lecture5-solving by the newton-raphson method

8/7/2019 Lecture5-Solving by the Newton-Raphson Method

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Power Systems I

The Power Flow Solution

l Most common and important tool in power systemanalysis

u also known as the “Load Flow” solution

u used for planning and controlling a system

u assumptions: balanced condition and single phase analysis

l Problem:

u determine the voltage magnitude and phase angle at each bus

u determine the active and reactive power flow in each line

u each bus has four state variables:

n voltage magnitude

n voltage phase angle

n real power injection

n reactive power injection



Power Systems I

The Power Flow Solution

u Each bus has two of the four state variables defined or given

l Types of buses:

u Slack bus (swing bus)

n voltage magnitude and angle are specified, reference bus

n solution: active and reactive power injections

u Regulated bus (generator bus, P-V bus)

n models generation-station buses

n real power and voltage magnitude are specified

n solution: reactive power injection and voltage angle

u Load bus (P-Q bus)n models load-center buses

n active and reactive powers are specified (negative values for loads)

n solution: voltage magnitude and angle



Power Systems I

Newton-Raphson PF Solution

l Quadratic convergence

u mathematically superior to Guass-Seidel method

l More efficient for large networks

u number of iterations required for solution is independent of

system size

l The Newton-Raphson equations are cast in natural powersystem form

u solving for voltage magnitude and angle, given real and reactivepower injections



Power Systems I

Newton-Raphson Method

l A method of successive approximation using Taylor’sexpansion

u Consider the function: f ( x) = c, where x is unknown

u Let x[0] be an initial estimate, then ∆ x[0] is a small deviation fromthe correct solution

u Expand the left-hand side into a Taylor’s series about x [0] yeilds

c x x f =∆+ ]0[]0[

( ) ( ) c xdx

f d x

dx

df x f =+∆

+∆

+ L

2]0[

2

2

21]0[]0[



Power Systems I


u Assuming the error, ∆ x[0], is small, the higher-order terms areneglected, resulting in

u where

u rearranging the equations

( ) ]0[]0[]0[]0[ xdx

df cc x

dx

df x f ∆

≈∆⇒≈∆

+

( )]0[]0[ x f cc −=∆

]0[]0[]1[

]0[

]0[

x x x

dx

df

c x

∆+=

∆

=∆



Power Systems I

Example

l Find the root of the equation: f (x ) = x 3 - 6x 2 + 9x - 4 = 0



Power Systems I


0 1 2 3 4 5 6-10

0

10

20

30

40

50

x

f(x) = x

3

-6x

2

+9x-4



Power Systems I

( )∑

∑∑

=

==

+∠−∠=−

=−

+∠==

n

j

jij jijiii

iiii

n

j

jij jij

n

j

jiji

V Y V Q jP

I V Q jP

V Y V Y I

1

*

11

δθδ

δθ

Power Flow Equations

l KCL for current injection

l Real and reactive power injection

l Substituting for I i yields:



Power Systems I

Power Flow Equations

( )

( )∑

∑

=

=

+−−=

+−=

n

j

jiijij jii

n

j

jiijij jii

Y V V Q

Y V V P

1

1

sin

cos

δδθ

δδθ

l Divide into real and reactive parts



Power Systems I

Newton-Raphson Formation

( )

( )

( ) ( )( )

=

=

=

+−−=

+−=

∑

∑

=

=

][

][

][

][

][

][

1

][][][][][

1

][][][][][

sin

cos

k

inj

k

injk

k

k

k

sch

inj

sch

inj

n

j

k

j

k

iijij

k

j

k

i

k

i

n

j

k

j

k

iijij

k

j

k

i

k

i

xQ

xP x f V

xQ

Pc

Y V V Q

Y V V P

δ

δδθ

δδθ

l Cast power equations into iterative form

l Matrix function formation of the system of equations



Power Systems I

Newton-Raphson Formation

( )

( )( )

( )dx

xdf

dx xdf

x f c x x

x x x f c

k

k

k k k

solutionsolution

][

][

][][]1[

]0[ of estimateinitial

−+=

==

+

l General formation of the equation to find a solution

l The iterative equation

l The Jacobian - the first derivative of a set of functions

a matrix of all combinatorial pairs



Power Systems I

The Jacobian Matrix

( )

∆

∆

∆

∆

=

∆

∆

∆

∆

∆∆

∂∂

∂∂

∂∂

∂∂

=

∆∆

⇒

−

−

∂∂

∂∂

∂∂

∂∂

∂ ∂∂∂∂∂∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

−

−

−

−−

−

−−

−−

−

−−

−

−−

−−

mn

n

V

Q

V

QQQ

V Q

V QQQ

V

P

V

PPP

V

P

V

PPP

mn

n

V

V

Q

Q

P

P

V V

QQ

V PP

Q

P

dx xdf

mn

mnmn

n

mnmn

mnn

mn

nn

n

nn

mnn

M

M

LL

MOMMOM

LL

LL

MOMMOM

LL

M

M

1

1

1

1

1

1

111

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

δ

δ

δ

δ

δ

δδ

δδ

δδ

δδ



Power Systems I

Jacobian Terms

( )

