Download - Glover 5e SI_Chapter 05
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CHAPTER 5:
Transmission
Lines: Steady-
State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 0
Chapter 5: Transmission Lines: Steady-State Operation
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Transmission Line Models
Previous lectures have covered how to calculate the
distributed inductance, capacitance and resistance of
transmission lines.
In this section we will use these distributed parameters
to develop the transmission line models used in power
system analysis.
Chapter 5: Transmission Lines: Steady-State Operation
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Transmission Line Equivalent Circuit
Our current model of a transmission line is shown below
Chapter 5: Transmission Lines: Steady-State Operation
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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!
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Derivation of V, I Relationships
Chapter 5: Transmission Lines: Steady-State Operation
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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
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Setting Up a Second Order Equation
Chapter 5: Transmission Lines: Steady-State Operation
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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 zyV
dxdx
d V xzyV
dx
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V, I Relationships, cont’d
Chapter 5: Transmission Lines: Steady-State Operation
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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
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Equation for Voltage
Chapter 5: Transmission Lines: Steady-State Operation
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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
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Real Hyperbolic Functions
For real x the cosh and sinh functions have the following
form:
Chapter 5: Transmission Lines: Steady-State Operation
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cosh( ) sinh( )sinh( ) cosh( )
d x d xx x
dx dx
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Complex Hyperbolic Functions
For x = + j the cosh and sinh functions have the
following form:
Chapter 5: Transmission Lines: Steady-State Operation
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cosh cosh cos sinh sin
sinh sinh cos cosh sin
x j
x j
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Determining Line Voltage
Chapter 5: Transmission Lines: Steady-State Operation
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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
Chapter 5: Transmission Lines: Steady-State Operation
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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
z
y
impedance
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Determining Line Current
Chapter 5: Transmission Lines: Steady-State Operation
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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
P
P
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Transmission Line Example
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 12
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 km
y = 7.75 10 = 7.75 10 90.0
j
j
km
Then
zy 2.036 88.9 /km
262.7 -1.1 c
z
y
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Transmission Line Example, cont’d
Chapter 5: Transmission Lines: Steady-State Operation
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*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
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Transmission Line Example, cont’d
Chapter 5: Transmission Lines: Steady-State Operation
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Lossless Transmission Lines
Chapter 5: Transmission Lines: Steady-State Operation
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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 l
jwc c
V
I
I Z V
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Lossless Transmission Lines
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 16
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.”
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 17
VS VR + +
- -
IS IR Transmission
Line
S
S
VWith
I
R
R
VA B
IC D
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Transmission Matrix Model, cont’d
Chapter 5: Transmission Lines: Steady-State Operation
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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
Chapter 5: Transmission Lines: Steady-State Operation
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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
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 20
'
' 2
' '1 '
2
' '
2 2
' ' ' '' 1 1
4 2
' '1 '
2
' ' ' '' 1 1
4 2
S RR R
S R R
S S R R
S R R
S R
S R
V V YV I
Z
Z YV V Z I
Y YI V V I
Z Y Z YI Y V I
Z YZ
V V
Z Y Z YI IY
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Equivalent Circuit Parameters
Chapter 5: Transmission Lines: Steady-State Operation
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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 1tanh
2 sinh 2
C
c c
B Z l Z
Z Yl
Y l l
Z l Z
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Simplified Parameters
Chapter 5: Transmission Lines: Steady-State Operation
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These values can be simplified as follows:
' sinh sinh
sinhwith Z zl (recalling )
' 1tanh tanh
2 2 2
tanh2 with Y
22
C
c
z l zZ Z l l
y l z
lZ zy
l
Y l y l y l
Z z l y
lY
yll
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Simplified Parameters
Chapter 5: Transmission Lines: Steady-State Operation
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For short lines make the following approximations:
sinh' (assumes 1)
' tanh( / 2)(assumes 1)
2 2 / 2
80 km 0.998 0.02 1.001 0.01
160 km 0.993 0.09 1.004 0.04
320 km 0.9
lZ Z
l
Y Y l
l
sinhγl tanh(γl/2)Length
γl γl/2
72 0.35 1.014 0.18
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Medium Length Line Approximations
Chapter 5: Transmission Lines: Steady-State Operation
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For shorter lines we make the following approximations:
sinh' (assumes 1)
' tanh( / 2)(assumes 1)
2 2 / 2
80 km 0.998 0.02 1.001 0.01
160 km 0.993 0.09 1.004 0.04
320
lZ Z
l
Y Y l
l
sinhγl tanh(γl/2)Length
γl γl/2
km 0.972 0.35 1.014 0.18
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Three Line Models
Chapter 5: Transmission Lines: Steady-State Operation
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(longer than 320 km)
tanhsinh ' 2use ' ,2 2
2
(between 80 and 320 km)
use and 2
(less than 80 km)
use (i.e., assume Y is zero)
ll Y Y
Z Zll
YZ
Z
Long Line Model
Medium Line Model
Short Line Model
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Power Transfer in Short Lines
Often we’d like to know the maximum power that could
be transferred through a short transmission line
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 26
V1 V2 + +
- -
I1 I1 Transmission
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
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Power Transfer in Lossless Lines
Chapter 5: Transmission Lines: Steady-State Operation
© 2012 Cengage Learning Engineering. All Rights Reserved. 27
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
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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°C.
– Trees grow, and will eventually hit lines if they are
planted under the line.
Chapter 5: Transmission Lines: Steady-State Operation
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Other Limits Affecting Power Transfer
Angle limits
– while the maximum power transfer occurs when line
angle difference is 90°, 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.
Chapter 5: Transmission Lines: Steady-State Operation
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