# EKC314: TRANSPORT PHENOMENA Core Course For B.Eng.(Chemical Engineering .

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EKC314: TRANSPORT PHENOMENACore Course forB.Eng.(Chemical Engineering)Semester II (2008/2009)Dr. Mohamad Hekarl Uzir-chhekarl@eng.usm.mySchool of Chemical EngineeringEngineering Campus, Universiti Sains MalaysiaSeri Ampangan, 14300 Nibong TebalSeberang Perai SelatanPenang.EKC314-SCE – p. 1/5

Syllabus1. Introduction and Concepts2. Momentum TransportViscosity and and mechanism of momentumtransportNewtonian and non-newtonian fluidsFlux and gradientsVelocity distribution in in laminar and turbulent flowsBoundary layerVelocity distribution with more than oneindependent variablesEKC314-SCE – p. 2/5

Syllabus3. Equation of changes in Isothermal Systems;Interphase transport in isothermal system.Macroscopic balances for isothermal systems.EKC314-SCE – p. 3/5

Introduction and ConceptsWhat are Transport Phenomena?1. Fluid dynamics2. Heat transfer3. Mass transferThey should be studied together since:1. they occur simultaneously2. basic equations that described the 3 transportphenomena are closely related3. mathematical tools required are very similar4. molecular mechanisms underlying various transportphenomena are very closely relatedEKC314-SCE – p. 4/5

Introduction and ConceptsMacroscopic levelWrite down balance equation;macroscopic balances.Equations should describe; mass,momentum, energy and angularmomentum within the system.No need to understand the details ofthe system.Microscopic levelExamination of the fluid mixture in asmall region within the equipment.Can write down the equation of changethat describe mass, momentum,energy and angular momentum changewithin the small regionAim is to get information aboutvelocity, temperature, pressure andconcentration profile within thesystem.Molecular levelTo seek a fundamentalunderstanding of themechanism of mass,momentum, energy and angularmomentum in terms ofmolecular structure andintermolecular forcesInvolve some theoreticalphysics and physical chemistryworkEKC314-SCE – p. 5/5

Introduction and ConceptsMixture of gasesMacroscopicHeat added tothe systemMolecularMicroscopicEKC314-SCE – p. 6/5

Introduction and ConceptsThe flow of fluid are studied in 3 different parts whichconsist of:flow of pure fluids at constant temperature-withemphasis on viscous and convective momentumtransportflow of pure fluids with varying temperature-withemphasis on conductive, convective and radiativeenergy transportflow of fluid mixtures with varying composition-withemphasis on diffusive and convective mass transportEKC314-SCE – p. 7/5

Introduction and ConceptsConservation Laws (mass):Consider two colliding diatomic molecules (N2 and O2 ).The conservation of mass can be written as;′′mN mO mN mOfor a system with no chemical reaction;′mN mNand′mO mOEKC314-SCE – p. 8/5

Introduction and ConceptsConservation Laws (momentum):According to the law of conservation of momentum, thesum of the momenta of all atoms before collision mustequal that after the collision.The conservation of momentum can be written as;mN1 ṙN1 mN2 ṙN2 mO1 ṙO1 mO2 ṙO2 m′N1 ṙ′N1 m′N2 ṙ′N2 m′O1 ṙ′O1 m′O2 ṙ′O2EKC314-SCE – p. 9/5

Introduction and ConceptsConservation Laws (momentum):rN1 is the position vector for atom 1 of molecule N andṙN1 is its velocity.It can be written as;r N1 r N RN1rN1 is the sum of the position vector for the centre ofmass and the position vector of the atom w.r.t thecentre of mass.EKC314-SCE – p. 10/5

Introduction and ConceptsConservation Laws (momentum):Also RN2 RN1Then the conservation equation can be simplified as;mN ṙN mO ṙO mN ṙ′N mO ṙ′OEKC314-SCE – p. 11/5

Introduction and ConceptsHow to study Transport Phenomena?Read the text with pencil and paper inhand-work through details of themathematical developmentsRefer back to any maths textbook justto brush up on culculus, differentialequations, vectors etc.Make a point to give a physicalinterpretation of key results-get intohabit of relating physical ideas to theequationsAsk whether results seem reasonable.If they do not agree with intuition, it isimportant to find out which is correctMake a habit to check the dimensionsof all results. This is a very good wayto locate errors in derivationsEKC314-SCE – p. 12/5

