Xcos examples from textbooks and for blocksHEADmaster

23 files changed, 532 insertions, 0 deletions

diff --git a/Working_Examples/83/CH5/EX5.1/example_5_1.sce b/Working_Examples/83/CH5/EX5.1/example_5_1.sce
new file mode 100755
index 0000000..d64c735
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.1/example_5_1.sce
@@ -0,0 +1,43 @@
+//Chapter 5
+//Example 5.1
+//page 132
+//To find sending-end voltage and voltage regulation
+clc;clear;
+
+load1=5000; //kW
+pf=0.707;
+Vr=10000; //receiving end voltage
+R=0.0195*20;
+X=2*%pi*50*0.63*10^-3*20;
+
+//to find sending end voltage and voltage regulation
+I=load1*1000/(Vr*pf);
+Vs=Vr+I*(R*pf+X*sin(acos(pf)));
+voltage_regulation=(Vs-Vr)*100/Vr;
+printf('\n\nReceiving current =I=%d A\n',I);
+printf('Sending end voltage =Vs=%d V\n',Vs);
+printf('Voltage Regulation=%0.2f %%',voltage_regulation);
+
+//to find the value of the capacitor to be connected in parallel to the load
+voltage_regulation_desi=voltage_regulation/2;
+Vs=(voltage_regulation_desi/100)*Vr+Vr;
+//by solving the equations (i) and (ii)
+pf=0.911;
+Ir=549;
+Ic=(Ir*(pf-%i*sin(acos(pf))))-(707*(0.707-%i*0.707));
+Xc=(Vr/imag(Ic));
+c=(2*%pi*50*Xc)^-1;
+printf('\n\nCapacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by :\n C = %d uF\n',c*10^6);
+
+//to find efficiency in both the cases
+//case(i)
+losses=I*I*R*10^-3;
+n=(load1/(load1+losses))*100;
+printf('\n Efficiency in : \nCase(i) \t n=%0.1f%%',n);
+//caase(ii)
+losses=Ir*Ir*R*10^-3;
+n=(load1/(load1+losses))*100;
+printf('\nCase(ii) \t n=%0.1f%%',n);
+
+
+
diff --git a/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt b/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt
new file mode 100755
index 0000000..99064b3
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt
@@ -0,0 +1,11 @@
+
+Receiving current =I=707 A
+Sending end voltage =Vs=12174 V
+Voltage Regulation=21.75 %
+
+Capacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by :
+ C = 87 uF
+
+ Efficiency in : 
+Case(i) 	 n=96.2%
+Case(ii) 	 n=97.7% 
diff --git a/Working_Examples/83/CH5/EX5.10/example_5_10.sce b/Working_Examples/83/CH5/EX5.10/example_5_10.sce
new file mode 100755
index 0000000..af6af96
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.10/example_5_10.sce
@@ -0,0 +1,39 @@
+//Chapter 5
+//Example 5.10
+//page 170
+//To determine the MVA rating of the shunt reactor
+clear;clc;
+v=275;
+l=400;
+R=0.035*l;
+X=2*%pi*50*1.1*l*10^-3;
+Z=R+%i*X;
+Y=2*%pi*50*0.012*10^-6*l*%i;
+A=1+(Y*Z/2);
+B=Z;
+Vs=275;
+Vr=275;
+r=(Vs*Vr)/abs(B);
+Ce=abs(A/B)*Vr^2;
+printf('Radius of the receiving-end circle=%0.1f MVA\n\n',r);
+printf('Location of the center of receiving-end circle= %0.1f MVA\n\n',Ce);
+printf('From the graph, 55 MVA shunt reactor is required\n\n');
+theta=180+82.5;
+x=-75:0.01:450;
+a=Ce*cosd(theta); //to draw the circle
+b=Ce*sind(theta); 
+y=sqrt(r^2-(x-a)^2)+b;
+x1=a:0.001:0;
+y1=tand(theta)*x1;
+plot(x,y,x1,y1);
+title('Circle diagram for example 5.10');
+xlabel('MW');
