Here are the steps to solve this problem: a) Given: Generator is 6 pole, 50 Hz Using the synchronous speed formula: nm = 120f/P nm = 120*50/6 = 1000 RPM b) Terminal voltage at different power factors: 1) Given load: Ia = 60 A, PF = 0.8 lagging Using phasor diagram: Vt = Ea - IaXs Ea = Vt + IaXs = 480 + 60*1 = 540 V Vt = 540*cos(cos-1(0.8)) = 480 V 2) PF = 1.0 Vt = Ea = 540 V 3) PF

1. Electrical Power Systems
Synchronous Generator
By: Mubarek Kurt

2.  A synchronous machine rotates at a constant speed in the steady
state.
 Unlike induction machines, the rotating air gap field and the rotor
in the synchronous machine rotate at the same speed, called the
synchronous speed.
 Sync. Machine are used primarily as generators of electrical power
(Hydro, nuclear or thermal power stations).
 Like most rotating machines, a synchronous machine can also
operate as both a generator and a motor.
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3.  An important feature of a synchronous motor is that it can
draw either lagging or leading reactive current from ac supply
system.
 Its rotor poles are excited by a DC current and its stator
winding (or armature winding) are connected to the AC supply.
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4.  The stator of the 3 phase sync. Machine has a 3 phase
distributed winding similar to that of the 3 phase induction
machine.
 The stator winding, which is connected to the ac supply
system, is sometimes called the armature winding. It is
designed for high voltage and current.
 The rotor has a winding called the field winding, which carries
direct current. The field winding on the rotating on the
structure is normally fed from an external dc source through
slip rings and brushes.
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5.  Synchronous machines can be broadly divided into 2 groups as
follows:
a) High speed machines with cylindrical (or non-salient pole)
rotors.
b) Low-speed machines with salient pole rotors.
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6. MubarekKurt

7.  Syn. Gen. are by definition sync., meaning that the electrical freq.
is locked in or synchronized with mechanical rate of rotation of the
gen.
 The electrical frequency is synchronized with the mechanical rate
of rotation
 Relationship between magnetic field speed and electrical
frequency,
nm is motor speed in RPM
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8.  Determine the rotor speed in RPM of the following 3 phase
synchronous machines
a) f=60Hz, number of poles=6
b) f=50Hz, number of poles=12
c) f=400Hz, number of poles=4
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9. The magnitude of the voltage induced in a
given stator phase is
This voltage depends on the flux, ɸ in the
machine, the freq. or speed of rotation, and the
machine construction. The equation can be
simplified as follow
Or
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10. • The voltage, Ea is not usually the voltage that
appears at the terminals, Vt of the generator.
• In fact, the only time the internal voltage Ea is
the same as the output voltage, Vt of a phase is
when there is no armature current flowing in
the machine.
• There are some factors that cause the diff.
between Ea and Vt
a) The distortion of the air gap magnetic field by
he current flowing in the stator, called
Armature reaction
b) Self inductance of the armature coilMubarekKurt

11.  We will explore the effects of the first three factors and derive
a machine model from them.
 The effect of a salient pole shape on the operation of a sync.
Machine will be ignored; in other words, all the machine in
this note are assumed to have non-salient or cyclindrical
rotors.
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12. The first effect mentioned, and normally the
largest one is armature reaction.
To understand armature reaction, please refer
below.
The figure shows a 2 pole rotor spinning inside
a 3 phase stator. There is no load connected to
the stator.
The rotor magnetic field BR produces an
internal generated voltage Ea.
With no load, thus Ea=Vt
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13.  Now suppose that the gen. is connected to a lagging load. Because
the load is lag. The peak current will occur at an angle behind the
peak voltage.
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14.  This current flowing in the stator windings produces a magnetic
field of its own, Bs and produces a voltage of its own in the stator,
Estat.
 With 2 voltages present in the stator winding, the total voltage in
a phase is just sum of internal generated voltage Ea and armature
reaction voltage Estat.
SRnet
statat
BBB
EEV


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15.  Since the angle of Ea and BR is the same and the angle of Estat
and Bs are the same, the resulting magnetic field will coincide
with net voltage, Vt.
 First note that, the voltage Estat lies at an angle 90 degree behind
current IA, thus the armature reactionn voltage can be expressed
as
Aat
Astat
jXIEV
jXIE


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16. MubarekKurt

17.  In addition to the effects of armature reaction, the stator coils have
a self inductance, XA and a resistance, RA. Thus,
 Where Xs=Syn. reactance
ASat
As
AAAAAat
IjXEV
XXX
IjRIjXjXIEV



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18. Full
equivalen
t circuit
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19. MubarekKurt

20. X=leakage reactance, Xa=armature reaction reactance, Xs=Xa+X=
synchronous reactance, Ra=armature resistance, Zs=synchronous
impedance, Vt=terminal voltages and Em=Magnetizing Voltage.
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21. MubarekKurt

