Turbulence and CFD models 1 Roadmap 1. Transition to turbulence in - - PowerPoint PPT Presentation

Turbulence and CFD models

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Roadmap

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• 1. Transition to turbulence in shear flows

Transition to turbulence in shear flows

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Transition to turbulence in shear flows

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Motivation for transition work

Transition to turbulence in shear flows

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Motivation for transition work

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Motivation for transition work

Control:

Added benefits: CO2 emission reductions and reduced operatings costs

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Motivation for transition work

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The usual picture

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Effect of roughness on skin friction

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• When/where/why/how do instabilities start?
• Why does roughness affect skin friction?
• What kind of waves are most likely to be amplified?
• Can they be controlled (eliminated, anticipated,

delayed)?

• How long does transition last?
• Once the turbulent flows sets in does it present a

universal character?

• Can we control turbulence?

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(analytical) (numerical)

The usual methodology starts with:

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( reduced frequency)

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Environmental conditions

The type(s) of disturbances which grow, their self- or mutual interactions, and the amount by which perturbations are amplified (in other words, the transition process) depend on the forcing conditions provided by the environment.

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Receptivity

Broadly speaking, the manner in which exogeneous perturbations [sound waves (irrotational), free-stream turbulence (rotational), leading edge curvature and/or vibrations, gusts, vortical structures, wall roughness, discontinuities in surface curvature at junction LE/flat plate,

• etc. …] enter the boundary layer and are filtered, eventually

turning into instability waves, determines the path to turbulence, the coherent flow structures arising, the ‘critical’

• r ‘transitional’ Reynolds number, the skin friction and heat

transfer to/from the wall.

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Morkovin, 1994

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Standard scenario for 2D boundary layer (A)

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Standard scenario for 2D boundary layer (A)

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Experiments: smoke and laser light sheet

K-type transition H-type transition Streaks-induced transition

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Turbulent spot (Matsubara & Alfredsson 2005) Sinuous instability Varicose instability

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mean and rms values ,

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mean and rms values , , turbulence intensity

• Flight conditions and

few wind tunnels: Tu < 0.1%

• Most wind tunnels:

Tu < 1%

• Turbines/compressors:

Tu > 10%

Wind tunnels can give trends opposite to flight

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Experiments versus theory (TS waves)

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… and DNS

Schlatter, 2009

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Luchini, 2000 (large free stream Tu)

Experiments versus theory (streaks)

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Zaki & Durbin, 2000 (large free stream Tu)

… and DNS

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The initial stages of transition

Except for the cases of transition scenarios D or E, small disturbances are initially filtered and amplified; this justifies focussing on the growth

• f infinitesimal perturbations: the

equations are thus linearized. Nonlinear interactions acquire importance only once the amplitude

• f the disturbances becomes large

enough.

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Recap on linear matrix algebra

𝑒𝒚 𝑒𝑢 = 𝒈[𝒚 𝑢 , 𝑢; 𝑠] Generic evolution system: x = state vector (N components, column vector) f = evolution function (another N-column vector) t = time r = control parameter 𝑒𝒚 𝑒𝑢 = 𝒈[𝒚 𝑢 ; 𝑠] Autonomous system:

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Recap on linear matrix algebra

𝑒𝒚 𝑒𝑢 = 𝒈[𝒚 𝑢 ; 𝑠] Statement of the problem: Predict the characteristics of the asymptotic state (t → ∞) as function of the initial conditions and the control parameter. Note: we will see later that the behavior of the system for small times is also of importance

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Recap on linear matrix algebra

𝑒𝒚0 𝑒𝑢 = 𝒈[𝒚0; 𝑠] Basic state: x0 that satisfies Perturbation: 𝜗x’(t) (𝜗 small amplitude) satisfying 𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′(𝑢)

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Recap on linear matrix algebra

𝑒𝒚0 𝑒𝑢 + 𝜗 𝑒𝒚′ 𝑒𝑢 = 𝒈 𝒚0 + 𝜗 𝒚′ = 𝒈(𝒚0) + 𝜗 ቚ

𝜖𝒈 𝜖𝒚 𝒚0

𝒚′ + O (𝜗2) = 𝒈(𝒚0) + 𝜗 𝑩 𝒚′ + O (𝜗2) 𝑩 = Jacobian matrix of coefficients (N x N ) 𝑢 = 0+ 𝒚 = 𝒚0 + 𝜗 𝒚′

