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Introduction to Transport PhenomenaCh406/Detailed Lecture Topics

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Colorado State UniversityDepartment of Chemical Engineering

Ch 406, “Introduction to Transport Phenomena”

Table of Contents

Momentum 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 analogies

Viscous 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


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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 viscometers

Creeping flow analysis of three viscometers

Parallel-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 torque

Rotating 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 torque

Cone-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 velocity

Dimensionless 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


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Terminal velocity of glass spheres (turbulent flow)Applications of hydrodynamic drag forces via f vs. Re to calibrate a rotameter

Macroscopic 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 tubes

Macroscopic 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 area

Macroscopic 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 requirements

Mass Transfer in Reactive and Non-reactive SystemsAccumulation rate processConvective mass transferMolecular mass transfer via Fick's 1st law of diffusionRates of production due to multiple chemical reactions

Mass transfer equationSteady state diffusion in stagnant media with no chemical reaction; Laplace’s equation

Rectangular coordinatesCylindrical coordinatesSpherical coordinates


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Analogy with steady state conductive heat transferRectangular coordinatesCylindrical coordinatesSpherical coordinates

The analogous problem in fluid dynamics which is described by Laplace's 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 integrals

Diffusion & 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 analogies

Momentum 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 time

Flux elevates the tensorial rank of a quantity by one unitScalars 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;

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∇ = δi∂∂xiι=1

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∑ = δx∂∂x

+ δy∂∂y

+ δz∂∂z


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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)

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∂∂t, ddt, DDt

Mathematical 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;

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∂ρ∂t

= −∇•ρv = −(ρ∇•v+ v•∇ρ)

via 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 on pages 222-3, TPfCRD.

Steady state form of the microscopic Equation of Continuity for incompressible fluidsComparision of steady state microscopic EOC,

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∇ •v = 0, with the steady statemacroscopic mass balance for (i) tube flow, and (ii) venturi & orifice meters forincompressible fluids, ρ1<v1>S1 = ρ2<v2>S2, where S = πD2/4 and ρ1 = ρ2. 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,

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∇ •v = 0, in 3 different coordinate systems


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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 involvedSubtract 2 for the "dot" operationSubtract 1 for the "cross" operationUse the following table to relate the resultant rank to a vector, tensor or scalar

Quantity Rank Examplesscalar 0 mass, pressure, temperature, energyvector 1 velocity, mass flux, ∇T & ∇p, ∇ x vnth-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 streamAverage velocity, <vz> = constant, via macroscopic mass balance with no changein 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

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1r∂∂r(rvr) = 0 , implies that rvr ≠ f(r), and Θ is the symmetry variable

(e) Diverging flow bounded by two stationary walls @ Θ = ±α; vr(r,Θ) = f(Θ)/r

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1r∂∂r(rvr) = 0 , implies that rvr ≠ f(r), z isn't important, planar flow


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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

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1rsinθ

∂∂θ(vθ sinθ) = 0 , implies that vΘ sinΘ ≠ f(Θ), φ is the symmetry variable.

(g) Transient radial flow in spherical coordinates, induced by an expanding bubble;

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1r2

∂∂r(r2vr ) = 0, yields vr(r,t) = f(t)/r 2, because r2vr ≠ f( r), Θ & φ are symmetry

variables, and the time-varying radius of the bubble precludes a steady stateanalysis.

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;

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∇•v =1r∂∂r

rvr( ) +∂vz∂z

= 0

Order-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


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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;

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∂vz∂z

≈1L

Qout −Qin

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-dimensionalflow 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;

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d rvr( ) = −∂vz∂z

rdr ≈

Qin −Qout

πR2L

rdr

d rvr( )rvr =0

Rvr r=R( )

∫ ≈Qin −Qout

πR2L

rdrr=0

r=R

∫

Rvr r = R( ) ≈ Qin −Qout

πR2L

12R2

Now, it is possible to (i) estimate the magnitude of the radial velocity component at thetube wall;

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vr r = R( ) ≈ Qin −Qout

2πRL

This “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 ifthe 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)?


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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 Newton's law of viscosity for Newtonian fluidsPressure contribution to momentum flux, only contributions are normal stresses

Each 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 SShear stresses act in the two coordinate directions which describe surface S

Surface S across which the stresses due to momentum flux actthe unit normal vector n is oriented in the coordinate direction identified by the 1st

subscript 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 coordinatesvr v θ vz in cylindrical coordinatesvr v θ v ϕ in spherical coordinates

It 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.


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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 Newton's law of viscosity, there is a linear relation between viscousstress and velocity gradients. All fluids do not obey Newton's 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. Fourier's 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.


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The binary molecular diffusion coefficient, DAB, 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/DAB = 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


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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 Newton's 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 volume

2 is the rate at which fluid momentum enters the control volume via momentumflux acting across the surfaces that bound the fluid within the control volume

3 is the rate at which fluid momentum leaves the control volume via momentumflux acting across the surfaces that bound the fluid within the control volme

4 is the sum of all external forces that act on the fluid within the control volume

It 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


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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

ddt

(ρvΔV)

is 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 ∂

∂t(ρv )


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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;

ρΔV∂v∂t

Remember 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


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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, ρvy, ρvz) by the three scalarcomponents of the velocity vector (vx, vy, vz). Using rigorous mathematical terminology,


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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 = δxvx + 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 ρvzvz

It 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.


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 00 p 00 0 p


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 Newton's 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 ρvivj

above. In rectangular coordinates, the matrix representation for viscous momentum fluxis written as follows;

τxx τxy τxz

τyx τyy τyz

τzx τzy τzz

The 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.


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;

ρgΔV

where 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


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).


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 forcesForces 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 surfacewhere 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 NLVAll forces or stresses due to total momentum flux act across the samesurface area, 2πRL;

Direction inConvective forces Viscous forces Pressure forces which the

forces actρvrvr τrr p δr

ρvrvΘ τrΘ - δΘρvrvz τrz - δz

Steady 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


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 volume

z-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 2πrLρvΘvz δΘ shear stress Ldrρvzvz δz normal stress 2πrdr

Viscous momentum fluxτrz δr shear stress 2πrLτΘz δΘ shear stress Ldrτzz δz normal stress 2πrdr

Use Newton's law of viscosity to determine which components of τ are importantOnly τrz = τzr survives if vz(r) and ∇ • v = 0The nine elements of τ can be represented by a symmetric 3x3 matrixThere are only 6 independent equations for τ via NLV, due to symmetryAll of the 2nd-rank tensors in NLV are symmetric

Lecture#5z-component force balance for one-dimensional laminar tube flow

Accumulation rate process is volumetric;

€

∂∂t(ρvzdV ) = ρdV ∂vz

∂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


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 dV

Surface area 2πrdr is perpendicular to the z-directionSurface areas 2πrL and 2π(r+dr)L are perpendicular r-direction

Accumulation = Input - Output + Sum of external forces

At 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 = 0

The 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+dr

Divide 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 )rrdr

=p(z = 0)− p(z = L)

L=ΔpL

Take 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;

€

1rddr(rτ rz ) =

ΔpL

"Divide and conquer", or "separate and integrate". Integrate the previous equation withrespect to independent variable r. The result is;

€

rτ rz =C1 +Δp2L

r2


24

€

τ rz (r) =C1r

+Δp2L

r

Since the range of r is from the centerline @ r=0 to the wall @ r=R, C1=0 because viscousstress is finite. Hence;

€

τ rz (r) =Δp2L

r

Now, relate τrz to the velocity gradient via Newton's law of viscosity if the fluid isNewtonian;

€

τ rz (r) = −µdvzdr

=Δp2L

r

Integration yields;

€

vz (r) =C2 −ΔpL

r2

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;

€

vz (r) =R2

4µ

ΔpL

(1−η2 )

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 satisfied

Consider the classic problem of 1-dimensional incompressible laminar flow through astraight tube with circular cross section


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

+ ρgz

Take 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 + ρgz

One 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., τzz≈0) through astraight tube in the entrance region, prior to the establishment of fully developed flow;


26

€

∂∂t(ρvz ) = −

∂∂z(ρvzvz) −

1r∂∂r(rτ rz ) −

∂p∂z

+ ρgz

Manipulate the accumulation and convective momentum flux terms with assistance fromthe Equation of Continuity;

€

∂∂t(ρvz ) +

∂∂z(ρvzvz) = ρ

∂vz∂t

+ vz∂ρ∂t

+ ρvz∂vz∂z

+ vz∂∂z(ρvz )

= vz∂ρ∂t

+∂∂z(ρvz)

+ ρ∂vz∂t

+ vz∂vz∂z

EOC for 1-dimensional flow (i.e., z-direction) of any type of fluid, with vr = vΘ = 0;

€

∂ρ∂t

+∂∂z(ρvz ) = 0

Hence;

€

∂∂t(ρvz ) +

∂∂z(ρvzvz) = ρ

∂vz∂t

+ vz∂vz∂z

= −1r∂∂r(rτ rz) −

∂p∂z

+ ρgz

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 theright 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 + ρg

Define 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 = - ∇Φ

For incompressible fluids, where the fluid density ρ is constant, the gravitational force perunit volume can be written as;


27

ρg = - ρ ∇Φ = - ∇ρΦ

Pressure and gravity forces in the Equation of Motion can be written as follows forincompressible fluids;

- ∇p + ρg = - ∇p - ∇ρΦ = - ∇(p + ρΦ) = - ∇P

where 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;

z-component;

€

0 = −1r∂∂r(rτ rz) −

∂P∂z

r-component: 0 = - ∂P/∂r, implies that P ≠ f(r)

Θ-component:

€

0 = −1r∂P∂Θ

, implies that P ≠ f(Θ)

Since 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), Newton's 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;

z-component;

€

1rddr(rτ rz ) = −

dPdz

= Constan t(C1)

via 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


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 / L

Now, one solves for the viscous shear stress distribution τrz(r) from;

€

1rddr(rτ rz ) = −

dPdz

= C1 =ΔPL

d rτ rz( ) =ΔPL

rdr

rτ rz =ΔP2L

r2 + C3

τ rz (r) =ΔP2L

r +

C3

r

Lecture#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 laminarregime 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;

€

τ rz(r) = −µdvzdr

=ΔP2L

r

vz (r) = C4 −ΔP4µL

r2

vz (r) =R2ΔP4µL

1−η2{ }


29

(b) One-dimensional (i.e., only vz) incompressible Newtonian fluid flow in the laminarregime between two concentric cylinders: axial annular flow at any orientationangle 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 wallsvz = 0 at r=Rinner and r=Router

€

τ rz (r) = −µdvzdr

=ΔP2L

r +

C3

r

vz(r) = C5 −C3

µln r − ΔP

4µL

r2

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 finalresult 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, anda2 that minimize the difference between the dimensionless “distorted”quadratic velocity profile, given by the previous equation after integrationconstants C3 and C5 are evaluated via two no-slip boundary conditions, andthe following dimensionless second-order polynomial;


30

€

vz r( ){ }AnnularFlow

vz ,max imum{ }TubeFlow= a0 + a1η+ a2η

2

where η = r/R. Results are summarized in the table below;

κ a0 a1 a2

0.01 0.31 1.60 -1.970.05 0.08 1.94 -2.060.10 -0.09 2.12 -2.070.20 -0.34 2.37 -2.050.30 -0.57 2.58 -2.030.40 -0.78 2.79 -2.020.50 -0.99 3.00 -2.010.60 -1.19 3.20 -2.010.70 -1.40 3.40 -2.000.80 -1.60 3.60 -2.000.90 -1.80 3.80 -2.000.95 -1.90 3.90 -2.000.99 -1.98 3.98 -2.00

The 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


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,

€

Q = S vz Average= vz (r)dS = vz (r)rdrdΘ ≈ ΔP{ }

1n

r=0

R∫Θ=0

2π∫

S∫∫

where 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 asthe 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;


32

Rectangular coordinate directions; x y zDifferential lengths in these directions; dx dy dz

Cylindrical coordinate directions; r Θ zDifferential lengths in these directions; dr r dΘ dz

Spherical coordinate directions; r Θ φDifferential lengths in these directions; dr r dΘ r sinΘ dφ

Construct 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 convection

Lecture#9Force-flow relations for Newtonian and non-Newtonian fluids through a straight tube withcircular cross-section;

€

logQ ≈1n

log ΔP( )

Apply 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. The exit tube has radius R, length L, and it is oriented at angle Θwith 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;

€

Q =πR4ΔP8µL

where;ΔP = Pinlet - Poutlet


33

At the tube outlet, which is chosen as the “zero of potential energy”, one writes;

Poutlet = pambient

At the tube inlet;Pinlet = { pambient + ρgh(t) } + ρgL cosΘ

Hence;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 Ω

Concentric cylinder viscometerInner solid cylinder at radius κR rotates at constant angular velocity Ωinner

Outer cylinder at radius R rotates at constant angular velocity Ωouter

Fast 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 spiralsThe Reynolds number is large

Slow rotation of the inner solid cylinder yields 1-dimensional flow (i.e., only vΘ)Centrifugal forces are negligibleStreamlines are represented by concentric circlesThe 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)


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 flowImportant velocity component vz(r) vΘ(r)Streamlines straight curvedSteady 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 Newton's law of viscosity, thefollowing information is useful to determine the important nonzero components ofviscous stress;

vr = 0 vz = 0 vΘ = f(r) ∇ • v = 0

The only nonzero components of τ are;

€

τ rΘ = τΘr = −µr ddr

vΘr

This illustrates an example where Newton's 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 radiusκR rotates at constant angular velocity Ωinner, and the outer cylinder at radius R rotates atconstant angular velocity Ωouter in the same direction.

€

vΘ(r) =1

1−κ 2Ωouter −κ

2Ωinner( )r − Ωouter −Ωinner( )κ2R2

r


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 = ΩvΘ = Ω r

(b) There is no inner cylinder; κ = 0vΘ = Ωouter r

In 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 τrΘ 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 slot

2B.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 bytranslation of the inner cylinder

Identify 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, let's consider the two components of secondary importance;


36

r-component:

€

−ρvΘ2

r= −

∂p∂r

+ρgr = −∂P∂r

z-component:

€

0 = −∂p∂z

+ρgz = −∂P∂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;

r-component:

€

∂p∂r

= ρvΘ2

r , due to centrifugal forces

z-component:

€

∂p∂z

= ρg , due to gravitational forces

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 τΘΘ. Certainly, conventional pumps are notequipped 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.

Θ-component of EOM in terms of τ:

€

0 = −1r2

ddr

r2τ rΘ( )

The general solution is:

€

τ rΘ =C1r2

Θ-component of EOM in terms of velocity gradients for incompressible Newtonian fluids;


37

€

0 = µddr

1rddr

rvΘ( )

1rddr

rvΘ( ) = C2

ddr

rvΘ( ) = C2r

rvΘ =12

C2r

2 + C3

vΘ (r) =12

C2r +

C3

r

Check for consistency between the separate solutions for τrΘ and vΘ using NLV;

€

τ rΘ(r) = −µr ddr

vΘr

=

C1r2

= −µr ddr

12

C2 +

C3

r2

=2µC3

r2

Therefore, 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 Router

ProblemConsider 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.


38

Review Problems for Exam#1

Problem#1Is the following functional form of the fluid velocity field valid or invalid for two-dimensional (i.e., vx and vy) laminar flow in the xy-plane for an incompressible non-Newtonian polymer solution at steady state? Provide quantitative support for youranswer.

vx = f(y) vy = g(x,y) vz = 0Answer:For two-dimensional flow of an incompressible fluid, the Equation of Continuity must besatisfied in the following form;

€

∇•v =∂vx∂x

+∂vy∂y

= 0

for both Newtonian and non-Newtonian fluids. Hence, if vx is not a function of x, asprovided in the problem statement, then the 1st term vanishes, and vy must not be afunction of independent variable y. The functional form of vy, provided in the problemstatement, contradicts this conclusion. The complete velocity field does not satisfy theEquation of Continuity for incompressible fluids, which implies that the microscopicbalance on overall fluid mass is not satisfied. Hence, the functional forms of vx & vy areinvalid when they are analyzed together.

Problem#2Consider fully-developed one-dimensional flow of an incompressible Newtonian fluid in thepolar direction (i.e., only vΘ) between two stationary concentric spheres of radius κR & R,where κ<1. The Equation of Continuity reveals that the functional form of the steadystate fluid velocity profile is;

€

vΘ(r,Θ) =f (r)sinΘ

and the flow configuration is illustrated on page#106 (Fig. 3B.4) in TransportPhenomena.

(a) Construct an integral expression for the volumetric flowrate Q. Be sure to includelimits of integration in your final answer.

Answer:The desired result for Q is obtained by integrating vΘ over the differential surfaceelement dS, which is constructed from a product of two differential lengths in the "no-


39

flow" directions (i.e., r & φ). One integrates in the radial direction from κR to R, and inthe φ-direction from 0 to 2π. Hence;

€

Q = vΘS∫∫ (r,Θ)dS =f (r)sinΘ

r sinΘdrdφ

κR

R∫0

2π∫

€

Q = 2π rf (r)dr ≠ g(Θ)κR

R∫

(b) At which polar angle Θ is the volumetric flowrate Q largest? [ ε ≤ Θ ≤ π-ε ]

Answer:Since the answer to part (a) does not depend on polar angle Θ, it should be obvious thatthe volumetric flowrate is constant for incompressible fluids at steady state, with oneinlet stream and one outlet stream.

(c) Identify all of the non-zero scalar components of total momentum flux which actacross the solid-liquid interface at r = R (i.e., the spherical shell), and thenindicate the coordinate direction in which each stress acts. Do not include anyquantities in your final answer which are identically zero.

Answer:The unit normal vector at the solid-liquid interface, where r=R, is oriented in the radialdirection. This implies that pressure stress acts in the radial direction, because it is anormal force, and the 1st subscript on the scalar components of convective and viscousmomentum flux is r. All of the possible stresses that act across the solid-liquid interfaceat r=R are tabulated below;Stress Coordinate Non-zero? Reason

DirectionPressure p r yes p > 0Convective momentum flux

ρvrvr r no vr = 0ρvrvΘ Θ no vr = 0ρvrvφ φ no vr = 0

Viscous momentum fluxτrr r no vr = 0, ∇ • v = 0τrΘ Θ yes ∂/∂r (vΘ/r) ≠ 0τrφ φ no vr = vφ = 0


40

Hence, the only non-zero forces due to total momentum flux acting across the solid-liquid interface at r=R are fluid pressure p, which acts in the radial direction, and viscousshear stress τrΘ, which acts in the Θ-direction.

Problem#3Consider one-dimensional (i.e., only vz) steady state laminar flow of an incompressibleNewtonian fluid through a straight channel with rectangular cross section. Independentvariables x & y are defined in the transverse plane, perpendicular to the flow, and zincreases along the axis of the channel in the primary flow direction. Fluid flow is drivenby a dynamic pressure gradient dP/dz in the primary flow direction.

(a) What important information is obtained from the Equation of Continuity for thisincompressible one-dimensional flow problem?

Answer:Since vx = 0 and vy = 0, the Equation of Continuity reduces to;

€

∇•v =∂vz∂z

= 0

which implies that vz is not a function of z.

(b) What is the functional form of the non-zero velocity component? vz = f(?).

Answer:Steady state analysis implies that vz is not a function of time, and the Equation ofContinuity reveals that vz ≠ f(z). Hence, vz is a function of x & y, and neither of thesetwo independent variables can be eliminated because velocity gradients exist in both thex and y directions. If the rectangular-shaped cross section has an extremely large aspectratio, which corresponds to either a large height or width, then gradients of vz in the"long" dimension are negligible with respect to similar gradients in the "short" direction.

(c) Write the 2nd-order differential equation which must be solved to calculate thevelocity profile, vz. Do not include any terms that are trivially zero in the mostimportant component of the Equation of Motion.

Answer:If vx = 0, vy = 0, and vz = f(x,y), then the x- and y-components of the Equation of Motionyield the following information;


41

x-component: 0 = - ∂P/∂x, therefore P ≠ f(x)

y-component: 0 = - ∂P/∂y, therefore P ≠ f(y)

These results imply that dynamic pressure is only a function of z at steady state. Now,the important component of the Equation of Motion in terms of velocity gradients forincompressible Newtonian fluids provides the 2nd-order partial differential equation whichallows one to calculate vz(x,y);

z-component:

€

0 = −dPdz

+ µ∂ 2vz∂x2

+∂ 2vz∂y2

If fluid flow is driven only by a pressure gradient in the z-direction, then gz = 0 and onecan replace the dynamic pressure gradient in the previous equation by the fluid pressuregradient, dp/dz. Separation of variables yields dp/dz = - Δp/L.

Numerical analysis directly related to Problem#3The axial velocity profile, driven by an imposed pressure drop, for laminar flow of anincompressible Newtonian fluid through a horizontal channel of rectangular cross sectionrepresents a complex flow problem. The dimensionless velocity profile is given below interms of the appropriate dimensionless independent variables.

v‡z =

2

(1−y‡2) + 4∑

n→∞

n=0

(−1)n+1 M−3n cosh-1 (Mn Ar) cosh(Mn Ar x

‡) cos(Mn y‡)

43 − 8

Ar∑n→∞

n=o

[ M−5n tanh(Mn Ar)]

where;

x‡ = xa y‡ = y

b Ar = ab Mn = (2n + 1)

π2

The flow cross section has a length of 2a in the x-direction and height of 2b in the y-direction, and the rectangular coordinate system is oriented such that the range of theindependent variables is;

-a ≤ x ≤ a-b ≤ y ≤ b


42

a) Write the differential equation with all of its boundary conditions that must besolved to obtain the complex velocity profile given above.

b) Obtain a numerical answer for the volumetric rate of flow in dimensionlessform by using Simpson's double integration rule based on second-order Legendreinterpolation polynomials when the aspect ratio Ar = 3. Numerical integrationshould be performed over a grid that contains a minimum of 49 points in eachcoordinate direction.

Problem#4Consider steady state incompressible creeping (i.e., very slow) flow of a Newtonian fluidin a single-screw extruder. The inner solid cylinder of radius κR in a concentric-cylinder configuration translates in the z-direction with a constant linear velocity Vand rotates at a constant angular velocity Ω. The outer cylinder of radius R is stationary.This motion of the inner cylinder transports fluid through the single-screw extruder.There is no gradient of dynamic pressure in either the Θ- or z-directions. Hint: Postulatethat vΘ = f(r) and vz = g(r).

(a) Use unit vectors in cylindrical coordinates and write a generic expression for thefluid velocity vector when the screw rotates and translates very slowly in thecreeping flow regime, such that centrifugal forces are negligible.

Answer:Flow occurs in two dimensions when the screw rotates slow enough such that vr = 0.The fluid velocity vector is;

v = δΘ vΘ(r) + δz vz(r)

(b) Identify all scalar components of the viscous stress tensor that are non-zero.

Answer:If vΘ = f(r) and vz = g(r), with ∇ • v = 0, then the non-zero scalar components of theviscous stress tensor for incompressible Newtonian fluids are τrΘ = τΘr and τrz = τzr.

(c) Write a detailed expression for each scalar velocity component that is non-zero ifthe fluid is incompressible and Newtonian, and centrifugal forces are negligible.Please do not derive these expressions.

Answer:The solution for tangential annular flow in cylindrical coordinates, due to slow rotation ofthe inner cylinder is obtained from the answer to Problem 3B.1(a) on page#105 in


43

Transport Phenomena, when the angular velocity of the outer cylinder (i.e., Ωouter) is zero.Hence;

€

vΘ(r) =ΩR κ 2

1−κ 2

Rr−rR

The solution for axial annular flow due to translation of the inner solid cylinder in theabsence of any gradient in dynamic pressure is obtained from the answer to 2B.7(a) onpage#65 in Transport Phenomena. Hence;

€

vz (r) =Vln r

R

lnκ


44

Continuation of Detailed Lecture Information

Torque vs. angular velocity relation for concentric cylinder viscometersConsider the solid-liquid interface at the rotating spindle, where r = κRCalculate the differential vector force exerted by the solid on the fluid, which is theopposite of the differential vector force exerted by the fluid on the solid, due to totalmomentum flux (i.e., ρvv, τ and p) acting across a differential surface element at r = κRUnit normal vector from the solid to the fluid is n = δr, in the positive radial directionHence, the 1st subscript on ρvv and τ must be r, and all forces exerted by the solidcylinder on the fluid across the surface at r=κR have positive signs.Identify all forces or stresses due to total momentum flux that act across the samedifferential surface area, κR dΘ dz (i.e., a product of 2 differential lengths in the Θ- and z-directions in cylindrical coordinates, because the simple surface is defined by a constantvalue of r=κR, and the unit normal vector is in the radial direction)

Stress due to; Coordinate Non-zero? ReasonDirection

Pressure p r yes p > 0

Stress due to; Coordinate Non-zero? ReasonDirection

Convective momentum fluxρvrvr r no vr = 0ρvrvΘ Θ no vr = 0ρvrvz z no vr = 0

Viscous momentum fluxτrr r no vr = 0, ∇ • v = 0τrΘ Θ yes ∂/∂r (vΘ/r) ≠ 0τrz z no vr = vz = 0

As a vector, the differential force exerted by the inner rotating solid cylinder on the fluidin contact with this spindle at r = κR, is given by a product of the non-zero stresses dueto total momentum flux acting across the surface at r = κR, including the appropriate unitvectors, and the differential surface element, κR dΘ dz;

dFSolid on Fluid = { δr p(r=κR) + δΘ τrΘ(r=κR) } κR dΘ dz


45

This differential force exerted by the solid cylinder on the fluid across the differentialsurface element at r=κR generates the following differential torque;

Differential Torque dT = [ Lever Arm ] x dFSolid on Fluid

where the Lever Arm is a position vector from the axis of rotation to the point on thedifferential surface element where stress is transmitted from solid to fluid. This positionvector must be perpendicular to the rotation axis. Hence;

Lever Arm = δr ( κR )

Torque calculations are more complex when the lever arm varies over the macroscopicsurface across which stress is transmitted from solid to fluid. For the concentric cylinderviscometer with constant lever arm, one calculates the differential torque as follows;

Differential Torque dT = [ δr ( κR ) ] x { δr p(r=κR) + δΘ τrΘ(r=κR) } κR dΘ dz

dT = [ δr x δΘ ] ( κR )2 τrΘ(r=κR) dΘ dz

dT = δz ( κR )2 τrΘ(r=κR) dΘ dz

Now, recall the basic information for this problem, which was generated primarily fromsolution of the Θ-component of EOM in terms of velocity gradients for incompressibleNewtonian fluids;

€

vΘ(r) =12C2r +

C3

r

(NLV )τ rΘ = −µr ddr

vΘr

=

C1r2

= −µr − 2C3

r 3

=2µC3

r2

with C1 = 2 µ C3. Hence, the product of r2 and τrΘ, evaluated at r=κR, can be replaced byC1, or 2µC3. The macroscopic torque vector is obtained by integrating the expression fordifferential torque;

€

Torque = dT = 2µC3 δzdΘdzz=0

L*

∫Θ=0

2π∫∫


46

Since the orientation of δz never changes on the lateral surface of the solid cylinder [i.e.,δz ≠ f(Θ,z), ∂δz/∂Θ = 0, and ∂δz/∂z = 0], it is constant and can be removed from theprevious integral expression;

€

Torque =δz2µC3 dΘ dz =δz4πµL *C3z=0

L*

∫Θ=0

2π∫

where L* represents a manufacturer's suggested corrected length of the rotating spindle,which empirically accounts for end effects. Calculate C3 via comparison of the generalsolution for vΘ(r) with the answer to Problem 3B.1 on page#105 in TransportPhenomena;

€

vΘ(r) =1

1−κ 2Ωouter −κ

2Ωinner( )r − Ωouter −Ωinner( )κ2R2

r

When the outer cylinder is stationary and Ωouter = 0, and Ωinner = Ω;

€

vΘ(r) =ΩR κ 2

1−κ 2

Rr−rR

=12C2r +

C3

rHence;

€

C3 =ΩR2 κ 2

1−κ 2

Torque =δz 4πµL*C3 = δzΩ( )4πµR2L* κ 2

1−κ 2

where the angular velocity vector for rotation of the inner solid cylinder is ( δz Ω ). Thetorque required to overcome viscous shear at the solid-liquid interface and spin the innercylinder at angular velocity Ω is co-linear with the angular velocity vector of the solidspindle. Think about the necessary modifications which must be implementedif the fluid does not obey Newton's law of viscosity.

ProblemConsider laminar flow of an incompressible Newtonian fluid on the shell side of a double-pipe heat exchanger and calculate the z-component of the vector force exerted by thefluid on the outer surface of the inner pipe. Prove that your answer is correct byevaluating this interfacial force at two strategically chosen values of the radius ratioκ and comparing your answers with known results for these special cases.


47

______________________________________________________________

Now, consider three viscometers described below, where very slow rotation of a solidsurface produces 1-dimensional fluid flow in the creeping flow regime. The entire leftside of the Equation of Motion can be neglected at very low Reynolds numbers, and thenon-zero velocity component depends on two spatial coordinates during steady stateoperation. If one calculates the fluid velocity at the interface with the rotating solid byinvoking solid-body rotation, then it is not necessary to any solve partial differentialequations to obtain an expression for the important component of the velocity vector.

