Salient pole motor excitation and reluctance power

Work in per unit: $V = 1\angle0^\circ$ pu, rated current $I_a = 1$ pu at 0.8 lagging, so $\vec I_a = 1\angle -36.87^\circ = 0.8 - j0.6$ pu.

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. $E_f = E' + (X_d - X_q)I_d$.

a) Excitation voltage

$$\begin{aligned} \vec E' &= 1 - j0.4(0.8 - j0.6) = 1 - 0.24 - j0.32 \\ &= 0.76 - j0.32 = 0.8246\angle -22.83^\circ\ \text{pu} \\ \delta &= 22.83^\circ,\quad \psi = 22.83^\circ - 36.87^\circ = -14.04^\circ \\ I_d &= 1 \times \sin(-14.04^\circ) = -0.2425\ \text{pu},\quad I_q = 0.9701\ \text{pu} \\ E_f &= 0.8246 + (0.8 - 0.4)(-0.2425) = 0.7276\ \text{pu} \end{aligned}$$

$E_f = 0.7276 \times 11 = 8.00$ kV (line), i.e. 4.62 kV per phase. $E_f < V$, as expected for an under-excited (lagging) motor.

b) Excitation power and reluctance power

$$P = \underbrace{\frac{E_fV}{X_d}\sin\delta}_{\text{excitation}} + \underbrace{\frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta}_{\text{saliency}}$$ $$\begin{aligned} P_{exc} &= \frac{0.7276 \times 1}{0.8}\sin22.83^\circ = 0.3529\ \text{pu} = 17.65\ \text{MW} \\ P_{rel} &= \frac{1}{2}(2.5 - 1.25)\sin45.67^\circ = 0.4471\ \text{pu} = 22.35\ \text{MW} \\ P &= 0.3529 + 0.4471 = 0.8\ \text{pu} = 40\ \text{MW} \end{aligned}$$

Check: input $= VI\cos\phi = 1 \times 1 \times 0.8 = 0.8$ pu. Shaft output $= 0.8 - 0.15 = 0.65$ pu $= 32.5$ MW.

c) Field current reduced to zero

With $E_f = 0$ only reluctance power remains:

$$P_{rel,max} = \frac{V^2}{2}\left(\frac{1}{X_q} - \frac{1}{X_d}\right) = \frac12(2.5 - 1.25) = 0.625\ \text{pu} \ (\text{at } \delta = 45^\circ)$$

The motor must take $0.8$ pu (0.65 pu load + 0.15 pu rotational loss). Since $0.625 < 0.8$, the reluctance torque cannot carry the load, so the machine falls out of synchronism. It would stay in step as a reluctance motor only if the shaft load were reduced so that total input is below 0.625 pu (load below 0.475 pu).

Answer: (a) $E_f = 0.728$ pu (8.0 kV line), $\delta = 22.8^\circ$; (b) excitation power 0.353 pu (17.65 MW), saliency power 0.447 pu (22.35 MW); (c) No, it loses synchronism.