Power Flow Analysis for a grid connected system

Power Flow Analysis [Document subtitle]

Rango [Date] [Course title]

Name: Power Flow Analysis Roll No.

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Abstract In this experiment, a power flow analysis of a power grid system has been studied

and simulated using Matlab. After that the reactive power compensation is

performed using STATCOM device. The system behavior under higher load

condition is investigated. Higher load at any buses also increases the transmission

losses. Therefore, the transmission line should be capable of handling maximum

load that can be introduced anytime and a specific time of duration.


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Table of Contents Abstract .................................................................................................................. 1

List of Figures ......................................................................................................... 4

List of tables ........................................................................................................... 4

1.0 Introduction .................................................................................................. 5

1.1 Aims and objectives.................................................................................... 5

2.0 System configuration..................................................................................... 6

3.0 Newton-Raphson Method .............................................................................. 7

3.1 Generator Reactive Limits .......................................................................... 7

3.2 Terminology ............................................................................................... 8

3.3 The Admittance Matrix ............................................................................. 10

3.4 The power flow equations ......................................................................... 13

3.5 Analytic statement of the power flow problem .......................................... 14

3.6 The Newton-Raphson Solution Procedure................................................. 16

3.6.1 Newton Raphson for the Scalar Case: ................................................ 16

3.6.2 A solution x(0) is guessed to the problem f(x)=0 (f(x(0))0). F(x) can be

expanded in a Taylor series, it becomes: ......................................................... 16

3.6.3 Newton-Raphson for the Multidimensional Case: ............................... 17

4.0 System simulation ....................................................................................... 20

4.1 5 buses and 7 transmission lines power grid system ................................ 20

4.1.1 One line diagram code ....................................................................... 20

4.1.2 System admittance calculation .......................................................... 21

4.1.3 Power flowing in and out of each transmission line ............................ 25

4.1.4 Amount of power drawn from the generator ....................................... 25

4.1.5 Power losses in each transmission line .............................................. 26

4.2 5 buses and 7 transmission lines power grid system where Main bus is consuming 50% more active power .................................................................... 26




4.3 5 buses and 7 transmission lines power grid system where Elm bus is

consuming 50% more active power .................................................................... 29




5.0 Reactive Power compensation using STATCOM ........................................... 31

5.1 Compensation Techniques ....................................................................... 33


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5.2 Shunt compensation ................................................................................ 33

5.3 Lake bus connected to the STATCOM and Reactive Compensation........... 34




6.0 Conclusion .................................................................................................. 40

7.0 Reference .................................................................................................... 41

8.0 Appendix ..................................................................................................... 42

8.1 Matlab code ............................................................................................. 42


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List of Figures Figure 1: Complete power grid system for simulation .............................................. 7

Figure 2: Generator Capability Curve and Approximate Reactive Limits

(Approximate Q-limits of each generator using dotted box) ..................................... 8

Figure 3: Single Line Diagram for Simple Power System.......................................... 9

Figure 4: Illustration of (a) positive injection, (b) negative injection, and (c) net

injection .................................................................................................................. 9

Figure 5: Network for Motivating Admittance Matrix ............................................. 11

Figure 6: Bus p Connected to Only Bus q ............................................................. 14

Figure 7: Flow Chart of the NR method ................................................................. 19

Figure 8: Two buses and one transmission line diagram model............................. 20

Figure 9: Matlab code for defining buses ............................................................... 20

Figure 10: Impedance of the one line system ........................................................ 21

Figure 11: System admittance matrix calculation ................................................. 22

Figure 12: Shunt Compensation ........................................................................... 33

Figure 13: Lake bus connected to the STATCOM and Reactive Compensation ...... 34

Figure 14: Modified Jacobian Matirx. .................................................................... 35

List of tables Table 1: Network connectivity and transmission line parameters ............................ 6

Table 2: Specified generation of active power .......................................................... 6

Table 3: Load Power ................................................................................................ 6

Table 4: Types of buses and base powers ................................................................ 6

Table 5: Power generation and consumption comparison for given system ............ 24

Table 6: Power flowing in and out of each transmission line ................................. 25

Table 7: power drawn from the generator .............................................................. 25

Table 8: Transmission line losses .......................................................................... 26

Table 9: Power generation and consumption comparison for given system with 50%

increase in Main bus ............................................................................................ 27

Table 10: Power flowing in and out of each transmission line ............................... 28

Table 11: power drawn from the generator ............................................................ 28

Table 12: Transmission line losses ........................................................................ 28

Table 13: Power generation and consumption comparison for given system with 50%

increase in Elm bus .............................................................................................. 30




Table 17: Power generation and consumption comparison for power grid system

with STATCOM device at Lake bus ........................................................................ 38





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1.0 Introduction Power flow analysis is known as one of the prime factors of power system analysis

and design. It is always required for planning, process, economic scheduling and

exchange of power between utilities. The principal topic of power flow analysis is to

find the magnitude and phase angle of voltage at each bus and the real and

reactive power flowing in each transmission lines. Connecting numerical analysis

applied to a power system is also significant tool. Iterative methods such as

Newton-Raphson, Gauss Seidel Newton-Raphson method are used in this study

due to there are no known analytical methods to solve the problem. For comprise

amply with step depend on the size of system to finish this analysis There are

methods of mathematical calculations. This procedure is demanding to perform

manually takes a lot of times. The purpose of this experiment, Newton-Rap son

method is analyzed and then steady state system has been solved for 5 buses

system where 2 generators and 4 loads are interconnected through 7 transmission

lines.

