Chapter 2 Canonical Turbulent Flows Before we consider two canonical turbulent flows we need a general description of turbulence. 2.1 A Brief Introduction to Turbulence One way of looking at turbulent flow is to consider it to be subject to two types of random movement: one on the scale of the thermal motion of molecules and one on the scale (macroscopic) of fluid velocity fluctuations (eddies). The main difference between these two motion types is: 1. Scale the turbulent fluctuation scale is typical molecular mean free path. 2. Continuity in a turbulent fluctuation some chunk of fluid must be pushed out of the way for a turbulent fluctuation to occur (and similarly, fluid must replace the void left by the mass of fluid carried in the turbulent fluctuation). How do we know a fluid is turbulent? Symptoms of turbulence: 1
2 Canonical Turbulent Flows Diffusive (more so than due to molecular diffusion, much much more so...). Unsteady Random (there is coherence in turbulence so it is not truly random). Wide range of scales of motion Consider the following: 2.1.1 The Energy Spectrum Big whirls have little whirls which feed on their velocity; and little whirls have lesser whirls, and so on to viscosity. L.F. Richardson, c. 1920 The Cascade Picture The energy of an eddy u 2 where u is the typical (read mean) velocity at a location.
Canonical Turbulent Flows 3 Let s assume that an eddy passes all of its energy to smaller scales in the time it takes the eddy to rotate approximately one revolution, known as the eddy turnover time. Then the rate of loss of energy to smaller scales u2 T e Where T e is the eddy turnover time L u eddy). (L is the diameter, or length scale, of the typical Therefore the rate of loss of energy to the smallest scales u3 L turbulent kinetic energy. = ε = dissipation rate of If the flow is in equilibrium (meaning the rate of the production of turbulence is equal to the rate of the dissipation of turbulence locally), this energy is passed on unattenuated by inviscid nonlinear interactions (e.g., vortex stretching, ω u where ω = u is the vorticity vector and u is the velocity vector) to smaller scales where it is ultimately dissipated by viscosity to heat, in which case we can write: u 3 L = ε = dissipation rate of turbulent kinetic energy (2.1) a classic scaling estimate for the dissipation rate. What is the smallest length scale of turbulence? It is set by the kinematic viscosity, ν, and the turbulent kinetic energy transfer rate = turbulent kinetic energy dissipation rate = ε in equilibrium flows. Applying dimensional analysis we find: l ε a ν b [L] = [L2 ] [L 2 ] [T 3 ] [T ] for [L] 1 = 2a +2b for [T ] 0 = 3a + b 1 = 4a a = 1 4 b = 3 4 [ ] ν l = ε 1/4 ν 3/4 3 1/4 = = η ε
4 Canonical Turbulent Flows Where η is known as the Kolmogorov length scale - the smallest length scale of a turbulent eddy. What is the ratio of the largest length scale, L, to the smallest, η? L η ( L ( ν 3 ) 1/4 Lu3/4 Lu ν 3/4 L 1/4 ν ε ) 3/4 Re 3/4 L E.g., to computationally resolve a turbulent flow (direct numerical simulation) we must resolve η x η where x is the computational grid spacing, over the whole flow domain L. Hence the number of grid points required in any one coordinate direction to fully resolve the flow is N = L x L η Re3/4 L. Therefore in 3-D we would need O(N 3 ) Re 9/4 L points. In Cayuga Lake L 1000m u 10 1 m/s ν 10 6 m/s. Therefore Re L 10 8 N 10 18 or 1 billion billion points! That is order 1 billion Gigabytes of memory just to store your data at one time step, or 1 million Terabytes or 1thousandPetabytes or what is known as an Exabyte. Not happening! Today (2017) the typical largest simulations are run on grids of 16384 3 points, or O(10 12 ) points, and this is extremely rare! Hence the only way to determine exactly how even a moderate Reynolds number flow behaves is to make measurements! And instead of looking at all this data in an instantaneous sense we take a stochastic approach.
