Two wattmeter method for unbalanced load

Principle: For any three-wire three-phase load, whether balanced or unbalanced, star or delta, the total power is the sum of the readings of two wattmeters.

Circuit diagram

 R ──[ W1: CC ]──────────────┬──► Z_R ─┐
        PC ──────┐           │         │
                 │           │         ● N / Δ
 Y ──────────────┴───────────┼──► Z_Y ─┤
        PC ──────┐           │         │
 B ──[ W2: CC ]──────────────┴──► Z_B ─┘

• The current coils (CC) of W₁ and W₂ are in lines R and B.
• The pressure coils (PC) of both are connected between their own line and the third line (Y).

Principle for an unbalanced load

Instantaneous power p = v_R i_R + v_Y i_Y + v_B i_B. In a three-wire system i_R + i_Y + i_B = 0, so i_Y = −i_R − i_B. Then

p = (v_R − v_Y) i_R + (v_B − v_Y) i_B = v_RY i_R + v_BY i_B

• W₁ reads the average of v_RY i_R = V_RY I_R cos(angle between V_RY and I_R)
• W₂ reads the average of v_BY i_B = V_BY I_B cos(angle between V_BY and I_B)

P = W₁ + W₂

This does not depend on any balance. No assumption about load equality is used. For an unbalanced load the two readings are generally unequal, and the simple formulas cos(30° ± φ) no longer apply, because the phase angles and the line currents differ.

Procedure

1. Connect the wattmeters as shown and read W₁ and W₂.
2. If a wattmeter deflects backwards (power factor below 0.5 in a balanced case, or a strongly reactive phase), reverse its pressure-coil leads and treat the reading as negative.
3. Total power = W₁ + W₂ (algebraic sum).

Limits

• It works only for a three-wire system. With a neutral wire and unbalanced load, three wattmeters are needed.
• For a balanced load, tan φ = √3 (W₁ − W₂)/(W₁ + W₂) gives the power factor, but this relation is not valid for unbalanced loads.