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1 - principles of electrical machines and power electronics p_c_sen

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one negative, in a wave winding (and this mini-
mum number is often used in small machines), in large machines more 
brush positions are used in order to decrease the current density in the 
brushes. 
In wave windings, the number of parallel paths (a) is always two and there 
may be two or more brush positions. 
Also note from Figs. 4.17a and 4.18a that when the coil ends pass the 
brushes, the current through the coil reverses. This process is known as 
commutation, and it happens when the coil sides are in the interpolar region. 
F 
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DC Machines 13 7 
During the time when two adjacent commutator bars make contact with a 
brush, one coil is shorted by the brush in the lap winding and p/2 coils in 
the wave winding. The effects of these short-circuited coils, undergoing 
commutation, will be discussed later. 
In small de motors, the armature is machine wound by putting the wire 
into the slots one turn at a time. In larger motors, the armature winding is 
composed of prefabricated coils that are placed in the slots. 
Because many parallel paths can be provided with a lap winding, it is 
suitable for high-current, low-voltage de machines, whereas wave windings 
having only two parallel paths are suitable for high-voltage, low-current de 
machines. 
4.2.4 ARMATURE VOLTAGE 
As the armature rotates in the magnetic field produced by the stator poles, 
voltage is induced in the armature winding. In this section an expression 
will be derived for this induced voltage. We can start by considering the r·· 
induced voltage in the coils due to change of flux linkage (Faraday's law) 
or by using the concept of "conductor cutting flux." Both approaches will 
provide the same expression for the armature voltage. 
The waveform of the voltage induced in a turn is shown in Fig. 4.12b, 
and because a turn is made of two conductors, the induced voltage in a turn 
a-b (Fig. 4.12) from Eq. 4.1 is 
e, = 2B(8)lwmr 
where l is the length of the conductor in the slot of the armature 
wm is the mechanical speed 
(4.4) 
r is the distance of the conductor from the center of the armature, 
that is, the radius of the armature 
The average value of the induced voltage in the turn is 
e, = 2B(8)lwmr 
Let 
Then 
From Eqs. 4.5 and 4.6, 
<P = flux per pole 
2rrrl A = area per pole = --
p 
- <Pp 
e =-w t m 
1T 
(4.5) 
(4.6) 
(4.7) 
138 chapter 4 DC Machines 
The voltages induced in all the turns connected in series for one parallel 
path across the positive and negative brushes will contribute to the average 
terminal voltage E •. Let 
Then 
N = total number of turns in the armature winding 
a = number of parallel paths 
From Eqs. 4.7 and 4.8, 
(4.8) 
(4.9) 
where Ka is known as the machine (or armature) constant and is given by 
or 
K = Zp 
a 2rra 
(4.10) 
(4.11) 
where Z is the total number of conductors in the armature winding. In the 
MKS system, if <I> is in webers and wm in radians per second, then E. is in 
volts. 
This expression for induced voltage in the armature winding is indepen-
dent of whether the machine operates as a generator or a motor. In the case 
of generator operation it is known as a generated voltage, and in motor 
operation it is known as back emf 
4.2.5 DEVELOPED (OR ELECTROMAGNETIC) TORQUE 
There are various methods by which an expression can be derived for the 
torque developed in the armature (when the armature winding carries cur-
rent in the magnetic field produced by the stator poles). However, a simple 
method is to use the concept of Lorentz force, as illustrated by Eq. 4.2. 
Consider the turn aa'b'b shown in Fig. 4.19, whose two conductors aa' 
and bb' are placed under two adjacent poles. The force on a conductor 
(placed on the periphery of the armature) is 
fc = B(O)l ic = B(O)l ~ 
a 
where ic is the current in the conductor of the armature winding 
I. is the armature terminal current 
(4.12) 
DC Machines 139 
a' 
.1J·· 
b 
FIGURE 4.19 Torque production in de machine. 
The torque developed by a conductor is 
Tc = fcr 
The average torque developed by a conductor is 
From Eqs. 4.6 and 4.14 
T = <f.Jp/a 
c 21Ta 
(4.13) 
(4.14) 
(4.15) 
All the conductors in the armature winding develop torque in the same 
direction and thus contribute to the average torque developed by the arma-
ture. The total torque developed is 
T= 2NTc ( 4.16) 
From Eqs. 4.15 and 4.16 
T = N<f.Jp I = K <f.Jl 1Ta a a a ( 4.17) 
In the case of motor action, the electrical power input (Eala) to the mag-
netic field by the electrical system must be equal to the mechanical power 
(Twm) developed and withdrawn from the field by the mechanical system. 
