Regulation and power angle of 1650 kVA generator

Given: star, 1650 kVA, 13.8 kV, $R_a = 1.6\ \Omega$, $X_s = 35\ \Omega$; load 1320 kW at 0.8 pf (= 1650 kVA, full load).

$$\begin{aligned} V_{ph} &= \frac{13800}{\sqrt3} = 7967.43\ \text{V} \\ I &= \frac{1320\times10^3}{\sqrt3\times13800\times0.8} = 69.03\ \text{A} \\ V\cos\phi + IR_a &= 6373.95 + 110.45 = 6484.40\ \text{V} \\ V\sin\phi &= 4780.46\ \text{V},\quad IX_s = 2416.09\ \text{V} \end{aligned}$$

The power angle $\delta$ is the angle between $E$ and $V$. If $\theta$ is the angle of $E$ from the current phasor, $\theta = \tan^{-1}\dfrac{V\sin\phi \pm IX_s}{V\cos\phi + IR_a}$, then $\delta = \theta - \phi$ for lagging pf and $\delta = \phi - \theta$ for leading pf.

(a) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{6484.40^2 + (4780.46 + 2416.09)^2} = \sqrt{6484.40^2 + 7196.55^2} = 9686.98\ \text{V} \\ \theta &= \tan^{-1}\frac{7196.55}{6484.40} = 47.98^\circ \\ \delta &= 47.98^\circ - 36.87^\circ = 11.11^\circ \\ \%\text{Reg} &= \frac{9686.98 - 7967.43}{7967.43}\times100 = 21.58\% \end{aligned}$$

(b) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{6484.40^2 + (4780.46 - 2416.09)^2} = \sqrt{6484.40^2 + 2364.37^2} = 6902.00\ \text{V} \\ \theta &= \tan^{-1}\frac{2364.37}{6484.40} = 20.03^\circ \\ \delta &= 36.87^\circ - 20.03^\circ = 16.84^\circ \\ \%\text{Reg} &= \frac{6902.00 - 7967.43}{7967.43}\times100 = -13.37\% \end{aligned}$$

(In both cases $E$ leads $V$, as it must for a generator.)

Answer: (a) regulation = 21.58%, power angle = 11.11°; (b) regulation = −13.37%, power angle = 16.84°.