ECE 476 Power System Analysis

Lecture 14: Power Flow

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Read Chapter 6, Chapter 12.4 and 12.5 • Quiz today on HW 6• HW 7 is 6.50, 6.52, 6.59, 12.20, 12.26; due October

22 in class (no quiz) • Power and Energy scholarships will be decided on

Monday; application on website; apply to Prof. Sauer; Grainger Awards due on Nov 1; application on website; apply to Prof. Sauer• energy.ece.illinois.edu

2

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

3

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

4

Modeling Voltage Dependent Load

So far we've assumed that the load is independent of

the bus voltage (i.e., constant power). However, the

power flow can be easily extended to include voltage

depedence with both the real and reactive l

Di Di

1

1

oad. This

is done by making P and Q a function of :

( cos sin ) ( ) 0

( sin cos ) ( ) 0

i

n

i k ik ik ik ik Gi Di ik

n

i k ik ik ik ik Gi Di ik

V

V V G B P P V

V V G B Q Q V

5

Voltage Dependent Load Example

22 2 2 2

2 22 2 2 2 2

2 2 2 2

In previous two bus example now assume the load is

constant impedance, so

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

Now calculate the power flow Jacobian

10 cos 10sin 4.0( )

10

V V

Q V V V

V VJ

x

x

x2 2 2 2 2sin 10cos 20 2.0V V V

6

Voltage Dependent Load, cont'd

(0)

22 2 2(0)

2 22 2 2 2

(0)

1(1)

0Again set 0, guess

1

Calculate

(10sin ) 2.0 2.0f( )

1.0( 10cos ) (10) 1.0

10 4( )

0 12

0 10 4 2.0 0.1667Solve

1 0 12 1.0 0.9167

v

V V

V V V

x

x

J x

x

7

Voltage Dependent Load, cont'd

Line Z = 0.1j

One Two 1.000 pu 0.894 pu

160 MW 80 MVR

160.0 MW120.0 MVR

-10.304 Deg

160.0 MW 120.0 MVR

-160.0 MW -80.0 MVR

With constant impedance load the MW/Mvar load at

bus 2 varies with the square of the bus 2 voltage

magnitude. This if the voltage level is less than 1.0,

the load is lower than 200/100 MW/Mvar

8

Dishonest Newton-Raphson

• Since most of the time in the Newton-Raphson iteration is spend calculating the inverse of the Jacobian, one way to speed up the iterations is to only calculate/inverse the Jacobian occasionally– known as the “Dishonest” Newton-Raphson– an extreme example is to only calculate the Jacobian for

the first iteration( 1) ( ) ( ) -1 ( )

( 1) ( ) (0) -1 ( )

( )

Honest: - ( ) ( )

Dishonest: - ( ) ( )

Both require ( ) for a solution

v v v v

v v v

v

x x J x f x

x x J x f x

f x

9

Dishonest Newton-Raphson Example

2

1(0)( ) ( )

( ) ( ) 2(0)

( 1) ( ) ( ) 2(0)

Use the Dishonest Newton-Raphson to solve

( ) - 2 0

( )( )

1(( ) - 2)

21

(( ) - 2)2

v v

v v

v v v

f x x

df xx f x

dx

x xx

x x xx

10

Dishonest N-R Example, cont’d

( 1) ( ) ( ) 2(0)

(0)

( ) ( )

1(( ) - 2)

2

Guess x 1. Iteratively solving we get

v (honest) (dishonest)

0 1 1

1 1.5 1.5

2 1.41667 1.375

3 1.41422 1.429

4 1.41422 1.408

v v v

v v

x x xx

x x

We pay a pricein increased iterations, butwith decreased computationper iteration

11

Two Bus Dishonest ROC

Slide shows the region of convergence for different initial

guesses for the 2 bus case using the Dishonest N-RRed regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution

12

Honest N-R Region of Convergence

Maximum of 15

iterations

13

Decoupled Power Flow

• The completely Dishonest Newton-Raphson is not used for power flow analysis. However several approximations of the Jacobian matrix are used.

• One common method is the decoupled power flow. In this approach approximations are used to decouple the real and reactive power equations.

14

Decoupled Power Flow Formulation

( ) ( )

( ) ( )( )

( )( ) ( ) ( )

( )2 2 2

( )

( )

General form of the power flow problem

( )( )

( )

where

( )

( )

( )

v v

v vv

vv v v

vD G

v

vn Dn Gn

P P P

P P P

P Pθθ V P x

f xQ xVQ Q

θ V

x

P x

x

15

Decoupling Approximation

( ) ( )

( )

( ) ( )( )

( ) ( ) ( )

Usually the off-diagonal matrices, and

are small. Therefore we approximate them as zero:

( )( )

( )

Then the problem

v v

v

v vv

v v v

P QV θ

P0

θ P xθf x

Q Q xV0V

1 1( ) ( )( )( ) ( ) ( )

can be decoupled

( ) ( )v v

vv v v

P Qθ P x V Q x

θ V16

Off-diagonal Jacobian Terms

Justification for Jacobian approximations:

1. Usually r x, therefore

2. Usually is small so sin 0

Therefore

cos sin 0

cos sin 0

ij ij

ij ij

ii ij ij ij ij

j

ii j ij ij ij ij

j

G B

V G B

V V G B

P

V

Qθ

17

Decoupled N-R Region of Convergence

18

Fast Decoupled Power Flow

• By continuing with our Jacobian approximations we can actually obtain a reasonable approximation that is independent of the voltage magnitudes/angles.

• This means the Jacobian need only be built/inverted once.

• This approach is known as the fast decoupled power flow (FDPF)

• FDPF uses the same mismatch equations as standard power flow so it should have same solution

• The FDPF is widely used, particularly when we only need an approximate solution

19

FDPF Approximations

ij

( ) ( )( )( ) 1 1

( ) ( )

bus

The FDPF makes the following approximations:

1. G 0

2. 1

3. sin 0 cos 1

Then

( ) ( )

Where is just the imaginary part of the ,

except the slack bus row/co

i

ij ij

v vvv

v v

V

j

P x Q xθ B V B

V VB Y G B

lumn are omitted

20

FDPF Three Bus Example

Line Z = j0.07

Line Z = j0.05 Line Z = j0.1

One Two

200 MW 100 MVR

Three 1.000 pu

200 MW 100 MVR

Use the FDPF to solve the following three bus system

34.3 14.3 20

14.3 24.3 10

20 10 30bus j

Y

21

FDPF Three Bus Example, cont’d

1

(0)(0)2 2

3 3

34.3 14.3 2024.3 10

14.3 24.3 1010 30

20 10 30

0.0477 0.0159

0.0159 0.0389

Iteratively solve, starting with an initial voltage guess

0 1

0 1

bus j

V

V

Y B

B

(1)2

3

0 0.0477 0.0159 2 0.1272

0 0.0159 0.0389 2 0.1091

22

FDPF Three Bus Example, cont’d

(1)2

3

i

i i1

(2)2

3

1 0.0477 0.0159 1 0.9364

1 0.0159 0.0389 1 0.9455

P ( )( cos sin )

V V

0.1272 0.0477 0.0159

0.1091 0.0159 0.0389

nDi Gi

k ik ik ik ikk

V

V

P PV G B

x

(2)2

3

0.151 0.1361

0.107 0.1156

0.924

0.936

0.1384 0.9224Actual solution:

0.1171 0.9338

V

V

θ V

23

FDPF Region of Convergence

24

“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

25

Power System Control

• A major problem with power system operation is the limited capacity of the transmission system– lines/transformers have limits (usually thermal)– no direct way of controlling flow down a transmission

line (e.g., there are no valves to close to limit flow)– open transmission system access associated with industry

restructuring is stressing the system in new ways

• We need to indirectly control transmission line flow by changing the generator outputs

26

DC Power Flow Example

27

DC Power Flow 5 Bus Example

slack

One

Two

ThreeFourFiveA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.000 pu 1.000 pu

1.000 pu

1.000 pu

1.000 pu 0.000 Deg -4.125 Deg

-18.695 Deg

-1.997 Deg

0.524 Deg

360 MW

0 Mvar

520 MW

0 Mvar

800 MW 0 Mvar

80 MW 0 Mvar

Notice with the dc power flow all of the voltage magnitudes are 1 per unit.

28

Indirect Transmission Line Control

What we would like to determine is how a change in

generation at bus k affects the power flow on a line

from bus i to bus j. The assumption isthat the changein generation isabsorbed by theslack bus

29

Power Flow Simulation - Before

•One way to determine the impact of a generator change is to compare a before/after power flow.•For example below is a three bus case with an overload

Z for all lines = j0.1

One Two

200 MW 100 MVR

200.0 MW 71.0 MVR

Three 1.000 pu

0 MW 64 MVR

131.9 MW

68.1 MW 68.1 MW

124%

30

Power Flow Simulation - After

Z for all lines = j0.1Limit for all lines = 150 MVA

One Two

200 MW 100 MVR

105.0 MW 64.3 MVR

Three1.000 pu

95 MW 64 MVR

101.6 MW

3.4 MW 98.4 MW

92%

100%

Increasing the generation at bus 3 by 95 MW (and hence

decreasing it at bus 1 by a corresponding amount), results

in a 31.3 drop in the MW flow on the line from bus 1 to 2.

31

Analytic Calculation of Sensitivities

• Calculating control sensitivities by repeat power flow solutions is tedious and would require many power flow solutions. An alternative approach is to analytically calculate these values

The power flow from bus i to bus j is

sin( )

So We just need to get

i j i jij i j

ij ij

i j ijij

ij Gk

V VP

X X

PX P

32

Analytic Sensitivities

1

From the fast decoupled power flow we know

( )

So to get the change in due to a change of

generation at bus k, just set ( ) equal to

all zeros except a minus one at position k.

0

1

0

θ B P x

θ

P x

P

Bus k

33

Three Bus Sensitivity Example

line

bus

12

3

For the previous three bus case with Z 0.1

20 10 1020 10

10 20 1010 20

10 10 20

Hence for a change of generation at bus 3

20 10 0 0.0333

10 20 1 0.0667

j

j

Y B

3 to 1

3 to 2 2 to 1

0.0667 0Then P 0.667 pu

0.1P 0.333 pu P 0.333 pu

34

Balancing Authority Areas

• An balancing authority area (use to be called operating areas) has traditionally represented the portion of the interconnected electric grid operated by a single utility

• Transmission lines that join two areas are known as tie-lines.

• The net power out of an area is the sum of the flow on its tie-lines.

• The flow out of an area is equal to

total gen - total load - total losses = tie-flow 35

Area Control Error (ACE)

• The area control error (ace) is the difference between the actual flow out of an area and the scheduled flow, plus a frequency component

• Ideally the ACE should always be zero.• Because the load is constantly changing,

each utility must constantly change its generation to “chase” the ACE.

int schedace 10P P f

36

Automatic Generation Control

• Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero.

• Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds.

37

Power Transactions

• Power transactions are contracts between generators and loads to do power transactions.

• Contracts can be for any amount of time at any price for any amount of power.

• Scheduled power transactions are implemented by modifying the value of Psched used in the ACE calculation

38