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Recap on linear matrix algebra

𝒚′ 𝑢 = 𝒚′ 0 + 𝑢 ฬ

𝑒𝒚′ 𝑒𝑢 𝑢=0

+

𝑢2 2

ฬ

𝑒2𝒚′ 𝑒𝑢2 𝑢=0

+ … Setting the eigenproblem

𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′, 𝑒2𝒚′ 𝑒𝑢2 = 𝑩2 𝒚′, … 𝑒𝑜𝒚′ 𝑒𝑢𝑜 = 𝑩𝑜 𝒚′

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Recap on linear matrix algebra

𝒚′ 𝑢 = 𝒚′ 0 + 𝑢 ฬ

𝑒𝒚′ 𝑒𝑢 𝑢=0

+

𝑢2 2

ฬ

𝑒2𝒚′ 𝑒𝑢2 𝑢=0

+ … Setting the eigenproblem

𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′, 𝑒2𝒚′ 𝑒𝑢2 = 𝑩2 𝒚′, … 𝑒𝑜𝒚′ 𝑒𝑢𝑜 = 𝑩𝑜 𝒚′

𝒚′ 𝑢 = 𝒚′ 0 + 𝑢 𝑩 𝒚′ 0 +

𝑢2 2 𝑩2 𝒚′ 0 + …

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Recap on linear matrix algebra

Setting the eigenproblem 𝒚′ 𝑢 = 𝒚′ 0 + 𝑢 𝑩 𝒚′ 0 +

𝑢2 2 𝑩2 𝒚′ 0 + …

= σ𝑜=0

∞ 𝑢𝑜𝑩𝑜 𝑜!

𝒚′ 0 Definition of the analytic function of a matrix:

𝑓𝑩𝑢 = ෍

𝑜=0 ∞

𝑢𝑩 𝑜 𝑜!

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Recap on linear matrix algebra

Setting the eigenproblem The solution of our disturbance problem is thus: and to assess the stability of the system it is useful to decompose the matrix A in the sum of products

• f left and right eigenvectors

𝒚′ 𝑢 = 𝑓𝑩𝑢 𝒚′ 0

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Recap on linear matrix algebra

Setting the eigenproblem The N eigenvalues of the matrix A are the solutions lk of the characteristic equation The right eigenvectors uk are non-trivial solutions, defined up to an arbitrary factor, of the system: det 𝑩 − lk𝑱 = 0 𝑩 𝒗𝑙 = lk 𝒗𝑙

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Recap on linear matrix algebra

Setting the eigenproblem The left eigenvectors 𝒘k are non-trivial solutions, defined up to an arbitrary factor, of the system: Note: the left eigenvectors of A are also the right eigenvectors of the conjugate transpose of A 𝒘𝑙

𝑈 ഥ

𝑩 = ഥ lk 𝒘𝑙

𝑈

ഥ 𝑩𝑈𝒘𝑙 = ഥ lk 𝒘𝑙

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Recap on linear matrix algebra

Definition of the scalar product between (in general complex) vectors: (𝒗𝑙, 𝒘𝑙) ≡ 𝒗𝑙

𝑈 𝒘𝑙

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Recap on linear matrix algebra

Definition of the scalar product between (in general complex) vectors: (𝒗𝑙, 𝒘𝑙) ≡ 𝒗𝑙

𝑈 𝒘𝑙

Definition of the adjoint matrix: 𝑩𝒗, 𝒘 = 𝑩𝒗𝑈𝒘 = ഥ 𝒗𝑈ഥ 𝑩𝑈 𝒘 = 𝒗, ഥ 𝑩𝑈𝒘 𝑩† ≡ ഥ 𝑩𝑈

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Recap on linear matrix algebra

Adjoint operators/matrices are important in many areas, including

• hydrodynamic stability, receptivity, sensitivity
• ptimal and robust control theory
• ptimal shape design
• inverse design
• data assimilation

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Recap on linear matrix algebra

𝑩† = 𝑩 If , the matrix A is self-adjoint In this case the matrix is a real, symmetric matrix, its eigenvalues are real and the eigenvectors form an orthogonal basis. Furthermore, left and right eigenvectors coincide. A non-self-adjoint matrix has, in general, complex eigenvalues, plus its conjugates.

