Load Flow Analysis Of Ieee14 Bus System Using MATLAB

DOI : 10.17577/IJERTV2IS50038

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Load Flow Analysis Of Ieee14 Bus System Using MATLAB

Load Flow Analysis Of Ieee14 Bus System Using MATLAB

P. Srikanth, O. Rajendra, A. Yesuraj, M. Tilak K.Raja

Students, Electrical & Electronics Engineering, Asst Professor, Electrical & Electronics Engineering,

Lingayas Institute of Management & Technology Lingayas Institute of Management & Technology Vijayawada, India. Vijayawada, India.

AbstractThe power system analysis and design is generally done by using power flow analysis .This analysis is carried out at the state of planning, operation, control and economic scheduling

.they are useful in determining the magnitude and phase angle of load buses, and active and reactive power flows over transmission lines, and active and reactive powers that are injected at the buses.

For this work the gauss-seidel method is used for numerical analysis. The objective of this project is to develop a MATLAB program to calculate voltages, active and reactive power at each bus for IEEE 14 bus systems. At first IEEE 5 bus system is calculated by using hand calculations and compared with MATLAB Program results and then IEEE 14 bus system MATLAB program is executed with the input data. This type of analysis is useful for solving the power flow problem in different power systems which will useful to calculate the unknown quantities.

Index Terms Power Flow Studies (PFS), Guass-Siedel (GS) method, and voltage magnitudes (V), active power (P) & reactive power (Q)

  1. INTRODUCTION

    Load flow studies are used to ensure that electrical power transfer from generators to consumers through the grid system is stable, reliable and economic. Conventional techniques for solving the load flow problem are iterative using the Gauss- Seidel methods Load flow analysis forms an essential prerequisite for power system studies. Considerable research has already been carried out in the development of computer programs for load flow analysis of large power systems. However, these general purpose programs may encounter convergence difficulties when a radial distribution system with a large number of buses is to be solved and, hence, development of a special program for radial distribution studies becomes necessary.

    There are many solution techniques for load flow analysis. The solution procedures and formulations can be precise or approximate, with values adjusted or unadjusted, intended for either on-line or off-line application, and designed for either single-case or multiple-case applications. Since an engineer is always concerned with the cost of products and services, the efficient optimum economic operation and planning of electric power generation system have always occupied an important position in the electric power industry. With large interconnection of the electric networks, the energy crisis in the world and continuous rise in prices, it is very essential to reduce the running charges of the

    electric energy. A saving in the operation of the system of a small percent represents a significant reduction in operating cost as well as in the quantities of fuel consumed. The classic problem is the economic load dispatch of generating systems to achieve minimum operating cost.

  2. POWER FLOW OVER VIEW

    • Power flow analysis is very important in planning stages of new networks or addition to existing ones like adding new generator sites, meeting increase load demand and locating new transmission sites.

    • The load flow solution gives the nodal voltages and phase angles and hence the power injection at all the buses and power flows through interconnecting power channels.

    • It is helpful in determining the best location as well as optimal capacity of proposed generating station, substation and new lines

    • It determines the voltage of the buses. The voltage level at the certain buses must be kept within the closed tolerances.

    • System transmission loss minimizes.

    • Economic system operation with respect to fuel cost to generate all the power needed

    • The line flows can be known. The line should not be overloaded, it means, we should not operate the close to their stability or thermal limits.

  3. POWER FLOW ANALYSIS

BUS CLASSIFICATION:

A bus is a node at which one or many lines, one or many loads and generators are connected. In a power system each node or bus is associated with 4 quantities, such as magnitude of voltage, phage angle of voltage, active or true power and reactive power in load flow problem two out of these 4 quantities are specified and remaining 2 are required to be determined through the solution of equation. Depending on the quantities that have been specified, the buses are classified into 3 categories.

Buses are classified according to which two out of the four variables are specified

  • Load bus:

No generator is connected to the bus. At this bus the real and reactive power are specified. It is desired to

E 1 ( PP jQP

n

n

Y E )

find out the voltage magnitude and phase angle through load flow solutions. It is required to specify

Ypp P

q 1 q p

pq q

p

p

E

E

*

*

only Pd and Qd at such bus as at a load bus voltage can be allowed to vary within the permissible values.

