Emf and regulation of 1500 kVA alternator

Given: star connected, $V_L$ = 12000 V, $R_a = 2\ \Omega$, $X_s = 25\ \Omega$ per phase, load 1250 kW at 0.75 pf.

$$\begin{aligned} V_{ph} &= \frac{12000}{\sqrt3} = 6928.2\ \text{V} \\ I &= \frac{1250\times10^3}{\sqrt3\times12000\times0.75} = 80.19\ \text{A} \\ \cos\phi &= 0.75,\quad \sin\phi = 0.6614 \\ V\cos\phi + IR_a &= 5196.15 + 80.19\times2 = 5356.53\ \text{V} \\ V\sin\phi &= 4582.58\ \text{V},\quad IX_s = 80.19\times25 = 2004.69\ \text{V} \end{aligned}$$

(Note: this load is $1250/0.75 = 1667$ kVA, which is above the 1500 kVA rating. The calculation follows the data given in the question.)

(i) 0.75 pf leading

$$\begin{aligned} E &= \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2} \\ &= \sqrt{5356.53^2 + (4582.58 - 2004.69)^2} \\ &= \sqrt{5356.53^2 + 2577.89^2} = 5944.57\ \text{V per phase} \\ E_L &= \sqrt3 \times 5944.57 = 10296\ \text{V} \\ \%\text{Reg} &= \frac{5944.57 - 6928.2}{6928.2}\times100 = -14.20\% \end{aligned}$$

(ii) 0.75 pf lagging

$$\begin{aligned} E &= \sqrt{5356.53^2 + (4582.58 + 2004.69)^2} \\ &= \sqrt{5356.53^2 + 6587.26^2} = 8490.26\ \text{V per phase} \\ E_L &= \sqrt3 \times 8490.26 = 14706\ \text{V} \\ \%\text{Reg} &= \frac{8490.26 - 6928.2}{6928.2}\times100 = 22.55\% \end{aligned}$$

Answer: (i) 0.75 lead: E = 5944.6 V/phase (10.30 kV line), regulation = −14.20%; (ii) 0.75 lag: E = 8490.3 V/phase (14.71 kV line), regulation = +22.55%.