diff options
Diffstat (limited to 'Working_Examples/2777')
132 files changed, 6449 insertions, 0 deletions
diff --git a/Working_Examples/2777/CH2/EX2.1/Ex2_1.sce b/Working_Examples/2777/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 0000000..990fbe9 --- /dev/null +++ b/Working_Examples/2777/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 2 : FORCES IN AN ELECTROMAGNETIC SYSTEMS
+
+// EXAMPLE : 2.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+A = 0.0001; // The Cross-sectional area of core in metre-square
+Mo = 4*%pi*(10)^(-7); // Permeability of air in Henre/metre
+Mr = 1000; // Relative permeability of core
+N1 = 10;N2=20;N3=10; // Number of turns
+I1 = 1.0;I2=0.5;I3=1.5; // Currents in Amphere
+d = 2.5; // Dimension of inner window in centimetre
+w = 1.0; // Each limb wide in centimeter
+
+
+// CALCULATIONS
+
+F = (N1*I1)+(N2*I2)-(N3*I3); // MMF in Amphere-turns (minus because third coil produces the flux in opposite direction to that of other to coils)
+L = ((d*4)+(I2*2*4))*10^-2; // Length of the Magnetic path in metre (4-is sides of the windows)(2-Going and returning of current I2)
+R = L/(Mr*Mo*A); // Reluctance of the Magnetic path in MKS unit of Reluctance
+phi = (F*10^3)/R; // Flux in milli-Weber
+B = phi/A; // Flux Density in Weber/metre Square
+H = F/L; // Magnetic Field Intensity in Amphere-turns/Metre
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 2.1 : SOLUTION :-") ;
+printf("\n (a) Flux in the core, phi = %.6f mWb ,\n",phi);
+printf("\n (b) Flux Density in the core, B = %.2f Wb/metre square \n",B);
+printf("\n (c) Magnetic Field Intensity in the core, H = %.2f At/m \n",H);
diff --git a/Working_Examples/2777/CH2/EX2.2/Ex2_2.sce b/Working_Examples/2777/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 0000000..d8992bb --- /dev/null +++ b/Working_Examples/2777/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,53 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 2 : FORCES IN AN ELECTROMAGNETIC SYSTEMS
+
+// EXAMPLE : 2.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N = 100; // Number of turns
+La = 0.3; // Mean arc length of material "a" is a Nickel-iron alloy in Metre
+Lb = 0.2; // Mean arc length of material "b" is a Steel in Metre
+Lc = 0.1; // Mean arc length of material "c" is a Cast Steel in Metre
+a = 0.001; // Area of the all Materials "a,b,c" in Metre-Square
+phi = 6*10^-4; // Magnetic Flux in Weber
+mue_0 = 4*%pi*10^ -7; // Permeability of the air in Henry/Meter
+
+
+// CALCULATIONS
+
+B = phi/a; // Flux Density in Telsa (Here Flux Density same for all the Materials "a,b,c" because Area of Cross Section is Same)
+Ha = 10; // Fileld Intensity in Amphere-Turn/Meter Correspounding to Flux density (B) of material "a" obtained from the Standard B-H curve
+Hb = 77; // Fileld Intensity in Amphere-Turn/Meter Correspounding to Flux density (B) of material "b" obtained from the Standard B-H curve
+Hc = 270; // Fileld Intensity in Amphere-Turn/Meter Correspounding to Flux density (B) of material "c" obtained from the Standard B-H curve
+F = (Ha*La)+(Hb*Lb)+(Hc*Lc); // The Total MMF Required in Amphere-Turns
+I = F/N; // Current flowing through the Coil in Amphere
+mue_r_a = B/(Ha*mue_0); // Relatative permeability of the Material "a"
+mue_r_b = B/(Hb*mue_0); // Relatative permeability of the Material "a"
+mue_r_c = B/(Hc*mue_0); // Relatative permeability of the Material "a"
+Ra = (Ha*La)/phi; // Relucatnce of the Material "a" in MKS unit
+Rb = (Hb*Lb)/phi; // Relucatnce of the Material "b" in MKS unit
+Rc = (Hc*Lc)/phi; // Relucatnce of the Material "c" in MKS unit
+L = (N*phi)/I; // Inductance of the Coil in Henry
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 2.2 : SOLUTION :-") ;
+printf("\n (a) The Total MMF , F = %.1f At \n ",F);
+printf("\n (b) Current flowing through the Coil , I = %.3f A \n",I);
+printf("\n (c.1) Relatative permeability of the Material a, mue_r_a = %.f \n ",mue_r_a);
+printf("\n (c.2) Relatative permeability of the Material b, mue_r_b = %.f \n ",mue_r_b);
+printf("\n (c.3) Relatative permeability of the Material c, mue_r_c = %.f \n ",mue_r_c);
+printf("\n (c.4) Relucatnce of the Material a, Ra= %.f MKS unit \n",Ra);
+printf("\n (c.5) Relucatnce of the Material b, Rb= %.1f MKS unit \n",Rb);
+printf("\n (c.6) Relucatnce of the Material c, Rc= %.f MKS unit \n",Rc);
+printf("\n (d) Inductance of the Coil , L = %.4f H \n",L);
diff --git a/Working_Examples/2777/CH2/EX2.3/Ex2_3.sce b/Working_Examples/2777/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 0000000..2821ad5 --- /dev/null +++ b/Working_Examples/2777/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,34 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivartava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 2 : FORCES IN AN ELECTROMAGNETIC SYSTEMS
+
+// EXAMPLE : 2.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+F = 35; // Total MMF in Amphere-Turns
+Lc = 0.1; // Inductance of The Material "c" in Henry
+a = 0.001; // Area of the all Materials "a,b,c" in Metre-Square
+
+
+// CALCULATIONS
+
+Hc = F/Lc; // Field Intensity in Amphere-Turns/Meter (Given that entire MMf apperas on Material "c" Because of the highest reluctance about 45000 MKS unit From Example 2.2)
+Bc = 0.65; // Flux density of material "c" in in Telsa obtained from the Standard B-H curve
+phi = Bc*a; // Flux in the core in Weber
+Ba = Bc; // Flux density of material "a" in in Telsa Same because Area of Cross Section is Same
+Bb = Bc; // Flux density of material "b" in in Telsabecause Area of Cross Section is Same
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 2.3 : SOLUTION :-") ;
+printf("\n (a) Flux in the core , phi = %.5f Wb \n ",phi);
+printf("\n (b) Flux density of material a,b,c , Ba = Bb = Bc %.2f T \n",Ba);
diff --git a/Working_Examples/2777/CH2/EX2.4/Ex2_4.sce b/Working_Examples/2777/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 0000000..8c46895 --- /dev/null +++ b/Working_Examples/2777/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,48 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 2 : FORCES IN AN ELECTROMAGNETIC SYSTEMS
+
+// EXAMPLE : 2.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer figure 2.7:- Page no. 41
+
+a = 0.0001; // Cross Sectional Area of the Core in Meter-Square
+Li = 0.158; // Total length of the Path abcdef in Meter (4.0*4.0 - 0.2 = 15.8cm = 0.158m)
+Lg = 0.002; // Length of the air gap in Meter
+mue_0 = 4*%pi*10^-7; // Permeability of the air in Henry/Meter
+mue_r = 10000; // Permeability of the core
+N = 10; // Number of Turns
+I = 1.0; // Current in the Coil in Amphere
+v = 50; // hall effect sensor generates volatge produces in milli volt per 1 Telsa
+Li_new = 0.16; // Length of the Flux path in Absence of the Air gap in Meter
+
+
+// CALCUALTIONS
+
+F = N*I; // MMF of the Coil in Amphere-turn
+Ri = Li/(mue_0*mue_r*a); // Relucatnce of the Iron Coil in MKS unit
+Rg = Lg/(mue_0*a); // Relucatnce of air gap in MKS unit
+R = Ri+Rg; // Total Reluctance in MKS unit
+phi = F/R; // Flux in the Core in Weber
+B = phi/a; // FLux density in the core(Presence of the Air gap) in Weber/Meter-Square
+HEV = B*50; // Output of the Hall effect Sensor device in Milli-Volt
+R_new = Li_new/(mue_0*mue_r*a) // Relucatance of the Magnetic Circuit in Absence of the Air gap
+phi_new = F/R_new; // New Flux in the Core in Weber
+B_new = phi_new/a; // New FLux density in the core in Weber/Meter-Square
+Ratio = B_new/B; // Ratio of the Flux Density in Absence of the Air gap and in the presence of the Air gap
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 2.4 : SOLUTION :-") ;
+printf("\n (a) Flux density in the core(Presence of the Air gap) , B = %.8f Wb/Meter-Square \n ",B);
+printf("\n (b) Output of the Hall effect Sensor device , HEV = %.7f mV \n",HEV);
+printf("\n (c) Ratio of the Flux Density in Absence of the Air gap and in the presence of the Air gap , Ratio = %.2f \n ",Ratio);
diff --git a/Working_Examples/2777/CH2/EX2.5/Ex2_5.sce b/Working_Examples/2777/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 0000000..068905a --- /dev/null +++ b/Working_Examples/2777/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,44 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 2 : FORCES IN AN ELECTROMAGNETIC SYSTEMS
+
+// EXAMPLE : 2.5
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer figure 2.3(a):- Page no. 36
+
+B = 1.0; // Flux Density in the Core in Weber/Meter-Square
+Liron = 0.55; // Mean length of the flux path of Iron in Meter
+Lair = 0.002; // Mean length of the flux path of Air Gap in Meter
+I = 20; // Coil Current in Amphere
+H = 200; // Field Intensity in Amphere-Turns/Meter
+mue_r = 20000; // Relative permeability of Ferrite core
+mue_0 = 4*%pi*10^-7; // Permeability of the air in Henry/Meter
+a = 0.0025; // Area of the Cross sectional of the core oin Metre-Square
+
+
+// CALCULATIONS
+
+phi = B*a; // Toatl Flux in the core in Weber
+Rair = Lair/(mue_0*a); // Relucatnce in the Air gap
+Fair = Rair*phi; // MMf in the Air gap in Amphere-Turns
+Firon = H*Liron; // MMf in the Iron core in Amphere-Turns
+F = Firon+Fair; // Total MMF in Amphere-Turns
+N = F/I; // Number of turns in the Coil
+F_new = B/(mue_0*mue_r); // Field Intensity in Amphere-Turns/Meter
+F_new_total = (Fair+F_new); // Total MMF in Amphere-Turns
+N_new = F_new_total/I; // Number of turns in the Coil
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 2.5 : SOLUTION :-") ;
+printf("\n (a) Number of turns in the Coil in air gap made of Silicon Steel having an field intensity 200At/m corresounds to 1.0 T Flux Density , N = %.2f appoximately 85 \n ",N);
+printf("\n (b) Number of turns in the Coil for a ferrite core of having Relative premeability of 20000 and magnetic Field Density corresponnds to 1.0 T , N_new = %.2f appoximately 82 \n",N_new);
diff --git a/Working_Examples/2777/CH3/EX3.1/Ex3_1.sce b/Working_Examples/2777/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 0000000..36f9aab --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,46 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Z = (0.05 + 0.05 * %i) * 100; // Transmission line parameters (impedance) in Ohms (multiplied by 100 because distance of the Transmission line is 100km)
+R = 0.05 * 100; // Transmission line Resistance in Ohms (multiplied by 100 because distance of the Transmission line is 100km)
+V1 = 220; // Terminal voltage in Volts
+V2 = 1 * 10 ^ 3; // Terminal volatge from Generator side in Volts
+P = 20 * 10 ^ 3; // Power in Watts
+
+
+// CACULATIONS
+
+I1 = P/V1; // Line current for 220V in Amphere
+I2 = P/V2; // Line current for 1kV in Amphere
+I1Z = Z*I1; // Voltage drop due to I1 in Volts
+I2Z = Z*I2; // Voltage drop due to I2 in Volts
+Loss1 = (I1 ^ 2) * R * 10 ^ -3; // Line loss for I1 in kW
+Loss2 = (I2 ^ 2) * R * 10 ^ -3; // Line loss for I2 in kW
+Vg1 = V1 + I1Z; // Input Voltages on Generator Terminal in Volts
+Vg2 = V2 + I2Z; // Input Voltages on Generator Terminal in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.1 : SOLUTION :-") ;
+printf("\n (a.1) Voltage drop due to I1 , I1Z = % .2f+j%.2f V \n ",real(I1Z),imag(I1Z));
+printf("\n (a.2) Voltage drop due to I2 , I2Z = % .f+j%.f V \n",real(I2Z),imag(I2Z));
+printf("\n (b.1) Line loss for I1 , Loss1 = %.2f kW \n ",Loss1);
+printf("\n (b.2) Line loss for I2 , Loss2 = % .2f kW \n",Loss2);
+printf("\n (c.1) Input Voltages on Generator Terminal from a load terminal , Vg1 = %.2f+j%.2f = %.2f V \n ",real(Vg1),imag(Vg1),abs(Vg1));
+printf("\n (c.2) Input Voltages on Generator Terminal from a Generating Station , Vg2 = % .f+j%.f = %.2f V \n",real(Vg2),imag(Vg2),abs(Vg2));
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) I1Z = (450.45)+j(450.45)V instead of (454.55)+j(454.55) V\n" );
+printf("\n (b) Vg1 = (670.45)+j(450.45) = 807.72 V instead of % .2f+j%.2f = %.2f V \n",real(Vg1),imag(Vg1),abs(Vg1) );
diff --git a/Working_Examples/2777/CH3/EX3.10/Ex3_10.sce b/Working_Examples/2777/CH3/EX3.10/Ex3_10.sce new file mode 100755 index 0000000..c46e524 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.10
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V1 = 440; // HV Side Voltage Rating of the Transformer in Volts
+V2 = 220; // LV Side Voltage Rating of the Transformer in Volts
+pf_o = 0.2; // No-load Power factor lagging
+pf_l = 0.8; // Load Power factor lagging
+I_o = 5; // No-load current in Amphere
+I_2 = 120; // Load current in Amphere
+
+// CALCULATIONS
+
+a = V1/V2; // Turns ratio of the two winding Transformers
+theta_o = acosd(pf_o); // No load power factor of the two winding Transformers in Degrees
+Io = I_o * exp(-(%i*theta_o*%pi/180)); // No load current of the two winding Transformers (minus because lagging) in Amphere
+theta = acosd(pf_l); // load power factor of the two winding Transformers in Degrees
+I2 = I_2 * exp(-(%i*theta*%pi/180)); // secondary load current of the two winding Transformers (minus because lagging) in Amphere
+I21 = I2/a; // Secondary referred to the primaryin Amphere
+I1 = Io + I21; // Primary current in Amphere
+I1_mag = abs(I1); // Primary current magnitude inj Amphere
+theta_1 = atand( imag(I1),real(I1)); // Primary current angle in Degree
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.10 : SOLUTION :-") ;
+printf("\n (a) Primary current , I1 = %.2f < %.1f A \n ",I1_mag,theta_1);
+
diff --git a/Working_Examples/2777/CH3/EX3.11/Ex3_11.sce b/Working_Examples/2777/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 0000000..b112baa --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,53 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.11
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 50; // kVA Rating of the Transformer
+f = 50; // Frequency in Hertz
+Wo = 190; // Meter Readings when HV Winding kept open in Watt
+Vo = 230; // Meter Readings when HV Winding kept open in Volts
+Io = 6.5; // Meter Readings when HV Winding kept open in Amphere
+R2 = 0.06; // Resistance of the LV Winding in Ohms
+V1 = 2300; // Voltage across the HV Side in Volts
+V2 = 230; // Voltage across the LV Side in Volts
+AC = 230; // Tranformer connected to AC mains in Volts
+
+
+// CALCULATIONS
+
+a = V1/V2; // Trasformation ratio of the Transformer
+Wc = Wo - ((Io ^ 2) * R2); // Core loss in Watts
+Po = Wc; // Core loss in Watts
+Pc = Wc; // Core loss in Watts
+cos_theta_o = Po/(Vo*Io); // No load power factor
+theta_o = acosd(cos_theta_o); // No load power factor angle in Degrees
+Ic = Io * cosd(theta_o);
+E = V1 - Io * exp(%i*(theta_o)*%pi/180);
+Rc = Pc/(Ic ^ 2 ); // Core loss Resistance in Ohms
+Im = Io * sind(theta_o); // Current through the Magnetizing branch in Amphre
+Xm = E/Im; // Magnetizing Reactance in Ohms
+Ift = (S * 10 ^ 3)/V2; // Full Load current in Amphere
+Ie = (Io/Ift)*100; // Percentage of the Exicting Current in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.11 : SOLUTION :-") ;
+printf("\n (a) Core loss , Wc = %.2f W \n ",Wc);
+printf("\n (b.1) No load power factor angle , theta_o = % .2f Degree \n",theta_o);
+printf("\n (b.2) No load power factor , cos(theta_o) = % .6f \n",cos_theta_o );
+printf("\n (c.1) Curent through Core loss Component , Ic = %.4f A \n ",Ic);
+printf("\n (c.2) Core loss Resistance , Rc = %.2f Ohms \n ",Rc);
+printf("\n (d) Current through the Magnetizing Component Xm , Im = % .2f A \n",Im);
+printf("\n (e) Percentage of the Exicting Current = % .2f Percent \n",Ie);
diff --git a/Working_Examples/2777/CH3/EX3.12/Ex3_12.sce b/Working_Examples/2777/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 0000000..05b3c12 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,57 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.12
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1000; // 1st Test at No-load condition f1 Frequency, Speed in RPM
+Vo1 = 250; // 1st Test at No-load condition f1 Frequency, Voltage in Volts
+Io1 = 0.5; // 1st Test at No-load condition f1 Frequency, Current in Amphere
+Wo1= 230; // 1st Test at No-load condition f1 Frequency, Power in Watts
+
+N2 = 900; // 2nd Test at No-load condition f2 Frequency, Speed in RPM
+Vo2 = 225; // 2nd Test at No-load condition f2 Frequency, Voltage in Volts
+Io2= 0.5; // 2nd Test at No-load condition f2 Frequency, Current in Amphere
+Wo2 = 200; // 2nd Test at No-load condition f2 Frequency, Power in Watts
+p = 6; // Number of poles of single phase alternator
+N = 220; // Number of the turns of single phase alternator
+R = 0.66; // Resistance of the single phase alternator in Ohms
+
+
+// CALCULATIONS
+
+f1 = (N1*p)/120; // 1st case Supply Frequency in Hertz
+Ratio1 = Vo1/f1; // 1st case Ratio of the Volatge and Frequency in Volts/Hertz
+f2 = (N2*p)/120; // 2nd case Supply Frequency in Hertz
+Ratio2 = Vo2/f2; // 2nd case Ratio of the Volatge and Frequency in Volts/Hertz
+
+c = (Wo1-(Io1^2)*R)/f1; // No-load corrected losses Eq 1 in Watts
+d = (Wo2-(Io2^2)*R)/f2; // No-load corrected losses Eq 2 in watts
+
+x = [ 1 f1 ; 1 f2 ]; // No-load corrected losses Eq 1 in watts
+y = [ c ; d ]; // No-load corrected losses Eq 2 in watts
+
+E = x\y; // Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
+A = E(1,1); // Solution of constant A in Watts/Hertz
+B = E(2,1); // Solution of constant B in watts/Hertz-Sqare
+Ph = f1*A; // Hysteresis loss at 50 Hertz in Watts
+Pe = (f1^2)*B; // Eddy current loss at 50 Hertz in Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.12 : SOLUTION :-") ;
+printf("\n (a) Hysteresis loss at %.f Hertz , Ph = %.3f W \n ",f1,Ph);
+printf("\n (b) Eddy current loss at %.f Hertz , Pe = % .2f W \n",f1,Pe);
+
+
+
diff --git a/Working_Examples/2777/CH3/EX3.13/Ex3_13.sce b/Working_Examples/2777/CH3/EX3.13/Ex3_13.sce new file mode 100755 index 0000000..898bb00 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,55 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1500; // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Speed in RPM
+Vo1 = 250; // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Voltage in Volts
+Wo1= 55; // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Power in Watts
+N2 = 1200; // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Speed in RPM
+Vo2 = 200; // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Voltage in Volts
+Wo2 = 40; // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Power in Watts
+p = 4; // Number of poles of single phase alternator
+
+
+// CALCULATIONS
+
+f1 = (N1*p)/120; // 1st case Supply Frequency in Hertz
+Ratio1 = Vo1/f1; // 1st case Ratio of the Volatge and Frequency in Volts/Hertz
+f2 = (N2*p)/120; // 2nd case Supply Frequency in Hertz
+Ratio2 = Vo2/f2; // 2nd case Ratio of the Volatge and Frequency in Volts/Hertz
+
+c = Wo1/f1; // No-load corrected losses Eq 1 in Watts
+d = Wo2/f2; // No-load corrected losses Eq 2 in watts
+
+x = [ 1 f1 ; 1 f2 ]; // No-load corrected losses Eq 1 in watts
+y = [ c ; d ]; // No-load corrected losses Eq 2 in watts
+
+E = x\y; // Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
+A = E(1,1); // Solution of constant A in Watts/Hertz
+B = E(2,1); // Solution of constant B in watts/Hertz-Sqare
+Ph1 = f1*A; // Hysteresis loss at 50 Hertz in Watts
+Pe1 = (f1^2)*B; // Eddy current loss at 50 Hertz in Watts
+Ph2 = f2*A; // Hysteresis loss at 40 Hertz in Watts
+Pe2 = (f2^2)*B; // Eddy current loss at 40 Hertz in Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.13 : SOLUTION :-") ;
+printf("\n (a.1) Hysteresis loss at %.f Hertz , Ph = %.f W \n ",f1,Ph1);
+printf("\n (a.2) Eddy current loss at %.f Hertz , Pe = %.f W \n",f1,Pe1);
+printf("\n (b.1) Hysteresis loss at %.f Hertz , Ph = %.f W \n ",f2,Ph2);
+printf("\n (b.2) Eddy current loss at %.f Hertz , Pe = %.f W \n",f2,Pe2);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Hysteresis loss at %.f Hertz , Ph = 25 W instead of %.f W \n ",f2,Ph2);
diff --git a/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce b/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce new file mode 100755 index 0000000..e0b0ab9 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,75 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 10 * 10 ^ 3; // Rating of the Single Transformer in VA
+f = 50; // Frequency in Hertz
+Pc = 110; // Required input no-load at normal voltage in Watts (Core loss)
+Psc = 120; // Required input Short-circuit at full-load current in Watts (copper loss or short circuit loss)
+
+
+// CALCUATIONS
+// case (a) for Unity power factor
+
+cos_theta1 = 1; // Unity Power factor
+K1 = 1.0; // Full load
+K2 = 0.5; // Half load
+eta_11 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+( K1 ^ 2 )*Psc); // Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+eta_12 = 100 * (K2*S*cos_theta1)/((K2*S*cos_theta1)+Pc+( K2 ^ 2 )*Psc); // Efficiency at unity factor and half load ( beacuse taken k2 = 0.5 ) in percentage
+
+// case (b) for 0.8 power factor lagging
+
+cos_theta2 = 0.8; // 0.8 power factor lagging
+eta_21 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+( K1 ^ 2 )*Psc); // Efficiency at 0.8 power factor lagging and full load ( beacuse taken k1 = 1 ) in percentage
+eta_22 = 100 * (K2*S*cos_theta2)/((K2*S*cos_theta2)+Pc+( K2 ^ 2 )*Psc); // Efficiency at 0.8 power factor lagging and half load ( beacuse taken k2 = 0.5 ) in percentage
+
+// Case (c) for 0.8 poer factor leading
+
+eta_31 = eta_21; // Efficiency at 0.8 power factor leading and full load will be same as the Efficiency at 0.8 power factor lagging and full load in percentage
+eta_32 = eta_22; // Efficiency at 0.8 power factor leading and half load will be same as the Efficiency at 0.8 power factor lagging and half load in percentage
+
+// Case (d) Maximum Efficiency assumed that unity power factor
+// Psc = Pc At Maximum Efficiency
+
+eta_41 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+Pc); // Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+
+// Case (e) Maximum Efficiency assumed that 0.8 power factor lagging
+// Psc = Pc At Maximum Efficiency
+
+eta_51 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+Pc); // Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+
+// Case (f) Maximum Efficiency assumed that 0.8 power factor leading
+// Psc = Pc At Maximum Efficiency
+
+eta_61 = eta_51; // Maximum Efficiency at 0.8 power factor leading and full load will be same as the Maximum Efficiency at 0.8 power factor lagging and full load in percentage
+out1 = K1*S*cos_theta1; // Output at which maximum efficiency occurs at unity power factor at full load in Watts
+out2 = K1*S*cos_theta2; // Output at which maximum efficiency occurs at 0.8 power factor lagging at full load in Watts
+out3 = K1*S*cos_theta2; // Output at which maximum efficiency occurs at unity power factor leading at full load in Watts
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.14 : SOLUTION :-") ;
+printf("\n (a.1) Efficiency at unity power factor and full load , eta = %.2f Percent \n ",eta_11);
+printf("\n (a.2) Efficiency at unity power factor and half load , eta= % .2f Percent \n",eta_12);
+printf("\n (b.1) Efficiency at 0.8 power factor lagging and full load , eta = %.2f Percent \n ",eta_21);
+printf("\n (b.2) Efficiency at 0.8 power factor lagging and half load , eta= % .2f Percent \n",eta_22);
+printf("\n (c.1) Efficiency at 0.8 power factor leading and full load , eta = %.2f Percent \n ",eta_31);
+printf("\n (c.2) Efficiency at 0.8 power factor leading and half load , eta= % .2f Percent \n",eta_32);
+printf("\n (d) Maximum Efficiency at unity power factor and full load , eta = %.2f Percent \n ",eta_41);
+printf("\n (e) Maximum Efficiency at 0.8 power factor lagging and full load , eta = %.2f Percent \n ",eta_51);
+printf("\n (f) Maximum Efficiency at 0.8 power factor leading and full load , eta = %.2f Percent \n ",eta_61);
+printf("\n (g) Output at which maximum efficiency occurs at unity power factor at full load = %.f W \n ",out1);
+printf("\n (h) Output at which maximum efficiency occurs at 0.8 power factor lagging at full load = %.f W \n ",out2);
+printf("\n (i) Output at which maximum efficiency occurs at 0.8 power factor leading at full load = %.f W \n ",out3);
+printf("\n IN THE ABOVE PROBLEM MAXIMUM EFFICIENCY AND THE OUTPUT AT WHICH THE MAXIMUM EFFICIENCY OCCURS IS NOT CALCULATED IN THE TEXT BOOK \n")
diff --git a/Working_Examples/2777/CH3/EX3.15/Ex3_15.sce b/Working_Examples/2777/CH3/EX3.15/Ex3_15.sce new file mode 100755 index 0000000..8cdda47 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.15
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer figure 3.17 page no. 101
+
+S = 500 * 10 ^ 6; // Rating of power transformer in VA
+V1 = 400 * 10^3; // HV side rating of the power transformer in Volts
+V2 = 131 * 10^3; // LV side rating of the power transformer in Volts
+pcu = 5; // Rated Copper loss in Percentage
+pi = 1; // Rated Core loss in Percentage
+
+
+// CALCULATIONS
+
+Pcu = S*(pcu/100); // Rated Copper loss in Watts
+Pi = S*(pi/100); // Rated Core loss in Watts
+kt = 0.25*3 + 0.75*3 + 1*3 + 0.5*3 + 1.0*3 + 0.25*6 + 1.0*3; // From graph figure 3.17 page no. 101
+out = S*kt; // Output energy in kilo-watt-hour
+kt2 = 0.54375; // From graph figure 3.17 page no. 101
+eloss = 24*Pi + S*kt2; // Energy required in losses in kilo-watt-hour { Energy required in losses = 24*Pi + sigma(copper loss * duration)}
+eta = 100*(out/(out+eloss)); // All day efficiency
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.15: SOLUTION :-");
+printf("\n (a) All day efficiency = %.2f percent \n",eta)
diff --git a/Working_Examples/2777/CH3/EX3.16/Ex3_16.sce b/Working_Examples/2777/CH3/EX3.16/Ex3_16.sce new file mode 100755 index 0000000..7286b30 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.16
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 20 * 10 ^ 3; // Rating of the Step-down Transformer in VA
+f = 50; // Frequency in Hertz
+V = 200; // Normally supplied Voltage of Step-down Transformer in Volts
+Vsc = 100; // Potential difference when Secondary being Short- Circuited in Volts
+Isc = 10; // Primary Current when Secondary being Short- Circuited in Amphere
+Cos_theta_sc = 0.28; // Power factor when Secondary being Short- Circuited
+
+
+// CALCULATIONS
+
+I = S/V; // Rated primary current in Amphere
+Wsc = Vsc * Isc * Cos_theta_sc; // Power loss when Secondary being Short- Circuited in Watts
+R = Wsc/(Isc ^ 2); // Resistance of Transformer referred to primary side in Ohms
+Z = Vsc/Isc; // Referred Impedence in Ohms
+X = sqrt((Z^2)-(R^2)); // Leakage Reactance referred to primary side in Ohms
+Er = (I*R)/V; // Per unit Resistance in Ohms
+Ex = (I*X)/V; // Per unit Reactance in Ohms
+Cos_theta1 = 1.0; // Unity Power factor
+Cos_theta2 = 0.6; // 0.6 Power factor Lagging
+Cos_theta3 = 0.6; // 0.6 Power factor Leading
+Sin_theta1 = 0.0; // Unity Power factor
+Sin_theta2 = 0.8; // 0.6 Power factor Lagging
+Sin_theta3 = 0.8; // 0.6 Power factor Leading
+E1 = (Er*Cos_theta1)+(Ex*Sin_theta1); // pu Regulation at Unity Power factor
+E2 = (Er*Cos_theta2)+(Ex*Sin_theta2); // pu Regulation at 0.6 Power factor Lagging
+E3 = (Er*Cos_theta3)-(Ex*Sin_theta3); // pu Regulation at 0.6 Power factor Leading
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.16 : SOLUTION :-") ;
+printf("\n (a) pu Regulation at Unity Power factor , E = %.1f \n ",E1);
+printf("\n (b) pu Regulation at 0.6 Power factor Lagging , E= % .2f \n",E2);
+printf("\n (c) pu Regulation at 0.6 Power factor Leading , E= % .2f \n",E3);
+
+
+
diff --git a/Working_Examples/2777/CH3/EX3.17/Ex3_17.sce b/Working_Examples/2777/CH3/EX3.17/Ex3_17.sce new file mode 100755 index 0000000..10e6b62 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,41 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.17
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 500; // Rating of the 3-Phase transformer in kVA
+V1 = 11 * 10 ^ 3; // Votage rating of the 3-Phase transformer on HV side in Volts
+V2 = 400; // Votage rating of the 3-Phase transformer on LV side in Volts
+f = 60; // Frequencyin Hertz
+eta = 98; // Maximum Efficency of the Transformer in Percentage Operating at 80% full load and Unity Power factor
+K = 0.8; // Beacuse 80% Full load
+x = 1.0; // Unity Power factor
+Ex = 4.5; // Percentage impedance
+
+
+// CALCULATIONS
+
+Out = S * K * x; // Output in KiloWatts at 80% full load and Unity Power factor
+Inp = Out/(eta/100); // Input in KiloWatts at full load and Unity Power factor
+Total_loss = Inp - Out; // Total loss at full load in KiloWatts
+Cu_loss = Total_loss/2; // Copper loss in KiloWatts at 80% full load and Unity Power factor
+Pcu = Cu_loss/(K ^2 ); // Full load Copper loss in KiloWatts
+Er = Pcu/S; // Per unit Resistance
+theta = atand((Ex/100)/Er); // Power factor angle at secondary terminal voltage is minimum in Degrees
+Pf = cosd(theta); // Load power factor for minimum volatge of the secondary terminal
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.17 : SOLUTION :-") ;
+printf("\n (a) Load power factor for minimum volatge of the secondary terminal , cos(theta) = %.4f lagging \n ",Pf);
diff --git a/Working_Examples/2777/CH3/EX3.19/Ex3_19.sce b/Working_Examples/2777/CH3/EX3.19/Ex3_19.sce new file mode 100755 index 0000000..f64cc2f --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.19/Ex3_19.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.19
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Sa = 200; // Rating of the TWO 1-Phase Transformer in kVA
+Z1 = 0.005 + 0.08 * %i // Equivalent Impedance of the Transformer-1 in Per-Unit
+Z2 = 0.0075 + 0.04 * %i // Equivalent Impedance of the Transformer-2 in Per-Unit
+P = 400; // Total load in kiloWatts
+Cos_theta = 1.0; // Unity power factor
+
+
+// CALCULATIONS
+
+kVA = P/Cos_theta; // kVA rating of the Transformer
+S = kVA; // kVA rating of the Transformer
+S1 = ( Z2/(Z1+Z2) )*S; // Load shared by Transformer-1 in kVA
+S2 = S - S1; // Load shared by Transformer-2 in kVA
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.19 : SOLUTION :-") ;
+printf("\n (a) Load shared by Transformer-1 , S1 = %.2f+j(%.2f) kVA \n ",real(S1),imag(S1));
+printf("\n (b )Load shared by Transformer-2 , S2 = %.2f+j%.2f kVA \n ",real(S2),imag(S2));
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) S1 = (-131.90)+j(38.47)kVA instead of %.2f+j(%.2f) kVA \n ",real(S1),imag(S1));
+printf("\n (b) S2 = (268.1)+j(.38047)kVA instead of %.2f+j%.2f kVA \n ",real(S2),imag(S2));
diff --git a/Working_Examples/2777/CH3/EX3.2/Ex3_2.sce b/Working_Examples/2777/CH3/EX3.2/Ex3_2.sce new file mode 100755 index 0000000..dbc0933 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,37 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+E1 = 6.6 * 10 ^ 3; // Primary voltage in Volts
+E2 = 220; // Secondary Voltage in volts
+f = 50; // Frequency in Hertz
+phi_m = 0.06; // Flux in Weber
+S = 50 * 10^6; // Rating of the single-phase transformer in VA
+
+// CALCULATIONS
+
+N1 = E1/(4.44*f*phi_m); // Number of turns in Primary
+vpn1 = E1/N1; // Voltage per turns in Primary in Volts/turn
+N2 = E2/(4.44*f*phi_m); // Number of turns in Secondary
+vpn2 = E2/N2; // Voltage per turns in Secondary in Volts/turn
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.2 : SOLUTION :-") ;
+printf("\n (a.1) Number of turns in Primary , N1 = %.1f Turns nearly 496 Turns \n ",N1);
+printf("\n (a.2) Number of turns in Secondary , N2 = %.1f Turns nearly 16 Turns \n",N2);
+printf("\n (b.1) Voltage per turns in Primary , vpn1 = %.1f Volts/turns \n ",vpn1);
+printf("\n (b.2) Voltage per turns in Secondary , vpn2 = %.2f Volts/turns \n ",vpn2);
+
diff --git a/Working_Examples/2777/CH3/EX3.20/Ex3_20.sce b/Working_Examples/2777/CH3/EX3.20/Ex3_20.sce new file mode 100755 index 0000000..6bef844 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.20/Ex3_20.sce @@ -0,0 +1,42 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.20
