State-space model of a hydroelectric dam

Normal tree, order, and variables

Now, we define a normal tree by overlaying it on the system graph in . There are six independent energy storage elements, making it a sixth-order (\(n=6\)) system. We define the state vector to be \[\begin{aligned} \bm{x} = \begin{bmatrix} P_{C_1} & P_{C_2} & Q_{L_1} & \Omega_J & i_{L_2} & v_{C_3} \end{bmatrix}^\top. \end{aligned}\] The input vector is defined as \(\bm{u} = \begin{bmatrix} Q_s & P_{s1} & P_{s2} \end{bmatrix}^\top\).

Elemental equations

Yet to be encountered is a turbine’s transduction. A simple model is that the torque \(T_2\) is proportional to the flowrate \(Q_1\), which are both through-variables, making it a transformer, so \[\begin{aligned} T_2 = -\alpha Q_1 \quad {and} \quad \Omega_2 = \frac{1} {\alpha} P_1, \end{aligned}\] where \(\alpha\) is the transformer ratio.

The other elemental equations have been previously encountered and are listed, below.

Continuity and compatibility equations

Continuity and compatibility equations can be found in the usual way—by drawing contours and temporarily creating loops by including links in the normal tree. We proceed by drawing a table of all elements and writing a continuity equation for each branch of the normal tree and a compatibility equation for each link.

State equation

The system of equations composed of the elemental, continuity, and compatibility equations can be reduced to the state equation. There is a substantial amount of algebra required to eliminate those variables that are neither state nor input variables. Therefore, we use the Mathematica package StateMint (Picone2018?). The resulting system model is: \[\begin{gathered} \frac{d \bm{x}}{d t} = A \bm{x} + B \bm{u}, \\ A = \begin{bmatrix} 0 & 0 & -1/C_1 & 0 & 0 & 0 \\ 0 & 0 & 1/C_2 & 0 & 0 & 0 \\ 1/L_1 & -1/L_1 & -(R_1+R_2)/L_1 & -\alpha/L_1 & 0 & 0 \\ 0 & 0 & \alpha/J & -B/J & -k_m/J & 0 \\ 0 & 0 & 0 & k_m/L_2 & -R_3/L_2 & -1/L_2 \\ 0 & 0 & 0 & 0 & 1/C_3 & -1/(R_4 C_3) \end{bmatrix}, \\ B = \begin{bmatrix} 1/C_1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1/L_1 & 1/L_1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{gathered}\]

The rub is estimating all these parameters.