a. dokhane, phys487, ksu, 2008 chapter3- thermal neutrons 1 news need to fix a date for the mid-term...

A. Dokhane, PHYS487, KSU, 2008

Chapter3- Thermal Neutrons

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NEWS

• Need to fix a date for the mid-term exam?

• The first week after vacation!



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Chapter3

Lecture 1

THERMAL NEUTRONS



3

1. Review

2. Neutron Reactions

3.3. Nuclear FissionNuclear Fission

4.4. Thermal NeutronsThermal Neutrons5. Nuclear Chain Reaction

6. Neutron Diffusion

7. Critical Equation



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• Introduction

• Energy distribution of thermal neutrons

• Effective cross section for thermal neutrons.

• The slowing down of reactor neutrons

• Scattering angles in L and C.M systems

• Forward scattering in L system

• Transport mean free path and scattering cross section

• Average logarithmic energy decrement

• Slowing-down power and moderating ratio

• Slowing-down density

• Slowing-down time

• Resonance escape probability

• The effective resonance integral

Lecture content:



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4.1 Introduction

Question: why it is necessary to reduce the neutron energies from 2 Mev to thermal energies?

Neutrons born in a fission 2 Mev

Answer: very high fission cross section for thermal neutrons as compared to that for the nonfission capture cross section.

A reactor which is designed that almost all neutron fissions occur with neutrons of thermal energies

Thermal Reactor

This type represents the main object of interest to us in this course



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4.1 Introduction

because the existence of large number of neutrons in reactor

To get some understanding of the physical processesphysical processes involved

it is necessary to employ a statistical approachstatistical approach very similar to that of the kinetic theory of gaseskinetic theory of gases.



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4.2 Energy Distribution of Thermal neutrons

Neutrons in nuclear reactors are not of uniform energynot of uniform energy but are distributed

over an energy range that extend from very slow to very fast neutrons of about 17 Mev.

See Figure 6.1, Page 135

Fast neutrons are continuously being produced by fission

Slow neutrons are continuously being absorbed leads to fission creation of fast neutron.



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Question: How the compensation for the steady loss of slow neutrons through absorption is made?

Answer: by rapid and efficient slowing down of the fast neutrons so as to maintain the supply of slow neutrons.

This process is called the moderation or thermalization

moderation process is achieved by a moderating material and called moderator which is incorporated in the reactor.



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Question: How the presence of moderator slows down the fast fission neutrons?

Answer: by elastic collisions between the moderator nuclei and the neutrons until the average kinetic energy of neutrons corresponds to that of the moderator nuclei.

See Figure 6.2, page 136



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Answer: Higher scattering cross section compared to the absorption cross section.

Question: what is the best moderator?

This allows a rapid reach of thermal energies and avoid the nonproductive absorption.



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When neutrons are thermalized their energy distribution will be approximately Maxwellian , corresponding to the temperature of the surrounding medium

For neutrons in thermal equilibium with the moderator velocity distribution will be given by the Maxwell-Boltzmann expression:

dvkT

mvv

mkT

ndvvndn

2

2

23

0 2

1

exp)/2(

4)(

(4.1)

Where 0n is the number of neutrons per cm3, m is the neutron mass, T the temperature,

and k the Boltzmann constant.

Number of neutrons whose velocities lie between v and v+dv is given by dn=n(v)dv



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dvkT

mvv

mkT

ndvvndn

2

2

23

0 2

1

exp)/2(

4)(

(4.1)

This distribution is shown in Figure 6.3, page 136.



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The energy distribution of the neutrons can be written in the form:

dEkT

EE

kT

ndEEndn

exp

)(

2)( 2

1

23

0

(4.2)

This distribution is shown in Figure 6.4, page137



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The energy distribution of the neutrons can be written in the form:

dEkT

EE

kT

ndEEndn

exp

)(

2)( 2

1

23

0

(4.2)

The distribution functions n(v) and n(E) are called “ density function”

Give: the number of neutrons per unit velocity interval and per unit energy interval, respectively.

What means??

dv

dnvn )( dE

dnEn )(and

Very important



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represented by the small shaded areas in figure 6.3 and 6.4

The velocity distribution function has a maximum for a value of the velocity vp

called most probable velocityprobable velocity

give the number of neutrons that are found within a small velocity or energy interval

Equations 4.1 and 4.2 dvkT

mvv

mkT

ndvvndn

2

2

23

0 2

1

exp)/2(

4)(

dEkT

EE

kT

ndEEndn

exp

)(

2)( 2

1

23

0



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Question: How the most probable velocity is calculated?

Answer: By differentiating (4.1) , and setting this equal to zero

dvkT

mvv

mkT

ndvvndn

2

2

23

0 2

1

exp)/2(

4)(

2

1

2

m

kTvp

(4.4) Can you prove this ????



