On the Boundary Condition Between Two Multiplying Media

Abstract. The transition region between two parts of a pile which have different compositions is investigated. In the case where the moderator is the same in both parts of the pile, it is found, that the diffusion constant times fast neutron density satisfies the usual pile equations everywhere, right to the boundary. More complicated formulae apply in a more general case. 

We shall first treat a case in which the moderator is the same 1 over the pile so that the diffusion constant for both fast and thermal neutrons remains constant throughout the whole system. It has been stated in CP-455 that, for such a system, Fermi s equations can be solved with the two group theory as described in CP-1461 and CP-1554. It will be shown that in this case, the two methods of calculation give the same result. The general case of different moderators in the two parts of the pile will be treated finally by means of the two group theory. 

r = ln(JB/ thermai) has a somewhat different definition from the usual one it is zero for neutrons, the energy E of which is thermal, (t t) is the transport cross section which may depend on the energy but does not depend on the position to, the value of this quantity for r = 0, i.e. for thermal neutrons the average logarithmic energy loss (independent of position) 7, the absorption cross section for thermal neutrons which depends on the position. q(r) is the density of fast neutrons per unit r (it is not Fermi s slowing down density Q), multiplied with the velocity, n the density of thermal neutrons times their velocity, /(r) dr is the number of fission neutrons per slow neutron captured in U, for which r is between r and r + dr. Finally pi is the chance of escaping resonance absorption and p2 the thermal utilization. The multiplication constant here 

The advantage of the above way of writing the ordinary pile equations is that it does not assume that the fission neutrons are monochromatic. 

If (Ta depends on the position, the ratio of n and q r) will not be constant over the pile and no simple equation will hold for either of them separately. This is natural, since transients occur both in n and in q which make their behavior quite complicated. The method of solution to be given consists of finding a quantity i/ in which the transients just cancel so that a simple equation shall hold for it. 

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