 2g Where Vfi is the velocity through the device or fitting number I and Ki is the corresponding loss coefficient. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 54 5.5 PIPELINE PROBLEMS 5.5.1 Pipes in Parallel When main pipeline at upstream point divides into two or more parallel pipes that again join downstream and continues as a main line, such pipes are said to be in parallel as shown in figure 5.3 below Q1, f1, D1, L1 Q Q Q2, f2, D2, L2 .Figure 5.3: A System of Pipes in Parallel The frictional losses are the same for all the branches while the discharge is the sum of the discharges in each of the branches. Therefore, h f =h f 1 = h f 2 5.21 Q = Q1 + Q 2 5.22 Consequently, equation 5.21 can be written as L1 V1 2 L 2 V2 2 f1 . . = f2. . D1 2g D 2 2g This equation yields the velocity ratio as : 5.23 1/ 2 V2 f1 L1 D 2 = . . V1 f 2 L 2 D1 Equation 5.22 can also be expressed in terms of flow velocities as V π π Q = D12 V1 + D 2 2 V2 or Q = V1 D12 + D 2 2 2 4 V1 4 Generally, if there are n pipes in parallel, equations 5.21 to 5.25 can be modified as L V2 hf = fi . i . i D i 2g Where I denote the ith pipe in the parallel which contains n pipe ( ) Q = Q1 + Q 2 + ....... + Q n Q= V V π 2 V1 D1 + D 2 2 2 + ..... + D n 2 n 4 V1 V1 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 5.24 5.25 5.26 5.27 5.28 55 1/ 2 V2 f1 L1 D 2 . = . V1 f 2 L 2 D1 V3 f1 L1 D 3 = . . V1 f 3 L 3 D1 . 1/ 2 5.29 1/ 2 Vn f1 L1 D n . = . V1 f n L n D1 5.5.2 Pipes in Series Consider a case where three pipes as shown in figure 5.4 are connected in series to two reservoirs. The length and diameters are L1, L2, L3, D1, D2 and D3 .Let D2 < D1 and D2 < D3 1 H 2 HA 3 HB Figure 5.4: Pipes in Series As the rate of flow through each pipes is same, therefore Q = A1V1 = A 2 V2 = A 3 V3 5.30 Q = V1D12 = V2 D 2 2 = D 2 3 V3 Also the difference in the liquid surface levels equals sum of various head losses in the pipe V3 2 H = H A − H B = he + hf1 + hc + hf 2 + hl + hf 3 + 5.31 2g If we assume that the minor losses are negligible compared to frictional losses, then H A − H B = h f1 + h f 2 + h f 3 5.32 L 3 V3 2 L1 V1 2 L 2 V12 H = H A − H B = f1 . . + f2. . + f3. . D1 2g D 2 2g D 3 2g Using equation 5.30 in equation 5.33 yields CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 5.33 56 4 4 L f . 1 . + f . L 2 . D1 + f . L 3 . D1 5.34 2 3 1 D1 D 2 D 2 D 3 D 3 5.5.3 Equivalent Pipe An equivalent pipe is defined as the pipe of uniform diameter having loss of head and discharge equal to the loss of head and discharge of a compound pipe consisting of several pipes of different lengths and diameter. The uniform diameter of the equivalent pipe is known as the equivalent diameter of the series or compound pipe. Consider the series of pipes in figure 5.4, the equivalent single pipe must have the same total losses with the system. h fe = h fs = H 5.35 The loss in the equivalent pipe can also be written as L e Ve 2 h fe = f e . . 5.36 D e 2g The equivalent length of the pipes is obtained as 2 4 4 V1 D e L1 L 3 D1 L 2 D1 + f 3 . 5.37 L e = f1 . . + f 2 . . . V f D D D D D 1 2 2 3 3 e e From equation 5.37,the length of a pipe equivalent to a pipe of length L1 and diameter D1 is V2 H= 1 2g 2 V D e L1 f1 . . L e = 1 Ve f e D1 D V Since 1 = e Ve D1 5.38 2 5 f D L e = L1 1 e 5.39 f e D1 5.5.4 Pipe Network In calculating the flow rates through the pipe network, the following conditions must be satisfied by the network: (i) Continuity must be satisfied at each junction of the network, that is the total flow into a junction must be equal to the total flow out of the junction. (ii) The algebraic sum of pressure head drops around each circuit must be zero. (iii) The Darcy-Weisbach equation (equation 5.7) or any equivalent exponential formula for head loss calculation must be satisfied in each pipe 5.6 Problems 1 Classify the following flows as laminar or turbulent a) Water at 200C flowing in a 0.30m diameter pipe at 1.1m/s. b) Oil of specific gravity 1.1, viscosity 0.05kg/m.s at 200C flowing in a 0.15m diameter pipe at 0.5m/s 2. A liquid of viscosity 3.2N.s/m2 and specific gravity 1.1 flows in a pipe of diameter 125mm.The pressure drop over a meter is 150N/m2. Calculate the frictional factor for the pipe if the flow is in the laminar regime. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 57 3 A pipe whose average roughness is 0.02mm is employed to raise water through a height of 10m.The pipe length is 100m and its diameter is 300mm.Calculate the power required for a discharge rate of 30,000l/min. 4 Three pipes of diameter 300mm, 200mm and 400mm and lengths 450mm, 255mm and 315mm respectively are connected in series. The difference in water surface levels in two tanks is 18m. Determine the rate of flow of water if coefficients of friction are 0.0075, 0.0078 and 0.0072 respectively considering cases of minor losses and neglecting minor losses. 5 When a sudden contraction is introduced in a horizontal pipe line from 500mm diameter to 250mm diameter, the pressure changes from 105kN/m2 to 69kN/m2.If the coefficient of contraction is assumed to be 0.65, calculate the water flow rate. Following this, if there is a sudden enlargement from 250mm to 500mm and if the pressure at the 250mm section is 69kN/m2, what is the pressure at the 500mm enlarged portion. