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Diffstat (limited to '3872/CH9')
| -rw-r--r-- | 3872/CH9/EX9.2/Ex9_2.jpg | bin | 0 -> 22519 bytes | |||
| -rw-r--r-- | 3872/CH9/EX9.2/Ex9_2.sce | 24 | ||||
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| -rw-r--r-- | 3872/CH9/EX9.3/Ex9_3.sce | 37 | ||||
| -rw-r--r-- | 3872/CH9/EX9.4/Ex9_4.jpg | bin | 0 -> 36371 bytes | |||
| -rw-r--r-- | 3872/CH9/EX9.4/Ex9_4.sce | 30 | ||||
| -rw-r--r-- | 3872/CH9/EX9.5/Ex9_5.jpg | bin | 0 -> 90878 bytes | |||
| -rw-r--r-- | 3872/CH9/EX9.5/Ex9_5.sce | 49 | ||||
| -rw-r--r-- | 3872/CH9/EX9.6/Ex9_6.jpg | bin | 0 -> 48332 bytes | |||
| -rw-r--r-- | 3872/CH9/EX9.6/Ex9_6.sce | 43 | ||||
| -rw-r--r-- | 3872/CH9/EX9.7/Ex9_7.sce | 54 | ||||
| -rw-r--r-- | 3872/CH9/EX9.7/Ex9_7_1.JPG | bin | 0 -> 42961 bytes | |||
| -rw-r--r-- | 3872/CH9/EX9.7/Ex9_7_2.JPG | bin | 0 -> 45302 bytes |
13 files changed, 237 insertions, 0 deletions
diff --git a/3872/CH9/EX9.2/Ex9_2.jpg b/3872/CH9/EX9.2/Ex9_2.jpg Binary files differnew file mode 100644 index 000000000..006a096af --- /dev/null +++ b/3872/CH9/EX9.2/Ex9_2.jpg diff --git a/3872/CH9/EX9.2/Ex9_2.sce b/3872/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..7c76719af --- /dev/null +++ b/3872/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,24 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.2
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+Xn=0.05 //motor neutral is grounded through reactance in per unit
+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
+Z1=%i*0.13893 //Positive sequence impedance in per unit
+
+
+If=Vf/Z1 //positive 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)]*[0;If;0] //subtransient fault current in each phase in per unit
+
+disp(abs(Isf),'The magnitude of fault current in each phase in per unit is given by :',);
+disp(atand(imag(Isf),real(Isf)),'The angle of fault current in each phase in degrees is given by:',);
+
+
+
diff --git a/3872/CH9/EX9.3/Ex9_3.jpg b/3872/CH9/EX9.3/Ex9_3.jpg Binary files differnew file mode 100644 index 000000000..a5eef04c5 --- /dev/null +++ b/3872/CH9/EX9.3/Ex9_3.jpg diff --git a/3872/CH9/EX9.3/Ex9_3.sce b/3872/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..b5b8775a5 --- /dev/null +++ b/3872/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,37 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.3
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+Xn=0.05 //motor neutral is grounded through reactance in per unit
+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
+
+If0=Vf/(Z0+Z1+Z2+(3*Zf)) //sequence line to ground fault current in per unit
+If1=If0;If2=If0; // Since If0=If1=If2
+If=[If0;If1;If2]
+Isf=3*If0 //subtransient fault current in per unit
+Ib2=Sb/(Vb*sqrt(3)) //base current at bus 2 in kA
+Ib22=Isf*Ib2
+Vsf=[0;Vf;0]-([Z0 0 0;0 Z1 0;0 0 Z2]*If) //sequence componenets of the voltages at the fault in per unit
+a=exp(%i*(120)*(%pi/180)); //operator a
+Vlg2=[1 1 1;1 (a^2) a;1 a (a^2)]*Vsf //line to ground voltages at faulted bus 2 in per unit
+for i=1:3 //This loop is included to avoid discrepancies in angle values when the voltage value is near to zero or zero
+ if abs(Vlg2(i))<1e-6 //For example, atand(0,0) gives 0 degree and atand(0,-0) gives 180 degree
+ Vlg2(i)=0;
+ end
+end
+
+printf('The magnitude of subtransient at bus 2 in is %.4f kA and its angle is %.4f degrees\n',abs(Ib22),atand(imag(Ib22),real(Ib22)));
+disp(abs(Vlg2),'The magnitude of line to ground voltages at faulted bus 2 in per unit is:');
+disp(atand(imag(Vlg2),real(Vlg2)),'The angle of line to ground voltages at faulted bus 2 is :');
+
+
