Sep 26, 2013 ... How does 2D inverse cascade saturate? For Ro ≪ 1 what is Re ≫ 1? 2 / 1.
Alexandros Alexakis, Martin Schrinner. Rotating Taylor Green ...
Rotating Taylor Green Flows Alexandros Alexakis, Martin Schrinner
September 26, 2013
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Rotating Flows A rotating world Stars, planets ... Earth’s atmosphere ... laboratory experiments ...
Weak Wave turbulence, is it there? What controls forward to backward cascade ratio? How does 2D inverse cascade saturate? For Ro 1 what is Re 1? 2/1
Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Taylor Green Forcing: A test case ∂t u + u · ∇u + 2Ω × u = −∇P + ν∇2 u + F
Ω f q=2
1 • F= Lz
Z Fdz = 0,
F = F0
ex sin(qx) cos(qy ) sin(qz) − ey cos(qx) sin(qy ) sin(qz) ez 0
1 • hF · ∇ × Fi = V
Z F · ∇ × FdV = 0
• u = F is not a solution of the Euler equations 3/1
Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Control Parameters
Re = UL/ν, U = kuk,
U=
Ro = U/2LΩ
p F0 L,
U = 1/3 L1/3
RoF =
FL 2ΩL
ReF =
√ FL3 ν
RoU =
kuk 2ΩL
ReU =
kukL ν
RoD =
1/3 2ΩL2/3
ReD =
1/3 L4/3 ν
√
Ω f q=2
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Parameter Space
••• Slow rotating 5/1
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Alexandros Alexakis, Martin Schrinner
•••
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Rotating Taylor Green
••• Fast rotating
The Runs
167 Runs Resolutions 643 − 5123 0 < ReD . 700 0.0...01 < RoD < ∞
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Different behaviors observed Energy evolution
“Laminar” Bursts
Quasi-2D Inverse cascade Quasi-isotropic turbulence
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Parameter Space
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Parameter Space
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Expansion
Ro 1 u → = RoF u, τ = RoF t and ReF−1 = λRoFn+1
∂t u + ez × u − ∇P = F − RoF
∂τ u + w × u − ∇P 0 + λRoFn ∇2 u. or
∂t u + L [u] = F + Ro P [−∂τ u + u × w] + λRo n ∇2 u P[u] = projection to incompressible fields P [ez × g]
L[g] =
u = u0 + Ro u1 + Ro2 u + . . . 10 / 1
Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
What we can show
For n = 0 O(ReF ) = O(RoF−1 ) Laminar solution, kuk ∼ F0 /Ω, (ReU = RoF ReF ) u=0 Nonlinearly (energy) stable for ReF < C1 RoF−1 . For n = 1 O(ReF ) = O(RoF−2 ) u=0 Laminar solution, kuk ∼ F0 /Ω, −2 Linearly stable for ReF < C2 RoF . For n = 3 O(ReF ) = O(RoF−4 ) Laminar solution, kuk ∼ F0 /Ω, u 6= 0 −4 2D-flow-unstable for ReF > C4 RoF . For n > 3 O(Re) > O(Ro −4 ) No Laminar solution exists.
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Parameter Space
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Energy dissipation Have we reached high enough Re yet?
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Rotating Taylor Green quasi 2D flows
vertical vorticity
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
Conclusions What is high Re for low Ro is not a trivial question. Do not extrapolate present results to other forcing functions (Different behavior for F 6= 0 is expected.)
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Alexandros Alexakis, Martin Schrinner
Rotating Taylor Green
