Electro-Mechanical Systems
EEEB413
TRANSFORMERS
Introduction
INTRODUCTION
• Applications, Types, and Construction of
Transformers
• Applications: Transfers Electric Energy, changing
the voltage level (or current level), through a
magnetic field (In our study)
• Other applications: e.g., voltage & current sampling
and measurement, impedance transformation
• It has two or more coils wrapped around a common
electromagnetic core
• Generally, flux in the core is common among the
coils
INTRODUCTION• One winding is connected to source of ac power, the 2nd (& 3rd )
supplies power to loads
• Winding connected to source named “Primary”
• Winding connected to load named “Secondary”
• If there is another one is called “Tertiary”
• Importance of Transformers:
• Main: to transfer electrical energy over long distances (from power plants to load centers)
• In modern power system electric energy is generated at voltages between 12 to 25 kV, Transformers step up voltage between 110 kV to 1200 kV for transmission over long distances with very small losses
• in Malaysia 500kV, 275kV and 132kV for transmission (frequency: 50 Hz)
• Then Transformers step down to 33 kV, 22 kV, 11kV, 6.6kV for local distribution & finally supply safely homes, offices & factories at voltages as low as 240 V (single phase)
Power Loss in Transmission Lines
RIPLoss
2
Transformers play a key role in the transmission of electric power.
Power Transmission
INTRODUCTION
• Transformers are classified, based on types
of core structure, into: (both use thin
laminations)
1- Core form,transformer windings
wrapped on two legs as shown
2- Shell form, transformer windings
Wrapped only on center leg as shown:
(leakage flux is minimized)
Single phase transformer construction
A) Core type B) Shell type
INTRODUCTION
INTRODUCTION• The primary and secondary windings are wrapped one on
top of the other to:
Reduce the leakage flux
• And the low-voltage winding innermost to :
Simplify insulating of the high-voltage winding
from the coreTypes of transformers :
• Step up/Unit transformers – located at output of a generator to step up the voltage level to transmit the power
• Step down/Substation transformers – Located at main distribution or secondary level transmission substations to lower the voltage levels for distribution 1st level purposes
• Distribution Transformers – located at small distribution substation. It lowers the voltage levels for 2nd level distribution purposes.
• Special Purpose Transformers - E.g. Potential Transformer (PT) , Current Transformer (CT)
Oil immersed Distribution
Transformers
Dry Type Distribution Transformers
Nameplate of Transformer
TWO WINDING TRANSFORMER
CONNECTION
• SINGLE PHASE
TRANSFORMER
CONSTRUCTION
Basic components of single phase transformer
N1 N
2Supply Load
Primary winding Secondary winding
Laminated iron core
Ideal Transformer
IDEAL TRANSFORMER
(SINGLE PHASE)
• a lossless transformer with an input winding
and an output winding in which magnetic core
has an infinite permeability
• Figure below shows: ideal transformer and
schematic symbols of a transformer
IDEAL TRANSFORMER
• Np : turns of wire on its primary side
• Ns : turns of wire on its secondary sides
• The relationship between the primary and secondary
voltage is as follows: (a: is the turns ratio)
• The relationship between primary and secondary
current is Np ip(t) = Ns is(t)
•
a
N
N
tv
tv
s
p
s
p
ati
ti
s
p 1
IDEAL TRANSFORMER
• In terms of Phasor quantities:
• Vp/Vs=a , Ip / Is=1/a
• while:
1- phase angles of Vp and Vs are the same
2- phase angles of Ip and Is are the same
• ideal transformer turn ratio affects the magnitude of
voltages & currents not their angles
• Now: given primary circuit voltage is positive at
specific end of coil, what would be the polarity of
secondary circuit’s voltage?
IDEAL TRANSFORMER
• It is possible to specify the secondary’s polarity
only if transformers were opened & it windings
examined
• To avoid requirement of this examination,
transformers employ a dot convention:• If the primary voltage is +ve at the dotted end of the winding wrt the
undotted end, then the secondary voltage will be positive at the dotted
end also. Voltage polarities are the same wrt the dots on each side of
the core.
