1
Instructor: Kai Sun
Fall 2014
ECE 421/599 Electric Energy Systems
5 – Line Model and Performance
2
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 l=80km l=250km R=r×l (Ω) XL=ωL×l (Ω) XC=1/(ωC) ×l (Ω)
0.8=0.027XL 2.5=0.027XL
30.2 94.3 3315.6=109.8XL 1061.0=11.3XL
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Short Line Model • Capacitance is ignored.
( )Z r j L l R jX
S R
S R
V VA BI C D I
=
1 0 1A B Z C D= = = =*
(3 ) 3S S SS V Iφ =
(3 ) (3 ) (3 )L S RS S Sφ φ φ= −(3 )
(3 )
R
S
PP
φ
φ
η =
10 1
S R R R
S R R
V V ZI VZI I I
+ = =
4
•Voltage Regulation (VR):
– 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
( ) ( )
( )
| | | |Percent 100%
| |
R NL R FL
R FL
V VVR
V−
= ×
ZIR ZIR ZIR
( )
( )
| | | | 100%
| |S R FL
R FL
V VV−
= ×
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Medium Line Model
• Model the total shunt admittance of the line by
– 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
( )Y g j C j Cω ω= + ≈
6
2L R RYI I V= +
( )2
S R L
R R R
V V ZIYV Z I V
= +
= + +
(1 )2S R R
ZYV V ZI= + +
2
( ) (1 )2 2 2
S L S
R R R R
YI I V
Y Y ZYI V V ZI
= +
= + + + +
12
(1 ) 14 2
S R R
S R R
ZY ZV V VA BI C D I ZY ZY IY
+ = = + +
12
ZYA B Z= + = (1 ) 14 2
ZY ZYC Y D= + = +
(1 ) (1 )4 2S R R
ZY ZYI Y V I= + + +
det 1A B
AD BCC D
= − =
SR
R S
VV D BI C A I
− = −
Linear, passive, bilateral two-port network (no source)
• Find VS, IS ↔ VR, IR
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Long Line Model
• Series impedance per unit length
• Shunt admittance per unit length
• Consider a small segment of ∆x at distance x from the receiving end
z r j L
y g j C
( ) ( ) ( )V x x V x z xI x+ ∆ = + ∆ ( ) ( ) ( )V x x V x zI xx
+ ∆ −=
∆
( ) ( )dV x zI xdx
=
( ) ( ) ( )I x x I x y xV x x+ ∆ = + ∆ + ∆ ( ) ( ) ( )I x x I x yV x xx
+ ∆ −= + ∆
∆
( ) ( )dI x yV xdx
=
2
2
( ) ( ) ( )d V x dI xz zyV xdx dx
= =
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Principal square root:
2
2
( ) ( )d V x zyV xdx
=2 zyγ = 2
22
( ) ( ) 0d V x V xdx
γ− =
1 2( ) x xV x A e A eγ γ−= +
( )( )j zy r j L g j Cγ α β ω ω∆
= + = = + + γ - Propagation constant α - Attenuation constant (≥0) β - Phase constant (≥0)
1 2 1 21 ( )( ) ( ) ( )x x x xdV x yI x A e A e A e A ez dx z z
γ γ γ γγ − −= = − = −
ZC = 𝑧/𝑦 - Characteristic impedance
If line losses are neglected, i.e. r=0 and g=0
2j LC j LCγ α β ω ω= + = − =
0, LCα β ω= =
with jz re ϕ π ϕ π= − < </2 jz re ϕ
∆
=
1 21 = ( )x x
C
A e A eZ
γ γ−−
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• Find A1 and A2: at the receiving end, x=0, V(x)=VR and I(x)=IR
1 2(0) RV V A A= = +
1 21(0) ( )R
C
I I A AZ
= = −
1 2R C RV Z IA +
=
2 2R C RV Z IA −
=
( ) 2 2 2 2
x x x xx xR c R R c R
R c RV Z I V Z I e e e eV x e e V Z I
γ γ γ γγ γ
− −−+ − + −
= + = +
1( ) 2 2 2 2
R RR R x x x x
x xc cR R
c
V VI Ie e e eZ ZI x e e V I
Z
γ γ γ γγ γ
− −−
+ −− +
= − = +
1 2( ) x xV x A e A eγ γ−= +
1 21( ) ( )x x
C
I x A e A eZ
γ γ−= −
cosh2
x xe ex
sinh2
x xe ex
|A1|>|A2| or
|A1|<|A2|?
