Notes on Transformer
Transcript of Notes on Transformer
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Chapter 4.
Transformer
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Transformer- Introduction
Two winding transformers Construction and principles Equivalent circuit Determination of equivalent circuit
parameters Voltage regulation Efficiency Auto transformer 3 phase transformer
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Transformer- Introduction
Varieties of transformers
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Transformer- Introduction
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Transformer- Introduction
Transformer is a device that makes use of the magnetically coupled coils to transfer energy
It is typically consists of one primary winding coil and one or more secondary windings
The primary winding and its circuit is called the Primary Side of the transformer
The secondary winding and its circuit is called the Secondary Side of the transformer
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Transformer- Introduction
If one of those winding, the primary, is connected to an alternating voltage source, an alternating flux will be produced. The mutual flux will link the other winding, the secondary, and will induced a voltage in it.
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Transformer- Introduction
Transformers are adapted to numerous engineering applications and may be classified in many ways:
Power level (from fraction of a volt-ampere (VA) to over a thousand MVA),
Application (power supply, impedance matching, circuit isolation),
Frequency range (power, audio, radio frequency (RF)) Voltage class (a few volts to about 750 kilovolts) Cooling type (air cooled, oil filled, fan cooled, water
cooled, etc.) Purpose (distribution, rectifier, arc furnace, amplifier
output, etc.).
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Transformer- Introduction
Power transmission
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Transformer- Introduction
Power transmission
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Transformer
4.1 Construction
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Transformer- construction
Basic components of single phase transformer
N1 N
2Supply Load
Primary winding Secondary winding
Laminated iron core
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Transformer- construction
Single phase transformer construction
A) Core type B) Shell type
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Transformer- construction
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Transformer- construction
Primary
Winding
Secondary
Winding
Multi-layer
Laminated
Iron Core
X1
X2H1 H
2
WindingTerminals
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4.2 Ideal Transformer
Transformer
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Transformer
i
e1
Φ
v1 v2
e2
The emf which induced in
transformer primary winding is
known as self induction emf as
the emf is induced due to to flux
which produced by the winding
itself.
While the emf which induced in
transformer secondary winding
is known as mutual induction
emf as the emf is induced due
to to flux which produced by the
other winding.
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Transformer
i
e1
Φ
v1 v2
e2
Acording to Faraday’s Law, the emf which induced in the primary winding is,
e1 =dt
dN
1
Since the flux is an alternating flux,
tmak sin e1 = dt
tdN mak )sin(
1
tN mak cos1
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Transformer
i
e1
Φ
v1 v2
e2
e1
where,
tfN mak cos21
tE cosmax1
max1E fN 2max1=
2
max1
1
EE rms fN max144.4
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Transformer
i
e1
Φ
v1 v2
e2
e2 =
Similarly it can be shown that,
dt
dN
2
E2 rmsfN max244.4
kN
N
fN
fN
E
E
1
2
max1
max2
1
2
44.4
44.4
k is transformation ratio
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Transformer
The voltage ratio of induced voltages on the secondary to primary windings is equal to the turn ratio of the winding turn number of the secondary winding to the winding turn number of the primary winding. Therefore the transformers can be used to step up or step down voltage levels by choosing appropriate number their winding turns. In power system it’s necessary to step up the output voltage of a generator which less than 30kV to up 500kV for long distance transmission. High voltage for long distance power transmission can reduce current flow in the transmission lines, thus line losses and voltage drop can be reduced.
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2
NE m1
1
2
NE m2
2
Current, voltages and flux in an unloaded ideal transformer
Transformer- Ideal Transformer
Winding resistances are zero, no leakage inductance and iron loss
Magnetization current generates a flux that induces voltage in both windings
N1 N
2
mI
m
V1
E1
E2 = V
2
2222
Transformer
i
e1
Φ
v1 v2
e2
Transformer on no load.