• If the primary current of the transformer flows into the dotted end of the
primary winding, the secondary current will flow out of the dotted end of
the secondary winding
IDEAL TRANSFORMER
• Power in Ideal Transformer
-Power supplied to Transformer:Pin= Vp Ip cosθp
-Power supplied to loads : Pout=Vs Is cosθs
• Since V & I angles unaffected by ideal transformer θp = θs=θ
• Using the turn ratio; Vp/Vs=a , Ip / Is=1/a
• Pout = Vp / a (a Ip) cosθ = Pin
• similiar relation for reactive power Q & S
• Qin= Vp Ip sinθp= Vs Is sinθs = Qout
• Sin= Vp Ip = Vs Is = Sout
• 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
IDEAL TRANSFORMER
IDEAL TRANSFORMER
Impedance Transformation
• Load impedance ZL = Vs/Is and apparent impedance of primary circuit: Z’L=Vp/Ip
• Vp = aVs
• Is = a Ip
• Z’L=Vp/Ip= aVs / Is /a= a² ZL
• With a transformer,
it is possible to match
magnitude of a load
impedance with source
impedance by picking proper turn ratio
IDEAL TRANSFORMER
• Analysis of CCT.s containing Ideal Transformer
In equivalent cct.
a) Voltages & impedances replaced by scaled values,
b) polarities reversed if the dots on one side of transformer windings are reversed compared to dots on the other side of transformer windings
• Example 1: A single phase power system consists of a 480 V, 50 Hz generator supplying a load Zload=4+j3 Ωthrough a transmission line of impedance: Zline=0.18+j0.24 Ω
• a) what is the voltage at load? What is the transmission losses?
IDEAL TRANSFORMER• b) a 1:10 step-up transformer placed at the
generator end of transmission line & a step
down transformer placed at load end of line.
What is load voltage ? What is transmission
losses?
IDEAL TRANSFORMER
• a) IG=Iline=Iload
• Iline = V / (Zline + Zload)=480 / [(0.18b+ j 0.24)+(4+j3)]
= 480 / (4.18 + j 3.24) = 480 / 5.29 =
90.8 A
• Load voltage : Vload= Iline Zload=(90.8 )(4+j3)=
454 V
• And the losses are :
• Pline=(Iline)² Rline=(90.8)²(0.18)=1484 W
00 0 8.37
8.378.37
9.0
IDEAL TRANSFORMER
b) need to convert the system to a common voltage
• Need two steps to be followed:1- eliminate T2 referring to load to Transmission line’s voltage
2-eliminate T1 by referring transmission line’s elements & equivalent load to
source side
step 1: Z’load=a² Zload = (10/1)² (4+j3)=400+j300Ω
Zeq=Zline+Z’load=400.18+j300.24Ω=500.3 Ω88.36
IDEAL TRANSFORMER
• Step 2: total impedance reflected cross T1 to source
side
• Z’eq=a² Zeq =a² (Zline+Z’load)
• =(1/10)²(0.18+j0.24+400+j300)=
• = 0.0018+j0.0024+4+j3=5.003 Ω88.36
IDEAL TRANSFORMER
• The generator’ current is :
• IG=480/ [5.003 ] =95.94
• Now it can be worked back to find Iline & Iload through T1
• Np1IG=NS1IlineIline = Np1/Ns1 IG =(1/10)(9.594 )
• Working back through T2:
• Np2IG=NS2Iline
• Iload = Np2/Ns2 Iline =(10/1)(95.94 )=95.94
• The load voltage:
• Vload =Iload Zload=(95.94 )(5 )=479.7
Volts
88.3688.36
88.36
88.3688.36
88.36 87.36 01.0
IDEAL TRANSFORMER
• The line losses are given:
Ploss= Iline² Rline = 9.594 ² 0.18 = 16.7 W
• Note: rising transmission voltage of power
system reduced transmission losses
by a factor of 90
Also voltage at load dropped much
less
Real/Practical
Transformer
REAL SINGLE PHASE
TRANSFORMER
• Operation of a real Transformer
• primary connected to ac source, secondary
open circuited
REAL SINGLE PHASE
TRANSFORMER
• The transformer’s hysteresis curve is shown
• Based on Faraday’s law:
• eind= dλ /dt
Where λ = ∑ φi (on N turn)
• Φav.=λ / N
• And e ind= N d Φav/dt
REAL SINGLE PHASE
TRANSFORMER
• Voltage Ratio of realizing the leakage flux in a real
Transformer
• φp=φm+φLp
• φS= φm+φLS
• Since φm >>φLS , φm >>φLp
• φm can be employed to determine the induced voltage
