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Larsen Lecture Notes
CHAPTER 1 : NUCLEAR REACTIONS IMPORTANT IN FISSION CHAIN
REACTIONS (Chapter 2, Duderstadt and Hamilton)
Spontaneous Radioactive Decay
 Rate Equations
 Half Life
Nuclear Collision Reactions
 Capture
 Scattering (n,)
 Fission (n,fission)
Microscopic Cross Sections
Macroscopic Cross Sections
 Number Density
 Mean Free Path
 Mean Free Time
 Mean Collision Frequency
Differential Scattering Cross Sections
 Scattering Angle
 Isotropic Scattering
 Linerly Anisotropic Scattering
Fission
 Fissile Nuclides
 Fissionable Nuclides
 Fission Cross Sections
 Fission Spectrum
 Fuels
 Fertile Nuclides
 Breeding
Resonances
Doppler Broading
1
Spontaneous Radioactive Decay (SRD)
Consider a certain type of atomic nucleus which undergoes SRD. Let:
N(t) = the probable number of these nuclei at time t
Q(t) = the rate (number per second) at which these nuclei are created by sources
Then, the general rate equation:
Rate of change of objects in a given state
= Rate at which objects enter the state
 Rate at which objects leave the state
implies
dN(t)dt
= Q(t) (rate of SRD)
The term on the left hand side is net rate of change. The first term and the second term on the right
hand side are gain term and loss term, respectively.
For N>>1, we have the experimental and physically intuitive result
probable rate of SRD = N(t),
where is the radioactive decay constant. Therefore,
dN(t)dt
= Q(t) N(t) (1)
dN
dt+ N(t) = Q
d(etN)dt
+ N(t) = Q
N(t) = N(0)et + t0
e(tt)Q(t) dt (2)
2
t
N(t)
2
1
1 > ( 2 )
If Q = 0, then
N(t) = N(0)et
We then have:
(probable) rate of SRD = N(t) = N(0)et
(probable) number of SRDs between t and t+ dt = = [ rate of SRD at timet ] dt = N(0)etdt
probability that a single nucleus will undergo SRD between t and t+dt=
= [N(0)etdt] / N(0) = [et]dt = p(t)dt (3)
Check:
0
p(t) dt =0
et dt = 1
We also have
probable (or mean) decay time =
= 10tp(t) dt
=0 te
tdt = 1
Inaddition,
t1/2 = half life = time required for the number of original nuclei to decrease by 1/2
3
N(t1/2) = 12N(0)
N(0)et1/2 = 12N(0)
et1/2 = 2
t1/2 = ln2 =0.693 = 0.693t
Now, supposewehavearadioactivedecaychain
[Here X is the chemical symbol for a type of nucleus.] Then, the governing equations are
dN1(t)dt
= 1N1(t) + Q1(t)dN2(t)
dt= 2N2(t) + 1N1(t) + Q2(t)
dN3(t)dt
= 3N3(t) + 2N2(t) + Q3(t)Nuclear Collision Reactions (neutronnucleus reactions)
Let
X= chemical symbol (i.e., H or C)
Z= atomic number
N= number of neutrons in a nucleus of X
A=Z+N= mass number of X
Then, the three important neutronnucleus reactions are:
1) Capture (n,)
n01 + XZA ( XZA+1 ) XZA+1 +
2) Scattering (n,n) or (n,n)
n10 + XAZ
n10 + XAZ [ elastic scattering (n,n)]
n10 + ( XAZ )
[ inelastic scattering (n,n)]
n10 + XAZ + [ inelastic scattering (n,n
)]
3) Fission (n,fission)
n01 + X1Z1A1
X2Z2A2 + X3Z3A3 + neutrons + energy
4
Here means excited state. In elastic scattering, the kinetic energy of the nucleus is altered, but the nucleusis not left in an excited state.
We need to describe these interactive processes mathematically, as we did with SRD. To do this, we intro
duce various types of material cross sections. These cross sections will basically play the same role as the
radioactive decay constants, but with the following difference:
radioactive decay constant = [mean time between events (SRD)]1
cross section = [mean distance between events (nuclear collisions)]1
We have already shown the first of these equalities; later we will show the second.
Microscopic Cross Sections Consider a uniform, monoenergetic, pencil beam of neutrons, normally
incident on a very thin target material. Let
I = the number of neutrons / cm2/ sec incident on the target
T = the number of target nuclei / cm2
R = the number of reactions / cm2 / sec
thin target
area = a
I
Then, we define the microscopic cross section by:
R = I T
# / cm2 sec = cm2 . # / cm2 sec . # / cm2(4)
5
Thus,
= microscopic cross section = R / (IT) = (R/T)/I
=number of reactions per target nucleus per sec
number of incident neutrons per cm2 per sec(5)
or
a=
number of reactions per target nucleus per sec
number of incident neutrons per sec
= probable number of reactions per target nucleus per incident neutron
We have the following microscopic scattering cross sections:
: capture
s : scattering
f : fission
e : elastic scattering
in : inelastic scattering
Then,
s = e + in
Also,
a = absorption cross section (any event other than scattering)
= f +
t = total cross section (any neutronnucleus event)
= s + a
= e + in + f
ne = nonelastic cross section (any event other than elastic scattering)
= t  e
These quantities all usually depend on the speed v (or, the kinetic energy E = mv2/2) of the incident
neutrons.
6
Macroscopic Cross Sections Now let us consider a normal beam of neutrons incident on a target of
finite thickness, and let us define
N = the number of target nuclei / cm3
= the number density of target nuclei
I(x) = the probable number of uncollided neutrons / cm2 / sec that passes through x
In a very thin part of this slab, between x and x+dx, we define
dR(x) = the number of reactions / cm2 / sec between x and x+dx
x
I(x)
x x+dx
dT = the number of target nuclei/ cm2 between x and x+dx
= N dx
Then, by Eqn.(4),
dR(x) = t . I . dT = t . I . Ndx (6)
However, we also have, by definition,
dR(x) = I(x) I(x+ dx) = I(x) [I(x) + I (x) dx] = I (x) dx
Combining these two equations, we get
dI
dx= N t I (7)
We now define
7
t = N t = macroscopic total cross section
=number of reactions / cm3
number of incident neutrons / cm2
Then the dimension of t is : [t] = length1
Eqn.(7) now implies
I(x) = I(0) et x
Therefore, Eqn.(6) gives
probable number of interactions between x and x+dx =
= t I(x) dx = t I(0) etx dx
and so
p(x) dx = probability that a single neutron undergoes a collision between x and x+dx
= t etx dx
and, as with the case of spontaneous radioactive decay,
0
p(x) dx = 1
0
x p(x) dx = mean free path between collisions =1t
If v is the speed of a neutron, with
dimension of v = cm / sec dimension of t = cm1
then
8
1 / (vt) = mean free time between collisions (dimension = sec)
We may define the following three statements
1) vt = mean collision frequency (dimension = sec1)
= collision rate (number of collisions / sec) of a neutron with speed v
2) If there are n neutrons in a system, then n(vt) = number of neutrons in the system undergoing
collisions per sec.
3) n(vt) dt = number of neutrons in the system that undergoes collision in a time interval dt.
f = N f = macroscopic fission cross section,
a = N a = macroscopic absorption cross section,
etc., so the additive formulas for also apply to :
f = e + in ,
a = + f , etc.
Also, if we have a homogenous mixture of nuclides X1, X2, ... with number densities N1, N2, ... and
microscopic cross sections 1, 2, ..., then for the mixture we have
= N1 1 + N2 2 + ...
where any subscript (t,a,s, etc.) can be used on and . In addition, if Ni depends on x and t, then
(x, t, E) = N1(x, t) 1(E) + N2(x, t) 2(E) + ...
so can depend on x and t.
A typical plot of t versus kinetic energy E of an incident neutron is given below:
9
E
(5)
(4)(3)
(2)
(1) ()T
(1) wavelength of neutron >> interatomic spacing ( = E1/2)
(2) wavelength of neutron = interatomic spacing. (diffraction effects)
(3) potential (billiard ball) scattering; = geometrical cross section of the nucleus
(4) resonance effects; incident neutron energy = lowest energy levels in nuclei
(5) wavelength of neutron
= v / v = (vx/v) i + (vy/v) j + (vz/v) k
= x i + y i + z k
= sin() cos() i + sin() sin() j + cos() k
=1 2 cos() i +
1 2 sin() j + cos()k
y
x
z
j
i
k
sin cos
cos
sin
sin sin
11
To compute the increment in area d caused by increments d and d, we have
y
x
z
area = d
sinsin
sin
+
+
d
d
d
d
Since d = sin()d,
then d = sin() d d
For any function of angle
f() = f(, ) = f(, )
we have
4
f() d = 2=0
=0
f(, ) sin() d d = 2=0
1=1
f(, ) d d
Also,
12
E =1/2 m v2 (m = neutron mass) and dE = m v dv
Thus, the variables vx, vy , vz can be replaced by
E, or E, , or E, ,
Now, consider a neutron with initial energy E and direction , which undergoes a scattering event. We
define
ps(E, E,) dE d = the probability that the scattered neutron has final energy within dEabout E, and final direction within d about
Then, we must have
4
0
ps(E, E,) dE d = 1
We now define the macroscopic differential scattering cross section
s(E, E,) = s(E) ps(E, E,)
Then the following three statements are equivalent:
1) v s(E, E,) dE d= v s(E) ps(E, E, dE)= the rate [probable number / sec] at which a neutron initially at state E, , is scattered within dE about
E and within d about .
2) If there are n neutrons in a system, then
n v s(E, E,) dE d= n v s(E) ps(E, E, dE)= the number of neutrons in the system that are scattered from E, to within dE and E and within d
about per second.
13
3) n v s(E, E,) dE d dt= n v s(E) ps(E, E, dE)= the number of neutrons in the system that are scattered from E, to within dE about E and within
d about in a time interval dt.
Also
s(E, E,) dE d =
sps(E, E,) dE d = s(E) (8)
This motivates the term differential scattering cross section; the integral of a derivative of a fuction is
the function, and the integral of a differential cross section is a cross section.
We may also define the microscopic differential cross section
s(E, E ,) = sps(E, E ,
)
Then
s(E, E ,
) dE
d
= s(E)
In most cases of interest, neutron scattering does not depend on the four variables that define and , but
rather on the single scalar variable = cos(0) = 0 where 0 is the scattering angle:
all on this cone
are equally probable
o
_
__
14
Hence, we can write
s(E, E ,) = s(E E , ) = s(E E , 0
If
s(E E , 0) = so(E E then scattering is isotropic. (9)
If
s(E E , 0) = so(E E + 3 0 s1(E E (10)
then scattering is linearly anisotropic. Otherwise, i.e in general, scattering is anisotropic.
