NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
48
VI. Implications of Neutron Conservation
Introduction
We are well aware that a reactor’s behavior depends on the gain and loss rates of neutrons in the
reactor. We also know that some gain and loss is due to neutron-nucleus reactions, and that
some is due to neutron leakage. In the previous chapter we developed mathematical expressions
for reaction rates and leakage rates. In this chapter we put these expressions into statements of
conservation to see what we can learn about how the neutron population changes with time in
reactors and how the neutrons distribute themselves in position and energy.
To do this we will examine conservation statements for different neutron populations in different
settings, ranging from simplistic to realistic:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
All neutrons in a given volume of an infinite uniform medium, pretending that all
neutrons have the same speed and that all fission neutrons are emitted promptly.
All neutrons in a given volume of an infinite uniform medium, pretending that all
fission neutrons are emitted promptly but not that all neutrons have the same speed.
Neutrons in a given volume and given energy interval in an infinite uniform medium,
pretending that all fission neutrons are emitted promptly.
All neutrons in a given volume of an infinite uniform medium, pretending that all
neutrons have the same speed but accounting for delayed neutrons.
Neutrons in a given volume and given energy interval in an infinite uniform medium,
accounting for delayed neutrons.
All neutrons in a finite reactor, pretending that all fission neutrons are emitted
promptly.
All neutrons in a finite reactor, accounting for delayed neutrons.
Neutrons in a given volume of a finite reactor, accounting for delayed neutrons.
Neutrons in a given energy interval in a given volume of a finite reactor, accounting
for delayed neutrons.
Neutrons in a given cone of directions in a given energy interval in a given volume
of a finite reactor, accounting for delayed neutrons.
In every case we will follow the same procedure. First we will note that
change rate = gain rate – loss rate
Second, for whatever neutron population we are considering, we will fill in a mathematical
expression for each of these three terms. Third, we will try to solve the resulting mathematical
equation and learn what it tells us about the behavior of the neutron population.
Neutron Conservation: Infinite uniform medium (Cases 1-5 above)
1. SETTING:
POPULATION:
One-speed neutrons; no delayed neutrons
All neutrons in volume
We solved this problem in Chapter III of these notes for the case in which there were no
“extraneous” sources spontaneously emitting neutrons in to the medium. Here we will allow for
the possibility of such sources. We shall continue to assume that the neutrons are distributed
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
49
uniformly and isotropically throughout the medium, and we shall continue to focus on a fixed
volume in the medium, which we call V. Now our gain mechanisms are:
* Emission from the fixed source
* Emission from neutron-induced fission
Our loss mechanism is:
* Absorption
This is the only loss mechanism, because there is no net leakage from any volume in our infinite
medium, because the neutrons are distributed uniformly in position and direction.
We define:
n(t)
= neutron density [n/cm3]
v
= neutron speed [cm/s]
Sext
= “fixed” or “extraneous” source-rate density
Then (see Eq. (5) of Chapter III) our conservation statement for neutrons in the volume V is:
(1)
The volume is constant and cancels out of the equation, leaving:
(2)
We can solve this equation (by using an integrating factor, for example). We obtain:
(3)
Once again, conservation plus a little bit of mathematics has told us a great deal about the
behavior of the neutron population of interest. Note:
1. If there is no fixed source, then the neutron population:
a.
grows exponentially if
b.
falls exponentially if
c.
stays steady if
2. If there is a fixed source, then the neutron population:
a.
eventually grows exponentially if
b.
reaches a steady value if
c.
grows linearly in time if
The last case, 2.c. – a critical medium with a fixed source – is slightly tricky mathematically.
There are two ways to find the result. One is to return to Eq. (2) and note that the fission and
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
50
absorption terms cancel, leaving a simple equation that says the time-derivative of n is the
constant Sext. The other is to use L’Hospital’s rule to evaluate the troublesome term in Eq. (3) –
the one that evaluates to 0/0 when νΣf=Σa.
2. SETTING:
POPULATION:
Distribution of neutron energies; no delayed neutrons
All neutrons in volume
We still have the same conservation statement, but now that we are addressing the reality that
neutrons have a whole spectrum of energies, each term in Eq. (1) involves an integral over
neutron energies:
(4)
As before, the volume V cancels out of the equation, because it is a constant.
However, this equation contains a problem: the fission and absorption terms are not just
That is, this equation is not of the form we had previously, which was
What can we do? There are at least two choices:
1. We can
2. We can examine conservation
Let’s take the first choice first. (We’ll explore the second choice in the next subsection.)
