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:HAPTER 2:HAPTER 2 OWER TRANSFORMEROWER TRANSFORMER
ur Diyana Kamarudin ur Diyana Kamarudin
EEE3233POWER SYSTEM
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IntroductionIntroduction A transformer is a static machinesstatic machines. The word transformer comes form the word transform. The word transformer comes form the word transform. Transformer is not an energy conversion devicenot an energy conversion device But is a device that changes AC electrical power at onedevice that changes AC electrical power at one
voltage level into AC electrical power at anothervoltage level into AC electrical power at anothervoltage level through the action of magnetic field,voltage level through the action of magnetic field,without a change in frequency.without a change in frequency.
Can raise or lower the voltage/current in ac circuitCan raise or lower the voltage/current in ac circuit
GenerationGenerationStationStation
T X 1 T X 1
DistributionsDistributions
T X 1
T X 1
Transmission SystemTransmission System
33/13.5kV33/13.5kV 13.5/6.6kV13.5/6.6kV
6.6kV/415V6.6kV/415V
Consumer Consumer
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TransformerConstruction
There are 3 basic parts of transformer:A primary coil/winding
receives energy from the ac sourceA secondary coil/winding
receives energy from primary winding &delivers it to the loadA core that supports the coils/windings.
provide a path for magnetic lines of flux
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rans ormerConstruction
The operation of transformer is based on the principal of mutual inductance
A transformer usually consists of two coils of wire wound onthe same core
The primary coil is the input coil while the secondary coil isthe output coil A changing in the primary circuit creates a changingmagnetic field
This changing magnetic field induces a changing voltage inthe secondary circuit
This effect is called mutual induction
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TransformerConstruction
Transformer can be either step-up or step-downtransformer
If the output voltage of transformer is greater than theinput voltage step-up transformer
If the output voltage of a transformer is less than theinput voltage step-down transformer By selecting appropriate numbers of turns, a transformerallows an alternating voltage to be stepped up bymaking Ns more than Np
Or stepped down by making Ns less than Np
Vs NsVp Np
=
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Example 1 There are 400 turns of wire in an iron-corecoil. If this coil is to be used as theprimary of a transformer, how manyturns must be wound on the coil to
form the secondary winding of thetransformer to have a secondaryvoltage of one volt if the primaryvoltage is 5 volts?
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Example 1 (solution)
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TransformerConstruction
Core characteristic: The composition of a transformer core depends on:
voltage, current, frequency, size limitations andconstruction costs
Commonly used core materials are air, soft iron, and steel Air-core transformers are used when the voltage source has
a high frequency (above 20 kHz) Iron-core transformers are usually used when the source
frequency is low (below 20 kHz) A soft-iron-core transformer is very useful where the
transformer must be physically small, yet efficient The iron-core transformer provides better power transfer
than does the air-core transformer
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TransformerConstruction
A transformer whose core is constructed of laminated sheets of steel dissipates heat readily; thus it provides for the efficienttransfer of power.
The purpose of the laminations is to reduce certain losses whichwill be discussed later in this part
Hollow-core construction
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TransformerConstruction
The most efficient transformer core is one that offers the bestpath for the most lines of flux with the least loss in magnetic andelectrical energy There are two main shapes of cores used in laminated-steel-core transformers:
Core-type transformers Shell-core transformers
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TransformerConstruction
Shell-core transformers: The most popular and efficient transformer core Each layer of the core consists of E- and I-shaped sections of
metal These sections are butted together to form the laminations The laminations are insulated from each other and then pressed
together to form the core.
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Advantages of Shell types Transformer
Better cooling facilityLess leakage reactanceGreater mechanical strengthLess magnetising currentLess magnetic loss
D isd va n ta g e s o f S h e ll typ e sTra n sfo rm e r
M ore d ifficu lt fo r m an u factu rin g G re a te r d ifficu lty in ca rryin g ou t re p a irs
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TransformerConstruction
Core - type construction: so named because the core is shaped with a hollow square
through the center the core is made up of many laminations of steel
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The basic difference between these two transformersare:
1)The core type has two limbs & shell type has three limbs.2)Core type has longer mean length of iron core & shorter
mean length of coil turn.
