CHAPTER 5

BEE3143:POWER SYSTEM ANALYSIS- Power flow solution- Newton-Raphson Expected Outcomes Able to solve power flow solution using Newton-Raphson technique

Newton-Raphson Power Flow Solution

• Newton-raphson method is found to be more practical and efficient for large power system.

• The number of iterations required to obtain a solution is independent of the system size, but more functional evaluations are required at each iteration.

... Newton-Raphson Power Flow Solution Rewrite the current entering bus i in a

typical bus power system in figure below in terms of the bus admittance matrix

∑=

=n

jjiji VYI

1

The typical element Yij is

ijijijijijijijijij jBGYjYYY +=+=∠= θθθ sincos

The voltage at a typical bus i

(1)

Expressing Equation (1) in polar form

jij

n

jjiji VYI δθ +∠= ∑

=1

∑=

=

=−n

jjiji

iiii

VYV

IVjQP

1

*

*

yin

yi1

yi2

Ii

Vi

V1

V2

Vn

yi0

( )iiiiii jVVV δδδ sincos +=∠=

The complex conjugate of the power injected at bus i is

...Newton-Raphson Power Flow Solution The complex conjugate of the power in polar form

ijij

n

jjiij

jij

n

jjijii

n

jjijiii

VVY

VYV

VYVjQP

δδθ

δθδ

−+∠=

+∠−∠=

=−

∑

∑

∑

=

=

=

1

1

1

*

Separating this equation into real and reactive parts

( )ijij

n

jjiiji VVYP δδθ −+= ∑

=1cos

( )ijij

n

jjiiji VVYQ δδθ −+−= ∑

=1sin

Equation (2) and (3) constitute a polar form of the power flow equations.

(2)

(3)

...Newton-Raphson Power Flow Solution These Eqs. constitute a set of non-linear algebraic equations in terms of the independent variables, voltage magnitude in per-unit, and phase angle in radians. Expanding (2) and (3) in Taylor’s series about the initial estimate and neglecting all higher order terms results in the following set of linear equations.

...Newton-Raphson Power Flow Solution

MISMATCHES

kn

k

kn

k

SCORRECTION

kn

k

kn

k

MATRIXJACOBIAN

k

n

nk

n

k

n

k

k

n

nk

n

k

n

k

k

n

nk

n

k

n

k

k

n

nk

n

k

n

k

Q

Q

P

P

V

V

VQ

VQ

JVQ

VQ

VP

VP

JVP

VP

QQJ

QQ

PPJ

PP

∆

∆

∆

∆

=

∆

∆

∆

∆

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

)(

)(2

)(

)(2

)(

)(2

)(

)(2

)()(

2

4

)(2

)(

2

2

)()(

2

2

)(2

)(

2

2

)()(

2

3

)(2

)(

2

2

)()(

2

1

)(2

)(

2

2

δ

δ

δδ

δδ

δδ

δδ

Newton-Raphson power flow equations:

...Newton-Raphson Power Flow Solution

• From the previous equation, bus 1 is assumed to be slack bus. • The Jacobian matrix gives the linearized relationship between small

changes in voltage angle and voltage magnitude with the small changes in real and reactive power.

• Elements of Jacobian matrix are the partial derivatives of (2) and (3), evaluated at small changes in voltage angle and voltage magnitude

...Newton-Raphson Power Flow Solution

• From the previous equation, bus 1 is assumed to be slack bus. • The Jacobian matrix gives the linearized relationship between small

changes in voltage angle and voltage magnitude with the small changes in real and reactive power.

• Elements of Jacobian matrix are the partial derivatives of (2) and (3), evaluated at small changes in voltage angle and voltage magnitude

• In short form, it can be written as:

∆∆

=

∆∆

VJJJJ

QP δ

43

21 (4)

Procedure of NR

• For load buses, is calculated from (2) and calculated from (3); • is calculated from: • is calculated from: • For PV buses, is calculated from (2) and calculated from • The element of Jacobian matrix (J1, J2, J3 and J4) are calculated as

follows:

)(kiP )(k

iQ)(k

iP∆ )()( ki

schi

ki PPP −=∆

)()( ki

schi

ki QQQ −=∆)(k

iQ∆)(k

iP∆)(kiP

)()( ki

schi

ki PPP −=∆

…Procedure of NR

The diagonal and off-diagonal elements of J1 are

∑≠

+−=∂∂

ijjiijijji

i

i YVVP )sin( δδθδ ( )ijijjiij

j

i VVYP

δδθδ

−+−=∂∂

sin

The diagonal and off-diagonal elements of J2 are

∑≠

+−+=∂∂

ijjiijijjiiiii

i

i YVYVVP )cos(cos2 δδθθ

( )ijijiijj

i VYVP δδθ −+=

∂∂ cos

…Procedure of NR

The diagonal and off-diagonal elements of J3 are

∑≠

+−=∂∂

ijjiijijji

i

i YVVQ )cos( δδθδ

)cos( jiijijjij

i YVVQ δδθδ

+−−=∂∂

The diagonal and off-diagonal elements of J4 are

∑≠

+−−−=∂∂

ijjiijijjiiiii

i

i YVYVVQ )sin(sin2

||δδθθ

)sin(|| jiijiji

j

i YVVQ δδθ +−−=

∂∂

…Procedure of NR

• The equation (4) is solved directly by optimally ordered triangular

factorization of Gaussian elimination • The new voltage and phase angle are computed from:

• The process is continued until the residuals and are less than specified accuracy

)(kiP∆ )(k

iQ∆

)()()1(

)()()1(

ki

ki

ki

ki

ki

ki

VVV ∆+=

∆+=+

+ δδδ