Transcript primary voltage side

EET 103
Chapter 5
(Lecture 1)
Single Phase Transformer
1
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.
2
Introduction to Transformer
•
Transformers are constructed of two coils or
more
placed
around
the
common
freeomagnetic core so that the charging flux
developed by one will link to the other.
•
The coil to which the source is applied is called the
primary coil.
•
The coil to which the load is applied is called the
secondary coil.
3
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).
4
5
Mutual Inductance
• Mutual inductance exits between coils of the
same or different dimensions.
• Mutual inductance is a phenomenon basic to
the operation of the transformer.
6
Mutual Inductance
• A transformer is constructed of 2 coils placed so that
the changing flux developed by one will link the other.
• The coil to which the source is applied is called
primary
• The coil which the load is applied is called secondary.
7
Mutual Inductance (Cont…)
8
9
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
i p (t )
1

i s (t ) a
N p i p ( t )  N s is ( t )
a = turns ratio of the transformer
10
Power in ideal transformer
Pout  Pin  V p I p cos
Qout  Qin  V p I p sin 
S out  Sin  Vs I s  V p I P
Where  is the angle between voltage and current
11
Impedance transformation
transformer
through
the
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
12
Non-ideal or actual transformer
Mutual flux
13
Losses in the transformer
• 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.
14
Losses in 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.
15
Transformer equivalent circuit
Ie
Ic
Im
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
16
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.
17
Exact equivalent circuit the actual transformer
a. The transformer model referred to primary side
b. The transformer model referred to secondary
side
18
Approximate equivalent circuit the actual
transformer
a. The
transformer
model referred to
primary side
b. The
transformer
model referred to
secondary side
19
Exact equivalent circuit of a transformer
refer to primary side
Ep = primary induced voltage
Vp = primary terminal voltage
Ip = primary current
Ie = excitation current
XM = magnetizing reactance
RC = core resistance
winding
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
Xp = primary leakage reactance
20
The Equation:
Primary side
Secondary side
I p  Ie  Is / 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
Vp
Ep
Is N p
a



Vs Es I p N s
E p  I M ( jX M )
E p  I e ( RC // jX M )
21
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
Vs
XM/a2
22
Approximate equivalent circuit of a transformer referred
to primary side
Ip
Reqp
jXeqp
Is/a
+
+
Reqp=Rp+a2Rs
Vp
Rc
aVs
jXM
-
Xeqp=Xp+a2Xs
-
Approximate equivalent circuit of a transformer referred to
secondary side
aIp
Reqs
+
jXeqs
Is
+
Reqs=Rp / a2+Rs
Vp/a
Rc/a2
-
jXM/a2
Vs
Xeqs=Xp / a2+Xs
23
Example 1
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?
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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
ZLoad=4+j3W
V=48000V
VLoad
Figure 1 (b)
-
25
Solution 5.1
(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

26
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  Rline
2
 (90.8 A) (0.18W)
2
 1484W
27
(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
28
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
29
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
4800V
IG 
 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
30
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
31
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
32
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.
33
Parameter determination of the transformer
Open circuit test
Provides magnetizing reactance and core loss
resistance
Obtain components are connected in parallel
34
Experiment Setup
In the open circuit test, transformer rated voltage
is applied to the primary voltage side of the
transformer with the secondary side left open.
Measurements of power, current, and voltage are
made on the primary side.
Since the secondary side is open, the input
current IOC is equal to the excitation current
through the shunt excitation branch. Because this
current is very small, about 5% of rated value,
the voltage drop across the secondary winding
and the winding copper losses are neglected.
35
Admittance
Yoc
I oc

Voc
Open circuit Power Factor
Poc
PF  cos  
Voc I oc
Open circuit Power Factor Angle
  cos
1
Poc
Voc I oc
Angle of current always lags angle of voltage by 
I oc
1
1
Yoc      GC  jBM   j
Voc
RC
XM
36
Short circuit test
– Provides combined leakage reactance and
winding resistance
– Obtain components are connected in series
37
Experiment Setup
In the short circuit test, the secondary side is
short circuited and the primary side is
connected to a variable, low voltage source.
Measurements of power, current, and voltage
are made on the primary side. The applied
voltage is adjusted until rated short circuit
currents flows in the windings.
This voltage is generally much smaller than the
rated voltage.
38
Impedances referred to the primary side
Z sc
Vsc

I sc
Power Factor of the current
Psc
PF  cos  
Vsc I sc
Angle Power Factor
1 Psc
  cos
Vsc I sc
Therefore
Z sc
Vsc 0
0
Vsc
0



0
I sc
I sc    

 
Z sc  Req  jX eq  R p  a Rs  j X p  a X s
2
2

39
The equivalent circuit impedances of a 20-kVA,
8000/240-V, 60-Hz transformer are to be
determined. The open circuit test and the short
circuit test were performed on the primary side of
the transformer and the following data were taken:
Open- circuit test (on Short- circuit test (on
primary)
primary)
Voc = 8000 V
Vsc = 489 V
Ioc = 0.214 A
Isc = 2.5 A
Poc = 400 W
Psc = 240 W
Find the impedances of the approximate
equivalent circuit referred to the primary side and
sketch that circuit
40
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
41
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.
42
Per Unit (pu) in Single Phase System
Pbase ,Qbase , Sbase  VbaseI base
Vbase
Z base 
I base
I base
Ybase 
Vbase
2
( Vbase )
Z base 
Sbase
43
Voltage Regulation (VR)
The voltage regulation of a transformer is defined
as the change in the magnitude of the secondary
voltage as the current changes from full load to
no load with the primary held fixed.
VR 
At no load,
V
VS  P
Vp
VR  a
VS , fl
Req
a
 VS , fl
VS , fl
VS ,nl  VS , fl
X 100%
Is
X 100%
Xeq
+
+
Vp/a
Vs
-
-
44
Phasor Diagram
Vp/a
jIsXeq
Vs
IsReq
Is
Vp/a
Lagging power factor
jIsXeq
Is
Vs
IsReq
Unity power factor
45
Vp/a
jIsXeq
Is
IsReq
Vs
Leading power factor
46
Efficiency
The efficiency of a transformer is defined as the ratio of the
power output (Pout) to the power input (Pin).
Pout

X 100%
Pin

Pout
Pout   P losses
X 100%
Vs I s cos 

X 100%
Vs I s cos   Pcu  Pcore
Pcore = Peddy current + Physteresis
And
Pcu= Pcopper losses
47
Pcu = Copper losses are resistive losses in the primary and
secondary winding of the transformer core. They are modeled by
placing a resistor Rp in the primary circuit of the transformer and
resistor Rs in the secondary circuit.
PCORE = Core loss is resistive loss in the primary winding of the
transformer core. It can be modeled by placing a resistor Rc in the
primary circuit of the transformer.
48
Example
A 15 kVA, 2400/240-V transformer is to be tested
to determine its excitation branch components,
its series impedances and its voltage regulation.
The following test data have been taken from the
primary side of the transformer
Open- circuit test
Short- circuit test
Voc = 2400 V
Vsc = 48 V
Ioc = 0.25 A
Isc = 6.0 A
Poc = 50 W
Psc = 200 W
49
The data have been taken by using the
connections of open circuit test and short circuit
test
a. Find the equivalent circuit of this transformer
referred to the high voltage side.
b. Find the equivalent circuit of this transformer
referred to the low voltage side.
c. Calculate the full load voltage regulation at
0.8 lagging power factor and 0.8 leading
power factor.
d. What is the efficiency of the transformer at full
load with a power factor of 0.8 lagging?
50