This presentation is about synchronous motor. it's basic principle and steady state operation has been described rigorously along with problem . additionally the power factor correction is also described . The starting methods of synchronous motor are also part of ppt.

1. Synchronous Motors
Zeeshan Ahmed Lodro
UG Student of EE 2k14
Department of
Electrical Engineering
Sukkur IBA University

2. 5.1 Basic Principles Of Motor
Operation
Two-pole synchronous motor
The basic principle of
synchronous motor operation is
that the rotor "chases" the
rotating stator magnetic field
around in a circle, never quite
catching up with it.

3. Synchronous Motor and Generator
Motor Generator
V  RAI A  jX sIA  EAV  RAI A  jX sIA  EA

4. 5.2 Steady-state Synchronous Motor
Operation
 This section explores the behavior of
synchronous motors under varying conditions
of load and field current.
 The following discussions will generally
ignore the armature resistance of the
motors for simplicity.

5. The Synchronous Motor Torque-Speed

6. The maximum or pullout torque
occurs when sigma= 90°
The Synchronous Motor Torque-Speed

7. The Effect of Load Changes on a
Synchronous Motor
X s
V : fixed (from electrical course)
EA : fixed (when I F fixed or using permanent magnets)
 A
V E
P  3V I cos   3  A
sin 
   (leading  lagging)
A A AP  (load increases)  sin   I  more heat (3I 2
R )

jXsIA P

8. Example 6.1
A 208 V,45 kVA, 0.8 PF leading, delta connected 60Hz synchronous motor has a
synchronous reactance of 2.5 and a negligible armature resistance. Its friction and
windage losses are 1.5 kW, and its core losses are 1 kW. Initially, the shaft is supplying 1
15 hp load, and the motor’s power factor is 0.8 leading.
(1) Sketch the phasor diagram of this motor, and find the values of IA, IL and EA.
(2) Assume that the shaft load is now increased to 30 hp. Sketch the behavior of phasor
diagram in response to this change.
(3) Find IA, IL and EA after the load change. What is the new motor power factor?

9. Solution [1]

10. Solution [2]

11. Solution [3]
B] As the power on the shaft is increased to 30 hp, the
shaft slows momentarily, and
the internal generated voltage EA swings out to a
larger angle a while maintaining a constant
magnitude.

12. Solution [4]
(c) After the load changes, the electric input power of the machine
becomes
= (30 hp)(0.746 kW/hp) + 1.5 kW + 1.0 kW + 0 kW
= 24.88 kW

13. Solution [5]

14. The Effect of Field Current Changes on a
Synchronous Motor
V  E A  RAI A  jX sI A
 E A  jX sI A


15. The Synchronous Motor and
Power-Factor Correction

16. 5.3 Starting Synchronous Motors
Three basic approaches can be used to safely start a
synchronous motor:
 Motor Starting by Reducing Electrical Frequency
 Motor Starting with an External Prime Mover
 Motor Starting by Using Amortisseur Windings