Internal emf and load angle of salient pole motor

Assumptions: the motor operates at rated voltage and rated current ($V = 1$ pu, $I_a = 1$ pu) with losses and $R_a$ neglected; bases are the machine ratings (11 kV line, 6.351 kV phase).

$\cos\phi = 0.98$ leading $\Rightarrow \phi = 11.48^\circ$, $\vec I_a = 1\angle 11.48^\circ = 0.98 + j0.199$ 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. $E_f = E' + (X_d - X_q)I_d$.

Load angle

$$\begin{aligned} \vec E' &= 1 - j0.8(0.98 + j0.199) = 1 + 0.159 - j0.784 \\ &= 1.1592 - j0.784 = 1.3994\angle -34.07^\circ\ \text{pu} \end{aligned}$$

So the load angle is $\delta = 34.07^\circ$.

Internal emf

$$\begin{aligned} \psi &= \delta + 11.48^\circ = 45.55^\circ \ (\text{leading current}) \\ I_d &= 1 \times \sin45.55^\circ = 0.7139\ \text{pu},\quad I_q = \cos45.55^\circ = 0.7003\ \text{pu} \\ E_f &= 1.3994 + (1.2 - 0.8)(0.7139) = 1.3994 + 0.2856 = 1.685\ \text{pu} \end{aligned}$$

In volts: $E_f = 1.685 \times 6.351 = 10.70$ kV per phase $= 18.53$ kV line.

 V = 1.0 ---------------------->
   \                  Ia (leads V by 11.5 deg)
    \ delta = 34.07
     \
      '--> Ef = 1.685 pu  (Ef > V: over-excited)

Check: $P = \dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)\sin2\delta = 0.787 + 0.193 = 0.98$ pu $= V I\cos\phi$.

Answer: Internal emf $E_f = 1.685$ pu (10.70 kV/phase, 18.53 kV line); load angle $\delta = 34.07^\circ$.