ECE 3600
Long-length Lines:
Lumped-Parameter Transmission Line Models
over 240 km (150 miles)
Need:
b
(over 200 mi in some texts)
Units
line length:
len , d
Resistance per unit length:
m or km
Ω
r
or
m
Inductance per unit length:
H
l
F
c
or
S
g
m
Units
H
OR Inductive reactance per unit length:
Ω
x
km
or
m
Conductance to ground:
Ω
km
m
Capacitance per unit length:
stick to the same unit length for all parameters
miles may also be used
F
OR Capacitance admittance per unit length:
S
y
km
or
km
S
Common assumption: g
0.
S
S
S
or
m
km
Ω
or
m
km
siemens
km
Find:
Units
Surge impedance:
Zc =
j. x r
j.y g
Propagation constant:
γ
( j.x
=
r) .( j.y
Ω
1
g)
or
m
1
km
If your calculator doesn't have hyperbolic trig functions
Series impedance
Z series
Y shunt
Shunt admittance:
= Z c. sinh( γ. len )
=
2
1 .
len
tanh γ.
Zc
2
OR
2 . Z shunt =
Shunt impedance:
e
= Z c.
γ. len
e
Ω
2
γ. len
2
=
γ. len
1 .e
Z c γ. len
2
e
e
e
γ. len
2
len
γ.
2
Zc
γ. len
=
1 . e
Zc
γ. len
e
e
e
( γ. len )
Ω
( γ. len )
S or
len
tanh γ.
2
1
Ω
If your calculator can't handle complex exponents
e
Model:
IS
(a
b. j )
a b. j
a
= e . e = e /b (in radians)
IR
Z series
VS
2 . Z shunt
VR
2 . Z shunt
neutral
ECE 3600
Transmission Line notes p6
Medium-length Lines:
80 - 240 km (50 to 150 miles)
Need:
(100 - 200 mi in some texts)
Units
line length:
len , d
Resistance per unit length:
m or km
Ω
r
Inductance per unit length:
H
l
km
F
c
S
g
Units
OR Inductive reactance per unit length:
x
km
F
or
m
Conductance to ground:
H
or
m
Capacitance per unit length:
Ω
or
m
stick to the same unit length for all parameters
miles may also be used
OR Capacitance admittance per unit length:
m
y
S
km
Common assumption: g
km
S
or
m
S
Ω
or
m
km
or
Ω
km
S
0.
km
Units
Find:
x
Ω
Surge Impedance:
Zc =
Series Resistance:
R line =
Series impedance
Z series =
(r
Y shunt
len
j . y.
2
S or
2
j . y. len
Ω
Shunt admittance:
Only needed if load is in terms of SIL
y
=
2
OR
2 . Z shunt =
Shunt impedance:
Ω
r. len
Ω
j . x ) . len
IS
IR
Z series
VS
VR
2 . Z shunt
2 . Z shunt
neutral
OR:
r. len
IS
j . x. len
IR
VS
VR
2
.
j y. len
2
.
j y. len
neutral
Short-length Lines:
less than 80km (50 mi)
r. len
IS
j . x. len
IR
(less than 100 mi in some texts)
Same as above but without the capacitors
ECE 3600
Transmission Line notes p7
VS
VR
neutral
1
Ω
ECE 3600
Transmission Line Examples
b
Ex1. A 500 kV transmission line is 500 km long and has the line parameters shown below. Use the long-length
model to find V S and IS if the line is loaded to 900 MVA and |VRLL| is 490 kV. Assume the phase angle of
VR is 0o and assume load pf = 1.
500. km
len
r
x
490. kV
V RLL
V RLL
VR
900. MVA
S 1φ
3
3
Ω
0.029.
km
Assume:
g
0.
S
km
Ω
0.326.
km
y
5.220. 10
Note: These are typical values
for a 500 kV transmission line
S
6.
km
Long-length line model:
Surge Impedance:
Zc
Propagation constant:
γ
j. x r
j. y g
( j. x
Z c = 250.151
r ) . ( j. y
γ = 5.797 10
g)
5
11.104j Ω
km
Series impedance:
Z series
Z c. sinh( γ. len )
Z series = 12.508 + 151.772j Ω
Shunt admittance:
(Not used in my solution)
Y shunt
2 .
len
tanh γ.
Zc
2
Y shunt
Shunt impedance:
Z shunt
IS
6
2 . Z shunt = 2.451
len
2 . tanh γ.
