KUNDU Fluid Mechanics 2 Episode 3 ppt

64 Kmematira A P mR m- B QC B' Q' C' 1 Ngurc3.13 DcformationoPclcmenllrina~lelshearflow. Theelement is strctchcdalongtheprincipal axis i1 and compressed along the principal axis &. -y, and that of BC is zero, giving -y/2 as the overall angular velocity (half the vorticity). The average value does not depend on which two mutually perpendicular elements in the x1 x2-plane are chosen to compute it. In contrast, the components of strain rate do depend on the orientation of the element. From Eq. (3.1 l), the strain rate tensor of an element such as ABCD, with the sides parallel to the XI XZ-axes, is 0 iyo e=[ iy 0 01, 0 00 which shows that there are only off-diagonal elements of e. Therefore, the element ABCD undergoes shear, but no normal strain. As discussed in Chapter 2, Section 12 and Example 2.2, a symmetric tensor with zero diagonal elements can be diagonalized by rotating the coordinate system through 45". It is shown there that, along these principal axes (denoted by an overbar in Figure 3.13), the strain rate tensor is 00 00 e= so that there is a linear extension rate of Z1 1 = y/2, a linear compression rate of E= = - y/2, and no shear. This can be understood physically by examining the deformation of an element PQRS oriented at 45", which deforms to P'Q'R'S'. It is clear that the side PS elongates and the side PQ contracts, but the angles between the sides of the clement remain 90'. In a small time interval, a small spherical element in this flow would become an ellipsoid oriented at 45" to the XI x2-coordinate system. Smmurizing, the element ABCD in a parallel shear flow undergoes only shear but no normal strain, wherear the element PQRS undergoes only normal but no shear strain. Both of these elements rotate at the same angular velocity. 1 I. Kinetnacic: Conxideratiom OJ Vorlm Flows Flows in circular paths arc callcd vot-texflows, some ba,ic forms of which are described in what lollows. Solid-Body Rotation Consider first the case in which the velocity is proportional to thc radius of the stream- lines. Such a flow can be generated by steadily rotating a cylindrical tank containing a viscous fluid and waiting until the transients die out. Using polar coordinates (rr H), the velocity in such a flow is ug u, =o, (3.27) where 041 is a constant equal to thc angular vclocity of revolution of each particle about the origin (Figure 3.14). We shall scc shortly that fN is also equal to the angular specd of roturion of each particle about its own center. The vorticity cornponcnts of a fluid clcment in polar coordinates are given in Appendix B. The component about thc z-axis is (3.28) 1. a 1 au, r ar r aH w: = (rue) - = 2~, whcrc wc' havc used thc vclocity distribution cquation (3.27). This shows that the angular velocity or each fluid element about its own centcr is a constant and qual to wg. This is evident in Figure 3.14, which shows the location af element ABCD at two succcssivc timcs. It is sccn that thc two mutually perpcndicular Ruid lincs AD and AB both rotatc countcrclockwisc (about the center ofthe elcment) wih speed q-,. Figure 3.14 Solid-hody rotation. Muid clcmcnls arc spinning about thcir own ccnkrs while they revolvc around the origin. There is no dcli)rmalion or the elements. 66 Kinemutiwv The time period for one mtation of the particle about its own center equals the time period for one revolution around the origin. It is also clear that the deformation of the fluid elements in this flow is zero, as each fluid particle retains its location relative to other particles. A flow defined by ue = wr is called a sok-body rotation as the fluid elements behave as in a rigid, rotating solid. The circulation around a circuit of radius r in this flow is 2 I’ = u=ds = uerde = 2arus = 2nr 00: s 1” (3.29) which shows that circulation equals vorticity 200 times area. It is easy to show (Exercise 12) that this is true of my contour in the fluid, regardless of