( )

( )

( ) jiY V V

P

Y V Y V

V

P

jiY V V P

Y V V P

jiijiji

j

i

i j

jiijij jiiiii

i

i

jiijij ji

j

i

i j

jiijij ji

i

i

≠+−=∂∂

+−+=∂

∂

≠+−−=∂∂

+−=∂∂

∑

∑

≠

≠

δδθ

δδθθ

δδθδ

δδθδ

cos

coscos2

sin

sin

l Real power w.r.t. the voltage angle

l Real power w.r.t. the voltage magnitude



Power Systems I

l Reactive power w.r.t. the voltage angle

l Reactive power w.r.t. the voltage magnitude

Jacobian Terms

( )

( )

( )

( ) jiY V V

Q

Y V Y V

V

Q

jiY V V Q

Y V V Q

jiijiji

j

i

i j

jiijij jiiiii

i

i

jiijij ji

j

i

i j

jiijij ji

i

i

≠+−−=∂∂

+−+−=∂

∂

≠+−−=∂∂

+−=∂∂

∑

∑

≠

≠

δδθ

δδθθ

δδθδ

δδθδ

sin

sinsin2

cos

cos



Power Systems I

Iteration process

l Power mismatch or power residuals

u difference in schedule to calculated power

l New estimates for the voltages

][][]1[

][][]1[

][][

][][

k i

k i

k i

k

i

k

i

k

i

k i

schi

k i

k

i

sch

i

k

i

V V V

QQQ

PPP

∆+=

∆+=

−=∆

−=∆

+

+δδδ



Power Systems I

Bus Type and the Jacobian Formation

l Slack Bus / Swing Bus

u one generator bus must be selected and defined as the voltageand angular reference

n The voltage and angle are known for this bus

n The angle is arbitrarily selected as zero degrees

n bus is not included in the Jacobian matrix formation

l Generator Bus

n have known terminal voltage and real (actual) power injection

n the bus voltage angle and reactive power injection are computed

n bus is included in the real power parts of the Jacobian matrix

l Load Bus

n have known real and reactive power injections

n bus is fully included in the Jacobian matrix



Power Systems I

Newton-Raphson Steps

1. Set flat start

u For load buses, set voltages equal to the slack bus or 1.0∠0°u For generator buses, set the angles equal the slack bus or 0°

2. Calculate power mismatch

u For load buses, calculate P and Q injections using the known andestimated system voltages

u For generator buses, calculate P injections

u Obtain the power mismatches, ∆P and ∆Q

3. Form the Jacobian matrix

u Use the various equations for the partial derivatives w.r.t. thevoltage angles and magnitudes



Power Systems I

Newton-Raphson Steps

4. Find the matrix solution (choose a or b)

u a. inverse the Jacobian matrix and multiply by the mismatchpower

u b. perform gaussian elimination on the Jacobian matrix with the b

vector equal to the mismatch powercompute ∆δ and ∆V

5. Find new estimates for the voltage magnitude and angle

6. Repeat the process until the mismatch (residuals) areless than the specified accuracy

ε

ε

≤∆

≤∆][

][

k

i

k i

Q

P



Power Systems I

Line Flows and Losses

l After solving for bus voltages and angles, power flowsand losses on the network branches are calculated

u Transmission lines and transformers are network branches

u The direction of positive current flow are defined as follows for a

branch element (demonstrated on a medium length line)u Power flow is defined for each end of the branch

n Example: the power leaving bus i and flowing to bus j

V jV i

y j0 yi0

yijBus i Bus j

I ij I ji I L

I j0 I i0



Power Systems I

Line Flows and Losses

l current and power flows:

l power loss:

V jV i

y j0 yi0

yijBus i Bus j

I ij I ji I L

I j0 I i0

( )

( )***

0

2*

00

jijiiijiijiij

ii jiiji Lij

V yV y yV I V S

V yV V y I I I

ji

−+==

+−=+=→

( )

( )***

0

2*

00

iij j jij j ji j ji

j ji jij j L ji

V yV y yV I V S

V yV V y I I I

i j

−+==

+−=+−=→

jiijij Loss SSS +=



Power Systems I

Example

j0.04

3

1

2

138.6 MW45.2 MVAR

256.6 MW110.2 MVAR

Slack Bus

V 1

= 1.05∠0°

j0.02 j0.025

l Using N-R method, find thephasor voltages at buses 2and 3

l Find the slack bus real

and reactive powerl Calculate line flows

and line losses

u 100 MVA base

lecture5-solving by the newton-raphson method

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