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Recap on linear matrix algebra

Property of orthogonality among eigenvectors Thus (lk- lℎ) 𝒘ℎ,𝒗𝑙 = 0, or 𝒘ℎ,𝒗𝑙 = a dhk a is some amplitude coefficient; if a = 1 left and right eigenvectors are orthonormalized 𝒘ℎ, 𝑩 𝒗𝑙 = 𝒘ℎ, lk𝒗𝑙 = lk 𝒘ℎ,𝒗𝑙 ഥ 𝑩𝑈𝒘ℎ, 𝒗𝑙 = ഥ lℎ 𝒘ℎ, 𝒗𝑙 = lℎ 𝒘ℎ,𝒗𝑙

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Recap on linear matrix algebra

Let us imagine that the N eigenvalues are distinct and the eigenvectors are linearly independent (so as to form a basis); at t = 0 we have 𝒚′(0) = ෍

𝑙=1 𝑂

𝒗𝑙 𝑑𝑙 𝒘ℎ, 𝒚′ = σ𝑙=1

𝑂

𝒘ℎ, 𝒗𝑙 𝑑𝑙 = σ𝑙=1

𝑂

𝑑𝑙 𝒘ℎ, 𝒗𝑙 = 𝑑ℎ 𝒘ℎ, 𝒗ℎ 𝑑ℎ = 𝒘ℎ, 𝒚′ 𝒘ℎ, 𝒗ℎ

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Recap on linear matrix algebra

Let us imagine that the N eigenvalues are distinct and the eigenvectors are linearly independent (so as to form a basis): at t = 0 we have 𝒚′ 0 = 𝒚′0 = ෍

𝑙=1 𝑂

𝒗𝑙 𝑑𝑙 𝒘ℎ, 𝒚′0 = σ𝑙=1

𝑂

𝒘ℎ, 𝒗𝑙 𝑑𝑙 = σ𝑙=1

𝑂

𝑑𝑙 𝒘ℎ, 𝒗𝑙 = 𝑑ℎ 𝒘ℎ, 𝒗ℎ 𝑑ℎ = 𝒘ℎ, 𝒚′0 𝒘ℎ, 𝒗ℎ

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Recap on linear matrix algebra

Let us assume that the eigenvectors are

• rthonormalized → , then

𝒚′0 = σ𝑙=1

𝑂

𝒗𝑙 𝒘𝑙, 𝒚′0 = σ𝑙=1

𝑂

𝒗𝑙 𝒘𝑙

𝑈𝒚′0 = 𝑱 𝒚′0

i.e. given that 𝒚′0 is any vector, the identity matrix can be retrieved from I = σ𝑙=1

𝑂

𝒗𝑙 𝒘𝑙

𝑈

𝑑𝑙 = 𝒘𝑙, 𝒚′0

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Recap on linear matrix algebra

The matrix A can thus be represented as 𝑩 = 𝑩 𝑱 = σ𝑙=1

𝑂

𝑩 𝒗𝑙 𝒘𝑙

𝑈 = σ𝑙=1 𝑂

lk 𝒗𝑙 𝒘𝑙

𝑈

i.e. A can be written as the sum of the product of eigenvalues and eigenvectors (of course, under the assumption that eigenvalues are distinct, and eigenvectors are linearly independent)

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Recap on linear matrix algebra

Let U be the matrix whose columns are the N right eigenvectors and V the matrix with the N left eigenvectors in the colums; assume eigenvalues to be

• distinct. When e-vectors are orthonormal:

𝒘ℎ, 𝒗𝑙 = 𝜀ℎ𝑙 → 𝒘ℎ

𝑈𝒗𝑙 = 𝜀ℎ𝑙 → ഥ

𝑾𝑈𝑽 = 𝑱 𝒘ℎ

𝑈𝑩 𝒗𝑙 = lk 𝒘ℎ 𝑈 𝒗𝑙 = lk𝜀ℎ𝑙

→ 𝑾𝑈𝑩 𝑽 = L (the diagonal matrix of e−values)

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Recap on linear matrix algebra

Let U be the matrix whose columns are the N right eigenvectors and V the matrix with the N left eigenvectors in the colums; assume eigenvalues to be

• distinct. When e-vectors are orthonormal:

𝒘ℎ, 𝒗𝑙 = 𝜀ℎ𝑙 → 𝒘ℎ

𝑈𝒗𝑙 = 𝜀ℎ𝑙 → ഥ

𝑾𝑈𝑽 = 𝑱 𝒘ℎ

𝑈𝑩 𝒗𝑙 = lk 𝒘ℎ 𝑈 𝒗𝑙 = lk𝜀ℎ𝑙

→ 𝑾𝑈𝑩 𝑽 = L (the diagonal matrix of e−values)