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V1 = 110; // Primary voltage of the Two Transformers the two primaries are connected in parallel in Volts
+I1 = 2.0; // Primary Current in Amphere
+P1 = 40; // Primary power intake in Watts
+V2 = 28; // secondary voltage of the Two Transformers the two secondary are connected in phase opposition in Volts
+I2 = 6.8; // secondary Current in Amphere
+P2 = 180; // secondary power intake in Watts
+a = 110/220; // Turn ratio of the Transformer
+
+
+// CALCULATIONS
+
+theta_o = acosd((a*P1)/(a*I1*V1)); // Primary Power factor angle in Degrees
+Io = 1.0 * (cosd(theta_o)-sind(theta_o)* %i); // No-load current in individual transformer in Amphere
+theta_sc = acosd((a*P2)/(a*I2*V2)); // Secondary Power factor angle in Degrees
+i_sc = I2 * ( cosd(theta_sc)-sind(theta_sc)* %i); // Secondary current in Amphere
+I_sc = (1/a)*i_sc; // referred Secondary current in each of the primary side in Amphere
+It1 = Io + I_sc; // LT winding current in the 1st Transformer in Amphere
+It2 = Io - I_sc; // LT winding current in the 2nd Transformer in Amphere
+In1 = It1 + It2; // The current flowing on paralel connected LT winding (This is same as total no-load current in the two Transforemer) in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.20 : SOLUTION :-") ;
+printf("\n (a) LT ( Primary ) winding current in the 1st Transformer , It1 = %.3f+j(%.4f) A \n ",real(It1),imag(It1));
+printf("\n (b) LT ( Primary ) winding current in the 2nd Transformer , It2= %.3f+j%.5f A \n",real(It2),imag(It2));
+printf("\n (c) LT winding are connected in parallel, the current flowing on paralel connected LT winding, In1 = %.3f+j(%.5f) A \n",real(In1),imag(In1));
diff --git a/Working_Examples/2777/CH3/EX3.21/Ex3_21.sce b/Working_Examples/2777/CH3/EX3.21/Ex3_21.sce new file mode 100755 index 0000000..807aae2 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,54 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.21
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer figures 3.31(a), 3.31(b) and 3.31(c):- Page no. 121
+
+VaH = 220; // HV side Voltage of the two winding Transformer in volts for case(a)
+VaL = 110; // LV side Voltage of the two winding Transformer in volts for case(a)
+VbH = 330; // HV side Voltage of the two winding Transformer in volts for case(b)
+VbL = 220; // LV side Voltage of the two winding Transformer in volts for case(b)
+VcH = 330; // HV side Voltage of the two winding Transformer in volts for case(c)
+VcL = 110; // LV side Voltage of the two winding Transformer in volts for case(c)
+S = 1.5; // Ratings of the the two winding Transformer in kVA
+I1 = 6.8; // Rated current in HV side in Amphere
+I2 = 13.6; // Rated current in LV side in Amphere
+
+
+// CALCULATIONS
+// for case(a):- figure 3.31(b) page no. 121
+
+IbH = I2; // Current of Auto-Transformer in HV side in Amphere
+IbL = I1 + I2; // Current of Auto-Transformer in LV side in Amphere
+KVA_b_L = (VbL*IbL)/1000; // LV side kVA rating of the Auto-Transformer in kVA
+KVA_b_H = (VbH*IbH)/1000; // HV side kVA rating of the Auto-Transformer in kVA
+
+// for case(b):- figure 3.31(c) page no. 121
+
+IcH = I1; // Current of Auto-Transformer in HV side in Amphere
+IcL = I1 + I2; // Current of Auto-Transformer in LV side in Amphere
+KVA_c_L = (VcL*IcL)/1000; // LV side kVA rating of the Auto-Transformer in kVA
+KVA_c_H = (VcH*IcH)/1000; // HV side kVA rating of the Auto-Transformer in kVA
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.21 : SOLUTION :-") ;
+printf("\n (a.1) Current of Auto-Transformer in HV side for case (b) , IH = %.1f A \n ",IbH);
+printf("\n Current of Auto-Transformer in LV side for case (b), IL= % .1f A \n",IbL);
+printf("\n (a.2) LV side kVA rating of the Auto-Transformer for case (b), KVAL = % .3f kVA \n ",KVA_b_L);
+printf("\n HV side kVA rating of the Auto-Transformer for case (b), KVAH= % .3f kVA \n",KVA_b_H);
+printf("\n (b.1) Current of Auto-Transformer in HV side for case (c) , IH = %.1f A \n ",IcH);
+printf("\n Current of Auto-Transformer in LV side for case (c) , IL= % .1f A \n",IcL);
+printf("\n (b.2) LV side kVA rating of the Auto-Transformer for case (c), KVAL = % .3f kVA \n ",KVA_c_L);
+printf("\n HV side kVA rating of the Auto-Transformer for case (c) , KVAH= % .3f kVA \n",KVA_c_H);
diff --git a/Working_Examples/2777/CH3/EX3.22/Ex3_22.sce b/Working_Examples/2777/CH3/EX3.22/Ex3_22.sce new file mode 100755 index 0000000..fce8b52 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.22/Ex3_22.sce @@ -0,0 +1,55 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.22
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 10 * 10 ^ 3; // Rating of the Two-winding Transformer in VA
+V1 = 2000; // HV side voltage of the Two-winding Transformer in Volts
+V2 = 200; // LV side voltage of the Two-winding Transformer in Volts
+V_A_H = 2200; // Two-winding Transformer is connected to auto transformer HV side in Volts
+V_A_L = 200; // Two-winding Transformer is connected to auto transformer LV side in Volts
+f = 50; // Frequency in Hertz
+
+
+// CALCULATIONS
+// for finding (a)
+
+I2 = S/V2; // Rated LV side current of winding for Step-up Auto transformer in Amphere
+I1 = S/V1; // Rated HV side current of winding for Step-up Auto transformer in Amphere
+IaH = I2; // The HV side current in the Auto-Transformer for Full-load in Amphere
+IaL = I2 + I1 ; // The LV side current in the Auto-Transformer for Full-load in Amphere
+VL = V1; // LV side voltage in Volts
+VH = V1 + V2; // HV side voltage in Volts
+KVA_a_L = (VL*IaL)/1000; // kVA rating of LV SIDE
+KVA_a_H = (VH*IaH)/1000; // kVA rating of HV SIDE
+
+// For finding (b)
+
+IbH = I1; // HV side Rated current through the Auto-Transformer in Amphere
+IbL = I1 + I2; // LV side Rated current through the Auto-Transformer in Amphere
+KVA_b_L = (V_A_L*IbL)/1000; // kVA rating of LV SIDE as output Auto-Transformer
+KVA_b_H = (V_A_H*IbH)/1000; // kVA rating of HV SIDE as output Auto-Transformer
+
+// case (c)
+
+V = V1; // Voltage on the Secondary, if the Commom windings are open
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.22 : SOLUTION :-") ;
+printf("\n (a.1) HV side Curent supplied by the common windings , IH = %.f A \n ",IaH);
+printf("\n (a.2) LV side Curent supplied by the common windings , IL= %.f A \n",IaL);
+printf("\n (b.1) KVA rating of LV SIDE as output Auto-Transformer , KVAL = %.f kVA \n ",KVA_b_L);
+printf("\n (b.2) KVA rating of HV SIDE as output Auto-Transformer , KVAH= %.f kVA \n",KVA_b_H);
+printf("\n (c) Voltage on the Secondary, if the Commom windings are open , V = %.f V \n ",V);
diff --git a/Working_Examples/2777/CH3/EX3.23/Ex3_23.sce b/Working_Examples/2777/CH3/EX3.23/Ex3_23.sce new file mode 100755 index 0000000..28ed6db --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.23/Ex3_23.sce @@ -0,0 +1,58 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.23
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 100; // Rating of the 3-Phase Transformer in kVA
+VH = 11; // HV side voltage in kilo-Volts
+VL = 440; // LV side voltage in Volts
+Vl = 400; // Line voltage in Volts
+ZA = 0.6; // Line impedance in line A in Ohms
+ZB = 0.6*(0.8 + 0.6 * %i); // Line impedance in line B in Ohms
+ZC = 0.6*(0.5 - 0.866 * %i); // Line impedance in line C in Ohms
+
+
+// CALCULATIONS
+
+Vp = Vl/sqrt(3); // Phase voltage in Volts
+VAB = Vl * exp( %i * 0 * %pi/180); // Line Voltage across line A and B in Volts
+VBC = Vl * exp( %i * (-120) * %pi/180); // Line Voltage across line B and C in Volts
+VCA = Vl * exp( %i * 120 * %pi/180); // Line Voltage across line C and A in Volts
+VAN = (Vl/sqrt(3)) * exp( %i * (-30) * %pi/180); // Phase Voltage across line A and Neutral in Volts
+VBN = (Vl/sqrt(3)) * exp( %i * (-150) * %pi/180); // Phase Voltage across line B and Neutral in Volts
+VCN = (Vl/sqrt(3)) * exp( %i * (90) * %pi/180); // Phase Voltage across line C and Neutral in Volts
+IA = VAN/ZA; // Line current in line A in Amphere
+IB = VBN/ZB; // Line current in line B in Amphere
+IC = VCN/ZC; // Line current in line C in Amphere
+IN = IA + IB + IC ; // Current in the Neutral in Amphere
+Y = (1/ZA)+(1/ZB)+(1/ZC); // Net Admittance in mho
+VN = IN/Y; // Neutral Potential in Volts
+VDA = VAN - VN; // Voltage drops across the ZA in Volts
+VDB = VBN - VN; // Voltage drops across the ZB in Volts
+VDC = VCN - VN; // Voltage drops across the ZC in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.23 : SOLUTION :-") ;
+printf("\n (a.1) Line current in line A , IA = %.f<%.f A \n ",abs(IA),atand(imag(IA),real(IA)));
+printf("\n (a.2) Line current in line B , IB = %.f<%.2f A \n",abs(IB),atand(imag(IB),real(IB)));
+printf("\n (a.3) Line current in line C , IC = %.f<%.f A \n ",abs(IC),atand(imag(IC),real(IC)));
+printf("\n (b.1) Phase Voltage across line A and Neutral , VAN = %.f<%.f V \n",abs(VAN),atand(imag(VAN),real(VAN)));
+printf("\n (b.2) Phase Voltage across line B and Neutral , VBN = %.f<%.f V \n ",abs(VBN),atand(imag(VBN),real(VBN)));
+printf("\n (b.3) Phase Voltage across line C and Neutral , VCN = %.f<%.f V \n",abs(VCN),atand(imag(VCN),real(VCN)));
+printf("\n (c) Neutral Potential , VN = %.1f<%.2f V \n ",abs(VN),atand(imag(VN),real(VN)));
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) IC = 385<-90.1 V instead of %.f<%.f A \n ",abs(IC),atand(imag(IC),real(IC)));
+printf("\n (b) VN = 230.5<78.17 V instead of %.1f<%.2f V \n ",abs(VN),atand(imag(VN),real(VN)) );
+printf("\n From Calculation of the IC, rest all the Calculated values in the TEXT BOOK is WRONG because of the IC value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH3/EX3.24/Ex3_24.sce b/Working_Examples/2777/CH3/EX3.24/Ex3_24.sce new file mode 100755 index 0000000..e37c3d1 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.24/Ex3_24.sce @@ -0,0 +1,85 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.24
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+VL = 11000; // Line-line voltage of the 3 identical 1-phase Transformer in Volts
+IL = 10; // Line current of the 3 identical 1-phase Transformer in Amphere
+a = 10; // Ratio of trun per phase of the 3 identical 1-phase Transformer
+
+
+// CALCULATIONS
+// For case (a) STAR-STAR
+
+VPp_a = VL/sqrt(3); // Primary phase volatge in Volts
+IPp_a = IL; // Primary phase current in Amphere
+VSp_a = VPp_a/a; // Secondary phase voltage in Volts
+ISp_a = a*IPp_a; // Secondary phase current in Amphere
+ISl_a = ISp_a; // Secondary line current in Amphere
+VSl_a = VSp_a*sqrt(3); // Secondary line voltage in Volts
+Out_a = sqrt(3)*VSl_a*ISl_a/1000; // Output in kVA
+
+// For case (b) STAR-DELTA
+
+VPp_b = VL/sqrt(3); // Primary phase volatge in Volts
+IPp_b = IL; // Primary phase current in Amphere
+VSp_b = VPp_a/a; // Secondary phase voltage in Volts
+ISp_b = a*IPp_b; // Secondary phase current in Amphere
+ISl_b = sqrt(3)*ISp_b; // Secondary line current in Amphere
+VSl_b = VSp_b; // Secondary line voltage in Volts
+Out_b = sqrt(3)*VSl_b*ISl_b/1000; // Output in kVA
+
+// For case (c) DELTA-DELTA
+
+VPp_c = VL; // Primary phase volatge in Volts
+IPp_c = IL/sqrt(3); // Primary phase current in Amphere
+VSp_c = VPp_c/a; // Secondary phase voltage in Volts
+ISp_c = a*IPp_c; // Secondary phase current in Amphere
+ISl_c = sqrt(3)*ISp_c; // Secondary line current in Amphere
+VSl_c = VSp_c; // Secondary line voltage in Volts
+Out_c = sqrt(3)*VSl_c*ISl_c/1000; // Output in kVA
+
+// For case (d) DALTA-STAR
+
+VPp_d = VL; // Primary phase volatge in Volts
+IPp_d = IL/sqrt(3); // Primary phase current in Amphere
+VSp_d = VPp_d/a; // Secondary phase voltage in Volts
+ISp_d = a*IPp_d; // Secondary phase current in Amphere
+ISl_d = ISp_d; // Secondary line current in Amphere
+VSl_d = sqrt(3)*VSp_d; // Secondary line voltage in Volts
+Out_d = sqrt(3)*VSl_d*ISl_d/1000; //Output in kVA
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.24 : SOLUTION :-") ;
+printf("\n For STAR-STAR Connection \n\n (a.1) Secondary line voltage = %.f V \n ",VSl_a);
+printf("\n (a.2) Secondary line current = % .f A \n",ISl_a);
+printf("\n (a.3) Primary phase current = %.f A \n ",IPp_a);
+printf("\n (a.4) Secondary phase current = %.f A \n",ISp_a);
+printf("\n (a.5) Output = %.2f kVA \n ",Out_a);
+printf("\n For STAR-DELTA Connection \n\n (b.1) Secondary line voltage = % .f V \n",VSl_b);
+printf("\n (b.2) Secondary line current = %.f A \n ",ISl_b);
+printf("\n (b.3) Primary phase current = %.f A \n",IPp_b);
+printf("\n (b.4) Secondary phase current = %.f A \n ",ISp_b);
+printf("\n (b.5) Output = % .2f kVA \n",Out_b);
+printf("\n For DELTA-DELTA Connection \n\n (c.1) Secondary line voltage = %.f V \n ",VSl_c);
+printf("\n (c.2) Secondary line current = %.2f A \n",ISl_c);
+printf("\n (c.3) Primary phase current = %.2f A \n ",IPp_c);
+printf("\n (c.4) Secondary phase current = %.1f A \n",ISp_c);
+printf("\n (c.5) Output = %.1f kVA \n ",Out_c);
+printf("\n For DELTA-STAR Connection \n\n (d.1) Secondary line voltage = % .2f V \n",VSl_d);
+printf("\n (d.2) Secondary line current = %.1f A \n ",ISl_d);
+printf("\n (d.3) Primary phase current = %.2f A \n",IPp_d);
+printf("\n (d.4) Secondary phase current = %.1f A \n ",ISp_d);
+printf("\n (d.5) Output = % .1f kVA \n",Out_d);
diff --git a/Working_Examples/2777/CH3/EX3.3/Ex3_3.sce b/Working_Examples/2777/CH3/EX3.3/Ex3_3.sce new file mode 100755 index 0000000..0dd28bb --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,29 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 50; // Frequency in Hertz
+N = 50; // Number of turns in Secondary
+E =220; // Induced voltage in Volts
+
+
+// CALCULATIONs
+
+phi_m = E/(4.44*f*N); // Maximum value of the Flux in Weber
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.3 : SOLUTION :-") ;
+printf("\n (a) Maximum value of the Flux , phi_m = % .7f Wb \n ",phi_m);
diff --git a/Working_Examples/2777/CH3/EX3.4/Ex3_4.sce b/Working_Examples/2777/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 0000000..337d1dc --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,31 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 1.5; // Transformer Rating in KVA
+E1 = 220; // HV side voltage in volts
+E2 = 40; // LV side voltage in volts
+
+
+// CALCULATION
+
+Ihv = (S * 10 ^ 3)/E1; // Rated HV side Curent in Amphere
+Ilv = (S * 10 ^ 3)/E2; // Rated lV side Curent in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.4 : SOLUTION :-") ;
+printf("\n (a) Rated HV side Curent , Ihv = %.2f A \n ",Ihv);
+printf("\n (b) Rated LV side Curent , Ilv = %.1f A \n",Ilv);
diff --git a/Working_Examples/2777/CH3/EX3.5/Ex3_5.sce b/Working_Examples/2777/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 0000000..a12f972 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,48 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.5
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Ai = 2.3 * 10 ^ -3; // Cross-Sectional area of the core in Meter-Square
+mue_0 = 4*%pi*10^ -7; // Permeability of the air in Henry/Meter
+Fe_loss = 2.6; // Iron loss at the working Flux density Watts/kg
+Fe_den = 7.8 * 10 ^ 3; // Density of the Iron in kg/Meter-Cube
+N1 = 800; // Number of Turns of the Primary winding
+L = 2.5; // Length of the Flux path in Meter
+mue_r = 1000; // Relative Permeability
+E = 400; // Primary Volatge of the Transformer in Volts
+f = 50; // Frequency in Hertz
+
+
+// CALCULATIONS
+
+Bm = E/(4.44*f*Ai*800); // Flux Density in Weber/Meter-Square
+phi_m = (Bm*Ai)*10^3; // Maximum Flux in the core in milli-Weber
+F = (L*Bm)/(mue_r*mue_0); // Magnetizing MMF in Amphere-turns
+Im = F/(N1*sqrt(2)); // Magnetizing Current in Amphere
+Vol = L*Ai; // Volume of the Core in Meter-Cube
+W = Vol * Fe_den; // Weight of the Core in kg
+Total_Fe_loss = Fe_loss * W; // Total Iron loss in Watt
+Ic = Total_Fe_loss/E; // Loss component of Current in Amphere
+Io= sqrt((Ic ^ 2)+(Im ^ 2)); // No load Current in Amphere
+pf_angle = atand(Io/Ic); // No load Power factor angle in degree
+pf = cosd(pf_angle); // No load Power factor
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.5 : SOLUTION :-") ;
+printf("\n (a) Maximum Flux in the core , phi_m = %.6f mWb \n ",phi_m);
+printf("\n (b) No load Current , I0 = %.3f A \n",Io);
+printf("\n (c) No load Power factor angle = %.3f degree \n ",pf_angle);
+printf("\n (d) No load Power factor = %.4f \n ",pf);
diff --git a/Working_Examples/2777/CH3/EX3.6/Ex3_6.sce b/Working_Examples/2777/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 0000000..9b91c0c --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,44 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.6
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 5; // Transformer Rating in kVA
+V1 = 220; // HV side voltage in volts
+V2 = 110; // LV side voltage in Volts
+P = 4 * 10 ^ 2; // Load of the Transformer
+pf = 0.8; // Power Factor (lagging)
+f = 50; // Frequency in Hertz
+
+
+// CALCULATIONS
+
+a = V1/V2; // Turn Ratio of the Transformer
+
+// case (a) At full load
+I1 = (S * 10 ^ 3)/V1; // Primary current at full load in Amphere
+I2 = (S * 10 ^ 3)/V2; // Secondary Current at full Load in Amphere
+
+// Case (b) At 4kW, 0.8 lagging pf load
+I11 = (4 * 10 ^ 3 * 0.8)/V1; // Primary current At 4kW, 0.8 lagging pf load in Amphere
+I22 = (4 * 10 ^ 3 * 0.8)/V2; // Secondary Current At 4kW, 0.8 lagging pf load in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.6 : SOLUTION :-") ;
+printf("\n (a) Turn Ratio of the Transformer , a = %.f \n ",a);
+printf("\n (b.1.1) Primary current at full load , I1 = %.2f A \n",I1);
+printf("\n (b.1.2) Secondary current at full load , I2 = %.2f A \n ",I2);
+printf("\n (b.2.1) Primary current at 4kW, 0.8 lagging pf load , I1 = %.3f A \n",I11);
+printf("\n (b.2.1) Secondary current at 4kW, 0.8 lagging pf load , I2 = %.3f A \n",I22);
diff --git a/Working_Examples/2777/CH3/EX3.7/Ex3_7.sce b/Working_Examples/2777/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 0000000..c61f784 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,30 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.7
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+a = 100/200; // Turn ratio of the Ideal transformer
+R = 1.0; // Resistance across the secondary side having 200 turns in Ohms
+
+
+// CALCULATIONS
+
+R1 = (a ^ 2)*R; // Referred value of the resistance from Primary side having 100 turns in Ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.7 : SOLUTION :-") ;
+printf("\n (a) Referred value of the %.f Ohm resistance from Primary side having 100 turns , R1 = %.2f ohms \n ",R,R1);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Referred value of the resistance from Primary side having 100 turns = 0.025 Ohms instead of %.2f Ohms \n ",R1);
diff --git a/Working_Examples/2777/CH3/EX3.8/Ex3_8.sce b/Working_Examples/2777/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 0000000..9b41b90 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.8
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 60; // Tranformer Rating in kVA
+V1 = 6600; // HV Side Voltage Rating of the Transformer in Volts
+V2 = 220; // LV Side Voltage Rating of the Transformer in Volts
+R1 = 7.8; // primary Resistances of the Transformer in Ohms
+R2 = 0.0085; // Secondary Resistances of the Transformer in Ohms
+
+
+// CALCULATIONS
+
+a = V1/V2; // Transformation Ratio
+Rp = R1+(a^2)*R2; // Resistance referred to Primary side in Ohms
+Rs = (R1/(a^2))+R2; // Resistance referred to Secondary side in Ohms
+Ip = (S*1000)/V1 // Current in Primary Side in Ampheres
+Cu_loss = Rp*(Ip^2); // Copper loss in Transformer in Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.8 : SOLUTION :-") ;
+printf("\n (a) Equlivalent Resistance as Referred to Primary Side, Rp = % .2f ohms \n",Rp)
+printf("\n (b) Equlivalent Resistance as Referred to Secondary Side, Rs = % .5f ohms \n",Rs)
+printf("\n (c) Total Copper Loss, Cu_loss = % .2f W \n",Cu_loss)
+
diff --git a/Working_Examples/2777/CH3/EX3.9/Ex3_9.sce b/Working_Examples/2777/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 0000000..57edaa3 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,30 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.9
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V1 = 11000; // HV Side Voltage Rating of the Transformer in Volts
+V2 = 440; // LV Side Voltage Rating of the Transformer in Volts
+R = 1.0; // Resistance across the secondary side having 11kV in Ohms
+
+
+// CALCULATIONS
+
+a = V1/V2; // Turns ratio of the ideal transformers
+R2 = (a ^ 2)*R; // Referred value of the resistance from Primary side having 440V in Ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.9 : SOLUTION :-") ;
+printf("\n (a) Referred value of the resistance from Primary side having 440V , R2 = %.f Ohms \n ",R2);
diff --git a/Working_Examples/2777/CH4/EX4.1/Ex4_1.sce b/Working_Examples/2777/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 0000000..9972057 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,36 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N = 600; // Speed of the driven Machine in RPM
+D = 2; // Diameter of the Machine in Meter
+L = 0.3; // Length of the Machine in Meter
+Bm = 1.0; // Flux Density in Weber per Meter-Square
+
+
+// CALCULATIONS
+
+n = N/60; // Revolution per second
+v = %pi * D * n; // Peripheral velocity in Meter per second
+E = Bm * v * L; // Maximum EMF induced in the Conducter in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.1 : SOLUTION :-") ;
+printf("\n (a) Maximum EMF induced in the Conducter , E = %.3f V \n ",E);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a)Induced EMF, E = 2.826 V instead of %.3f A \n ",E);
+printf("\n From Calculation of the peripheral velocity(v), rest all the Calculated values in the TEXT BOOK is WRONG because of the peripheral velocity(v) value is WRONGLY calculated and the same used for the further Calculation part \n")
+
diff --git a/Working_Examples/2777/CH4/EX4.10/Ex4_10.sce b/Working_Examples/2777/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 0000000..359b3c9 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,37 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.10
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 6; // Number of the Poles
+P = 100 * 10 ^ 3; // Power rating of the DC machine in KiloWatts
+V = 440; // Voltage rating of the DC machine in Volts
+Z = 500; // Total number of the Armature Conductor
+Ig = 1.0 * 10 ^ -2; // Interpolar Air gap in Meter
+Bi = 0.28; // Interpolar Flux Densist in Weber per Meter-Square
+mue_0 = 4*%pi*10^ -7; // Permeability of the air in Henry/Meter
+
+
+// CALCULATIONS
+
+Ia = P/V; // Full load current in Amphere
+a = p; // Number of the Parallel path (Equal to p because LAP WINDING)
+ATi = (Z*Ia)/(2*a*p)+((Bi*Ig)/mue_0); // Amphere turns for each Interpole
+Nc = ATi/Ia; // Number of turns per pole of interpole
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.10 : SOLUTION :-") ;
+printf("\n (a) Amphere turns for each Interpole, ATi = %.2f AT \n",ATi);
+printf("\n (b) Number of turns per pole of interpole, Nc = %.2f turns per pole nearly %.f turns per pole \n",Nc,Nc);
diff --git a/Working_Examples/2777/CH4/EX4.11/Ex4_11.sce b/Working_Examples/2777/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 0000000..f3ca8cc --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,35 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.11
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 4.11 : \n\n Given Data between the Field current and Open-Circuit EMF generated by DC shunt wound Generator \n");
+printf("\n If(A) 0 1 2 3 4 5 6 \n");
+printf("\n Voc(v) 10 90 170 217.5 251 272.5 281 \n");
+N = 1000; // Speed of an DC Shunt wound generator on open circuit in RPM
+Rf = 50; // Shunt field resistance in Ohms
+
+
+// CALCULATIONS
+// Refer Figure 4.20:- Page no. 180
+
+Vt = 277.17; // Terminal Voltage in Volts from Figure 4.20 (The slope of the Resistance line Rf cuts the OCC at this Voltage [point])
+Voc_r = 90; // Critical Open circuit voltage in Volts from Figure 4.20 page no. 180
+If_r = 1.0; // Critical Field current in Amphere from Figure 4.20 page no. 180
+Rc = Voc_r/If_r; // Crictical field Resistance in Ohms
+
+
+// DISPLAY RESULTS
+
+printf(" \n\n SOLUTION :-\n") ;
+printf("\n (a) Crictical field Resistance, Rc = %.f Ohms \n",Rc);
diff --git a/Working_Examples/2777/CH4/EX4.12/Ex4_12.sce b/Working_Examples/2777/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 0000000..8618778 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.12
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 4.12 : \n\n Given Data between the Field current and Open-Circuit EMF generated by DC Machine \n");
+printf("\n If(A) 0 0.25 0.5 1.0 1.5 2.0 2.5 3.0 \n");
+printf("\n Voc(v) 8 43 77 151 198 229 253 269\n");
+N = 600; // Speed of an DC Shunt wound generator on open circuit in RPM
+Rf1 = 100; // Shunt field resistance in Ohms
+Rf2 = 125; // Shunt field resistance in Ohms
+
+
+// CALCULATIONS
+// Refer Figure 4.21:- Page no. 181
+
+Vt1 = 253.33; // Terminal Voltage in Volts correspounding to field resistance of 100 Ohms from Figure4.21 Page no. 181 (The slope of the Resistance line Rf cuts the OCC at this Voltage [point])
+Vt2 = 213.33; // Terminal Voltage in Volts correspounding to field resistance of 125 Ohms from Figure 4.21 Page no. 181 (The slope of the Resistance line Rf cuts the OCC at this Voltage [point])
+Voc_r = 151; // Critical Open circuit voltage in Volts from Figure 4.20
+If_r = 1.0; // Critical Field current in Amphere from Figure 4.20
+Rc = Voc_r/If_r; // Crictical field Resistance in Ohms
+
+
+// DISPLAY RESULTS
+
+printf(" \n\n SOLUTION :-\n") ;
+printf("\n (a) Crictical field Resistance, Rc = %.f Ohms \n",Rc);
+printf("\n (b.1) Terminal Voltage correspounding to field resistance of 100 Ohms is %.2f V \n ", Vt1);
+printf("\n (b.1) Terminal Voltage correspounding to field resistance of 125 Ohms is %.2f V \n ", Vt2);
+
diff --git a/Working_Examples/2777/CH4/EX4.13/Ex4_13.sce b/Working_Examples/2777/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 0000000..477bad0 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,35 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1200; // Rotation speed of the Separately excited Generator in RPM at case (1)
+Ia1 = 100; // Current supplied by the Generator in Amphere
+V1 = 220; // Opearting Volatge of the Generator in Volts
+Ra = 0.08; // Armature Resistance in Ohms
+N2 = 1000; // Rotation speed of the Separately excited Generator in RPM at case (2)
+Vb = 2.0; // Total Brush drop in Volts
+
+
+// CALCULATIONS
+
+RL = V1/Ia1; // Load resistance in Ohms
+E1 = V1 + Vb + (Ra * Ia1); // Back EMF at case (1) in Volts
+E2 = (N2/N1)*E1; // Back EMF at case (2) in Volts (Excitation is Constant)
+Ia2 = (E2 - Vb)/(RL + Ra); // New load current in Amphere for case (2)
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.13 : SOLUTION :-") ;
+printf("\n (a) New load current at %.f RPM , Ia2 = %.2f A \n",N2,Ia2);
diff --git a/Working_Examples/2777/CH4/EX4.14/Ex4_14.sce b/Working_Examples/2777/CH4/EX4.14/Ex4_14.sce new file mode 100755 index 0000000..72fe191 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+I = 50; // Curent supplied by the Separately Excitated Generator in Amphere
+V = 250; // Dc bus bar in Volts
+phi_1 = 0.03; // Useful Flux in Weber
+Ra = 0.5; // Armature Resistance in Ohms
+phi_2 = 0.029; // New(Changed) Flux in Weber
+
+
+// CALCULATIONS
+
+Vd = I * Ra; // Voltage drop in the Armature in Volts
+E1 = V + Vd; // EMF Generated in Volts
+E2 = (phi_2/phi_1)*E1; // EMF Generated in Volts immediately after flux changes but speed will remains same
+Ia = (E2 - V)/Ra; // Armature Current in Amphere immediately after flux changes
+perct = 100 * (( phi_1 - phi_2)/phi_2); // Percenatge change in the speed of the machine that is required to restore the original Armature current but EMF raised to the original value and its Proportional to the speed and flux
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.14 : SOLUTION :-") ;
+printf("\n (a) Armature Current immediately after flux changes, Ia = %.1f A \n",Ia);
+printf("\n (b) Percenatge change in the speed of the machine (that is required to restore the original Armature current) is %.2f Percenatge \n",perct);
+
+
diff --git a/Working_Examples/2777/CH4/EX4.15/Ex4_15.sce b/Working_Examples/2777/CH4/EX4.15/Ex4_15.sce new file mode 100755 index 0000000..69a032f --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,67 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.15
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 500 * 10 ^ 3; // Rating of the Generator-1 and Generator-2
+VI = 800 * 10 ^ 3' // Actual load
+
+
+// CALCULATIONS
+//For Case (a)
+
+Voc_a = 500; // Open-circuit EMF Generator-1 and Generator-2 in Volts
+I = 1000; // Full load current in Amphere
+perct_1a = 2/100; // Percenatge fall of the Voltage in Generator-1
+perct_2a = 3/100; // Percenatge fall of the Voltage in Generator-2
+V1a = Voc_a - (perct_1a * Voc_a); // Voltage in the Generator-1 in Volts when it falls to 2% at fully loaded
+V2a = Voc_a - (perct_2a * Voc_a); // Voltage in the Generator-2 in Volts when it falls to 3% at fully loaded
+// From Chacteristics can be assumed linear as, for Generator 1 is V = 500 + ((500-490)*I1)/(0-1000), V = -0.01*I1+500 and for Generator 2 is V = 500 + ((500-485)*I2)/(0-1000), V = 0.015*I2+500
+// When sharing load of 800KVA at voltage, the load current will be I = I1+I2 = (800*1000)/V
+// From above equations we get I1 = 1.5*I2 thus, 2.5*I2 = (800*1000)/V
+// Putting the above equations in the Generator 2 equation we get V = -0.015*((800*1000)/(2.5*V))+500 solving we get, 25*V^2 - 12500V + 120000 = 0
+V_a = poly ([120000 -12500 25],'x','coeff'); // Expression for the load Voltage in Quadratic form
+r_a = roots (V_a); // Value of the load Voltage in Volts (neglecting lower value)
+I_a = VI/r_a(1,1);
+I2_a = I_a/2.5;
+I1_a = 1.5*I2_a;
+
+// For Case (b)
+
+perct = 2/100; // Percenatge fall of the Voltage in Generator-1and Generator-2
+Voc_1b = 500; // Open-circuit EMF Generator-1 in Volts
+Voc_2b = 505; // Open-circuit EMF Generator-2 in Volts
+I = 1000; // Full load current in Amphere
+V1 = Voc_1b - (perct * Voc_1b); // Voltage in the Generator-1 in Volts when it falls to 2% at fully loaded
+V2 = Voc_2b - (perct * Voc_2b); // Voltage in the Generator-2 in Volts when it falls to 2% at fully loaded
+// From Chacteristics can be assumed linear as, for Generator 1 is V = 500 + ((500-490)*I1)/(0-1000), V = -0.01*I1+500 and for Generator 2 is V = 505 + ((505-494.5)*I2)/(0-1000), V = -0.0101*I2+505
+// When sharing load of 800KVA at voltage, the load current will be I = I1+I2 = (800*1000)/V
+// From above equations we get V = -0.01*I1 + 500, I1 = -V/0.01 + 500/0.01 = 50000 - 100*V, V = -0.0101*I2 + 505 and I2 = 505/0.0101 - V/.0101 = 50000-99.0099*V
+// Putting the above equations in the Current I equation we get I = I1+I2 = (800*1000)/V = 2*50000-199.0099*V solving we get, 199.0099*V^2 - 100000V + 800000 = 0
+V_b = poly ([800000 -100000 199.0099],'x','coeff'); // Expression for the load Voltage in Quadratic form
+r_b = roots (V_b); // Value of the load Voltage in Volts (neglecting lower value)
+I_b = VI/r_b(1,1);
+I1_b = 50000-100*r_b(1,1)
+I2_b = 50000-99.0099*r_b(1,1)
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.15: SOLUTION :-");
+printf("\n For case (a) Having open-circuit EMfs of 500V but their voltage falls to 2 percent and 3 percent when fully loaded Load Voltage,\n\n Load Voltage = %.2f V \n\n Load current = %.2f A \n\n Individual currents are %.2f A and %.2f A \n",r_a(1,1),I_a,I1_a,I2_a)
