Question:

In the tank-level control system shown, the tank cross-sectional area is $A$. The flow controllers are perfect. The level controller (LC) is PI. For an integral time $\tau_I$, the controller gain $K_c$ is tuned for critical damping. The value of $\dfrac{K_c\tau_I}{A}$ is \(\underline{\hspace{2cm}}\) (rounded off to nearest integer).

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Critical damping for second-order systems requires $a^2 = 4b$. Apply this directly to PI-controlled tank systems.
Updated On: Jan 13, 2026
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Correct Answer: 4

Solution and Explanation

The linearized tank mass balance: \[ A\frac{dh}{dt}= q_{in}-q_{out}. \] With perfect flow controllers, the flow dynamics vanish. Closed-loop PI controller transfer function: \[ G_c(s)=K_c\left(1+\frac{1}{\tau_I s}\right). \] Tank transfer function: \[ G_p(s)=\frac{1}{As}. \] Open-loop transfer function: \[ L(s)=\frac{K_c}{A}\left(1+\frac{1}{\tau_I s}\right)\frac{1}{s}. \] Closed-loop characteristic equation: \[ s^2 + \frac{K_c}{A} s + \frac{K_c}{A\tau_I} = 0. \] For critical damping: \[ \left(\frac{K_c}{A}\right)^2 = 4\left(\frac{K_c}{A\tau_I}\right). \] Simplifying: \[ \frac{K_c\tau_I}{A}=4. \] Rounded to nearest integer: \[ 4. \]
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