DC Generator

1
DC Generator
Types of Generators Generators are usually
classified according to the way in which their
fields are excited. The classification of DC
generator is shown as follows
2
Generators may be divided into Separately-excited
generator Separately-excited generators are
those whose field magnets are energized from an
independent external source of DC current. It is
shown diagrammatically in Fig. 26.41 ( or
equivalent circuit).
Self-excited generator Self-excited generators
are those whose field magnets are energized by
the current produced by the generators
themselves. Due to the residual magnetism, there
is always present some flux in the poles. When
the armature is rotated, some e.m.f. and hence
some induced current is produced which is partly
or fully passed through the field coils thereby
strengthening the residual pole flux.
3
There are three types of self-exited generators
named according to the manner in which their
field coils (or winding) are connected to the
armature.
(a) Shunt Wound Generator The field windings are
connected across or in parallel with the armature
conductors and have the full voltage of the
generator applied across them. It is shown
diagrammatically in Fig. 26.42.
(b) Series Wound Generator In this case, the
field windings are joined in series with the
armature conductors. It is shown diagrammatically
in Fig. 26.43.
4
Compound Wound Generator It is a combination of
a few series and as a few shunt windings. In a
compound generator shunt field is stronger than
the series field. Compound wound generators are
two types
(a) Short-Shunt where shunt field and armature
are connected in parallel and that connection is
in series with the series field.
(b) Long-Shunt where series filed and armature
are connected in series and that connection is in
parallel with the shunt field.
5
When series field aids the shunt the shunt field,
the generator is said to be commutative-compound.
On the other hand, if series field opposes the
shunt field, the generator is said to be,
differentially compound.
Depending on load characteristics and the
relative additional aiding flux produced by the
series field, the cumulative compound generator
(whether long-shunt or short-shunt) are three
types. These types are called (i) over-compound,
(2) flat-compound, and (iii) under-compound.
Most commercial compound dc dynamos are normally
supplied by the manufacturer as over-compound
machine. The degree of compounding (over, flat,
and under) may be adjusted by means of diverter
that shunts the series field.
6
Diverter A diverter is a variable resistance
shunting the series field of compound generator
to adjust the degree of compounding to produce a
desired voltage regulation. A diverter is used
to control and produce a sufficient voltage rise
at the generator to compensate for the voltage
drop in the lines at full load.
The following figures show the connection of
diverter in the long-shunt and short-shunt
cumulative compound generator.
7
Brush Contact Drop
In the voltage drop over the brush contact
resistance when current passes from commutator
segments to brushes and finally to the external
load. Its value depends on the amount of current
and the value of contact resistance. This drop is
usually small and includes brushes of both
polarities. However, in practice, the brush
contact drop is assumed to have following
constant values for all loads. 0.5 V for
metal-graphite brushes. 2.0 V for carbon brushes.
8
Generated E.M.F. or E. M. F. Equation of a
Generator
Let, ?flux/pole in weber, Z total number of
armature conductors or Z No. of slots ? No. of
conductors/slot, P No. of generator poles, A
No. of parallel paths in armature, For lap
winding, AmP, For wave winding 2m, m The
multiplicity (such m3 for triplex winding) N
armature rotation in revolutions per minute
(rpm) E e.m.f. induced in any parallel path in
armature Eg e.m.f generated in any one of the
parallel paths i.e. E.
9
No. of revolution/second N/60 So, time for one
revolution, dt60/N second Hence, according to
Faradays Law of Electromagnetic Induction,
No. of conductors (in series) per parallel path
Z/A
For a simplex lap-wound generator m1 and AmPP
then
10
For a duplex lap-wound generator m2 and AmP2P
then
For a triplex lap-wound generator m3 and
AmP3P then
For a simplex wave-wound generator m1 and
A2m2 then
For a duplex wave-wound generator m2 and A2m4
then
For a triplex wave-wound generator m3 and
A2m6 then
11
For a given DC machine, Z, P, and A are
constant. Hence, putting KaZP/A, we get,
It is seen from the above equation that the
generated emf is directly proportional to the
flux (?) and the speed (N).
