Combined Load with a Synchronous Motor — Power Factor: 22-Elec-A6 Power Systems and Machines, December 2018, Q4

Question 4: Combined Load with a Synchronous Motor — Power Factor (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 lagging factory load in parallel with a leading-p.f. synchronous motor.

Given data
Factory load	$S_1 = 100\ \text{kVA}$ at 0.45 p.f. lagging
Synchronous motor	0.2 p.f. leading, real intake $P_2 = 10\ \text{kW}$
Motor power angle (c)	$\delta_1 = 35^{\circ}$ at rated voltage
Change (c)	supply frequency −5%, load −10%

Find. (a) total real power; (b) overall power factor; (c) the new power angle after the frequency and load change.

Approach. Resolve each load into real ($P$) and reactive ($Q$) components — lagging vars positive, leading vars negative — add them, and recombine into an overall apparent power and power factor. For (c), use the synchronous-machine power relation $P=\frac{V E_a}{X_s}\sin\delta$ and track how each factor scales with frequency.

Factory load components. $$P_1 = S_1\cos\theta_1 = 100(0.45) = 45\ \text{kW},\qquad Q_1 = S_1\sin\theta_1 = 100\sqrt{1-0.45^2} = 89.3\ \text{kvar (lagging)}.$$
Synchronous-motor components. From its real intake and leading power factor, $$S_2 = \frac{P_2}{\cos\theta_2} = \frac{10}{0.2} = 50\ \text{kVA},\qquad Q_2 = S_2\sin\theta_2 = 50\sqrt{1-0.2^2} = 48.99\ \text{kvar (leading)}.$$
Total real power (a). $$P_{tot} = P_1 + P_2 = 45 + 10 = 55\ \text{kW}.$$
Total reactive power. The leading motor vars partially cancel the lagging factory vars: $$Q_{tot} = Q_1 - Q_2 = 89.3 - 48.99 = 40.31\ \text{kvar (net lagging)}.$$
Overall apparent power and power factor (b). $$S_{tot} = \sqrt{P_{tot}^2 + Q_{tot}^2} = \sqrt{55^2 + 40.31^2} = 68.2\ \text{kVA},$$ $$\boxed{\text{p.f.} = \frac{P_{tot}}{S_{tot}} = \frac{55}{68.2} = 0.807\ \text{lagging}.}$$ The synchronous motor has improved the plant power factor from 0.45 to 0.81 while still doing 10 kW of useful work.
New power angle after the change (c). For the synchronous motor, $P = \dfrac{V E_a}{X_s}\sin\delta$. Holding the terminal voltage at its rated value and the field constant, the internal EMF scales with frequency ($E_a \propto f$) and so does the synchronous reactance ($X_s = 2\pi f L \propto f$); these two effects cancel in $VE_a/X_s$, leaving the electrical loadability unchanged. The power angle then adjusts only to the new load: $$\sin\delta_2 = \frac{P_2'}{P_2}\sin\delta_1 = 0.90\,\sin 35^{\circ} = 0.90(0.5736) = 0.5162,$$ $$\delta_2 = \sin^{-1}(0.5162) = 31.1^{\circ}.$$ The angle falls from $35^{\circ}$ to about $31^{\circ}$ because the shaft load has been reduced by 10%.

Check: Part (c) as printed is garbled in the source ("…if the frequency and torque the supply frequency is reduced by 5% and the load by 10% if the."). It is solved under the stated "rated voltage" condition with the field held constant, so $E_a\propto f$ and $X_s\propto f$ cancel and the $-5\%$ frequency change does not alter the sync-motor loadability; the power angle then responds only to the $-10\%$ load, giving $\delta_2\approx31^{\circ}$. If instead a constant-$V/f$ policy is assumed, $VE_a/X_s$ scales by $0.95$ and $\sin\delta_2 = (0.90/0.95)\sin35^{\circ}$, giving $\delta_2\approx32.9^{\circ}$. If the 10% load cut is read as a torque reduction (the printed wording mentions torque), the shaft power also falls with the 5% lower synchronous speed, so $\sin\delta_2 = (0.90)(0.95)\sin35^{\circ}$ and $\delta_2\approx29.4^{\circ}$. Every reading lands between about $29^{\circ}$ and $33^{\circ}$, so the conclusion is the same: the power angle decreases from $35^{\circ}$.

Final results — Question 4
Quantity	Value
(a) Total real power $P_{tot}$	55 kW
(b) Overall power factor	0.807 lagging ($S_{tot}=68.2$ kVA)
(c) New power angle $\delta_2$	≈ 31° (rated-V basis)

Topic: Power-factor correction with a synchronous motor and the synchronous power–angle relation.

Key relations: $P = S\cos\theta$, $Q = S\sin\theta$ (lag +, lead −); $S_{tot}=\sqrt{P_{tot}^2+Q_{tot}^2}$; $\text{p.f.}=P_{tot}/S_{tot}$; $P=\frac{V E_a}{X_s}\sin\delta$.

Why this works: An over-excited synchronous motor draws leading current, so it supplies reactive power to the system just like a capacitor while simultaneously delivering mechanical power. Adding its leading vars to the plant's lagging vars reduces the net reactive demand, raising the overall power factor. Because real and reactive powers each add algebraically, resolving every load into a $P+jQ$ pair and summing is always valid. In the power–angle relation the angle $\delta$ is set by how hard the machine is loaded relative to its pull-out capability $VE_a/X_s$.

Common pitfall: Adding apparent powers (kVA) arithmetically instead of adding $P$ and $Q$ as phasors; or sign-confusing the leading motor vars. In part (c), forgetting that both $E_a$ and $X_s$ move with frequency — if only one is scaled the loadability change is mis-estimated.

Source: Chapman, §5.4 (synchronous motor power factor) and §4.5 (power–angle characteristic); Sen, ch. 6.

Question 4 of 5: Combined Load with a Synchronous Motor — Power Factor

Question 4: Combined Load with a Synchronous Motor — Power Factor (20 marks)