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authorpriyanka2015-06-24 15:03:17 +0530
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-rwxr-xr-x83/CH6/EX6.3/example_6_3.sce64
-rwxr-xr-x83/CH6/EX6.3/result_example_6_3.txt42
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diff --git a/83/CH6/EX6.3/example_6_3.sce b/83/CH6/EX6.3/example_6_3.sce
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+//Chapter 6
+//Example 6.3
+//page 201
+//To find an approximate load flow solution
+clear;clc;
+
+/////////////////////////////////////////////////////////////////////////////////
+//Realdemand Reactive demand Real generation Reactive generation Bus
+/////////////////////////////////////////////////////////////////////////////////
+ Pd1=1; Qd1=0.5; Pg1=0; Qg1=0;//initialization 1
+ Pd2=1; Qd2=0.4; Pg2=4; Qg2=0;//initialization 2
+ Pd3=2; Qd3=1; Pg3=0; Qg3=0;//initialization 3
+ Pd4=2; Qd4=1; Pg4=0; Qg4=0;//initialization 4
+
+Pg1=Pd1+Pd2+Pd3+Pd4-Pg2;
+
+//Ybus matrix from the network
+Ybus=[-21.667*%i 5*%i 6.667*%i 10*%i;
+ 5*%i -21.667*%i 10*%i 6.667*%i;
+ 6.667*%i 10*%i -16.667*%i 0;
+ 10*%i 6.667*%i 0 -16.667*%i];
+printf('Ybus matrix of the system is given by \nYbus=');disp(Ybus);
+//as given in the text book using approximate load flow equations and simplifying (ii),(iii),(iv)
+//delta matrix(x) is of the from A*x=B
+A=[-5 21.667 -10 -6.667;
+ -6.667 -10 16.667 0;
+ -10 -6.667 0 16.667
+ 1 0 0 0];
+
+B=[3; -2; -2;0];
+
+delta=inv(A)*B; //solving for delta
+printf('\nDelta of the system is given by \ndelta(rad)=');disp(delta);
+
+Q1=-5*cos(delta(2,1))-6.667*cos(delta(3,1))-10*cos(delta(4,1))+21.667;
+Q2=-5*cos(delta(2,1))-10*cos(delta(3,1)-delta(2,1))-6.667*cos(delta(4,1)-delta(2,1))+21.667;
+Q3=-6.667*cos(delta(3,1))-10*cos(delta(3,1)-delta(2,1))+16.667;
+Q4=-10*cos(delta(4,1))-6.667*cos(delta(4,1)-delta(2,1))+16.667;
+
+Q=[Q1;Q2;Q3;Q4];
+printf('\nInjected reactive power at the buses is given by \nQi(in pu)=');disp(Q);
+
+Qg1=Q1+Qd1;
+Qg2=Q2+Qd2;
+Qg3=Q3+Qd3;
+Qg4=Q4+Qd4;
+
+Qg=[Qg1;Qg2;Qg3;Qg4];
+printf('\n Reactive power generation at the four buses are \nQgi(in pu)=');disp(Qg);
+Qd=[Qd1;Qd2;Qd3;Qd4];
+Ql=sum(Qg)-sum(Qd);
+printf('\nReactive power losses are QL=%0.5f pu',Ql);
+
+printf('\n\nLine Flows are given as:\n');
+P13=(abs(Ybus(1,3)))*sin(delta(1,1)-delta(3,1));P31=-P13;printf('\nP13=-P31=%0.3f pu',P13);
+P12=(abs(Ybus(1,2)))*sin(delta(1,1)-delta(2,1));P21=-P12;printf('\nP12=-P21=%0.3f pu',P12);
+P14=(abs(Ybus(1,4)))*sin(delta(1,1)-delta(4,1));P41=-P14;printf('\nP14=-P41=%0.3f pu',P14);
+
+Q13=abs(Ybus(1,3))-(abs(Ybus(1,3)))*cos(delta(1,1)-delta(3,1));Q31=-Q13;printf('\n\nQ13=-Q31=%0.3f pu',Q13);
+Q12=abs(Ybus(1,2))-(abs(Ybus(1,2)))*cos(delta(1,1)-delta(2,1));Q21=-Q12;printf('\nQ12=-Q21=%0.3f pu',Q12);
+Q14=abs(Ybus(1,4))-(abs(Ybus(1,4)))*cos(delta(1,1)-delta(4,1));Q41=-Q14;printf('\nQ14=-Q41=%0.3f pu',Q14);
+
+
+ \ No newline at end of file
diff --git a/83/CH6/EX6.3/result_example_6_3.txt b/83/CH6/EX6.3/result_example_6_3.txt
new file mode 100755
index 000000000..ec1cc1025
--- /dev/null
+++ b/83/CH6/EX6.3/result_example_6_3.txt
@@ -0,0 +1,42 @@
+
+-->exec('/home/samyak/scilab_working_directory/chapter_6/ex_3/example_6_3.sce', -1)
+Ybus matrix of the system is given by
+Ybus=
+ - 21.667i 5.i 6.667i 10.i
+ 5.i - 21.667i 10.i 6.667i
+ 6.667i 10.i - 16.667i 0
+ 10.i 6.667i 0 - 16.667i
+
+Delta of the system is given by
+delta(rad)=
+ 0.
+ 0.0769217
+ - 0.0738455
+ - 0.0892280
+
+Injected reactive power at the buses is given by
+Qi(in pu)=
+ 0.0727367
+ 0.2200358
+ 0.1316085
+ 0.1315939
+
+ Reactive power generation at the four buses are
+Qgi(in pu)=
+ 0.5727367
+ 0.6200358
+ 1.1316085
+ 1.1315939
+
+Reactive power losses are QL=0.55597 pu
+
+Line Flows are given as:
+
+P13=-P31=0.492 pu
+P12=-P21=-0.384 pu
+P14=-P41=0.891 pu
+
+Q13=-Q31=0.018 pu
+Q12=-Q21=0.015 pu
+Q14=-Q41=0.040 pu
+-->diary(0);