Synchronous Machines-Part 2

3-Ph Synchronous Machine

1. Tdo’ – D axis Open Circuit (arm) transient Time Constant

This time constant indicates the decay of field current with arm open circuited

Time constant=Inductance/ resistance

Xmd

Arm

term

inal

s

OPE

N

xf

xa

For Inductance see from this side

OR

Time constants from Equivalent Circuits

1. Tdo’ – D axis Open Circuit (arm) transient Time Constant

This time constant indicates the decay of field current with arm open circuited

Time constant=Inductance/ resistance

Xmd

Arm

term

inal

s

OPE

N

xa

Time constants from Equivalent Circuits

mdffo XxX '

3-Ph Synchronous Machine

fXx

L mdff 2

f

fdo r

LT '

fXx

rT mdf

fdo 2

1'

xf

2. Td’ – D axis Short Circuit transient Time Constant

This time constant indicates the decay of field current with arm short circuited

Xmd

Arm

term

inal

s

Sho

rted

xf

xa

amd

amdff xX

xXxX

'

xf

amd

amdff xX

xXxf

L21

amd

amdf

fd xX

xXxfr

T211'


amd

amdf

fd xX

xXxfr

T211'

amd

amdafmdf

f xXxXxxXx

fr 211

)(12

11mdfamdf

df

XxxXxXfr

)()(

12

11mdfa

mdf

mdf

df

XxxXx

XxXfr

'1d

d

XX

)( mdf Xx frf 2

11

Xmd

xa xf

Xd

Xmd

xa xf

Xd’

OPE

N

Xmd

xa xf

Tdo’

d

ddod X

XTT'

''

3. Tdo” – D axis Open Circuit Subtransient Time

Constant

This time constant indicates the decay of d axis damper bar current with arm open circuited

Xmd

Arm

term

inal

s

OPE

N

xf

xa

mdf

mdfkdkdo Xx

XxxX

'

xkd


mdf

mdfkd

kddo Xx

Xxx

frT

211"

Are so rahe ho kya? udhar dekho

AAAAAAAA?Sir, This is the answer

Very Easy Subject

4. Td” – D axis Short Circuit Subtransient Time Constant

This time constant indicates the decay of d axis damper bar current with arm short circuited

Xmd

Arm

term

inal

s

OPE

N

xf

xa

fmda

kdkd

xXx

xX 1111'

xkd


Arm

term

inal

s

Sho

rted

fmda

kdkd

d

xXx

xfr

T 1111

211"

'

"""

d

ddod X

XTTthatshownbecanIt

Sir, very difficult, three branches are in parallel

4. Tqo” – Q axis Open Circuit Subtransient Time

Constant

Xmq

Arm

term

inal

s

OPE

N

xa

mqkqkqo XxX "

xkq


mqkqkq

qo Xxfr

T 211"

This time constant indicates the decay of q axis damper bar current with arm open circuited

Soma write down very easy

5. Tq” – Q axis Short Circuit Subtransient Time

Constant

Xmq

Arm

term

inal

s

OPE

N

xa

mqa

mqakqkq Xx

XxxX

"

xkq


This time constant indicates the decay of q axis damper bar current with arm short circuited

Arm

term

inal

s

Sho

rted

mqa

mqakq

kqq Xx

Xxx

frT

211"

Don’t laugh

K & KWrite

down

mqa

mqakq

kqq Xx

Xxx

frT

211"

mqa

mqamqkqakq

kq XxXxXxxx

fr 211

"'

1q

q

XXfrkq 2

11

Xmq

xa xkq

Xq’

Xmq

xa xkqXq

”

OPE

N

Xmq

xa xkq

Tqo”

'

"""

q

qqoq X

XTT

)(12

11' kqmqamqkq

qkq

xXxXxXfr

)()(

12

11' kqmqa

kqmq

mqkq

qkq

xXxxX

XxXfr

)( kqmq xX

6. Ta – Armature Time Constant

2211

""qd

aa

XXfr

T

This time constant indicates the decay of DC component

or Second harmonic component of arm short circuit current.The flux produced by DC component is stationary

Since the field is rotating, at one instant DC flux comes along d-axis

and at other instant that comes along q-axis The reluctance , inductance and reactance

offered are that of Xd” and Xq”

If resistances and reactances are given in puThen 2πf =1 in pu in all above time constants equations.

Synchronous machine connected to Infinite Bus

Infin

ite B

us

Vt,

f

Ef Iara

Xs Ef Ia

XsIf ra =0,

Vt

Iara

Vt

θδ

Ia

Ef

jIaXs

Single Line Diagram

Phasor Diagram

IaZs Synchronous generator supplies electrical power at constant volatge Vt, since it is connected to Infinite Bus

Ef=MdωIf, can be varied by varying excitationPe=VtIacosθ, (elect power) can be controlled by varying mechanical power of PM

For generatorMechanical power is converted into electrical power

Mechanical power = Equivalent of Mech. Power converted to Elect. Power Pm

Power being converted from Mechanical to Electrical form Pm

a) Cylindrical rotor generator =3VtIacosθ =3Vt • Ia (dot product) for single phase

=VtIacosθ +Ia2ra =VtIacosθ +Ia

2Zscosθ

=Vt • Ia +Ia2

• Zs =(Vt +Ia Zs) • Ia

=Ef • Ia

Pe

Pm

As Ef=Vt +Ia Zs,s

tfa Z

VEI

=Vt • Ia

+ Constant loss (iron loss + f & w loss)

= Electrical power Pe + Arm copper loss

S A M E

z 0

z

For Pe, take Vt as a referenceSince Pe=Vt • Ia

s

tfa Z

VEI

a

sz r

Xanglewhere 1tan

Iara

Vt

θδ

Ia

Ef

jIaXs

IaZs

s

zt

s

zfa Z

VZ

EI

s

ztt

s

zftate Z

VVZ

EVIVP

zs

tz

s

fte Z

VZEV

P cos)cos(2

0

z


s

tfa Z

VEI

a

sz r

Xanglewhere 1tan

s

zt

s

zfa Z

VZ

EI

s

ztt

s

zftate Z

VVZ

EVIVP

zs

tz

s

fte Z

VZEV

P cos)cos(2

If resistance is negligible,

ssz XZandthen 090

sins

fte X

EVP

For maximum Pe

0 z)(8580 approxtoz

For maximum Pe, power angle = impedance angle

Electromagnetic Power Pem

emP

Pe

90 (ra=0)80 to 85 (ra≠0)

0

z

For Pm, take Ef as a reference

Now Pm=Ef • Ia

s

tfa Z

VEI

Iara

Vt

θδ

Ia

Ef

jIaXs

IaZs

)cos(cos2

zs

tfz

s

fm Z

VEZE

P

sins

tfe X

VEP

Pe

90 (ra=0)80 to 85 (ra≠0)

0

z

For P, take Ef as a reference

Now Pm=Ef • Ia

s

tfa Z

VEI

)cos(cos2

zs

tfz

s

fm Z

VEZE

P

For maximum Pm

180 z

z 180

95


ss

z

XZandthen

090

sins

tfm X

VEP

Thus if resistance is neglected,

Pm=Pe=Pem Electromagnetic Power

Otherwise Pm= Pe + Ia2ra

aa ritoeqivalent 2,10

emP

b) Salient Pole generator

Ef

IaraVt

Ia

jIdXd

jIqXq

θδ

jIa X

qEf

’

Consider similaritySo, Xs=Xq,

Zs=Zq

Zq=ra+jXq

zs

tz

s

fte Z

VZEV

P cos)cos(2

qq

tq

q

fte Z

VZEV

P cos)cos(2'

