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 = E

_{bph }and I_{aph }is also ψ. The mechanical per phase power developed is given by,

In triangle OBD,

BD = OB cosψ = I

_{a }Zs cosψ OD = OB sin ψ = Ia Zs sin

Now BD = CD - BC = AE - BC

Substituting in (2),

I

_{a }Zs cosψ = V_{ph }cos (θ-δ) - E_{b }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 V

_{ph }and constant E_{ph }i.e. excitation.

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