Program Development
Develop 3-Phase Radial Load-flow Program & set up Feeder Model
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)
- 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.
- 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
- 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
- Perform Radial Load-flow with Embedded Solar.
- 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
- 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.
- Perform Radial Load-flow with Embedded Solar & Energy Storage Units.
- Code for integration:
The maximum load reduction is 28.95kVA. the storage capacity of the storage system should be greater than 250 kJ.
- 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.
- 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
