Direct and Quadrature Axis Synchronous Reactance

Blondel's Two Reaction Theory (Theory of Salient Pole Machine) : part 2 
We know that, the armature reaction flux ΦAR has two components, Φd along direct axis and Φq along quadrature axis. These fluxes are proportional to the respective m.m.f. magnitudes and the permeance of the flux path oriented along the respective axes.
...                          Φd  = Pd Fd
            where        Pd = permeance alomng the direct axis
       Permeance is the reciprocal of reluctance and indicates ease with which flux can travel along the path.
       But                 Fd = m.m.f. = Kar Id in phase with  Id 
       The m.m.f. is always proportional to current. While Kar is the armature reaction coefficient.
...                          Φd  = Pd Kar Id 
       Similarly          Φq = Pq Kar Iq
       As the reluctance along direct axis is less than that along quadrature axis, the permeance  Pd along direct axis is more than that along quadrature axis, (Pd < Pq ).
        Let Ed and Eq be the induced e.m.f.s due to the fluxes Φ and Φq respectively. Now Ed lags Φd  by 90o while E lags Φq by 90o .

       where K = e.m.f. constant of armature winding
       The resultant e.m.f. is the phasor sum of Ef, E and Eq.
       Substituting expressions for Φ and Φ
       Now       Xard  = Equivalent reactance corresponding to the d-axis component of armature reaction
                               = K P Kar 
       and         Xarq    = Equivalent reactance corresponding to the q-axis component of armature reaction
                                =  K P Kar 
       For a realistic alternator we know that the voltage equation is,
        where       Vt = terminal voltage
                         XL  = leakage reactance
       Substituting in expression for ĒR ,

        where  X   = d-axis synchronous reactance = XL  + Xard                                   .............(2)
        and      X   = q-axis synchronous reactance =  XL  +  Xarq                                     .........(3)
It can be seen from the above equation that the terminal voltage Vt is nothing but the voltage left after deducing ohmic drop Ia Ra, the reactive drop Id Xd in quadrature with Id and the reactive drop Iq Xq  in quadrature with Id, from the total e.m.f. Ef.
The phasor diagram corresponding to the equation (1) can be shown as in the Fig. 1. The current Ia lags terminal voltage Vt by Φ. Then add Ia Ra in phase with Ia to Vt. The drop Id Xd leads Id by 90o as in case purely reactive circuit current lags voltage by 90o i.e. voltage leads current by 90o . Similarly the drop Iq Xq  leads Xq  by 90o . The total e.m.f. is Ef.

Read : Part 1  , Part 3

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