Ybus&powerflow

8/6/2019 Ybus&powerflow

1/7

6 Power Flow Notes on Power System Analysis 1

Chapter 6: Power Flow

Network Matrices

Network SolutionsPower Flow Equations

Newton-Raphson MethodFast Decoupled Method


Bus Admittance Matrix

LetI = vector of currents injected into

nodes V = vector of node voltages Y bus = bus admittance matrixThen: I = Y bus V


Example

==2

1

2

1

V

V

I

IVI

=2

1

2221

1211

2

1

V

V

YY

YY

I

I

VYI =6 Power Flow Notes on Power System Analysis 4

Constructing Ybus

Ybus is symmetric unless the circuithas phase shifters or active devices

Diagonal term Y ii is the sum of alladmittances connected to bus i

Off-diagonal term Y ij is the negativeof the sum of all admittancesdirectly connecting bus i to bus j


Network Solution

Given current injections, find nodevoltages For small systems, simply invert the

Ybus matrix: V = Y bus -1 I = Z bus I For medium, solve the set of linear

equations using Gaussian elimination For large systems, the Gaussian

elimination is best done by triangular(LU) factorization


Triangular factorization

Triangular matrices L and U areobtained so that Y bus = L U The triangular factors L and U are

saved to be used on othercalculations

Note that L is a lower triangularmatrix and U is an upper triangularmatrix


2/7


Triangular factorization

Once the factors are known, solve:

forward substitution I = L V' back substitution V' = U V


Triangular factorization Forward substitution gives V' by

solution of set of (lower) triangularequations that can be solvedsequentially starting at 1

Back substitution gives V bysolution of a set of (upper)triangular equations that are solvedsequentially starting at n


Kron Reduction

If one bus has current injection of zero, that bus can be eliminated:

=3

2

1

333231

232221

131211

2

1

V

V

V

YYY

YYY

YYY

0

I

I

k j,in,,2,1 j,iY

YYYY

kk kjik ij)new(ij == L


Kron reduction

The Kron reduction method isequivalent to performing a generalwye-delta conversion to eliminate anode in the network It applies to any node that has a zero

current injection It can be generalized to give a partial

inverse (see Mathcad worksheet)


Bus Impedance Matrix

V = Z bus I, so Z bus = Y bus -1 The bus impedance matrix can be

formed by inverting the busadmittance matrix or by thebuilding it one bus at a time


Power Flow Analysis

Power Flow EquationsPower Flow Problem

Several Iterative SolutionsNewton-Raphson andDecoupled Methods

Control of Power Flow


3/7


Example

SG2SG1 SG3 V1

V2

V3

V4 V5

SD1 SD3

SD4 SD5


Power Flow Equations

Complex power injection:Si = S Gi S Di = generation load

Si = k Sik where k = 1, , n and i =1, , n

Current injection (phasor)Ii = I Gi I Di = k Iik = k (Y ik Vk )


Power Flow Equations

Ii = I Gi I Di = k Iik = k (Y ik Vk )Si = V i Ii* = V i k (Y ik * Vk *)Si = k |V i| |V k | e jik (Gik j B ik )

Vk = |V k | e jk

ik = i k Y ik = G ik + jB ik


Power flow equations

Break the complex power flowequation into real and imaginaryparts:Si = P i + j Q i = k |V i| |V k | e jik (Gik j B ik )givesPi = k |V i| |V k | [G ik cos( ik ) + B ik sin( ik )]

andQi = k |V i| |V k | [G ik sin( ik ) - B ik cos( ik )]


Power Flow Problem

Assumptions: At most generator buses, the active power P G

is controlled (by speed governor) and thevoltage magnitude is controlled (by thevoltage regulator). Treat these as known.

At most load buses, a reasonableapproximation is that the load active andreactive power demand P D and Q D are known.

At one generator bus, leave the active poweras a variable (to make up system losses).


Power flow problem: bus type

The slack bus: the one generator busthat has a variable P G. Bus number 1 PV buses: those having known P and

|V| (mostly all other generator buses).buses number 2, ... , m

PQ buses: those having known P and Q(mostly load buses, but also fixedgenerators). buses number m+1, ... , n


4/7


Power flow problem

Statement: Given: (|V 1|, 1), (P 2, |V 2|),...,(P m, |V m|)

(Pm+1 , Q m+1 ), ..., (P n, Q n) Find: (P 1, Q 1), (Q 2, 2), ..., (Q m, m),

(|V m+1 |, m+1 ), ..., (|V n|, n) Note:

P1 and Q 1, ..., Q m are calculated once allvoltages and phase angles are known


Remarks: Seek an iterative solution The equations are nonlinear algebraic

equations There are n-1 unknown phase angles

(all but the slack bus, which is givenas 0)

There are n-m unknown voltagemagnitudes (all but the PV and slack buses, which total m)


Remarks: Seek an iterative solution The equations are nonlinear algebraic

equations Multiple solutions are possible Solutions may fail to exist Numerical method may fail Well-designed system will usually have

only one realistic solution


Iterative Solutions Gauss iteration:

Solve x = h(x) by using initial guess x 0to compute x 1 = h(x 0), then iteratexp+1 =h(x p) where p is iteration

Vector case, use Gauss-Seidel:x1p+1 =h(x 1p, x2p , x 3p, ..., x np)x

2

p+1 =h(x1

p+1 , x2

p , x3

p, ..., xn

p)x3p+1 =h(x 1p+1 , x 2p+1 , x 3p, ..., x np)...