+printf("\n For case (b) Having open-circuit EMfs of 500V and 505V but their governors have identical speed regulation of 2 percent when fully loaded Load Voltage,\n\n Load Voltage = %.2f V \n\n Load current = %.2f A \n\n Individual currents are %.2f A and %.2f A \n",r_b(1,1),I_b,I1_b,I2_b)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- For case(b) Load voltage = 493.35 V A instead of %.2f V \n ",r_b(1,1));
+printf("\n Load current = 1634.73 A instead of %.2f A \n ",I_b)
+printf("\n Individual currents 665 A and 1153.5 A instead of %.2f A and %.2f \n ",I1_b,I2_b)
+printf("\n For Case (b):- From Calculation of the Load Voltage (V), rest of all the Calculated values in the TEXT BOOK is WRONG because of the value Load Voltage (V) is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH4/EX4.16/Ex4_16.sce b/Working_Examples/2777/CH4/EX4.16/Ex4_16.sce new file mode 100755 index 0000000..04e11d4 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,50 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.16
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Out_hp = 20; // Output of the Motor in HP
+eta = 90/100; // Full load efficiency of the Motor
+V = 220; // Motor voltage in Volts
+ns = 5; // Number of the step of Starter
+Rf = 220; // Field Resistance in Ohms
+cr = 1.8; // Lowest Current rating is 1.8 times of the Full load current
+Cu = 5/100; // Total Copper loss is 5% of the Input
+
+
+// CALCULATIONS
+
+Out = 20 * 746; .. // Output of the Motor in Watt
+Inp = (Out/eta); // Input of the Motor in KiloWatt
+I = Inp/Rf; // Full-Load Current in Amphere
+Cu_l = Inp*Cu; // Total Copper loss in Watts
+olf = (V ^ 2)/Rf; // Ohmic loss in the Fiels in the Watts
+Acu = Cu_l - olf; // Armature Copper loss in Watts
+Ra = Acu/(I * I); // Armature Resistance in Ohms
+I2 = I * cr; // Lower Current in Amphere
+n = ns - 1; // Number of the Resistance
+gama = ( (I2 * Ra)/Rf ) ^ (1/(n + 1)); // Current Ratio
+I1 = I2/gama; // Initial Current in amphere
+R1 = V/I1; // Initial Resistance in Ohms
+R2 = gama * R1; // Initial Resistance in Ohms
+r1 = R1 - R2; // Graded Resistance in Ohms
+R3 = gama * R2; // Initial Resistance in Ohms
+r2 = gama * r1; // Graded Resistance in Ohms
+r3 = gama ^ 2 * r1; // Graded Resistance in Ohms
+r4 = gama ^ 3 * r1; // Graded Resistance in Ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.16 : SOLUTION :-") ;
+printf("\n (a) Graded Resistances are %.4f Ohms, %.4f Ohms, %.4f Ohms and %.4f Ohms \n",r1,r2,r3,r4);
diff --git a/Working_Examples/2777/CH4/EX4.17/Ex4_17.sce b/Working_Examples/2777/CH4/EX4.17/Ex4_17.sce new file mode 100755 index 0000000..14005fd --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.17
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+I = 30; // Initial starting Current in Amphere
+ns = 5; // Number of Steps of the starter
+V = 500; // Operating Voltage of the DC Shunt Motor in Volts
+I1 = 50; // Peak(Upper) Current limit in Amphere
+Ra = 1.0; // Armature Circuit resistance in Ohms
+
+
+// CALCULATIONS
+
+R1 = V / I; // Initial Resistance in Ohms
+gama = ( Ra/R1) ^ (1/(ns-1)); // Current Ratio
+I2 = gama * I1; // Lower Current limit in Amphere
+r1 = R1 * (1-gama); // Graded Resistances in Ohms
+r2 = gama * r1; // Graded Resistances in Ohms
+r3 = gama * r2; // Graded Resistances in Ohms
+r4 = gama * r3; // Graded Resistances in Ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.17 : SOLUTION :-") ;
+printf("\n (a)Graded Resistances are %.2f Ohms, %.4f Ohms, %.4f Ohms and %.4f Ohms \n",r1,r2,r3,r4);
+
diff --git a/Working_Examples/2777/CH4/EX4.18/Ex4_18.sce b/Working_Examples/2777/CH4/EX4.18/Ex4_18.sce new file mode 100755 index 0000000..6168cde --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,41 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.18
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 500; // Operating voltage of the DC series motor in Volts
+P_hp = 10; // Operating Power in HP
+Il = 40; // Lower currents limit in Amphere
+Ih = 60; // Higher currents limit in Amphere
+f = 0.5/100; // Motor flux rises by 0.5% per amphere
+Rt = 0.8; // Motor terminal resistance in Ohms
+eta = 90/100; // Motor efficiency
+
+// CALCULATIONS
+
+E1 = V-Il*Rt; // Induced EMF E1 in Volts
+// Induced EMF, E2 = 500-60(0.8+r4) = 500 - 60*R4 where r4 is the fourth-step resistance, and R4 = 0.8+r4 and E1 = 1.1*E2 , 500 - 40*0.8 = 1.1*(500-60(0.8+r4)), 500-32 = 550-66*R4 thus we get, R4 = (550-500+32)/66 refer page no. 197
+R4 = (V-(E1/1.1))/Ih;
+r4 = R4 - Rt; // Fourth-step resistance in ohms
+R3 = (V-((V-Il*R4)/1.1))/Ih;
+r3 = R3 - R4; // Third-step resistance in ohms
+R2 = (V-((V-Il*R3)/1.1))/Ih;
+r2 = R2 - R3; // Second-step resistance in ohms
+R1 = (V-((V-Il*R2)/1.1))/Ih;
+r1 = R1 - R2; // First-step resistance in ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.18: SOLUTION :-");
+printf("\n (a) The resistance steps in series motor stater are %.3f Ohms,%.4f Ohms, %.3f Ohms and %.2f Ohms \n",r1,r2,r3,r4)
diff --git a/Working_Examples/2777/CH4/EX4.19/Ex4_19.sce b/Working_Examples/2777/CH4/EX4.19/Ex4_19.sce new file mode 100755 index 0000000..79960a5 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.19
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Vac = 250; // Operating AC Voltage in Volts
+V = 220; // Operating Voltage of the separately excited DC motor in Volts
+fa = 30; // Firing Angle in Degree
+Out_hp = 20; // DC Motor Output in HP
+La = 20 * 10 ^ -3; // Armature Inducatnce in Henry
+Ra = 0.15; // Armature Resistance in Ohms
+E_cons = 0.2; // EMF Constant in Volts/RPM
+eta = 90/100; // Motor Operating Efficiency
+N = 1000; // Rotational Speed of the Motor in RPM
+
+
+// CALCULATIONS
+
+out = 20 * 746; // DC Motor Output in Watt
+Vt = ((Vac*2*sqrt(2))/%pi)*cosd(fa); // Average Terminal volatge in Volts
+Ia = out/(V*eta); // Rated Current in Amphere
+E = Vt - ( Ia * Ra ); // Back EMF in Volts
+n = E/E_cons; // Speed of the Motor in RPM
+e_cons = (E_cons*60)/ ( 2 * %pi); // EMF Constant in Volts-Second per radians
+T = e_cons * Ia; // Devolped Torque in Newton-Meter
+pi = (E*Ia)+(Ia^2*Ra); // Power intake in Watts
+pi_v = Vt * Ia; // Power intake in Watts (Verification)
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.19 : SOLUTION :-") ;
+printf("\n (a) Speed of the Motor, N = %.2f RPM \n",n);
+printf("\n (b) Devolped Torque, T = %.2f N-m \n",T);
+printf("\n (b) Power intake at Rated current and Firing angle of %.f deg, VI = %.1f W \n",fa,pi);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS IS :- (a) T = 114.07 N-m instead of 143.91 N-m \n" );
+
+
+
+
+
diff --git a/Working_Examples/2777/CH4/EX4.2/Ex4_2.sce b/Working_Examples/2777/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 0000000..4b36006 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,33 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 8; // Number of the poles in Dc machine
+a = 8; // Number of the Parallel path
+N = 500; // Rotation per minute in RPM
+phi = 0.095; // Average flux in air gap in Weber per meter
+Za = 1000; // Total number of the Conductor in Armature
+
+
+// CALCUALTIONS
+
+n = N/60; // Rotation (Revolution) per Second
+E = (p/a)*n*phi*Za; // EMF induced in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.2 : SOLUTION :-") ;
+printf("\n (a) EMF induced , E = %.1f A \n ",E);
+
diff --git a/Working_Examples/2777/CH4/EX4.20/Ex4_20.sce b/Working_Examples/2777/CH4/EX4.20/Ex4_20.sce new file mode 100755 index 0000000..c5d4dc9 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,59 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.20
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N = 1500; // Speed of the separately excited DC Motor in RPM
+Out_hp = 100; // Output of the DC Motor in HP
+V = 500; // Motor operating Volatge in Volts
+VL = 440; // 3-phase Line-line Voltage in Volts
+f = 50; // Frequency in Hertz
+Ra = 0.0835; // Armature Resistance in Ohms
+La = 5.7 * 10 ^ -3; // Armature Inductance in Henry
+eta = 89/100; // Operating Efficiency of the Motor
+E_cons = 0.35; // EMF constant in Volts per RPM
+
+
+// CALCULATIONS
+// For case (a)
+
+Out = Out_hp * 746; // Output of the DC Motor in Watts
+Ia = Out/(V*eta); // Rated Current in Amphere
+Vph = VL/sqrt(3); // Phase Voltage in Volts
+a = (3*Vph*sqrt(6)) / %pi; // Constant
+E = N * E_cons; // Back EMF at Rated speed
+V = E + (Ia * Ra); // Terminal Volatge in Volts
+alpa = acosd(V/a); // Firing Angle
+
+// For case (b)
+// Assumed that No load current is about 12% of full load current
+
+Io = ( 0.12 * Ia ); // No load current in Amphere
+V_b1 = a * cosd(0); // Terminal Voltage at Firing Angle 0 deg
+E_b1 = V_b1 + (Io * Ra); // Back EMF at Firing Angle 0 deg
+N_b1 = E_b1/E_cons; // No load speed at Firing Angle 0 deg
+V_b2 = a * cosd(30); // Terminal Voltage at Firing Angle 30 deg
+E_b2 = V_b2 + (Io * Ra); // Back EMF at Firing Angle 30 deg
+N_b2 = E_b2/E_cons; // No load speed at Firing Angle 30 deg
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.20 : SOLUTION :-") ;
+printf("\n (a) Firing Angle at Rated speed and Rated Motor Current , alpa = %.2f deg \n",alpa);
+printf("\n (b.1) No load speed at Firing Angle 0 deg, N = %.1f RPM \n",N_b1);
+printf("\n (b.2) No load speed at Firing Angle 30 deg, N = %.1f RPM \n",N_b2);
+
+
+
+
diff --git a/Working_Examples/2777/CH4/EX4.21/Ex4_21.sce b/Working_Examples/2777/CH4/EX4.21/Ex4_21.sce new file mode 100755 index 0000000..09b149d --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,37 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.21
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Given that Back EMF is Zero Because Motor is at Standstill
+
+V = 250; // DC supply Voltage to separately excited DC Motor in Volts
+Ra = 1.0; // Armature Resistance in Ohms
+La = 30 * 10 ^ -3; // Armature Inductance in Henry
+E_cons = 0.19; // Motor (EMF) Constant in Volts per RPM
+Ia = 25; // Average Armature Current in Amphere
+
+
+// CALCULATIONS
+
+V1 = Ia * Ra; // Minimum Terminal Volatge in Volts
+alpa_mini = Ia/V; // Minimum Duty Cycle
+alpa_max = 1.0; // Maximum Duty Cycle
+V2 = V; // Maximum Terminal Volatge in Volts when Duty cycle (alpa) is 1.0
+E2 = V2 - (V1 * alpa_max); // Back EMF at Maximum Duty cycle ( i.e alpa = 1.0) in Volts
+N = E2/E_cons; // Speed of the Motor
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.21 : SOLUTION :-") ;
+printf("\n (a) Range of the Speed is from 0 RPM to %.2f RPM and Range of the Duty Cycle is %.1f to %.1f \n",N,alpa_mini,alpa_max);
diff --git a/Working_Examples/2777/CH4/EX4.22/Ex4_22.sce b/Working_Examples/2777/CH4/EX4.22/Ex4_22.sce new file mode 100755 index 0000000..2fa4afc --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.22
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1000; // Speed of the DC shunt Motor in RPM
+Out_hp = 20; // Output of the DC shunt Motor in HP
+V = 220; // Motor operating Volatge in Volts
+Ra = 0.9; // Armature Resistance in Ohms
+Rf = 200; // Field Resistance in Ohms
+eta = 89/100; // Operating Efficiency of the Motor
+Ra_a = 0.2; // Resistance inserted to the armature circuit
+
+
+// CALCULATIONS
+
+out = Out_hp * 746; // Output of the DC Motor in watts
+I = out/(V * eta); // Rated current in Amphere
+If = V/Rf; // Field current in Amphere
+Ia1 = I - If; // Armature current in Amphere
+E1 = V - (Ia1 * Ra); // Back EMF in Volts
+// Assuming that Torque and Armature current is constant
+E2 = V - ( Ra + Ra_a ) * Ia1; // New Back EMF in Volts
+N2 = N1*(E2/E1); // New speed in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.22: SOLUTION :-") ;
+printf("\n (a) New Speed of the Motor , N2 = %.2f RPM \n",N2);
diff --git a/Working_Examples/2777/CH4/EX4.23/Ex4_23.sce b/Working_Examples/2777/CH4/EX4.23/Ex4_23.sce new file mode 100755 index 0000000..9157d1c --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.23/Ex4_23.sce @@ -0,0 +1,43 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.23
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 600; // Speed of the DC shunt Motor in RPM
+Out_hp = 10; // Output of the DC shunt Motor in HP
+V = 220; // Motor operating Volatge in Volts
+Ra = 1.5; // Armature Resistance in Ohms
+Rf = 250; // Field Resistance in Ohms
+eta = 88/100; // Operating Efficiency of the Motor
+Rf_a = 50; // Resistance inserted to the field circuit
+
+
+// CALCULATIONS
+
+out = Out_hp * 746 // Output of the DC Motor in watts
+I = out/(V * eta); // Rated current in Amphere
+If1 = V/Rf; // Field current in Amphere
+Ia1 = I - If1; // Aramature current in Amphere
+E1 = V - Ra*Ia1; // Back EMF in Volts
+If2 = V/(Rf+Rf_a); // New Field current in Amphere after 50 Ohms Resistance inserted to the field circuit
+
+// Refer page no. 217 we have T1 = K*If1*Ia1 proportional to 1/W1^2 and T1 = K*If2*Ia2 proportional to 1/W2^2 thus T1/T2 = (If1*Ia1)/(If2*Ia2) = (W2^2)/(W1^2) = (N2^2)/(N1^2), Ia2 = (If1*Ia1*W1^2)/(If1*W1^2) = (0.88*37.65*N2^2)/(0.733*600*600)
+// Now New EMF E2 is E2 = V - Ia2*Ra, E1/E2 = (k*If1*N1)/(k*If2*N2), E2 = (0.733*N2)/(0.88*600) = 220 - (0.88*37.65*1.5*N2^2)/(0.733*600*600) Thus we have 0.001388*N2^2 = 220 - 1.833*10^-4*N2
+N2 = poly ([-220 0.001388 1.833*10^-4],'x','coeff'); // Expression for the new speed of the motor in Quadratic form
+r = roots (N2); // Value of the New speed of the motor in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.23 : SOLUTION :-") ;
+printf("\n (a) New speed of the motor, N2 = %.2f RPM nearly %.f RPM \n",r(2,1),r(2,1));
diff --git a/Working_Examples/2777/CH4/EX4.24/Ex4_24.sce b/Working_Examples/2777/CH4/EX4.24/Ex4_24.sce new file mode 100755 index 0000000..ff41d23 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.24/Ex4_24.sce @@ -0,0 +1,37 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.24
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1000; // Speed of the DC shunt Motor in RPM
+Out_hp = 10; // Output of the DC shunt Motor in HP
+V = 220; // Motor operating Volatge in Volts
+Ra = 0.5; // Armature Resistance in Ohms
+Rf = 100; // Field Resistance in Ohms
+eta = 90/100; // Operating Efficiency of the Motor
+
+
+// CALCULATIONS
+
+out = Out_hp * 746; // Output of the DC Motor in watts
+I = out/(V * eta); // Rated current in Amphere
+If = V/Rf; // Field current in Amphere
+Ia = I-If; // Armature current in Amphere
+E = V - (Ia*Ra); // Back EMF of the Motor in Volts
+Rd = E/I; // Resistance at Dynamic Braking in Ohms
+Rc = (V+E)/I; // Resistance at Counter Current Braking in Ohms
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.24 : SOLUTION :-") ;
+printf("\n (a) Resistance at Dynamic Braking, Rd = %.2f Ohms \n",Rd);
+printf("\n (b) Resistance at Counter Current Braking, Rc = %.1f Ohms \n",Rc);
diff --git a/Working_Examples/2777/CH4/EX4.25/Ex4_25.sce b/Working_Examples/2777/CH4/EX4.25/Ex4_25.sce new file mode 100755 index 0000000..00b4846 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.25/Ex4_25.sce @@ -0,0 +1,44 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.25
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // Motor operating Volatge in Volts
+Ra = 1.0; // Armature Resistance in Ohms
+Rf = 220; // Field Resistance in Ohms
+Ia1 = 20; // Armature Current in Amphere
+N1 = 800; // Motor drving speed in RPM
+N2 = 1000; // To be obtained speed in RPM
+
+
+// CALCULATIONS
+
+If = V/Rf; // Field Current in Amphere
+E1 = V - ( Ia1 - If ) * Ra; // Back EMF E1 at N1 Speed in Volts
+// Now we have Back EMF E2 at N2 Speed, E2 = 220-Ia2*1.0 = 220-Ia2 and the field flux be proportional to the field current, since torque is constant we get, If2*Ia2 = If1*Ia1 = 20
+// Thus (220-Ia2)/201 = (If2*N2)/(If1*N1) = If2*(1000/(800*1.0)), 220-Ia2 = 201*(10/8)*(20/Ia2) = 5000/Ia2 solving this we get Ia2^2 - 220Ia2 + 2000 = 0
+Ia2 = poly ([5000 -220 1],'x','coeff'); // Expression for the new Armature current in Quadratic form
+r = roots (Ia2); // Value of the New Armature current in Amphere
+If2 = If*(Ia1/r(2,1)); // New field current in Amphere when New Armature current is 39.29A
+Rfn = V/If2; // New field resistance in ohms
+ERf = Rfn - Rf; // Extra resistance in Ohms
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.25 : SOLUTION :-") ;
+printf("\n (a) Extra resistance should be added in the field circuit for raising the speed to %.f RPM is = %.2f Ohms \n",N2,ERf);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Ia2 = 39.29 A and 180.71 A instead of %.2f A and %.2f A \n ",r(1,1),r(2,1));
+printf("\n (b) Extra resistance required is 212.22 Ohms instead of %.2f Ohms \n ",ERf);
+printf("\n From Calculation of the New armature current (Ia2), rest all the Calculated values in the TEXT BOOK is WRONG because of the New armature current (Ia2) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH4/EX4.26/Ex4_26.sce b/Working_Examples/2777/CH4/EX4.26/Ex4_26.sce new file mode 100755 index 0000000..4d284bc --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.26/Ex4_26.sce @@ -0,0 +1,47 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.26
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+R = 2.0; // DC series Motor Resistance between the terminals in Ohms
+V1 = 220; // Motor Operating voltage in Volts
+N1 = 500; // Rotation Sped of the DC series Motor in RPM
+I1 = 22; // Current in Motor in Amphere
+N2 = 600; // New Rotation Sped of the DC series Motor in RPM
+
+
+// CALCULATIONS
+// For case (a) when magnetic circuit is Unsaturated
+
+E1 = V1 - (I1 * R); // Back EmF at N1 Speed in Volts
+I2_a = (N2/N1)*I1; // Current in Motor at N2 speed in Amphere
+E2_a = (E1*I2_a*N2)/(I1*N1); // Back EmF at N2 Speed in Volts
+V2_a = E2_a + (I2_a * R); // Applied Voltage at N2 Speed in Volts
+
+// For case (b) when magnetic circuit is Saturated
+
+I2_b = ((N2/N1)^2)*I1; // Current in Motor at N2 speed in Amphere
+E2_b = (N2/N1)*E1; // Back EmF at N2 Speed in Volts
+V2_b = E2_b + (I2_b * R); // Applied Voltage at N2 Speed in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.26 : SOLUTION :-") ;
+printf("\n (a.1) Applied Voltage when magnetic circuit is Unsaturated,V2 = %.2f V \n",V2_a);
+printf("\n (a.2) Current in Motor when magnetic circuit is Unsaturated, I2 = %.1f A \n",I2_a);
+printf("\n (b.1) Applied Voltage when magnetic circuit is Saturated,V2 = %.2f V \n",V2_b);
+printf("\n (b.2) Current in Motor when magnetic circuit is Saturated, I2 = %.2f A \n",I2_b);
+
+
+
diff --git a/Working_Examples/2777/CH4/EX4.27/Ex4_27.sce b/Working_Examples/2777/CH4/EX4.27/Ex4_27.sce new file mode 100755 index 0000000..8279509 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.27/Ex4_27.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.27
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // DC series Motor operating Volatge in Volts
+Ra = 1.0; // Armature Resistance in Ohms
+Rf = 1.0; // Field Resistance in Ohms
+I1 = 20; // Armature Current in Amphere
+N1 = 1800; // Motor drving speed in RPM
+If = 20; // Armature Current in Amphere
+Rd = 0.5; // Diverter Resistance in Ohms
+
+
+// CALCULATIONS
+
+E1 = V - ( Ra + Rf ) * I1; // Back EMF in Volts
+I2 = sqrt(3)*I1; // New Armature current in Amphere
+If2 = ( Rd * I2 )/(Ra + Rd); // New field Current in Amphere
+E2 = V - ( Ra + (1/3))*I1; // New BAck EMF in Volts
+N2 = (N1*E2*If)/(E1*If2); // New Rotation speed of the Motor in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.27 : SOLUTION :-") ;
+printf("\n (a) New Rotation speed of the Motor at torque remains constant, N2 = %.f RPM \n",N2);
+printf("\n (b.1) New Armature Current at torque remains constant, I2 = %.2f A \n",I2);
+printf("\n (b.2) New Field Current at torque remains constant, If2 = %.2f A \n",If2);
+
diff --git a/Working_Examples/2777/CH4/EX4.28/Ex4_28.sce b/Working_Examples/2777/CH4/EX4.28/Ex4_28.sce new file mode 100755 index 0000000..765cff0 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.28/Ex4_28.sce @@ -0,0 +1,53 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.28
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // DC shunt Motor operating Volatge in Volts
+Ra = 1.0; // Armature Resistance in Ohms
+Rf = 220; // Field Resistance in Ohms
+In1 = 5; // No-Load Current in Amphere
+N1 = 1000; // Motor drving speed in RPM
+inp = 10 * 10 ^ 3; // Motor input in Watts
+
+
+// CALCULATIONS
+
+If = V/Rf; // Field Current in Amphere
+Ian1 = In1 - If; // No load Armature Current in Amphere
+E1 = V - (Ian1 * Ra); // Back EMF in Volts
+Iin = inp/V; // Motor Input Current in Amphere
+Ia = Iin - If; // Armature current in Amphere
+E2 = V - (Ia * Ra); // New Back EMF in Volts
+N2 = (N1*E2)/E1; // New Rotation speed of the Motor in RPM
+Pa = E2 * Ia; // Developed Armature Power in Watts
+T = Pa/((2*%pi*N2)/60); // Developed Torque in Newton-Meter
+Pi = V * In1; // No-Load input Power in Watts
+Pa_cu = Ian1 ^ 2 * Ra; // No-Load Armature Copper loss in Watts
+F_loss = Pi - Pa_cu; // Fixed losses in Watts
+Pa_cu_load = Ia ^ 2 * Ra; // Loaded Armature Copper loss in Watts
+Total_loss = F_loss + Pa_cu_load; // Total losses in loaded conditions in Watts
+out = inp - Total_loss; // Shaft output in Watts
+Ts = out/((2*%pi*N2)/60); // Shaft torque in Newton-Meter
+eta = (out/inp)*100;
+ // Efficiency in Percentage
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.28 : SOLUTION :-") ;
+printf("\n (a) New Rotation speed of the Motor , N2 = %.f RPM \n",N2);
+printf("\n (b.1) Developed Torque, T = %.1f N-m A \n",T);
+printf("\n (b.2) Shaft torque, Ts = %.2f N-m \n",Ts);
+printf("\n (c) Efficiency in Percentage, eta = %.2f percent \n",eta);
+
+
diff --git a/Working_Examples/2777/CH4/EX4.29/Ex4_29.sce b/Working_Examples/2777/CH4/EX4.29/Ex4_29.sce new file mode 100755 index 0000000..6d7823a --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.29/Ex4_29.sce @@ -0,0 +1,62 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.29
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // Shunt Motor operating Line Volatge in Volts
+Ra = 0.2; // Armature Resistance in Ohms
+Iam = 72; // Motor Armature current in Amphere
+I = 12; // Line Current in Amphere
+Ifm = 1; // Motor field Current in Amphere
+Ifg = 1.5; // Generator field Current in Amphere
+
+
+// CALCULATIONS
+
+Iag = Iam - I; // Geneartor Armature current in Amphere
+Pfm = V * Ifm; // Loss in Motor Field winding in Watts
+Pfg = V * Ifg; // Loss in Geneartor Field winding in Watts
+loss_ma = Iam ^ 2 * Ra; // Loss in Motor Armature circuit in Watts
+loss_ga = Iag ^ 2 * Ra; // Loss in Generator Armature circuit in Watts
+Em = V - Iam * Ra; // Motor EMF in Volts
+Eg = V + Iag * Ra; // Generator EMF in Volts
+T_loss = (V*I) - (Ra*Iam^2 + Ra*Iag^2); // Total Iron and Rotational Loss in Watts
+Pim = (V*Iam)+(V*Ifm); // Motor input in Watts
+Wc = 0.5 * T_loss; // Total Iron and Rotational Loss in each Machine in Watts
+Wm = Wc+(Ra*Iam^2)+V*Ifm; // Motor losses in Watts
+Pom = Pim - Wm; // Motor output in Watts
+eta_m = (1-(Wm/Pom))*100; // Motor Efficiency in Percentage
+Pog = V*Iag; // Generator output in Watts
+Wg = Wc+(Ra*Iag^2)+V*Ifg; // Generator losses in Watts
+Pin = Pog + Wg; // Generator input power in Watts
+eta_g = (1-(Wg/Pin))*100; // Generator Efficiency in Percentage
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.29 : SOLUTION :-") ;
+printf("\n (a) Motor Efficiency , eta = %.2f Percentage \n ",eta_m);
+printf("\n (b) Generator Efficiency , eta = %.2f Percantage \n ",eta_g);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Total Iron and Rotational Loss = 720 W instead of %.1f W \n ",T_loss);
+printf("\n (b) Pim = 15912 W instead of %.f W \n ",Pim);
+printf("\n (c) Wm = 1371.4 Winstead of %.1f W \n ",Wm);
+printf("\n (d) Pom = 14540.6 W instead of %.1f W \n ",Pom);
+printf("\n (e) eta_m = 90.54 Percentage instead of %.2f Percentage \n ",eta_m);
+printf("\n (f) eta_g = 93.22 Percentage instead of %.2f Percentage \n ",eta_g);
+printf("\n From Calculation of the Total Iron and Rotational Loss in each Machine (Wc), rest all the Calculated values in the TEXT BOOK is WRONG because of the Total Iron and Rotational Loss in each Machine (Wc) value is WRONGLY calculated and the same used for the further Calculation part \n")
+
+
+
+
+
diff --git a/Working_Examples/2777/CH4/EX4.3/Ex4_3.sce b/Working_Examples/2777/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 0000000..d1848ac --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,31 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+E = 420; // EMF induced in Volts
+N = 900; // Rotation speed in RPM
+phi = 0.06; // Flux per pole in Weber per pole
+Two_p = 4; // Total number of poles
+
+
+// CALCULATIONS
+
+n = N/60; // Revolution Per second
+Zc = E/(Two_p*phi*n); // Number of the Conductor in Parallel Path
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.3 : SOLUTION :-") ;
+printf("\n (a) Number of the Conductor in Parallel Path , Zc = %.2f Conductors nearly 117 conductors \n ",Zc);
diff --git a/Working_Examples/2777/CH4/EX4.30/Ex4_30.sce b/Working_Examples/2777/CH4/EX4.30/Ex4_30.sce new file mode 100755 index 0000000..92dedb0 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.30/Ex4_30.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.30
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Vg = 110; // Generator operating Volatge in Volts
+Vm = 102; // Motor operating Volatge in Volts
+Vs = 274; // Supply Volatge in Volts
+Ra = 1.0; // Armature Resistance in Ohms for both the Machines
+Rf = 0.82; // Field Resistance in Ohms for both the Machines
+N = 1440; // Speed of the Set in RPM
+Ig = 17.5; // Generator current in Amphere
+Im = 9.5; // Motor current in Amphere
+
+
+// CALCULATIONS
+
+Pi = Vs * Im; // Input power in Watts
+Pg = Vg * Ig; // Output power in Watts
+Pim = Vm * Im; // Power Input to the Motor in Watts
+Pl = Pi - Pg; // Losses in the entire set in Watts
+Pcu = Im^2*(Ra+2*Rf) + Ig^2*Ra; // Total Copper loss for both the Machines in Watts
+P_l = Pi - Pg - Pcu; // Frictional, Windage and core losses of the both Machines in Watts
+Po = P_l/2; // Frictional, Windage and core loss of each Machines in Watts
+eta_m = (1 - ((Po + Im^2*(Ra+Rf))/Pim))*100; // Motor Effiicency in Percentage
+Pig = Pg + Po + Ig^2*Ra + Im^2*Rf; // Generator input in Watts
+eta_g = (Pg / Pig)*100; // Generator Effiicency in Percentage
+T = (Vg*Ig *60)/(2*%pi*N); // Torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.30 : SOLUTION :-") ;
+printf("\n (a) Motor Efficiency , eta_m = %.2f percentage \n ",eta_m);
+printf("\n (b) Generator Efficiency , eta_g = %.2f Percentage \n ",eta_g);
+printf("\n (c) Torque , T = %.2f N-m \n ",T);
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Generator input = 2307.5 W instead of %.f W \n ",Pig);
+printf("\n (b) eta_g = 83.42 Percenatge instead of %.2f Percentage \n ",eta_g);
+printf("\n From Calculation of the Generator input, rest all the Calculated values in the TEXT BOOK is WRONG because of the Generator input value is WRONGLY calculated and the same used for the further Calculation part \n")
+
+
diff --git a/Working_Examples/2777/CH4/EX4.4/Ex4_4.sce b/Working_Examples/2777/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 0000000..104c136 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,33 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+L = 0.3; // Length of the Machine in Meter
+Ia = 10; // Current through The Conductors in Ampheres
+N = 10; // Number of the Conductors in each Slot
+Za = 24; // Number of the Slots
+Bav = 0.6; // Average Flux Density in Telsa
+D = 0.1; // Machine Daimeter in Meter
+
+
+// CALCULATIONS
+
+F = N*Ia*Bav*L; // Force due to the Single Slot in Newton
+T = (Bav*L*Ia*N*D*Za)/2 // Torque produced in the Machine in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.4 : SOLUTION :-") ;
+printf("\n (a) Torque produced in the Machine, T = %.1f N-m \n",T);
diff --git a/Working_Examples/2777/CH4/EX4.5/Ex4_5.sce b/Working_Examples/2777/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 0000000..2a41ee2 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,42 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.5
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 4; // Number of the Poles in the DC machine
+Nt = 100; // Number of the turns in the Dc machine
+N = 600; // Rotation speed of the DC machine in RPM
+E = 220; // EMF generated in open circuit in Volts
+Z = 200; // Total number of the Conductor in armature
+
+
+// CALCUALTIONS
+// For case (a) Lap Connected
+
+a = 4; // Number of the Poles in the DC machine
+n = N/60; // Revolution per second
+phi_a = (E*a)/(p*Z*n); // Useful flux per pole when Armature is Lap connected in Weber
+
+// For case (b) Wave Connected
+
+a = 2; // Number of the Poles in the DC machine
+phi_b = (E*a)/(p*Z*n); // Useful flux per pole when Armature is Wave connected in Weber
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.5 : SOLUTION :-") ;
+printf("\n (a) Useful flux per pole when Armature is Lap connected , phi = %.1f Wb \n ",phi_a);
+printf("\n (B) Useful flux per pole when Armature is Lap connected , phi = %.3f Wb \n ",phi_b);
+
+
diff --git a/Working_Examples/2777/CH4/EX4.6/Ex4_6.sce b/Working_Examples/2777/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 0000000..cc4f2f7 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,33 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.6
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 6; // Number of the pole in DC Motor
+Ia = 20; // Armature Current in Amphere
+Z = 1000; // Number of the Conductors
+a = 6; // Number of the Parallel paths
+phi = 25 * 10 ^ -3; // Flux per pole in Weber
+
+
+// CALCULATIONS
+
+T = (p/a)*((Z*Ia*phi)/(2*%pi)); // Deleloped Torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.6 : SOLUTION :-") ;
+printf("\n (a) Developed Torque in an Six-pole DC Motor , T = %.1f N-m \n ",T);
+
+
diff --git a/Working_Examples/2777/CH4/EX4.7/Ex4_7.sce b/Working_Examples/2777/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 0000000..d444b1c --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,41 @@ + +// ELECTRICAL MACHINES +// R.K.Srivastava +// First Impression 2011 +// CENGAGE LEARNING INDIA PVT. LTD + +// CHAPTER : 4 : DIRECT CURRENT MACHINES + +// EXAMPLE : 4.7 + +clear ; clc ; close ; // Clear the work space and console + + +// GIVEN DATA + +p = 2; // Number of the Pole +N = 1000; // Rotation speed of the Armature in RPM +Ia = 20; // Armature Current in Amphere +CS = 36; // Commutator Segments +BW = 1.4; // Brush width is 1.4 times of the Commutator Segments +L = 0.09 * 10 ^ -3; // Inducatnce of the each Armature Coil + + +// CALULATIONS + +a = p; // Number of the Parallel paths (Equal to number of poles because Lap Connected Armature) +n = N/60; // Revoultion per second +I = Ia/2; // Current Through the each Conductor in Amphere +v = n * CS; // Peripheral Velocity of Commutator in Commutator segments per Seconds +Tc = BW/v; // Time of the Commutation in Seconds +Er = (L*2*I)/Tc; // Reactance voltage in Volts + + +// DISPLAY RESULTS + +disp("EXAMPLE : 4.7 : SOLUTION :-") ; +printf("\n (a) Reactance voltage assuming Linear Commutation , Er = %.4f V \n",Er); +printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" ); +printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Tc = 0.014 s instead of %.4f s \n",Tc); +printf("\n (b) Er = 1.2857 V instead of %.4f V\n",Er); +printf("\n From Calculation of the Time of commutation (Tc), rest all the Calculated values in the TEXT BOOK is WRONG because of the Time of commutation (Tc) value is WRONGLY calculated and the same used for the further Calculation part \n") diff --git a/Working_Examples/2777/CH4/EX4.8/Ex4_8.sce b/Working_Examples/2777/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 0000000..3ee5f34 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,33 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.8
