Complete information about Transformers

1. EVEN SEMESTER
P. Maria Sheeba
ASSISTANT PROFESSOR /ECE
MOUNT ZION COLLEGE OF ENGINEERING AND TECHNOLOGY
PUDUKKOTTAI
1
BE8254 – BASIC ELECTRICAL AND INSTRUMENTATION
ENGINEERING

2. .
A transformer is a device for increasing or decreasing an ac voltage.
Introduction2.1
2

3. • A transformer consists of a primary coil and a
secondary coil, both wound on an iron core.
• The changing magnetic flux produced by the
current in the primary coil induces an emf in
the secondary coil.
Introduction2.1
3

4. • A transformer is a
device that changes ac
electric power at one
voltage level to ac
electric power at
another voltage level.
Introduction2.1
4

5. 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
Introduction2.1
5

6. • In terms of number of windings
– Conventional transformer: two windings
– Autotransformer: one winding
– Others: more than two windings
• In terms of number of phases
– Single-phase transformer
– Three-phase transformer
Transformer Classification
6

7. • Depending on the voltage level at which the winding is
operated
– Step-up transformer: primary winding is a low voltage (LV)
winding
– Step-down transformer : primary winding is a high voltage
(HV) winding
Transformer Classification2.1
7

8. • It is based on principle of MUTUAL
INDUCTION.
Principle of Operation
2.2
8

9. MUTUAL INDUCTION
• . When an e.m.f. is induced in a coil when
current in the neighbouring coil changes
2.2
9

10. Working of a transformer
1. When current in the
primary coil changes
being alternating in
nature, a changing
magnetic field is
produced
.
2.2
10

11. 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
Working of a transformer
2.2
11

12. Power Transmission
Transformers play a key role in the transmission
of electric power.
2.2
12

13. • There are two basic parts of transformer
1. Transformer core
2. Transformer coil
Construction2.2
13

14. Transformer winding
• There are three types of transformer winding
• 1. Shell type.
• 2.Core type.
• 3.Berry type.
2.2
14

15. Shell type
• Windings are wrapped around the center leg of a
laminated core.
2.2
15

16. Core type
• Windings are wrapped around two sides of a laminated square
core.
2.2
16

17. Sectional view of transformers
Note:
High voltage conductors are smaller cross section conductors
than the low voltage coils
2.2
17

18. Core type
Coil and laminations of core
type transformer
Various types of cores
2.2
18

19. 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
Sandwich windings
2.2
19

20. Cut view of transformer2.2
20

21. 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
2.2
21

22. Ideal Transformers
• The core has infinite permeability
1. Reluctance of the core is zero
2. Negligible current is required to establish
magnetic flux
• Loss-less magnetic core
No hysteresis or eddy currents
2.2
22

23. 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
2.2
23

24. Let
• N1 = Number of turns in primary windings.
• N2 = Number of turns in second windings.
• ∅m =flux.
• ∅m = Maximum flux in the core in Webbers.
EMF equation of Transformer2.3
24

25. • ∅m = Bm.A,
• f = Frequency of A.C
input in Hz.
• E1 = e.m.f induced in
primary.
• E2 = e.m.f induced in
secondary.
EMF equation2.3
25

26. • As shown in fig- flux increases from its zero
value to maximum value Øm in one quarter of
the cycle i.e. in ¼ second.
• Average rate of change of flux = (Øm / ¼f.)
• = 4f Øm Wb/s or volt.
EMF equation2.3
26

27. • Average e.m.f /per turn = 4f Øm volt.
• If flux Øm varies sinusoidally, then r.m.s value of
induced .e.m.f is obtained by multiplying the
average value with form factor.
EMF equation
27
2.3

28. • Form factor =r.m.s value /
Average value =1.11
• r.m.s value of e.m.f/turn = 1.114 f Øm
= 4.44f Øm volt
• now r.m.s value of the induced e.m.f in the whole primary
winding.=
( induced e.m.f/turn)*number of primary turns
EMF equation2.3
28

29. • E1=4.44fN1Øm…………………………(1)
• E1 = 4.44fN1BmA.
(Øm= BmA)
Similarly, r.m.s value of the e.m.f. induced in
secondary is,
• E2 = 4.44fN2 Øm
EMF equation2.3
29

30. • E2 = 4.44fN2BmA.……..(2)
(Øm= BmA)
It’s seen from (1) and (2) that
E1/N1=E2/N2= = 4.44f Øm.
EMF equation2.3
30

31. Ratio of transformer
• 1.Voltage ratio.
• 2.Current Ratio.
• 3.Volt Amp Ratio.
Voltage ratio
2.3
31

32. • Transformation Ratio=K
2.3
32

33. • Current ratio:
• Volt Amp Ratio:
33
2.3

34. • The maximum flux density in the core of 240/2400 V, 50 Hz,
Single phase transformer is 1 Wb/m3. If the e.m.f. per turn is 8
V, Determine:
1. Primary and secondary turns.
2. Area of the core.
Solution:
Example 1
34
2.3

