Lesson 12 Objectives
•
•
•
•
Time dependent solutions
Derivation of point kinetics equation
Derivation of diffusion theory from
transport theory (1D)
Simple view of 1st order perturbation
theory
12-1
Time dependent BE
•
Moving back to a previous slide (Lecture 2), the
last time we had time dependence the BE looked
like this:
ˆ , E, t )
1  (r , 
ˆ  (r , 
ˆ , E , t )    r , E , t  (r , 
ˆ , E, t )

t
v
t

ˆ , E , t )  dE  d  ( r , 
ˆ 
ˆ , E   E, t ) ( r , 
ˆ , E , t )
 qex ( r , 
s
 
4
0

ˆ , E , t )
   r , E   dE   d  f ( r , E , t ) ( r , 
0
•
4
Since I am too lazy to write all that, I will shorten it
to:
12-2
Time dependent BE (2)
1  ( P, t ) ˆ
   ( P, t )   t  P, t  ( P, t ) 
v
t
   s ( P  P, t ) ( P, t )    P    f ( P, t ) ( P, t )
P  (r , , E )
P  (r , , E )
(I took out the fixed source since time-dependent
problems usually involve reactors)
• We will look at 3 techniques for attacking this
equation (given that we can readily solve both the
forward and adjoint STATIC BE)
12-3
Time dependent BE (3)
•
It gets a little more complicated in practice
because we have to account for the delayed
neutrons:
1  ( P, t ) ˆ
   ( P, t )   t  P, t  ( P, t ) 
v
t
   s ( P  P, t ) ( P, t )  1      P    f ( P, t ) ( P, t )
+ i Ci  r , t 
i
Ci (r , t )
  i   f ( P, t ) ( P, t )  i Ci  r , t  , i  1, 6 (usually)
t
12-4
Method 1: Explicit
•
For the explicit approach, we employ a forward
finite difference for the time derivative:
1  ( P, t  t )  ( P, t ) ˆ
   ( P, t )   t  P, t  ( P, t ) 
v
t
   s ( P  P, t ) ( P, t )  1      P    f ( P, t ) ( P, t )
+ i Ci  r , t 
i
Ci (r , t  t )  Ci (r , t )
 i   f ( P, t ) ( P, t )  i Ci  r , t 
t
•
and solve for the future values:
12-5
Method 1: Explicit (2)

 ( P, t  t )   ( P, t )  vt 


 s ( P  P, t ) ( P, t )
 1      P    f ( P, t ) ( P, t )+ i Ci  r , t 
i

ˆ
   ( P, t )   t  P, t  ( P, t ) 

Ci (r , t  t )  Ci (r , t )  t i   f ( P, t ) ( P, t )  iCi  r , t 
12-6
Method 1: Explicit (2)
•
This is a standard forward differencing, which
tends to require VERY small time steps, although it
always converges to the right solution (if you have
the patience)
• Notice the velocity term
• And it has the added advantage (shared by
several methods) that you can determine the
bracketed term (production rate – loss rate)
BEFORE you pick the time step
• Therefore, for example, you can restrict the
timestep so that no value increases more than
a certain percentage
12-7
Method 2: Implicit
•
For the implicit approach, we employ a backward
finite difference for the time derivatives on Slide
12-4 to get:
ˆ  ( P, t  t )    P, t  t   1  ( P, t  t ) 

 t
vt 
   s ( P  P, t  t ) ( P, t )  1      P    f ( P, t ) ( P, t  t )
1
+ i Ci  r , t  t  
 ( P, t )
vt
i
•
•
This uses the standard static solution
I did not include the precursor solution since it is
usually much easier (and explicit works fine)
12-8
Method 3: Quasi-static
•
The quasi-static approach rests on an assumption
that the ABSOLUTE MAGNITUDE of the flux
changes much more rapidly than the SHAPE (in
space, direction, and energy) of the flux
•
•
•
We employ the point kinetics equation to solve for the
rapidly-changing flux magnitude over a fairly large time
step
Every so often we use the current cross sections (which
have changed due to temperature and material
movement) in a new STATIC flux calculation to get the
flux shape
This usually allows for extremely large times steps
(compared to the other two methods)
12-9
Derivation of point kinetics equation
•
The transport equation can be reduced to the time
(only) dependent point kinetics equation
dn  t    t   
=
n  t    i Ci  r , t 
dt

i
dCi (t ) i
 n  t   i Ci  t  , i  1, 6
dt

• The equation itself can be solved implicitly or
explicitly (or using fancier time-dependent solutions
like Runge-Kutta)
• It is the DEFINITION of the parameters from the
current flux that is of prime importance in this
derivation
12-10
Derivation of point kinetics equation (2)
•
To get there, I am going to integrate the static and
transient equations, returning to the brack-et
notation for integration over all variables (except
time):


 1  ( P, t ) ˆ





(
P
,
t
)


P
,
t

(
P
,
t
)



t
 v

t


    s ( P  P, t ) ( P, t )  1      P    f ( P, t ) ( P, t )  dr d  dE


 + i Ci  r , t 

i


 Ci (r , t )






