Steady State Load Flow Analysis for Power Systems

Steady State Load Flow Analysis for PowerSystems

Praveen Tripathy

Department of Electronics & Electrical EngineeringINDIAN INSTITUTE OF TECHNOLOGY GUWAHATI

INDIA

Presentation at IIT Guwahati, India

Steady State Load Flow Analysis for Power Systems ... P Tripathy IIT Guwahati, INDIA 1 / 34

Outline

Why load flow ?Numerical solution techniquesNewton Raphson Load flow (NRLF)NRLF in rectangular formDecoupled Newton load flowFast Decoupled load flowDC load flow


Why Load Flow ? (Contd...)

Power System Analysis:Load Flow AnalysisFault AnalysisSecurity AnalysisStability Analysis


Why Load Flow ?

Why we need reference bus ?

0 5 10 15 20 25 30−1

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0 5 10 15 20 25 30−1

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G G

0

POWER SYSTEM PV Bus or Generator bus

PQ Bus or load bus

Slack bus

V θ

V θ

L L

G GV

Load Flow Analysis:off-line method of calculating the voltage and angle at the bussolve the set of nonlinear power balance equations



Load flow is root finding problemThis root finding problem is converter to optimization problem


Numerical solution techniques

For solution of non-linear algebraic equations

y1 = f1(x1,x2, ...,xN)

y2 = f2(x1,x2, ...,xN)

.

.

yN = fN(x1,x2, ...,xN)

(1)

Gauss Iterative method & Gauss SeidalNewton Raphson method


Gauss Iterative method & Gauss seidal

Step 1: Transform the given set of equation to express each variable asa non-linear function of other variable

x1 = F1(x1,x2, ...,xN)

x2 = F2(x1,x2, ...,xN)

.

.

.

xN = FN(x1,x2, ...,xN)

(2)


Gauss Iterative method & Gauss Seidal ...

Step 2: Starting from initial Guess, use successive displacement method( GI or GS) to obtain better estimate of variables.In Gauss Iterative (GI) method, 1st iteration is given by

x(1)1 = F1(x(0)1 ,x(0)2 , ...,x(0)N )

x(1)2 = F2(x(0)1 ,x(0)2 , ...,x(0)N )

.

.

x(1)N = FN(x(0)1 ,x(0)2 , ...,x(0)N )

(3)

Second iteration

x(2)1 = F1(x(1)1 ,x(1)2 , ...,x(1)N )

.

x(2)N = FN(x(1)1 ,x(1)2 , ...,x(1)N )

(4)Steady State Load Flow Analysis for Power Systems ... P Tripathy IIT Guwahati, INDIA 9 / 34


Step 3: The number of iteration continues, till

|x(k+1)i − x(k)i | ≤ ε (5)

Where ε→ tolerence value.Normally for 100 MVA base value,ε is of the order of10−5or10−6p.u.



Gauss Seidal Method utilizes latest value of the variable

x(1)1 = F1(x(0)1 ,x(0)2 , ...,x(0)N )

x(1)2 = F2(x(1)1 ,x(0)2 , ...,x(0)N )

.

x(1)N = FN(x(1)1 ,x(1)2 , ...,x(1)N−1,x

(0)N )

(6)

(r+1)th iteration

x(r+1)1 = F1(x

(r)1 ,x(r)2 , ...,x(r)N )

x(r+1)2 = F2(x

(r+1)1 ,x(r)2 , ...,x(r)N )

.

x(r+1)i = Fi(x

(r+1)1 ,x(1)2 , ...x(r+1)

i−1 ,x(r)i , ...,x(r)N )

x(r+1)N = FN(x

(r+1)1 ,x(r+1)

2 , ...,x(r)N )


Gauss Seidal

The (r+1)th iteration using the GS method, is given by

x(r+1)i cal = Fi(x

(r+1)1 ,x(1)2 , ...x(r+1)

i−1 ,x(r)i , ...,x(r)N )

Using the acceleration factor the updated variable is

x(r+1)i = x(r)i +α(x(r+1)

i − xri )

Typical value of α = 1.2 to 1.6


Newton Raphson method

y1 = f1(x1,x2, ...,xN)

y2 = f2(x1,x2, ...,xN)

.

