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Chapter 4.
Transformer
1
Transformer
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Introduction
Two winding transformers
Construction and principles
Equivalent circuit
Determination of equivalent circuit
parameters
Voltage regulation
Efficiency
Auto transformer
3 phase transformer
2
Transformer-
Introduction
Varieties of transformers
3
Transformer-
Introduction
4
Transformer
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Introduction
Transformer is a device that makes use of the
magnetically coupled coils to transfer energy
It is typically consists of one primary winding
coil and one or more secondary windings
The primary winding and its circuit is called
the Primary Side of the transformer
The secondary winding and its circuit is called
the Secondary Side of the transformer
5
Transformer
Introduction
If one of those winding, the primary, is
connected to an alternating voltage
source, an alternating flux will be
produced. The mutual flux will link the
other winding, the secondary, and will
induced a voltage in it.
6
Transformer-
Introduction
Transformers are adapted to numerous engineering
applications and may be classified in many ways:

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Power level (from fraction of a volt-ampere (VA) to over a
thousand MVA),
Application (power supply, impedance matching, circuit
isolation),
Frequency range (power, audio, radio frequency (RF))
Voltage class (a few volts to about 750 kilovolts)
Cooling type (air cooled, oil filled, fan cooled, water
cooled, etc.)
Purpose (distribution, rectifier, arc furnace, amplifier
output, etc.).
7
Transformer
Introduction
Power transmission
8
Transformer
Introduction
Power transmission
9
Transformer
4.1 Construction
10
TransformerPrimary winding
Supply
N1
construction
Secondary winding
N2
Load
Laminated iron core
Basic components of single phase transformer
11
Transformer-
A) Core type
construction
B) Shell type
Single phase transformer construction
12
Transformer-
construction
13
Transformer-
construction
Primary
Winding
Multi-layer
Laminated
Iron Core
Secondary
Winding
H1 H2
X1
X2
Winding
Terminals
14
Transformer
4.2 Ideal Transformer
15
Transformer
Φ
The emf which induced in
transformer primary winding is
known as self induction emf as
the emf is induced due to to flux
which produced by the winding
itself.
i
v1
e1
v2
e2
While the emf which induced in
transformer secondary winding
is known as mutual induction
emf as the emf is induced due
to to flux which produced by the
other winding.
16
Transformer
Φ
Acording to Faraday’s Law,
the emf which induced in
the primary winding is,
i
v1
d
e1 = N1
dt
e1
v2
e2
Since the flux is an
alternating flux,
   mak sin  t
e1 = N1
d ( mak sin  t )
dt
 N1mak cos  t
17
Transformer
Φ
e1  N1mak 2f cos  t
 E1max cos  t
where,
E1max = N1max 2f
E1rms 
E1max
i
v1
e1
v2
e2
 4.44 N1max f
2
18
Transformer
Φ
e2 = N 2
d
dt
Similarly it can be shown
that,
E2 rms  4.44 N 2max f
E2 4.44 N 2max f N 2


k
E1 4.44 N1max f N1
i
v1
e1
v2
e2
k is transformation ratio
19
Transformer
The voltage ratio of induced voltages on the secondary to
primary windings is equal to the turn ratio of the winding turn
number of the secondary winding to the winding turn number
of the primary winding. Therefore the transformers can be
used to step up or step down voltage levels by choosing
appropriate number their winding turns. In power system it’s
necessary to step up the output voltage of a generator which
less than 30kV to up 500kV for long distance transmission.
High voltage for long distance power transmission can reduce
current flow in the transmission lines, thus line losses and
voltage drop can be reduced.
20
Transformer

