Synchronous Motor,electrical machine, Study notes of Electric Machines

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Synchronous Motor

Synchronous Motor-General

A synchronous motor is electrically identical with an alternator or AC

y

y

C

generator.A given alternator ( or synchronous machine) can be used as a motor,when driven electrically.Some characteristic features of a synchronous motor are as follows:1. It runs either at synchronous speed or not at all

i.e.

while running it

maintains a constant speed. The only way to change its speed is to varyth

l

f

(b

N

f/P

the supply frequency (because

N

S

f/P

2. It is not inherently self-starting. It has to be run up to synchronous(or

near

synchronous)

speed

by

some

means

before

it

can

be

(or

near

synchronous)

speed

by

some

means,

before

it

can

be

synchronized to the supply.3 It is capable of being operated under a wide range of power factors3. It is capable of being operated under a wide range of power factors,both lagging and leading. Hence, it can be used for power correctionpurposes, in addition to supplying torque to drive loads.

Now, consider the condition shown in

Fig. 36.

(a) where the stator

and rotor poles are attracting each other. Suppose that the rotor is notstationary, but it is rotating clockwise, with such a speedthat turns through one pole-pitch by the time the statorpitch by the time the statorpoles

interchange

their

positions, as shown in Fig.36 3(b)36.3(b). Here, again the stator and rotor poles attract each other.It means that if the rotor poles also shift their positions along with the

t t

l

th

th

ill

ti

l

i

idi

ti

l

stator poles, then they will continuously experience a unidirectionaltorque

i.e

. clockwise torque, as shown in Fig. 36.3.

Method of Starting

There are several methods to start the synchronous motor such as:

y

(a) Auxiliary drive (induction motor or dc motor),(b) Induction start (using damper winding) etc(b) Induction start (using damper winding), etc

General Starting Procedure:

The rotor (which is as yet

unexcited) is speeded up to synchronous or near synchronous speed byunexcited) is speeded up to synchronous or near synchronous speed bysome arrangement and then excited by the DC source.

The moment this (near) synchronously rotating rotor is excited, it

y

y

g

is magnetically locked into position with the stator

i.e.

the rotor poles are

engaged with the stator poles and both run synchronously in the samedirection.direction.

It is because of this inter-locking of stator and rotor poles that the

motor has either to run synchronously or not at all.

However,

it

is

important

to

understand

that

the

arrangement

between the stator and rotor poles is

not an absolutely rigid one

Procedure for Starting a Synchronous Motor

Using the Damper WindingUsing the Damper Winding

While starting a modern synchronous motor provided withdamper windings following procedure is adopteddamper windings, following procedure is adopted.1.

First, main field winding is short circuited.

Reduced

voltage

with

the

help

of

auto-transformers

is

applied across stator terminals. The motor starts up.

When it reaches a steady state speed (as judge by its sound),

y

p

j

g

y

a weak DC excitation is applied by removing the short-circuit on the main field winding. If excitation is sufficient,then the machine will be pulled into synchronismthen the machine will be pulled into synchronism.

Full supply voltage is applied across stator terminals bycutting out the auto-transformers.

Th

b

d

d

i

d

f

b

The motor may be operated at any desired power factor bychanging the DC excitation.

Motor on Load With Constant Excitation

In a synchronous machine, a back emf

E

b

(as like DC motor) is set up in the armature

a sy c

o ous

ac

e, a bac

e

b

(as

e

C

o o ) s se up

e a

a u e

(stator) by the rotor flux which opposes the applied voltage

V

.

This back emf depends on rotor excitation only (and not on speed as in DC motors).The net voltage in armature (stator) is the

vector difference

(not arithmetical, as in DC

motors) of

V

and

E

b .

Armature current is obtained by dividing this

vector

difference of voltages by

Armature current is obtained by dividing this

vector

difference of voltages by

armature impedance (not resistance as in DC machines). Fig. 36.

shows the condition when the motor (properly synchronized to the supply) is

running on

no

load

and has

no

losses

and is having fixed excitation which makes

running on

no

-load

and

has

no losses

and

is

having

fixed

excitation

which

makes

E

b

=V

.

It is seen that vector difference of

E

b

and

V

is zero

and so is the armature current.Motor intake is zero, as there is neither load nor

,

losses to be met by it.In other words, the motor just floats.

Equivalent Circuit of a Synchronous MotorFig.

36.

(a)

shows

the

equivalent

circuit model for one armature phaseof

a

cylindrical

rotor

synchronous

It is seen from

Fig. 36.9(b)

that the

phase applied voltage

V

is the vector sum

of reversed back emf

i.e.

E

b

and the

of

a

cylindrical

rotor

synchronous

motor.

b

impedance drop

I

a

Z

. s

In other words,

V

=

(

-

E

b

+

I

a

Z

). s

The angle

α

between the phasor for

V

and

E

b

is called the

load angle

or

power

angle

of the synchronous motor.

