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Electrical Power System Analysis
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ECE 476 Power System Analysis
Lecture 9: Transformers
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
Announcements
• Please read Chapter 3• H4 is 4.34, 4.41, 5.2, 5.7, 5.16
• It should be turned in on Sept 24 (hence no quiz this week)
2
Lossless Transmission Lines
2
( ) cosh sinh
( ) cosh sinh
( )( )
V(x)Define as the surge impedance load (SIL).
Since the line is lossless this implies
( )
( )
R R
R R
c
c
R
R
V x V x V x
I x I x I x
V xZ
I x
Z
V x V
I x I
If P > SIL then line consumes
vars; otherwise line generates vars.
3
Tree Trimming: Before
4
Tree Trimming: After
5
Transformers Overview
• Power systems are characterized by many different voltage levels, ranging from 765 kV down to 240/120 volts.
• Transformers are used to transfer power between different voltage levels.
• The ability to inexpensively change voltage levels is a key advantage of ac systems over dc systems.
• In this section we’ll development models for the transformer and discuss various ways of connecting three phase transformers.
6
Transmission to Distribution Transfomer
7
Transmission Level Transformer
8
Ideal Transformer
• First we review the voltage/current relationships for an ideal transformer– no real power losses– magnetic core has infinite permeability– no leakage flux
• We’ll define the “primary” side of the transformer as the side that usually takes power, and the secondary as the side that usually delivers power.– primary is usually the side with the higher voltage, but
may be the low voltage side on a generator step-up transformer.
9
Ideal Transformer Relationships
1 1 2 2
1 21 1 2 2
1 2 1 1
1 2 2 2
Assume we have flux in magnetic material. Then
= turns ratio
m
m m
m m
m
N N
d d d dv N v N
dt dt dt dtd v v v N
adt N N v N
10
Current Relationships
'1 1 2 2
'1 1 2 2
'1 1 2 2
'1 1 2 2
To get the current relationships use ampere's law
mmf
length
length
Assuming uniform flux density in the core
lengtharea
d N i N i
H N i N i
BN i N i
N i N i
H L
11
Current/Voltage Relationships
'1 1 2 2
1 2 1 2'
1 2 12
1 2
1 2
If is infinite then 0 . Hence
1or
Then
0
10
N i N i
i N i NN i N ai
av v
i ia
12
Impedance Transformation Example
•Example: Calculate the primary voltage and current for an impedance load on the secondary
21
21
0
10
a vvvi
Za
21 2 1
21
1
1 vv av i
a Zv
a Zi
13
Real Transformers
• Real transformers– have losses– have leakage flux– have finite permeability of magnetic core
• Real power lossesresistance in windings (i2 R)
core losses due to eddy currents and hysteresis
14
Transformer Core losses
Eddy currents arise because of changing flux in core.
Eddy currents are reduced by laminating the core
Hysteresis losses are proportional to area of BH curve
and the frequency
These losses are reduced by using material with a
thin BH curve15
Effect of Leakage Flux
2
22
1 1 1
2 2 2
'1 1 1 2 2
11 1 1 1 1
''
2 2 2 2
Not all flux is within the transformer core
Assuming a linear magnetic medium we get
v
v
l m
l m
l l l l
ml
ml
N
N
L i L i
ddir i L N
dt dt
di dr i L N
dt dt
16
Effect of Finite Core Permeability
m
1 1 2 2 m
m 21 2
1 1
2 m1 2 m
1 1
Finite core permeability means a non-zero mmf
is required to maintain in the core
N
This value is usually modeled as a magnetizing current
where im
i N i
Ni i
N N
Ni i i
N N
17
Transformer Equivalent Circuit
Using the previous relationships, we can derive an
equivalent circuit model for the real transformer
' 2 '2 2 1 2
' 2 '2 2 1 2
This model is further simplified by referring all
impedances to the primary side
r e
e
a r r r r
x a x x x x
18
Simplified Equivalent Circuit
19
Calculation of Model Parameters
• The parameters of the model are determined based upon – nameplate data: gives the rated voltages and power– open circuit test: rated voltage is applied to primary with
secondary open; measure the primary current and losses (the test may also be done applying the voltage to the secondary, calculating the values, then referring the values back to the primary side).
