Radial Distribution System Forward Backward Sweep Algorithm

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.

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 

5.        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; 8^th 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; 8^th 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; 14^th October, 2018

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