Power Flow within a Synchronous Motor

Let

R

a

= armature resistance/phase;

e

a

a

a u e es s a ce/p ase;

X

S

= synchronous reactance/phase

Then,

Z

S

=

R

a +jX

S

;

Hence, the power available at the shaftwould

be

less

than

the

developed

power by this amount.

S

b

R S

a

Z

E

V

E Z

I

=

=

b

Z

I

E

V

Obviously

Out of the input power/phase

VI

a

cos

φ

,

and

amount

I

a

2 R

a

is

wasted

in

armature,

the

rest

(

VI

a cos

φ

I

a

2 R

a

)

S

a

b

Z

I

E

V

−

=

Obviously

,

The angle

θ

(known as

internal angle

) by

which

I

a

lags

behind

E

R

is

given

by

θ

/

f

i^

li ibl

h

θ

90

a

a

a

appears as mechanical power in rotor;out of it, iron, friction and excitationlosses are meet and the rest is available

tan

θ

=

X

S /R

a

. If

R

a

is negligible, then

θ

=

o .

Motor input =

VI

a cos

φ

-per phase, here

V

is

applied voltage/phase.

at the shaft.If power input/phase of the motor is

P

then:

P=P

m

+ I

a 2

R

a

.

pp

g

p

Total input for a star-connected 3-phasemachine is,

P=

3

V

L

I

L

cos

φ

m

a

a

Or mechanical power in rotor

P

m

=P-

I

a

2 R

a

-per phase

The mechanical power developed, somewould go to meet iron and fraction andexcitation losses.

For

three

phases

P

m

=

3

V

L

I

L

cos

φ

-

3

I

a

2 R

a

.

Power Developed by a Synchronous Motor

The

armature

resistance

of

a

synchronous

motor

is

negligible

as

compared

to

its

The

armature

resistance

of

a

synchronous

motor

is

negligible

as

compared

to

its

synchronous reactance.Hence the equivalent circuit for the motor becomes as shown in

Fig. 36.

(a).

From the phasor diagram of

Fig. 36.10(b)

, it

is seen that

φ

=

=

cos

sin

S X a I b E

AB

sin

cos

Or,

V b E S X a

VI

= φ

Now,

VI

a cos

φ

= motor power input/phase

phase

per

sin

−

=

∴

α

S X

V b E

in P

phase

three

for

sin

3

−

=

∴

α

S X

V b E

in P

Since

Stator

Cu

losses

have

been

neglected

P

also

3

V

E

Since

Stator

Cu

losses

have

been

neglected

,^

P

in

also

represents

the

gross

mechanical

power

( P

m

)

developed by the motor.

α

sin

3

S X

V b E

in P

m P

=

=

∴

The gross torque developed by the motor is

T

P

/N

N

m

N

in rpm

The

gross torque developed by the motor is

T

g

P

m

/N

S

N

-m

N

S

in

rpm.

m

N

sin

(^65).

28

sin

3

(^55). 9

(^55). 9

− = × = = ∴

S X S N

V b E S X S N

V b E

S N

m P

g T

Example 38.

A 75 kW, 3-phase, Y-connected, 50 Hz, 440 V cylindrical rotor

synchronous motor operates at rated condition with 0.8 power factor leading. Themotor efficiency excluding field and stator losses, is 95% and

XS

=2.5 ohm. Calculate:

o o

e

c e cy e c ud

g

e d a d s a o

osses, s 95% a d

S

.5 o

. Ca cu a e:

(

i ) mechanical power developed, (

ii

) armature current, (

iii

) back emf, (

iv

) power angle,

and (

v

) maximum or pull-out torque of the motor.

S l

i^

N

120 50/

1500

25

S

olution

: N

S

= 120

×

50/

= 1500 rpm = 25 rps.

W

(^950) ,

78

(^95). 0 /

103

75

/

) (

=

×

=

=

=

out P

m P

in P

i (ii) Si

i^

t i

k

k

th t

(ii)

Since power input is known, we know that

A

129

(^8). 0

440

3

(^950) ,

78

W;

(^950) ,

78

(^8). 0

440

3

cos

3

= × × = = × × × = φ ∴

a I a I a I L V

(iii) A

li d

l

/ h

2 4

440

3

2 4

∠

0

h

i

i^

36 11

(iii)

Applied voltage/phase=

254

440

3

=

×

Let V= 254

∠

0

o as shown in

Fig. 36.11.

° = − = φ + = (^9).

36

) (^8). 0 ( 1

cos

and

Now,

S

jIX

b E

V

° − ∠ = ° ∠ × ° ∠ − ° ∠ = − =

30

516

90

(^5). 2

(^9).

36

129

0

254

or,

S

j IX

V

b E

S

j

b

°

−

∴

30

,

angle

Power

(iv)

(v) Maximum or Full-out torque occurs when

α

=

o .

W

(^8).

