# Reactor Kinetics

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```					REACTOR KINETICS:

• The study of time-dependent phenomena in nuclear reactor systems is
referred to as reactor kinetics or reactor dynamics.

• Until now, we have considered critical systems. We now examine systems
which are super-critical or sub-critical.

• Some neutrons emitted as a result of fission appear in the system seconds
after fission (as a result of fission product decay) resulting in a dominant
influence on kinetic behavior of the system.

In general, a neutron produced by fission will experience scattering
interactions which will decrease its energy. Eventually it will either be
absorbed or it will leak out of the system.

The average time a neutron exists in an infinite system can be
estimated by noting that the average distance of travel without being
1
absorbed is the mean free path,  a 
a

If the neutron has an average speed, it will take a time, l, to interact.
a   1
l      
V V a
Here l is the prompt neutron lifetime. The lifetime is often broken
into slowing-down and thermal lifetime with the average results:

lSD ~ 10-7 sec
lTh ~ 10-4 sec

Where lSD -fast reactor

lTh - thermal reactor

l = lSD+ lTh ~ lTh = 10-4 sec

1
In a finite system, if a fraction of the neutrons leak from the system,
this will result in a decrease of neutron lifetime.

l
l finite 
D
1  B2
a
D
B2      -leakage fraction
a

(for large reactors, leakage fraction is only a few percent)

The average neutron lifetime is the characteristic multiplication time
or generation time for a prompt neutron system. If there are N(t) Neutrons
in the system at a given time, t, at a time, l, later there will be N(t+l)
neutrons. The magnitude of N(t+l) will depend on keff .

N(t+l) = N(t)keff or

dN  t 
N  t  l )  N  t   l                             N  t  keff
dt

Rearranging the above (Taylor expanded) expression for N(t+l) gives:

dN  t          keff  1              keff
               t                   t
dt               l                    l

Where keff is the excess multiplication excess over the critical
multiplication factor keff = 1.

This differential equation form has a general solution:

 keff   

 t       

N  t   N  0 e                

For neutron population, N(t), at time t, based on an initial population,
1
N(0) at arbitrary time t=0. When t                                       , the population will change by a
keff
factor of e; this quantity is called the period:

2
1
Prompt Neutron Period = TP 
keff

3
For a thermal reactor where l ~ 10-4 an excess multiplication of 0.1%
(keff = 1.00; keff = 1.001-1.00 = 0.001) will give

1 x104 sec
TP                0.1sec
1 x103
or
t
N  t   N  0 e   0.1sec

so that one second
1.0sec
N  t   N  0 e   0.1sec

 N  0 e10  22,026 N  0

An increase in power of a factor of over 20,000 in one second.

For fast reactors where l ~ 10-7, the power increase would, of course, be
substantially more rapid. On the basis of such typical "time constants",
reactor systems would be very difficult to control if only prompt neutrons
were available.

4
DELAYED NEUTRONS:

Recall that of the v neutrons associated with each fission event, a
small fraction result from fission product decay, denoting this fraction by
the symbol ; we then have:

Delayed Neutrons/fission V

Prompt Neutrons/fission (1-V

Then, considering the average time between the fissions and the
absorption (or leakage) of the neutrons, we have an effective lifetime based
on both prompt and delayed neutrons of:
I
i
leff   1    l  
j 1  i

where i is the fraction of the v neutrons associated with each of the I
precursor decay chains.

KINETIC EQUATIONS:

Recall that the general neutron balance is

[net increase] = [production] - [absorption] - [leakage]

Using the diffusion theory approximation theory for the leakage term
and assuming that the spatial and time dependents are separable, we can
write:
dN  t 
 S  t   DB 2   t      t 
dt

Here the parameters are arranged over space, energy and the source
term will be composed of prompt and delayed neutron parts:
6
S  t   1   V  t   t     i C i  t 
i 1

5
where concentration of the i precursor CI(t) must be determined from the
equation.

