Alternator

Alternator
Chapter Five
Synchronous Generators
5.1 Introduction
Three phase synchronous generators are the primary source of
all the electrical energy we consume. These machines are the
largest energy converters in the world. They convert mechanical
energy into electrical energy, in powers ranging up to 1500 MW.
In this chapter we will study the construction and characteristics of
these large, modern generators. They are based upon the
elementary principles.
5.2 Commercial Synchronous Generators
Commercial synchronous generators are built with either a
stationary or a rotating DC magnetic field.
A stationary field synchronous generator has the same outward
appearance as a DC generator. The salient poles create the DC
field, which is cut by a revolving armature. The armature possesses
a 3-phase winding whose terminals are connected to three
slip-rings mounted on the shaft. A set of brushes, sliding on the
slip-rings, enables the armature to be connected to an external
3-phase load. The armature is driven by a gasoline engine, or some
other source of motive power. As it rotates, a 3-phase voltage is
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induced, whose value depends upon the speed of rotation and upon
the DC exciting current in the stationary poles. The frequency of
the voltage depends upon the speed and the number of poles on the
field. Stationary-field generators are used when the power output is
less than 5 kVA. However, for greater outputs, it is cheaper, safer,
and more practical to employ a revolving DC field.
A revolving field synchronous generator has a stationary
armature called a stator. The 3-phase stator winding is directly
connected to the load, without going through large, unreliable
slip-rings and brushes. A stationary stator also makes it easier to
insulate the windings because they are not subjected to centrifugal
forces. Fig.5.1 is a schematic diagram of such a generator,
sometimes called an alternator. The field is excited by a DC
generator, usually mounted on the same shaft. Note that the
brushes on the commutator have to be connected to another set of
brushes riding on slip-rings to feed the DC current IX into the
revolving field.
Fig.5.1Schematic diagram and cross-section view of a typical 500
MW synchronous generator.
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5.3 Number of Poles
The number of poles on a synchronous generator depends upon
the speed of rotation and the frequency we wish to produce.
Consider, for example, a stator conductor that is successively
swept by the N and S poles of the rotor. If a positive voltage is
induced when an N pole sweeps across the conductor, a similar
negative voltage is induced when the S pole speeds by. Thus, every
time a complete pair of poles crosses the conductor, the induced
voltage goes through a complete cycle. The same is true for every
other conductor on the stator; we can therefore deduce that the
alternator frequency is given by
f =
Pn
120
(5.1)
f =frequency of the induced voltage [Hz].
P = number of poles on the rotor.
n = speed of the rotor [r/min].
Example 5.1 A hydraulic turbine turning at 200 r/min is connected to a synchronous generator. If the induced voltage has a
frequency of 60 Hz, how many poles does the rotor have?
Solution:
From Eqn.(5.1), we have :
P= 120 * 60/200
= 36 poles, or 18 pairs of N and S poles.
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5.4 Main Features Of The Stator
From an electrical standpoint, the stator of a synchronous
generator is identical to that of a 3-phase induction motor. It is
composed of a cylindrical laminated core containing a set of slots
that carry a 3-phase lap winding. The winding is always connected
in wye and the neutral is connected to ground. A wye connection is
preferred to a delta connection because:
1. The voltage per phase is only 1 / 3 or 58% of the voltage
between the lines. This means that the highest voltage between a
stator conductor and the grounded stator core is only 58% of the
line voltage. We can therefore reduce the amount of insulation in
the slots that in turn, enables us to increase the cross section of the
conductors. A larger conductor permits us to increase the current
and, hence, the power output of the machine.
2. When a synchronous generator is under load, the voltage
induced in each phase becomes distorted, and the waveform is no
longer sinusoidal. The distortion is mainly due to an undesired
third harmonic voltage whose frequency is three times that of the
fundamental frequency. With a wye connection, the distorting
line-to-neutral harmonics do not appear between the lines because
they effectively cancel each other. Consequently, the line voltages
remain sinusoidal under all load conditions. Unfortunately, when a
delta connection is used, the harmonic voltages do not cancel, but
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257
add up. Because the delta is closed on itself, they produce a
third-harmonic circulating current, which increases the I 2 R losses.
