KUNDU Fluid Mechanics 2 Episode 13 ppt

and any tensor a Ui /ax is equal to the product of uiui and symmetric part of a Vi /hj, namely, Eij ; this is proved in Chapter 2, Section 11 .) If the mean flow is given by U (y), then W(aUi/axj) = E(dU/dy). We saw in the preceding scction that is likely to be negative if dU/dy is positive. The fifth term uiu,(aUi/axj) js therefore likely to be negative in shear flows. By analogy with the fourth term, it must represcnt an energy loss to the agency that generates turbulent stress, namely the fluctuating field. Indced, we shall see in the following section that this term appears on the right-hand side of an equation for the rate of change of turbulent kinetic energy, but with the sign reversed. Therefore, this term generally results in a loss of mean kinetic energy and a gain of turbulent kinetic energy. We shall call this term the shearproduction of turbulence by the interaction of Reynolds stresses and the mcan shear. The sixth term represents the work done by gravity on the mean vertical motion. For example, an upward mean motion results in a loss of mean kinetic energy, which is accompanied by an increase in the potential energy of the mean field. Thc two viscous terms in Eq. (13.32), namcly, the viscous transport 2ua(Ui Eij)/axj and the viscous dissipation -2uEijEij, are smallin afully turbulent flow at high Reynolds numbers. Compare, for example, the viscous dissipation and the shear production terns: where U is the scale for mean velocity, L is a length scale (for example, thc width of the boundary layer), and u,, is the rms value of the turbulent fluctuation; we have also assumed that urms and U are of the same order, since experiments show that urms is a substantial fraction of U. The direct influence of viscous terms is therefore negligible on the meun kinetic energy budget. We shall see in the following section that this is not true for the turbulent kinetic energy budget, in which the viscous terms play a major role. What happens is the following: The mean flow loses energy to the turbulent field by means of the shear production; the furbulent kinetic energy so generated is then dissipated by viscosity. 7. Kinelic Knew Budget of Turbulcnl How An equation for the turbulent kinetic energy is obtained by first finding an equation for aulat and taking the scalar product with u. The algebra becomes compact if we use the “comma notation,” introduced in Chapter 2, Section 15, namely, that a comma denotes a spatial derivative: 3A A,i - axi ’ where A is any variable. (This notation is very simple and handy, but it may take a little practice to get used to it. It is used in this book only if the algebra would become cumbersome otherwise. There is only one other place in the book where this notation has been applied, namely Section 5.6. With a little initial patience, the reader will quickly see the convenience of this notation.) Equations of motion for the total and mean flows are, respectively, a at -(Ui + Ui) + (Uj + uj)(Vi + Ui).,j 1 Po = (I' + p),i - gll - u(T + T' - q1)]Si:3 + u(U; + ~i),jj, Subtracting, we obtain thc equation of motion for the turbulent vclocity u;: dlli 1 - + UjUi.j + ~jUi.j + ~j~i~j - (W),j = pp.; + gaT'G. 13 + vUi,jj- at Po (13.33) The equation for the turbulent kinctic energy is obtained by multiplying this cquation by ui and averaging. The first two terms on the left-hand side of Eq. (13.33) give i)ui a I- at at 2 ui- = - ( ;) I -u;@Jq).j = -ii;@Jq).j = 0, where we have used thc continuity equation uiqi = 0 and Ui = 0. The first and second terms on the right-hand si& of Eq. (13.33) givc 1 1 -ui -~,i = GT)j 7 Po Po ujgaT'Si3 = gawT'. - Thc last term on the right-hand side of Eq. (13.33) gives vu;ui.jj = v{uiui.jj + $(ui,j + uj,i)(ui,j - uj,;)}: where we have added the doubly contracted product of a symmetric tensor (U;J +uj,i) and an antisymmetric tensor (ui,~ - uj,i), such a product bcing zero. In the first term on thc right-hand side, we can write ui~j = (U~J + uj~),j because ofihe