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EE 0308 POWER SYSTEM ANALYSIS

Chapter 2

POWER FLOW ANALYSIS



UNIT-II

Primitive network, Formation of Bus admittance

matrix by inspection method and singular method

Bus classification –

Formulation of Power Flowproblems – Power flow solution using Gauss Seidel

method and Newton Raphson method,

Comparison between these methods.

Handling of Voltage controlled buses, Off nominal

transformer ratios and phase shifting transformer.



Primitive network

A power network is essentially an interconnection of several two-terminal

components such as generators, transformers, transmission lines, motors and

loads. Each element has an impedance. The voltage across the element is called

element voltage and the current flowing through the element is called the element

current. A set of components when they are connected form a Primitive network.

A representation of a power system and the corresponding oriented graph are

shown in Fig. 2.1.

Fig. 2.1 A power system and corresponding oriented graph

7

6

5 4

3

21

1 2 3

4

0

24

3

1



1 2 3 4 5 6 7

Connectivity various elements to form the network can be shown by the bus

incidence matrix A. For above system, this matrix is obtained as

A =

-1 1

-1 1 -1 1

-1 -1

-1 1 -1

Fig. 2.1 A power system and corresponding oriented graph

4

3

2

1

(2.1)

7

6

5 4

3

21

1 2 3

4

0



Element voltages are referred as v1, v2, v3, v4, v5, v6 and v7. Element currents are

referred as i1, i2, i3, i4, i5, i6 and i7. In power system network, bus voltages and bus

currents are of more useful. For the above network, the bus voltages are V1, V2, V3

and V4. The bus voltages are always measured with respect to the ground bus.

The bus currents are designated as I1, I2, I3, and I4. The element voltages are

related to bus voltages as:

7

6

5 4

3

2

1

1 2 3

4

0

v1 = - V1

v2 = - V2

v3 = - V4

v4 = V4 – V3

v5 = V2 - V3

v6 = V1 – V2

v7 = V2 – V4



I2I3

I1I4

Expressing the relation in matrix form

7v

v

v

v

v

v

v

6

5

4

3

2

1

=

Thus v = AT Vbus (2.3)

The element currents are related to bus currents as:

-1

-1

-1

-1 1

1 -1

1 -1

1 -1

V1

V2

V3

V4

(2.2)

I1 = - i1 + i6

I2 = - i2 + i5 – i6 + i7

I3 = - i4 – i5

I4 = - i3 + i4 – i7

7

6

5 4

3

21

1

2 3 4

0



Expressing the relation in matrix form

4

3

2

1

I

I

I

I

=

Thus Ibus = A i (2.4)

-1 1

-1 1 -1 1

-1 -1

-1 1 -1

I1 = - i1 + i6

I2 = - i2 + i5 – i6 + i7

I3 = - i4 – i5

I = - i + i – i

i1

i2

i3

i4

i5

i6

i7



The element voltages and element impedances are related as:

7v

v

v

vv

v

v

6

5

4

3

2

1

=

77767574737271

67666564636261

57565554535251

47464544434241

37363534333231

27262524232221

17161514131211

zzzzzzz

zzzzzzz

zzzzzzz

zzzzzzzzzzzzzz

zzzzzzz

zzzzzzz

7i

i

i

ii

i

i

6

5

4

3

2

1

(2.5)

Here zii is the self impedance of element i and zij is the mutual impedance

between elements i and j. In matrix notation the above can be written as

v = z i (2.6)

Here z is known as primitive impedance matrix. The inverse form of above is

i = y v (2.7)

In the above y is called as primitive admittance matrix. Matrices z and y are

inverses of each other.



v = z i (2.6)

i = y v (2.7)

Similar to the above two relations, in terms of bus frame

Vbus = Zbus Ibus (2.8)

Here Vbus is the bus voltage vector, Ibus is the bus current vector and Zbus is the

bus impedance matrix. The inverse form of above is

Ibus = Ybus Vbus (2.9)

Here Ybus is known as bus impedance matrix. Matrices Zbus and Ybus are inversesof each other.



Derivation of bus admittance matrix

It was shown that

v = AT Vbus (2.3)

Ibus = A i (2.4)

i = y v (2.7)

Ibus = Ybus Vbus (2.9)

Substituting eq. (2.7) in eq. (2.4)

Ibus = A y v (2.10)

Substituting eq. (2.4) in the above

Ibus = A y AT Vbus (2.11)

Comparing eqs. (2.9) and (2.11)

Ybus = A y AT (2.12)

This is a very general formula for bus admittance matrix and admits mutual

coupling between elements.



In power system problems mutual couplings will have negligible effect and often

omitted. In that case the primitive impedance matrix z and the primitive

admittance matrix y are diagonal and Ybus can be obtained by inspection. This is

illustrated through the seven-elements network considered earlier. When mutual

couplings are neglected

77

66

55

44

33

22

11

y

yy

y

y

y

y

y (2.13)

Ybus = A y AT

= A

77

66

55

44

33

22

11

y

y

y

y

y

yy

-1-1

-1

-1 1

1 -1

1 -1

1 -1



=-1 1

-1 1 -1 1-1 -1

-1 1 -1

-y11

-y22

-y33

-y44 y44

y55 -y55

y66 -y66

y77 -y77

y11 + y66 - y66 0 0

- y66

y22 + y55 + y66+y77 - y55

- y77

0 - y55 y44 + y55 - y44

0 - y77 - y44 y33 + y44 + y77

Ybus =

1

1

2

2

3

3

4

4



The rules to form the elements of Ybus are:

The diagonal element Yii equals the sum of the admittances directly

connected to bus i.