Parallel-disk viscometerNewtonian fluids; Torque vs. Ω, Problem 3B.5 on page#106 in Transport Phenomena

Begin with a picture of the flow configuration and identify vΘ as the only nonzerocomponent of the fluid velocity vector. The Equation of Continuity in cylindricalcoordinates reveals that;

€

∇•v =1r∂vΘ∂Θ

= 0

Hence, one concludes that vΘ ≠ f(Θ) based on the overall mass balance. One also arrivesat this conclusion, qualitatively, by invoking symmetry in cylindrical coordinates. For aparallel disk viscometer in which the fluid of interest is placed in the narrow gap betweena stationary plate and a rotating plate, the tangential velocity vΘ on the rotating plate isΩr, via solid body rotation. One obtains this result very easily by taking the vector crossproduct of the angular velocity vector (i.e., δzΩ) with the variable lever arm (i.e., δrr).This reveals the radial dependence of vΘ at any axial position z between the rotating &stationary plates. If one moves into the fluid in the z-direction from the moving plate atconstant r, and a separation of variables solution to the Θ-component of the Equation ofMotion is valid, then the r-dependence shouldn't change. Hence, at steady state;

vΘ = f(r) g(z) = r g(z)g(z) = Ω on the rotating plate at z=B

g(z) = 0 on the stationary plate at z=0

Notice how the postulated "separation-of-variables" form of the fluid velocity profile,based on solid body rotation at the interface between the fluid and the rotating plate,agrees with the Equation of Continuity. For Newtonian fluids, this one-dimensional flowfield in the parallel-disk viscometer generates the following important scalar componentsof the viscous stress tensor;


48

€

τ rΘ = τΘr = −µr ∂∂r

vΘr

= 0

τΘz = τ zΘ = −µ∂vΘ∂z

= −µr dgdz

Based on the postulated separation-of-variables form for vΘ, which accounts for solidbody rotation at the interface between the fluid and the rotating plate, the modifiedform of Newton's law of viscosity for flow problems with curved streamlines eliminatesany contribution of solid body rotation to viscous stress. Hence, the r-Θ component ofthe viscous stress tensor vanishes because the radial dependence of vΘ was adoptedfrom solid body rotation of the upper plate. As a general rule, if there is only oneimportant scalar component of the fluid velocity vector which depends on two spatialvariables, such as vi(xk,xm), then the following scalar components of the viscous stresstensor will be nonzero; τik, τki, τim, and τmi. However, if the dependence of vi on xk is basedon rigid body rotational characteristics within the fluid, then τik = τki = 0.

Force balances for the parallel-disk viscometer. Since the Θ-component of theEquation of Motion is most important, a consideration of the r- and z-components of theforce balance will provide information about the fluid pressure distribution.

r-component:

€

−ρvΘ2

r= −

∂P∂r

= −∂p∂r (i.e., gr = 0)

z-component:

€

0 = −∂P∂z

= −∂p∂z

+ρgz

Once again, fluid pressure depends on radial position r, due to centrifugal forces, andaxial position z, due to gravity. Since there are no restraining walls in the parallel-diskviscometer, it is necessary to operate the device in the creeping flow regime such that(i) centrifugal forces are negligible, and (ii) surface tension provides a restraining barrierto prevent fluid from moving in the radial direction. In the creeping flow regime, oneneglects the entire left side of the Equation of Motion which scales as the square of thefluid velocity. Now, fluid pressure depends only on axial position z, due to gravity. Thefluid pressure distribution in the parallel-disk viscometer is the same in the creeping flowregime as it is in the hydrostatic situation when there is no flow (i.e., p = pambient + ρgh).It is physically impossible to induce a pressure gradient in the Θ-direction if there are nonormal viscous stresses like τΘΘ, and gΘ = 0 if the viscometer is placed on a horizontal


49

surface such that gravity acts only in the z-direction. The Θ-component of the Equationof Motion for incompressible Newtonian fluids is;

Θ-component:

€

0 = µ∂∂r

1r∂∂r

rvΘ( )

+∂ 2vΘ∂z2

The 1st-term on the right side of the previous equation vanishes if vΘ = r g(z). This isequivalent to the fact that τrΘ is not important, as illustrated above, because the r-dependence of vΘ emulates solid body rotation which generates no viscous stress. Now,one solves for g(z) from the Θ-component of the force balance as follows;

Θ-component:

€

0 = µ∂ 2vΘ∂z2

= µr d2gdz2

Hence, the function g(z) must be linear in z, and the general solution for vΘ is;

vΘ(r,Θ) = r ( C1 z + C2 )

Integration constant C2 must be zero because the no-slip boundary condition suggeststhat there is no fluid motion at the stationary solid plate @ z=0 for all r>0. The boundarycondition at the rotating plate (i.e., z=B) is consistent with the postulated form of thevelocity profile, and yields an expression for C1. The final result is;

€

vΘ(r,Θ) =Ωr zB

Differential force due to total momentum flux which acts across the interfacebetween the fluid and the rotating solid plate. Identify all forces or stresses due to totalmomentum flux that act across the differential surface with unit normal vector in the z-direction. In particular, the unit normal vector from the solid to the fluid is in thenegative z-direction. Consequently, all forces transmitted across this interface from thesolid to the fluid must include a negative sign. The differential surface area underinvestigation is dS = r dr dΘ (i.e., a product of 2 differential lengths in the r- & Θ-directions in cylindrical coordinates, because the simple surface is defined by a constantvalue of z=B, and the unit normal vector is in the z-direction). The first subscript on anyimportant scalar components of convective and viscous momentum flux must be z.


50

Stress due to; Coord. Direction Non-zero? ReasonPressure -p z yes p > 0

Convective momentum flux-ρvzvr r no vz = 0-ρvzvΘ Θ no vz = 0-ρvzvz z no vz = 0

Viscous momentum flux-τzr r no vr = vz = 0-τzΘ Θ yes ∂vΘ/∂z ≠ 0-τzz z no vz = 0, ∇ • v = 0

As a vector, the differential force exerted by the upper rotating plate on the fluid incontact with this surface at z=B, is given by a product of the non-zero stresses due tototal momentum flux acting across the surface at z=B, including the appropriate unitvectors, and the differential surface element, dS = r dr dΘ;

dFSolid on Fluid = - { δΘ τzΘ(z=B) + δz p(z=B) } r dr dΘ

This differential force generates the following differential torque;

Differential Torque dT = [ Lever Arm ] x dFSolid on Fluid

where the Lever Arm is a position vector from the axis of rotation to the point on thedifferential surface element where stress is transmitted from solid to fluid. Hence;

Lever Arm = δr r

Notice that the Lever Arm is not constant. One calculates the differential torque asfollows;

Differential Torque dT = - [ δr r ] x { δΘ τzΘ(z=B) + δz p(z=B) } r dr dΘ

dT = - { [ δr x δΘ ] r2 τzΘ(z=B) + [ δr x δz ] r2 p(z=B) } dr dΘ

It is acceptable to neglect the pressure contribution to the relation between torque andangular velocity, even though the pressure force does not extend through the axis ofrotation. Detailed vector calculus analysis of the effect of pressure reveals that its


51

contribution is zero. As a general rule, forces due to fluid pressure will never contributeto the relation between torque and angular velocity. Hence;

dT = - δz r2 τzΘ(z=B) dr dΘ

Now, recall the basic information for this problem, which was generated primarily fromsolution of the Θ-component of the Equation of Motion in terms of velocity gradients forincompressible Newtonian fluids;

€

vΘ(r) =Ωr zB

(NLV )τ zΘ = −µ∂vΘ∂z

= −µΩrB

The macroscopic torque T is obtained via integration of dT over the entire interfacebetween the rotating solid and the fluid;

€

Torque = dT = dΘ −δzr2τ zΘ(z = B){ }dr =

2πµΩB

δzr3dr

r=0

R∫r=0

R∫Θ=0

2π∫∫

Since the orientation of δz never changes on the surface of the rotating solid plate atz=B, [i.e., δz ≠ f(r,Θ), ∂δz/∂r=0], it is constant and can be removed from the previousintegral expression;

€

Torque = δzΩ( ) 2πµB

r 3dr = δzΩ( ) πµR4

2Br=0

R∫

Once again, the macroscopic torque vector is co-linear with the angular velocity vector(i.e., δz Ω), which can be considered as a general rule.

Rotating sphere viscometerNewtonian fluids; Torque vs. Ω, see pages#95-96 in Transport Phenomena

Statement of the problem, and evaluation of the fluid velocity at the solid-liquidinterface via solid-body rotation. A solid sphere of radius R is suspended from a wire androtates very slowly at constant angular velocity Ω about the long axis of the wire in anincompressible Newtonian fluid. The fluid is quiescent far from the sphere. For a rotatingsphere viscometer, solid-body rotation at the fluid-solid interface suggests that thetangential fluid velocity vφ on the surface of the sphere is ΩRsinΘ. This result is obtained


52

by analyzing rigid body rotation of a solid sphere about the z-axis of a Cartesiancoordinate system and calculating the velocity vector at the fluid-solid interface byinvoking the "no-slip" condition;

v = [ Ω x { Lever Arm} ]r=R

The angular velocity vector is oriented in the z-direction (i.e., Ω = Ω δz), and the variablelever arm from the axis of rotation (i.e., along the wire) to any point on the surface ofthe solid sphere is;

Lever Arm = R sinΘ { δr sinΘ + δΘ cosΘ }

where Θ is the polar angle measured from the z-axis. Upon taking the cross product, oneobtains;

v = Ω R sinΘ { [ δz x δr ] sinΘ + [ δz x δΘ ] cosΘ }

Trigonometric relations between unit vectors in rectangular & spherical coordinates yieldthe following expression for δz (see p.#828, Transport Phenomena, 2nd edition, by RBBird, WE Stewart & EN Lightfoot). Hence;

δz = δr cosΘ - δΘ sinΘ δz x δr = δφ sinΘ δz x δΘ = δφ cosΘ

If the sphere rotates very slowly and centrifugal forces do not induce flow in the radialdirection or the Θ-direction, then one calculates the fluid velocity at the fluid-solidinterface via solid body formalism summarized above. Vector algebra reveals that thisproblem is described by one-dimensional flow in the φ-direction, because;

v = Ω R sinΘ { δφ sin2Θ + δφ cos2Θ } = δφ Ω R sinΘ = δφ vφ

This calculation from solid body rotation reveals the angular dependence of vφ at anyradial position, because if one moves into the fluid at larger r and constant Θ, and aseparation of variables solution to the φ-component of the Equation of Motion is valid,then the sinΘ dependence shouldn't change. Hence;

vφ(r,Θ) = f(r) g(Θ) = f(r) sinΘf(r) = ΩR @ r = Rf(r) → 0 as r → ∞

This functional form of the fluid velocity profile satisfies the Equation of Continuity,because when rotation is slow enough and flow occurs only in one direction, the balanceon overall fluid mass for incompressible fluids is;


53

€

∇•v =1

r sinΘ∂vφ∂φ

= 0

which stipulates that vφ cannot be a function of the azimuthal angle φ.

Creeping flow analysis of the Equation of Motion. In the creeping flow regimewhere centrifugal forces are negligible, one sets the entire left side of the Equation ofMotion to zero and considers only a balance between viscous, pressure and gravityforces. Since the φ-component of the vector force balance is most important, the r- andΘ-components yield the following information about dynamic pressure;

r-component: 0 = - ∂P/∂r P ≠ f(r)

Θ-component:

€

0 = −1r∂P∂Θ P ≠ f(Θ)

Steady state analysis implies that dynamic pressure is not time-dependent, and intuitionsuggests that it is almost impossible to induce a pressure gradient in the primary flowdirection, unless normal viscous stresses like τφφ are nonzero. Since the φ-component ofgravity vanishes if the sphere is suspended vertically from a wire, one analyzes the φ-component of the Equation of Motion in the creeping flow regime when dynamic pressureis constant, similar to a hydrostatic situation. Fluid flow is induced by (i) rotation of thesolid and (ii) viscous shear which is transmitted across the solid-liquid interface. Asexpected, the φ-component of the force balance yields useful information to calculate vφ.The only terms which survive in the φ-component of the Equation of Motion are;

€

0 =1r2

∂∂r

r2∂vφ∂r

+

∂∂Θ

1sinΘ

∂∂Θ

vφ sinΘ( )

Now, one calculates f(r) from the previous equation by letting vφ = f(r) sinΘ . Forexample;


54

€

∂∂Θ

vφ sinΘ( ) = 2 f (r)sinΘcosΘ

1sinΘ

∂∂Θ

vφ sinΘ( ) = 2 f (r)cosΘ

∂∂Θ

1sinΘ

∂∂Θ

vφ sinΘ( )

= −2 f (r)sinΘ

The φ-component of the Equation of Motion reduces to Euler’s differential equation;

€

sinΘr2

ddr

r2 dfdr

− 2 f (r)

= 0

If one adopts a trial solution of the form f(r) ≈ rn, or rnln r if two values of n are the samefor this 2nd-order ordinary differential equation, then both terms in brackets {} areproportional to rn. Substitution of f(r) ≈ rn into the previous equation yields the followingquadratic polynomial with two roots for the exponent n;

sinΘ [ n ( n + 1 ) - 2 ] rn-2 = 0

The solution is; n = -2, 1. The φ-component of the Equation of Motion is a lineardifferential equation in the creeping flow regime, because one neglects terms that scaleas the square of fluid velocity. Hence, the general solution for the φ-component of thefluid velocity vector is obtained by adding both solutions for f(r) and including theappropriate integration constants;

€

vφ (r,Θ) = Ar +Br2

sinΘ

The solution for n = 1 must be discarded because the fluid is stagnant at large r. Hence,A = 0. The boundary condition at the fluid-solid interface yields B = ΩR3. The finalcreeping flow solution is;

€

vφ (r,Θ) =ΩR3 sinΘ

r2

Since vφ is a function of r and Θ , one predicts that there are four nonzero scalarcomponents of the viscous stress tensor. τrφ = τφr is important because vφ depends on r,


55

and this functional dependence does not conform to solid body rotation. In fact, the rφ-component of τ is solely responsible for the torque/angular-velocity relation. Since vφ

depends on Θ , at first glance, one expects that τφΘ = τΘφ should yield a nonzerocontribution to the state of viscous stress. However, the Θ-dependence of vφ wasconstructed from solid body rotation and detailed calculations of the Θφ- and φΘ-components of τ indicate that no viscous forces are generated from the sinΘ functionaldependence of vφ. In the parallel disk viscometer, the rΘ- and Θr-components of τ vanishbecause no viscous forces result from the linear dependence of vΘ on r due to solid bodyrotation. In summary, when the important nonzero velocity component for one-dimensional fluid flow depends on two independent spatial variables, and the functionaldependence on one of these spatial variables is postulated to match solid body rotationat the fluid-solid interface, the viscous shear stress based on this solid-body-typefunctional dependence vanishes (i.e., r for the parallel disk viscometer, sinΘ for therotating sphere viscometer). Hence, the state of viscous stress in the fluid is simplifiedbecause the fluid adopts some, but not all, aspects of solid body rotation.

Differential vector force due to total momentum flux transmitted across the fluid-solid interface at r=R. Now, one focuses on the fluid-solid interface and calculates thedifferential vector force dFSolid On Fluid exerted by the solid on the fluid. Begin by identifyingthe unit normal vector from the solid to the fluid across the surface at r=R; n = + δr. The1st subscript on each scalar component of convective and viscous momentum flux is r,each non-zero component of total momentum flux is preceeded by a positive sign, & thedifferential surface element is dS = R2 sinΘ dΘ dφ. Construct the following table;

Stress Coord. direction Non-zero? ReasonPressure p r yes p > 0

Convective momentum fluxρvrvr r no vr = 0ρvrvΘ Θ no vr = 0ρvrvφ φ no vr = 0

Stress Coord. direction Non-zero? ReasonViscous momentum flux

τrr r no vr = 0, ∇ • v = 0τrΘ Θ no vr = vΘ = 0τrφ φ yes ∂/∂r (vφ/r) ≠ 0


56

Hence, there is a normal pressure force which acts in the r-direction, and a viscous shearforce which acts in the φ-direction. The total differential vector force which acts acrossthe solid-liquid interface at r=R is;

dFSolid On Fluid = { δr p + δφ τrφ }r=R R2 sinΘ dΘ dφ

The differential vector torque dT that arises from dFSolid On Fluid acting across the solid-liquid interface @ r=R is calculated by performing the following cross-product operation;

dT = { Lever Arm } x dFSolid On Fluid

where the lever arm was calculated on page#32. Hence;

dT = R sinΘ { δr sinΘ + δΘ cosΘ } x { δr p + δφ τrφ }r=R R2 sinΘ dΘ dφ

= { δr τrφ cosΘ - δΘ τrφ sinΘ - δφ p cosΘ }r=R R3 sin2Θ dΘ dφ

As a general rule, neglect the pressure contribution to dT because detailed vectorcalculus analysis will reveal that it vanishes for any type of viscometer. The quantity ofinterest reduces to;

dT = { δr cosΘ - δΘ sinΘ } R3 τrφ(r=R) sin2Θ dΘ dφ

Now, (i) evaluate the important scalar component of the viscous stress tensor at thesolid-liquid interface, (ii) rewrite the three spherical coordinate unit vectors in terms ofconstant unit vectors in rectangular coordinates (i.e., see page#828 in TransportPhenomena, 2nd edition, by RB Bird, WE Stewart & EN Lightfoot), and (iii) integrate theprevious expression to calculate the macroscopic torque/angular-velocity relation fromwhich the Newtonian viscosity µ can be determined via measurements of torque vs. Ω.

€

τ rφ (r = R,Θ)= −µ r ∂∂r

vφr

r=R

= 3µΩsinΘ

δr = δx sinΘ cosφ + δy sinΘ sinφ + δz cosΘ

δΘ = δx cosΘ cosφ + δy cosΘ sinφ - δz sinΘ

δφ = - δx sinφ + δy cosφ


57

If the solution for n=1 in the expression for vφ were not discarded based on qualitativephysical arguments, and integration constant A ≠ 0, then the form of Newton’s law ofviscosity for τrφ eliminates any contribution from the term Ar in the torque/angular-velocity relation because Ar emulates solid body rotational characteristics within thefluid. Macroscopic torque T is obtained via integration of dT over the surface of thesolid sphere, where, for example, Θ ranges from 0 to π, and φ ranges from 0 to 2π. Thereare no contributions to T in the x- & y-coordinate directions because, in all cases, oneintegrates either sinφ or cosφ over the complete period of these trigonometric functions.Hence, this provides quantitative justification for the claim that fluid pressure does notcontribute to the relation between torque and angular velocity in this particularviscometer. It is only necessary to consider terms in the z-direction due to δr and δΘ.These are;

€

Torque = dT = 3µΩR3 δz cos2Θ+δz sin2Θ( )sin3ΘdΘdφΘ=0

π

∫φ=0

2π∫∫

=δz6πµΩR3 sin3ΘdΘ= 8πµR3 δzΩ( )Θ=0

π

∫

Once again, the macroscopic torque vector is co-linear with the angular velocity vector ofthe solid sphere, and both of these vectors act in the z-direction. These results areuniversal for all viscometers if the z-direction is vertical. The origin of R3 in the finalexpression for torque vs. angular velocity is (i) one factor of R from the lever arm, and(ii) one factor of R2 from both the differential and macroscopic interfacial surface areabetween the solid sphere and the fluid. The origin of sin3Θ in the integral expression fortorque is (i) one factor of sinΘ from the lever arm, (ii) one factor of sinΘ from thedifferential surface area, and (iii) one factor of sinΘ when the important scalar shearcomponent of the viscous stress tensor is evaluated at the fluid-solid interface.

Problem#1Determine the exponents a, b & c in the following scaling relation for the macroscopictorque in the rotating sphere viscometer;

magnitude of the torque ≈ µa Ωb Rc

Answera=1, b=1 & c=3.

Problem#2How do your answers for the scaling law exponents a & b relate to the fact that theconstitutive relation between viscous stress and velocity gradients is linear via Newton'slaw of viscosity?


58

AnswerThe values for a & b in the previous scaling law are a direct consequence of the fact thattorque is linearly proportional to viscous shear stress, and Newton's law of viscosity is alinear constitutive relation between viscous stress & viscosity (i.e., a=1) and viscousstress & velocity gradients (i.e., b=1), where the velocity gradient can be approximatedby the angular velocity Ω of the solid sphere.

Problem#3Estimate the scaling law exponent b if the fluid were non-Newtonian with power-law indexn in the classic Ostwald-de Waele model as described on page#241 in TransportPhenomena, 2nd edition, by RB Bird, WE Stewart & EN Lightfoot.

AnswerFor power-law fluids, viscous stress is proportional to the nth-power of the shear rate,which represents the magnitude of the rate-of-strain tensor. Since torque scales linearlywith viscous shear stress, and shear rate scales linearly with angular velocity, it followsdirectly that torque scales as the nth-power of Ω. Hence, b = n.___________________________________________________________________

Cone-and-plate viscometerAll fluids; Torque vs. ΩSee Problems 2B.11 on page#67-68, and 8C.1 on page#261 in Transport Phenomena.

An incompressible Newtonian fluid is placed within the gap between a stationary plateand a solid cone whose apex touches the plate. The radius of the cone and the plate isR, and the conical surface is described by a constant value of polar angle Θ = Θ1, whereΘ1 is approximately 89.50. Hence, the test fluid resides in a very narrow gap betweenthe cone and the plate. The cone rotates very slowly at constant angular velocity Ωabout the z-axis, and surface tension is sufficient to retain the fluid within the gapbecause centrifugal forces are negligible in the creeping flow regime at very slowrotational speeds of the cone. One-dimensional fluid flow occurs at low Ω, whereas eithertwo- or three-dimensional flow is appropriate at high Ω. The r- and φ-components of thefluid velocity vector are important in the equatorial plane at Θ = π/2 when the rotationalspeed of the cone is high enough such that the Reynolds number is sufficiently largerthan unity. For creeping flow, analysis of solid-body rotation on the conical surfaceallows one to obtain the fluid velocity at the solid-liquid interface (i.e., Θ = Θ1) byinvoking the "no-slip" condition. Hence;

v = [ Ω x { Lever Arm} ] @ Θ = Θ1


59

The angular velocity vector is oriented in the z-direction (i.e., Ω = Ω δz), and the leverarm from the axis of rotation (i.e., z-axis) to any point on the conical surface is;

Lever Arm = r sinΘ1 { δr sinΘ1 + δΘ cosΘ1 }

where Θ1 is the polar angle measured from the z-axis to any point on the surface of thecone. Upon taking the cross product, one obtains;

v = Ω r sinΘ1 { [ δz x δr ] sinΘ1 + [ δz x δΘ ] cosΘ1 }

Trigonometric relations between unit vectors in rectangular and spherical coordinatesyield the following expression for δr, δΘ, and δφ (see p.#828, Transport Phenomena, 2nd

edition, by RB Bird, WE Stewart & EN Lightfoot). Hence;

δr = δx sinΘ cosφ + δy sinΘ sinφ + δz cosΘ

δΘ = δx cosΘ cosφ + δy cosΘ sinφ - δz sinΘ

δφ = δx (-sinφ) + δy cosφ

The cross products of interest are;

δz x δx = δy δz x δy = - δx δz x δz = 0

Vector algebra reveals that this problem is described by one-dimensional flow in the φ-direction, because the velocity vector at the fluid/rotating-solid interface is;

v = Ω r sinΘ1 { δy sin2Θ1 cosφ - δx sin2Θ1 sinφ + δy cos2Θ1 cosφ - δx cos2Θ1 sinφ }

= Ω r sinΘ1 { - δx sinφ + δy cosφ } = δφ Ω r sinΘ1 = δφ vφ

This calculation from solid body rotation reveals that vφ = Ωr sinΘ1. Hence, the radialdependence of vφ is a linear function of r at any angle Θ between the rotating cone @ Θ1

and the stationary plate @ Θ=π/2, because if one moves into the fluid in the Θ-directionfrom the rotating cone toward the stationary plate at constant radial position r, and aseparation of variables solution to the φ-component of the Equation of Motion is valid,then the r-dependence shouldn't change. Hence;

vφ(r,Θ) = f(r) g(Θ) = r g(Θ)


60

g(Θ) = Ω sinΘ1 @ Θ = Θ1

g(Θ) = 0 @ Θ = π/2

If one postulates vφ(r,Θ) to agree with the boundary condition at the interface betweenthe fluid and the rotating cone, then the Equation of Continuity is satisfied for anincompressible fluid in which vr = vΘ = 0 when centrifugal forces are negligible in thecreeping flow regime, because;

€

∇•v =1

r sinΘ∂vφ∂φ

= 0

Hence, the important nonzero component of the fluid velocity vector is not a function ofthe azimuthal angle φ in spherical coordinates.

Important nonzero components of the viscous stress distribution. If the fluid isNewtonian, then Newton's law of viscosity allows one to estimate the importance of thesix independent scalars that summarize the state of viscous stress. For one-dimensionalflow of an incompressible fluid in the φ-direction, the important quantities are;

€

τΘφ = τφΘ = −µsinΘr

∂∂Θ

vφsinΘ

= h Θ( )

τφr = τ rφ = −µr ∂∂r

vφr

= 0

The rφ-component of the viscous stress distribution vanishes because the r-dependenceof vφ resembles solid-body rotation (i.e., it is a linear function of r), for which there is noviscous stress. Mathematically, this concept is included in Newton's law when thestreamlines are curved. Hence, it is only necessary to consider the Θφ- and φΘ-components of the viscous stress tensor in the Equation of Motion.

Creeping flow analysis of the Equation of Motion. The force balances in terms ofτ reveal that the important component of the viscous stress tensor is essentiallyconstant within the gap between the rotating cone and the stationary plate. Since the φ-component of the Equation of Motion provides the most important information about vφ

and τΘφ, the other two components of the force balance are considered first;


61

r-component:

€

−ρvφ2

r= −

∂P∂r

If the z-axis about which the cone rotates is vertical, then the fluid experiences acentrifugal force that is horizontally outward. Hence, the r-component of the centrifugalforce is ρ(vφ)2sinΘ divided by the radius of curvature at position r and Θ, which is r{sinΘ}.This centrifugal force appears on the left side of the r-component of the Equation ofMotion in spherical coordinates. In the creeping flow regime, dynamic pressure is not afunction of radial coordinate r because centrifugal forces are negligible. Surface tensionacts in the radial direction and provides the restraining wall that opposes radial fluidmotion. However, surface tension is operative only at the boundary of the flow problem.At most, one should account for surface tension in the boundary condition at r=R.

Θ-component:

€

−ρvφ2

rcotΘ = −

1r∂P∂Θ

If the fluid experiences a centrifugal force which is horizontally outward, then the Θ-component of this force is ρ(vφ)2cosΘ divided by the radius of curvature (i.e., r sinΘ), asgiven by the term on the left side of the previous equation. In the creeping flow regime,dynamic pressure is not a function of polar angle Θ . Since azimuthal angle φ is thesymmetry variable in spherical coordinates, qualitative arguments suggest that P isindependent of φ. This is reasonable because it is almost impossible to impose a pressuregradient in the primary flow direction when the streamlines are curved in a curvilinearcoordinate system. One exception is axial flow through a helical cooling tube. For steadystate analysis, where P does not exhibit time dependence, dynamic pressure is constantand the flow is not driven by gravity or a pressure gradient. This is typical forviscometers which contain a rotating solid surface. The φ-component force balanceyields the following information;

φ-component:

€

0 = −1

r sinΘ∂∂Θ

τΘφ sinΘ{ }−τΘφrcotΘ = −

1r∂τΘφ∂Θ

− 2τΘφrcotΘ

Since the viscous stress tensor is symmetric and τΘφ is only a function of polar angle Θ,separation of variables allows one to obtain the viscous stress distribution as follows;


62

€

dτΘφτΘφ

= −2 cosΘsinΘ∫∫ dΘ

lnτΘφ = −2 ln sinΘ{ }+C1

τΘφ Θ( ) =C2

sin2Θ

Since the angle between the rotating cone and the stationary plate is approximately 0.5degree (i.e., from 89.50 to 900), fluids experience essentially constant shear stress in thecone-and-plate viscometer, which makes this device attractive for both Newtonian andnon-Newtonian fluids. Torque is proportional to τΘφ, and shear rate or velocity gradient isgiven by the linear velocity of the cone ΩrsinΘ1 divided by the spacing between the coneand the plate, r(π/2 - Θ1), via an arc-length calculation at radial position r. Hence, thevelocity gradient is ΩsinΘ1/(π/2 - Θ1), and sinΘ1 is approximately equal to unity becausethe angle of the cone is very close to 900. Now, one calculates the viscosity of any fluidvia the ratio of τΘφ and Ω/{π/2 - Θ1}, where the analysis below reveals that the magnitudeof the torque transmitted by the fluid to the stationary plate is;

Magnitude of the Torque(@ Θ=π/2) = (2/3) π R3 τΘφ(Θ=π/2)

One obtains this expression for the magnitude of the torque rather quickly because thevariable lever arm is rsinΘ ≈ r and the force transmitted across the fluid-solid interface atΘ = π/2 is given by the product of the viscous shear stress τΘφ = C2 and the differentialsurface element rdrdφ. Upon integraton over the entire surface of the stationary solidplate;

€

Torque = δrr[ ]x0

R

∫0

2π

∫ δφτΘφ Θ = π / 2( )[ ]rdrdφ = −δΘτΘφ Θ = π / 2( ) 23πR3

one relates viscous shear stress to the magnitude of the torque. It should beemphasized that -δΘ everywhere on the surface of the stationary plate corresponds to δz,which is not a function of spatial coordinates r and φ in the previous integral expression.