12
Example 26.3 A shunt generators delivers 450 A at
230 V and the resistance of the shunt field and
armature are 50? and 0.03? respectively.
Calculate the generated e.m.f..
Solution Generator circuit is shown in Fig.
26.46.
Given, Vt230V, IL 450A, Rsh50?, Ra0.03?.
Eg? Current through shunt field winding is
Ish230/504.6 A. Load current IL 450 A
So, armature current, IaILIsh 4504.6454.6
A Armature voltage drop, IaRa 454.6?0.0313.6
V Now, e.m.f. generated in the armature Eg
terminal voltage (Vt) armature drop (IaRa)
23013.6243.6 V
13
Example 26.4 A long-shunt compound generator
delivers a load current of 50A at 500V and has
armature, series field and shunt field
resistances of 0.05?, 0.03? and 250?
respectively. Calculate the generated voltage and
the armature current. Allow 1 V per brush for
contact drop.
Solution Generator circuit is shown in Fig.
26.47.
Given, Vt500V, IL 50A, Ra0.05?, Rse0.03?,
Rsh250? and total brush drop 2?12 V. Eg? and
Ia?. Current through shunt field winding is
Ish500/2502 A. Current through the armature and
series field is Ia50252 A
Voltage drop on armature and series field
winding
52(0.050.03)4.16 V Voltage drop
at brushes 2?1 2 V Now, Eg Vt(IaRa Series
drop) brush drop5004.162506.16 V
14
Example 26.5 A short-shunt compound generator
delivers a load current of 30A at 220 V, and has
armature, series field and shunt field
resistances of 0.05 ohm, 0.3 ohm and 200 ohm
respectively. Calculate the induced e.m.f. and
the armature current. Allow 1.0 V per brush for
contact drop.
Solution Generator circuit is shown in Fig.
26.48.
Given, Vt220V, IL 30A, Ra0.05 ohm, Rse0.3
ohm, Rsh200 ohm and total brush drop 2?12 V.
Eg? and Ia?. Voltage drop in series winding
30?0.39V Voltage
drop across shunt winding
2209229 V
Ish229/2001.145 A Ia 301.14531.145
A IaRa 31.145 ? 0.05 1.56 V Brush drop
2?12 V Now, Eg VtIaRa series drop brush
drop Eg 220 1.5692 232.56 V
15
Example 26.6 In a long-shunt compound generator,
the terminal voltage is 230 V when generator
delivers 150 A. Determine (i) induced e.m.f.,
(ii) total power generated, and (iii)
distribution of this power. Given that shunt
field, series field, divertor and armature
resistances are 92 ohm, 0.015 ohm, 0.03 ohm and
0.032 ohm respectively.
Solution Generator circuit is shown in Fig.
26.49.
Given, Vt230V, IL 150A, Rsh92 ohm, Rse0.015
ohm, Rd0.03 ohm, Ra0.032 ohm. Eg? Total power
generated? and Distribution of the geneated
power?. Ish 230/92 2.5 A Ia1502.5152.5
A Since series field resistance and divertor
resistance are in parallel their combined
resistance is 0.03?0.015/(0.030.015) 0.01
ohm. Total resistance is 0.0320.010.042
ohm Voltage drop 152.5?0.0426.4 V (i) voltage
generated by armature, Eg 2306.4236.4 V (ii)
total power generated in armature
EgIa236.4?152.536,051 W
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(iii) Total loss power lost in armature (Ia2Ra)
power lost in series field and divertor
(152.52?0.01) Power dissipated in shunt winding
( VtIsh) Power delivered to load
(230?150) 152.52?0.032152.52?0.01230?0.01230
?150 36,051 W
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Example 26.7 The following information is given
for a 300 kW, 600 V, long-shunt compound
generator, the shunt field resistance75?,
armature resistance including brush resistance
0.03?, commutating field winding resistance
0.011?, series field resistance 0.012?,
divertor resistance 0.036 ?. When the machine is
delivering full load, calculate the voltage and
power generated by the armature.
