Top Banner
1 BEE 3133 ELECTRICAL POWER SYSTEMS Chapter 4 Line Model and Performance Rahmatul Hidayah Salimin
45
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Ch5 Power System Slide08

1

BEE 3133 ELECTRICAL POWER

SYSTEMS

Chapter 4Line Model and Performance

Rahmatul Hidayah Salimin

Page 2: Ch5 Power System Slide08

2

Introduction Analyze the performance of single-phase

and balanced three-phase transmission lines under normal steady-state operating conditions.

Expression of voltage and current at any point along the line are developed, where the nature of the series impedance and shunt admittance is taken into account.

The performance of transmission line is measured based on the voltage regulation and line loadability.

Page 3: Ch5 Power System Slide08

3

Transmission Line Representation

ABCD

+

VR

-

+

Vs

-

Is IR

A line is treated as two-port network which the ABCD parameters and an equivalent π circuit are derived.

Page 4: Ch5 Power System Slide08

4

Transmission Line Representation To facilitate the performance calculations

relating to a transmission line, the line is approximated as a series–parallel interconnection of the relevant parameters.

Consider a transmission line to have: A sending end and a receiving end; A series resistance and inductance; and A shunt capacitance and conductance

Page 5: Ch5 Power System Slide08

5

Transmission Line Representation The relation between sending–end and

receiving–end quantities of the two–port network can be written as:

R

R

S

S

RRS

RRS

I

V

DC

BA

I

V

DICVI

BIAVV

Page 6: Ch5 Power System Slide08

6

Transmission Line Representation Short Line Model < 80 km in length Shunt effects are neglected.

Medium Line Model Range from 80–240 km in length Shunt capacitances are lumped at a few

predetermined points along the line. Long Line Model >240 km in length. Uniformly distributed parameters. Shunt branch consists of both capacitance and

conductance.

Page 7: Ch5 Power System Slide08

7

Short Line Model

l

VRVS

IRIS R XL

Z

Page 8: Ch5 Power System Slide08

8

Short Line Model

length line

inductance phase-per

resistance phase-per

:where

L

r

jXR

LjrzZ

L

Page 9: Ch5 Power System Slide08

9

Short Line Model Thus, the ABCD parameters are easily

obtained from KVL and KCL equations as below:

SCZBpuDA

I

VZ

I

V

II

ZIVV

R

R

S

S

RS

RRS

0;;1

10

1

Page 10: Ch5 Power System Slide08

10

Complex Power Sending end power

Receiving end power

lineRlineRR

phaseRphaseRR

IVS

or

IVS

*3

*3

3

3

lineSlineSS

phaseSphaseSS

IVS

or

IVS

*3

*3

3

3

phaseline VV 3

Remember!

Page 11: Ch5 Power System Slide08

11

Transmission Line Efficiency Total Full–Load Line Losses

Transmission Line Efficiency

Note that only Real Power are taken into account!

333 RSL SSS

100%

3

3

3

3

S

R

S

R

P

P

P

P

Page 12: Ch5 Power System Slide08

12

Voltage Regulation ABCD parameters can be used to describe

the variation of line voltage with line loading.

Voltage regulation is the change in voltage at the receiving end of the line when the load varies from no–load to a specified full–load at a specified power factor, while the sending end is held constant.

Page 13: Ch5 Power System Slide08

13

Voltage Regulation

RFLRS

NLR VVA

VV )()(

100%)(

)()(

FLR

FLRNLR

V

VVVR

No–load receiving–end voltage

Full–load receiving–end voltage

Page 14: Ch5 Power System Slide08

14

??

21

VV

VV

V

;

0:

AVV

SRNL

SRNL

RNL

RNLS

LineLong

ZYLineMedium

LineShortA

V

Thus

IConditionLoadNo

BI

s

R

R

Page 15: Ch5 Power System Slide08

15

Voltage Regulation The effect of load power factor on voltage

regulation is illustrated in phasor diagram. The phasor diagrams are graphical

representation of lagging, unity and leading power factor.

