chapter 17: synchronous motors - oakland universityfrick/ee4220-em_dynamics/lecture2… · chapter...

3/28/00 Electromechanical Dynamics 1

Chapter 17: Synchronous Motors


Starting a Synchronous Motor

• A synchronous motor can not start by itself– the motor is equipped with a squirrel case winding so as to

start as an induction motor

– during starting, the dc field winding is short circuited

– when the motor has accelerated close to synchronous speed, the dc excitation is then applied to produce the field flux

• Pull-in torque– if the poles on the rotor at the moment the exciting current is

applied happen to be facing poles of opposite polarity on the stator, a strong magnetic attraction is set up between them

• the mutual attraction locks the rotor and stator poles together

• the rotor is literally yanked into step with the revolving field


Motor under Load

• At no-load conditions, the rotor poles are directly opposite the stator poles and their axes coincide

• As mechanical load is applied, the rotor poles fall slightly behind the stator poles, but continues to turn at synchronous speed– greater torque is developed

with increase separation angle

– there is a limit when the mechanical load exceeds the pull-out torque; the motor will stall and come to a halt

– the pull-out torque is a function of the dc excitation current and the ac stator current


Motor under Load

δ∠=

°∠=

−−=

=−=

00

0

0

0

EE

EE

X

EEjI

EIjX

EEE

S

XS

X


Motor under Load

• Example– a 500 hp, 720 rpm synchronous motor connected to a 3980 V,

3-phase line generates an excitation voltage, E0 of 1790 V line-to-neutral when the dc exciting current is 25 A

• the synchronous reactance is 22 ohms

• the torque angle between E0 and E is 30°

– find• the value of EX

• the ac line current

• the power factor of the motor

• the developed horsepower

• the developed torque


Power and Torque

• When a synchronous machine operates as a motor under load, the converted power is given by the same equation used for the synchronous generator

• As far as torque is concerned, it is directly proportional the the mechanical power because of the fixed rotor speed

δsin0

SD X

EEP =

S

DD n

PT

55.9=


Maximum Torque

• The power equation shows that the mechanical power increases with the torque angle– its maximum value is reached when δ is 90°

– the poles of the rotor are then midway between the north and south poles of the stator

SX

EEP 0

max =


Power and Torque

• Example– 150 kW, 460 V, 1200 rpm, 60 Hz motor has a synchronous

reactance of 0.8 Ω per phase

– the excitation voltage is fixed at 300 V per phase

– determine the following:• the power versus the torque angle curve

• the torque versus the torque angle curve

• the pull out torque of the motor


Excitation and Reactive Power

• Consider a wye-connected synchronous motor connected to a power system with fixed line voltage VL

– the line current I produces a mmf in the stator

– the dc field current produces a dc mmf in the rotor

– the total flux Φ is created by the combined actions of the two mmf’s

• The total flux Φ induces the voltage Ea in the stator– neglecting the very small voltage drop IRa, Ea = VL

– because VL is fixed, the flux Φ is also fixed, as in a transformer

– the constant flux Φ may be produced either by the stator or the rotor or by both


Excitation and Reactive Power

• If the rotor exciting current Ix is zero– all the flux has to be produced by the stator

– the stator circuit absorbs considerable reactive power

• If the rotor exciting current is increased– the rotor mmf helps produce part of the flux

– less reactive power is drawn from the ac power system

• Eventually by raising the rotor exciting current gradually– the rotor produces all of the required flux

– the stator circuit draws no reactive power (unity power factor)

• If the exciting current exceeds this critical level– the stator delivers reactive power to the ac power system


Effects of Excitation

VT E

I

j XS

VT

E

I

j XS Iφδ

constantcos

constantsin

sinsin

0

00

==

==

==

T

TS

S

T

TS

V

PI

V

PXE

X

EVP

V

PXE

φ

φ

φφ

φ

δ

δδ

φφ

φδφδ

φδ

φ

φ

sin

cos

cossin0:

sincos:

0

0

0

0

IVQ

IVP

XIE

XIEV

XIjEV

T

T

S

ST

ST

=

=−=ℑ

+=ℜ∠+∠=°∠



φm

VT

E0 Im j XS Im δ

VT

E0

Im

j XS Im

φm

δ

VT

E0

Im

j XS Im δ

Unity Power Factor

Lagging Power Factor Leading Power Factor

Constant Power Locus

Constant Power Locus


Power Factor Rating

• Most synchronous machines are designed to operate at unity power factor– may be operated at full-load with a 0.8 leading power factor

– this is equivalent to a 0.6 leading reactive power factor

– the motor can deliver a reactive power equal to 75% of the rated mechanical power


V-Curves

• Consider a synchronous motor operating at rated mechanical load– examine the behavior as the excitation is varied

• mechanical power remains constant

• at unity power factor the motor current is at a minimum

• at unity power factor the total power equals the active power

• as excitation increases or decreases

– the motor current increases

– the total power increases

– by varying the excitation, a plot of total power, S, with respect to the excitation voltage E0 is generated for a fix load

• the family of active power curves are shaped as the letter V


V-Curves



• Example– 3000 kW, 200 rpm, 6600 V synchronous motor operates at

full-load at a 80% leading power factor• synchronous reactance is 11 Ω

– calculate the following• the apparent power of the motor

• the ac line current

• the value and phase angle of the induced voltage, E

• draw the phasor diagram

• determine the torque angle, δ


Stopping the Synchronous Motor

• Synchronous motors with their loads have large inertia– may take several hours to stop after power has been

disconnected from the power line

– to stop faster, electrical or mechanical braking can be applied• maintain full dc excitation on rotor and short the 3-phase

armature windings (stator windings), or

• maintain full dc excitation on rotor and connect the armature (stator windings) to a bank of external resistors, or

• apply mechanical braking

– with electrical braking, the motor slows because the stored energy is dissipated into the resistive elements of the circuit

– mechanical braking is usually applied only after the motor has reached half speed or less


Stopping the Synchronous Motor

• Example– a 1500 kW, 4600 V, 600 rpm motor is stopped by using the

short-circuit method• E0 = 2400, XS = 16 Ω and RA = 0.2Ω, per phase

• moment of inertia = 275 kg m2

– calculate• the power dissipated in the armature at 600 rpm

• the power dissipated in the armature at 150 rpm

• the kinetic energy at 600 rpm

• the kinetic energy at 150 rpm

• the time required for the speed to fall from 600 rpm to 150 rpm


Machine Comparison

• Induction machines have excellent properties– when speeds are above 600 rpm

– simple construction and maintenance

– at lower speeds induction machines become heavy and costly with relatively low power factors and efficiencies

• Synchronous motors are particularly attractive for low-speed drives– power factor can always be adjusted to 1.0 with high

efficiencies and reduced weight and costs

– can improve the power factor of a plant while carrying its rated load

– can be designed to deliver a higher starting torque


Machine Comparison

a squirrel-cage induction motor and a synchronous motor, both rated at 4000 hp, 1800 r/min, 6.9 kV, 60 Hz.

comparison of the efficiency

comparison of the starting torque


Synchronous Condenser

• A synchronous condenser (synchronous capacitor) is a synchronous motor running at no load– only purpose is to absorb or deliver reactive power in order to

stabilize the system voltage

– the machine acts as an enormous 3-phase capacitor or inductor

– the reactive power is varied by changing the dc field excitation


Synchronous Condenser

• Example– a synchronous condenser is rated at 160 MVar, 16 kV, and

1200 rpm, and is connected to 16 kV line

– the machine has a synchronous reactance of 0.8 Ω per phase

– calculate the value of E0 so that the machine • absorbs 160 Mvar

• delivers 120 Mvar


Synchronous Motors

• Homework– 17-14, 17-15, 17-19, and 17-20

chapter 17: synchronous motors - oakland universityfrick/ee4220-em_dynamics/lecture2… · chapter...

Documents