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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,γ ) - 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

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• CHAPTER 1 : NUCLEAR REACTIONS IMPORTANT IN FISSION CHAIN

REACTIONS (Chapter 2, Duderstadt and Hamilton)

- 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

1

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

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

[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 (neutron-nucleus 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 neutron-nucleus 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 neutron-nucleus 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

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: U-233, U-235, Pu-

239, Pu-241

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 high-energy (1 Mev) neutrons : Th-232, U-238, Pu-240,

Pu-242

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, neutron-rich. 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 fission-produced 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 half-lives of 55.0,

22.0, 6.0, 2.0, 0.5, and 0.2 sec.

i = decay constant of the i-th precursor group

i = fraction of all fission neutrons emitted from i-th 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 beta-decay 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 : U-233, U-235, Pu-239, Pu-241

Fissionable : Th-232, U-238, Pu-240, Pu-242

Only U-235 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 steady-state chain reaction is maintained if

1 =1f1a1

=1f1a1

= 1

For a mixture of N types of nuclei, a steady-state 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 stedy-state 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 Breit-Wigner formula (derived using quantum me-

chanics) describes this analytically.

23

• CHAPTER 2 : FISSION CHAIN REACTIONS (AND NUCLEAR REACTORS)

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 (steady-state 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)

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 six-factor and four-factor 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 six-factor 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 six-factor and four-factor 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 mass-number 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 (30-40

times as much as in thermal reactors) is needed to maintain chain reaction.

2. Heavy, high mass-number 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 Energy-Averaged , , 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 3-1.

2.2 D H, p. 100, problem 3-3.

2.3 D H, p. 100, problem 3-6.

2.4 D H, p. 100, problem 3-9.

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 non-leakage 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-

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

Time-Dependent Neutron Transport Equation With Delayed Neutrons

- Initial and Boundary Conditions

Steady-State Transport in a Purely Absorbing Medium - Steady-State 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

One-Group Approximation

- Angular Discretization

SN (Dissrete-Ordinates)

PN (Spherical Harmonics)

- Representation of Macroscopic Scattering Cross Section

Scattering Ratio, c

Mean Scattering Cosine, 0

- Spatial Discretization

- Iterative Methods

Special Geometries

- One-Dimensional Slab Geometry

1

• Integral Transport Equation

SN Equations

PN Equations

Closure Relations

- One-Dimensional Spherical Geometry

- Two-Dimensional X,Y-Geometry

N-th Collided Flux Equations

One-Group Time-Dependent Diffusion Equation

- Initial Condition

- Boundary Conditions

Prescribed Incident Flux

Reflecting

Extrapolated Endpoint (Extrapolation Distance)

- Interface Conditions

Energy-Dependent 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

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

, and d3r

(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

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

= [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

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 j-th 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 j-th 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 j-th precursor group

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 time-dependent 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) Neutron-neutron 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) non-reentrant boundary

3) incident flux does not depend on the exiting flux

D

exterior of D= vacuum

1) non-convex

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)

Steady-State Transport Equation Without Delayed Neutrons:

+ t =

s dEd

+

x

4

f dE

d

+

14

Q (24)

+ boundary condition (3.21)

Steady-State Transport Equation in a Purely Absorbing Medium: Set s = f = 0 ; then

+ t = 14Q (25)

+ boundary condition (3.21)

Steady-State 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 Steady-State 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 first-order 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)

= 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 delta-fuction 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 r0|2(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 free-streaming neutrons that enter D through its outer

boundary, and there is no exponential attenuation away from the boundary.

One-Dimensional Half-Space: Let D consist of the half-space 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

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 time-dependent problem reduces to solving a steady-state 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 steady-state 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

• One-Group Approximation:

(r,) + t(r) (r,) =4

s( ) (r,) d + 1

4Q(r) (43)

3) Angle (SNandPN ) : There are two widely-used 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 discrete-ordinates 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 forward-scattering 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.

Special Geometries: The general steady-state one-group 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 one-dimensional slab geometry in detail, and then we will very briefly describe one-dimensional

spherical geometry and two-dimensional x, y-geometry.

26

• One-Dimensional 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 slab-geometry 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 non-vacuum 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.

Discrete-Ordinates 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 Gauss-Legendre 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 discrete-ordinates 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 well-accepted, and we will derive

them in the next chapter. However, boundary conditions for the general PN equations with N odd and geq3

are controversial.

One-Dimensional 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 non-radial 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 (discrete-ordinates) 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.

Two-Dimensional X,Y-Geometry: 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,y-plane

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.

N-th 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 = once-collided flux, ....., n = n-th 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 One-Group Diffusion Equation: Let us consider the following general-geometry one-group 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 steady-state 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 time-dependent 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

• n
• 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 transport-corrected 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 One-Group Diffusion Theory by the Method of Successive Approximations:

To conclude this section on the one-group 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 zero-th 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 order-result. 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 Energy-Dependent Diffusion Equation: Now let us consider the energy-dependent 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 energy-dependent 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 one-group 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 energy-dependent 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 one-group diffusion equation.

59

• CHAPTER 4 : ONE-GROUP DIFFUSION THEORY (Chapter 5, Duderstadt and Hamilton)

Analytic One-Group 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

- k-Eigenvalue

- Reactivity

- Perturbation Theory

Inner Product

Perturbation of k-Eigenvalue Problems

Numerical One-Group Diffusion Theory

- Fixed Source Problems

Spatial Mesh

Tridiagonal Matrix

Diagonally Dominant Matrix

Gaussian Elimination

- k-Eigenvalue (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 one-group 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 delta-function 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) = e|rr0|/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

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