4.1 GENERAL
This
chapter examines what goes on inside a generator. First we will examine the normal ‘ohmic’
voltage drop due to load current in the main windings. Also we shall consider how the emf which is
generated in the armature by the rotating field system undergoes certain changes. They cause it to be less than the emf due to
the pure field system and so reduce still further the voltage that appears at
the generator terminals.
Take
first the simpler case of a d.c.
generator.
4.2 D.C. GENERATOR
In a d.c.
generator the current which flows from the machine to the external load and
back again also flows through the internal resistance of the generator, as
shown in Figure 4.1.
FIGURE 4.1
D.C. GENERATOR - INTERNAL
RESISTANCE
If the
excitation causes an emf E in the
generator armature, the open-circuit voltage V0 will be equal to E. But if a load current I flows, it flows also through the
generator’s internal resistance r causing
an internal voltage drop equal to I.r. When the load is applied therefore the
terminal voltage drops from V0
to a lower value V, where
V
= V0 - I.r ….(i)
The
reduction of terminal voltage (V0 -
V) is the ‘regulation’ of the
generator under the stated load conditions.
Its value is given by the drop of voltmeter reading when the load is
applied (assuming no change in the exciting field).
Since V0 = E, equation (i) can also
be written
V = E - I.r
or E = V + I.r ….(ii)
4.3 A.C. GENERATOR
A
similar situation occurs with a.c. generators, but the effect is more
complicated, as shown
in Figure 4.2.
FIGURE 4.2
A.C.
GENERATOR - INTERNAL IMPEDANCE
In this
case the ‘load’ may be an impedance which draws both active and reactive
current, and the internal resistance of the d.c. generator of Figure 4.1 is
replaced by a generator internal impedance z,
which consists of both resistance r and
reactance x.
The same
principles apply as for the d.c.
generator - namely that the terminal voltage V0 is reduced to a lower value V by the internal ‘voltage drop’ caused by the load current flowing
through the internal impedance z.
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but
these are now all vector quantities.
The
numerical difference between no-load terminal voltage and the terminal voltage
on load, namely V0 - V, is still the regulation of the
generator under the stated load condition.
Its value is given by the drop of voltmeter reading when the load is
applied (assuming no change in the excitation).
When
considering what goes on inside a generator it would be logical to take first
the generated emf E; then to examine
the voltage drop due to the load current flowing through the generator’s
internal impedance and finally to arrive at the terminal voltage V. However,
because the emf of a machine cannot actually be measured (except on open
circuit), it would be more convenient to start with the terminal voltage
V - for that alone is what the
external load sees - and then to work backwards through the internal voltage
drop, adding it to V, to arrive at
the emf E needed to produce the
required terminal voltage V. Expressed as an equation (similar to
equation (ii)):
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 |
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E
= V + I.z (vectorially) ….(iii)
This
gives a measure of the excitation needed to produce a given voltage at the
terminals for any given load current, and, in particular, to find the maximum
excitation required from the AVR and exciter to maintain full rated terminal
voltage at maximum rated load.
It is
known that, if an a.c. voltage V is
applied across a resistance R, a
current equal to V/R (by Ohm’s Law)
will flow through it and will be in phase with V. Similarly, if a voltage V is applied across a pure reactance X,
a current equal to V/X (by Ohm’s Law
for a.c.) will flow
through it and will lag 90o on V.
FIGURE 4.3
COMMON CURRENT IN
RESISTOR AND REACTOR
If,
instead of starting with a common voltage V,
a common current I had been
flowing through both a resistance R and
a pure reactance X, as shown in Figure 4.3, then the voltage VR
equal to I.R and
developed across R, will be in phase
with I.
The voltage Vx, equal to I.X
and developed across X, will lead the current I, since the current must lag on the voltage. This is shown in Figure 4.3, where the
voltage vectors are in blue, and the currents in red. The voltages VR and VX
are respectively the voltage drops in R and
X due to the common current I
flowing through them.
