Combining a Lagging Load With a Leading Synchronous Motor, and the Effect of Frequency on Power Angle: 16-Mechatronics-B10, December 2019, Q4

Question 4: Combining a Lagging Load With a Leading Synchronous Motor, and the Effect of Frequency on Power Angle (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.

Quantity	Value
Existing factory load, S₁	100 kVA at 0.45 p.f. lagging, 60 Hz
Added synchronous motor input power, P₂	10 kW at 0.2 p.f. leading
Rated power angle (at rated V, f, torque), δ_rated	35°
Frequency change	−5%
Load (torque) change	−10%

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

Check

Part (c) is read as: at rated voltage/frequency/torque the motor's power angle is 35°; find the new power angle when frequency drops 5% and torque (shaft load) drops 10%, holding terminal voltage V and field excitation E_f fixed (no proportional V/f adjustment is stated). This is the only reading consistent with the question's own phrase "for rated voltage, frequency and torque."

Approach. Resolve both loads to real/reactive power (a leading load subtracts vars), sum for (a)/(b); for (c), use the round-rotor power equation P = (VE_f/X_s)sinδ with X_s ∝ f and torque T = P/ω_s, ω_s ∝ f, to get sinδ ∝ Tf².

Existing plant load, resolved. $$P_1=S_1\cos\varphi_1=100(0.45)=45.0\ \text{kW}$$ $$Q_1=S_1\sin\varphi_1=100\sin(\cos^{-1}0.45)=\boxed{89.30\ \text{kVAR (lagging)}}$$
Synchronous-motor contribution (leading ⇒ supplies vars). $$Q_2=-P_2\tan(\cos^{-1}0.2)=-10(4.899)=\boxed{-48.99\ \text{kVAR}}$$
(a) Total real power. $$P_T=P_1+P_2=45.0+10=\boxed{55.0\ \text{kW}}$$
(b) Overall power factor. $$Q_T=Q_1+Q_2=89.30-48.99=40.31\ \text{kVAR (net lagging)}$$ $$S_T=\sqrt{P_T^2+Q_T^2}=\sqrt{55.0^2+40.31^2}=68.19\ \text{kVA}$$ $$\text{pf}_T=\frac{P_T}{S_T}=\boxed{0.807\ \text{lagging}}$$
(c) New power angle after −5% frequency, −10% torque. With X_s=2πfL, V and E_f fixed: $$\sin\delta \propto P\,X_s \propto (T f)f = T f^2$$ $$\frac{\sin\delta_{new}}{\sin\delta_{rated}}=\left(\frac{T_{new}}{T_{rated}}\right)\left(\frac{f_{new}}{f_{rated}}\right)^2=(0.9)(0.95)^2=0.8123$$ $$\sin\delta_{new}=\sin(35^{\circ})(0.8123)=0.5736(0.8123)=\boxed{0.4659}$$ $$\delta_{new}=\sin^{-1}(0.4659)=\boxed{27.8^{\circ}}$$

Quantity	Value
(a) Total real power, P_T	55.0 kW
(b) Overall power factor	0.807 lagging
(c) New power angle, δ_new	27.8° (was 35°)

Concept & Theory

Topic: Complex-power combination of mixed-p.f. loads; synchronous-motor power-angle sensitivity to frequency and torque.

Key relations: P=S cosφ, Q=S sinφ (sign by lagging/leading); P=(VE_f/X_s)sinδ, X_s∝f, T=P/ω_s, ω_s∝f.

Why this works: leading-p.f. loads (over-excited synchronous motors, capacitor banks) act as local reactive-power sources, cancelling part of the lagging vars drawn by induction-motor-heavy plant loads — this is exactly the power-factor correction service an over-excited synchronous motor sells. The power-angle relation follows directly from the round-rotor synchronous machine's power equation once the frequency dependence of X_s and the torque–power–speed link are substituted in.

Common pitfall: adding kVA or p.f. values directly instead of resolving to P and Q first (only P and Q add algebraically for loads in parallel); dropping the sign that a leading load subtracts reactive power rather than adding it.

Source: Chapman, Electric Machinery Fundamentals, Ch. 4 §4.13 (power-factor correction) and Ch. 6 (synchronous-motor power-angle equation).

Question 4 of 5: Combining a Lagging Load With a Leading Synchronous Motor, and the Effect of Frequency on Power Angle

Question 4: Combining a Lagging Load With a Leading Synchronous Motor, and the Effect of Frequency on Power Angle (20 marks)