in the windings and approximately : Vp(t)/Vs(t)=Np/NS=a
• As smaller the leakage fluxes, the better ideal
transformer turn ratio approximate the real transformer
turn ratio
REAL SINGLE PHASE
TRANSFORMER
• Magnetization Current in a Real Transformer
• ac source even when the secondary is open
circuited supply a current to produce flux in real
ferromagnetic core (as seen in chapter One)
• There are two components in the current:
(a) magnetization current iM, required to produce
flux
(b) core-loss current ih+e supplies hysteresis & eddy
current losses of core
REAL SINGLE PHASE
TRANSFORMER
• Magnetization curve of a typical transformer
core can be considered as a saturation curve
REAL SINGLE PHASE
TRANSFORMER
• Knowing the flux in the core magnitude of magnetization current can be found from curve
• Ignoring the leakage flux in the core:
• φav = 1/Np∫ vp(t) dt
• If vp(t) = Vm cos ωt φav= 1/Np∫ Vm cos ωt dt =
= Vm/(ω Np) sin ωt
• If current required to produce a given flux determined at different times from the magnetization curve (above), the magnetization current can be found
REAL SINGLE PHASE
TRANSFORMER
• Finding magnetization current
REAL SINGLE PHASE
TRANSFORMER
• Magnetizing current (another example)
REAL SINGLE PHASE
TRANSFORMER
• Note (Magnetization Current):
1 – magnetization current is nonsinusoidal
2 - once peak flux reaches the saturation point, a small increase in peak flux results in a very large increase in magnetization current
3 - fundamental component of magnetization current lags the voltage applied by 90◦
4 - higher harmonics (odd one) are present in the magnetization current and may have relatively large amount compared to the fundamental &
as core driven further into saturation, larger the harmonic components become
REAL SINGLE PHASE
TRANSFORMER
• Other components of no-load current of
transformer
• is required to supply the hysteresis and eddy
current losses in the core
• assuming sinusoidal flux in the core , eddy
current loss in core proportional to dφ/dt and is
largest when flux pass 0
• Eddy and hysteresis loss shown in Fig 1 and
the total current required to produce flux in the
core shown in Fig 2
REAL SINGLE PHASE
TRANSFORMER• Exciting Current (components: e+h & m)
Fig 1 Fig 2
REAL SINGLE PHASE
TRANSFORMER
• Current Ratio & Dot Convention
• A current flowing into dotted
end of winding produces
a pos. mmf, while current
flowing to undotted end of winding
proguces neg. mmf
• Two current flowing into dotted ends of their respective windings produce mmfs that add
• If one current flows into a dotted end of a winding and one flows out of dotted end, then mmfs will subtract each other
REAL SINGLE PHASE
TRANSFORMER
• In this situation as shown in last figure
• ζP =NPIP , ζS=-NSIS
• ζnet= NPIP-NSIS
• The net mmf produce net flux in core
• ζnet= NPIP-NSIS = φ R
• Where R; reluctance of transformer core
• Since R of well designed is very small until core saturate ζnet= NPIP-NSIS ≈ 0
• Therefore until core unsaturated NPIP ≈ NSIS
• IP/IS ≈ NP /NS =1/a
REAL SINGLE PHASE
TRANSFORMER
• To convert a real transformer to an ideal transformer following assumptions are required:
1- core must have no hysteresis or eddy current
2- magnetization curve must have shape shown (infinite permeabilty before satuartion)ζnet=0 and
NPIP=NSIS
3- leakage flux in core must be zero, implying all flux in core couples both windings
4- resistance of transformer windings must be zero
REAL SINGLE PHASE
TRANSFORMER
Practical Transformer
Equivalent Circuit
mI
1
V1
V2
l1
l2
R1
I2
R2
N1
N2
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).
EQUIVALENT CIRCUIT
• 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.