)cos(cos2

zs

tfz

s

fm Z

VEZE

P )cos(cos'2'

qq

tfq

q

fm Z

VEZ

EP

θz= θq

a

qq r

X1tan

Cyl. Rotor Salient Pole Rotor

Iara

Vt

θδ

Ia

EfjIa X

s

But if resistance is neglected, Pm=Pe ≠ Pem for salient pole rotor Pe=VtIacosθ

Active power =Voltage component • In phase current component Vt

Ia

θ

Iacosθ Ef

Vt

Ia

jIdXd

jIqXq

θδ

Pe=VtsinδxId +VtcosδxIq

Id

Iq

Vtsinδ Vtcosδ

Now obtain Id and Iq

costfdd VEXI

d

tfd X

VEI

cos

sintqq VXI

q

tq X

VI sin

d

tfd X

VEI

cos

q

tq X

VI sin

q

tt

d

tft X

VVXVE

V

sincoscos

sin

qt

d

t

d

ft

XV

XV

XEV sincoscossinsin 2

2

2sin112

sin2

dq

t

d

tfe XX

VX

VEP

Pe=VtsinδxId +VtcosδxIq

Electromagnetic Power Pem+ Reluctance PowerElectrical Power=(Power due to saliency)

dq XX11

P

Sin

9080 to 85 2Sin

Power-AngleCharacteristics ofSalient Pole Machine

mP relem PP

Present, Ef=0

02sin112

sin2

dq

t

d

tf

XXV

XVE

dd

ddP

Determine the δ, for maximum power

P

Sin

9080 to 85 2Sin


02cos11cos 2

dqt

d

tf

XXV

XVE

1cos22cos, 2 Now

thenanda

acbbObtain ,2

4cos,2

δ is less than 900.

Synchronous MotorElectrical power is converted into mechanical power

Electrical power Pe = Equivalent of Elect. Power converted to Mech. Power Pm

Power being converted from Electrical to Mechanical form Pm

a) Cylindrical rotor Motor =3VtIacosθ =3Vt • Ia (dot product) for single phase

=VtIacosθ - Ia2ra =VtIacosθ - Ia

2Zscosθ

=Vt • Ia - Ia2

• Zs =(Vt - Ia Zs) • Ia

=Ef • Ia

Pe

Pm

As Ef=Vt - Ia Zs,s

fta Z

EVI

=Vt • Ia

+ Arm copper loss

= Output power Po or Shaft power Psh

S A M E

+ Constant loss (iron loss + f & w loss)

z0

z


a

sz r

Xanglewhere 1tan

Iara

Ef

θδ

Ia

Vt

jIaXs

IaZs

s

zf

s

zta Z

EZ

VI

s

zft

s

zttate Z

EVZ

VVIVP

)cos()cos(2

zs

ftz

s

te Z

EVZVP

s

fta Z

EVI

0

z


a

sz r

Xanglewhere 1tan

s

zf

s

zta Z

EZ

VI

s

zft

s

zttate Z

EVZ

VVIVP

)cos(cos2

zs

ftz

s

te Z

EVZVP

s

fta Z

EVI


ss

z

XZandthen

090

sins

fte X

EVP

For maximum Pe

180 z

)(10095180

approxtoz

For maximum Pe, power angle =180 - impedance angle

emP

Pe

90 (ra=0)

0

z


Now Pm=Ef • Ia

s

fta Z

EVI

Iara

Ef

θδ

Ia

Vt

jIaXs

IaZs

zs

fz

s

tfm Z

EZ

VEP cos)cos(

2

sins

tfe X

VEP

95 to 100 (ra≠0)

Pemax

Pe

90 (ra=0)

0

z


Now Pm=Ef • Ia

s

fta Z

EVI

zs

fz

s

tfm Z

EZ

VEP cos)cos(

2

For maximum Pm

0 z

z

)(8580 approxtoz


ss

z

XZandthen

090

sins

tfm X

VEP

Thus if resistance is neglected,

Pm=Pe=Pem Electromagnetic Power

Otherwise Pe= Pm + Ia2ra

aa ritoeqivalent 2,10

emP

PmMax 80 to 85

95 to 100 (ra≠0)

PeMax

b) Salient Pole motor

Ef

Iara

Ef’

Ia

jIdXd

jIqXq

θδ

jIa X

q

VtConsider similarity

So, Xs=Xq,Zs=Zq

Zq=ra+jXq

)cos(cos'2

qq

ftq

q

te Z

EVZVP

qq

fq

q

tfm Z

EZ

VEP cos)cos(

2''

θz= θq

a

qq r

X1tan

Cyl. Rotor Salient Pole Rotor

Iara

Ef

θ

δ

Ia

VtjIa X

s

)cos(cos2

zs

ftz

s

te Z

EVZVP

zs

fz

s

tfm Z

EZ

VEP cos)cos(

2

But if resistance is neglected, Pm=Pe ≠ Pem for salient pole rotor Pe=VtIacosθ

Active power =Voltage component x In phase current component Vt

Ia

θ

Iacosθ Pe=-VtsinδxId VtcosδxIq

Now obtain Id and Iq

ftdd EVXI cos

d

ftd X

EVI

cos

sintqq VXI

q

tq X

VI sin

Ef

Vt

Ia

jIdXd

jIqXq

θδ

Id

Iq

Vtsinδ

Vtcosδ

d

ftd X

EVI

cos

q

tq X

VI sin

q

tt

d

tft X

VVXVE

V

sincoscos

sin

qt

d

t

d

ft

XV

XV

XEV sincoscossinsin 2

2

2sin112

sin2

dq

t

d

tfe XX

VX

VEP

Pe=-VtsinδxId +VtcosδxIq

Electromagnetic Power Pm+ Reluctance PowerElectrical Power=(Power due to saliency)

dq XX11

P

Sin

9080 to 85 2Sin


mP relem PP

02sin112

sin2

dq

t

d

tf

XXV

XVE

dd

ddP

Determine the δ, for maximum power

P

Sin

9080 to 85 2Sin


02cos11cos 2

dqt

d

tf

XXV

XVE

1cos22cos, 2 Now

thenanda

acbbObtain ,2

4cos,2

δ is less than 900.

Reactive power Generator

Ef

Vt

Ia

jIdXd

jIqXq

θδ

Iq

Vt

θδ

Ia

EfjIaXs

Salient pole Cylindrical rotor

Reactive power, Q= at IVImag OR=Voltage x

Quadrature lagging component of arm current

=Vtcosδ x Id

Id

-Vtsinδ x Iq

Vtcosδ

Vtsinδ

d

tfd X

VEINow

cos,

q

tq X

VIand sin

at IVImag

Ef

Vt

Ia

jIdXd

jIqXq

θδ

Iq

Vt

θδ

Ia

EfjIaXs


Reactive power, Q= OR=Voltage x


=Vtcosδ x Id

Id

-Vtsinδ x Iq

Vtcosδ

Vtsinδ

d

tfd X

VEINow

cos,

d

tq X

VIand sin

Duster

Reactive power, Q

q

t

d

tft X

VXVE

V2sincos

cos

q

t

d

t

d

tf

XV

XV

XVE 2222 sincoscos

222

sin11cos

dqt

d

t

d

tf

XXV

XV

XVE

For cyl rotor, Xd=Xq=Xs

Qd

t

d

tf

XV

XVE 2

cos tfd

t VEXV

cos

For maximum Q, the δ can be obtained from dQ/dδ=0

For generator, +ve value indicates that Q flows out of generator.

