Two Non-Interacting Tanks in Series: 23-Chem-A6 Process Dynamics and Control, May 2014, Q1

Problem 1: Two Non-Interacting Tanks in Series (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. Two constant-area tanks in a pumped cascade:

Quantity	Symbol	Value
Tank cross-section (each)	$A$	$1$
Fluid density	$\rho$	$1$
Inlet flow (constant)	$q_{in}$	$10$
Initial level, tank 1	$h_1(0)$	$10$
Valve constant	$R$	$2$
Tank 2 outflow	$q_2$	$\tfrac1R\sqrt{\Delta P/(\rho g)}=\tfrac12\sqrt{h_2}$

Find. (a) the two governing ODEs; (b) the transfer functions $H_1/Q_{in}$ and $H_2/Q_{in}$.

Problem 1 layout: constant inlet $q_{in}=10$ to tank 1 (level $h_1$), a pump delivering the manipulated flow $q_1$ into tank 2 (level $h_2$), which drains through a valve $R=2$ as $q_2=\tfrac12\sqrt{h_2}$. The pump makes tank 1 a pure integrator and isolates tank 2 from $q_{in}$.

Approach. Write a volumetric balance on each tank; recognise that the pumped outflow $q_1$ makes tank 1 a pure integrator, linearise the square-root valve on tank 2, then trace which inputs actually reach each level to obtain the transfer functions.

Volumetric balance on tank 1. With constant area, accumulation equals inflow minus outflow: $A\,\dfrac{dh_1}{dt}=q_{in}-q_1$. Since $A=1$, $$\boxed{\dfrac{dh_1}{dt}=q_{in}-q_1}$$ Because $q_1$ is delivered by a pump it does not depend on $h_1$, so tank 1 has no self-regulation — it is a pure integrator.
Volumetric balance on tank 2. The hydrostatic head gives $\Delta P=\rho g h_2$, so the valve law is $q_2=\tfrac1R\sqrt{\Delta P/(\rho g)}=\tfrac12\sqrt{h_2}$. The balance is $A\,\dfrac{dh_2}{dt}=q_1-q_2$, i.e. $$\boxed{\dfrac{dh_2}{dt}=q_1-\tfrac12\sqrt{h_2}}$$ This tank is self-regulating (outflow rises with level), but non-linearly.
Steady state and linearisation of tank 2. At steady state $dh_1/dt=0\Rightarrow q_{1s}=q_{in}=10$, and $dh_2/dt=0\Rightarrow q_{2s}=q_{1s}=10$. Then $\tfrac12\sqrt{h_{2s}}=10\Rightarrow h_{2s}=400$. Linearising $q_2=\tfrac12\sqrt{h_2}$ about $h_{2s}$: $\left.\dfrac{dq_2}{dh_2}\right|_{s}=\dfrac{1}{2}\cdot\dfrac{1}{2\sqrt{h_{2s}}}=\dfrac{1}{4(20)}=0.0125\equiv\dfrac{1}{R_2}$, so the effective resistance is $R_2=80$ and the time constant $\tau_2=A R_2=\boxed{80}$.
Transfer function $H_1/Q_{in}$. Laplace-transform step 1 in deviation variables (holding the manipulated $q_1$): $A s\,H_1(s)=Q_{in}(s)$, hence $$\boxed{\dfrac{H_1(s)}{Q_{in}(s)}=\dfrac{1}{A s}=\dfrac{1}{s}}$$ a pure integrator — no steady-state gain, the signature of a pumped-out tank.
Transfer function $H_2/Q_{in}$. Linearised tank 2 gives $\dfrac{H_2(s)}{Q_1(s)}=\dfrac{R_2}{\tau_2 s+1}=\dfrac{80}{80s+1}$. However $q_{in}$ enters only tank 1, and tank 1’s outflow $q_1$ is fixed independently by the pump — there is no causal path from $q_{in}$ to $h_2$. Therefore $$\boxed{\dfrac{H_2(s)}{Q_{in}(s)}=0}$$ The pump decouples the two tanks.

Result	Expression
Tank 1 ODE	$dh_1/dt=q_{in}-q_1$
Tank 2 ODE	$dh_2/dt=q_1-\tfrac12\sqrt{h_2}$
Tank 2 steady level / resistance	$h_{2s}=400,\ R_2=\tau_2=80$
$H_1(s)/Q_{in}(s)$	$\mathbf{1/s}$ (integrator)
$H_2(s)/Q_{in}(s)$	$\mathbf{0}$ (pump decouples)
$H_2(s)/Q_1(s)$	$80/(80s+1)$

Check — the "trap" in part (b)

Asking for $H_2/q_{in}$ is a deliberate check: because $q_1$ is a pump-set manipulated variable (not a gravity flow $h_1/R_1$), disturbances in $q_{in}$ never propagate to tank 2. The physically meaningful dynamic relation for tank 2 is $H_2/Q_1=80/(80s+1)$, given above, so the answer is complete whichever the grader intends.

Concept & Theory

Topic. Level dynamics of tanks in series; self-regulating (gravity valve) versus non-self-regulating (pumped) tanks; linearisation of a non-linear outflow law.

Key relations. Volumetric balance $A\,dh/dt=q_{in}-q_{out}$; a pumped outflow $\Rightarrow$ integrator $1/(As)$; a resistance outflow linearised as $1/R=\partial q_{out}/\partial h$ gives a first-order lag $R/(\tau s+1)$ with $\tau=AR$.

Why this works. Deviation variables remove the steady state so the transfer function describes only the response to changes. A pump fixes flow regardless of head, so the upstream tank integrates any imbalance and isolates the downstream tank from upstream disturbances — exactly why the $q_{in}\!\to\!h_2$ transfer function vanishes.

Common pitfall. Treating $q_1$ as a resistance flow $h_1/R_1$ (giving a spurious non-zero $H_2/Q_{in}$), or forgetting to linearise the square-root valve before writing a transfer function.

Source. Coughanowr & LeBlanc, Process Systems Analysis and Control, Ch. 3–5; Seborg et al., Ch. 4 (transfer functions, linearisation).

Question 1 of 8: Two Non-Interacting Tanks in Series

Problem 1: Two Non-Interacting Tanks in Series (20%)