UNIT 3 DC MACHINES
DC Machines
Structure
3.1
Introduction
Objectives
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Principles of Electromechanical Energy Conversion
3.2.1
Basics of Electromagnetism
3.2.2
Generation of Electromotive Force (EMF) or Voltage
3.2.3
Torque Developed in DC Machine
Construction of DC Machine
3.3.1
Field System
3.3.2
Armature
3.3.3
Commutator
3.3.4
Armature Windings
Effect of Armature Current
3.4.1
Armature Reaction
3.4.2
Commutation
Characteristics of DC Generator
3.5.1
Separately Excited DC Generator
3.5.2
Self Excited DC Generators
DC Motor
3.6.1
Back EMF
3.6.2
Speed Regulation
3.6.3
Power Equation and Torque
3.6.4
Power Flow and Losses
3.6.5
Motor Current and Voltage Equations
Characteristics of DC Motors
3.7.1
DC Series Motor
3.7.2
DC Shunt Motor
3.7.3
DC Compound Motors
Starting, Speed Control and Application sof DC Motors
3.8.1
Starting of DC Motors
3.8.2
Speed Control
3.8.3
Applications
Summary
3.10 Answers to SAQs
3.1 INTRODUCTION
Rotating electrical machines are electromechanical energy conversion devices. When these
devices change electrical energy into mechanical then the device is said to function as a
motor and when the device converts mechanical energy into electrical, it is said to act as a
generator. In this unit, we will get an understanding of construction and working of DC
machines. Further, we will discuss different excitation schemes used in DC machines.
Objectives
After studying this unit, you should be able to
•
identify and list the various components of energy loss,
51
Electrical Technology
•
give an elementary description of DC armature winding and the commutator,
•
explain the various excitation schemes used in DC machines,
•
calculate induced emf and terminal voltage of DC generators,
•
know characteristics and applications of DC generators,
•
calculate torque and speed of DC motors,
•
describe characteristics, starting and speed control of DC motors, and
•
know applications of DC motors.
3.2 PRINCIPLES OF ELECTROMECHANICAL
ENERGY CONVERSION
3.2.1 Basics of Electromagnetism
In DC machines, there is an electromechanical energy conversion. In DC motor, the input
is electrical energy and the output is mechanical energy. In a DC generator, it is viceversa. An electromechanical conversion device interrelates electric and magnetic fields on
one hand and mechanical force and motion on the other.
For energy conversion, the principle of energy conservation is applicable. In DC machines,
the input power is converted into output power and some heat loss. Thus,
PInput = POutput + PLoss
. . . (3.1)
DC machine contains one or more than one magnetic circuit depending on the number of
poles. Before discussing magnetic circuit of a DC machine, let us recall the basic laws of
magnetic circuits.
In any magnetic circuit, the magnetomotive force (MMF) depends on number of turns in
coil N and the current I flowing through it, i.e.
MMF = N . I (AT)
...
(3.2)
The intensity of magnetic force H is given by
H=N.I/l =
MMF
= (AT/m)
l
...
(3.3)
where l is length of magnetic path.
This MMF sets a flux φ in the magnetic materials. There is a reluctance (S), which
opposes the setting of flux. The reluctance is similar to resistance in electrical circuits and
depends on the length l, area A and permeability µ of magnetic path, i.e.
S=
l
µA
...
S=
l
µo µr A
...
(3.4)
or
(3.5)
where µo is absolute permeability, and
µr is relative permeability.
52
The relative permeability of a magnetic material can be obtained by plotting its
B – H curve. The relative permeability is the ratio of flux density B and magnetic field
intensity H of material. It can be determined by taking ratio of B and H in linear region of
B – H curve. But actually the value of permeability depends on the value of flux density
(B) as shown in Figure 3.1. It is known as permeability curve.
DC Machines
Permeability
B
Figure 3.1 : Permeability Curve
A DC machine has magnetic circuits which are not exactly in parallel and the flux passes
through poles, air gap, armature teeth and magnetic core. The magnetic materials used for
different parts have different permeability. This type of magnetic circuit is known as
composite circuit.
So, the total MMF required to setup the flux φ in this magnetic circuit is
Total MMF = H p × l p + H g × lg + H t × lt + H c × lc + H y × l y
. . . (3.6)
where Hp and lp are field intensity and length of path, respectively, for the poles.
Hg and lg are field intensity and length of path, respectively, for air gap.
Similarly t, c and y refer to teeth, coils and yoke or magnetic core, respectively, of
DC machine.
3.2.2 Generation of Electromotive Force (EMF) or Voltage
EMF induced in rotating machines is dynamically induced emf.
According to Faraday’s law of dynamically induced emf, if a conductor having length l is
moving in a magnetic field of flux density B, with a velocity v m/s and θ as angle between
B and direction of motions then induced emf e is given by
e = B l v sin θ
...
(3.7)
The direction of induced emf can be determined by Fleming’s right hand rule for emf.
The induced emf can also be calculated by taking statically induced emf relations like
Faraday’s law. The induced emf in a conductor is directly proportional to rate of change of
flux linked with conductor
e=
dφ
dt
...
(3.8)
Consider a DC machine having the following :
P = Number of poles,
N = Speed of rotation of armature in RPM,
Zt = Number of total conductors in armature,
53
Electrical Technology
φ = Flux per pole in Webers, and
A = Number of parallel paths of armature conductors.
The number of parallel paths depends on type of armature winding.
For Wave wound armature
A=2
For Lap wound armature
A=P
Now, the flux linked with a conductor in one revolution ⇒ dφ = φ . P .
Time taken in one revolution = dt =
60
sec onds.
N
So emf induced per conductor in one revolution
e=
d φ P .φ P φ N
=
=
60
dt
60
N
...
(3.9)
The average emf induced in DC machine is always equal to emf across each parallel path.
So average emf across each parallel path
eav = e × No. of conductors in parallel path
=
P φ N Zt
volts
×
60
A
...
(3.10)
For Wave wound armature
eav =
P φ N Zt
volts and
×
60
2
...
(3.11)
for Lap wound armature
eav =
P φ N Zt φ N Zt
×
=
volts
60
P
60
...
(3.12)
In case of generator the average emf is known as generated emf (eg) and in case of motor it
is known as back emf (eb).
3.2.3 Torque Developed in DC Machine
When a conductor carrying current is placed at right angles to an uniform magnetic field, a
force is setup between the conductor and magnetic field, given by
F = B l Ia
. . . (3.13)
here B is flux density, l is length of conductor and Ia is current carried by conductor.
The direction of force can be obtained by Fleming’s Left hand rule. In DC motors, when
the current carrying coil of armature is placed in uniform flux then a torque is developed.
Average force on any conductor is given by
I 
Fav = Bav ×  a  l newtons
 A
...
(3.14)
where Bav is average flux density and A is number of parallel paths. For an armature of
diameter D, average torque contributed by each conductor is
I 
 D
Tav = Fav × armature radius =   Bav ×  a  l
2
 A
54
(3.15)
...
DC Machines
Total Torque due to all conductors,
I 
D
T = Tav × Zt = Zt   Bav l =  a  Nm
2
 A
...
(3.16)
But
Bav =
Pφf
πDl
...
(3.17)
 1  P
 .   Z t φ f I a Nm
T = 
 2π   A 
∴
T = Ka φ f I a
...
(3.18)
 P Zt 

K a = 
 2π A 
where
3.3 CONSTRUCTION OF DC MACHINE
Construction-wise both DC generator and DC motor are same. A DC machine has four
essential parts :
(a)
Field System
(b)
Armature
(c)
Commutator
(d)
Carbon Brushes
3.3.1 Field System
In DC machines, the field magnets provide uniform magnetic field surrounding the
armature. Generally, electromagnets are preferred in DC machines in place of permanent
magnets. The field magnet consists of four parts
(a)
Yoke or Frame
(b)
Pole Cores and Pole Shoes
(c)
Magnetic Coils
The Yoke or Frame
(a)
Generally, cylindrical yoke is used in DC machines.