+ylabel('MVAR');
+plot(a,b,'markersize',150);
+xgrid(2)
+set(gca(),"grid",[0,0])
+get("current_axes");
+xstring (-75,25,'55 MVAR');
+xstring(-75,-25,'83.5 deg');
+xstring(-20,-300,'487.6 MVA');
+xstring(300,-100,'544.3 MVA');
\ No newline at end of file
diff --git a/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt b/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt
new file mode 100755
index 0000000..abd9fbc
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt
@@ -0,0 +1,7 @@
+ 
+Radius of the receiving-end circle=544.3 MVA
+
+Location of the center of receiving-end circle= 487.6 MVA
+
+From the graph, 55 MVA shunt reactor is required
+
diff --git a/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg b/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg
new file mode 100755
index 0000000..522057e
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg
diff --git a/Working_Examples/83/CH5/EX5.11/example_5_11.sce b/Working_Examples/83/CH5/EX5.11/example_5_11.sce
new file mode 100755
index 0000000..19bd34f
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.11/example_5_11.sce
@@ -0,0 +1,19 @@
+//Chapter 5
+//Example 5.11
+//page 172
+//To determine sending-end voltage.maximum power delivered
+clear;clc;
+
+A=0.93*(cosd(1.5)+%i*sind(1.5));
+B=115*(cosd(77)+%i*sind(77));
+Vr=275;
+Ce=abs(A/B)*Vr^2;
+printf('Centre of the receiving end circle is = %0.1f MVA\n\n',Ce);
+CrP=850;Vs=CrP*abs(B)/Vr;
+printf('(a) From the diagram,\n\tCrP=%d \n \tSending end voltage|Vs|= %0.1f kV\n\n',CrP,Vs);
+Vs=295; //given
+r=(Vs*Vr)/abs(B);
+Pr_m=556; //from the diagram
+printf('(b) Radius of the circle diagram = %0.1f MVA\n\t PR_max=%d MW\n\n',r,Pr_m);
+Ps=295; //from the diagram;
+printf('(c) Additional MVA to be drawn from the line is = P''S=%d MVAR\n\n',Ps);
\ No newline at end of file
diff --git a/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt b/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt
new file mode 100755
index 0000000..322c788
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt
@@ -0,0 +1,12 @@
+ 
+
+Centre of the receiving end circle is = 611.6 MVA
+
+(a) From the diagram,
+	CrP=850 
+ 	Sending end voltage|Vs|= 355.5 kV
+
+(b) Radius of the circle diagram = 705.4 MVA
+	 PR_max=556 MW
+
+(c) Additional MVA to be drawn from the line is = P'S=295 MVAR
diff --git a/Working_Examples/83/CH5/EX5.2/example_5_2.sce b/Working_Examples/83/CH5/EX5.2/example_5_2.sce
new file mode 100755
index 0000000..925e68d
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.2/example_5_2.sce
@@ -0,0 +1,28 @@
+//Chapter 5
+//Example 5.2
+//page 134
+//To find voltage at the bus at the power station end
+clc;clear;
+
+base_MVA=5;
+base_kV=33;
+pf=0.85;
+cable_impedance=(8+%i*2.5);
+cable_impedance=cable_impedance*base_MVA/(base_kV^2);
+
+transf_imp_star=(0.06+%i*0.36)/3; //equivalent star impedance of winding of the transformer
+Zt=(transf_imp_star*5/(6.6^2))+((0.5+%i*3.75)*5/(33^2));
+total=cable_impedance+2*Zt;
+
+load_MVA=1;
