 In this lecture, we shall continue with our discussion of the sustaining mechanism of turbulence, but by employing spectral analysis and vorticity dynamics. Both these terms need explanation in the course of this lecture, you will understand what they mean. I must want that the subject matter is somewhat mathematical and therefore, very close attention would be required. So, we recall that in the previous lecture, we discovered three characteristic length scales of turbulence. One was the largest, the integral length scale which was obtained by integrating spatial correlation coefficient L int, intermediate Taylor micro scale, L f and G f being longitudinal and G being transverse length scale and then the very smallest scales, which we call Kolmogorov scales L sub epsilon, where energy dissipation takes place because of the strong influence of viscosity. Spectral analysis goes a little further and it explains how turbulence energy is distributed among the range of wave scales and how the energy exchange between eddies of different scales takes place. In this sense, it is a much more continuous description of energy transfer from larger scales to the smaller scale. To do this, we use the notion of spectra. The spectra are decompositions of any non-linear function into waves of different wavelengths or they can even be periods. The value of the spectrum at a given wavelength or frequency in case of time dependent function is the mean energy in that wave. Spectra analysis leads to the understanding that turbulence receives its energy at the large scales and while its energy dissipates at the very small scales, there also exist waves within a range of wavelengths called the inertial sub range, which are not directly affected by the sustenance mechanism of turbulence. That is the overall story that we wish to narrate by going through a somewhat tedious mathematics. You will recall the spatial correlation tensor B ij at a vector position r del x 1 del x 2 del x 3 is related to the spectral tensor phi ij as a function of wavelength vector k via the 3D Fourier transform and the Fourier transform is defined as B ij at vector r equal to minus infinity to plus infinity 3 times over phi ij k vector k exponential of i k dot r d k i is the complex number under root minus 1. The inverse transform of this is given by phi ij k vector k equal to integral from minus infinity to plus infinity 3 times B ij r exponential of minus i k dot r d r. Therefore, spectral interpretation of the Reynolds stress which is of the 1 point correlation tensor rho u i dash u j dash both u i dash and u j dash are measured at the same point. Therefore, minus rho u i dash u j dash will become minus rho B ij r equal to 0 equal to minus rho minus infinity to plus infinity 3 times over and since r is 0 exponential of that would be 1 and you get phi j k vector d k. Now, since u i dash u i j dash time average determine the energy in the various velocity components of the fluctuations, the value of phi ij k gives the division of this energy in different eddy sizes or wave numbers. The small value of wave number corresponds to a large eddy and vice versa. So, a large wave number wavelength corresponds to very small eddies and very small wavelength corresponds to large eddies. So, what the Fourier transform does is it gives you the division of the energy of the fluctuations in different eddy sizes. Consequently, phi ij k vector is called the energy spectrum tensor. Further, if we take the sum of the diagonal components of the tensor gives the turbulent kinetic energy spectrum at a given wave number. So, B i i r equal to 0 u i dash u i dash will be 2 times the kinetic energy E equal to minus infinity to plus infinity 3 times over phi i i k d k. E here is the turbulent kinetic energy. The spectral tensor phi i k is a function of 3 wave number components k 1 k 2 and k 3 because k is a vector. This makes interpretation of this expression somewhat difficult. Therefore, in order that physical interpretation becomes easier, it is customary to remove directional dependence by integrating phi i k bar over a spherical shell of radius k, where k is square root of k 1 square plus k 2 square plus k 3 square. In other words, the total the magnitude of the vector k is taken as the radius on this spherical shell. So, we form a sphere of radius k and if we take an element d A on this sphere, then the energy spectrum of E k is half times integral phi i k d A and E would be equal to integral 0 to infinity E k d k. So, this is what I have shown here energy at wave length k is one half area integral of phi i k bar d A and that gives E from 0 to infinity E k d k is the kinetic energy E. The function E k is called the scalar kinetic energy spectrum. To derive the transport equation for E k, an equation for B i j k is first derived for a non homogenous anisotropic and steady turbulent flow. Now, as you know in a steady turbulent flow, d u dash k d x k plus d u k by d x k is 0 from continuity and the instantaneous form of the Navier-Stokes equation would be given by d u i dash d by d t plus u cap k d u cap i by d k d x k equal to the pressure gradient term and the diffusion term. Now, what I am going to do is explain how the second term is modified. So, u k d u i by d x k is written as u k plus u dash k d by d x k by d x k is equal