Condition For Maximum Power Developed In Synchronous Motor

The value of ? for which the mechanical power developed is maximum can be obtained as,

Note : Thus when R_a is negligible, ? = 90^o for maximum power developed. The corresponding torque is called pull out torque.

1.1 The Value of Maximum Power Developed

The value of maximum power developed can be obtained by substituting ? =? in the equation of P_m.

When R_a is negligible,     ? = 90^o  and cos (?) = 0 hence,
.^..               R_a = Z_s cos?   and X_s = Z_s sin?

Substituting   cos? = R_a/Z_s in equation (6b) we get,

Solving the above quadratic in E_b we get,

As E_b  is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step.

1.2 Condition for Excitation When Motor Develops (P_m ) R_max

Let us find excitation condition for maximum power developed. The excitation controls E_b. Hence the condition of excitation can be obtained as,

Assume load constant hence ? constant.

but    ? = ?  for P_m = (P_m)_max

         Substitute            cos? = R_a/Zs

This is the required condition of excitation.

Note : Note that this is not maximum value of but this is the value of foe which power developed is maximum.

The corresponding value of maximum power is,