ECE 476 Power System Analysis

Lecture 17: OPF, Symmetrical Faults

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Special Guest Lecturer: TA Won Jang

Announcements

• Read Chapter 7• HW 8 is 12.19, 12.20, 12.26, 12.28; due October 29 in

class (no quiz) – P. 12.26: PW case 12.8– P. 12.28: PW case 12.28– Both posted on the webpage

• Second exam is Thursday Nov. 12 during class. Closed book, closed notes, one note sheet and calculators allowed

• Abbott Power Plant tour is Tuesday Nov. 10 during class

2

“DC” Power Flow

• The “DC” power flow makes the most severe approximations:– completely ignore reactive power, assume all the voltages

are always 1.0 per unit, ignore line conductance

• This makes the power flow a linear set of equations, which can be solved directly

1θ B P

3

Optimal Power Flow (OPF)

• OPF functionally combines the power flow with economic dispatch

• Minimize cost function, such as operating cost, taking into account realistic equality and inequality constraints

• Equality constraints– bus real and reactive power balance– generator voltage setpoints– area MW interchange

4

OPF, cont’d

• Inequality constraints– transmission line/transformer/interface flow limits– generator MW limits– generator reactive power capability curves– bus voltage magnitudes (not yet implemented in

Simulator OPF)

• Available Controls– generator MW outputs– transformer taps and phase angles

5

Two Example OPF Solution Methods

• Non-linear approach using Newton’s method– handles marginal losses well, but is relatively slow and

has problems determining binding constraints

• Linear Programming – fast and efficient in determining binding constraints, but

can have difficulty with marginal losses.– used in PowerWorld Simulator

6

LP OPF Solution Method

• Solution iterates between– solving a full ac power flow solution

• enforces real/reactive power balance at each bus• enforces generator reactive limits• system controls are assumed fixed • takes into account non-linearities

– solving a primal LP• changes system controls to enforce linearized

constraints while minimizing cost

7

Two Bus with Unconstrained Line

Total Hourly Cost :

Bus A Bus B

300.0 MWMW

197.0 MWMW 403.0 MWMW300.0 MWMW

8459 $/hr Area Lambda : 13.01

AGC ON AGC ON

13.01 $/MWh 13.01 $/MWh

Transmission line is not overloaded

With no overloads theOPF matchesthe economicdispatch

Marginal cost of supplyingpower to each bus (locational marginal costs)

8

Two Bus with Constrained Line

Total Hourly Cost :

Bus A Bus B

380.0 MWMW

260.9 MWMW 419.1 MWMW300.0 MWMW

9513 $/hr Area Lambda : 13.26

AGC ON AGC ON

13.43 $/MWh 13.08 $/MWh

With the line loaded to its limit, additional load at Bus A must be supplied locally, causing the marginal costs to diverge.

9

Three Bus (B3) Example

• Consider a three bus case (bus 1 is system slack), with all buses connected through 0.1 pu reactance lines, each with a 100 MVA limit

• Let the generator marginal costs be – Bus 1: 10 $ / MWhr; Range = 0 to 400 MW– Bus 2: 12 $ / MWhr; Range = 0 to 400 MW– Bus 3: 20 $ / MWhr; Range = 0 to 400 MW

• Assume a single 180 MW load at bus 2

10

Bus 2 Bus 1

Bus 3

Total Cost

0.0 MW

0 MW

180 MW

10.00 $/MWh

60 MW 60 MW

60 MW

60 MW120 MW

120 MW

10.00 $/MWh

10.00 $/MWh

180.0 MW

0 MW

1800 $/hr

120%

120%

B3 with Line Limits NOT Enforced

Line from Bus 1to Bus 3 is over-loaded; all buseshave same marginal cost

B3 with Line Limits Enforced

Bus 2 Bus 1

Bus 3

Total Cost

60.0 MW

0 MW

180 MW

12.00 $/MWh

20 MW 20 MW

80 MW

80 MW100 MW

100 MW

10.00 $/MWh

14.00 $/MWh

120.0 MW

0 MW

1920 $/hr

100%

100%

LP OPF redispatchesto remove violation.Bus marginalcosts are nowdifferent.

12

Bus 2 Bus 1

Bus 3

Total Cost

62.0 MW

0 MW

181 MW

12.00 $/MWh

19 MW 19 MW

81 MW

81 MW100 MW

100 MW

10.00 $/MWh

14.00 $/MWh

119.0 MW

0 MW

1934 $/hr

81%

81%

100%

100%

Verify Bus 3 Marginal Cost

One additional MWof load at bus 3 raised total cost by14 $/hr, as G2 wentup by 2 MW and G1went down by 1MW

13

Why is bus 3 LMP = $14 /MWh

• All lines have equal impedance. Power flow in a simple network distributes inversely to impedance of path. – For bus 1 to supply 1 MW to bus 3, 2/3 MW would take

direct path from 1 to 3, while 1/3 MW would “loop around” from 1 to 2 to 3.

