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Ch. VII. Neutron Diffusion
Equation Chapter 7 Section 1
VII. Neutron Diffusion
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. We have learned how to compute reaction rates using the
neutron scalar flux and how to compute net leakage rates using the neutron current. We have
studied the implications of various neutron conservation equations, considering different settings
and writing conservation statements for different neutron populations. In the real-world setting
of a finite heterogeneous reactor, when we applied conservation to the sub-population of
neutrons in some sub-volume of the reactor, some interval in energy, and some cone of
directions, we obtained the following transport equation for the “angular flux,” ψ:
(1)
where we recall that
.
(2)
Equation (1) is the transport equation. It is essentially exact, and it has a unique solution that
tells us everything we need to know about the neutron distribution, but it is difficult to solve.
We also derived the accompanying equation for the delayed-neutron precursor concentrations:
.
(3)
Before we derived the transport equation, we derived a slightly less detailed conservation
statement by looking at the neutron population in some spatial sub-volume and some energy
interval, but not distinguishing among neutrons travelling in different directions. This produced
the following as an alternative to Eq. (1):
(4)
We recognized that while this equation is simpler than the transport equation, it is not solvable,
because it has four unknown functions: φ, Jx, Jy, and Jz. Thus, to make use of this equation, we
will need to augment it with equations for Jx, Jy, and Jz.
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
85
In this chapter we develop an approximate equation that relates the net current density (J) to the
scalar flux (φ). These approximate equations lead to
which is an approximate but very useful equation. After we derive the diffusion equation we
will solve it for various problems that will help us understand out how neutrons distribute
themselves in reactors.
Fick’s Law and the Neutron Diffusion Equation
In this section we will obtain “more equations” that relate the neutron current density J to the
scalar flux φ. These equations, which can be written as one vector equation, will take the form:
J(r)
(5)
This equation, known as Fick’s Law, is
However, in many cases of practical interest, it is a reasonable approximation. For a particularly
relevant example, this approximation works quite well for large homogeneous reactors.
The idea that current flows in a direction opposite the gradient of the scalar flux is intuitively
appealing. If you imagine a box of neutrons, with a higher concentration in one corner than
elsewhere, you can easily imagine a net flow from the higher-concentration area to the lower.
Mono-energetic case
We can derive Fick’s Law if we look at a simple enough problem. We assume that:
a)
b)
c)
d)
e)
the medium (containing the neutrons)
the medium is uniform –– cross sections
the extraneous source varies very little over the distance of a mean-free path;
scattering is
scalar flux, φ(r), is a slowly-varying function of position
f)
g) all neutrons have the same energy.
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Ch. VII. Neutron Diffusion
86
The idea now is to compute the net current density, J, as a function of the scalar flux, φ. We
recall that
ez⋅J dxdy
=
= net n/s crossing
(6)
Recall also that
net n/s crossing
dxdy in +z direction
=
≡
(7)
[Jz+ and Jz– are called “partial currents” or “one-way currents” in the + and – z directions.]
Given our simplifying assumptions, we can approximately calculate these upward and downward
crossing rates. Consider the figure:
Note: all neutrons that cross dxdy must have streamed to this area directly after being emitted by
the extraneous source or directly from a scattering event.
Now we figure out the rate at which neutrons will cross dxdy after either scattering or being
“born” in d3r. First, the rate at which neutrons in d3r scatter is:
(8)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
87
and the rate at which they are “born” in d3r is:
(9)
We combine these by defining a total source:
(10)
Second, the fraction that head toward dxdy is
(11)
(If dxdy were tangent to the surface of a sphere centered at r, the fraction would just be dxdy /
area of sphere. The cosθ factor is because it is not tangent.)
Third, the fraction that actually get to dxdy without colliding is
(12)
If we put this all together, we find
n/s that
cross dxdy after
scattering in d3r
=
(13)
The next step is to integrate over all differential volumes (d3r) that are above the x-y plane,
thereby obtaining the total neutrons/second that flow downward through dxdy. When we try this,
however, we encounter a difficulty: we don’t know φ(r)! We get around that by invoking our
assumptions that neither Sext nor φ changes much over a distance of a few mean-free paths. We
use a truncated Taylor-series expansion:
Stot(x,y,z) ≈ Stot(0,0,0) +
(14)
We then assume that
Next is simply integration over the volume above x-y plane. We find that
Jz–(0) =
(15)
The upward part is computed similarly:
Jz+(0) =
(16)
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Ch. VII. Neutron Diffusion
88
Thus, we obtain the z-component of the net current density:
Jz(0) = Jz+(0) – Jz–(0) =
(17)
The other components are, of course, similar:
Jx(0) =
,
Jy(0) =
,
or, in shorthand,
J(r) =
(18)
We dropped indication of evaluation at the origin, since the location of the origin of coordinates
in this derivation is arbitrary. This equation is valid at any point in the medium.
Now we attempt to express Stot in terms of φ. Recall the balance equation:
,
or
or
Given our assumption that Stot is slowly varying, the second term in the last expression is much
smaller than the first, and we have
(19)
This approximation yields:
J(r) =
(20)
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Ch. VII. Neutron Diffusion
We obtained this result by considering a very simple situation and making some simplifying
approximations. A similar result can be shown to hold (approximately, anyway) under more
general conditions:
J(r) =
(21)
D(r) = diffusion coefficient [cm].
(22)
Here
Energy-dependent case
The generalization to the energy-dependent case is simple to write down:
J(r,E) =
(23)
where
D(r,E) = energy-dependent diffusion coefficient [cm].
