lec4 power system

8/18/2019 lec4 power system

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POWER SYSTEMS ILecture 4

06-88-590-68

Electrical and Computer Engineering

University of Windsor

Dr. Ali Tahmasebi


2/31

1

Transmission Line Models

l Transmission lines have distributed inductance,

capacitance and resistance, which can be calculated

based on the conductor material and line structure.

l In this section we will use these distributed parameters to develop the transmission line models

used in power system analysis.


3/31

2

Transmission Line Equivalent Circuit

Our current model of a transmission line is shown

below for a section of line with the length = dx.

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

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

w w

w


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3

Derivation of V, I Relationships

We can then derive the following relationships:

( )( ) ( )

dV I z dx

dI V dV y dx V y dxdV x dI x

z I yV dx dx

=

= + »

= =

z and y are the

values for the entire

line.


5/31

4

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 x z I yV

dx dx

d V x dI x z zyV

dxdx

d V x zyV

dx

= =

= =

- =


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5

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

g

g a b

a

b

g g g

= = +

=

=

- = - + =

V, I Relationships, cont’d

[m-1]


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6

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

g g

g g g g

g g g g

g g

-

- -

- -

= +

+ -= + + -

= + = -

+ -= +

= +

(voltage at position x):

( x is measured from

receiving end of line)


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7

Real Hyperbolic Functions

For real x the cosh and sinh functions have the

following form:

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

d x d x x x

dx dx

g g g g g g = =


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8

Complex Hyperbolic Functions

For x = a + jb the cosh and sinh functions have thefollowing form

cosh cosh cos sinh sin

sinh sinh cos cosh sin

x j

x j

a b a b

a b a b

= +

= +


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9

Determining Line Voltage

The voltage along the line is determined based on the

current/voltage relationships at the terminals. Assuming

we know V and I at one end (usually at the “receiving

end” with VR and IR where x = 0) we can determine theconstants K 1 and K 2, and hence the voltage at any point of

the line.


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10

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 R R

R R c

V x K x K x

V V K K

K V

dV x zI K x K x

dx

zI I z zK I

y yzV x V x I Z x

z

y

g g

g g g g

g g g

= +

= = +

= = Þ =

= = +

Þ = = =

= +

= = impedance

x = 0

[W]


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11

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

R R

c

out

in

V I x I x x

Z

P

P

g g

h

= +

=


13/31

12

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 87.8mile

y = 7.75 10 = 7.75 10 90

j

j - -

= +

WÐ °

´ ´ Ð .0

Then

zy 2.036 88.9 / mile

262.7 -1.1c

mile

z

y

g

°

= = Ð °

Z = = Ð °W

W


14/31

13

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 kV

3

(2000 1000) 101688 26.6 A

3 441.7 0 10

( ) cosh( ) sinh( )441,700 0 cosh(

R

R

R R c

V

j I

V x V x I Z xg g

= = Ð °

é ù+ ´= = Ð - °ê ú

´ Ð °́ë û

= += Ð ° 2.036 88.9 )

443,440 27.7 sinh( 2.036 88.9 )

x

x

´ Ð °+

Ð - °́ ´ Ð °


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14

Transmission Line Example, cont’d


16/31

15

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

V A B

I C D

é ù é ùé ù=ê ú ê úê ú

û ûû


17/31

16

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

V A B

I C D

V V l Z I l

V I I l l

Z

l Z l A B

l lC D Z

g g

g g

g g

g g

é ù é ùé ù=ê ú ê úê ú

ë ûë ûë û

= +

= +

é ùé ù ê ú= =ê ú ê úë ûê úë û

T


18/31

17

Equivalent Circuit Model

The common representation is the equivalent circuitp

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

parameters Z' and Y'.


