Glover 5e SI_Chapter 05

CHAPTER 5:

Transmission

Lines: Steady-

State Operation

© 2012 Cengage Learning Engineering. All Rights Reserved. 0

Chapter 5: Transmission Lines: Steady-State Operation

Transmission Line Models

Previous lectures have covered how to calculate the

distributed inductance, capacitance and resistance of

transmission lines.

In this section we will use these distributed parameters

to develop the transmission line models used in power

system analysis.



Transmission Line Equivalent Circuit

Our current model of a transmission line is shown below



For operation at frequency , let z = r + j L

and y = g +j C (with g usually equal 0)

Units on

z and y are

per unit

length!

Derivation of V, I Relationships



We can then derive the following relationships:

( )

( ) ( )

dV I z dx

dI V dV y dx V y dx

dV x dI xz I yV

dx dx

Setting Up a Second Order Equation



2

2

2

2

( ) ( )

We can rewrite these two, first order differential

equations as a single second order equation

( ) ( )

( )0

dV x dI xz I yV

dx dx

d V x dI xz zyV

dxdx

d V xzyV

dx

V, I Relationships, cont’d



2 2

Define the propagation constant as

where

the attenuation constant

the phase constant

Use the Laplace Transform to solve. System

has a characteristic equation

( ) ( )( ) 0

yz j

s s s

Equation for Voltage



1 2

1 2 1 2

1 1 2 2 1 2

1 2

1 2

The general equation for V is

( )

Which can be rewritten as

( ) ( )( ) ( )( )2 2

Let K and K . Then

( ) ( ) ( )2 2

cosh( ) sinh( )

x x

x x x x

x x x x

V x k e k e

e e e eV x k k k k

k k k k

e e e eV x K K

K x K x

Real Hyperbolic Functions

For real x the cosh and sinh functions have the following

form:



cosh( ) sinh( )sinh( ) cosh( )

d x d xx x

dx dx

Complex Hyperbolic Functions

For x = + j the cosh and sinh functions have the

following form:



cosh cosh cos sinh sin

sinh sinh cos cosh sin

x j

x j

Determining Line Voltage



R R

The voltage along the line is determined based upon

the current/voltage relationships at the terminals.

Assuming we know V and I at one end (say the

"receiving end" with V and I where x 0) we can

1 2determine the constants K and K , and hence the

voltage at any point on the line.

Determining Line Voltage, cont’d



1 2

1 2

1

1 2

2

c

( ) cosh( ) sinh( )

(0) cosh(0) sinh(0)

Since cosh(0) 1 & sinh(0) 0

( )sinh( ) cosh( )

( ) cosh( ) sinh( )

where Z characteristic

R

R

R RR

R R c

V x K x K x

V V K K

K V

dV xzI K x K x

dx

zI I z zK I

yyz

V x V x I Z x

z

y

impedance

Determining Line Current



By similar reasoning we can determine I(x)

( ) cosh( ) sinh( )

where x is the distance along the line from the

receiving end.

Define transmission efficiency as

RR

c

out

in

VI x I x x

Z

P

P

Transmission Line Example



R

6 6

Assume we have a 765 kV transmission line with

a receiving end voltage of 765 kV(line to line),

a receiving end power S 2000 1000 MVA and

z = 0.0201 + j0.535 = 0.535 87.8 km

y = 7.75 10 = 7.75 10 90.0

j

j

km

Then

zy 2.036 88.9 /km

262.7 -1.1 c

z

y

Transmission Line Example, cont’d



*6

3

Do per phase analysis, using single phase power

and line to neutral voltages. Then

765 441.7 0 kV3

(2000 1000) 101688 26.6 A

3 441.7 0 10

( ) cosh( ) sinh( )

441,700 0 cosh(

R

R

R R c

V

jI

V x V x I Z x

2.036 88.9 )

443,440 27.7 sinh( 2.036 88.9 )

x

x

Transmission Line Example, cont’d



Lossless Transmission Lines



c

c

c

For a lossless line the characteristic impedance, Z ,

is known as the surge impedance.

Z (a real value)

If a lossless line is terminated in impedance

Z

Then so we get...

R

R

R c R

jwl l

jwc c

V

I

I Z V

Lossless Transmission Lines



2

( ) cosh sinh

( ) cosh sinh

( )

( )

V(x)Define as the surge impedance load (SIL).

Since the line is lossless this implies

( )

( )

R R

R R

c

c

R

R

V x V x V x

I x I x I x

V xZ

I x

Z

V x V

I x I

If P > SIL then line consumes

vars; otherwise line generates vars.