(24)
Note: If you are not told otherwise, use the following expression for the diffusion coefficient:
D(r,E) =
where
=
the average
Also, if you are not told anything about , just state that you are assuming isotropic scattering,
which means
=0 and Σtr(r,E) = Σt(r,E).
The conservation equation, Eq. (4), is exact. That is, the true scalar flux φ and the true net
current density J in a reactor
will always satisfy this equation.
However, Fick’s Law (the “diffusion approximation”) is not exact. That is, in general the true
scalar flux and net current density
However, in many cases of practical interest this diffusion approximation is
We will use it extensively in this course.
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Ch. VII. Neutron Diffusion
90
If we insert the diffusion approximation (23) into our balance equation (4), we obtain:
(25)
(Here I is the number of types of delayed-neutron precursors.) This is the
energy-dependent neutron diffusion equation.
It is not exact, but for most of this course it is the model that we will use to describe the behavior
of neutrons in reactors.
If we return for a moment to the mono-energetic balance equation and use the mono-energetic
diffusion approximation in it, we obtain:
(26)
This is the
mono-energetic, or “one-speed” diffusion equation.
It is not exact, but we will study it quite a bit to develop an understanding of how neutrons
distribute themselves spatially.
Two-Group Diffusion
Suppose we divide all the neutrons in a reactor into two “groups” according to their energies:
What would the conservation equation look like for each group?
(27)
(28)
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Ch. VII. Neutron Diffusion
91
Here we have defined “group-averaged” absorption cross sections (Σa,1 and Σa,2), groupaveraged speeds (v1 and v2), group-to-group scattering cross sections (Σs,1→2 and Σs,2→1), and
the fraction of fission neutrons that is born in each group (χp1, χdn,1, χp2, and χdn,2).
If we made the diffusion approximation for the net current density in each group:
, g=1 and 2,
(29)
then the resulting 2-group diffusion equations would be:
(30)
(31)
We will worry later about how to obtain the group-averaged cross-sections. For now, we simply
note that:
the 2-group diffusion equations are
Boundary Conditions for the Neutron Diffusion Equation
The steady-state diffusion equation is a
We know that differential equations are usually satisfied by
It takes
to determine which solution is the right one for a given problem.
With our steady-state diffusion equation, we must specify at each point on the problem boundary
one of the following:
1)
2)
3)
(You will sometimes see en⋅∇φ written as dφ/dn. The two expressions mean the same thing ––
the derivative of φ in the outward direction. Here en is the outward unit normal.)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
92
The Usual Boundary Conditions
Vacuum boundary.
If part of our problem domain is surrounded by something we are pretending is a perfect
vacuum, we usually will use the “extrapolated boundary” condition. That is, we assume that the
scalar flux is decreasing as we move toward such a boundary, and that it is decreasing at such a
rate that if we extrapolated it, it would extrapolate to zero at “d” cm from the boundary:
1) General extrapolated boundary condition:
2) 1D slab extrapolated b. c. at right face (x=a/2):
3) 1D slab extrapolated b. c. at left face (x=–a/2):
4) 1D sphere extrapolated b.c. at outer surface (r=R):
Of course, the scalar flux does not really go to zero out there. We are just saying that it is
decreasing fast enough that it would hit zero if we extrapolated it. See the figure below.
Rigorous calculations (using transport theory) tell us that the best linear extrapolation distance to
use, at least in certain simple problems, is
dlinear = (0.7104...)(3D),
where D is the diffusion coefficient in the problem domain.
It is sometimes much more convenient mathematically to specify the distance at which the
scalar-flux function, instead of a linear extension of it, goes to zero. We will almost always use
d=
where D is the diffusion coefficient in the problem domain, for this purpose.
When we say the scalar-flux function goes to zero at some distance, we are
We do this because it
See figure!
(32)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
93
Reflecting (or “mirror”) boundary.
There is no net flow of neutrons across a reflecting boundary. No net flow means
In the diffusion approximation, this means
(33)
That is the reflecting boundary condition in general. In specific,
1) 1D slab reflecting b. c. at right face (x=a/2, e.g.):
2) 1D slab reflecting b. c. at left face (x=0, e.g.):
Interface between different materials.
At the interface between materials we have continuity of two quantities:
1) neutron density
2) net flow across interface
This translates to:
1) General interface at ri, with normal en,i:
2) 1D slab interface at xi:
Other helpful truths
When properly specified, the boundary conditions of a diffusion problem uniquely determine the
solution. However, sometimes we can use common-sense shortcuts to save us a bit of algebra.
That is, we know that the scalar flux:
1) is
everywhere that the diffusion equation is valid
2) is
except possibly at singular points of a source distribution.
Recognizing these facts often allows us to eliminate certain potential solutions before we even
apply the boundary conditions.
Note that we stated “everywhere the diffusion equation is valid.” For example, we might find in
some problem that φ(x) = ex/L for x in the interval (0,10). This function goes to infinity as x goes
to infinity, but that doesn’t matter, because it is valid only for x in (0,10).
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Ch. VII. Neutron Diffusion
Math: the ∇ and ∇⋅ Operators
In various coordinate systems the derivative operators appearing in the diffusion equation take
different forms, as shown in the following table. Here we use f to denote any scalar function
(such as scalar flux) and v to denote any vector function (such as net current density).
Geometry
∇f [“gradient”]
∇⋅v [“divergence”]
∇⋅∇f ≡ ∇2f
3D Cartesian
1D Cartesian (slab)
1D spherical
1D cylindrical
Steady-State One-Speed Diffusion Solutions in Homogeneous Media
In this section we shall use the diffusion equation to compute the scalar flux as a function of
position in several simple problems. In each problem we shall assume that
material properties are
(That’s what homogeneous means.) We shall also assume mono-energetic neutrons and no
time dependence. Finally, we shall consider only
media
for now.
Under these assumptions our diffusion equation is
[SS, 1-speed, homogeneous]
(34)
We often rewrite this as
(35)
where
L ≡
(36)
In one-dimensional slabs this becomes:
[SS, 1-spd, homog., slab] (37)
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Ch. VII. Neutron Diffusion
The solution of the diffusion equation is
[the diffusion approximation of] the
Once we have the scalar flux, we can get
[the diffusion approximation of] the
by using
Remember, we need the flux & current because:
a) reactor behavior depends on neutron production & loss rates,
b) production & loss rates depend on reaction rates & leakage rates,
c) reaction-rate densities depend on scalar flux,
d) leakage rates depend on net current density.
We will now solve the steady-state one-speed diffusion equation in a few homogeneous
problems, to illustrate how its done.
Plane Source, infinite medium
Suppose at x=0 in an infinite homogeneous medium there is an infinitesimally thin plane source
of neutrons emitting
S0 [n/cm2-s].
This is a 1D problem –– nothing varies in the y or z directions –– so Eq. (34) applies, as long as
we can figure out how to write Sext(x). We can:
[plane source, ∞ med]
(38)
where
δ(x) =
Properties of the Dirac delta:
1)
2) δ(x)dx is
3)
=
4) δ(x) is
This delta-function information is not really required for this problem; it is introduced here
because you’ll need it later, and because I wanted to scare you a little bit.
Away from x=0, we have
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NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
(39)
The general solution of this ODE is
(40)
Let us consider x>0 first. By our finiteness requirement we see that
Thus, all that remains is finding the constant “A.”
To do this, we take a close look at
neutron leakage rates
(i.e., the rate at which neutrons cross imaginary surfaces) near the source plane. Let us look at
the net neutrons per cm2 per sec crossing the plane at x=ε, where ε is a small positive number.
We know that the net rate to the right is just
(41)
We also know that this is just
crossing rate/cm2 to right – crossing rate/cm2 to left
= [S0/2 +
–
As ε → 0, the second and third terms become equal in magnitude, by symmetry. Thus,
(42)
This is often called a “source condition.” As far as our ODE is concerned, it serves as a
boundary condition:
= S0/2 ⇒
=
which means
A=
(43)
Thus, for x>0 our diffusion solution is
φ(x) =
(44)
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Ch. VII. Neutron Diffusion
If we consider x<0, everything works the same way. Finiteness tells us A=0. Leakage rate at
x=–ε leads to the source condition
.
(45)
This leads to the solution
φ(x) =
(46)
which by now is obvious from symmetry. We can combine Eqs. (44) and (46):
(47)
Sketch of Diffusion Solution, Plane Source, ∞ Homogeneous Medium.
Point Source in Infinite Homogeneous Medium
Suppose at r=0 in an infinite homogeneous medium there is an infinitesimally small source of
neutrons emitting
Stot [n/s].
This is a 1D problem in spherical coordinates. Away from r=0, the problem is source-free, and
we have:
for r≠0.
[point source, ∞ med]
(48)
The general solution of this homogeneous ODE is
(49)
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Ch. VII. Neutron Diffusion
(You should verify this!!!) This solution was found by making the substitution
into Eq. (48), which results in a slab-like diffusion equation for w.
By our finiteness requirement (and by physical reasoning) we see that
Thus, all that remains is finding A.
To do this, we take a close look at the neutron leakage rates (i.e., the rate at which neutrons cross
imaginary surfaces) near the source point. Consider the net neutrons per sec crossing a spherical
shell of radius r, in the outward direction. We know that this is just
(50)
We also know that this is just
crossing rate outward – crossing rate inward .
As r approaches 0 the second term vanishes, because the surface area gets vanishingly small.
Thus,
(51)
We can use this to determine A. In spherical coordinates we have
(52)
Thus,
=
=
which means
A=
(53)
Thus,
(54)
Comparing this to the plane-source result, you will see that the scalar flux decreases more rapidly
in the point-source case. This makes physical sense –– the neutrons are now
as well as getting absorbed.
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
99
Plane Source in center of Finite Slab
We consider now a plane source at x=0 in a slab that goes from x=–a to x=+a. This problem is
much like the plane-source-infinite-medium problem, except that the finiteness conditions are
now replaced by boundary conditions. For x≠0 we still have
φ(x) = Aexp(–x/L) + Cexp(x/L),
(55)
but this time we cannot say C=0. We say instead that
φ(a+d) =
(56)
where d=extrapolation distance. (This is extrapolation with curvature.) The source condition for
x>0 is still
= S0/2 ,
(57)
but now φ has two terms instead of just one. Equations (56) and (57) are two equations for the
two constants A and C; a bit of algebra produces them.
The x<0 half of the problem is handled in the same way. The final solution is
(58)
This solution is shown below, with the infinite-medium solution shown also for comparison.
(Parameters used for plot: a=L or 2L, d=0.1L, S0L/2D = 1.)
Hints on 1D Diffusion Solutions:
When solving 1D slab diffusion equations with a given extraneous source, we will generally
have (in each region) a
particular solution and a “homogeneous” solution.
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Ch. VII. Neutron Diffusion
The “homogeneous” solution (the general solution of the equation without a source) will be a
linear combination of
These can be written, for example, as
exp(x/L) and exp(–x/L)
or
Which you choose can dramatically affect how much algebra you do to find your constants.
