1200 kVA alternator: voltage at leading pf

Given: 1200 kVA, 6600 V (line), star, $R_a = 0.4\ \Omega$, $X_s = 6\ \Omega$ per phase, full-load current, 0.8 pf lagging at rated voltage. Find the terminal voltage for the same excitation and current at 0.8 pf leading.

Rated quantities

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}, \qquad I = \frac{1200\times10^3}{\sqrt3\times6600} = 104.97\ \text{A}$$ $$Z_s = 0.4 + j6\ \Omega$$

Excitation emf (0.8 lagging)

Taking $V_{ph}$ as reference, $\vec I = 104.97\angle -36.87^\circ$ A:

$$\vec E = \vec V + \vec I Z_s = 3810.5 + 104.97\angle-36.87^\circ(0.4 + j6)$$ $$IR = 41.99\ \text{V},\quad IX_s = 629.84\ \text{V}$$ $$E = \sqrt{(3810.5 + 41.99\times0.8 + 629.84\times0.6)^2 + (629.84\times0.8 - 41.99\times0.6)^2}$$ $$E_{ph} = 4249.05\ \text{V} \;(\text{line } 7359.6\ \text{V})$$

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

Now $\vec I = 104.97\angle+36.87^\circ$ A. The drop $\vec I Z_s$ has magnitude $104.97\times 6.0133 = 631.23$ V at an angle $36.87^\circ + 86.19^\circ = 123.06^\circ$ with respect to $V$. For the same excitation, $|\vec V + \vec I Z_s| = E$, so

$$V^2 + 2V(631.23)\cos 123.06^\circ + 631.23^2 = 4249.05^2$$ $$V^2 - 688.62\,V - 17.656\times10^{6} = 0$$ $$V_{ph} = \frac{688.62 + \sqrt{688.62^2 + 4\times17.656\times10^{6}}}{2} = 4560.3\ \text{V}$$ $$V_L = \sqrt3\times 4560.3 = 7898.7\ \text{V}$$

Answer: terminal voltage at 0.8 leading $\approx 7899$ V (line), i.e. 4560 V per phase, higher than rated because of the leading current (Ferranti-type effect; armature reaction aids the field).