Question 4 of 6: Synchronous Generator on an Infinite Bus

Question 4: Synchronous Generator on an Infinite Bus (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Y-connected, 3-phase synchronous generator, 14 kV (line) / 10 MVA / 60 Hz, 2 poles, $X_S=20.0\ \Omega$, $R_A=2.0\ \Omega$ per phase, on an infinite bus at 0.85 lagging.

Find. (a) rotor speed; (b) internal generated (excitation) voltage $E_A$ at rated power; (c) power angle $\delta$; (d) maximum power at constant field (neglect $R_A$); (e) current and power factor at that maximum.

Approach. Speed comes from the pole/frequency relation. Rated per-phase quantities give the armature current; $E_A=V_\phi+I(R_A+jX_S)$ gives the excitation voltage and $\delta$. With field (hence $|E_A|$) fixed, maximum power occurs at $\delta=90^\circ$; the current and pf follow from the phasor equation with $R_A$ neglected.

1. Speed (part a). $n_s=\dfrac{120f}{p}=\dfrac{120(60)}{2}=\boxed{3600\ \text{rpm}}$.
2. Rated per-phase quantities. $V_\phi=\dfrac{14\,000}{\sqrt3}=8083\ \text{V}$; rated current $I_L=\dfrac{S}{\sqrt3\,V_L}=\dfrac{10\times10^6}{\sqrt3\,(14\,000)}=412.4\ \text{A}$, at $\theta=\cos^{-1}0.85=31.8^\circ$ lagging, so $I=412.4\angle{-}31.8^\circ$ A.
3. Excitation voltage (part b). With $Z_S=R_A+jX_S=2+j20=20.10\angle84.3^\circ\ \Omega$, $$E_A=V_\phi+I\,Z_S=8083+ (412.4\angle{-}31.8^\circ)(20.10\angle84.3^\circ)=13\,129+j6577$$ $$\boxed{E_A=14.68\ \text{kV per phase}\ (=25.43\ \text{kV line})}$$
4. Power angle (part c). $\delta=\angle E_A=\tan^{-1}\dfrac{6577}{13\,129}=\boxed{26.6^\circ}$.
5. Maximum power at constant field (part d). Holding the field constant fixes $|E_A|=14.68$ kV. Neglecting $R_A$, the three-phase power is $P=\dfrac{3V_\phi E_A}{X_S}\sin\delta$, maximum at $\delta=90^\circ$: $$\boxed{P_{max}=\frac{3V_\phi E_A}{X_S}=\frac{3(8083)(14\,683)}{20}=17.80\ \text{MW}}$$
6. Current and pf at maximum (part e). At $\delta=90^\circ$, $E_A=14\,683\angle90^\circ$ V. With $R_A$ neglected, $$I=\frac{E_A-V_\phi}{jX_S}=\frac{(-8083+j14\,683)}{20\angle90^\circ}=838\angle{+}28.8^\circ\ \text{A}$$ $$\boxed{I=838\ \text{A},\qquad \text{pf}=\cos(28.8^\circ)=0.876\ \text{leading}}$$ Check: $P=3V_\phi I\cos\theta=3(8083)(838)(0.876)=17.80\ \text{MW}$ ✓, matching part (d).