The cylindrical-rotor construction is used for two- and four-pole turbine generators.
The salient-pole construction is best adapted to multipolar, slow-speed, hydroelectric
generators and to most synchronous motors.
The dc power required for excitation is supplied by the excitation system.
In older machines, the excitation current was typically supplied through slip rings from a dc
machine, the exciter, which was often mounted on the same shaft as the synchronous machine.
In modern systems, the excitation is supplied from ac exciters and solid-state rectifiers (either
simple diode bridges or phase-controlled rectifiers).
When a synchronous generator (G) is connected to a large interconnected system containing
many other synchronous generators, the voltage and frequency at its armature terminals are
fixed by the system. G cannot significantly affect the system voltage or frequency. It is usual
to represent the remainder of the system as a constant-frequency, constant-voltage source,
referred to as an infinite bus.
Torque in a synchronous machine:
T R F f sin RF
R : resultant air-gap flux per pole.
F f : mmf of dc field winding.
RF : torque angle, angle between magnetic axes of R and F f .
Torque-angle characteristic; rotor mmf and resultant airgap flux are assumed constant.
Positive values of torque represent generator action, corresponding to positive values of RF for which
the rotor mmf wave leads the resultant air-gap flux.
loss of synchronism (or pulling out of step):
An increase in prime-mover torque will result in a corresponding increase in the torque angle.
When RF becomes 90°, the electromechanical torque reaches its maximum value, known as
the pull-out torque. Any further increase in prime-mover torque cannot be balanced by a
corresponding increase in synchronous electromechanical torque. The result is that
synchronism will no longer be maintained and the rotor will speed up.
Under these conditions, the generator is disconnected from the external electrical system by
the automatic operation of circuit breakers, and the prime mover is quickly shut down to
prevent dangerous overspeed.
Synchronous Machine Inductances
Flux linkage of phase-a: a L aaia L abib L acic L af i f
Stator-to rotor mutual inductances:
L af L fa Laf cos me
Rotor rotating at synchronous speed s , m s t 0 0 : rotor position at t 0.
me ( P / 2)m et e0
Flux linkage of phase-a due to field:
L af L fa Laf cos(e t e 0 )
af Laf I f cos(et e0 )
L aa L bb L cc Laa Laa 0 Lal Laa 0 : self-inductance due to fundamental air-gap flux,
Lal : armature leakage flux
Mutual inductances between armature phases:
L ab L ba L ac L ca L bc L cb Laa cos(120 ) Laa 0
Resultant phase-a flux linkage:
a ( Laa 0 Lal )ia Laa 0 (ib ic ) af
Balanced currents ia ib ic 0 ib ic ia
a ( Laa 0 Lal )ia af
Ls : synchronous inductance
Terminal voltage (phase-a).
d a di
va Ra ia Ra ia Ls a eaf
eaf : voltage generated by field winding flux (excitation voltage)
eaf e Laf I f sin(et e 0 )
e Laf I f
rms value Eaf
e Laf I f
As an rms phasor Eaf j e je 0
Phase-a terminal voltage eqn. in phasor form:
ˆ ˆ ˆ ˆ
Va Ra I a jX s I a Eaf (motor reference direction for current)
X s Ls : synchronous reactance.
ˆ ˆ ˆ ˆ
Eaf Ra I a jX s I a Va (generator reference direction for current)
(a) motor reference direction (b) generator reference direction.
X s e Ls e Lal e Laa 0 X al X A
X al : armature leakage reactance
X A : armature reaction reactance
ER : air-gap voltage
A 60-Hz, three-phase synchronous motor is observed to have a terminal voltage of 460 V (line-line) and a
terminal current of 120 A at a power factor of 0.95 lagging. The field-current under this operating condition is 47
A. The machine synchronous reactance is equal to 1.68 Ω (0.794 per unit on a 460-V, 100-kVA, 3-phase base).
Assume the armature resistance to be negligible.
