Question 5.9: The reactances Xd and Xq of a salient-pole synchronous gener......

First, the phase \hat{E}_{af}  \text{must be found so that}  \hat{I}_a can be resolved into its direct- and quadrature- axis components. The phasor diagram is shown in Fig. 5.25. As is commonly done for such problems, the terminal voltage \hat{V}_a will be used as the reference phasor, i.e., \hat{V}_a = V_a  e^{j0.0°} = V_a.

In per unit

\hat{I}_a = I_a  e^{j\mathcal{\phi}} = 0.80  –  j0.60 = 1.0  e^{-j36.9°}

The quadrature axis is located by the phasor

\hat{E}^′ = \hat{V}_a  + jX_q\hat{I}_a = 1.0  +  j0.60(1.0  e^{-j36.9°}) = 1.44  e^{j19.4°}

Thus, δ = 19.4°, and the phase angle between \hat{E}_{af}  \text{and}  \hat{I}_a  \text{is}  δ-\mathcal{\phi} = 19.4°  –  (-36.9°) = 56.3°. Note, that although it would appear from Fig. 5.25 that the appropriate angle is δ + \mathcal{\phi}, this is not correct because the angle \mathcal{\phi} as drawn in Fig. 5.25 is a negative angle. In general, the desired angle is equal to the difference between the power angle and the phase angle of the terminal current.

The armature current can now be resolved into its direct- and quadrature-axis components.

Their magnitudes are

I_d = |\hat{I}_a| \sin (δ – \mathcal{\phi}) = 1.00  \sin (56.3 °) = 0.832

and

I_q =|\hat{I}_a| \cos (δ – \mathcal{\phi}) = 1.00  \cos (56.3°) = 0.555

As phasors,

\hat{I}_d = 0.832  e^{j(-90°+19.4°)} = 0.832  e^{j70.6°}

and

\hat{I}_q=0.555  e^{j19.4°}

We can now find E_{af} from Eq. 5.59

and we see that E_{af}= 1.77 per unit. Note that, as expected, ∠\hat{E}_{af} = 19.4 ° =δ , thus confirming that \hat{E}_{af} lies along the quadrature axis.