ECE 421/521
Electric Energy Systems
Power Systems Analysis I
5 – Line Model and Performance
Instructor:
Kai Sun
Fall 2013
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
l=250km
2.5=0.027XL
94.3
3315.6=109.8XL
2
1061.0=11.3XL
Short Line Model
• Capacitance is ignored.
Z  ( r  j L)l  R  jX
VS   A B 
 I  = C D 

 S 
=
A 1=
B Z=
C 0=
D 1
S S (3φ ) = 3VS I S*
S=
S S (3φ ) − S R (3φ )
L (3φ )
VR 
I 
 R
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
4
ZIR
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
Y
I=
I R + VR
L
2
V=
VR + ZI L
S
Y
=
VR + Z ( I R + VR )
2
VS   A B  VR 
 I  = C D   I 
  R
 S 
ZY
VS =
(1 +
)VR + ZI R
2
ZY
A=
1+
B=
Z
2
ZY
ZY
C=
Y (1 +
) D=
1+
4
2
Y
I=
I L + VS
S
2
Y
Y
ZY

)VR + ZI R 
=( I R + VR ) + (1 +
2
2
2

A B
det 
= AD − BC = 1

C D 
VR   D
 I  =  −C
 R 
ZY
ZY
I S = Y (1 +
)VR + (1 +
)IR
4
2
6
−B
A 
VS 
 
IS 
Linear, passive, bilateral two-port network
(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)
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
d 2V ( x)
2
0
−
γ
V ( x) =
2
dx
γ = zy
2
∆
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
Principal square root:
γ =α + j β = −ω LC =jω LC
2
=
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
• Find A1 and A2: at the receiving end, x=0, V(x)=VR and I(x)=IR
VR + Z C I R
2
V − ZC I R
A2 = R
2
V (0)
= V=
A1 + A2
R
A1 =
1
( A1 − A2 )
I (0)
= I=
R
ZC
|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
9
e x  e x
sinh x 
2
=
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 
ZY
VS =
(1 +
)VR + ZI R
2
I S = Y (1 +
1
Z ′Y ′
)
sinh γ=
 Y ′(1 +
4
ZC
D = cosh γ 
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
B=
Z
ZY
ZY
C=
Y (1 +
) D=
1+
4
2
Z ′Y ′
A = cosh γ  = 1 +
2
B = Z C sinh γ  = Z ′
C=
Compared to the medium line π model:
Z ′ Z=
=
C sinh γ 
=Z
z
sinh γ 
zy 
y
γ
sinh γ 
γ
Y ′ cosh γ  − 1 cosh γ  − 1
= =
2
Z′
Z C sinh γ 
γ
tanh
2
= =
ZC
11
γ
γ
tanh
tanh
y
2 Y
2
zy  =
z
γ
2 γ
2
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 )
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 Ω/mi
ZC = 𝑧/𝑦=387.3∠-6.05o Ω/mi ≈ Real number
γl =225 𝑧𝑧=0.4688∠83.95o=0.0494+j0.466
2sinhγl=eγl - e-γl=e0.0494 ej0.466-e0.0494 e-j0.466=1.051∠0.466rad-0.952∠-0.466rad
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
Vs =
cosh γ VR + Z c sinh γ I R =
89.28∠19.39 kV
1
Is =
sinh γ VR + cosh γ xI R = 162.42∠14.76° A
Zc
13
=
η
IR=184.1∠ 18.195o
Pload
13.33
=
= 92%
Re(VS I S* ) 14.45
Voltage and Current Waves
V ( x) =A1eγ x + A2 e −γ x =A1eα x e j β x + A2 e −α x e − j β x
• 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 ) 
v=
(t , x) v1 (t , x) + v2 (t , x)
v1 (t , x )
v2 ( t , x )
2 | A1 | eα x cos(ωt + β x + ∠A1 )
2 | A2 | e −α x cos(ωt − β x + ∠A2 )
Incident wave (ampl. ↑ when x ↑)
Reflected wave (ampl. ↑ when x ↓)
Let ∠VR=0. ∠A1≈0 and ∠A2≈0 for high-voltage transmission line
v1 (t , x) ≈ 2 | A1 | eα x cos(ωt + β x)
v2 (t , x) ≈ 2 | A2 | e −α x cos(ωt − β 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)
15
Receiving
t(s)
Animation in MATLAB
16
Traveling Waves
1.5
• v1(t, x): S→R
x1 lags x2
x2 lags x3
x 10
5
v1(t, x) (V)
1
0.5
x1=0
0
x2=300km
-0.5
-1
• v2(t, x): R→S
x1 leads x2
x2 leads x3
x3=600km
-1.5
0
0.002
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
0.018
0.02
t(s)
4
2.5
x 10
2
v2(t, x) (V)
1.5
• Velocity of propagation
300km/0.001s=3×108m/s
0.004
1
0.5
x1=0
0
x2=300km
-0.5
-1
-1.5
x3=600km
-2
-2.5
0
0.002
17
0.004
0.006
0.008
0.01
t(s)
0.012
0.014
0.016
Velocity and Wavelength of Propagation
• Consider
=
v2 ( t , x )
2 | A2 | cos(ωt − β x )
constant
– For a point on the traveling wave: ωt − β x =
– Its moving speed (velocity of propagation) and the wavelength
2π
dx ω 2π f =
=
λ
v
/
f
=
= =
v
β
dt β
β
– If α=0, β = ω LC
– 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
18
λ
3
1
3×4
r /=
r ' e= 1.09
1
60 µ0ε 0
= 5000km
Losses Lines
Vs
cosh γ VR + Z c sinh γ I R
Is
1
sinh γ VR + cosh γ xI R
Zc
• For a lossless line:
γ = jβ
e j x  e j x
cosh  x  cosh j x 
2
 cos  x
sinh  x  sinh j x  j sin  x
=
V ( x ) cos β xVR + jZ C sin β xI R
=
I ( x) j
1
sin β xVR + cos β xI R
ZC
19
Impacts on the Sending End Due to the Open/Short
Circuit at the Receiving End
• Assume at the sending end (x=l), VS is constant
=
VS cos β VR + jZ C sin β I R
IS
1
j
sin β VR + cos β I R
ZC
β=
 ω LC ⋅ =