(a) the generated voltage Eaf in volts,
(b) the magnitude of the field-to-armature mutual inductance Laf, and
(c) the electrical power input to the motor in kW and in horsepower.
For motor reference direction for the current
ˆ ˆ ˆ
Va Eaf jX s I a ˆ ˆ ˆ
Eaf Va jX s I a
ˆ 460 265.6 V line-to-neutral I cos1 (0.95) 18.2 ( sign is for lagging p.f.)
I a 120 e j18.2 A ˆ
Eaf 265.6 j1.68(120 e j18.2 ) 278.8e j 43.4 V line-to-neutral
The field-to-armature mutual inductance
2 Eaf 2 278.8
Laf 22.3 mH
e I f 120 47
The three-phase power input to the motor, P 3Va I a cosa 3 265.6 120 0.95 90.83 kW 122 hp
Assuming the input power and terminal voltage for the motor of Example 5.1 remain constant, calculate (a) the
phase angle of the generated voltage and (b) the field current required to achieve unity power factor at the
a. Unity power factor current in phase with voltage
Pin 90.83 ˆ
Ia 114 A I a 1140 A
3Va 3 265.6
ˆ ˆ ˆ
Eaf Va jX s I a 265.6 j1.68 114 327.45e j 35.8 V line-to-neutral 35.8
2 Eaf 2 327.45
b. If 55.2 A
e Laf 120 22.3 103
OPEN- AND SHORT-CIRCUIT CHARACTERISTICS
Open-circuit Saturation Characteristic
The air-gap line represents the machine open-circuit
voltage characteristic corresponding to unsaturated
Deviation from this curve is a measure of the degree of
saturation in the machine.
The open-circuit characteristic can be used to measure the
field-to-armature mutual inductance Laf.
An open-circuit test performed on a three-phase, 60-Hz synchronous generator shows that the rated
open-circuit voltage of 13.8 kV is produced by a field current of 318 A. Extrapolation of the air-gap
line from a complete set of measurements on the machine shows that the field current corresponding to
13.8 kV on the air-gap line is 263 A. Calculate the saturated and unsaturated values of Laf.
13.8 2 Eaf 2 7.97 103
Eaf 7.97 kV Saturated value of Laf 94 mH
3 e I f 120 318
2 7.97 103
Unsaturated value of Laf 114 mH
The mechanical power required to drive the synchronous machine during the open-circuit test is used
to measure the no-load rotational losses.
no-load rotational losses friction and windage losses + core loss (due to flux at no load)
The friction and windage losses at synchronous speed are constant.
The open-circuit core loss is a function of the flux (which is proportional to the open-circuit voltage).
open-circuit core-loss curve
Open- and short-circuit characteristics SCC:
SCC is obtained with the machine driven at synchronous
speed. The field current is increased and the corresponding
armature current values are measured.
The machine operates in unsaturated condition. The short-
circuit armature current is directly proportional to the field
current over the range from zero to above rated armature
At If = Of , Ia,sc = O b
line-to-neutral voltage corresponding to Oa = Va,ag
unsaturated synchronous reactance X s ,u a ,ag
I a ,sc
Phasor diagram for short-circuit conditions.
Armature short-circuited ˆ ˆ ˆ
Va 0 Eaf I a ( Ra jX s )
Air-gap voltage: ˆ ˆ
ER I a ( Ra jX al )
The saturated value of the synchronous reactance at rated
voltage Va,rated is
The short-circuit ratio (SCR):
Of : field current required for rated voltage on open circuit.
