EE 369
POWER SYSTEM ANALYSIS
Lecture 7
Transmission Line Models
Tom Overbye and Ross Baldick
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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  jC (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
6
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.
Define transmission efficiency as 
Pout

;
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 = 0.53587.8 
y = j 7.75  106
mile
= 7.75  106 90.0 S
mile
Then
3
 
zy
 2.036  10
Zc 
z
y
 262.7 -1.1 
88.9 / mile
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Transmission Line Example, cont’d
Do per phase analysis, using single phase power
and line to neutral voltages. Then
 441.70 kV
 765
VR
3
6 *
 (2000  j1000)  10
 1688  26.6 A
 
IR
3 
 3  441.70 10 
V ( x)  VR cosh( x)  I R Z c sinh( x)
 441,7000 cosh( x  2.036  103 88.9) 
443,440  27.7  sinh( x  2.036  103 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
jC
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 
 S
Y ' 1 
 R
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 '  ZC 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
0.998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
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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
the steady-state stability limit.
P12Max
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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
31
Tree Trimming: After
32
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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