Power flow problem [1]-[2] is a very renowned problem where voltage magnitudes

and angles for one set of buses are desired In the field of power systems

engineering,, given that voltage magnitudes and power levels for another set of

buses are acknowledged and that a model of the network configuration (unit

commitment and circuit topology) is useful. To solve the power flow problem a

power flow solution procedure is a numerical method that is employed. For solving

the load flow problem unsurprising techniques are iterative, using the Newton-

Raphson or the Gauss-Seidel methods [3]-[4]. A power flow program is a computer

code that apparatus a power flow solution method. The voltages and angles at all

buses is contained by the power flow solution, and from this information, it may

compute the real and reactive generation and load levels at all buses and the real

and reactive flows across all circuits. The word load substituted for power, is

often used with the above terminology i.e., load flow problem, load flow solution

procedure, load flow program, and load flow solution. However, the former

expressions are favored as one normally does not think of load as something that

flows.

1.1 Aims and objectives Aims and objectives of this experiment are as follows:

a. Understanding the power system structure and power flow analysis

technique.

b. Understanding the iterative technique to solve the power flow analysis of a

power grid system. Such system contains generators, loads, buses, and

transmission lines.

c. Understanding Newton-Raphson method to solve the power flow in a given

power grid system that comprises 2 generators, 4 local loads, 5 buses, and 7

transmission lines.

d. Understanding the development of power grid system solution technique

such as Newton-Raphson technique.

e. Write Matlab code to solve the power flow for the given system

f. Simulate the Matlab code and analyze the results.

g. Modify the Matlab code to analyze the ractive power compensation technique

or the evaluation of STATCOM.


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h. Understand the assumption made during power flow analysis.

2.0 System configuration For the power grid system diagram given bus system are as follows:

Table 1: Network connectivity and transmission line parameters

Table 2: Specified generation of active power

Table 3: Load Power

Table 4: Types of buses and base powers

Complete system is drawn according to the information given as follows:


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Figure 1: Complete power grid system for simulation

3.0 Newton-Raphson Method

3.1 Generator Reactive Limits The maximum reactive power the maximum reactive power capability resembles to

that the generator may produce when operating with a lagging power factor. The

maximum reactive power the generator may absorb corresponds to the minimum

reactive power capability to when operating with a leading power aspect. A

function of the real power output of the generator is these limitations that is, the

reactive power limitations move closer to zero as the real power increases.


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Pmax

Qmax

Qmin

P

Q

leading

operation

lagging

operation

Figure 2: Generator Capability Curve and Approximate Reactive Limits (Approximate Q-limits of each generator using dotted box)

3.2 Terminology It is suitable to typify power system networks using the pretended one-line diagram,

which can be considered of as the circuit diagram of the per-phase equivalent, but

without the neutral conductor. The one-line diagram of a small transmission

system is illustrated in figure 3.

An injection is the power, that is being injected into or withdrawn from a bus by an

element having its other terminal (in the per-phase equivalent circuit) connected to

ground is either real or reactive. It would be either a generator or a load. It can be

thought as a positive injection as one where power is flowing from the element into

the bus (i.e., into the network); when power is flowing from the bus (i.e., from the

network) into the element is regarded as a negative injection. Figure 4 (a), (b), and

(c) demonstrates positive power injection, negative power injection, and net power

(sum of all injections).


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Figure 3: Single Line Diagram for Simple Power System

Pk=100 Qk=30

(a)

Pk= - 40 Qk= -20

(b)

Pk=100+(-40)=60 Qk=30+(-20)=10

(c)

Figure 4: Illustration of (a) positive injection, (b) negative injection, and (c) net injection

An arbitrary bus numbered k is considered. The four variables are real and

reactive power injection, Pk and Qk, , and voltage magnitude and angle, |Vk | and

k, are the four variable which are real and reactive.

PV Buses: it is acknowledged as Pk and |Vk | but not Qk or k SD For type

PV buses. Because of the capability to state (and therefore to know) the

voltage magnitude of this bus these buses fall under the category of voltage

controlled buses. Most generator buses fall into this group, enlightened of

whether it also has load; exceptions are buses that have reactive power

inoculation at either the generators upper limit (Qmax) or its lower limit


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(Qmin), and (2) the system swing bus (it has been described the swing bus

below).

PQ Buses: It is known that Pk and Qk but not |Vk | or k for type PQ buses.

All load buses fall into this class, as well as buses that have not either load

or generation.

The swing bus is referred as the third type of bus. There are two other communal

terms for this bus are slack bus and reference bus. There is only one swing bus,

and it can be labelled by the engineer to be any generator bus in the system. For

the swing bus, it is known as |V| and. The voltage angle may be designated to be

any angle, but normally it is designated as 0o.

3.3 The Admittance Matrix A network characterized in a hybrid fashion using one-line diagram illustration for

the nodes (buses 1-5) and circuit demonstration for the branches connecting the

nodes and the branches to ground Is shown in figure 5. The nodes denote lines are

associated by the branches. The branches to ground signify any shunt elements at

the buses, including the charging capacitance at either end of the line. All branches

are denoted with their admittance values Yij for a branch connecting bus i to bus j

and Yi for a shunt element at bus i. Ii indicates the current injections at each bus i.


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North (1)

South (2)

Lake (5)

Main (4)

Elm (3)

40

MW

20

MW

10M

Var

60

MW

10

MV

ar

40MW5MVar

45MW15MVar

10

0MW

Y1

Y12

Y2

Y15

Y24

Y45

Y34

Y23

Y5

Y4

Y3

Figure 5: Network for Motivating Admittance Matrix

The current injected into bus 1 may be written as (Kirchoffs Current Law (KCL)):

I1=(V1-V2)Y12 + (V1-V5)Y15 + V1Y1 (1)

Bus 1 is connected to bus 3, and 4 through an infinite impedances considered to

be completed, which implies that the resultant admittance Y13, and Y14 is zero.