  • Generator bus or voltage controlled bus: Here the voltage magnitude corresponding to the generator voltage and real power Pg corresponds to its rating are specified. It is required to find out the reactive power generation Qg and phase angle of the bus voltage.

  • Slack (swing) bus: For the Slack Bus, it is assumed that the voltage magnitude |V| and voltage phase

p 1,2,….., n

p s

This involves only bus voltages as variables. Formulating the load flow problem in this manner results in asset of non-linear equations that can be solved by an iterative method. A significant reduction in the computing time for a solution will be obtained by performing as many arithmetic operations as possible before initiating the iterative calculation .Letting

are known, whereas real and reactive powers Pg and Qg are obtained through the load flow solution.

1

Ypp

Lp

Equation (4) can be written

Gauss iterative method using Ybus:-

( p jQ )L n

The solution of the load flow problem is initiated by assuming voltages for all buses except the slack bus, where the voltage

Ep p p p

E*

YpqLp Eq

is specified and remains fixed. Then currents are calculated for all buses except slack bus s from the loading equation

p q1

q p

E

E

*

*

Ip Pp jQp

p

p 1,2,….., n

p s

Letting

Where n is the number of buses in the network .The performance of the network can be obtained from the equation

IBUS = YBUS EBUS

Pp jQp Lp KLp

And

Ypq LPQ = YLPQ

Then, the bus voltage equation (5) becomes

Selecting the ground as the reference bus, a set of (n-1)

n

KLp

KLp

Ep

E*

YL pq Eq

simultaneous equations can be written in the form

p q 1

q p

p

p

p

p

E 1 (I

Ypp

n

n

q1 q p

YpqEq )

The normal procedure for a load flow study is to assume a balanced system and to ue a single-phase representation

p 1,2,….., n

p s

The bus currents calculated from equation (1), the slack bus voltage, and the estimated bus voltages are substituted in to EQ (3) to obtain a new set of bus voltages. The new voltages are used inEQ (8.3.1) to re calculate bus currents from a subsequent solution of EQ (3). This process is continued until changes in all bus voltages are negligible. After the voltage solution has been obtained, the power at the slack bus and line flows can be calculated.

The network EQ (3) and the bus loading EQ (1) can be

equivalent to the positive sequence network. Since there is no mutual coupling, the bus admittance matrix can be formed by inspection and many of its elements will be zero. Selecting bus 2 as the slack bus in the system shown in the fig IEEE 5 BUS SYSTEM, the formulas for the gauss iterative solutions are Gauss-sidle iterative method using Ybus The bus voltage EQ (6) also can be solved by the gauss-seidle iterative method (glimn and Stagg 1957) in this method the new calculated voltage immediately replaces and is used in the solution of the subsequent equations. For the system shown in fig of IEEE 5 BUS SYSTEM(8.1)

E

E

Ek 1 KL1 YL E YL Ek YL Ek

combining to obtain

1 k 12 2

1

13 3

14 4

E2 Specified

fixed

value

E

E

k 1 3

KL3

3

3

(Ek )*

YL Ek

YL Ek

To verify the effectiveness of MATLAB program,both MATLAB results and hand calculations results are compared

.It is found that ,These two results are equal.The input data for sample 5 bus system is given in appendix.

Ek 1

KL4

YL Ek YL Ek

31 1

31 1

35 5

35 5

The MATLAB results & Hand calculations are

4

Ek 1

(Ek )*

4

4

KL5

41 1

YL E

46 6

YL Ek

5

Ek 1

(Ek )*

5

5

KL6

52 2

YL E

53 3

YL Ek

6

6

6 (Ek ) *

62 2

64 3

IV CASE STUDIES

Figure 1: Sample 5-Bus System.

A sample 5 bus system is used for this case,the input data of 5 bus is taken from the reference[].The hand calculations are done by using guass-seidel equations to calculate voltages

,active and reactive powers at each bus for every iteration.

A matlab program is written to calculate the voltages ,active power and reactive power at every bus in the system for every iteration.The input data for the program is given with input file name.

