Salient pole generator phasor diagram in per unit

Using the two-reaction (Blondel) theory: $X_d$ = 0.6 pu, $X_q$ = 0.45 pu, $R_a$ = 0.015 pu, $V$ = 1 pu, $I$ = 1 pu, pf = 0.8 lagging ($\phi = 36.87^\circ$).

                      E (on q-axis)
                     /:
                    / : j Id (Xd - Xq)
                   /  :
                  / .-+ E' = V + I(Ra + jXq)
                 /.' delta
                O-------------- V
                 \  phi
                  \ I (psi = phi + delta from q-axis)

(i) Load angle $\delta$

$$\begin{aligned} \tan\psi &= \frac{V\sin\phi + IX_q}{V\cos\phi + IR_a} = \frac{0.6 + 0.45}{0.8 + 0.015} = \frac{1.05}{0.815} = 1.2883 \\ \psi &= 52.18^\circ \\ \delta &= \psi - \phi = 52.18^\circ - 36.87^\circ = 15.31^\circ \end{aligned}$$

(ii) $I_d$ and $I_q$

$$\begin{aligned} I_d &= I\sin\psi = \sin 52.18^\circ = 0.790\ \text{pu} \\ I_q &= I\cos\psi = \cos 52.18^\circ = 0.613\ \text{pu} \end{aligned}$$

(iii) Open-circuit voltage (excitation emf)

$$\begin{aligned} E &= V\cos\delta + I_qR_a + I_dX_d \\ &= \cos 15.31^\circ + 0.613\times0.015 + 0.790\times0.6 \\ &= 0.9645 + 0.0092 + 0.4740 = 1.448\ \text{pu} \end{aligned}$$

(iv) Voltage regulation

$$\%\text{Reg} = \frac{E - V}{V}\times100 = \frac{1.448 - 1}{1}\times100 = 44.8\%$$

Answer: δ = 15.31°, I_d = 0.790 pu, I_q = 0.613 pu, E = 1.448 pu, regulation = 44.8%.