EE 369POWER SYSTEM ANALYSIS

Lecture 11Power Flow

Tom Overbye and Ross Baldick

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AnnouncementsStart reading Chapter 6 for lectures 11 and 12.Homework 9 is: 3.47, 3.49, 3.53, 3.57, 3.61,

6.2, 6.9, 6.13, 6.14, 6.18, 6.19, 6.20; due November 7. (Use infinity norm and epsilon = 0.01 for any problems where norm or stopping criterion not specified.)

Read Chapter 12, concentrating on sections 12.4 and 12.5.

Homework 10 is 6.23, 6,25, 6.26, 6.28, 6.29, 6.30 (see figure 6.18 and table 6.9 for system), 6.31, 6.38, 6.42, 6.46, 6.52, 6.54; due November 14.

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Wind Blade Failure

Photo source: Peoria Journal Star

Several years ago, a 140 foot, 6.5 ton blade broke off from a Suzlon Energy wind turbine.The wind turbine is located in Illinois. Suzlon Energy isone of the world’s largest wind turbine manufacturers; its shares fell 39% followingthe accident. No one was hurtand wind turbines failuresare extremely rare events. (Vestas and Siemens turbineshave also failed.) 3

Thermal Plants Can Fail As Well: Another Illinois Failure, Fall 2007

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Springfield, Illinois City Water, Light and Power Explosion, Fall

2007

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Gauss Two Bus Power Flow Example•A 100 MW, 50 MVAr load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 MVAr on each end (100 MVA base). •Also, there is a 25 MVAr capacitor at bus 2. •If the generator voltage is 1.0 p.u., what is V2?

SLoad = 1.0 + j0.5 p.u. 6

Gauss Two Bus Example, cont’d2

2 bus

The unknown is the complex load voltage, .

To determine we need to know the ,

which is a 2 2 matrix. The capacitors have

susceptances specified by the reactive power

at the rated voltage.

Line

V

V

Y

bus

11 22

1 1series admittance = 5 15.

0.02 0.06

5 14.95 5 15Hence .

5 15 5 14.70

( Note: 15 0.05; 15 0.05 0.25).

jZ j

j j

j j

B B

Y

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Gauss Two Bus Example, cont’d1 1

2

*2

2 2 2*22 1, 22

( 1)2

2

Note that =1.0 0 is specified, so we do not update .

We only consider one entry of ( ), namely ( ).

1 SEquation to solve: ( ).

1 1 0.5Update:

5 14.70 (

n

k kk k

V V

h V h V

V Y V h VY V

jV

j V

( )

(0)2

( ) ( )2 2

( 5 15)(1.0 0))*

Guess 1.0 0 (this is known as a flat start)

0 1.000 0.000 3 0.9622 0.0556

1 0.9671 0.0568 4 0.9622 0.0556

2 0.9624 0.0553

v v

j

V

v V v V

j j

j j

j

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Gauss Two Bus Example, cont’d

2

* *1 1 11 1 12 2

ˆFixed point: 0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values can

be determined, including the generator powers and

the line flows.

ˆ( ) 1.023 0.239

In actua

V j

S V Y V Y V j

1 1

22

l units 102.3 MW, 23.9 MVAr

The capacitor is supplying 25 23.2 MVAr

P Q

V

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Slack BusIn previous example we specified S2 and V1 and

then solved for S1 and V2. We can not arbitrarily specify S at all buses

because total generation must equal total load + total losses.

We also need an angle reference bus.To solve these problems we define one bus as the

“slack” bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection. In the previous example, this was bus 1.

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Gauss for Systems with Many Buses

*( 1) ( )

( )*1,

( ) ( ) ( )1 2

( 1)

With multiple bus systems we could calculate

new values of the voltages as follows:

S1

( , ,..., )

But after we've determined , it is

i

i

nv vii ik kv

ii k k i

v v vi n

vi

V

V Y VY V

h V V V

V

( )

a better estimate

of the voltage at bus than , so it makes sense to use

this new value. Using the latest values is known as the

Gauss-Seidel iteration.

vii V

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Gauss-Seidel Iteration

( 1) ( ) ( ) ( )2 12 2 3

( 1) ( 1) ( ) ( )3 13 2 3

( 1) ( 1) ( 1) ( ) ( )4 14 2 3 4

( 1) ( 1) (1 2 3

Immediately use the new voltage estimates:

( , , , , ) (bus 1 is slack),

( , , , , )

( , , , , )

( , ,

v v v vn

v v v vn

v v v v vn

v v vn n

V h V V V V

V h V V V V

V h V V V V V

V h V V V

1) ( 1) ( )

4, , )

Gauss-Seidel usually works better than the Gauss, and

is actually easier to implement.

Gauss-Seidel is used in practice instead of Gauss.

v vnV V

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Three Types of Power Flow BusesThere are three main types of buses:

– Load (PQ), at which P and Q are fixed; goal is to solve for unknown voltage magnitude and angle at the bus.

