Chapter 4 Transmission.ppt

CHAPTER 4
Transmission Line
BEF 23803 – Polyphase Circuit Analysis

BEF 23803 – Polyphase Circuit Analysis – Chapter 4
2
Module Outline
 Introduction
 Types of Power Lines
 Short Line
 Medium Line
 Long Line

3
Introduction
Distribution System
Transmission System
Generation System

4
Introduction – Transmission Line
 The equivalent model is on a “per-phase” basis,
i.e. VL-N, and Ip.
 Two port networks theory is used to express the
voltage and current relations.
 Short, medium, and long line models are
considered as well as the regulation and losses.

5
Type of Power Lines
Transmission Line Model
Short
Line
≤80km
Medium
Line
≤250km
Long
Line
≥250km

6
Short Line
 Definition: ≤ 80 km or ≤ 69 kV.
 Multiplying series impedance per unit length
(r + jL) by the line length (ℓ).
Z = (r + jL)ℓ = R + jX
Z = R + jX
SR
VS
+
-
+
-
VR
IS IR

7
Short Line
 Consider a 3Ф load with apparent power SR(3Ф)
is connected at the end of the transmission line,
the receiving end current is obtained by
 The sending end voltage is
VS = VR + ZIR
 Since the shunt capacitance is neglected, we
have
IS = IR
*
R
)
(3
*
R
R
3V
S
I

 * means conjugate, says S=(2+j3),
thus S* becomes (2-j3)

8
Short Line
 Two port network (ABCD) representation:
VS = AVR + BIR
IS = CVR + DIR
or in matrix form
ABCD
+
-
+
-
VS VR
IR
IS



















R
R
S
S
I
V
D
C
B
A
I
V



















R
R
S
S
I
V
1
0
Z
1
I
V

9
Short Line
 It is obvious that for short line,
A=1 B=Z C=0 D=1
 Voltage regulation is defined as the % change in
voltage at the receiving end in going from no-
load to full-load:
 At no-load, IR=0, thus,
100%
X
V
V
V
VR
%
R(FL)
R(FL)
R(NL) 

A
V
V S
R(NL) 

10
Short Line
 For short line, A=1 and VR(NL)=VS.
 Voltage regulation is measure of line voltage
drop and depends on the power factor (cos θ).
 Voltage regulation is poorer at low lagging
power factor loads (inductive).
 Voltage regulation become negative with leading
power factor loads (capacitive).
 VR(FL)
RIR
jXIR
VS
IR
Lagging pf
VR=+ve

11
Short Line
 Sending-end power,
 The total line loss is given by
SL(3Ф)=SS(3Ф) – SR(3Ф)
 Transmission line efficiency is given by
*
S
S
)
S(3 I
3V
S 

)
S(3
)
R(3
P
P


 
VR(FL)
RIR
jXIR
VS
IR
Leading pf
VR=-ve

12
Short Line
Example 4.1
A 220-kV, three-phase transmission line is 40 km long.
The resistance per phase is 0.15  per km and the
inductance per phase is 1.3263 mH per km. The shunt
capacitance is negligible. Use the short line model to find
the voltage and power at the sending end and the
voltage regulation and the efficiency when the line is
supplying a three-phase load of
a. 381 MVA at 0.8 power factor lagging at 220 kV.
b. 381 MVA at 0.8 power factor leading at 220 kV.

13
Short Line
Solution
a. The series impedance per phase is (f = 60 Hz)
Z=(r+jL)ℓ =(0.15+j2x60x1.3263x10-3)40
=6+j20 
The receiving end voltage per phase is
The apparent power is
SR(3Ф)= 381cos-10.8
= 381 36.87° = 304.8+j228.6 MVA
kV
0
127
3
0
220
VR(LN) 






14
Short Line
The current per phase is
The sending end voltage is
The sending end line-to-line voltage magnitude
A
36.87
1000
0
127
x
3
x10
36.87
381
3V
S
I
3
*
R(LN)
*
)
R(3
)
R(1 












kV
4.93
144.33
)
)(10
36.87
-
j20)(1000
(6
0
127
ZI
V
V -3
)
R(1
R(LN)
S(LN)











 
kV
250
V
3
V S(LN)
L)
-
S(L 


15
Short Line
The sending end power is
Voltage regulation is
Transmission line efficiency is
MVA
41.8
433
Mvar
j288.6
MW
322.8
10
x
36.87
1000
x
4.93
144.33
x
3
I
3V
S -3
*
)
S(1
S(LN)
)
S(3









 

13.6%
100%
x
220
220
-
250
R
V
% 

94.4%
100%
x
8
.
322
8
.
304
P
P
)
S(3
)
R(3







16
Short Line
b. The current for 381 MVA with 0.8 leading pf is
The sending end voltage is
The sending end line-to-line voltage magnitude
A
36.87
1000
0
127
x
3
x10
36.87
381
3V
S
I
3
*
R(LN)
*
)
R(3
R(p) 









kV
9.29
121.39
)
)(10
36.87
j20)(1000
(6
0
127
ZI
V
V -3
)
R(1
R(LN)
S(LN)











 
kV
210.26
V
3
V S(LN)
L)
-
S(L 


17
Short Line
The sending end power is
Voltage regulation is
Transmission line efficiency is
MVA
58
.
27
-
364.18
Mvar
j168.6
MW
322.8
10
x
36.87
1000
x
29
.
9
121.39
x
3
I
3V
S -3
*
)
S(1
S(LN)
)
S(3










