Lecture 13 Newton-Raphson Power Flow Professor Tom Overbye Department of Electrical and Computer...

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Lecture 13Newton-Raphson Power Flow

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

Announcements

Homework 6 is 2.38, 6.8, 6.23, 6.28; you should do it before the exam but need not turn it in. Answers have been posted.

First exam is 10/9 in class; closed book, closed notes, one note sheet and calculators allowed. Last year’s tests and solutions have been posted.

Abbott power plant and substation field trip, Tuesday 10/14 starting at 12:30pm. We’ll meet at corner of Gregory and Oak streets.

Be reading Chapter 6; exam covers up through Section 6.4; we do not explicitly cover 6.1.

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

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.

10 10busj j

Two Bus Example, cont’d

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

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

Two Bus Example, cont’d

22 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

Two Bus Example, First Iteration

2 2(0)2

2 2 2(0)

2 2 2 2

0Set 0, guess

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

x1 2.0 0.2

1.0 0.9

Two Bus Example, Next Iterations

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

(2) (3)

0.0145 0.236)

0.0190 0.8554

0.0000906f( ) Done! V 0.8554 13.52

0.0001175

Two Bus Solved Values

Line Z = 0.1j

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

Two Bus Case Low Voltage Solution

2 2(0)2

This case actually has two solutions! The second

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

0Set 0, guess

Calculate

(10sin ) 2.0f( )

( 10cos ) (10) 1.0

2 2 2(0)

2 2 2 2

10 cos 10sin 2.5 0( )

10 sin 10cos 20 0 5

Low Voltage Solution, cont'd

(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

f x x x

Line Z = 0.1j

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

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

Using the Power Flow: Example 1

SLACK345

SLACK138

RAY345

RAY138

FERNA69

DEMAR69

BLT138

BOB138

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

1.02 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

1.02 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 pu1.03 pu

1.02 pu

LYNN138

1.02 pu

1.00 pu

218 MW 54 Mvar

21 MW 7 Mvar

45 MW 12 Mvar

140 MW 45 Mvar

13 Mvar

12 MW 5 Mvar

150 MW 0 Mvar

13 Mvar

15 MW 5 Mvar

2 Mvar

42 MW 2 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 30 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

16.0 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

0.0 Mvar

106 MW 8 Mvar

20 MW 8 Mvar

150 MW 0 Mvar

17 MW 3 Mvar

0 MW 0 Mvar

14 MW 4 Mvar

Usingcasefrom Example6.13

Three Bus PV Case Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

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

For this three bus case we have

( ) ( ) 0

x f x x

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

oad. This

is done by making P and Q a function of :

( cos sin ) ( ) 0

( sin cos ) ( ) 0

i k ik ik ik ik Gi Di ik

V V G B P P V

V V G B Q Q V

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( )

Q V V V

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

Voltage Dependent Load, cont'd

22 2 2(0)

2 22 2 2 2

0Again set 0, guess

Calculate

(10sin ) 2.0 2.0f( )

1.0( 10cos ) (10) 1.0

10 4( )

0 10 4 2.0 0.1667Solve

1 0 12 1.0 0.9167

Voltage Dependent Load, cont'd

Line Z = 0.1j

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

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

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

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

x x J x f x

Dishonest Newton-Raphson Example

1(0)( ) ( )

( ) ( ) 2(0)

( 1) ( ) ( ) 2(0)

Use the Dishonest Newton-Raphson to solve

( ) - 2 0

( )( )

1(( ) - 2)

(( ) - 2)2

df xx f x

x x xx

Dishonest N-R Example, cont’d

( 1) ( ) ( ) 2(0)

( ) ( )

1(( ) - 2)

Guess x 1. Iteratively solving we get

v (honest) (dishonest)

1 1.5 1.5

2 1.41667 1.375

3 1.41422 1.429

4 1.41422 1.408

x x xx

We pay a pricein increased iterations, butwith decreased computationper iteration

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

Honest N-R Region of Convergence

Maximum of 15

iterations

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.

Decoupled Power Flow Formulation

( ) ( )

( ) ( )( )

( )( ) ( ) ( )

( )2 2 2

General form of the power flow problem

( )( )

vv v v

vn Dn Gn

P Pθθ V P x

f xQ xVQ Q

Decoupling Approximation

( ) ( )

( ) ( )( )

( ) ( ) ( )

Usually the off-diagonal matrices, and

are small. Therefore we approximate them as zero:

( )( )

Then the problem

P QV θ

θ P xθf x

Q Q xV0V

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

can be decoupled

( ) ( )v v

vv v v

P Qθ P x V Q x

Off-diagonal Jacobian Terms

Justification for Jacobian approximations:

1. Usually r x, therefore

2. Usually is small so sin 0

Therefore

cos sin 0

ii ij ij ij ij

ii j ij ij ij ij

V V G B

Decoupled N-R Region of Convergence

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

FDPF Approximations

( ) ( )( )( ) 1 1

( ) ( )

The FDPF makes the following approximations:

1. G 0

3. sin 0 cos 1

( ) ( )

Where is just the imaginary part of the ,

except the slack bus row/co

P x Q xθ B V B

V VB Y G B

lumn are omitted

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

FDPF Three Bus Example, cont’d

(0)(0)2 2

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 0.0477 0.0159 2 0.1272

0 0.0159 0.0389 2 0.1091

FDPF Three Bus Example, cont’d

1 0.0477 0.0159 1 0.9364

1 0.0159 0.0389 1 0.9455

P ( )( cos sin )

0.1272 0.0477 0.0159

0.1091 0.0159 0.0389

nDi Gi

k ik ik ik ikk

P PV G B

0.151 0.1361

0.107 0.1156

0.1384 0.9224Actual solution:

0.1171 0.9338

FDPF Region of Convergence

“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

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

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

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

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

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.

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

i j ijij

Analytic Sensitivities

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.

θ B P x

Three Bus Sensitivity Example

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

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

Lecture 13 Newton-Raphson Power Flow Professor Tom Overbye Department of Electrical and Computer...

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