2
Solve circuit:
= 4.49 10
3
+ 1.353 10 j
S
2
Zc
S 1φ
IR
1
3
+ 1.306 10 j
VR
738.924j Ω
(Not complex in this case because pf = 1
otherwise include a phase angle calculated
from the pf or load other information)
I R = 1060.4 A
Z series
VS
V R = 282.902 kV
IL
I Zshunt
2 . Z shunt
RL
2 . Z shunt
I Zshunt
neutral
VR
2 . Z shunt
I Zshunt = 1.27 + 382.852j A
IL
I Zshunt
VS
VR
I ZshuntS
IS
IR
I L. Z series
VS
2 . Z shunt
I ZshuntS
IL
3
I L = 1.062 10 + 382.852j
5
A
5
V S = 2.381 10 + 1.659 10 j
V
I ZshuntS = 223.48 + 322.934j A
I S = 838.23 + 705.786j A
V S = 290.192 kV
arg V S = 34.874 deg
3 . V S = 502.628 kV
I S = 1096 A
ECE 3600
arg I S = 40.097 deg
Transmission Line notes p8
ECE 3600
Transmission Line notes p9
Ex 2. A 345 kV transmission line is 220 km long and has the line parameters shown below.
Find V S and IS if the line is loaded to 0.9 SIL with pf = 95% lagging. |VRLL| is 510 kV.
220. km
len
V RLL
510. kV
VR
V RLL
x
Ω
0.037.
km
0.376.
Assume:
0.
g
of VR is 0 o if VR is given
S
Note: These are typical values
for a 345 kV transmission line
km
Ω
0.95
Assume the phase angle
3
r
pf
6 S
4.518. 10 .
km
y
km
Medium-length line model:
Surge Impedance:
x
Zc
Z c = 288.5 Ω
y
Z series
(r
Shunt admittance:
Not used in my solution
Y shunt
j. y. len
Shunt impedance:
Z shunt
Figure out 1 SIL:
SIL
3.
VR
Z series = 8.14 + 82.72j Ω
j. x) . len
Series impedance:
Y shunt
1
3
2 . Z shunt = 2.012 10 j
j. y. len
2
Actual:
SIL = 902 MVA
S 1φ
Zc
Solve circuit:
S 1φ
IR
.e
j. acos( pf )
0.9 . SIL
Ω
( 0.9 SIL loading)
3
(Negative phase angle
because the pf is lagging)
VR
I R = 872.7
IS
= 496.98j µS
2
286.8i A
Z series
V R = 294.449 kV
IL
2 . Z shunt
VS
2 . Z shunt
I Zshunt
neutral
I Zshunt
VR
2 . Z shunt
IL
I Zshunt
VS
VR
I ZshuntS
IS
IR
I L. Z series
VS
2 . Z shunt
I ZshuntS
ECE 3600
IL
ZL
I Zshunt = 146.335j A
I L = 872.68
140.501j A
5
4
V S = 3.132 10 + 7.104 10 j
V
V S = 321.132 kV
Line voltage:
arg V S = 12.781 deg
3 . V S = 556.216 kV
I ZshuntS = 35.308 + 155.641j A
I S = 837.372 + 15.141j A
Transmission Line notes p9
I S = 838 A
arg I S = 1.036 deg
ECE 3600
Ex3. A 230 kV transmission line has the
following length and line parameters.
150. km
len
Medium-length
line model:
Ω
0.06.
km
r
Ω
0.5 .
x
Transmission Line notes p10
0.
g
km
y
km
Z series
(r
Shunt admittance:
Y shunt
j. y. len
Z shunt
1
.
j y. len
6 S
4 . 10 .
km
Z series = 9 + 75j Ω
j. x) . len
Series impedance:
Shunt impedance:
S
Y shunt
= 0.3j mS
2
2 . Z shunt = 3.333j kΩ
a) The load is 250Ω with a power factor of 0.87, leading. Find the line current, ILine .
IS
VS
IR
Z series
230. kV
(9
75. j ) . Ω
ZL
3
I Line
2 . Z shunt
V S = 132.79 kV
.5. Y shunt
3.33. jkΩ
Assume the phase angle
ZL
0.3j. mS
of VS is 0 o if VS is given
I Zshunt
neutral
pf
ZL
250. Ω
0.87
Z L. e
Z L = 217.5
j. acos( pf )
123.263j Ω
= 250Ω / 29.54 o
Z
Z series
1
Y shunt
1
2
ZL
Z = 210.467
I Line
VS
56.544j Ω
= 219.7Ω / -15.04o
I Line = 588.459 + 158.096j A = 609.3A / 15.04 o
Z
I Line. Z series = 6.561 + 45.557j kV
b) Find the load line voltage.
VR
VS
I Line. Z series
V R = 139.352
45.557j kV = 146.6kV / -18.1 o
Receiving line voltage =
3 . V R = 253.9 kV
Notice that |V R| is bigger than |V S| , this can happen when the receiving-end power factor is leading.
δ =
c) What is the "power angle" (δ)?
arg V R = 18.104 deg
d) How much power is delivered to the load?
IR
VR
= 3 . V R . I R. pf = 224.4 MW
PL
ZL
Power estimate for the same |V R| and
~ 3.
|V S| , but neglecting the line resistance:
V S . V R . sin( 18.1. deg )
e) Express this loading in terms of SIL
Surge Impedance:
Zc
= 240 MW
Z series
x
y
Z c = 353.6 Ω
ECE 3600
Zc
= 1.414
SIL load
ZL
Transmission Line notes p10