Gauss iteration (scalar case)x

xx0

h(x)h(x 0)

x1

h(x 1)

Note slow convergence, dependingon shape of h(x)


Gauss-Seidel power flow

After rearranging, the power flowequation gives a form that can beused for a Gauss or Gauss-Seideliterative solution

n,,3,2iVYV

SY1

Vn

ik 1k

k ik *i

*i

iii L==

=


5/7


Gauss-Seidel power flow

Results of the Gauss-Seidel powerflow: acceptable for small systems, but it

takes many iterations to converge number of iterations for convergence

increases as the number of buses inthe system increase

not suitable for medium and largesizes of systems


Newton-Raphson

Newtons method (scalar):Equations in this form: f(x) = 0

where x0

is an initial guessLinearize:

f(xp + x) = f(x p) + f (x p) x 0Solve for x = - f '(x p)-1 f(xp)


Newtons method (scalar)


Newton-Raphson Newton-Raphson (vector):

Vector function of vector: f ( x ) = 0where x 0 = [x 10,..., x n0]T is initial guessLinearize:

f ( x p+ x ) = f ( x p) + J ( x p) x 0where J ( x p) is the Jacobian matrix:

a matrix of partial derivatives

Jij(x p) = f i/x j evaluated at x p p = iteration number


Newton-Raphson

Newton-Raphson (vector):Linearize:

f ( x p+ x ) = f ( x p) + J ( x p) x 0Solve for x = - J ( x p) -1 f ( x p)J ( x p) is the Jacobian matrix:

Jij(x p) = f i/x j evaluated at x p.


Newton Raphson Power Flow

Define a vector of unknowns x = [ 2, 3,..., n, |V m+1 |, |V m+2 |,..., |V n|] T

Net power injections computedPi( x ) = k [|V i| |V k | [G ik cos( i-k )

+ B ik sin( i-k )]Qi( x ) = k [|V i| |V k | [G ik sin( i-k )

- Bik cos( i-k )]


6/7


Newton Raphson Power Flow

Power mismatches (= specified value computed value)

Solution is obtained when mismatchesgo to zeroP( x ) = [P 2 P 2( x ), ... , P n P n( x )] T

Q( x ) = [Q m+1 Q m+1 ( x ), ... , Q n Q n( x )] T


The solution will be achieved whenthe mismatches are driven to zero

Solve the following linear equationfor the updates and | V | :

= QP

V

JJJJ

||22211211

After each iteration gives and |V |: p+1 = p + and | V | p+1 = |V | p + |V |

mismatch vector is updated for next iteration


J 11 is a matrix of P ( x )/ :Pi / j =|V i||V j|[G ijsin( i- j)-Bijcos( i- j)Pi / i = - Q i - Bii | V i| 2

J 12 is a matrix of P ( x )/ | V | :Pi / |V| j =|V i| [G ijcos( i- j) + B ijsin( i- j)Pi / |V| i = P i /|V i| + G ii |V i|


J 21 is a matrix of Q( x )/ :Qi / j= -|V i||V j|[G ijcos( i- j)

+ B ij sin( i- j)]Qi / i = P i - Gii | V i| 2

J 22 is a matrix of Q( x )/ | V | :Qi / |V| j=|V i|[G ijsin( i- j)

- Bij cos( i- j)]Qi / |V| i = Q i /|V i| - B ii |V i|


Jacobian elements

Elements of J 12 are relatively small Elements of J 21 are relatively small Elements of J 11 and J 22 are relatively

large


Decoupled Power Flow

Neglect the relatively small termsin the Jacobian, but calculate thepower mismatches exactly

J 1 1 = P andJ 22 | V | = Q


7/7


Decoupled Power Flow

Fast decoupled method makesconstant Jacobian approximation

|V i| 1 pu andi > G ij


Fast Decoupled Method

-B = P' and -B | V | = Q'where Pi' = Pi / | Vi| and Qi' = Qi / | Vi|The approximations do not affect the

mismatches. If the solution converges,we get the exact solution

This may take more iterations but lesscomputation time than the Newton-Raphson.


Control Implications

P primarily depends on , changeactive power flow by controllingangles change Pg which drives the local

angle ahead or operate phase shifting

transformers


Control Implications

Q primarily depends on V, changereactive power flow by controllingvoltage magnitudes change voltage regulator settings or operate tap changers on

transformers


Large Systems

Large systems typically exhibit asparse Ybus matrix A sparse matrix is best stored as a

linked list only the non-zero elements of the

matrix are stored in the linked list pointers are used to indicate the

position of the next non-zero entry

Ybus&powerflow

Documents

Transcript of Ybus&powerflow