ZC = 𝑧/𝑦 - Characteristic impedance
10
( ) cosh sinhR c RV x xV Z xIγ γ= +
1( ) sinh coshR Rc
I x xV xIZ
γ γ= +
11
• At the sending end, x=l, V(l)=VS, I(l)=IS
cosh sinhs R c RV V Z Iγ γ= +
1 sinh coshs R Rc
I V xIZ
γ γ= +
S R
S R
V VA BI C D I
=
coshsinhC
AB Z
γγ
==
1 sinh
coshC
CZ
D
γ
γ
=
=
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ZYA B Z= + =
(1 ) 14 2
ZY ZYC Y D= + = +
Compared to the medium line π model:
1 2
Z Y
Z
′ ′= +
′=
(1 ) 4
12
Z YY
Z Y
′ ′′= +
′ ′= +
(1 )2S R R
ZYV V ZI= + +
(1 ) (1 )4 2S R R
ZY ZYI Y V I= + + +
sinhsinh
sinh
CzZ Z zyy
Z
γγ
γγ
γ
′ = =
=
tan
t
hcosh 1 cosh 1 22 sinh
2
anh2
2
C
Y y zyZ Z z
Y
γγ γ
γγ
γ
γ′ − −= = =
′
=
A=D, AD-BC=1
Linear, passive, bilateral two-port network (no source)
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Equivalent π Model for Long Length Lines
( )( )zy r j L g j Cγ ω ω= = + +
( )Z zl r j L l
( )Y yl g j C l
|γl| (pu)
Z’/Z = sinh(γl)/(γl)
Y’/Y = tanh(γl/2)/(γl/2)
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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 γ=α+jβ= 𝑧𝑦=0.136 ×10-6 +j1.29×10-6 γl =225 𝑧𝑦=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
cosh sinh 89.28 19.39 kVs R c RV V Z Iγ γ= + = ∠
1 sinh cosh 162.42 14.76 As R Rc
I V xIZ
γ γ= + = ∠ °
*
13.33 92%Re( ) 14.45
load
S S
PV I
η = = =
|A1|=72.1kV
|A2|=1.57kV
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Voltage and Current Waves
•Instantaneous voltage as a function of t and x Incident wave: amplitude ↑ when x ↑
Reflected wave: amplitude ↓ when x ↑
1 2 1 2( ) x x x j x x j xV x A e A e A e e A e eγ γ α β α β− − −= + = +
( ) ( )1 2
1 2
( , ) 2 Re 2 Re
( )) ,( ,
x j t x x j t xv t x A e e A
v t x
e e
v t x
α ω β α ω β+ − − = + = +
1 1 1( , ) 2 | | cos( )xv t x A e t x Aα ω β= + +∠
2 2 2( , ) 2 | | cos( )xv t x A e t x Aα ω β−= − +∠
( ) 2 | | cos( )vv t V tω θ= +
| | vV V θ= ∠
15 t(s) x(mi)
v(t, x) (V)
Receiving
Sending
t(s) x(mi)
v2(t, x) (V)
Receiving
Sending
t(s) x(mi)
v1(t, x) (V)
Receiving
Sending
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Velocity and Wavelength of Propagation • Consider
– For a point on the traveling wave: – Its moving speed (velocity of propagation) and the wavelength
– If α=0,
– GMRL≈GMRC
2 2( , ) 2 | | cos( )v t x A t xω β= −
constantt xω β− =
2dx fvdt
ω πβ β
= = =2/v f πλβ
= =
LCβ ω= CLZC
= Surge impedance
1vLC
=1
f LCλ =
8
7 120 0
1 1 3 10 m/4 10 8.85 10
v sµ ε π − −
= = ×× × ×
0 0
1 5000km60
λµ ε
=
13 3 4For 3 bundled conductors: / / ' 1.09C LGMR GMR r r e ×= = =
0
0
1 ln2C
c
GMDZGMR
µπ ε
60lnC
GMDGMR
0.2 ln mH/kmL
GMDLGMR
=0.0556 F/km
lnC
C GMDGMR
µ=
coming from the receiving end (x=0)
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Lossless Lines
cosh cosh cos2
j x j xe ex j x x
sinh sinh sin2
j x j xe ex j x j x
( ) cos sinR C RV x xV jZ xIβ β= +
1( ) sin cosR RC
I x j xV xIZ
β β= +
( ) cosh sinhR c RV x xV Z xIγ γ= +
1( ) sinh coshR Rc
I x xV xIZ
γ γ= +
cos sinS R C RV V jZ Iβ β= +
1 sin cosS R RC
I j V IZ
β β= +
Sending end
jγ β=
( ) cosR nl SV V β= 0RI SV≥• Open circuit at the receiving end:
sinS C RV jZ Iβ=
cosS RI Iβ=
0RV =• Short circuit at the receiving end:
0β ≈
( )R nl SV V≈
S RI I≈ → ∞
For short lines
1cos , sin , sinCC
A D B jZ C jZ
β β β= = = =
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Surge Impedance Loading
• When ZL=ZC
– For a lossless line, ZC is purely resistive.
