Two-wattmeter method calculator

Total power, reactive power and power factor of a balanced three-wire load from its two wattmeter readings.

Wattmeter readings
kW
kW

Type a negative reading with its minus sign.

Find the readings for a known load
V
kW

Line current: 75.18 A

Higher reading: 35.83 kW · Lower reading: 14.17 kW

Balanced load. Lagging or leading only changes which meter reads higher, not the two values.

Result

Power factor
0.866
Total power, P
60kW
Reactive power, |Q|
34.64kVAR
Apparent power, S
69.28kVA
Power-factor angle, |φ|
30°

Balanced three-wire load. The readings alone do not say whether the power factor is lagging or leading: that depends on the phase sequence.

What it calculates

The total power, reactive power and power factor of a balanced three-phase, three-wire load measured with two wattmeters. Each meter has its current coil in one line and its potential coil from that line to the third line, the one with no current coil. For a balanced load with power-factor angle φ, the two meters read:

W1=VLILcos⁡(30∘−φ)W_1 = V_L I_L \cos(30^\circ - \varphi)W2=VLILcos⁡(30∘+φ)W_2 = V_L I_L \cos(30^\circ + \varphi)

Their sum is the total power, and their difference, times √3, is the reactive power. The ratio of the two gives the power-factor angle without knowing the voltage or the current:

P=W1+W2P = W_1 + W_2Q=3 (W1−W2)Q = \sqrt{3}\,(W_1 - W_2)
tan⁡φ=3 (W1−W2)W1+W2\tan\varphi = \dfrac{\sqrt{3}\,(W_1 - W_2)}{W_1 + W_2}pf=cos⁡φ\text{pf} = \cos\varphi

Worked example

The meters read 40 kW and 20 kW. The calculator opens on this case:

P=40+20=60 kWQ=3 (40−20)=34.64 kVARtan⁡φ=34.6460=0.5774φ=30.0∘pf=cos⁡30.0∘=0.866\begin{aligned} P &= 40 + 20 = 60\ \text{kW} \\[4pt] Q &= \sqrt{3}\,(40 - 20) = 34.64\ \text{kVAR} \\[4pt] \tan\varphi &= \dfrac{34.64}{60} = 0.5774 \quad \varphi = 30.0^\circ \\[4pt] \text{pf} &= \cos 30.0^\circ = 0.866 \end{aligned}

When one meter reads negative

Below a power factor of 0.5 the angle in the lower meter's reading passes 90°, so that meter reads negative. Keep the sign. With readings of 50 kW and −10 kW:

P=50+(−10)=40 kWQ=3 (50−(−10))=103.9 kVARφ=tan⁡−1103.940=68.95∘pf=0.359\begin{aligned} P &= 50 + (-10) = 40\ \text{kW} \\[4pt] Q &= \sqrt{3}\,\bigl(50 - (-10)\bigr) = 103.9\ \text{kVAR} \\[4pt] \varphi &= \tan^{-1}\dfrac{103.9}{40} = 68.95^\circ \quad \text{pf} = 0.359 \end{aligned}

At exactly 0.5 power factor (φ = 60°) the lower meter reads zero. At unity power factor the two readings are equal.

Readings from a known load

To go the other way, from a load to the readings you should see, open the helper under the inputs. It finds the line current from the total power, the line voltage and the power factor:

IL=P3 VL pfI_L = \dfrac{P}{\sqrt{3}\,V_L\,\text{pf}}

and puts it into the two reading formulas above. For a 50 kW load at 480 V and 0.80 power factor, the line current is 75.18 A and the meters read 35.83 kW and 14.17 kW, which add back to 50 kW. To get line current from other combinations of kW, kVA and power factor, see the three-phase power calculator.

Assumptions

  • The load and the supply are balanced. The sum of the readings is the total power of any three-wire load, but the reactive power and power factor formulas hold only for a balanced one.
  • The circuit is three-wire. A four-wire load with current in the neutral needs three wattmeters, one per phase.
  • The meters are ideal: no burden on the circuit and no ratio or phase-angle error from instrument transformers.
  • The result gives the size of the angle and of the reactive power. Whether the load is lagging or leading depends on the phase sequence and on which meter reads higher, which the two numbers alone do not show.

Questions

Why does one wattmeter read negative?
Each meter reads V_L × I_L × cos(30° ± φ). Once the power-factor angle passes 60°, which is a power factor below 0.5, the angle in the lower meter's reading passes 90° and its cosine turns negative. The meter is not miswired. Enter the reading with its minus sign, and the total power is still the sum of the two.
Can the readings tell me whether the power factor is lagging or leading?
Not on their own. Which meter reads higher depends on the phase sequence as well as on the sign of the angle. With abc sequence, current coils in lines a and c and potential coils to line b, a lagging load makes the line-a meter read lower. This calculator gives the size of the angle and of the reactive power, not their sign.
Does the two-wattmeter method work for an unbalanced load?
The sum of the two readings is the total power of any three-wire load, balanced or not (Blondel's theorem). The reactive power and power factor formulas assume a balanced load. A four-wire load with current in the neutral needs three wattmeters.

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