Transport Delay in a Pipe Feeding an Integrating Tank: 23-Chem-A6 Process Dynamics and Control, December 2013, Q6

Problem 6: Transport Delay in a Pipe Feeding an Integrating Tank (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. Pipe $L=10\ \mathrm{m}$, liquid speed $u=1\ \mathrm{m/s}$; tank area $A=1\ \mathrm{m^{2}}$, level $h$; the tank has an inflow only (no drain specified), so it integrates.

Find. (a) the delay $\theta$; (b) the ODE and $H(s)/U(s)$; (c) the closed-loop $H/H_{sp}$ under proportional control.

Problem 6: liquid enters the pipe at velocity $u$ and, after a transport delay $\theta=L/u$, discharges into the tank (area $A$, level $h$). With no outflow the tank integrates the delayed inflow, so the open-loop path is a pure delay in series with an integrator.

Approach. The pipe is a pure transport delay $\theta=L/u$; the tank with no outlet is an integrator. Combine them for the open loop, then close the proportional loop with the standard $L/(1+L)$ formula.

(a) Transport delay. A fluid element traverses the pipe in $$\boxed{\theta=\dfrac{L}{u}=\dfrac{10}{1}=10\ \mathrm{s}.}$$
(b) Model and open-loop transfer function. The flow reaching the tank is the inlet flow delayed by $\theta$; with the pipe cross-section absorbed into the input (or unit pipe area), the volumetric balance on the outlet-free tank is $A\,\dfrac{dh}{dt}=u(t-\theta)$. In deviation/Laplace form ($A=1$): $$\boxed{\dfrac{H(s)}{U(s)}=\dfrac{e^{-\theta s}}{A s}=\dfrac{e^{-10s}}{s}}$$ — a pure integrator preceded by a $10\,$s dead time (no steady-state gain; level is non-self-regulating).
(c) Closed-loop transfer function. With $G_c=K_c$ and $G_p=e^{-10s}/s$, $$\dfrac{H(s)}{H_{sp}(s)}=\dfrac{K_cG_p}{1+K_cG_p}=\boxed{\dfrac{K_c\,e^{-10s}}{s+K_c\,e^{-10s}}.}$$
Stability note. An integrator-plus-delay under pure proportional control is only conditionally stable: the phase $-90^\circ-10\omega$ reaches $-180^\circ$ at $\omega_{pc}=\pi/20=0.157\ \mathrm{rad/s}$, where $|G_p|=1/\omega$, so the ultimate gain is $K_{cu}=\omega_{pc}=0.157$. The loop is stable only for $0<K_c<0.157\ \mathrm{s^{-1}}$ — the dead time imposes a hard speed limit.

Check — modelling assumption

The problem states an inlet velocity $u$ but no pipe diameter or tank outflow. We take the volumetric inflow proportional to $u$ (unit effective pipe area) and the tank as outlet-free, which is the standard reading of this classic dead-time-plus-integrator exercise. If a finite outlet resistance $R$ were intended, the integrator $1/s$ would become a first-order lag $R/(A R s+1)$ and $H/U=R\,e^{-10s}/(ARs+1)$; the delay term and its stability implications are unchanged.

Result	Expression
Transport delay	$\theta=L/u=10\ \mathrm{s}$
Tank ODE	$A\,dh/dt=u(t-\theta)$
Open-loop TF	$H/U=e^{-10s}/s$
Closed-loop TF	$H/H_{sp}=K_ce^{-10s}/(s+K_ce^{-10s})$
Stability limit	$0<K_c<0.157\ \mathrm{s^{-1}}$

Concept & Theory

Topic. Transport (dead-time) lag; integrating (non-self-regulating) level process; proportional closed loop.

Key relations. Pipe delay $\theta=L/u$, transfer $e^{-\theta s}$; outlet-free tank $A\,\dot h=q_{in}\Rightarrow H/Q=1/(As)$; closed loop $H/H_{sp}=K_cG_p/(1+K_cG_p)$; integrator+delay ultimate gain $K_{cu}=\pi/(2\theta)$.

Why this works. Plug flow means every element takes the same time $L/u$ to cross the pipe, a pure delay. With no drain the tank accumulates indefinitely (an integrator), so the process has infinite DC gain and can only be stabilised by feedback — but the delay caps how much loop gain that feedback can use.

Common pitfall. Adding a spurious self-regulating outflow (turning the integrator into a lag) when none is given, or forgetting that the delay makes even this simple loop go unstable at high gain.

Source. Stephanopoulos, Ch. 10 & 17 (dead-time systems, stability); Coughanowr & LeBlanc, Ch. 7 & 18 (transport lag, integrating processes).

Question 6 of 8: Transport Delay in a Pipe Feeding an Integrating Tank

Problem 6: Transport Delay in a Pipe Feeding an Integrating Tank (20%)