Combined Load with a Synchronous Motor — Power Factor: 22-Elec-A6 Power Systems and Machines, Undated paper, 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) whether the synchronous motor still sits at a $35^{\circ}$ power angle once the supply frequency falls 5% and the load falls 10% — i.e. the new power angle.

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 pull-out capability unchanged. The power angle then adjusts only to the new load: $$\sin\delta_2 = \frac{P'}{P}\sin\delta_1 = 0.90\,\sin 35^{\circ} = 0.90(0.5736) = 0.5162,$$ $$\boxed{\delta_2 = \sin^{-1}(0.5162) = 31.1^{\circ} \ne 35^{\circ}.}$$ So the answer to the question as posed is no — the motor will not still sit at $35^{\circ}$. The angle falls to about $31^{\circ}$, because a synchronous machine holds its speed rigidly and signals a lighter shaft load only by letting its rotor swing closer into line with the stator field.

Check — assumptions behind part (c). The printed sentence has lost a verb ("…and the load by 10% if the synchronous motor has a power angle of 35 for rated voltage, frequency and torque"), so two modelling choices must be declared, exactly as the paper's own rubric invites. (1) Excitation and terminal voltage: the baseline is stated at rated voltage, so $V$ is held at rated and the field current is unchanged; then $E_a\propto f$ and $X_s = 2\pi f L\propto f$ cancel in the pull-out term $VE_a/X_s$ and the $-5\%$ frequency change does not alter loadability. (2) What "the load" means: taken as the motor's stated real intake (10 kW), so $P'/P = 0.90$ and $\delta_2 = 31.1^{\circ}$ — the value boxed above.

The alternative readings are all close and none changes the verdict. Reading "load" as a torque reduction (the baseline does say "rated…torque") makes $P' \propto T\omega = 0.90\times0.95$, giving $\delta_2 = 29.4^{\circ}$. Assuming instead a constant-$V/f$ supply policy scales $VE_a/X_s$ by 0.95, giving $\sin\delta_2 = (0.90/0.95)\sin35^{\circ}$ and $\delta_2 = 32.9^{\circ}$. Every reading lands between $29^{\circ}$ and $33^{\circ}$, i.e. below the original $35^{\circ}$, so the answer — the power angle does not stay at $35^{\circ}$, it decreases — is robust to the ambiguity. State whichever assumption you adopt and carry it through.

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 $\delta_2$	31.1° — not 35° (rated-V basis; 29.4°–32.9° across the alternative assumptions)

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)