Then, it has:

I1=(V1-V2)Y12 + (V1-V3)Y13 + (V1-V4)Y14+ (V1-V5)Y15+ V1Y1 (2)

Rearranging eq. 2, it has:

I1= V1(Y1 + Y12 + Y13 + Y14+ Y15) + V2(-Y12)+ V3(-Y13) + V4(-Y14) + V5(-Y15)(3)

likewise, it may be developed the current injections at buses 2, 3, 4 and 5 as:


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I2= V1(-Y21) + V2(Y2 + Y21 + Y23 + Y24+ Y25) + V3(-Y23) + V4(-Y24)+ V5(-Y25) (4)

I3= V1(-Y31)+ V2(-Y32) + V3(Y3 + Y31 + Y32 + Y34+ Y35) + V4(-Y34)+ V5(-Y35) I4= V1(-Y41)+ V2(-Y42) + V3(-Y43)+ V4(Y4 + Y41 + Y42 + Y43+ Y45)+ V5(-Y45) I5= V1(-Y51)+ V2(-Y52) + V3(-Y53)+V4(-Y54)+V5(Y5 + Y51 + Y52 + Y53+ Y54)

That the admittance of the circuit from bus k to bus i is the same as the

admittance from bus i to bus k can be distinguished, i.e., Yki=Yik From eqs. (3) and

(4),

1 1 12 13 14 15 12 13 14 15

2 21 2 21 23 24 25 23 24 25

3 31 32 3 31 32 34 35 34 35

4 41 42 43 4 41 42 43 45 45

5 51 52 53 54 5 51 52 53 54

I Y Y Y Y Y Y Y Y Y

I Y Y Y Y Y Y Y Y Y

I Y Y Y Y Y Y Y Y Y

I Y Y Y Y Y Y Y Y Y

I Y Y Y Y Y Y Y Y Y

1

2

3

4

5

V

V

V

V

V

(5) The matrix containing the network admittances in eq. (5) is the admittance matrix,

also recognized as the Y-bus, and denoted as:

1 12 13 14 15 12 13 14 15

21 2 21 23 24 25 23 24 25

31 32 3 31 32 34 35 34 35

41 42 43 4 41 42 43 45 45

51 52 53 54 5 51 52 53 54

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

(6)

Eq.(6) can be rewritten by Denoting the element in row i, column j, as Yij, as:

11 12 13 14 15

21 22 23 24 25

31 32 33 34 35

41 42 43 44 45

51 52 53 54 55

Y Y Y Y Y

Y Y Y Y Y

Y Y Y Y Y Y

Y Y Y Y Y

Y Y Y Y Y

(7)

Where the terms ijY are not admittances but rather elements of the admittance

matrix. Therefore, eq. (6) becomes:

1 11 12 13 14 15 1

2 21 22 23 24 25 2

3 31 32 33 34 35 3

4 41 42 43 44 45 4

5 51 52 53 54 55 5

I Y Y Y Y Y V

I Y Y Y Y Y V

I Y Y Y Y Y V

I Y Y Y Y Y V

I Y Y Y Y Y V

(8)

By using eq. (7) and (8), and eq (8) can be rewritten by defining the vectors V and I,

in compact form according to:


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1 1

2 2

3 3

4 4

5 5

,

V I

V I

V V I I

V I

V I

VYI (9)

3.4 The power flow equations Sk might expressed by Drawing on the acquainted relation for complex power, :

Sk=VkIk* (10)

From eq. (8),

j

N

j

kjk VYI

1

(11)

Substitution of eq. (11) into eq. (10) yields:

*

1

*

*

1

j

N

j

kjkj

N

j

kjkk VYVVYVS

(12)

Since Vk=|Vk|k and Ykj=Gkj+jBkj , eqn (12) can be written as ;

N

j

kjkjjkjk

N

j

kjkjjjkk

jj

N

j

kjkjkkjj

N

j

kjkjkkj

N

j

kjkk

jBGVVjBGVV

VjBGVVjBGVVYVS

11

1

*

1

**

1

*

)()()(

)()(

(13)

Since V=|V|=|V|{cos+jsin}, eq. (13) can be rewritten according to Euler

relation

N

j

kjkjjkjkjk

N

j

kjkjjkjkk

jBGjVV

jBGVVS

1

1

)()sin()cos(

)()(

(14)

Equating real and imaginary part:

N

j

jkkjjkkjjkk

N

j

jkkjjkkjjkk

BGVVQ

BGVVP

1

1

)cos()sin(

)sin()cos(

(15)

These two equations of (15) are recognized as the power flow equations.


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It seems amazing to consider the case of eq. (15) if bus k, relabeled as bus p, is

only joined to one other bus, named bus q. Then the bus p injection is the same as

the flow into the lineup. The situation is illustrated in figure 6.

