Fission Product Buildup and Decay

Abstract

Nuclear fission products are the atomic fragments left after a large atomic nucleus undergoes nuclear fission. Typically, a nucleus that has a large atomic mass like uranium could fission by splitting into two smaller nuclei, along with a few neutrons. This process results in release of heat energy such as kinetic energy of the nuclei and gamma rays. The fission products themselves are often unstable and radioactive, due to being relatively neutron-rich for their high atomic number, and many of them quickly undergo beta decay. This releases additional energy in the form of beta particles, antineutrinos, and gamma rays. Thus, fission events normally result in beta radiation and antineutrinos, even though these particles are not produced directly by the fission event itself.

Problems

Problem 15.1

A thermal reactor using uranium-235 as fissile materials has been operating for some time (about 6.7 h) at an average flux ϕ = 2 × 10¹⁴ neutron/(cm²)(s): How long after shutdown will the xenon poisoning reach a maximum, and what is the poisoning at this time?

Assume that the time to attain the maximum concentration of xenon (Xe) after shutdown is t_max, and it is given by:

$$ {t}_{\mathrm{max}}=\frac{1}{\lambda_X-{\lambda}_I}\ln \frac{\lambda_x}{\lambda_I}\left(1-\frac{\lambda_X-{\lambda}_I}{\lambda_I}\cdot \frac{X_0}{I_0}\right) $$

(a)

where

• λ_I = Decay constant of iodine-135

• λ_X = Decay constant of xenon-135

• X₀ = Xenon concentration at equilibrium

• I₀ = Iodine concentration at equilibrium

In addition, the poisoning ψ(t) in general of a reactor is defined as the ratio of the number of thermal neutrons absorbed by the poison to those absorbed in fuel, hence we can write:

$$ \psi \left({t}_s\right)=\frac{X\left({t}_s\right){\sigma}_X}{\sum_u} $$

(b)

where

• X(t_s) = Xenon concentration at time t_s after shutdown

• σ_X = Absorption cross section of xenon-135

• ∑_u = Fuel macroscopic absorption cross section

Additionally, the equilibrium concentration, I₀, attained after reactor has been operating for some time, is obtained by:

$$ {I}_0=\frac{\gamma_I{\Sigma}_f\phi }{\lambda_I+{\sigma}_I\phi}\approx \frac{\gamma_I{\Sigma}_f\phi }{\lambda_I} $$

(c)

where

• Σ_f = Macroscopic fission cross section

• γ_I = Fission yield for iodine-135

• ϕ = Neutron flux

• σ = Absorption cross section of iodine-135

And, similarly, the equilibrium concentration, X₀, of xenon is obtained from the following equation as well:

$$ {X}_0=\frac{\lambda_I{I}_0+{\gamma}_X{\Sigma}_f\phi }{\lambda_X^{\ast }}=\frac{\lambda_I\left(\frac{\gamma_I{\Sigma}_f\phi }{\lambda_I}\right)+{\gamma}_X{\Sigma}_f\phi }{\lambda_X^{\ast }}=\frac{\left({\gamma}_I+{\gamma}_X\right){\Sigma}_f\phi }{\lambda_X^{\ast }} $$

(d)

where ${\lambda}_X^{\ast }={\lambda}_X+{\sigma}_X\phi$; because the absorption cross section of xenon-135 is so high, ${\lambda}_X^{\ast }$ is appreciably greater than λ_X and γ_X is fission yield for xenon-135.

We can also write the xenon-135 concentration X(t_s) at time t_s after shutdown as:

$$ X\left({t}_s\right)=\frac{\lambda_I}{\lambda_X-{\lambda}_I}{I}_0\left({e}^{-{\lambda}_I{t}_s}-{e}^{-{\lambda}_X{t}_s}\right){X}_0{e}^{-{\lambda}_X{t}_s} $$

(e)

Problem 15.2

Let I and X denote the concentrations of the iodine and xenon isotopes. We then have

$$ \frac{d}{dt}I(t)={\gamma}_I{\Sigma}_f\phi -{\lambda}_II(t) $$

(a)

and

$$ \frac{d}{dt}X(t)={\gamma}_X{\Sigma}_f\phi -{\lambda}_II(t)-{\lambda}_XX(t)-{\sigma}_{aX}X(t)\phi $$

(b)

Note that no neutron absorption term, σ_aII(t)ϕ, in the Eq. (a) here, since even at high flux levels, iodine absorption is insignificant compared to its decay.