whether or not it contains the center. Irrotational vortex Circular streamlines, however, do not imply that a flow should have vorticity every- where. Consider the flow around circular paths in which the velocity vector is tan- gential and is inversely proportional to the radius of the streamline. That is, C r ue = - u, =o. (3.30) Using Q. (3.28), the vorticity at any point in the flow is 0 r 0- = This shows that the vorticity is zero everywhere except at the origin, where it canuot be determined from this expression. However, the vorticity at the origin can be deter- mined by considering the circulation around a circuit enclosing the origin. Around a contour of radius r, the circulation is I’ = 6” uer de = 2aC. This shows that r is constant, independent of the radius. (Compare this with the case of solid-body rotation, for which Eq. (3.29) shows that I‘ is proportional to r2.) In fact, the circulation around a circuit of any shape that encloses the origin is ~JcC. Now consider the implication of Stokes’ theorem (3.31) for a contour enclosing the origin. The left-hand side of Eq. (3.31) is nonzero, which implies that o must be nonzero somewhere within the area enclosed by the contour. Because r in this flow is independent of r, we can shrink the contour without altering the left-hand side of Eq. (3.31). In the limit the area approaches zero, so that the vorticity at the origin must be infinite in order that o SA may have a finite nonzero limit at the origin. We have therefore demonstrated that thcjhw represented by (7 u e- r Figure 3.15 where e: se. Irrotational vortex. Vorticity of a Ruid element is iniinite at the origin and zero every- ue = C/r is irrotutional everywhere except at th.e origin, where the vortici1.y is iqlinire. Such a flow is called an imtatianul or potentiul vortex. Although the circulation around a circuit containing the origin in an irrotational vortex is nonzero, that around a circuit not contaiajng the originis zero. The circulation around any such conlour ABCD (Figure 3.15) is Because thc linc intcgrals of u ds around BC and DA are 72~0, wc obtain FAUCI) = -uor A0 + (ug + Auo)(r + Ar) AQ = 0, where we have noted that thc line integral along AB is negative bccause u and ds arc oppositcly directed, and we have used ugr = const. A zero circulation around ABCD is expected becausc of Stokes' theorem, and the fact that vorticity vanishes everywhere within ABCD. Rankine Vortex Real vortices, such as a bathtub vortex or an atmospheric cyclone, havc a core thal rotates nearly likc a solid body and an approximately irrotational far field (Figure 3.16a). A rotational core must exist bccduse the tangential vclocity in an irrotational vortcx has an infinite velocity jump at the origin. An idcalizalion of such a behavior is called the Runkine vortex, in which the vorticity is assumed uniform within a corc ol'radius R and zero outside the core (Figurc 3.16b). Figure 3.16 Vclocity and vorticiy distributions in a rcal vortex and a Rankine vorh: (a) real vorm; (b) Rankine vortex. 12. Om-, lluo-, and ~!e-l)imc?n~ional~~ow~ A truly one-dimensional Jlow is one in which all flow characteristics vary in one direction only. Few real flows are strictly one dimensional. Consider the flow in a conduit (Figure 3.17a). The flow characteristics here vary both along the direction of flow and over the cross section. However, for somc purposes, the analysis can be simpliiied by assuming that the flow variables are uniform over the cross section (Figure 3.1 7b). Such a simplification is called a one-dimensional approximation, and is satisfactory if one is interested in the overall effects at a cross section. A t wo-dimensional or plane flow is one in which the variation of flow charac- teristics occurs in two Cartesian directions only. The flow past a cylinder of arbikary cross section and infinite length is an example of plane flow, (Note that in this contcxt the word “cylinder” is used for describing any body