Question 3.2: Calculate the initial conditions of the induction machine op......
Calculate the initial conditions of the induction machine operating in steady state in Example 3-1 using the voltage equations Eq. (3-28), Eq. (3-29), Eq. (3-31), and Eq. (3-32). Also calculate the load torque.
\nu _{s d}=R_s i_{s d}+\frac{d}{d t} \lambda_{s d}-\omega_d \lambda_{s q}\hspace{30 pt} \text{(3-28)}
\nu _{s q}=R_s i_{s q}+\frac{d}{d t} \lambda_{s q}+\omega_d \lambda_{s d}.\hspace{30 pt} \text{(3-29)}
\nu _{r d}=R_r i_{r d}+\frac{d}{d t} \lambda_{r d}-\omega_{d A} \lambda_{r q}\hspace{30 pt} \text{(3-31)}
\nu _{r q}=R_r i_{r q}+\frac{d}{d t} \lambda_{r q}+\omega_{d A} \lambda_{r d},\hspace{30 pt} \text{(3-32)}
In a balanced steady state, with \omega_d chosen as the synchronous speed \omega_\text{{syn}}, all dq winding variables are dc quantities and \omega_{dA}=s\omega_\text{{syn}}. Therefore, their time derivatives are zero in Eq. (3-28), Eq. (3-29), Eq. (3-31), and Eq. (3-32), resulting in the following equations:
\nu _{s d}=R_s i_{s d}-\omega_\text{{syn}} \lambda_{s q}\hspace{30 pt} \text{(3-66)}
\nu _{s q}=R_s i_{s q}+\omega_\text{{syn}} \lambda_{s d}\hspace{30 pt} \text{(3-67)}
0=R_r i_{r d}-s \omega_\text{{syn}} \lambda_{r q}\hspace{30 pt} \text{(3-68)}
0=R_r i_{r q}+s \omega_\text{{syn}} \lambda_{r d}.\hspace{30 pt} \text{(3-69)}
Substituting in the above equations for flux linkages from Eq. (3-19) through Eq. (3-22)
\lambda _{sd}=L_si_{sd}+L_mi_{rd}\hspace{30 pt} \text{(3-19)}
\lambda _{rq}=L_ri_{rq}+L_mi_{sq},\hspace{30 pt} \text{(3-22)}
\begin{bmatrix}\nu _{s d} \\\nu _{s q} \\0 \\0\end{bmatrix}=\underbrace{\begin{bmatrix}R_s & -\omega_{ \text{syn} } L_s & 0 & -\omega_{ \text{syn} } L_m \\\omega_{ \text{syn} } L_s & R_s & \omega_{ \text{syn} } L_m & 0 \\0 & -s \omega_{ \text{syn} } L_m & R_r & -s \omega_{ \text{syn} } L_r \\s \omega_{ \text{syn} } L_m & 0 & s \omega_{ \text{syn} } L_r & R_r\end{bmatrix}}_{[A]}\begin{bmatrix}i_{s d} \\i_{s q} \\i_{r d} \\i_{r q}\end{bmatrix}.\quad \text{(3-70)}
The machine currents can be calculated from Eq. (3-70) by inverting matrix [A]:
\left[\begin{array}{l}i_{s d} \\i_{s q} \\i_{r d} \\i_{r q}\end{array}\right]=[A]^{-1}\left[\begin{array}{c}\nu _{s d} \\\nu _{s q} \\0 \\0\end{array}\right].\hspace{30 pt} \text{(3-71)}
Once the dq winding currents are calculated, the flux linkages can be calculated from Eq. (3-61).
\left[\begin{array}{c}\lambda_{s d} \\\lambda_{s q} \\\lambda_{r d} \\\lambda_{r q}\end{array}\right]=\underbrace{\left[\begin{array}{cccc}L_s & 0 & L_m & 0 \\0 & L_s & 0 & L_m \\L_m & 0 & L_r & 0 \\0 & L_m & 0 & L_r\end{array}\right]}_{[M]}\left[\begin{array}{l}i_{s d} \\i_{s q} \\i_{r d} \\i_{r q}\end{array}\right].\hspace{30 pt} \text{(3-61)}
The results from a MATLAB file EX3_2.m on the accompanying website are as follows for currents, with flux linkages and the load torque exactly as in Example 3-1.
\begin{aligned}& i_{s d}(0)=5.34 A \\& i_{s q}(0)=-3.7 A \\& i_{r d}(0)=-5.5 A \\& i_{r q}(0)=0.60 A.\end{aligned}