Question 10.14: Thorium-232 is a fertile material the may be transmuted to f......
Thorium-232 is a fertile material the may be transmuted to fissile uranium-233 through the following reaction {}_{90}^{232}\mathrm{Th} \overset{\mathrm{n}}{\underset{}{\longrightarrow}}{}_{90}^{233}\mathrm{Th} \overset{\mathrm{\beta}}{\underset{22 \ \mathrm{min}}{\longrightarrow}}{}_{91}^{233}\mathrm{Pa} \overset{\mathrm{\beta}}{\underset{27.4 \ \mathrm{days}}{\longrightarrow}}{}_{92}^{233}\mathrm{U} where the half lives are indicated. Assume that a fresh core is put into operation containing only thorium-232 and uranium-235. Thereafter neutron capture in thorium takes place at a constant rate of \Sigma_{a}^{t h}\overline{{{\phi}}};
a. Assuming that the half-life _{90}^{233}T h can be ignored, write down and solve the differential equation for the concentration of _{91}^{233}Pa
b. Write down and solve the differential equation for the concentration of _{92}^{233}U
Part a. Assume that N^{02}(t)\approx N^{02}(0)
{\frac{d}{d t}}N^{13}(t)=\sigma_{\gamma}^{02}N^{02}(0)\phi-\lambda^{13}N^{13}(t)
Using an integrating factor of \exp(\lambda^{13}t)\,,\,\mathrm{and}\ N^{13}(0)=0\,, we obtain
N^{13}(t)=\frac{\sigma_{\gamma}^{02}N^{02}(0)\phi}{\lambda^{13}}\Bigl[1-\exp(-\lambda^{13}t)\Bigr]
Part b:
{\frac{d}{d t}}N^{23}(t)={\lambda}^{13}N^{13}(t)-\sigma_{a}^{23}\phi(t)N^{23}(t)
Substituting in { N}^{13}(t), we obtain
\frac{d}{d t}\,N^{23}(t)=\sigma_{\gamma}^{02}N^{02}(0)\phi\!\left[1-\exp(-\lambda^{13}t)\right]-\sigma_{a}^{23}\phi(t)N^{23}(t)
Employing an integrating factor of \mathrm{\exp}(\sigma_{a}^{23}\phi t)\mathrm{~and~}N^{23}(t)=0, we have,
N^{23}(t)=\exp(-\sigma_{a}^{23}\phi t)\sigma_{\gamma}^{02}N^{02}(0)\phi{\int_{0}}^{t}~\left\{\exp(\sigma_{a}^{23}\phi t^{\prime})-\exp[(\sigma_{a}^{23}\phi-\lambda^{13})t^{\prime}]\}d t^{\prime}\right\}
N^{23}(t)=\exp(-\sigma_{a}^{23}\phi t)\sigma_{\gamma}^{02}N^{02}(0)\phi\left\{\frac{\exp(\sigma_{a}^{23}\phi t)-1}{\sigma_{a}^{23}\phi}-\frac{\exp[(\sigma_{a}^{23}\phi-\lambda^{13})t]-1}{\sigma_{a}^{23}\phi-\lambda^{13}}\right\}
N^{23}(t)=(\sigma_{\gamma}^{02}/\sigma_{a}^{23})N^{02}(0)\left\{1-\frac{\sigma_{a}^{23}\phi\exp(-\lambda^{13}t)-\lambda^{13}\exp(-\sigma_{a}^{23}\phi t)}{\sigma_{a}^{23}\phi-\lambda^{13}}\right\}
Note that
N^{23}(\infty)/N^{02}(0)=(\sigma_{\gamma}^{02}\;/\,\sigma_{a}^{23}) is independent of the flux.