Transcript Approximate equivalent circuit of the actual

ELECTRICAL MACHINE
DET 204/3
JIMIRAFIZI BIN JAMIL
CHAPTER 1
Transformer
Introduction
A transformer is a device that changes ac
electric energy at one voltage level to ac
electric energy at another voltage level
through the action of a magnetic field.
The most important
transformers are:
tasks
performed
by
• changing voltage and current levels in
electric power systems.
• matching source and load impedances for
maximum power transfer in electronic and
control circuitry.
• electrical isolation (isolating one circuit from
another or isolating dc while maintaining ac
continuity between two circuits).
It consists of two or more coils of wire wrapped
around a common ferromagnetic core. One of
the transformer windings is connected to a
source of ac electric power – is called primary
winding and the second transformer winding
supplies electric power to loads – is called
secondary winding.
Ideal Transformer
An ideal transformer is a lossless device
with an input winding and output winding.
v p (t )
v s (t )

Np
Ns
a
a = turns ratio of
the transformer
a = N2 = V2
N1
V1
Sp
SS
 1 lossless
N p i p ( t )  N s is ( t )
i p (t )
i s (t )

1
a
Power in ideal transformer
Output Power
Pout  Pin  V p I p cos
Reactive Power
Qout  Qin  V p I p sin 
Apparent Power
S out  S in  V p I p sin 
Impedance transformation through the
transformer
The impedance of a device – the ratio of the
phasor voltage across it in the phasor current
flowing through it:
ZL 
VL
IL
Z L'  a2Z L
The equivalent circuit of a transformer
The major items to be considered in the construction of such a model
are:
• Copper (I2R) losses: Copper losses are the resistive heating in the
primary and secondary windings of the transformer. They are
proportional to the square of the current in the windings.
• Eddy current losses: Eddy current losses are resistive heating losses
in the core of the transformer. They are proportional to the square of
the voltage applied to the transformer.
• Hysteresis losses: Hysteresis losses are associated with the
arrangement of the magnetic domain in the core during each half
cycle. They are complex, nonlinear function of the voltage applied to
the transformer.
• Leakage flux: The fluxes ΦLP and ΦLS which escape the core and
pass through only one of the transformer windings are leakage fluxes.
These escaped fluxes produce a self inductance in the primary and
secondary coils, and the effects of this inductance must be accounted
for.
Nonideal or actual transformer
Mutual flux
Nonideal or actual transformer
Ep = primary induced voltage
Es = secondary induced voltage
Vp = primary terminal voltage
Vs = secondary terminal voltage
Ip = primary current
Is = secondary current
Ie = excitation current
IM = magnetizing current
XM = magnetizing reactance
IC = core current
RC = core resistance
Rp = resistance of primary winding
Rs = resistance of the secondary winding Xp = primary leakage reactance
Xs = secondary leakage reactance
Exact equivalent circuit of the actual
transformer
a) The transformer model referred to primary side
b) The transformer model referred to secondary side
Approximate equivalent circuit of the actual
transformer
a) The transformer model referred to primary side
b) The transformer model referred to secondary side
Per unit System
The per unit value of any quantity is defined as
Actual Quantity
Per Unit, pu 
Base value of quantity
Quantity – may be power, voltage, current or
impedance
Two major advantages in using a per unit
system:
1. It eliminates the need for conversion of the
voltages, currents, and impedances across
every transformer in the circuit; thus, there is
less chance of computational errors.
2. The need to transform from three phase to
single phase equivalents circuits, and vise
versa, is avoided with the per unit quantities;
hence, there is less confusion in handling and
manipulating the various parameters in three
phase system.
Per Unit (pu) in Single Phase System
Pbase , Sbase ,Qbase  VbaseI base
Vbase
Z base 
I base
I base
Ybase 
Vbase
( Vbase )2
Z base 
Sbase
Example
 An ideal transformer is rated at 2400/120 V, 9.6 kVA
and has 50 turns on the secondary side. Calculate;
a) Turns ratio
b) The number of turns on the primary side
c) The current ratings for the primary and secondary
windings
Solution
a) This is step-down transformer, since V1 = 2,400 V > V2 = 120 V
a = V2 = 2,400 = 20
V1
120
b) N1 = ?
a = N2
N1
20 =
50
N1
N1 = 50
20
= 2.5 turns
c) S = V1 I1 = V2 I2 = 9.6kVA.
I1 = 9,600
V1
= 9,600
2,400
= 4A
I2 = 9,600
V2
= 9,600
120
@
= 80 A
I2 = I1
a
Assignment 2
Consider an ideal, single phase 2400/240 V
transformer. The primary is connected to a
2200 V source and the secondary is
connected to an impedance of 236.9 o  .
a) Find the secondary output current and
voltage.
b) Find the primary input current.
c) Find the load impedance as seen from the
primary side.
d) Find the input and output apparent powers.
e) Find the output power factor.
Try It !