continuity equation. Then wc can writc = v{ui(ui,j + .j:i),j + (ui,,j + Uj:i)(.i,,j - ;Ui.,j - $uj:i)I = v{[ui(ui.j + .j,i)~:j - i(.i.j + ujyiI2I- Defining the fluctuating strain rate by we finally obtain Collecting terms, the turbulent energy equation becomes Iranrpnrt - -uiujUi,j + ~LIWT' - 2veijeii. (1 3.34) shear pmd buoyant pd ~iiscous dins The fmt three terms on the right-hand side are in Ihc flux divergence form and con- sequently represent the spatial transport of turbulent kinetic energy. The first two terms represent the transport by turbulence itself, whereas the third lerm is viscous lrdnsport . The fourth term -Ui,j also appears in the kinetic energy budgct of the mean flow with its sign rcversed, as seen by comparing Eq. (13.32) and Eq. (13.34). As argued in the preceding section, -WUi,j is usually positive, so that this term rep- resents a loss of mean kinetic energy and a gain of turbulent kinetic energy. It must then represent the rate of generation of turbulent kinetic energy by thc interaction of the Reynolds stress with the mean shear Ui,j. Therefore, -aUi Shear production = -uiuj - axj (13.35) The fifth term gawT' can have either sign, depending on the nature of the back- ground temperature distribution T(z). In a stable situation in which the background temperature increases upward (as found, e.g., in the atmospheric bounddry layer at night), rising fluid elements are likely to be associated with a negative temperdture fluctuation, resulting in wT' e 0, which means a downward turbulent heat flux. In such a stable situation gam represents the rate of turbulent energy loss by work- ing against the stable background density gradient. In the opposite CSLSC, when the background density profile is unstable, the turbulent heat flux wT' is upward, and convective motions cause an increase of turbulent kinetic energy (Figure 13.9). We shall call gam thc buoyantproduction of turbulent kinetic energy, keeping in mind that it can also be a buoyant "destruction" if the turbulcnt heat flux is downward. Therefore, 1 Buoyant production = gawT'. (1 3.36) -I \ E = Viscous dissipation = 2um. convection (1 3.37) Figarc 13.Y Heat flux in an unstdblc environment? gencraling turhulenl kioctic cncrgy and lowcring the mcan potential cncrgy. The buoyant generation of turbulent kinetic energy lowcrs the potential cnergy of thc mean field. This can be understood from Figure 13.9, where it is seen that he heavier fluid has moved downward in the final state as a rcsult of the heat flux. This can also be demonslrated by dcriving an equalion for thc mean potential cnergy, in which thc term gcrwT’appears with a negutive sign on the right-hand side. Thcrefore, the huoyunt generution of turbulent kinetic energy by the upward heat flux occurs at thc expense of the mean potenrial energy. This is in contrast to the shear pmduction of turbulent kinetic energy, which occurs at lhe expensc or the mean kineric energy. The sixth knn 2vm is the viscous dissipation of turbulent kinetic energy, and is usually denoted by E: This term is nor negligible in thc turbulent kinetic encrgy equation, allhough an analogous term (namely 2uE;) is negligible in the mean kinetic energy equation, as discussed in thc preceding section. In fact, the viscous dissipation E is of the ordcr of thc turbulence production terms (11IUICJi.j or gcrwT‘) in most locations. 8. 7MultmCre Prociuclion and Cwcack! Evidence suggests that the largc eddies in a turbulent flow arc anisotropic, in the scnse that thcy are “aware” of thc direction of mean shear or of background density gradient. In a complctcly isotropic field the off-diagonal components of the Reynolds stress cliuj are zero (see Section 5 here), as is the upward heat flux wT‘ because there is no prcltrence between thc upward and downward dircctions. In such an isolropic Fip 13.10 Large eddics oriented dong the principal dirations or a parallel shear flow. Note thal he vortcx aligned wih the a-axis has a posilive u when M is negalive and a ncgative u when