POWER SYSTEM MODELING & ANALYSIS – 1

Overview of power system modeling for various studies – Distinction between

steady state, quasi steady state and transient modeling of power systems –

Generation system planning – transmission system planning – steady state and

transient analysis – load forecasting.

Overview – mathematics for basic power system analysis – algebraic equations –

differential algebraic equations – differential equation, numerical solution of

algebraic equations – Gauss elimination method and bifactorization method –

sparsity techniques for large system – sparsity oriented network solution.

Bus classification, Power flow model using Y bus – computational aspects of

power flow problem – Gauss Seidel iterative technique – Newton Raphson

method – Fast decoupled power flow method – Multi area power flow analysis

with tie line control – contingency and sensitivity analysis.

Special purpose power flow studies – Harmonic Power flow, three phase load

flow, distribution power flow. Optimal power flow using Newton’s method &

interior point method.

Symmetrical short circuit analysis, symmetrical components and sequence

impedances. Algorithm for symmetrical fault analysis using Z bus –

Unsymmetrical fault analysis using symmetrical components – Algorithm for

unsymmetrical fault analysis using Z bus – limitations.

Physical description of synchronous machine – Mathematical description of a

synchronous machine – dqo transformation – Per unit representation –

Equivalent circuit – Steady state analysis – transient performance characteristics

– Magnetic saturation – simplified model with damper neglected – classical model

– constant flux linkage model including the effect of sub-transient circuits.

Study using simulation softwares for above problems – power flow, short circuit,

transient stability studies of power systems.

REFERENCE BOOKS

1. Stagg, G.W. & Abiad, A.H.. “Computer methods in Power System Analysis”,

McGraw Hill International Editions, 1968

2. Elgerd, Olle L., “Electrical Energy System theory - An introduction” Tata Mc

Graw Hill Edition, 1982.

3. George L. Kusic,”Computer Aided Power System Analysis” Prentice Hall of

India Ltd., 1986

4. John, J. Grainger & Stevenson, D, “Power System Analysis”, Mc Graw Hill

International Editions, 1994.

5. Singh, L. P.,” Advanced Power System Analysis”, Wiley Eastern Ltd., 1986.

6. Kundur, P,” Power System Stability and Control”, McGraw Hill, Inc. 1994.

POWER FLOW ANALYSIS

1 INTRODUCTION

Power flow analysis is the most fundamental study to be performed in a

power system both during the Planning and Operational phases. It

constitutes the major portion of electric utility. The study is concerned with

the normal steady state operation of power system and involves the

determination of bus voltages and power flows for a given network

configuration and loading condition.

The results of power flow analysis help to know

1 the present status of the power system, required for continuous

monitoring.

2 alternative plans for system expansion to meet the ever increasing

demand.

The mathematical formulation of the power flow problem results in a

system of non-linear algebraic equations and hence calls for an iterative

technique for obtaining the solution. Gauss-Seidel method and Newton

Raphson ( N.R.) method are commonly used to get the power flow solution.

Decoupled N.R. method, that uses two sub-matrices of Jacobian matrix, is

significant improvement over Newton Raphson method. The coefficient

matrices are constant in Fast Decoupled Power Flow( FDPF) method. Since

factorization is done only once, FDPF method is well suited for large scale

power systems. These methods are discussed now.

Description of the problem

With reasonable assumptions and approximations, a power system may be

modeled as shown in Fig.1 for purpose of steady state analysis.

Fig. 1 Typical power system network

Static Capacitor

4

44 jQDPD

2

3

G

33 V33 jQGPG

G

1

5

11 jQGPG

11 jQDPD

G 55 jQGPG

a:1

22 jQDPD

The model consists of a network in which a number of buses

representing either generating station or switching stations are

interconnected by means of lines which may either transmission lines or

power transformers.

The generators and loads are simply characterized by the complex powers

flowing into and out of buses respectively. The omission of dynamic of

the generators and frequency and voltage dependency of the loads are

justified since the problem is only Steady State in nature.

Each transmission line is characterized by a lumped impedance and a line

charging capacitance. Static capacitors or reactors may be located at

certain buses either boosting or bucking the load-bus voltages at

times of need.

The ultimate objective in the steady state analysis of a power system is to

obtain the generation schedule for which total cost of generation is

minimum which can meet the given load schedule keeping the line flows

and bus voltages within the allowable limits. This complete problem is termed

as Optimal Power Flow problem.

If the economic aspect of the above problem is dropped and if search is

limited only to a generation schedule, then we have what is termed as a

Feasible Power Flow problem or simply a Power Flow problem.

Thus the Power Flow problem may be stated as follows:

Given the network configuration and the loads at various buses, determine

a schedule of generation so that the bus voltages and hence line flows

remain within security limits.

A more specific statement of the problem will be made subsequently after

taking into consideration the following three observations.

1 For a given load, we can arbitrarily select the schedules of all the

generating buses, except one, to lie within the allowable limits of the

generation. The generation at one of the buses, called as the slack

bus, cannot be specified beforehand since the total generation should

be equal to the total demand plus the transmission losses, which is

not known unless all the bus voltages are determined.

2 Once the complex voltages at all the buses are known, all other

quantities of interest such as line flows, transmission losses and

generation at the buses can easily be determined. Hence the foremost

aim of the power flow problem is to solve for the bus voltages.

3 It will be convenient to use the Bus Power Specification which is

defined as the difference between the specified generation and load at a

bus. Thus for the kth bus, the bus power specification Sk is given by

Sk = PIk + QIk = (PGk + j QGk) - (PDk + j QDk)

= (PGk – PDk) + j(QGk – QDk) (1)

In view of the above three observations Power Flow Problem may be

defined as that of determining the complex voltages at all the buses, given

the network configuration and the bus power specifications at all the buses

except the slack bus.

2 CLASSIFICATION OF BUSES

There are four quantities associated with each bus. They are PI, QI, |V| and δ.

Here PI is the real power injected into the bus

QI is the reactive power injected into the bus

|V| is the magnitude of the bus voltage

δ is the phase angle of the bus voltage

Any two of these four may be treated as independent variables ( i.e.

specified ) while the other two dependent variables may be computed by

solving the power flow equations. Depending on which of the two variables

are specified, buses are classified into three types. Three types of buses

classified based on practical requirements are shown in Fig. 2.

|Vm|, δm PIm, QIm |Vk|, δk PIk, QIk PIs, QIs

? ? ? ? ? ?

Fig. 2 Three types of buses

|Vs|, δs

Slack bus

In a power system with N buses, power flow problem is primarily concerned

with determining the 2N bus voltage variables, namely the voltage

magnitude and phase angles. These can be obtained by solving the 2N

power flow equations provided there are 2N power specifications. However

as discussed earlier the real and reactive power injection at the SLACK

BUS cannot be specified beforehand.

This leaves us with no other alternative but to specify two variables |V| and

δ arbitrarily for the slack bus so that 2( N-1 ) variables can be solved from

2( N-1 ) known power specifications.

Incidentally, the specification of |Vs| helps us to fix the voltage level of the

system and the specification of δs serves as the phase angle reference for

the system.

Thus for the slack bus, both |V| and δ are specified and PI and QI are to

be determined. PI and QI can be computed at the end, when all the |V|s

and δs are solved for.

Generator bus

In a generator bus, it is customary to maintain the bus voltage magnitude

at a desired level which can be achieved in practice by proper reactive

power injection. Such buses are termed as Voltage Controlled Buses or P –

V buses. At these buses PI and |V| are specified and QI and δ are to be

solved for.

Load bus

The buses where there is no controllable generation are called as Load

Buses or P – Q buses. At the load buses, both PI and QI are specified and

|V| and δ are to be solved for.

3 DEVELOPMENT OF POWER FLOW MODEL

Development of power flow model

The power flow model will comprise of a set of simultaneous non-linear

algebraic equations. These equations relate the complex power injection to

complex bus voltages. The solution of this model will yield all the bus

voltages.

There are three types of equations namely

i) Network equations

ii) Bus power equations

iii) Line flow equations

The first two equations are used for the development of power flow model.

The solution of power flow model will yield all the bus voltages. Once the

bus voltages are known, the line and transformer flows can be determined

using the line flow equations.

Network equations

Network equations can be written in a number of alternative forms. Let us

choose the bus frame of reference in admittance form which is most

economical one from the point of view of computer time and memory

requirements.

The equations describing the performance of the network in the bus

admittance form is given by

I = Y V (2)

Here I is the bus current vector,

V is the bus voltage vector and

Y is the bus admittance matrix

In expanded form these equations are

N

2

1

I

I

I

=

NNN2N1

2N2221

1N1211

YYY

YYY

YYY

N

2

1

V

V

V

(3)

Typical element of the bus admittance matrix is

jiY = jiY jiθ = jiY cos jiθ + j jiY sin jiθ = jiG + j jiB (4)

Voltage at a typical bus i is

iV = iV iδ = iV ( cos iδ + j sin iδ ) (5)

The current injected into the network at bus i is given by

n

N

1nni

nNi22i11ii

VY

VYVYVYI

(6)

Bus power equations

In addition to the linear network equations given by eqn. (3), bus power

equations should also be satisfied in the power flow problem. These bus

power equations introduce non- linearity into the power flow model.

The complex power entering at the thk bus is given by

*kkkk IVQjP (7)

This power should be the same as the bus power specification. Thus

*kkkk IVQIjPI (8)

sk

N,1,2,k

Equations (6) and (8) can be suitably combined to obtain the power flow

model.

The following two methods are used to solve the power flow model.

1. Gauss-Seidel method 2. Newton Raphson method

Gauss-Seidel method

Gauss-Seidel method is used to solve a set of algebraic equations.

Consider

NNNN2N21N1

2N2N222121

1N1N212111

yxaxaxa

yxaxaxa

yxaxaxa

N,1,2,k

xaya

1xgivesThisxayxaThus

yxaxaxaxalySpecifical

m

N

km1m

kmkkk

k

N

km1m

mkmkkkk

kNkNkkk2k21k1

;

Initially, values of N21 x,,x,x are assumed. Updated values are calculated

using the above equation. In any iteration 1h , up to ,1km values of mx

calculated in 1h iteration are used and for 1km to N , values of

mx calculated in h iteration are used. Thus

1k

1m

N

1km

hmkm

1hmkmk

kk

1hk xaxay

a

1x (9)

4 GAUSS-SEIDEL METHOD FOR POWER FLOW SOLUTION

In this method, first an initial estimate of bus voltages is assumed. By

substituting this estimate in the given set of equations, a second estimate,

better than the first one, is obtained. This process is repeated and better

and better estimates of the solution are obtained until the difference

between two successive estimates becomes lesser than a prescribed

tolerance.