(b)
Material of yoke possess high permeability and it provides the path to the
flux.
(c)
It is made of forged or cast steel or iron. Cast steel has very good magnetic
properties. The yoke provides mechanical protection to the internal parts of a
DC machine.
Pole Cores and Pole Shoes
Pole core has circular section and it carries the field magnets. The pole shoes are
attached to pole core and act as support to the field coils. It also spreads out the flux
uniformly over the armature periphery. Usually pole cores are made of iron and
casted with yoke. The pole shoes are formed by laminated sheets of steel and bolted
to the pole cores.
Magnetic Coils
55
Electrical Technology
To set up the flux in magnetic circuit of DC machine the magnetomotive force or
MMF is required. An electromagnet is formed by field or magnetic coils, which are
supported by the pole core. The magnetic flux produced by MMF is developed in
these magnetic coils.
Typical Magnetic
Flux Lines
Yoke
N
S
Field
Windings
Pole Shoes
S
Slotted
Armature
N
Figure 3.2 : Sectional View of Field System
In series field machines, the field coil is made of thick wire of copper with less
turns. For a shunt field coil, many turns of fine wire are used. After proper winding
of coils these are dipped in insulating varnish to provide mechanical strength and
better insulating properties.
3.3.2 Armature
Armature is the drum shaped rotating part of DC machine. The armature conductors are
fixed at the upper surface of drum in slots. There is a small air gap between armature and
pole shoes of field magnets to avoid any rubbing in DC machine. This air gap should be
kept minimum. Usually armature is made of
0.3 mm to 0.6 mm thick laminated stampings of high grade steel to reduce hysteresis and
eddy current losses. On the outer periphery the slots are formed by die cut or punch as
shown in Figure 3.3. Also some air ducts are provided for proper ventilation.
Key Way
Slot
Tooth
Ventilating Ducts
Figure 3.3 : Armature Lamination
The armature conductors carry the current and are insulated using several layers of paper
or mica insulation. Figure 3.4(a) depicts a slot containing two coil sides, each consisting of
a single conductor. Two sides of one coil are housed approximately one pole pitch apart.
One side occupies top layer and another side occupies the bottom layer of the respective
slots. In a multiturn coil, each coil side consists of as many conductors as the number of
turns in the coil.
56
DC Machines
Sectional View
Slot Wedge
Top
Coil Side
Overhang
Insulation
Bottom
Coil Side
Active Length
or
Length of Armature Core
Conductors
Start of Coil
(a) Coil Sides in Slot
Finish of Coil
(b) Coil Viewed From Top
Figure 3.4 : Arrangements of Coils in Slots
3.3.3 Commutator
Commuator is an essential part of DC machines. It is placed between armature and the
external circuit. The armature coils are connected with the commutator, which in turn gets
connected to external circuit through carbon brushes sliding on commutator. Its serves the
following purpose :
(i)
It completes electrical circuit by connecting the rotating armature coils and
stationary electrical circuit.
(ii)
In generating action it works as a rectifier which converts the generated AC
voltage into DC voltage.
(iii)
In motoring action it reverses the direction of DC current to maintain the
torque in same direction.
Commutator is made of wedge shaped segments of drop forged and hard drawn copper. A
thin sheet of mica is used to separate or laminate the segments from each other.
It has cylindrical shape and approximately same diameter as armature. The winding ends
of armature are directly soldered to the commulator segments.
Slotted Armature
Mica
Shaft
Segments
Figure 3.5 : The Commutator
3.3.4 Armature Windings
Many different types of armature windings are employed in dc machines depending on
requirements. The most common of these is the simplex lap, though the simplex wave is
used in high voltage machines. (More elaborate windings, referred to as duplex and
multiplex are also sometimes used). In this section, we will confine ourselves to a
description of the simplex lap winding and refer in passing to the simplex wave.
57
Electrical Technology
Discussion of armature windings is greatly facilitated by using armature winding
diagrams. Figure 3.6 is the armature winding diagram of a two-pole dc machine having 8
slots, 8 coils and 8 commutator segments operating in generator made. The diagram is a
conventionalized representation of the cylindrical armature surface assuming that it is cut
along an axial line and unrolled flat on to a plane. The 8 slots will then appear at equal
distances from each other. Each slot has a top coil side and a bottom coil side. In the
Figure 3.6, the top coil sides in the active length of the armature are represented by firm
straight lines numbered 1 to 8 corresponding to the 8 slots. The bottom coil sides are
indicated by dotted straight lines close to the firm lines.
Since the machine is meant for two poles, there are 4 slots per pole. A coil whose top and
bottom layers are 4 slot pitches is said to be a full-pitched coil. We will use full pitch coils.
In this case, if the top layer of a coil is in slot number x, the bottom layer will be in slot
number (x + 4). The triangular connections at the top are meant to indicate the
interconnection of coil sides and belong to the overhang of the winding on the side away
from the commutator. Thus, the firm line in slot 1 is connected to the broken line in slot 5
to represent coil number 1 etc. In an armature meant for four or more poles, top coil sides
in slots 5, 6, 7 and 8 would in fact be connected to bottom coil sides in slots 9, 10, 11 and
12. Since our armature is for two poles and has only 8 slots, these coil sides would be
connected to the bottom layers in slots 1, 2, 3 and 4. Thus, each side of a coil falls under
separate poles.
The segmented rectangular strip at the bottom represents the commutator. As seen from
Figure 3.5. The representation in Figure 3.6 is the developed view of an equivalent
commutator having the same diameter as the armature. In a lap winding, the ends of a coil
are connected to adjacent commutator segments. Thus each commutator segment will be
connected to top coil side of one coil and to the bottom coil side of a different coil. We will
adopt the convention that the segment number is the same as that of the slot to whose top
layer it is connected. In a lap-winding, the ends of coil 1 will be connected across segments
1 and 2, etc. The same type of diagram is used even for multiturn coils).
Flux Directions
N-pole
Domain
Velocity
1
2
3
4
5
6
7
P
Commutator
7
8
1
Slot 1
Slot 8
S-pole
Domain
8
Q
2
3
4
B
5
6
7
B
Figure 3.6 : Winding Diagram
3.4 EFFECT OF ARMATURE CURRENT
3.4.1 Armature Reaction
The distribution of flux under main poles is affected by magnetic field set up by the
armature current; as a result the flux distribution or flux wave shape in air gap is distorted
and demagnetised. This phenomenon is known as armature reaction.
58
When the field is energized and there is no current in armature, the flux distribution is
rectangular in space along armature periphery with axis of magnetisation through the
centre of poles.
DC Machines
When a armature carries current then there is some MMF due to number of turns in slots
and it works as an electromagnet. The axis of magnetization differs for the various coils so
the resulting axis of magnetization for the complete winding passes between active
conductors and brushes are placed in interpolar region. In generating action, when the
armature rotates in clockwise direction, the field produced by armature current acts
opposite to the direction of rotation, so, in air gap flux is weakened under leading pole tips
and strengthened under the trailing pole tips. This distorts the original rectangular flux
wave shape leading to shift in resultant axis magnetisation and hence the brushes do not
now fall on interpolar axis, thereby leading to poor commutation and speaking.
The effect of armature MMF is summarized as The cross magnetising effect of
armature reaction. Due to cross magnetizing effect DC machines have poor
commutation.
Compensating Winding
In large capacity DC machines a compensating winding is used to neutralize effect
of armature reaction. It maintains uniform flux distribution under pole faces. A
copper winding is placed in the slots provided in pole shoes. It is connected in series
with armature winding in such a way that the direction of current flowing through it
is reverse of direction of armature current. This in effect neutralizes field produced
by armature current and hence minimises cross-magnetising effect of armature
reactions. The armature reaction is severe in large capacity and high speed
machines. So compensating winding is used with inter poles arrangement.