 

4.43%
-
100%
x
220
220
-
210.26
R
V
% 

94.4%
100%
x
8
.
322
8
.
304
P
P
)
S(3
)
R(3







18
Medium Line
 Definition: 80 km ≤ length ≤ 250 km.
 Shunt capacitance of the line is included and is
divided into two equal parts placed at the
sending and receiving ends of the line to form
the so-called nominal  model.
Z = R + jX
VS
+
-
+
-
VR
IS IR
IL
2
Y
2
Y

19
Medium Line
 Total shunt admittance
Y = (g +jC)ℓ
 The shunt conductance per unit length, g is
negligible, C = line to neutral capacitance per km,
and ℓ = line length.
R
R
L V
2
Y
I
I 

L
R
S ZI
V
V 
 R
R
S ZI
V
2
ZY
1
V 








S
L
S V
2
Y
I
I 
 R
R
S I
2
ZY
1
V
4
ZY
1
Y
I 
















20
Medium Line
 Representing into the two-port network:
 A and D are dimensionless and equal each
other if the line is the same when viewed from
either end.
 The dimensions of B and C are ohms and mhos,
respectively. The determinant of the line matrix
is unity, i.e.,
AD – BC = 1








2
ZY
1
A Z
B  







4
ZY
1
Y
C 







2
ZY
1
D

21
Medium Line
 We can find VR and IR if VS and IS are known.
 In matrix form (inverse matrix),
BC
AD
BI
DV
V S
S
R



BC
AD
CV
AI
I S
S
R



S
S
R BI
DV
V 

S
S
R CV
AI
I 






















S
S
R
R
I
V
A
C
B
D
I
V

22
Medium Line
 At no-load, i.e. IR is zero and thus A is the ratio
VS/VR.
 If the receiving end is short-circuited, i.e. VR is
zero and thus B is the ratio VS/IR.
 The constant A is useful in computing voltage
regulation. If VR(FL) is the receiving end voltage
at full load for a sending end voltage of VS,
100%
X
V
V
A
V
VR
%
R(FL)
R(FL)
S 


23
Medium Line
Example 4.2
A 345 kV, three-phase transmission line is 130
km long. The resistance per phase is 0.036 
per km and the inductance per phase is 0.8 mH
per km. The shunt capacitance is 0.0112 F per
km. The receiving end load is 270 MVA with 0.8
power factor lagging at 325 kV. Use the medium
line model to find the voltage and power at the
sending end and the voltage regulation.

24
Medium Line
Solution
The series impedance per phase is (assume f = 60 Hz)
Z=(0.036+j2x60x0.8x10-3)130=4.68+j39.207 
Y=(0+j2x60x0.0112x10-6)130
=j0.000548899 siemens
Z = R + jX
VS
+
-
+
-
VR
IS IR
IL
2
Y
2
Y Load
270MVA
325 kV
0.8 lagging

25
Medium Line
0012844
.
0
j
98924
.
0
2
ZY
1
D
A 










j39.207
4.68
Z
B 


5
j0.0005459
3.5251x10
4
ZY
1
Y
C 7










 
kV
0
187.64
3
0
325
V (LN)
R 





MVA
j162
216
MVA
36.87
270
0.8
cos
270
S -1
)
R(3 








26
Medium Line
A
36.87
479.64
0
187.64
x
3
x10
36.87
270
3V
S
I
3
*
R(LN)
*
)
R(3
)
1
R( 












)
R(1
R(LN)
)
S(1 DI
CV
I 
 





 012
.
4
19
.
199
BI
AV
V )
R(1
R(LN)
S(LN) 





 36.79
474.42
j284.0929
379.9522



 012
.
4
01
.
345
3
x
V
V S(LN)
S(LL)
0.7570
)]
angle(I
-
)
V
cos[angle(
factor
power )
S(1
S(LN) 
 

27
Medium Line





 36.79
x0.47442
4.012
3x199.19
xI
3xV
S *
)
S(1
S(LN)
)
S(3 

Mvar
j185.25
MW
214.6
40.802
283.5 




0,
I
load,
-
no
During R 
0
V
2
ZY
1
V R(LL)
S(LL) 









kV
76
.
348
98924
.
0
01
.
345
A
V
V
S(LL)
R(LL) 


7.3108%
x100%
325
325
-
348.76
R
V
% 



28
Long Line
 Definition: length  250 km.
Zx
VS
+
-
+
-
VR
IS
IR
I(x)
IS(x+x)
+
-
+
-
V(x+ x) V(x)
yx yx
x x
l

29
Long Line
)
(
)
(
,
0
x
zI
dx
x
dV
x
zI(x)
Δx
Δx)-V(x)
V(x
xI(x)
z
V(x)
Δx)
V(x









)
(
)
(
,
0
)
(
)
(
)
(
)
(
)
(
)
(
x
yV
dx
x
dI
x
x
x
yV
x
x
I
x
x
I
x
x
xV
y
x
I
x
x
I

