to of u k plus u dash k sorry u i divided by u dash i and that gives me the first term as u k plus u dash k d by d x k of u i plus u k plus u dash k of d u dash i by d x k. So, this term remains as it is plus u k d u dash i by d x k plus u dash k d u dash i by d x k and this equals u k d u dash i by d x k plus d u dash i by d x k plus u dash k by d x k minus u dash i by d u dash k minus by d x k. Now, of course, that term is 0 because of continuity and you will see this is what I have written here. So, you have three terms u k plus u dash k d u i by d x k plus u k d u dash by d x k d u dash k u dash i by d x k equal to all that. Now, if I subtract from this equation the Rans equation and you will recall what is the Rans equation? The Rans equation is u k d u i by d x k equal to minus 1 over rho d p by d x i plus d by d x k u i into nu minus d u dash i u dash k by d x k. So, if I subtract this equation from this equation that you see here, then you will notice that I would get the following equation d u i dash i by d u dash k d u i by d x k plus u k d u dash i by d x k into d by d x k u dash k u dash i minus u dash k u dash i average time average minus 1 over rho d p dash by d x i plus nu d 2 u dash i by d x l d x l. So, this is simply subtracting the mean Rans equation from the instantaneous form of the momentum equation. If I say this is the equation for u dash i at position r 1, I could also write a similar equation for u dash j at another position r 2 and that would read as, so I have let us say position r 1 where I wrote an equation for u dash i and position r 2 where I wrote an equation for u dash j and that equation would again read very much like the first one with the d p dash by d x j here and these terms. Now, to do further development, we multiply the first equation this equation at position r 1 by u dash j at r 2 and the second equation by u dash i at r 1. We then add the two equations and time average which would yield the required equation for b i j which I will show on the x hand side with two independent variables psi k which is the difference between these. So, psi k equal to x k at r 2 minus x k at r 1 and x k mean equal to half of x k at r 2 plus x k at r 1. So, you have the midpoint m and the difference psi. So, we have two independent variables in this equation. This may be little bit unfamiliar to you, but it is a straight forward algebra to show that this is indeed the case. So, the total equation for non homogenous anisotropic turbulence would read like this d b i j by d t plus b k j d u i by d x k at r 1 plus b i k d u j by d x k at r 2 plus half of d u k r 1 by d u k r 2 into this b i j d x k at mean point plus u k r 2 minus u k r 1 d b i j by d psi k equal to minus half of t i k j t i k j. Now, t i k j here is defined as u i dash at r 1 multiplied by u dash j at r 2 multiplied by u dash k at r 2 and t i k j the second term is defined as u i dash at r 1 u dash k at r r 1 multiplied by u dash j at r 2. So, you get that transfer term minus d by d psi k based on the difference of the between the two points then minus 1 over rho 2 rho d C p j by d x i at m and C p j is the product of pressure fluctuation at r 1 multiplied by u dash j at r 2 that is the C p j and C p i likewise is the pressure fluctuation at r 2 multiplied by u dash i at r 1 and b i j of course, is u dash i at r 1 into u dash j at r 2 is the tensor b i j plus nu times 1 by 2 b i j d x these are the diffusion terms and this. So, to make further progress now we postulate the equation of the previous slide represents complete equation for non-homogeneous turbine flow. The equation is not tractable because it has two types of independent variables as you can see psi k and x k. So, midpoint and the difference between the two points it is not tractable, but we can make it tractable somewhat by postulating homogeneous turbulence and as you will recall in homogeneous turbulence derivatives of correlations with x k vanish, but are finite with respect to psi k the difference and therefore, you will get all these terms involving d x k of b i j would vanish sorry. So, would this vanish and you will get this equation d b i j by d t equal to psi l times d u x k by d x l into d b i j by d x k. Now, that is the mean convection term this term because this is already 0 it is this term u k r 2 minus u k r 1 d b i j by d psi k. So, that mean convection plus production remember the product of stress multiplied by rate of strain mean strain it gives you the production term. So, that becomes the production term d by d psi k tau i k j minus tau i k j is as you will see it is this term and that is the diffusion of b i j due to velocity fluctuation. So, we say that is v diffusion and then the pressure diffusion terms which are given by these are the pressure diffusion terms these of course, vanish, but these survive because these are derivatives with respect to psi and then finally, this term survives because it is the derivative with respect to psi l psi l that is 2 nu times d 2 b i j d psi l which is really the dissipation due to viscosity. So, you have fairly complex equation now the mean convection terms d u i k d x l d a b i j d psi k is really the u k r 2 minus u k r 1 d b i j by d psi k as I explained. However, u k r 2 minus u k r 1 will be psi l d u k into d x by d x l at m and therefore, we