This eqn shows that:

tfd

t VEXV

cos

1. If Ef cosδ = Vt , Q = 0,Generator neither delivers nor absorbs Q Gen is operating at normal excitation, and power factor is unity

2. If Ef cosδ > Vt , Q = +ve,Generator delivers Q Gen is over excited, and pf is Lagging Demagnetizing action

3. If Ef cosδ < Vt , Q = -ve,Generator absorbs Q Gen is under excited, and pf is Leading Magnetizing action

Ef

Vt

jIaXs

Ia

δ

Ef

Vt

jIaXs

Ia

δθ

Q

Ef Vt

jIaXs

Iaδθ

Ef cosδ

Ef cosδ

Reactive power Motor

Ef

Vt

Ia

jIdXd

jIqXq

θ

δ

Iq

Vt

θ

δ

Ia

Ef

jIaXs




=Vtcosδ x Id

Id

+Vtsinδ x Iq

VtcosδVtsinδ

d

ftd X

EVINow

cos,

d

tq X

VIand sin

MotorReactive power

Ef

Vt

Ia

jIdXd

jIqXq

θ

δ

Iq

Vt

θ

δ

Ia

Ef

jIaXs




=Vtcosδ x Id

Id

+Vtsinδ x Iq

VtcosδVtsinδ

d

ftd X

EVINow

cos,

q

tq X

VIand sin

Duster

MotorReactive power, Q

q

t

d

ftt X

VX

EVV

2sincoscos

q

t

d

tf

d

t

XV

XVE

XV 2222 sincoscos

222

sin11cos

dqt

d

t

d

tf

XXV

XV

XVE

For cyl rotor, Xd=Xq=Xs

d

t

d

tf

XV

XVE 2

cos cosftd

t EVXV

For maximum Q, the δ can be obtained from dQ/dδ=0

For motor, +ve value indicates that Q flows in to the motor.

Q


tftd

t EVXV cos

1. If Ef cosδ = Vt , Q = 0,Motor neither absorbs nor delivers Q Motor is operating at normal excitation, and power factor is unity

2. If Ef cosδ > Vt , Q = -ve,Motor delivers Q Motor is over excited, and pf is Leading Demagnetizing action

3. If Ef cosδ < Vt , Q = +ve,Motor absorbs Q Motor is under excited, and pf is Lagging Magnetizing action

Ef Vt

jIaXs

Ia

δ

Ef

Vt

jIaXs

Ia

δ

θ

Q

Ef Vt jIaXs

Iaδ

θ

Ef cosδ

Ef cosδ


tftd

t EVXV cos

1. If Ef cosδ = Vt , Q = 0,Motor neither absorbs nor delivers Q Motor is operating at normal excitation, and power factor is unity

2. If Ef cosδ > Vt , Q = -ve,Motor delivers Q Motor is and pf is Leading Demagnetizing action

3. If Ef cosδ < Vt , Q = +ve,Motor absorbs Q Motor is under excited, and pf is Lagging Magnetizing action

Ef Vt

jIaXs

Ia

δ

Ef

Vt

jIaXs

Ia

δ

θ

Q

Ef Vt jIaXs

Iaδ

θ

Ef cosδ

Ef cosδ

over excited,

In general, the conclusions for Synchronous Machine are:

1. If Ef cosδ > Vt ,


tfd

t VEXV

cos

1. If Ef cosδ = Vt , Q = 0,Generator neither delivers nor absorbs Q Gen is operating at normal excitation, and power factor is unity

2. If Ef cosδ > Vt , Q = +ve,GeneratorGen is and pf is Lagging Demagnetizing action

3. If Ef cosδ < Vt , Q = -ve,Generator absorbs Q Gen is under excited, and pf is Leading Magnetizing action

Ef

Vt

jIaXs

Ia

δ

Ef

Vt

jIaXs

Ia

δθ

Q

Ef Vt

jIaXs

Iaδθ

Ef cosδ

Ef cosδ

GENERATOR

over excited, delivers Q

In general, the conclusions for Synchronous Machine are:

1. If Ef cosδ > Vt , Machine is over excited,

Machine produces or delivers or exports reactive power to infinite bus.

2. If Ef cosδ < Vt , Machine is under excited,

Machine consumes or absorbs or imports reactive power from infinite bus.