Bus q Bus p Series

admittance G-jB

Bus injection Pp and Qp Line flow

Ppq and Qpq

Figure 6: Bus p Connected to Only Bus q

Fig. 6, eqn. (15) Become:

)cos()sin(

)sin()cos(

2

2

qppqqpqppqqppppp

qppqqpqppqqppppp

BVVGVVBVQ

BVVGVVGVP

(16)

3.5 Analytic statement of the power flow problem A power system network is regarded as having N buses, NG of which is voltage-

regulating generators. The swing bus must be one of these. Thus there are NG-1

type PV buses, and N-NG type PQ buses. It is unspecified that the swing bus is

numbered bus 1, the type PV buses are numbered 2,, NG, and the type PQ buses

are numbered NG+1,,N. The assumptions are following:

1. The admittances of all series and shunt elements

2. The voltage magnitudes Vk, k=1,,NG, at all NG generator buses,

3. The real power injection of all buses except the swing bus, Pk, k=2,,N

4. The reactive power injection of all type PQ buses, Qk, k=NG+1, , N

5. The relationship holds (N-1) + (N-NG) =2N-NG-1

The power flow equations are:

NNkBGVVQ

NkBGVVP

G

N

j

jkkjjkkjjkk

N

j

jkkjjkkjjkk

,...,1 , )cos()sin(

,...,2 , )sin()cos(

1

1

(17)

It is the following information about the network which is needed to be found:

a. The angles for the voltage phasors at all buses except the swing bus (it is 0

at the swing bus), i.e., k, k=2,,N

b. The magnitudes for the voltage phasors at all type PQ buses, i.e., |Vk|,

k=NG+1, , N

Is the angle of unknown vector (an underline beneath the variable means it is a

vector or a matrix) and the vector of unknown voltage magnitudes is |V|.


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||V

||V

||V

|V |

N

N

N

N

G

G

2

1

3

2

,

(18)

Vector x is defined as the composite vector of unknown angles and voltage

magnitudes.

12

1

1

2

1

2

1

3

2

x

G

G

G

NN

N

N

N

N

N

N

N

x

x

x

x

x

x

||V

||V

||V

|V|

(19)

Similarly eqn. (17) can be rewritten as

,...,NNkxQQ

,...,NkxPP

Gkk

kk

1 , )(

2 , )(

(20)

In eqn. (20), the specified injections are Pk and Qk (known constants) while the

right-hand sides are functions of the elements in the unknown vector x. Bringing

the left-hand side over to the right-hand side, it can be written that

,...,NNkQxQ

,...,NkPxP

Gkk

kk

1 , 0)(

2 , 0)(

(21)

Now a vector-valued function f(x) can be defined as:

0

0

0

0

0

)(

)(

)(

)(

)(

)(

)(

)(

)(

1

2

11

22

12

1

1

N

N

N

NN

NN

NN

NN

N

N

Q

Q

P

P

QxQ

QxQ

PxP

PxP

xf

xf

xf

xf

xf

GGG

G

(22)

Equation (22) executes in the form of f(x) =0, where f(x) is a vector-valued function

and 0 is a vector of zeros.


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3.6 The Newton-Raphson Solution Procedure Two basic methods can be used to solve the power flow problem: Gauss-Siedel (GS)

and Newton-Raphson (NR) [3]. These methods are both iterative root finding

schemes.

The GS and NR methods are often classified as root finding schemes because they

are geared towards solving equations like f(x) =0 (or f(x)=0). The solution to such an

equation, call it x* (or x*), is clearly a root of the function f(x) (or f(x)).

3.6.1 Newton Raphson for the Scalar Case:

3.6.2 A solution x(0) is guessed to the problem f(x)=0 (f(x(0))0). F(x) can be expanded in a

Taylor series, it becomes:

0...))((''2

1)(')()( 2)0()0()0()0()0()0()0( xxfxxfxfxxf (23)

It is reasonable to approximate eq. (23) as

0)(')()( )0()0()0()0()0( xxfxfxxf (24)

,

, it can be written Taking f(x(0)) to the right hand side,

)()(' )0()0()0( xfxxf (25)

eq. (25) might be easily solved for x(0) according to:

)()(' )0(1)0()0( xfxfx (26)

Because f (x(0)) in eq. (26) is scalar, using simple division its inverse is very easily

evaluated so that:

)('

)()0(

)0()0(

xf

xfx

(27)

The basis for the update formula to be used in the first iteration of the scalar NR

method is provided in equation (26). This update formula is:

)('

)()0(

)0()0()0()0()1(

xf

xfxxxx

(28)

And from eq. (26), the update formula might be interfered for any particular

iteration as:


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)('

)()(

)()()()()1(

j

jjjjj

xf

xfxxxx

(29)

3.6.3 Newton-Raphson for the Multidimensional Case: It is assumed that there are n nonlinear algebraic equations and n unknowns

characterized by f(x) =0, and that a solution x(0) is guessd. Then f(x(0))0 because x(0)

is just a supposition. But there must be some x(0) which will make f(x(0) + x(0))=0.

Again, the function f(x) is expanded in a Taylor series, as follows:

0...))((''2

1)(')()( 2

)0()0(

1

)0()0(

1

)0(

1

)0()0(

1 xxfxxfxfxxf

0...))((''2

1)(')()( 2

)0()0(

2

)0()0(

2

)0(

2

)0()0(

2 xxfxxfxfxxf (32)

0...))((''2

1)(')()( 2

)0()0()0()0()0()0()0( xxfxxfxfxxf nnnn

Equations (30) may be written more neatly as

0...))((''2

1)(')()( 2

)0()0()0()0()0()0()0( xxfxxfxfxxf (31)

Assuming the guess is a good one such that x(0) is small, then the higher order

terms are also small and it can be printed

0)(')()()0()0()0()0()0( xxfxfxxf (32)

A derivative can be computed for each individual function with respect to each

individual unknown as there are n functions and n variables, like fk(x)/xj, which

give the derivative of the kth function with respect to the jth unknown. Thus, it is

suitable to store all of these derivatives in a matrix. This matrix has become quite

renowned as the Jacobian matrix, and it is often denoted using the letter J.

