Power Flow

Semester 2 15/16 POWER SYSTEMS I EEPB353 POWER FLOW ASSIGNMENT

NAME: MUHD. SYAHMI AZRI BIN SANSUL BAHRI ID: EP092693 SECTION: 3 LECTURER: MR. JOHN STEVEN NAVAMANY

Contents

No . 1 2 3 4 5 6

Introduction

Title

Page

Introduction Load flow application Simulation results Discussions Conclusions References

1 1 4-16 17 18 18

Power flow studies or also known as load flow is an important part of power system analysis[1]. Normally, numerical methods are used to solve the load flow. These are important for planning, economic scheduling, as well as control of an existing system and for future expansion. Load flow focusses on voltage and voltages angle at each bus, and real power and reactive power for lines and loads. Computer software such as PSS®E Xplore 34 and MATLAB are very essential to calculate the desired parameters. By running the simulation tools in the software, one can simulate and solve the power system in real life with real conditions. To solve complex and multiple circuits with many diagrams, it is convenient to solve using a computer software. One also can draw the diagram using the software mention above. For this assignment, PSS®E Xplore 34 will be used.

Load Flow Applications

Generally, load flow solutions are mostly used for the power system design, planning and its operations. These are some another essential applications for any electrical power engineers in power flow analysis [2]:

1. 2. 3. 4. 5. 6.

Operation planning for transmission division expansion. Operation planning for distribution expansion planning. Industrial/Commercial distribution system planning, operational planning Network interconnection, Grid interconnection studies Evaluation of energy used. Sizing of transformers, cables, overhead lines, transformer tap ranges, shunt capacitors, shunt reactors, reactive power management, HVDC operation

SIMULATION RESULTS

Impedance Calculations Base values: SB = 100 MVA VB1 = 13.8 kV VB2 = 13.8 kV VB3 = 13.8 kV VB4 = 13.8 kV VB5 = 13.8(230/13.8) = 230 kV VB6 = 230 kV VB7 = 230 kV VB8 = 230 kV VB9 = 230 kV VB10 = 230 kV VB11 = 13.8 kV

Generators Impedances: XG1 = 0.1 (100MVA/100MVA) = 0.1 pu XG2 = 0.1 (100MVA/200MVA) = 0.05 pu XG3 = 0.1 (100MVA/500MVA) = 0.02 pu XG4 = 0.1 (100MVA/300MVA) = 0.03333 pu XG5 = 0.1 (100MVA/100MVA) = 0.0125 pu Transformers Impedances:

Load in per units

XT15 = 0.1 (100MVA/100MVA) = 0.1 pu XT26 = 0.1 (100MVA/100MVA) = 0.1 pu XT37 = 0.1 (100MVA/200MVA) = 0.05 pu XT48 = 0.1 (100MVA/200MVA) = 0.05 pu XT11,10 = 0.1 (100MVA/200MVA) = 0.05 pu

S5 = (70MW+ j15MVAR)/100 MVA = 0.7 + j0.15 pu S6 = (50MW+ j10MVAR)/100 MVA = 0.5 + j0.1 pu S7 = (80MW+ j20MVAR)/100 MVA = 0.8 + j0.2 pu S8 = (60MW+ j10MVAR)/100 MVA = 0.6 + j0.1 pu S9 = (250MW+j20MVAR)/100MVA = 2.5 + j0.4 pu

Lines Impedances: ZBL = (230kV)2/100MVA = 529 Ω Z56 = (0.05 + j0.6 Ω/km) x 15km/529 Ω = 0.001418 + j0.017 pu Z57 = (0.05 + j0.7 Ω/km) x 60km/529 Ω = 0.005671 + j0.0794 pu Z59 = (0.05 + j0.9 Ω/km) x 30km/529 Ω = 0.002836 + j0.05103 pu Z68 = (0.06 + j0.8 Ω/km) x 60km/529 Ω = 0.006805 +j0.09074 pu Z6,10 = (0.06 + j0.1 Ω/km) x 30km/529 Ω = 0.003402 + j0.005671 pu Z78 = (0.07 + j0.8 Ω/km) x 15km/529 Ω = 0.001985 + j0.02268 pu Scenario 1

Z79 = (0.07 + j0.9 Ω/km) x 30km/529 Ω = 0.003970 + j0.05104 pu Z8,10 = (0.05 + j0.6 Ω/km) x 15km/529 Ω = 0.004537 + j0.005671 pu

Bus: Bus Number

Base (kV)

Voltage (pu)

Angle (deg)

Normal Vmin (pu) 0.9

Emergency Vmax (pu)

Emergency Vmin (pu)

