1500 kVA alternator: excitation emf, voltage at leading pf

Given: 1500 kVA, 6600 V, star, $R_a = 0.4\ \Omega$, $X_s = 6\ \Omega$ per phase, full-load at 0.8 pf lagging, rated voltage.

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}, \qquad I = \frac{1500\times10^3}{\sqrt3\times6600} = 131.22\ \text{A}$$

(i) Excitation emf (0.8 lagging)

Taking $V_{ph}$ as reference, $\vec I = 131.22\angle-36.87^\circ$, $Z_s = 0.4 + j6$:

$$IR_a = 52.49\ \text{V}, \quad IX_s = 787.30\ \text{V}$$ $$\vec E = 3810.5 + 131.22\angle-36.87^\circ(0.4+j6)$$ $$E = \sqrt{(3810.5 + 52.49\times0.8 + 787.30\times0.6)^2 + (787.30\times0.8 - 52.49\times0.6)^2}$$ $$E_{ph} = 4366.07\ \text{V}, \qquad E_L = \sqrt3\times4366.07 = 7562.3\ \text{V}, \qquad \delta = 7.88^\circ$$

(ii) Terminal voltage at 0.8 leading (same $E$ and $I$)

$\vec I = 131.22\angle+36.87^\circ$. The drop $\vec IZ_s$ has magnitude $131.22\times6.0133 = 789.1$ V at $36.87^\circ + 86.19^\circ = 123.06^\circ$ from $V$. With $|\vec V + \vec IZ_s| = E$:

$$V^2 + 2V(789.1)\cos123.06^\circ + 789.1^2 = 4366.07^2$$

Solving this quadratic gives (taking the positive root)

$$V_{ph} = 4746.1\ \text{V}, \qquad V_L = \sqrt3\times4746.1 = 8220.5\ \text{V}$$

Answer: (i) $E_{ph} = 4366$ V (line 7562 V); (ii) terminal voltage $\approx 8220$ V line (4746 V per phase). The voltage rises because the leading current's armature reaction is magnetising.