Al-Balqa Applied
University
1
Dr.Audih
In previous lectures have covered how to calculate the parameters
inductance, capacitance and resistance of transmission lines.
In this section we will use these parameters to develop the
transmission line models used in power system analysis.
 Transmission line take power from generators, transmit it to
location where it will be used ( loads center).
 The power capability of a transmission line is proportional
to the square of the voltage on the line.
Dr.Audih
2
Therefore, very high voltage levels are used to transmit
power over long distances.
 Once the power reaches the area where it will be used, it
is stepped down to a lower voltages in distribution
substations, and then delivered to customers through
distribution lines.
 An overhead transmission line usually consists of three
conductors or bundles of conductors containing the three
phases of the power system.
 The conductors are usually aluminum cable steel
reinforced (ACSR), which are steel core (for strength)
and aluminum wires .
Dr.Audih
3
1-Short length model : Overhead transmission lines(OHL) shorter than
80 km can be modeled as a
series
resistance
and
inductance, since the shunt
capacitance can be neglected
over short distances.
2- Medium length model: OHL between (80-240 km), can be modeled
as shunt capacitance should
be
considered
by
two
capacitors of a half at both
ends of the line.
3- Long length model: OHL longer than 240 km are long transmission
lines and will be discussed later ,
Dr.Audih
4
In the single line equivalent circuit of
a short line ,the current sending
IS and current received IR are :
IS  IR
Neglected the shunt admittance and using KVL , the relation
between sending end voltage VS and receiving end voltages
VR is:
VS  VR  ZI  VR   R  jX L  I
VR  VS   R  j L  I
which is very similar to the equation derived for a synchronous generator.
Dr.Audih
5
1- Phasors Diagrams of TL:
Load with lagging power factor.
Load with unity power factor.
Load with leading power factor.
Dr.Audih
6
2- Voltage regulation of a transmission line:
It’s define as the
percentage change in voltage at the receiving end of the transmission line in
case of loads and no loads.
VR% 
VNL  VFL
VFL
VS
 VR
A
100% 
100=
VR
where :VNL and VFL are the no-load and full-load voltages at the line output and
constant A=1 for short module( will discuss later).
 If the load at terminals of the line is lagging (inductive) , the voltage at the
end of the transmission line decreases ( large positive VR).
 If unity-PF (resistive) loads, the voltage at the end of the transmission line
decreases slightly (small positive VR).
 If the load at terminals of the line is leading (capacitive) , the voltage at the
end of the transmission line increases (negative VR).
Dr.Audih
7
3. Power flow in a transmission line:
The real ,reactive and apparent power input and output of a 3-phase transmission
line can be computed as:
Psending  3VLL , S I S cos  S
Precived  3VLL , R I R cos  R
Qsending  3VLL , S I S sin  S
QRecived  3VLL , R I R sin  R
S sending  3VLL , S I S
4. Transmission line efficiency:
Srecived  3VLL, R I R
Pout
Pout
  100%
100%
Pin
Pout P
Dr.Audih
8
5-Maximum power transfer: Consider the following diagram
IR
IS
+
VS
Ssend
-
+
Transmission
Line with
Impedance Z
-
*
VR
*
 V 
 VS  VR 
*
S sen d  V S I S  V S 
  VS 

Z
Z




w ith V S  V S   , V R  V R  0 o , Z  Z   Z
o
S sen d
*
 VS    VR  0 
 VS   
 
Z Z


VS

Z
2
VS VR
      Z  
  0o   Z
Dr.Audih
Z
9
S send 
VS
2
Z
VS VR
 Z 
Z
If w e assum e a line im pedance is jX
L
   Z
and
just interested in real
pow er transfer then  = 90 0 ,thuse :
Psend  jQ send