Shell type has shorter mean length of iron core &longer mean length of coil turn.3) In core type transformers the LV(low voltage) coil is wound
next to the core & HV(high voltage) coil is wound onthe LV coil after the insulation layer. In Shell type
transformers the LV & HV windings are sandwichedbetween each other.
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f
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TransformerConstruction
An ideal transformer is a transformer which has no loseswhich has no loses,i.e. its winding has no ohmic resistance, no magneticleakage, and therefore no I 2R and core loses.
However, it is impossibleimpossible to realize such a transformer inpractice.
Yet, the approximate characteristic of ideal transformer willapproximate characteristic of ideal transformer willbe used in characterized the practical transformer.be used in characterized the practical transformer.
VV11 VV22
NN11 : N: N 22
EE 11 EE 22
II11 II22
VV11 Primary Voltage Primary VoltageVV22 Secondary Voltage Secondary VoltageEE 11 Primary induced Voltage Primary induced VoltageEE 22 secondary induced Voltage secondary induced VoltageNN11 :N:N22 Transformer ratio Transformer ratio
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TransformerConstruction
No-load condition: is said to exist when a voltage is applied to the primary,
but no load is connected to the secondary Because of the open switch, there is no current flowing in
the secondary winding. With the switch open and an ac voltage applied to theprimary, there is, however, a very small amount of current called EXCITING CURRENT flowing in the primary
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TransformerConstruction
With-load condition: When a load device is connected across the secondary
winding of a transformer, current flows through thesecondary and the load
The magnetic field produced by the current in thesecondary interacts with the magnetic field produced bythe current in the primary
This interaction results from the mutual inductancebetween the primary and secondary windings.
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Transformer Equation For an ac sources,
Let V(t) = V m sin ti(t) = i m sin t
Since the flux is a sinusoidal function;
Then:
Therefore:
Thus:
t t m sin)( =
t N dt
t d N Emf V
m
mind ind
cos
sin
=
==
maxmind ind fN N Emf V === 2(max)
maxmm
rmsind fN fN N
Emf =
=
= 44.42
22
)(
m m B x A =
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Transformer Equation For an ideal transformer,
In the equilibrium condition, both the input power will beequaled to the output power, and this condition is said toideal condition of a transformer.
From the ideal transformer circuit, note that,
1 1 2 2V I V I =
1 2
2 1
V I V I
=
(i)
2211 V E and V E ==
max
max
fN E
fN E = =
= =
22
11
44.4
44.4
Input power = output power
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Transformer Equationn
a I I
N N
E E
Therefore ===1
2
2
1
2
1,
aa = Voltage Transformation RatioVoltage Transformation Ratio ;which will determine whether the transformer is goingto be step-up or step-down
EE 22 > E> E 11For a >1For a >1
For a
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Transformer Equation Transformer rating is normally writtenwritten in terms of
Apparent Power Apparent Power. Apparent power is actually the product of its ratedits rated
current and rated voltagecurrent and rated voltage.
2211 I V I V VA == Where,
I1 and I 2 = rated current on primary and secondary
winding. V1 and V 2 = rated voltage on primary and secondarywinding.
**** Rated currents are actually the full load currents inRated currents are actually the full load currents intransformer transformer
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Example 1 1.5kVA single phase transformer has ratedvoltage of 144/240 V. Finds its full loadcurrent.