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Recap on linear matrix algebra for the problem let us take

𝒚′ = 𝑽 𝒓 𝑢 ,

𝑒𝒚′ 𝑒𝑢 = 𝑽 ሶ

𝒓, 𝑽 ሶ 𝒓 = 𝑩 𝑽 𝒓, ሶ 𝒓 = 𝑽−1𝑩 𝑽 𝒓, ሶ 𝒓 = L 𝒓 → 𝒓 𝑢 = 𝑓L𝑢𝒓(0) ഥ 𝑾𝑈𝑩 𝑽 = L → 𝑩 = U L ഥ 𝑾𝑈 𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′(𝑢)

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Recap on linear matrix algebra

so that the solution of is 𝒚′ = 𝑽 𝑓L𝑢 𝑽−1 𝒚′ 0 = 𝑽 𝑓L𝑢 ഥ 𝑾𝑈 𝒚′ 0

𝒓 𝑢 = 𝑽−1𝒚′ = 𝑓L𝑢 𝒓 0 = 𝑓L𝑢 𝑽−1 𝒚′ 0 𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′(𝑢)

𝒚′ = 𝑴 𝒚′(0)

propagator of the initial condition

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Recap on linear matrix algebra

𝑩 = ෍

𝑙=1 𝑂

lk 𝒗𝑙 𝒘𝑙

𝑈

𝑩2 = 𝑩𝑩 = ෍

𝑙=1 𝑂

lk 𝒗𝑙 𝒘𝑙

𝑈 ෍ ℎ=1 𝑂

lh 𝒗ℎ 𝒘ℎ

𝑈 =

σ𝑙=1

𝑂

σℎ=1

𝑂

lklh𝒗𝑙 𝒘𝑙

𝑈𝒗ℎ 𝒘ℎ 𝑈 =

σ𝑙=1

𝑂

σℎ=1

𝑂

lklh𝒗𝑙dℎ𝑙 𝒘ℎ

𝑈 =

σℎ=1

𝑂

lh

𝟑 𝒗ℎ 𝒘ℎ

𝑈

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Recap on linear matrix algebra

The matrix A𝑜 has the same eigenvectors as A, and eigenvalues which are lk

𝑜 . In general, for a linear

combination 𝑕 of powers of A 𝑩𝑜 = ෍

𝑙=1 𝑂

lk

𝑜 𝒗𝑙 𝒘𝑙

𝑈

𝑕(𝑩) = ෍

𝑙=1 𝑂

𝑕(lk) 𝒗𝑙 𝒘𝑙

𝑈

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Recap on linear matrix algebra

and in particular The solution of our linear problem reads also: with the left eigenvectors weighting the initial condition 𝑓𝑩 = ෍

𝑙=1 𝑂

𝑓lk 𝒗𝑙 𝒘𝑙

𝑈

𝒚′ 𝑢 = 𝑓𝑩𝑢 𝒚′ 0 = σ𝑙=1

𝑂

𝑓lkt 𝒗𝑙 (𝒘𝑙, 𝒚′ 0 )

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𝒚′ 𝑢 = σ𝑙=1

𝑂

𝑓lkt 𝒗𝑙 𝑑𝑙 Stability conditions

the eigenvalues lk define the asymptotic growth/decay

• f the disturbance.

Should there be a double eigenvalue, terms of the form t 𝑓lkt would appear in the expansion of the solution (resulting in a linear time growth of the disturbance even when Re(lk) < 0, for all k ).

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Stability conditions

An autonomous system, with evolution matrix A equipped with N distict eigenvalues is:

• Asymptotically stable is all eigenvalues of A have

negative real part

• Marginally stable if one (or more) eigenvalues

have real part equal to zero (and the others have negative real part)

• Unstable if at least one eigenvalue has real part

larger than zero

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Stability conditions

Near the marginal stability conditions, typically a single (1!) eigenvalue crosses the stability boundary, i.e. for 𝑢 → ∞ we have 𝒚′ 𝑢 ~ 𝑓l1t 𝒗1 (𝒘1, 𝒚 0 ) = 𝑓l1t 𝒗1 𝑑1 all other modes (k = 2, 3, 4, … N ) being damped.