15
Nuclear Fission
Fissile nuclides undergo fission when struck by low energy (about 1 ev) neutrons: U233, U235, Pu
239, Pu241
E (ev)10 1010
10
10
10
10
212
0
1
2
3
107
f (E) (barns)
Note that f for small E can be several orders of magnitude larger than f for large E.
Fissionable nuclides undergo fission when struck by highenergy (1 Mev) neutrons : Th232, U238, Pu240,
Pu242
E (ev)1. 10 2. 10 6. 10
1
2
7 7 7
thresholdenergy
f (E) (barns)
16
For fissionable nuclei f = 1 barn for large E; this is much less than f for fissile nuclei for low E. Only
fissile nuclei are capable, by themselves, of supporting a chain reaction.
Recall that
a + + f
Thus, for a neutron that is absorbed, we can define
(E) = (E) / f (E) = (rate of capture) / (rate of fission) = capture to fission ratio
E10
5
10 10721
()
Fission events produce :
(a) fissioned nuclei
(b) neutrons
(c) gammas, betas, neutrinos
(d) energy
In more detail,
(a) Fissioned Nuclei : charged, very energetic, neutronrich. 80% of yhe energy release consists of kinetic
energy of these nuclei. decay accounts for another 4% after a delay time.
17
(b) Neutrons : about 99% appear within 1014 sec (prompt)
about 1% appear within 0.2 to 55.0 sec (prompt)
Let us define
E = the probable total number of neutrons ( prompt and delay ) released in a fission reaction
initiated by a neutron with energy E.
(11)
(Mev)5 10 15
3
4
5
()
() = + 0 1
Also, let us define
(E) dE = the probability that a prompt, fissionproduced neutron has energy between E and E+dE.(12)
( 0
(E) dE = 1)
18
The prompt fission spectrum (E) is independent of the energy of the neutron that caused the fission.
1 2 3 E (Mev)
() (prompt)
Note that prompt fissionproduced neutrons are fast, about 1 Mev.
Delayed neutrons are produced by a more complicated process:
X + X + prompt neutrons + neutrinos + A A1
0 1 2 3
An + X
1 2 3
X A
X A
2
2
4
4
11n+
prompt
decay ( 0.2 to 55.0 )
(10 sec)14
+ + k
19
It is customary to lump delayed neutrons into six precursor groups characterized by halflives of 55.0,
22.0, 6.0, 2.0, 0.5, and 0.2 sec.
i = decay constant of the ith precursor group
i = fraction of all fission neutrons emitted from ith precursor group
= i = total fraction of delayed neutrons
E (Mev)0.4 1
(composite delayed) ()
(c) Gammas, Betas, Neutrinos :
gammas : 4% of energy (prompt)
betas : 4% of energy (delayed)
neutrinos : 5% of energy (lost)
20
(d) Energy:
Product % of energy Range Time delay
KE of fission products 80 < 0.01 cm prompt
prompt fast neutrons 3 10  100 cm prompt
decay 4 100 cm prompt
fission product betadecay 4 short delayed
neutrinos 5 infinite
other nonfission reactions 4 100 cm delayed
Note: About 97% of recoverable fission energy is deposited in the fuel material.
Fission Fuels:
Fissile : U233, U235, Pu239, Pu241
Fissionable : Th232, U238, Pu240, Pu242
Only U235 is found in nature. But, fissile fuel can be produced by transmutation of fertile nuclei:
U + n U +
Np
Pu
(23 min)
(2.3 days)
1 239
239
239
238
Th + n Th +
Pa
U
(22 min)
(27 days)
232 1 233
233
233
(protactinium)
21
Let us define
= average number of neutrons produced (by fission) per neutron absorbed in fuel.
Then, for a single fuel isotope (nuclide 1)
1 =1f1a1
=1f1
f1 + 1=
11 + 1f1
=1
1 + 1
For an infinite, homogenous medium of this material, a steadystate chain reaction is maintained if
1 =1f1a1
=1f1a1
= 1
For a mixture of N types of nuclei, a steadystate chain reaction is maintained if
=
Nj=1 (f )jN
j=1 aj= 1
If 1 > 1, then excess neutrons are available for transmutation of fertile nuclei. For example, consider a
mixture of nuclide 1 and a fertile nuclide 2. Then, the mixture is stedystate if
1f1a1 + a2
= 1, then 1f1a1 = a2
1 = 1 + a2a1
Now, if 1 = 2 , then a2 = a1
so fissile fuel is generated as fast as it is depleted.
If 1 > 2, then a2 > a1
and the system breeds fissile fuel. This is the principle behind the design of fast breeder reactors.
Resonances, Doppler Broadening: Various cross sections have resonance regions characterized by large
changes in for small changes in E. As the temperature of a material increases, incident neutrons see nuclei
with a larger range of kinetic energies, and the relative velocity between the neutron and nucleus becomes less
specifically tied to the energy of the neutron. Therefore, the fine details of the resonance become smeared out.
22
E
(,) (,)
EE+ +
1
2
< 1 2
This process is known as Doppler Broadening. The BreitWigner formula (derived using quantum me
chanics) describes this analytically.
23
CHAPTER 2 : FISSION CHAIN REACTIONS (AND NUCLEAR REACTORS)
(Chapter 3, Duderstadt and Hamilton)
Multiplication Factor,k
 Subcritical Reactor
 Critical Reactor
 Supercritical Reactor
Six Factor Formula
 Resonance Escape Probability
 Thermal Utilization Factor
 Fast Fission Factor
 Four Factor Formula
Conversion Ratio
Breeding Ratio
Thermal Reactors
Fast Reactors
Suppose an incident neutron causes a fission. The results are :
a) , , particles,
b) fissioned nuclei,
c) prompt and delayed neutrons (occuring in a certain fission generation). Some of these (prompt and
delayed) neutrons are eventually absorbed, and some eventually leak out of the system. The rest, after pos
sibly numerous scattering events, will lead to new fission neutrons occuring in the next fission generation.
We define
k = multilication factor =number of neutrons in a given generation
number of neutrons in the previous generation
Then:
k < 1 : subcritical (number of neutrons 0 in time)
1
k = 1 : critical (steadystate solution exists)
k > 1 : supercritical (number of neutrons in time)
A more practical definition is
k0 =rate of neutron gain in reactor
rate of neutron loss in reactor=
G(t)L(t)
If neutrons are lost at a rate L(t), and if there are N(t) neutrons, then
l =N(t)L(t)
= mean neutron lifetime
Using the general rate equation
rate of change = rate of gain  rate of loss
we obtain
dNdt = G(t) L(t) = k0L(t) L(t) = (k0 1)L(t) = k01l N(t)
N(t) = N(0) ek01
l t
N(t+ l) = N(0) ek01
l (t+l) = N(0) ek01
l tek01 = N(t) ek01
Hence,
k =N(t+ l)N(t)
= ek01
We have for k0 = 1,
k = ek01 = 1 + (k0 1) + 12 (k0 1)2 + ..... = k0 + O(k0 1)2
Also,
k > 1 if k0 > 1
k = 1 if k0 = 1
k < 1 if k0 < 1
Therefore, for criticality, k0 and k operationally mean the same, are equal for k = 1, and for k0 = 1
2
we have
k0 = k, N(t) = ek01
l t
The Six Factor Formulas : In thermal teactors, fission neutrons (= 107 ev) are produced that must slow
down to thermal energies (= 0.1 ev) before they are likely to create the next generation of fission neutrons.
The sixfactor and fourfactor formulas describe qualtitavely how various physical processes affect the mul
tiplication factor k.
Fast Neutron
Leaks outDoes notleak out
absorbed whileslowing down
slows down tothermal energies
leaks outof system
absorbed insystem
absorbed innonfuel
absorbed infuel
capturefission(thermal)
capture(fast)fission
neutronsproduced
1 PP
1 pp
P 1 P
P 1 P
P 1 P
FNL
TNLTNL
AF AF
FF
FNL
PFNL = probability that a fast neutron does not leak out of the system,
p = probability that a fast neutron is not absorbed while slowing down = resonance escape probability,
3
PTNL = probability that a thermal neutron does not leak out of the system,
PAF = probability that a thermal neutron that is absorbed is absorbed in the fuel
= fuelf / ( fuelf +
nonfuelf ) = f = thermal utilization factor
PF = probability that athermal neutron absorbed in the fuel produces a fission
= fuelf / fuela ( PF = fuel)
= probable number of fast fission produced per fast neutron.
Then, if N1 = the number of neutrons in one generation and N2 = the number in the next generation,
N2 = N1 [ PFNL p PTNL PAF PF + ]
so
k = N2/N1 = PFNL p PTNL f fuel +
= PFNL p PTNL ffuel [ PF NL p PT NL f fuel +
PF NL p PT NL f fuel]
Thus we have the sixfactor formula
k = PFNL p PTNL f fuel
where
= fast fission factor =probable number of neutrons from thermal + fast fissions
probable number of neutrons from thermal fissions
For an infinite medium, PFNL = PTNL = 1
and we get the four factor formula : k = p f fuel
The sixfactor and fourfactor formulas show roughly how various different physical phenomena affect the
criticality of the reactor.
Conversion and Breeding; Thermal and Fast Reactors : We know return to the concept of creating fis
sile fuel by transmutation:
238U + 1n ... 239Pu [ works best for fast neutrons]
4
232Th + 1n ... 233U [ works best for thermal neutrons]
We define:
ConversionRatio = CR =rate of fissile atom production
rate of fissile atom consumption
BreedingRatio = BR = CR if CR > 1
Now, consider an infinite hoogenous mixture of a fissile nuclide (1) and a fertile nuclide (2). Suppose the
mixture is critical. To obtain a formula that determines whether the mixture will breed, we write the criti
cality condition as
=1 f1
a1 + a2= 1
Then, the system breeds if
CR =a2a1
> 1 or a2 > a1
or 1 f1 = a1 + a2 > 2a1
or
1 =1 f1a1
=1 f1
f1 + 1=
11 + 1
> 2
Thermal Reactors : In thermal reactors, an average energy is comparable to the thermal neutron energy
(= 0.1 ev). Some characteristics of these reactors:
1. f is largest for thermal neutrons, so it is relatively easy to maintain a chain reaction. Not much fuel
needed.
2. Light, low massnumber materials are used in the construction of these reactors to help moderate
(slow down) the fast neutrons. (In elastic collisions with light nuclei, neutrons lose much more kinetic energy
than they do with heavy nuclei.)