(
P
,
t
)

(
P
,
t
)


C
r
,
t


i 
f
i i
 t
 dr


12-11
Derivation of point kinetics equation (3)
•
Yielding:
ˆ       

t
s
E 

  f
E 

(static)
d 
ˆ       
 
t
s
dt v
d Ci
dt
E 
 1      f
E 
  i Ci
i
 i  f   i Ci
(time-dependent)
12-12
Derivation of point kinetics equation (4)
•
This can be cleaned up a bit. The scattering terms
can be reduced to:
 s
E 


ˆ 
ˆ , E   E , t ) (r , 
ˆ , E , t )
  dr  dE  d   dE   d  s (r , 
V
0
4
0
4


ˆ
ˆ
ˆ , E , t )
  dr  dE   d    dE  d  s (r ,   , E   E , t )  (r , 
V
0
4
 0 4



ˆ , E , t )
  dr  dE   d  s (r , E , t ) (r , 
V
0
4
  s
12-13
Derivation of point kinetics equation (5)
•
And the fission term can be reduced to:
  f
E 


ˆ , E , t )
  dr  dE  r , E   d   dE   d  f (r , E , t ) (r , 
V
4
0
0
4



ˆ , E , t )
  dr   dE  r , E     d    dE   d  f (r , E , t ) (r , 
V
4
0
  4
0

ˆ , E , t )
  dr  dE   d  f (r , E , t ) (r , 
V
0
4
  f 
12-14
Derivation of point kinetics equation (6)
•
Leaving us with (slightly rearranged):
ˆ        

t
s
 f

(static)
d 
ˆ         1         C
 
t
s
f
i
i
dt v
i
d Ci
dt
 i  f   i Ci
(time-dependent)
12-15
Derivation of point kinetics equation (7)
•
Substituting the static into the time-dependent, the
latter becomes:
 f
d 

 1     f    i Ci
dt v

i
d Ci
dt
•
 i  f   i Ci
Finally, if we define:
n(t ) 

v
; Ci (t )  Ci ;  


v
 f
v

 f



1
v
 f
12-16
Derivation of point kinetics equation (8)
•
this becomes:
dn  t 
=
 t   
n  t    i Ci  r , t 
dt

dCi (t ) i
 n  t   i Ci  t 
dt

•
i
if we define reactivity as:
 t   1 
1


keff  1
keff
12-17
Derivation of diffusion theory (1D)
•
Avoiding the much more complicated 3D form, we
begin with the 1D slab BE with first order Legendre
scattering:

  ( x,  )   t ( x) ( x,  )  S0 ( x,  )   s 0 ( x)0 ( x)  3 s1 ( x)1 ( x) 
x
1
•
1
First we integrate the entire equation using  d  :
2 1
1

1 1

d 1
   ( x,  )d     t ( x)    ( x,  )d  
dx  2 1

 2 1

1 1

1 1

1 1

 S0 ( x)   d     s 0 ( x)0 ( x)   d    3 s1 ( x)1 ( x)    d  
 2 1 
 2 1 
 2 1

12-18
Derivation of diffusion theory (2)
•
Or:
d
1  x    t ( x)0  x   S0 ( x)   s 0 ( x)0  x 
dx
1
•
1
Next we integrate the entire equation using   d 
2 1
1
1



d 1
1
2
    ( x,  )d     t ( x)    ( x,  )d  
dx  2 1

 2 1

1 1

1 1

1 1 2 
 S0 ( x)    d     s 0 ( x)0 ( x)    d    3 s1 ( x)1 ( x)    d  
 2 1

 2 1

 2 1

12-19
Derivation of diffusion theory (3)
•
To get:
1

d 1
2
    d     t ( x)1 ( x)   s1 ( x)1 ( x)
dx  2 1

•
We get rid of the remaining integral by assuming
that the second LEGENDRE moment of the flux is
zero:
1
1
1
1  3 2  1 
1  3 2 
1 1 

 d   0   
  d      d 

2 1  2 
2 1  2 
2 1  2 
1
1
3
11
1
2


d



d


0  x 


4 1
2 2 1
2
1
1
1
2


d


0  x 

2 1
3
12-20
Derivation of diffusion theory (4)
•
Substituting this leaves us with the coupled
equations:
d
1  x    t ( x)0  x   S0 ( x)   s 0 ( x)0  x 
dx
1 d
0 ( x)   t ( x)1 ( x)   s1 ( x)1 ( x)
3 dx
•
The second equation can be solved for 1 ( x) to
get:
1
d
d
1 ( x)  
0 ( x)   D 0 ( x)
3  t ( x)   s1 ( x)  dx
dx
12-21
Derivation of diffusion theory (5)
•
Substituting this into the first equation gives us the
familiar diffusion equation:
d2
 D 2 0  x    a ( x)0  x   S0 ( x)
dx
where
 a ( x)   t ( x)   s ( x)
12-22