Download UNIT – 4 Explain the construction of Single phase transformer with

UNIT – 4
1. Explain the construction of Single phase transformer with neat diagrams.
ANS:
TRANSFORMER CONSTRUCTION:
There are two general types of transformers
1. Core type transformer
2. Shell type transformer
These two differ by the manner in which the windings are wound around the magnetic core.
The magnetic core is a stack of thin silicon-steel laminations about 0.35 mm thick for 50 Hz
transformer. In order to reduce the eddy current losses, these laminations are insulated from
one another by thin layers of varnish. In order to reduce the core losses, transformers have
their magnetic core made from cold-rolled grainoriented sheet steel (C.R.G.O). This material,
when magnetized in the rolling direction, has low core loss and high permeability.
CORE TYPE TRANSFORMER:
In the core-type,
the
windings
surround
a
considerable part
of steel core as
shown in fig (a).
The
core
type
transformers
require
more
conductor
material and less
iron
when
compared
to
shell-type. The vertical portions of the core are usually called limbs or legs and the top and
bottom portions are called the yoke. For single phase transformers, core-type has two legged
core. In order to reduce leakage flux, half of the L.V. winding is placed over one leg and other
half over other leg. For H.V. winding also, half of the winding is placed over one leg and the
other half over the other leg. L.V. winding is placed adjacent to the steel core and H.V. winding
outside, in order to minimize the amount of insulation required.
Shell Type Transformer:
In the core-type,
the
steel
core
surrounds
a
considerable part of
the windings as
shown in fig (b).
Shell-type
transformer
has
three legged core.
The L.V. and H.V.
windings
are
wound
on
the
central limb. In order to reduce leakage flux, the windings are interleaved or sandwiched. The
shell type transformers require more iron and less conductor material when compared to coretype. There are two types of windings employed for transformers.
2. Explain the principle of operation of a Transformer in detail with neat diagrams.
ANS:
WORKING PRINCIPLE OF A TRANSFORMER
A transformer is a static device which transfers electric energy from one circuit to another
circuit without changing the frequency of the system. It works on electromagnetic induction
principle. According to this principle, an e.m.f. is induced in a coil if it links a changing flux.
Consider two coils 1 and 2 wound
on a simple magnetic circuit as
show in fig.1.1. These two coils are
insulated from each other and there
is no electrical connection, but
magnetically coupled. The two coils
posses high mutual inductance. If
one coil is connected to a source of
alternating voltage, an alternating
flux is set up in the laminated core,
most of which is linked with the
other coil, in which it produces
mutually induced e.m.f. according
to Faraday’s law of electromagnetic
induction. If the secondary coil
circuit is closed a current flows in it
and so an electrical energy transfers
from the first coil to the second coil.
Coil 1 which receives energy from
the source of ac supply is called the
primary coil or primary winding and coil 2, which is connection to load and delivers energy to
the load is called the secondary coil or secondary winding. The symbolic representation of a two
winding transform is as shown in fig.1.1. The two vertical lines are used to represent magnetic
core, which signify the tight magnetic coupling between the windings.
3. Write the classifications of Transformer.
ANS:
TRANSFORMER CLASSIFICATION
1. According to number of phases
(a)Single phase transformer
(b)Three phase transformer
2. According to construction
(a) Core type transformer
(b) Shell type transformer
(c) Berry type transformer
3. According to Function
(a) Power transformer
(i)Step-up transformer
(ii)Step-down transformer
(b) Distribution transformer
(c) Instrument transformer
(i) Current transformer (CT)
(ii) Potential transformer (PT)
4. According to type of cooling
(a) Air cooled transformer
(b) Oil cooled transformer
5. According to frequency
(a) Low frequency (50 Hz)
(b) High frequency (HFT’s pulse transformer)
6. Special purpose transformer
(a) Scott connection transformer
(b)V-connection transformer
(c) Pulse transformer
(d)Auto transformer
(e)Welding transformer
7. According to location
(a) Indoor transformer
(b) Outdoor transformer
8. According to rating
(a) Low rating transformer
(b) Medium rating transformer
(c) High rating transformer
4. Derive the EMF equation of a Transformer.
ANS:
E.M.F. EQUATION OF A TRANSFORMER
When the primary winding is excited by an alternating
voltage, it circulates
alternating current and hence alternating flux is
produced in the core.
Let Ф = Фm Sin ωt.
Where Фm = Maximum value of flux
f= Frequency of supply voltage
N1= Number of primary winding turns
N2= Number of secondary winding turns
E1= r.m.s value of the primary induced e.m.f.
E2= r.m.s. value of the secondary induced e.m.f.
According to Faraday’s Law of electromagnetic induction, E.M.F induced is given by
eN
d
d
  N  m Sin t    N  Cost  N  m  Sin (t  90 0 )
dt
dt
It is clear from the above equation that maximum value of induced e.m.f. is
Emax  N m 
The r.m.s. value induced e.m.f. is
E rms 
E max
2

