# Sign of 3-phase instantaneous reactive power and phase sequence

For three-phase systems, in p-q theory instantaneous reactive power is

$$q = \frac{(v_a-v_b)i_c + (v_b-v_c)i_a + (v_c-v_a)i_b}{\sqrt{3}}$$

(see, e.g.,  Eq. (5))

If we substitute in expressions for a balanced inductive load, in which the current lags the voltage by 90 degrees: \begin{align} v_a &= V \sin(\omega t)\\ v_b &= V \sin(\omega t - 2\pi/3)\\ v_c &= V \sin(\omega t + 2\pi/3)\\ i_a &= I \sin(\omega t - \pi/2)\\ i_b &= I \sin(\omega t - \pi/2 - 2\pi/3)\\ i_c &= I \sin(\omega t - \pi/2 + 2\pi/3)\\ \end{align}

we find (I used Sympy to verify this): $$q = \frac{3VI}{2} > 0 \quad\text{(if V, I > 0)}$$ which matches up with, say, a phasor-based computation of total reactive power for a pure inductive load.

However, if we reverse the phase sequence by swapping phases b and c, this equation becomes \begin{align} q' &= \frac{(v_a-v_c)i_b + (v_c-v_b)i_a + (v_b-v_a)i_c}{\sqrt{3}}\\ &=\frac{-(v_c-v_a)i_b - (v_b-v_c)i_a - (v_a-v_b)i_c}{\sqrt{3}}\\ &=\frac{-(v_a-v_b)i_c -(v_b-v_c)i_a -(v_c-v_a)i_b}{\sqrt{3}}\\ &=-q\\ \end{align}

and so the instantaneous reactive power changes sign.

My question: to relate p-q theory 3-phase instantaneous reactive power to "conventional" reactive power, must we take the phase sequence into account?

 "p-q Theory Power Components Calculations", Afonso, Freitas, and Martins. DOI 10.1109/ISIE.2003.1267279, available here

• Somebody else will explain it much better, but here's a quick test: i.stack.imgur.com/EIM6J.png. The 3 lower plots are the currents from the grid and on the loads. Each plot has two overlapping traces and they coincide; the voltages change from abs to acb. That's with [200,230,230] V on the grid, [10,10,5] A and -[30,60,30] deg on the loads. The upper plot, though, is the p after filtering. Because the compensation has the Clarke transform with quadrature output, so the signs of vb*ib change: i.stack.imgur.com/TmaeK.png. Oct 9, 2021 at 12:38