The converse is true for generator action. This is confirmed from Eqs. 4. 9 
and 4.17. Electrical power, 
mechanical power (4.17a) 
EXAMPLE 4.1 
A four-pole de machine has an armature of radius 12.5 em and an effective 
length of 25 em. The poles cover 75% of the armature periphery. The arma-
ture winding consists of 33 coils, each coil having seven turns. The coils are 
accommodated in 33 slots. The average flux density under each pole is 0. 75T. 
140 chapter 4 DC Machines 
1. If the armature is lap-wound, 
(a) Determine the armature constant K •. 
(b) Determine the induced armature voltage when the armature ro-
tates at 1000 rpm. 
(c) Determine the current in the coil and the electromagnetic torque 
developed when the armature current is 400 A. 
(d) Determine the power developed by the armature. 
2. If the armature is wave-wound, repeat parts (a) to (d) above. The current 
rating of the coils remains the same as in the lap-wound armature. 
Solution 
1. Lap-wound de machine 
(a) K = Np = !:__ll_ 
a 1Ta 2a 1T 
Z = 2 X 33 X 7 = 462, 
K = 462 X 4 = 73.53 
a 2X4Xrr 
a =p = 4 
(b) P l A _ 2rr X 0.125 X 0.25 X 0.75 o e area, P- 4 
(c) 
(d) 
or 
= 36.8 X 10-3 m 2 
<I>= AP X B = 36.8 X 10-3 X 0.75 
= 0.0276Wb 
1000 Ea = K.<Pwm = 73.53 X 0.0276 X~ 
X 2rr = 212.5 V 
I .1 = ~ = 400 = 1 00 A em a 4 
T = K.<PI. = 73.53 X 0.0276 X 400 = 811.8 N · m 
P. = E.I. = 212.5 X 400 = 85.0 kW 
1000 
= Twm = 811.8 X~ X 2rr = 85.0kW 
2. Wave-wound de machine. 
p =4, a= 2, z = 462 
(a) 
(b) 
(c) 
(d) 
K = 462 X 4 = 147.06 
a 2X2X7T 
1000 
Wm =~X 27T = 104.67 rad/sec 
DC Machines 141 
Ea = 147.06 X 0.0276 X 104.67 = 425 V 
/coil= 100 A 
la = 2 X 100 = 200 A 
T = 147.06 X 0.0276 X 200 = 811.8 N · m 
Pa = 425 X 200 = 85.0kW • 
4.2.6 MAGNETIZATION (OR SATURATION) CURVE OF A 
DC MACHINE 
A de machine has two distinct circuits, a field circuit and an armature circuit. 
The mmf's produced by these two circuits are at quadrature-the field mmf 
is along the direct axis and the armature mmf is along the quadrature axis. 
A simple schematic representation of the de machine is shown in Fig. 4.20. 
The flux per pole of the machine will depend on the ampere turns Fr 
provided by one or more field windings on the poles and the reluctance rzJt 
of the magnetic path. The magnetic circuit of a two-pole de machine is 
shown in Fig. 4.21a. The flux passes through the pole, air gap, rotor teeth, 
rotor core, rotor teeth, air gap, and opposite pole and returns through the 
yoke of the stator of the machine. The magnetic equivalent circuit is shown 
in Fig. 4.21b, where different sections of the magnetic system in which 
the flux density can be considered reasonably uniform are represented by 
separate reluctances. 
The magnetic flux <I> that crosses the air gap under each pole depends on 
the magnetomotive force Fr (hence the field current) of the coils on each 
pole. At low values of flux <I> the magnetic material may be considered to 
have infinite permeability, making the reluctances for magnetic core sections 
zero. The magnetic flux in each pole is then 
d-axis 
-n 
Field circuit 
(4.18) 
q-axis 
Armature circuit FIGURE 4.20 DC machine representation. 
......... 
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.. ,.,, 
142 chapter 4 DC Machines 
(a) 
(b) 
FIGURE 4.21 Magnetic circuit. (a) Cross-sectional view. (b) Equiva-
lent circuit. 
If