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Stability conditions

lk = 𝜏𝑙 + 𝑗 𝜕𝑙 growth rate angular frequency The eigenvalue problem is 𝑩 𝒗𝑙 = l𝑙 𝒗𝑙 and in hydrodynamic stability analysis the 𝒗𝑙’s are called normal modes

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Stability conditions 𝐹 𝑢 = (𝒚′, 𝒚′) E rE rG rc r

unstable region

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Stability conditions

Stable Conditionally stable (III) Globally stable (II) Monotonically stable (I)

E rE rG rc r

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A simple example

Consider the very simple linear system (A in the chosen example is called a Jordan block): A has a double eigenvalue (l1 = l2 = 0) to which is associated the double eigenvector 𝒗1 = 𝒗2 =

1

𝑒𝒚′ 𝑒𝑢 = 𝑩 𝒚′ 𝑢 𝑩 =

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Example

If 𝒚′(0) =

𝑦10 𝑦20

is the initial condition (at t = 0) then the solution is 𝒚′ = 𝑦10 + 𝑢 𝑦20 𝑦20 i.e. the disturbance vector grows linearly in time (algebraic growth), despite the fact that the eigenvalues have vanishing real part.

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Example

Consider now the perturbed system with A = eigenvalues: l1 = −ϵ , l2= −2ϵ e-vectors:

𝒗1 = 1 0 , 𝒗2= 1 −ϵ , 𝒘1 = 1 1/ϵ , 𝒘2= −1/ϵ

and the solution is: 𝒚′ = 𝑑1 𝑓−𝜗𝑢 𝒗1 + 𝑑2 𝑓−2𝜗𝑢 𝒗2

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Example The question is: how do the two solutions (linear growth and exponential decrease) match as ϵ decreases to become ϵ = 0 ?

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Example The question is: how do the two solutions (linear growth and exponential decrease) match as ϵ decreases to become ϵ = 0 ? To answer we must focus on the energy of the disturbance, E(t) = (𝒚′, 𝒚′). For large times (when ϵt >> 1) the exponential behavior of the previous slide holds and eventually at large times the solution goes like 𝒚′~ 𝑓−𝜗𝑢 𝒗1 . What about short times?

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Example

For short times (𝜗t << 1) a Taylor series of the energy gives: and a linear growth in time is possible if the factor of (𝜗t) is negative. This growth is related to the fact that the two eigenvectors 𝒗1 and 𝒗2 are not orthogonal to one another (in fact, they are almost parallel !) 𝐹 𝑢 = 𝑑1 + 𝑑2 2 + 𝜗2𝑑2

2

− 2𝑑1

2 + 4 1 + 𝜗2 𝑑2 2 + 6𝑑1𝑑2

𝜗𝑢 + O 𝜗2𝑢2

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Example

𝜗t growth, then decay monotonic decay

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Example

The optimal initial condition It is easy to see that the initial condition which yields the largest gain, ratio of final to initial energy, for t large enough, is the first left eigenvector.

𝜗t growth, then decay monotonic decay

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Another simple example

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Another simple example

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Another simple example

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Another simple example

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Shear flow problems

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Shear flow problems

For the simple problems above, the flow is parallel or quasi- parallel and it is a good approximation to consider the velocity profile as

𝒗0 = 𝑉 𝑧 𝒋

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Hydrodynamic stability of // shear flows

• n solid boundaries

𝒗 = 𝒗0 + 𝜗 𝒗1 + 𝜗2𝒗2 + … 𝑞 = 𝑞0 + 𝜗 𝑞1 + 𝜗2𝑞2 + … 𝜗 ≪ 1 𝛼 . 𝒗 = 0

𝜖𝒗 𝜖𝑢 + (𝒗 .𝛼)𝒗 = − 𝛼 𝑞 + 1 𝑆𝑓 𝛼2𝒗

𝒗 𝒚, 0 assigned 𝒗 𝒚, 𝑢 = 0

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Hydrodynamic stability

𝛼 . 𝒗0 = 0 (𝒗0 .𝛼)𝒗0 = − 𝛼 𝑞0 +

1 𝑆𝑓 𝛼2𝒗0

𝛼 . 𝒗1 = 0

𝜖𝒗1 𝜖𝑢 + (𝒗1 .𝛼)𝒗0 + (𝒗0 .𝛼)𝒗1= − 𝛼 𝑞1 + 1 𝑆𝑓 𝛼2𝒗1

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Hydrodynamic stability

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Hydrodynamic stability

into the equations and collecting terms of O(𝜗):

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Hydrodynamic stability

The solution of the linear O(𝜗) problem relies crucially on the streamlines being parallel, i.e. U is independent of x. This makes the solution separable, e.g. ………………………………… . More particularly, it can be shown that the solutions are the sum of exponential functions of the invariant directions, x and t (Fourier modes). A normal mode has the form: ; this is a x-travelling wave, with a the streamwise wavenumber and w the circular frequency.

