23-Chem-A6 Process Dynamics and Control · May 2018
Question 3 of 8: Two Interacting Tanks — Nonlinear ODEs and Linear State-Space Model
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion, state-space models and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order tank/thermal dynamics, step/ramp response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 3: Two Interacting Tanks — Nonlinear ODEs and Linear State-Space Model (20%)
Given. Two tanks in series with an interacting (head-difference) inter-tank flow and a square-root outlet:
Quantity
Symbol
Relation
Inter-tank flow
$F_1$
$\beta_1\sqrt{h_1-h_2}$
Outlet flow
$F_2$
$\beta_2\sqrt{h_2}$
Tank areas
$A_1,A_2$
constant
Input
$F$
steady value $F_s$
Find. (1) the two nonlinear level ODEs; (2) their linearisation $\dot{\mathbf x}=\mathbf A\mathbf x+\mathbf B\,u$ about $F=F_s$.
Problem 3: tank 1 receives $F$ and discharges to tank 2 through an interacting connection driven by the head difference $h_1-h_2$; tank 2 discharges through a square-root outlet $\beta_2\sqrt{h_2}$.
Approach. Write a volume balance on each tank (nonlinear because of the square-root flows), locate the steady state from $F=F_s$, then linearise each nonlinear term with a first-order Taylor expansion to obtain constant state and input matrices.
(1) Nonlinear balances. Volume balance on each tank (constant density): $$A_1\frac{dh_1}{dt}=F-\beta_1\sqrt{h_1-h_2},\qquad A_2\frac{dh_2}{dt}=\beta_1\sqrt{h_1-h_2}-\beta_2\sqrt{h_2}.$$ These are the two governing ODEs; they are coupled and nonlinear through the radicals.
Steady state at $F=F_s$. Setting the derivatives to zero, $F_1=F_2=F_s$, so $$\beta_1\sqrt{h_{1s}-h_{2s}}=F_s\Rightarrow h_{1s}-h_{2s}=\Big(\tfrac{F_s}{\beta_1}\Big)^2,\qquad \beta_2\sqrt{h_{2s}}=F_s\Rightarrow h_{2s}=\Big(\tfrac{F_s}{\beta_2}\Big)^2.$$
Linearise the radicals. Define deviation variables $x_1=h_1-h_{1s}$, $x_2=h_2-h_{2s}$, $u=F-F_s$. Using $\dfrac{d}{dh}\sqrt{h}=\dfrac{1}{2\sqrt h}$, the Taylor slopes at steady state are $$\frac{\partial F_1}{\partial h_1}=\frac{\beta_1}{2\sqrt{h_{1s}-h_{2s}}}=\frac{\beta_1^2}{2F_s}\equiv\alpha=-\frac{\partial F_1}{\partial h_2},\qquad \frac{\partial F_2}{\partial h_2}=\frac{\beta_2}{2\sqrt{h_{2s}}}=\frac{\beta_2^2}{2F_s}\equiv\gamma.$$
Assemble the linear ODEs. Substituting the linearised flows, $$A_1\dot x_1=u-\alpha(x_1-x_2),\qquad A_2\dot x_2=\alpha(x_1-x_2)-\gamma x_2.$$
(2) State-space form. Writing $\dot{\mathbf x}=\mathbf A\mathbf x+\mathbf B u$ with $\mathbf x=[x_1,x_2]^\mathsf T$: $$\boxed{\mathbf A=\begin{bmatrix}-\dfrac{\alpha}{A_1} & \dfrac{\alpha}{A_1}\\[4pt] \dfrac{\alpha}{A_2} & -\dfrac{\alpha+\gamma}{A_2}\end{bmatrix},\qquad \mathbf B=\begin{bmatrix}\dfrac{1}{A_1}\\[4pt]0\end{bmatrix},\qquad \alpha=\frac{\beta_1^2}{2F_s},\ \gamma=\frac{\beta_2^2}{2F_s}.}$$ Both eigenvalues of $\mathbf A$ have negative real parts (the trace is negative and the determinant $\alpha\gamma/(A_1A_2)>0$), so the linearised interacting system is stable about $F_s$.