have written that as that this is u k r 2 minus u k r 1 the v diffusion and p diffusion terms represent diffusion of energy due to velocity and pressure fluctuation respectively. Now, in order to study the transfer process each term of this equation is Fourier transform. So, as to yield an equation for phi i j k bar so, we will have a differential equation in Fourier space of phi i j d by phi i j k bar if we then set i equal to j we would get an equation phi i i k bar and further to achieve directional independence each term is integrated over a spherical shell and I left out all the algebra here to yield d e k by d t equal to p k minus d t t k by d k minus d k where p k is production, d k is dissipation and t t k is the transfer term and this due to convection due to v diffusion and due to p diffusion that is the diffusion due to pressure fluctuation diffusion due to velocity fluctuation and this is diffusion due to mean convection. So, essentially then we get a an equation in scalar k space it is a one dimensional equation because there is only one independent variable in wave number space k d k and it is a transient equation d e k by d t equal to production minus d by d k tau t t k minus d k which is the dissipation. We can look upon this equation which is the spectral form of the turbulent kinetic energy equation for homogeneous turbulence and it is nothing but a one dimensional equation representing the energy balance over a control volume d k in the wave number space. The p k term comprises spectral functions arising out of b k j and b i k and multiplied by the mean velocity gradients and is not expected to be large at high wave numbers where the velocity fluctuations have almost died down and viscosity has taken over. So, the p k term would largely dominate at small wave number the gradient t t k term would vanish when integrated from k equal to 0 to k infinity giving d e by d t equal to d by d t integral 0 to infinity e k d k equal to 0 to infinity p k minus d k d k. This transfer term simply redistributes energy both directionally in the components as well as along the wave numbers that is along the sizes and therefore, it is t t means total transfer term. The dissipation term after Fourier transforming of this term would be written as 2 nu k square e k and the presence of k square confirms that it would be significant only at very very high wave numbers k square suggest that it would be significant at very very high wave numbers. Now, of course, I have left out the algebra as I said because it is considerably long, but the essential ideas that we had we had seen in Rance equations from which we derived the turbulent kinetic energy in physical space and the equation we have now derived in the wave number space have very similar characteristics that there is a rate of change of energy is equal to the rate of its production minus net rate of its transfer to smaller scales and minus dissipation. If you consider a flow in which homogeneous shear flow in which v and w are 0 the mean velocities are 0 and d u by d y is a constant then here are the results computed at Reynolds number turbulence Reynolds number based on transverse Taylor micro scale of about 100 look something like this. You see at the loss of loss means negative contribution to energy balance gain means positive contribution to energy balance and you see that the production term is indeed very high at small wave numbers this is the wave number axis at small wave numbers. In fact, it peaks at very very small wave numbers and therefore, it is called large eddies as I said very very large eddies at which the production dominates, but then it declines as the wave number increases. In this range the transfer term is negative which means energy is being sucked out from large eddies and being pushed towards smaller eddies energy is continuously being fed to smaller eddies. A point is reached where the production almost dies down and so does the transfer almost died down and there can be a region called the inertial sub range about which I will talk in a minute, but then dissipation takes over as you can see it is shown as negative dissipation which occurs at very very large wave numbers and the transfer term becomes positive at large wave numbers and again dies down and of course, as we would expect the area under the negative sign here of the transfer term must equal the area under the positive part of d t by d x k so that the net area is 0 in this case. The energy itself of course, rises to very very large value E k and peaks at some wave number which is designated by k sub E and then begins to decline. So, energy itself is goes on declining then it falls here at the rate proportional to the k to the power of minus 5 by 3 and then finally, k very very rapidly to k to the power of minus 7. What does this indicate? So, the decay term dominates at very high k which means large wavelengths or Kolmogorov small eddies say of the power of minus 0.1 divided by L sub epsilon k to 1 divided by L sub epsilon that is the range of wave numbers in this part here 0.1 divided by 1 over L epsilon to 1 over L epsilon energy is mainly supplied by the transfer term d t by d k and the energy extraction from the mean motion is minimal as we saw here. The energy for dissipation is essentially supplied by the transfer term