Two-Wattmeter Measurement and Power-Factor Correction: 22-Elec-A6 Power Systems and Machines, May 2014, Q1

Question 1: Two-Wattmeter Measurement and Power-Factor 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. A three-phase, three-load system at 600 V line-to-line, 60 Hz, measured by the two-wattmeter method (W1 current coil in line a with its voltage coil across a–b; W2 current coil in line c with its voltage coil across c–b).

Given data
Load	Rating	Real power P	Reactive power Q
A (heating)	20 kW, unity pf	20.00 kW	0
B (25 hp motor)	0.65 pf lagging	from rating	lagging
C	24 kVA, 18.6 kVAR lag	from triangle	18.60 kVAR

Find. The two wattmeter readings for the uncorrected system, the capacitive VARs per phase needed for unity power factor, and the corrected wattmeter readings.

[Figure not reproduced: Figure 1 (redrawn): two-wattmeter connection W1 (line a, voltage a–b) and W2 (line c, voltage c–b) feeding the three parallel loads A, B, C. See the official exam paper.]

Check: No efficiency is given for the 25 hp motor, so its rating is taken as the real electrical power drawn, P_B = 25 × 746 = 18.65 kW — the standard convention when only the horsepower and power factor are stated.

Approach. Resolve each load into (P, Q), sum to obtain the total complex power, convert to a line current and load angle, then apply the two-wattmeter formulas; for correction, cancel the total lagging Q with capacitors.

Resolve each load to real and reactive power. With $\theta_B=\cos^{-1}0.65 = 49.46^\circ$, $$P_B = 25\times746 = 18.65\ \text{kW},\qquad Q_B = P_B\tan\theta_B = 21.80\ \text{kVAR}.$$ For Load C the power triangle gives $P_C=\sqrt{S_C^2-Q_C^2}=\sqrt{24^2-18.6^2}=15.17\ \text{kW}$, with $Q_C=18.60$ kVAR.
Total complex power. Summing, $$P_T = 20+18.65+15.17 = 53.82\ \text{kW},\qquad Q_T = 0+21.80+18.60 = 40.40\ \text{kVAR}.$$ $$S_T=\sqrt{P_T^2+Q_T^2}=67.30\ \text{kVA},\quad \theta=\tan^{-1}\frac{Q_T}{P_T}=36.90^\circ,\quad \text{pf}=\cos\theta=0.800\ \text{lag}.$$
Line current. $$I_L=\frac{S_T}{\sqrt3\,V_L}=\frac{67\,300}{\sqrt3\,(600)} = 64.76\ \text{A}.$$
Wattmeter readings (uncorrected). For this connection the meters read $$W_1 = V_L I_L\cos(30^\circ+\theta),\qquad W_2 = V_L I_L\cos(30^\circ-\theta).$$ $$\boxed{W_1 = 15.24\ \text{kW},\qquad W_2 = 38.57\ \text{kW}}$$ Check: $W_1+W_2 = 53.82$ kW $=P_T$ and $\sqrt3\,(W_2-W_1)=40.40$ kVAR $=Q_T$.
Capacitors for unity power factor. Unity pf requires the capacitors to supply the entire lagging reactive power, $Q_C^{cap}=Q_T=40.40$ kVAR, so $$\boxed{Q_{cap,\ \text{per phase}} = \frac{Q_T}{3} = 13.47\ \text{kVAR/phase}.}$$
Corrected wattmeter readings. At unity pf $\theta=0$, the line current falls to $I_L'=P_T/(\sqrt3 V_L)=51.79$ A and the two meters read equally, $$W_1'=W_2'=V_L I_L'\cos 30^\circ = \tfrac{1}{2}P_T = \boxed{26.91\ \text{kW each}.}$$

Final results — Question 1
Quantity	Value
Total real / reactive power	53.82 kW / 40.40 kVAR
System pf, line current	0.800 lag, 64.76 A
W1, W2 (uncorrected)	15.24 kW, 38.57 kW
Capacitor VARs per phase	13.47 kVAR
W1, W2 (corrected, unity pf)	26.91 kW each

Topic: Three-phase power measurement by the two-wattmeter method and shunt capacitive power-factor correction.

Key relations: $P_T=\sqrt3 V_LI_L\cos\theta$; $W_{1,2}=V_LI_L\cos(30^\circ\pm\theta)$; $W_1+W_2=P_T$, $\sqrt3(W_2-W_1)=Q_T$; $Q_{cap}=V^2/X_C$.

Why this works: In a balanced three-wire system, Blondel’s theorem says two wattmeters measure total power exactly. Each meter’s line current lags its own line-to-line voltage reference by $30^\circ\pm\theta$, so their sum recovers $P_T$ while their difference is proportional to $Q_T$. Capacitors supply leading VARs that cancel the load’s lagging VARs without changing real power, so at unity pf both meters read half the (unchanged) total real power.

Common pitfall: Forgetting the $30^\circ$ reference shift and equating each meter to a single phase’s power; or correcting to unity by matching total kVA instead of cancelling only the reactive component.

Source: Chapman, Electric Machinery Fundamentals, ch. 1 (three-phase power); Wildi, Electrical Machines, ch. 8 (two-wattmeter method, pf correction).

Question 1 of 6: Two-Wattmeter Measurement and Power-Factor Correction

Question 1: Two-Wattmeter Measurement and Power-Factor Correction (20 marks)