186
5 Miscellaneous Calculations
Since the cross-sections will vary in time due to the varying abundance fractions,
ν must also vary; it has to be computed afresh at each timestep. ν will increase with
time since the fission cross-section will decrease due to consumption of
235 U.
The next step is to set up expressions for how many nuclei of isotope iso undergo a
given process during time t. Nuclei of a given isotope can be created from neutron
capture by a nucleus of lower weight (if applicable), while simultaneously being
lost due to fission and capturing neutrons themselves to produce fission products or
isotopes of greater weight. Now, from above, the number of neutrons released over
Δt days will be νR Δt. If all of these neutrons are involved in some event over time
t, then the number of events that correspond to some process will be given by the
total number of events involved multiplied by the ratio of the total cross-section for
that process to the total available cross-section. Hence, for a given isotope we can
write
N
iso
(t + t) = N
iso
(t) +
ν R t
σ total
N
F
lower
σ
lower
c
− F
iso
σ
iso
f − F
iso
σ
iso
c
.
(5.19)
The simulation actually tracks fractional abundances, that is, (5.19) divided by N:
F
iso
(t + t) = F
iso
(t) +
ν R t
σ total
F
lower
σ
lower
c
− F
iso
σ
iso
f − F
i
σ
iso
c
.
(5.20)
We will need to know N, however, as it remains in σ total through (5.15). For
simplicity in determining N, it will be assumed that the fuel is initially composed
entirely of
238 U. Given that the fuel in most reactors is enriched to only a few percent
235 U, this will not be a drastic approximation.
What of the fission products? “Product 1” accumulates from fissions of the three
fissile isotopes in the simulation,
235 U,
239 Pu, and
241 Pu, but is lost according as
its own abundance and capture cross-section for neutrons. In terms of fractional
abundances,
F
Prod−1
(t + t) = F
Prod−1
(t)
+
ν R t
σ total
F
235
σ
235
f
+ F
239
σ
239
f
+F
241
σ
241
f
− F
Prod−1
σ
Prod−1
c
.
(5.21)
Similarly, product 2 accumulates from neutron capture by product 1; there is no
loss mechanism for product 2:
F
Prod−2
(t + t) = F
Prod−2
(t) +
ν R t
σ total
F
Prod−1
σ
Prod−1
c
.
(5.22)
5 Miscellaneous Calculations
Since the cross-sections will vary in time due to the varying abundance fractions,
ν must also vary; it has to be computed afresh at each timestep. ν will increase with
time since the fission cross-section will decrease due to consumption of
235 U.
The next step is to set up expressions for how many nuclei of isotope iso undergo a
given process during time t. Nuclei of a given isotope can be created from neutron
capture by a nucleus of lower weight (if applicable), while simultaneously being
lost due to fission and capturing neutrons themselves to produce fission products or
isotopes of greater weight. Now, from above, the number of neutrons released over
Δt days will be νR Δt. If all of these neutrons are involved in some event over time
t, then the number of events that correspond to some process will be given by the
total number of events involved multiplied by the ratio of the total cross-section for
that process to the total available cross-section. Hence, for a given isotope we can
write
N
iso
(t + t) = N
iso
(t) +
ν R t
σ total
N
F
lower
σ
lower
c
− F
iso
σ
iso
f − F
iso
σ
iso
c
.
(5.19)
The simulation actually tracks fractional abundances, that is, (5.19) divided by N:
F
iso
(t + t) = F
iso
(t) +
ν R t
σ total
F
lower
σ
lower
c
− F
iso
σ
iso
f − F
i
σ
iso
c
.
(5.20)
We will need to know N, however, as it remains in σ total through (5.15). For
simplicity in determining N, it will be assumed that the fuel is initially composed
entirely of
238 U. Given that the fuel in most reactors is enriched to only a few percent
235 U, this will not be a drastic approximation.
What of the fission products? “Product 1” accumulates from fissions of the three
fissile isotopes in the simulation,
235 U,
239 Pu, and
241 Pu, but is lost according as
its own abundance and capture cross-section for neutrons. In terms of fractional
abundances,
F
Prod−1
(t + t) = F
Prod−1
(t)
+
ν R t
σ total
F
235
σ
235
f
+ F
239
σ
239
f
+F
241
σ
241
f
− F
Prod−1
σ
Prod−1
c
.
(5.21)
Similarly, product 2 accumulates from neutron capture by product 1; there is no
loss mechanism for product 2:
F
Prod−2
(t + t) = F
Prod−2
(t) +
ν R t
σ total
F
Prod−1
σ
Prod−1
c
.
(5.22)