El-Hawary M.E. Electrical Energy Systems
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Figure 5.23 A Cascade of Two two-Port Networks.
=
M
M
s
s
I
V
DC
BA
I
V
11
11
,
=
r
r
M
M
I
V
DC
BA
I
V
22
22
From which, eliminating (V
M
, I
M
), we obtain
=
r
r
s
s
I
V
DC
BA
DC
BA
I
V
22
22
11
11
Thus the equivalent ABCD parameters of the cascade are
2121
CBAAA
+=
(5.116)
2121
DBBAB
+=
(5.117)
2121
CDACC
+=
(5.118)
2121
DDBCD
+=
(5.119)
If three networks or more are cascaded, the equivalent ABCD parameters can be
obtained most easily by matrix multiplications as was done above.
5.6 TRANSMISSION LINE MODELS
The line parameters discussed in the preceding sections were obtained
on a per-phase, per unit length basis. We are interested in the performance of
lines with arbitrary length, say l. To be exact, one must take an infinite number
of incremental lines, each with a differential length. Figure 5.24 shows the line
with details of one incremental portion (dx) at a distance (x) from the receiving
end.
The assumptions used in subsequent analyses are:
1. The line is operating under sinusoidal, balanced, steady-state
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Figure 5.24 Incremental Length of the Transmission Line.
conditions.
2. The line is transposed.
With these assumptions, we analyze the line on a per phase basis. Application
of Kirchhoff’s voltage and current relations yields
xyxVI
xzxIV
∆=∆
∆=∆
)(
)(
Let us introduce the propagation constant
υ
defined as
zy
=
υ
(5.120)
The series impedance per-unit length is defined by
LjRz
ω
+= (5.121)
The shunt admittance per-unit length is defined by
CjGy
ω
+=
(5.122)
R and L are series resistance and inductance per unit length, and G and C are
shunt conductance and capacitance to neutral per unit length.
In the limit, as 0
→∆
x , we can show that
V
dx
Vd
2
2
2
υ
=
(5.123)
I
dx
Id
2
2
2
υ
= (5.124)
Equation (5.123) can be solved as an ordinary differential equation in
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V. The solution turns out to be
)exp()exp()(
21
xAxAxV
υυ
−+=
(5.125)
Now taking the derivative of V with respect to x to obtain I(x) as
c
Z
xAxA
xI
)exp()exp(
)(
21
υυ
−−
=
(5.126)
Here we introduce
y
z
Z
c
=
(5.127)
Z
c
is the characteristic (wave) impedance of the line.
The constants A
1
and A
2
may be evaluated in terms of the initial
conditions at x = 0 (the receiving end). Thus we have
21
21
)0(
)0(
AAIZ
AAV
c
−=
+=
from which we can write
[][]
{}
)exp()0()0()exp()0()0(
2
1
)( xIZVxIZVxV
cc
υυ
−−++= (5.128)
−
−+
+=
)exp(
)0(
)0()exp(
)0(
)0(
2
1
)( x
Z
V
Ix
Z
V
IxI
cc
υυ
(5.129)
Equations (5.128) and (5.129) can be used for calculating the voltage
and current at any distance x from the receiving end along the line. A more
convenient form of these equations is found by using hyperbolic functions.
We recall that
2
)exp()exp(
cosh
2
)exp()exp(
sinh
θθ
θ
θθ
θ
−+
=
−−
=
By rearranging Eqs. (5.128) and (5.129) and substituting the hyperbolic function
for the exponential terms, a new set of equations is found. These are
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© 2000 CRC Press LLC
xIZxVxV
c
υυ
sinh)0(cosh)0()(
+=
(5.130)
and
x
Z
V
xIxI
c
υυ
sinh
)0(
cosh)0()(
+=
(5.131)
We define the following ABCD parameters:
xxA
υ
cosh)(
=
(5.132)
xZxB
c
υ
sinh)(
=
(5.133)
x
Z
xC
c
υ
sinh
1
)(
=
(5.134)
xxD
υ
cosh)(
=
(5.135)
As a result, we have
)0()()0()()(
)0()()0()()(
IxDVxCxI
IxBVxAxV
+=
+=
For evaluation of the voltage and current at the sending end x = l, it is
common to write
)0(
)0(
)(
)(
II
VV
lII
lVV
r
r
s
s
=
=
=
=
Thus we have
rrs
BIAVV
+=
(5.136)
rrs
DICVI
+=
(5.137)
The subscripts s and r stand for sending and receiving values, respectively. We
have from above:
llAA
υ
cosh)(
==
(5.138)
lZlBB
c
υ
sinh)(
==
(5.139)
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l
Z
lCC
c
υ
sinh
1
)(
==
(5.140)
llDD
υ
cosh)(
==
(5.141)
It is practical to introduce the complex variable
θ
in the definition of
the ABCD parameters. We define
ZYl
==
υθ
(5.142)
As a result,
θ
cosh
=
A
(5.143)
θ
sinh
c
ZB
=
(5.144)
θ
sinh
1
c
Z
C
=
(5.145)
AD
=
(5.146)
Observe that the total line series impedance and admittance are given by
zlZ
=
(5.147)
ylY
=
(5.148)
Evaluating ABCD Parameters
Two methods can be employed to calculate the ABCD parameters of a
transmission line exactly. Both assume that
θ
is calculated in the rectangular
form
21
θθθ
j
+=
The first method proceeds by expanding the hyperbolic functions as follows:
()
()
22
22
11
11
2
1
2
sinh
2
1
2
θθ
θ
θθ
θθ
θθ
θθ
θθ
−∠−∠=
−
=
−∠+∠=
+
=
−
−
−
−
ee
ee
ee
ee
A
(5.149)
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θ
sinh
Y
Z
B
=
(5.150)
θ
sinh
Z
Y
C
=
(5.151)
Note that
θ
2
is in radians to start with in the decomposition of
θ
.
