Maximum steady state power with Vt held

When the terminal voltage $|V_t|$ is held constant (by a fast voltage regulator), the generator terminal bus transfers power to the infinite bus through the series reactance $X_e$:

$$P = \frac{|V_t|\,|V|}{X_e}\sin\theta$$

where $\theta$ is the angle between $V_t$ and $V$. The power is largest when $\theta = 90^\circ$. The angle between $E$ and $V$ is then more than $90^\circ$, which is acceptable because the regulator raises $E$ as the load angle grows.

Data: $X_g = 0.4$ pu, $X_e = 1.0$ pu (series reactance), $|V_t| = 1.1$ pu, $V = 1.0\angle0^\circ$ pu.

Step 1: Maximum power

$$P_{max} = \frac{|V_t|\,|V|}{X_e} = \frac{1.1\times 1.0}{1.0} = 1.1\ \text{pu}$$

Step 2: Check that this point is feasible

At the limit, $V_t = j1.1$ pu. The current is

$$I = \frac{V_t - V}{jX_e} = \frac{j1.1 - 1}{j1.0} = 1.1 + j1.0\ \text{pu}$$

so $P = \text{Re}(V I^*) = 1.1$ pu. The required internal emf is

$$E = V_t + jX_gI = j1.1 + j0.4(1.1 + j1.0) = -0.4 + j1.54 = 1.591\angle104.6^\circ\ \text{pu}$$

Check: $\dfrac{EV}{X_g + X_e}\sin\delta = \dfrac{1.591\times 1}{1.4}\sin104.6^\circ = 1.10$ pu.

Comparison with constant $E$

If $E$ were fixed and the limit were taken at $\delta = 90^\circ$, then $V_t = 1.1$ pu would need $E = 1.487$ pu, giving only $P_{max} = EV/X = 1.487/1.4 = 1.062$ pu. If $E$ were simply taken as 1.1 pu, the limit would be $1.1/1.4 = 0.786$ pu. Holding $V_t$ constant with the regulator raises the limit to $|V_t||V|/X_e$.

Answer: Maximum steady-state power capability $P_{max} = 1.1$ pu (at $\theta = 90^\circ$ between $V_t$ and $V$, with $E = 1.59$ pu).