Question 3 of 8: Linear vs. Nonlinear-Valve Draining-Tank Dynamics

Problem 3: Linear vs. Nonlinear-Valve Draining-Tank Dynamics (20% total)

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 independent tanks, each cross-section $A=1\ \text{m}^2$, common initial steady state $h_0=1\ \text{m}$, $q_0=1\ \text{m}^3/\text{s}$; Case I outlet $q=h/R$, $R=1$; Case II outlet $q=K\sqrt{h}$ ($K$ from the IC); inlet flow steps $q_i:1\to2\ \text{m}^3/\text{s}$ at $t=0$.

Find. (a) $\delta H(s)/\delta Q_i(s)$ for each case; (b) $\delta h(t)$ for each case for the given inlet step.

Approach. Write the unsteady mass balance $A\,dh/dt=q_i-q$ for each tank; Case I's outlet is already linear so the transfer function is exact, while Case II's outlet must first be Taylor-linearized about $h_0$ before Laplace-transforming; both reduce to a standard first-order lag, then apply the known step response with $M=\delta q_i=1\ \text{m}^3/\text{s}$.

1. (a) Case I transfer function. $A\dfrac{dh}{dt}=q_i-\dfrac{h}{R}\Rightarrow AR\dfrac{dh'}{dt}+h'=Rq_i'$, so with $A=R=1$: $$\boxed{\frac{H'(s)}{Q_i'(s)}=\frac{R}{ARs+1}=\frac{1}{s+1}\qquad(\tau_I=AR=1\ \text{s},\ K_I=R=1).}$$
2. (a) Case II: fix $K$, then linearize. From the IC, $q_0=K\sqrt{h_0}\Rightarrow K=q_0/\sqrt{h_0}=1/\sqrt1=1$. Linearizing $q=K\sqrt h$ about $h_0$: $\left.\dfrac{dq}{dh}\right|_{h_0}=\dfrac{K}{2\sqrt{h_0}}=\dfrac12=0.5$. The balance becomes $A\dfrac{dh'}{dt}=q_i'-0.5\,h'$, so $$\boxed{\frac{H'(s)}{Q_i'(s)}=\frac{1}{As+0.5}=\frac{2}{2s+1}\qquad(\tau_{II}=A/0.5=2\ \text{s},\ K_{II}=1/0.5=2).}$$
3. (b) Apply the standard first-order step response, $M=1\ \text{m}^3/\text{s}$ (the $1\to2\ \text{m}^3/\text{s}$ inlet step). $h'(t)=K M(1-e^{-t/\tau})$ for each case: $$\boxed{\delta h_I(t)=1-e^{-t}\ \text{m},\qquad \delta h_{II}(t)=2\left(1-e^{-t/2}\right)\ \text{m}\qquad(t\ \text{in s}).}$$
4. Sample values and a note on the linearization. At $t=1\ \text{s}$: $\delta h_I=0.632\ \text{m}$, $\delta h_{II}=0.443\ \text{m}$ (Case II starts slower, $\tau_{II}=2\tau_I$). As $t\to\infty$: $\delta h_I\to1\ \text{m}$ (new level $h=2\ \text{m}$, exact, since $q=h/R$ is already linear and $q_i=2=h/1\Rightarrow h=2$ exactly). $\delta h_{II}\to2\ \text{m}$ (linearized new level $h\approx3\ \text{m}$); the exact nonlinear new steady state solves $q_i=2=K\sqrt h=\sqrt h\Rightarrow h=4\ \text{m}$, i.e. $\delta h=3\ \text{m}$ — the linearized model under-predicts the true new level by $1\ \text{m}$ because the step ($\delta q_i/q_0=100\%$) is large relative to the point of linearization.