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Armature Reaction in Alternator at Lagging Power Factor Load
Consider a 2-pole alternator as shown in Figure-1. Suppose that the alternator is loaded with an inductive load of zero power factor lagging. In this case, the phase current πΌπ , πΌπ and πΌπ΅ will be lagging with their respective phase voltages πΈπ , πΈπ and πΈπ΅ by 90°. The phasor diagram of the alternator is shown in Figure2.
Now, at time t = 0, the instantaneous values of currents and fluxes are given by,
$$\mathrm{π_{π } = 0;\:\:φ_{π } = 0}$$
$$\mathrm{π_{π} = πΌ_{π}\:sin(−120°) = −\frac{\sqrt{3}}{2}πΌ_{π};\:\:\:φ_{π} = −\frac{\sqrt{3}}{2}φ_{π}}$$
$$\mathrm{π_{π΅} = πΌ_{π}\:sin(120°) = \frac{\sqrt{3}}{2}πΌ_{π};\:\:\:φ_{π΅} = \frac{\sqrt{3}}{2}φ_{π}}$$
The space diagram of the magnetic fluxes is shown in Figure-3. By the parallelogram law, the resultant armature reaction flux is given by,
$$\mathrm{π_{π΄π } =\sqrt{φ^{2}_{π} + φ^{2}_{B} + 2 φ_{π}\:φ_{π΅}\:cos\:60°}}$$
$$\mathrm{\Rightarrow\:φ_{π΄π } =\sqrt{(\frac{\sqrt{3}}{2}φ_{π})^{2}+(\frac{\sqrt{3}}{2}φ_{π})^{2}+2 \times (\frac{\sqrt{3}}{2}φ_{π})\times (\frac{\sqrt{3}}{2}φ_{π}) \times (\frac{1}{2})}}$$
$$\mathrm{∴\:π_{π΄π } = 1.5\:φ_{π}}$$
Here, it can be seen that the direction of the armature reaction flux φπ΄π is opposite to the main field flux. Therefore, it will oppose and weaken the main field flux. This effect of the armature reaction is said to be demagnetising. Again it can also be shown that for successive positions of the rotor, the armature reaction flux φπ΄π remains constant in magnitude equal to 1.5 ππ and rotates at synchronous speed. Also, the EMF induced in each phase by the armature reaction flux lag the respective phase currents by 90°.
Therefore, when the alternator supplies a load at lagging power factor, then the effect of the armature reaction is partly demagnetising and partly cross-magnetising.
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