3. Relatively simple to build.
5
4. = 2 in fuel, so it is difficult to breed.
Fast Reactors : In fast reactors, an average neutron energy is much higher than in thermal reactors
(= 105ev). Some characteristics:
1. f is small, so it is relatively difficult to maintain a chain reaction. Considerably more fuel (3040
times as much as in thermal reactors) is needed to maintain chain reaction.
2. Heavy, high massnumber materials are used in the construction of these reactors to inhibit the slowing
down of neutrons.
3. relatively complicated to build.
4. > 2 in fuel, so it is possible to breed.
6
Comparison of EnergyAveraged , , andf in Thermal and Fast Reactors :
Thermal (LWR) Fast (LMFBR)
U235 Pu239 U235 Pu239
2.4 2.9 2.6 3.1
2.0 1.9 2.1 2.6
f 580 790 1.9 1.8
reflector
core
blanket U238
(fast) core Pu
(thermal)
thermal reactorfast reactor
7
CHAPTER 2 PROBLEMS
2.1 D H, p. 100, problem 31.
2.2 D H, p. 100, problem 33.
2.3 D H, p. 100, problem 36.
2.4 D H, p. 100, problem 39.
2.5 Measurements on a critical thermal reactor indicate that for every 100 neutrons emitted in fission,
10 leak out while slowing down, 10 are absorbed while slowing down, and 10 leak out afyter being thermal
ized. Also, 70% of the thermal absoptions occur in the fuel, and 2 fission neutrons are emitted for every
thermal absorption in the fuel. Calculate:
a) the fast nonleakage probability,
b) the resonance escape probability,
c) the thermal utilization factor,
d) the fast fission factor,
e) the infinite medium multiplication factor.
8
CHAPTER 3 : NEUTRON TRANSPORT AND DIFFUSION EQUATIONS (Chapter 4, Dud
erstadt and Hamilton)
Density Fuctions
Angular Neutron Density
 probable number of neutrons in an infinitesimal element of phase space
 probable rate at which neutrons pass through an infinitesimal element of surface area
Delayed Neutrons
 Precursor Groups
TimeDependent Neutron Transport Equation With Delayed Neutrons
 Initial and Boundary Conditions
SteadyState Transport Equation
SteadyState Transport in a Purely Absorbing Medium  SteadyState Transport in a Vacuum
 Method of Characteristics
 Decay from a Point Source
Angular Flux
Scalar Flux
Current
Discretized Representation of the Transport Equation
 Time Discretization
 Energy Discretization
Multigroup Approximation
OneGroup Approximation
 Angular Discretization
SN (DissreteOrdinates)
PN (Spherical Harmonics)
 Representation of Macroscopic Scattering Cross Section
Scattering Ratio, c
Mean Scattering Cosine, 0
 Spatial Discretization
 Iterative Methods
Special Geometries
 OneDimensional Slab Geometry
1
Integral Transport Equation
SN Equations
PN Equations
Closure Relations
 OneDimensional Spherical Geometry
 TwoDimensional X,YGeometry
Nth Collided Flux Equations
OneGroup TimeDependent Diffusion Equation
 Initial Condition
 Boundary Conditions
Prescribed Incident Flux
Reflecting
Extrapolated Endpoint (Extrapolation Distance)
 Interface Conditions
EnergyDependent Diffusion Equation
2
Density Functions : Consider a bar consisisting of a mixture of materials that vary with position x. We
define the mass density function of the rod by:
(x) dx = the mass between x and x+dx,
then, x2x1
(x) dx = the mass between x1 and x2,
and has units of mass per length. The concept of a density function plays a fundamental role in the
mathematical description of neutron transport and diffusion processes.
Angular Neutron Density : We define the angular density function n(r, E,, t) by:
n(r, E,, t) d3rdEd (1)
= the probabale number of neutrons in d3r about , having energy within dE about E,
travelling with direction in the solid angle d about , at time t.
x
y
z
x
y
z
E E+dE
energy
dy dx
dz
d 3r = dx dy dz
r
d
Then, for example, if D is some domain in physical space, we have
[0 n(r, E,, t) d
3r ] dE d = the probable number of neutrons in D, having energies within dE about
E and directions d about , at time t.
and
3
4
E2E1
0n(r, E,, t) d3r dE d = the total number of neutrons in D, with energies between E1 and
E2, at time t.
Also, let us consider a surface with an area increment dA and a unit normal vector n at the point r:
r
n
area=dA
y
x
z
We wish to compute the rate at which neutrons at ( r, E,, t) pass through dA. To do this, let us consider
neutrons at r, travelling in the direction , with energy E [or speed v = (2E/m)1/2]. In time dt, the neutrons
travel a distance d = v dt:
n
r
volume=dV=dAds
ds
area=dA
d
dsd = cos =  n
dV = dA dS = dA  n d = dA  n v dt
4
Since neutrons travel a distance d in time dt, then every neutron in the volume dV at time t, having
direction and speed v, will exit through dA between t and t+dt. Therefore,
number of neutrons that pass through dA in time dt = number of neutrons initially
in the volume ( and in dE about E, d about ) = ( n dE d ) dV = ( n dE d ) dA n v dt
and
v  n n(r, E,, t) dE d dA= the rate at which neutrons at r, within dE about E, within d about , pass through dA
(2)
Now let us consider an arbitrary volume D with surface D. At each point r D letn= n(r) be the unitouter normal vector. Then for fixed , E, and t,
D : that part of D for which n >< 0
n
n
[D
v n n(r, E,, t) dA ] dE d= [
D+ v n n(r, E,, t) dA ] dE d [
D v n n(r, E,, t) dA ] dE d
= [ the rate at which neutrons flow out of D ] [ the rate at which neutrons flow into D ]or
[D
v n n(r, E,, t) dA ] dE d= the net rate at which neutrons within dE about E and d about leak out of D = net leakage rate
(3)
Thus, integral can be positive or negative. If it is positive, then the rate of flow out of D is greater than the
rate of flow in. If it is negative, then the rate of flow in of D is greater than the rate of flow out.
5
Similarly, if S is a surface with a continuously varying normal vector n(r):
n
n
then
[S
v n n(r, E,, t) dA ] dE d (4)
= the net rate at which neutrons within dE about E and d about flow through S.
This integral can be positive or negative. If it is positive, then the net neutron flow is in the direction
of the normal vectors; otherwise, it is in the positive direction.
If the above expression is integrated over , we also have:
[4
S
v n n(r, E,, t) dA d ] dE (5)
= the net rate at which neutrons within dE about E flow through S.
Again, this integral can be positive or negative, with the same meaning as before.
Now we shall derive an equation for n(x,E,, t). We have, for any volume D, and for fixed E and ,
the general rate equation:
Rate of change of the neutrons for [r D, energies within dE about E, and directions d about ]= Rate of gain of neutrons [ r D, energies within dE about E, and directions d about ] Rate of loss of neutrons [ r D, energies within dE about E, and directions d about ]
(6)
However,
6
Rate of change of neutrons [r D, energies within dE about E, and directions d about ]
=d
dt[D
n(r, E,, t) dr3 ] dE d] = [D
n
t(r, E,, t) dr3 ] dE d ] (7)
Also
Rate of loss of neutrons in dE about E and direction d about =
Rate at which neutrons in dE about E and direction d about undergo collisions
+ leakage rate of neutrons in dE about E and direction d about out of D
= [D
v T (E) n(r, E,, t) d3r] dE d + [D
v n n(r, E,, t) d2r] dE d (8)
Here we have set dr2 = dA , dr3 = dV
However, the Divergence Theorem (or Greens Theorem) gives for a general vector function f(r):
D n f(r)d2r =
D f(r)d3r ,
where
= i x + j y + k z = gradient operator
Therefore,
D
n [ v n ] dr2 = D v n dr3 =
Dv n dr3 ,
so Eqn.(3.8) can be written
Rate of loss of neutrons in dE about E and direction d about
= [D
[ v n(r, E,, t) + v T (E) n(r, E,, t)] d3r ] dE d (9)
Next,
7
Rate of gain into dE about E and direction d about
= Rate of gain into dE about E and direction d about due to scattered neutrons
+ Rate of gain into dE about E and direction d about due to prompt fission neutrons
+ Rate of gain into dE about E and direction d about due to delayed fission neutrons
+ Rate of gain into dE about E and direction d about due to interior sources
(10)
However,
Rate of gain due to scattered neutrons
= [D
[4
0
vs(E
E, ) n(r, E , , t) dE d ] d3r] dE d (11)
Note : see Eqn.(1.9). Also,
Rate of gain due to prompt fission neutrons
= [D
p(E)4
[4
0
[ 1 (E) ] (E) v f (E) n(r, E ,, t) dE
d
] d3r] dE d (12)
Notes:
vf (E
) n(r, E
,
, t) dE
d
d3r = the fisison rate in dE
about E
, about d
about
, and d3r
about r,
(E) = the total number of neutrons (prompt and delayed) produced in a fission event that is caused
by a neutron with energy E,
(E) = the fraction of neutrons in a fission event, caused by a neutron with energy E
, that are delayed,
[1 (E )] (E) v f (E) n(r, E ,, t) dE
d
] d3r = the rate at which prompt fission neutrons
are created by neutrons in dEabout E
, about d
about
, and d3r about r,
p(E) = prompt fission spectrum [0
p(E)dE = 1]
p(E)4 dE d = probability that a prompt neutron is emitted in dE about E, about d about .
8
Rate of gain due to delayed fission neutrons
= [D
Qd(r, E, t) d3r] dE d (Qd to be determined) (13)
Rate of gain due to interior sources
= [14
D
Q(r, E, t) d3r] dE d (14)
Note: Interior sources are usually isotropic. The factor 4 is included as a normalization factor, so that
[4
0
14
D
Q(r, E, t) d3r] dE d = 0
D
Q(r, E, t) d3r dE
= the total rate at which source neutrons are introduced in D.
Combining Eqns.(3.10)(3.14), we get
Rate of gain into dEabout E
, about d
about
, and d3r about r,
= [D
[
vs n dE
d
+
p(E)4
(1 ) v f n dE d + Qd + 14Q] d
3r] dE d (15)
Finally, combining Eqns.(3.6), (3.7), (3.9), and (3.15), we obtain
D
[dn
dt+ vn + vTn
v
sndE
d
p4
(1 )vfndEd Qd 14Q]d
3r = 0
Because D is arbitrary, the integrand [...] = 0. Thus, defining
(r, E,, t) = v n(r, E,, t) = angular neutron flux, (16)
we obtain
1v
t+ + t =
s dE
d
+ p
4
(1 ) f dE d + Qd + 14Q (17)
Now we shall derive an expression for Qd:
Qd(r, E, t) d3r dE d = the rate at which delayed neutrons are emitted in dEabout E
, about d
about , and d3r about r.