Canonical Turbulent Flows 5 2.1.2 Reynolds Decomposition Let u = u + u (2.2) where u is the temporal mean component and u is the fluctuating component. This decomposition is known as the Reynolds decomposition after Osborne Reynolds, who first proposed it as a statistical approach to aid in the study of turbulence. The definition of u is Therefore u = 1 T T 0 udt (2.3) u = 0 u = u uu = 0 But u u 0 O.K., to get a feel for the analysis of a turbulent signal let s consider some data. The signal below is the axial-component of velocity measured downstream from a turbulent round jet on the jet axis. The signal was collected with a one-component laser Doppler velocimeter (LDV), which basically measures the Doppler shift of light due to the motion of small seed particles in the direction of a light ray. This is somewhat analogous to an acoustic Doppler velocimeter (ADV),which we will discuss in detail. The signal was sampled at f s = 100 Hz for 194.56 seconds. Only 5 seconds of the signal is shown along with the u = 0.1502 m/s (dashed line). Following Reynolds decomposition for this signal we can determine u 2 = 0.0029 m 2 /s 2. The square root of this value, u 2 = 0.0542 m/s, is often reported as a measure of the mean turbulence intensity level for a single velocity component. For those of you who recall your statistics this is essentially
6 Canonical Turbulent Flows 0.4 0.3 u (m/s) 0.2 0.1 0 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 t (s) the standard deviation hence we expect about 63% of the velocity fluctuations to be less than 5.42 cm/s and 95% of the velocity fluctuations to be less than 10.8 cm/s if the statistics are normally distributed which is a good assumption for the turbulent round jet. Looking at the figure we see this looks about right for the limited portion of the data we can see. In fact, let s verify the assumption of a normally distributed data, how? Look at the histogram! Counts 1200 1000 800 600 400 Data Gaussian 200 0 0.1 0 0.1 0.2 0.3 0.4 u (m/s) Now, let s look at the spectrum (determined by averaging records of length of 512 points) Note for completeness I have plotted the entire spectrum (S uu ) where the negative frequencies are on the right half of the plot (e.g., between 50 and 100 Hz - since the plot is log-log the plot does not appear symmetric but the data is!). Normally we would only plot G uu (the positive frequencies) so the symmetric negative frequency data would not be shown.
Canonical Turbulent Flows 7 10 2 10 3 S uu (m 2 /s 3 ) 10 4 10 5 10 6 10 1 10 0 10 1 10 2 f (Hz) 10 2 10 3 G uu (m 2 /s 3 ) 10 4 10 5 10 6 10 1 10 0 10 1 f (Hz) Of course, just to be confusing, it is entirely common to plot just the positive frequencies of S uu (i.e., without the factor of two on the amplitudes of G uu ) henceyoumust be careful to understand what other researchers have done when you are looking at spectra! We see a well developed -5/3 region, known as the inertial subrange, which connects the low frequency region where the turbulence is produced to the high frequency region where the turbulence is dissipated. Now our data set does not really indicate the turbulence dissipation region as either instrument noise contaminates the signal at the highest frequencies - burying the actual turbulence signal which is decaying, or we have not sampled fast enough to resolve the dissipative frequencies. Looking at our data it appears to be a bit of both as there is evidence of the spectra beginning to flatten out at the highest frequencies, an indication of noise starting to dominate. Note that the integral of the spectrum yields the variance (0.0029 m 2 /s 2 ). We can get a rough estimate of where
8 Canonical Turbulent Flows the dissipation region occurs by estimating the Kolmogorov length scale as η = 0.1mm (typical smallest η in water in laboratory flows we will discuss how to better estimate this value later) and noting that the mean flow is moving at about 0.15 m/s therefore the smallest scales are crossing the measurement volume in a time τ = η/u =0.001 mm/0.15 m/s = 0.0067 s, or at a rate of 150 Hz. We would need to sample at twice this rate to resolve 150 Hz, right?! (Nyquist criteria). Hence by sampling at 100 Hz (and resolving only down to 50 Hz due to Nyquist criteria) we would expect to see perhaps just the beginning of the dissipation region but not much. The autocorrelation function (averaging records of length of 512 points) is shown below and on the top of the next page. We see that for the first 0.1 s or so the signal is strongly correlated to itself and this correlation then rapidly decays. After this, the signal slowly decays to a weakly negative correlation (second plot calculated with just 2 ensembles - N/2=9728 points, or 97.28 second length records). This all suggests that the turbulence forgets its history on the order of 0.1-1 second and behaves essentially randomly after this point. In fact, we define the autocorrelation time scale as the integral of the autocorrelation function. Integrating the second figure we find a time scale of 0.31 s. This is important information if one is interested in designing an efficient system for determining the mean or other moments as data that is truly independent with respect to each point will converge toward the true statistical value the most rapidly hence we are often interested in knowing the autocorrelation time scale. R uu /R uu (0) 1 0.8 0.6 0.4 0.2 0 0.2 0 0.5 1 1.5 2 2.5 t (s)
Canonical Turbulent Flows 9 1 R uu /R uu (0) 0.8 0.6 0.4 0.2 0 0.2 0 10 20 30 40 t (s) 2.1.3 The Reynolds Stress u v If you apply Reynolds decomposition to the governing equations for fluid mechanics (the Navier-Stokes equations) and then average the equations over some time scale (which could be an infinite time in an ergodic flow) you can develop an expression forthefeatures of the mean quantities (e.g., the mean velocity and pressure fields). The linear terms in the Navier-Stokes equations just become the mean over the time scale of interest. However, decomposing the non-linear terms and averaging yields new terms. Let s take a look at why. Consider the two instantaneous quantities A and B which we decompose to A = A + A where the overline indicates a temporal average over the time scale of interest. Based on our definition of the Reynolds decomposition we have: A =0 Proof A = (A + A )=A + A = A + A A = 0 (2.4) and importantly the nonlinear product yields: AB = (A + A )(B + B )=(AB + AB + A B + A B ) (2.5) = AB + A B + AB + A B = AB + A B + A B + A B = AB + A B Where A B is the correlation of fluctuating quantities. These types of terms are the source of turbulent transport and stirring (and great modeling challenges!) in turbulent flows.
