7. Magnetic Induction and the DC
Generator
• Faraday’s Law e = N dΦ / dt
– e = the induced voltage in volts (V)
– N = the number of series-connected turns of wire
in turns (t)
– dΦ/dt = rate of change in flux in Webers/second
(Wb/s)
• e = B L v
B = the flux density in teslas (T)
L = the length of the conductor that is in the magnetic
field in meters (m)
v = the relative velocity between the wire and the
flux, in meters/second (m/s)
13. 13
•The structure of an electric
machine has two major
components, stator and rotor,
separated by the air gap.
• Stator:
Does not move and normally is
the outer frame of the machine.
• Rotor:
Is free to move and normally is
the inner part of the machine.
•Both rotor and stator are made of
ferromagnetic materials.
a'
b
c'
a
b'
c
Stator
Rotor
Electric Machines
Basic Structure
Y’
B
Y
B’
Stator
Rotor
R
R’
N
S
14. Determining the Direction of Induced
Electromotive Force (e.m.f)
Notes:-
• If the direction of motion is reversed keeping
flux direction same then the direction of
induced e.m.f. and hence the direction of
current is reversed.
if flux direction is reversed keeping direction
of motion same then the direction of induced
e.m.f. and hence the direction of current is
reversed.
16. Type of DC machine
• Generator action:An emf(voltage) is induced in
a conductor if it moves through a magnetic field.
• •Motor action:A force is induced in a conductor
that has a current going through it and placed in
a magnetic field
17. Generator
• Conversion from mechanical to electrical
Practical Generator
1 Magnetic Frame or Yoke
2. Pole-Cores and Pole-Shoes
3. Pole Coils or Field Coils
4. Armature Core
5. Armature Windings or Conductors
6. Commutator
7. Brushes and Bearings
18.
19. D.C. GENERATORS
An electrical generator is a machine which converts
mechanical energy (or power) into electrical energy (or
power).
The energy conversion is based on the principle of the production of
dynamically or motionally) induced e.m.f. As seen rom Fig. 26.1, whenever a
conductor cuts magnetic flux, dynamically induced e.m.f. s produced in it
according to Faraday’s Laws of Electromagnetic Induction. This e.m.f. causes
a current to flow if the conductor circuit is closed. Hence, two basic essential
parts of an electrical generator are (i) a magnetic field and (ii) a conductor or
conductors which can so move as to cut the flux.
20. D.C. GENERATORS
are joined to two slip-rings ‘a’ and ‘b’ which are insulated from each
other and from the central shaft. Two collecting brushes (of carbon or
copper) press against the slip-rings. Their function is to collect the
current induced in the coil and to convey it to the external load
resistance R.
21. Working
the coil assumes successive positions in the field, the flux linked
with it changes. Hence, an e.m.f. is induced in it which is proportional
to the rate of change of flux linkages (e = NdΦdt). When the plane of
the coil is at right angles to lines of flux i.e. when it is in position, 1,
then flux linked with the coil is maximum but rate of change of flux
linkages is minimum. It is so because in this position, the coil sides
AB and CD do not cut or shear the flux, rather they slide along them
i.e. they move parallel to them. Hence, there is no induced e.m.f. in
the coil. Let us take this no-e.m.f. or vertical position of the coil as the
starting position. The angle of rotation or time will be measured from
this position.
22. Working
As the coil continues rotating further, the rate of change of flux
linkages (and hence induced e.m.f. in it) increases, till position 3 is
reached where θ = 90º. Here, the coil plane is horizontal i.e. parallel
to the lines of flux. As seen, the flux linked with the coil is minimum
but rate of change of flux linkages is maximum. Hence, maximum
e.m.f. is induced in the coil when in this position (Fig. 26.3).
In the next quarter revolution i.e. from 90º to 180º, the flux linked with
the coil gradually increases but the rate of change of flux linkages
decreases. Hence, the induced e.m.f. Decreases gradually till in
position 5 of the coil, it is reduced to zero value. For making the flow
of current unidirectional in the external circuit, the slip-rings are
replaced by split-rings (Fig. 26.4). The split-rings are made out of a
conducting cylinder which is cut into two halves or segments
insulated from each other by a thin sheet of mica or some other
nsulating material
(Fig. 26.5).