The Jacobian matrix can be written as:

n

nnn

n

n

x

xf

x

xf

x

xf

x

xf

x

xf

x

xf

x

xf

x

xf

x

xf

J

)()()(

)()()(

)()()(

)0(

2

)0(

1

)0(

)0(

2

2

)0(

2

1

)0(

2

)0(

1

2

)0(

1

1

)0(

1

(33)

it is found by taking f(x(0)) to the right hand side In eq. (32),

)()(')0()0()0(

xfxxf (34)


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or,it can be written in terms of the Jacobian matrix J,

)()0()0(

xfxJ (35)

Solving eq. (35) it is found for x(0),

)()()(' )0(1)0(1)0()0( xfJxfxfx (36)

Equation (36) gives the basis for the update formula to be used in the first iteration

of the multi-dimensional case. This update formula is:

)()0(1)0()0()0()1(

xfJxxxx

(37)

And from eq. (37), the inform formula might be interfered for any exacting iteration

as:

)()(1)()()()1( iiiii

xfJxxxx

(38)

To see how to avoid matrix inversion the update formula can be stated differently.

To do this, it become eq. (38) as

)()()1( iii

xxx

(39)

Where x(i) is found from

)()()( ii

xfxJ (40a)

Equation (40a) is a very simple relation. Observing that J is just a constant nn

matrix, x(i) is an n1 vector of unknowns, and f(x(i)) is an n1 vector of knowns, it

is seen that eq. (40a) is just the linear matrix equation

Az=b (40b)

(40b) can be solved in various ways c. These notes will later cover this topic. First,

however, for a multi-dimensional case Newton-Raphson procedure must be

illustrated. As much as simple multi-dimensional case must be applied, a two-

variable problem.

The stopping criteria

Type 1 stopping criterion: the maximum change must be experienced in the

solution elements from one iteration to the next, and if this maximum change is

smaller than a certain predefined tolerance, then stop. This means to compare

the maximum complete value of elements in x against a small number, call it

1.


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Type 2 stopping criterion: The maximum value must be tested in the function

elements of the most current iteration f(x), and it is stopped if this maximum

value of elements in f(x) is smaller than a certain predefined tolerance. This

means to compare the maximum absolute value of elements in f(x) against a

small number, call it 1. For power flow solutions this is the most common

stopping criterion, and the value of each element in the function is referred to

as the power mismatch for the bus corresponding to the function. It is tested

for both real and reactive power mismatches for type PQ. Real power

mismatches is only tested for type PV buses,.

Figure 7: Flow Chart of the NR method


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4.0 System simulation

4.1 5 buses and 7 transmission lines power grid system

4.1.1 One line diagram code

R

iB/2

jB/2

jXSend

ing

Bu

s

Re

ceiv

ing

Bu

s

Figure 8: Two buses and one transmission line diagram model

Figure 7 shows the system where two buses and one transmission line are

considered. Consider the sending bus is 1 and the receiving bus is 2. For

configuring the buses, the bus 1 is considered as the PV bus and the bus 2 is

considered as the PQ buses.

Figure 9: Matlab code for defining buses

Since there is no generator and load in the system, the active power and reactive

power is zero for the buses.

For impedance the transmission line is modeled as -model. Therefore, the total

susceptance is divided by two and assigned to each end. The admittance matrix for

the system is following:


21 | P a g e

Figure 10: Impedance of the one line system

4.1.2 System admittance calculation According to the eqn. 6 the system admittance matrix is following (details analysis

is performed in section 3.3):

1 12 13 14 15 12 13 14 15

21 2 21 23 24 25 23 24 25

31 32 3 31 32 34 35 34 35

41 42 43 4 41 42 43 45 45

51 52 53 54 5 51 52 53 54

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y Y Y Y

If there is no direct transmission line between two buses the admittance between

these buses is zero. The admittance that exists between the bus and the ground is

the sum of the half of the admittance of all transmission line connected to that bus

(-model). Since the impedance of each transmission line is given, the inverse of

impedance will be the admittance between these two lines. According to the above

information Matlab code is written to find the system admittance matrix.


22 | P a g e

Figure 11: System admittance matrix calculation

At first each admittance is defined and then finally using eqn.6 the system

admittance is calculated.

For the power flow analysis the Matlab code is written according to the lab

instruction [6]. The complete code is attached in appendix.

Each bus Voltages and angles are as follows:


23 | P a g e

Algotithm is converged after 10 iterations. The Slack bus (North bus) voltage is 1.06

p.u and angle is zero as specified. It will be unchanged. The PV bus (South) bus

voltage is 1 p.u as specified at first and it will be unchanged. The angle is -2.06123

p.u. The voltage of the other buses is lower than one as these buses are consuming

the power. Due to the losses through the transmission line(s), other buses voltage

is lower than one.

For transmission losses calculation, Matlab code is written. The Transmission

losses summary is as follows:


24 | P a g e

0.0

20

.06

j

0.06/2j

0.06/2j

North (1)

South(2)

0.08

0.24j Lake (5)

0.0

5/2

j

0.05

/2j

0.0

60

.18

j

0.04/2j

0.04/2j

0.06 0.18j

0.0

4/2

j

0.0

4/2

j

Main (4)

0.0

1

0.02/2j

0.02/2j

0.03

j

Elm (3)

0.0

8

0.05/2j

0.05/2j

0.24

j

0.04

0.0

3/2

j

0.0

3/2

j

0.12j

40

MW

20

MW

10

MV

ar

60M

W1

0MV

ar

40MW5MVar

45MW15MVar

13

1MW

2.4

9M

W+

1.0

9M

var

1.52MW-0.69Mvar

0.3

6M

W-2

.87

Mva

r

0.46MW-2.55Mvar

1.22MW+0.73Mvar

0.0

4M

W-1

.82

Mva

r

0.0

4M

W-4

.65

Mva

r

To verify the equalization of the total active power generation and consumption, the

power generated by Slack bus and PV bus are added, similarly, power consumption

and transmission losses are summed up. If these two sum is equal the equalization

is hold.

