Fluid Flow for Chemical and Process Engineers

In preparing the second edition of this book, the authors have been concerned to maintain or expand those aspects of the subject that are specific to chemical and process engineering.


Professor F, A. Holland, Dr R. Bragg


375 Pages

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Chemical Engineering

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  • Professor F, A. Holland, Dr R. Bragg   
  • 375 Pages   
  • 12 Feb 2015
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    Fluid Flow for Chemical Engineers .............................. IIII II I IIIIIIII I II III II II II I I l_ Second edition Professor F, A. Holland Overseas Educational Development Office University of Salford Dr R. Bragg Department of Chemical Engineering University of Manchester Institute of Science and Technology I~E I N E M A N N OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI read more..

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    Butterworth-Heinemann Linacre House, Jordan Hill, Oxford OX2 8DP 225 Wildwood Avenue, Wobum MA 01801-2041 A division of Reed Educational and Professional Publishing Ltd -~~A member of the Reed Elsevier plc group First published in Great Britian 1973 First published by Edward Arnold 1995 Reprinted 1999 Transferred to digital printing 2002 9 F. A. Holland and R. Bragg i995 All rights reserved. No part of this publication may be reproduced in any material form (including photocopying or storing in read more..

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    Contents List of examples Preface to the second edition Nomenclature 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 Fluids in motion Units and dimensions Description of fluids and fluid flow Types of flow Conservation of mass Energy relationships and the Bernoulli equation Momentum of a flowing fluid Stress in fluids Sign conventions for stress Stress components Volumetric flow rate and average velocity in a pipe Momentum transfer in laminar flow Non-Newtonian behaviour Turbulence and read more..

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    vi CONTENTS 2.9 2.10 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 6 6.1 6.2 Universal velocity distribution for turbulent flow in a pipe Flow in open channels Flow of incompressible non-Newtonian fluids in pipes Elementary viscometry Rabinowitsch-Mooney equation Calculation of flow rate-pressure drop relationship for laminar flow using r data Wall shear stress--flow characteristic curves and scale-up for laminar flow Generalized read more..

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    CONTENTS vii 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 Isothermal flow of an ideal gas in a horizontal pipe Non-isothermal flow of an ideal gas in a horizontal pipe Adiabatic flow of an ideal gas in a horizontal pipe Speed of sound in a fluid Maximum flow rate in a pipe of constant cross-sectional area Adiabatic stagnation temperature for an ideal gas Gas compression and compressors Compressible flow through nozzles and constrictions 195 199 200 202 203 205 206 209 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7 read more..

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    viii CONTENTS 10.3 10.4 10.5 Time for a solid spherical particle to reach 99 per cent of its terminal velocity when falling from rest in the Stokes regime Suddenly accelerated plate in a Newtonian fluid Pressure surge in pipelines 311 312 317 Appendix I The Navier-Stokes equations 322 Appendix II Further problems 332 Answers to problems 345 Conversion factors 348 Friction factor charts 349 Index 3 51 read more..

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    List of examples Chapter 1 1.1 Application of Bernoulli's equation to a circulating liquid 1.2 Calculation of discharge rate from nozzle 1.3 Determination of direction of forces acting on a pipe with a reducer 1.4 Calculation of reaction on bend due to fluid momentum 1.5 Determination of contraction of a jet 1.6 Determination of forces acting on a nozzle 1.7 Application of force balance to determine the wall shear stress in a pipe 1.8 Determination of radial variation of shear stress for flow in read more..

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    X LiST OF EXAMPLES Chapter 5 5.1 Calculation of power for a turbine agitator in a baffled tank 5.2 Calculation of power for a turbine agitator in an unbaffled tank 179 180 Chapter 6 6.1 Calculation of pipe diameter for isothermal compressible flow in a pipeline 6.2 Calculation of work in a compressor 6.3 Calculation of flow rate for compressible flow through a converging nozzle 198 207 216 Chapter 7 7.1 Calculation of pressure drop using the homogeneous model for gas-liquid two-phase flow 7.2 read more..

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    Preface to the second edition In preparing the second edition of this book, the authors have been concerned to maintain or expand those aspects of the subject that are specific to chemical and process engineering. Thus, the chapter on gas-liquid two-phase flow has been greatly extended to cover flow in the bubble regime as well as to provide an introduction to the homogeneous model and separated flow model for the other flow regimes. The chapter on nonoNewtonian flow has also been extended to read more..

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    Nomenclature a a A b r C C C C Cd C,, d D De g E E EO f F F Fr g G h H H He blade width, m propagation speed of pressure wave in equation 10.39, m/s area, m 2 width, m speed of sound, m/s couple, N m Chezy coefficient (2g/f) 1/2, ml/2/s constant, usually dimensionless solute concentration, kg/m ~ or kmol/m 3 drag coefficient or discharge coefficient, dimensionless specific heat capacity at constant pressure, J/(kg K) specific heat capacity at constant volume, J/(kg K) diameter, m equivalent read more..

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    NOMENCLATURE xlll IT J Jl J k k K K K K, K' KE L e In log m M Ma tl tt r N NPSH P P PA Ps PE Po q q Q r r rl R R R' Re RMM tank turnovers per unit time in equation 5.8, s- volumetric flux, m/s basic friction factor jr -- f/2, dimensionless molar diffusional flux in equation 1.70, kmol/(m 2 s) index of polytropic change, dimensionless proportionality constant in equation 5.1, dimensionless consistency coefficient, Pa s ~ number of velocity heads in equation 2.23 proportionality constant in read more..

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    xlv NOMENCLATURE $ s s S So t T To u UG ,L Ut UT U V V W W We X X Y Y Z ,y e E r p Or" T distance, m scale reading in equation 8.39, dimensionless slope, sin0, dimensionless cross-sectional flow area~ m 2 surface area per unit volume, m- time, s temperature, K stagnation temperature in equation 6.85, K volumetric average velocity, m/s characteristic velocity in equation 7.29, m/s terminal velocity, m/s tip speed, m/s internal energy per unit mass, J/kg or m2/s 2 point velocity, m/s volume, read more..

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    NOMENCLATURE xv oJ square root of two-phase multiplier, dimensionless pressure function in equation 6.108, dimensionless correction factor in equation 9.12, dimensionless angular velocity, rad/s vorticity in equation A26, s-l Subscripts Q a A b r d f G i L m mf M N 0 P r $ sh t t T v V tV tV Y referring to apparent referring to accelerative component referring to agitator referring to bed or bubble referring to coarse suspension, coil, contraction or critical referring to discharge side read more..

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    1 Fluids in motion 1.1 Units and dimensions Mass, length and time are commonly used primary units, other units being derived from them. Their dimensions are written as M, L and T respectively. Sometimes force is used as a primary unit. In the Syst~me International d'Unit~s, commonly known as the SI system of units, the primary units are the kilogramme kg, the metre m, and the second s. A number of derived units are listed in Table I.I. 1.2 Description of fluids and fluid flow 1.2.1 read more..

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    2 FLUID FLOW FOR CHEMICAL ENGINEERS Table 1.1 i i Quantify Derived unit Relationship to Symbol primary units Force Work, energy, quantity of heat Power Area Volume Density Velocity Acceleration Pressure Surface tension Dynamic viscosity newton N kg m/s 2 joule J N m watt W J/s = N m/s square metre m 2 cubic metre m 3 kilogramme per cubic kg/m 3 metre metre per second m/s metre per second per m/s 2 second pascal, or newton per Pa N/m 2 square metre newton per metre N/m pascal second, or Pa s N read more..

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    FLUIDS IN MOTION $ homogeneous fluid is one whose properties are the same at all locations and this is usually true for single--phase flow. The flow of gas--liquid mixtures and of sofid-fluid mixtures exemplifies heterogeneous flow problems. A material is isotropic if its properties are the same in all directions. Gases and simple liquids are isotropic but liquids having complex, chain-like molecules, such as polymers, may exhibit different properties in different directions. For example, read more..

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    4 FLUID FLOW FOR CHEMICAL ENGINEERS done by experimental flow visualization or by calculating the velocity field. The terms 'path', 'streakline' and 'streamline' have different meanings. Consider a flow visualization study in which a small patch of dye is injected instantaneously into the flowing fluid. This will 'tag' an element of the fluid and, by following the course of the dye, the path of the tagged element of fluid is observed. If, however, the dye is introduced con- tinuously, a read more..

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    FLUIDS IN MOTION iS observed. In the first type, shown schematically in Figure 1.1(a), the streaklines are straight and the dye remains intact. The dye is observed to spread very slightly as it is carried through the tube; this is due to molecular diffusion. The flow causes no mixing of the dye with the surrounding water. In this type of flow, known as laminar or streamline flow, elements of the fluid flow in an orderly fashion without any macroscopic intermixing with neighbouring fluid. In this read more..

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    8 FLUID FLOW FOR CHEMICAL ENGINEERS Q. 2 "0 ulent ..,,,. Flow rote Figure 1.2 The relationship between pressure drop and flow rate in a pipe promoting rapid mixing and enhances convective heat and mass transfer. The penalty that has to be paid for this is the greater power required to pump the fluid. Measurements with different fluids, in pipes of various diameters, have shown that for Newtonian fluids the transition from laminar to turbulent flow takes place at a critical value of the read more..

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    FLUIDS IN MOTION 7 are much more compressible than liquids but if the pressure of a flowing gas changes little, and the temperature is sensibly constant, then the density will be nearly constant. When the fluid density remains constant, the flow is described as incompressible. Thus gas flow in which pressure changes are small compared with the average pressure may be treated in the same way as the flow of liquids. When the density of the gas changes significantly, the flow is described as read more..

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    $ FLUID FLOW FOR CHEMICAL ENGINEERS Q Ul Figure 1.3 | U2 Flow through a pipe of changing diameter PlQI ---- p2Q2 + V OPm, (1.4) c3t where V is the constant volume of the section between planes 1 and 2, and po~ is the density of the fluid averaged over the volume V. This equation represents the conservation of mass of the flowing fluid: it is frequently called the 'continuity equation' and the concept of 'continuity' is synony- mous with the principle of conservation of mass. In the case of read more..

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    FLUIDS IN MOTION 9 This is the form of the Continuity Equation that will be used most frequently but it is valid only when there is no accumulation. Although Figure 1.3 shows a pipe of circular cross section, equations 1.4 to 1.7 are valid for a cross section of any shape. 1.5 Energy relationships and the Bemoulll equation The total energy of a fluid in motion consists of the following components: internal, potential, pressure and kinetic energies. Each of these energies may be considered with read more..

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    10 FLUID FLOW FOR CHEMICAL ENGINEERS where each term has the dimensions of force times distance per unit mass, ie (ML/T2)L/M or L2/T 2. Consider fluid flowing from point 1 to point 2 as shown in Figure 1.4. Between these two points, let the following amounts of heat transfer and work be done per unit mass of fluid: heat transfer q to the fluid, work Wi done on the fluid and work Wo done by the fluid on its surroundings. Wi and Wo may be thought of as work input and output. Assuming the read more..

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    FLUIDS IN MOTION 11 For an inviscid fluid, ie frictionless flow, and no pump, equation (1.10) becomes ( ( z2g + --- + = zig +--- +-- P2 T p! (1.]1) Equation 1.11 is known as Bernoulli's equation. Dividing throughout by g, these equations can be written in a slightly different form. For example, equation 1. I0 can be written as z z + --'-O z g + ~ ] z l + Pig" + "~g + --g - ---g (I. 12) In this form, each term has the dimensions ot ~ length. The terms z, Pl(pg) and v2/(2g) are known as read more..

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    12 FLUID FLOW FOR CHEMICAL ENGINEERS Bernoulli's equation is based on the principle of conservation of energy and, in the form in which the work terms are zero, it states that the total mechanical energy remains constant along a streamline. Fluids flowing along different streamlines have different total energies. For example, for laminar flow in a horizontal pipe, the pressure energy and potential energy for an element of fluid flowing in the centre of the pipe will be virtually identical to read more..

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    FLUIDS IN MOTION 18 m 2~ ) st tl ~ Arbitrarily chosen base line Figure 1.5 Diagrammatic representation of heads in a liquid flowing through a pipe The method of calculating frictional losses is described in Chapter 2. It may be noted here that losses occur as the fluid flows through the plain pipe, pipe fittings (bends, valves), and at expansions and contractions such as into and out of vessels. A slightly more general case is incompressible flow through an inclined pipe having a change of read more..

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    14 FLUID FLOW FOR CHEMICAL ENGINEERS Denoting the total pressure drop (Pl- Pz) by AP, it can be written as M ~ = APa, + M'~o + M~t (1.17) where kP, h, AP=, APf are respectively the static head, accelerative and frictional components of the total pressure drop given in equation 1.16. Equation 1.16 shows that each component of the pressure drop is equal to the corresponding change of head multiplied by pg. An important application of BernouUi's equation is in flow measure- ment, discussed in read more..

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    FLUIDS IN MOTION 15 For a recirculating flow like this, the fluid's destination is the same as its origin so the two locations can be chosen to be the same, for example the point marked X. In this case equation 1.14 reduces to - hf showing that the pump is required simply to overcome the losses. There is no change in the potential, pressure and kinetic energies of the liquid because it ends with a height, pressure and speed identical to those with which it started. An alternative is to choose read more..

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    16 FLUID FLOW FOR CHEMICAL ENGINEERS that there is no pump in the section, Bernoulli's equation reduces to I'2 u2 2 P, u 2 +___=. +-.- pg 2e pg 2g Thus u 2 - u 2 = 2(P, - Pz)/P The fluid pressure P2 at the exit plane is the atmospheric pressure, ie zero gauge pressure. Therefore u 2 -u 2 = 2(2 x 105 Pa)/(1000 kg/m 3) = 400 m2/s 2 By continuity = therefore u2 = ul~/~ = 9ul Thus 80U 2 = 400 m2/s 2 and hence ui = 2.236 ntis and u2 - 20.12 m/s The volumetric discharge rate can be calculated from read more..

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    FLUIDS IN MOTION 17 Consider the case of incompressible, horizontal flow. Equation 1.11 shows that if a flowing element of fluid is brought to rest (~2 - 0), the pressure P2 is given by P2 - Pl + 2 (1.19) In coming to rest without losses, the fluid's kinetic energy is converted into pressure energy so that the pressure P2 of the stopped fluid is greater than the pressure P l of the flowing fluid by an amount 89 For a fluid having a pressure P and flowing at speed ~, the quantity 89 2 is known as read more..

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    18 FLUID FLOW FOR CHEMICAL ENOINEERS net force - rate of change of momentum applies to an element of fluid, it is difficult to follow the motion of such an element as it flows. It is more convenient to formulate a version of Newton's law that can be applied to a succession of fluid elements flowing through a particular region, for example flowing through the section between planes I and 2 inFigure 1.3. To understand how an appropriate momentum equation can be derived, consider first a stationary read more..

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    FLUIDS IN MOTION 19 When each mass is brought to rest its momentum is destroyed and a corresponding impulse is thereby imposed on the tank. As the input of a succession of masses increases towards a steady stream, the impulses merge into a steady force. This is also the case with the stream of liquid: the fluid's momentum is destroyed at a constant rate and by Newton's second law there must be a force acting on the fluid equal to the rate of change of its momentum. If there is no accumulation of read more..

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    20 FLUID FLOW FOR CHEMICAL ENGINEERS Thus a force equal to M(u2- u~) must be applied to the fluid. This force is measured as positive in the positive x-direction. These equations are valid when there is no accumulation of momentum within the section. When accumulation of momentum occurs within the section, the momentum equation must be written as net force acting on the fluid - rate of change of momentum = momentum flow rate out of section - momentum flow rate into secton + rate of accumulation read more..

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    FLUIDS IN MOTION 21 By continuity But M2u2 -Mlus - M(u2 - ul) plUlSl ---- P2u2S2 Pl = P2 and $2 <$1 Therefore u2>ul Thus, the rate of change of momentum is positive and by Newton's law a positive force must act on the fluid in the section, ie a force in the positive x-direction. If the flow were reversed, the force would be reversed. The above example shows the effect of a change in pipe diameter, and therefore flow area, on the momentum flow rate. It is clear that for steady, fully read more..

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    22 FLUID FLOW FOR CHEMICAL ENGINEERS ul Yt__ x Figure 1.8 Reaction components acting on a pipe bend due to the change in fluid momentum y-component: y-momentum flow rate out = 0 y-momentum flow rate in = M(-ul) = -Mul Note: the minus sign arises from the fact that the fluid flows in the negative y-direction. Thus the rate of change ofy-momentum is Mu~ and the force acting on the fluid in the y-direction is equal to Mul. There is therefore a reaction Ry of magnitude Mu~ acting on the pipe in the read more..

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    FLUIDS IN MOTION 23 1.6.1 Laminar flow The cases considered so far are ones in which the flow is turbulent and the velocity is nearly uniform over the cross section of the pipe. In laminar flow the curvature of the velocity profile is very pronounced and this must be taken into account in determining the momentum of the fluid. The momentum flow rate over the cross sectional area of the pipe is easily determined by writing an equation for the momentum flow through an infinitesimal element of area read more..

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    24 FLUID FLOW FOR CHEMICAL ENGINEERS air as a jet from the end of the tube. What is the relationship between the diameters of the jet and the tube? Calculations It is assumed that the Reynolds number is sufficiently high for the fluid's momentum to be dominant and consequently the momentum flow rate in the jet will be the same as that in the tube. On emerging from the tube, there is no wall to maintain the liquid's parabolic velocity profile and consequently the jet develops a uniform velocity read more..

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    FLUIDS IN MOTION 26 Example 1.6 Figure 1.9 illustrates a nozzle at the end of a hose-pipe. It is convenient to align the x-coordinate axis along the axis of the nozzle. The y-axis is perpendicular to the x-axis as shown and the x--y plane is vertical. It is necessary first to define the region or 'control volume' for which the momentum equation is to be written. In this example, it is convenient to select the fluid within the nozzle as that control volume. The control volume is defined by read more..

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    26 FLUID FLOW FOR CHEMICAL ENGINEERS As before, the rate of change of the fluid's x-component of momentum is M(u2- Ul), SO the net force acting on the fluid in the x-direction is equal to M(u2 - ul). There is no change of momentum in the y or z directions. The momentum equation can now be written but it must include the unknown reaction between the fluid and the nozzle. The unknown reaction of the nozzle on the fluid is denoted by Fx and for convenience (to show it acting on the fluid across the read more..

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    FLUIDS IN MOTION 27 that this force is required by the fact that the exterior control surface cuts through the bolts of the coupling. Similarly, if there were a restraining bracket the force exerted by it on the control volume would be incorpo- rated in the force-momentum balance. In a case such as this, the force of the atmosphere on the surface of the nozzle can be simplified by using a cylindrical control volume shown by the dotted line in Figure 1.9(b). Assuming the thickness of the nozzle read more..

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    28 FLUID FLOW FOR CHEMICAL ENGINEERS --ts h J. n •• h ~ ............. , ) ) (a) Figure 1.10 Shearing of a solid material (a) Sample of area A Co) Cross-section showing displacement /: F plate as shown in Figure 1.10, the solid material is deformed until the resulting internal forces balance the applied force. In order to keep the lower plate stationary it is necessary that a restraining force of magnitude F acting in the opposite direction be provided by the fixture. The direction of the read more..

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    FLUIDS IN MOTION 29 'i iI / / / / / "q= F -,,.' \ ...... ~, -~,,' \ _/ Figure 1.11 Sheafing of an element of material 8y i_._. X internal force acting on the upper surface of the slice acts in the direction of the applied force. Below the slice, the material is displaced less far and this lower material therefore exerts a force in the opposite direction. If the slice at distance y from the fixed plate is displaced a distance x from its unstressed position, the shear strain ~, is equal to read more..

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    30 FLUID FLOW FOR CHEMICAL ENGINEERS 1.7.2 Newton's law of viscosity In contrast to the behaviour of a solid, for a normal fluid the shear stress is independent of the magnitude of the deformation but depends on the rate of change of the deformation. Gases and many liquids exhibit a simple linear relationship between the shear stress 9 and the rate of shearing: d~ T.= ~-~ = g.9 (1.32) This is a statement of Newton's law of viscosity and the constant of proportionality /~ is known as the read more..

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    FLUIDS IN MOTION 31 X _f _f "- "f= d"y "~Upper plate Figure 1.12 Shearing of a fluid showing the velocity gradient In general, the velocity profile will be curved but as equation 1.33 contains only the local velocity gradient it can be applied in these cases also. An example is shown in Figure 1.13. Clearly, as the velocity profile is curved, the velocity gradient is different at different values of y and by equation 1.32 the shear stress ~-must vary with y. Flows generated read more..

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    32 FLUID FLOW FOR CHEMICAL ENGINEERS =...__ v X d~ " ~ $ = d-7 / ] __2' s element Figure 1.13 The varying shear rate for a curved velocity profile rL__ X Figure 1.14 Typical velocity profile for flow in a pipe--the annular element shown is used in the analysis of such flows will carry its momentum to that layer and tend on average to accelerate the slower layer. Similarly, when a molecule diffuses from a slow layer to a faster one it will retard the faster layer. This molecular diffusion read more..

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    FLUIDS IN MOTION 33 pipe flow shown in Figure 1.14, the flow is caused by the imposition of a higher pressure upstream (ie at the left) than downstream. By virtue of the no-sfip condition at the pipe wall, the fluid there must be stationary but layers of fluid closer to the centre line have successively higher velocities. For the element shown, the fluid nearer the wall retards the element while that closer to the centre drags the element in the direction of flow. A steady state is achieved when read more..

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    34 FLUID FLOW FOR CHEMICAL ENGINEERS The upstream pressure P~ acts over the cross-sectional area of the element, so that the force acting on the element in the direction of flow is given by force acting in flow direction - ~Tr~iPl (].34) where the radius of the element is the same as the inside radius ri of the pipe. The downstream pressure P2 acts on the element against the flow, as does the drag of the pipe wall on the fluid. The shear stress at the wall is called the wall shear stress and is read more..

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    FLUIDS IN MOTION 35 in the x-direction. The first subscript, here r, indicates that the stress component acts on the surface normal to the r-coordinate direction. (Specifying the normal to a surface is the easiest way of defming the orientation of the surface.) The shear stress component ~-,~ is caused by the shearing of the liquid such that the velocity component v,, varies in the r-direction. The wall shear stress ~-,~ is just the value of ~-,,, at the wall of the pipe. Example 1.8 How does read more..

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    38 FLUID FLOW FOR CHEMICAL ENGINEERS balance it is necessary for the shear stress r,x to be proportional to r. (Note that AP is uniform over the whole cross section.) It is important to note that in deriving the shear stress distribution no assumption was made as to whether the fluid was Newtonian or whether the flow was laminar. In the case of turbulent flow, it is the time-averaged values of Tr~ and r,o that are given by equations 1.40 and 1.41. In Section 1.13 these time-averaged stresses read more..

