Three-Phase Power: 22-Elec-A6 Power Systems and Machines, December 2015, Q1

Question 1: Three-Phase Power (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.

Part (a) — Induction-motor input power, apparent power, line current

Given. A three-phase induction motor: rated mechanical output 40 hp, line voltage V_L = 460 V, efficiency η = 0.936, power factor 0.83 lagging.

Find. (i) active (real) input power in kW, (ii) apparent input power, (iii) full-load line current.

Approach. The nameplate horsepower is the shaft output; divide by efficiency for electrical input real power, then divide by power factor for apparent power, then use the three-phase relation $S=\sqrt3\,V_L I_L$.

Convert output power and find input real power. The shaft output is $P_{out}=40\times746=29\,840\text{ W}$. Since efficiency is output over input, $$P_{in}=\frac{P_{out}}{\eta}=\frac{29\,840}{0.936}=\boxed{31.88\text{ kW}}$$
Apparent power from the power factor. Real power is apparent power times the power factor, so $$S=\frac{P_{in}}{\text{pf}}=\frac{31\,880}{0.83}=38.41\text{ kVA}.$$
Line current from the three-phase apparent-power relation. For a balanced three-phase load $S=\sqrt3\,V_L I_L$, hence $$I_L=\frac{S}{\sqrt3\,V_L}=\frac{38\,410}{\sqrt3\,(460)}=48.2\text{ A}.$$

Question 1(a) — results
Quantity	Value
Active (real) input power $P_{in}$	31.88 kW
Apparent input power $S$	38.41 kVA
Full-load line current $I_L$	48.2 A

Part (b) — Parallel Y and Δ loads; capacitor for unity power factor

Given. Balanced 3-wire 208 V, 60 Hz supply feeding two parallel loads:

Given data — part (b)
Item	Value
Line voltage $V_L$	208 V
Phase voltage (Y) $V_\phi=V_L/\sqrt3$	120.1 V
Load 1 (Y), per phase	$Z_1=10\angle45^\circ\ \Omega$
Load 2 (Δ), per phase	$8\ \Omega + \text{16 mH}$
Angular frequency $\omega=2\pi f$	376.99 rad/s

Find. The capacitance (per phase) of a shunt bank that raises the combined power factor to unity.

One-line diagram: the Y load, the Δ load and the correction capacitors all share the 208 V bus.

Approach. Resolve each load into real power P and reactive power Q using $S=3\,V_\phi^{2}/Z^{*}$ (Y, phase voltage) or $S=3\,V_L^{2}/Z^{*}$ (Δ, line voltage). Add them; the capacitor bank must supply reactive power equal to the total inductive $Q$ so that the net reactive power is zero.

Load 1 (Y) complex power. With phase voltage across each 10∠45° impedance, $$S_1=3\,\frac{V_\phi^{2}}{Z_1^{*}}=3\,\frac{(120.1)^2}{10\angle{-45^\circ}} =3(1442)\angle45^\circ=3059+j3059\ \text{VA}.$$ So $P_1=3.06\text{ kW}$, $Q_1=+3.06\text{ kvar}$ (inductive).
Load 2 (Δ) branch impedance. The inductor reactance is $X_L=\omega L=376.99(0.016)=6.03\ \Omega$, so each Δ branch is $$Z_2=8+j6.03=10.02\angle37.0^\circ\ \Omega.$$
Load 2 (Δ) complex power. A Δ branch sees the full line voltage, so $$S_2=3\,\frac{V_L^{2}}{Z_2^{*}}=3\,\frac{(208)^2}{10.02\angle{-37.0^\circ}} =3(4318)\angle37.0^\circ=10\,345+j7797\ \text{VA}.$$ So $P_2=10.34\text{ kW}$, $Q_2=+7.80\text{ kvar}$.
Combine the two loads. $$P=P_1+P_2=13.40\text{ kW},\qquad Q=Q_1+Q_2=+10.86\text{ kvar (lagging).}$$
Size the capacitor bank. For unity power factor the capacitors must supply $Q_C=10.86\ \text{kvar}$. For a Δ-connected bank each unit sees $V_L$ and supplies $Q_C=3\,\omega C_\Delta V_L^{2}$, so $$C_\Delta=\frac{Q_C}{3\,\omega V_L^{2}} =\frac{10\,856}{3(376.99)(208)^2}=\boxed{222\ \mu\text{F per phase.}}$$ A Y-connected bank instead sees $V_\phi$ and needs $C_Y=Q_C/(\omega V_L^{2})=666\ \mu\text{F per phase.}$

Question 1(b) — results
Quantity	Value
Total real power $P$	13.40 kW
Total reactive power $Q$ (before correction)	+10.86 kvar
Correction capacitance, Δ bank	222 µF / phase
Correction capacitance, Y bank	666 µF / phase

Part (c) — Two-wattmeter readings, before and after correction

Given. The combined load of part (b): before correction $P=13.40\text{ kW}$, $Q=+10.86\text{ kvar}$; after correction the same real power at unity power factor.

