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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., [1] 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?

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

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  • \$\begingroup\$ 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. \$\endgroup\$ Oct 9, 2021 at 12:38

2 Answers 2

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Phase sequence does not have an effect on P or Q when P or Q is averaged over an integral number of cycles. When thus averaged, Q will be zero and P can be positive, negative or zero. P is negative when the source is absorbing power rather than supplying power.

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The problem is that pq theory misinterprets the power phenomea. It calls "reactive power" to something not generally related to the physical true meaning of reactive power, i.e., the energy stored in magnetic (inductor) or electric fields (Capacitor). Only in balanced and single phase systems, there is one to one correspondence, but not in non-sinusoidal or multiphase systems.

The true meaning can only be attached to Maxwell equations through the Poynting Vector. So the result is that you find a "strange" behaviour for q.

PQ theory is widely used for current compensation and it does a good job, but it cannot be used for a physical interpretation of power flow.

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  • \$\begingroup\$ Thanks - so, to my original question: "for balanced 3-phase systems, is the sign of instantaneous reactive power as defined by p-q theory is function of phase sequence", the answer is "yes" ? Is the idea you're presenting that reactive power is actually a vector, and so this sign dependence is a symptom of a larger problem? I see in onlinelibrary.wiley.com/doi/pdf/10.1002/9780470667057.app2 that (1) Akagi's q is presented as having opposite sign to what I've shown (paragraph below (B.30)), and (2) that there is an extension of p-q theory where q is a vector (see (B.49)). \$\endgroup\$
    – Rory Yorke
    Nov 9, 2021 at 7:40
  • \$\begingroup\$ Yes, as you already showed, the sign of the term q in Akagi theory changes depending on the phase sequence. So, this evidence clearly suggests that it doesn't represent the true meaning of reactive power. Moreover, using vectors is a convenient way to better understand the power flow from an economical point of view. In my opinion, Geometric Algebra is now the more powerful tool for this. \$\endgroup\$ Nov 15, 2021 at 7:31

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