Consider now the voltage drops due to a load current I flowing through the internal resistance r and reactance x of a generator armature.
FIGURE 4.4
INTERNAL VOLTAGE
DROPS
In
Figure 4.4 a generator is supplying an external load Z (resistance R and reactance
X) with a current I. The generator is being excited to an emf E,
and it has an internal resistance r and
reactance x. The terminal voltage
while load current is flowing is V.
0
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The external impedance Z causes the load current / to lag by an angle φ on the terminal voltage, cos φ being the power factor of the load. The load current I is then flowing through both r and x (x is assumed typically to be larger than r).
FIGURE 4.5
TERMINAL
VOLTAGE AND EMF IN A GENERATOR
As already explained, the voltage drop vr due to current I flowing through r is I.r and is in phase
with I; it is represented by the
vector OP. The voltage drop vx
due to current I flowing through x is
I.x and leads I by 90o. It is
represented by the vector OQ. |
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The total voltage drop I.z in the generator is the vector sum of vx and vr namely (I.x + I.r)
=OQ + OP.
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Since E = V+ I.z (equation (iii)),
E
= V + OQ + OP
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|
 |
=
V + CS
+ ST (Since CS = OQ and ST = OP)
 |
and is
so shown in Figure 4.5, the resultant vector E being in
red.
This
shows that, to achieve a terminal voltage V
on load, the emf must not only be greater than V numerically but must also in general lead on V in phase. This angle of
lead is called the ‘power angle’ or ‘load angle’ for that particular load and
is given the symbol λ. It represents the
advance in the rotor’s pole position as compared with its no-load
position. The power angle λ should not
be confused with the load’s phase angle φ.
It
should be noted that r and x are the
resistance and reactance (in ohms) per phase in a 3-phase system. This assumes a star connection, and V in the formulae is the phase voltage
(line voltage divided by √3).
Figure 4.5 gives the relationship between the emf E and the terminal
voltage V for a given load current.
It was evolved by starting with V and
evaluating the corresponding E, but
equally it shows that, given an emf E, at the stated load current the
terminal voltage will be V - that is,
it can be used either way.
For
example, if the terminal voltage has to be maintained at a steady value V at all loads, then the emf (and so the
excitation) can be determined for any stated load current, and in particular
for the maximum rated load. From this
the size of the excitation system can be deduced.
FIGURE 4.6
TERMINAL VOLTAGE AND
EMF IN A GENERATOR
AT CONSTANT NO-LOAD
EXCITATION
A
special case of this is at no load, where there is no load current I and therefore no internal voltage
drops I.r and I.x. The vector E
then lies on top of V and is equal to
it; that is to say, at no load the terminal voltage is identical with the emf,
as shown in Figure 4.6(a). If V0 is the terminal voltage at
no load and E0 the corresponding emf at no load,
then V0
= E0. This is the only
situation where an emf can be actually measured.
As
before, V and E are the phase
voltage and phase emf.
4.4 REGULATION
If the
no-load terminal voltage V0 is the system voltage (as it
normally will be), then E0 is also equal to the system
voltage. If, as load is applied, the emf
were maintained at its no-load value E0 (that is, if there
were no change in the excitation), then the situation would be as shown in
Figure 4.6(b), which is a repetition of Figure 4.5 but everywhere
reduced in scale. Since the magnitude E is unaltered from its no-load value E0,
the on-load terminal voltage vector V, the
current vector and both the volt-drop vectors will be reduced in proportion but
will retain their relative sizes and positions.
In
Figure 4.6(b), therefore, V0 is
the no-load terminal voltage and V the
terminal voltage when a stated load current is applied. The numerical difference between V0
and V is the drop of terminal
voltage, as measured by voltmeter, when the load is applied with no change in
the excitation; it is the regulation of the generator at that load.
The
regulation of a generator is usually expressed as a percentage of the system
voltage when the generator is supplying full rated load current at rated power
factor. Then the percentage regulation
is
at rated
output.