The off-diagonal element Yij equals the negative of the admittance

connected between buses i and j. If there is no element between buses i

and j, then Yij equals to zero.

y11 + y66 - y66 0 0

- y66 y22 + y55 + y66+y77 - y55 - y77

0 - y55 y44 + y55 - y44

0 - y77 - y44 y33 + y44 + y77 Ybus =

1

1

2

2

3

3

4

4

7

6

5 4

3

21

1 2 3

4

0



Bus admittance matrix can be constructed by adding the elements one by one.

Separating the entries corresponding to the element 5 that is connected between

buses 2 and 3 the above Ybus can be written as

It can be inferred that the effect of adding element 5 between buses 2 and 3 is to

add admittance y55 to elements Ybus(2,2) and Ybus(3,3) and add – y55 to elements

Ybus(2,3) and Ybus(3,2). To construct the bus admittance matrix Ybus, initially all the

elements are set to zero; then network elements are added one by one, each time

four elements of Ybus are modified.

y11 + y66 - y66 0 0

- y66 y22 + y66+y77 0 -y77

0 0 y44 - y44

0 -y77 - y44 y33 + y44 + y77

0 0 0 0

0 y55 - y55 0

0 - y55 y55 0

0 0 0 0

Ybus =

1

1

2

2

3

3

4

4

+

1

1

2

2

3

3

4

4



Example 2.1

Consider the power network shown in Fig. 2.2. The ground bus is marked as 0.

Grounding impedances at buses 1, 2, and 4 are j0.6 Ω, j0.4 Ω and j0.5 Ω

respectively. Impedances of the elements 3-4, 2-3, 1-2 and 2-4 are j0.25 Ω, j0.2 Ω,

0.2 Ω and j0.5 Ω. The mutual impedance between elements 2-3 and 2-4 is j0.1 Ω.

Obtain the bus admittance matrix of the power network.

Fig. 2.2 Power network – Example 2.1

24

3

1

Solution



z = j

1 2 3 4 5 6 7

j 0.5

j 0.2

j 0.2 j 0.25

0.5 j 0.4

j 0.1

1 2

7

5

3

1

2

3

4

5

6

7

Solution

The oriented graph of the network, with impedances marked is shown in Fig. 2.3.

Primitive impedance matrix is:

0.60.4

0.5

0.25

0.2 0.1

0.2

0.1 0.5

Fig. 2.3 Data for Example 2.1

j 0.6

1 2 3

4

0

6

4



y = - j

A =

1 2 3 4 5 6 7

1

2

3

4

5

6

7

Inverting this

Bus incidence matrix A is:

1.66672.5

2.0

4.0

5.5556 -1.1111

5.0

-1.1111 2.2222

-1 1

-1 1 -1 1

-1 -1

-1 1 -1



Ybus = - j A

Ybus = - j

Bus admittance matrix Ybus = A y AT

1.6667

2.52.0

4.0

5.5556 -1.1111

5.0

-1.1111 2.2222

-1

-1

-1

-1 1

1 -1

1 -1

1 -1

-1.6667

- 2.5

- 2.0

- 4.0 4.0

4.4444 - 5.5556 1.1111

5.0 - 5.0

1.1111 1.1111 - 2.2222

-1 1

-1 1 -1 1

-1 -1

-1 1 -1

- j6.6667 j5.0 0 0

j5.0 - j13.0556 j4.4444 j1.1111

0 j4.4444 - j9.5556 j5.1111

0 j1.1111 j5.1111 - j8.2222

Ybus =

1

1

2

2

3

3

4

4

E l 2 2



- j2.0

- j 5.0

- j5.0 - j 4.0

- j 2.0- j 2.51 2

7

5

3

Example 2.2

Neglect the mutual impedance and obtain Ybus for the power network described in

example 2.1.

Solution

Admittances of elements 1 to 7 are

- j1.6667, - j2.5, - j2.0, - j4.0, - j5.0, - j5.0 and – j2.0. They are marked blow.

- j6.6667 j5.0 0 0

j5.0 - j14.5 j5.0 j2.0

0 j5.0 - j9.0 j4.0

0 j2.0 j4.0 - j8.0

Ybus =

1

1

2

3

4

2 3 4

- j 1.6667

1 2 3

4

0

6

4



Example 2.3

Repeat previous example by adding elements one by one.

Solution

Initially all the elements of Ybus are set to zeros.

Add element 1: It is between 0-1 with admittance – j1.6667

Add element 2: It is between 0-2 with admittance – j2.5

- j1.6667 0 0 0

0

0 0

00 0 0 0

0 0 0 0

- j1.6667 0 0 0

0 - j2.5 0 0

0 0 0 0

0 0 0 0

Ybus =

1

1

2

3

4

2 3 4

Ybus =

1

1

2

3

4

2 3 4



Add element 3: It is between 0-4 with admittance – j2


- j1.6667 0 0 0

0 - j2.5 0 0

0 0 0 0

0 0 0 - j2.0

- j1.6667 0 0 00 - j2.5 0 0

0 0 - j4.0 j4.0

0 0 j4.0 - j6.0

Ybus =

1

1

2

3

4

2 3 4

Ybus =

1

1

2

3

4

2 3 4

Add element 5: It is between 2 3 with admittance j5





Add element 7: It is between 2-4 with admittance

–

j2.Final bus admittance matrix

- j1.6667 0 0 0

0 - j7.5 j5.0 0

0 j5.0 - j9.0 j4.0

0 0 j4.0 - j6.0

- j6.6667 j5.0 0 0

j5.0 - j12.5 j5.0 0

0 j5.0 - j9.0 j4.0

0 0 j4.0 - j6.0

- j6.6667 j5.0 0 0

j5.0 - j14.5 j5.0 j2.0

0 j5.0 - j9.0 j4.0

0 j2.0 j4.0 - j8.0

Ybus =

1

1

2

3

2 3 4

4

Ybus =

1

1

2

3

2 3 4

4

Ybus =

1

1

2

3

2 3 4

4



NETWORK REDUCTION

For a four node network, the performance equations in bus frame using the

admittance parameter can be written as:

44434241

34333231

24232221

14131211

Y Y Y Y

Y Y Y Y

Y Y Y Y Y Y Y Y

4

3

2

1

V

V

VV

=

4

3

2

1

I

I

II

Suppose current I4 = 0, the node 4 can be eliminated and the network equations

can be written as:

'

33

'

32

'

31

'

23

'

22

'

21

'

13

'

12

'

11

Y Y Y

Y Y Y

Y Y Y

3

2

1

V

V

V

=

3

2

1

I

I

I

Consider the first set of equations. The equation corresponding to node 4 is

Y41 V1 + Y42 V2 + Y43 V3 + Y44 V4 = 0

Thus, V4 = -44

41

Y

Y V1 -

44

42

Y

Y V2 -

44

43

Y

Y V3



Thus, V4 = -44

41

Y

Y V1 -

44

42

Y

Y V2 -

44

43

Y

Y V3

Substituting the above in the equation of node 1

Y11 V1 + Y12 V2 + Y13 V3 + Y14 (-44

41

Y

Y V1 -

44

42

Y

Y V2 -

44

43

Y

Y V3) = I1 i.e.

( Y11 -44

4114

Y

Y Y) V1 + ( Y12 -

44

4214

Y

Y Y) V2 + ( Y13 -

44

4314

Y

Y Y ) V3 = I1

Comparing the above with the first equation in the second set of equations

Y11’ = Y11 -

44

4114

Y

Y Y

Y12’ = Y12 -

44

4214

Y

Y Y

Y13’ = Y13 -

44

4314

Y

Y Y



Y11’ = Y11 -

44

4114

Y

Y Y

Y12’ = Y12 -

44

4214

Y

Y Y

Y13’ = Y13 -44

4314

Y

Y Y

Thus in general, when node k is eliminated, the modified elements can be

calculated as

Yij’ = Yij -

kk

kjik

Y Y Y

where i = 1, 2, …, N 1 ≠ k and j = 1, 2, …, N 1 ≠ k



Example

Solve the equations

0.83330.50.33330

0.50.750.250

0.33330.251.08330.5

000.50.625

4

3

2

1

V

V

V

V

=

4

0

2

0

for the node voltages using network reduction.

Solution

Eliminating node 1, we get

0.83330.50.3333

0.50.750.25

0.33330.250.6833

4

3

2

V

V

V

=

4

0

2



Eliminating node 1, we get

0.83330.50.3333

0.50.750.25

0.33330.250.6833

4

3

2

V

V

V

=

4

0

2

Eliminating node number 3, we get

0.50.5

0.50.6

4

2

V

V =

4

2

On solving the above, V2 = 60 and V4 = 68



0.83330.50.33330

0.50.750.250

0.33330.251.08330.5

000.50.625

4

3

2

1

V

V

V

V

=

4

0

2

0

The node 1 equation of first set gives

0.625 V1 – 0.5 V2 = 0 Thus V1 = 480.625

60x0.5

The node 3 equation of first set gives

- 0.25 V2 + 0.75 V3 - 0.5 V4 = 0 Thus V3 = 65.33330.75

68x0.560x0.25

Thus

4

3

2

1

V

V

V

V

=

68

65.3333

60

48



Formulation of Power Flow problem

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.



With reasonable assumptions and approximations, a power system may be

modeled as shown in Fig. 2.4 for purpose of steady state analysis.

Fig. 2.4 Typical power system network

Static Capacitor

4

44 jQD PD

2

3

22 jQD PD

G

33 V

33 jQG PG

G

1

5

11 jQG PG

11 jQD PD

G55 jQG PG

a:1



The model consists of a network in which a number of buses 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. 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 to boost or buck the load-bus voltages at times of

need.

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 beequal 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 thk bus, the bus power specification kS is given by

)k

QDk

QG( j)k

PDk

PG(

)k

QD jk

PD()k

QG jk

PG(

kQI j

KPI

kS

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 busesexcept the slack bus.

(2.14)

Cl ifi ti f b



Classification of buses

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

Here PI is the real power injected into the bus

QI is the reactive power injected into the bus

ΙVΙ 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 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 bus classification

based on practical requirements are shown below.

ss QI,PI ss δ,V k k QI,PI k V , k δ mm QI,PI mV , mδ

? ? ? ? ? ?

Fig. 2.5 Three types of buses



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 2Npower 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 sV and s

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 ofsV helps us to fix the voltage level of the

system and the specification ofs

δ 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 δ s 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 the Voltage Controlled Buses

or P – V buses. In 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.



Iterative solution for solving power flow model

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

algebraic equations. 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

yxaxaxayxaxaxa



Specifically

N,1,2,k

xaya

1xgivesThis

xayxaThus

yxaxaxaxa

m

N

km1m

kmk

kk

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 ofmx calculated in h iteration are used. Thus

1k

1m

N

1km

h

mkm

1h

mkmk

kk

1h

k xaxaya

1x (2.15)



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.

First let us consider a power system without any P-V bus. Later, the modification

required to include the P-V busses will be discussed. This means that given the

net power injection at all the load bus, it is required to find the bus voltages at all

the load busses.

