Power angle depends on real power

Consider a cylindrical-rotor synchronous generator with internal emf $E_f\angle\delta$, terminal voltage $V_t\angle 0^\circ$, synchronous reactance $X_s$ and negligible armature resistance.

Derivation

Armature current:

$$I_a = \frac{E_f\angle\delta - V_t\angle 0}{jX_s}$$

Complex power delivered at the terminal:

$$\begin{aligned} S &= V_tI_a^* = V_t\left[\frac{E_f\angle -\delta - V_t}{-jX_s}\right] = \frac{jV_t\left(E_f\cos\delta - jE_f\sin\delta - V_t\right)}{X_s} \\ &= \frac{V_tE_f\sin\delta}{X_s} + j\,\frac{V_tE_f\cos\delta - V_t^2}{X_s} \end{aligned}$$

So

$$P = \frac{E_fV_t}{X_s}\sin\delta,\qquad Q = \frac{V_t(E_f\cos\delta - V_t)}{X_s}$$

Showing δ is due to real power alone

Write $I_a = I_a\cos\phi - jI_a\sin\phi$ (lagging). Then

$$E_f = V_t + jX_sI_a = (V_t + X_sI_a\sin\phi) + j\,X_sI_a\cos\phi$$ $$\tan\delta = \frac{X_sI_a\cos\phi}{V_t + X_sI_a\sin\phi}$$

• The imaginary (quadrature) part of $E_f$, which creates the angle $\delta$, is $X_sI_a\cos\phi = X_sP/V_t$, i.e. proportional only to real power.
• The reactive current $I_a\sin\phi$ adds only to the in-phase part of $E_f$, changing its magnitude, not producing any angle by itself.

Two limiting cases make this clear:

1. P = 0 (pure reactive load, $\cos\phi = 0$): $E_f\sin\delta = 0$, so $\delta = 0$ however large Q is; $E_f$ and $V_t$ are in phase and differ only in magnitude.
2. Q = 0 (unity pf): $E_f = V_t + jX_sI_a$, and $\delta = \tan^{-1}(X_sI_a/V_t)$, a non-zero angle set by P.

Hence the power angle $\delta$ between $E_f$ and $V_t$ exists only because real power is delivered ($\delta = 0 \Leftrightarrow P = 0$), while Q affects mainly the magnitude of $E_f$. This is the basis of P–δ (P–f) and Q–V decoupling.