Chemegate

GATE 2025 Chemical Engineering, Q60 · Process Control

The correct answer is 5.6 to 5.7.

Question

Consider a process with transfer function Gp=2es/(5s+1)2.G_{p} = 2e^{−s} / (5s+1)². A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let τmτ_{m} and θmθ_{m} denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of τm/θmτ_{m}/θ_{m} is ____ (rounded off to 2 decimal places). Given (as shown in the figure): for G =1/(τs+1)2,= 1/(τs+1)², the unit step output response y(t) =1(1+t/τ)et/τ,= 1−(1+t/τ)e^{−t/τ}, dy(t)/dt =(t/τ2)et/τ,= (t/τ²)e^{−t/τ}, and d2y(t)/dt2=(1/τ2)(1t/τ)et/τ.d²y(t)/dt² = (1/τ²)(1−t/τ)e^{−t/τ}.

Figure for Q60 (Process Control, FOPDT Modeling): Consider a process with transfer function G_p = 2e^(−s) / (5s+1)². A first-order plus dead time (FOP…

Official answer: 5.6 to 5.7

Why

GATE 2025 official key: 5.60 to 5.70. The inflection point occurs at t =θ0+τ=6,= θ₀+τ = 6, with maximum slope Ke1/τ.K\cdot e⁻¹/τ. Constructing the tangent line there gives θp2.41θ_{p} \approx 2.41 and τp13.59,τ_{p} \approx 13.59, so τp/θp5.64.τ_{p}/θ_{p} \approx 5.64.

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