Sequence components decoupled for power flow

The complex power of a three-phase system splits into three separate terms, one for each sequence, with no cross terms between different sequences. So each sequence network carries its own power independently; this is what "decoupled for power flow" means.

Proof

Complex power in phase quantities:

$$S = V_aI_a^* + V_bI_b^* + V_cI_c^* = V_p^TI_p^*$$

With $V_p = AV_s$, $I_p = AI_s$, where $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$:

$$S = (AV_s)^T(AI_s)^* = V_s^T\,A^TA^*\,I_s^*$$

$A$ is symmetric, so $A^T = A$; and $a^* = a^2$, so $A^* = \begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$. Using $1+a+a^2 = 0$, $a^3 = 1$:

$$A^TA^* = \begin{bmatrix} 3 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3&0&0\\0&3&0\\0&0&3 \end{bmatrix}$$

Hence

$$S = 3V_s^TI_s^* = 3V_{a0}I_{a0}^* + 3V_{a1}I_{a1}^* + 3V_{a2}I_{a2}^*$$

Why this shows decoupling

• Each term contains a voltage and a current of the same sequence only. Terms like $V_{a1}I_{a2}^*$ or $V_{a0}I_{a1}^*$ do not appear, because the off-diagonal elements of $A^TA^*$ are zero.
• Physically, a positive sequence voltage acting with a negative (or zero) sequence current produces only an oscillating power whose sum over the three phases is zero; it gives no net power.
• Therefore the power delivered by the positive sequence network ($S_1 = 3V_{a1}I_{a1}^*$), negative sequence network ($S_2$) and zero sequence network ($S_0$) can be calculated separately and simply added:

$$S = S_0 + S_1 + S_2$$

• Together with the diagonal sequence impedance matrix of symmetrical equipment (voltage of one sequence depends only on current of the same sequence), this shows the three sequence networks are fully decoupled for both voltage drop and power.

Example

A balanced system has only positive sequence: $S = 3V_{a1}I_{a1}^*$. A generator feeding an unbalanced load delivers useful power mainly through the positive sequence network; power in the negative sequence network appears as losses (for example, extra rotor heating).

The factor 3 appears because $A$ is not unitary. With the power-invariant form $\frac{1}{\sqrt3}A$, $S = V_s^TI_s^*$ exactly.