9
Recalling that delayed neutrons arise from fissioned nuclei that undergo spontaneous radioactive decay,
we define:
Cj(r, t)d3r = the probable number of fissioned nuclei in precursor group j, in d3r about r at time t. (The
decay constant for these nuclei is j .)
j(E) = the spectrum of neutrons emitted from precursor group j (0
jdE = 1)
j(E) = the fraction of all fission neutrons, caused by a neutron with energy E, that are emitted from
the jth precursor group.
=6
j=1 j(E) = the total fraction of delayed neutrons in a fission event caused by a neutron with
energy E.
Then, nuclei are introduced into jth precursor group in d3r about r at the rate
[ 0
4
j(E) (E
) v
f (E
) n(r, E
,
, t) dE
d
] d3r]
= [ 0
4
j(E) (E
) f (E
) (r, E
,
, t) dE
d
] d3r]
Hence,
tCj(r, t) + j Cj(r, t) =
0
4
j(E) (E
) v
f (E
) n(r, E
,
, t) d
dE
(18)
The rate at which precursor group j nuclei spontaneously decay is j Cj , and hence this is the rate at which
group j neutrons are produced. Since the spectrum for these neutrons is j(E) and
j(E)4 dEd = probability that a group j delayed fission neutron is emitted in dE about E and d about .
[0
4
j(E)4 dEd =
0
j(E)dE = 1]
then
10
j(E)4 jCj(r, t)d
3rdEd = the rate at which delayed neutrons are emitted from the jth precursor group
into d3r, dE about E and d about at time t.
Hence,
Qd(r, E, t) =6
j=1j(E)4 jCj(r, t) ,
and Eqn.(3.17) becomes:
1v
t(r, E,, t) + (r, E,, t) + t(r, E,, t) =
0
4
s(E E, )(r, E,, t)dEd
+ p(E)4
0
4
(1 (E))f (E)(r, E,, t)dEd + Qd(r, E, t) =6
j=1
j(E)4
jCj(r, t) +14
Q(r, E, t)
(19)
This is the full timedependent neutron transport equation with delayed neutrons; it is coupled with
Eqn.(3.18), which govern the precursor densities. (Note: in some formulations of these equations, the
4 factors are absent, but then p(E), j(E), and Q(r,E,t) have different normalizations.)
Some physics which is omitted from these equations :
1) Certain quantum mechanical effects,
2) Motion of the host material,
3) Statistical fluctuations in the neutron density n,
4) Neutronneutron and ather rare interactions,
5) Forces (for example, gravity) on neutrons,
6) Temperature feedback (t depends on temperature which depends on ).
Initial and Boundary Conditions : Eqns.(3.18) and (3.19) do not, by themselves, describe ; we also need
initial and boundary conditions. Physically, we expect that given all (interior and boundary) sources of
neutrons, the initial values of the angular flux and precursor densities, we should be able to determine
and Cj uniquely. This expectation is correct, and it gives us the appropriate initial and boundary conditions.
Interior source of neutrons: 14 Q(r, E, t) r D, 0 < E < , t > 0
This known term is already in Eqn.(3.19).
11
Initial source of neutrons : we prescribe
(r, E,, 0) = i(r, E,) r D, 0 < E < ,  = 1, (20)
where i is known.
Boundary source of neutrons: we prescribe
(r, E,, 0) = b(r, E,, t) r D, 0 < E < , t > 0, n < 0, (21)
where b is known. Note that since n is the unit outer normal, n < 0 corresponds to all directions pointing into the spatial domain D. Therefore, we are free to prescribe the incoming or incident flux,
corresponding to n < 0, but we cannot prescribe the outgoing or exiting flux, corresponding to n < 0.
Initial values of the precursor densities:
Cj(r, 0) = Cij(r) r D, 1 < j < 6 (22)
where Cij are all known.
Eqns.(3.18)(3.22) uniquely determine and Cj in a given physical system D.
Note: an assumption implicit in this formulation is that nrutrons entering D through its outer boundary can
be arbitirarily chosen and are independent of the exiting fluxes. That is true only if the boundary of D is
convex:
12
D
exterior of D = vacuum
1) convex
2) nonreentrant boundary
3) incident flux does not depend on the exiting flux
D
exterior of D= vacuum
1) nonconvex
2) reentrant boundary
3) incident flux does depend on the exiting flux
This problem can be cured by enlarging our definition of D to include some exterior points, so that the
new boundary is convex:
newboundary
13
Neutron transport problems are always solved in physical systems that are convex.
Transport Equation Without Delayed Neutrons: Set j = = Cj = 0. Then, Eqn.() becomes
1v
t+ + t =
s dE
d
+
x
4
f dE
d
+
14
Q (23)
+ initial condition (3.20) + boundary condition (3.21)
SteadyState Transport Equation Without Delayed Neutrons:
+ t =
s dEd
+
x
4
f dE
d
+
14
Q (24)
+ boundary condition (3.21)
SteadyState Transport Equation in a Purely Absorbing Medium: Set s = f = 0 ; then
+ t = 14Q (25)
+ boundary condition (3.21)
SteadyState Transport Equation in a Vacuum:
= 0 (26)
+ boundary condition (3.21)
Equations (3.25) and (3.26) can be explicitly solved because these equations do not couple angle or en
ergy. Eqn.() couples angle and energy, can be explicitly solved only in very special idealized cases.
Solution of the SteadyState Transport Equation in a Purely Absorbing Homogenous Medium: We consider
a convex homogenous domain D and
(r, E,, t) + t(r, E,, t) = 14Q(r, E, t) r D,  = 1
(r, E,) = b(r, E,) r D, n < 0 (27)
Eqn.(3.27) can be written
14
x
x+ y
y+ z
z+ T =
14
Q (28)
This is a firstorder differential equation that can be solved by the method of characteristics. To do this, we
define a curve
r(s) = x(s) i + y(s) j + z(s) k
and
(s, E,) = [ r(s), E, ] = [ x(s), y(s), z(s), E, ]
Q(s, E) = Q[ r(s), E ] = Q[ x(s), y(s), z(s), E ]
Then,
s=
s [ x(s), y(s), z(s), E, ] =
dx
ds
x+
dy
ds
y+
dz
ds
z
Let us require x(s), y(s), and z(s) to satisfy:
dx
ds= x,
dy
ds= y,
dz
ds= z , (29)
Then, s = xx + y
y + z
z
Hence, using Eqn.(3.28), we obtain the system of equations (3.29) and
s(s, E,) + t(E) (s, E,) =
14
Q(s, E,) (30)
We must impose initial conditions for this system. Let r0 = x0i+ y0j + z0k be any point on the boundary
of D for which n < 0. Then, we set
x(0) = x0, y(0) = y0, z(0) = z0 (31)
and
(0, E,) = b(ro, E,) (32)
15
Solving Eqns.(3.29) and (3.31), we get
x(s) = x0 +x , y(s) = y0 +y , z(s) = z0 +z
so r(s) = r0 +s
Thus, for s > 0, r(s) tracks into D. In fact, the characteristic line r(s) for s > 0 traces the physical path of a
neutron as it enters the system at r0 and propagates inward in the direction .
n
s
D
.
dD
r +so
Eqns.(3.30) and (3.32) now give
s+ t =
14
Q
sets =
14
Q ets
(s) ets (0) = s0
Q(s)
4ets
ds
(s) = (0) ets + s0
Q(s)
4et(ss
) ds
= (0) ets + s0
Q(s )4
et d
16
or
(r0 + s, E,) = b(r0, E,) et(E)s +
s0
Q(r0 + s , E,)4
et(E)d (33)
Let us now change notation a bit. For s fixed point r in domain D and direction , let
r0 = the intersection of D with the line r + , < 0
s =  r r0  (34)
s
ro
dD
r
Then, making contact with our previous notation, we have
r = r0 + s , and Eqn.(3.33) becomes
(r, E,) = b (r0, E,) et(E)s + s0
Q(r , E,)4
et(E) d (35)
Note that the above equation gives the exiting flux from a system.
Thus, uncollided neutrons decay exponentially with rate t(E). (We predicted this earlier, in Chapter
1 of these notes.) Now for some special cases:
17
Vacuum Boundary Condition: b = 0. Eqn.(3.35) becomes
(r, E,) = s0
Q(r , E,)4
et(E) d
Therefore
(r, E) = scalar flux =4
s0
Q(r , E,)4
et(E) d
The variables s, constitute a set of polar ccodinates with origin at the point r. Let us convert this set to
Cartesian ccordinates. We take
r= r s (variable of integration)
= r r  (radius)d3r
= 2 d d = r r 2 d d
and then we obtain
(r, E) =rD
Q(r, E)
4et(E) rr

 r r 2 d3r
(36)
Isotropic Point Source: If Q(r,E) is a deltafuction source at r = r0, E = E0, i.e.,
Q(r, E) = (x x0) (y y0) (z z0) (E E0),
then
(r, E) =et(E0) rr0
4 r r02(E E0)
Therefore, the decay of the scalar flux away from a point source is as etr/r2, where r is the distance to the
point source.
Pure Streaming in a Vacuum: (t = Q = 0). Eqn.(3.35) reduces to
(r, E,) = b (r0, E,)
18
r
ro
dD
D
Thus, the angular flux inside D consists of sloy of the freestreaming neutrons that enter D through its outer
boundary, and there is no exponential attenuation away from the boundary.
OneDimensional HalfSpace: Let D consist of the halfspace z > 0, and let all quantities be indepen
dent of x and y. Then for > 0, Eqn.(3.35) becomes
z
s=z/
r=xi + yj + zkz
z/s =cos
=
(z, E, ) = (0, E, ) et(E)s + s0
Q(z )4
et(E) d
19
= b (E, ) et(E)z/ + z /0
Q(z )4
et(E) d
= b (E, ) et(E)z/ + z0
Q(z )4
et(E) / d
(37)
For < 0, z = , so
(z, E, ) = 0
Q(z + , E)
4et(E)
/
 d
Some definitions:
n(r, E,, t) = angular neutron density
(r, E,, t) = v n(r, E,, t) = angular flux
(r, E, t) =4
(r, E,, t) d = scalar flux
J(r, E, t) =4 (r, E,, t) d = current
If e is a unit vector, then
J(r, E, t) =e0  e (r, E,, t) d = partial current
Note that if e is perpendicular to an area element dA, then
J+(r, E, t)dA = rate at which neutrons at (r,E,t) flow through dA in the direction of e.
