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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```