Analysis of the important nonzero component of the fluid velocity vector. If thefluid is Newtonian and one accounts for the slight dependence of viscous shear stress onpolar angle Θ, then the following equations are combined to calculate vφ;


63

€

vφ r,Θ( ) = rg Θ( )

τΘφ = −µsinΘr

∂∂Θ

vφsinΘ

= −µ sinΘ ddΘ

g Θ( )sinΘ

=C2sin2Θ

Since g(Θ) vanishes on the surface of the stationary plate via the "no-slip" condition, oneintegrates the previous equation from π/2 to Θ. The result is;

€

dg Θ( )sinΘ

0

Θ

∫ = −C2µ

dxsin3 xπ /2

Θ

∫

g Θ( ) = −C2

µsinΘ dx

sin3 xπ /2

Θ

∫

The following integral theorems with m ≠ 1 are useful to complete the development andcalculate the fluid velocity profile;

€

dxsinm ax∫ = −

cos ax( )a m −1[ ]sinm−1 ax( )

+m − 2m −1

dxsinm−2 ax∫

When m=3 and a=1, one obtains;

€

dxsin3 x∫ = −

cos x2sin2 x

+12

dxsin x∫

Finally;

€

dxsin x∫ = csc x( ) csc x − cot xcsc x − cot x

dx∫ = ln csc x − cot x{ }

Since the integral of 1/sin3x vanishes when x=π/2, it is only necessary to evaluate theresult at x=Θ. Hence;

€

vφ r,Θ( ) = rg Θ( ) = −C2

µr sinΘ dx

sin3 xπ /2

Θ

∫ =C2r2µ

cotΘ− sinΘ ln cscΘ− cotΘ( ){ }

=C2r2µ

cotΘ− 12sinΘ ln 1− cosΘ

1+ cosΘ


64

Integration constant C2 is evaluated via the "no-slip" condition at the interface betweenthe fluid and the rotating cone, where Θ=Θ1;

€

vφ r,Θ1( ) =Ωr sinΘ1 = rg Θ1( )

g Θ1( ) =ΩsinΘ1 =C2

2µTrigFunction Θ1( )

TrigFunction Θ( ) = cotΘ− 12sinΘ ln 1− cosΘ

1+ cosΘ

The final result for the fluid velocity profile can be expressed as;

€

vφ r,Θ1( ) =Ωr sinΘ1TrigFunction Θ( )TrigFunction Θ1( )

This profile is essentially a linear function of polar angle Θ when the conical surface isdescribed by values of Θ1 between 470 and 89.90.

Problem#1Consider the following flux terms that appear either in the Equation of Motion or in theEquation of Continuity:

(a) 1r sin(θ)

∂∂θ

(ρvθsin{θ})

(b) µ ∂∂r 1

r∂∂r

(rvr)

(i) Be specific and identify the type of transport and the mechanism in each case.Answer: (a) Convective mass flux in the EOC; spherical coordinates

(b) Normal viscous stress in the EOM; cylindrical coordinates

(ii) Use differential elements of length and identify the surface across which each fluxacts.Answer: (a) Perpendicular to the Θ-direction; dS = rsinΘdrdφ

(b) Perpendicular to the r-direction; dS = rdΘdz

(iii) Identify the coordinate direction in which each flux acts.Answer: (a) Convective mass flux in the Θ-direction; spherical coordinates


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(b) Viscous momentum flux in r-direction; cylindrical coordinates

(iv) Identify the unit normal vector to the surface across which each flux acts.Answer: (a) Normal vector in the Θ-direction; spherical coordinates

(b) Normal vector in the r-direction; cylindrical coordinates

(v) Obtain an analytical solution for the function ω(r,Θ), as described by the followingfluid dynamics equations that describe the flow field in the cone-and-plateviscometer;

dz(θ)

dθ = −2z

cos(θ)

sin(θ)

[1]

z(θ) = −µ

sin(θ)

r

∂∂θ

ω(r,θ)

sin(θ)

[2]

ω(r,θ) = rf(θ) [3]

B.C. θ = π2 f = 0 [4]

θ = θ1 f = Ωsin(θ1) [5]

(v) Identify the variable z(Θ) in terms of fluid dynamics nomenclatureAnswer: z(Θ) = τΘφ (i.e., the Θφ-component of the viscous stress tensor)

(vi) Identify equations [1] and [2] above in terms of the generalized equations ofchange for momentum transport problems, and be extremely specific in yourdescription.Answer: [1] is the φ-component of EOM in terms of τ; spherical coordinates

[2] is Newton’s law of viscosity for the Θφ-component of τ

(viii) Graph the analytical solution to the problem described in (v) above as follows;

ω(r,θ)ΩRsin(θ1) vs. Θ at three (3) constant values of r/R = 0.2, 0.5, 0.9; and six (6)

different values of the parameter Θ1 = 600, 750, 850, 880, 890, 89.50. The rangeof the independent variable Θ is; Θ1 ≤ Θ ≤ 900. When Θ1 = 600, 750, scale thehorizontal axis from 600 to 900. When Θ1 = 850, 880, 890, 89.50, scale thehorizontal axis from 850 to 900. You should generate six (6) graphs, one for each


66

value of Θ1. Three solutions should be plotted on each graph, corresponding tothe three values of r/R = 0.2, 0.5, 0.9

Answer: The velocity profile is a linear function of polar angle Θ for cone angles Θ1

that are larger than ≈ 50 degrees. Miraculously, the complex trigonometricfunction derived above for vφ yields a straight line. When the cone angle Θ1

is less than 54 degrees, deviations of the velocity profile from linearity aregreatest near the surface of the cone. For example;

Θ1 (degrees) % deviations from linearity54 031 1029 1328 1527 179 74

When the profile for vφ(r,Θ) is a linear function of polar angle Θ from the surface of therotating cone to the stationary plate, one postulates the following expression thatconforms to the boundary conditions at both solid surfaces;

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vφ r,Θ( ) ≈ Ωr sinΘ1π / 2 −Θπ / 2 −Θ1

Consistent with this linear velocity profile when the cone angle is ≈ 890, one equates themagnitude of the Θφ-component of the velocity gradient tensor, which is not asymmetric 2nd-rank tensor, with the shear rate in the cone-and-plate viscometer. Forexample;

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∇v( )Θφ =1r∂vφ∂Θ

=ΩsinΘ1ddΘ

π / 2 −Θπ / 2 −Θ1

≈ −

Ωπ / 2 −Θ1

As mentioned above, one calculates the viscosity of any fluid in the cone-and-plateviscometer via division of the viscous shear stress τΘφ at the surface of the stationaryplate, obtained experimentally from the measured torque, by the shear rate which isgiven by the magnitude of the previous calculation.


67

(ix) Compare your graphical results of

ω(r,θ)ΩRsin(θ1) vs. Θ for a Newtonian fluid within

the gap of a cone-and-plate viscometer with three (3) other viscometerproblems;

(a) steady state velocity profile vx(y) for a fluid contained between twoparallel plates; the lower plate is stationary and the upper platemoves unidirectionally in the x-direction with a constant velocity V,

(b) velocity profile vΘ(r) for a Newtonian fluid in the Couette (i.e.,concentric cylinder) viscometer,

(c) velocity profile vΘ(r,z) for a Newtonian fluid in the parallel-diskviscometer.

One of the four fluid flow examples mentioned above does not conform to theother three profiles. Identify the non-conformer.

Problem#2Solve the following ordinary differential equation for the function y(x), subject to theboundary condition; y = 15 when x = π/2 radians.

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dydx

=4ytan x

Problem#3Solve the following ordinary differential equation for y as a function of x, subject to theboundary condition; y = 10 when x = 0 radians.

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dydx

= 3y tan x( )

Problem#4Write an integral expression for the average fluid velocity in the cone-and-plateviscometer. Be sure to include the limits of integration.

Answer: Since fluid flow occurs in the φ-direction, the differential surface element dS isconstructed from a product of two differential lengths in the r- and Θ-


68

directions. Hence, dS = rdrdΘ. vφ is averaged over the flow cross-section asfollows;

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vφ Average=

vφ r,Θ( )dS∫∫dS∫∫

=

rg Θ( )0

R

∫Θ1

π /2

∫ rdrdΘ

rdrdΘ0

R

∫Θ1

π /2

∫=

13 R

3 g Θ( )dΘΘ1

π /2

∫12 R

2 π / 2 −Θ1( )=

2R3 π / 2 −Θ1( )

g Θ( )dΘΘ1

π /2

∫

Differential vector force transmitted by the fluid to the stationary plate. Tosimplify the calculation of the macroscopic torque/angular-velocity relation, oneconsiders the force transmitted by the fluid to the stationary solid plate at Θ = π/2.Calculations are simplified on the stationary plate because the normal vector alwayspoints in the z-direction, and the variable lever arm is completely in the r-direction. Inspherical coordinates, the unit normal vector from the fluid to the solid is oriented in thepositive Θ-direction. Hence, the 1st subscript on the scalar components of convectiveand viscous momentum flux which act across this surface is Θ, pressure forces act in theΘ-direction, and all scalar components of total momentum flux which act across thissurface contain a positive sign. The differential surface element is constructed from aproduct of two differential lengths in the r- and φ-directions. Hence, dS = r sinΘ dr dφ,which reduces to r dr dφ on the surface of the stationary plate, where Θ = π/2. Considerthe following table;

Stress Coord. direction Non-zero? ReasonPressure p Θ yes p > 0

Convective momentum fluxρvΘvr r no vΘ = 0ρvΘvΘ Θ no vΘ = 0ρvΘvφ φ no vΘ = 0

Viscous momentum fluxτΘr r no vr = 0, vΘ = 0τΘΘ Θ no vΘ = 0, ∇ • v = 0τΘφ φ yes ∂/∂Θ(vφ/sinΘ)≠0

Hence, the normal pressure force acts in the Θ-direction, and viscous shear acts in the φ-direction. The differential vector force which acts across the solid-liquid interface at Θ =π/2 is;


69

dFFluid On Solid = { δΘ p + δφ τΘφ }Θ=π/2 r dr dφ

Macroscopic relation between torque and angular velocity. The differential vectortorque dT that arises from dFFluid On Solid acting across the solid-liquid interface @ Θ = π/2 iscalculated as follows;

dT = { Lever Arm } x dFFluid On Solid

= { δr r } x { δΘ p + δφ τΘφ }Θ=π/2 r dr dφ

Performing the cross-product operation, neglecting the pressure contribution to thetorque, and integrating the result over the entire surface of the stationary plate yields;

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Torque = dT∫ = δr xδφ[ ]0

R

∫0

2π

∫ τΘφ Θ = π / 2( )r2drdφ

= −δΘτΘφ Θ = π / 2( ) dφ0

2π

∫ r2dr0

R

∫ =δz23πR3τΘφ Θ = π / 2( )

where (i) δr x δφ = - δΘ = + δz on the stationary plate, (ii) the viscous shear stress is givenby C2 when Θ = π/2, and (iii) and the constant C2 is calculated from the "no-slip"boundary condition on the surface of the rotating cone;

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τΘφ Θ = π / 2( ) =C2 =2µΩsinΘ1

TrigFunction Θ1( )The final result is;

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Torque =δz43 πR

3µΩsinΘ1

TrigFunction Θ1( )

This relation allows one to calculate the viscosity of a Newtonian fluid via measurementof torque and angular velocity. If the fluid does not obey Newton's law of viscosity, thenone calculates the viscous shear stress from the magnitude of the macroscopic torque asfollows;

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τΘφ Θ = π / 2( ) =3 Torque{ }2πR3

The fluid viscosity is obtained via division of the previous equation by Ω/{π/2 - Θ1},where Ω is the angular velocity of the cone. Non-Newtonian fluids exhibit a shear-rate-


70

dependent viscosity. However, when the gap between the cone and plate is very narrow(i.e., Θ1 ≈ 89.50), the fluid experiences essentially one viscous shear stress, and thevelocity profile is a linear function of polar angle Θ, with a constant shear rate given byΩ/{π/2 - Θ1}. The effect of shear rate on the viscosity of a non-Newtonian fluid isobtained, in practice, by incrementing Ω and repeating the experiment in the cone-and-plate viscometer several times. Then, one constructs a graph of;

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3 Torque{ } π / 2 −Θ1( )2πΩR3

vs. Ωπ / 2 −Θ1( )

Dimensionless Momentum Transfer Correlations

Forces exerted by moving fluids on stationary solid surfaces viamacroscopic dimensionless momentum transfer correlations between thefriction factor and the Reynolds number. The magnitude of the dynamic forceexerted across a fluid-solid interface represents the focal point for these correlations.For one-dimensional tube flow of an incompressible Newtonian fluid, with vz(r), τrz(r) andP(z), the differential vector force exerted by the fluid on the wall at r=R, due to totalmomentum flux, is;

dFFluid on Solid = { δr p + δz τrz }r=R R dΘ dz

where p is fluid pressure and P is dynamic pressure. This can be verified rather easily,because the unit normal vector from the fluid to the stationary solid wall is n = δr, whichimplies that the first subscript on all of the important nonzero components of totalmomentum flux is r and pressure forces across this surface act in the r-direction. Now,the component of dFFluid on Solid in the primary flow direction is obtained by performing thescalar product operation of dFFluid on Solid with δz, which yields τrz dS. Hence;

δz • dFFluid on Solid = { dFFluid on Solid }z-component = τrz(r=R) R dΘ dz

For laminar or turbulent flow of incompressible fluids, the rz-component of the viscousstress tensor, evaluated at the tube wall, is;

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τ rz r = R( ) =RΔP2L

Integration of the differential expression for the z-component force exerted by the fluidon the stationary wall yields;


71

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FFluid⇒Solid{ }z−component = dFFluid⇒Solid{ }z−component∫ =12

dΘ R2ΔPd zL

0

1

∫0

2π

∫ = πR2ΔP

By definition, the magnitude of the dynamic vector force exerted by fluids on stationarysolid surfaces for any flow problem is given by;

| { FFluid on Solid }z-component | = (1/2) ρV2 (Shear Area) f

where V represents an average velocity. The previous equation can be considered as anoperational definition of the friction factor f. If one employs previous results for spatiallyaveraged properties to calculate V (or <vz>) for laminar flow of incompressible Newtonianfluids, then the previous two equations yield f = 16/Re, where the shear area, or lateralsurface area of the tube, is 2πRL, the Reynolds number is defined by Re = ρ<vz>2R/µ,and the correlation is valid when Re is below 2100.

A word of caution is important, here, if one employs dimensionless momentumtransfer correlations between f and Re that were developed elsewhere. The followinginformation must be available;

(1) How is the Shear Area defined? (2) What is the characteristic length in the definition of Re? (3) How is the Reynolds number defined? (4) Over what range of Re is the correlation valid?

Generalized interpretation of f vs. Re; what is the physicalsignificance of these correlations? When the mass flowrate and the Reynoldsnumber increase, the friction factor typically decreases, except for high Reynolds numberflow around submerged objects. However, the dynamic force transmitted across a fluid-solid interface increases at higher Reynolds numbers in all flow regimes. The generalizedcorrelations are;

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FFluid⇒Solid{ }Pr imaryFlowDirection =12ρV 2 ShearArea( ) f

f =C1Rea

;Re =ρVL

µ

where L is the characteristic length in the definition of Re, and the exponent "a" is thenegative slope of f vs. Re on log-log coordinates. The dependence of the dynamic forcein the primary flow direction on density, viscosity and average fluid velocity is;


72

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FFluid⇒Solid{ }Pr imaryFlowDirection ≈ µ aρ1−aV 2−a

Intuitively, this interfacial force should increase for fluids with (i) higher viscosity in thecreeping or laminar flow regime, (ii) higher density in the turbulent regime, and (iii) higherflowrates in general. Hence, the acceptable range of the exponent "a" is;

0 ≤ a ≤ 1

Consider the following correlations for two-dimensional flow of incompressible Newtonianfluids around solid spheres, where the shear area is πR2 (not 4πR2) and the characteristiclength in the definition of the Reynolds number is the sphere diameter.

(i) f = 24 / Re Re ≤ 0.5

(ii) f ≈ 18.5 / Re0.6 2 < Re < 500

(iii) f ≈ 0.44 500 < Re < 2 x 105

Correlation (i) in the creeping flow regime represents an exact solution to the steadystate Equation of Motion when its entire left-hand side has been neglected becauseforces due to convective momentum flux are negligible. The latter two experimentalcorrelations [i.e., (ii) and (iii)] have been obtained at higher Reynolds numbers whereforces due to convective momentum flux are important, and an exact solution of the r-and Θ-components of the Equation of Motion, together with the Equation of Continuity ismuch more difficult, if not impossible. In the creeping flow regime where a=1;

| { FFluid on Solid }Primary Flow Direction | ≈ µ V

At much higher flowrates in the turbulent flow regime, where the exponent "a"approaches zero;

| { FFluid on Solid }Primary Flow Direction | ≈ ρ V2

Turbulent flow past a solid sphere is appropriate to describe the flight of a baseball orgolfball, for example, and these correlations reveal that the hydrodynamic drag forcescales as V2. Furthermore, density, not viscosity, is the most important fluid propertythat should be considered. The dimples on a golf ball reduce hydrodynamic drag in thehighly turbulent flow regime by approximately a factor of two because the fluidstreamlines moving past this rough surface follow the contour of the ball for a longer


73

distance beyond the stagnation point relative to a smooth surface. This phenomenonhas been demonstrated by attaching sandpaper to the front of a bowling ball, at itsstagnation point, and plunging the ball into a container of liquid. The fluid streamlines aremore efficient at following the contour of the rough bowling ball because the turbulenteddies that are generated by the rough surface allow momentum to be transferred fromthe free stream into the boundary layer adjacent to the fluid-solid interface. Hence, fluidparticles attempting to traverse a streamline on the backside of the submerged objecthave a higher probability of penetrating regions of higher pressure as their tangentialvelocity component decreases in magnitude. If these fluid particles are not successful intraversing a streamline on the backside of a submerged object, then they are redirectedtoward the free stream and a chaotic low-pressure wake develops behind the object.This phenomenon is described as “boundary layer separation”, and the consequence isthat the submerged object experiences significant hydrodynamic drag. “Streamlinedobjects”, whose surface contours match that of a dolphin, for example, are designed toreduce hydrodynamic drag by minimizing boundary layer separation

Practical examples of hydrodynamic drag. Consider the followingsituations where (i) air or water flow around submerged streamlined objects has beendesigned to reduce hydrodynamic drag forces, or (ii) hydrodynamic forces are primarilyresponsible for the direction of flight of an object.

1) Aerobars on a bicycle

2) Bladed bicycle spokes, that look like linguine, which “slice” the air instead ofcylindrical spokes that “chop” the air as wheels rotate

3) $900 Disk wheels on a bicycle that shield spokes from "cutting" the wind---diskwheels are extremely unstable while riding in a strong crosswind

4) Air flow transverse to tear-drop bicycle tubes instead of cylindrical tubes

5) New designs for bicycle helmets that include an extended tail---Greg LeMond wonthe 1989 Tour de France by 8 seconds over Laurent Fignon after more than 100hours and 2031 miles of racing in three weeks!!! LeMond rode the final 15-miletime trial from Versailles to Paris in 26 minutes 57 seconds (34 miles per hour)

6) Recumbent bicycles that are equipped with wind shields (the world hour record ina human-powered vehicle is 47 miles in 1 hour at sea level on one of these bikes)

7) Bicycle racers forming a paceline where the lead rider does most of the work


74

8) Canadian geese flying in a V-pattern as they travel from City Park to a golf coursesouth of Fort Collins. It is not known whether it’s one goose, one vote, or somestrange version of the electoral college, but however they decide, the goose thatstarts out at the point of the “V-pattern” is most likely the strongest flier. Asthe lead goose fatigues, it moves back and is replaced by a bird that continues tofly strong. There’s a tremendous force of wind resistance that the lead bird isbreaking, which is extremely power-demanding. Goose-flying formations areactually lessons in physics and fluid dynamics. As a bird flies, vortices of air spiraloff its wings, creating slipstreams. When one bird flies directly behind anotherbird, there is some resistance from this slipstream of air. However, the secondbird experiences a suction effect when it flies behind the lead bird’s wingtips.Studies estimate that geese flying in a “V-pattern” can continue to flyapproximately 70% farther than one goose could fly alone.

9) Polymer solutions used to coat ship hulls for reduced drag in the ocean---theoverall objective is to recapture America's cup

10) Take-off gear on airplane wings to minimize power requirements for departures

11) Designs for Indy 500 racing cars that travel faster than 200 miles per hour---drinking a quart of milk in the winner’s circle never hurts anyone

12) The air foil on the roof of a semi-cab prevents wind from crashing into the bluntfront surface of its trailer on Interstate 25

13) Dimpled golf balls that travel much farther than smooth ones, particularly at theMount Massive golf course near Leadville (elevation ≈ 9800 feet above sea level).In July 1969, Edwin "Buzz" Aldrin followed Neill Armstrong down the steps of theEagle, which landed the previous day in the Sea of Tranquility, and hit a golf ballout of sight in the "thin" lunar atmosphere, coupled with a gravitational field thatis 6-fold weaker than the one on Earth.

14) Hot wax applied to the base of Nordic skating skis to enhance "slip" at the wall

15) Downhill skiers testing various "tuck" positions in a wind tunnel trying to reducetheir time (in seconds) in the first or second decimal place on the giant slalomcourse at the 2002 scandal-plagued Olympics in Salt Lake City

16) Swimmers who shave their head for the freestyle event in quest of a gold medal


75

17) Skyscraper designs in large cities to minimize the effects of gale force winds

18) Roofing designs to prevent property destruction in Boulder during wind storms

19) Dandy Don Meredith and Frank Gifford say that it's easier to kick long-distancefield goals during Monday night football games at Mile High stadium in Denverrelative to any other professional football stadium in the country. Jason Elamtied Tom Dempsey’s record with a 63-yard field goal at Invesco Field.

20) If the placekick holder (i.e., most likely, a 2nd- or 3rd-string quarterback) positionsthe football with the laces facing right, then field goal attempts will be "wideright" because of the hydrodynamic force imbalance. Remember when ScottNorwood missed a last-second field goal attempt in the 1991 Super Bowl and theNew York Giants defeated the Buffalo Bills, 20-19? On more than one occasion,the Florida State Seminoles field goal specialist was "wide right" in the finalminutes of the game against Miami, and the Hurricanes won the game and thenational championship of collegiate football.

21) If a pitcher throws a curve ball, then it breaks more near sea level at YankeeStadium in the Bronx (i.e., a borough of New York City) than it does at CoorsField in Denver. That might explain why the Colorado Rockies don’t generatemuch offense during “away” games.

22) Relative to baseball games played in humid conditions, there aren't many homeruns hit in domed stadiums where water vapor is removed from the air via "airconditioning". Hank Aaron played many baseball games in muggy Atlanta and hit755 lifetime homers. However, the Seattle Mariners set a new record for mosthome runs hit by an entire team during the 162-game season in 1997, and halfof their games were played in the Kingdome where there is a 20 mile-per-hourwind blowing from home plate toward the outfield. Hey, humid air is lighter.I can ignore it if you only say it once per week. However, you have nowmentioned it twice this week and I feel compelled to correct your error. Contraryto what you are telling your readers, humid air is actually lighter than dry air.That’s right, with all other things being equal, particularly altitude, curveballs“break” less and batted balls travel farther in humid air relative to dry air. If youdon’t believe me (and why should you?), then contact someone in theDepartment of Atmospheric Science at Colorado State University.


76

23) Major league baseball scores of games played at Coors Field resemble footballscores in defensive-minded struggles. Look what happened at the All-Star gamein 1998! (i.e., the commissioner let the players finish the game and the AmericanLeague won by the score of 13-8).

24) At the highest airport on the planet in La Paz, Bolivia, planes must achieve veryfast take-off velocities to create the required lift because, at 13,000 feet abovesea level, the density of air is reduced considerably relative to that at sea level.

25) Speed-skating revolution. When someone asked Michael Jordan, “Is it in theshoes?”, they had the correct question but the wrong sport. Long-track speedskaters definitely will break records at the winter Olympic games, thanks not totheir abilities but to a new skate that shaves off a second per lap. The noisy new“klap” (Dutch for “slap”) skate allows a longer and smother stride. Since theirdebut at international events, they have helped tie or break more than 16 worldrecords. Every medal contender wears them at international speed skatingevents. Fast-forward a few years. At first glance, they seem insignificant;tiny strips of rubber roughly the size of a seam, hardly noticeable on a speedskater’s skin-tight suit. However, these racing stripes (if that’s the correct wordfor them, the concept is so new that no one is sure what their official nameshould be) are the latest weapon in the technological tussle to increase skater’sspeeds. Amazingly, the stripes actually transform the “klap” skate into asecondary issue. It used to be that one would simply wake up in the morning andskate. Now, there’s all of this tinkering going on. Theory dictates that thestripes provide an aerodynamic edge by reducing hydrodynamic drag (i.e., windresistance) when skaters reach speeds of 40 mph. Does it work? The Dutchthink so because they tested the concept in wind tunnels and surprised thecompetition at Olympic events by obtaining permission to attach a few squigglystripes to each leg of their racing suits, running from knee to ankle, and anotherstripe was attached to their hood.

Identify two more practical examples of drag reduction, or situations wherehydrodynamic drag has a major effect on the flight of a submerged object movingthrough air or water.

Use of f vs. Re to analyze pressure drop vs. flowrate in tubes.Laminar flow of an incompressible Newtonian fluid through a straight tube with radius Rand length L corresponds to f = 16/Re and log Q ≈ log ΔP, where f is the friction factor,Re is the Reynolds number based on the tube diameter, Q is the volumetric flowrate and


77

P is dynamic pressure. Determine the scaling law exponent α for turbulent flow of anincompressible Newtonian fluid through the same tube;

log Q ≈ α log {ΔP}

where f = 0.0791/Re1/4. Hint: The z-component of the dynamic force exerted by thefluid on the wall at r=R is πR2ΔP.

AnswerPrior to solving this problem, it is instructive to consider the underlying fundamentalsrelated to the hint provided above. In terms of forces or stresses due to totalmomentum flux that act across the solid-liquid interface, the differential vector forceexerted by the fluid on the tube wall is;

dFFluid on Solid = { δr ( τrr + p ) + δΘ τrΘ + δz τrz }r=R R dΘ dz

The z-component of dFFluid on Solid is obvious from the previous expression. Rigorously, it isobtained via the following scalar dot product operation;

δz • dFFluid On Solid = {τrz(r=R)} R dΘ dz

Integration of the previous equation over the complete lateral surface (i.e., 0≤Θ≤2π,0≤z≤L) for incompressible Newtonian fluids yields;

€

FFluid⇒Solid{ }z−component = δz •d∫ FFluid⇒Solid = dΘ0

2π

∫ −µdvzdr

r=R

Rdz

0

L

∫

For one-dimensional flow in the z-direction, where vz(r) is not a function of spatialcoordinates within the lateral surface, the final expression for the macroscopic dynamicforce simplifies considerably because τrz is also independent of the lateral surfacecoordinates. Hence;

{ FFluid on Solid }z-component = { - µ (dvz/dr)r=R } 2π R L = π R2 ΔP

This result is verified rather easily for laminar flow in terms of the microscopic fluidvelocity gradient at the tube wall. For steady state one-dimensional flow through astraight tube in any regime, a combination of the macroscopic mass and momentumbalances yields the same result, as given by the previous equation. The solution to thisproblem begins by employing the macroscopic momentum transfer correlation, which


78

includes the definition of the friction factor, to evaluate the z-component of the dynamicforce exerted by the fluid on the tube wall, with shear area given by 2πRL. For example;

{ FFluid on Solid }z-component = π R2 ΔP = (1/2) ρ <vz>2 { 2 π R L } f

Now, use the dimensionless correlation for f vs. Re, where the Reynolds number isdefined in terms of the tube diameter;

Re = ρ <vz> ( 2R ) / µ

In terms of the scaling law for dynamic force;

{ FFluid on Solid }z-component = π R2 ΔP ≈ µa ρ(1-a) <vz>(2-a)

where a = 1/4 in the turbulent flow regime. For tube flow, average velocity <vz> andvolumetric flowrate Q are related by the cross-sectional area for flow (i.e., πR2). Hence,the dynamic pressure drop ΔP scales as Q taken to the (2-a) power. In other words;

ΔP ≈ Q(2-a)

Therefore, the scaling law exponent which relates Q to ΔP is, α = 1/(2-a) = 4/7. Thecomplete result for laminar or turbulent flow of an incompressible Newtonian fluidthrough a straight tube of radius R and length L is;

€

Q2−a =2aπ 2−aR5−aΔPC1ρ

1−aµ aL

when the dimensionless momentum transfer correlation is f = C1/Rea. In the laminar flowregime, where C1 = 16 and a = 1, the previous equation reduces to the classic Hagen-Poiseuille law;

€

Q =πR4ΔP8µL

The solution to this problem reveals that Q and ΔP do not follow a linear relation forturbulent flow of an incompressible Newtonian fluid through a tube. Hence, if one hasdata for Q vs. ΔP that correspond to flow of an incompressible Newtonian fluid through astraight tube with radius R and length L, then the exponent "a" in the experimental f vs.Re correlation can be obtained from the slope of a log-log graph of Q vs. ΔP, provided


79

that the data do not span more than one flow regime. From the viewpoint of linear leastsquares analysis for the pressure-drop/flowrate relation in straight tubes, the followingprocedure should be employed to calculate the experimental scaling law exponent;

(i) polynomial model; y(x) = a0 + a1x

(ii) independent variable; xi = log Q

(iii) dependent variable; yi = log ΔP;

(iv) intercept, or zeroth-order coefficient;

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a0 = log C1ρ1−aµ aL

2aπ 2−aR5−a

(v) slope, or 1st-order coefficient; a1 = 2 - a

Analysis of terminal velocities for submerged objects. A solid sphere ofradius Rsphere and density ρsphere falls through an incompressible Newtonian fluid that isquiescent far from the sphere. The viscosity and density of the fluid are µfluid and ρfluid,respectively. The Reynolds number is 50, based on the physical properties of the fluid,the diameter of the sphere and its terminal velocity. The following scaling lawcharacterizes the terminal velocity of the sphere in terms of geometric parameters andphysical properties of the fluid and solid;

log vterminal ≈ α log Rsphere + β log { ρsphere - ρfluid } + γ log µfluid + δ log ρfluid

(i) Calculate the scaling law parameters α, β, γ and δ in the previous equation. Fournumerical answers are required, here.