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Transformer- Ideal Transformer
Loaded transformer
i
Φ
V1E2E1
V2 ZL
I2
Φ2
N1 N2
Φ1
When a load is connected to the secondary output terminals of a transformer as shown in Figure 4.5, a current I2 flows into the load and into transformer secondary winding N2. The current I2
which flowing in N2 produces flux Φ2 which opposite –by Lenz’s law- to the main magnetic flux Φ in the transformer core. This will weaken or slightly reduce the main flux Φ to Φ’.
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Transformer- Ideal Transformer
Loaded transformer
i
Φ
V1E2E1
V2 ZL
I2
Φ2
N1 N2
Φ1
The reduction of main flux Φ –by Faraday’s law- could also reduce the induced voltage in primary winding E1. Consequently E1 is now smaller than the supply voltage V1, then the primary current would be increased due to that potential differences. Therefore on loaded transformer, the primary current has an additional current of I1’.
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Transformer- Ideal Transformer
Loaded transformer
i
Φ
V1E2E1
V2 ZL
I2
Φ2
N1 N2
Φ1
The extra current I1’ which flowing in the primary winding N1
produces flux Φ1 which naturally react according to Lenz’s law, demagnetize the flux Φ2. Therefore the net magnetic flux in the core is always maintained at original value, it is the main flux Φ (the flux which produced by the magnetizing current).
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Transformer- Ideal Transformer
Loaded transformer
i
Φ
V1E2E1
V2 ZL
I2
Φ2
N1 N2
Φ1
The magneto motive force (mmf) source N2I2 at the secondary winding produces flux Φ2, while the mmf N1I1‘ produces flux Φ1. Since the magnitude of Φ1 equal to magnitude of Φ2 and the reluctance seen by these two mmf sources are equal, thus
N1I1‘ = N2I2
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Currents and fluxes in a loaded ideal transformer
Transformer- Ideal Transformer
Loaded transformer
E2
Load
V2
I2
2
1
m
Im
+ I1
V1
E1
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Transformer- Ideal Transformer
Turn ratio
If the primary winding has N1 turns and secondary winding has N2 turns, then:
The input and output complex powers are equal
1
2
2
1
2
1
I
I
E
E
N
Na
**IESSIE 222111
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Transformer- Ideal Transformer
Functional description of a transformer:
When a = 1 Isolation Transformer
When | a | < 1 Step-Up Transformer Voltage is increased from Primary side to secondary side
When | a | > 1 Step-Down Transformer Voltage is decreased from Primary side to secondary side
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Transformer- Ideal Transformer
Transformer Rating Practical transformers are usually rated
based on:
Voltage Ratio (V1/V2) which gives us the turns-ratio
Power Rating, small transformers are given in Watts (real power) and Larger ones (Power Transformers) are given in kVA (apparent power)
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Transformer- Ideal Transformer
Example 4.1
Determine the turns-ratio of a 5 kVA 2400V/120V Power Transformer
Turns-Ratio = a = V1/V2 = 2400/120 = 20/1 = 20
This means it is a Step-Down transformer
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Transformer- Ideal Transformer
Example 4.2
A 480/2400 V (r.m.s) step-up ideal transformer delivers 50 kW to a resistive load. Calculate:
(a) the turns ratio, (0.2)(b) the primary current, (104.17A)(c) the secondary current. (20.83A)
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Transformer- Ideal Transformer
Nameplate of transformer
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Transformer- Ideal Transformer
Equivalent circuit
I2
I1 = I
2 /T
E2 = V
2
V1 = E
1 = T E
2
V1
E1
T
Equivalent circuit of an ideal transformer
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Transformer- Ideal Transformer
Transferring impedances through a transformer
2
22
2
2
1
11
I
V
I
V
I
VZ a
a
a
Equivalent circuit of an ideal transformer
loada ZZ2
1
Vac
Zload
T
V1
V2
I1
I2
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Vac a2ZloadV1
I1
Vac
/k Zload
V2
I2
a) Equivalent circuit when
secondary impedance is
transferred to primary side and
ideal transformer eliminated
b) Equivalent circuit when
primary source is transferred to
secondary side and ideal
transformer eliminated
Thévenin equivalents of transformer circuit
Transformer- Ideal Transformer
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Transformer- practical transformer
Practical Transformer
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4.3 Equivalent Circuits
Transformer
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Transformer- equivalent circuit
mI
1
V1
V2
l1
l2
R1
I2
R2
N1
N2
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Transformer- equivalent circuit
I1
R1
V1
X1
I2
R2
V2
X2
N1:N
2
Development of the transformer equivalent circuits
The effects of winding resistance and leakage flux are
respectively accounted for by resistance R and leakage
reactance X (2πfL).
