Get Instant Help From 5000+ Experts For

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost

a) Develop a radial load flow program and set up a uniformly distributed load feeder model of 10 equal loads.  Total feeder load = 100 kVA (3-phase), at peak, operating at 0.9 power factor lagging.  Total feeder length = 500 m.  The cable impedance = 0.315 +j0.365 ohms/phase/kilometre.  The peak load refers to the residential load curve at about 5 pm (refer Addendum).

b)Run your load flow.  Are current and voltage constraints being exceeded?  Current constraint = 200 A/phase; voltage constraint = 415 V +/- 5%.   If the supply end voltage was raised above 415 V would this solve any peak load?

c)Now, run the load flow for light-load conditions = 30% of peak load conditions (refer to the residential load curve at about 11 am).  Are voltage constraints being exceeded?  Record all your feeder current (supply end) and voltage drops. If the supply volts is raised as in 1(b) are voltages too high in light-load conditions?

d)The load at node #10 is a large induction motor and is a fluctuating load.  Its starting current is 5 times its normal running load (as in part 1(b)) and starting power factor is 0.4 lagging.  Re-run your load flow and determine the change in voltage drop, at both node #9 and node #10.  Is the voltage drop at node #10 > 10%?  Is the voltage drop at node #9 (the “common point of coupling”) > 4%?

## Program Development

Linecurrent(upperNode,IntermediateNodes(i,1))=(conj(complex(busdata(IntermediateNodes(i,1),2),busdata(IntermediateNodes(i,1),3))/Busvoltage(IntermediateNodes(i,1),1)));%current equal to load or generation connected to intermediate node

for i1=1:size(Connection,1)%current to intermediate node is equal to currents going from inter node to all connected downstream nodes + current of node itself if it exists

if Connection(IntermediateNodes(i,1),i1)~=0 && i1~=IntermediateNodes(i,1)

Linecurrent(upperNode,IntermediateNodes(i,1))=Linecurrent(upperNode,IntermediateNodes(i,1))+Linecurrent(IntermediateNodes(i,1),i1);

%if converged break

if Busvoltage(referencebus,1)<=1+epsilon && Busvoltage(referencebus,1)>=1-epsilon

converged=1;

end

if converged==1

break;

end

%%%%%%%%%%%%%%%%%

%%forward sweep%%

%%%%%%%%%%%%%%%%%

Busvoltage(referencebus,1)=1;

for i=1:size(Connection,1)

for j=1:size(Connection,1)

if Connection(i,j)~=0

Busvoltage(j,1)=Busvoltage(i,1)-Linecurrent(i,j)*Connection(i,j);

end

end

end

iteration=iteration+1;

end

%%

%%%%%%%%%%

%%losses%%

%%%%%%%%%%

for i=1:size(linedata,1)

PowerLoss(i,1)=linedata(i,1);%from bus

PowerLoss(i,2)=linedata(i,2); %to bus

PowerLoss(i,3)=Busvoltage(linedata(i,1))*conj(Linecurrent(linedata(i,1),linedata(i,2)))+Busvoltage(linedata(i,2))*conj(-1*Linecurrent(linedata(i,1),linedata(i,2))); %and the loss in p.u.

end

currents=[];

b=1;

f=2;

for a=1:9;

currents(b,1)=Linecurrent(a,f);

b=b+1;

f=f+1;

end

currents=[currents;0+0*j];

Phase_currents=415.^2/1e5.*abs(currents)

Busvoltage=415.*abs(Busvoltage)

1. Results of the program:

Phase_currents =

2.3073

2.1041

1.8863

1.6536

1.4061

1.1447

0.8708

0.5866

0.2952

0

Busvoltage =

1.0e+03 *

0.4150

0.2741

0.3329

0.4933

0.6616

0.8126

0.9383

1.0350

1.1006

1.1336

If the supply end voltage was raised above 415 V would this solve any peak load?

It would not solve the problem. Since a larger current will be flowing the system.

1. Yes the constraints are exceeded.

Phase_currents =

1.7856

1.6229

1.4499

1.2667

1.0737

0.8716

0.6614

0.4448

0.2236

0

Busvoltage =

415.0000

322.9582

341.0636

431.7459

541.6766

646.1511

735.3562

804.8060

852.1489

876.1086

1. Results:

Re-run your load flow and determine the change in voltage drop, at both node #9 and node #10.  Is the voltage drop at node #10 > 10%?  Is the voltage drop at node #9 (the “common point of coupling”) > 4%?

Yes the voltage drop at node number 9 is greater than 4 percent. The voltage drop is less than 10 percent.

