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- 1. Colorado State University Department of Chemical Engineering Ch 406, “Introduction to Transport Phenomena” Table of ContentsMomentum Transport and Fluid DynamicsDefinitions; vectors, tensors, and scalarsDel operator and various time derivativesMathematical statement of the Equation of Continuity for incompressible fluidsSolid body rotation; example of the cross product of two vectors; v = Ω x rApplications of the Equation of Continuity for one-dimensional flow problemsTwo-dimensional incompressible fluid flow analysis via the Equation of ContinuityPhysical properties and transport analogiesViscous stress and momentum fluxThe fundamental balances that describe momentum transportThe accumulation rate processRate processes due to momentum fluxMomentum rate processes due to external body forcesGeneral objectives for solving problems in fluid dynamicsOne-dimensional laminar tube flow of incompressible Newtonian fluidsMomentum shell balancesz-Component of the force balance for τrz(r)Partial differential nature of the Equation of MotionDefinition of dynamic pressureAnalyzing all three components of the Equation of Motion for laminar tube flowDynamic pressure and viscous stress distributionVelocity profiles for boundary value problemsTube flow and flow between two tubesSpatially averaged properties; volumetric flowrateDifferential surface and volume elements in three coordinate systemsMethods to induce fluid flowForce-flow relations for Newtonian and non-Newtonian fluids through a straight tubesThe Hagen-Poiseuille lawA note of caution about the momentum shell balance approach 1 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 2. Couette flowAnalysis of Torque vs. angular velocity for Newtonian fluidsComparison between tube flow and Couette flowModification of Newton’s law for fluids that exhibit solid body rotational characteristicsEffect of centrifugal and gravitational forces on the fluid pressure distributionViscous stress and velocity distributionTorque vs. angular velocity relation for concentric cylinder viscometersCreeping flow analysis of three viscometersParallel-disk viscometersNewtonian fluids; Torque vs. angular velocityForce balances for the parallel-disk viscometerDifferential force due to total momentum flux which acts across the interfaceThe corresponding torqueRotating sphere viscometersNewtonian fluids; Torque vs. Ω, see pages#95-96 in Transport PhenomenaEvaluation of the fluid velocity at the solid-liquid interface via solid-body rotationCreeping flow analysis of the Equation of MotionDifferential vector force due to total momentum flux transmitted across the interfaceThe corresponding torqueCone-and-plate viscometersAll fluids; Torque vs. angular velocityImportant nonzero components of the viscous stress distributionCreeping flow analysis of the Equation of MotionAnalysis of the important nonzero component of the fluid velocity vectorGraphical and analytical solution for the one-dimensional velocity profileDifferential vector force transmitted by the fluid to the stationary plateMacroscopic relation between torque and angular velocityDimensionless momentum transfer correlationsForces exerted by moving fluids on stationary solid surfacesGeneralized interpretation of friction factors vs. Reynolds numbersPractical examples of hydrodynamic dragUse of friction factors vs. Re to analyze pressure drop vs. flowrate in tubesAnalysis of terminal velocities for submerged objectsLinear least squares analysis of terminal velocities for spheres of different radii 2 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 3. Terminal velocity of glass spheres (turbulent flow)Applications of hydrodynamic drag forces via f vs. Re to calibrate a rotameterMacroscopic mass balanceImportant Considerations in the Development and Use of the macroscopic mass balanceTransient and steady state analysis of the dilution of salt solutionsTransient analysis of draining incompressible Newtonian fluids from a spherical bulbCapillary viscomety for the determination of momentum diffusivitiesDetailed evaluation of the capillary constant and comparison with experimental resultsDraining power-law fluids from a right circular cylindrical tank via a tilted capillary tubeComparision of efflux and half-times for incompressible Newtonian fluids in capillary tubesMacroscopic momentum balanceImportant considerations and assumptions in the development and use of the macroscopicmomentum balanceUnsteady state applications of the macroscopic momentum balance to rocket propulsionSteady state applications of the macroscopic momentum balanceOne-dimensional flow through tilted tubes with no change in flow cross-sectional areaMacroscopic mechanical energy balance (the Bernoulli equation)Important concepts in the development of the macroscopic mechanical energy balanceUnique pump designOrifice meter analysis via the non-ideal Bernoulli equation with frictional energy lossMass flowrate vs. pressure drops in Venturi metersApplication of the fluid flow meter equation for laminar tube flowViscous flow through a parallel configuration of orifice and venturi metersFrictional energy loss and pump horsepower requirementsMass Transfer in Reactive and Non-reactive SystemsAccumulation rate processConvective mass transferMolecular mass transfer via Ficks 1st law of diffusionRates of production due to multiple chemical reactionsMass transfer equationSteady state diffusion in stagnant media with no chemical reaction; Laplace’s equation Rectangular coordinates Cylindrical coordinates Spherical coordinates 3 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 4. Analogy with steady state conductive heat transfer Rectangular coordinates Cylindrical coordinates Spherical coordinatesThe analogous problem in fluid dynamics which is described by Laplaces equation.Mass transfer coefficients and Sherwood numbersSteady state film theory of interphase mass transferSteady state diffusion and conduction across flat interfacesEffect of curvature for radial diffusion and conduction across cylindrical interfacesSteady state radial diffusion and conduction in spherical coordinatesGeneral strategy to calculate interphase transfer coefficientsLeibnitz rule for differentiating one-dimensional integralsDiffusion & chemical reaction across spherical gas-liquid interfacesReactive distillationLaminar flow heterogeneous catalytic “tube-wall” reactors; two-dimensional modelConversion vs. time for variable-volume batch reactors; gas phase production of CH3OHTransport analogiesMomentum Transport and Fluid DynamicsLecture#1Nomenclature; dependent (v, τ, p) & independent variables [ (x,y,z); (r,Θ,z); (r,Θ,φ) ]Definition of a vector: associates a scalar with each coordinate direction in a givencoordinate systemVelocity vector---examples in rectangular, cylindrical & spherical coordinatesMass flux vector for pure fluids; ρv, with dimensions of mass per area per timeThe flux of any quantity has dimensions of that quantity per area per timeFlux elevates the tensorial rank of a quantity by one unit Scalars have a rank of zero (i.e., mass, pressure, temperature, energy) Vectors have a rank of 1 (i.e., velocity, mass flux, temp. & pressure grad.) nth-order tensors have a rank of n (i.e., momentum flux, for n=2)"Del" or gradient operator in rectangular coordinates (i.e., spatial rates of change), withdimensions of inverse length; 3 ∂ ∂ ∂ ∂ ∇ = ∑ δi = δ x + δy + δ z ι=1 ∂x i ∂x ∂y ∂z 4 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics €
- 5. Examples of gradients; temperature (∇T), pressure (∇p) and velocity (∇v) gradientsBalance on overall fluid mass---statement of the Equation of Continuity in words(each term has dimensions of mass per time) “Rate of accumulation of overall fluid mass within a differential control volume (CV)= Net rate at which mass enters the control volume due to mass flux acting across all of the surfaces which bound fluid within the control volume”Accumulation rate processes:time rate of change that depends on the nature of the control volume;(partial derivative for stationary CV, total or substantial derivative for moving CV) ∂ d D , , ∂t dt DtMathematical statement of the Equation of Continuity---analogy w/ macroscopic massbalance (i.e., accumulation = rate of input – rate of output, or net rate of input)After division by the size of the control volume, each term in EOC has dimensions ofmass per volume per time;€ ∂ρ = −∇ • ρv = −(ρ∇ • v + v • ∇ρ ) ∂tvia the product rule for differentiation involving the “del” operator and the product of ascalar with a vector. The Equation of Continuity is derived using general vector notationin Problem#8.1 € pages 222-3, TPfCRD. onSteady state form of the microscopic Equation of Continuity for incompressible fluidsComparision of steady state microscopic EOC, ∇ • v = 0 , with the steady statemacroscopic mass balance for (i) tube flow, and (ii) venturi & orifice meters forincompressible fluids, ρ 1<v1>S1 = ρ 2<v2>S2, where S = πD2/4 and ρ 1 = ρ 2. If the flowcross-sectional area does not change from inlet to outlet, then S1 = S2 and the averagefluid velocity is the same across the inlet € plane and the outlet plane.Hence, <v1> = <v2>. An equivalent statement of the balance on overall fluid mass at themicroscopic level, for steady state operation with one-dimensional flow in the z-direction,is ∂vz/∂z = 0, in cylindrical coordinates.Lecture#2Expressions for incompressible EOC, ∇ • v = 0 , in 3 different coordinate systems 5 Introduction to Transport Phenomena € Ch406/Detailed Lecture Topics