J(r, E, t)dA = rate at which neutrons at (r,E,t) flow through dA in the direction of e.
ee
dAdA
J J+
20
J > 0 [and J are not vectors]
J can not be determined without first specifying e.
J e dA = (J+ J) dA = net rate at which neutrons flow through dA (+ if net flow is in direction ofe,  if net flow is in direction of e)
Discretized Representation of the Transport Equation:
1) Time: The following scheme is stable for all t. Also, if Q is independent of t, it is accurate for
very large t.
n n1v t
+ n + t n =
s n dEd
+
14
Q
or
n + (t + 1v t
) n
s n dEd
=
14
Q +n1
v t
Thus, the timedependent problem reduces to solving a steadystate problem within each time step. For
very large t,
n + t n
s n dEd
=
14
Q
and we obtain the correct equation for the steadystate solution.
2) Energy: (Multigroup approximation) We assume Emin E Emax and divide this large intervalinto G smaller energy groups:
Emin = Ea < Ea1 < ........ < E0 = Emax
Then, taking the transport equation
+ t(E) = EmaxEmin
4
s(E E, ) d dE + 1
4Q
=G
g=1
4
Eg1
Eg
s(E E, ) dE d + 1
4Q,
21
we operate by Eg1Eg
(.) dE and we get
( EgEg+1
dE ) + (
EgEg+1
t(E) dE EgEg+1
dE) EgEg+1
dE
=G
g=1
4
(
EgEg+1
EgE
g+1
s(E E, ) dE dE
EgEg+1
dE) EgEg+1
dEd
+
EgEg+1
Q dE
Thus, if we define
g (r,) = EgEg+1
(r, E,) dE, (38)
Qg(r) = EgEg+1
Q(r, E) dE, (39)
tg(r) =
EgEg+1
t(E) dE EgEg+1
dE(40)
s,gg(r, ) =
EgEg+1
Eg
Eg+1
s(E E, ) dE
EgEg+1
dE(41)
where (E) is a suitably approximate shape function, then we get the multigroup transport equations
g + t(E) g =G
g=1
4
s,gg( ) g (r,) d
+
14
Qg, g = 1, 2, ..., G(42)
Note: These equations are exact if for each g and g
t(E) = constant [= tg] for Eg+1 < E < Eg,
s(E E, ) = function of )
[= s,gg( )Eg ] for Eg+1 < E < Eg, Eg+1 < E < Eg ,
because the shape fuction then cancels out of the expression for tg and s,gg. They are also exact if
(r, E,) = (r,) (E),
where Psi(E) is the shape function, because then psi(E) cancels out of the expression for tg and s,gg.
This shows that to get good accuracy, the shape fuction Psi(E) must be representative of the energy
dependence of (r, E,).
22
OneGroup Approximation:
(r,) + t(r) (r,) =4
s( ) (r,) d + 1
4Q(r) (43)
3) Angle (SNandPN ) : There are two widelyused ways to discretize in angle. The first is the discrete
ordinates or SN approximation, in which beutrons are assumed to travel only in discrete directions m, m
= 1,2,....,N. With each of these discrete directions (or ordinates) we associate a section of the unit sphere
wm, and we make the approximation
f() d
Nm=1
f(m) wm
where, by the definition of the wm, 4 =N
m=1 wm
The set mwm is a quadrature set, and Eqn.(3.42) becomes
g + t(E) g =G
g=1
Nm=1
s,gg(m m) gm wm +14
Qg , 1 g G , 1 m N (44)
There are three discreteordinates or SN equations. Eqn.(3.44) holds in Cartesian geometries only. Boundary
conditions are obvious.
An alternative to SN is the spherical harmonics or PN method. Consider the various spherical harmonic
functions
lm() , l m l , l = 0, 1, 2, ..... ,N
It is known that any reasonable function f() can be expanded as a linear combination of these func
tions:
f() =l=0
lm=l
flmlm()
Let us multiply Eqn.() by lm() and integrate over . This produces a system of
23
[1 + 3 + 5 + ...... + (2N+1)] G = (2N + 1)2 G equations.
Next we insert the approximate expansion
g(r,) l=0
lm=l
flmg(r)lm()
into this system. [Note that there are (2N + 1)2 G unknowns, flmg(r).] One then has a system with the
same number of equations as unknowns. [We will explicitly carry out this derivation later, in slab geometry.]
Note: Unlike the SN method, the derivation of boundary conditions for the PN equations is somewhat
problematical.
Explicit Representation for s( ) : In a given group, for the important linearly anisotropic scat
tering, we set
s( ) = 1
4(s0 + 3
s1) (45)
To define s0 and s1, we shall need the identities
4
d = 4,4
d =
4
d = 0,
4
( )2d =
4
( )2d = 4
3,
We then have:
4
s( ) d = 1
4
4
s0 d = s0
and
4
( ) s(
) d = 14
4
[ ( ) s0 + 3 (
)2] d = s1 (46)
Now let () be the angular flux at a fixed spatial point in a given group. Then,
[4
s( ) () d ] d = rate at which neutrons are scattered into d about ,
24
so
4
[4
s( ) () d ] d = rate at which neutrons are scattered
=4
[4
s( ) d] () d =
4
s0 () d
= s0
4
() d
Dividing this equation by t4
()d
, we get
s0t
=
s( ) () d d
t4 (
) d
=rate at which neutrons are scattered
rate at which neutrons undergo collisions
= mean number of scattered neutrons per collision = c. (47)
Next, we define
= mean scattering cosine
=
( ) s(
) () d d s(
) () d d
=[[(
) s0 + 3 ( )2 s1] d] (
) d
s0(
) d
, for = 0,
=s1s0
Therefore, if s( ) can be written in the form of Eqn.(3.45), then
s0 = c t
s1 = 0 s0 = c 0 t
s( ) = c t
4(1 + 3 0
) (48)
25
Usually, > 0, which is a consequence of the fact that forwardscattering is more probable than back
scattering.
4) Spatial Discretizations: A lengthy topic. See NE 542.
5) Efficient Iterative Methods for Solving the Fully Discretized SN Equations: Another lenghty topic.
See also NE 542.
Special Geometries: The general steadystate onegroup transport equation depends on five independent
variables: x, y, z, and . This can be require an enormous amount of computer storage, even for prob
lems with moderate numbers of discrete values of x, y, z, and . However, in specialgeometries for which
symmetries occur, the number of independent variables can be reduced to a manageable number. We will
now describe onedimensional slab geometry in detail, and then we will very briefly describe onedimensional
spherical geometry and twodimensional x, ygeometry.
26
OneDimensional Slab Geometry: = (z, ). Let us consider a situation in which all geometrical quan
tities and all boundary conditions are independent of x and y, and all boundary conditions depend only on
the polar angle and not on the azimuthal angle :
kz
x,yplane
=1 2cosi +
1 2sinj + k
= cos , is independent of x, y, and
We have Eqn.(3.43):
+ t(E) = 144
s0 + 3 s1 d
+
Q
4(49)
However,
= (x
x+ y
y+ z
z) (z, ) =
z(z, )
and
14
4
(s0 + 3 s1) d
=14
1=1
2=0
(0 + 3 [ sqrt1 2 sqrt1 2 (cos cos + sin sin) + ] s1) (z, ) d d
27
=12
1=1
(s0 + 3 s1) (z, ) d
Therefore, Eqn.(3.49) becomes
z(z, ) + t (z, ) =
12
1
1( s0 + 3 s1 ) (z, ) d + Q(z)4 (50)
It is customary to define
(z, ) = 20
(z, ) d = 2 (z, ),
and then
(z, ) d dz = the total number of neutrons in d about , in dz about z, per unit area in the x, yplane,
and Eqn.(3.50) becomes
z
(z, ) + t (z, ) =12[1
1( s0 + 3 s1 ) (z, ) d + Q(z) ] (51)
For the case of general anisotropic scattering, this equation generalizes to
z
(z, ) + t (z, ) =12(1
1[n=0
(2n+ 1) Pn() Pn() sn ] (z,
d
+ Q(z)) (52)
where Pn() are the Legendre Polynomials:
P0() = 1 , P1() = , P2() = 12 (32 1) , . . . . . .
which satisfy the recursion relations
Pn() = n2n+1Pn1() +n+12n+1Pn+1()
and the orthogonality relations
11
Pn() Pm() d =2
2n+ 1nm (53)
28
Eqns.(3.51) and (3.52) typically hold on a slab Z < z < Z. Using
> 0 < 0
z
(neutrons flow to the right)(neutrons flow
to the left)
we see that incident boundary conditions for must be assigned as follows:
Z Z
< 0
(,) ( ,)
> 0
so
(Z, ) = b () > 0(Z, ) = b () < 0
(54)
If b = 0 on either edge, then that edge is termed a vacuum boundary.
29
z
1
+1
ZZ
boundary conditions imposedon edges denoted by
Other boundary conditions, based on physical symmetries, are also possible. For example, suppose that
b = 0, Q(z) = Q(z), t(z) = t(z), s0(z) = s0(z), and s1(z) = s1(z) :
z = Z z = 0 z = Z
.. .z z
Q(z) = Q(z) (z) = (z)
Then, must have a similar type of symmetry about z = 0:
(z, ) = (z,) [ (0, ) = (0,) ] (55)
30
z = Z z = 0 z = Z
.. .z z
Thus, we can reformulate the problem for the smaller slab 0 < z < Z and assign reflecting boundary
conditions on the left edge:
0
vacuum boundary :reflecting boundary :
(,) = 0 ,1 < < 0
(0,) = (0,)
(0, ) = (0,) [reflecting boundary](z, ) = 0 , 1 < 0 [vacuum boundary]
(56)
This reduces by a factor of two the amount of storage and arithmetic required to solve a problem.