Without approximation, we can multiply and divide the absorption term by the energy-integrated
neutron density:
(5)
Now we recognize that the term in brackets is a
with weight function n(E,t). We use a bracket notation to denote this average:
(6)
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NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
We define a similar average in the fission term:
(7)
Finally, we define:
(8)
(9)
With these definitions, our conservation equation can be rewritten as:
(10)
This appears to be of the same form as Eq. (2), which we have already solved. So have we
managed to solve the problem despite the issue identified above?
No, not really, although we have made some progress. The main issue is that
Even if we are given exactly what ν(E), Σa(E), and Σf(E) are, we cannot say what their nweighted average values are, because we do not know n(E,t). Further, the solution of Eq. (10)
cannot tell us the energy distribution of n(E,t) – it knows only about the energy-integrated
neutron density, ntot(t).
This is where approximations could enter the picture. Suppose we have a guess for the energy
distribution of the neutron density. For example, suppose we guess that
(11)
and we also make a guess for the function f. Then our guess would produce numbers for the
needed averages:
(12)
.
(13)
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NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
This allows us to solve our equation:
.
(14)
However, this is approximate – the < > quantities are only approximations of the correct values!
There are several lessons to learn from this exercise:
1.
Even though we don’t know the exact values of the <bracketed> constants in Eq.
(14), the equation still shows a lot about how the total neutron density changes with
time in the infinite-medium problem we are studying. In fact, the conclusions we
stated just after Eq. (3) are still valid, as long as we replace νΣf and Σa by their
<average> values.
2.
When a term in our conservation equation depends on neutron sub-populations for
which we haven’t written a detailed conservation equation, we have a problem. In
the current example, we wrote conservation for the total neutron density, but our
production and loss terms depend on the neutron density in each energy interval.
This is what forced us to resort to approximations, such as inventing Eq. (11) and
inventing a function f for Eqs. (12) and (13).
These lessons apply quite generally, as we shall see.
3. SETTING:
POPULATION:
Distribution of neutron energies; no delayed neutrons
Only the neutrons in volume V that have energies in ΔE
Here we continue to study an infinite medium and continue to pretend that there are no delayed
neutrons. Now that we are writing our conservation statement only for neutrons in a certain
energy range, though, we have an additional gain mechanism, because neutrons can scatter from
other energies into the chosen energy interval. Now our gain mechanisms are:
* Emission from the fixed source
* Emission from neutron-induced fission
* Scattering
Similarly, neutrons can scatter out of the energy interval, so our loss mechanisms are:
* Absorption
* Scattering
Again, there is no net leakage from any volume in our infinite medium, because the neutrons are
distributed uniformly in position and direction – gain from inleakage cancels with loss from
outleakage.
We define:
n(E,t)
= energy-dependent neutron density
v(E)
= neutron speed [cm/s]
Sext(E) = energy-dependent “extraneous” source-rate density
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
53
χ(E)dE = fraction of fission neutrons emitted with energies in dE at E.
The distribution function χ is called the
Now our conservation statement for neutrons that are in the volume V and that have energies in
the interval ΔE is:
(15)
The volume is constant and cancels out of the equation. Also, we note that Σa+Σs=Σt:
(16)
We recognize that every term involves an integral over the energy interval ΔE, so we group the
terms into a single integral:
(17)
Note that this equation holds for any ΔE that we pick. This means that the function inside the {}
has a zero integral over every possible range of integration. What does that tell us about the
function inside the {}?
Thus we have:
(18)
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Ch. VI. Implications of Neutron Conservation
54
We can write this in terms of the energy-dependent scalar flux instead of the neutron density:
(19)
This is an
for the energy-dependent scalar flux in an infinite uniform medium. This equation is well-posed
and has a unique solution. If the source is known and the cross sections are known, then this
problem can be solved to whatever accuracy we require.
That is, since we have now explicitly recognized conservation for every sub-population of
neutrons that affects gain and loss (i.e., the neutrons in every spatial volume in every energy
interval), we are not forced to make approximations about neutron distributions. This is in sharp
contrast to the situation in the previous subsection, where we did not recognize conservation in
every energy interval and ended up with an unsolvable equation.