whosc shape is invariant along the length of the body. It can have an arbitrary cross section. A cylinder with a circlslar (a) 0 E’igure 3.17 Flow through a conduit and its onc-dimcnsional approximation: (a) real flow; (b) onedimensional approximation. cross section is a special case. Sometimes, however, the word “cylinder” is used to dcswibe circular cylinders only.) Around bodies of revolution, the flow variables are identical in planes containing the axis of the body. Using cylindrical polar coordinatcs (R, q, x), with x along the axis of the body, only two coordinates (R and x) are neccssary to describe motion (see Figure 6.27). The flow could therefore be called “two dimensional” (although not plane), but it is customary to describe such motions as three-dimensional uxispunerric JlOlVS. 13. The Mmzmjmclion The description of incompressible two-dimensional flows can be considerably sim- plified by dcfining a function that satisfies the law of conscrvation of mass for such flows. Although the conservation laws are derived in the following chapter, a simple and allernalive derivation of the mass conservation equation is given here. We proceed from thc volumetric slrain rate given Eq. (3.9), namely, The D/.; signifies that a specific fluid particle is followed, so thal the volume of a particle is inversely proportional to its density. Subslituling 6T o( p-’ , we obtain (3.32) This is called the CCJ~ZU~Q equation because it assumes that the fluid flow has no voids in it; the name is somewhat mislcading because all laws of continuum mechanics makc this assumption. The density of Ruid particles docs not change appreciably along the fluid path under certain conditions, the most importanl of which is that the flow spccd should be small compared with the spccd of sound in the medium. This is callcd the Boussinesq approximation and is discussed in more detail in Chapter 4, Section 18. The condition holds in most flows of liquids, and in flows of gases in which the speeds are less than 70 KILCIIZ~GB about 100m/s. In these flows p-' Dp/Dt is much less than any o€lhe derivatives in aui laxi, under which condition the conlinuity equation (steady or unsteady) becomes pq - =o. In many cases the continuity equation consists of two terms only, say au av ax ay -+-=o. (3.33) This happens if w is not a function of E. A planc flow with tu = 0 is the most common example of such two-dimensional flows. If a function +(x, y, t) is now defined such that a@ aY u -, (3.34) a$ ax ' v E then Eq. (3.33) is automatically satisfied. Therefore, a streamfunction JI can be defined whenever Eq. (3.33) is valid. (A similar streamfunction can be defincd for incom- pressible misymmetric flows in which the continuity equation involvcs R and x coor- dinates only; for compressible flows a streamfunction can be defined if the motion is two dimensional und steady (Exercisc 2).) The streamlines of the flow are given by dx dy = - uv Substitution of Eq. (3.34) into Eq. (3.35) shows (3.35) which says that d@ = 0 along a streamline. The instanlaneous streamlines in a flow are therefore givcn by the curves @ = const., a different value of the constant giving a different streamline (Figure 3.18). Consider an arbitrary line element dx = (dx, dy) in the flow of Figure 3.18. Here we have shown a case in which both dx and dy are positive. The volume rate of flow across such a line element is showing that the volume flow rate between a pair of streamlines is numerically equal to the difference in their + values. Thc sign oF $ is such that, facing the direction of motion, II. increases to the left. This can also be seen hm the defmition equation (3.34), according to which the dcrivative of @ in a certain direction gives the vclocity Figure 3.18 Flow thmugh a pair of streamlines. component in a direction 90" clockwise from the direction of differentiation. This requires that e in Figure 3.18 must increase downward if the flow is from right to left. One