CHAPTER TWO
TRANSFORMERS
Abbreviations inside this lecture
x’mer = transformer
ckt = circuit
[NB: these abbreviations are not standard]
Prepared by: Eskeziyaw A.
1
Transformer is a device that changes ac electric
power at one voltage level to ac electric power
at another volt age level through the action of a
magnetic field.
Transformers are extremely versatile devices that
can be used to either step up and step down AC
voltages or to step up and step down AC current.
They only allow AC to pass; and block DC.
The only connection between the coils is the
common magnetic flux present within the core.
Prepared by: Eskeziyaw A.
2
2.1 Principle of Operation of Transformer
When two coils are placed very close to each other, the changing
magnetic flux line produced by the first coil will cut through the second
coil. The two coils are said to be magnetically linked or coupled. As a
result, a voltage is induced.
The basic of transformer operation, i.e mutual induction, can be
expressed using Faraday’s law:
d
eind 
dt
N
   i
i 1
Prepared by: Eskeziyaw A.
3
Let flux through the core be
Induced voltage at primary winding becomes:
Peak value of induced voltage is:
Prepared by: Eskeziyaw A..
4
rms value of primary winding is
E1 
E1m
2

2
2
fN1 m
 2 fN1 m
 4.44 fN1 m
If primary winding resistance is negligible , so
[Do the same thing for the secondary winding]
Prepared by:
5
Advantages of x’mer besides their principal
purpose:
Impedance matching
Electrical isolation between source and load
DC isolation
Voltage sampling (e.g PT)
Current sampling (e.g CT)
.
6
2.2 CONSTRUCTION OF TRANSFORMERS
Two types of x’mer Construction:
1. Core type:
Core has two legs on which windings are wrapped.
7
2. Shell type:
Shape of the core is 3-legged.
Winding is wrapped in the middle leg(limb)
8
TRANSFORMER CORE CONSTRUCTION:
9
Comparison between core and shell type transformer
Core type
Shell type
1
The winding encircles the core
The core encircles most parts of the winding
2
It has single magnetic circuit
It has a double magnetic circuit
3
The core has two limbs
The core has three limbs
4
The cylindrical coils are used
The multi layer disc or sandwich type coils are used
5
The winding are uniformly distributed on two
The natural cooling does not exist as the windings
limbs hence natural cooling is effective.
are surrounded by the core.
The coils can be easily removed from
The coils can not be removed easily.
6
maintenance point of view
7
Preferred to low voltage transformers.
Preferred to high voltage transforme
68
THE IDEAL TRANSFORMER
For a transformer to be an ideal, the various
assumptions are as follows:
1. winding resistances are negligible.
2. all of the flux is confined to the magnetic core.
3. core losses (hysteresis and eddy current losses)
are negligible.
.
11
Transformer Losses
The losses that occur in transformers are:
1.
Copper (I2R) losses. Copper losses are the resistive heating losses in the primary and secondary
windings of the transformer. They are proportional to the square of the current in the windings.
2.
Eddy current losses. Eddy current losses are resistive heating losses in the core of the
transformer. They are proportional to the square of the voltage applied to the transformer.
3. Hysteresis losses. Hysteresis losses are associated with the rearrangement of the magnetic
domains in the core during each half-cycle. They are a complex, nonlinear function of the voltage
applied to the transformer.
4.
Leakage flux. The fluxes which escape the core and pass through only one of the transformer
windings are leakage fluxes. These escaped fluxes produce a self-inductance in the primary and
secondary coils.
87
Voltage and current relationships in ideal x’mer
Consider the following fig,
12
Transfer of Voltage and current in ideal x’mer
• affects only magnitudes of V & I
• Phase angles of V & I are not affected.
• Power remains constant.
13
Power in an Ideal Transformer
Active power:
Since phase is not affected,  p   s  
Reactive power:
Apparent power:
14
Impedance Transformation through a Transformer
Impedance of any device is:
Impedance of load connected to the secondary
winding is
15
Primary impedance of x’mer
But
and
So, primary impedance
becomes:
The above formula
shows that it is possible
to match load
impedance of load to
source impedance by
using the turns ratio a.
16
2.3 Ideal and Practical Models for Single Phase
Transformer
Consider the following figure
17
The basis of x’mer operation is mutual induction;
Faraday’s law:
eind
d

