ECE 476 Power System Analysis

Lecture 19: Ground, Symmetrical Components

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Read Chapters 8 and 9• HW 9 is 7.6, 7.13, 7.19, 7.28, 8.4; will be covered by

an in-class quiz on Nov 5• Second exam is Thursday Nov. 12 during class.

Closed book, closed notes, one new note sheet, first exam note sheet, and calculators allowed

• Abbott Power Plant tour is Tuesday Nov. 10 during class (we’ll be meeting at Abbott on the north side, NW corner of Oak and Armory)

• Design project is assigned today; due on Nov 19; details are on the website

2

Design Project: Due on Nov 19

• Goal is to optimally add new transmission lines and/or transformers to fix existing problems and problems caused by the addition of a new wind farm.

slack

Island Electric Company (IEC)

350 MW

505 MW

396 MW

268 MW 128 Mvar

115 MW 25 Mvar

150 MW 70 Mvar

110 MW 30 Mvar

150 MW 39 Mvar

60 MW 15 Mvar

55 MW 15 Mvar

130 MW 30 Mvar

175 MW 40 Mvar

140 MW 32 Mvar 176 MW

15 Mvar

165 MW 30 Mvar

132 MW

15 Mvar

160 MW

35 Mvar

112 MW 40 Mvar

200 MW 60 Mvar

95 MW

23 Mvar

75 MW 15 Mvar

198 MW 35 Mvar

87 MW 19 Mvar

161 MW

21 Mvar

135 MW 20 Mvar

140 MW 20 Mvar

88 MW 11 Mvar

130 MW 45 Mvar

128 MW

28 Mvar

45%

22%

33%

35%

4%

13%

23%

14%

13%

28%

35%

8%

17%

13%

57%

28%

6%

3%

6%

16%

30%

22%

10%

27%

12%

30%

44%

9%

5%

46%

78%

45%

53%

47%

19%

900 MW

70 Mvar

51 Mvar

78 Mvar

47%

3%

10%

19%

16%

22%

48%

RobinEagle

Rook

SparrowBluebird

Hawk

Parrot

Turkey

Mallard

Condor

Hen

Dove

Piper

Cardinal

24%

1120 MW

2%

2%

29%

8%

36%

7%

18%

7%

11%

24%

17%

38% 39%

30%

7%

78%

14%

83 Mvar

300 MWSystem Losses: 35.22 MW

0.000 pu

NewWind

0 MW

56 Mvar

0.996 pu

1.018 pu

1.018 pu

1.015 pu

1.019 pu

1.015 pu

1.002 pu

0.997 pu

0.996 pu

1.004 pu

0.988 pu

1.001 pu

1.010 pu

1.020 pu

0.999 pu

0.989 pu

0.993 pu

0.996 pu

1.014 pu0.993 pu

1.010 pu

1.015 pu

1.023 pu

1.015 pu

1.010 pu1.006 pu

0.999 pu

1.018 pu1.020 pu

0.993 pu

1.012 pu

1.013 pu

0.992 pu

0.991 pu

1.005 pu

1.014 pu

1.015 pu0.994 pu

1.025 pu

57 Mvar

Ostrich

Crow

Peacock

Lark Finch

Oriole

Heron

Finch

Owl

Woodpecker

Pheasant

Canary

Flamingo Assignmentwill countas threeregularhomeworks

3

Symmetrical Faults: Extension to Larger Systems

bus

bus

The superposition approach can be easily extended

to larger systems. Using the we have

For the second (2) system there is only one voltage

source so is all zeros except at the fault loca

Y

Y V I

I tion

0

0

fI

I

However to use thisapproach we need tofirst determine If

4

Determination of Fault Current

bus

1bus bus

(2)1

(2)11 1 2

(2)1 1

(2)

(1)f

Define the bus impedance matrix as

0

Then

0

For a fault a bus i we get -I

bus

n

f

n nn n

n

ii f i

V

Z Z V

I

Z Z V

V

Z V V

Z

Z Y V Z I

5

Determination of Fault Current

(1)

f

ii

ij

(1) (2) (1)i

Hence

I

Where

Z driving point impedance

Z ( ) transfer point imepdance

Voltages during the fault are also found by superposition

V are prefault values

i

ii

i i i

VZ

i j

V V V

6

Three Gen System Fault Example

g1 2 3

For simplicity assume the system is unloaded

before the fault with

E 1.05 0

Hence all the prefault currents are zero.

g gE E

7

Three Gen Example, cont’d

15 10 0

10 20 5

0 5 9bus j

Y

115 10 0

10 20 5

0 5 9

0.1088 0.0632 0.0351

0.0632 0.0947 0.0526

0.0351 0.0526 0.1409

busZ j

j

8

Three Gen Example, cont’d

1

(2)