AnswerSince there is no longer any acceleration when submerged objects achieve terminalvelocity, the sum of all forces acting on the object must be zero. Hence, there is abalance between buoyancy, gravity, and hydrodynamic drag. The gravity force actsdownward, and the buoyant and drag forces act in the opposite direction. In general, thehydrodynamic drag force acts (i) in the opposite direction of the motion of the objectwhen the fluid is stationary, (ii) in the same direction as the motion of the fluid when theobject is stationary, or (iii) in the direction of the relative motion of the fluid with respectto the object when neither one is stationary. Each force is calculated as follows;


80

Gravitational force: (4/3) π R3sphere ρsolid g

Buoyant force: (4/3) π R3sphere ρfluid g

Hydrodynamic drag force: (1/2) ρfluid v2terminal ( π R2

sphere ) f

For flow around spheres in any regime, the dimensionless momentum transfer correlationadopts the following form;

f = C1 / Rea Re = ρfluid vterminal ( 2 Rsphere ) / µfluid

Now, the hydrodynamic drag force can be expressed explicitly in terms of physicalproperties of the fluid and solid;

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HydrodynamicDragForce=πC121+a

µ fluida ρ fluid

1−a Rspherevtermin al( )2−a

Rearrangement of the above-mentioned force balance yields the following solution forvterminal;

€

vtermin al2−a =

23+a

3C1

Rsphere1+a ρsolid −ρ fluid( )g

µ fluida ρ fluid

1−a

Therefore, the scaling law parameters are;

€

α =1+ a2 − a

;β =12 − a

;γ =−a2 − a

;δ = −1− a2 − a

where a = 3/5 in the intermediate (i.e., laminar) flow regime. With reference to acreeping flow falling sphere viscometer, one measures the terminal velocity of a solidsphere that falls slowly through an incompressible Newtonian fluid. In the creeping flowregime, the dimensionless momentum transfer correlation for solid spheres is f = 24/Re,which corresponds to C1 = 24 and a = 1. Hence;

€

vtermin al =2Rsphere

2 ρsolid −ρ fluid( )g9µ fluid


81

One estimates the fluid viscosity by rearranging the previous equation. This prediction isaccurate if the Reynolds number is smaller than 0.5. For consistency, one measures theterminal velocity of a solid sphere that falls through an incompressible Newtonian fluid,rearranges the previous equation to calculate the fluid viscosity, and uses the physicalproperties of the fluid, together with vterminal, to demonstrate that the Reynolds numbercorresponds to creeping flow. See page#61 in the 2nd edition of Transport Phenomenaby RB Bird, WE Stewart & EN Lightfoot.

(ii) A different sphere of the same density with radius 2Rsphere falls through the sameincompressible Newtonian fluid. Now, the Reynolds number is greater than 50,but less than 500, because the diameter of the sphere has increased by a factorof 2. Does the terminal velocity of the sphere increase, decrease, or remainunchanged?

AnswerSince the scaling law exponent α > 0, one achieves larger terminal velocity if the size ofthe sphere increases.

(iii) By how much, or by what factor, does vterminal change in part (ii)? For example, ifthe terminal velocity of the sphere remains unchanged, then it changes by afactor of one.

AnswerThe scaling law in part (a) can provide both qualitative and quantitative results. If thesphere radius increases by a factor of 2, then vterminal increases by {2}(1+a)/(2-a), whichcorresponds to {2}(8/7) in the intermediate flow regime where a = 3/5.

(iv) How does the scaling law for terminal velocity change if a non-deformable bubbleof radius Rbubble rises with constant velocity through the same incompressibleNewtonian fluid in the same flow regime (i.e., 50 ≤ Re ≤ 500)?

AnswerFirst, one must replace Rsphere by Rbubble, but this is a minor change. Secondly, and mostimportantly, the hydrodynamic drag force acts downward when bubbles rise. Now, theupward buoyant force is counterbalanced by gravity and hydrodynamic drag.Consequently, one must replace ( ρ solid - ρ fluid ) by ( ρfluid - ρbubble ) in the scaling law forvterminal, as presented in part (i).


82

Linear least squares analysis of terminal velocities for spheres ofdifferent radii in various flow regimes

Algebraic rearrangement of the physical model for terminal velocity in the previoussection yields;

€

log vtermin al( ) =12 − a

log23+a g ρsolid −ρ fluid( )3C1µ fluid

a ρ fluid1−a

+1+ a2 − a

log Rsphere( )

= a0 + a1 log Rsphere( )

a0 =12 − a

log23+a g ρsolid −ρ fluid( )3C1µ fluid

a ρ fluid1−a

a1 =1+ a2 − a

Creeping flow:Re < 0.5, C1 = 24, a = 1;

€

a0 = log2g ρsolid −ρ fluid( )

9µ fluid

a1 = 2Laminar flow:2 ≤ Re ≤ 500, C1 ≈ 18.5, a = 0.6;

€

a0 =57log

23.6g ρsolid −ρ fluid( )3 18.5( )µ fluid

0.6 ρ fluid0.4

a1 =87

Turbulent flow:500 ≤ Re ≤ 2x105, C1 ≈ 0.44, a = 0;

€

a0 =12log

8g ρsolid −ρ fluid( )3 0.44( )ρ fluid

a1 =12


83

Terminal velocity of glass spheres (turbulent flow)

This program analyzes buoyant forces, gravity forces, and hydrodynamic drag forces onglass spheres that fall through an incompressible Newtonian fluid in the turbulent flowregime. The fluid is carbon tetrachloride at 200C with known density and viscosity. Theobjective is to achieve a terminal velocity of 65 centimeters per second. The steadystate force balance on glass spheres that don't accelerate is solved for the spherediameter that will accomplish this task. It is necessary to calculate the Reynolds numberand verify that fluid flow is actually turbulent and corresponds to the Newton's Lawregime.

density of carbon tetrachloride at 200 C, g/cm3

ρfluid = 1.59

density of glass spheres, g/cm3

ρsolid = 2.62

terminal velocity that is required, cm/secvterminal = 65

viscosity of CCl4, gram per cm per second, Poise at 200Cµfluid = 0.00958

gravitational acceleration constant, cm/sec2

gravity = 980.665

volume of the glass sphereVolumeSphere = (4/3)π(DSphere/2)3

projected area of the sphere, as seen by the approaching fluidAreaProjection = π(DSphere/2)2

VolumeSphere/AreaProjection = (2/3)DSphere

Steady state force balance that accounts for gravity, buoyancy, andhydrodynamic drag(ρsolid - ρfluid)gravity{VolumeSphere/AreaProjection} = (1/2)ρfluidv2

terminal{friction factor}

check the Reynolds number to verify the flow regimeReynolds number = ρfluidvterminalDSphere/µfluid


84

Hint: this flow problem corresponds to the turbulent flow regime;500 < Re < 2 x 105

friction factor = 0.44

No numerical solution exists when f = 24/Re and Re is restricted to be less than 0.5No numerical solution exists when f = 18.5/Re3/5 and 2 < Re < 500Hence, there are no solutions to this problem in the creeping and laminar flow regimes.

SolutionAreaProjection = 3.78e+0 [cm2]DSphere = 2.19e+0 [cm]friction factor = 4.40e-1 [dimensionless]gravity = 9.81e+2 [cm/sec2]µfluid = 9.58e-3 [g/cm-sec]Reynolds number = 2.37e+4 [dimensionless]ρfluid = 1.59e+0 [g/cm3]ρsolid = 2.62e+0 [g/cm3]VolumeSphere/AreaProjection = 1.46e+0 [Volume/Area, cm]VolumeSphere = 5.54e+0 [cm3]vterminal = 6.50e+1 [cm/sec]

Applications of hydrodynamic drag forces via f vs. Re to calibrate arotameter when the test fluid is different from the calibration fluid. Arotameter consists of a vertical conical tube that contains a float of higher density thanthat of the fluid passing through the meter. The tube diameter is not constant, but itincreases linearly as the float moves to higher positions in the conical tube. This featureallows the rotameter to measure a wide range of mass flow rates. When the rotameter iscalibrated for a particular fluid, it is very straightforward to measure mass flow rates forthat fluid in terms of the height of the float under steady state conditions. You are givena rotameter calibration curve for water which illustrates that mass flow rate is linearlyproportional to float height. However, experiments on a distillation column require thatyou must measure the mass flow rates of alcohols using the rotameter that wascalibrated for water. Devise a strategy and use that strategy to modify the rotametercalibration curve for water so that one can measure the mass flow rate of an alcoholusing the same rotameter.

Your final answer should include strategies when a log-log plot of friction factorvs. Reynolds number for flow through a conical tube that contains a submerged object(i.e, the float);


85

(i) is a straight line with a slope of -1.0(ii) is a straight line with a slope of -0.5(iii) is a straight line with zero slope

AnswerResults from the previous section provide a generalized expression for the terminalvelocity of solids or bubbles in stationary fluids. The same results describe the averagefluid velocity in the vicinity of a stationary submerged object, like the rotameter float.Of course, the shear area and volume of a solid sphere or gas bubble are well defined interms of the radius of the submerged object. The corresponding shear area and volumeof the rotameter float can be measured, but they can't be expressed in terms of onesimple geometric parameter. Fortunately, these quantities don't change when a differentfluid passes through the rotameter. The strategy below, which focuses on the followingscaling law for the average fluid velocity in the vicinity of the float;

€

vfluid2−a

≈ρsolid −ρ fluid( )µ fluida ρ fluid

1−a

reveals that the shear area, float volume, and gravitational acceleration constant do notaffect the rotameter correction factor. One obtains the corresponding mass flow ratefrom the previous scaling law via multiplication by the fluid density and the cross-sectional area. Since the rotameter correction factor compares mass flowrates for twodifferent fluids when the float height is the same, the flow cross section does not appearin the final result because it remains constant. Hence, it is only necessary to multiply<v>fluid by ρfluid. Therefore, when the float is at the same position, the mass flowrate ofany fluid through the same rotameter scales as;

€

ρ fluid v fluid ≈ρ fluid ρsolid −ρ fluid( ){ }

1/(2−a )

µ fluida /(2−a)

The quantity on the right side of the previous equation must be evaluated for the testfluid and the calibration fluid. This ratio (i.e., test fluid/calibration fluid) represents thecalibration factor which one must multiply by the mass flowrate of the calibration fluid ata given rotameter float height to obtain the mass flowrate of the test fluid when thefloat is in the same position.

For part (i), a = 1 and the mass flowrate for each fluid scales as;


86

€

MassFlowRate ≈ ρ fluid v fluid ≈ρ fluid ρsolid −ρ fluid( )

µ fluid

For part (ii), a = 0.5 and the mass flowrate for each fluid scales as;

€

MassFlowRate ≈ ρ fluid v fluid ≈ρ fluid ρsolid −ρ fluid( ){ }

2/3

µ fluid1/3

For part (iii), a = 0 and the mass flowrate for each fluid scales as;

Mass flowrate ≈ { ρfluid ( ρsolid - ρfluid ) }(1/2)

(iv) At 200C, the density of water is 1.00 g/cm3 and the density of methanol is 0.79g/cm3. The float density is 3.95 g/cm3. Compare the mass flow rates of waterand methanol through the same rotameter at 200C when the float rests at thesame position in the rotameter. In both cases, the dimensionless momentumtransport correlation is f ≈ constant in the high Reynolds number regime.

AnswerEvaluate the mass flowrate scaling factor for water and methanol via the prescriptionfrom part (iii), because a = 0. Then, construct the ratio of these scaling factors tocompare the mass flowrates of the two fluids. For example;

(1) Mass flowrate of water ≈ { ρwater ( ρfloat - ρwater ) }(1/2)

(2) Mass flowrate of methanol ≈ { ρmethanol ( ρfloat - ρmethanol ) }(1/2)

The ratio of (1) to (2) is 1.09, which indicates that the mass flowrate of water is 9%larger than that of methanol.

(v) How does your comparision of the mass flow rates of water and methanol at200C from part (iv) change if the float density is only 1.35 g/cm3?

AnswerUse the scaling laws in part (iv) for water and methanol, but reduce the float densityfrom 3.95 g/cm3 to 1.35 g/cm3. Now, the ratio of (1) to (2) is 0.89, which indicatesthat the mass flowrate of water is about 11% less than that of methanol.


87

(vi) In the highly turbulent regime, the mass flow rates of water and methanol will bethe same @ 200C when a particular float rests at the same position in therotameter. What float density is required for this statement to be true?

AnswerEquate the scaling laws in part (iv) for water and methanol and solve for ρfloat;

{ ρwater ( ρfloat - ρwater ) }(1/2) = { ρmethanol ( ρfloat - ρmethanol ) }(1/2)

ρfloat = 1.8 g/cm3.


88

Review Problems for Exam#2

Problem#1Use one or two sentences and describe the phrase "very low Reynolds numberhydrodynamics". Do not include any equations in your description.

Problem#2 Sketch velocity profiles in 4 viscometers(a) Concentric cylinder viscometer

Sketch vΘ vs. r when the inner cylinder rotates very slowly at angular velocity Ω,and the outer cylinder is stationary.

(b) Parallel-plate viscometerSketch vΘ vs. z when r/R = 0.5. The lower plate at z=0 is stationary and theupper plate at z=B rotates very slowly at anglular velocity Ω.

(c) Rotating sphere viscometerSketch vφ vs. r at Θ = π/2. The sphere rotates very slowly at angular velocity Ωand the fluid is stationary far from the sphere.

(d) Cone-and-plate viscometerSketch vφ vs. Θ at r/R = 0.5, when the angle of the cone Θ1 is 800, 850 & 890.Put all three curves on one set of axes and indicate the value of Θ1 thatcorresponds to each curve. The cone rotates very slowly at angular velocity Ω,and the plate at Θ = π/2 is stationary.

(e) The fluid velocity profile in the cone-and-plate viscometer is "similar" to the fluidvelocity profile in one of the other three viscometers listed above. Identify thisother viscometer and explain briefly why the fluid velocity profiles are "similar".

Problem#3Use information in Transport Phenomena on pages#55-56 in the 2nd edition and page#53in the 1st edition to develop a relation between the friction factor and the Reynoldsnumber for laminar flow of an incompressible Newtonian fluid between two concentriccylinders, where flow in the z-direction vz(r) is driven by a gradient in dynamic pressure.The radius ratio of the inner to the outer cylinder is κ = Rinner/Router < 1. It is necessary toconsider dynamic forces exerted by the fluid on both solid surfaces, but it is notnecessary to simplify your algebraic expression to obtain a concise result for f vs. Re.The characteristic length, or effective diameter, in the definition of the Reynolds number


89

for flow on the shell side of the double-pipe heat exchanger is the difference betweenthe outer and inner cylindrical diameters, 2Router(1-κ).

Problem#4The volumetric flowrate, Q = πR2<vz>, of an incompressible Newtonian fluid through asmooth straight tube of radius R and length L increases by a factor of 4. This 4-foldincrease in Q causes the Reynolds number to increase from 5 x 103 to 2 x 104.

(a) Does the z-component of the dynamic interfacial shear force exerted by the fluidon the stationary wall at r=R increase, decrease, remain constant, or is it toocomplex to obtain a quantitative estimate of the change in this interfacial force?

(b) By what factor does the z-component of the dynamic interfacial shear force frompart (a) change? A factor of 1 indicates no change, a factor less than 1 indicatesa decrease, and a factor greater than 1 indicates an increase in the interfacialforce. A numerical answer is required, here.

Problem#5It should be obvious that the terminal velocity of a bowling ball in air is much larger thanthe terminal velocity of a popped corn in air. However, in both cases, a steady stateforce balance on the object that accounts for buoyancy, gravity, and hydrodynamic dragreveals that;

log { vterminal } ≈ ζ log ( ρsolid - ρair )

where ζ is the scaling law exponent, and ρsolid corresponds to either the bowling ball orthe popped corn.

(a) What is the value of ζ if ReBowlingBall ≈ 200,000?

(b) What is the value of ζ if RePoppedCorn ≈ 0.1?

Problem#6Consider a baseball that is 3 inches in diameter (i.e., 0.25 ft) moving at a velocity of 100miles/hr. (i.e., 147 ft/sec) through stagnant air at 200C having a kinematic viscosity (ormomentum diffusivity) of 1.6 x 10-4 ft2/sec off the bat of Roger Maris in the 6th inning ofthe 6th game of the 1964 World Series (for sports trivia fans, the St. Louis Cardinals beatthe New York Yankees, 4 games to 3, and right-handed pitcher Bob Gibson, who wongames #5 and #7 with 22 strikeouts in those two games, was the MVP of the series).The Reynolds number is 2.3 x 105, based on the sphere diameter. Provide a qualitativeranking of the distance that the baseball will travel if the game is played at each of the


90

following locations. The temperature of the air, wind conditions, initial velocity of thebaseball and its angle of inclination are the same in each location.

(a) Texas Rangers stadium in Arlington, Texas near Dallas where Nolan Ryan pitchedwithout air conditioning and sweated profusely,

(b) Minute-Maid Park in Houston, Texas where the domed stadium is air conditionedto remove unwanted moisture,

(c) Leadville Giants minor league stadium which sits at an elevation of 10,150 feet atthe base of Mount Elbert,

(d) Coors' Field in Denver at an elevation of 5280 feet (don't account for the factthat this stadium and major league baseball did not exist in Denver during theRoger Maris era).

Problem#7A crazy cyclist was riding a bicycle in the horrendous westerly winds when Channel 9weather experts estimated that the intensity of convective momentum flux was "off ofthe charts". Use arrows to represent vectors and illustrate the motion of a cyclist in acrosswind, where the wind is blowing perpendicular to the path of the cyclist. Then, drawa vector that illustrates the direction in which the fluid (i.e., air) exerts a hydrodynamicdrag force on the submerged object (i.e., the cyclist). Assume that the wind is blowingfrom the west at 30 miles per hour, and that the cyclist is riding north at 15 miles perhour.

These are additional topics in fluid dynamics which were not discussedduring the first 7 or 8 weeks of the course, but they could be mentioned inconnection with analogous topics in mass transfer

(a) Why is the Reynolds number important?(b) Why is the Reynolds number calculated for all momentum transport problems?

Dimensional analysis of the equations governing momentum transportDynamic similarity and scaling concepts

Macroscopic mass balances(a) Application of the unsteady state mass balance and the Hagen-Poiseuille law to

calculate the time required to drain capillary tubes with cylindrical or sphericalbulbs


91

(b) Parameters that effect the capillary constant in the rheology laboratoryexperiments

Macroscopic Mass Balance

Important Considerations in the Development and Use of the Unsteady StateMacroscopic Mass Balance

(1) Accumulation rate process is balanced by the net rate of input, since there are nosources or sinks of overall fluid mass.

(2) Each term in the macroscopic mass balance has dimensions of mass per time.

(3) The accumulation rate process on the left-hand side of the equation requires atime derivative for unsteady state analysis. Ordinary differential equations mustbe solved to analyze the transient behaviour of a system.

(4) Mass flux provides a convective mechanism in the macroscopic mass balance dueto fluid flow across the inlet/outlet fictitious planes that bound the system.

(5) The divergence of convective mass flux, via the “del” operator (i.e., whom Mr. &Mrs. Shannon named their only son after), accounts for the net rate of output(i.e., output – input) on the right-hand side of the macroscopic mass balance.

(6) One equates rates of input to rates of output at steady state, after the transientresponse of the system decays to zero. Steady state response of a system isanalyzed by solving algebraic equations.

(7) If a system initially contains a given amount of mass, then this quantity appearsin the integrated form of the unsteady state macroscopic mass balance via theinitial condition, most likely at t=0, but not as a rate of input or a rate of output.

(8) One should only analyze the transient response of a system during a period oftime when there are no abrupt changes in rates of input or rates of output. If theinput and/or output streams change smoothly, then it is acceptable to analyzethe system over a timescale that includes these smooth changes.


92

(9) If it is necessary to analyze the transient response of a system over severaldifferent time intervals because rates of input and/or output change abruptly, ordiscontinuously, then one must be sure that the mass of the system iscontinuous from one time interval to the next. No principles are violated if theslope of total system mass vs. time changes abruptly from the end of one timeinterval to the beginning of the next interval, in response to an abrupt change inrates of input and/or output.

(10) Total system mass achieves an extremum (i.e., maximum or minimum) when thesum of all inlet mass flowrates equals the sum of all outlet mass flowrates.

(11) For a system that is well mixed in an agitated tank, for example, the overall massdensity within the tank and in the exit stream might be time-dependent but it willnot depend on spatial coordinates due to the mixing process. Hence, onemanipulates the accumulation rate process on the left-hand side of themacroscopic mass balance as follows;

€

ddtmtotal =

ddt

ρdV =ddtV

∫∫∫ ρV{ } = ρdVdt

+V dρdt

(12) The system volume will change with time and, hence, dV/dt will be nonzero if thesum of all inlet volumetric flow rates is different from the sum of all outletvolumetric flow rates. The time-rate-of-change of total system volume iscalculated from the difference between the sum of all inlet volumetric flow ratesand the sum of all outlet volumetric flow rates. These statements are valid whenthe system corresponds to fluid in a tank, not fluid plus “empty space”, forexample, as the tank is drained.

(13) The most generalized form of the unsteady state macroscopic mass balance forsystems with variable volume, multiple inlet/outlet streams, and no sources orsinks of overall fluid mass is;

€

ddtmtotal = ρ

dVdt

+V dρdt

= ρQ{ }iinletstreami

∑ − ρQ{ } joutletstream

j

∑

where ρ is fluid density and Q represents volumetric flow rate with respect to theappropriate fictitious inlet or outlet plane.


93

(14) The time constant for the transient response of any system with inlet and outletstreams is the residence time τ. When the sum of all inlet volumetric flow rates isbalanced by the sum of all outlet volumetric flow rates and the total systemvolume remains constant, approximately five residence times (i.e., 5τ) arerequired for the transient response to decay to within 99.7% of the new steadystate response.

Transient and Steady State Analysis of the Dilution of a SaltSolution in a Well-Mixed Vessel

A storage vessel contains 1000 Litres of salt solution with a mass of 1.05x103

kg, and the initial salt concentration is 50 grams per Litre. At time t=0, salt-free waterenters the vessel at a rate of 9 Litres per minute. The outlet valve is opened at t=0 andsalt solution exits the vessel at a rate of 10 Litres per minute. Hence, the volume offluid in the tank decreases uniformly at a rate of 1 Litre per minute. Perfect mixinghomogenizes the concentration of salt in the liquid within the tank. There is also a largeexcess of solid salt at the bottom of the tank that dissolves into solution at a steadyrate of 5 grams per minute throughout the entire operation of the system.

a) Write all of the equations that are required to calculate the salt mass fraction inthe fluid phase within the vessel and in the exit stream. The salt mass fraction isdefined as ρSalt/ρSolution, where each density ρ is expressed with respect to thesolution volume.

b) Use an ODE solver to generate a quantitative graph of the mass fraction of salt inthe tank and in the exit stream as a function of time.

c) Calculate the steady state mass fraction of salt in the exit stream.

d) Does the system reach 99.9% of its steady state operating point before thevessel is drained? If your answer to this question is YES, then calculate the timerequired for the salt mass fraction in the exit stream to reach 99.9% of itssteady state value. In other words, find the value of time t that satisfies thefollowing equation.

€

ωSalt,Initial −ωSalt t( )ωSalt,Initial −ωSalt t > 5τ( )

= 0.999


94

where ωSalt = ρSalt/ρSolution is the salt mass fraction.

(e) Perform steady state analysis of this well-mixed 1000-Litre vessel when Qin = Qout

= Q = 10 Litre/minute and the rate at which solid salt dissolves into the fluidwithin the vessel is given by;

€

Dissolution_ Rate[grams /min]= 2 ρSalt,Interface −ρSalt t( ){ }

where ρSalt,Interface [= grams/Litre] represents the equilibrium liquid phase massdensity (i.e., solubility) of salt at the solid-liquid interface via thermodynamicconsiderations. This dissolution rate is time-dependent because ρSalt(t) decreasesduring the dilution process. There is a large excess of salt that does not dissolvecompletely into the liquid phase during the time frame of the dilution process. Isthe following expression for the steady state salt mass fraction correct?

€

ωSalt t⇒∞( ) =1

1+ 6ρH2O

ρSalt,Interface

The factor of 6 in the denominator of the previous equation is derived from thesum of 1 + Q/2, and 2 is the coefficient of the mass density difference (i.e.,driving force) on the right side of the dissolution rate expression.

Answer to part (e):Begin by writing unsteady state mass balances for the overall solution and the salt,separately. Then, set the accumulation terms to zero at steady state and solve twocoupled linear algebraic equations for ρSolution and ρSalt when the transient behaviourvanishes. For example, when the inlet and outlet flowrates are equal and the overallsolution volume remains constant;

€

VSolutiondρSolutiondt

=Q ρH2O − ρSolution{ } + kMTCS ρSalt,int erface − ρSalt{ }⇒ 0

VSolutiondρSaltdt

= 0 −QρSalt + kMTCS ρSalt ,int erface − ρSalt{ }⇒ 0


95

Obtain an expression for the steady state mass density of salt in solution from thesecond equation;

€

ρSalt =kMTCS

Q+ kMTCS

ρSalt ,int erface

Now, use this result to solve for ρSolution from the overall mass balance (i.e., the firstequation) when t ⇒ ∞. One obtains;

€

ρSolution = ρH2O +kMTCSQ

ρSalt,int erface −ρSalt{ }

The ratio of these steady state mass densities yields the required expression for themass fraction of salt in the well-mixed vessel and in the exit stream. This answer canbe verified numerically by using specific parametric values for VSolution, Q, kMTCS, andρSalt,interface, together with any reasonable initial conditions, solving both unsteady statemass balances, and inspecting the numerical solution to these coupled ODE’s whentime t is greater than five residence times

€

ωSalt t⇒∞( ) =ρSaltρSolution

=ρSalt

ρH2O +kMTCSQ

ρSalt ,int erface − ρSalt{ }

=1

ρH2O

ρSalt+kMTCSQ

ρSalt,int erfaceρSalt

−1

=1

ρH2O

ρSalt ,int erface1+

QkMTCS

+kMTCSQ

1+Q

kMTCS−1

=1

1+ρH2O

ρSalt ,int erface1+

QkMTCS

(f) For typical operating conditions in this vessel-draining problem, why will ρSalt,Interface

be greater than the initial mass density of salt in solution, ρSalt(t=0)?


96

Transient analysis of draining an incompressible Newtonian fluidfrom a spherical bulb with a tilted capillary tube to simulate theperformance of capillary viscometers for the determination of

momentum diffusivities

This problem combines the unsteady state macroscopic mass balance and theHagen-Poiseuille law for laminar tube flow, together with the volume of fluid in a partiallyfilled sphere. The overall objectives are to (i) predict the capillary constant “b”, basedsolely on geometric parameters of the viscometer, and (ii) compare this prediction withexperimental values obtained by calibrating a capillary viscometer using fluids with knownviscosity and density. The system is defined as fluid within the bulb plus the capillary,and one seeks the time required to drain only the bulb above a capillary that is orientedat angle Θ with respect to gravity. Hence, this is an example of the unsteady statemacroscopic mass balance where the fictitious inlet plane “floats” on the upper surfaceof liquid in the bulb such that the average velocities of the fluid and the surface areequal. Consequently, there is no contribution from convective mass transfer across theinlet plane. Fluid flow across the stationary outlet plane at the exit from the capillary isdescribed by the Hagen-Poiseuille law for incompressible Newtonian fluids. Themacroscopic mass balance for an incompressible fluid with time-varying system volume,no inlet contribution, and one stationary outlet plane reduces to;

€

ρdVsystemdt

= −ρQHP = −ρπRTube

4 ΔP8µL

where the capillary has radius RTube and length L, and P represents dynamic pressure.Laminar flow occurs through a cylindrical capillary tube of length L, regardless of whetherthe capillary is vertical or tilted at angle Θ with respect to gravity. The angle of tilt isconsidered in the dynamic pressure difference ΔP from tube inlet to tube outlet. If h(t)describes the height of fluid within the spherical bulb above the capillary at any time t,and the “zero of potential energy” is placed arbitrarily at the exit from the capillary, thenfluid pressure at the capillary entrance is pambient + ρgh(t), based on approximatehydrostatic conditions in the bulb, and dynamic pressure at the capillary entrance is givenby the sum of fluid pressure and gravitational potential energy per unit volume of fluid.Since the capillary entrance is at higher elevation than the capillary exit, by a distanceLcosΘ, one evaluates dynamic pressure at the capillary inlet as follows;

€

Pinlet = pambient +ρgh t( ) +ρgL cosΘ


97

There is no contribution from gravitational potential energy to dynamic pressure at thecapillary exit because it coincides with the potential energy reference plane. Ambientpressure exists on the upper surface of liquid in the bulb and at the capillary exit. Hence,the dynamic pressure difference ΔP = Pinlet - Poutlet is given by ρg{h+LcosΘ}. One mustsolve the following time-dependent ODE to relate momentum diffusivity to efflux time;

€

dVsystemdt

=ddt

VPartiallyFilledSphere +πRTube2 L[ ] =

ddtVPartiallyFilledSphere = −

πRTube4 g

8 µ /ρ( )Lh t( ) + L cosΘ{ }

Fluid volume within the capillary tube is constant during the analysis of efflux timesbecause one measures the time required to drain the bulb, not the capillary. The nexttask is to evaluate the volume of fluid in a partially filled sphere of radius RSphere when thefluid achieves height h(t). This calculation is performed in cylindrical coordinates bystacking an infinite number of cylinders with infinitesimal thickness dz and radius ω(z),such that ω(z) vanishes when z = 0 and z = 2Rsphere, but ω(z) = RSphere when the sphere is50% filled. Let the spherical bulb sit on the origin of an xyz-coordinate system such thatthe center of the sphere is found at a distance z = RSphere upward from the origin in the z-direction. If the sphere is filled with fluid to height z that can be greater than or lessthan the sphere radius, then the liquid surface is circular and the following relation allowsone to predict the radius ω(z) of the circular surface of liquid;

€

ω z( ){ }2

+ z − Rsphere( )2 = RSphere2

ω z( ){ }2

= 2zRsphere − z2

Now, calculate the volume of an infinite number of cylinders with radius ω(z) andthickness dz stacked upon each other using a differential volume element in cylindricalcoordinates. When fluid achieves height h(t) in this partially filled sphere, one evaluatesthe following triple integral to obtain the liquid volume;

€

VPartiallyFilledSphere = dV = dΘ0

2π

∫∫∫∫ dz0

h t( )

∫ rdr0

ω z( )

∫ = π ω z( ){ }2dz

0

h t( )

∫

= π 2zRSphere − z2{ }dz =

0

h t( )

∫ π RSphereh2 t( )− 13

h3 t( )

As expected, the liquid volume vanishes when h=0, it achieves the normal volume of asphere [i.e., (4/3)πR3

Sphere] when h=2RSphere, and it achieves 50% of the normal sphere


98

volume when h=RSphere. The time-rate-of-change of system volume in the unsteady statemass balance is obtained via differentiation of the volume of this partially filled spherewith respect to time, because the fluid height h(t) is time-dependent as the spheredrains. One obtains the following result via separation of variables;

€

ddtVPartiallyFilledSphere = π 2Rsphereh t( )− h2 t( ){ } dhdt = −

πRTube4 g

8 µ /ρ( )Lh t( ) + L cosΘ{ }

µρ

=gRTube

4 dt0

tefflux

∫

8Lh 2Rsphere − h( )h+ L cosΘ( )

dh0

2Rsphere

∫= btefflux

with the following integration limits; h = 2RSphere initially at t = 0, and h = 0 at the effluxtime required to drain the bulb. The results of this analysis yield the functionaldependence of the capillary constant “b”;

€

b =gRTube

4

8Lh 2Rsphere − h( )h+ L cosΘ( )

dh0

2Rsphere

∫

The capillary constant depends on the (i) dimensions of the capillary tube, (ii) orientationof the capillary with respect to gravity, (iii) volume (or radius) of the spherical bulb, and(iv) strength of the gravitational field. The capillary constant does not depend ontemperature or the physical properties of the fluid, provided that the fluid isincompressible and Newtonian.