Let us now find the required equations for calculating the bus voltages.

From the network equations, we have

m

N

km1m

kmkkkm

N

1mkmk

NkNkkk2k21k1k

VYVYVYIThus

VYVYVYVYI

(10)

From the bus power specifications, we have

sk

N,1,2,k

IVQIjPI *kkkk

(11)

sk

N,1,2,k

IVQIjPI *kkkk

(11)

kI is the intermediate variable. This can be eliminated. Taking conjugate of

above equation

*k

kkkk

*kkk V

QIjPIITherefore;IVQIjPI

(12)

From equations (10) and (12) we get

m

N

km1m

kmkkk VYVY*k

kk

V

QIjPI

sk

N,1,2,k

VY

Y

V

1

Y

QIjPI

k

N,1,2,k

VYV

jQIPI

Y

1VThus

m

N

km1m kk

km*

kkk

kk

N

km1m

mkm*k

kk

kkk

s

(13)

A significant reduction in computing time for a solution can be achieved

by performing as many arithmetic operations as possible before initiating

the iterative calculation. Let us define

kkk

kk AY

QIjPI

(14)

kmkk

km BY

Yand (15)

Having defined kmk BandA equation (13) becomes

skN,1,2,kVBV

AV

N

km1m

mkm*k

kk

(16)

When Gauss-Seidel iterative procedure is used, the voltage at the thk bus

during th1h iteration, can be computed as

sk

N,1,2,k

VBVBV

AV

N

1km

hmkm

1k

1m

1hmkm*h

k

k1hk

(17)

m k

ykm’ ykm

’

Ikm

5 LINE FLOW EQUATIONS

Knowing the bus voltages, the power in the lines can be computed as

shown below.

Current Ikm = (Vk – Vm) ykm + Vk y’km (18)

Power flow from bus k to bus m is *kmkkmkm IVQjP (19)

Substituting equation (18) in equation (19)

]yVy)VV([VQjP *'km

*k

*km

*m

*kkkmkm (20)

Similarly, power flow from bus m to bus k is

]yVy)VV([VQjP *'km

*m

*km

*k

*mmmkmk (21)

The line loss in the transmission line mk is given by

)QjP()QjP(P mkmkkmkmmkL (22)

Total transmission loss in the system is

linesthealloverji

jiLL PP (23)

kmy

Fig. 3 Circuit for line flow calculation

6 REPRESENTATION OF OFF NOMINAL TAP SETTING TRANSFORMER

A transformer with no tap setting arrangement can be represented similar to a short transmission line. However off nominal tap setting transformer which has tap setting facility at the HT side calls for different representation.

Consider a transformer 110 / 11 kV having tap setting facility. Its nominal ratio is 10 : 1 i.e. its nominal turns / voltage ratio = 10. Suppose the tap is adjusted so that the turns ratio becomes 11 : 1. This can be thought of connecting two transformers in cascade, the first one with ratio 11 : 10 and the second one with the ratio 10 : 1 (Nominal ratio). In this case the off nominal turns ratio is 11 / 10 = 1.1.

Suppose the tap is adjusted so that the turns ratio becomes 9 : 1. This is same as two transformers in cascade, the first with a turn ratio 9 : 10 and the second with the ratio 10 : 1 ( Nominal ratio ). In this case off nominal turn ratio is 9 / 10 = 0.9.

A transformer with off nominal turn ratio can be represented by its impedance or admittance, connected in series with an ideal autotransformer as shown in Fig. 4.

C B

An equivalent π circuit as shown in Fig. 5, will be useful for power flow studies. The elements of the equivalent π circuit can be treated in the same manner as the line elements.

Let ‘a’ be the turn ratio of the autotransformer. ‘a’ is also called as off nominal turns ratio. Usually ‘a’ varies from 0.85 to 1.15.

Fig. 4 Representation of off nominal transformer

a:1

qI

pqy

pI pqi

p q t

I q I p

Fig. 5 Equivalent circuit of off nominal transformer

A p q

The parameters of the equivalent π circuit shown in Fig. 5 can be derived

by equating the terminal current of the transformer with the corresponding

current of the equivalent π circuit.

It is to be noted that

aI

ianda

E

E

p

qt

t

p (24)

qqp

p2

qp

qpqp

qpqtqt

p

Ea

yE

a

y

y)Ea

E(

a

1y)EE(

a

1

a

iI

Further

qqppqp

qpp

qqptqq

EyEa

y

y)a

EE(y)EE(I

(25)

(26)

a:1

qI

pqy

pI pqi

p q t

a

y qp

qpy)1a

1(

a

1 qpy)

a

11(

p

Referring to the π equivalent circuit

qppqpp EAE)BA(BEA)E(EI and

qpqpqq E)CA(EACEA)EE(I

Comparing the coefficients of Ep and Eq in eqns. (25) to (28), we get

qp

qp

2

qp

yCAa

yA

a

yBA

qpqp

qp

qpqp

2

qp

qp

y)a

11(

a

yyC

y)1a

1(

a

1

a

y

a

yB

a

yAThus

The equivalent π circuit will then be as shown in Fig. 6

(28)

(27)

I q I p

q

Fig. 6 Equivalent circuit of off nominal transformer

When the off nominal turns ratio is represented at bus p for transformer

connected between p and q is included, the following modifications are

necessary in the bus admittance matrix.

a

yYY

a

yYY

yY

y)a

11(

a

YYY

a

yY

y)1a

1(

a

1

a

yYY

qpoldpqnewpq

qpoldqpnewqp

qpoldqq

qpqp

oldqqnewqq

2

qpoldpp

qpqp

oldppnewpp

(29)

Example 1

Obtain the bus admittance matrix of the transmission system with the following data.

Line data

Line

No.

Between

buses Line Impedance HLCA

Off nominal

turns ratio

1 1 – 4 0.08 + j 0.37 j 0.007 ---

2 1 – 6 0.123 + j 0.518 j 0.010 ---

3 2 – 3 0.723 + j 1.05 0 ---

4 2 – 5 0.282 + j 0.64 0 ---

5 4 – 3 j 0.133 0 0.909

6 4 – 6 0.097 + j 0.407 j 0.0076 ---

7 6 – 5 j 0.3 0 0.976

Shunt capacitor data

Bus No. 4 Admittance j 0.005

% % Bus admittance matrix including off nominal transformer % Shunt parameter must be in ADMITTANCE % Reading the line data % Nbus = 6; Nele = 7; Nsh = 1; % % Reading element data % Edata=[1 1 4 0.08+0.37i 0.007i 1;2 1 6 0.123+0.518i 0.01i 1;… 3 2 3 0.723+1.05i 0 1; 4 2 5 0.282+0.64i 0 1; ... 5 4 3 0.133i 0 0.909;6 4 6 0.097+0.407i 0.0076i 1; ... 7 6 5 0.3i 0 0.976]; % % Reading shunt data % Shdata = [1 4 0.005i]; % % Displaying data % disp([' Sl.No From bus To bus Line Impedance HLCA ONTR']) Edata if Nsh~=0 disp([' Sl.No. At bus Shunt Admittamce']) Shdata end

%

% Formation of Ybus matrix % Ybus = zeros(Nbus,Nbus); for k = 1:Nele p = Edata(k,2); q = Edata(k,3); yele = 1/Edata(k,4); Hlca = Edata(k,5); offa = Edata(k,6); offaa = offa*offa; Ybus(p,p) = Ybus(p,p) + yele/offaa + Hlca; Ybus(q,q) = Ybus(q,q) + yele + Hlca; Ybus(p,q) = Ybus(p,q) - yele/offa; Ybus(q,p) = Ybus(q,p) - yele/offa; end if Nsh~=0 for i = 1:Nsh q = Shdata(i,2); yele = Shdata(i,3); Ybus(q,q) = Ybus(q,q) + yele; end end disp([' *** BUS ADMITTANCE MATRIX ***']) Ybus

Answer

j7.6341

0.988

j3.4153j2.3209

0.554100

j1.8275

0.4339

j3.4153j4.6418

0.576500

j1.3085

0.57650

j2.3249

0.5541

0j13.9869

1.1124j8.27150

j2.5820

0.5583

00j8.2715j8.1649

0.4449

j0.6461

0.44490

0j1.3085

0.57650

j0.6461

0.4449

j1.9546

1.02140

j1.8275

0.4339

0j2.5820

0.558300

j4.3925

.99920

Example 2

For a power system, the transmission line impedances and the line

charging admittances in p.u. on a 100 MVA base are given in Table 1. The

scheduled generations and loads on different buses are given in Table 2.

Taking the slack bus voltage as 1.06 + j 0.0 and using a flat start perform

the power flow analysis and obtain the bus voltages, transmission loss and

slack bus power.

Table 1 Transmission line data:

Sl. No. Bus code

k - m Line Impedance

kmz HLCA

1 1 – 2 0.02 + j 0.06 j 0.030 2 1 – 3 0.08 + j 0.24 j 0.025 3 2 – 3 0.06 + j 0.18 j 0.020 4 2 – 4 0.06 + j 0.18 j 0.020 5 2 – 5 0.04 + j 0.12 j 0.015 6 3 – 4 0.01 + j 0.03 j 0.010 7 4 – 5 0.08 + j 0.24 j 0.025

Table 2 Bus data:

Solution

Flat start means all the unknown voltage magnitude are taken as 1.0 p.u.

and all unknown voltage phase angles are taken as 0.