The nominal line voltage of a synchronous generator depends
upon its kVA rating. In general, the greater the power rating, the
higher the voltage. However, the nominal line-to-line voltage
seldom exceeds 25 kV because the increased slot insulation takes
up valuable space at the expense of the copper conductors.
5.5 Main Features Of The Rotor
Synchronous generators are built with two types of rotors:
salient pole rotors and smooth, cylindrical rotors. Salient pole
rotors are usually driven by low-speed hydraulic turbines, and
cylindrical rotors are driven by high-speed steam turbines.
1. Salient pole rotors. Most hydraulic turbines have to turn at
low speeds (between 50 and 300 r/min) in order to extract the
maximum power from a waterfall. Because the rotor is directly
coupled to the waterwheel, and because a frequency of 50 Hz or 60
Hz is required, a large number of poles are required on the rotor.
Low-speed rotors always possess a large diameter to provide the
necessary space for the poles. The salient poles are mounted on a
large circular steel frame which is fixed to a revolving vertical
shaft. To ensure good cooling, the field coils are made of bare
copper bars, with the turns insulated from each other by strips of
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mica. The coils are connected in series, with adjacent poles having
opposite polarities.
In addition to the DC field winding, we often add a squirrel-cage
winding, embedded in the pole faces. Under normal conditions,
this winding does not carry any current because the rotor turns at
synchronous speed. However, when the load on the generator
changes suddenly, the rotor speed begins to fluctuate, producing
momentary speed variations above and below synchronous speed.
This induces a voltage in the squirrel-cage winding, causing a large
current to flow therein. The current reacts with the magnetic field
of the stator, producing forces which dampen the oscillation of the
rotor. For this reason, the squirrel-cage winding is sometimes
called a damper winding.
The damper winding also tends to maintain balanced 3-phase
voltages between the lines, even when the line currents are unequal
due to unbalanced load conditions.
2. Cylindrical rotors. It is well known that high-speed steam
turbines are smaller and more efficient than low-speed turbines.
The same is true of high-speed synchronous generators. However,
to generate the required frequency we cannot use less than 2 poles,
and this fixes the highest possible speed. On a 60 Hz system it is
3600 r/min. The next lower speed is 1800 r/min, corresponding to a
4-pole machine. Consequently, these steam-turbine generators
possess either 2 or 4 poles.
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The rotor of a turbine-generator is a long, solid steel cylinder
which contains a series of longitudinal slots milled out of the
cylindrical mass. Concentric field coils, firmly wedged into the
slots and retained by high-strength end-rings serve to create the N
and S poles.
The high speed of rotation produces strong centrifugal forces,
which impose an upper limit on the diameter of the rotor. In the
case of a rotor turning at 3600 r/min, the elastic limit of the steel
requires the manufacturer to limit the diameter to a maximum of
1.2 m. On the other hand, to build the powerful 1000 MVA to 1500
MVA generators the volume of the rotors has to be large. It follows
that high-power, high-speed rotors have to be very long.
5.6 Field Excitation And Exciters
The DC field excitation of a large synchronous generator is an
important part of its overall design. The reason is that the field
must ensure not only a stable ac terminal voltage, but must also
respond to sudden load changes in order to maintain system
stability. Quickness of response is one of the important features of
the field excitation. In order to attain it, two DC generators are
used: a main exciter and a pilot exciter Static exciters that involve
no rotating parts at all are also employed.
The main exciter feeds the exciting current to the field of the
synchronous generator by way of brushes and slip-rings. Under
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normal conditions the exciter voltage lies between 125 V and 600
V. It is regulated manually or automatically by control signals that
vary the current Ic, produced by the pilot exciter (Fig.5.1).
The power rating of the main exciter depends upon the capacity
of the synchronous generator. Typically, a 25 kW exciter is needed
to excite a 1000 kVA alternator (2.5% of its rating) whereas a 2500
kW exciter suffices for an alternator of 500 MW (only 0.5% of its
rating).
Under normal conditions the excitation is varied automatically.
It responds to the load changes so as to maintain a constant ac line
voltage or to control the reactive power delivered to the electric
utility system. A serious disturbance on the system may produce a
sudden voltage drop across the terminals of the alternator. The
exciter must then react very quickly to keep the ac voltage
constant. For example, the exciter voltage may have to rise to twice
its normal value in as little as 300 to 400 milliseconds. This
represents a very quick response, considering that the power of the
exciter may be several thousand kilowatts.