u is positivc, resulting in W e 0. case no turbulent energy can be extracted from the mean field. Therefore, turbulence must dcvelop anisotropy if it has to sustain itself against viscous dissipation. A possible mechanism of generating anisotropy in a turbulent shear flow is dis- cussed by Tennekes and Ludey (1 972, p. 41). Consider a parallcl shear flow U (y) shown in Fiprc 13.10, in which the fluid elements translate, rotate, and undergo shearing deformation. The nature of deformation of an elemcnt depcnds on the ori- entation of the clement. An element oriented parallel to the xy-axes undergoes only a shear strain rate Ex,, = f dU/dy, but no linear strain rate (Exx = Eyp = 0). The strain rate tensor in the xy-coordinate system is therefore I- O $dU/dy idU/dy 0 As shown in Chapter 3, Section 10, such a symmetric tensor can be diagonalized by rotating the coordinate system by 45". Along thesc principal axes (denoted by a and /I in Figure 13. IO), the strain rate tensor is E=[ gdU/dy 0 0 -idU/dy so that there is a linear extension rate of Emu = f dU/dy, a linear compression rate of Epp = -: dU/dy, and no shear (Eap = 0). The kinematics of strctching and compression along the principal directions in a parallel shear flow is discussed further in Chapter 3, Section 10. The large eddies with vorticity oriented along the a-axis intensify in stren,@h due to the vortex stretching, and the ones with vorticity oriented along the &axis decay in strength. The net effect of the mean shear on the turbulent field is therefore to cause a predominance of eddics with vorticity oriented along the a-axis. As is evident in Figure 13.10, thesc cddies are associated with apositive u when u is negative, and with a negative u whcn u is positivc, resulting in a positive value for the shear production The largest cddies are of order of the width of the shear flow, for examplc the diameter of a pipe or the width of a boundary layer along a wall or along the uppcr surface of thc ocean. Thew eddies extract kinetic energy from the mean field. The eddies that are somewhal smaller than thcse are straincd by the velocity field of the largest cddies, and exhact energy from he larger eddics by the same mechanism of vortcx stretching. The much smaller eddies arc cssenliaUy advectcd in the velocity field of the large eddics, as the scales of the strain rate field of the large cddies are much larger than the size of a small eddy. Thcrdore, the small eddies do not interact with either thc large eddics or the mean ficld. The kinetic energy is therefore cascuded down J.om large to snzall eddies in n series of snwll steps. This process of energy criscude is essentially inviscid, as the vortex stretching mechanism arises jmna the nonlinear terms of the equations af motion. h B fully turbulent shcar flow (i.e., for large Reynolds numbers), therefoE, the viscosity of the fluid does not alTect the shci production, if all other variablcs are held constant. The viscosity does, howevcr, determine thc scales at which turbulent cnergy is dissipated into hcat. From the expression E = 2ueijeij, it is clcar that the encrgy dissipation is effective only at very small scales, which have high fluctuating strain rates. The continuous strclching and cascade generate long and thin filaments, somewhat like “spaghetti.” When these fi lamcnls become thin enough, molecular diffusive effects arc able to smear out their vclocity gradients. These arc the small- est scales in a turbulent flow and are responsible for the dissipation of the lurbulent kinclic energy. Figure 13.1 1 illustrates the deformation of a fluid particle in a tur- bulent motion, suggesting that molecular effccls can act on thin filaments generatcd by continuous stretching. The large mixing rates in a turbulent flow, therefore, are essentially a result of the turbulent fluctuations generating thc large suijiuces on which the molecular diffusion finally acts. It is clear that E docs not depend on u, but is dctermined by the inviscid properties of the