Internal emf, no-load emf and regulation

Given: star, 50 kVA, 440 V, $R_a = 0.25\ \Omega$, $X_s = 3.2\ \Omega$, leakage reactance $X_l = 0.5\ \Omega$ per phase; rated load at unity pf.

• The internal (air-gap) emf $E_i$ is found using only the leakage impedance $R_a + jX_l$.
• The no-load emf $E_0$ (excitation emf) uses the full synchronous impedance $R_a + jX_s$, since $X_s = X_l + X_a$ includes armature reaction.

$$\begin{aligned} V_{ph} &= \frac{440}{\sqrt3} = 254.03\ \text{V} \\ I &= \frac{50\times10^3}{\sqrt3\times440} = 65.61\ \text{A}\quad(\text{in phase with } V) \end{aligned}$$

(i) Internal emf

$$\begin{aligned} \bar E_i &= V + I(R_a + jX_l) = (254.03 + 65.61\times0.25) + j(65.61\times0.5) \\ &= 270.44 + j32.80 \\ E_i &= 272.42\ \text{V per phase}\quad(471.8\ \text{V line}) \end{aligned}$$

(ii) No-load emf

$$\begin{aligned} \bar E_0 &= V + I(R_a + jX_s) = 270.44 + j(65.61\times3.2) \\ &= 270.44 + j209.95 \\ E_0 &= 342.36\ \text{V per phase}\quad(593.0\ \text{V line}) \end{aligned}$$

(iii) Voltage regulation

$$\%\text{Reg} = \frac{E_0 - V}{V}\times100 = \frac{342.36 - 254.03}{254.03}\times100 = 34.77\%$$

              E0
             /|
            / |  I Xa (armature reaction)
           /  |
          / Ei+
         / /  |  I Xl
        O-----+  V + I Ra
        O-------> I  (unity pf)

Answer: (i) $E_i$ = 272.4 V/phase (471.8 V line); (ii) $E_0$ = 342.4 V/phase (593.0 V line); (iii) regulation = 34.77%.