1. Solve the load flow for power system data given in Figure 1 and Tables I & II. The base power is Sbase=100 MVA. From the load flow analysis, calculate the power losses in the transmission lines and the total system losses. Hint: bus 1 is slack bus, bus 2 is PV bus, and the rest of buses are PQ bus.
2) Find the point of minimum system losses for phase shift transformer a34 within ±20° of zero phase shift (i.e. a=ejπθ/180 , -20°<θ<20°) by adjusting a34 from unity tap setting (i.e. a=ej0=1) by steps of ±1° shifts.
Section One:
Calculations:
Section One:
Code:
clear all;close all;clc;
S=1e5; %%base power for the power system.
% BusNo RealPower(pu) ReactivePower(pu) bdata stands for bus data
bdata=[ 1 0 0
2 0.55 0.13
3 0.1 0.2
4 0.30 0.18
5 0.5 0.5
6 0.4 0.7
7 0.55 0.8
];
% Inbus Outbus Resistance(pu) Reactance(pu) ldata stands for line data
ldata=[ 1 4 0.08 0.37
1 6 0.123 0.5180
2 3 0.7230 1.05
2 5 0.282 0.64
3 4 0.05 0.133
4 6 0.097 0.407
5 6 0 0.30
];
%Forward Backward Sweep algorithm for Radial Distribution System
%function [] = radialforwardbackwardsweep ()
c = input('Enter the central bus no.: ');
Iterations = input('Enter the no. of iterations : ');
% ldat;
% bdat;
sizl=size(ldata);
sizb=size(bdata);
totalbus=sizb(1,1); %total no. of buses
%Find Terminal Buses by using the fact that a Terminal bus is never an inbus
for i=1:totalbus-1
Terminalbuses(i,1)=0;
Intermediatebuses(i,1)=0;
end
co=0;in=0;
for i=1:sizl(1,1)
for j=1:sizl(1,1)
if (i~=j)&&(ldata(i,2)==ldata(j,1))
co=co+1;
end
end
if co==0
in=in+1;
Terminalbuses(in,1)=ldata(i,2);
else
in=in+1;
Intermediatebuses(in,1)=ldata(i,2);
end
co=0;
end
%Initializing Voltage and Current matrices
for i=1:totalbus
Busvoltage(i,1)=0;
end
for i=1:sizl(1,1)
Linecurrent(i,1)=0;
end
for i=1:totalbus-1
if(Terminalbuses(i,1)~=0)
Busvoltage(Terminalbuses(i,1),1)=1;
end
end
for i=1:Iterations
%forward sweep
t=0;int=0;
for i=1:sizl(1,1)
for j=1:totalbus-1
if ldata(i,2)== Terminalbuses(j,1)
t=t+1;
end
end
if(t~=0)
Linecurrent(i,1)=conj(complex(bdata(ldata(i,2),2),bdata(ldata(i,2),3))/Busvoltage(ldata(i,2),1));
end
t=0;
end
for i=1:sizl(1,1)
for j=1:totalbus-1
if ldata(i,2) == Intermediatebuses(j,1)
int=int+1;
end
end
if(int~=0)
Busvoltage(ldata(i,2),1)=Busvoltage(ldata(i+1,2),1)+((Linecurrent(i+1,1))*(complex(ldata(i+1,3),ldata(i+1,4))));
Linecurrent(i,1)=(conj(complex(bdata(ldata(i,2),2),bdata(ldata(i,2),3))/Busvoltage(ldata(i,2),1)))+ Linecurrent(i+1,1);
end
int=0;
end
%backward sweep
Busvoltage(c,1)=1;
for i=1:sizl(1,1)
if (ldata(i,1)==c)
Busvoltage(ldata(i,2),1)=Busvoltage(c,1)-(Linecurrent(i,1)*(complex(ldata(i,3),ldata(i,4))));
else
Busvoltage(ldata(i,2),1)=Busvoltage(ldata(i,1),1)-(Linecurrent(i,1)*(complex(ldata(i,3),ldata(i,4))));
end
end
end
Busvoltage=Busvoltage.*415;
Linecurrent=abs(Linecurrent).*1400;
display(Linecurrent);
Busvoltage=abs(Busvoltage);
display(Busvoltage);
display(Iterations);
% end
Results:
Power losses in per unit =
1.0e-01 *
2.1831
1.8749
0.2857
0.9068
2.2290
1.8749
1.8749
Total losses= 0.17589 MVA
Section Two:
Code:
Results:
power_losses =
1.0e+03 *
2.6294
2.2041
2.3885
2.1508
2.6988
2.2041
2.2041
Busvoltage =
0
415.0000
768.8354
683.1217
98.2750
277.3165
0
Iterations =
3
Total losses =0.2538MVA.
- Simulation:
Section One:
The IEEE 6 bus power system was constructed on matlab using Simpower tab.
The component values were adjusted to values as given in table I and II. We then added the power module to the Simulink model.
To run the simulation we double clicked on the powergui>>Tools>>Load flow>>compute>>generate report.
Results:
The results were obtained as given below:
The Newto- Raphson method principles was adopted in iterations to achieve the results.
The Load Flow converged in 2 iterations!
