Direction of active and reactive power flow

Consider two buses joined by a short line of reactance $X$ (resistance neglected, as $X \gg R$). Bus 1 voltage $V_1\angle\delta_1$, bus 2 voltage $V_2\angle\delta_2$, and $\delta = \delta_1 - \delta_2$.

  V1/_d1        jX         V2/_d2
   o-----------/\/\/----------o
   |      P, Q  -->           |
  Bus 1                     Bus 2

Derivation

$$I = \frac{V_1\angle\delta_1 - V_2\angle\delta_2}{jX}$$

Power received at bus 2:

$$S_2 = V_2\angle\delta_2\,I^* = \frac{V_1V_2\sin\delta}{X} + j\,\frac{V_1V_2\cos\delta - V_2^2}{X}$$

So

$$P = \frac{V_1V_2}{X}\sin\delta,\qquad Q = \frac{V_2}{X}\left(V_1\cos\delta - V_2\right)$$

Active power: from leading to lagging angle

• $P$ depends on $\sin\delta$. If $\delta_1 > \delta_2$ (bus 1 leads), $\sin\delta > 0$ and $P > 0$: power flows from 1 to 2.
• If bus 2 leads, $P < 0$: power flows from 2 to 1.
• $P$ flows even if $V_1 = V_2$, as long as there is an angle difference.

Reactive power: from higher to lower magnitude

In practice $\delta$ is small, so $\cos\delta \approx 1$:

$$Q \approx \frac{V_2(V_1 - V_2)}{X}$$

• If $|V_1| > |V_2|$, $Q > 0$: reactive power flows from 1 to 2.
• If $|V_1| < |V_2|$, $Q$ reverses.
• It hardly depends on $\delta$.

Phasor illustration

Case A: P flow           Case B: Q flow
  V1 (leads)               V1 (longer)
   /                       ------------>
  / delta                  V2 (shorter, same angle)
 O-------> V2              ------->
 P: 1 -> 2                 Q: 1 -> 2

Example

$X = 10\ \Omega$, $V_1 = 11\angle5^\circ$ kV, $V_2 = 10.5\angle0^\circ$ kV: $P = \frac{11\times10.5}{10}\sin5^\circ = 1.01$ MW (1 → 2), $Q = \frac{10.5}{10}(11\cos5^\circ - 10.5) = 0.48$ MVAr (1 → 2). If instead $V_2 = 11.5\angle0^\circ$ kV, $P$ still flows 1 → 2, but $Q = \frac{11.5}{10}(11\cos5^\circ - 11.5) = -0.62$ MVAr, i.e. Q flows 2 → 1.

Hence the statement is verified: P follows the phase angle, Q follows the voltage magnitude.