Thus initial solution is

0j1.0VVVV

0j1.06V(0)5

(0)4

(0)3

(0)2

1

STEP 1

For the transmission system, the bus admittance matrix can be calculated as

Bus code

k

Generation Load Remark MWinPGk MVARinQGk MWinPDk MVARinQDk

1 --- --- 0 0 Slack bus 2 40 30 20 10 P – Q bus 3 0 0 45 15 P – Q bus 4 0 0 40 5 P – Q bus 5 0 0 60 15 P – Q bus

1 2 3 4 5

1 --- --- --- --- ---

2 -5 +j15 10.833–j32.415 -1.6667 + j5 -1.6667+j5 -2.5 + j7.5

Y = 3 -1.25 + j3.75 -1.6667 + j5 12.9167–j38.695 -10 + j30 0

4 0 -1.6667 + j5 -10 + j30 12.9167-j38.695 -1.25 + j3.75

5 0 -2.5+j7.5 0 -1.25+j3.75 3.75-j11.21

STEP 2

Calculation of elements of A vector and B matrix.

manner.similaraincalculatedbecanBmatrixofelementsOther

0.00036j0.4626332.415j10.833

5j5

Y

YB

.calculatedareAandA,ASimilarly

0.0037j0.007432.415j10.833

0.2j0.2

Y

QIjPIA

0.2j0.2QIjPI

0.2j0.2)20j20(100

1QIjPI

Y

YBand

Y

QIjPIA

22

2121

543

22

222

22

22

kk

kmkm

kk

kkk

Thus

1 -----

2 0.00740 + j 0.00370

A = 3 -0.00698 - j 0.00930

4 -0.00427 - j 0.00891

5 -0.02413 - j 0.04545

B =

--- --- --- --- --- - 0.46263

+ j 0.00036

--- - 0.15421

+ j 0.00012 - 0.15421

+ j 0.00012 - 0.23131

+ j 0.00018 - 0.09690

+ j 0.00004 - 0.12920

+ j 0.00006

--- - 0.77518

+ j 0.00033

0

0 - 0.12920 + j 0.00006

- 0.77518 + j 0.00033

---

- 0.09690 + j 0.00004

0 - 0.66881 + j 0.00072

0

- 0.33440 + j 0.00036

---

STEP 3

Iterative computation of bus voltage can be carried out as shown.

New estimate of voltage at bus 2 is calculated as:

0.00290j1.03752

)0.0j1.0()0.00018j0.23131(

)0.0j1.0()0.00012j0.15421()0.0j1.0()0.00012j0.15421(

)00.0j1.06()0.00036j0.46263(0j1.0

0.00370j0.00740

VBVBVBVBV

AV )0(

525)0(

424)0(

323121*)0(2

2)1(2

This value of voltage )1(2V will replace the previous value of voltage )0(

2V

before doing subsequent calculations of voltages.

The rate of convergence of the iterative process can be increased by

applying an ACCELERATION FACTOR to the approximate solution

obtained. For example on hand, from the estimate )1(2V we get the change

in voltage

0.00290j0.03752

)0j1.0()0.00290j1.03752(VVVΔ )0(2

)1(22

The accelerated value of the bus voltage is obtained as

0.00406j1.05253

)0.00290j0.03752(1.4)0j1.0(V

1.4assumingBy

ΔVVV

)1(2

2)0(

2)1(

2

This new value of voltage )1(2V will replace the previous value of the bus

voltage )0(2V and is used in the calculation of voltages for the remaining

buses. In general

)VV(VV hk

1hk

hk

1haccldk (30)

The process is continued for the remaining buses to complete one

iteration. For the next bus 3

0.00921j1.00690

)0j1.0()0.00033j0.77518(

)0.00406j1.05253()0.00006j0.12920(

)0j1.06()0.00004j0.09690(0j1.0

0.00930j0.00698

VBVBVBV

AV )0(

434)1(

232131*)0(3

3)1(3

The accelerated value can be calculated as

0.01289j1.00966

)0.00921j0.00690(1.4)0j1.0(

)VV(VV )0(3

)1(3

)0(3

)1(accld3

Continuing this process of calculation, at the end of first iteration, the bus

voltages are obtained as

0.07374j1.02727V

0.02635j1.01599V

0.01289j1.00966V

0.00406j1.05253V

0.0j1.06V

)1(5

)1(4

)1(3

)1(2

1

If and are the acceleration factors for the real and imaginary

components of voltages respectively, the accelerated values can be

computed as

)ff(βff

)ee(eehk

1hk

hk

1haccldk

hk

1hk

hk

1haccldk

(31

CONVERGENCE

The iterative process must be continued until the magnitude of change of

bus voltage between two consecutive iterations is less than a certain level

for all bus voltages. We express this in mathematical form as

hk

1hk VV <

If 1ε and 2ε are the tolerance level for the real and imaginary parts of

bus voltages respectively, then the convergence criteria will be

hk

1hk ee < 1ε and h

k1h

k ff < 2ε

For the problem under study 0.0001εε 21

The final bus voltages obtained after 10 iterations are given below.

0.10904j1.01211V

0.09504j1.01920V

0.08917j1.02036V

0.05126j1.04623V

0.0j1.06V

5

4

3

2

1

sk

N..,1,2,......kforε

(32)

sk

N..,1,2,......kfor

(33)

ε

COMPUTATION OF LINE FLOWS AND TRANSMISSION LOSS

Line flows can be computed from

)0.086j0.888(

]})0.03j0.0(0)j1.06({

)15j5(})0.05126j1.04623()0j1.06({[)0j(1.06

]yVy)VV([VQjP

]yVy)VV([VQjP

*'12

*1

*12

*2

*111212

*'km

*k

*km

*m

*kkkmkm

)0.062j0.874(

]})0.03j(0.0)0.05126j1.04623({

15)j5}()0j1.06(0.05126)j1.04623({[)0.05126j1.04623(

yVy)VV([VQjP

Similarly*'

12*2

*12

*1

*222121

Power loss in line 1 – 2 is

)0.024j0.014( )QjP()QjP(P 2121121221L

Power loss in other lines can be computed as

0.051j0.0P

0.019j0.0P

0.002j0.011P

0.029j0.004P

0.033j0.004P

0.019j0.012P

54L

43L

52L

42L

32L

31L

Total transmission loss = (0.045 - j 0.173) i.e.

Real power transmission loss = 4.5 MW

Reactive power transmission loss = 17.3 MVAR (Capacitive)

Total transmission loss can also computed as the total generation minus total

load.

7 VOLTAGE CONTROLLED BUS

In voltage controlled bus k net real power injection kPI and voltage

magnitude kV are specified. Normally minmax QIandQI will also be specified

for voltage controlled bus. Since kQI is not known, kA given by kk

kk

Y

QIjPI

cannot be calculated. An expression for kQI can be developed as shown

below.

We know *kkkkm

N

1mmkk IVQIjPIandVYI

Denoting haveweδVVandθYY iiijijiji

N

1mmkmkmkmkk

mkmkmk

N

1mmk

mmkm

N

1mmkkkkk

)θδδ(sinYVVQIThus

θδδYVV

δθVYδVQIjPI

(34)

The value of kV to be used in equation (34) must satisfy the relation

specifiedkk VV

Because of voltage updating in the previous iteration, the voltage

magnitude of the voltage controlled bus might have been deviated from the

specified value. It has to be pulled back to the specified value, using the

relation

Adjusted voltage hk

hk

hkh

k

hk1h

khkspecifiedk

hk fjeVtaking)

e

f(tanδwhereδVV

Using the adjusted voltage hkV as given in eqn. (35), net injected reactive

power hkQI can be computed using eqn. (34). As long as h

kQI falls within

the range specified, hkV can be replaced by the Adjusted h

kV and kA can

be computed.

(35)

In case if hkQI falls beyond the limits specified, h

kV should not be

replaced by Adjusted hkV , h

kQI is set to the limit and kA can be

calculated. In this case bus k is changed from P – V to P – Q type.

Once the value of kA is known, further calculation to find 1hkV will be

the same as that for P – Q bus.

Complete flow chart for power flow solution using Gauss-Seidel method is

shown in Fig. 7. The extra calculation needed for voltage controlled bus is

shown between X – X and Y – Y.

=

8 NEWTON RAPHSON METHOD OF SOLVING A SET OF NON-LINEAR

ALGEBRAIC EQUATIONS

Let the non-linear equations to be solved be

1k)x,,x,(xf n211

2n212 k)x,,x,(xf

nn21n k)x,,x,(xf

Let the initial solution be (0)n

(0)2

(0)1 x,,x,x

If 0)x,,x,(xfk (0)n

(0)2

(0)111

0)x,,x,(xfk (0)n

(0)2

(0)122

0)x,,x,(xfk (0)n

(0)2

(0)1nn

then the solution is reached.

(36)

Let us say that the solution is not reached. Assume n21 Δx,,Δx,Δx are the

corrections required on (0)n

(0)2

(0)1 x,,x,x respectively. Then

1k)]Δx(x ,), Δx(x),Δx[(x f n(0)n2

(0)21

(0)11

2n

(0)n2

(0)21

(0)12 k)]Δx(x,),Δx(x),Δx[(xf

nn(0)n2

(0)21

(0)1n k)]Δx(x ,),Δx(x),Δx[(xf

Each equation in the above set can be expanded by Taylor’s theorem

around (0)n

(0)2

(0)1 x,,x,x . For example, the following is obtained for the first

equation.

(0)n

(0)2

(0)11 x,,x,(xf

1

1

x

f)

2

11 x

fΔx

n

12 x

fΔx

11n kφΔx

where 1φ is a function of higher derivatives of 1f and higher powers of

n21 Δx,,Δx,Δx . Neglecting 1φ and also following the same for other

equations, we get

(37)

0

0

0

0

0 0 0

0 0 0

0 0 0

(39)

(0)n

(0)2

(0)11 x,,x,(xf

1

1

x

f)

2

11 x

fΔx

n

12 x

fΔx

1n kΔx

(0)n

(0)2

(0)12 x,,x,(xf

1

2

x

f)

2

21 x

fΔx

n

22 x

fΔx

1n kΔx

(0)n

(0)2

(0)1n x,,x,(xf

1

n

x

f)

2

n1 x

fΔx

n

n2 x

fΔx

nn kΔx

The matrix form of equations (38) is

)x,,x,x(fkΔxx

f

x

f

x

f (0)n

(0)2

(0)1111

n

1

2

1

1

1

)x,,x,x(fkΔxx

f

x

f

x

f (0)n

(0)2

(0)1222

n

2

2

2

1

2

)x,,x,x(fkΔxx

f

x

f

x

f (0)n

(0)2

(0)1nnn

n

n

2

n

1

n

0 0

0

0

=

0 (38)

This set of linear equations need to be solved for the correction vector

n

2

1

Δx

Δx

Δx

ΔX

In eqn.(40) )X(F (0)' is called the JACOBIAN MATRIX and the vector

)X(FK (0) is called the ERROR VECTOR. The Jacobian matrix is also

denoted as J.

Solving eqn. (40) for ΔX

ΔX = 1)0(' )X(F )X(FK )(0 (41)

Then the improved estimate is

ΔXXX )0(1)(

Generalizing this, for th)1h( iteration

ΔXXX )h()1h( where (42)

)X(FK)X(FΔX )h(1)h(' (43)

i.e. ΔX is the solution of

)X(FKΔX)X(F )h()h(' (44)

)X(FKΔX)X(F )h()h(' (44)

Thus the solution procedure to solve K)X(F is as follows :

(i) Calculate the error vector )X(FK (h)

If the error vector zero, convergence is reached; otherwise formulate

)X(FKΔX)X(F (h)(h)'

(ii) Solve for the correction vector ΔX

(iii) Update the solution as

ΔXXX (h)1)(h

Values of the correction vector can also be used to test for convergence.