diff --git a/3872/CH9/EX9.4/Ex9_4.jpg b/3872/CH9/EX9.4/Ex9_4.jpg Binary files differnew file mode 100644 index 000000000..97e80aca8 --- /dev/null +++ b/3872/CH9/EX9.4/Ex9_4.jpg diff --git a/3872/CH9/EX9.4/Ex9_4.sce b/3872/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..afd70f2fa --- /dev/null +++ b/3872/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,30 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.4
+//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
+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
+I0=0 //Zero sequence current in per unit
+
+I1=Vf/(Z1+Z2+Zf) //sequence fault current in per unit
+Isfb=-%i*sqrt(3)*I1 //subtransient fault current in phase b in per unit
+Ib2=(Sb/(Vb*sqrt(3)))*Isfb //subtransient fault current at phase b in kA
+Isfc=-Isfb; //subtransient fault current at phase c in pu
+Ic=-Ib2 //subtransient fault current at phase c in kA
+
+printf('The magnitude of subtransient fault current in phase b in per unit is :%.4f pu and its angle is:%.4f degrees\n',abs(Isfb),(180/%pi)*atan(imag(Isfb),real(Isfb)));
+printf('The magnitude of subtransient fault current in phase b in kA is %.4f kA and its angle is %.4f degrees\n',abs(Ib2),(180/%pi)*atan(imag(Ib2),real(Ib2)));
+
+
+printf('The magnitude of sequence fault current in phase c in per unit is %.4f pu and its angle is %.4f degrees\n',abs(Isfc),(180/%pi)*atan(imag(Isfc),real(Isfc)));
+printf('The magnitude of sequence fault current in phase c in kA is %.4f kA and its angle is %.4f degrees\n',abs(Ic),(180/%pi)*atan(imag(Ic),real(Ic)));
+
+
diff --git a/3872/CH9/EX9.5/Ex9_5.jpg b/3872/CH9/EX9.5/Ex9_5.jpg Binary files differnew file mode 100644 index 000000000..65958489b --- /dev/null +++ b/3872/CH9/EX9.5/Ex9_5.jpg 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:');
+
+
diff --git a/3872/CH9/EX9.6/Ex9_6.jpg b/3872/CH9/EX9.6/Ex9_6.jpg Binary files differnew file mode 100644 index 000000000..bc0b3841b --- /dev/null +++ b/3872/CH9/EX9.6/Ex9_6.jpg diff --git a/3872/CH9/EX9.6/Ex9_6.sce b/3872/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..4b0ca7e58 --- /dev/null +++ b/3872/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,43 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.6
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+
+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
+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
+Ilead1=Iline1*exp(%i*(30)*(%pi/180)) //Positive sequence fault current from the line leads by 30 degree 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
+Ilag2=Iline2*exp(%i*(-30)*(%pi/180)) //Negative sequence fault current from the line lags by 30 degree in per unit
+Imotor2=(Znl/(Znr+Znl))*I2 //Negative sequence motor current from the motor in per unit
+a=exp(%i*(120)*(%pi/180)) //operator a
+Iline=[1 1 1;1 (a^2) a;1 a (a^2)]*[0;Ilead1;Ilag2] //transforming the line currents to the phase domain
+Ilineb=Iline*0.41837 //transforming the line currents to the phase domain with base currents of 0.41837 kA
+
+disp(abs(clean(Iline,1e-10)),'The magnitude of transforming the line currents to the phase domain in per unit for each phase is given by:');
+disp(atand(imag(Iline),real(Iline)),'The angle of transforming the line currents to the phase domain in degreess for each phase is given by:');
+disp(abs(clean(Ilineb,1e-10)),'The magnitude of transforming the line currents to the phase domain in kA for each phase is given by:');
+disp(atand(imag(Ilineb),real(Ilineb)),'The angle of transforming the line currents to the phase domain in degreess for each phase is given by:');
+
+
+
+
diff --git a/3872/CH9/EX9.7/Ex9_7.sce b/3872/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..3e613327f --- /dev/null +++ b/3872/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,54 @@ +//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.7
