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Transcript
ELECTRICAL MACHINE
 140070110029
 140070110046
 140070110022
 140070110006
 140070110064
RUSHABH MEHTA
UDIT PATEL
JAIMIN PUJ
BHARGAV KAKADIYA
MAHENDRA SHIYAL
 BRANCH
ELECTRONICS ENGINEERING
Guided By:J.C.Baria
Single Phase
Transformer
TRANSFORMER
•
•
•
•
A transformer is a static device.
The word ‘transformer’ comes form the word ‘transform’.
Transformer is not an energy conversion device, but it is device that changes
AC electrical power at one voltage level into AC electrical power at another
voltage level through the action of magnetic field but with a proportional
increase or decrease in the current ratings., without a change in frequency.
It can be either to step-up or step down.
WORKING
The main principle of operation of a transformer is mutual inductance between two
circuits which is linked by a common magnetic flux. A basic transformer consists of two
coils that are electrically separate and inductive, but are magnetically linked through a
path of reluctance. The working principle of the transformer can be understood from
the figure below
WORKING OF TRANSFORMER
As shown above the transformer has primary and secondary windings.
Both the coils have high mutual inductance. A mutual electro-motive force
is induced in the transformer from the alternating flux that is set up in
the laminated core, due to the coil hat is connected to a source of
alternating voltage. Most of the alternating flux developed by this coil is
linked with the other coil and thus produces the mutual induced electromotive force. The so produced electro-motive force can be explained with
the help of Faraday’s laws of Electromagnetic Induction as
e=M*dI/dt
If the second coil circuit is closed, a current flows in it and thus electrical
energy is transferred magnetically from the first to the second coil.
The alternating current supply is given to the first coil and hence it can be
called as the primary winding. The energy is drawn out from the second
coil and thus can be called as the secondary winding.
Types of Transformer
 CORE Type Transformer:
 It has a single magnetic circuit.
 In single phase type, the core has




two limbs.
The windings encircle the core.
The cylindrical type coil is used.
As winding are distributed natural
cooling is effective.
It is preferred for Low Voltage
transformer.
The low voltage windings are placed nearer to the core as it is
the easiest to insulate. The effective core area of the
transformer can be reduced with the use of laminations and
insulation
Shell Type Transformer:
 It has double magnetic circuit.
 In single phase type, the core




has three limbs.
The core encircle most part of
winding.
Multilayer disc type or sandwich
coils are used.
As winding are surround by core
the natural cooling is not
possible.
The construction is for high
voltage transformer.
OPEN CIRCUIT TEST
 The purpose of this test is to find the iron loss(core loss).
 A wattmeter, voltmeter and an ammeter are connected in




low voltage side.
Secondary side of transformer we will keep open.
We give the normal voltage applied to the primary, normal
flux will be set up in the core, hence normal iron losses will
occur which are recorded by the wattmeter.
As the primary no-load current is measured by ammeter
and its value is small, Cu-loss is negligibly small in primary
and nil in secondary (it being open).
Hence, the wattmeter reading represents practically the
CORE LOSS under no-load condition.
SHORT CIRCUIT TEST
 A wattmeter, voltmeter and an ammeter are connected in
high voltage side.
 The low voltage side of transformer is short circuited.
 The applied voltage is a small percentage of the normal
voltage.
 Hence the core losses are very small with the result of the
wattmeter reading represent the full-load
CU-LOSS or
loss for the whole transformer .
Transformer Efficiency
 To check the performance of the device, by
comparing the output with respect to the input.
 The higher the efficiency, the better the system.
Efficiency, 
Output Power
100%
Input Power
Pout

100%
Pout  Plosses
 ( fullload) 
V2 I 2 cos 

100%
V2 I 2 cos   Pc  Pcu
 (load n ) 
Where Pcu = Psc
Pi = Poc
kVAcos  1000
(kVAcos  1000)  Pc  Pcu
nkVAcos  1000
(nkVAcos  1000)  Pc  n 2 Pcu
Where, if ½ load, hence n = ½ ,
¼ load, n= ¼ ,
90% of full load, n =0.9
Transformer Losses
 Copper Loss:
 The total power loss in winding resistance of transformer is copper loss.
 Copper Loss=Power loss in primary resistance + Power loss in
secondary resistance
 Copper Losses
Pcopper  Pcu  ( I 1) 2 R1  ( I 2) 2 R2  Pshort circuit
or if referred , Pcu  ( I 1) 2 R01  ( I 2) 2 R02
 Copper loss should be kept as low as possible to increase efficiency.
Iron loss
 Pi = Hysteresis loss + eddy current loss
 Iron Losses
Piron  Pc  ( I c) 2 Rc  Popen circuit
 Both hysteresis loss and eddy current loss are dependent
on frequency.
 Hence iron loss Pi of the total loss is dependent on
frequency but the copper loss Pcu is constant irrespective
of frequency.