ECE 421/599
Electric Energy Systems
5 – Line Model and Performance
Instructor:
Kai Sun
Fall 2014
1
Line Models
• Short Line Model
– 80km (50 miles) or less, 69kV or lower
– Ignoring capacitance
• Medium Line Model
– 80km (50 miles) ~ 250km (150 miles)
– Lumped line parameters
• Long Line Model
– 250km (150 miles) or longer
– Distributed line parameters
• Example: L=1mH/km, C=0.01µF/km and r=0.01Ω/km
ω=2π×60=377Hz
R=r×l
(Ω)
XL=ωL×l (Ω)
XC=1/(ωC) ×l (Ω)
l=80km
0.8=0.027XL
30.2
3315.6=109.8XL
l=250km
2.5=0.027XL
94.3
1061.0=11.3XL
2
Short Line Model
• Capacitance is ignored.
Z  ( r  j L)l  R  jX
VS   A B 
 I  = C D 

 S 
VR 
I 
 R
VS  VR + ZI R  1 Z  VR 
=
 I  =
 0 1   I 
I
 R
R
 
 S 
=
A 1=
B Z=
C 0=
D 1
S S (3φ ) = 3V I
*
S S
S=
S S (3φ ) − S R (3φ )
L (3φ )
η=
PR (3φ )
PS (3φ )
3
•Voltage Regulation (VR):
=
Percent VR
=
| VR ( NL ) | − | VR ( FL ) |
| VR ( FL ) |
| VS | − | VR ( FL ) |
| VR ( FL ) |
×100%
× 100%
– It is a measure of line voltage drop
– Depends on the load power factor: VR is poorer at low lagging power factor
– Perhaps, VR<0 for a leading power factor (i.e. |VS|<|VR|). See Example 5.1
ZIR
ZIR
ZIR
4
Medium Line Model
• Model the total shunt admittance of the line by
Y=
( g + jωC ) ≈ jωC 
– g, the shunt conductance per unit length, represents the leakage current
over the insulators is negligible under normal condition.
– C is the line to neutral capacitance per unit length
• Nominal π model:
– Half of C is considered to be lumped at each end of the line
5
• Find VS, IS ↔
I=
IR +
L
VR, IR
Y
VR
2
V=
VR + ZI L
S
Y
=
VR + Z ( I R + VR )
2
ZY
VS =
(1 +
)VR + ZI R
2
ZY


