synchronous machines-part 2
DESCRIPTION
Synchronous Machines-Part 2.pptTRANSCRIPT
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'
For Inductance see from this side
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
For Inductance see from this side
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
For Inductance see from this side
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
For Inductance see from this side
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
For Inductance see from this side
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
For Pe, take Vt as a referenceSince Pe=Vt • Ia
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
If resistance is negligible,
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
tq
q
fte Z
VZEV
P cos)cos(2'
)cos(cos2
zs
tfz
s
fm Z
VEZE
P )cos(cos'2'
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
Power-AngleCharacteristics ofSalient Pole Machine
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
For Pe, take Vt as a referenceSince Pe=Vt • Ia
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
For Pe, take Vt as a referenceSince Pe=Vt • Ia
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
If resistance is negligible,
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
For Pm, take Ef as a reference
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
For Pm, take Ef as a reference
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
If resistance is negligible,
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
ftq
q
te Z
EVZVP
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
Power-AngleCharacteristics ofSalient Pole Machine
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
Power-AngleCharacteristics ofSalient Pole Machine
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
Salient pole Cylindrical rotor
Reactive power, Q= OR=Voltage x
Quadrature lagging component of arm current
=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
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
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
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
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
This eqn shows that:
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
This eqn shows that:
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 ,
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,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
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
Vt
Ef
Leading
Lagging
ω
Rotor
Stator
Sf Nf
Sa Na
Φf
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
Vt
Ef
Leading
Lagging
ω
Rotor
Stator
Sf Nf
Sa Na
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
Vt
Ef
Leading
Lagging
ω
Rotor
Stator
Sf Nf
Sa Na
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
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
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
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
Rotor try to move clockwise
Φf
Sf
Φa
ω
Te is opposite to 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
Increase θ VIcosθ decreases
δ decreases
ΦfNf
Te
Y1
Y2
B1
B2
R1
R2
Sa
Na
Ψ<180
Te
Te
ω
Φf
Rotor try to move clockwise
Φf
Sf
Φa
ω
Te is opposite to 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
Increase θ VIcosθ decreases
δ decreases
ΦfNf
Te
Y1
Y2
B1B2
R1
R2
Sa
Na
Ψ<180
Te
Te
ω
Φf
Rotor try to move clockwise
Φf
Sf
Φa
ω
Te is opposite to 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
Rotor try to move clockwise
Φf
Sf
Φa
ω
Te is opposite to 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
Again Increase θ upto ≈ 90
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
Te is opposite to 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
Again Increase θ upto ≈ 90
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
Repulsion on BOTH sides
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
Repulsion on BOTH sides
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
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
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
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
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
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
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
Again Increase θ > 90 cos θ is –ve, δ = -ve
Vt leads Ef
MOTOR
ΨΨ<180
Repulsion, rotor moves anticlockwise
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
rpsrawph nIKfT 2/)cos(44.43
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
3 rpsrar nIE 2/)cos(3
rpsrawph nIKfT 2/)cos(44.43
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
3 rpsrar nIE 2/)cos(3
rpsrawph nIKfT 2/)cos(44.43
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
Increase excitation to If4
If1 If2 If3
L
F
G
I KA
Ia
G
A
J
O
Vt
θz
C
B D
Vt /Zs
Ef /Zs
Pm=0E Excitation CircleF
H I
J
K
L
Increase excitation to If4
MN Increase excitation to If5
Increase excitation to If6
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
Pm=0E Excitation CircleF
H I
J
K
L
Increase excitation to If4
If1 If2 If3
M
If4
N Increase excitation to If5
If5
Increase excitation to If6
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
Pm=0E Excitation CircleF
H I
J
K
L
Increase excitation to If4
MN Increase excitation to If5
Increase excitation to If6
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 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 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 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 Line1
2
At pt 3, Ia=O3, If=A3
3
At pt D, Ia=OD, If=AD=If5
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 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 Line1
2
3
At pt 3, Ia=O3, If=A3
At pt D, Ia=OD, If=AD=If5
Unity Power factor Line
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
Excitation LineUPF Line
For Pm1, Iamin= O1
1
, Iamax= O22
Ifmin= AE
Pm2
3Iamin= O3
4
Iamax= O4Ifmin= AH Ifmax= AM
Minimum and Maximum Ia and If
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
Excitation LineUPF Line
For Pm1, Iamin= O1
1
, Iamax= O22
Ifmin= AE
Pm2
For Pm2, 3
Iamin= O3
4
Iamax= O4Ifmin= AH Ifmax= AM
Minimum and Maximum Ia and If
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
Minimum and Maximum Ia and If
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
Minimum and Maximum Ia and If
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
a) No load operation
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
a) No load operation
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
a) No load operation
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
a) No load operation
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
Load of Gen 1 Load of Gen 2 O
AOA=50.5HzFrequency
BOB=50.0Hz
x y z
ge 700kW 48Hzd c 48.5Hz
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
Load of Gen 1 Load of Gen 2 O
AOA=50.5HzFrequency
BOB=50.0Hz
x y z
ge 700kW 48Hzd c 48.5Hz
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
Load of Gen 1 Load of Gen 2 O
AOA=50.5HzFrequency
BOB=50.0Hz
x y z
ge 700kW 48Hzd c 48.5Hz
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
Consider motorConnected to infinite bus and on no load
Vt
Ef
Vt
NS N S δ=0
Ef
jIaXs
Ia
Apply some load on shaft gradually.
Speed decreases gradually.
Consider motorConnected to infinite bus and on no load
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
For Pm, take Ef as a reference
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