23-Chem-A6 Process Dynamics and Control · December 2013
Question 4 of 8: Thermocouple Lag — Response to a Triangular Input
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2013 — 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 paper is almost entirely quantitative — linearisation, transfer functions, step/pulse/ramp responses, Routh and frequency-response (Bode/Nyquist) stability, and IMC design — and every boxed number.
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — linearisation, transfer functions, frequency response, IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh array, Bode and Nyquist stability, dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order dynamics and step/pulse/ramp response. Standard control conventions are used (deviation variables about a steady state; unity valve/sensor gains unless stated).
Problem 4: Thermocouple Lag — Response to a Triangular Input (20%)
Find. (a) $T(s)/T_L(s)$; (b) $T(t)$ registered for the triangular liquid profile.
Approach. A lumped energy balance on the bead gives a unity-gain first-order lag; get $\tau$ from $mC/hA$ (watch the hour→second conversion). Then treat the triangular input as a sum of ramps and use the known first-order ramp response by superposition.
(a) Energy balance and time constant. $mC\,\dfrac{dT}{dt}=hA\,(T_L-T)$, a unity-gain lag with $\tau=\dfrac{mC}{hA}=\dfrac{0.25\times1}{60\times1}=4.167\times10^{-3}\ \mathrm{h}$. Converting, $\tau=4.167\times10^{-3}\times3600=\boxed{15\ \mathrm{s}}$, so $$\dfrac{T(s)}{T_L(s)}=\dfrac{1}{15s+1}.$$
Decompose the input into ramps. The triangle has slope $+1\ \mathrm{^{\circ}C/s}$ then $-1\ \mathrm{^{\circ}C/s}$: $T_L(t)=t-2(t-300)\mathcal{U}(t-300)+(t-600)\mathcal{U}(t-600)$. The first-order response to a unit-slope ramp is $r(t)=t-\tau(1-e^{-t/\tau})$.
(b) Response by superposition. $T(t)=r(t)-2\,r(t-300)+r(t-600)$. Explicitly, for $0\le t\le300$: $T(t)=t-15+15e^{-t/15}$; for $300\le t\le600$: $T(t)=615-t+15e^{-t/15}-30e^{-(t-300)/15}$; for $t\ge600$ the polynomial part cancels and $T(t)=15e^{-t/15}-30e^{-(t-300)/15}+15e^{-(t-600)/15}\to0$.
Interpret — a 15 s tracking lag. Because $\tau=15\ \mathrm{s}\ll300\ \mathrm{s}$, after the first few $\tau$ the reading tracks the ramp offset downward by $\tau\times\text{slope}=15\ \mathrm{^{\circ}C}$. At $t=300\,$s (liquid at its 300 °C apex) the thermocouple reads only $\boxed{285\ \mathrm{^{\circ}C}}$. It keeps rising briefly after the liquid turns down, meeting the descending liquid line at the reading’s own peak of $289.6\ \mathrm{^{\circ}C}$ at $t=310.4\,$s, then lags the fall by the same $15\ \mathrm{^{\circ}C}$ (see figure).
Problem 4(b): liquid temperature (red) and thermocouple reading (blue). The bead tracks the triangular input with a $\approx15\ \mathrm{s}$ lag — reading $285\ \mathrm{^{\circ}C}$ when the liquid is at its $300\ \mathrm{^{\circ}C}$ apex, peaking at $289.6\ \mathrm{^{\circ}C}$ at $t=310\,$s, and trailing the descent symmetrically.
