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

Question 1: Two-Wattmeter Method 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. Three parallel 3-phase loads on a 440 V (line-to-line) system, measured by the two-wattmeter connection of Figure 1 (W1: current coil in line a, voltage coil a–c; W2: current coil in line b, voltage coil b–c; line c carries no current coil).

Find. (a) the two wattmeter readings W1 and W2; (b) the per-phase capacitive reactance to raise the overall power factor to unity, and the two wattmeter readings after correction.

[Figure not reproduced: Figure 1 (redrawn): W1 current coil in line a , voltage coil a–c ; W2 current coil in line b , voltage coil b–c (line c is the common line). See the official exam paper.]

Approach. Resolve every load into real (P) and reactive (Q) power, sum to a system complex power, obtain the line current and overall angle, then apply the two-wattmeter identities for the Figure 1 connection (common line c, abc sequence): W1 = V_LI_Lcos(30°−θ), W2 = V_LI_Lcos(30°+θ). Unity correction supplies the total lagging vars with capacitors.

1. Resolve each load into P and Q. For A, $S_A=P_A/\text{pf}=21.5/0.6=35.83\ \text{kVA}$, so $Q_A=\sqrt{S_A^2-P_A^2}=28.67\ \text{kVAR}$. For B, $Q_B=\sqrt{22^2-11^2}=\sqrt{363}=19.05\ \text{kVAR}$. For C, $\tan\theta_C=0.6/0.8=0.75$, so $P_C=Q_C/\tan\theta_C=17.6/0.75=23.47\ \text{kW}$. All are lagging (inductive), so the vars add.
2. Sum to the system complex power. $$P_T=21.5+11+23.47=55.97\ \text{kW},\qquad Q_T=28.67+19.05+17.60=65.32\ \text{kVAR}$$ $$\boxed{S_T=\sqrt{P_T^2+Q_T^2}=86.02\ \text{kVA},\quad \theta=\tan^{-1}\!\frac{65.32}{55.97}=49.4^\circ\ (\text{pf}=0.651\ \text{lag})}$$
3. Line current. For a balanced three-phase system $I_L=\dfrac{S_T}{\sqrt{3}\,V_L}=\dfrac{86.02\times10^3}{\sqrt3\,(440)}=112.9\ \text{A}.$
4. Wattmeter readings (part a). In Figure 1, W1 measures $I_a$ against $V_{ac}$ and W2 measures $I_b$ against $V_{bc}$. With positive (abc) sequence and $V_{an}=V_\phi\angle0^\circ$: $V_{ac}=V_L\angle{-}30^\circ$, $I_a=I_L\angle{-}\theta$, and $V_{bc}=V_L\angle{-}90^\circ$, $I_b=I_L\angle({-}120^\circ-\theta)$, so $$W_1=V_L I_L\cos(\theta-30^\circ)=440(112.9)\cos(19.4^\circ)=46.84\ \text{kW}$$ $$W_2=V_L I_L\cos(30^\circ+\theta)=440(112.9)\cos(79.4^\circ)=9.13\ \text{kW}$$ $$\boxed{W_1=46.84\ \text{kW},\qquad W_2=9.13\ \text{kW}}$$ Check: $W_1+W_2=55.97\ \text{kW}=P_T$ ✓. The large split reflects the low 0.651 power factor. (If the supply sequence were acb, the two readings would swap meters; the pair of values is unchanged.)
5. Capacitor sizing for unity pf (part b). To reach unity the bank must supply the entire lagging reactive power, $Q_C=65.32\ \text{kVAR}$, i.e. $21.77\ \text{kVAR}$ per phase. Taking the capacitors delta-connected across the 440 V lines (each capacitor sees $V_L$): $$X_C=\frac{V_L^2}{Q_{C,\text{ph}}}=\frac{440^2}{21.77\times10^3}=\boxed{8.89\ \Omega\ \text{per phase}}\quad\Big(C=\tfrac{1}{2\pi f X_C}=298\ \mu\text{F}\Big)$$
6. Wattmeter readings after correction. At unity pf $\theta=0$ and $Q_T=0$, so the line current falls to $I_L'=\dfrac{P_T}{\sqrt3\,V_L}=73.4\ \text{A}$. Both meters now read equally: $$W_1'=W_2'=V_L I_L'\cos(30^\circ)=\tfrac{1}{2}P_T=\boxed{27.98\ \text{kW each}}$$