The expression for net power injection being Vk Ik*, the equations to be solved are

Vk Ik* = PIk + j QIk for

sk

N,1,2,k

(2.16)



Vk Ik* = PIk + j QIk for

sk

N,1,2,k

In the above equations bus currents Ik are the intermediate variables that are to

be eliminated. Taking conjugate of the above equations yields

Vk* Ik = PIk - j QIk (2.17)

Therefore Ik = *

k

kk

VQI jPI

From the network equations

N

2

1

I

I

I

=

NNN2N1

2N2221

1N1211

Y Y Y

Y Y Y

Y Y Y

N

2

1

V

V

V

(2.19)

(2.16)

(2.18)

we can write



we can write

Ik = Yk1 V1 + Yk2 V2 + ….. + Ykk Vk + …..+ YkN VN

= Ykk Vk +

N

km1m

mkm V Y (2.20)

Combining equations (2.20) and (2.18), we have

sk

N,1,2,k

V Y

Y

V

1

Y

QI jPI

sk

N,1,2,k

V YV

jQIPI

Y

1VThus

m

N

km1m kk

km

*

kkk

kk

N

km1m

mkm*

k

kk

kk

k

*

k

kkm

N

km1m

kmkkkV

QIPIV YV Y

(2.21)



m

N

km1m kk

km

*

kkk

kkk V

Y

Y

V

1

Y

QI jPIV

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

k

kk

kk A Y

QI jPI

(2.22)

km

kk

km B Y Yand (2.23)

Having defined kmk BandA equation (2.21) becomes

skN,1,2,kVBV

AV

N

km 1m

mkm*

k

kk

(2.24)

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

during th1h iteration, can be computed as

skN;........2.1,kfor VBVB

V

AV

N

1km

hmkm

1k

1m

1hmkm*h

k

k1hk

(2.25)

(2.21)

Line flow equations



mk

ykm ykm

Ikm

Line flow equations

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

shown below.

'kmkkmmkkm yVy)VV(ICurrent

(2.26)

Power flow from bus k to bus m is

*

kmkkmkm IVQ jP (2.27)

Substituting equation (2.26) in equation (2.27)

]yVy)VV([VQ jP *'

km

*

k

*

km

*

m

*

kkkmkm (2.28)

kmy

Fig. 2.6 Circuit or line flow calculation



]yVy)VV([VQ jP *'km

*k

*km

*m

*kkkmkm (2.28)

Similarly, power flow from bus m to bus k is

]yVy)VV([VQ jP *'

km

*

m

*

km

*

k

*

mmmkmk (2.29)

The line loss in the transmission line mk is given by

)Q jP()Q jP(P mkmkkmkmmkL (2.30)

Total transmission loss in the system is

linestheallover

ji

jiLL PP (2.31)



Example 2.4

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 0.0204 2 – 4 0.06 + j 0.18 j 0.020

5 2 – 5 0.04 + j 0.12 j 0.0156 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

0 j1.0VVVV

0 j1.06V

(0)5

(0)4

(0)3

(0)2

1

Bus

codek

Generation Load

RemarkMWinPGk 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 10 P – Q bus

STEP 1



For the transmission system, the bus admittance matrix is to be calculated.

Sl.No.

Buscodek - m

Line Impedance

kmz

Line admittanceykm

HLCA

1 1 – 2 0.02 + j 0.06 5 - j 15 j 0.030

2 1 – 3 0.08 + j 0.24 1.25 – j 3.75 j 0.0253 2 – 3 0.06 + j 0.18 1.6667 – j 5 j 0.020

4 2 – 4 0.06 + j 0.18 1.6667 – j 5 j 0.020

5 2 – 5 0.04 + j 0.12 2.5 – j 7.5 j 0.015

6 3 – 4 0.01 + j 0.03 10 – j 30 j 0.010

7 4 – 5 0.08 + j 0.24 1.25 – j 3.75 j 0.025

Y22 = (5 – j15) + (1.6667 – j5) + (1.6667 – j5) + (2.5 – j7.5) + j 0.03 + j 0.02 + j 0.02 + j 0.015

= 10.8334 – j 32.415

Similarly Y33 = 12.9167 – j 38.695 ; Y44 = 12.9167 – j 38.695 ; Y55 = 3.75 – j 11.21

1 2 3 4 5

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

2 -5 +j15 10.8334 – 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.similar aincalculatedbecanBmatrixof elementsOther

0.00036 j0.4626332.415 j10.83345 j5

Y YB

.calculatedareAandA,ASimilarly

0.0037 j0.007432.415 j10.8334

0.2 j0.2

Y

QI jPIA

0.2 j0.2QI jPI

0.2 j0.2)20 j20(100

1QI jPI

Y

YBand

Y

QI jPIA

22

2121

543

22

222

22

22

kk

kmkm

kk

kkk

1



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:

0.00290 j1.03752

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

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

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

0.00370 j0.00740

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.

f f



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

0.00290 j0.03752

)0 j1.0()0.00290 j1.03752(VVVΔ )0(2

)1(22

The accelerated value of the bus voltage is obtained as

0.00406 j1.05253

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

(2.32)



The process is continued for the remaining buses to complete one

iteration. For the next bus 3

0.00921 j1.00690

)0 j1.0()0.00033 j0.77518()0.00406 j1.05253()0.00006 j0.12920(

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

0.00930 j0.00698

VBVBVBV

AV )0(

434

)1(

232131*)0(

3

3)1(

3

The accelerated value can be calculated as

0.01289 j1.00966

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

)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

0.07374 j1.02727V

0.02635 j1.01599V

0.01289 j1.00966V

0.00406 j1.05253V

0.0 j1.06V

)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

)f f (βf f

)ee(ee

h

k

1h

k

h

k

1h

accldk

h

k

1h

k

h

k

1h

accldk

(2.33)



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 hk

1hk f f

< 2ε

sk

N..,1,2,......kfor ε

(2.34)

sk

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

(2.35)

