Consider the phasor diagram of a synchronous motor running on leading power factor cosΦ as shown in the Fig. 1.
Fig. 1 |
The line CD is drawn at an angle θ to AB.
The lines AC and DE are perpendicular to CD and AE.
and angle between AB = Ebph and Iaph is also ψ.
In triangle OBD,
BD = OB cosψ = Ia Zs cosψ
OD = OB sin ψ = Ia Zs sin
Now BD = CD - BC = AE - BC
Substituting in (2),
Ia Zs cosψ = Vph cos (θ-δ) - Eb cosθ
All values are per phase values
Substituting (3) in (1),
This is the expression for the mechanical power developed interms of the load angle δ and the internal machine angle θ, for constant voltage Vph and constant Eph i.e. excitation.
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