purpose of defining a streamfunction is to be able to plot streamlines. A more theoretical reason, however, is that it decreases thc number of simultaneous equations to be solved. For example, it will be shown in Chapter 10 that the momentum and mass conservation equations for viscous flows near a planar solid boundary are given, respectively, by au au a2u u- + u- = u-: ax ay ay2 au av ax ay -+-=o. (3.36) (3.37) The pair of simultaneous equations in u and u can be combined into a single equation by defining a streamfunction, when the momentum equation (3.36) becomes a+ a2$ a+a2$ a3$ ay axay ax ay2 ay3 = v We now have a single unknown function and a single differential equation. The continuity equation (3.37) has been satisfied automatically. Sum.m.arizing, a streamfunction can be defined whenever the continuity equation consists of two ms. The flow can otherwise be completely general, for example, it can be rotational, viscous, and so on. The lines $ = C are thc instantaneous streamlines, and the flow rate between two streamlines equals d@. This concept will be generalized following our derivation of mass conservation in Chapter 4, Section 3. 14. I'olur Cmrdinatca It is sometimes easier to work with polar coordinates, especially in problems involv- ing circular boundaries. In fact, we often select a coordinate system to conform to the shape of the body (boundary). It is customary to consult a reference source for expressions of various quantities in non-Cartesian coordinates, and this practice is perfectly satisfactory. However, it is good to know how an equation can be trans- formed from Cartesian into other coordinates. Here, we shall illustrate the procedure by transforming the Laplace equation to plane polar coordinates. Cartesian and polar coordinatcs are related by x = r cose y = r sin e H = tan-'(.y/x), r = ,/=. (3.38) Let us fust determine the polar velocity components in terms a€ the streamfunction. Bccause rl. = f(x, y), and x and y are themselves functions of r and 8, the chain rule of partial differentiation gives (Z)* = (:)y (E)() + ($)x ($)/ Omitting parentheses and subscripts, we obtain (3.39) Figure 3.19 shows that ug = vcose - u sine, so that Eq. (3.39) implies a$/&- = -u6. Similarly, we can show that a$/% = Tur. Themfore, the polar velocity components are related to the streamfunction by 1 a$ r ae' w ar ur = ug = This is in agreement with our previous observation that the derivative of $ gives the velocity component in a direction 90 clockwise €om the direction of differentiation. Now let us write the Laplace equation in polar coordinatcs. The chain rule gives a* a+ar a*ase a$ shea$ -= + =case- - ax ar ax ae ax ar r ae' Differentiating this with respect to x, and following a similar rule, we obtain [ a$ sinO;;]. case- - ar r (3.40) X Hpre 3.19 Relation d vclocily components in Cartesian and plane polar coordinates. Tn a similar manncr, (3.41) The addition of Eqs. (3.40) and (3.41) leads to which completes the transformation. lhmises 1. A two-dimensional steady flow has velocity components u=y v=x. Show that thc streamlines are rectangular hyperbolas 2 x - y2 = const. Sketch the flow pattern, and convince yourself that it represents an irrotational flow in a 90" comer. 2. Consider a steady axisymmetric flow of a compressible fluid. The equation of continuity in cylindrical coordinates (R, p: x) is [...]... are shown u r21+7,dXZ l 1h2 2 1 h2-7 & 1 Figure 46 Torqw on an clcmcnt + centraid axis 4 " 86 Cunrrpsva~ion Iuuw rotational equilibrium then requires (ti2 + P - t 2 1 ) dxl d ~ = - dxl d ~ z ( d x ? dx;) h3, 2 12 that is, t 12 P - t21 = -(dxl 2 12 + d ~2 2& ) As dxl and dx2 go to zero,the preceding condition can be satisfied only if t l 2 = t21 In general, El tij =tji (4.14) See Exercise 3 at the end... w = x 2 z = x2y, Consider the fluid region inside a spherical volume x 2 validity of Gauss’ theorem + y2 + z2 = u2 Verify the by inlegrating over the sphere 10 Show that the vorticity field for any flow satisfies v*w=o 1 1 A flow field on h e xy-plane has the velocity components u=3x+y u=2x-3y + Show that the circulation around the circle (x - 1 )2 (y - 6 )2 = 4 is 417 12 Consider the solid-body rotation... Pedlosky (1987), and Holton (1979) Consider (Figure 4. 