The second method uses two well-known identities to arrive at the
parameter of interest.
)cosh(
21
θθ
jA
+=
2121
sinsinhcoscoshcosh
θθθθθ
j
+=
(5.152)
We also have
2121
sincoshcossinhsinh
θθθθθ
j
+=
(5.153)
Example 5.9
Find the exact ABCD parameters for a 235.92-mile long, 735-kV, bundle-
conductor line with four subconductors per phase with subconductor resistance
of 0.1004 ohms per mile. Assume that the series inductive reactance per phase
is 0.5541 ohms per mile and shunt capacitive susceptance of 7.4722
×
10
-6
siemens per mile to neutral. Neglect shunt conductance.
Solution
The resistance per phase is
ohms/mile0251.0
4
1004.0
==
r
Thus the series impedance in ohms per mile is
ohms/mile5541.00251.0 jz
+=
The shunt admittance is
lesiemens/mi104722.7
6
−
×=
jy
For the line length,
$
$
90107628.1)92.235)(104722.7(
41.8786.130)92.235)(5541.00251.0(
36
∠×=×==
∠=+==
−−
jylY
jzlZ
We calculate
θ
as
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© 2000 CRC Press LLC
()( )
[]
4802.00109.0
90107628.141.8786.130
21
3
j
ZY
+=
∠×∠=
=
−
$$
θ
Thus,
4802.0
0109.0
2
1
=
=
θ
θ
We change
θ
2
to degrees. Therefore,
$
5117.27
180
)4802.0(
2
=
=
π
θ
Using Eq. (5.149), we then get
()
$
$$
3242.08870.0
5117.275117.27
2
1
cosh
0109.00109.0
∠=
−∠+∠=
−
ee
θ
From the above,
$
3242.08870.0
∠==
AD
We now calculate sinh
θ
as
()
$
$$
8033.884621.0
5117.275117.27
2
1
sinh
0109.00109.0
∠=
−∠−∠=
−
ee
θ
We have
()
$
$
$
295.146.272
59.217.74234
90107628.1
41.8786.130
21
21
3
−∠=
−∠=
∠×
∠
==
−
Y
Z
Z
c
As a result,
$
508.87904.125
sinh
∠=
=
θ
c
ZB
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Also,
$
098.9010696.1
sinh
1
3
∠×=
=
−
θ
c
Z
C
Let us employ the second method to evaluate the parameters. We find
the hyperbolic functions:
2
0109.00109.0
1
0109.00109.0
1
1009002.1
2
sinh
000059.1
2
)0109.0cosh(cosh
−
−
−
×=
−
=
=
+
=
=
ee
ee
θ
θ
(Most calculators have build-in hyperbolic functions, so you can skip the
intermediate steps). We also have
4619566.0
180
)4802.0(sinsin
8869.0
180
)4802.0(coscos
2
2
=
=
=
=
π
θ
π
θ
Therefore, we have
$
$
801.884620851.0
)4619566.0)(000059.1()8869.0)(1009002.1(
sincoshcossinhsinh
32527.08869695.0
)4619566.0)(1009002.1()8869.0)(000059.1(
sinsinhcoscoshcosh
2
2121
2
2121
∠=
+×=
+=
∠=
×+=
+=
−
−
j
j
j
j
θθθθθ
θθθθθ
These results agree with the ones obtained using the first method.
Example 5.10
Find the voltage, current, and power at the sending end of the line of Example
5.9 and the transmission efficiency given that the receiving-end load is 1500
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© 2000 CRC Press LLC
MVA at 700 kV with 0.95 PF lagging.
Solution
We have the apparent power given by
VA101500
6
×=
r
S
The voltage to neutral is
V
3
10700
3
×
=
r
V
Therefore,
A19.1818.1237
95.0cos
3
10700
3
101500
1
3
6
$
−∠=
−∠
×
×
=
−
r
I
From Example 5.9 we have the values of the A, B, and C parameters. Thus the
sending-end voltage (to neutral) is obtained as
()
()( )
kV66.190938.439
19.1818.1237508.87904.125
3
10700
3253.08870.0
3
$
$$
$
∠=
−∠∠+
×
∠=
+=
rrs
BIAVV
The line-to-line value is obtained by multiplying the above value by
3
, giving
kV533.760
=
L
S
V
The sending-end current is obtained as
()
()( )
$
$$
$
49.1805.1100
19.1818.12373253.0887.0
3
10700
098.9010696.1
3
3
∠=
−∠∠+
×
∠×=
+=
−
rrs
DICVI
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The sending-end power factor is
99979.0)17.1cos(
)49.1866.19cos(cos
==
−=
s
φ
As a result, the sending-end power is
MW1077.1448
)99979.0)(05.1100)(100938.439(3
6
3
×=
×=
s
P
The efficiency is
9836.0
1077.1448
95.0101500
6
6
=
×
××
=
=
s
r
P
P
η
Lumped Parameter Transmission Line Models
Lumped parameter representations of transmission lines are needed for
further analysis of interconnected electric power systems. Their use enables the
development of simpler algorithms for the solution of complex networks that
involve transmission lines.
Here we are interested in obtaining values of the circuit elements of a
π
circuit, to represent accurately the terminal characteristics of the line given by
rrs
rrs
DICVI
BIAVV
+=
+=
It is easy to verify that the elements of the equivalent circuit are given in terms
of the ABCD parameters of the line by
BZ
=
π
(5.154)
and
B
A
Y
1
−
=
π
(5.155)
The circuit is shown in Figure 5.25.