Regulation of 1.5 MVA generator and zero-regulation pf

Given: star, 1.5 MVA, 11 kV, $R_a = 1.2\ \Omega$, $X_s = 25\ \Omega$; load 1.4375 MVA at 0.8 pf.

$$\begin{aligned} V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\ I &= \frac{1.4375\times10^6}{\sqrt3\times11000} = 75.45\ \text{A} \\ V\cos\phi + IR_a &= 5080.68 + 90.54 = 5171.22\ \text{V} \\ V\sin\phi &= 3810.51\ \text{V},\quad IX_s = 1886.23\ \text{V} \end{aligned}$$

(i) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{5171.22^2 + (3810.51 + 1886.23)^2} = 7693.79\ \text{V} \\ \%\text{Reg} &= \frac{7693.79 - 6350.85}{6350.85}\times100 = 21.15\% \end{aligned}$$

(ii) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{5171.22^2 + (3810.51 - 1886.23)^2} = 5517.64\ \text{V} \\ \%\text{Reg} &= \frac{5517.64 - 6350.85}{6350.85}\times100 = -13.12\% \end{aligned}$$

Power factor for zero regulation

For $E = V$, with $Z_s = 1.2 + j25 = 25.029\angle 87.25^\circ\ \Omega$ and current angle $\phi$ (leading positive):

$$\begin{aligned} \cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{75.45\times25.029}{2\times6350.85} = -0.1487 \\ \theta_z + \phi &= 98.55^\circ \\ \phi &= 98.55^\circ - 87.25^\circ = 11.30^\circ \\ \text{pf} &= \cos 11.30^\circ = 0.981\ \text{leading} \end{aligned}$$

Answer: (i) 21.15%; (ii) −13.12%; zero regulation at pf = 0.981 leading.