Lecture 14Newton-Raphson Power Flow

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Announcements

Homework 6 is due now Be reading Chapter 6 Homework 7 is 6.12, 6.19, 6.22, 6.45 and 6.50.

Due date is October 25 Design Project 2 from the book is assigned today

(page 345 to 348). It will be due Nov 29

3

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

4

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

5

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

6

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

7

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

8

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

9

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

10

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

11

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 atbus 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

12

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

13

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

14

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

15

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

16

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

17

Two Bus Dishonest ROC

Slide shows the region of convergence for different initialguesses for the 2 bus case using the Dishonest N-R

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

18

Honest N-R Region of Convergence

Maximum of 15

iterations

19

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.

20

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

21

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 V

22

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

23

Decoupled N-R Region of Convergence

24

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

25

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

26

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

27

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

28

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

29

FDPF Region of Convergence

30

“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

31

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

32

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

33

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%

34

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

35

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

36

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

37

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