Detailed evaluation of the capillary constant and comparison withexperimental results. The next task is to evaluate the complex integral expression inthe previous equation for the capillary constant. Begin with the following substitution sothat the denominator of the integrand can be rewritten in terms of only one variable Ψ.Let Ψ = h + LcosΘ. Integration proceeds as follows;

€

h 2Rsphere − h( )h+ L cosΘ( )

dh0

2Rsphere

∫ =Ψ − L cosΘ( ) 2Rsphere −Ψ + L cosΘ( )

ΨdΨ

L cosΘ

2Rsphere+L cosΘ

∫


99

The integrand reduces to a simple function of Ψ that can be integrated rather easily;

€

Ψ − L cosΘ( ) 2Rsphere −Ψ + L cosΘ( )Ψ

= 2Rsphere −Ψ + 2L cosΘ−L cosΘ 2Rsphere + L cosΘ( )

Ψ

Integration from LcosΘ to 2Rsphere+LcosΘ yields the following expression;

€

2 Rsphere + L cosΘ( )−Ψ −L cosΘ 2Rsphere + L cosΘ( )

Ψ

dΨ

L cosΘ

2Rsphere+L cosΘ

∫

= 2 Rsphere + L cosΘ{ } 2Rsphere + L cosΘ− L cosΘ{ }− 12 2Rsphere + L cosΘ( )2 − L cosΘ( )2{ }−L cosΘ 2Rsphere + L cosΘ( ) ln

2Rsphere + L cosΘL cosΘ

= 2Rsphere Rsphere + L cosΘ( )− L cosΘ 2Rsphere + L cosΘ( ) ln 1+2RsphereL cosΘ

Finally, the capillary constant can be written in terms of the gravitational accelerationconstant and several geometric parameters that characterize the spherical bulb and thetilted capillary tube;

€

b =

gRTube4

8L

2Rsphere Rsphere + L cosΘ( )− L cosΘ 2Rsphere + L cosΘ( ) ln 1+2RsphereL cosΘ

Geometric parameters and capillary constants are summarized below for two differentCannon-Fenske capillary viscometers. If longer efflux times are desirable to minimizeerrors associated with end effects and experimental reproducibility, then one should usea viscometer with a smaller capillary constant.

Geometric Characteristics Size#100 Size#150Bulb volume, assumed to be spherical (mL) 8 8

Bulb radius, RSphere (cm) 1.24 1.24Capillary length, L (cm) 7.6 6.7

Capillary radius, RTube (cm) 0.041 0.05Capillary tilt angle with respect to gravity (degrees) 15 15

Capillary constant, predicted (cm2/sec2) 1.53x10-4 3.45x10-4

Capillary constant, experimental (cm2/sec2) 1.5x10-4 3.5x10-4


100

Draining Power-law fluids from a right circular cylindrical tank via atilted capillary tube

The capillary viscometer in the previous section is re-analyzed when anincompressible power-law fluid is drained from a cylindrical tank instead of a sphericalbulb. The unsteady state mass balance with no inlet stream and one outlet is analogousto the previous development, except that it is necessary to (i) modify the time-varyingsystem volume and (ii) use a generalized expression for the volumetric flowrate of non-Newtonian fluids through straight tubes with radius RTube and length L in the laminarregime. The dynamic pressure difference from capillary inlet to capillary outlet in thegeneralized Hagen-Poiseuille law for tilted tubes is identical to that in the previoussection if h(t) represents the variable height of fluid in a cylindrical tank. Hence, thestarting point for this analysis, to drain the tank but not the capillary tube, is;

€

dVsystem

dt=ddt

VPartiallyFilledTank + πRTube2 L[ ] = πRTank

2 dhdt

= −n

1+ 3nπRTube

3+ 1/ n( ) ρg2mL

h t( ) + LcosΘ{ }

1/ n

If the initial height of fluid in the tank is H (i.e., h=H at t=0), then one defines the half-time t1/2 and the efflux time tefflux as h=H/2 at t=t1/2 and h=0 at t=tefflux, respectively. Theremainder of this analysis compares half-times and efflux times for incompressibleNewtonian fluids, when n=1 and m=µ. The overall objective is to prove, unequivocally,that the efflux time is greater than twice the half-time for any set of initial conditionsand viscometer geometries, including all orientations (i.e., angle Θ) of the exit capillarywith respect to gravity. For fluids that obey Newton’s law of viscosity, the previousexpression reduces to;

€

RTank2 dh

dt= −

gRTube4

8 µ /ρ( )Lh t( ) + L cosΘ{ }

This unsteady state macroscopic mass balance for incompressible Newtonian fluids yieldsa much simpler result for the momentum diffusivity via the half-time or the efflux time,relative to the final expression for µ/ρ from the previous section when a spherical bulb isdrained. From a practical viewpoint, there are two geometric parameters (i.e., H andRTank) that must be related to the volume of the bulb above the capillary tube (i.e.,VolumeBulb ≈ πR2

TankH). In contrast, when the bulb volume is modeled as a sphere insteadof a right circular cylinder, one identifies the sphere radius via VolumeBulb = (4/3)πR3

Sphere.Hence, even though two parameters (i.e., H and RTank) are related by one equation (i.e.,


101

VolumeBulb ≈ πR2TankH), one predicts the momentum diffusivity for this “tank-draining”

problem as follows;

€

µρ

=gRTube

4 dt0

tefflux

∫

8RTank2 L dh

h+ L cosΘ0

H

∫=

gRTube4 dt

0

t1/2

∫

8RTank2 L dh

h+ L cosΘ12H

H

∫

Obviously, one can predict momentum diffusivities for incompressible Newtonian fluidsvia laboratory measurements of efflux times or half-times. The capillary constant basedon efflux times is smaller than the capillary constant based on half-times, because theproduct of the appropriate capillary constant and either the half-time or the efflux timeyields the momentum diffusivity which is insensitive to the time required to drain eitherone-half of the total volume of the tank (or bulb) or the total volume of fluid above thecapillary tube. The rather simple relation between half-time and efflux time, based on theprevious equation, is;

€

teffluxt1/2

=

dhh+ L cosΘ0

H

∫dh

h+ L cosΘ12H

H

∫=ln H + L cosΘ

0+ L cosΘ

ln H + L cosΘ12 H + L cosΘ

> 2

Numerical substitutions for the (i) initial height H of fluid in the cylindrical tank, (ii) lengthL of the capillary tube, and (iii) angle of tilt Θ with respect to gravity reveal that the ratioof tefflux to t1/2 is always greater than 2. In fact, this ratio (i.e., tefflux/t1/2) becomessignificantly greater than 2 when H is larger, L is smaller, and Θ approaches π/2. Whenthe capillary tube is horizontal (i.e., Θ = π/2), it is important to emphasize that the half-time is finite;

€

t1/2 =8 µ /ρ( )LRTank2

gRTube4 ln H + L cosΘ

12 H + L cosΘ

⇒Θ=

π2

8 µ /ρ( )LRTank2 ln 2{ }gRTube

4

but an infinite amount of time is required to drain the total volume of fluid in the tank.These trends can be rationalized in terms of a dynamic pressure difference from capillaryinlet to capillary outlet that decreases at longer times because the hydrostatic pressureat the capillary inlet is directly proportional to the instantaneous height of fluid in thereservoir.


102

Macroscopic Momentum Balance

Important Considerations and Assumptions in the Development and Use of theMacroscopic Momentum Balance; Steady State and Unsteady State Analysis

(1) The system is defined as fluid within control volume V, bounded by solid surfacesand fictitious inlet and outlet planes.

(2) The total system mass within V, which is a scalar, is defined by;

€

mtotal = ρdVV∫∫∫

(3) The total momentum within V, which is a vector, is defined by;

€

Ptotal = ρvdVV∫∫∫

(4) The average fluid velocity <v> is a scalar in the contributions from convectivemomentum flux.

(5) Pressure and viscous forces exerted by the fluid on all solid surfaces in contactwith the fluid are represented by FFluid on Solid, which contains all of the importantcontributions due to the viscous stress tensor.

(6) Normal pressure forces are much more important than normal viscous forcesacross all of the fictitious inlet and outlet planes that bound fluid within thecontrol volume.

(7) The gravitational acceleration vector g is essentially constant within the system.Hence;

€

ρgdV = g ρdV = mtotal gV∫∫∫V∫∫∫

This is the only external body force in the macroscopic momentum balance.Forces due to electric and magnetic fields are not considered.

(8) Across the fictitious inlet and outlet planes which bound fluid within the system,one identifies unit normal vectors ninlet and noutlet, respectively, which are oriented


103

in the direction of primary fluid flow, not necessarily pointing outward from thesystem to the surroundings.

(9) A qualitative statement of the macroscopic momentum balance is given by;

1 = 2 – 3 + 4 – 5 – 6 + 7where

1 is the accumulation of fluid momentum within control volume V2 represents the rate of input due to convective momentum flux across surface S3 represents the rate of output due to convective momentum flux across surface S4 represents normal pressure forces acting across the fictitious inlet planes5 represents normal pressure forces acting across the fictitious outlet planes6 represents pressure and viscous forces exerted by the fluid on all solid surfaces7 represents the external body force due to gravity

(10) A quantitative statement of the macroscopic momentum balance, with termsthat correspond to those numbered from 1 to 7 above, is given by;

€

dPtotal

dt= ρv(v − vSurface )S ninlet −

inletplanes

∑ ρv(v − vSurface )S noutletoutletplanes

∑

+ pSinletplanes

∑ ninlet − pS noutletoutletplanes

∑

−FFluid→Solid + mtotal g

A word of caution is needed about the fact that the macroscopic momentumbalance requires absolute pressure, not gauge pressure, because normal forcesdue to pressure stress are constructed from a product of pressure and surfacearea across which pressure acts. In particular, for inlet and outlet planes of acontrol volume that correspond to a reduction in flow cross-section, pressureforces written in terms of gauge pressure will be incorrect. When one evaluatesdifferences in pressure forces across fictitious inlet and outlet planes that havethe same area for unidirectional flow, the use of absolute or gauge pressuresyields the same difference because the surface areas are the same and thepressure forces are subtracted. However, when the inlet and outlet planes aredescribed by different surface areas, or when the flow is not unidirectional (i.e.,


104

the U-tube is an example where pressure forces across the inlet and outlet planesare additive), absolute pressure is required to evaluate pressure forces due to thevector nature of the linear momentum balance and the fact that pressure stressmust be multiplied by the appropriate surface area.

Unsteady state applications of the macroscopic momentumbalance to rocket propulsion

Consider a propulsion vehicle, like a rocket, with solid mass mrocket and initial liquidfuel mass mfuel,Initial at time t0 that is traveling vertically upward at constant velocityvpropulsion. The system is chosen as the solid rocket and its liquid fuel. Unsteady stateanalysis is required because, even though the system does not experience anyacceleration at constant rocket velocity, the mass of the system decreases at longertime as fuel is burned to provide the required propulsion that maintains constant velocity.Exhaust gases from burned fuel can potentially escape the system at supersonicvelocities on the order of 2000 meters/sec through a converging-diverging nozzle if theexit (i.e., back) pressure is low enough. Relative to a stationary frame of reference, theexhaust manifold with flow cross-sectional area Soutlet moves upward at velocity vpropulsion.Since there are no inlet planes, and mass is neither depleted nor produced during thecombustion reaction, the overall unsteady state macroscopic mass balance yields thefollowing expression for the instantaneous mass of fuel in the system at time t whencombustion gases are ejected continuously across the converging-diverging nozzle atconstant volumetric flowrate Qexhaust = vexhaustSoutlet;

€

ddtmsystem = −ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet

msystem = mrocket +mfuel t( ) = mrocket +mfuel,initial −ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet t − t0( )

where ρfuel,exhaust is the mass density of the exhaust gas. The mass of the solid rocketmrocket remains constant during the analysis if booster stages do not separate from thespace capsule. The total momentum of the system, and its time derivative, arecalculated to evaluate the accumulation term on the left side of the unsteady statemacroscopic momentum balance for systems that lose mass, in the form of liquid orgaseous fuel, at constant velocity;


105

€

Ptotal = msystemv propulsion = mrocket + mfuel{ }v propulsiondPtotal

dt= v propulsion

dm fuel

dt= −ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet v propulsion

There are no inlet planes across which mass enters the system, and there is only onefictitious outlet plane that provides an escape route for burned exhaust gases that couldexceed “Mach 1”. Relative to a stationary frame of reference, the nozzle in the fictitiousoutlet plane moves upward at the velocity of the system, vpropulsion, whereas the exitstream moves downward at velocity vexhaust. Consequently, the relative velocity of theexit stream is enhanced with respect to motion of the fictitious outlet plane in theopposite direction. The unsteady state macroscopic momentum balance reduces to;

€

dPtotal

dt= −ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet v propulsion =

−ρ fuel,exhaustvexhaust vexhaust + vpropulsion( )Soutlet noutlet − pexhaustSoutlet noutlet − FFluid→Solid + msystem g

On the right side of the previous equation, the interfacial frictional force exerted by fluidon all of the solid surfaces within the system and at its boundaries is typically neglectedrelative to the thruster force due to convective momentum flux, the “back-pressure”force (i.e., pexhaustSoutlet), and the gravitational force. This approximation allows one todesign propulsion vehicles under “ideal” conditions when hydrodynamic drag forces canbe neglected. For interplanetary travel, space vehicles that escape the Earth’satmosphere and gravitation field do not experience forces due to back-pressure, gravity,or hydrodynamic drag, and no thruster force is required to maintain effortless constant-velocity motion. In realistic situations closer to home, the final expressions below for therequired initial mass of liquid fuel and the exhaust gas flowrate represent lower limits toobtain a given propulsion velocity. The gravitational force (i.e., last term on the rightside of the previous equation) contains variable system mass because fuel is burnedcontinuously to provide propulsion. Since the mass of liquid fuel decreases linearly withtime to maintain constant exhaust mass flowrate of the gaseous combustion products,msystem in the gravitational force term is approximated by the average mass of the solidrocket and liquid fuel during the time frame of operation (i.e., t-t0) of the system.Hence;


106

€

msystem ⇒ mstysten average

= mrocket +12mfuel,initial + mfuel,initial − ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet t − t0( )[ ]{ }

= mrocket + mfuel,initial −12ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet t − t0( )

For the specific example where the propulsion velocity is vertically upward (i.e., in thepositive z-direction) and exhaust gases are ejected vertically downward in the samedirection as the gravitation acceleration vector, the z-component of the macroscopicmomentum balance for constant-velocity motion in the absence of frictional forcesreduces to;

€

−2ρ fuel,exhaustQexhaustv propulsion − ρ fuel,exhaustv propulsion2 Soutlet = ρ fuel,exhaustvexhaust

2 Soutlet

+pexhaustSoutlet − mrocket + mfuel,initial −12ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet t − t0( )

g

because the unit normal vector at the outlet plane, oriented in the primary direction ofexhaust gas flow, points in the negative z-direction. This z-component force balanceprovides a relation between several design variables for rocket propulsion. If oneidentifies the mass flow rate of exhaust gases ωexhaust, with respect to a stationaryreference frame, as;

€

ωexhaust = ρ fuel,exhaustQexhaust = ρ fuel,exhaustvexhaustSoutlet

then the z-component of the unsteady state macroscopic momentum balance yields thefollowing ideal rocket propulsion design equation;

€

1ρ fuel,exhaustSoutlet

ωexhaust2 + 2vpropulsion +

12g t − t0( )

ωexhaust

+pexhaustSoutlet + ρ fuel,exhaustv propulsionSoutlet v propulsion +12g t − t0( )

− mrocket + mfuel,initial{ }g = 0

If the system operates for time, t-t0, and ejects exhaust gases continuously at constantvolumetric flowrate Qexhaust, then the initial mass of liquid fuel must be greater than;

€

mfuel,initial ≥ ρ fuel,exhaust vexhaust + vpropulsion( )Soutlet t − t0( )


107

Solution of the nonlinear equation for the mass flowrate of exhaust gases ωexhaust requiresthe positive square root in the quadratic formula. When the initial mass of liquid fuel isgreater than its minimum value, as required for continuous rocket propulsion, one obtainsthe following expression for Qexhaust to maintain constant-velocity motion during the timeframe of system operation, t-t0, if frictional forces are negligible;

€

Qexhaust =Soutlet2

− 2vpropulsion +12g t − t0( )

+

2vpropulsion +12g t − t0( )

2

+4

ρ fuel,exhaustSoutlet

mrocket + mfuel,initial( )g − pexhaustSoutlet−ρ fuel,exhaustv propulsionSoutlet v propulsion +

12g t − t0( )

The system of equations searches for positive values of vexhaust, Qexhaust, and ωexhaust. Thisoccurs when the following inequality is satisfied;

€

mrocket + mfuel,initial( )g > pexhaustSoutlet + ρ fuel,exhaustv propulsionSoutlet v propulsion +12g t − t0( )

ProblemDesign propulsion vehicles with a solid mass of 1250 kg that move vertically upwardfrom sea level at constant velocity until they achieve an altitude of 3000 meters. Liquidfuel is vaporized and burned at 500K. The gaseous combustion products (i.e., primarilyCO2 and H2O), with an average molecular weight of 35 daltons, are ejected at 1atmosphere total pressure (i.e., 101 kiloPascals = 1.01 x 105 Newtons per square meter)through a converging-diverging nozzle that has a diameter of 1.5 feet at the exhaustmanifold. It is acceptable to use the ideal gas law with Rgas = 0.082 Litre-atm/mol-K toestimate the exit gas molar density with dimensions of gram-moles per Litre. Since thereare 103 grams per kilogram, and 103 Litres per cubic meter, mass densities in grams perLitre are equivalent to those with dimensions of kg/m3. Use the MKS system of units tocorrelate the following quantities in tabular and graphical form for propulsion velocitiesbetween 10 meters per second and 40 meters per second, in increments of 1 meter persecond.

(a) minimum initial mass of liquid fuel (kg)


108

(b) average velocity of exhaust gases through the exit manifold (m/s)(c) volumetric flowrate of exhaust gases through the exit manifold (m3/s)(d) mass flowrate of exhaust gases through the exit manifold (kg/s)

10 14 18 21 25 29 32 36 409

19

29

39

49

60

70

80

90

100

Vpropulsion [meters/ sec]

Vex

haus

t [m

eter

s/se

c]Effect of Propulsion Velocity on Exhaust Velocity

Solid Rocket Mass = 1250 kg

10 14 18 21 25 29 32 36 40890

975

1060

1145

1230

1315

1400

Vpropulsion [meters/ sec]

Mas

sFue

lini

tial

[k

g]

Effect of Propulsion Velocity on Initial Mass of Liquid Fuel

Solid Rocket Mass = 1250 kg


109

Steady state applications of the macroscopic momentum balance

(11) Steady state analysis of the macroscopic momentum balance is accomplished byneglecting the left hand side of the previous equation. Hence, the sum of allforces acting on the system must vanish. This is reasonable if the system doesnot (i) accelerate, (ii) gain mass, or (iii) lose mass, because each of theseprocesses gives rise to a time rate of change of total system momentum.

(12) For incompressible fluids with constant density ρ, stationary inlet and outletplanes such that vSurface=0, and fluid pressure which does not change much acrossthe entire inlet plane or the outlet plane (but fluid pressure usually decreasesfrom inlet to outlet), one rearranges the steady state macroscopic momentumbalance to calculate the pressure and viscous forces exerted by the fluid incontact with all of the solid surfaces;

€

FFluid→Solid = ρ v 2S ninlet −inletplanes

∑ ρ v 2S noutletoutletplanes

∑

+ pSinletplanes

∑ ninlet − pSnoutletoutletplanes

∑

+mtotal g

(13) Consider steady state one-dimensional flow through a straight horizontal tubewith radius R and length L. If there is no change in the flow cross-sectional areaS from inlet to outlet, then <v>inlet = <v>outlet via the steady state macroscopicmass balance for incompressible fluids. Since the fluid does not changedirections from inlet to outlet, the unit normal vectors in the primary flowdirection across the fictitious inlet and outlet planes are oriented in the positivez-direction. Hence, contributions from convective momentum flux across theinlet and outlet planes cancel because (i) the fluid density doesn’t change if thefluid is incompressible, (ii) the cross-sectional area S remains constant, and (iii)the average fluid velocity across the inlet and outlet planes is the same at steadystate. The only surviving terms in the z-component of the previous steady statemacroscopic momentum balance are, with gz = 0;

€

FFluid→Solid{ }z−component = pinlet − poutlet( )S = πR2Δp


110

(14) Analysis of the vector force per unit area exerted by a moving fluid on thestationary wall of a tube at r=R, with unit normal vector in the +r-direction, yieldscontributions from pressure and viscous forces, only. There is no contributionfrom convective momentum flux at the fluid-solid interface (i.e., @ r=R) becausethe “no-slip” boundary condition states that the fluid velocity vector must vanishat r=R if the wall is stationary. Furthermore, the tube wall at r=R is the only solidsurface in contact with the fluid. The fictitious inlet and outlet planes are notsolid surfaces. Hence, pressure stress acts in the r-direction and the 1st subscripton the important scalar components of the viscous stress tensor is r;

€

VectorForce /Area{ }Fluid→Solid@r=R = δ r p+ τ rr( ) +δΘτ rΘ +δ zτ rz{ }r=R

(15) It should seem reasonable to equate the z-component of the vector force perunit area exerted by the fluid on the solid surface from (14), multiplied by thelateral surface area (i.e., 2πRL), and the z-component of the pressure and viscousforces (actually 100% due to τ) exerted by the fluid in contact with all of thesolid surfaces via the steady state macroscopic momentum balance in (13).Hence;

€

2πRLτ rz (r = R) = πR2Δp

τ rz (r = R) =RΔp2L

=12R −

dpdz

Steady state analysis of the z-component of the macroscopic momentum balance hasbeen performed together with a microscopic analysis of forces exerted by fluids onsimple solid surfaces, due to the momentum flux tensor. One obtains a relation, orbalance, between viscous shearing forces on the lateral surface of the tube and normalpressure forces that act across the fictitious inlet and outlet planes at z=0 and z=L. It isimportant to emphasize that macroscopic analysis provides an evaluation of τrz only atthe tube wall (i.e., r=R). Microscopic analysis of the linear momentum balance, known asthe Equation of Motion, yields the viscous shear stress distribution throughout the fluidwhich is consistent with the previous equation when τrz(r) is evaluated at r=R. Eventhough there is an unlimited number of functions of radial position r that yield R whenthey are evaluated at the tube wall (i.e., r=R), the z-component of the microscopic linearmomentum balance (i.e., Equation of Motion) reveals that the viscous shear stress profileis;

€

τ rz (r) =12r − dp

dz


111

One-dimensional flow through tilted tubes with no change in flowcross-sectional area. Let’s revisit steady state one-dimensional incompressible flowthrough the same straight tube of radius R and length L, as described in (13) through(15) above. Now, the flow configuration is oriented at angle Θ (i.e., 0≤Θ≤π) with respectto vertical, such that gravity plays a role when one balances forces in the primarydirection of fluid flow, which is identified as the z-direction in cylindrical coordinates. Thez-axis and the tube axis are aligned such that the tube wall is a simple surface atconstant value of the radial coordinate. The unit normal vector at the fluid-solidinterface is oriented in the radial direction at all points on the solid surface. Once again,z-component normal forces due to convective momentum flux acting across the fictitiousinlet and outlet planes cancel because the average fluid velocity is the same (i.e., <vz>inlet

= <vz>outlet) when the flow cross-sectional area remains constant. The steady statemacroscopic momentum balance yields the following expression for the z-component ofthe total vector force exerted by the fluid on the tube wall, which is due completely toviscous shear at the fluid-solid interface because pressure forces at the tube wall actsolely in the radial direction;

€

FFluid→Solid{ }z−component = pinlet − poutlet( )S +mtotalgcosΘ

= πR2Δp+ρ fluidπR2LgcosΘ = πR2 Δp+ρ fluid gL cosΘ{ }

The effect of gravitational potential energy per unit volume of fluid on the driving forcefor flow can be added to fluid pressure (i.e., they have the same dimensions) via thedefinition of a quantity called dynamic pressure P;

€

P = p+ρ fluidgh

where p represents fluid pressure, h is a position variable that increases as one movesvertically upward from an arbitrarily chosen zero of potential energy, and ρfluidgh is thegravitational potential energy per unit volume of fluid. The horizontal plane thatrepresents the zero of potential energy can be chosen arbitrarily because driving forcesfor fluid flow are expressed as differences in dynamic pressure, so the absolute value ofpotential energy at any position within the fluid has no effect on the final solution to themomentum balance. For example, when forced convective flow is driven by acombination of gravity and a decrease in fluid pressure, as described in this section, andthe zero of potential energy is placed at the fictitious outlet plane of the tilted tube suchthat the inlet plane is either LcosΘ higher (i.e., 0≤Θ≤π/2) or lower (i.e., π/2≤Θ≤π) than


112

the outlet plane, the previous result in this section for viscous shear at the fluid-solidinterface reduces to;

€

FFluid→Solid{ }z−component = 2πRLτ rz r = R( )

= πR2 pinlet +ρ fluidgL cosΘ[ ]− poutlet + 0[ ]{ } = πR2ΔP

with assistance from statement (14) above to evaluate the z-component of theinterfacial force exerted by the fluid on the solid surface due to the viscous stresstensor. Rearrangement of this equation allows one to evaluate τrz at the tube wall;

€

τ rz r = R( ) =RΔP2L

=12R ΔP

L

=12R −

dPdz

This generalized result for straight tilted tubes is essentially the same as the one instatement (15) above for horizontal tubes, except that fluid pressure p must be replacedby dynamic pressure P when gravity assists or hinders fluid flow. Furthermore, it ispossible to extrapolate the previous expression for tilted tubes to obtain the viscousshear stress distribution throughout the fluid;

€

τ rz r( ) =12r − dP

dz

which is applicable for incompressible flow through straight tubes with constant cross-sectional area, oriented at any angle with respect to gravity. For Newtonian fluids withconstant viscosity that cannot store elastic energy upon deformation due to viscousstress, the dynamic pressure gradient in the direction of flow is constant, such that–dP/dz = ΔP/L. As a consequence of this analysis of one-dimensional flow throughstraight tilted tubes, it is acceptable to modify the generalized macroscopic momentumbalance for incompressible fluids and account for gravitational forces via dynamicpressure. Hence, one neglects mtotalg in the force balance and replaces fluid pressure pby dynamic pressure P. The result is;

€

dPtotal

dt= ρv(v− vSurface )S ninlet −

inletplanes

∑ ρv(v− vSurface )S noutletoutletplanes

∑

+ PSinletplanes

∑ ninlet − PS noutletoutletplanes

∑ −FFluid→Solid


113

Macroscopic Mechanical Energy Balance

Important Concepts in the Development of the Macroscopic MechanicalEnergy Balance (i.e., the Bernoulli Equation)

1) The mechanical energy balance, which is a scalar, is developed from firstprinciples via the (i) balance on overall fluid mass, and the (ii) linear momentumbalance. The units of each term in the Bernoulli equation are energy per time,which is equivalent to power.

2) The system is defined as the fluid contained within the control volume V(t),which is not necessarily stationary. Moving control volumes are described bytime-dependent boundaries.