12) a frame of reference ( X I , x2, x3) rotating at a uniform , angular velocity 51 with respect to a fixed frame ( X IXzl X3) Any vector P is represented in the rotating frame by n Figure 4. 12 Coordinate liamc (XI x2 X:d 3 ( X I x2 x3) rotating at angular velocity S with respect to a fixcd frame 2 ... (Figure 3 la), assuming thatuo=Uatr=R 8 Take a planc polar elemcnt of fluid of dimensions dr and r de Evaluate the right-hand si& of Stokes’ t e r m hoe and thcreby show that the expression for vorticity in polar coordinates is Also, find the cxpressions for w, and we in polar coordinates in a similar manner 9 The velocity field of a certain flow is givcn by u = zr1 .2 + 2xz2, 2: w = x 2 z = x2y, Consider... x-component of the momentum principle (4 .22 ) reduces to D = &Iou', (4 . 23 ) where f i j O u l is the net outflow rate of x-momentum through the boundaries of the region There is no flow of momentum through the central hole in Figure 4.8 Outflow rates of x-momentum through PS and QR are -1, b ME = U,(pU,dy) Lb 1 (4 .24 ) h b hQR = = -2bpU&, u(pu d y ) = p -b u2 d y (4 .25 ) An important point is that there is... + A2 + & Here A 1 is the outer surface, A2 is wrappcd around the body like a tight-fittingrubber glove with dA2 pointing outwardsfrom the fluid volume and, therefore, into the body, and Ag is the connection surface between the outcr A, and thc inner A2 Now L (puu - t) dA3 +0 as A3 + 0, bccause il may be taken as the bounding surface of an cvancscent thread On the surfacc or a solid body, u d A 2 =... surface Here t dA2 is the rorce the body exerts on the fluid from our definition of t Then the force the fluid exerts on the body is Fe = - J, t - dAz =- J,,(puu - - dA, t) (4 .28 ) Using similar arguments, mass conscrvationcan be written in the form J,,pu.dA' =o (4 .29 ) Equations (4 .28 ) and (4 .29 ) can bc used to solve Example 4.1 Of course, the same final result is obtained when t 2 conslant pressure... if Y (4 .36 ) =LL- Only two constants p and A, of the original 8 1 , have therefore survived under the restrictions of material isotropy and stress symmety Substitution of Eq (4 .35 ) into the constitutive equation (4 .34 ) gives aij =2 p e i j + kern,,, i j , S - where emrn V u is the volumetric strain rate (explained in Chapter 3, Scction 6) = The complete stress tensor (4 .33 ) then becomes rij + 2 p e i... 82 6 Stnx.sa~alhbit 84 7 Conwnwhn rfMomerilum 86 8 Womnhrn Pririujil(?Ji)r(1 Ikcd Ih!.urne 88 Examplc4.1 89 9 Angular :MommtumI’rincipleJi,r (I F h d K h n e 92 Kxamplt: 4 .2 93 IO Gn,s iru&icer;i t;) n.$r :Vewtoniun Fluid 94 Nori-R’cwtnnianFluids 97 1I :Vm*k4ok(!.!.u KqimLori 97 Cnmmrxitlls or1 the hcouii X:im 98 12 Ikituhg I k m e 99 Effm or C m i h g d I h t 1 02. .. (4 .20 ) where Fb is the net body force acting over the entire volume The volume integral of the €ourth term in Eq (4.17) gives, after applying Gauss' theorem, (4 .21 ) wherc F, is the ne1 surface force at tbe boundary of V Tf we define F = Fb thc sum or all forces, then the volume integral of Eq.(4.17) finally gives I dM F = dr +Mo", + F, as (4 .22 ) where Eqs (4.18X4 .21 ) have been used Equation (4 .22 ) . field of a certain flow is givcn by 2 u = zr1 .2 + 2xz2, 2: = x2y, w=x z. Consider the fluid region inside a spherical volume x2 + y2 + z2 = u2. Verify the validity of Gauss’. equation (3. 36) becomes a+ a2$ a+a2$ a3$ ay axay ax ay2 ay3 = v We now have a single unknown function and a single differential equation. The continuity equation (3. 37) has been. that a@ aY u -, (3. 34) a$ ax ' v E then Eq. (3. 33) is automatically satisfied. Therefore, a streamfunction JI can be defined whenever Eq. (3. 33) is valid. (A similar