2
VS
X
 90  
L
VS VR
X
  o  90 
L
M athm atically ,w e get :
2
VS
P

sending
X
cos 90  
L
VS VR
X
cos( o  90 ) 
L
Dr.Audih
V
S
X
V
R sin  o
L
10
Hence the maximum power transfer is meet when   90 o , then :
Psend Max 
VS V R
XL
and for 3 system
Psend Max  3
VS V R
XL
The power supplied by a transmission line depends on the angle(δ) between the
phasors (input and output voltages). The maximum power is called steady-state
stability limit of the transmission line.
The maximum power of a transmission line is inversely proportional to its series
reactance, which may be a serious problem for long transmission lines.
Some very long lines include series capacitors to reduce the total series
reactance and thus increase the total power capability of the line.
*Note: The angle  controls the power flowing through the line. It is possible to
control power flow by placing a phase-shifting transformer at one end of the line
and varying voltage phase.
Dr.Audih
11
6. Transmission line limits:
a) Power losses : Is the main limiting factors in transmission line
operation (since its resistivity). The maximum steady-state current
must be limited to prevent the overheating of the transmission line.
Ploss  3I L2 R
b) Voltage drop: Its the ratio of the magnitude of the receiving end
voltage to the magnitude of the sending end voltage ,should be
limited to approximately 5%,
VR
VS
 0.95
c) Stability limits: The angle  in a transmission line should typically
be  300 to the static stability limit .
Dr.Audih
12
Example:
An over-head 3-phase transmission line delivers 5 MW at 22 kV at 0·8 p.f.
lagging. The resistance and reactance of each conductor is 4Ω and 6Ω
respectively. Determine :
(i)sending end voltage (ii)percentage regulation (iii)transmission efficiency.
Solution
Load power factor cos  0.8 lagging    36.869o
Receiving end voltage per phase VR1 
22 103
 12.7 kV
3
impedance per phase Z  R  jX  4  j6  7.21156.3o 
5 106
o
Line current I 

164

36.86
A
3
3 22 10  0.8
Taking VR0o of single phase as the reference phasor
VR = 12.70o kV
Dr.Audih
13
(i) Sending end voltage per phase is
VS (1 )  VR(1 ) + I.Z=127000o + 164  36.86o    7.21156.3o   =
=13.8211.63o kV
VS (3 )  3 13.821  23.938 kV
(ii) % age Regulation =
**
VS  VR
VR
13.821 12.700
100=
100  8.83 %
12.700
note : the %VR equation here is valid for short module only since A  1
.
(iii) Line losses = 3R.I2 = 3 4  (164)2  322 752kW
5000
Transmission efficiency =
100  93.94%
5000  322 752
.
Dr.Audih
14
Example:
A 3-phase line delivers 3600 kW at a p.f. 0·8 lagging to a load. If the
sending end voltage is 33 kV, determine (i) the receiving end voltage
(ii) line current (iii) transmission efficiency.
The resistance and reactance of each conductor are 5·31Ω and 5·54 Ω
respectively.
Solution.
Resistance of each conductor, R  5·31 
Reactance of each conductor, X L  5·54 
Load power factor, cos  0·8 (lagging)
Sending end voltage per phase, VS  33/ 3  19.052 kV
Let VR be the phase voltage at the receiving end.
P3
3600
power delivered per phase 

 1200 kW
3
3
Dr.Audih
15
I per phase
1200 103 15 105



VR(1 ) .cos 
VR  0.8
VR
P(1 )
(i) Using approximation expression for VS , we get;
15 105
15 105
VS =VR  IR cos   IX L sin   VR 
 5.31 0.8 
 5.54  0.6 
VR
VR
 VR2  19,052VR  1,13,58,000  0  VR  18.435kV
VS ( LL )  3 18.435  31.93kV
15 105
(ii)line current I 
 81.36 A
18, 435
(iii)line losses  3RI 2  3 5.31 (81.36)2  105.447kW
Pout
3600
The efficiency  
100 
100  97%
Pout  Plosses
3600  105.447
Dr.Audih
16
Example:
220kV ,60 Hz ,three phase transmission line with 40km long and
R=0.15Ω/km/phase , L=1.3263mH/km/phase, if C is negligible ,find
voltage sending ,voltage regulation and efficiency when supplying
a load of:
a) 381MVA at 0.8 lagging at 220kV.
b) 381MVA at 0.8 leading at 220kV.
Solution:
The total impedance per km length is :
Z  z.l  (r  j L).l  (0.15  j 2  60 1.3263 103 )  40  6  j 20 
VR (3 ) 2200o
VR (1 ) 

 1270o kV
3
3
S R  381 cos1 (0.8)  38136.87o  304.8  j 228.6 MVA
*
 S R  381106   36.87o
o
IR  