zy ⋅ 
• Open circuit at the receiving end:
IR  0
VR ( nl )
VR ( nl ) ≈ VS
VS
=
≥ VS
cos β 
β ≈ 0
• Short circuit at the receiving end:
VR = 0
IS ≈ IR → ∞
VS = jZ C sin β I R
I S = cos β I R
20
Surge Impedance Loading
• When ZL=ZC
ZL
V
IR = R
ZC
– For a losses line, ZC is purely resistive.
– Surge impedance loading (SIL): for 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
21
( PF=1 for
any x)
Observations from SIL
• |V(x)|=|VS|=|VR|
ZL
• |I(x)|=|IS|=|IR|
• 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 overvoltage issues at the receiving end
22
23
(Source: Kundur’s book)
24
Complex Power Flow Through Transmission Lines
VS   A B  VR 
 I  = C D   I 
  R
 S 
VR   D
 I  =  −C
 R 
IR
VS  VS 
D  A  A  A
B  B B
VS − AVR | VS | ∠δ − | A || VR | ∠θ A
=
B
| B | ∠θ B
S R (3φ ) =PR (3φ ) + jQR (3φ ) =3V I
*
=
R R
=
IS
− B  VS 
A   I S 
VR  VR 0
| 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
DVS − VR | A || VS | ∠θ A + δ − | VR | ∠0
=
B
| B | ∠θ B
S S (3φ ) =PS (3φ ) + jQS (3φ ) =3VS I S*
25
Q
• Sending end:
| A || VS ( L− L ) |2
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|
Sending end
circle
cos(θ B + δ )
δ
sin(θ B + δ )
(PS, QS)
θB-θA P
S(3φ)max
Radius
PR(3φ)max
PR (3φ )
| A || VR ( L− L ) |2
|V
|| V
|
=
−
cos(θ B − θ A ) + S ( L− L ) R ( L− L ) cos(θ B − δ )
|B|
|B|
QR (3φ )
| A || VR ( L− L ) |2
|V
|| V
|
=
−
sin(θ B − θ A ) + S ( L− L ) R ( L− L ) sin(θ B − δ )
|B|
|B|
P
δ
(PR, QR)
Receiving end
circle
• For a lossless line, B=jX’, θA=0, θB=90o, and A=cosβl
P3φ =
QR 3φ
| VS ( L − L ) || VR ( L − L ) |
sin δ
X′
| VS ( L − L ) || VR ( L − L ) |
| VR ( L − L ) |2
cos δ −
cos β 
X′
X′
P=
PS (3φ ) − PR (3φ )
L (3φ )
Q=
QS (3φ ) − QR (3φ )
L (3φ )
26
Sending & Receiving End Power Circle Diagram
CS
R=
| A || VS ( L− L ) |2
|B|
CR
∠θ B − θ A =
| A || VR ( L− L ) |2
|B|
Q
∠θ B − θ A + π
Sending end
circle
| VS ( L− L ) || VR ( L− L ) |
CS
|B|
δ
• Can two circles intersect? (PR=PS and QR=QS)
| CS | + | CR |≤ 2 R
θB-θA
2 | VS ( L − L ) || VR ( L − L ) |
| A|
2
2
(| VS ( L − L ) | + | VR ( L − L ) | ) ≤
|B|
|B|
| A |≤
2 | VS ( L − L ) || VR ( L − L ) |
| VS ( L − L ) | + | VR ( L − L ) |
2
2
≤1
| A=| | cosh γ =| | cosh(α  + j β )=| | cosh( zy ⋅ ) |
• Lossless line:
|A|=|cosβl|≤1. Two circles may intersect, e.g. if |VS|=|VR|
A special case is when PS=PR=SIL and QS=QR=0
27
(PS, QS)
P
(PR, QR)
δ
2nd
The
“=” holds if
and only if |VS|=|VR|
R
R
CR
Receiving end
circle
28
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-carry capacity of the conductor (Ithermal) provided by
the manufacturer, the thermal loading limit is
Sthermal = 3Vφ rated I theramal
• Steady-state stability limit (ignoring losses)
– Theoretical limit
P3φ max
X ′ = Z c sin β 
| VS ( L − L ) || VR ( L − L ) |
| VS ( L − L ) || VR ( L − L ) |

