Inverse Heat Conduction Problem using TC deconvolution and IR measurements: Application to heat flux estimation in a Tokamak Application to heat flux estimation in a Tokamak JL. Gardarein, J. Gaspar IUSTI Laboratory Provence University JL. Gardarein, J. Gaspar 1 METTI-2011 Tutorial 11
Summarize 1. Experimental Set-Up Problematic 2. Description of the method Application to a 1D Inverse Heat Conduction Problem 3. Application to a 2D experimental case (If we have time) JL. Gardarein, J. Gaspar 2 METTI-2011
Plasma at T ~ 100 millions of degrés Lawson Critera: (Density) x (T ° ) x (Confinement Time) > 10 21 KeV.s.m -3 The tokamaks… The confinement is insuring with 2 magnetic fields : – an axial field produced with the toroidal coils – a poloidal field created with the plasma current – a poloidal field created with the plasma current The resulting magnetic fields are helicoïdal JL. Gardarein, J. Gaspar 3 METTI-2011
Plasma Wall Interaction: Heat flux of about 10MW/m2 DSMF : Dernière surface magnétique fermée Divertor Limiteur (Last Closed Magnetic Surface) SOL : Scrape off layer T ore S upra ( TS ) Cadarache (France) J oint E uropean T orus ( JET ) Culham (UK) Heat flux of about 10MW/m 2 Heat flux intercepted by the Plasma Facing Component JL. Gardarein, J. Gaspar 4 METTI-2011
Divertor de JET JET Divertor lignes de champ magnétiques lignes de champ magnétiques JET : Diagnostics and Components � Resolution ~ 8-10mm -1.3 -1.3 -1.4 -1.4 � Spectral Range: [3-5 µm] côté côté côté côté lignes de champ magnétiques lignes de champ magnétiques extérieur extérieur intérieur intérieur Z(m) Z(m) -1.5 -1.5 -1.3 -1.3 � F acquisition = 50 Hz � Type K -1.6 -1.6 -1.4 -1.4 TC TC TC TC côté côté côté côté � Observation of each Divertor’Side extérieur extérieur intérieur intérieur � F acq = 20Hz Z(m) Z(m) -1.7 -1.7 -1.5 -1.5 -1.8 -1.8 -1.6 -1.6 TC TC TC TC 2.4 2.4 2.6 2.6 2.8 2.8 3.0 3.0 -1.7 -1.7 r(m) r(m) -1.8 -1.8 -1.8 -1.8 2.4 2.4 2.6 2.6 2.8 2.8 3.0 3.0 ����� ����� ����� ����� r(m) r(m) 300 ���� ° �� 200 ����� ����� ����� ����� 100 [Gautier, EPS 1997] JL. Gardarein, J. Gaspar 5 METTI-2011
Example of Experimental Temperatures x time (s) ( ° C) IR data T T ( TC (1 cm) time (s) 100 200 300 400 500 JL. Gardarein, J. Gaspar 6 METTI-2011
Objectives : Heat flux computation on the plasma facing components Why ? • Critical Heat Flux and temperature of the PFC (10 MW/m², 1200 ° ° ° ° C) - Components destruction - Water leak in water-cooled machines - Water leak in water-cooled machines • Better understanding of the plasma physic. Problem • Is the plasma component perfectly known ? - Dimensions - Thermal properties JL. Gardarein, J. Gaspar 7 METTI-2011
Problems • IR data: - Direct computation (surface temperature measurement) - Unknown Thermal Properties • TC data: - Spatial resolution • TC data: - Spatial resolution - Inverse Problem We have to use the thermocouple data => Solve an inverse Problem JL. Gardarein, J. Gaspar 8 METTI-2011
A simplified version of our problem : 1D, Linear,Semi-Infinite Wall z k = 1 W/mK Cp = 1000 J/kg.m 3 ρ = 2500 kg/m 3 Direct Problem Direct Problem It’s possible to have the temperature field T(z,t) using : Temperature • Numerical Simulation + • Analytic Solution Noise • Semi-Analytic Solution JL. Gardarein, J. Gaspar 9 METTI-2011
Knowing the thermal properties, is it possible to estimate the heat flux with a temperature measurement ? z k = 1 W/mK Cp = 1000 J/kg.m 3 + ρ = 2500 kg/m 3 e Inverse Problem => Q(t) ??? JL. Gardarein, J. Gaspar 10 METTI-2011