23. Working
As before, the coil ends are joined to these segments on which rest the
carbon or copper brushes. It is seen [Fig. 26.6 (a)] that in the first half
revolution current flows along (ABMNLCD) i.e. The brush No. 1 in contact
with segment ‘a’ acts as the positive end of the supply and ‘b’ as the negative
end. In the next half revolution [Fig. 26.6 (b)], the direction of the induced
current in the coil has reversed. But at the same time, the positions of
segments ‘a’ and ‘b’ have also reversed with the result that brush No. 1
comes in touch with the segment which is positive i.e. segment ‘b’ in this
case. Hence, current in the load resistance again flows from M to L. The
waveform of the current through the external circuit is as shown in Fig. 26.7.
This current is unidirectional but not continuous like pure direct current.
24. Working
As before, the coil ends are joined to these segments on which rest the
carbon or copper brushes. It is seen [Fig. 26.6 (a)] that in the first half
revolution current flows along (ABMNLCD) i.e. The brush No. 1 in contact
with segment ‘a’ acts as the positive end of the supply and ‘b’ as the negative
end. In the next half revolution [Fig. 26.6 (b)], the direction of the induced
current in the coil has reversed. But at the same time, the positions of
segments ‘a’ and ‘b’ have also reversed with the result that brush No. 1
comes in touch with the segment which is positive i.e. segment ‘b’ in this
case. Hence, current in the load resistance again flows from M to L. The
waveform of the current through the external circuit is as shown in Fig. 26.7.
This current is unidirectional but not continuous like pure direct current.
25. Types of Generators
(a)separately-excited generators
(b) self-excited generators.
(a) Separately-excited generators are those whose field
magnets are energised from an independent
external source of d.c. current.
27. 26.4. Yoke
The outer frame or yoke serves double
purpose :
(i) It provides mechanical support for the
poles and acts as a protecting cover for the
whole machine and
(ii) It carries the magnetic flux produced by
the
poles.
26.5. Pole Cores and Pole Shoes
(i) they spread out the flux in the air gap and also, being of larger cross-
section, reduce the reluctance of the magnetic path
(ii) they support the exciting coils (or field coils) as shown in Fig. 26.14.
There are two main types of pole construction.
28. 26.6. Pole Coils
The field coils or pole coils, which consist of copper wire or strip, are former-
wound for the correct dimension (Fig. 26.13). Then, the former is removed
and wound coil is put into place over the
core as shown in Fig. 26.14.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39. Self-excited generators
There are three types of self-excited generators named
according to the manner in which their field coils (or
windings) are connected to the armature
(i) Shunt wound
The field windings are connected across or in parallel with
the armature conductors and have the full voltage of the
generator applied
(ii) Series Wound
In this case, the field windings are joined in series with the armature
conductors
Eg = terminal voltage + armature drop
42. Compound Wound
It is a combination of a few series and a
few shunt windings and can be either short-shunt or long-
shunt as shown in Fig. 26.44 (a) and (b) respectively. In a
compound generator, the shunt field is stronger than the
series field.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52. Example
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 voltage
and the armature current. Allow 1 V per
brush for contact drop.
53. Example
A short-shunt compound generator delivers
a load current of 30 A at 220 V, and has
armature, series-field and shunt-field
resistances of 0.05 Ω, 0.30 Ω and 200 Ω
respectively. Calculate the induced e.m.f.
and the armature current. Allow 1.0 V per
brush for contact drop.
54.
55.
56.
57. Example
A shunt generator delivers 450 A at 230 V
and the resistance of the shunt field and
armature are 50 Ω and 0.03 Ω respectively.
Calculate the generated e.m.f.
58.
59.
60.
61.
62.
63.
64. D.C. MOTOR
DC motor is any of a class of electrical machines that converts direct
current electrical power into mechanical power. electrical power into
mechanical power.
65.
66.
67.
