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The balance on tanks (1 & 2) induding linear rexutance wiol give:\begin{array}{l}\text { at skady stater } \left.\begin{array}{rl}\left(q-q_{12}\right) & =\left(A_{1}\right)(0)=0 \\\& & \left(q-q_{2 s}\right)=0\end{array}\right\} \\\end{array}scolving all equations givatTank-ti \left(Q-Q_{1}\right)=\left(A_{1}\right)\left(d H_{1} / d t\right)-1valve-t?value-2, \quad Q_{2}=\left(\mathrm{H}_{2} \mid \mathrm{R}_{2}\right)Taking Laplace tranforms of (1) (2), (3) & (4) give:We have doriveds\frac{H_{2}(l)}{Q(e)}=\left[\left(T_{1} T_{2}\right) x^{2}+\left(T_{1}+T_{2}+A_{1} R_{2}\right) s+1\right]where, \left(\tau_{1}=A_{1} R_{1}\right) ;\left(\tau_{2}=A_{2} R_{2}\right)given,\begin{array}{l}A_{1}=1 \mathrm{~m}^{2} ; \quad A_{2}=0.5 \mathrm{~m}^{2} ; \quad R_{1}=0.5 \mathrm{hm}^{-2} ; \quad R_{2}=2 \mathrm{hm}^{-2} \\T_{1}=A_{1} R_{1}=(1 \times 0.5)=0.5 \mathrm{~h} \\T_{2}=A_{2} R_{2}=(0.5 \times 2)=\underline{1.0 h} \\\Leftrightarrow\left[\frac{\overline{H_{2}(s)}}{\bar{Q}(s)}=\frac{(2)}{\left[(0.5) s^{2}+(3.5) s+1\right]}\right. \\\end{array}Transfer function\overline{Q(8)} ...