Convolution / Deconvolution Theory of the Linear System s(t) e(t) Linear Syst. OUTPUT INPUT t ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ = = = = + + + + ⊗ ⊗ ⊗ ⊗ = = = = + + + + τ τ − − − − τ τ τ τ s s t t = = = = s s t t + + + + e e t t ⊗ ⊗ ⊗ ⊗ i i t t = = = = s s t t + + + + e e i i t t − − − − d d ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) 0 0 0 0 0 q(t) ? T(t) i(t) = Impulse response of the System δ ( t ) i(t) Linear Syst. t z JL. Gardarein, J. Gaspar 11 METTI-2011
Convolution / Deconvolution • Duhamel’Theorem: t ∫ ∫ ∫ ∫ T t − − − − T = = = = Q t ⊗ ⊗ ⊗ ⊗ i t = = = = Q τ i t − − − − τ d τ ( ) ( ) ( ) ( ) ( ) 0 0 • Step Response: H ( t ) u ( t ) ∂ − τ − + − − u t u F f u F f ( ) ( 1 ) ( ) − τ = ≈ i t t ( ) ∂ t ∂ ∆ ∆ τ τ t u(t) = Step Response of the System (response to the Heaviside Function) System • Discretization: F = Number of step time F F ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∆ T = = = = T − − − − T = = = = u F − − − − f + + + + − − − − u F − − − − f q f = = = = ∆ u q f ( ( 1 ) ( )) ( ) ( ) F F F − − − − f 0 = = = = f = = f = = 1 1 12 JL. Gardarein, J. Gaspar METTI-2011
F ∑ ∑ ∑ ∑ ∆ T = = = = ∆ u q f ( ) F F − − − − f F = Number of step time = = f = = 1 ∆ = = + + − − ≤ ≤ ≤ ≤ − − u k = = u k + + − − u k ≤ ≤ k ≤ ≤ F − − ∆ ∆ T u q � � � ( 1 ) ( ) 0 1 0 0 1 0 1 ∆ T ∆ u ∆ u q � � � 0 2 1 0 2 � � � � � . . 0 − − − − = = = = ∆ T T q u F = . 1 0 1 0 � � � � � � � . T − − − − T = = = = q ∆ u + + + + q ∆ u � � � � � � � . 2 0 1 1 2 0 ∆ ∆ ∆ ∆ T u u u q � � − − − − = = = = ∆ + + + + ∆ + + + + ∆ . T T q u q u q u F F − F − F 1 2 0 3 0 1 2 2 1 3 0 . Convolution × × � T � T . . = = Q Q X X − − = = ∆ + + ∆ + + ∆ + + + + ∆ T − − T = = q u + + q u + + q u + + + + q u ....... F F − − − − F − − − − F − − − − F 0 1 1 2 2 3 3 0 The matrix X can be inverted if X is well conditioned Deconvolution = = = = − − − − � T 1 If K(X) is low => X is well conditioned Q X . = − K X X X 1 ( ) If K(X) >> 1 => X is ill conditioned JL. Gardarein, J. Gaspar 13 METTI-2011
z k = 1 W/mK Cp = 1000 J/kg.m 3 + ρ = 2500 kg/m 3 e • Step response computation: analytical solution τ π q − x x x 2 ² τ = = = = − − − − − − − x erfc � T 0 ( , ) exp ατ ατ π ατ b 4 2 2 • The temperatures are produced with a FEM code We can solve this problem with Excel or Matlab JL. Gardarein, J. Gaspar 14 METTI-2011
Deconvolution of the temperature at z=5mm: Application with Excel and Matlab JL. Gardarein, J. Gaspar 15 METTI-2011
− = • Why doesn’t it work ? Q X � T 1 . => The matrix is ill conditioned because the problem is ill posed => The solution doesn’t respect the Stability condition: Q is very sensitive to measurement errors contained in deltaT. • How to find a solution ? => Need to use a regularization procedure => Regularization with a penalisation => We choose the Tikhonov operator JL. Gardarein, J. Gaspar 16 METTI-2011
~ 2 • Without regularization, the function to minimize is: γ = = = = − − − − � T J X R X Q ( , , ) . • With regularization, the new J is: J is the function to minimise ~ ~ 2 γ = = − − + + γ = = − − + + � T J ( X , R , ) X . Q I . Q R is the regularization operator γ is the regularization parameter ~ ~ Q Q • In our case, we want to limit the norm of , so R=I, this is a 0 order regularization. • To minimise J is equivalent to have the following expression of Q ( ( ( ( ) ) ) ) ~ − − − − 1 = = = = t + + + + γ t t I I � T Q X X X . ~ • One can note that is not the exact heat flux but an estimated Q heat flux; it is a biased value of Q JL. Gardarein, J. Gaspar 17 METTI-2011
How to choose γ ? => γ is chosen to have the best compromise between an exact solution and a stable solution ~ − � T − → → − − → → => Exact solution means that X . Q 0 ~ → => Stable means that → → → Q min JL. Gardarein, J. Gaspar 18 METTI-2011
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