Back emf and new current of 9 kVA motor

Assumption: star connection (not stated); $R_a$ neglected.

Given: $V_L = 320$ V, $X_s = 7.2\ \Omega$, $P = 4.6$ kW at unity pf.

(i) Back emf and power angle

$$\begin{aligned} V &= \frac{320}{\sqrt3} = 184.75\ \text{V per phase} \\ I_a &= \frac{4600}{\sqrt3 \times 320 \times 1} = 8.299\ \text{A} \\ I_aX_s &= 8.299 \times 7.2 = 59.76\ \text{V} \end{aligned}$$

Motor: $\vec E_f = \vec V - j X_s \vec I_a$, with $I_a$ in phase with $V$:

$$\begin{aligned} \vec E_f &= 184.75 - j59.76 \\ E_f &= \sqrt{184.75^2 + 59.76^2} = 194.18\ \text{V per phase} \\ \delta &= \tan^{-1}\frac{59.76}{184.75} = 17.92^\circ \end{aligned}$$

Line value of $E_f = \sqrt3 \times 194.18 = 336.3$ V.

  V = 184.75 -------------->
   \ delta = 17.92 deg      |
    \                       | jXs Ia = 59.76
     \--------------------->|
          Ef = 194.18
  Ia in phase with V

(ii) Load angle increased by 25 %

$$\delta' = 1.25 \times 17.92^\circ = 22.40^\circ, \qquad \vec E_f = 194.18\angle -22.40^\circ = 179.52 - j74.01\ \text{V}$$ $$\begin{aligned} \vec I_a' &= \frac{\vec V - \vec E_f}{jX_s} = \frac{(184.75 - 179.52) + j74.01}{j7.2} \\ &= \frac{5.23 + j74.01}{j7.2} = 10.279 - j0.727 \\ &= 10.30\angle -4.04^\circ\ \text{A} \end{aligned}$$

Power factor $= \cos 4.04^\circ = 0.9975$ lagging.

Check: new input $= 3 \times 184.75 \times 10.30 \times 0.9975 = 5.70$ kW, equal to $\dfrac{3VE_f}{X_s}\sin22.40^\circ$.

Answer: (i) $E_f = 194.2$ V/phase (336.3 V line), $\delta = 17.92^\circ$. (ii) $I_a = 10.30$ A at 0.9975 lagging pf.