3) Total kinetic energy within the system is defined in terms of fluid density ρ andthe square of the fluid velocity, as follows;

€

Ktotal = 12V ( t )∫∫∫ ρvfluid

2 dV

4) Total potential energy within the system is defined in terms of fluid density ρ andheight h, which is measured vertically upward relative to the “zero of potentialenergy”.

€

Φ total = ρgh{ }V ( t)∫∫∫ dV

5) Inlet plane S1 and outlet plane S2 represent fictitious surfaces that are not solid.Fluid flow crosses these fictitious planes. The rate at which kinetic energy entersand leaves the system due to convective flux, or bulk fluid flow, is proportional tothe third power of the fluid velocity. There is no molecular flux of kinetic energy.

Net rate of convective input of kinetic energy =

€

12 ρ1 v1

3 S1 − 12 ρ2 v2

3 S2

6) The rate at which potential energy enters and leaves the system due toconvective flux is given by a product of (i) the potential energy per unit mass offluid, (ii) overall convective mass flux, and (iii) the surface area normal to fluidflow across the inlet or outlet planes. Once again, there is no molecular flux ofpotential energy. Molecular fluxes exist for (i) viscous transport of momentumvia Newton’s law, (ii) conduction of thermal energy via Fourier’s law, (iii) diffusion


114

of species mass via Fick’s law, and (iii) entropy via concepts based on irreversiblethermodynamics. The net rate of convective input of potential energy to thesystem is given by;

Net rate of convective input of potential energy =

€

gz1ρ1 v1 S1 − gz2ρ2 v2 S2

7) The Bernoulli equation contains several work-related terms which are (i) positivewhen work is done by the surroundings on the fluid, or (ii) negative when work isdone by the fluid on the surroundings. For example, the rate at which work isdone on the fluid due to pressure forces is given by a product of (i) theisothermal Gibbs free energy of the fluid, (ii) overall convective mass flux, and(iii) the surface area normal to fluid flow across the inlet or outlet planes. This isequivalent to the fact that the rate at which work is performed (i.e., power) isgiven by the scalar dot product of a normal pressure force and the fluid velocityvector. Hence;

Net rate of work done on the fluid by pressure forces =

€

p1 v1 S1 − p2 v2 S2

8) The rate at which work is done on the fluid by viscous forces across the fictitiousinlet and outlet planes is negligible, relative to the work done by pressure forces,as outlined in (7).

9) The rate at which work is done by the fluid on moving solid surfaces Smoving due topressure forces is given by a surface integral of the product of fluid pressure andthe normal component of the fluid velocity vector evaluated at the fluid-solidinterface. The fluid velocity is the same as the velocity of the moving solidsurface at the point of contact between the fluid and the solid. The unit normalvector in the expression below is directed from the control volume that containsthe fluid toward the moving solid surface. This term does not appear in theBernoulli equation if the solid surfaces are stationary.

€

dWdt

moving,pressure

= p n•vfluid{ }Smoving∫∫

@SmovingdS

10) The rate at which work is done by the fluid on moving solid surfaces Smoving due toviscous forces is given by a surface integral of a complex dot product involving (i)the unit normal vector directed from the fluid toward the moving solid surface,


115

(ii) the fluid velocity vector, and (iii) the viscous stress tensor. Once again, thisterm vanishes at any solid surface that is stationary.

€

dWdt

moving,viscous

= n• τ •v[ ]{ }Smoving∫∫

@SmovingdS

11) The horsepower requirement of a pump, which is part of the system, is given bythe following sum; (dW/dt)pump = – (dW/dt)moving,pressure – (dW/dt)moving,viscous, withunits of energy per time. Horsepower requirements originate from the rate atwhich work is done on the fluid due to pressure and viscous forces acting acrossmoving solid surfaces that are in contact with the fluid. Pumps in series act likevoltage sources in series. They experience the same flow rates or currents, andthe total increase in fluid pressure or voltage is additive. A series configurationof pumps is recommended when each one delivers the correct flow rate, butgreater increase in fluid pressure is required. Pumps in parallel function like aparallel configuration of voltage sources. They produce the same pressure orvoltage increase, and the total flow rate or current is additive. A parallelconfiguration of pumps is recommended when each one delivers the correctincrease in fluid pressure, but higher flow rates are required. Please refer to (18)below for the design of pumping configurations to meet desired specifications.

12) The non-ideal Bernoulli equation contains an irreversible rate of conversion ofmechanical energy to thermal energy due to viscous dissipation, Ev. Other namesfor this process are (i) friction loss, (ii) viscous heating, (iii) viscous dissipation,and (iv) the degradation of mechanical energy to thermal energy. The secondlaw of thermodynamics for irreversible processes dictates the path, or direction,by which this process occurs. In fact, the degradation of mechanical or kineticenergy to thermal energy always causes the fluid temperature to increase. If thefluid temperature decreased and all of this thermal energy were convertedcompletely to mechanical energy, then one could construct perpetual motionmachines of the second kind. These devices are prohibited by the second law ofthermodynamics. The irreversible degradation of mechanical energy to thermalenergy, written below using vector-tensor notation, is always positive forNewtonian fluids that cannot store elastic energy.

€

Ev = − τ :∇v( )V (t )∫∫∫ dV


116

13) Simple forms for the irreversible conversion of mechanical energy to thermalenergy in various fluid flow configurations are summarized below;

Flow in straight channels with arbitrary cross-section;

€

Ev = ρ vz S{ } 12 vz

2 LTubeLengthRHydraulicRadius

fFrictionFactor

Flow through valves, fittings, and other obstacles with dimensionless friction lossfactor ev;

€

Ev = ρ vz S{ } 12 vz

2ev

14) The complete macroscopic mechanical energy balance, based on all of theconcepts mentioned above, for a control volume with one inlet stream (i.e.,subscript 1) and one outlet stream (i.e., subscript 2) is;

€

ddt

Ktotal +Φ total{ } = 12 ρ1 v1

3 S1 − 12 ρ2 v2

3 S2

+gz1ρ1 v1 S1 − gz2ρ2 v2 S2+p1 v1 S1 − p2 v2 S2

+dWdt

Pump

−Ev

15) For systems that operate at steady state with no moving parts (i.e., no pumpwork term) and negligible friction loss, the macroscopic mechanical energybalance reduces to the ideal Bernoulli equation. Fluids that obey theserestrictions are classified as ideal, inviscid, irrotational and isentropic. Flowoccurs at constant entropy and the fluid experiences no increase in temperature,because the irreversible conversion of mechanical energy to thermal energy isinsignificant. The ideal Bernoulli equation can be summarized as follows; the sumof kinetic energy, potential energy, and isothermal Gibbs free energy (i.e.,pressure work term) remains constant everywhere within the fluid. Theappropriate mechanical energy balance is;


117

€

12 ρ1 v1

3 S1 + p1 v1 S1 + gz1ρ1 v1 S1= 12 ρ2 v2

3 S2 + gz2ρ2 v2 S2 + p2 v2 S2

16) Analysis of the steady state ideal Bernoulli equation is performed after division ofthe previous equation by the total mass flow rate, which can be written asρ1<v1>S1 or ρ2<v2>S2, because both expressions for the total mass flow rate arethe same at steady state with one inlet stream and one outlet stream. Thestarting point for the analysis of inviscid fluids with no moving parts via the idealBernoulli equation is;

€

12

v13

v1+ gz1 +

p1ρ1

= 12

v23

v2+ gz2 +

p2ρ2

In terms of the steady state Bernoulli equation with one inlet stream and oneoutlet stream, where each term has dimensions of energy per mass, and overallmass flow rate is constant, the pressure terms that are typically subtractedcorrespond to pV-work across the fictitious inlet and outlet planes, and nosurface areas appear because one divides the equation with dimensions of energyper time by total mass flow rate which does not change from inlet to outlet atsteady state. Hence, predictions from the steady state Bernoulli equation withone inlet and one outlet do not depend on the use of absolute or gauge pressure,but absolute pressure is required in the momentum balance due to the vectornature of that equation and the fact that pressure stress must be multiplied bythe appropriate surface area.

17) For incompressible fluids, in which fluid density ρ is a very weak function ofpressure, the difference between gz + p/ρ at the inlet plane S1 and outlet planeS2 can be written as (ΔP)/ρ, where the dynamic pressure difference ΔP is P1 minusP2.

18) Application of the steady state non-ideal Bernoulli equation around a pumpreveals that the downstream pressure, after fluid passes through the pump, isgreater than the upstream pressure. If the inlet and outlet planes arecharacterized by the same flow cross-sectional areas such that there is nochange in kinetic energy, then the rate at which work must be done on the fluidby pressure and viscous forces acting across moving solid surfaces (i.e., impeller)


118

in contact with fluid inside of the pump is related to the increase in dynamicpressure via the following equation;

€

Efficiency[ ] dWdt

PumpRequired( )

=dWdt

PumpDelivered

toFluid

= v S p2 +ρgz2( )− p1 +ρgz1( ){ }

= v S P2 −P1{ } = v SΔP ⇒horizontal

configuration

v SΔp =QΔp

All frictional energy losses within the pump are accounted for by the pump efficiency.Six pumps are available with flowrate specifications at a particular horsepower rating[i.e., (dW/dt)Pump,Required] when each pump operates very close to its optimum efficiency.

Pump Volumetric flowrate, Q Δp = [Efficiency{dW/dt}Pump,Required]/Q# (gallons/minute, gpm) (psi)A 10 20B 15 15C 20 20D 30 50E 30 30F 40 35

(i) If each pump can be chosen only once and it is desired to use them when theyoperate near optimum efficiency, then design a pumping configuration thatdelivers 30 gallons/minute with a 100-psi increase in fluid pressure downstreamfrom the pumps. Remember that pumps in series should operate at matchedflowrates and pumps in parallel should operate at matched Δp.

Answer:Put D (30 gpm; 50 psi) and E (30 gpm; 30 psi) in series with a parallel configuration ofA (10 gpm; 20 psi) and C (20 gpm; 20 psi)

(ii) If each pump can be chosen only once and it is desired to use them when theyoperate near optimum efficiency, then design a pumping configuration thatdelivers 60 gallons/minute with a 50-psi increase in fluid pressure downstreamfrom the pumps. Remember that pumps in series should operate at matchedflowrates and pumps in parallel should operate at matched Δp.


119

Answer:Put E (30 gpm; 30 psi) in series with a parallel configuration of A (10 gpm; 20 psi) andC (20 gpm; 20 psi). Then, put this entire configuration in parallel with D (30 gpm; 50psi)

Orifice meter analysis via the ideal and non-ideal Bernoulli equationwith frictional energy loss

The discharge coefficient for an orifice meter, Cv, is defined as the ratio of the actualmass flow rate with frictional energy loss (i.e., Ev>0) to the ideal mass flow rate withoutfriction loss (i.e., Ev=0). If the frictional energy loss Ev, with units of energy per unit massof fluid, for a sharp-edged orifice meter is described by:

€

Ev =12vtube

2ev

and the dimensionless friction loss factor ev in the turbulent regime is approximated by:

€

ev = λ 1−β( )1−β 4( )β 4

where β = D2/D1 is defined as the ratio of the orifice diameter D2 to the tube diameter D1,then calculate the discharge coefficient, Cv, when β = 0.75

Answer:Begin with the steady state nonideal Bernoulli equation and define the fictitious inlet andoutlet planes that provide the necessary surfaces which identify the control volume andthe system as the fluid between these planes. Inlet plane “1” is upstream from theorifice meter, within the tube, and outlet plane “2” is placed at the throat of the meterwhere maximum velocity and minimum pressure exist. There are no moving solidsurfaces within the control volume, so the pump work term vanishes. All remaining termsin the steady state Bernoulli equation, including frictional energy loss, are divided by thetotal mass flow rate which remains constant for steady state analysis. The kineticenergy correction factor α is defined as the ratio of the cube of the average velocityrelative to the average of the velocity cubed. In other words;


120

€

α =

1Scross−sec tion

vzdSScross−sec tion

∫∫

3

1Scross−sec tion

vz3dS

Scross−sec tion

∫∫=vz

3

vz3

The primary objective is to predict the mass flow rate either in the tube or in the throatof the meter via the following equation;

€

v12

2α+ gz1 +

p1ρ1

=v2

2

2α+ gz2 +

p2ρ2

+Ev

The steady state mass balance for incompressible fluids allows one to relate <v1> and<v2> via flow cross-sectional areas S1 in the tube and S2 at the contraction. Hence;

€

v1 S1 = v2 S2

v1 =S2S1

v2 = β2 v2

Now, the nonideal Bernoulli equation is solved for average fluid velocity <v2> at themeter;

€

v22

2α1−β 4 +αβ 4ev{ } =

1ρ

p1 − p2( ) + g z1 − z2( )

v2 =

2α 1ρ

p1 − p2( ) + g z1 − z2( )

1−β 4 +αβ 4ev=

2α 1ρP1 −P2( )

1−β 4 +αβ 4ev

Multiplication of the previous expression for <v2> by ρS2 allows one to predict the totalmass flow rate for both real (i.e., ev > 0) and ideal (i.e., ev = 0) fluids. The ratio of thesemass flow rates yields the coefficient of discharge Cv. It is important to emphasize thatboth mass flow rates are evaluated at the same dynamic pressure difference P1 – P2 = ΔPon the far right side of the previous equation for orifice meters that are either horizontalor tilted at any angle with respect to gravity. The appropriate equation for Cv is;


121

€

Cv =ρS2 v2 ev > 0( )ρS2 v2 ev = 0( )

=1−β 4

1−β 4 +αβ 4ev=

1

1+αβ 4

1−β 4

ev

Two important analyses follow. First, the previous equation is inverted to express thedimensionless friction loss factor ev in terms of the coefficient of discharge Cv. Theresult is;

€

ev =1Cv2 −1

1−β 4

αβ 4

Hence, experimental determination of the coefficient of discharge via (i) measurement ofthe real mass flow rate and (ii) use of the ideal Bernoulli equation to predict the idealmass flow rate allows one to estimate frictional energy loss for an orifice meter. Thedynamic pressure drop across the meter is given by;

€

ΔP =12ρ vtube

2ev =12ρ vtube

2 1Cv2 −1

1−β 4

αβ 4

via application of the nonideal Bernoulli equation with no change in flow cross-sectionalarea upstream and downstream from the meter. The second important calculation in thissection employs the dimensionless friction loss model from the previous page toestimate the coefficient of discharge for sharp-edged orifice meters. One obtains thefollowing result;

€

Cv =1

1+αβ 4

1−β 4

ev

=1

1+αλ 1−β( )

where β is the diameter ratio for the orifice meter, λ is a parameter in the model for thedimensionless friction loss factor, and α is approximately unity (i.e., 0.945) for turbulentflow.


122

Mass Flow Rate vs. Pressure Drop in Venturi Meters

Obtain an expression for the actual, not the ideal, mass flow rate through thetilted venturi meter illustrated below, where the primary direction of flow makes an angleθ with respect to the horizontal. An incompressible fluid with density ρ1 flows down theincline from left to right. The manometer contains mercury with density ρ2. Mercury,with density ρHg, rises to height z4 in the left leg of the manometer, and it rises to heightz3 in the right leg of the manometer. It is important to realize that the pressure taps atpositions #1 and #2 are at elevations z1 and z2, respectively, in the gravitational field,where z1 > z2.

Answer:Since models for frictional energy loss (i.e., eV) are not readily available for Venturimeters, except for the following empirical relation between ev and the coefficient ofdischarge CV;

€

eV =1Cv2 −1

1−β 4

αβ 4

the strategy involves starting with the steady state ideal Bernoulli equation. One definesthe fictitious inlet and outlet planes that provide the necessary surfaces which identify (i)the control volume and (ii) the system as the fluid between these planes. Inlet plane “1”at position z1 is upstream from the orifice meter, within the tube, and outlet plane “2” atposition z2 is placed at the throat of the meter where maximum velocity and minimumpressure exist. There are no moving solid surfaces within the control volume, so thepump work term vanishes. All remaining terms in the ideal Bernoulli equation are dividedby the total mass flow rate that remains constant for steady state analysis. The primaryobjective is to predict the mass flow rate either in the tube or in the throat of the metervia the following equation without frictional energy loss;


123

€

v12

2α+ gz1 +

p1ρ1

=v2

2

2α+ gz2 +

p2ρ2

The steady state mass balance for incompressible fluids allows one to relate <v1> and<v2> via flow cross-sectional areas S1 in the tube and S2 at the contraction. Hence;

€

v1 S1 = v2 S2

v1 =S2S1

v2 = β2 v2

Now, the steady state ideal Bernoulli equation for incompressible fluids (i.e., ρ1 = ρ2 = ρ)is solved for average fluid velocity <v2> at the meter;

€

v22

2α1−β 4{ } =

1ρ

p1 − p2( ) + g z1 − z2( )

v2 =

2α 1ρp1 − p2( ) + g z1 − z2( )

1−β 4

Multiplication of the previous expression for <v2> by ρS2 allows one to predict the totalmass flow rate for ideal fluids (i.e., eV = 0). Then, multiplication of ρ<v>S2 by thecoefficient of discharge CV yields the actual mass flowrate through the Venturi meter.Hence;

€

ActualMassFlowrate=CVρ v2 S2 =CVρπ4dContraction2 2α p1 − p2( ) +ρg z1 − z2( ){ }

ρ 1−β 4( )

The coefficient of discharge CV for Venturi meters is approximately 0.92-0.95

The venturi meter is equipped with a very dangerous and outdated mercurymanometer to measure the pressure difference between positions #1 and #2 forsubsequent calculation of the pressure drop Δp = p1 – p2. If the difference between theheight of mercury in the two legs of the manometer is the same when the venturi is


124

tilted at angle Θ and when the venturi is horizontal, then in which orientation is the massflow rate smaller?

Answer: Both flow rates are the SAME.

When the meter is horizontal, z1 = z2 and the dynamic pressure difference in the previousequation for the actual mass flowrate reduces to the actual fluid pressure difference.However, upon applying hydrostatics within the manometer for tilted meters, in general,and equating fluid pressure at elevation z4 in both legs, one obtains the following relation;

€

p1 +ρg z1 − z4( ) = p2 +ρg z2 − z3( ) +ρHgg z3 − z4( )p1 − p2 = ρg z2 − z3( )−ρg z1 − z4( ) +ρHgg z3 − z4( )

p1 − p2 +ρg z1 − z2( ) = g z3 − z4( ) ρHg −ρ{ }

ActualMassFlowrate=CVρπ4dContraction2 2αg z3 − z4( ) ρHg −ρ{ }

ρ 1−β 4( )

where ρ is the fluid density of interest and ρHg is the density of mercury in the bottom ofthe manometer. As indicated by the previous equation for the actual mass flowrate intilted Venturi meters, elevations z1 and z2 at the inlet and outlet planes do not affect thefinal answer, so one obtains the same mass flowrate in tilted and horizontal meters whenthe difference between the height of mercury in both legs of the manometer (i.e., z3 –z4) is the same. Why does this happen? The hydrostatic contribution to fluid pressure ineach leg of the manometer cancels the effect of gravitational potential energydifferences at the inlet and outlet planes.

The venturi meter is equipped with 21st-century pressure transducers at positions#1 and #2 to measure the pressure drop Δp = p1 – p2. If the pressure transducersmeasure the same Δp when the venturi is tilted at angle Θ and when the venturi ishorizontal, then in which orientation is the mass flow rate larger?

Answer: The tilted Venturi, in which flow proceeds downhill, exhibits thelarger mass flowrate.

Now, hydrostatic calculations within each leg of the manometer do not contribute to fluidpressure at the inlet and outlet planes because the transducers provide a direct measureof fluid pressure. When z1 ≠ z2, gravitational potential energy differences at the inlet and


125

outlet planes cannot be neglected and no cancellation occurs. Hence, if Δp = p1 – p2 isthe same for horizontal and tilted meters, then the tilted meter exhibits a larger flowratewhen z1 > z2 for downhill flow, and the horizontal meter exhibits a larger flowrate when z1

< z2 for uphill flow.

Application of the fluid flow meter equation for laminar tube flow

Frictional energy loss, with units of energy per unit mass of fluid, for one-dimensional fluid flow through a straight tube with circular cross-section is;Ev=(1/2)<vtube>2ev, where the dimensionless friction loss factor for a tube with diameterD=2R and length L is; ev=(4L/D)f, and the friction factor for laminar flow of anincompressible Newtonian fluid is; f=16/Re when Re={ρ<vtube>D/µ} < 2100. Use the fluidflow meter equation to calculate <vtube> when β=1. Remember that Δp≠0 when β=1. Youshould obtain a classic result in fluid mechanics.

Answer:Begin with the fluid flow meter equation from the nonideal Bernoulli equation for either<vtube> or <vmeter>, because both are the same when the diameter ratio β = 1, andcombine fluid pressure and gravitational potential energy via the definition of dynamicpressure;

€

v =

2α 1ρp1 − p2( ) + g z1 − z2( )

1−β 4 +αβ 4eV=

2α P1 −P2{ }ρ 1−β 4 +αβ 4eV( )

⇒β=1

2αΔPραeV

Now, use the dimensionless friction loss factor expression for straight tubes in thelaminar flow regime, where the kinetic energy correction factor α = 1/2. One obtains thefollowing result;

€

v ⇒β=1

2αΔPραeV

=2ΔP

ρ4LD

f Re( )=

2ΔP

ρ4L2R

16µρ v 2R

=R2 v ΔP8µL

v =R2ΔP8µL

;This_ is _ the_HP _ Law!!!


126

Viscous flow through a parallel configuration of orifice and venturimeters

Water at ambient temperature flows into a branch point with an inlet volumetric flowrateof 103 cubic centimeters per second (i.e., 1 Litre per second). The flow configurationsplits into two segments at the branch point. The first horizontal segment contains 100cm of smooth 5-cm inner diameter straight tubing and a sharp-edged orifice meter with adiameter ratio of 0.504. The second horizontal segment contains 100 cm of smooth 5-cm inner diameter straight tubing and a venturi meter with a diameter ratio of 0.316.The coefficients of discharge are 0.62 for the orifice meter and 0.90 for the venturi.Calculate the fraction of the inlet stream (i.e., Qinlet = 103 cm3/sec.) that flows througheach branch.

The incorrect approach to solve this problemThe steady state macroscopic mass balance at the junction point, with one inlet streamand two outlet streams for incompressible fluids, reveals that the inlet volumetric flowrate of 103 cm3/sec must be balanced by the sum of volumetric flowrates through thefirst branch that contains the orifice meter and the second branch that contains theventuri. The fluid flow meter equation provides a route to calculate volumetric flowrates,because z1 = z2 for horizontal configurations, and the coefficient of discharge CV and thediameter ratio β are provided for each meter. Hence;

€

Qinlet =103cm3 /sec =Qorifice +Qventuri

Qorifice =CV ,orificeπ4dorifice2 2αorifice p1 − p2( )

ρ 1−βorifice4( )

;Qventuri =CV ,venturiπ4dventuri2 2αventuri p1 − p2( )

ρ 1−βventuri4( )

Since the meters are arranged in parallel, the pressure drop across each one (i.e., p1–p2)should be the same. A second relation between volumetric flowrates through bothmeters is obtained by rearranging the flow meter equations and equating p1–p2. It isassumed that the flow regime is the same in each branch and that the kinetic energycorrection factor α will not appear in the final result. If, for example, laminar flow occursthrough the orifice meter and turbulent flow occurs through the venturi, then αorifice =1/2, αventuri = 0.945, and cancellation does not occur. Upon equating pressure drops, oneobtains;

€

p1 − p2 =16ρ

2αorificeπ2dTube

orifice

4

Qorifice2

CV ,orifice2

1βorifice4 −1

=

16ρ2αventuriπ

2dTubeventuri

4

Qventuri2

CV ,venturi2

1βventuri4 −1


127

Why is the previous expression incorrect?The pressure drop in the fluid flow meter equation corresponds to the differencebetween upstream pressure and fluid pressure at the contraction, where minimumpressure exists. When the fluid subsequently experiences a gradual (i.e., venturi) orsudden (i.e., orifice) expansion and exits the meter, the streamlines recover from theobstacle (i.e., meter) in the line of flow and fluid pressure increases to some extent, butit doesn’t increase beyond the upstream pressure. The permanent pressure drop acrosseach meter, defined by the difference between upstream pressure and downstreampressure, is significantly less than p1–p2 in the fluid flow meter equation. Branches inparallel experience the same overall, or permanent, pressure drop. Hence, it is notcorrect to equate p1–p2 for both meters because p1 corresponds to the upstreampressure, but p2 represents minimum fluid pressure at the vena contracta. Furthermore,it is necessary to add the pressure drop across smooth straight tubing to the permanentpressure drop across each meter before one should equate the overall pressure dropacross each branch in parallel.

The preferred approach to solve this problemFrictional energy loss across any obstacle in the line of flow can be correlated usingdimensionless friction loss factors eV and the square of the average fluid velocityupstream from the obstacle, but within the appropriate branch. This methodology wasadopted earlier in the discussion of the Macroscopic Mechanical Energy Balance to obtaina relation between dimensionless friction loss factors and coefficients of discharge CV fororifice and venturi meters. The desired relation between CV and eV for any meter withdiameter ratio β is;

€

eV =1CV2 −1

1−β 4

αβ 4

where the kinetic energy correction factor α is flow-regime-specific. Application of thenon-ideal Bernoulli equation to a control volume in which the fictitious inlet plane liesupstream from the meter and the fictitious outlet plane is at least eight tube diametersdownstream from the meter allows one to estimate the dynamic pressure differenceacross the obstacle. For horizontal configurations, the dynamic pressure difference isidentical to the permanent fluid pressure drop because there is no change in potentialenergy from inlet to outlet. Furthermore, there are no moving solid surfaces within thecontrol volume, and there is no change in kinetic energy from inlet to outlet if theupstream tube diameter is the same as the downstream tube diameter. Hence;


128

€

Pupstream −Pdownstream ⇒

horizontalconfiguration

pupstream − pdownstream =12ρ vTube

2eV

If the control volume within each branch of the parallel configuration contains smoothstraight tubing in addition to the appropriate meter, then application of the non-idealBernoulli equation yields the following result;

€

Pupstream −Pdownstream ⇒

horizontalconfiguration

pupstream − pdownstream =12ρ vTube

2 1CV2 −1

1−β 4

αβ 4+ 4 L

dTubef ReTube( )

It is reasonable to add permanent pressure drops across obstacles in series, includingsmooth straight tubing in which f=16/Re for laminar flow and f≈0.08/Re0.25 for turbulentflow. The correct strategy to analyze partitioned flow in a parallel configuration of orificeand venturi meters is;

(1) Apply the steady state overall macroscopic mass balance at the junction pointand relate the inlet volumetric flowrate to the sum of flowrates through bothmeters (i.e., Qinlet = Qorifice + Qventuri).

(2) Express the average tube velocity in each branch in terms of the correspondingvolumetric flowrate and cross-sectional area.

(3) Assume turbulent flow to evaluate the kinetic energy correction factor (i.e., α ≈0.945) and identify the appropriate friction factor correlation. It will benecessary to verify the flow regime by inspecting the Reynolds number when thevolumetric flowrates are determined.

(4) Use the appropriate discharge coefficient and geometric parameters for eachmeter and equate the permanent pressure drop across each branch, as given bythe previous expression.

(5) This problem is completely defined, with no degrees of freedom, and it can besolved implicitly by non-linear algebraic equation solvers. A summary of theappropriate equations and numerical results is provided on the following page.


129


130

Frictional Energy Loss and Pump Horsepower Requirements

Calculate the horsepower requirement of a pump to transport water at 680F fromreservoir A to tank B at the same elevation in the gravitational field. Reservoir A is atambient pressure (i.e., 1 atmosphere) and tank B is at an absolute pressure of 3atmospheres. The flow configuration contains 100 feet of hydraulically smooth 2-inchinner diameter tubing with four 900 rounded elbow, one globe valve fully opened, and onesharp-edged orifice meter that has a diameter ratio of 0.5 and the following coefficientof discharge (i.e., CV ≈ 0.62). The volumetric flowrate is 100 gallons per minute, and thepump efficiency is 30%. The physical properties of water at 680F and some importantconversion factors are provided for your calculations.

