23-Chem-A6 Process Dynamics and Control · December 2018
Question 8 of 8: Draining Tank — Step and Impulse Response of the Level
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: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion, dead-time systems and state-space models; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order tank/thermal dynamics, step/impulse 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 8: Draining Tank — Step and Impulse Response of the Level (20%)
Given. A self-regulating tank with a linear outlet resistance:
Quantity
Symbol
Value
Cross-sectional area
$A$
$1\ \mathrm{m^2}$
Initial level
$h_0$
$7$ m
Outlet law
$F_1$
$R_1h$, $R_1=4\ \mathrm{m^2/min}$
Initial steady inflow
$F_{0s}$
$R_1h_0=28\ \mathrm{m^3/min}$
Find. (a) $\delta h(t)$ for a unit step in $F_0$; (b) $\delta h(t)$ for a unit impulse in $F_0$.
Problem 8: the tank is a first-order lag with $\tau=A/R_1=0.25$ min. A step in inflow lifts the level to a new steady value $0.25$ m; an impulse jumps the level to $1$ m and it decays back to the original steady state.
Approach. Write the volume balance, linearise about the initial steady state (already linear here), form the first-order transfer function $\delta H/\delta F_0=1/(As+R_1)$, and invert for the step and impulse inputs.
Volume balance and transfer function. $A\dfrac{dh}{dt}=F_0-R_1h$. In deviation variables about $h_0=7$ ($F_{0s}=R_1h_0=28$): $A\dfrac{d\,\delta h}{dt}=\delta F_0-R_1\,\delta h$, so $$\frac{\delta H(s)}{\delta F_0(s)}=\frac{1}{As+R_1}=\frac{1}{s+4}=\frac{0.25}{0.25\,s+1},$$ i.e. gain $K=1/R_1=0.25\ \mathrm{m\,per\,(m^3/min)}$ and time constant $\tau=A/R_1=0.25$ min.
(a) Unit step. $\delta F_0(s)=1/s$, so $\delta H(s)=\dfrac{1}{s(s+4)}=\dfrac14\Big(\dfrac1s-\dfrac{1}{s+4}\Big)$, giving $$\boxed{\delta h(t)=0.25\left(1-e^{-4t}\right)\ \text{m}.}$$ The level climbs from $0$ toward a new steady rise of $0.25$ m (reaching $63\%$, i.e. $0.158$ m, after one time constant $\tau=0.25$ min).
(b) Unit impulse. $\delta F_0(s)=1$, so $\delta H(s)=\dfrac{1}{s+4}$, giving $$\boxed{\delta h(t)=e^{-4t}\ \text{m}.}$$ The impulse instantaneously adds $1\ \mathrm{m^3}$ of liquid, raising the level by $1/A=1$ m at $t=0^+$; it then drains back to the original steady state with the same time constant $\tau=0.25$ min.
Consistency check. The impulse response is the time-derivative of the step response: $\dfrac{d}{dt}[0.25(1-e^{-4t})]=e^{-4t}$, matching (b) — as it must, since an impulse is the derivative of a step.