+
Z
1
 VR 
VS   A B  VR  
2
=
  
 I  =
 I  
C
D
ZY
ZY
  R  Y (1 +
 S 
 IR 
) 1+

4
2 
ZY
ZY
ZY
Y (1 +
) D=
1+
A=
1+
B=
Z C=
4
2
2
Y
I=
I
+
VS
S
L
2
Y
Y
ZY

)VR + ZI R 
=( I R + VR ) + (1 +
2
2
2

ZY
ZY
)VR + (1 +
)IR
I S = Y (1 +
4
2
A B
det 
= AD − BC = 1

C D 
VR   D
 I  =  −C
 R 
−B
A 
VS 
 
IS 
Linear, passive, bilateral two-port network
6
(no source)
Long Line Model
• Series impedance per unit length
z  r  j L
• Shunt admittance per unit length
y  g  j C
• Consider a small segment of ∆x at distance x from the receiving end
V (x + ∆
=
x) V ( x) + z ∆xI ( x)
V ( x + ∆x) − V ( x)
= zI ( x)
∆x
I ( x + ∆=
x ) I ( x ) + y∆xV ( x + ∆x )
I ( x + ∆x) − I ( x)
= yV ( x + ∆x)
∆x
dV ( x)
= zI ( x)
dx
dI ( x)
= yV ( x)
dx
d 2V ( x)
dI ( x)
=
z= zyV ( x)
dx 2
dx
7
d 2V ( x)
= zyV ( x)
2
dx
γ 2 = zy
d 2V ( x)
2
V ( x) =
0
−
γ
2
dx
∆
γ - Propagation constant
γ =α + j β = zy = (r + jω L)( g + jωC )
α - Attenuation constant (≥0)
β - Phase constant (≥0)
If line losses are neglected, i.e. r=0 and g=0
γ =α + j β = −ω LC =jω LC
2
Principal square root:
=
z re jϕ with − π < ϕ < π
∆
z = re jϕ /2
=
α 0,=
β ω LC
V=
( x) A1eγ x + A2 e −γ x
1 dV ( x ) γ
= ( A1eγ x − A2e −γ x ) =
I ( x) =
z dx
z
=
1
( A1eγ x − A2e −γ x )
ZC
y
( A1eγ x − A2e −γ x )
z
ZC = 𝑧/𝑦 - Characteristic impedance
8
V=
( x) A1eγ x + A2 e −γ x
1
=
I ( x)
( A1eγ x − A2 e −γ x )
ZC
ZC = 𝑧/𝑦 - Characteristic impedance
• Find A1 and A2: at the receiving end, x=0, V(x)=VR and I(x)=IR
V (0)
= V=
A1 + A2
R
1
( A1 − A2 )
I (0)
= I=
R
ZC
VR + Z C I R
2
V − ZC I R
A2 = R
2
A1 =
|A1|>|A2| or
|A1|<|A2|?
VR + Z c I R γ x VR − Z c I R −γ x eγ x + e −γ x
eγ x − e − γ x
V ( x) =
e +
e
VR + Z c
IR
=
2
2
2
2
VR
VR
+ IR
− IR
−γ x
−γ x
γx
γx
−
+
e
e
e
e
1
Zc
Z
−
γx
γ
x
=
I ( x) =
e − c
e
VR +
IR
Zc
2
2
2
2
e x  e x
cosh x 
2
e x  e x
sinh x 
2
9
=
V ( x) cosh γ xVR + Z c sinh γ xI R
1
sinh γ xVR + cosh γ xI R
I ( x)
=
Zc
10
• At the sending end, x=l, V(l)=VS, I(l)=IS
Vs
cosh γ VR + Z c sinh γ I R
Is
1
sinh γ VR + cosh γ xI R
Zc
VS   A B 
 I  = C D 