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N = 800; // Rotation speed of the Commutator in RPM
+D = 50; // Diameter in Centimeter
+BW = 1.5; // Brush Width in Centimeter
+
+
+// CALCULATIONS
+
+r = D/2; // Radius in Centimeter
+n = N/60; // Revoultion per second
+w = (2 * %pi)*n; // Angular velocity
+v = w*r; // Peripheral Speed in centimeter per second
+Tc = (BW/v)*1000; // Time of the Commutation in Second
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.8 : SOLUTION :-") ;
+printf("\n (a) Time of the Commutation , Tc = %.4f ms \n",Tc);
diff --git a/Working_Examples/2777/CH4/EX4.9/Ex4_9.sce b/Working_Examples/2777/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 0000000..f92ae73 --- /dev/null +++ b/Working_Examples/2777/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 4 : DIRECT CURRENT MACHINES
+
+// EXAMPLE : 4.9
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 4; // Number of the pole in the Generator
+Ia = 100; // supplying Current by the Generator in Amphere
+Za = 500; // Armature conductor
+beta = 8; // Brush shift in degrees
+If = 5; // Current in the Seperately exicted field winding
+ratio = 0.7; // Ratio of Pole arc to Pole pitch
+
+
+// CALCULATIONS
+// For case (a) Lap winding
+
+a_a = p; // Number of the Parallel Paths
+AT_a = (Za*Ia)/(2*a_a*p); // Amphere turns per pole
+ATd_a = (beta*Za*Ia)/(360*a_a); // Demagnetizing Armature Amphere turns per pole
+ATc_a = ((1/p)-(beta/180))*((Za*Ia)/(2*a_a)); // CrossMagnetizing Armature Amphere turns per pole
+ATw_a = ratio*AT_a; // Amphere turns of Compensating winding
+
+// For case (b) Wave winding
+
+a_b = p/2; // Number of the Parallel Paths
+AT_b = (Za*Ia)/(2*a_b*p); // Amphere turns per pole
+ATd_b = (beta*Za*Ia)/(360*a_b); // Demagnetizing Armature Amphere turns per pole
+ATc_b = ((1/p)-(beta/180))*((Za*Ia)/(2*a_b)); // CrossMagnetizing Armature Amphere turns per pole
+ATw_b = ratio*AT_b; // Amphere turns of Compensating winding
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 4.9 : SOLUTION :-") ;
+printf("\n For LAP winding \n\n (a.1) Amphere turns per pole, AT = %.1f AT \n",AT_a);
+printf("\n (a.2) Demagnetizing Armature Amphere turns per pole, ATd = %.1f AT \n",ATd_a);
+printf("\n (a.3) Cross-Magnetizing Armature Amphere turns per pole, ATc = %.1f AT \n",ATc_a);
+printf("\n (a.4) Amphere turns of Compensating winding, ATw = %.1f AT \n",ATw_a);
+printf("\n For WAVE winding \n\n (b.1) Amphere turns per pole, AT = %.f AT \n",AT_b);
+printf("\n (b.2) Demagnetizing Armature Amphere turns per pole, ATd = %.2f AT \n",ATd_b);
+printf("\n (b.3) Cross-Magnetizing Armature Amphere turns per pole, ATc = %.f AT \n",ATc_b);
+printf("\n (b.4) Amphere turns of Compensating winding, ATw = %.1f AT \n",ATw_b);
diff --git a/Working_Examples/2777/CH5/EX5.1/Ex5_1.sce b/Working_Examples/2777/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 0000000..7a60ea9 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,66 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// For Case (a)
+
+S_a = 30; // Total number of Slots
+m_a = 3; // Total number of Poles
+p_a = 2; // Total number of Phases
+
+// For Case (b)
+
+S_b = 60; // Total number of Slots
+m_b = 3; // Total number of Poles
+p_b = 4; // Total number of Phases
+
+// For Case (c)
+
+S_c = 24; // Total number of Slots
+m_c = 3; // Total number of Poles
+p_c = 4; // Total number of Phases
+
+// For Case (d)
+
+S_d = 12; // Total number of Slots
+m_d = 3; // Total number of Poles
+p_d = 2; // Total number of Phases
+
+
+// CALCULATIONS
+// For Case (a)
+
+spp_a = S_a/(p_a*m_a); // Slot per poles per phase
+
+// For Case (b)
+
+spp_b = S_b/(p_b*m_b); // Slot per poles per phase
+
+// For Case (c)
+
+spp_c = S_c/(p_c*m_c); // Slot per poles per phase
+
+// For Case (d)
+
+spp_d = S_d/(p_d*m_d); // Slot per poles per phase
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.1 : SOLUTION :-") ;
+printf("\n For case (a) Slot per poles per phase, spp = %.f \n ",spp_a);
+printf("\n For case (b) Slot per poles per phase, spp = %.f \n ",spp_b);
+printf("\n For case (c) Slot per poles per phase, spp = %.f \n ",spp_c);
+printf("\n For case (d) Slot per poles per phase, spp = %.f \n ",spp_d);
+
+
diff --git a/Working_Examples/2777/CH5/EX5.10/Ex5_10.sce b/Working_Examples/2777/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 0000000..d15b4a4 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,32 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.10
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles in Induction Motor
+f = 50; // Frequency in Hertz
+s = 0.03; // Slip
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+Nr = (1-s)*Ns; // Rotor Speed in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.10 : SOLUTION :-");
+printf("\n (a) Rotor Speed , Nr = %.f RPM \n",Nr)
diff --git a/Working_Examples/2777/CH5/EX5.11/Ex5_11.sce b/Working_Examples/2777/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 0000000..ca8b07f --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.11
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+s = 0.03; // Slip
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+Nr = (1-s)*Ns; // Rotor Speed in RPM
+Nf = Ns - Nr; // Speed of Forward Rotating magnetic fields with respect to stator and rotor in RPM
+Nb = Ns + Nr; // Speed of Backward Rotating magnetic fields with respect to stator and rotor in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.11 : SOLUTION :-");
+printf("\n (a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM \n",Nf)
+printf("\n (b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM \n",Nb)
+
+
+
diff --git a/Working_Examples/2777/CH5/EX5.12/Ex5_12.sce b/Working_Examples/2777/CH5/EX5.12/Ex5_12.sce new file mode 100755 index 0000000..3541e45 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,35 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.12
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+s = 0.05; // Slip
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+Nr = (1-s)*Ns; // Rotor Speed in RPM
+Nf = s*Ns; // Speed of Forward Rotating magnetic fields with respect to stator and rotor in RPM
+Nb = (p-s)*Ns; // Speed of Backward Rotating magnetic fields with respect to stator and rotor in RPM
+fr = (p-s)*f; // Backward rotating magnetic field induces a current of frequency in Hertz
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.12 : SOLUTION :-");
+printf("\n (a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM \n",Nf)
+printf("\n (b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM \n",Nb)
diff --git a/Working_Examples/2777/CH5/EX5.13/Ex5_13.sce b/Working_Examples/2777/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 0000000..88850d4 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,45 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+s = 0.05; // Slip
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+fr = s*f; // Rotor-induced Frequency of forward field in Hertz
+Nfr = s*Ns; // Speed of Forward Rotating magnetic fields with respect to rotor surface in RPM
+f2r = s*f; // Rotor-induced Frequency of Backward field in Hertz
+Nbr = -(s*Ns); // Speed of Backward Rotating magnetic fields with respect to rotor surface in RPM
+Nr = (1-s)*Ns; // Rotor running in Forward direction in RPM
+Nfs = Nr+(s*Ns); // Speed of Forward Rotating magnetic fields with respect to stator surface in RPM
+Nbs = Nr-(s*Ns); // Speed of Backward Rotating magnetic fields with respect to stator surface in RPM
+Nbs_new = -(0.5*Ns)+(1-0.5)*Nr; // Speed of Backward Rotating magnetic fields with respect to stator for 50% of slip in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.13 : SOLUTION :-");
+printf("\n (a.1) Speed of Forward Rotating magnetic fields with respect to rotor surface is equal to + %.f RPM \n",Nfr)
+printf("\n (a.2) Speed of Backward Rotating magnetic fields with respect to rotor surface is equal to + %.f RPM \n",Nbr)
+printf("\n (b.1) Speed of Forward Rotating magnetic fields with respect to stator surface is equal to + %.f RPM \n",Nfs)
+printf("\n (b.2) Speed of Backward Rotating magnetic fields with respect to stator surface is equal to + %.f RPM \n",Nbs)
+printf("\n (c) Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to %.1f RPM \n",Nbs_new)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to 0 RPM instead of %.1f RPM \n ",Nbs_new);
diff --git a/Working_Examples/2777/CH5/EX5.14/Ex5_14.sce b/Working_Examples/2777/CH5/EX5.14/Ex5_14.sce new file mode 100755 index 0000000..d967110 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,31 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 50; // Stator Frequency of Inductor Motor in Hertz
+fr = 10; // Rotor Frequency of Inductor Motor in Hertz
+p = 2; // Number of poles
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed of Induction Motor in RPM
+s = fr/f; // Slip of the Induction Motor
+Nr = (1-s)*Ns; // Rotor Speed of the Induction Motor
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.14 : SOLUTION :-");
+printf("\n (a) Rotor Speed of Induction Motor, Nr = %.f RPM \n",Nr)
diff --git a/Working_Examples/2777/CH5/EX5.15/Ex5_15.sce b/Working_Examples/2777/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 0000000..2eb9ed8 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,79 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.15
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 5.15 : \n\n Given Data No-load test : 440V, 30A, 4.5KW \n");
+printf("\n Blocked rotor test : 90V,50Hz, 120A, 16KW \n");
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+V = 440; // Operating voltage of the Induction motor in Volts
+out_hp = 100; // Output of the Induction motor in Horse-Power
+R = 0.15; // Average dc Resistance in Ohms
+Wsc = 16000; // Power at Blocked Rotor test in Watts
+Vsc = 90; // Voltage at Blocked Rotor test in Volts
+Isc = 120; // Current at Blocked Rotor test in Amphere
+W0 = 4500; // Power at No-load test in Watts
+V0 = 440; // Voltage at No-load test in Volts
+I0 = 30; // Current at No-load test in Amphere
+s = 0.05; // Slip
+f = 50; // Frequency in Hertz
+
+
+// CALCULATIONS
+
+R1 = R/2; // DC winding Resistance per phase in Ohms
+Rac = Wsc/(3*Isc^2); // AC Resistance referred to stator from locked rotor test at supply frequency in Ohms
+R_2 = Rac - R1; // Per phase Rotor Resistance to Stator in Ohms
+Zsc = Vsc/(sqrt(3)*Isc); // Per phase Impedance from locked rotor test in Ohms
+Xs = sqrt((Zsc^2)-(Rac^2)); // Per phase leakage Reactance referred to stator in Ohms
+theta_0 = acosd(W0/(V0*I0*sqrt(3))); // No-load power factor angle in degree
+Im = I0*sind(theta_0); // Reactive component of no-load current in Amphere
+Xm = V0/(Im*sqrt(3)); // Magnetizing Reactance in Ohms
+Pc = W0 - 3*I0^2*R1; // Total Core loss in Watts
+Rc = (V0/sqrt(3))^2*(3/Pc); // Per phase core loss Resistance in Watts
+Vph = V0/sqrt(3); // Per phase Voltage in Volts
+Ic = Vph/Rc; // Core loss current in Amphere
+I_m = Vph/(%i * Xm); // Magnetizing Current in Amphere
+I_o = Ic + I_m; // No-load current in Amphere
+I_2 = Vph/(R1+(R_2/s)+(%i*Xs)); // Current in Amphere
+I1 = I_o + I_2; // Input Current in Amphere
+Pf = cosd(atand(imag(I1)/real(I1))); // Power factor
+P1 = (3*(abs(I_2)^2*R_2)/s)/1000; // 3-phase air gap power or Rotor intake Power in Kilo-Watts
+Po = P1*(1-s); // Output Power in Kilo-Watts
+Ws = 2*%pi*((120*f/p)*(1/60)); // Angular Roatation in Radians per Seconds
+T = P1*1000/Ws; // Torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a.1) DC winding Resistance per phase, R1 = %.3f Ohms \n",R1)
+printf("\n (a.2) AC Resistance referred to stator from locked rotor test at supply frequency = %.4f Ohms \n",Rac)
+printf("\n (a.3) Per phase Rotor Resistance to Stator, R2 = %.4f Ohms \n",R_2)
+printf("\n (a.4) Per phase Impedance from locked rotor test,Zsc = %.3f Ohms \n",Zsc)
+printf("\n (a.5) Per phase leakage Reactance referred to stator, Xs = %.4f Ohms \n",Xs)
+printf("\n (a.6) No-load power factor angle, theta_O = %.2f Degree \n",theta_0)
+printf("\n (a.7) Reactive component of no-load current, Im = %.1f A \n",Im)
+printf("\n (a.8) Magnetizing Reactance, Xm = %.2f Ohms \n",Xm)
+printf("\n (a.9) Total Core loss, Pc = %.1f W \n",Pc)
+printf("\n (a.10) Per phase core loss Resistance, Pc = %.f Ohms \n",Rc)
+printf("\n (a.11) Per phase Voltage, Vph = %.f V \n",Vph)
+printf("\n (a.12) Core loss current, Ic = %.2f < %.f A \n",abs(Ic),atand(imag(Ic),real(Ic)))
+printf("\n (a.13) Magnetizing Current, Im = %.1f < %.f A \n",abs(I_m),atand(imag(I_m),real(I_m)))
+printf("\n (a.14) No-load current, I0 = %.2f < %.2f A \n",abs(I_o),atand(imag(I_o),real(I_o)))
+printf("\n (a.15) Current, I2 = %.2f < %.2f A \n",abs(I_2),atand(imag(I_2),real(I_2)))
+printf("\n (b) Input current, I1 = %.2f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)))
+printf("\n (c) Power Factor, Pf = %.4f Lagging \n",Pf)
+printf("\n (d) Output Power, P0 = %.1f kW \n",Po)
+printf("\n (e) Torque, T = %.2f NM \n",T)
diff --git a/Working_Examples/2777/CH5/EX5.17/Ex5_17.sce b/Working_Examples/2777/CH5/EX5.17/Ex5_17.sce new file mode 100755 index 0000000..ea44107 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,59 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.17
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 5.17 : \n\n Given Data No-load test : 440V, 3.0A, 500KW, 50Hz \n");
+printf("\n Blocked rotor test at rated frequency : 110V, 18A, 2500W, 50Hz \n");
+printf("\n DC test on Stator per phase : 10V, 15A \n");
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+out_hp = 20; // Motor Power Rating in Horse-Power
+Vdc = 10; // DC Voltage in Volts
+Idc = 15; // DC Current in Amphere
+Wsc = 2500; // Power at Blocked Rotor test rated frequency in Watts
+Wsc_red = 2050; // Power at Blocked Rotor test reduced frequency in Watts
+Vsc = 110; // Voltage at Blocked Rotor test rated frequency in Volts
+Isc = 18; // Current at Blocked Rotor test rated frequency in Amphere
+Wo = 500; // Power at No-load test in Watts
+Vo = 440; // Voltage at No-load test in Volts
+Io = 4.0; // Current at No-load test in Amphere
+fsc = 50; // Rated Frequency at blocked rotor test in Hertz
+fo = 50; // Rated Frequency at no-load test in Hertz
+fsc1 = 15; // Reduced Frequency at blocked rotor in Hertz
+Pfw = 200; // Friction and Windage loss in Watts
+
+
+// CALCULATIONS
+
+R1dc = Vdc/Idc; // DC winding Resistance per phase in Ohms
+Rac = Wsc/(3*Isc^2); // AC Resistance from Locked rotor test at supply frequency
+Rac_red = Wsc_red/(3*Isc^2); // AC Resistance from Locked rotor test at reduced frequency
+R1ac = (Rac/Rac_red)*R1dc; // Corrected Value of AC stator winding Resistance in Ohms
+R2dc = Rac_red - R1dc; // Second rotor parameter, rotor resistance referred to stator is at low frequency in Ohms
+Zsc = Vsc/(sqrt(3)*Isc); // Per phase Impedance from locked rotor test at power frequency in Ohms
+Xs = sqrt((Zsc^2)-(Rac^2)); // Per phase leakage Reactance referred to stator in Ohms
+theta_0 = acosd(Wo/(Vo*Io*sqrt(3))); // No-load power factor angle in degree
+Im = Io*sind(theta_0); // Reactive component of no-load current in Amphere
+Xm = Vo/(Im*sqrt(3)); // Magnetizing Reactance in Ohms
+Pc = Wo - 3*Io^2*R1ac-Pfw; // Total Core loss in Watts
+Rc = (Vo/sqrt(3))^2*(3/Pc); // Per phase core loss Resistance in Watts
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a) Magnetizing Reactance of Equivalent circuit, Xm = %.1f Ohms \n",Xm)
+printf("\n (b) Per phase core loss Resistance, Pc = %.f Ohms \n",Rc)
diff --git a/Working_Examples/2777/CH5/EX5.18/Ex5_18.sce b/Working_Examples/2777/CH5/EX5.18/Ex5_18.sce new file mode 100755 index 0000000..20f0e97 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,54 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.18
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// From Previous problem data (Example 5.17)
+
+R1ac = 0.8127; // Corrected Value of AC stator winding Resistance in Ohms
+R2dc = 1.4433; // Second rotor parameter, rotor resistance referred to stator is at low frequency in Ohms
+Xs = 2.42; // Per phase leakage Reactance referred to stator in Ohms
+Xm = 64.4; // Magnetizing Reactance in Ohms
+Rc = 742; // Per phase core loss Resistance in Watts
+s = 0.035; // Slip
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+out_hp = 20; // Motor Power Rating in Horse-Power
+
+
+// CALCULATIONS
+
+Vph = V/sqrt(3); // Per phase Voltage in Volts
+Ic = Vph/Rc; // Core loss current in Amphere
+I_m = Vph/(%i * Xm); // Magnetizing Current in Amphere
+I_o = Ic + I_m; // No-load current in Amphere
+I_2 = Vph/(R1ac+(R2dc/s)+(%i*Xs)); // Current in Amphere
+I1 = I_o + I_2; // Input Current in Amphere
+Pf = cosd(atand(imag(I1)/real(I1))); // Power factor
+P1 = 3*(abs(I_2)^2*R2dc)/s; // 3-phase air gap power or Rotor intake Power in Watts
+Po = P1*(1-s); // Output Power in Watts
+Ws = 2*%pi*((120*f/p)*(1/60)); // Angular Roatation in Radians per Seconds
+T = P1/Ws; // Torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.18 : SOLUTION :-");
+printf("\n (a) Input current, I1 = %.2f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)))
+printf("\n (b) Power Factor, Pf = %.3f Lagging \n",Pf)
+printf("\n (c) Output Power, P0 = %.2f W \n",Po)
+printf("\n (d) Torque, T = %.2f NM \n",T)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) T = 4340.82 Nm instead of %.2f Nm \n ",T);
+printf("\n\n IN TEXT BOOK, CALCULATION OF TORQUE IS NOT DONE \n ");
diff --git a/Working_Examples/2777/CH5/EX5.19/Ex5_19.sce b/Working_Examples/2777/CH5/EX5.19/Ex5_19.sce new file mode 100755 index 0000000..5fc9518 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.19/Ex5_19.sce @@ -0,0 +1,58 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.19
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+s = 0.05; // Slip
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operrating Voltage of the Inductuon Motor
+R1 = 0.10; // Circuit Parameter in Ohms
+R2 = 0.11; // Circuit Parameter in Ohms
+X1 = 0.35; // Circuit Parameter in Ohms
+X2 = 0.40; // Circuit Parameter in Ohms
+pf = 0.2; // Power factor (Lagging)
+Pr = 900; // Rotational Loss in Watts
+Psc = 1000; // Stator core loss in Watts
+I = 15; // Line current draws by the motor in Amphere
+
+
+// CALCULATIONS
+
+Vph = V/sqrt(3); // Per phase Voltage in Volts
+I_2 = Vph/(R1+(R2/s)+(%i*(X1+X2))); // Current in Amphere
+Io = I * exp(-( %i * acosd(pf) * %pi/180)); // No-load current in Amphere
+I1 = Io + I_2; // Input line Current in Amphere
+PF = cosd(atand(imag(I1)/real(I1))); // Power factor
+Ws = 2*%pi*((120*f/p)*(1/60)); // Angular Roatation in Radians per Seconds
+Pg = (3*(abs(I1)^2*R2))/s; // 3-phase air gap power or Rotor intake Power in Watts
+T = Pg/Ws; // Torque in Newton-Meter
+Po = Pg*(1-s)-Pr; // Output Power in Watts
+Po_HP = Po/746; // Output Power in Horse-Power
+eta = (Po/(Po+Psc+Pr))*100; // Efficiency in Percentage
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.19 : SOLUTION :-");
+printf("\n (a) Input line current, I1 = %.1f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)))
+printf("\n (b) Power Factor, Pf = %.4f Lagging \n",PF)
+printf("\n (c) Output Power, P0 = %.1f HP \n",Po_HP)
+printf("\n (d) Torque, T = %.2f Nm \n",T)
+printf("\n (e) Efficiency, eta = %.1f Percenatge \n",eta)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) I1 = 114.2<-24.68 A instead of %.1f<%.2f A \n ",abs(I1),atand(imag(I1),real(I1)));
+printf("\n (b) T = 548.24 Nm instead of %.2f Nm \n ",T);
+printf("\n (c) Po = 108.4 HP instead of %.1f HP \n ",Po_HP);
+
diff --git a/Working_Examples/2777/CH5/EX5.2/Ex5_2.sce b/Working_Examples/2777/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 0000000..4a53f79 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,91 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// For Case (a)
+
+S_a = 54; // Total number of Slots
+m_a = 3; // Total number of Poles
+p_a = 8; // Total number of Phases
+
+// For Case (b)
+
+S_b = 32; // Total number of Slots
+m_b = 3; // Total number of Poles
+p_b = 4; // Total number of Phases
+
+// For Case (c)
+
+S_c = 30; // Total number of Slots
+m_c = 3; // Total number of Poles
+p_c = 4; // Total number of Phases
+
+
+// CALCULATIONS
+// For Case (a)
+
+spp_a = S_a/(p_a*m_a); // Slot per poles per phase
+l_a = 0 * spp_a; // Phase allociation Series
+m_a = 1 * spp_a; // Phase allociation Series
+n_a = 2 * spp_a; // Phase allociation Series
+o_a = 3 * spp_a; // Phase allociation Series
+p_a = 4 * spp_a; // Phase allociation Series
+d_a = 0; // d_a = l_a (Rounding off)
+e_a = 2; // e_a = m_a (Rounding off)
+f_a = 4; // f_a = n_a (Rounding off)
+g_a = 6; // g_a = o_a (Rounding off)
+h_a = 9; // h_a = p_a (Rounding off)
+R_a = e_a - d_a; // Phase allociation
+Y_a = f_a - e_a; // Phase allociation
+B_a = g_a - f_a; // Phase allociation
+R1_a = h_a - g_a; // Phase allociation
+
+// For Case (b)
+
+spp_b = S_b/(p_b*m_b); // Slot per poles per phase
+l_b = 0 * spp_b; // Phase allociation Series
+m_b = 1 * spp_b; // Phase allociation Series
+n_b = 2 * spp_b; // Phase allociation Series
+o_b = 3 * spp_b; // Phase allociation Series
+d_b = 0; // d_b = l_b (Rounding off)
+e_b = 2; // e_b = m_b (Rounding off)
+f_b = 5; // f_b = n_b (Rounding off)
+g_b = 8; // g_b = o_b (Rounding off)
+R_b = e_b - d_b; // Phase allociation
+Y_b = f_b - e_b; // Phase allociation
+B_b = g_b - f_b; // Phase allociation
+
+// For Case (c)
+
+spp_c = S_c/(p_c*m_c); // Slot per poles per phase
+l_c = 0 * spp_c; // Phase allociation Series
+m_c = 1 * spp_c; // Phase allociation Series
+n_c = 2 * spp_c; // Phase allociation Series
+d_c = 0; // d_b = l_b (Rounding off)
+e_c = 2; // e_b = m_b (Rounding off)
+f_c = 5; // f_b = n_b (Rounding off)
+R_c = e_c - d_c; // Phase allociation
+Y_c = f_c - e_c; // Phase allociation
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.2 : SOLUTION :-") ;
+printf("\n For Case (a) Slot per poles per phase , spp = %.3f \n ",spp_a);
+printf("\n Phase allociation series is %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f,slots are allocated respectively to R, Y, B, R, Y, B, R, Y, B....... phase in Sequence\n ",R_a,Y_a,B_a,R1_a,R_a,Y_a,B_a,R1_a,R_a);
+printf("\n By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible \n");
+printf("\n For Case (b) Slot per poles per phase , spp = %.3f \n ",spp_b);
+printf("\n Phase allociation series is %.f, %.f, %.f \n",R_b,Y_b,B_b);
+printf("\n By seeing Sequence its Slot per pole per phase are not Integer therefore R-phase will have 8 slots whereas Y-phase and B-phase will have 12 slots \n");
+printf("\n For Case (c) Slot per poles per phase , spp = %.1f \n ",spp_c);
+printf("\n Phase allociation series is %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f slots are allocated respectively to R, Y, B, R, Y, B, R, Y, B, R, Y, B....... phase in Sequence\n ",R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c);
+printf("\n By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible \n");
diff --git a/Working_Examples/2777/CH5/EX5.20/Ex5_20.sce b/Working_Examples/2777/CH5/EX5.20/Ex5_20.sce new file mode 100755 index 0000000..1971ca4 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.20/Ex5_20.sce @@ -0,0 +1,60 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.20
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+R1 = 0.25; // Circuit Parameter in Ohms
+R2 = 0.25; // Circuit Parameter in Ohms
+X1 = 0.75; // Circuit Parameter in Ohms
+X2 = 0.75; // Circuit Parameter in Ohms
+Xm = 1000; // Circuit Parameters in Ohms
+Rc = 100; // Circuit Parameters in Watts
+s = 0.025; // Slip
+Pr = 450; // Rotational Loss in Watts
+Psc = 800; // Stator core loss in Watts
+
+
+// CALCULATIONS
+
+Vph = V/sqrt(3); // Per phase Voltage in Volts
+I_2 = Vph/(R1+(R2/s)+(%i*(X1+X2))); // Current in Amphere
+Ic = Vph/Rc; // Core loss current in Amphere
+I_m = Vph/(%i * Xm); // Magnetizing Current in Amphere
+I_o = Ic + I_m; // No-load current in Amphere
+I1 = I_o + I_2; // Input Current in Amphere
+Pf = cosd(atand(imag(I1)/real(I1))); // Power factor
+Ns = (120*f)/p; // Synronous Speed in RPM
+Pg = 3*(abs(I_2)^2*R2)/s; // 3-phase air gap power or Rotor intake Power in Watts
+Pm = Pg*(1-s); // Output Power in Watts
+Ws = 2*%pi*Ns*(1/60); // Angular Roatation in Radians per Seconds
+T = Pg/Ws; // Torque in Newton-Meter
+Po = Pm-Pr; // Output Power in Watts
+Po_HP = Po/746; // Output Power in Horse-Power
+eta = (Po/(Po+Psc+Pr))*100; // Efficiency in Percentage
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.20 : SOLUTION :-");
+printf("\n (a) Input line current, I1 = %.f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)))
+printf("\n (b) Power Factor, Pf = %.4f Lagging \n",Pf)
+printf("\n (c) Output Power, P0 = %.2f HP \n",Po_HP)
+printf("\n (d) Torque, T = %.1f Nm \n",T)
+printf("\n (e) Efficiency, eta = %.1f Percenatge \n",eta)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) I1 = 26.8-j3.584 {27<-7.62} A in instead of (%.1f)+(j%.3f) {%.f<%.2f} A \n ",real(I1),imag(I1),abs(I1),atand(imag(I1),real(I1)));
+printf("\n (b) pf = 0.9885 Lagging instead of %.4f Lagging \n ",Pf);
diff --git a/Working_Examples/2777/CH5/EX5.21/Ex5_21.sce b/Working_Examples/2777/CH5/EX5.21/Ex5_21.sce new file mode 100755 index 0000000..3ff9730 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.21/Ex5_21.sce @@ -0,0 +1,41 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.21
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+s = 0.05; // Slip
+f = 50; // Frequency in Hertz
+Tm = 500; // Maximum Torque in Newton-Meter
+Tst = 200; // Starting Torque in Newton-Meter
+sst = 1.0; // Starting Slip
+
+
+// CALCULATONS
+
+p1 = poly([1 -5 1],'sm','c'); // Slip at Maximum Torque (obtained from Equation Tst = (2*Tm)/((sst/sm)+(sm+sst))
+a = roots(p1); // Value of slip at Maximum Torque (obtained from Equation Tst = (2*Tm)/((sst/sm)+(sm+sst))
+sm = a(2,1); // Slip at Maximum Torque (obtained from Equation Tst = (2*Tm)/((sst/sm)+(sm+sst)) { 1st root is 4.8 so its out of range because slip value is lies between 0-1 so its neglected and second root will be slip }
+T = (2*Tm)/((s/sm)+(sm/s)); // Torque at 0.05 slip
+Ns = (120*f)/p; // Synchronous Speed in RPM
+Wr = (2*%pi)*(1-s)*(Ns/60); // Angular Velocity in Radians-per-Second
+P = T * Wr; // Power Output in Watts
+P_HP = P/746; // Power Output in Horse-Power
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.21 : SOLUTION :-");
+printf("\n (a) Torque at 0.05 slip, T = %.2f Nm \n",T)
+printf("\n (b) Power Output at 0.05 slip, P = %.1f W = %.2f HP \n",P,P_HP)
diff --git a/Working_Examples/2777/CH5/EX5.22/Ex5_22.sce b/Working_Examples/2777/CH5/EX5.22/Ex5_22.sce new file mode 100755 index 0000000..6103f5f --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.22/Ex5_22.sce @@ -0,0 +1,76 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.22
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Wsc = 1000; // Power at Blocked Rotor test in Watts
+Vsc = 56; // Voltage at Blocked Rotor test in Volts
+Isc = 18; // Current at Blocked Rotor test in Amphere
+Woc = 52; // Power at No-load test in Watts
+Voc = 220; // Voltage at No-load test in Volts
+Ioc = 2.6; // Current at No-load test in Amphere
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+V = 220; // Operating voltage of the Induction motor in Volts
+f = 50; // Frequency in Hertz
+s = 0.05; // Slip
+R = 0.65; // Per phase stator Resistance in Ohms
+
+
+// CALCULATIONS
+
+Vph = Voc/sqrt(3); // Per phase Voltage in Volts
+Wo = Woc/m; // Per phase No-load loss in Watts
+theta_0 = acosd(Wo/(Voc*Ioc*sqrt(3))); // No-load power factor angle in degree
+VSC = Vsc/sqrt(3); // Per phase locked rotor Voltage in Volts
+WSC = Wsc/m; // Per phase locked rotor loss in Watts
+theta_sc = acosd(WSC/(VSC*Isc)); // No-load power factor angle in degree
+ISC = Isc*(Voc/Vsc); // locked rotor current at full Voltage in Amphere
+Re = WSC/Isc^2; // Resistance in Ohms
+R1 = R*1.1; // Per phase AC stator Resistance in Ohms
+R_2 = Re - R1; // Per phase rotor Resistance in Ohms
+Zsc = VSC/Isc; // Per phase impedance in Ohms
+Xs = sqrt((Zsc^2)-(Re^2)); // Leakage Reactance in Ohms
+I_2 = (Voc/sqrt(3))/sqrt((R1+(R_2/s))^2+(Xs^2)); // Current in Amphere
+pf = cosd(atand(Xs/(R1+(R_2/s)))); // Power Factor
+Ws = 2*%pi*((120*f/p)*(1/60)); // Rotational Speed in Radians per Seconds
+Pg = (3*(abs(I_2)^2*R_2))/s; // 3-phase air gap power or Rotor intake Power in Watts
+T = Pg/Ws; // Torque in Newton-Meter
+// CALCULATIONS OR DATA OBTAINED FROM CIRCLE DIAGRAM FIGURE 5.35 and PAGE NO:-303
+OA = 2.60; // Correspounding Current in Amphere at 87' from Y-axis (from Circle diagram)
+OE = 70.70; // Correspounding Current in Amphere at 55' from Y-axis (from Circle diagram)
+OP = 17.77; // Current in Amphere (from Circle diagram)
+OV = Voc/sqrt(3); // Phase Voltage in No-load test or value obatined from circle diagram in Volts
+PK = 11.6; // Correspounding Value from Circle diagram
+JK = 0.8; // Correspounding Value from Circle diagram
+PJ = 10.8; // Correspounding Value from Circle diagram
+PM = 11.6; // Correspounding Value from Circle diagram
+Pir = 3*OV*PK; // Total Rotor intake in Watts
+Plr = 3*OV*JK; // Total Rotor loss in Watts
+Po = 3*OV*PJ; // Total Mechanical power output in Watts
+T_c = (3*OV*PK)/Ws; // Total Torque in Newton-Meter
+s_c = JK/PK; // Slip obtained from Circle diagram
+s_pc = 100*s_c; // Slip in percentage
+eta = 100*(PJ/PM); // Eifficiency in Percentage
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.22 : SOLUTION :-");
+printf("\n (a) Input line current, I2 = %.2f A \n",I_2)
+printf("\n (b) Power Factor, Pf = %.3f \n",pf)
+printf("\n (c) Torque, T = %.2f Nm \n",T)
+printf(" \n Verification Results from Circle Diagram :-\n");
+printf("\n (a) Efficency, eta = %.2f Percent \n",eta)
+printf("\n (b) slip, s = %.3f = %.f percent \n",s_c,s_pc)
+printf("\n (c) Torque, T = %.2f Nm \n",T_c)
diff --git a/Working_Examples/2777/CH5/EX5.23/Ex5_23.sce b/Working_Examples/2777/CH5/EX5.23/Ex5_23.sce new file mode 100755 index 0000000..5e298d4 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.23/Ex5_23.sce @@ -0,0 +1,92 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.23
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+R1 = 0.2; // Circuit Parameter in Ohms
+R2 = 0.4; // Circuit Parameter in Ohms
+X1 = 1.0; // Circuit Parameter in Ohms
+X2 = 1.5; // Circuit Parameter in Ohms
+m = 3; // Total Number of phase in Induction Motor
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+
+
+// CALCULATIONS
+
+Ws = 2*%pi*f; // Synchronous angular speed in Radians per second
+Z = (R1+R2)+((%i)*(X1+X2)); // At slip s=1, the impedance seen from the terminals in Ohms
+s = 1; // Slip
+
+// For Case(a) Winding is connected in star
+
+Isy_a = V/(abs(Z)*sqrt(3)); // Current in Amphere
+Tsy_a = (3*Isy_a^2*R2)/(s*Ws); // Torque in Newton-Meter
+
+// Winding is connected in delta
+
+Isd_a = (V*sqrt(3))/abs(Z); // Current in Amphere
+Tsd_a = (3*(Isd_a/sqrt(3))^2*R2)/(s*Ws); // Torque in Newton-Meter
+I_R = Isd_a/Isy_a; // Ratio of the line current
+T_R = Tsd_a/Tsy_a; // Ratio of the Torque
+
+// For Case(b) Machine is started using auto-transfromer and voltage is 50% reduced
+
+Isy_b = (0.5*V)/(abs(Z)*sqrt(3)); // Current in Amphere when Winding is connected star
+Tsy_b = (3*Isy_b^2*R2)/(s*Ws); // Torque in Newton-Meter when Winding is connected star
+Isd_b = (0.5*V*sqrt(3))/abs(Z); // Current in Amphere when Winding is connected delta
+Tsd_b = (3*(Isd_b/sqrt(3))^2*R2)/(s*Ws); // Torque in Newton-Meter when Winding is connected delta
+
+// For Case(c) Both Voltage and Frequency are reduced to 50%
+
+f_new = (10/100)*f; // New Frequency
