EE 369 POWER SYSTEM ANALYSIS Lecture 7 Transmission Line Models Tom Overbye and Ross Baldick.
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Transcript EE 369 POWER SYSTEM ANALYSIS Lecture 7 Transmission Line Models Tom Overbye and Ross Baldick.
EE 369
POWER SYSTEM ANALYSIS
Lecture 7
Transmission Line Models
Tom Overbye and Ross Baldick
1
Announcements
For lecture 7 to 10 read Chapters 5 and 3.
HW 6 is problems 5.2, 5.4, 5.7, 5.9, 5.14, 5.16,
5.19, 5.26, 5.31, 5.32, 5.33, 5.36; case study
questions chapter 5 a, b, c, d, is due Thursday,
10/15.
Homework 7 is 5.8, 5.15, 5.17, 5.24, 5.27,
5.28, 5.29, 5.34, 5.37, 5.38, 5.43, 5.45; due
10/22.
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Transmission Line Models
Previous lectures have covered how to calculate
the distributed series inductance, shunt
capacitance, and series resistance of
transmission lines:
– That is, we have calculated the inductance L,
capacitance C, and resistance r per unit length,
– We can also think of the shunt conductance g per
unit length,
– Each infinitesimal length dx of transmission line
consists of a series impedance rdx + jωLdx and a
shunt admittance gdx + jωCdx,
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 model of an infinitesimal length of
transmission line is shown below:
dx
dx
Ldx
Units on
z and y are
per unit
length!
For operation at frequency , let z r j L
and y g jC (with g usually equal to 0)
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Derivation of V, I Relationships
dx
Ldx
dx
We can then derive the following relationships:
dV I ( x ) z dx
dI
(V ( x ) dV ) y dx V ( x ) y dx,
on neglecting the dVdx term,
dV
dI
( x) z I ( x)
( x ) yV ( x )
dx
dx
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Setting up a Second Order Equation
dV
dI
( x) z I ( x)
( x ) yV ( x )
dx
dx
We can rewrite these two, first order differential
equations as a single second order equation
d 2V
dI
( x ) z ( x ) zyV ( x )
2
dx
dx
d 2V
( x ) zyV ( x ) 0
2
dx
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V, I Relationships, cont’d
Define the propagation constant as
yz j
where
the attenuation constant
the phase constant
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Equation for Voltage
The general equation for V is
V ( x ) k1e x k2 e 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 )
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Real Hyperbolic Functions
For real x, the cosh and sinh functions have
the following form:
cosh( x)
d cosh( x)
sinh( x)
dx
sinh( x )
dsinh( x)
cosh( x)
dx
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Complex Hyperbolic Functions
For complex x ˆ j ˆ
cosh( x )
cosh ˆ cos ˆ j sinh ˆ sin ˆ
sinh ˆ cos ˆ j cosh ˆ sin ˆ
sinh( x )
Make sure your calculator handles sinh and
cosh of complex numbers.
You will need this for homework and for the
mid-term!
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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.
We will mostly be interested in the voltage and
current at the other, "sending end," of the line.
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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 ( x ) 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 Z c
z
y
characteristic impedance
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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.
Pout
Define transmission efficiency as
;
Pin
that is, efficiency means the real power out (delivered)
divided by the real power in.
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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 S R 2000 j1000 MVA and
z = 0.0201 + j 0.535 = 0.53587.8
y = j 7.75 106
mile
= 7.75 106 90.0 S
mile
Then
zy
2.036 103 88.9 / mile
Zc
z
y
262.7 -1.1
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Transmission Line Example, cont’d
Do per phase analysis, using single phase power
and line to neutral voltages. Then
441.70 kV
765
VR
3
6 *
(2000 j1000) 10
1688 26.6 A
IR
3
3 441.70 10
V ( x) VR cosh( x) I R Z c sinh( x)
441,7000 cosh( x 2.036 103 88.9)
443,440 27.7 sinh( x 2.036 103 88.9)
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Transmission Line Example, cont’d
Squares and crosses show real and reactive power flow, where
a positive value of flow means flow to the left.
