Line Models and SIL
1.0 Introduction
In these notes, I present different line
models that are used, and I also make some
comments on Examples 4.2 and 4.3 leading
to discussion of surge impedance loading,
and finally I give a hint for problem 4.21.
2.0 Simplified models (Section 4.5)
We recall two things. First, we have the socalled “exact” transmission line equations:
V1  V2 coshl  ZC I 2 sinh l
(1)
V2
I1  I 2 cosh l 
sinh l
(2)
ZC
Second, we may represent a transmission
line using a π-equivalent model, shown
below, if we use Z’ and Y’/2, where
IZ
I1
I2
Z’
IY1
IY2
Y’/2
Y’/2
V1
Fig. 1
1
V2
sinh l
Z '  Z C sinh l  Z
(3)
l
2
l
tanh( l / 2)
Y'
tanh  Y
(4)
ZC
2
l / 2
Note that the two are equivalent, i.e., use of
the π-equivalent transmission model with Z’
and Y’ is equivalent to using eqs. (1), (2).
Question: When is it OK to use the πequivalent transmission model with Z and Y
(instead of Z’ and Y’)?
(Recall Z=zl, Y=yl where l is line length).
Let’s look at eqs. (3), (4) in more detail.
They tell us that Z’≈Z and Y≈Y’ when
sinh l
1
(5)
l
tanh( l / 2)
1
(6)
l / 2
2
To see when this happens, let’s look up in a
good math table how to express sinh(x) and
tanh(x) as a Taylor’s series. I used [1, p. 5859] to find that:
x3 x5 x7
sinh x  x 



3! 5! 7!
x 3 2 x 5 17 x 7
tanh x  x 



3
15
315
2
π
x2 
4
Using these in eqs. (5) and (6), we get:
3
5
7

l  l  l 
l  



3!
5!
l 
7!
1
3
5
7

l / 2 2l / 2 17l / 2
l / 2 



3
15
l / 2
315
(7)
1
(8)
For both of these equations, they become
true as the higher-order terms in the
numerator get small relative to the first term
in the numerator. This happens for small |γl|,
which occurs for small line length l.
So when |γl| is small, it is quite reasonable to
use Z’=Z=zl and Y’=Y=yl.
3
Consider a lossless line, i.e., a line for which
r=0 in z=r+jx and g=0 in y=g+jb. (Note
that for inductive series elements and
capacitive shunt elements, that x and b will
be positive numbers when defined with
positive signs in z and y). Then
  zy  (r  jx)( g  jb)  rg  j (rb  gx)  xb
For transmission lines, g=0 always. So
  jrb  xb
In the lossless case, r=0, and we get
   xb  j xb  j
(9)
(10)
For a lossless line (γ=jβ), it is possible to
show that γ=jβ=j0.0000013/meter [2, pg
211] is quite typical for most transmission
lines. For a 100 mile-long line:
l 
0.0000013 1609meters

 100miles  0.2092
meter
mile
Then:
sinh( j 0.2092)  j 0.2077
tanh( j 0.2092 / 2)  tanh( j 0.1046)  j 0.105
which shows that
 sinh(γl)≈γl
 tanh(γl/2)≈ γl/2
4
as required.
The loss of accuracy from the approximation
in these cases can be seen from:
sinh l 0.2077

 0.9928
l
0.2092
tanh( l / 2) 0.105

 1.0038
l / 2
0.1046
The text recommends that lines longer than
150 miles should use Long-Line model, but
[2] recommends that lines longer than about
100 miles should use the Long-Line model.
Lines below about 100 miles may use Z=zl
and Y=yl. Doing so results in the MediumLength model, sometimes also referred to as
the nominal π-equivalent model.
A final model suggested by the text is the
short-length model, for lines shorter than 50
miles. This is the same as the MediumLength model except Y is neglected
altogether. This makes sense from the point
of view that the “parallel-plate capacitor” in
5
this case, can be considered to have short
length, and thus a small area of the “plates.”
3.0 Surge impedance loading
Recall our definition of characteristic
impedance ZC as:
z
y
ZC 
(11)
Example 4.2 considers a transmission line
terminated in its characteristic impedance,
per Fig. 2 (the long-line model is used).
IZ
I1
I2
Z’
IY1
Y’/2
IY2
Y’/2
V1
V2
ZC 
Fig. 2
One result of the analysis in Example 4.2 is
to show that the “complex power gain” (the
ratio of the power flowing out of the line to
the power flowing into the line) is given by:
6
z
y
 S21  P21