Integral Transport Equation: Let us consider the following slabgeometry problem with t = 1 and s1 = 0 :
z(z, ) + (z, ) =
c
2
1
1(z, ) d
+
q(z)2
0 < z < L (57)
(0, ) = 0 , 0 < 1(L, ) = 0 , 1 < 0
31
Defining the scalar flux
(z) =1
1 (z, ) d
(58)
we can write Eqn.(3.57) as
z +
1 =
12 (c + q)
z e
z/ = ez/
2 (c + q)
For > 0, z0 ()dz
gives
(z, ) ez/ (0, ) = z0
ez/
2 [ c (z
) + q(z
) ] dz
for (0, ) = 0 ,
(z, ) = z0
e (zz)/
2 [ c (z
) + q(z
) ] dz
(59)
For < 0, Lz()dz gives
(L, ) eL/ (z, ) = Lz
ez/
2 [ c (z
) + q(z
) ] dz
for
(L, ) = 0 ,
(z, ) = Lz
e(zz)/
2 [ c (z) + q(z
) ] dz
(60)
Combining Eqns.(3.58)(3.60), we obtain
(z) = 10
(z, ) d + 01
(z, ) d
= z0
[ 10
e(zz)/
2 d] [ c (z
) + q(z
) ] dz
+ Lz
[ 01
e(zz)/
2 d] [ c (z) + q(z
) ] dz
However, for z< z,
10
e(zz)/
2 d =
1
e(zz)t(
dtt
) = 1
e(zz)t
t dt
32
= E1( z z) = E1(  z z  ) (61)
and for z> z,
01
e(zz)/
2 d = 01
e(zz)/s
s(ds) =
10
e(zz)/s
sds
= E1( z z ) = E1(  z z  )
Combining the last three equations, we find
(z) =12
L0
E1(  z z  ) [ c (z) + Q(z)] dz (62)
This is the integral transport or Peierls equation for the scalar flux. It can be derived in any geometry and
with inhomogenous media, and nonvacuum boundary conditions. The only requirements is that scattering
be isotropic.
The Peierls equation has the advantage that is is removed. However, it has the disadvantage that every
spatial point is explicitly coupled to every other point; the equation is spatially global, unlike the integro
differential form of the transport equation, which is spatially local. Therefore, the Peierls equation is normally
used in computer codes only for small physical systems for which ona can safely make the approximation
that scattering is isotropic.
DiscreteOrdinates or SN Equations : Consider an angular quadrature set consisting of angles m and
angular weights wm, 1 m N , satisfying
Nm=1 wm = 2 , wm > 0
m = N+1m , wm = wN+1m (symmetry)wN/2 + ....... + wm1 < m < wN/2 + ....... + wm N/2 + 1 m N
. . . .
1 2
1 2
N1 N
NN1
The GaussLegendre Quadrature Sets satisfy these conditions; some of these sets are given at the end of this
33
chapter. For m, we can make the approximations
(z, ) (z, m) m(z)
and
11
(z, ) d
(z, 1) w1 + . . . . . + (z, N ) wN =N
n=1
(z, n) wn
Therefore, the problem (3.57) can be approximated by
mdmdz
(z) + m(z) =12[c
Nn=1
n(z) wn + q(z)] 0 < z < L (63)
m(0) = 0 1 m N/2 (m > 0)m(L) = 0 N/2 + 1 m N (m < 0)
which is termed the discreteordinates or SN approximation. This approximation can be developed for any
transport geometry. Physically, it amounts to constraining neutrons to travel in only a finite set of directions
1, ...., N , rather than an infinite set 1 1.
Spherical Harmonics or PN Equations : We now make explicit use of the Legendre polynomials defined in
Eqn.(3.53):
P0() = 1 , P1() = , P2() = 12 (32 1) , .........
Pn() = n2n+1Pn1() +n+12n+1Pn+1() n 1 [P1 = 0]
11 Pn()Pm()d =
22n+1nm n,m 0
It is known that for a reasonable function f(), one has the following Legendre polynomial expansion
f() =n=0
2n+ 12
fn Pn()
where
fn = 11
Pn() f() d = expansion coefficients
34
Let us now define the angular flux moments
n(z) = 11
Pn() (z, ) d, n 0
Note that 0 = scalar flux and 1 = current.
We multiply Eqn.(3.57) by Pn() and obtain
Pn()
z+ Pn() = Pn()
12[ c (z) + q(z)]
or
z[
n
2n+ 1Pn1() +
n+ 12n+ 1
Pn+1() ] + Pn() = Pn()12[ c (z) + q(z)]
Now we operate by 11()d. For n = 0, we obtain
d
dz1(z) + (1 c) 0(z) = q(z)
For 1 n N , we obtain
n
2n+ 1dn1dz
(z) +n+ 12n+ 1
dn+1dz
(z) + n(z) = 0
These are N+1 equations in N+2 unknowns 0, 1, ...., N+1. The standart closure relation is simply to take
N+1 = 0.
Then, we obtain the standart slab geometry PN equations, which have the same number of equations as
unknowns:
d1dz
+ (1 c) 0 = qn
2n+ 1dn1dz
+n+ 12n+ 1
dn+1dz
+ n = 0 1 n N 1N
2N + 1dN1dz
+ N = 0 (64)
Normally, N is odd, so there are an even number of equations and unknowns 0, 1, ...., N .
In principle, one solves these equations, and then is obtained from
(z, ) N
n=0
2n+ 12
Pn() n(z)
35
However, in practice, one is usually only interested in 0(x) and 1(x), and the higher order moments are
usually discarded at the end of the calculation.
P3 Equations:
d1dz + (1 c) 0 = q
13
d0dz +
23
d2dz + 1 = 0
25
d1dz +
35
d3dz + 2 = 0
37
d2dz + 3 = 0
(65)
Eliminating 1 and 3, we obtain two coupled diffusion equations:
13
d20dx2 + (1 c) 0 = q + 23 d
22dx2 ,
1121
d22dx2 + 2 =
215
d20dx2
(66)
P1 Equations:
d1dz + (1 c) 0 = q ,
13d0dz + 1 = 0
(67)
Eliminating 1, we obtain the single diffusion equation:
13
d20dx2
+ (1 c) 0 = q
P1 or diffusion theory is a very common and useful approximation to transport theory, and later we will
discuss it in detail.
Note: Boundary conditions for the diffusion (P1) equations are known and wellaccepted, and we will derive
them in the next chapter. However, boundary conditions for the general PN equations with N odd and geq3
are controversial.
OneDimensional Spherical Geometry : We take Eqn.(3.49)
+ t(E) = 144
(s0 + 3 s1) d
+
Q
4(68)
36
with = (r, )
where
r = x i + y j + z k = spatial position
r =  r  = ( x2 + y2 + z2 )1/2 = distance to origin (69)
= direction of neutron travel
= cos = ( r/r ) = ( x x + y y + z z )/r (70)
z
x
y
r
r / r
We note that if is fixed and r is varied in any nonradial direction, then varies. Therefore, an uncollided
particle moves along a path in which r and both change. We have
(r, ) = r
( r) +
( ) (71)
However,
r2 = x2 + y2 + z2
37
2r rx = 2x 2rry = 2y 2r
rz = 2z
rx =
xr
ry =
yr
rz =
zr
so
r = x rx + y ry + z rz= x xr + y
yr + z
zr =
Also, by Eqn.(3.70),
x =
xx
+ yy
+ zz
r ==
x
r x
x+y
y+ z
z
r2xr =
x
r xr2y = . . . . . . . . . =
y
r yr2z = . . . . . . . . . =
z
r zr2so
= x x + y y + z z= x(
x
r xr2 ) + y(yr yr2 ) + z(
z
r zr2 )= 1r r
xx
+ yy
+ zz
r =12
r
Hence, Eqn.(3.71) becomes
(r, ) = r
+1 2
r
(72)
Next, we consider the scattering term
14
( s0 + 3
s1) (r,
r/r)d
The value of this integral cannot depend on our choice of angular coordinate system. Therefore, we can
choose this coordinate system to simplify the evaluation of the integral. In particular, for r fixed, we set
k = r/r
and we choose i and j in any suitable way. Then,
38
=1 2cosi +
1 2sinj + k
=
1 ()2cos i + 1 ()2sinj + k
and d
= dd
so
14
(s0 + 3
s1) (r,
r/r) d
= 14 1=1
2=0 (0 + 3 [ sqrt1 2 sqrt1
2(coscos + sin
sin) +
]s1) (z,
) d
d
= 12=1 1(s0 + 3
s1) (z,
) d
(73)
Using Eqns.(3.72) and (3.73), Eqn.(3.68) becomes:
r+
1 2r
+ t =
12
11
(s0 + 3 s1) d + Q(r)4 (74)
Again, it is customary to define
(r, ) = 20 (r, )d = 2 (r, )
so that
(r, )ddV = (r, )d(4r2)dr = the total number of neutrons in d about , in dr about r,
and Eqn.(3.74) becomes
r
(r, ) +1 2
r
+ t (r, ) =12[1
1(s0 + 3 s1) (r, ) d + Q(r)] (75)
This equation holds for 0 < r < R and 1 0 corresponds to directions pointing toward the center ofthe sphere, and 0 < 1 corresponds to directions pointing away from the center of the sphere. Boundaryconditions to go with Eqn.(3.75) are therefore:
(R,) = b() , 1 0 (76)
[This prescribes the incident angular flux at the outer boundary.]
39
rR
1
0
1
boundary conditionsare applied here
Note: SN (discreteordinates) is trickier to apply here because of the angular derivative term. However, PN
(spherical harmonics) is just as easy as to apply. As before, P1 theory leads to a standart diffusion equation.
TwoDimensional X,YGeometry: We again take Eqn.(3.49):
+ t(E) = 144
s0 + 3 s1 d
+
Q
4
= (x, y,) [independent of z]
and (x, y,) = (x, y,r)
y
x
z
r
cos
cos k
k
40
= sin() cos() i + sin() sin() j + cos() k
r = sin() cos() i + sin() sin() j cos() k= reflection of across the x,yplane
Then,
x
x+ y
y+ t =
14
4
s0 + 3 s1 d
+
Q
4(77)
where
x = sin() cos() =
y = sin() sin() =
( z = cos() = )
(78)
The surface of the unit sphere is defined by
1 = f( , , ) = 2 + 2 + 2
and the unit outer normal for a point on the sphere is
n =f
 f  =2 i + 2 j + 2 k2 (2 + 2 + 2)1/2
= i + j + k
z
x
y
n
=
d d
d
i j k+ +
41
In the plane generated by n and k, we have
kn
d
d d
Therefore,
d dd = cos = k n = =
1 2 2
Thus, d = d d1 2 2
and Eqn.(3.77) becomes:
x+
y+ t
=14
[ Sigmas0 + 3 (
+
+
) s1 ]
dd
1 ()2 ()2
+14
[ Sigmas0 + 3 (
+
+
) s1 ]
dd
1 ()2 ()2 +
Q
4
=12
[ Sigmas0 + 3 (
+
) s1 ]
dd
1 ()2 ()2 +
Q
4(79)
It is customary to define
(x, y, , ) = 2 (x, y, , )
so that
42
(x, y, , ) dx dy = d d1 2 2 = the number of neutrons in d about , d about , dx about
x, dy about y, per unit length in z,
and Eqn.(3.79) becomes:
x
(x, y, , ) + y
(x, y, , ) + t (x, y, , )
= [12
[ Sigmas0 + 3 (
+
) s1 ] (x, y,
,
)
dd
1 ()2 ()2 + Q(x, y) ] (80)
This equation holds for X < x < X and Y < y < Y . Using
> 0 > 0
> 0 < 0
< 0 < 0
< 0 > 0
x
y
43
we find that incident boundary conditions for must be assigned as follows:
(0,0) < 0 > 0
< 0
> 0
( ,),)
( ,)
,)
YX X,y,
x,Y,
b
b
X, y,
( b
x,Y,
(b
The function b is defined at each point on the boundary, and for each incoming direction (, ). Then we
set = b for all such points and directions.