+Ws_c = 2*%pi*f_new; // Synchronous angular speed in Radians per second
+Z_c = ((R1+R2)+((%i)*(X1+X2))*(f_new/f)); // At slip s=1, the impedance seen from the terminals in Ohms
+Isy_c = (0.1*V)/(abs(Z_c)*sqrt(3)); // Current in Amphere when Winding is connected star
+Tsy_c = (3*Isy_c^2*R2)/(s*Ws_c); // Torque in Newton-Meter when Winding is connected star
+Isd_c = (0.1*V*sqrt(3))/abs(Z_c); // Current in Amphere when Winding is connected delta
+Tsd_c = (3*(Isd_c/sqrt(3))^2*R2)/(s*Ws_c); // Torque in Newton-Meter when Winding is connected delta
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.23 : SOLUTION :-");
+printf("\n For Case (a.1) Winding is connected in star \n");
+printf("\n (a.1.1) Per phase impedance seen from the terminals in Ohms, Z = %.3f < %.1f Ohms \n",abs(Z),atand(imag(Z),real(Z)));
+printf("\n (a.1.2) Initial Starting Current , Isy = %.2f A \n",Isy_a)
+printf("\n (a.1.3) Starting Torque , Tsy = %.1f Nm \n",Tsy_a)
+printf("\n For Case (a.2) Winding is connected in delta \n" );
+printf("\n (a.2.1) Initial Starting Current , Isd = %.2f A \n",Isd_a)
+printf("\n (a.2.2) Starting Torque , Tsd = %.2f Nm \n",Tsd_a)
+printf("\n For Case (b) Machine is started using auto-transfromer and voltage is 50 percentage reduced :- (b.1) Winding is connected in star \n ")
+printf("\n (b.1.1) Per phase impedance seen from the terminals in Ohms, Z = %.3f<%.1f Ohms \n",abs(Z),atand(imag(Z),real(Z)));
+printf("\n (b.1.2) Initial Starting Current , Isy = %.1f A \n",Isy_b)
+printf("\n (b.1.3) Starting Torque , Tsy = %.2f Nm \n",Tsy_b)
+printf("\n For Case (b.2) Winding is connected in delta \n" );
+printf("\n (b.2.1) Initial Starting Current , Isd = %.2f A \n",Isd_b)
+printf("\n (b.2.2) Starting Torque , Tsd = %.f Nm \n",Tsd_b)
+printf("\n For Case (c) Both Voltage and Frequency are reduced to 50 percentage :- (c.1) Winding is connected in star \n ");
+printf("\n (c.1.1) Per phase impedance seen from the terminals in Ohms, Z = %.2f<%.2f Ohms \n",abs(Z_c),atand(imag(Z_c),real(Z_c)));
+printf("\n (c.1.2) Initial Starting Current , Isy = %.2f A \n",Isy_c)
+printf("\n (c.1.3) Starting Torque , Tsy = %.2f Nm \n",Tsy_c)
+printf("\n For Case (c.2) Winding is connected in delta \n" );
+printf("\n (c.2.1) Initial Starting Current , Isd = %.2f A \n",Isd_c)
+printf("\n (c.2.2) Starting Torque , Tsd = %.2f Nm \n",Tsd_c)
+printf('\nComparing the Calculated values of starting current and torque eid rated frequency and rated voltage\n")
+printf("\n star delta\n")
+printf("\n 440V,50Hz 44V,5Hz 440V,50Hz 44V,5Hz \n")
+printf("\n starting current %.2f A %.f A %.f A %.2f A \n",Isy_a,Isy_c,Isd_a,Isd_c)
+printf("\n starting Torque %.1f Nm %.2f Nm %.f Nm %.2f Nm \n",Tsy_a,Tsy_c,Tsd_a,Tsd_c)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- For Case (a.2) Winding is connected in delta :- (a) Initial Starting Current Isy = 254.01 A instead of %.2f A \n\n ",Isd_a);
diff --git a/Working_Examples/2777/CH5/EX5.24/Ex5_24.sce b/Working_Examples/2777/CH5/EX5.24/Ex5_24.sce new file mode 100755 index 0000000..bd0bf31 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.24/Ex5_24.sce @@ -0,0 +1,74 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.24
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating voltage of the Induction Motor in Volts
+R1 = 0.2; // Circuit Parameter in Ohms
+R2 = 0.4; // Circuit Parameter in Ohms
+X1 = 1.0; // Circuit Parameter in Ohms
+X2 = 1.5; // Circuit Parameter in Ohms
+Rc = 150; // Circuit Parameter in Ohms
+Xm = 30; // Circuit Parameter in Ohms
+
+
+// CALCULATIONS
+
+V1 = V/sqrt(3); // Rated Voltage in Volts
+Zdol = (R1+%i*X1)+(Rc*%i*Xm*(R2+%i*X2))/(Rc*%i*Xm+Rc*(R2+%i*X2)+(%i*Xm)*(R2+%i*X2)); // Equivalent impedance per phase in DOL starter in Ohms
+I = V1/Zdol; // Starting Current in DOL starter in Amphere
+
+// For Case(a) A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit
+
+Zsr = (0.5+R1+%i*X1)+((Rc*%i*Xm*(R2+%i*X2))/((Rc*%i*Xm+Rc*(R2+%i*X2)+(%i*Xm)*(R2+%i*X2)))); // Total impedance seen from the terminals in Ohms
+Isr = V1/Zsr; // Starting Current in DOL starter in Amphere
+
+// For Case(b) A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit here assumed that stator to rotor turn ratio is 1.0
+
+Zrr = (R1+%i*X1)+((Rc*%i*Xm*(0.5+R2+%i*X2))/(Rc*%i*Xm+Rc*(0.5+R2+%i*X2)+(%i*Xm)*(0.5+R2+%i*X2))); // Total impedance seen from the terminals in Ohms
+Irr = V1/Zrr; // Starting Current in DOL starter in Amphere
+
+// For Case(c) When applied Voltage reduced to 50%
+
+I_c = (0.5*V1)/Zdol; // Starting Current in DOL starter in Amphere
+
+// For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10%) and the reduced frequency is 5Hz
+
+f_n = 5; // Reduced Frequency in Hertz
+X1_n = (f_n/f)*X1; // Changed Circuit Parameter in Ohms
+X2_n = (f_n/f)*X2; // Changed Circuit Parameter in Ohms
+Xm_n = (f_n/f)*Xm; // Changed Circuit Parameter in Ohms
+Zdol_n = (R1+%i*X1_n)+((Rc*%i*Xm_n*(R2+%i*X2_n))/(Rc*%i*Xm_n+Rc*(R2+%i*X2_n)+(%i*Xm_n)*(R2+%i*X2_n))); // Equivalent impedance per phase in DOL starter in Ohms
+I_n = (V1*0.1)/Zdol_n; // Starting Current in DOL starter in Amphere
+Ratio = abs(I_n)/abs(I); // Ratio of the Starting Current witha the rated Voltage and frequency to the reduced Voltage and frequency
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.24 : SOLUTION :-");
+printf("\n Normal Initial Starting Current in DOL starter, I = %.1f <%.1f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n For Case(a) A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit \n")
+printf("\n Initial Starting Current in DOL starter, I = %.1f <%.2f A \n",abs(Isr),atand(imag(Isr),real(Isr)))
+printf("\n For Case(b) A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit \n")
+printf("\n Initial Starting Current in DOL starter, I = %.2f <%.1f A \n",abs(Irr),atand(imag(Irr),real(Irr)))
+printf("\n For Case(c) When applied Voltage reduced to 50 percentage \n")
+printf("\n Initial Starting Current in DOL starter, I = %.2f <%.1f A \n",abs(I_c),atand(imag(I_c),real(I_c)))
+printf("\n For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz \n")
+printf("\n Initial Starting Current in DOL starter, I = %.1f <%.1f A \n",abs(I_n),atand(imag(I_n),real(I_n)))
+printf("\n By reducing volatge as well as the frequency, the peak starting current at the instant os starting is reduced by a fector of %.4f of the starting current with the reted volatge and frequency \n",Ratio)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz, I = 24.1 < 25.6 A instead of %.1f < (%.2f) A \n ",abs(I_n),atand(imag(I_n),real(I_n)));
+printf("\n Ratio of the Starting Current with the rated Voltage and frequency to the reduced Voltage and frequency, Ratio = 0.2518 instead of %.4f \n ",Ratio);
+
diff --git a/Working_Examples/2777/CH5/EX5.25/Ex5_25.sce b/Working_Examples/2777/CH5/EX5.25/Ex5_25.sce new file mode 100755 index 0000000..98245e2 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.25/Ex5_25.sce @@ -0,0 +1,34 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.25
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m1 = 3; // Total Number of phase in 1st Induction Motor
+p1 = 6; // Total number of Poles of 1st Induction Motor
+f = 50; // Frequency in Hertz
+m2 = 3; // Total Number of phase in 2nd Induction Motor
+p2 = 10; // Total number of Poles of 2nd Induction Motor
+
+
+// CALCULATIONS
+
+Ns1 = (120*f)/p1; // Synchronous speed of 1st Induction Motor in RPM
+Ns2 = (120*f)/p2; // Synchronous speed of 2nd Induction Motor in RPM
+Nscu = (120*f)/(p1+p2); // Speed during cumalative casade in RPM
+Ndiff = (120*f)/(p2-p1); // Speed during cumalative casade in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.25 : SOLUTION :-");
+printf("\n (a) Range of speed is %.f - %.f - %.f - %.f RPM \n",Nscu,Ns2,Ns1,Ndiff)
diff --git a/Working_Examples/2777/CH5/EX5.26/Ex5_26.sce b/Working_Examples/2777/CH5/EX5.26/Ex5_26.sce new file mode 100755 index 0000000..33e3732 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.26/Ex5_26.sce @@ -0,0 +1,68 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.26
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+R1 = 0.25; // Circuit Parameter in Ohms
+R2 = 0.5; // Circuit Parameter in Ohms
+X1 = 1.5; // Circuit Parameter in Ohms
+X2 = 1.5; // Circuit Parameter in Ohms
+
+
+// CALCULATIONS
+
+Vph = V/sqrt(3); // Per phase Voltage in Volts
+Ns = (120*f)/p; // Synchoronous Speed in RPM
+Ws = (2*%pi*Ns)/60; // Roatation Speed in Radians per Seconds
+
+// For Case (a) Machine running at, N = 1400 RPM
+
+N_a = 1400; // Machine running in RPM
+s_a = (Ns-N_a)/Ns; // Slip
+I_2_a = Vph/(R1+(R2/s_a)+(%i*(X1+X2))); // Rotor per phase Current referred to the stator side in Amphere
+Pg_a = 3*(abs(I_2_a)^2*R2)/s_a; // 3-phase air gap power or Rotor intake Power in Watts
+T_a = Pg_a/Ws; // Torque in Newton-Meter
+
+// For Case (b) Machine running at, N = 1600 RPM
+
+N_b = 1600; // Machine running in RPM
+s_b = (Ns-N_b)/Ns; // Slip
+I_2_b = Vph/(R1+(R2/s_b)+(%i*(X1+X2))); // Rotor per phase Current referred to the stator side in Amphere
+Pg_b = 3*(abs(I_2_b)^2*R2)/s_b; // 3-phase air gap power or Rotor intake Power in Watts
+T_b = Pg_b/Ws; // Torque in Newton-Meter
+
+// For Case (b) Machine running at, N = -100 RPM
+
+N_c = -100; // Machine running in RPM
+s_c = (Ns-N_c)/Ns; // Slip
+I_2_c = Vph/(R1+(R2/s_c)+(%i*(X1+X2))); // Rotor per phase Current referred to the stator side in Amphere
+Pg_c = 3*(abs(I_2_c)^2*R2)/s_c; // 3-phase air gap power or Rotor intake Power in Watts
+T_c = -Pg_c/Ws; // Torque in Newton-Meter (minus sign because its counter opposing torque)
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.26 : SOLUTION :-");
+printf("\n For Case (a) Machine running at, N = 1400 RPM \n ")
+printf("\n (a.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A \n",abs(I_2_a),atand(imag(I_2_a),real(I_2_a)))
+printf("\n (a.2) Developed Torque, T = %.2f Nm \n",T_a)
+printf("\n For Case (b) Machine running at, N = 1600 RPM \n ")
+printf("\n (a.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A \n",abs(I_2_b),atand(imag(I_2_b),real(I_2_b)))
+disp(" ( angle -157.52 + 180 = 22.48 ) ")
+ printf("\n (a.2) Developed Torque, T = %.2f Nm \n",T_b)
+printf("\n For Case (c) Machine running at, N = -100 RPM \n ")
+printf("\n (c.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A \n",abs(I_2_c),atand(imag(I_2_c),real(I_2_c)))
+printf("\n (c.2) Developed Torque, T = %.2f Nm \n",T_c)
diff --git a/Working_Examples/2777/CH5/EX5.27/Ex5_27.sce b/Working_Examples/2777/CH5/EX5.27/Ex5_27.sce new file mode 100755 index 0000000..4375b97 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.27/Ex5_27.sce @@ -0,0 +1,56 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.27
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor
+R1 = 0.25; // Circuit Parameter in Ohms
+R2 = 0.25; // Circuit Parameter in Ohms
+X1 = 0.75; // Circuit Parameter in Ohms
+X2 = 0.75; // Circuit Parameter in Ohms
+out_hp = 50; // Output of the induction motor in HP
+
+
+// CALCULATIONS
+
+V1 = V/sqrt(3); // Phase Voltage in Volts
+I = (out_hp*746)/(V*sqrt(3)); // Rated Current in Amphere
+sm = R2/(sqrt(R1^2+(X1+X2)^2)); // Slip at Maximum torque both its in Positive and negative sign
+Ws = 2*%pi*((120*f/p)*(1/60)); // Angular Roatation in Radians per Seconds
+Tm = (3*V1^2)/((2*Ws)*(R1+sqrt((R1^2)+(X1+X2)^2))); // Maximum torque during motoring in Newton-Meter
+Tg = -(3*V1^2)/((2*Ws)*(-R1+sqrt((R1^2)+(X1+X2)^2))); // Maximum torque during generating in Newton-Meter
+
+// For Case (a) slip = 0.05
+
+s_a = 0.05; // Slip
+T_a = (2*Tm)/((s_a/sm)+(sm/s_a)); // Torque in Newton-Meter
+
+// For Case (b) slip = -0.05
+
+s_b = -0.05; // Slip
+T_b = (2*Tg)/((s_b/sm)+(sm/s_b)); // Torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.27 : SOLUTION :-");
+printf("\n Maximim Torque during Motoring, Tm = %.f N-m \n",Tm)
+printf("\n Maximim Torque during Generating, Tm = %.2f N-m \n",Tg)
+printf("\n For Case (a) slip = 0.05 \n ")
+printf("\n (a.1) Torque, T = %.2f Nm \n",T_a)
+printf("\n For Case (b) slip = -0.05 \n ")
+printf("\n (b.1) Torque, T = %.2f Nm \n",T_b)
+
diff --git a/Working_Examples/2777/CH5/EX5.28/Ex5_28.sce b/Working_Examples/2777/CH5/EX5.28/Ex5_28.sce new file mode 100755 index 0000000..5eda796 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.28/Ex5_28.sce @@ -0,0 +1,43 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.28
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of phase in Induction Motor
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor in Volts
+R0 = 0.5; // Circuit Parameter in Ohms
+Ri = 0.05; // Circuit Parameter in Ohms
+X0 = 0.2; // Circuit Parameter in Ohms
+Xi = 0.9; // Circuit Parameter in Ohms
+s = 0.07; // Slip
+
+
+// CALCULATIONS
+
+Ws = 2*%pi*f; // Synchronous speed in Radins per second
+v = V/sqrt(3); // Phase Voltage in Volts
+Io = v/(R0+%i*X0); // Starting Current in the outer cage in Amphere
+Ii = v/(Ri+%i*Xi); // Starting Current in the inner cage in Amphere
+Tst = ((3*abs(Io)^2*R0)/Ws)+((3*abs(Ii)^2*Ri)/Ws); // Starting torque i.e at standstill, s=1
+Ios = v/((R0/s)+(%i*X0)); // Current in the outer cage at slip = 0.07
+Iis = v/((Ri/s)+(%i*Xi)); // Current in the outer cage at slip = 0.07
+T = ((3*abs(Ios)^2*R0)/(s*Ws))+((3*abs(Iis)^2*Ri)/(s*Ws)); // Starting torque at s=0.07 in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.28 : SOLUTION :-");
+printf("\n (a) Starting torque, Tst = %.2f Nm \n",Tst)
+printf("\n (b) Running torque at slip = 0.07, T = %.2f Nm \n",T)
diff --git a/Working_Examples/2777/CH5/EX5.29/Ex5_29.sce b/Working_Examples/2777/CH5/EX5.29/Ex5_29.sce new file mode 100755 index 0000000..573903d --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.29/Ex5_29.sce @@ -0,0 +1,48 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.29
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 4; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 440; // Operating Voltage of the Inductuon Motor in Volts
+out = 25*1000; // Power rating of the Induction motor in Watts
+R0 = 2.5; // Circuit Parameter in Ohms
+Ri = 0.5; // Circuit Parameter in Ohms
+X0 = 1.0; // Circuit Parameter in Ohms
+Xi = 5.0; // Circuit Parameter in Ohms
+Rc = 500; // Circuit Parameter in Ohms
+R1 = 0.2; // Circuit Parameter in Ohms
+Xm = 50; // Circuit Parameter in Ohms
+X123 = 2.0; // Circuit Parameter in Ohms
+s = 0.05; // Slip
+
+
+// CALCULATIONS
+
+Ws = (2*%pi*120*f)/(p*60); // Synchronous speed in Radins per second
+Zo = (R0/s)+(%i*X0); // Outer cage impedance at slip = 0.05 in Ohms
+Zi = (Ri/s)+(%i*Xi); // Inner cage impedance at slip = 0.05 in Ohms
+Z = (R1+%i*X123)+((Zo*Zi)/(Zo+Zi)); // Total impdance in Ohms
+I = V/Z; // Current in the Cage winding in Amphere
+Io = (I*((Zo*Zi)/(Zo+Zi)))/Zo; // Current in the outer cage in Amphere
+Ii = (I*((Zo*Zi)/(Zo+Zi)))/Zi; // Current in the inner cage in Amphere
+T = ((3*abs(Io)^2*R0)/(s*Ws))+((3*abs(Ii)^2*Ri)/(s*Ws)); // Starting torque in Newton-Meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.29 : SOLUTION :-");
+printf("\n (a) Torque at slip %.2f, T = %.2f Nm \n",s,T)
+
+
diff --git a/Working_Examples/2777/CH5/EX5.3/Ex5_3.sce b/Working_Examples/2777/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 0000000..de6886f --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,46 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+s = 24; // Total number of the pole
+p = 4; // Total number of the poles in the Alternator
+
+
+// CALCULATIONS
+// For Case (a) Short pitching by one Slots
+
+spp = s/p; // Slot per pole
+E_a = ((180*2)/24)*(4/2); // Slot angle in Electrical
+kp_a = cosd(E_a/2); // Pitch Factor
+kp5_a = cosd((5*E_a)/2); // Pitch Factor
+kp7_a = cosd((7*E_a)/2); // Pitch Factor
+
+// For Case(b) Short pitching by two Slots
+
+E_b = 2*((180*2)/24)*(4/2); // Slot angle in Electrical
+kp_b = cosd(E_b/2); // Pitch Factor
+kp5_b = cosd((5*E_b)/2) // Pitch Factor
+kp7_b = cosd((7*E_b)/2); // Pitch Factor
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.3 : SOLUTION :-") ;
+printf("\n For Case (a) Short pitching by one Slots:- Pitch Facor , kp = %.4f \n ",kp_a);
+printf("\n kp5 = %.4f \n ",kp5_a);
+printf("\n kp7 = %.4f \n ",kp7_a);
+printf("\n For Case (a) Short pitching by Two Slots:- Pitch Facor , kp = %.4f \n ",kp_b);
+printf("\n kp5 = %.4f \n ",kp5_b);
+printf("\n kp7 = %.4f \n ",kp7_b);
+
diff --git a/Working_Examples/2777/CH5/EX5.30/Ex5_30.sce b/Working_Examples/2777/CH5/EX5.30/Ex5_30.sce new file mode 100755 index 0000000..70ec500 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.30/Ex5_30.sce @@ -0,0 +1,50 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.30
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 1; // Total Number of phase in Induction Motor
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+V = 220; // Operating Voltage of the Inductuon Motor in Volts
+R1 = 10; // Circuit Parameter in Ohms
+R2 = 11; // Circuit Parameter in Ohms
+X1 = 12; // Circuit Parameter in Ohms
+X2 = 12; // Circuit Parameter in Ohms
+Xm = 125; // Circuit Parameter in Ohms
+s = 0.03; // Slip
+
+
+// CALCULATIONS
+
+Zf = ((%i*Xm/2)*((R2/(2*s))+(%i*X2/2)))/((%i*Xm/2)+(R2/(2*s))+(%i*X2/2)); // Impedance offered by the forward field in Ohms
+Zb = ((%i*Xm/2)*((R2/(2*(2-s)))+(%i*X2/2)))/((%i*Xm/2)+(R2/(2*(2-s)))+(%i*X2/2)); // Impedance offered by the backward field in Ohms
+Z = (R1+%i*X1)+Zf+Zb; // Total Impedance in Ohms
+I = V/Z; // Total input current in Amphere
+pf = cosd(atand(imag(I),real(I))); // Power Factor (lagging)
+Vf = I*Zf; // Forward Volatge at slip 0.03 in Volts
+Vb = I*Zb; // Backward Volatge at slip 0.03 in Volts
+If = Vf/(0.5*R2/s); // Forward Current in Amphere
+Ib = Vb/(0.5*R2/(2-s)); // Forward Current in Amphere
+Ws = 2*%pi*f; // Synchronous Speed in radians per second
+T = ((0.5*If^2*R2)/(s*Ws))-((0.5*Ib^2*R2)/((2-s)*Ws)); // Starting torque
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.30 : SOLUTION :-");
+printf("\n (a) Input Current, I = %.2f < %.f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n (b) Power factor, pf = %.2f Lagging \n",pf)
+printf("\n (c) Developed Torque, T = %.3f Nm \n",T)
+
+
diff --git a/Working_Examples/2777/CH5/EX5.31/Ex5_31.sce b/Working_Examples/2777/CH5/EX5.31/Ex5_31.sce new file mode 100755 index 0000000..42e3428 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.31/Ex5_31.sce @@ -0,0 +1,63 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.31
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Wsc = 900; // Power at Blocked Rotor test in Watts
+Vsc = 200; // Voltage at Blocked Rotor test in Volts
+Isc = 5.0; // Current at Blocked Rotor test in Amphere
+Wo = 60; // Power at No-load test in Watts
+Vo = 220; // Voltage at No-load test in Volts
+Io = 1.5; // Current at No-load test in Amphere
+m = 1; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+V = 220; // Operating voltage of the Induction motor in Volts
+f = 50; // Frequency in Hertz
+s = 0.07; // Slip
+R1 = 12; // Resistance of the main primary winding in Ohms
+
+
+// CALCULATIONS
+
+Zsc = Vsc/Isc; // Impedance in Blocked Rotor test in Ohms
+Rsc = Wsc/(Isc^2); // Resistance in Blocked Rotor test in Ohms
+Xsc = sqrt((Zsc^2)-(Rsc^2)); // Reactance in Blocked Rotor test in Ohms
+Xl1 = Xsc/2; // Leakage reactance of stator and rotor to be equal in Ohms
+Xl2 = Xsc/2; // Leakage reactance of stator and rotor to be equal in Ohms
+R2 = Rsc-R1; // Equivalent resistance of rotor referred to stator in Ohms
+Z0 = Vo/Io; // Impedance in Blocked Rotor test in Ohms
+R0 = Wo/(Io^2); // Resistance in Blocked Rotor test in Ohms
+X0 = sqrt((Z0^2)-(R0^2)); // Reactance in Blocked Rotor test in Ohms
+Wloss = Wo - ((Io^2)*(R1+R2)); // Loss in Watts
+Xm_half = X0-Xl1-Xl2/2;
+R2f = (R2/s)+((%i*Xl2)/2); // Forward resiatance in Ohms
+Zf = ((%i*Xm_half)*R2f)/(%i*Xm_half+R2f); // Total Forward impedance in Ohms
+R2b = (R2/(2-s))+((%i*Xl2)/2); // Backward resiatance in Ohms
+Zb = ((%i*Xm_half)*R2b)/(%i*Xm_half+R2b); // Total Backward impedance in Ohms
+Z = Zf+Zb+(R1+%i*Xl1); // Total impedance in Ohms
+I = V/Z; // Motor Current in Amphere
+pf = cosd(atand(imag(I),real(I))); // Power Factor (lagging)
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.31 : SOLUTION :-");
+printf("\n Circuit Parameters are \n\n (a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = %.2f Ohms \n",Xl1)
+printf("\n (b) Equivalent resistance of rotor referred to stator, R2 = %.f Ohms \n",R2)
+printf("\n (c) Total Forward impedance, Zf = %.1f < %.2f Ohms \n",abs(Zf),atand(imag(Zf),real(Zf)))
+printf("\n (c) Total Backward impedance, Zb = %.2f < %.2f Ohms \n",abs(Zb),atand(imag(Zb),real(Zb)))
+printf("\n (d) Total impedance, Z = %.2f < %.2f Ohms \n",abs(Z),atand(imag(Z),real(Z)))
+printf("\n (e) Input Current, I = %.2f < %.2f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n (f) Power factor, pf = %.2f Lagging \n",pf)
+
+
diff --git a/Working_Examples/2777/CH5/EX5.32/Ex5_32.sce b/Working_Examples/2777/CH5/EX5.32/Ex5_32.sce new file mode 100755 index 0000000..7741ec1 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.32/Ex5_32.sce @@ -0,0 +1,72 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.32
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Wsc = 600; // Power at Blocked Rotor test in Watts
+Vsc = 125; // Voltage at Blocked Rotor test in Volts
+Isc = 15.0; // Current at Blocked Rotor test in Amphere
+Wo = 360; // Power at No-load test in Watts
+Vo = 220; // Voltage at No-load test in Volts
+Io = 6.5; // Current at No-load test in Amphere
+m = 1; // Total Number of phase in Induction Motor
+p = 4; // Total number of Poles of Induction Motor
+V = 220; // Operating voltage of the Induction motor in Volts
+f = 50; // Frequency in Hertz
+s = 0.05; // Slip
+R1 = 1.2; // Resistance of the main primary winding in Ohms
+
+
+// CALCULATIONS
+
+Zlr = Vsc/Isc; // Impedance in Blocked Rotor test in Ohms
+Rlr = Wsc/(Isc^2); // Resistance in Blocked Rotor test in Ohms
+Xlr = sqrt((Zlr^2)-(Rlr^2)); // Reactance in Blocked Rotor test in Ohms
+Xl1 = Xlr/2; // Leakage reactance of stator and rotor to be equal in Ohms
+Xl2 = Xlr/2; // Leakage reactance of stator and rotor to be equal in Ohms
+R2 = (Rlr-R1); // Equivalent resistance of rotor referred to stator in Ohms
+R2_half = R2/2; // Equivalent resistance of rotor referred to stator in Ohms
+Z0 = Vo/Io; // Impedance in Blocked Rotor test in Ohms
+R0 = Wo/(Io^2); // Resistance in Blocked Rotor test in Ohms
+X0 = sqrt((Z0^2)-(R0^2)); // Reactance in Blocked Rotor test in Ohms
+Wloss = Wo - ((Io^2)*(R1+R2)); // Loss in Watts
+Xm_half = X0-Xl1-Xl2/2;
+R2f = (R2/(2*s))+((%i*Xl2)/2); // Forward resiatance in Ohms
+Zf = ((%i*Xm_half)*R2f)/(%i*Xm_half+R2f); // Total Forward impedance in Ohms
+R2b = (R2/(2*(2-s)))+((%i*Xl2)/2); // Backward resiatance in Ohms
+Zb = ((%i*Xm_half)*R2b)/(%i*Xm_half+R2b); // Total Backward impedance in Ohms
+Z = Zf+Zb+(R1+%i*Xl1); // Total impedance in Ohms
+I = V/Z; // Motor Current in Amphere
+pf = cosd(atand(imag(I),real(I))); // Power Factor (lagging)
+Vf = I*Zf; // Voltage across forward impedance in Volts
+If = Vf/R2f; // Forward current producing torque in Amphere
+Tf = ((abs(If)^2)*R2)/(2*s); // Forward torque in synchronous Watts
+Vb = I*Zb; // Voltage across Backward impedance in Volts
+Ib = Vb/R2b; // Backward current producing torque in Amphere
+Tb = ((abs(Ib)^2)*R2)/(2*(2-s)); // Backward torque in synchronous Watts
+T = Tf-Tb; // Net torque in Synchronous Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.32 : SOLUTION :-");
+printf("\n Circuit Parameters are \n\n (a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = %.2f Ohms \n",Xl1)
+printf("\n (b) Equivalent resistance of rotor referred to stator, R2 = %.2f Ohms \n",R2)
+printf("\n (c) Total Forward impedance, Zf = %.1f < %.2f Ohms \n",abs(Zf),atand(imag(Zf),real(Zf)))
+printf("\n (c) Total Backward impedance, Zb = %.2f < %.2f Ohms \n",abs(Zb),atand(imag(Zb),real(Zb)))
+printf("\n (d) Total impedance, Z = %.2f < %.2f Ohms \n",abs(Z),atand(imag(Z),real(Z)))
+printf("\n (e) Input Current, I = %.2f < %.f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n (f) Power factor, pf = %.4f Lagging \n",pf)
+printf("\n (g) Forward torque, Tf = %.2f Synchronous Watts \n",Tf)
+printf("\n (h) Backward torque, Tb = %.2f Synchronous Watts \n",Tb)
+printf("\n (i) Net torque, T = %.2f Synchronous Watts \n",T)
diff --git a/Working_Examples/2777/CH5/EX5.4/Ex5_4.sce b/Working_Examples/2777/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 0000000..927a99f --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,31 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+s = 60; // Total number of Slot
+m = 3; // Total number of Phase
+p = 4; // Total number of Pole
+
+
+// CALCULATIONS
+
+M = s/(m*p); // Slot per pole per Phase
+sigma = 180/m; // Phase Spread in angle (deg)
+Ka = sind((M*sigma)/2)/(M*sind(sigma/2)); // Distribution Factor
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.4 : SOLUTION :-");
+printf("\n (a) Distribution Factor, Ka = %.1f \n",Ka)
+
diff --git a/Working_Examples/2777/CH5/EX5.7/Ex5_7.sce b/Working_Examples/2777/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 0000000..214e31c --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,36 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.7
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 50; // Frequency of the 2-pole Induction Motor
+p = 2; // Total Number of Poles
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+Ns5 = -(120*f)/(5*p); // Synchronous Speed of 5th order space harmonic in RPM
+N5 = -(120*5*f)/p; // Synchronous Speed of 5th order time harmonic in RPM
+Ns7 = (120*f)/(7*p); // Synchronous Speed of 7th order space harmonic in RPM
+N7 = (120*7*f)/p; // Synchronous Speed of 7th order time harmonic in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.7 : SOLUTION :-");
+printf("\n (a.1) Synchronous Speed of 5th order space harmonic, Ns5 = %.f RPM \n",Ns5)
+printf("\n (a.2) Synchronous Speed of 5th order time harmonic, N5 = %.f RPM \n",N5)
+printf("\n (b.1) Synchronous Speed of 7th order space harmonic, Ns7 = %.2f RPM \n",Ns7)
+printf("\n (b.2) Synchronous Speed of 7th order time harmonic, N7 = %.f RPM \n",N7)
+
+
diff --git a/Working_Examples/2777/CH5/EX5.8/Ex5_8.sce b/Working_Examples/2777/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 0000000..b8b1079 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,34 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.8
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p_a = 6; // Total number of Poles in the Alternator
+p_m = 4; // Total number of Poles of Induction Motor
+N_a = 900; // Running Speed of the Alternator in RPM
+N_m = 1250; // Running Speed of the Induction Motor in RPM
+m = 3; // Total Number of phase in Induction Motor
+
+
+// CALCULATIONS
+
+f = (N_a*p_a)/120; // Frequency of the 6-pole Alternator running at 900 RPM in Hertz
+Ns = (120*f)/p_m; // Synchronous Speed of 4-pole Induction Motor in RPM
+s = (Ns-N_m)/Ns; // Slip
+fr = s*f; // Frequency of the Rotor Current in Hertz
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.8 : SOLUTION :-");
+printf("\n (a) Frequency of the Rotor Current, fr = %.2f Hz \n",fr)
+
diff --git a/Working_Examples/2777/CH5/EX5.9/Ex5_9.sce b/Working_Examples/2777/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 0000000..ebc4ea3 --- /dev/null +++ b/Working_Examples/2777/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,35 @@ +// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.9
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 2; // Total number of Poles of Induction Motor
+f = 50; // Frequency in Hertz
+Nr = 2800; // Running Speed of the Induction Motor in RPM
+m = 3; // Total Number of phase in Induction Motor
+V = 400; // Operating Voltage of Induction Motor in Volts
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous Speed in RPM
+s = 100*((Ns-Nr)/Ns); // Slip in Percentage
+fr = (s/100)*f; // Frequency of the Rotor Current in Hertz
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.9 : SOLUTION :-");
+printf("\n (a) Slip, s = %.2f percent \n",s);
+printf("\n (b) Frequency of the Rotor Current, fr = %.2f Hz \n",fr)
+
+
diff --git a/Working_Examples/2777/CH6/EX6.1/Ex6_1.sce b/Working_Examples/2777/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 0000000..8eace61 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,47 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 50; // Generating Frequency in Hertz
+
+
+// CALCULATIONS
+// For Case(a)
+
+Ns_a = 3000; // Synchronous speed in RPM
+p_a = (120*f)/Ns_a; // Number of poles
+
+// For Case(b)
+
+Ns_b = 1000; // Synchronous speed in RPM
+p_b = (120*f)/Ns_b; // Number of poles
+
+// For Case(c)
+
+Ns_c = 300; // Synchronous speed in RPM
+p_c = (120*f)/Ns_c; // Number of poles
+
+// For Case(d)
+
+Ns_d = 40; // Synchronous speed in RPM
+p_d = (120*f)/Ns_d; // Number of poles
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.1 : SOLUTION :-") ;
+printf("\n For Case(a) Ns = %.f, p = %.f \n ",Ns_a,p_a);
+printf("\n For Case(b) Ns = %.f, p = %.f \n ",Ns_b,p_b);
+printf("\n For Case(c) Ns = %.f, p = %.f \n ",Ns_c,p_c);
+printf("\n For Case(d) Ns = %.f, p = %.f \n ",Ns_d,p_d);
diff --git a/Working_Examples/2777/CH6/EX6.10/Ex6_10.sce b/Working_Examples/2777/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 0000000..d8e74d0 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,44 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.10
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3; // Total Number of Phase in Alternator
+p = 2; // Total number of Poles of Alternator
+V = 11*10^3; // Operating voltage of the Alternator in Volts
+VA = 10*10^6; // VA rating of the Alternator in Volts-Amphere
+f = 50; // Operating Frequency of the alternator in Hertz
+pf = 0.8; // Power factor (lagging)
+Vf = 12*10^3; // Operating field voltage of the Alternator in Volts
+If = 160; // Field Current in Amphere
+Ra = 0.05; // Armature Resistance per phase in Ohms
+Xs = 1.5; // Winding leakage reactance per phase in Ohms
+A = 150; // The armature MMF at rated current is equivalent to Field Current in Amphere
+
+
+// CALCULATIONS
+
+Vt = V/sqrt(3); // Rated per phase Voltage in Volts
+Ia = VA/(sqrt(3)*V); // Rated Armature Current in Amphere
+pfa = acosd(pf); // Power factor angle in degree
+Er = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*(Ra+%i*Xs); // Induced EMF in Volts
+R_a = 90 + atand(imag(Er),real(Er)); // Angle of R in Degree
+R = 160 * exp( %i * (R_a) * %pi/180); // (Line-line Voltage) Er = 11902.40V will get R from Air gap Characteristics
+A_n = A * exp( %i * (-pfa) * %pi/180);
+F = R - A_n; // Field Current required to produce the excitation EMF in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.10: SOLUTION :-");
+printf("\n (a) Field Current required to produce the excitation EMF, F = %.2f A \n",abs(F))
diff --git a/Working_Examples/2777/CH6/EX6.11/Ex6_11.sce b/Working_Examples/2777/CH6/EX6.11/Ex6_11.sce new file mode 100755 index 0000000..442c25c --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,81 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.11
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.11 : \n\n Given Data \n");
+printf("\n Voc( V) 12 13 13.8 14.5 15.1 \n");
+printf("\n If(A) 175 200 225 250 275 \n\n");
+V = 11*10^3; // Operating voltage of the Synchronous generator in Volts
+VA = 50*10^6; // VA rating of the Synchronous generator in Volts-Amphere
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+N = 1500; // Speed of the Synchronous generator in RPM
+If_scc = 200; // SCC test field Rated current in Amphere at rated Short circuit current