large cddies, which supply thc cnergy to thc dissipating scales. Suppose 1 is a typical length scale of the large eddies (which may bc taken equal to the integral length -E(d U/dy). (Kolmogorov microscale) Ayre 13.11 thc scale bccomcs of thc odcr of thc Kolmogorov microscalc. Successivc &limnations ol‘a marked hid cleinenl. Di flusivc cll’cc~s causc smearing whcn 520 'liub~lec scale defined hm a spatial correlation function, analogous to the integral time scale defined by Eq. (13.10)), and u' is a typical scale of the fluctuating velocity (which may be taken equal to the rms fluctuating speed). Then the time scale of large eddies is of order l/d. Observations show that the large eddies lose much of their energy during the time they turn over one or two times, so that the rate of energy transferred from large eddies is proportional to un times their frequency u'/Z. The dissipation rate must then be of order (13.38) signifying that the viscous dissipation is determined by the inviscid large-scale dynamics of the turbulent ficld. Kolmogorov suggested in 1941 that the size of the dissipating eddies depends on hose parameters that are relevant to the smallest eddies. These parameters are the rate E at which energy has to be dissipated by the eddies and the diffusivity u that does the smearing out of the vclocity gradients. As the unit of E is cm2/s3, dimensional reasoning shows that the lcngth scale formed from E and u is (1 3.39) which is called the Kolmogomv micmscale. A decrease afv merely decreases the scale at which viscous dissipation takes place, and not the rate of dissipatian E. Estimates show that is of the order of millimeters in the ocean and the atmosphere. Tnlaboratory flows the Kolmogorov microscale is much smaller because of the largerrate of viscous dissipation. Landahl and Mollo-Christensen (1986) give a nice illustration of this. Suppose we are using a 100-W household mixer in 1 kg of water. As all the power is used to generate the turbulence, the rate of dissipation is E = 100 W/kg = 100 m2/s3. Using u = m2/s for water, we obtain q = mm. 9. Speci.rum oJ Turbulence in lizertial Subrangc In Section 4 we &fined the wavenumber spectrum S(K), representing turbulent kinetic energy a, a function of the wavenumber vector K. Tf the turbulence is isotropic, then the spectrum becomes independent of the orientation a€ the wavenumber vector and depends on its magnitude K only. In that case we can wrile oc - u3 =$ S(K)dK. In this section we shall derive the Form d S(K) in a certain rangc of wavenumbers in which he turbulence is nearly isotropic. Somewhat vaguely, wc shall associate a wavenumber K with an eddy of size K-' . Small cddies are therefore represented by large wavenumbers. Suppose I is the scale of thc large eddics, which may bc he width of the boundary laycr. A1 the relativcly small scales represented by wavenumbers K >> 1-I, there is no direct interaction between thc turbulence and the motion of the large, encrgy-containing eddies. This is because the small scalcs have been gencrakd by a long series of small steps, losing information at each stcp. The spectrum in this range oJfarge wtsvenumhers is nerrrly isotropic, as only thc large eddies are aware of the directions of mcan gradients. Thc spcctruin here docs not depend on how much encrgy is present at large scales (wherc most 01 the energy is contained), or the scales at which most of thc cnergy is present. The spectrum in this range depends only on the parametcrs that determine thc nature o€ he small-scale flow, so that we can write 1- lo" S 10-2 S = S(K, E, u) K >> I-'. - - . . Thc range of wavenumbers K >> 1-' is usually called the equifihri-ium rcmge. The dissipating wavenumbers with K - rj-I, beyond which the spectrum falls off very rapidly, form the high end of thc equilibrium range (Figure 13.12). The lower cnd of this range, for which 1-' << K << q-'. is called the inertial subrange, as only the n-dnsfer of encrgy by inertial forces (vortex stmtching) takes place in this rangc. Both production and dissipation are small in the inertial subrangc. The production of energy by large eddics causes a peak of S at a ccrlah K 2 1 '-I, and the dissipation of energy causes a sharp drop of S for K =- I)'-'. The question is, how does S vary with K between the two limits in the inertial subrange? IO-' . . . . 