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40     CHAPTER 4 Solutions of the point reactor kinetics equations




                                             1              q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                                                                       2
                                       s 1 ,s 2 ¼    β  ρ + λΛÞ   ð β  ρ + λΛÞ +4λρΛ
                                                  ð
                                             2Λ
                                                            s 1 t
                                                                  s 2 t
                         The general solution has the form  PtðÞ  ¼ Ae + Be .
                                                     P 0ðÞ
                            The solution for the example case is given by
                                              PtðÞ      0:0088t     603:09t
                                                  ¼ 1:111e   0:111e                     (4.10)
                                              P 0ðÞ
                         Note that the solution for reactor power contains two exponential terms, one of the
                         exponents is positive and the other is negative. The term with the negative exponent
                         goes to zero rapidly, leaving the term with the positive exponent to define the con-
                         tinuing transient. The reactor period for this example is given by T ¼ 1/
                         0.0088 ¼ 113.6 s.
                            Now consider the response to a negative step change in reactivity. The following
                         equation is the solution for the same reactor as above for a   10¢ reactivity step.
                                             PtðÞ
                                                 ¼ 0:909e  0:0073t  +0:0909e  737:07t   (4.11)
                                             P 0ðÞ
                         Note that for the positive reactivity step change, one of the exponents is positive and
                         the other is negative; and for the negative reactivity step change, both the exponents
                         are negative. As shown in Section 4.8, for a six-delayed neutron group model, there is
                         one positive exponential and six negative exponentials for a positive step and seven
                         negative exponentials for a negative step.
                            The behavior described above can be deduced by examining the equation for neu-
                         tron density.
                                                               6
                                                  dn  ð ρ βÞ  X
                                                    ¼      n +   λ i C i                (4.12)
                                                  dt    Λ
                                                              i¼1
                         Note that the first term on the right is negative for ρ < β and the second term is
                         always positive. Thus, delayed neutrons are essential to maintain an increasing neu-
                         tron density if ρ < β.If ρ > β, the rate of increase is positive even if there were no
                         delayed neutron contribution.




                         4.5 Solutions for small perturbations

                         The small perturbation form of the zero-power reactor equations is sometimes
                         used. But the reader must remember that it is applicable only for small perturbations
                         with respect to a nominal. Fig. 4.5 shows simulations performed with the full, six-
                         delay group model and the small perturbation model (also with six delay groups).
                         The discrepancies in the results obtained with the small perturbation model are
                         apparent.
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