.

yN = fN(x1,x2, ...,xN)

(8)

X =

x1x2.

xN

or Y = f (X), Here the vector Y is already known.


Newton Raphson method ...

Using Taylor series enpension

Ys = f (X0)+∂f∂X|X=X0

∆X+12

∂2f∂X2 |X=X0

∆X2+ ... (9)

Where, ∆X = (X− X0)If one assume that initial guess X0 is close to the final value of X, then ∆X =small quantity.With this assumption, one can neglect 2nd & higher order derivatives of theTaylor series, i.e., we can write

Ys = f (X0)+∂f∂X|X=X0

∆X


Newton Raphson method ...

Ys =⇒ Specified value of YYs− Y0 = [J]∆XWhere, f (X0) =⇒ Computed value of Y

∆Y = [J][∆X]

∆X = [J]−1[∆Y]

(10)

Here,∆X =⇒ correction vectorJ =⇒ Jacobian∆Y =⇒ Mismatch Vector|∆Yi| ≤ ε

State Update is given by X(1) = X0 +∆X


Load flow equations

SLFE in Complex Form

~S∗i =~V∗iN

∑j=1

~Yij~Vj

SLFE in Polar Form

Pi =N

∑j=1

ViVjYijcos(δi−δj−θij), Qi =N

∑j=1

ViVjYijsin(δi−δj−θij)

(11)

SLFE in Rectangular form

Pi =N

∑j=1

[Gij(eiej + fifj)+Bij(ejfi− fjei)]

Qi =N

∑j=1

[Gij(ejfi− fjei)−Bij(eiej + fifj)]


Bus classifications

Reference bus,Slack bus,Swing bus Here δ = 0 and V prescribed. Loadflow results provide the value of Pg and Qg.P-V bus (also called Generator bus) if a load bus having SVC,V-controland suppose load is constant voltage type , it is also called P-V typeP-Q bus (load bus) Pg and Qg are prescribed. Load flow results providethe value of V & δ.


Newton Raphson load flow (NRLF) method

∴ Total number of unknown variable :

= (m−1)+2(N−m)

= 2N−m−1(13)


Newton Raphson load flow (NRLF) method ...

For generator bus, only Pi is specified, thus we have (m−1) number ofequation for which Pi are specifiedFor PQ bus, both Pi & Qi is specified, thus we have 2(N−m) numberof equation for which Pi & Qi are specifiedThus we have a total of 2N−2m+m−1 = 2N−M−1 =⇒ thenumber of unknown variables is equal to known number of equation,hence, we can solve the non-linear algebraic equation.




Newton Raphson load flow (NRLF) method

[∆P∆Q

]=

[∂P∂δ

∂P∂V V

∂Q∂δ

∂Q∂V V

][∆δ∆VV

]=

[H(N−1)(N−1) L(N−1)(N−m)

M(N−M)(N−1) N(N−m)(N−m)

][∆δ∆VV

]The jacobian elements are given as:

Hij = Nij = ViVj(Gij sin δij−Bij cos δij)

Hii =−Qi−BiiV2i

Nii = Qi−BiiV2i

Lij =−Mij = ViVj(Gij cos δij +Bij sin δij)

Lii = Pi +GiiV2i

Mii = Pi−BiiV2i

(14)



The state update equation are:

∆δ(k+1) = δ

k +∆δk

∆V(k+1) = Vk +∆Vk

= Vk(1+∆Vk

Vk ) (15)

(16)

For convergence the infinity norm of mismatch vector should be less than ε,i.e,

‖∆P∆Q‖∞ < ε


About NRLF Method:

1 More reliable convergence as compared to GSLF (ie. more robust).2 No of iterations are independent of system size, it takes 2-7 iterations

for converge.3 It has good convergence even in ill conditioned situations:

(i) High RX ratio of line, Normally in transmission line, it is about (0.1 to

0.001).(ii) Presence of negative reactance e.g series capacitor.

(iii) Heavily loaded system.

Due to these ill condition, the jacobian element will have weak diagonalelement. The [J]−1 will be very large.