(^276) ,

157

90

sin

(^5). 2

516

254

3

sin

3

(max)

= ° × × = =

S X

V b E

m P

∴

Pull-out torque=9.

×

157,276.8/1500=1001.328 N-m

The value of

and back emf

E

b

can be found with the help of vector

diagrams for various power factors. (i) L

i

P

F

A

f

Fi

(i)

Lagging Power Factor

: As seen from Fig.

36.14(a)

φ − θ + φ − θ − = + =

2 )]

sin(

[

2 )]

cos(

[

2

2

2

R E

R E

V

BC

AB

AC

φ − θ

−

−

φ − θ + φ − θ − = =

)

sin(

1

tan

1

tan

2 )]

sin(

[

2 )]

cos(

[

S Z a I

BC

S Z a I S Z a I V

AC

b

E α

a R

s

X 1

tan

−

=

θ

b

R

=

E

V

E

φ − θ − = =

)

cos(

tan

tan

S Z a I

V

AB

)

cos(

2

2

2

−

=

R

VE

R E V b E )

sin

]

cos

[

)]

o

180

sin(

)

o

180

[cos(

) o 0

sin

o 0

(cos

b jE

b E

V

R

j b E j V R

b

R

−

=

− + − + + =

E E

R

b

α

sin

:

is

and

between

Angle

b E

V

R

E

α

α

φ

θ

cos sin

1

tan

b E

V

b E −

−

=

(ii) Leading Power Factor

: As seen from Fig. 36.14(b)

2

2

⎞⎟ ⎟ ⎠

⎛⎜ ⎜ ⎝

φ − θ − ° − φ − θ − ° − =

φ + θ − ° + φ + θ − ° + =

)}

(

180

cos{

)}

(

180

sin{

1

tan

2

)}]

(

180

sin{

[

2

)}]

(

180

cos{

[

S Z a I

V

S Z a I

S Z a I S Z a I V b

E α

Where,

α

=load angle

⎠

⎝

⎞

⎛^ ⎜

φ

θ

−

=

φ + θ + φ + θ − =

)

sin(

1

tan

2 )]

sin(

[

2 )]

cos(

[

S Z a I

S Z a I S Z a I V b

E α

⎠

⎜ ⎜ ⎝^

φ

θ

−

=

)

cos(

tan

S Z a I

V

S

)

cos(

2

2

2

−

=

R

VE

R E

V

b E

)]

o

180

sin(

)

o

180

[cos(

) o 0

sin

o 0

(cos

j b E j V R

b

R

− + − + + =

=

E

E

V

E

)

sin

]

cos

[

)]

(

)

[

(

)

(

b jE

b E

V

R

j

b

j

R

−

=

E

:

angle

factor

Power

a R

s

X 1

tan

g

−

=

θ

Example

A

25-hp

220V

60Hz

four

pole

Y-connected

synchronous motor is rotating with a light load. The angle between

y

g

g

g

the rotor and stator fields is 3

o

. The excitation is adjusted for a

generated armature voltage per phase of 110V. (a) what is theresultant armature voltage per phase? (b) what is the angle betweenresultant armature voltage per phase? (b) what is the angle betweenresultant voltage (

E

R

) and terminal voltage (

V

Solution

: Terminal voltage per phase,

V

3=127V

g

p

p

o

. Cos

=0.9986 and sin

We know that

)

i

]

[

jE

E

V

E

We know that,

)

sin

]

cos

[

b jE

b E

V

R

−

=

E

2 )]

sin

[

2 ]

cos

[

) (

Thus,

b E b E V R E a

−

=

V

R E

(^95).

17

2 ]

(^0523). 0

(^110) [

2 ]

(^9986). 0

110

(^127) [^

= × + × − =

i

o (^7).

18

(^9986). 0

110

127

(^0523). 0

110

1

tan

cos sin

1

tan

) (

=

×

−

× − = − − =

b E

V

b E

b

Effect of Increased Load with Constant Excitation

The effect of increased load on a synchronous motor will be studied under

diti

f

l

d

d

it ti

(i

i

th

ff

t^

f

t

conditions of normal, under and over-excitation (ignoring the effects of armaturereaction).With

normal-excitation,

E

b

=

V

,^

with

under-excitation,

E

b

V

. The value of excitation would be kept

constant

.

R

a

is considered negligible as compared to

X

S

so that phase angle between

E

R

and

I

a

i .e

.^

θ

=

o

.

i.e.

θ

Normal Excitation, (

E

b

=

V

)

Fig. 36.15(a) shows the condition when motoris running with light load so that (a) torqueangle

α

1

is small, (b) so

E

R

1

is small, (c) hence

I

a 1

is small and (d)

φ

1

is small so that cos

φ

1

is

large. Now, suppose that load on the motor is

increased

as shown in Fig. 36.15(b).

,^

pp

g

( )

For meeting this extra load, motor must develop more torque by drawing morearmature current.