Identifying the “wavemaker” of fluid/structure instabilities

Identifying the “wavemaker” of fluid/structure instabilities

Olivier Marquet1 & Lutz Lesshafft2

1 Department of Fundamental and Experimental Aerodynamics2 Laboratoire d’Hydrodynamique, CNRS-Ecole Polytechnique

24th ICTAM

21-26 August 2016, Montreal, Canada

Context

2

Flow-induced structural vibrations

Civil engineering Aeronautics Offshore-marine industry

Stability analysis of the fluid/structure problem

A tool to predict the onset of vibrations

Model problem: spring-mounted cylinder flow

3

�� = ����

Reynolds number

= �⁄Structural frequency

�� = 40� = 0.010.4 < � < 1.2 � = 10�, 200, 10(�� = 0.73)

�Structural damping

� = � ���Density ratio

One spring - cross-stream

Stability analysis of the coupled fluid/solid problem

4

�(��) !��"#!� (� , �)

$�$ = (% + '�) (� 00 )

$�$

$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid

components

Growth rate/frequency

Cossu & Morino (JFS, 2000)

Steady solution – No solid component

Diagonal operators: intrinsinc dynamics

5

�(��) !��"#!� (� , �)

$�$ = (% + '�) (� 00 )

$�$

$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid

components

Growth rate/frequency

Cossu & Morino (JFS, 2000)

Steady solution – No solid component

Solid operator

(damped harmonic oscillator)

Fluid operator

Off-diagonal operators: coupling terms

6

�(��) !��"#!� (� , �)

$�$ = (% + '�) (� 00 )

$�$

$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid

components

Growth rate/frequency

Mougin & Magaudet (IJMF, 2002), Jenny et al. (JFM 2004)

Steady solution – No solid component

Coupling operator

(boundary condition + non-inertial terms)

Coupling operator

Weighted fluid force

Stability analysis at large density ratio

7

� = 0.75 0.4 < � < 1.2Structural Mode (� = 0.75)

Wake Mode

� = 10�

Methods to identify structural and wake modes

- Vary the structural frequency

- Look at the vertical displacement/velocity (not the fluid component)

Stability analysis at smaller density ratio

8

SM WMEffect of decreasing the density ratio

� = 200� = 10� � = 10

- Stronger interaction between the two branches

- The two branches exchange their « nature » for small �- Coalescence of modes (not seen here)

SM is destabilized WM is destabilizedStable

Zhang et al (JFM 2015), Meliga & Chomaz (JFM 2011)

Objective and outlines

9

Objective:

- Identify the « wavemaker » of a fluid/solid eigenmode

- Quantify the respective contributions of fluid and solid

dynamics to the eigenvalue of a coupled eigenmode

Outlines:

1 - Presentation of the operator/eigenvalue decomposition

2 - The infinite mass ratio limit ( � = 10�)3 - Finite mass ratio ( � = 200and � = 10)

State of the art for the method

10

• Energetic approach of eigenmodes (Mittal et al, JFM 2016)

- Transfer of energy from the fluid to the solid / Growth rate

- Does not identify the « wavemaker » region in the fluid

• Wavemaker analysis (Giannetti & Luchini, JFM 2008, …)

- Structural sensitivity analysis of the eigenvalue problem.

- Largest eigenvalue variation induced by any perturbation of the

operator ?

- Output of this analysis is an inequality. We would like an identify !

• Operator/Eigenvalue decomposition

From operator to eigenvalue decomposition

11

$ = 4 + 5 $ = 6$Operator decomposition

In general, $ is not an eigenmode of 4 or 5 , so

4$ = 64$ + 74 5$ = 65$ + 75

Eigenvalue decomposition

with residuals 74 ≠ 0, 75 ≠ 0 but 74 = −75 = 7

64 + 65 = 6

How to compute the eigenvalue contributions 64/65 ?

Computing eigenvalue contributions

12

Expansion of the residual on the set of other eigenmodes $;

Orthogonal projection on the mode $using the adjoint mode $.

$.<( 4$) = 64($.<$) +=7;($.<$;);= 1 Bi-orthogonality= 0Normalisation

4$ = 64$ +=7;$;;

7 = =7;$;;

64 = $.<( 4$) 65 = $.<( 5$)

Adjoint mode-based decomposition

6 = 64 + 65

Why this particular eigenvalue decomposition?

13

For an identical decomposition of the operator,

other eigenvalue decompositions are possible

6>4/5 = 64/5 ±=7;($<$;);

≠ 0

Non-orthogonal projection on the mode $

6>4 = $<( 4$) 6>5 = $<( 5$)

Direct mode-based decomposition

6 = 6>4 + 6>5

4$ = 64$ +=7;$;;

But it includes contributions from other eigenmodes

M.Juniper (private communication)

5$ = 65$ −=7;$;;

Application to the spring-mounted cylinder flow

14

6 = 6� + 6Adjoint mode-based decomposition

6� = $�.<( �$� + !�$)Fluid contribution

6 = $.<( $ + �"#!� $�)Solid contribution

� !��"#!�

$�$ = 6 (� 00 )

$�$

6� = @ $�.∗(+) ⋅ ( �$� + !�$) + C+D

Local contributions

Infinite mass ratio - Fluid Modes

15

6 = 6� + 6Adjoint mode-based decomposition

6� = $�.<( �$� + !�$)Fluid contribution

6 = $.<( $ + �"#!� $�)Solid contribution

Fluid ModesEF = G

6� = $�.< �$� = 6

� !�G

$�$ = 6 (� 00 )