- 6. See B.4 on page#846 in BSL’s Transport PhenomenaComment about the additional geometric factors for EOC in cylindrical (1/r) and spherical(1/r2) coordinates.Determine the rank of a vector/tensor operation; Sum the rank of each quantity involved Subtract 2 for the "dot" operation Subtract 1 for the "cross" operation Use the following table to relate the resultant rank to a vector, tensor or scalar Quantity Rank Examples scalar 0 mass, pressure, temperature, energy vector 1 velocity, mass flux, ∇T & ∇p, ∇ x v nth-order tensor n momentum flux (i.e., n=2)For solid-body rotation, the velocity vector of the solid is given by v = Ω x r, where Ω isthe angular velocity vector and r is the position vector from the axis of rotation. Thevorticity vector is defined by (1/2) ∇ x v = (1/2) ∇ x [Ω x r] = Ω.Applications of the microscopic EOC for steady state one-dimensional flow problems withno time dependence (i.e., ∂/∂t = 0)(a) Tube flow---vz(r); w/ 1 inlet stream & 1 outlet stream Average velocity, <vz> = constant, via macroscopic mass balance with no change in the flow cross sectional area ∂vz/∂z = 0, implies that vz ≠ f(z), and Θ is the symmetry variable(b) Flow on the shell side of the double-pipe heat exchanger; ∂vz/∂z=0 & vz(r) EOC is not affected by the boundaries. Same result for tube flow.(c) Tangential flow between two rotating concentric solid cylinders; vΘ(r) ∂vΘ/∂Θ = 0, implies that vΘ ≠ f(Θ), & vΘ ≠ f(z) if end effects are negligible(d) Radial flow, horizontally, between two parallel circular plates; vr(r,z) = f(z)/r 1∂ (rv r ) = 0 , implies that rvr ≠ f(r), and Θ is the symmetry variable r ∂r(e) Diverging flow bounded by two stationary walls @ Θ = ±α; vr(r,Θ) = f(Θ)/r 1∂ € (rv r ) = 0 , implies that rvr ≠ f(r), z isnt important, planar flow r ∂r € 6 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 7. Note: this is a two-dimensional flow problem in rectangular coordinates with vx & vy, butonly one-dimensional flow in polar or cylindrical coordinates (i.e., only vr)(f) Polar flow between two stationary concentric solid spheres; vΘ(r,Θ) = f(r)/sinΘ 1 ∂ (vθ sin θ ) = 0 , implies that vΘ sinΘ ≠ f(Θ), φ is the symmetry variable. r sinθ ∂θ(g) Transient radial flow in spherical coordinates, induced by an expanding bubble; 1 ∂ 2€ 2 (r v r ) = 0 , yields vr(r,t) = f(t)/r 2, because r2vr ≠ f( r), Θ & φ are symmetry r ∂r variables, and the time-varying radius of the bubble precludes a steady state analysis.€For analysis of simple 2-dimensional and 3-dimensional flows via the Equation ofContinuity, see Problem#8-16 (p. 235, TPfCRD) and Problem#8-17 (p. 236, TPfCRD),respectively. Other complex examples of 2-dimensional flow can be found on pp. 284-287 and pp. 304-305. Example of 2-dimensional incompressible fluid flow analysis for hollowfiber ultrafiltration via the Equation of Continuity. Consider axial flow through atube with a permeable wall such that the radial component of the fluid velocity vectorcannot be neglected. Cylindrical coordinates is most appropriate to exploit the symmetryof the macroscopic boundaries. Axial and radial flow occur, where the former is ofprimary importance and the latter is much smaller in magnitude relative to vz, but vr isextremely important for membrane separations that employ hollow filbers. Constant fluiddensity within the tube is reasonable, particularly when another component external tothe capillary is transported into the tube across the porous wall. Application of theEquation of Continuity for this problem yields; 1∂ ∂v ∇•v = (rvr ) + z = 0 r ∂r ∂zOrder-of-magnitude analysis for ∂vz/∂z allows one to rearrange the previous equation andestimate the magnitude of the radial velocity component at the tube wall. Furthermore,multiplication of vr(r=R) by the lateral surface area of the porous section of the tube €(i.e., 2πRL) provides a calculation of the volumetric flowrate across the permeable wallthat agrees with the steady state macroscopic mass balance. If fluid escapes across thetube wall, then the inlet volumetric flowrate Q(z=0) = Qin will be reduced in the exitstream at z=L (Qin > Qout). Hence, unlike steady state flow through a straight tube withconstant cross-sectional area and an impermeable wall, vz depends on axial coordinate z 7 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 8. and the second term in the Equation of Continuity is estimated in “finite-difference”fashion using grid points at the tube inlet and tube outlet; ∂vz 1 Qout − Qin ≈ ∂z L πR2 This “back-of-the-envelope” approximation for ∂vz/∂z is used to estimate vr(r=R) viarearrangement and integration of the Equation of Continuity for 2-dimensional flow incylindrical coordinates. This approach illustrates a general strategy for 2-dimensional €flow problems, based solely on the Equation of Continuity. If one of the two velocitycomponents can be approximated with reasonable accuracy, then the other importantcomponent of the fluid velocity vector is calculated such that one does not violate thebalance on overall fluid mass. The calculation proceeds as follows; ∂v Q − Q d (rvr ) = − z rdr ≈ in 2 out rdr ∂z πR L Rvr (r=R ) Qin − Qout r=R ∫ d (rvr ) ≈ πR2 L ∫ rdr r=0 rvr =0 Q − Q 1 Rvr (r = R) ≈ in 2 out R 2 πR L 2Now, it is possible to (i) estimate the magnitude of the radial velocity component at thetube wall; € Qin − Qout vr (r = R) ≈ 2πRLThis “semi-quantitative” prediction from the microscopic balance on overall fluid massagrees completely with the steady state macroscopic mass balance that equates rates ofinput to rates of output. Since mass flowrates can be replaced by volumetric flowrates if €the fluid density does not change appreciably, the rate of input is Qin, whereas the rate ofoutput is the sum of Qout and 2πRLvr(r=R).ProblemWhat coordinate system is most appropriate to describe the basic information of fluidmechanics for incompressible laminar flow through a tapered tube (i.e., the tube radius isa linear function of axial position)? 8 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 9. Lecture#3Forces due to momentum flux that act across surfaces with orientation defined by nConvective momentum flux, ρvv is obtained from a product of the mass flux vector, ρv,and the ratio of momentum to mass (i.e., v) Viscous momentum flux---use Newtons law of viscosity for Newtonian fluids Pressure contribution to momentum flux, only contributions are normal stressesEach of the 9 scalar components of ρvv and τ needs 2 subscriptsNormal stress vs. shear stress---diagonal vs. off-diagonal matrix elements Normal stresses act in the direction of the unit normal vector n to surface S Shear stresses act in the two coordinate directions which describe surface SSurface S across which the stresses due to momentum flux actthe unit normal vector n is oriented in the coordinate direction identified by the 1stsubscript on ρvv or τ, and the force or stress acts in the coordinate direction given bythe 2nd subscript Viscous Stress and Momentum Flux in Fluid Dynamics The fluid velocity vector is one of the most important variables in fluidmechanics. Remember that a vector is best described as a quantity that has magnitudeand direction. A more sophisticated description identifies a vector as a mathematicalentity that associates a scalar with each coordinate direction in a particular coordinatesystem. Hence, there are three scalar velocity components that constitute the velocityvector, and they are typically written in the following manner in three differentcoordinate systems: vx vy vz in rectangular cartesion coordinates vr vθ vz in cylindrical coordinates vr vθ vϕ in spherical coordinatesIt is important to mention here that each flow problem is solved in only one coordinatesystem---the coordinate system that best exploits the symmetry of the macroscopicboundaries. At the introductory level, the problems will be simple enough that thestudent should identify a primary direction of flow, and only consider the velocitycomponent that corresponds to this flow direction. Hence, it will be acceptable toassume one-dimensional flow and disregard two of the three velocity components forsimple problems in fluid dynamics. 