 S 
1
Z ′Y ′
)
sinh γ=
 Y ′(1 +
4
ZC
D = cosh γ 
ZY
VS =
(1 +
)VR + ZI R
2
I S = Y (1 +
Z ′Y ′
= 1+
2
A=D, AD-BC=1
Linear, passive, bilateral two-port network
(no source)
ZY
ZY
)VR + (1 +
)IR
4
2
ZY
A=
1+
2
VR 
I 
 R
Z ′Y ′
A = cosh γ  = 1 +
2
B = Z C sinh γ  = Z ′
C=
Compared to the medium line π model:
B=
Z
ZY
ZY
C=
Y (1 +
) D=
1+
4
2
=
Z ′ Z=
C sinh γ 
=Z
sinh γ 
z
zy 
y
γ
sinh γ 
γ
Y ′ cosh γ  − 1 cosh γ  − 1
= = =
2
Z′
Z C sinh γ 
=
Y
2
y
zy 
z
γ
2
γ
tanh
γ
2
γ
2
tanh
11
Equivalent π Model for Long Length Lines
Z  zl  (r  j L)l
Y  yl  ( g  jC )l
γ=
zy =
(r + jω L)( g + jωC )
Y’/Y = tanh(γl/2)/(γl/2)
Z’/Z = sinh(γl)/(γl)
|γl| (pu)
12
Example 4.1 in Bergen and Vittal’s Book
• A 60-Hz 138kV 3-phase transmission line is 225 mi long. The distributed line
parameters are r=0.169Ω/mi, L=2.093mH/mi, C=0.01427µF/mi, g=0. The transmission
line delivers 40MW at 132kV with 95% power factor lagging.
– Find the sending-end voltage and current.
– Find the transmission line efficiency
Solution:
ω=2π×60=377rad/s
z=r+jωL=0.169+j377×2.093 ×10-3= 0.169+j0.789 = 0.807∠77.9o Ω/mi
y=jωC=j377×0.01427 ×10-6 = j5.38×10-6 =5.38×10-6 ∠90o S/mi
ZC =
𝑧/𝑦=387.3∠-6.05o
Ω/mi ≈ Real number
|A1|=72.1kV
|A2|=1.57kV
γ=α+jβ= 𝑧𝑧=0.136 ×10-6 +j1.29×10-6
γl =225 𝑧𝑧=0.4688∠83.95o=0.0494+j0.466 ≈ Imaginary number
2sinhγl=eγl - e-γl=e0.0494 ej0.466-e-0.0494 e-j0.466=1.051∠0.466 rad -0.952∠-0.466 rad
sinhγl=0.452∠84.4o. Similarly, coshγl=0.8950∠1.42o
Let ∠VR=0. VR=132 ×103/ 3=76.2kV
Pload=0.95|VR||IR|=40/3=13.33MW θ=cos-1(0.95)=18.195o
IR=184.1∠-18.195o A
Pload
13.33
Vs =
cosh γ VR + Z c sinh γ I R =
89.28∠19.39 kV
=
η
=
= 92%
*
1
Re(
V
I
)
14.45
S S
Is =
sinh γ VR + cosh γ xI R = 162.42∠14.76° A
Zc
13
Voltage and Current Waves
=
v (t )
V ( x) =A1eγ x + A2 e −γ x =A1eα x e j β x + A2 e −α x e − j β x
2 | V | cos(ωt + θ v )
=
V | V | ∠θ v
•Instantaneous voltage as a function of t and x
v (t , x )
2 Re  A1eα x e j (ωt + β x )  + 2 Re  A2e −α x e j (ωt − β x ) 
v1 (t , x )
v1 (t , x )
+ v2 ( t , x )
2 | A1 | eα x cos(ωt + β x + ∠A1 )
Incident wave: amplitude ↑ when x ↑
v2 ( t , x )
2 | A2 | e −α x cos(ωt − β x + ∠A2 )
Reflected wave: amplitude ↓ when x ↑
14
v2(t, x) (V)
v1(t, x) (V)
Sending
Sending
x(mi)
x(mi)
t(s)
t(s)
Receiving
Receiving
