Power Flow Analysis

Well known as : Load Flow

The Power Flow Problem

• Power flow analysis is fundamental to the study of power

systems.

• In fact, power flow forms the core of power system analysis.

• power flow study plays a key role in the planning of additions

or expansions to transmission and generation facilities.

• A power flow solution is often the starting point for many other

types of power system analyses.

• In addition, power flow analysis is at the heart of contingency

analysis and the implementation of real-time monitoring

systems.

Problem Statement

For a given power network, with known complex power

loads and some set of specifications or restrictions on

power generations and voltages, solve for any unknown

bus voltages and unspecified generation and finally for

the complex power flow in the network components.

Network Structure

Power Flow Study Steps

1. Determine element values for passive network components.

2. Determine locations and values of all complex power loads.

3. Determine generation specifications and constraints.

4. Develop a mathematical model describing power flow in the network.

5. Solve for the voltage profile of the network.

6. Solve for the power flows and losses in the network.

7. Check for constraint violations.

Formulation of the Bus Admittance Matrix

• The first step in developing the mathematical model

describing the power flow in the network is the

formulation of the bus admittance matrix.

• The bus admittance matrix is an n*n matrix (where n is

the number of buses in the system) constructed from the

admittances of the equivalent circuit elements of the

segments making up the power system.

• Most system segments are represented by a combination

of shunt elements (connected between a bus and the

reference node) and series elements (connected between

two system buses).

Bus Admittance Matrix

Formulation of the bus admittance matrix follows two

simple rules:

1. The admittance of elements connected between node

k and reference is added to the (k, k) entry of the

admittance matrix.

2. The admittance of elements connected between nodes

j and k is added to the (j, j) and (k, k) entries of the

admittance matrix.

• The negative of the admittance is added to the (j, k)

and (k, j) entries of the admittance matrix.

Bus Admittance Matrix

Bus Admittance Matrix

Node-Voltage Equations

Applying KCL at each node yields: Defining the Y’s as

The Y-Bus

The current equations reduced to

In a compact form

Where,

Gauss Power Flow

*

* * *i

1 1

* * * *

1 1

*

*1 1,

*

*1,

We first need to put the equation in the correct form

S

S

S

S1

i i

i

i

n n

i i i ik k i ik kk k

n n

i i i ik k ik kk k

n ni

ik k ii i ik kk k k i

ni

i ik kii k k i

V I V Y V V Y V

V I V Y V V Y V

Y V Y V Y VV

V Y VY V

Difficulties

• Unless the generation equals the load at every bus, the

complex power outputs of the generators cannot be

arbitrarily selected.

• In fact, the complex power output of at least one of the

generators must be calculated last, since it must take up

the unknown “slack” due to the uncalculated network

losses.

• Further, losses cannot be calculated until the voltages are

known.

• Also, it is not possible to solve these equations for the

absolute phase angles of the phasor voltages. This simply

means that the problem can only be solved to some

arbitrary phase angle reference.

Difficulties

• For a 4- bus system, suppose that SG4 is arbitrarily allowed to float

or swing (in order to take up the necessary slack caused by the

losses) and that SG1, SG2, SG3 are specified.

Remedies

• Now, with the loads known, the equations are seen as

four simultaneous nonlinear equations with complex

coefficients in five unknowns. (V1, V2, V3, V4 and SG4).

• Designating bus 4 as the slack bus and specifying the

voltage V4 reduces the problem to four equations in four

unknowns.

Remedies

• The slack bus is chosen as the phase reference for all

phasor calculations, its magnitude is constrained, and

the complex power generation at this bus is free to take

up the slack necessary in order to account for the

system real and reactive power losses.

• Systems of nonlinear equations, cannot (except in rare

cases) be solved by closed-form techniques.

Load Flow Solution

• There are four quantities of interest associated with each

bus:

1. Real Power, P

2. Reactive Power, Q

3. Voltage Magnitude, V

4. Voltage Angle, δ

• At every bus of the system, two of these four quantities will

be specified and the remaining two will be unknowns.

• Each of the system buses may be classified in accordance

with which of the two quantities are specified

Bus Classifications

Slack Bus — The slack bus for the system is a single bus for which the voltage

magnitude and angle are specified.

• The real and reactive power are unknowns.

