Question 10.8: A reactor has operated for several weeks at constant power, ......
A reactor has operated for several weeks at constant power, reaching the equilibrium concentrations of iodine and xenon I_0 and X_0 given by Eqs. (10.14) and (10.15). At t = 0 the power is cut back, dropping the flux level from \phi to \tilde{\phi} . Solve Eqs. (10.12) and (10.13) and show that the iodine and xenon concentrations following the power reduction are
I(t)=\frac{\gamma_{I}}{\lambda_{I}}\overline{{{\Sigma}}}_{f}\Bigl[\tilde{\phi}+(\phi-\tilde{\phi})e^{-\lambda_{I}t}\Bigr]
and
X(t)=\frac{(\gamma_{I}+\gamma_{X})}{\lambda_{_X}+\sigma_{a X}\phi}\overline{{{{\Sigma}}}}_{f}\phi e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t}+\frac{(\gamma_{I}+\gamma_{X})}{\lambda_{_X}+\sigma_{a X}\tilde{\phi}}\overline{{{\Sigma}}}_{f}\tilde{\phi}\Big[1-e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t}\Big]
+\frac{\gamma_{I}}{\lambda_{X}-{\lambda_{I}}+\sigma_{a X}\tilde{\phi}}\overline{\Sigma}_{f}(\phi-\tilde{\phi})\biggl[e^{-\lambda_{I}t}-e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t}\biggr]
Apply the integrating factor \mathrm{exp}(\lambda_{I}t) to Eq. (10.12):
\frac{d}{d t}\Bigl[{I}(t)\exp({\lambda}_{I}t)\Bigr]=\gamma_{I}\Sigma_{f}\tilde{\phi}\exp({\lambda}_{I}t) and integrate between 0 and t:
{I}(t)\exp(\lambda_{I}t)-{I}(0)=\gamma_{I}\Sigma_{f}\tilde{\phi}\int_0^t\ \exp(\lambda_{I}t^{\prime})d t^{\prime}=\gamma_{I}\Sigma_{f}\tilde{\phi}\frac{{\exp}(\lambda_{I}t)-1}{\lambda_{I}}
Using Eq. (10.14) for the initial condition, we have
{I}(t)=\frac{\gamma_{I}\Sigma_{f}\phi}{\lambda_{I}}\exp(-\lambda_{I}t)+\frac{\gamma_{I}\Sigma_{f}\tilde{\phi}}{\lambda_{I}}[1-\exp(-\lambda_{I}t)]
which reduces to the required equation.
Apply the integrating factor \exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t] to Eq. (10.13):
\frac{d}{d t}\Bigl\{X(t)\exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t]\Bigr\}=\Bigl[\gamma_{X}\Sigma_{f}\tilde{\phi}+\lambda_{I}I(t)\Bigr]\exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t]
Inserting I(t):
\frac{d}{d t}\{X(t)\exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t]\}=\Bigl[(\gamma_{X}+\gamma_{I})\Sigma_{f}\tilde{\phi}+\gamma_{I}\Sigma_{f}(\phi-\tilde{\phi})\exp(-\lambda_{I}t)\Bigr]\exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t]
Integrate between 0 and t:
X(t)\exp[(\lambda_{X}+\sigma_{a X}{\tilde{\phi}})t]-X(0)
=\int_{0}^{t}\ \left\{\left[(\gamma_{X}+\gamma_{I})\Sigma_{f}\tilde{\phi}+\gamma_{I}\Sigma_{f}(\phi-\tilde{\phi})\exp(-\lambda_{I}t^{\prime})\right]\exp[(\lambda_{X}+\sigma_{a X}\tilde{\phi})t^{\prime}]\right\}d t^{\prime}
=\frac{(\gamma_{X}+\gamma_{I})\Sigma_{f}\tilde{\phi}(e^{(\lambda_{X}+\sigma_{aX}\tilde{\phi})t}-1)}{\lambda_{X}+\sigma_{a X}\tilde{\phi}}+\frac{\gamma_{I}\Sigma_{f}(\phi-\tilde{\phi})(e^{-(\lambda_{X}-\lambda_{I}+\sigma_{aX}\tilde{\phi})t}-1)}{\lambda_{X}-\lambda_{I}+\sigma_{a X}\tilde{\phi}}
Replace X(t) with Eq. (10.15) and solve:
X(t)=\frac{(\gamma_{X}+\gamma_{I})\Sigma_{f}\phi}{\lambda_{X}+\sigma_{a X}\phi}e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t}
=\frac{(\gamma_{X}+\gamma_{I})\Sigma_{f}\tilde{\phi}}{\lambda_{X}+\sigma_{a X}\tilde{\phi}}[1-e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t}]+\frac{\gamma_{I}\Sigma_{f}(\phi-\tilde{\phi})}{\lambda_{X}-\lambda_{I}+\sigma_{a X}\tilde{\phi}}(e^{-\lambda_{I}t}-e^{-(\lambda_{X}+\sigma_{a X}\tilde{\phi})t})
Which is the required equation.