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    FLUIDS IN MOTION 37 force acting in the direction of motion and this can be done in two equivalent ways: (i) net force = sum of forces acting in direction of motion (ii) net force = propulsive forces- retarding forces In (i) all forces are taken as acting in the direction of motion so the propulsive forces are positive but the retarding forces must be entered as negative quantifies. In (ii), the fact that the retarding forces act opposite to the direction of motion has been incorporated into the read more..

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    38 FLUID FLOW FOR CHEMICAL ENOINEER$ With the positive sign convention, Newton's law of viscosity is express- ed as Positive convention: d~ *'Y" = ~ dy (1.44a) but with the negative sign convention it must be expressed as Negative convention: d~x 9 yx - -~ dy (1.44b) r is the shear stress component associated with the velocity gradient dvfldy and corresponding equations hold for other components. In cylindrical coordinates, the velocity gradient d~fldr generates the shear stress component read more..

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    FLUIDS IN MOTION 39 Derivations The velocity profile must have a form like that shown in Figure 1.17. The velocity is zero at the pipe wall and increases to a maximum at the centre. From Example 1.8, it is known that the shear stress vanishes on the centre-line r = 0, so from Newton's law of viscosity (equation 1.45) the velocity gradient must be zero at the centre. The general arrangement of a representative element of fluid is shown in Figure 1.17. A cylindrical shell of fluid of length L has read more..

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    40 FLUID FLOW FOR CHEMICAL ENGINEERS On letting & tend to zero, mo( / l d (r+ &)r,~l,+8, rr _~ r& r dr -- (rT~x) (1.48) therefore 1 d PI-P2 r dr (rr = L (1.491 Integrating Pl -P2) r As *'~ = L T +-'-r (1.5o) The shear stress must remain finite at r = 0 so A] - 0. Thus the shear stress distribution is given by r ~'~ - L "2" - 2 (1.511 as found in Example 1.8. The upstream pressure P~ is greater than P2 so 9 ,x is positive, indicating that the shear stress components read more..

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    FLUIDS IN MOTION 41 and the velocity profile is (1.54) This is the equation of a parabola. A slightly different procedure is to substitute for ~'rx in equation 1.49 using Newton's law of viscosity. If this is done and the resulting equation integrated twice, equations 1.55 and 1.56 are obtained: dvx (Pl-P2) r A2 = L T +-r (1.55) PI-P2) r 2 -I~vx = .... L "' ~- +A21nr+B (1.56) There are two constants of integration in equation 1.56 so two boundary conditions are required. The first is the read more..

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    42 FLUID FLOW FOR CHEMICAL ENGINEERS This leads to and 1 d PI-P2 - -- --(rr, x) = (1.58) r dr L -~',x = L T = T (1.59) Equations 1.58 and 1.59 are equivalent to equations 1.49 and 1.51. It will be noted that the only difference is the sign of each term containing the shear stress r,~. P1 being greater than P2, equation 1.59 shows that r~ is negative, which indicates that the shear stress components act physically in the opposite directions to those employed in the sign convention. This is in read more..

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    FLUIDS IN MOTION 43 centre-line, only the outer surface, with ~,~ taken as positive in the negative x-direction remains. With the positive sign convention, ~-,,~ on that surface is taken as positive in the positive x-direction. This may be the easiest way to remember the sign conventions. 1.9 Stress components In the preceding section, only one stress component was considered and that component was the only one of direct importance in the simple flow considered. The force acting at a point in a read more..

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    44 FLUID FLOW FOR CHEMICAL ENGINEERS clarity, the stress components acting on the face normal to the x-axis are omitted from Figure 1.18. When using the positive sign convention, the direction of each stress component that is taken as positive is the opposite of that shown in Figure 1.18. Independent of the sign convention used, the stress components can be classified into two types: those that act tangentially to the face of the element and those that act normal to the face. Tangential read more..

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    FLUIDS IN MOTION 45 1.10 Volumetric flow rate and average velocity in a pipe The volumetric flow rate is determined by writing the equation for the volumetric flow rate across an infinitesimal element of the flow area then integrating the equation over the whole flow area, ie the cross-sectional area of the pipe. It is necessary to use an infinitesimal element of the flow area because the velocity varies over the cross section. Over the infinite- simal area, the velocity may be taken as uniform, read more..

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    46 FLUID FLOW FOR CHEMICAL ENGINEERS 8~ (1.64) This last result is known as the Hagen-Poiseuille equation. The volumetric average velocity u could be determined from equation 1.63 but as the expression for Q has already been found it is more convenient to determine u by dividing equation 1.64 by the flow area ~r~" /.i2 u = -~ \L] 3-~\-L--] (1.65) (1.66) On putting r = 0 in equation 1.54, the maximum velocity is '.'m,x = Thus, for a parabolic velocity profile in a pipe, the volumetric read more..

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    FLUIDS IN MOTION 47 Heat conduction is described by Fourier's law and diffusion by Fick's first law: heat flux dT k d qy - -k-~?.. = (pCpT) (1.69) dy uy molar diffusional flux JAy - -~ dCA dy (1.70) Equation 1.70 shows that the molar diffusional flux of component A in the y-direction is proportional to the concentration gradient of that compo- nent. The constant of proportionality is the molecular diffusivity ~. Similarly, equation 1.69 shows that the heat flux is proportional to the gradient of read more..

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    48 FLUID FLOW FOR CHEMICAL ENGINEERS Table 1.2 Viscosities of fluids at 25 ~ Fluid ~ (Pa s) v (m2/s) Air 1.8 x 10 -s 1.5 x 10 -~ Water 1.0 x 10- 3 1.0 x 10 -6 Mercury 1.5 x 10 -3 1.1 x 10 -7 Castor oil 0.99 1.0 x 10- 3 , , ,,,, ,,, i ,, 1.12 Non-Newtonian behaviour For a Newtonian fluid, the shear stress is proportional to the shear rate, the constant of proportionality being the coefficient of viscosity. The viscosity is a property of the material and, at a given temperature and pressure, is read more..

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    FLUIDS IN MOTION 49 / B/ingham plas S tt h Cea r thinning ~ Newtonian , ~ Shear thickening Shear rate .......... "- (~ • Bingham plastic ~ Shear thinning Newtonian ~kening r Shear rate (b) Figure 1.19 Flow curves for time-independent fluids (a) Shear stress against shear rate (b) Apparent viscosity against shear rate point such as that indicated, and Figure 1.19(b) shows the apparent viscosity for each type of behaviour corresponding to the curves in Figure 1.19(a). Fluids for which the read more..

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    50 FLUID FLOW FOR CHEMICAL ENGINEERS guide, dilute and moderately concentrated suspensions and solutions of macromolecules exhibit shear thinning behaviour, the suspended matter or the molecules tending to become aligned with the flow. Shear thicken- ing behaviour occurs most commonly with highly concentrated suspen- sions, in which progressively stronger interactions occur between the suspended particles as the shear rate increases. Examples of shear thinning fluids are polymer solutions and read more..

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    FLUIDS IN MOTION 81 the ~'-5' curve at the origin, ie a kink in the curve, for values of n other than unity. More importantly, with many fluids it is impossible to fit a single value of n tO the flow curve over a wide range of shear rates, partly because most fluids tend to Newtonian behaviour at very low deformation rates. Nevertheless, the flow behaviour of some materials can be repre- sented quite well by the power law and, provided it is appreciated that it is merely a curve-fitting tool, read more..

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    52 FLUID FLOW FOR CHEMICAL ENGINEERS 3; Lower Newtonian region Po Upper Newtonian region p| log (shear rate) v Figure 1.20 Variation of apparent viscosity with shear rate for a polymer In equation 1.75, ~ and/z| are the values of the apparent viscosity for the lower and upper Newtonian regions respectively. The constant ~,~ is the shear rate evaluated at the mean apparent viscosity (/~o +/~| The second category, time-dependent behaviour, is common but dif- ficult to deal with. The best known read more..

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    FLUIDS IN MOTION 63 J Shear rate ,,,..._ v Figure 1.21 Flow curves for a thixotropic fluid That recovery which does occur reflects thixotropy: true thixotropy is reversible. Many food preparations and some paints are deliberately designed to be thixotropic so that the solid matter remains in suspension when the product is standing but, on being shaken, the apparent viscosity falls and the product can be poured. The opposite behaviour, increasing apparent viscosity during shear is called rheopexy read more..

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    54 FLUID FLOW FOR CHEMICAL ENGINEERS The rapid stopping may be due merely to the higher apparent viscosity but the recoil is a manifestation of solid-like behaviour: a purely viscous material cannot recoil. The simplest model that can show the most important aspects of viscoelastic behaviour is the Maxwell fluid. A mechanical model of the Maxwell fluid is a viscous element (a piston sliding in a cylinder of oil) in series with an elastic element (a spring). The total extension of this mechanical read more..

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    FLUIDS IN MOTION 56 By inspecting equations 1.79 and 1.80, it is clear that purely viscous behaviour corresponds to .A = 0 and purely elastic behaviour is approached as A--, oo. The relaxation time has a physical origin, being related to the time taken for molecules or particles to change orientation in response to the applied stress. In Chapter 3 it is shown that the response of a viscoelastic fluid depends on how rapidly it is deformed in relation to the relaxation time: when the deformation read more..

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    56 FLUID FLOW FOR CHEMICAL ENGINEERS The word 'eddy' is simply a convenient term to denote an identifiable group of fluid elements having a common motion, whether that motion be shearing, stretching or rotation. The eddies have a wide range of sizes: in pipe flow the largest eddies are comparable in size to the diameter of the pipe, while the size of the smallest eddies will be typically 1 per cent of the pipe diameter. The various sizes of eddy also have different characteristic speeds and read more..

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    FLUIDS IN MOTION 57 The properties of the turbulence are different at the two extremes of the scale of turbulence. The largest eddies, known as the macroscale turbu- lence, contain most of the turbulent kinetic energy. Their motion is dominated by inertia and viscosity has little direct effect on them. In contrast, at the microscale of turbulence, the smallest eddies are domin- ated by viscous stresses, indeed viscosity completely smooths out the microscale turbulence. 1.13.1 Velocity read more..

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    58 FLUID FLOW FOR CHEMICAL ENGINEERS component v, at a point can be represented as the sum of the mean velocity ~ at that point and the instantaneous fluctuation v~ from the mean: v~ - Ox + v~ (1.82) For turbulent flow near the axis of a pipe, the fluctuation v~ will not exceed about + 10 per cent of the mean value. The time-averaged value ~ can be calculated from 1 I r+Au2 = Vx dt (1.83) ~3x A't T-At/2 This equation is an ordinary definition of an average but it is for the average at time T at read more..

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    FLUIDS IN MOTION 59 two velocity components. It follows from equation 1.88 that turbulent flow possesses more kinetic energy than it would w~th the same mean velocity but no fluctuations. Just as the velocity fluctuations give turbulent flow extra kinetic energy, so they generate extra momentum transfer. Consider the transfer of x-component momentum across a plane of area 8ySz perpendicular to the x-coordinate direction. The momentum flow rate is the product of the mass flow rate (ffo~SySz) read more..

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    80 FLUID FLOW FOR CHEMICAL ENGINEERS Writing the instantaneous velocity components vx, vy as the sums of the mean values and fluctuations, and taking the time average gives the mean momentum flux as: mean x-momentum flux in y-direction = p(oy + v.D(o + v;,) = p~y~3,, + po~v~ (1.93) In general, the time-averaged value of the product of the fluctuations is non-zero so there is an additional flux of x-momentum in the y-direction I due to the velocity fluctuations v~ and ~y. This momentum flux is read more..

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    FLUIDS IN MOTION 61 Yl / l/ / __/ 7 / x Figure 1.23 Velocity fluctuation in the y-direction across the gradient of the mean velocity When the mean velocity gradient dOx/dy is negative, a velocity fluctuation v~ will produce a fluctuation ~ of the same sign and the Reynolds stress will be positive. It follows that if an element of fluid moves in they-direction in a region where the mean velocity gradient d0~/dy is zero, a fluctuation ~ gives rise, on average, to a zero fluctuation ~. The read more..

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    62 FLUID FLOW FOR CHEMICAL ENGINEERS desirable to attempt to relate the mean values of the turbulent fluxes to the corresponding gradients of the mean profiles. In order to do this, it is necessary to introduce the concept of eddy diffusivities. By analogy wit.h the flux equations for purely molecular transfer, equations 1.68 to 1.70, mean turbulent flux equations can be written as momentum flux ~y~ - -~v+ e)d~ dy (1.95) -- 0i" heat flux qy = -oCp(a + en)-~y (1.96) dCA molar flux JAy = - (~ read more..

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    FLUIDS IN MOTION 63 Laminar Figure 1.24 Comparison of the time-averaged velocity profile for turbulent flow and the profile for laminar flow at the same volumetric flow rate length of 8 rn is 14 000 Pa. For water, the value of the kinematic viscosity v is 1 x 10 -6 mZ/s. It may be assumed that the mean velocity profile is described by Prandtl's l/7th power law over most of the cross section. Calculations The eddy kinematic viscosity e is defined by 9 ,~ = -p(v+ e) dex from (1.95) dr Prandd's read more..

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    64 FLUID FLOW FOR CHEMICAL ENGINEERS m Trx 1" -- =- from (1.41) r~ ri Using these equations, Table 1.3 can be constructed. Table 1.3 i i r/ri ~ (Pa) dO~/dr (s-l) ~ (m2/s) e/v 0.1 4.38 -19.1 0.23 x 10 -3 230 0.2 8.75 -21.1 0.41 x 10 -3 410 0.4 17.50 -27.0 0.65 x 10 -3 650 0.6 26.25 -38.2 0.68 x 10 -3 680 0.8 35.00 -69.3 0.51 x 10 -3 510 0.9 39.38 - 125 0.31 x 10 -3 310 , l lll i , These values of e may be compared with the value of the molecular kinematic viscosity v (1 x 10 -6 mZ/s), e is read more..

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    FLUIDS IN MOTION 6S this is known as a boundary layer. As a result, it is possible to treat the turbulent flow as two regions: the boundary layer where viscosity has a significant effect, and the region outside the boundary layer, known as the free stream, where viscosity has no direct influence on the flow. This artificial segregation allows considerable simplification in the analysis of turbulent flow. Figure 1.25 shows the boundary layer that develops over a flat plate placed in, and aligned read more..

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    66 FLUID FLOW FOR CHEMICAL ENGINEERS laminar boundary layer. After the transition, the structure of the bound- ary layer is more complex: the flow in most of that part of the boundary layer is turbulent and hence it is called a turbulent boundary layer. However, in the turbulent boundary layer there is a very thin layer of fluid adjacent to the solid surface where turbulent stresses are negligible and the flow is dominated by viscous stresses: this is known as the 'viscous sublayer'. The viscous read more..

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    FLUIDS IN MOTION 67 moving downstream slowly, compared with the surrounding fluid, erupts away from the wall into the main part of the turbulent boundary layer, ie the viscous sublayer appears to burst. An inrush is the opposite of a burst: relatively rapidly moving fluid rushes in towards the wall. More recently, 'horseshoe' or 'hairpin' vortices have been observed. The general picture is that elongated patches of fluid (streaks) having a low mean velocity, but with large fluctuations, appear read more..

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    68 FLUID FLOW FOR CHEMICAL ENGINEERS It was noted earlier that the Reynolds stress is negative when the mean velocity gradient is positive and vice versa, consequently turbulent energy production is always positive. Very close to the wall, the Reynolds stress f f PVx~r is small, while far from the wall the Reynolds stress and the mean velocity gradient d#~ldr are small, so in both these regions there is very little production of turbulent energy. In the buffer or generation zone, neither read more..

  • Page - 86

    FLUIDS IN MOTION 69 The distribution of the shear stress components is shown schematically in Figure 1.26. For clarity, the magnitude of the viscous stress is exaggerated in Figure 1.26. References Cross, M.M., Rheology of non-Newtonian fluids: a new flow equation for pseudoplastic systems,Journal of CoUoM Solace, 20, pp. 417-37 (1965). Kline, S.J., Reynolds, W.C., Schraub, F.A. and Runstadler, P.W., The structure of turbulent boundary layers,yournal of FluidMechanics, 30, pp. 741-73 (1967). read more..

  • Page - 87

    Flow of incompressible Newtonian fluids in pipes and channels 2.1 Reynolds number and flow patterns in pipes and tubes As mentioned in Chapter 1, the first published work on fluid flow patterns in pipes and tubes was done by Reynolds in 1883. He observed the flow patterns of fluids in cylindrical tubes by injecting dye into the moving stream. Reynolds correlated his data by using a dimensionless group later known as the Reynolds number Re: Re- C1.3) p In equation 1.3, p is the density, ~ the read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 71 2.2 Shear stress in a pipe Consider steady, fully developed flow in a straight pipe of length L and internal diameter d~. As shown in Example 1.8, a force balance on a cylindrical element of the fluid can be written as ~n~AP/- 2~L~-,~ = 0 (2.3) where APf is the frictional component of the pressure drop over the pipe length L. In the case of fully developed flow in a horizontal pipe ~LP/is the only component of the pressure drop, see equations 1.16 and read more..

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    72 FLUID FLOW FOR CHEMICAL ENGINEERS Owing to its complexity, turbulent flow does not admit of the simple solutions available for laminar flow and the approach to calculating the pressure drop is based on empirical correlations. It was noted in Section 1.3 that the frictional pressure drop for turbulent flow in a pipe varies as the square of the flow rate at very high values of Re. At lower values of Re the pressure drop varies with flow rate, and therefore with Re, to a slightly lower power read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 73 Fanning friction factor, which is defined by equation 2.10, equation 2.9 may be written as (LtpuZ 2JLpu2 AP/= 4f ~ 2 = d, (2.13) This is the basic equation from which the frictional pressure drop may be calculated. It is valid for all types of fluid and for both laminar and turbulent flow. However, the value off to be used does depend on these conditions. Although it is unnecessary to use the friction factor for laminar flow, exact solutions being read more..

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    74 FLUID FLOW FOR CHEMICAL ENGINEERS Figure 2.1 Friction factor chart for Newtonian fluids. (See Friction Factor Charts on page 349.) greater the relative roughness, the higher the value of f for a given value of Re. At high values of Re, the friction factor becomes independent of Re; this is true for the region of the chart above and to the right of the broken line. The reason for this behaviour is discussed at the end of Section 2.9. In the region of transition between laminar and turbulent read more..

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    FLOW OF iHCOMPRES$16LE NEWTONIAN FLUIO$ 75 fi/2 = 4.06 log + 2.16 A very useful correlation has been given by Haaland (1983): (2.18) 1 It e till 09] fi/2 -- --3.6 log 3.7di +'~e (2.19) Equation 2.19 has the advantages of giving f explicitly and being adequately accurate over the whole range of turbulent flow. Use of the friction factor chart or a correlation such as equation 2.19 enables calculation of the frictional pressure drop for a specified flow rate from equation 2.13. The inverse problem read more..

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    76 FLUID FLOW FOR CHEMICAL ENGINEERS 101o lO' 103 10" 105 e = Absolute roughness, m d~ = Inside pipe diameter, m ,0005 ,0001 e/d, lo a lo' 107 lo 6 ~ lO' lO3 2x102 10 s 10" 10 s 1(# ~ runberRe Figure 2.2 Plot of ~ fRe t against Reynolds number lo 7 lo s lo s 10` Calculations Q mean velocity u = 7rd2/4 From the given values (from 1.6) read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 77 4 (3.142)(0.0526 m) 2 = 0.002173 m 2 9.085 m3/h 0 = 3600 s/h = 0.002 524 m% Therefore 0.002 524 m3/s u =0.002 i73 m 2 = 1.160 m/s The Reynolds number is given by pud Re-~ /z Substituting the given values (1200 kg/m3)(1.160 rrds)(0.0526 m) Re= 0.01 Pas = 7322 (1.3) Relative roughness is given by e 0.000045 m d; 0.0526 m = 0.000856 From the graph of f against Re in Figure 2.1, f= 0.0084 for Re = 7322 and e/di = 0.000 856. The frictional pressure drop is read more..

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    78 FLUID FLOW FOR CHEMICAL ENGINEERS Example 2.2 Estimate the steady mean velocity for a commercial steel pipe with the following characteristics: length L inside diameter di wall roughness e frictional pressure drop AP liquid dynamic viscosity/~ liquid density p = 30.48 m = 0.0526 m = 0.000045 m = 15720 N/m 2 =0.01 Pas = 1200 kg/m 2 Calculations fRe 2 = 2L~ 2 Substituting the given values d, p f 2L/~ 2 (0.0526 m)3(1200 kgtm3)[15 720 kg/(s z m)] 2(30.48 mX0.01 Pa s) 2 (2.21) = 4.503 x l0 s read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 79 Given a suitable algebraic correlation such as equation 2.19, the friction factor chart might be considered obsolete. Both f and fRe z can be represented algebraically as functions of Re allowing both types of calculations to be done. In the case of the inverse problem, that is the calculation of the flow rate for a specified pressure drop, an alternative is to use an iterative calculation, a procedure that is particularly attractive with a pocket read more..

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    80 FLUID FLOW FOR CHEMICAL ENGINEERS This quantity gives a measure of convergence to the specified value. In this case, the calculation might be stopped after the fourth step, when the error is 0.0026, ie 0.26 per cent. The calculation converges to a value of Re (and hence u) very close to the value in Example 2.2. There is no point iterating beyond a discrepancy of about 1 per cent because the correlations are no better than this. It should be borne in mind that frictional pressure drop read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 81 If the frictional losses were expressed as the head loss, hi= AP/Ipg, then the quantity 4fL,/d; would multiply uZ/2g. Thus 4fL,/di is the total number of velocity heads lost. Consequently, an alternative presentation of frictional losses for fittings is in terms of the number of velocity heads K lost for each fitting. In this case, the total frictional pressure drop may be calculated as ,u, AP/= 4f --~ + Y.,K 2 = (2.23) In equation 2.23, the first term read more..

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    82 FLUID FLOW FOR CHEMICAL ENGINEERS Q | li Figure 2.3 Row through a sudden expansion These expressions for the losses due to sudden expansion and sudden contraction can be derived by application of Bernoulli's equation and the momentum equation. Figure 2.3 shows a sudden expansion. The control surface is taken as shown on the inside surface between planes I and 2. It might be thought that plane 1 should be placed upstream of the expansion but if this were done it would be necessary to include a read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 8S equation 2.26, the pressure drop can be eliminated between equations 2.26 and 2.27 and the result rearranged to give Ill -- (Ul -- U2)212g (2.28) Now AP, = pgh/ (2.29) so equation 2.24 is obtained. Note that if plane I had been placed inside the pipe just upstream of the expansion, the pressure being the same as at the expansion, the momen- tum equation would be written as PISI + PI(S2 -- SI) - P2S2 -- M(u2 - Ul) (2.30) where the first term is the read more..

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    84 FLUID FLOW FOR CHEMICAL ENGINEERS the pressure distribution is unknown in the region of the contraction. However, the analysis for the sudden expansion can be applied beyond the vena contracta but it is necessary to have experimental values of the contraction ratio. 2.4.1 Pressure drop in coils A number of equations have been proposed for use in the calculation of pressure drop in coils of constant curvature [Srinivasan ~t al (1968)]. The latter are known as helices. For laminar flow, Kubair read more..