9 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 10. Viscous stress is an extremely important variable, and this quantity is identifiedby the Greek letter, τ. Viscous stress represents molecular transport of momentum thatis analogous to heat conduction and diffusion. All molecular transport mechanismscorrespond to irreversible processes that generate entropy under realistic conditions.When fluids obey Newtons law of viscosity, there is a linear relation between viscousstress and velocity gradients. All fluids do not obey Newtons law of viscosity, butalmost all gases and low-molecular-weight liquids are Newtonian. In this course, we willdiscuss problems in Newtonian fluid dynamics, as well as non-Newtonian fluid dynamics.The problems in non-Newtonian fluid dynamics relate directly to the rheology laboratoryexperiments next semester. Fluid pressure is the third important variable, and it is designated by the letter p.The force balances that we will generate contain fluid pressure because pressure forcesare exerted across surfaces, and there are, at most, six surfaces that completely enclosefluid within a so-called control volume. The fluid within the control volume is the system.Our force balances will apply to fluids in motion---hence, the name, fluid dynamics.However, our balances will be completely general to describe the situation when fluidsare at rest. In other words, the force balances will be applicable to describe hydrostaticswhen the velocity vector and τ vanish. Physical properties and transport analogies Physical properties of a fluid can be described within the context of transportanalogies for all of the transport processes. Numerical solutions to fluid dynamicsproblems require that the viscosity µ and the density ρ are known. If the fluid isNewtonian and incompressible, then both of these physical properties are constants thatonly depend on the fluid itself. The viscosity µ is the molecular transport property thatappears in the linear constituitive relation that equates the molecular transport ofmomentum with velocity gradients. The ratio of viscosity to density is called thekinematic viscosity, υ = µ/ρ, or momentum diffusivity with units of (length)2/time. Numerical solutions to simple thermal energy transport problems in the absenceof radiative mechanisms require that the viscosity µ, density ρ, specific heat Cp, andthermal conductivity k are known. Fouriers law of heat conduction states that thethermal conductivity is constant and independent of position for simple isotropic fluids.Hence, thermal conductivity is the molecular transport property that appears in the linearlaw that expresses molecular transport of thermal energy in terms of temperaturegradients. The thermal diffusivity α is constructed from the ratio of k and ρCp. Hence, α= k/ρCp characterizes diffusion of thermal energy and has units of (length)2/time. 10 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 11. The binary molecular diffusion coefficient, D AB, has units of (length)2/time andcharacterizes the microscopic motion of species A in solvent B, for example. DAB is alsothe molecular transport property that appears in the linear law that relates diffusionalfluxes and concentration gradients. In this respect, the same quantity, DAB, represents amolecular transport property for mass transfer and a diffusion coefficient. This is not thecase for the other two transport processes. Before we leave this section on physical properties, it is instructive to constructthe ratio of the diffusivities for thermal energy transfer and mass transfer with respectto momentum transport. In doing so, we will generate dimensionless numbers thatappear in correlations for heat and mass transfer coefficients. The ratio of momentumdiffusivity υ to thermal diffusivity α is equivalent to the Prandtl number, Pr = υ/α = µCp/k.The Prandtl number is simply a ratio of physical properties of a fluid. However, a verylarge value of the Prandtl number means that diffusion of thermal energy away from ahot surface, for example, is poor relative to the corresponding diffusion of momentum.This implies that the thermal boundary layer which contains all of the temperaturegradients will remain close to the surface when the fluid flow problem is fully developed.Convective transport parallel to a hot surface maintains thin thermal boundary layers by"sweeping away" any thermal energy that diffuses too far from the surface. Fullydeveloped laminar flow in a straight tube of circular cross-section means that themomentum boundary layer (containing all of the velocity gradients) next to the surfaceon one side of the tube has grown large enough to intersect the boundary layer from thesurface on the other side of the tube. It should be no surprise that these boundarylayers will meet in the center of the tube when fully developed flow is attained, and thethickness of the momentum boundary layer is actually the radius of the tube. Hence, avery large Prandtl number means qualitatively that under fully developed laminar flowconditions when the momentum boundary layer has filled the cross-section of the tube,the thermal energy or temperature boundary layer hugs the wall. As a consequence, highrates of heat transfer are prevalent because transport normal to a surface is inverselyproportional to the thickness of the boundary layer adjacent to the surface in question.This boundary layer contains all of the gradients that generate molecular transport. Analogously, the ratio of momentum diffusivity υ to mass diffusivity DAB isequivalent to the Schmidt number, Sc = υ/D AB = µ/ρDAB. It follows directly from thediscussion in the previous paragraph that for very large values of the Schmidt number,mass transfer boundary layers remain close to the adjacent surface and high rates ofmass transfer are obtained. Hence, the Schmidt number is the mass transfer analog ofthe Prandtl number. The momentum transport analog of the Schmidt or Prandtl numbersis 1, because we take the ratio of momentum diffusivity to momentum diffusivity. Theconsequence of this statement is that if a heat transfer correlation containing the Prandtl 11 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 12. number can be applied to an analogous momentum transport problem, then the Prandtlnumber is replaced by 1 to calculate the friction factor. Of course, if the heat transferproblem is completely analogous to a posed mass transfer problem, then the Prandtlnumber in the heat transfer correlation is replaced by the Schmidt number to calculatethe mass transfer coefficient. The fundamental balances that describe momentum transport In this section, we discuss the concepts that one must understand to constructforce balances based on momentum rate processes. The fluid, the specific problem, andthe coordinate system are generic at this stage of the development. If the discussionwhich follows seems quite vague, then perhaps it will become more concrete whenspecific problems are addressed. The best approach at present is to state the forcebalance in words, and then focus on each type of momentum rate process separately. The strategy for solving fluid dynamics problems begins by putting a controlvolume within the fluid that takes advantage of the symmetry of the boundaries, andbalancing the forces that act on the system. The system is defined as the fluid that iscontained within the control volume. Since a force is synonymous with the time rate ofchange of momentum as prescribed by Newtons laws of motion, the terms in the forcebalance are best viewed as momentum rate processes. The force balance for an opensystem is stated without proof as, 1 = 2 - 3 + 4 , where:1 is the rate of accumulation of fluid momentum within the control volume2 is the rate at which fluid momentum enters the control volume via momentum flux acting across the surfaces that bound the fluid within the control volume3 is the rate at which fluid momentum leaves the control volume via momentum flux acting across the surfaces that bound the fluid within the control volme4 is the sum of all external forces that act on the fluid within the control volumeIt should be emphasized that force is a vector quantity and, hence, the force balancedescribed qualitatively above is a vector equation. A vector equation implies that threescalar equations must be satisfied. This is a consequence of the fact that if two vectorsare equal, then it must be true that they have the same x-component, the same y-component, and the same z-component, for example, in rectangular coordinates. At theintroductory level, it is imperative that we choose the most important of the three scalarequations that represent the vector force balance. The most important scalar equation 12 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 13. is obtained by balancing forces in the primary direction of flow--and since we assumeone-dimensional flow, it should be relatively straightforward to identify the importantflow direction and balance forces in that direction. The accumulation rate process We are ready to associate mathematical quantities with each type of momentumrate process that is contained in the vector force balance. The fluid momentum vector isexpressed as ρv, which is equivalent to the overall mass flux vector. This is actually themomentum per unit volume of fluid because mass is replaced by density in the vectorialrepresentation of fluid momentum. Mass is an extrinsic property that is typically a linearfunction of the size of the system. In this respect, mv is a fluid momentum vector thatchanges magnitude when the mass of the system increases or decreases. This change influid momentum is not as important as the change that occurs when the velocity vectoris affected. On the other hand, fluid density is an intrinsic property, which means that itis independent of the size of the system. Hence, ρv is the momentum vector per unitvolume of fluid that is not affected when the system mass increases or decreases. Weare ready to write an expression for the rate of accumulation of fluid momentumwithin the control volume. This term involves the use of a time derivative todetect changes in fluid momentum during a period of observation that is consistent withthe time frame during which the solution to the specified problem is required. If ΔVrepresents the size of the control volume, then d (ρv ΔV) dtis the mathematical representation of the accumulation term with units of momentumper time--hence, rate of momentum. A few comments are in order here before weproceed to the form of the other terms in the force balance. If the control volume isstationary, or fixed in space, then the spatial coordinates of ΔV are not functions of time.Consequently, the control volume ΔV can be moved to the left side of the derivativeoperator because ΔV is actually a constant and the total time derivative can be replacedby the partial time derivative. Hence, terms of type 1 in the force balance can