Two-Wattmeter Measurement and Capacitor Sizing
At a 480 V, three-phase, three-wire industrial service, a technician uses the two-wattmeter method to measure a balanced motor load. With current coils in lines A and C and voltage coils connected A-B and C-B, the meters read W1 = 12.5 kW and W2 = 4.0 kW. The plant engineer wants to add a 100 kvar capacitor bank at the same bus and asks whether the existing 200 A feeder and 150 kVA transformer have margin. The technician notes that W2 fluctuates between 3.8 and 4.3 kW and that the CTs are 200:5 with a 5 A meter. No power factor meter is installed. The load is assumed balanced and sinusoidal.
1. Compute the total real power, reactive power, and power factor from the two wattmeter readings. State the formula and the sign convention.
Read the worked answer
Use the two-wattmeter relations:
- Total real power: P = W1 + W2 = 12.5 + 4.0 = 16.5 kW.
- Reactive power: Q = √3 (W1 - W2) = √3 (8.5) = 14.72 kvar.
- Apparent power: S = √(P² + Q²) = √(16.5² + 14.72²) = 22.11 kVA.
- Power factor: pf = P/S = 16.5/22.11 = 0.746, lagging because W1 > W2.
The sign convention matters: if the larger reading is assigned to W1, the positive Q result means lagging current. Reversing the meter leads or swapping which reading is called W1 changes the sign of the calculated Q, not the physical load. The 0.746 pf is consistent with a partly loaded induction motor.
2. Determine the feeder current and transformer loading before and after adding the capacitor bank. Is the 200 A feeder adequate?
Read the worked answer
Before the bank, line current is I = S/(√3 V) = 22.11×10³/[√3(480)] = 26.6 A.
After adding 100 kvar, the net reactive power is Q = 14.72 - 100 = -85.28 kvar (leading). With P = 16.5 kW,
S = √(16.5² + 85.28²) = 86.9 kVA, and I = 86.9×10³/[√3(480)] = 104.5 A.
The 200 A feeder is adequate in both cases. The 150 kVA transformer is loaded to 14.7% before and 57.9% after. However, the 100 kvar bank is much larger than the 14.7 kvar load requirement; it overcorrects to a leading pf of about 0.19. A bank near 15 kvar would correct to unity without overvoltage or leading-power concerns. The engineer should not size the bank from transformer margin alone.
3. What does the fluctuation in W2 suggest, and how should the technician verify the measurement?
Read the worked answer
The fluctuation in W2 is too large to ignore if the motor load is steady. Likely causes are a loose voltage-coil connection, wrong CT polarity or ratio, CT burden too high for the meter, or an actual load variation that also should appear in W1. The technician should:
- Compare both wattmeter readings and line currents with a clamp-on meter.
- Verify CT ratio, polarity, and secondary burden against the meter data.
- Check phase angles or use a power-quality analyzer to confirm the two-wattmeter connection.
- Repeat the measurement at a stable load.
Because the calculation depends on the difference W1 - W2, a small error in W2 produces a large error in reactive power and power factor. The measurement should be validated before using it to size the capacitor bank.