3. TORQUE IN A
SYNCHRONOUS
MOTOR
Torque causes an a object to acquire angular acceleration.
Torque is a measure of the force the can cause an object to rotate
about an axis.
Torque is a vector quantity. The direction of the torque vector depend
on the direction of the force on the axis.
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4. TORQUE IN A
SYNCHRONOUS MOTOR
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The mechanical power developed (Pm) by any
synchronous motor can be expressed as −
Pm=2πNSτg60Watts
Also, the mechanical power developed (Pm) is,
Pm=VIaCos(δ−φ),
Where, NS is the synchronous speed in RPM.
τgis the gross-torque in N-m.
5. TORQUE IN A
SYNCHRONOUS MOTOR
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Therefore, the gross torque of the synchronous
motor is given by,
τg=602π*PmNS = 9.55×PmNSN−m
And the shaft torque is given by,
τsh=9.55×PoNSN−m
6. Types of Torque in a Synchronous Motor
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Starting Torque
Running Torque
Pull-in Torque
Pull-out Torque
7. Torque angle of a synchronous motor.
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• A synchronous motor is meant to rotate at synchronous speed i.e, speed
of the rotating field.
• If the motor is loaded, the rotor lags behind the synchronous speed. Thus
the rotor goes out of sync with the rotating resultant field.
• Angle between rotor and the rotating resultant field is called Torque
angle.
• With the help of torque angle, the stability of the motor can be analyzed.
Similarly, when the motor is suddenly unloaded the rotor slips ahead of the
rotating field.
• At any instant, separation between rotor and rotating field is calculated
by an angle called Torque Angle.
8. Blondel diagram in synchronous
motor.
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The Blondel diagram of a synchronous motor is an extension of a simple phasor diagram of a
synchronous motor.
For a synchronous motor, the power input to the motor per phase is given by,
Pin = Vph Iph cos
The gross mechanical power developed per phase will be equal to the difference
between Pin per phase and the per phase copper losses of the winding.
Copper loss per phase = (Iaph)2 Ra
Pm = Vph Iph cos? – (Iaph)2 Ra
9. Presentation title 9
For mathematical convenience let Vph = V and
Iaph = I,
Pm = VI cos – I2 Ra
I2 Ra – VI cos + Pm = 0
fd
10. Thus if excitation is varied while the power is kept constant, then
working point B while move along the circle of constant power.
Let O’B = Radius of circle = r
OO’ = Distant d
Applying cosine rule to triangle OBO’,
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12. 12
One of the most important characteristics of a synchronous motor is that, by
changing the field excitation of the motor, its power factor can be made
both lagging and leading. The change in the power factor of the
synchronous motor with the change in excitation can be explained with the
help of its phasor diagram.
Pi= Vef/Xs Sinδ = 3VIaCosφ
Since V and XS are constant for a given synchronous motor, thus for constant
power output,
EfSinδ=Constant
IaCosφ=Constant
Effect of Changing Field Excitation on
Synchronous Motor at Constant Load
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•If the synchronous motor is under-excited, it has a lagging power
factor.
•If the synchronous motor is normally-excited, it has unity power
factor.
•If the synchronous motor is over-excited, it has a leading power
factor.
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Effect of Load Change on a Synchronous Motor
In induction motors and DC motors, when the mechanical load attached to the shaft of the motor is
increased, the speed of the motor decreases. The decrease in the speed reduces the back EMF so
that additional current is drawn from the source to carry the increased load at a reduced speed. But,
this action cannot take place in a synchronous motor, since it runs at a constant speed (i.e.,
synchronous speed) at all loads.
15. Presentation title 15
For a synchronous motor, the armature current per phase is
given by,
Ia=V−Ef/Zs=Er/Zs
When the load on a synchronous motor is increased, then
•The motor continues to run at synchronous speed
• The torque angle (δ) increases
•The magnitude of the excitation voltage (Ef) remains constant
•The armature current (Ia) drawn from the supply increases.
16. Presentation title 16
When the load on a synchronous motor is decreased,
then
•The motor continues to run at synchronous speed.
•The torque angle (Ia) decreases.
•The magnitude of the excitation voltage (Ef) remains constant.
•The armature current (Ia) drawn from the supply decreases.