besimplified as follows when the control volume is fixed in space; ∂( ) ΔV ρv ∂t 13 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 14. It should be obvious that this term is volumetric, meaning that the accumulationmechanism applies to the entire system contained within the control volume. Thestipulation that the control volume is stationary did simplify the mathematics to someextent, but the final form of the force balance does not depend on details pertaining tothe movement of the control volume. Possibilities for this motion include a controlvolume that is stationary with fixed spatial coordinates, a control volume that moves atevery point on its surface with the local fluid velocity, or a control volume that moveswith a velocity that is different from the local fluid velocity. Of course, we chose thesimplest case, but as I indicated above, the final form of all of the balances is not afunction of this detail. Finally, if the fluid is a liquid, then the assumption ofincompressibility is typically invoked. An incompressible liquid is characterized by adensity that is not a strong function of pressure based on equation of state principles.Hence, for isothermal liquid systems, one assumes that density changes are negligibleand the accumulation mechanism can be simplified further as; ∂v ρΔV ∂tRemember that this "unsteady state" term is unimportant and will be neglected when weseek solutions that are independent of time---the so-called steady state behaviour of thesystem. In practice, we must wait until the transients decay and the measurablequantities are not functions of time if we hope to correlate steady state predictions withexperimental results. Rate processes due to momentum flux The terms identified by 2 and 3 in the force balance are unique because they aresurface-related and act across the surfaces that bound the fluid within the controlvolume. Surface-related is a key word, here, which indicates that flux is operative. Theunits of momentum flux are momentum per area per time. There are three contributions to momentum flux that have units of momentumper area per time. Since these units are the same as force per unit area, one of the fluxmechanisms is pressure. Remember that pressure is a scalar quantity, which means thatthere is no directional nature to fluid pressure. In other words, fluid pressure actssimilarly in all coordinate directions. However, pressure forces are operative in a fluid,and they act perpendicular to any surface that contacts the fluid. These forces act alongthe direction of the unit normal vector that characterizes the orientation of the surfaceand, for this reason, pressure forces are classified as normal forces. In general, a normal 14 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 15. force is one that acts perpendicular to the surface across which the force is transmitted.Choose any well-defined simple surface in one of the coordinate systems mentionedabove (rectangular, cylindrical, or spherical) and the student should be able to identifytwo orthogonal coordinate directions within the surface, and one coordinate that isnormal to the surface. Consider the walls, floor, or ceiling of a room in rectangularcoordinates, for example. An alternative viewpoint is a follows--as one moves on asimple surface, two coordinates change and one remains fixed. This simple surface istypically defined as one with a constant value of the coordinate that remains fixed in thesurface. The coordinate that remains fixed is also in the direction of the unit normalvector. In summary, forces due to momentum flux act across surfaces and can beclassified as normal forces or shear forces. As mentioned above, normal forces actperpendicular to a surface along the unit normal vector. Shear forces act parallel to thesurface along one of the two coordinate directions that make up the surface. Hence,momentum flux initially identifies a simple surface with a unit normal vector that iscoincident with one of the unit vectors of an orthogonal coordinate system. Thenmomentum flux identifies a vector force per unit area that acts across this surface, andthis vector force has three scalar components. One of these scalar force componentsacts co-linear with the unit normal vector to the surface, and this force is designated asa normal force. The other two scalar force components act along coordinate directionswithin the surface itself, and these forces are called shear forces because the surfacearea across which the force acts is parallel to the direction of the force. Another important contribution to momentum flux is due to convective fluidmotion, and this mechanism is called convective momentum flux---designated by ρvv.The mathematical form of convective momentum flux is understood best by initiallyconstructing the total mass flux vector for a pure or multicomponent fluid, and thengenerating the product of mass flux with momentum per unit mass. Mass flux is avectorial quantity that has units of mass per area per time, and ρv is the mathematicalrepresentation of the total mass flux vector. Of course, ρv also represents themomentum vector per unit volume of fluid as introduced above for the accumulation rateprocess. The total mass flux vector represents an important contribution to the balanceon overall fluid mass. If one accepts ρv as a vectorial representation of the convectiveflux of overall fluid mass, then it is possible to construct the product of ρv with themomentum vector per unit mass of fluid, the latter of which is analogous to the velocityvector v. This product of ρv and v is not the scalar ("dot") product or the vector("cross") product that the student should be familiar with from vector calculus.Convective momentum flux is a quantity that generates nine scalars. This should beobvious if one chooses the rectangular coordinate system for illustration and multipliesthe three scalar components of the mass flux vector (ρvx, ρv y, ρv z) by the three scalarcomponents of the velocity vector (vx, vy, vz). Using rigorous mathematical terminology, 15 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 16. convective momentum flux ρvv is a second-rank tensor that associates a vector witheach coordinate direction. Since there are three orthogonal coordinate directions that areidentified by the unit vectors of the chosen coordinate system, ρvv identifies a vectorwith each of the three coordinate directions. Remember that a vector associates a scalarwith each of three coordinate directions, also in the chosen coordinate system. Hence,the student should rationalize that there are nine scalars that one can generate fromthree distinct vectors, and these nine scalars constitute a second-rank tensor such asconvective momentum flux. At this stage in our discussion of momentum flux, which isunique to the study of fluid dynamics, it is instructive to write all nine scalars of ρvv andcomment about the subscripts on the scalar velocity components. It should beemphasized that the discussion which follows is applicable to the most complex flowproblems and actually contains much more detail than that which is necessary to analyzeone-dimensional flow. This claim is substantiated by the fact that eight of the ninescalars of ρvv are identically zero for a simple one-dimensional flow problem with avelocity vector given by v = δ xvx + δy(0) + δz(0). If fluid motion is restricted t o the x-direction in rectangular coordinates as illustrated above, then the only non-vanishingscalar of convective momentum flux is ρvxvx, which has units of momentum per time perarea or force per unit area. Hence, ρvv represents force per unit area that is transmittedacross the surfaces that bound fluid within the control volume, and terms of this naturedue to convection motion of a fluid must be included in a force balance. As mentionedabove, one must construct the product of each of the nine scalars generated from thissecond-rank tensor with the surface area across which the force (or stress) istransmitted. Information about these surfaces and the coordinate direction in which theforce acts is contained in the subscripts of the velocity components. For the mostgeneral type of fluid flow in rectangular coordinates, the nine scalars that one cangenerate from convective momentum flux are given below; ρvxvx ρvxvy ρvxvz ρvyvx ρvyvy ρvyvz ρvzvx ρvzvy ρvzvzIt should be obvious that the nine scalars illustrated above for convective momentum fluxfit nicely in a three-by-three matrix. All second-rank tensors generate nine scalars, and itis acceptable to represent the tenor by the matrix of scalars. If the matrix is symmetric,then the tensor is classified as a symmetric tensor. This is true for convectivemomentum flux because the product of two velocity components vivj does not change ifthe second component is written first. Another positive test for symmetry is obtainedby interchanging the rows and columns of the three-by-three matrix to generate asecond matrix that is indistinguishable from the original matrix. 