Measurement of three-phase power and power factor by the two-wattmeter method
Measuring three-phase power and power factor with two wattmeters: connections, why a reading goes negative below 0.5 pf, the tan φ calculation, and viva questions with answers.
Aim
To measure the power and power factor of a three-phase balanced load using two wattmeters, and to verify that the sum of the two readings gives the total three-phase power.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| Three-phase balanced load | Resistive and inductive banks, 415 V | 1 each |
| Wattmeter (UPF) | 600 V, 10 A | 2 |
| Voltmeter (MI) | 0–600 V | 1 |
| Ammeter (MI) | 0–10 A | 1 |
| Three-phase variac | 415 V, 10 A | 1 |
Theory
Blondel's theorem says that the power in an n-wire system can be measured with n−1 wattmeters. A three-wire three-phase system therefore needs only two, whatever the load and whether or not it is balanced — because the three line currents must sum to zero, so knowing two of them is knowing all three.
Each wattmeter has its current coil in one line and its pressure coil connected from that line to the third line, the one carrying no current coil. The sum of the two readings is the total three-phase power, regardless of power factor.
The individual readings, however, depend strongly on the power factor. At unity power factor the two are equal. As the power factor falls the readings diverge; at exactly 0.5 one of them reads zero, and below 0.5 that one goes negative and its pointer tries to deflect backwards.
That divergence is useful. For a balanced load, tan φ = √3 (W₁ − W₂)/(W₁ + W₂), so the two readings alone give the power factor without any voltmeter or ammeter. The difference also gives the reactive power directly: Q = √3 (W₁ − W₂).
The power-factor formula assumes a balanced load. The total power W₁ + W₂ does not — it is valid for any three-wire load, balanced or not.
Circuit connections
Check every point below against your board before switching on. There is no diagram here on purpose — a wrong diagram is worse than none, and this is the list a demonstrator actually walks through with you.
- Current coil of W₁ in line R, current coil of W₂ in line Y. Line B carries no current coil.
- Pressure coil of W₁ from R to B; pressure coil of W₂ from Y to B. Both M terminals to the supply side of their own current coils.
- Ammeter in one line and voltmeter across two lines, for the cross-check.
- Three-phase load connected in star or delta as required, downstream of both wattmeters.
Procedure
- 1Complete the connections with the variac at zero and have them checked.
- 2Switch on and raise the supply to the rated line voltage.
- 3Connect a purely resistive load first. Record W₁, W₂, V_L and I_L — the two wattmeter readings should be nearly equal.
- 4Add inductive load in steps to lower the power factor, recording a full set at each step.
- 5When one pointer tries to deflect backwards, switch off, reverse that wattmeter's pressure-coil connections, switch on again and record its reading as negative.
- 6Compute the total power and the power factor from the two readings, and cross-check the total against √3·V_L·I_L·cos φ.
Work out your readings
Type in the numbers off the meters. This fills the tabular column, works the calculation through step by step, plots the characteristic — and tells you when a reading cannot physically be right, which is the part a manual can't do. Everything stays on this device, and it works with the network off.
Nameplate and machine data
Enter a negative value if the pointer reversed
Fill in the machine data to see the results, the worked calculation and the curve.
Precautions
- Reverse the pressure coil, never the current coil, when a wattmeter deflects backwards — and record the reading with a minus sign.
- Both wattmeter pressure coils must go to the same line, the one without a current coil. Connecting them to different lines gives readings that mean nothing.
- Apply the multiplying factor to each wattmeter separately; the two may be on different ranges.
- Do not exceed the current or voltage rating of either coil — a wattmeter can be overloaded on current while its pointer sits mid-scale, because deflection depends on the power factor as well.
Sources of error
Every record asks for these, and every record gets the same three lines copied from the one before. These are the errors this particular experiment actually has.
- Pressure-coil power consumption is included in the reading; on a light load it is a noticeable fraction.
- The tan φ formula assumes a balanced load. Any unbalance puts an error into the power factor, though the total power stays correct.
- At low power factor both readings are far apart and one is small, so the percentage error in the difference — which the power factor depends on — is much larger than the error in either reading.
- Harmonic distortion in the load current, which the formula does not account for.
Viva questions with answers
Why are only two wattmeters needed for a three-phase system?
Blondel's theorem: an n-wire system needs n−1 wattmeters. In a three-wire system the three line currents sum to zero, so the third measurement carries no new information.
At what power factor does one wattmeter read zero?
At exactly 0.5. Below that the same meter reads negative, and above it both read positive. At unity power factor the two readings are equal.
The total power is W₁ + W₂. Is that true for an unbalanced load?
Yes — the sum is valid for any three-wire load, balanced or not. What is not valid for an unbalanced load is the power-factor formula tan φ = √3(W₁−W₂)/(W₁+W₂), which assumes balance.
How do you find reactive power from the two readings?
Q = √3 (W₁ − W₂). The difference of the readings is proportional to the reactive power, just as the sum is equal to the active power.
Can this method be used on a four-wire system?
Not with two wattmeters. A four-wire system needs three, by the same theorem — unless the load is known to be balanced, in which case one wattmeter and a multiplication by three will do.
Why should the pressure coil be reversed rather than the current coil?
Reversing either reverses the deflection, but the current coil carries the line current and breaking it interrupts the load. The pressure coil is a high-resistance branch that can be safely reversed at the terminals.