Data: µ=6.72x10-4 lbmass/foot-second (viscosity of water @ 680F)ρ=62.4 lbmass per cubic foot (density of water @ 680F)g=32.174 feet/second2 (gravitational acceleration constant)1 gallon = 0.13368 cubic feet1 cubic foot = 28.316 Litres1 atmosphere = 68105.9 lbmass/foot-second2 = 14.7 lbforce/inch2

1 lbforce = 32.174 lbmassfoot/second2

1 BTU/second = 778.223 foot-lbforce/second1 Horsepower = 0.7068 BTU per second = 550 ft-lbforce per second

This sequence of calculations analyzes the power requirements of a pump via thenonideal Bernoulli equation which includes frictional energy losses through variousobstacles, straight tubing, and pump inefficiencies

Enter the fluid densitydensity = 62.4 {lb-mass per cubic foot for water @ 680F}

Enter the fluid viscosityviscosity = 6.72 x 10-4 {lb-mass per foot per second for water @ 680F}

Enter the required volumetric flowrateVolumeFlow = 100 {gallons per minute}

Convert the volumetric flowrate from gallons per minute to cubic feet per second1 gallon = 0.13368 cubic feetcfs = VolumeFlow*0.13368/60

Enter the inner diameter of the tube through which fluid flows, feet


131

diameter = 2/12 {2-inch inner diameter tubing}

Fluid velocity in feet per secondVelocity = 4*cfs/(pi*diameter**2)

Reynolds number, turbulent flowRe# = density*Velocity*diameter/viscosity

Kinetic energy correction factor for turbulent flowalpha = 0.945

f vs. Re correlation for hydraulically smooth tubes, turbulent flowfriction = 0.0791/Re#**0.25

Length of straight tubing in feetLength = 100

Friction loss factor for contraction from an infinite reservoirSuddenContraction = 0.45

Friction loss factor for expansion into an infinite reservoir, turbulent flowSuddenExpansion = 1/alpha

Enter the number of rounded 90-degree elbows, eV = 0.75Elbows90 = 4

Enter the number of gate valves, completely opened, eV = 0.17GateValves = 0

Enter the number of globe valves, completely opened, eV = 8.0GlobeValves = 1

Enter the number of 180-degree return bends, eV = 1.5ReturnBends180 = 0

Enter the number of orifice or venturi meters, either 1 or 0Meter = 1

Enter the diameter ratio for the orifice/venturi meter, smaller/largerDiameterRatio = 0.5


132

Enter the coefficient of discharge for the orifice/venturi meter, highly turbulentflowCV = 0.62

Friction loss in the meterMeterLoss = (1/(CV**2)-1)*(1-DiameterRatio**4)/(alpha*(DiameterRatio**4))

EvStraightTube = 0.5*Velocity**2*(4*Length*friction/diameter) {ft2/sec2}

EvObstacles =0.5*Velocity**2*(SuddenContraction+SuddenExpansion+0.75*Elbows90+0.17*GateValves+8.0*GlobeValves+1.5*ReturnBends180+Meter*MeterLoss)

Total frictional energy loss, ft2/sec2

EvHat=EvStraightTube+EvObstacles

Fluid pressure in reservoir#1, prior to the pump, atmospheresp1 = 1

lb-mass/ft-sec2, 1 atmosphere = 68105.9 lb-mass/ft-sec2

p1Convert = p1*68105.9

Fluid pressure in reservoir#2, downstream from the pump, atmospheresp2 = 3

lb-mass/ft-sec2, 1 atmosphere = 14.7 lb-force/square-inchp2Convert = p2*68105.9

Height of reservoir#1 relative to the zero of potential energy, feetz1 = -250

Height of reservoir#2 relative to the zero of potential energy, feetz2 = -250 {pressurizing liquid fuel tanks in Death Valley, California}

Gravitational acceleration constant, feet per square secondgravity = 32.174

Dynamic pressure in reservoir#1, lb-mass/ft-sec2

scriptP1 = p1Convert+density*gravity*z1


133

Dynamic pressure in reservoir#2, lb-mass/ft-sec2

scriptP2 = p2Convert+density*gravity*z2

Power delivered by the pump to the fluid in units of foot lb-force/secondThere are 32.174 lb-mass foot per square second in 1 lb-forceWpumpDelivered = density*cfs*((scriptP2-scriptP1)/density+EvHat)/gravity

The pump is 30% efficientPumpEfficiency = 0.30

Required pump power, ft lb-force/secondWpumpRequired = WpumpDelivered/PumpEfficiency

Calculate the horsepower delivered to the fluid and required by the inefficientpumpThe following conversions are useful;1 BTU per second = 778.223 ft lb-force per second1 Horsepower = 0.7068 BTU per second = 550 ft lb-force per second}HorsepowerDelivered = WpumpDelivered/(778.223*0.7068)HorsepowerRequired = WpumpRequired/(778.223*0.7068)


134

Mass Transfer in Non-reactive Systems

Begin with the mass transfer equation (MTE) for species i; #1 = #2 + #3 + #4,which is a scalar equation with dimensions of moles per volume per time, for isothermalproblems. The system is defined as the fluid within the control volume (CV). Since thecontrol volume is differentially thick in all coordinate directions, this mass balance yields apartial differential equation. Consider each mass transfer rate process, as describedbelow;

#1; Accumulation rate processAccumulation of the moles of species i within a stationary control volume; ∂Ci/∂tThis term appears on the left side of the mass transfer equation.

#2; Convective mass transferNet rate at which the moles of species i enters a stationary control volume due toconvective mass flux acting across all of the surfaces that surround fluid within the CV;

- ∇ • Civ = - { Ci ∇ • v + v • ∇Ci }

where v represents the mass-average velocity vector from fluid dynamics. The net rateof input is given by the negative of the divergence (i.e., convergence) of convectivemass flux. For an incompressible fluid, the Equation of Continuity reveals that ∇ • v = 0.Hence, the contribution from convective mass transfer on the right side of the masstransfer equation for incompressible liquids reduces to;

- ∇ • Civ = - v • ∇Ci

#3; Molecular mass transfer via Fick's 1st law of diffusionNet rate at which the moles of species i enters a stationary control volume (CV) due todiffusional mass flux acting across all of the surfaces that surround fluid within the CV,where the diffusional flux is given by Fick's 1st law (i.e., - Di,Mixture∇Ci);

- ∇ • ( - Di,Mixture ∇Ci ) = Di,Mixture ∇ • ∇Ci + ∇Ci • ∇Di,Mixture

The net rate of input is given by the negative of the divergence (i.e., convergence) ofdiffusional mass flux. For non-reactive or chemically reactive mixtures with constantphysical properties, like density ρ and mass diffusivity Di,Mixture, only the first term on theright side of the previous equation survives. Hence, the term that characterizesmolecular mass transfer via diffusion on the right side of the mass transfer equation isrepresented by;


135

Di,Mixture ∇ • ∇Ci = Di,Mixture ∇2Ci

where ∇2Ci corresponds to the Laplacian of the molar density of species i. Diffusionoccurs, at most, in three coordinate directions. Partial differential equations must besolved for problems that are described by two- or three-dimensional diffusion. Whenone-dimensional diffusion is sufficient to describe a mass transfer problem, the governingequation (i.e., MTE) corresponds to an ODE. One obtains the following forms for theLaplacian of a scalar (i.e., molar density of species i), which represents one-dimensionaldiffusion in three important coordinate systems;

Rectangular coordinates (z-direction);

€

∇2Ci =d2Ci

dz2

Cylindrical coordinates (r-direction);

€

∇2Ci =1rddr

r dCi

dr

Spherical coordinates (r-direction);

€

∇2Ci =1r2

ddr

r2 dCi

dr

The additional factors of r in cylindrical coordinates and r2 in spherical coordinatesrepresent curvature correction factors because the surface area normal to radial massflux scales as either r or r2, respectively.

#4; Rates of production due to multiple chemical reactionsIf species i participates in several chemical reactions, and its stoichiometric coefficient inthe jth reaction is υ ij, then one must consider each reaction to account for the rate ofproduction of the moles of species i. The final result, which includes information fromchemical kinetics to construct the appropriate rate law for the jth chemical reaction, is;

Σj υij Rj

where Rj represents the intrinsic rate of the jth reaction. For simple nth-order chemicalkinetics, where the rate law is only a function of the molar density of one species;

€

Rj = knj T( )Cin j

where nj is the overall order of the jth reaction with nth-order kinetic rate constant kn.


136

The Mass Transfer Equation for species i that will be analyzed with variouscombinations of mass transfer rate processes in an incompressible fluid is;

€

∂Ci

∂t= −v•∇Ci +Di,Mixture∇

2Ci + υijRjj∑

Usually, the 1st-term on the right side of the previous equation appears on the left sidewith a positive sign because its dimensional scaling factor is the same as that for theaccumulation rate process. Basic information at the microscopic continuum level isobtained by solving the previous equation (i.e., MTE) for the molar density of species i ina mixture.

Steady state diffusion in a stagnant medium with no chemicalreaction. Molecular mass transfer via one-dimensional diffusion is considered for thesesimple steady state problems, which correspond to the film theory of interphase masstransfer with a constant film thickness that contains all of the concentration gradients.The molar density of species i must satisfy the following equation;

€

Di,Mixture∇2Ci = 0

This is Laplace's equation, which stipulates that the Laplacian of molar density mustvanish. One obtains the following generalized molar density profiles for species i in threeimportant coordinate systems, as dictated by the following steady state mass balances;

Rectangular coordinates (x,y,z)3-dimensional diffusion (see page#832 in Transport Phenomena, 2nd-edition);

€

∇2Ci =∂ 2Ci

∂x2+∂ 2Ci

∂y2+∂ 2Ci

∂z2= 0

1-dimensional diffusion in the z-direction, Ci(z);

€

d2Ci

dz2= 0

1-dimensional solution; Ci(z) = a0 + a1 z


137

Cylindrical coordinates (r,Θ,z)3-dimensional diffusion (see page#834 in Transport Phenomena, 2nd-edition);

€

∇2Ci =1r∂∂r

r ∂Ci

∂r

+1r2∂ 2Ci

∂Θ2 +∂ 2Ci

∂z2= 0

1-dimensional diffusion in the radial direction, Ci(r);

€

1rddr

r dCi

dr

= 0

1-dimensional solution; Ci(r) = a2 + a3 ln r

Spherical coordinates (r,Θ,φ)3-dimensional diffusion (see page#836 in Transport Phenomena, 2nd-edition);

€

∇2Ci =1r2

∂∂r

r2 ∂Ci

∂r

+

1r2 sinΘ

∂∂Θ

sinΘ∂Ci

∂Θ

+

1r2 sin2Θ

∂ 2Ci

∂φ2= 0

1-dimensional diffusion in the radial direction, Ci(r);

€

1r2

ddr

r2 dCi

dr

= 0

1-dimensional solution; Ci(r) = a4 + a5 / r

Analogy with steady state conductive heat transfer in a stagnantmedium with no viscous dissipation, no radiation, and no work terms due tocompression or expansion of the system. The thermal energy balance is rathercomplex, but when molecular transport of thermal energy in a pure material viaconduction is the only rate process that must be considered, the following equationdescribes the process;

∇ • q = 0

Convective transport of thermal energy, reversible compression/expansion work,irreversible conversion of mechanical energy to thermal energy, and radiation are notincluded in the previous balance. The molecular flux of thermal energy, given by q, isrelated to temperature gradients via Fourier's law of heat conduction in pure materials;


138

q = - kTC ∇ T

with thermal conductivity kTC of an isotropic medium. The steady state thermal energybalance for conductive heat transfer reduces to;

∇ • ( - kTC ∇ T ) = - kTC ∇ • ∇ T = - kTC ∇2T = 0

Once again, this is Laplace's equation, and the Laplacian of temperature must vanish forsteady state heat conduction in a pure material. If one replaces Ci by T, then molardensity profiles in three important coordinate systems can be adopted from the previoussection as the solution to Laplace's equation for the corresponding temperature profiles.Hence, the one-dimensional and three-dimensional steady state heat conductionequations are summarized below;

Rectangular coordinates (x,y,z)3-dimensional conduction (see page#832 in Transport Phenomena, 2nd-edition);

€

∇2T =∂ 2T∂x2

+∂ 2T∂y2

+∂ 2T∂z2

= 0

1-dimensional conduction in the z-direction, T(z);

€

d2Tdz2

= 0

1-dimensional solution; T(z) = a0 + a1 z

Cylindrical coordinates (r, Θ, z)3-dimensional conduction (see page#834 in Transport Phenomena, 2nd-edition)

€

∇2T =1r∂∂r

r ∂T∂r

+1r2∂ 2T∂Θ2 +

∂ 2T∂z2

= 0

1-dimensional conduction in the radial direction, T(r);

€

1rddr

r dTdr

= 0


139

1-dimensional solution; T(r) = a2 + a3 ln r

Spherical coordinates (r,Θ,φ)3-dimensional conduction (see page#836 in Transport Phenomena, 2nd-edition);

€

∇2T =1r2

∂∂r

r2 ∂T∂r

+

1r2 sinΘ

∂∂Θ

sinΘ∂T∂Θ

+

1r2 sin2Θ

∂ 2T∂φ2

= 0

1-dimensional conduction in the radial direction, T(r);

€

1r2

ddr

r2 dTdr

= 0

1-dimensional solution; T(r) = a4 + a5 / r

The analogous problem in fluid dynamics which is described byLaplace's equation. The cross product of the "del" operator with the velocity vector,yields the vorticity vector [i.e., (1/2) ∇ x v] with dimensions of inverse time. Fluid flowproblems that exhibit rotational characteristics, like all of the viscometers discussed inprevious sections of this document, are described by nonzero vorticity. When fluid flowoccurs far from a high-shear no-slip solid-liquid interface, or in the vicinity of a zero-shearperfect-slip gas-liquid interface, it is reasonable to invoke no vorticity. Hence;

∇ x v = 0

This is the realm of irrotational, potential, inviscid, isentropic, or ideal flow, where there isno tendency for an object placed within the fluid to undergo any type of rotation. Forcomparison, if the velocity vector of a rigid solid which experiences solid-body rotation isgiven by;

v = Ω x r

then the vorticity vector corresponds to the angular velocity vector. For 2-dimensionalflow in cylindrical coordinates, with vr(r,Θ) and vΘ(r,Θ), the volume-averaged vorticityvector, defined by;


140

€

1V

12 ∇xv[ ]

V∫ dV =

12πR2L

δr1r∂vz∂Θ

−∂vΘ∂z

+δΘ∂vr∂z

−∂vz∂r

+δz1r∂ rvΘ( )∂r

−1r∂vr∂Θ

V∫ rdrdΘdz

=δz2πR2

1r∂ rvΘ( )∂r

−1r∂vr∂Θ

0

R

∫0

2π

∫ rdrdΘ

only contains a non-trivial z-component which simplifies considerably to the averageangular velocity vector of the fluid. Potential flow in liquids implies that there are norotational tendencies within the fluid, especially near a boundary. The microscopicdescription of potential flow requires that the vorticity vector must vanish. Themacroscopic description of potential flow requires that there is no large-scale vorticity,which implies that the volume-averaged vorticity vector must vanish. From amathematical viewpoint based on the microscopic description, the vorticity vector willvanish if one identifies any scalar velocity potential Φ (not to be confused with thegravitational energy per unit mass of fluid), such that;

v = ∇ Φbecause;

∇ x ∇ Φ = 0

via Stokes' theorem if Φ is an exact differential. This is true for any multivariable scalarfunction that is analytic and path-independent, because the order of mixed 2nd partialdifferentiation can be reversed without affecting the final result. Hence, the requirementof no vorticity at the microscopic level, which is consistent with irrotational flow,suggests that the fluid velocity vector can be expressed as the gradient of a scalarvelocity potential. However, the requirement of no vorticity does not provide a uniquefunction for Φ because any scalar that is an exact differential will satisfy the previousequation. The unique scalar velocity potential for a particular ideal flow problem iscalculated by invoking incompressibility. Hence;

∇ • v = ∇ • ∇ Φ = ∇ 2 Φ = 0

which is Laplace's equation. Potential flow solutions in n-dimensions (i.e, 1≤n≤3) areobtained by solving one 2nd-order partial differential equation (i.e., Laplace's equation) forΦ in terms of n independent spatial variables. This is one of the most straightforwardroutes to calculate 3-dimensional flows.


141

Mass transfer coefficients and Sherwood numbers that are consistentwith the steady state film theory of interphase mass transfer. The film theorydescribes steady state one-dimensional diffusion across an interface and into a stagnantmedium. The examples presented below are specific to planar, cylindrical, and sphericalinterfaces. The overall objective is to develop relations between mass transfercoefficients, molecular transport properties, and film thicknesses. Curvature correctionfactors are required when radial diffusion occurs across cylindrical and sphericalinterfaces. Analogies between heat and mass transfer are invoked to obtain thecorresponding heat transfer coefficients. General molar density and temperature profileshave been obtained in three important coordinate systems from the solution of Laplace'sequation in one dimension. Two boundary conditions are required to evaluate theintegration constants in each case. In terms of the molar density of mobile componentA, equilibrium is achieved at the interface (i.e., CA,equilibrium), which corresponds to thesolubility of A in the phase of interest. As one moves away from the interface in thedirection of the unit normal vector and travels through the stagnant film which containsall of the concentration gradients, bulk fluid conditions are achieved (i.e., CA,bulk) at theedge of the film. Three separate mass transfer problems are considered below.

Steady state diffusion and conduction across flat interfaces in rectangularcoordinates;If z is the independent variable measured normal to the interface and the stagnant film isdescribed by a constant thickness L, then the boundary conditions are;

(1) At z = 0, CA = CA,equilibrium

(2) At z = L, CA = CA,bulk

The two integration constants in the linear molar density profile are calculated as follows;

(1) CA,equilibrium = a0

(2) CA,bulk = a0 + a1L

Hence, a1 = -∆CA/L, where ∆CA = CA,equilibrium – CA,bulk represents the overall concentrationdriving force for mobile component A. The basic information for mass transfer at themicroscopic continuum level, consistent with steady state one-dimensional diffusion intoa stagnant medium, is;

€

CA,equilibrium −CA z( )ΔCA

=zL

The corresponding temperature profile for steady state one-dimensional conduction in anisotropic solid or stagnant fluid is;


142

€

TSurface −T z( )TSurface −TBulk

=zL

where TSurface is analogous to CA,equilibrium, and TBulk is analogous to CA,bulk. If film thickness Lis interpretted as a hydrodynamic factor that decreases at higher Reynolds numbers forthe phase where CA(z) and T(z) were determined, then this simple model exhibits someof the characteristics of more complex heat and mass transfer problems, even thoughconvective transport was neglected.

Effect of curvature for radial diffusion and conduction across cylindricalinterfaces;Results from the previous section must be modified slightly when radial diffusion occursacross a curved interface, like the lateral surface of a long cylinder. The logarithmicprofile accounts for the fact that the surface area normal to diffusional flux in the radialdirection is not constant (i.e., it scales linearly with radial position r). If a stagnant film ofradius RFilm surrounds a cylindrical solid of radius RSolid, then the appropriate "film theory"boundary conditions are;

(1) At r = RSolid, CA = CA,equilibrium

(2) At r = RFilm, CA = CA,bulk

The film thickness is RFilm – RSolid, and the integration constants in the molar density profileare calculated from the following equations;

(1) CA,equilibrium = a2 + a3 ln RSolid

(2) CA,bulk = a2 + a3 ln RFilm

One obtains;

€

a3 =−ΔCA

ln RFilm /RSolid( );a2 =CA,bulk +ΔCA

lnRFilm

ln RFilm /RSolid( )

The molar density profile for radial diffusion across a cylindrical interface is given by;

€

CA r( )−CA,bulk

ΔCA

=ln RFilm / r( )

ln RFilm /RSolid( )


143

The corresponding temperature profile is;

€

T r( )−TBulkTSurface −TBulk

=ln RFilm / r( )

ln RFilm /RSolid( )

Steady state radial diffusion and conduction in spherical coordinates;Now, the surface area normal to diffusional flux in the radial direction scales as thesquare of radial position r, and this effect of curvature yields a molar density ortemperature profile that depends inversely on r. A stagnant film of radius RFilm surroundsa bubble, liquid droplet, or spherical solid pellet of radius RSolid. The constant filmthickness is RFilm – RSolid. This is the only coordinate system that will produce reasonableresults when the thickness of the stagnant film becomes infinitely large (i.e., when RFilm ⇒∞). The appropriate boundary conditions are;

(1) At r = RSolid, CA = CA,equilibrium

(2) At r = RFilm, CA = CA,bulk

Integration constants a4 and a5 are calculated as follows;

(1) CA,equilibrium = a4 + a5 / RSolid

(2) CA,bulk = a4 + a5 / RFilm

Hence;

€

a5 =ΔCA

1RSolid

−1

RFilm

;a4 =CA,bulk −ΔCA

RFilm1

1RSolid

−1

RFilm

The final results at the microscopic continuum level for molecular transport in the radialdirection across a spherical interface are;

€

CA r( )−CA,bulk

ΔCA

=

1r−

1RFilm

1RSolid

−1

RFilm

; T r( )−TBulkTSurface −TBulk

=

1r−

1RFilm

1RSolid

−1

RFilm

When the stagnant film is infinitely thick, which implies that one must travel an infinitedistance away from the spherical interface to achieve bulk conditions, the right side ofthe previous two equations for CA(r) and T(r) reduces to RSolid/r.


144

General strategy to calculate interphase transfer coefficients;(i) Draw a picture of the problem and identify the coordinate system that exploits

the symmetry of the macroscopic boundary. Now, the interface underinvestigation should conform to a “simple surface”.

(ii) Solve a simplified version of the mass transfer equation that includes only themost important mass transfer rate processes, together with the supportingboundary conditions. The objective is to obtain the molar density profile ofmobile component A that is transported across the interface. When the masstransfer Peclet number is sufficiently large, it is reasonable to neglect moleculartransport (i.e., conduction, diffusion, or viscous stress) in the primary flowdirection. However, one should never neglect molecular transport normal to theinterface.

(iii) Identify the unit normal vector n, which is perpendicular to the interface andpoints in the direction of interphase transport.

(iv) Construct the scalar "dot" product of n with the molecular flux of species Awhose molar density profile within the mass transfer boundary layer was obtainedin the previous sections for interfaces with rectangular, cylindrical, or sphericalsymmetry.

(v) Define the interphase transfer coefficient in terms of the normal component ofmolecular flux, evaluated at the interface;

{ n • Molecular flux }Evaluated at the Interface = [ Transfer Coefficient ] [ Driving Force ]

Rectangular coordinates;The flat interface is defined by z=0, and the unit normal vector, oriented in the directionof interphase mass transfer, is n = δz. The molar density profile for mobile component A,which is consistent with one-dimensional diffusion in the z-direction, was developed in aprevious section;

€

CA,equilibrium −CA z( )ΔCA

=zL

The normal component of the diffusional mass flux vector of species A, evaluated at theinterface, is;

€

n• −DA,mixture∇CA{ }z=0 = −DA,mixturedCA

dz

z=0

= DA,mixtureΔCA

L= kCΔCA


145

Hence, the simplest mass transfer coefficient for steady state one-dimensional diffusionacross a flat interface into a stagnant fluid with a constant film thickness L, is given by;

€

kC =DA,mixture

L

This corresponds to a Sherwood number of unity, when the characteristic length is L inthe definition of the Sherwood number.

Unsteady state diffusion with no chemical reaction(a) Fick's 2nd law(b) Introduction to boundary layer analysis

Fluid dynamics p.#115Heat transfer Eq.#11.2-10 on p.#375, also p.#338Mass transfer Eq.#19.1-18 on p.#585, also p.#613

Leibnitz rule for differentiating one-dimensional integrals when the limits ofintegration are not constant. This theorem is based on the following integral expression,where the integrand and the limits of integration depend on the independent variable.For example, if;

€

Γ t( ) = f x,t( )dxx=u t( )

w t( )

∫then, in general;

Γ(t) = Γ{t,u(t),w(t)} = Γ(t,u,w)

The hierarchy is that Γ depends directly on t, u, and w; whereas u and w both depend ont. Hence, the total differential of Γ is;

€

dΓ =∂Γ∂t

u,wdt +

∂Γ∂u

t ,wdu +

∂Γ∂w

t ,udw

Therefore;

€

dΓdt

=∂Γ∂t

u,w

+∂Γ∂u

t,w

dudt

+∂Γ∂w

t ,u

dwdt


146

The 1st term on the right side of the previous equation is obtained by taking the partialderivative of f(x,t) with respect to t "inside the integral", because the limits ofintegration are treated as constants. Now, it is necessary to obtain expressions for(∂Γ/∂u)t,w and (∂Γ/∂w)t,u. This is achieved by introducing a new function Ξ(x,t), whichrepresents the indefinite integral of f(x,t) with respect to x. In other words;

€

Ξ x,t( ) = f x,t( )∫ dx

Γ t,u,w( ) = f x, t( )dxx=u t( )

w t( )

∫ =Ξ w,t( )−Ξ u,t( )

The "fundamental theorem of calculus" states that;

€

∂Ξ∂x

t

= f x, t( )

Therefore, if one replaces x by either u or w, then;

€

∂Ξ∂u

t

= f u,t( ); ∂Ξ∂w

t

= f w,t( )

The relation between Γ and Ξ above yields the following partial derivatives of interest;

€

∂Γ∂u

t,w

= −∂Ξ∂u

t

= − f u,t( )

∂Γ∂w

t,u

= +∂Ξ∂w

t

= + f w,t( )

The final result is;

€

dΓdt

=∂f x,t( )∂t

x=u t( )

w t( )

∫x

dx+ f w,t( ) dwdt

− f u,t( ) dudt

_______________________________________________________

Convection and diffusion through permeable membranes(a) Plug flow models(b) Analysis of blood capillaries


147

Review for Exam#3

Problem#1Consider steady state heat conduction across a solid-liquid interface with sphericalsymmetry. A solid sphere of radius RSolid is suspended by a metal wire in a stagnant fluid.An electrical current generator maintains the solid sphere at constant temperature, givenby TSolid, everywhere throughout the solid. Thermal energy is transported from the solidto the stagnant fluid solely by radial conduction, and the stagnant film which surroundsthe solid sphere is infinitely thick, such that TBulk, which is less than TSolid, is achieved as r⇒ ∞. There is no source of thermal energy within the fluid. Calculate the heat transfercoefficient for this steady state process. Then, modify your result for a very thinstagnant film in the liquid phase surrounding the sphere, such that RFilm ≈ RSolid.

Answer:Begin with the temperature profile that represents the solution to Laplace's equation forsteady state conduction exclusively in the radial direction in spherical coordinates;

€

T r( )−TBulkTSurface −TBulk

=

1r−

1RFilm

1RSolid

−1

RFilm

Use Fourier's law and evaluate the molecular flux of thermal energy, normal to the solid-liquid interface in the radial direction. By definition, this conductive heat flux, evaluatedat the interface, is given by the product of a heat transfer coefficient and a temperaturedriving force;

€

n• −kTC∇T{ }r=RSolid = −kTCdTdr

r=RSolid

= −kTC TSurface −TBulk( ) −1RSolid2

11

RSolid−

1RFilm

= hHTC TSurface −TBulk( )

hHTC =kTCRSolid

1

1− RSolidRFilm

Now, evaluate the result for the heat transfer coefficient when the steady state filmthickness (i.e., RFilm – RSolid) is infinitely large (i.e., RFilm >> RSolid). The final result for this“stagnant film theory” problem in spherical coordinates is;


148

hHTC = kTC / RSolid

which corresponds to a Nusselt number of Nu = 2 when the characteristic length is thediameter of the sphere. When the stagnant film is very thin, the previous heat transfercoefficient expression is manipulated by defining ε = RFilm/RSolid, expanding thedenominator of the curvature correction factor in spherical coordinates about ε = 1, andtruncating the Taylor series after the linear term because ε is very close to unity;

€

1− RSolidRFilm

=1− 1ε

= 0+ε −11!

ddε

1− 1ε

ε=1

+ ...≈ ε −1

CurvatuveCorrectionFactor =1

1− RSolid

RFilm

≈1

ε −1≈

RSolid

RFilm − RSolid

One obtains the following expression for the steady state heat transfer coefficient that isconsistent with radial conduction into a stagnant medium when the thermal boundarylayer thickness, δT = RFilm – RSolid, is very small and localized at the solid-liquid interface;

€

hHTC =kTCRSolid

1

1− RSolid

RFilm

≈kTCRSolid

RSolidδT

≈kTCδT

This is the heat-transfer analog of steady state diffusion across a “locally flat” interface(see pages 132-133 where L represents the mass transfer boundary layer thickness)because the curvature correction factor in spherical coordinates is negligible for thinboundary layers.

Problem#2(a) A cold solid sphere of radius RSolid at initial temperature TSolid is submersed in a

large container of liquid. A heater maintains this liquid bath at a temperaturewhich is much hotter than the initial temperature of the solid sphere, andvigorous stirring of the liquid within the container eliminates all temperaturegradients in the liquid phase such that T = TSurface at r = RSolid throughout the entireanalysis. There is no source of thermal energy within the solid sphere. Write theappropriate differential equation that must be solved to calculate the localtemperature profile T within the solid, and don't include any unnecessary terms.


149

Answer:This problem is classified as unsteady state heat conduction in a solid sphere. Sincethere is no convective transport in solids, and viscous dissipation is nonexistent, onemust solve the unsteady state heat conduction equation;

€

∂T∂t

=α∇•∇T =α∇2T

where α = kTC/(ρCp) is the thermal diffusivity of the solid. If heat conduction occursexclusively in the radial direction, then the previous equation requires the contribution inthe r-direction for the divergence of the gradient of a scalar in spherical coordinates.This is equivalent to the contribution in the r-direction for the Laplacian of a scalar.Hence, one must account for the fact that the surface area normal to the molecular fluxof thermal energy in the radial direction is variable, and scales as the square of radialposition. The partial differential equation for T(r,t) is;

€

∂T∂t

=α1r2

∂∂r

r2 ∂T∂r

(b) Include all of the boundary conditions that are required to obtain a uniquesolution to your thermal energy balance in part (a).

Answer:One condition is required on independent variable t;

(1) t = 0, for all r < RSolid; T = TSolid (initial condition)

Two conditions are required on independent spatial variable r for heat conduction;

(2) r = RSolid, for all t > 0; T = TSurface (interfacial condition)

At short times, this problem can be treated as a boundary layer problem via combinationof variables. Hence, if δT represents the thermal boundary layer thickness measuredinward from the fluid-solid interface at r = RSolid, then;

(3a) r < RSolid - δT, for finite t; T = TSolid (boundary layer BC)

However, if one seeks the solution for T(r,t) from the initial condition until the solidequilibrates with the constant temperature bath, then separation of variables is the


150

method of choice, and symmetry at the center of the sphere is appropriate. In otherwords, the molecular flux of thermal energy vanishes at r=0 due to symmetry, andFourier's law stipulates that;

(3b) r = 0; dT/dr = 0 (symmetry at the center of the sphere)

Another possible condition is T = TSurface at long times for all values of 0≤r≤RSolid, but thisobvious result might be satisfied automatically by the exponentially decreasing functionof time in the separation-of-variables solution for T(r,t).

