Page 112 - Fundamentals of Magnetic Thermonuclear Reactor Design
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96     Fundamentals of Magnetic Thermonuclear Reactor Design


               It is clear that a detailed description of EM transients is only possible through
            a solution of coupled electrodynamic and thermal problems taking into account
            the system configuration.
               Many tokamak conductive components present structures with walls that
            are thin compared with other linear dimensions [32]. For ITER, they are the
            vacuum vessels, thermal shield and cryostat. EM loads occurring in such struc-
            tures often become the key design constraint.
               In ITER, the following procedure is accepted for the analysis of EM tran-
            sients. The R&D stage for the reactor involves axisymmetric DINA simulations
            of plasma scenarios (computer codes other than DINA can also be used). Then
            the results of solving a related EM problem are used as initial data for 3D com-
            putations.
               For a qualitative description of transient processes as per, for instance [28]
            one can use existing analytical solutions to the problem of a field penetration
            into a conductive medium. Such a solution was first found [28,33] in describing
            a uniform thermal field diffusion in an infinitely long flat layer with a thick-
            ness of d. In our case, boundary conditions take the form: B(0, t) = B (t), B(d,
                                                                    1
            t) = B (t), with the initial condition B(x,0) = f(x).
                 2
               The x coordinate is directed normally from the medium boundary inwards
            the layer. At zero boundary conditions, B  = B  = 0 and
                                             1
                                                 2
                                                 /
 fx=B sinπx/d                       fx     0   π ( x d )
                                     () = B sin
 0
                                                   2
                            ( t,
                                          /
 2
                                                         0
                                   0
                                                     0
 Bx,t=B sinπx/dexp−π ρ t/µ d 2  Bx  ) = B sin  π ( xd )exp (− πρ t /  µ d 2 )
 0
 0
 0
               As seen, the field decay occurs without any spatial re-distribution. The char-
                                                       2
 τ =µ d /π ρ 0  acteristic decay time for the first harmonic is τ =  µ d /  2 ρ . Higher harmon-
 2
 2
                                                             0
                                                     0
                                                 0
 0
 0
 τ 0        ics tend to decay more rapidly. For a 60-mm-thick steel wallτ  is 3 ms.
                                                              0
               Similar analytical solutions for a problem involving field diffusion through
 τ 0        a flat layer contain the same characteristic time  τ  and linear dimension d.
                                                      0
            They are used in numerical analyses to assess spatial and temporal steps for FE
            models.
               Solving a diffusion problem for a field induced by external currents that run
            parallel to the layer, one can obtain the asymptotic values of a tangential compo-
                      ,
 BF(−d,t)   nent  B ( −d t)  behind the conducting layer with boundaries x = 0, x =−d. One
                  F
            example is related to current filaments. If the depth of field penetration into the
            layer is much smaller than its thickness and the characteristic linear dimensions,
            the solution for a current filament loop lying in plane x = h can be reduced to the
            principal term of the expansion in powers with respect to the minor parameter
 ρ t/µ d 2   ρ t /  µ d 2  . This solution is the more accurate the smaller t is. For an instant
              0
                  0
 0
 0
            current step from 0 to i , the tangential field [28]
                               0
                                                                  2
                                  h
                                    i
              B  (−d t,  ) = −∂8  b () ∂⋅ ⋅  t /  µ d 2 ) 3 2  ⋅d /  π ⋅ exp(− µ d /4 ρ ) t .
                              0/
 2
 BF−d,t=−8∂bF0/∂h⋅i ⋅ρ t/µ d 32  F  F  0  ρ ( 0  0              0      0
 0
 0
 0
 2
 ⋅d/π⋅exp−µ d /4ρ t.
 0
 0
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