3.4.2 Commutation
In DC machines, the EMF induced in each coil of armature alternates with a frequency
NP

corresponding to product of number of poles and speed  f =
 . To obtain the direct
120 

current in external circuit it is necessary to reverse periodically the connections of each
armature coil with the external circuit. This reversal should take place at the instant when
emf induced in coil is zero. For this purpose the commutator and carbon brushes are used.
The reversal of current at this instant is affected by the following :
Variable Contact Resistance
The contact resistance between carbon brushes and commutator is not constant but
largely depends on current density at contact surface. Also voltage drop across
brush contact varies but this variation is very small compared to resistance
variations.
Current Density under the Brush
When the brush thickness is less than one commutator segment, then reversal of
current follows the simple linear law i.e. known as straight line commutation.
A sparkles commutation is achieved by the following methods :
(a)
By using high resistance carbon brush.
(b)
By shifting carbon brushes in inter pole region.
(c)
By using inter poles or commutation poles.
(d)
By using equalizer rings.
59
Electrical Technology
3.5 CHARACTERISTICS OF DC GENERATOR
3.5.1 Separately Excited DC Generator
In separately excited DC generators, the field coils are excited by the external DC supply
by using field regulating resistance. The open circuit characteristic or magnetizing
characteristic is obtained by driving the machine at normal speed and varying the exciting
current in steps from zero to maximum. Main circuit of machine is kept open and a
voltmeter connected across the terminals. The excitation current is entirely independent of
the load current in the armature. Equivalent circuit for mathematical calculations gives :
Ia = I = Load current
...
(3.19)
V = Eg – Ia Ra – Voltage drop across Carbon brush
Power Input = Eg Ia
...
Power Output = V . Ia
...
(3.20)
(3.21)
If
A
FF
I
Rheostat
+
Ia
Load
E
V
Eg Ra
FF
–
Figure 3.7 : Separately Excited DC Generator
Magnetizing Characteristic
Magnetizing characteristic or open circuit characteristic (OCC) is known as no load
saturation curve which is obtained by method discussed above. The curve between
the field current and armature terminal voltage is drawn when armature current is
zero, i.e. at no load. Generally, it starts from zero but in self excited DC generator it
has residual voltage and starts at slightly higher value. For DC generators,
magnetizing curve is shown in Figure 3.8, its shape can be explained using theory of
domain alignment.
Eg
Self Excited
Separately Excited
c
b
Residual
Voltage
a
If
O
Figure 3.8 : Magnetic Characteristic
Internal Curve
60
Internal curve gives the relation between the emf actually generated in the
armature and the armature current Ia.
DC Machines
External Curve
External curve also known as load curve is plotted between terminal voltage
V and armature current Ia.
For a separately excited DC generator these curves are shown in Figure 3.9.
Eg,V
Eg
Ideal (Generated emf)
V
Inter
nal C
urve
Exte
rna
l Cu
rve
Armature Reaction Drop
Ohmic Drops (Ia Ra)
Ia
Figure 3.9 : Internal and External Curves for Separately Excited DC Generators
The internal curve is obtained by subtracting the drops due to armature
reaction from generated emf (Eg). Also, it is a combination of external
characteristic and Ia Ra drops. So, internal characteristic can be achieved by
adding Ia Ra drops to external characteristic.
Separately excited generators operate in stable condition with any excitation
and generally used in laboratories.
3.5.2 Self Excited Generators
Self excited machines have their field winding connected in series or parallel with the
armature or main circuit. These are classified as :
(a)
Series wound
(b)
Shunt wound
(c)
Compound wound
(i)
Long shunt compound
(ii)
Short shunt compound
To build up the voltage in self excited DC generators, following conditions should be met :
(i)
Presence of residual magnetism in pole curve, and
(ii)
Forward direction of rotation.
When the armature of a self excited DC generator rotates at rated speed, the voltage across
terminals increases up to rated voltage. Initially, due to residual magnetism there is a small
flux which induces small residual voltage, which increases the field current. Finally,
terminal voltage reaches the rated value.
If the direction of rotation is reversed then induced emf due to residual magnetism
demagnetizes the residual flux so there is no voltage built up in self excited generators.
61
Electrical Technology
E
Eg , E, V
Magnetic Characteristic And
Armature Reaction Drop
Eg
Drops Due to Armature and Field
V
B’
External Characteristic
Ia
O B
Stable Operating Point
Figure 3.10 : Characteristics of Self Excited DC Generator
External characteristics can be obtained by subtracting armature and field winding
resistances drops from internal characteristic. External characteristic shows that first
voltage increases with increase in load current, at max value of load current it decreases.
This is known as dropping region. Under dropping region terminal voltage may fall to
zero. Usually, this type of generators are used for welding purpose.
Critical load resistance is that resistance of load above which a series wound generator
fails to built up the voltage. The slope at external curve represent critical load resistance.
Series Wound DC Generator
In series wound DC generator, the field winding is series connected with main
circuit and it delivers load current same as armature current. The equivalent circuit
of DC series generator is shown in Figure 3.11.
Rse
Ia
Ise
IL
Eg R
a
V
Figure 3.11 : Equivalent Circuit of DC Series Generator
If series field resistance is Rse then
Ia = Ise = IL
V = Eg – Ia Ra – Ia Rse – voltage drop across carbon brush.
(3.22)
...
V = Eg – Ia (Ra + Rse) – voltage drop due to carbon brush.
(3.23)
...
Power Input = Eg . Ia
(3.24)
...
Power Output = V . Ia
(3.25)
...
Characteristics of DC Series Generator
In a series wound dc generator the voltage cannot build up if the load terminals are
open. To built up voltage first connect the load across output terminals of DC
62
generator and rotate the armature in forward direction. Due to residual magnetism,
there is small voltage across armature, which results in a small current in field
winding. This small current developes a flux in field and voltage across armature
increases due to increase in flux. This loop continues till the saturation. Now,
voltage across the armature is rated. This given by magnetizing characteristic. Due
to armature reaction there is some voltage drop, by subtracting it from magnetic
characteristic, internal characteristic can be obtained.
DC Machines
Shunt Wound Generator
If field winding of a DC generator is connected across the main circuit then its
known as DC shunt generator. The resistance of shunt field is very high as
compared to armature resistance. The equivalent circuit is shown in Figure 3.12.
IL
If
Ia
Rsh
Eg
V
Ra
Load
Figure 3.12 : Equivalent Circuit of DC Shunt Generator
Rsh is the shunt field resistance then field current I f =
Here,
V
.
Rsh
Ia = IL + If
Eg = V + Ia Ra + voltage drop due to carbon brush.
Power Input = Eg . Ia
Power Output = V . IL
Voltage Built Up
The emf induced by the armature winding due to its rotation through the
residual magnetic flux sends a current in field winding. This current should
be in such a direction as to tend to increase the magnetic flux. With proper
connections and direction of rotation the magnetic field will be gradually
strengthened and as a result the induced emf increases. This emf increases
exciting current and so the voltage builds up until steady state conditions are
obtained. Usually, shunt wound generators are connected with load after the
rated voltage is built up. Generally the load resistance is less than field
resistance otherwise the generator may fail to build up the rated voltage.
On other land if field resistance is very high or more than critical field
resistance then a dc generator fails to build up the voltage. The critical field
resistance can be obtained by drawing the tangent at open circuit
characteristic of DC shunt generator.
Characteristics of DC Shunt Generator
Magnetic characteristic is plotted between If and generated EMF. For a given
magnetic flux, the emf generated varies in direct proportion to the speed of rotation.
Then magnetization curve for various speeds plotted at the same scale gives
different critical resistances at different speed. The shunt field resistance needs to be
less than this critical value.
Slopes OA, OB and OC give critical resistances at speeds N3, N2 and N1
respectively, here N3 > N2 > N1.