Solution:Solution:
A I
A I
FL
FL
62401500
42.10144
1500
2
1
==
==
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Example 2 A single phase transformer has 400 primary and
1000 secondary turns. The net cross-sectionalarea of the core is 60m2. If the primary winding
is connected to a 50Hz supply at 520V,calculate: The induced voltage in the secondary winding The peak value of flux density in the core
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Example 2 (solution)N1=400 V1=520V A=60m2 N2=1000
V2=?a) We know that,n
n
n
b) Emf,n
nn
n
2
520
1000
400
V =
2
1
2
1
V
V
N
N a
==V V 1300
2=
[ ]
[ ]
2
21
/0976.0
)60)()(400)(50(44.4520
44.4
1300,520,
44.4
44.4
mmWb B
B
A B fN E
V E V E known
A B fN
fN E
m
m
m
m
m
=
=
=
==
=
=
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Example 3 A 25kVA transformer has 500 turns on the
primary and 50 turns on the secondarywinding. The primary is connected to3000V, 50Hz supply. Find:
a) Full load primary currentb) The induced voltage in the secondary
windingc) The maximum flux in the core
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Equivalent Circuit of
Practical TransformerEquivalent circuit: Often given to explain the operation of a complicated or
unfamiliar device
If a circuit is truly an equivalent circuit, the original devicecan be removed from a system & replaced with itsequivalent circuit without changing the behavior orperformance of the system
For purposes of analysis the transformer may be representedby a 1:1 turns ratio equivalent circuit This circuit is based on the following assumptions: Primary and secondary turns are equal in number. One
winding is chosen as the reference winding; theother is the referred winding.
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Equivalent Circuit of
Practical Transformer Core loss may be represented by a resistance across theterminals of the reference winding.
Core flux reactance may be represented by a reactanceacross the terminals of the reference winding. Primary and secondary IR and IX voltage drops may belumped together; the voltage drops in the referredwinding are multiplied by a factor derived at the end of this section, to give them the correct equivalent value .
Equivalent reactance and resistances are linear.
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Equivalent Circuit of PracticalTransformer
V1 = primary supply voltageV2 = 2 nd terminal (load) voltageE1 = primary winding voltageE2 = 2 nd winding voltageI1 = primary supply currentI2 = 2 nd winding currentI1= primary winding currentIo = no load current
Ic = core currentIm = magnetism currentR
1= primary winding resistance
R2= 2 nd winding resistanceX1= primary winding leakage reactanceX2= 2 nd winding leakage reactanceR c= core resistanceXm= magnetism reactance
VV 11
II 11 R R 11XX11
R R CC
II cc
XXmm
II mm
II oo
EE 11 EE 22VV22
II 11
NN 11: N: N 22R R 22
XX 22
LoadLoad
II 22
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Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary)
Actual MethodActual Method
22
22
2
2
12 '' Ra ROR R N
N R =
=
22
22
2
2
12 '' X a X OR X
N
N X =
=a
I I
aV V ORV N N
V E
22
2222
1'21
'
'
=
=
==
VV 11
II 11 R R 11 XX 11
R R CC
II cc
XX mm
II mm
II oo
EE 11 EE 22 VV 22
II 22 NN 11: N: N 22
R R 22 XX 22
Load
II 22
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Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary)
Approximate MethodApproximate Method
VV11
II 11 R R 11XX 11
R R CC
II cc
XX mm
II mm
II oo
EE 11 EE 22 VV22
II 22 NN 11 : N: N 22R R 22
XX22
Load
II 22
22
22
2
2
12 '' Ra ROR R N
N R =
=
22
22
2
2
12 '' X a X OR X
N
N X =
= a I
I
aV V ORV N N
V E
22
2222
1'21
'
'
=
=
==
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Equivalent Circuit of Practical Transformer
Single Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary) Approximate MethodApproximate Method
22
22
2
2
12 '' Ra ROR R N
N R =
=
22
22
2
2
12 '' X a X OR X N
N X =
= '
'
2101
2101
X X X
R R R
+=+=
2222
1'2 ' aV V ORV N
N V =
=
In some application, the excitation
branch has a small current comparedto load current, thus it may beneglected without causing seriouserror.
