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diff --git a/3872/CH9/EX9.5/Ex9_5.sce b/3872/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..ab7f9c5e7 --- /dev/null +++ b/3872/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,49 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.5
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+Sb=100 //Base value of system in MVA
+Vb=13.8 //Base voltage of system in kV
+Vf=1.05 //Prefault voltage in per unit
+Z0=%i*0.250 //Zero sequence impedance in per unit
+Z1=%i*0.13893 //Positive sequence impedance in per unit
+Z2=%i*0.14562 //Negative sequence impedance in per unit
+Zf=0 //Fault through impedance in per unit
+Zpr=0.20 //The positive sequence thevenin motor impedance at bus 2
+Zpl=0.455 //The positive sequence thevenin line impedance at bus 2
+Znr=0.21 //The negative sequence thevenin motor impedance at bus 2
+Znl=0.475 //The negative sequence thevenin line impedance at bus 2
+
+
+I1=Vf/(Z1+((Z0*Z2)/(Z0+Z2))) //Positive sequence fault current in per unit
+I2=-I1*(Z0/(Z0+Z2)) //Negative sequence fault current in per unit
+I0=-I1*(Z2/(Z0+Z2)) //Zero sequence fault current in per unit
+a=exp(%i*(120)*(%pi/180)) //operator a
+Isf=[1 1 1;1 (a^2) a;1 a (a^2)]*[I0;I1;I2] //Subtransient fault current in each phase in per unit
+Ib2=Isf*((Sb)/(Vb*sqrt(3))) //Using the base current 4.1837kA at bus 2 in kA
+In=3*I0 //Neutral fault current in per unit
+In2=In*((Sb)/(Vb*sqrt(3))) //Neutral fault current in kA
+Iline0=0 //Zero sequence fault current from the line in per unit
+Imotor0=I0 //Zero sequence motor current from the motor in per unit
+Iline1=(Zpr/(Zpr+Zpl))*I1 //Positive sequence fault current from the line in per unit
+Imotor1=(Zpl/(Zpr+Zpl))*I1 //Positive sequence motor current from the motor in per unit
+Iline2=(Znr/(Znr+Znl))*I2 //Negative sequence fault current from the line in per unit
+Imotor2=(Znl/(Znr+Znl))*I2 //Negative sequence motor current from the motor in per unit
+Iline=[1 1 1;1 (a^2) a;1 a (a^2)]*[Iline0;Iline1;Iline2] //transforming to the phase domain for the line
+Ilineb=Iline*(0.41837) //Transforming to the phase domain with base currents of 0.41837 kA for the line in kA
+Imotor=[1 1 1;1 (a^2) a;1 a (a^2)]*[Imotor0;Imotor1;Imotor2] //transforming to the phase domain for the motor
+Imotorb=Imotor*((Sb)/(Vb*sqrt(3))) //Transforming to the phase domain with base currents of 4.1837 kA for the motor in kA
+
+disp(abs(clean(Ib2,1e-10)),'The magnitude of subtransient fault current in each phase in kA is given by:');
+disp(atand(clean(imag(Ib2),1e-10),clean(real(Ib2),1e-10)),'The angle of subtransient fault current in each phase in degrees is given by:');
+printf('The magnitude neutral fault current is %.4f kA and its angle is %.4f degree\n',abs(In2),atand(imag(In2),real(In2)));
+disp(abs(clean(Ilineb,1e-10)),'The magnitude of fault current contribution from the line in kA for each phase is given by: ');
+disp(atand(clean(imag(Ilineb),1e-10),clean(real(Ilineb),1e-10)),'The angle of fault current contribution from the line in degrees for each phase is given by:');
+disp(abs(clean(Imotorb,1e-10)),'The magnitude of fault current contribution from motor in kA for each phase is given by:');
+disp(atand(clean(imag(Imotorb),1e-10),clean(real(Imotorb),1e-10)),'The angle of fault current contribution from motor in degrees for each phase is given by:');
+
+
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