ε



For the problem under study 0.0001εε 21

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

0.10904 j1.01211V

0.09504 j1.01920V

0.08917 j1.02036V

0.05126 j1.04623V

0.0 j1.06V

5

4

3

2

1

COMPUTATION OF LINE FLOWS AND TRANSMISSION LOSS

Line flows can be computed from

)0.086 j0.888(

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

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

]yVy)VV([VQ jP

]yVy)VV([VQ jP*'

12*

1*12

*2

*111212

*'km

*k

*km

*m

*kkkmkm

Similarly



)0.062 j0.874(

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

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

yVy)VV([VQ jP *'12

*2

*12

*1

*222121

Power loss in line 1 – 2 is

)0.024 j0.014(

)Q jP()Q jP(P 2121121221L

Power loss in other lines can be computed as

0.051 j0.0P0.019 j0.0P

0.002 j0.011P

0.029 j0.004P

0.033 j0.004P

0.019 j0.012P

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 )

COMPUTATION OF SLACK BUS POWER



Slack bus power can be determined by summing up the powers flowing

out in the lines connected at the slack bus.

Thus in this case slack bus power is

)0.075 j1.295()0.086 j0.888()0.011 j0.407(

)Q jP()Q jP(Q jP 12121313ss

Thus the power supplied by the slack bus:

Real power = 129.5 MW

Reactive power = 7.5 MVAR (Capacitive )

P13 + j Q13

Ps + j Qs G

31

2

P12 + j Q12

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

QI jPI cannot be calculated. An expression for kQI can be developed

as shown below.

We know *kkkkm

N

1mmkk IVQI jPIandV YI

Denoting haveweδVVandθ Y Y iii ji ji ji

N

1m

mkmkmkmkk

mkmkmk

N

1m

mk

mmkm

N

1mmkkkkk

)θδδ(sin YVVQIThus

θδδ YVV

δθV YδVQI jPI

(2.36)



The value ofkV to be used in equation (2.36) 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 f jeVtaking)

e

f (tanδwhereδVV

Using the adjusted voltage hkV as given in eqn. (2.37), net injected reactive

power hkQI can be computed using eqn. (2.36). As long as hkQI falls within

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

kV and kA can

be computed.

(2.37)



In case if

h

kQI 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 1h

kV 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. 2.7. The extra calculation needed for voltage controlled bus

is shown between X – X and Y – Y.

START

READ LINE DATA & BUS DATA



READ LINE DATA & BUS DATAFORM Y MATRIX

ASSUME sk N k V k ,,2,1)0(

COMPUTE k A FOR P – Q BUSES

COMPUTE km B SET h = 0

SET k = 1 AND 0.0 MAX V

k : s YES

X -------------------------------------------- NO----------------------------------------- NO k : V.C. Bus X

YES

COMPUTEh

k ADJUSTED VOLTAGE,h

k QI

MAX k

h

k QI QI : MIN k

h

k QI QI :

> <

REPLACEh

k QI BY MAX k QI REPLACEh

k QI BY MIN k QI REPLACEh

k V BY ADJh

k V

Y COMPUTE k A Y

-----------------------------------------------------------------------------------------------------------------

COMPUTE ;1hk V COMPUTE

h

k

h

k V V V 1

YES

MAX V V SET V V MAX NO

REPLACEh

k V BY1h

k V

SET k = k + 1

k : N IS MAX V < YES

NOSET h = h + 1

COMPUTE LINE FLOWS, SLACK BUS POWERPRINT THE RESULTS

STOP

>



Representation of off nominal tap setting transformer

A transformer with no tap setting arrangement can be represented similar to ashort 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 OR 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. 2.10.

An equivalent ∏ circuit as shown in Fig. 2.11, will be useful for power flow studies.The elements of the equivalent ∏ 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. 2.10 Representation of off nominal transformer

a:1

q I

pq y

p I pqi

p qt



The parameters of the equivalent ∏ circuit shown in Fig. 2.11 can be derived by

equating the terminal current of the transformer with the corresponding current of

the equivalent ∏ circuit.

It is to be noted that aIianda

EE

p

qt

t

p (2.86)

q

qp

p2

qp

qpq

p

qpqt

qt

p

Ea

y

Ea

y

y)Ea

E(

a

1y)EE(

a

1

a

iI

qqpp

qp

qp

p

qqptqq

EyEa

y

y)a

EE(y)EE(I Also

q I

pq y

p I pqi

p qt

(2.87)

(2.88)



Referring to the ∏ equivalent circuit

qppqpp EAE)BA(BEA)E(EI and

qpqpqq E)CA(EACEA)EE(I

Comparing the coefficients of pE and qE in eqns. (2.87) to (2.90), we get

qp

qp

2

qp

yCA

a

yA

a

y

BA

qp

qp

qp

qp

qp

2

qp

qp

y)a

11(

a

yyC

y)1a

1(

a

1

a

y

a

yB

a

y

AThus

(2.90)

(2.89)

QP

CB

A

IqIP

Fig. 2.11

y

The equivalent ∏ circuit will then be as shown in Fig. 2.12



a

y qp

qpy)1a

1(

a

1 qpy)

a

11(

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

yYy)a

11(YY

a

yYy)1

a

1(

a

1

a

yYY

qp

oldpqnewpq

qp

oldqpnewqp

qpoldqqqpoldqqnewqq

2

qp

oldppqp

qp

oldppnewpp

(2.91)

I qI p

q

Fig. 2.12 Equivalent circuit of off nominal transformer

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

data



data.

Line data

Line

No.

Between

busesLine 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



LineNo.

