Question 10.9: Under load following conditions a reactor operates each day ......

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

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

Integrate between 0 and t:

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

Thus

I(t)=I(0)e^{-\lambda_{I}t}+\frac{\gamma_{I}\Sigma_{f}\phi}{\lambda_{I}}(1-e^{-\lambda_{I}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

I(t)=I(12)e^{-\lambda_{I}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:

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

Solve for I(0):

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

Thus

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

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

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