Introduction and ConceptsRecap: Vector CalculusIf given a vector of a form;a [a1 , a2 , a3 ] a1 i a2 j a3 kconsists of the real vector space R3 with vectoraddition defined as;[a1 , a2 , a3 ] [b1 , b2 , b3 ] [a1 b1 , a2 b2 , a3 b3 ]and scalar multiplication defined by;c[a1 , a2 , a3 ] [ca1 , ca2 , ca3 ]EKC314-SCE – p. 13/5

Introduction and ConceptsThe dot product of two vectors is defined by;a · b a b cosγ a1 b1 a2 b2 a3 b3where γ is the angle between a and b. This gives thenorm or length a of a with a formula;q a a · a a21 a22 a33If a · b 0 therefore a and b orthogonal.EKC314-SCE – p. 14/5

Introduction and ConceptsExample of a dot product is given by;W p·dwhich is work done by a force p in a displacement d.EKC314-SCE – p. 15/5

Introduction and ConceptsThe cross product v a b is a vector of length; a b a b sinγand perpendicular to both a and b such that a, b, vform a right-handed triple.This can also be written in the form of; i j ka b a1 a2 a3 b1 b2 b3The cross product is anticommutative a b b aand not associative.EKC314-SCE – p. 16/5

Introduction and ConceptsFor a vector function given byv(t) [v1 (t), v2 (t), v3 (t)] v1 (t)i v2 (t)j v3 (t)kThen the derivative is;dvv(t t) v(t)v lim t 0dt t′therefore;v′ [v1′ , v2′ , v3′ ] v1′ i v2′ j v3′ kEKC314-SCE – p. 17/5

Introduction and ConceptsVector function r(t) can be used to represent a curve Cin space.Then r(t) associates with each t t0 in some intervala t b the point of C with position vector r(t0 ).The derivative r′ (t) is a tangent vector of C.If a vector in a Cartesian coordinate is given by;v(x, y, z) [v1 (x, y, z), v2 (x, y, z), v3 (x, y, z)] v1 (x, y, z)i v2 (x, y, z)j v3 (x, y, z)kEKC314-SCE – p. 18/5

Introduction and ConceptsTherefore, the partial derivative of v can be obtainedby; v1 v2 v3 v ,, x x x x v2 v3 v1i j k x x xEKC314-SCE – p. 19/5

Introduction and ConceptsGradient of a function f can be written as; f f f,,grad f f x y zTherefore, using the operation, the directionalderivative of f in a direction of a unit vector b can beobtained by;dfDb f b · fdsEKC314-SCE – p. 20/5

Introduction and ConceptsDivergence of a vector function v can be written as; v1 v2 v3 div v · v x y zAnd the curl of v is given as; curl v v ijk x y zv1v2 v3 EKC314-SCE – p. 21/5

Introduction and ConceptsSome basic formula for grad, div and curl;Grad: (f g) f g g f (f /g) (1/g 2 )(g f f g)Div:div(f v) f div v v · fdiv(f g) f 2 g f · gEKC314-SCE – p. 22/5

Introduction and ConceptsDiv and Grad (Laplacian): 2 f div( f ) 2 (f g) g 2 f 2 f · g f 2 gDiv and Curl:curl(f v) f v f curl vdiv(u v) v · curl u u · curl vEKC314-SCE – p. 23/5

Introduction and ConceptsExtra:curl( f ) 0div(curl v) 0EKC314-SCE – p. 24/5

Introduction and ConceptsExample 1:Let a particle A of mass M be fixed at point P0 and let aparticle B of mass m be free to take up various positions Pin space. Then A attracts B. According to Newton’s Lawof Gravitation, the corresponding gravitational force p isdirected from P to P0 and its magnitude is proportional to1, where r is the distance between P and P0 . This can ber2written as;c p 2rEKC314-SCE – p. 25/5

Introduction and ConceptsHence p defines a vector field in space. Using Cartesiancoordinates, such that P0 (x0 , y0 , z0 ) and P (x, y, z), thereforethe distance r can be determined as;pr (x x0 )2 (y y0 )2 (z z0 )2For r 0 in vector form;r [x x0 , y y0 , z z0 ] (x x0 )i (y y0 )j (z z0 )kEKC314-SCE – p. 26/5