Solution Generator circuit is shown in Fig.
26.50.
Given, Vt600V, Output power VtIL 300kW,
Rsh75?, Ra0.03 ohm, Rcom0.011 ohm, Rse0.012?,
Rd0.036 ?,. Eg? Power generated Eg Ia ?
Output current, IL 300000W/600V500A Ish
600/758A Ia5008508 A Since the series field
resistance and divertor resistance are in
parallel their combined resistance is
(0.012?0.036)/0.0480.009 ?. Total armature
circuit resistance 0.030.0110.0090.05
? Voltage drop 508?0.05 25.4 V Voltage
generated by armature 600 25.4625.4 V Power
generated 625.4?508317,700 W 317.7 kW.
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Example 26.8 A four-pole generator having
wave-wound armature winding has 51 slots, each
slot containing 20 conductors. What will be the
voltage generated in the machine when driven at
1500 rpm assuming the flux per pole to be 7.0 mWb
and what will be the generated voltage if the
generator is the triplex lap wonding?
Solution Given, ?07.0 mWb7?10-3 Wb
Z51?201020 A2m2 ?12 P4 N1500 rpm
We know for the lap winding that, AmP3?412
Thus for triplex lap-winding generator
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Example 26.11 An 8 pole DC shunt generator with
778 armature conductors and running at 500 rpm.
Supplies a load 12.5 ohm resistance at terminal
of 250 V. The armature resistance is 0.24? and
the field resistance is 250?. Find the armature
current, the induced e.m.f. and the flux per pole
for (a) wave-connected winding and (b) Triplex
lap-connected winding.
Solution Generator circuit is shown in Fig.
26.53.
Given, P8, Z778, N 500 rpm, Vt250V, Ra0.24
ohm, Rsh250?, Ia?, Eg? ?? Load current, IL
Vt/Ra 250/12.5 20A Shunt current, Ish
Vt/Rsh 250/250 1 A. Armature current,
IaILIsh20121 A Induced e.m.f.
250(21?0.24)255.04 V
(a) For wave-connected winding, A2m 2 (m1),
Thus
(b) For triplex lap-connected winding, AmP
3?824, Thus
20
Example 26.12 A separately excited generator,
when running 1000 rpm supplied 200 A at 125 V.
What will the load current when the speed drops
to 800 rpm if If (field current) is unchanged?
Given that the armature resistance 0.04? and
brushes drop 2V.
Solution Generator circuit is shown in Fig.
26.54.
Given, N1 1000 rpm, Vt125V, IL200A, Ra0.04?,
Brushes drop 2V, Eg2 (at 800 rpm)? The load
resistance, RL 125/2000.625? Eg1(at 1000
rpm)125200?0.042 135 V N1 1000 rpm
21
Thus, Eg2(at 800 rpm) 135?800/1000108 V If IL2
is the new load current, then terminal voltage is
given by Vt2 Eg-(IL2Ra Brushes drop)108-0.04
IL2-2 106-0.04 IL2. So, IL2 Vt2 /RL(106-0.04
IL2)/0.625 0.625IL2106-0.04 IL2
(0.6250.04)IL2106 0.665IL2106
IL2106/0.665 IL2 159.398 A
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Backward Force or Magnetic Drag
In the case of DC generator as shown in Fig.
29.2, it is seen that the flux due the armature
current carrying conductor a force is
produced. This force is in a direction opposite
to that of armature rotation. Hence, it is known
as backward force or magnetic drag on the
conductors. It is against this drag action on all
armature conductors that the prime mover has to
work. The work done in overcoming this opposition
is converted into electrical energy.
23
Armature reaction
The current in the armature produces a flux. So,
the interaction between this flux and the main
field flux is called armature reaction. The
armature magnetic field has two effects (i) it
demagnetizes or weakens the main flux which leads
to reduced generated voltage, and (ii) it
cross-magnetizes, which leads to the sparking at
the brushes, or distorts it. When there is no
load connected to the generator, the current in
the armature conductors is zero. Under these
conditions there is only one magnetic field in
the generator, and that field is produced by the
main-field poles of the generator.
