Chapter 6 SOLUTION OF VISCOUS-FLOW PROBLEMS

1 Chapter 6 SOLUTION OF VISCOUS-FLOW IntroductionTHE previous Chapter contained derivations of the relationships for the con-servation of mass and momentum|theequations of motion|in rectangular,cylindrical, and spherical coordinates. All the experimental evidence indicatesthat these are indeed the most fundamental equations of uid mechanics, and thatin principle they govern any situation involving the ow of a Newtonian , because of their all-embracing quality, their SOLUTION in analyticalterms is di cult or impossible except for relatively simple situations. However,it is important to be aware of these \Navier-Stokes equations," for the followingreasons:1. They lead to the analytical and exact SOLUTION of some simple, yet importantproblems, as will be demonstrated by examples in this They form the basis for further work in other areas of chemical If a few realisticsimplifying assumptionsare made, they can often lead toapproximate solutions that are eminently acceptable for many engineeringpurposes.

2 Representative examples occur in the study of boundary layers,waves, lubrication, coating of substrates with lms, and inviscid (irrotational) With the aid of more sophisticated techniques, such as those involving powerseries and asymptotic expansions, andparticularlycomputer-implemented nu-merical methods, they can lead to the SOLUTION of moderately or highly ad-vanced PROBLEMS , such as those involving injection-molding of polymers andeven the incredibly di cult problem of weather following sections present exact solutions of the equations of motion forseveral relatively simple PROBLEMS in rectangular, cylindrical, and spherical coor-dinates. Throughout, unless otherwise stated, the ow is assumed to besteady,laminarandNewtonian,withconstan tdensity and viscosity. Although these as-sumptions are necessary in order to obtain solutions , they are nevertheless realisticin many of the examples in this Chapter are characterized by low Reynolds is, the viscous forces are much more important than the inertial forces, |Introduction273are usually counterbalanced by pressure or gravitational e ects.

3 Typical applica-tions occur at low ow rates and in the ow of high-viscosity polymers. Situationsin which viscous e ects are relativelyunimportantwill be discussed in Chapter generalprocedurefor solving each problem in-volves the following steps:1. Make reasonable simplifying assumptions. Almost all of the cases treatedhere will involvesteady incompressible owofaNewtonian uid in asinglecoordinate direction. Further,gravitymay or may not be important, and acertain amount ofsymmetrymay be Write down theequations of motion|both mass (continuity) and momentumbalances|and simplify them according to the assumptions made previously,striking out terms that are zero. Typically, only a very few terms|perhapsonly one in some cases|will remain in each di erential equation. The simpli- ed continuity equation usually yields information that can subsequently beused to simplify the momentum simpli ed equations in order to obtain expressions for the de-pendent variables such as velocities and pressure.

4 These expressions will usu-ally contain some, as yet, arbitrary constants|typicallytwofor the velocities(since they appear in second-order derivatives in the momentum equations)andonefor the pressure (since it appears only in a rst-order derivative).4. Invoke theboundary conditionsin order to evaluate the constants appearingin the previous step. Forpressure,such a condition usually amounts to aspeci ed pressure at a certain location|at the inlet of a pipe, or at a freesurface exposed to the atmosphere, for example. For thevelocities, theseconditions fall into either of the following classi cations:(a) Continuity of the velocity, amounting to ano-slipcondition. Thus, thevelocity of the uid in contact with a solid surface typically equals thevelocity of that surface|zero if the surface is , for thefew cases in which one uid (A, say) is in contact with another immiscible uid (B), the velocity in uid A equals the velocity in uidBat thecommon interface.

5 (b) Continuity of the shear stress, usually between two uidsAandB, leadingto the product of viscosity and a velocity gradient having the same valueat the common interface, whether in uidAorB. If uidAis a liquid, and uidBis a relatively stagnant gas, which|because of its low viscosity|is incapable of sustaining any signi cant shear stress, then the commonshear stress is e ectively At this stage, the problem is essentially solved for the pressure and , if desired, shear-stress distributions can be derived by di erentiating1In a few exceptional situations there may be lack of adhesion between the uid and surface, in which caseslipcan 6| SOLUTION of VISCOUS-FLOW Problemsthe velocities in order to obtain the velocitygradients; numerical predictionsof process variables can also be broad classes of viscous ow will be illustrated in ow,in which an applied pressure di erence causes uid motionbetween stationary ow,in which a moving surface drags adjacent uid along with it andthereby imparts a motion to the rest of the , it is possible to have both types of motion occurring simultaneously,as in the screw extruder analyzed in Example SOLUTION of the Equations of Motion in Rectangular CoordinatesThe remainder of this Chapter consists almost entirely of a series of workedexamples, illustrating the above steps for solving viscous - ow |Flow Between Parallel PlatesFig.

