Chemegate

GATE 2024 Chemical Engineering, Q63 · Process Control

The correct answer is 0.32 to 0.34.

Question

Consider the surge drum in the figure. Initially the system is at steady-state with a hold-up =5m3,= 5 m³, which is 50% of full tank capacity, Vfull,V_{full}, and volumetric flow rates F̄_in = F̄_out =1m3/h.= 1 m³/h. The high hold-up alarm limit Vhigh=0.8×VfullV_{high} = 0.8 \times V_{full} while the low hold-up alarm limit Vlow=0.2×Vfull.V_{low} = 0.2 \times V_{full}. A proportional (P-only) controller manipulates the outflow to regulate the hold-up V as Fout=Kc(VVˉ)+F_{out} = K_{c}(V−V̄) + F̄_out. At t=0, FinF_{in} increases as a step from 1m3/h1 m³/h to 2m3/h.2 m³/h. Assume linear control valves and instantaneous valve dynamics. Let Kc,minK_{c,min} be the minimum controller gain that ensures V never exceeds Vhigh.V_{high}. The value of Kc,min,K_{c,min}, in h1,h⁻¹, rounded off to 2 decimal places, is _________

Figure for Q63 (Process Control, Level Control): Consider the surge drum in the figure. Initially the system is at steady-state with a hold-up V̄ = 5…

Official answer: 0.32 to 0.34

Why

GATE 2024 official key: 0.32 to 0.34.Vfull=10m3,0.34. V_{full}=10 m³, so Vhigh=8,V_{high}=8, giving allowable rise VhighVˉ=3m3.V_{high}−V̄=3 m³. The step response V(t)−V̄ =(ΔFin/Kc)(1e(Kc= (\Delta F_{in}/K_{c})(1−e^(−K_{c} t)) approaches a maximum offset of ΔFin/Kc=1/Kc\Delta F_{in}/K_{c} = 1/K_{c} as t.t→\infty . Requiring 1/Kc31/K_{c} \leq 3 gives Kc,min=1/30.33h1.K_{c,min} = 1/3 \approx 0.33 h⁻¹.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.