VV 11
II 11R R 0101 XX0101
aVaV 22
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Equivalent Circuit of PracticalTransformer
Single Phase transformer (referred to Secondary)Sing le Phase transformer (referred to Secondary) Actual MethodActual Method
21
11
2
1
21 '' a
R ROR R
N N
R =
= a
V V ORV
N N
V 1111
21 '' =
=
21
11
2
1
21 ''
a
X X OR X
N
N X =
=
I1 R 1X1
R C
Ic
Xm
Im
IoI2 R 2
X2
VV 22
aV 1
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Equivalent Circuit of
Practical Transformer )Single Phase transformer (referred to Secondary)Sing le Phase transformer (referred to Secondary) Approximate MethodApproximate Method
21
11
2
1
21 ''
a
X X OR X
N
N X =
=
2102
2102
'
'
X X X
R R R
+=+=
II 11 R R 0202 XX 0202
aV 1
21
11
2
1
21 '' a
R ROR R
N N
R =
=
aV
V ORV N N
V 1111
21 '' =
=
11 ' aI I =
Neglect the excitation branch,
VV 22
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Example 4 For the parameters obtained from the test of
20kVA 2600/245 V single phase transformer,refer all the parameters to the high voltageside if all the parameters are obtained atlower voltage side.Rc = 3.3 , X m =j1.5 , R 2 = 7.5 , X 2 =
j12.4
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Example 4 (solution)Given : R c = 3.3 , X m =j1.5 , R 2= 7.5 , X 2 = j12.4
i) Refer to H.V side (primary)
R2=(10.61) 2 (7.5) = 844.65 ,
X2=j(10.61) 2 (12.4) = j1.396k
Rc=(10.61) 2 (3.3) = 371.6 ,Xm =j(10.61) 2 (1.5) = j168.9
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Example 5 A 10 kVA single phase transformer 2000/440V
has primary resistance and reactance of 5.5and 12 respectively, while the resistanceand reactance of secondary winding is 0.2
and 0.45 respectively. Calculate:i) The parameter referred to high voltageside and draw the equivalent circuit
ii) The approximate value of secondaryvoltage at full load of 0.8 lagging powerfactor, when primary supply is 2000V.
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Example 5 (solution)R1=5.5 X 1=j12 R 2 =0.2 X 2=j0.45
i) Refer to H.V side (primary)
R2 =(4.55) 2 (0.2) = 4.14 ,X2 =j(4.55) 2(0.45) = j9.32
Therefore,
R01 =R 1+R 2 =5.5 + 4.13 = 9.64X01 =X 1 +X 2 =j12 + j9. 32 = j21.32
55.4440
2000
2
1
2
1 ====V
V
E
E a
V1 aV 2
R01 X01
21.329.64
I1
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Example 5 (solution)ii) Secondary voltage
p.f = 0.8cos = 0.8
=36.87 o
Full load,
From cct eqn.,
o
oo
oo
V
V j
aV I jX RV
8.06.422
)55.4()87.365)(32.2164.9(02000
))((0
2
2
2101011
=++=
++=
AV VA I FL 52000
10103
1==
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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
%100cos
cos
%100
%100,
22
22 ++=
+=
=
cuc
lossesout
out
P P I V I V
P P P
Power Input Power Output
Efficiency
%100cos
cos
%100cos
cos
2)(
)(
++=
++=
cucnload
cucload full
P n P nVAnVA
P P VAVA
Where, if load, hence n = , load, hence n = , load, n= , load, n= ,
90% of full load, n =0.990% of full load, n =0.9Where P cu = P scP c = P oc
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Voltage Regulation The voltage regulation of the transformer is the
percentage change in the output voltage fromno-load to full-load
Voltage Regulation can be determined based on 3methods: Basic Definition Short circuit Test Equivalent Circuit
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Voltage Regulation
Basic Definition In this method, all parameters are being referred
either to primary or secondary side. It can be represented in either
Down voltage Regulation
Up Voltage Regulation
%100. = NL
FL NL
V V V
RV
%100. = FL
FL NL
V V V
RV
Voltage Regulation
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Voltage Regulation
Short-circuit test In this method, direct formula can be used.
( )%100
cos.
1
. =V
V RV f p sc sc
If referred to primaryside
( )%100
cos.
2
. =V
V RV f p sc sc
If referred to secondaryside
Note that:Note that:
is for Lagging power factor + is for Leading power factor Isc must equal to IFL
P sc = V sc I sc cos sc
V l R l i
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Voltage Regulation( Equivalent Circuit )
Equivalent circuit In this method, the parameters must be referred to
primary or secondary
[ ]%100
sincos.
1
.01.011
= V X R I
RV f p f p If referred to
primary side
If referred tosecondary side
Note that:Note that: + is for Lagging power factor is for Leading power factor
j terms ~0
[ ]%100
sincos.