1


36.87
kA

3
o
3 127 10 0
Dr.Audih
 3VR 
17
VS (1 )  VR  Z .I R  1270o  (20.8873.3o 1  36.87o ) 
 144.334.93o kV (note is greater than VR )
VS (3 )  3 VS (1 )  3 144.33  250 kVLL
SS (3 )  3.Ss (1 )  3.Vs I s*  3 144.334.93o 136.87o 
 322.8MW  j 288.6MVAr  43341.8o MVA
VR% 
VSLL  VRLL
VRLL
100 
250  220
220
100  13.6%
Pout
PR
304.8

100%  100% 
100%  94.4%
Pinp
PS
322.8
Dr.Audih
18
b) for leading P.F .
I  136.87 o kA
VS (1 )  VR  Z .I R  1270o  (20.8873.3o 136.87 o ) 
 121.399.29o kV (note is less than VR )
VS (3 )  3 VS (1 )  3 121.39  210.26 kV
S S (3 )  3.S s (1 )  3.Vs I s*  3  121.399.29o  1  36.87 o 
 322.8MW  j168.6 MVAr  364.17  27.58o MVA
210.26  220
VR % 
 100  4.43%
220
(negative)
PR
304.8
  100% 
100%  94.4%
PS
322.8
Dr.Audih
19
A transmission line can be represented by a two-port as shown. If the
network is linear, the relationship between the sending and receiving end
voltages and currents are:
VS  AVR  BI R
I S  CVR  DI R
Vs   A B VR 
as matrix    
. 

Is  C D IR 
ABCD constants are sometimes called TL constants .
- A represents the effect of a change in the sending end voltage
per the receiving end voltage.
- B represents the effect of a change in the sending end
Dr.Audih
voltage per the receiving end current
20
-
C denotes the effect of a change in the sending end
current per the receiving end voltage.
D is the effect of a change in the sending end current
per the receiving end current.
VS
 VR
A
VR % 
 100
VR
For short model IS = IR = I, and the ABCD constants are
A=1
B=Z
VS  VR  ZI  AVR  BI R
I S  I R  CVR  DI R
C=0
D=1
Where:
A and D are dimensionless, B has
units of , and C is measured in
Siemens.
Dr.Audih
21
Considering medium-length lines (80 to 240km long), the shunt
admittance must be included in calculations. However, the total
admittance is usually modeled ( model) as two capacitors of equal values
(each is a half of total admittance) placed at the sending and receiving
ends.
The current of the receiving end capacitor (IC2) is:
The current of the series impedance elements is
I ser  I C 2  I R  V R
Y
 IR
Dr.Audih
2
22
From the Kirchhoff’s voltage law, the sending end voltage is
VS  V VR  ZIser VR  Z  IC2  IR  VR 
ZY
 Y

 YZ 
 Z VR  IR  VR  VR
 ZIR VR   1VR  ZIR
2
 2

2 
The source of current will be
Y
Y
IS  IC1  Iser  IC1   IC2  IR   VS VR  IR 
2
2
 YZ 
Y
Y
YZ Y
Y
Y
Y
  1VR  ZIR  VR  IR  VR . VR.  ZIR. VR  IR
2
2 2
2
2
2
 2 
2
 Y 2Z
Y
Y 
Y
  ZY 
 ZY 
VR. VR    ZIR.  IR   Y  1VR   1 IR
VR
4
2
2 
2
  4 
 2 

Dr.Audih
23
Therefore, the ABCD constants of a medium-length π of transmission
line are:
ZY
A  D 
1
2
B  Z
 ZY

C Y 
 1
 4

VS  AVR  BIR
IS  CVR  DIR
 ZY
1

Vs 
2
 I     ZY 
 s  Y
1

  4


Z 
VR 
. 
ZY
IR 


1

2
If the shunt capacitance of the line is ignored (Y=0) , the ABCD
constants are the constants for a short transmission line.
Dr.Audih
24
Example 1: (general case of TL)
The ABCD constant of a three phase short length of TL are:
A  D  0.936  j0.016  0.9360.98o
o
B  33.5  j138  14276.4 
6
C  (5.18  j914) 10 S
The load at the receiving end is 50MW at 220kV with a power
factor of 0.9 lagging. Find the magnitude of the sending end voltage
and the voltage regulation. Assume the magnitude of the sending
end voltage remains constant.
25
Dr.Audih
Solution:
P

3.V LL .cos 
IR 
50  10 6
 145.8 [A]
3
3  220  10  0.9
cos  1 (0.9)  25.84 o and since its lagging then the angle is negative,thus;
I R  145.8   25.84 o
[ A]
V LL 220

