Analysis of Phasor Diagram

Analysis of Phasor Diagram

Consider a phasor diagram with normal excitation i.e. such a current through field winding which will produce flux that will adjust magnitude of E_bph same as V_ph.

Let δ be the load angle corresponding to the load on the motor. So from the exact opposing position of E_bph with respect to V_ph. Ebpb gets displaced by angle δ.

Vector difference of  E_bph and V_ph, gives the phasor which represents I_a Z_s, called E_Rph-

Now Z_s = R_a + j X_s |Ω| = |Z_S| ∠θΩ

where R_a = Resistance of stator per phase

X_s = Synchronous reactance of stator per phase

i.e θ ta ^-1 X_s / R_a

and |Z_S|  = R_a²  + X₅² Ω

This angle ' θ ' is called internal machine angle or an impedance angle.

The significance of ' θ ' is that it tells us that phasor I_aph lags behind E_Rph i.e. I_a Z_s by angle θ. Current always lags in case of inductive impedance with respect to voltage drop across that impedance. So phasor Iapb can be shown lagging with respect to ERpb by angle θ. Practically R_a is very small compared to X_a and hence θ tends to 90°.

Key Point The power factor at which motor is running, gets decided by the angle between V_ph and I_aph, shown. This angle is denoted as  ϕ  and called power factor angle.

ϕ = V_ph Λ I_aph

= Power factor angle

and cos ϕ = Power factor at which motor is working

The nature of this p.f. is lagging if I_aph lags V_ph by angle ϕ.Phasor diagram indicating all the details is shown in the Fig.4.9.1.