**Blondel's Two Reaction Theory (Theory of Salient Pole Machine) : part 2**We know that, the armature reaction flux Φ

_{AR}has two components, Φ_{d}along direct axis and Φ_{q}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.**.**Φ

^{.}._{d}= P

_{d}F

_{d}

where P

_{d}= permeance alomng the direct axis Permeance is the reciprocal of reluctance and indicates ease with which flux can travel along the path.

But F

_{d }= m.m.f. = K_{ar}I_{d}in phase with I_{d}_{} The m.m.f. is always proportional to current. While K

_{ar}is the armature reaction coefficient.**.**Φ

^{.}._{d}= P

_{d }K

_{ar }I

_{d}

Similarly Φ

_{q}= P_{q }K_{ar }I_{q} As the reluctance along direct axis is less than that along quadrature axis, the permeance P

_{d }along direct axis is more than that along quadrature axis, (P_{d }< P_{q }). Let E

_{d }and E_{q }be the induced e.m.f.s due to the fluxes Φ_{d }and Φ_{q}respectively. Now E_{d }lags Φ_{d}by 90^{o }while E_{q }lags Φ_{q}by 90^{o }. where K

_{e }= e.m.f. constant of armature winding The resultant e.m.f. is the phasor sum of E

_{f}, E_{d }and E_{q}. Substituting expressions for Φ

_{d }and Φ_{q } Now X

_{ard }= Equivalent reactance corresponding to the d-axis component of armature reaction = K

_{e }P_{d }K_{ar }_{} and X

_{arq }_{ }= Equivalent reactance corresponding to the q-axis component of armature reaction = K

_{e }P_{q }K_{ar } For a realistic alternator we know that the voltage equation is,

where V

_{t}= terminal voltage X

_{L}= leakage reactance Substituting in expression for

**Ē**_{R }, where X

_{d }_{ }= d-axis synchronous reactance = X_{L}+ X_{ard }.............(2) and X

_{q }_{ }= q-axis synchronous reactance = X_{L}+ X_{arq }_{ }.........(3)It can be seen from the above equation that the terminal voltage V

_{t}is nothing but the voltage left after deducing ohmic drop I_{a}R_{a}, the reactive drop I_{d}X_{d}in quadrature with I_{d}and the reactive drop I_{q}X_{q}in quadrature with I_{d}, from the total e.m.f. E_{f}.The phasor diagram corresponding to the equation (1) can be shown as in the Fig. 1. The current I

_{a}lags terminal voltage V_{t}by Φ. Then add I_{a}R_{a}in phase with I_{a}to V_{t}. The drop I_{d}X_{d}leads I_{d}by 90^{o }as in case purely reactive circuit current lags voltage by 90^{o }i.e. voltage leads current by 90^{o }. Similarly the drop I_{q}X_{q}leads X_{q}by 90^{o }. The total e.m.f. is E_{f}.**Read**: Part 1 , Part 3**Sponsored Articles**
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