Regulation of 10 kVA round-rotor generator

Given: star, 10 kVA, 230 V, $R_a = 0.5\ \Omega$, $X_s = 1.2\ \Omega$ per phase.

$$\begin{aligned} V_{ph} &= \frac{230}{\sqrt3} = 132.79\ \text{V},\quad I = \frac{10000}{\sqrt3\times230} = 25.10\ \text{A} \\ V\cos\phi + IR_a &= 106.23 + 12.55 = 118.78\ \text{V} \\ V\sin\phi &= 79.67\ \text{V},\quad IX_s = 30.12\ \text{V} \end{aligned}$$

(a) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{118.78^2 + (79.67 + 30.12)^2} = 161.76\ \text{V} \\ \%\text{Reg} &= \frac{161.76 - 132.79}{132.79}\times100 = 21.81\% \end{aligned}$$

(b) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{118.78^2 + (79.67 - 30.12)^2} = 128.70\ \text{V} \\ \%\text{Reg} &= \frac{128.70 - 132.79}{132.79}\times100 = -3.08\% \end{aligned}$$

(c) Power factor for zero regulation

Zero regulation means $E = V$. With $Z_s = R_a + jX_s = 1.3\angle 67.38^\circ\ \Omega$ and current $I\angle\phi$ ($\phi$ positive for leading):

$$\begin{aligned} E^2 &= V^2 + 2VIZ_s\cos(\theta_z + \phi) + I^2Z_s^2 = V^2 \\ \cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{25.10\times1.3}{2\times132.79} = -0.1229 \\ \theta_z + \phi &= 97.06^\circ \\ \phi &= 97.06^\circ - 67.38^\circ = 29.68^\circ\ \text{(leading)} \\ \text{pf} &= \cos 29.68^\circ = 0.869\ \text{leading} \end{aligned}$$

Answer: (a) 21.81%; (b) −3.08%; (c) zero regulation at pf = 0.869 leading.