1 0-' 1V* IV' 1 m Figure 13.12 A typical wavenumbcr spectrum observed in Lhc ocean, plotted on a log-log scalc. The unit of S is arbitrary, and the dots rcpresent hypothetical dah. S = AE2f3K-j/3 1-1 << K << q-', ~ ~ ~ _______ where A 21 1.5 has been €ound to be a universal constant, valid for all turbulent flows. Quation (13.40) is usually called Kalmoguraw's K-5/3 law. If the Reynolds number of the flow is large, then the dissipating eddics are much smaller than the energy-containing eddies, and the inertial subrange is quite broad. Because very large Reynolds numbcrs are difficult to generate in the laboratory, the Kolmogorov spectral law was not verified for many years. In fact, doubts were being raised about its theoretical validity. The first confirmation of the Kolmogorov law camc from the oceanic observations of Grant etaf. (1 962), who obtained a vdocity spectrum in a tidal flow through a narrow passage between two islands near the west coast of Canada. The velocity fluctuations were measured by hanging a hot film anemometer from the bottom of a ship. Based on the depth of water and the average flow velocity, the Reynolds number was of order lo8. Such large Reynolds numbers are typical of gcophysical flows, since the length scales are very large. The K-s/3 law has since been verified in the ocean over a wide range of wavenumbers, a typical behavior being sketched in Figure 13.12. Notc that the spectrum drops sharply at Kq - 1, where viscosity begins to affect the spectral shape. The figure also shows that the spectrum departs fom the K-'I3 law for small values of the wavenumber, where thc turbulence production by large eddies bcgins to affect the spectral shape. Laboratory cxperimcnts are also in agreemcnt with the Kolmogorov spectral law, although in a namwcr range of wavenumbers because the Reynolds number is not as large as in geophysical flows. The K 'l3 law has become one of the most important results of turbulence theory. (1 3.40) Nearly parallel shear flows are divided into two classes-wall-free shear flows and wall-bounded shear flows. In this section we shall examine some aspects of turbulent flows hat are free of solid boundaries. Common examples of such flows are jets, wakes, and shear layers (Figure 13.1 3). For simplicity we shall consider only plane two-dimensional Bows. hisymmetric flows are discussed in Townsend (1 976) and Tennekes and Lumlcy (1 972). Intermittency Consider aturbulent flow confined to a limited region. To be specific we shall consider the example of a wake (Figm 13.13b), but our discussion also applies to a jet, a shear layer, or the outer part of a boundary layer on a wall. The fluid outside the turbulent region is either in irrotational motion (as in the case of a wake or H boundary layer), OT nearly static (as jn the case of a jet). Observations show that the instantaneous interface between thc turbulent and nonturbulent fluid is very sharp. In fact, the thickness of the interface must equal the size of he smallest scalcs in the flow, namely the Kolmogorov [...]... statc, the mean velocity at various downstream n viscous eddies irrotational fluid 4 irrotational fluid turbulent fluid turbulent fluid Figure 13. 14 htrainmcnt of a nonturbulent fluid and its assimilation into turhuleni fluid by viscous aciion at lhc interke distanccs is given by U - = f($) u, (jet), (13. 41) u-U I - f u -u = 2 1 (5) (shcar layer) Here S(x) is thc width offlow, Uc(x)is thc centerline... cnsemblc averdgc and not time avcragc We can write d dXu -(Xi) = 2xu- (13. 75) dr dt ’ where we have used the commutation rule (13. 3) of averaging and differentiation Defining Uu dX, =- dt ’ as the Lugrungian velocity component of afluidparticle at timet, Eq.