SUMMARY for sub network No 1
Total generation: P= 17.91 MW Q= 6.71 Mvar
Total PQ load : P= -0.00 MW Q= -0.00 Mvar
Total Zshunt load: P= 17.90 MW Q= 6.33 Mvar
Total ASM load : P= 0.00 MW Q= 0.00 Mvar
Code:
Total losses : P= 0.01 MW Q= 0.37 Mvar
1: generator bus V= 1.000 pu/11kV 0.00 deg ; Swing bus
Generation: P= 14.11 MW Q= 0.48 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= -0.00 MW Q= 0.00 Mvar
--> Zone A : P= 14.11 MW Q= 0.48 Mvar
2: zone A V= 0.999 pu/11kV -1.21 deg
Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= -0.00 MW Q= -0.00 Mvar
Z_shunt : P= 4.74 MW Q= 1.56 Mvar
--> Generator bus: P= -14.10 MW Q= -0.18 Mvar
--> Zone B : P= 9.36 MW Q= -1.38 Mvar
3: zone B V= 1.000 pu/11kV -1.62 deg
Generation: P= 1.90 MW Q= 3.74 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.58 MW Q= 2.39 Mvar
--> Zone A : P= -9.36 MW Q= 1.44 Mvar
--> Zone C : P= 4.68 MW Q= -0.09 Mvar
4: zone C V= 1.000 pu/11kV -1.72 deg
Generation: P= 1.90 MW Q= 2.49 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.58 MW Q= 2.39 Mvar
--> Zone B : P= -4.68 MW Q= 0.10 Mvar
5: zone B V= 1.000 pu/11kV -1.62 deg
Generation: P= 1.90 MW Q= 3.74 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.58 MW Q= 2.39 Mvar
--> Zone A : P= -9.36 MW Q= 1.44 Mvar
--> Zone C : P= 4.68 MW Q= -0.09 Mvar
6: zone C V= 1.000 pu/11kV -1.72 deg
Generation: P= 1.90 MW Q= 2.49 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.58 MW Q= 2.39 Mvar
--> Zone B : P= -4.68 MW Q= 0.10 Mvar
Section Two:
To obtain the minimum point of system power loss the center tap of transformers one and two were varied according the questions. We then ran the simulation as explained in section one.
The results were obtained as given below:
Results:
The Load Flow converged in 1 iteration!
SUMMARY for subnetwork No 1
Total generation: P= 17.44 MW Q= 6.85 Mvar
Total PQ load : P= 0.00 MW Q= -0.00 Mvar
Total Zshunt load: P= 17.43 MW Q= 6.16 Mvar
Results:
Total ASM load : P= 0.00 MW Q= 0.00 Mvar
Total losses : P= 0.01 MW Q= 0.69 Mvar
1: generator bus V= 1.000 pu/11kV 0.00 deg; Swing bus
Generation: P= 17.44 MW Q= 6.85 Mvar
PQ_load : P= 0.00 MW Q= 0.00 Mvar
Z_shunt : P= -0.00 MW Q= 0.00 Mvar
--> Zone A : P= 17.44 MW Q= 6.85 Mvar
2: zone A V= 0.990 pu/11kV -1.50 deg
Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= 0.00 MW Q= -0.00 Mvar
Z_shunt : P= 4.65 MW Q= 1.53 Mvar
--> Generator bus: P= -17.43 MW Q= -6.33 Mvar
--> Zone B : P= 12.78 MW Q= 4.80 Mvar
3: zone B V= 0.986 pu/11kV -2.06 deg
Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= 0.00 MW Q= -0.00 Mvar
Z_shunt : P= 6.39 MW Q= 2.32 Mvar
--> Zone A : P= -12.78 MW Q= -4.66 Mvar
--> Zone C : P= 6.38 MW Q= 2.33 Mvar
4: zone C V= 0.985 pu/11kV -2.21 deg
Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= -0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.38 MW Q= 2.32 Mvar
--> Zone B : P= -6.38 MW Q= -2.32 Mvar
- Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= 0.00 MW Q= -0.00 Mvar
Z_shunt : P= 6.39 MW Q= 2.32 Mvar
--> Zone A : P= -12.78 MW Q= -4.66 Mvar
--> Zone C : P= 6.38 MW Q= 2.33 Mvar
6: zone C V= 0.985 pu/11kV -2.21 deg
Generation: P= 0.00 MW Q= 0.00 Mvar
PQ_load : P= -0.00 MW Q= 0.00 Mvar
Z_shunt : P= 6.38 MW Q= 2.32 Mvar
--> Zone B : P= -6.38 MW Q= -2.32 Mvar
Conclusion
It was observed that the total active and reactive power loss was 0.01 and 0.37 respectively. From the report it can also be observed that the voltage was about -1.72 across the power system which meets the demands. From the analysis it can be concluded that;
Different faults bring different fault currents; single line to ground double line to ground, line to line and symmetrical 3 phase.
Position of the fault affects the fault current in the system.
Newto- Raphson method principles was adopted in iterations to achieve the results.
Change in resistances or reactance in the system affect the fault current in the system.
Power lost in transmission line
Zone A: P= -9.36 MW
Zone B: P= -4.68 MW
Total losses: P= 0.01 MW
The results for section two can be summarized as given below:
|
0.9 p.f. lagging |
||||||
|
bus |
PR |
QR |
VR (pu) |
δ (deg) |
PS |
QS |
|
(MW) |
(MVAr) |
(MW) |
(MVAr) |
|||
|
1-4 |
0 |
0.05 |
1.01 |
-0.03 |
0 |
-14.7 |
|
1-6 |
50 |
-24.22 |
0.98 |
-2.34 |
50.24 |
12.3 |
|
4-6 |
99.99 |
-48.53 |
0.95 |
-4.8 |
101.07 |
45.29 |
|
4-3 |
150 |
-76.65 |
0.91 |
-7.46 |
152.73 |
90.55 |
|
3-2 |
200 |
-96.85 |
0.88 |
-10.39 |
205.19 |
135.87 |
|
2-5 |
249.9 |
-121.01 |
0.83 |
-13.79 |
259.12 |
200.11 |
|
5-6 |
300 |
-145.3 |
0.77 |
-18.07 |
315.55 |
289.48 |
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