• The bus selected as the slack bus must have a source of both real and

reactive power, since the injected power at this bus must “swing” to take

up the “slack” in the solution.

• The best choice for the slack bus (since, in most power systems, many

buses have real and reactive power sources) requires experience with the

particular system under study.

• The behavior of the solution is often influenced by the bus chosen.

Bus Classifications

Load Bus (P-Q Bus) : A load bus is defined as any bus of the system for which the real and reactive power are specified.

• Load buses may contain generators with specified real and reactive power outputs;

• however, it is often convenient to designate any bus with specified injected complex power as a load bus.

Voltage Controlled Bus (P-V Bus) : Any bus for which the voltage magnitude and the injected real power are specified is classified as a voltage controlled (or P-V) bus.

• The injected reactive power is a variable (with specified upper and lower bounds) in the power flow analysis.

• (A P-V bus must have a variable source of reactive power such as a generator.)

Solution Methods

• The solution of the simultaneous nonlinear power flow

equations requires the use of iterative techniques for

even the simplest power systems.

• There are many methods for solving nonlinear

equations, such as:

- Gauss Seidel.

- Newton Raphson.

- Fast Decoupled.

Guess Solution

• It is important to have a good approximation to the

load-flow solution, which is then used as a starting

estimate (or initial guess) in the iterative procedure.

• A fairly simple process can be used to evaluate a good

approximation to the unknown voltages and phase

angles.

• The process is implemented in two stages: the first

calculates the approximate angles, and the second

calculates the approximate voltage magnitudes.

Gauss Iteration Method

Gauss Iteration Example

( 1) ( )

(0)

( ) ( )

Example: Solve - 1 0

1

Let k = 0 and arbitrarily guess x 1 and solve

0 1 5 2.61185

1 2 6 2.61612

2 2.41421 7 2.61744

3 2.55538 8 2.61785

4 2.59805 9 2.61798

v v

v v

x x

x x

k x k x

Stopping Criteria

=

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 0.05 p.u on each end (100 MVA base). Also, there is a

0.25 p.u. capacitance at bus 2. If the generator voltage is

1.0 p.u., what is V2?

100+j50 0.25 p.u.

Z = 0.02 + j0.06

V= 1 0

Y-Bus

2

2 bus

bus

22

The unknown is the complex load voltage, V .

To determine V we need to know the .

15 15

0.02 0.06

5 14.95 5 15Hence

5 15 5 14.70

( Note - 15 0.05 0.25)

jj

j j

j j

B j j j

Y

Y

Solution

*2

2 *22 1,2

2 *2

(0)2

( ) ( )2 2

1 S

1 -1 0.5( 5 15)(1.0 0)

5 14.70

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

n

ik kk k i

v v

V Y VY V

jV j

j V

V

v V v V

j j

j j

.9624 0.0553j

Solution (cont.)

Gauss-Seidel Iteration

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

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

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

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

Immediately use the new voltage estimates:

( , , , , )

( , , , , )

( , , , , )

( , , , ,

v v v vn

v v v vn

v v v v vn

v v vv 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 V V

)

The Gauss-Seidel works better than the Gauss, and

is actually easier to implement. It is used instead

of Gauss.

Newton-Raphson Power Flow

The General Form of the Load-Flow Equations

• In Practice, bus powers Si is specified rather than the bus currents Ii .

• As a result, we have

)(||)(1 1

**

ininniin

N

n

N

n

niniiiii VVYVYVIVjQP

ijijijijijijijijij jBGYjYYY sin||cos||||

i

i*

V

SI

Load-Flow Equations

• These are the static power flow equations. Each equation is complex,

and therefore we have 2n real equations. The nodal admittance matrix

current equation can be written in the power form:

)(||)(1 1

**

ininniin

N

n

N

n

niniiiii VVYVYVIVjQP

ijijijijijijijijij jBGYjYYY sin||cos||||

Let,

Load-Flow Equations

• Finally,

N

inn

inininniiiii YVVGVP1

2 )cos(||||

N

inn

inininniiiii YVVBVQ1

2 )sin(||||

o This is known as NR (Newton – Raphson) formulation

Newton-Raphson Algorithm

• The second major power flow solution method is the

Newton-Raphson algorithm.

