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1 - principles of electrical machines and power electronics p_c_sen

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Conse-
\-1 
100 
-- Terminal 
ill. 80 AVAR ,-c ...... k ~ with no 
~ armature 
"0 60 reaction 
~ Operating point Terminal 
*- 40 characteristic 
with armature 
reaction 
20 
% rated current 
FIGURE 4.29 Terminal characteristic of a sepa-
rately excited de generator. 
c 
r 
,,, 
148 chapter 4 DC Machines 
(a) 
-- --
--
--
(b) 
B 
(c) 
FIGURE 4.30 Armature reaction effects. 
quently, flux density under the pole increases in one half of the pole and 
decreases under the other half of the pole. If the increased flux density 
causes magnetic saturation, the net effect is a reduction of flux per pole. 
This is illustrated in Fig. 4.30c. 
To have a better appreciation of the mmf and flux density distribution in 
a de machine, consider the developed diagram of Fig. 4.3la. The armature 
mmf has a sawtooth waveform as shown in Fig. 4.31b. For the path shown 
by the dashed line, the net mmf produced by the armature current is zero 
because it encloses equal numbers of dot and cross currents. The armature 
mmf distribution is obtained by moving this dashed path and considering 
the dot and cross currents enclosed by the path. The flux density distribution 
p 
DC Generators 149 
produced by the armature mmf is also shown in Fig. 4.31b by a solid curve. 
Note that in the interpolar region (i.e., near the q-axis), this curve shows a 
dip. This is due to the large magnetic reluctance in this region. In Fig. 4.31c 
the flux density distributions caused by the field mmf, the armature mmf, 
and their resultant mmf are shown. Note that 
• Near one tip of a pole, the net flux density shows saturation effects 
(dashed portion). 
• The zero flux density region moves from the q-axis when armature cur-
rent flows. 
• If saturation occurs, the flux per pole decreases. This demagnetizing 
effect of armature current increases as the armature current increases. 
At no load Ua = 11 = 0) the terminal voltage is the same as the generated 
voltage (V10 = Ea0 ). As the load current flows, if the flux decreases because 
.------...., 
I I i J •i ~ J s! L [t_ 
m I m I m 
••• 0 8 ® ® .,1® ® ® ® 0 ••• • 1 •••• 0 ® ® ® ® L ______ _j 
(a) 
(b) 
(c) 
B (resultant)- effect of 
saturation 
FIGURE 4.31 MMF and flux density distribution. 
150 chapter 4 DC Machines 
FIGURE 4.32 Effect of armature re-
action. 
of armature reaction, the generated voltage will decrease (Eq. 4.21). The 
terminal voltage will further decrease because of the I.R. drop (Eq. 4.24). 
In Fig. 4.32, the generated voltage for an actual field current Ir(actuai) is Eao. When the load current I a flows the generated voltage is E. = Vt + laRa. If E. < E.0 , the flux has decreased (assuming the speed remains unchanged) because of armature reaction, although the actual field current / 1(actuaD in the field winding remains unchanged. In Fig. 4.32, the generated voltage E. is produced by an effective field current Ir(eff). The net effect of armature reac-
tion can therefore be considered as a reduction in the field current. The difference between the actual field current and effective field current can be considered as armature reaction in equivalent field current. Hence, 
/r(eff) = /!(actual) - /f(AR) (4.25) 
where Ir(AR) is the armature reaction in equivalent field current. 
Compensating Winding 
The armature mmf distorts the flux density distribution and also produces 
the demagnetizing effect known as armature reaction. The zero flux density 
region shifts from the q-axis because of armature mmf (Fig. 4.31), and this 
causes poor commutation leading to sparking (Section 4.3.5). Much of the 
rotor mmf can be neutralized by using a compensating winding, which is fitted in slots cut on the main pole faces. These pole face windings are so 
arranged that the mmf produced by currents flowing in these windings 
opposes the armature mmf. This is shown in the developed diagram of Fig. 4.33a. The compensating winding is connected in series with the armature 
winding so that its mmf is proportional to the armature mmf. Figure 4.33b 
shows a schematic diagram and Fig. 4.33c shows the stator of a de machine having compensating windings. These pole face windings are expensive. Therefore they are used only in large machines or in machines that are 
subjected to abrupt changes of armature current. The de motors used in 
steel rolling mills are large as well as subjected to rapid changes in speed 
and current. Such de machines are always provided with compensating 
windings. 
