Analysis of Synchronous Machines, 2nd Edition

Combining the two current components to obtain the phasor stator current yields

${\tilde{I}}_s=I_s\text{ cos }\varphi -j{I}_s\text{ sin }\varphi$

(4.115)

whereupon

${\tilde{I}}_s=V_s\text{sinδ cos δ}\left(\frac{1}{x_{qs}}-\frac{1}{x_{ds}}\right)+\frac{E_i\text{ sin δ}}{x_{ds}}-j\frac{V_s}{x_{ds}}-j{V}_s{\text{sin}}^2\text{δ}\left(\frac{1}{x_{qs}}-\frac{1}{x_{ds}}\right)+j\frac{E_i\text{ cos δ}}{x_{ds}}$

(4.116)

This equation can be rearranged to form

${\tilde{I}}_s=V_s \text{sin}\text{δ}(\text{cos δ}-j \text{sinδ})\left(\frac{1}{x_{qs}}-\frac{1}{x_{ds}}\right)+\frac{V_s}{j{x}_{ds}}-\frac{E_i}{j{x}_{ds}}\left(\text{cos δ}-j\sin \text{δ}\right)$

(4.117)

However,

$\text{sinδ}(\text{cos δ}-j\sin \text{δ})=\sin \text{δ}(e^{-j\text{δ}})=\frac{\text{ sin δ}}{e^{j\text{δ}}}$

(4.118)

$=\frac{\text{ sin δ}}{\text{ cos δ}+j\text{ sin δ}}=\frac{1}{\frac{1}{\text{ tan δ}}+j1}$

(4.119)

The phasor equation for ${\widehat{I}}_s$ can now be written as

${\tilde{I}}_s=\frac{V_s-{\tilde{E}}_i}{j{x}_{ds}}+\frac{V_s}{\frac{x_{ds}{x}_{qs}}{x_{ds}-x_{qs}}\frac{1}{\text{ tan δ}}+j\frac{x_{ds}{x}_{qs}}{x_{ds}-x_{qs}}}$

(4.120)

where ${\tilde{E}}_i=E_i{e}^{-j\text{δ}}$.

The circuit described by this equation is shown in Figure 4.13, where the stator resistance has now again been introduced to complete the result. The similarity of the equivalent ...