+If_zpf = 400; // ZPF test field Rated current in Amphere at rated voltage and rated current
+pf = 0.8; // Power factor (lagging)
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.30 & Page no:-413
+
+Vt = V/sqrt(3); // Rated per phase Voltage in Volts
+Ia = VA/(sqrt(3)*V); // Rated Armature Current in Amphere
+pfa = acosd(pf); // Power factor angle in degree
+O = 13000; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
+Xs = O/(sqrt(3)*Ia); // Synchronous reactance per phase in Ohms
+BC = 4000; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
+Xl = BC/(sqrt(3)*Ia ); // Per phase leakage reactance in Ohms
+
+// For Case (a) General (ZPF) Method
+
+Er_a = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*(%i*Xl); // Induced EMF in Volts
+R_a = 220; A_a = 200; //From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 13776V will get R_a & A_a value Respectively from SCC (Figure6.30 & Page no:-403)
+angle_a = 140.3; // Angle between R_a & A_a = 90'+13.43'+36.87' = 140.3'
+F_a = sqrt((R_a^2)+(A_a^2)-(2*R_a*A_a*cosd(angle_a))); // From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere
+Eo_a = 20000; // Corresponding to field current F_a = 470.90 A the open circuit EMF from OCC is 20000 V (Figure6.30 & Page no:-413)
+r_a = 100*((Eo_a-V)/V); // Percentage regulation
+
+
+// For Case(b) EMF Method
+
+Er_b = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*(%i*Xs); // Induced Voltage in Volts
+F_b = 500; //From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 21404 V will get 500A from SCC (Figure6.15 & Page no:-386)
+
+// For Case (c) MMF Method
+
+Er_c = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*0; // Induced Voltage in Volts ( Zero is multipied because Armature reistance is zero (not mentioned))
+R_c = 160; A_c = 200; //From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 11000 V will get R_c & A_c value Respectively from SCC (Figure6.30 & Page no:-413)
+angle_c = 126.27; // Angle between R_c & A_c = 90'-0'+36.87' = 126.27' {can refer figure 6.21a at page no:-400}
+F_c = sqrt((R_c^2)+(A_c^2)-(2*R_c*A_c*cosd(angle_c))); // From phasor diagram {can refer figure 6.21a at page no:-400} the neccessary field excitation in Amphere
+
+
+// For Case (d) ASA Method
+
+Er_d = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*(%i*Xl); // Induced Voltage in Volts
+R_d = 220; A_d = 200; //From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 13800 V will get R_d & A_d value Respectively from SCC (Figure6.30 & Page no:-413)
+angle_d = 126.87; // Angle between R_d & A_d = 90'+36.87' = 126.87'{can refer figure 6.22a at page no:-40}
+F_d1 = sqrt((R_d^2)+(A_d^2)-(2*R_d*A_d*cosd(angle_d))); // from Phasor diagram {can refer figure 6.2a at page no:-400 The neccessary field excitation in Amphere
+F_d = F_d1 + 30; // from Phasor diagram {can refer figure 6.2a at page no:-400 The Total neccessary field excitation in Amphere
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a) Leakage Reactance, Xl = %.2f Ohms \n",Xl)
+printf("\n (b) Synchronous Reactance, Xs = %.2f Ohms \n",Xs)
+printf("\n For Case (a) General (ZPF) Method \n Field Current required for maintaing the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = %.2f A \n",pf,F_a)
+printf("\n For Case (a) EMF Method \n Field Current required for maintaing the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = %.f A \n",pf,F_b)
+printf("\n For Case (a) MMF Method \n Field Current required for maintaing the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = %.2f A \n",pf,F_c)
+printf("\n For Case (a) ASA Method \n Field Current required for maintaing the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = %.f A \n",pf,F_d)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- For Case (a) General (ZPF) Method \n (a) Field Current required for maintaining the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = 470.90 A instead of %.2f A \n",pf,F_a);
diff --git a/Working_Examples/2777/CH6/EX6.12/Ex6_12.sce b/Working_Examples/2777/CH6/EX6.12/Ex6_12.sce new file mode 100755 index 0000000..d4b040a --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.12
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+printf("\n EXAMPLE : 6.12 : \n\n Given Data \n");
+printf("\n Voc(V) 175 250 280 300 330 350 370 380 \n");
+printf("\n If(A) 10 17 20 23 30 38 50 60 \n");
+printf("\n Vzpf(V) - - - 0 130 210 265 280 \n\n");
+V = 433; // Operating voltage of the Alternator in Volts
+N = 3000; // speed of the Alternator in RPM
+VA = 20*10^3; // VA rating of the Alternator in Volts-Amphere
+f = 50; // Operating Frequency of the Alternator in Hertz
+pf = 0.8; // Power factor (lagging)
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.35 & Page no:-420
+
+Vt = V/sqrt(3); // Rated per phase Voltage in Volts
+Ia = VA/(sqrt(3)*V); // Rated Armature Current in Amphere
+pfa = acosd(pf); // Power factor angle in degree
+O = 298; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
+Xs = O/(sqrt(3)*Ia); // Synchronous reactance per phase in Ohms
+BC = 70; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
+Xl = BC/(sqrt(3)*Ia ); // Per phase leakage reactance in Ohms
+E = Vt+Ia*(cosd(pfa)-%i*sind(pfa))*(%i*Xs); // Induced EMF in Volts using EMF Method
+c = 380-60; // The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation
+y = 694.50; // The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation
+// Extrapolation Equation is y = (x*(380-370)/(60-50))*c
+x = y - c; // The required field current in Amphere
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a) Leakage Reactance, Xl = %.2f Ohms \n",Xl)
+printf("\n (b) Synchronous Reactance, Xs = %.2f Ohms \n",Xs)
+printf("\n (c) Field Current required for maintaing the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = %.2f A \n",pf,x)
+
+
diff --git a/Working_Examples/2777/CH6/EX6.13/Ex6_13.sce b/Working_Examples/2777/CH6/EX6.13/Ex6_13.sce new file mode 100755 index 0000000..b159320 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,71 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+V = 400; // Operating voltage of the Synchronous generator in Volts
+VA = 60*10^3; // VA rating of the Synchronous generator in Volts-Amphere
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+xd = 1.5; // Direct axis reactances in Ohms
+xq = 0.6; // Quadrature axis reactances in Ohms
+
+
+// CALCULATIONS
+
+I = VA/(sqrt(3)*V); // Rated current in Amphere
+v = V/sqrt(3); // Rated Phase Votage in Volts
+
+// For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)
+
+pf_a = 0.8; // Power factor
+pfa_a = acosd(pf_a); // Power factor angle in deg
+pa_a = atand((I*xq*cosd(pfa_a))/(v+I*xq*sind(pfa_a))); // Power angle in deg
+Iq_a = I*cosd(pfa_a+pa_a); // Current in Amphere
+Id_a = I*sind(pfa_a+pa_a); // Current in Amphere
+Eo_a = sqrt((v+Id_a*xd*cosd(pa_a)-Iq_a*xq*sind(pa_a))^2 + (Id_a*xd*sind(pa_a)+Iq_a*xq*cosd(pa_a))^2); // Induced EMF in Volts
+pr_a = ((Eo_a-v)/v)*100; // Percentage regulation
+
+// For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)
+
+pf_b = 1.0; // Power factor
+pfa_b= acosd(pf_b); // Power factor angle in deg
+pa_b = atand((I*xq*cosd(pfa_b))/(v+I*xq*sind(pfa_b))); // Power angle in deg
+Iq_b = I*cosd(pfa_b+pa_b);
+Id_b = I*sind(pfa_b+pa_b);
+Eo_b = sqrt((v+Id_b*xd*cosd(pa_b)-Iq_b*xq*sind(pa_b))^2 + (Id_b*xd*sind(pa_b)+Iq_b*xq*cosd(pa_b))^2); // Induced EMF in Volts
+pr_b = ((Eo_b-v)/v)*100; // Percentage regulation
+
+// For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)
+
+pf_c = 0.8; // Power factor
+pfa_c = acosd(pf_c); // Power factor angle in deg
+pa_c = atand((I*xq*cosd(pfa_c))/(v-I*xq*sind(pfa_c))); // Power angle in deg
+Iq_c = I*cosd(pfa_c-pa_c);
+Id_c = I*sind(pfa_c-pa_c);
+Eo_c = sqrt((v-Id_c*xd*cosd(pa_c)-Iq_c*xq*sind(pa_c))^2 + (-Id_c*xd*sind(pa_c)+Iq_c*xq*cosd(pa_c))^2); // Induced EMF in Volts
+pr_c = ((Eo_c-v)/v)*100; // Percentage regulation
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.13: SOLUTION :-");
+printf("\n For Case (a) 0.80 lagging Power factor \n Induced EMF, EMF = %.2f V \n",Eo_a)
+printf("\n Power angle = %.3f degree \n",pa_a)
+printf("\n Percenatge Regulation, R = %.1f Percenatge \n",pr_a)
+printf("\n For Case (b) Unity Power factor \n Induced EMF, EMF = %.2f V \n",Eo_b)
+printf("\n Power angle = %.2f degree \n",pa_b)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",pr_b)
+printf("\n For Case (c) 0.80 leading Power factor \n Induced EMF, EMF = %.2f V \n",Eo_c)
+printf("\n Power angle = %.2f degree \n",pa_c)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",pr_c)
diff --git a/Working_Examples/2777/CH6/EX6.14/Ex6_14.sce b/Working_Examples/2777/CH6/EX6.14/Ex6_14.sce new file mode 100755 index 0000000..54830d7 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,67 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+v = 1.0; // Operating voltage of the Synchronous generator in pu
+xd = 1.0; // Direct axis reactances in pu
+xq = 0.5; // Quadrature axis reactances in pu
+I = 1.0; // Rated current in pu
+
+
+// CALCULATIONS
+
+// For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)
+
+pf_a = 0.8; // Power factor
+pfa_a = acosd(pf_a); // Power factor angle in deg
+pa_a = atand((I*xq*cosd(pfa_a))/(v+I*xq*sind(pfa_a))); // Power angle in deg
+Iq_a = I*cosd(pfa_a+pa_a);
+Id_a = I*sind(pfa_a+pa_a);
+Eo_a = sqrt((v+Id_a*xd*cosd(pa_a)-Iq_a*xq*sind(pa_a))^2 + (Id_a*xd*sind(pa_a)+Iq_a*xq*cosd(pa_a))^2); // Induced EMF in Volts
+pr_a = ((Eo_a-v)/v)*100; // Percentage regulation
+
+// For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)
+
+pf_b = 1.0; // Power factor
+pfa_b= acosd(pf_b); // Power factor angle in deg
+pa_b = atand((I*xq*cosd(pfa_b))/(v+I*xq*sind(pfa_b))); // Power angle in deg
+Iq_b = I*cosd(pfa_b+pa_b);
+Id_b = I*sind(pfa_b+pa_b);
+Eo_b = sqrt((v+Id_b*xd*cosd(pa_b)-Iq_b*xq*sind(pa_b))^2 + (Id_b*xd*sind(pa_b)+Iq_b*xq*cosd(pa_b))^2); // Induced EMF in Volts
+pr_b = ((Eo_b-v)/v)*100; // Percentage regulation
+
+// For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)
+
+pf_c = 0.8; // Power factor
+pfa_c = acosd(pf_c); // Power factor angle in deg
+pa_c = atand((I*xq*cosd(pfa_c))/(v-I*xq*sind(pfa_c))); // Power angle in deg
+Iq_c = I*cosd(pfa_c-pa_c);
+Id_c = I*sind(pfa_c-pa_c);
+Eo_c = sqrt((v-Id_c*xd*cosd(pa_c)-Iq_c*xq*sind(pa_c))^2 + (-Id_c*xd*sind(pa_c)+Iq_c*xq*cosd(pa_c))^2); // Induced EMF in Volts
+pr_c = ((Eo_c-v)/v)*100; // Percentage regulation
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.14: SOLUTION :-");
+printf("\n For Case (a) 0.80 lagging Power factor \n Induced EMF, EMF = %.4f V \n",Eo_a)
+printf("\n Power angle = %.1f degree \n",pa_a)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",pr_a)
+printf("\n For Case (b) Unity Power factor \n Induced EMF, EMF = %.2f V \n",Eo_b)
+printf("\n Power angle = %.2f degree \n",pa_b)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",pr_b)
+printf("\n For Case (c) 0.80 leading Power factor \n Induced EMF, EMF = %.4f V \n",Eo_c)
+printf("\n Power angle = %.2f degree \n",pa_c)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",pr_c)
diff --git a/Working_Examples/2777/CH6/EX6.15/Ex6_15.sce b/Working_Examples/2777/CH6/EX6.15/Ex6_15.sce new file mode 100755 index 0000000..277dbeb --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,36 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.15
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+If = 1.25; // Given that rated voltage at air gap line for this field current in pu
+IF = 0.75; // Rated current in SC test for this field current in pu
+Ia = 1.0; // Rated current in Per unit
+pf = 0.8; // Power factor
+V = 1.0; // Rated Volatge in pu
+
+
+// CALCULATIONS
+
+pfa = acosd(pf); // Power factor angle in deg
+Voc = (V*IF)/If; // Open circuit volatge in pu
+xs = Voc/Ia; // Syncronous reactance in pu
+E = V + Ia*(cosd(pfa)-(%i)*sind(pfa))*(%i*xs); // Induced EMF in pu
+a = abs(E)*If;
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.15: SOLUTION :-");
+printf("\n Induced EMF, E = %.2f < %.2f pu \n ",abs(E),atand(imag(E),real(E)))
+printf("\n The field current required for %.2f pu voltage on air gap line %.1f pu \n",abs(E),a)
diff --git a/Working_Examples/2777/CH6/EX6.16/Ex6_16.sce b/Working_Examples/2777/CH6/EX6.16/Ex6_16.sce new file mode 100755 index 0000000..a7e20a6 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.16/Ex6_16.sce @@ -0,0 +1,51 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.16
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+V = 440; // Operating voltage of the alternator in Volts
+VA = 20*10^3; // VA rating of the alternator in Volts-Amphere
+f = 50; // Operating Frequency of the alternator in Hertz
+N = 3000; // Rotation of the alternator in RPM
+Ra = 0.0; // Armature reistance in Ohms
+xl = 0.6; // Armature reactances in Ohms
+pf = 0.8; // Power factor lagging
+pfa = acosd(pf); // ower factor angle in deg
+p = (120*f)/N; // Number of poles
+w = (2*%pi*f); // Rotation speed in Radians per second
+v = V/sqrt(3); // Rated phase voltage in Volts
+I = VA/(sqrt(3)*V); // Rated curent in Amphere
+If = I; // Given field current = armature current from SCC test in Amphere
+E = 16*If; // Open-circuit EMF at field current in Volts given from Equation E = 16If refer page no. 431
+xs = E/(If*sqrt(3)); // Synchronous reactance in Ohms
+Eo = sqrt((v+I*xs*sind(pfa))^2 + (I*xs*cosd(pfa))^2); // Induced EMF in Volts
+pa = atand(193.98/399.49); // From above equation Eo
+pr = ((Eo-v)/v)*100; // Percent regulation
+P = (3*v*Eo*sind(pa))/(xs*1000); // Power inKilo-Watts
+T = (P*1000)/w; // Torque devolped in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.16: SOLUTION :-");
+printf("\n Induced EMF, EMF = %.f V \n",Eo)
+printf("\n Power angle = %.2f degree \n",pa)
+printf("\n Power, P = %.3f kW \n",P)
+printf("\n Counter Torque, T = %.2f N-m \n",T)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- \n (a) Induced EMF, EMF = 471 V instead of %.f V \n",Eo)
+printf("\n (b) Power angle = 18.05 degree instead of %.2f degree \n",pa)
+printf("\n (c) Power, P = 12.003 kV instead of %.3f kW \n",P)
+printf("\n (d) Counter Torque, T = 38.23 N-m instead of %.2f N-m \n",T)
+printf("\n From Calculation of the Induced EMF(E), rest all the Calculated values in the TEXT BOOK is WRONG because of the Induced EMF(E) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH6/EX6.17/Ex6_17.sce b/Working_Examples/2777/CH6/EX6.17/Ex6_17.sce new file mode 100755 index 0000000..16e28f5 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.17/Ex6_17.sce @@ -0,0 +1,59 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.17
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+V = 440; // Operating voltage of the Synchronous Motor in pu
+E = 200; // Induced voltage in Volts
+xs = 8.0; // Synchronous reactance in Ohms
+f = 50; // Frequency in Hertz
+pa = 36; // Power angle in degree
+
+
+// CALCULATIONS
+
+v = V/sqrt(3); // Rated phase voltage in Volts
+ws = 2*%pi*f; // Synchronous speed in Radians per second
+// To calculate the power factor angle refer page no 438 n figure 6.50
+// Since E*cos(delta) < v so Power factor is lagging, let power factor angle be theta from ohasor diagram figure 6.50:- page no. 438
+// v = E*cos(delta) + I*xs*sin(theta), I*sin(theta) = (254-0.809*200)/8 = 11.525
+// Similarly, E*sin(delta) = I*xs*cos(theta), I*cos(theta) = (200*0.59)8 = 14.70
+// From above two equations, tan(theta) = 0.784
+theta = -38.1; // Power factor angle in degree (minus sign because of lagging)
+pf = cosd(theta); // Power factor lagging
+I = 14.7/cosd(theta); // Line current in Amphere (I*cos(theta) = 14.7)
+p = 3 * v * 14.7; // Input to motor in watts ( p = 3*V*I*cos(theta), I*cos(theta) = 14.7)
+P = (3*E*v*sind(pa))/(xs*1000); // Power in Kilo-watts
+T = (P*1000)/ws; // Torque in Newton-meter
+// For Power factor unity
+// let the current will be I2, thus 3*v*I2 = 3*v*I*cos(theta) , I2 = I*cos(theta) = 14.10 A
+// let ecitation will be E2, thus v = E2*cos(delta2) and E2*sin(delta2) = I2*xs, E2*cos(delta2) = 254 and E2*sin(delta2) = 117.60, by solving these two equations we get E2 = sqrt(254^2+117.6^2) = 279.90 V and delta2 = atand(117.6/254) = 24.84 degree
+E2 = 279.90;
+delta2 = 24.84;
+P_1 = (3*v*E2*sind(delta2))/(xs*1000); // Power in kilo-watts
+T_1 = (P_1*1000)/ws; // Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.17: SOLUTION :-");
+printf("\n (a) Line current, I = %.2f A \n",I)
+printf("\n (b) Power factor angle = %.1f degree \n",theta)
+printf("\n (c) Power , P = %.3f kW \n",P)
+printf("\n (d) Torque , T = %.2f N-m \n",T)
+printf("\n (e) Power factor = %.2f lagging \n",pf)
+printf("\n To make the Power factor to UNITY requirements are:- \n (a) Excitation EMF, E = %.2f V \n",E2)
+printf("\n (b) Power angle = %.2f degree \n",delta2)
+printf("\n (c) Power , P = %.3f kW \n",P_1)
+printf("\n (d) Torque , T = %.2f N-m \n",T_1)
diff --git a/Working_Examples/2777/CH6/EX6.18/Ex6_18.sce b/Working_Examples/2777/CH6/EX6.18/Ex6_18.sce new file mode 100755 index 0000000..aec7ef6 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.18/Ex6_18.sce @@ -0,0 +1,51 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.18
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+v = 1100; // Operating voltage of the Synchronous Motor in Volts
+p = 4; // Total number of Poles
+m = 3; // number of phase
+xs = 5.0; // Synchrouons reactances in Ohms
+f = 50; // Frequency in Hertz
+delta = 9; // Power angle in degree
+p_hp = 150; // Motor delivering power in HP
+eta = 89/100; // Efficiency of motor
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+ws = (4*%pi*f)/p; // Synchronous speed in Radians per second
+// We have (746*150)/0.89) = 125730.34 W = sqrt(3)*1100*I*cos(theta) refer page no. 440, thus we get I*cos(theta) = 12530.34/(1100*sqrt(3)) = 65.99 and E*sin(delta) = I*xs*cos(theta)
+E = (xs*65.99)/sind(delta); // Exitation EMF per phase in Volts
+// since E*cos(delta) > V, therefore the machine is over excited and power factor is leading, thus we get V = E*cos(delta) + I*xs*sin(theta), I*sin(theta) = (635.1-2109.2*cos(9)/5 = -289.586 and we have I*cos(theta) = 65.99 thus by solving these two equations we get theta = atand(-286.586/65.99) = 77.16 degre
+theta = 77.16; // Power factor angle in degree
+I = 65.99/cosd(theta); // Current in Amphere
+pf = cosd(theta); // Power factor leading
+P = (3*V*E*sind(delta))/(xs*1000); // Power in kilo-Watts
+T = (P*1000)/ws; // torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.18: SOLUTION :-");
+printf("\n (a) Excitation EMF, E = %.1f V \n",E)
+printf("\n (b) Line current, I = %.2f A \n",I)
+printf("\n (c) Power factor = %.3f leading \n",pf)
+printf("\n (d) Power , P = %.4f kW \n",P)
+printf("\n (e) Torque , T = %.2f N-m \n",T)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- \n (a) Power , P = 13.0667 kW instaed of %.4f kW \n",P)
+printf("\n (b) Torque , T = 83.22 N-m instaed of %.2f N-m \n",T)
+printf("\n From Calculation of the Power(P), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power(P) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH6/EX6.19/Ex6_19.sce b/Working_Examples/2777/CH6/EX6.19/Ex6_19.sce new file mode 100755 index 0000000..c80914d --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.19/Ex6_19.sce @@ -0,0 +1,47 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.19
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 440; // Operating voltage of the Synchronous Motor in Volts
+p = 6; // Total number of Poles
+m = 3; // Number of phase
+xs = 5; // Synchrouons reactances per phase in Ohms
+f = 50; // Frequency in Hertz
+p_hp = 10; // Motor delivering power in HP
+loss = 1000; // Total iron,copper and friction losses in Watts
+pf = 0.8; // Power factor lagging
+I = 10; // Motor drawing current in Amphere at 0.8 PF lagging
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+ws = (4*%pi*f)/p; // Synchronous speed in Radians per second
+theta = acosd(pf); // Power factor angle in degree
+Po = p_hp*746; // Output power in Watts
+Pi = Po+loss; // Input power in Watts
+eta = (Po/Pi)*100; // Efficiency
+// we have V = E*cos(delta) - I*xs*sin(theta), 254 = E*cos(delta) - 5*10*0.6, so E*cos(delta) = 254 + 30 = 284 and E*sin(delta) = I*xs*cos(theta) = 5*10*0.8 = 40 by solving these two equations we get delta = atand(40/284) = 8.01 degree
+delta = 8.01; // Power angle in degree
+E = 40/sind(delta); // Induced EMF per phase in Volts
+P = (3*V*E*sind(delta))/(xs*1000); // Power in Kilo-watts
+T = (P*1000)/ws; // Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.19: SOLUTION :-");
+printf("\n (a) Efficiency, eta = %.2f Percent \n",eta)
+printf("\n (b) Induced EMF, E = %.f V per phase and \n\n Power (Torque) angle = %.2f degree \n",E,delta)
+printf("\n (c) Power , P = %.4f kW \n",P)
+printf("\n (d) Torque , T = %.2f N-m \n",T)
diff --git a/Working_Examples/2777/CH6/EX6.2/Ex6_2.sce b/Working_Examples/2777/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 0000000..4884b1f --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,30 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 60; // Generating Frequency in Hertz
+Ns = 1200; // Synchronous speed in RPM
+Ns_r = 1000; // Alternator running speed in RPM
+
+
+// CALCULATIONS
+
+p = (120*f)/Ns; // Total number of poles
+f_r = (p*Ns_r)/120; // Alternator running frequency in Hertz
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.2 : SOLUTION :-") ;
+printf("\n Alternator running frequency, f = %.f Hz \n ",f_r);
diff --git a/Working_Examples/2777/CH6/EX6.20/Ex6_20.sce b/Working_Examples/2777/CH6/EX6.20/Ex6_20.sce new file mode 100755 index 0000000..d3f8b1a --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.20/Ex6_20.sce @@ -0,0 +1,73 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.20
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+v = 11*10^3; // Operating voltage of the Synchronous Motor in Volts
+p = 4; // Total number of Poles
+m = 3; // number of phase
+xs = 7; // Synchrouons reactances per phase in Ohms
+f = 50; // Frequency in Hertz
+KVA = 1500; // KVA rating (whole)
+kva = 500; // Each case KVA rating
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+ws = (4*%pi*f)/p; // Synchronous speed in Radians per second
+I = (sqrt(3)*kva)/v; // Phase Current in Amphere
+
+// For Case (a) 0.8 pf lagging
+
+pf_a = 0.8; // Power factor lagging
+pfa_a = acosd(pf_a); // Power factor angle in degree
+// we have E*cos(delta) = V - I*xs*sin(theta) = 6351-78.73*7*0.6 = 6020.334 and E*sin(delta) = I*xs*cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6036.46 V and delta = atand(440.888/6020.334) = 4.19 degree
+E_a = 6036.46; // Induced Voltage in Volts
+delta_a = 4.19; // Power angle in degree
+P_a = (3*V*E_a*sind(delta_a))/(xs*10^6); // Power in Mega-Watts
+T_a = (P_a*10^6)/ws; // Torque in Newton-meter
+
+// For Case (b) 0.8 pf leading
+
+pf_b = 0.8; // Power factor lagging
+pfa_b = acosd(pf_b); // Power factor angle in degree
+// we have E*cos(delta) = V + I*xs*sin(theta) = 6351+78.73*7*0.6 = 6681.666 and E*sin(delta) = I*xs*cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6696.2 V and delta = atand(440.888/6681.666) = 3.78 degree
+E_b = 6696.2; // Induced Voltage in Volts
+delta_b = 3.78; // Power angle in degree
+P_b = (3*V*E_b*sind(delta_b))/(xs*10^6); // Power in Mega-Watts
+T_b = (P_b*10^6)/ws; // Torque in Newton-meter
+
+// For Case (c) UPf
+
+pf_c = 1.0; // Power factor lagging
+pfa_c = acosd(pf_c); // Power factor angle in degree
+// we have E*cos(delta) = V = 6351 and E*sin(delta) = I*xs = 78.73*7 = 551.11 thus we get by sloving these two equatins E = 6374.9 V and delta = atand(551.11/6351) = 4.96 degree
+E_c = 6374.9; // Induced Voltage in Volts
+delta_c = 4.96; // Power angle in degree
+P_c = (3*V*E_c*sind(delta_c))/(xs*10^6); // Power in Mega-Watts
+T_c = (P_c*10^6)/ws; // Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.20: SOLUTION :-");
+printf("\n For Case (a) 0.80 pf lagging :- \n (a) Induced EMF, E = %.2f V \n",E_a)
+printf("\n (b) Power , P = %.1f MW \n",P_a)
+printf("\n (c) Torque , T = %.2f N-m \n",T_a)
+printf("\n For Case (b) 0.80 pf leading :- \n (a) Induced EMF, E = %.1f V \n",E_b)
+printf("\n (b) Power , P = %.3f MW \n",P_b)
+printf("\n (c) Torque , T = %.2f N-m \n",T_b)
+printf("\n For Case (a) UPf :- \n (a) Induced EMF, E = %.1f V \n",E_c)
+printf("\n (b) Power , P = %.2f MW \n",P_c)
+printf("\n (c) Torque , T = %.f N-m \n",T_c)
diff --git a/Working_Examples/2777/CH6/EX6.21/Ex6_21.sce b/Working_Examples/2777/CH6/EX6.21/Ex6_21.sce new file mode 100755 index 0000000..de2ad26 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.21/Ex6_21.sce @@ -0,0 +1,55 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.21
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 440; // Operating voltage of the Synchronous Motor in Volts
+f = 50; // Operating Frequency of the Synchronous Motor in Hertz
+xd = 10; // Direct axis reactances in Ohms
+xq = 7.0; // Quadrature axis reactances in Ohms
+p = 6; // Total number of Poles
+pf = 0.8; // Power factor lagging
+i = 10; // Motor drawing current in Amphere
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+ws = (4*%pi*f)/p; // Synchronous speed in Radians per second
+theta = acosd(pf); // Power factor angle in degree
+I = 10*(cosd(theta)+(%i*sind(theta))); // Motor drawing current in Amphere at 0.8 PF leading
+delta = atand((i*xq*cosd(theta))/(V+i*xq*sind(theta))); // Power angle for motoring mode in degree
+Iq = i*cosd(theta+delta); // Current in Amphere
+Id = i*sind(theta+delta); // Current in Amphere
+Eo = V*cosd(delta) + Id*xd; // Induced EMF in Volts
+P = ((3*V*Eo*sind(delta))/xd)+(3*V^2*((1/xq)-(1/xd))*sind(2*delta))/2; // Power in Watts
+T = ((3*V*Eo*sind(delta))/(xd*ws))+(3*V^2*((1/xq)-(1/xd))*sind(2*delta))/(2*ws); // Torque in Newton-meter
+
+// when the machine is running as alternator, the magnitude of induced EMF = 323.38V. Let the new current will be Inew at lagging power factor thetanew. Now torque angle is 10.71 deg from phasor diagram Figure 6.51 and page no. 444 we get V+Id*xd*cos(delta)-Iq*xq*sin(delta) = Eo*cos(delta), 254+9.825*Id-1.3Iq = 317.75, 9.825*Id-1.3*Iq = 63.75, 7.56*Id-Iq = 49 and we have Id*xd*sin(delta)+Iq*xq*cosdelta) = Eo*sin(delta), 1.85*Id+6.88*Iq = 60.1, 0.27*Id+Iq = 8.74 by solving these two equations we get Idnew = 123.85/10.095 = 12.27A and Iqnew = 5.43A
+Iqnew = 5.43; // New current in Amphere
+Idnew = 12.27; // New current in Amphere
+Inew = sqrt(Idnew^2 + Iqnew^2); // New total Current in Amphere
+// We know that torque angle, tan(delta) = (I*xd*cos(theta))/(V+I*Xq*sin(theta)) so by calutaion for new power factor angle thetanew we get, tan(10.17) = (13.42*7*cos(thetanew))/(254+13.42*7*sin(thetanew)), 0.189(254+13.42*7*sin(thetanew) = 13.42*7*cos(thetanew), 48-93.94cos(thetanew)+17.75*sin(thetanew) = 0 by solving this equatuon we gwt thetanew = 49.5 lagging
+thetanew = 49.5; // New power factor angle in degree
+pfnew = cosd(thetanew); // Power factor lagging
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.21: SOLUTION :-");
+printf("\n (a) Induced EMF, E = %.2f V \n",Eo)
+printf("\n (b) Power (Torque) angle = %.2f degree \n",delta)
+printf("\n Power , P = %.2f W \n",P)
+printf("\n Torque , T = %.2f N-m \n",T)
+printf("\n (c) when the machine is running as alternator requirements are:- \n\n New Current = %.2f A\n",Inew)
+printf("\n Power factor = %.3f lagging \n",pfnew)
diff --git a/Working_Examples/2777/CH6/EX6.22/Ex6_22.sce b/Working_Examples/2777/CH6/EX6.22/Ex6_22.sce new file mode 100755 index 0000000..2714e32 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.22/Ex6_22.sce @@ -0,0 +1,42 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.22
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+E1 = 1100 + (%i*0); // EMFs of two identicel synchronous Generators in Volts per phase
+E2 = 1100*(cosd(5)-(%i*sind(5))); // EMF in Volts per phase
+Zl = 1.0 + (%i*1.0); // Load impedance in Ohms per phase
+Zs1 = 0.15 + (%i*2.1); // Synchronous impedance in Ohms per phase
+Zs2 = 0.2 + (%i*3.3); // Synchronous impedance in Ohms per phase
+f = 50; // Frequency in Hertz
+
+
+// CALCULATONS
+
+Ys1 = 1/Zs1; // Synchronous Admittance in Ohms per phase
+Ys2 = 1/Zs2; // Synchronous Admittance in Ohms per ohase
+Yl = 1/Zl; // Load Admittance in Ohms per ohase
+V = ((E1*Ys1)+(E2*Ys2))/(Yl+Ys2+Ys1); // Terminal Voltage in Volts per phase (From Millman's Theorem)
+I1 = (E1-V)/Zs1; // Individual current in Amphere per phase
+I2 = (E2-V)/Zs2; // Individual current in Amphere per phase
+P1 = abs(V)*abs(I1)*cosd(atand(imag(V),real(V))-atand(imag(I1),real(I1))); // Per phase actice power in Watts
+P2 = abs(V)*abs(I2)*cosd(atand(imag(V),real(V))-atand(imag(I2),real(I2))); // Per phase actice power in Watts
+Ic = (E2-E1)/(Zs1+Zs2); // No-load circulating current in Amphere per phase
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.22 : SOLUTION :-");
+printf("\n (a) Terminal Voltage per phase, V = %.2f < %.1f V \n",abs(V),atand(imag(V),real(V)))
+printf("\n (b) Individual currents per phase, I1 = %.f < %.1f A \n\n I2 = %.1f < %.1f A \n",abs(I1),atand(imag(I1),real(I1)),abs(I2),atand(imag(I2),real(I2)))
+printf("\n (c) Per phase Active Power , P1 = %.f W \n\n P2 = %.1f W \n ",P1,P2)
+printf("\n (d) No-load current per phase, Ic = %.2f < %.2f A \n",abs(Ic),atand(imag(Ic),real(Ic)))
diff --git a/Working_Examples/2777/CH6/EX6.23/Ex6_23.png b/Working_Examples/2777/CH6/EX6.23/Ex6_23.png Binary files differnew file mode 100755 index 0000000..f8cb7f6 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.23/Ex6_23.png diff --git a/Working_Examples/2777/CH6/EX6.23/Ex6_23.sce b/Working_Examples/2777/CH6/EX6.23/Ex6_23.sce new file mode 100755 index 0000000..0fb8aba --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.23/Ex6_23.sce @@ -0,0 +1,49 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.23
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 4; // Number of the poles in the Alternator
+f = 50; // Frequency in Hertz
+pkw = 500; // Alternator delivering load in kilo-watts
+pkwinc = 1000; // Generator increases its share of the common elictrical in kilo-watts
+Kj = 1.5; // Inertia acceleration coefficient for the combined prime mover-alternator in N-m/elec deg/second square
+Kd = 12; // Damping torque coefficient in N-m/elec deg/second
+delta1 = 9; // Initial value of the Power angle in degree
+
+
+// CALCULATIONS
+
+delta2 = (pkwinc/pkw)*delta1; // Final value (maximum value) of the Power angle in degree (considering Linear variation)
+ws = (4*%pi*f)/p; // Rotational speed in Radians per second
+Ts = (pkw*1000)/ws; // Synchornizing torque at 500kW in N-m
+Ks = Ts/delta1; // Synchornizing torque cofficient at 500kW in N-m/elec-deg
+// Laplace transform of the swing Equation can be written as :- s^2 + ((Kd/Kj)*s) + (Ks/Kj) = 0, s^2 + (12/1.5)s + (353.86/1.5) = 0 and compring with the standard equation s^2 + s(2*zeta*Wn) + Wn^2 = 0 we get:- mentined below (refer page no. 454 and 455)
+Wn = sqrt(Ks/Kj); // Natural frequency of oscillations in Radians per second
+fn = Wn/(2*%pi); // Frequency of natural oscillations in Hertz
+zeta = (1*Kd)/(2*Wn*Kj); // Damping ratio
+Wd = Wn*(sqrt(1-zeta^2)); // Frequency of damped oscillations in radians/s
+fd = Wd/(2*%pi); // Frequency of damped oscillations in Hertz
+ts = 5/(zeta*Wn); // Settling time in second
+deltamax = delta1 + 1.42*(delta2-delta1); // The maximum overshoot for damping ratio of 0.2604 is about 42% the maximum appoximate value of the overshoot in terms of 1% tolearance band in Electrical degree
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.23: SOLUTION :-");
+printf("\n (a.1) Final value (maximum value) of the Power angle (considering Linear variation), delta2 = %.f degree \n",delta2)
