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AMME5202 Lecture 9 Summary: Turbulence Turbulence occurs when hydrodynamic instability prevents a steady flow from occurring. Energy which is contained in large scale motions with a size L is passed down to smaller length scales via an energy cascade until the scales are small enough for viscosity to dissipate their energy. One critical feature of this energy cascade is that it relies on three dimensional motion. Two dimensional simulations cannot correctly capture the turbulent energy cascade. The amount of turbulence in a flow is related to the Reynolds number, Re = uL ν . Different situation call for various choices of velocity and length scale used in this cal- culation. Turbulence presents a significant problem for computational fluid dynamics. In order to simulate a turbulent flow, the simulation must use a sufficient grid resolution in order to capture all turbulent length scales of the flow. It is also impossible to use a two dimensional simulation so even flows which appear two dimensional such as the ducts, flows over steps and over airfoils that we have studied must be run using a third dimension. Also the simulation will never reach a steady state, generally will need to run until a large number of time samples are calculated in order to capture time averaged quantities of interest. The smallest length scales are given by the Kolmogorov length scale l K . The ratio between the Kolmogorov length scale and the energy containing scales is, L l K ∼ Re 3 4 In order to illustrate the computational requirements for direct numerical simulation of turbulent flows, let us consider a duct with a typical engineering scale Reynolds number of 100,000. Assuming that we have a grid size equal to the Kolmogorov length scale and a domain size at least equal to the energy containing scale then the number of grid points is roughly proportional to Re 3 4 . Now since we need that many cells in three dimensions, the total number of cells is, n ∼ Re 9 4 = 1 . 8 × 10 11 . If each cell requires the storage of four variables using 8 bytes of computer memory each then that is roughly 5 . 8 × 10 12 bytes or 5.8 TB of memory required. This scale of simulation is still not feasible on the worlds largest supercomputers. This is why turbulence modelling is an important area of research. 1

Tensor notation In the remainder of this lecture we will introduce tensor notation. Tensors are a gen- eralized way to represent scalars, vectors and matrices as well as higher rank tensors which have no equivalent. A scalar is a rank zero tensor, a vector is a rank one tensor and a matrix is a rank two tensor. In this course we will use Greek subscripts to denote tensors, for example a vector is written as, u ˜ = u α , where α can take three values, 1, 2 and 3 corresponding to the x , y , and z directions. In tensor notation summation is implied over repeated indices, for example we would write the dot product as, u ˜ · v ˜ = u α v α = u x v x + u y v y + u z v z . The del operator ∇ ˜ , a vector of derivatives, is written as, ∇ ˜ = ∂ α . The gradient of a scalar is written using the del operator, ∇ ˜ φ = ∂ α φ. The divergence of a vector is written as, ∇ ˜ · u ˜ = ∂ α u α . A matrix vector product can be written as, A αβ u α =   A xx u x + A yx u y + A zx u z A xy u x + A yy u y + A zy u z A xz u x + A yz u y + A zz u z   . A commonly used tensor is the Kronecker delta, δ αβ = 1 : α = β 0 : otherwise The Kronecker delta is equivalent to the identity matrix. Multiplication by the Kronecker delta where one index is repeated is equivalent to replacing the repeated index by the other one, A αβ δ βγ = A αγ The trace of a matrix is a scalar, it may be written as, A αβ δ αβ = A xx + A yy + A zz 2

The cross product makes use of the Levi-Civita symbol of rank and dimension three. It’s value is 0 if any indices are repeated, 1 if the indices are all unique and form a cyclic permutation of { 1 , 2 , 3 } and - 1 if the indices are unique and form an anti-cyclic permutation of { 1 , 2 , 3 } , ε αβγ =      +1 for { α, β, γ } = { 1 , 2 , 3 } or { 2 , 3 , 1 } or { 3 , 1 , 2 } - 1 for { α, β, γ } = { 3 , 2 , 1 } or { 2 , 1 , 3 } or { 1 , 3 , 2 } 0 otherwise The definition of a cross product is, u ˜ × v ˜ = ε αβγ u β v γ . Notice that two of the indices are repeated and hence summed over, one index is not repeated so the result is a vector. The convective form of the Navier-Stokes equations can be written as, ∂ α ρu α = 0 ∂ t u α + u β ∂ β u α = - 1 ρ ∂ α P + ν∂ β ∂ β u α Reynolds-averaged Navier-Stokes equations In order to model the effects of turbulence, the flow variables u α and P are split into mean and fluctuating quantities, ¯ u α = 1 t 1 - t 0 Z t 1 t 0 u α d t Now the fluctuating component is, u 0 α = u α - ¯ u α . The same definitions are used for mean and fluctuating pressure. These definitions are substituted into the Navier-Stokes equations, ∂ α u α = 0 ∂ t u α + ∂ β u α u β = - 1 ρ ∂ α P + ν∂ β ∂ β u α and the time average of the entire set of equations is taken resulting in, ∂ α (¯ u α + u 0 α ) = 0 ∂ t (¯ u α + u 0 α ) + ∂ β ( ¯ u α + u 0 α )( ¯ u β + u 0 β ) = - 1 ρ ∂ α ( ¯ P + P 0 ) + ν∂ β ∂ β (¯ u α + u 0 α ) . 3

