3.3 kV synchronous motor back emf

Given: 3.3 kV, 3-phase star motor, $Z_s = 0.2 + j2.2\ \Omega$ per phase, 0.5 pf leading, $I_L = 100$ A.

$$V_{ph} = \frac{3300}{\sqrt3} = 1905.26\ \text{V}, \qquad I = 100\ \text{A}$$

$\cos\phi = 0.5$ leading, so $\phi = 60^\circ$ and $\vec I = 100\angle +60^\circ$ A with $V_{ph}$ as reference.

For a motor, the back emf is

$$\vec E_b = \vec V - \vec I Z_s$$

Impedance drop:

$$\vec I Z_s = 100\angle60^\circ\times(0.2 + j2.2) = 100\angle 60^\circ\times 2.209\angle 84.8^\circ = 220.9\angle144.8^\circ = -180.53 + j127.32\ \text{V}$$ $$\vec E_b = 1905.26 - (-180.53 + j127.32) = 2085.79 - j127.32\ \text{V}$$ $$E_b = \sqrt{2085.79^2 + 127.32^2} = 2089.66\ \text{V}, \qquad \delta = \tan^{-1}\!\left(\frac{-127.32}{2085.79}\right) = -3.49^\circ$$

The line value is $\sqrt3\times 2089.66 = 3619.4$ V.

Phasor diagram

         / I (leads V by 60 deg)
       /
     /
   /
 O------------------------> V
  `-._                     \
      `-._    Eb            \ I Zs = 220.9 V at 144.8 deg
          `-._               \ (from tip of Eb to tip of V)
              `-._            \
                  `-._         \
                      `-._      \
                          `-._   \
                              `-._*  tip of Eb

(Not to scale. $\vec V = \vec E_b + \vec I Z_s$; $E_b$ lags $V$ by $3.49^\circ$ and $I$ leads $V$ by $60^\circ$. The motor is over-excited because $E_b > V$.)

Answer: back emf per phase $= 2089.7$ V (line 3619 V), load angle $3.49^\circ$ ($E_b$ lagging $V$).