22. MubarekKurt

23. MubarekKurt

24. - RA assumed to be zero
-  Torque angle
- Maximum torque due to
max power when sin  is
1
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25. Basic torque equation:
From power expression Pout = Pconv =
indm
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26.  A 3 phase, 5KVA, 208 V, 4 pole, 60 Hz, star connected sync.
Machine has negligible stator winding resistance and a sync.
Reactance of 8 ohms per phase at rated terminal voltage.
 The machine is first operated as a generator in parallel with a
3phase, 208 V, 60 Hz
a) Determine the excitation voltage and the power angle when the
machine is delivering rated KVA at a 0.8 lagging. Draw the
phasor diagram for this condition.
b) If the field excitation current is now increased by 20 %(without
changing the prime mover), find the stator current , power
factor and reactive KVA supplied by the machine.
c) With the field current as in (a) the prime mover power is slowly
increased. What is the steady state stability limit? What are
the corresponding values of the stator (or armature) current,
PF, and reactive power at the max. power transfer condition?
Draw the phasor diagram for this condition.
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27. Purpose of test is to determine these parameters:
a) Field current and flux relationship (and therefore between the
field current and Ea)
b) Synchronous reactance
c) Armature resistance
Open circuit and short circuit tests should be performed in order to
get all parameters
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28. Procedures:
1) Generators is rotated at the rated speed
2) No load is connected at the terminals and field current, If is
set to zero.
3) Field current is increased from 0 to maximum and the
terminal Voltage Vt is measured at each step along the way.
With terminal open IA=0, Ea=Vt
4) Record values and plot graph of the terminal voltage and
field current value. This plot is called open circuit
characteristic (OCC) of a generator.
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29. IA = 0, so EA = V, possible to plot EA or VT vs IF
graph. It is possible to find internal generated
voltage for any given field current
The iron saturated, mmf
getting slow down due to
increasing reluctance of the
iron
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30. Procedures of short circuit test:
1) Generator is rotated at rated speed
2) Adjust field current to 0
3) Short circuit the terminals through a set of ammeters.
4) Measure armature current, Ia or line current as the field
current is increased and plot the graph. This graph is
known as short-circuit characteristic (SCC).
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31. The net magnetic field is very small, the
iron is not saturated, so the relationship
is linearMubarekKurt

32. From the both tests. EA from OCC while IA
from SCC
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33.  The resistance is usually determined by applying a DC voltage
to 2 of the stator terminals.
 For Y connection: Ra=R/2=Vdc/(2Idc)
 For Delta connection: Ra=3R/2 =3Vdc/(2Idc)
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34. Where,± (+ for lag. Pf, - for lead Pf)
%%
)sin()cos( 22
Vfl
VflVnl
VR
VIaXsVIaRaVnl flfl


 
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35.  A 3 phase syn. Gen. rated at 50 kVA, 220 V, 60 Hz is Y
connected. The test results are given below. Find Ra, Zs and
Xs.
 Resistance test: V=2, I=22
 SCT: I1=I2=I3=rated current, If=22 A
 OCT: If=22 A, V=95 V
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36.  A 200kVA, 480 V, 50 Hz, Y-connected syn. Gen. with a rated
field current of 5 A was tested, and the following data were
taken:
 Vt,oc at the rated If was measured to be 540 V.
 IL,sc at the rated If was found to be 300 A.
 When a dc voltage of 10 V was applied to 2 of the terminals, a
current of 25 A was measured. Find Ra & Xs
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37. A 2300-V 1000-kVA 0.8-PF-lagging 60-Hz two-pole Y-connected
synchronous generator has a synchronous reactance of 1.1 Ω and an
armature resistance of 0.15 Ω. At 60 Hz, its friction and windage
losses are 24 kW, and its core losses are 18 kW. The field circuit has
a dc voltage of 200 V, and the maximum IF is 10 A. The resistance of
the field circuit is adjustable over the range from 20 to 200 Ω. The
OCC of this generator is shown in Figure.
(a) How much field current is required to make VT equal to 2300 V
when the generator is running at no load?
(b) What is the internal generated voltage of this machine at rated
conditions?
(c) How much field current is required to make VT equal to 2300 V
when the generator is running at rated conditions?
(d) How much power and torque must the generator’s prime mover
be capable of supplying?
(e) Construct a capability curve for this generator.
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38. MubarekKurt