5. SETTING:
POPULATION:
Distribution of neutron energies; accounting for delayed neutrons
Only the neutrons in volume V that have energies in ΔE
Here we continue to study an infinite medium but this time we make no significant
approximations to the physics. We account for the true distribution of neutrons in energy and
for the true physics of delayed neutrons. Now our gain mechanisms are:
* Emission from the fixed source
* Emission of prompt neutrons from neutron-induced fission
* Scattering into ΔE from other energies
* Emission of delayed neutrons from neutron-induced fission
Our loss mechanisms are:
* Absorption
* Scattering from energies within ΔE to other energies
Again, there is no net leakage from any volume in our infinite medium, because the neutrons are
distributed uniformly in position and direction – gain from inleakage cancels with loss from
outleakage.
We define:
Sdn(E,t) = energy-dependent delayed-neutron source-rate density
χp(E)
= fission spectrum of prompt neutrons
νp(E)
= average number of prompt neutrons emitted
from a fission caused by a neutron of energy E
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NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
Everything proceeds just as in the previous subsection, and we end up with:
(20)
The big difference here is that the delayed-neutron source, Sdn, is not known – it is something we
have to solve for, just like the scalar flux φ. How do we do this?
Delayed neutrons
To figure this out, let us review the processes that lead to emission of delayed neutrons.
Consider the decay scheme for Bromine-87:
(2.6% of
Br87→d.n.)
Figure. Decay scheme for Bromine-87.
The Br-87 atoms that eventually decay to metastable Kr-87 (and thus yield a neutron) are
called
(These are only ≈ 2.6% of the Br-87 atoms.) Notice that the precursor does not actually expel a
neutron. Instead, it decays to an excited state of another nuclide, which then emits the neutron.
Also note that the precursor does not have to be a direct fission product, but may simply be in the
decay chain of a direct fission product. Bromine-87, for example, is produced by As-87 → Se87 → Br-87. The delayed neutron in question is delayed by the sum of the decay times of the
precursor and its parents.
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
56
Approximately 270 delayed-neutron precursors have been identified to date. Of these, only a
few dozen are of much practical significance.
We seek an equation or set of equations that will give us Sdn(E,t), the energy-dependent delayedneutron emission rate density, which appears in our equation for the energy-dependent scalar
flux in our infinite-medium problem. Let us define some helpful quantities:
Ci(t) = expected density of type-i precursors at time t.
λi
= decay constant of type-i precursors.
χi(E) = energy spectrum of neutrons emitted after decay of type-i precursors.
From these definitions it follows that
(21)
This is fine, but it still doesn’t help us unless we know Ci(t) for each precursor type, i. How do
we find these precursor densities?
As usual, we turn to our favorite equation:
Change rate = gain rate – loss rate
For what population of things should we write this equation?
We already know the loss rate from decay of the precursors – it’s just λi times the precursor
population. But what is the gain rate?
Note that each precursor had its ultimate origin in fission. That is, each precursor is either a
fission product or came from the decay of a fission product. So we know that the production rate
density of precursors will be tied to
Also note that
We define
νdi(E) = expected number of type-i precursors produced from
This is also the expected number of delayed neutrons that will ultimately be emitted from type-i
precursors, per fission caused by a neutron of energy E.
With these definitions we can now write our conservation equation for precursors of type i:
(22)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
57
Let us collect what we have now:
(23)
(24)
We see that
The scalar flux depends on
The delayed-neutron precursors concentrations depend on
Thus, we have a
of equations to solve. While this may look like a complicated system that could be difficult to
solve, it is at least a well-posed set of equations that has a unique solution (given initial
conditions for the scalar flux and the precursor concentrations).
So once again conservation has come through for us! That is, we have written “change rate =
gain rate – loss rate” in mathematical terms, for both the neutron population and the precursor
populations, and arrived at a well-posed system of equations with a unique solution. The
solution, if we can find it, tells us in detail how the neutrons and precursors behave in the setting
we have considered.
I hope you see that conservation is a powerful tool with far-reaching implications!
Now let us see what we can figure out about solutions of Eqs. (23) and (24). We begin by
integrating the first equation over all neutron energies:
(25)
Here we have recognized that the integral of probability distribution functions, such as χ(E) and
P(E’→E), is unity. Now we subtract the scattering-rate density from both sides of the equation
and change dummy variable E’ to E:
(26)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
58
Rewrite in terms of neutron density:
(27)
Recall previous definitions:
(28)
,
(29)
.
(30)
.
(31)
and add another:
.