– short circuit test: with secondary shorted, apply voltage to primary to get rated current to flow; measure voltage and losses.
20
Transformer Example
Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data:
open circuit: 20 amps, with 10 kW losses
short circuit: 30 kV, with 500 kW losses
Determine the model parameters.
21
Transformer Example, cont’d
e
2sc e
2 2e
2
e
100 30500 , R 60
200 500
P 500 kW R 2 ,
Hence X 60 2 60
2004
10
200R 10,000 10,000
20
sc e
e sc
c
e m m
MVA kVI A jX
kV A
R I
kVR M
kW
kVjX jX X
A
From the short circuit test
From the open circuit test
22
Residential Distribution Transformers
Single phase transformers are commonly used in
residential distribution systems. Most distribution
systems are 4 wire, with a multi-grounded, common
neutral.
23
Per Unit Calculations
• A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer
impedances to the different sides of the transformers
• This problem is avoided by a normalization of all variables.
• This normalization is known as per unit analysis.
actual quantityquantity in per unit
base value of quantity
24
Per Unit Conversion Procedure, 1f
1. Pick a 1 VA base for the entire system, SB
2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.
3. Calculate the impedance base, ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unitNote, per unit conversion on affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
25
Per Unit Solution Procedure
1. Convert to per unit (p.u.) (many problems are already in per unit)
2. Solve
3. Convert back to actual as necessary
26
Per Unit Example
Solve for the current, load voltage and load power
in the circuit shown below using per unit analysis
with an SB of 100 MVA, and voltage bases of
8 kV, 80 kV and 16 kV.
Original Circuit27
Per Unit Example, cont’d
2
2
2
80.64
100
8064
100
162.56
100
LeftB
MiddleB
RightB
kVZ
MVA
kVZ
MVA
kVZ
MVA
Same circuit, withvalues expressedin per unit.
28
Per Unit Example, cont’d
L
2*
1.0 00.22 30.8 p.u. (not amps)
3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
LL L L
G
Ij
VS V I
ZS
29
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
LActual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 16 kV 13.7 30.8 kV
0.189 0 100 MVA 18.9 0 MVA
0.22 30.8 100 MVA 22.0 30.8 MVA
100 MVAI 1250 Amps
80 kV
I 0.22 30.8 Amps 275 30.8
V
S
S
30
Per Unit Change of MVA Base
• Parameters for equipment are often given using power rating of equipment as the MVA base
• To analyze a system all per unit data must be on a common power base
2 2
Hence Z /
Z
base base
OriginalBase NewBasepu actual pu
OriginalBase NewBasepu puOriginalBase NewBase
BaseBase
NewBaseOriginalBase NewBaseBasepu puOriginalBase
Base
Z Z Z
V VZ
SS
SZ
S
31
Per Unit Change of Base Example
•A 54 MVA transformer has a leakage reactance of 3.69%. What is the reactance on a 100 MVA base?
1000.0369 0.0683 p.u.
54eX
32
Transformer Reactance
• Transformer reactance is often specified as a percentage, say 10%. This is a per unit value (divide by 100) on the power base of the transformer.
• Example: A 350 MVA, 230/20 kV transformer has leakage reactance of 10%. What is p.u. value on 100 MVA base? What is value in ohms (230 kV)?
2
1000.10 0.0286 p.u.
350
2300.0286 15.1
100
eX
33
Three Phase Transformers
• There are 4 different ways to connect 3 transformers
Y-Y D-D
Usually 3 transformers are constructed so all windings
share a common core34
3f Transformer Interconnections
D-Y Y-D
35
Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
1, ,An AB A
an ab a
V V Ia a
V V I a
36