N m 
2

N  m 2f
2
 4.44 N f  m
RMS value of e.m.f. induced in the primary winding
Similarly
RMS value of e.m.f. induced in the secondary winding
E1= 4.44 N1 f Фm Volts
E2= 4.44 N2 f Фm Volts
5. Explain the Transformer action when at NO-Load and When Loaded. Also draw the phasor
diagram for different loaded conditions.
ANS:
PRACTICAL TRANSFORMER ON NO-LOAD
In the case of ideal transformer, losses are neglected, but practically it is not possible. When
the primary of a transformer is connected to ac supply source, and the secondary is open, the
transformer is said to be on no-load. When the transformer is on no-load, the primary current
is not completely reactive. But it has to supply iron losses in the core and a very small amount
of copper losses in primary (there being no copper loss in secondary as it is open). Hence, the
no load primary input current I0 is not at 900 behind V1 but lags it by an angle Ф0 <900.
No load input power of the
transformer is W0 =V1I0 CosФ0
The phasor diagram under noload condition is drawn as
shown in fig(b).
From the pharos diagram Io
has two components
i) Working
or
active
component, Iw =
I0CosФ0 , which is in
phase with V1
ii) Magnetizing component,
Im = I0CosФ0 , which is in quadrature with V1
 I 0  I m2  I w2
PRACTICAL TRANSFORMER ON LOAD
When the transformer is loaded, the secondary current, I 2 is set up. I2 will be in phase with V2
if the load is resistive, it lags V2 if the load is inductive, it leads V2 if the load is capacitive. The
secondary current I2 sets up its own m.m.f.(=N2I2) and hence it produces flux Ф2, which is in
opposition to the main primary flux Ф , which is due to I0. Secondary flux Ф2 weakens the main
flux Ф momentarily and hence primary back e m f E1 tends to be reduced. For a moment V1
gains the upper hand over E1 and hence causes additional current I2’ to flow in primary and
hence flux Ф2’ (due to m.m.f. N1 I2’ ) which counter balance the secondary flux Ф2.
 2'   2
N1 I 2'  N 2 I 2
 I 2' 
where K = Transformation Ratio
N2
I2  K I2
N1
I 2' is known as load component of primary current. This current is in anti-phase with I1.
I 1  I 0  I 2'
Here
The phasor diagram of a transformer under
load condition can be drawn as shown below.
6. Draw the Equivalent circuit of the Transformer when referred to primary and secondary.
ANS:
EQUIVALENT CIRCUIT OF A TRANSFORMER
In transformers, the problems concerning voltages and currents can be solved by use of phasor
diagrams. However, it is more convenient to represent the transformer by an equivalent circuit.
The term equivalent circuit of a transformer means the combination of fixed and variable
resistances and reactances, which exactly simulates performance and working of the machine.
If an equivalent circuit is available, the computations can be done by the direct application of
circuit theory. The transformer shown diagrammatically in fig. can be resolved into an
equivalent circuit in which the resistance and leakage reactance of the transformer are
imagined to be external to the windings whose function is only to transform the voltage.
For a transformer, no-load primary current I0 has two components
I m  I 0 Sin 0  Magnetizin g Component
I w  I 0 Cos 0  Working Component
Im produces the flux and is assumed to flow through reactance X 0 called no-load reactance
while Iw is active component representing core losses hence is assumed to flow through
resistance R0. R0 and X0 are connected in parallel across the primary circuit as shown in fig.
Equivalent Circuit of a Transformer
From the equivalent circuit we can write
R0 
E1
Iw
and
X0 
E1
Im
To make the calculations simpler, it is preferable to transfer voltage, current and impedance
to the one side either primary or secondary of transformer.
(i) Equivalent circuit when referred to Primary
Let us see how to transfer secondary values to primary side:
E 2'  The primary equivalent of secondary E.M.F. =
E2
 E1
K
V2'  The primary equivalent of secondary terminal voltage 
i.e. I1 V2'  I 2 V2  V2' 
V2
K
I2
V
V2  2
I1
K
I 2'  The primary equivalent of the sec ondary current  K I 2
i.e. V1 I 2'  V2 I 2  I 2' 
V2
I2  K I2
V1
R2
K2
X
X 2'  The primary equivalent of secondary reactance  22
K
Z
Z 2'  The primary equivalent of secondary Impedance  22
K
R2'  The primary equivalent of secondary resistance 
R2
X
;
X 01  X 1  22 ;
2
K
K
 R01  j X 01
R01  R1 
 Z 01
where K 
V2
E
N
I
 2  2  1  Transformation Ratio
V1
E1
N1
I2
Thus the equivalent circuit of the transformer when referred to primary is as shown in the
below fig.
Equivalent circuit when referred to primary
Approximate equivalent circuit of a transformer:
The no-load current I0 usually less than 5% of the full load primary current. The voltage drop
produced by I0 in (R1+jX1) is negligible for practical purposes. Therefore, it is immaterial that
the shunt branch R0//X0 is connected before the primary series impedance (R1+jX1) or after it.
The currents Im and Iw are not much affected. Therefore, the equivalent circuit can be further
simplified by shifting the no load branch containing R 0 and X0 to the supply terminals as
shown in fig. By doing this, we are creating an error that the drop across R1 and X1 due to I0 is
neglected. Hence equivalent circuit is called approximate equivalent circuit.
Approximate equivalent circuit when referred to primary
The equivalent circuit can be further simplified by combining parameters R1 and
and combining parameters X1 and
R21 as R01
X 2' as X01 Where
R2
X
;
X 01  X 1  22 ;
2
K
K
 R01  j X 01
R01  R1 
 Z 01
The circuit is shown in the fig.
Approximate equivalent circuit when referred to primary
In the similar fashion the approximate equivalent circuit referred to secondary can also be
obtained
(ii) EQUIVALENT CIRCUIT WHEN REFERRED TO SECONDARY
Similarly if all the primary quantities are transferred to secondary, we get the
equivalent circuit of transformer referred to secondary as shown in fig.
R1'  R1 K 2
X 1'  X 1 K 2
Z1'  Z1 K 2
I
I
I 1'  1
I 0'  0
E1'  E1 K
K
K
2
2
R02  R2  K R1
X 02  X 2  K X 1
Z 02  R02  j X 02
1
1
Similarly the exciting circuit parameters also gets transformed to secondary as R0 and X 0
The equivalent circuit referred to secondary is as shown in fig.
Equivalent circuit when referred to secondary
(a)
(b)
Equivalent circuit when referred to secondary
(c)
7. Write the efficiency equation of a transformer.
ANS:
EFFICIENCY OF A TRANSFORMER
Efficiency of a transformer at a given load and power factor is defined as the ratio of output
power to the input power, both the quantities expressed in the same units.
 Efficiency, 

