Introduction to Turbulence and Turbulence Modeling

Transcription

1 Introduction to Turbulence and Turbulence Modeling Part I Venkat Raman The University of Texas at Austin Lecture notes based on the book Turbulent Flows by S. B. Pope

2 Turbulent Flows Turbulent flows Commonly found in nature Relevant to industrial devices Contain random fluid motion

3 Importance of Turbulent Flows Naturally occurring in practical flows Necessary for many applications - Enhanced fuel/air mixing in internal combustion and aircraft engines Reduces size of engines drastically May not be desired in others - Increases drag on surfaces Reduced efficiency Key theoretical issues How do we characterize/study turbulence? What are the essential features of turbulent flow?

4 Equations of Motion Fluid motion governed by Navier-Stokes equations ρu i + ρu ju i = p + τ ij, t x j x i x j ( Ui τ ij = µ + U ) j 2 x j x i 3 µδ U j ij x j Solved along with continuity equation ρ t + ρu j =0 x j Need to solve energy equation (if needed) along with ideal gaslaw to close the system Density assumed to be constant here

5 Reynolds Number Reynolds number is a key non-dimensional quantity Re = UL µ = U(U/L) µu/l 2 Ratio of inertial to viscous forces - Alternate definition based on length scales (to be discussed) High Re associated with turbulence Exact number at which transition occurs is dependent on flow conditions - Possible to maintain laminar flow at very high Re numbers as well

6 Scales of Turbulence Turbulent flows involve a range of length-scales Characteristic lengths that are associated with certain flow features A wide range of length-scales possible One representative length-scale

7 Scales of Turbulence Turbulent motion composed of eddies of different sizes Let eddy of size l have a velocity u(l) Associated timescale: Let the largest eddies be characterized by lengthscale Comparable to the flow scale τ(l) =l/u(l) Velocity of these eddies proportional to turbulence intensity u (2/3k) 1/2 Reynolds number for this size: Re 0 u 0 l 0 /ν = u l 0 /ν What happens to these large eddies? L l 0

8 Energy Cascade Large eddies are unstable and break up into smaller eddies Also transfer their energy to the smaller eddies When Reynolds number is high, the viscous forces are very small compared to the inertial forces - Viscous dissipation is small at large eddy sizes Energy cascade Eddies break up until the characteristic size and velocity decrease - Note: Re(l) =u(l)l/ν - Viscous dissipation finally kills the turbulent kinetic energy

9 Energy Dissipation Rate Large-scale eddies have energy Dissipation rate u 2 0 ϵ = u 2 0/(l 0 /u 0 )=u 3 0/l 0 Large-scales determine the dissipation rate Break up of these eddies start the cascade process Dissipation rate is independent of viscosity - However, viscosity itself is the cause of dissipation Verified from experiments Valid only for high Reynolds number flows

10 Kolmogorov Hypotheses Hypothesis of local isotropy At sufficiently high Reynolds number, the small-scale turbulent motions ( ) are statistically isotropic - Implies that all geometric information is lost as energy is transferred through the length-scales - All information about the boundaries and geometry lost - Consider length scale l EI defined such that Anisotropic for Isotropic for Roughly, l<<l 0 l>l EI l<l EI l EI l 0 /6

11 Kolmogorov Hypotheses First similarity hypothesis In every turbulent flow at sufficiently high Reynolds number, the statistics of small-scale motions have a universal form that is uniquely determined by and - Leads to the following length, velocity, time scales Kolmogorov length Velocity Timescale u η (ϵν) 1/4 τ η (ν/ϵ) 1/2 η ( ν 3 /ϵ ) 1/4 - Velocity and length scales decrease with eddy size - Kolmogorov length is comparable to the size of the smallest eddies that are found in a turbulent flow ν ϵ

12 Reynolds Number Reynolds number also represents information about the range of scales Re 3/4 l 0 As Re increases, the range of scales increases u u 0 Re 1/4 Velocity-scale at smallest scales decreases w.r.t large scale velocity 0 Re 1/2 Time-scale at the small scales also decrease w.r.t large scale

13 Kolmogorov Hypotheses Second similarity hypothesis At sufficiently high Reynolds numbers, the statistics of the motions of scale l in the range l 0 >> l >> η have a universal form that is uniquely determined by ϵ, independent of ν - This universal range referred to as the inertial subrange - Motions determined by inertial forces Dissipation range Inertial Subrange Energy-containing range η l DI l EI l 0 L