– Slack at which the voltage magnitude and angle are fixed; iteration solves for unknown P and Q injections at the slack bus

– Generator (PV) at which P and |V| are fixed; iteration solves for unknown voltage angle and Q injection at bus:special coding is needed to include PV buses in the Gauss-

Seidel iteration.

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Inclusion of PV Buses in G-S

* *

1

( ) ( )* ( )

1

To solve for at a PV bus we must first make a

guess of using the power flow equation:

Hence Im is an

estimate of the reactive power injectio

k

i

i

n

i i ik k i ik

nv v vi i ik

k

V

Q

S V Y V P jQ

Q V Y V

( ) ( )

n.

For the Gauss iteration we use the known value

of real power and the estimate of the reactive power: v vi i iS P jQ

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Inclusion of PV Buses, cont'd( 1)

( )*( 1) ( )

( )*1,

( 1) ( 1)

( 1)

( 1)

Tentatively solve for

1

In update, set .

But since is specified, replace by .

That is, set

i

vi

v nv vii ik kv

ii k k i

vi i

vi i i

i i

V

SV Y V

Y V

V V

V V V

V V

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Two Bus PV Example

Bus 1

(slack bus)

Bus 2V1 = 1.0 V2 = 1.05

P2 = 0 MW

z = 0.02 + j 0.06

Consider the same two bus system from the previousexample, except the load is replaced by a generator

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Two Bus PV Example, cont'd( ) ( )* ( )

22 21

( ) ( )* ( ) ( )*21 221 2 2 2

( )* ( )*( 1) ( ) ( )2 2

2 212 1( )* ( )*22 221, 22 2

(0)2

( ) ( 1) ( 1)2 2 2

Im ,

Im[ ]

1 1

Guess 1.05 0

0 0 0.457

k

nv v

kk

n

k kk k

v v v

Q V Y V

Y V V Y V V

S SV Y V Y V

Y YV V

V

v S V V

j

1.045 0.83 1.050 0.83

1 0 0.535 1.049 0.93 1.050 0.93

2 0 0.545 1.050 0.96 1.050 0.96

j

j

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Generator Reactive Power LimitsThe reactive power output of generators varies

to maintain the terminal voltage; on a real generator this is done by the exciter.

To maintain higher voltages requires more reactive power.

Generators have reactive power limits, which are dependent upon the generator's MW output.

These limits must be considered during the power flow solution.

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Generator Reactive Limits, cont'dDuring power flow once a solution is obtained,

need to check if the generator reactive power output is within its limits

If the reactive power is outside of the limits, then fix Q at the max or min value, and re-solve treating the generator as a PQ bus– this is know as "type-switching"– also need to check if a PQ generator can again regulate

Rule of thumb: to raise system voltage we need to supply more VArs.

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Accelerated G-S Convergence

( 1) ( )

( 1) ( ) ( ) ( )

(

Previously in the Gauss-Seidel method we were

calculating each value as

( )

To accelerate convergence we can rewrite this as

( )

Now introduce "acceleration parameter"

v v

v v v v

x

x h x

x x h x x

x

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

With = 1 this is identical to standard Gauss-Seidel.

Larger values of may result in faster convergence.

v v v vx h x x

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Accelerated Convergence, cont’d

( 1) ( ) ( ) ( )

Consider the previous example: 1 0

(1 )

Matlab code: alpha=1.2;x=x0;x=x+alpha*(1+sqrt(x)-x).

Comparison of results with different values of

1 1.2 1.5 2

0 1 1 1 1

1 2 2.20 2.5 3

2

v v v v

x x

x x x x

2.4142 2.5399 2.6217 2.464

3 2.5554 2.6045 2.6179 2.675

4 2.5981 2.6157 2.6180 2.596

5 2.6118 2.6176 2.6180 2.626 21

Gauss-Seidel Advantages

Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system).

Relatively easy to program.

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Gauss-Seidel Disadvantages

Tends to converge relatively slowly, although this can be improved with acceleration.

Has tendency to fail to find solutions, particularly on large systems.

Tends to diverge on cases with negative branch reactances (common with compensated lines)

Need to program using complex numbers.

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Newton-Raphson AlgorithmThe second major power flow solution

method is the Newton-Raphson algorithmKey idea behind Newton-Raphson is to use

sequential linearizationGeneral form of problem: Find an such that

( ) 0

x

f x

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Newton-Raphson Method (scalar)

( )

( )

( ) ( ) ( ) ( )

2 2( ) ( )2

1. Represent by a Taylor series about the

current guess . Write for the deviation

from :

( ) ( ) ( )

1( )

2

higher order terms.

v v v v

v v

f

x x

x

dff x x f x x x

dx

d fx x

dx

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Newton-Raphson Method, cont’d

( ) ( ) ( ) ( ) ( )

( )

1( ) ( ) ( )

2. Approximate by neglecting all terms

except the first two

( ) ( ) ( )

3. Set linear approximation equal to zero

and solve for

( ) ( )

4. Sol

v v v

v

v v v

f

dff x x f x x x

dx

x

dfx x f x

dx

( 1) ( ) ( )

ve for a new estimate of solution:v v vx x x 26

Newton-Raphson Example

2

1( ) ( ) ( )

( ) ( ) 2( )

( 1) ( ) ( )

( 1) ( ) ( ) 2( )

Use Newton-Raphson to solve ( ) 0,

where: ( )= 2.