and the production term itself contributes nothing to energy dissipation. The transfer term is negative at small k and positive at high k indicating that energy is indeed transferred out of low k region into high k region and therefore, the integral of d t by d k d k is 0. The dominance of p k in the low k region indicates that most of the energy production is brought about by large eddies. As you saw here, most of the production is brought about by large eddies that is why I have said large eddies here. As k tends to 0, very large eddies dominate and the e k spectrum is not expected to be universal at all being influenced by the mean velocity gradients. Also, d e k by d t is small and d t k by d k therefore, equals production which means all the energy produced is simply goes into its transfer to smaller and smaller eddies or bigger and bigger wavelengths. This region therefore, is called the region of rapid distortion. This region is called the region of rapid distortion. The e k is maximum near k sub e which characterizes the most energy containing eddies and the l integral that we saw in the previous lecture is largely determined by these eddies. So, the characteristic eddy of this region is we call the energy containing eddies. This is where the l integral corresponds somewhere to k sub e. If k dissipation is much greater than the energy containing wave number k sub e then an inertial sub range exists at k less than 0.1 divided by l sub epsilon. So, as you can see if k dissipation is much bigger than k sub e then a range exists where as you can see here k less than 0.1 divided by l sub epsilon identified with Taylor micro scale exists in which the conditions of isotropy of the small k eddies and of independence of turbulent structure from energy containing eddies are simultaneously satisfied. In other words this range of wave number is small enough for isotropy to prevail is a range in which the isotropy of the large scale and the independence of the structure from energy containing eddies are simultaneously satisfied and in this range e k is proportional to minus phi by 3 Now, this is very important many experimentalists always want to ensure is their turbulent sufficiently vigorous so that they can take it as fully turbulent or not which means are the large wave numbers where dissipation takes place and the small wave numbers the energy containing wave numbers where the production takes place are they sufficiently separated or not because if they are separated then an inertial sub range must exist with k raised to minus phi by 3 as the energy spectrum. If it exists they would declare such turbulence as being truly representative of vigorous turbulence the existence or otherwise of the inertial sub range has considerable significance for the near wall turbulence this is a matter which is used employed in turbulence modeling which we may use a little later finally at very high wave numbers again where k is greater than 1 over epsilon the energy spectrum varies as e k k raised to minus 1 and at this point d k is maximum as we saw earlier. So, spectral analysis essentially gives you how continuously the energy is transferred from large eddies to small eddies or from small wave numbers to large wave numbers. Vorticity dynamics is another way of explaining the same phenomenon and that is by the process of breakdown of eddies. So, for example, consider a three dimensional cubic element here this is direction 1 and this is direction 2 and this is direction 3 and imagine that this element is being stretched because of the strain rate in direction 1 then if the turbulence was very high in the large where viscosity would have very little influence then you will see that the plane of cross section must shrink as I have shown here because of the stretching in x direction the cross section in y and z direction or the length scales associated with y and z direction must decrease. So, figure a shows a 3D cubic element which is stretched in the direction of the linear strain S 1 1 then the cross section in a plane perpendicular to the strain will become smaller as shown in figure b. Similarly, in figure c I now consider a vortex element rather than a cubic element the vortex in the direction of strain S 1 1 becomes smaller in cross section as you can see here this is S 1 1 and the vortex is being stretched. So, it becomes smaller in cross section but while the cross section normal to the strain becomes larger as shown in figure d. So, the cross section in this direction would become smaller would become larger while in the stretch direction it would become smaller from this to there. Now, intuitively of course this is understandable, but it is useful to consider equation for vorticity fluctuations on the next slide. I will return to 3D cubic element after this explanation. So, consider large eddy structure where effects of viscosity are small then vorticity equation will read as d omega i dash by d t equal to omega dash i into S i j. Again vorticity equation is simply I have not derived here the full form of it, but when viscosity goes to 0 you could very well take it as that at the moment for the purposes of discussion. Now, for simplicity consider a 2D strain field in which only S 1 1 is equal to minus S 2 2 S 1 1 being stretching tensile S 2 2 being compressive minus