Ideal Transformer
Winding resistances are zero, no leakage
inductance and iron loss
Magnetization current generates a flux that
induces voltage in both windings
N1  m 
E1 
2
N2 m 
E2 
2
m
Im
V1
E1
N1
N2
E2 = V2
Current, voltages and flux in an unloaded ideal transformer
21
Transformer
Φ
Transformer on no load.
i
v1
e1
v2
e2
22
Transformer
Loaded transformer
Ideal Transformer
Φ2
Φ1
Φ
I2
i
V1
N1
E2
E1
V2 ZL
N2
When a load is connected to the secondary output terminals of
a transformer as shown in Figure 4.5, a current I2 flows into the
load and into transformer secondary winding N2. The current I2
which flowing in N2 produces flux Φ2 which opposite –by Lenz’s
law- to the main magnetic flux Φ in the transformer core. This
will weaken or slightly reduce the main flux Φ to Φ’.
23
Transformer
Loaded transformer
Ideal Transformer
Φ2
Φ1
Φ
I2
i
V1
N1
E2
E1
V2 ZL
N2
The reduction of main flux Φ –by Faraday’s law- could also
reduce the induced voltage in primary winding E1.
Consequently E1 is now smaller than the supply voltage V1,
then the primary current would be increased due to that
potential differences. Therefore on loaded transformer, the
primary current has an additional current of I1’.
24
Transformer
Loaded transformer
Ideal Transformer
Φ2
Φ1
Φ
I2
i
V1
N1
E2
E1
V2 ZL
N2
The extra current I1’ which flowing in the primary winding N1
produces flux Φ1 which naturally react according to Lenz’s
law, demagnetize the flux Φ2. Therefore the net magnetic
flux in the core is always maintained at original value, it is
the main flux Φ (the flux which produced by the magnetizing
current).
25
Transformer
Loaded transformer
Ideal Transformer
Φ2
Φ1
Φ
I2
i
V1
N1
E2
E1
V2 ZL
N2
The magneto motive force (mmf) source N2I2 at the
secondary winding produces flux Φ2, while the mmf N1I1‘
produces flux Φ1. Since the magnitude of Φ1 equal to
magnitude of Φ2 and the reluctance seen by these two mmf
sources are equal, thus
N 1I 1‘ = N 2I 2
26
Transformer
Ideal Transformer
Loaded transformer
Im + I
V1
E1
I2
m
1
1
2
E2
V2
Load
Currents and fluxes in a loaded ideal transformer
27
Transformer
Ideal Transformer
Turn ratio

If the primary winding has N1 turns and
secondary winding has N2 turns, then:
N1 E1 I 2
a


N 2 E 2 I1

The input and output complex powers are equal
E1 I1*  S1  S 2  E2 I*2
28
Transformer
Ideal Transformer
Functional description of a transformer:
When a = 1
Isolation Transformer
When | a | < 1
Step-Up Transformer Voltage
is increased from Primary
side to secondary side
When | a | > 1
Step-Down Transformer
Voltage is decreased from
Primary side to secondary
side
29
Transformer
Ideal Transformer
Transformer Rating

Practical transformers are usually rated
based on:
 Voltage Ratio (V1/V2) which gives us the
turns-ratio
 Power Rating, small transformers are
given in Watts (real power) and Larger
ones (Power Transformers) are given in
kVA (apparent power)
30
Transformer
Ideal Transformer
Example 4.1



Determine the turns-ratio of a 5 kVA
2400V/120V Power Transformer
Turns-Ratio = a = V1/V2 = 2400/120 =
20/1 = 20
This means it is a Step-Down transformer
31
Transformer
Ideal Transformer
Example 4.2
A 480/2400 V (r.m.s) step-up ideal
transformer delivers 50 kW to a resistive
load. Calculate:
(a) the turns ratio, (0.2)
(b) the primary current, (104.17A)
(c) the secondary current. (20.83A)
32
Transformer
Ideal Transformer
Nameplate of transformer
33
Transformer
Ideal Transformer
Equivalent circuit
I1 = I2 /T
T
V1
E1
I2
E2 = V2
V1 = E1 = T E2
Equivalent circuit of an ideal transformer
34
Transformer
Ideal Transformer
Transferring impedances through a transformer
Z1 
V1 a V2
V