2
.02.022 =V
X R I RV f p f p
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Example 6 Determine the Voltage regulation by using
down voltage regulation and equivalentcircuit in example 5.
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Example 6 (solution) By using e quivalent circuit,
I1=5A R 01 =9.64 X 01 = 21.32 V 1=2000V, 0.8 lagging p.f
[ ][ ]
%12.5
%1002000
)6.0(32.21)8.0(64.95
%100sincos.1
.01.011
=
+=
=V
X R I RV f p f p
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Example 7 A short circuit test was performed at the
secondary side of 10kVA, 240/100Vtransformer. Determine the voltageregulation at 0.8 lagging power factor if :
Vsc =18V Isc =100 Psc=240W
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Example 7 (solution)( )
o
sc sc
sc sc
sc sc sc sc
o f p
f p sc sc
I V P
I V P that Know
Hence
f pGiven
V V RV
34.82)100)(18(
240cos
cos
cos,
87.368.0cos,
8.0.
%100cos.
1
1
1.
2
.
=
=
=
=
===
=
( )
%62.12
%100100
87.3634.82cos18.
=
=oo
RV
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Example 8 The following data were obtained in test on 20kVA
2400/240V, 60Hz transformer:
Vsc =72V Isc =8.33APsc=268W Poc=170W
The measuring instrument are connected inthe primary side for short circuit test. Determine
the voltage regulation for 0.8 lagging p.f. (use all3 methods), full load efficiency and half loadefficiency.
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Example 8 (solution)( )
.72.786.34.6364.8
64.833.8
72
4.63)33.8)(72(
268cos
cos
cos,
87.368.0cos,
8.0.
%100cos
.
0101
1
1
1.
2
.
side primarytoconnected because jX R j Z
I V
Z
I V P
I V P that Know
Hence
f pGiven
V
V RV
o sc
sc
sc sc
o
sc sc
sc sc
sc sc sc sc
o f p
f p sc sc
+=+==
===
=
=
=
=
===
=
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Example 8 (solution)
( )
%68.2
%100240
58.233240
%100.
79.058.233
2402400
4.6364.887.36240020000
02400
,.3
2
2
20111
=
=
=
=
+
=
+=
NL
FL NL
o
ooo
V
V V
RV
V V
V
aV Z I V
Defination Basic
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Open circuit testq is conducted to determine magnetism
parameter, Rc and Xm and core lossesRc and Xm and core losses.q Also known as no-load test.
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With the secondary open, the primary voltage wasincreased from zero to rated voltage, where therated voltage is the name plate stamp.
A digital multimeter was used as an ammeter tomeasure the open circuit current. A wattmeter
was used to measure the open circuit power. Thepower measured was the power dissipated in R m ,the core losses.
Pin(W) = core loss+ copper loss
Copper loss is neglected because of small no loadcurrent.
Wattmeter only shows the reading of core loss.
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Open-Circuit Test
( )22
1
,
,,
sin
cos
,
cos
cos
mcoc
m
ocmc
occ
mc
ococm
ococc
ococ
ococ
ococococ
I I I
I V
X I V
R
X and RThen
I I
I I
Hence
I V P
I V P
+===
==
=
=
Rc XmVoc
Ic Im
Voc
Ioc cos oc
Ioc
Voc
Ic
Im
Ioc sin oc
oc
Note:If the question asked parameters referred to lowquestion asked parameters referred to lowvoltage sidevoltage side , the parameters (R c and X m) obtainedneed to be referred to low voltage sideneed to be referred to low voltage side
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Short circuit test
is conducted to determine the copperparameter depending where the test isperformed. If performed at primary, hencethe parameters are RR0101 and XX0101 and vice-vice-versaversa.
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Short-Circuit Test Normally, measurement at lower voltage sideat lower voltage side If the given test parameters are taken on primary side,taken on primary side,
RR 0101 and Xand X 0101 will be obtained &will be obtained & vvice-versa.