Drawback of NRLF:

1 Require more memory space to store [J] and [J]−1

2 Also require more CPU time per iteration, due to formation andinversion of Jacobian.

To overcome them1 Use sparsity technique2 To develop decouple method


Decoupled Newton load flow method:(By B. Stott, IEEE onPAS, 1972)

[∆P∆Q

]=

[∂P∂δ

∂P∂V V

∂Q∂δ

∂Q∂V V

][∆δ∆VV

][

∆P∆Q

]=

[∂P∂δ

00 ∂Q

∂V V

][∆δ∆VV

]Thus the decoupled equation are given by

∴ ∆P =∂P∂δ

∆δ

and

∆Q =∂Q∂V

∆V

Normally we follow 1δ−1V update.

1 time per iteration ≈ half.2 No. of iteration� NRLF.3 poor convergence for ill condition.


FAST decoupled load flow (FDLF):

[∆P∆Q

]=

[H 00 N

][∆δ∆VV

]

∆P = [H]∆δ and ∆Q = [N][

∆VV

]

Element of H and N :

Hii =−Qi−BiiV2i

Hij = ViV− j(Gijsinδij−Bijcosδij)

Nii =+Qi−BiiV2i

Nij = Hij


Physically justifiable assumption:

1 δij = δi−δj is quite small in a physical system, cos(δij)≈ 1 andsin(δij)≈ δij

2 RX ratio of line in transmission network is small. Then we can sayGijsinδij� Bijcosδij ' Bij

3 Qi� BiiV2i

Hii = Vi(−Bii)Vi

Nii = Vi(−Bii)Vi

Hij = Vi(−Bij)Vj

Nij = Vi(−Bij)Vj

(17)

∆P = [V ′B′V][∆δ]

∆Q = [V ′B′′V][∆VV

]

(18)


Physically justifiable assumption: ...

Where B’ and B” are derived from negative Susceptance part of [Ybus]

∆PV

= [B′V][∆δ]

∆QV

= [B′′V][∆VV

]

(19)

Normally taking V=1;

∆PV

= [B′][∆δ]

∆QV

= [B′′][∆V]

(20)


Additional assumption: (IEEE Transection on PAS May1974 by B. stott and O. ALSAE

1 While forming [B’] omit the effect of all these element whichpre-dominatingly affect MVAR flow such as shunt reactance (line shunt+ bus shunt) and nominal in transformer setting.

2 While forming [B’] matrix, neglect the series resistance of the line.3 While forming [B”] , omit the phase shifting transformer (as it

pre-dominatingly affect MW flow)Final model of FDLF:

[∆PV

]= [B′][∆δ][

∆QV

]= [B′′][∆V]

(21)

This method is as Robust as NRLF, and it also uses less CPU time.


Construction of B’ and B” matrix:

(i) Ybus =Actual Y-bus(ii) Y ′bus = neglect the resistance, tap=1, shunt reactance

(iii) Y ′′bus = neglect the phase shifting; B’=-{Susceptance part of Y’ bus };B”=-{Susceptance part of Y” bus}


DC load flow methods: (Real power of calculations)

Assumption:1 Line resistance and all the shunt are neglected2 Transformer tap are set at 1 pu3 voltage magnitude at all the buses set at 1 pu4 angle difference of voltage vector of two buses are small δi−δj ≈ 0



Pij =ViVj

Xijsin(δi−δj)

using DC load flow assumption Vi = V=1pu as δi−δj is small ∴sinδi−δj ≈ δi−δj

Pij =1

Xij(δi−δj)

it is similar to DC expression , thus we call it as DC load flow.

Pi =n

∑j=1

Pij =n

∑j=1

1Xij

(δi−δj) =1

Xijδi−

1Xij

δj)



let,

[P]=

P1P2..

Pn

;[

δ]=

δ1δ2..

δn

[P] = [Bx][δ]

(Bx)ii =n

∑j=16=i

1Xij

(Bx)ij =−1

Xij

∴ δ = [Bx]−1[P]


Thank you


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Steady State Load Flow Analysis for Power Systems

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