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Hydrodynamic stability

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Hydrodynamic stability

Boundary conditions

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Hydrodynamic stability

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Hydrodynamic stability

.

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Stability conditions

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The inviscid problem

Rayleigh inflection point theorem

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The inviscid problem

Rayleigh inflection point theorem The imaginary part of the equation above (a real) is ad this relation is satisfied for 𝑑𝑗 ≠ 0 only when the integral vanishes, which occurs only if 𝑉′′ ≡ 0 or 𝑉′′ changes sign at least once in

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The inviscid problem

Rayleigh inflection point theorem: a necessary, but not sufficient, condition for instability is that the velocity profile have an inflection point

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The inviscid problem

Fjørtoft’s theorem Let there be an inflection point at and let If 𝑑𝑗 ≠ 0 then The real part of the expression derived previously is

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The inviscid problem

Fjørtoft’s theorem Adding up leads to: a necessary, but not sufficient, condition for instability is that < 0, somewhere in the flow

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The inviscid problem

Fjørtoft’s theorem

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The inviscid problem

Howard’s semi-circle theorem the complex phase velocity lies inside, or on, the semi- circle centred on

𝑽𝒏𝒃𝒚 + 𝑽𝒏𝒋𝒐 𝟑

• f radius

𝑽𝒏𝒃𝒚 − 𝑽𝒏𝒋𝒐 𝟑

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The viscous problem

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The viscous problem

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The viscous problem: PPF

Re = 10000, a = 1

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The viscous problem: Blasius

Re = 500, a = 0.2

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The viscous problem: Blasius

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The viscous problem

Typical wind tunnel experiments say that transition occurs around Re = 2000 in PPF and around Re = 400 in the Blasius boundary layer … Could this be a 3D effect? Or something else?

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(ax + bz – wt)

The viscous problem: 3D disturbances

Consider 3D disturbances and replace by

ො 𝑣, ො 𝑤, ෝ 𝑥, ො 𝑞 = 𝑣 𝑧 , 𝑤 𝑧 , 𝑥 𝑧 , 𝑞(𝑧)

to end up with the OS and Squire equations

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The viscous problem: 3D disturbances



෥  = 𝑗 𝛾 ෤ 𝑣 − 𝑗 𝛽 ෥ 𝑥

mode shape of the normal vorticity

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The 3D viscous problem

In discrete form the temporal problem is a generalized eigenvalue problem of the form:

𝑩𝟐 𝟏 𝑫 𝑩𝟑 𝒘

 = 𝜕

𝑪𝟐 𝟏 𝟏 𝑪𝟑 𝒘

 𝑩 𝒚′ = 𝜕 𝑪 𝒚′

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The 3D viscous problem

There are two families of solutions of the Orr-Sommerfeld and Squire problems OS modes: Squire modes:

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The 3D viscous problem: Squire theorem

It is easy to show that (i) Squire modes are always damped and (ii) for each 3D OS mode there is a 2D OS mode of lower Reynolds number This means that the search of the critical Reynods number (smaller value of Re at which wi becomes positive for the first time) can be carried out looking at 2D OS modes only.

critical Reynolds number

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The 3D viscous problem: Squire theorem

It is easy to show that (i) Squire modes are always damped and (ii) for each 3D OS mode there is a 2D OS mode of lower Reynolds number. This means that the search of the critical Reynods number (smaller value of Re at which wi becomes positive for the first time) can be carried out looking at 2D OS modes only. So, if 3D disturbances are not the answer, what happens?

critical Reynolds number

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The forced problem

Let us imagine that an oscillating source term forces our system

• f equations (before Laplace-Fourier transforming them):

𝜖𝑔 𝜖𝑢 = 𝑀 𝑔 + 𝑡𝑓𝜏𝑡𝑢

f is some state function, L is a linear evolution operator, s is the spatial distribution of the forcing signal and 𝜏𝑡 its growth rate.