+load_voltage=6/6.6;
+load_current=1/load_voltage;
+
+Vs=load_voltage+load_current*(real(total)*pf+imag(total)*sin(acos(pf)));
+Vs=Vs*6.6;
+printf('\n\nCable impedance= (%0.3f+j%0.4f) pu\n',real(cable_impedance),imag(cable_impedance));
+printf('\nEquivalent star impedance of 6.6kV winding of the transformer =(%0.2f+j%0.2f) pu\n',real(transf_imp_star),imag(transf_imp_star));
+printf('\nPer unit transformer impedance,Zt=(%0.4f+j%0.3f) pu\n',real(Zt),imag(Zt));
+printf('\nTotal series impedance=(%0.3f+j%0.3f) pu\n',real(total),imag(total));
+printf('\nSending end Voltage =|Vs|=%0.2fkV (line-to-line)',Vs);
+
diff --git a/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt b/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt
new file mode 100755
index 0000000..f9b1d06
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt
@@ -0,0 +1,10 @@
+
+Cable impedance= (0.037+j0.0115) pu
+
+Equivalent star impedance of 6.6kV winding of the transformer =(0.02+j0.12) pu
+
+Per unit transformer impedance,Zt=(0.0046+j0.031) pu
+
+Total series impedance=(0.046+j0.073) pu
+
+Sending end Voltage =|Vs|=6.56kV (line-to-line) 
diff --git a/Working_Examples/83/CH5/EX5.3/example_5_3.sce b/Working_Examples/83/CH5/EX5.3/example_5_3.sce
new file mode 100755
index 0000000..195c0b5
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.3/example_5_3.sce
@@ -0,0 +1,24 @@
+//Chapter 5
+//Example 5.3
+//page 135
+//problem with mixed end condition
+clc;clear;
+Vr=3000; //receiving end voltage
+pfs=0.8; //sending end power factor
+Ps=2000*10^3; //sending end active power
+z=0.4+%i*0.4; //series impedance
+Ss=Ps/pfs; //sending end VA 
+Qs=Ss*sqrt(1-pfs^2); //sending end reacive power
+
+//by substituting all the values to the equation (iii)
+deff('[y]=fx(I)',"y=(Vr^2)*(I^2)+2*Vr*(I^2)*(real(z)*((Ps-real(z)*(I^2))/Vr)+imag(z)*((Qs-imag(z)*(I^2))/Vr))+(abs(z))^2*(I^4)-(Ss^2)");
+I=fsolve(100,fx);
+
+pfR=(Ps-real(z)*(I^2))/(Vr*I); //Cos(phi_r)
+Pr=Vr*I*pfR;
+Vs=(Ps/(I*pfs));
+
+printf('\nLoad Current |I|= %0.2f A',I);
+printf('\nLoad Pr=%d W',Pr);
+printf('\nReceiving end power factor=%0.2f',pfR);
+printf('\nSupply Voltage=%0.2fV',Vs);
diff --git a/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt b/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt
new file mode 100755
index 0000000..537ce19
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt
@@ -0,0 +1,6 @@
+ 
+Load Current |I|= 733.14 A
+Load Pr=1785001 W
+Receiving end power factor=0.81
+Supply Voltage=3409.99V 
+
diff --git a/Working_Examples/83/CH5/EX5.4/example_5_4.sce b/Working_Examples/83/CH5/EX5.4/example_5_4.sce
new file mode 100755
index 0000000..7a4015d
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.4/example_5_4.sce
@@ -0,0 +1,24 @@
+//Chapter 5
+//Example 5.4
+//page 138
+//to find sending end voltage and voltage regulation of a medium transmission line system
+clear;clc;
+D=300;
+r=0.8;
+L=0.461*log10(D/(0.7788*r));
+C=0.0242/(log10(D/r));
+R=0.11*250;
+X=2*%pi*50*L*0.001*250;