5.7 Brushless Excitation
Due to brush wear and carbon dust, we constantly have to clean,
repair, and replace brushes, slip-rings, and commutators on
conventional DC excitation systems. To eliminate the problem,
brushless excitation systems have been developed. Such a system
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261
consists of a 3-phase stationary field generator whose ac output is
rectified by a group of rectifiers. The DC output from the rectifiers
is fed directly into the field of the synchronous generator (Fig.5.2).
Fig.5.2 Brushless exciter system.
The armature of the ac exciter and the rectifiers are mounted on
the main shaft and turn together with the synchronous generator. In
comparing the excitation system of Fig.5.2 with that of Fig.5.1, we
can see they are identical, except that the 3-phase rectifier replaces
the commutator, slip rings, and brushes. In other words, the
commutator (which is really a mechanical rectifier) is replaced by
an electronic rectifier. The result is that the brushes and slip-rings
are no longer needed.
The DC control current I c from the pilot exciter regulates the
main exciter output I x , as in the case of a conventional DC exciter.
The frequency of the main exciter is generally two to three times
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the synchronous generator frequency (50 Hz). The increase in
frequency is obtained by using more poles on the exciter than on
the synchronous generator. Static exciters that involve no rotating
parts at all are also employed.
5.8 Factors Affecting The Size Of Synchronous
Generators
The prodigious amount of energy generated by electrical utility
companies has made them very conscious about the efficiency of
their generators. For example, if the efficiency of a 1000 MW
generating station improves by only 1 %, it represents extra
revenues of several thousand dollars per day. In this regard, the
size of the generator is particularly important because its efficiency
automatically improves as the power increases. For example, if a
small 1 kilowatt synchronous generator has an efficiency of 50%, a
larger, but similar model having a capacity of 10 MW inevitably
has an efficiency of about 90%. This improvement in efficiency
with size is the reason why synchronous generators of 1000 MW
and up possess efficiencies of the order of 99%.
Another advantage of large machines is that the power output
per kilogram increases as the power increases. For example, if a 1
kW generator weighs 20 kg (yielding 1000 W/20 kg = 50 W/kg), a
10 MW generator of similar construction will weigh only 20 000
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kg, thus yielding 500 W/kg. From a power standpoint, large
machines weigh relatively less than small machines; consequently,
they are cheaper. Section 16.24 at the end of this chapter explains
why the efficiency and output per kilogram increase with size.
Everything, therefore, favors the large machines. However, as
they increase in size, we run into serious cooling problems. In
effect, large machines inherently produce high power losses per
unit surface area (W/m2); consequently, they tend to overheat. To
prevent an unacceptable temperature rise, we must design efficient
cooling systems that become ever more elaborate as the power
increases. For example, a circulating cold air system is adequate to
cool synchronous generators whose rating is below 50 MW but
between 50 MW and 300 MW we have to resort to hydrogen
cooling. Very big generators in the 1000 MW range have to be
equipped with hollow, water-cooled conductors. Ultimately, a
point is reached where the increased cost of cooling exceeds the
savings made elsewhere, and this fixes the upper limit to size.
To sum up, the evolution of big alternators has mainly been
determined by the evolution of sophisticated cooling techniques.
Other technological breakthroughs, such as better materials, and
novel windings have also played a major part in modifying the
design of early machines.
As regards speed, low-speed generators are always bigger than
high-speed machines of equal power. Slow-speed bigness
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simplifies the cooling problem; a good air-cooling system,
completed with a heat exchanger, usually suffices. For example,
the large, slow-speed 500 MVA, 200 r/min synchronous generators
installed in a typical hydropower plant are air-cooled whereas the
much smaller high-speed 500 MVA, 1800 r/min units installed in a
steam plant have to be hydrogen-cooled.
5.9 No-Load Saturation Curve
Fig.5.3 shows a 2-pole synchronous generator operating at
no-load. It is driven at constant speed by a turbine. The leads from
the 3-phase, wye-connected stator are brought out to terminals A,
B, C, N, and a variable exciting current IX produces the flux in the
air gap.
Fig.5.3 Generator operation at no-load.