 127  0 o kV and
3
3
 AV R (1 )  BI R (1 )  (0.936  0.98 o  (127  10 3  0 o ) 
V R (1 ) 
V s (1 )
 (142  76.4 o  145.8   25.84 o )  133.23 7.77 o k V
V S LL 
V R ( nL ) 
3  133.23  230.8 kV
V S LL
A

230.8
 246.5 kV LL
0.9361
(% ) regulation 
V R ( nL )  V R (FL)
V R (FL )
 100% 
246.5  220
 100%  12%
220
26
Dr.Audih
Example 2:
Find the ABCD constants of a π circuit having a 600Ω resistor for
shunt branch al sending end, a 1kΩ resistor for the shunt branch at
the receiving end and an 80Ω resistor for series branch.
Solution:
V s  V R   V  V R  (80  I L )
from KCL
I L  I R  I (1000  )
VR
 IR  (
)
1000 
27
Dr.Audih
V s  V R  (80  I L )  V R  (80  ( I R  (
 V R  80 I R 
VR
)) 
1000 
80V R
 V R  80 I R  0.08V R  1.08V R  80 I R
1000 
 V s  1.08V R  80 I R
and since
A  1.08 and
V s  AV R  BI R then ;
B  80
And for C and D from KCL Is  IL  I 600
VR
Vs
VR
1.08VR  80IR
Is  IL  I600  (IR 
)  ( )  (I R 
) (
)
1000 600
1000
600
IR  0.001VR  0.0018VR  0.133IR  (0.001 0.0018)VR  (1 0.133)IR
And since Is  CVR  DIR  0.0028VR 1.133IR 
C  0.0028 and D1.133
28
Dr.Audih
Example 3:
Determine the voltage regulation of 3-phase,150km long,50Hz
transmission line delivering 20MW at p.f. of 0.8 lagging and 66kV to at
balance load. Resistance of the line is 0.075Ω/km,1.5cm outside diameter,
placed equilaterally 2 meters between center. Use normal π method.
Solution :
R  r .l  0.075  150  11.25 
d 1.5

 0.75 cm
2
2
D  2  10 2  200 cm
radius of the condctor is r 
D istance
D
T he inductance is L  2  10  7 . ln    l 
 r 
 200 cm 
7
3
 2  10 . ln 
  (150  10 ) m  0.1675 H enry
 0.75 cm 
X L  2  f . L  2  50  0.1675  52 .62 
Dr.Audih
29
2 o
2  8.854 1012 F / m
Now Cphase 
.l 
 (150 103 )m  1.49 F
 D
 200 cm 
ln  
ln 

r
 
0.75
cm


Y  jC  j 2 fC  j 2  50 1.49 106  j 468.1106 mho
Y
 234 106 90o mho
2
Z  R  jX L  11.25  j52.62  53.80977.9o 
ZY
Now
 0.01259167.9o  (0.0123  j0.00264)
2
and B  Z then;
Dr.Audih
30
P
20106  cos1(0.8)
o
I


218.7

36.87
A
3
3V.cos
3  6610  0.8
66000
And VR 
 38104 V
3
 ZY 
VS  AVR  BIR  1 VR  ZIR 
2 

 1 0.01259167.9o   381050o   53.80977.9o  218.7 36.87o  
 47.1679.554o kV or
VS (3 )  3  47.15  81.696kV
VR% 
VR(nL)  VR(FL)
VR(FL)
.100 
VS
 VR
A
VR

81.696
 66
0.9877
66
%  25.3%
Dr.Audih
31
Example 4:
A three-phase,60-Hz,completely transposed 345-kV,200-km
,and the following .positive-sequence line constants:
z  0.032  j 0.35  / km
y  j 4.2 10
6
S / km
Full load at the receiving end of the line is 700 MW at 0.99
p.f. leading and at 95% of rated voltage. Assuming a
medium-length line, determine the following:
a. ABCD parameters of the nominal π circuit
b. Sending-end voltage VS, current IS, and real power PS
c. Percent voltage regulation
d. Transmission-line efficiency at full load
Dr.Audih
32
Solution :
a)The total series impedance and shunt admittance values are:
Z = z.l  (0.032  j0.35)  (200)  70.2984.78o 
Y  y.l  j4.2 106  200  8.4 104 90o S
 70.2984.78o  8.4 104 90o 
ZY
A  D  1  1 