Considering now, following reactor start-up, both iodine and xenon concentrations build from zero to equilibrium values over a period of several half-lives. Since the half-lives are in hours, after a few days equilibrium is achieved. Find the final form of both Eqs. (a) and (b) for t → ∞. Variables definitions are as what are given in Problem 15.1.

Problem 15.3

Let I₀ and X₀ be the concentrations of the iodine and xenon at the time of reactor shutdown. If the reactor is put on a large negative period, to first approximation, we may assume that the shutdown is instantaneous compared to the time spans of hours over which the iodine and xenon concentration evolves. Find the solution of Eqs. (a) and (b) that are given by Problem 15.2. Variables definitions are as what are given in Problem 15.1.

Problem 15.4

Prove that the following relation is valid, if the reactor has been running for several days – long enough for iodine and xenon in order to reach equilibrium.

$$ X(t)={\Sigma}_f\phi \left[\frac{\left({\gamma}_I+{\gamma}_X\right)}{\lambda_X\_{\sigma}_{aX}\phi }{e}^{-{\lambda}_Xt}+\frac{\gamma_I}{\lambda_I-{\lambda}_X}\left({e}^{-{\lambda}_Xt}-{e}^{-{\lambda}_It}\right)\right] $$

(a)

Variables definitions are as what are given in Problem 15.1. Hint: Use the results that are given by solution of Problems 15.2 and 15.3.

Problem 15.5

Prove that for a reactor operating at a very high flux level, the maximum xenon-135 concentration takes place at approximately 11.3 h following shutdown.

Variables definitions are as what are given in Problem 15.1. Hint: Use result of solution for Problem 15.4.

Problem 15.6

Make a logarithmic plot of the effective half-life of xenon-135 over the flux range of 10¹⁰ ≤ ϕ ≤ 10¹⁵ n = cm² = s. Use the relation of λ′ = λ+σ_aϕ, where σ_a is absorption cross section, λ radioactive decay constant, and neutron flux ϕ. Note that the effective half-life xenon is given by t_1/2 = 0.693/λ.

Problem 15.7

A thermal reactor fueled with uranium has been operating at constant power for several days. Make a plot of the ratio of concentration of xenon-135 to uranium-235 atoms in the reactor versus its average flux. Determine the maximum value in this ratio. Note that the effective half-life xenon is given by t_1/2 = 0.693/λ. Hint: Use the result of solution of Problem 15.2 for xenon. Note also that uranium-235 has date of γ_I = 0.0639 and γ_X = 0.00237 and ${\sigma}_f^{25}=2.65\times {10}^6$ barns.

Problem 15.8

If the rate of formation of neutron designated by q and is taken to be constant outside the rod and zero inside, then the fraction of neutrons absorbed in single rod δk, which may be taken as the total worth of all the rods as it can be seen in figure below, and it may be written as following equation (Fig. 15.9):

Fig. 15.9

Poison cell for control rod calculation

$$ {\displaystyle \begin{array}{l}\delta k=\frac{\left(\mathrm{Exposed}\ \mathrm{primeter}\ \mathrm{of}\ \mathrm{rod}\right)\ D{\bigtriangledown}^2\phi\ \left(\mathrm{at}\ \mathrm{rod}\ \mathrm{surface}\right)}{q\ \left(\mathrm{Area}\ \mathrm{of}\ \mathrm{source}\ \mathrm{region}\right)}\\ {}\kern1em \approx \frac{4l}{{\left(m-2a\right)}^2}\cdot \frac{1}{h{\Sigma}_a+\frac{1}{L}\coth \left(\frac{m-2a}{2L}\right)}\end{array}} $$

(a)

where

• h = Linear extrapolation distance

• m = Poison cell dimension

• a = Half-thickness of cruciform control rod

• l = Half-length of arm of cruciform rod

• L = Diffusion length

• Σ_a = Macroscopic absorption cross section

Now consider that in a water-moderated reactor, in which 32 cruciform control rods are distributed evenly throughout the core, the following dimensions are applicable: l = 10.0 cm, a = 0.336 cm, and m = 19.5 cm. The thermal-neutron diffusion length in the core is 1.8 cm, and Σ_a = 0.114 cm⁻¹. Estimate the total worth of the system of control rods.