Betweenbuses

Line admittance HLCA ypq / a ypq / a2

1 1 4 0.5583 – j2.582 j 0.007 --- ---

2 1 6 0.4339 –

j1.8275 j 0.01 ---- ---3 2 3 0.4449 – j 0.6461 0 ---- ----

4 2 5 0.5765 – j 1.3085 0 ---- ----

5 4 3 - j 7.5188 0 - j8.2715 - j 9.09966 4 6 0.5541 – j 2.3249 j 0.0076 ---- ----

7 6 5 - j 3.3333 0 - j 3.4153 - j 3.4992

Y11 = 0.5583 – j 2.582 + j 0.007 + 0.4339 – j 1.8275 + j 0.01 = 0.9922 – j 4.3925

Y22 = 0.4449 – j0.6461 + 0.5765 – j1.3085 = 1.0214 – j 1.9546

Y33 = 0.4449 – j0.6461 – j7.5188 = 0.4449 – j 8.1649

Y44 = 0.5583 –

j 2.582 + j 0.007 –

j 9.0996 + 0.5541 –

j 2.3249 + j 0.0076 + j0.005

= 1.1124 – j 13.9869 Y55 = 0.5765 – j1.3085 – j 3.3333 = 0.5765 – j 4.6418

Y66 = 0.4339 – j 1.8275 + j 0.01+ 0.5541 – j 2.3249 - j 3.4992 = 0.988 – j 7.634



%

% 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 LineImpedance HLCA ONTR'])Edataif Nsh~=0

disp([' Sl.No. At bus Shunt Admittamce'])

Shdataend

%

% 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;

endif Nsh~=0

for i = 1:Nshq = Shdata(i,2);yele = Shdata(i,3);Ybus(q,q) = Ybus(q,q) + yele;

endenddisp([' *** BUS ADMITTANCE MATRIX ***'])Ybus

Answer



j7.6341

0.988

j3.4153 j2.3209

0.554100

j1.8275

0.4339

j3.4153 j4.6418

0.576500

j1.3085

0.57650

j2.3249

0.5541

0 j13.9869

1.1124 j8.27150

j2.5820

0.5583

00 j8.2715 j8.1649

0.4449

j0.6461

0.44490

0 j1.3085

0.57650

j0.6461

.4449

j1.9546

1.02140

j1.8275

0.4339

0 j2.5820

0.558300

j4.3925

.99920



Introduction to Power Flow Analysis by Newton Raphson method

Gauss-Seidel method of solving the power flow has simple problem formulation and hence

easy to explain However, it has poor convergence characteristics. It takes large number of

iterations to converge. Even for the five bus system discussed in Example 2.4, it takes 10

iterations to converge.

Newton Raphson (N.R.) method of solving power flow is based on the Newton Raphson

method of solving a set of non-linear algebraic equations.

N. R. method of solving power flow problem has very good convergence characteristics.

Even for large systems it takes only two to four iterations to converge.

Power flow model of Newton Raphson method



The equations describing the performance of the network in the bus admittance

form is given by

I = Y V (2.38)

In expanded form these equations are

N

2

1

I

I

I

=

NNN2N1

2N2221

1N1211

YYY

YYY

YYY

N

2

1

V

V

V

(2.39)

Typical element of the bus admittance matrix is

jiY = ji

Y jiθ = ji

Y cos jiθ + j ji

Y sin jiθ = jiG + j jiB (2.40)

Voltage at a typical bus i is

iV =iV iδ =

iV ( cos iδ + j sin iδ ) (2.41)

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

n

N

1nni

nNi22i11ii

VY

VYVYVYI

(2.42)

In addition to the linear network equations given by eqn. (2.39), bus power

equations should also be satisfied in the power flow problem These 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 the network at bus i is given by

*

iiii IVQ jP (2.43)

Bus power equations can be obtained from the above two equations (2.42) and (2.43) by

eliminating the intermediate variable iI . From eqn. (2.43)

i

IVQ jP*

iii =

*

iV

n

N

1nni VY

=iV

i

δ

N

1nni

Y ni

θ nV

n

δ

=

N

1n

iV nV niY niθ + nδ - iδ

Separating the real and imaginary parts, we obtain

iP =

N

1n

iV nV niY cos ( niθ + nδ - iδ ) (2.44)

iQ = -

N

1n

iV nV niY sin ( niθ + nδ - iδ ) (2.45)



The real and reactive powers obtained from the above two equations are referred as

calculated powers. During the power flow calculations, their values depend on the latestbus voltages. These calculated powers should be equal to the specified powers. Thus the

non-linear equations to be solved in power flow analysis are

N

1n

iV nV niY cos ( niθ + nδ - iδ ) = iPI (2.46)

-

N

1n

iV nV niY sin ( niθ + nδ - iδ ) = iQI (2.47)

It is to be noted that equation (2.46) can be written for bus i, only if real power injection at

bus i is specified.

Similarly, equation (2.47) can be written for bus i, only if reactive power injection at bus

i is specified.



Of the N total number of buses in the power system, let the number of P-Q buses be

1N , P-V buses be 2N . Then 1NNN 21 . Basic problem is to find the

i) Unknown phase angles δ at the 21 NN number of P-Q & P-V buses and

ii) Unknown voltage magnitudes V at the 1N number of P-Q buses.

Thus total number of unknown variables = 21 NN2

We can write 21 NN real power specification equations (eqn.2.46) and 1N reactive

power specification equations (eqn.2.47).

Thus total number of equations = 21 NN2 .

Therefore Number of equations = Number of variables = 21 NN2

Thus in power flow study, we need to solve the equations



N

1n


for i = 1, 2, …….., N

i ≠ s

and

-

N

1n


for i = 1, 2, …….., N

i ≠ s

i ≠ P – V buses

for the unknown variables δi i = 1,2,…….., N, i ≠ s and

|Vi| i = 1,2,…....., N, i ≠ s , i ≠ P – V buses

The unknown variables are also called as state variables.



Example 2.5

In a 9 bus system, bus 1 is the slack bus, buses 2,5 and 7 are the P-V buses. List the state

variables. Also indicate the specified power injections.

