Aside from energy production, it is important to evaluate the potential of hybrid reactors for transmutation, i. e. to what extent they produce excess neutrons. A standard reactor can be viewed as a device producing energy and neutrons. Both energy and neutrons are primarily produced by fission. Fission releases about 200 MeV and 2.5 neutrons. It follows that one may

say that 80 MeV are needed to produce one neutron. The spallation process requires only 30 MeV to produce one neutron. Should the fission 200 MeV be available for proton acceleration one would, then, get more than nine neutrons per fission (2.5 ± 6.6)! True enough, no usable energy would be produced. In fact, assuming a thermodynamic efficiency of 40% and an accelerating efficiency of 40%, one finds that about 6GeV are needed to accelerate a proton to 1 GeV. It would then be possible to obtain about 3.5 neutrons per fission, still without producing usable energy. For more realistic scenarios one sees that an accelerator allows an increase of the number of neutrons available for transmutation at the expense of usable energy. It is interesting to see if, as far as neutron availability is concerned, hybrid reactors are more or less efficient than the association of a critical reactor and an accelerator. The number of neutrons produced in the hybrid reactor is

N0

N = ^; (44)

while the number of fissions is

On average a fission is produced by (<rF ± <rc)/aF neutrons. The total number of neutrons needed to produce NF fissions is

where ] is the number of neutrons produced following the capture of an initial neutron by a fissile nucleus. The total number of neutrons available for transmutation is therefore

NDhyb = N — Nnf = 1 — k). (4.7)

We now consider a critical reactor coupled to an accelerator. NDr is the number of neutrons available when using a reactor producing NF fissions, in addition to the N0 spallation neutrons. The number of neutrons necessary per fission is

= 1 + a (4.8)

°F

while the number of neutrons produced per fission is v. It follows that the number of neutrons available per fission is v — 1 — a. The total number of neutrons available in the reactor is then

and the total number of neutrons available for the reactor + accelerator system is

N°- = N»(>+*rh)(v -1 — a)) = (’ — k)’ (4Л0)

Thus

NDhyb = NDr — (411)

It follows that the choice of a specific value of k is irrelevant as far as the transmutation capabilities are concerned. Whatever the method of coupling between the fission reactor and the accelerator, the number of available neutrons is

Nd = N0 + Nf(v — 1 — a). (4.12)

From the preceding, it is seen that using 10% of the available energy allows us to obtain about 0.1 additional neutrons per fission. Although small, this number has to be compared with the number of neutrons which are effectively available in reactors. We know that the maximum number of available neu­trons per fission amounts to v — 1 — a. In practice the real number is smaller than this value due to captures in structural materials and to transmutations of fertile nuclei. Let the number of such capture neutrons be vc. The number of available neutrons is then v — 1 — a — vc. Captures in structural materials cannot be much less than 0.2 neutron per fission, particularly as reactivity changes are counterbalanced by the presence of consumable neutronic poisons. For each fissioning nucleus a fissile nuclei suffer neutron capture leading, in general, to a fertile nucleus. If one requires regeneration of the nuclear fuel, one sees that vc = 0.2 + 1 + a at least. The number of available neutrons amounts to v — 2(1 + a) — 0.2. We consider four cases.

1. The thermal 238U-239Pu system. Then, v = 2.871, a = 0.36. The number of available neutrons is 2.871 -(2 x 1.36)- 0.2 = —0.05. Regeneration is not possible and no neutrons are available for transmutation. The 0.1 additional neutrons made available by the use of an accelerator would allow regeneration.

2. The thermal 232Th-233U system. In this case v = 2.492, a = 0.09. The number of available neutrons is 2.492 -(2 x 1.09)- 0.2 = 0.11. Regenera­tion is possible and 0.1 neutrons are available for transmutation. The addi­tional number of neutrons that an accelerator would bring is significant.

3. The fast 238U-239Pu system. In this case v = 2.98, a = 0.14. The number of available neutrons is 2.98 -(2 x 1.14)- 0.2 = 0.5. Regeneration is easy. The advantage of an accelerator is not compelling.

4. The fast 232Th-233U system. In this case v = 2.492, a = 0.093. The number of available neutrons becomes 2.492 -(2 x 1.093)- 0.2 = 0.10. Regen­eration is possible. The additional number of neutrons that an accelerator would bring is significant.