Power-Factor Correction with a Synchronous Motor: 22-Elec-A6 Power Systems and Machines, May 2018, Q4

Question 4: Power-Factor Correction with a Synchronous Motor (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. An existing factory load in parallel with a newly added synchronous motor (running over-excited, hence leading). Real and reactive powers add as phasors on the complex-power plane.

Given data
Load	Apparent / real power	Power factor
Factory	$S_1=100$ kVA	0.45 lagging
Synchronous motor	$P_2=10$ kW	0.2 leading

Find. (a) the total (real, and apparent) power; (b) the overall power factor; (c) the new rotor power angle when $f$ is reduced 5% and the load (torque) is reduced 10%, starting from $\delta_1=35^\circ$ at rated conditions.

Complex-power (P–Q) diagram: the lagging factory load and the leading synchronous-motor load add head-to-tail; the leading vars of the motor partly cancel the lagging vars of the factory, shrinking the resultant apparent power and improving the overall power factor.

Approach. Resolve each load into $P$ and $Q$ (lagging positive, leading negative), sum component-wise, then recombine into $S$ and $\cos\theta$. For part (c), express the synchronous-motor torque in terms of $V$, $f$ and $\delta$ and solve for the new angle.

Resolve the factory load. $P_1=S_1\cos\theta_1=100(0.45)=45\ \text{kW}$; with $\sin\theta_1=\sqrt{1-0.45^2}=0.893$, $$Q_1=S_1\sin\theta_1=100(0.893)=+89.3\ \text{kvar (lagging).}$$
Resolve the synchronous motor. $S_2=P_2/\cos\theta_2=10/0.2=50\ \text{kVA}$; $\sin\theta_2=\sqrt{1-0.2^2}=0.980$, and leading so $Q_2$ is negative: $$Q_2=-S_2\sin\theta_2=-50(0.980)=-48.99\ \text{kvar (leading).}$$
(a) Total real, reactive and apparent power. $$P=P_1+P_2=45+10=55\ \text{kW},\qquad Q=Q_1+Q_2=89.3-48.99=40.31\ \text{kvar.}$$ $$S=\sqrt{P^{2}+Q^{2}}=\sqrt{55^{2}+40.31^{2}}=\boxed{68.2\ \text{kVA}\ (P=55\ \text{kW}).}$$
(b) Overall power factor. Still lagging (net $Q$ positive): $$\cos\theta=\frac{P}{S}=\frac{55}{68.2}=\boxed{0.807\ \text{lagging.}}$$ The motor has raised the plant power factor from 0.45 to 0.81 while doing 10 kW of useful work.
(c) New power angle after $f\downarrow5\%$, load$\downarrow10\%$. For a round-rotor synchronous machine the developed torque is $$T=\frac{3\,V\,E_a}{\omega_s X_s}\sin\delta.$$ With the terminal voltage held at rated ($V$ constant), constant field excitation so $E_a\propto f$, and $X_s=\omega L\propto f$, $\omega_s\propto f$, the frequency factors combine to $T\propto \dfrac{V\sin\delta}{f}$. Taking the ratio of the new state ($T_2=0.90\,T_1$, $f_2=0.95\,f_1$, $V$ unchanged) to the rated state: $$\frac{T_2}{T_1}=\frac{\sin\delta_2}{\sin\delta_1}\cdot\frac{f_1}{f_2}\ \Rightarrow\ 0.90=\frac{\sin\delta_2}{\sin 35^\circ}\cdot\frac{1}{0.95}.$$ $$\sin\delta_2=0.90\times0.95\times\sin 35^\circ=0.490\ \Rightarrow\ \boxed{\delta_2=29.4^\circ.}$$

Check: Question 4(c) is worded ambiguously in the source (a double "if" clause). It is solved here on the physical reading that "load reduced by 10%" means the mechanical torque falls to $0.90\,T$ while the terminal voltage stays at rated and the field is unchanged, giving $T\propto V\sin\delta/f$ and $\delta_2=29.4^\circ$. If instead only the electrical power is scaled with $E_a/X_s$ held constant ($P\propto V\sin\delta$, frequency immaterial), then $\sin\delta_2=0.90\sin35^\circ\Rightarrow\delta_2=31.1^\circ$. Both readings agree that the angle decreases (the machine is more lightly loaded), differing only by the treatment of the 5% frequency change.

Final Results — Question 4
Quantity	Result
Total real power $P$	55 kW
Total reactive power $Q$	40.3 kvar (lagging)
Total apparent power $S$	68.2 kVA
Overall power factor	0.807 lagging
New power angle $\delta_2$	29.4° (torque reading)

Topic: Complex-power addition, power-factor correction by an over-excited synchronous motor, and the synchronous power-angle relation.

Key relations: $S=P+jQ$; $Q=S\sin\theta$ (lagging +, leading −); $\cos\theta=P/S$; synchronous-motor torque $T=\dfrac{3VE_a}{\omega_sX_s}\sin\delta$.

Why this works: Real and reactive powers are conserved separately, so parallel loads add as vectors on the P–Q plane. An over-excited synchronous motor draws leading current, injecting vars that cancel part of the lagging load; the plant does more real work yet draws less apparent power, which is exactly why synchronous motors and condensers are used for power-factor correction. The power-angle expression shows how the rotor angle re-settles when supply conditions change: reducing torque lowers $\sin\delta$, while lowering frequency at fixed voltage over-fluxes the machine and raises the torque per degree, so both effects push $\delta$ down here.

Common pitfall: giving the leading vars the wrong sign (which would worsen the computed power factor), or reporting the overall p.f. as leading — the net $Q$ is still positive here, so it remains lagging. In part (c), holding $E_a$ constant instead of $\propto f$ drops the frequency dependence entirely.

Source: Glover et al., Power System Analysis and Design, 6th ed., §2.5 (complex power); Chapman, 5th ed., §4.5 (synchronous-motor power and torque).

Load	Apparent / real power	Power factor
Factory	\(S_1=100\) kVA	0.45 lagging
Synchronous motor	\(P_2=10\) kW	0.2 leading

Quantity	Result
Total real power \(P\)	55 kW
Total reactive power \(Q\)	40.3 kvar (lagging)
Total apparent power \(S\)	68.2 kVA
Overall power factor	0.807 lagging
New power angle \(\delta_2\)	29.4° (torque reading)

Question 4 of 5: Power-Factor Correction with a Synchronous Motor

Question 4: Power-Factor Correction with a Synchronous Motor (20 marks)