+Z=R+%i*X;
+Y=%i*2*%pi*50*C*0.000001*250;
+Ir=((25*1000)/(132*sqrt(3)))*(cosd(-36.9)+%i*sind(-36.9));
+Vr=(132/sqrt(3));
+A=(1+(Y*Z/2));
+Vs=A*Vr+Z*Ir*10^(-3);
+printf('\n\nVs(per phase)=(%0.2f+%0.2f)kV',real(Vs),imag(Vs));
+Vs=abs(Vs)*sqrt(3);
+printf('\n\n|Vs|(line)=%d kV',Vs);
+Vr0=Vs/abs(A);
+printf('\n\n|Vr0|(line no load)=%0.1fkV',Vr0);
+Vol_regu=(Vr0-132)/132;
+printf('\n\nVoltage Regulation=%0.1f%%\n\n',Vol_regu*100);
diff --git a/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt b/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt
new file mode 100755
index 0000000..4a47cad
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt
@@ -0,0 +1,12 @@
+ 
+
+Vs(per phase)=(82.26+7.46)kV
+
+|Vs|(line)=143 kV
+
+|Vr0|(line no load)=148.4kV
+
+Voltage Regulation=12.4%
+
+ 
+
diff --git a/Working_Examples/83/CH5/EX5.5/example_5_5.sce b/Working_Examples/83/CH5/EX5.5/example_5_5.sce
new file mode 100755
index 0000000..8c62731
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.5/example_5_5.sce
@@ -0,0 +1,29 @@
+//Chapter 5
+//Example 5.5
+//page 147
+//to find maximum permissible length and and frequency
+clc;clear;
+R=0.125*400;
+X=0.4*400;
+Y=2.8*(10^-6)*400*%i;
+Z=R+X*%i;
+
+//(i) At no-load
+A=1+(Y*Z/2);
+C=Y*(1+Y*Z/6);
+VR_line=220000/abs(A);
+Is=abs(C)*VR_line/sqrt(3);
+printf('\n\n |VR|line = %d kV',VR_line/1000);
+printf('\n |Is| = %d A',Is);
+
+//(ii) to find maximum permissible length
+//By solving the equations shown in the book,we get
+l=sqrt((1-0.936)/(0.56*10^(-6)));
+printf('\n\n Maximum permissible length of the line = %d km',l);
+
+//(iii) to find maximum permissible frequency for the case(i)
+//By solving the equations shown in the book,we get
+f=sqrt(((1-0.88)*50*50)/(0.5*1.12*10^-3*160));
+printf('\n\n Maximum permissible frequency = %0.1f Hz\n\n',f);
+
+
diff --git a/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt b/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt
new file mode 100755
index 0000000..d027f9c
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt
@@ -0,0 +1,11 @@
+ 
+
+
+ |VR|line = 241 kV
+ |Is| = 151 A
+
+ Maximum permissible length of the line = 338 km
+
+ Maximum permissible frequency = 57.9 Hz
+
+
diff --git a/Working_Examples/83/CH5/EX5.6/example_5_6.sce b/Working_Examples/83/CH5/EX5.6/example_5_6.sce
new file mode 100755
index 0000000..21b98e4
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.6/example_5_6.sce
@@ -0,0 +1,35 @@
+//Chapter 5
+//Example 5.6
+//page 149
+//to find incident and reflected voltages
+clear;clc;
+
+R=0.125;
+X=0.4;
+y=%i*2.8*10^(-6);
+z=R+%i*X;
+
+r=sqrt(y*z); //propogation constant
+a=real(r); //attenuation constant
+b=imag(r); //phase constant
+
+//(a) At the receiving-end;
+Vr=220000;
+Inci_vol=Vr/(sqrt(3)*2);
+Refl_vol=Vr/(sqrt(3)*2);
+printf('\n\nIncident Vvoltage=%0.2f kV',Inci_vol/1000);
+printf('\nReflected Vvoltage=%0.2f kV',Refl_vol/1000);
+
+//(b) At 200km from the receiving-end
+x=200;
+Inci_vol=Inci_vol*exp(a*x)*exp(%i*b*x);