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The forced problem

The adjoint system is −

𝜖𝑕 𝜖𝑢 = 𝑀+ 𝑕

g is the adjoint function, L+ is the adjoint operator. This equation runs backward in time, i.e. it is integrated from t = T to t = 0. It arises easily from the definition of the inner product: 𝑏, 𝑐 = න

𝑈

න

𝑊

ത 𝑏 𝑐 𝑒𝑊 𝑒𝑢

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The forced problem

𝑕, 𝜖𝑔 𝜖𝑢 = 𝑕, 𝑀𝑔 + 𝑕, 𝑡𝑓𝜏𝑡𝑢 − 𝜖𝑕 𝜖𝑢 , 𝑔 + න

𝑊

ҧ 𝑕 𝑔|0

𝑈 𝑒𝑊 =

𝑀+𝑕, 𝑔 + 𝑕, 𝑡𝑓𝜏𝑡𝑢

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The forced problem

− 𝜖𝑕 𝜖𝑢 , 𝑔 + න

𝑊

ҧ 𝑕 𝑔|0

𝑈 𝑒𝑊 =

𝑀+𝑕, 𝑔 + 𝑕, 𝑡𝑓𝜏𝑡𝑢 the solution of the direct problem at the final time is: න

𝑊

ҧ 𝑕 𝑔|𝑢=𝑈 𝑒𝑊 = න

𝑊

ҧ 𝑕 𝑔|𝑢=0 𝑒𝑊 + 𝑕, 𝑡𝑓𝜏𝑡𝑢

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The forced problem

As terminal condition of the adjoint problem choose g(T) = d(x − x’) so that the direct solution at T is i.e. the adjoint solution weights both the initial condition and the source term in determining the solution at the final time T . 𝑔(𝑦′, 𝑈) = න

𝑊

ҧ 𝑕 𝑔|𝑢=0 𝑒𝑊 + 𝑕, 𝑡𝑓𝜏𝑡𝑢

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The forced problem

If we solve the linear stability problem for a plane shear flow it is very easy to show that the eigenfunctions of the adjoint

• perator act as receptivity functions, both for the temporally

evolving case and for the eigenproblem. In particular, adjoint e- functions provide inflow/wall/source receptivity coefficients.

Hill, 1995 Luchini & Bottaro, 1998

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Sensitivity analysis

Let us go back to the discrete world, and imagine that the matrices A and B are perturbed, for example by disturbances in the boundary conditions or by a noisy base flow. This will produce perturbations in both the eigenvalues and the eigenvectors. 𝑩 𝒚′ = 𝜕 𝑪 𝒚′ 𝑩 + 𝜀𝑩 𝒚′ + 𝜀𝒚′ = 𝜕 + 𝜀𝜕 𝑪 + 𝜀𝑪 𝒚′ + 𝜀𝒚′ and to first order (for small variations): 𝑩 𝜀𝒚′ + 𝜀𝑩 𝒚′ = 𝜕 𝑪 𝜀𝒚′ + 𝜕 𝜀𝑪 𝒚′ +𝜀𝜕 𝑪 𝒚′

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Sensitivity analysis

The left eigenproblem is: 𝒛𝑈 ഥ 𝑩 = ഥ 𝜕 𝒛𝑈 ഥ 𝑪 𝒛𝑈 𝑩 = 𝜕 𝒛𝑈 𝑪 Left multiply 𝑩 𝜀𝒚′ + 𝜀𝑩 𝒚′ = 𝜕 𝑪 𝜀𝒚′ + 𝜕 𝜀𝑪 𝒚′ +𝜀𝜕 𝑪 𝒚′ by 𝒛𝑈 to obtain the eigenvalue drift:

𝜀𝜕 =

𝒛, 𝜀𝑩 𝒚′ 𝒛, 𝑪 𝒚′

− 𝜕 𝒛, 𝜀𝑪 𝒚′

𝒛, 𝑪 𝒚′

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𝜗-pseudospectrum

Small variations in 𝑩 and 𝑪 with respect to their ideal behavior, linked to noise or imperfect knowledge of base flow and/or boundary conditions, can destabilize (and modify the frequency)

• f a nominally stable flow. The 𝜗-pseudospectrum of a matrix C

is the set of all eigenvalues which are 𝜗-close to C C :