The iterative update is:

( ) ( )

1(( ) 2)

2

1(( ) 2).

2

v v v

v vv

v v v

v v vv

f x

f x x

dfx x f x

dx

x xx

x x x

x x xx

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Newton-Raphson Example, cont’d( 1) ( ) ( ) 2

( )

(0)

( ) ( ) ( )

3 3

6

1(( ) 2)

2Matlab code: x=x0; x = x-(1/(2*x))*(x^2-2).

Guess 1. Iteratiting, we get:

( )

0 1 1 0.5

1 1.5 0.25 0.08333

2 1.41667 6.953 10 2.454 10

3 1.41422 6.024 10

v v vv

v v v

x x xx

x

x f x x

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Sequential Linear Approximations

Function is f(x) = x2 - 2.Solutions to f(x) = 0 are points wheref(x) intersects x axis.

At each iteration theN-R methoduses a linearapproximationto determine the next valuefor x

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Newton-Raphson Comments

• When close to the solution the error decreases quite quickly -- method has what is known as “quadratic” convergence: – number of correct significant figures roughly

doubles at each iteration.

• f(x(v)) is known as the “mismatch,” which we would like to drive to zero.

• Stopping criteria is when f(x(v)) <

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Newton-Raphson Comments

• Results are dependent upon the initial guess. What if we had guessed x(0) = 0, or x(0) = -1?

• A solution’s region of attraction (ROA) is the set of initial guesses that converge to the particular solution.

• The ROA is often hard to determine.

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Multi-Variable Newton-Raphson

1 1

2 2

Next we generalize to the case where is an -

dimension vector, and ( ) is an -dimensional

vector function:

( )

( )( )

( )

Again we seek a solution of ( ) 0.n n

n

n

x f

x f

x f

x

f x

x

xx f x

x

f x

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Multi-Variable Case, cont’di

1 11 1 1 2

1 2

1

1 21 2

The Taylor series expansion is written for each f ( )

( ) ( ) ( ) ( )

( ) higher order terms

( ) ( ) ( ) ( )

( ) higher order terms

nn

n nn n

nn

n

f ff x x f x x x x x

x x

fx x

x

f ff x x f x x x x x

x x

fx x

x

x

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Multi-Variable Case, cont’d

1 1 1

1 21 1

2 2 22 2

1 2

1 2

This can be written more compactly in matrix form

( ) ( ) ( )

( )

( ) ( ) ( )( )( )

( )( ) ( ) ( )

n

n

nn n n

n

f f fx x x

f xf f f

f xx x x

ff f fx x x

x x x

x

x x xxf x +Δx

xx x x

higher order terms

nx

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Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The by matrix of partial derivatives is known

as the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )

n

n

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 xJ x

x x x

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Multi-Variable N-R ProcedureDerivation of N-R method is similar to the scalar case

( ) ( ) ( ) higher order terms

( ) ( ) ( )

To seek solution to ( ) 0, set linear

approximation equal to zero: 0 ( ) ( ) .

f x x f x J x x

f x x f x J x x

f x x

f x J x x

x 1

( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

( ) ( )

( ) ( )

Iterate until ( )

v v v

v v v v

v

J x f x

x x x

x x J x f x

f x36

Multi-Variable Example1

2

2 21 1 2

2 22 1 2 1 2

1 1

1 2

2 2

1 2

Solve for = such that ( ) 0 where

( ) 2 8

( ) 4

First symbolically determine the Jacobian

( ) ( )

( ) =( ) ( )

x

x

f x x x

f x x x x x

f fx x

x x

f fx x

x x

x f x

J x

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Multi-variable Example, cont’d1 2

1 2 1 2

11 1 2 1

2 1 2 1 2 2

4 2( ) =

2 2

4 2 ( )Then

2 2 ( )

Matlab code: x1=x10; x2=x20;

f1=2*x1^2+x2^2-8;

f2=x1^2-x2^2+x1*x2-4;

J = [4*x1 2*x2; 2*x1+x2 x1-2*x2;

[x1;x2] =

x x

x x x x

x x x f

x x x x x f

J x

x

x

[x1;x2]-inv(J)*[f1;f2].

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Multi-variable Example, cont’d(0)

1(1)

1(2)

( )

1Initial guess

1

1 4 2 5 2.1

1 3 1 3 1.3

2.1 8.40 2.60 2.51 1.8284

1.3 5.50 0.50 1.45 1.2122

At each iteration we check ( ) to see if it

x

x

x

f x

(2)

is

0.1556below our specified tolerance : ( )

0.0900

If = 0.2 then done. Otherwise continue iterating.

f x

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