v(t, x) (V)
Sending
x(mi)
t(s)
Receiving
15
Velocity and Wavelength of Propagation
• Consider
=
v2 ( t , x )
2 | A2 | cos(ωt − β x )
coming from the receiving end (x=0)
ωt − β x =
constant
– For a point on the traveling wave:
– Its moving speed (velocity of propagation) and the wavelength
2π
dx ω 2π f =
=
λ
v
/
f
=
= =
v
β
dt β
β
β = ω LC
– If α=0,
– GMRL≈GMRC
ZC 
1
µ0ε 0
1
2π
=
L
C
Surge impedance
1
λ=
f LC
1
LC
v=
v
ZC =
0.0556
µ F/km
C=
GMD
GMD
L = 0.2ln
mH/km
ln
GMRL
GMRC
For 3 bundled conductors: GMRC / GMR
=
L
1
4π × 10 × 8.85 × 10
−7
µ0 GMD
GMD
ln
 60ln
ε 0 GMRc
GMRC
−12
= 3 × 108 m/ s
λ
3
1
3×4
r /=
r ' e= 1.09
1
60 µ0ε 0
= 5000km
16
Lossless Lines
V ( x) cosh γ xVR + Z c sinh γ xI R
I ( x)
1
sinh γ xVR + cosh γ xI R
Zc
γ = jβ
e j x  e j x
cosh  x  cosh j x 
 cos  x
2
=
V ( x ) cos β xVR + jZ C sin β xI R
I ( x)
j
1
sin β xVR + cos β xI R
ZC
1
j
sin β VR + cos β I R
ZC
A= D= cos β ,
VS = jZ C sin β I R
I S = cos β I R
B= jZ C sin β ,
C= j
1
sin β 
ZC
VR ( nl ) ≈ VS
VR ( nl ) = VS cos β  ≥ VS
• Short circuit at the receiving end:
VR = 0
=
VS cos β VR + jZ C sin β I R
Sending
end
=
IS
• Open circuit at the receiving end:
IR  0
e j x  e j x
 j sin  x
sinh  x  sinh j x 
2
For short lines
β ≈ 0
IS ≈ IR → ∞
17
Surge Impedance Loading
• When ZL=ZC
ZL
V
IR = R
ZC
– For a lossless line, ZC is purely resistive.
• Surge impedance loading (SIL) is the loading when ZL=ZC at rated voltage
3 | VR |2 3 | VLrated / 3 |2 ( kVLrated ) 2
=
SIL 3=
V I
=
=
MW
ZC
ZC
ZC
*
R R
=
V ( x ) cos β xVR + jZ C sin=
β xI R (cos β x + j sin β x )VR = VR ∠β x
I ( x)
j
1
sin β xVR + cos=
β xI R (cos β x + j sin β x ) I R = I R ∠β x
ZC
18
Observations from SIL
V ( x=
) VR ∠β x
I ( x )= I R ∠β x
ZL
• |V(x)|=|VS|=|VR|, |I(x)|=|IS|=|IR|
• PF=1 for any x
• QS=QR=0: Q losses due to line inductance are exactly offset by
Q supplied by shunt capacitance, i.e.
2
 L I R  C VR
2
•SIL is a useful measure of transmission line capacity:
– For load >>SIL, shunt capacitors may be needed to minimize voltage
drop along the line
– For load <<SIL, shunt inductors may be needed to avoid over-voltage
issues at the receiving end
19
20
(Source: Kundur’s book)
21
Complex Power Flow Through Transmission Lines
VS   A
 I  = C
 S 
B  VR 
=
A D, AD − BC
= 1