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The main field is represented by an arrow, which
indicates the direction of the magnetic flux from
the north pole to south pole as shown in Fig. 6.1.
A load is now connected to the generator, and of
course current flow exists.
The current to the load is the current in the
armature conductors and is equal to the sum of
the currents from the parallel paths in the
armature.
Consider Fig. 6.2, which allows the armature
rotating in the magnetic field and the resulting
armature current when a load is connected to the
generator.
When the current flows through a conductor a
magnetic field is set up around the conductor as
indicated in Fig. 6.2.
25
The flux from the conductors on the left side of
the armature and the flux from the conductors on
the right side of the armature cause a resultant
flux in the center of the armature that is
downward in direction.
This resultant flux can be represented by an
arrow as indicated, noting that the arrow passes
through both top and bottom brushes.
There are now two fluxes inside the generator,
one produced by the main field poles of the
generator and the other by the current in the
armature conductors.
These two fluxes now combine to form a new
resultant flux as shown in Fig. 6.3.
This new resultant flux is not in the same
direction as the original main field flux but
runs from the tip of one of the poles, across the
armature, to the tip of the other pole.
26
The armature conductors are now cutting this new
resultant flux that is not the same direction as
the main field flux which conductors were
originally cutting.
The brushes are supposed to be located at the
point of minimum flux, which of courses at right
angles to the direction of the flux.
Since the brushes were at right angles to the
main-field flux, they certainly cannot be at
right angles to the new resultant flux. With the
brushes in their present location they will be
short-circuiting coils in which there is a
voltage induced, thereby producing sparking at
the brushes, undue brushes wear, and other
unfavorable conditions.
27
Effect of Brush Shifting
If the brushes are no longer at the points of
minimum flux, or magnetic neutral, as the points
of minimum flux are known, it might appear to be
a simple solution to shift the brushes until they
do fall on the magnetic neutral, and then the
brushes will once again be at the points of
minimum flux. It has just been seen that the
direction of resultant flux depends upon both the
flux from the main-field poles and the flux
produces by the current in the armature
conductors. The flux from the main-field poles is
fairly constant and will remain constant even
though the generator is supplying current to a
load. If the load current is small, the armature
current will be small and the flux produced by
the armature conductors will be small hence the
shift in the resultant flux will be small as
compared with the main-field flux. The greater
the current delivered by the generator, the
greater the current in the armature conductors,
and therefore the greater the flux produced by
the armature conductors, ending with greater
shift in the direction of the resultant
flux. Hence if the brushes are to be moved to a
new neutral position, the new position will
depend upon the load. With the load on a
generator constantly varying, it would be
impossible to preset the position of the brushes
and expect satisfactory result.
28
Fig. 6.4 shows a new position of brushes at the
minimum flux point.
Referring Fig. 6.4, it is seen by Flemings
right-hand rule that the conductors under the
north pole carry current away from the observer
and the conductors under the south pole carry
current toward the observer.
The flux from these conductors is indicated on
the diagram, and of course the combined flux from
all the conductors is still in a direction from
the top brush to the bottom brush. But the flux
from the armature is not at right angles to the
flux from thee main-field poles.
The effects of the armature flux in the new
position of brushes is illustrated in Fig. 6.5.
29
It is seen from Fig. 6.5 that there have two
components of armature flux that are at right
angles to each other.
One component is at right angles to the main
field, and because this component crosses the
main-field flux, it is known as cross-magnetizing
component of the armature flux. The second
component is in the same plane as the main-field
flux. The direction of this component is opposite
to the direction of the main-field flux, with the
result that it tends to reduce the effect of the
main-field flux. This component of the armature
flux is known as the demagnetizing component of
the armature flux.
It now appears that the shifting of the brushes
has not improved the situation. In fact, it seems
to have become worse.