Table 5: Power generation and consumption comparison for given system

Power Generation Value (MW) Power Consumtion

Value (MW)


25 | P a g e

Slack bus to South (1->2) 89.33138 South bus load 20

Slack bus to Lake (1->5) 41.79085 Lake bus load 45

PV bus (South bus) generator 40 Main bus load 40

Elm bus load 60

Transmission loss (1->2) 2.48587







Total Generation 171.12223 Total Consumption 171.12224

Since total generation and total consumption of active power is similar, the Matlab

Coding is competent.

4.1.3 Power flowing in and out of each transmission line For calculating the power flowing in and out of each transmission line, the power

flow output is considered. For example, 1-> 2 is the power in to the transmission

line between and opposite sign of 2->1 is the power out of that transmission line.

Table 6: Power flowing in and out of each transmission line

Bus no

Transmission line

Power in Power Out

Active Power (MW)

Reactive power (MVAR)

Active Power (MW)


1 1 -> 2 89.33138 73.99518 86.84551 72.60839

2 1 -> 5 41.79085 16.82034 40.27302 17.5125

3 2 -> 5 24.47266 -2.51849 24.11315 0.3523

4 2 -> 4 27.713 -1.72391 27.25215 0.83056

5 2 -> 3 54.65985 5.55794 53.44485 4.82921

6 5 -> 4 19.38618 2.8648 19.34611 4.68775

7 4 -> 3 6.59825 0.51832 6.55515 5.17079

4.1.4 Amount of power drawn from the generator For calculating the amount of power drawn from the generator is calculating by

summing the total active power from the bus it is attached. A generator is attached

at Slack bus and its active power is varied. A constant 40 MW generator is attached

at PV bus.

Table 7: power drawn from the generator

Power Generation Value (MW) Total (MW)


26 | P a g e

Slack bus to South (1->2) 89.33138 131.12223

Slack bus to Lake (1->5) 41.79085

PV bus (South bus) generator 40 40

Total Generation 171.12223

4.1.5 Power losses in each transmission line The difference between power in and out of a transmission line is the power losses

in each transmission line.

Table 8: Transmission line losses

Bus no

Transmission line

Power in Power Out Transmission Loss

Active Power (MW)


Active Power (MW)


Active Power (MW)


1 1 -> 2 89.33138 73.99518 86.84551 72.60839 2.48587 1.38679

2 1 -> 5 41.79085 16.82034 40.27302 17.5125 1.51783 -0.69216

3 2 -> 5 24.47266 -2.51849 24.11315 0.3523 0.35951 -2.87079

4 2 -> 4 27.713 -1.72391 27.25215 0.83056 0.46085 -2.55447

5 2 -> 3 54.65985 5.55794 53.44485 4.82921 1.215 0.72873

6 5 -> 4 19.38618 2.8648 19.34611 4.68775 0.04007 -1.82295

7 4 -> 3 6.59825 0.51832 6.55515 5.17079 0.0431 -4.65247

4.2 5 buses and 7 transmission lines power grid system where Main bus is

consuming 50% more active power Each bus Voltages and angles are as follows:







27 | P a g e

is lower than one. The load at Main bus is increased. To support this extra load,

the current flow through the transmission line that feeds the Main bus increases.

As a result the voltage at the Main bus is lower than previous case.



To support this extra load, the current flow through the transmission line that

feeds the Main bus increases. As a result the transmission line loss for the line that

deliver power to the Main bus in increased.

Table 9: Power generation and consumption comparison for given system with 50% increase in Main bus


Value (MW)




Elm bus load 60












28 | P a g e





Bus no Transmission line Power in Power Out

Active Power Reactive power Active Power Reactive power

1 1 -> 2 103.66076 69.85414 100.79364 67.6236

2 1 -> 5 48.85806 16.99562 46.87918 16.28348

3 2 -> 5 28.91445 -1.41174 28.41262 1.01514

4 2 -> 4 34.01442 -0.58883 33.31904 1.24157

5 2 -> 3 57.86477 6.30474 56.50107 5.1243

6 5 -> 4 30.2918 2.29862 30.19572 3.93487

7 4 -> 3 3.51476 0.17645 3.49893 4.8757




at PV bus.










Bus no

Transmission line


Active Power (MW)


Active Power (MW)


Active Power (MW)


1 1 -> 2 103.66076 69.85414 100.79364 67.6236 2.86712 2.23054


29 | P a g e

2 1 -> 5 48.85806 16.99562 46.87918 16.28348 1.97888 0.71214

3 2 -> 5 28.91445 -1.41174 28.41262 1.01514 0.50183 -2.42688

4 2 -> 4 34.01442 -0.58883 33.31904 1.24157 0.69538 -1.8304

5 2 -> 3 57.86477 6.30474 56.50107 5.1243 1.3637 1.18044

6 5 -> 4 30.2918 2.29862 30.19572 3.93487 0.09608 -1.63625

7 4 -> 3 3.51476 0.17645 3.49893 4.8757 0.01583 -4.69925

4.3 5 buses and 7 transmission lines power grid system where Elm bus is consuming

50% more active power Each bus Voltages and angles are as follows:






is lower than one. The load at Main bus is increased. To support this extra load,

the current flow through the transmission line that feeds the Elm bus increases. As

a result the voltage at the Elm bus is lower than previous case.