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 58 MODULE 6 COMPRESSIBLE FLOW When a compressible fluid, i.e a gas, flows from a region of high pressure to one of low pressure it expands and its density decreases. It is necessary to take this variation of density into account in compressible flow calculations. In a pipe of constant cross-sectional area, the falling density requires that the fluid accelerate to maintain the same mass flow rate. Consequently, the fluid’s kinetic energy increases. It is found convenient to base compressible flow calculations on an energy balance per unit mass of fluid and to work in terms of the fluid’s specific volume V rather than the density ρ. The specific volume is the volume per unit mass of fluid and is simply the reciprocal of the density: 6.0 6.1 6.1 Energy Relationships The total energy E per unit mass of fluid is given by the following equations: 6.2 Or equation 6.2 can also be written as 6.3 Where are the internal, potential, pressure and kinetic energies per unit mass respectively. Consider unit mass of fluid flowing in steady state from a point 1 to a point 2. Between these two points, let a net amount of heat energy q be added to the fluid and let a net amount of work W be done on the fluid. An energy balance for unit mass of fluid can be written either as 6.4 For steady flow in a pipe or tube the kinetic energy term can be written as u /(2α) where u is the volumetric average velocity in the pipe or tube and α is a dimensionless correction factor which accounts for the velocity distribution across the pipe or tube. Fluids that are treated as compressible are almost always in turbulent flow and α is approximately 1 for turbulent flow. Thus for a compressible fluid flowing in a pipe or tube, equation 6.4 can be written as 2 6.5 where in SI units each term is in J/kg. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 59 Since the enthalpy per unit mass of a fluid H is defined by the equation 6.6 Equation 6.5 can be written in the alternative form 6.7 The work term W in equations 6.5 and 6.7 is positive if work is done on the fluid by a pump or compressor. W is negative if the fluid does work in a turbine. W is often referred to as shaft work since it is transmitted into or out of a system by means of a shaft. The differential form of equation 6.5 is 6.8 For a reversible change, the first law of thermodynamics can be expressed by the equation 6.9 where dU is the increase in internal energy per unit mass of fluid and PdV is the work of expansion on the fluid layers ahead for a net addition of heat dq to the system. In flow, energy is required to overcome friction. The effect of friction is to generate heat in a system by converting mechanical to thermal energy. Thus where friction is involved, equation 6.9 can be written as 6.10 where dF is the energy per unit mass required to overcome friction. Substituting equation 6.10 into equation 6.8 gives 6.11 Equation 6.11 can be integrated between states 1 and 2 to give 6.12 where in SI units each term is in J/kg. Equations 6.5, 6.7 and 6.12 all relate to the energy changes involved for a fluid in steady turbulent flow. The most appropriate equation is selected for each particular application: equation 6.12 is a convenient form from which a basic flow rate-pressure drop equation will be derived. Due to the change in the average velocity u, it is more convenient in calculations for compressible flow in pipes of constant cross-sectional area to work in terms of the mass flux G. This is the mass flow rate per unit flow area and is sometimes called the mass velocity. If the mass flow rate is constant, as will usually be the case, then G is constant when the area is constant. In SI units G is in kg/(m2s). The relationship between G and u is given by 6.13 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 60 Writing equation 6.11 in terms of G, noting that d(V2/2) = VdV, gives 6.14 Given that the pressure drop ∆Pf due to friction in a pipe of length L and inside diameter di is given by 6.15 where f is the Fanning friction factor. For an element of length dx of the pipe, equation 6.15 can be written as 6.16 The corresponding energy required to overcome friction is df = VdPf . Thus equation 6.16 gives dF as 6.17 where advantage has been taken of equations 6.1 and 6.13. Substituting for dF in equation 6.14 gives 6.18 Dividing equation 6.18 throughout by V2 and integrating between states 1 and 2 over a length L of pipe gives 6.19 In integrating the frictional term it has been assumed that the value of the friction factor is constant: this is a good approximation because the Reynolds number will usually be very high, a condition for which f is independent of Re. In almost all cases, the change in potential energy will be negligible for gas flow. Also, it is convenient to treat a compressor separately from flow in the pipe, ie equation 6.19 will be applied to a section of pipe in which no shaft work is done. Consequently, 6.19 can be written in a reduced form: 6.20 Equation 6.20 is that required for most calculations involving compressible flow in a pipe. The three terms represent, respectively, changes in pressure energy, kinetic energy and the conversion of mechanical energy to thermal energy by frictional dissipation. The terms in the square brackets are necessarily positive so the pressure energy term must be negative: this reflects the fact that the pressure falls in the direction of flow. In most cases the kinetic energy term will be negligible compared with the frictional term. This is useful when calculating the pressure drop for a given flow rate because in this case one of the pressures and therefore the corresponding specific volume will be unknown. An approximate calculation can be made neglecting the kinetic energy term then, when the pressures are known, the value of that term can be calculated to check whether it was in fact negligible. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 61 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 In equation 6.20 , and all other equations in this module, P denotes the absolute pressure.In order to make use of equation 6.20 or equation 6.19 it is necessary to know the relationship between the pressure P and the specific volume V so that terms such as relationship between P and V is known as the equation of state. can be evaluated. The 6.2 Equations of State An ideal or perfect gas obeys the equation 6.21 where R is the universal gas constant, T the absolute temperature and RMM the relative molecular mass converson factor for the gas. In SI units R = 8314.3 J/(kmol K) and T is in K. The conversion factor RMM has the numerical value of the relative molecular mass and the units kg/kmol in the SI system. Equation 6.21, which is a combination of Boyle's and Charles's laws, will be more familiar in the form 6.22 where is the molar volume of the gas. The relative molecular mass has to be introduced in equation 6.21 because V is the specific volume, ie the volume per unit mass. It is convenient to define a specific gas constant R' by 6.23 so that equation 6.21can be written as 6.24 It is essential to remember that in equation 6.24 both V and R' are values per unit mass of gas and they must not be confused with the molar equivalents. The value of R' is different for gases of different relative molecular masses. Many gases obey equation 6.24 up to a few atmospheres pressure. At high pressures it is necessary to modify equation 6.24 by introducing the compressibility factor Z: 6.25 The compressibility factor is a function of the reduced pressure Pr, and the reduced temperature Tr, of the gas. Pr, is the ratio of the actual pressure P to the critical pressure Pc of the gas: 6.26 and Tr, is the ratio of the actual temperature T to the critical temperature Tc, of the gas: 6.27 Plots of Z against P, at constant T, are available [Perry (1984), Smith and Van Ness (1987)l. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 62 When ideal gases are compressed or expanded they obey the following general equation: 6.28 Thus, for two states 1 and 2, equation 6.28 gives 6.29 6.30 Combining equation 6.30 with 6.24 gives the relationship between pressure and temperature: 6.31 Equation 6.31 shows that, in general, expansion or compression of a gas is accompanied by a change of temperature. A change of state according to equation 6.28 is called a polytropic change. Two special cases are the isothermal change and the adiabatic change.As the name implies, an isothermal change takes place at constant temperature. This requires that the process be relatively slow and heat transfer between the gas and the surroundings be rapid. An isothermal change corresponds to k = 1 and equation 6.28 for an ideal gas becomes: 6.32 The other extreme case is the adiabatic change, which occurs with no heat transfer between the gas and the surroundings. For a reversible adiabatic change, where , the ratio of the specific heat capacities at constant pressure (Cp,) and at constant volume (Cv,). For areversible adiabatic change of an ideal gas, equation 6.28 becomes 6.33 Then, 6.30 becomes 6.34 And 6.31 becomes 6.35 In a reversible adiabatic change the entropy remains constant and therefore this type of change is called an isentropic change. Although not rigorously valid for irreversible changes, equations 6.33 to 6.35 are good approximations for these conditions. Approximate values of at ordinary temperatures and pressures are 1.67 for monatomic gases such as helium and argon, 1.40 for diatomic gases such as hydrogen, carbon monoxide and nitrogen, and 1.30 for triatomic gases such as carbon dioxide. Gases and vapours of complex molecules can have significantly lower values of , for example 1.05 for n-heptane and 1.03 for n-decane. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION 63 Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 6.3 Isothermal flow of an ideal gas in a horizontal pipe For steady flow of a gas between points 1 and 2, distance L apart, in a horizontal pipe of constant cross-sectional area in which no shaft work is done, the energy relationships are given by equation 6.20. Using Isothermal equation of state in equation 6.20 gives the following working equation for isothermal flow of an ideal gas: 6.36 As noted previously, the first term is negative, as must be the case because P1 > P 2 . Equation 6.36 is the basic form of the energy equation to be used for isothermal conditions, however it is instructive to write the equation in a slightly different form that allows easy comparison with incompressible flow. The pressure energy term can be written in the following form: 6.37 where Pm, = (P2 + P1)/2 is the arithmetic mean pressure in the pipe. Then 6.38 So that 6.39 where Vm, is the specific volume at the mean pressure P,. Thus, equation 6.36 can be written as 6.40 As noted previously, the kinetic energy term is usually negligible compared with the frictional term and this is certainly true when the pressure drop ∆P' = P1 - P2 is small compared with P1. In this case, equation 6.40 can be approximated by 6.41 Or 6.42 where ρm and um, are the density and average velocity at the mean pressure Pm,. Equation 6.42 will be recognized as being of the same form as incompressible flow, except that it is written in terms of average properties. Thus, when the pressure drop is small compared with the meanpressure in the pipe, the gas flow may be treated as incompressible flow. For large values of the pressure drop it is necessary to use equation 6.36. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 64 In order to maintain isothermal flow it is necessary for heat to be transferred across the pipe wall. From equation 6.7, for flow in a section with no shaft work and negligible change in elevation, the energy equation takes the form 6.43 For an ideal gas under isothermal conditions, the enthalpy remains constant and hence it follows from equation 6.43 that the required heat leak into the pipe is equal to the increase in kinetic energy. This is usually a small quantity and therefore flow in long, uninsulated pipes will be virtually isothermal. Example 6.1 Hydrogen is to be pumped from one vessel through a pipe of length 400 m to a second vessel, which is at a pressure of 20 bar absolute. The required flow rate is 0.2 kg/s and the allowable pressure at the pipe inlet is 25 bar. The flow conditions are isothermal and the gas temperature is 25°C. If the friction factor may be assumed to have a value of 0.005, what diameter of pipe is required? Solution : Recall that For Isothermal conditions Equation of state Substituting these values into equation 6.20 gives CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 65 It may be anticipated that 0.223 << 4/di, i.e di <<17.9 m. Thus the calculation may be simplified by neglecting the kinetic energy term, so that 6.4 Adiabatic Flow of an Ideal Gas in a Horizontal Pipe For adiabatic flow in a horizontal pipe with no shaft work, equation 6.7 reduces to 6.44 The differential form of equation 6.44 with u expressed in terms of G and V is 6.45 The enthalpy change can be found from the following two fundamental thermodynamic relationships which, in the case of ideal gases, are valid for irreversible processes as well as reversible ones: 6.46 From equation 6.6 6.47 Thus 6.48 And 6.50 Then 6.51 Thus equation 6.45 can be written as 6.52 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 66 Integrating equation 6.52 gives the desired relationship between P and V: 6.53 Thus, 6.54 and the integral in equation 6.20 is readily found by integrating equation 6.54: 6.55 Substituting this result in equation 6.20 gives 6.56 Where from equation 6.53 6.57 Calculations for adiabatic flow require the use of equations 6.56 and 6.57. For example, if the upstream conditions P1 and V1are known, and G and d, are specified, C can be calculated from equation 6.57 then V2 from equation 6.57. Substituting this value of V2 in equation 6.57 gives P2. If the logarithmic term is not negligible, an iterative calculation will be needed to determine V2 from equation 6.56. 6.5 Gas Compression and Compressors Compressors are devices for supplying energy or pressure head to a gas. For the most part, compressors like pumps can be classified into centrifugal and positive displacement types. Centrifugal compressors impart a high velocity to the gas and the resultant kinetic energy provides the work for compression. Positive displacement compressors include rotary and reciprocating compressors although the latter are the most important for high pressure applications. From equation 6.12, the shaft work of compression W required to compress unit mass of gas from pressure P1 to pressure P2 in a reversible frictionless process, in which changes in potential and kinetic energy are negligible, is 6.58 Although isothermal compression is desirable, in practice the heat of compression is never removed fast enough to make this possible. In actual compressors only a small fraction of the heat of compression is removed and the process is almost adiabatic. When ideal gases are compressed under reversible adiabatic conditions they obey equation 6.33, which can be written as 6.59 so that the specific volume is given by CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 67 6.60 Substituting V into equation 6.58 and integrating gives 6.61 Equation 6.60 gives the theoretical adiabatic work of compression from pressure PI to pressure P2. Compression is often done in several stages with the gas being cooled between stages. For two-stage compression from P1 to P2to P3, with the gas cooled to the initial temperature T1 at constant pressure, equation 6.61 becomes 6.62 In the case of compression from pressure P1 to pressure P2 through n stages each having the same pressure ratio (P2/Pl)1/n the compression work is given by 6.63 Equations 6.61 to 6.63 give the work required to compress unit mass of the gas. It should be noted that the work required depends on the pressure ratio so that compression from 1 bar to 10 bar requires as much power as compressing the same mass of gas, with the same initial temperature, from 10 bar to 100 bar. In practice it is possible to approach more nearly isothermal compression by carrying out the compression in a number of stages with cooling of the gas between stages. When ideal gases are compressed under reversible adiabatic conditions the temperature rise from T1 to T2 is given by equation 6.35: 6.64 So far only reversible adiabatic compression of an ideal gas has been considered. For the irreversible adiabatic compression of an actual gas, the shaft work W required to compress the gas from state 1 to state 2 can be obtained from equation 6.7, which in this case becomes 6.65 where H is the enthalpy per unit mass of gas.The actual work of compression is greater than the theoretical work because of clearance gases, back leakage and friction. Example 6.2 Calculate the theoretical work required to compress 1 kg of a diatomic ideal gas initially at a temperature of 200 K adiabatically from a pressure of 10000Pa to a pressure of 100000Pa in (i) a single stage, (ii) a compressor with two equal stages and (iii) a compressor with three equal stages. The relative molecular mass of the gas is 28.0 and the ratio of specific heat capacities is 1.40. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 68 Solution (i) For a single stage compression, 6.61 From given values Therefore Equation of state Therefore Also Substituting these values into 6.61 gives (ii) For adiabatic compression of an ideal gas in n equal stages For n = 2 CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 69 Since And as before It follows that (iii) Repeating the above calculation for n = 3 gives 6.6 Problems 1 2 3 4 An ideal gas in which the pressure P is related to the volume V by the equation PV = 75 m2/s2 flows in steady isothermal flow along a horizontal pipe of inside diameter di, = 0.02 m. The pressure drops from 20000 Pa to 10000 Pa in a 5 m length. Calculate the mass flux assuming that the Fanning function factorf= 9.0 X10-3. Ethylene flows through a pipeline 10 km long to a receiving station A. At a point 3 km from A, a spur leads off the main pipeline and runs 5 km to a receiving station B. The internal diameter of the main pipeline is 0.20 m and that of the spur is 0.15 m. The flow rates into A and B are regulated by valves at these locations. If the pressure immediately upstream of valve A is 3.88 bar (absolute) and that at B is 3.69 bar when the flow rate into B is 0.63 kg/s, calculate the pressure at the beginning of the main pipeline, assuming that flow in the pipeline is isothermal at a temperature of 20°C. Data: specific volume of ethylene at 200C, 1 bar = 0.870 m3/kg, Fanning friction factor = 0.0045. Calculate the air velocity in m / s required to cause a temperature drop of 1 K on a conventional thermometer given that for the air at atmospheric pressure and 373 K, the thermal capacity per unit mass at constant pressure Cp, = 1006 J/(kg K). An ideal gas flows in steady state adiabatic flow along a horizontal pipe of inside diameter di, = 0.02 m. The pressure and density at a point are P = 20000 Pa and ρ = 200 kg/m3 respectively. The density drops from 200kg/m3 to 100kg/m3 in a 5m length. Calculate the mass flux assuming that the Fanning friction factor f = 9.0 x 10-3 and the ratio of heat capacities at constant pressure and constant volume = 1.40. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 70 MODULE 7 7.0 PUMPS Pumps are devices for supplying energy or head to a flowing liquid in order to overcome head losses due to friction and also, if necessary, to raise the liquid to a higher level. The head imparted to a flowing liquid by a pump is known as the total head ∆h. If a pump is placed between points 1 and 2 in a pipeline, the heads for steady flow are related by equation: 7.1 In equation 7.1, z, P/(ρg), and u /(2gα) are the static, pressure and velocity heads respectively and hf is the head loss due to friction. The dimensionless velocity distribution factor α is 0.5 for laminar flow and approximately 1 for turbulent flow. For a liquid of density ρ flowing with a constant mean velocity u through a pipeline of circular cross section and constant diameter between points 1 and 2 separated by a pump, equation 7.1 can be written as 2 7.2 There exist a wide variety of pumps that are designed for various specific applications. However, most of them can be broadly classified into two categories: positive displacement and centrifugal. The most significant characteristics of each of these will be described later. 7.1 System Heads The important heads to consider in a pumping system are the suction, discharge, total and available net positive suction heads. The following definitions are given in reference to the typical pumping system shown in Figure 7.1 where the arbitrarily chosen base line is the centre-line of the pump. Figure 7.1: Typical Pumping System CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 71 Sunction Head: 7.3 Discharge Head: 7.4 In equation 7.3, hfs is the head loss due to friction, zs is the static head and Ps is the gas pressure above the liquid in the tank on the suction side of the pump. If the liquid level on the suction side is below the centre-line of the pump, zs, is negative. In equation 7.4, hfd is the head loss due to friction, zd is the static head and Pd is the gas pressure above the liquid in the tank on the discharge side of the pump. hs and hd are the values of at the suction flange and at the discharge flange respectively. Equations 7.3. and 7.4 are obtained by applying Bernoulli’s equation between the supply tank and the suction flange, and between the discharge flange and the receiving tank, respectively. On the suction side, the frictional loss hfs reduces the total head at the suction flange but on the discharge side, hfd increases the head at the discharge flange. The total head ∆h which the pump is required to impart to the flowing liquid is the difference between the discharge and suction heads: 7.5 Equation 7.5 can be written in terms of 7.3 and 7.4 as: 7.6 The head losses due to friction are given be equations : 7.7 And 7.8 where are the total equivalent lengths on the suction and discharge sides of the pump respectively. The suction head hs decreases and the discharge head hd increases with increasing liquid flow rate because of the increasing value of the friction head loss terms hfs and hfd . Thus the total head ∆h which the pump is required to impart to the flowing liquid increases with the liquid pumping rate. It is clear from equation 7.3 that the suction head hs can fall to a very low value, for example when the suction frictional head loss is high and the static head z, is low. If the absolute pressure in the liquid at the suction flange falls to, or below, the absolute vapour pressure Pv of the liquid, bubbles of vapour will be formed at the pump inlet. Worse still, even if the pressure at the suction flange is CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 72 slightly higher than the vapour pressure, Cavitation - the formation and subsequent collapse of vapour bubbles will occur within the body of the pump because the pressure in the pump falls further as the liquid is accelerated. In order that cavitation may be avoided, pump manufacturers specify a minimum value by which the total head at the suction flange must exceed the head corresponding to the liquid's vapour pressure. The difference between the suction head and the vapour pressure head is known as the Net Positive Suction Head, NPSH: 7.9 Substituting for hs from equation 7.3, the available NPSH is given by 7.10 The available NPSH given by equations 7.9 and 7.10 must exceed the value required by the pump and specified by the manufacturer. The required NPSH increases with increasing flow rate. 7.2 Positive Displacement Pumps The term positive displacement