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Energy corresponding to most probable velocity is

kTmvE pp 2

2

1

Neutron temperature is k

mv

k

ET

pkin

2

2

1

The energy distribution equation (4.2) leads to the most probable energy E0 by the same method derivation and make it equal to zero

dEkT

EE

kT

ndEEndn

exp

)(

2)( 2

1

23

0

kTE2

10 Can you prove this?



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Conclusion:the most probable energy E0 is not equal to the energy corresponding to the most probable velocity Ep but it is half of it

02EEp

The average energy is E is given by kinetic theory:

kTvmE2

3

2

1 2 21

21

2 3

m

kTvv rms

In combination with 2

1

2

m

kTv p prms vv

21

2

3

The average velocity is related to vp pvv2

1

4

Prove this???

Prove this???



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Summary

2

1

2

m

kTv p kTmvE pp 2

2

1

pvv2

1

4

kTvmE2

3

2

1 2

21

21

2 3

m

kTvv rms

kTE2

10



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The energy distribution of neutrons in a reactor is not exactly Maxwelllian, it corresponds to a temperature which is slightly higher than that of the moderator material. WHY????

Reality:

Existence of a steady influx of high-energy neutrons from fission

a steady absorption of the low-energy neutrons by the fissionable material

++

Consequence



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4.2 Energy Distribution of Thermal neutronsThe energy distribution of neutrons in a reactor is not exactly Maxwelllian, it corresponds to a temperature which is slightly higher than that of the moderator material. WHY????

Effect of raising the high-energy end of the Maxwellian distribution and depressing the low-energy portion of the distribution.

Consequence




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The actual neutron distribution will therefore correspond to an effective temperature that is somewhat higher than that of the actual reactor material.

Consequence

effective rise of neutron temperature effect is called: spectrum or thermal hardening



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The elevation of the neutron temperature Tn above that of the moderator temperature T can be calculated by the relation:

s

an ATTTT 89.0 valid for A<25 and 5.0

s

aA

Where A is the atomic mass number of the moderator atoms and a and s

are

their cross sections at temperature T.

For heavy moderator atoms we have to substitute 0.89 by 0.6

Example 6.1, page 140



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4.3 Effective Cross Section for Thermal Neutrons

We have seen that the absorption cross sections for slow neutrons are strongly dependent on the neutron energies.

In many cases, this dependence follows the 1/v law

Answer: Because we have not only one speed of neutrons in the thermal region but a range. Then we have to find an average value that is the effective one

Example: reaction (n, alpha) with boron is a typical example for the 1/v behaviour

Question: why effective cross section فعلي ??مقطع



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However, Not all absorption reactions for slow neutrons follow strictly the 1/v law

a correction factor is introduced called “not 1/v” or f-factor . Ranges from 0.85 (for Gadolinium) to 1.5 (for Somarium).

See Table6.1, page 140



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For more rigorous calculations, f-factors should be considered temperature dependent.

Because thermal neutrons have a spread in a wide range of speed

the corresponding cross section is also has spread over a wide range

we use an effective cross section which is defined as:

Value of the cross section that results when the total number of absorption per second per unit volume is averaged over the neutron flux

Therefore

0

0

)(

)()(

vdvvn

dvvvvn



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If neutron absorption Xsec obeys the 1/v law 0tan)( ctconsvv

0

00

)(

)(

vdvvn

dvvnc

However, the average speed is defined as

0

0

)(

)(

dvvn

vdvvnv

v

c0

the effective cross section is equal to the cross section for neutrons with speeds equal to the average speed for thermal neutron distribution



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How is calculated?? 0c

conventionally taken with reference to the most probable velocity vp for the thermal neutrons

ppvc 0 with p is cross section for neutrons of speed vp

ppp

v

v

2

21

Note that with a factor of = 0.886 p 2

21



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The thermal neutron cross sections are tabulated based on the following condition:

speed=2200 meters/sec most probable speed of a Maxwellian neutrondistribution at 293.6K and corresponds to a neutron energy of 0.0253 ev

Question: how the effective neutron cross section at a temperature T is obtained from the tabulated value?

Answer:

1. multiply the value by 22

1

2. multiply by 21

6.293T

3. multiply by not 1/v factor

0)2200(

21

21

"/1"6.293

2 pT factorvnot

T



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Example 6.2, page 142



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4.4 The Slowing Down of Reactor Neutrons

Question: the rapid slowing down and thermalization of the fast fission neutrons are an important phase in the design and operation of a nuclear reactor. Why??

Question: How neutrons slow down in a nuclear reactor?

Answer: it is necessary to reduce neutron loss due to nonfission and resonance absorption and to increase the fissioning rate

Answer: The most important contribution to slowing down of neutrons in a nuclear reactor are elastic collisions with the moderator nuclei.

whereby a neutron transfers a portion of its kinetic energy to its collision partner.