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+Vf=1.05 //Prefault voltage in per unit
+Zf=0 //Fault through impedance in per unit
+
+
+Ybus0=-%i*[20 0;0 4] //zero sequence bus admittance matrix in per unit
+Zbus0=inv(Ybus0) //zero sequence bus impedance matrix in per unit
+Ybus1=-%i*[9.9454 -3.2787;-3.2787 8.2787] //Positive sequence bus admittance matrix in per unit
+Zbus1=inv(Ybus1) //Positive sequence bus admittance matrix in per unit
+Ybus2=-%i*[9.1611 -3.2787;-3.2787 8.0406] //Negative bus admittance matrix in per unit
+Zbus2=inv(Ybus2) //Negative sequence bus admittance matrix in per unit
+Z110=%i*0.05 //zero sequence impedance Z110 find out from the Zbus0
+Z111=%i*0.11565 //positive sequence impedance Z111 find out from the Zbus1
+Z112=%i*0.12781 //negative sequence impedance Z112 find out from the Zbus2
+I10=Vf/(Z110+Z111+Z112) //zeor sequence fault current at bus 1 in per unit
+I11=I10 //positive sequence fault current at bus 1 in per unit
+I12=I11 //Negative sequence fault current at bus 1 in per unit
+a=exp(%i*120*%pi/180) //operator a
+Isf1=[1 1 1;1 (a^2) a;1 a (a^2)]*[I10;I11;I12] //Subtransient fault current in each phase at bus 1 in per unit
+Z220=%i*0.25 //zero sequence impedance Z220 find out from the Zbus0
+Z221=%i*0.13893 //positive sequence impedance Z221 find out from the Zbus1
+Z222=%i*0.14562 //negative sequence impedance Z222 find out from the Zbus2
+I20=Vf/(Z220+Z221+Z222) //zeor sequence fault current at bus 1 in per unit
+I21=I20 //positive sequence fault current at bus 1 in per unit
+I22=I21 //Negative sequence fault current at bus 1 in per unit
+Isf2=[1 1 1;1 (a^2) a;1 a (a^2)]*[I20;I21;I22] //Subtransient fault current in each phase at bus 2 in per unit
+V1=[0;Vf;0]-[Z110 0 0;0 Z111 0;0 0 Z112]*[I10;I11;I12] //The sequence components of the line to ground voltages at bus 1 during tha fault at bus 1 with k=1 and n=1 in per unit
+V1lg=[1 1 1;1 (a^2) a;1 a (a^2)]*[V1] //The line to ground voltages at bus 1 during tha fault at bus 1 in per unit
+Z210=%i*0.05 //zero sequence impedance Z210 find out from the Zbus0
+Z211=%i*0.11565 //positive sequence impedance Z211 find out from the Zbus1
+Z212=%i*0.12781 //negative sequence impedance Z212 find out from the Zbus2
+V2=[0;Vf;0]-[Z210 0 0;0 Z211 0;0 0 Z212]*[I10;I11;I12] //The sequence components of the line to ground voltages at bus 1 during tha fault at bus 2 with k=2 and n=1in per unit
+V2lg=[1 1 1;1 (a^2) a;1 a (a^2)]*[V2] //The line to ground voltages at bus 1 during tha fault at bus 1 in per unit
+
+
+
+disp(clean(Zbus0,1e-10),'The zero sequence bus impedance matrix is:');
+disp(clean(Zbus1,1e-10),'The positive sequence bus impedance matrix is:');
+disp(clean(Zbus2,1e-10),'The negative sequence bus impedance matrix is:');
+disp(clean(Isf1,1e-10),'The Subtransient fault current in pu in each phase during fault at bus 1 are:');
+disp(clean(Isf2,1e-10),'The Subtransient fault current in pu in each phase during fault at bus 2 are:');
+disp(abs(clean(V1lg,1e-10)),'The magnitude of the line to ground voltages at bus 1 in pu during fault at bus 1 :',);
+disp(atand(clean(imag(V1lg),1e-10),clean(real(V1lg),1e-10)),'The angle of the line to ground voltages at bus 1 in degrees during fault at bus 1 :',);
+disp(abs(clean(V2lg,1e-10)),'The magnitude of the line to ground voltages at bus 2 in pu during fault at bus 1 :',);
+disp(atand(clean(imag(V2lg),1e-10),clean(real(V2lg),1e-10)),'The angle of the line to ground voltages at bus 1 in degrees during fault at bus 1 :',);
+
+
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