S equal to a l S 1 is equal to S and that is equal to constant. These are strain fields for all times t greater than 0 and S 1 2 is equal to 0 that is the strain rate in the then if omega dash naught is the vorticity at time t equal to 0 then from this equation we say that d omega dash 1 by d t would be equal to omega dash 1 S because S 1 1 is positive which would give the solution omega dash 1 divided by omega dash naught equal to exponential of plus S t. And likewise omega dash 2 divided by d t will be equal to minus omega dash S equal to omega dash or omega minus equal to exponential of minus S t. And therefore, if I were to consider omega 1 square plus omega 2 square as the total vorticity that would be equal to omega naught square exponential of 2 S t plus exponential of minus 2 S t. And therefore, the total vorticity increases with S multiplied by t at large values of S t omega dash 1 in the direction of stretching increases very rapidly because of the plus and omega dash 2 in the direction of compression decreases slowly because of the minus. What does this tell us? This tells us that the eddies are thus stretched at a rapid rate into smaller eddies their growth to larger eddies. However, occurs at a much smaller rate resulting in the net reduction in their size. Now, this is a very important observation to make from vorticity dynamics. Now, for the same case when effect of mu is small then the angular momentum must be conserved or omega dash square are equal to constant. So, if we now go back to cubic element and I am stretching it in x 1 direction and therefore, it shrinks in the y z plane, but since angular momentum must be conserved reduced length scale which means increase intensity of vorticity would mean v dash and w dash must now increase compared to their state before stretching whereas, u dash of course, is doing the stretching. So, in figure b element is stretched in x direction then the kinetic energy of rotation in the y z plane increases at the expense of the kinetic energy of the velocity component u dash which does the stretching and therefore, the length scales of motion in y z plane decrease and hence v dash and w dash increase. Now, this increase v dash and w dash will bring about further stretching in y z directions and so on. So, you can see stretching in x direction brings about intensification of velocities in y and z direction which do further stretching. So, intensification in y will do stretching will increase the intensity in z and x direction while reducing the scale. Similarly, intensified w dash would bring about reduction in size in x and y direction and so on and so forth. At each stretching however, the length scale of the element will go on decreasing and this is called the breakdown of the eddies. So, in summary then I can say the tree demonstrates that the stretching in x direction intensifies motions in y and z directions producing smaller scale stretching in these directions and intensifying motions in x, y and z directions at the end of the second stage and so on to the further stretches. As the length scales are progressively reduced the effects of mean motion are weakened and small eddies tend towards a universal structure that is homogeneous and isotropic. Despite the fact that the mean flow and large scale structure are an isotropic and inhomogeneous. The breakdown of eddies would continue indefinitely if it were not for the action of viscosity which finally kills the fluctuations and maintains the fluid continuum. Now, this is the notion which we started with. We questioned whether the fluid will continuum survives in our previous lecture the first slide and whether the fluid will actually split up and by observing the scales at the molecular levels and the scales at the turbulence level being so far separate that we concluded there that the turbulence in no way influences the molecular motion structure, molecular motions and vice versa and therefore, there is no question of continuum being brought under question continuum indeed prevails as the analysis shows at small in turbulence and therefore, as we showed here the breaking down of the eddies would continue indefinitely if it were not for the action of viscosity which kills all fluctuations and maintains the fluid continuum. Now, this idea is very well captured in a beautiful poem by Richardson and I will big walls have little walls which feed on their velocity and little walls have lesser walls and so on to viscosity. This poem was written in 1922. The turbulence models while appreciating the poem very much because it explains to a layman how turbulence sustains itself. They often point out that the existence of the inertial sub range is not revealed in this poem, but then this was a poem and not science and the existence of inertial sub range in which you have independence from energy containing eddies and the isotropy of the dissipating eddies is not made explicit in this poem. That is well taken, but nonetheless to a layman at least and it is a good take home message to remember that the big walls have little walls which feed on their velocity and the little walls have lesser walls and so on to viscosity. So, big walls indicating large scale eddies which create small scale eddies and small scale eddies create smaller scale eddies and so on to viscosity.