=
sin 90
X′
X′ =
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=
29
1
 5000km if f =60Hz
f LC
Example 5.6
P3φ =
| VSpu || VRpu | SIL
sin(2π  / λ )
30
sin δ
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
31
Line Loadability Curves
• Assume VR≈VS=400kV, Ithermal=3000A, SIL=499.83MW and δmax=30o
– SThermal =2078MW
– Line loadability curve vs. Line length:
5000
4500
4000
Loading Limit (MVA)
Theoretical limit
Loading Limit (MVA)
5000
0
3500
3000
-5000
0
500
1000
2500
2000
2078
Thermal Limit
1500
1000
0
Practical line
loadability
SIL
500
50
1500
2000
2500
Line Length (km)
100
150
200
Line Length (km)
250
32
300
3000
3500
4000
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
33
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)
34
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).
35
Long line
Shunt Reactors
• Use a reactor of XLsh to limit the
receiving end open-circuit voltage
to VR
VR
IR =
jX Lsh
Z
=
VS VR (cos β  + C sin β )
X Lsh
X Lsh =
sin β 
VS
− cos β 
VR
ZC
• For VS=VR
X Lsh =
sin β 
ZC
1 − cos β 
VS and VR are in phase
(no real power is transmitted over the line)
1
sin β VR + cos β I R
ZC
1
=
(−
sin β X Lsh + cos β ) I R
ZC
=
IS j
= −IR
What does that mean?
Prove, at the midspan:
Vm =
VR
cos
36
β
2
Im = 0
37
Long line
Series Capacitors
| 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
38
Homework
• Read through Saadat’s Chapter 5
• ECE521: 5.8-5.13, and Prove Vm and Im in slide 36
• ECE421: 5.8-5.13
• Due date: 10/24 (Thursday), to Weihong@MK207 and
Nan@MK205, or by email
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