 a2 2
I1  I 2 
I2
 
a
I1
Vac
Z1  a 2 Z load
T
V1
I2
V2
Zload
Equivalent circuit of an ideal transformer
35
Transformer-
Ideal Transformer
I1
Equivalent circuit when
secondary
impedance
is
transferred to primary side and
ideal transformer eliminated
a)
Vac
V1
a2Zload
I2
Vac/k
V2
Zload
b)
Equivalent circuit when
primary source is transferred to
secondary
side
and
ideal
transformer eliminated
Thévenin equivalents of transformer circuit
36
Transformer
practical transformer
Practical Transformer
37
Transformer
4.3 Equivalent Circuits
38
TransformerR1
equivalent circuit
m
I1
I2
l2
l1
V1
N1
R2
V2
N2
39
Transformer-
equivalent circuit
Development of the transformer equivalent circuits
The effects of winding resistance and leakage flux are
respectively accounted for by resistance R and leakage
reactance X (2πfL).
X1
R1
V1
X2
I1
R2
I2
V2
N1:N2
40
Transformer

practical equivalent circuit
In a practical magnetic core having finite
permeability, a magnetizing current Im is
required to establish a flux in the core.
This effect can be represented by a
magnetizing inductance Lm. The core loss can
be represented by a resistance Rc.
41
TransformerI1
V1
I’1
X1
R1
practical equivalent circuit
X2
I0
Ic
Rc
R2
I2
Im
V2
Xm
N1:N2
Rc :core loss component,
Xm : magnetization component,
R1 and X1 are resistance and reactance of the primary winding
R2 and X2 are resistance and reactance of the secondary winding
42
Transformer
practical equivalent circuit
The impedances of secondary side such
as R2, X2 and Z2 can be moved to
primary side and also the impedances
of primary side can be moved to the
secondary side, base on the principle
of:
The power before transferred = The
power after transferred.
43
Transformer-
practical equivalent circuit
The power before transferred = The power
after transferred.
I22R2 = I1’ 2R2’
Therefore R2’= (I2/ I1’ ) 2 R2
= a2R2
44
Transformer-
practical equivalent circuit
The turns can be moved to the right or left by referring
all quantities to the primary or secondary side.
I1
V1
I’1
X1
R1
X’2
R’2
I2
I0
Ic
Rc
Im
Xm
V2’
V2
N1:N2
V2' = a V2 , I1' = I2/a
X2' = a2 X2 , R2' = a2 R2
a = N1/N2
The equivalent circuit with secondary side moved to the primary.
45
Transformer-
Approximate equivalent
circuit

For convenience, the turns is usually not
shown and the equivalent circuit is drawn
with all quantities (voltages, currents, and
impedances) referred to one side.
R1
I1
X1
X’2
I’1
R’2
I0
Ic
V1 R
c
Im
Xm
V2’
Z2’
N
46
Transformer
equivalent circuit
Example 4.3
A 100kVA transformer has 400 turns on the primary
and 80 turns on the secondary. The primary and
secondary resistance are 0.3 ohm and 0.01 ohm
respectively and the corresponding leakage
reactances are 1.1 ohm and 0.035 ohm respectively.
The supply voltage is 2200V. Calculate:
(a) the equivalent impedance referred to the primary
circuit (2.05 ohm)
(b) the equivalent impedance referred to the
secondary circuit
47
Transformer
4.4 Determination of
Equivalent Circuit
Parameter
48
Transformer


o/c-s/c tests
The equivalent circuit model for the actual
transformer can be used to predict the behavior of
the transformer.
The parameters R1, X1, Rc, Xm, R2, X2 and N1/N2
must be known so that the equivalent circuit model
can be used.
These parameters can be directly and more easily
determined by performing tests:
1. No-load test (or open-circuit test).
2. Short-circuit test.
49
Transformer

o/c-s/c tests
No load/Open circuit test
 Provides magnetizing reactance (Xm) and core
loss resistance (RC)
 Obtain components are connected in parallel
Short circuit test
 Provides combined leakage reactance and
winding resistance
 Obtain components are connected in series
50
Transformer
open circuit test
No load/Open circuit test
P oc
A
X1
R1
X2
R2
W
I oc
V
X
Rc
m
V oc
Equivalent circuit for open circuit test, measurement at the primary side.
Poc
A
W
V
Simplified equivalent
circuit
Ioc
Xm
Rc
Voc
51
Transformer
open circuit test
Open circuit test evaluation
 Poc 

 0  cos 
 Voc I oc 
Voc2
Rc 
Poc
1
Q  Voc I oc sin  0
Voc2
Xm 
Q
52
Transformer
short circuit test
Short circuit test