(1 3.75) becomes d -(X:) = 2X,u, = 2 [ g t u , ( t l ) d r r ] u, dt =2 I’ U, (t’)U,(t)dr’, (13. 76) where we havc used the commutation rule ( 1 3.4) of... then becomes - g ( t )= 2Zt I’ 5) (1 - r,(t) d t (13. 79) f i o limiling cases are examined in what follows BehaviorJor small t: If t is small compared to the correlation scale of r , ( t ) , then r,(t) 2: 1 throughout the intcgral in Q (13. 78) (Figure 13 .28 ).This gives - - x:(t) 21 u y Taking the square root of both sides, we obtain (1 3.80) 3Figurc 13 .28 Small and large valucs or time o n a plot of... more convenient to match thcir gradients (The derivation given here closely follows Tennekes and Lumley (19 72) .) From Eqs (13. 50) and (1 3. 52) , the vclocity grddicnts in the inner and outcr regions are given by e dU _ - u: d.f dy v dY+!+' dU = -u,dF dy S de (13. 53) (13. 54) Equating (13. 53)and (13. 54) and multiplying by y / u * , we obtain d 6- F dt = y dJ 1 =dy+ k' (1 3.55) e valid for large y+ and... particle, Eq (13. 76) becomes d(z) 2 2 = dr lr r,(t’ - t ) dt’ (13. 77) where we have changed the integration variable from f’ to t = t - t’ Integrating, we obtain (13. 78) which shows how the variance of the particle position changes with time Anotheruseful form of Eq (13. 78)is obtained by inugrating it by p a t We have =t l’ r r n ( td t ) - I’ t’rrn(r’) dr’ Equation (1 3.78) then becomes - g ( t )= 2Zt I’... laycr Figure 13 .22 is a photograph from Klinc et al (1967), showing the top view of thc flow within the viscous sublayer rm at a distancc y = 2. 7 f o the wall (Here x is thc direction of flow, and z is the %panwisc” direction.! The wire producing the hydrogen bubbles in the figure was Figure l 2 Top view of near-wall structure (at y+ = 2. 7) i a turbulent boundary layer on a horizontal 32 n flat plate... large i Jf t is large compared with the correlation scale of r , ( t ) then : t / r in Eq (13. 79) is negligible, giving - X;(t) 2 2 : ~3-t: (1 3. 82) where 3i ( 9 ru(t)dt, is the integral tirnc scale detcrmined f o tbe Lagrangian corrclation r , ( t ) Taking rm the square root, Eq (13. 82) gives (1.3.83) The t ‘ j 2 behavior or Eq (1 3.83) at large times is similar to the behavior in a random wulk, in... will be considered in this infleclional profile n X lifted and stretchcd vortex elemcnt 2 F i g m 13 .23 Mechanics orslrcak brcak up S J Kline el a/.:Journal ofFluidMechmim 3 :741-773, 0 1967 wd icpr;.ntedwith the permission oTCmbridge University Prcss section Further discussion can be found in Tennckcs and Lumley (19 72) , Phillips ( 1 9771, and Panorsky and Dutton (1984) As is customary in gcophysicalliterature,... ,using the logarithmic velocity distribution (13. 60), from which clU/dz = u , / k t (Notc that we are now using z for distances perpendicular to the surface.) Using UW = u: because of the near uniformity of stress in the logarithmic iayer, Eq (13. 68) becomcs z Rf = - (13. 72) LM As Kf is the ratio of buoyant destruction to shear production of turbulence, (13. 72) shows that LM is Ihc hcight at which these... x-momentum That is M=d_ 00 U 2dy = independcnt oix: (13. 42) where M is the momentum flux of thc jct (=integral of inass flux p U d y times velocity U) The momentum flux is the basic externally controlled parameter for a jet and is known from an evaluation of Eq. (13. 42) at the orifice opening The mass flux p l U dg across thc jet must incrcasc because of entrainment of the surrounding fluid The assumption of . eddies irrotational fluid irrotational fluid turbulent fluid turbulent fluid 4 Figure 13. 14 htrainmcnt of a nonturbulent fluid and its assimilation into turhuleni fluid by viscous. y/S < 0 .2. Thc logarithmic law, herefore, holds accuratcly in a rather small percentagc ( -20 2) of thc total bound- ary layer thickness. The gcneral defect law (13. 52) , where F(e). the cross-stream direction y, and 3 is the intensity in the z-direction; q2 = (u2 + Y~ + w2) /2 is the turbulent kinctic encrgy per unit mass. The Reynolds stress is zero