Physical Explanation of Torque

R1

R2

Y1

Y2

B1

B2

Sa

Na

SfNf Φf

Φa

ω

Repulsion

Ψ=90+δ+θ

Generator

Φf

Φa

IaraVtθ

δIa

EfjIaXs

ω

Ψ

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf

Rotor try to move clockwise

Φf

SfNf

Φa

ω

Te

Te is opposite to ω

Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Φf

Ψ

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf


Φf

SfNf

Φa

ω

Te


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

Φf

Ψ

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf


Φf

SfNf

Φa

ω

Te


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Φf

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

Ψ

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf


Φf

SfNf

Φa

ω

Te


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Φf

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

Ψ

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf


Φf

SfNf

Φa

ω

Te


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Φf

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

Te

180 Ψ

ΨΨ<180

R1

R2

Y1

Y2

B1

B2

Sa

Na

Te

ω

Φf


Φf

SfNf

Φa

ω

Te


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Φf

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

180 Ψ

Ψ

Increase θ VIcosθ decreases

δ decreases Ψ<180

Te

Te

ω

Φf


Φf

Sf

Φa

ω


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

180 Ψ

Ψ


δ decreases

ΦfNf

Te

Y1

Y2

B1

B2

R1

R2

Sa

Na

Ψ<180

Te

Te

ω

Φf


Φf

Sf

Φa

ω


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

180 Ψ

Ψ


δ decreases

ΦfNf

Te

Y1

Y2

B1B2

R1

R2

Sa

Na

Ψ<180

Te

Te

ω

Φf


Φf

Sf

Φa

ω


Repulsion

Φf

Φa

IaraVtθ

δIa

EfjIaXs

Ψ=90+δ+θ

Generator

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

180

Ψ

Again Increase θ upto ≈ 90

VIcosθ =0, δ =0

ΦfNf

Te

Y1

Y2

B1B2

R1

R2

Sa

Na

Ψ<180Ψ=180

Ψ

Te

Te

ω

Φf


Φf

Sf

Φa

ω


Repulsion

Φf

Φa

IaraVtθ Ia

EfjIaXs

Ψ=90+δ+θ

Generator

Vt

Ef

Leading

Lagging

ω

Rotor

Stator

Sf Nf

Sa Na

180

Ψ


VIcosθ =0, δ =0

ΦfNf

Te

Y1

Y2B1

B2

R1R2

Sa

Na

Ψ=180

Ψ

In Phase with Ef

In Phase with Vt

Te

Te

ω

ΦfΦf

Sf

Φa

ω


Repulsion on BOTH sides

Φf

Φa

IaraVtθ Ia

EfjIaXs

Ψ=90+δ+θ

No Gen, No Motor

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

180

Ψ


VIcosθ =0, δ =0

ΦfNf

Te

Y1

Y2B1

B2

R1R2

Sa

Na

Resultant Te=0

Ψ=180

Generator

Ψ

In Phase with Ef

In Phase with Vt

Te

ω

ΦfΦf

Sf

Φa

ω

Φf

Φa

IaraVtθ Ia

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

180

Ψ

Again Increase θ > 90

ΦfNf

Y1

Y2B1

B2

R1R2

Sa

Na

Resultant Te=0

Φa Ia

Vt

cos θ is –ve,

No Gen, No Motor

δ = -ve

δ

Ψ

Vt leads Ef

Ψ=180


MOTOR

ΨΨ<180

In Phase with Ef

In Phase with Vt

G

M

ω

ΦfΦf

Sf

Φa

ω

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

180

Again Increase θ > 90

ΦfNf

Y1

Y2B1

B2

R1R2

Sa

Na

Φa Ia

Vt

cos θ is –ve, δ = -ve

δ

Ψ

Vt leads Ef


MOTOR

ΨΨ<180

In Phase with Ef

In Phase with Vt

G

M

ω

ΦfΦf

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

180 Ψ

Y1

Y2B1

B2

R1R2

Sa

Na

Φa Ia

Vt

Sf

Φa

ωΦf

Nf

Again Increase θ > 90 cos θ is –ve, δ = -ve

Vt leads Ef

MOTOR

ΨΨ<180

Repulsion, rotor moves anticlockwise

δ

In Phase with Ef

In Phase with Vt

G

M

ω

ΦfΦf

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

Te

180 Ψ

Y1

Y2

B1

B2

R1

R2

Sa

Na

Φa Ia

Vt

Te

SfNf


Vt leads Ef

MOTOR

ΨΨ<180


δ

In Phase with Ef

In Phase with Vt

Φa

ωΦf

Te

G

M

ω

ΦfΦf

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

Te

180 Ψ

Y1

Y2

B1

B2

R1

R2

Sa

Na

Φa Ia

Vt

SfNf


Vt leads Ef

MOTOR

ΨΨ<180


δ

In Phase with Ef

In Phase with Vt

Φa

ωΦf

Te

Te

G

M

ω

ΦfΦf

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

Te

180 Ψ

Y1

Y2

B1

B2

R1

R2

Sa

Na

Φa Ia

Vt

SfNf


Vt leads Ef

MOTOR

ΨΨ<180


δ

In Phase with Ef

In Phase with Vt

Φa

ωΦf

Sa

Leading

Lagging Thus for GEN or for MOTOR

Ψ=90+δ+θ <180

Te

Te

G

M

ω

ΦfΦf

Φf

Iara

θ

EfjIaXs

Ψ=90+δ+θ

Vt

Ef

ω

Rotor

Stator

Sf Nf

Sa Na

Te

180 Ψ

Y1

Y2

B1

B2

R1

R2

Sa

Na

Φa Ia

Vt

SfNf


Vt leads Ef

MOTOR

ΨΨ<180


δ

Φa

ωΦf

Sa

Leading

Lagging Thus for GEN or for MOTOR

Ψ=90+δ+θ <180

Te

Te

G

M

Expression for Torque GENERATOR

NmradMechinSpeedRotor

PDevelopedPowerMechanicalT me .sec/..

,

rps

m

nP2

rps

af

nIE

23

rpsaf nIE 2/)cos(3

Ef jIaXs

δθ Ia

VtTm is opposite to Te

Cyl Rotor

araf IEIE

Er

δr

rps

are n

IET2

3 rpsrar nIE 2/)cos(3

rpsrawph nIKfT 2/)cos(44.43

GENERATOR

rpsaf nIE 2/)cos(3 '

Salient Pole Rotor

araf IEIE 'Er

δr rps

afe n

IET

23 '

rpsrar nIE 2/)cos(3


Ef

IaraVtIa

jIdXd

jIqXq

θδ

jIa X

qEf

’

MOTOR

rps

afe n

IET

23

rpsaf nIE 2/)cos(3

Tm is in same direction as that of Te

Cyl Rotor

araf IEIE

Erδr

rps

are n

IET2



θ

δ

Ia

Ef

jIaXsVt

MOTOR

rps

afe n

IET

23 '

rpsaf nIE 2/)cos(3 '

Salient Pole Rotor

araf IEIE '

Er

δr

rps

are n

IET2



Ef

Ef’

Ia

jIdXd

jIqXq

θ

δjI

a Xq

Vt

Circle Diagram of a Syn Motor Circle diagram gives very good idea about steady state performance.

saft ZIEV

This is obtained by varying mechanical load and excitation.

Consider only cylindrical rotor motor with constant ra and Xs.

1. Excitation Circle

This gives the locus of armature current Ia whenexcitation voltage Ef and load angle δ are varied.

The voltage equation is

s

f

s

ta Z

EZVI

If Vt is taken as reference, then )()( zs

fz

s

ta Z

EZVI

zs

t

ZV

θz

Vt

Ef

δ

CB

O AEf /Zs

Vt /Zsδδ

Ia

)( zs

f

ZE

aI

If load is changed, δ will change for a constant Ef

θz

Vt

Ef

δ

CB

O AEf /Zs

Vt /Zsδδ

Ia

Then Ia follows the path of CIRCLE.

This locus is known as Excitation Circle.

If Ef >Vt , then for constant δ, Ia is equal to OD.

D

zs

t

ZV )( z

s

f

ZE

aI

Ef >Vt

Ia MAX

θz

For MAX power, δ=θz

If Ef <Vt , then for constant δ,

Pf leading

If load is changed, δ will change for a constant Ef

θz

Vt

O AEf /Zs

Then Ia follows the path of CIRCLE.

This locus is known as Excitation Circle.

If Ef >Vt , then for constant δ, Ia is equal to OD.

D

zs

t

ZV )( z

s

f

ZE

aI

Ef >Vt

For MAX power, δ=θz

If Ef <Vt , then for constant δ,

Pf leading

Ef

δ

CB

Vt /Zsδδ

Ia

Ia MAX

Pf lagging

Ef <Vtθz

mechanical power developed Pm and power factor angle θ are varied.

2. Power CircleThis gives the locus of armature current Ia when

The equation can be written as

OR

0cos2 a

ma

a

ta r

PIrVI

Pm = Output power Po or Shaft power Psh + Constant loss

Pm = Pe - Copper loss aaat rIIV 2cos

0cossincos 2222 a

ma

a

taa r

PIrVII

Let x=Iasinθ and y=Iacosθ, then

022 a

m

a

t

rPy

rVyx

This is a equation of a circle

Radius of a circle is

General eqn of circle is

a

t

rV2

,0

a

m

a

t

rP

rV

2

2

and radius=

022 a

m

a

t

rPy

rVyx Centre of a circle is

Centre = (-g, -f)

02222 cfygxyx

cfg 22

Radius of a circle is

a

t

rV2

,0

a

m

a

t

rP

rV

2

2

022 a

m

a

t

rPy

rVyx Centre of a circle is

Ox

y

Vt

Vt/ra

Vt/2ra

x=Iasinθ

y=IacosθIa

θ

C

This is a power circle

A

aaa IIIOA 22 cossin

LaggingCOAand ,

Thus any point on circle gives Ia and its pf angle θ.

If Pm =0, then radius isa

t

rV2

Zero power circle

Ox

y

Vt

Vt/ra

Vt/2ra

x=Iasinθ

y=IacosθIa

θ

C

A

Zero power circle

As power increases, radius decreasesPoint C is the maximum power point.

02

max

2

aa

t

rP

rV and

aa

t IOCr

Vradius 2

a

t

rVP4

2

max

cosat IVMax power Input

12

a

tt r

VV

a

t

rV2

2

Max power output, Pmaxa

t

rV4

2

So effn is 50%50% Loss More temp rise

Motor never operated at that point

Ox

y

Vt

Vt/ra

Vt/2ra

x=Iasinθ

y=IacosθIa

θ

C

A

Zero power circle

If power equation is solved, TWO arm currents are obtained

Ia1

Ia2

Ia1 and Ia2

P

Ia1 Ia2

Out of these, LOWER current Ia1 is selected

O

Vt

θz

C

V Curves

A

B D

Vt /Zs

Ef /Zs

If

Ia

For point A, Ef /Zs=0, If=0, Ia=OA

O

A

Increase excitationPm=0

O

Vt

θz

C

V Curves

A

B D

Vt /Zs

Ef /Zs

If

Ia

For point A, Ef /Zs=0, If=0, Ia=OA

O

A

Increase excitationPm=0E

AE=If1

Excitation Circle

If1

F

FG

G

Again increase excitation to If2

O

Vt

θz

C

V CurvesB D

Vt /Zs

Ef /ZsFor point A, Ef /Zs=0, If=0, Ia=OAIncrease excitationPm=0

EAE=If1

Excitation CircleFAgain increase excitation to If2

H I

J

If1 If2

Ia

A

GA

If O

F

G

I

J

Increase excitation to If3

O

Vt

θz

C

B D

Vt /Zs

Ef /Zs

If O

Pm=0E Excitation CircleF

H I

J

K

L


If1 If2 If3

L

F

G

I KA

Ia

G

A

J

O

Vt

θz

C

B D

Vt /Zs

Ef /Zs


H I

J

K

L


MN Increase excitation to If5


A

If

Ia

O

A

FG

G

I K

If1 If2 If3 If4

N

If5 If6

JL

O

Vt

θz

C

B D

Vt /Zs

Ef /Zs


H I

J

K

L


If1 If2 If3

M

If4

N Increase excitation to If5

If5


If6

Dotted and thick line complete, is ‘O’ Curve

A

If

Ia

O

A

FG

G

I

J

K

L

N

V Curve

Inverted V Curve

O

Vt

θz

C

B D

Vt /Zs

Ef /Zs


H I

J

K

L


MN Increase excitation to If5


Dotted and thick line complete, or as whole, is ‘O’ Curve

A

If

Ia

O

A

FG

G

I

J

K

LIf1 If2 If3 If4

N

If5 If6

V Curve

Inverted V Curve

O

Vt

θz

C

A

B D

Vt /Zs

Ef /Zs

If

Ia

O

A

Pm=0E F

FG

G

H I

JI

J

K

KL

LIf1 If2 If3

M

If4

N

N

If5 If6

Excitation LineConsider AB At pt A, Ia=OA At pt E, Ia=OE At pt H, Ia=OH At pt C, Ia=OC At pt M, Ia=OM At pt B, Ia=OB

