Vs from P+jQ and reactive power flow direction

Derivation

Short line, per phase impedance $Z = R + jX$. Take $V_R = |V_R|\angle0^\circ$ and the receiving-end demand $S_R = P + jQ$ (per phase).

Since $S_R = V_RI^*$,

$$I = \frac{P - jQ}{V_R}$$

Then

$$\begin{aligned} V_S &= V_R + (R + jX)\frac{P - jQ}{V_R} \\ &= \left(V_R + \frac{RP + XQ}{V_R}\right) + j\left(\frac{XP - RQ}{V_R}\right) \end{aligned}$$

So

$$|V_S| = \sqrt{\left(V_R + \frac{RP + XQ}{V_R}\right)^2 + \left(\frac{XP - RQ}{V_R}\right)^2}$$

The first bracket is the in-phase drop $\Delta V = (RP + XQ)/V_R$ and the second is the quadrature drop $\delta V = (XP - RQ)/V_R$.

Reactive power flows from higher to lower voltage

For transmission lines $X \gg R$ and the quadrature part changes the magnitude very little. Then

$$|V_S| - |V_R| \approx \frac{RP + XQ}{V_R} \approx \frac{XQ}{V_R}$$ $$\Rightarrow\ Q \approx \frac{V_R\,(|V_S| - |V_R|)}{X}$$

• If $|V_S| > |V_R|$: $Q > 0$, reactive power flows from sending end to receiving end.
• If $|V_S| < |V_R|$: $Q < 0$, reactive power flows from receiving end to sending end.
• If $|V_S| = |V_R|$: almost no reactive power transfer.

So reactive power always flows from the bus with the higher voltage magnitude to the bus with the lower voltage magnitude, whatever the phase angles. (Active power, on the other hand, follows the angle: $P \approx V_SV_R\sin\delta/X$.)