Download 32-Synchronous Generators Part 1

11/30/2012
Overview
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32-Synchronous Generators Part 1
text: 7.7 to 7.9
ECEGR 450
Electromechanical Energy Conversion
Excitation
Induced frequency
Induced EMF
Equivalent Circuit
Armature Reaction
Power Relationship
Approximate Power Relationship
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Dr. Louie
Excitation
Excitation
• Synchronous generators (motors) require
revolving magnetic field
a b c
 Permanent magnet
 Field winding (dc)
stator
i
• Exciter: supplies current to field winding (if )
commutator
N
 DC generator
 Brushless generator
 Power rating: <3% of generator rating
N
if
shaft
Pilot
exciter
field coil
S
slip rings
dc generator
• Field current is related to fp by kf
Voltage: 125 to 600VDC
Automatically controlled
(terminal voltage magnitude,
reactive power)
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field coil
S
stator
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Brushless Excitation
Induced Frequency
• 2-pole synchronous generator
• Balanced three-phase voltage induced
a b c
stator
i
3-phase rotor
N
N
if
x
shaft
Pilot
exciter
rectifier
x
xx
S
x
S
stator
x
5
x
x
Two-pole machine:
electrical frequency =
mechanical frequency
x
N
x
x
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x
x
x
field coil
S
Stationary field
AC generator
x
x
field coil
x
Tm, wm
x
x
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Induced Frequency
Induced EMF
• 4-pole synchronous generator
• Each coil “sees” two Norths and two Souths per
rotation
• We next examine the induced emf in a
synchronous generator
• Flux linking a single stator coil (fc):
 Two electrical sinewaves for each mechanical
rotation
• In general:
fc  fpkp cos(wt)
• Induced voltage in an Nc turn coil is:
f P N P
f m  m
2
120
ec  Ncfpwkp sin(wt)
S
N
N
S
• Maximum induced voltage:
Em  Nckpfpw
Tm, wm
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Induced EMF
Induced EMF
• RMS value of the induced emf:
Em  Nckpfpw
| Ec |
• If a generator has “a” parallel paths and P poles,
then the emf per phase is:
N P
f  m
120
1
Em  4.44fNckpfp
2
| Ea |
• Induced voltage in a phase group, accounting for
the number of coils in series, pitch factor and the
distribution factor, is:
| Epg | nk dEc  4.44nNckpk dffp
Ne
P
4.44nNck wffp
a
PnNck w
a
 Ne: effective turns per phase
• We can then write:
| Epg | 4.44nNck wffp
kw
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| Ea | 4.44Neffp
kpk d (winding factor)
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Dr. Louie
Example
Example
Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10
turns, and the flux per pole is 0.025Wb. The rotor
rotates at 375 rpm. The pitch factor is 0.94, and
the distribution factor is 0.96. There are two
parallel paths.
Compute:
frequency of the induced voltage
RMS value of the per-phase voltage
RMS value of the line-line voltage
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Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10
turns, and the flux per pole is 0.025Wb. The rotor
rotates at 375 rpm. The pitch factor is 0.94, and
the distribution factor is 0.96. There are two
parallel paths.
Induced frequency:
f
375  16
 50 Hz
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Example
Example
Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10
turns, and the flux per pole is 0.025Wb. The rotor
rotates at 375 rpm. The pitch factor is 0.94, and
the distribution factor is 0.96. There are two
parallel paths.
Induced RMS phase voltage:
Computing generator parameters:
PnNck w 16  3  10  0.902