EQUIVALENT CIRCUIT
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
I1
R1
V1
X1
I2
R2
V2
X2
N1:N
2
I’1
Rc
Xm
Ic
I0
Im
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
EQUIVALENT CIRCUIT
I22R2 = I1’
2R2’
Therefore R2’= (I2/ I1’ ) 2 R2
= a2R2
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.
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
EQUIVALENT CIRCUIT
• Example 2
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
EQUIVALENT CIRCUIT
Determination of
Equivalent Circuit
Parameters
• 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:
1. No-load test (or open-circuit test).
2. Short-circuit test.
Transformer- o/c-s/c tests
• 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
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
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
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.
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
Transformer- short circuit test
• Short circuit test
V
A W
Xe1Re1
I sc
Vsc
P sc
Simplified circuit for calculation of series impedance
22
11 RaRReq
22
11 XaXX eq
Primary and secondary
impedances are
combined
Transformer- short circuit test
• Short circuit test evaluation
21
211
121
eeeq
sc
sceq
sc
sceq
RZX
I
VZ
I
PR
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.
Transformer- o/c-s/c tests
• Example 3
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, Re=0.21ohm,
Xe=0.31 ohm)
Voltage Regulation
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.
ZV2
I2’ I2
V12’=V20
R1’ R2
Ze2
X1’
Rc’ Xm’
X2
Ze2 = R1’ + R2 + jX1’ + jX2
= Re2 + jXe2
Transformer – voltage regulation
ZV2
I2’ I2
V12’=V20
R1’ R2
Ze2
X1’
Rc’ Xm’
X2
Applying KVL, V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
Or V2 = V20 - I2 (Ze2 )
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
OA
θ2
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
A
B
θ2
V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Voltage drop = AM = OM – OA
= AD + DN + NM
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
AD = I2 Re2 cosθ2
DN=BL= I2 Xe2 sinθ2
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Applying Phytogrus theorem to OCN triangle.
(NC)2 = (OC)2 – (ON)2
= (OC + ON)(OC - ON) ≈ 2(OC)(NM)
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Therefore NM = (NC)2/2(OC)
NC = LC – LN = LC – BD
= I2 Xe2 cosθ2 - I2 Re2 sinθ2
Transformer – voltage regulation
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 Recosθ2 + I2 Xe2 sin θ2 +
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
thus,
votage regulation = (AM)/V20 per unit
In actual practice the term NM is negligible since its value is very small
compared with V2. Thus the votage regulation formula can be reduced
to:
Transformer – voltage regulation
I2Xe2
I2Re2
I2
V2
O
V20
I2Re2
I2Xe2
C
MNDA
B
θ2
L
θ2
θ2
Voltage regulation =
20
222222 sincos
V
XIRI ee
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)
Transformer – voltage regulation
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)
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
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
Transformer- voltage regulation
• Example 4
Based on Example 3 calculate the voltage
regulation and the secondary terminal
voltage for full load having a power factor
of
(i) 0.8 lagging (0.0336pu,14.8V)
(ii) 0.8 leading (-0.0154pu,447V)
Transformer’s Efficiency
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
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
Transformer- Efficiency
• Efficiency on full load
where S is the apparent power (in volt amperes)
scocFLFL
FLFL
PPS
S
cos
cos
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
Transformer- Efficiency
• Example 5
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%)
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
Transformer- Efficiency
Autotransformer
Transformer- Autotransformer
• It is a transformer whose primary and
secondary coils are in a single winding
Autotransformer
Transformer- Autotransformer
• 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
Transformer- Autotransformer
• 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
Transformer- Autotransformer
• A Step Down Autotransformer:
and
Transformer- Autotransformer
• A Step Up Autotransformer:
and
Transformer- Autotransformer
• Example 6
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.
Transformer- Autotransformer
• Solution
3 Phase Transformer
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
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
Transformer -3 phase transformer
• Analyses of the grounded wye / delta transformer
IA
IB
IC
N
IAN
ICN
IBN
Iab
Ibc
Ica
Ib
Ia
Ic
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
Transformer -3 phase transformer
107
• Three phase transformer
Transformer
Transformer
Transformer
Transformer
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.
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
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
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
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
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