Power angle and regulation of 10 kVA alternator

Given: star, 10 kVA, 400 V, 50 Hz, $R_a = 0.5\ \Omega$, $X_s = 10\ \Omega$; rated load at 0.8 pf lagging.

Phase voltage and current

$$\begin{aligned} V_{ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\ I &= \frac{10\times10^3}{\sqrt3\times400} = 14.43\ \text{A} \\ \cos\phi &= 0.8,\quad \sin\phi = 0.6,\quad \phi = 36.87^\circ \end{aligned}$$

Excitation emf

$$\begin{aligned} V\cos\phi + IR_a &= 184.75 + 14.43\times0.5 = 191.97\ \text{V} \\ V\sin\phi + IX_s &= 138.56 + 14.43\times10 = 282.90\ \text{V} \\ E &= \sqrt{191.97^2 + 282.90^2} = 341.89\ \text{V per phase}\quad(592.2\ \text{V line}) \end{aligned}$$

Power angle

The angle of $E$ from the current phasor is

$$\begin{aligned} \theta &= \tan^{-1}\frac{282.90}{191.97} = 55.84^\circ \\ \delta &= \theta - \phi = 55.84^\circ - 36.87^\circ = 18.97^\circ \end{aligned}$$

                       E = 341.9 V
                      /
                     /  delta = 18.97 deg
                    O-------------- V = 230.9 V
                     \  phi = 36.87 deg
                      \ I = 14.43 A

Voltage regulation

$$\%\text{Reg} = \frac{E - V}{V}\times100 = \frac{341.89 - 230.94}{230.94}\times100 = 48.04\%$$

(The regulation is high because $X_s = 10\ \Omega$ is large compared with the rated impedance $V/I = 16\ \Omega$, i.e. $X_s \approx 0.625$ pu.)

Answer: power angle δ = 18.97°, voltage regulation = 48.04% (E = 341.9 V per phase).