16 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 17. As an illustrative example, let us focus on the element in the first row and secondcolumn, ρvxvy, for the matrix representation of the convective momentum flux tensor.The subscript x on the first velocity component indicates that ρvxvy is a force per unitarea acting across a simple surface oriented with a unit normal vector in the ±x-direction.The subscript y on the second velocity component tells us that this force acts in the y-direction. If we perform this analysis for all nine components in the matrix for ρvv above,then the three entries in the first row represent x-, y-, and z-components, respectively,of the force per unit area that is transmitted across the simple surface defined by aconstant value of the x coordinate, which means that the unit normal vector to thesurface is co-linear with the x-direction. Likewise, the three entries in the second row ofthe matrix represent x-, y-, and z-components, respectively, of the force per unit areathat is transmitted across the simple surface defined by a constant value of the ycoordinate, which means that the unit normal vector to the surface is co-linear with they-direction. Finally, the three entries in the third row of the matrix represent x-, y-, andz-components, respectively, of the force per unit area that is transmitted across thesimple surface defined by a constant value of the z coordinate, which means that theunit normal vector to the surface is co-linear with the z-direction. Notice that the matrixcomponents on the main diagonal from upper left to lower right have the same twosubscripts and can be written in general as ρ(vi)2. These forces satisfy the requirementfor normal forces. Each one acts in the ith coordinate direction (where i = x, y, or z) andthe unit normal vector to the surface across which the force is transmitted is also in theith direction. In summary, when a momentum flux tensor is expressed in matrix form, themain diagonal entries from upper left to lower right represent forces per unit area thatact along the direction of the normal vector to the surface across which the force istransmitted. The off-diagonal elements represent shearing forces because these forcesact in one of the two coordinate directions that define the surface across which the forceis transmitted. Using this formalism, it is also possible to represent the pressure contribution tomomentum flux in matrix notation. However in this case, all of the entries have the samemagnitude (i.e., p) and they lie on the main diagonal from upper left to lower right.There are no off-diagonal components because fluid pressure generates surface forcesthat act in the direction of the unit normal vector to the surface across which the forceor stress is transmitted---they are all normal forces. In each coordinate system, thematrix representation of the pressure contribution to momentum flux can be written inthe following form; p 0 0 0 p 0 0 0 p 17 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 18. Before we depart from the discussion of rate processes due to momentum flux, itis necessary to consider the molecular mechanism that relates viscous stress to linearcombinations of velocity gradients via Newtons law of viscosity, if the fluid is classifiedas a Newtonian fluid. Viscous momentum flux is also a second-rank tensor that identifiesa vector force per unit area with each of the three coordinate directions. These forcesare not due to inertia or bulk fluid motion like those that are generated from ρvv, butthey are best viewed as frictional forces that arise when fluid parcels on adjacentstreamlines slide past one another because they are moving at different relative speeds.A simple analogy of the shearing forces generated by viscous momentum flux is theaction that one performs with a piece of sandpaper to make a wood surface smoothe.The wood surface is analogous to the wall of a pipe, for example, and the motion of thesandpaper is representative of the fluid layers that are adjacent to the wall. The surfaceforces under consideration definitely meet the requirements of shearing forces becausethe surface is oriented parallel to the direction of fluid motion, the latter of whichcoincides with the direction of the force. In polymer processing operations, if the fluidviscosity is large enough and the flow is fast enough, then thermal energy will begenerated by frictional shear at the interface between the fluid and the wall. This isanalogous to the fact that a wood surface is slightly warmer after it is sanded, and thesurface temperature is higher when the sanding is performed more vigorously. When wegenerate the matrix representation of viscous momentum flux τ, it is necessary to puttwo subscripts on the letter τ to facilitate the row and column for each entry. Unlike theconvective momentum flux tensor where we inserted a single subscript on each velocitycomponent in the product vv, we must now put both subscripts on τ . However, it isacceptable to analyze two subscripts on τ in the same manner that we analyzed ρvivjabove. In rectangular coordinates, the matrix representation for viscous momentum fluxis written as follows; τxx τxy τxz τyx τyy τyz τzx τzy τzzThe interpretation of these nine scalars follows directly from the discussion of the ninescalars that are generated by ρvv. The only difference is that the forces result from amolecular mechanism that is analogous to heat conduction and mass diffusion, ratherthan bulk fluid motion. For example, the second row of scalars represents x-, y-and z-components, respectively, of the viscous force per unit area that is transmitted acrossthe simple surface defined by a constant value of the y coordinate, which means that theunit normal vector to the surface is co-linear with the y-direction. 18 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 19. We have now introduced a total of 21 scalar quantities; 9 from ρvv, 9 from τ ,and 3 from the pressure contribution to momentum flux; that identify all of the possiblesurface force components which can be generated from the total momentum flux tensor.When each of these scalars is multiplied by the surface area across which the force acts,a quantity with units of momentum per time is obtained that represents a term of type 2or 3 in the force balance. Inputs are identified as type 2, and outputs are classified astype 3. It should be no surprise that the 21 scalar surface forces are distributed equallyamong the three scalar balances that constitute the total vector force balance. Basedon the introductory statements above with respect to the double-subscript nature of thescalar forces generated by ρvv and τ, the student should be able to identify the surfaceacross which the force acts from the first subscript, and the direction in which the forceacts from the second subscript. Momentum rate processes due to external body forces All terms in the momentum balance have units of momentum per unit time, whichis synonymous with the units of force. In this respect, it is necessary to include terms oftype 4 in the force balance because they account for all of the external forces that acton the fluid within the control volume. These terms are different than those catagorizedby types 2 and 3 because they do not act across surfaces that bound the fluid withinthe control volume. Type 4 forces are usually called body forces because they actvolumetrically like the accumulation rate process, which means that each fluid parcelwithin the system experiences the same effect due to a body force. The primary bodyforce that we will encounter is gravity. The external vector force due to gravity is writtenintrinsically via the fluid density in the following manner; ρgΔVwhere g is the gravitational acceleration vector. Once again, we have used the fluiddensity instead of the total fluid mass within the system to insure that our external forcedoes not have to be modified if the mass or size of the system changes. It is true thatthe size of the control volume ΔV could change in response to an increase or decrease insystem mass at constant density. However, one of the last steps in deriving the forcebalance for a specific problem is division by the control volume which generates acompletely intrinsic equation that is independent of system size or mass. There areother types of external body forces in addition to gravity that should be included in acompete study of fluid dynamics. For example, fluid particles that have permanentelectric dipoles will experience body forces in the presence of an electric field, andparticles with magnetic moments experience forces and torques due to magnetic fields.These forces are important and must be considered in a study of ferrohydrodynamics and 19 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 20. magnetohydrodynamics. Unfortunately, fluid flow in the presence of electric andmagnetic fields is rarely covered in undergraduate as well as graduate courses becausethe complexity of the problems increases several-fold, limiting discussion to the simplestexamples for which exact solutions require the use of advanced mathematical techniques.Even though surface tension forces cannot be classified as body forces, they play animportant role in the operation of viscosity-measuring devices, like the parallel-plate andcone-and-plate viscometers, where a thin film of fluid is placed between two closelyspaced horizontal surfaces---the lower surface being stationary and the upper onerotating at constant angular velocity. In the absence of surface tension, the test fluidwould spread and completely wet the solid surfaces in response to rotation whichgenerates centrifugal forces. Then the fluid would "fall off the table" since there are norestraining walls. Of course, surface tension plays the role of restraining walls and keepsthe fluid from exiting the viscometer if the rotational speeds are slow enough. General objectives for solving problems in fluid dynamics Most flow problems that we will encounter involve a fluid in motion adjacent to astationary wall---the wall