In
practice, of course, the situation of Figure 4.6(b) would be immediately
detected by the AVR, which senses only the terminal voltage. Finding V
reduced below the system voltage, it would raise the excitation and so the
emf E. This would increase the scale of Figure
4.6(b) again until it approached that of Figure 4.5 and the terminal voltage V
once more reached the system voltage. As
will be seen from Figure 4.5, the emf will then be much greater than
the system voltage.
4.5 PRACTICAL APPLICATION
The
foregoing is the full treatment for determining the emf and the regulation of a
generator, taking into account the power factor of the load current and both
the internal resistance and reactance of the machine.
FIGURE 4.7
DIAGRAM NEGLECTING INTERNAL RESISTANCE
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The vector CS, equal to I.x, is
then the principal element in determining the difference in magnitude between E and V, and so in determining the regulation.
In Figure 4.7 the angle between the vector CS
and the horizontal is also φ, so
that the numerical difference between E and
V is approximately CY, or CS sin φ. So
the difference between the emf and the terminal voltage is approximately I.x sin φ. But I.sin φ is the quadrature component (Iq) of the load current, so the regulation can be taken
as Iq.x.
YS, the other component of CS, namely CS cos φ
(= I.x cos φ), has virtually no
effect on the voltage difference between E
and V, but it does affect the
power angle λ. But I cos φ is the in-phase component (Ip) of the load current, so the power angle depends
mainly on the in-phase, or active power, component of the load current. The power angle is given approximately by
tan λ
|
=
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SY |
OY
|
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=
|
CS.cos φ |
|
OC +
CY
|
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=
|
I.x cos φ
|
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V + I.x sin φ
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||
=
|
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As before, x is the reactance (in ohms) per phase and V is the phase voltage.
To sum
up, with the approximations made (that r may
be neglected), the regulation or voltage drop within a loaded generator depends
principally on the quadrature component of the load current - that is, on the
kvar loading only - and on the generator reactance.
The
in-phase component of the load current - that is, the kW loading - affects
principally the power angle and has little effect on the voltage drop.
A
notable case of this occurs when starting large motors. On switching on, not only is the starting
current heavy but it is also drawn at a low power factor. In other words a very large reactive load is
thrown onto the system - a situation causing the maximum voltage drop in the
generators. This will cause a noticeable
dip as evidenced by the dimming of lights until the AVR has had time to take charge
and restore the voltage.
4.6 SYNCHRONOUS REACTANCE
There is
one further matter which has a very marked effect on the value of excitation
needed to produce the required emf. It
was shown in Figure 2.4 of Chapter 2 that, when a load current - particularly a
reactive load current - passes through the armature of a generator, the
magnetic field of that current opposes that of field poles and demagnetises
them; this is the phenomenon of ‘armature reaction’. Consequently, with this effect present, the net
field flux is reduced and, for a given exciting current, less emf is generated.
FIGURE 4.8
ARMATURE REACTION AND SYNCHRONOUS REACTANCE
All that
has been said regarding the relationship between the emf E and the terminal voltage V still
stands. Up until now the emf E has been considered to reflect exactly
the field exciting current, but, because of demagnetisation, this is no longer
true and the net flux is now less than that produced by the excitation. Consequently the emf E actually generated by the reduced flux is less than would occur
with the given exciting current alone.
This effect is shown in Figure 4.8(a).
Eex represents the
emf that would be generated by the exciter field alone, but E is the reduced emf due to the
demagnetisation of the field. This
results in a smaller V and a smaller I. The heavy reduction in V is as if the exciter emf Eex
had remained unaltered but the generator had had a much higher reactance Xd
than its real one x. This is
indicated by the blue vectors in Figure 4.8(a).
Indeed
the effect of armature reaction is taken into account by assuming just this, as
shown in Figure 4.8(b). The generator is
considered to have a larger, false reactance whose effect is to restore the
relationship between a new and much increased excitation emf Eex (not the actual E) and the terminal voltage to that of
Figure 4.7 by using a higher value of x.