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Now we will make use of some identities, ( a + b ) = ¯ a + ¯ b. The time average of a time averaged quantity is just the time average, ¯ a = ¯ a. The time average of a fluctuating quantity is zero, a 0 = a - ¯ a = ¯ a - ¯ a = 0 . Therefore the continuity equation for the mean velocity is, ∂ α ¯ u α = 0 . Similarly these identities reduce the momentum equation to, ∂ t ¯ u α + ∂ β ( ¯ u α + u 0 α )( ¯ u β + u 0 β ) = - 1 ρ ∂ α ¯ P + ν∂ β ∂ β ¯ u α We will simplify the non-linear term by making use of another identity, ¯ ab = ¯ a ¯ b. Since the mean quantity ¯ a doesn’t vary with time, it can be taken outside of the time integral used to take the average. Now we can simplify the non-linear term, ∂ β ( ¯ u α + u 0 α )( ¯ u β + u 0 β ) = ∂ β ¯ u α ¯ u β + ∂ β u 0 α ¯ u β + ∂ β ¯ u α u 0 β + ∂ β u 0 α u 0 β = ∂ β ¯ u α ¯ u β + ∂ β u 0 α u 0 β Substituting into the momentum equations and moving the new term to the right hand side we have, ∂ t ¯ u α + ∂ β ¯ u α ¯ u β = - 1 ρ ∂ α ¯ P + ν∂ β ∂ β ¯ u α - ∂ β u 0 α u 0 β | {z } Reynolds stress The equations satisfied by the mean velocity and pressure fields are almost identical to the Navier-Stokes equations satisfied by the full time varying fields however with the addition of the Reynolds stress terms which are the result of velocity fluctuations around the mean. 4

Mixing length turbulence model The Reynolds averaged Navier-Stokes ( rans ) equations are, ∂ t ¯ u α + ¯ u β ∂ β ¯ u α = - 1 ρ ∂ α ¯ P α + ν∂ β ∂ β ¯ u α - ∂ β u 0 α u 0 β | {z } Reynolds stress . The eddy viscosity model assumes that the Reynolds stress term may be modelled as, - u 0 α u 0 β = ν t ( ∂ α ¯ u β + ∂ β ¯ u α ) , The mixing length turbulence model, for a wall normal to the y direction drops all derivatives in the eddy viscosity approximation except for ∂ y ¯ u . In other words, the Reynolds stress tensor in two dimensions is, - u 0 u 0 = - v 0 v 0 = 0 , - u 0 v 0 = - v 0 u 0 = ν t ∂ y ¯ u x . In order to calculate the eddy viscosity, the mixing length turbulence model assumes that the eddy viscosity is proportional to the mixing length l mix squared multiplied by the magnitude of the velocity gradient normal to the wall. ν t = l 2 mix · | ∂ y ¯ u | The mixing length is generally not known. As is common in turbulence modelling, some factors need to be determined experimentally. The mixing length may be approximated using the Nikuradse formula in terms of the channel half height δ as, l mix δ = 0 . 14 - 0 . 08 1 - y wall δ 2 - 0 . 06 1 - y wall δ 4 Here y wall is the distance from the nearest wall. Writing out the two dimensional rans equations in full, ∂ t ¯ u + ¯ u∂ x ¯ u + ¯ v∂ y ¯ u = - 1 ρ ∂ x ¯ P + ν∂ x ∂ x ¯ u + ν∂ y ∂ y ¯ u - ∂ x u 0 u 0 - ∂ y u 0 v 0 , ∂ t ¯ v + ¯ u∂ x ¯ v + ¯ v∂ y ¯ v = - 1 ρ ∂ y ¯ P + ν∂ x ∂ x ¯ v + ν∂ y ∂ y ¯ v - ∂ x v 0 u 0 - ∂ y v 0 v 0 . Now substituting the mixing length parameters, ∂ t ¯ u + ¯ u∂ x ¯ u + ¯ v∂ y ¯ u = - 1 ρ ∂ x ¯ P + ν∂ x ∂ x ¯ u + ν∂ y ∂ y ¯ u + ∂ y ( l 2 mix · | ∂ y ¯ u | · ∂ y ¯ u ) , ∂ t ¯ v + ¯ u∂ x ¯ v + ¯ v∂ y ¯ v = - 1 ρ ∂ y ¯ P + ν∂ x ∂ x ¯ v + ν∂ y ∂ y ¯ v + ∂ x ( l 2 mix · | ∂ y ¯ u | · ∂ y ¯ u ) . 5

In your code for the assignment, you may find it easier to calculate the gradient of the Reynolds stress term in two steps, first calculating, - u 0 v 0 = l 2 mix · | ∂ y ¯ u | · ∂ y ¯ u at all grid points, then calculating, - ∂ y u 0 v 0 and - ∂ x v 0 u 0 during a second pass. You should use a central approximation for the derivatives in these calculations. Mixing length model for a one dimensional duct Consider the rans equations with a mixing length model in one dimension, the wall normal direction. In this case there are no variations along the duct. This is a valid assumption for fully developed flow. In this case the problem is simplified to an ordinary differential equation. ∂ t ¯ u = - 1 ρ ∂ x ¯ P + ν∂ y ∂ y ¯ u + ∂ y ( l 2 mix · | ∂ y ¯ u | · ∂ y ¯ u ) . This flow is not driven by an inlet boundary condition but rather by an imposed pressure gradient which is equivalent to an applied body force f , ∂ t ¯ u = f + ν∂ y ∂ y ¯ u + ∂ y ( l 2 mix · | ∂ y ¯ u | · ∂ y ¯ u ) . (1) In order to understand the equivalence between the body force and the wall shear, consider the balance of forces on the fluid in a rectangular prism duct with half height δ . fV = τA, where A is the area of the wall and V is the volume of the duct. Thus the body force can be calculated from your duct half height and wall shear, f = τ δ . It is interesting to obtain the fully developed one–dimensional turbulent duct flow by solving the ordinary differential equation above, and comparing to the solution obtained by solving the Reynolds averaged Navier-Stokes equations in Assignment 3. 6

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