9. Solved Problems

Example 2: Tank Dynamics for Step Change in In-flow (Different Solution Methods) Consider a cylindrical tank of cross sectional area 2 m\(^2\). Steady inflow of liquid to the tank is 0.015 m\(^3\)/s. Outflow (\(q_o\)) is related to the head (\(h\), in m) of liquid in the tank as \[q_o = 0.01\sqrt{h}\] At time \(t=0\), the inflow valve is closed and so there is no inflow for \(t\ge 0\). Find the time necessary to empty the tank to half the original head. Solve by: (i) direct analytical solution of differential equation, and by (ii) Laplace transform method with linearized \(h\).

Solution: At initial steady state, \(q=q_o\). Therefore, \[0.015 = 0.01\sqrt{h_s} \qquad \Longrightarrow \quad h_s = 2.25\text{ m}\] From mass balance for the constant density systems (applicable for liquids), \[\begin{aligned} q - q_o &= A\frac{dh}{dt}\\ \text{For $t\ge 0$, $q=0$. Therefore,}\\ -q_o &=A\frac{dh}{dt}\\ \text{Substituting for $q_o$ and $A$, we get}\\ -0.01\sqrt{h} &= 2\frac{dh}{dt} \end{aligned}\]

Rearranging, and integrating the above we get \[\begin{aligned} \int_{h_s}^{h_s/2}\frac{dh}{\sqrt{h}} &= -0.005\int_0^{t}dt \\ \left[\frac{h^{[(-1/2) + 1}]}{1/2} \right]_{h_s}^{h_s/2} &= 0.005 t \\ \left[2h^{1/2}\right]_{h_s}^{h_s/2} &= 0.005 t\\ \text{Substituting for $h_s=2.25$ m, we get}\\ 2\left[1.125^{1/2} - 2.25^{1/2}\right] &= 0.005 t \\ \Longrightarrow \quad t &= 175.74 \text{ s} \end{aligned}\] This result is obtained by direct solution of differential equation.

From mass balance for the constant density systems (applicable for liquids), \[\begin{align*} q - q_o &= A\frac{dh}{dt}\\ \text{At initial state,}\\ 0.015 - 0.01\sqrt{h_s} &= 0\\ \text{Subtracting the above two equations,}\\ (q-0.015) - (q_o-0.01\sqrt{h_s}) &= A\frac{d(h-h_s)}{dt}\\ \text{Using deviation variables,}\\ Q - Q_o &= A\frac{dH}{dt} \tag*{(1)}\\ \text{where $Q=q-q_s = q-0.015$, $Q_o = q_o-q_{os}=q_o-0.01\sqrt{h_s}$ and, $H=h-h_s$.} \end{align*}\]

In the above, \(Q_o = q_o - 0.01\sqrt{h_s} = 0.01\sqrt{h} - 0.01\sqrt{h_s}\) i.e., \[Q_o =0.01(\sqrt{h}-\sqrt{h_s}) \tag*{(2)}\] Since \(Q_o\) is having non-linear relation with \(h\), i.e., \(Q\propto \sqrt{h}\), we have to linearize this function before taking Laplace transform. From Taylor series expansion, for the variable \(f(x)\), around \(x_o\), and considering terms upto \(f'(x_o)\), we get \[f(x) = f(x_o) + f'(x_o)(x-x_o)\] Here, \(f(x)=\sqrt{h}\); and, \(f'(x)=\dfrac{1}{2\sqrt{h}}\). Hence, \[\sqrt{h} = \sqrt{h_o} + \frac{1}{2\sqrt{h_o}}(h-h_o)\]

For \(\sqrt{h}\) around \(h_s\), we get \[\sqrt{h} = \sqrt{h_s} + \frac{1}{2\sqrt{h_s}}(h-h_s)\]

Using this in Eqn.(2), we get \[Q_o = 0.01\left[\frac{1}{2\sqrt{h_s}}(h-h_s) \right]\] We know that \(h_s=2.25\) m. Therefore, \[Q_ o = 0.01\left[\frac{1}{2\sqrt{2.25}}(h-h_s) \right] = \frac{0.01}{3}(h-h_s) = \frac{H}{300}\] where \(H=h-h_s\).

Substituting for \(Q_o\) from above in Eqn.(1), we get \[Q - \frac{H}{300} = A\frac{dH}{dt}\] Since, \(A=2\) m\(^2\), we get \[\begin{align*} 2\frac{dH}{dt} + \frac{H}{300} &= Q \\ 600\frac{dH}{dt} + H &= 300Q\\ \text{Taking Laplace transform,} \frac{H(s)}{Q(s)} &= \frac{300}{600s+1} = \frac{K_p}{\tau_p s+1} \tag*{(2)} \end{align*}\] where \(K_p=R=300\) s/m\(^2\); and, \(\tau_p=AR=2\times300=600\) s.

For \(t\ge0\), \(Q(t)=q(t) - q_s = 0-0.015\) m\(^3\)/s. Hence, \(Q(s) = -0.015/s\). Substituting this in Eqn.(2), and taking inverse Laplace transform, we get \[\begin{aligned} H(t) &= -0.015\times300\times(1-e^{-t/600})\\ \text{Since $H(t) = h(t) - h_s = h(t) - 2.25$, we get}\\ h(t) &= 2.25-4.5\times(1-e^{-t/600})\\ \text{For $h(t)=h_s/2=2.25/2=1.125$ m, we get}\\ 1.125 &= 2.25 - 4.5\times(1-e^{-t/600}) \\ \Longrightarrow \quad t &= 172.61 \text{ s} \end{aligned}\]

Note: The time obtained by this Laplace transform method (i.e., 172.61 s) is slightly different from that obtained by direct solution method (i.e., 175.74 s). This is because of the approximation involved in Taylor series expansion.

The function \(q=0.01\sqrt{h}\) is approximated around its initial steady state of \(h_s=2.25\) m, as \(q=\frac{0.01}{2\sqrt{h_s}}h\), i.e., \(q=0.00333h\).