• Key idea behind Newton-Raphson is to use sequential

linearization

General form of problem: Find an x such that

( ) 0ˆf x

Newton-Raphson Method

( )

( ) ( )

( )( ) ( )

2 ( ) 2( )

2

1. For each guess of , , define ˆ

-ˆ

2. Represent ( ) by a Taylor series about ( )ˆ

( )( ) ( )ˆ

1 ( )higher order terms

2

v

v v

vv v

vv

x x

x x x

f x f x

df xf x f x x

dx

d f xx

dx

Newton-Raphson Method

( )( ) ( )

( )

1( )( ) ( )

3. Approximate ( ) by neglecting all terms ˆ

except the first two

( )( ) 0 ( )ˆ

4. Use this linear approximation to solve for

( )( )

5. Solve for a new estim

vv v

v

vv v

f x

df xf x f x x

dx

x

df xx f x

dx

( 1) ( ) ( )

ate of x̂

v v vx x x

Example

2

1( )( ) ( )

( ) ( ) 2

( )

( 1) ( ) ( )

( 1) ( ) ( ) 2

( )

Use Newton-Raphson to solve ( ) - 2 0

The equation we must iteratively solve is

( )( )

1(( ) - 2)

2

1(( ) - 2)

2

vv v

v v

v

v v v

v v v

v

f x x

df xx f x

dx

x xx

x x x

x x xx

Example Solution

( 1) ( ) ( ) 2

( )

(0)

( ) ( ) ( )

3 3

6

1(( ) - 2)

2

Guess x 1. Iteratively solving we get

v ( )

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 v

v

v v v

x x xx

x f x x

Comments

• When close to the solution the error decreases

quite quickly -- method has quadratic convergence

• Stopping criteria is when f(x(v)) <

• Results are dependent upon the initial guess. What

if we had guessed x(0) = 0, or x (0) = -1?

Multi-Variable Newton-Raphson

1 1

2 2

Next we generalize to the case where is an n-

dimension vector, and ( ) is an n-dimension function

( )

( )( )

( )

Again define the solution so ( ) 0 andˆ ˆ

n n

x f

x f

x f

x

f x

x

xx f x

x

x f x

x ˆ x x

Multi-Variable Case, cont’d

i

1 11 1 1 2

1 2

1

n nn n 1 2

1 2

n

The Taylor series expansion is written for each f ( )

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order terms

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order terms

nn

nn

x xx x

xx

x xx x

xx

x

x xx x

x

x xx x

x

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

( ) ( ) ( )

( )( ) ( ) ( )

( )( )ˆ

( )( ) ( ) ( )

n

n

nn n n

n

f f f

x x xf x

f f ff x

x x x

ff f f

x x x

x x x

xx x x

xf x

xx x x

higher order terms

nx

Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The n by n matrix of partial derivatives is known

as the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

f f f

x x x

f f f

x x x

f f f

x x x

J x

x x x

x x x

J x

x x x

Multi-Variable N-R Procedure

1

( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

Derivation of N-R method is similar to the scalar case

( ) ( ) ( ) higher order termsˆ

( ) 0 ( ) ( )ˆ

( ) ( )

( ) ( )

Iterate until ( )

v v v

v v v v

v

f x f x J x x

f x f x J x x

x J x f x

x x x

x x J x f x

f x

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

Solution

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

Solution, 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.

NR Application to Power Flow

*

* * *i

1 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

=

=

=

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

NR 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

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

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

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 f

x x x

f f f

x x x

f f f

x x x

J x

x x x

x x x

J x

x x x

Power Flow Jacobian Matrix,

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

Two Bus, 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. Assume that bus one is the slack and SBase = 100 MVA.

2

2

10 10

10 10bus

j j

V j j

x Y

Two Bus Example, cont’d

i1

i1

2 2 1 2

22 2 1 2 2

General power balance equations

P ( cos sin )

Q ( sin cos )

Bus two power balance equations

P (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

Q V V V

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

First Iteration

(0)

2 2(0)

22 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

x

12.0 0.2

1.0 0.9

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

Two Bus Solved Values

Line Z = 0.1j

One Two 1.000 pu 0.855 pu

200 MW

100 MVR

200.0 MW

168.3 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

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

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 MVR170.0 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

N-R 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