Voltage Equation of an Alternator

If E_ph is the induced e.m.f. per phase in the alternator, there are following voltage drops occur in an alternator.

i) The drop across armature resistance I_a R_a both I_a and R_a are per phase values.

ii) The drop across synchronous reactance  I_a X_s, both  I_a and X_s are per phase values.

       After supplying these drops, the remaining voltage of E_ph is available as the terminal voltage V_ph.

Note : Now drop I_a R_a is always in phase with I_a due to a resistive drop while current I_a lags by 90^o with respect to drop  I_a X_s as it is a drop across purely inductive reactance.

       Hence all these quantities can not be added or subtracted algebraically but must be added or subtracted vectorially considering their individual phases. But we can write a voltage equation in its phasor from as,

       This is called voltage equation of an alternator.

       From this equation, we can draw the phasor diagram for various load power factor conditions and establish the relationship between E_ph and V_ph, in terms of armature current i.e. load current and the power factor cos(Φ).

For drawing the phasor diagram consider all per phase values and remember following steps.

Steps to draw the phasor diagram :

1. Choose current as a reference phasor.

2. Now if load power factor is cosΦ it indicates that angle between V_ph and I_a is Φ as V_ph is the voltage available to the load.

       So show the phasor V_ph in such a way that angle between V_ph and I_a is Φ. For lagging 'Φ', I_a should lag V_ph and for leading 'Φ', I_a should lead V_ph. For unity power factor load Φ is zero, so V_ph and I_a are in phase.

3. Now the drop I_a R_a is a resistive drop and hence will always be in phase with I_a. So phasor I_a R_a direction will be always same as I_a, i.e. parallel to I_a. But as it is to be added to V_ph, I_a R_a phasor must be drawn from the tip of the V_ph phasor drawn.

4. The drop I_a X_s is drop across purely inductive reactance. In pure inductance, current lags voltage by 90^o. So 'I_a X_s' phasor direction will be always such that I_a will lag I_a X_s phasor by 90^o. But this phasor is to be drawn from the tip of the  I_a R_aphasor to complete phasor addition of V_ph, I_a R_a and I_a X_s.

5. Joining the starting point to the terminating point, we get the phasor E_ph.

       Whatever may be the load power factor, I_a R_a is a resistive drop, will be in phase with I_a while I_a X_s is purely inductive drop and hence will be perpendicular to I_a in such a way that I_a will lag  I_a X_s by 90^o. This is shown in the Fig. 1.

Fig. 1

       By using the above steps, the phasor diagrams for various load power factor conditions can be drawn.

1.1 Lagging Power Factor Load

       The power factor of the load is cosΦ lagging so I_a lags V_ph by angle Φ. By using steps discussed above, phasor diagram can be drawn as shown in the Fig. 2.

Fig. 2  Phasor diagram for leading p.f. load

       To derive the relationship between E_ph and V_ph, the perpendicular are drawn on the current phasor from points A and B. These intersect current phasor at points D and E respectively.

.^..        (E_ph)² = (OD + DE)² + (BE - BC)²

.^..        (E_ph)² = (V_ph cosΦ + I_a R_a)² + (V_phsinΦ - I_a X_s)²

       It can be observed that the sign of the I_a X_s is negative as against its positive sign for lagging p.f. load. This is because X_s consists of X_ar i.e. armature reaction reactance. Armature reaction is demagnetising for lagging while magnetising for leading power factor loads. So sign of I_a X_s is opposite for lagging and leading p.f. conditions.

1.3 Unity Power Factor Load

       The power factor of the load is unity i.e. cosΦ = 1. So Φ = 0, which means V_ph is in phase with I_a. So phasor diagram can be drawn as shown in the Fig. 3.

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

       Consider ΔOBC, for which we can write,

         (OC)² = (OB)² + (BC)²

.^..      (E_ph)² = (OA + AB)² + (BC)²

.^..      (E_ph)² = (V_ph + I_a R_a)² + (I_a Xs)²

       As cosΦ = 1, so sinΦ = 0 hence does not appear in the equation.

Note : The phasor diagrams can be drawn by considering voltage V_ph as a reference phasor. But to derive the relationship, current phasor selected as a reference makes the derivation much more simplified. Hence current is selected as a reference phasor.

       It is clear from the phasor diagram that V_ph is less than E_ph for lagging and unity p.f. conditions due to demagnetising and cross magnetising effects of armature reaction. While V_ph is more than E_ph for leading p.f. condition due to the magnetising effect of armature reaction.

       Thus in general for any power factor condition,

(E_ph)² = ( V_ph cos + I_a R_a)² + (V_ph sin I_a X_s)²

+ sign for lagging p.f. loads

- sign for leading p.f. loads

and  V_ph = per phase rated terminal voltage

        I_a = per phase full load armature current