Page 16: Ch5 Power System Slide08

17

Voltage Regulation In practice, transmission line voltages

decrease when heavily loaded and increase when lightly loaded.

EHV lines are maintained within ±5% of rated voltage, corresponding to about 10% voltage regulation.

10% voltage regulation for lower voltage lines also considered good operating practice.

Page 17: Ch5 Power System Slide08

19

Example 1 :Short TL A 220-kV, 50 Hz, three-phase transmission line

is 40 km long. The resistance per phase is 0.15 Ω/km and the inductance per phase is 1.5915 mH/km. The shunt capacitance is negligible. Use the line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying a three-phase load of

a) 381 MVA at 0.8 pf lagging at 220 kVb) 381 MVA at 0.8 pf leading at 220 kV

Page 18: Ch5 Power System Slide08

20

Example 2 :Short TL A 220-kV, 50 Hz, three-phase transmission line

is 40 km long. The resistance per phase is 0.15 Ω/km and the inductance per phase is 1.5915 mH/km. The shunt capacitance is negligible. Use the line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying a three-phase load of

a) 381 MVA at 0.8 pf lagging at 220 kVb) 381 MVA at 0.8 pf leading at 220 kV

Page 19: Ch5 Power System Slide08

21

Solution (a) Given

R = 0.15 Ω/km , L = 1.5915 mH/kmS =381 MVA with pf 0.8 lagVR(line)=220 kV

+

Vs

_

+

VR

_

Is IR

R jXL

Z=R+jωL Ω

Page 20: Ch5 Power System Slide08

22

206

405915.150215.0

Z

phase;per impedance series The

40km

j

mj

lLjr

kV

kV

VV

o

o

LineRphaseR

0127

3

0220

3

RR

RRS

I and Z,,V find Therefore,

ZIVV voltage,end sending Find

Page 21: Ch5 Power System Slide08

23

A

kV

MVA

MjMWMVAS

Thus

o

o

o

oR

o

87.361000

01273

87.36381

3V

S I

3V

S I

I3VS

var6.2288.30487.36381

,

87.368.0cos MVA, 381S

*R(Phase)

*R

R

R(Phase)

R*R

*RR(Phase)R

-1

Page 22: Ch5 Power System Slide08

24

250V

144.33

3

93.4144.3

87.3610002060127

VV

Therefore,

PhaseRS(Phase)

PhaseSLineS

o

oo

R

VV

kV

AjkV

ZI

Page 23: Ch5 Power System Slide08

25

SISS 3V S Power, end-Sending Find

MVA

MjMW

AV

AII

o

o

oRS

8.41433

var6.2888.322

87.3610004.93144.33 3

I3VS

87.361000

o

*SS(Phase)S

Page 24: Ch5 Power System Slide08

26

%6.13

100220

220250

100 %VR

RFL

RFLRNL

V

VV

Voltage Regulation,

%4.94

1008.322

8.304

100 %

S

R

P

P

Effiency,η

Page 25: Ch5 Power System Slide08

27

Medium Line Model – Nominal π Circuit

l

VR

IRIS R XL

Z

VS Y/2 Y/2

Page 26: Ch5 Power System Slide08

28

Medium Line Model Shunt capacitor is considered. ½ of shunt capacitor considered to be

lumped at each end of the line – π circuit Total shunt admittance, Y

length line

kmper econductanc line

kmper ecapacitanc neutral toline

:where

g

C

CjgY

Page 27: Ch5 Power System Slide08

29

Medium Line ModelUnder normal condition, shunt conductance per unit length (the

leakage current) over the insulators and due to corona is negligible

Thus, g = 0

Page 28: Ch5 Power System Slide08

30

Medium Line Model To obtain ABCD parameters, the current in

the series branch is denoted as IL.

Using KCL and KVL, the sending–end voltage is:

3..2

1

2

2 and 1 From

2..2

1..

