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

3

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−

= ×

5

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

7

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

= =

8

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−−

9

• 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

=

=

12

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)

12

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)

13

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

16

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)

17

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= = = =

18

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= ∠

19

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

24

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)