Excitation Line

Unity Power factor Line

O

Vt

θz

C

A

B D

Vt /Zs

Ef /Zs

If

Ia

O

A

Pm=0E F

FG

G

H I

JI

J

K

KL

LIf1 If2 If3

M

If4

N

N

If5 If6


Excitation Line

Unity Power factor LineConsider OD At pt O, Ia=0, If=AO=AM=If4

1

At pt 1, Ia=O1, If=A1At pt 2, Ia=O2, If=A2

2

At pt C, Ia=OC, If=AC=If3

O

Vt

θz

C

A

B D

Vt /Zs

Ef /Zs

If

Ia

O

A

Pm=0E F

FG

G

H I

JI

J

K

KL

LIf1 If2 If3

M

If4

N

N

If5 If6


Excitation Line1

2

At pt 3, Ia=O3, If=A3

3

At pt D, Ia=OD, If=AD=If5


O

Vt

θz

C

A

B D

Vt /Zs

Ef /Zs

If

Ia

O

A

Pm=0E F

FG

G

H I

JI

J

K

KL

LIf1 If2 If3

M

If4

N

N

If5 If6


Excitation Line1

2

3

At pt 3, Ia=O3, If=A3

At pt D, Ia=OD, If=AD=If5


O

Vt

θz

C

B DEf /Zs

Pm=0E

H

M

A

If

Ia

O

A

Minimum and Maximum Ia and If

for different power

Pm1

Excitation LineUPF Line

For Pm1, Iamin= O1

1

, Iamax= O22

Ifmin= AE Ifmax= AE’

E’

O

Vt

θz

C

B DEf /Zs

Pm=0E

H

M

A

If

Ia

O

A

Pm1


For Pm1, Iamin= O1

1

, Iamax= O22

Ifmin= AE

Pm2

3Iamin= O3

4

Iamax= O4Ifmin= AH Ifmax= AM


for different power

Ifmax= AE’

E’

There are bends in thses two lines

O

Vt

θz

C

B DEf /Zs

Pm=0E

H

M

A

If

Ia

O

A

Pm1


For Pm1, Iamin= O1

1

, Iamax= O22

Ifmin= AE

Pm2

For Pm2, 3

Iamin= O3

4

Iamax= O4Ifmin= AH Ifmax= AM


for different power

Ifmax= AE’

E’

‘O’ Curve for different powers

Pm=0

Pm1

Pm2

For Pm2, if excitation is reduced ie less than AH Strength of Magnetic locking decreases Motor oscillates Meter pointer oscillates Motor is unable to drive load This is called as loosing synchronism or out of step

There are bends in thses two lines

O

Vt

θz

C

B DEf /Zs

Pm=0E

H

M

A

If

Ia

O

A

Pm11

2

Pm2

3

4


for different power E’ Meter

Pm=0

Pm1

Pm2

For Pm2, if excitation is reduced ie less than AH Strength of Magnetic locking decreases Motor oscillates Meter pointer oscillates Motor is unable to drive load This is called as loosing synchronism or out of step

Leadinglagging

From M decrease excitationUpto H stable opn

After H unstable opn

O

Vt

θz

C

B DEf /Zs

Pm=0E

H

M

A

If

Ia

O

A

Pm11

2

Pm2

3

4


for different power E’

Pm=0

Pm1

Pm2

Max Excitation

s

f

ZE

AB max ODa

t

rV

sa

tf Z

rVE max

If

Ia

O

A Excitation LineUPF Line

‘V’ and inverted ‘V’ curves

Pm=0

Pm1

Pm2

pf

1

Pm=0

Pm1=0.25puPm1=0.75pu

Leading pflagging pf

V Curves

Inverted V Curves

If

Ia

O

A Excitation LineUPF Line

‘V’ and inverted ‘V’ curves

Pm=0

Pm1

Pm2

pf

1

Pm=0

Pm1=0.25puPm1=0.75pu

Leading pf

V Curves

Inverted V Curves

lagging pf

Synchronization,

Parallel Operation of TWO AlternatorsReasons and methods

Two important characteristics of syn machine1. Constant speed at constant frequency

2. Ability to operate both at leading and lagging power factor.

1. Effect of Changing Mechanical Torque

After Synchronization,

The Mechanical Driving Torque can be varied by controlling

Gate opening in case of hydro-generators

Throttle opening in case of turbo-generatorsFor simplicity consider cyl rotor generator

But the results are applicable to both types

First consider no load operation

a) No load operation

Zs1Zs2

Ef1 Ef2

Vt

Two alternators in parallelUnloaded

Externally Ef1 and Ef2 are in phase

External Circuit

Ef1Ef2

In local circuit Ef1 and Ef2 are in phase opposition

Ef1

Ef2

Local Circuit


Ef1Ef2

Ef1

Ef2

Local Circuit

Increase driving torque of Gen 1Speed of Gen 1 increases.Ef1 gets ahead of Ef2

Resultant voltage appears between Ef1 & Ef2

Due to this there is circulating current Ic between Gen 1 & 2.

Ic lags Ec by 900 which is given by

0

21

211 90tan

aa

ss

rrXX

External Circuit

β

β

Ec

Ec

Since circulating current Ic is flowing in local circuit, this circuit gives better idea.

21 ffc EEE

21

21

ss

ffc ZZ

EEI

Zs1Zs2

Ef1 Ef2

Vt

Ef1

Ic

Ic

Ic

1500rpm1520


Ef1Ef2

Ef2

Local Circuit

Increase driving torque of Gen 1Speed of Gen 1 increases.Ef1 gets ahead of Ef2

Resultant voltage appears between Ef1 & Ef2

Due to this there is circulating current Ic between Gen 1 & 2.

Ic lags Ec by 900 which is given by

0

21

211 90tan

aa

ss

rrXX

External Circuit

β

β

Ec

Ec

Since circulating current Ic is flowing in local circuit, this circuit gives better idea.

Ef1

θ1

θ2

1500rpm1520


Ef1Ef2

Ef2

Local CircuitExternal Circuit

β

β

Ec

Ec

Ef1

θ1

θ2

External Circuit

For Gen 1

Ef1Iccosθ1=+ve

So Gen 1 operates as GEN

Speed of Gen 1 decreases

Ef2Iccosθ2=-ve

So Gen 2 operates as MOTORSpeed of Gen 2 increases

For Gen 2

1500rpm1520 1510

1510

Thus there is AUTO-Equalization of speed or auto synchronizing action


Ef1Ef2

Ef2

Local CircuitExternal Circuit

α

Ec

Ec

Ef1

External Circuit

Now consider Zs1 & Zs2 purely resistive.

Ic will be in phase with Ec

Ef1Iccosα= Ef2Iccosα

Therefore there is no AUTO-Equalization of speed or auto synchronizing action

αBoth generators operate as a generators

Thus reactance is useful for auto synchronizing action But bad for voltage regulation

a) On load operation

Zs1Zs2

Ef1 Ef2

Vt

LOADIa1 Ia2

IL

Assume that

Ef1=Ef2 Ia1=Ia2 Total current =IL= Ia1+Ia2= 2Ia1=2Ia2

Power factor=cosθ

On load operation depends on PM’s speed-load characteristics

PM

1

PM

2

load

Load of Gen 1 Load of Gen 2

Speed or freq

P1 P2

f

P2= P2

Now increase mech. driving torque of Gen 1, it’s speed becomes more.

f ’In order to maintain freq. constant, decrease the mech. driving torque of Gen 2.