dt
N
Where
   i
i 1
Total flux linkage in a coil is not Nɸ but average flux
per turn can be defined as
Therefore,
18
If Vp(t) is input voltage to primary winding;
This flux is produced at primary winding, but the
whole flux will not reach secondary winding.
It can be divided into two:
1. Mutual flux (effective flux)
2. Leakage flux
19
Average primary flux can be expressed as;
20
for the secondary winding, average flux can be
expressed as;
21
Illustration of primary and secondary leakage fluxes
22
Faraday's law for the primary circuit can be
expressed as
Faraday’s law for the secondary winding can be
expressed as
23
24
Since in a well-designed transformer,
mutual flux>>leakage flux; leakage flux can be
neglected.
Thus,
25
The total no-load current in the core of x’mer is
called excitation current.
It consists of two components:
1. magnetization current, im- current required
to produce flux in the transformer core.
2. core-loss current, ic - current required to
make up for hysteresis and eddy current losses
26
Real x’mer with load connected to its secondary
winding
The net mmf of the ckt is
28
But, reluctance of a well-designed x’mer core is very small
(nearly zero), hence
Therefore, in order for the net mmf to be nearly zero, current
must flow into one dotted end and out of the other dotted end.
29
Example 1
A 100-kVA, 2400/240-V, 60-Hz step-down transformer (ideal) is used between a
transmission line and a distribution system.
a)
Determine turns ratio.
b)
What secondary load impedance will cause the transformer to be fully loaded,
and what is the corresponding primary current?
c)
Find the load impedance referred to the primary.
The following assumptions must be
considered to convert a real x’mer into
ideal one:
1. Hysteresis & eddy current losses are zero
2. Resultant mmf is zero
3. Winding losses(copper losses) are zero
4. Leakage flux must be zero
31
In real x’mer, losses must be considered to get
accurate model.
Major terms to be considered for real x’mer model:
1. Copper losses-are resistive heating losses in x’mer
windings.

Proportional to square of current
1. Eddy current losses-are resistive heating losses in
the core.
 proportional to square of voltage applied
3. Hysteresis losses
4. Leakage flux- produces self-inductance
32
Therefore, it is possible to construct an equivalent
circuit that takes into account all the major
imperfections in real transformers.
Equivalent Circuit of a Transformer
 Copper losses are modeled by Rp in primary, and
Rs in secondary circuit
 Leakage flux produces voltage in both windings:
33
Since much of flux passes through air, and air has
constant reluctance, flux directly proportional to
current;
34
35
Leakage flux is modeled by inductor
 magnetization current, im -is a current
proportional to the voltage applied to the core
and lagging the applied voltage by 90°, so it can
be modeled by a reactance XM connected across
the primary voltage source.
36
 core- loss current ic is a current proportional to
the voltage applied to the core that is in phase
with the applied voltage, so it can be modeled by
a resistance Rc connected across the primary
voltage source.
The resulting equivalent circuit of a real x’mer can
be drawn as shown below:
37
38
To simplify analysis, there is a process of replacing
one side of a transformer by its equivalent at the
other side's voltage level which is known as referring
one side of the transformer to the other side.
Referring is simply done using the turns ratio , a , of
transformer.
So, the above equivalent ckt can be redrawn by
referring to either the primary side or the secondary
side.
39
Primary impedance of x’mer:
But
and
So, primary impedance
becomes:
These formula are used
to refer transformer
parameters either to the
primary or secondary.
40
Equivalent ckt of x’mer referred to primary side
41
Equivalent ckt of x’mer referred to secondary side
42
The transformer models shown before are more
complex than necessary in order to get good results
in practical engineering applications.
Hence, the following assumptions are taken to
simplify the ckt so that an approximate equivalent
ckt can be drawn.
1. Excitation branch has a very small current
compared to load current.
2. Excitation branch is moved to the front, and
primary & secondary impedances are left in series
with each other.
3. Sometimes this branch is neglected entirely.
43
Approximate transformer models. (a) Referred to the
primary side; (b) referred to the secondary side;
T.
44
Approximate transformer models.
(c) with no excitation branch, referred to the primary side;
(d) with no excitation branch, referred to the secondary side.
45
Example
SOLUTION
2.4 Parameter Testing
This testing is helpful to determine the values of
components in the transformer model.
It is possible to experimentally determine the values
of the inductances and resistances in the transformer
model.
This can be done using two tests only:
(i) open-ckt (no-load) test
(ii) Short-ckt test
46
Open –Circuit Test
-secondary winding is open-circuited, and its primary
winding is connected to a full-rated line voltage.
-as a result, all the input must be flowing trough
excitation branch
- Rp and Xp Are too small compared to Rc and Xm, so
all the input voltage is dropped at the excitation
branch.
 Input voltage, current and power are measured.
 So it is possible to determine:
• PF of input current
• Magnitude and phase angle of excitation impedance.
.
47
Connection diagram for transformer open-circuit test.
48
The easiest way to calculate Rc and Xm is:
Conductance of core loss resistor is:
Susceptance of magnetizing inductor is:
Total admittance is:
49
The magnitude of the excitation admittance (referred
to the primary circuit) can be found from the opencircuit test voltage and current:
The open-circuit power factor (PF) is given by
50
PF is always lagging for a real transformer, so the
angle of the current always lags the angle of the
voltage by  degrees.
Therefore, the admittance
Hence, by comparing admittance values of the
above eqn. and its magnitude which is calculated
earlier, it is possible to determine values of
Rc and Xm directly from open-ckt test.
51
Short-Circuit Test
• secondary terminals of x’mer are short-circuited,
• primary terminals are connected to a fairly
low-voltage source,
• input voltage is adjusted until the current in the
short-circuited windings is equal to its rated value.
• input voltage, current, and power are measured.
52
Connection diagram for transformer short-circuit test.
.
53
• Since input voltage is so low during the
short-circuit test, negligible current flows through
the excitation branch.
• The magnitude of the series impedances referred
to the primary side of the transformer is
• power factor [lagging] of the current is given by
54
• This shows that it is possible to determine the
total series impedance referred to the primary
side by using this technique, but there is no easy
way to split the series impedance into primary and
secondary components.
55
Example: We need to determine the equivalent circuit
impedances of a 20 kVA, 8000/240 V, 60 Hz transformer. The
open-circuit and short-circuit tests led to the following data:
Open-ck measurements
Short-ckt measurements
VOC = 8000 V
VSC = 489 V
IOC = 0.214 A
ISC = 2.5 A
POC = 400 W
PSC = 240 W
The power factor during the open-circuit test is
POC
400
PF  cos  

