Question 10.10: Under load following conditions a reactor operates each day ......

Apply the integrating factor \mathrm{exp}(\lambda_{P}t) to Eq. (10.12)

\left[{\frac{d}{d t}}P(t)+\lambda_{P}P(t)\right]\exp(\lambda_{I}t)={\frac{d}{d t}}\{P(t)\exp(\lambda_{P}t)\}=\gamma_{P}\Sigma_{f}\phi\exp(\lambda_{P}t)

Integrate between 0 and t:

P(t)\exp(\lambda_{P}t)-P(0)=\int_0^{t}\gamma_{P}\Sigma_{f}\phi\exp(\lambda_{P}t^{\prime})d t^{\prime}={\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}}[\exp(\lambda_{P}t)-1]

Thus

P(t)=P(0)e^{-\lambda_{I}t}+\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}(1-e^{-\lambda_{P}t})\ \ \mathrm{for}\ 0\lt t\lt 12\mathrm{~hr}.

Let t^{\prime} = t  –  12  \mathrm{hr}. Then while the reactor is shut down, no iodine is produced and thus

P(t)=P(12)e^{-\lambda_{P}t^{\prime}}\quad\qquad{\mathrm{for~}}12 \ \ \mathrm{h}\mathrm{r.\lt }t\lt 24\mathrm{~h}\mathrm{r.}

Now using the periodic boundary condition:

P(0)=P(24)=P(12)e^{-\lambda_{P}12}=\left\{P(0)e^{-\lambda_{I}12}+\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}(1-e^{-\lambda_{P}12})\right\}e^{-\lambda_{P}12}\ .

Solve for P(0):

P(0)=\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}\frac{(1-e^{-\lambda_{P}12})}{(1-e^{-\lambda_{P}24})}e^{-\lambda_{P}12}=\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}\frac{1}{(e^{\lambda_{P}12}+1)}

Thus

P(t)=\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}\frac{1}{\left(e^{\lambda_{P}12}+1\right)}e^{-\lambda_{P}t}+\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}(1-e^{-\lambda_{P}t})\ \ \mathrm{for}\ 0\lt t\lt 12\mathrm{~hr}.

P(t)=\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}\frac{e^{\lambda_{P}1{2}}}{\left(e^{\lambda_{P}12}+1\right)}e^{-\lambda_{P}t^{\prime}}\quad\mathrm{for}\,0\lt t^{\prime}\lt 12\,\mathrm{h}\mathrm{r}.,\,\,\,\,\mathrm{or}\,

P(t)=\frac{\gamma_{P}\Sigma_{f}\phi}{\lambda_{P}}\frac{e^{\lambda_{P}24}}{\left(e^{\lambda_{P}12}+1\right)}e^{-\lambda_{P}t}\quad\mathrm{for~}12\ \mathrm{hr.}\lt t\lt 24\mathrm{~hr}.