Synchronous Motor Power Equation & Efficiency

In order to drive the Synchronous Motor Power Equation & Efficiency, we have to assume some conditions and neglect some losses. Assume that synchronous motor is driving a constant torque load. The active power converted by the machine is constant, since the power, the voltage, and the motor speed are constant.

Synchronous Motor Power Equation

Synchronous Motor Power Equation circuit diagram

Except for very small machines, the armature resistance of a synchronous motor is relatively insignificant compared to its synchronous reactance, so that Equation to be approximated to

VT =  E+ IaZs

The equivalent-circuit and phasor diagram corresponding to this relation is shown in Fig given below.

Synchronous Motor Power Equation phasor diagram

These are normally used for analyzing the behavior of synchronous motor, due to changes in load and/or changes in field excitation. From this phasor diagram, we have,

I_aX_s\cos { \theta_i =-E_f\sin { \delta } }

Multiplying through by VT and rearranging terms we have,

V_TI_a\cos { \phi_i =\frac { -V_TE_f }{ X_s } } \sin { \delta }

Since the left side of the above equation is an expression for active power input and as the winding resistance is assumed to be negligible this power input will also represent the electromagnetic power developed, per phase, by the synchronous motor.

P_{in,ph}=V_TI_a\cos { \phi_i }

P_{in,ph}=\frac { -V_TE_f }{ X_s } \sin { \delta }

Thus, for a three-phase synchronous motor,

P_{in}=3\ast V_TI_a\cos { \phi_i }

P_{in}=3\ast \frac { -V_TE _f}{ X _s} \sin { \delta }

This equation called the synchronous machine power equation expresses the electromagnetic power developed per phase by a cylindrical-rotor motor, in terms of its excitation voltage and power angle. Assuming a constant source voltage and constant supply frequency.

P\quad \propto \quad I_a\cos { \theta }

P\quad \propto \quad E_f\sin { \delta }

Synchronous Motor Losses and Efficiency

The flow of power through a synchronous motor, from stator to rotor, and then to shaft
output. As indicated in the power-flow diagram, the total power loss for the motor is given by

Ploss = Pscl + Pcore + Pfcl + Pf,w + Pstray W

where:
Pscl = stator-copper loss
Pfcl = fie1d-copper.loss
Pcore = core loss Pf,w= friction and windage loss
Pstray = stray load loss

Except for the transient conditions that occur when the field current is increased or decreased the total energy supplied to the field coils is constant and all of it is consumed as  I2R losses in the field winding. Just as in the case of the synchronous generator, the overall efficiency of a synchronous motor is given by

\eta =\frac { P_{shaft} }{ P_{in}+P_{field} } =\frac { P_{shaft} }{ P_{shaft}+P_{loss} }

The nameplates of synchronous motors and manufacturers’ specification sheets customarily provide the overall efficiency for rated load and few load conditions only.

power flow diagram of synchronous motor

Hence, only the total losses at these loads can be determined. The separation of losses into the components needs a very involved test procedure in the laboratory. However, a closer approximation of the mechanical power developed can be calculated by subtracting the copper losses of the armature and field winding if these losses can be calculated. The shaft power can then be calculated subtracting the mechanical losses from the mechanical power developed.

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