GATE 2024 CH – Question 63

Instrumentation and Process Control · Controller modes, control valves and tuning · 2 marks · Numerical answer

Consider the surge drum in the figure. Initially the system is at steady-state with a hold-up $\bar{V} = 5$ m$^3$, which is 50% of full tank capacity, $V_{full}$, and volumetric flow rates $\bar{F}_{in} = \bar{F}_{out} = 1$ m$^3$ h$^{-1}$. The high hold-up alarm limit $V_{high} = 0.8V_{full}$ while the low hold-up alarm limit $V_{low} = 0.2V_{full}$. A proportional (P-only) controller manipulates the outflow to regulate the hold-up $V$ as $F_{out} = K_c(V - \bar{V}) + \bar{F}_{out}$. At $t = 0$, $F_{in}$ increases as a step from 1 m$^3$ h$^{-1}$ to 2 m$^3$ h$^{-1}$. Assume linear control valves and instantaneous valve dynamics. Let $K_c^{min}$ be the minimum controller gain that ensures $V$ never exceeds $V_{high}$. The value of $K_c^{min}$, in h$^{-1}$, rounded off to 2 decimal places, is _________

a surge drum with the inlet $F_{in}$ at the top and the outlet $F_{out}$ through a control valve operated by a level controller LC. The levels $V_{high}$, $\bar{V}$ and $V_{low}$ are marked, with $\bar{V} = 5$ m³.

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Correct answer: 0.32 to 0.34

Explanation

The full capacity is $V_{full} = \frac{5}{0.5} = 10$ m³, so $V_{high} = 8$ m³, which is 3 m³ above $\bar{V}$. The balance is $\frac{dV}{dt} = F_{in} - F_{out} = 2 - K_c(V - \bar{V}) - 1 = 1 - K_c(V - \bar{V})$. This first-order response approaches its final value $V - \bar{V} = \frac{1}{K_c}$ without any overshoot. The hold-up never exceeds $V_{high}$ if $\frac{1}{K_c} \le 3$, so $K_c^{min} = \frac{1}{3} = 0.33$ h⁻¹.