of a tube, for example---or a fluid that is set in motion by amoving surface---this is the case in a viscosity-measuring device. In general, the bulkfluid and the solid surface are moving at different relative speeds, and this generatesvelocity gradients and viscous stress at the interface. Macroscopic correlations in fluiddynamics focus on the fluid-solid interface and calculate the force exerted by the fluid onthe solid, or vice versa, via the fluid velocity gradient "at the wall". These macroscopicmomentum transport correlations contain the friction factor and the Reynolds number.Hence, one calculates the Reynolds number from the characteristics of the flow problemand uses these dimensionless correlations to determine the value of the friction factor.Frictional energy losses in straight sections of a tube are estimated from the frictionfactor. The size of a pump required to offset all of the dissipative processes that reducefluid pressure can be estimated from the non-ideal macroscopic mechanical energybalance (i.e., Bernoulli equation) that incorporates friction loss via the friction factor. Insome cases, macroscopic momentum transfer correlations relate torque and angularvelocity for the viscosity-measuring devices, allowing one to calculate the viscosities ofNewtonian and non-Newtonian fluids from measurements of torque vs. angular velocity.Example:The lateral surface of a straight tube with radius R and length L. Considerone dimensional flow in the z-direction induced by some combination of apressure gradient and gravitational forces. Hence, steady state applicationof the Equation of Continuity yields, vz = f(r). 20 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 21. The unit normal is n = δr. Hence, the 1st subscript on ρvv and τ must be r. Pressure force acts in the r-direction, δrp pressure forces are normal forces Forces due to convective momentum flux vanish, if no-slip is a valid assumption The only nonzero force due to ρvv is ρvzvz, which acts across a surface where n = δz. For 3-dimensional flow, there are δrρvrvr, δΘρvrvΘ, and δzρvrvz The primary force due to viscous momentum flux is τrz, if vz(r) via NLV All forces or stresses due to total momentum flux act across the same surface area, 2πRL; Direction inConvective forces Viscous forces Pressure forces which the forces act ρvrvr τrr p δr ρvrvΘ τrΘ - δΘ ρvrvz τrz - δzSteady state one-dimensional fluid flow of an incompressible Newtonian fluidIf the functional dependence of the fluid velocity profile is vi(xj), then balance forces inthe ith-coordinate direction. Hence, the 2nd subscript on ρvv and τ is i. The controlvolume which contains the fluid should be differentially thick only in the jth-coordinatedirection. This yields an ordinary differential equation for the dependence of τji on xj viadτji/dxj. If the velocity profile is vi(xj), then for incompresible Newtonian fluids, one shouldexpect the following nonzero elements of the viscous stress tensor; τij and τji.Qualitative statement of the momentum shell balance, or the force balanceThere are 3 scalar components because this is a vector equation Accumulation of fluid momentum in control volume (CV) = + Input due momentum flux acting across surfaces which bound fluid in the CV - Output due to momentum flux acting across surfaces which bound fluid in the CV + Sum of all external forces acting on the fluid in the CV (i.e., gravitational force)Lecture#4>Forces are represented by arrows>By convention, draw all arrows in the positive coordinate directions; + forces>Inputs due to momentum flux act across surfaces defined by smaller values of theindependent variable that remains constant within the surface>Outputs due to momentum flux act across surfaces defined by larger values of theindependent variable that remains constant within the surface 21 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 22. Steady state analysis; laminar flow of an incompressible Newtonian fluid through a tube;The classic problem in fluid dynamics; vz(r); tube is horizontal, no gravity forces in z-dir.The control volume is represented by a cylindrical shell at radius r with length LSize of the control volume is 2πrLdr; differentially thick in the radial direction, onlyThere are four different surfaces that bound fluid within the control volumeThe system is chosen as the fluid that occupies the differential control volumez-Component of the force balance---the objective is to obtain an ODE for τrz(r)z-component forces are represented by arrows. Only consider forces or stress due tototal momentum flux that act in the z-direction. The 2nd subscript on ρvv & τ is z. Theonly stresses that must be considered are; Unit normal vector to the surfaceStress across which the stress acts Surface areaPressure p δz normal stress 2πrdrConvective momentum flux ρvrvz δr shear stress 2πrL ρvΘvz δΘ shear stress Ldr ρvzvz δz normal stress 2πrdrViscous momentum flux τrz δr shear stress 2πrL τΘz δΘ shear stress Ldr τzz δz normal stress 2πrdrUse Newtons law of viscosity to determine which components of τ are important Only τrz = τzr survives if vz(r) and ∇ • v = 0 The nine elements of τ can be represented by a symmetric 3x3 matrix There are only 6 independent equations for τ via NLV, due to symmetry All of the 2nd-rank tensors in NLV are symmetricLecture#5z-component force balance for one-dimensional laminar tube flow ∂ ∂v Accumulation rate process is volumetric; (ρvz dV ) = ρdV z ∂t ∂t Input due to convective momentum flux occurs at z=0; [ ρvzvz 2πr dr ]z=0 Input due to fluid pressure occurs at z=0; [ p 2πr dr ]z=0 Input due to viscous momentum flux occurs at r; [ τrz 2πrL ]r € 22 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 23. Output due to convective momentum flux occurs at z=L; [ ρvzvz 2πr dr ]z=L Output due to fluid pressure occurs at z=L; [ p 2πr dr ]z=L Output due to viscous momentum flux occurs at r+dr; [ τrz 2πrL ]r+dr External force due to gravity is volumetric; ρgz dVSurface area 2πrdr is perpendicular to the z-direction Surface areas 2πrL and 2π(r+dr)L are perpendicular r-directionAccumulation = Input - Output + Sum of external forcesAt steady state, Accumulation = 0Input = forces due to momentum flux exerted by the surroundings on the systemOuptut = forces due to momentum flux exerted by the system on the surroundingsInput = Output, for convective momentum flux, because vz ≠ f(z) via EOCThere is no horizontal z-component of the force due to gravity; gz = 0The z-component force balance represents a balance between pressure and viscousforces; [ p 2πr dr ]z=0 + [ τrz 2πrL ]r = [ p 2πr dr ]z=L + [ τrz 2πrL ]r+drDivide the previous equation by the size of the control volume, dV = 2πrLdr, which isdifferentially thick in the radial direction, rearrange the equation and combine terms; (rτ rz )r+dr − (rτ rz )r p(z = 0) − p(z = L) Δp = = rdr L LTake the limit as dr → 0, and recognize the definition of the 1st derivative. The result isthe z-component of the Equation of Motion in cylindrical coordinates in terms of viscousstress; € 1 d Δp (rτ rz ) = r dr L"Divide and conquer", or "separate and integrate". Integrate the previous equation withrespect to independent variable r. The result is; € Δp rτ rz = C1 + r 2 2L 23 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics €
- 24. C1 Δp τ rz (r) = + r r 2L Since the range of r is from the centerline @ r=0 to the wall @ r=R, C1=0 because viscousstress is finite. Hence; € Δp τ rz (r) = r 2L Now, relate τ rz to the velocity gradient via Newtons law of viscosity if the fluid isNewtonian; € dvz Δp τ rz (r) = −µ = r dr 2L Integration yields; Δp r 2 vz (r) = C2 − L 4µ €and integration constant C2 is evaluated from the “no-slip” boundary condition at thestationary wall, where vz=0. Finally, one obtains the quadratic velocity profile forincompressible flow of a Newtonian fluid through a straight tube of radius R and length L; € R 2 Δp vz (r) = (1− η 2 ) 4 µ L where η = r/R. The fluid velocity vz is maximum along the centerline of the tube andminimum at the stationary solid wall (i.e., no slip). Viscous shear stress τrz is maximum atthe solid wall and minimum at the centerline (i.e., r=0) where symmetry is invoked. €Lecture#6Equation of Motion (EOM) represents the vector force balance in fluid dynamicsA vector equation implies that 3 scalar equations must be satisfiedConsider the classic problem of 1-dimensional incompressible laminar flow through astraight tube with circular cross section 24 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 25. Make the control volume differentially thick in both the r-direction and the z-directionReplace the inlet plane at z=0 with an inlet plane at zReplace the outlet plane at z=L with an outlet plane at z + dzNow, dV = 2πrdrdz (i.e., replace L by dz)The "shell balance" approach for the z-component force balance for fluids in which onlyτrz is important yields; Accumulation = Input - Output + Sum of external forces ∂/∂t { ρvz dV } = + [ ρvzvz 2πr dr ]z - [ ρvzvz 2πr dr ]z+dz + [ τrz 2πr dz ]r - [ τrz 2πr dz ]r+dr + [ p 2πr dr ]z - [ p 2πr dr ]z+dz + ρgz dVDivide by dV = 2πr dr dz; ∂/∂t { ρvz } = { + [ ρvzvz 2πr dr ]z - [ ρvzvz 2πr dr ]z+dz } / 2πr dr dz { + [ τrz 2πr dz ]r - [ τrz 2πr dz ]r+dr } / 2πr dr dz { + [ p 2πr dr ]z - [ p 2πr dr ]z+dz } / 2πr dr dz + ρgzTake the limit as both dr → 0 and dz → 0; ∂/∂t { ρvz } = { + [ ρvzvz ]z - [ ρvzvz ]z+dz } / dz { + [ r τrz ]r - [ r τrz ]r+dr } / r dr { + [ p ]z - [ p ]z+dz } / dz + ρgzOne obtains the z-component of the Equation of Motion (EOM) for 1-dimensional laminarflow of any type of fluid