Regulation of delta connected 1600 kVA alternator

Given: delta connected, so $V_{ph} = V_L = 13.5$ kV; $R_a = 1.5\ \Omega$, $X_s = 3\ \Omega$ per phase; load 1.28 MW at 0.8 pf, i.e. $1.28/0.8 = 1.6$ MVA (full load).

$$\begin{aligned} I_{ph} &= \frac{1.6\times10^6}{3\times13500} = 39.51\ \text{A} \\ V\cos\phi + IR_a &= 13500\times0.8 + 39.51\times1.5 = 10800 + 59.26 = 10859.26\ \text{V} \\ V\sin\phi &= 8100\ \text{V},\quad IX_s = 39.51\times3 = 118.52\ \text{V} \end{aligned}$$

(i) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{10859.26^2 + (8100 + 118.52)^2} = \sqrt{10859.26^2 + 8218.52^2} \\ &= 13618.65\ \text{V} \\ \%\text{Reg} &= \frac{13618.65 - 13500}{13500}\times100 = 0.879\% \end{aligned}$$

(ii) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{10859.26^2 + (8100 - 118.52)^2} = \sqrt{10859.26^2 + 7981.48^2} \\ &= 13476.93\ \text{V} \\ \%\text{Reg} &= \frac{13476.93 - 13500}{13500}\times100 = -0.171\% \end{aligned}$$

The regulation is small because the given impedance (3 Ω) is very small compared with the phase voltage.

Answer: (i) +0.88% at 0.8 pf lagging; (ii) −0.17% at 0.8 pf leading.