2
2


 0.97060.159o
B  Z  70.2984.78o
4
o
o


8.4

10

90

70.29

84.78
 ZY 
4
o
C  Y  1 8.4 10 90 
1 
4
 4



 8.277 104 90.08o
Dr.Audih
33
b ) sending end voltage :
V R  345  95%(from rated voltage)  327.8kV LL
V R (3 )
327.8
V R (1 ) 

 189.20o kV Ln
3
3
3
P
700

10
IR 
. cos1  
 cos1 (0.99) 
3.V LL .cos 
3  327.8  0.99
 1.245 8.110 kA (leading )
The sending quantities are
o
o

V S (1 )  AV R  BI R   0.97060.159   189.20   
 70.2984.78o   1.2458.110    199.526.15o kV Ln


V S (3 )  3 199.5  345.5 kV LL
Dr.Audih
34
I S  CVR  DI R  (8.277  10 4  90.08 o  189.2  0 o ) 
 (0.9706  0.159 o  1.2458.110 )  1.240 o 15.447 o kA
and the real power delivered to the sending end is :
Ps  3VLL , S I S cos( v   i ) 

i 
v
0
o
 3  345.5  1.240 cos(26.15 o  15.447 o )  729.77MW
Dr.Audih
35
(c) voltage regulation
VR % 
VR ( nl )  VR ( Fl )
V
R ( Fl )

VS
 VR ( Fl )
A
 100 
 100 
VR ( Fl ) 
345.8
 327.8
0.9706

 100  8.6%
327.8
(d )
Pout
700

100% 
 100%  96%
Pin
729.77
Plosses  Pin  Pout  729.77  700  30MW
Dr.Audih
36
inverse matrix :
If we need to find VR and IR
1
VR  A B 
then inverse matrix is used    

IR  C D 
 D 
 AD CB 




  C 


  AD CB 
Vs 
.  
Is 
 B  

 V
 AD CB    s   D B Vs 
.   
. 

 A   Is  C A  Is 


 AD CB  
Dr.Audih
37
It’s longer than 240 km. The single-line diagram of a long transmission line
is shown in Fig.
.
ΔI
The length of the line is l . Let us consider a small strip Δx that is at a
distance x from the receiving end
Dr.Audih
38
V = value of voltage just before entering the element Δx.
I = value of current just before entering the element Δx.
V+ΔV = voltage leaving the element Δx.
I+ΔI = current leaving the element Δx.
ΔV = voltage drop across element Δx.
zΔx = series impedance of element Δx
yΔx = shunt admittance of element Δx
The voltage drop across the infinitely small element Δx is given
by:
V
V  zx.I  I.z 
x
or
dV
 I.z 
dx
1
Now to determine the current ΔI through the capacitor branch, from A to B we
apply KCL to node A.
Dr.Audih
39
I  (V  V ) yx  Vyx  Vyx 
Since the term Vyx is the product of 2 infinitely small
values, we ignore it for making easier calculation. 
2
I
I  Vyx

 Vy
x
or
dI
 yV
.
dx
3
Now derivate both sides of eq (1)by d/dx
d
dV
d
I .z .

.
dx
dx dx
d 2V
dI

 z.
2
dx
dx
Dr.Audih
4
40
Now substituting 3 in 4 we get:
2
dV
 z .y V
. or
2
dx
2
dV
 z .y V
. 0
2
dx
5
The solution of the above second order differential equation for V is
given by.
x yz
 x yz
6
1
2
V ( x )  A .e
 A .e
Taking derivative of (6) with respect to (x) we get
dV (x)
 A1
dx
y z .e x
yz
y z .e  x
 A2
yz
7
Where A1 and A2 are arbitrary constants ,if we combining (1) with (7) we get:
1  dV ( x ) 
I ( x)  

z  dx 
A1
.e x
z/ y
yz

A2
.e  x
z/y
Dr.Audih
yz
8
41
Let us define the following two quantities
z
 Z C ( ch a ra cteristic im p ed a n ce ,  )
y
yz      j 
9
( p ro p a g a tio n co n sta n t )
Substituting 9 into equation 6 and 8, and written in terms
of the characteristic impedance and propagation
constant we get:
V ( X )  A1 .e
x
 A2 .e
 x
A1  x
A2   x
I(X ) 
.e 
.e
ZC
ZC
10
Dr.Audih
42
assume that x = 0 not (X). Then V (x)= VR and I(x) = IR
V ( X )  A1 .e  x  A 2 .e   x
A1  x
A2   x
I(X ) 
.e 
.e
ZC
ZC
Note X with R and e0=1 we get:
V R  A1  A2
and
A1
A2
IR 

