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23-Chem-A6 Process Dynamics and Control · December 2014

Question 5 of 8: Ultimate Gain of a Third-Order Loop with Sensor Delay

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams / EGBC — December 2014 — 04-Chem-A6 Process Dynamics & Control. Three-hour open-book examination; any non-communicating calculator is permitted. Eight problems are printed and any five constitute a complete paper (each worth 20%); all eight are solved below for completeness. The parts are quantitative throughout — two-capacitance thermal modelling, state-space transfer functions, IMC design with a right-half-plane zero, Nyquist stability of an open-loop-unstable plant, ultimate gain with sensor dead time, second-order damping regimes, a non-linear CSTR, and Bode gain-margin design — and every requested plot (step response, IMC servo response, Nyquist locus, damping family, Bode diagram) is drawn as a real figure.

Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist and Bode stability, dead-time systems and controller design; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first- and second-order dynamics, linearisation of non-linear balances. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout.

Problem 5: Ultimate Gain of a Third-Order Loop with Sensor Delay (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Loop $L=\dfrac{k_c\,H}{(s+1)^3}$; (a) $H=1$, (b) $H=e^{-0.7s}$.

Find. The ultimate (maximum) gain $k_{c,\max}$ in each case (phase-crossover condition $\angle L=-180^\circ$, $|L|=1$).

Approach. Find the phase-crossover frequency $\omega_{co}$ where $\angle L=-\pi$, then set $|L(j\omega_{co})|=1$ to solve for $k_c$; the pure delay adds phase but no magnitude.

  1. (a) $H=1$. Phase $-3\arctan\omega=-\pi\Rightarrow\arctan\omega=60^\circ\Rightarrow\omega_{co}=\sqrt3$. There $|(1+j\omega)^3|=(\sqrt{\omega^2+1})^3=(\sqrt4)^3=8$, so $|L|=1$ gives $$\boxed{k_{c,\max}=8.}$$
  2. (b) $H=e^{-0.7s}$ — phase-crossover equation. The delay adds $-0.7\omega$, so $3\arctan\omega+0.7\omega=\pi$. This is transcendental; solve $g(\omega)=3\arctan\omega+0.7\omega-\pi=0$ (bisection/Newton), showing only the first three iterations:
    • Iteration 1: $\omega=1.00$: $g=3(0.7854)+0.700-\pi=-0.085$ (too small).
    • Iteration 2: $\omega=1.05$: $g=3(0.8098)+0.735-\pi=+0.023$ (too large).
    • Iteration 3: $\omega=1.04$: $g=3(0.8058)+0.728-\pi=+0.003$ — converged to $\omega_{co}\approx1.039$.
  3. (b) Ultimate gain. The delay leaves the magnitude unchanged, so $|L|=\dfrac{k_c}{(\sqrt{\omega_{co}^2+1})^3}=1$ gives $$\boxed{k_{c,\max}=(\omega_{co}^2+1)^{3/2}\approx3.0.}$$ The 0.7-unit sensor delay slashes the ultimate gain from 8 to about 3 — nearly threefold — because it drives the phase to $-180^\circ$ at a lower frequency where the process gain is still large.
Case$\omega_{co}$$k_{c,\max}$
(a) $H=1$$\sqrt3=1.732$$8$
(b) $H=e^{-0.7s}$$1.039$$\approx3.0$