8. Solved Problems

Example 1: Tank Dynamics for Step Change in In-flow


Solution: From balance on volumetric flow rate, \[F_i - F_o = \frac{d(Ah)}{dt} = A\frac{dh}{dt}\] i.e., \[A\frac{dh}{dt} + F_o = F_i\] Given: \(F_o= F_{\text{out}}=0.1h\) m\(^3\)/min. And, \\ \(A=V_{\text{total}}/h_{\text{total}}=0.25/1=0.25\) m\(^2\). Therefore, the above equation becomes \[0.25\frac{dh}{dt} + 0.1\,h = F_i \label{eqn200862one}\]


At initial steady state, \(dh/dt=0\) and, \(F_i=F_o=0.1\,h_o\).

Given: \(F_i=0.05\) m\(^3\)/min; Therefore, \(h_o=0.05/0.1=0.5\) m.

Rewriting the Eqn.[eqn(1)] as below: \[2.5\frac{dh}{dt} + h = 10\,F_i \label{eqn200862two}\] At the initial steady state, the above equation becomes, \[0 + h_o = 10\times F_{io}\] i.e., \[0 + 0.5 = 10\times0.05 \label{eqn200862three}\]

Eqn.[eqn(2)]\(-\) Eqn.[eqn(3)] \(\Longrightarrow\) \[2.5\frac{dh}{dt} + (h-0.5) = 10(F_i-0.05)\] Writing \((h-0.5)\) as \(\bar{h}\), and \((F_i-0.05)\) as \(\bar{F_i}\) we have \[2.5\frac{d\bar{h}}{dt} + \bar{h} = 10\bar{F_i} \qquad \left(\text{Note: }\frac{dh}{dt} = \frac{d\bar{h}}{dt} \qquad \text{ as } \quad (h-10)=\bar{h} \right)\] Taking Laplace transform for the above equation, \[2.5\,s\bar{h}(s) + \bar{h}(s) = 10\bar{F_i}(s)\] i.e., \[\frac{\bar{h}(s)}{\bar{F_i}(s)} = \frac{10}{2.5s+1} = \frac{K_p}{\tau_p s+1}\] Here, \(K_p=10\) and \(\tau_p=2.5\).

For a step change in input of magnitude \(A\), we get the response as \[\bar{h}(t) = AK_p\bigl(1-e^{-t/\tau}\bigr)\] Here, \(A=0.15-0.05=0.1\) m\(^3\)/min. Therefore, \[\bar{h}(t) = 0.1\times10\bigl(1-e^{-t/2.5}\bigr) =1-e^{-t/2.5}\] i.e., \[h-0.5=1-e^{-t/2.5}\] The tank gets filled when \(h\) reaches the \(h_{\text{total}}\) of 1 m. i.e., \[1-0.5=1-e^{-t/2.5}\] i.e, \[e^{-t/2.5} = 0.5\]

Taking logarithms on both sides, we get \[-t/2.5 = \ln(0.5) \qquad \Longrightarrow \quad t=1.733 \text{ min}\] The tank gets filled and starts to overflow, after 1.733 min from the start of change of flow rate to 0.15 m\(^3\)/min from its initial value of 0.05 m\(^3\)/min.