Here are some tips on choosing. In general,
1) Use sinh & cosh in finite regions. For one thing, sinh has a zero, which is very nice
for dealing with extrapolated boundary conditions. For another, cosh is an even
function of x, which is nice in problems whose solutions are symmetric.
2) Use exp(±x/L) in infinite media. One exponential goes to zero as x goes to infinity,
and the other blows up –– this can make it very easy to determine that one constant
is zero, for example.
3) sinh(c + x/L) and cosh(c + x/L) are just as valid as sinh(x/L) and cosh(x/L). In some
problems you can use this to great advantage. For example, if you have a reflecting
boundary at x=b, you might try sinh[(x–b)/L] and cosh[(x–b)/L].
The Diffusion Length (L)
The diffusion length L, defined as
L≡
does have a physical interpretation. It is proportional to an average straight-line distance
between neutron birth point and absorption point, at least in an infinite medium.
Recall that if Stot neutrons/second are born at a point in an infinite medium, the scalar flux a
distance r always is (in the diffusion approximation) given by:
(59)
a result we derived a few pages ago. The absorption-rate density a distance r from the source is
therefore
A.R.D. =
and the absorption rate between r and r+dr is:
.
(60)
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Ch. VII. Neutron Diffusion
Thus, the probability of absorption between r and r+dr, where r is the distance from a point
source in an infinite medium, is just
(61)
Note:
General definition of an average:
In our case,
(62)
The denominator =1 –– the probability that a neutron gets absorbed somewhere in the infinite
medium is unity:
.
If we let f(r) = r2, we get
(63)
This
is the
average squared
between birth and absorption points in an infinite homogeneous medium. We see from Eq. (63)
that
(64)
So: The diffusion length, L, tells us how far away neutrons get, on average, before they die.
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Ch. VII. Neutron Diffusion
Steady-State One-Speed Diffusion Solutions in Heterogeneous Media
In this section we consider problem domains composed of more than one homogeneous region.
The figure below shows, for example, a two-region slab that is infinite in the y and z directions:
We still use the diffusion equation in each region. In a problem with regions F and M, for
example, we will have (given steady state and the one-speed assumption):
,
(65)
and
(66)
We write down the general solution in each region; each will contain unknown constants. We
will have to use boundary conditions at problem boundaries and
at each interface between regions. In a two-region slab, this means
Example
Consider the slab shown above. In the left region let Sext be SF n/(cm3-s); in the right let Sext be
SM n/(cm3-s). Let the boundaries at x=0 and x=b be reflecting (“mirror”). Compute the
(diffusion approximation of the) scalar flux in the slab.
1) Our diffusion equations are:
and
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Ch. VII. Neutron Diffusion
.
Our general solutions are:
,
(67)
(68)
You will soon see why we’ve chosen this form for our general solutions. Note that all of the
above means that our scalar flux is:
(69)
Our boundary conditions are:
1) reflecting at x=0:
2) reflecting at x=b:
Our interface conditions are:
1) continuity of scalar flux:
2) continuity of normal component of current:
Now let us determine our four constants (AF, CF, AM, CM) using our four conditions (two
boundary, two interface). Reflecting at x=0 tells us that
⇒
(70)
Reflecting at x=b tells us that
⇒
Thus, so far we have:
(71)
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Ch. VII. Neutron Diffusion
.
(72)
Now we apply the interface conditions. Continuity of scalar flux tells us that:
;
(73)
continuity of Jx tells us that:
(74)
These are two equations for our two unknown constants AF and AM. Note that all the cosh and
sinh terms in these two equations are
That is, there is no “x” anywhere. This is just a simple system of two equations [(73) and (74)]
and two unknowns (AF, AM). You know how to solve this! It has the form
α1 AF + β1 AM = q1 ,
α2 AF + β2 AM = q2 .
Don’t let the ugly α’s and β’s intimidate you –– they are just numbers!
Criticality (k-eigenvalue) Problems
So far when we have solved the diffusion equation to give us a scalar flux distribution we have
ignored fission. In this section we will treat fission, and we will solve for the multiplication
factor and the neutron distribution in simple reactors.
Suppose a bare reactor is just critical and operating in steady state, which means the loss rate
due to leakage and absorption exactly equals the production rate from fission. The one-speed (or
one-group) diffusion description of this reactor is just:
(75)
(76)
Here V denotes the volume of the reactor, ∂V is its boundary,
en is the outward unit normal
on the boundary, and
d is the
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Ch. VII. Neutron Diffusion
105
Suppose now that a similar reactor is not critical, but that we can multiply the loss terms by some
number so that
(loss rate)*(number) = production rate.
If (number) < 1, the reactor is subcritical; if (number) >1, the reactor is supercritical. If we call
this number “k”, and divide through by it, we obtain:
(77)
This equation says that for some scalar flux function φ(r) and some number k,
(78)
Question: Can a k-eigenvalue problem have a time variable or a time-derivative term?
Answer:
Math
Equation (77) is an eigenvalue equation. The number “k” is an
and the function φ is an
Note that eigenvalue and eigenfunction mean
“characteristic” value and “characteristic” function.
There is a large set of {eigenvalue, eigenfunction} pairs that satisfy (77) and the boundary
condition (76). In fact, there are infinitely many! Only one eigenfunction in the set is
(This is not obvious, but with some math we can prove it.) The eigenvalue corresponding to this
special eigenfunction is, in fact, the
eigenvalue in the set. We give it a special name:
In fact, this value, keff, is the
of the reactor. To see this, let’s assume we know the eigenfunction that goes with the eigenvalue
keff –– call it φfund(r) –– and then let’s integrate the equation over the reactor volume:
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
106
or
keff =
(79)
=
(80)
This agrees with our earlier “definition” of multiplication factor. The really precise definition of
multiplication factor, for one-group or one-speed homogeneous diffusion problems, is
multiplication factor = .