• Surge impedance loading (SIL) is the loading when ZL=ZC at rated voltage
RR
C
VIZ
=ZL
2 2 2* 3 | | 3 | / 3 | ( )3 MWR Lrated Lrated
R RC C C
V V kVSIL V IZ Z Z
= = = =
(cos sin ) Rx j x Vβ β= +( ) cos sinR C RV x xV jZ xIβ β= +
1( ) sin cosR RC
I x j xV xIZ
β β= +
RV xβ= ∠
(cos sin ) Rx j x Iβ β= +RI xβ= ∠
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Observations from SIL
• |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.
•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
2 2R RL I C V
ZL
( ) RV x V xβ= ∠ ( ) RI x I xβ= ∠
20
21
(Source: Kundur’s book)
22
Complex Power Flow Through Transmission Lines
S R
S R
V VA BI C D I
=
0R RV V S SV V BB B AA A
| | | || | | | ( ) | || | ( )| | | |
S R S R A S B R A BR
B
V AV V A V V A VIB B B
δ θ δ θ θ θθ
− ∠ − ∠ ∠ − − ∠ −= = =
∠
SR
R S
VV D BI C A I
− = −
| || | | | 0 | || | ( ) | | = | | | | | |
S R S A R S A B R BS
B
DV V A V V A V VIB B B B
θ δ θ θ δ θθ
− ∠ + − ∠ ∠ − + ∠−= = −
∠
*(3 ) (3 ) (3 ) 3R R R R RS P jQ V Iφ φ φ= + =
2| || | | || |3 ( ) 3 ( )| | | |S R R
B B AV V A V
B Bθ δ θ θ= ∠ − − ∠ −
2( ) ( ) ( )| || | | || |
( ) ( )| | | |
S L L R L L R L LB B A
V V A VB B
θ δ θ θ− − −= ∠ − − ∠ −
2( ) ( ) ( )*
(3 ) (3 ) (3 )
|| | ( ) | || | ( )3
| | | | S L L B A S L L R L L B
S S S S S
A V V VS P jQ V I
B Bφ φ φ
θ θ θ δ− − −∠ − ∠ += + = = −
, 1A D AD BC= − =
( 1) /C AD B= −
• Define
θA≈0, θB≈90o
23
2( ) ( ) ( )
(3 )
| || | | || |cos( ) cos( )
| | | |S L L S L L R L L
S B A B
A V V VP
B Bφ θ θ θ δ− − −= − − +
2( ) ( ) ( )
(3 )
| || | | || |sin( ) sin( )
| | | |S L L S L L R L L
S B A B
A V V VQ
B Bφ θ θ θ δ− − −= − − +
2( ) ( ) ( )
(3 )
| || | | || |cos( ) cos( )
| | | |R L L S L L R L L
R B A B
A V V VP
B Bφ θ θ θ δ− − −= − − + −
2( ) ( ) ( )
(3 )
| || | | || |sin( ) sin( )
| | | |R L L S L L R L L
R B A B
A V V VQ
B Bφ θ θ θ δ− − −= − − + −
Radius
• For a lossless line, B=jX’, θA=0, θB=90o, and A=cosβl
( ) ((3 )
)(3 ) 3
| || |sinS L L R L
RL
S
VP P
VP
X φφφ δ− −= = =′
2( ) ( ) ( )
(3 )
| || | | |cos cosS L L R L L R L L
R
V V VQ
X Xφ δ β− − −= −′ ′
(3 ) (3 ) (3 )L S RP P Pφ φ φ= −
(3 ) (3 ) (3 )L S RQ Q Qφ φ φ= −
• Sending end:
• Receiving end: P
Q Sending end circle (SL)
Receiving end circle (SR)
θB-θA
δ
PR(3φ)max
PS(3φ)max (PS, QS)
(PR, QR) δ
0
θA≈0, θB≈90o
2( ) ( ) ( )
(3 )
| || | | |cos cosS L L R L L S L L
S
V V VQ
X Xφ δ β− − −= − +′ ′
π+θB
θB
θA
θA
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Sending & Receiving End Power Circle Diagram
• Can two circles intersect? (PR=PS and QR=QS)
P
Q Sending end
circle
Receiving end circle
θB-θA
δ
(PS, QS)
(PR, QR) δ
CS
CR
R
R
2( )| || |
| |S L L
S B A
A VC
Bθ θ−= ∠ −
2( )| || |
| |R L L
R B A
A VC
Bθ θ π−= ∠ − +
( ) ( )| || || |
S L L R L LV VR
B− −=
| | | | 2S RC C R+ ≤ ( ) ( )2 2( ) ( )
