The phasor diagram of a synchronous motor is shown below. From the phasor diagram,
Let
V = Supply voltage / phase
Ia = Armature current / phase
Ra = Armature resistance / phase
α = Load angle
φ = Power factor angle
Input Power to Motor :
Motor input power / phase = V Ia Cos φ
Total input power for 3-φ star-connected motor,
P
= √ 3 VL IL
Cos φ
= 3 Vph Iph
Cos φ
Where
VL and IL are line values
Vph and Iph are phase values
Power Developed by Motor :
The mechanical power developed / phase,
Pm = Back emf * Armature current * Cosine of the angle between Eb and Ia
= Eb Ia
Cos ( α - φ ) for lagging p.f
= Eb Ia
Cos ( α + φ ) for leading p.f
The copper loss in a synchronous motor takes place in the armature windings.
Therefore,
Armature copper loss / phase = Ia2 Ra
Total copper loss = 3 Ia2 Ra
By subtracting the copper loss from the power input, we obtain the mechanical power developed by a synchronous motor as
Pm = P -
Pcu
For three-phase,
Pm = √ 3
IL IL Cos φ – 3 Ia2 Ra
Power Output of the Motor :
To obtain the power output we subtract the iron, friction, and excitation losses from the power developed.
Therefore,
Net output power, Pout
= Pm - iron, friction, and excitation losses.
The above two stages can be shown diagrammatically called as Power Flow Diagram of a Synchronous Motor
The power developed in a synchronous motor as follows.
Motor Input Power, P
1. Stator ( Armature ) copper loss Pcu
2. Mechanical power developed, Pm
a. Iron, friction, and excitation losses
b. Output power, Pout
Net Power Developed by a Synchronous Motor :
The expression for power developed by the synchronous motor in terms of α, θ, V, Eb, and Zs are as follows :
Let
V = Supply voltage
Eb = Back emf / phase
α = Load angle
θ = Internal or Impedance angle = Tan-1 ( Xr / Zs
)
Ia = Armature current / phase = Er / Zs
Zs
= Ra + J Xs = Synchronous impedance
Mechanical power developed / phase,
The armature resistance is neglected
If Ra is neglected, then Zs ≈ Xs and θ = 90°. substituting these values in the above equation