Example 3

Using Newton-Raphson method, solve for 1x and 2x of the non-linear

equations

4 2x sin 1x = - 0.6; 4 22x - 4 2x cos 1x = - 0.3

Choose the initial solution as (0)1x = 0 rad. and (0)

2x = 1. Take the precision index

on correction vector as 310 .

Solution

Errors are calculated as

- 0.6 – (4 2x sin 1x ) = - 0.6

- 0.3 – (4 22x - 4 2x cos 1x ) = - 0.3

The error vector

0.3

0.6is not small.

It is noted that 1f = 4 2x sin 1x 2f = 4 22x - 4 2x cos 1x

1x

0x

2

1

1x

0x

2

1

Jacobian matrix is: J =

2

2

1

2

2

1

1

1

x

f

x

fx

f

x

f

=

1212

112

xcos4x8xsinx4

xsin4xcosx4

Substituting the latest values of state variables 1x = 0 and 2x = 1

J =

40

04 ; Its inverse is 1J =

0.250

00.25

Correction vector is calculated as

2

1

xΔ

xΔ =

0.250

00.25

0.3

0.6 =

0.075

0.150

The state vector is updated as

2

1

x

x =

1

0 +

0.075

0.150 =

0.925

0.150

This completes the first iteration.

Errors are calculated as

- 0.6 – (4 2x sin 1x ) = - 0.047079

- 0.3 – (4 22x - 4 2x cos 1x ) = - 0.064047

The error vector

0.064047

0.04709

is not small.

Jacobian matrix is: J =

1212

112

xcos4x8xsinx4

xsin4xcosx4

=

3.4449160.552921

0.5977533.658453

Correction vector is calculated as

2

1

xΔ

xΔ =

3.4449160.552921

0.5977533.658453

0.064047

0.047079 =

0.021214

0.016335

- 1

0.925x

rad.0.15x

2

1

0.925x

rad.0.15x

2

1

0.925x

rad.0.15x

2

1

The state vector is updated as

2

1

x

x =

0.925

0.150 +

0.021214

0.016335 =

0.903786

0.166335

This completes the second iteration.

Errors are calculated as

- 0.6 – (4 2x sin 1x ) = = - 0.001444

- 0.3 – (4 22x - 4 2x cos 1x ) = - 0.002068

Still error vector exceeds the precision index.

Jacobian matrix is: J =

1212

112

xcos4x8xsinx4

xsin4xcosx4

=

3.2854950.598556

0.6622763.565249

0.903786x

rad.0.166335x

2

1

0.903786x

rad.0.166335x

2

1

0.903786x

rad.0.166335x

2

1

Correction vector is calculated as

2

1

xΔ

xΔ =

3.2854950.598556

0.6622763.565249

0.002068

0.001444 =

0.000728

0.000540

These corrections are within the precision index. The state vector is updated as

2

1

x

x =

0.903786

0.166335 +

0.000728

0.000540 =

0.903058

0.166875

Errors are calculated as

- 0.6 – (4 2x sin 1x ) = = - 0.000002

- 0.3 – (4 22x - 4 2x cos 1x ) = - 0.000002

Errors are less than 10-3.

The final values of 1x and 2x are - 0.166875 rad. and 0.903058 respectively.

- 1

0.903058x

rad.0.166875x

2

1

0.903058x

rad.0.166875x

2

1

The results can be checked by substituting the solution in the original equations:

4 2x sin 1x = - 0.6 4 22x - 4 2x cos 1x = - 0.3

In this example, we have actually solved our first power flow problem by N.R.

method. This is because the two non-linear equations of this example are the

power flow model of the simple system shown in Fig. 8.

Here bus 1 is the slack bus with its voltage 1V 1δ = 1.0 00 p.u. Further 1x

represents the angle 2δ and 2x represents the voltage magnitude 2V at bus 2.

We now concentrate on the application of Newton-Raphson procedure in the

power flow studies.

Ιv2 Ι 2δ

Fig. 8 Two bus power flow model

p.u.)0.3j0.6(QP 2d2d j 6.0 p.u.

j 0.25 p.u. G

1 2 Ιv1 Ι 1δ

9 POWER FLOW SOLUTION BY NEWTON RAPHSON METHOD

As discussed earlier, taking the bus voltages and line admittances in polar form,

in power flow study we need to solve the non-linear equations

N

1niV nV niY cos ( niθ + nδ - iδ ) = iPI (45)

-

N

1niV nV niY sin ( niθ + nδ - iδ ) = iQI (46)

Separating the term with in we get

ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ ) = iPI (47)

ii

2

i BV

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) = iQI (48)

In a compact form, the above non-linear equations can be written as

PI)Vδ,(P (49)

QI)Vδ,(Q (50)

On linearization, we get

ΔQ

ΔP

VΔ

Δδ

V

Q

δ

QV

P

δ

P

(51)

where

PIΔP computed value of )Vδ,(P corresponding to the present solution.

QIΔQ computed value of )Vδ,(Q corresponding to the present solution.

To bring symmetry in the elements of the coefficient matrix, V

VΔ is taken

as problem variable in place of VΔ . Then eqn. (61) changes to

ΔQ

ΔP

V

VΔ

Δδ

VV

Q

δ

Q

VV

P

δ

P

(62)

In symbolic form, the above equation can be written as

ΔQ

ΔP

V

VΔ

Δδ

LM

NH

(63)

The matrix H N is known as JACOBIAN matrix.

M L

The dimensions of the sub-matrices will be as follows:

H )N(N 21 x )N(N 21

N )N(N 21 x 1N

M 1N x )N(N 21 and

L 1N x 1N

where 1N is the number of P-Q buses and 2N is the number of P-V buses.

Consider a 4 bus system having bus 1 as slack bus, buses 2 and 3 as P-

Q buses and bus 4 as P-V bus for which real power injections

432 PI&PI,PI and reactive power injections 32 QI&QI are specified. Noting

that 32432 VandV,δ,δ,δ are the problem variables, linear equations that

are to be solved in each iteration will be

32432 VVδδδ

2P 33

22

2

2

4

2

3

2

2

2 VV

PV

V

P

δ

P

δ

P

δ

P

2Δδ 2ΔP

3P 33

32

2

3

4

3

3

3

2

3 VV

PV

V

P

δ

P

δ

P

δ

P

3Δδ 3ΔP

4P 33

42

2

4

4

4

3

4

2

4 VV

PV

V

P

δ

P

δ

P

δ

P

4Δδ = 4ΔP (64)

2Q 2

2

δ

δQ

3

2

δ

Q

4

2

δ

Q

22

2 VV

Q

33

2 VV

Q

2

2

V

VΔ 2ΔQ

3Q 2

3

δ

δQ

3

3

δ

Q

4

3

δ

Q

22

3 VV

Q

33

3 VV

Q

3

3

V

VΔ 3QΔ

The following is the solution procedure for N.R. method of power flow analysis.

1 Read the line data and bus data; construct the bus admittance matrix.

2 Set k = 0. Assume a starting solution. Usually a FLAT START is assumed in

which all the unknown phase angles are taken as zero and unknown voltage

magnitudes are taken as 1.0 p.u.

3 Compute the mismatch powers i.e. the error vector. If the elements of error

vector are less than the specified tolerance, the problem is solved and hence

go to Step 7; otherwise proceed to Step 4.

k

4 Compute the elements of sub-matrices H, N, M and L. Solve

ΔQ

ΔP

V

VΔ

Δδ

LM

NH

for

V

VΔ

Δδ

5 Update the solution as

V

δ =

V

δ +

VΔ

Δδ

6 Set k = k + 1 and go to Step 3.

7 Calculate line flows, transmission loss and slack bus power. Print the

results and STOP

k+1

k

k

Calculation of elements of Jacobian matrix

We know that the equations that are to be solved are

ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ ) = iPI (65)

ii

2

i BV

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) = iQI (66)

i.e. ii PI)V,(δP (67)

ii QI)Vδ,(Q (68)

The suffix i should take necessary values.

Jacobian matrix is

LM

NH where

VV

QLand

δ

QM;V

V

PN;

δ

PH

Here

iP = ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ ) (69)

iQ = ii

2

i BV

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) (70)

Diagonal elements:

i

iii δ

PH

=

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) = ii

2

ii BVQ (71)

ii

2

iii

iii GV2V

V

PN

+

N

in1n

iV nV niY cos ( niθ + nδ - iδ )

= iP + ii

2

i GV (72)

i

iii δ

QM

=

N

in1n

iV nV niY cos ( niθ + nδ - iδ ) = iP - ii

2

i GV (73)

ii

iii V

V

QL

= ii

2

i BV2 -

N

in1n

iV nV niY sin ( niθ + nδ - iδ )

= iQ ii

2

i BV (74)

Off-diagonal elements:

We know that

iP = ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ ) (75)

iQ = ii

2

i BV

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) (76)

jijij

iji YVV

δ

PH

sin ( jiθ + jδ - iδ ) (77)

j

j

iji V

V

PN

= jiji YVV cos ( jiθ + jδ - iδ ) (78)

j

iji δ

QM

= jiji YVV cos ( jiθ + jδ - iδ ) (79)

j

j

iji V

V

QL

= jiji YVV sin ( jiθ + jδ - iδ ) (80)

Summary of formulae

iiH ii

2

ii BVQ

iiN iP + ii

2

i GV

iiM iP ii

2

i GV

iiL iQ ii

2

i BV

jiH jiji YVV sin ( jiθ + jδ - iδ )

jiN jiji YVV cos ( jiθ + jδ - iδ )

jiM jiji YVV cos ( jiθ + jδ - iδ )

jiL jiji YVV sin ( jiθ + jδ - iδ )

It is to be noted that in general, the sub-matrices of Jacobian matrix as well as the

Jacobian matrix are not symmetrical.

(81)

k

START

READ LINE DATA COMPUTE Y MATRIX ASSUME FLAT START

SET k = 0

FOR ALL P-V BUSES COMPUTE iQ

IF iQ VIOLATES THE LIMITS SET

iQ = LimitiQ AND TREAT BUS i AS A P-Q BUS

COMPUTE MISMATCH POWERS ΔQ&ΔP

COMPUTE MATRICES H,N,M & L

FORM

ΔQ

ΔP

V

VΔΔδ

LM

NH ; SOLVE FOR

V

VΔΔδ

AND UPDATE

V

δ =

V

δ +

VΔ

Δδ

YES

SET k = k + 1

COMPUTE LINE FLOWS, TRANSMISSION LOSS & SLACK BUS POWER PRINT THE RESULTS

STOP

NO

k+1 k

ELEMENTS OF ΔQ&ΔP < ε ?

Example 4

Perform power flow analysis for the power system with the data given

below, using Newton Raphson method, and obtain the bus voltages.

Line data ( p.u. quantities )

Bus data ( p.u. quantities )

Bus

No Type

Generator Load V δ minQ maxQ

P Q P Q

1 Slack --- --- 0 0 1.0 0 --- ---

2 P - V 1.8184 --- 0 --- 1.1 --- 0 3.5

3 P - Q 0 0 1.2517 1.2574 --- --- --- ---

0.2j0313

0.2j0322

0.1j0211

impedancesLinebusesBetweenNo.Line

Solution

The bus admittance matrix can be obtained as

1 2 3 1 2 3

Y =

j10j5j5

j5j15j10

j5j10j15

= j

1055

51510

51015

This gives

Y

1055

51510

51015

and θ =

000

000

000

909090

909090

909090

In this problem

1.2574QI

1.2517PI

1.8184PI

3

3

2

and unknown quantities =

3

3

2

V

δ

δ

1

2

3

1 2 3

1

2

3

1 2 3

1 2

3

1

2

3

With flat start 01 01.0V

02 01.1V

03 01.0V

We know that

iP = ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ )

iQ = ii

2

i BV

N

in1n

iV nV niY sin ( niθ + nδ - iδ )

Substituting the values of bus admittance parameters, expressions for 2P , 3P

and 3Q are obtained as follows

2P = )δδθ(cosYVV)δδθ(cosYVVGV 233232322112121222

2

2

= 0 + )δδ90(cosVV5)δδ90(cosVV10 23322112

= )δδ(sinVV5)δδ(sinVV10 23322112

Similarly

3P = 5 3V 1V )δδ(sin 31 5 3V 2V )δδ(sin 32

Likewise

33

2

33 BVQ 3V 1V )δδ90(sinY 3113 3V 2V )δδ(90sinY 3223

= 5V102

3 3V 1V 5)δδ(cos 31 3V 2V )δδ(cos 32

To check whether bus 2 will remain as P-V bus, 2Q need to be calculated.

10V15Q2

22 2V 1V 5)δδ(cos 21 2V 3V )δδ(cos 23

= ( 15 x 1.1 x 1.1 ) – ( 10 x 1.1 x 1 x 1 ) – ( 5 x 1.1 x 1 x 1 ) = 1.65

This lies within the Q limits. Thus bus 2 remains as P – V bus.

ii

2

iiii

iiiiii

ii

2

iiii

BVQL

PM;PN

BVQH

)δδ(sinYVVM

)δδ(sinYVVN

)δδ(cosYVVH

ijjijiji

ijjijiji

ijjijiji

Since 0,δδδ 321 we get 0PP 32

3Q = ( 10 x 1 x 1 ) – ( 5 x 1 x 1 ) – ( 5 x 1 x 1.1 ) = - 0.5

Mismatch powers are: 1.818401.8184PPIΔP 222

1.251701.2517PPIΔP 333

0.75740.51.2574QQIΔQ 333

333332

333332

232322

332

LMM

NHH

NHH

Vδδ

3

3

3

2

V

VΔΔδ

Δδ

=

3

3

2

ΔQ

ΔP

ΔP

For this problem, since iiG are zero and jiθ are 090

2P

3P

3Q

Linear equations are

22

2

2222 BVQH - 1.65 + ( 1.1 x 1.1 x 15 ) = 16.5

10.5100.5BVQH 33

2

3333

0PM;0PN 333333

9.5100.5BVQL 33

2

3333

32H 2V 3V )δδ(cosY 2332 = 5.51x5x1x1.1 and 23H = 5.5

32N 2V 3V 0)δδ(sinY 2332

32M 3V 2V 0)δδ(sinY 3223

Thus

9.500

010.55.5

05.516.5

3

3

3

2

V

VΔΔδ

Δδ

=

0.7574

1.2517

1.8184

Solving the above

3

3

3

2

V

VΔΔδ

Δδ

=

0.0797

0.07449

0.08538

Therefore

0)1(2 4.89rad.0.085380.085380δ

0)1(3 4.27rad.0.074490.074490δ

0.92030.07971.0V)1(

3

Thus 01 01.0V

02 4.891.1V

03 4.270.9203V

This completes the first iteration.

Second iteration:

2Q (15 x 1.1 x 1.1) - (10 x 1.1 x 1.0 cos 04.89 ) - (5 x 1.1 x 0.9203 cos 09.16 )

= 2.1929

This is within the limits. Bus 2 remains as P-V bus.

2P (10 x 1.1 x 1.0 sin 04.89 ) + ( 5 x 1.1 x 0.9203 sin 09.16 ) = 1.7435

3P - ( 5 x 0.9203 x 1.0 sin 4.27 0 ) – ( 5 x 0.9203 x 1.1 sin 9.16 0 ) = -1.1484

3Q = 10x 0.9203 x 0.9203 - (5 x 0.9203 x 1.0 cos 4.27 0 ) - (5 x 0.9203 x 1.1 cos 9.16 0 )

= - 1.1163

2ΔP 1.8184 – 1.7435 = 0.0749

3ΔP - 1.2517 + 1.1484 = - 0.1033

3ΔQ - 1.2574 + 1.1163 = -0.1444

22H - 2.1929 + ( 1.1 x 1.1 x 15 ) = 15.9571

33H 1.1163 + (0.9203 x 0.9203 x 10) = 9.5858

33N = - 1.1484; 33M = - 1.1484

33L - 1.1163 + ( 0.9203 2 x 10 ) = 7.3532

23H - 1.1 x 0.9203 x 5 cos 9.16 0 = - 4.9971

32H = - 4.9971

23N 1.1 x 0.9203 x 5 sin 9.16 0 = 0.8058

32M = 0.8058

The linear equations are

7.35321.14840.8058

1.14849.58584.9971

0.80584.997115.9571

3

3

3

2

V

VΔΔδ

Δδ

=

0.1444

0.1033

0.0749

Its solution is

3

3

3

2

V

VΔΔδ

Δδ

=

0.021782

0.012388

0.001914

3VΔ - 0.9203 x 0.02178 = - 0.02

0)2(2 5.00rad.0.087290.0019140.08538δ

0)2(3 4.98rad.0.086880.0123880.07449δ

0.90320.020.9232V)2(

3

Thus at the end of second iteration

01 01.0V

02 5.001.1V

03 4.980.9032V

Continuing in this manner the final solution can be obtained as

01 01.0V

02 51.1V

03 50.9V

Once we know the final bus voltages, if necessary, line flows, transmission

loss and the slack bus power can be calculated following similar procedure

adopted in the case of Gauss – Seidel method.

10 DECOUPLED / FAST DECOUPLED POWER FLOW METHOD

In Newton Raphson method of power flow solution, in each iteration,

linear equations

ΔQ

ΔP

V

VΔ

Δδ

LM

NH

(82)

are to be solved for the correction vector

V

VΔ

Δδ

. When the power

system has N1 number of P-Q buses and N2 number of P-V buses the

size of the Jacobian matrix is 2N1 + N2 . This will not exceed 2 x ( N-1 )

where N is the number of buses in the power system under study. Even

though factorization method can be adopted to solve such large size linear

algebraic equations, factorization has to be carried out in each iteration

since the elements of the Jacobian matrix will change in values in each

iteration. This results in enormous amount of calculations in each iteration.

In practice, however, the Jacobian matrix is often recalculated only every

few iterations and this speeds up the overall solution process. The final

solution is obtained, of course, by the allowable power mismatches at the

buses.

When solving large scale power systems, an alternative strategy for

improving computational efficiency and reducing computer storage

requirements, is the FAST DECOUPLED POWER FLOW METHOD, which

makes use of an approximate version of the Newton Raphson procedure.

The principle underlying the decoupled approach is based on a few

approximations which are acceptable in large practical power systems.

As a first step, the following two observations can be made:

1 Change in voltage phase angle at a bus primarily affects the flow of

real power in the transmission lines and leaves the flow of the reactive

power relatively unchanged.

2 Change in the voltage magnitude at a bus primarily affects the flow of

reactive power in the transmission lines and leaves the flow of the real

power relatively unchanged.

The first observation states essentially that the elements of the Jacobian

sub-matrix H are much larger than the elements of sub-matrix M, which we

now consider to be approximately zero.

The second observation means that the elements of sub-matrix L are much

larger than the elements of sub-matrix N which are also considered to be

approximately zero.

Incorporation of these two approximations in equation (82) yields two

separated systems of equations

ΔPΔδH (83)

ΔQV

VΔL (84)

The above two equations are DECOUPLED in the sense that the voltage

phase angle corrections Δδ are calculated using only real power mismatches

ΔP, while voltage magnitude corrections ΔΙVΙ are calculated using only ΔQ

mismatches.

However, the coefficient matrices H and L are still interdependent because

the elements of matrix H depend on voltage magnitudes, being solved in

eqn. (84), whereas the elements of matrix L depend on voltage phase

angles that are computed from eqn. (83). Of course, the two sets of

equations could be solved alternately, using in one set the most recent

solution from the other set.

The power flow method that uses the decoupled equations is known as

DECOUPLED POWER FLOW METHOD. But this scheme would still require

evaluation and factorizing of the coefficient matrices at each iteration. The

order of the two equations to be solved will not be more than N-1. As

compared to Newton Raphson method, wherein the order on equations to

be solved will be about 2 x ( N-1), Decoupled Power Flow method requires

less computational effort.

If the coefficient matrices do not change in every iteration, factorization

need to be done only once and this will result in considerable reduction in

the calculations. To achieve this we introduce further simplifications, which

are justified by the physics of transmission line power flow. This leads to

FAST DECOUPLED POWER FLOW METHOD in which the coefficient

matrices become constant matrices. These matrices are factorized only once.

During different iteration, only mismatch powers are recalculated and the

solution is updated easily.

In a well designed and properly operated power transmission system:

1 The differences )δδ( qp between two physically connected buses of the

power system are usually so small that

)δδ(cos qp )δδ()δδ(sinand;1 qpqp (85)

2 The line susceptances Bpq are many times larger than the line

conductances Gpq so that

)δδ(sinG qppq << )δ(δcosB qppq (86)

3 The reactive power Qp injected into any bus p of the system during

normal operation is much less than the reactive power which would flow

if all lines from that bus were short circuited to reference bus. That is

Qp << pp

2

p BV (87)

The above approximations can be used to simplify the elements of

Jacobian sub-matrices H and L.