1.05For a fault at bus 1 we get I 9.6

0.1088

0.1088 0.0632 0.0351 9.6

0.0632 0.0947 0.0526 0

0.0351 0.0526 0.1409 0

1.05 0

0.60 0

0.337 0

fj Ij

j

j

V

9

Three Gen Example, cont’d

1.05 0 1.05 0 0 0

1.05 0 0.606 0 0.444 0

1.05 0 0.337 0 0.713 0

V

10

PowerWorld Example 7.5: Bus 2 Fault

slack

One

Two

ThreeFourFive

0.724 pu 0.579 pu

0.000 pu

0.687 pu

0.798 pu 0.000 deg 0.000 deg

0.000 deg

0.000 deg

0.000 deg

7 pu 11 pu

11

Problem 7.28

slack

SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHIMKO69

ROGER69

UIUC69

PETE69

HISKY69

TIM69

TIM138

TIM345

PAI69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

JO138 JO345

0.79 pu

0.70 pu

0.79 pu

0.83 pu

0.50 pu

0.59 pu0.52 pu

0.43 pu

0.50 pu

0.61 pu

0.61 pu

0.32 pu0.24 pu

0.58 pu

0.56 pu

0.78 pu

0.61 pu

0.64 pu

0.52 pu

0.564 pu

0.56 pu

0.62 pu

0.68 pu

0.70 pu0.75 pu

0.64 pu

0.35 pu 0.62 pu

0.77 pu

0.77 pu 0.77 pu

0.84 pu 0.91 pu

0.64 pu

LYNN138

0.82 pu

0.00 pu

5 pu

9 pu

3 pu

2 pu

0 pu 3 pu

9 pu

3 pu

0 pu

12

Analysis of Unsymmetric Systems

• Except for the balanced three-phase fault, faults result in an unbalanced system.

• The most common types of faults are single line-ground (SLG) and line-line (LL). Other types are double line-ground (DLG), open conductor, and balanced three phase.

• System is only unbalanced at point of fault! • The easiest method to analyze unbalanced system

operation due to faults is through the use of symmetrical components

• First we’ll go brief coverage of ground calculations13

Grounding

• When studying unbalanced system operation how a system is grounded can have a major impact on the fault flows

• Ground current does not come into play during balanced system analysis (since net current to ground would be zero).

• Becomes important in the study of unbalanced systems, such as during most faults.

14

Grounding, cont’d

• Voltages are always defined as a voltage difference. The ground is used to establish the zero voltage reference point– ground need not be the actual ground (e.g., an airplane)

• During balanced system operation we can ignore the ground since there is no neutral current

• There are two primary reasons for grounding electrical systems1. safety

2. protect equipment

15

How good a conductor is dirt?

• There is nothing magical about an earth ground. All the electrical laws, such as Ohm’s law, still apply.

• Therefore to determine the resistance of the ground we can treat it like any other resistive material:

8

8

onductor lengthResistance

cross sectional area

2.65 10 -m for aluminum

1.68 10 -m for copper

where is the resistivity

cR

16

How good a conductor is dirt?

8

16

2.65 10 -m for aluminum

5 10 -m for quartz (insulator!)

160 -m for top soil

900 -m for sand/gravel

20 -m for salt marsh

What is resistance of a mile long, one inch diameter,

circular wir

8

2

e made out of aluminum ?

2.65 10 1609 R= 0.083 mile0.0128

17

How good a conductor is dirt?

62

What is resistance of a mile long, one inch diameter,

circular wire made out of topsoil?

160 1609 R= 500 10 mile0.0128

In order to achieve 0.08 with our dirt wiremilewe would need a cross sectio

6 2

nal area of

160 16093.2 10 (i.e., a radius of about 1000 m)

0.08But what the ground lacks in , it makes up for in A!

m

18

Calculation of grounding resistance

• Because of its large cross sectional area the earth is actually a pretty good conductor.

• Devices are physically grounded by having a conductor in physical contact with the ground; having a fairly large area of contact is important.

• Most of the resistance associated with establishing an earth ground comes within a short distance of the grounding point.

• Typical substation grounding resistance is between 0.1 and 1 ohm; fence is also grounded, usually by connecting it to the substation ground grid.

19

Calculation of grounding R, cont’d

• Example: Calculate the resistance from a grounding rod out to a radial distance x from the rod, assuming the rod has a radius of r:

In general we have

xcross sectional area

but now area changes

with length.

R

2

ˆln

2 2ˆ

x

r

dxdR

length x

dx xR

length x length r

20

Calculation of grounding R, cont’d

For example, if r 1.5 inches, length = 10 feet,

and 160 we get the following values as

a function of x (in meters)

160ln

2 3.05 0.038x R

1 m 27.2

10 m 46.4

100 m 65.6

100 km 83.4

m

xR

The actual values will besubstantially less since we’ve assumed no currentflowing downward intothe ground

21

Symmetric Components

• The key idea of symmetrical component analysis is to decompose the system into three sequence networks. The networks are then coupled only at the point of the unbalance (i.e., the fault)

• The three sequence networks are known as the– positive sequence (this is the one we’ve been using)– negative sequence– zero sequence

• Presented in paper by Charles .L Fortescue in 1918 (judged as most important power paper of 20th century)

Heydt, G. T.; Venkata, S. S.; Balijepalli, N. (October 24, 2000). "High Impact Papers in Power Engineering, 1900-1999"  Proceedings 2000 North American Power Symposium, vol. 1, October 2000. North American Power Symposium (NAPS). Waterloo, Ontario. 22

Positive Sequence Sets

• The positive sequence sets have three phase currents/voltages with equal magnitude, with phase b lagging phase a by 120°, and phase c lagging phase b by 120°.

• We’ve been studying positive sequence sets

Positive sequencesets have zeroneutral current

23

Negative Sequence Sets

• The negative sequence sets have three phase currents/voltages with equal magnitude, with phase b leading phase a by 120°, and phase c leading phase b by 120°.

• Negative sequence sets are similar to positive sequence, except the phase order is reversed

Negative sequencesets have zeroneutral current

24

Zero Sequence Sets

• Zero sequence sets have three values with equal magnitude and angle.

• Zero sequence sets have neutral current

25

Sequence Set Representation

• Any arbitrary set of three phasors, say Ia, Ib, Ic can be represented as a sum of the three sequence sets

0

0

0

0 0 0

where

, , is the zero sequence set

, , is the positive sequence set

, , is the negative sequence set

a a a a

b b b b

c c c c

a b c

a b c

a b c

I I I I

I I I I

I I I I

I I I

I I I

I I I

26