+printf("\n (a.2) Natural frequency of oscillations, Ns = %.2f radians/s \n",Wn)
+printf("\n (a.3) Damping ratio, zeta = %.4f \n",zeta)
+printf("\n (a.4) Frequency of damped oscillations, Wd = %.2f radians/s \n",Wd)
+printf("\n (a.5) Settling time, ts = %.2f seconds \n",ts)
+printf("\n (b) The maximum overshoot for damping ratio of 0.2604 is about 42 percent the maximum appoximate value of the overshoot in terms of 1 percent tolearance band is, deltamax = %.2f degree \n",deltamax)
+printf("\n\n FOR CASE (C) CANNOT BE DO IT IN THIS BECAUSE AS IT REQUIRES MATLAB SIMULINK \n")
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\ No newline at end of file diff --git a/Working_Examples/2777/CH6/EX6.25/Ex6_25.sce b/Working_Examples/2777/CH6/EX6.25/Ex6_25.sce new file mode 100755 index 0000000..1259182 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.25/Ex6_25.sce @@ -0,0 +1,46 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.25
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 440; // Operating voltage of the Synchronous generator in Volts
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+m = 3; // Total number of Phase
+pf = 0.8; // Power factor lagging
+Il = 100; // Motor drawing current in Amphere
+xs = 2; // Synchronous reactances in Ohms
+delta = 20; // Power angle in degree
+P = 50*10^3; // Total Power developed by the motor in Watts
+Ppp = (50*10^3)/3; // Power developed by the motor per phase in Watts
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+Eo = (Ppp*xs)/(3*V*sind(delta)); // Per phase Induced voltage in Volts
+// Let us assume thetam is Power factor angle in degree and Im is the Motor current now, from phasor diagram figure 6.67 page no. 465 we get, Eo*cosd(delta) = V+Im*xs*sind(thetam), Im*sind(thetam) = ((383.84*cosd(20))-254.03)/2 = 53.35 and Im*xs*cosd(thetam) = Eo*sin(delta), Im*cosd(theta) = ((383.84*sind(20))/2 = 65.60 by sloving these two equations we get Im = sqrt(65.60^2 + 53.35^2) = 84.56 A and thetam = atand(53.35/65.60) = 39.13 degree
+Im = sqrt(65.60^2 + 53.35^2); // Motor current in Amphere
+thetam = atand(53.35/65.60); // Power factor angle in degree
+kVA = (sqrt(3)*V*Im*sind(thetam))/1000; // Rective kVA of the motor in kVAR
+thetal = acosd(pf); // Load power factor angle in degree
+thetaR = atand((Im*sind(thetam)-Il*sind(thetal))/(Im*cosd(thetam)+Il*cosd(thetal))); // Resultant Power factor angle in degree
+ovpf = cosd(thetaR); // Overall Power factor lagging
+IR = sqrt((Im*sind(thetam)-Il*sind(thetal))^2 + (Im*cosd(thetam)+Il*cosd(thetal))^2); // Resultant (magnitude) current in Amphere refer phasor diagram figure 6.69 page no. 467
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.25: SOLUTION :-");
+printf("\n (a) Rective kVA of the motor = %.3f kVAR \n",kVA)
+printf("\n (b) Overall Power factor of the load and motor = %.4f lagging and \n",ovpf)
+printf("\n Resultant (magnitude) current = %.2f A \n",IR)
diff --git a/Working_Examples/2777/CH6/EX6.27/Ex6_27.sce b/Working_Examples/2777/CH6/EX6.27/Ex6_27.sce new file mode 100755 index 0000000..e2a3646 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.27/Ex6_27.sce @@ -0,0 +1,52 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.27
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 440; // Operating voltage of the Synchronous generator in Volts
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+m = 3; // Total number of Phase
+xs = 5; // Synchronous reactances in Ohms
+Eo = 500; // Indduced Voltage in Volts per phase
+R1 = 0.1; // Circuit Parameter in Ohms
+R2 = 0.1; // Circuit Parameter in Ohms
+X1 = 1.55; // Circuit Parameter in Ohms
+X2 = 1.55; // Circuit Parameter in Ohms
+s = 0.03; // Slip
+P = 30*10^3; // Total Power developed by the motor in Watts
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Phase voltage in Volts
+Ii = V/sqrt((R1+R2/s)^2 + (X1+X2)^2); // Per phase induction motor current in Amphere
+thetal = atand((X1+X2)/(R1+R2/s)); // Power factor angle of the induction motor in degree
+pf = cosd(thetal); // Power factor of the induction motor lagging
+// Let us assume thetam is leading Power factor angle in degree and Im is the synchronous Motor current now, from phasor diagram figure 6.70 page no. 469
+delta = asind((xs*P)/(3*V*Eo)); // Power angle in degree
+// From phasor diagram figure 6.70 page no. 469 we have, Im*xs*cos(thetam) = Eo*sin(delta), Im*cos(delta) = ((500*sind(23.18))/5 = 39.37 and Eo*cosd(delta) = V+Im*xs*sind(thetam), Im*sind(thetam) = ((500*cosd(23.18))-254.03)/5 = 41.12 by sloving these two equations we get Im = sqrt(39.37^2 + 41.12^2) = 56.93 A and thetam = atand(41.12/39.37) = 46.25 degree
+Im = sqrt(39.37^2 + 41.12^2); // Motor current in Amphere
+thetam = atand(41.12/39.37); // Power factor angle in degree
+kVA = (sqrt(3)*V*Im*sind(thetam))/1000; // Rective kVA of the motor in kVAR
+II = Ii * exp( %i * (-thetal) * %pi/180); // Induction Motor current in Amphere
+Im = Im * exp( %i * thetam * %pi/180); // Synchronous Motor current in Amphere
+It = II + Im; // Total per phase current in Amphere
+ovpf = cosd(atand(imag(It),real(It))); // Overall Power factor leading
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.27: SOLUTION :-");
+printf("\n (a) Reactive kVA of the motor = %.3f kVAR \n",kVA)
+printf("\n (b) Individual currents:- \n Induction Motor current, II = %.2f + i(%.2f) A \n\n Synchronous Motor current, Im = %.2f + i(%.2f) A \n",real(II),imag(II),real(Im),imag(Im))
+printf("\n (c) Resultant (overall) current = %.2f < %.2f A \n",abs(It),atand(imag(It),real(It)))
+printf("\n (d) Overall Power factor = %.4f leading \n",ovpf)
diff --git a/Working_Examples/2777/CH6/EX6.28/Ex6_28.sce b/Working_Examples/2777/CH6/EX6.28/Ex6_28.sce new file mode 100755 index 0000000..264cf05 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.28/Ex6_28.sce @@ -0,0 +1,38 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.28
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+V = 400; // Operating voltage of the Synchronous generator in Volts
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+xd = 12; // Direct axis reactances in Ohms
+xq = 5; // Quadrature axis reactances in Ohms
+delta = 15; // Power(Torque) angle in degree
+p = 2; // Number of the poles
+m = 3; // Number of the phase
+
+
+// CALCULATIONS
+
+v = V/sqrt(3); // Rated Phase Votage in Volts
+Ns = (120*f)/p; // Operating speed in RPM
+Ws = (2*%pi*f)/(p/2); // Synchronous speed in radians/s
+T = (3*v^2*sind(2*delta)/(2*Ws))*((1/xq)-(1/xd)); // Developed Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.28: SOLUTION :-");
+printf("\n (a) Operating speed, Ns = %.f RPM \n",Ns)
+printf("\n (b) Developed Torque , T = %.2f N-m \n",T)
diff --git a/Working_Examples/2777/CH6/EX6.29/Ex6_29.sce b/Working_Examples/2777/CH6/EX6.29/Ex6_29.sce new file mode 100755 index 0000000..ca9535b --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.29/Ex6_29.sce @@ -0,0 +1,46 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.29
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 400; // Operating Frequency of the Synchronous generator in Hertz
+Ld = 50*10^-3; // Direct axis reactances in Henry
+Lq = 15*10^-3; // Quadrature axis reactances in Henry
+delta = 15; // Power(Torque) angle in degree
+p = 2; // Number of the poles
+m = 3; // Number of the phase
+I = 10; // Operating current in Amphere
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Operating speed in RPM
+Ws = (2*%pi*f)/(p/2); // Synchronous speed in radians/s
+xd = 2*%pi*f*Ld; // Direct axis reactances in reactance
+xq = 2*%pi*f*Lq; // Quadrature axis reactances in reactance
+E1 = 0; // Induced EMF in Volts (Its ZERO beacuse when field winding current is zero)
+v = xq*I; // Applied voltage in Volts
+T = (3*v^2*sind(2*delta)/(2*Ws))*((1/xq)-(1/xd)); // Developed Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.29: SOLUTION :-");
+printf("\n (a) Operating speed, Ns = %.f RPM \n",Ns)
+printf("\n (b) Developed Torque , T = %.5f N-m \n",T)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) xd = 12.56 instead of %.2f Ohms \n ",xd);
+printf("\n (b) xq = 3.768 instead of %.2f Ohms \n ",xq);
+printf("\n (c) v = 36.68 instead of %.2f V \n ",v);
+printf("\n (d) T = 0.07875 instead of %.4f N-m \n ",T);
+printf("\n From Calculation of the d-axis and q-axis reactance (xd and xq respectively), rest all the Calculated values in the TEXT BOOK is WRONG because of the d-axis and q-axis reactance (xd and xq respectively) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH6/EX6.3/Ex6_3.sce b/Working_Examples/2777/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 0000000..2bb72bf --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,53 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+p = 2; // Total number of poles
+
+
+// CALCULATIONS
+// For Case(a)
+
+f_a = 10; // Frequency in Hertz
+Ns_a = (120*f_a)/p; // Synchronous speed in RPM
+
+// For Case(b)
+
+f_b = 50; // Frequency in Hertz
+Ns_b = (120*f_b)/p; // Synchronous speed in RPM
+
+// For Case(c)
+
+f_c = 60; // Frequency in Hertz
+Ns_c = (120*f_c)/p; // Synchronous speed in RPM
+
+// For Case(d)
+
+f_d = 100; // Frequency in Hertz
+Ns_d = (120*f_d)/p; // Synchronous speed in RPM
+
+// For Case(e)
+
+f_e = 400; // Frequency in Hertz
+Ns_e = (120*f_e)/p; // Synchronous speed in RPM
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.3 : SOLUTION :-");
+printf("\n For Case (a) When f = %.f, Synchronous speed, Ns = %.f RPM \n",f_a,Ns_a)
+printf("\n For Case (b) When f = %.f, Synchronous speed, Ns = %.f RPM \n",f_b,Ns_b)
+printf("\n For Case (c) When f = %.f, Synchronous speed, Ns = %.f RPM \n",f_c,Ns_c)
+printf("\n For Case (d) When f = %.f, Synchronous speed, Ns = %.f RPM \n",f_d,Ns_d)
+printf("\n For Case (e) When f = %.f, Synchronous speed, Ns = %.f RPM \n",f_e,Ns_e)
diff --git a/Working_Examples/2777/CH6/EX6.30/Ex6_30.sce b/Working_Examples/2777/CH6/EX6.30/Ex6_30.sce new file mode 100755 index 0000000..d2172dd --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.30/Ex6_30.sce @@ -0,0 +1,34 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.30
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+f = 50; // Operating Frequency of the Synchronous generator in Hertz
+p = 2; // Number of the poles
+Pt = 800; // Total loss in Watts
+Pr = 10; // Rotational loss in Watts
+
+
+// CALCULATIONS
+
+Ws = (4*%pi*f)/p; // Synchronous speed in radians/s
+Ph = Pt-Pr; // Hysteresis loss refered to stator in Watts
+Th = Ph/Ws; // Torque at the shaft in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.30: SOLUTION :-");
+printf("\n (a) Power at the shaft, Ph = %.f W \n",Ph)
+printf("\n )b) Torque at the shaft , Th = %.2f N-m \n",Th)
+
diff --git a/Working_Examples/2777/CH6/EX6.31/Ex6_31.sce b/Working_Examples/2777/CH6/EX6.31/Ex6_31.sce new file mode 100755 index 0000000..7cd049b --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.31/Ex6_31.sce @@ -0,0 +1,75 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.31
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Pi = 2*10^6; // Power input in Volt-Amphere
+v = 6.6*10^3; // Operating voltage in Volts
+
+
+// CALCULATIONS
+
+I = Pi/(v*sqrt(3)); // Rated current in Amphere
+V = v/sqrt(3); // Phase voltage in Volts
+xs = v/(I*sqrt(3)); // Synchronous reactance in Ohms
+
+// For case (a) 0.8 pf lagging
+
+pf_a = 0.8; // Power factor
+pfa_a = acosd(pf_a); // Power factor angle in degree
+a_a = (V + (I*xs*sind(pfa_a)));
+b_a = (I*xs*cosd(pfa_a));
+E_a = sqrt(a_a^2 + b_a^2); // Induced EMF in Volts
+delta_a = atand(b_a/a_a); // Torque (power) angle in degree
+P_a = (3*V*E_a*sind(delta_a))/(xs*10^6); // Power developed in MVA
+
+// For case (b) 0.8 pf leading
+
+pf_b = 0.8; // Power factor
+pfa_b = acosd(pf_b); // Power factor angle in degree
+a_b = (V - (I*xs*sind(pfa_a)));
+b_b = (I*xs*cosd(pfa_b));
+E_b = sqrt(a_b^2 + b_b^2); // Induced EMF in Volts
+delta_b = atand(b_b/a_b); // Torque (power) angle in degree
+P_b = (3*V*E_b*sind(delta_b))/(xs*10^6); // Power developed in MVA
+
+// For case (c) UPF
+
+pf_c = 1.0; // Power factor
+pfa_c = acosd(pf_c); // Power factor angle in degree
+a_c = V;
+b_c = I*xs;
+E_c = sqrt(a_c^2 + b_c^2); // Induced EMF in Volts
+delta_c = atand(b_c/a_c); // Torque (power) angle in degree
+P_c = (3*V*E_c*sind(delta_c))/(xs*10^6); // Power developed in MVA
+
+
+disp("EXAMPLE : 6.31: SOLUTION :-");
+printf("\n For Case (a) 0.80 lagging Power factor \n Induced EMF, EMF = %.2f V \n",E_a)
+printf("\n Power (Torque) angle = %.2f degree \n",delta_a)
+printf("\n Power developed, P = %.1f MVA \n",P_a)
+printf("\n For Case (b) 0.80 leading Power factor \n Induced EMF, EMF = %.f V \n",E_b)
+printf("\n Power (Torque) angle = %.2f degree \n",delta_b)
+printf("\n Power developed, P = %.3f MVA \n",P_b)
+printf("\n For Case (c) Unity Power Factor \n Induced EMF, EMF = %.1f V \n",E_c)
+printf("\n Power (Torque) angle = %.2f degree \n",delta_c)
+printf("\n Power developed, P = %.1f MVA \n",P_c)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- xs = 20.14 instead of %.2f Ohms \n ",xs);
+printf("\n For Case (a) 0.80 lagging Pf (a.1) E = 6561.42 instead of %.2f V \n ",E_a);
+printf("\n (a.2) delta = 25.45 instead of %.2f degree \n ",delta_a);
+printf("\n For Case (b) 0.80 leading Pf (b.1) E = 3290 instead of %.1f V \n ",E_b);
+printf("\n (b.2) delta = 58.98 instead of %.2f degree \n ",delta_b);
+printf("\n (b.3) Power developed = 1.617 instead of %.3f MVA \n ",P_b);
+printf("\n For Case (c) UPF (c.1) E = 5190.2 instead of %.2f V \n ",E_c);
+printf("\n (c.2) delta = 42.77 instead of %.2f degree \n ",delta_c);
+printf("\n In all the three cases from Calculation of the Synchronous reactance (xs), rest all the Calculated values in the TEXT BOOK is WRONG because of the Synchronous reactance (xs) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH6/EX6.32/Ex6_32.sce b/Working_Examples/2777/CH6/EX6.32/Ex6_32.sce new file mode 100755 index 0000000..1642d8a --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.32/Ex6_32.sce @@ -0,0 +1,41 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.32
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer phasor diagram figure 6.76 and page no. 476
+
+pf = 0.8; // Power factor lagging
+pa = acosd(pf); // Power factor angle in degree
+v = 1.0 * exp( %i * pa * %pi/180); // Operating voltage of the alternator in pu
+xd = 0.8; // Direct axis reactances in pu
+xq = 0.4; // Quadrature axis reactances in pu
+I = 1.0; // Current in pu taking this as reference
+
+
+// CALCULATIONS
+
+A = v + (%i*xq*I);
+delta = atand(imag(A),real(A))-pa; // Power angle in degree
+Iq = I * cosd(atand(imag(A),real(A))); // d-axis currents in Amphere
+
+Id = I * sind(atand(imag(A),real(A))); // q-axis currents in Amphere
+E = abs(v)*cosd(delta) + Id*xd; // Induced EMF per phase in Per unit
+pr = ((abs(E)-abs(v))/abs(v))*100; // Percentage regulation
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.32: SOLUTION :-");
+printf("\n (a) Induced EMF per phase, E = %.4f < %.2f pu \n",E,delta)
+printf("\n (b) Power angle = %.2f degree \n",delta)
+printf("\n (C) Percenatge Regulation, R = %.2f Percent \n",pr)
+printf("\n\n IN THIS PROBLEM PERCENTAGE REGULATION IS NOT CALCULATED IN THE TEXT BOOK\n")
diff --git a/Working_Examples/2777/CH6/EX6.33/Ex6_33.sce b/Working_Examples/2777/CH6/EX6.33/Ex6_33.sce new file mode 100755 index 0000000..27c1e6a --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.33/Ex6_33.sce @@ -0,0 +1,44 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.33
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 6.6*10^3; // Operating voltage of the Synchronous motor in Volts
+P = 5*10^6; // Operating power of the Synchronous motor in Watts
+pf = 1.0; // Power factor
+xd = 3.0; // Direct axis reactances in Ohms
+xq = 1.0; // Quadrature axis reactances in Ohms
+eta = 0.98; // OPerating efficiency
+
+
+// CALCULATIONS
+
+V = v/sqrt(3); // Per phase voltage in Volts
+I = P/(eta*v*sqrt(3)); // Line current in Amphere
+delta = atand((xq*I)/v); // power angle in degree
+E = v*cosd(delta) + xd*I*sind(delta); // Induced EMF in Volts
+Tmax = ((3*E*V*sind(90))/xd) + ((3*V^2*sind(180))/2)*((1/xq)-(1/xd)); // Maximum electromagnetic torque in N-m
+T = ((3*E*V*sind(delta))/xd) + ((3*V^2*sind(2*delta))/2)*((1/xq)-(1/xd));
+ // Actual electromagnetic torque in N-m
+Ratio = Tmax/T; // Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.33: SOLUTION :-");
+printf("\n (a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is %.2f \n",Ratio)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) delta = 2.41 instead of %.2f degree \n ",delta);
+printf("\n (b) E = 6379 instead of %.2f V \n ",E);
+printf("\n (c) Ratio = 10.84 instead of %.2f \n ",Ratio);
+printf("\n From Calculation of the Power angle (delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle (delta) value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH6/EX6.4/Ex6_4.sce b/Working_Examples/2777/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 0000000..bf5422e --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,112 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.4 : \n\n Given Data \n");
+printf("\n Voc(v) 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n");
+printf("\n If(A) 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n\n");
+m = 3; // Total Number of phase
+p = 6; // Total number of Poles
+V = 400; // Operating voltage in Volts
+I = 13.5; // Operating current in Amphere
+N = 1000; // Speed in RPM
+Ia_scc = 13.5; // SCC test Armature current in Amphere at If = 9.5 A
+If_scc = 9.5; // SCC test field Rated current in Amphere
+Ia_zpf = 13.5; // ZPF test Armature current in Amphere at If = 24 A
+If_zpf = 24; // ZPF test field Rated current in Amphere
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
+Ra = 1.0; // Armature resistance in Ohms
+
+// For case (a)
+
+BC = 120; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386
+Xl = BC/(sqrt(3)*Ia_scc); // Per phase leakage reactance in Ohms
+
+
+// For Case (b.1) 0.8 pf Lagging
+
+pfa_b1 = acosd(0.8); // Power factor angle in degree
+Er_b1 = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b1)-%i*sind(pfa_b1))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b1 = 10; A_b1 = 9.5; //From OCC the field current required for Er_b1 (Should be in Line-line Voltage) Er_b1 = 379.12V will get R_b1 & A_b1 value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b1 = 136.35; // Angle between R_b1 & A_b1 (figure 6.16(a) & Page no:-388) = 90'+9.48'+36.87' = 136.35'
+F_b1 = sqrt((R_b1^2)+(A_b1^2)-(2*R_b1*A_b1*cosd(angle_b1))); // From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere
+Eo_b1 = 525; // Corresponding to field current F_b1 = 18.12 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
+r_b1 = 100*((Eo_b1-V)/V); // Percentage regulation
+
+
+// For Case (b.2) 0.8 pf Leading
+
+pfa_b2 = acosd(0.8); // Power factor angle in degree
+Er_b2 = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b2)+%i*sind(pfa_b2))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b2 = 8.3; A_b2 = 9.5; //From OCC the field current required for Er_b2 (Should be in Line-line Voltage) Er_b1 = 363.71 V will get R_b2 & A_b2 value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b2 = 70.61; // Angle between R_b2 & A_b2 (figure 6.16(b) & Page no:-388) = 90'+17.48'-36.87' = 70.61'
+F_b2 = sqrt((R_b2^2)+(A_b2^2)-(2*R_b2*A_b2*cosd(angle_b2))); // From phasor diagram in figure 6.16(b) & Page no:-388 the neccessary field excitation in Amphere
+Eo_b2 = 338; // Corresponding to field current F_b2 = 10.36 A the open circuit EMF from OCC is 338 V (Figure6.15 & Page no:-386)
+r_b2 = 100*((Eo_b2-V)/V); // Percentage regulation
+
+
+// For Case (b.3) Unity pf Leading
+
+pfa_b3 = acosd(1.0); // Power factor angle in degree
+Er_b3 = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b3)-%i*sind(pfa_b3))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b3 = 13; A_b3 = 9.5; //From OCC the field current required for Er_b3 (Should be in Line-line Voltage) Er_b1 = 440.30 V will get R_b3 & A_b3 value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b3 = 105.81; // Angle between R_b3 & A_b3 (figure 6.16(c) & Page no:-388) = 90'+15.81' = 105.81'
+F_b3 = sqrt((R_b3^2)+(A_b3^2)-(2*R_b3*A_b3*cosd(angle_b3))); // From phasor diagram in figure 6.16(c) & Page no:-388 the neccessary field excitation in Amphere
+Eo_b3 = 520; // Corresponding to field current F_b2 = 18.10 A the open circuit EMF from OCC is 520 V (Figure6.15 & Page no:-386)
+r_b3 = 100*((Eo_b3-V)/V); // Percentage regulation
+
+
+// For Case (b.4) ZPF Lagging
+
+pfa_b4 = acosd(0); // Power factor angle in degree
+Er_b4 = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b4)-%i*sind(pfa_b4))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b4 = 18; A_b4 = 9.5; //From OCC the field current required for Er_b4 (Should be in Line-line Voltage) Er_b4 = 521 V will get R_b4 & A_b4 value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b4 = 177.57; // Angle between R_b4 & A_b4 = 90'-2.43'+90' = 177.57'
+F_b4 = sqrt((R_b4^2)+(A_b4^2)-(2*R_b4*A_b4*cosd(angle_b4))); // The neccessary field excitation in Amphere
+Eo_b4 = 570; // Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
+r_b4 = 100*((Eo_b4-V)/V); // Percentage regulation
+
+
+// For Case (b.4) ZPF Lagging
+
+pfa_b5 = acosd(0); // Power factor angle in degree
+Er_b5 = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b5)+%i*sind(pfa_b5))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b5 = 6.0; A_b5 = 9.50; //From OCC the field current required for Er_b5 (Should be in Line-line Voltage) Er_b5 = 280.70 V will get R_b5 & A_b5 value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b5 = 4.77; // Angle between R_b5 & A_b5 = 90'-4.77'-90' = 4.77'
+F_b5 = sqrt((R_b5^2)+(A_b5^2)-(2*R_b5*A_b5*cosd(angle_b5))); // The neccessary field excitation in Amphere
+Eo_b5 = 135; // Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 135 V (Figure6.15 & Page no:-386)
+r_b5 = 100*((Eo_b5-V)/V); // Percentage regulation
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a) Per phase leakage reactance, Xl = %.2f Ohms \n",Xl)
+printf("\n For Case (b.1) 0.8 pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_b1)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b1)
+printf("\n For Case (b.2) 0.8 pf Leading \n Open circuit EMF, EMF = %.f V \n",Eo_b2)
+printf("\n Percenatge Regulation, R = %.1f Percenatge \n",r_b2)
+printf("\n For Case (b.3) Unity pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_b3)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_b3)
+printf("\n For Case (b.4) ZPF Lagging \n Open circuit EMF, EMF = %.f V\n",Eo_b4)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b4)
+printf("\n For Case (b.5) ZPF Leading \n Open circuit EMF, EMF = %.f V \n",Eo_b5)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n\n",r_b5)
+disp(" Calculated Answer in Tabular Column")
+printf("\n Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n")
+printf("\n Open circuit EMF (V) %.f %.f %.f %.f %.f \n",Eo_b1,Eo_b2,Eo_b3,Eo_b4,Eo_b5)
+printf("\n Percenatge Regulation %.2f %.1f %.f. %.2f %.2f \n",r_b1,r_b2,r_b3,r_b4,r_b5)
diff --git a/Working_Examples/2777/CH6/EX6.5a/Ex6_5a.sce b/Working_Examples/2777/CH6/EX6.5a/Ex6_5a.sce new file mode 100755 index 0000000..ab6c379 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.5a/Ex6_5a.sce @@ -0,0 +1,118 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.5a
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.5a( Data is same as Example 6.4): \n\n Given Data \n");
+printf("\n Voc(v) 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n");
+printf("\n If(A) 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n\n");
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+V = 400; // Operating voltage of the Induction motor in Volts
+I = 13.5; // Operating current of the Induction motor in Amphere
+N = 1000; // speed of the Induction motor in RPM
+Ia_scc = 13.5; // SCC test Armature current in Amphere at If = 9.5 A
+If_scc = 9.5; // SCC test field Rated current in Amphere
+Ia_zpf = 13.5; // ZPF test Armature current in Amphere at If = 24 A
+If_zpf = 24; // ZPF test field Rated current in Amphere
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
+Ra = 1.0; // Armature resistance in Ohms
+v = V/sqrt(3); // Rated phase Voltage in Volts
+
+// For case (a)
+
+EMF_a1 = 345; // From OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386 open-circuit line-line voltage per phase is 345vVfor If = 9.50A in Volts
+Zs_a1 = EMF_a1/(Ia_zpf*sqrt(3)); // Unsaturated synchronous impedance at If=9.50A in Ohms
+Xs_a1 = sqrt((Zs_a1^2)-(Ra^2)); // Synchronous reactance at If =9.50A in Ohms
+Ia_a2 = 15.75; // Current from SCC in Figure6.15 & Page no:-386 is 15.75A for correspounding to the rated Voltage in Volts
+Zs_a2 = V/(Ia_a2*sqrt(3)); // Unsaturated synchronous impedance at If=9.50A in Ohms
+Xs_a2 = sqrt((Zs_a2^2)-(Ra^2)); // Synchronous reactance at If =9.50A in Ohms
+
+// For Case (b.1) 0.8 pf Lagging
+
+pfa_b1 = acosd(0.8); // Power factor angle in degree
+real_b1 = (v+Ia_zpf*Ra*cosd(pfa_b1)+Ia_zpf*Xs_a1*sind(pfa_b1));
+imag_b1 = (Ia_zpf*Xs_a1*cosd(pfa_b1)-Ia_zpf*Ra*sind(pfa_b1));
+E_b1 = sqrt(real_b1^2+imag_b1^2); // Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for lagging pf
+del_b1 = atand(imag_b1/real_b1); // Power angle in degree
+r_b1 = 100*(E_b1-v)/v; // Percantage regulation
+
+
+// For Case (b.2) 0.8 pf Leading
+
+pfa_b2 = acosd(0.8); // Power factor angle in degree
+real_b2 = (v+Ia_zpf*Ra*cosd(pfa_b2)-Ia_zpf*Xs_a1*sind(pfa_b2));
+imag_b2 = (Ia_zpf*Xs_a1*cosd(pfa_b2)+Ia_zpf*Ra*sind(pfa_b2));
+E_b2 = sqrt(real_b2^2+imag_b2^2); // Induced Voltage pr phase in Volts from Figure6.19 (b) & Page no:-394 shows the phasor diagram for leading pf
+del_b2 = atand(imag_b2/real_b2); // Power angle in degree
+r_b2 = 100*(E_b2-v)/v; // Percantage regulation
+
+
+// For Case (b.3) Unity pf
+
+pfa_b3 = acosd(1.0); // Power factor angle in degree
+real_b3 = (v+Ia_zpf*Ra);
+imag_b3 = (Ia_zpf*Xs_a1);
+E_b3 = sqrt(real_b3^2+imag_b3^2); // Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for unity pf
+del_b3 = atand(imag_b3/real_b3); // Power angle in degree
+r_b3 = 100*(E_b3-v)/v; // Percantage regulation
+
+// For Case (b.4) ZPF pf Lagging
+
+pfa_b4 = acosd(0); // Power factor angle in degree
+real_b4 = (v+Ia_zpf*Xs_a1);
+imag_b4 = (-Ia_zpf*Ra);
+E_b4 = sqrt(real_b4^2+imag_b4^2); // Induced Voltage pr phase in Volts ZPF for lagging pf
+del_b4 = atand(imag_b4/real_b4); // Power angle in degree
+r_b4 = 100*(E_b4-v)/v; // Percantage regulation
+
+// For Case (b.5) ZPF pf Leading
+
+pfa_b5 = acosd(0); // Power factor angle in degree
+real_b5 = (v-Ia_zpf*Xs_a1);
+imag_b5 = (Ia_zpf*Ra);
+E_b5 = sqrt(real_b5^2+imag_b5^2); // Induced Voltage pr phase in Volts ZPF for lagging pf
+del_b5 = atand(imag_b5/real_b5); // Power angle in degree
+r_b5 = 100*(E_b5-v)/v; // Percantage regulation
+
+
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n (a.1) Synchronous reactance for rated current at If = %.2f, Xs = %.2f Ohms \n",If_scc,Xs_a1)
+printf("\n (a.2) Synchronous reactance for rated per phase Voltage at v = %.f, Xs = %.2f Ohms \n",v,Xs_a2)
+printf("\n For Case (b.1) 0.8 pf Lagging \n Induced EMF per phase , EMF = %.2f V \n",E_b1)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b1)
+printf("\n Power angle = %.2f degree \n",del_b1)
+printf("\n For Case (b.2) 0.8 pf Leading \n Induced EMF per phase, EMF = %.2f V \n",E_b2)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b2)
+printf("\n Power angle = %.2f degree \n",del_b2)
+printf("\n For Case (b.3) Unity pf Lagging \n Induced EMF per phase, EMF = %.2f V \n",E_b3)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b3)
+printf("\n Power angle = %.2f degree \n",del_b3)
+printf("\n For Case (b.4) ZPF Lagging \n Induced EMF per phase, EMF = %.2f V\n",E_b4)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b4)
+printf("\n Power angle = %.1f degree \n",del_b4)
+printf("\n For Case (b.5) ZPF Leading \n Induced EMF per phase, EMF = %.2f V \n",E_b5)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b5)
+printf("\n Power angle = %.2f degree \n\n",del_b5)
+disp(" Calculated Answer in Tabular Column")
+printf("\n Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n")
+printf("\n Open circuit EMF (V) %.2f %.2f %.2f %.2f %.2f \n",E_b1,E_b2,E_b3,E_b4,E_b5)
+printf("\n Percenatge Regulation %.2f %.2f %.2f. %.2f %.2f \n",r_b1,r_b2,r_b3,r_b4,r_b5)
+printf("\n Power angle %.2f %.2f %.2f. %.1f %.2f \n",del_b1,del_b2,del_b3,del_b4,del_b5)
diff --git a/Working_Examples/2777/CH6/EX6.6/Ex6_6.sce b/Working_Examples/2777/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 0000000..ec33d7a --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,106 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.6
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.4 ( Data is same as Exaple 6.4 ): \n\n Given Data \n");
+printf("\n Voc(v) 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n");
+printf("\n If(A) 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n\n");
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+V = 400; // Operating voltage of the Induction motor in Volts
+I = 13.5; // Operating current of the Induction motor in Amphere
+N = 1000; // speed of the Induction motor in RPM
+Ia_scc = 13.5; // SCC test Armature current in Amphere at If = 9.5 A
+If_scc = 9.5; // SCC test field Rated current in Amphere
+Ia_zpf = 13.5; // ZPF test Armature current in Amphere at If = 24 A
+If_zpf = 24; // ZPF test field Rated current in Amphere
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
+Ra = 1.0; // Armature resistance in Ohms
+v = V/sqrt(3); // Rated phase voltage in Volts
+
+
+// For Case (a) 0.8 pf Lagging
+
+pfa_a = acosd(0.8); // Power factor angle in degree
+E_a = v+(Ia_scc*(cosd(pfa_a)-%i*sind(pfa_a))*Ra); // Induced Voltage in Volts
+R1_a = 11.8; A_a = 9.50; //From OCC the field current required for E_a (Should be in Line-line Voltage) E_a = 419.05V will get R1_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_a = 124.95; // Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'-1.92'+36.87' = 124.95'
+F_a = sqrt((R1_a^2)+(A_a^2)-(2*R1_a*A_a*cosd(angle_a))); // From phasor diagram in figure 6.21(a) & Page no:-400 the neccessary field excitation in Amphere
+Eo_a = 538; // Corresponding to field current F_a = 18.94 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)
+r_a = 100*((Eo_a-V)/V); // Percentage regulation
+
+
+// For Case (b) 0.8 pf Leading
+
+pfa_b = acosd(0.8); // Power factor angle in degree
+E_b = v+(Ia_scc*(cosd(pfa_b)+%i*sind(pfa_b))*Ra); // Induced Voltage in Volts
+R1_b = 11.80; A_b = 9.50; //From OCC the field current required for E_b (Should be in Line-line Voltage) E_b = 419.10V will get R1_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b = 55.07; // Angle between R1_b & A_b (Figure6.21b & Page no:-400) = 90'-1.92'-36.87' = 55.07'
+F_b = sqrt((R1_b^2)+(A_b^2)-(2*R1_b*A_b*cosd(angle_b))); // From phasor diagram in figure 6.21(b) & Page no:-400 the neccessary field excitation in Amphere
+Eo_b = 382; // Corresponding to field current F_b = 10.10 A the open circuit EMF from OCC is 382 V (Figure6.15 & Page no:-386)
+r_b = 100*((Eo_b-V)/V); // Percentage regulation
+
+// For Case (c) Unity pf
+
+pfa_c = acosd(1); // Power factor angle in degree
+E_c = v+(Ia_scc*(cosd(pfa_c)+%i*sind(pfa_c))*Ra) // Induced Voltage in Volts
+R1_c = 12.10; A_c = 9.50; //From OCC the field current required for E_c (Should be in Line-line Voltage) E_c = 423.50V will get R1_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_c = 90; // Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'