 e  2l  
S12
P12
(12)
where α is the attenuation constant
(γ=α+jβ).
This says that when a line is terminated in
ZC, the complex power gain (actually loss) is
purely real.
The implication of this is that the line (when
terminated in ZC), only affects the real
power (decreases it) but does not affect the
reactive power at all.
Consider reactive power implication:
 Whatever reactive power flows out of the
line (and into the load) also flows into the
line. So a line terminated in ZC has a very
special character with respect to reactive
power: the amount of reactive power
consumed by the series X is exactly
compensated by the reactive power supplied
by the shunt Y, for every inch of the line!
7
Another result of Example 4.2 is:
V2
 e  l
(13)
V1
It is the case that α is always non-negative.
This means that eαl>1 and 0<e-αl<1.0, and
therefore |V2|<|V1|. So when the line is
terminated in ZC, |V2| will always be less
than or equal to |V1|.
But what if the line is lossless? As shown by
eq. (10), we see that α=0.
Referring back to eq. (13), this means that:
V2
1
(14)
V1
This is a remarkable thing. If we terminate a
lossless line in ZC, the voltage profile along
the line will be flat! In other words, the
voltage along the line will everywhere have
the same magnitude.
8
So |V2|=|V1| for lossless line terminated in
ZC.
Note also that lossless implies
ZC 
jl ind

j c
z

y
l ind

c
l ind l
L

cl
C (15)
So ZC is purely real for the lossless case.
So for lossless line terminated in ZC, since
ZC is purely real, then only real power is
delivered to it, and since the line is lossless,
this same real power is delivered from the
source. Therefore:
2
P12 
2
V2
V
 1
ZC
ZC
(16)
The characteristic impedance is clearly an
important parameter. It is also commonly
referred to in the industry as the surge
impedance. And the power flowing into it,
per eq. (16), is called the surge impedance
loading, denoted by PSIL, i.e.,
2
PSIL
V1

ZC
(17)
9
Equation (17) gives SIL in terms of perphase power (and line-to-neutral voltage).
The text also gives it in terms of 3-phase
power (and line-to-line voltage) as:
3
PSIL

V1ll
2
(18)
ZC
Surge impedance and SIL are given for
typical overhead 60 Hz three-phase
transmission lines in Table 1 [3, 4].
Table 1
2
PSIL  Vrated
/ ZC
Vrated
ZC  L / C
(kV)
(MW)
(ohms)
69
366-400
12-13
115
380*
35
138
366-405
47-52
161
380*
69
230
365-395
134-145
345
280-366
325-425
500
233-294
850-1075
765
254-266
2200-2300
1100
231
5238
* Estimated
10
4.0 Line limits (Section 4.9)
Fig. 3 is a well-known conceptual curve that
captures some attributes of transmission
lines (note that the numbers are typical and
limits for any particular line may vary).
Note the vertical axis is given as a
percentage of SIL, simply because SIL
provides a convenient characteristic of a
transmission line that captures an attribute
related to its power handling capability as a
function of its physical construction.
However, SIL does not capture the influence
of length on power handling capability.
Fig. 3
11
These attributes are:
 Power limit decreases with line length
 Short lines are limited mainly by thermal
problems.
 Medium length lines tend to be limited by
voltage-related problems.
 Very long lines tend to be limited by
stability problems.
5.0 Complex power expression
This material combines Sec. 4.6, 4.8, 4.9.
Consider the long transmission line of Fig.
4. The voltages at the ends are specified as:
V1  V1 e j1
V2  V2 e j 2
(19)
The series impedance is
Z   Z  e j Z 
(20)
IZ
I1
I2
Z’
IY1
Y’/2
IY2
Y’/2
V1
Fig. 4
12
V2
ZL
The complex power transferred into the line
from bus 1 is:
2
*
*
S12  V1 I Z  V1 Y  / 2
(21)
But IZ can be expressed as:
V1  V2
IZ 
(22)
Z
Substitution of (22) into (21) yields:
*
V

V
2
 1
*
2 
S12  V1 
  V1 Y  / 2
(23)
 Z 
Distributing the V1 through, we obtain:
V1V1*  V1V2*
2
*



S12 

V
Y
/
2
1
Z *
V1
V1 V2 e j 1 2
2
* (24)