14 Turbulent Energy Spectrum Energy spectrum provides the distribution of energy among the different lengthscales Plotted as a function of wavenumber k = 2π l Shape deduced from theory confirmed by experiments

15 Numerical Simulation of Turbulence Direct numerical simulation (DNS) Solve the Navier-Stokes equations on a computational grid DNS needs to resolve all time and length scales of turbulence Grid-width should be comparable to Kolmogorov length scale Computational domain proportional to integral length scale Number of grid points (in each direction) depends on Reynolds number N / l 0 =) N / Re3/4

16 DNS Cost Timestep also proportional to Reynolds number Total time of integration comparable to integral time scale Time-step based on Courant-number restriction u 0 t x 1 0 l 0 /u 0 N t = 0 t l 0/ x Re 3/4 Total cost = number of grid points X number of time steps Cost / Re 3

17 DNS Uses DNS is often used to test turbulence models (to be discussed next) Isotropic Turbulence However, DNS is often performed at low Re numbers Computational limitations

18 Example DNS using 4096^3 grid points Performed at the Earth Simulation Center in Japan Estimated computational time ~ 240 hrs/eddy turnover time Computed on 2000 processors

19 Turbulence Modeling Direct simulation of governing equations is not practical Reduced descriptions sought - Leads to turbulence modeling Two main approaches Reynolds averaged Navier-Stokes (RANS) approach Large eddy simulation (LES) approach Other approaches Partially-averaged NS (PANS) Hybrid RANS/LES

20 RANS Approach Main motivation For engineering applications, only statistics of the flow field required - E.g. average velocity and its variance RANS formulates equations for moments of the flow variables RANS formulation based on one-point one-time probability density function (PDF) of flow variables Key properties such as dissipation are two point quantities and require modeling

21 Ensemble Averaging Consider a turbulent flow prescribed using a set of boundary and initial conditions These conditions are insufficient to describe the flow completely Each realization thus produces variations due to the chaotic nature of turbulence Ensemble averages are defined as NX hu i i = 1 N j=1 U j i U i = U i + u i Y i = Y i + y i T = T + T i

22 RANS Profiles DNS RANS

23 Favre-Averaging For variable density flows, Reynolds-averaging leads to terms with fluctuating density ρut = ρ U T + ρ ut + ρ u T + ρ T U + ρ ut Instead, Favre averages are defined U = Ũ + u ρu =0 ρu = ρũ + ρu = ρũ ρut = ρũ T + ρ u T

24 Reynolds Stresses Turbulence modeling essentially refers to closure models for the Reynolds stress tensor Arise from the nonlinear term in the momentum equation ρu i t + ρu ju i x j = p x i + τ ij x j, Reynolds stress hu i U j i = hu i ihu j i + hu i u j i Reynolds stresses cannot be obtained directly from the mean velocity field Signifies transfer of momentum by fluctuating field

25 Properties of Reynolds Stresses Second-order tensor Symmetric Diagonal terms are normal stresses, off-diagonal are shear stresses Turbulent kinetic energy is half the trace of tensor k = 1 2 hu iu i i Two approaches to closing Reynolds stresses Turbulent viscosity based models Reynolds-stress transport equations

26 Turbulent Viscosity based Models u i u j requires closure Most common closure approach is based on the turbulentviscosity hypothesis Introduced by Boussinesq (1877) Relates the deviatoric part of the Reynolds stress to the mean rate of strain ρ u i u j = µ t ( Ui x j + U j x i ) 2 3 ρδ ijk k = 1 2 u ju j Need models for and µ t k - Several models have been proposed

27 Example: Turbulent Round Jet Round jet issuing into quiescent surroundings Exhibits self-similar behavior quiescent surroundings x 0.2 x/d = 30 W U V r U /U J U J θ 0.1 d nozzle x/d = 60 x/d = 100 fluid supply r/d

28 Round Jet: Self-similar Profile Normalized based on half-width of jet hu(x, r 1/2, 0)i = 1 hu(x, 0, 0)i U /U J J x/d = 30 x/d = 30 U U U J U 0 (x) x/d = 60 x/d = r/r 1/ x/d x/d = x/d = 100 r/d r/d