)
(
)
(
)
(
2
2
x
zyV
dx
x
dI
z
dx
x
V
d



30
Long Line
0
)
(
)
(
)
(
)
(
2
2
2
2
2
2
2





x
V
dx
x
V
d
x
V
dx
x
V
d
zy



If we take Second order differential equation:
x
x
e
A
e
A
x
V 
 

 2
1
)
(
where
length)
unit
per
(radian
constant
phase
constant,
n
attenuatio
)
)(
(
zy
constant
n
propagatio

















C
j
g
L
j
r
j

31
Long Line
 
   
 
y
z
z
e
A
e
A
z
x
I
or
e
A
e
A
z
y
e
A
e
A
z
x
I
e
A
e
A
z
dx
x
dV
z
x
I
c
x
x
c
x
x
x
x
x
x















impedance
stic
characteri
1
)
(
)
(
1
)
(
1
)
(
2
1
2
1
2
1
2
1












32
Long Line
R
R
I
x
I
V
x
V
x



)
(
,
)
(
,
0
when
2
2
2
)
(
1
)
(
1
1
2
2
2
2
2
1
2
1
R
c
R
R
c
R
c
R
R
c
c
R
R
I
z
V
A
I
z
V
A
z
A
V
A
A
V
z
A
A
z
I
A
A
V













To find the constant A1 and A2

33
Long Line
x
R
c
R
x
R
c
R
x
R
c
R
x
R
c
R
c
x
R
c
R
x
R
c
R
e
I
z
V
e
I
z
V
x
I
e
I
z
V
e
I
z
V
z
x
I
e
I
z
V
e
I
z
V
x
V


















 







2
2
)
(
2
2
1
)
(
2
2
)
(

34
Long Line
R
x
x
c
R
x
x
x
R
c
x
R
c
x
R
x
R
I
e
e
Z
V
e
e
x
V
e
I
Z
e
I
Z
e
V
e
V
x
V







 








 









2
2
)
(
2
2
2
2
)
(








Re-arrange the equations we have,
R
x
x
R
x
x
c
x
R
x
c
R
x
R
x
c
R
I
e
e
V
e
e
Z
x
I
e
I
e
Z
V
e
I
e
Z
V
x
I







 








 









2
2
1
)
(
2
2
2
2
)
(









35
Long Line
Hyperbolic function,
R
R
c
R
c
R
xI
xV
Z
x
I
xI
Z
xV
x
V




cosh
sinh
1
)
(
sinh
cosh
)
(




Setting x=l, V(l)=Vs, I(l)=Is
R
R
c
s
R
c
R
s
I
V
Z
I
I
Z
V
V








cosh
sinh
1
sinh
cosh






cosh

 D
A


sinh
c
Z
B 


sinh
1
c
Z
C 
1

 BC
AD

36
Long Line
R
R
s I
Z
V
Y
Z
V '
2
'
'
1 








R
R
s I
Y
Z
V
Y
Z
Y
I 















2
'
'
1
4
'
'
1
'
Comparing B constant with hyperbolic function,











sinh
sinh
sinh
sinh
' Z
z
z
y
z
Z
Z c 



Nominal  representation for long line,
Method 2

37
Long Line
2
tanh
2
'
sinh
1
cosh
2
tanh
,
sinh
1
cosh
2
'
1
cosh
2
'
sinh
cosh
2
'
'
1


































ZY
where
ZY
Y
Z
Y
Z
To obtain the Y’/2, compare the A constant,

38
Long Line
2
tanh
2
' 
 

Z
Y

2
tanh
2
tanh






z
y
z
zy


2
tanh
1
2
' 

c
Z
Y

2
tanh
2
tanh






y
y
y



39
Long Line
2
tanh
2
' 



Y
Y

2
2
tanh
2 



Y

VS
+
-
+
-
VR
IS IR
IL
2
2
tanh
2
2
Y'




Y

2
Y'




sinh
' Z
Z 
Equivalent  model for
long length line:

40
Long Line
Example 4.3:
250 km, 500 kV transmission line has per
phase,
z = (0.045 + j0.4) /km
y = j4. 0 S/km
Find ABCD for a  model of the long
transmission line.

41
Long Line
Solution:
76
.
17
7
.
316
10
4
4
.
0
045
.
0
6
j
j
j
y
z
Zc 




 
001267
.
0
10
104
.
7
)
10
4
)(
4
.
0
045
.
0
( 5
6
j
j
j
zy 





 


36
.
98
88
.
10
)
sinh(
' j
Z
Z c 

 

001
.
0
2
2
tanh
1
' j
Z
Y
c


















42
Long Line
0055
.
0
j
9504
.
0
2
Y'
Z'
1
D
A 










36
.
98
88
.
10
Z'
B j



j0.00098
4
Y'
Z'
1
Y'
C 









Chapter 4 Transmission.ppt

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Chapter 4 Transmission.ppt