63
Eg
Electrical Technology
pe
Slo
N3
N2
A
N1
B
C
If
O
Figure 3.13 : DC Shunt Generator Magnetic Characteristic at Different Speeds
Ideally, the terminal voltage of any dc shunt generator should be constant. But due
to armature reaction, load voltage decreases with increasing load current. It further
decreases due to Ia Ra drops.
Example 3.1
The emf induced in the armature of a 450 kW, 250 volt shunt generator is 258.8
volt, when the field current is 20.0 amp and the generator is supplying power to a
load at rated terminal voltage. The armature circuit resistance is 0.005 ohm.
Determine (i) load current, (ii) power generated, (iii) power output, and (iv)
electrical efficiency. Neglect brush contact drop.
Solution
Let the load current be I amp
Armature current = Load current + field current
or
Ia = I + 20
Induced emf = Terminal voltage + Ia Ra drop
or
258.8 = 250 + (I + 20) × 0.005
or
I =
8.7
= 1740 amp
0.005
(i)
Load current = 1740 amp
(ii)
Power generated = E . Ia = 258.8 (1740 + 20) watts
= 455.488 kW
(iii)
Power output V . I = 250 × 1740 watts
= 435 kW
(iv)
Electrical efficiency =
435
× 100 = 95.5%
455.488
SAQ 1
A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal
voltage of 220 volts. Its armature has 150 single turn coils and has a resistance of
0.01 ohm. The air gap flux per pole is 0.02 weber. Shunt field resistance is
50 ohm. Calculate the speed at which it is being driven. Neglect brush contact drop.
64
DC Machines
DC Compound Generator
In compound generators the field winding is connected in series as well as parallel to
the armature circuit. In long shunt generators, the shunt winding is connected across
armature and series field but in short stunt it covers only armature winding. If the
flux produced by series and shunt windings is additive, the dc generator is known as
cumulative compound generator and if fluxes are deductive then it is known as
differential compound generator.
For a long shunt generator as shown in Figure 3.14,
Ia = IL + If
...
(3.26)
If =
V
Rsh
...
(3.27)
V = E g − I a ( Ra + Rse ) − voltage drop due to carbon brush
...
(3.28)
Input Power = Eg . Ia
...
(3.29)
Output Power = V . IL
...
(3.30)
IL
If
Rse
Ia
V
Rsh
Eg R
a
Load
Figure 3.14 : Long Shunt DC Compound Generator
In a short shunt generator as shown in Figure 3.15,
Ise = IL
Ia = IL + If
...
(3.31)
If =
V + I se Rse
Rsh
...
(3.32)
V = Eg – Ia Ra – IL Rse – voltage drop due to carbon brush
(3.33)
Power Input = Eg Ia
...
...
(3.34)
65
Electrical Technology
Power Output = V . IL
...
(3.35)
Ise
Rse
If
IL
Ia
Rsh
V
Eg R
b
Load
Figure 3.15 : Short Shunt DC Compound Generator
The degree of compounding depends on flux shared by series or shunt windings.
SAQ 2
(a)
A long shunt compound generator delivers a load current of 50 A at 500 V
and has armature, series-field and shunt field resistances of 0.05 Ω, 0.03 Ω
and 250 Ω respectively. Calculate the generated emf and the armature
current. Allow 1.0 volt per brush for contact drop.
Also draw the circuit diagram of this arrangement.
(b)
A 4 pole, lap wound, 11.5 kW, 230 volts, dc shunt generator has armature
resistance of 0.2 ohm and field resistance of 100 ohms. Calculate (i) emf
generated if the brush contact drop is one volt per brush, and (ii) flux per pole
if the machine is driven at 1000 rpm and 1200 conductors are in the armature
slots with generator supply rated load at rated voltage.
Characteristics of Compound Generators
Usually the series winding is arranged to assist the shunt winding and the terminal
voltage variation with increasing load is determined by the relative strengths of two
windings. Under suitable conditions terminal voltage may remain constant from no
load to full load. By using relatively strong series windings the terminals voltage
may increase with load. Its known as over compounded generator. If the shunt
winding is strong then terminal voltage decreases with rise in load. It is under
compound generator. If two fluxes are subtracting then the external curve goes
down sharply, as shown in Figure 3.16.
66
DC Machines
V
Over Compounded
Flat Compounded
Under Compounded
Differential
Compounded
O
IL
Figure 3.16 : Characteristics of Compound Generators
3.6 DC MOTOR
DC motor converts electrical energy into mechanical energy. When a current carrying
conductor is placed in a magnetic field, a force acts on the conductor and conductor moves
in the direction of force. When the DC machine is connected to DC supply a current
passes through the armature winding. When conductors of armature winding carry
outward current under north and incoming current under south pole then those conductors
experience a force in clockwise direction according to fleming’s left hand rule. Due to this
force, conductors move in clockwise direction. The direction of current is reversed by
commutator, which causes the moving conductor coming under different pole to carry
reverse current. This causes the force on the conductor to be again in the same direction as
flux and current both change direction simultaneously. Thus armature conductors always
experience force in same direction.
3.6.1 Back EMF
When the armature of a DC motor rotates, an emf is induced in armature conductors
known as back emf which opposes the applied voltage.
Eb =
φZNP
volts
60 A
...
(3.36)
If Ra is armature resistance then
V = Eb + I a Ra
...
(3.37)
here Ia is armature current and V is applied voltage
so
Ia =
V − Eb
Ra
...
(3.38)
Back emf makes a dc motor self regulating. When speed is low then back emf will less and
armature current will be large.
3.6.2 Speed Regulation
The back emf in armature is
Eb =
Pφ Z N
60 A
...
(3.39)
67
Electrical Technology
N =
so
Eb 60 A
PφZ
...
(3.40)
since P, Z and A are constant for a machine then
or
N∝
Eb
φ
N∝
V − I a Ra
φ
...
(3.41)
For a shunt motor φ is constant then
N ∝ Eb or N ∝ V − I a Ra
...
(3.42)
For a series motor φ is proportional to Ia
so
Nα
Eb
Ia
...
(3.43)
percentage speed regulation =
No − N f
Nf
× 100
...
(3.44)
here No is speed at no load in RPM and Nf is speed at full load in RPM.
Example 3.2
A shunt generator delivers 50 kW at 250 volts and 400 rpm. The armature and field
resistances are 0.02 and 50 ohms respectively. Calculate the speed of the machine
running as a shunt motor and taking 50 kW input at 250 volts. Allow one volt per
brush for contact drop.
Solution
As Generator :
Line current I =
50 × 103
= 200 amp
250
Shunt field current I sh =
250
= 5 amp
50
Armature current I a = I + I sh = 200 + 5 = 205 amp
Armature drop = I a Ra = 205 × 0.02 = 4.1 volts
Induced emf
E g = 250 + armature drop + brush contact drop
E g = 250 + 4.1 + 2 × 1 [There are two brushes]
= 256.1 volts
As Motor :
Armature current I a = I − I sh = 200 − 5 = 195 amp
Armature drop = I a Ra = 195 × 0.02 = 3.9 volts
Back emf
68
Eb = 250 − armature drop − brush contact drop
DC Machines
Eb = 250 − 3.9 − 2 × 1 = 244.1 volts
or
Since field current is constant
E g ∝ N1 and Eb ∝ N 2
where N1 and N2 are respectively the speeds of the machine as a generator and
as a motor.
Eg
Eb
=
E N
N1
244.1 × 400
or N 2 = b 1 =
= 381.257 rpm
N2
Eg
256.1
SAQ 3
A shunt wound motor runs at 500 rpm from a 200 volt supply. Its armature
resistance is 0.5 ohm and the shunt field resistance is 100 ohms and takes
32 amperes line current from the supply. What resistance must be reduced to 300
rpm, the armature and field currents remaining the same? Neglect brush drop.
3.6.3 Power Equation and Torque
The voltage equation for motor is
V = Eb + I a Ra
...
(3.45)
By multiplying Ia in this equation by Ia we get
V I a = Eb I a + I a2 Ra
...
