23-Chem-A6 Process Dynamics and Control · December 2019
Question 4 of 8: First-Order Thermometer — Response to a Ramping Bath
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
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, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 4: First-Order Thermometer — Response to a Ramping Bath (20%)
Find. (a) the error $e=T_b-T_m$ at $t=0.1$ and $10$ min; (b) the maximum error and its timing; (c) the ultimate time lag.
Problem 4(c): the bath rises as a straight ramp; the thermometer reading is a parallel ramp displaced downward and to the right. The vertical gap grows to the steady error $a\tau=0.2,{}^ rc$C, equivalent to a constant time lag of $\tau=0.2$ min.
Approach. For a first-order lag the tracking error to a ramp has the closed form $e(t)=a\tau\!\left(1-e^{-t/\tau}\right)$; evaluate it at the two times, take its monotone asymptote for the maximum, and read the ultimate lag from the steady offset divided by the slope.
Error transfer function. The bath is the input $T_b=at$, so $T_b(s)=a/s^2$. The reading follows $T_m(s)=\dfrac{1}{\tau s+1}T_b(s)$, hence the error $$E(s)=T_b(s)-T_m(s)=\Big(1-\tfrac{1}{\tau s+1}\Big)\frac{a}{s^2}=\frac{a\tau}{s(\tau s+1)}.$$
Invert to the time domain. Partial fractions give $\dfrac{a\tau}{s(\tau s+1)}=\dfrac{a\tau}{s}-\dfrac{a\tau^{2}}{\tau s+1}$, so $$\boxed{e(t)=a\tau\left(1-e^{-t/\tau}\right)}=0.2\left(1-e^{-t/0.2}\right).$$ The error starts at zero and rises monotonically toward $a\tau$.
(a-i) At $t=0.1$ min. $e=0.2\left(1-e^{-0.1/0.2}\right)=0.2\left(1-e^{-0.5}\right)=0.2(0.3935)=\boxed{0.0787,{}^ rc\mathrm C}$ — the reading trails the bath by about $0.08,{}^ rc$C.
(a-ii) At $t=10$ min. $e=0.2\left(1-e^{-10/0.2}\right)=0.2\left(1-e^{-50}\right)\approx\boxed{0.200,{}^ rc\mathrm C}$; fifty time constants have elapsed, so the transient is dead and the error sits at its asymptote.
(b) Maximum deviation. Because $e(t)=a\tau(1-e^{-t/\tau})$ increases monotonically (its derivative $ae^{-t/\tau}$ is always positive), the largest deviation is the limit $$\boxed{e_{\max}=a\tau=0.2,{}^ rc\mathrm C}\quad\text{approached as } t\to\infty.$$ It is not a finite-time peak — the ramp keeps the sensor permanently behind, and the offset saturates at $a\tau$ (already reached to four figures by $t=10$ min).
(c) Ultimate time lag. At steady state the reading is $T_m=at-a\tau=a(t-\tau)$: an identical ramp shifted right by a fixed time. The lag is the offset divided by the slope, $$\boxed{\Delta t=\frac{a\tau}{a}=\tau=0.2\ \text{min}.}$$ The thermometer forever reports the temperature the bath had $0.2$ min earlier.