that exhibits no normal viscous stress (i.e., τ zz≈0) through astraight tube in the entrance region, prior to the establishment of fully developed flow; 25 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 26. ∂ ∂ 1∂ ∂p ( ρv z ) = − (ρv zv z ) − (rτ rz ) − + ρgz ∂t ∂z r ∂r ∂zManipulate the accumulation and convective momentum flux terms with assistance fromthe Equation of Continuity; € ∂v ∂v ∂ ∂ ∂ρ ∂ ( ρv z ) + (ρv zv z ) = ρ z + v z + ρv z z + v z ( ρv z ) ∂t ∂z ∂t ∂t ∂z ∂z ∂ρ ∂ ∂v ∂v = v z + (ρv z ) + ρ z + v z z ∂t ∂ z ∂t ∂z EOC for 1-dimensional flow (i.e., z-direction) of any type of fluid, with vr = vΘ = 0; € ∂ρ ∂ + ( ρv z ) = 0 ∂t ∂zHence; ∂ ∂ ∂v ∂v 1∂ ∂p ( ρv z ) + (ρv zv z ) = ρ z + v z z = − (rτ rz ) − + ρgz ∂t ∂z ∂t ∂z r ∂r ∂z €The previous equation is not restricted to incompressible fluids with constant density,even though ρ is outside of the derivative operators. Consider the last two terms on the €right side of the previous z-component force balance; z-component forces due to pressure and gravity = - ∂p/∂z + ρgz In vector notation, these two forces can be expressed as - ∇p + ρgDefine the gravitational potential energy per unit mass Φ, such that Φ = gh, where g isthe gravitational acceleration constant and h is a spatial coordinate which increases inthe coordinate direction opposite to gravity (i.e., h = z, measured vertically upward).The gravitation acceleration vector can be written as; g = g ( - δh ) = - δh ∂Φ/∂h = - ∇ΦFor incompressible fluids, where the fluid density ρ is constant, the gravitational force perunit volume can be written as; 26 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 27. ρg = - ρ ∇Φ = - ∇ρΦPressure and gravity forces in the Equation of Motion can be written as follows forincompressible fluids; - ∇p + ρg = - ∇p - ∇ρΦ = - ∇(p + ρΦ) = - ∇Pwhere P represents "dynamic pressure", which is a combination of fluid pressure andgravitational potential energy.Lecture#7As a consequence of the previous vector algebra, it is acceptable to neglect thegravitational force in each scalar component of the Equation of Motion and replace fluidpressure p by dynamic pressure P. The previous steady state z-component force balancefor fully developed 1-dimensional laminar tube flow of an incompressible fluid [i.e., vz(r)]reduces to; 1∂ ∂P z-component; 0=− (rτ rz ) − r ∂r ∂z r-component: 0 = - ∂P/∂r, implies that P ≠ f(r) € 1 ∂P Θ-component: 0=− , implies that P ≠ f(Θ) r ∂ΘSince steady state analysis implies that P ≠ f(t), it is reasonable to consider that thefunctional dependence of dynamic pressure is P = f(z), and one replaces ∂P/∂z by dP/dz. €Furthermore, for 1-dimensional flow with vz = f(r), Newtons law of viscosity reveals thatτrz is only a function of r. Hence, one replaces ∂/∂r (rτrz) by d/dr (rτrz) if the fluid isNewtonian. Now, the z-component force balance can be written as; 1 d dP z-component; (rτ rz ) = − = Constan t(C1) r dr dzvia separation of variables, because the viscous stress term involving τrz is only a functionof r, whereas the dynamic pressure term is only a function of z. If; € - dP/dz = Constant (C1)then; P(z) = - C1 z + C2 27 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 28. represents a linear dynamic pressure distribution along the tube axis. Integrationconstants C1 and C2 can be determined from the following boundary conditions; z=0 P = P0, which implies that C2 = P0 z=L P = PL, which implies that C1 = ( P0 - PL ) / L = ΔP / LNow, one solves for the viscous shear stress distribution τrz(r) from; 1 d dP ΔP (rτ rz ) = − = C1 = r dr dz L ΔP d ( rτ rz ) = rdr L ΔP rτ rz = r 2 + C3 2L ΔP C τ rz (r) = r + 3 2L rLecture#8Boundary value problems; €The previous generic result for the viscous shear stress distribution is valid for;(a) One-dimensional (i.e., only vz) incompressible Newtonian fluid flow in the laminar regime through a tube at any orientation angle with respect to gravity Boundary conditions based on the range of the independent variable, 0 ≤ r ≤ R; τrz is finite along the symmetry axis of the tube @ r=0. Hence, C3 = 0. "No slip" at the solid-liquid interface, vz(@r=R) = 0. Let η = r/R; dv z ΔP τ rz (r) = −µ = r dr 2L ΔP 2 v z (r) = C4 − r 4 µL R 2ΔP 2 v z (r) = {1− η } 4 µL 28 Introduction to Transport Phenomena € Ch406/Detailed Lecture Topics
- 29. (b) One-dimensional (i.e., only vz) incompressible Newtonian fluid flow in the laminar regime between two concentric cylinders: axial annular flow at any orientation angle with respect to gravity. Boundary conditions based on the range of the independent variable, Rinner ≤ r ≤ Router; "No slip" at the inner and outer stationary tube walls vz = 0 at r=Rinner and r=Router dv z ΔP C3 τ rz (r) = −µ = r + dr 2L r C ΔP 2 v z (r) = C5 − 3 ln r − r µ 4 µL This is essentially a “distorted” quadratic profile for vz(r) due to the presence of thelogarithmic term when axial flow occurs between two concentric cylinders. The final €result for vz(r) in these previous two examples could have been obtained directly via twointegrations of the z-component of the Equation of Motion in terms of velocity gradientsfor incompressible Newtonian fluids, without solving for the viscous shear stressdistribution as an intermediate step. It is only justified to use the tabulated form of theEquation of Motion on page#848 when the fluid is incompressible and Newtonian, and theflow is laminar. The Equation of Motion in terms of τ on page#847 should be employedfor compressible fluids and non-Newtonian fluids.ProblemConsider radius ratio κ = Rinner/Router = 0.1 and 0.9 for laminar flow of an incompressibleNewtonian fluid on the shell side of the double pipe heat exchanger. At which of thesetwo radius ratios does a quadratic approximation to the actual velocity profile provide abetter fit to the exact solution obtained by solving the z-component of the Equation ofMotion?Answer: For each value of the radius ratio κ, one finds the best values for a0, a1, and a2 that minimize the difference between the dimensionless “distorted” quadratic velocity profile, given by the previous equation after integration constants C3 and C5 are evaluated via two no-slip boundary conditions, and the following dimensionless second-order polynomial; 29 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 30. {vz (r )} AnnularFlow = a0 + a1η + a2η 2 {vz,maximum }TubeFlow where η = r/R. Results are summarized in the table below; € κ a0 a1 a2 0.01 0.31 1.60 -1.97 0.05 0.08 1.94 -2.06 0.10 -0.09 2.12 -2.07 0.20 -0.34 2.37 -2.05 0.30 -0.57 2.58 -2.03 0.40 -0.78 2.79 -2.02 0.50 -0.99 3.00 -2.01 0.60 -1.19 3.20 -2.01 0.70 -1.40 3.40 -2.00 0.80 -1.60 3.60 -2.00 0.90 -1.80 3.80 -2.00 0.95 -1.90 3.90 -2.00 0.99 -1.98 3.98 -2.00The effects of curvature are much less significant when κ = 0.9 relative to smaller valuesof the radius ratio. In fact, the axial quadratic velocity profile is not significantly differentfrom the “distorted” quadratic profile when κ = 0.9 because flow between two concentriccylinders with very similar radii is essentially the same as flow between two flat plates,the latter of which is given by a quadratic profile.Numerical AnalysisConsider the axial velocity profile for flow of an incompressible Newtonian fluid on theshell side of the double-pipe heat exchanger. The expression for vz is given as equation2.4-14 on page 55 [i.e., Transport Phenomena, 2nd edition, RB Bird, WE Stewart, ENLightfoot (2002)], and the corresponding volumetric flow rate, Q, is given as equation2.4-17 on the same page.ProblemUse the Newton-Raphson numerical root-finding technique to calculate the dimensionlessradial position, η = r/R, in the annular flow configuration where the streamline velocity isa specified fraction (β) of the centerline (i.e., maximum) fluid velocity for tube flow. Use 30 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 31. double precision and report your answer as ηroot that satifies the following convergencecriterion; F(ηroot) < 10-6, for the cases given below; [ β = 1 – η2 + (1 – κ2)ln(η)/ln(1/κ) ] a) κ = 0.20 β = 0.30 (Two solutions; η = 0.41 and η = 0.70) b) κ = 0.20 β = 0.35 (No real solutions exist; β > βMaximum = 0.341) c) κ = 0.30 β = 0.25 (Two solutions; η = 0.57 and η = 0.66) d) κ = 0.30 β = 0.30 (No real solutions exist; β > βMaximum = 0.254) e) κ = 0.40 β = 0.15 (Two solutions; η = 0.55 and η = 0.81) f) κ = 0.40 β = 0.20 (No real solutions exist; β > βMaximum = 0.184)As radius ratio κ increases, the maximum value of β (see Eq. 2.4-15, BSL) decreases.ProblemUse the Newton-Raphson numerical root-finding technique to calculate the required radiusratio, κ, such that the volumetric rate of flow on the shell side of the double-pipe heatexchanger is a specified fraction (β) of the volumetric flow rate for tube flow. Usedouble precision and report your answer as κroot that satifies the following convergencecriterion; G(κroot) < 10-6, for the three cases given below; a) β = 0.20 (κ = 0.405) b) β = 0.50 (κ = 0.147) c) β = 0.80 (0.006 ≤ κ ≤ 0.007) [ β = 1 – κ4 – (1 – κ2)2/ln(1/κ) ]Spatially averaged properties;Volumetric flowrate for one-dimensional axial flow though a straight tube, 2π R 1 Q = S vz Average = ∫∫ v (r)dS = ∫ ∫ z Θ=0 v (r)rdrdΘ ≈ {ΔP} r=0 z n Swhere S is the flow cross-sectional area (i.e., πR2) and n is the power-law index for thepower-law model. When n=1 for Newtonian fluids in the laminar flow regime, the relationbetween volumetric flowrate Q and dynamic pressure difference ΔP is known classically as €the Hagen-Poiseuille law. dS is a differential surface element normal to the flow direction,which is constructed from