Principle of operation of dc machine

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DC Machine
Principle of Operation of dc Machine
Classification of dc Machines
Efficiency
Figure 12–7
The spatial relationship between magnetic flux and a single armature coil.
Figure 12–8
One complete rotation of the coil produces one complete sine-wave cycle of voltage.
Figure 12–9
(a) Instantaneous coil flux condition at the 0º instant (vertical instant).
Figure 12–9 (continued)
(b) Instantaneous coil flux condition at the 30º instant.
Figure 12–9 (continued)
(c) Instantaneous coil flux condition at the 60º instant.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
Figure 12–9 (continued)
(d) Instantaneous coil flux condition at the 90º instant.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
Figure 12–10
Slip-rings and brushes for getting the electrical energy out of the generator to the external load.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
Figure 12–11
Splitting the slip-ring into segments for rectifying the ac to dc.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
Figure 12–12
Placing a second coil on the rotor makes the output waveform smoother, with higher average voltage.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
Figure 12–13
Converting a dc machine into a motor.
Timothy J. Maloney
Modern Industrial Electronics, Fourth
Edition
Copyright ©2001 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
THE EMF EQUATION
Let us define the following symbols to determine emf
by using the physical parameters of a DC machine.
p : number of field poles
 : flux per pole, Wb
n : speed of rotation of the armature, rpm
Z : number of active conductors on the armature
a : number of parallel paths in the armature winding
For a lap winding a = p
For a wave winding a = 2
DERIVING THE EMF EQUATION
np
p
flux cut by
one conductor
in a rotation
flux cut by
one conductor
in n rotation (rpm)
n p
60
flux cut per second
by one conductor
number of conductors in series
n  p  Z   E
a
60  a 
Ea ka m
flux cut per second
by Z/a conductor
ka : the armature constant, the emf constant, torque constant.
THE ELECTROMAGNETIC TORQUE
Assumptions ! no saturation, e = f() only, no loss
P E Ia k a   m Ia
Te 


 k a  Ia
m m
m
Zp
ka 
2 a
SEPARATELY EXCITED DC GENERATOR
Ia
IL
+
Rfw
If
+
Vt
Ea
-
Rfc
Ra
RL
-
Equivalent circuit model at steady-state
+
Vf
-
Ea = Ra Ia + Vt
Vf = (Rfc + Rfw) If
Ea = ka m
SHUNT EXCITED DC GENERATOR
If
IL
Ia
Ra
Rfw
+
Rfc
+
Vt
RL
Ea
-
-
The generator itself provides its own field excitation.
Ea = Ra Ia + Vt
Vt = Vf = Rf If = (Rfc + Rfw) If = RL IL
Ea = ka m
Ia = IL +If
Questions
 How can you increase the generated voltage:
• By generator design?
• During operation?
 How can you increase the generated torque:
• By generator design?
• During operation?
EXAMPLE
Determine the voltage induced in the armature of a DC machine running at 1750
rpm and having four poles. The flux per pole is 25 mWb, and the armature is
lap-wound with 728 conductors.
If the armature carries a current of 123 A. Calculate the electromagnetic torque
developed by the armature.
SOLUTION :
p=4
= 25 mWb
728 conductors in total
a = p (lap-winding)
Z = number of active conductor
= 728
3 4  728 
 Z

  530.8 V
Ea  n  p    1750 25x10


60  a 
60
 4 
Te  k a  Ia 
Zp
728 x 4
 Ia 
0.025 x 123  356.3 Nm
2 a
2 4
Classification of DC Machines
Two main types of DC machines
 Separately excited machines
 Self excited machines
Shunt machines
 Series machines
 Compound machines

