ece 476 power system analysis lecture 8: transmission line parameters, transformers prof. tom...
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ECE 476 Power System Analysis
Lecture 8: Transmission Line Parameters,
Transformers
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
Special Guest Lecturer: TA Won Jang
Announcements
• Please read Chapters 5 and then 3• Quiz today on HW 3• H4 is 4.34, 4.41, 5.2, 5.7, 5.16
• It should be turned in on Sept 24 (hence no quiz next week)
2
Leidos Engineering
Transmission Line Equivalent Circuit
•Our current model of a transmission line is shown below
For operation at frequency , let z = r + j L
and y = g +j C (with g usually equal 0)
Units on
z and y are
per unit
length!
4
Derivation of V, I Relationships
We can then derive the following relationships:
( )
( ) ( )
dV I z dx
dI V dV y dx V y dx
dV x dI xz I yV
dx dx
5
Setting up a Second Order Equation
2
2
2
2
( ) ( )
We can rewrite these two, first order differential
equations as a single second order equation
( ) ( )
( )0
dV x dI xz I yV
dx dx
d V x dI xz zyVdxdx
d V xzyV
dx
6
V, I Relationships, cont’d
2 2
Define the propagation constant as
where
the attenuation constant
the phase constant
Use the Laplace Transform to solve. System
has a characteristic equation
( ) ( )( ) 0
yz j
s s s
7
Equation for Voltage
1 2
1 2 1 2
1 1 2 2 1 2
1 2
1 2
The general equation for V is
( )
Which can be rewritten as
( ) ( )( ) ( )( )2 2
Let K and K . Then
( ) ( ) ( )2 2
cosh( ) sinh( )
x x
x x x x
x x x x
V x k e k e
e e e eV x k k k k
k k k k
e e e eV x K K
K x K x
8
Real Hyperbolic Functions
•For real x the cosh and sinh functions have the following form:
cosh( ) sinh( )sinh( ) cosh( )
d x d xx x
dx dx
9
Complex Hyperbolic Functions
•For x = + j the cosh and sinh functions have the following form
cosh cosh cos sinh sin
sinh sinh cos cosh sin
x j
x j
10
Determining Line Voltage
R R
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 V and I where x 0) we can
1 2determine the constants K and K , and hence the
voltage at any point on the line.
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Determining Line Voltage, cont’d
1 2
1 2
1
1 2
2
c
( ) cosh( ) sinh( )
(0) cosh(0) sinh(0)
Since cosh(0) 1 & sinh(0) 0
( )sinh( ) cosh( )
( ) cosh( ) sinh( )
where Z characteristic
R
R
R RR
R R c
V x K x K x
V V K K
K V
dV xzI K x K x
dx
zI I z zK I
yyz
V x V x I Z x
zy
impedance
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Determining Line Current
By similar reasoning we can determine I(x)
( ) cosh( ) sinh( )
where x is the distance along the line from the
receiving end.
Define transmission efficiency as
RR
c
out
in
VI x I x x
Z
PP
13
Transmission Line Example
R
6 6
Assume we have a 765 kV transmission line with
a receiving end voltage of 765 kV(line to line),
a receiving end power S 2000 1000 MVA and
z = 0.0201 + j0.535 = 0.535 87.8 mile
y = 7.75 10 = 7.75 10 90
j
j
.0
Then
zy 2.036 88.9 / mile
262.7 -1.1 c
mile
zy
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Transmission Line Example, cont’d
*6
3
Do per phase analysis, using single phase power
and line to neutral voltages. Then
765 441.7 0 kV3
(2000 1000) 101688 26.6 A
3 441.7 0 10
( ) cosh( ) sinh( )
441,700 0 cosh(
R
R
R R c
V
jI
V x V x I Z x
2.036 88.9 )
443,440 27.7 sinh( 2.036 88.9 )
x
x
15
Transmission Line Example, cont’d
16
Lossless Transmission Lines
c
c
c
For a lossless line the characteristic impedance, Z ,
is known as the surge impedance.
Z (a real value)
If a lossless line is terminated in impedance
Z
Then so we get...
R
R
R c R
jwl ljwc c
VI
I Z V
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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.
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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 VR
+ +
- -
IS IR
Transmission
Line
S
S
VWith
IR
R
VA B
IC D
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Transmission Matrix Model, cont’d
S
S
VWith
I
Use voltage/current relationships to solve for A,B,C,D
cosh sinh
cosh sinh
cosh sinh
1sinh cosh
R
R
S R c R
RS R
c
c
c
VA B
IC D
V V l Z I l
VI I l l
Z
l Z lA B
l lC DZ
T
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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'.
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Equivalent Circuit Parameters
'' 2
' '1 '
2
' '2 2
' ' ' '' 1 1
4 2
' '1 '
2' ' ' '
' 1 14 2
S RR R
S R R
S S R R
S R R
S R
S R
V V YV I
ZZ Y
V V Z I
Y YI V V I
Z Y Z YI Y V I
Z YZ
V V
Z Y Z YI IY
22
Equivalent circuit parameters
We now need to solve for Z' and Y'. Using the B
element solving for Z' is straightforward
sinh '
Then using A we can solve for Y'
' 'A = cosh 1
2' cosh 1 1
tanh2 sinh 2
C
c c
B Z l Z
Z Yl
Y l lZ l Z
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Simplified Parameters
These values can be simplified as follows:
' sinh sinh
sinhwith Z zl (recalling )
' 1tanh tanh
2 2 2
tanh 2 with Y2
2
C
c
z l zZ Z l l
y l z
lZ zy
l
Y l y l y lZ z l y
lY
yll
24
Medium Length Line Approximations
For shorter lines we make the following approximations:
sinh' (assumes 1)
' tanh( / 2)(assumes 1)
2 2 / 2
50 miles 0.998 0.02 1.001 0.01
100 miles 0.993 0.09 1.004 0.0
lZ Z
l
Y Y ll
sinhγl tanh(γl/2)Length
γl γl/2
4
200 miles 0.972 0.35 1.014 0.18
25
Three Line Models
(longer than 200 miles)
tanhsinh ' 2use ' ,2 2
2 (between 50 and 200 miles)
use and 2
(less than 50 miles)
use (i.e., assume Y is zero)
ll Y Y
Z Zll
YZ
Z
Long Line Model
Medium Line Model
Short Line Model
26
Power Transfer in Short Lines
•Often we'd like to know the maximum power that could be transferred through a short transmission line
V1 V2
+ +
- -
I1 I1Transmission
Line with Impedance Z
S12 S21
1
** 1 2
12 1 1 1
1 1 2 2 2
21 1 2
12 12
with , Z
Z Z
V VS V I V
Z
V V V V Z Z
V V VS
Z Z
27
Power Transfer in Lossless Lines
21 1 2
12 12 12
12 12
1 212 12
If we assume a line is lossless with impedance jX and
are just interested in real power transfer then:
90 90
Since - cos(90 ) sin , we get
sin
Hence the maximu
V V VP jQ
Z Z
V VP
X
1 212
m power transfer is
Max V VP
X
28
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.
29
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.
30
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.
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Transmission to Distribution Transfomer
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Transmission Level Transformer
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