68. Torque
By the term torque is meant the turning or twisting moment of a force about
an axis. It is measured by the product of the force and the radius at which this
force acts. Consider a pulley of radius r metre acted upon by a
circumferential force of F Newton which causes it to rotate at N r.p.m. (Fig.
29.10).
T = F × r Newton-metre (N - m)
Work done by this force in one revolution
= Force × distance = F × 2πr Joule
Power developed = F × 2 πr × N Joule/second or Watt
= (F × r) × 2π N Watt
Now 2 πN = Angular velocity ω in radian/second and F ×r = Torque T
72. Speed of a D.C. Motor
From the voltage equation of a motor
It shows that speed is directly proportional to back e.m.f. Eb and inversely to
the flux Φ on N ∝ Eb/Φ.
For Series Motor
73. Speed Regulation
The term speed regulation refers to the change in speed of a motor with
change in applied load torque, other conditions remaining constant. By
change in speed here is meant the change which occurs under these
conditions due to inherent properties of the motor itself and not those
changes which are affected through manipulation of rheostats or other
speed-controlling devices.
defined as the change in speed when the load on the motor is reduced
from rated value to zero, expressed as percent of the rated load speed
74. Torque and Speed of a D.C.
Motor
It will be proved that though torque of a motor is admittedly a function of flux
and armature current,. In fact, it is the speed which depends on torque and
not. It has been proved earlier that
Suppose that the flux of a motor is decreased by decreasing the field current.
81. Compound Motors
(a) Cumulative-compound Motors
(b) Differential-compound Motors
Since series field opposes the shunt field, the flux is decreased as load is
applied to the motor. This results in the motor speed remaining almost
constant or even increasing with increase in load (because, N ∝ Eb/(Φ). Due
to this reason, there is a decrease in the rate at which the motor torque
increases with load. Such motors are not in common use. But because they
can be designed to give an accurately constant speed under all conditions,
they find limited application for experimental and research work
82.
83.
84.
85.
86.
87. SPEED CONTROL OF
D.C. MOTORS
Speed Control of Shunt motors
(i) Variation of Flux or Flux Control Method
It is seen from above that N ∝ 1/ΦThe flux of a d.c. motor
can be changed by
changing Ish with help of a shunt field rheostat (Fig. 30.1).
Since
Ish is relatively small, shunt field rheostat has to carry only
a small current, which means I2R loss is small.
88. so that rheostat is small in size. This method is, therefore, very efficient. In
non-interpolar machine, the speed can be increased by this method in the
ratio 2 : 1. Any further weakening of flux Φ adversely affects the
communication and hence puts a limit to the maximum speed obtainable with
the method. In machines fitted with interpoles, a ratio of maximum to minimum
speed of 6 : 1 is fairly common.
89. (ii) Armature or Rheostatic Control Method
This method is used when speeds below the no-load speed are required. As
the supply voltage is normally constant, the voltage across the armature is
varied by inserting a variable rheostat or resistance (called controller
resistance) in series with the armature circuit as shown in Fig. 30.4 (a).
As controller resistance is increased, p.d.
across the armature is decreased,
thereby decreasing the
armature speed. For a load
constant torque, speed is
approximately proportional to the
p.d. across the
armature. From the speed/armature
current characteristic [Fig. 30.4 (b)],
it is seen that greater the
resistance in the armature circuit,
greater is the fall in the speed.
90.
91.
92.
93.
94. (iii) Voltage Control Method
(a) Multiple Voltage Control
In this method, the shunt field of the motor is connected permanently to a
fixed exciting voltage, but the armature is supplied with different voltages by
connecting it across one of the several different voltages by means of suitable
switchgear. The armature speed will be approximately proportional to these
different voltages. The intermediate speeds can be obtained by adjusting the
shunt field regulator. The method is not much used, however
(b) Ward-Leonard System
This system is used where an unusually wide (upto 10 : 1) and very sensitive
speed control is required as for colliery winders, electric excavators, elevators
and the main drives in steel mills and blooming and paper mills. The
arrangement is illustrated in Fig. 30.9..