2 | || || | (| | | | )| | | |
S L L R L LS L L R L L
V VA V VB B
− −− −+ ≤
( ) ( )2 2
( ) ( )
2 | || || |
| | | |S L L R L L
S L L R L L
V VA
V V− −
− −
≤+
| | | cosh | | cosh( ) | | cosh( ) | 1A j zyγ α β= = + = ⋅ ≤
≤1 (=1 iff |VS|=|VR|)
• Lossless line:
|A|=|cosβl|≤1. Two circles may intersect, e.g. when |VS|=|VR|
A special case is when PS=PR=SIL and QS=QR=0
A necessary condition:
π+θB
θB
θA
θA
25
26
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
•Steady-state stability limit (ignoring losses) – Theoretical limit: δ=90o
– Practical line loadability: δ<30o~45o
( ) ( ) ( ) ( )3 max
| || | | || |sin 90
sinS L L R L L S L L R L L
c
V V V VP
X Zφ β− − − −= =
′
3thermal rated theramalS V Iφ=
sin
sin( )
cX Z
L LCC
β
ω
′ =
= ⋅
2( ) ( )
3
| | | | sin( )( )( )sin
S L L R L L rated
rated rated C
V V VPV V Zφ
δβ
− −=
| || |sin
sin(2 / )Spu RpuV V SIL
δπ λ
=
sin| || |sinSpu RpuV V SIL δ
β=
1 5000km if =60Hzff LC
λ =
27
Example 5.6
3
| || |sin
sin(2 / )Spu RpuV V SIL
P φ δπ λ
=
28
Sending & Receiving End Power Circle Diagram
R=1167 (MVA) = P3φ(max)
CS=0+j1196 (MVA)
CR= -j969 (MVA)
Assume δ<30o
Practical line loadability =583.5MW
-1500 -1000 -500 0 500 1000 1500-2500
-2000
-1500
-1000
-500
0
500
1000
1500
2000
2500
P (MW)
Q (
Mva
r)
CS
CR
R 30o
30o
-700 -600 -500 -400 -300 -200 -100 0 100 200 300 400 500 600 700-300
-200
-100
0
100
200
300
400
P (MW)
Q (M
var)
1167
583.5
29
Line Loadability Curves • Assume VR≈VS=400kV, Ithermal=3000A, SIL=499.83MW and δmax=30o
– SThermal =2078MW – Line loadability curve vs. Line length:
50 100 150 200 250 3000
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
Line Length (km)
Load
ing
Lim
it (M
VA
)
SIL
Thermal Limit
Theoretical limit
Practical line loadability
2078
0 500 1000 1500 2000 2500 3000 3500 4000-5000
0
5000
Line Length (km)Lo
adin
g Li
mit
(MV
A)
30
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
31
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)
32
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).
33
Shunt Reactors • Use XLsh to limit the receiving end
open-circuit voltage to VR
• If VR=VS
Long line
RR
Lsh
VIjX
=
(cos sin )CS R
Lsh
ZV VX
β β= +
VS and VR are in phase
(no real power is transmitted over the line)
sin
cosLsh C
S
R
X ZVV
β
β=
−
sin1 cosLsh CX Zβ
β=
−
1( sin cos )S Lsh RC
I X IZ
β β= − + RI= −
What does IS= -IR mean?
Prove, at the mid-point of the line (x=l/2):
cos2
Rm
VV β=
0mI =
cos sinS R C RV V jZ Iβ β= +
1 sin cosS R RC
I j V IZ
β β= +
34
35
Series Capacitors
• 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 If fs=60Hz, fr= 30Hz for 25% compensation
Long line
( ) ( )3
| || |sinS L L R L L
Cser
V VP
X Xφ δ− −=′ −
% Compensation 100%CsrXX
= ×′
1Cserr s s
ser
Xf f fX L C
= =′ ′
36
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)