– Likewise, for bus 2 to supply 1 MW to bus 3, 2/3MW would go from 2 to 3, while 1/3 MW would go from 2 to 1to 3.

14

Why is bus 3 LMP $ 14 / MWh, cont’d

• With the line from 1 to 3 limited, no additional power flows are allowed on it.

• To supply 1 more MW to bus 3 we need – Pg1 + Pg2 = 1 MW– 2/3 Pg1 + 1/3 Pg2 = 0; (no more flow on 1-3)

• Solving requires we up Pg2 by 2 MW and drop Pg1 by 1 MW -- a net increase of $14.

15

Both lines into Bus 3 Congested

Bus 2 Bus 1

Bus 3

Total Cost

100.0 MW

4 MW

204 MW

12.00 $/MWh

0 MW 0 MW

100 MW

100 MW100 MW

100 MW

10.00 $/MWh

20.00 $/MWh

100.0 MW

0 MW

2280 $/hr

100% 100%

100% 100%For bus 3 loadsabove 200 MW,the load must besupplied locally.Then what if thebus 3 generator opens?

16

Fault Analysis

• The cause of electric power system faults is insulation breakdown

• This breakdown can be due to a variety of different factors– lightning– wires blowing together in the wind– animals or plants coming in contact with the wires– salt spray or pollution on insulators

17

Fault Types

• There are two main types of faults– symmetric faults: system remains balanced; these faults

are relatively rare, but are the easiest to analyze so we’ll consider them first.

– unsymmetric faults: system is no longer balanced; very common, but more difficult to analyze

• The most common type of fault on a three phase system by far is the single line-to-ground (SLG), followed by the line-to-line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults

18

Lightning Strike Event Sequence

1. Lighting hits line, setting up an ionized path to ground 30 million lightning strikes per year in US! a single typical stroke might have 25,000 amps, with a

rise time of 10 s, dissipated in 200 s. multiple strokes can occur in a single flash, causing the

lightning to appear to flicker, with the total event lasting up to a second.

2. Conduction path is maintained by ionized air after lightning stroke energy has dissipated, resulting in high fault currents (often > 25,000 amps!)

19

Lightning Strike Sequence, cont’d

3. Within one to two cycles (16 ms) relays at both ends of line detect high currents, signaling circuit breakers to open the line nearby locations see decreased voltages

4. Circuit breakers open to de-energize line in an additional one to two cycles breaking tens of thousands of amps of fault current is no

small feat! with line removed voltages usually return to near normal

5. Circuit breakers may reclose after several seconds, trying to restore faulted line to service

20

Worldwide Lightning Strike Density

Source: http://science.nasa.gov/science-news/science-at-nasa/2001/ast05dec_1/

Units are Lightning Flashes per square km per year; Florida istop location in the US; very few on the West Coast, or HI, AK. Thisis an important consideration when talking about electric reliability!

21

Fault Analysis

• Fault currents cause equipment damage due to both thermal and mechanical processes

• Goal of fault analysis is to determine the magnitudes of the currents present during the fault– need to determine the maximum current to insure devices

can survive the fault– need to determine the maximum current the circuit

breakers (CBs) need to interrupt to correctly size the CBs

22

RL Circuit Analysis

• To understand fault analysis we need to review the behavior of an RL circuit

( )

2 cos( )

v t

V t

Before the switch is closed obviously i(t) = 0.When the switch is closed at t=0 the current willhave two components: 1) a steady-state value2) a transient value

23

RL Circuit Analysis, cont’d

ac

2 2 2 2

1. Steady-state current component (from standard

phasor analysis)

2 cos( )i ( )

where ( )

ac

V tt

Z

Z R L R X

VI

Z

24

RL Circuit Analysis, cont’d

dc 1

1

ac dc 1

1

2. Exponentially decaying dc current component

i ( )

where T is the time constant,

The value of is determined from the initial

conditions:

2(0) 0 i ( ) i ( ) cos( )

2

tT

tT

Z

t C e

LT RC

Vi t t t C e

Z

VC

Z

cos( ) which depends on Z 25

Time varying current

26

RL Circuit Analysis, cont’d

dc

1

Hence i(t) is a sinusoidal superimposed on a decaying

dc current. The magnitude of i (0) depends on when

the switch is closed. For fault analysis we're just

2concerned with the worst case:

( )

VC

Zi t

ac dci ( ) i ( )

2 2( ) cos( )

2(cos( ) )

tT

tT

t t

V Vi t t e

Z Z

Vt e

Z

27

RMS for Fault Current

2 2RMS

22 2

2The function i(t) (cos( ) ) is not periodic,

so we can't formally define an RMS value. However,

as an approximation define

I ( ) ( ) ( )

2

This function has a maximum va

tT

ac dc

tT

ac ac

Vt e

Z

t i t i t

I I e

lue of 3

Therefore the dc component is included simply by

multiplying the ac fault currents by 3

acI

28

Generator Modeling During Faults

• During a fault the only devices that can contribute fault current are those with energy storage

• Thus the models of generators (and other rotating machines) are very important since they contribute the bulk of the fault current.