BUS VOLTAGES AT THE END OF 10 ITEATIONS E(1)=(1.060000)+j(0.000000)

E(2)=(1.046249)+j(-0.051279)

E(3)=(1.020376)+j(-0.089174)

E(4)=(1.019215)+j(-0.095041)

E(5)=(1.012128)+j(-0.109042)

The Line Power Flows are:

Ppq(1)(2)=88.822114 MW Qpq(1)(2)=-8.684865 MVAr Pqp(2)(1)=-87.412785 MW Qqp(2)(1)=6.250251 MVAr Ppq(1)(3)=40.696820 MW Qpq(1)(3)=1.126016 MVAr Pqp(3)(1)=-39.506563 MW Qqp(3)(1)=-2.987044 MVAr Ppq(2)(3)=24.674725 MW Qpq(2)(3)=3.539738 MVAr Pqp(3)(2)=-24.323822 MW Qqp(3)(2)=-6.779800 MVAr Ppq(2)(4)=27.925725 MW Qpq(2)(4)=2.963475 MVAr Pqp(4)(2)=-27.484747 MW Qqp(4)(2)=-5.930740 MVAr Ppq(2)(5)=54.822293 MW Qpq(2)(5)=7.360576 MVAr Pqp(5)(2)=-53.697096 MW Qqp(5)(2)=-7.185329 MVAr Ppq(3)(4)=18.930215 MW Qpq(3)(4)=-5.155906 MVAr Pqp(4)(3)=-18.894450 MW Qqp(4)(3)=3.166250 MVAr Ppq(4)(5)=6.340671 MW Qpq(4)(5)=-2.277924 MVAr Pqp(5)(4)=-6.309887 MW Qqp(5)(4)=-2.840041 MVAr

The line losses are:

Ppq(1)(2)=156.317898 MW Qpq(1)(2)=-20.255152 MVAr Pqp(2)(1)=-152.051712 MW Qqp(2)(1)=27.431249 MVAr Ppq(2)(3)=73.087538 MW Qpq(2)(3)=3.578016 MVAr Pqp(3)(2)=-70.773582 MW Qqp(3)(2)=1.545299 MVAr

Ploss(1)=1.409328 MW

Qloss(1)=-2.434615 MVAr

Ppq(2)(4)=55.967986 MW

Qpq(2)(4)=-2.296184 MVAr

Ploss(2)=1.190257 MW

Qloss(2)=-1.861028 MVAr

Pqp(4)(2)=-54.301066 MW

Qqp(4)(2)=3.371402 MVAr

Ploss(3)=0.350903 MW

Qloss(3)=-3.240062 MVAr

Ppq(1)(5)=75.313365 MW

Qpq(1)(5)=3.492425 MVAr

Ploss(4)=0.440978 MW

Qloss(4)=-2.967264 MVAr

Pqp(5)(1)=-72.567026 MW

Qqp(5)(1)=2.519567 MVAr

Ploss(5)=1.125196 MW

Qloss(5)=0.175247 MVAr

Ppq(2)(5)=41.385188 MW

Qpq(2)(5)=0.749092 MVAr

Ploss(6)=0.035765 MW

Qloss(6)=-1.989657 MVAr

Pqp(5)(2)=-40.488427 MW

Qqp(5)(2)=-1.637334 MVAr

Ploss(7)=0.030784 MW

Qloss(7)=-5.117965 MVAr

Ppq(3)(4)=-23.397752 MW

Qpq(3)(4)=2.785299 MVAr

Pqp(4)(3)=23.770984 MW

Qqp(4)(3)=-5.392707 MVAr

Slack Bus Real Power Generation=129.518934 MW Slack Bus Reactive Power Generation=-7.558849 MAVr

The IEEE 14 bus system is simulated for voltages ,active and reactive powers at each bus.

The standard IEEE14 bus system is shown in figure.