a product of two differential lengths in the "no-flow"directions.Differential surface and volume elements; 31 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 32. Rectangular coordinate directions; x y z Differential lengths in these directions; dx dy dz Cylindrical coordinate directions; r Θ z Differential lengths in these directions; dr r dΘ dz Spherical coordinate directions; r Θ φ Differential lengths in these directions; dr r dΘ r sinΘ dφConstruct dV via the product of differential lengths in all three coordinate directions.Construct dS for the flow cross-sectional area or the differential surface element at asolid-liquid interface from the product of two differential lengths (i.e., choose the correctones).There are at least 4 methods to induce fluid flow, excluding capillary or surface tensionforces: (1) Impose a pressure gradient---forced convection (2) Take advantage of gravity---forced convection (3) Move a solid surface that contacts the fluid---viscous shear (4) Impose a temperature gradient---free convectionLecture#9Force-flow relations for Newtonian and non-Newtonian fluids through a straight tube withcircular cross-section; 1 logQ ≈ log(ΔP ) nApply a quasi-steady state model for laminar flow through an exit tube to analyze theunsteady state behaviour associated with draining a cylindrical tank or spherical bulb witha cylindrical exit tube. € exit tube has radius R, length L, and it is oriented at angle Θ Thewith repect to gravity. The height of fluid in the tank or bulb above the tube is h(t). Forincompressible Newtonian fluids, the volume rate of flow through the exit tube, withradius R, is given by the Hagen-Poiseuille law; πR 4 ΔP Q= 8µLwhere; ΔP = Pinlet - Poutlet € 32 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 33. At the tube outlet, which is chosen as the “zero of potential energy”, one writes; Poutlet = pambientAt the tube inlet; Pinlet = { pambient + ρgh(t) } + ρgL cosΘHence; ΔP = ρg [ h(t) + L cosΘ ]The previous relations illustrate an example where dynamic pressure is calculated for theapproximate hydrostatic conditions in a very large tank or bulb (i.e., p = pambient + ρgh),and the resulting dynamic pressure difference between the tube inlet and the tube outletis employed in a hydrodynamic relation to calculate the volumetric flowrate forincompressible Newtonian fluids through the exit tube.A note of caution---the momentum shell balance approach should not be used toanalyze fluid flow problems with curved streamlines because centrifugal and coriolisforces will not be accounted for properly in cylindrical and spherical coordinates. Hence,one should apply the vector force balance in fluid dynamics, in general, by considering allthree components of the Equation of Motion for a particular problem in only onecoordinate system.Couette flow---analysis of the rheology experiment in Ch306 for Newtonian and non-Newtonian fluids; Torque T vs. angular velocity ΩConcentric cylinder viscometer Inner solid cylinder at radius κR rotates at constant angular velocity Ωinner Outer cylinder at radius R rotates at constant angular velocity ΩouterFast rotation of the inner solid cylinder produces 2-dimensional flow (i.e., vΘ & vr) Centrifugal forces are important, which are repsonsible for vr Streamlines are represented by outward spirals The Reynolds number is largeSlow rotation of the inner solid cylinder yields 1-dimensional flow (i.e., only vΘ) Centrifugal forces are negligible Streamlines are represented by concentric circles The Reynolds number is small (i.e., <1). This is the creeping flow regime. It is acceptable to neglect the entire left side of the Equation of Motion.Steady state analysis implies that vΘ ≠ f(t) 33 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 34. The Equation of Continuity yields; (1/r) ∂vΘ/∂Θ = 0. Hence, vΘ ≠ f(Θ)Analysis of the flow problem far from the ends of the rotating cylinders suggests thatend effects are not important. Hence, vΘ ≠ f(z), but there is no balance or law whichprovides this result. If end effects are important in practice, then the actual length ofthe rotating cylinder is usually increased empirically in the final calculations.One conlcudes that vΘ = f(r), which is reasonable because the tangential fluid velocitychanges considerably as one moves from the inner cylinder to the outer cylinder.Comparison between tube flow and Couette flow; Consideration Tube flow Couette flow Important velocity component vz(r) vΘ(r) Streamlines straight curved Steady state analysis vz ≠ f(t) vΘ ≠ f(t) Symmetry vz ≠ f(Θ) --- Neglect end effects --- vΘ ≠ f(z) Equation of Continuity vz ≠ f(z) vΘ ≠ f(Θ)For Newtonian fluids in the Couette viscometer that obey Newtons law of viscosity, thefollowing information is useful to determine the important nonzero components ofviscous stress; vr = 0 vz = 0 vΘ = f(r) ∇•v =0The only nonzero components of τ are; d vΘ τ rΘ = τ Θr = −µr dr r This illustrates an example where Newtons law must be modified for flow systems withcurved streamlines so that solid body rotational characteristics within the fluid produceno viscous stress. For example, consider the final result when the inner cylinder at radius €κR rotates at constant angular velocity Ωinner, and the outer cylinder at radius R rotates atconstant angular velocity Ωouter in the same direction. 1 κ 2 R2 vΘ (r) = (Ωouter − κ Ωinner )r − (Ωouter − Ωinner ) 2 1− κ 2 r € 34 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 35. Two limiting cases (i.e., see Problem 3B.1(a) on page#105); (a) Both cylinders rotate at the same angular velocity; Ωinner = Ωouter = Ω vΘ = Ω r (b) There is no inner cylinder; κ = 0 vΘ = Ωouter rIn both cases, fluid motion is the same as solid body rotation, where v = Ω x r, andthere is no viscous stress if the angular velocity is constant. If there is no viscous stress,then there is no torque/angular-velocity relation, and the device does not function as aviscosity-measuring device. In the complete expression for vΘ above, the term whichscales as r does not contribute to τ rΘ or the torque. The term which scales as 1/r issolely responsible for the torque/angular-velocity relation.Look at the following problems;2B.3 page#63 Laminar flow through a rectangular slot2B.6 page#64 Laminar flow of a falling film on the outside of a tubeIdentify the important non-zero component of viscous stress. Identify the radialposition where viscous shear stress τrz is maximum. Then calculate the maximum valueof viscous shear stress.2B.7 page#65 Axial annular flow in the laminar regime induced by translation of the inner cylinderIdentify the important non-zero component of viscous stress. Identify the radialposition where viscous shear stress τrz is maximum. Then calculate the maximum valueof viscous shear stress.3B.10 page#108 Radial flow between two parallel disks.Calculate the functional form of the pressure distribution for radial flow between twoparallel disks of circular geometry and sketch your answer.Lecture#10 Continuation of the Couette viscometer problemConsider all three components of the Equation of Motion in cylindrical coordinates for anytype of fluid (i.e., EOM in terms of τ), with vΘ(r). Since the most important component ofEOM is the Θ-component, lets consider the two components of secondary importance; 35 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 36. 2 vΘ ∂p ∂P r-component: −ρ = − + ρgr = − r ∂r ∂r ∂p ∂P z-component: 0=− + ρgz = − ∂z ∂z €If the z-axis of the concentric cylinder configuration is vertical and independent variable zincreases downward, such that gz = g and both gr and gΘ vanish, then; € 2 ∂p vΘ r-component: = ρ , due to centrifugal forces ∂r r ∂p z-component: = ρg , due to gravitational forces ∂z €Intuitively, one invokes symmetry and writes that p ≠ f(Θ), because it is difficult toenvision how one could impose a pressure gradient in the Θ-direction without theexistence of normal viscous stress, like τ ΘΘ. Certainly, conventional pumps are not €equipped to perform this task. Hence, fluid pressure depends on radial position r, due tocentrifugal forces, and axial position z, due to gravity. One obtains the fluid pressuredistribution p(r,z) via "partial integration" of the previous two equations after solving forvΘ(r).Consider the most important component of the Equation of Motion in cylindricalcoordinates for, vΘ = f(r), τrΘ = g(r), and P ≠ h(Θ). All other quantities are zero. 1 d 2 Θ-component of EOM in terms of τ: 0=− r 2 dr (r τ rΘ ) C1 The general solution is: τ rΘ = r2 €Θ-component of EOM in terms of velocity gradients for incompressible Newtonian fluids; € 36 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics
- 37. d 1 d 0=µ ( rvΘ ) dr r dr 1 d ( rvΘ ) = C2 r dr d ( rvΘ ) = C2 r dr 1 rvΘ = C2 r 2 + C3 2 1 C vΘ (r) = C2 r + 3 2 rCheck for consistency between the separate solutions for τrΘ and vΘ using NLV; € d vΘ C1 τ rΘ (r) = −µr = dr r r 2 d 1 C 2µC = −µr C2 + 23 = 2 3 dr 2 r rTherefore, C1 = 2 µ C3. €Boundary Conditions (to calculate C2 & C3); @ Rinner @ Router (a) Stationary inner cylinder vΘ = 0 vΘ = Ω Router (b) Stationary outer cylinder vΘ = Ω Rinner vΘ = 0 (c) Both cylinders are rotating vΘ = Ωinner Rinner vΘ = Ωouter RouterProblemConsider tangential annular flow of an incompressible Newtonian fluid between tworotating cylinders. Obtain an expression for the important velocity component whenboth cylinders are rotating at the same angular velocity Ω, but in opposite directions. 37 Introduction to Transport Phenomena Ch406/Detailed Lecture Topics

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