L𝜗 𝑫 = l ∈ ℂ | ∃ 𝒚 ∈ ℂ𝑜\ 0 , ∃ 𝑭 ∈ ℂ𝑜×𝑜: 𝑫 + 𝑭 𝒚 = l𝒚, | 𝑭 |2 ≤ 𝜗

PPF, Re = 10 000, a = 1 ci cr

1 2

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𝜗-pseudospectrum

The 𝜗−pseudospectrum is particularly useful to understand non-normal matrices and their eigenvectors, i.e. matrices which do not commute with their conjugate traspose, and for which the eigenvectors are not orthogonal to one another 𝑫 ഥ 𝑫𝑈 ≠ ഥ 𝑫𝑈 𝑫

Clearly non-normal matrices are not self-adjoint. The OS operator/matrix is strongly non-normal

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Transient growth

Damped e-vectors in time can produce a disturbance f = F1 – F2 whose amplitude is initially/transiently amplified

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Transient growth

PPF, Re = 1000, a = 1

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Transient growth

Butler & Farrell, 1992

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Transient growth

The optimal transient growth (i.e. that producing the largest energy gain) transforms streamwise elongated vortices (present at t = 0 or x = 0) into streamwise elongated streaks at the final time/position. The mechanism is inviscid. Let us take the OS/Squire system, in symbolic form 𝑩 𝒚′ = 𝜕 𝑪 𝒚′ and let’s go one step backwards, i.e. before the Laplace transform: 𝑩 𝒚′ = −𝑗 𝑪

𝑒𝒚′ 𝑒𝑢 ,

𝒚′ 𝑢 = 0 = 𝒚0

′

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Transient growth

This equation also reads:

𝑒𝒚′ 𝑒𝑢 = 𝑫 𝒚′, with 𝑫 = 𝑗 𝑪−1𝑩

We have already seen that in this case we can decompose the solution in the sum of eigenvectors, i.e. 𝒚′ = 𝑽 𝑓L𝑢 ഥ 𝑾𝑈𝒚0

′ = 𝑴 𝒚0 ′

The energy of the disturbance is 𝐹(𝑢) = 𝒚′, 𝒚′ and the gain of the disturbance at a generic time T is

G(T) = 𝐹(𝑈)

𝐹(0) = 𝑴𝒚0

′ , 𝑴𝒚0 ′

𝒚0

′ , 𝒚0 ′

=

ത 𝑴𝑈𝑴𝒚0

′ , 𝒚0 ′

𝒚0

′ , 𝒚0 ′ propagator of the IC

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Transient growth

This is called Rayleigh quotient

G(T) =

ത 𝑴𝑈𝑴𝒚0

′ , 𝒚0 ′

𝒚0

′ , 𝒚0 ′

and the initial condition 𝒚0

′ which yields the largest gain is easily

identified by power iterations (adjoint looping)

𝑴𝒚0

′ = 𝒚𝑈 ′

𝒛𝑈 = 𝒚𝑈

′

ത 𝑴𝑈𝒛𝑈 = 𝒛0 𝒚0

′ = 𝒛0

direct iterations adjoint iterations t = 0 t = T

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Case closed?

Despite the fact that non-normality (and – as a direct consequence – transient growth) is an important concept, it is not sufficient to explain the breakdown to turbulence observed in experiments.

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Case closed?

Despite the fact that non-normality (and – as a direct consequence – transient growth) is an important concept, it is not sufficient to explain the breakdown to turbulence observed in experiments.

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Non-linearities matter!

• waves are coupled
• growth is due to linear mechanism
• nonlinear terms redistribute kinetic energy among modes

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Weakly non-linear approach

Consider a simple 1D model system:

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Weakly non-linear approach

Consider a simple 1D model system: = =

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Weakly non-linear approach

Thus, for each mode k and if we only considered three modes, k = -1, 0, 1, we would have:

𝑒𝑏0 𝑒𝑢 = 0, 𝑏0 is the mean flow correction

𝑒𝑏1 𝑒𝑢 + 𝑗𝛽𝑉𝑏1 + 𝜉𝛽2𝑏1 = −𝑗𝛽𝑏0𝑏1 𝑒𝑏−1 𝑒𝑢 − 𝑗𝛽𝑉𝑏−1 + 𝜉𝛽2𝑏−1 = 𝑗𝛽𝑏0𝑏−1

= -

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Fully non-linear analysis

• Prof. Joel Guerrero!

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M.C. Escher, Angels and Demons https://www.youtube.com/watch?v=YWVFIz4f5qk