 0.234lagging
VOC I OC 8000  0.214
The excitation admittance is
YE 
I OC
0.214
1
1
  cos 1 PF 
  cos 1 0.234  0.0000063  j 0.0000261 
j
VOC
8000
RC
XM
59
Therefore,
RC 
1
1
 159k ; X M 
 38.3k 
0.0000063
0.0000261
The power factor during the short-circuit test is
PF  cos  
PSC
240

 0.196lagging
VSC I SC 489  2.5
The series impedance is given by
Z SE 
VSC
489
 cos 1 PF 
78.7
I SC
2.5
 38.4  j192
Req  38.3; X eq  192
.
60
Example
A 50-kVA 2400:240-V, 60-Hz distribution transformer has a
leakage impedance of 0.72 + j0.92  in the high-voltage
winding and 0.0070 + j0.0090  in the low-voltage winding.
At rated voltage and frequency, the impedance of the shunt
branch accounting for the exciting current is 6.32 + j43.7 
when viewed from the low-voltage side.
Draw the equivalent circuit referred to
(a) the high-voltage side; and
(b) the low-voltage side, and label the impedances
numerically.
61
Ans. (a) referred to high-voltage side
(b) referred to low-voltage side
62
2.7 Voltage Regulation and Efficiency
Voltage Regulation
Since a real transformer contains series impedances, the transformer’s
output voltage varies with the load even if the input voltage is constant.
To compare transformers in this respect, the quantity called a full-load
voltage regulation (VR) is defined as follows:
VR 
Vs , nl  Vs , fl
Vs , fl
100% 
V p a  Vs , fl
Vs , fl
100%
In a per-unit system:
VR 
V p , pu  Vs , fl , pu
Vs , fl , pu
100%
Where Vs,nl and Vs,fl are the secondary no load and full load voltages.
Note,
Note,the
theVR
VRofofan
anideal
idealtransformer
transformerisiszero.
zero.
76
The transformer phasor diagram at different loading conditions
A transformer operating at a lagging power factor:
It is seen that Vp/a > Vs, VR > 0
A transformer operating at
a unity power factor:
It is seen that VR > 0
A transformer operating at a
leading power factor:
If the secondary current is leading,
the secondary voltage can be higher
than the referred primary voltage;
VR < 0.
77
.
Figure. Derivation of the approximate equation for
T.
Vp
a
78
The transformer efficiency
The efficiency of a transformer is defined as:
Pout
Pout

