Two Tanks in Series with an Inter-tank Pump: 23-Chem-A6 Process Dynamics and Control, December 2016, Q4

Problem 4: Two Tanks in Series with an Inter-tank Pump (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. $A=1$, $\rho=1$, $R=2$, $q_{in}=10$ (constant), $q_1$ pumped (independent), tank-2 outlet $q_2=\tfrac1R\sqrt{h_2}$.

Find. (a) the two level ODEs; (b) $\delta h_1/\delta q_{in}$ and $\delta h_2/\delta q_{in}$.

A pump sits between the tanks and sets $q_1$ independently of either level. Consequently tank 1 is a pure integrator on $q_{in}-q_1$, and tank 2 is driven entirely by the pump — the pump decouples $h_2$ from the inlet disturbance.

Approach. Write a volume balance on each tank, linearise the square-root outlet of tank 2, and note that because $q_1$ is a pump command (not a function of $h_1$), the inlet sees only the tank-1 integrator.

(a) Tank-1 balance. $A\,\dfrac{dh_1}{dt}=q_{in}-q_1$, and with $A=1$ $$\boxed{\frac{dh_1}{dt}=q_{in}-q_1.}$$ There is no self-regulation: the outflow $q_1$ is set by the pump, not by $h_1$.
(a) Tank-2 balance. $A\,\dfrac{dh_2}{dt}=q_1-q_2$ with $q_2=\tfrac1R\sqrt{h_2}$: $$\boxed{\frac{dh_2}{dt}=q_1-\tfrac1R\sqrt{h_2}.}$$
Steady state & linearised resistance. At steady state $q_2=q_1=q_{in}=10$, so $\sqrt{\overline{h_2}}=R\,\overline{q_2}=20\Rightarrow\overline{h_2}=400$. Linearising $q_2$: $\left.\dfrac{dq_2}{dh_2}\right|_{ss}=\dfrac{1}{2R\sqrt{\overline{h_2}}}$, so the effective resistance is $R_L=2R\sqrt{\overline{h_2}}=2(2)(20)=80$ and $\tau_2=A\,R_L=80$.
(b) $\delta h_1/\delta q_{in}$. The pump fixes $q_1$, so $\delta q_1=0$ when $q_{in}$ moves: $s\,\delta h_1=\delta q_{in}$, giving a pure integrator $$\boxed{\frac{\delta h_1}{\delta q_{in}}=\frac{1}{s}.}$$
(b) $\delta h_2/\delta q_{in}$. Tank 2 is fed only by the pumped $q_1$, which does not respond to $q_{in}$. The inlet disturbance therefore never reaches tank 2: $$\boxed{\frac{\delta h_2}{\delta q_{in}}=0.}$$
Manipulable path (context). Through the actual handle $q_1$, tank 2 is a first-order lag $\dfrac{\delta h_2}{\delta q_1}=\dfrac{R_L}{\tau_2 s+1}=\dfrac{80}{80s+1}$ — this is what a controller would use to regulate $h_2$.

Quantity	Result
Tank-1 ODE	$\dot h_1=q_{in}-q_1$
Tank-2 ODE	$\dot h_2=q_1-\tfrac1R\sqrt{h_2}$
$\delta h_1/\delta q_{in}$	$1/s$ (integrator)
$\delta h_2/\delta q_{in}$	$0$ (pump-decoupled)
$\delta h_2/\delta q_1$ (context)	$80/(80s+1)$

Concept & Theory

Topic. Level dynamics; non-self-regulating (pump) outflow; linearised hydraulic resistance; decoupling.

Key relations. Volume balance $A\dot h=q_{in}-q_{out}$; a pump outlet gives $H/Q=1/(As)$ (integrator); a valve outlet gives $R_L=2R\sqrt{\overline h}$ and $\tau=AR_L$.

Why this works. When the outflow is set by a pump it is independent of head, so the tank integrates the flow imbalance rather than self-regulating; and because the pump between the tanks fixes $q_1$, any change in $q_{in}$ only accumulates in tank 1 and cannot propagate downstream.

Common pitfall. Treating the pump flow as head-dependent (it is not), or linearising $\sqrt{h_2}$ about the wrong steady level. The decoupled $\delta h_2/\delta q_{in}=0$ is the whole point of the problem.

Source. Seborg, Edgar, Mellichamp & Doyle, Process Dynamics and Control (4th ed., Wiley) Ch. 2 & 6 (dynamic models, interacting vs non-interacting tanks); Coughanowr & LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) Ch. 4–7.

Question 4 of 8: Two Tanks in Series with an Inter-tank Pump

Problem 4: Two Tanks in Series with an Inter-tank Pump (20%)