Post on 07-Jul-2018
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POWER SYSTEMS ILecture 4
06-88-590-68
Electrical and Computer Engineering
University of Windsor
Dr. Ali Tahmasebi
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Transmission Line Models
l Transmission lines have distributed inductance,
capacitance and resistance, which can be calculated
based on the conductor material and line structure.
l In this section we will use these distributed parameters to develop the transmission line models
used in power system analysis.
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Transmission Line Equivalent Circuit
Our current model of a transmission line is shown
below for a section of line with the length = dx.
For operation at frequency , let z = r + j L
and y = g +j C (with g usually equal 0)
w w
w
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Derivation of V, I Relationships
We can then derive the following relationships:
( )( ) ( )
dV I z dx
dI V dV y dx V y dxdV x dI x
z I yV dx dx
=
= + »
= =
z and y are the
values for the entire
line.
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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 x z I yV
dx dx
d V x dI x z zyV
dxdx
d V x zyV
dx
= =
= =
- =
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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
g
g a b
a
b
g g g
= = +
=
=
- = - + =
V, I Relationships, cont’d
[m-1]
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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
g g
g g g g
g g g g
g g
-
- -
- -
= +
+ -= + + -
= + = -
+ -= +
= +
(voltage at position x):
( x is measured from
receiving end of line)
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Real Hyperbolic Functions
For real x the cosh and sinh functions have the
following form:
cosh( ) sinh( )sinh( ) cosh( )
d x d x x x
dx dx
g g g g g g = =
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Complex Hyperbolic Functions
For x = a + jb the cosh and sinh functions have thefollowing form
cosh cosh cos sinh sin
sinh sinh cos cosh sin
x j
x j
a b a b
a b a b
= +
= +
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Determining Line Voltage
The voltage along the line is determined based on the
current/voltage relationships at the terminals. Assuming
we know V and I at one end (usually at the “receiving
end” with VR and IR where x = 0) we can determine theconstants K 1 and K 2, and hence the voltage at any point of
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 R R
R R c
V x K x K x
V V K K
K V
dV x zI K x K x
dx
zI I z zK I
y yzV x V x I Z x
z
y
g g
g g g g
g g g
= +
= = +
= = Þ =
= = +
Þ = = =
= +
= = impedance
x = 0
[W]
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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
R R
c
out
in
V I x I x x
Z
P
P
g g
h
= +
=
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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 87.8mile
y = 7.75 10 = 7.75 10 90
j
j - -
= +
WÐ °
´ ´ Ð .0
Then
zy 2.036 88.9 / mile
262.7 -1.1c
mile
z
y
g
°
= = Ð °
Z = = Ð °W
W
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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 kV
3
(2000 1000) 101688 26.6 A
3 441.7 0 10
( ) cosh( ) sinh( )441,700 0 cosh(
R
R
R R c
V
j I
V x V x I Z xg g
= = Ð °
é ù+ ´= = Ð - °ê ú
´ Ð °́ë û
= += Ð ° 2.036 88.9 )
443,440 27.7 sinh( 2.036 88.9 )
x
x
´ Ð °+
Ð - °́ ´ Ð °
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Transmission Line Example, cont’d
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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
I
R
R
V A B
I C D
é ù é ùé ù=ê ú ê úê ú
û ûû
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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
V A B
I C D
V V l Z I l
V I I l l
Z
l Z l A B
l lC D Z
g g
g g
g g
g g
é ù é ùé ù=ê ú ê úê ú
ë ûë ûë û
= +
= +
é ùé ù ê ú= =ê ú ê úë ûê úë û
T
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Equivalent Circuit Model
The common representation is the equivalent circuitp
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 1
4 2
S R R R
S R R
S S R R
S R R
S R
S R
V V Y V I
Z
Z Y V V Z I
Y Y I V V I
Z Y Z Y I Y V I
Z Y Z
V V
Z Y Z Y I I Y
-- =
æ ö= + +ç ÷
è ø
= + +
æ ö æ ö= + + +ç ÷ ç ÷
è ø è ø
é ù+
ê úé ù é ù= ê úê ú ê úæ ö æ ö ë ûë û ê ú+ +ç ÷ ç ÷
ê úè ø è øë û
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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 Y l
Y l l
Z l Z
g
g
g g g
= =
= +
-= =
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Simplified Parameters
These values can be simplified as follows:
' sinh sinh
sinh with Z zl (recalling )
' 1tanh tanh
2 2 2
tanh2 with Y
22
C
c
z l z Z Z l l
y l z
l Z zyl
Y l y l y l
Z z l y
lY
yll
g g
g g g
g g
g
g
= =
= =
= =
=
=
=
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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
l Z Z
l
Y Y ll
g
g
g g
= »
= »
Ð ° Ð - °
Ð ° Ð -
sinhγl tanh(γl/2)Length
γl γl/2
4
200 miles 0.972 0.35 1.014 0.18
°
Ð ° Ð - °
(this approximation is applied to lines between 80 to 250 km)
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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 Z ll
Y Z
Z
g g
g g = =
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
V1 V2
+ +
- -
I1 I1Transmission
Line withImpedance 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 V S V I V
Z
V V V V Z Z
V V V S
Z Z
q q q
q q q
-æ ö= = ç ÷
è ø
= Ð = Ð = Ð
= Ð - Ð +(where q12 =
q1-q2)
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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 l
jwc c
V
I
I Z V
= = W
=
=
For quick, hand calculations, we can assume R=0 (so
z=jw l ), which is called a lossless line.
Which is known as surge impedance.
= = ( )( ) = = [m-1]
(gwould be a purely imaginary number)
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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 x Z
I x
Z
V x V
I x I
g g
g g
= +
= +
=
=
=If P > SIL then line consumes
vars; otherwise line generates vars.
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Surge Impedance Loading
If a lossless line is terminated by (or connected to) a load
equal to the surge impedance Z C , the power delivered is
called surge impedance loading (SIL).
= → = →
() = cosh + sinh = cosh + sinh ()
= =
→ () =
® Voltage profile at SIL condition is flat.
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Surge Impedance Loading, cont’d
() = cosh +
sinh = cosh + sinh ()
=
=
→ () =
Current at position x, at surge impedance loading:
Complex power flowing at any point x:
() = () + () = ()()∗ =
∗
=
∗
=
= SIL ® constant and real
number
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ABCD Parameters of Lossless Line
= = cosh = cosh = +
2 = cos
sinh = sinh = − 2
= sin
→ = sinh = sin =
sin [W]
=sinh ()
=
sin ()
[W-1]
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Wavelength of a Transmission Line
Wavelength is defined as the distance on the transmission
line required to change the phase of the voltage or current by
360°(2p radians). If we assume a lossless line:
() = + = cos + sin ()
() = + = sin ()
+ cos ()
These two equations change phase by 360°when =
→ wavelength = l =
=
=
[m]
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Wavelength of a Transmission Line, cont’d
Velocity of propagation of voltage and current waves on
the line:
l =
1
[m/s]
For overhead lines, typically
≅ 3 ×10 m/s
→ l ≅ 3 ×10
60 = 5000 = 3100
Typical line are usually much shorter than this.