Transcript ECE 310 - University of Illinois at Urbana–Champaign

ECE 476
POWER SYSTEM ANALYSIS
Lecture 8
Transmission Lines, Transformers, Per Unit
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements




Start reading Chapter 3.
HW 2 is due now.
HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class.
“Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit
a coal power plant, a coal mine, a wind farm and a bio-diesel
processing plant. Sponsored by Students for Environmental
Concerns. Cost isn’t finalized, but should be between $10
and $20. Contact Rebecca Marcotte at
[email protected] for more information or to sign up.
1
V, I Relationships, cont’d
Define the propagation constant  as
  yz    j 
where
  the attenuation constant
  the phase constant
Use the Laplace Transform to solve. System
has a characteristic equation
( s 2   2 )  ( s   )( s   )  0
2
Equation for Voltage
The general equation for V is
V ( x)  k1e x  k2e  x
Which can be rewritten as
e x  e x
e x  e  x
V ( x)  (k1  k2 )(
)  (k1  k2 )(
)
2
2
Let K1  k1  k2 and K 2  k1  k2 . Then
e x  e x
e x  e  x
V ( x)  K1 (
)  K2 (
)
2
2
 K1 cosh( x)  K 2 sinh( x)
3
Real Hyperbolic Functions
For real x the cosh and sinh functions have the
following form:
d cosh( x)
  sinh( x)
dx
d sinh( x)
  cosh( x)
dx
4
Complex Hyperbolic Functions
For x =  + j the cosh and sinh functions have the
following form
cosh x
sinh x
 cosh  cos   j sinh  sin 
 sinh  cos   j cosh  sin 
5
Determining Line Voltage
The voltage along the line is determined based upon
the current/voltage relationships at the terminals.
Assuming we know V and I at one end (say the
"receiving end" with VR and I R where x
0) we can
determine the constants K1 and K 2 , and hence the
voltage at any point on the line.
6
Determining Line Voltage, cont’d
V ( x)  K1 cosh( x)  K 2 sinh( x)
V (0)  VR
 K1 cosh(0)  K 2 sinh(0)
Since cosh(0)  1 & sinh(0)  0  K1  VR
dV ( x)
dx
 zI  K1 sinh( x)  K 2 cosh( x)
 K2 
zI R

IR z
z

 IR
y
yz
V ( x)  VR cosh( x)  I R Z c sinh( x)
where Zc 
z
y
 characteristic impedance
7
Determining Line Current
By similar reasoning we can determine I(x)
VR
I ( x)  I R cosh( x)  sinh( x)
Zc
where x is the distance along the line from the
receiving end.
Define transmission efficiency as 
Pout

Pin
8
Transmission Line Example
Assume we have a 765 kV transmission line with
a receiving end voltage of 765 kV(line to line),
a receiving end power SR  2000  j1000 MVA and
z = 0.0201 + j0.535 = 0.53587.8  mile
y = j 7.75  10 6 = 7.75  10 6 90.0
mile
Then
 
zy  2.036 88.9 / mile
c 
z
y
 262.7 -1.1 
9
Transmission Line Example, cont’d
Do per phase analysis, using single phase power
and line to neutral voltages. Then
VR
 765
3
 441.70 kV
6 *
 (2000  j1000)  10
IR
 
 1688  26.6 A
3 
 3  441.70  10 
V ( x)  VR cosh( x)  I R Z c sinh( x)
 441, 7000 cosh( x  2.03688.9) 
443, 440  27.7  sinh( x  2.03688.9)
10
Transmission Line Example, cont’d
11
Lossless Transmission Lines
For a lossless line the characteristic impedance, Zc ,
is known as the surge impedance.
Zc 
jwl
l

 (a real value)
jwc
c
If a lossless line is terminated in impedance
VR
Zc 
IR
Then I R Z c  VR so we get...
12
Lossless Transmission Lines
V ( x)  VR cosh  x  VR sinh  x
I ( x)  I R cosh  x  I R sinh  x
V ( x)
 Zc
I ( x)
2
V(x)
Define
as the surge impedance load (SIL).
Zc
Since the line is lossless this implies
V ( x)  VR
I ( x)  I R
If P > SIL then line consumes
vars; otherwise line generates vars.
13
Transmission Matrix Model
Oftentimes we’re only interested in the terminal
characteristics of the transmission line. Therefore we
can model it as a “black box”.
+
VS
IS
IR
Transmission
Line
-
+
VR
-
 VS   A B  VR 
With    
 I 
I
C
D
 R
 S 
14
Transmission Matrix Model, cont’d
 VS   A B  VR 
With    
 I 
I
C
D
 R
 S 
Use voltage/current relationships to solve for A,B,C,D
VS  VR cosh  l  Z c I R sinh  l
VR
I S  I R cosh  l  sinh  l
Zc
 cosh  l
A B
1
T  


 sinh  l
C
D


 Z c
Z c sinh  l 

cosh  l 

15
Equivalent Circuit Model
The common representation is the  equivalent circuit
Next we’ll use the T matrix values to derive the
parameters Z' and Y'.
16
Equivalent Circuit Parameters
VS  VR
Y'
 VR  I R
Z'
2
Z 'Y ' 