Solution

Buses 3,4,6,8 and 9 are P-Q buses.

9864398765432 VandV,V,V,V,δ,δ,δ,δ,δ,δ,δ,δ are the state variables.

9864398765432 QIandQI,QI,QI,QI,PI,PI,PI,PI,PI,PI,PI,PI are the specified power

injections.

Newton Raphson method of solving a set of non-linear equations



=

Let the non-linear equations to be solved be

1k )x,,x,(xf

n211

2n212 k )x,,x,(xf

(2.50)

nn21n k )x,,x,(xf

Let the initial solution be(0)

n

(0)

2

(0)

1 x,,x,x

If 0)x,,x,(xf k (0)

n

(0)

2

(0)

111

0)x,,x,(xf k (0)

n

(0)

2

(0)

122

0)x,,x,(xf k (0)

n

(0)

2

(0)

1nn

then the solution is reached. Let us say that the solution is not reached.



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[(xf n

(0)

n2

(0)

21

(0)

11

2n

(0)

n2

(0)

21

(0)

12 k )]Δx(x,),Δx(x),Δx[(xf

(2.51)

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

1

1x

f Δx

n

1

2x

f Δx

11n k φΔx



0

0

0 0

0

(0)

n

(0)

2

(0)

11 x,,x,(xf

1

1

x

f )

2

1

1x

f Δx

n

1

2x

f Δx

11n k φΔx

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

(0)

n

(0)

2

(0)

11 x,,x,(xf

1

1

x

f

)

2

1

1 x

f

Δx

n

1

2 x

f

Δx

1n k Δx

(0)

n

(0)

2

(0)

12 x,,x,(xf

1

2

x

f )

2

2

1x

f Δx

n

2

2x

f Δx

2n k Δx

(0)

n

(0)

2

(0)

1n x,,x,(xf

1

n

x

f )

2

n

1x

f Δx

n

n

2x

f Δx

nn k Δx

(2.52)

00

0

0

The matrix form of equations (2.52) is



0 0 0

0 0 0

0 0 0

2.53

)x,,x,x(f k Δxx

f

x

f

x

f (0)

n

(0)

2

(0)

1111

n

1

2

1

1

1

)x,,x,x(f k Δxxf

xf

xf (0)

n(0)2

(0)1222

n

2

2

2

1

2

)x,,x,x(f k Δxx

f

x

f

x

f (0)

n

(0)

2

(0)

1nnn

n

n

2

n

1

n

The above equation can be written in a compact form as

)X(FK ΔX)X(F(0)(0)' (2.54)

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

n

2

1

Δx

Δx

Δx

ΔX

=



In eqn.(2.54) )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. (2.54) for ΔX

ΔX =1)0('

)X(F )X(FK )(0 (2.55)

Then the improved estimate is

ΔXXX)0(1)(

Generalizing this, forth)1h( iteration

ΔXXX)h()1h(

where (2.56)

)X(FK )X(FΔX

)h(1)h('

(2.57)

i.e. ΔX is the solution of

)X(FK ΔX)X(F)h()h(' (2.58)



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(FK ΔX)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 2.6



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

4 2x sin 1x = - 0.6

4 2

2x - 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 1 x ) = - 0.6

- 0.3 – (42

2x - 4 2x cos 1x ) = - 0.3

The error vector

0.3

0.6is not small.

1x

0x

2

1

1x

0x

2

1

It is noted that 1f = 4 2x sin 1x



2f = 4

2

2x - 4 2x cos 1x

Jacobian matrix is: J =

2

2

1

2

2

1

1

1

x

f

x

f

x

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

1

J

=

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.




- 0.6 – (4 2x sin 1x ) = = - 0.047079

- 0.3 – (42

2x - 4 2x cos 1x ) = - 0.064047

The error vector

0.064047

0.04709is not small.


1212

112

xcos4x8xsinx4

xsin4xcosx4

=

3.4449160.552921

0.5977533.658453

0.925x

rad.0.15x

2

1

0.925x

rad.0.15x

2

1

0.925x

rad.0.15x

2

1


xΔ 0 5977533 658453 0 047079 0 016335

- 1



2

1

xΔ

xΔ =

3.4449160.552921

0.5977533.658453

0.064047

0.047079 =

0.021214

0.016335

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.


- 0.6 – (4 2x sin 1x ) = = - 0.001444

- 0.3 – (42

2x - 4 2x cos 1x ) = - 0.002068

Still error vector exceeds the precision index.

0.903786x

rad.0.166335x

2

1

0.903786x

rad.0.166335x

2

1




1212

112

xcos4x8xsinx4

xsin4xcosx4

=

3.2854950.598556

0.6622763.565249


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

0.903786x

rad.0.166335x

2

1

- 1




- 0.6 – (4 2x sin 1

x ) = = - 0.000002

- 0.3 – (42

2x - 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.

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

4 2x sin 1x = - 0.6

42

2x - 4 2x cos 1

x = - 0.3

0.903058x

rad.0.166875x

2

1

0.903058x

rad.0.166875x

2

1

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



This is because the two non linear equations of this example are the power flow model of

the simple system shown in Fig. 2.8.

1V 1δ 2V 2δ

Here bus 1 is the slack bus with its voltage 1V 1δ = 1.0 0

0 p.u. Further 1x represents

the angle 2δ and 2x represents the voltage magnitude2V at bus 2.

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

studies.

Fig. 2.8 Two bus power flow model

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

j 6.0 p.u.

j 0.25 p.u.

G

1 2

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

1n


-

N

1n


Separating the term with in we get

ii

2

i GV +

N

in1n


ii

2

i BV

N

in1n


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



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

PI)Vδ,(P (2.63)

QI)Vδ,(Q (2.64)

On linearization, we get

ΔQ

ΔP

VΔ

Δδ

V

Q

δ

Q

V

P

δ

P

(2.65)

where

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

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



The matrix H N is known as JACOBIAN matrix.