Output Power
Output power

Input Power
Output power  Losses
V2 I 2 * power factor
V2 I 2 * power factor  I 12 R1  I 22 R2  Wi
It may appear that efficiency can be determined by directly loading the transformer and
measuring the input power and output power. However, this method has the following
drawbacks:



Since the efficiency of a transformer is very high, even 1% error in each wattmeter
(output and input) may give ridiculous results. This test, for instance, may give
efficiency higher than 100%.
Since the test is performed with transformer on load, considerable amount of power is
wasted. For large transformers, the cost of power alone would be considerable.
The test gives no information about the proportion of various losses.
Due to these drawbacks, direct loading method is seldom used to determine the efficiency of a
transformer. In practice, the efficiency can be calculated by determining iron loss from open
circuit test and copper loss from short circuit test.
8. Explain briefly about the regulation of an Alternator.
ANS:
REGULATION OF A TRANSFORMER
When the transformer is loaded its terminal voltage falls from no load to full load. The change
in secondary terminal voltage from no load to full load and expressed as secondary no load
voltage is known as regulation down.
 V  V2 
 * 100
 Voltage Re gulation down   0 2
V
 0 2 
If the change in voltage is expressed as secondary full-load voltage then it is known as
regulation up
 V  V2 
 * 100
 Voltage Re gulation up   0 2
 V2

