Power System Analysis
Lecture 9: Transformers
Announcements
• Please read Chapter 3
• HW 4 is 4.32, 4.41, 5.1, 5.7, 5.14; this one must be
turned in on Sept 22 (hence there will be no quiz
that day)
1
Three Line Models
Long Line Model (longer than 200 miles)
l
tanh
sinh  l Y ' Y
2
use Z '  Z
,

l
2 2 l
2
Medium Line Model (between 50 and 200 miles)
Y
use Z and
2
Short Line Model (less than 50 miles)
use Z (i.e., assume Y is zero)
2
Power Transfer in Short Lines
•Often we'd like to know the maximum power that
could be transferred through a short transmission line
V1
+
-
I1
I1
Transmission
Line with
Impedance Z
S12
+
S21
-
V2
*
V1  V2 

*
S12  V1I1  V1 

 Z 
with V1  V1 1 , V2  V2  2
Z  Z  Z
2
S12
V1
V1 V2

 Z 
 Z  12
Z
Z
3
Power Transfer in Lossless Lines
If we assume a line is lossless with impedance jX and
are just interested in real power transfer then:
2
P12  jQ12
V1
V1 V2

90 
90  12
Z
Z
Since - cos(90  12 )  sin 12 , we get
V1 V2
P12 
sin 12
X
Hence the maximum power transfer is
P12Max
V1 V2

X
4
Limits Affecting Max. Power Transfer
• Thermal limits
–
–
–
–
limit is due to heating of conductor and hence depends
heavily on ambient conditions.
For many lines, sagging is the limiting constraint.
Newer conductors limit can limit sag. For example, in
2004 ORNL working with 3M announced lines with a
core consisting of ceramic Nextel fibers. These lines can
operate at 200 degrees C.
Trees grow, and will eventually hit lines if they are
planted under the line.
5
Other Limits Affecting Power Transfer
• Angle limits
–
while the maximum power transfer occurs when line
angle difference is 90 degrees, actual limit is substantially
less due to multiple lines in the system
• Voltage stability limits
–
as power transfers increases, reactive losses increase as
I2X. As reactive power increases the voltage falls,
resulting in a potentially cascading voltage collapse.
6
Example 5.8
765.0 kV
0.0 Deg
816.7 kV
-11.4 Deg
847.3 kV
-19.5 Deg
857.8 kV
-27.3 Deg
A
A
A
MVA
MVA
MVA
A
A
A
MVA
MVA
MVA
A
A
A
MVA
MVA
MVA
A
A
A
MVA
MVA
MVA
A
A
MVA
MVA
slack
9000 MW
-2912 Mvar
9000 MW
0 Mvar
7
Transmission Line Series
Compensation
• One way to increase the transmission capacity of a
transmission line that is limited by its reactance is
to add series compensation
–
Capacitors are placed in series with the transmission line
(covered in Example 5.10)
Image shows
BPA series
capacitors in
a 500 kV
line
Image: https://www.bpa.gov/news/newsroom/Pages/Chief-engineers-reunite-reminisce-for-BPAs-75th.aspx
Transmission Line Series
Compensation
• Amount of series compensation is expressed as a
percentage of the total line reactance (e.g., 50%)
• The series capacitance is usually setup so that it
can be bypassed sometimes
–
There can be excessive reactive power generation on the
system during light loads, like at night
• There can be a concern with sub-synchronous
interactions (SSI)
fn 
XC
1
 f0
XL
LC
Example 5.10
765.0 kV
765.0 kV
slack
2200.0 MW
YES
-0.2 Mvar
YES
Line Angle Difference: -21.4 Deg
2200 MW
0 Mvar
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.
11
Transmission to Distribution
Transfomer
12
Transmission Level Transformer
13
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.
14
Ideal Transformer Relationships
Assume we have flux m in magnetic material. Then
1  N1m
d 1
2  N 2m
d 2
d m
d m
v1 
 N1
v2

 N2
dt
dt
dt
dt
d m
v1
v2
v1
N1




 a = turns ratio
dt
N1
N2
v2
N2
15
Current Relationships
To get the current relationships use ampere's law
mmf 

H idL  N1i1  N 2i2'
H  length  N1i1  N 2i2'
B  length

 N1i1  N 2i2'
Assuming uniform flux density in the core
  length
 N1i1  N 2i2'
  area
16
Current/Voltage Relationships
If  is infinite then 0  N1i1  N 2i2' . Hence
i1
N2
 
or
'
N1
i2
i1
N2 1


i2
N1 a
Then
v1 
i 
 1
a 0  v
 2



1  
0
  i2 

a
17
Impedance Transformation Example
•Example: Calculate the primary voltage and current
for an impedance load on the secondary
a

 v1 


i 
0
 1

v1  a v2
i1
0   v2 
1   v2 



Z
a
1 v2

aZ
v1
 a2 Z
i1
18
Real Transformers
• Real transformers
–
–
–
have losses
have leakage flux
have finite permeability of magnetic core
• Real power losses
resistance in windings (i2 R)
core losses due to eddy currents and hysteresis
19
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 curve
20
Effect of Leakage Flux
Not all flux is within the transformer core
1  l1  N1m
2  l 2  N 2m
Assuming a linear magnetic medium we get
l1 ≜ Ll1i1
l 2 ≜ Ll 2i 2'
di1
dm
v1  r1i1  Ll1  N1
dt
dt
'
v 2  r2i 2  Ll 2
di2'
d m
 N2
dt
dt
21
Effect of Finite Core Permeability
Finite core permeability means a non-zero mmf
is required to maintain m in the core
N1i1  N 2i2   m
This value is usually modeled as a magnetizing current
 m N 2
i1 

i2
N1
N1
i1
N2
 im 
i2
N1
 m
where i m 
N1
22
Transformer Equivalent Circuit
Using the previous relationships, we can derive an
equivalent circuit model for the real transformer
This model is further simplified by referring all
impedances to the primary side
r2'  a 2 r2
re  r1  r2'
x2'  a 2 x2
xe  x1  x2'
23
Simplified Equivalent Circuit
24
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.
25
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.
26
Transformer Example, cont’d
From the short circuit test
100 MVA
30 kV
I sc 
 500 A, R e  jX e 
 60 
200kV
500 A
2
Psc  Re I sc
 500 kW  R e  2 ,
Hence X e  602  22  60 
From the open circuit test
200 kV 2
Rc 
 4M 
10 kW
200 kV
R e  jX e  jX m 
 10, 000 
20 A
X m  10, 000 
27
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.
28