 * 

V
Y
/
2
1
*


Z
Z
where θ12= θ1- θ2. Eq (24) is the same as eq.
4.29, pg. 104 in the text except eq. (24) also
includes the effect of the sending-end linecharging, as represented by the last term.
2
13
6.0 Circle diagrams
If we ignore the last term in eq. (24), which
means assuming no charging capacitance,
and if we assume Z’=Z, i.e., for the “shortline model,” eq. (24) becomes:
2
V1
V1 V2 e
S12  * 
Z
Z*
j12
Using Z*=|Z|e-j∟Z, we get
2
V1 jZ V1 V2 jZ j
S12 
e 
e e (24a)
|Z |
|Z |
12
We may repeat this same process to obtain
power from the other direction as
2
V2 jZ V1 V2 jZ  j
S 21 
e 
e e
|Z |
|Z |
12
The power flowing into the receiving bus
from the line is then just –S21, or
2
V2 jZ V1 V2 jZ  j
 S 21  
e 
e e (24b)
|Z |
|Z |
12
14
We can then define the following:
2
V1 jZ
C1 
e
|Z|
2
V2 jZ
C2  
e
|Z|
V1 V2 jZ
B
e
|Z|
which are all constants if we assume the
voltages are fixed.
Then eqs. (24a) and (24b) become:
S12  C1  Be j
(24c)
12
 S 21  C2  Be
 j 12
(24d)
Eqs. (24c) and (24d) characterize sending
end and receiving end complex power as a
function of the angle between the buses, θ12.
C1 and C2 are the centers of circles having
radii |B|=|V1||V2|/|Z|.
Figure 4.7 illustrates circle diagrams. The
text makes 6 observations about these that I
encourage you to review, among which are:
#5: Transmission is strengthened by
increasing
voltages
(generator
field
windings) and decreasing reactance (series
compensation)
15
#6: Strong coupling between P-flow and θ12
and between Q-flow and |V|.
7.0 An alternative development
We want simple expressions for P12 and Q12.
Recall eq. (3) above, repeated here:
sinh l
Z '  Z C sinh l  Z
(3)
l
In the lossless case, γ=jβ, so eq. (3) is
e jl  e  jl
Z '  Z C sinh j l  Z C
2
cos  l  j sin  l  cos  l  j sin  l
 ZC
2
 Z   jZ C sin  l
(25)
Likewise, we can derive, in lossless case:
tan(  l / 2)
Y '  j C
(26)
l / 2
Here, C is total line capacitance. Eq. (25),
(26) are same as (4.46), (4.45) in text, p.
115, and we see that Z’ and Y’ are pure
reactances and susceptances, respectively,
i.e., Z’=jX and Y’=jB.
16
Conjugating eq. (25) and (26), we get:
Z *   jZ C sin  l
(27)
tan(  l / 2)
Y '   j C
l / 2
*
(28)
Recall eq. (24):
2
V1
V1 V2 e j
2
*

S12  * 
 V1 Y / 2 (24)
*
Z
Z
Substitution of eqs. (27), (28) into the
expression for S12, eq. (24), we get:
12
V1
V1 V2 e j 12
2 tan(  l / 2)
S12 

 jC V1
(29)
 jZ C sin  l  jZ C sin  l
l
2
Bringing the j of the first term into the
numerator and canceling the two negative
signs of the second term results in:
j V1
V1 V2 e j 12
2 tan(  l / 2)
S12 

 jC V1
(30)
ZC sin  l jZ C sin  l
l
2
8.0 Real power
The first and last terms of eq. (30) are purely
imaginary and so do not affect real power.
Let’s look more closely at the second term:
17
V1 V2 e j 12 V1 V2 e j 12 e  j 90 V1 V2 e j ( 12  90)


jZ C sin  l
Z C sin  l
Z C sin  l

V1 V2 cos(12  90)  j sin(12  90)
Z C sin  l
V1 V2 sin12  j cos12 

Z C sin  l
(31)
Of this term, we see that the real part is the
sinθ12 term. Since this is the only real part of
eq. (30), it must be true that:
P12 
V1 V2 sin12
ZC sin  l
(32)
It is generally the case that voltage
magnitudes at either end of a line are not
very different (and when the line is lossless
and terminated in ZC, they are
exactly the same – see eq. (14) above). If we
assume that |V2|=|V1|, then:
V sin12
P12  1
ZC sin  l
2
(33)
Recalling eq. (17), repeated here:
2
PSIL
V1