The precise definition of the multiplication factor in general is:
multiplication factor =
By the way: eigenvalue problems
never contain extraneous sources!
They also never contain fixed incoming partial currents on boundaries. Eigenvalue problems tell
us
characteristics of the
They couldn’t do this if sources were present, because sources “force” the system behavior
instead of letting it behave as it wants to.
Please do not forget this, or bad things will happen to you!
Here ends the math part, for now.
Some Reactor Physics
Suppose a reactor is just critical and operating in steady state, which means the loss rate due to
leakage and collisions exactly equals the production rate from fission. What would be the
diffusion equation for thermal neutrons? The net outleakage rate density would be
where Dth is the thermal diffusion coefficient (i.e., a diffusion coefficient appropriately averaged
over all neutron energies less than an eV or so). The absorption rate density would be
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where Σa,th is the thermal absorption cross section (i.e., the absorption cross section
appropriately averaged over all neutron energies less than an eV or so). The outscattering rate
density would be
where Σs,th→fast is the cross section for thermal neutrons scattering to energies higher than the
boundary that we’ve defined for “thermal” neutrons (usually around an eV or so). Note that this
“upscattering” term will usually be very small compared to the absorption term.
What would be the production rate density? Note the following:
1) The mechanism for creating thermal neutrons is
2) The downscattering rate density at position r is approximately
This is only an approximation because
The approximation is reasonably accurate if
3) The fast-neutron production rate density at r is
It follows that the production rate density of thermal neutrons at r is approximately
Thus, if a reactor is just critical and operating in steady state, the diffusion equation for thermal
neutrons is approximately:
.
(81)
Thus, the k-eigenvalue problem, not making any silly assumptions about neutrons all moving
with the same speed, can be written approximately as follows:
(82)
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Ch. VII. Neutron Diffusion
108
For the remainder of this chapter, let us agree to the following:
• D means Dth, the thermal diffusion coefficient.
• Σa means Σa,th – Σs,th→fast, the thermal absorption + upscattering cross section.
• Σf means Σf,th, the thermal fission cross section.
• ν means
• φ means th, the thermal flux.
Then our physically realistic k-eigenvalue problem can be written exactly as we wrote it when
we were making the (terribly unphysical) approximation that all neutrons move with the same
speed:
(83)
(84)
Solving k-Eigenvalue Problems
Bare Homogeneous Slab
We consider first a bare, homogeneous slab reactor (infinite in y and z directions), of width a in
the x direction. Bare means it is surrounded by
nothing but
Homogeneous means
all material properties are
We want to know whether this reactor is critical, and also what the neutrons are doing in it. We
have to solve the following eigenvalue problem:
(85)
(86)
where
=
(87)
where d is the extrapolation distance. (Unless you are told otherwise, use d=2D. Remember
this!)
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Ch. VII. Neutron Diffusion
109
Let us rewrite our equation:
(88)
and define the number B2:
(89)
which gives us
for x ∈ (–a/2, a/2).
(90)
Note and remember: there is a simple one-to-one relationship between B2 and the eigenvalue k.
In fact, we can think of B2 as an eigenvalue itself.
Eigenvalue means
Our equation can’t be satisfied by just any old B2; only certain characteristic values will work.
Let’s solve the problem. It looks like φ is a linear combination of sines and cosines, at least if B2
is positive (which we can prove, but for now will simply assume):
(91)
[Exercise: see what happens if you assume B2 <0!] Let’s apply the boundary conditions:
(92)
If these equations had to be true for all constants B, the only solution would be
(Adding the equations gives C=0; subtracting gives A=0.)
We are not interested in zero solutions. Instead, we seek those characteristic values of B that
allow nontrivial solutions. If we add the boundary-condition equations (92), we obtain
= 0.
(93)
Either C or the cosine term must be zero. The cosine term will be zero for the following values
of B:
Bn =
(94)
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Ch. VII. Neutron Diffusion
110
These are some of the eigenvalues of our slab problem. The associated eigenfuntions are
where Cn is any constant (and n is odd).
If we subtract the boundary-condition equations (92) we obtain
(95)
Either A or the sine term must be zero. The sine term will be zero for the following values of B:
Bn =
(96)
These are the rest of the eigenvalues of our slab problem. The associated eigenfuntions are
where An is any constant (and n is even).
That’s it. We have completely solved the eigenvalue problem specified by Eqs. (85) and (86).
Now we will discuss the solution at some length.
“Modes”
The nth eigenfunction is often called the
n-th mode.
The one eigenfunction that is positive throughout the problem domain is called the
In our problem the fundamental mode is
fundamental mode = φ1(x) =
(97)
A very important special case is source-free steady-state. We know that if a source-free reactor
is operating in steady state, then its multiplication factor is exactly 1. If the reactor is operating
in steady state, then we know that the scalar flux satisfies the steady-state equation, which is the
same as the k-eigenvalue equation with k=1. Thus, we can conclude that the scalar flux function
that satisfies the steady-state source-free equation must be the fundamental mode! So remember
this:
In a source-free steady state reactor,
Note that eigenvalue problems tell us only the shapes of the modes. Even if we were told that
our reactor is source-free and steady-state, we would still need more information to determine
This information might be the total reactor power, the value of the peak scalar flux, the total net
outleakage rate, etc. – something that pins down the amplitude of the flux.