The diagonal elements of sub-matrices H and L are given in eqns. (74) and (77).

They now become

pp2

ppppp BVLH (88)

The off-diagonal elements of sub-matrices H and L are given in eqns. (78)

and (83). They now become

pqqppqpq BVVLH (89)

For a 4 bus system having bus 1 as slack bus, buses 2 and 3 as P-Q

buses and bus 4 as P-V bus the linear equations to be solved in N.R. are

shown in eqn. (71). Incorporating the above two equations the decoupled

equations become

24422332222

2 BVVBVVBV 2Δδ 2ΔP

3443332

33223 BVVBVBVV 3Δδ 3ΔP (90)

442

443344224 BVBVVBVV 4Δδ 4ΔP

32 VV

2332222

2 BVVBV 2

2

V

VΔ 2ΔQ

332

33223 BVBVV 3

3

V

VΔ 3ΔQ

Q2

Q3

(91)

δ2 δ3 δ4

P4

P3

P2

=

Equation (90) can be rearranged as

432 δδδ

244233222 BVBVBV 2Δδ 2

2

V

ΔP

344333322 BVBVBV 3Δδ = 3

3

VΔP

(92)

444433422 BVBVBV 4Δδ 4

4

V

ΔP

To make the above coefficient matrix to be independent of bus voltage

magnitude, 32 V,V and 4V are set to 1.0 per unit in the left hand side

expression. Then the above equation becomes

242322 BBB 2Δδ 2

2

V

ΔP

343332 BBB 3Δδ = 3

3

V

ΔP (93)

444342 BBB 4Δδ 4

4

V

ΔP P4

P3

P2

P4

P3

P2

δ2 δ3 δ4

This can be written in a compact form as

V

ΔPΔδB ' (94)

Now equation (91) can be rearranged as

32 VV

2322 BB 2VΔ 2

2

V

ΔQ

3332 BB 3VΔ 3

3

V

ΔQ

This can be written in a compact form as

V

ΔQVΔB" (96)

In a large power network, the bus admittance matrix is symmetrical and

sparse. Separating the real and imaginary parts, it can be written as

BjGY

= (95)

Q2

Q3

The constant matrix B’ is obtained from matrix B

(1) deleting the row and column corresponding to the slack bus and

(2) changing the sign of all the elements.

The constant matrix B” is obtained from matrix B’ by deleting the rows and

columns corresponding to all the P-V buses.

One typical solution strategy is to;

1 Calculate the initial mismatches V/ΔP for all buses except slack bus

2 Solve eqn. (94) for Δδ

3 Update the angles δ and use them to calculate mismatches V/ΔQ for

all P-Q buses

4 Solve eqn. (96) for VΔ and update the magnitudes V and

5 Return to eqn. (94) to repeat the iteration until all mismatches are

within specified tolerances.

The Fast Decoupled Power Flow method uses the constant matrices B’ and

B” that are factorized only once. During different iterations repeat solution

is obtained corresponding to the present mismatch power vectors V

ΔP and

V

ΔQ. Thus tremendous amount of computational simplifications are

achieved in Fast Decoupled Power Flow method and hence it is ideal for

large scale power systems.

Example 5

Consider the power system described in Example 4. Determine the bus

voltages at the end of second iteration, employing Decoupled Power Flow

method.

Solution

Referring to the calculations in Example 4, the decoupled equations with

change in phase angles as variables are

10.55.5

5.516.5

3

2

Δδ

Δδ =

1.2517

1.8184

Solving this 0.08538Δδ2

0.07449Δδ3

Therefore 0)1(2 4.89rad.0.085380.085380δ

0)1(3 4.27rad.0.074490.074490δ

Thus the bus voltages are

03

02

01

4.271.0V

4.891.1V

01.0V

Reactive power at bus 3 is calculated as

[{Q3 5 x 1.0 x 1.0 cos ( 04.27 )} + { 5 x 1.0 x 1.1 cos ( 09.16 )}- ( 10 x 1.0 x 1.0)]

= - 0.4160

0.84140.41601.2574QQIΔQ 333

233

2

3333 1.0(0.4160BVQL x 10 ) = 9.584

Decoupled equation with change in voltage magnitude as variable is

0.8414V

VΔ9.584

3

3

3VΔ = - 0.8414 x 1.0 / 9.584 = - 0.08779

Therefore )(1

3V 1.0 - 0.08779 = 0.9122

At the end of first iteration, bus voltages are

03

02

01

4.270.9122V

4.891.1V

01.0V

Second iteration:

2Q {( 10 x 1.1 x 1.0 cos )4.890 ( 15 x )1.12 + ( 5 x 1.1 x 0.9122 cos )}9.160

= 2.2369

This is within the limits. Bus 2 remains as P-V bus.

2P ( 10 x 1.1 x 1.0 sin 0)4.890 ( 5 x 1.1 x 0.9122 sin )9.16 0 = 1.7364

3P { 5 x 0.9122 x 1.0 sin ( )}4.270 + { 5 x 0.9122 x 1.1 sin ( 0)}9.160

= - 1.1383

0.0821.73641.8184ΔP2 0.11341.13831.2517ΔP3

[{Q3 5 x 0.9122 x 1.0 cos ( )}4.270 + { 5 x 0.9122 x 1.1 cos ( )}9.160

( 10 x )]0.91222 = - 1.1804

(2.2369H22 1.1 x 1.1 x 15 ) = 15.9131

233 0.9122(1.1804H x 10 ) = 9.5015

23H 1.1 x 0.9122 x 5 cos ( )9.160 4.9531

4.9531H32

Decoupled equations with change in phase angles as variables are

9.50154.9531

4.953115.9131

3

2

Δδ

Δδ =

0.1134

0.082

3

2

Δδ

Δδ =

126.6651

1

15.91314.9531

4.95319.5015

0.1134

0.082 =

0.01104

0.001717

0)2(2 4.99rad.0.087100.0017170.08538δ

0)2(3 4.90rad.0.085530.011040.07449δ

Therefore

01 01.0V

02 4.991.1V

03 4.900.9122V

Reactive power at bus 3 is calculated as

3Q [{ 5 x 0.9122 x 1.0 cos ( )}4.900 + { 5 x 0.9122 x 1.1 cos ( )}9.890

( 10 x )]0.91222 = 1.1658

0.09161.16581.2574ΔQ3

233 0.9122(1.1658L x 10 ) = 7.1553

Decoupled equation with change in voltage magnitude as variable is

7.1553 0.0916V

VΔ

3

3

0.012807.1553

0.0916

V

VΔ

3

3

3VΔ 0.01280 x 0.9122 0.01167 0.90050.011670.9122V)2(

3

Thus at the end of second iteration, bus voltages are

01 01.0V

02 4.991.1V

03 4.900.9005V

B =

Example 6

Consider the power system described in Example 4. Determine the bus

voltages at the end of second iteration, employing Fast Decoupled Power

Flow method.

Solution

Susceptance matrix of the power network is

10553

515102

510151

321

The constant matrices are

1053

5152

32

Initial solution is

03

02

01

01.0V

01.1V

01.0V

'B and B” = 10

As in example 4, 1.8184ΔP2 and 1.2517ΔP3

Therefore 1.2517V

ΔP;1.6531

V

ΔP

3

3

2

2

Thus V

ΔPΔδB ' yields

105

515

3

2

Δδ

Δδ =

1.2517

1.6531

On solving the above

3

2

Δδ

Δδ =

125

1

155

510

1.2517

1.6531 =

0.08408

0.08218

0)1(2 4.71rad.0.082180.082180δ

0)1(3 4.82rad.0.084080.084080δ

This gives

03

02

01

4.821.0V

4.711.1V

01.0V

Reactive power at bus 3 is calculated as

3Q [{ 5 x 1.0 x 1.0 cos ( )}4.820 +{ 5 x 1.0 x 1.1 cos ( )}9.530 - ( 10 x 1.0 x 1.0)]

= 0.4064

0.8510.40641.2574QQIΔQ 333

Thus V

ΔQVΔB" yields

10 3VΔ = - 0.851 i.e. 3VΔ = - 0.0851

This gives 0.91490.08511.0V )1(3

At the end of first iteration, bus voltage

03

02

01

4.820.9149V

4.711.1V

01.0V

Second iteration:

2Q {( 10 x 1.1 x 1.0 cos )4.710 ( 15 x )1.12 + ( 5 x 1.1 x 0.9149 cos )}9.530

= 2.2246

This is within the limits. Bus 2 remains as P-V bus.

2P ( 10 x 1.1 x 1.0 sin 0)4.710 ( 5 x 1.1 x 0.9149 sin )9.530 = 1.7363

3P { 5 x 0.9149 x 1.0 sin ( )}4.820 + { 5 x 0.9149 x 1.1 sin ( 0)}9.530

= - 1.2175

0.08211.73631.8184ΔP2

0.03421.21751.2517ΔP3

0.03738V

ΔP;0.07464

V

ΔP

3

3

2

2

Equation with 2Δδ and 3Δδ as variables are

105

515

3

2

Δδ

Δδ =

0.03738

0.07464

On solving this

3

2

Δδ

Δδ =

0.0015

0.004476

0)2(2 4.97rad.0.086660.0044760.08218δ

0)2(3 4.90rad.0.085580.00150.08408δ

This gives

03

02

01

4.900.9149V

4.971.1V

01.0V

Reactive power at bus 3 is calculated as

3Q [{ 5 x 0.9149 x 1.0 cos ( )}4.900 +{ 5 x 0.9149 x 1.1 cos ( )}9.870

( 10 x 20.9149 )]

= 1.1448

0.1231V

ΔQ;0.11261.14481.2574ΔQ

3

33

Thus V

ΔQVΔB" yields

10 3VΔ = - 0.1231 i.e. 3VΔ = - 0.01231

This gives 0.90260.012310.9149V )2(3

At the end of second iteration, bus voltages are

03

02

01

4.900.9026V

4.971.1V

01.0V

Comparison of Newton Raphson, Decoupled and Fast Decoupled power flow methods

Newton Raphson power flow method is accurate and no approximations are

involved. In each iteration equations of the type

J Δx = Δy

are to be solved. The coefficient matrix J keeps changing in every iteration. When

the system size is large, computational effort required is very large. In general

matrix J is not symmetric.