T 
I I Stator I I 
~~ 
000000000E!'JE!'J®E!'JE!'J$®0 
Armature 
(a) 
(c) 
DC Generators 151 
Compensating 
windings 
Armature 
Shunt windings 
n 
(b) 
FIGURE 4.33 Compensating winding. (a) Developed diagram. (b) Schematic dia-
gram. (c) Photograph. (Courtesy of General Electric Canada Inc.) 
EXAMPLE 4.2 
A 12 kW, 100 V, 1000 rpm de shunt generator has armature resistance 
R. = 0.1 !1, shunt field winding resistance Rrw = 80 !1, and N1 = 1200 turns per 
pole. The rated field current is 1 ampere. The magnetization characteristic at 
1000 rpm is shown in Fig. 4.24. 
The machine is operated as a separately excited de generator at 1000 rpm 
with rated field current. 
(a) Neglect the armature reaction effect. Determine the terminal voltage 
at full load. 
(b) Consider that armature reaction at full load is equivalent to 0.06 
field amperes. 
(i) Determine the full-load terminal voltage. 
(ii) Determine the field current required to make the terminal voltage 
V1 = 100 Vat full-load condition. 
( 
r 
.. 
152 chapter 4 DC Machines 
Note: In this book, to avoid confusion, rating data on a de machine is 
considered to apply to the armature, whether the machine is used as a 
generator or motor. 
Solution 
In a de shunt generator, the main field winding is the shunt field winding. 
Also, from the data on the machine, 
Rated armature voltage Ealrated = 100 V 
Rated armature power (or full load) = 12 kW 
Rated armature current (or full load) Ialrated = 12,0001100 = 120 A 
Rated speed = 1000 rpm 
Rated field current Irlrated = 1 A 
(a) 
= 100- 120 X 0.1 
= 88V 
(b) (i) From Eq. 4.25 
lr(eff) = 1 - 0.06 
= 0.94A 
From Fig. 4.24, at this field current 
(ii) 
Ea = 98 V 
= 98- 120 X 0.1 
= 86V 
= 100 + 120 X 0.1 
= 112V 
From Fig. 4.24, the effective field current required is 
/f(eff) = 1.4 A 
From Eq. 4.25, 
/f(actual) = 1.4 + 0.06 
= 1.46 A • 
+ 
DC Generators 15 3 
FIGURE 4.34 Schematic of a shunt or self-excited 
de machine. 
4.3.2 SHUNT (SELF -EXCITED) GENERATOR 
In the shunt or self-excited generator the field is connected across the arma-
ture so that the armature voltage can supply the field current. Under certain 
conditions, to be discussed here, this generator will build up a desired 
terminal voltage. 
The circuit for the shunt generator under no-load conditions is shown in 
Fig. 4.34. If the machine is to operate as a self-excited generator, some 
residual magnetism must exist in the magnetic circuit of the generator. 
Figure 4.35 shows the magnetization curve of the de machine. Also shown 
in this figure is the field resistance line, which is a plot of Rrlr versus Ir. A 
simplistic explanation of the voltage buildup process in the self-excited de 
generator is as follows. 
Assume that the field circuit is initially disconnected from the armature 
circuit and the armature is driven at a certain speed. A small voltage, Ear, 
will appear across the armature terminals because of the residual magnetism 
in the machine. If the switch SW is now closed (Fig. 4.34) and the field 
circuit is connected to the armature circuit, a current will flow in the field 
winding. If the mmf of this field current aids the residual magnetism, eventu-
ally a current In will flow in the field circuit. The buildup of this current will 
depend on the time constant of the field circuit. With In flowing in the field 
circuit, the generated voltage is £.1-from the magnetization curve-but the 
Ea2 
Eal 
E., 
0 
Field resistance line 
~~~------------------------_.~