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    FLOW OF INCOMPRESSlDLE NEWTONIAN FLUIDS 85 Consider a pipe of circular cross section with an inside and an outside diameter of d~ and do respectively. Let this pipe be placed symmetrically inside a larger pipe having an inside diameter of D; and let a fluid flow through the annulus. Since the shear stress resisting the flow of fluid acts on both walls of the annulus, the appropriate flow perimeter required to calculate the equivalent diameter of the annulus d~ is (r r There- fore 4[( r / 4) - ( read more..

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    86 FLUID FLOW FOR CHEMICAL ENGINEERS Note that the volumetric average velocity u is exactly half the maximum velocity. 2.7 Kinetic energy in laminar flow At a point where the velocity is Vx, the kinetic energy per unit volume is equal to pv~2/2. The volumetric flow rate through an element of area of infinitesimal width #r is equal to 2r Thus the flow rate of kinetic energy 8(KE) through the element of area is given by 8(KE) = 2 r = r (2.36) The total flow rate of kinetic energy through the whole read more..

  • Page - 104

    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 87 cross section, corresponding to equation 1.67 for laminar flow, may be written as ~x (i r)I/7 (2r) 1/7 ....... = -- = I- (2.40) f~max ri dii Equation 2.40 is an empirical equation known as the one-seventh power velocity distribution equation for turbulent flow. It fits the experimentally determined velocity distribution data with a fair degree of accuracy. In fact the value of the power decreases with increasing Re and at very high values of Re it falls read more..

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    88 FLUID FLOW FOR CHEMICAL ENGINEERS The volumetric average velocity is defined by Q = ~rri2u, so 49 u = 60 #max 0.82#max (2.45) The average kinetic energy per unit mass can also be found as for laminar flow. The kinetic energy flow rate is given by KE = 7tO (O~)3rdr (2.37) o Substituting for O~ from equation 2.40, the kinetic energy flow rate is given by KE ~pf r' (Omax)3(l r) 3`7 = - - rdr (2.46) o ri This integral can be evaluated "conveniently by changing variables using equations 2.41 read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 89 2.9 Universal velocity distribution for turbulent flow in a pipe Consider a fully developed turbulent flow through a pipe of circular cross section. A turbulent boundary layer will exist with a thin viscous sublayer immediately adjacent to the wall, beyond which is the buffer or generation layer and finally the fully turbulent outer part of the boundary layer. In the viscous sublayer, the magnitude of the time-averaged value of the shear stress ~ is read more..

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    90 FLUID FLOW FOR CHEMICAL ENGINEERS and y+ = v,y (2.57) /2 so equation 2.55 can also be written as v + =y+ (2.58) The dimensionless distance y+ has the form of a Reynolds number. Equation 2.58 fits the experimental data in the range 0 -< y+ -< 5. In the viscous sublayer, the velocity increases linearly with distance from the wall. Conditions in the fully turbulent outer part of the turbulent boundary layer are quite different. In a turbulent fluid, the shear stress q is given by read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 91 fluctuation in the positive y-direction will retard the faster fluid, thus producing a negative fluctuation. Prandtl assumed further that the fluctuations in the x and y-directions are of the same order of magnitude, which is now known to be true for this type of flow. Consequently, the magnitude of the Reynolds stress is given by (d~3~) 2 dy - Vo~,v~ ~pr (2.62) Comparing equations 2.59 and 2.62, it follows that the eddy kinematic viscosity can be read more..

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    92 FLUID FLOW FOR CHEMICAL ENGINEERS where C 2 is another constant. Equation 2.68 is not applicable near the wall because it neglects the viscous shear stress and consequently gives ~5~ - -oo instead of ~ x = 0 at y = 0. Rewriting equation 2.68 in terms of v + = Oily, and y+ = pv,y/~ gives v+ 1 = ~, ln y + + C (2.69) where C is a constant. Equation 2.69 fits the experimental data for turbulent flow in smooth pipes of circular cross section for y+> 30 when I/K and C are given the values 2.5 read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 93 for smooth pipes is followed and the wall is said to be hydraulically smooth. If the roughness is large enough to protrude through the viscous sublayer into the buffer zone, the protuberances will be subject to form drag as well as the viscous drag. The proportion of form drag will increase as the roughness protrudes further and eventually, when it reaches right into the fully turbulent zone, form drag on the protuberances will be dominant. Under these read more..

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    94 FLUID FLOW FOR CHEMICAL ENGINEERS 2.10 Flow in open channels Consider a liquid flowing in an open channel of uniform cross section under the influence of gravity. The liquid has a free surface subjected only to atmospheric pressure. If the flow is steady, the depth of the liquid is uniform and the hydraulic slope of the free liquid surface is parallel to the slope of the channel bed. Consider a length AL in Figure 2.6 in which the ~L Ah t --r Figure 2.6 Flow in an open channel frictional head read more..

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    FLOW OF INCOMPRESSIBLE NEWTONIAN FLUIDS 95 where the mean velocity u is proportional to the square root of the channel slope s. Equation 2.74 is frequently written in the form u = C ~/~ (2.75) which is known as the Chezy formula. The Chezy coefficient C is C= 2/~ (2.76) Manning and others gave values of C for various types of surface roughness [Barna (1969)]. A typical value for C when water flows in a concrete channel is 100 m t/2/s. In general, liquids such as water which commonly flow in open read more..

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    Flow of incompressible non-Newtonian fluids in pipes Among other characteristics, non-Newtonian fluids exhibit an apparent viscosity that varies with shear rate. Consequently, the determination of the shear stress-shear rate curve must be an initial consideration. Although the apparent viscosity of a thixotropic or "a rheopectic fluid changes with the duration of shearing, meaningful measurements may be made if the change is relatively slow. Viscoelastic fluids also exhibit behaviour that read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 97 C_ t_~ ta f ,\, .... I / I-----~r ! Plate (disc) ,., ;/t,'//,',;," Figure 3.1 Cone and p/ate viscometer plate. If the cone is rotated at a constant angular rate, the fluid is sheared with every element of fluid describing a horizontal circular path. The tangential velocity component vo varies linearly from zero at the lower plate to the speed of the cone at the cone's surface. At a radial distance r, the cone's tangential speed is fir read more..

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    98 FLUID FLOW FOR CHEMICAL ENOINEERS The couple acting on either the cone or the plate may be measured as they are equal but act in opposite directions. Thus ~o in equation 3.3 is strictly the magnitude of the shear stress. Dividing equation 3.3 by equation 3.1 gives an expression for the apparent viscosity: 3C a ~~ = 2~rR ~ II (3.4) Note that a must be in radians and ~1 in radians per second. If the rotational speed is measured as N revolutions per minute (rpm), then the required conversion is read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 99 Various arrangements at the bottom of the inner cylinder are available: in Figure 3.2 an indentation is provided so that an air gap is formed and shearing in the sample below the inner cylinder is negligible. Another arrangement is to make the bottom of the inner cylinder a cone. When one of the cylinders is rotated, a Couette flow is generated with fluid particles describing circular paths. The only non-zero velocity component is vo and it read more..

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    100 FLUID FLOW FOR CHEMICAL ENGINEERS Again, fl is in rad/s and may be calculated from equation 3.5. From equation 3.6, the magnitude of the shear stress can be calculated from the measured couple C on either cylinder: C C ~"~ ~ 2 ~rR ~ h ~ 2 ~rR Z h (3.10) Assuming that the shear stress is calculated for the inner cylinder, the apparent viscosity is given by C(Ro-R~) t~,, ~ 2~rR~hfl (3.11) Equations 3.4 and 3.11 are unnecessary for the calculation of/~o because both r and ~ will be read more..

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    FLOW OF INCOMPRE881BLE NON-NEWTONIAN FLUIDS IN PIPES 101 under test is forced through the tube either by driving a ram through the reservoir or by applying gas pressure. In the case of testing a molten polymer, the reservoir might be a vertical heated barrel with a capillary tube fitted into its end. The capillary tube may be only about 50 mm long and have a diameter of 0.5 mm or less. A piston driven down the barrel extrudes the viscous molten polymer through the capillary tube. In contrast, read more..

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    102 FLUID FLOW FOR CHEMICAL ENGINEERS determined. In the case of flow in a vertical tube, the measured pressure drop must be corrected for the static head. In order to find the shear stress-shear rate relationship, ie to be able to plot a ~'-;/curve, it is necessary to know the shear rate at the wall. Thus, ~,, can be plotted against corresponding values of ~,~. The difficulty that arises is that ~,, is equal to the gradient of the velocity profile at the wall and the shape of the velocity read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 103 The volumetric flow rate through an annular element of area perpen- dicular to the flow and of width 8r is given by 8Q = 2m,'Sr.v,~ (1.61) and, consequently, the flow rate through the whole tube is I: Q-- 2r re, dr (1.62) Integrating by parts gives 9 rZ ~ Idr} Q-2~[~~'~];+f:'2( - dr, _ (3 14) Provided there is no slip at the tube wall, the first term in equation 3.14 vanishes. Now, the velocity gradient is equal to the shear rate ~, so read more..

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    104 FLUID FLOW FOR CHEMICAL ENGINEERS non-Newtonian fluid, as well as for a Newtonian fluid, the flow character- istic 8u/d~ is a unique function of the wall shear stress ~. The shear rate ~, can be extracted from equation 3.17 by differentiating with respect to ~. Moreover, if a definite integral is differentiated wrt the upper limit (here ~'w), the result is the integrand evaluated at the upper limit. It is convenient first to multiply equation 3.17 by ~ throughout, then differentiating wrt ~ read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 105 3 Plot ln(Su/d;) against ln~'w and measure the gradient at various points on the curve. [log(Su/di) and log~-~ may be used if more convenient.] Alternatively, calculate the gradient from the differences between the successive values of these quantities. 4 Calculate the true wall shear rate from equation 3.20 with the derivative determined in 3. In general, the plot of ln(Su/d~) against lnT-,~ will not be a straight line and the gradient read more..

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    106 FLUID FLOW FOR CHEMICAL ENGINEERS Using the given values 8u 32M di "rr(6 x 10 .3 m)3(870 kg/m 3) = 54200M s -l where the mass flow rate M is in kg/s. Using these expressions for r~, and 8u/di enables the values in the first two columns of Table 3.2 to be calculated. This provides the shear stress-flow characteristic curve. In order to determine the true shear rate at the wall it is necessary to use the Rabinowitsch-Mooney equation: 8u[~ 1 d ln(Su/d,)] (3.20) By plotting the calculated read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 107 120 d' 30 8u/d (s -~ ) 0 100 200 300 400 500 .... i ....... I ............. ! " ~ I~,~" 600 I I I I I 100 200 300 400 500 I~wl (s-') Figure 3.4 9 w - 8u/d~ and Tw - ~, curves for Example 3. I 600 and hence the derivative in equation 3.20; however, the reader should be aware that graphical differentiation is notoriously inaccurate and conse- quently requires great care. Other methods include fitting a low order polynomial equation read more..

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    108 FLUID FLOW FOR CHEMICAL ENGINEERS 3.3 Calculation of flow rate-pressure drop relationship for laminar flow using ~-# data Flow rate-pressure drop calculations for laminar non-Newtonian flow in pipes may be made in various ways depending on the type of flow information available. When the flow data are in the form of flow rate and pressure gradient measured in a tubular viscometer or in a pilot scale pipeline, direct scale-up can be done as described in Section 3.4. When the data are in the read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 109 d~AP (2.5) r,~= 4L (37 • 10 -3 m)(1100 Pa/m) = 10.18 Pa The volumetric average velocity u is given by equation 3.17 as 8u 4 f,, 2(_~)dr (3.17) It is necessary to evaluate the integral from r = 0 to r = 10.18 Pa. This can be done by calculating r 2~ for each of the values given in the table and plotting r2~ against I-. The area under the curve between r = 0 and r = 10.18 Pa can then be measured. An alternative, which will be used here, is read more..

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    110 FLUID FLOW FOR CHEMICAL ENGINEERS The above is the general method but in this case the viscometric data can be well represented by Thus = 0.749~ ~176 Pa ~, = 1.62,/.1.667 $-1 This allows the integral in equation 3.17 to be evaluated analytically. f~" /'~ ~ .~7 3s-, ~2~,d~ = 1.62 d~= 17510 Pa 0 This agrees with the value found by numerical integration and would give the same value for u. Note that the values of the apparent viscosity po were not used; they were provided to show that the read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 111 8u r -- = -- (3.22) Recall that the wall shear rate for a Newtonian fluid in laminar flow in a tube is equal to -Su/d~. In the case of a non-Newtonian fluid in laminar flow, the flow characteristic is no longer equal to the magnitude of the wall shear rate. However, the flow characteristic is still relateduniquely to ~ because the value of the integral, and hence the fight hand side of equation 3.17, is determined by the value of ,,~. If read more..

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    112 FLUID FLOW FOR CHEMICAL ENGINEERS t t J Figure 3.5 Decreasing diameter d. d+= d~ Turbulent flow I Laminar flow 8uld+ Shear stress at the pipe wall against flow characteristic for a non-Newtonian fluid flowing in a pipe Given a wall shear stress-flow characteristic curve such as that in Figure 3.5, the flow rate-pressure drop relationship can be found for any diameter of pipe provided the flow remains laminar and is within the range of the graph. For example, if it is required to calculate read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 113 --= y I tangent = n' I I I I I .... ~ ! : - Figure 3.6 Logarithmic plot of wall shear stress against flow characteristic: the gradient at a point defines n' In general, both K' and n' have different values at different points along the curve. The values should be found at the point corresponding to the required value of 1-,~. In some cases, the curve in Figure 3.6 will be virtually straight over the range required and a single value may be read more..

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    114 FLUID FLOW FOR CHEMICAL ENGINEERS smaller magnitude than the Newtonian value 5'wN. The value of the correction factor varies from 2.0 for n' = 0.2 to 0.94 for n'= 1.3. 3.5 Generalized Reynolds number for flow in pipes For Newtonian flow in a pipe, the Reynolds number is defined by ~d Re = ~----:' (1.3) /z In the case of non-Newtonian flow, it is necessary to use an appropriate apparent viscosity. Although the apparent viscosity ~o is defined by equation 1.71 in the same way as for a read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 118 = ' (3.27) The pipe flow apparent viscosity, defined by equation 3.31, is given by /'LAP = 8u/d~ (3.33) When this equation for /~op is substituted into equation 3.32, the generalized Reynolds number takes the form pu 2-"' d~" Re' = 8n, IK , (3.34) Use of this generalized Reynolds number was suggested by Metzner and Reed (1955). For Newtonian behaviour, K' = ~ and n' = 1 so that the generalized Reynolds number reduces to the read more..

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    provides another way of calculating the pressure gradient for a given flow rate for laminar non-Newtonian flow, instead of using the methods of Sections 3.3 and 3.4. 3.6.1 Laminar-turbulent transition 2500 A stability analysis made by Ryan and Johnson (1959) suggests that the transition from laminar to turbulent flow for inelastic non-Newtonian fluids occurs at a critical value of the generalized Reynolds number that depends on the value of n'. The results of this analysis are shown in Figure read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 117 ~0.01 0.001 _-I J !l ! i ! ..... 9 [F ,,, J ! ~i ~ i.,"..~ ~/g~~ ----"1 :'xperimental r;gio~s I " "il'[ Exptrapolated regions- 1000 r irt~o~o i::::i:!i:i~:~l~!~,oooo ~ Generalized Reynolds number, Re' Figure 3.8 Friction factor chart for purely viscous non-Newtonian fluids. (See Friction Factor Charts on page 349.) Source: D. W. Dodge and A. B. Metzner, AIChEJournalS, pp. 189-204 (1959) confirm the findings of Dodge read more..

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    118 FLUID FLOW FOR CHEMICAL ENOINEER8 Re' pudi Pap The flow characteristic is given by 0.32) 8u = 8(2.0 m/s) = 210 s -l d~ 0.0762 m and --1 = 210 (~176 = 0.0237 s ~ Hence ~p = (1.48 Pa s~ s ~ = 0.0351 Pa s and (0.0762 m)(2.0 m/s)(961 kg/m 3) Re' = = 4178 (0.0351 Pa s) From Figure 3.8, the Fanning friction factor f has a value 0.0047. Therefore the pressure drop is given by APf= 4f 2 = d, 2(0.0047)(3.048 m)(961 kg/m3)(2.0 m/s) z (0.0762 m) (2.13) = 1445 Pa 3.7 Power law fluids The methods read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 119 must not be confused with equation 1.72a defining a power law fluid. The relationship between n' and n, and K' and K will now be demonstrated. For the conditions at the pipe wall, denoted by the subscript w, the equation of the power law fluid can be written as r~ = K(-#~)n = K - -d-r-r (3.37) The minus sign has been placed inside the parentheses recognizing the fact that the shear rate y (equal to dvJdr) is negative. J'w is the true shear read more..

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    120 FLUID FLOW FOR CHEMICAL ENGINEERS Combining equations 2.4 and 3.41 gives the velocity gradient: dvx (APf) l`" dr = - 2LK rl/" (3.42) On integrating equation 3.42 with the boundary conditions v~ = 0 at r = ri, the velocity profile is found as v =(4KLAP;diI]/"( n r) ("+l The volumetric flow rate is readily calculated from Q- 2~" rv~dr (1.62) with vx given by equation 3.43. The result is n (3.44) and the volumetric average velocity u is equal to Q/wr/2. Consequently, read more..

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    FLOW OF INCOMPRESSIDLE NON-NEWTONIAN FLUIDS IN PIPES 121 It can be seen from equaton 3.45 that 7)ma x 3n + 1 = ------ (3.46) u n+l All the results for a power law fluid reduce to the corresponding ones for a Newtonian fluid on putting n = I and K =/~. 3.7.2 Velocity profile for turbulent flow in a pipe Dodge and Metzner (1959) deduced the velocity profile from their measurements of flow rate and pressure gradient for turbulent flow of power law fluids in pipes. For the turbulent core, the read more..

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    122 FLUID FLOW FOR CHEMICAL ENGINEERS This is in excellent agreement with equation 2.70, considering that equation 3.47 was derived from flow rate and pressure gradient measure- ments rather than from direct measurement of the velocity profiles. For Newtonian fluids the velocity profile in the viscous sublayer adjacent to the wall is V + =y+ (2.58) The corresponding equation for power law fluids is [Dodge and Metzner (1959)] v + = (y+)~/" (3.50) 3.7.3 Expansion and contraction losses for read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 123 3.8 Pressure drop for Bingham plastics in laminar flow The behaviour of a Bingham plastic is described by r- ~y = -//~ for r -> Ty and (1.73) q/=0 for "r < Ty As the shear stress for flow in a pipe varies from zero at the centre-line to a maximum at the wall, genuine flow, ie deformation, of a Bingham plastic occurs only in that part of the cross section where the shear stress is greater than the yield stress Zy. In the part read more..

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    124 FLUID FLOW FOR CHEMICAL ENGINEERS Govier (1959) developed a method for solving equation 3.55 for r and the pressure gradient for a given value of the Reynolds number. He defined a modified Reynolds number ReB in terms of/3: Ren = ~ (3.56) /3 It was pointed out in Section 1.12 that the coefficient of rigidity/3 is equal to the apparent viscosity at infinite shear rate. Gorier also defined a dimensionless yield number Y by y= ryd~ (3.57) /3u Dividing equation 3.57 by equation 3.56 and using read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 125 1C ~ 9 -~ 10- 10- 10 ~2 46 10 ~'2 46 1042 4 6 105 2 46 10 e 2 46 107 Reynolds number, Re. = pud~/~ Figure 3.10 Friction factor chart for laminar flow of Bingham plastic materials. (See Friction Factor Charts on page 349.) Source: D. G. Thomas, AIChEJoumal6, pp. 631-9 (1960) 3.9 Laminar flow of concentrated suspensions and apparent slip at the pipe wall The flow properties of suspensions are complex. The apparent viscosity at a given shear read more..

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    126 FLUID FLOW FOR CHEMICAL ENGINEERS usual approach is to treat the suspension as being uniform up to the wall but assuming that slip occurs at the wall. In reality there must be a very high velocity gradient but the thickness of the layer over which this occurs is unknown, although it may be expected to be of the order of the particle diameter. By supposing slip to occur, this high velocity gradient over a small distance is replaced by a discontinuity in the velocity at the wall. 3.9.1 read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 127 8u 8,), 4 Iv" di - di + ~ rz(-7)dr (3.66) jo Equations 3.65 and 3.66 reduce to equation 3.17 when v, = 0. It is important to remember that in these equations Q is the measured total flow rate and u is calculated from Q. When trying to determine the flow behaviour of a material suspected of exhibiting wall slip, the procedure is first to establish whether slip occurs and how significant it is. The magnitude of slip is then determined read more..

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    128 FLUID FLOW FOR CHEMICAL ENGINEERS u against r,,,, all the results for different tube diameters will fall on, or close to, a single line. If this is the case, the genuine flow behaviour cannot be determined but for pipe flow calculations the u-r,,, plot gives the informa- tion that is required for scale-up, assuming that the slip behaviour is the same in the large pipe. Having established that wall slip occurs but is not dominant, the procedure is to estimate the value of v, and hence read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 129 ! =l'er ~ Tw4 ~ Twl _._g.-.--- 1/~ Figure 3.12 Plot of apparent #uidity against I/~ to determine the slip velocity (Mooney's method) Q~ --- Q-Q, (3.71) The corrected volumetric average velocity u~ is given by Q~ (3.72) u.= This must be done for each of a range of values of the wall shear stress ~,~. The standard Rabinowitsch--Mooney equation can then be used with the corrected values of uc: 8uc[ 4 l d ln(8ucldi)] (3..73) This enables the read more..

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    130 FLUID FLOW FOR CHEMICAL ENGINEERS 8v, 8c. d,',-w = ,/2 C3.75) If this type of behaviour occurs, a plot of apparent fluidity against l/d~ at constant T,~ will be a straight line. The gradient of the line is equal to 8Cy for the corresponding value of T,~. Hence, v~ can be calculated from equation 3.74 and then the corrected flow rate as before. Several workers [see for example Cheng (1984)] have generalized Jastrzebski's method, writing the slip term in equation 3.67 in the form C/d'/'~. A read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 131 allow a good estimate of the genuine bulk flow behaviour to be obtained but use of these methods for calculating the flow rate in a pipe requires great caution, with the possible exception of very large pipes in which the effect of slip is negligible. The safest procedure is to make measurements in pipes as large as possible, made of the same material and having the same wall characteristics as the full-scale pipe. Not all suspensions will read more..

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    132 FLUID FLOW FOR CHEMICAL ENGINEERS stress pushes radially outwards. Consequently, when the liquid emerges from the end of the tube it swells under the influence of the normal stresses. This phenomenon, known as die swell, is in contrast to the contraction of the Newtonian liquid under similar conditions (see example 1.5). Another weft-known phenomenon is the Weissenberg effect, which occurs when a long vertical rod is rotated in a viscoelastic liquid. Again, the shearing generates a tension read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 133 Integrating Vx = kx + C (3.77) If Vx is negligibly small at x = 0, then C = 0. Noting that Vx = dx/dt, equation 3.77 may be integrated to give the position x of a material point: x - e kt (3.78) Thus, even when the elongation rate, as defined by equation 3.76, is constant, the separation of two material points increases exponentially with time. As stress relaxation occurs exponentially, it is clear that at high elongation rates the stress read more..