 216.48
a
2
| Ea | 4.44fNefp  1201.5V
Ne 
k w  kpk d  0.902
n
Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10
turns, and the flux per pole is 0.025Wb. The rotor
rotates at 375 rpm. The pitch factor is 0.94, and
the distribution factor is 0.96.
144
 3 coils/pole/phase
16  3
| EL | 1201.5  3  2081V
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Equivalent Circuit
Equivalent Circuit
• Generator terminal voltage (Va) of a synchronous
generator depends upon the load
 Terminal voltage may be greater or smaller than
induced emf
 Will be higher when the power factor is leading
 Assumes generator is not grid-connected
 Ra: per-phase armature resistance (Ohm)
 Xa: armature leakage reactance (Ohm)
Rf
 Armature resistance voltage drop
 Armature leakage reactance voltage drop
 Armature reaction
+
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Equivalent Circuit
jXf
armature circuit
vf
Ra
fp
if
-
Ea
jXa
Ia
Va
Ea  Ia(R a  jXa )  Va
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Equivalent Circuit
Phasor diagrams (compare magnitude of Ea, Va)
Draw the phasor diagram for a synchronous
generator with a leading PF
Ea  Ia(R a  jXa )  Va
Ea
Ea
IaRa
• Equivalent circuit ignoring armature reaction
field circuit
• Terminal voltage is affected by:
Ia
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jIaXa
Ia
fPF V
a
IaRa
jIaXa
Va
Unity power factor
Lagging power factor
Va: reference
Ia: in phase Va (unity PF)
IaRa: in phase with Ia
jIaXa: 90o out of phase from Ia
E a > Va
Va: reference
Ia: lags Va (by fPF)
IaRa: in phase with Ia
jIaXa: 90o out of phase from Ia
Ea > Va
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Va
Leading power factor
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Equivalent Circuit
Armature Reaction
Draw the equivalent circuit for a leading PF
Consider a load with unity power factor
 Ea lags fp by 90 degrees
 Ia is in phase with Va
 fe = fp + far
Ea
Ia
• fe: effective flux per pole
• far: armature reaction flux per pole
jIaXa IaRa
fPF
 fe induces an emf the armature winding
Va
• Ear armature reaction emf
• Ear lags far by 90 degrees
Leading power factor
Va: reference
Ia: leads Va (by fPF)
IaRa: in phase with Ia
jIaXa: 90o out of phase from Ia
Ea < Va
Possible for induced voltage
to be larger than terminal
voltage
 Ee = Ea + Ear
• Effective per-phase emf
 Ee = Va + Ia(Ra + jXa)
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Armature Reaction
Armature Reaction
• Armature reaction can be modeled by placing a
voltage source in series with the induced emf
• Equivalent circuit
• Phasor Diagram (unity PF load)
 Armature reaction reduces the effective flux per pole
 Terminal voltage is smaller than the induced voltage
Ea
Rf
+
-
Nf
vf
Ear
+
Ea
-
if
-
+
Ra
+
Ee
-
Ia
fe = fp + far
+
Va
-
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Ea
fe = fp + far
Ia
IaRa
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Armature Reaction
• Phasor Diagram (leading PF load)
 Armature reaction increases the effective flux per pole
 Terminal voltage is greater than the induced voltage
Ear
Ee
jIaXa
Ee  Ea  Ear
• Phasor Diagram (lagging PF load)
 Armature reaction reduces the effective flux per pole
 Terminal voltage is smaller than the induced voltage
far
fPF Va
IaRa
Va
Unity power factor
Armature Reaction
fe
fp
Ea  Ear + Ee = Ear  Ia(R a  jXa )  Va
Ee  Ea + Ear
Dr. Louie
Ee
fe
Ia
Ea  Ear + Ia(R a  jXa )  Va
fp
Ear
far
jXa
far
jIaXa
fe = fp + far
fp
fe
Ia
Ea
fPF
Ear
Ee
jIaXa IaRa
Va
Leading power factor
Lagging power factor
Ea  Ear + Ee = Ear  Ia(R a  jXa )  Va
Ea  Ear + Ee = Ear  Ia(R a  jXa )  Va
Ee  Ea  Ear
Ee  Ea  Ear
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Armature Reaction
Equivalent Circuit
• Per phase equivalent circuit:
• Note: if must be adjusted with changing load to
keep terminal voltage constant
• Armature reaction can be expressed as a
magnetizing reactance
 Xs is used instead of Ear, Xa
 Ear = -jIaXm (emf uses active sign convetion)
• Xm and Xa can be combined together as the
synchronous reactance, Xs
Rf
 Xs = Xm + Xa
+
• Synchronous impedance of the generator is
 Zs = Ra + jXs
vf
4
1.06
d
Ia
Ia
fPF
Ea
IaRa
jIaXs
Lagging power factor
Unity power factor
Ea
fPF Va
Ea as a function of
power factor.
Terminal voltage
held constant at
10kV.
Induced Voltage (V)
Ea
Va
x 10
lagging
1.04
1.02
1
0.98
leading
0.96
0
jIaXs IaRa
d
0.2
d is measured from Va to Ea
d is positive for generators
Va
Leading power factor
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Equivalent Circuit
 d: angle between Ea and Va (induced voltage and
terminal voltage), known as the power angle or
torque angle
Ia
Va
Ia
Dr. Louie
• Phasor diagrams of new per-phase circuit
jIaXs
jXs
Ea  Ia(R a  jXs )  Va
Equivalent Circuit
IaRa
Ea
if
-
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d
Ra
jXf
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0.4
0.6
Power Factor
0.8
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Example
1
Example
A synchronous generator has a per-phase
synchronous impedance of 0.2 + j4. The generator
supplies a per-phase load current of 100A at a
lagging power factor of 0.866 lagging. The perphase terminal voltage is 10kV.
• Per phase armature current:
Ia  100  30 A
• Solving the circuit:
Ea  10.21.8 kV
• Power angle: 1.8 degrees
Compute the per-phase induced voltage.
Compute the power angle.
0.2
Ea
j4
100 kV
Ea  Ia(R a  jXs )  Va
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Voltage Regulation
Power Relationships
• Similar to dc generators, the voltage regulation of
a synchronous generator is:
VR 
• Mechanical power supplied to the shaft of a
synchronous generator by the prime mover




