Forced magnetic reconnection and its applications B. H. Oh 1 1 - - PowerPoint PPT Presentation
Forced magnetic reconnection and its applications B. H. Oh 1 1 Seoul National University, Department of Nuclear Engineering supervised by Prof. T. S. Hahm 1 KSTAR Conference 2014, 24-26 Feb 2014, Gangwon-do, Korea 1 Introduction Motivation
1Seoul National University, Department of Nuclear Engineering
supervised by Prof. T. S. Hahm1
KSTAR Conference 2014, 24-26 Feb 2014, Gangwon-do, Korea
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Motivation
Toroidal magnetostatic axisymmetric equilibrium Toroidal magnetostatic equilibrium with boundary perturbation
· 0
Resonant surface The equilibrium with the original topology(nested tori) and they have surface currents on this resonant surface The equilibrium with magnetic islands on the resonant surface of the original equilibrium, but they have no surface currents Two class of equilibrium solutions for perturbing boundary Magnetic islands?
Surface current? Solution I Solution II
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The geometry of the probelm
̂ ,
In the work by Hahm and Kulsrud, they treated a simple classic model problem suggested by Taylor to determine which one is the correct solution. The displacement of the boundary is . Perfect conducting wall at x=-a Perfect conducting wall at x=a
Sheared magnetic field
The displacement of the boundary is .
Toroidal magnetic field
Sheared magnetic field Resonant surface
̂ ̂ . 2 , cos ,
Introducing the flux function, [Hahm and Kulsrud, Phys. Fluids 28, 2412 (1985)]
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Review of fundamentals Current sheet leads to discrete jump
~ 0
half width of magnetic islands Current sheet (outward direction) Magnetic islands If 0 = 0, there are no magnetic islands at x=0. If 0 0, there are magnetic islands at x=0.
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First, consider the MHD equilibrium equations given by
, 4, · 0.
· · · 0, From the curl of the force balance equation
→
, . With boundary condition sinh /sinh , cosh / cosh . Equilibrium (I) Equilibrium (II) Two class of equilibrium solutions are possible. cosh sinh tanh sinh sinh . A general solution is combination of Equilibrium (I) and Equilibrium (II)
it has mirror symmetry 0 0, No island formation
sinh . 0 0 2/sinh Here has a finite jump corresponding to a surface current. 0 / cosh , 0 is finite so the topology has changed and equilibrium (II) has magnetic islands.
cosh . 0 0 0 There is no surface current since is continuous at 0.
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4 .
sin , = cos .
The resistive MHD equations which will be used for analysis
cosh
.
Recall the equilibrium solution phases A, B phase C phase D
~ // ≫ // ≪ //
time scale for analysis
̂ , ̂ · ,
where
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Hydromagnetic time scale
/.
It is invariant under transformation → , → .
4 0 4
2
sin
~ / 4 2
We linearize the resistive MHD equations and neglect the resistivity because it takes time for the resistivity to play a role in dynamics, then we obtain these equations Physical meaning of a similarity solution Current distribution get more peaked and confined within the narrower width at the center, asymptotes to a singular sheet.
Combine these equation into where A similarity solution, , .
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/).
0 2
After current is confined within a layer which is narrow enough at the center, resistivity begins to act. But, in the early phase of resistive evolution, 0 and profiles do not deviate much from the ideal solution. (0) from ideal solution in phase A
∝ .
(half width of magnetic islands) ∝ t
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By using Laplace transformation and introducing dimensionless variables, we obtain
4 Ψ.
4,
≡ Ψ Ω 4Ψ . Combine these equation into Asymptotic matching consists of: Magnetic induction equation Vorticity equation lim
→ lim →
lim
→
Ψ Ψ
The exterior solution could be obtained from lim
→
Ψ 0 tanh . cosh
.
Thus, we obtain a solution Ψ 0
1 / tanh / 2/ , ̅.
where The interior solution as → ∞ is where 2
The exterior solution as → 0 is where
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cosh Δ/2 sinh .
Δ ≪ 1,
The constant approximation. Δ 1 / 12Ω
2 /
/ / /
.
≫ //
Interior solution is can be reduced to a real form and evaluated numerically. 0, ~
82/ 5 Γ 3 4 / , ≪ 1 0, ~
4 / , ≫ 1 An important point is that a time which is long compared to the tearing mode time scale // is required to reach equilibrium (II).
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/
/ /,
,
If ≳ ∆ , nonlinear theory must be employed. ~/,
∆ ~
/ / ,
∆
/
≡
Performing the standard matching procedure of [P .H. Rutherford, Phys. Fluids 16, 1903 (1973)]. Solving ∆ from
cosh Δ/2 sinh . where
1, /,
/
,
/
. And restoring dimensions to the variables, we have
→ but
tanh . z reaches 1, which corresponds to equilibrium (II), on a time scale ∼ / /. Asymptotic limits are 0, ~
; 0, ~
≪ , ≫ , 1 3 , tanh . Thus, the overshooting of the island width disappears in nonlinear evolution.
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[E Lazzaro and L Comisso, Plasma Phys. Control. Fusion 53, 054012 (2011)] Nonlinear two-fluid simulation results addressing the effects of RF driven current on the magnetic islands in the classical model problem RF driven current on the magnetic islands shrinks magnetic islands driven by boundary perturbation. And, there is a narrow current on both sides of the magnetic islands in relation to the work by Hahm and Kulsrud.
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~ Width of RF driven current Magnetic islands width A narrow current on both sides of the magnetic islands does not vanish. A narrow current on both sides of the magnetic islands nearly vanishes as time goes on. Nonlinear two-fluid simulations (E Lazzaro and L Comisso, 2011) show different results depending on the magnitude of with respect to .
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0 cosh ∓ sinh tanh 1 sinh sinh ,
S : Lundquist number.
~
Consider a single harmonic only, Current sheets at / for the time harmonic forcing RF driven current term (E Lazzaro and L Comisso, 2011) We will extend this work to more realistic situation. The tendency of the simulation results could be explained by analytical results(Hahm and Kulsrud, 1985) by including additional RF driven current term.
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Future plans
Linear analysis with RF driven current term , 0 , , where , is constant. Analysis with RF driven current term which has the following properties.
(RF driven current term should depend on )
cos /2 sin , , where /.
From the ideal MHD equations, we obtain the time evolution of
0 4, / .
Transition from ideal MHD to resistive evolution of The time scale of 0 is long compare to the tearing mode time scale
/ /.
The Induced eddy current opposes to and competes with . The reason 0, is constant is that the vorticity at x=0 is zero. wave length in x ∝ /.
amplitude of ∝ /.
(For / ≫ 1)
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