ZC
ZC
11
Solving (11) we get the following values for A1 and A2 .
VR  ZC I R
A1 
2
VR  ZC I R
and A2 
2
Dr.Audih
12
43
Also note that for x = l we have V(x) = Vs and I(x) = IS . Therefore
replacing x by l and substituting the values of A1 and A2 in (12) in (10) we
get
VR ZCIR l VR ZCIR l
Vs 
.e 
.e
2
2
VR / ZC  IR l VR / ZC IR l
Is 
.e 
.e
2
2
13
14
And noting from hyperbolic equation that
e( l )  e(  l )
cosh( l ) 
2
and
e( l )  e(  l )
sinh( l ) 
2
Dr.Audih
15
44
Then We can rewrite (13) and (14) using hyperbolic equation(15) we get:
V S  V R c o s h  l  Z c I R s in h  l
IS
VR

s in h  l  I R c o s h  l
Zc
The ABCD parameters of the long transmission line can be written as:
A  D  cosh   l  , B  Zc sinh( l )
sinh( l )
C
Zc
,
D  cosh ( l )
Dr.Audih
45
hyperbolic and exponential equation
cosh( x )  cosh( x ) cos(  x )  j sinh( x ) sin(  x )
sinh( x )  sinh( x ) cos(  x )  j cosh( x ) sin(  x )
 x   x  j x
or as exponential equation
e ( x )  e (   x ) 1  x
cosh( x ) 
 (e   x  e   x    x )
2
2
e ( x )  e (   x ) 1  x
sinh( x ) 
 (e   x  e   x    x )
2
2
e (  x )  e ( x ) .e j (  x )  e (  x )   ( x )
( radian )
e (   x )  e  (  x ) .e  j (  x )  e   x    x ( radian )
Dr.Audih
46
surge impedance loading
The term surge impedance loading or SIL is often used to indicate the
nominal capacity of the line. The surge impedance is the ratio of voltage
and current at any point along an infinitely long line. The
term SIL or natural power is a measure of power delivered by a
transmission line when terminated by surge impedance and is given by
SIL  3V R I R*  3
VR
ZC
2
2

3 V R ( 3 ) /
3
ZC
If P > SIL then line consumes vars; otherwise line generates vars.
Dr.Audih
47
Example 5:
Consider a 500 km long line for which the per kilometer line impedance
and admittance are given respectively by z = 0.1 + j 0.5145 Ω and
y = j 3.1734 x 10 -6 mho. Find A B C D constants of TL.
Solution :
 79o 90o 
z
0.52413
o

.



406.403


5.5
 and


6
y
3.1734  10
2 
 2
Zc 

 79o 90o  
6
 l  zy .l   500 0.5241 3.1734  10  



2 
 2

 0.644884.5o  0.0618  j 0.6419   l  0.0618 and  l  j 0.6419


use the following two formulas for evaluating the hyperbolic forms
c o s h (  l )  c o s h ( l ) c o s (  l ) o  j s in h ( l ) s in (  l ) o
s in h (  l )  s in h ( l ) c o s (  l ) o  j c o s h ( l ) s in (  l ) o
( l)o 
1 8 0 o  0 .6 4 1 9

 3 6 .7 8 o
Dr.Audih
48
Applied the equations and noting that sin and cos are in degree as well
as cosh and sinh in rad
co sh  l  co sh ( l  j  l )  co sh (0 .0 6 1 8 ) co s(3 6 .7 8 o ) 
j sin h (0 .0 6 1 8 ) sin (3 6 .7 8 o )  0 .8 0 2 5  j 0 . 0 3 7
si n h  l  sin h ( l  j  l )  sin h (0 .0 6 1 8) co s(3 6 .7 8 o ) 
j co sh (0 .0 6 1 8) sin (3 6 .7 8 o )  0 .0 4 9 5  j 0 . 5 9 9 8
Therefore the ABCD parameters of the system can be written as
A  D  c o s h  l