Of : field current required for rated armature current on
( X s ) pu
Va ,rated Of
Xs . From the short-circuit characteristic: a ,rated slope of SCC
Of I a Of Of
Va ,rated Of Va ,rated Of Va ,rated Xs Of 1
Xs . . Z base ( X s ) pu
Of I a ,rated I a ,rated Of I a ,rated Z base Of SCR
The following data are taken from the open- and short-circuit characteristics of a 45-kVA, three-phase, Y-
connected, 220-V (line-to-line), six-pole, 60-Hz synchronous machine. From the open-circuit characteristic:
Line-to-line voltage = 220 V Field current = 2.84 A
From the short-circuit characteristic:
Armature current, A 118 152
Field current, A 2.20 2.84
From the air-gap line:
Field current = 2.20 A Line-to-line voltage = 202 V
Compute the unsaturated value of the synchronous reactance, its saturated value at rated voltage in accordance
and the short-circuit ratio. Express the synchronous reactance in ohms per phase and in per unit on the machine
rating as a base.
air-gap line: If = 2.20 A Va,ag = 202 / 3 = 116.7 V
for the same field current the armature current on short circuit is Ia,sc = 118 A
Xs,u = 116.7 / 118 = 0.987 Ω / phase.
rated armature current
I a ,rated 118 A
Ia,sc = 1.0 per unit Va,ag = 202 / 220 = 0.92 per unit
Xs,u = 0.92 / 1.00 = 0.92 per unit
saturated synchronous reactance:
Va ,rated 220 / 3
Xs 0.836 /phase
In per unit I a 152 /118 1.29 Xs = 1.00 / 1.29 = 0.775 per unit
the short-circuit ratio SCR = 2.84 / 2.20 = 1.29 Xs = 1 / SCR = 1 / 1.29 = 0.775 per unit
STEADY-STATE POWER-ANGLE CHARACTERISTICS
Synchronous machine is connected to an external system represented as an impedance in series with a
E1 E1 e j ˆ
Z R jX Z e jZ Z tan 1
The power P2 delivered through the impedance to the load-end voltage source E 2 :
ˆ ˆ j
P2 E2 I cos ˆ E E2 E1 e E2 E1 e j ( Z ) E2 e jZ
Z Z e jZ Z Z
ˆ ˆ E1 E
I I e j I cos jI sin Re( I ) I cos cos( Z ) 2 cos Z
E1 E2 E2
P2 cos( Z ) 2 cos Z
R E1 E2 E2R E E E 2R
cos Z P2 cos( Z ) 2 2 1 2 sin( Z ) 2 2
Z Z Z Z Z
where Z 90 Z
Power P1 at source end E1 of the impedance:
E1 E2 E2R
1 sin( Z ) 1 2
If the resistance R is negligible, Z 90 Z 0
1 sin the power-angle characteristic for a synchronous machine, : power angle
Maximum power transfer occurs when 90
ˆ ˆ ˆ ˆ ˆ ˆ
If > 0 E1 leads E 2 and the power flows from source E1 to source E 2 . If < 0 E 2 leads E1 and the
power flows from source E 2 to source E1 .
For a synchronous machine with generated voltage Eaf and synchronous reactance Xs connected to a
system whose Thevenin equivalent is a voltage source VEQ in series with a reactive impedance j XEQ,
X s X EQ
power transferred from the synchronous
machine to the system. is the phase
angle of Eaf with respect to VEQ .
A three-phase, 75-MVA, 13.8-kV synchronous generator with saturated synchronous reactance Xs = 1.35 per
unit and unsaturated synchronous reactance Xs,u = 1.56 per unit is connected to an external system with
equivalent reactance XEQ = 0.23 per unit and voltage VEQ 1.0 per unit, both on the generator base. It achieves
rated open-circuit voltage at a field current of 297 amperes.
(a) Find the maximum power Pmax (in MW and per unit) that can be supplied to the external system if the
internal voltage of the generator is held equal to 1.0 per unit.
(b) Using MATLAB, plot the terminal voltage of the generator as the generator output is varied from zero
to Pmax under the conditions of part (a).
(c) Now assume that the generator is equipped with an automatic voltage regulator which controls the
field current to maintain constant terminal voltage. If the generator is loaded to its rated value, calculate
the corresponding power angle, per-unit internal voltage, and field current. Using MATLAB, plot per-
unit Ear as a function of per-unit power.