(3.46)
It is a power equation, V Ia is input power, Eb I is power developed in armature and Ia2 Ra
represents power losses in armature. So mechanical power developed by the motor is
Pm = Eb I a = V I a − I a2 Ra
...
(3.47)
or
Pm = input power – losses
Differentiate the Eq. (3.47) with respect to Ia we get
dPm
= V − 2 I a Ra
dI a
...
(3.48)
For max power output
dPm
= 0.
dI a
or
V = 2 I a Ra
but
(3.49)
I a Ra = V − Eb
so
V = 2 (V − Eb )
...
69
Electrical Technology
or
Eb =
V
2
...
(3.50)
In any motor mechanical power developed in armature is maximum when back emf is half
of applied voltage.
3.6.4 Power Flow and Losses
Electrical Losses in Core/Iron Parts
In the iron part of machine some electrical losses occur in the form of hysteresis
and eddy current losses.
Hysteresis losses occur due to magnetic reversals caused by the rotating armature.
Hysteresis losses are directly proportional to the number of magnetic reversal per
second.
hysteresis loss Pn = n (Bmax)x f V watts
these losses occur in armature core and teeth of the dc machine. To reduce the
hysteresis loss armature core is made of silicon steel.
When armature core rotates in magnetic fields of poles which induce emf in
armature core and yoke. Due to this induced emf eddy currents circulate in
armature core, the eddy current losses mainly depend on thickness of material.
Pe = K Bmax f 2 V t2 watt
...
(3.51)
To minimize eddy current losses the armature core is made of laminated
stampings. Hysteresis and eddy current losses are known as core losses and are
about 20% to 30% of full load losses.
Mechanical Losses
Due to friction of bearings, air friction or windage some losses occur in dc
machines. These are known as mechanical losses. The brush friction losses are quite
large. These losses are about 10% to 20% of full load losses.
Losses and Efficiency
In electrical machines, the efficiency is always less then one. It means that the
output is less than the input.
For any machine, efficiency =
Output
Input
...
(3.52)
In electrical machine input power is sum of output power and power loss i.e.
Power (Input) = Power (Output) + Power loss
So,
efficiency =
Output Power
Input Power
=
Input Power − Power loss
Input Power
=1−
Power loss
Input Power
...
(3.53)
70
The various machine losses may be classified as electrical losses and mechanical
losses.
DC Machines
Electrical Losses
In DC machines electrical losses occur in several parts of machine. The maximum
electrical losses occur due to I2R losses because a large current flows through
various machine windings. In addition to I2R losses there is brush contact loss at the
contacts between the brushes and commutator. These losses are known as copper
losses and amount to 40% to 60% of the full load losses.
The power input of any DC machine is distributed as shown in Figure 3.17 below.
POWER INPUT
Total Losses
Useful Output
Copper Losses
Armature
losses
Iron Losses
Field Winding Eddy
Losses
Current
Loss
Mechanical Losses
Hysterisis
Loss
Friction
Loss
Windage
Loss
Figure 3.17 : Power Input Distribution in DC Machine
Power Flow Analysis
In DC machines the sequence of energy conversion is necessary in analysis of the
operation and characteristics of the machine.
Power Flow in DC Generator
Armature Copper Loss
DC Generator
Mechanical Power
Input from Turbine, Engine etc.
Mechanical Power Converted
to Electrical Power
Mechanical
Power Loss and Hystirisis
and Eddy Current Loss
Output Power
Power Lost
in Shunt Field
Figure 3.18 : Power Flow in DC Generator
Power Flow for DC Motor
DC Motor
Electrical
Power Input
Loss in Shunt
Field Winding
Power Converted In
Mechanical Power
Copper Losses
in Armature
Output
Core and
Mechanical Losses
Figure 3.19 : Power Flow for DC Motor
Torque
In a DC motor, output power is converted to torque.
If at a wheel of radius r metre, a force F acts on circumference then
71
Electrical Technology
Torque T = F . r
...
(3.54)
work done per revolution = F . 2π r joules
work done per second = F . 2π r . n
(3.55)
here, n =
...
N
and N is speed of rotation in RPM.
60
r
F
Figure 3.20
Now power developed, P = work done per second
or
P = F r. 2π n
or
(3.56)
P=T.ω
where,
T = Torque, and
...
ω = angular velocity.
here
ω = 2π n =
2π N
60
...
(3.57)
Power developed in armature is Eb . Ia
So,
(3.58)
Eb Ia = T . ω
...
T=
Eb I a
ω
or
T =
PφZ N
Ia
60 A
or
T =
PφZ
I
. I a = 0.159 φ Z P a
2π A
A
or
...
(3.59)
2π N
60
...
(3.60)
or
T = 9.55
Eb I a
N
...
(3.61)
Since Z, P and A are constant so
T ∝ φ Ia
. . (3.62)
72
.
DC Machines
For DC shunt motor φ is constant so T α Ia and for dc series motor φ α Ia,
so T α Ia2.
Example 3.3
A 125 V, dc shunt motor at its rated conditions develops 1 kW at 1800 rpm. Its line
current is 10.67 A. The motor has a field resistance of 110 ohms and armature
circuit resistance of 1.23 ohms. If the motor torque is increased by 20% determine
(i) its probable new speed, and (ii) line current.
Solution
Shunt field current I sh =
125
V
=
= 1.136 amp
Rsh 110
Under rated conditions I a1 = I − I sh = 10.67 − 1.136 = 9.534 amp
Torque ∝ φ Ia
For a shunt motor, φ is constant
∴
(3.63)
T ∝ Ia
...
When the motor torque is increased by 20%, the new value of torque becomes 1.2 T
and let the new value of armature current be Ia2. We have,
1.2 T ∝ Ia2
...
(3.64)
Dividing Eq. (ii) by Eq. (i), we have
I
1 .2 T
= a 2 or I a 2 = 9.534 × 1.2 = 11.44 amp
T
9.534
Under rated conditions,
Eb1 = V − I a1 Ra = 125 − 9.534 × 1.23
or
Eb1 = 113.27 volts
But back emf ∝ speed [φ being constant]
∴
(3.65)
E b1 ∝ N1
...
with increased torque
Eb 2 = V − I a 2 Ra = 125 − 11.44 × 1.23
or
Eb 2 = 110.93
Let the new speed be N2, then
Ebe ∝ N2
...
(3.66)
Dividing Eqs. (3.66) by Eq. (3.65), we have
110.93
N
110.93 × 1800
= 2 or N 2 =
= 1762.8 rpm
113.27 1800
113.27
New speed = 1762.8 rpm and
New line current = Ia2 + Ish
73
Electrical Technology
= 11.44 + 1.136
= 12.576 amp
Example 3.4
The flux per pole of a 4 pole, 220 volt dc series motor is 0.025 weber for a load
current of 52 amperes. The armature is wave wound with 500 conductors. Calculate
: (i) the gross torque, (ii) the speed, (iii) the output torque, and
(iv) the efficiency.
The corresponding iron, friction and windage losses total 860 watt. The armature
and field resistances are 0.21 ohm and 0.15 ohm respectively.
Solution
(a)
P
Gross torque T = 0.159 φ Z IA  
 A
4
T = 0.159 × 0.025 × 500 × 52  
2
or
= 206.7 N-m
(b)
Back emf
Eb = V − I a ( Ra + Rse )
Eb = 220 − 52 (0.21 + 0.15) = 201.28 volts
But
or
Eb = φ
(c)
0.025 × 500 × N 4
×
60
2
201.28 =
or
ZN P
×
60
A
N = 483.072 rpm
Torque lost in friction windage etc.
T fw =
=
860
2π N
60
860 × 60
= 17 N-m
2 π × 483.072
∴ Output torque = T – 17 = 206.7 – 17 = 189.7 N-m
(d)
Efficiency =
=
Output Eb I a − 860
=
Input
V Ia
201.28 × 52 − 860
= 83.97%
220 × 52
SAQ 4
A dc series motor develops 30 kW and takes a current of 80 amp when running at
1200 rpm. Find the starting torque if at starting armature current is 120 Amperes.