19/31

18

Equivalent Circuit Parameters

'

' 2

' '1 '

2

' '

2 2

' ' ' '' 1 1

4 2

' '1 '

2

' ' ' '' 1 1

4 2

S R R R

S R R

S S R R

S R R

S R

S R

V V Y V I

Z

Z Y V V Z I

Y Y I V V I

Z Y Z Y I Y V I

Z Y Z

V V

Z Y Z Y I I Y

-- =

æ ö= + +ç ÷

è ø

= + +

æ ö æ ö= + + +ç ÷ ç ÷

è ø è ø

é ù+

ê úé ù é ù= ê úê ú ê úæ ö æ ö ë ûë û ê ú+ +ç ÷ ç ÷

ê úè ø è øë û


20/31

19

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 1 tanh2 sinh 2

C

c c

B Z l Z

Z Y l

Y l l

Z l Z

g

g

g g g

= =

= +

-= =


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20

Simplified Parameters

These values can be simplified as follows:

' sinh sinh

sinh with Z zl (recalling )

' 1tanh tanh

2 2 2

tanh2 with Y

22

C

c

z l z Z Z l l

y l z

l Z zyl

Y l y l y l

Z z l y

lY

yll

g g

g g g

g g

g

g

= =

= =

= =

=

=

=


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21

Medium Length Line Approximations

For shorter lines we make the following approximations:

sinh' (assumes 1)

' tanh( / 2)(assumes 1)2 2 / 2

50 miles 0.998 0.02 1.001 0.01

100 miles 0.993 0.09 1.004 0.0

l Z Z

l

Y Y ll

g

g

g g

= »

= »

Ð ° Ð - °

Ð ° Ð -

sinhγl tanh(γl/2)Length

γl γl/2

4

200 miles 0.972 0.35 1.014 0.18

°

Ð ° Ð - °

(this approximation is applied to lines between 80 to 250 km)


23/31

22

Three Line Models

(longer than 200 miles)

tanhsinh ' 2use ' ,2 2

2 (between 50 and 200 miles)

use and 2

(less than 50 miles)

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

ll Y Y

Z Z ll

Y Z

Z

g g

g g = =

Long Line Model

Medium Line Model

Short Line Model


24/31

23

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 I1Transmission

Line withImpedance 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 V S V I V

Z

V V V V Z Z

V V V S

Z Z

q q q

q q q

-æ ö= = ç ÷

è ø

= Ð = Ð = Ð

= Ð - Ð +(where q12 =

q1-q2)


25/31

24

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

= = W

=

=

For quick, hand calculations, we can assume R=0 (so

z=jw l ), which is called a lossless line.

Which is known as surge impedance.

= = ( )( ) = = [m-1]

(gwould be a purely imaginary number)


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25

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 x Z

I x

Z

V x V

I x I

g g

g g

= +

= +

=

=

=If P > SIL then line consumes

vars; otherwise line generates vars.


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26

Surge Impedance Loading

If a lossless line is terminated by (or connected to) a load

equal to the surge impedance Z C , the power delivered is

called surge impedance loading (SIL).

= → = →

() = cosh + sinh = cosh + sinh ()

= =

→ () =

® Voltage profile at SIL condition is flat.


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27

Surge Impedance Loading, cont’d

() = cosh +

sinh = cosh + sinh ()

=

=

→ () =

Current at position x, at surge impedance loading:

Complex power flowing at any point x:

() = () + () = ()()∗ =

∗

=

∗

=

= SIL ® constant and real

number


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28

ABCD Parameters of Lossless Line

= = cosh = cosh = +

2 = cos

sinh = sinh = − 2

= sin

→ = sinh = sin =

sin [W]

=sinh ()

=

sin ()

[W-1]


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29

Wavelength of a Transmission Line

Wavelength is defined as the distance on the transmission

line required to change the phase of the voltage or current by

360°(2p radians). If we assume a lossless line:

() = + = cos + sin ()

() = + = sin ()

+ cos ()

These two equations change phase by 360°when =

→ wavelength = l =

=

=

[m]


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30

Wavelength of a Transmission Line, cont’d

Velocity of propagation of voltage and current waves on

the line:

l =

1

[m/s]

For overhead lines, typically

≅ 3 ×10 m/s

→ l ≅ 3 ×10

60 = 5000 = 3100

Typical line are usually much shorter than this.

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