39.  A 480V, 50 Hz, Y-connected, 6 pole syn. Gen. has a per-phase syn. Reactance of
1 ohm. Its full load armature current is 60 A at 0.8 PF lagging. This gen. has
friction and windage losses of 1.5 kW and core losses of 1 kW at 60 Hz full
load. Since the armature resistance is being ignored, assume that i2R
negligible. The field current has been adjusted so that the terminal voltage is
480 V at no load.
a) What is the speed of this generator.
b) What is the terminal voltage of this generator if the following are true?
1. It is loaded with the rated current at 0.8 PF lagging.
2. It is loaded with the rated current at 1.0 PF
3. It is loaded with the rated current at 0.8 PF leading.
c) What is the efficiency of this generator when it is operating at the rated
current and 0.8 PF leading?
d) How much shaft torque must be applied by the prime mover at full load?
How large is the induced countertorque.
e) What is the voltage regulation of this generator at 0.8 PF lagging?at 1
PF?at 0.8 PF leading.
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40.  A three syn. Gen has the following data;
a) The rated kVA=1250V
b) Rated voltage= 6000V
c) Mode of connection of the armature winding is star.
d) Ra=0.45 ohm, Xs=6.5 ohm.
The machine supplies full load current at 0.85 lagging at normal
rated voltage. Find the terminal voltage at the same excitation
and the load current at 0.85 leading.
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41.  At a particular current in the field winding, the short-circuit
armature current of the syn. Gen. attains the value 255 A,
when the open circuit generated nemf becomes 1550 V.
Determine the terminal potential difference when the load
current is 225 A at the 6.63 kV and lagging power factor is 0.8.
assume the armature resistance top be 2.5 ohm
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42.  A 1000 kVA, 1200 V 3 phase alternator is Y connected. Its
resistance per phase is 0.12 ohm and the reactance per phase
is 1.5 ohm, Find its voltage regulation if it supplies rates load
at:
a) PF=1
b) PF=0.9 lagging
c) PF = 0.9 Leading
d) What would the no-load line voltage be in part (c)
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43.  Repeat previous example but the alternator is delta connected
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44.  There are 2 factors that determine the power limits of electric
machines.
 One is mechanical torque on the shaft of the machine, and the
other is the heating of the machine’s windings.
 In all practical syn. Motors and generators, the shaft is strong
enough mechanically to handle a much larger steady state
power than the machine is rated for, so the practical steady
state limits are set by heating in the machine windings.
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45.  The stator and rotor heat limits, together with any external
limits on a syn. Gen. can be expressed in graphical form by a
gen. capability diagram.
 A capability diagram is a plot of complex power, S=P+jQ. It is
derived from the phasor diagram of the generator., assuming
that the V is constant at the machine rated voltage.
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46. MubarekKurt

47. Problem statement;
 Notice that for some possible current angles the required EA
exceeds EA,max. If the generator were operated at the rated
armature current and these power factors, the field winding would
burn up.
 Based upon these limits, there is a need to plot the capability of
the synchronous generator. This is so that it can be shown
graphically the limits of the generator.
 A capability diagram is a plot of complex power S=P+jQ. The
capability curve can be derived back from the voltage phasor of the
synchronous generator.
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48.  
SS X
V
V
X
V
Q 

 33

S
A
E
X
VE
D 3

On the voltage axes, the origin of the phasor diagram is at -
Vf on the horizontal axis, so the origin on the power
diagram is at:
The field current is proportional to the machine’s flux,
and the flux is proportional to EA = Kfw. The length
corresponding to EA on the power diagram is:
The armature current IA is proportional to XSIA , and the
length corresponding to XSIA on the power diagram is 3VfI-
A.
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49. MubarekKurt

50. Reasons for operating in parallel:
a) Handling larger loads.
b) Maintenance can be done w/t power disruption.
c) Increasing system reliability.
d) Increased efficiency.
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51. Condition required:
1) RMS line voltage must be equal.
2) Both have same phase sequence.
3) Output phase angles are same.
4) Must have a slightly higher frequency for new generator. It
will change slowly.
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52.  Frequency rating: the frequency at the system.
 Voltage rating: voltage generated that depend on flux and
speed.
 Apparent power rating: maximum power with maximum
armature current.
 Power factor rating:
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53. A 480-V 400-kVA 0.85-PF-lagging 50-Hz four-pole Δ-connected
generator is driven by a 500-hp diesel engine and is used as a standby
or emergency generator. This machine can also be paralleled with the
normal power supply (a very large power system) if desired.
(a) What are the conditions required for paralleling the emergency
generator with the existing power system? What is the generator’s rate
of shaft rotation after paralleling occurs?
(b) If the generator is connected to the power system and is initially
floating on the line, sketch the resulting magnetic fields and phasor
diagram.
(c) The governor setting on the diesel is now increased. Show both by
means of house diagrams and by means of phasor diagrams what
happens to the generator. How much reactive power does the generator
supply now?
(d) With the diesel generator now supplying real power to the power
system, what happens to the generator as its field current is increased
and decreased? Show this behavior both with phasor diagrams and with
house diagrams.
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54. A 13.8-kV 10-MVA 0.8-PF-lagging 60-Hz two-pole Y-connected
steam-turbine generator has a synchronous reactance of 12 Ω per
phase and an armature resistance of 1.5 Ω per phase. This
generator is operating in parallel with a large power system
(infinite bus).
(a) What is the magnitude of EA at rated conditions?
(b) What is the torque angle of the generator at rated conditions?
(c) If the field current is constant, what is the maximum power
possible out of this generator? How much reserve power or torque
does this generator have at full load?
(d) At the absolute maximum power possible, how much reactive
power will this generator be supplying or consuming? Sketch the
corresponding phasor diagram. (Assume IF is still unchanged.)
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