Identifying the wavemaker of fluid/structure instabilities Olivier - - PowerPoint PPT Presentation

Identifying the “wavemaker”

• f fluid/structure instabilities

Olivier Marquet1 & Lutz Lesshafft2

1 Department of Fundamental and Experimental Aerodynamics 2 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 = 0.01 0.4 < < 1.2 = 10, 200, 10 ( = 0.73)

• Structural damping

=

• Density ratio

One spring - cross-stream

Stability analysis of the coupled fluid/solid problem

4

() !

• "#!

(, ) $ $ = (% + ') (

• )

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

Fluid/solid components Growth rate/frequency

Cossu & Morino (JFS, 2000)

Steady solution – No solid component

Diagonal operators: intrinsinc dynamics

5

() !

• "#!

(, ) $ $ = (% + ') (

• )

$ $ $′(+, ,) = ($, $) + -./ 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

() !

• "#!

(, ) $ $ = (% + ') (

• )

$ $ $′(+, ,) = ($, $) + -./ 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 < < 1.2

Structural 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 WM

Effect 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 destabilized Stable 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 ( = 200 and = 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
• perator ?
• 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 $ + 7

4

5 $ = 65 $ + 7

5 Eigenvalue decomposition with residuals

7

4 ≠ 0, 7 5 ≠ 0 but 7 4 = −7 5 = 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

= 0

Normalisation

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,

• ther 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

+ 6

Adjoint mode-based decomposition

6

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

6 = $

.<( $ + "#! $) Solid contribution

• !
• "#!
• $

$ = 6 (

• )

$ $ 6

= @ $ .∗(+) ⋅ ( $ + ! $) + C+ D

Local contributions

Infinite mass ratio - Fluid Modes

15

6 = 6

+ 6

Adjoint mode-based decomposition

6

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

6 = $

.<( $ + "#! $) Solid contribution

Fluid Modes

EF = G 6

= $ .< $ = 6

• !
• G
• $

$ = 6 (

• )

$ $ "# = 0 "# = 0 6 = 0 OK

Infinite mass ratio - Structural Mode

16

6 = 6

+ 6

Adjoint mode-based decomposition

6

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

6 = $

.<( $ + "#! $) Solid contribution

Structural Mode

6

= 0

• !
• G
• $

$ = 6 (

• )

$ $ "# = 0 "# = 0 6 = $

.< $ = 6

OK EH

. = G

Infinite mass ratio – Direct mode-based decomposition

17

6 = 6 >

+ 6

>

Direct mode-based decomposition

6 >

= EH I ( $ + ! $) Fluid contribution

6 > = EF

I( $ + "#! $) Solid contribution

Structural Mode

6 >

= 6 ($ < $)

• !
• $

$ = 6 (

• )

$ $ "# = 0 "# = 0 6 > = 6 ($

< $)

NOT OK $ + !

$ = 6$

Infinite mass ratio – Direct mode-based decomposition

18

6 = 6 >

+ 6

>

Direct mode-based decomposition

6 >

= EH I ( $ + ! $) Fluid contribution

6 > = EF

I( $ + "#! $) Solid contribution

Structural Mode

6 >

= 6 ($ < $)

• !
• $

$ = 6 (

• )

$ $ "# = 0 "# = 0 6 > = 6 ($

< $)

NOT OK

(large fluid response)

(6) − )$ = !

$

Mass ratio J = KGG: Structural Mode

19

SM WM

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 Fluid Solid 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

SM WM

Frequency (WM)

For small , ∼ - For large , ∼

Growth rate (WM)

Wake Mode: frequency decomposition

23

Frequency Solid Fluid

= 10

For small : ∼ = frequency selection by the fluid Unstable range : ∼ For large : ∼ = frequency selection by the solid

Wake Mode: growth rate decomposition

24

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 !

• P

$ $

Fluid modes

6∗ Q Q = O< !

• <

P< Q Q

Direct Adjoint Solid modes

6 $ = O $ $ = 0 6 $ = P $ (6) − O) $ = !

$

6

• ∗Q

R = O<Q R

6

• ∗) − P< Q

R = !

• <Q

R

6

∗Q R = P<Q R

Q

R = 0

Projection of coupled problem on pure fluid modes

29

Titre présentation

(6 ) − O) $ = !

$

(6) − P)$ = "#! $ Q

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

6∗Q

R − O<Q R <

$ + 6∗Q

R − P<Q R − !

• <Q

R <

$ = "#Q

R<! $

6 − 6

(Q R<$) + 6 − 6 (Q R<$) = "#Q R<! $

6 − 6

=

"#Q

R<! $

(Q

R<$ + Q R<$)

Projection of coupled problem on pure solid modes

30

Titre présentation

(6 ) − O) $ = !

$

(6) − P)$ = "#! $ Q

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

6∗Q

R − O<Q R <

$ + 6∗Q

R − P<Q R − !

• <Q

R <

$ = "#Q

R<! $

(6 − 6)(Q

R<$) = "#Q R<! $

6 − 6 = "#Q

R<! $

Q

R<$

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