Negative Wattmeter Reading on a Grinding-Machine Motor
During a load survey in a vocational workshop, a 7.5 kW, 415 V, 4-pole squirrel-cage motor driving a grinding machine is tested by the two-wattmeter method. The wattmeter in the R line reads 2.4 kW; the wattmeter in the B line reads −0.4 kW. The instruments are moving-coil dynamometer types fitted with a reversing switch, and the negative reading was confirmed by reversing the current-coil connections and re-reading, so the polarity is genuine and not a wiring error. A clip-on ammeter shows a line current of 7.4 A and the line voltage is 412 V. The nameplate full-load current is 14.5 A, the motor is protected by a 16 A MCB and a thermal overload relay set at 15 A, and the machine has done the same job for years. The supervisor, seeing the pointer below zero, concludes that the wattmeter is defective and wants it replaced.
1. Calculate the total power, the phase angle and the power factor from the two readings, and state whether a negative reading is a fault.
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Total power
P = W1 + W2 = 2.4 + (−0.4) = 2.0 kW
Phase angle and power factor
For the two-wattmeter method:
tan φ = √3 (W1 − W2) / (W1 + W2) = 1.732 × (2.4 − (−0.4)) / 2.0 = 1.732 × 2.8 / 2.0 = 2.425
φ = tan⁻¹ 2.425 = 67.6°, so power factor = cos 67.6° = 0.38 lagging
Is the negative reading a fault? No. One wattmeter reads negative whenever φ exceeds 60°, i.e. whenever the power factor falls below 0.5. The pointer simply goes below zero because the current coil's current is more than 90° out of phase with its voltage. Reversing the current coil to read the magnitude (as was done) is the correct procedure; the reading is then subtracted. Replacing the wattmeter would change nothing.
2. Cross-check the result against the measured voltage and current, and from that check explain how heavily the motor is actually loaded.
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Reactive power from the two readings
Q = √3 (W1 − W2) = 1.732 × 2.8 = 4.85 kVAr
Apparent power
S = √(P² + Q²) = √(2.0² + 4.85²) = 5.25 kVA
Line current implied
I = S / (√3 × V) = 5250 / (1.732 × 412) = 7.36 A
This matches the measured 7.4 A within instrument error, which confirms both the readings and the calculation - there is no meter fault and no wiring error.
Loading
7.36 A against a nameplate full-load current of 14.5 A is about 51 % of full-load current, but current is a poor guide to shaft load at low loads. Input power is only 2.0 kW; allowing roughly 85 % efficiency, the shaft output is about 1.7 kW against a 7.5 kW rating - around 23 % loaded. An induction motor that lightly loaded has a large magnetising component and therefore a very poor power factor, which is exactly what the 0.38 shows. The motor is heavily oversized for the grinding job.
3. Recommend what the workshop should actually do, and calculate the capacitor kVAr that would correct the power factor to 0.90 at the present load.
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Recommendations, in order of preference
- Right-size the drive. Replace the 7.5 kW motor with a 2.2–3.7 kW unit of the same frame and speed. This raises the power factor at source, reduces losses and costs less to run than any capacitor scheme. The overload relay is set at 15 A for a motor drawing 7.4 A, so it cannot protect the motor against a sustained overload either - it must be reset to the new nameplate current.
- If the motor must stay, connect a capacitor bank switched with the motor.
Capacitor calculation
Qc = P (tan φ1 − tan φ2)
- tan φ1 = 2.425
- cos φ2 = 0.90 → tan φ2 = 0.484
Qc = 2.0 × (2.425 − 0.484) = 3.88 kVAr, i.e. a standard 4 kVAr capacitor.
Caution: the capacitor must not exceed the motor's no-load reactive demand, otherwise the motor self-excites when switched off and generates overvoltage. Connect the capacitor on the motor side of the contactor so it is switched with the motor, and check that the machine does not run for long periods at very light load, where 4 kVAr would over-correct.