RR

RRRS

RRL

LRS

ZIVZY

VY

IZVV

VY

II

ZIVV

Page 29: Ch5 Power System Slide08

31

Medium Line Model

Page 30: Ch5 Power System Slide08

32

Medium Line Model

Page 31: Ch5 Power System Slide08

33

Medium Line Model Using KCL to obtain equation for sending–

end current:

5..2

14

1

221

2

4 into 3 and 2 Substitute

4..2

RR

RRR

RS

SLS

IYZ

VYZ

Y

YZIV

YZYVII

VY

II

Page 32: Ch5 Power System Slide08

34

Medium Line Model Thus, the ABCD parameters can be

obtained from equation [3] and [5];

SZY

YCZBpuZY

DA

I

V

ZYZYY

ZZY

I

V

R

R

S

S

41;;

21

21

41

21

Page 33: Ch5 Power System Slide08

35

Medium Line Model ABCD constant are complex since π model

is a symmetrical two-port networkA = D

The determinant of the transmission matrix is unity(1)

AD – BC = 1 (Prove this!)

Page 34: Ch5 Power System Slide08

36

Medium Line Model The receiving and quantities can be

expressed in terms of the sending end quantities

If, ignore the shunt capacitance of the TL, the shunt admittance, Y=0, it become the short transmission line constant.

S

S

R

R

I

V

AC

BD

I

V

Page 35: Ch5 Power System Slide08

37

Example 2 : Medium TL A 345-kV, 60 Hz, three-phase transmission line

is 130 km long. The resistance per phase is 0.036 Ω/km and the inductance per phase is 0.8 mH/km. The shunt capacitance is 0.0112 μF/km. Use the medium line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying a three-phase load of

a) 325 MVA at 0.8 pf lagging at 325 kVb) 381 MVA at 0.8 pf leading at 325 kV

Page 36: Ch5 Power System Slide08

38

Medium Line Model – Nominal

T Circuit

l

VR

IRIS

VS Y

Z/2 Z/2

Find the ABCD Parameters for this circuit using KVL and KCL

Page 37: Ch5 Power System Slide08

39

Long Line Model

l

VR

IRISZ’

VS Y’/2 Y’/2

Page 38: Ch5 Power System Slide08

40

Long Line Model The shunt capacitance and series

impedance must be treated as distributed quantities

The ‘V’ and ‘I’ on the line must be found by solving the differential equation of the transmission line.

Page 39: Ch5 Power System Slide08

41

Long Line Model

2tanh

1

2

2tanh

22

'

sinhsinh

'

c

c

c

Z

YY

ZZZ

y

zZzy

CjgyLjRz

γ = propagation constant

Zc = characteristic impedance

Page 40: Ch5 Power System Slide08

42

Long Line Model If γl <<0 sinh (γl )/( γl ) & tanh (γl /2)/ (γl /2) ≈ 1.0The ABCD parameters:

12

'' D 1

4

'''

' 12

''

YZYZYC

ZBYZ

A

I

V

DC

BA

I

V

R

R

S

S

Page 41: Ch5 Power System Slide08

43

ABCD Parameters

ABCD Parameters

A B

C D

Short Line

1 Z

0 1

Mediumπ

MediumT

LongLine

Page 42: Ch5 Power System Slide08

44

Surge Impedance Loading When the line is loaded by being

terminated with an impedance equal to its characteristic impedance, the receiving end current is

For a lossless line, Zc is purely resistive. The load corresponding to the surge impedance at rated voltage is known as the surge impedance loading (SIL).

C

RR Z

VI impedanceSurge

C

LZC ;

Page 43: Ch5 Power System Slide08

45

Surge Impedance Loading

Since VR = VLrated/√3, SIL in MVA becomes

3

32

C

RRR Z

VIVSIL

MW

2

C

Lrated

Z

kVSIL

Page 44: Ch5 Power System Slide08

46

Surge Impedance Loading SIL is useful measure of transmission line

capacity as it indicates a loading where the line’s reactive requirement are small.

For loads significantly above SIL, shunt capacitor may be needed to minimize voltage drop along the line.

While for light loads significantly below SIL, shunt inductors may be needed.

Page 45: Ch5 Power System Slide08

47

Power Transmission Capability Power handling ability of a line is limited

by: Thermal loading limit Stability limit

Thermal loading limit:

3 thermalratedthermal IVS