23-Chem-A6 Process Dynamics and Control · May 2017
Question 4 of 8: Liquid-Level Dynamics — Resistance Valve vs. Exit Pump
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2017 — 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 entirely quantitative — dynamic modelling, transfer functions, step/ramp responses, Routh and Nyquist stability, frequency-response (Bode) design and IMC — and qualitative sketches are drawn as real figures wherever 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 4: Liquid-Level Dynamics — Resistance Valve vs. Exit Pump (20%)
Given. Two single tanks with a common initial steady state:
Quantity
Symbol
Value
Cross-sectional area
$A$
$1\,\mathrm{m^2}$
Initial level / flow
$h_s,\,q_s$
$1\,$m, $1\,\mathrm{m^3/s}$
Valve coefficient (Case I)
$R$
$1$
Inlet step
$\Delta q_i$
$1\to2$, i.e. $+1\,\mathrm{m^3/s}$
Find. (a) $\delta h/\delta q_i$ for Case I (resistance valve) and Case II (pump); (b) the level response $\delta h(t)$ for each.
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Approach. Write the unsteady mass balance $A\,dh/dt=q_i-q$ for each tank, linearise the outlet law about the initial steady state to get a transfer function, then invert the unit-step response.
(a) Case I — self-regulating tank. The balance is $A\,dh/dt=q_i-R\sqrt{h}$. Linearising the outlet about $h_s=1$: $q\approx q_s+\dfrac{R}{2\sqrt{h_s}}\,\delta h=1+0.5\,\delta h$. In deviation form $A\,\dfrac{d\,\delta h}{dt}=\delta q_i-0.5\,\delta h$, giving $$\boxed{\frac{\delta h}{\delta q_i}=\frac{1}{As+R/(2\sqrt{h_s})}=\frac{1}{s+0.5}=\frac{2}{2s+1}}\quad(K=2,\ \tau=2\,\text{s}).$$ The valve resistance provides negative feedback ($+0.5\,\delta h$ opposes a rise), so the tank has a finite gain and settles.
(a) Case II — integrating tank. The pump fixes the outlet at its set value, $q=q_s$ (constant), so $\delta q=0$ and $A\,\dfrac{d\,\delta h}{dt}=\delta q_i$. Hence $$\boxed{\frac{\delta h}{\delta q_i}=\frac{1}{As}=\frac{1}{s}},$$ a pure integrator — there is no level feedback at all.
(b) Case I response. With $\delta q_i=1/s$, $\delta h=\dfrac{2}{s(2s+1)}$, so $$\delta h_I(t)=2\left(1-e^{-t/2}\right)\ \text{m}.$$ The level climbs from $1\,$m and approaches a new steady state $h=1+2=3\,$m (reaching $63\,\%$ of the rise, $\delta h=1.26\,$m, at $t=\tau=2\,$s).
(b) Case II response. With $\delta q_i=1/s$, $\delta h=\dfrac{1}{s^2}$, so $$\delta h_{II}(t)=t\ \text{m}.$$ The level ramps upward at $1\,$m/s indefinitely: with the pump removing exactly the original $1\,\mathrm{m^3/s}$, the extra $1\,\mathrm{m^3/s}$ inlet simply accumulates and the tank eventually overflows. See the figure.