(32)
Define the
as follows:
(33)
Rewrite our equation in terms of these defined quantities:
(34)
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Ch. VI. Implications of Neutron Conservation
59
Now define
= average time from prompt-neutron birth to absorption (35)
That is,
Also define
(36)
In terms of these quantities, our equation for ntot can be written:
(37)
or
(38)
We define
(39)
(40)
With these definitions our equation for the neutron density can be written in a form that has
become standard for time-dependent neutronics calculations:
(41)
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Ch. VI. Implications of Neutron Conservation
60
Similarly, we can write our precursor equation in standard form:
(42)
where we have defined
(43)
We have now written our equations in the form of
We can learn a lot about the behavior of the neutron population by studying and solving these
equations!
We observe that:
*
The quantities ρ, β, βi, and Λ are all defined in terms of n(E,t), which of course
depends on time. Thus, in general, these quantities all depend on time.
*
It turns out that in many interesting problems, ρ, β, βi, and Λ are all constants.
Thus, it is very useful for us to study the solution of our coupled first-order ODEs in
this interesting case.
Equations (41) and (42) are known as the
They are used extensively to study the time-dependent behavior of nuclear reactors. Let us see
what they tell us.
We consider the case in which. In this case the Point-Reactor Kinetics Equations (PRKEs) are a
set of coupled 1st-order ODEs with constant coefficients. We know from our previous
mathematical studies that:
1.
The solution of each equation is the sum of
2.
The no-source solution of equation is
Given our assumption that Stot and the coefficients are constants, it is easy to show that constant
particular solutions satisfy the PRKEs. You should verify that the following solutions do satisfy
the equations:
(44)
(45)
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Ch. VI. Implications of Neutron Conservation
61
(Actually, you can see that there is a problem with this particular solution if the reactivity, ρ, is
zero. This is a special case that we will address later.) Now let’s address the “homogeneous”
solution – the solution that satisfies the equations with Stot removed. If we define
I = number of types of delayed-neutron precursors that we are tracking,
(46)
then the PRKEs are a set of I+1 coupled ODEs (with constant coefficients in the case we are
studying). We know from our previous mathematics studies that each “homogeneous” solution
(ntot and all of the Ci’s) will be
and that the time constants in the exponentials are
Thus, our PRKE solutions have the following form:
(47)
(48)
But how do we find the constants sj, j=1,…,I+1? This is not terribly complicated: we insert the
solutions into the equations and find the s values that allow the solutions to be non-trivial (i.e.,
the solutions that can have non-zero Aj and Ci,j coefficients in front of the exponentials in the
solutions). After a bit of algebra we find that each of the s values must satisfy the following
equation:
(49)
There are exactly I+1 distinct values of s that will satisfy this equation. This is called the
It is well-known among nuclear engineers. It gives us the s values, which determine how quickly
or slowly each part of the solution changes with time. Note that the s values have units of
In the early days of reactor analysis, people often used “inverse hours” as the unit for the s
values. This led to the name “in-hour,” which the equation still bears.
It is not difficult to show that the I+1 values of s have the following properties:
*
I of them are
*
The other one (call it s1)
*
If ρ=0, then
*
If ρ<0, then
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
62
That is, I values of s will always be real and negative; the other value, which is the largest
algebraically, is real and has the same sign as the reactivity, ρ. This simple information tells us a
great deal about the time variation of the neutron population in a reactor. We explore this for
several interesting cases in the subsections that follow.
Subcritical Reactor with Fixed Source
If the reactor is subcritical, then ρ < 0 and, according to the discussion above, all exponential
time constants {sj} are negative. It follows that for large t, the exponentials become vanishingly
small, and we are left with the “particular” part of the solution:
(50)
That is,
In a source-driven subcritical system,
The steady-state neutron density is proportional to the source strength, proportional to the
prompt-neutron lifetime, and inversely proportional to 1–k. The factor of 1–k in the denominator
expresses the phenomenon of
Note that for k arbitrarily close to unity, the steady-state neutron population can be arbitrarily
large, even with a “small” source and a subcritical reactor.
Note further that for k close to unity, ρ is close to zero and thus s1 will be close to zero. This
means that one exponential will be slow to decay away, which means it will take a long time for
the neutron population to attain its steady-state level.
Supercritical Reactor with Fixed Source
This case is relatively simple. The particular solution is a constant, I of the exponentials decay
with time, and one of the exponential terms has a positive s value. The increasing exponential
eventually dominates all other terms, and
(51)
That is,
In a source-driven supercritical system,
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
63
Critical Reactor with Fixed Source
If the reactor is critical, then ρ = 0 and, according to the discussion above, all exponential time
constants {sj} are negative except for one that is zero. However, we find that a constant
“particular” solution does not satisfy the equations when ρ = 0. (The 1/ρ factor in Eq. (45) is a
hint that this doesn’t work!) We try the next-simplest function we can imagine, namely one that
is linear in t. We find then that the following particular solution does satisfy the equations with
ρ=0:
(52)
and
,
(53)
where
= a positive constant.