+Refl_vol=Refl_vol*exp(-a*x)*exp(-%i*b*x);
+printf('\n\nIncident voltage=%0.2f @ %0.1f deg kV',abs(Inci_vol)/1000,atand(imag(Inci_vol)/real(Inci_vol)));
+printf('\nReflected voltage=%0.2f @ %0.1f deg kV',abs(Refl_vol)/1000,atand(imag(Refl_vol)/real(Refl_vol)));
+
+//(c) Resultant voltage at 200km from the receiving-end
+res=Inci_vol+Refl_vol;
+printf('\n\nResultant line-to-line voltage at 200km =%0.2f kV',abs(res)*sqrt(3)/1000);
+
+
+
diff --git a/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt b/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt
new file mode 100755
index 0000000..cd94d77
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt
@@ -0,0 +1,10 @@
+ 
+
+
+Incident Vvoltage=63.51 kV
+Reflected Vvoltage=63.51 kV
+
+Incident voltage=65.62 @ 12.3 deg kV
+Reflected voltage=61.47 @ -12.3 deg kV
+
+Resultant line-to-line voltage at 200km =215.09 kV 
diff --git a/Working_Examples/83/CH5/EX5.7/example_5_7.sce b/Working_Examples/83/CH5/EX5.7/example_5_7.sce
new file mode 100755
index 0000000..dbd47d2
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.7/example_5_7.sce
@@ -0,0 +1,83 @@
+//Chapter 5
+//Example 5.7
+//page 138
+//to tabulate characteristics of a system using different methods
+clear;clc;
+
+Z=40+125*%i;
+Y=%i*10^(-3);
+Ir=((50*10^6)/(220000*0.8*sqrt(3)))*(cosd(-36.9)+%i*sind(-36.9));
+Vr=220000/sqrt(3);
+
+//(a) Short line approximation
+Vs=Vr+Ir*Z;
+Vs_line1=Vs*sqrt(3);
+Is1=Ir;
+pfs1=cos(atan(imag(Vs)/real(Vs))+acos(0.8));
+Ps1=sqrt(3)*abs(Vs_line1)*abs(Is1)*pfs1;
+
+//(b) Nominal pi method
+A=1+Y*Z/2;
+D=A;
+B=Z;
+C=Y*(1+Y*Z/4);
+Vs=A*Vr+B*Ir;
+Is2=C*Vr+D*Ir;
+Vs_line2=sqrt(3)*Vs;
+pfs2=cos(atan(imag(Is2)/real(Is2))-atan(imag(Vs)/real(Vs)));
+Ps2=sqrt(3)*abs(Vs_line2)*abs(Is2)*pfs2;
+
+//(c) Exact transmission line equations
+rl=sqrt(Z*Y); //propogation constant
+Zc=sqrt(Z/Y); //characteristic impedance
+A=cosh(rl);
+B=Zc*sinh(rl);
+C=sinh(rl)/Zc;
+D=cosh(rl);
+Vs=A*Vr+B*Ir;
+Is3=C*Vr+D*Ir;
+Vs_line3=sqrt(3)*Vs;
+pfs3=cos(atan(imag(Is3)/real(Is3))-atan(imag(Vs)/real(Vs)));
+Ps3=sqrt(3)*abs(Vs_line3)*abs(Is3)*pfs3;
+
+//(d) Approximation
+A=(1+Y*Z/2);
+B=Z*(1+Y*Z/6);
+C=Y*(1+Y*Z/6);
+D=A;
+Vs=A*Vr+B*Ir;
+Is4=C*Vr+D*Ir;
+Vs_line4=sqrt(3)*Vs;
+pfs4=cos(atan(imag(Is4)/real(Is4))-atan(imag(Vs)/real(Vs)));
+Ps4=sqrt(3)*abs(Vs_line4)*abs(Is4)*pfs4;
+
+//converting all the values to their standard form before writing it to table
+
+//voltage to kV
+Vs_line1=abs(Vs_line1)/1000;
+Vs_line2=abs(Vs_line2)/1000;
+Vs_line3=abs(Vs_line3)/1000;
+Vs_line4=abs(Vs_line4)/1000;
+
+//Current to kA
+Is1=Is1/1000;
+Is2=Is2/1000;
+Is3=Is3/1000;
+Is4=Is4/1000;
+
+//power to MW5
+Ps1=Ps1/1000000;
+Ps2=Ps2/1000000;
+Ps3=Ps3/1000000;
+Ps4=Ps4/1000000;
+
+//preparinf table
+printf("\n\n_______________________________________________________________________________________________________________________");
+printf('\n        \t\tShort line \t\t       Nominal Pi \t\t Exact     \t\t Approximation');