| Ea |  | Va |
100
| Va |
 Ea: induced emf, also the no-load terminal voltage
 Va: terminal voltage at full load (V)
steam turbine
combustion turbine
dc motor
others
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Power Relationships
Power Relationships
• Mechanical power in:
• Power output: Po  3 | Va || Ia | cos fPF  3 Re{VaIa}
Pinm  Tsws
 Requires knowledge (usually computation) of
armature current
 Ts: shaft torque (Nm)
 w s: shaft speed (rad/s)
• Desired to have an equivalent expression of
generator power output without having to
compute armature current
• Total power in:
Pin  Tsws  v fif
• Electrical power out:
Po  3 | Va || Ia | cos fPF  3 Re{VaIa}
• Copper2 losses:
Pcu  3 | Ia | R a
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Power Expressions
Power Expressions
• From the equivalent circuit:
Ia 
Continuing:
Ea  Va Ea  Va

R a  jXs
Zs
Pd  3 Re{
• Power developed:
*
Po  3 Re{VaIa}  3 Re{
 3 Re{
VaEa  | Va |2
}
Z*s
VaE*aZs | Va |2 Zs
V E* Z | V |2 R
| V |2 X

}  3 Re{ a a 2 s  a 2 a  j a 2 s }
| Zs |2
| Zs |2
| Zs |
| Zs |
| Zs |
Zs | Zs | z
| Zs |    z
| Z | 
Zs
1
1 Z*s


 s 2 z
Z*s Z*s Z*s | Zs |2   2z
| Zs |
| Zs |2
Dr. Louie
VaE*aZs | Va |2 R a
| V |2 X

 j a 2 s}
| Zs |2
| Zs |2
| Zs |
 3 Re{
VaE*aZs | Va |2 R a
}
| Zs |2
| Zs |2
 3 Re{
VaE*a (R a  jXs ) | Va |2 R a
}
| Zs |2
| Zs |2
Note:
Ea | Ea | d
Va | Va | 0 | Va |
|V| |E |(cos d  jsin d )(R a  jXs ) | Va |2 R a
 3 Re{ a a
}
| Zs |2
| Zs |2
 Above expansion uses:
Z*s | Zs |   z
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Dr. Louie
Recall that dividing by a phasor
means dividing by the magnitude
and subtracting the angle
35

3 | V|a|Ea|
| V |2 R
(R a cos d  Xs sin d )  a 2 a
| Zs |2
| Zs |
Dr. Louie
Important result
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Power Relationships
Power Relationship
• Power balance equation:
• Generator efficiency:
Pin  Tsws  if v f  3 | Va || Ia | cos fPF  3 | Ia |2 R a  if v f  Pr  Psl

