Download Chapter 6: Single Phase Transformer

BASIC ELECTRICAL TECHNOLOGY
DET 211/3
Chapter 6: Single Phase
Transformer
Introduction to
Transformer
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 tasks performed by transformers are:
•
•
•
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 )

i p (t )
Np
Ns
1

i s (t ) a
Sp
SS
1
lossless
a
N p i p ( t )  N s is ( t )
a = turns ratio
of the
transformer
Power in ideal transformer
Pout  Pin  V p I p cos
Qout  Qin  V p I p sin 
S out  S in  V p I p sin 
Where  is the angle between voltage and
current
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
ZL'  a2ZL
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
Nonideal or actual transformer
Transformer equivalent circuit, with secondary impedances referred to
the primary side
Ep = primary induced voltage
Vp = primary terminal voltage
Ip = primary current
Ie = excitation current
XM = magnetizing reactance
RC = core resistance
Rs = resistance of the secondary winding
Xs = secondary leakage reactance
Es = secondary induced voltage
Vs = secondary terminal voltage
Is = secondary current
IM = magnetizing current
IC = core current
Rp = resistance of primary winding
Xp = primary leakage reactance
Nonideal or actual transformer
Transformer equivalent circuit
Dot convention
1. If the primary voltage is positive at the dotted
end of the winding with respect to the
undotted end, then the secondary voltage will
be positive at the dotted end also. Voltage
polarities are the same with respect to the
dots on each side of the core.
2. 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.
Exact equivalent circuit the actual transformer
a) The transformer model referred to primary side
b) The transformer model referred to secondary side
Approximate equivalent circuit the actual
transformer
a) The transformer model referred to primary side
b) The transformer model referred to secondary side
Exact equivalent circuit of a transformer
Ep = primary induced voltage
Vp = primary terminal voltage
Ip = primary current
Ie = excitation current
XM = magnetizing reactance
RC = core resistance
Rs = resistance of the secondary winding
Xs = secondary leakage reactance
Es = secondary induced voltage
Vs = secondary terminal voltage
Is = secondary current
IM = magnetizing current
IC = core current
Rp = resistance of primary winding
Xp = primary leakage reactance
Primary side
Secondary side
I p  Ie  I s / a
ES  I s ( Rs  jX s )  Vs
Ie  IC  I M
Vs  I s Z L
V p  I p ( R p  jX p )  E p
E p  I C RC
E p  I M ( jX M )
E p  I e ( RC // jX M )
Vp
Ep
Is N p
a



Vs Es I p N s
Exact equivalent circuit of a transformer referred to
primary side
Rp
Ip
a2Xs
Xp
Is/a
a2Rs
Ie
Vp
Ep
aVs
Exact equivalent circuit of a transformer referred to
secondary side
Rp/a2
aIp
Xp/a2
Rs
Is
Xs
aIe
Vp/a
aIc
Rc/a2
aIm
Ep/a = Es
XM/a2
Vs
Approximate equivalent circuit of a transformer
referred to primary side
Ip
Reqp
jXeqp
Is/a
+
Reqp=Rp+a2Rs
+
Xeqp=Xp+a2Xs
Vp
Rc
aVs
jXM
-
-
Approximate equivalent circuit of a transformer
referred to secondary side
aIp
Reqs
+
jXeqs
Is
+
Reqs=Rp/a2+Rs
Xeqs=Xp/a2+Xs
Vp/a
Rc/a2
-
jXM/a2
Vs
-
Example
A single phase power system consists of a 480V 60Hz
generator supplying a load Zload=4+j3W through a
transmission line ZLine=0.18+j0.24W. Answer the following
question about the system.
a) If the power system is exactly as described below
(figure 1(a)), what will be the voltage at the load be?
What will the transmission line losses be?
b) Suppose a 1:10 step-up transformer is placed at the
generator end of the transmission line and a 10:1 step
down transformer is placed at the load end of the line
(figure 1(b)). What will the load voltage be now?
What will the transmission line losses be now?
Example
ILine
IG
ZLoad=0.18+j0.24W
+
ILoad
VLoad
V=48000V
ZLoad=4+j3W
-
Figure 1 (a)
T1
IG
1:10
ILine
T2
ILoad
ZLine=0.18+j0.24W
+
10:1
V=48000V
VLoad
Figure 1 (b)
-
Solution Example
(a) From figure 1 (a) shows the power system without
transformers. Hence IG = ILINE = ILoad. The line current in
this system is given by:
I line 
V
Z line  Z load
4800V

(0.18W  j 0.24W)  (4W  j 3W)
4800
4800


4.18  j 3.24W 5.2937.8
 90.8  37.8
Solution Example
Therefore the load voltage is:
Vload  I lineZ load
 (90.8  37.8 A)( 4W  j 3W)
 (90.8  37.8 A)(536.9W)
 454  0.9
and the line losses are
Ploss  ( I line ) 2 Rline
 (90.8 A) 2 (0.18W)
 1484W
Solution Example
(b) From figure 1 (b) shows the power system with the
transformers. To analyze the system, it is necessary to
convert it to a common voltage level. This is done in two
steps:
i) Eliminate transformer T2 by referring the load over to
the transmission’s line voltage level.
ii) Eliminate transformer T1 by referring the transmission
line’s elements and the equivalent load at the
transmission line’s voltage over to the source side.
The value of the load’s impedance when reflected to the
transmission system’s voltage is
Z 'load  a 2 Z load
10 2
 ( ) ( 4W  j 3W)
1
 400W  j 300W
Solution Example
The total impedance at the transmission line level is now:
Z eq  Z line  Z 'load
 400.18W  j 300.24W
 500.336.88W
The total impedance at the transmission line level
(Zline+Z’load) is now reflected across T1 to the source’s
voltage level:
Z 'eq  a 2 Z eq
 a 2 ( Z line  Z 'load )
1 2
 ( ) (0.18W  j 0.24W  400W  j 300W)
10
 (0.0018W  j 0.0024W  4W  j 3W)
 5.00336.88W
Solution Example
Notice that Z’’load = 4+j3 W and Z’line =0.0018+j0.0024 W.
The resulting equivalent circuit is shown below. The
generator’s current is:
IG 
4800V
 95.94  36.88 A
5.00336.88W
a) System with the load
referred to the transmission
system voltage level
b) System with the load and
transmission line referred to
the generator’s voltage level
Solution Example
Knowing the current IG, we can now work back and find Iline
and ILoad. Working back through T1, we get:
N p1 I G  N s1 I line
I line 
N p1
N S1
IG
1

(95.94  36.88 A)
10
 9.594  36.88 A
Solution Example
Working back through T2 gives:
N p 2 I line  N s 2 I load
I load 
N p2
Ns2
I line
10
 ( )(9.594  36.88 A)
1
 95.94  36.88 A
It is now possible to answer the questions. The load
voltage is given by
Vload  I load Z load
 (9.594  36.88 A)(536.87W)
 479.7  0.01V
Solution Example
and the line losses are given by:
Ploss  ( I line ) 2 Rline
 (9.594 A) 2 (0.18W)
 16.7W
Notice that raising the transmission voltage of the power
system reduced transmission losses by a factor of nearly
90. Also, the voltage at the load dropped much less in the
system with transformers compared to the system without
transformers.