98. Transformer
An A.C. device used to change high voltage
low current A.C. into low voltage high current
A.C. and vice-versa without changing the
frequency
1. Transfers electric power from one circuit to
another
2. It does so without a change of frequency
3. It accomplishes this by electromagnetic
induction
4. Where the two electric circuits are in mutual
inductive influence of each other.
99. Principle of operation
It is based on
principle of MUTUAL
INDUCTION.
According to which
an e.m.f. is induced
in a coil when
current in the
neighbouring coil
changes.
100. Constructional detail : Shell type
• Windings are wrapped around the center leg of a
laminated core.
104. Core type
Fig1: Coil and laminations of
core type transformer
Fig2: Various types of cores
105. Shell type
• The HV and LV
windings are split
into no. of sections
• Where HV winding
lies between two
LV windings
• In sandwich coils
leakage can be
controlled
Fig: Sandwich windings
108. Working of a transformer
1. When current in the primary coil
changes being alternating in
nature, a changing magnetic field
is produced
2. This changing magnetic field gets
associated with the secondary
through the soft iron core
3. Hence magnetic flux linked with
the secondary coil changes.
4. Which induces e.m.f. in the
secondary.
109.
110. Ideal Transformers
• Zero leakage flux:
-Fluxes produced by the primary and secondary currents
are confined within the core
• The windings have no resistance:
- Induced voltages equal applied voltages
• The core has infinite permeability
- Reluctance of the core is zero
- Negligible current is required to establish magnetic
flux
• Loss-less magnetic core
- No hysteresis or eddy currents
111. Ideal transformer
V1 – supply voltage ; I1- noload input current ;
V2- output voltgae; I2- output current
Im- magnetising current;
E1-self induced emf ; E2- mutually induced emf
112. EMF equation of a transformer
• Worked out on board /
• Refer pdf file: emf-equation-of-tranformer
114. Transformer on load assuming no
voltage drop in the winding
Fig shows the Phasor diagram of a transformer
on load by assuming
1. No voltage drop in the winding
2. Equal no. of primary and secondary turns
115. Transformer on load
Fig. a: Ideal transformer on load
Fig. b: Main flux and leakage
flux in a transformer
121. • The effect of circuit parameters shouldn’t be changed
while transferring the parameters from one side to
another side
• It can be proved that a resistance of R2 in sec. is
equivalent to R2/k2 will be denoted as R2’(ie. Equivalent
sec. resistance w.r.t primary) which would have caused
the same loss as R2 in secondary,
2
2
2
2
1
2
'
2
2
2
2
'
2
2
1
k
R
R
R
I
I
R
I
R
I
123. Equivalent circuit referred to
secondary side
•Transferring primary side parameters to secondary side
Similarly exciting circuit parameters are also transferred to
secondary as Ro’ and Xo’
126. Transformer Tests
Electrical Machines
•The performance of a transformer can be calculated on the basis of
equivalent circuit
•The four main parameters of equivalent circuit are:
- R01 as referred to primary (or secondary R02)
- the equivalent leakage reactance X01 as referred to primary
(or secondary X02)
- Magnetising susceptance B0 ( or reactance X0)
- core loss conductance G0 (or resistance R0)
•The above constants can be easily determined by two tests
- Oper circuit test (O.C test / No load test)
- Short circuit test (S.C test/Impedance test)
•These tests are economical and convenient
- these tests furnish the result without actually loading the
transformer
127. In Open Circuit Test the transformer’s secondary winding is open-circuited, and
its primary winding is connected to a full-rated line voltage.
• Usually conducted on
H.V side
• To find
(i) No load loss or core
loss
(ii) No load current Io
which is helpful in
finding Go(or Ro ) and Bo
(or Xo )
2
0
2
0
0
2
0
oc
0
0
2
0
oc
0
0
o
0
0
0
2
2
0
0
0
m
0
0
w
c
0
0
0
0
0
0
B
e
susceptanc
Exciting
&
V
W
G
e
conductanc
Exciting
;
G
V
W
Y
;
Y
V
I
sin
I
I
cos
I
I
cos
cos
loss
Core
G
Y
V
I
-I
I
I
or
I
or
I
V
W
I
V
W
w
oc
oc
Open-circuit Test
0
0
0
0
0
0
0
0
V
I
B
V
I
G
I
V
X
I
V
R
w
w
128. Short-circuit Test
In Short Circuit Test the secondary terminals are short circuited, and the
primary terminals are connected to a fairly low-voltage source
The input voltage is adjusted until the current in the short circuited windings
is equal to its rated value. The input voltage, current and power is
measured.