If b = 0 along a section of the boundary, then this section is termed a vacuum boundary.
Other boundary conditions are possible, based on symmetry considerations. For example, suppose the
outer boundaries of the system are vacuum and
X
Y
.
. .
.(x,y) (x,y)
(x,y) (x,y)
44
Q(x,y) = Q(x,y) = Q(x,y) = Q(x,y), and similarly for t,s0,s1.
Then, must have a similar type of symmetry:
(x, y, , ) = (x,y, ,) = (x, y,, ) = (x,y,,)
. .
..
Y
X(x,y)
(x,y)(x,y)
(x,y)
(,) (,)
(,)(,)
Thus, we can reformulate the problem in the firts quadrant and assign reflecting boundary conditions on the
left and bottom edges:
y
x
(vacuum)
(vacuum)
(reflecting)
(reflecting)
( ,)
(
((
= (,)
,) = ,)( ,)
= 0
= 0 < 0
< 0Y
X
x, Y,
X, y,
x, 0, x, 0,, )
0, y,
0, y,
45
This reduces by a factor of four the amount of storage and arithmetic required to solve this problem.
Finally, Sn and PN versions of Eqn.(3.80) are relatively easy to formulate.
Nth Collided Flux Equations: Suppose we wish to solve the transport problem
z+ t =
s02
11
(z, ) d
+
Q(z)2
0 < z < L (81)
(0, ) = l() 0 < 1(L, ) = r() 1 < 0
Let us consider the following recursive sequence of problems for functions 0, 1, 2, .....
We define 0 by the problem
0z
+ t 0 =Q(z)2
0 < z < L (82)
0(0, ) = l() 0 < 10(L, ) = r() 1 < 0,
and for n 1, we define n in terms of n1 by
nz
+ t n =s02
11
n1 d
0 < z < L (83)
n(0, ) = 0 0 < 1n(L, ) = 0 1 < 0
Then, adding up all of Eqns.(3.82) and (3.83), we obtain
z [ 0 + 1 + 2 + . . . . . . ]
+ t [ 0 + 1 + 2 + . . . . . . . ]
= Q2 +s02
11 [ 0 + 1 + . . . . . . . ] d
c
Likewise,adding up all of the first boundary conditions of Eqns.(3.82) and (3.83), we obtain
[0(0, ) + 1(0, ) + . . . . ] = l() 0 < 1,
46
and adding up all of the second boundary conditions of Eqns.(3.82) and (3.83), we obtain
[ 0(L, ) + 1(L, ) + . . . . ] = r() 1 < 0
Therefore, if we define
(z, ) =n=0
n(z, ) , (84)
then satisfies the original problem (3.81). We have from problems (3.82) and (3.83):
n = the angular flux due to neutrons that have undergone exactly n collisions.
Thus, 0 = uncollided flux, 1 = oncecollided flux, ....., n = nth collided flux. These interpretations
are consistent with Eqn.(3.85), because then is the sum of the angular fluxes of neutrons that have under
gone all possible numbers of collisions.
The OneGroup Diffusion Equation: Let us consider the following generalgeometry onegroup transport
problem:
1v
t+ + t = 14
4
(s0 + 3 s1) d
+
Q
4
=s04
d
+
3 s14
d
+
Q
4
=s04
+3 s14
J + Q4
=14
(s0 + 3 s1 J + Q) (85)
where
= d
= scalar flux
J = d
= current
(86)
Eqn.(3.86) holds for points r in a spatial domain D. The initial and boundary conditions to go with Eqn.(3.86)
are
(r,, 0) = i(r,) (87)
47
(r,, t) = b(r,, t) , r D , n < 0 (88)
We wish to derive the full P1 approximation to this general problem. To do this, we need the identities
d = 4
d = 0
d = 43 I d = 0
(89)
Now, we first operate on Eqn.(3.86) by()d and get
1v
t+ J + (t s0) = Q (90)
This is the balance or conversation equation. If we multiply Eqn.(3.91) by a small volume element d3r,
then each term has the following physical interpretation:
1v
td3r + J d3r + ( t s0) d3r = Q d3r
 First term on the left: rate of change of in d3r
 Second term on the left: rate of leakage out of d3r
 Last term on the left: rate of absorption in d3r
 Term on the right: production rate due to source in d3r
Next, we operate on Eqn.(3.86) by()d and get
1v
tJ +
d + ( t s1 ) J = 0 (91)
Eqns.(3.91) and (3.92) are exact. To close them (so that there are the same number of equations as un
knowns) we must approximate the integral in Eqn.(3.92) using and J . We do this using
(r, J, t) 14
[ (r, t) + 3 J(r, t) ] (92)
This approximates in terms of spherical harmonic functions of order 0 and 1 in such a way that Eqn.()
are preserved. [Thus, this approximation sets to zero all spherical harmonic moments of of order 2.]
48
Introducing Eqn.(3.93) into the integral in Eqn.(3.92), we obtain
d ( + 3 J ) d , for J = 0 ,= 14 ( 14
d) = 14 (43 I)
= 13 I = 13
Hence, Eqn.(3.92) becomes
1v
tJ +
13 + ( t s1 ) J = 0
Defining
a = t s0 ( absorption cross section )tR = t s1 ( transport cross section )
(93)
we obtain the general geometryP1equations:
1v
t + J + a = Q , (94)
1v
tJ +
13 + tr J = 0 (95)
The approximation that lead to these equations is valid provided is nearly isotropic (of the form of Eqn.()
with J  ), absorption is low, and space and time derivatives are weak:
= 0(1)
 J  = 0()  = 0()/t = 0(2)a = 0(2)
Q = 0(2) all for 1
(96)
This scaling is such that each term in Eqn.(3.95) is 0(2). However, applying this scaling to Eqn.(3.96), wefind
tr J = 0()13 = 0()1v
t J = 0(3)
49
Thus, we can delete J/t in Eqn.() [note that this is not an approximation for steadystate problems] and
obtain
J = 13 tr
= D (97)
where
D(r) =1
3 tr(r)= diffusioncoefficient (98)
Now, using Eqn.(3.98) to eliminate J from Eqns.(3.95) and (3.96), we obtain the timedependent diffusion
equation
1v
t D + a = Q (99)
[note that the second term on the left side of the above equation is approximate.]
and
(r,, t) 14
[ (r, t) 1tr
(r, t) ] (100)
Note that the scaling (3.97) implies that each term in Eqn.(3.100) is O(2), and that the second term onthe right side of (3.101) is smaller than the first.
Eqn.(3.100) does not fully determine ; initial and boundary conditions must also be derived. To obtain
the initial condition, we operate on Eqn.(3.88) by()d and obtain
(r, 0) =
i (r,, 0) d (101)
Next, let us attempt to solve Eqn.(3.89) with the approximatio (3.101), i.e.,
b (r,, t) =14
[ (r, t) 1tr
(r, t) ] r D , n < 0 (102)
This boundary condition clearly cannot be satisfied, in general, for every direction . Therefore, we shall
approximately satisfy it: we operate by
50
where r is the reflection of of the boundary:
n
(.
(
.(
n)n
.n)n
n)n
r
r = [ ( n) n ] ( n) n = 2 ( n) n
r n = n (107)
Then, the transport current at this point satisfies
n Jtr =n d
=n>0 [ n () + n r (r) ] d
=n>0 [ n () n () ] d = 0
Requiring the diffusion current to also satisfy this, we get, from (3.98), the following reflecting boundary
condition:
n (r, t) = 0 r D (108)
52
Finally, if the domain D has material discontinuities across an interface :
then is continuous across , so [by Eqn.(3.87)] transport and J transport will be continuous:
material 1 material 2
rr
+. .
transport (r+, t) = transport (r, t)
J transport (r, t) = J transport (r, t)
We also require the diffusion scalar flux and current to be continuous across an interface.
Thus, using Eqn.(3.98), we obtain the interface conditions:
(r+, t) = (r, t)
D(r+) (r+, t) = D(r) (r, t)
Therefore, across an interface, will be continuous, but will be discontinuous:
Q = 0 Q = 0
Q > 0
We have already stated that b = 0, then the boundary condition (3.105) can be replaced by the more
accurate condition (3.106). These can be written as
0 = (r, t) + n (r, t) (110)
where
=
2/(3 tr) standart diffusion theory
0.7104/tr transportcorrected diffusion theory
Using
(r + , t) (r) + n (r, t) + 22 (n )2 (r) + . . . .=
2
2 (n )2 (r) = 0(2) [ see Eqn. (3.97) ] ,
we see that Eqn.(3.110) can be approximated by the extrapolated endpoint condition:
0 = (r + , t) r D , n = unit outer normal (111)
true boundaries
extrapolated boundaries
>0 , =
>0 ,
=
Remarks: Because Eqn.(3.109) cannot be satisfied, the diffusion solution will be inaccurate at material
boundaries. However, a few mean free paths away from boundaries, it is often much more accurate.
transport solution
diffusion solution
vacuum boundary
dDD
z
Therefore, if a system is many mean free paths thick, the diffusion solution can be quite accurate in most
of the system (i.e., away from boundaries and interfaces, where transport effects can dominate).
Derivation of OneGroup Diffusion Theory by the Method of Successive Approximations:
To conclude this section on the onegroup diffusion equation, we shall discuss the validity of the P1 method
for deriving this equation. The idea is to derive the diffusion equation by a different method that more
clearly demonstrates the validity of the assumptions used in the P1 method. We begin this derivation with
the slab geometry transport equation
z(z, ) + t (z, ) =
12[ s0
11
(z, ) d
+ Q(z) ] (112)
and we assume that the leakage term is small compared to the collision and scattering terms, i.e.,
z Sigmat or t
z 1 (113)
Then, just using the inequality (3.114), we can derive the standart diffusion approximation
55
ddz
13t
d
dz(z) + a (z) = Q(z) (114)
to Eqn.(3.113) using the method of successive approximations. This derivation is of onterest because it
provides insight into the P1 method.