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Ch. VII. Neutron Diffusion
111
What the modes look like. Let’s draw the first few eigenfunctions of our slab reactor:
As promised, only the fundamental mode is positive throughout the domain (which in this
picture goes from –0.46 to +0.46). Note that this mode is “less leaky” than the others: it has a
smaller gradient at the boundaries.
Some important concepts, terms, and definitions
Here we define some terms that nuclear engineers find very useful in analyzing, designing, and
discussing reactors. You should understand them all.
Geometric buckling
Recall the eigenvalue equation that we have been working on, and write it for the nth
{eigenvalue, eigenfunction) pair:
.
Divide through by φn(x):
(98)
Note:
1) Bn2 is a measure of
2) If Bn2 and φn are both positive, then
We have a name for the Bn2 associated with the fundamental mode:
Fundamental-mode
≡
(99)
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Ch. VII. Neutron Diffusion
This will always be the smallest of the B2 eigenvalues. It is a measure of how much the
fundamental mode is curved, or “buckled.” In a bare slab we have
=
[geom. buckling, bare slab] (100)
If you forget this, bad things will happen! [Build your intuition: how does solution curvature
change as the reactor width increases?]
Connection with k
The problem we wrote down originally was
(101)
.
(102)
[We have written it here with the index n explicitly denoting the nth (k,φ) pair.] We then wrote it
in a slightly simpler-looking form by defining
(103)
This is the relationship between the B-eigenvalues and the k-eigenvalues. Solve for kn:
(104)
The kn associated with the fundamental mode has a special name:
Fundamental kn ≡
(105)
It is hopefully obvious that we have:
(106)
Materials Buckling
If our reactor is critical, then keff =1 and
=
[only if critical] (107)
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Ch. VII. Neutron Diffusion
113
The term on the left is geometric buckling, which depends only on reactor geometry (shape and
size). (In a bare slab, the only “geometry” information is the extrapolated slab width.) The term
on the right depends only on
We assign this term a special name:
≡
(108)
If the two bucklings are equal, that is
if the material properties and the reactor geometry
then the reactor is critical. In general,
<
⇔
=
⇔
>
⇔
The critical relationship is easy to remember. To easily remember the others, imagine what must
happen when Σf gets larger or smaller (which makes the materials buckling larger or smaller).
The above relations apply to any bare homogeneous reactor, regardless of shape. Only the
definition of
, which of course depends on geometry, will change.
Bare Homogeneous Sphere
We consider next a bare, homogeneous spherical reactor. Bare means it is surrounded by
nothing but vacuum. Homogeneous means all material properties are constant. We want to
know whether or not this reactor is critical, and also what the neutrons are doing in it. We must
solve the following eigenvalue problem:
(109)
(110)
where
= extrapolated radius =
(111)
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Ch. VII. Neutron Diffusion
114
where d is the extrapolation distance. (Unless you are told otherwise, use d=2D.) We have
recognized, by using the subscript n, that our eigenvalue problem will have many {eigenvalue,
eigenfunction} pairs as solutions.
If we bring all terms to the left side of Eq. (109) and divide through by (–D), we obtain
(112)
where we have defined as before:
.
(113)
Note that there is a one-to-one relationship between Bn2 and the eigenvalue kn. In fact, we can
think of Bn2 as an eigenvalue itself.
Eigenvalue means characteristic value. Our equation can’t be satisfied by just any old B2; only
certain characteristic values will work.
Let’s solve the problem. You can verify that the following satisfies our equation:
φn(r) =
(114)
If we apply our finiteness condition [see Eq. (110)], we see that
Our boundary condition then looks like:
(115)
If this equation had to be true for all constants B, the only solution would be the trivial solution
An=0.
We are not interested in zero solutions. Instead, we seek those certain values of B that allow
nontrivial solutions. The sine term will be zero for the following values of B:
Bn =
(116)
These are the eigenvalues of our spherical problem. The associated eigenfuntions are:
where An is any constant.
That’s it. We have completely solved the eigenvalue problem specified by Eqs. (109) and (110).
Now we will discuss the solution.
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Ch. VII. Neutron Diffusion
The one eigenfunction that is positive throughout the problem domain is called the fundamental
mode. In our problem the fundamental mode is
fundamental mode = φ1(r) =
(117)
If a bare reactor has been left alone for “a long time” and there are no extraneous sources, the
scalar flux will be in the fundamental mode. This just describes the
shape
of the flux. In a given problem, the magnitude of the flux must be determined by other
conditions. For example, we may be told the reactor is operating in steady state at a certain
power level. Then
P = known power =
(118)
What the modes look like. Let’s draw a picture of the first few eigenfunctions of our spherical
reactor:
As promised, only the fundamental mode is positive throughout the domain (which in this
picture goes from r/R = 0 to 0.98).
Geometric buckling. Recall that:
The
associated with the fundamental mode ≡
(119)
It is a measure of how much the fundamental mode is curved, or “buckled.” In a bare sphere we
have
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Ch. VII. Neutron Diffusion
=
[geom. buckling, bare sphere] (120)
If you forget this, bad things will happen!
Connection with k. This is just like in slab geometry, or any other bare geometry for that matter.
We still have:
.
(121)
And of course, we still have:
(122)
And again, these relations apply to any bare homogeneous reactor, regardless of shape. Only the
definition of
, which of course depends on geometry, will change.
Bare Shoebox
In a bare homogeneous rectangular parallelepiped (brick- or shoebox-shaped) reactor, our
eigenvalue equations look like:
(123)
, (124)
where the ~ quantities are extrapolated widths & heights & lengths:
,
(125)
and d is the extrapolation distance. (Unless you are told otherwise, use d=2D.)