29 Reynolds Stresses in Round Jet Stress tensor 2 hu 2 i huvi 0 4huvi hv 2 i hw 2 i u i u j U u 2 w 2 v 2 U uv r/r 1/ η r/r 1/2 3.0

30 Turbulent Viscosity Turbulent viscosity is also self-similar T (x, r) =U 0 (x)r 1/2 (x)b T ( ) Turbulent viscosity in terms of length scale T = u 0 l 0.15 ν ^ T l r 1/ r/r 1/ r/r 1/ η

31 Transport of Kinetic Energy Transport of kinetic energy is important to understand for modeling purposes Derived from governing equations (no i i = P T i = 1 2 hu iu j u j i + hu i p 0 i/ s ij = j i 2 hu j s ij i Production and dissipation terms are important P = hu i u j j =2 hs ij s ij i

32 Budget of Kinetic Energy Equation Production and dissipation P = Turbulent transport hu i u j j T i = 1 2 hu iu j u j i + hu i p 0 i/ i =2 hs ij s ij i 2 hu j s ij i Gain Loss r/r 1/2 Production Mean-flow convection Turbulent transport Dissipation

33 Model for Kinetic Energy Production Turbulent viscosity hypothesis 2 a ij = hu i u j i 3 k ij = j i Relates anisotropy tensor to strain rate Production related to anisotropy tensor P = a ij S ij S ij = 1 2 a ij = 2 T S j i P =2 T S ij S ij According to the model, production is always positive

34 Transport Equation for Kinetic Energy Turbulent transport modeled using gradient diffusion hypothesis T i = 1 2 hu iu j u j i + hu i p 0 i/ 2 hu j s ij i Modeled kinetic energy equation T i = T Dissipation rate still needs to be modeled - Focus now shifts to modeling this @x i, k i i + P

35 One-equation Model k related to turbulent viscosity using mixing-length T = ck 1/2 l m Coefficient is roughly 0.55 l m is the mixing length and needs to be specified by the user needs to be obtained separately k ρk t + U j k x j = x j ( [µ + µ t ] k x j ) ( Ui + µ t + U ) j Ui x j x i x j ϵ ϵ nows needs to be modeled - Represents energy dissipation rate = C Dk 3/2 l m

36 k ϵ Turbulence Model Dissipation rate can be solved using a transport equation Equation based on very strong closure assumptions ρϵ t + ρ U j ϵ x j = x j ( µt ( Ui P k = µ t + U ) j Ui x j x i x j µ t = ρc µ k 2 Several modeling constants σ ϵ ϵ x j σ ϵ =1.3 C ϵ1 =1.44 C ϵ2 =1.92 ϵ ) + C ϵ1 P k ϵ k ρc ϵ2 ϵ 2 k

37 k Turbulence Model In most RANS viscosity-based models, kinetic energy is set as one variable A second variable is needed to obtain length scale Several choices possible Treat = /k as the i + C 1 P k C 2 2 Expression for turbulent viscosity same as before Numerical issue when treating cases with non-turbulent regions (k=0)

38 Reynolds-Stress Transport Models Alternate approach to turbulent-viscosity methods Directly solves transport equations for the Reynolds stress i u j kihu i u j k T kij = P ij + R ij ij Production term (P) appears closed Turbulent transport (T), pressure-rate-of-strain tensor (R) and dissipation tensor ( ij ) need to be modeled Dissipation rate still needed Modeled using equation similar to the k-e model Except that the production term is directly obtained from the Reynolds stresses

39 Reynolds Stress Model Directly solves transport equations for the Reynolds stresses No need to use turbulent-viscosity hypothesis In three-dimensional flows, six additional equations have to be solved In addition, a transport equation for dissipation has to be solved Found to be more accurate than the two-equation closures for a variety of complex flows - Swirling flows, flows with stagnation points etc. However, many closures required leading to higher degree of model uncertainty

40 Issues in RANS Modeling RANS performance highly sensitive to turbulence models Models are based on simple flows Coefficients are tuned based on canonical flows and are not universal Modeling burden is too high with information only on mean quantities available - In spite of decades of developments, models are still not sufficiently accurate for complex flows RANS unable to utilize the growing power of computers RANS accuracy is not related to the availability of computing power