Introduction and ConceptsTherefore; r rand ( 1/r)r is a unit vector in the direction of p. The minussign indicates that p is directed from P to P0 . Thus, 1cx x0y y0z z0p p r 3 r ci c 3 j c 3 k3rrrrrEKC314-SCE – p. 27/5

Introduction and ConceptsExample 2:From the former example, by the Newton’s Law ofGravitation, the force of attraction between 2 particles isgiven as; x x0y y0z z0ci j kp 3 r c333rrrrwhere r is a distance between two particles P0 and P ofthe given coordinates. Thus,pr (x x0 )2 (y y0 )2 (z z0 )2EKC314-SCE – p. 28/5

Introduction and ConceptsThe important observation now is 1 2(x x0 )x x0 2223/2 x r2[(x x0 ) (y y0 ) (z z0 ) ]r3Similarly; y y0 1 3 y rrand 1z z0 3 z rrEKC314-SCE – p. 29/5

Introduction and ConceptsFrom here, it is observed that p is the gradient of the scalarfunction;cf (x, y, z) rand f is a potential of that gravitational field. ApplyingLaplace Equation of the form; 2f 2f 2f 2 2 02 x y zEKC314-SCE – p. 30/5

Introduction and ConceptsApplying this to the unit vector, gives for the individualcomponent; 2 113(x x0 )2 3 2 xrrr5 3(y y0 )211 3 2 yrrr5 2 113(z z0 )2 3 2 zrrr52EKC314-SCE – p. 31/5

Introduction and ConceptsCombining the 3 gives; 2 22 c 2 2 02 x y zror it can be written in the form of; 2 f 0which is normally termed as the Laplacian operator of afunction.EKC314-SCE – p. 32/5

Introduction and ConceptsExample 3:From Example 2, it is known that the gravitational force p isthe gradient of the scalar function f (x, y, z) rc whichsatisfies Laplace’s equation 2 f 0. Therefore,div p 0for r 0.EKC314-SCE – p. 33/5

Introduction and ConceptsExample 4:Consider a motion of a fluid in a region R having nosources and sinks in R (no points at which the fluid isproduced or disappeared). Assuming that the fluid iscompressible and it flows through a small rectangular boxW of dimension x, y and z with edges parallel to thecoordinate axes. W has a volume of V x y z.Let v is the velocity vector of the form given by;v [v1 , v2 , v3 ] v1 i v2 j v3 kEKC314-SCE – p. 34/5

Introduction and ConceptsSettingu ρv [u1 , u2 , u3 ] u1 i u2 j u3 kwhere ρ is the density of the fluid. Consider the flow of fluidout of the box W is through the left face whose area is x z and the components v1 and v3 are parallel to thatface and contribute nothing to that flow. Thus, the mass offluid entering through that face during a short time interval t is approx. given by;(ρv2 )y x z t (u2 )y x z tEKC314-SCE – p. 35/5

Introduction and ConceptsAnd the mass of fluid leaving the opposite face of the boxW, during the same time interval is approx. given by;(u2 )y y x z tTherefore, the difference is in the form of; u2 u2 x z t V t ywhere u2 (u2 )y y (u2 )yEKC314-SCE – p. 36/5

Introduction and ConceptsTwo other pairs in x and z directions are taken andcombined in the form given by; u1 u2 u3 V t x y zThe loss of mass in W is caused by the rate of change ofthe density and therefore equals to; ρ V t tEKC314-SCE – p. 37/5

Introduction and ConceptsEquating both equations gives; u1 u2 u3 ρ V t V t x y z tDiving the above equation by V t and let x, y, zand t approaching 0 leads to; u1 u2 u3 ρ x y z tEKC314-SCE – p. 38/5

Introduction and ConceptsOr in its simplified form; ρdiv u tor ρdiv(ρv) twhich consequently becomes; ρ div(ρv) 0 twhich is the condition for the conservation of mass orthe continuity equation of a compressible fluid flow!EKC314-SCE – p. 39/5

Introduction and ConceptsIf the flow is steady, or it is independent of time, thus; ρ 0 tand therefore the continuity equation reduces intodiv(ρv) 0If the density ρ is constant, which means the fluid isincompressible, therefore the continuity equation becomes;div v 0EKC314-SCE – p. 40/5