Phase_currents =

4.6005

4.4048

4.1757

3.9024

3.5691

3.1530

2.6215

1.9365

1.1133

0

Busvoltage =

1.0e+03 *

0.4150

0.8911

1.3887

1.8684

2.3196

2.7336

3.0998

3.4042

3.6273

3.7457

2. Phase_currents =

1.7747

1.6129

1.4418

1.2614

1.0723

0.8755

0.6729

0.4689

0.2769

0

Busvoltage =

1.0e+03 *

0.4150

0.5044

0.6333

0.7679

0.8936

1.0035

1.0940

1.1628

1.2083

1.2295

1. Phase_currents =

2.7638

2.5396

2.2875

2.0104

1.7129

1.4004

1.0788

0.7554

0.4500

0

Busvoltage =

1.0e+03 *

0.4150

0.3222

0.4601

0.6686

0.8717

1.0497

1.1954

1.3057

1.3784

1.4123

From the results the voltage drop across the feeder reduces because of the low flow of current. The voltage profile across the feeder depreciates with time. This is because the supply of power from the embedded system decreases with time.

2. Code for integration:

The maximum load reduction is 28.95kVA. the storage capacity of the storage system should be greater than 250 kJ.

1. Results for distributed storage facility:

Phase_currents =

4.0221

3.8197

3.5863

3.3139

2.9921

2.6078

2.1461

1.5946

0.9582

0

Busvoltage =

1.0e+03 *

0.4150

0.4779

0.8308

1.2193

1.5925

1.9347

2.2354

2.4841

2.6693

2.7796

Observation.

It can be observed that the voltage profile the feeder improved.

1. Results for storage facility at different location.

Node 1:

Phase_currents =

3.8065

3.5956

3.3540

3.0749

2.7499

2.3692

1.9235

1.4078

0.8328

0

Busvoltage =

1.0e+03 *

0.4150

0.3739

0.6870

1.0489

1.3953

1.7100

1.9833

2.2063

2.3698

2.4656

Node 2:

Phase_currents =

3.8084

3.6445

3.4036

3.1248

2.7991

2.4163

1.9660

1.4421

0.8546

0

Busvoltage =

1.0e+03

0.4150

0.3889

0.7062

1.0728

1.4245

1.7447

2.0234

2.2512

2.4187

2.5171

Node 7:

Phase_currents =

4.0747

3.8691

3.6318

3.3544

3.0262

2.6336

2.2644

1.6888

1.0151

0

Busvoltage =

1.0e+03 *

0.4150

0.5018

0.8642

1.2587

1.6369

1.9833

2.2873

2.5492

2.7454

2.8625

Node 10:

Phase_currents =

4.4008

4.2013

3.9687

3.6934

3.3615

2.9539

2.4453

1.8093

1.0516

0

Bus voltage =

1.0e+03 *

0.4150

0.6778

1.1075

1.5474

1.9671

2.3533

2.6949

2.9787

3.1887

3.3070

Conclusion

From the results the best position for the storage unit is at node 10 since it causes the profile of the feeder network to be almost flat.

From the simulation results it can be observed that the voltage profile in a LV feeder depends on the impedance of the cables, the power drawn by the loads connected to the feeder.  Embedded solar-PV and battery storage units helps to almost maintain a flat load curve. The storage units should be placed at the extreme node of the feeder away from the source. It will make the system to appear to be in a ring hence the effects of peak loading of the feeder reduced.

Reference

Grigsby, L. L. (2012). Power system stability and control. Boca Raton, Taylor & Francis. Available from; https://www.crcnetbase.com/isbn/9781439883204. Date of Access; 8th September, 2018

Musirin, I., & Sulaiman, S. I. (2015). Recent trends in power engineering: selected, peer reviewed papers from the 2015 9th International Power Engineering and Optimization Conference (PEOCO 2015), March 18-19, 2015, Melaka, Malaysia. Pfaffikon, Trans Tech Publications. Available from; https://search.ebscohost.com/login.aspx?direct=true&scope=site&db=nlebk&db=nlabk&AN=1060570. Date of Access; 8th September, 2018

Nagsarkar, T. K., & Sukhija, M. S. (2016). Power System Analysis: Power System Analysis. New Delhi, Oxford University Press India. Available from; https://app.knovel.com/hotlink/toc/id:kpPSAE0003/power-system-analysis. Date of Access; 8th September, 2018

Cite This Work

"Develop 3-Phase Radial Load-flow Program & Set Up Feeder Model." My Assignment Help, 2021, https://myassignmenthelp.com/free-samples/300197-power-systems-planning-and-economics/feeder-reduced.html.

My Assignment Help (2021) Develop 3-Phase Radial Load-flow Program & Set Up Feeder Model [Online]. Available from: https://myassignmenthelp.com/free-samples/300197-power-systems-planning-and-economics/feeder-reduced.html
[Accessed 07 August 2024].

My Assignment Help. 'Develop 3-Phase Radial Load-flow Program & Set Up Feeder Model' (My Assignment Help, 2021) <https://myassignmenthelp.com/free-samples/300197-power-systems-planning-and-economics/feeder-reduced.html> accessed 07 August 2024.

My Assignment Help. Develop 3-Phase Radial Load-flow Program & Set Up Feeder Model [Internet]. My Assignment Help. 2021 [cited 07 August 2024]. Available from: https://myassignmenthelp.com/free-samples/300197-power-systems-planning-and-economics/feeder-reduced.html.

Get instant help from 5000+ experts for

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost