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Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

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Page 1: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

Lecture 13Power Flow

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

Page 2: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

2

Announcements

Be reading Chapter 6, also Chapter 2.4 (Network Equations). HW 5 is 2.38, 6.9, 6.18, 6.30, 6.34, 6.38; do by October 6 but

does not need to be turned in. First exam is October 11 during class. Closed book, closed

notes, one note sheet and calculators allowed. Exam covers thru the end of lecture 13 (today)

An example previous exam (2008) is posted. Note this is exam was given earlier in the semester in 2008 so it did not include power flow, but the 2011 exam will (at least partially)

Page 3: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

3

Multi-Variable Example

1

2

2 21 1 2

2 22 1 2 1 2

1 1

1 2

2 2

1 2

xSolve for = such that ( ) 0 where

x

f ( ) 2 8 0

f ( ) 4 0

First symbolically determine the Jacobian

f ( ) f ( )

( ) =f ( ) f ( )

x x

x x x x

x x

x x

x f x

x

x

x x

J xx x

Page 4: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

4

Multi-variable Example, cont’d

1 2

1 2 1 2

11 1 2 1

2 1 2 1 2 2

(0)

1(1)

4 2( ) =

2 2

Then

4 2 ( )

2 2 ( )

1Arbitrarily guess

1

1 4 2 5 2.1

1 3 1 3 1.3

x x

x x x x

x x x f

x x x x x f

J x

x

x

x

x

Page 5: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

5

Multi-variable Example, cont’d

1(2)

(2)

2.1 8.40 2.60 2.51 1.8284

1.3 5.50 0.50 1.45 1.2122

Each iteration we check ( ) to see if it is below our

specified tolerance

0.1556( )

0.0900

If = 0.2 then we wou

x

f x

f x

ld be done. Otherwise we'd

continue iterating.

Page 6: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

6

Possible EHV Overlays for Wind

AEP 2007 Proposed Overlay

Page 7: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

7

NR Application to Power Flow

** * *

i1 1

We first need to rewrite complex power equations

as equations with real coefficients

S

These can be derived by defining

Recal

i

n n

i i i ik k i ik kk k

ik ik ik

ji i i i

ik i k

V I V Y V V Y V

Y G jB

V V e V

jl e cos sinj

Page 8: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

8

Real Power Balance Equations

* *i

1 1

1

i1

i1

S ( )

(cos sin )( )

Resolving into the real and imaginary parts

P ( cos sin )

Q ( sin cos

ikn n

ji i i ik k i k ik ik

k k

n

i k ik ik ik ikk

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik

P jQ V Y V V V e G jB

V V j G jB

V V G B P P

V V G B

)k Gi DiQ Q

Page 9: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

9

Newton-Raphson Power Flow

i1

In the Newton-Raphson power flow we use Newton's

method to determine the voltage magnitude and angle

at each bus in the power system.

We need to solve the power balance equations

P ( cosn

i k ik ikk

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n

i k ik ik ik ik Gi Dik

B P P

V V G B Q Q

Page 10: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

10

Power Flow Variables

2 2 2

n

2

Assume the slack bus is the first bus (with a fixed

voltage angle/magnitude). We then need to determine

the voltage angle/magnitude at the other buses.

( )

( )

G

n

P P

V

V

x

x f x

2

2 2 2

( )

( )

( )

D

n Gn Dn

G D

n Gn Dn

P

P P P

Q Q Q

Q Q Q

x

x

x

Page 11: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

11

N-R Power Flow Solution

( )

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedure

discussed last time:

Set 0; make an initial guess of ,

While ( ) Do

( ) ( )

1

End While

v

v

v v v v

v

v v

x x

f x

x x J x f x

Page 12: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

12

Power Flow Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The most difficult part of the algorithm is determining

and inverting the n by n Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

f f fx x x

f f fx x x

f f fx x x

J x

x x x

x x x

J x

x x x

Page 13: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

13

Power Flow Jacobian Matrix, cont’d

i

i

i1

Jacobian elements are calculated by differentiating

each function, f ( ), with respect to each variable.

For example, if f ( ) is the bus i real power equation

f ( ) ( cos sin )n

i k ik ik ik ik Gik

x V V G B P P

x

x

i

1

i

f ( )( sin cos )

f ( )( sin cos ) ( )

Di

n

i k ik ik ik iki k

k i

i j ik ik ik ikj

xV V G B

xV V G B j i

Page 14: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

14

Two Bus Newton-Raphson Example

Line Z = 0.1j

One Two 1.000 pu 1.000 pu

200 MW 100 MVR

0 MW 0 MVR

For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.

2

2

10 10

10 10busj j

V j j

x Y

Page 15: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

15

Two Bus Example, cont’d

i1

i1

2 1 2

22 1 2 2

General power balance equations

P ( cos sin )

Q ( sin cos )

Bus two power balance equations

(10sin ) 2.0 0

( 10cos ) (10) 1.0 0

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

V V

V V V

Page 16: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

16

Two Bus Example, cont’d

2 2 2

22 2 2 2

2 2

2 2

2 2

2 2

2 2 2

2 2 2 2

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

Now calculate the power flow Jacobian

P ( ) P ( )

( )Q ( ) Q ( )

10 cos 10sin

10 sin 10cos 20

V

Q V V

VJ

V

V

V V

x

x

x x

xx x

Page 17: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

17

Two Bus Example, First Iteration

(0)

2 2(0)2

2 2 2

2 2 2(0)

2 2 2 2

(1)

0Set 0, guess

1

Calculate

(10sin ) 2.0 2.0f( )

1.0( 10cos ) (10) 1.0

10 cos 10sin 10 0( )

10 sin 10cos 20 0 10

0 10 0Solve

1 0 10

v

V

V V

V

V V

x

x

J x

x1 2.0 0.2

1.0 0.9

Page 18: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

18

Two Bus Example, Next Iterations

(1)2

(1)

1(2)

0.9(10sin( 0.2)) 2.0 0.212f( )

0.2790.9( 10cos( 0.2)) 0.9 10 1.0

8.82 1.986( )

1.788 8.199

0.2 8.82 1.986 0.212 0.233

0.9 1.788 8.199 0.279 0.8586

f(

x

J x

x

(2) (3)

(3)2

0.0145 0.236)

0.0190 0.8554

0.0000906f( ) Done! V 0.8554 13.52

0.0001175

x x

x

Page 19: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

19

Two Bus Solved Values

Line Z = 0.1j

One Two 1.000 pu 0.855 pu

200 MW 100 MVR

200.0 MW168.3 MVR

-13.522 Deg

200.0 MW 168.3 MVR

-200.0 MW-100.0 MVR

Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values,such as the line flows and the generator reactive power output

Page 20: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

20

Two Bus Case Low Voltage Solution

(0)

2 2(0)2

2 2 2

This case actually has two solutions! The second

"low voltage" is found by using a low initial guess.

0Set 0, guess

0.25

Calculate

(10sin ) 2.0f( )

( 10cos ) (10) 1.0

v

V

V V

x

x

2 2 2(0)

2 2 2 2

2

0.875

10 cos 10sin 2.5 0( )

10 sin 10cos 20 0 5

V

V V

J x

Page 21: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

21

Low Voltage Solution, cont'd

1(1)

(2) (2) (3)

0 2.5 0 2 0.8Solve

0.25 0 5 0.875 0.075

1.462 1.42 0.921( )

0.534 0.2336 0.220

x

f x x x

Line Z = 0.1j

One Two 1.000 pu 0.261 pu

200 MW 100 MVR

200.0 MW831.7 MVR

-49.914 Deg

200.0 MW 831.7 MVR

-200.0 MW-100.0 MVR

Low voltage solution

Page 22: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

22

Two Bus Region of Convergence

Slide shows the region of convergence for different initialguesses of bus 2 angle (x-axis) and magnitude (y-axis)

Red regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution

Page 23: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

23

PV Buses

Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x or write the reactive power balance equations– the reactive power output of the generator varies to

maintain the fixed terminal voltage (within limits)– optionally these variations/equations can be included by

just writing the explicit voltage constraint for the generator bus

|Vi | – Vi setpoint = 0

Page 24: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

24

Three Bus PV Case Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

One Two 1.000 pu 0.941 pu

200 MW 100 MVR

170.0 MW 68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW 63 MVR

2 2 2 2

3 3 3 3

2 2 2

For this three bus case we have

( )

( ) ( ) 0

V ( )

G D

G D

D

P P P

P P P

Q Q

x

x f x x

x

Page 25: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

25

400 MVA15 kV

400 MVA15/345 kV

T1

T2800 MVA345/15 kV

800 MVA15 kV

520 MVA

80 MW40 Mvar

280 Mvar 800 MW

Line 3 345 kV

Lin

e 2

Lin

e 1345 kV

100 mi345 kV 200 mi

50 mi

1 4 3

2

5

Single-line diagram

The N-R Power Flow: 5-bus Example

Page 26: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

26

Bus Type

V

per unit

degrees

PG

per

unit

QG

per

unit

PL

per

unit

QL

per

unit

QGmax

per

unit

QGmin

per

unit

1 Swing 1.0 0 0 0

2 Load 0 0 8.0 2.8

3 Constant voltage

1.05 5.2 0.8 0.4 4.0 -2.8

4 Load 0 0 0 0

5 Load 0 0 0 0

Table 1. Bus input data

Bus-to-Bus

R’

per unit

X’

per unit

G’

per unit

B’

per unit

Maximum

MVA

per unit

2-4 0.0090 0.100 0 1.72 12.0

2-5 0.0045 0.050 0 0.88 12.0

4-5 0.00225 0.025 0 0.44 12.0

Table 2. Line input data

The N-R Power Flow: 5-bus Example

Page 27: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

27

Bus-to-Bus

R

per

unit

X

per

unit

Gc

per

unit

Bm

per

unit

Maximum

MVA

per unit

Maximum

TAP

Setting

per unit

1-5 0.00150 0.02 0 0 6.0 —

3-4 0.00075 0.01 0 0 10.0 —

Table 3. Transformer input data

Bus Input Data Unknowns

1 V1 = 1.0, 1 = 0 P1, Q1

2 P2 = PG2-PL2 = -8

Q2 = QG2-QL2 = -2.8

V2, 2

3 V3 = 1.05

P3 = PG3-PL3 = 4.4

Q3, 3

4 P4 = 0, Q4 = 0 V4, 4

5 P5 = 0, Q5 = 0 V5, 5

Table 4. Input data and unknowns

The N-R Power Flow: 5-bus Example

Page 28: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

28

Time to Close the Hood: Let the Computer Do the Math! (Ybus Shown)

Page 29: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

29

Ybus Details

02321 YY

unitperjjjXR

Y 91964.989276.01.0009.0

11'24

'24

24

unitperjjjXR

Y 83932.1978552.105.00045.0

11'25

'25

25

22

11 '25

'24

'25

'25

'24

'24

22

Bj

Bj

jXRjXRY

2

88.0

2

72.1)83932.1978552.1()91964.989276.0( jjjj

unitperj 624.845847.284590.2867828.2

Elements of Ybus connected to bus 2

Page 30: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

30

Here are the Initial Bus Mismatches

Page 31: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

31

And the Initial Power Flow Jacobian

Page 32: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

32

])0()0(cos[){0()()0( 211212122222 VYVPxPPP])0()0(cos[]cos[ 233232322222 VYVY

])0()0(cos[ 2442424 VY]})0()0(cos[ 2552525 VY

)624.84cos()0.1(5847.28{0.10.8 )143.95cos()0.1(95972.9 )}143.95cos()0.1(9159.19

unitper99972.7)1089.2(0.8 4

])0()0(sin[)0()0()0(1 2442424224 VYVJ

]143.95sin[)0.1)(95972.9)(0.1( unitper91964.9

And the Hand Calculation Details!

Page 33: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

33

Five Bus Power System Solved

slack

One

Two

ThreeFourFiveA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.000 pu 0.974 pu

0.834 pu

1.019 pu

1.050 pu 0.000 Deg -4.548 Deg

-22.406 Deg

-2.834 Deg

-0.597 Deg

395 MW

114 Mvar

520 MW

337 Mvar

800 MW 280 Mvar

80 MW 40 Mvar

Page 34: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

34

37 Bus Example Design Case

slack

Metropolis Light and Power Electric Design Case 2SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1.02 pu

1.03 pu

1.03 pu

1.01 pu

1.00 pu1.01 pu

1.00 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu

1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.00 pu

1.00 pu

1.02 pu

0.99 pu

0.99 pu

1.00 pu

1.02 pu

1.00 pu1.01 pu

1.01 pu

1.00 pu 1.00 pu

1.01 pu

1.02 pu 1.02 pu

1.02 pu 1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

1.00 pu

A

MVA

System Losses: 10.70 MW

220 MW 52 Mvar

12 MW 3 Mvar

20 MW 12 Mvar

124 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 0 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

38 MW 3 Mvar

45 MW 0 Mvar

25 MW 36 Mvar

36 MW 10 Mvar

10 MW 5 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 28 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

15.9 Mvar 18 MW 5 Mvar

58 MW 40 Mvar

60 MW 19 Mvar

14.2 Mvar

25 MW 10 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.9 Mvar

7.3 Mvar

12.8 Mvar

28.9 Mvar

7.4 Mvar

0.0 Mvar

55 MW 25 Mvar

39 MW 13 Mvar

150 MW 0 Mvar

17 MW 3 Mvar

16 MW -14 Mvar

14 MW 4 Mvar

KYLE69A

MVA

Page 35: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

35

Good Power System Operation

• Good power system operation requires that there be no reliability violations for either the current condition or in the event of statistically likely contingencies• Reliability requires as a minimum that there be no transmission

line/transformer limit violations and that bus voltages be within acceptable limits (perhaps 0.95 to 1.08)

• Example contingencies are the loss of any single device. This is known as n-1 reliability.

• North American Electric Reliability Corporation now has legal authority to enforce reliability standards (and there are now lots of them). See http://www.nerc.com for details (click on Standards)

Page 36: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

36

Looking at the Impact of Line Outages

slack

Metropolis Light and Power Electric Design Case 2SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1.02 pu

1.03 pu

1.03 pu

1.01 pu

1.00 pu1.01 pu

1.00 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu

1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.01 pu

1.00 pu

1.02 pu

0.90 pu

0.90 pu

0.94 pu

1.01 pu

0.99 pu1.00 pu

1.00 pu

1.00 pu 1.00 pu

1.01 pu

1.01 pu 1.02 pu

1.02 pu 1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

1.00 pu

A

MVA

System Losses: 17.61 MW

227 MW 43 Mvar

12 MW 3 Mvar

20 MW 12 Mvar

124 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 4 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

38 MW 9 Mvar

45 MW 0 Mvar

25 MW 36 Mvar

36 MW 10 Mvar

10 MW 5 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 40 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

16.0 Mvar 18 MW 5 Mvar

58 MW 40 Mvar

60 MW 19 Mvar

11.6 Mvar

25 MW 10 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.9 Mvar

7.2 Mvar

12.8 Mvar

28.9 Mvar

7.3 Mvar

0.0 Mvar

55 MW 32 Mvar

39 MW 13 Mvar

150 MW 4 Mvar

17 MW 3 Mvar

16 MW -14 Mvar

14 MW 4 Mvar

KYLE69A

MVA

80%A

MVA

135%A

MVA

110%A

MVA

Opening one line (Tim69-Hannah69) causes an overload. This would not be allowed

Page 37: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

37

Contingency Analysis

Contingencyanalysis providesan automaticway of lookingat all the statisticallylikely contingencies. Inthis example thecontingency setIs all the single line/transformeroutages