Magnetic circuit remains unsaturated.
74
DC Machines
3.6.5 Motor Current and Voltage Equations
Separately Excited DC Motor
Eb = V – Ia Ra
...
(3.67)
Ia = IL,
If =
E
Rf
Power drawn from supply
...
P = V IL
(3.68)
Mechanical power developed
Pm = Eb Ia = Ia (V – Ia Ra)
...
(3.69)
= V Ia – Ia2 Ra
= V I L − I L2 Ra = Powers drawn from supply
– I2 R losses in armature
...
(3.70)
IL
+
If
Ia
F
A
Eb
V
Ra
Rf
E
Battery
AA
FF
–
Figure 3.21 : Separately Excited DC Motor
Series Wound DC Motor
Eb = V − I a ( Ra + Rse )
...
(3.71)
Power drawn from supply
⇒ P = V Ia
...
Pm = V I a − I a2 Ra − I a2 Rse
...
Pm = I a (V − I a ( Ra + Rse ) )
...
Pm = I a Eb
...
(3.72)
(3.73)
or
(3.74)
(3.75)
75
F
Electrical Technology
FF
+
IL
RSE
Ia
A
Eb
Ra
AA
–
Figure 3.22 : Series Wound DC Motor
Shunt Wound DC Motor
...
I L = I a + I sh
(3.76)
V
Rsh
...
Eb = V − I a Ra
...
I sh =
(3.77)
(3.78)
P = V . IL
Pm = V . I L − V I sh − I a2 Ra
...
(3.79)
= V ( I L − I sh ) − I a2 Ra = V I a − I a2 Ra
= I a (V − I a Ra )
...
(3.80)
...
Pm = I a Eb
(3.81)
IL
Ish
Ia
F
A
Eb
Rsh
Ra
V
AA
FF
e
Figure 3.23 : Shunt Wound DC Motor
Direction of rotation of DC motor can be reversed by reversing the direction of
current either in armature or in field windings. If current through both windings will
reverse, then direction of rotation will remain same.
3.7 CHARACTERISTICS OF DC MOTORS
To determine performance and stability of a dc motor following are drawn :
76
(a)
Torque – Armature current : Electrical characteristics
(b)
Torque – Speed : Mechanical characteristics
(c)
Speed – Armature current : Speed characteristics
DC Machines
3.7.1 DC Series Motor
For series motors the flux is directly proportional to armature current. From speed
equation, we get
N µ
Eb
V − I a ( Ra + Rse )
=
φ
φ
. . . (3.82)
T ∝ φ Ia
T ∝ I a2
N∝
N∝
and
V
− ( Ra + Rse )
Ia
V
k T
− ( Ra + Rse ) (in saturated region)
T ∝ Ia
N∝
V − I a ( Ra + Rse )
k2
N∝
V − T ( Ra + Rse )
(in saturated region)
k2
A curve between armature current and speed will be rectangular hyperbola before
saturation. Series DC motor have variable speed but on no load speed is dangerously high.
To start a series DC motor, first we must put a mechanical loading on it.
Torque
N
N
T
Ia
(a)
T
(b)
Figure 3.24 : DC Series Motor Characteristics (a) Torque-current Characteristics of Armature
Current Speed Characteristics, and (b) Speed-torque Characteristics
It is suitable for gear drive because gear provides some load on account of frictional
resistances. Before saturation, flux (φ) is proportional to Ia, so torque (T) is proportional
to Ia2 but after saturation flux is constant so torque is directly proportional to armature
current.
From torque current characteristic (Figure 3.24(a)) we see that the series motor develops
high starting torque for heavy loads. So these are used for electric traction. From speed
torque characteristic (Figure 3.24(b)) we see that the speed are sharply falls with increase
in torque. So another application of DC series is in fans whose speed falls with load.
3.7.2 DC Shunt Motor
DC shunt motors are constant speed motors, there is only slight variation in speed from no
load to full load. These are suitable for constant speed requirement. Torque-armature
current characteristic is a straight line according to its speed torque characteristic, shunt
motor is used for medium starting torque i.e. in pumps, fans, conveyors, shapers, milling
etc. Here since φ is constant
77
Electrical Technology
N ∝ Eb = V − I a Ra
T ∝ φ Ia α Ia
N ∝ V − R3 Ra T
No Load Speed
N
Torque
Speed
N
T
T
Ia
(a)
(b)
Figure 3.25 : DC Shunt Motor Characteristics (a) Torque-current Characteristics of Armature
Current Speed Characteristics, and (b) Speed-torque Characteristics
3.7.3 DC Compound Motors
Differential compound motors are rarely used. Cumulative compound motors have
combined characteristic of series and shunt motors. Shunt field provides constant speed
and series field provides high starting torque. They are used in – rolling mills, shearing
machines, lifts, mine hoist, etc.
l
tia
ren
e
f
Dif
Speed
Shunt
Co
mm
ula
tive
Com
mula
tive
N
T
Torque
Differential
Ia
Figure 3.26 : DC Compound Motor Characteristics
3.8 STARTING, SPEED CONTROL AND
APPLICATIONS OF DC MOTORS
3.8.1 Starting of DC Motors
When the armature of a dc motor is at standstill, the back emf is zero as N = 0.

PφZ N 
∵ Eb =
 . If a dc motor is connected directly to the supply then a heavy current
60 A 


V − Eb 
 because Ra is very small
will flow through armature conductors ∵ I a =
Ra 

(i.e. less than 1 Ω). This heavy current may damage the armature windings. Thus, for the
protection of the motor against the flow of large current during starting period starters are
used. Starter provides high resistance at the time of starting and removes this additional
resistance when motor attains its normal speed. Starters are :
78
DC Machines
(a)
Two point starter – used for DC series motors.
(b)
Three point starter – used for DC shunt and compound motors.
(c)
Four point starter – for DC shunt and compound motors when speed
variation is achieved by flux control method.
Here we will discuss three-point starter only.
Three Point Starter
It consists of series starting resistance in several sections, connected on brass studs.
A spiral spring is placed on starter arm to bring this arm to OFF position when
holding coil is de-energised. Under normal running condition the arm is held at ON
position by holding coil which is magnetized by field current.
Operation
To start motor, the starter arm is moved gradually from OFF position to ON
position. At position A, total starting resistance is inserted in series with
armature and field is directly connected across supply through Brass arc and
holding coil, hereby magnetising the holding coil. On moving starter arm
clockwise further the resistance decreases in steps. At ON position, starting
resistance is totally out of circuit. The soft iron on starter arm latches with
the magnetised holding coil and hence keeps the starter arm at ON position.
Starting Resistance
C
D
A
Holding Coil or
No Volt Release
ON
OFF
Brass Arc
Soft Iron
Starter Arm
Spring
Overload Release
L
F
A
+
220 VDC
F
A
–
FF
AA
Figure 3.27 : Three-point Starter
No Volt Release
No volt release coil is an electromagnet which is series connected with the
field winding. In the abnormal position (i.e. supply failure and field winding
open) the holding coil is demagnetized and starter arm goes back to OFF
position due to spring force.
Over Load Release Coil
To provide overload protection to motors overload release is used. This coil is
connected in series with armature winding when motor draws heavy currents
79
Electrical Technology
then coil magnetises it to such an extent that it pulls lever and closes contact
L of overload release coil. This short circuits the No volt release coil. The No
volt release coil is demagnetized and releases the starting arm which goes
back to OFF position.
3.8.2 Speed Control
Speed of any dc motor is directly proportional to back emf and inversely proportional to
magnetic flux.
Eb
φ
i.e.
N∝
∵
Eb = V − I a Ra
so
N∝
V − I a Ra
φ
...
(3.83)
Thus the speed of any DC motor can be controlled by adjusting the following :
(a)
Applied voltage across armature, V.
(b)
Voltage drop across armature, Ia Ra.