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Goal: to get an indicator of the amount of energy transferred from the neutron to the moderator nucleus in an elastic collision.?

the collision process needs to be studied in more detail

Hence

Collision between two particles can be described either in:

1. Laboratory system (L system) the target nucleus (moderator) is initially at rest

2. Center of Mass system (CM system) the center of mass of the colliding particles is at rest initially and remains throughout the collision




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by definition of the CM of two particles: The total linear momentum of the colliding particles is zero in the CM system

0 MVmvc

Where m is neutron mass, M is moderator nucleus mass,

vc is neutron velocity and –V is the velocity of M

That is why CM system is often more convenient for purpose of calculation

However, L system is the frame in which all experimental measurements are made.

It is important to establish relations that will permit us to relate the motion of a particle in one system to its motion in the other system.



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See Figure 6.6, page 143.

It is clear that the velocity of the target nucleus in the CM system is equal and opposite to the velocity of the CM in the Lab system.

Consider now, the V

the velocity of the CM in L system and 0v

the velocity of a particle

in the L system and cv

its velocity in the CM system, the three velocities are related as

follow:

Vvv c

0 (4.19)

Applying the condition that the total linear momentum of the colliding particles must be zero in the CM system

0 MVmvc

Where m is neutron mass, M is moderator nucleus mass and –V is the velocity of M.

(4.20)



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See Figure 6.7, page 144 .

0vMm

mV

0vmM

Mvc

and

Combining and we can find Vvv c

0

0 MVmvc



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Conservation of energy: speeds after collision are not changed in the CM system (see Figure 6.8)

they must be also collinear to give zero linear momentum

The result of the collision in the CM system is a rotation of the particle system by an angle of



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See Figure 6.9, page 145 the collision in the two systems

Where v1 and v2 are the velocities of m and M, respectively, after the collision in the L system



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The following geometrical relations can be deduced from the figure:

cos22221 cc VvvVv (4.22)

sin

sin

2

1 v

v(4.23)

=scattering angle in CM system and =scattering angle in L system



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(4.25)

By introducing the mass number A of the moderator nucleus, and the mass number 1 for the neutron, we can write with only a negligible error

AM

m 1

from (4.20) we can get and A

vV

1

10

A

Avvc

10

Consider: E0 and E1 is the neutron energy in the L system before and after the collision respectively.

by combining Equations 4.22 and 4.25 the relationship between E0 and E1

2

2

20

21

0

1

1

cos21

A

AA

v

v

E

E



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2

2

2

21

0

1

1

cos21

A

AA

v

v

E

E

c

For 0 , 01 EE which means that no energy is transferred from the neutron to the

moderator nucleus

For 2

2

0

1

1

1

A

A

E

E

This type of collision causes the maximum energy transfer from neutron to the moderator nucleus

i.e, largest possible energy loss for the neutron in a single collision



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Let’s set

2

2

1

1

A

A

Maximum energy loss:

11 0

max0

10max10max E

E

EEEEE

The maximum fractional energy loss is

1

max0E

E

For all intermediate angles of scattering 0< <

The fractional energy loss lies between 0 and 1

100E

Efor 0< <



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Hence: cos

2

1

2

1

0

1

E

E(4.31)

The maximum fractional energy loss depends through on the atomic mass number of the moderator nucleus A. by expanding

....32

14

1

2

max0

AAA

E

E

Important conclusion Maximum fractional energy loss is the greatergreater mass of the moderator nucleus is the smallermoderator nucleus is the smaller

light nuclei to be more effective moderators than heavy nuclei.

See Examples 6.3 and 6.4page147



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4.4 The Slowing Down of Reactor NeutronsSee Examples 6.3 and 6.4

page147



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4.5 Scattering Angles in L system and CM system

cos21

sin1

sin1cos

2

22

22

AA

A

cos21

cos12

2

AA

A

2

12 cos21

cos1cos

AA

A

In the case when A>>1 coscos .



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4.5 Scattering Angles in L system and CM system

neutron energy after scattering, E1, is given by cos

2

1

2

1

0

1

E

E

relates this energy as measured in the L system to the scattering angle in the CM system

it is very useful in this form because the angular dependence of the scattering in the CM system is very simple

A relation between the scattering angle and the scattering angle as measured in the L system can be established by combining Equations. 4.23 and 4.26

2

21

20

21

2

2

2

1sin

sin

A

A

v

v

v

vc

cos211 2

22

1

0

AA

A

A

A

E

E

cos21

sinsin

2

222

AA

A



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Homework

• Problems: 1, 3, 4, 6, 7, 10 and 14 of Chapter 6 in Text Book, Pages 168

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a. dokhane, phys487, ksu, 2008 chapter3- thermal neutrons 1 news need to fix a date for the mid-term...

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