pump is quite descriptive, because such pumps are designed to displace a more or less fixed volume of fluid during each cycle of operation. They include piston, diaphragm, screw, gear, progressing cavity, and other pumps. The volumetric flow rate is determined by the displacement per cycle of the moving member (either rotating or reciprocating) times the cycle rate (e.g., rpm). The flow capacity is thus fixed by the design, size, and operating speed of the pump. The pressure (or head) that the pump develops depends upon the flow resistance of the system in which the pump is installed and is limited only by the size of the driving motor and the strength of the parts. Consequently, the discharge line from the pump should never be closed off without allowing for recycle around the pump or damage to the pump could result. In general positive displacement pumps have limited flow capacity but are capable of relatively high pressures. Thus these pumps operate at essentially constant flow rate, with variable head. They are appropriate for high pressure requirements, very viscous fluids, and applications that require a precisely controlled or metered flow rate. 7.3 Centrifugal Pumps The term ‘‘centrifugal pumps’’ is also very descriptive, because these pumps operate by the transfer of energy (or angular momentum) from a rotating impeller to the fluid, which is normally inside a casing. A sectional view of a typical centrifugal pump is shown in Figure 7.2. The fluid enters at the axis or ‘‘eye’’ of the impeller (which may be open or closed and usually contains radial curved vanes) and is discharged from the impeller periphery. The kinetic energy and momentum of the fluid are increased by the angular momentum imparted by the high-speed impeller. This kinetic energy is then converted to pressure energy (‘‘head’’) in a diverging area (the ‘‘volute’’) between the impeller discharge and the casing before the fluid exits the pump. The head that these pumps can develop depends upon the pump design and the size, shape, and speed of the impeller and the flow capacity is determined by the flow resistance of the system in which the pump is installed. Thus, these pumps operate at approximately constant head and variable flow rate, within limits, of course, determined by the size and design of the pump and the size of the driving motor.Centrifugal pumps can be operated in a ‘‘closed off’’ condition (i.e., closed discharge line), because the liquid will recirculate within the pump without causing damage. However, such conditions should be avoided, because energy dissipation within the pump could result in excessive heating of the fluid and/or the pump or unstable CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 73 operation, with adverse consequences. Centrifugal pumps are most appropriate for ‘‘ordinary’’ (i.e.,low to moderate viscosity) liquids under a wide variety of flow conditions and are thus the most common type of pump. Figure 7.2 Sectional view of a typical centrifugal pump. The performance of a centrifugal pump for a particular rotational speed of the impeller and liquid viscosity is represented by plots of total head against capacity, power against capacity, and required NPSH against capacity. These are known as characteristic curves of the pump. The relationship between ∆P and ∆h is given by : 7.11 Combining equations 7.6,7.7 and 7.8 gives the total head as 7.12 The mean velocity u of the liquid is related to the volumetric flow rate or capacity Q by CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 74 7.13 Using 7.13 in 7.12 gives 7.14 For laminar flow, the Fanning friction factor f is given by equation: 7.15 Substituting for fin equation 7.14, the total head for laminar flow can be written as 7.16 Or 7.17 The system ∆h against Q curve can be plotted using equation 7.14 to calculate the values of the system total head ∆h at each volumetric flow rate of liquid or capacity Q. Equation 7.17 shows that for laminar flow the total head ∆h increases linearly with capacity Q. Thus for laminar flow, the system ∆h against Q curve is a straight line. The available NPSH in a system can be calculated from equation 7.10 having substituted for hfs: 7.18 Equation 7.18 shows that the available NPSH in a system decreases as the liquid throughput increases because of the greater frictional head losses. Example 4.1 Calculate the values for a system total head against capacity curve for the initial conditions of the system shown in Figure 7.1 given the following data: CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 75 Solution From f against Re graph Repeating the calculations for other values of u gives the following results: CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 76 Table 7.1 7.4 Centrifugal pump relations Holland and Chapman ( 1966)] developed an approach through dimensional analysis to express the power PE required in an ideal centrifugal pump as a function of the liquid density ρ, the impeller diameter D and the rotational speed of the impeller N. 7.19 7.20 7.21 7.22 Also, Consider a centrifugal pump with an impeller diameter D1 operating at a rotational speed N1 and developing a total head ∆hl. Consider an homologous pump with an impeller diameter D2 operating at a rotational speed N2 and developing a total head ∆h2. Equations 7.19 to 7.22 can be manipulated to 7.23 Similarly’ 7.24 and by analogy with equation 7.23 the net positive suction heads for the two homologous pumps can be related by the equation 7.25 Equations 7.23 to 7.25 are the affinity laws for homologous centrifugal pumps. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 77 For a particular pump where the impeller of diameter D1 is replaced by an impeller with a slightly different diameter D2, the following equations hold [Holland and Chapman (1966): 7.26 If the characteristic performance curves are available for a centrifugal pump operating at a given rotation speed, equations in 7.26 enable the characteristic performance curves to be plotted for other operating speeds and for other slightly different impeller diameters. Example 7.2 A volute centrifugal pump with an impeller diameter of 0.02 m has the following performance data when pumping water at the best efficiency point: Evaluate the performance characteristics of a homologous pump with twice the impeller diameter operating at half the impeller speed. Solution Let subscripts 1 and 2 refer to the first and second pumps respectively. Given: The ratio of impeller spreeds N1/ N2 = 2 and the ratio of impeller diameters D1/D2 = 1/2. The ratio of capacities is given by CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 78 7.5 Pumping Efficiencies The liquid power PE can be defined as the rate of useful work done on the liquid. It is given by the equation 7.27 If the volumetric flow rate Q is in m3/s and the pressure developed by the pump ∆P is in Pa or N/m2, the liquid power PE is in Nm/s or W. The pressure developed by the pump ∆P is related to the total head developed by the pump ∆h by equation 7.28 7.29 Or CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 79 7.30 since M = ρQ is the mass flow rate. If M is in kg/s, ∆h is in m and the gravitational accelerating g = 9.81m/s2, PE is in W. The brake power PB can be defined as the actual power delivered to the pump by the prime mover. It is the sum of liquid power and power lost due to friction and is given by the equation 7.31 is the mechanical efficiency expressed in per cent. The mechanical efficiency decreases as the liquid viscosity and hence the frictional losses increase. The mechanical efficiency is also decreased by power losses in gears, bearings, seals etc. In rotary pumps contact between the rotor and the fixed casing increases power losses and decreases the mechanical efficiency. These losses are not proportional to pump size. Relatively large pumps tend to have the best efficiencies whilst small pumps usually have low efficiencies. Furthermore, high speed pumps tend to be more efficient than low speed pumps. In general, high efficiency pumps have high NPSH requirements. Sometimes a compromise may have to be made between efficiency and NPSH. Another efficiency which is important for positive displacement pumps is the volumetric efficiency. This is the delivered capacity per cycle as a percentage of the true displacement per cycle. If no slip occurs, the volumetric efficiency of the pump is 100 per cent. For zero pressure difference across the pump, there is no slip and the delivered capacity is the true displacement. The volumetric efficiency of a pump is reduced by the presence of entrained air or gas in the pumped liquid. It is important to know the volumetric efficiency of a positive displacement pump when it is to be used for metering. 7.6 Factors In Pump Selection The selection of a pump depends on many factors which include the required rate and properties of the pumped liquid and the desired location of the pump. In general, high viscosity liquids are pumped with positive displacement pumps. Centrifugal pumps are not only very inefficient when pumping high viscosity liquids but their performance is very sensitive to changes in liquid viscosity. A high viscosity also leads to high frictional head losses and hence a reduced available NPSH. Since the latter must always be greater than the NPSH required by the pump, a low available NPSH imposes a severe limitation on the choice of a pump. Liquids with a high vapour pressure also reduce the available NPSH. If these liquids are pumped at a high temperature, this may cause the gears to seize in a close clearance gear pump. If the pumped liquid is shear thinning, its apparent viscosity will decrease with an increase in shear rate and hence pumping rate. It is therefore an advantage to use high speed pumps to pump shear thinning liquids and in fact centrifugal pumps are frequently used. In contrast, the apparent viscosity of a shear thickening liquid will increase with an increase in shear rate and hence pumping rate. It is therefore an advantage to use large cavity positive displacement pumps with a low cycle speed to pump shear thickening liquids. Some liquids can be permanently damaged by subjecting them to high shear in a high speed pump. For example, certain liquid detergents can be broken down into two phases if subject to too much shear. Even though these detergents may exhibit shear thinning characteristics they should be pumped with relatively low speed pumps. Wear is a more serious problem with positive displacemeni pumps than with centrifugal pumps. Liquids with poor lubricating qualities increase the wear on a pump. Wear is also caused by corrosion and by the pumping of liquids containing suspended solids which are abrasive. where CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 80 In general,-centrifugal pumps are less expensive, last longer and are more robust than positive displacement pumps. However, they are unsuitable for pumping high viscosity liquids and when changes in viscosity occur. 