-2.62

Normal Vmax (pu) 1.1

1

13.8

1.0000

1.1

0.9

2 3 (swing bus) 4

13.8 13.8

1.0000 1.0000

-2.53 0.00

1.1 1.1

0.9 0.9

1.1 1.1

0.9 0.9

13.8

0.9994

0.14

1.1

0.9

1.1

0.9

5

230

0.9872

-4.36

1.1

0.9

1.1

0.9

6

230

0.9877

-4.27

1.1

0.9

1.1

0.9

7 8

230 230

0.9897 0.9907

-2.90 -2.75

1.1 1.1

0.9 0.9

1.1 1.1

0.9 0.9

Machines: Bus number 1 2 3 4

P gen (MW) 30 30 100.177 7 100

P max (MW) 50 50 150

P min (MW) 0 0 0

Q gen (Mvar) 13.2777 12.7478 23.2107

Q max (Mvar) 30 30 50

Q min (Mvar) -30 -30 -50

M base (MVA) 100 100 100

X source (pu) 0.1 0.05 0.02

200

0

20.0000

20

-20

100

0.03333

Loads: Bus number 5 6 7 8

Branch

P load (MW) 70 50 80 60

Q load (Mvar) 15 10 20 10

From bus 5 5 6 7

To bus 6 7 8 8

Line R (pu) 0.001418 0.005671 0.006805 0.001985

Line X (pu) 0.017 0.07940 0.090740 0.022680

2 winding transformers: From bus

To bus

Specified X (pu)

Winding (MVA) base

1 2 3 4

5 6 7 8

0.1 0.1 0.05 0.05

100 100 100 100

Winding 1 Nominal (kV) 13.8 13.8 13.8 13.8

Winding 2 Nominal (kV) 230 230 230 230

Rating (MVA)

Y-admittance matrix 4,

4,

0.00000000000000

, -20.0000000000000

100 100 200 200

4, 2, 2, 1, 1, 5, 5, 5, 5, 8, 8, 8, 8, 7, 7, 7, 7, 6, 6, 6, 6,

Power Flow Diagram

Scenario 2

8, -0.00000000000000 2, 0.00000000000000 6, -0.00000000000000 1, 0.00000000000000 5, -0.00000000000000 5, 5.76764273643494 1, -0.00000000000000 7,-0.894970655441284 6, -4.87267208099365 8, 4.65151566267014 4, -0.00000000000000 7, -3.82966136932373 6,-0.821854293346405 7, 4.72463202476501 5,-0.894970655441284 8, -3.82966136932373 3, -0.00000000000000 6, 5.69452637434006 2, -0.00000000000000 5, -4.87267208099365 8,-0.821854293346405

, , , , , , , , , , , , , , , , , , , , ,

20.0000000000000 -10.0000000000000 10.0000000000000 -10.0000000000000 10.0000000000000 -80.9476242065430 10.0000000000000 12.5305366516113 58.4170875549316 -74.7153940200806 20.0000000000000 43.7565307617188 10.9588632583618 -76.2870674133301 12.5305366516113 43.7565307617188 20.0000000000000 -79.3759508132935 10.0000000000000 58.4170875549316 10.9588632583618

Bus: Bus Number

Base (kV)

Voltage (pu)

Angle (deg)

Normal Vmin (pu) 0.9

Emergency Vmax (pu)

Emergency Vmin (pu)

-5.83

Normal Vmax (pu) 1.1

1

13.8

0.9793

1.1

0.9

2 3 (swing bus) 4

13.8 13.8

0.9823 1.0000

-5.27 0.00

1.1 1.1

0.9 0.9

1.1 1.1

0.9 0.9

13.8

0.9596

1.11

1.1

0.9

1.1

0.9

5 6 7 8 9

230 230 230 230 230

0.9501 0.9531 0.9560 0.9549 0.9347

-8.91 -8.33 -6.37 -5.15 -11.70

1.1 1.1 1.1 1.1 1.1

0.9 0.9 0.9 0.9 0.9

1.1 1.1 1.1 1.1 1.1

0.9 0.9 0.9 0.9 0.9

Machines: Bus number 1 2 3 4

P gen (MW) 50 50 212.095 200

P max (MW) 50 50 150 200

P min (MW) 0 0 0 0

Q gen (Mvar) 30.0 30.0 99.7437 20.0

Q max (Mvar) 30 30 50 20

Q min (Mvar) -30 -30 -50 -20

M base (MVA) 100 100 100 100

X source (pu) 0.1 0.05 0.02 0.03333

Loads: Bus number 5 6 7 8 9

P load (MW) 70 50 80 60 250

Q load (Mvar) 15 10 20 10 20

Branch From bus 5 5

To bus 6 7

Line R (pu) 0.001418 0.005671

Line X (pu) 0.017 0.07940

6 7 5 7

8 8 9 9

0.006805 0.001985 0.002836 0.003970

0.090740 0.022680 0.051040 0.051040

2 winding transformers: From bus

To bus

Specified X (pu)