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    134 FLUID FLOW FOR CHEMICAL ENGINEERS 0.010 ~.. 0.006 ~ 0.0o5[ 1~ O.OO3- "E 0.00'2- 0.11111 turbulent I .... I \ I 5000 10000 20000 Reynolds number, Re ! 10ppm Figure 3.13 Friction factors for turbulent flow of dilute solutions of poly ethylene oxide Source: R. W. Patterson and F. H. Abemathy, Journal of Fluid Mechanics, 51, pp. 177-85 (1972) 70000 Drag reducing polymers are susceptible to degradation and consequent- ly find only limited application. However, suspensions of fibres, particu- read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 135 As noted in Chapter 1, purely viscous behaviour corresponds to ,~ = 0, while purely elastic behaviour is approached as ,~ ---, oo. It will now be shown that the response of a viscoelastic fluid to unsteady sheafing depends on how rapidly it is deformed compared with the rate at which it can relax. Imagine a Maxwell liquid placed between two parallel plates and sheared by moving the upper plate in its own plane. However, instead of moving read more..

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    136 FLUID FLOW FOR CHEMICAL ENGINEERS Equation 3.85 is valid after initial transients have died away. Comparing equation 3.85 with equations 3.80 and 3.81 shows that the first term in parentheses is in phase with the rate of strain, while the second term is in phase with the strain. Thus the first term represents the viscous part of the fluid's response and the second term the elastic part. It can be seen from equation 3.85 that and 9 -, -p.A~ocos~ot as Ao-,O (3.86) va ,'-, - --- sin ~ot = -GA read more..

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    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 137 It is interesting to consider the response of a Maxwell fluid to an arbitrary shear rate history. Denoting the shear rate as j(t), an arbitrary function of time, the equivalent of equation 3.83 is f! :tet/%. = -~ et/^~(t)dt (3.89) where the shearing at all times in the past is considered. In the integral of equation 3.89, t is a dummy variable so it can be changed to another variable t' without changing the value of the integral provided read more..

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    138 FLUID FLOW FOR CHEMICAL ENGINEERS The function O(t-t') may be interpreted as a memory function having a form as shown in Figure 3.14. For an elastic solid, 0 has the value unity at all times, while for a purely viscous liquid 0 has the value unity at the current time but zero at all other times. Thus, a solid behaves as if it 'remembers' the whole of its deformation history, while a purely viscous liquid responds only to its instantaneous deformation rate and is unin- fluenced by its read more..

  • Page - 156

    FLOW OF INCOMPRESSIBLE NON-NEWTONIAN FLUIDS IN PIPES 139 Ryan, N.W. and Johnson, M.M., Transition from laminar to turbulent flow in pipes, AIChEJournal, 5, pp. 433-5 (1959). Schofield, R.K. and Scott Blair, G.W., The influence of the proximity of a solid wall on the consistency of viscous and plastic materials, Journal of Physical Chon/stry, 34, No. 1, pp. 248-62 (1930). Skelland, A.H.P., Non-Nezvtonian Flow and Heat Transfer, New York, John Wiley and Sons Inc, 1967. Tanner, R., Engine~ng read more..

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    pumping of liquids 4.1 Pumps and pumping Pumps are devices for supplying energy or head to a flowing liquid in order to overcome head losses due to friction and also, i~ 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 Ah. If a pump is placed between points 1 and 2 in a pipeline, the heads for steady flow are related by equation 1.14 zz+----+2ga2P2g - ~" +--- ~ = Ah- hf (1.14) In equation 1.14, z, Pl(pg), and read more..

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    PUMPING OF LIQUIDS 141 ..F. ', 1 z~ Zs Arbitrary chosen base line , a, . Discharge side Suction side .... T FigUre 4.1 Typical pumping system Suction head: Discharge head" P$ h, = z, +--- - h/-, (4.2) pg hd f zd + P--Aa + hfd (4.3) Pg In equation 4.2, hf~ is the head loss due to friction, z, is the static head and P, 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, z, is negative. read more..

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    142 FLUID FLOW FOR CHEMICAL ENGINEERS The total head Ah which the pump is required to impart to the flowing liquid is the difference between the discharge and suction heads: Ah = hd- h, (4.4) Equation 4.4 can be written in terms of equations 4.2 and 4.3 as pg The head losses due to friction are given by the equations f(Y.L,, u 2 h:,= . ) and (4.5) (4.6) 4f(2L~a~ u z hfa= " ~, d~ /'-~g (4.7) where 2Le, and Y,L,a are the total equivalent lengths on the suction and discharge sides of the pump read more..

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    PUMPING OF LIQUIDS 143 Substituting for h, from equation 4.2, the available NPSH is given by P$ ~ PV NPSH --- ~, + ------- - hf, (4.9) Pg The available NPSH given by equations 4.8 and 4.9 must exceed the value required by the pump and specified by the manufacturer. The required NPSH increases with increasing flow rate as discussed below. 4.3 Centrifugal pumps In centrifugal pumps, energy or head is imparted to a flowing liquid by centrifugal action. The most common type of centrifugal pump is read more..

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    144 FLUID FLOW FOR CHEMICAL ENGINEERS ensure a smooth flow of liquid. The velocity head imparted to the liquid is gradually converted into pressure head as the velocity of the liquid is reduced. The efficiency of this conversion is a function of the design of ~he impeller and casing and the physical properties of the liquid. 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 read more..

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    PUMPING OF LIQUIDS 148 t 0---.~ Figure 4.3 Total head against capacity characteristic curve for a volute centrifugal pump (P P) ~h = (zd- z,) + ,-a- -,, + (h~d + h~,) (4.5) Pg Combining equation 4.5 with equations 4.6 and 4.7, which give the frictional head losses hf, and hfd respectively, allows the total head to be written as Ah = (zd-- zs) + (Pd- Ps) + 4f[(XLe, + Y.L.d) ] u 2 ....... d, (4.J]) The mean velocity u of the liquid is related to the volumetric flow rate or capacity Q by Q u = read more..

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    146 FLUID FLOW FOR CHEMICAL ENGINEERS For laminar flow, the Fanning friction factor fis given by equation 2.15 16 f= Re (2.15) Substituting for f in equation 4.12, the total head for laminar flow can be written as Ah = (za - z, ) + (Pd - ) + \-~ d~ u (4.13) Pg or as Ah = (zd- z,) + (Pd-P,) pg +\~_.~){32~ [(~,L~,+ ~,Lr (1rdQ/4) The system Ah against Q curve shown in Figure 4.4 can be plotted using equation 4.12 to calculate the values of the system total head Ah at each volumetric flow rate of read more..

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    PUMPING OF LIQUIDS 147 s imm s m = Im mm m m s m m G,m m m = "m = = = I 7= ~ -- in the system s/~ ,' by the pump / / ,,, , , ,t ,,,, , .... Figuro 4.,5 Available and required net positive suction heads against capacity in a pumping system In the above discussion it is assumed that the available NPSH in the system is adequate to support the flow rate of liquid into the suction side of the pump. If the available NPSH is less than that required by the pump, cavitation occurs and the normal read more..

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    148 FLUID FLOW FOR CHEMICAL ENGINEERS Normal pump curve for adequate suction condition System curve I I I I I I I I I Normal operating point ' Pump curve for insufficient I I available NPSH i LI 0"'4" Figure 4.6 Effect of insufficient NPSH on the performance of a centrifugal pump A centrifugal pump will operate normally at a point on its total head against capacity characteristic curve until the available NPSH falls below the required NPSH curve. Beyond this point, the total head read more..

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    PUMPING OF LIOUID8 140 Pump curve /////~ / System curve ---... / ~ with throttling //" ~rmal // J point Normol system curve o-.----.~ Figure 4.7 Effect of throttling the discharge valve on the operating point of a centrifugal pump I I I I / / I / I / I / / I / /// I t I ,,./ I i / I ," tl ol " u I o "-3 I go o[.I , , , | 0..--~ Figure 4.8 Effect of adding a safety factor to the system total head against capacity curve read more..

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    150 FLUID FLOW FOR CHEMICAL ENGINEERS Example 4.1 Calculate the values for a system total head against capacity curve for the initial conditions of the system shown in Figure 4.1 given the following data: dynamic viscosity of liquid density of liquid static head on suction side of pump static head on discharge side of pump inside diameter of pipe pipe roughness gas pressure above the liquid in the tank on the suction side of the pump gas pressure above the liquid in the tank on the discharge read more..

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    PUMPING OF LIQUID8 181 e f=0.0112 for Re-1578 and ~i=0.000856 (XL, + XL,d) 4.9 m + 63.2 m = = 1294.7 di 0.0526 m u 2 (1.0 m/s) 2 = 0.050 97 m ....- ---- 2g (2)(9.81 m/s 2) total head Ah = (zd- z,) + (Pa- P,) + 4f[(XL,, + XL,a) ] u2 = 4 m + 4(0.0112)(1294.7)(0.05097 m) -- 6.671 m Q mean velocity u = r / 4 ,rd 2 (3.142)(0.0526 m) z 4 4 ,rd, capacity Q = u .... 4 = 0.002173 m 2 = (1.0 m/s)(O.O02173 m 2) = 0.002173 m3/s = 0.00217 m3/s Repeating the calculations for other values of u gives the read more..

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    182 FLUID FLOW FOR CHEMICAL ENGINEERS 4.4 Centrifugal pump relations The power PE required in an ideal centrifugal pump can be expected to be a function of the liquid density p, the impeller diameter D and the rotational speed of the impeller N. If the relationship is assumed to be given by the equation PE = Cp~ bD ~ (4.16) then it can be shown by dimensional analysis [Holland and Chapman (1966)] that PE = CIP N3D5 (4.17) where C l is a constant which depends on the geometry of the system. The read more..

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    PUMPING OF LIQUIDS 163 A more satisfactory definition is that of a dimensionless specific speed N ~. The above deficiency can be removed by replacing equation 4.20 by equation 4.22: gab = C '4NZD z (4.22) where C ~ is a dimensionless constant. The dimensionless specific speed N ~ is given by N'~ = (gAb)3/4 (4.23) The value of N', is a unique number provided that consistent units are used. In SI units, the units are N in rev/s, Q in m~/s, h in m, and g has the value 9.81 m/s 2. The specific speed read more..

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    154 FLUID FLOW FOR CHEMICAL ENGINEERS ,,,,s. NPSH2 (~- (4.27) Equations 4.24 to 4.27 are the affinity laws for homologous centrifugal pumps. For a particular pump where the impeller of diameter D l is replaced by an impeller with a slightly different diameter De, the following equations hold [Holland and Chapman (1966)]: Q2 Ah, =/NI/2/DI '~2 Ah2 \~2 ] \D2 / (4.29) and PEI = ( Nl l3 ( Dl l 3 PE2 \~2 ] \~2 / (4.30) If the characteristic performance curves are available for a centrifugal pump read more..

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    PUMPING OF LIQUIDS 166 The ratio of impeller spreeds N~/Nz = 2 and the ratio of impeller diameters DI/1)2 -- 1/2. The ratio of capacities is given by o. Qz (4.24) = (2)(~) -- t The capacity of the second pump is Q2 = 4Ql = (4)(0.012 m%) = 0.048 m3/s The ratio of total heads is Ahz (~2 (4.25) = (4)(t) = 1 The total head of the second pump is &hz = Ahl = 70 m The ratio of powers is PEZ = (8)(~) = (4.26) Assume Then Pal P~l Psz P~2 Psi 1 --- ._. PB2 4 (This is equivalent to assuming that the read more..

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    lS6 FLUID FLOW FOR CHEMICAL ENGINEERS The ratio of required net positive suction heads is NPSHn- (N~)2~DI / 2 NPSH2 \'~z/ = (4X~) = ] Therefore net positive suction head of second pump NPSH2 = NPSHI = 18 m (4.27) 4.5 Centrifugal pumps in sedes and in parallel Diskind (1959) determined the operating characteristics for centrifugal pumps in parallel and in series using a simple graphical method. Consider two centrifugal pumps in parallel as shown in Figure 4.9. The L. I I I Or Pump curve (I) A~ read more..

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    PUMPING OF LIQUIDS 157 total head for the pump combination Ahr is the same as the total head for each pump, ie Ahr = Ahl = Ah2 (4.31) The volumetric flow rate or capacity for the pump combination Qr is the sum of the capacities for the two pumps, ie QT = QI + Q2 (4.32) The operating characteristics for two pumps in parallel are obtained as follows. 1 Draw the Ah against Q characteristic curves for each pump together with the system Ah~ against Q~ curve on the same plot as shown in Figure 4.9. 2 read more..

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    188 FLUID FLOW FOR CHEMICAL ENGINEERS Ah r I I p---Z~h, .... ~ ,~h 2 I I I i I I ,9 , .~- : o, i I L$h r Ah2 Ah~ Pump curve (I) curve (2) System curve Figure 4.10 Operating point for centrifugal pumps in series I I I I I -.I 9 % I I I I I I Qr 1 Draw the Ah against Q characteristic curves for each pump together with thesystem Ah, against Qs curve on the same plot as shown in Figure 4.10. 2 Draw a vertical constant capacity line in Figure 4.10 which intersects the two pump curves at total heads read more..

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    PUMPING OF LIQUIDS 169 An alternative to this trial and error procedure for two pumps in series is to calculate Ahr from equation 4.33 for various values of the capacity from known values of Ah] and Ahz at these capacities. The operating point for stable operation is at the intersection of the Ahr against QT curve with the Ah, against Q~ curve. The piping and valves may be arranged to enable two centrifugal pumps to be operated either in series or in parallel. For two identical pumps, series read more..

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    160 FLUID FLOW FOR CHEMICAL ENGINEERS the power from the motor. The idler gear runs free. As the rotating gears unmesh they create a partial vacuum which causes liquid from the suction line to flow into the pump. Liquid is carried through the pump between the rotating gear teeth and the fixed casing. The meshing of the rotating gears generates an increase in pressure which forces the liquid into the outlet line. In principle an external gear pump can discharge liquid either way depending on the read more..

  • Page - 178

    PUMPING OF LIQUIOS 161 kP = pAhg Substituting equation 4.10 into equation 4.35 gives /'E = pQzxhg which can also be written as (4.10) (4.36) P~ = M Ah g (4.37) since M = oQ is the mass flow rate. If M is in kg/s, Ah is in rn and the gravitational accelerating g = 9.81 m/s 2, PE is in W. The brake power Ps 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 (100) Pa = PE -~- (4.38) read more..

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    162 FLUID FLOW FOR CHEMICAL ENOINEERS 4.8 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. e In general, high viscosity liquids are pumped with positive displace- ment 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 read more..

  • Page - 180

    PUMPING OF LIQUIDS 163 Refer~ces Diskind, T., Solve muliple hookups graphically, Chemical Engineering, 66 p. 102 (2 Nov 1959). Holland, F.A. and Chapman, F.S., Pumping of Liquids, New York, Reinhold Publishing Corporation (1966). read more..

  • Page - 181

    5 Mixing O f liquids in tanks 5.1 Mixers and mixing Mixing may be defined as the 'intermingling of two or more dissimilar portions of a material, resulting in the attainment of a desired level of uniformity, either physical or chemical, in the final product' [Quillen (1954)]. Since natural diffusion in liquids is relatively slow, liquid mixing is most commonly accomplished by rotating an agitator in the liquid confined in a tank. It is possible to waste much of this input of mechanical energy if read more..

  • Page - 182

    MIXING OF LIQUIDS IN TANKS 165 where k is a dimensionless proportionality constant for a particular system. For a liquid mixed in a tank with a rotating agitator, the shear rate is greatest in the immediate vicinity of the agitator. In fact the shear rate decreases exponentially with distance from the agitator [Norwood and Metzner (1960)]. Thus the shear stresses and strain rates vary greatly throughout an agitated liquid in a tank. Since the dynamic viscosity of a Newtonian liquid is read more..

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    f I t I j/ Figure 5.1 Six-blade flat blade turbine Figure 5.2 Marine propeller read more..

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    /J k\ t Figure 5.3 Radial flow pattern produced by a flat blade turbine t Ii Figure 5.4 Axial flow pattern produced by a marne propeller read more..

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    168 FLUID FLOW FOR CHEMICAL ENGINEERS b ~-- Baffle Driving shaft Tank "_'2-.'-..': .- ':.:=-, -. ----" ----'------" , 6-blade flat ~,~ blade turbine I I_ F .... Or ....... Figure 5.5 Standard tank configuration .,-. ~._-_--.--_- --. immediately adjacent to the wall have a width b. The agitator has a blade width a and blade length r and the blades are mounted on a central disc of diameter s. A typical turbine mixing system is the standard configuration defined by the following read more..

  • Page - 186

    MIXING OF LIQUIDS IN TANKS 169 ur- IrDAN Tip speed ranges for turbine agitators are recommended as follows: 2.5 to 3.3 m/s for low agitation 3.3 to 4.1 m/s for medium agitation and (5.3) 4.1 to 5.6 m/s for high agitation If turbine or marine propeller agitators are used to mix relatively low viscosity liquids in unbaflied tanks, vortexing develops. In this case the liquid level falls in the immediate vicinity of the agitator shaft. Vortexing increases with rotational speed N until eventually the read more..

  • Page - 187

    170 FLUID FLOW FOR CHEMICAL ENGINEERS where V is the tank volume and IT is the number of turnovers per unit time. To get the best mixing, IT should be at a maximum. For a given tank volume V, this means that th .,. qlating capacity QA should have the highest possible value for the minirr n consumption of power. The head developed by the rotatih,, agitator hA can be written as hA = CIN2D~ (5.9) where C~ is a constant. Equation 5.9 is analogous to equation 4.20 for centrifugal pumps. Combining read more..

  • Page - 188

    MIXIN6 OF LIQUID8 IN TANK8 171 Figure 5.6 Gate type anchor agitator operate within close proximity to the tank wall. The shearing action of the anchor blades past the tank wall produces a continual interchange of fiquid between the bulk fiquid and the fiquid film between the blades and the wall [Holland and Chapman (1966)]. Anchors have successfully been used to mix liquids with dynamic viscosities up to 100 Pa s, [Brown et al. (1947), Uhl and Voznick (1960)]. For heat transfer applications, read more..

  • Page - 189

    172 FLUID FLOW FOR CHEMICAL EHOIHEERS Figure 5.7 Flow pattern in a baffled helical screw system unbaltled tank is nearly motionless. Baffles set away from the tank wall create turbulence and facilitate the entrainment of liquid in contact with the tank wall. The flow pattern in a baffled helical screw system is shown in Figure 5.7. Baffles are not required if the helical screw is placed in an off-centred position since in this case the system becomes self-baffling. However, off-centred helical read more..

  • Page - 190

    MIXIN6 OF LIQUIDS IN TANKS 173 5.4 Dimensionless groups for mixing In the design of liquid mixing systems the following dimensionless groups are of importance. The power number PA Po = (5.13) pN DI The Reynolds number for mixing ReM represents the ratio of the applied to the opposing viscous drag forces. pND~ ReM = ------- (5.14) The Froude number for mixing FrM represents the ratio of the applied to the opposing gravitational forces. FrM - (5.15) g The Weber number for mixing W'eM represents read more..

  • Page - 191

    174 FLUID FLOW FOR CHEMICAL ENGINEERS In liquid mixing systems, baffles are used to suppress vortexing. Since vortexing is a gravitational effect, the Froude number is not required to describe baffled liquid mixing systems. In this case the exponent y in equations 5.17 and 5.18 is zero and Fr~ = 1. Thus for non-vortexing systems equation 5.18 can be written either as ck = Po = CReXM (5.19) or as log Po = log C + x log Rein (5.20) The Weber number for mixing Were is only Of importance when read more..

  • Page - 192

    MIXING OF LIQUIDS IN TANKS 175 ~. Turbulent ~_ Viscous ,L -J-- - - Transition range - 9 ranoe -]- -I- ronoe , l | ! .=, =-- 10 e -_ A t 10 ! , _ ~-- 6 E C ! ..I .... I..l,I J .J..J,l , i.l,I.., j I I~l I ~ ,~1 ~o o ~o ~ Io ~ - I0 ~ - Io 4 Io ~ Figure 5.8 Power curve for the standard tank configuration For the transition flow region BCD which extends up to ReM = 10000, the parameters C and x in equation 5.20 vary continuously. In the fully turbulent flow region DE, the curve becomes horizontal read more..

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    176 FLUID FLOW FOR CHEMICAL ENGINEERS For the unbaffled system, ~ = Po at ReM<300 and ~ = Po/Fr~M at ReM > 300. A plot of Po against ReM on log-log coordinates for the unbaffled system gives a family of curves at ReM > 300. Each curve has a constant Froude number for mixing FrM. A plot of Po against FrM on log-log coordinates is a straight line of slope y at a constant Reynolds number for mixing ReM. A number of fines can be plotted for different values of ReM. A plot ofy against log read more..

  • Page - 194

    MIXING OF LIQUIDS IN TANKS 177 "A 10 ~ Viscous .L range 7-" t tO t Transition range _J_ Turbulent_~ range ! t i l 1 iJ lo 0 lo' i I IIi C~ D [,,I Ill i i i I] l , ,~ 10 = 10 s 10 4 ReM Figure 5.9 Power curve for the standard tank configuration without baffles liquids. Metzner and Otto (1957) used this procedure to obtain the dimensionless proportionality constant k in equation 5.1 and a non- Newtonian power curve for a particular system geometry. ~/,,, = kN (5.1) The procedure is as read more..

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    178 FLUID FLOW FOR CHEMICAL ENGINEERS 10 2 Line of Rushton, Costich and Everett ..... Line of Metzner ond Otto l 10 t _-- 15" PA ~v3~ % I i , n , ~J I .... I ~ n I I l|ll ,to ~ 10 4 tO z ReM Figure 5.10 Deviation from Newtonian power curve for shear thinning liquids J 9 speeds N beyond the laminar flow region calculate values of shear rate ~, from equation 5.1. Read the corresponding values of apparent viscosity /~o from the log-log plot of/~o against ~ and calculate the Reynolds number read more..

  • Page - 196

    MIXING OF LIQUIDS IN TANKS 179 Nienow and Elson (1988) have reviewed work done mainly by them and their co-workers on the mixing of non-Newtonian liquids in tanks. The above approach for inelastic, shearing thinning liquids has been largely substantiated but considerable doubt has been cast over using this method for dilatant, shear thickening materials. In the case of highly elastic liquids mixed by a Rushton turbine, flow reversal may occur in the low Reynolds number region, ReM < 30, read more..

