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Mechanical and Electrical Systems
SKAA 2032
Transformers
Dr. Asrul Izam Azmi
Faculty of Electrical Engineering
Universiti Teknologi Malaysia
Transformers
• A transformer is a static machine which change
voltage: step up or down.
• There is no electromechanical energy conversion.
• The transfer of energy takes place through the
magnetic field and all currents and voltages are
AC
• Transformer can be categorized as:
The ideal transformers
Practical transformers
Special transformers
Three phase transformers
Transformers
Transformers
Transformers
Applications of the transformer
• A typical power system consists of generation,
transmission and distribution
• Power from plant/station is generated around 11-1320-30kV (depending upon manufacturer and demand)
• This voltage is carried out at a distance to reach for
utilization through transmission line system by step up
transformer at different voltage levels depending upon
distance and losses
• The distribution is made through step down
transformer according to the consumer demand.
Functions of transformer
• Raise or lower voltage or current in AC circuit
• Isolate circuit from each other
• Increase or decrease the apparent value of a
capacitor, inductor or resistor
• Enable transmission of electrical energy over great
distances
• Distribute safely in homes and factories
Transformer is one of the most useful electrical
devices ever invented
Principles of Transformer
• A transformer consists of two electric circuits called
primary and secondary
• A magnetic circuit provides the link between
primary and secondary
Principles of Transformer
The basic idea:
• AC voltage is applied to the primary circuit, AC
current Ip flows into the primary circuit.
• Ip sets up a time-varying magnetic flux Ф in the core
• An AC voltage is induced to the secondary circuit, Vs
according to the Faraday’s law
Core Types of Transformer
• The magnetic (iron) core is made of thin laminated
steel sheet. The reason of using laminated steel is to
minimize the eddy current loss by reducing
thickness
• There are two common cross
section of core which include
square or (rectangular) for small
transformers and circular (stepped)
for the large and 3 phase
transformers.
Configuration of Single phase
Primary
Winding
Multi-layer
Laminated
Iron Core
Secondary
Winding
H1 H2
X1
X2
Winding
Terminals
Three Phase Transformer
The three phase transformer iron core has three legs
A phase winding is placed in each leg
Construction of Transformer
• Core (U/I) Type: Is constructed from a stack of U
and I shaped laminations. In a core-type
transformer, the primary and secondary windings
are wound on two different legs of the core
• Shell Type: Is constructed from a stack E and I
shaped laminations. In a shell-type transformer, the
primary and secondary windings are wound on the
same leg of the core, as concentric windings, one on
top of the other.
Construction of Transformer
• Core and Shell Type Transformer
Shell Type
Core Type
Construction of a Small Transformer
Iron core
Insulation
Secondary
winding
Terminals
Transformer with Cooling System
High voltage
bushing
Oil tank
Bushing
Low voltage
bushing
Steel
tank
Iron core
behind the steel
bar
Cooling
radiators
Winding
Insulation
Radiator
The Ideal Transformer
For an ideal transformer, we assume:
• No losses
• Flux produced by the primary is completely linked
by the secondary and vice versa
• Core is infinitely permeable
• No leakage flux of any kind
The Ideal Transformer
• Primary and secondary posses N1 and N2 turns
respectively
• Primary is connected to a sinusoidal source Eg
• Magnetizing current Im creates a flux of Φm
The Ideal Transformer
• The flux is completed linked by the primary and
secondary windings – mutual flux
• Flux varies sinusoidally and reaches a peak value of Φm
The Ideal Transformer
• A transformer with more turns in its primary than
its secondary coil will reduce voltage and is called a
step-down transformer
• One with more turns in the secondary than the
primary is called a step-up transformer
The Ideal Transformer
• The sinusoidal current Im produces a sinusoidal mmf
(magnetomotive force )= NIm which in turn creates a
sinusoidal flux. The flux induces an effective voltage
E across the terminals of the coil
Induces Voltages:
• The effective induced emf in primary winding is
E1  4.44 fN1m
• Where N1 is the number of winding turns in primary
winding, Фm the maximum (peak) flux and f is the
frequency of the supply voltage
The Ideal Transformer
• This equation shows that for a given frequency and
a given number of turns, Фm varies in proportion to
the applied voltage, Eg
• This means that if Eg is kept constant, the peak flux
must remain constant
• Similarly, the effective induced emf in secondary
winding:
E2  4.44 fN2m
The Ideal Transformer - at No Load
E1 N1

a
E2 N 2
a = Turn ratio
E1 = Voltage induced in the primary [V]
E2 = Voltage induced in the secondary [V]
N1 = Number of turns on the primary
N = Number of turns on the primary
2
Example problem
A transformer has 500 primary turns and 3000
secondary turns. If the primary voltage is 240V,
determine the secondary voltage, assuming an
ideal transformer
N1=500, N2=3000, V1=240V
For ideal transformer, V1=E1 and V2=E2
E1 V1 N1