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In a practical magnetic core having finite permeability, a magnetizing current Im is required to establish a flux in the core.
This effect can be represented by a magnetizing inductance Lm. The core loss can be represented by a resistance Rc.
Transformer- practical equivalent circuit
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Rc :core loss component,
Xm : magnetization component,
R1 and X1 are resistance and reactance of the primary winding
R2 and X2 are resistance and reactance of the secondary winding
Transformer- practical equivalent circuit
I1
R1
V1
X1
I2
R2
V2
X2
N1:N
2
I’1
Rc
Xm
Ic
I0
Im
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Transformer- practical equivalent circuit
The impedances of secondary side such as R2, X2 and Z2 can be moved to primary side and also the impedances of primary side can be moved to the secondary side, base on the principle of:
The power before transferred = The power after transferred.
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Transformer- practical equivalent circuit
The power before transferred = The power after transferred.
I22R2 = I1’
2R2’
Therefore R2’= (I2/ I1’ ) 2 R2
= a2R2
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Transformer- practical equivalent circuit
I1
R1
V1
X1 I
2R’2
V2
X’2
N1:N
2
I’1
Rc
Xm
Ic
I0
Im
V2’
V2' = a V2 , I1' = I2/a
X2' = a2 X2 , R2' = a2 R2a = N1/N2
The turns can be moved to the right or left by referring
all quantities to the primary or secondary side.
The equivalent circuit with secondary side moved to the primary.
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Transformer- Approximate equivalent
circuit
For convenience, the turns is usually not shown and the equivalent circuit is drawn with all quantities (voltages, currents, and impedances) referred to one side.
I1
R1
V1
X1 R’2X’
2
N
V2’
Z2’
I0
Rc
Xm
Ic Im
I’1
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Transformer- equivalent circuit
Example 4.3
A 100kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and 0.035 ohm respectively. The supply voltage is 2200V. Calculate:(a) the equivalent impedance referred to the primary circuit (2.05 ohm)(b) the equivalent impedance referred to the secondary circuit
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4.4 Determination of
Equivalent Circuit
Parameter
Transformer
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1. No-load test (or open-circuit test).
2. Short-circuit test.
Transformer- o/c-s/c tests
The equivalent circuit model for the actual transformer can be used to predict the behavior of the transformer.
The parameters R1, X1, Rc, Xm, R2, X2 and N1/N2 must be known so that the equivalent circuit model can be used.
These parameters can be directly and more easily determined by performing tests:
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No load/Open circuit test
Provides magnetizing reactance (Xm) and core
loss resistance (RC)
Obtain components are connected in parallel
Short circuit test
Provides combined leakage reactance and
winding resistance
Obtain components are connected in series
Transformer- o/c-s/c tests
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Transformer- open circuit test
No load/Open circuit test
V
A
X1 R1
X m R c
X2 R2
W
V oc
I oc
P oc
Equivalent circuit for open circuit test, measurement at the primary side.
Simplified equivalent
circuitV
A
Xm Rc
W
Voc
Ioc
Poc
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Transformer- open circuit test
Open circuit test evaluation
Q
VX
P
VR
IVQIV
P
ocm
oc
occ
ococ
ococ
oc
22
0
1
0 sincos
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Transformer- short circuit test
Short circuit test Secondary (normally the LV winding) is shorted,
that means there is no voltage across secondary terminals; but a large current flows in the secondary.
Test is done at reduced voltage (about 5% of rated voltage) with full-load current in the secondary. So, the ammeter reads the full-load current; the wattmeter reads the winding losses, and the voltmeter reads the applied primary voltage.