Winding (MVA) base

1 2 3 4

5 6 7 8

0.1 0.1 0.05 0.05

100 100 100 100

Winding 1 Nominal (kV) 13.8 13.8 13.8 13.8

Winding 2 Nominal (kV) 230 230 230 230

Rating (MVA)

Admittance Matrix 9,

9,

2.60007083415985

, -39.0068264007568

9, 9, 4, 4,

5, -1.08529078960419 7, -1.51478004455566 4, 0.00000000000000 8, -0.00000000000000

, 19.5321731567383 , 19.4746532440186 , -20.0000000000000 , 20.0000000000000

100 100 200 200

2, 2, 1, 1, 5, 5, 5, 5, 5, 8, 8, 8, 8, 7, 7, 7, 7, 7, 6, 6, 6, 6,

Power Flow Diagram

2, 0.00000000000000 6, -0.00000000000000 1, 0.00000000000000 5, -0.00000000000000 5, 6.85293352603912 9, -1.08529078960419 1, -0.00000000000000 7,-0.894970655441284 6, -4.87267208099365 8, 4.65151566267014 4, -0.00000000000000 7, -3.82966136932373 6,-0.821854293346405 7, 6.23941206932068 9, -1.51478004455566 5,-0.894970655441284 8, -3.82966136932373 3, -0.00000000000000 6, 5.69452637434006 2, -0.00000000000000 5, -4.87267208099365 8,-0.821854293346405

, , , , , , , , , , , , , , , , , , , , , ,

-10.0000000000000 10.0000000000000 -10.0000000000000 10.0000000000000 -100.479797363281 19.5321731567383 10.0000000000000 12.5305366516113 58.4170875549316 -74.7153940200806 20.0000000000000 43.7565307617188 10.9588632583618 -95.7617206573486 19.4746532440186 12.5305366516113 43.7565307617188 20.0000000000000 -79.3759508132935 10.0000000000000 58.4170875549316 10.9588632583618

Based on results above, the P gen obtained is 212.095 MW and Q gen obtained is 99.7437 Mvar. Both P gen and Q gen are more than the generator’s limit. This means, the generator 3 must produce more power to fulfill the demand of new load even though all other 3 plants are operating at maximum capacity. Therefore, the power network is not capable of handling the new load.

Scenario 3 Bus: Bus Number

Base (kV)

Voltage (pu)

Angle (deg)

Normal Vmax (pu)

Normal Vmin (pu)

Emergency Vmax (pu)

Emergency Vmin (pu)

1

13.8

1.0000

-0.26

1.1

0.9

1.1

0.9

2 3 (swing bus) 4

13.8 13.8

1.0000 1.0000

1.11 0.00

1.1 1.1

0.9 0.9

1.1 1.1

0.9 0.9

13.8

0.9892

2.33

1.1

0.9

1.1

0.9

5 6 7 8 9 10 11

230 230 230 230 230 230 230

0.9740 0.9791 0.9765 0.9813 0.9579 0.9842 1.0000

-3.21 -1.82 -3.01 -1.51 -6.98 -1.39 3.86

1.1 1.1 1.1 1.1 1.1 1.1 1.1

0.9 0.9 0.9 0.9 0.9 0.9 0.9

1.1 1.1 1.1 1.1 1.1 1.1 1.1

0.9 0.9 0.9 0.9 0.9 0.9 0.9

Machines: Bus number 1 2 3 4 11

P gen (MW) 50 50 102.485 1 130 180

P max (MW) 50 50 150

P min (MW) 0 0 0

Q gen (Mvar) 27.2927 22.1775 49.7330

Q max (Mvar) 30 30 50

Q min (Mvar) -30 -30 -50

M base (MVA) 100 100 100

X source (pu) 0.1 0.05 0.02

200 400

0 0

20.0000 39.8275

20 50

-20 -50

100 100

0.03333 0.0125

Loads: Bus number 5 6 7 8 9

P load (MW) 70 50 80 60 250

Q load (Mvar) 15 10 20 10 20

Branch From bus 5 5 6 7 5 7

To bus 6 7 8 8 9 9

Line R (pu) 0.001418 0.005671 0.006805 0.001985 0.002836 0.003970

Line X (pu) 0.017 0.07940 0.090740 0.022680 0.051040 0.051040

6 8

10 10

0.003403 0.004537

0.005671 0.005671

2 winding transformers: From bus

To bus

Specified X (pu)