100% 
100%
Pin
Pout  Ploss
Note: the same equation describes the efficiency of motors and generators.
Maximum efficiency occurs when the iron losses equal the copper losses.
Considering the transformer equivalent circuit, we notice three types of losses:
1. Copper (I2R) losses – are accounted for by the series resistance
2. Hysteresis losses – are accounted for by the resistor Rc.
3. Eddy current losses – are accounted for by the resistor Rc.
Since the output power is
Pout  Vs I s cos  s
The transformer efficiency is

Vs I s cos 
100%
PCu  Pcore  Vs I s cos 
79
Example: A 15 kVA, 2300/230 V transformer was tested to by open-circuit
and closed-circuit tests. The following data was obtained:
VOC = 2300 V
VSC = 47 V
IOC = 0.21 A
ISC = 6.0 A
POC = 50 W
PSC = 160 W
a. Find the equivalent circuit of this transformer referred to the high-voltage side.
b. Find the equivalent circuit of this transformer referred to the low-voltage side.
c. Calculate the full-load voltage regulation at 0.8 lagging power factor, at 1.0 power
factor, and at 0.8 leading power factor.
d. Plot the voltage regulation as load is increased from no load to full load at power
factors of 0.8 lagging, 1.0, and 0.8 leading.
e. What is the efficiency of the transformer at full load with a power factor of 0.8
lagging?
80
a. The excitation branch values of the equivalent circuit can be determined as:
Poc
50
1
 oc  cos
 cos
 84
Voc I oc
2300  0.21
1
The excitation admittance is:
YE 
I oc
0.21
  84 
  84  0.0000095  j 0.0000908S
Voc
2300
The elements of the excitation branch referred to the primary side are:
1
 105k 
0.0000095
1

 11k 
0.0000908
Rc 
XM
.
81
From the short-circuit test data, the short-circuit impedance angle is
 SC  cos 1
PSC
160
 cos 1
 55.4
VSC I SC
47  6
The equivalent series impedance is thus
V
47
Z SE  SC  SC 
55.4  4.45  j 6.45
I SC
6
The series elements referred to
the primary winding are:
Req  4.45; X eq  6.45
The equivalent circuit
82
b. To find the equivalent circuit referred to the low-voltage side, we need to
divide the impedance by a2. Since a = 10, the values will be:
RC  1050
X M  110
Req  0.0445
X eq  0.0645
The equivalent circuit will be
83
c. The full-load current on the secondary side of the transformer is
I S ,rated 
S rated 15, 000VA

 65.2 A
VS ,rated
230V
Vp
 VS  Req I S  jX eq I S
a
1
At PF = 0.8 lagging, current I s  65.2  cos (0.8)  65.2  36.9 A
Since:
and
Vp
a
 2300  0.0445   65.2  36.9   j 0.0645   65.2  36.9   234.850.40V
The resulting voltage regulation is, therefore:
VR 
V p a  VS , fl
VS , fl
100%
234.85  230
100%
230
 2.1%

84
At PF = 1.0, current
and
Vp
a
I s  65.2 cos 1 (1.0)  65.20 A
 2300  0.0445   65.20   j 0.0645   65.20   232.941.04V
The resulting voltage regulation is, therefore:
VR 
V p a  VS , fl
VS , fl
100% 
232.94  230
100%  1.28%
230
85
At PF = 0.8 leading, current
and
Vp
a
I s  65.2 cos 1 (0.8)  65.236.9 A
 2300  0.0445   65.236.9   j 0.0645   65.236.9   229.851.27V
The resulting voltage regulation is, therefore:
VR 
V p a  VS , fl
VS , fl
100% 
229.85  230
100%  0.062%
230
86
e. To find the efficiency of the transformer, first calculate its losses.
The copper losses are:
PCu  I S2 Req  65.2 2  0.0445  189W
The core losses are:
Pcore 
Vp a 
RC
2
234.852

 52.5W
1050
The output power of the transformer at the given Power Factor is:
Pout  VS I S cos   230  65.2  cos 36.9  12000W
Therefore, the efficiency of the transformer is
Pout

100%  98.03%
PCu  Pcore  Pout
87
2.8 Three Phase Transformers
The majority of the power generation/distribution systems in the world are 3-phase
systems. The transformers for such circuits can be constructed either as a 3-phase
bank of independent identical transformers (can be replaced independently) or as a
single transformer wound on a single 3-legged core (lighter, cheaper, more efficient).
95
We assume that any single transformer in a three-phase
transformer (bank) behaves exactly as a single-phase
transformer.
The impedance, voltage regulation, efficiency, and other
calculations for three-phase transformers are done on a perphase basis, using the techniques studied previously for singlephase transformers.
Four possible connections for a three-phase transformer
bank are:
1. Y-Y
2. Y-
3. - 
4. -Y
96
Voltage , Current relationships in star and delta connection
For star (Y) connection
V 
VL
3
I  I L
For delta () connection
V  VL
I 
IL
3
97
1. Y-Y connection:
The primary voltage on each phase of
the transformer is
1. Y-Y connection:
V P 
VLP
3
The secondary phase voltage is
VLS  3V s
The overall voltage ratio is
3V p
VLP

a
VLS
3V p
98
The Y-Y connection has two very serious problems:
1. If loads on one of the transformer circuits are unbalanced, the voltages on the
phases of the transformer can become severely unbalanced.
2. The third harmonic issue. The voltages in any phase of a Y-Y transformer are
1200 apart from the voltages in any other phase. However, the third-harmonic
components of each phase will be in phase with each other. Nonlinearities in
the transformer core always lead to generation of third harmonic! These
components will add up resulting in large (can be even larger than the
fundamental component) third harmonic component.
Both problems can be solved by one of two techniques:
1. Solidly ground the neutral of the transformers (especially, the primary side).
The third harmonic will flow in the neutral and a return path will be
established for the unbalanced loads.
2. Add a third -connected winding. A circulating current at the third harmonic
will flow through it; suppressing the third harmonic in other windings.
.
99
2. Y- connection:
The primary voltage on each
phase of the transformer is
V P 
VLP
3
The secondary phase voltage is
VLS  V S
The overall voltage ratio is
3V P
VLP

 3a
VLS
V S
.
100
The Y- connection has no problem with third harmonic
components due to circulating currents in .
It is also more stable to unbalanced loads since the  partially
redistributes any imbalance that occurs.
One problem associated with this connection is that the
secondary voltage is shifted by 300 with respect to the primary
voltage. This can cause problems when paralleling 3-phase
transformers since transformers secondary voltages must be
in-phase to be paralleled. Therefore, we must pay attention to
these shifts.
In the U.S., it is common to make the secondary voltage to lag
the primary voltage. The connection shown in the previous
slide will do it.
101
3.  -Y connection:
The primary voltage on each
phase of the transformer is
V P  VLP
The secondary phase voltage is
VLS  3V S
The overall voltage ratio is
V P
VLP
a


VLS
3V S
3
The same advantages and the same
phase shift as the Y- connection.
102
4.  -  connection:
The primary voltage on
each phase of the
transformer is
V P  VLP
The secondary phase
voltage is
VLS  V S
The overall voltage ratio is
VLP V P

a
VLS V S
No phase shift, no problems with
unbalanced loads or harmonics.
103
Transformer Ratings
Transformers have the following major ratings:
1. Apparent power;
3. Voltage;
2. Current;
4. Frequency.
Example: A typical name plate rating of a single phase transformer is as
follows: 20 KVA, 3300/220V, 50Hz.
 20 KVA is rated output at the secondary terminal.
Note that the rated output is expressed in kilo-volt-ampere (KVA) rather than in
kilowatt (KW).
This is due to the fact that rated transformer output is limited by the heating and
hence by losses in the transformer. The two types of losses in a transformer are
core loss and ohmic (I2R) loss.
The core loss depends on transformer voltage and ohmic loss on transformer
current.
As these losses depend on transformer voltage (V) and current (I) and are almost
unaffected by the load power factor, the transformer rated output is expressed in
VA (VI) or in kVA and are not in kW.
112
 3300/220 V refers to the design voltages of the two windings.
 3300 V is rated primary voltage and refers to the voltage applied to
the primary winding.
 220V is rated secondary voltage and refers to the voltage
developed between output terminals at no-load with rated voltage
applied to the primary terminals.
 Rated primary and secondary currents are calculated from the
rated KVA and the corresponding rated voltages.
 Rated frequency refers to the frequency for which the
transformer is designed to operate.
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