electrical power and machines- sync motor

8/9/2019 Electrical Power and Machines- sync motor

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Chapter 17: Synchronous

MotorsThree-phase, unity power factor synchronous

motor rated 3000 hp (2200 kW), 327 r/min,

4000 , !0 "# dri$in% a compressor used in a

pumpin% station on the Trans &anada pipe'inerush'ess e*citation is pro$ided +y a 2 kW,

20 a'ternator/rectifier, which is mounted on

the shaft +etween the +earin% pedesta' and the

main rotor (&ourtesy of .enera' 'ectric)


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&onstruction

ynchronous motors are identica' in

construction to sa'ient-po'e ac

%enerators1 tator is composed of a s'otted

ma%netic core, which carries a 3-

phase 'ap windin%1 otor has a set of sa'ient po'es1 & current e*citer otor of a 0 "# to ! 2/3 "# freuency

con$erter used to power an e'ectric rai'way

The 4-po'e rotor at the 'eft is associated with

a sin%'e-phase a'ternator rated 7000 k5, !

2/3 "#, 6 89 The rotor on the ri%ht is for a

!:00 k5, 0 "#, 3-phase, :09 6

synchronous motor which dri$es the sin%'e-

phase a'ternator oth rotors are euipped

with suirre'-ca%e windin%s Today, these

machines are rep'aced +y so'id-state

freuency con$erters (ee ection 2:!)

(&ourtesy of 5)


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- dc contro' source

2- stationary e*citer po'es

3 - a'ternator (3-phase e*citer'

4 - 3-phase &onnection

- +rid%e rectifier ! - dc 'ine

7 - rotor of synchronous motor

8 - stator of synchronous motor

: - 3-phase input to stator

Diagram showing the main components of

a brushless exciter for a synchronous

motor. It is similar to that of a synchronous

generator.


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Starting a SynchronousMotor

1 A synchronous motor can not start by itself the motor is equipped with a squirrel cage 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

1 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 strongmagnetic attraction is set up between them

o the mutual attraction locks the rotor and stator poles together

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


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Motor under Load

1 At no-load conditions, the rotor poles

are directly opposite the stator poles

and their axes coincide

1 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; themotor will stall and come to a halt

the pull-out torque is a function of the

dc excitation current and the ac stator

current


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Motor under Load


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Motor under Load

Example

1 a 5 hp, !" rpm synchronous motor

connected to a #$% &, #-phase line

generates an excitation voltage, ' of (!$

& line-to-neutral when the dc exciting

current is "5 A

o the synchronous reactance is "" ohms

o the torque angle between ' and ' is

#)

1 find

o the value of EX

o

the ac line currento the power factor of the motor

o the developed horsepower

o the developed torque


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Motor under Load

Example





current is "5 A



#)

1 find

o the value of EX

o



o the developed torqueThus, phasor * has a $a'ue of !8 and it

'eads phasor +y 0;


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Motor under Load

Example





current is "5 A



#)

1 find

o the value of EX

o



o the developed torque


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Motor under Load

Example





current is "5 A



#)

1 find

a* the value of EX

b* the ac line currentc* the power factor of the motor

d* the developed horsepower

e* the developed torque


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Maximum Torque

1 he power equation shows that the

mechanical power increases with

the torque angle

its maximum value is reached when d

is $)

the poles of the rotor are then midway

between the north and south poles of

the stator


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Power and Torque

Example

(5 k+, . &, (" rpm, . /0

motor has a synchronous

reactance of *% Ω per phase*

he excitation voltage is fixed at# & per phase* 1etermine the

following2

a* the power versus the torque

angle curve

b* the torque versus the torqueangle curve

c* the pull out torque of the

motor


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Power and Torque

Example

(5 k+, . &, (" rpm, . /0

motor has a synchronous

reactance of *% Ω per phase*

he excitation voltage is fixed at# & per phase* 1etermine the

following2

a* the power versus the torque

angle curve

b* the torque versus the torqueangle curve

c* the pull out torque of the

motor

The actua' pu''-out torue is 3 times as %reat

(2400 =m) +ecause this is a 3-phase machine

imi'ar'y, the power and torue $a'ues %i$en in

the a+o$e %raph must a'so +e mu'tip'ied +y 3

&onseuent'y, this 0 kW motor can de$e'op a

ma*imum output of 300 kW, or a+out 400 hp


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Excitation and eacti!ePower

13onsider a wye-connected synchronous motorconnected to a power system with fixed line

voltage V 4

o the line current I produces a mmf in the stator

o the dc field current produces a dc mmf in the

rotor

o the total flux is created by the combinedactions of the two mmf’ s

1 he total flux Φ induces the voltage E a in the stator

neglecting the very small voltage drop IRa , E a

= V 4

because V 4 is fixed, the flux Φ is also fixed, asin a transformer

the constant flux Φ may be produced either by

the stator or the rotor or by both

The total flux Φ is due to the mmf

produced by the rotor (Ur) plus themmf produced by the stator (Ua).