P1’ P2’ P1 +P2=2P

P1’ +P2’=2P f’ >f, P1’ >P1

P2’ <P2 Thus increase in mech. driving torque, increases frequency and load sharing of that generator.

For internal behaviour of gen, draw phasor diagram,

Vt

Ef1 = E

f2

jIa1Xs = jIa2Xs

Ia1= Ia2

δ

θ

IL

Power shared by Alt r

sin1

s

tf

XVE

P1 =P2

sin2

s

tf

XVE

Now increase mech driving torque of Alt 1Speed of Alt 1 increases, freq f increases to f ’ and δ increases to δ1

Power P1 increases to P1’ 11 sins

tf

XVE

For freq f , decrease mech driving torque of Alt 2

δ1

δ2

Power P2 decreases to P2’ 22 sins

tf

XVE

Ef1

Ef2

cos1at IV cos2at IV

Between Ef1 and Ef2, there is a voltage difference Ec

Vt

Ef1 = E

f2

jIa1Xs = jIa2Xs

Ia1= Ia2

δ

θ

IL

δ1

δ2

Ef1

Ef2

Ec

Due to Ec,the Ic circulates between two alternators.

Ic

Ic flows from Ef1 to Ef2.

Ic is added to Ia1

Ia1 + Ic = Ia1’

Ia1’Ic is subtracted from Ia2

Ia2 - Ic = Ia2’

Ia2’

θ1

θ2

But again Ia1’+ Ia2

’ =IL

Pf angle decreases, pf increases

Pf angle increases, pf decreases

Thus it is concluded that if mech driving torque is increased

1. Ia increases. 2. pf increases or improves 3. Load sharing increases

21 ffc EEE

21

21

ss

ffc ZZ

EEI

Zs1Zs2

Ef1 Ef2

Vt

Ic

2. Effect of Changing ExcitationThe Excitation can be controlled with the help of Field RheostatField current effects reactive power and power factorActually Ia changes, Ia

2ra loss changes so active power changesBut this change is negligible.a) No load operation

Zs1Zs2

Ef1 Ef2

Vt

Ef1

Ef2

Local circuit phasor diagram is useful

Increase excitation of Gen 1Ef1 Increases

Ef1’

Ec=Ef1’- Ef1Ec

IcIc lags to Ef1

’

DemagnetizingEf1

’ decreases to Ef1”

Ef1”

Ic leads to Ef2 Magnetizing

Ef2 increases to Ef2

” Ef2

”

Ic

2. Effect of Changing ExcitationThus there is Auto-Equalization of emf or voltages.

Zs1Zs2

Ef1 Ef2

Vt

Ef1

Ef2

Ef1’

Ec

Ic

Ef1”

Ef2”

Ic

230v

240v

235v

235v

2. Effect of Changing ExcitationThus there is Auto-Equalization of emf or voltages.

Zs1Zs2

Ef1 Ef2

Vt

New terminal voltages Vt= Ef1 - Ic Zs1 and Vt= Ef2 + Ic Zs2

221 ff

t

EEV

b) On load operation

LOADIa1 Ia2

IL Again assume that

Ef1=Ef2 Ia1=Ia2

Total current =IL= Ia1+Ia2

=2Ia1=2Ia2

Power factor=cosθ

Vt

Ef1= Ef2jIa1Xs = jIa2Xs

Ia1= Ia2

δ

θ

IL

Power shared by Alt r

sin1

s

tf

XVE

P1 =P2

sin2

s

tf

XVE

δ1

δ2

Ef1’

Ef2’

cos1at IV cos2at IV

Due to excitation, power remains constantstf X and V constant for constant,E sin

V constant for constant,I ta cos

constantE f sin

Now increase excitation of Alt 1 to Ef1’

For maintaining Vt constant,

Decrease excitation of Alt 2 to Ef2’

sin1

s

tf

XVE

1sin'

1 s

tf

XVE

2

'2 sin

s

tf

XVE

Vt

Ef1= Ef2jIa1Xs = jIa2Xs

Ia1= Ia2

δ

θ

IL

δ1

δ2

Ef1’

Ef2’

Now Ec=Ef1’- Ef2

’

Ic lags to Ec by 900

Thus if excitation is increased, Ia increases, Q increases

Ic

Ia1’=Ia1+ Ic

Ia1’θ1

jIa1’Xs

pf angle increases, pf decreases

Ia2’=Ia2- Ic Ia2

’

θ2

jIa2’Xs

Thus if excitation is decreased, Ia decreases, Q decreasespf angle decreases, pf increases

constantE f sin

Load DivisionZL=load impedance

1

11

s

tfa Z

VEI

Zs1Zs2

Ef1 Ef2

Vt

LOADIa1 Ia2

Laat ZIIV 21

111 saft ZIEV

222 saft ZIEV

2

22

s

tfa Z

VEI

Ls

tf

s

tft Z

ZVE

ZVE

V

2

2

1

1

On simplification2

2

1

1

21

111

s

f

s

f

Lsst Z

EZE

ZZZV

Put Ia1 and Ia2 in eqn (1)

(1)

Zs1Zs2

Ef1 Ef2

Vt

LOADIa1 Ia2

021 Laa ZII111 saf ZIE

222 saf ZIE

2

11122

s

saffa Z

ZIEEI

2

11

11

s

LsLs

fa

ZZZZZ

EI

Putting (3) in (1) gives

21

21

ss

ffc ZZ

EEI

021 Laa ZII

(1)

(2)

(1) – (2) gives

21 ff EE 2211 sasa ZIZI

(3)

L

ssss

ff

ZZZZZ

EE

2121

21

IcNo load condition, ZL=α

ExampleSketch the phasor diagrams of a 3-phase syn machine(a) At the moment of synchronizing

(b) When working as a motor(c) When working as a generatorDraw equivalent circuit diagrams

(a) At the moment of synchronizing: Ef=Vt Both are in phaseδ is zero, Power is zero, Ia is zero,

Ef Vt

VtEf

+

-

Xs +

-

Ia=0

(b) When working as a motor, Vt=Ef+jIaXs

Vt Ef

VtEf

+

-

Xs +

-

IajIaXs

δ

Iaθ

(cb) When working as a generator, Ef=Vt+jIaXs

Vt Ef

VtEf

+

-

Xs +

-

IajIaXs

δ

Iaθ

ExampleIt is desirable that the incoming machine should be a little too fast at the time of synchronizing. Explain.

Consider Gen 1 is on load and supplying Ia1 & power P1

Ef1=Ef2 Incoming machine is Gen 2.

Ef1

Ef2

Gen 2 -slowerFalls back

Ia1

Ec

Ic

Ia1’

P=Ef1Ia1’cos θ1>P1

θ1

Current of Gen 1 increases to Ia1

’

For Gen 2

θ2

P=Ef2Iccos θ2=-ve Operates as Motor

Both operations are undesirableLoad on Gen 1 increases

ExampleIt is desirable that the incoming machine should be a little too fast at the time of synchronizing. Explain.

Consider Gen 1 is on load and supplying Ia1 & power P1

Ef1=Ef2 Incoming machine is Gen 2.

Ef1

Ef2

Gen 2 –little fasterEf2 gets ahead

Ia1

Ec

Ic

Ia1’

P=Ef1Ia1’cos θ1>P1

θ1Current of Gen 1 decreases to Ia1

’

For Gen 2

θ2

P=Ef2Iccos θ2=+ve Operates as Generator

Both operations are desirableLoad on Gen 1 decreases

Gen 1 is relieved

ExampleTwo similar alternators operating in parallel have data:Alt 1: Capacity 700kW, freq 50Hz at no load & drops at 48.5Hz at full loadAlt 2: Capacity 700kW, freq 50.5Hz at no load & drops at 48Hz at full loadSpeed regulation of PM is linear

a) Calculate how a total load of 1200kW is shared by each alternator Also find the operating bus bar frequency at this load.b) Calculate the maximum load that these two units can

deliver without overloading either of them.