7.7 Problems 1 Calculate the available net positive section head NPSH in a pumping system if the liquid density ρ = 1200 kg/m3. The liquid dynamic viscosity µ = 0.4 Pa.s, the mean velocity u = 1 m/s, the static head on the suction side zs = 3m, the inside pipe diameter di, = 0.0526m, the gravitational acceleration g = 9.81 m/s2 , and the equivalent length on the suction side 2 The liquid is at its normal boiling point. Neglect entrance and exit losses. A volute centrifugal pump has the following performance data at the best efficiency point: Evaluate the performance of an homologous pump which operates at an impeller speed of 29.2 rev/s but which develops the same total head ∆h and requires the same NPSH. CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 81 MODULE 8 FLOW MEASUREMENT 8.1 The Pitot Tube As previously discussed, the volumetric flow rate of a fluid through a conduit can be determined by integrating the local (‘‘point’’) velocity over the cross section of the conduit: 8.1 If the conduit cross section is circular, this becomes 8.2 The pitot tube is a device for measuring v(r), the local velocity at a given position in the conduit, as illustrated in Figure 8.1 The measured velocity is then used to determine the flow rate in equation 8.2. It consists of a differential pressure measuring device (e.g., a manometer, transducer, or DP cell) that measures the pressure difference between two tubes. One tube is attached to a hollow probe that can be positioned at any radial location in the conduit, and the other is attached to the wall of the conduit in the same axial plane as the end of the probe. The local velocity of the streamline that impinges on the end of the probe is v(r). The fluid element that impacts the open end of the probe must come to rest at that point, because there is no flow through the probe or the DP cell; this is known as the stagnation point. The Bernoulli equation can be applied to the fluid streamline that impacts the probe tip: 8.3 where point 1 is in the free stream just upstream of the probe and point 2 is just inside the open end of the probe (the stagnation point). Since the friction loss is negligible in the free stream from 1 to 2, and v2 =0 because the fluid in the probe is stagnant, Equation 8.3 can be solved for v1 to give 8.4 The measured pressure difference ∆P is the difference between the ‘‘stagnation’’ pressure in the velocity probe at the point where it connects to the DP cell and the ‘‘static’’ pressure at the corresponding point in the tube connected to the wall. Figure 8.1 : The Pitot Tube CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 82 8.2 The Venturi And Nozzle There are other devices, however, that can be used to determine the flow rate from a single measurement. These are sometimes referred to as obstruction meters, because the basic principle involves introducing an ‘‘obstruction’’ (e.g., a constriction) into the flow channel and then measuring the pressure drop across the obstruction that is related to the flow rate. Two such devices are the venturi meter and the nozzle, illustrated in Figure. 8.2 and 8.3 respectively. In both cases the fluid flows through a reduced area, which results in an increase in the velocity at that point. The corresponding change in pressure between point 1 upstream of the constriction and point 2 at the position of the minimum area (maximum velocity) is measured and is then related to the flow rate through the energy balance. The velocities are related by the continuity equation,and the Bernoulli equation relates the velocity change to the pressure change: 8.5 Figure 8.2: Venturi Meter Figure 8.3 : Nozzle For constant density 8.6 And the Bernoulli Equation is 8.7 where plug flow has been assumed. Using Equation 8.6 to eliminate V1 and neglecting the friction loss, Equation 8.7 can be solved for V2: CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 83 8.8 Where (where d2 is the minimum diameter at the throat of the venturi or nozzle). To account for the inaccuracies introduced by assuming plug flow and neglecting friction, Equation 8.8 is written 8.9 where Cd is the ‘‘discharge’’ or venturi (or nozzle) coefficient and is determined by calibration as a function of the Reynolds number in the conduit. 8.3 The Orifice Meter The simplest and most common device for measuring flow rate in a pipe is the orifice meter, illustrated in Figure 8.4 This is an ‘‘obstruction’’ meter that consists of a plate with a hole in it that is inserted into the pipe, and the pressure drop across the plate is measured. The major difference between this device and the venturi and nozzle meters is the fact that the fluid stream leaving the orifice hole contracts to an area considerably smaller than that of the orifice hole itself. This is called the vena contracta, and it occurs because the fluid has considerable inward radial momentum as it converges into the orifice hole, which causes it to continue to flow ‘‘inward’’ for a distance downstream of the orifice before it starts to expand to fill the pipe. If the pipe diameter is D, the orifice diameter is d, and the diameter of the vena contracta is d2, the contraction ratio for the vena contracta is defined as For highly turbulent flow, Cc≈ 0:6. The complete Bernoulli equation, as applied between point 1 upstream of the orifice where the diameter is D and point 2 in the vena contracta where the diameter is d2, is 8.10 Figure 8.4 :Orifice Meter As for the other obstruction meters, when the continuity equation is used to eliminate the upstream velocity from Equation 8.10, the resulting expression for the mass flow rate through the orifice is CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 84 8.11 where is the orifice coefficient: Co is obviously a function of β and the loss coefficient Kf (which depends on Re). For incompressible flow,equation 8.11 becomes 8.12 8.13 While equation 8.11 becomes 8.14 for compressible flow 8.14 It is more convenient to express this result in terms of the ratio of Equation 8.14 to the corresponding incompressible equation, Equation 8.13, which is called the expansion factor Y: 8.15 where the density ρ1 is evaluated at the upstream pressure (P1). For convenience, the values of Y are shown as a function of ∆P/P1 and β for the square-edged orifice, nozzles, and venturi meters for values of of 1.3 and 1.4 . CHE 241 – FUNDAMENTALS OF FLUID MECHANICS 2010/2011 SESSION Department of Chemical Engineering, University of Ilorin. Dr. O.O.Ogunleye and Mr. E.O.Ajala - 2010 85