5 MVA alternator: regulation and pf for zero regulation

Given: 5 MVA, 11 kV, star, $R_a = 0.12\ \Omega$, $X_s = 2\ \Omega$ per phase, full load at 0.9 pf lagging.

$$V_{ph} = \frac{11000}{\sqrt3} = 6350.85\ \text{V}, \qquad I = \frac{5\times10^6}{\sqrt3\times11000} = 262.43\ \text{A}$$ $$IR_a = 31.49\ \text{V}, \qquad IX_s = 524.86\ \text{V}$$

Voltage regulation at 0.9 lagging

$\cos\phi = 0.9$, $\sin\phi = 0.4359$:

$$E = \sqrt{(V + IR_a\cos\phi + IX_s\sin\phi)^2 + (IX_s\cos\phi - IR_a\sin\phi)^2}$$ $$= \sqrt{(6350.85 + 28.34 + 228.78)^2 + (472.38 - 13.73)^2} = \sqrt{6607.97^2 + 458.65^2}$$ $$E_{ph} = 6623.9\ \text{V}$$ $$\text{Regulation} = \frac{6623.9-6350.85}{6350.85}\times100 = 4.30\%$$

Power factor for zero regulation

For zero regulation $E = V$ at full load current. The current must lead. Let $\vec I = I\angle\theta$ with $\theta > 0$ leading, $Z_s = 2.003\angle86.57^\circ$ ($|Z_s| = \sqrt{0.12^2+2^2}= 2.0036\ \Omega$):

$$|\vec V + \vec IZ_s|^2 = V^2 \;\Rightarrow\; 2VI|Z_s|\cos(\theta + 86.57^\circ) + I^2|Z_s|^2 = 0$$ $$\cos(\theta+86.57^\circ) = -\frac{I|Z_s|}{2V} = -\frac{262.43\times2.0036}{2\times6350.85} = -0.0414$$ $$\theta + 86.57^\circ = 92.37^\circ \;\Rightarrow\; \theta = 5.81^\circ \text{ (leading)}$$ $$\cos\theta = 0.9949$$

Answer: regulation at 0.9 lagging $= 4.30\%$ (E = 6623.9 V per phase). Regulation is zero at 0.995 leading power factor.