Step and Impulse Response of a Linear Draining Tank: 20-Bio-A2 Process Dynamics and Control, December 2019, Q7

Problem 7: Step and Impulse Response of a Linear Draining Tank (20% total)

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. Single tank, cross-section $A=1\ \text{m}^2$, initial (steady-state) level $h_0=7\ \text{m}$, outlet law $F_1=R_1h$ (already linear — no linearization needed), $R_1=4\ \text{m}^2/\text{min}$; initial steady state has $F_0=F_1=R_1h_0=28\ \text{m}^3/\text{min}$.

Find. $\delta h(t)$ for (a) a unit step in $F_0$; (b) a unit impulse in $F_0$.

Single draining tank: inlet $F_0$ (manipulated/disturbance), linear outlet resistance $F_1=R_1h$, level $h$, cross-section $A$.

Approach. Write the unsteady-state mass balance $A\,dh/dt=F_0-F_1$, substitute $F_1=R_1h$ (already linear, so this is an exact model, not a linearization), Laplace-transform in deviation variables to get a standard first-order lag, then apply the known step and impulse responses.

Mass balance and transfer function. $A\dfrac{dh}{dt}=F_0-R_1h\Rightarrow A\dfrac{dh'}{dt}+R_1h'=F_0'$ (deviation variables), so $$\boxed{\frac{H'(s)}{F_0'(s)}=\frac{1}{As+R_1}=\frac{1/R_1}{(A/R_1)s+1}=\frac{0.25}{0.25s+1},\qquad \tau=\frac{A}{R_1}=0.25\ \text{min},\ \ K=\frac{1}{R_1}=0.25\ \frac{\text{min}}{\text{m}^2}.}$$
(a) Unit step in $F_0$. Standard first-order step response $h'(t)=K(1-e^{-t/\tau})$: $$\boxed{h'(t)=0.25\left(1-e^{-4t}\right)\ \text{m}\qquad(t\ \text{in min}).}$$ At $t=0.25\ \text{min}$ (one time constant): $h'=0.25(1-e^{-1})=0.158\ \text{m}$; as $t\to\infty$: $h'\to0.25\ \text{m}$ (new steady level $=7.25\ \text{m}$ for a $1\ \text{m}^3/\text{min}$ step increase in $F_0$).
(b) Unit impulse in $F_0$. Standard first-order impulse response $h'(t)=(K/\tau)e^{-t/\tau}$: $$\boxed{h'(t)=\frac{0.25}{0.25}e^{-4t}=e^{-4t}\ \text{m}\qquad(t\ \text{in min}).}$$ At $t=0$: $h'=1\ \text{m}$ (the impulse instantaneously adds a unit volume of liquid, raising the level by $1\ \text{m}$ since $A=1\ \text{m}^2$, before it drains away); at $t=0.25\ \text{min}$: $h'=e^{-1}=0.368\ \text{m}$; as $t\to\infty$: $h'\to0$ (returns exactly to the original $7\ \text{m}$ steady level, since an impulse carries no net change to the long-run inflow).

Result	Value
Time constant	$\tau=A/R_1=0.25\ \text{min}$
Steady-state gain	$K=1/R_1=0.25\ \text{min/m}^2$
Step response	$h'(t)=0.25(1-e^{-4t})\ \text{m}$
Impulse response	$h'(t)=e^{-4t}\ \text{m}$

Concept & Theory

Topic. First-order dynamic response of a single liquid-level tank with a linear (Ohm's-law-type) outlet resistance.

Key relations. Unsteady mass balance $A\,dh/dt=F_0-F_1$; linear resistance $F_1=R_1h$; standard first-order step response $K(1-e^{-t/\tau})$ and impulse response $(K/\tau)e^{-t/\tau}$.

Why this works. Because the given outlet law $F_1=R_1h$ is already linear in $h$ (unlike the more common $F_1=C_v\sqrt{h}$), the mass balance is an exact first-order linear ODE with no Taylor-series linearization needed — the deviation-variable transfer function is exact, not an approximation about a steady state.

Common pitfall. Reflexively linearizing $F_1=R_1h$ as if it were a square-root valve law (introducing an unnecessary $\partial F_1/\partial h|_{h_0}$ step); also, forgetting that a "unit impulse" produces an instantaneous level jump of $K/\tau=1/A$, not a step change.

Source. Seborg et al., Process Dynamics and Control, Ch. 4–5 (liquid-level process modelling, first-order step and impulse responses).

Question 7 of 8: Step and Impulse Response of a Linear Draining Tank

Problem 7: Step and Impulse Response of a Linear Draining Tank (20% total)