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Transformer- short circuit test
Short circuit test
R2
I scV
A W
X1R1 X2P sc
Vsc
Equivalent circuit for short circuit test, measurement at the primary side
V
A W
a2R2
X1R1 a2X2
I sc
P sc
Vsc
Simplified equivalent circuit for short circuit test
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Transformer- short circuit test
Short circuit test
V
A W
Xe1Re1
I sc
Vsc
P sc
Simplified circuit for calculation of series impedance
2
2
11 RaRRe
2
2
11 XaXXe
Primary and secondary impedances are combined
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Transformer- short circuit test
Short circuit test evaluation
2
1
2
11
121
eee
sc
sce
sc
sce
RZX
I
VZ
I
PR
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Transformer- o/c-s/c tests
Equivalent circuit obtained by measurement
Xm Rc
X e1 R e1
Equivalent circuit for a real transformer resulting from the
open and short circuit tests.
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Transformer- o/c-s/c tests
Example 4.4
Obtain the equivalent circuit of a 200/400V, 50Hz
1-phase transformer from the following test data:-
O/C test : 200V, 0.7A, 70W - on L.V. side
S/C test : 15V, 10A, 85W - on H.V. side
(Rc =571.4 ohm, Xm=330 ohm, Re1=0.21ohm, Xe1=0.31 ohm)
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Transformer – voltage regulation
Voltage Regulation
Most loads connected to the secondary of a transformer are designed to operate at essentially constant voltage. However, as the current is drawn through the transformer, the load terminal voltage changes because of voltage drop in the internal impedance.
To reduce the magnitude of the voltage change, the transformer should be designed for a low value of the internal impedance Zeq
The voltage regulation is defined as the change in magnitude of the secondary voltage as the load current changes from the no-load to the loaded condition.
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ZV2
I2’ I2
V12’=V20
R1’ R2
Ze2
X1’
Rc’ Xm’
X2
Ze2 = R1’ + R2 + jX1’ + jX2
= Re2 + jXe2
Transformer – voltage regulation
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ZV2
I2’ I2
V12’=V20
R1’ R2
Ze2
X1’
Rc’ Xm’
X2
Applying KVL, V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
Transformer – voltage regulation
Or V2 = V20 - I2 (Ze2 )
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I2Xe2
I2Re2
I2
V2
OA
θ2
Transformer – voltage regulation
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
A
B
θ2
Transformer – voltage regulation
V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
Voltage drop = AM = OM – OA
= AD + DN + NM
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
AD = I2 Re2 cosθ2
DN=BL= I2 Xe2 sinθ2
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
Applying Phytogrus theorem to OCN triangle.
(NC)2 = (OC)2 – (ON)2
= (OC + ON)(OC - ON) ≈ 2(OC)(NM)
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
Therefore NM = (NC)2/2(OC)
NC = LC – LN = LC – BD
= I2 Xe2 cosθ2 - I2 Re2 sinθ2
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
20
2
222222
2
sincos
V
RIXI ee
20
2
222222
2
sincos
V
RIXI ee
NM =
AM = AD + DN + NM= I2 Re2 cosθ2 + I2 Xe2 sin θ2 +
Transformer – voltage regulation
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
thus,votage regulation = (AM)/V20 per unit
In actual practice the term NM is negligible since its value isvery small compared with V2. Thus the votage regulationformula can be reduced to:
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I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Transformer – voltage regulation
Voltage regulation =
20
222222 sincos
V
XIRI ee
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Transformer- voltage regulation
The voltage regulation is expressed as follows:
NL
LNL
V
VVregulationVoltage
2
22
V2NL= secondary voltage (no-load condition)
V2L = secondary voltage (full-load condition)
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Transformer- voltage regulation
For the equivalent circuit referred to the primary:
1
21
V
VVregulationVoltage
'
V1 = no-load voltage
V2’ = secondary voltage referred to the primary (full-load condition)
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Transformer- voltage regulation
Consider the equivalent circuit referred to the secondary,
I2' R
1'
V2NL
X1' R
2X
2
V2 Z
2
I2
Re2
Xe2
NL
ee
V
sinXIcosRIregulationVoltage
2
222222
(-) : power factor leading
(+) : power factor lagging
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Transformer- voltage regulation
Consider the equivalent circuit referred to the primary,
1
211211
V
sinXIcosRIregulationVoltage ee
I1
R1
V1
X1 R
2'X
2'
Z’2
I1'
V2
'
Re1
Xe1
(-) : power factor leading
(+) : power factor lagging
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Transformer- voltage regulation
Example 4.5
Based on Example 4.3 calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of
(i) 0.8 lagging (0.0336pu,425V)
(ii) 0.8 leading (-0.0154pu,447V)
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Transformer- Efficiency
Losses in a transformer
Copper losses in primary and secondary windings
Core losses due to hysteresis and eddy current. It depends on maximum value of flux density, supply frequency and core dimension. It is assumed to be constant for all loads
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Transformer- Efficiency
As always, efficiency is defined as power output to power input ratio
The losses in the transformer are the core loss (Pc) and copper loss (Pcu).