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    180 FLUID FLOW FOR CHEMICAL ENGINEERS pND~ Re, w =----- Substituting the given values (1000 kg/m3)(0.2 rev/s)(9.0 m 2) Re, w = 1.0 Pas Re, w = 1800 From the graph of ~b against ReM in Figure 5.8 d~ = Po = 4.5 The theoretical power for mixing is PA = PopN 3D~ = (4.5)(1000 kg/m3)(O.O08 rev3/s3)(243 m ~) = 8748 W (5.14) Example 5.2 Calculate the theoretical power for a six-blade flat blade turbine agitator with diameter DA = 0.1 m running at N = 16 rev/s in a tank system without baffles but read more..

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    MIXING OF LIQUIDS IN TANKS 181 Now with and Therefore y _. a- log ReM a = 1.0 and B = 40.0 log 1800 = 3.2553 Substituting known values -2.2553 = -0.056 38 40 So N2DA (256 rev2/s2)(0.1 m) , g 9.81 m/s 2 = 2.610 ( N2DA)y = 2.610 (-o.05638) = 0.9479 g Therefore PA = (2.2)(900 kg/m3)(4096 rev3/s3)(0.00001 m ~) (0.9479) = 76.88 W (5.25) 5.6 Scale-up of liquid mixing systems The principle of similarity [Holland (1964), Johnstone and Thring (1957)] together with the use of dimensionless groups is the read more..

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    182 FLUID FLOW FOR CHEMICAL ENGINEERS the other. Hence, geometrical similarity exists between two pieces of equipment of different sizes when both have the same shape. Kinematic similarity exists between two systems of different sizes when they are not only geometrically similar but when the ratios of velocities between corresponding points in one system are equal to those in the other. Dynamic similarity exists between two systems when, in addition to being geometrically and kinematically read more..

  • Page - 200

    MIXIN6 OF LIQUID8 IN TANK8 183 conflict. In order to scale up with accuracy, it is often necessary to design pilot equipment so that the effects of certain dimensionless groups are deliberately suppressed in favour of a particular dimensionless group [Holland and Chapman (1966)]. For example, baffles can be used to eliminate vortexing so that the Froude number need not be considered. Frequently it is not possible to achieve the desired similarity when scaling up from small to large scale units. read more..

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    184 FLUID FLOW FOR CHEMICAL ENGINEERS for a constant tip speed ur DAI DA2 = ~ (5.37) N, N2 for a constant ratio of circulating capacity to head QA/hA and N 32 3D2 ~Dal = N2 A2 (5.38) for a constant power per unit volume PA/V. The scale-up rules given by equations 5.32 to 5.34 and 5.36 to 5.38 are all mutually conflicting. In practice, the process result and corresponding agitator speeds can be obtained in three small geometrically similar tank systems of different sizes. These data can then be read more..

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    MIXING OF LIOUIDS IN TANKS 185 nzrND~ QA = ------ (5.7) 4 For scale-up from system 1 to system 2, equation 5.7 can be written in the form ~] QA2 = (5.42) nlNID~l ~72N2D~2 Combining equations 5.41 and 5.42 gives QA 2 = -~l -~2 QA , (5.43) Equation 5.43 shows that the circulation capacity of low speed square pitch propellers greatly exceeds that of high speed propellers for the same power consumption and Reynolds number [Holland and Chapman (1966)]. 5.7 The purging of stirred tank systems In read more..

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    186 FLUID FLOW FOR CHEMICAL ENGINEERS Equation 5.45 can also be written in the form Clt = Cl0 e- ~' (5.46) where a = QIV is the reciprocal of the nominal holding time for the liquid in the tank. The fraction x of the original solute which has been purged from the tank after a time t is given by the equation mo-m Clo-Cl, x = - = 1 - e- ~,t (5.47) mo el0 For a second tank of the same size in series~ the rate of change of solute concentration is given by the equation vdCet ~t = Q(Ctt- C2,) (5.48) read more..

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    MIXING OF LIQUIDS IN TANKS 187 V(C~, + C2, + C~, +... + C.,) a2t2 a.-lt~-l ] = VClo e-"* 1 + at + ~ +... + (n - 1)'----~. (5.54) Equation 5.54 gives the total amount of solute remaining in a system of n equal size tanks after a time t where at time t = 0 the concentration in the first tank was C~o and the concentration in all other tanks was zero. The amount of solute purged from the system after time t is given by the equation { [ on,it mt= V CIO -- Clo e- at 1 + at + ~ +... + (n- read more..

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    188 FLUID FLOW FOR CHEMICAL ENOINEER$ Rushton, J.H., Costich, E.W. and Everett, H.J., Power characteristics of mixing impellers, Chemical Engineering Progress, 46, pp. 395-404 (1950). Uhl, V.W. and Voznick, H.P., Anchor agitator, Chemical Engineering Progress, 56, pp. 72-7 (1960). Weber, A.P., Selecting propeller mixers, Chemical Engineering, 70, pp. 91-8 (2 Sept 1963). read more..

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    Flow of compressible fluids in conduits When a compressible fluid, ie 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 calcula- tions. In a pipe of constant cross-sectional area, the falling density requires that the fluid accelerate to maintain the same mass flow rate. Consequent- ly, the fluid's kinetic energy increases. It is found convenient to base read more..

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    190 FLUID FLOW FOR CHEMICAL ENGINEERS or as An energy balance for unit mass of fluid can be written either as E~ + q + W = E2 (6.3) (U2 - Ul ) + (z2 - zl )g + (P2V2 - Pl Vl ) + = q + W (6.4) For steady flow in a pipe or tube the kinetic energy term can be written as u21(2a) where u is the volumetric average velocity in the pipe or tube and a is a dimensionless correction factor which accounts for the velocity distribution across the pipe or tube. Fluids that are treated as compressi- ble are read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 191 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 dq = dU + PdV- dF (6.10) where (iF is the energy per unit mass required to overcome friction. Substituting equation 6.10 into equation 6.8 gives gdz+ VdP+ d(~) + dF = dW (6.11) Equation 6.11 can be integrated between states 1 and 2 to give (z2 read more..

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    192 FLUID FLOW FOR CHEMICAL ENGINEERS 2f(dx ] ou2 alP/- \-d-~ ] (6.15) The corresponding energy required to overcome friction is df = Vdl~/. Thus equation 6.15 gives dF as dE---2~d--~x)u2--2f(d~ii)G2V2 "\ di (6.16) where advantage has been taken of equations 6.1 and 6.13. Substituting for dF in equation 6.14 gives gdz + VdP + G2VdV+ 2f(d~ii )GzV2 = dW (6.17) Dividing equation 6.17 throughout by V 2 and integrating between states 1 and 2 over a length L of pipe gives f2 g f2dP (~1) 2fG2L read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 193 tion 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. In equation 6.19, and all other equations in this chapter, P denotes the absolute pressure. In order to make use of equation 6.19 or equation 6.18 it is necessary to know the relationship between the pressure P and the specific volume V so that terms such as f 2 dP/V can be read more..

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    194 FLUID FLOW FOR CHEMICAL ENOINEER8 high pressures it is necessary to modify equation 6.23 by introducing the compressibility factor Z: PV = ZR'T (6.24) The compressibility factor is a function of the reduced pressure P, and the reduced temperature 7", of the gas. P, is the ratio of the actual pressure P to the critical pressure Pc of the gas: P P~ = p--~ (6.25) and 7', is the ratio of the actual temperature. T to the critical temperature T, of the gas: T 7', = ~,~ (6.26) Plots of Z read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 195 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 6.27 becomes PV = constant (6.31) for an ideal gas. 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, read more..

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    196 FLUID FLOW FOR CHEMICAL ENGINEERS --~+G In + =0 (6.19) In the case of isothermal flow of an ideal gas, the equation of state can be written as PV = PI VI (6.35) and evaluation of the integral in equation 6.19 gives i2dp i f 2 p~_p2 P dP = (6.36) I ~V ~ PiVI t 2PIVl Often the upstream pressure P l will be unknown but for an isothermal change PiVl can be replaced by any known value of PV at the same temperature, for example the downstream conditions P2V2 if P2 is specified. In the kinetic read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 197 Pm(P2-Pt) P2-PI PIVI ffi Vm (6.41) where Vm is the specific volume at the mean pressure Pro. Thus, equation 6.38 can be written as v. +G2L \P2 +-~i =0 (6.42) As noted previously, the kinetic energy term is usually negligible com- pared with the frictional term and this is certainly true when the pressure drop AP = P] - P2 is small compared with P]. In this case, equation 6.42 can be approximated by P2-PI 2fLG 2 t ffi 0 (6.43) v. di or lkP.f ---- P I - read more..

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    198 FLUID FLOW FOR CHEMICAL ENGINEERS 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~ If the friction factor may be assumed to have a value of 0.005, what diameter of pipe is required.~ Calculations ,-'~-+'G 2 In ~l +-~/ =0 For isothermal read more..

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    FLOW OF COMPRESSIBLE FLUIOS IN COliOUIT$ 199 It may be anticipated that 0.223<<4/d~, ie da<<17.9m. Thus the calculation may be simplified by neglecting the kinetic energy term, so that dS 4 x 6.485 • 10 -2 = 5.025x 10 ~ -- 5.162x 10 -Tm s and d~ = 0.0553 m 6.4 Non-isothermal flow of an ideal gas In a horizontal pipe For steady flow of an ideal gas between points 1 and 2, distance L apart, in a pipe of constant cross-sectional area in which no shaft work is done, the energy read more..

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    200 FLUID FLOW FOR CHEMICAL EHOINEER$ 6.5 Adiabatic flow of an ideal gas in a horizontal pipe Equation 6.19 is the basic equation relating the pressure drop to the flow rate. The difficulty that arises in the case of adiabatic flow is that the equation of state is unknown. The relationship, PV ~ = constant, is valid for a reversible adiabatic change but flow with friction is irreversible. Thus a difficulty arises in determining the integral in equation 6.19: an alternative method of finding an read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 201 ie Thus, equation 6.51 can be written as ("Y)d(PV)+G2VdV=O (6.59) Integrating equation 6.59 gives the desired relationship between P and V: (,1) PV+ ,, '27' G2V2 = constant - C (6.60) Thus (6.58) Ide C (~-1)a~l V dV = -V'J- 7 2 V (6.61) and the integral in equation 6.19 is readily found by integrating equation 6.61: (6.62) = ~;I v ~, ~ T ~, V 2 Substituting this result in equation 6.19 gives (, ,) C V1 V~ + G2 y In ~ +-d7 =0 (6.63) where, from read more..

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    202 FLUID FLOW FOR CHEMICAL ENGINEERS is common practice to assume isothermal conditions, any departure providing a small bonus. 6.6 Speed of sound in a fluid The speed u,~ with which a small pressure wave propagates through a fluid can be shown [Shapiro (1953)] to be related to the compressibility of the fluid Op/OP by equation 6.65: u,~ = (6.65) Assuming that the pressure wave propagates through the fluid polytropi- cally, then the equation of state is PV k = constant = K (6.66) from which read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 203 when u is greater than c, ie supersonic flow. It is therefore useful to define the Mach number Ma: u Ma = - (o./~) c Subsonic flow corresponds to Ma < 1 and supersonic flow to Ma > 1. The conditions of incompressible flow are approached as Ma-, O. 6.7 Maximum flow rate in a pipe of constant cross-sectional area Consider the case of steady polytropic flow in a horizontal pipe described by equation 6.49: - +-~-0 (6.49) If the upstream pressure P~ read more..

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    204 FLUID FLOW FOR CHEMICAL ENGINEERS achieved when the gas speed reaches the speed at which a small pressure wave propagates through the gas. (This is why the subscript w has been used to denote this condition.) If the downstream pressure Pz is reduced further, there can be no increase in the flow rate and the flow is said to be choked. A simple interpretation of this choking condition is as follows. The gas flows as a result of the pressure difference P~- P2. When the gas speed reaches the read more..

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    FLOW OF COMPRESSIBLE FLUIOS IN CONOUITS 205 6.8 Adiabatic stagnation temperature for an ideal gas For adiabatic flow with negligible change of elevation and no shaft work, the energy equation reduces to ui-u 2 H2- H~ + = 0 (6.50) 2 which may be written in differential form as dH+d(~) = O (6.81) Substituting for dH using equation 6.53 gives CrdT+ d(~) = O (6.82) Equations 6.81 and 6.82 show that as the velocity rises the kinetic energy increases at the expense of the enthalpy and consequently the read more..

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    206 FLUID FLOW FOR CHEMICAL ENGINEERS where Tm is the measured temperature. For thermometers of conventional design, r/= 0.88, Barna (1969). 6.9 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 read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 207 between stages. For two-stage compression from Pm to Pz to P3, with the gas cooled to the initial temperature TI at constant pressure, equation 6.89 becomes i /,'v' - ] + - 1 (6.90) In the case of compression from pressure P] to pressure P2 through n stages each having the same pressure ratio (P,IPI) TM, the compression work is given by W- ~,-'1 P~V~ - 1 (6.91) Equations 6.89 to 6.91 give the work required to compress unit mass of the gas. It should be read more..

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    208 FLUID FLOW FOR CHEMICAL ENGINEERS stages. The relative molecular mass of the gas is 28.0 and the ratio of specific heat capacities 3, is 1.40. Calculations (i) For a single stage compression, ~,- i lyv_ 1] From given values P2 --=10 P1 Therefore = 10 0.2857= 1.931 Equation of state PV= R'T= RT RMM Therefore PiVl =--.----. RTI RMM 8314.3 J/(kmol K) x 200 K 28.0 kg/kmol = 5.939 x 104 J/kg (6.89) Also 3' = 3.5 3,-1 Substituting these values into equation 6.89 W (1 stage) = (3.5)(5.939 • 104 read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 209 IF= y,1 PlVl -1 (6.91) For n = 2 = 100"1429- 1.389 Since and as before ny .... =7.0 y-I Pl V] = 5.939 • 104 J/kg it follows that IF(2 stages) = (7.0)(5.939 • 10 4 J/kg)(1.389- 1) IF(2 stages) = 1.617 • l0 s J/kg = 161.7 kJ/k8 (iii) Repeating the above calculation for n = 3 gives W (3 stages) = 152.8 kJ/kg 6.10 Compressible flow through nozzles and constrictions High speed gas flow through nozzles and other constrictions is essentially read more..

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    210 FLUID FLOW FOR CHEMICAL EHOINEER8 Po-----~.- & P. Po I I I t Figure 6.1 Distance along nozzle Regime 1 t Regime 2 Pressure profiles for compressible flow through a convergent nozzle the discharge reservoir, is gradually reduced below Po, the gas flow through the nozzle will gradually increase. This is illustrated by conditions (a) to (c). In each case the pressure PE at the exit plane is equal to the back pressure PB. Flow is subsonic throughout the nozzle. This type of behaviour in read more..

  • Page - 228

    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 211 illustrated in condition (e). The profiles of pressure and gas speed are identical to those of condition (d) up to the exit plane. For condition (e) the exit plane pressure P~ is equal to P, and it is necessary for the pressure to change to the imposed back pressure PB: this occurs in an oblique shock wave outside the nozzle, indicated by the jagged line. The profile of the gas speed is a mirror image of the pressure profile: the speed increases where read more..

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    212 FLUID FLOW FOR CHEMICAL ENOINEER$ pressure Po constant, flow in the converging section is as discussed above for a convergent nozzle. When the back pressure is only slightly lower than the supply pressure, as in condition (a) in Figure 6.2, the pressure passes through a minimum value at the throat, where the gas speed is a maximum, but pressure recovery occurs in the diverging section as the gas decelerates. This type of behaviour, denoted as regime l, is observed until the back pressure is read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 213 instructive to see how the constraints of continuity, energy and the equation of state allow both subsonic and supersonic flow in a nozzle. From equation 1.7, continuity can be expressed as puS = constant (6.93) Writing this in differential form and dividing throughout by puS gives dS du dp -- +--- +-- = 0 (6.94) S u p For isentropic flow with negligible change of elevation and no shaft work, equation 6.11 reduces to VdP + udu = 0 (6.95) Thus du VdP dP read more..

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    214 FLUID FLOW FOR CHEMICAL ENGINEERS (Ma > 1). If flow in a converging section is subsonic, it accelerates but if it were then to become supersonic, equation 6.100 shows that it would decelerate. Thus the maximum speed in a converging section is the sonic speed and this is reached at the throat where dS/S - O. As equation 6.95 shows, for isentropic flow with negligible change in elevation (potential energy) and no shaft work, there is an interchange between only two forms of energy: read more..

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    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 215 M 2 $2( 2T ) -- V ~ ~-1 (poVo - P v) (6. ~06) Using the equation of state and rearranging the result enables the mass flow rate to be given as M = (3,PolVo)l/2S~ (6.107) where { [ } 10, The pressure-dependence of the flow is contained entirely within the term ~,. The quantity V' ~'Po/Vo is constant for specified upstream conditions. (It is equal to the mass flux the gas would have if flowing at the sonic speed Co = ~/3'Po Vo corresponding to the read more..

  • Page - 233

    218 FLUID FLOW FOR CHEMICAL ENGINEERS convergent--divergent nozzles, the pressure at the throat cannot be lower than P. but if a low back pressure is imposed a shock wave will occur somewhere downstream of the throat. Example 6.3 Nitrogen contained in a large tank at a pressure P = 200000 Pa and a temperature of 300 K flows steadily under adiabatic conditions into a second tank through a converging nozzle with a throat diameter of 15 mm. The pressure in the second tank and at the throat of the read more..

  • Page - 234

    FLOW OF COMPRESSIBLE FLUIDS IN CONDUITS 217 RT PV=R'T= RMM (8314.3 J/kmol K)(300 K) - 28.02kg/kmol - 8.904 x 104 J/kg Thus, for Po - 2 x l0 s Pa, V0 - 0.4451 m~/kg. Substituting these values into equation 6.107 M = (1.39x 2 x 105 Pa)l/2 0.445'1 m3/kg ' (1.767 x 10 -4 m2)(0.5407) M = 0.0755 kg/s At the throat, Pt = 1.4 x l0 s Pa and the specific volume there is given by = = 0.4451 = 0.5753 m3/kg The gas speed at the throat is given by MVt (0.0755 kg/s)(0.5753 m3/kg) St 1.767 x 10 -4 m 2 = 245.8 read more..

  • Page - 235

    218 FLUID FLOW FOR CHEMICAL EtiGINEERS 6.111, applies to isentropic, ie constant entropy flow. It therefore applies to the smooth curves but not across a shock wave. When a normal shock wave occurs in regime 2, the flow is isentropic up to the shock wave, then isentropic but at a different entropy downstream of the shock wave. The value of the product SO is constant throughout the flow if no shock wave occurs but is different upstream and downstream of a shock when one Occurs. A shock wave from read more..

  • Page - 236

    7 Gas--liquid two-phase flow The flow of gas-liquid mixtures in pipes and other items of process equipment is common and extremely important. In some cases the quality, that is the mass fraction of gas in the two-phase flow, will vary very little over a large distance. An example of this is the flow in many gas-oil pipelines. In other cases, boiling or condensation occurs and the quality may change very significantly although the total mass flow rate remains constant. It is important to read more..

  • Page - 237

    220 FLUID FLOW FOR CHEMICAL ENGINEERS Figure 7.1 Flow regimes in vertical gas-liquid flow most of the cross section there is a churning motion of irregularly shaped portions of gas and liquid. Further increase in the gas flow rate causes a degree of separation of the phases, the liquid flowing mainly on the wall of the tube and the gas in the core. Liquid drops or droplets are carried in the core: it is the competing tendencies for drops to impinge on the liquid film and for droplets to be read more..

  • Page - 238

    GAS-LIQUID TWO-PHASE FLOW 221 Figure 7.2 Flow regimes in horizontal gas-liquid flow As the gas flow rate is increased further, the interracial shear stress becomes sufficient to generate waves on the surface of the liquid produc- ing the wavy flow regime. As the gas flow rate continues to rise, the waves, which travel in the direction of flow, grow until their crests approach the top of the pipe and, as the gas breaks through, liquid is distributed over the wall of the pipe. This is known as the read more..

  • Page - 239

    222 FLUID FLOW FOR CHEMICAL ENGINEERS lOS: "- I lO' lO 3 lo' NO "'e O. 10 10-1 =,,. I "l " i Annular / / I I / f Wispy annular Chum ~.--I I Bubble / I flow / t / \ / Slug % / / I J ..,!. I I 10 10 ~ I(P 10' ~0 s PL/~ [ko/( s= m)] lO e Figure 7.3 Flow regime map for vertical gas-liquid flow Source: G. F. Hewitt and D. N. Roberts, Studies of two-phase flowpattems by simultaneous X-ray and flash photography, Report AERE-M 2159 (London: HMSO, 1969) gradual and the classification read more..

  • Page - 240

    OAS-LIQUIO TWO-PHASE FLOW 223 "E 10 Wav~,~ Stratified flow 10 Dispersed flow .... . _~ Annular flow ( /\ \ Slug flow Bubble flow , P,,ug,ow ""~* Ioo ~ooo Figure 7.4 Flow regime map for horizontal gas-liquid flow Soume: O. Baker, Oil and Gas Journal53, pp. 185--95 (26 July, 1954) 10000 the same as the superficial velocity.) In addition to allowing the flow regime for a specified combination of gas and liquid flow rates to be determined, the diagram shows how changes of operating read more..

  • Page - 241

    224 FLUID FLOW FOR CHEMICAL ENGINEERS atmosphere; consequently A and q~ have the value unity for the air-water system under these conditions. One of the problems with two-phase flow is that a significant distance may be required for the flow regime to become established and the flow regime may be changed by flow through pipe fittings and bends. When a change of phase occurs several different flow regimes may be obtained in a short distance as demonstrated by the schematic representation of flow read more..

  • Page - 242

    OA$-LIQUIO TWO-PHASE FLOW 226 Figure 7.6 Element of two-phase flow The terms represent the contributions from the total pressure gradient, the frictional drag of the pipe wall and the hydrostatic head of the two-phase mixture. The rate of change of momentum is (Mo+ 8MG)(UG+ 6UG) + (ML + SMz.)(UL + 6Uz.) - MGUO -- MLUZ. (7.7) where the first two terms are the momentum flow rate of the fluid leaving the element and the third and fourth terms are the momentum flow rates into the element. For steady read more..

  • Page - 243

    226 FLUIO FLOW FOR CHEMICAL ENGINEERS d"x = + + (7. l I) a #h In principle, this is the same as for single-phase flow. For example in steady, fully developed, isothermal flow of an incompressible fluid in a straight pipe of constant cross section, friction has to be overcome as does the static head unless the pipe is horizontal, however there is no change of momentum and conscquendy the accelerative term is zero. In the case of compressible flow, the gas expands as it flows from high read more..

  • Page - 244

    6AS-LIQUIO TWO-PHASE FLOW 227 It is also conventional to work in terms of the mass flux, sometimes called the mass velocity, G: M G = --- (7.16) S so, using equations 7.14 and 7.15 MG = wGS and ML = (1 - w)GS (7.17) Using the specific volumes of the gas Va and liquid VL, the gas and liquid velocities can be written in the following forms: MoVo wGVo ua -- .... -- .... (7.18) aS a and MLV,~ (I -w)GVL - = ...... (7.19) It is left as an exercise in using this notation for the reader to show that the read more..