35. 35
2.3

36. 2.3
36

37. Phasor diagram: Transformer on No-load
2.4
37

38. 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
2.4
38

39. Transformer on load
Fig. a: Ideal transformer on load
Fig. b: Main flux and leakage flux
in a transformer
2.5
39

40. Phasor diagram of transformer with UPF load
2.5
40

41. Phasor diagram of transformer with lagging p.f load2.5
41

42. Phasor diagram of transformer with leading p.f load
2.5
42

43. Equivalent circuit of a transformer
𝐼 𝑚 = 𝐼0 𝑆𝑖𝑛∅0
𝐼𝑐 = 𝐼0 𝐶𝑜𝑠∅0
𝑅0 = 𝑉1
𝐼𝑐
𝑉0 = 𝑉1
𝐼 𝑚
2.6
43

44. Referred to primary and secondary sides
respectively
2.5
44

45. Contd.,
• 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
2.6
45

46. Contd.,
• Equivalent sec. resistance w.r.t primary- which would have
caused the same loss as R2 in secondary,
2
2
2
'
2
2
1 RIRI 
2
2
1
2'
2R R
I
I







2
2
k
R

2.6
46

47. Transferring secondary parameters to primary
side
2.6
47

48. Contd.,
2.6
48

49. Contd.,
2.6
49

50. Equivalent circuit referred to secondary side
• Transferring primary side parameters to secondary
side
2.6
50

51. Contd.,
• Similarly exciting circuit parameters are also
transferred to secondary as Ro’ and Xo’
2.6
51

52. Contd.,
2.6
52

53. equivalent circuit w.r.t primary
2.6
53

54. 2.6
54

55. Approximate equivalent circuit
• Since the noload current is 1% of the full load current, the
nolad circuit can be neglected
2.6
55

56. • A 6600/400 V single phase transformer has primary resistance
of 2.5 ohms and secondary resistance of 0.01 ohms. Calculate
the total equivalent resistance with respect to primary and
secondary.
• Solution :
The given values are
Example 2
56
2.6

57. 2.6
57

58. 2.6
58

59. 2.6
59

60. Transformer test
• 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)
2.7
60

61. • - Magnetising susceptance B0 ( or reactance X0)
• - core loss conductance G0 (or resistance R0)
• The above constants can be easily determined by
two tests
1. OC Test
2. SC Test
2.7
61

62. • - Open 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
2.7
62

63. Open-circuit Test
• In Open Circuit Test the transformer’s secondary winding is
open-circuited, and its primary winding is connected to a full-
rated line voltage.
2.7
63

64. 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 )
2.7
64

65. Short-circuit Test
• Secondary terminals are short circuited
• Primary terminals are connected to a fairly low-voltage
source.
2.7
65

66. • 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
2.7
66

67. 2
01
2
0101
01
2
sc
01
01
2
sc
X
W
R
WlossculoadFull
RZ
I
V
Z
I
RI
sc
sc
sc
sc




2.7
67

68. Losses in transformer
• Core loss (iron loss)
- due to hysteresis.
• Copper loss (I2R loss)
-Resistive heating.
2.8
68

69. Core loss (iron loss)
• Copper losses are the resistive heating losses
in the primary and secondary windings of the
transformer.
• Hysteresis loss =
2.8
69

70. Copper loss (I2R loss)2.8
70

71. • Total loss =iron loss+copper loss
• Iron loss- Due to voltage
• Copper loss-Due to current
2.8
71

72. Regulation of transformer
• Full load Voltage Regulation is a quantity that compares the
output voltage at no load with the output voltage at full load.
voltageload-no
voltageload-fullvoltageload-no
regulationVoltage


2.8
72

73. • E2-Secondary terminal voltage on no load
• V2-Secondary terminal voltage onGiven load
Then
73

74. • I2 – Full load secondary current
. V2- No load secondary Voltage.
2.8
74

75. • R2e- Equivalent resistance refer to secondary.
• X2e- Equivalent reactance refer to secondary.
• Cos∅ – Power factor
2.8
75

76. 2.8
76

77. 2.8
77

78. Efficiency
• Transformer efficiency is defined as (applies to motors,
generators and transformers):
%100
in
out
P
P

%100


lossout
out
PP
P

2.8
78

79. -
• Therefore, for a transformer, efficiency may be calculated
using the following:
%100
cos
cos
x
IVPP
IV
SScoreCu
SS





CopperlossPCu 
CorelossPcore 
2.8
79

80. Condition for maximum efficiency
2.8
80

81. 2.8
81

82. The Condition for maximum efficiency
• The Condition for maximum efficiency is
Copper loss = Core loss
2.8
82

83. All day efficiency
hours)24(
kWhinInput
kWhinoutput
in wattsinput
in wattsputout
efficiencycommercialordinary
day forall 


•All day efficiency is always less than the
commercial efficiency
2.9
83

84. Auto transformer
• .
84
2.10

85. Introduction
• An autotransformer (sometimes
called autostep down transformer) is an
electrical transformer with only one Winding.
85
2.10

86. • portions of the same winding act as both
the primary and secondery sides of the
transformer.
86
2.10
Introduction

87. Operation
• An autotransformer has a single
winding with two end terminals.
• The primary voltage is applied
across two of the terminals.
• The secondary voltage taken from
two terminals,
• Almost always having one
terminal in common with the
primary voltage.
87
2.10

88. Advantages of Autotransformers
• Its efficiency is more when compared with
theconventional one.
• Its size is relatively very smaller.
• Voltage regulation of autotransformer is much better.
• Lower cost
• Low requirements of excitation current.
• Less copper is used in its design and construction
88
2.10

89. Applications:
1. Used in both synchronous motor and
industrial motors..
2. Used in electrical apparatus testing labs since
the voltage can be smoothly and
continuously varied.
3. They find application as boosters in
AC feeders to increase the voltage levels.
89
2.10

90. Revision
• Construction of transformer
• Types of transformer
• Equivalent circuit of transformer
• Efficiency of transformer
90