Viscosity and the Mechanisms ofMomentum TransportConsider a parallel plates with area A separated bydistance Y with a type of fluid;The system is initially at restAt time t 0, the lower plate is set in motion in thepositive x direction at a constant velocity V .As time proceeds, the fluid gains momentum and thelinear steady-state velocity is established.At steady motion, a constant force, F is required tomaintain the motion of the lower plate and this can beexpressed as;VF µAYEKC314-SCE – p. 41/5

Viscosity and the Mechanisms ofMomentum TransportYt 0Fluid initially at restt 0Lower plate set in motionVvx(y,t)Small tVelocity build up in unsteady flowLarge tFinal velocity distribution in steady flowVvx(y)yxVEKC314-SCE – p. 42/5

Viscosity and the Mechanisms ofMomentum TransportConsider a parallel plates with area A separated bydistance Y with a type of fluid;Force in the direction of x perpendicular to the ydirection given by F/A is replaced by τyx .The term V /Y is replaced by dvx /dy and thereforethe equation becomes;τyxdvx µdyThis means that the shearing force per unit area isproportional to the negative of the velocity gradient i.e.Newton’s Law of Viscosity.EKC314-SCE – p. 43/5

Viscosity and the Mechanisms ofMomentum TransportSimilarly, let the angle between the fixed plate and themoving boundary δγ and the distance from the origin isδx.Therefore,δxtan δγ Yfor a very small angle,δx δγYEKC314-SCE – p. 44/5

Viscosity and the Mechanisms ofMomentum TransportButδx V δtthus,V δtδγ YTaking limit on on the above terms;Vdvδγ limδt 0 δtYdyEKC314-SCE – p. 45/5

Viscosity and the Mechanisms ofMomentum TransportHence;δγdv δtdyButThus,γ̇ τdvτ dyEKC314-SCE – p. 46/5

Viscosity and the Mechanisms ofMomentum TransportWhich then becomesdvτ µdywith µ as the proportionality constant.The minus sign represents the force exerted by thefluid of the lesser Y on the fluid of greater Y.EKC314-SCE – p. 47/5

Viscosity and the Mechanisms ofMomentum TransportτShear thinningNewtonianShear thickeningdγ/dtEKC314-SCE – p. 48/5

Viscosity and the Mechanisms ofMomentum TransportGeneralisation of Newton’s Law of Viscosity.Consider a fluid moving in 3-dimensional space withrespect to time t.Therefore, the velocity components are given asfollows;vx vx (x, y, z, t) vy vy (x, y, z, t) vz vz (x, y, z, t)In the given situation, there will be 9 stresscomponents, τij .EKC314-SCE – p. 49/5

Viscosity and the Mechanisms ofMomentum TransportBased on the given diagram:The pressure force-always perpendicular to theexposed surface.Thus, the force per unit area on the shaded surface willbe a vector pδx (pressure multiplied by the unit vectorδx in the x direction)Same goes to the y and z directions.EKC314-SCE – p. 50/5

Viscosity and the Mechanisms ofMomentum TransportThe velocity gradients within the fluid are neitherperpendicular to the surface element nor parallel to it,but rather at some angle to the surface.The force per unit area, τx exerted on the shaded areawith components (τxx , τxy and τxz ).This can be conveniently represented by standardsymbols which include both types of stresses(pressure and viscous stresses);πij pδij τijwhere i and j may be x, y or z.EKC314-SCE – p. 51/5

Viscosity and the Mechanisms ofMomentum TransportHere, δij is the Kronecker delta, which is 1 if i j andzero if i 6 j.The term πij can be defined as;force in the j direction on a unit area perpendicularto the i direction, where it is understood that thefluid in the region of lesser xi is exerting the forceon the fluid of greater xi .flux of j-momentum in the positive i direction – thatis, from the region of lesser xi to that of greater xi .EKC314-SCE – p. 52/5

Viscosity and the Mechanisms ofMomentum TransportSummary of the components of the molecular stresstensor.Dir.Components of forcesVector forcex-componenty-componentz-componentxπx pδx τxπxx p τxxπxy τxyπxz τxzyπy pδy τyπyx τyxπyy p τyyπyz τyzzπz pδz τzπzx τzxπzy τzyπzz p τzzEKC314-SCE – p. 53/5

What are Transport Phenomena? 1. Fluid dynamics 2. Heat transfer 3. Mass transfer They should be studied together since: 1. they occur simultaneously 2. basic equations that described the 3 transport phenomena are closely related 3. mathematical tools required are very similar 4. molecular mechanisms underlying various transport

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