  • Page - 245

    228 FLUID FLOW FOR CHEMICAL ENGINEERS ,jtT oOb ~ o Oob: 0 o 0 o o o 0 0 b o 0 t> 0 0 o o 0 l Oo o 0 o 0 o o X 0 ^ i - " Uc~ ---.--.- um#~ Figure 7.7 Schematic representation of a bubble column with cocurrent flow of gas and liquid case cocurrent flow occurs. Alternatively, the liquid may be introduced at the top of the column and be removed at the bottom thus providing counter-current flow of the two phases. In bubble columns the static head of the fluid is the dominant component of the read more..

  • Page - 246

    OAS-LIQUIO TWO-PHASE FLOW 22tl For counter-current flow the value of Qt. must be taken as negative. For many dispersed systems (gas bubbles in fiquids, fiquid droplets in another liquid, solid particles in a liquid), it has been found that the slip velocity is related to the terminal velocity ut of a single bubble, droplet or particle by the equation slip velocity = ut(1 - a)"- l (7.25) This result follows from the Richardson-Zaki equation. In their original work, Richardson and Zaki (1954) read more..

  • Page - 247

    230 FLUID FLOW FOR CHEMICAL ENGINEERS where the bubble Reynolds number Re=pt.utD,1l~t is based on the terminal rise velocity of the bubble and its equivalent diameter, that is the diameter of a sphere having the same volume as the bubble. Equation 7.25 shows that the velocity of rise of a bubble swarm relative to the liquid is lower than the terminal rise velocity of a single, isolated bubble of the same size in the same liquid. It follows that the presence of neighbouring bubbles increases the read more..

  • Page - 248

    6AS-LIQUID TWO-PHASE FLOW 231 Ue,L / ' I ,,.,..~ 0 Figure 7.8 Wallis plot for flow in a bubble column ,uta(1 - a) "-1 (1 - Q)-~o Q~ o~ ,..._.. S fraction. Using this Wallis plot, it is easy to see how the void fraction must change to accommodate variations of the gas and liquid flow rates. As an example, Figure 7.9 shows the effect of reducing the liquid flow rate (l to 3) when the gas flow rate is kept constant. The flow is still cocurrent. As the liquid flow rate is reduced the void read more..

  • Page - 249

    232 FLUID FLOW FOR CHEMICAL ENGINEERS oo ..,.., s (-~./S), (-OJSh (-QJSh (-OJS), 0 al o+ + o+ 0...-..I,, Figure 7.10 9 Wallis plot for counter-current flow o; 1 Figure 7.10. The sequence I to 4 corresponds to increasing the magnitude of the liquid flow for a fixed gas flow rate. Cases 1 and 2 are typical and demonstrate the fact that in general two values of the void fraction are possible for counter-current flow, however, in practice the higher value is difficult to obtain owing to coalescence read more..

  • Page - 250

    GAS-LIQUID TWO-PHASE FLOW 233 For bubble columns of moderate height the static head is sufficiently low for the expansion of the rising bubbles to be negligible. Consequently, the gas density, the volumetric flow rate of the gas, and the void fraction are sensibly constant. In this case, equation 7.31 is readily integrated to give the pressure drop over a dispersion of height H: ~ = [a06 + (1 - ~)OL] g dx = [a0~ + (1 - ~)OL] gtt (7.32) o Note that the quantity o~ l-a + (l - ' )PL = WG + VL- read more..

  • Page - 251

    234 FLUID FLOW FOR CHEMICAL ENGINEERS Table 7.1 Peebles and Garber correlation , ,, =,,, , ,,, Terminal velocity L Range of applicability ZR$.(OL-po)g U t = .... 9#L Ut = 0.33g 0"76 VL0"52R~ 28 ~ ( g(r 10.25 ut - 1.18\p-~/ i| i Re<2 2 <Re< 4.02Gi -~ 4 n9~_-0.214 -0.25 .v~vm <Re<3.1OG~ or 16.32G ~ 5.75 3. l OG~ ~ < Re or 5.75 <G2 The bubble Reynolds number Re is defined as PLut(2Rb,) pLutDb, Re = ~ = ..... (7.34) P'L #L where Db, = 2Rb, is the diameter of a sphere read more..

  • Page - 252

    OAS-LIQUID TWO-PHASE FLOW 235 over a very wide range of conditions, and Wallis (1974) has given a comprehensive correlation for the rise velocity of bubbles and drops in both pure and contaminated liquids. It should be noted that the bubble rise velocity is independent of the bubble diameter over an extremely wide range of bubble size; this corresponds to the fourth region in the Peebles and Oarber correlation. This region is likely to encompass almost all conditions in bubble columns so read more..

  • Page - 253

    236 FLUID FLOW FOR CHEMICAL ENGINEERS (a) Figure 7.11 Definition of bubble diameters (a) For spherical cap bubble 03) For a gas slug -•Uquid -----Gas slug ---- Dp .~Uquid {b} where Dp is the diameter of the pipe or tube. The diameter of the slug is only slightly smaller than the tube diameter, the liquid film between the slug and the tube wall being very thin. It may appear surprising that the shear force in this film is negligible but this must be so because the gas has such a low viscosity read more..

  • Page - 254

    GAS-LIQUID TWO-PHASE FLOW 237 slugs rising through Newtonian liquids of low viscosity. Corrections for the effects of surface tension and higher viscosities have been given by White and Beardmore (1962). The constant 0.35 is replaced by a constant kl" u, = 1.2(jG "t-jL) + kl ~ (7.43) The value of hi depends on the E6tvos number, E~, and the Morton number, Gn, as shown in Figure 7.12. 0.40 i ii=' .... i i I i I lli I I i i~1'i 0.30 "-'..---'--- / i. t ~o,/ _ ' 1031 S 0.10 , 104 ~~i read more..

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    238 FLUID FLOW FOR CHEMICAL ENGINEERS 7.4.1 Voidfraction For the gas flow, continuity gives Q6 = u, aS (7.45) or (20 JG m usS us (7.46) Substituting for u, from equation 7.43, the void fraction is obtained as JG a = .... (7.47) +A) + x/gOp 7.4.2 Pressuredrop In many cases the frictional and accelerative components of the pressure gradient are negligible so only the static head component need be considered: (--~) = -[aPo + (l - a)pL]g (7.48) sh If necessary the frictional component of the read more..

  • Page - 256

    6AS-LIQUID TWO-PHASE FLOW 239 7.5 The homogeneous model for two-phase flow The complex nature of the churn flow regime does not lend itself to the relatively simple analyses used for the bubble flow and slug flow regimes. Friction will certainly be significant and acceleration may be, particularly when a change of phase occurs. Although the annular flow regime appears to be simpler and Wallis (1969) has suggested an approach using an interracial friction factor, the details of annular flow are read more..

  • Page - 257

    240 FLUID FLOW FOR CHEMICAL ENGINEERS where u is the velocity of both phases and/J is the average density of the two-phase mixture. By definition, the average density is the reciprocal of the average specific volume of the mixture, so that 1 -- = f:= wvo+(]-W)VL (7.51) The momentum equation takes a simple form, which may be derived from the general form given in equations 7.9 by putting uc = UL = U (7.52) and S~pc + SLpL = S~ (7.53) Thus the momentum equation becomes dP dF du -S S~g sin 0 = M read more..

  • Page - 258

    GAS-LIQUID TWO-PHASE FLOW 241 (dP) - ~x = d~ (7.58) which is equivalent to equation 2.7, the latter being written for a finite length of pipe. Using the Fanning friction factor, the wall shear stress is given by ~-,~ -- 89 and the frictional component of the pressure gradient is then (dP.)f 2f/mz _ 2f GzIT - ~x - d, - d, (7.59) This equation is equivalent to equation 2.13. The velocity, which generally varies with position along the pipe, has been rewritten in terms of the constant mass flux G read more..

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    242 FLUID FLOW FOR CHEMICAL ENOINEER$ vapour with a much higher specific volume Vo so the mean specific volume increases and by continuity the flow must accelerate. Substituting equation 7.61 into equation 7.60 allows the accelerative component of the pressure gradient to be written as (7.62) Static head component The static head component is simply - - ~g sin 0 = (7.63) ,h V Total pressure gradient Summing the three components of the pressure gradient in equation 7.55 gives the total pressure read more..

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    6AS-LIQUID TWO-PHASE FLOW 243 Simplifications that may be applicable can be summarized as follows. I[G" dVo -w-~ << 1 ie gas compressibility negligible. 2 f, 17," remain constant over the length of integration. 3 With the above simplifications it is possible to integrate equation 7.65 analytically for the special case of evaporation with dwldx constant. This condition occurs when the heat flux is uniform along the tube length. If the liquid is saturated at x=0 and has a quality read more..

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    244 FLUID FLOW FOR CHEMICAL ENOINEERS 3 Use a friction factor for a corresponding single-phase flow. As an example of method 3, in bubbly flow with a low quality it would be appropriate to calculate the friction factor based on the properties of the liquid. The frictional component of the pressure gradient for the actual two-phase flow is given by (dP): 2fG29 - - C7.59) while if the whole of the two-phase flow were liquid the frictional component of the pressure gradient would be given by --(--~ read more..

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    6AS-LIQUIO TWO-PHASE FLOW 246 flow. Again this reference flow has the same total flow rate. Writing the equivalents of equations 7.68, 7.69 and 7.70 will give (~xx)f - I? ---[w+ l- VL (7.72) The concept in this method is to replace the actual two-phase flow by a corresponding single-phase flow, for which the frictional pressure gradient is readily calculated. The relationship between the frictional pressure gradient for the two-phase flow and that for the reference single-phase flow is then read more..

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    246 FLUID FLOW FOR CHEMICAL ENOIHEER$ S = .... ~r~ = 3.142 x 10 -4 m 2 4 G Mo +Mz. (0.008 + 0.4) kg/s = S = "3.142 x 10 -4 m z = 1299 kg/(mZs) (7.16) MG w = M o + ML = 0.0196 (7.14) For isothermal expansion, PVo = constant. Therefore dVc V~ g dP P From thermodynamic tables, P, ir = 1.1984 kg/m s at 20 ~ 1 atm (1.01325 bar). Therefore dVo dP = -8.455 x 10 -6 mS/(kg Pa) at 1 bar and dVa G2w--~ = 12992 x 0.0196 x (-8.455 x 10 -6) = -0.2796 Thus the denominator of equation 7.65 has the value read more..

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    6AS-LIQUID TWO-PHASE FLOW 247 ft.o is determined for the whole flow as liquid, for which the Reynolds number is Gdi [1299 kg/(m2s)] (20 • 10 -3 m) ReLo = = ....... = 2.598 x 10 4. /tL I X 10 -3 Pa s From the friction factor chart, Figure 2.1, for a smooth tube and this value of the Reynolds number, fz.o = 0.0058. Therefore (~xx) =(1"687x105 Pa/m)(0.0058) 978.5 Pa/m LO Making the approximation that the value of the friction factor for the two-phase flow is equal to fLo, the frictional read more..

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    248 FLUID FLOW FOR CHEMICAL ENGINEERS the above calculations. If the pressure gradient were constant over the length of the tube, the pressure drop over the 1.3 m length would be AP = (2.461 x 104 Pa/m)(1.3 m) = 3.199 x 10 4 Pa = 0.3199 bar This estimated pressure drop is a significant fraction of the outlet pressure (1 bar) so various quantifies calculated above will vary through the tube; this is particularly true for Vo and consequently I?. The calculation accuracy can be increased by read more..

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    OAS-LIQUIO TWO-PHASE FLOW 249 cent. In high pressure systems the pressure drop will generally be a very small fraction of the average pressure and pressure-dependent variations will be less significant. An example of the use of the homogeneous flow model for a case in which boiling occurs, and equation 7.66 is used, is given as part of Example 7.2. 7.6 Two-phase multiplier In the separated flow models presented in Sections 7.7 and 7.8, the method of calculating the frictional component of the read more..

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    280 FLUID FLOW FOR CHEMICAL ENGINEERS determined using a suitable correlation, the two-phase frictional pressure gradient is readily calculated from equation 7.74. The case of a wholly gas reference flow is similar. When the reference flow is only the liquid in the two-phase flow, the equations are slightly different because the liquid flow rate and not the total flow rate must be used for the reference. The 'only liquid' frictional pressure gradient is given by dP ) = 2fL(l -- w)2G ell (7.76) read more..

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    OA$-LIOUIO TWO-PHASE FLOW 251 7.7 Separated flow models In these models the phases are treated as if they are separate and flow in well defined but unspecified parts of the cross section. Only the simplest case, in which the phases are allowed to have different but uniform velocities, will be considered here. An overall momentum equation will be given and it will be seen that merely allowing the gas and liquid velocities to differ leads to considerable complexity. Two empirical correlations from read more..

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    282 FLUID FLOW FOR CHEMICAL ENGINEERS Comparing equation 7.81 with its homogeneous model equivalent, equation 7.65, it is clear that merely allowing the phases to have different velocities leads to a considerable increase in complexity. In both equa- tions, the middle term in the numerator derives from the accelerative component of the pressure gradient and represents acceleration due to a change of phase. Also, in both equations, the term involving G 2 in the denominator originates in the read more..

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    GAS-LIQUID TWO-PHASE FLOW 253 component vanishes. Consequently the frictional pressure gradients in the two phases were assumed to be equal. Lockhart and Martinelli (1949) used 'only liquid' and 'only gas' reference flows and, having derived equations for the frictional pressure gradient in the two-phase flow in terms of shape factors and equivalent diameters of the portions of the pipe through which the phases are assumed to flow, argued that the two-phase multipliers ~b~ and ~ could be read more..

  • Page - 271

    254 FLUID FLOW FOR CHEMICAL ENGINEERS 100 -0- t.." (D 1 0.01 0.10 1.00 10.0 100 1.00 t- ._o 0.10 0 > 0.01 Martinelli parameter, X Figure 7.13 Void fraction and square root of two-phase multiplier against Martine/li parameter X Source: R. W. Lockhart and R. C. Martinelli, Chemical Engineering Progress 45, pp. 39--46 (1949) Use of the correlation is very simple. First the frictional pressure gradients are calculated for only the liquid flowing in the pipe, and for only the gas" (dP) read more..

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    OAS-LIQUIO TWO-PHASE FLOW 25S The value of the square root of the two-phase multiplier is read from Figure 7.13, or calculated from equation 7.85 or 7.86, and the two-phase frictional pressure gradient calculated from or Obviously, it is better to use the smaller multipfier. The curves for each combination of flow regimes cross at X = I, as they must from the defmition of X, so the 'only gas' reference is preferable for X < 1 and the 'only liquid' reference for X> I. The reader is read more..

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    256 FLUID FLOW FOR CHEMICAL ENGINEERS If both the gas and the liquid are in laminar flow or both in turbulent flow, KL = K~ and m = n. Consequently W / \-~G PL In practice, the flow regimes of both phases will be turbulent in most cases. Using the value n = 0.20 ; )o, This provides a simple method of determining the ratio of the 'only liquid' and 'only gas' frictional pressure gradients without evaluating both pressure gradients. At high values of the Martinelli parameter X, the gas-liquid flow read more..

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    GAS-LIQUID TWO-PHASE FLOW 257 /LO , Consequently, from the definitions of the Two-Phase Multipliers, equa- tions 7.74 and 7.77 (dP) /(dP) = d~(l _ w),.s (7.95) The Lockhart-Martinelli correlation provides the relationship between ~b~. and the Martinelli parameter Xt,. Therefore, use of equation 7.95 enables the relationship between r and Xt, at low pressures to be found. At the other pressure extreme, namely the critical pressure, the phases are indistinguishable so it follows that = 1 X, _ read more..

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    268 FLUID FLOW FOR CHEMICAL ENGINEERS 100 - . -; .... . .; ,.' , . bar ...... ----- ~_ ,. . . _.....~-- ~- : : i11.01~_~:~--~ ........ 9 :/ '6.89 " '~--" ,, 10 10~ P 1.0~ ----------V-- '0 20 40 60 80 100% Mass quality, w Figure 7.14 Two-phase multiplier as a function of mass quality Source: R. C. Martinelli and D. B. Nelson, Transactions of ASME, 70, No. 6, pp. 695-702 (1948) The definition of r2 used by Mardnelli and Nelson is different from that used here: theirs is the present read more..

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    1.0 GAS-LIQUID TWO-PHASE FLOW 280 0.8 i /lo3 ..... / 207" .2 ...... / i/ O- 20 40 - 60 80 100% Mass quality, w Figure 7.15 Void fraction as a function of mass quality Source: R. C. Martinelli and D. B. Nelson, Transactions of ASME, 70, No. 6, pp. 695--702 (1948) The original Martinelli-Nelson correlation was based on relatively limited data. Thom (1964) has derived revised values of the quantities shown in Figures 7.14 to 7.17 from extensive measurements. One of the disadvantages of the read more..

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    260 FLUID FLOW.FOR CHEMICAL ENGINEERS 1000 100 10 [Z 1 2 1 "10 ,,,3 "1# i ! ! | ! ii=-'i II Ilmlll IlIII Illll IIIII IIIII IIIII inRiil| Immlll rimIIII mllll PIHIII lUlIII IIIII IIIII IIIIII IImilII i~nemml IiilIl liIlll IIllll IIIIII IIIIII IIIIII IIIIII Pressure (bar) Figure 7.16 Mean value of the two-phase multiplier as a function of absolute pressure Source: R. C. Martinelli and D. B. Nelson, Transactions of ASME, 70, No. 6, pp. 695-702 (1948) Chisholm (1973) presented a convenient read more..

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    OAS-LIQUID TWO-PHASE FLOW 261 100 4- ~1~ ~o II 100% Exit quality 0.1 1 2 5 10 20 50 100 200 Pressure (bar) Figure 7.17 Factor r~ as a function of absolute pressure Source: R. C. Martinelli and D. B. Nelson, Transactions of ASME, 70, No. 6, pp. 695-702 (1948) a single increment. For the conditions specified, the water entering the tubes is at its boiling temperature. From the data and using steam tables, the following values are found: VL - 1.302 x 10 -3 m3/kg, Vo - 3.563 x 10 -2 m3/kg (at inlet read more..

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    262 FLUID FLOW FOR CHEMICAL ENGINEERS latent heat, A = 1605 kJ/kg /~r = 9.9 x 10 -5 Pas S = 1.227 x 10 -4 m 2 M = 500/3600 = 0.1389 kg/s G = 1132 kg/(m2s) (i) Homogeneous model From steam tables dVo (0.03944- 0.03244) m3/kg dP ~ ' " 10' Pa ' = 7.0 x 10 -9 m3/(kg Pa) As the water is saturated at the tube inlet, all the heat transfer results in boiling (neglecting changes in kinetic and potential ener&,y), consequently the rate of vaporization is 50 kW 1605 kJ/kg = 0.031 15 kg/s and the read more..

  • Page - 280

    GAS-LIQUID TWO-PHASE FLOW 263 Relative roughness = 0.005 mm/12.5 mm= 0.0004. Hence, from the friction factor chart, Figure 2.1, fzo = 0.0044 ~ f. The frictional pressure drop is given by 2~, , 2VLL 1+ (2)(0.0044X11322)(1.302 x 10-')(2.5)[ 0.2243 (26 37)] Pa = i2.5x10 -3 ...... 1+ 2 " = 11.62 kPa Accelerative term APa -- G 2VL \-~L / gOe -- G 2VLGWr = [1132 kg/(m2s)]2(3.433 x 10 -2 m3/kg) x 0.2243 Pa = 9.867 kPa Static head term AP,, = VLow, In 1+ w, (2.5)(9.81) - (3.433 • ~o )t 9 '~ read more..

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    264 FLUID FLOW FOR CHEMICAL ENGINEERS Using these values shows that G 2 dVo + << 1 de ~w (l--a) 2 a2 It is therefore possible to use equation 7.83: o (I -w,) 2 l-a + ~ + dw gO e 0 V L ,] Frictional term From the values in part (i) 2ft.oG2VLL (2)(0.0044)(11322)(1.302 x 10-3)(2.5) = Pa di 12.5 x 10 -3 = 2.936 x 103 Pa From Figure 7.16, for P = 55.05 bar and we = 0.2243 Therefore l fo'~2odw 6.2 AP/= (2.936 x 103 Pa)(6.2) = 1.821 x 104 Pa = 18.21 kPa Accelerative term From Figure 7.17 APa -- read more..

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    OAS-LIQUID TWO-PHASE FLOW 266 Using Figure 7.15, a curve of a against w at 55 bar can be estimated. Reading off values of a at various values of w (Table 7.4) allows the above integrand to be evaluated as a function of w: Table 7.4 ,, ii i ...................... i I . I I I IIII I I L a |'~ g II,. (kg/m 3) II II I II I II I 0 0 768 0.02 0.40 472 0.04 0.53 376 0.O6 0.56 354 0.08 0.61 317 0.10 0.63 302 0.15 0.72 235 0.2O 0.76 206 0.2243 0.77 198 , ,,,,, Using these values to evaluate the integral read more..

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    266 FLUID FLOW FOR CHEMICAL ENOINEER$ Further reading The presentation of the material in Sections 7.3 and 7.4 has been greatly influenced by the work of Wallis (1969), while the remainder of this chapter closely follows the treatment of Collier (1972). These two books represent excellent starting points for anyone seeking further reading. It should be noted that it is customary in the two-phase flow literature to use the symbol x to denote mass quality and z to denote the axial coordinate. In read more..

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    GAS-LIQUID TWO-PHASE FLOW 267 in Chemical Engineering, Volume 8,255-368. Eds. Drew, T. B., Cokelet, G. R., Hoopes, J. W., Jr. and Vermeulen, T. (1970). Lockhart, R.W. and Martinelli, R.C., Proposed correlation of data for isothermal two-pha~, two-component flow in pipes, Chemical Engineering Progress, 45, pp. 39--46 (1949). McAdams, W.H., Woods, W.K. and Heroman, L.C. Jr., Vaporization inside horizontal tubes----ll--Benzene--oil mixtures, Transactions of ASME, 64, pp. 193--200 (1942). read more..

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    8 Flow measurement 8.1 Flowmeters and flow measurement The flow of fluids is-most commonly measured using head flowmeters. The operation of these flowmeters is based on the Bernoulli equation. A constriction in the flow path is used to increase the flow velocity. This is accompanied by a decrease in pressure head and since the resultant pressure drop is a function of the flow rate of fluid, the latter can be evaluated. The flowmeters for dosed conduits can be used for both gases and liquids. The read more..

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    FLOW MEASUREMENT 21111 Figure 8.1 U-tube manometer (a) Manometer reading ~ when P~ > Pt (o) A vertical differerce in the location of the manometer taps does not affect the reading which can be written as P~ - P2 = (p,,, - p)A,e',,, g (8.2) If p and p, are in kg/m 3, ~u~, is in m, and g is 9.81 m/s z, the pressure differential across the primary element Pl - Pz is in N/m z or Pa. The head differential across the primary element Ak based on the flowing fluid is Ah = PI-P2 (8.3) pg Combining read more..