• Usually conducted on L.V side
• To find
(i) Full load copper loss – to pre determine the
efficiency
(ii) Z01 or Z02; X01 or X02; R01 or R02 - to predetermine the
voltage regulation
130. Transformer Voltage Regulation
and Efficiency
Electrical Machines
The output voltage of a transformer varies with the load even if the input
voltage remains constant. This is because a real transformer has series
impedance within it. Full load Voltage Regulation is a quantity that compares
the output voltage at no load with the output voltage at full load, defined by
this equation:
%
100
down
Regulation
%
100
up
Regulation
,
,
,
,
,
,
nl
S
fl
S
nl
S
fl
S
fl
S
nl
S
V
V
V
V
V
V
%
100
/
down
Regulation
%
100
/
up
Regulation
V
V
k
noload
At
,
,
,
,
p
s
x
V
V
k
V
x
V
V
k
V
nl
S
fl
S
P
fl
S
fl
S
P
Ideal transformer, VR = 0%.
132. Transformer Phasor Diagram
10/3/2023 132
Electrical Machines
Aamir Hasan Khan
To determine the voltage regulation of a transformer, it is necessary
understand the voltage drops within it.
133. Transformer Phasor Diagram
10/3/2023 133
Electrical Machines
Aamir Hasan Khan
Ignoring the excitation of the branch (since the current flow through the
branch is considered to be small), more consideration is given to the series
impedances (Req +jXeq).
Voltage Regulation depends on magnitude of the series impedance and the
phase angle of the current flowing through the transformer.
Phasor diagrams will determine the effects of these factors on the voltage
regulation. A phasor diagram consist of current and voltage vectors.
Assume that the reference phasor is the secondary voltage, VS. Therefore the
reference phasor will have 0 degrees in terms of angle.
Based upon the equivalent circuit, apply Kirchoff Voltage Law,
S
eq
S
eq
S
P
I
jX
I
R
V
k
V
134. Transformer Phasor Diagram
10/3/2023 134
Electrical Machines
Aamir Hasan Khan
For lagging loads, VP / a > VS so the voltage regulation with lagging loads is > 0.
When the power factor is unity, VS is lower than VP so VR > 0.
135. Transformer Phasor Diagram
10/3/2023 135
Electrical Machines
Aamir Hasan Khan
With a leading power factor, VS is higher than the referred VP so VR < 0
136. Transformer Phasor Diagram
Electrical Machines
For lagging loads, the vertical components of Req and Xeq will partially
cancel each other. Due to that, the angle of VP/a will be very small, hence
we can assume that VP/k is horizontal. Therefore the approximation will
be as follows:
138. Transformer Efficiency
Electrical Machines
Transformer efficiency is defined as (applies to motors, generators and
transformers):
%
100
in
out
P
P
%
100
loss
out
out
P
P
P
Types of losses incurred in a transformer:
Copper I2R losses
Hysteresis losses
Eddy current losses
Therefore, for a transformer, efficiency may be calculated using the following:
%
100
cos
cos
x
I
V
P
P
I
V
S
S
core
Cu
S
S
139. Losses in a transformer
Core or Iron loss:
Copper loss:
143. Transformers
• Figure 34.2 shows the general arrangement of a
transformer. A steel core C consists of laminated sheets,
about 0.35–0.7 mm thick, insulated from one another.
The purpose of laminating the core is to reduce the
eddy-current loss.