E2 V2 N 2
240 500

V2
3000
V2  1440V
Ideal Transformer Under Load: Current Ratio
Let us connect a load Z across the secondary of the ideal
transformer. A secondary current I2 will immediately flow.
1) In an ideal transformer the primary and secondary windings
are linked by the mutual flux, Фm, consequently voltage ratio
will be the same as at no load
2) If the supply voltage Eg is kept fixed, then the primary induced
voltage E1 remain fixed. Consequently, Фm also remains
fixed. It follows that E2 also remain fixed
We conclude that E2 remains fixed whether a load is connected
or not
Ideal Transformer Under Load: Current Ratio
• Let us now examine the mmf created by the primary
and secondary windings. First, current I2 produces a
secondary mmf N2I2. If it acted alone, this mmf
would produce a profound change in the Фm. But we
just saw that Фm does not change under load.
• We conclude that flux Фm can only remain fixed if
the primary develops a mmf which exactly
counterbalances N2I2 at every instant. Thus, a
primary current I1 must flow so that
N1 I1  N 2 I 2
Ideal Transformer Under Load: Current Ratio
To obtain the required instant-to-instant bucking effect,
current I1 and I2 must increase and decrease at the
same time
(a)Ideal transformer under load
(b) Phasor relationships under load
Ideal Transformer Under Load: Current Ratio
I1
I 2 N1

a
I1 N 2
I2
Secondary
Primary
I 1 = Primary current [A]
Single phase transformer
I 2 = Secondary current [A]
N1
N2
= Number of turns on the primary
= Number of turns on the secondary
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformer Impedance
Z L' 
VP
IP
Primary Impedance:
Primary impedance in terms of
Z L'  a 2 Z L
secondary impedance :
VP  aVS
Primary Voltage :
I
I

Primary Current :
a
S
P
INSPIRING CREATIVE AND INNOVATIVE MINDS
Example problem
An ideal transformer has a turns ratio of 8:1 and
the primary current is 3A when it is supplied at
240V. Calculate the secondary voltage and
current.
E1 V1 N1


E2 V2 N 2
N2
V2  V1
N1
1
V2  240   30V
8
Example problem
N1 I 2

N 2 I1
N1
8
I 2  I1
 3   24 A
N2
1
Example problem
A transformer coil possesses 4000 turns and
links an ac flux having a peak value of 2 mWb. If
the frequency is 60 Hz, calculate the effective
value of the induced voltage E.
E  4.44 fN m
 4.44(60)(4000)(0.002)
 2131V
Example problem
A coil having 90 turns is connected to a 120V, 60 Hz
source. If the effective value of the magnetizing
current is 4 A, calculate the following:
a. The peak value of flux
b. The peak value of the mmf
c. The inductive reactance of the coil
d. The inductance of the coil.
Real Transformer
Leakage Flux: Not all of the flux produced by the
primary current links the winding, but there is leakage
of some flux into air surrounding the primary. Similarly,
not all of the flux produced by the secondary current
(load current) links the secondary, rather there is loss of
flux due to leakage. These
effects are modeled as
leakage reactance in the
equivalent circuit
representation.
Voltage Regulation
• Voltage Regulation is defined as the change in the
magnitude of the secondary voltage as the load
current changes from the no-load to full load
• The primary side voltage is always adjusted to meet
the load changes; hence V’s and Vs are kept
constant
VNL  VFL
%VR 
100
VNL
 Vp  aVs  aVs 100


 Vp  Vs' Vs' 100
Efficiency of Transformer
• As always, efficiency is defined as power output to
power input ratio
  Pout Pin 100%
Pin  Pout  Pcore  Pcopper
• Pcopper represents the copper losses in primary and
secondary windings. There are no rotational losses.
Example Problem
A not-quite-ideal transformer having 90 turns on the
primary and 2250 turns on the secondary is connected
to a 120 V, 60 Hz source. The coupling between the
primary and the secondary is perfect but the
magnetizing current is 4 A. calculate:
a. The effective voltage across the secondary terminals
b. The peak voltage across the secondary terminals.
c. The instantaneous voltage across the secondary when
the instantaneous voltage across the primary is 37 V.
Ans: 3000V, 4242 V, 925 V.
Example Problem
An ideal transformer having 90 turns on the primary
and 2250 turns on the secondary is connected to a 200
V, 50 Hz source. The load across the secondary draws a
current of 2 A at a power factor of 80 per cent lagging.
Calculate :
• The effective value of the primary current
• The instantaneous current in the primary when the
instantaneous current in the secondary is 100 mA.
• The peak flux linked by the secondary winding.
Ans: 50 A, 2.5 A, 10 mWb
Example Problem
A 125 kVA transformer has 500 turns on the primary
and 80 turns on the secondary. The primary and
secondary resistances are 0.5 Ω and 0.025 Ω
respectively, and the corresponding leakage reactances
are 2.5 Ω and 0.025 Ω respectively. The supply voltage
is 2.2 kV. Calculate the voltage regulation and the
secondary terminal voltage for full load having a power
factor of 0.85 lagging
Example Problem
The primary and secondary windings of a 400 kVA
transformer have resistances of 0.3 Ω and 0.0015 Ω
respectively. The primary and secondary voltages are
15 kV and 0.4 kV respectively. If the core loss is 2.5 kW
and the power factor of the load is 0.80, calculate the
efficiency of the transformer on full load.
Summary
At the end of this topic, you should know:
• The functions of a transformer in electrical
distribution system
• The working principles of a transformer
• The voltage and current ratios
• The difference between the ideal and real
transformer