D  IR 
=
C ( AD − 1) / B
VR   D − B  VS 
 I  =  −C A   I 
  S
 R 
• Define
IR
=
VR  VR 0
VS  VS 
B  B B
VS − AVR | VS | ∠δ − | A || VR | ∠θ A | VS | ∠(δ − θ B )− | A || VR | ∠(θ A − θ B )
=
=
| B | ∠θ B
|B|
B
S R (3φ ) =PR (3φ ) + jQR (3φ ) =3V I
*
=
R R
=
IS
A  A  A
| VS ( L− L ) || VR ( L− L ) |
|B|
∠(θ B − δ ) −
| VS || VR |
| A || VR |2
∠(θ B − δ ) − 3
∠(θ B − θ A )
3
|B|
|B|
| A || VR ( L− L ) |2
|B|
∠(θ B − θ A )
θA≈0, θB≈90o
DVS − VR | A || VS | ∠θ A + δ − | VR | ∠0 | A || VS | ∠(θ A − θ B + δ ) | VR | ∠ − θ B
=
=
−
B
| B | ∠θ B
|B|
|B|
S S (3φ ) =PS (3φ ) + jQS (3φ )
2
A
||
V
|
∠(θ B − θ A ) | VS ( L − L ) || VR ( L − L ) | ∠(θ B + δ )
S ( L− L)
*
=3VS I S =
−
|B|
|B|
22
• Sending end:
2
| A || VS ( L − L ) |
PS (3φ )
|B|
| A || VS ( L − L ) |2
QS (3φ )
|B|
cos(θ B − θ A ) −
| VS ( L − L ) || VR ( L − L ) |
sin(θ B − θ A ) −
| VS ( L − L ) || VR ( L − L ) |
• Receiving end:
|B|
|B|
Radius
cos(θ B + δ )
Q
Sending end
circle (SL)
π+θB
sin(θ B + δ )
θA
θA≈0, θB≈90o
θB-θA
0
2
PR (3φ )
| A || VR ( L − L ) |
| VS ( L − L ) || VR ( L − L ) |
=
−
cos(θ B − θ A ) +
cos(θ B − δ )
|B|
|B|
θA
θB
2
δ
δ
(PS, QS)
PR(3φ)max
PS(3φ)max
P
(PR, QR)
| A || VR ( L − L ) |
|V
|| V
|
QR (3φ ) =
−
sin(θ B − θ A ) + S ( L − L ) R ( L − L ) sin(θ B − δ )
|B|
|B|
• For a lossless line, B=jX’, θA=0,
P=
P=
S (3φ )
R (3φ )
QR (3φ )
QS (3φ )
θB=90o,
| VS ( L − L ) || VR ( L − L ) |
=
δ P3φ
sin
X′
| VS ( L − L ) || VR ( L − L ) |
| VR ( L − L ) |2
cos δ −
cos β 
X′
X′
| VS ( L − L ) || VR ( L − L ) |
| VS ( L − L ) |2
=
−
cos δ +
cos β 
′
′
X
X
and A=cosβl
Receiving end
circle (SR)
P=
PS (3φ ) − PR (3φ )
L (3φ )
Q=
QS (3φ ) − QR (3φ )
L (3φ )
23
Sending & Receiving End Power Circle Diagram
| A || VS ( L− L ) |2
CS
|B|
R=
CR
∠θ B − θ A =
| A || VR ( L− L ) |2
|B|
Q
∠θ B − θ A + π
Sending end
circle
π+θB
| VS ( L− L ) || VR ( L− L ) |
CS
|B|
θA
• Can two circles intersect? (PR=PS and QR=QS)
2 | VS ( L − L ) || VR ( L − L ) |
| A|
(| VS ( L − L ) |2 + | VR ( L − L ) |2 ) ≤
|B|
|B|
| CS | + | CR |≤ 2 R
| A |≤
2 | VS ( L − L ) || VR ( L − L ) |
| VS ( L − L ) | + | VR ( L − L ) |
2
2
≤1 (=1 iff |VS|=|VR|)
A necessary condition:
=
| A | | cosh=
γ  | | cosh(α  + j=
β ) | | cosh( zy ⋅ ) |≤ 1
• Lossless line:
δ
θB-θA
θA
R
(PS, QS)
P
(PR, QR)
δ
R
CR θB
Receiving end
circle
|A|=|cosβl|≤1. Two circles may intersect, e.g. when |VS|=|VR|
A special case is when PS=PR=SIL and QS=QR=0
24
25
Power Transmission Capacity
•Thermal loading limit:
– Conductors are stretched if its temperature increases due to real power
loss, which will increase the sag between transmission towers
– With the current-carrying capacity (Ithermal) of the conductor provided by the
manufacturer, the thermal loading limit is
Sthermal = 3Vφ rated I theramal
•Steady-state stability limit (ignoring losses)
– Theoretical limit: δ=90o
P3φ max
X ′ = Z c sin β 
| VS ( L − L ) || VR ( L − L ) |
| VS ( L − L ) || VR ( L − L ) |
=
sin 90
X′
Z c sin β =
L
sin(ω LC ⋅ )
C
– Practical line loadability: δ<30o~45o
2
| VS ( L − L ) | | VR ( L − L ) | Vrated
sin δ
sin δ
P3φ = (
)(
)(
)
=| VSpu || VRpu | SIL
Vrated
Vrated
Z C sin β 
sin β 
=
| VSpu || VRpu | SIL
sin(2π  / λ )
sin δ
λ=
1
 5000km if f =60Hz
f LC
26
Example 5.6
P3φ =
| VSpu || VRpu | SIL
sin(2π  / λ )
sin δ
27
Sending & Receiving End Power Circle Diagram
2500
R=1167
2000
(MVA) = P3φ(max)
CS=0+j1196 (MVA)
1500
CR= -j969 (MVA)
CS
Assume δ<30o
1000
30o
R
Practical line loadability =583.5MW
0
400
1167
300
30o
-500
200
Q (Mvar)
Q (Mvar)
500
-1000
CR
0
-100
-1500
583.5
-200
-300