30 | P a g e

To support this extra load, the current flow through the transmission line that

feeds the Elm bus increases. As a result the transmission line loss for the line that

deliver power to the Elm bus in increased.

Table 13: Power generation and consumption comparison for given system with 50% increase in Elm bus


Value (MW)




Elm bus load 90















Bus no Transmission line Power in Power Out

Active Power Reactive power Active Power

Reactive power

1 1 -> 2 115.00476 66.67351 111.77722 63.3617

2 1 -> 5 48.79316 16.37946 46.8359 15.73939

3 2 -> 5 25.28554 -1.20085 24.90155 1.58532

4 2 -> 4 29.78206 -0.4367 29.24841 1.88603

5 2 -> 3 76.70962 9.35141 74.30877 5.03216

6 5 -> 4 26.73745 2.32471 26.66256 4.03096

7 4 -> 3 15.91097 0.91699 15.69123 4.96784


31 | P a g e




at PV bus.










Bus no

Transmission line


Active Power (MW)


Active Power (MW)


Active Power (MW)


1 1 -> 2 115.00476 66.67351 111.77722 63.3617 3.22754 3.31181

2 1 -> 5 48.79316 16.37946 46.8359 15.73939 1.95726 0.64007

3 2 -> 5 25.28554 -1.20085 24.90155 1.58532 0.38399 -2.78617

4 2 -> 4 29.78206 -0.4367 29.24841 1.88603 0.53365 -2.32273

5 2 -> 3 76.70962 9.35141 74.30877 5.03216 2.40085 4.31925

6 5 -> 4 26.73745 2.32471 26.66256 4.03096 0.07489 -1.70625

7 4 -> 3 15.91097 0.91699 15.69123 4.96784 0.21974 -4.05085

5.0 Reactive Power compensation using STATCOM In AC circuits energy is stored temporarily in inductive and capacitive elements,

which results in the periodic turnaround of the direction of flow of energy between

the source and the load. The real power is the average power after the completion of

one whole cycle of the AC waveform, and this is the usable energy of the system

and is used to do work, whereas the portion of power flow which is temporarily


32 | P a g e

stored in the form of magnetic or electric fields and flows back and forth in the

transmission line due to inductive and capacitive network elements is known as

reactive power. In order to transmit power. This is the unused power which the

system has to incur.

In an inductive circuit, we know the instantaneous power to be:

max m

max m max m

I cos cos

I Icos 1 cos 2 sin sin 2

2 2 2 2

ax

ax ax

p V t t

V Vp t t

(41)

The instantaneous reactive power is given by:

max mI

sin sin 22 2

axV t (42)

max

max

:

p = instantaneous power

V = Peak value of the voltage waveform

I = Peak value of the current waveform

= Angular frequency

= 2 f where f is the frequency of the waveform.

t = Time period

Where

= Angle by which the current lags the voltage in phase

From here, it can be terminated that the instantaneous reactive power pulsates at

twice the system occurrence and its average value is zero and it gives the maximum

instantaneous reactive power:

Q = |V| |I| sin (43)

It does not necessarily mean that no energy is flowing by the zero average, but the

actual amount that is flowing for half a cycle in one direction, is coming back in the

next half cycle.


33 | P a g e

5.1 Compensation Techniques

5.2 Shunt compensation

Figure 12: Shunt Compensation

A source V1, a power line and an inductive load without any type of compensation

comprises the figure 12. The above part shows the phasor diagram. The active

current Ip is in phase with the load voltage V2. Here, the load is inductive and

hence it requires reactive power for its appropriate operation and this has to be

supplied by the source, thus increasing the current from the generator and through

the power lines. Instead of the lines carrying this, the line current can be

minimized if the reactive power can be supplied near the load, reducing the power

losses and improving the voltage regulation at the load terminals. This can be

completed in three ways: 1) A voltage source. 2) A current source. 3) A capacitor.

In this case, to compensate Iq a current source device is used, which is the reactive

component of the load current. Similarly, the voltage regulation of the system is

improved and the reactive current constituent from the source is concentrated or

almost eliminated. This is in case of lagging compensation. , we need an inductor

For leading compensation.


34 | P a g e

5.3 Lake bus connected to the STATCOM and Reactive Compensation

0.0

20

.06

j

0.06/2j

0.06/2j

North (1)

South(2)

0.08

0.24j Lake (5)

0.0

5/2

j

0.0

5/2

j

0.0

60

.18

j

0.04/2j

0.04/2j

0.06 0.18j

0.0

4/2

j

0.0

4/2

j

Main (4)

0.0

1

0.02/2j

0.02/2j

0.03

j

Elm (3)

0.0

8

0.05/2j

0.05/2j

0.24

j

0.04

0.0

3/2

j

0.0

3/2

j

0.12j

40

MW

20

MW

10

MV

ar

60

MW

10

MV

ar

40MW5MVar

45MW15MVar

13

1MW

2.4

9M

W+

1.0

9M

var

1.52MW-0.69Mvar

0.3

6M

W-2

.87M

var

0.46MW-2.55Mvar

1.22MW+0.73Mvar

0.0

4M

W-1

.82

Mva

r

0.0

4M

W-4

.65

Mva

r

STATCOM Device

Figure 13: Lake bus connected to the STATCOM and Reactive Compensation

STATCOM device is attached to Lake bus to achieve the bus voltage 1 p.u. using

available technique. For this purpose the Jacobian matrix and the updated bus

voltage should be changed.