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Let us gradually increase the exciting current while observing
the ac voltage Eo between terminal A, say, and the neutral N. For
small values of IX, the voltage increases in direct proportion to the
exciting current. However, as the iron begins to saturate, the
voltage rises much less for the same increase in IX. If we plot the
curve of Eo versus IX, we obtain the no-load saturation curve of
the synchronous generator.
Fig.5.4 shows the actual no-load saturation curve of a 36 MW,
3-phase generator having a nominal voltage of 12 kV (line to
neutral). Up to about 9 kV, the voltage increases in proportion to
the current, but then the iron begins to saturate. Thus, an exciting
current of 100 A produces an output of 12 kV, but if the current is
doubled, the voltage rises only to 15 kV.
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Fig.5.5 is a schematic diagram of the generator showing the
revolving rotor and the three phases on the stator.
Fig.5.5 Electric circuit representing the generator of Fig.5.3
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5.10 Synchronous Reactance Equivalent Circuit Of
An Ac Generator
Consider a 3 phase synchronous generator having terminals A,
B, C feeding a balanced 3 phase load. The generator is driven by
a turbine, and is excited by a DC current IX. The machine and its
load are both connected in wye, yielding the circuit of Fig.5.6.
Although neutrals N1 and N2 are not connected, they are at the same
potential because the load is balanced. Consequently, we could
connect them together (as indicated by the short dash line) without
affecting the behavior of the voltages or currents in the circuit.
The field carries an exciting current which produces a flux φ . As
the field revolves, the flux induces in the stator three equal voltages
Eo that are 120 degrees out of phase (Fig.5.7). Each phase of the
stator winding possesses a resistance R and a certain inductance L.
Because this
Fig.5.6 Electric circuit representing the alternator connected
with 3-phase load.
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Fig.5.7 Voltage and impedances in a 3-phase generator and its
connected load
The synchronous reactance of a generator is an internal
impedance, just like its internal resistance R. The impedance is
there, but it can neither be seen nor touched. The value of X S is
typically 10 to 100 times greater than R; consequently, we can
always neglect the resistance, unless we are interested in efficiency
or heating effects.
We can simplify the schematic diagram of Fig.5.7 by showing
only one phase of the stator. In effect, the two other phases are
identical, except that their respective voltages (and currents) are
out of phase by 120 degrees. Furthermore, if we neglect the
resistance of the windings, we obtain the very simple circuit of
Fig.5.8. A synchronous generator can therefore be represented by
an equivalent circuit composed of an induced voltage Eo in series
with a reactance X S .
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269
Fig.5.8 Equivalent circuit of a 3-phase generator, showing only
one phase.
In this circuit the exciting current IX produces the flux
φ which
induces the internal voltage Eo . For a given synchronous
reactance, the voltage E at the terminals of the generator depends
upon Eo and the load Z. Note that Eo and E are line-to-neutral
voltages and I is the line current.
Example 5.2 A 3-phase synchronous generator produces an
open-circuit line voltage of 6928 V when the do exciting current is
50 A. The ac terminals are then short-circuited, and the three line
currents are found to be 800 A.
a. Calculate the synchronous reactance per phase.
b. Calculate the terminal voltage if three 12 Ω resistors are
connected in wye across the terminals.
Solution:
a. The line-to-neutral induced voltage is
Eo = E L / 3 = 6928 / 3 = 4000V
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When the terminals are short-circuited, the only impedance
limiting the current flow is that due to the synchronous reactance.
Consequently,
X S = Eo / I = 4000 / 80 = 5Ω
The synchronous reactance per phase is therefore 5 fl.
b. The equivalent circuit per phase is shown in Fig.5.9a.
The impedance of the circuit is:
Z = R 2 + X S2 = 12 2 + 52 = 13 Ω
The current is :
I = Eo / Z = 4000 / 13 = 308 A
The voltage across the load resistor is
E = IR = 308 *12 = 3696 V
The line voltage under load is:
E L = 3E = 3 * 3696 = 6402 V
The schematic diagram of Fig.5.9b helps us visualize what is
happening in the actual circuit.
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Fig.5.9. See Example 5.2. b. Actual line voltages and currents.
