Transcript Distributed Power Systems ELCT 908

Modeling of Power
Transformers
A Static Device
Transformers

The transformer enables us to utilize different voltage
levels across the system for the most economical
value.

Stepping up the generator voltage to high voltage, is
done through power transformers to minimize losses
and increase the transmission capacity of the lines.

This transmission voltage level is then stepped down in
many stages for distribution and utilization purposes.
General Theory

A transformer contains two or more windings linked by a
mutual field.

The primary winding is connected to an alternating
voltage source.

The input current results in an alternating flux whose
magnitude depends on the voltage and number of turns
of the primary winding.

The alternating flux links the secondary winding and
induces a voltage in it with a value that depends on the
number of turns of the secondary winding.
Transformers
Primary winding
Supply
NP
Secondary winding
NS
Laminated iron core
Basic components of single phase transformer
Load
Transformers
Primary
Winding
Multi-layer
Laminated
Iron Core
Secondary
Winding
H1 H2
X1
X2
Winding
Terminals
5.2 Single phase transformer arrangement
Transformers
H2
X2
H2
X1
Vp
Vs
Vp
Vs
H1
X1
H1
X2
100
100
Vp ( t )
Vs( t )
Vp ( t )
0
100
Vs( t )
0
5
10
15
20
0
100
0
5
10
t
t
milli  s
milli  s
(b)
(a)
Polarity for transformer
15
20
Transformers
Iron core
Insulation
Secondary
winding
Terminals
Small transformer construction
a) Lamination
b) Iron core with winding
Transformers
Dry-type three-phase transformer
Transformers
Bushing
Steel
tank
Iron core
behind the steel
bar
Winding
Insulation
Radiator
Oil Insulated and cooled transformer
Power Transformers
Transformation ratio
Primary (supply)
Secondary (Load)
Transformers at no load
Ic
IF
E1
Qc
Im
f
The no load current If is needed to supply the no load
losses and to magnetize the transformer core.
E1
If
IF
Ic Im
Loaded Transformer
Z2’ is the load impedance
referred to the primary
Transformer losses

The transformer losses are divided into electrical losses
(copper losses) and Magnetic losses (Iron losses).

Copper losses in both the primary and secondary windings.
I12 R1  I 22 R2

Magnetic losses, these losses are divided into eddy current
losses and hysteresis losses.
Pmag  Peddy  Physterises  V1I m
Equivalent circuit
V1: Primary voltage (supply)
I1 : Primary current.
V2: Secondary voltage (load)
I2: : Secondary current
Exact Circuit
Approximate Circuit
(a)
(b)
The no load current ranges from 1% to 3% of the full load current.
Therefore, the circuit can be simplified to circuit (b).
Phasor Diagram
'
V1  V2
'
 I 2 ( Req
 jX eq )
Performance Measures

The percent regulation

The transformer efficiency
Voltage Regulation
VR  V1  V2'  I 2' Req cosf  I 2' X eq sin f
Example
A 100-kVA, 400/2000 V, single-phase transformer has the following
parameters
R1 = 0.01
R2 = 0.25 ohms
X1 = 0.03 ohms
X2 = 0.75 ohms
The transformer supplies a load of 90 kVA at 2000 V and 0.8 PF lagging.

Calculate the primary voltage and current using the simplest
equivalent circuit.
 Find also the V.R. and efficiency for the transformer