Excitation emf of 6600 V salient pole motor

Assumption: the power factor is not stated as lagging or leading; the usual textbook version of this problem takes 0.8 leading (over-excited motor). The lagging result is given at the end. Star connection, all losses neglected, so input = 2 MW.

$$\begin{aligned} V &= \frac{6600}{\sqrt3} = 3810.51\ \text{V per phase} \\ I_a &= \frac{2\times10^6}{\sqrt3 \times 6600 \times 0.8} = 218.69\ \text{A} \\ \vec I_a &= 218.69\angle 36.87^\circ = 174.95 + j131.22\ \text{A} \end{aligned}$$

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

Load angle

$$\begin{aligned} jX_q\vec I_a &= j3.1(174.95 + j131.22) = -406.77 + j542.36 \\ \vec E' &= 3810.51 - (-406.77 + j542.36) = 4217.28 - j542.36 \\ &= 4252.0\angle -7.33^\circ\ \text{V} \end{aligned}$$

So $\delta = 7.33^\circ$.

Excitation emf

$$\begin{aligned} \psi &= \delta + \phi = 7.33^\circ + 36.87^\circ = 44.20^\circ \\ I_d &= 218.69\sin44.20^\circ = 152.46\ \text{A},\quad I_q = 156.79\ \text{A} \\ E_f &= 4252.0 + (5 - 3.1)(152.46) = 4541.7\ \text{V per phase} \end{aligned}$$

Line value $= \sqrt3 \times 4541.7 = 7866$ V.

Check: $3\left[\dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right)\sin2\delta\right] = 2.000$ MW. Correct.

          Ia (leads V by 36.87 deg)
         /
        /
  -----+-------------------> V = 3811 V
        \  delta = 7.33 deg
         '----------> Ef = 4542 V

If the pf is 0.8 lagging: $\vec E' = 3446.7\angle -9.05^\circ$, $\psi = -27.82^\circ$, $I_d = -102.05$ A, $E_f = 3446.7 - 1.9 \times 102.05 = 3252.8$ V per phase (5634 V line).

Answer (0.8 leading): $E_f = 4542$ V per phase $= 7.87$ kV line, load angle $\delta = 7.33^\circ$.