Identifying the “wavemaker” of fluid/structure instabilities
Transcript of Identifying the “wavemaker” of fluid/structure instabilities
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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
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Context
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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
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Model problem: spring-mounted cylinder flow
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�� = ����
Reynolds number
= �⁄Structural frequency
�� = 40� = 0.010.4 < � < 1.2 � = 10�, 200, 10(�� = 0.73)
�Structural damping
� = � ���Density ratio
One spring - cross-stream
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Stability analysis of the coupled fluid/solid problem
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�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 00 )
$�$
$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid
components
Growth rate/frequency
Cossu & Morino (JFS, 2000)
Steady solution – No solid component
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Diagonal operators: intrinsinc dynamics
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�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 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
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Off-diagonal operators: coupling terms
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�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 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
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Stability analysis at large density ratio
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� = 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)
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Stability analysis at smaller density ratio
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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)
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Objective and outlines
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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)
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State of the art for the method
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• 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
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From operator to eigenvalue decomposition
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$ = 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 ?
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Computing eigenvalue contributions
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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
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Why this particular eigenvalue decomposition?
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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;$;;
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Application to the spring-mounted cylinder flow
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6 = 6� + 6Adjoint mode-based decomposition
6� = $�.<( �$� + !�$)Fluid contribution
6 = $.<( $ + �"#!� $�)Solid contribution
� !��"#!�
$�$ = 6 (� 00 )
$�$
6� = @ $�.∗(+) ⋅ ( �$� + !�$) + C+D
Local contributions
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Infinite mass ratio - Fluid Modes
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6 = 6� + 6Adjoint mode-based decomposition
6� = $�.<( �$� + !�$)Fluid contribution
6 = $.<( $ + �"#!� $�)Solid contribution
Fluid ModesEF = G
6� = $�.< �$� = 6
� !�G
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6 = 0OK
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Infinite mass ratio - Structural Mode
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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
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Infinite mass ratio – Direct mode-based decomposition
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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$�
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Infinite mass ratio – Direct mode-based decomposition
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6 = 6>� + 6>Direct mode-based decomposition
6>� = EHI( �$� + !�$)Fluid contribution
6> = EFI( $ + �"#!� $�)Solid contribution
Structural Mode
6>� = 6($�<$�)
� !�0
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6> = 6($<$)NOT OK(large fluid response)
(6) − �)$� = !�$
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Mass ratio J = KGG: Structural Mode
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SMWM
Growth rate (SM)
Frequency (SM)
• The frequency is quasi-equal to �• The growth rate gets positive for �close ��
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Structural Mode: frequency/growth rate decomposition
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Frequency
Fluid
Solid
FluidSolid
Growth rate
Very large (resonance) and opposite contributions
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Spatial distribution of the fluid component
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Growth rate
Local fluid contributions
The solid contribution
induces
destabilisation
The fluid contribution
induces
destabilisation
Phase change between and $�. �$�Schmid & Brandt (AMR 2014)
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Mass ratio J = LG: Wake Mode
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SMWMFrequency (WM)
For small �, � ∼ �� - For large �, � ∼ �
Growth rate (WM)
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Wake Mode: frequency decomposition
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Frequency SolidFluid� = 10
For small �: �� ∼ � = frequency selection by the fluid
Unstable range : �� ∼ �
For large �: � ∼ � = frequency selection by the solid
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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
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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
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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
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Conclusion & perspectives
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• 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
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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
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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<$� + QR<$)
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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<!� $�QR<$
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Direct-based decomposition of the unstable mode
31 Titre présentation
Frequency
Fluid
Solid
Growth rate
� = 200
Solid
Fluid
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Free oscillation
�� = 40; � = 50
No oscillation� = 0.60
� = 0.66 Weak oscillation
� = 0.90 Strong oscillation
� = 1.10 No oscillation
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Solid displacement – Fluid fields
Multiple solutions