Megger a Guide to Transformer Winding Resistance Measurements
1 Chapter 4. Transformer. 2 Transformer- Introduction Two winding transformers Construction and...
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Transcript of 1 Chapter 4. Transformer. 2 Transformer- Introduction Two winding transformers Construction and...
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1
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 N2Supply 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
PrimaryWinding
SecondaryWinding
Multi-layerLaminatedIron Core
X1X
2H1 H2
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
max11
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
ratiok is transformation ratio
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TransformerThe 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 N2
mIm
V1
E1
E2 = V
2
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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
E 2
Load
V 2
I2
2 1
mIm + I 1
V 1E 1
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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
I2I1 = I2 /T
E2 = V2
V1 = E1 = T E2
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
VIV
I
VZ a
a
a
Equivalent circuit of an ideal transformer
loada ZZ 21
Vac Zload
T
V1 V2
I1 I2
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Vac a2ZloadV1
I1
Vac/k ZloadV2
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
mI1
V1
V2
l1l2
R1
I2
R2
N1
N2
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Transformer- equivalent circuit
I1
R1
V1
X1
I2
R2
V2
X2
N1:N2
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 windingR2 and X2 are resistance and reactance of the secondary winding
Transformer- practical equivalent circuit
I1
R1
V1
X1
I2
R2
V2
X2
N1:N2
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 I2R’2
V2
X’2
N1:N2
I’1
Rc Xm
Ic
I0
ImV2
’
V2' = a V2 , I1' = I2/aX2' = a2 X2 , R2' = a2 R2
a = 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.I
1R
1
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
AX1 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 circuit
V
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
ococococ
oc
22
01
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
a2R2X1R1 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 scVsc
P sc
Simplified circuit for calculation of series impedance
22
11 RaRRe
22
11 XaXX e
Primary and secondary impedances are combined
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Transformer- short circuit test
Short circuit test evaluation
21
211
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 4.4Obtain the equivalent circuit of a 200/400V, 50Hz1-phase transformer from the following test
data:-
O/C test : 200V, 0.7A, 70W - on L.V. sideS/C test : 15V, 10A, 85W - on H.V. side
(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm, Xe=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
2222222
2
sincos
V
RIXI ee
20
2222222
2
sincos
V
RIXI ee
NM =
AM = AD + DN + NM = I2 Recosθ2 + 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 is very small compared with V2. Thus the votage regulation formula 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' R1'
V2NL
X1' R2X2
V2 Z2
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
I1R1
V1
X1 R2'X2
'
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,14.8V) (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
222222
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 efficiency occurs when
If this condition is applied, the condition for maximum efficiency is
that is, core loss = copper loss.
02
dI
d
222 ec RIP
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Transformer- Efficiency
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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 whichmay 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
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Three phase transformer
Transformer
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Three phase transformer
Transformer
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Three phase transformer
Transformer
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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
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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
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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