35 | P a g e

Each time Jacobian matrix is evaluated using the Slack bus voltage 1.06 p.u. and

PV bus voltage 1.00 p.u. Now another bus is maintaining 1 p.u. voltage. Therefore,

during Jacobian matrix calculation this situation is considered.

Figure 14: Modified Jacobian Matirx.

The Lake bus number is 5. When these bus appears the Jacobian matrix

calculation replace the voltage with 1.0 p.u.

When only the Lake bus is appeared,

Similarly for other Jacobian calculation, the code is modified.


36 | P a g e


37 | P a g e

Each bus Voltages and angles are as follows:




p.u. The voltage of the other buses except the Lake bus (bus no. 5) is lower than

one. In this result, the bus voltage is lower than one, the STATCOM device will

provide additional reactive power to make the bus voltage 1 p.u.




38 | P a g e

Since the Lake bus voltage is 1.0 p.u. the transmission loss is different than that of

the previous configuration.

Table 17: Power generation and consumption comparison for power grid system with STATCOM device at Lake bus

Power Generation Value (MW) Total (MW) Power Consumtion

Value (MW)


South bus load 20


PV bus (South bus) generator 40 40 Main bus load 40

Elm bus load 60















Bus Transmission Power in Power Out


39 | P a g e

no line Active Power Reactive power Active Power Reactive power

1 1 -> 2 89.33138 73.99518 86.84551 72.90839

2 1 -> 5 41.79085 16.82034 40.27302 17.5125

3 2 -> 5 24.47266 -2.51849 24.11315 0.3523

4 2 -> 4 27.713 -1.72391 27.25215 0.83056

5 2 -> 3 54.65985 5.55794 53.44485 4.82921

6 5 -> 4 19.38618 2.8648 19.34611 4.68775

7 4 -> 3 6.59825 0.51832 6.55515 5.17079




at PV bus.










Bus no

Transmission line


Active Power (MW)


Active Power (MW)


Active Power (MW)


1 1 -> 2 89.33138 73.99518 86.84551 72.90839 2.48587 1.08679

2 1 -> 5 41.79085 16.82034 40.27302 17.5125 1.51783 -0.69216

3 2 -> 5 24.47266 -2.51849 24.11315 0.3523 0.35951 -2.87079


40 | P a g e

4 2 -> 4 27.713 -1.72391 27.25215 0.83056 0.46085 -2.55447

5 2 -> 3 54.65985 5.55794 53.44485 4.82921 1.215 0.72873

6 5 -> 4 19.38618 2.8648 19.34611 4.68775 0.04007 -1.82295

7 4 -> 3 6.59825 0.51832 6.55515 5.17079 0.0431 -4.65247

6.0 Conclusion To ensure best operation of a current power grid system as well as to expand the

system, power flow analysis is very important. The power flow analysis summarizes

the whole power grid system by providing voltage and angle of each bust associated

with power flow through each line. In this experiment, power flow analysis is

performed for a power grid system of 5 buses and 7 transmission lines. Prior to this,

popular power flow technique known as Newton-Raphson method is studied well.

The steady state voltage and angle of each bus and power flow of each line is found

by writing a matlab program. The voltage and angle of each bus is measured by

varying active loading of different buses and compared with that of previous

scenarios. From the findings, it is accomplished that raising the active loading

resulted in an increased Transmission losses. The calculations are complex In

Newton-Rahpson method, but the number of iterations is low even when the

number of buses is high. For any power grid system where any bus is drawing

more power than the specified limit can be identified easily using load flow analysis.

Therefore, load flow analysis is also useful for fault analysis in the system.

STATCOM device is assumed to be attached with a bus so that the voltage of that

bus is maintained 1.0 p.u. following reactive compensation. To achieve this,

Jacobian matrix is modified and the result is analyzed.


41 | P a g e

7.0 Reference [1] A.E. Guile and W.D. Paterson, Electrical power systems, Vol. 2, (Pergamon

Press, 2nd edition, 1977).

[2] W.D. Stevenson Jr., Elements of power system analysis, (McGraw-Hill,4th

edition, 1982).

[3] W. F. Tinney, C. E. Hart, "Power Flow Solution by Newton's Method," IEEE

Transactions on Power Apparatus and systems, Vol. PAS-86, pp. 1449-1460,

November 1967.

[4] W. F. Tinney, C. E. Hart, "Power Flow Solution by Newton's Method, "IEEE

TRANS. POWER APPARATUS AND SYSTEMS, Vol. PAS-86, pp. 1449-1460,

November 1967.

[5] O. Elgerd, Electric Energy Systems Theory, McGraw-Hill, New York, 1982.

[6] Lab Manual, ELEC5525-Power Flow Analysis


42 | P a g e

8.0 Appendix

8.1 Matlab code

This is the type of buses that existis in a power grid systems.

No. of iteration and stopping criteria is set here. If the stooping criteria is smaller

enough it is good for accuracty but it will take time to converge.

Buses are defined with initial values.


43 | P a g e

Transmission lines are described.

Calculating the total buses and no. of lines.

Power system admittance matrix is prepared here.


44 | P a g e

Variable initialization for output and intermediate temp variable.

Initialization of powers and number to the buses.


45 | P a g e

Sizing Jacobian Matrix and initialization.


46 | P a g e

Main body of the iteration.


47 | P a g e


48 | P a g e

Updating Jacobian sub-Matrix.


49 | P a g e

Updating Jacobian Matix and other variable with convergence checking

Printing output.

Output formatting.


50 | P a g e

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Power Flow Analysis for a grid connected system

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