0 .8 0 2 5  j 0 .0 3 7
B  Z c s in h ( l )  4 3 .4  j 2 4 0 .7 2 
s in h ( l )
C 
  2 .0 1  1 0  5  j 0 .0 0 1 5
Zc
Dr.Audih
49
Example 6:
Find Vs for 765 kV transmission line with a receiving end voltage
of 765 kV(line to line), a receiving end power SR = 2000+ j1000 MVA
and
z=0.53587.8  / m and
y=7.75106 90.0   m
z
Solution:   zy  2.03688.9 / m ,and c 
 262.7 -1.1 
y
VR(1 )  765
3
 441.70 kV
*
6 *
 S   (2000  j1000) 10 
IR(3 )  
   3 441.70103   1688 26.6 A
 3.V  

Vs (x)  VR cosh( x)  IR Zc sinh( x)  441,7000 cosh(x  2.03688.9) 
(1688 26.6 262.7 -1.1sinh(x  2.03688.9)
where x is the lenght in m
Dr.Audih
50
EXAMPLE 7:
A three-phase 765-kV, 60-Hz, 300-km, completely transposed line has
the following positive-sequence impedance and admittance using
exponential method , find ABCD parameters of the line.
z  0.0165  j0.3306  0.331087.14o  / km
y  j 4.674 106
S / km
Solution:
z
0.331087.14o
4
o
ZC 


7.082

10

2.86

6
o
y
4.674 10 90
 266.11.43o 
 l  zy .l  (0.331087.14 o )  (4.674  10 6 90o  (300) 
 1.547  10 6 177.14 o  (300)  0.373188.57 o
Dr.Audih
51
o
 l   l  j  l  0.373188.57 
 0.00931  j 0.3730 (deg.)
l
l
l
0.00931
e  e . e
j l
l
 e  l (radian)
j 0.3730
0.00931
e e
. e
e
0.3730 
 1.00940.3730 rad .
Transfer radian in degree
o
deg. 
180  rad

o

180  0.3730

Dr.Audih
 21.37
o
52
l
e in degree
l
o
e  1.009421.37  0.9400  j 0.3678 same
e
(  x )
e
 l
e
e
 ( x )
.e
0.00931
in degree e
 j( x)
. e
 l
e
 j 0.3730
 x
   x (radian) then;
 0.9907  0.3730 rad .
 0.9226  j 0.3610
l
 l
e  e
Then ;
cosh( l ) 

2
(0.9400  j 0.3678)  (0.9226  j 0.3610)


2
 0.93230.209o
Dr.Audih
53
e ( l )  e (   l )
sinh( l ) 

2
(0.9400  j 0.3678)  (0.9226  j 0.3610)


2
 0.3645  88.63 o
A = D = cosh(  l)=0.9313  0.209 o
B  Z c sinh( l )  266.1  1.43 o  0.3645  88.63 o 
 97.0  87.2 o
sinh( l ) 0.3645  88.63 o
3
o
C 


1.37

10

9
0.06
Zc
266.1  1.43 o
Dr.Audih
54
Example 8:
Determine the theoretical maximum power in MW and in per unit of
SIL,that the line in example 7 can deliver.AssumeVS =VR =765kV
S o lu tio n :
Z  z .l  ( 0 . 0 1 6 5  j 0 . 3 3 0 6 )  3 0 0 
 9 9  8 7 .1 4 o , th e n ;
Z  99 
and

Z
 8 7 .1 4 o
a ls o
A  0 .9 3 1 3  0 .2 0 9 o th e n ;
A  0 .9 3 1 3 a n d
Z
C
 2 6 6 .1 

A
 0 .2 0 9 o a n d
Vs
A 
 V S  A .V R
VR
Dr.Audih
55
Using powe r transfer equation we get;
V S 2 A.V R2
PR (max) 

.cos( Z   A )
Z
Z
 Z  Impedance angle and  A is the angle of conta nt A
Since V R  V S  765 kV then ;
 765 
2
2
0.9313   765 
PR (max) 

.cos(87.14 o  0.209 o ) 
99
99
 5911.36  294.74  5616.62 M W
Fo r SIL=
V
2
ZC
765 2

 2199 M W
266.1
PR (max)
5616.62
The max. power in per unit PR(max) 

 2.55 p .u .
SIL
2199
This value is about 4% less than that found in example
befor where losses were neglected.
Dr.Audih
56
Dr.Audih
57