lossesP
P
)P(powerinput
)P(poweroutput
out
out
in
out
2
2
2222
222
ec RIPcosIV
cosIV
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Transformer- Efficiency
Efficiency on full load
where S is the apparent power (in volt amperes)
scocFLFL
FLFL
PPS
S
cos
cos
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Transformer- Efficiency
Efficiency for any load equal to n x full load
where corresponding total loss =
scocFLFL
FLFL
PnPSn
Sn
2cos
cos
scoc PnP 2
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Transformer- Efficiency
Example 4.6
The following results were obtained on a 50 kVA transformer: open circuit test – primary voltage, 3300 V; secondary voltage, 400 V; primary power, 430W.Short circuit test – primary voltage, 124V;primary current, 15.3 A; primary power, 525W; secondary current, full load value. Calculate the efficiency at full load and half load for 0.7 power factor.
(97.3%, 96.9%)
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Transformer- Efficiency
For constant values of the terminal voltage V2 and load power factor angle θ2 , the maximum efficiencyoccurs when
If this condition is applied, the condition for maximum efficiency is
that is, core loss = copper loss.
02
dI
d
2
2
2 ec RIP
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2
2
2222
222
cos
cos
ec RIPVI
VI
22
2
22
22
cos
cos
ec RI
I
PV
V
From transformer efficiencyformula,
Efficiency =
= (4.11)
Transformer-Maximum Efficiency
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If V2 is approximately constant, hence for a load of givenpower factor, the efficiency is a maximum when thedenominator of Equation 4.11 is a minimum, it is when
0cos 22
2
22
2
RI
I
PV
dI
d c
022
2
e
c RI
P
Or I22Re2 = Pc (4.12)
Hence the efficiency is a maximum when the variable lossI22Re2 (copper loss) is equal to the constant core loss.
Transformer-Maximum Efficiency
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Transformer- Efficiency
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Example 4.7
From Example 4.6: [The following results were obtained on a 50 kVA transformer: open circuit test – primary voltage, 3300 V; secondary voltage, 400 V; primary power, 430W.Short circuit test –primary voltage, 124V;primary current, 15.3 A; primary power, 525W; secondary current, full load value]. Calculate the maximum efficiency for 0.7 power factor. Also find the load at maximum efficiency.
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Assignment 2 SEE2053
Question:
Explain in detail the occurrence of hysteresis loss and eddy current loss in a transformer core and how these losses can be reduced.
Please submit this assignment before or on 24th October 2010.
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Transformer- Auto transformer
It is a transformer whose primary and secondary coils are in a single winding
Autotransformer
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Transformer- Auto transformer
Same operation as two windings transformer
Physical connection from primary to secondary
Sliding connection allows for variable voltage
Higher kVA delivery than two windings connection
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Transformer- Auto transformer
Advantages: A tap between primary and secondary sides which
may be adjustable to provide step-up/down capability
Able to transfer larger S apparent power than the two winding transformer
Smaller and lighter than an equivalent two-winding transformer
Disadvantage: Lacks electrical isolation
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Transformer- Auto transformer
A Step Down Autotransformer:
and
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Transformer- Auto transformer
A Step Up Autotransformer:
and
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Transformer- Auto transformer
Example 4.7
An autotransformer with a 40% tap is supplied by a 400-V, 60-Hz source and is used for step-down operation. A 5-kVA load operating at unity power factor is connected to the secondary terminals.
