Receiving end complex power and power circle

Derivation of receiving-end complex power

For any line, $V_S = AV_R + BI_R$. Take

$$V_R = |V_R|\angle0^\circ,\ V_S = |V_S|\angle\delta,\ A = |A|\angle\alpha,\ B = |B|\angle\beta$$

Then

$$I_R = \frac{V_S - AV_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta)$$

Receiving-end complex power (per phase; with line voltages it is three-phase):

$$\begin{aligned} S_R &= V_R I_R^* \\ &= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha) \end{aligned}$$

In rectangular form:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$

Construction of the receiving-end power circle

Write the result as

$$\left(P_R + \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)\right)^2 + \left(Q_R + \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)^2 = \left(\frac{|V_S||V_R|}{|B|}\right)^2$$

This is a circle in the $P$–$Q$ plane with:

• Centre $N$: $\left(-\frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha),\ -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)$, fixed for a fixed $|V_R|$.
• Radius $NM = \frac{|V_S||V_R|}{|B|}$.

Steps:

1. Draw $P$ (horizontal) and $Q$ (vertical) axes with origin $O$.
2. From $O$ draw line $ON$ of length $|A||V_R|^2/|B|$ at angle $(\beta-\alpha)$ below the negative $P$-axis; $N$ is the centre.
3. With centre $N$ draw an arc of radius $|V_S||V_R|/|B|$.
4. From $O$ draw the load line at angle $\phi_R$ above the $P$-axis (lagging); it cuts the circle at the operating point $M$. $M$ gives $P_R$ and $Q_R$.
5. Angle between $NM$ and the horizontal through $N$ is $(\beta-\delta)$, so $\delta$ can be read off.

          Q
          |        . M (P_R, Q_R)
          |     .   /
          |  .     /  load line (phi_R)
  --------O-------/------------- P
         /|     .
        / |  .   radius = |Vs||Vr|/|B|
       /  |.
      N (centre)
   angle (beta-alpha) below -P axis

The maximum $P_R$ is at the point where the radius $NM$ is horizontal ($\delta = \beta$). Circles for different $|V_S|$ are concentric about $N$.