22-Elec-A6 Power Systems and Machines · December 2016
Question 1 of 6: Three-Phase Plant Loads and Two-Wattmeter Method
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, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).
Question 1: Three-Phase Plant Loads and Two-Wattmeter Method (20 marks)
Given. A 480 V (line-to-line), 60 Hz three-phase bus supplying the four parallel loads tabulated below.
Given data
Load
Real power P
Power factor
Induction motor 1
250 kW
0.9 lagging
Induction motor 2
350 kW
0.75 lagging
Synchronous motor
300 kW
0.9 lagging (then 0.9 leading)
Δ impedance bank
Z = 2.5∠30° Ω per branch (line voltage across each)
Find. Total P and Q, total line current (magnitude and angle), then—with the synchronous motor at 0.9 leading—the synchronous line current, the new total line current and power factor, and the two-wattmeter readings.
[Figure not reproduced: Figure 1 (redrawn): two wattmeters W1, W2 in lines a and b; the four loads tap all three lines. See the official exam paper.]
Approach. Resolve every load into complex power $S=P+jQ$, add them (power adds arithmetically for parallel loads), then obtain line current from $|S|=\sqrt{3}\,V_{LL} I_L$ and the wattmeter readings from the standard two-wattmeter relations.
Reactive power of each motor load. For a load $Q=P\tan\theta$ with $\theta=\cos^{-1}(\text{pf})$:
$$Q_1 = 250\tan(\cos^{-1}0.9)=121.1\ \text{kVAR},\quad Q_2 = 350\tan(\cos^{-1}0.75)=308.7\ \text{kVAR}$$
$$Q_{SM}=300\tan(\cos^{-1}0.9)=145.3\ \text{kVAR (lagging)}.$$
Delta impedance bank. Each branch sees the full line voltage $V_{LL}=480$ V, so the three-phase complex power is
$$S_\Delta = 3\,\frac{V_{LL}^{2}}{Z^{*}} = 3\,\frac{480^{2}}{2.5\angle{-30^\circ}} = 276.5\angle 30^\circ\ \text{kVA},$$
giving $P_\Delta = 239.4$ kW and $Q_\Delta = 138.2$ kVAR (inductive, since $Z$ is inductive).
Totals (part a). Summing all four:
$$\boxed{P_{tot}=250+350+300+239.4 = 1139.4\ \text{kW},\qquad Q_{tot}=121.1+308.7+145.3+138.2 = 713.3\ \text{kVAR}.}$$
The apparent power is $S_{tot}=\sqrt{P_{tot}^2+Q_{tot}^2}=1344.3$ kVA at $\text{pf}=1139.4/1344.3=0.848$ lagging.
Total line current (part b). From $|S|=\sqrt{3}\,V_{LL}I_L$:
$$I_L=\frac{S_{tot}}{\sqrt3\,V_{LL}}=\frac{1\,344\,300}{\sqrt3\,(480)} = 1617\ \text{A}.$$
The current lags the phase voltage by $\theta=\cos^{-1}0.848 = 32.1^\circ$, so $\boxed{I_L = 1617\angle{-32.1^\circ}\ \text{A}}$.
Synchronous motor at 0.9 leading — its line current (part c). Its apparent power is unchanged, $S_{SM}=300/0.9=333.3$ kVA, so
$$I_{SM}=\frac{S_{SM}}{\sqrt3\,V_{LL}}=\frac{333\,333}{\sqrt3\,(480)}=401\ \text{A},$$
now leading: $\boxed{I_{SM}=401\angle{+25.8^\circ}\ \text{A}}$.
New plant totals (part d). Only the synchronous-motor reactive power changes sign ($+145.3\to-145.3$ kVAR):
$$Q_{tot}'=121.1+308.7-145.3+138.2 = 422.7\ \text{kVAR},\quad S_{tot}'=\sqrt{1139.4^2+422.7^2}=1215.3\ \text{kVA}.$$
Hence $\text{pf}'=1139.4/1215.3=0.938$ lagging and
$$\boxed{I_L'=\frac{1\,215\,300}{\sqrt3\,(480)}=1462\angle{-20.4^\circ}\ \text{A}.}$$
Two-wattmeter readings (part e). With the current lagging by $\theta'=20.4^\circ$,
$$W_1=V_{LL}I_L'\cos(30^\circ+\theta')=480(1462)\cos 50.4^\circ = 447.7\ \text{kW},$$
$$W_2=V_{LL}I_L'\cos(30^\circ-\theta')=480(1462)\cos 9.6^\circ = 691.7\ \text{kW}.$$
Check: $W_1+W_2=1139.4$ kW $=P_{tot}$ and $\sqrt3(W_2-W_1)=422.7$ kVAR $=Q_{tot}'$. $\boxed{W_1=447.7\ \text{kW},\ W_2=691.7\ \text{kW}.}$