Secondary (normally the LV winding) is shorted,
that means there is no voltage across secondary
terminals; but a large current flows in the
secondary.
Test is done at reduced voltage (about 5% of
rated voltage) with full-load current in the
secondary. So, the ammeter reads the full-load
current; the wattmeter reads the winding losses,
and the voltmeter reads the applied primary
voltage.
53
Transformer
short circuit test
Short circuit test
P sc
A
V sc
R1
X1
R2
X2
W
I sc
V
Equivalent circuit for short circuit test, measurement at the primary side
P
A
V sc
sc
R1
X1
a2R2
a2X2
W
V
I
sc
Simplified equivalent circuit for short circuit test
54
Transformer
short circuit test
Short circuit test
P
A
V sc
sc
Re1
X e1
W
I
V
sc
Simplified circuit for calculation of series impedance
Re1  R1  a R2
2
X e1  X 1  a X 2
2
Primary and
secondary
impedances are
combined
55
Transformer
short circuit test
Short circuit test evaluation
Psc
Re1  2
I sc
X e1 
2
Z e1
Z e1

Vsc

I sc
2
Re1
56
Transformer
o/c-s/c tests
Equivalent circuit obtained by measurement
X e1
Xm
R e1
Rc
Equivalent circuit for a real transformer resulting from the
open and short circuit tests.
57
Transformer
o/c-s/c tests
Example 4.4
Obtain the equivalent circuit of a 200/400V, 50Hz
1-phase transformer from the following test data:O/C test : 200V, 0.7A, 70W
S/C test : 15V, 10A, 85W
- on L.V. side
- on H.V. side
(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm,
Xe=0.31 ohm)
58
Transformer –

voltage regulation
Voltage Regulation



Most loads connected to the secondary of a transformer are
designed to operate at essentially constant voltage. However,
as the current is drawn through the transformer, the load
terminal voltage changes because of voltage drop in the
internal impedance.
To reduce the magnitude of the voltage change, the
transformer should be designed for a low value of the
internal impedance Zeq
The voltage regulation is defined as the change in
magnitude of the secondary voltage as the load current
changes from the no-load to the loaded condition.
59
Transformer –
voltage regulation
Ze2
I2’
V12’=V20
R1’
Rc’
X1’
Xm’
R2
X2
I2
V2
Z
Ze2 = R1’ + R2 + jX1’ + jX2
= Re2 + jXe2
60
Transformer –
voltage regulation
Ze2
I2’
V12’=V20
Rc’
Applying KVL,
Or
R1’
X1’
Xm’
R2
X2
I2
V2
Z
V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
V2 = V20 - I2 (Ze2 )
61
Transformer –
voltage regulation
I2Xe2
A
O
θ2
V2
I2Re2
I2
62
Transformer –
voltage regulation
I2Xe2
V20
I2Xe2
A
O
θ2
I2Re2
V2
I2Re2
B
I2
V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
63
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
I2Re2
D
N
M
V2
I2Re2
B
L
θ2
I2
Voltage drop = AM = OM – OA
= AD + DN + NM
64
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
I2Re2
D
M
V2
I2Re2
B
I2
N
L
θ2
AD = I2 Re2 cosθ2
DN=BL= I2 Xe2 sinθ2
65
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
I2Re2
D
M
V2
I2Re2
B
I2
N
L
θ2
Applying Phytogrus theorem to OCN triangle.
(NC)2 = (OC)2 – (ON)2
= (OC + ON)(OC - ON) ≈ 2(OC)(NM)
66
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
D
N
M
V2
I2Re2
I2Re2
B
I2
L
θ2
Therefore NM = (NC)2/2(OC)
NC = LC – LN = LC – BD
= I2 Xe2 cosθ2 - I2 Re2 sinθ2
67
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
I2Re2
D
M
V2
I2Re2
B
I2
N
L
θ2
2