The successive approximations idea is simple: It says that if the leakage term is small, then it can be
approximated more crudely than the other terms, and in thr first approximation, we can ignore it altogether.
Doing this, Eqn.(3.113) reduces to
t (z, ) =12[ s0
11
(z, ) d
+ Q(z) ] (115)
Thus, defining the scalar flux
(z) = 11
(z, ) d
we get from Eqn.(3.116)
t = s + Q a = Q , (116)
and using this to eliminate Q from Eqn.(3.116), we obtain
t =12[ s + a ] =
12[ t ]or
(z, ) = 12 (z)
(117)Eqns.(3.117) and (3.118) are the zeroth order (infinite medium) result.
To obtain a better approximation, we introduce the result (3.118) into the leakage term in Eqn.(3.113)
and obtain
d
dz
2+ t =
12[ s + Q ] (118)
Integrating over , ( 11 ()d), we obtain
t = s + Q a = Q (119)
and using this to eliminate Q in Eqn.(3.119), we get
d
dz
2+ t =
12[ s + a ] =
12t
56
Thus,
(z, ) =12[ (z)
td
dz(z) ] (120)
This is the first orderresult. We note that Eqns.(3.117) and (3.120) agree, but Eqn.(3.121) contains a (small)
correction term to (3.118).
To do netter, we use the result (3.121) in the leakage term in Eqn.(3.113) and get:
2d
dz(
td
dz) + t =
12[ s + Q ] (121)
Integrating over , we obtain
ddz
13 t
d
dz(z) + a = Q , (122)
and using this result to eliminate Q from Eqn.(3.122), we get
2d
dz(
td
dz) + t =
12[ s d
dz
13 t
d
dz+ a =
12[ t d
dz
13t
d
dz]
or
=12[
td
dx+
32 13
(1t
d
dx) (
1t
d
dx) ] (123)
Eqn.(3.123) is the standart diffusion equation, and using
P0() = 1 , P1() = , P2() = 3212
We see that Eqn.(3.124) is a Legendre polynomial expansion of the angular flux, with the coefficient of
P2 small compared to that of P1, and the coefficient of P1 small compared to that of P0. This justifies the
closure relation used in the P1 method; when space derivatives are weak, the P2 moment of the angular flux
is small, and it is consistent to set it to zero.
Finally, operating on Eqn.(3.124) by 11 () d ,
we obtain
J(z) = 11
(z,
) d
= 1
3 td
dz(z) ,
57
which is just Ficks Law.
In summary, we have shown that in regions where space derivatives are weak; i.e., where scattering dom
inates leakage, the assumptions underlying the P1 approximation are valid, and diffusion theory should be a
good approximation to transport theory.
The EnergyDependent Diffusion Equation: Now let us consider the energydependent transport equa
tion
1v
t+ + t
=14
[ s0(E
E) + 3 s1(E E) ] (r, E , , t) d dE + 14 Q(r, E, t) (124)
We wish to derive the P1 approximation for this more general equation. As before, we operate by() d
and () d to obtain exactly
1v
t+ J + t =
[ s0(E
E) (r, E , t) dE + Q(r, E , t) (125)
1v
J
t+
d + t(E) J =
[ s1(E
E) J(r, E , t) dE (126)
To close these equations, we take
(r, E,
, t) 1
4[ (r, E, t) + 3 J((r, E, t) ]
so that, as before,
d 13
and then we obtain the energydependent P1 equations:
1v
t(r, E, t) + J(r, E, t) + t (r, E, t) =
s0(E
E) (r, E , t) dE + Q(r, E, t) , (127)
1v
J
t(r, E, t) +
13 (r, E, t) + t(E) J(r, E, t) =
s1(E
E) J(r, E , t) dE (128)
To proceed, approximations to terms (A, the first term in the left hand side of Eqn.(3.129)) and (B, the
58
right hand side of Eqn.(3.129)) are introduced:
(A): As in the onegroup case, it is customary to set
1v
J
t= 0
( B): Here, one takes
s1(E E) s1(E) (E E) (129)
where
s1(E) =
s1(E
E) dE (130)Then,
s1(E
E) J(r, E , t) dE
s1(E) (E
E) J(r, E , t) dE = s1(E) J(r, E, t)
Although this is approximate,() dE of this equation is exact because of Eqn.(3.131). Eqn.(3.129) now
becomes
13 + t J = s1 J
so, once again, we obtain Ficks law
J(r, E, t) = D(r, E) (r, E, t) (131)D(r, E) =
13 [ t ( D(r, E) ) s1 ( D(r, E) ) ] , (132)
and Eqn.(3.128) becomes the standart time and energydependent diffusion equation:
1v
t(r, E, t) D(r, E) (r, E, t) + t(r, E) (r, E, t)
=
s0(r, E E) (r, E , t) dE + Q(r, E, t) (133)
Note that to derive this equation, we need to make an approximation [Eqn.(3.130)] that is not required for
the derivation of the onegroup diffusion equation.
59
CHAPTER 4 : ONEGROUP DIFFUSION THEORY (Chapter 5, Duderstadt and Hamilton)
Analytic OneGroup Diffusion Theory
 Diffusion Coefficient
 Diffusion Length
 Jump Condition (across a delta source)
 Greens Functions
Plane Source, Infinite Medium
Point Source, Infinite Medium
 Geometric Buckling
 Material Buckling
 kEigenvalue
 Reactivity
 Perturbation Theory
Inner Product
Adjoint Operator
Adjoint Boundary Conditions
SelfAdjoint Operator
Perturbation of kEigenvalue Problems
Numerical OneGroup Diffusion Theory
 Fixed Source Problems
Spatial Mesh
Tridiagonal Matrix
Diagonally Dominant Matrix
Gaussian Elimination
 kEigenvalue (Criticality) Problems
Inverse Power Iteration Method
1
In this chapter we shall consider analytical and numerical solution methods for fixed source and eigen
value problems in the onegroup diffusion approximation derived in the previous chapter. First we shall work
out some fixed source problems that can be solved analtically.
Example 1: Plane source in an infinite homogeneous medium
Let us consider a deltafunction source at x = x0, of strength S0:
D d2
dx2(x) + (x) = S0 (x x0)
or
(x) 1
L2= S0
D(x x0), (1)
where
D =1
3 tr= diffusion coefficient (2)
L =
D
a=
13 tr a
= diffusion length (3)
At x = x0, we want (x) to be continuous, but we shall allow (x) to be discontinuous. Then, operating
on Eqn.(1) by x0+x0 () dx
we obtain
(x0+ ) (x0 ) 1
L2
x0+x0
(x) dx = S0D
Now, letting 0, we obtain the jump condition
(x0) (x0) = S0
D= 0 (4)
The diffusion problem can now be written
(x) 1
L2= 0 x = x0 (5)
(x0) (x0) = 0 (6)
(x0) (x0) = S0
D(7)
2
() = 0 (infinite medium) (8)
Solutions of Eqn.(5) are = ex/L. Therefore, Eqns.(5),(6),and (8) imply
(x) =
A e(xx0)/L x > x0
A e(x0x)/L x < x0
Then
(x) =
AL e(xx0)/L x > x0AL e
(xx0)/L x < x0
so Eqn.(7) implies
( AL ) (AL ) = S0D or A = S0L2Dand
(x) =S0L
2De xx0 /L (9)
xxo
If we have a source which is a sum of plane delta functions:
S(x) =I
i=1 Si (x xi) then (x) = L2DI
i=1 Si e xxi /L
3
Likewise, if the source is distributed,
S(x) = S(x
) (x x) dx
then (x) = L2D S(x) e
 xx /L dx
= G(x, x
) S(x
) dx
where
G(x, x) =
L
2De xx
/L (10)
is the infinite medium Greens function for a plane source.
Physical Interpretation of the Jump Condition: Using
J(x) = D ddx
(x) i = current,
the jump condition (4) can be written
[ D ddx (x0 + 0) ] + [ D ddx (x0 0) ] = S0or [ i J(x0 + 0) ] + [ i J(x0 0) ] = S0
or J+(x0 + 0) + J(x0 + 0) S0 (J = partial currents)
i i
x xxo
o o +
Thus, the rate at which neutrons are produced by the source (S0) equals the rate at which neutrons leak
away from the point x0.
4
Example 2: Point source in a finite homogeneous slab: (See Chapter 4 problems.)
Example 3: Point source in an infinite medium:
1r2
d
drD r2
d
dr+ a = S0
(r)4r2
(11)
() = 0 (12)
In spherical geometry, dV = 4r2dr. Therefore, for all R > 0,
R0
S0(r)4r2
dV = S0 R0
(r) dr = S0
so we have a point source of strength S0. Operating on Eqn.(11) by 0() dV , we get
4 D r2 ddr 0 + 4 0a r2 dr = S0
Letting 0, we obtain
S0 = limr0
(4r2)[ D ddr
(r) ] = limr0
(ar)J+(r) (13)
This says that the rate (S0) at which neutrons are produced by the delta function source at r=0 equals the
rate at which neutrons flow out of the sphere centered at r=0 with infinitely small radius. The problem thus
can be rewritten as
1r2
d
drr2
d
dr 1
L2 = 0 0 < r < (14)
limr(r) = 0 (infinite medium) (15)
limr0
r2d
dr= S0
4D(limr0
4r2J+ = S0) (16)
Solutions of Eqn.(15) are:
(r) =er/L
r
B er/L
r , r > 0
A er/Lr , r < 0
5
To satisfy Eqn.(16), we must ignore the solution that grows as r and take
(r) =er/L
r
Then,
r2 ddr = r2 A [ 1
r2 er/L 1rL er/L ]= A (1 + rL) er/L A forr = 0
so A = S04Dor
(r) =S04D
er/L
r(17)
In cartesian geometry, with a delta function source of unit strength at r = r0, we obtain from Eqn.(18)
(r) = err0/L4D r r0 = G(r, r0) = infinite medium Greens function for a point source
Remark 1: Let us consider one neutron emitted per second, S0 = 1. Then, the probability that the neutron
is absorbed in dr about r is
a (r) dV = a (r) 42 dr = a ( 14Der/L
r ) 4r2 dr
= aD r er/L dr = rL2 e
r/L dr ( L =
Da
)
Thus, the root mean square distance between emission and absorption, < r2 >1/2, satisfies
< r2 > =0
r2 [ rL2 er/L ] dr = L2
0
( rL)3 er/L d( rL )
for t = rL ,
= L20
t3 et dt = L2 (4) = 6 L2
so