Draining Tank — Step and Impulse Response: 23-Chem-A6 Process Dynamics and Control, May 2016, Q6

Problem 6: Draining Tank — Step and Impulse 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
Tank cross-section	$A$	$1\ \mathrm{m^2}$
Initial level	$h_0$	$1\ \mathrm{m}$
Outflow law	$F_1$	$R_1 h,\ R_1=1\ \mathrm{m^2/min}$
Inlet pipe length / area	$L,\ A_p$	$1.0\ \mathrm{m},\ 0.01\ \mathrm{m^2}$

Find. The deviation level $\delta h(t)=h(t)-h_0$ for (a) a unit step and (b) a unit impulse in $F_0$.

Problem 6: single gravity-drained tank, imposed inflow $F_0$ through the inlet pipe, level $h$ ($A=1\ \mathrm{m^2}$, $h_0=1\ \mathrm{m}$), linear outflow $F_1=R_1 h$ through valve $R_1=1\ \mathrm{m^2/min}$.

Check — role of the inlet-pipe dimensions

The forcing is specified as a change in the inlet flow $F_0$ itself, so $F_0$ is imposed directly on the tank and the level dynamics are first order; the pipe length and area do not enter this response. They would matter only if the forcing were an upstream pressure step, in which case the liquid slug in the pipe adds a fluid inertance $I=\rho L/A_p=1000(1.0)/0.01=1\times10^{5}$ (SI), giving a second-order inertial mode. We solve the problem as posed (flow-forced, first order) and note the inertance for completeness.

Approach. Write the linear tank balance, reduce it to standard first-order form to read off $\tau$ and gain $K$, then apply the standard step and impulse solutions.

Model and standard form. $A\,\dfrac{dh}{dt}=F_0-R_1 h$. In deviation variables about the initial steady state ($F_{0s}=R_1 h_0=1\times1=1\ \mathrm{m^3/min}$): $\dfrac{dh'}{dt}=F_0'-h'$. This is standard first order with $$\tau=\dfrac{A}{R_1}=1\ \mathrm{min},\qquad K=\dfrac{1}{R_1}=1,$$ so $\dfrac{H'(s)}{F_0'(s)}=\dfrac{K}{\tau s+1}=\dfrac{1}{s+1}.$
(a) Unit-step response. For $F_0'=1$, the first-order step response is $$\delta h(t)=K\!\left(1-e^{-t/\tau}\right)=\boxed{1-e^{-t}}\ \mathrm{m}.$$ The level rises from $1\ \mathrm{m}$ toward $h(\infty)=2\ \mathrm{m}$ with a $1\ \mathrm{min}$ time constant (~99% settled in $5\tau=5\ \mathrm{min}$).
(b) Unit-impulse response. A Dirac impulse has $F_0'(s)=1$, so $\delta H(s)=\dfrac{1}{s+1}$ and $$\delta h(t)=\dfrac{K}{\tau}e^{-t/\tau}=\boxed{e^{-t}}\ \mathrm{m}.$$ The unit-area flow impulse injects a volume $\int F_0'\,dt=1\ \mathrm{m^3}$ instantaneously, so the level jumps by $\Delta h=V/A=1\ \mathrm{m}$ to $h=2\ \mathrm{m}$ at $t=0^+$, then decays back to $1\ \mathrm{m}$.
Consistency check. The impulse response is the time-derivative of the step response: $\dfrac{d}{dt}\!\left[1-e^{-t}\right]=e^{-t}$ — as it must be for a linear system. Both share the $\tau=1\ \mathrm{min}$ decay.

Problem 6(a): unit-step response. The level climbs from $1\ \mathrm{m}$ to the new steady state $2\ \mathrm{m}$ as $1+(1-e^{-t})$, effectively settled after $\approx5\ \mathrm{min}$.

Problem 6(b): unit-impulse response. The injected $1\ \mathrm{m^3}$ raises the level instantly to $2\ \mathrm{m}$, which then decays as $1+e^{-t}$ back to $1\ \mathrm{m}$.

Quantity	Value
Time constant $\tau=A/R_1$	$1\ \mathrm{min}$
Gain $K=1/R_1$	$1$
(a) Step response	$\delta h(t)=1-e^{-t}$; $h(\infty)=2\ \mathrm{m}$
(b) Impulse response	$\delta h(t)=e^{-t}$; peak $h=2\ \mathrm{m}$ at $t=0^+$
Inlet-pipe inertance (unused)	$I=\rho L/A_p=1\times10^{5}$ SI

Concept & Theory

Topic. First-order self-regulating tank; step and impulse (Dirac) forcing.

Key relations. Standard first-order lag $\tau\,dy/dt+y=Ku$; step response $y=K\Delta u(1-e^{-t/\tau})$; impulse response $y=(K/\tau)e^{-t/\tau}$ (the derivative of the step response); an area-$a$ flow impulse adds a volume $a$ instantly, jumping the level by $a/A$.

Why this works. A Dirac input has unit Laplace transform, so the output transform equals the transfer function and its inverse is the impulse response — the system’s fundamental fingerprint. Because the system is linear and time-invariant, the impulse response is exactly $d/dt$ of the step response.

Common pitfall. Treating the inlet-pipe dimensions as part of a flow-forced response (they are not), or forgetting that a flow impulse causes an instantaneous level jump rather than a gradual rise.

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

Question 6 of 8: Draining Tank — Step and Impulse Response

Problem 6: Draining Tank — Step and Impulse Response (20%)