-700 -600 -500 -400 -300 -200 -100 0 100 200 300 400 500 600 700
P (MW)
-2000
-2500
-1500
100
-1000
-500
0
P (MW)
500
1000
1500
28
Line Loadability Curves
• Assume VR≈VS=400kV, Ithermal=3000A, SIL=499.83MW and δmax=30o
– SThermal =2078MW
– Line loadability curve vs. Line length:
Loading Limit (MVA)
5000
5000
4500
4000
0
Loading Limit (MVA)
Theoretical limit
3500
-5000
3000
0
500
1000
1500
2000
2500
Line Length (km)
3000
3500
4000
2500
2000
2078
Thermal Limit
1500
1000
SIL
500
0
Practical line
loadability
50
100
150
200
Line Length (km)
250
300
29
Line Compensation
•Voltage Improvement:
•A long transmission line loaded at its SIL has no net Mvar flow
into or out of the line, and has approximately a flat voltage
profile along its length.
– A light load << SIL may cause high voltage at the receiving
end
– A heavy load >>SIL may cause low voltage at the receiving
end
– A reactor or capacitor may be installed at the receiving end to
improve voltage profiles
•Other purposes of line compensation:
– Changing the impedance of a line
30
Use of Capacitors and Reactors
•Can be designed to be a permanent part of the system (fixed)
or be switched in and out of service via circuit breakers or
switchers
– Shunt capacitors: supply Mvar to the system at a location
and increase voltages near that location.
– Shunt reactors: absorb excessive Mvar from the system at a
location and reduce voltages near that location.
– Series capacitors: reduce the impedance of the path by
adding capacitive reactance (to improve stability and reduce
reactive losses).
– Series reactors: increase the impedance of the path by
adding inductive reactance (to limit fault currents or reduce
power oscillations between generators)
31
Shunt Capacitors
•Locations:
– Connected directly to a bus bar or to the tertiary winding of
a main transformer
•Advantage:
– Low cost and flexibility of installation and operation
•Disadvantage:
– Reactive power output Q is proportional to its V2, and is hence
reduced at low voltages (when it is likely to be needed most)
– For example, if a 25 Mvar shunt capacitor normally rated at 115
kV is operated at 109 kV (0.95pu) the output of the capacitor is
22.5 Mvar or 90% of the rated value (Q=0.952=0.90pu).
32
Shunt Reactors
• Use XLsh to limit the receiving end
open-circuit voltage to VR
Long line
VR
IR =
jX Lsh
VS
Z
VR (cos β  + C sin β )
X Lsh
=
VS cos β VR + jZ C sin β I R
=
IS
VS and VR are in phase
j
1
sin β VR + cos β I R
ZC
(no real power is transmitted over the line)
X Lsh =
sin β 
VS
− cos β 
VR
• If VR=VS
ZC
1
(−
sin β X Lsh + cos β ) I R = − I R
IS =
ZC
What does IS= -IR mean?
Prove, at the mid-point of the line (x=l/2):
X Lsh =
sin β 
ZC
1 − cos β 
Vm =
VR
cos
β
2
Im = 0
33
34
Series Capacitors
Long line
| VS ( L− L ) || VR ( L− L ) |
P3φ =
sin δ
X ′ − X Cser
% Compensation
=
X Csr
× 100%
′
X
• Advantage:
– “Self-regulating” nature: unlike a shunt capacitor, series capacitors produce
more reactive power with heavier power current flows
• Disadvantage:
– Sub-synchronous resonance (SSR) is often caused by the series-resonant circuit
X Cser
1
=
f r f=
f
s
s
X′
L′Cser
If fs=60Hz, fr= 30Hz for 25% compensation
35
Homework #6
•Read through Saadat’s Chapter 5
•ECE421: 5.8-5.13, and draw the sending & receiving end power
circles for Example 5.6 (slide 27) with VS=VR=1pu and indicate on
both circles the operating points with SIL, the practical line
loadability with δ=45o and the theoretical maximum power
transfer.
•ECE599: plus proving Vm and Im in slide 33
•Due date: 10/31 (Friday)
36