Where 0V2 = No-load secondary voltage=KV1
V2= Secondary voltage on load
Unless stated otherwise, regulation is to be taken as regulation down. Lesser the regulation
better the transformer performance. It may be noted that percentage voltage regulation of the
transformer will be the same whether primary or secondary side.
From the phasor diagram considering OE ≈OF (by neglecting EF)
∴OC = OE =OA + AD + DE
0V2 = V2+ AD+ DE
AD
From Δle ADB,
AB
AD AB *Cos2  I 2 R02 Cos 2
Cos 2 
From Δle BCG, Cos (90   2 ) 
BG
BC
 Sin  2 
DE
BC
 DE  I 2 X 02 Sin  2
∴ From above equations,
0
V2  V2  AD  DE
 V2  I 2 R02 Cos  2  I 2 X 02 Sin  2
 0V2  V2  I 2 R02 Cos  2  I 2 X 02 Sin  2
In general the approximate voltage drop is given by 0 V2  V2  I 2 R02 Cos 2  I 2 X 02 Sin 2
Note: + ve sign for lagging p.f & - ve sign for leading p.f
 V  V2 
I R Cos 2  I 2 X 02 Sin 2
 * 100  2 02
 Voltage Re gulation   0 2
* 100
0 V2
 0 V2 
9. What is ALL DAY EFFICIENCY?
ANS:
ALL-DAY OR ENERGY EFFICIENCY
The ordinary or commercial efficiency of a transformer is defined as the ratio of output power to
the input power i.e.,
 Commercial Efficiency, 
Output Power
Output power

