Two-Wattmeter Method with a Backward Deflection
A technician commissions a 7.5 kW, 415 V, 50 Hz, 4-pole cage induction motor using the two-wattmeter method with two 5 A, 150 V dynamometer wattmeters connected directly in the lines, with no CTs or PTs. At the measured shaft load, wattmeter 1 reads 1.9 kW. Wattmeter 2 deflects backwards; the technician reverses its potential-coil connection, reads 0.6 kW, and records the motor input as 1.9 + 0.6 = 2.5 kW. The nameplate gives 7.5 kW, 415 V, 14.3 A, power factor 0.83 and efficiency 88%. A clamp meter on the same line reads 6.3 A. The technician concludes the motor is running at about one-third load with 'far too low' an efficiency and proposes to derate the feeder cable. The plant engineer asks you to check the power measurement before any cable work is authorised.
1. State the correct motor input power and power factor, explain why one wattmeter deflected backwards, and quantify the technician's error.
Read the worked answer
In the two-wattmeter method the readings are W1 = VL IL cos(30° − φ) and W2 = VL IL cos(30° + φ), so:
P = W1 + W2 and Q = √3 (W1 − W2)
The sign of W2 is part of the measurement, not a fault. When φ > 60° (power factor below 0.5) the angle (30° + φ) exceeds 90°, the cosine goes negative and the wattmeter tries to read below zero. Reversing the potential coil makes the pointer readable but inverts the number's meaning: the reading must be subtracted, not added.
So the true input power is P = 1.9 + (−0.6) = 1.30 kW.
Reactive power: Q = √3 (1.9 − (−0.6)) = √3 × 2.5 = 4.33 kvar.
Apparent power S = √(1.30² + 4.33²) = 4.52 kVA, giving power factor 1.30/4.52 = 0.288, i.e. φ = 73.3° - comfortably beyond the 60° threshold, which is exactly why the needle reversed.
The technician's 2.5 kW is 1.2 kW too high, an error of +92%. At that figure the implied power factor would be (2500/4528) = 0.55, above the 0.5 threshold - and in that case wattmeter 2 would not have reversed at all. That contradiction by itself should have stopped the record.
2. Cross-check the power measurement against the 6.3 A line current and the nameplate data, showing what readings a genuine 2.5 kW input would have produced.
Read the worked answer
With VL = 415 V and IL = 6.3 A, the product VL IL = 415 × 6.3 = 2614.5 W. Using φ = 73.3°:
W1 = 2614.5 cos(30° − 73.3°) = 2614.5 cos(−43.3°) = 2614.5 × 0.728 = 1903 W ✓ (measured 1.9 kW)
W2 = 2614.5 cos(30° + 73.3°) = 2614.5 cos(103.3°) = 2614.5 × (−0.230) = −601 W ✓ (measured −0.6 kW)
The two meters and the clamp meter agree to better than 1%. This is the strongest possible confirmation that 1.30 kW is right.
Now invert the argument: if the input really were 2.5 kW at 6.3 A, the power factor would be 0.552, φ = 56.5°, and the meters would have read W1 = 2614.5 cos(−26.5°) = 2.34 kW and W2 = 2614.5 cos(86.5°) = +0.16 kW - both positive, with W1 far from the observed 1.9 kW. Neither number matches the bench.
Sanity check against the nameplate: full-load current is 7500/(√3 × 415 × 0.83 × 0.88) = 14.3 A, matching the plate. At 6.3 A the motor is at about 44% of full-load current but only about 17% of full-load power, and a lightly loaded induction motor draws its magnetising reactive current almost unchanged while the active current collapses - precisely the condition that produces pf ≈ 0.29 and a backward wattmeter.
3. What do you recommend to the plant engineer? Address the measurement practice, the efficiency conclusion and the cable-derating proposal.
Read the worked answer
Do not derate the cable. The feeder carries 6.3 A on a motor rated 14.3 A, i.e. 44% of full-load current, and the measured 1.30 kW input corresponds to roughly 17% of rated power. The cable is not stressed; the proposal was based entirely on the mis-recorded 2.5 kW.
Correct the measurement method. For any load that can operate below 0.5 power factor, the two-wattmeter method must allow a negative reading to be carried with its algebraic sign - by using a wattmeter with a zero-at-centre scale, by reversing the coil and recording the number as negative, or by using a digital power analyser that reports signed per-phase quantities. Recording |W2| and adding it converts a correct measurement into a 92% error. The instrument range is also worth noting: 150 V and 5 A dynamometer wattmeters are near their limits on a 415 V, 14.3 A motor and would not have coped at full load - a CT/PT set or a clip-on power analyser with a 600 V, 20 A range is the appropriate instrument here.
Fix the efficiency conclusion. A 7.5 kW motor at 17% load does run inefficiently and at very low power factor - that is normal behaviour, not a fault, and the nameplate efficiency applies at full load only. If the duty really is light, the engineering answer is to replace the motor with a correctly sized unit or run it on a variable-speed drive; a 92% reading error, however, is not evidence about efficiency at all. Re-measure the shaft output (torque and speed) before any claim about efficiency is made, and log power factor over a duty cycle if power-factor correction is being considered.