6
For convenience, the twenty-or-so known precursor nuclides are often
represented by six "effective" delayed neutron groups:
6
i
leff   1    l  
i 1   i

Where the number of delayed neutrons per fission, and the precursor
half-lives are shown below for the important fissile nuclides. To calculate
leff it is convenient to convert as per the table for U235 with the result that for
a thermal system with l ~ 10-4

leff = (1-0.0065)x10-4 + 0.084 ~ 0.084 sec

Using leff rather than l as a characteristic multiplication time will
greatly enhance the stability of the system to small changes in
multiplication.

dC i  t 
  iV    t    C i  t    ai   t 
dt

Fission Product                    Decay Rate Absorption Rate

Due to the fact that these nuclides normally have small absorption
cross sections, the last term can usually be neglected. By using the
definitions of k00 and keff from diffusion theory along with the following:

  t   vN  t 
keff  1

keff

l  1 
Neutron Generation  l                                   
k  k v  a 

7
We can rewrite the equations for neutron and precursor population in
a simplified form.

dN  t                         6
         N  t     iCi  t 
dt              l           i 1

dC i  t          i
      N  t    i Ci  t 
dt             l

This form is known as the point kinetic. Seven linear, coupled
differential equations which can be solved in a straight forward manner.

One delayed neutron group:

Some insight into the kinetic behavior of a reactor can be gained by
considering the effect of a single "average" group of delayed neutrons,
where:

i  

i           6

 i
i 1   i

dN  t   
   N  t   C  t 
dt       l
dC  t  
  N t   C t 
dt     l

Using a general solution to form e  wt for both N(t) and C(t) and an
assumption of an initially critical system with "step-change" of reactivity.
We have:

      
                  t 
             
  
           TP   
N  t   N  0       e             
       e     l 
            
  

                

     l   
TP     
         
delayed                     prompt

8
Note that the equation can be very different depending on the
relationship between  and .

1.  The second term will be positive as will the exponential; l x
is small, the increase will be rapid, out-stripping the delayed
neutron effect in the first term.

: This system is said to be prompt critical (critical on prompt
neutrons only.)

: This system is prompt super critical since the relationship
between  and  is significant, a reactivity unit, the dollar was
defined:

\$1 


so    \$1 is prompt critical

1 
yes,        \$0.01 = 1 cent
100 

2.  The first term will dominate falling a prompt-jump
due to the second term.



The system in not critical on prompt neutrons alone (the additional
prompt neutrons produce less than one net prompt neutron in the
next generation and must wait for the delayed neutrons to continue
the supercritical increase. This system will establish a stable period
characteristic of the delayed neutron period.

3. 

The second term causes a prompt-drop due to decreased fission rate;
the first term then dominates since the delayed precursor nuclides
must decay out for the neutron population to die out completely.

In this case the period of decay is that of the one delayed group.

9
SIX DELAYED GROUPS:

With all six groups effective groups, the procedures and results are
similar.

1.  prompt critical

2.  established a stable period, TP, after
the transient terms (6 ??) dies out.

l   6
i
      
TP i 1 1   iTP

3.  stable period of decrease established
by the longest precursor half-life:

t1/2 = 56 seconds          TP = 80

10
PROBLEM 1:

It has been determined that the reactor is in a condition such that keff is
equal to 0.95. Do you have to add or subtract reactivity to make the reactor
critical?

SOLUTION:

 k  1
k

How much reactivity do you have to add to make the reactor critical?

k = 0.95                               
 k  1   0.95  1  0.053
k          0.95

k=1             

k
   0   0.053    0.053
k

11
PROBLEM 2:

Reactor power is 100 watts and increasing on a 30 second period.
How long will it take for the reactor power to reach 500 kilowatts?

   t 
   
 t 
    
N  t   N  0     e P  
T
N  t   N  0 e    TP 
        
           

t
P  P0 e T
t
P
 eT
P0
t      P
 ln
TP      P0
P
t  TP ln
P0

= 30 ln(500,000 / 100)
= 255 sec
= 4.25 min

12
PROBLEM 3:

Calculate the reactor period for the given conditions (assume that
eff = 0.007.)

keff = 1.0001 for a prompt critical reactor


k   eff    1
 0.0001
k
keff                     k

l
T                                      


l
TP                                     l*-provided @ (0.0001 sec)


0.0001

0.0001
 1sec

What is the reactor period for the same reactivity insertion if the reactor is
not Prompt Critical?