Only the two sub-matrices of Jacobian matrices are involved in the decoupled

power flow solution. Thus in each iteration equations of the type

H Δx = Δy

L Δx’ = Δy’

are to be solved.

The size of matrices H and L are much smaller as compared to matrix J. In each

iteration, matrices H and L are to be recomputed. In general matrices H and L are

not symmetric.

Certain practical approximations are there in developing fast decoupled power

flow solution. In each iteration equations of the type

B’ Δδ = ΔP

B” Δ|V| = ΔQ

are to be solved. The size of B’ and B” are much smaller and they do not vary in

different iterations. Further as they are derived from the bus susceptance matrix,

they are symmetric matrices.

11 FACTORIZATION OF SYMMETRIC MATRIX

For power flow analysis of large scale system, in each iteration, it is required to

solve algebraic equations of the type

Ax = b

where A may be symmetrical or unsymmetrical. Factor matrix method can be

followed to solve the above set of linear algebraic equations.

Factors of symmetric matrix

Let us consider a set of linear algebraic equations involving 4 variables,

such as

44434241

43333231

42322221

41312111

YYYY

YYYY

YYYY

YYYY

4

3

2

1

V

V

V

V

=

4

3

2

1

I

I

I

I

(97)

The coefficient matrix Y can be written as the product of three matrices

tLandDL, .

Then tLDLY (98)

Matrix L is a lower triangular matrix of the form

1

01

001

0001

L

434241

3231

21

(99)

and matrix D is

44

33

22

11

d000

0d00

00d0

000d

D (100)

The matrix Lt is the transpose of matrix L.

Using eqns. (99) and (100) in eqn. (98)

1

01

001

0001

434241

3231

21

44

33

22

11

d000

0d00

00d0

000d

1000

100

10

1

43

4232

413121

=

44434241

43333231

42322221

41312111

YYYY

YYYY

YYYY

YYYY

After multiplying first two matrices

44334322421141

3322321131

221121

11

dddd

0ddd

00dd

000d

1000

100

10

1

43

4232

413121

=

44434241

43333231

42322221

41312111

YYYY

YYYY

YYYY

YYYY

(101)

Considering element 3-3 of the above equation

3333322232311131 Yddd

i.e. 222

32112

313333 ddYd

Generalizing the above, we get

kk

1i

1k

2kiiiii dYd

(102)

Extracting element 4-3 in equation (101)

433343322242311141 Yddd

i.e. 333222423111414343 d/)ddY(

Generalizing this we have

jjkjkk

1j

1kkijiji d/)dY(

(103)

The elements of the factor matrices L and D can be obtained using eqns.

(102) and (103) For matrix of order 4, these elements are calculated in the

order

;,d;,,d;,,,d 433342322241312111 and d44 Generalizing this, for matrix of

order N, elements of factor matrices L and D are calculated in the order

iNi2ii1iii ,,,,d for i = 1,2,….., N (104)

Solving Y V = I using factor matrices

Once the coefficient matrix Y is factored, it can be written as

tLDLY (105)

Hence, the problem is to solve for vector V for a given value of vector I, using

IVLDL t (106)

Let us define two intermediate dummy vectors V’ and V” as

't VVL (107)

and "' VVD (108)

Then IVL " (109)

To solve for the unknown voltage vector V, the above three equations are

to be solved in the order eqn. (109), eqn.(108) and eqn,(107).

For a 4 bus system eqn. (109) is

1

01

001

0001

434241

3231

21

"4

"3

"2

"1

V

V

V

V

=

4

3

2

1

I

I

I

I

(110)

From the above equation

3"

3"

232"

131 IVVV i.e. "232

"1313

"3 VVIV

Generalizing the above we have

1i

1k

"kkii

"i VIV i = 1,2,……..,N (111)

The elements of V” vectors are calculated using the above equation.

Now eqn. (108) can be written as

44

33

22

11

d000

0d00

00d0

000d

'4

'3

'2

'1

V

V

V

V

=

"4

"3

"2

"1

V

V

V

V

(112)

From the above equation, it is clear that

ii"

i'

i d/VV i = 1,2,………,N (113)

The elements of V’ vector are calculated using the above equation.

Finally, equation (107) will be of the form

1000

100

10

1

43

4232

413121

4

3

2

1

V

V

V

V

=

'4

'3

'2

'1

V

V

V

V

(114)

From the above equation

'24423322 VVVV

i.e. 442332'

22 VVVV

Generalizing the above, we get

N

1ikkik

'ii VVV i = N, N-1,…….,1 (115)

It is to be noticed that, while calculating elements of vector V from the

above equation, Vi s are computed in the order i = N, N-1,…….,1

Example 7

Using the factor matrices, solve the equation

20j020j00

020j0020j

20j036j16j0

0016j26.2j10j

020j010j30j

5

4

3

2

1

V

V

V

V

V

=

5

4

3

2

1

I

I

I

I

I

for the bus voltages, when the bus currents are

5

4

3

2

1

I

I

I

I

I

=

1

1

0

0

0

Solution

Calculations are carried out in 7 decimal accuracy and recorded in 4 decimal

accuracy.

Elements of factor matrices are computed from

kk

1i

1k

2kiiiii dYd

and jjkjkk

1j

1kkijiji d/)dY(

1111 Yd = - j 30

112121 d/Y j 10 / ( -j 30 ) = - 0.3333

0d/Y 113131

114141 d/Y = j 20 / ( -j 30) = -0.6667

0d/Y 115151

j22.8667)j30(0.3333)(26.2jdYd 211

2212222

0.6997)j22.8667(/j16d/)dY( 222111313232

0d/)dY(

0.2915)j22.8667(/)0.3333()j30()0.6667(d/)dY(

222111515252

222111414242

j24.8047)j22.8667()0.6997(0j36ddYd 222

23211

2313333

333222423111414343 d/)ddY(

0.1881)j24.8047(/})0.6997()j22.8667()0.2915(00{ 0.8063j24.8047)(/)00j20(d/)ddY( 333222523111515353

332

43222

42112

414444 dddYd

)j24.8047(0.1881)(j22.8667)(0.2915)(j30)(0.6667)(j20 222

= - j 3.8458

444333534222524111515454 d/)dddY(

= { 0 – 0 – 0 - ( - 0.8063 ) ( - j24.8047 ) ( - 0.1881 ) } / ( - j 3.8458 ) = - 0.9780

442

54332

53222

52112

515555 ddddYd

j0.1956

)j3.8458()0.9780()j24.8047()0.8063(00j20 22

Thus the factor matrices are

L =

10.97800.806300

010.18810.29150.6667

0010.69970

00010.3333

00001

D =

j0.19560000

0j3.8458000

00j24.804700

000j28.86670

0000j30

Now the solution can be obtained in three step. Elements of vector "V are

calculated from

1i

1k

"kkii

"i VIV i = 1,2,……..,N

0.0220

)0.9780(0001VVVVIV

10001VVVIV

0000VVIV

000VIV

0IV

"454

"353

"252

"1515

"5

"343

"242

"1414

"4

"232

"1313

"3

"1212

"2

1"1

Elements of vector V’ are calculated from ii"

i'

i d/VV i = 1,2,……..,N

j0.1125)0.1956j(/)0.0220(V

j0.2600)j3.8458(/1V

0VVV

'5

'4

'3

'2

'1

Finally elements of vector V are obtained from

N

1ikkik

'ii VVV i = N, N-1,…….,1

j0.10)j0.15()0.6667(000

VVVVVV

00)j0.1500()0.2915()j0.0625()0.6997(0

VVVVV

j0.0625

)j0.1125()0.8063()j0.1500()0.1881(0VVVV

j0.1500)j0.1125()0.9780(j0.2600VVV

j0.1125VV

551441331221'11

552442332'22

553443'33

554'44

'55

Thus, the bus voltages are

5

4

3

2

1

V

V

V

V

V

=

j0.1125

j0.15

j0.0625

0

j0.1

12 FACTORIZATION OF UNSYMMETRIC MATRIX

Let

A X = b (116)

be the set of linear equations to be solved. In the first phase, the coefficient

matrix A is factorized such that

L U = A (117)

where L is a lower triangular matrix and U is a upper triangular matrix. In the

second phase, the unknown vector X is obtained from

L U X = b (118)

Triangular Factorization

Triangular factorization is illustrated through an example.

Example 8

For the matrix

A =

22.514.28.55.0

10.112.67.44.0

3.93.06.33.0

0.60.40.22.0

obtain the factor matrices L and U. Verify L U = A

Solution

Step 1

Create a matrix M as discussed. Copy the first column of A matrix. Divide elements a12, a13 and a14 by the pivotal element a11 and enter them as m12, m13 and m14. Retain the other elements as they are and write the M matrix as

M =

22.514.28.55.0

10.112.67.44.0

3.93.06.33.0

0.30.20.12.0

Elements in the first column and first row must be retained without any change.

Step 2

Consider the 3 x 3 matrix obtained by canceling the first column and the first row

of matrix M. Modify the elements as

m’jk = mjk – mj1 m1k for j and k = 2,3,4 (119)

Then we get

M =

21.013.28.05.0

8.911.87.04.0

3.02.46.03.0

0.30.20.12.0

Consider the recalculated 3 x 3 matrix. Copy its first column. Divide the elements

m23 and m24 by the pivotal element m22. Retain the other elements as they are

and write matrix M as

M =

21.013.28.05.0

8.911.87.04.0

0.50.46.03.0

0.30.20.12.0

Elements in the first two columns and the first two rows must be retained without

any change.

Step 3

Consider the 2 x 2 matrix obtained by canceling the first two columns and first

two rows of matrix M. Recalculate the elements as

m’jk = mjk – mj2 m2k for j and k = 3,4 (120)

Thus

M =

17.010.08.05.0

5.49.07.04.0

0.50.46.03.0

0.30.20.12.0

Consider the recalculated 2 x 2 matrix. Copy the first column. Divide the element

m34 by the pivotal element m33. Retain the other element as it is and write matrix

M as

M =

1710.08.05.0

0.69.07.04.0

0.50.46.03.0

0.30.20.12.0

Step 4

Consider the single element obtained by canceling the first three columns and

three rows of matrix M. Recalculate its value

m’jk = mjk – mj3 m3k for j and k = 4 (121)

Then

M =

11.010.08.05.0

0.69.07.04.0

0.50.46.03.0

0.30.20.12.0

The lower triangular matrix L consists of all elements in the lower triangular

portion including the diagonal elements. The upper triangular matrix U consists

of all elements in the upper triangular portion with all diagonal elements as one.