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    270 FLUID FLOW FOR CHEMICAL ENOINEERS tic flowmeters have the advantages of no restriction in a conduit and no moving parts. 8.2 Head flowmeters in closed conduits The primary element of an orifice meter is simply a flat plate containing a drilled hole located in a pipe perpendicular to the direction of fluid flow as shown in Figure 8.2. Figure 8.2 Orifice meter Equation 1.15 is the modified Bernoulli equation for steady flow in a pipe with no pump in the section. ( ( z2+ ~ ~ + = zl+ + - P2g read more..

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    FLOW MEASUREMENT 271 uZ (1_ fezU! ) (P, - Pz) (8.5) Zaz alul = p Consider points 1 and 2 in Figure 8.2. At point 1 in the pipe, the fluid flow is undisturbed by the orifice plate. The fluid at this point has a mean velocity um and a cross-sectional flow area St. At point 2 in the pipe the fluid attains its maximum mean velocity u2 and its smallest cross-sectional flow area $2. This point is known as the vena contracta. It occurs at about one half to two pipe diameters downstream from the orifice read more..

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    272 FLUID FLOW FOR CHEMICAL ENGINEERS given by equation 8.10. Viscous frictional effects retard the flowing fluid. In addition, boundary layer separation occurs on the downstream side of the orifice plate resulting in a substantial pressure or head loss from form friction. This effect is a function of the geometry of the system. In practice, the volumetric flow rate Q is given by equation 8.11 - / 2(PI'P2) Q = S~ )4] (8.11) In equation 8.11, which is analogous to equation 8.10, Ca is the read more..

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    FLOW MEASUREMENT 273 difference. Equations 8.13 and 8.14 reduce to equations 8.11 and 8.12, respectively, when z~ = z2. It is essential to appreciate that the pressure difference measured by .a manometer automatically eliminates the static head difference. This is shown in Figure 8. l(b). The static head pg(zl-z2) in the pipe is exactly balanced by the extra static head above the right hand limb of the manometer. Consequently, if ~h is calculated from ~Lz,~ using equation 8.4, no further read more..

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    274 FLUID FLOW FOR CHEMICAL ENGINEERS Figure 8.4 OdFce discharge coefficients Source: J. H. Perry, Chemical Engineers' Handbook (Sixth edition, New York: McGraw- Hill, 1984) p. 5--15 [ < o)2j 1- (8.15) Orifice plates are inexpensive and easy to install since they can readily be inserted at a flanged joint. Figure 8.5 shows a Venturi meter. The theory is the same as for the orifice meter but a much higher proportion of the pressure drop is recoverable than is the case with orifice meters. The read more..

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    FLOW MEASUREMENT 27S Figure 8.5 Venturi meter Figure 8.6 shows a flow nozzle. This is a modified and less expensive type of Venturi meter. The theoretical treatment of head flowmeters in this section is for incompressible fluids. The flow of compressible fluids through a constric- tion in a pipe is treated in Chapter 6. Orifice meters, Venturi meters and flow nozzles measure volumetric flow rate Q or mean velocity u. In contrast the Pitot tube shown in a horizontal pipe in Figure 8.7 measures a read more..

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    276 FLUID FLOW FOR CHEMICAL ENGINEERS Figure 8.7 Pitot tube read more..

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    FLOW MEASUREMENT 277 Combining equations 8.4 and 8.16 gives o. V'- p (s.17) Equation 8.17 gives the point velocity v in terms of the difference in level between the two arms of the manometer Azm, the density of the flowing fluid p, the density of the immiscible manometer liquid pm and the gravitational acceleration g. Most Pitot tubes consist of two concentric tubes parallel to the direction of fluid flow. The inner tube points into the flow and the outer tube is perforated with small holes read more..

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    278 FLUID FLOW FOR CHEMICAL ENOINEERS Ah = <p''- P)~" p = (13.6- 1)(0.254 m) = 3.20 m The volumetric flow rate is given by 2gAh Q SoCd il - (do/d,)'] Given that So = ~d2 : (3.142)(0.10 m) 2 4 4 - 0.007 855 m 2 g = 9.81 m/s 2 Ah = 3.20 m 2gAb do O.lO 1 a-;: o.i5 : 1_(d~ \a, / /(2)(9.81 m/s2)(3.20 m) V 0.8025 Ca = 0.60 = 8.845 m/s it follows that = (0.007855 m2)(0.60)(8.845 m/s) = 0.0417 m3/s (8.4) (8.]2) 8.3 Head flowmeters in open conduits Weirs are commonly used to measure the flow rate read more..

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    FLOW MEASUREMENT 279 z2g +--- + -- 'el Pl "2" P2 ~ g+----+ =0 (8.]8) Consider a liquid flowing over a sharp crested weir as shown in Figure 8.8. Let the upstream level of the liquid be Zo above the level of the weir crest. As the liquid approaches the weir, the liquid level gradually drops and the flow velocity increases. Downstream from the weir, a jet is formed. This is called the nappe and it is ventilated underneath to enable it to spring free from the weir crest [Barna (1969)]. read more..

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    280 FLUID FLOW FOR CHEMICAL ENGINEERS Equation 8.20 gives the point velocity v at a height z above the weir crest. If the width of the conduit is b at this point, the volumetric flow rate through an element of cross-sectional flow area of height dz is dQ = by dz (8.21) Substituting equadon 8.20 into equation 8.21 gives dQ = b (X/~) *o + ~ - * dz (8.22) Let so that h = zo + ~ - z (8.23) dh = -dz (8.24) On substituting equations 8.23 and 8.24 into equation 8.22, the volumetric flow rate through read more..

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    FLOW MEASUREMENT 281 channel bed. A typical value is Ca = 0.65 for z,,o/zo = 2 and zo = 0.3 m. Viscosity has a negligible influence on Ca. Thus the flow rate can be determined from a measurement of zo. In addition to rectangular weirs, V-notch or triangular weirs are commonly used with a cross-sectional flow area as shown in Figure 8.9. In Figure 8.9 Cross-sectional flow area in a V-notch weir this case the width of the conduit b is variable and at any height z above the bottom of the weir is 0 read more..

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    282 FLUID FLOW FOR CHEMICAL ENGINEERS Integrating equation 8.34 gives 0 (~ 2) Q = -2 tan -~ (~ ,oh 3'2-g h "2 + C (8.35) Noting that Q = 0 for, - 0 (ie h = *o), then 8 0 (~ ~ros/2 (8.36) C -== -i~ tan - ~ The full flow rate occurs for,-- Zo, ie h = 0, when the first term on the right hand side of equation 8.35 vanishes. Thus the flow rate is given by 8 o 2 (8.37) Q = E tan -~ Equation 8.37 gives the volumetric flow rate Q through a V-notch weir when there is no friction in the system. In read more..

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    FLOW MEASUREMENT 283 Figure 8.10 Rotameter These usually consist of a non-magnetic casing, a rotor, and an electro- magnetic pickup. The rotor is either a propeller or turbine freely suspended on ball bearings in the path of the flowing fluid with the axis of rotation in line with the flow. The rotor turns in the fluid flow stream at a rate proportional to the flow rate. As the rotor turns it cuts through the lines of force of an electric field produced by an adjacent induction coil. The read more..

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    284 FLUID FLOW FOR CHEMICAL ENGINEERS the fluid velocity. This voltage is detected by the electrodes and is then amplified and transmitted to either readout instruments or recorders. The magnetic field is usually alternating so that the induced voltage is also alternating. Electromagnetic flowmeters can only be used to meter fluids which have some electrical conductivity. They cannot be used to meter hydrocarbons. Although electromagnetic flowmeters are expensive they are especially suitable for read more..

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    FLOW MEASUREMENT 285 All head flowmeters are used with a square root scale in which the volumetric flow rate Q is proportional to the square root of the scale reading s. Q --- k2 $1/2 (8.41) where k2 is a constant. Differentiating equation 8.41 gives dQ k2 k~ = 2-~ = 2Q (8.42) The per cent flow rate error "Ae can be defined as Ae = 100 A Q Q (8.43) where AQ is the absolute error in the volumetric flow rate. Let As be the indicator or recorder error. Then, for small errors AQ -.. d__QQ As ds read more..

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    286 FLUID FLOW FOR CHEMICAL ENGINEERS 5000 As Ae = (8.49) Q2 The square root scale is more accurate than the linear scale at flow rates near the maximum of the scale. The linear scale is more accurate than the square root scale at flow rates much less than the maximum of the scale. Thus head flowmeters are unsuitable for measuring flow rates which vary widely. Example 8.2 A flowmeter is inherently accurate at all points to 0.5 per cent of the full range. Calculate the per cent flow rate error read more..

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    FLOW MEASUREMEHT 287 A -!5000)(0.5) (lO) 2 = 25 per cent for a flow rate 10 per cent of Qmax A~ ---- (5O0O)(0.5) (25) 2 = 4 per cent for a flow rate 25 per cent of Qmax (5ooo)(0.5) Ae = (50) 2 - I per cent for a flow rate 50 per cent of Qm,~ Ae -- (5OOO)(0.5) (100)2 -- 0.25 per cent for Qmax Ro s Barna, P.S., Fluid Mechanics for Engineers, London, Butterworths (1969). Buzzard, W., InStrument scale error study throws new fight on flowmeter accuracy, Chemical Eng/neer/ng, 66, pp. 147-50 (9 Mar read more..

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    9 Fluid motion in the presence of solid particles 9.1 Relative motion between a fluid and a single particle Consider the relative motion between a particle and an infinitely large volume of fluid. Since only the relative motion is considered the following cases are covered: 1 a stationary particle in a moving fluid; 2 a moving particle in a stationary fluid; 3 a particle and a fluid moving in opposite directions; 4 a particle and a fluid both moving in the same direction but at different read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 289 Figure 9.1 Streamlines for flow around a sphere Upper half: intermediate Rep giving a laminar boundary layer Lower half: high Rep giving a turbulent boundary layer Section 1.5). In flowing round the sphere, the fluid has to accelerate and therefore, by Bernoulli's equation, the pressure falls towards the mid- point of the sphere's surface. At very low values of Rep, when the flow is dominated by viscous stresses, the fluid creeps round the rear read more..

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    290 FLUID FLOW FOR CHEMICAL EHOINEERS On increasing the Reynolds number further, a point is reached when the boundary layer becomes turbulent and the point of separation moves further back on the surface of the sphere. This is the case illustrated in the lower half of Figure 9.1 with separation occurring at point C. Although there is still a low pressure wake, it covers a smaller fraction of the sphere's surface and the drag force is lower than it would be if the boundary layer were laminar at read more..

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    FLUIO MOTION IN THE PRESENCE OF SOLIO PARTICLES 291 As a result of the changing flow patterns described above, the drag coefficient Ca is a function of the Reynolds number. For the streamline flow range of Reynolds numbers, Rep < 0.2, the drag force F2 is given by. F2 = 3"rrdp/.tut (9.6) and consequently, from equation 9.3, the drag coefficient is 24 From equations 9.5 and 9.7, the terminal falling velocity ut for the . streamline flow range of Reynolds numbers is given by d (pp-p)g u, read more..

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    292 FLUID FLOW FOR CHEMICAL ENGINEERS coefficient Cd against the particle Reynolds number for spheres, discs and cylinders. Equation 9.5 gives the terminal settling velocity for a spherical particle. For a non-spherical particle, equation 9.5 can be written in the modified form /46~,Pp-P)g ut=~l ~ 3C.p (9.12) For a spherical particle the dimensionless correction factor ~- 1 and equations 9.5 and 9.12 become identical. 9.2 Relative motion between a fluid and a concentration of particles So far read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 293 fluid and large particles sink. Particles of the same size but of different densities can also be separated by settling or elutriation. Consider two spherical particles 1 and 2 of the same diameter but of different densities settling freely in a fluid of density p in the streamline Reynolds number range Rep<0.2. The ratio of the terminal settling velocities utl/ut2 is given by equation 9.8 rewritten in the form ut_. L _ ppl - p (9.14) ut2 read more..

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    294 FLUID FLOW FOR CHEMICAL ENGINEERS A very small particle may still be in laminar flow in a centrifugal separator. In this case ro, 2 may be written in place of g in equ.ation 9.8 to give u, = d op- p)r 18~ (9.18) In addition to hydrodynamic interactions between solid particles and a fluid, physico-chemical forces may act between pairs of particles. These forces tend to form a structure which prevents the particles from settling out [Gheng (1970)]. If the forces are sufficiently strong a read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLE8 296 4e d,- ,-.~- ~.~,- (9.20) t1--~)3o The velocity calculated by dividing the volumetric flow rate by the whole cross-sectional area of the bed is known as the superficial velocity u. The mean velocity within the interstices of the bed is then ub - u/~. A Reynolds number for flow through a packed bed can be defined as Reb = pubd, (9.21) which when combined with equation 9.20 can be written as 4pu Reb-- ~(l- '~)So (9.22) An alternative Reynolds read more..

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    298 FLUID FLOW FOR CHEMICAL ENGINEERS - ,ou~ = (1- e~opU 2 (9.28) where fb is a dimensionless friction factor for flow through a packed bed. Various other definitions of friction factors for flow in packed beds have been used [Longwell (1%6)]. For laminar flow where Re'b <~ 2 fb 5 - (9.29) 2 Re'b The transition to turbulent flow is gradual. Turbulence commences initially in the largest channels and eventually extends to the smaller channels. For the complete range of Reynolds number Carman read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 297 ~kPf-- (Kcla,L) [ (1-F')2S2~ ] e3 - u (9.34) Equation 9.33 is the Carman-Kozeny equation [Carman (1937)]. The parameter K~ has a value which depends on the particle shape, the porosity and particle size range. The value lies in the range 3.5 to 5.5 but the value most commonly used is 5. For spherical particles So --" 6/dp and equation 9.34 can be written as ~J~/= (180/d.,) L e3d] u (9.35) where K~ = 5.0. Example 9.1 A gas of density p = read more..

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    298 FLUID FLOW FOR CHEMICAL ENGINEERS it follows that 9 P/= (180)(1.5 x I0 -s Pa s) (3.0 m)(12)(0.03 m/s) ......... (2.5x 10 -s m 2) = 116.6 Pa 9.4 Fluidization If a fluid is passed upwards in laminar flow through a packed bed of solid particles the superficial velocity u is related to the pressure drop AP by equation 9.33: - e)2S ] (9.33) As the fluid velocity is increased the drag on the particles increases and a point is reached where the pressure drop balances the effective weight of bed read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 299 regime may be compared with the churn-flow regime in gas-liquid two-phase flow (see Section 7.1). In tall, narrow beds the gas voids may 'coalesce' producing a slugging bed. This condition is generally undesirable owing to its unsteady nature and the difficulty of scale-up. At very high gas velocities the particles are carried out of the top of the bed. This is known as fast fluidization and is a type of pneumatic conveying. Fast fluidization read more..

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    300 FLUID FLOW FOR CHEMICAL ENGINEERS rate for the packed bed can be seen. The fluid velocity at the transition is taken as urn/. On fluidizing the bed again, the latter type of behaviour with no peak may be observed. If it is necessary to predict the minimum fluidization velocity the following correlation [Grace (1982)] may be used for gas-solid systems. where pumfdp _ (C,2 + 0.0408Ar)1/2_C (9.36) p, At = pgd~(pp- p)/ft 2 (9.37) is the Archimedes (or Galileo) number. Some doubt exists regarding read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 301 slurries can be pumped through a pipeline either in laminar or turbulent flow. Compared with non-settling slurries, settling slurries contain larger solid particles at lower concentrations. Settling slurries are essentially two-phase heterogeneous mixtures. The liquid and the solid particles exhibit their own characteristics. Thus in contrast to non-settling slurries, the solid particles in settling slurries do not alter the viscosity of the read more..

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    302 FLUID FLOW FOR CHEMICAL ENGINEERS 2.5 x 10 -2 to 3 • 10 -~ m. In equations 9.38 and 9.39,/~ is the dynamic viscosity of the transporting liquid and p, pp, and p, are the densities of the liquid, solid particles and slurry mixture respectively. The last is given by the equation p,,, = a(pp- p) + p (9.40) where a is the volume fraction of the solids in the slurry. Empirical equations are also available to calculate the pressure drop for slurries flowing through pipelines [Condolios and read more..

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    FLUID MOTION IN THE PRESENCE OF SOLID PARTICLES 303 9.6 Filtration When a slurry flows through a filter, the solid particles become entrapped by the filter medium which is permeable only to the liquid. Either of two mechanisms are used: cake filtration or depth filtration. In cake filtration, the filter medium acts as a strainer and collects the solid particles on top of the initial layer. A filter cake is formed and the flow obeys the Carman-Kozeny equation for packed beds. Depth filtration is read more..

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    304 FLUID FLOW FOR CHEMICAL ENGINEERS Perry, J.H., Chemical Engineers' Handbook. Sixth edition, New York: McGraw- Hill Book Company Inc, pp. 5-53 and 5-54 (1984). Pettyjohn, E.S. and Chrisfiansen, E.B., Effect of particle shape on free-settling rates of isomeric particles, Chemical Engineering Progress, 44, pp. 157-72 (1948). Richardson, J.F. and Zaki, W.N., Sedimentation and fluidisation: Part I, Transactions of the Institution of Chemical Eng/neers, 32, pp. 35-52 (1954). Spells, K.E., read more..

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    0 Introduction to unsteady flow Four aspects of unsteady fluid flow will be considered in this chapter: quasi-steady flow as in the filling or emptying of vessels, incremental calculations, start-up of shearing flow, and pressure surge in pipelines. 10.1 Quasi-steady flow Very often processing operations change only slowly with time and at any instant may be treated as if conditions were steady. However, in predict- ing the course of the operation, it is necessary to recognize that conditions read more..

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    306 FLUIO FLOW FOR CHEMICAL ENOINEER$ ---- - Dr =- .- .'--- --_-_- .- ,--_-- -- -- - ---.-- .-.-- 8 Figure 10.1 Liquid flowing from a tank In equation 10.2, hf is the head loss due to friction and is given by the equation In equation 10.3, 2L, is the equivalent length of the outlet pipe (including the contraction at its inlet) and di is the pipe's inside diameter; u is the mean velocity in the outlet pipe. It has been assumed in writing equations 10.1 and 10.2 that the fluid's velocity in the read more..

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    INTRODUCTION TO UNSTEADY FLOW 307 ~r D~Sz= r -By-- -T Td~uSt (10.6) Note that, for this case of emptying a tank, 8z and 8V are negative. Rearranging equation 10.6 and going to the limit #t-, 0 gives the rate of change of level in the tank as d~ d, ~ ~7 -- -u~ (10.7) Substituting for u in equation 10.7 using equation 10.5 and integrating from time t~ to t2, when the liquid levels are z~ and z2, gives f., d, d i,, dt ., [Z. + (PA -- PB)I(pg)]I/2 - -(2g )1/2 ~ ', [lla + 4f(~Leldi)] 1/2 (10.8) read more..

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    308 FLUIO FLOW FOR CHEMICAL ENGINEERS Caiculations Substituting the given values 4j~LJd~ = 40 In this example, the value of l/a is small compared with 4)~L~/di and could be ignored. When this is not the case it will be necessary to know whether the flow in the outlet pipe is laminar or turbulent, so that the appropriate value of a can be used. The following calculation shows how this can be done. From equation 10.5, with PA = PB, the velocity u is given by ,s [ 2g ] 1/2 u = i/a + 4f(2L./di) ZI'2 read more..

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    INTRODUCTION TO UNSTEADY FLOW 309 where y+ 1)/~,]} 1/2 (6.108) As the gas flows from the first tank, to which there is no feed, the pressure P0 falls and consequently so does the flow rate. Thus it takes progressively longer for each unit mass of gas to flow from the tank. In principle, this problem could be treated in a manner similar to that in Section 10.1; however, the complexity of equation 6.108 makes this impracticable. A suitable method of calculation is to divide the period of flow into read more..

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    310 FLUID FLOW FOR CHEMICAL ENGINEERS For the first increment Po falls from 200000 Pa to 198 000 Pa. The mean pressure in the first vessel during the first increment is 15o = 199000 Pa. For isentropic expanson, the mean specific volume is given by /Too \ = 0.4451 | -Z=~-~ } = 0.44670 m3/kg \ 199 / consequently, (~/~0/]~ro) 1/2 ----" 786.9 kg/(m2s) and Pt 140 -- = = 0.7035 Po 199 From equation 6.108 From equation 6.107 r 0.5390 M = 0.07495 kg/s At the end of increment, Po - 198 000 Pa read more..

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    INTRODUCTION TO UNSTEADY FLOW 311 tank being evaluated at the mean pressure of 197000 Pa. Repeating the calculation until the pressure has fallen to 180000 Pa gives the results shown in Table 10.1. Table 10.1 Po M W' AW At kPa kg/s kg kg s 2.16 4.36 6.60 8.88 11.22 13.59 16.03 18.52 21.07 23.68 initially 22.467 199 0.07495 22.305 0.162 2.16 197 0.07381 22.143 0.162 2.20 195 0.07265 21.980 0.163 2.24 193 0.07146 21.817 0.163 2.28 191 0.07023 21.653 0.164 2.34 189 0.06898 21.489 0.164 2.38 187 read more..

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    312 FLUID FLOW FOR CHEMICAL ENGINEERS equation 10.11 can be written as dt 1- g 1-- Ut Integrating equation 10.12 gives (10.12) (lO.13) where C is a constant. The initial condition is u = 0 at t = 0, therefore C = 0. Consequently, equation 10.13 becomes Pp ut (lO.14) which can be written as U=lu, -exp [- (I. ~) gt In, (10.15) Neglecting the trivial case ut = 0 when pp = p, equation 9.8 can be used to substitute for ut so that equation I0.15 becomes u = l_exp(_ 18/~t ) u-~ d~p, (10.16) 10.4 read more..

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    INTRODUCTION TO UNSTEADY FLOW 313 directions in which the stress physically acts in this case), the shear forces acting on the lower and upper surfaces of the element are (l)r ly acting in the positive x-direction and (1)~-y,, [y+Ss acting in the negative x-direction respectively. As the flow is the same at all values of x, the momentum tlow rates into and out of the element are equal and the rate of change of momentum is just the rate of change of momentum of the fluid instan- taneously in the read more..

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    314 FLUIO FLOW FOR CHEMICAL EHOINEERS ordinary differential equation. This is applicable when the problem has an open range, in this case y > 0. The method of combination of variables requires that a suitabJe combination ofy and t can be found. Dimensionally, equation 10.19 can be written as = (10.21) This suggests that the group y/V~) can be used as a combined variable. For convenience, a factor of 2 may be introduced and the combined variable ~ defined as Y ~=~ (10.22) It is necessary to read more..

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    INTRODUCTION TO UNSTEADY FLOW 318 This simple ordinary differential equation can be solved by making the substitution so that equation 10.26 becomes dvx 4, = (10.27) d,? d~b -- + 2'r/~b = 0 (10.28) dn Integrating equation 10.28 gives Therefore lnd, = -7/2 +Cl (10.29) dvx = ~b = C2e -'t (10.30) dT/ Integrating again gives the velocity Vx as Vx = C2f e -'~ dT/+ C3 (10.31) The integral in equation 10.31 cannot be evaluated analytically but it can be written in terms of the error function eft(v/) read more..