Solution

Load of Gen 1 Load of Gen 2 O

AOA=50.5HzFrequency

BOB=50.0Hz

x y z

xyz is line for load =1200kW xy+yz =1200kW

ge 700kW 48Hzd c 48.5Hz

700kW

From similar triangles, Bcd and Byx,

Bccd

Byxy

OcOBOyOBxy

700

5.4850700

50

fxy

fxy 505.1

700

Solution


AOA=50.5HzFrequency

BOB=50.0Hz

x y z


700kW

From similar triangles, Aeg and Ayz,Aeeg

Ayyz

OeOAOyOAyz

700

485.50700

5.50

fyz

fyz 5.505.2

700

xyz is line for load =1200kW xy+yz =1200kW

Solution


AOA=50.5HzFrequency

BOB=50.0Hz

x y z


700kW

f505.1

7001200= f5.50

5.2700

Hzf 58.48

kWyz 6.53758.485.505.2

700

andkWxy ,4.66258.48505.1

700

Solution


AOA=50.5HzFrequency

BOB=50.0Hz

x y z


700kW

OeOAeg

OcOAcc

'

b) From A to O, Alt 1 is loaded first for 700kW,

HzatkW 5.481260560700

kWcc 56025.2

700'

ie de line at 48.5Hz.Extend c to c’

c’

485.50700

5.485.50'

cc

Maximum possible load =

Synchronous Machine connected to Infinite Bus

If syn gen is connected in parallel with another gen then

voltage can be changed by excitation and

But if syn machine is connected in parallel with infinite bus, then

frequency can be changed by driving torque.

voltage & frequency remain constant due to infinite bus.

Ef Ia

Xs

Vt, f

Gate opening in case of hydro-generatorsThrottle opening in case of turbo (or steam) -generatorsFor simplicity consider cyl rotor generator

But the results are applicable to both types

First consider no load operation

1. Effect of Changing Mechanical Torque

The Mechanical Driving Torque can be varied by controlling

a) No load operationUnloaded δ=0, Ef in phase with Vt

The voltage equation

Ef

Vt

Ef=Vt+jIaXs No Active Power is transferred to infinite bus

Ia

jIaXs

PM merely supplies gen losses.Gen supplies reactive power.

b) ON load operationLoaded, load angle is δ and pf lagging

Ef

Vt

Ef=Vt+jIaXs

jIaXs

θ

δ

cossin ats

tf IVX

VEP

If driving torque increases, δ increases, power increases. The maximum stable value of δ is 900.

90

Ia1

θ1

Ef

jI a1X s

Power factor is leadingVt & f remain constantIf δ is decreased, δ becomes zeroNo load condition

Ef

Vt

Ia

jIaXs

Ia

b) ON load operationLoaded, load angle is δ and pf lagging

Ef

Vt

Ef=Vt+jIaXs

jIaXs

θ

δ

cossin ats

tf IVX

VEP

If driving torque increases, δ increases, power increases. The maximum stable value of δ is 900.

90

Ia1

θ1

Ef

jI a1X s

Power factor is leadingVt & f remain constantIf δ is decreased, δ becomes zeroNo load conditionAfter that if PM is decoupledEf lags Vt, δ is -ve

Ia

Ef

Vt

jIaXs

θ

-δ

Ia

Gen operates as Motor Vt=Ef+jIaXs

2. Effect of Changing ExcitationThe Excitation can be controlled with the help of Field RheostatField current effects reactive power and power factorActually Ia changes, Ia

2ra loss changes so active power changesBut this change is negligible.a) No load operation

Ef1If losses are neglected,δ is zero, power is zero

Increase Excitation to Ef1

Ef1

Now decrease excitation to Ef2

Ic

Ic lags to Ef1

Demagnetizing

Ic leads to Ef2

Magnetizing

jIcXs

If Ef=Vt, no current flows neither to, or from infinite busSynchronous machine is said to be on the busFLOATING

Vt Vt

s

tfc X

VEI

ft EV Ef2

jIcXs

Vt

ft EV

Delivers Q

Absorbs Q

Thus alternator voltage = Vt

b) ON load operationSuppose Gen is supplying power with angle δ and pf Unity

With Pm constant, cossin ats

tf IVX

VEP

Increase Ef to Ef1

Power factor is lagging

Supply Q

= Constant

Vt Ef

jIaXs

δ θ1

δ1

Ef1

constantE f sin

ConstantIE af cossin

11 sinsin ff EE

Ia1

Ia

jIa1 X

s Ia increases to Ia1=Ia+Ic1

Ia =active component

Ic1=demagnetizing component Ic1

aaa III 11 coscos

b) ON load operationSuppose Gen is supplying power with angle δ and pf Unity

With Pm constant, cossin ats

tf IVX

VEP

Decrease Ef to Ef2

Power factor is leading

Absorb Q

= Constant

Vt Ef

jIaXs

δ θ1

δ1

δ2

Ef1

Ef2

constantE f sin

ConstantIE af cossin

22 sinsin ff EE

Ia1

Ia

jIa1 X

s Ia increases to Ia2=Ia+Ic2

Ia =active component

Ic2=magnetizing componentIc1

jI a2X s

Ic2

Ia2θ2

aaa III 22 coscos

Active power P remains constant

Ia, pf, and Q change

b) ON load operation

Vt Ef

jIaXs

δ θ1

δ1

δ2

Ef1

Ef2

constantE f sin

Ia1

Ia

jIa1 X

s

Ic1

jI a2X s

Ic2

Ia2θ2

Ia

If

UPF

Leading Lagging

pf

If

UPF line

Leading Lagging

ExampleA 3-ph star connected cyl rotor alternator with synchronousreactance 5Ω per phase, is supplying 240A at unity pf to a 11kV infinite bus.

a) If the excitation emf of the altr is increased by 25% without changing its driving torque, calculate the new values of current, pf and δ.Will there be any change in power supplied to the infinite bus?Explain

b) With the increased excitation of part (a), held fixed, at what power power output would the alternator break from synchronism? Find the

corresponding values of current and power factor.

Per phase infinite bus voltage

Ef1

Vt

Vb=11000/√3=6351V

jIa1Xs

θ

δ

Pf =1

Ef=Vt+jIaXs

δ1

a) New value of excitation Ef1 =1.25x6463.4

Ef12=(Vt+Ia1Xssinθ)2+(Ia1Xscosθ)2

=406.22

Ia1

Ef

jIaXs

Ia

=6351+j240x5 =6463.4V

=8079.3VIa1cosθ=Ia =240A

θIa1sinθ=327.74A

Ia1= √{(Ia1sinθ)2+ (Ia1cosθ)2}Pf=cosθ=0.591 Lagging

Initial angle δ=tan -1(IaXs /Vt)=10.70

11 sinsin ff EE 0

1 54.8

Since driving torque is unchanged, Power P remains unchanged

b) With constant Ef, maximum power output takes place

Ef1

Vt

when δ is 900

jIa1Xs

θ

δ

Max power output

90

Ia2δ1

Ef1

jI a2X s

Ia2=2055.3A

Ia1

Ef

jIaXs

Ia

θ

90sin1

s

tf

XVE

phaseperkW3.10262

2212, tfsa VEXIFigFrom 67.10276

8079.3V

6351V

Power factor is leading

θ2

Power factor =2at IV

P

3.2055351.63.10262

786.0

ExampleAn alternator is supplying 60% of its rated power to an infinite bus at rated voltage and frequency. The excitation voltage ismade equal to rated voltage. The per unit reactance is 0.8

a) Determine the power angle, armature current and pf of the machine

b) If the excitation is increased by 40% with prime mover driving torqueUnchanged. Find the new values of power angle, arm current and pf.

a) 28.690, 0.6194, 0.9687

b) 20.050, 0.7177, 0.836

HuntingSatisfactory operation: Pm=Pe with losses neglected andRelative speed between stator and rotor field =0

Pm

ω

Pe

Generator

Pm

ωPe

Motor

If any power in changed then Pm≠Pe

and Pm-Pe=Pa Accelerating power

Consider motorConnected to infinite bus and on no load

Vt

Ef

Vt

NS N S δ=0

Ef

Ia

jIaXs


Vt

Ef

Vt

NS N S δ=0

Ef

jIaXs

Ia

Apply some load on shaft gradually.