+F_c = sqrt((R1_c^2)+(A_c^2)-(2*R1_c*A_c*cosd(angle_c))); // From phasor diagram in figure 6.21(c) & Page no:-400 the neccessary field excitation in Amphere
+Eo_c = 480; // Corresponding to field current F_c = 15.38 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)
+r_c = 100*((Eo_c-V)/V); // Percentage regulation
+
+
+// For Case (d) ZPF Lagging
+
+pfa_d = acosd(0.0); // Power factor angle in degree
+E_d = v+(Ia_scc*(cosd(pfa_d)-%i*sind(pfa_d))*Ra) // Induced Voltage in Volts
+R1_d = 11.20; A_d = 9.50; //From OCC the field current required for E_d (Should be in Line-line Voltage) E_d = 400.80V will get R1_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_d = 179.40; // Angle between R1_d & A_d = 90'-0.6'+90' = 179.40'
+F_d = sqrt((R1_d^2)+(A_d^2)-(2*R1_d*A_d*cosd(angle_d))); // From phasor diagram the neccessary field excitation in Amphere
+Eo_d = 545; // Corresponding to field current F_d = 18.12 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)
+r_d = 100*((Eo_d-V)/V); // Percentage regulation
+
+// For Case (d) ZPF Lagging
+
+pfa_e = acosd(0.0); // Power factor angle in degree
+E_e = v+(Ia_scc*(cosd(pfa_e)+%i*sind(pfa_e))*Ra) // Induced Voltage in Volts
+R1_e = 11.20; A_e = 9.50; //From OCC the field current required for E_e (Should be in Line-line Voltage) E_d = 400.80V will get R1_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_e = 0.60; // Angle between R1_e & A_e = 90'+0.6'-90' = 0.60'
+F_e = sqrt((R1_e^2)+(A_e^2)-(2*R1_e*A_e*cosd(angle_e))); // From phasor diagram the neccessary field excitation in Amphere
+Eo_e = 63; // Corresponding to field current F_e = 1.70 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)
+r_e = 100*((Eo_e-V)/V); // Percentage regulation
+
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n For Case (a) 0.8 pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_a)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_a)
+printf("\n For Case (b) 0.8 pf Leading \n Open circuit EMF, EMF = %.f V \n",Eo_b)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b)
+printf("\n For Case (c) Unity pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_c)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_c)
+printf("\n For Case (d) ZPF Lagging \n Open circuit EMF, EMF = %.f V\n",Eo_d)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_d)
+printf("\n For Case (e) ZPF Leading \n Open circuit EMF, EMF = %.f V \n",Eo_e)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n\n",r_e)
+disp(" Calculated Answer in Tabular Column")
+printf("\n Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n")
+printf("\n Open circuit EMF (V) %.f %.f %.f %.f %.f \n",Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)
+printf("\n Percenatge Regulation %.2f %.2f %.f %.2f %.2f \n",r_a,r_b,r_c,r_d,r_e)
diff --git a/Working_Examples/2777/CH6/EX6.7/Ex6_7.sce b/Working_Examples/2777/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 0000000..6ca6d00 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,108 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.7
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.7 ( Data as same as Example 6.4 ) : \n\n Given Data \n");
+printf("\n Voc(v) 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n");
+printf("\n If(A) 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n\n");
+m = 3; // Total Number of phase in Induction Motor
+p = 6; // Total number of Poles of Induction Motor
+V = 400; // Operating voltage of the Induction motor in Volts
+I = 13.5; // Operating current of the Induction motor in Amphere
+N = 1000; // speed of the Induction motor in RPM
+Ia_scc = 13.5; // SCC test Armature current in Amphere at If = 9.5 A
+If_scc = 9.5; // SCC test field Rated current in Amphere
+Ia_zpf = 13.5; // ZPF test Armature current in Amphere at If = 24 A
+If_zpf = 24; // ZPF test field Rated current in Amphere
+
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
+Ra = 1.0; // Armature resistance in Ohms
+BC = 120; // Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386
+Xl = BC/(sqrt(3)*Ia_scc); // Per phase leakage reactance in Ohms for this referring to example6.4 & page no:- 386
+
+// For Case (a) 0.8 pf Lagging
+
+pfa_a = acosd(0.8); // Power factor angle in degree
+Er_a = (V/sqrt(3))+(Ia_scc*(cosd(pfa_a)-%i*sind(pfa_a))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_a = 9.8; A_a = 9.5; //From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 479.60V will get R_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_a = 126.87; // Angle between R_a & A_a (Figure6.22(a) & Page no:-403) = 90'+36.87' = 126.87'
+F_a = sqrt((R_a^2)+(A_a^2)-(2*R_a*A_a*cosd(angle_a))); // From phasor diagram in figure 6.22(a) & Page no:-403 the neccessary field excitation in Amphere
+Eo_a = 560; // Corresponding to field current ( OF'=OF+FF'), F_a = 17.28 + 6.2 = 23.46 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)
+r_a = 100*((Eo_a-V)/V); // Percentage regulation
+
+
+// For Case (b) 0.8 pf Leading
+
+pfa_b = acosd(0.8); // Power factor angle in degree
+Er_b = (V/sqrt(3))+(Ia_scc*(cosd(pfa_b)+%i*sind(pfa_b))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_b = 9.8; A_b = 9.5; //From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 363.90 V will get R_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_b = 53.13; // Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'-36.87' = 53.13'
+F_b = sqrt((R_b^2)+(A_b^2)-(2*R_b*A_b*cosd(angle_b))); // From phasor diagram in figure 6.22(b) & Page no:-403 the neccessary field excitation in Amphere
+Eo_b = 380; // Corresponding to field current ( OF'=OF+FF') F_b = 8.62+1.5=10.12 A the open circuit EMF from OCC is 380 V (Figure6.15 & Page no:-386)
+r_b = 100*((Eo_b-V)/V); // Percentage regulation
+
+
+// For Case (c) Unity pf Leading
+
+pfa_c = acosd(1.0); // Power factor angle in degree
+Er_c = (V/sqrt(3))+(Ia_scc*(cosd(pfa_c)-%i*sind(pfa_c))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_c = 9.8; A_c = 9.5; //From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 440.11 V will get R_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_c = 90; // Angle between R_c & A_c (Figure6.22c & Page no:-403) = 90' = 90'
+F_c = sqrt((R_c^2)+(A_c^2)-(2*R_c*A_c*cosd(angle_c))); // From phasor diagram in figure 6.22(c) & Page no:-403 the neccessary field excitation in Amphere
+Eo_c = 510; // Corresponding to field current ( OF'=OF+FF') F_c = 13.65+3.0=16.65A the open circuit EMF from OCC is 510 V (Figure6.15 & Page no:-386)
+r_c = 100*((Eo_c-V)/V); // Percentage regulation
+
+
+// For Case (d) ZPF Lagging
+
+pfa_d = acosd(0); // Power factor angle in degree
+Er_d = (V/sqrt(3))+(Ia_scc*(cosd(pfa_d)-%i*sind(pfa_d))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_d = 9.8; A_d = 9.5; //From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 570.20 V will get R_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_d = 180.0; // Angle between R_d & A_d = 90'+90' = 180'
+F_d = sqrt((R_d^2)+(A_d^2)-(2*R_d*A_d*cosd(angle_d))); // The neccessary field excitation in Amphere
+Eo_d = 600; // Corresponding to field current ( OF'=OF+FF') F_d = 19.3+16=35.30 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
+r_d = 100*((Eo_d-V)/V); // Percentage regulation
+
+
+// For Case (e) ZPF Lagging
+
+pfa_e = acosd(0); // Power factor angle in degree
+Er_e = (V/sqrt(3))+(Ia_scc*(cosd(pfa_e)+%i*sind(pfa_e))*(Ra+%i*Xl)); // Induced Voltage in Volts
+R_e = 9.8; A_e = 9.50; //From OCC the field current required for Er_e (Should be in Line-line Voltage) Er_e = 281.10 V will get R_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)
+angle_e = 0.0; // Angle between R_e & A_e = 90'-90' = 0.0'
+F_e = sqrt((R_e^2)+(A_e^2)-(2*R_e*A_e*cosd(angle_e))); // The neccessary field excitation in Amphere
+Eo_e = 5; // Corresponding to field current ( OF'=OF+FF') F_e = 0.0+0.30=0.30 A the open circuit EMF from OCC is 5 V (Figure6.15 & Page no:-386)
+r_e = 100*((Eo_e-V)/V); // Percentage regulation
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n Per phase leakage reactance, Xl = %.2f Ohms \n",Xl)
+printf("\n For Case (a) 0.8 pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_a)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_a)
+printf("\n For Case (b) 0.8 pf Leading \n Open circuit EMF, EMF = %.f V \n",Eo_b)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_b)
+printf("\n For Case (c) Unity pf Lagging \n Open circuit EMF, EMF = %.f V \n",Eo_c)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_c)
+printf("\n For Case (d) ZPF Lagging \n Open circuit EMF, EMF = %.f V\n",Eo_d)
+printf("\n Percenatge Regulation, R = %.f Percenatge \n",r_d)
+printf("\n For Case (e) ZPF Leading \n Open circuit EMF, EMF = %.f V \n",Eo_e)
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n\n",r_e)
+disp(" Calculated Answer in Tabular Column")
+printf("\n Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n")
+printf("\n Open circuit EMF (V) %.f %.f %.f %.f %.f \n",Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)
+printf("\n Percenatge Regulation %.f %.f %.1f %.f %.2f \n",r_a,r_b,r_c,r_d,r_e)
diff --git a/Working_Examples/2777/CH6/EX6.9/Ex6_9.sce b/Working_Examples/2777/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 0000000..a20cbc0 --- /dev/null +++ b/Working_Examples/2777/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,71 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.9
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+printf("\n EXAMPLE : 6.9 : \n\n Given Data \n");
+printf("\n Voc(kV) 10 10.80 11.50 12.10 12.60 13 14 14.50 14.80 \n");
+printf("\n If(A) 175 200 225 250 275 300 400 450 500 \n\n");
+p = 6; // Total number of Poles of Alternator
+V = 11*10^3; // Operating voltage of the Alternator in Volts
+N = 1500; // speed of the Alternator in RPM
+Ia_scc = 2099; // SCC test Armature current in Amphere at If = 200 A
+If_scc = 200; // SCC test field Rated current in Amphere
+Ia_pt = 2099; // Pottier test Armature current in Amphere at If = 450 A
+If_pt = 450; // Pottier test field Rated current in Amphere
+VA = 40*10^6; // VA rating of the Alternator in Volts-Amphere
+f = 50; // Operating Frequency of the Alternator in Hertz
+pf = 0.8; // Power factor (lagging)
+
+// CALCULATIONS
+// Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407
+
+v = V/sqrt(3); // Rated phase Voltage in Volts
+I = VA/(sqrt(3)*V); // Full-load phase current in Amphere
+Xl = 0.4481; // Leakage reactance in Ohms From OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407
+
+
+// For Case(a) General Method
+
+pfa_a = acosd(pf); // Power factor angle in degree
+Er_a = (V/sqrt(3))+(Ia_scc*(cosd(pfa_a)-%i*sind(pfa_a))*Xl); // Induced Voltage in Volts
+R_a = 208.4; A_a = 200; //From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 11043.66 V will get R_a & A_a value Respectively from SCC (Figure6.24 & Page no:-407)
+angle_a = 131.93; // Angle between R_a & A_a (Figure6.25(a) & Page no:-408) = 90'+5.06'+36.87' = 131.93'
+F_a = sqrt((R_a^2)+(A_a^2)-(2*R_a*A_a*cosd(angle_a))); // From phasor diagram in figure 6.25(a) & Page no:-408 the neccessary field excitation in Amphere
+Eo_a = 13720; // Corresponding to field current, F_a = 373 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)
+r_a = 100*((Eo_a-V)/V); // Percentage regulation
+
+
+// For Case (b) ASA Method
+
+pfa_b = acosd(pf); // Power factor angle in degree
+Er_b = (V/sqrt(3))+Ia_scc*(cosd(pfa_b)-%i*sind(pfa_b))*Xl; // Induced Voltage in Volts
+R_b = 160; A_b = 200; //From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 11043.66 V will get R_b & A_b value Respectively from SCC (Figure6.24 & Page no:-407)
+angle_b = 126.87; // Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'+36.87' = 126.87'
+F_b = sqrt((R_b^2)+(A_b^2)-(2*R_b*A_b*cosd(angle_b))); // From phasor diagram in figure 6.25(b) & Page no:-408 the neccessary field excitation in Amphere
+Eo_b = 13660; // Corresponding to field current ( OF'=OF+FF') F_b = 337.88+15.38=337.88 A the open circuit EMF from OCC is 13660 V (Figure6.15 & Page no:-386)
+r_b = 100*((Eo_b-V)/V); // Percentage regulation
+
+
+// DISPLAY RESULTS
+
+disp(" SOLUTION :-");
+printf("\n For Case (a) General(ZPF) Method \n Induced EMF, EMF = %.f < %.2f V \n",abs(Er_a),atand(imag(Er_a),real(Er_a)))
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_a)
+printf("\n For Case (b) ASA Method \n Induced EMF, EMF = %.f < %.2f V \n",abs(Er_b),atand(imag(Er_b),real(Er_b)))
+printf("\n Percenatge Regulation, R = %.2f Percenatge \n",r_b)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- For Case (a) General(ZPF) Method (a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f \n ",abs(Er_a),atand(imag(Er_a),real(Er_a)))
+printf("\n For Case (b) ASA Method (a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f \n\n ",abs(Er_b),atand(imag(Er_b),real(Er_b)))
+printf(" CALCULATION OF THE POWER ANGLE IS NOT CALCULATED IN THE TEXT BOOK FOR THIS PROBLEM\n ")
+printf("\n INDUCED EMF AND PERCENTAGE REGULATION IS APPROXIMATED VALUE BECACUSE IN THE TEXT BOOK, CALCULATED INDUCED EMF IS WRONGLY PRINTED")
diff --git a/Working_Examples/2777/CH7/EX7.1/Ex7_1.sce b/Working_Examples/2777/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 0000000..afbd574 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,31 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.1
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+D = 35*10^-2; // Outer diameter of the conducting disk in Meter
+d = 10*10^-2; // Inner diameter of the conducting disk in Meter
+B = 1.0; // Axial magnetic field in Telsa
+N = 900; // Rotating shaft running in RPM
+
+
+// CALCULATIONS
+
+Wr = (2*%pi*N)/60; // Rotational angular speed in radians/s
+Er = ((D^2-d^2)*B*Wr)/8; // EMF induced in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.1: SOLUTION :-");
+printf("\n (a) Induced EMF in the outer and inner rims of the disk, Er = %.4f V \n",Er)
diff --git a/Working_Examples/2777/CH7/EX7.10/Ex7_10.sce b/Working_Examples/2777/CH7/EX7.10/Ex7_10.sce new file mode 100755 index 0000000..97b59e5 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,41 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.10
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+L = 1.0; // Length in Meter
+S = 60; // Number of slots
+f = 50; // Frequency in Hertz
+v = 440; // Operating Volage of the Motor in Volts
+V = 11.5; // Running speed of the motor in Meter/second
+m = 3; // Number of phases
+p = 8; // Total number of Poles
+
+
+// CALCULATIONS
+
+Vs = (2*L*f)/p; // Synchronous linear speed in Meter/second
+s = (Vs-V)/Vs; // Linear slip
+Vph = v/sqrt(3); // Phase Voltage in Volts
+Z1 = 6.0 + %i*5; // Impedance in Ohms refer figure and page no. 526
+Z2 = ((100*%i)*(5*%i+8.2/s))/(100*%i+5*%i+8.2/s); // Impedance in Ohms refer figure and page no. 526
+Z = Z1 + Z2; // Total Impedance in Ohms
+I = Vph/Z; // Per phase Current when Machine is running at 11.5 m/s in Amphere
+pf = cosd(atand(imag(I),real(I))); // Power factor lagging
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.10 : SOLUTION :-") ;
+printf("\n (a) Synchronous linear speed, Vs = %.1f m/s \n ",Vs);
+printf("\n (b) Per phase current when Machine is running at 11.5 m/s, I = %.2f < %.2f A \n",abs(I),atand(imag(I),real(I)))
diff --git a/Working_Examples/2777/CH7/EX7.11/Ex7_11.sce b/Working_Examples/2777/CH7/EX7.11/Ex7_11.sce new file mode 100755 index 0000000..f079848 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,29 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.11
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+s = 9; // Degree per step of the stepper motor
+N = 200; // Rotation Speed of the Stepper motor in RPM
+
+
+// CALCULATIONS
+
+spr = 360/s; // Steps Per Revolution (360 is full revolution)
+pps = (N*spr)/60; // Input pulse rate in pulses per second
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.11: SOLUTION :-");
+printf("\n (a) Input pulse rate is %.2f pulses per second \n",pps)
diff --git a/Working_Examples/2777/CH7/EX7.13/Ex7_13.sce b/Working_Examples/2777/CH7/EX7.13/Ex7_13.sce new file mode 100755 index 0000000..cacb284 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.13/Ex7_13.sce @@ -0,0 +1,34 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+L1 = 1.1; // Inductance in Henry
+L2 = 1.07; // Inductance in Henry
+dtheta = 1; // Rotor rotation in Mechanical degree
+r = 0.10; // Radius of the rotor in Meter
+I = 20; // Coil Current in Amphere
+
+
+// CALCULATIONS
+
+dl = L1-L2; // Change in Inductance of one of its stator coils in Henry
+F = (I^2*dl)/(2*dtheta); // Force on the single rotor pole in Newton
+T = F*r; // Instantaneous Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.13: SOLUTION :-");
+printf("\n (a) Instantaneous Torque, T = %.1f N-m \n\n",T)
+printf("The force is a motoring force since inductance of the coil is rising")
diff --git a/Working_Examples/2777/CH7/EX7.14/Ex7_14.sce b/Working_Examples/2777/CH7/EX7.14/Ex7_14.sce new file mode 100755 index 0000000..2b3a13c --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.14/Ex7_14.sce @@ -0,0 +1,37 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+Va = 220 * exp( %i * 0 * %pi/180); // Three phase in Volts
+Vb = 230 * exp( %i * (-115) * %pi/180); // Three phase in Volts
+Vc = 250 * exp( %i * (-245) * %pi/180); // Three phase in Volts
+
+
+// CALCUALTIONS
+// We know that operator :-
+
+alpha = 1 * exp( %i * 120 * %pi/180);
+alpha2 = 1 * exp( %i * (-120) * %pi/180);
+Va0 = (Va+Vb+Vc)/3 // Zero sequence Voltage in Volts
+Va1 = (Va+alpha*Vb+alpha2*Vc)/3 // Positive sequence Voltage in Volts
+Va2 = (Va+alpha2*Vb+alpha*Vc)/3 // Negative sequence Voltage in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.14 : SOLUTION :-") ;
+printf("\n (a) Zero sequence Voltage, Va0 = %.2f < %.2f V \n",abs(Va0),atand(imag(Va0),real(Va0)))
+printf("\n (b) Positive sequence Voltage, Va1 = %.3f < %.2f V \n",abs(Va1),atand(imag(Va1),real(Va1)))
+printf("\n (c) Negative sequence Voltage, Va1 = %.2f < %.1f V \n",abs(Va2),atand(imag(Va2),real(Va2)))
+
diff --git a/Working_Examples/2777/CH7/EX7.2/Ex7_2.sce b/Working_Examples/2777/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 0000000..8e518d8 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,37 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.2
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+D = 0.120; // Outer Radius of the Printed Circuit Motor in meter
+d = 0.060; // Inner Radius of the Printed Circuit Motor in meter
+ B = 0.7; // Axial Flux Density in Telsa
+V = 12; // Volage Supplied to the Motor in Volts
+R = 2700; // Motor Speed in RPM
+n = 0.65; // Efficiency of Motor
+p = 94; // Output Power of Motor in Watts
+I = 12; // Motor current in Ampheres
+
+
+// CALCULATIONS
+
+w = ((2*(%pi))*R)/60; // The Angular Velocity in Radians/second
+T = p/w; // Torque in Newton-Meter
+N = (8*T)/((D^2-d^2)*B*I) // Total Number of Conductors
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.2 : SOLUTION :-") ;
+printf("\n (a) Torque, T = %.2f N-m \n ",T);
+printf("\n (b) Total Number of Conductors, N = %.2f nearly 30 \n",N);
diff --git a/Working_Examples/2777/CH7/EX7.3/Ex7_3.sce b/Working_Examples/2777/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 0000000..f214279 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,54 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.3
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 2; // Total number of phase in servo Motor
+f = 50; // Frequency in Hertz
+V = 220; // Operating Voltage of the servo Motor in Volts
+R1 = 250; // Circuit Parameter in Ohms
+R2 = 750; // Circuit Parameter in Ohms
+X1 = 50; // Circuit Parameter in Ohms
+X2 = 50; // Circuit Parameter in Ohms
+Xm = 1000; // Circuit Parameter in Ohms
+s = 0.6; // Slip
+Va = 220; // Unbalanced Voltage in Volts
+Vb = 150 * exp(%i * (-60) * %pi/180); // Unbalanced Voltage in Volts
+
+
+// CALCULATIONS
+
+Va1 = (Va + %i*Vb)/2; // Positive sequence voltage in Volts
+Va2 = (Va - %i*Vb)/2; // Negative sequence voltage in Volts
+Z11 = (R1+%i*X1);
+Z12 = (((%i*Xm)*(R2/s+%i*X2))/(%i*Xm+R2/s+%i*X2));
+Z1 = Z11 + Z12 ; // Positive sequence impedance in Ohms
+Z2 = (R1+%i*X1) + (((%i*Xm)*(R2/(2-s)+%i*X2))/(%i*Xm+R2/(2-s)+%i*X2)); // Negative sequence impedance in Ohms
+Ia1 = Va1/Z1; // Positive sequence current in Amphere
+I12 = (Ia1*Z12)/(R2/s); // Positive sequence current in Amphere
+Ia2 = Va2/Z2; // Negative sequence current in Amphere
+I22 = (Ia2*Z2)/(R2/(2-s)); // Negative sequence current in Amphere
+T1 = 2*(abs(I12)^2)*R2/s; // Positive sequence torque in Newton-meter
+T2 = 2*(abs(I22)^2)*R2/(2-s); // Negative sequence torque in Newton-meter
+T = T1 - T2; // Resultant torque in Newton-meter
+Ia = Ia1 + Ia2; // Line current in Amphere
+Ib = (-%i*Ia1) + (%i*Ia2); // Line current in Amphere
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.3: SOLUTION :-");
+printf("\n (a) Resultant torque, T = %.2f N-m \n",T)
+printf("\n (b) Phase currents (line currents), Ia = %.2f < %.2f A \n\n Ib = %.2f < %.2f \n",abs(Ia),atand(imag(Ia),real(Ia)),abs(Ib),atand(imag(Ib),real(Ib)))
+printf("\n\n IN THE ABOVE PROBLEM ALL THE VALUES PRINTED IN THE TEXT BOOK ARE NOT ACCURATE, SO VALUE OF THE TORQUE AND LINE CURRENTS ARE DIFFERING, WHEN WE COMPARED TO THE TEXT BOOK ANSWERS FOR THE SAME. \n\n")
+printf("\n IN EVERY CALCULATED PARAMETER IN THE TEXT BOOK SLIGHT VARIATION IS THERE AS WE COMPARED TO MANUAL CALCULATION ITS FROM POSITIVE SEQUENCE VOLTAGE (Va1) \n")
diff --git a/Working_Examples/2777/CH7/EX7.4/Ex7_4.sce b/Working_Examples/2777/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 0000000..ddb5252 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,51 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.4
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 2; // Total number of phase in AC drag-cup servo Motor
+p = 2; // Number of poles
+Va = 220; // Operating Voltage of the servo Motor in Volts
+R1 = 350; // Circuit Parameter in Ohms
+R2 = 250; // Circuit Parameter in Ohms
+X1 = 60; // Circuit Parameter in Ohms
+X2 = 50; // Circuit Parameter in Ohms
+Xm = 900; // Circuit Parameter in Ohms
+s = 0.3; // Slip
+p = 0.8; // Ratio of the control winding and main winding voltage
+
+
+// CALCULATIONS
+
+Va1 = (Va*(1+p))/2; // Positive sequence voltage in Volts
+Va2 = (Va*(1-p))/2; // Negative sequence voltage in Volts
+Z11 = (R1+%i*X1);
+Z12 = (((%i*Xm)*(R2/s+%i*X2))/(%i*Xm+R2/s+%i*X2));
+Z1 = Z11 + Z12 ; // Positive sequence impedance in Ohms
+Z2 = (R1+%i*X1) + (((%i*Xm)*(R2/(2-s)+%i*X2))/(%i*Xm+R2/(2-s)+%i*X2)); // Negative sequence impedance in Ohms
+Ia1 = Va1/Z1; // Positive sequence current in Amphere
+I12 = (Ia1*Z12)/(R2/s); // Positive sequence current in Amphere
+Ia2 = Va2/Z2; // Negative sequence current in Amphere
+I22 = (Ia2*Z2)/(R2/(2-s)); // Negative sequence current in Amphere
+T1 = 2*(abs(I12)^2)*R2/s; // Positive sequence torque in Newton-meter
+T2 = 2*(abs(I22)^2)*R2/(2-s); // Negative sequence torque in Newton-meter
+T = T1 - T2; // Resultant torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.4: SOLUTION :-");
+printf("\n (a) Resultant torque, T = %.2f N-m \n",T)
+printf("\n\n IN THE ABOVE PROBLEM ALL THE VALUES PRINTED IN THE TEXT BOOK ARE NOT ACCURATE, SO VALUE OF THE TORQUE AND LINE CURRENTS ARE DIFFERING WHEN WE COMPARED TO THE TEXT BOOK ANSWERS FOR THE SAME. \n\n")
+printf("\n IN EVERY CALCULATED PARAMETER IN THE TEXT BOOK SLIGHT VARIATION IS THERE AS WE COMPARED TO MANUAL CALCULATION ITS FROM POSITIVE SEQUENCE IMPEDANCE (Z1) \n")
+
diff --git a/Working_Examples/2777/CH7/EX7.5/Ex7_5.sce b/Working_Examples/2777/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 0000000..9711507 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,35 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.5
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+R = 15; // Resistance of the fractional horse power AC series motor in Ohms
+V = 230; // AC supply voltage in Volts
+f = 50; // Frequency in Hertz
+I = 1.2; // Motor current in Amphere
+NDC = 2500; // Rotating speed of the Motor during DC Operation in RPM
+L = 0.5; // Total inductance in Henry
+
+
+// CALCUALTIONS
+
+X = 2*%pi*f*L; // Reactance in Ohms
+NAC = NDC * (sqrt(V^2-(I*X)^2)-(I*R)) / (V-(I*R)); // Rotating speed of the Motor during AC Operation in RPM
+pf = sqrt(1-((I*X)/V)^2); // Power factor lagging
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.5: SOLUTION :-");
+printf("\n When the Motor operting at AC 230V, 50 Hz \n\n (a) NAC = %.f RPM \n",NAC)
+printf("\n (b) Power factor = %.4f lagging \n",pf)
diff --git a/Working_Examples/2777/CH7/EX7.6/Ex7_6.sce b/Working_Examples/2777/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 0000000..4bf4f07 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,54 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.6
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+R = 1.4; // Total Resistance of the AC series motor in Ohms
+V = 115; // supply voltage in Volts
+f = 50; // Frequency in Hertz
+N = 5000; // Rotating speed in RPM
+X = 12; // Total reactance in Ohms
+P = 250; // Electrical power output in Watts
+loss = 18; // Rotational losses in Watts
+
+
+// CALCULATIONS
+
+Pd = P + loss; // Mechanical power developed in Watts
+// We know that Er = Pd/I and from phasor diagram in figure 7.11 page no. 501 V^2 = (Er+I*R)^2+(I*X)^2, 115^2 = (268/I-1.4*I)^2+(12*I)^2, 13225*I^2 = 71824+2.036*I^4-750.4*I^2+144*I^2, solving this we get 2.036*I^4-13831.4*I^2+71824 = 0, I^4-6793.42*I^2+3577 = 0 this gives I = 2.28A or 82.38A (The above calculation part is wrong )
+i = poly ([3577 0 -6793.42 0 1],'x','coeff'); // Expression for the Current in Quadratic form
+a = roots (i); // 4-Value of the current in Amphere
+I = a(4,1); // Curent in Amphere neglecting higher value and negative value
+
+pf_a = sqrt(1-((I*X)/V)^2); // Power factor lagging
+Er_a = sqrt(V^2-(I*X)^2)-(I*R); // Rotational Voltage in Volts
+T_a = (Er_a*I)/(2*%pi*N/60); // Developed torque in Newton-meter
+Ih = I/2; // Current halved in Amphere
+pf_b = sqrt(1-((Ih*X)/V)^2); // Power factor lagging when load current halved
+Er_b = sqrt(V^2-(Ih*X)^2)-(Ih*R); // Rotational Voltage in Volts when load current halved
+N2 = (N*Er_b*I)/(Er_a*Ih); // New speed in RPM when load current halved
+T_b = (Er_b*Ih)/(2*%pi*N2/60); // Developed torque in Newton-meter when load current halved
+eta = 100*(Er_b*Ih)/(V*Ih*pf_b); // Efficiency when load current halved
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.6: SOLUTION :-");
+printf("\n At rated condition, \n\n (a.1) Current, I = %.2f A \n",I)
+printf("\n (a.2) Power factor = %.3f lagging \n",pf_a)
+printf("\n (a.3) Developed torque = %.2f N-m \n",T_a)
+printf("\n When load current halved (reduced to half), \n\n (b.1) Speed, N2 = %.f RPM \n",N2)
+printf("\n (b.2) Power factor = %.4f lagging \n",pf_b)
+printf("\n (b.3) Developed torque = %.2f N-m \n",T_b)
+printf("\n (b.4) Efficiency = %.1f percenatge \n",eta)
+printf("\n From Calculation of the Current(I), rest all the Calculated values in the TEXT BOOK is WRONG because of the Current equation and its value both are WRONGLY calculated and the same used for the further Calculation part, so all the values are in the TEXT BOOK IS WRONG \n")
diff --git a/Working_Examples/2777/CH7/EX7.7/Ex7_7.sce b/Working_Examples/2777/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 0000000..79bd9e0 --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,51 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.7
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // supply voltage in Volts
+f = 50; // Frequency in Hertz
+p = 4; // Number of poles
+Xm = 50; // Mutual reactance in Ohms
+Rs = 0.4; // Resistance of stator windings in Ohms
+Xs = 2.5; // Leakage reactance of stator windings in Ohms
+Ra = 2.2; // Resistance of Armature windings in Ohms
+Xa = 3.1; // Leakage reactance of armature windings in Ohms
+loss = 30; // Rotational losses in Watts
+N = 2000; // Motor running speed in RPM
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous speed in RPM
+s = N/Ns; // Speed ratio
+I1 = V/(2*Rs + 2*%i*Xs + 2*%i*Xm + (%i*Xm^2)*((s-%i)/(Ra+%i*Xa+%i*Xm)));// line current in Amphere
+pf = cosd(atand(imag(I1),real(I1))); // Power factor lagging
+I2 = (s-%i)*(%i*Xm*I1)/(Ra+%i*Xa+%i*Xm); // line current in Amphere
+P1 = V*abs(I1)*cosd(atand(imag(I1),real(I1))); // Input power in Watts
+Pm = P1 - 2*(abs(I1)^2)*Rs - (abs(I2)^2)*Ra; // Mechanical power developed in Watts
+Po = Pm - loss; // output power in Watts
+eta = 100*(Po/P1); // Efficiency
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.7: SOLUTION :-");
+printf("\n (a) Line currents, I1 = %.2f < %.2f A and I2 = %.2f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)),abs(I2),atand(imag(I2),real(I2)))
+printf("\n (b) Power factor = %.2f lagging \n",pf)
+printf("\n (c) Efficiency = %.2f percentage \n",eta)
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) I1 = 3.37 < -42.78 A instead of %.2f < j(%.2f) A \n ",abs(I1),atand(imag(I1),real(I1)));
+printf("\n (b) I2 = 5.26 < -77.34 A instead of %.2f < j(%.2f) A \n ",abs(I2),atand(imag(I2),real(I2)));
+printf("\n (b) eta = 81.53 percent instead of %.2f percent \n ",eta)
+printf("\n From Calculation of the I1, rest all the Calculated values in the TEXT BOOK is WRONG because of the I1 value is WRONGLY calculated and the same used for the further Calculation part \n")
diff --git a/Working_Examples/2777/CH7/EX7.8/Ex7_8.sce b/Working_Examples/2777/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 0000000..0c7e9ea --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,40 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.8
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220; // supply voltage in Volts
+f = 50; // Frequency in Hertz
+p = 4; // Number of poles
+Xm = 60; // Mutual reactance in Ohms
+Rs = 1.0; // Resistance of stator windings in Ohms
+Xs = 6.0; // Leakage reactance of stator windings in Ohms
+Ra = 2.5; // Resistance of Armature windings in Ohms
+Xa = 6.0; // Leakage reactance of armature windings in Ohms
+P_hp = 1; // Output power in HP
+N = 1400; // Motor running speed in RPM
+alpha = 15; // Brush displacement from the low-impedance position in degree
+
+// CALCULATIONS
+
+Ns = (120*f)/p; // Synchronous speed in RPM
+s = N/Ns; // Speed ratio
+I = V / (Rs + %i*(Xs+Xm) + (%i*Xm^2*cosd(alpha))*(s*sind(alpha)-(%i*cosd(alpha)))/(Ra+%i*(Xa+Xm))); // Curent in Amphere
+pf = cosd(atand(imag(I),real(I))); // Power factor lagging
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.8: SOLUTION :-");
+printf("\n (a) Currents, I = %.2f < %.2f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n (b) Power factor = %.4f lagging \n",pf)
diff --git a/Working_Examples/2777/CH7/EX7.9/Ex7_9.sce b/Working_Examples/2777/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 0000000..2afaf6b --- /dev/null +++ b/Working_Examples/2777/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,32 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.9
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+E2 = 100; // Per phase standstill EMF in Volts
+Z2s = 0.025 + %i*0.08; // Rotor circuit impedance at standstill
+E = 50; // Injected EMF in Volts
+
+
+// CALCULATIONS
+
+I2 = 0; // Assuming Current is zero
+s1 = (E/E2)+(I2*Z2s)/E2; // Slip when injected EMF is opposite to the E2
+s2 = (-E/E2)+(I2*Z2s)/E2; // Slip when injected EMF is phase with E2
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.9: SOLUTION :-");
+printf("\n (a) Slip when injected EMF is opposite to the E2, s = %.1f \n",s1)
+printf("\n (b) Slip when injected EMF is phase with E2, s = %.1f \n",s2)
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