Question 1 of 7: Feeder Analysis and Power-Factor 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 1: Feeder Analysis and Power-Factor Correction (20 marks)
Given. A single-phase (per-phase) feeder delivering to load L1, with a second parallel load L2 added for correction.
Given data — Question 1
Load L1 real power
P1 = 1500 kW
Load L1 power factor
0.75 lagging
Load-terminal voltage
VL = 2400 V (rms, reference)
Line impedance
Zline = 0.05 + j0.5 Ω
Added load L2
+100 kW, overall PF → 0.95 lagging
Find. (a) load current; (b) line loss; (c) source voltage; then with L2 added, (d) L2's reactive power and PF, and (e) the new combined current, line loss, and source voltage with VL held at 2400 V.
Figure 1.1/1.2 — single-phase feeder: source Vg through the line impedance to load L1, with L2 added in parallel for power-factor correction.
Approach. Resolve each load into complex power S = P + jQ, take the load-terminal voltage as the phase reference, obtain current from I = (S/V)*, then walk the line drop to the source with Vg = VL + I·Zline.
Load current from apparent power. The apparent power is $S_1=\dfrac{P_1}{\text{pf}}=\dfrac{1500}{0.75}=2000\text{ kVA}$, and the phase angle is $\theta_1=\cos^{-1}0.75=41.41^\circ$ (current lags voltage). With $V_L=2400\angle 0^\circ$, $$\boxed{I_1=\frac{S_1}{V_L}\angle{-\theta_1}=\frac{2\,000\,000}{2400}\angle{-41.41^\circ}=833.3\angle{-41.41^\circ}\text{ A}.}$$
Transmission-line loss. Only the line resistance dissipates power: $$P_{\text{loss}}=|I_1|^2R_{\text{line}}=(833.3)^2(0.05)=34.72\text{ kW}.$$
Source voltage. Adding the series drop, with $Z_{\text{line}}=0.05+j0.5=0.5025\angle 84.29^\circ$, $$V_g=V_L+I_1Z_{\text{line}}=2400+ (833.3\angle{-41.41^\circ})(0.5025\angle 84.29^\circ).$$ The drop is $418.8\angle 42.88^\circ=306.8+j284.9$, so $$\boxed{V_g=2706.8+j284.9=2722\angle 6.01^\circ\text{ V}.}$$
Reactive power of the added load L2. With L2 the total is $P_T=1600$ kW at 0.95 lag, so $Q_T=P_T\tan(\cos^{-1}0.95)=1600(0.3287)=525.9$ kvar. The original demand is $Q_1=1500\tan(\cos^{-1}0.75)=1322.9$ kvar, hence L2 supplies $$\boxed{Q_2=Q_T-Q_1=525.9-1322.9=-797.0\text{ kvar (i.e. 797 kvar capacitive).}}$$
Power factor of L2. With $P_2=100$ kW and $Q_2=-797$ kvar, $S_2=\sqrt{100^2+797^2}=803.3$ kVA, so $\text{pf}_2=\dfrac{100}{803.3}=0.1245$ leading — L2 is essentially a capacitor bank with a small real load.
Combined current with VL held at 2400 V. $S_T=1600+j525.9$ kVA, $|S_T|=1684$ kVA, so $$\boxed{|I|=\frac{|S_T|}{V_L}=\frac{1\,684\,200}{2400}=701.8\text{ A}\ (\angle{-18.2^\circ}).}$$
New line loss and source voltage. $P_{\text{loss}}'=(701.8)^2(0.05)=24.62$ kW, and $V_g'=2400+I\,Z_{\text{line}}=2543+j322=2563\angle 7.23^\circ$ V. Correction cut the current 833→702 A, the loss 34.7→24.6 kW, and the source voltage 2722→2563 V.