X01R01
For a case referred toPrimary side 2
012
0101
01
1
012
,
cos
cos
X R Z
I
V Z
Hence
I V
P
R I P
I V P
sc
sc
sc
sc sc
sc sc
sc sc
sc sc sc sc
+=
=
=
==
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Example 9 Given the test on 500kVA 2300/208V are as follows:
Poc = 3800W P sc = 6200WVoc = 208V V sc = 95VIoc = 52.5A I sc = 217.4A
Determine the transformer parameters and drawequivalent circuit referred to high voltage side. Alsocalculate full load efficiency, half load efficiency andvoltage regulation, when power factor is 0.866 lagging.
[1392 , 517.2 , 0.13 , 0.44 , 97.74%, 97.59%, 3.04%]
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EMPS_BEE2133_NJ
Example 9 (solution)
A
I I
A
I I
I V P
o
ococm
o
ococc
ooc
ococococ
2.49
6.69sin5.52
sin
26.18
6.69cos5.52
cos
6.69
)208)(5.52(
3800cos
cos
1
======
=
=
=
Ioc cos oc
Ioc
Voc
Ic
Im
Ioc sin oc
oc
From Open Circuit Test,
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===
===
23.421.49
208
39.1126.18
208
m
ocm
c
occ
I V
X
I V
R
Since V oc =208Vall reading are taken on the secondary side
Parameters referred to high voltage side,
=
=
=
= =
=
21.517208
230023.4'
1392208
230039.11'
22
2
1
22
2
1
E E
X X
E E R R
mm
cc
Example 9 (solution)
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EMPS_BEE2133_NJ
Example 9 (solution)
AV VA
I FL 4.217230010500 3
11
===
o sc
sc sc sc sc I V P
53.72)4.217)(95(
6200cos
cos
1 =
=
=
From Short Circuit Test,First, check theFirst, check the II scsc
Since IFL1 =Isc , all reading are actually taken on the primary side
+=
=
=
=
42.013.0
53.7244.053.724.217
95
01
j
I V
Z
oo
sc sc
sc
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Example 9 (solution)Equivalent circuit referred to high voltage side,
VV22 =aV=aV 22VV11 RR cc13921392
XXmm517.21517.21
RR 01010.130.13
XX01010.420.42
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Example 9 (solution)Efficiency,
%59.97
%1003800)5.0)(6200()866.0)(10500)(5.0(
)866.0)(10500)(5.0(
%100cos
cos
%74.97
%10038006200)866.0)(10500(
)866.0)(10500(
%100cos
cos
23
3
22
1
3
3
=
++
=
++=
=
++ =
++
=
oc sc L
oc sc FL
P P nnVAnVA
P P VAVA
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Example 9 (solution)Voltage Regulation,
[ ]
[ ]
%04.3
%1002300
3053.72cos)95(
%100cos
.1
=
=
= E
V RV pf sc sc
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Example 10 Data obtained from short-circuit and open-circuit test of a
75-kVA, 4600-230V, 60Hz transformer are
Open-Circuit Test Short-Circuit Test(Low-Side Data) (High-Side Data)Poc = 521W P sc = 1200WVoc = 230V V sc = 160.8VIoc = 13.04A I sc = 16.3A
Determine a) the magnetizing reactance andequivalent core-loss resistance; b) the resistance,reactance and impedance of the transformer windings;c) the voltage regulation when operating at rated loadand 0.75 power-factor lagging.
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Summary Transformers convert AC electricity from one voltage to another with
little loss of power Transformers work only with AC and this is one of the reasons why
mains electricity is AC Step-up transformers increase voltage, step-down transformers
reduce voltage Most power supplies use a step-down transformer to reduce thedangerously high mains voltage (230V in UK) to a safer lowvoltage.
The input coil is called the primary and the output coil is called thesecondary
There is no electrical connection between the two coils, instead theyare linked by an alternating magnetic field created in the soft-iron core of the transformer
The two lines in the middle of the circuit symbol represent the core.
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Summary The two lines in the middle of the circuit symbol represent the core Transformers waste very little power so the power out is (almost)
equal to the power in Note that as voltage is stepped down current is stepped up.
The ratio of the number of turns on each coil, called the turnsratio , determines the ratio of the voltages A step-down transformer has a large number of turns on its primary
(input) coil which is connected to the high voltage mains supply,and a small number of turns on its secondary (output) coil togive a low output voltage.