144. • When the secondary is on open circuit, its terminal
voltage is the same as the induced e.m.f. The primary
current is then very small, so that the applied voltage V1
is practically equal and opposite to the e.m.f. induced in
P. Hence:
• Since the full-load efficiency of a transformer
is nearly 100 per cent
• When a load is connected across the secondary
terminals, the secondary current – by Lenz’s law –
produces a demagnetizing effect
154. Phasor diagram for a transformer on no load
It is most convenient to commence the phasor diagram with the phasor
representing the quantity that is common to the two windings, namely the flux
Φ. This phasor can be made any convenient length and may be regarded
merely as a reference phasor, relative to which other phasors have to be
drawn.
that the e.m.f. Induced by a sinusoidal flux leads the flux by a quarter of a
cycle
Consequently the
e.m.f. E1 induced in the primary winding is represented by a phasor drawn
90° ahead of Φ
The e.m.f. E2 also leads the flux by 90°,
but the effect which this produces
at the terminals of the transformer depends
on the manner in which the secondary winding
The no-load current, I0, taken by the primary
consists of two components:
155.
156.
157.
158. Phasor diagram for an ideal loaded transformer
it follows that the secondary terminal voltage V2 is the same as the e.m.f. E2
induced in the secondary, and the primary applied voltage V1 is equal to the
e.m.f. E1 induced in the primary winding. Also, if we again assume equal
number of turns on the primary and secondary windings, then E1 = E2
Let us consider the general case of a load having a lagging power factor
cos φ2; hence the phasor representing the secondary current I2 lags V2 by an
angle φ2, as shown in Fig. 34.5.
Phasor I2′ represents the component of the
primary current to neutralize the
Demagnetizing effect of the secondary
current and is drawn equal and opposite
to I2.
I0 is the no-load current of the transformer,
already discussed
in section 34.5. The phasor sum of I2′ and
I0 gives the total current I1 taken
from the supply, and the power factor on
the primary side is cos φ1, where φ1
is the phase difference between V1 and I1.
In Fig. 34.5 the phasor representing I0 has,
for clarity, been shown far
larger relative to the other current phasors
than it is in an actual transformer.
175. Open-circuit and short-circuit tests on a transformer
The primary current
on no load is usually less than 5 per cent of
the full-load current, so that the
I2R loss on no load is less than 1/400 of
the primary I2R loss on full load and
is therefore negligible compared with the
core loss. Hence the wattmeter
reading can be taken as the core loss of
the transformer.
(b) Short-circuit test
The secondary is short-circuited through a suitable ammeter A2, as shown in
Fig. 34.23 and a low voltage is applied to the primary circuit. This voltage
should, if possible, be adjusted to circulate full-load currents in the primary
and secondary circuits. Assuming this to be the case, the I2R loss in the
windings is the same as that on full load. On the other hand, the core loss is
negligibly small, since the applied voltage and therefore the flux are only
about one-twentieth to one-thirtieth of the rated voltage and flux, and
the core loss is approximately proportional to the square of the flux. Hence
the power registered on wattmeter W can be taken as the I2R loss in the
windings.
176.
177.
178.
179.
180. Three-phase core-type transformers
Modern large transformers are usually of the three-phase core type
shown in Fig. 34.24. Three similar limbs are connected by top and
bottom yokes, each limb having primary and secondary windings,
arranged concentrically. In Fig. 34.24 the primary is shown star-
connected and the secondary delta connected. Actually, the windings
may be connected star–delta, delta–star, star–star or delta–delta,
depending upon the conditions under which the transformer is to be
used.
181.
182. The nearer the ratio of transformation is to unity, the greater is the
economy of conductor material. Also, for the same current density in the
windings and the same peak values of the flux and of the flux density, the I2R
loss in the auto-transformer is lower and the efficiency higher than in the two
winding transformer. Auto-transformers are mainly used for (a)
interconnecting systems that are operating at roughly the same voltage and
(b) starting cage-type induction motors (section 38.8). Should an auto-
transformer be used to supply a low-voltage system from a high-voltage
system, it is essential to earth the common connection, for example, B in Fig.
34.25, otherwise there is a risk of serious shock. In general, however, an
auto-transformer should not be used for interconnecting high-voltage and low-
voltage systems.