Figure 12–1
Subclasses of dc motors.
CLASSIFICATION OF DC MACHINES
SELF EXCITED DC MACHINES
Shunt
Series
A1
SEPARATELY EXCITED
DC MACHINES
Compound
Armature
A1
A1
Field
F1
F1
Field F2
F1
F2
Cumulative
Compound
A2
A2
Differential
Compound
A1
Series
winding
F1
F2
S1
A2
S2
Series
winding
A1
F1
F2
S2
A2
S1
A2
F2
Figure 12–16
(a) Schematic drawing of a dc motor with field winding in parallel (in shunt) with
armature winding. (b) Physical wiring of shunt motor. To avoid crowding, this
drawing does not show the wires leading between the armature coil-ends and the
commutator segments.
Example 12-1 page 551.
As they are illustrated in the previous slide, in separately excited DC
machines, the armature and field windings are electrically isolated. Each
winding is excited from a separate DC source. Permanent magnet DC
machines are equivalent to separately excited machine having constant
excitation.
If the armature and the field voltages are the same, these machines can be
connected in parallel (SHUNT MACHINE).
In series DC machines The armature and field windings are electrically
connected in series. The series field carries the same current as the armature
(load current) and consists of only a few turns of thick wire.
The compound DC machine has two field windings: one connected in series
(series field) with the armature, and the other is connected in parallel (shunt
field) with the armature.
SEPARATELY EXCITED DC MOTOR
Ia
+
Rfw
Ra
Vt
Ea
-
Rfc
+
DC voltage source 1
m
If
+ Vf
-
DC voltage source 2
1 It is the most flexible DC motor connection.
2 Suitable for feedback control.
3 The demagnetising, armature reaction and the voltage
drop effects can be compensated by increasing field excitation.
Ra
Rfc
Ea
Vt
-
-
Vt = Ea + Ra Ia
m
If
Vf = (Rfc + Rfw) If = Rf If
+
+
Rfw
Ia
+ Vf
Assuming :
-
No saturation
No armature reaction
If = constant
T = ka a = ka kf fa = kT a
Ea = ka m = ka kf fm = kT m
Motor Constant: kT = ka kf f
Nm/A
SHUNT EXCITED DC MOTOR
Used for constant speed applications, centrifugal pumps, fans
and machine tools.
If
It
Ia
Vf = (Rfw + Rfc) If
+V
Ra
Rfw
Rfc
Ea = ka m
+
Ea
-
T = ka a
-
If = 0 ? (what is going to happen)
Example 12-2 page 552
t
Vt = Ea + Ra Ia
Vt  I a R a
m
ka 
SHUNT EXCITED DC MOTOR
(Torque-speed characteristic)
Vt = Ea + Ra Ia
T = ka a
Ea = ka m
m 
1
ka 
Vt 
Ra
k a  
2
T
m, n
0
Torque, T
If Vt and constant , for a given load, the speed decreases linearly
with a torque
! In a practical machine, the denominator term also decreases due
to the armature reaction.  CONSTANT SPEED APPLICATIONS
Example 12-3 page 553
SERIES DC MOTOR
Used in electric traction applications and overhead cranes , due to their high
starting torque characteristics.
Ia
Vt = Ea + (Ra + Rs ) Ia
Ea = ka m = ka ksam = k am
T = ka a = ka ks aa = k a2
+
Rs
Vt
Ra
+
Ea
-
In the unsaturated region :
If
m
-
COMPOUND DC MOTOR
(cumulative compound motor)
Ia
If
+ It
Ra
+
Rfw
Rs
Vt
Ea
-
Rfc
-
Vt = Ea + Ra Ia + Rs It
It = Ia + If
Ea = ka m
Te = ka a
PM DC MOTORS
Permanent Magnet
Ia
+
Rfw
Ea
Ra
+ Vf
+
m
If
+
Vt
-
Rfc
Ia
Ea
Ra
+
Vt
-
-
They create the fields
-
Ia
+
Ra
Ea
Ea = kc m
Vt
back emf or
torque
constant
torque
-
Vt = Ea + Ra Ia
+
T = kc a
Vt  k c  m
T  kc
Ra
Linear T-n
Vt
T  Tstart  k c
Ra
Vt(3) Vt(2)
0
0
Vt(1)
speed
Vt
 max 
kc
PM DC MOTORS
ADVANTAGES:
1. Smaller in size than…
2. Higher efficiency (no
field coil power loss)
3. High stall (starting)
torque
4. Linear speed-torque
characteristics
DISADVANTAGES:
1. Absence of field control
c.f. …
2. Demagnetisation might
occur
3. Normally require a high
number of poles to
avoid demagnetisation
with low magnet
volumes
EFFICIENCY
Pinput = Pelectrical
= V t IL
Power flow in a DC motor
ROTATIONAL LOSSES:
ELECTRICAL AND
COPPER LOSSES :
Armature copper loss
Shunt field copper loss
Series field copper loss
Brush electrical loss
Magnetic or core loss
Mechanical losses
Stray load losses
Poutput
Poutput
(%)
100 
100
Pinput
Poutput   Plosses
Poutput
= Pmechanical
= Tshaft m
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