Transmission Matrix Model

Oftentimes we’re only interested in the terminal

characteristics of the transmission line. Therefore we

can model it as a “black box.”



VS VR + +

- -

IS IR Transmission

Line

S

S

VWith

I

R

R

VA B

IC D

Transmission Matrix Model, cont’d



S

S

VWith

I

Use voltage/current relationships to solve for A,B,C,D

cosh sinh

cosh sinh

cosh sinh

1sinh cosh

R

R

S R c R

RS R

c

c

c

VA B

IC D

V V l Z I l

VI I l l

Z

l Z lA B

l lC DZ

T

Equivalent Circuit Model



The common representation is the equivalent circuit

Next we’ll use the T matrix values to derive the

parameters Z' and Y'.

Equivalent Circuit Parameters



'

' 2

' '1 '

2

' '

2 2

' ' ' '' 1 1

4 2

' '1 '

2

' ' ' '' 1 1

4 2

S RR R

S R R

S S R R

S R R

S R

S R

V V YV I

Z

Z YV V Z I

Y YI V V I

Z Y Z YI Y V I

Z YZ

V V

Z Y Z YI IY

Equivalent Circuit Parameters



We now need to solve for Z' and Y'. Using the B

element solving for Z' is straightforward

sinh '

Then using A we can solve for Y'

' 'A = cosh 1

2

' cosh 1 1tanh

2 sinh 2

C

c c

B Z l Z

Z Yl

Y l l

Z l Z

Simplified Parameters



These values can be simplified as follows:

' sinh sinh

sinhwith Z zl (recalling )

' 1tanh tanh

2 2 2

tanh2 with Y

22

C

c

z l zZ Z l l

y l z

lZ zy

l

Y l y l y l

Z z l y

lY

yll

Simplified Parameters



For short lines make the following approximations:

sinh' (assumes 1)

' tanh( / 2)(assumes 1)

2 2 / 2

80 km 0.998 0.02 1.001 0.01

160 km 0.993 0.09 1.004 0.04

320 km 0.9

lZ Z

l

Y Y l

l

sinhγl tanh(γl/2)Length

γl γl/2

72 0.35 1.014 0.18

Medium Length Line Approximations



For shorter lines we make the following approximations:

sinh' (assumes 1)

' tanh( / 2)(assumes 1)

2 2 / 2

80 km 0.998 0.02 1.001 0.01

160 km 0.993 0.09 1.004 0.04

320

lZ Z

l

Y Y l

l

sinhγl tanh(γl/2)Length

γl γl/2

km 0.972 0.35 1.014 0.18

Three Line Models



(longer than 320 km)

tanhsinh ' 2use ' ,2 2

2

(between 80 and 320 km)

use and 2

(less than 80 km)

use (i.e., assume Y is zero)

ll Y Y

Z Zll

YZ

Z

Long Line Model

Medium Line Model

Short Line Model

Power Transfer in Short Lines

Often we’d like to know the maximum power that could

be transferred through a short transmission line



V1 V2 + +

- -

I1 I1 Transmission

Line with

Impedance Z S12 S21

1

** 1 2

12 1 1 1

1 1 2 2 2

21 1 2

12 12

with , Z

Z Z

V VS V I V

Z

V V V V Z Z

V V VS

Z Z

Power Transfer in Lossless Lines



21 1 2

12 12 12

12 12

1 212 12

If we assume a line is lossless with impedance jX and

are just interested in real power transfer then:

90 90

Since - cos(90 ) sin , we get

sin

Hence the maximu

V V VP jQ

Z Z

V VP

X

1 212

m power transfer is

Max V VP

X

Limits Affecting Max. Power Transfer

Thermal limits

– limit is due to heating of conductor and hence depends

heavily on ambient conditions.

– For many lines, sagging is the limiting constraint.

– Newer conductors limit can limit sag. For example, in

2004 ORNL working with 3M announced lines with a

core consisting of ceramic Nextel fibers. These lines

can operate at 200°C.

– Trees grow, and will eventually hit lines if they are

planted under the line.



Other Limits Affecting Power Transfer

Angle limits

– while the maximum power transfer occurs when line

angle difference is 90°, actual limit is substantially

less due to multiple lines in the system

Voltage stability limits

– as power transfers increases, reactive losses increase

as I2X. As reactive power increases the voltage falls,

resulting in a potentially cascading voltage collapse.



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