Eaf VEQ 1
Pmax Eaf 1.0 pu, VEQ 1.0 pu Pmax 0.633 pu 47.5 MW
X s X EQ 1.35 0.23
the generator terminal current,
Eaf VEQ Eaf e j VEQ e j 1
j ( X s X EQ ) j ( X s X EQ ) j1.58
The generator terminal voltage
ˆ ˆ ˆ e j 1 j
Va VEQ jX EQ I a 1.0 j 0.23 0.8544 0.1456 e
generator loaded to its rated value P = 1.0 per unit
P sin t 4.35sin t ˆ
t : angle of the terminal voltage with respect to VEQ
1.0 t sin1 (1 / 4.35) 13.3
Va e j VEQ e j13.3 1
Ia 1.007e j 6.65
jX EQ j 0.23
ˆ ˆ ˆ
Eaf VEQ j ( X s X EQ ) I a 1.78e j 62.7 Eaf 1.78 I f 1.78 297 529 A
A 2000-hp, 2300-V, unity-power-factor, three-phase, Y-connected, 30-pole, 60-Hz synchronous motor has a
synchronous reactance of 1.95 Ω/phase. For this problem all losses may be neglected.
(a) Compute the maximum power and torque which this motor can deliver if it is supplied with power directly
from a 60-Hz, 2300-V infinite bus. Assume its field excitation is maintained constant at the value which
would result in unity power factor at rated load.
(b) Instead of the infinite bus of part (a), suppose that the motor is supplied with power from a three-phase, Y-
connected, 2300-V, 1500-kVA, two-pole, 3600 r/min turbine generator whose synchronous reactance is 2.65
Ω/phase. The generator is driven at rated speed, and the field excitations of generator and motor are adjusted
so that the motor runs at unity power factor and rated terminal voltage at full load. Calculate the maximum
power and torque which could be supplied corresponding to these values of field excitation.
This machine is of the salient-pole type. We will solve the problem by simple cylindrical-rotor theory.
Rated kVA = 2000 × 0.746 = 1492 kVA, three-phase
= 497 kVA/phase
Rated voltage = 1328 V line-to-neutral
Rated current = 374 A
From the phasor diagram,
Eafm Va2 ( I a X sm )2 1515 V
Pmax 1032 kW/phase 3096 kW 3-phase
X sm 1.95
In per unit, Pmax = 3096/1492 = 2.07 per unit
2 2 P 3096 103
Synchronous speed, s e 2 60 8 rad/sec Tmax max 123.2 kN-m
P 30 s 8
Va = 1328 V, Eafm =1515 V Eafg Va2 ( I a X sg )2 1657 V
Eafg Eafm 1657 1515 1638
Pmax 546 kW/phase 1638 kW 3-phase. In per unit, Pmax 1.10
X sg X sm 4.60 1492
Pmax 1638 103
Tmax 65.2 kN-m
STEADY-STATE OPERATING CHARACTERISTICS
Consider a synchronous generator delivering power at
constant frequency and rated terminal voltage to a load
whose power factor is constant. The curve showing the
field current required to maintain rated terminal voltage as
the constant-power-factor load is varied is known as a
Synchronous generators are usually rated in terms of the
maximum apparent power (kVA or MVA) load at a
specific voltage and power factor (often 80, 85, or 90
percent lagging) which they can carry continuously without overheating. The real power output of the
generator is usually limited to a value within the apparent power rating by the capability of its prime
When the real-power loading and voltage are fixed, the allowable reactive-power loading is limited by
either armature- or field-winding heating.
capability curves for a large, hydrogen-cooled turbine generator:
Capability curves of an 0.85 power factor, 0.80 short-circuit ratio, hydrogen-cooled turbine generator. Base
MVA is rated MVA at 0.5 psig hydrogen.
For a given real-power loading, the power factor at
which a synchronous machine operates, and hence its
armature current, can be controlled by adjusting its field
excitation. The curve showing the relation between
armature current and field current at a constant terminal
voltage and with a constant real power is known as
a V curve.