VS  1 
VR  Z ' I R
2 

Y'
Y'
I S  VS  VR  I R
2
2
Z 'Y ' 
Z 'Y ' 


I S  Y ' 1 
VR  1 
 IR
4 
2 


 1  Z 'Y '

Z
'

 VR 
VS 
2
 I     Z 'Y '   Z 'Y '    I 
 R
 S
Y ' 1 

1

 

 
4  
2  
17
Equivalent circuit parameters
We now need to solve for Z' and Y'. Using the B
element solving for Z' is straightforward
B  ZC sinh  l  Z '
Then using A we can solve for Y'
Z 'Y '
A = cosh l  1 
2
Y'
cosh  l  1 1
l

 tanh
2
Z c sinh  l Z c
2
18
Simplified Parameters
These values can be simplified as follows:
Z '  Z C sinh  l

zl z
sinh  l
yl z
sinh  l
Z
with Z zl (recalling   zy )
l
Y'
1
l

tanh
2
Zc
2

yl y
l
tanh
zl y
2
l
tanh
Y
2 with Y

2 l
2
yl
19
Simplified Parameters
For short lines make the following approximations:
sinh  l
Z'  Z
(assumes
 1)
l
Y' Y
tanh( l / 2)

(assumes
 1)
2 2
l /2
sinhγl
tanh(γl/2)
Length
γl
γl/2
50 miles
0.9980.02 1.001  0.01
100 miles
0.9930.09
1.004  0.04
200 miles
0.9720.35
1.014  0.18
20
Medium Length Line Approximations
For shorter lines we make the following approximations:
sinh  l
Z'  Z
(assumes
 1)
l
Y' Y
tanh( l / 2)

(assumes
 1)
2 2
l /2
sinhγl
tanh(γl/2)
Length
γl
γl/2
50 miles
0.9980.02 1.001  0.01
100 miles
0.9930.09
1.004  0.04
200 miles
0.9720.35
1.014  0.18
21
Three Line Models
Long Line Model (longer than 200 miles)
sinh  l Y ' Y
use Z '  Z
,

l
2 2

l
tanh
l
2
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)
22
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
S12
I1
Transmission
Line with
Impedance Z
S21
+
-
V2
*
V1  V2 

S12 
 V1 

 Z 
with V1  V1 1 , V2  V2  2
V1I1*
Z  Z  Z
2
S12
V1
V1 V2

 Z 
 Z  12
Z
Z
23
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
Max
P12
V1 V2

X
24
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.
25
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.
26
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.
27
Transmission to Distribution
Transfomer
28
Transmission Level Transformer
29
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.
30
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
31
Current Relationships
To get the current relationships use ampere's law
mmf 

H dL  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
32
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
33
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
34
Real Transformers

Real transformers
–
–
–
have losses
have leakage flux
have finite permeability of magnetic core
1. Real power losses
–
–
resistance in windings (i2 R)
core losses due to eddy currents and hysteresis
35
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
36
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'
dm
di1
v1  r1i1  Ll1  N1
dt
dt
v 2  r2i 2  Ll 2
'
di 2'
dm
 N2
dt
dt
37
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
38
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'
39
Simplified Equivalent Circuit
40
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.
41
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.
42
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
R e  jX e  jX m
200 kV

 10, 000 
20 A
X m  10, 000 
43
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.
44
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 quantity
quantity in per unit 
base value of quantity
45
Per Unit Conversion Procedure, 1
1.
2.
3.
4.
5.
Pick a 1 VA base for the entire system, SB
Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.
Calculate the impedance base, ZB= (VB)2/SB
Calculate the current base, IB = VB/ZB
Convert actual values to per unit
Note, 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)
46
Per Unit Solution Procedure
1.
2.
3.
Convert to per unit (p.u.) (many problems are
already in per unit)
Solve
Convert back to actual as necessary
47
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 Circuit
48
Per Unit Example, cont’d
Z BLeft
8kV 2

 0.64
100 MVA
Middle
ZB
Z BRight
80kV 2

 64
100 MVA
16kV 2

 2.56
100 MVA
Same circuit, with
values expressed
in per unit.
49
Per Unit Example, cont’d
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8p.u.
VL I L*
50
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
V LActual  0.859  30.8 16 kV  13.7  30.8 kV
S LActual  0.1890 100 MVA  18.90 MVA
SGActual  0.2230.8 100 MVA  22.030.8 MVA
I Middle

B
100 MVA
 1250 Amps
80 kV
I Actual
Middle  0.22  30.8 Amps  275  30.8
51