25 MVA salient pole motor maximum power

Given: 25 MVA, 11 kV, star, $X_d = 48\ \Omega$, $X_q = 3.5\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$\begin{aligned} V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\ I_a &= \frac{25\times10^6}{\sqrt3 \times 11000} = 1312.16\ \text{A}\ (\text{in phase with } V) \end{aligned}$$

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

$$\begin{aligned} \vec E' &= V - jX_qI_a = 6350.85 - j(3.5 \times 1312.16) = 6350.85 - j4592.56 \\ &= 7837.4\angle -35.87^\circ\ \text{V} \quad \Rightarrow\ \delta = 35.87^\circ \\ \psi &= \delta - 0 = 35.87^\circ \ (\text{upf}) \\ I_d &= 1312.16\sin35.87^\circ = 768.90\ \text{A},\quad I_q = 1063.28\ \text{A} \\ E_f &= E' + (X_d - X_q)I_d = 7837.4 + 44.5 \times 768.90 \\ &= 42053\ \text{V per phase} = 72.84\ \text{kV line} \end{aligned}$$

Check: $P$ at $\delta = 35.87^\circ$ from the power-angle equation $= 25.00$ 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 $a = \dfrac{E_fV}{X_d} = 5.5641\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 10.6835\times10^6$ (per phase).

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

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.5888 \\ \delta_m &= 53.93^\circ \\ P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\ &= 13.49 + 15.25 = 28.75\ \text{MW} \end{aligned}$$

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

Answer: (i) $E_f = 42053$ V per phase (72.84 kV line), load angle $35.87^\circ$; (ii) maximum power angle $\delta_m = 53.93^\circ$, $P_{max} = 28.75$ MW.