There is a fundamental difference between this and our previous examples: our equation is now
a
differential equation. How do we solve it? We guess a separable solution –– that is, we guess
that our φ(x,y,z) is composed of products of one-dimensional functions:
(126)
We put this into our equation:
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
117
(127)
and then divide through by XYZ:
(128)
We also divide through by –D, and define B2 ≡ [(1/k) (νΣf/Σa) –1]/L2, as before. Then:
(129)
Note:
• the first term depends only on
• the second term depends only on
• the third term depends only on
• the right-hand side doesn’t depend on x, y, or z.
Thus, as x varies across its allowed range (–a/2,+a/2), the second and third and fourth and righthand terms don’t change. But if they don’t change, that means the first term didn’t change either
(because all these terms are in an equation together). This (and similar reasoning for y and z)
means that:
≡
(130)
≡
(131)
≡
(132)
If we add these equations, we see that we must have the following relationship between these
constants and the B2 above:
(133)
Equations (130)-(132) are just three separate eigenvalue problems of the type we have already
solved. For example, the X problem looks like:
(134)
with
(135)
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Ch. VII. Neutron Diffusion
118
The Y and Z problems are similar. We have already solved these problems! Each is just a slabgeometry problem! The solutions are:
(Bx)i =
(136)
Xi(x) =
(137)
(By)j =
(138)
Yj(y) =
(139)
(Bz)k =
(140)
Zk(z) =
(141)
Recall that our original equation was
,
(142)
where
(143)
We see now that φijk(x,y,z) and
satisfy this equation if
(i,j,k)th mode: φijk(x,y,z) =
(i,j,k)th B2 eigenvalue:
(144)
=
(145)
What is the fundamental mode, and what is the geometric buckling?
fundamental mode: φfund(x,y,z) =
(146)
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
geometric buckling:
=
119
(147)
geometric buckling, bare homog. rectangular parallelepiped (shoebox)
What is keff?
(148)
Bare Homogeneous Infinite Cylinder
We consider next a bare, homogeneous infinitely-long cylindrical reactor. We want to know
whether or not this reactor is critical, and also what the neutrons in it are doing. We have to
solve the following eigenvalue problem:
(149)
(150)
where
=
(151)
where d is the extrapolation distance. (Use d=2D unless told otherwise.)
Let us rewrite our equation:
(152)
where we’ve defined B2 as usual:
(153)
To solve this system, we rewrite the equation (again) in a form that may ring a bell:
(154)
[We just carried out the d/dr operation using the product rule, and then added a zero term to put
the equation in a ‘standard’ form.] This is
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Ch. VII. Neutron Diffusion
(but he’ll let you use it). If
> 0, the solution is a linear combination of
of the first and second kind
φn(r) =
(155)
[Or see Appendix D of Stacey’s book, or see a differential equations book.] If <0, the solution
is a linear combination of modified Bessel functions of the first and second kind [AnI0(Bnr) +
CnK0(Bnr)]. It is not terribly difficult to prove that >0, so each of our eigenfunctions will
satisfy Eq. (155).
The Y0 function is
while the J0 function is bounded everywhere. Thus we must have:
or
φn(r) =
(156)
Our extrapolated boundary condition tells us that either
[which doesn’t interest us] or
Thus, the eigenvalues Bn are the numbers such that
Bn
=
(157)
This is analagous to Bn being the nth ‘zero’ of the sine function in our sphere problem. The
zeroes of the sine are simple –– nπ for n = 1,2,3,.... The zeroes of J0 are not that easy to write
down, but the concept is the same and the zeroes are known.
The smallest ‘zero’ of J0 is
The geometric buckling is therefore
[geom. buckling, infinite cylinder] (158)
The fundamental mode is
φfund(r) =
[fund. mode, infinite cylinder] (159)
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Ch. VII. Neutron Diffusion
121
Question: In terms of material properties and reactor dimensions, what is the multiplication
factor of an infinitely long bare homogeneous cylinder?
Answer: keff
Question: In terms of material properties and reactor dimensions, what is the multiplication
factor of a bare homogeneous sphere?
Answer: keff =
Question: In terms of material properties and reactor dimensions, what is the multiplication
factor of a bare homogeneous brick?
Answer: keff =
Question: In terms of material properties and reactor dimensions, what is the multiplication
factor of a bare homogeneous finite cylinder?
Answer: This will be an exercise.
:-)
Time Dependent Diffusion: Brief Discussion
Consider the diffusion equation for a bare, source-free, homogeneous reactor:
(160)
subject to the following initial and boundary conditions:
(161)
Here we are defining
∂V+
V+
=
=
extrapolated boundary of the domain V,
domain V extended to ∂V+ .
We assume now that we know the eigenvalues and eigenfunctions that satisfy:
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Ch. VII. Neutron Diffusion
(162)
We further assume that we have ordered the eigenvalues:
(163)
We will not prove it, but it turns out that
the eigenfunctions {ψ} are orthogonal, and
Because the eigenfunctions are orthogonal, we can normalize them such that:
.
(164)
Because the eigenfunctions are complete, we can expand our initial condition and scalar flux:
,
(165)
(166)
where the known coefficients {φ0m} and the unknown coefficients {φm(t)} are given by:
,
(167)
(168)
Note: each eigenfunction satisfies the boundary condition, so the composite solution will also.
Thus, we do not have to explicitly impose it below.
Let us insert our expansions into the diffusion equation:
(169)
or, using the fact that ψn is an eigenfunction of the leakage operator, with eigenvalue Bn2,
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Ch. VII. Neutron Diffusion
(170)
Now, because the eigenfunctions are linearly independent, we have simply
,
(171)
where
(172)
Thus, each “mode” evolves independently – φ2(t) doesn’t depend on φ3(t) for example.
