Question 9.4: Consider the case of a symmetrical two-phase motor such as i......
Consider the case of a symmetrical two-phase motor such as is discussed in Section 9.4.1. In this case, Eqs. 9.25 through 9.37 simplify with equal self and mutual inductances and resistances for the two windings. Using the notation of Section 9.4.1, ‘α’ and ‘β’ replacing ‘main’ and ‘aux’, the flux-linkage/current relationships of Eq. 9.39 and 9.40 become
\left[\begin{array}{c}\lambda_{\text {main }} \\ \lambda_{\text {aux }} \\ \lambda_{\mathrm{r}1} \\ \lambda_{\mathrm{r}2}\end{array}\right]=\left[\begin{array}{cccc}L_{\text {main }} & 0 & \mathcal{L}_{\text {main}, \mathrm{r}1}\left(\theta_{\text {me}}\right) & \mathcal{L}_{\text {main }, \mathrm{r}2}\left(\theta_{\text {me }}\right) \\0 & L_{\text {aux}} & \mathcal{L}_{\text {aux}, \mathrm{r}1}\left(\theta_{\text {me}}\right) & \mathcal{L}_{\text {aux }, \mathrm{r}2}\left(\theta_{\text {me }}\right) \\ \mathcal{L}_{\text {main,r}1}\left(\theta_{\text {me }}\right) & \mathcal{L}_{\text {aux }, \mathrm{r}1}\left(\theta_{\text {me }}\right) & L_{\mathrm{r}} & 0 \\ \mathcal{L}_{\text {main }, \mathrm{r} 2}\left(\theta_{\text {me}}\right) & \mathcal{L}_{\text {aux }, \mathrm{r} 2}\left(\theta_{\text {me }}\right) & 0 & L_{\mathrm{r}}\end{array}\right]\left[\begin{array}{c}i_{\text {main }} \\i_{\text {aux }} \\i_{\mathrm{r}1} \\i_{\mathrm{r}2}\end{array}\right]
(9.25)
\mathcal{v}_{r2}=0=i_{r2}R_r + \frac{dλ_{r2}}{dt} \quad \quad \quad (9.37)
\hat{λ}_{main}= [L_{main} – jL_{main,r}^2(\hat{K}^+ +\hat{K}^-)] \hat{I}_{main}+ L_{main,r} L_{aux,r} (\hat{K}^+ -\hat{K}^-)\hat{I}_{aux} \quad \quad \quad (9.39)
\hat{λ}_{aux}= -L_{main ,r} L_{aux,r} (\hat{K}^+ – \hat{K}^-)\hat{I}_{main} + [L_{aux} – jL_{aux,r}^2 (\hat{K}^+ + \hat{K}^-)]\hat{I}_{aux} \quad \quad (9.40)
\hat{λ}_α= [L_α – jL_{α,r}^2(\hat{K}^+ +\hat{K}^-)] \hat{I}_α+ L_{α,r}^2 (\hat{K}^+ -\hat{K}^-)\hat{I}_β \\ \hat{λ}_β= -L_{α ,r}^2 (\hat{K}^+ – \hat{K}^-)\hat{I}_α + [L_α – jL_{α,r}^2 (\hat{K}^+ + \hat{K}^-)]\hat{I}_β
and the voltage equations (Eqs. 9.43 and 9.44) become
\hat{V}_{main}=\hat{I}_{main} R_{main} +j ω_e\hat{λ}_{main} \quad \quad (9.43) \\ \hat{V}_{aux}=\hat{I}_{aux} R_{aux} +j ω_e\hat{λ}_{aux} \quad \quad (9.44)
\hat{V}_α=\hat{I}_α R_α +j ω_e\hat{λ}_α \\ \hat{V}_β=\hat{I}_β R_α +j ω_e\hat{λ}_β
Show that when operated from a positive sequence set of voltages such that \hat{V}_β = -j V_α the single-phase equivalent circuit is that of the forward-field (positive-sequence) equivalent circuit of Fig. 9.12a.
Substitution of the positive-sequence voltages in the above equations and solution for the impedance Z_{\alpha}=\hat{V}_{\alpha} / \hat{I}_{\alpha} gives
\begin{aligned}Z_{\alpha} & =R_{\alpha}+j \omega_{\mathrm{e}} L_{\alpha}+\frac{\left(\omega_{\mathrm{e}} L_{\alpha, \mathrm{r}}\right)^{2}}{\left(R_{\mathrm{r}} / s+j \omega_{\mathrm{e}} L_{\mathrm{r}}\right)} \\ & =R_{\alpha}+j X_{\alpha}+\frac{X_{\alpha, \mathrm{r}}^{2}}{\left(R_{\mathrm{r}} / s+j X_{\mathrm{r}}\right)}\end{aligned}
This equation can be rewritten as
Z_{\alpha}=R_{\alpha}+j\left(X_{\alpha}-X_{\alpha, \mathrm{r}}\right)+\frac{j X_{\alpha, \mathrm{r}}\left[j\left(X_{\mathrm{r}}-X_{\alpha, \mathrm{r}}\right)+R_{\mathrm{r}} / s\right]}{\left(R_{\mathrm{r}} / s+j X_{\mathrm{r}}\right)}
Setting R_{\alpha} \Rightarrow R_{1},\left(X_{\alpha}-X_{\alpha, \mathrm{r}}\right) \Rightarrow X_{1}, X_{\alpha, \mathrm{r}} \Rightarrow X_{\mathrm{m}},\left(X_{\mathrm{r}}-X_{\alpha, \mathrm{r}}\right) \Rightarrow X_{2}, and R_{\mathrm{r}} \Rightarrow R_{2}, we see that this equation does indeed represent an equivalent circuit of the form of Fig. 9.12a.