Nuclear Fuel Cycle

(a) Explain the terms “fission product inventory” and “decay heat”, and explain the importance of each in determining the hazards posed by irradiated fuel.

(b) The differential equation used to calculate the amount of a fission product $i$ in a reactor is:

$\frac{dN_i}{dt} = \sum_p \lambda_{pi} N_p + \sum_j \sigma_{ji} \phi N_j + \sum_k y_{ki} \sigma_{fk} \phi N_k - \lambda_i N_i - \sigma_i \phi N_i$

Explain how this equation would be simplified if:

(i) the reactor is shut down,

(ii) the half-life of nuclide $i$ is very long.

Solution

(a) Definitions and importance:

Fission product inventory is the total amount (expressed as activity in Bq, or as number of atoms, or as mass) of each radioactive fission product species present in the irradiated fuel at a given time. The inventory depends on the reactor power level, the duration of irradiation, the fission yields of individual isotopes, and the time since shutdown (if applicable).

Decay heat is the thermal energy released by the radioactive decay of fission products (and actinides) in the fuel after the fission chain reaction has ceased. Even after shutdown, the fuel continues to generate significant heat because the accumulated fission products continue to undergo beta and gamma decay.

Importance for hazards:

• The fission product inventory determines the radiological hazard posed by the fuel. Isotopes with high activity, volatile nature, and biological significance (e.g. $^{131}$I concentrating in the thyroid, $^{90}$Sr in bone) define the potential consequences of a release.
• Decay heat determines the cooling requirements for the fuel after shutdown. If cooling is lost, the fuel temperature will rise, potentially leading to fuel damage, cladding failure, and release of the fission product inventory. This is why irradiated fuel must be continuously cooled during storage --- in water pools or by passive air circulation in dry stores.
• Together, the inventory and decay heat dictate the shielding, cooling, containment, and emergency planning requirements at every stage of the back end of the fuel cycle.

(b)(i) After shutdown ($\phi = 0$):

When the reactor is shut down, the neutron flux $\phi$ drops to zero. All terms containing $\phi$ vanish:

$\frac{dN_i}{dt} = \sum_p \lambda_{pi} N_p - \lambda_i N_i$

This is the standard radioactive decay chain equation. Nuclide $i$ is only produced from the decay of its precursor(s), and is only removed by its own radioactive decay. There is no production from fission and no burn-up.

(b)(ii) If the half-life of nuclide $i$ is very long:

If $T_{1/2}$ is very long, then $\lambda_i$ is very small, and the terms involving $\lambda_i$ become negligible. The decay removal term $\lambda_i N_i \approx 0$ and, for the same reason, production from precursor decay of $i$ is negligible (since $i$ itself barely decays). If we further apply the standard simplifications (direct fission production only, neglect burn-up), the equation reduces to:

$\frac{dN_i}{dt} = R_i$

where $R_i = F \cdot Y_i$ (fission rate times yield). This integrates directly to:

$N_i(t) = R_i \cdot t$

The number of atoms builds up linearly with time. The corresponding activity is:

$A_i(t) = \lambda_i N_i(t) = R_i \cdot \lambda_i \cdot t$

For such a long-lived isotope, the activity also builds up steadily (linearly) with irradiation time.