The solution of Eq. (171) is simply
(173)
Thus, the solution of our diffusion problem is:
(174)
or
,
which implies that
(175)
Thus, the
determines the long-time behavior of the solution. This is the α associated with
The solution approaches the shape of the corresponding eigenfunction, which is called
Example
Consider a bare homogeneous slab of width a and extrapolated width a+4D, where D is the
diffusion coefficient. You are told that the solution is symmetric about the center of the slab.
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Ch. VII. Neutron Diffusion
At t=0 the scalar flux is a linear function of |x|, going from a value of 1 at x=0 to a value of 0 at
the extrapolated boundaries:
.
How does the scalar flux evolve over time? As you will show in a homework problem, the
solution is
,
where
,
.
What does this solution look like? We’ll plot it for the following case:
a = 130 cm, ν = 2.43, Σf = 0.2 cm−1, Σa = 0.4859 cm−1, D = 0.15 cm, v = 220,000 cm/s.
At t=0 we just have a straight line. At t = 0.0001 seconds, it is not far from a straight line:
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
125
The fundamental mode is also plotted for comparison. At 0.001 seconds the peak has dropped a
bit:
At 0.005 seconds you can tell that the shape is approaching that of the fundamental mode:
NUEN 301 Course Notes, Marvin Adams, Fall 2009
Ch. VII. Neutron Diffusion
At 0.01 seconds it is even closer. The higher modes are dying out quickly:
At 0.05 seconds the higher modes are virtually gone:
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Ch. VII. Neutron Diffusion
127
If we look at 5 seconds we can see that the solution is growing – the reactor is significantly
supercritical. We could tell that this would happen from the beginning by noting that the largest
α is >0.
Warning: In this exercise we have ignored something extremely important:
Some neutrons that are released by fission are released with a time delay. This significantly
changes the time behavior of reactors. However, it is still true that a reactor left on its own will
have a flux distribution that pretty quickly approaches the fundamental mode and then grows or
decays exponentially in time. This will continue until something causes a change in the:
of the reactor.
Note in particular that temperature changes will cause changes in material properties. Why?
Because temperature affects nucleus motion, and cross sections are averaged over nucleus
motion. This is a very important “feedback mechanism” in reactors! It keeps power from
growing exponentially without bound, for example.
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Ch. VII. Neutron Diffusion
128
Summary
Our basic goal in this course is to gain an understanding of how reactors behave and why they
behave that way. This requires, at the very least, an understanding of neutron production and
loss rates in reactors. In this chapter we went a long way toward that. We found that:
1)
The neutron balance equation (either one-speed or energy-dependent) comes from
simply counting neutrons. It is essentially exact –– the things that it ignores are not
important in reactors.
2)
The neutron balance equation has too many unknowns –– another equation is
needed to relate the net current density (J) and the scalar flux (φ).
3)
Fick’s Law, or the diffusion approximation, can be used to relate J and φ:
It is an approximation. It is a reasonably good approximation if all of the
following conditions are met:
•
•
•
•
away from localized sources,
away from material interfaces (including problem boundaries),
when Σa << Σs,
when φ is not changing ‘quickly’ in time.
You must know when the diffusion approximation is reasonable!
4)
If we approximate J in the balance equation using the diffusion approximation, we
get the diffusion equation. (Either one-speed or energy-dependent.)
5)
Each term in the diffusion equation is a physically-meaningful ‘rate density’:
•
= net outleakage rate density [n/cm3-s],
• Σa(r)φ(r) = absorption rate density [n/cm3-s],
• νΣf(r)φ(r) = fission-neutron production rate density [n/cm3-s],
• Sext(r) = “extraneous” or “fixed” source rate density [n/cm3-s].
Similar interpretations hold for the energy-dependent diffusion equation, although
in that case each term is a density in energy as well as space, and thus has units
n/(cm3-s-MeV).
6)
On the boundary of a diffusion problem we must specify:
• φ,
•
•
en•∇φ, or
a combination of φ and en•∇φ
in order to have a completely specified (‘well-posed’) problem.
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Ch. VII. Neutron Diffusion
7)
At an interface between different materials, we have continuity of
• scalar flux,
• normal component (en•J = – en•D∇φ) of net current density.
8)
We studied two different kinds of diffusion problems:
a) fixed-source (both homogeneous and multi-region),
b) k-eigenvalue (criticality).
9)
To solve a fixed-source diffusion problem, we
i) write the diffusion equation for each homogeneous region,
ii) write all interface, boundary, and finiteness conditions,
iii) find a particular solution and the no-source solution; add them to get
the general solution,
iv) use interface, boundary, and finiteness conditions to solve for the
constants in the general solution.
10) In a homogeneous diffusion k-eigenvalue problem, we
i) define B2 so that problem looks like ∇2φ + B2φ = 0,
ii) find general solution of that equation,
iii) find values of B that give non-trivial solution,
iv) find geometric buckling (
= smallest B2),
v) find multiplication factor (largest k = keff = νΣf/[Σa + D
vi) find fundamental mode (φ that goes with
).
]),
11) Materials buckling, keff, and various criticality conditions are:
• Bm2 = [νΣf – Σa]/D,
• Bm2 {<,=,>} Bg2 means reactor is {subcritical, critical, supercritical},
• keff = νΣf/[Σa + DBg2],
• keff {<,=,>} 1 means reactor is {subcritical, critical, supercritical}.
12) This was not a complete list of everything you need to know from this chapter.
(But it’s a good start!)