 Pr: rotational losses (W)
 Psl: stray load losses (W)
3 | Va || Ia | cos fPF
3 | Va || Ia | cos fPF  3 | Ia |2 R a  Pc
• For maximum efficiency:
• Constant losses grouped as:
3 | Ia |2 R a  Pc
Pc  if vf  Pr  Psl
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Power Relationship
Approximate Power Relationship
• Armature resistance is small
• Common to ignore it
Pd = Re{EaIa*}
power (p.u)
Po
jXs
Ea
Ia
d
Va
Ia
0
50
100
d (deg.)
150
fPF
Ea
jIaXs
Va
Example lagging PF load
200
Armature copper
losses negligible
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Dr. Louie
Approximate Power Relationship
Approximate Power Relationship
Computing the real power output:
• Synchronous generator power output
(approximate)
Ea | Ea | d | Ea | cos d  j | Ea | sin d
Ia | Ia |   fPF | Ia | cos fPF  j | Ia | sin fPF
Euler’s Identity
Va | Va | 0 | Va |  j0
Ia 

d
Ea  Va | Ea | cos d  | Va | j | Ea | sin d  0


jXs
jXs
jXs
| Ea | sin d
| E | cos d  | Va |
j a
Xs
Xs
| Ia | cos fPF 
Ia
fPF
Po  3 | Va || Ia | cos fPF 
Ea
jIaXs
3 | Va || Ea | sin d
Xs
3 | Va || Ea | sin d
Xs
• Assumes:
Va
Example lagging PF load
Va  Ea  jIaXs
| Ea | sin d
(equating real parts)
Xs
Po  3 | Va || Ia | cos fPF 
40




Armature resistance is zero
Constant speed
Constant field current
Cylindrical rotor
Important result
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Approximate Power Relationship
Power Relationship
• Power-angle relationship:
Po 
• Torque developed (approximate):
3 | Va || Ea | sin d
Xs
Td 
• Maximum power:
3 | Va || Ea |
Pdm 
Xs
Po (approximate)
Pd
ws
3 | Va || Ea | sin d
Xs
• Maximum torque (approximate):
Tdm 
power (p.u)

3 | Va || Ea |
Xsws
• Maximum power and torque occur at d = 90o
0
50
100
d (deg.)
150
200
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Dr. Louie
Example
44
Example
A 2-pole synchronous generator has a per-phase
terminal voltage of 7.5 kV, a per-phase induced
voltage of 7.9 kV and a synchronous reactance of
1W. If the power angle is 15 degrees, compute the
total real power delivered to the load. Assume the
rotational losses are 1MW.
A 2-pole synchronous generator has a per-phase
terminal voltage of 7.5 kV, a per-phase induced
voltage of 7.9 kV and a synchronous reactance of
1W. If the power angle is 15 degrees, compute the
total real power delivered to the load. Assume the
rotational losses are 1MW.
Po 
3 | Va || Ea | sin d
 46MW
Xs
Rotational losses are not electric, so we do not need
to subtract them.
Dr. Louie
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Dr. Louie
Power Expressions
Power Relationship Summary
Pin = Tsws +vfif (total input power)
Several different forms of round-rotor power
output:
vfif (field winding loss)
Po  3 | Va || Ia | cos fPF
=3Re{VaIa}

46
Pinm=Tsws (input mechanical power)
3 | Ea || Va |
3 | Va |2 R a
(R a cos d  X s sin d) 
| Zs |2
| Zs |2
Pr + Psl (rotational and stray load loss)
Pd  3Re{EaI*a}  developed electrical power 
3 | Va || Ea | sin d
Po 
(valid only if R a can be ignored)
Xs
Pcu  3 | I2a | R a (copper loss in armature)
Po  3 | Va || Ia | cos fPF =3Re{VaIa} (output electrical power)

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3 | Ea || Va |
3 | Va |2 R a
(R a cos d  X s sin d) 
| Zs |2
| Zs |2
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Power Relationship Example
Summary
Pin = Tsws +vfif = 44.21MW
Let:
vf=400V
if = 250A
Pr + Psl= 2MW
Zs = 0.2 + j4W
d = 30o
Va = 10kV
|Ea| = 11kV
Po 
• Exciters are used to supply DC current to the
rotor of synchronous generators
• Frequency of induced voltage increases with the
number of poles for a fixed mechanical speed
• Leakage reactance and armature reaction can be
combined into Xs, the synchronous reactance
• Approximate power delivered by a synchronous
generator is:
Vfif = 0.1MW
Pinm=Tsws = 44.11MW
Pr + Psl = 2MW
Pd  3Re{EaI*a}  42.11MW
Pcu  3 | I2a | R a  1.14MW
Po  3 | Va || Ia | cos fPF 
3 | Ea || Va |
3 | Va |2 R a
(R a cos d  Xs sin d) 
 40.97MW
| Zs |2
| Zs |2
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3 | Va || Ea | sin d
Xs
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