(c)
Flux per pole.
Speed Control of DC Shunt Motor
Field or Flux Control Method
+
I
If
Ia
A
V
F
Eb
AA
FF
–
Figure 3.28 : Field Control for DC Shunt Motor
Speed variation is accomplished by means of a variable resistance inserted in
series with shunt field. By controlling field current the speed of DC shunt
motor is controlled. This method is independent of load on the motor.
Drawbacks of this system are :
(i)
Creeping speeds cannot be obtained.
(ii)
Top speeds are only obtained at reduced torque due to very weak field.
(iii)
At higher speeds very weak field increases armature current required
for developing the desired torque.
SAQ 5
The speed of a 500 V shunt motor is raised from 700 rpm to 1000 rpm by field
weakening, the total torque remaining unchanged. The armature and field resistance
are 0.8 Ω and 750 Ω respectively, and the line current at lower speed is 12 amps.
Calculate the additional shunt field resistance required, assuming the magnetic
circuit to be unsaturated and neglecting all losses.
80
DC Machines
Armature Control Method
Speed variation of shunt motors by armature control requires applied voltage to
change without altering the field current. So speed is adjusted by armature control
method using :
(a)
Armature resistance control.
(b)
Shunted armature control.
(c)
Terminal voltage control impressed upon armature circuit.
Armature Resistance Control
Ia
Ish
F
+
V
A
–
Eb
FF
AA
Figure 3.29 : Armature Resistance Control
When a resistance is connected in series with armature it increases
armature drop so voltage applied across armature is reduced. Thus, the
speed is reduced, which is directly proportional to voltage drop,
without altering shunt field current. In this method, large amount of
power Ia2 R (R is resistance in series with armature) is wasted so this
method is not used for long operation. Its used for short period in
printing machines, hoists, cranes, fans and blowers.
Shunted Armature Method
I
Ish
+
Series
Resistance
V
F
–
Ia
A
Divider
FF
Eb
AA
Figure 3.30 : Shunted Armature Control
Speed variation in armature resistance control method also depends on
load current. This double dependence makes speed sensibly constant on
rapidly changing loads. For more stable operation a divertor is used
across armature. It is generally applied in low power rating machines.
Armature Voltage Control
Speed variation is obtained by using variable supply voltage which is
provided by the adjustable electronic rectifier. This gives large speed
range with good speed regulation and efficiency.
81
Electrical Technology
SAQ 6
The load torque and the loss torque of a dc shunt motor are constant and
independent of speed. When the motor is connected directly to the 440 V supply and
the speed is steady, the armature current is 20 A. The resistance of the armature is
1.5 Ω. Find the resistance of a starter to limit the armature starting current to 40 A.
The starter contact is to be held on the first active stud until the motor acquires a
steady speed. The contact is then to be moved to the second stud and the initial
current is to be limited to 40 A. Find the resistance between the first and second
studs of the starter. Assume that the field flux is constant.
Ward Leonard Method
Speed control by means of an adjustable voltage generator connected across
armature terminals of the motor is called Ward Leonard system.
R
Y
B
M1
G1
R
M2
A
S
RS
G2
B
Rheostat
Figure 3.31 : Ward Leonard Method
M1 is work motor powered by generator G1 which is driven by a synchronous or
Induction motor M2 which gives constant speed drive to generator G1, and exciter G2
mounted on the same shaft as the generator. The field current for the work motor M1
and the generator G1 is obtained from exciter E. The Ward Leonard set starts the
driving motor. Variable voltage across the terminals of the generator or across
motor is obtained by varying the exciting current of the generator G1 by means of
shunt
regulator R. Switch S, at positions A and B, is used to reverse direction of voltage of
G1 and hence rotation of M1. By controlling the field current of G1 through R, the
armature voltage of M1 and hence its speed can be controlled.
Advantage
82
(a)
Very fine speed control over whole range from zero to normal speed
in both directions.
(b)
Uniform acceleration can be obtained.
(c)
DC Machines
Speed regulation is good.
Disadvantage
(a)
Arrangement is costly due to three extra machines.
(b)
Low overall efficiency of the system.
Speed Control of DC Series Motor
Armature Control
Armature Resistance Control
Speed variation is obtained by varying armature voltage drops. The
maximum range of speed control will be available on the load. This is
most economical for constant torque drives.
F
FF
Control
Resistance
+
+
A
Eb
V
–
– AA
Figure 3.32 : DC Series Motor : Armature Resistance Control
Shunted Armature Control
This method gives slow speeds at light loads. There is speed control both by
lowering armature voltage and flux variation. Armature voltage is varied by series
rheostat R1 and flux is varied by rheostat R2.
F
FF
I
+
R1
A
Eb
V
R2
AA
–
Figure 3.33 : DC Series Motor : Shunted Armature Control
Armature Voltage Control
Speed varies by varying supply voltage using adjustable electronic rectifiers.
Field Control Method
Field Divertor
Flux can be reduced by shunting a portion of field winding. This method
gives speeds above normal speed. This method is economical and provides
speed control in range not exceeding 2 : 1.
83
F
Electrical Technology
FF
A
+
Eb
Diverter
V
–
AA
Figure 3.34 : Field Divertor Method
Tapped Field Control
This is another method of increasing speed by reduced flux. It is obtained by
reducing number of turns of field winding. This method is generally used in
traction.
F
FF
I
–
A
Eb
Diverter
AA
–
Figure 3.35 : Tapped Field Control
Paralleling Field Coil Method
By regrouping the field coils in series or parallel the variable speed is
obtained. It is used in electric traction.
3.8.3 Applications
DC Series Motors
(a)
Electric Traction
•
High starting torque and reduced torque at high speeds.
•
Large tractive effort, so a number of motors in series.
•
 Power 
 ratio ∴ Motors robust in construction.
High η, large 
 Weight 
(b)
Hoists, cranes, excavations, electric vehicles, streetcars, battery – powered
portable tools, automotive starter motors : all because of high starting torque.
(c)
Drive fan load
• Torque requirement increases with the square of speed.
(d)
Battery Operated Vehicles
• Cars and other battery – powered vehicles have traction characteristics.
• Speed control can be carried out by thyristers.
DC Shunt Motors
84
(a)
Constant speed applications.
(b)
Applications where a wide range of speed control is employed e.g. in lathes,
in paper industry etc.
(c)
As a separately – excited motor when field winding is disconnected from
armature and connected to an external voltage source.
DC Machines
DC Compound Motors
(a)
Rolling Mills : To improve characteristic and have higher starting torque
for the lower rolling motor of the twin-drive. Cummulative compound
motors are better suited than shunt motors. In conjunction with flywheel,
they can take sudden temporary loads and are ideal for rolling mills and
coal-cutting machines.
(b)
Punching Press.
(c)
Milling Machine.
(d)
Traction Motors : Only where supply voltage is likely to vary considerably.
(e)
Hoisting and Lowering of Loads (along with regenerative braking).
(f)
to (v) were for cumulatively compounded motor.
(g)
Differential compound motors find only rare application as in research and
experimental work.
Example 3.5
A DC generator has the following magnetization characteristics at 800 rpm.
Generated emf (V)
27.5
53
75
88
95
106
112
Field Current (A)
1
2
3
4
5
6
7
If the machine is shunt excited, determine the induced emf for a field circuit
resistance of 19 ohm. Calculate the load current for a terminal voltage of
76 volts. The armature resistance in 0.1 ohm. Neglect the armature reaction.
Solution
The magnetization characteristic at 800 rpm is plotted in Figure 3.36, the field
resistance line OA for 19 Ω is drawn as follow :
Take any current say 3 amp, multiply it by 19 Ω, we get 57 V. Locate
point B (3 amp, 57 volts) and draw a line joining the origin (0, 0), and the point B.
This gives 19 Ω, resistance line. It cuts the induced emf corresponding to this point
C is 100 volts. Hence the generator will develop 100 volts corresponding to field
circuit resistance of 19 Ω.