+printf("\n_______________________________________________________________________________________________________________________");
+printf('\n|Vs|line\t\t%0.2fkV        \t\t %0.2fkV\t\t %0.2fkV \t\t %0.2fkV',Vs_line1,Vs_line2,Vs_line3,Vs_line4);
+printf('\nIs      \t\t%0.3f@%0.1fdeg kA \t\t%0.2f@%0.1fdeg kA\t\t%0.4f@%0.1fdeg kA\t%0.2f@%0.1fdeg kA',abs(Is1),tand(imag(Is1)/real(Is1)),abs(Is2),tand(imag(Is2)/real(Is2)),abs(Is3),tand(imag(Is3)/real(Is3)),abs(Is4),tand(imag(Is4)/real(Is4)));
+printf('\npfs     \t\t%0.3f lagging       \t\t%0.3f leading \t\t%0.3f leading \t\t%0.3f leading',pfs1,pfs2,pfs3,pfs4);
+printf('\nPs      \t\t%0.2f MW          \t\t%0.2f MW \t\t%0.2f MW \t\t%0.2f MW',Ps1,Ps2,Ps3,Ps4);
+printf("\n_______________________________________________________________________________________________________________________\n\n\n");
+
diff --git a/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt b/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt
new file mode 100755
index 0000000..bb57c69
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt
@@ -0,0 +1,14 @@
+
+
+
+_______________________________________________________________________________________________________________________
+        		Short line 		       Nominal Pi 		 Exact     		 Approximation
+_______________________________________________________________________________________________________________________
+|Vs|line		251.34kV        		 238.08kV		 237.40kV 		 237.28kV
+Is      		0.164@-0.0deg kA 		0.13@0.0deg kA		0.1289@0.0deg kA	0.13@0.0deg kA
+pfs     		0.746 lagging       		0.988 leading 		0.987 leading 		0.987 leading
+Ps      		53.23 MW          		52.19 MW 		52.28 MW 		52.24 MW
+_______________________________________________________________________________________________________________________
+
+
+ 
diff --git a/Working_Examples/83/CH5/EX5.8/example_5_8.sce b/Working_Examples/83/CH5/EX5.8/example_5_8.sce
new file mode 100755
index 0000000..6479e6a
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.8/example_5_8.sce
@@ -0,0 +1,56 @@
+//Chapter 5
+//Example 5.8
+//page 162
+//to estimate the torque angle and station powerfactor
+clear;clc;
+Sd1=15+%i*5;
+Sd2=25+%i*15;
+//case(a) cable impedance=j0.05pu
+r=0;
+x=%i*0.05;
+PG1=20;
+PG2=20;
+Ps=5;Pr=5;
+V1=1;
+V2=1;
+d1=asind(Ps*abs(x)/(V1*V2)); //delta1
+V1=V1*(cosd(d1)+%i*sind(d1));
+Qs=((abs(V1)^2)/abs(x))-((abs(V1)*abs(V2))*cosd(d1)/(abs(x)));
+Qr=(((abs(V1)*abs(V2))*cosd(d1)/(abs(x)))-(abs(V1)^2)/abs(x));
+Ql=Qs-Qr;
+Ss=Ps+%i*Qs;
+Sr=Pr+%i*Qr;
+Sg1=Sd1+Ss;
+Sg2=Sd2-Sr;
+pf1=cos(atan(imag(Sg1)/real(Sg1)));
+pf2=cos(atan(imag(Sg2)/real(Sg2)));
+printf('\n\nCase(a)\nTotal load on station1=%d+j%0.3f pu',real(Sg1),imag(Sg1));
+printf('\nPower factor of station1=%0.3f pu lagging',pf1);
+printf('\n\Total load on station2=%d+j%0.3f pu',real(Sg2),imag(Sg2));
+printf('\nPower factor of station2=%0.3f pu lagging',pf2);
+//case(b) cable impedance=0.005+j0.05;
+r=0.005;
+PG1=20;
+V1=1;V2=1;
+Ps=5;