Speed decreases gradually.


Vt

Ef

Vt

NS N S δ=0

Ef

jIaXs

Ia

Apply some load on shaft gradually.

Speed decreases gradually. δ increases Power increases

Torque increases to drive load

δ

sins

tfem X

VEPP

No oscillation of rotor

Ef

Ia1

Vt

Ef

Vt

NS N S δ=0

Ef

jIaXs

Ia

Apply some load on shaftSpeed decreases suddenly. δ increases to δ1 > δ

δ1

SUDDENLY.

δ1 = δ +Δδ sin

s

tfm X

VEP 1sin

s

tfe X

VEP Speed increases

δ

δ decreases to δ2 < δ or δ2 = δ –Δδ, Pm > Pe Speed decreases

δ2

Pm < Pe Speed increases Pm > Pe Speed decreasesRotor attains its equilibrium position Pm = Pe

s

fta X

EVIMore

1

s

fta X

EVILess

2

Vt

Ef

Ia

Ef

Vt

NS

N

S

N S

Mechanical oscillations

Electrical oscillations

The phenomena involving rotor oscillations about its finalequilibrium position is called HUNTING.During hunting the orientation of phasor of Ef changes relative Vt , therefore hunting is also called as PHASE-SWINGING.

Due to sudden application of load, rotor has to search for or hunt for

new equilibrium space position. So the name hunting.

In laboratory, hunting can be observed by wattmeter and ammeter

or by stroboscope, light falling on rotor shaft. At syn speed light appearsAt syn speed light appears stationary.

In case of generator, if mechanical driving torque or electrical load is changed suddenly, then there are rotor oscillations or hunting.

Hunting is objectionable.

The BAD effects are

1. It may cause the machine to fall out of step.

2. For generator, the output voltage fluctuates.

3. It causes great surges in current and power flow.

4. It increases machine losses and temperature.

5. It increases mechanical stresses and fatigue of the shaft.

The CAUSES of hunting1. A sudden change in LOAD.

2. A FAULT in the supply system

3. A sudden change in FIELD current.

4. Load torque or prime Mover containing HARMONICS.

It can be GUARDED by

1. Using a FLYWHEEL.

2. Designing m/c with suitable synchronizing POWER coefficientor STIFFNESS factor.

3. Providing DAMPER or ammortisseur winding.

Incomplete Type,Non-connected orOpen Type

Damper WindingConsists ofLow resistance copper, brass or aluminium bars

embeded in slot of pole faces and connected to short circuiting ring on both sides similar to SCIM.

Complete Type orConnected TypeGenerally used

The salient pole circuit ie damper wdg, bolts and iron take partin damping out the rotor hunting, is called amortisseur circuit.Or these short circuited bars are also known as amortisseur

Winding.

This wdg serves dual purpose: 1. starting and 2. hunting.

A motor started on the principle of induction motor by means of

damper wdg is known as Synduction Motor

For zero relative speed, no damping torque is developed.

Damping torque is developed when speed departs from syn speed.

Relative speed =+10rpm.

Nr=1490rpm

Ns=1500rpm

Due to +10rpm, current is induced in damper bars, suppose dot.

This current produces damping torqueThis is induction motor torque

Speed increases, relative speed becomes=0

Nr=1510rpm

Ns=1500rpm

Due to -10rpm, current is induced in damper bars, suppose cross.

This current produces damping torque

This is induction generator torque

Speed decreases, relative speed becomes=0

Thus damper wdg damps the rotor oscillations or hunting

Xd Can be obtained by

1. Slip Test

Phasor diagram of salient pole syn genr with ra=0

Now, Vt sinδ=IqXq

Under short circuit condition, Vt=0

2. Isc oscillogram3. OCC and SCC

Ef

VtIa

jIdXd

jIqXq

θδ

Id

Iq

,0qX ,0 qI

Isc=Id+jIq=Id

From Fig. Ef=XdId

sc

f

d

fd I

EIE

X

Ef jIdXd

Id=Isc

fsc

ffd IsameforICurrentCircuitShort

IgivenforEVoltageCircuitOpenX

Phasor diagram under Short circuit condn

Open ckt voltage = Ef

A

fsc IKI 1d

fdsc L

IMKI

Voltage

If

OCCIsc

SCC

If1

1fId AB

ACX

B

C

O

Short Circuit Ratio (SCR)SCR is defined as the ratio of Field Current

required to generate rated voltage (Ef) on open circuit, to theField Current required to circulate rated armature current Iasc

on a 3-phase short circuit.

A

Voltage

If

OCC Isc

SCC

If1

B

C

D

EIasc

rated

Ef

ratedascf

ff

IcurrentSCratedforIEvoltageOCratedforI

SCR

ODOA

O

Δ OAB & ODE are similar

DEAB

ODOASCR

1

,fI

d ABACXNow

Thus SCR is equal to the reciprocal of per unit value of Xd

A

Voltage

If

OCC Isc

SCC

If1

B

C

D

EIasc

rated

Ef

rated

O

1fId AB

ACpuinX

ImpedanceBase1

1fIABAC

f

asc

EI

1fIABAC

ACDE

ABDE

OAOD

ODOA1

SCR1

SCR is

1. Useful in obtaining an estimate of operating characteristics2. A measure of physical size of m/c

Significance of SCRa) LOW SCR

means high Xd

1. Ef= Vt+jIdXd More voltage variation,Poor voltage regulationMore If is required to maintain voltage constant.

2. Less Power and lower stability limit

3. Low synchronizing torque under disturbance

4. Low value of short circuit current (advantage)

b) HIGH SCR means low Xd

1. Better voltage regulation2. More Power and better stability limit3. High synchronizing torque under disturbance 4. High value of short circuit current (disadvantage)

Physical Size and CostFor more air gap, Xd is less and SCR is more.

So, SCR is directly proportional to air gap length. If air gap is doubled, Xd is halved For constant voltage, double If is requiredDiameter of field winding is doubled.More copper, more cost and more sizeThus overall size and cost are directly proportional to SCR

There is no effect on Iasc

Md is halved

d

fd

d

fasc L

IMXE

I

fd

d ILM

fIK

Same field current is required for rated Iasc.But for same If , more turns of arm wdg are required for contant voltage

as per wphf KTfE 44.4Thus more overall size and cost

Typical values of SCR

0.5 to 0.8 for turbo-generator

1 to 1.4 for hydro-generator

0.4 for synchronous condensor

Example

as

fz

s

tfi r

ZE

ZVE

P 2

2

)sin(

With the help of phasor diagram show that power input to a Cylindrical rotor alternator, at lagging power factor is given by

where s

az X

r1tan and Zs=ra+jXs

z Iara

Vt

θδ

Ia

Ef

jIaXs

IaZs

Solution:

Pm is input power to generatorSo derive the eqn for Pm

z Iara

Vt

θδ

Ia

Ef

jIaXs

IaZs

0

z


Pm=Ef • Ia

s

tfa Z

VEI

)cos(cos2

zs

tfz

s

fm Z

VEZE

P

90 zz

)cos(cos2

zs

tfz

s

fm Z

VEZE

P

)90cos(2

zs

tf

s

a

s

fm Z

VEZr

ZE

P

)sin(2

2

zs

tfa

s

fim Z

VEr

ZE

PP

Synchronous Machines-Part 2

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