III. Chain Reactions and Criticality Introduction We know that neutron production and loss rates determine the behavior of a nuclear reactor. In this chapter we introduce some terms that help us describe production rates, loss rates, and how well they are balanced. Multiplication Factor: Definitions Suppose we could remove all the neutrons from a source-free reactor, then initiate exactly enough fissions to create N fission neutrons. Let us define these to be a generation of neutrons. Some neutrons in any given generation will ultimately cause fission. The neutrons resulting from such fissions are defined to be in the next generation. We now define (for the time being) k = multiplication factor = (1) Then we define: k < 1 neutron population is decreasing k = 1 neutron population is steady k > 1 neutron population is increasing Another definition of multiplication factor is: k = (2) P(t). L(t) Now that we have defined k, we note that unless the neutrons are in their fundamental mode (fundamental distribution) for the reactor in question. In that case, but only then, the two definitions are correct and equivalent. Later we will discuss this fundamental mode (and other modes) and develop a more rigorous definition of k. For now, we note simply that: k is a property of
23 A Simple Problem Consider an infinite uniform sea of some material. Pretend that and that all neutrons have the same speed, v, all cross sections (appropriately averaged over nuclear motion) are known. [Neutrons don t really all have the same speed, but if we pretend that they do then we can illustrate useful concepts more readily.] Let us assume that some neutrons are introduced in to the medium, uniformly and isotropically. What will happen? Any time we want to learn something about a problem like this, we should turn to our favorite fundamental truth: In this problem, we write this as: or where = = (3) ν [that s the Greek letter nu ] Question: Why do we not consider loss from leakage? Answer: Let us consider some fixed finite volume, V, of our infinite medium. Is there any net leakage into or out of this volume? Thus, the above conservation statement holds in every sub-volume of the infinite medium. Let us write our conservation statement in mathematical notation. First, let n = v = Note that our conservation statement involves reaction rates, for which we need the scalar flux. What is the scalar flux in this problem? Answer: φ = path-length rate per unit volume = [n s per unit volume]*[path length per n per s] = (4)
24 Now we can write our conservation statement mathematically: Now we can do lots of neat stuff. First, note that the volume divides out of each term, and we have: which means, in words, that: change-rate density = production-rate density absorption-rate density. We know how to solve this little equation for the neutron density, n. We use an integrating factor (exp{[σ a νσ f ]vt}), and we get Pretty remarkable all we did was recognize conservation, and now we ve figured out everything else about the problem! Question: What is k in the infinite-medium one-speed problem? Answer: Question: What must be true about the material cross sections to make a steady-state neutron population possible, given one-speed neutrons in a source-free infinite medium? Answer: (5) (6) (7) Multiplication Factor: Formulas We shall develop here a formula for the multiplication factor, k, in a thermal reactor. We do so by considering the life history of a neutron from birth to death, describing each fork in the road with a mathematical statement of probability. (We assume that the neutrons are in the fundamental-mode distribution.) First we define: P FNL P TNL p = fast non-leakage probability = probability that a = thermal non-leakage probability = probability that a thermal neutron does not leak, = resonance-escape probability = probability that a fast neutron given that it didn t leak,
25 f u F P TAF P FAF ν T ν F = thermal utilization = probability that given that it was absorbed = fast utilization = probability that given that it was absorbed while fast = prob. that thermal absorption in fuel = probability that = average number of neutrons = average number of neutrons We now sketch a neutron s history in a thermal reactor:
26 It follows that the multiplication factor must be: k = (8) It is customary to define ε fast fission factor = expected # of neutrons from P FNL p P TNL f P TAF ν T + P FNL (1 p) u F P FAF ν F = P FNL p P TNL f P TAF ν T = 1 +, and η T thermal reproduction factor = expected # of neutrons = It is not yet customary, but I also define: η F fast reproduction factor = expected # of neutrons produced per fast absorption in fuel. = This gives us: k = (9a) k = (9b) This is the six-factor formula. In an infinite medium nothing leaks, and we have the four-factor formula: k = (10)
27 Note that Example Recall that a neutron in a reactor will ultimately either leak or get absorbed. Suppose that in a certain reactor, for every 1000 neutrons that are born from fission, the following is how many leak and how many get absorbed: 20 leak after they become thermal 80 leak while they are still fast 200 are absorbed in fuel while they are fast 400 are absorbed in fuel after they become thermal 100 are absorbed in non-fuel materials while fast 200 are absorbed in non-fuel materials while thermal Suppose further that of every 400 thermal neutrons that are absorbed in fuel, 320 actually cause fission, and that on average there are 2.5 neutrons emitted from these thermally-induced fissions. Likewise, suppose that of every 200 fast neutrons absorbed in fuel, 60 actually cause fission, and that on average there are 3.0 fast neutrons emitted from these fast fissions. Question: What are the six factors in the six-factor formula, and what is the multiplication factor? Solution. First, forget all the formula stuff and use common sense to get the multiplication factor. Then you can check later to make sure your factors give the right answer. The commonsense approach to k goes like this: How many fission neutrons will result from 1000 initial fission neutrons? Well, we know that there are 320 thermal fissions, with 2.5 neutrons emitted from each, which gives us (2.5)(320) = 800 new neutrons. Also, there are 60 fast fissions, producing (3)(60) = 180 more new neutrons. So the multiplication factor is (800+180)/1000 = 0.98. Now we ll do the six factors: P FNL = (1000-80)/1000 = 920/1000 p = P TNL = (620-20)/620 = 600/620 f = η T = 2.5 320/400 = 800/400 That s five of them. For the fast-fission factor, ε, we ll get some intermediate results: (1-p) = (920-620)/920 = 300/920 u F = 200/(200+100) = 200/300 η F = 3.0 60/200 = 180/200 Now we can get the fast-fission factor:
28 P FNL p P TNL f P TAF ν T + P FNL (1 p) u F P FAF ν F ε =, P FNL p P TNL f P TAF ν T 800/1000 + 180/1000 = -------------------------- = 800/1000 It is easy to see that if we multiply the six factors, we obtain 980/1000, as we should. Note that because the probabilities are conditional, the numerator of one factor typically cancels the denominator of another! Summary The multiplication factor, k, tells us the basic state of a reactor: k <1 / =1 / >1 means the reactor is sub-critical/critical/super-critical. In a thermal reactor, k is given by the six-factor formula, a product of six terms. If we set the non-leakage probabilities to unity we get the four-factor formula for k. k, the multiplication factor for an infinitely large reactor, is an upper bound for k. Later we will calculate each term in the six-factor formula for k.