$�$�"# = 0

�"# = 0

6 = 0OK

Infinite mass ratio - Structural Mode

16

6 = 6� + 6Adjoint mode-based decomposition

6� = $�.<( �$� + !�$)Fluid contribution

6 = $.<( $ + �"#!� $�)Solid contribution

Structural Mode

6� = 0

� !�G

$�$ = 6 (� 00 )

$�$�"# = 0

�"# = 0

6 = $.< $ = 6OK

EH. = G

Infinite mass ratio – Direct mode-based decomposition

17

6 = 6>� + 6>Direct mode-based decomposition

6>� = EHI( �$� + !�$)Fluid contribution

6> = EFI( $ + �"#!� $�)Solid contribution

Structural Mode

6>� = 6($�<$�)

� !�0

$�$ = 6 (� 00 )

$�$�"# = 0

�"# = 0

6> = 6($<$)NOT OK

�$� + !�$ = 6$�

Infinite mass ratio – Direct mode-based decomposition

18

6 = 6>� + 6>Direct mode-based decomposition

6>� = EHI( �$� + !�$)Fluid contribution

6> = EFI( $ + �"#!� $�)Solid contribution

Structural Mode

6>� = 6($�<$�)

� !�0

$�$ = 6 (� 00 )

$�$�"# = 0

�"# = 0

6> = 6($<$)NOT OK(large fluid response)

(6) − �)$� = !�$

Mass ratio J = KGG: Structural Mode

19

SMWM

Growth rate (SM)

Frequency (SM)

• The frequency is quasi-equal to �• The growth rate gets positive for �close ��

Structural Mode: frequency/growth rate decomposition

20

Frequency

Fluid

Solid

FluidSolid

Growth rate

Very large (resonance) and opposite contributions

Spatial distribution of the fluid component

21

Growth rate

Local fluid contributions

The solid contribution

induces

destabilisation

The fluid contribution

induces

destabilisation

Phase change between and $�. �$�Schmid & Brandt (AMR 2014)

Mass ratio J = LG: Wake Mode

22

SMWMFrequency (WM)

For small �, � ∼ �� - For large �, � ∼ �

Growth rate (WM)

Wake Mode: frequency decomposition

23

Frequency SolidFluid� = 10

For small �: �� ∼ � = frequency selection by the fluid

Unstable range : �� ∼ �

For large �: � ∼ � = frequency selection by the solid

Wake Mode: growth rate decomposition

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Growth rate Solid Fluid� = 10

Small �destabilization due to the solid

Large �destabilization due to the fluid

Wake Mode: growth rate decomposition

25

Growth rate Solid Fluid� = 10

Small �destabilization due to the solid

Large �destabilization due to the fluid

Wake Mode: growth rate decomposition

26

Growth rate Solid Fluid� = 10

Small �destabilization due to the solid

Large �destabilization due to the fluid

Conclusion & perspectives

27

• The adjoint-based eigenvalue decomposition enables to discuss

- the frequency selection/ the destabilization of coupled modes

- The localization of this process in the fluid

• Comparison with structural sensitivity (not shown here)

Identification of the same spatial regions

• Use this decomposition in more complex fluid/structure problem

Cylinder with a flexible splitter plate (Jean-Lou Pfister)

Thank you to

Pure modes (infinite mass ratio)

28 Titre présentation

6 $�$ = O !�0 P

$�$

Fluid

modes

6∗ Q�Q = O< 0!�< P<

Q�Q

Direct Adjoint

Solid

modes

6$� = O$�$ = 0

6$ = P$(6) − O)$� = !�$

6�∗Q�R = O<Q�R

6�∗) − P< QR = !�<Q�R

6∗QR = P<QRQ�R = 0

Projection of coupled problem on pure fluid modes

29 Titre présentation

(6) − O)$� = !�$(6) − P)$ = �"#!� $�

Q�R< 6) − O $� + QR< 6) − P $ − Q�R<!� $= �"#QR<!�$�6∗Q�R − O<Q�R

<$� + 6∗QR − P<QR − !�<Q�R<$ = �"#QR<!�$�

6 − 6� (Q�R<$�) + 6 − 6� (QR<$) = �"#QR<!� $�

6 − 6� = �"#QR<!�$�(Q�R<$� + QR<$)

Projection of coupled problem on pure solid modes

30 Titre présentation

(6) − O)$� = !�$(6) − P)$ = �"#!� $�

Q�R< 6) − O $� + QR< 6) − P $ − Q�R<!� $= �"#QR<!�$�6∗Q�R − O<Q�R

<$� + 6∗QR − P<QR − !�<Q�R<$ = �"#QR<!�$�

(6 − 6)(QR<$) = �"#QR<!�$�

6 − 6 = �"#QR<!� $�QR<$

Direct-based decomposition of the unstable mode

31 Titre présentation

Frequency

Fluid

Solid

Growth rate

� = 200

Solid

Fluid

Free oscillation

�� = 40; � = 50

No oscillation� = 0.60

� = 0.66 Weak oscillation

� = 0.90 Strong oscillation

� = 1.10 No oscillation

Solid displacement – Fluid fields

Multiple solutions