L16

12/09/26 Sat 10:40am

Question 1:

A two tank system has the following parameters

L15-2026-09-10-08.11.34.excalidraw

$A_1$ = $1m^2$
$A_2$ = 1$m^2$
$R_2=R_1=2s/m^2$
$q_{in}=0.5m^3/s$

model equation:

$$q_{12}=\frac{h_1-h_2}{R_1},q_{out}=\frac{h_2}{R_2}$$ $$A_1\frac{\delta h_1}{\delta t}=q_{in}-q_{12}$$ $$A_2\frac{\delta h_2}{\delta t}=q_{12}-q_{out}$$

Compute:

1. State equation
2. steady state values
3. eigenvalues
4. stability of the system
5. $h_1\&h_2\text{after two euler steps with}\Delta T=0.5s$

Answer:

Substitue known values into the model equation

$$\frac{\delta h_1}{\delta t}=0.5-\frac{h_1-h_2}{2}$$ $$\frac{\delta h_2}{\delta t}=\frac{h_1-h_2}{2}-\frac{h_2}{2}$$

1. Now model it into the state equation:

$$\left[\begin{array}{c}\dot{h_1}\\ \dot{h_2}\end{array}\right]={\left[\begin{array}{c}A\end{array}\right]}_{2\times 2}\left[\begin{array}{c}{h}_1\\ h_2\end{array}\right]+{\left[\begin{array}{c}B\end{array}\right]}_{2\times 1}\left[\begin{array}{c}{q}_{in}\end{array}\right]$$ $$\boxed{\dot{h}=\left[\begin{array}{cc}-0.5& 0.5\\ 0.5& -1\end{array}\right]\left[\begin{array}{c}h\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{c}0.5\end{array}\right]}$$

2. The steady state values of h1, h2
   can be found by equating $q_{in}=q_{12}=q_{out}=0.5$

we know from $q_{out}=\frac{h_2}{R_2}$

$$0.5=\frac{h_2}{2}⟹h_2=1$$

and from $q_{12}=\frac{h_1-h_2}{R_1}$
we get

$$0.5=\frac{h_1-1}{2}⟹h_1=2$$ $$\boxed{h_1=2,h_2=1}$$

3.Next for eigenvalues

$$|[\lambda I-A]|=0$$ $$|\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}-0.5& 0.5\\ 0.5& -1\end{array}\right]|=0$$

simplifying that we get

$$|\left[\begin{array}{cc}\lambda +0.5& -0.5\\ -0.5& \lambda +1\end{array}\right]|=0$$

and the determinant simplifies to

$$(\lambda +0.5)(\lambda +1)-(-0.5)(-0.5)=0$$ $$⟹{\lambda }^2+\frac{3}{2}\lambda +\frac{1}{2}-\frac{1}{4}=0$$ $$⟹{\lambda }^2+\frac{3}{2}\lambda +\frac{1}{4}=0$$ $$\lambda =\frac{-\frac{3}{2}\pm \sqrt{\frac{9}{4}-1}}{2}$$ $$⟹\frac{-\frac{3}{2}\pm \sqrt{\frac{5}{4}}}{2}=-3\pm \sqrt{5}$$ $$\boxed{\lambda =-1.31,-0.19}$$

4. Since both the values are on the left half plane, the system is stable.

5. Euler Simulation ($\Delta T=0.5s$) $x_{new}=x_{old}+\Delta t.\frac{\delta x}{\delta t}$

Step	$h_1$	$h_2$	$\frac{\delta h_1}{\delta t}=0.5-\frac{h_1-h_2}{2}$	$\frac{\delta h_2}{\delta t}=\frac{h_1-h_2}{2}-\frac{h_2}{2}$	new values of $h_1,h_2$
1	0	0	$\frac{\delta h_1}{\delta t}$ = 0.5	$\frac{\delta h_2}{\delta t}$ = 0	$h_1=0.25,h_2=0$
2	0.25	0	$\frac{\delta h_1}{\delta t}$ = 0.375	$\frac{\delta h_2}{\delta t}$ = 0.125	$h_1=0.4375,h_2=0.0625$
3	0.4375	0.0625	$\frac{\delta h_1}{\delta t}$ = 0.3125	$\frac{\delta h_2}{\delta t}$ = 0.156	$h_1=0.75,h_2=0.219$