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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

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.

1. 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

1.        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

Reference

Conference Board of Canada. (2004). Electricity restructuring: opening power markets. Ottawa, Conference Board of Canada.

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; 14th October, 2018

World Bank. (2013). Vietnam Power Sector Generation Options. DC, Washington. Available from; https://hdl.handle.net/10986/12860. Date of Access: 14th October 2018

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