Winding (MVA) base

1 2 3 4 11

5 6 7 8 10

0.1 0.1 0.05 0.05 0.05

100 100 100 100 100

Winding 1 Nominal (kV) 13.8 13.8 13.8 13.8 13.8

Winding 2 Nominal (kV) 230 230 230 230 230

Rating (MVA) 100 100 200 200 200

Y admittance matrix 11, 11, 10, 10, 10, 10, 9, 9, 9, 4, 4, 2, 2, 1, 1, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8, 7, 7, 7, 7, 7, 6,

11, 0.00000000000000 10, -0.00000000000000 10, 163.817749023438 11, -0.00000000000000 8, -86.0182647705078 6, -77.7994842529297 9, 2.60007083415985 5, -1.08529078960419 7, -1.51478004455566 4, 0.00000000000000 8, -0.00000000000000 2, 0.00000000000000 6, -0.00000000000000 1, 0.00000000000000 5, -0.00000000000000 5, 6.85293352603912 9, -1.08529078960419 1, -0.00000000000000 7,-0.894970655441284 6, -4.87267208099365 8, 90.6697804331779 10, -86.0182647705078 4, -0.00000000000000 7, -3.82966136932373 6,-0.821854293346405 7, 6.23941206932068 9, -1.51478004455566 5,-0.894970655441284 8, -3.82966136932373 3, -0.00000000000000 6, 83.4940106272697

, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

-20.0000000000000 20.0000000000000 -257.168655395508 20.0000000000000 107.518096923828 129.650558471680 -39.0068264007568 19.5321731567383 19.4746532440186 -20.0000000000000 20.0000000000000 -10.0000000000000 10.0000000000000 -10.0000000000000 10.0000000000000 -100.479797363281 19.5321731567383 10.0000000000000 12.5305366516113 58.4170875549316 -182.233490943909 107.518096923828 20.0000000000000 43.7565307617188 10.9588632583618 -95.7617206573486 19.4746532440186 12.5305366516113 43.7565307617188 20.0000000000000 -209.026509284973

6, 6, 6, 6,

Power Flow Diagram

10, -77.7994842529297 2, -0.00000000000000 5, -4.87267208099365 8,-0.821854293346405

, , , ,

129.650558471680 10.0000000000000 58.4170875549316 10.9588632583618

Discussions

For scenario 3, the problem of lack of power supply in scenario 2 can be solved by building a new power plant. All generators can operate under their maximum capacity with all loads operating. Generator 3 expected to generate 102.4851 MW of real power and 49.733 Mvar reactive power which are less than its limits. Therefore, the proposed new power plant can fulfill the needs of new power system.

For scenario 1, the generators can support all loads given without having to run at maximum capacity. If the generators are generating more power than needed, this means the excessive power will go to the swing bus and thus the power is wasted. One can check this result with the software by setting generators 1,2 and 4 at maximum capacity. As a result, a negative sign will be shown at bus 3. Meanwhile, power system at scenario 2 has a problem with overloading.

Line impedances is also important for a power system. The higher the impedance value, the more power losses in the line, thus reducing efficiency. The length of the transmission line affects the power losses. The longer the transmission length, the higher the power losses. Therefore, the utility company usually use the short route to transmit the electricity. Keeping them closer is a good idea to reduce power losses in the transmission line cables.

In real life, the generators are usually not run at maximum capacity. They usually run at lower rate than its limit. This is very crucial to ensure that the generators can operate for longer time. Besides that, the power plants are interconnected between each other with national grid network [3]. When a power plant is under maintenance, the particular area still can receive power supply as the other generators still producing and sending the power via the transmission lines. The power plants don’t operate independently as this will increase the operating costs.

There are some important factors for planning a power network. The most important is the reliability and the capacity of generators to supply enough power to the big loads. Usually the load demand in a particular area expands per year. So, a utility company must consider and do a forecast on the increasing number of electricity demands every year.

Conclusions

To plan, design, construct and operate a power system correctly, load flow study is necessary to ensure that power is supplied correctly to the load with good efficiency. It is more convenient to use a computer software for the calculations using Gauss Seidel method. The software also allows us to determine emergency parameters. By using this PSSE software, one or an engineer is able to save time and cost in order to perform accurate calculations. The generators are not operated at their maximum capacity so that they can be used for a long period of time. A generator also is not isolated with other generators as they should be interconnected with national grid network. Last but not least, the reliability of generator to handle such huge number of loads is very crucial when designing a good power system network.

References [1] Hadi Saadat, “Power flow solution” in Power System Analysis, 2th ed. New York: McGrawHill, 1999, ch. 6, pp. 208. [2] “Load Flow or Power Flow or Contigency Ranking and Evaluation” [Online]. Retrieved from: http://www.powerapps.org/PAES_LoadFlow.aspx [3] Abdullah M. Al-Shaalan, “Essential aspects of power system planning in developing countries”, Journal of King Saud University – Engineering Sciences, vol. 23, pp. 27-32,2011.