initial commit / add all books

12 files changed, 264 insertions, 0 deletions

diff --git a/83/CH2/EX2.1/example_2_1.sce b/83/CH2/EX2.1/example_2_1.sce
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index 000000000..a05dcb8c0
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@@ -0,0 +1,16 @@
+//Chapter 2
+//Example 2.1
+//page 56
+//To find GMD of the conductor
+//From the given the text book,leaving out the factor of "r",we have the seven possible distances
+clear;clc;
+D1=0.7788*2*2*(2*sqrt(3))*4*(2*sqrt(3))*2;
+//since there are 7 identical conductors,the above products remains same dor all D's
+D2=D1;
+D3=D1;
+D4=D1;
+D5=D1;
+D6=D1;
+D7=D1;
+Ds=(D1*D2*D3*D4*D5*D6*D7)^(1/(7*7));
+printf("\n GMD of the conductor is %0.4fr",Ds);
diff --git a/83/CH2/EX2.1/result_example_2_1.txt b/83/CH2/EX2.1/result_example_2_1.txt
new file mode 100755
index 000000000..0c9d80cff
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+++ b/83/CH2/EX2.1/result_example_2_1.txt
@@ -0,0 +1,3 @@
+
+
+ GMD of the conductor is 2.2578r 
diff --git a/83/CH2/EX2.2/example_2_2.sce b/83/CH2/EX2.2/example_2_2.sce
new file mode 100755
index 000000000..c8680e6b9
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+++ b/83/CH2/EX2.2/example_2_2.sce
@@ -0,0 +1,35 @@
+//Chapter 2
+//Example 2.2
+//page 57
+//To find reactance of the conductor
+clear;clc;
+f=50; //frequency
+D=5.04; //diameter of the entire ACSR
+d=1.68; //diameter of each conductor
+Dsteel=D-2*d; //diameter of steel strand
+//As shown in fig
+D12=d;
+D13=(sqrt(3)*d);
+D14=2*d;
+D15=D13;
+D16=D12;
+//neglecting the central sttel conductor,we have the 6 possibilities
+D1=(0.7788*d)*D12*D13*D14*D15*D16;
+//we have total of 6 conductors,hence
+D2=D1;
+D3=D1;
+D4=D1;
+D5=D1;
+D6=D1;
+Ds=(D1*D2*D3*D4*D5*D6)^(1/(6*6));//GMR;
+//since the spacing between lines is 1m=100cm
+l=100;
+L=0.461*log10(l/Ds); //Inductance of each conductor
+Ll=2*L; // loop inductance
+Xl=2*%pi*f*Ll*10^(-3);//reactance of the line
+printf("\n\nInductance of each conductor=%0.4f mH/km\n\n",L);
+printf("Loop Inductance=%0.4f mH/km\n\n",Ll);
+printf("Loop Reactance=%f ohms/km\n\n",Xl);
+
+
+
diff --git a/83/CH2/EX2.2/result_example_2_2.txt b/83/CH2/EX2.2/result_example_2_2.txt
new file mode 100755
index 000000000..68706e8c8
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+++ b/83/CH2/EX2.2/result_example_2_2.txt
@@ -0,0 +1,9 @@
+ 
+
+
+Inductance of each conductor=0.7667 mH/km
+
+Loop Inductance=1.5334 mH/km
+
+Loop Reactance=0.481723 ohms/km
+
diff --git a/83/CH2/EX2.3/example_2_3.sce b/83/CH2/EX2.3/example_2_3.sce
new file mode 100755
index 000000000..d3f3e39a3
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+++ b/83/CH2/EX2.3/example_2_3.sce
@@ -0,0 +1,48 @@
+//Chapter 2
+//Example 2.3
+//page 58
+//To find inductance of each side of the line and that of the complete line
+clear;clc;
+//to find mutual GMD
+D14=sqrt(8*8+2*2);
+D15=sqrt(8*8+6*6);
+D24=sqrt(8*8+2*2);
+D25=sqrt(8*8+2*2);
+D34=sqrt(8*8+6*6);
+D35=sqrt(8*8+2*2);
+//sixth root of six mutual distances
+Dm=(D14*D15*D24*D25*D34*D35)^(1/6);//mutual GMD between lines
+
+//to find GMR of Side A conductors
+D11=0.7788*2.5*10^(-3);
+D22=D11;
+D33=D11;
+D12=4;
+D21=D12;
+D13=8;
+D31=8;
+D23=4;
+D32=D23;
+//ninth root nine distances in Side A
+Da=(D11*D12*D13*D21*D22*D23*D31*D32*D33)^(1/9);
+
+//to find GMR of Side A conductors
+D44=0.7788*5*10^(-3);
+D45=4;
+D54=D45;
+D55=D44;
+//fourth root of four distances in Side B
+Db=(D44*D45*D54*D55)^(1/4);
+
+La=0.461*log10(Dm/Da);//inductance line A
+Lb=0.461*log10(Dm/Db);//inductance line B
+
+L=La+Lb; //loop inductance 
+
+printf("\n\nMutual GMD between lines = %0.4f m\n\n",Dm);
+printf("GMR of Side A conductors = %0.4f m\n\n",Da);
+printf("GMR of Side B conductors = %0.4f m\n\n",Db);
+printf("Inductance of line A = %0.4f mH/km\n\n",La);
+printf("Inductance of line B = %0.4f mH/km\n\n",Lb);
+printf("Loop Inductance of the lines = %0.4f mH/km\n\n",L);
+
diff --git a/83/CH2/EX2.3/result_example_2_3.txt b/83/CH2/EX2.3/result_example_2_3.txt
new file mode 100755
index 000000000..bec7c7d42
--- /dev/null
+++ b/83/CH2/EX2.3/result_example_2_3.txt
@@ -0,0 +1,12 @@
+
+Mutual GMD between lines = 8.7937 m
+
+GMR of Side A conductors = 0.3670 m
+
+GMR of Side B conductors = 0.1248 m
+
+Inductance of line A = 0.6359 mH/km
+
+Inductance of line B = 0.8519 mH/km
+
+Loop Inductance of the lines = 1.4878 mH/km
diff --git a/83/CH2/EX2.5/example_2_5.sce b/83/CH2/EX2.5/example_2_5.sce
new file mode 100755
index 000000000..dc380bdf8
--- /dev/null
+++ b/83/CH2/EX2.5/example_2_5.sce
@@ -0,0 +1,34 @@
+//Chapter 2
+//Example 2.5
+//page 63
+//To find flux linkages with neutral and voltage induced in neutral
+//To find voltage drop in each of three-phase wires
+
+clear;clc;
+Ia=-30+%i*50;
+Ib=-25+%i*55;
+Ic=-(Ia+Ib);
+
+//(a) to find flux linkages with neutral and voltage induce in it
+Dan=4.5;
+Dbn=3;  //from figure
+Dcn=1.5;
+Phi_n=2*10^(-7)*(Ia*log(1/Dan)+Ib*log(1/Dbn)+Ic*log(1/Dcn));
+Vn=%i*2*%pi*50*Phi_n*15000; //voltage induced for 15km long TL
+Vn=abs(Vn) ;
+printf("\nFlux linkages of the neutral wire = %f Wb-T/m\n\n",Phi_n);
+printf("Voltage induced in the neutral = %d\n\n",Vn);
+
+//(b) to find voltage drop in each phase
+Phi_a=2*10^(-7)*(Ia*log(1/(0.7788*0.005))+Ib*log(1/1.5)+Ic*log(1/3));
+Phi_b=2*10^(-7)*(Ib*log(1/(0.7788*0.005))+Ia*log(1/1.5)+Ic*log(1/1.5));
+Phi_c=2*10^(-7)*(Ic*log(1/(0.7788*0.005))+Ib*log(1/1.5)+Ia*log(1/3));
+
+delta_Va=%i*2*%pi*50*Phi_a*15000; //like we did for neutral voltage
+delta_Vb=%i*2*%pi*50*Phi_b*15000;
+delta_Vc=%i*2*%pi*50*Phi_c*15000;
+
+printf("The Voltage drop of phase a(in volts) =");disp(delta_Va);
+printf("\n\nThe Voltage drop of phase b(in volts) =");disp(delta_Vb);
+printf("\n\nThe Voltage drop of phase c(in volts) =");disp(delta_Vc);
+
diff --git a/83/CH2/EX2.5/result_example_2_5.txt b/83/CH2/EX2.5/result_example_2_5.txt
new file mode 100755
index 000000000..1b290864f
--- /dev/null
+++ b/83/CH2/EX2.5/result_example_2_5.txt
@@ -0,0 +1,15 @@
+
+Flux linkages of the neutral wire = 0.000010 Wb-T/m
+
+Voltage induced in the neutral = 99
+
+The Voltage drop of phase a(in volts) = 
+  - 349.1594 - 204.26943i  
+
+
+The Voltage drop of phase b(in volts) = 
+  - 308.62198 - 140.28272i  
+
+
+The Voltage drop of phase c(in volts) = 
+    621.85121 + 328.22026i  
diff --git a/83/CH2/EX2.6/example_2_6.sce b/83/CH2/EX2.6/example_2_6.sce
new file mode 100755
index 000000000..8ec5b5006
--- /dev/null
+++ b/83/CH2/EX2.6/example_2_6.sce
@@ -0,0 +1,13 @@
+//Chapter 2
+//Example 2.6
+//page 65
+//To find mutual inductance between power line and telephone line and voltage induced in telephone line
+
+clear;clc;
+D1=sqrt(1.1*1.1+2*2); //from figure 2.14
+D2=sqrt(1.9*1.9+2*2);  //from figure 2.14
+Mpt=0.921*log10(D2/D1);  //mutual inductance
+Vt=abs(%i*2*%pi*50*Mpt*10^(-3)*100);//when 100A is flowing in the power lines
+
+printf("\n\nMutual inductance between power line and telephone line = %f mH/km\n\n",Mpt);
+printf("\n\nVoltage induced in the telephone circuit = %.3f V/km\n\n",Vt);
