23-Chem-A6 Process Dynamics and Control · December 2015
Question 6 of 8: Transport Delay + Integrating Tank — Level Control
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2015 — 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. Most parts are quantitative (dynamic modelling, transfer functions, step/pulse/ramp responses, Routh and Nyquist stability, and IMC design); qualitative sketches are drawn as real figures where the paper asks for them.
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: An Introduction to Theory and Practice (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.
Problem 6: Transport Delay + Integrating Tank — Level Control (20%)
Given. Pipe $L=10\ \mathrm m$, inlet speed $u=1\ \mathrm{m/s}$ (manipulated), tank area $A=1\ \mathrm{m^2}$; pipe cross-section $a$; no tank outlet.
Find. (a) $\theta$; (b) $H/U$; (c) $H/H_{sp}$ under P control.
Problem 6: liquid enters at speed $u$ and travels the pipe in $\theta=L/u$ before reaching the tank; with no outlet the tank integrates the delayed inflow $a\,u(t-\theta)$, so $H/U$ is a delayed integrator.
Approach. Get the pure transport lag from length over velocity, write a volumetric balance treating the tank (no outlet) as an integrator fed by the delayed inflow, then close the proportional loop and reduce.
(a) Transport delay. The residence (plug-flow) time in the pipe is $$\boxed{\theta=\frac{L}{u}=\frac{10}{1}=10\ \mathrm s}.$$ A change in inlet velocity appears at the pipe outlet exactly $\theta$ seconds later.
(b) Model. The volumetric inflow to the tank is (pipe area)×(velocity), delayed by $\theta$: $q_{in}(t)=a\,u(t-\theta)$. With no outlet, a volume balance gives $A\dfrac{dh}{dt}=a\,u(t-\theta)$. In deviation variables and Laplace domain, $As\,H=a\,e^{-\theta s}U$, so $$\boxed{\frac{H(s)}{U(s)}=\frac{a}{A}\,\frac{e^{-\theta s}}{s}=\frac{a}{A}\,\frac{e^{-10s}}{s}}$$ — a delayed integrator (non-self-regulating: no steady-state level, since the tank never stops filling for a sustained inflow).
(c) Close the proportional loop. With $u=K_c(h_{sp}-h)$ (unity valve/sensor), the loop transfer is $L(s)=K_c\dfrac{a}{A}\dfrac{e^{-\theta s}}{s}$. The set-point closed loop is $\dfrac{H}{H_{sp}}=\dfrac{L}{1+L}$: $$\boxed{\frac{H(s)}{H_{sp}(s)}=\frac{K_c\frac{a}{A}\,e^{-10s}}{s+K_c\frac{a}{A}\,e^{-10s}}}.$$ Writing the loop gain $K\equiv K_c\,a/A$ compresses this to $\dfrac{K e^{-10s}}{s+K e^{-10s}}$.
Stability limit (interpretation). For an integrator-plus-delay under P control, the phase reaches $-180^\circ$ when $\omega\theta=\pi/2$, i.e. $\omega_c=\dfrac{\pi}{2\theta}$; at that frequency $|L|=K/\omega_c=1$ gives the ultimate loop gain $$K_u=\frac{\pi}{2\theta}=\frac{\pi}{20}=0.157\ \mathrm s^{-1}.$$ The loop is stable for $K_c\,a/A<0.157$; because of the delay, arbitrarily high proportional gain is not admissible even though the plant is a bare integrator.