Page 38: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

38

Power Flow And Design

• One common usage of the power flow is to determine how the system should be modified to remove contingencies problems or serve new load• In an operational context this requires working with the

existing electric grid• In a planning context additions to the grid can be considered

• In the next example we look at how to remove the existing contingency violations while serving new load.

Page 39: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

39

An Unreliable Solution

slack

Metropolis Light and Power Electric Design Case 2SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.02 pu

1.01 pu

1.02 pu

1.03 pu

1.01 pu

1.00 pu1.01 pu

1.00 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu

1.01 pu

1.01 pu

1.01 pu

1.02 pu

0.99 pu

1.00 pu

1.02 pu

0.97 pu

0.97 pu

0.99 pu

1.02 pu

1.00 pu1.01 pu

1.01 pu

1.00 pu 1.00 pu

1.01 pu

1.02 pu 1.02 pu

1.02 pu 1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

1.00 pu

A

MVA

System Losses: 14.49 MW

269 MW 67 Mvar

12 MW 3 Mvar

20 MW 12 Mvar

124 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 1 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

38 MW 4 Mvar

45 MW 0 Mvar

25 MW 36 Mvar

36 MW 10 Mvar

10 MW 5 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 40 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

15.9 Mvar 18 MW 5 Mvar

58 MW 40 Mvar

60 MW 19 Mvar

13.6 Mvar

25 MW 10 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.9 Mvar

7.3 Mvar

12.8 Mvar

28.9 Mvar

7.4 Mvar

0.0 Mvar

55 MW 28 Mvar

39 MW 13 Mvar

150 MW 1 Mvar

17 MW 3 Mvar

16 MW -14 Mvar

14 MW 4 Mvar

KYLE69A

MVA

96%A

MVA

Case now has nine separate contingencies with reliability violations

Page 40: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

40

A Reliable Solution

slack

Metropolis Light and Power Electric Design Case 2SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1.01 pu

1.02 pu

1.03 pu

1.01 pu

1.00 pu1.01 pu

1.00 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu

1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.00 pu

0.99 pu

1.02 pu

0.99 pu

0.99 pu

1.00 pu

1.02 pu

1.00 pu1.01 pu

1.01 pu

1.00 pu 1.00 pu

1.01 pu

1.02 pu 1.02 pu

1.02 pu 1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

A

MVA

System Losses: 11.66 MW

266 MW 59 Mvar

12 MW 3 Mvar

20 MW 12 Mvar

124 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 1 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

38 MW 4 Mvar

45 MW 0 Mvar

25 MW 36 Mvar

36 MW 10 Mvar

10 MW 5 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 38 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

15.8 Mvar 18 MW 5 Mvar

58 MW 40 Mvar

60 MW 19 Mvar

14.1 Mvar

25 MW 10 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.9 Mvar

7.3 Mvar

12.8 Mvar

28.9 Mvar

7.4 Mvar

0.0 Mvar

55 MW 29 Mvar

39 MW 13 Mvar

150 MW 1 Mvar

17 MW 3 Mvar

16 MW -14 Mvar

14 MW 4 Mvar

KYLE69A

MVA

Kyle138A

MVA

Previous case was augmented with the addition of a 138 kV Transmission Line

Page 41: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

41

Generation Changes and The Slack Bus

• The power flow is a steady-state analysis tool, so the assumption is total load plus losses is always equal to total generation• Generation mismatch is made up at the slack bus

• When doing generation change power flow studies one always needs to be cognizant of where the generation is being made up• Common options include system slack, distributed across

multiple generators by participation factors or by economics

Page 42: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

42

Generation Change Example 1

slack

SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69

HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

0.00 pu

-0.01 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu-0.01 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu

-0.01 pu0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

-0.002 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu0.00 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu

0.00 pu0.00 pu

0.00 pu0.00 pu

A

MVA

-0.01 pu

A

MVA

A

MVA

LYNN138

A

MVA

0.00 pu

A

MVA

0.00 pu

A

MVA

162 MW 35 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-157 MW -45 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW 2 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

0 MW 3 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 4 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-0.1 Mvar 0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

-0.1 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

-0.1 Mvar

0.0 Mvar

-0.1 Mvar

-0.2 Mvar

0.0 Mvar

0.0 Mvar

0 MW 51 Mvar

0 MW 0 Mvar

0 MW 2 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

Display shows “Difference Flows” between original 37 bus case, and case with a BLT138 generation outage; note all the power change is picked up at the slack

Page 43: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

43

Generation Change Example 2

slack

SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69

HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

0.00 pu

-0.01 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu0.00 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu

-0.01 pu-0.01 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

-0.003 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu0.00 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu

0.00 pu0.00 pu

0.00 pu0.00 pu

A

MVA

0.00 pu

A

MVA

A

MVA

LYNN138

A

MVA

0.00 pu

A

MVA

0.00 pu

A

MVA

0 MW 37 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-157 MW -45 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

42 MW -14 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

99 MW -20 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-0.1 Mvar 0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

-0.1 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

0.0 Mvar

0.0 Mvar

-0.1 Mvar

-0.2 Mvar

-0.1 Mvar

0.0 Mvar

19 MW 51 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

Display repeats previous case except now the change in generation is picked up by other generators using a participation factor approach

Page 44: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

44

Voltage Regulation Example: 37 Buses

Display shows voltage contour of the power system, demo will show the impact of generator voltage set point, reactive power limits, and switched capacitors

slack

SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69

HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVAA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1.01 pu

1.02 pu

1.03 pu

1.01 pu

1.00 pu1.00 pu

0.99 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.00 pu

1.00 pu

1.02 pu

0.997 pu

0.99 pu

1.00 pu

1.02 pu

1.00 pu1.01 pu

1.00 pu

1.00 pu 1.00 pu

1.01 pu

1.02 pu1.02 pu

1.02 pu 1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

1.00 pu

A

MVA

219 MW 52 Mvar

21 MW 7 Mvar

45 MW 12 Mvar

157 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 0 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

38 MW 3 Mvar

45 MW 0 Mvar

58 MW 36 Mvar

36 MW 10 Mvar

0 MW 0 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 9 Mvar

23 MW 7 Mvar

33 MW 13 Mvar 15.9 Mvar 18 MW

5 Mvar

58 MW 40 Mvar 51 MW

15 Mvar

14.3 Mvar

33 MW

10 Mvar

15 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.8 Mvar

7.2 Mvar

12.8 Mvar

29.0 Mvar

7.4 Mvar

20.8 Mvar

92 MW 10 Mvar

20 MW 8 Mvar

150 MW 0 Mvar

17 MW 3 Mvar

0 MW 0 Mvar

14 MW 4 Mvar

1.010 pu 0.0 Mvar

System Losses: 11.51 MW

Page 45: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

45

Solving Large Power Systems

The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning

the amount of computation increases with the cube of the size size

– this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix

– using sparse matrix methods results in a computational order of about n1.5.

– this is a substantial savings when solving systems with tens of thousands of buses

Page 46: Lecture 13 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

46

Newton-Raphson Power Flow

Advantages– fast convergence as long as initial guess is close to solution– large region of convergence

Disadvantages– each iteration takes much longer than a Gauss-Seidel iteration– more complicated to code, particularly when implementing

sparse matrix algorithms

Newton-Raphson algorithm is very common in power flow analysis