(54)
Equation (52) says the “particular” portion of the solution for ntot is a linearly increasing
function. What about the “homogeneous” portion? It has I exponentials that decay away in
time, so they soon become unimportant. The other exponential is exp(0), because
We therefore have
(55)
[Question: Why were we able to replace Λ with lp?
Answer: Because
]
Thus, we have now found that
In a source-driven critical system,
Subcritical Reactor with No Source
In a source-free reactor the particular solution is zero in all cases. If the reactor is subcritical,
then ρ < 0 and, as discussed above, all exponential time constants {sj} are negative. If we
continue to let s1 represent the algebraically largest s (the one closest to zero, in this case), then
the other exponential terms decay more rapidly than the s1 term and we have
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
64
(56)
That is,
In a source-free subcritical system,
Important: At the beginning of a transient (for example, immediately after rapid insertion of
control rods) the population may drop much faster than this, because of the other rapidlyvarying exponentials that are part of the solution. But the population eventually approaches this
single decaying exponential with time constant s1.
Recall that
in a subcritical system. Also note that λmin is associated with the longest-living delayed-neutron
precursors, which have half-lives of around 55 seconds, so
(57)
This means it takes the neutron population at least 80 seconds to decrease by a factor of e. This
is not a very fast decrease! But remember, the population may decrease much faster in the early
stages of a transient.
Supercritical Reactor with No Source
In this case one of the exponential terms has a positive s value while all other exponential terms
are decaying. The increasing exponential eventually dominates the others, and
(58)
where s1>0. That is,
In a source-free supercritical system,
At the beginning of a transient the population may grow more slowly or more rapidly than this,
or even decrease for a little while (!), but it eventually approaches this single increasing
exponential.
Critical Reactor with No Source
If the reactor is critical, then ρ = 0 and, as discussed above, all exponential time constants {sj}
are negative except for one that is zero. It follows that
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
65
(59)
That is,
In a source-free critical system,
This is just what we would expect—the definition of a critical reactor is that it is able to sustain
a steady chain reaction, without help from “extra” sources of neutrons. However, note that the
population could change significantly while it is on its way to the steady value – even in a
source-free critical reactor the neutron population may change for a little while before it settles
out.
Summary of PRKE solutions
We have derived the PRKEs for an infinite uniform medium. We did not make any significant
approximations – neutrons have energy distributions and some fission neutrons are delayed,
exactly as in nature.
As we shall see later, we obtain exactly the same equations for a finite non-uniform reactor – the
PRKEs are quite general! So our work here applies far beyond the infinite uniform medium
that we started with.
We have studied PRKE solutions for situations in which Stot (fixed source), ρ (reactivity), β
(delayed-neutron fraction), βI (type-i delayed-neutron fraction), and Λ (mean generation time)
are all constants. This is a realistic case that does arise in nature! We studied six cases, which
are the combinations of subcritical, supercritical, and critical reactors with and without fixed
sources.
Interestingly, we find that while the presence of delayed neutrons can cause interesting and
complicated behavior early in a transient, eventually the neutron population behaves quite
predictably:
1. If there is no fixed source, then the neutron population:
a.
eventually grows exponentially if
b.
eventually falls exponentially if
c.
eventually stays steady if
2. If there is a fixed source, then the neutron population:
a.
eventually grows exponentially if
b.
eventually reaches a steady value if
c.
eventually grows linearly in time if
If you compare this against what we found in Case 1 early in the chapter, you see similarities,
and you may be tempted to conclude that delayed neutrons are not important. Not so – delayed
neutrons are extremely important! They have a dramatic effect on the time constants that appear
in the exponentials mentioned above!
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Ch. VI. Implications of Neutron Conservation
If there were no delayed neutrons, then it is easy to see (by looking at the inhour equation with
β=0) that
(60)
In a typical commercial reactor, lp is on the order of 0.0001 s. Now consider a slightly
supercritical reactor, with k=0.001. How much would the neutron population (and thus the
power) change in 1 second if there were no delayed neutrons?
(61)
Imagine how difficult it would be to control a reactor if its power could change this quickly!
Delayed neutrons slow this down enormously. They make it relatively easy to control reactors
that are slightly supercritical or subcritical. There will be exercises in which you will quantify
this for yourself.
