Excitation emf of salient pole generator

Use the two-reaction theory. Given: star, 500 V, $X_d = 0.1\ \Omega$, $X_q = 0.075\ \Omega$, $R_a = 0.1\ \Omega$, $I = 100$ A at 0.8 pf lagging.

Step 1: Voltage behind $R_a + jX_q$

$$\begin{aligned} V &= \frac{500}{\sqrt3} = 288.68\ \text{V},\quad \bar I = 100\angle -36.87^\circ = 80 - j60\ \text{A} \\ \bar E' &= \bar V + \bar I(R_a + jX_q) = 288.68 + (80 - j60)(0.1 + j0.075) \\ &= 288.68 + (8 + 4.5) + j(6 - 6) = 301.18 + j0\ \text{V} \end{aligned}$$

So the load angle $\delta = 0^\circ$ (the q-axis lies along $V$ here) and $\psi = \phi + \delta = 36.87^\circ$.

Step 2: d- and q-axis currents

$$\begin{aligned} I_d &= I\sin\psi = 100\sin 36.87^\circ = 60\ \text{A} \\ I_q &= I\cos\psi = 100\cos 36.87^\circ = 80\ \text{A} \end{aligned}$$

Step 3: Excitation emf

$$\begin{aligned} E &= E' + I_d(X_d - X_q) = 301.18 + 60(0.1 - 0.075) \\ &= 301.18 + 1.5 = 302.68\ \text{V per phase} \\ E_L &= \sqrt3\times302.68 = 524.25\ \text{V} \end{aligned}$$

Answer: excitation emf = 302.68 V per phase (524.25 V line), load angle ≈ 0°.