Receiving end
Sending end
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Lossless Transmission Lines
For a lossless line the characteristic impedance, Z c ,
is known as the surge impedance.
Zc
j L
jC
L
(a real value)
C
If a lossless line is terminated in impedance Z c then:
VR
Zc
IR
Then I R Z c VR so we get...
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Lossless Transmission Lines
V ( x) VR cosh( x) I R Z c sinh( x),
VR cosh x VR sinh x,
VR (cosh x sinh x).
VR
I ( x) I R cosh( x) sinh( x)
Zc
VR
VR
cosh x sinh x,
Zc
Zc
VR
(cosh x sinh x)
Zc
V ( x)
That is, for every location x,
Zc .
I ( x)
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Lossless Transmission Lines
Since the line is lossless this implies that for every location x,
the real power flow is constant. Therefore:
Real power flow (V ( x ) I ( x )*) ( Z c I ( x ) I ( x )*)
( Z c | I ( x ) |2 ) Z c | I ( x ) |2 is constant and equals Z c | I (0) |2 .
Therefore, at each x, I ( x) I (0) I R .
V ( x)
So, since
Zc ,
I ( x)
2
V ( x ) V (0) VR and
V ( x)
Define
to be the "surge impedance loading" (SIL).
Zc
If load power P > SIL then line consumes VArs;
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otherwise, the line generates VArs.
Transmission Matrix Model
•Often we are only interested in the terminal
characteristics of the transmission line.
Therefore we can model it as a “black box:”
+
VS
IS
Transmission
Line
-
IR
+
VR
-
VS A B VR
With
,
I S C D I R
where A, B, C , D are determined from general equation
noting that VS and I S correspond to x equaling length of line.
Assume length of line is l.
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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 V (l ) VR cosh l Z c I R sinh l
VR
I S I (l ) I R cosh l sinh l
Zc
cosh l
A B
1
T
sinh l
C
D
Zc
Z c sinh l
cosh l
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Equivalent Circuit Model
A circuit model is another "black box" model.
We will try to represent as a equivalent circuit.
To do this, we’ll use the T matrix values to derive the
parameters Z' and Y' that match the behavior of the
equivalent circuit to that of the T matrix.
We do this by first finding the relationship between
sending and receiving end for the equivalent circuit.
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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
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Equivalent circuit parameters
We now need to solve for Z ' and Y '.
Solve for Z ' using B element:
B Z C 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
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Simplified Parameters
These values can be simplified as follows:
Z ' Z C sinh l
sinh l
Z
with Z
l
Y'
1
l
tanh
2
Zc
2
l
tanh
Y
2
2 l
2
z l zl
sinh l
y l zl
zl (recalling zy )
y l yl
l
tanh
z l yl
2
with Y
yl
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Simplified Parameters
For medium 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
0.9980.02 1.001 0.01
100 miles 0.9930.09 1.004 0.04
200 miles
0.9720.35
1.014 0.18
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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)
The long line model is always correct.
The other models are usually good approximations
for the conditions described.
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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
-
I1
S12
I2
Transmission
Line with
Impedance Z
S21
+
-
V2
*
V1 V2
S12
V1
Z
with V1 V1 1, V2 V2 2 , Z Z Z
V1I1*
2
S12
V1
V1 V2
Z
( Z 12 )
Z
Z
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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
V1 V2
,
X
This power transfer limit is called
P12Max
the steady-state stability limit.
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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/materials limit can limit sag.
– Trees grow, and will eventually hit lines if they are
planted under the line,
– Note that thermal limit is different to the steadystate stability limit that we just calculated:
– Thermal limits due to losses,
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– Steady-state stability limit applies even for lossless line!
Tree Trimming: Before
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Tree Trimming: After
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Other Limits Affecting Power Transfer
Angle limits
– while the maximum power transfer (steady-state
stability limit) occurs when the line angle
difference is 90 degrees, actual limit is
substantially less due to interaction of 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.
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