For a given !" the flux Φ is

essentially fixed.


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E"ects o# Excitation


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%&Cur!es1 3onsider a synchronous motor operating at rated mechanical load

examine the behavior as the excitation is varied

o mechanical power remains constant

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

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

o 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 E is generated for a fix load

o the family of active power curves are shaped as the letterV


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%&Cur!es


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Example

1 # k+, " rpm, .. &

synchronous motor operates at

full-load at a %: leading power

factor* ynchronous reactance is

((Ω* 3alculate the following

a* the apparent power of the

motor

b* the ac line current

c* the value and phase angle ofthe induced voltage, E

d* draw the phasor diagram

e* determine the torque angle, δ


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E"ects o#Excitation

Example

1 # k+, " rpm, .. &






motor






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Example

1 # k+, " rpm, .. &






motor





e The torue an%'e is 2!;


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Stopping the SynchronousMotor

1 ynchronous 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 #-phase armaturewindings 7stator windings8, or

"* maintain full dc excitation on rotor and connect the armature 7stator

windings8 to a bank of external resistors, or

#* apply mechanical braking*

+ith electrical braking, the motor slows because the stored energy is

dissipated into the resistive elements of the circuit

9echanical braking is usually applied only after the motor has reached half

speed or less


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Stopping the Synchronous Motor

Example

1 a (5 k+, . &, . rpm motor is

stopped by using the short-circuit method

o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase

o moment of inertia < "!5 kg m"

calculate

a* the power dissipated in the armature at

. rpm

b* the power dissipated in the armature at

(5 rpm

c* the kinetic energy at . rpm

d* the kinetic energy at (5 rpm

e* the time required for the speed to fall

from . rpm to (5 rpm


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Example



o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase


calculate


. rpm


(5 rpm






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Example



o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase


calculate


. rpm


(5 rpm






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Example



o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase


calculate


. rpm


(5 rpm






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Stopping theSynchronous Motor

Example



o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase


calculate


. rpm


(5 rpm






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Stopping theSynchronous Motor

Example



o E = 2400, X = 16 Ω and RA = 0.2

Ω, per phase


calculate


. rpm


(5 rpm





This ener%y is 'ost as heat in the armature

resistance The time for the speed to drop from !00r/min to 0 r/min is %i$en +y


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Machine Comparison

1 6nduction machines have excellent properties when speeds are above . rpm

simple construction and maintenance

at lower speeds induction machines become heavy and costly with

relatively low power factors and efficiencies

1 ynchronous motors are particularly attractive for lowspeed

drives

power factor can always be ad=usted to (* 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


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Machine Comparison

1 a squirrel-cage induction motor and a synchronous motor, both

rated at hp, (% r>min, .*$ k&, . /0*

comparison of

the efficiency

comparison o#the starting

torque


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Synchronous Condenser

1 A synchronous condenser 7synchronous capacitor8 is a

synchronous motor running at no load

only purpose is to absorb or deliver reactive power in order to stabili0e

the system voltage

the machine acts as an enormous #-phase capacitor or inductor the reactive power is varied by changing the dc field excitation


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Example

A synchronous condenser is

rated at (. 9&ar, (. k&, and

(" rpm, and is connected to

(. k& line* he machine has a

synchronous reactance of *% Ω

per phase* 3alculate the value of

E so that the machine

a* absorbs (. 9var

b* delivers (" 9var


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Example








a* absorbs (. 9var

b* delivers (" 9var


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SynchronousCondenser

Example








a* absorbs (. 9var

b* delivers (" 9var


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SynchronousCondenser

ExampleA synchronous condenser is




synchronous reactance of *% Ω per phase* 3alculate the value of


a* absorbs (. 9var

b* delivers (" 9var

The e*citation $o'ta%e (4 800 ) is

now considera+'y %reater than the 'ine

$o'ta%e (:20 )


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electrical power and machines- sync motor

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