Draining Tank — Step and Pulse Response: 23-Chem-A6 Process Dynamics and Control, May 2013, Q2

Problem 2: Draining Tank — Step and Pulse Response (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A gravity-drained tank with a linear resistance:

Quantity	Symbol	Value
Cross-section area	$A$	$1\ \mathrm{m^2}$
Initial level	$h_0$	$7\ \mathrm{m}$
Outflow law	$F_1$	$R_1 h,\ R_1=4\ \mathrm{m^2/min}$
Pulse duration	$t_p$	$5\ \mathrm{min}$

Find. The deviation level $h'(t)=h(t)-h_0$ for (a) a unit step and (b) a unit pulse of width $t_p$ in $F_0$.

Problem 2: single gravity-drained tank, inflow $F_0$, level $h$ ($A=1\ \mathrm{m^2}$, $h_0=7\ \mathrm{m}$), linear outflow $F_1=R_1 h$ through valve $R_1=4\ \mathrm{m^2/min}$.

Approach. Write the linear tank balance, reduce it to standard first-order form to read off $\tau$ and gain $K$, then apply the step solution directly and build the pulse as the superposition of a step and a delayed negative step.

Model and standard form. $A\,\dfrac{dh}{dt}=F_0-R_1 h$. In deviation variables about the initial steady state ($F_{0s}=R_1 h_0=4\times7=28\ \mathrm{m^3/min}$): $\dfrac{dh'}{dt}=F_0'-4h'$. Dividing by $R_1$ puts it in standard form with $$\tau=\dfrac{A}{R_1}=0.25\ \mathrm{min},\qquad K=\dfrac{1}{R_1}=0.25,$$ so $\dfrac{H'(s)}{F_0'(s)}=\dfrac{K}{\tau s+1}=\dfrac{0.25}{0.25 s+1}=\dfrac{1}{s+4}.$
(a) Unit-step response. For $F_0'=1$ (a $1\ \mathrm{m^3/min}$ step), the first-order step response is $$h'(t)=K\!\left(1-e^{-t/\tau}\right)=0.25\left(1-e^{-4t}\right).$$ The level rises from $7\ \mathrm{m}$ toward $\boxed{h(\infty)=7.25\ \mathrm{m}}$ with time constant $0.25\ \mathrm{min}$ (~99% settled in $5\tau\approx1.25\ \mathrm{min}$).
(b) Unit-pulse response — superposition. A unit pulse of width $t_p$ is a unit step at $t=0$ minus a unit step at $t=t_p$. By linearity, for $0\le t\le t_p$ the response equals the step result, $h'(t)=0.25(1-e^{-4t})$; for $t>t_p$ the delayed negative step is added: $$h'(t)=0.25\Big[e^{-4(t-5)}-e^{-4t}\Big],\qquad t>5\ \mathrm{min}.$$
Interpret the pulse. Because $\tau=0.25\ \mathrm{min}\ll t_p=5\ \mathrm{min}$, the level essentially reaches the new steady state during the pulse: at $t=t_p$, $h'(5)=0.25(1-e^{-20})\approx\boxed{0.25\ \mathrm{m}}$ (peak, $h\approx7.25\ \mathrm{m}$). After the pulse ends the tank drains back to $7\ \mathrm{m}$ with the same $0.25\ \mathrm{min}$ time constant — a near-rectangular rise-and-decay bump of height $0.25\ \mathrm{m}$.

Quantity	Value
Time constant $\tau=A/R_1$	$0.25\ \mathrm{min}$
Gain $K=1/R_1$	$0.25$
(a) Step response	$h(t)=7+0.25(1-e^{-4t})$; $h(\infty)=7.25\ \mathrm{m}$
(b) Pulse peak at $t=5$	$h'\approx0.25\ \mathrm{m}$, then decays to $0$ as $0.25\,e^{-4(t-5)}$

Concept & Theory

Topic. First-order self-regulating tank; response to step and pulse forcing by superposition.

Key relations. Standard first-order lag $\tau\,dy/dt+y=Ku$; step response $y=K\Delta u(1-e^{-t/\tau})$; a pulse of width $t_p$ = step $-$ (step delayed by $t_p$), so $y_{\text{pulse}}(t)=y_{\text{step}}(t)-y_{\text{step}}(t-t_p)\,\mathcal{U}(t-t_p)$.

Why this works. The system is linear and time-invariant, so any input decomposed into shifted steps yields a response that is the same combination of shifted step responses. When $\tau\ll t_p$ the tank fully "charges" during the pulse, giving a flat-topped bump; when $\tau\gtrsim t_p$ the level would never reach $K\Delta u$ before the pulse ends.

Common pitfall. Using the square-root valve law here — the outflow is explicitly linear ($F_1=R_1 h$), so no linearisation is needed — or forgetting the delayed-step term when writing the post-pulse decay.

Source. Coughanowr & LeBlanc, Ch. 4–6 (first-order response, forcing functions); Seborg et al., Ch. 5 (dynamic behaviour of first-order systems).

Question 2 of 8: Draining Tank — Step and Pulse Response

Problem 2: Draining Tank — Step and Pulse Response (20%)