\ No newline at end of file
diff --git a/83/CH2/EX2.6/result_example_2_6.txt b/83/CH2/EX2.6/result_example_2_6.txt
new file mode 100755
index 000000000..cfce102d1
--- /dev/null
+++ b/83/CH2/EX2.6/result_example_2_6.txt
@@ -0,0 +1,8 @@
+ 
+
+
+Mutual inductance between power line and telephone line = 0.075774 mH/km
+
+
+
+Voltage induced in the telephone circuit = 2.381 V/km
diff --git a/83/CH2/EX2.7/example_2_7.sce b/83/CH2/EX2.7/example_2_7.sce
new file mode 100755
index 000000000..7d7a737fe
--- /dev/null
+++ b/83/CH2/EX2.7/example_2_7.sce
@@ -0,0 +1,43 @@
+//Chapter 2
+//Example 2.7
+//page 69
+//To find inductive reactance of for the three phase bundled conductors
+clear;clc;
+r=0.01725; //radius of each conductor
+//from the figure we can declare the distances
+d=7;
+s=0.4;
+//Mutual GMD between bundles of phases a and b
+Dab=(d*(d+s)*(d-s)*d)^(1/4);
+//Mutual GMD between bundles of phases b and c
+Dbc=Dab ; //by symmetry
+//Mutual GMD between bundles of phases c and a
+Dca=(2*d*(2*d+s)*(2*d-s)*2*d)^(1/4);
+//Equivalent GMD is calculated as
+Deq=(Dab*Dbc*Dca)^(1/3);
+//self GMD is given by
+Ds=(0.7788*1.725*10^(-2)*0.4*0.7788*1.725*10^(-2)*0.4)^(1/4);
+//Inductive reactance per phase is given by
+Xl=2*%pi*50*10^(-3)*0.461*log10(Deq/Ds); //10^(-3) because per km is asked
+printf("\n\nMutual GMD between bundles of phases a and b = %0.3fm\n\n",Dab);
+printf("Mutual GMD between bundles of phases b and c = %0.3fm\n\n",Dbc);
+printf("Mutual GMD between bundles of phases c and a = %0.3fm\n\n",Dca);
+printf("Equivalent GMD = %0.3fm\n\n",Deq);
+printf("Self GMD of the bundles = %0.3fm\n\n",Ds);
+printf("Inductive reactance per phase = %0.3f ohms/km\n\n",Xl);
+
+//now let us compute reactance when center to centerr distances are used
+Deq1=(d*d*2*d)^(1/3);
+Xl1=2*%pi*50*0.461*10^(-3)*log10(Deq1/Ds);
+printf("\n When radius of conductors are neglected and only distance between conductors are used, we get below results:\n\n");
+printf("Equivalent mean distance is = %f\n\n",Deq1);
+printf("Inductive reactance per phase = %0.3f ohms/km\n\n",Xl1);
+
+//when bundle of conductors are replaced by an equivalent single conductor 
+cond_dia=sqrt(2)*1.725*10^(-3); //conductor diameter for same cross-sectional area
+Xl2=2*%pi*50*0.461*10^(-3)*log10(Deq1/cond_dia);
+printf("\nWhen bundle of conductors are replaced by an equivalent single conductor:\n\n");
+printf("Inductive reactance per phase = %0.3f ohms/km\n\n",Xl2) ;
+percentage_increase=((Xl2-Xl1)/Xl1)*100;
+printf("This is %0.2f higher than corresponding value for a bundled conductor line.",percentage_increase);
+
diff --git a/83/CH2/EX2.7/results_example_2_7.txt b/83/CH2/EX2.7/results_example_2_7.txt
new file mode 100755
index 000000000..ab9b3234a
--- /dev/null
+++ b/83/CH2/EX2.7/results_example_2_7.txt
@@ -0,0 +1,28 @@
+ 
+
+
+Mutual GMD between bundles of phases a and b = 6.994m
+
+Mutual GMD between bundles of phases b and c = 6.994m
+
+Mutual GMD between bundles of phases c and a = 13.997m
+
+Equivalent GMD = 8.814m
+
+Self GMD of the bundles = 0.073m
+
+Inductive reactance per phase = 0.301 ohms/km
+
+
+ When radius of conductors are neglected and only distance between conductors are used, we get below results:
+
+Equivalent mean distance is = 8.819447
+
+Inductive reactance per phase = 0.301 ohms/km
+
+
+When bundle of conductors are replaced by an equivalent single conductor:
+
+Inductive reactance per phase = 0.515 ohms/km
+
+This is 71.04 higher than corresponding value for a bundled conductor line.