Find: (a) the secondary voltage,(b) the secondary current, (c) the primary current.
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Transformer- Auto transformer
Solution
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Three phase transformers
The three-phase transformer can be built by:
the interconnection of three single-phase transformers
using an iron core with three limbs
The usual connections for three-phase transformers are:
wye / wye seldom used, unbalance and 3th harmonics problem
wye / delta frequently used step down.(345 kV/69 kV)
delta / delta used medium voltage (15 kV), one of the transformer can be removed (open delta)
delta / wye step up transformer in a generation station
For most cases the neutral point is grounded
Transformer -3 phase transformer
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Transformer -3 phase transformer
Analyses of the grounded wye / delta transformer
Each leg has a
primary and a
secondary winding.
The voltages and
currents are in phase
in the windings
located on the same
leg.
The primary phase-to-
line voltage generates
the secondary line-to-
line voltage. These
voltages are in phase
A B C
VAN VB N VC N
VabVbc Vca
N
a b c
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Transformer -3 phase transformer
Analyses of the grounded wye / delta transformer
IA
IB
IC
N
IAN
ICN
IBN
Iab
Ibc
Ica
Ib
Ia
Ic
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Transformer -3 phase transformer
Analyses of the grounded wye / delta transformer
VCA
VAB
N
VA N
VC N
VBN
Vbc
Vbc
Vab
Vca
Vab
A
C
B
a
c
b
VB C Vbc
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Three phase transformer
Transformer -3 phase transformer
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Three phase transformer
Transformer
100
Three phase transformer
Transformer
101
Three phase transformer
Transformer
102
Three phase transformer
Transformer
103
Transformer Three phase transformer
Transformer Construction
Iron Core
The iron core is made of thin
laminated silicon steel (2-3 %
silicon)
Pre-cut insulated sheets are
cut or pressed in form and
placed on the top of each
other .
The sheets are overlap each
others to avoid (reduce) air
gaps.
The core is pressed together
by insulated yokes.
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Transformer Three phase transformer
Transformer Construction Winding
The winding is made of copper or
aluminum conductor, insulated with
paper or synthetic insulating material
(kevlar, maylard).
The windings are manufactured in
several layers, and insulation is
placed between windings.
The primary and secondary windings
are placed on top of each others but
insulated by several layers of
insulating sheets.
The windings are dried in vacuum
and impregnated to eliminate
moisture.
Small transformer winding
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Transformer Three phase transformer
Transformer Construction
Iron Cores
The three phase transformer iron
core has three legs.
A phase winding is placed in
each leg.
The high voltage and low voltage
windings are placed on top of
each other and insulated by
layers or tubes.
Larger transformer use layered
construction shown in the
previous slides.
A B C
Three phase transformer iron core
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Transformer Three phase transformer
Transformer Construction
The dried and treated
transformer is placed in a steel
tank.
The tank is filled, under vacuum,
with heated transformer oil.
The end of the windings are
connected to bushings.
The oil is circulated by pumps
and forced through the radiators.
Three phase oil transformer
107
Transformer Three phase transformer
Transformer Construction
The transformer is equipped with
cooling radiators which are
cooled by forced ventilation.
Cooling fans are installed under
the radiators.
Large bushings connect the
windings to the electrical system.
The oil is circulated by pumps
and forced through the radiators.
The oil temperature, pressure
are monitored to predict
transformer performance.
Three phase oil transformer
108
Transformer Three phase transformer
Transformer Construction
Dry type transformers are used
at medium and low voltage.
The winding is vacuumed and
dried before the molding.
The winding is insulated by
epoxy resin
The slide shows a three phase,
dry type transformer.
Dry type transformer