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    318 FLUID FLOW FOR CHEMICAL ENGINEERS The last statement represents both conditions (i) and (iii). Substituting the boundary conditions, using the properties of the error function given above, shows that -A -- B = Uo. The velocity field is therefore (10.34) It is clear from equation 10.34 that the time taken for the fluid at a given distance y from the surface of the plate to reach a specified fraction of the plate's velocity is proportional to y2 and inverselyproportional to the fluid's read more..

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    INTRODUCTION TO UNSTEADY FLOW 317 cent of the plate's velocity after 25 s and requires 2500 s to attain 88.75 per cent of it. 10.5 Pressure surge in pipelines Consider the consequences of closing the flow control valve V in the pipeline shown in Figure 10.3. The momentum of the liquid in a length L _N L ............ V Figure 10.3 Pipeline with check valve and flow control valve of pipe is equal to ~i2Lpu if the flow is assumed to be turbulent with volumetric average velocity u and to fill the read more..

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    318 FLUID FLOW FOR CHEMICAL ENGINEERS than gases, liquids transmit sound at higher speeds. The speed of sound in water is about 1400 m/s, compared with the value of 340 m/s for dry air under ambient conditions. It is convenient to introduce the bulk modulus K of the fluid, defined by t: - - v( ~ ) \ aV , (10.36) Replacing the specific volume V by l/p, equation 10.36 can be written as K = P (10.37) $ Thus, tile speed of sound is related to the bulk modulus by c = X/~p (10.38) When a pressure wave read more..

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    INTRODUCTION TO UNSTEADY FLOW 319 volume is p(um - u2). The pressure wave passes through the fluid at a speed a so that the rate of change of momentum per unit area is equal to p(ul- u2)a. Consequently the pressure rise is given by AP = p(ul - u2)a (10.40) When the valve is suddenly closed completely, u2 = 0 and equation 10.40 becomes AP = pula (10.41) Equation 10.41 shows that although the pressure rise is finite it may be very large. For water (p = 1000kg/m 3) flowing at 2 m/s, taking a as read more..

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    320 FLUID FLOW FOR CHEMICAL ENGINEERS Example 10.3 In order to protect a relief valve from corrosive materials, it is sometimes preceded by a bursting disc in the vent pipe. Similarly, two bursting discs may be used in series. It is known that, with these arrangements, if the first disc fails the resulting pressure surge will cause the second device to relieve even though the pressure may be significantly below its rated pressure. On the basis of experiments done with air (~, = 1.40 for dry read more..

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    INTRODUCTION TO UNSTEADY FLOW 321 P, ( 2 ) ~/(r~) The total pressure P+ is therefore related to the upstream pressure Po by ( 2 ) "/(v--') P+ = Po(1 + 7) 7+ ] (10.45) Equation 10.45 shows that P+/Po is a very weak function of 7. If the second device is to be on the point of just relieving when the upstream disc ruptures, the required ratings of the two devices must be in the ratio PoIP +. From equation 10.45 this ratio has the following values at various values of ~. 7 1.10 1.20 1.40 Po/P + read more..

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    Appendix I The Navier--Stokes equations Rather than setting up a force-momentum balance for a particular flow problem as was done in Chapter 1, general equations, known as the Navier-Stokes equations, may be formulated. Before discussing the Navier-Stokes equations, it is necessary to consider some related matters. Differentiation following the flow Let ~k represent some property of the flow, for example the velocity, temperature or density of the fluid. In general ~k is a function of the time t read more..

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    APPENDIX I THE NAVIER-STOKES EQUATIONS 323 D$ 05 05 Od~ = -- + ~)x + Vy 0~ + -- (A.3) Dt Ot ~x -~y v*Oz In equation A.3, Vx, vy, v, are the velocity components of the fluid. Thus, DcblDt gives the rate of change of ~b for a material element as it flows along. This is known as differentiation following the flow. In cylindrical coordinates (r,O,z) the substantive derivative is given by DO Ock + +_ __ + Dt - Ot V:~r r O0 v*-~z (A.4) Continuity equation A material balance on a fixed rectangular read more..

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    324 FLUID FLOW FOR CHEMICAL ENGINEERS Expanding equation A. 7, Op Or,, Ovy Or, - +Vx ~ + +~,,~ + +~,,~ + = o at Ox ~x Oy ~y Oz ~ (A.9) which can be written as 1 Dp OVx ovv Or, --~+ = + " + .... =0 (A.10) p Dt Ox Oy Oz Equation A. 10 shows that the fractional rate of change of the density of a fluid element is equal to minus its net rate of expansion. In cylindrical coordinates (r,O,z) the continuity equation is ap 1 O 1 o o ---+- --- (r/m,) +- "77-~ (gvo) +-7(/rv,) = 0 (A.I1) 0t r 0r read more..

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    APPENDIX I THE NAVIER-STOKES EQUATIONS 326 Substituting equations A.13 and A.14 into equation A.12 gives the result = ...... (A.15) Dt 8x 6y 8z Now the momentum of a fluid element depends on its velocity and not on the spatial distribution of its mass, so p can be taken outside the derivative. In the limit 8x, 6y, &--, 0, equation A. 15 becomes Dv~ 0~'x, 07"yx 0%x P'-~ = Pgx Ox ~y Oz (A.16) Equation A. 16 is valid for any fluid. It is convenient to decompose the stress component ~-~j read more..

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    326 FLUID FLOW FOR CHEMICAL ENGINEERS Equation A.20 is a generalization of equation 1.44b. Substituting for the stress components in equation A. 16 using equations A. 19 and A.20 gives Dv, OP [Ozv~ O2v, 02~)O[Ov, Ovy Or,) P'~t = ~" - -~x + "i'~x 2 + ~yz + "~ + la'~i Ox +-' + (A.21) Oy Oz By continuity, the last term in equation A.21 vanishes for incompressible flow and equation A.21 can be written as ,,(o,, + o,, +,~ + _~ " ~ Ot ~ Ox Y Oy v. O, : os.-g+.i-~x,+--+ oy 2 o~ ] read more..

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    APPENDIX I THE NAVlER-STOKES EQUATIONS 327 There is no general solution of the Navier-Stokes equations, which is due in part to the non-linear inertial terms. Analytical solutions are possible in cases when several of the terms vanish or are negligible. The skill in obtaining analytical solutions of the Navier-Stokes equations lies in recognizing simplifications that can be made for the particular flow being analysed. Use of the continuity equation is usually essential. Some of the read more..

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    328 FLUID FLOW FOR CHEMICAL ENGINEERS Potential flow Consider the two-dimensional flow show~n in Figure A. 1. If the velocity gradient o%,lay is positive it tends to c~use the element to rotate in the clockwise direction. Similarly, if a~ylax is positive it tends to cause rotation in the anti-clockwise direction. Thus, the quantity avylax- a~,,lay gives the net rate of rotation in the anti-clockwise direction as viewed. It is the clockwise direction about a line parallel to the z- coordinate as read more..

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    APPENDIX I THE NAVIER-STOKES EQUATIONS 329 seen. Laminar flow in a pipe is rotational everywhere except on the centre-line. The only non-zero velocity gradient for fully-developed flow is OvJOr, which has a maximum magnitude at the wall and falls to zero on the centre-line. A small neutrally-buoyant sphere placed in the fluid would be seen to rotate owing to the higher velocity of the fluid nearer the centre-line than that nearer the wall. Turbulent flow outside boundary layers often exhibits a read more..

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    330 FLUID FLOW FOR CHEMICAL ENGINEERS oh g" = -g Ox (A.31) where h is the height above an arbitrary datum, equation A.30 can be written as + +- +gh = 0 (A.32) at p For steady flow, Ov~/Ot = 0 so that the quantity v2/2+P/p+gh must be independent of x. Similar equations can be written for the y and z components of the velocity so it can be concluded that v2 p -- +-- + gh - constant (A.33) 2 p This is a statement of Bernoulli's theorem: the quantity vz/2+P/p+gh is constant throughout the read more..

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    APPENDIX I THE NAVIER-STOKE8 EQUATIONS SSl The function ~ is known as the velocity potential. In equation A.36 the minus sign is arbitrary but is usually incorporated so that flow is from a high value of the velocity potential to a low value. Substituting for the velocity components in equation A.8 using equation A.36 gives 02(I) 02@ 02@ o x -~ + 7y ~ + --F: = o ( ^ . 3 7 ) which shows that the velocity potential satisfies Laplace's equation in the case of incompressible flow. For potential read more..

  • Page - 349

    Appendix II Further problems (The numbers refer to the relevant chapter.) 1-1 1-2 An incompressible fluid flows upwards in steady state in a cylin- drical pipe at an angle 0 with the horizontal. Assume that the head loss due to friction is negh'gible. (a) Derive an expression for the pressure gradient in the pipe. (b) Derive an expression for the length of pipe over which the pressure is reduced by half. (c) Calculate the length of pipe L over which the pressure is reduced by half if the read more..

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    1-3 2-1 2-2 2-3 APPENDIX II FURTHER PROBLEMS 333 (e) Assuming the pressure drop to be negligible, calculate the moment on the double bend caused by the steam's pressure. Data: density of water= 1000kg/m 3, density of 20atm steam = 10.0 kg/m3 A corrosive liquid is to be transferred from one tank to a higher tank without using a pump but by pressurizing the space above the liquid in the lower tank. The frictional head loss in the pipe-work is equal to 1000 velocity heads [ie hf = l O00(u2/2g)] and read more..

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    334 FLUID FLOW FOR CHEMICAL ENGINEERS 2-4 2-5 2-6 2-7 3-1 p = 1200 kg/m 3, the liquid dynamic viscosity/t = 0.001 Pa s and the mean velocity u = 2 m/s. A liquid flows in a steady state in a cylindrical pipe of inside diameter di = 0.05 m at a flow rate Q = 2 x 10 -3 m3/s. Calculate the head loss and the pressure drop for a sudden expansion to a pipe of inside diameter 0.1 m, if the liquid density p = 1000 kg/m 3. A Newtonian liquid flows in steady state in a cylindrical pipe. (a) Calculate point read more..

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    APPENDIX II FURTHER PROBLEMS 33S 3--2 For Newtonian fluids in which the dynamic viscosity ~t is a function only of temperature, ie ~ = f(T), the expression ~ =/a~//z raised to some power is used to correct isothermal equations for non-isothermal conditions. Suggest an analogous correction for non-Newtonian power law fluids flowing in pipes in which the apparent dynamic viscosity/zo is a function of shear stress r shear rate ~, and temperature T, ie /r = f0", ~', T). The laminar flow read more..

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    336 FLUID FLOW FOR CHEMICAL ENGINEERS 3-6 4-1 4-2 4-3 If the material whose viscometric properties were determined in question 3-4 were pumped through a 25 mm diameter pipe so that the wall shear stress had the value corresponding to the .last measurement in that question, what would be the volumetric average velocity and what value of pressure gradient would be required? A plastic film is to be produced by extruding it through a narrow slit of depth 2h. The width of the slit is much greater read more..

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    APPENDIX II FURTHER PROBLEMS 337 4-4 4-5 the discharge and suction sides 2L,a and gL,, respectively, and the volumetric flow rate Q. A system total head against mean velocity curve for a particular power law liquid in a particular pipe system can be represented by the equation Ah = (0.03)(100n)(u n) + 4.0 for u~ 1.5 m/s where Ah is the total head in m u is the mean velocity in m/s and n is the power law index. A centrifugal pump operates in this particular system with a total head against mean read more..

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    338 FLUID FLOW FOR CHEMICAL ENGINEERS 4-6 Two centrifugal pumps are connected in series in a given pumping system. Plot total head Ah against capacity Q pump and system curves and determine the operating points for (a) only pump 1 running (b) only pump 2 running (c) both pumps running on the basis of the following data: operating data for pump 1 Ah]m, 50.0 49.5 48.5 48.0 46.5 44.0 42.0 39.5 36.0 32.5 28.5 O m 3/h, 0 25 50 75 100 125 150 175 200 225 250 operating data for pump 2 Ah2m, 40.0 39.5 read more..

  • Page - 356

    APPENDIX II FURTHER PROBLEMS 339 5-1 5-2 5-3 (c) (d) If very hot fluid is pumped with a gear pump, what difficulty might occur? Gear pumps can be small liquid cavity high speed pumps or large liquid cavity low speed pumps. Which type would you use to pump (i) a shear thinning liquid? (ii) a shear thickening liquid? (iii) a slurry? Calculate the theoretical power in watts for a 0.25 m diameter, six-blade fiat blade turbine agitator rotating at N = 4 rev/s in a tank system with a power curve given read more..

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    340 FLUID FLOW FOR CHEMICAL ENGINEERS 6--1 6-2 6-3 6--4 6-5 equal size tanks are used in series, the removal of solute is n times more effective after a time t where n and t are related by the equation V[(2n - 1)1/2 _ 1] t-- Q An ideal gas in which the pressure P is related to the volume V by the equation PV = 75 m2/s 2 flows in steady isothermal flow along a horizontal pipe of inside diameter d~ = 0.02 m. The pressure drops from 20000 Pa to 10000 Pa in a 5 rn length. Calculate the mass flux read more..

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    APPENDIX II FURTHER PROBLEMS 341 6-6 6-7 6-8 7-1 7-2 four equal stages where 7 is the ratio of heat capacities at constant pressure and constant volume. Air flows from a large reservoir where the temperature and pressure are 25~ and 10 atm, through a convergent-divergent nozzle and discharges to the atmosphere. The area of the nozzle's exit is twice that of its throat. Show that under these conditions a shock wave must occur. (y = 1.4.) Air at a pressure of 5 bar in a closed tank is to be vented read more..

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    342 FLUID FLOW FOR CHEMICAL ENGINEERS 7-3 7-4 8-1 8-2 8-3 0.215 bar. What is the value of the friction factor? Assume isothermal conditions. Data: at 9.90 bar, Vo = 8.28 x 10 -2 m3/kg, Vt. = 1.02 x 10 -3 m~/kg. A mixture of gas and liquid flows through a pipe of internal diameter da = 0.02 m at a steady total flow rate of 0.2 kg/s. The pipe roughness e = 0.000 045 m. The dynamic viscosities of the gas and liquid are ~ = 1.0 x 10 -5 and ~z. = 3.0x 10 -3 Pa s respec- tively. The densities of the read more..

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    APPENDIX II FURTHER PROBLEMS 343 8-4 8-5 9-1 10-1 10-2 be the manometer reading and what is the pressure difference between the two taps? The specific gravity of mercury is 13.6. Calculate the volumetric flow rate in m3/s through a V-notch weir when the height of liquid above the weir is 0.15 m given that the notch angle 0 = 20 ~ and the discharge coefficient Cd = 0.62. Show that a flowmeter with a square root scale has an error 50/Q times that for a linear scale where the maximum volumetric read more..

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    Answers to problems 1-1 1-2 1-3 2-1 2-2 2-3 2-4 2-5 2-7 (a) (PI-P2)IL = pgsinO (b) L = Pl/(2 pg sin 0) (c) 17 m (a) (b) (c) (d) (e) (a) (b) (a) (b) (a) (b) (i) R~ = 31.4 N, Ry = -31.4 N (ii) Rx = -31.4 N, Ry = 31.4 N (iii) R~ = Ry = 0 Total tension = 62.8 N (31.4 N on each flange) Same values because pu 2 has the same value 37.7 N m (anticlockwise) 18150 N m (anticlockwise) PI-P2 --" 0.91 bar /~ pd; 3 )]4/7 u = (~T) I(--~) (0.1584/~2 1.10 rn/s 286.5 Pa/m -2 1584 Pa/m 293 Pa Confirm laminar read more..

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    346 FLUID FLOW FOR CHEMICAL ENGINEERS 3-1 3-2 3-5 4-2 4-3 ~. ~. (a) 4-5 5-1 6-1 6-2 6-3 6--4 6--6 6--7 6-8 (b) (c) 1000 Pa/m ~b = (#~ = (-~-~) [(n~ 1)] (gop)m (m--at mean stress, b---at bulk temperature) At highest speed, ~-- 25.3 Pa, j,- 280 s -~ u = 0.82 m/s, AP/L - 4042 Pa/m 1.038 m Ah = kl + k2Q 1"75 where the constant h l = (Zd -- Z,) + (Pal -- P,)/(Pg) and the constant k2 = (~'L" + ~'L'dXO'239) (p ) (i) Ah = 6.88 m, u = 0.96 m/s (ii) Ah = 6.28 m, u = 1.21 m/s (iii) Ah = 5.60 m, read more..

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    7-1 7-2 7-3 7-4 8-1 8-2 8-3 8--4 9-I (a) (b) (c) ANSWERS TO PROBLEMS 347 0.12 0.18 0.35 f = 0.00461 2775 Pa/m (APa = 239 Pa, APf = 21261 Pa) M ~ = 2.25 kPa (fLo = 0.0061) 1.99 m/s 5.42 x 10 -2 m~/s Az,,, = 0.234 m, M ~ = 31730 Pa 2.25 • 10 -3 m3/s 2.315 • 10 -2 m/s 10-1 4510 s, 1.253 h read more..

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    Conversion factors area density dynamic viscosity energy flow rate, mass per unit time flow rate, volume per unit time heat capacity per unit mass kinematic viscosity length linear velocity mass mass flux pressure pressure gradient power specific volume surface tension temperature difference volume 1 fi2 1 lb/fi 3 1 Ib/UK gal 1 Ib/US gal I cP 1 lb/(h fi) 1 lb/(s fi) 1 Btu 1 ft pdl 1 lb/h I ft3/s I ft3/min 1 UK gal/min 1 US gal/min 1 Btu/(lb F) I ft2/s lft I ft/s 1 lb 1 lb/(h ft 2) 1 atm 1 pdl/ft read more..

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    Friction factor flow charts Friction factor chart for Newtonian Fluids Reynolds number Re read more..

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    350 FLUID FLOW FOR CHEMICAL ENGINEERS 0.01 8 ;F- ~_ "1 1 "" "/'~';" Exptrapo!ated regions-t--~-j,.,-H ..... "; ;i" 0.001 1000 10000 Generalized Reynolds number, Re' Friction factor chart for purely viscous non-Newtonian fluids I, 100000 10~ ~~04 -~ i0.~ .--~i0 e-- --~10 ~. -L-~ i0e "1-~109--"~101~ Hedstr~m no.- 4 - o \,,,\ \ ,, 10- i ..... \ k 9 .9 [- 9 9 "t 9 ,,,.- \ \ ~ ~ .... ~L- ~ ~ \J ~ ,, 0 4 "~ 2 f '16[.+' ~ ~,~ " 10-I read more..

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    Index ACCELERATED plate 312-7 Agitators anchor 170-1 calculation of power for 179, 180 classification of 164 helical screw 171-2 large blade low speed 170 marine propeller 165, 166, 169 circulating capacity of 169 small blade high speed 165 tip speed of 168-9 turbine 165, 166 Apparent fluidity 128 Archimedes number 300 BAFFLES in mixing tanks 165, 167, 168, 172, 175 Bernoulli equation 9-12, 330-1 applications of 12-16, 82, 268, 270, 275-6, 278-9, 305, 331 Bingham plastic see Non-Newtonian fluids read more..

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    352 INDEX Drew equation 74 Drift flux 230 EDDY kinematic viscosity 62, 90-1 Elastic solid 27-9 Elutriation 292 Energy balance 10, 190 cascade 56 conversion between types 11 internal 9, 10, 67, 189-91,200 kinetic 9, 86, 88, 189, 192 distribution correction factor .see Velocity distribution correction factor potential 9, 192 pressure 9, 192 relationships 9-10, 189-91 total 9 turbulent, production of 56, 67 Enthalpy 190, 197, 200, 205, 207 E6tvos number 237 Equation of motion 324-5, see also read more..

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    I N 0 E X 353 rotameter 282-3 scale errors in 284-6 Venturi 14, 275 volumetric flow rate in 277 weir 278-82 Flow rate calculation of, for compressible flow 216-7 calculation of, for incompressible flow 15-6, 78-80, 277-8 calculation of, for incompressible non-Newtonian fluids 108-9, 117-8 errors in measurement of 284 mass 7-8, 70, 19 l, 203, 21 4-5, 223, 308-11 volumetric 7,45 in an annulus 103 in an orifice . 271-2, 277 in a pipe 45, 85 over a weir 280 Fluid, difference from solid 29 read more..

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    364 INDEX HAALAND equation 75 Hagen-Poiseuille equation 46 Head developed by an agitator 170 differential in a manometer 269 discharge 141 flowmeters 268, 270-82 loss due to friction 1 I, 80, 140, 306 net positive suction 142-3, 147, 161,162 potential 11, 140 pressure 11 static 14, 140 suction 141 system 140 total (of pump) 11, 140, 144 velocity 11,140 Heat capacity, specific 195, 200 Heat transfer I0, 189-90, 197 Hedstr6m number 124 Homogeneity 2 Hydraulic slope 94 IDEAL fluids 10, 235 read more..

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    INDEX 3SS scale-up for flow in pipes 11 0-1, 130-1 shear thickening 49-50, 294 shear thinning 49-50, 116, 179, 294 suspensions 125-31, thixotropic 52 time-dependent 48 time-independent 48 turbulent flow of 115-7, 124, 133-4 viscoelastic 48, 53-5, 131-8, 179 PACKED beds 294 Particles non-spherical 291-2, 294 spherical 288-91,292-4, 311-2 Path 3-4 Peebles and Gather correlation 233-4 Pipes and pipelines characteristic time of 319 elasticity of 318 entrance and exit losses for 81 force acting on read more..

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    356 INOEX RABINOWITSCH-MOONEY equation 102-5, ll3, ll9, 123, 129 Recovery factor 205-6 Relative molecular mass 193 Relaxation time 54, 134-5, 136 Reynolds number 6 critical value in a coil 84 critical value for non-Newtonian fluids in pipes I 16 for boundary layers 66 for bubbles 230, 234 for flow round a particle 288 for head flowmeters 273 for mixing 173 for non-Newtonian fluids 114-5, 124 for packed beds 295 for pipes and tubes 70 Reynolds stresses 57, 91 Rheopectic fluids see Non-Newtoman read more..

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    INDEX 367 Turbulence eddies in 55-7 eddy diffusivities of 62 eddy viscosity, calculation of 62-4 energy cascade in 56 fluctuations 57-61 Reynolds stresses in 57, 60-1, 67-8 scales of 56 time averaging of 57-60 transport properties 61-2 turbulent energy production 56, 67 Turbulent flow see Flow, and Turbulence Two-phase multiplier 249-50, see also Gas-liquid two-phase flow UNITS 1 VALVES closing of 317-9 head losses due to 80 throttling 83, 148-9, 317 Velocity angular 293 average see mean read more..

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    360 INDEX Weber number 173 Weissenberg effect 132 Work done on, or by, a fluid 10, 189-90 shaft 190 YIEI~ number 124 Yield stress 51, 123, 179 ZERO slip boundary condition 29, 39, 40, 103, 331 read more..

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