10 Canonical Turbulent Flows Now, when you average the continuity equation and the Navier-Stokes equations you have linear terms x A = x A (2.6) where the temporal mean just passes through to the variable, why? Since the averaging process is a time integration we can move the spatial derivative outside the integral. e.g., 1 T t0 +T t 0 A x dτ = 1 x T t0 +T t 0 Adτ = A x Now we continue down Osborne Reynolds path and apply our decomposition to the Navier-Stokes equations under the assumption of ergotic incompressible flow. It is left as an exercise for the student to show that continuity just looks like: And the Navier Stokes equations become: u x + v y + w z = 0 (2.7) u u x + v u y + w u z u v x + v v y + w v z u w x + v w y + w w z = 1 ( P 2 ρ x + ν u x + 2 u 2 y + 2 u 2 z 2 ( 2 v = 1 P ρ y + ν = 1 P ρ z + ν ) u 2 ) x + 2 v 2 y + 2 v u v 2 z 2 x ( ) 2 w x + 2 w 2 y + 2 w 2 z 2 x u v y u w z v 2 y v w z (2.8) (2.9) u w x v w w 2 y z g (2.10) These equations, known as the Reynolds Averaged Navier Stokes equations, orrans equations, look essentially the same as the original equation with two exceptions: The unsteady temporal terms have been averaged out and a new set of termsemergesonthe right-hand-side. Looking more carefully at the new terms they actually arose from the nonlinear terms on the left-hand side but it is customary to move them to the right hand side as they can be viewed as the turbulent analog to the viscous stress terms, why? Let s look at these terms, known as the Reynolds stress terms, a bit more carefully. These terms are of the form u w. Let s try to get a physical sense for what these sort of terms are doing.
Canonical Turbulent Flows 11 Consider a turbulent shear flow with u/ z >0. A shear flow is any flow with a mean gradient. A picture of our flow might look like: If a parcel sitting at the point (x o,z o ) has a fluctuating vertical velocity w > 0 it has high probability that u < 0 since it is low momentum fluid moving to a region of high momentum. Now consider hundreds and thousands of such parcels, onaverageweex- pect u w < 0. If parcels have w < 0 then they have high probability that u > 0 since high momentum fluid is moving to a region of lower mean momentum u w < 0 again. Therefore u w < 0 in a turbulent shear flow with mean gradient u/ z >0. What does this mean? Consider that the w term is an advection term - meaning it is responsible for transporting stuff with it up or down (positive or negative). What is the stuff? It is u in our example what is u? Well, what if we multiply by the density, ρ? Wegetthatw is transporting ρu horizontal momentum! Thus we can think of u w as the turbulent vertical advection of streamwise turbulent momentum, or more simply the vertical flux of streamwise momentum. In general terms like u w are known as momentum flux terms. What would u c be (where c is a scalar concentration fluctuation)? A scalar flux - the turbulent streamwise flux of mass! If you make the same back-of-the-envelope analysis for u/ z <0youfindthatu w > 0.
12 Canonical Turbulent Flows Thus the sign of the momentum flux terms tell whether the flux is inducing a net increase or decrease in momentum (positive and negative, respectively). The momentum flux terms are frequently referred to as the Reynolds stress terms as they play an analogous role in the governing equations to the viscous stress. The sign convention is such that the Reynolds stress is most properly written in the form ρu v but frequently, the terms of the form u v will be referred to as the Reynolds stress. Looking at our Reynolds Averaged Navier Stokes (RANS) equations (equations 2.8-2.10) we see that there are six independent Reynolds stress terms, three tangential stress terms ρu v, ρu w,and ρv w and three normal stress terms, ρu 2, ρv 2,and ρw 2.We should note that for modelers this is a problem, we have introduced no new equations yet we have six new terms, i.e., six new unknowns! This is referred to as a closure problem as the RANS equations are not a closed set of equations. It is solved by developing models for the new terms based on resolved quantities. The best known of these models is perhaps the k ε model which developed the models for the Reynolds stress terms based on the turbulent kinetic energy k = 1 2 (u 2 + v 2 + w 2 )andε and includes two new equations to track k and ε. This is part of a broad class of solutions to the closure problem known as 2-equation models since two new equations are introduced. Significantly, experimentalist have no closure problem - we can directly measure the Reynolds stresses!