I
X
cos


I
R
sin

2
e
2
2
2
e
2
2
NM =
2V20
AM = AD + DN + NM
I 2 X e 2 cos 2  I 2 Re 2 sin  2 2
= I2 Recosθ2 + I2 Xe2 sin θ2 +
2V20
68
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
I2Re2
D
M
V2
I2Re2
B
thus,
votage regulation = (AM)/V20 per unit
I2
N
L
θ2
In actual practice the term NM is negligible since its value is
very small compared with V2. Thus the votage regulation
formula can be reduced to:
69
Transformer –
voltage regulation
I2Xe2
C
V20
θ2
I2Xe2
A
O
θ2
N
M
V2
I2Re2
I2Re2
B
L
θ2
I2
Voltage regulation =
D
I 2 Re 2 cos 2  I 2 X e 2 sin  2 
V20
70
Transformer
voltage regulation
The voltage regulation is expressed as
follows:
Voltage regulation 
V2 NL  V2 L
V2 NL
V2NL= secondary voltage (no-load condition)
V2L = secondary voltage (full-load condition)
71
Transformer
voltage regulation
For the equivalent circuit referred to the
primary:
V1  V2'
Voltage regulation 
V1
V1 = no-load voltage
V2’ = secondary voltage referred to the primary (full-load condition)
72
Transformer
voltage regulation
Consider the equivalent circuit referred to the
secondary,
I2'
V2NL
R1'
X1'
X2
Xe2
I2
R2
V2
Z2
Re2
I 2 Re 2 cos 2  I 2 X e 2 sin 2
Voltage regulation 
V2 NL
(-) : power factor leading
(+) : power factor lagging
73
Transformer
voltage regulation
Consider the equivalent circuit referred to the primary,
I1
R1
X2'
X1
R2' I1'
Xe1
V1
V2'
Z’2
Re1
Voltage regulation 
I1 Re1 cos 2  I1 X e1 sin 2
V1
(-) : power factor leading
(+) : power factor lagging
74
Transformer
voltage regulation
Example 4.5
Based on Example 4.3 calculate the
voltage regulation and the secondary
terminal voltage for full load having a
power factor of
(i) 0.8 lagging (0.0336pu,14.8V)
(ii) 0.8 leading (-0.0154pu,447V)
75
Transformer
Efficiency
Losses in a transformer
 Copper losses in primary and secondary
windings
 Core losses due to hysteresis and eddy
current. It depends on maximum value of
flux density, supply frequency and core
dimension. It is assumed to be constant for
all loads
76
Transformer

Efficiency
As always, efficiency is defined as power output to
power input ratio
output power ( Pout )

input power ( Pin )
Pout

Pout  losses
The losses in the transformer are the core loss
(Pc) and copper loss (Pcu).
V2 I 2 cos 2

2
V2 I 2 cos 2  Pc  I 2 Re 2
77
Transformer
Efficiency
Efficiency on full load
S FL cos  FL

S FL cos  FL  Poc  Psc
where S is the apparent power (in volt amperes)
78
Transformer
Efficiency
Efficiency for any load equal to n x full load
n  S FL cos  FL

2
n  S FL cos  FL  Poc  n  Psc
where corresponding total loss =
Poc  n  Psc
2
79
Transformer
Efficiency
Example 4.6
The following results were obtained on a 50 kVA
transformer: open circuit test – primary voltage,
3300 V; secondary voltage, 400 V; primary power,
430W.Short circuit test – primary voltage,
124V;primary current, 15.3 A; primary power,
525W; secondary current, full load value. Calculate
the efficiency at full load and half load for 0.7
power factor.
(97.3%, 96.9%)
80
Transformer
Efficiency
For constant values of the terminal voltage V2 and
load power factor angle θ2 , the maximum efficiency
occurs when
d
0
dI 2

If this condition is applied, the condition for
maximum efficiency is
2
c
2 e2
P I R
that is, core loss = copper loss.
81
Transformer-
Efficiency
82
Transformer- Auto transformer

It is a transformer whose primary and
secondary coils are in a single winding
Autotransformer
83
Transformer- Auto transformer




Same operation as two windings
transformer
Physical connection from primary to
secondary
Sliding connection allows for variable
voltage
Higher kVA delivery than two windings
connection
84
Transformer- Auto transformer

Advantages:




A tap between primary and secondary sides which
may be adjustable to provide step-up/down
capability
Able to transfer larger S apparent power than the
two winding transformer
Smaller and lighter than an equivalent twowinding transformer
Disadvantage:

Lacks electrical isolation
85
Transformer- Auto transformer

A Step Down Autotransformer:
and
86
Transformer- Auto transformer

A Step Up Autotransformer:
and
87
Transformer- Auto transformer

Example 4.7
An autotransformer with a 40% tap is supplied by a
400-V, 60-Hz source and is used for step-down
operation. A 5-kVA load operating at unity power
factor is connected to the secondary terminals.
Find:
(a) the secondary voltage,
(b) the secondary current,
(c) the primary current.
88
Transformer- Auto transformer

Solution
89
Transformer -3 phase transformer
Three phase transformers

The three-phase transformer can be built by:

the interconnection of three single-phase transformers

using an iron core with three limbs

The usual connections for three-phase transformers are:

wye / wye
seldom used, unbalance and 3th harmonics
problem


wye / delta
frequently used step down.(345 kV/69 kV)

delta / delta
used medium voltage (15 kV), one of the
transformer can be removed (open delta)

delta / wye
step up transformer in a generation station
For most cases the neutral point is grounded
90
Transformer -3 phase transformer

Analyses of the grounded wye / delta transformer
A
VAN
Vab
a
B
VB N
Vbc
b
C
VC N
N

Each leg has a
primary and a
secondary winding.

The voltages and
currents are in phase
in the windings
located on the same
leg.

The primary phase-toline voltage generates
the secondary line-toline voltage. These
voltages are in phase
Vca
c
91
Transformer -3 phase transformer

Analyses of the grounded wye / delta transformer
Ia
IA
IAN
IBN
IB
Iab
ICN
Ica I
c
N
Ibc
IC
Ib
92
Transformer -3 phase transformer

Analyses of the grounded wye / delta transformer
A
VA N
VBN
VAB
C
VC N
a
Vab
Vab
N
B
VB C
Vbc
VCA
c
Vbc
Vca
Vbc
b
93
Transformer -3 phase transformer

Three phase transformer
94
Transformer

Three phase transformer
95
Transformer

Three phase transformer
96
Transformer

Three phase transformer
97
Transformer

Three phase transformer
98
Transformer

Three phase transformer
Transformer Construction
Iron Core

The iron core is made of thin
laminated silicon steel (2-3 %
silicon)

Pre-cut insulated sheets are
cut or pressed in form and
placed on the top of each
other .

The sheets are overlap each
others to avoid (reduce) air
gaps.

The core is pressed together
by insulated yokes.
99
Transformer

Three phase transformer
Transformer Construction Winding

The winding is made of copper or
aluminum conductor, insulated with
paper or synthetic insulating material
(kevlar, maylard).

The windings are manufactured in
several layers, and insulation is
placed between windings.

The primary and secondary windings
are placed on top of each others but
insulated by several layers of
insulating sheets.

The windings are dried in vacuum
and impregnated to eliminate
moisture.
Small transformer winding
100
Transformer

Three phase transformer
Transformer Construction
Iron Cores
Three phase transformer iron core
The three phase transformer iron
core has three legs.



A phase winding is placed in
each leg.
The high voltage and low voltage
windings are placed on top of
each other and insulated by
layers or tubes.
Larger transformer use layered
construction shown in the
previous slides.
A
B
C
101
Transformer

Three phase transformer
Transformer Construction

The dried and treated
transformer is placed in a steel
tank.

The tank is filled, under vacuum,
with heated transformer oil.

The end of the windings are
connected to bushings.

The oil is circulated by pumps
and forced through the radiators.
Three phase oil transformer
102
Transformer

Three phase transformer
Transformer Construction

The transformer is equipped with
cooling radiators which are
cooled by forced ventilation.

Cooling fans are installed under
the radiators.

Large bushings connect the
windings to the electrical system.

The oil is circulated by pumps
and forced through the radiators.

The oil temperature, pressure
are monitored to predict
transformer performance.
Three phase oil transformer
103
Transformer

Three phase transformer
Dry type transformer
Transformer Construction

Dry type transformers are used
at medium and low voltage.

The winding is vacuumed and
dried before the molding.

The winding is insulated by
epoxy resin

The slide shows a three phase,
dry type transformer.
104