30
Before the brushes were shifted, the armature
flux was at right angles to the main-field flux
and therefore produced only a cross-magnetizing
field. With the brushes shifted to the new
position, there is still a cross-magnetizing
field, some what reduced in magnitude, but in
addition there is now a demagnetizing field which
tends to reduce the main-field flux, resulting in
a lower generated voltage. This demagnetizing
component was obtained only after the brushes
were shifted, and the brushes were shifted
because of the change in direction of the
resultant flux, which was due to the armature
conductors carrying current. The application of
some means to prevent the shift of the resultant
flux would eliminate the necessity of shifting
the brushes, and hence no demagnetizing field
would be produced.
31
Demagnetizing and Cross-magnetizing Conductors
The exact conductors which produce these
distorting and demagnetizing effects are shown in
Fig. 27.6 where the brush axis has been given a
forward lead of ? so as to lie along the new
position of magnetic neutral axis (M. N. A).
All conductors lying with in the angles
AOCBOD2? at the top and bottom of the armature,
are carrying current in such a direction as to
send the flux through the armature from right to
left. It is these conductors which act in direct
opposition to the main-field and are hence called
the demagnetizing armature conductors.
Now consider the remaining armature conductors
lying between angles AOD and COB as shown in Fig.
27.7.
These conductors carry current in such a
direction as to produce a combined flux pointing
at right angles to the main-field flux. This
results in distortion of the main field. Hence,
these conductors are known as cross-magnetizing
conductors and constitute distorting ampere
conductors.
32
Without shifting the brush the armature effect
can be minimized by using the following
methods (a) High-Reluctance Pole Tips (b)
Horizontal Slots in Main-Field Pole (c)
Compensating Windings
High-Reluctance Pole Tips
It can be seen that the flux from the current in
the armature conductors causes the main-field
flux to shift from the center of the main-field
poles to the tips of the poles. Applying the
knowledge that most of the flux flows the path of
least reluctance, poles are designed where the
reluctance at the ends of the poles is greater
than the reluctance at the center.
33
The variation of reluctance is obtained by
constructing the poles that the greater the
distance from the center of the pole, the greater
the air gap between the pole and the armature.
The greater the air gap, the greater the
reluctance.
With no current in the armature conductors, the
flux will be concentrated at the center of the
pole, and when current flows in the armature
conductors, the flux will tend to shift the end
of the pole.
The air gap will offer an increase in reluctance
to the flux as it moves from the center of the
pole, thereby tending to keep the flux in the
same original position.
The shape of this pole is shown in Fig. 6.6.
The reluctance of the center of pole can be
reduced by using lamination.
34
Horizontal Slots in Main-Field Pole
The high reluctance pole-tip construction reduced
the effects of armature reaction by not allowing
the flux to shift.
The shift in the resultant flux was caused by the
armature flux.
If the armature flux could be reduced to a
negligible value, then its cross-magnetizing
effect upon the main-field flux would be small
and the brushes would not have to be shifted.
Fig. 6.8 shows that part of the path of the
magnetic flux is through the field poles.
By cutting horizontal slots in the poles, several
air gaps are introduced in the path of the flux.
These slots increase the reluctance to the
armature flux while having very little effect on
the main-field flux.
The armature is materially reduced, and the
brushes need not be shifted.
35
Compensating Windings
The compensating windings are placed in the pole
faces of the filed pole can run parallel to the
armature conductors. A connection is made from
one of the brushes to one end of this winding, so
that current from armature must first pass
through the winding before going to the load. The
direction of current through the winding is
opposite to that of the current in the armature
conductors under that pole.
The location of the compensating winding and of
the connection to the winding is shown in Fig.
27.8.
Figure 27.8 indicates that the current in the
armature conductors located under the north pole
carry current away from the observer therefore
the direction of current flowing in that part of
the compensating winding situated in the north
pole is toward the observer.
Since the current in the compensating winding is
opposite in direction to the current in the
armature conductors, the flux produced by the
current in the compensating winding will be
opposite in direction to the armature flux. The
compensating flux, being opposite in direction to
the armature flux, tends to cancel the armature
flux. If the armature current increase, the
compensating current increases, so that the
armature flux is canceled for al load conditions.