Thus the factor matrices are

L =

11.010.08.05.0

09.07.04.0

006.03.0

0002.0

and U =

1.0000

0.61.000

0.50.41.00

0.30.20.11.0

Let us multiply these two matrices

L U =

11.010.08.05.0

09.07.04.0

006.03.0

0002.0

x

1.0000

0.61.000

0.50.41.00

0.30.20.11.0

=

22.514.28.55.0

10.112.67.44.0

3.93.06.33.0

0.60.40.22.0

Thus it can be seen that L U = A

Obtaining the solution

Our primary objective is to solve the equation

A X = b (122)

Once matrix A is factorized, we can write

L U X = b (123)

We need to solve for the vector X for a given vector b. Let us define a vector p

as

U X = p (124)

Substituting the above in eqn. (123), we get

L p = b (125)

Knowing the matrix L and vector b, vector p can be calculated from eqn. (125)

through forward substitution. Once the vector p is known, using eqn.(124) the

unknown vector X can be calculated through backward substitution. This

procedure is illustrated in the following example.

Example 9

Using triangular factorization, solve the equation

22.514.28.55.0

10.112.67.44.0

3.93.06.33.0

0.60.40.22.0

4

3

2

1

x

x

x

x

=

41.0

25.5

16.2

4.8

Solution

As seen in the previous example, factor matrices of the coefficient matrix are:

L =

11.010.08.05.0

09.07.04.0

006.03.0

0002.0

and U =

1.0000

0.61.000

0.50.41.00

0.30.20.11.0

Let us solve L p = b i.e.

11.010.08.05.0

09.07.04.0

006.03.0

0002.0

4

3

2

1

p

p

p

p

=

41.0

25.5

16.2

4.8

p1 = 4.8 / 2.0 = 2.4

p2 = [ 16.2 – ( 3.0 x 2.4 )] / 6.0 = 1.5

p3 = [ 25.5 – ( 4.0 x 2.4 ) – ( 7.0 x 1.5 ) ] / 9.0 = 0.6

p4 = [ 41.0 – ( 5.0 x 2.4 ) – ( 8.0 x 1.5 ) – ( 10.0 x 0.6 ) ] / 11 = 1.0

Now let us solve U X = p i.e.

1.0000

0.61.000

0.50.41.00

0.30.20.11.0

4

3

2

1

x

x

x

x

=

1.0

0.6

1.5

2.4

The calculations are to be carried out in backward direction.

x4 = 1.0

x3 = 0/6 – ( 0.6 x 1.0 ) = 0

x2 = 1.5 – ( 0.4 x 0.0 ) – ( 0.5 x 1.0 ) = 1.0

x1 = 2.4 – ( 0.1 x 1.0 ) – ( 0.2 x 0.0 ) – ( 0.3 x 1.0 ) = 2.0

Thus the solution vector is

X =

1.0

0

1.0

2.0

PROBLEM 1

For the transmission network with the following data, determine the bus admittance matrix.

Sl.

No.

Bus Code

k - m

Line Impedance

kmz HLCA

1 1 - 2 0.02 + j 0.06 j 0.03

2 1 -3 0.08 + j 0.24 ---

3 2 – 3 0.06 + j 0.18 j 0.02

4 2 – 4 0.06 + j 0.18 j 0.02

5 2 – 5 0.04 + j 0.12 ---

6 3 – 4 0.01 + j 0.025 j 0.015

7 4 – 5 0.08 + j 0.24 ---

Answer

j11.8104.224j4.3101.7240j7.52.50

4.310j1.724j43.75817.184j34.48313.793j51.6670

0j34.48313.793j43.19816.71j51.667j3.751.25

j7.52.5j51.667j51.667j32.4310.834j155

00j3.751.25j155j18.726.25

PROBLEM 2

Obtain the bus admittance matrix of the transmission system with the following data.

Line data

Line

No.

Between

buses Line Impedance HLCA

Off nominal

turns ratio

1 1 – 4 0.08 + j 0.37 j 0.007 ---

2 1 – 6 0.123 + j 0.518 j 0.010 ---

3 2 – 3 0.723 + j 1.05 0 ---

4 2 – 5 0.282 + j 0.64 0 ---

5 4 – 3 j 0.133 0 0.909

6 4 – 6 0.097 + j 0.407 j 0.0076 ---

7 6 – 5 j 0.3 0 0.976

Shunt capacitor data

Bus No. 4 Admittance j 0.005

Answer

j7.6341

0.988

j3.4153j2.3209

0.554100

j1.8275

0.4339

j3.4153j4.6418

0.576500

j1.3085

0.57650

j2.3249

0.5541

0j13.9869

1.1124j8.27150

j2.5820

0.5583

00j8.2715j8.1649

0.4449

j0.6461

0.44490

0j1.3085

0.57650

j0.6461

0.4449

j1.9546

1.02140

j1.8275

0.4339

0j2.5820

0.558300

j4.3925

.99920

PROBLEM 3

Consider the power system with the following data:

Line data Bus data

0.15j0.0543

0.30j0.1042

0.45j0.1532

0.30j0.1031

0.15j0.0521

impedanceLinecodeBus

0.10.34

0.51.03

0.20.52

1

QIPINo.Bus

All the buses other than the slack bus are P – Q type. Assuming a flat

voltage start and slack bus voltage as 001.04 , find all the bus voltages

at the end of first Gauss – Seidel iteration.

Answer

0.00923j1.02505V

0.08703j1.02802V

0.04636j1.01909V

0.0j1.04V

(1)4

(1)3

(1)2

1

PROBLEM 4

In the previous problem, let bus 2 be a P – V bus with 2V = 1.04 p.u.

Compute

4322 VandV,V,QI at the end of first Gauss – Seidel iteration. Assume flat

start and 1.0QI0.2 2 p.u.

Answer

0.0149j1.03968V

0.08929j1.03387V

0.03388j1.05129V

0.20799QI

4

3

2

2

PROBLEM 5

Repeat the previous problem taking the limits on reactive powers as

.p.u1.0QI0.25 2

Answer

0.01540j1.04118V

0.08949j1.034473V

0.03278j1.05460V

0.25QI

4

3

2

2

PROBLEM 6

Perform power flow analysis for the power system with the data given

below, using Newton Raphson method, and obtain the bus voltages.

Line data ( p.u. quantities )

0.2j0313

0.2j0322

0.1j0211

impedancesLinebusesBetweenNo.Line

Bus data ( p.u. quantities )

Bus

No Type

Generator Load V minQ maxQ

P Q P Q

1 Slack --- --- 0 0 1.0 0 --- ---

2 P - V 5.3217 --- 0 --- 1.1 --- 0 3.5

3 P - Q 0 0 3.6392 0.5339 --- --- --- ---

Answer

01 01.0V 0

2 151.1V 03 150.9V

PROBLEM 7

It is required to conduct power flow analysis, using Newton Raphson method, on the power system shown.

Here, bus 1 is the slack bus with its voltage 11 δV = 1.0 00 p.u. Obtain the non-

linear equations that are to be solved taking the state variable x 1 representing the voltage

phase angle 2δ and x 2 representing the voltage magnitude 2V .

Answer

4 2x sin 1x = - 0.6 and 4 22x - 4 2x cos 1x = - 0.3

j 0.25 p.u.

p.u.)0.3j0.6(QP 2d2d j 6.0 p.u.

G

1 2

PROBLEM 8

Figure shows the one line diagram of a three bus power system with generators at buses 1

and 3. The magnitude of voltage at bus 1 is adjusted to 1.05 p.u. Voltage magnitude at

bus 3 is fixed at 1.04 p.u. with a real power generation of 200 MW. A load comprising of

400 MW and 250 MVAR is taken from bus 2. Line impedances are marked in p.u. on 100

MVA base. Perform two iterations of power flow solution by the Newton Raphson

method. Give the linear equations to be solved in each iteration and the bus voltages at the

end of second iteration.

.1.04p.uV3

200 MW

0.0125 + j 0.025 0.01 + j 0.03

0.02 + j 0.04

G

G

1 2

3

400 MW

250 MVAR

Slack bus p.u.01.05V 0

1

Answer

2

2

3

2

V

VΔδΔ

δΔ

49.7216.6427.14

16.6466.0433.28

24.8633.2854.28

=

0.22

1.4384

2.86

2

2

3

2

V

VΔδΔ

δΔ

46.824317.402528.5380

14.970765.655932.9811

20.736131.765251.7240

=

0.050132

0.02145

0.099074

1V = 1,05 00 p.u.

2V = 0.97168 02.696 p.u.

3V = 1.04 00.499 p.u.

PROBLEM 9

Using the triangular factorization method solve the linear equations

2002000

0200020

20036160

001626.210

02001030

5

4

3

2

1

V

V

V

V

V

=

1

1

0

0

0

for 4321 V,V,V,V and 5V .

Answer

5

4

3

2

1

V

V

V

V

V

=

0.1125

0.15

0.0625

0

0.1

PROBLEM 10

For the power system described in Problem 8, perform two iterations of power flow solution by decoupled Newton Raphson method.

Answer

1.4384

2.86

δΔ

δΔ

66.0433.28

33.2854.28

3

2 51.22431 2

2

V

VΔ = -1.72431

0.52308

0.67265

δΔ

δΔ

65.607132.9233

31.315950.9990

3

2 45.7404 2

2

V

VΔ = 0.3179

1V = 1,05 00 p.u.

2V = 0.97306 02.574 p.u.

3V = 1.04 00.508 p.u.

PROBLEM 11

For the power system described in Problem 8, perform two iterations of power flow solution by fast decoupled power flow method.

Answer

1.3831

2.86

δΔ

δΔ

6232

3252

3

2 52 2VΔ = -1.79477

0.50128

0.83730

δΔ

δΔ

6232

3252

3

2 52 2VΔ = 0.38785

1V = 1.05 00 p.u.

2V = 0.97295 02.531 p.u.

3V = 1.04 00.492 p.u.

PROBLEM 12

For the power system described in Problem 8, the final bus voltages are

1V = 1,05 00 p.u.

2V = 0.97168 02.696 p.u.

3V = 1.04 00.4988 p.u.

Calculate the power flows and power losses and mark the values in one-line diagram.

Answer

Power flows and power loss in MW or MVAR are marked in the figure shown.

38.867 140.858

218.407

148.079

167.772

(19.693 )

( 9.846 ) 229.009

238.855

101.969

( 16.786 ) 118.755

170.947

21.575 22.124 ( 0.548 )

179.340

39.045

146.187 200 G

G

1 2

3

400

( 0.183 )

( 8.393 ) ))))