This higher, false reactance is called the ‘synchronous reactance’ of
the generator. Its symbol is Xd. If used instead of the actual reactance x,
it gives the same relationship between the excitation emf and terminal voltage
as in Figure 4.7.
In fact,
taking this artificial synchronous reactance into account, which is normally
several times greater than the actual reactance x, many large generators
have excitation systems some two to three times that required for the no-load
value in order to maintain full terminal voltage at full rated output. Compare
the relative sizes of Eex
and V(= E0) in Figure
4.8(b). It also causes a larger power angle
λ.
The whole
concept of ‘synchronous reactance is merely a device for taking into account
the armature reaction of a generator when loaded. The voltage drop within a generator is caused
partly by the loss of emf due to this armature reaction, and partly by the
normal impedance drop due to load current in the windings. By using this wholly fictitious synchronous
reactance, the total drop is considered to be due to an increased impedance
drop alone. Its use enables currents and
voltages to be calculated by ordinary methods as if no demagnetisation were
taking place.
It is
always used in network ‘steady-state’ problems such as power flow and load
sharing. It is also used in generator design to establish what excitation
system will be needed - the sizes of the exciters and main field system, rotor
cooling, etc.
4.7 PERCENTAGE REACTANCE
Throughout
this chapter reactances per phase have been used directly (x or Xd)
and are measured in ohms. In practice an alternative method is generally used,
namely to express them as ‘percentage reactances’.
Instead
of using Ohm’s Law I = V/X
to determine currents, the generator is considered to be short-circuited
and the excitation reduced so that the short-circuit current just equals the
rated full-load current. The reduced
excitation is expressed as a fraction (or percentage) of the excitation at full
rated load; percentage is then the ‘percentage reactance’ of the generator.
For example, if the rated full-load current of a generator were 1 000A, and on short-circuit this current were achieved with
only 25% excitation, the generator would have a ‘percentage
reactance’ of 25%. If the full-load current is divided by the
percentage reactance (expressed as a fraction), it gives the current on
short-circuit with full excitation. Thus
the short-circuit current in this case 
If the generator is
rated 5 000kVA at full
load, then on short-circuit it will deliver 
or 20MVA.
If the generator is
rated 5 000kVA at full
load, then on short-circuit it will deliver or 20MVA.
This
method of expressing reactances is very useful, for it avoids having to deal
with actual voltages and currents for any particular generator. Most drawings give reactances only in
percentage form.
In
Chapter 9 percentage synchronous, transient and subtransient reactances are
given for a typical large generator, together with formulae for converting from
percentage to ohmic reactance and from ohmic to percentage. These conversions, however, require knowledge
of the rated voltage and current.
The very
large synchronous reactances which arise as explained in para. 4.6, result in
percentage synchronous reactances being well above 100% - typically nearer
200%. This causes the current of a
short-circuited generator to settle (in the 200% case) at only one-half the
normal full-load current - that is to say, the generator current is actually
reduced by a short circuit (assuming that the excitation does not change) - a
result which perhaps may be surprising to some.
A
consequence of this is that normal overcurrent protection of the generator may
actually fail to work on short circuit.
This is investigated further in the manual ‘Electrical Protection’.
4.8 DIRECT AXIS AND QUADRATURE AXIS REACTANCES
In the
foregoing discussion it was assumed that the machine concerned had a uniform
air gap and was thus of the cylindrical rotor type. All reactances therefore were ‘direct axis
It was
explained in Chapter 2 that salient-pole generators, which form the majority in
offshore and onshore installations, do not have a uniform air gap. In consequence their reactances are of two
types - direct axis and quadrature axis.
In
practice, omitting the quadrature-axis quantities from the calculation has
little effect on the magnitude of the excitation E needed to sustain a given load as
described in the foregoing paragraphs.
The quadrature-axis reactance does, however, have a considerable
influence on the power angle λ for that load.
Omitting it leads to a calculated angle smaller than actually occurs and
may present stability problems.
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