Find. The two wattmeter readings $W_1,\,W_2$ in each case.

Approach. In the two-wattmeter method $W_1=V_L I_L\cos(30^\circ+\theta)$ and $W_2=V_L I_L\cos(30^\circ-\theta)$, where $\theta$ is the load angle. Find $\theta$ and $I_L$ from the combined $P$ and $Q$.

Check: the wattmeters are taken to measure the combined parallel load of part (b) at the common bus ("the loads of part (b)"), with the correction capacitors connected on the load side of the meters. This is the reading that changes with the capacitors and gives the intended before/after contrast.

Load angle and line current before correction. $$\theta=\tan^{-1}\!\frac{Q}{P}=\tan^{-1}\!\frac{10.86}{13.40}=39.0^\circ,\qquad I_L=\frac{\sqrt{P^2+Q^2}}{\sqrt3\,V_L}=\frac{17\,248}{\sqrt3\,(208)}=47.9\text{ A}.$$
Wattmeter readings before correction. $$W_1=V_L I_L\cos(30^\circ+\theta)=208(47.9)\cos 69.0^\circ=3.57\text{ kW},$$ $$W_2=V_L I_L\cos(30^\circ-\theta)=208(47.9)\cos 9.0^\circ=9.84\text{ kW}.$$ Check: $W_1+W_2=13.4\text{ kW}=P$ ✓.
After correction (unity power factor). The capacitors carry no real power, so $P$ is unchanged while $\theta=0$ and the line current falls to $$I_L=\frac{P}{\sqrt3\,V_L}=\frac{13\,404}{\sqrt3\,(208)}=37.2\text{ A}.$$ With $\theta=0$ both meters read the same, $$W_1=W_2=V_L I_L\cos 30^\circ=\tfrac12 P=6.70\text{ kW each.}$$

Question 1(c) — results
Condition	$W_1$	$W_2$	$I_L$
Before capacitors (pf 0.776 lag)	3.57 kW	9.84 kW	47.9 A
After capacitors (pf 1.0)	6.70 kW	6.70 kW	37.2 A

Topic. Balanced three-phase power accounting, power-factor correction, and the two-wattmeter method.

Key relations. $S=\sqrt3\,V_L I_L$; complex power $S=3V^2/Z^{*}=P+jQ$; capacitor reactive supply $Q_C=\omega C V^2$ per unit; two-wattmeter $W_{1,2}=V_L I_L\cos(30^\circ\pm\theta)$, with $P=W_1+W_2$ and $Q=\sqrt3\,(W_2-W_1)$.

Why this works. Real and reactive powers are conserved and add arithmetically across parallel branches, so each load is reduced independently to its $P$ and $Q$ and the totals are summed. An ideal capacitor supplies leading vars without consuming watts, so it cancels the inductive $Q$ at constant $P$; the line current then drops in proportion to the improved power factor while the metered real power is unchanged. The two-wattmeter algebra follows from writing each meter as a line-to-line voltage times a line current at the appropriate 30° offset.

Common pitfall. Applying the phase voltage to a Δ branch (or the line voltage to a Y branch); each Δ branch sees the full line voltage. A second trap is assuming the capacitors change the wattmeter sum — they do not, because they add no real power.

Source. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed., Ch 8 (three-phase circuits and the two-wattmeter method) and Ch 25 (power-factor correction).

Quantity	Value
Active (real) input power \(P_{in}\)	31.88 kW
Apparent input power \(S\)	38.41 kVA
Full-load line current \(I_L\)	48.2 A

Item	Value
Line voltage \(V_L\)	208 V
Phase voltage (Y) \(V_\phi=V_L/\sqrt3\)	120.1 V
Load 1 (Y), per phase	\(Z_1=10\angle45^\circ\ \Omega\)
Load 2 (Δ), per phase	\(8\ \Omega + \text{16 mH}\)
Angular frequency \(\omega=2\pi f\)	376.99 rad/s

Quantity	Value
Total real power \(P\)	13.40 kW
Total reactive power \(Q\) (before correction)	+10.86 kvar
Correction capacitance, Δ bank	222 µF / phase
Correction capacitance, Y bank	666 µF / phase

Question 1 of 5: Three-Phase Power

Question 1: Three-Phase Power (20 marks)

Part (a) — Induction-motor input power, apparent power, line current

Part (b) — Parallel Y and Δ loads; capacitor for unity power factor

Part (c) — Two-wattmeter readings, before and after correction