Input Power
Output power  Losses
There are certain types of transformers whose performance cannot be judged by this efficiency.
For instance, distribution transformers used for supplying lighting loads have their primaries
energized all the 24 hours in a day but the secondaries supply little or no load during the
major portion of the day. It means that a constant loss (i.e., iron loss) occurs during the whole
day but copper loss occurs only when the transformer is loaded and would depend upon the
magnitude of load. Consequently, the copper loss varies considerably during the day and the
commercial efficiency of such transformers will vary from a low value (or even zero) to a high
value when the load is high. The performance of such transformers is judged on the basis of
energy consumption during the whole day (i.e., 24 hours). This is known as all-day or energy
efficiency.
All-day efficiency is defined as the ratio of output in kWh (i.e output energy) to the input in
kWh (i.e input energy) of a transformer over a 24-hour period. i.e.
 all day 
kWh Output in 24 hrs
kWh Input in 24 hrs
This efficiency is always less than the commercial efficiency of a transformer. To find the allday efficiency, we have to know the load cycle on the transformer i.e., how much and how long
the transformer is loaded during 24 hours.
All-day efficiency is of special importance for those transformers whose primaries are never
open-circuited but the secondaries carry little or no load much of the time during the day. In
the design of such transformers, efforts should be made to reduce the iron losses which
continuously occur during the whole day.
Note: Efficiency of a transformer means commercial efficiency unless stated otherwise.
10. What are the different tests to be performed on the transformer to find out IRON and COPPER
losses in a transformer.
ANS:
TRANSFORMER TESTS
The circuit constants, efficiency and voltage regulation of a transformer at any load and p.f.
condition can be predetermined without actually loading the transformer. But the equivalent
circuit parameters of a transformer are determined by conducting two simple tests (i) opencircuit test and (ii) short-circuit lest. These tests give more accurate results than those
obtained by taking measurements on fully loaded transformers. Further, the power required to
carry out these tests is very small as compared with full-load output of the transformer. These
tests consist of measuring the input voltage, current and power to the primary first with
secondary open-circuited (open-circuit test) and then with the secondary short-circuited (short
circuit test).
OPEN CIRCUIT TEST
This test is conducted to determine iron loss (or core loss) and no load current, I0 which is
helpful in finding R0 and X0. In this test, the rated voltage is applied to the primary (usually
low-voltage winding) while the secondary is left open circuited. A wattmeter ‘W’ a voltmeter ‘V’
and an ammeter ‘A’ are connected in the low voltage side i.e primary winding in the present
case. The applied primary voltage V1 is measured by the voltmeter, the no load current I0 by
ammeter and no-load input power W0 by wattmeter as shown in fig. With normal voltage
applied to primary, normal flux will set up in the core and hence normal iron losses will occur,
which are recorded by the wattmeter ‘W’. As the primary no load current I0 (2 to 10% of rated
current) is small, copper loss is negligibly small in the primary and nill in the secondary (being
opened). Hence the O.C test gives core losses alone practically (i.e wattmeter reading) and is
same for all loads.
Iron loss, Wi= Wattmeter reading= W0
No-load current= Ammeter reading=I0
Applied voltage=Voltmeter reading=V1
Input power, W0  V1 I 0 Cos 0
no  load p. f  Cos 0 
I m  I 0 Sin 0 ,
R0 
V1
,
Iw
W0
V1 I 0
I w  I 0 Cos 0
X0 
V1
Im
Thus open-circuit test enables us to determine iron losses and parameters R 0 and X0 of the
transformer.
Point to be remembered: The OC test is always conducted on LV side of a transformer. In OC
test, the equipments are connected in LV side, since we need to get rated voltage. If it is
connected in HV side the value of voltage will be more and current will be less. Hence the
ordinary instruments cannot measure this value of current. The man aim of connecting meters
on LV side is to use low range meters and low voltage supply, which are easily available and
safe.
SHORT CIRCUIT TEST
This test is conducted to determine R 01 (or R02), X01 (or X02) and full-load copper losses of the
transformer. In this test, the secondary (usually low-voltage winding) is short circuited by a
thick conductor and variable low voltage is applied to the primary as shown in fig. A wattmeter
‘W’ a voltmeter ‘V’ and an ammeter ‘A’ are connected in the high voltage winding i.e primary
winding in the present case.
A low voltage (usually 5 to 10% of normal primary voltage) is applied through a variac to the
primary and is gradually increased till the ammeter ‘A’ indicates full load current I1 in the
primary. Since the applied voltage is very low, so flux produced is very small. Hence, the iron
losses are so small that these can be neglected, with the result, the wattmeter ‘W’ reads total
full load copper loss of the transformer.
Full load Cu loss, Wcu= Wattmeter reading= Wsc
Full load primary current= Ammeter reading=Isc
Applied voltage=Voltmeter reading=Vsc
2
2
2
2
WSC  I SC
R1  I SC
R2  I SC
( R1  R2 )  I SC
R01
Total impedance when referred to primary, Z 01 
 R01 
WSC
2
I SC
VSC
I SC
Total leakage reactance when referred to primary, X 01 
2
2
Z 01
 R01
Thus short-circuit lest gives full-load Cu loss, R01 and X01
Note: The short-circuit test will give full-load Cu loss only if the applied voltage Vsc is such so
as to circulate full-load currents in the windings. If, in a short circuit test, current value is
other than full-load value, the Cu loss will be corresponding to that current value.
Point to be remembered: The SC test is always conducted on HV side of a transformer. If the
measurements were made on LV side, the voltage needed would be inconveniently low and the
current would be inconveniently high.
11. Pre-determination of efficiency and Voltage Regulation of a transformer.
ANS:
Predetermination of efficiency and voltage regulation
Knowing the equivalent resistance and reactance referred to primary (or secondary) from Short
Circuit test, the voltage regulation of the transformer at any p.f. can be determined by
 Voltage Re gulation 

I 1 R01 Cos   I 1 X 01 Sin 
* 100
V1
I 2 R02 Cos  2  I 2 X 02 Sin  2
* 100
0 V2
By performing open circuit and short circuit test we can find transformer total losses
If W0 =Input power in watts from OC test
=Iron losses, Wi
Wsc =Input power in watts from SC test with full load current
=Full load copper losses, Wcu
Then the total losses on full load=Wi + Wcu
Full load efficiency is given by
 full load 
Full load kVA * Cos 
Full load kVA * Cos   Wi  Wcu
Efficiency at any load is given by
 at any load 
x S * Cos 
x S * Cos   Wi  x 2 Wcu
Where x=Fraction of F.L at which the transformer is working
S=Full load kVA of the transformer