A MATLAB program is used for IEEE 14 bus system with standard IEEE 14 bus system data input file.After simulating the program ,it didplays results ,i.e voltage ,active & Reactive powers and line losses .The results of MATLAB program for IEEE 14 bus system are

BUS VOLTAGES AT THE END OF 48 ITEATIONS E(1)=(1.060000)+j(0.000000) E(2)=(1.041071)+j(-0.090419)

E(3)=(0.985314)+j(-0.221865)

E(4)=(1.002309)+j(-0.181984)

E(5)=(1.008447)+j(-0.155282)

E(6)=(1.037393)+j(-0.261998)

E(7)=(1.033411)+j(-0.244615)

E(8)=(1.060686)+j(-0.251088)

E(9)=(1.020843)+j(-0.271512)

E(10)=(1.015241)+j(-0.273020)

E(11)=(1.022244)+j(-0.269053)

E(12)=(1.019093)+j(-0.273569)

E(13)=(1.014097)+j(-0.273800)

E(14)=(0.995706)+j(-0.285303)

The Line Power Flows are:

Ppq(4)(5)=-61.046827 MW Qpq(4)(5)=15.619484 MVAr Pqp(5)(4)=61.560360 MW Qqp(5)(4)=-15.330086 MVAr Ppq(5)(6)=43.904029 MW Qpq(5)(6)=12.856263 MVAr Pqp(6)(5)=-43.904029 MW Qqp(6)(5)=-8.455568 MVAr Ppq(4)(7)=27.926820 MW Qpq(4)(7)=-9.401528 MVAr Pqp(7)(4)=-27.926820 MW Qqp(7)(4)=11.075128 MVAr Ppq(7)(8)=0.010177 MW Qpq(7)(8)=-16.900100 MVAr Pqp(8)(7)=-0.010177 MW Qqp(8)(7)=17.346207 MVAr Ppq(4)(9)=16.024493 MW Qpq(4)(9)=-0.309868 MVAr Pqp(9)(4)=-16.024493 MW Qqp(9)(4)=1.602588 MVAr Ppq(7)(9)=28.061205 MW Qpq(7)(9)=5.825814 MVAr Pqp(9)(7)=-28.061205 MW Qqp(9)(7)=-5.024598 MVAr Ppq(9)(10)=5.244181 MW Qpq(9)(10)=4.309034 MVAr Pqp(10)(9)=-5.231047 MW Qqp(10)(9)=-4.274147 MVAr Ppq(6)(11)=7.332290 MW Qpq(6)(11)=3.470636 MVAr Pqp(11)(6)=-7.277693 MW Qqp(11)(6)=-3.356303 MVAr Ppq(6)(12)=7.769332 MW Qpq(6)(12)=2.503193 MVAr Pqp(12)(6)=-7.697799 MW Qqp(12)(6)=-2.354312 MVAr Ppq(6)(13)=17.727774 MW Qpq(6)(13)=7.175947 MVAr Pqp(13)(6)=-17.516427 MW Qqp(13)(6)=-6.759739 MVAr Ppq(9)(14)=9.452723 MW Qpq(9)(14)=3.661815 MVAr Pqp(14)(9)=-9.335661 MW Qqp(14)(9)=-3.412809 MVAr Ppq(10)(11)=-3.748790 MW Qpq(10)(11)=-1.536134 MVAr Pqp(11)(10)=3.760975 MW Qqp(11)(10)=1.564657 MVAr Ppq(12)(13)=1.612438 MW Qpq(12)(13)=0.733450 MVAr Pqp(13)(12)=-1.606212 MW Qqp(13)(12)=-0.727817 MVAr Ppq(13)(14)=5.628515 MW Qpq(13)(14)=1.689408 MVAr Pqp(14)(13)=-5.575015 MW Qqp(14)(13)=-1.580481 MVAr

The line losses are:

Ploss(1)=4.266185 MW Qloss(1)=7.176097 MVAr Ploss(2)=2.313956 MW Qloss(2)=5.123314 MVAr Ploss(3)=1.666920 MW Qloss(3)=1.075218 MVAr Ploss(4)=2.746339 MW Qloss(4)=6.011992 MVAr Ploss(5)=0.896762 MW Qloss(5)=-0.888243 MVAr Ploss(6)=0.373232 MW Qloss(6)=-2.607408 MVAr Ploss(7)=0.513533 MW Qloss(7)=0.289398 MVAr Ploss(8)=-0.000000 MW Qloss(8)=4.400695 MVAr Ploss(9)=0.000000 MW Qloss(9)=1.673601 MVAr Ploss(10)=0.000000 MW Qloss(10)=0.446107 MVAr Ploss(11)=0.000000 MW Qloss(11)=1.292721 MVAr Ploss(12)=0.000000 MW Qloss(12)=0.801215 MVAr Ploss(13)=0.03133 MW Qloss(13)=0.034887 MVAr Ploss(14)=0.054597 MW Qloss(14)=0.114333 MVAr Ploss(15)=0.071533 MW Qloss(15)=0.148881 MVAr Ploss(16)=0.211347 MW Qloss(16)=0.416208 MVAr Ploss(17)=0.117062 MW Qloss(17)=0.249006 MVAr Ploss(18)=0.012185 MW Qloss(18)=0.028523 MVAr Ploss(19)=0.006226 MW Qloss(19)=0.005633 MVAr Ploss(20)=0.053500 MW Qloss(20)=0.108928 MVAr

Slack Bus Real Power Generation=231.631263 MW Slack Bus Reactive Power Generation=-16.762727 MAVr

Figure 2: IEEE 14-Bus System.

In this paper, The IEEE 14 bus system is analyzed by using guass-seidel method. This is verified by calculating hand calculations by using the guass-seidel equations and MATLAB program for 5 bus sample system. Both these results are found equal .so this type of MATLAB programming is very useful for solving load flow problems. This MATLAB program can be applicable for any number of buses. The standard IEEE 14 bus input data is used for IEEE14bus system and sample 5 bus input data is used for 5 bus system. The future scope for this project can be extended with Newton-Raphson method and Fast Decoupled methods.

REFERENCES

  1. Glenn W Stagg, and I.Stagg, Computer Methods in Power System Analysis.

  2. J W.D. Stevenson Jr., Elements of power system analysis, (McGraw-Hill, 4th edition, 1982).

  3. H. Dommel, "Digital methods for power system analysis" (in German), Arch. Elektrotech., vol. 48, pp. 41-68, February 1963 and pp. 118-132, April 1963.

  4. Carpentier Optimal Power Flows, Electrical Power and Energy Systems, Vol.1, April 1979, pp 959-972.

APPENDIX

IEEE 5-Bus System input data

Bus code p

Assumed Bus Voltage

Generation

Load

MW

MVARs

MW

MVARs

1

1.06+j0.0

0

0

0

0

2

1.0+j0.0

40

30

20

10

3

1.0+j0.0

0

0

45

15

4

1.0+j0.0

0

0

40

5

5

1.0+j0.0

0

0

60

10

Bus Code p-q

Impedance

Zpq

Line Charging

Y' /2

pq

1-2

0.02+j0.06

0.0+j0.030

1-3

0.08+j0.24

0.0+j0.025

2-3

0.06+j0.18

0.0+j0.020

2-4

0.06+j0.18

0.0+j0.020

2-5

0.04+j0.12

0.0+j0.015

3-4

0.01+j0.03

0.0+j0.010

4-5

0.08+j0.24

0.0+j0.025

Bus code p

Assumed Bus Voltage

Generation

Load

MW

MVARs

MW

MVARs

1

1.06+j0.0

0

0

0

0

2

1.0+j0.0

40

30

20

10

3

1.0+j0.0

0

0

45

15

4

1.0+j0.0

0

0

40

5

5

1.0+j0.0

0

0

60

10

Bus Code p-q

Impedance

Zpq

Line Charging

Y' /2

pq

1-2

0.02+j0.06

0.0+j0.030

1-3

0.08+j0.24

0.0+j0.025

2-3

0.06+j0.18

0.0+j0.020

2-4

0.06+j0.18

0.0+j0.020

2-5

0.04+j0.12

0.0+j0.015

3-4

0.01+j0.03

0.0+j0.010

4-5

0.08+j0.24

0.0+j0.025

IEEE 14-Bus System input data

Line Data of IEEE 14-Bus System

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