5.11 Synchronous Generator Under Load
The behavior of a synchronous generator depends upon the type
of load it has to supply. There are many types of loads, but they
can all be reduced to two basic categories:
1. -Isolated loads, supplied by a single generator
2.-Consider a 3-phase generator that supplies power to a load
having a lagging power factor. Fig.5.10 represents the equivalent
circuit for one phase. In order to construct the phasor diagram for
this circuit, we list the following facts:
Fig.5.10 Equivalent circuit of a generator under load.
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1. Current I lags behind terminal voltage E by an angle θ .
2. Cosine θ = power factor of the load.
3. Voltage EX across the synchronous reactance leads current I by
90 degrees. It is given by the expression E x = jIX S .
4. Voltage Eo generated by the flux
φ is equal to the phasor sum
of E plus EX.
5. Both Eo and EX are voltages that exist inside the synchronous
generator windings and cannot be measured.
6. Flux φ is that produced by the DC exciting current I x .
The resulting phasor diagram is given in Fig.5.11. Note that Eo
leads E by
δ degrees. Furthermore, the internally-generated
voltage Eo is greater than the terminal voltage, as we would
expect.
In some cases the load is somewhat capacitive, so that current I
leads the terminal voltage by an angle
θ . What effect does this
have on the phasor diagram? The answer is found in Fig.5.11. The
voltage E X across the synchronous reactance is still 90 degrees
ahead of the current. Furthermore, EO is again equal to the phasor
sum of E and E X . However, the terminal voltage is now greater
than the induced voltage EO , which is a very surprising result. In
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273
effect, the inductive reactance X S enters into partial resonance
with the capacitive reactance of the load. Although it may appear
we are getting something for nothing, the higher terminal voltage
does not yield any more power.
If the load is entirely capacitive, a very high terminal voltage can
be produced with a small exciting current. However, in later
chapters, we will see that such under-excitation is undesirable.
Fig.5.11 Phasor diagram for a lagging power factor load.
Fig5.12 Phasor diagram for a leading power factor load.
Example 5.3 A 36 MVA, 20.8 kV, 3-phase alternator has a synchronous reactance of 9Ω and a nominal current of 1 kA. The
no-load saturation curve giving the relationship between Eo and I,
is given in Fig.5.5. If the excitation is adjusted so that the terminal
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voltage remains fixed at 21 kV, calculate the exciting current
required and draw the phasor diagram for the following
conditions:
a. No-load
b. Resistive load of 36 MW
c. Capacitive load of 12 Mvar
Solution:
We shall immediately simplify the circuit to show only one phase.
The line-to-neutral terminal voltage for all cases is fixed at
E = 20.8 / 3 = 12 kV
a.
At no-load there is no voltage drop in the synchronous
reactance; consequently,
Eo = E = 12 kV
The exciting current is :
I x = 100 A
(See Fig.5.5)
In no load E and Eo will be in phase
With a resistive load of 36 MW:
b. The power per phase is
P=36/3=12 MW
The full load line current is
I = P / E = 12 *106 / 12 000 = 1000 A
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275
The current is in phase with the terminal voltage.
The voltage across X S is:
E x = jX S = j1000 * 9 = 9 kV ∠90o
The voltage Eo generated by Ix is equal to the phasor sum of E
and E x Referring to the phasor diagram, its value is given by:
Eo = E 2 + E x2 = 12 2 + 9 2 = 15 kV
The required exciting current is
I x = 200 A
(See Fig.5.5)
The phasor diagram is given in Fig.5.13.
Fig.5.13 Phasor diagram with a unity power factor load.
With a capacitive load of 12 Mvar:
c. The reactive power per phase is
Q=12/3=4 Mvar.
The line current is
I = Q / E = 4 *10 6 / 12000 = 333 A
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The voltage across XS is
E x = jX S = j 333 * 9 = 3 ∠90o kV
As before E x leads I by 90 degrees. (Fig.5.14)
Fig.5.14 Phasor diagram with a capacitive load.
The voltage Eo generated by Ix is equal to the phasor sum of Ex
and E.
Eo = E + E x = 12 + (− 3) = 9kV
The corresponding exciting current is
I x = 70 A
(See Fig.5.5)
Note that Eo is again less than the terminal voltage E.
The phasor diagram for this capacitive load is given in Fig.5.14.
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