ZC
(17)
18
we see that eq. (33) can be re-written as:
P12  PSIL
sin12
sin  l
(34)
Although approximate, eq. (34) is very
useful for getting a “back-of-the-envelope”
sense of what voltages are required in order
to accommodate a given power transfer
level, as observed in problem 4.21 of the
homework assignment. If you take a
position as a transmission planner, you
might find this expression handy.
Problem 4.21:
Transient instability problems can occur
when the angular separation between
voltage phasors at either end of a
transmission line, denoted by θ12=θ1-θ2
becomes large.
So when does θ12 become large?
19
Note that in eq. (34) that PSIL and sinβl are
constants for an already-constructed
transmission line.
P12  PSIL
sin12
sin  l
(34)
So θ12 directly determines (or is directly
determined by) P12.
As θ12 increases from 0° to 90°, the real
power transfer P12 gets larger. Or, more
properly, we can see that as P12 gets larger,
θ12 increases from 0° to 90°.
So answer to question of When does θ12
become large? is: when P12 becomes large.
Problem 4.21 indicates that a practical limit
for transmission lines on θ12 is 45° (actually,
this is probably too large, a 30° limit is
better, but we will use 45°).
It also indicates that the desired transfer
level on a 300 mile-long line is 500 MW.
20
Question is: what voltage levels can we
consider?
This is a design (planning) problem.
The problem also indicates β=0.002/mile.
Solving eq. (34) for PSIL yields:
PSIL  P12
sin  l
sin( 0.002  300)
 500
 400 MW
sin 12
sin 45
This is 3-phase power. Referring to Table 1
above, we see we need voltage level of at
least 345 kV to accommodate this.
9.0 Reactive power
Consider again eq. (30), repeated here:
j V1
V1 V2 e j 12
2 tan(  l / 2)
S12 

 jC V1
(30)
ZC sin  l jZ C sin  l
l
2
Now let’s consider the imaginary part. In
doing so, we need to remember that the
middle term contributes a real part, but it
21
also contributes an imaginary part, as
indicated by eq. (31), repeated here:
V1 V2 e j 12 V1 V2 sin12  j cos12 

jZ C sin  l
Z C sin  l
(31)
Substituting eq. (31) into eq. (30), we have:
j V1
V V sin12  j cos12 
2 tan(  l / 2)
S12 
 1 2
 jC V1
ZC sin  l
ZC sin  l
l
2
(35)
Now taking only the imaginary part of (35):
V1
V V cos 12
2 tan(  l / 2)
Q12 
 1 2
 C V1
Z C sin  l
Z C sin  l
l
2
(36)
Recall that the denominator of the first two
terms is just |Z’| in the lossless case (see eq.
(25) above). Therefore, we can write:
V
V V cos  12
2 tan(  l / 2)
Q12  1  1 2
 C V1
Z'
Z
l
2
(37)
Now let’s consider eq. (37) for the short line
(and lossless) model. In this case,
 Z’≈Z, and for lossless, |Z|=X
 C≈0.
So eq. (37) becomes:
22
V1
V1 V2 cos  12
Q12 

X
X
2
(38)
This should be familiar expression to you
from EE 303.
10.0
Voltage instability (Section 4.7)
We can also write the real power equation
(32) for the short line model, as we have just
done for the reactive power equation, using
Z’=Z and |Z|=X, to get:
V1 V2 sin12 V1 V2 sin12
P12 

Z C sin  l
X
(39)
Section 4.7 applies eqs. (38) and (39) to
consider how the receiving end voltage
magnitude varies with load.
Assuming the (normalized) sending-end
voltage is 1.0, the text derives:
V2
2

1  PD  1  PD ( PD  2 )
2
where
23
(40)
|V2| is the normalized voltage at the
receiving end,
 PD is the receiving end real power load
 β=tanφ, where pf=cosφ (note β here is
different from our previous use of it).
Equation (40) can be used to plot receiving
end voltage |V2| as a function of receiving
end demand PD.

Fig. 4
24
Fig. 4 is similar to Fig. E4.9 in the text. This
figure is a very basic figure for transmission
planners and operators, who constantly
worry about the implications of this figure.
Why?
[1] R. Burington, “Handbook of Mathematical Tables and
Formulas,” 5th Edition, McGraw-Hill.
[2] O. Elgerd, “Electric Energy Systems Theory: An Introduction,”
2nd edition, McGraw-Hill, 1982.
[3] J. Glover and M. Sarma, “Power System Analysis and Design,”
PWS Publishers, 1987.
[4] “Transmission Line Reference Book: 345 kV and Above,”
Electric Power Research Institute, 2nd Edition, 1987.
25