Problem#3An incompressible Newtonian fluid flows past a rectangular solid plate that is soluble inthe liquid. Hence, mobile component A is transported from the solid to the liquid. Theinterface between the solid and the liquid is locally flat, and laminar flow is appropriate todescribe flow of the liquid parallel to the interface. The x-direction is parallel to theinterface, and the y-direction is perpendicular to the interface.

(a) If the mass transfer Peclet number is very large, then write the simplified versionof the steady state microscopic mass transfer equation which must be solved tocalculate the molar density profile of species A, CA(x,y), within the incompressibleliquid phase for this non-reactive problem. Include all three boundary conditionsthat are required to obtain a unique solution for CA(x,y).

Answer:Begin with the most general form of the microscopic mass transfer equation for speciesA;

€

∂CA

∂t= −v•∇CA +DA,Mixture∇

2CA + υAjRjj∑

For steady state analysis of non-reactive systems, the MTE reduces to;

€

v•∇CA = DA,Mixture∇2CA

Convection and diffusion occur, at most, in three coordinate directions. For two-dimensional transport in rectangular coordinates, the MTE for CA(x,y) reduces to;

€

vx∂CA

∂x+ vy

∂CA

∂y= DA,Mixture

∂ 2CA

∂x2+∂ 2CA

∂y2


151

Since the primary direction of fluid flow coincides with the x-direction, and the masstransfer Peclet number is large, it is reasonable to neglect molecular transport bydiffusion in the x-direction relative to convective mass transfer in the x-direction. Ifconvective mass transfer occurs in both the x- and y-directions, where flow in the y-direction is of secondary importance, then one obtains the solution for CA from thefollowing partial differential equation;

€

vx∂CA

∂x+ vy

∂CA

∂y= DA,Mixture

∂ 2CA

∂y2

If convective mass transfer occurs exclusively in the x-direction, then the appropriateform of the mass balance for CA(x,y) is;

€

vx∂CA

∂x= DA,Mixture

∂ 2CA

∂y2

(b) The solution to part (a) for the molar density profile of mobile component A isgiven by;

€

CA,equilibrium −CA η( )CA,equilibrium −CA,bulk

= P η( ) =1

Γ 43( )

exp −u3( )0

η

∫ du

η =y

δC x( )

where Γ(4/3) is the gamma function evaluated when the argument is 4/3rds andδC(x) represents the mass transfer boundary layer thickness that increases atlarger values of x along the interface, with δC=0 at x=0. Calculate the local masstransfer coefficient, kC,local, and indicate its functional dependence on theappropriate spatial coordinate or coordinates.

Answer:Since there is no convective flux at a high-shear no-slip interface, use Fick’s 1st-law ofdiffusion and calculate the y-component of diffusional mass flux at y=0, for all x>0 (i.e.,η=0). The unit normal vector at the interface, oriented in the direction that masstransfer occurs, is n = δ y. Hence, one performs the following scalar dot productoperation and equates the result to the product of a local mass transfer coefficient,kC,local, and the overall concentration driving force, CA,equilibrium - CA,bulk;


152

€

n• −DA,Mixture∇CA{ }y=0= −DA,Mixture

∂CA

∂y

y=0

= −DA,MixturedCA

dPdPdη

η=0

∂η∂y

x

= DA,Mixture CA,equilibrium −CA,bulk( ) dPdη

η=0

1δC x( )

= kC ,local CA,equilibrium −CA,bulk( )

Hence, the local mass transfer coefficient is;

€

kC ,local x( ) = DA,MixturedPdη

η=0

1δC x( )

The Leibnitz rule indicates that the derivative of the dimensionless molar density profilewith respect to the combined variable η, evaluated at the solid-liquid interface for all x >0 (i.e., η = 0), is;

€

dPdη

η=0

=1

Γ 43( )exp −η3( ){ }

η=0=

1Γ 4

3( )

The final expression for the local mass transfer coefficient, which depends on spatialcoordinate x measured parallel to the interface, is;

€

kC ,local x( ) =1

Γ 43( )DA,Mixture

δC x( )

(c) Without using any equations, describe how the local mass transfer coefficientchanges as one moves along the solid-liquid interface in the primary flow direction(i.e., increasing x), when the Reynolds number remains constant. Does kC,local

increase, decrease, remain constant, or is it too complex to determine how thelocal mass transfer coefficient changes? Provide a qualitative explanation foryour answer in one or two sentences.

Answer:The local mass transfer coefficient decreases as one moves along the interface in thedirection of flow (i.e., increasing x) because the resistance to mass transfer increases,due the fact that the boundary layer thickness increases. The answer to part (b) revealsthat kC,local depends inversely on the thickness of the mass transfer boundary layer, δC.


153

Helpful hint: The Leibnitz rule for differentiating a one-dimensional integral, where theintegrand and both limits of integration depend on the independent variable, is;

€

Γ t( ) = f x, t( )dxx=u t( )

w t( )

∫

dΓdt

=∂f x,t( )∂t

x=u t( )

w t( )

∫x

dx+ f w,t( ) dwdt

− f u,t( ) dudt

__________________________________________________________________

Diffusion and Chemical Reaction Across Spherical Gas-LiquidInterfacessee Chapter#13 in TPfCRD and Chapter#19 in BSL’s Transport Phenomena.

Radial diffusion and 1st-order chemical reaction in spherical coordinatesChemical reaction enhancement of mass transfer coefficientsProblem 19B.6, pp. 607-8 (BSL); transient, steady state and quasi-steady state analyses

Convection, Diffusion & Chemical Reaction in Multiphase Reactorssee Chapter#24 in TPfCRD

Chlorination of benzene in a gas-liquid continuous stirred tankGas phase mass balances with interphase mass transferLiquid phase mass balances with chemical reaction and interphase transfer

Chemical reaction enhancement of interphase mass transfer coefficientsEquilibrium at the gas-liquid interfaceMinimal gas phase resistance in the mass transfer boundary layerInterfacial area for gas-liquid mass transferTime constants for three important mass transfer rate processesDimensionless mass balances in the gas and liquid phasesMolecular diffusion in liquids

Coupled Heat and Mass Transfer in Non-Isothermal Liquid PhaseTubular Reactors with Strongly Exothermic Chemical Reactionsee Chapter#4 in TPfCRD

Thermal Runaway, Parametric Sensitivity and Multiple Stationary StatesStrategies to control thermal runaway


154

Plug flow mass balance at high mass transfer Peclet numbersThermal energy balance at high heat transfer Peclet numbersThermodynamics of multicomponent mixturesConductive heat transfer across the lateral surface of the reactor

Adiabatic reactorsConstant heat flux across the wallConstant outer wall temperature

Manipulating the outer wall temperatureManipulating the surface-to-volume ratio of the reactor

Coupled heat and mass transfer in PFR's with cocurrent coolingManipulating the flowrate of a cocurrent cooling fluid

Parametric sensitivity analysis in non-isothermal tubular reactorsExothermic/endothermic reactions with cocurrent coolingConcentric double-pipe configurations that are not insulatedCountercurrent cooling in concentric double-pipe configurationsMultiple stationary states in PFR's with countercurrent cooling

Examples of multiple stationary states

Review for Exam#4

Problem#1Consider steady state mass transfer via convection and diffusion through a blood vesselat very large mass transfer Peclet numbers with no chemical reaction.

(a) Obtain an expression for the molar density profile of mobile component A withinthe blood vessel, CA(z), as a function of axial coordinate z, which increases in theprimary flow direction through the permeable capillary. The well-mixedconcentration of species A outside of the capillary is approximately constant (i.e.,CA,bulk), and CA at the capillary inlet (i.e., z=0) is zero.

(b) Sketch CA vs. z in the laminar flow regime when the Reynolds number, based onthe capillary diameter, is 200 and 1000. Put both curves on one graph, with CA

on the vertical axis and z on the horizontal axis. Label each curve with theappropriate value of the Reynolds number.

(c) If the Reynolds number is 500 in the laminar flow regime, then sketch CA vs. z at350C and 400C. Put both curves on one graph, with CA on the vertical axis and zon the horizontal axis. Label each curve with the appropriate temperature.


155

Problem#2(a) Consider one-dimensional radial diffusion and 1st-order irreversible chemical

reaction in the liquid phase external to a gas bubble of radius R. Obtain anexpression for the molar density profile of reactant A, CA(r), if the steady statefilm that surrounds the bubble is infinitely thick. Hint: Adopt a solution in termsof exponential functions, not hyperbolic functions.

Answer:In dimensional notation for the molar density of species A, the spherical coordinate masstransfer equation in the liquid phase with radial diffusion and 1st-order irreversiblechemical reaction yields the following 2nd-order ODE;

€

DA,Mixture1r2

ddr

r2 dCA

dr

= k1CA

The following transformation [i.e., CA = (1/r)ΨA] allows one to rewrite the previous ODEwith variable coefficients as a frequently occurring 2nd-order ODE with constantcoefficients;

€

dCA

dr=1rdΨA

dr−ΨA

r2

r2 dCA

dr= r dΨA

dr−ΨA

ddr

r2 dCA

dr

= r

d2ΨA

dr2

DA,Mixture1r2

ddr

r2 dCA

dr

= DA,Mixture

1rd2ΨA

dr2= k1CA = k1

1rΨA

DA,Mixtured2ΨA

dr2= k1ΨA

If η = r/R, then the solution to this ODE for ΨA(η) is written as follows in terms ofexponential functions;

€

ΨA η( ) = Aexp ΛAη( ) + Bexp −Λ Aη( )

ΛA2 =

k1R2

DA,Mixture


156

where Λ2A is the Damkohler number for species A, which represents a ratio of the rate of

chemical reaction to the rate of molecular mass transfer via diffusion. The generalsolution for the molar density of reactant A is;

€

CA =1rΨA η( ) =

1rAexp ΛAη( ) + Bexp −Λ Aη( ){ }

(b) If the boundary conditions are;

(i) At r = R (i.e., η=1), CA = CA,equilibrium

(ii) As r ⇒ ∞, CA = 0

then calculate the integration constants in your expression for CA(r) from part (a)and express your final answer for CA(r) in terms of the Damkohler number forreactant A

Answer:The exponential term with the positive argument in the general solution for CA increasesin unbounded fashion as radial position r (or η) tends toward infinity. This can be verifiedvia one application of l’Hopital’s rule. Hence, one sets integration constant A to zero, asrequired by boundary condition (ii). Integration constant B is evaluated using boundarycondition (i);

€

CA,equilibrium =BRexp −Λ A( )

CA r( ) =CA,equilibriumRr

exp −ΛAη( )exp −ΛA( )

(c) Obtain an expression for the chemical-reaction-enhanced mass transfercoefficient kC,Liquid on the liquid side of the gas-liquid interface, which exhibitsdependence on the Damkohler number that is slightly different from the resultsdiscussed in class.

Answer:Use Fick’s 1st law and evaluate the diffusional flux of reactant A in the radial direction onthe liquid side of the gas-liquid interface, at r=R. Then, equate this result to the productof a liquid-phase mass transfer coefficient and the concentration difference, or drivingforce, CA,equilibrium – 0. This generalized procedure yields the following expression for kC,Liquid;


157

€

−DA,MixturedCA

dr

r=R

= −DA,MixtureCA,equilibriumR

exp −ΛA( )ddr

1rexp −ΛAr

R

r=R

= −DA,MixtureCA,equilibriumR

exp −ΛA( )−1r2exp −ΛAr

R

−

ΛA

Rrexp −ΛAr

R

r=R

=DA,Mixture

R1+ΛA[ ] CA,equilibrium − 0( ) = kC ,Liquid CA,equilibrium − 0( )

kC ,Liquid =DA,Mixture

R1+ΛA[ ]

The second term in brackets [] for kC,Liquid dominates for diffusion-controlled chemicalreactions because the Damkohler number is very large (i.e., 1+ΛA ≈ ΛA), whereas the firstterm in brackets [] is more important for reaction-controlled situations (i.e., 1+ΛA ≈ 1) orwhen no reaction occurs (i.e., ΛA ⇒ 0). This latter case reduces to steady state diffusionor heat conduction across a spherical interface into a stagnant medium in which theboundary layer thickness is infinitely large and curvature effects cannot be neglected.

Problem#3 Reactive distillationPure liquid B is flowing at steady state from left to right across a perforated tray in adistillation column and bubbles of gas A rise through the liquid. Gas A is soluble in liquidB, and A reacts irreversibly with B only in the liquid phase. Due to the high concentrationof B in the liquid phase, the "method of excess" suggests that the kinetic rate law ispseudo-first-order with respect to the liquid phase molar density of solubilized gas A.The rising motion of the bubbles produces a "well-stirred" liquid mixture of A and B, butthe two streams do not leave the tray in equilibrium with each other. At most,equilibrium is established at the spherical gas-liquid interface.

(a) Consider mass transfer rate processes and their corresponding time constants todescribe the conditions that must exist if the outlet liquid stream contains asignificant fraction of species A, realizing that the inlet stream contains pureliquid B. Do not use any equations.

Answer:The time constant for interphase mass transfer of species A must be significantly smallerthan either of the time constants for (i) chemical reaction in the liquid phase or (ii)convective mass transfer of the liquid phase across the tray in the distillation column(i.e., residence time).


158

(b) In the diffusion-limited regime, perform a macroscopic balance on the liquid phaseand obtain an algebraic equation that relates the outlet liquid phase molar densityof reactant A, CA,outlet, to the following quantities;

CA,equil. equilibrium molar density of species A on the liquid side of the gas-liquidinterface (i.e., equilibrium solubility of gas A in liquid B, g-mol/cm3)

DA,Liquid diffusion coefficient of species A in liquid B, (cm2/sec)k1 pseudo-first-order kinetic rate constant in the liquid phase, (1/sec)q volumetric flow rate of the liquid, (cm3/sec)VL liquid phase volume on the tray, (cm3)τ liquid phase residence time on the tray, VL/q (sec)aL interfacial area per unit volume of liquid, (1/cm)

Answer:The liquid phase can be analyzed as a well-mixed CSTR operating at steady state. Hence,one equates rates of input to rates of output for reactant A in its liquid phase massbalance. Since the inlet liquid stream contains pure component B, there is nocontribution from convective mass transfer across the inlet plane.

Rate of input due to interphase mass transfer = {k1DA,Liquid}1/2[CA,equilibrium - CA,outlet]aLVL

(using a chemical-reaction-enhanced mass transfer coefficient in the diffusion-limitedregime, where curvature effects are negligible for thin mass transfer boundary layers)

Rate of output due to convective mass transfer = qCA,outlet

Rate of disappearance of reactant A due to 1st-order irreversible reaction = k1CA,outletVL

The steady state liquid-phase CSTR mass balance for reactant A, with dimensions ofmoles per time, is;

€

k1DA,Liquid CA,equilibrium −CA,outlet( )aLVL = qCA,outlet + k1CA,outletVL

CA,outlet =CA,equilibrium

aL k1DA,Liquid

aL k1DA,Liquid + k1 +1τ


159

(c) The Arrhenius activation energy for diffusion of solubilized gas A in liquid B,Eac/Diffusion, is much smaller than the Arrhenius activation energy for the chemicalreaction, Eact/ChemicalReaction. Hence,

€

Eact /Diffusion << Eact /ChemicalReaction

ddTlnDA,Liquid =

Eact /Diffusion

RT 2 > 0

ddTln k1 =

Eact /ChemicalReaction

RT 2 > 0

Describe how a decrease in temperature T will affect the outlet liquid phase molardensity of reactant A. Will CA increase, decrease, remain unchanged, or is it toocomplex to determine how CA will change?

Problem#4Use numerical methods (i.e., finite difference calculus) to solve the steady statemicroscopic mass transfer equation for convective diffusion in heterogeneous catalytic“tube-wall” reactors with circular cross-section in the laminar flow regime forincompressible Newtonian fluids. Chemical reaction at the catalytic surface (i.e., r=R) isirreversible and first-order with respect to reactant A. Let the tube radius R be thecharacteristic length in the definitions of the Damköhler (i.e., β) and mass transfer Peclet(i.e., PeMT) numbers, and consider the regime where PeMT is large enough to justify theneglect of axial diffusion.

Answer:The appropriate mass transfer equation is given in Step#5 of Problem 23-7 on page#649of TPfCRD, and the laminar flow velocity profile is provided in Step#7 on page#650.Hence, the primary objective of this exercise is to calculate the molar density of reactantA, CA(r,z), from the following partial differential equation and its boundary conditions incylindrical coordinates, with variable coefficients and chemical reaction at the boundaryof the flow configuration;

€

2 vz Average1−η2{ }∂CA

∂z= DA

1r∂∂r

r ∂CA

∂r

= DA∂ 2CA

∂r2+1r∂CA

∂r

CA =CA,inlet@z = 0,r < R∂CA

∂r

r=0

= 0;−DA∂CA

∂r

r=R

= k1,SurfaceCA r = R, z( )


160

The zero-flux boundary condition along the tube axis at r=0 is a consequence ofsymmetry, and the radiation boundary condition at the catalytic surface (i.e., r=R)represents a balance between diffusion and chemical reaction. Radial and axial positionsare dimensionalized using tube radius R. Hence, η = r/R and ζ = z/R. Reactant molardensity is dimensionalized via the inlet condition, ΨA(η,ζ) = CA(r,z)/CA,inlet. In terms of theimportant dimensionless numbers that govern the solution to this problem;

€

Damkohler#;β =k1,SurfaceRDA,ordinary

Peclet#;PeMT =vz Average

RDA,ordinary

the mass transfer equation and its boundary conditions can be written as follows usingdimensionless variables;

€

2PeMT 1−η2{ }∂ΨA

∂ζ=∂ 2ΨA

∂η2+1η∂ΨA

∂η

ΨA =1@ζ = 0,η <1∂ΨA

∂η

η=0

= 0; ∂ΨA

∂η

η=1

= −βΨA η =1,ζ( )

Problem 23-7 in TPfCRD provides an asymptotically exact mass transfer boundary layersolution for CA(r,z) in the inlet region (i.e., z>0) for heterogeneous catalytic tubularreactors (see Step#16 on page#652). A much simpler approach is adopted below toinitiate the numerical algorithm by applying the radiation boundary condition at z=0 andr=R (or ζ=0 and η=1) to estimate the molar density of reactant A at the wall near theinlet plane. For example;

€

∂ΨA

∂η

η=1

≈ΨA η =1,ζ = 0( )−1

Δη= −βΨA η =1,ζ = 0( )

ΨA η =1,ζ = 0( ) ≈ 11+βΔη


161

which exhibits the correct trend, because reactant molar density at the catalytic surfacedecreases when the rate of reaction is faster and the Damköhler number increases. In aneffort to check the validity of the finite-difference solutions to the microscopic masstransfer equation, one poses the following question; Do the microscopic results satisfythe quasi-macroscopic mass balance? Hence, it is necessary to evaluate the bulk molardensity of reactant A at each axial position, given by equation (23-19) in TPfCRD,explicitly for tubular reactors. Analogous to equation (23-51) for rectangular ducts (seepage#632 in TPfCRD), the dimensionless bulk reactant molar density in tubular reactorsis;

€

CA,bulk z( ) =vz r( )CA r, z( )rdrdΘ

S∫∫

πR2 vz Average

= 4 CA r, z( ) 1−η2{ }η=0

1

∫ ηdη

ΨA,bulk ζ( ) = 4 ΨA η,ζ( ) 1−η2{ }η=0

1

∫ ηdη

Finally, the quasi-macroscopic mass balance for heterogeneous catalytic reactors withfirst-order irreversible chemical reaction at the boundary, as described on pages 634-636(TPfCRD), is analyzed completely for uniform catalyst activity on the inner wall of tubesin Problem 23-6 on pages 647-648. Hence, the second equation on page#648 ofTPfCRD is dimensionalized as follows;

€

vz AverageπR2 −

dCA,bulk

dz

= 2πRk1,SurfaceCA r = R, z( )

−dΨA,bulk

dζ=2βPeMT

ΨA η =1,ζ( )

These equations are analyzed via the following finite-difference algorithmthat can be implemented in conjunction with a linear equation solver. Anonlinear equation solver is required if the chemical kinetics are not first-order.

Important parameters that govern the solution to the convective diffusion mass transferequation for laminar flow tube-wall reactorsβ = 150 Damköhler number; heterogeneous reaction rate wrt diffusion ratePeMT = 25 Mass transfer Peclet number; rate of convection wrt diffusion rate


162

Numerical grid parameters that determine the total number of grid points and mesh sizeNR = 101 number of discretized points in the radial directionΔη = 1/(NR–1) step size in the radial directionΔζ = 0.001 step increment in axial position

Establish the dimensionless inlet molar density profile of reactant AΨA(j,k=0) = 1; 1≤j≤ NR–1 no conversion in the feed stream at ζ=0ΨA(NR,k=0) = 1/(1+βΔη) approximate molar density at the wall via the BC at η=1

Evaluate the dimensionless laminar flow velocity profile at each radial mesh pointη(j) = (j–1)Δη; 1≤j≤NR

v*Z(j) = 2{1–[η(j)]2}

Initiate a counter and calculate the dimensionless axial positionk=1***ζ = kΔζ use a loop and return to this statement each time counter k is incremented

Symmetry boundary condition at the center of the tube (i.e., η=0)Second-order-correct forward difference representation for first derivatives, Eq. (23-35)

€

12Δη

−ΨA 3,k( ) + 4ΨA 2,k( )− 3ΨA 1,k( ){ } = 0

Radiation boundary condition at the catalytic wall (i.e., η=1)Second-order-correct backward difference representation for first derivatives, Eq. (23-40)

€

12Δη

3ΨA NR ,k( )− 4ΨA NR −1,k( ) +ΨA NR − 2,k( ){ } = −βΨA NR ,k( )

Implicit finite-difference representation of the convective diffusion mass transferequation within the tube; 1st derivative with respect to axial position ζ is first-ordercorrect; 1st and 2nd spatial derivatives with respect to radial position η are second-ordercorrect, Eq. (23-24)

€

2 ≤ j ≤ NR −1

PeMTvz∗ j( )

ΨA j,k( ) −ΨA j,k −1( )Δζ

=ΨA j +1,k( ) − 2ΨA j,k( ) + ΨA j −1,k( )

Δη( )2+ΨA j +1,k( ) −ΨA j −1,k( )

η j( )2Δη


163

Calculate the bulk molar density of reactant A via the trapezoidal rule

€

ΨA,bulk k( ) = 2 Δη2

2η j( )vz∗ j( )ΨA j,k( )j=2

NR −1

∑

Verify that the finite-difference solution of the microscopic convective diffusion equationalso satisfies the quasi-macroscopic mass balance, using the trapezoidal rule

€

ΨA,bulk k −1( )−ΨA,bulk k( )Δζ

=2βPeMT

ΨA NR ,k( )

Increment the counter, return to the step denoted by 3 asterisks ***, solve the systemof linear algebraic equations at the next axial step, calculate the bulk molar density ofreactant A and verify that the finite-difference solution also satisfies the quasi-macroscopic mass balancek = k+1Go To ***

Calculate the dimensionless tube length ζ = z/R that is required to achieve50% conversion of reactant A to products when the Damköhler number is150 and the mass transfer Peclet number is 25. Hint: Graph ΨA,bulk vs. ζ toobtain the answer.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.4

0.5

0.6

0.7

0.8

0.9

1.0

Dimensional Axial Position, zeta=z/RDim

ensio

nles

s B

ulk

Mol

ar D

ensi

ty

Heterogeneous Catalytic Tube-Wall Reactor; 1st-Order Kinetics

Damkohler Number = 150Mass Transfer Peclet # = 25

Axial "Conversion" Profile

50% Conversion when z/R = 3. 5


164

Predict the thickness of the mass transfer boundary layer δMTBLT(ζ), measuredinward from the catalytically active surface toward the centerline of thetube, as a fraction of the tube radius R when ζ = 1, β = 150, and PeMT = 25.Hint: Graph ΨA(η,ζ=1) vs. η. Within the mass transfer boundary layer;ΨA(R-δMTBLT,ζ=1) ≤ 0.98

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.00.10.20.30.40.50.60.70.80.91.0

Dimensionless Radial Position (r/R)

Dim

ensio

nles

s M

olar

Den

sity

Radial Profile; Beta=150, PeMT=25, zeta=1

Mass Transf er Boundary Layer Thickness

≈ 70% of the Tube Radius

Problem#5Develop a plug-flow version of the differential thermal energy balance that must besolved to calculate the temperature profile within a tubular reactor T(z) as a function ofindependent variable z which increases in the direction of flow of the reactive fluidthrough a straight cylindrical tube of length L and radius R. There is one pseudo-first-order homogeneous irreversible chemical reaction that converts reactants to products.The control volume is dV = πR2dz, the wall of the tube at radius r=R is not insulated fromthe surroundings, pressure effects are negligible, and the heat transfer Peclet number isonly 10 because the Reynolds number is very small in the creeping flow regime. Theunits of each term in the thermal energy balance should be "energy per volume pertime".___________________________________________________________________

Coupled Heat & Mass Transfer in Batch Reactorssee Chapter#6 in TPfCRD

Isothermal analysis of calorimetric rate dataFormalism for multiple chemical reactions


165

Adiabatic operation with one chemical reactionComparison of adiabatic temperature rise

Non-isothermal analysis of a constant volume batch reactor

Batch Reactor Problem

Conversion vs. time for a variable-volume batch reactor thatproduces methanol in the gas phase at constant T and p

A stoichiometric feed of carbon monoxide and hydrogen is injected into a batchreactor that operates isothermally at 298K and isobarically at low pressure (i.e., 50 torr).The overall objective is to produce methanol, and simulate the time dependence of theconversion of the key limiting reactant, carbon monoxide. This gas phase reaction iselementary, reversible, and it requires low operating pressures to quench the initiation ofundesirable side reactions. The normal boiling point of methanol is 338K at 1atmosphere pressure. Of course, methanol boils at a much lower temperature when thepressure is 50 torr, so it is acceptable to operate the reactor at 298K and be assuredthat methanol remains in the gas phase during the course of the reaction. At 298K and50 torr, the compressibility factor for the gas mixture is Z = 0.1 at 298K and 50 torr.The kinetic rate constant for the forward reaction, based on partial pressures in the gasphase, is described by the following Arrhenius parameters; the pre-exponential factor is2x105 g-mol/cm3-min-(atm)3 and the activation energy divided by the gas constant R is5000K. Thermodynamic data at 298K reveal that the standard state free energy offormation is –32,800 cal/g-mol for carbon monoxide, -38,700 cal/g-mol for methanol,and 0 for hydrogen because gaseous H2 is the standard state for hydrogen at 298K. Thefeed to this batch reactor contains 5 grams of gaseous carbon monoxide. Useful valuesof the universal gas constant R are 1.987 cal/mol-K and 0.082 litre-atm/mol-K.

(a) Generate graphs that illustrate the time dependence of the following quantitiesfor the operating conditions described above: (i) conversion of carbon monoxide,(ii) mole fraction of each component in the gas mixture, and (iii) reactor volume.

(b) How long must the batch reactor operate at 298K and 50 torr to achieve 50%conversion of carbon monoxide?

(c) What is the maximum possible conversion of CO that can be achieved for theoperating conditions described above?

(d) Design the size of the reactor. Provide only one numerical answer in litres.


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(e) Demonstrate that it is possible to achieve 50% conversion of CO in half of thetime based on your answer from part (b), and quantitatively describe twodifferent sets of operating conditions that will achieve this goal.

(f) Qualitatively, describe at least 2 strategies that could be implemented toincrease the maximum possible conversion of CO in this batch reactor. Then,provide quantitative evidence that both strategies are feasible.

Polymath solution to this batch reactor problem for the production ofmethanol from a stoichiometric feed of carbon monoxide and hydrogend(x)/d(t)=kinfinity*exp(-Eactivation/T)*RxRate*ReactorVolume*(10**3)/InitialMolesAyCO=(1-x)/(1+ThetaH2-2*x)yH2=(ThetaH2-2*x)/(1+ThetaH2-2*x)yCH3OH=(x)/(1+ThetaH2-2*x)RxRate=(p**3)*(yCO*yH2**2-(p**(-2))*yCH3OH/Kequilibrium)Kequilibrium=exp(5900/(1.987*T))ReactorVolume=(0.1*0.082*T/p)*InitialMolesA*(1+ThetaH2-2*x)kinfinity=200000Eactivation=5000ThetaH2=2InitialMolesA=5/28T=298p=50/760

Problem focusing on the microscopic approach to mass transfer withchemical reaction in tubular reactorsThe appropriate description of a realistic mass transfer problem is posed in terms of thefollowing second-order, non-linear, partial differential equation with variable coefficients;

Vz, max 1 −

rR

2

∂ CA∂ z

= D AB

∂2 CA

∂ r2 + 1r ∂ CA∂ r

+ kr(CA)

32

1 2 3 4

a) Identify a mass transfer mechanisms for each of the four terms in theequation

b) Is the mass transfer model written explicitly for a reactant or a product?

c) What is the apparent order of the chemical reaction?


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d) Is the reaction reversible or irreversible?

e) Does the mass transfer equation represent a plug-flow model?

f) Write an expression for the control volume, dV = ?, that was used to generatethe equation above.

g) In what coordinate direction or directions must the control volume bedifferentially thick to obtain the governing mass balance?

h) Will residence time distributions affect the performance of this model in a tubularreactor? Provide a brief explanation.

Problem on Transport AnalogiesComplete the table below which focuses on analogies between heat, mass, andmomentum transport for non-reactive systems

Momentum Transport Heat Transfer Mass Transfer

Re is the dimensionlessscaling factor in the ? ?Equation of Motion

The friction factor vs. Reis a dimensionless correlationobtained by focusing on the ? ?interface and calculating forcesthat are exerted by the fluid ona stationary solid surface

? ? Schmidt Number

? ? Peclet Number

? Fourier's Law of Heat ?Conduction


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