6 Shows the ow of a uid of viscosity , which ows in thexdirectionbetween two rectangular plates, whose width is very large in thezdirection whencompared to their separation in theydirection. Such a situation could occur in adie when a polymer is being extruded at the exit into a sheet, which is subsequentlycooled and solidi ed. Determine the relationship between the ow rate and thepressure drop between the inlet and exit, together with several other quantities plateBottom plateInletExitFluidxyzCoordinate systemLargewidthNarrowgap, 2dFig. Geometry for ow through a rectangular duct. Thespacing between the plates is exaggerated in relation to their situation is analyzed by referring to a crosssection of the duct, shown in Fig. , taken at any xed value ofz. Let thedepth be 2d( dabove and below the centerline or axis of symmetryy= 0), andthe lengthL. Note that the motion is of thePoiseuilletype, since it is caused bythe applied pressure di erence (p1 p2).

7 Make the following realisticassumptionsabout the | SOLUTION of the Equations of Motion in Rectangular Coordinates2751. As already stated, it is steady and Newtonian, with constant density andviscosity. (These assumptions will often be taken for granted, and not restated,in later PROBLEMS .)2. There is only one nonzero velocity component|that in the direction of ow, ,vy=vz= Since, in comparison with their spacing, 2d, the plates extend for a very longdistance in thezdirection, all locations in this direction appear essentiallyidentical to one another. In particular, there is no variation of the velocity inthezdirection, so that@ Gravity acts vertically downwards; hence,gy= gandgx=gz= The velocity is zero in contact with the plates, so thatvx=0aty= by examining the general continuity equation, ( ):@ @t+@( vx)@x+@( vy)@y+@( vz)@z=0;(5:48)which, in view of the constant-density assumption, simpli es to Eqn.

8 ( ):@vx@x+@vy@y+@vz@z=0:(5:52)But sincevy=vz=0:@vx@x=0;(E6:1:1)sovxis independent of the distance from the inlet, and the velocity pro le willappear the same forallvalues ofx. Since@vx=@z= 0 (assumption 3), it followsthatvx=vx(y) is a function ofyonly. Axis ofsymmetryLExitVelocity profileInletWallWallxydd vx p 2p 1 Fig. Geometry for ow through a rectangular 6| SOLUTION of VISCOUS-FLOW ProblemsMomentum the stated assumptions of a Newtonian uidwith constant density and viscosity, Eqn. ( ) gives thex,y, andzmomentumbalances: @vx@t+vx@vx@x+vy@vx@y+vz@vx@z = @p@x+ @2vx@x2+@2vx@y2+@2vx@z2 + gx; @vy@t+vx@vy@x+vy@vy@y+vz@vy@z = @p@y+ @2vy@x2+@2vy@y2+@2vy@z2 + gy; @vz@t+vx@vz@x+vy@vz@y+vz@vz@z = @p@z+ @2vz@x2+@2vz@y2+@2vz@z2 + gz:Withvy=vz= 0 (from assumption 2),@vx=@x= 0 [from the simpli edcontinuity equation, ( )],gy= g,gx=gz= 0 (assumption 4), and steady ow (assumption 1), these momentum balances simplify enormously, to: @2vx@y2=@p@x;(E6:1:2)@p@y= g;(E6:1:3)@p@z=0:(E6:1:4)Pressure last of the simpli ed momentum balances,Eqn.

9 ( ), indicates no variation of the pressure across the width of the system(in thezdirection), which is hardly a surprising result. When integrated, thesecond simpli ed momentum balance, Eqn. ( ), predicts that the pressurevaries according to:p= gZdy+f(x)= gy+f(x):(E6:1:5)Observe carefully that since apartialdi erential equation is being integrated, weobtain not a constant of integration, but afunctionof integration,f(x).Assume|to be veri ed later|that@p=@xis constant, so that the centerlinepressure (aty= 0) is given by a linear function of the form:py=0=a+bx:(E6:1:6)The constantsaandbmay be determined from the inlet and exit centerline pres-sures:x=0:p=p1=a;(E6:1:7)x=L:p=p2=a +bL;(E6:1:8) | SOLUTION of the Equations of Motion in Rectangular Coordinates277leading to:a=p1;b= p1 p2L:(E6:1:9)Thus, the centerline pressure declineslinearlyfromp1at the inlet top2at the exit:f(x)=p1 xL(p1 p2);(E6:1:10)so that the complete pressure distribution isp=p1 xL(p1 p2) gy:(E6:1:11)That is, the pressure declines linearly, both from the bottom plate to the top plate,andalsofrom the inlet to the exit.

10 In the majority of applications, 2d L, andthe relatively small pressure variation in theydirection is usually ignored. Thus,p1andp2, although strictly thecenterlinevalues, are typically referred to astheinlet and exit pressures, pro , from Eqn. ( ),vxdoes not depend onx,@2vx=@y2appearing in Eqn. ( ) becomes atotalderivative, so this equationcan be rewritten as: d2vxdy2=@p@x;(E6:1:12)which is a second-order ordinary di erential equation, in which the pressure gra-dient will be shown to beuniformbetween the inlet and exit, being given by: @p@x=p1 p2L:(E6:1:13)[A minus sign is used on the left-hand side, since@p=@xis negative, thus renderingboth sides of Eqn. ( ) as positive quantities.]Equation ( ) can be integrated twice, in turn, to yield an expression forthe velocity. After multiplication through bydy, a rst integration gives:Zd2vxdy2dy=Zddy dvxdy dy=Z1 @p@x dy;dvxdy=1 @p@x y+c1:(E6:1:14)A second integration, of Eqn.