85
A
Electrical Technology
112
106
100
Generated e.m.f. (V)
95
88
75
B
53
52
I
If
+
Ia
Rse
27
E
Rsh
20
V
Ra
–
10
1
2 3
4
5
6
7
Field current (A)
Figure 3.36
Given terminal voltage = 76 volts
V
76
=
= 4 amp
Ra 19
Field current I f =
Induced emf E corresponding to If = 4 amp from the given OCC is 88 volts.
∴
Voltage drop in the armature Ia Ra = E − V
or
I a Ra = 88 − 76 = 12 volts
∴
Ia =
∴
Load current I = Ia − If
12 12
=
= 120 amp
Ra 0.1
= 120 – 4
= 116 amp.
3.9 SUMMARY
You learnt about principles of electromechanical energy conversion and calculations of
emf and force or torque experienced by current carrying conductor. After a brief
introduction to the constructional features of dc machine in Section 3.3, you learnt the
calculations of emf for generators, torque and speed regulation for motor. After study of
characteristics of generator and motors you were able to decide the applications of DC
machines. You also learnt the starting and different speed control schemes for DC motors
which are used for industrial purpose in Section 3.8.
3.10 ANSWERS TO SAQs
SAQ 1
Load current I =
86
W 45 × 103 2250
=
=
= 204.55 amp
V
220
11
Field current I sh =
DC Machines
V
220
=
= 4.4 amp
Rsh
50
Armature current I a = I + I sh = 204.55 + 4.4 = 208.95 amp
Armature drop I s Rs = 208.95 × 0.01 = 2.0895 volts
Induced emf E = V + I a Ra = 220 + 2.0895 = 222.0895 volts
But
E=
or
N =
φZ N P
×
60
A
E × 60 × A
222.0895 × 60 × 2
=
φZ P
0.02 × (150 × 2) × 4
= 1110.45 rpm
SAQ 2
The circuit diagram for the given system is shown in Figure 3.37.
A
120
110
100
Generated e.m.f. (V)
(a)
90
80
70
B
60
+
If
Ia
50
40
E
30
Ra
V
Rf
20
–
10
1
2
3
4
5
6
7
Field current (A)
Figure 3.37
Current in the shunt field
I sh =
500
= 2 amp
250
Armature current = I a = I + I sh
or
I a = 50 + 2 = 52 amp
Voltage drop in the armature and series resistance
= I a ( Ra + Rse )
= 52 (0.05 +0.03)
= 4.16 volts
Generated emf
E = Terminal voltage
+ I a ( Ra + Rse )
+ brush drop
87
E = 500 + 4.16 + 2
Electrical Technology
= 506.16 volts
(b)
Load current I =
P 11.5 × 1000
=
= 50 amp
V
230
Field current I sh =
V
230
=
= 2.3 amp
Rsh 100
∴ Armature current I a = I + I sh
or
I a = 50 + 2.3 = 52.3 amp
Armature drop I a Ra = 52.3 × 0.2 = 10.46 v
Brush drop = 1 × 2 = 2 volts
Generated emf = 230 + 10.46 + 2 = 242.46 volts
But
E=
φZ N P
×
60
A
For lap wound generator A = P
∴
E=
φZ N
60 E
or φ =
60
ZN
or
φ=
60 × 242.46
= 0.012123 webers/pole
1200 × 1000
SAQ 3
Line current I = 32 amp
Field current I sh =
V
200
=
= 2 amp
Rsh 100
Armature current I a = I − I sh = 32 − 2 = 30 amp
Back emf Eb1 = V − I a Ra = 200 − 30 × 0.5 = 185 volts
We know that back emf ∝ a flux per pole × speed. It is given that field current is
same, therefore, flux per pole remains constant
Back emf ∝ speed
∴
Eb1 ∝ N1
. . . (i)
Let R ohm be the external resistance to be added in the armature circuit to reduce
the speed to N2 = 300 rmp. At this speed let the back emf be Eb2.
∴
E b2 ∝ N 2
Dividing Eq. (ii) by Eq. (i), we have
Eb 2 300
=
or Eb 2 = 111 volts
185 500
Since armature current remains same
88
∴
Eb 2 = V − I a ( Ra + R)
or
111 = 200 – 30 (0.5 + R)
or
R = 2.467 Ω
. . . (ii)
∴
DC Machines
Required resistance to be added in the armature circuit = 2.467 Ω.
SAQ 4
Output of dc series motor = Eb I a = 30 × 103 watts
Output is also equal to
2π NT
1200
= 2π ×
×T
60
60
∴
2π ×
1200
× T = 30 × 103
60
or
T=
750
N-m
π
For a series motor with unsaturated field : T α I a2
T = k I a2
750
= k (80) 2
π
. . . (i)
At starting Ia = 120 amp. Hence torque at starting,
Ts = k (120)2
. . . (ii)
Dividing Eq. (ii) by Eq. (i) we get,
Ts
 120 
=

750  80 
π
2
2
750
 120 
Ts = 
 ×
80
π


2
750
3
Ts =   ×
π
2
Ts =
9 750
×
4
π
Ts = 537.15 N-m.
SAQ 5
In the first case :
I f1 =
500 2
= amp
750 3
I a1 = 12 −
2 34
=
amp
3
3
Eb1 = V − I a Ra = 500 −
or
Eb1 = 500 −
34
× 0.8
3
27.2
= 490.93 volts
3
89
Ish
Electrical Technology
IL
+
Ia
0.03 Ω
Eb
750 Ω
0.8 Ω
500 V
–
Let the new shunt field resistance be R. Hence in the second case,
If2 =
V
500
=
R
R
Let the new armature current be Ia2
∴
Eb 2 = V − I a 2 Ra = 500 − I a 2 × 0.8
. . . (i)
As torque is the same in two cases
∴
T ∝ φ1 I a1 ∝ φ 2 I a 2 or φ1 I a1 = φ 2 I a 2
Since saturation is neglected, hence flux is proportional to field current
∴
φ1 ∝ I f 1 ∝
2
500
and φ 2 ∝ I f 2 ∝
3
R
Substituting the values in Eq. (ii)
∴
2 34 500
×
=
× Ia2
3 3
R
or
I a2 =
2 × 34 R
= 15.1 R × 10 − 3
3 × 3 × 500
Substituting in Eq. (i), we have
Eb 2 = 500 − 0.0151 × 0.8 R
= 500 − 0.01208 R
Using the relation
N 2 Eb 2 φ1
=
×
, we get
N1
Eb1 φ 2
2
1000 500 − 0.01208 R
=
× 3
500
700
490.93
R
or
10 (500 − 0.01208 R) 2
R
=
× ×
7
490.93
3 500
or
R (500 – 0.01208 R) = 525996.42
or
0.01208 R2 – 500 R + 525996.42 = 0
Solving this equation, we get
R = 1080.18 Ω
Hence additional shunt field resistance required
= 1080.18 – 750
90
. . . (ii)
= 330.18 Ω
DC Machines
SAQ 6
Let the total resistance required (including armature resistance) in the armature
circuit to limit armature current to 40 A be R
R=
440
= 11 Ω [as Eb = 0 at starting, i.e. N = 0]
40
External resistance to be inserted in the armature circuit
R − Ra = 11 − 1.5 = 9.5 Ω
With this resistance present in the armature circuit the back emf developed under
steady speed.
Eb = 440 − I a R = 440 − 20 × 11 = 220 volts
[Since T α Ia for shunt motor and it remains constant due to constant load torque, Ia
at steady state remains same, i.e. 20 A].
Let the resistance present between the first and the second stud be r Ω. When this
resistance is cut out by moving the arm to the second stud, the resistance remaining
in the armature circuit = (R – r)
∴
V − Eb
440 − 220
= 40 amp (given ) or ( R − r) =
= 5.5
R−r
40
∴
r = R – 5.5 = 11 – 5.5 = 5.5 Ω
91