20 MVA salient pole motor maximum power

Given: 20 MVA, 11 kV, star, $X_d = 50\ \Omega$, $X_q = 3\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$\begin{aligned} V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\ I_a &= \frac{20\times10^6}{\sqrt3 \times 11000} = 1049.73\ \text{A}\ (\text{in phase with } V) \end{aligned}$$

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

(i) Excitation voltage

$$\begin{aligned} \vec E' &= V - jX_qI_a = 6350.85 - j(3 \times 1049.73) = 6350.85 - j3149.18 \\ &= 7088.8\angle -26.38^\circ\ \text{V} \quad \Rightarrow\ \delta = 26.38^\circ \\ \psi &= \delta - 0 = 26.38^\circ \ (\text{upf}) \\ I_d &= 1049.73\sin26.38^\circ = 466.34\ \text{A},\quad I_q = 940.45\ \text{A} \\ E_f &= E' + (X_d - X_q)I_d = 7088.8 + 47 \times 466.34 \\ &= 29007\ \text{V per phase} = 50.24\ \text{kV line} \end{aligned}$$

Check: $P$ at $\delta = 26.38^\circ$ from the power-angle equation $= 20.00$ MW $=$ rated input.

(ii) Maximum power and its power angle

$$P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right]$$

Let $a = \dfrac{E_fV}{X_d} = 3.6844\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 12.6378\times10^6$ (per phase).

$\dfrac{dP}{d\delta} = 0 \Rightarrow 2b\cos^2\delta + a\cos\delta - b = 0$:

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.6380 \\ \delta_m &= 50.36^\circ \\ P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\ &= 8.51 + 18.63 = 27.14\ \text{MW} \end{aligned}$$

The maximum occurs below $90^\circ$ because of the reluctance term.

Answer: (i) $E_f = 29007$ V per phase (50.24 kV line), load angle $26.38^\circ$; (ii) maximum power angle $\delta_m = 50.36^\circ$, $P_{max} = 27.14$ MW.