Expression for Back E.M.F or Induced E.M.F. per Phase in S.M.

Case i) Under excitation, E_bph < V_ph .

       Z_s = R_a + j X_s = | Z_s |  ∟θ Ω

       θ = tan^-1(X_s/R_a)

       E_Rph  ^ I_aph = θ, I_a lags always by angle θ.

       V_ph = Phase voltage applied

       E_Rph = Back e.m.f. induced per phase

       E_Rph = I_a x Z_s V            ... per phase

       Let p.f. be cosΦ, lagging as under excited,

       V_ph  ^ I_aph = Φ

       Phasor diagram is shown in the Fig. 1.

Fig. 1 Phasor diagram for under excited condition

       Applying cosine rule to Δ OAB, 

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph x (V_ph ^ E_Rph)

       but V_ph ^ E_Rph = x = θ - Φ

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph x (θ - Φ)                   ......(1)

       where E_Rph =  I_aph x Z_s

       Applying sine rule to Δ OAB,

       E_bph/sinx = E_Rph/sinδ

        So once E_bph is calculated, load angle δ can be determined by using sine rule.

Case ii) Over excitation, E_bph > V_ph

       p.f. is leading in nature.

       E_Rph  ^ I_aph = θ

       V_ph  ^ I_aph = Φ

       The phasor diagram is shown in the Fig. 2.

Fig.2  Phasor diagram for overexcited condition

       Applying cosine rule to Δ OAB,

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph  E_Rph x cos(V_ph ^ E_Rph)

       V_ph ^ E_Rph = θ + Φ

.^..    (E_bph)² = (V_ph)² + (E_Rph)² - 2 V_ph   E_Rph cos(θ + Φ) .......(3)

       But θ + Φ is generally greater than  90^o

.^..    cos (θ + Φ) becomes negative, hence for leading p.f., E_bph > V_ph .

       Applying sine rule to Δ OAB,

       E_bph/sin( E_Rph ^ V_ph) = E_Rph/sinδ

       Hence load angle δ can be calculated once E_bph is known.

Case iii) Critical excitation

       In this case E_{bph ≈} V_ph, but p.f. of synchronous motor is unity.

.^..         cos = 1   .^..    Φ = 0^o

       i.e. V_ph and I_aph are in phase

       and  E_Rph ^ I_aph = θ

       Phasor diagram is shown in the Fig. 3.

Fig. 3  Phasor diagram for unity p.f. condition

       Applying cosine rule to OAB,

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph cos θ            ............(5)

       Applying sine rule to OAB,

       E_bph/sinθ = E_Rph/sinδ

       where   E_Rph = I_aph x Zs V