There follow a series of very important sketches that summarize what we found above. Study
these. Be able to explain them to a high-school student. Practice explaining them to each other.
Work all of this into your intuition about the way neutrons behave in reactors.
n(t)/n(0) for CRITICAL reactor
3
With fixed source
Without fixed source
2.5
2
1.5
1
0.5
0
0
2
4
6
8
10
12
14
16
18
time
Figure. Behavior of neutron population in a CRITICAL reactor.
20
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Ch. VI. Implications of Neutron Conservation
n(t)/n(0) for SUBCRITICAL reactor
1
0.9
With fixed source
0.8
Without fixed source
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
100
200
300
400
500
600
time
Figure. Behavior of neutron population in a SUB-CRITICAL reactor.
n(t)/n(0) for SUPERCRITICAL reactor
With fixed source
10
9
Without fixed source
8
7
6
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
18
time
Figure. Behavior of neutron population in a SUPER-CRITICAL reactor.
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Ch. VI. Implications of Neutron Conservation
68
FEEDBACK
A few words are in order about an important phenomenon that we have not yet described:
In a nuclear reactor, feedback happens when changes in the neutron population cause changes in
the reactor properties. These changes in properties then alter the way that the neutron population
behaves. Most of the feedback in a nuclear reactor is caused by
in the reactor material. Remember that temperature can be strongly influenced by the
which of course is determined by the
which of course is determined by the
Also recall that
depend upon the
and that
is a function of the cross sections.
Illustration:
*
Consider a reactor that is critical at a very low power level, such that fission is not
causing the fuel to be noticeably hotter than its surroundings.
*
Now suppose that control rods are pulled out so that the reactivity becomes positive
(because the absorption cross section has been reduced). As we have seen, the
neutron population will soon be on an exponential increase. Thus, the fission rate
will increase exponentially with time.
*
Eventually this will cause the fuel to heat up significantly. This will change the
cross sections in the fuel (remember – the cross sections are averaged over nucleus
motion, and temperature determines the motion). In fact, in almost all reactors this
will cause the absorption cross section to increase more than the fission cross
section, which causes the multiplication factor to decrease.
*
As temperature increases further, the multiplication factor decreases further. This
continues until k reaches 1 and the reactor settles into steady state.
The above is an example of
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
69
This is what we want. Negative feedback gives us
which means that fluctuations in neutron population are damped out in time. Positive feedback,
on the other hand, would amplify changes. Imagine a reactor in which increased temperature
caused an increase in k and ρ. In such a reactor, a small increase in neutron population would
produce an increase in reactivity, which would cause the population to increase, which would
further increase reactivity, etc. The reactor power would spiral upward until something (like
melting fuel) introduced negative feedback. Not good.
So we design our reactors with negative feedback.
Now we turn from our infinite-medium cases to the finite-reactor cases.
6. SETTING:
POPULATION:
finite reactor; no delayed neutrons
All neutrons in the whole reactor
For this population our gain mechanisms are:
* Emission from the fixed source
* Emission from neutron-induced fission
*
Our loss mechanisms are:
* Absorption
*
We take the difference [outleakage – inleakage] to create a net outleakage loss term. Then we
know how to write all of the terms in our conservation equation:
(62)
We now encounter the same difficulty that we had previously in CASE 2:
Some terms in our equation depend on neutron sub-populations
for which we haven’t written a conservation equation.
As we found in CASE 2, we have a couple of choices. We can:
1)
Define some average quantities, which we cannot calculate, and continue to try to
learn something;
2)
Express conservation for a smaller neutron sub-population and hope that this leads
to a well-posed problem.
70
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VI. Implications of Neutron Conservation
Here we will explore the first path; later we will explore the second. Learning from our previous
experience, we define:
ntot(t) ≡
= total neutrons in reactor at time t,
Stot(t) ≡
(63)
(64)
,
(65)
(66)
We also need to do something with the net outleakage term. The ratio we need is clear – it is just
the term itself divided by ntot – but we need a name and symbol for this ratio that reminds us of
its physical meaning. Its physical meaning can be deduced by noting that if we multiply it by dt,
we get the probability that a neutron in the reactor leaks in time dt.
That is, this ratio plays the same role as a decay constant in a conservation equation for a
radionuclide. We will therefore adopt the definition:
(67)
Question: What are the units of the bracketed quantities and λleak?
Answer:
With these definitions, our conservation equation becomes:
(68)
If we pretend for a moment that the quantities inside the [ ] are constant, then we can solve this
equation to obtain:
(69)