The matrix H N is known as JACOBIAN matrix.

M L

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

H )N(N21

x )N(N21

N )N(N21

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



32432 VVδδδ

2P 3

3

2

2

2

2

4

2

3

2

2

2

VV

P

VV

P

δ

P

δ

P

δ

P

2Δδ 2ΔP

3P 3

3

3

2

2

3

4

3

3

3

2

3 VV

PV

V

P

δ

P

δ

P

δ

P

3Δδ 3ΔP

4P 3

3

4

2

2

4

4

4

3

4

2

4 VVPV

VP

δP

δP

δP

4Δδ = 4ΔP (2.68)

2Q

2

2

δ

δQ

3

2

δ

Q

4

2

δ

Q

2

2

2 VV

Q

3

3

2 VV

Q

2

2

V

VΔ 2ΔQ

3Q

2

3

δ

δQ

3

3

δ

Q

4

3

δ

Q

2

2

3 VV

Q

3

3

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



k

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

Step 7; otherwise proceed to Step 4.

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


ii

2

i BV

N

in1n


i.e. ii PI)V,(δP (2.71)

ii QI)Vδ,(Q (2.72)

The suffix i should take necessary values.

Jacobian matrix is

LMNH where

VV

QLand

δ

QM;V

V

PN;

δ

PH

Here

iP =ii

2

i GV +N

iV nV niY cos ( niθ + nδ - iδ ) (2.73)



i iii

in1n

i n ni( ni n i ) ( )

iQ = ii

2

i BV

N

in 1n

iV nV niY sin ( niθ + nδ - iδ ) (2.74)

Diagonal elements:

i

i

iiδ

PH

=

N

in1n

iV nV niY sin ( niθ + nδ - iδ ) = ii

2

ii BVQ (2.75)

ii

2

ii

i

iii GV2V

V

PN +

N

in1n

iV nV niY cos ( niθ + nδ - iδ )

= iP + ii

2

i GV (2.76)

i

i

ii

δ

QM

=

N

in1n

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

2

i GV (2.77)

i

i

iii V

V

QL

= ii

2

i BV2 -

N

in1n

iV nV niY sin ( niθ + nδ - iδ )

= iQii

2

i BV (2.78)



Summary of formulae



iiH ii

2

ii BVQ

ii

N i

P +ii

2

i

GV

iiM iP ii

2

i GV

iiL iQ ii

2

i BV

jiH ji ji YVV sin ( jiθ + jδ - iδ )

jiN ji ji YVV cos ( jiθ + jδ - iδ )

jiM ji ji YVV cos ( jiθ + jδ - iδ )

jiL ji ji YVV sin ( jiθ + jδ - iδ )

Flow chart for N.R. method of power flow solution is shown below.

(2.85)

START

READ LINE d BUS DATA



k

READ LINE and BUS 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 OFΔQ&ΔP < ε ?

Example 2.7

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

NoType

Generator LoadV 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 --- --- --- ---

0.2 j0313

0.2 j0322

0.1 j0211

impedancesLinebusesBetweenNo.Line

Solution

The bus admittance matrix can be obtained as



The bus admittance matrix can be obtained as

1 2 3

Y =

j10 j5 j5

j5 j15 j10

j5 j10 j15

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

With flat start 0

1 01.0V

001 1V



0

2 01.1V

0

3 01.0V

We know that

iP = ii

2

i GV +

N

in1n

iV nV niY cos ( niθ + nδ - iδ )

iQ = ii2

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.



Since 0,δδδ321

we get 0PP32

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

Mismatch powers are: 1.818401.8184PPIΔP222

1.251701.2517PPIΔP333

0.75740.51.2574QQIΔQ333

333332

333332

232322

332

LMM

NHH

NHH

Vδδ

3

3

3

2

VVΔ

Δδ

Δδ

=

3

3

2

ΔQ

ΔP

ΔP

2P

3P

3Q

Linear equations

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



ii

2

iiii

iiiiii

ii

2

iiii

BVQL

PM;PN

BVQH

)δδ(sinYVVM

)δδ(sinYVVN

)δδ(cosYVVH

i j ji ji ji

i j ji ji ji

i j ji ji ji

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

32H2V 3V )δδ(cosY

2332 = 5.51x5x1x1.1 and 23H = 5.5

32N2V 3V 0)δδ(sinY 2332

32M 3V 2V 0)δδ(sinY 3223

Thus

9 500

010.55.5

05.516.5

3

2

VΔ

Δδ

Δδ

=

0 7574

1.2517

1.8184



9.500

3

3

V

VΔ 0.7574

Solving the above

3

3

3

2

V

VΔ

ΔδΔδ

=

0.0797

0.074490.08538

Therefore

0)1(2 4.89rad.0.085380.085380δ

0)1(

3 4.27rad.0.074490.074490δ

0.92030.07971.0V)1(

3

Thus

0

1 01.0V

0

2 4.891.1V

0

3 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 4.270 ) – ( 5 x 0.9203 x 1.1 sin 9.16

0 ) = -1.1484

3Q = 10 x 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

H 1 1163 + (0 9203 x 0 9203 x 10) = 9 5858



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 9.160 = - 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

VΔ

Δδ

Δδ

=

0.1444

0.1033

0.0749



Its solution is

3

3

3

2

V

VΔ

Δδ

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



Thus at the end of second iteration

0

1 01.0V

0

2 5.001.1V

0

3 4.980.9032V

Continuing in this manner the final solution can be obtained as

01 01.0V

0

2 51.1V

0

3 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.

lighting network

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