T
                                     = 0.01



 0.0075  0.0001 
 0.1 0.0001
 690sec

13
PROBLEM 4:

At t=0 the reactor power at 100 kw. Thirty seconds later the power
level is 150 kw. What is the reactivity?

t
P  t   P  0 e   T

t       P t  
 ln 
T       P  0 

        
 P t  
T  t / ln 
 P  0 

        

30

 150 
ln      
 100 
 74sec

               1
            from  P  
1  T               TP
0.0075

1  0.1  74  
k
 0.00089
k

14
PROBLEM 5:

Calculate the final stable reactor period if a rod is inserted with a
k
reactivity of 0.005      .
k

   0.0065  0.005
TP         
    0.1  0.005 
 23sec

Not So!

= -80 seconds!

15
PROBLEM 6:

A reactor is made subcritical by a large negative reactivity insertion.

(a) Determine the stable negative period that is reached during shut-
down.

Reactor neutron level is maintained by the decay of delayed neutron
sources during shutdown. After the shorter-lived fission products
decay away, neutron population is maintained by the decay of Br-87
with a 55.72 second half-life. This decay causes neutron level to
decline on an apparent period that is determined by:
t        0.693 t

P  P0 e  P0 e
T          T2

t      0.693 t

eT  e     T2

t
 0.693T2
T
T2
T
0.693
  55.72sec 

0.693
 80sec

(b) How long does it take to establish the stable T=-80 sec after the
rods are inserted?

The next longest-lived precursor is I-137 with a 24.4 second half-life.
Using the seven half-life rule, I-137 will decay away to less than 1%
of its steady state operating value 7x24.4 = 171 sec ~`3 min.

16
REACTIVITY CONTROL:

Objective: to control neutron population; introduce material that
absorbs neutrons. The system may use one or more of the following to
provide absorption:

•control rods

•soluble poison (boric acid)

•burnable poison

Short Term Changes:

1. Fuel temperature

2. Moderate and/or coolant temperature

3. Fuel motion

Each of these is often quantified in terms of a reactivity coefficient.

17
TEMPERATURE EFFECTS:

•Dominant absorption effect is known as the Doppler:

1-1000 ev

•Coolant and Moderator Effects:

•Changes due to:

•temperature

•pressure

•density *

•void contact

*Decrease Density

•increase leakage

•decrease moderation

•decrease moderator absorption

•Fuel movement:

•for example, Bowing

18
Limiting Conditions
Reactivity and Period

Reactivity Limit             Reactor Period       Reactor Condition
k/k                        (sec)
       > eff                          *
T l /             Prompt critcal

       eff > >            T   eff   /      Super critical

                               T                   Critical

0 > eff/1-80             T                 Sub critical

    eff/1-80                T                 Sub critical

19
REACTOR SHUTDOWN:

• reactor period limited by the radioactive decay of the longest lived
neutron precursor Br-80

Show that the shortest period for shutdown is -80 sec.
.693t                        t
N t   N0 e            t 1/2
 N  0 e       T

for this to be true then

 . 693t t

t 1/2   T

and T
t 1/2
T
. 693

by substituting Br-80 half-life 55.72 sec results in

55. 72 sec
T                    80 sec
.693

20
XENON AND SAMARIUM POISONING:

• All fission products absorb neutrons.
• The accumulation of fission products reduce the multiplication
factor k
• This reduction can be described with reactivity

 aP
f

p

In general the mass balance for any fission fragment being produced
in a reactor is:
accumulation = production - destruction

production = fission product + decay product
fi-1

          destruction = decay + absorption
a

Therefore

  f   N i1    a  N
dN
dt

21
For Xenon

 (11sec )       (6.7hour)            (9.2hour)         (10 6 years )
Te135   I135    Xe135    135   Ba 135 (stable)
                      Cs     

fission                 fission               fission

While the reactor is in operation and combining the fission yields for I and
Te the following system of equations results

dI
  I  f   I
dt
  x  f   I   x   aX X
dX
dt

At equilibrium:

  
I  I f
I
 I   X  f 
X 
 X   aX 

After reactor shutdown

X  t   X 0 e t X 
 II0
 I  X       
e t X  e t I      

22
For Samarium

By Simplifying the Sm chain, During reactor operation the concentration of
Sm and Pm are:

dP
  P f    PP
dt
dS
  P P   aS S
dt

at equilibrium:

 Pf 
P 
P
 
S  P f
 aS

After Shutdown

St   S 0  P 0 1  e tP   

23
TEMPERATURE EFFECTS ON REACTIVITY:

d
T 
dT

Nuclear Doppler Effect

• absorption cross sections vary with temperature
• nuclei are in atoms which are continual motion due to their
thermal energy
As a result of these thermal motions even of beam of monoenergetic
neutrons appear to have a continuous spread of energies. Therefore the
resonance peak widens with temperature. (shape changes but the area under
the curve remains essentially constant

i     1 
 prompt        ln     
2 T p(300) 

Moderator coefficent
• thermal cross sections change
• physical density of the moderator changes

 mod   T  f    T (p)   T ( P)

Void coefficient
In a liquid moderator and/or cooled reactor when the liquid
boils the volume occupied by the vapor is a void. These voids effect
reacivity.

24
d
v 
dx
EXAMPLE:
Determine the limiting temperature of a reactor that has a excess reactivity
of 0.50 % at 80 F and a temperature coefficient of reactivity of -0.000040/F.
Assume the coefficient is constant throughout the temperature range
involved

d
T 
dT
Rearrange

T 
T
Substitute
. 0050
T             125F
. 000040
Therefore the reactor will become subcritical if the temperature rises above
205F (80+125)

25
EXAMPLE:
A triga operating at 100 kW is inadvertantly exposed to a positive reactivity
of \$.90 for 5 seconds when the operator recognizes the unplanned reactivity
increase and scrams the reactor decreasing the power level to 10% of its
original power. What reacivity was inserted to scram the reactor?

t1                                 t2
P1      1        T1         P2     1                T2
         e        and              e
P 0 1  1 \$                P1 1  2 \$

by substitution
t2 t1

P2      1        1              T 2 T1
                  e               e
P 0 1  1 \$ 1  2 \$

Solve for p2\$

t1       t2
P        1  T T
2 \$  1  0              e 1 e 2
P2       1  1 \$ 


Identify Knowns

1 \$  0. 95
t 1  5sec
t 2  60 sec Scram = fastest period a reactor can shut down is -80 sec
therefore T2=-80 sec

Still need T1

26
1  1 \$
T1 
1 \$

By substitution

1. 90
T1                  1.39 sec
. 08 . 90 

Now solve for p2\$

5   60
100  1  1.39 80
2 \$  1            e e     1723
10 1. 90 

What is P1
t1                              5
1           T1                   1
P1           P0 e                 P1         100e 1.39
1  1 \$                           1. 95

P1=72986 kW

27
CONTROL RODS:

Black rod - strongly absorbing rod
Grey rod - less strongly absorbing rod

For a Black Rod the reactivity can be expressed by

7. 43M2                                   1
T               R  d 
w                   0.116  ln          
    0 T      
1  B 2 M 2 R 2           2. 405a  a 


Incremental reactivity for control rods

 x 1     2 x 
w (x )   w (H)     sin      
H 2      H 

Typical Geometry’s
Cluster Control Rods
Cruciform Control Rods

Chemical Shim

Affects the thermal utilization factor

f = absorption in the fuel divided by the absorption in the
fuel + everything else.

 aB
reactivity is w  1  f 0 
 aM

where f0 is the thermal utilization factor without the chemical

28

```
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