Internal emf and armature reaction reactance

Given: star, 60 kVA, 400 V, $R_a = 0.25\ \Omega$, $X_s = 3.5\ \Omega$, $X_l = 0.75\ \Omega$ per phase; rated load at 0.8 pf lagging.

Phase voltage and current

$$\begin{aligned} V_{ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\ I &= \frac{60\times10^3}{\sqrt3\times400} = 86.60\ \text{A} \end{aligned}$$

(i) Internal emf $E_i$

The internal (air-gap) emf is the voltage behind the leakage impedance $R_a + jX_l$:

$$\begin{aligned} V\cos\phi + IR_a &= 184.75 + 86.60\times0.25 = 206.40\ \text{V} \\ V\sin\phi + IX_l &= 138.56 + 86.60\times0.75 = 203.52\ \text{V} \\ E_i &= \sqrt{206.40^2 + 203.52^2} = 289.86\ \text{V per phase} \\ E_{i,L} &= \sqrt3\times289.86 = 502.1\ \text{V} \end{aligned}$$

(ii) Armature reactance (armature reaction)

Since $X_s = X_l + X_a$:

$$X_a = X_s - X_l = 3.5 - 0.75 = 2.75\ \Omega\ \text{per phase}$$

              E0
             /|
            / |  j I Xa  (armature reaction)
           /  |
          /   + Ei
         /   /|  j I Xl
        O---+-+ V + I Ra
         `-.
            `-. I (0.8 lag)

(For reference, the no-load emf using $X_s$ would be $E_0 = \sqrt{206.40^2 + (138.56 + 303.11)^2} = 487.5$ V per phase.)

Answer: (i) $E_i$ = 289.9 V per phase (502.1 V line); (ii) $X_a$ = 2.75 Ω per phase.