22-Elec-A6 Power Systems and Machines · May 2016

Question 2 of 7: Three-Phase Y Load, Motor Addition and PF Correction

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.

Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.

Question 2: Three-Phase Y Load, Motor Addition and PF Correction (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.

Given data — Question 2
Y-load impedance / phase	Z = 17∠62° Ω
Distribution-line impedance	Z_L = 0.1 + j0.3 Ω
Load voltage (line-to-line)	600 V ⇒ V_ph = 346.4 V
Induction motor	23 hp out, η = 83%, pf = 0.85 lag
Frequency	60 Hz

Find. (a) source line current and voltage; (b) load P and Q; (c) motor P, Q, new line current and source voltage; (d) capacitor vars and Y phase impedance for unity pf; (e) line current after correction; (f) phasor diagram.

Figure 2 — one-line diagram: source through the distribution line to the 600 V load bus feeding the Y load, the induction motor, and (parts d–e) the correction capacitor bank.

Approach. Work per phase with the load phase voltage as reference. Get the Y-load current directly from Ohm's law, add the motor as a complex-power load at the bus, and carry the total current through Z_L for the source voltage. Size the capacitor bank to cancel the total reactive power.

Line current and source voltage (Y load alone). With $V_{\text{ph}}=\dfrac{600}{\sqrt3}=346.4\angle 0^\circ$ V, $$I_a=\frac{V_{\text{ph}}}{Z}=\frac{346.4\angle 0^\circ}{17\angle 62^\circ}=20.38\angle{-62^\circ}\text{ A}.$$ The source phase voltage is $V_{s}=V_{\text{ph}}+I_aZ_L=346.4+ (20.38\angle{-62^\circ})(0.316\angle 71.57^\circ)=352.8\angle 0.17^\circ$ V, so $$\boxed{I_{\text{line}}=20.38\text{ A},\qquad V_{s,LL}=\sqrt3(352.8)=611.0\text{ V}.}$$
Load real and reactive power. Using $S=3V_{\text{ph}}I=3(346.4)(20.38)=21.18$ kVA at 62°, $$\boxed{P_Y=3|I|^2R=9.94\text{ kW},\qquad Q_Y=3|I|^2X=18.70\text{ kvar}.}$$ (R = 17cos62° = 7.98 Ω, X = 17sin62° = 15.01 Ω.)
Induction-motor power. The mechanical output is $23\times746=17\,158$ W, so the electrical input is $P_m=\dfrac{17\,158}{0.83}=20.67$ kW, with $Q_m=P_m\tan(\cos^{-1}0.85)=12.81$ kvar and per-phase current $I_m=\dfrac{|S_m|}{3V_{\text{ph}}}=\dfrac{24.32\text{ kVA}}{3(346.4)}=23.41$ A.
New line current and source voltage (load + motor). Totals at the bus: $P_T=9.94+20.67=30.61$ kW, $Q_T=18.70+12.81=31.51$ kvar. The line current magnitude is $$\boxed{|I_{\text{line}}|=\frac{|S_T|}{3V_{\text{ph}}}=\frac{43.94\text{ kVA}}{3(346.4)}=42.27\text{ A}\ (\angle{-45.83^\circ}),}$$ and $V_{s}=346.4+I_{\text{line}}Z_L=358.5\angle 0.93^\circ$ V → $V_{s,LL}=620.9$ V.
Capacitor bank for unity power factor. To cancel all reactive power the bank must supply $Q_C=Q_T=31.51$ kvar (three-phase). Per phase $Q_{C,\text{ph}}=10.50$ kvar, so for a Y bank at $V_{\text{ph}}=346.4$ V, $$\boxed{X_C=\frac{V_{\text{ph}}^2}{Q_{C,\text{ph}}}=\frac{346.4^2}{10\,503}=11.43\ \Omega\ (\text{i.e. } Z=-j11.43\ \Omega/\text{phase}).}$$
Line current after correction. At unity pf the bus draws only real power, so $$\boxed{I_{\text{line}}=\frac{P_T}{\sqrt3\,V_{LL}}=\frac{30\,614}{\sqrt3(600)}=29.46\text{ A}.}$$ The current falls from 42.27 A to 29.46 A.
Phasor diagram (part f). Before correction the line current lags the load voltage by 45.83°; after correction it is in phase with V and shorter. See the figure below.

Figure 2(f) — line-current phasors: 42.3 A lagging 45.8° (uncorrected) versus 29.5 A in phase with V (corrected to unity).

Final results — Question 2
Quantity	Value
(a) Line current / source voltage (Y load)	20.38 A / 611.0 V (LL)
(b) Y-load power	P = 9.94 kW, Q = 18.70 kvar
(c) Motor power / current	P = 20.67 kW, Q = 12.81 kvar, I_m = 23.41 A
(c) Line current / source voltage (with motor)	42.27 A / 620.9 V (LL)
(d) Capacitor bank	31.51 kvar; Y phase X_C = 11.43 Ω
(e) Line current at unity pf	29.46 A

Topic: Balanced three-phase per-phase analysis, complex-power aggregation, and shunt VAR compensation.

Key relations: $V_{\text{ph}}=V_{LL}/\sqrt3$; $S_{3\phi}=3V_{\text{ph}}I^*=\sqrt3\,V_{LL}I_L$; parallel loads add in S; $Q_C=V_{\text{ph}}^2/X_C$ per phase; $\eta=P_{\text{out}}/P_{\text{in}}$.

Why this works: A balanced system is fully described by one phase, so the per-phase circuit carries all the physics while the line-to-line and three-phase totals follow by the $\sqrt3$ and factor-3 relations. Because every load shares the bus voltage, their complex powers add directly; the capacitor supplies leading vars that exactly offset the lagging vars, leaving only real current in the line.

Common pitfall: Using $V_{LL}$ where $V_{\text{ph}}$ is required (a $\sqrt3$ error), or sizing the Y capacitor with the line voltage instead of the phase voltage.

Source: Glover, Power System Analysis and Design §2.5–2.6; Chapman, Electric Machinery Fundamentals §1.

← Question 1 Next: Question 3 →