• Generators can be approximated as a constant voltage behind a time-varying reactance

'aE

29

3 bal. windings (a,b,c) – stator

Field winding (fd) on rotor

Damper in “d” axis

(1d) on rotor

2 dampers in “q” axis

(1q, 2q) on rotor

30

Synchronous Machine Modeling

Generator Modeling, cont’d

"d

'd

d

The time varying reactance is typically approximated

using three different values, each valid for a different

time period:

X direct-axis subtransient reactance

X direct-axis transient reactance

X dire

ct-axis synchronous reactance

31

Generator Modeling, cont’d

'

"

''

ac

" '

"d

For a balanced three-phase fault on the generator

terminal the ac fault current is (see page 386)

1 1 1

i ( ) 2 sin( )1 1

where

T direct-axis su

d

d

tT

d dda t

T

d d

eX XX

t E t

eX X

'd

btransient time constant ( 0.035sec)

T direct-axis transient time constant ( 1sec)

32

Generator Modeling, cont'd

'

"

''

ac

" '

'

DC "

A

The phasor current is then

1 1 1

1 1

The maximum DC offset is

2I ( )

where T is the armature time constant ( 0.2 seconds)

d

d

A

tT

d dda t

T

d d

tTa

d

eX XX

I E

eX X

Et e

X

33

Generator Short Circuit Currents

34

Generator Short Circuit Currents

35

Generator Short Circuit Example

• A 500 MVA, 20 kV, 3 is operated with an internal voltage of 1.05 pu. Assume a solid 3 fault occurs on the generator's terminal and that the circuit breaker operates after three cycles. Determine the fault current. Assume

" '

" '

A

0.15, 0.24, 1.1 (all per unit)

0.035 seconds, 2.0 seconds

T 0.2 seconds

d d d

d d

X X X

T T

36

Generator S.C. Example, cont'd

2.0

ac0.035

ac

6

base ac3

0.2DC

Substituting in the values

1 1 11.1 0.24 1.1

( ) 1.051 1

0.15 0.24

1.05(0) 7 p.u.0.15

500 10I 14,433 A (0) 101,000 A

3 20 10

I (0) 101 kA 2 143 k

t

t

t

e

I t

e

I

I

e

RMSA I (0) 175 kA37

Generator S.C. Example, cont'd

0.052.0

ac 0.050.035

ac

0.050.2

DC

RMS

Evaluating at t = 0.05 seconds for breaker opening

1 1 11.1 0.24 1.1

(0.05) 1.051 1

0.15 0.24

(0.05) 70.8 kA

I (0.05) 143 kA 111 k A

I (0.05

e

I

e

I

e

2 2) 70.8 111 132 kA

38

Network Fault Analysis Simplifications

• To simplify analysis of fault currents in networks we'll make several simplifications:– Transmission lines are represented by their series

reactance– Transformers are represented by their leakage reactances– Synchronous machines are modeled as a constant voltage

behind direct-axis subtransient reactance– Induction motors are ignored or treated as synchronous

machines– Other (nonspinning) loads are ignored

39

Network Fault Example

For the following network assume a fault on the terminal of the generator; all data is per unitexcept for the transmission line reactance

2

19.5Convert to per unit: 0.1 per unit

138100

lineX

generator has 1.05

terminal voltage &

supplies 100 MVA

with 0.95 lag pf

40

Network Fault Example, cont'd

Faulted network per unit diagram

*'a

To determine the fault current we need to first estimate

the internal voltages for the generator and motor

For the generator 1.05, 1.0 18.2

1.0 18.20.952 18.2 E 1.103 7.1

1.05

T G

Gen

V S

I

41

Network Fault Example, cont'd

f

The motor's terminal voltage is then

1.05 0 - (0.9044 - 0.2973) 0.3 1.00 15.8

The motor's internal voltage is

1.00 15.8 (0.9044 - 0.2973) 0.2

1.008 26.6

We can then solve as a linear circuit:

1I

j j

j j

.103 7.1 1.008 26.60.15 0.5

7.353 82.9 2.016 116.6 9.09

j j

j

42