+//from the eq(i) in the textbook,we can calculate d1
+z=r+x;
+theta=atand(imag(z)/real(z));
+z=abs(z);
+d1=acosd(z*(V1^2*cosd(theta)/z-Ps)/(V1*V2))-theta;
+Qs=(V1^2*sind(theta)/z)-(V1*V2*sind(theta+d1)/z);
+Qg1=5+Qs;
+Pr=(V1*V2*cosd(theta-d1)/z)-(V1^2*cosd(theta)/z);
+Pg2=25-Pr;
+Qr=(V1*V2*sind(theta-d1)/z)-(V1^2*sind(theta)/z);
+Qg2=15-Qr;
+Ss=Ps+%i*Qs;
+Sr=Pr+%i*Qr;
+Sg1=Sd1+Ss;
+Sg2=Sd2-Sr;
+pf1=cos(atan(imag(Sg1)/real(Sg1)));
+pf2=cos(atan(imag(Sg2)/real(Sg2)));
+printf('\n\nCase(b)\nTotal load on station1=%d+j%0.3f pu',real(Sg1),imag(Sg1));
+printf('\nPower factor of station1=%0.3f pu lagging',pf1);
+printf('\n\Total load on station2=%d+j%0.3f pu',real(Sg2),imag(Sg2));
+printf('\nPower factor of station2=%0.3f pu lagging\n\n',pf2);
diff --git a/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt b/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt
new file mode 100755
index 0000000..b7f4dee
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt
@@ -0,0 +1,15 @@
+
+
+Case(a)
+Total load on station1=20+j5.635 pu
+Power factor of station1=0.963 pu lagging
+Total load on station2=20+j15.635 pu
+Power factor of station2=0.788 pu lagging
+
+Case(b)
+Total load on station1=20+j5.132 pu
+Power factor of station1=0.969 pu lagging
+Total load on station2=20+j16.119 pu
+Power factor of station2=0.781 pu lagging
+
+
diff --git a/Working_Examples/83/CH5/EX5.9/example_5_9.sce b/Working_Examples/83/CH5/EX5.9/example_5_9.sce
new file mode 100755
index 0000000..7ca7bf2
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.9/example_5_9.sce
@@ -0,0 +1,33 @@
+//Chapter 5
+//Example 5.9
+//page 165
+//to determine power,voltage,compensating equipment rating
+clear;clc;
+A=0.85;
+B=200;
+
+//case(a)
+Vs=275000;
+Vr=275000;
+a=5;b=75; //alpha and beta
+Qr=0;
+//from equation 5.62
+d=b-asind((B/(Vs*Vr))*(Qr+(A*Vr^2*sind(b-a)/B))); //delta
+Pr=(Vs*Vr*cosd(b-d)/B)-(A*Vr^2*cosd(b-a)/B);
+printf('\n\ncase(a)\nPower at unity powerfactor that can be received =%0.1f MW',Pr/10^6);
+
+//case(b)
+Pr=150*10^6;
+d=b-acosd((B/(Vs*Vr))*(Pr+(A*Vr^2*cosd(b-a)/B))); //delta
+Qr=(Vs*Vr*sind(b-d)/B)-(A*Vr^2*sind(b-a)/B);
+Qc=-Qr;
+printf('\n\ncase(b)\nRating of the compensating equipment = %0.2f MVAR',Qc/10^6);
+printf('\ni.e the compensating equipment must feed positive VARs into the line');
+
+
+//case(c)
+Pr=150*10^6;
+Vs=275000;
+//by solving the two conditions given as (i) and (ii), we get
+Vr=244.9*10^3;
+printf('\n\ncase(c)\nReceiving end voltage = %0.1f kV',Vr/1000);
diff --git a/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt b/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt
new file mode 100755
index 0000000..18b9c60
--- /dev/null
+++ b/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt
@@ -0,0 +1,11 @@
+
+
+case(a)
+Power at unity powerfactor that can be received =117.6 MW
+
+case(b)
+Rating of the compensating equipment = 27.40 MVAR
+i.e the compensating equipment must feed positive VARs into the line
+
+case(c)
+Receiving end voltage = 244.9 kV